Bruce K. Driver

Analysis Tools with Examples

March 10, 2004 File:tempanal.tex

Springer

Berlin Heidelberg NewYork Hong Kong London Milan Paris Tokyo

Contents

Part I Back Ground Material

1 Introduction / User Guide . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 1.1 Topology beginnings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 1.2 A Better Integral and an Introduction to Measure Theory …. 3

2 Set Operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 2.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

3 A Brief Review of Real and Complex Numbers . . . . . . . . . . . . 11

3.1 The Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

3.1.1 The Decimal Representation of a Real Number . . . . . . . . 16 3.2 The Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 3.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

4 Limits and Sums. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 4.1 Limsups, Liminfs and Extended Limits . . . . . . . . . . . . . . . . . . . . . 21 4.2 Sums of positive functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 4.3 Sums of complex functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 4.4 Iterated sums and the Fubini and Tonelli Theorems . . . . . . . . . . 32 4.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

4.5.1 Limit Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 4.5.2 Dominated Convergence Theorem Problems . . . . . . . . . . 35

5 ℓ^{p} – spaces, Minkowski and Holder Inequalities . . . . . . . . . . . . 39 5.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

Part II Metric and Banach Space Basics 6 Metric Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

6.1 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 6.2 Completeness in Metric Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 6.3 Supplementary Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

6.3.1 Word of Caution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 6.3.2 Riemannian Metrics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

6.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

7 Banach Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 7.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 7.2 Bounded Linear Operators Basics . . . . . . . . . . . . . . . . . . . . . . . . . 62 7.3 General Sums in Banach Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 66 7.4 Inverting Elements in L(X) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68 7.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

8 The Riemann Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 8.1 The Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . . . . . . 77 8.2 Integral Operators as Examples of Bounded Operators . . . . . . . 81 8.3 Linear Ordinary Differential Equations . . . . . . . . . . . . . . . . . . . . . 83 8.4 Classical Weierstrass Approximation Theorem. . . . . . . . . . . . . . . 87 8.5 Iterated Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94 8.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

9 Hölder Spaces as Banach Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 9.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

Part III Topological Spaces

10 Topological Space Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 10.1 Constructing Topologies and Checking Continuity . . . . . . . . . . . 110 10.2 Product Spaces I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116 10.3 Closure operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119 10.4 Countability Axioms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 10.5 Connectedness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123 10.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127

10.6.1 General Topological Space Problems . . . . . . . . . . . . . . . . . 127 10.6.2 Connectedness Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . 128 10.6.3 Metric Spaces as Topological Spaces . . . . . . . . . . . . . . . . . 129

11 Compactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131 11.1 Metric Space Compactness Criteria . . . . . . . . . . . . . . . . . . . . . . . . 132 11.2 Compact Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139 11.3 Local and σ – Compactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140 11.4 Function Space Compactness Criteria . . . . . . . . . . . . . . . . . . . . . . 142 11.5 Tychonoff’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146 11.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149

11.6.1 Ascoli-Arzela Theorem Problems . . . . . . . . . . . . . . . . . . . . 150 11.6.2 Tychonoff’s Theorem Problem . . . . . . . . . . . . . . . . . . . . . . 151

12 Locally Compact Hausdorff Spaces . . . . . . . . . . . . . . . . . . . . . . . . 153 12.1 Locally compact form of Urysohn’s Metrization Theorem . . . . . 158 12.2 Partitions of Unity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160 12.3 C_{0}(X) and the Alexanderov Compactification . . . . . . . . . . . . . . . 165 12.4 Stone-Weierstrass Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168 12.5 *More on Separation Axioms: Normal Spaces . . . . . . . . . . . . . . . 172 12.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175

13 Baire Category Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179 13.1 Metric Space Baire Category Theorem . . . . . . . . . . . . . . . . . . . . . 179 13.2 Locally Compact Hausdorff Space Baire Category Theorem … 180 13.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186

14 Hilbert Space Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187 14.1 Hilbert Space Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195 14.2 Some Spectral Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202 14.3 Compact Operators on a Hilbert Space . . . . . . . . . . . . . . . . . . . . . 208

14.3.1 The Spectral Theorem for Self Adjoint Compact

Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210

14.4 Weak Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214 14.5 Supplement 1: Converse of the Parallelogram Law . . . . . . . . . . . 216 14.6 Supplement 2. Non-complete inner product spaces . . . . . . . . . . . 218 14.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 220

Part IV Calculus and Ordinary Differential Equations in Banach Spaces

15 Ordinary Differential Equations in a Banach Space . . . . . . . . 223 15.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223 15.2 Uniqueness Theorem and Continuous Dependence on Initial

Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225 15.3 Local Existence (Non-Linear ODE) . . . . . . . . . . . . . . . . . . . . . . . . 227 15.4 Global Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 230 15.5 Semi-Group Properties of time independent flows . . . . . . . . . . . . 236 15.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 238

16 Banach Space Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241 16.1 The Differential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241 16.2 Product and Chain Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243 16.3 Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247 16.4 Higher Order Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249 16.5 Inverse and Implicit Function Theorems . . . . . . . . . . . . . . . . . . . . 253 16.6 Smooth Dependence of ODE’s on Initial Conditions* . . . . . . . . 260 16.7 Existence of Periodic Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263 16.8 Contraction Mapping Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . 265 16.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267

16.9.1 Alternate construction of g. To be made into an exercise.269

Part V Lebesgue Integration Theory

17 Introduction: What are measures and why “measurable” sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273 17.1 The problem with Lebesgue “measure” . . . . . . . . . . . . . . . . . . . . . 274

18 Measurability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 279 18.1 Algebras and σ – Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 279 18.2 Measurable Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284

18.2.1 More general pointwise limits . . . . . . . . . . . . . . . . . . . . . . . 291 18.3 σ – Function Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292 18.4 Product σ – Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 299

18.4.1 Factoring of Measurable Maps . . . . . . . . . . . . . . . . . . . . . . 302 18.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302

19 Measures and Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 305

19.1 Example of Measures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 308

19.1.1 ADD: Examples of Measures . . . . . . . . . . . . . . . . . . . . . . . . 310 19.2 Integrals of Simple functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 310 19.3 Integrals of positive functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 312 19.4 Integrals of Complex Valued Functions . . . . . . . . . . . . . . . . . . . . . 320 19.5 Measurability on Complete Measure Spaces . . . . . . . . . . . . . . . . . 329 19.6 Comparison of the Lebesgue and the Riemann Integral . . . . . . . 330 19.7 Determining Classes of Measures . . . . . . . . . . . . . . . . . . . . . . . . . . 333 19.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 336

20 Multiple Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 339 20.1 Fubini-Tonelli’s Theorem and Product Measure . . . . . . . . . . . . . 340 20.2 Lebesgue Measure on ℝ^{d} and the Change of Variables Theorem348 20.3 The Polar Decomposition of Lebesgue Measure . . . . . . . . . . . . . . 359 20.4 More proofs of the classical Weierstrass approximation

Theorem 8.34 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 362 20.5 More Spherical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365 20.6 Sard’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 369 20.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 373 21 L^{p}-spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 377

21.1 Jensen’s Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 381 21.2 Modes of Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 385 21.3 Completeness of L^{p} – spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 389

21.3.1 Summary: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393 21.4 Converse of Hölder’s Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393 21.5 Uniform Integrability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 399 21.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 406

22 Approximation Theorems and Convolutions . . . . . . . . . . . . . . . 409 22.1 Density Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 409 22.2 Convolution and Young’s Inequalities . . . . . . . . . . . . . . . . . . . . . . 416

22.2.1 Smooth Partitions of Unity . . . . . . . . . . . . . . . . . . . . . . . . . 426 22.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 427

23 L^{2} - Hilbert Spaces Techniques and Fourier Series . . . . . . . . . 431 23.1 L^{2}-Orthonoramal Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 431 23.2 Hilbert Schmidt Operators. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 433 23.3 Fourier Series Considerations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 438

23.3.1 Dirichlet, Fejér and Kernels . . . . . . . . . . . . . . . . . . . . . . . . . 439 23.3.2 The Dirichlet Problems on D and the Poisson Kernel . . 444

23.4 Weak L^{2}-Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 447 23.5 *Conditional Expectation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 449 23.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 452 23.7 Fourier Series Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 454 23.8 Conditional Expectation Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 458

Part VI Appendices

A Multinomial Theorems and Calculus Results. . . . . . . . . . . . . . . 5 A.1 Multinomial Theorems and Product Rules . . . . . . . . . . . . . . . . . . 5 A.2 Taylor’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

B Zorn’s Lemma and the Hausdorff Maximal Principle . . . . . . 11

C Nets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

D Study Guides . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

D.1 Study Guide For Math 240A: Fall 2003 . . . . . . . . . . . . . . . . . . . . . 19 D.1.1 Basic things you should know about numbers and limits 19 D.1.2 Basic things you should know about topological and

measurable spaces: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 D.1.3 Basic things you should know about Metric Spaces …. 21 D.1.4 Basic things you should know about Banach spaces …. 21 D.1.5 The Riemann integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 D.1.6 Basic things you should know about Lebesgue

integration theory and infinite sums . . . . . . . . . . . . . . . . . 22

D.2 Study Guide For Math 240B: Winter 2004 . . . . . . . . . . . . . . . . . . 23 D.2.1 Basic things you should know about Multiple Integrals: 23 D.2.2 Basic things you should know about L^{p} – spaces . . . . . . 23 D.2.3 Additional Basic things you should know about

topological spaces: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 D.2.4 Things you should know about Locally Compact

Hausdorff Spaces: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 D.2.5 Approximation Theorems and Convolutions. . . . . . . . . . . 24 D.2.6 Things you should know about Hilbert Spaces . . . . . . . . 25

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

Part I

Back Ground Material Introduction / User Guide

Not written as of yet. Topics to mention.

1. A better and more general integral. a) Convergence Theorems

b) Integration over diverse collection of sets. (See probability theory.)

c) Integration relative to different weights or densities including singular weights.

d) Characterization of dual spaces.

e) Completeness.

2. Infinite dimensional Linear algebra.

3. ODE and PDE.

4. Harmonic and Fourier Analysis.

5. Probability Theory

*** Beginning of WORK material. ***

1.1 Topology beginnings

Recall the notion of a topology by extrapolating from the open sets on ℝ^{2}. Also recall what it means to be continuous, namely f : X → ℝ is continuous at x if for all ε > 0 there exists V ∈ τ_{x} such that

f (V) ⊂ f (x) + (−ε, ε) .

1.2 A Better Integral and an Introduction to Measure Theory

Let a, b ∈ ℝ with a < b and let

I^{0}(f) := ∫_{a}^{b} f(t)dt for all f ∈ C ([a, b])

denote the Riemann integral. Also let H denote the smallest linear subspace of bounded functions on [a, b] which is closed under bounded convergence and contains C ([a, b]) . Such a space exists since we can take the intersection over all such spaces of functions.

Theorem 1.1. There is an extension I of I^{0} to H such that I is still linear and lim_{n→∞} I (f_{n}) = I (f) for all f_{n} ∈ H with f_{n} → f boundedly. Moreover this extension is unique and is positive in the sense that I (f) ≥ 0 if f ∈ H and f ≥ 0.

Proof. We will only prove the uniqueness here. Suppose that J and I are two such extensions and let

K := {f ∈ H : J (f) = I (f)} .

Then K is a linear subspace closed under bounded convergence which contains C ([a, b]) and hence K = H. The existence of I is the hard part. The positivity of I can be seen from the existence construction.

Example 1.2. Here are some examples of functions in H and their integrals:

1. Suppose [α, β] ⊂ [a, b] , then 1_{[α,β]} ∈ H and I (1_{[α,β]}) = β − α.(Draw a picture.)

2. I (1_{{α}}) = 0.

3. The space H is an algebra, i.e. if f, g ∈ H then fg ∈ H. To prove this, first assume that f ∈ C ([a, b]) and let

H_{f} := {g ∈ H : fg ∈ H} .

Then H_{f} is closed under bounded convergence and contains C ([a, b]) and hence H_{f} = H, i.e. the product of a continuous function and an element in H is back in H. Now suppose that f ∈ H and again let H_{f} be as above. By the same reasoning we may show again that H_{f} = H and this proves the assertion.

4. If f ∈ H and ϕ ∈ C (ℝ) , then ϕ ∘ f ∈ H. This a consequence of the Weierstrass approximation Theorem 8.34. In particular |f| ∈ H and f_{±} := \frac{|f|±f}{2} ∈ H if f ∈ H.

5. If f_{n} ∈ H, f_{n} ≥ 0 and f = ∑_{n=1}^{∞} f_{n} is a bounded function, then f ∈ H and

I (f) = ∑_{n=1}^{∞} I (f_{n}) .  (1.1)

To prove Eq. (1.1) we have

∑_{n=1}^{∞} I (f_{n}) = lim_{N→∞} I (∑_{n=1}^{N} f_{n}) = I (f) .

6. As an example of item 4., 1_{ℚ∩[a,b]} = ∑_{n=1}^{∞} 1_{{α_{n}}} ∈ H and I (1_{ℚ∩[a,b]}) = 0. Here {α_{n}}_{n=1}^{∞} is an enumeration of the rational numbers in the interval [a, b].

7. Let M := {A ⊂ [a, b] : 1_{A} ∈ H} and for A ∈ M let m (A) := I (1_{A}) . Then M and m have the following properties:

a) ∅, [a, b] ∈ M and m (∅) = 0 and m ([a, b]) = b−a. Moreover m (A) ≥ 0 for all A ∈ M.

b) If A ∈ M then A^{c} ∈ M and m (A^{c}) = b − a − m (A) . This follows from the fact that 1_{A^{c}} = 1 − 1_{A}.

c) If A, B ∈ M, then A ∩ B ∈ M since if 1_{A∩B} = 1_{A} · 1_{B} and H is an algebra. Definition: a collection of sets M satisfying a) – c) is called an algebra of subsets of [a, b] .

d) More generally if A_{n} ∈ M then ∩A_{n} ∈ M since 1_{∩A_{n}} = lim_{N→∞} 1_{A_{1}} ⋯ 1_{A_{N}} and the convergence is bounded. Definition: a collection of sets M satisfying a) – d) is called an σ – algebra.

e) If A_{n} ∈ M, then ∪A_{n} ∈ M. Indeed we know ∪A_{n} ∈ M iff (∪A_{n})^{c} ∈

M. But

(∪A_{n})^{c} = ∩A_{n}^{c} ∈ M

by item d. above.

f) If A_{n} ∈ M are pairwise disjoint, then

m (∪A_{n}) = ∑_{n=1}^{∞} m (A_{n}) .

To prove this it suffices to observe that 1_{∪A_{n}} = ∑_{n=1}^{∞} 1_{A_{n}}.

g) M is not 2^{[a,b]}, i.e. M is not all subset of [a, b]. This is not obvious and it is not possible to really write down an “explicit” subset [a, b] which is not in M. We will prove the existence of such sets later.

8. Fact: M is the smallest σ – algebra on [a, b] which contains all subintervals of [a, b].

9. Fact: A bounded function f : [a, b] → ℝ is in H iff {f > α} ∈ M for all α ∈ ℝ.

10. Fact: The integral I may be recovered from the measure m by the formula

I (f) = lim_{mesh→0} ∑_{0<α_{1}<α_{2}<α_{3}<...}^{∞} α_{i}m ({x ∈ [a, b] : α_{i} < f (x) ≤ α_{i}}) .

We will prove items 8. – 10. later in the course. The proof if Items 9. and 10. is not so hard and the energetic reader may wish to give them a try.

Notation 1.3 The collection of sets M is called the Borel σ – algebra on [a, b] and the function m : M → ℝ is called Lebesgue measure. We will usually write I (f) as ∫_{[a,b]} fdm and I (f) will be called the Lebesgue integral of f. This integral may be extended to all positive functions f such that f1_{|f|≤M} ∈ H for all M by

I (f) = lim_{M→∞} I (f1_{|f|≤M}) .

Again, we will come back to all of this again later.

*** End of WORK material. ***

Set Operations

Let ℕ denote the positive integers, ℕ_{0} := ℕ∪ {0} be the non-negative integers and ℤ = ℕ_{0} ∪ (−ℕ) – the positive and negative integers including 0, ℚ the rational numbers, ℝ the real numbers (see Chapter 3 below), and ℂ the complex numbers. We will also use 𝔽 to stand for either of the fields ℝ or ℂ.

Notation 2.1 Given two sets X and Y, let Y^{X} denote the collection of all functions f : X → Y. If X = ℕ, we will say that f ∈ Y^{ℕ} is a sequence with values in Y and often write f_{n} for f (n) and express f as {f_{n}}_{n=1}^{∞} . If X = {1, 2, …, N}, we will write Y^{N} in place of Y^{{1,2,...,N}} and denote f ∈ Y^{N} by f = (f_{1}, f_{2}, …, f_{N}) where f_{n} = f(n).

Notation 2.2 More generally if {X_{α} : α ∈ A} is a collection of non-empty sets, let X_{A} = ∏_{α∈A} X_{α} and π_{α} : X_{A} → X_{α} be the canonical projection map

defined by π_{α}(x) = x_{α}. If If X_{α} = X for some fixed space X, then we will write ∏_{α∈A} X_{α} as X^{A} rather than X_{A}.

Recall that an element x ∈ X_{A} is a “choice function,” i.e. an assignment x_{α} := x(α) ∈ X_{α} for each α ∈ A. The axiom of choice (See Appendix B.) states that X_{A} ≠ ∅ provided that X_{α} ≠ ∅ for each α ∈ A.

Notation 2.3 Given a set X, let 2^{X} denote the power set of X – the collection of all subsets of X including the empty set.

The reason for writing the power set of X as 2^{X} is that if we think of 2 meaning {0, 1} , then an element of a ∈ 2^{X} = {0, 1}^{X} is completely determined by the set

A := {x ∈ X : a(x) = 1} ⊂ X.

In this way elements in {0, 1}^{X} are in one to one correspondence with subsets of X.

For A ∈ 2^{X} let

A^{c} := X \ A = {x ∈ X : x ∉ A}

and more generally if A, B ⊂ X let

B \ A := {x ∈ B : x ∉ A} = A ∩ B^{c}.

We also define the symmetric difference of A and B by

A△B := (B \ A) ∪ (A \ B) .

As usual if {A_{α}}_{α∈I} is an indexed collection of subsets of X we define the union and the intersection of this collection by

∪_{α∈I}A_{α} := {x ∈ X : ∃ α ∈ I ∋ x ∈ A_{α}} and
∩_{α∈I}A_{α} := {x ∈ X : x ∈ A_{α} ∀ α ∈ I }.

Notation 2.4 We will also write ∐_{α∈I} A_{α} for ∪_{α∈I}A_{α} in the case that {A_{α}}_{α∈I} are pairwise disjoint, i.e. A_{α} ∩ A_{β} = ∅ if α ≠ β.

Notice that ∪ is closely related to ∃ and ∩ is closely related to ∀. For example let {A_{n}}_{n=1}^{∞} be a sequence of subsets from X and define

{A_{n} i.o.} := {x ∈ X : # {n : x ∈ A_{n}} = ∞} and
{A_{n} a.a.} := {x ∈ X : x ∈ A_{n} for all n sufficiently large}.

(One should read {A_{n} i.o.} as A_{n} infinitely often and {A_{n} a.a.} as A_{n} almost always.) Then x ∈ {A_{n} i.o.} iff

∀N ∈ ℕ ∃ n ≥ N ∋ x ∈ A_{n}

and this may be expressed as

{A_{n} i.o.} = ∩_{N=1}^{∞} ∪_{n≥N} A_{n}.

Similarly, x ∈ {A_{n} a.a.} iff

∃ N ∈ ℕ ∋ ∀ n ≥ N, x ∈ A_{n}

which may be written as

{A_{n} a.a.} = ∪_{N=1}^{∞} ∩_{n≥N} A_{n}.

Definition 2.5. A set X is said to be countable if is empty or there is an injective function f : X → ℕ, otherwise X is said to be uncountable.

Lemma 2.6 (Basic Properties of Countable Sets).

1. If A ⊂ X is a subset of a countable set X then A is countable.

2. Any infinite subset Λ ⊂ ℕ is in one to one correspondence with ℕ.

3. A non-empty set X is countable iff there exists a surjective map, g : ℕ → X.

4. If X and Y are countable then X × Y is countable.

5. Suppose for each m ∈ ℕ that A_{m} is a countable subset of a set X, then A = ∪_{m=1}^{∞}A_{m} is countable. In short, the countable union of countable sets is still countable.

6. If X is an infinite set and Y is a set with at least two elements, then Y^{X} is uncountable. In particular 2^{X} is uncountable for any infinite set X.

Proof. 1. If f : X → ℕ is an injective map then so is the restriction, f|_{A}, of f to the subset A. 2. Let f (1) = min Λ and define f inductively by

f(n + 1) = min Λ \ {f(1), …, f(n)} .

Since Λ is infinite the process continues indefinitely. The function f : ℕ → Λ defined this way is a bijection. 3. If g : ℕ → X is a surjective map, let

f(x) = min g^{−1} ({x}) = min {n ∈ ℕ : f(n) = x} .

Then f : X → ℕ is injective which combined with item 2. (taking Λ = f(X)) shows X is countable. Conversely if f : X → ℕ is injective let x_{0} ∈ X be a fixed point and define g : ℕ → X by g(n) = f^{−1}(n) for n ∈ f (X) and g(n) = x_{0} otherwise. 4. Let us first construct a bijection, h, from ℕ to ℕ × ℕ. To do this put the elements of ℕ × ℕ into an array of the form

( \begin{matrix}(1, 1) & (1, 2) & (1, 3) & … \\ (2, 1) & (2, 2) & (2, 3) & … \\ (3, 1) & (3, 2) & (3, 3) & … \\ ⋮ & ⋮ & ⋮ & ... \end{matrix} )

and then “count” these elements by counting the sets {(i, j) : i + j = k} one at a time. For example let h (1) = (1, 1) , h(2) = (2, 1), h (3) = (1, 2), h(4) = (3, 1), h(5) = (2, 2), h(6) = (1, 3), etc. etc. If f : ℕ →X and g : ℕ →Y are surjective functions, then the function (f × g) ∘ h : ℕ →X × Y is surjective where (f × g) (m, n) := (f (m), g(n)) for all (m, n) ∈ ℕ × ℕ. 5. If A = ∅ then A is countable by definition so we may assume A ≠ ∅. With out loss of generality we may assume A_{1} ≠ ∅ and by replacing A_{m} by A_{1} if necessary we may also assume A_{m} ≠ ∅ for all m. For each m ∈ ℕ let a_{m} : ℕ →A_{m} be a surjective function and then define f : ℕ × ℕ → ∪_{m=1}^{∞}A_{m} by f(m, n) := a_{m}(n). The function f is surjective and hence so is the composition, f ∘ h : ℕ → X × Y, where h : ℕ → ℕ×ℕ is the bijection defined above. 6. Let us begin by showing 2^{ℕ} = {0, 1}^{ℕ} is uncountable. For sake of contradiction suppose f : ℕ → {0, 1}^{ℕ} is a surjection and write f (n) as (f_{1} (n) , f_{2} (n) , f_{3} (n) , …) . Now define a ∈ {0, 1}^{ℕ} by a_{n} := 1 − f_{n}(n). By construction f_{n} (n) ≠ a_{n} for all n and so a ∉ f (ℕ) . This contradicts the assumption that f is surjective and shows 2^{ℕ} is uncountable. For the general case, since Y_{0}^{X} ⊂ Y^{X} for any subset Y_{0} ⊂ Y, if Y_{0}^{X} is uncountable then so is Y^{X}. In this way we may assume Y_{0} is a two point set which may as well be Y_{0} = {0, 1} . Moreover, since X is an infinite set we may find an injective map x : ℕ → X and use this to set up an injection, i : 2^{ℕ} → 2^{X} by setting i (a) (x_{n}) = a_{n} for all n ∈ ℕ and i (a) (x) = 0 if x ∉ {x_{n} : n ∈ ℕ} . If 2^{X} were countable we could find a surjective map f : 2^{X} → ℕ in which case f ∘ i : 2^{ℕ} → ℕ would be surjective as well. However this is impossible since we have already seed that 2^{ℕ} is uncountable.

We end this section with some notation which will be used frequently in the sequel.

Notation 2.7 If f : X → Y is a function and E ⊂ 2^{Y} let

f^{−1}E := f^{−1} (E) := {f^{−1}(E)|E ∈ E}.

If G ⊂ 2^{X}, let

f_{∗}G := {A ∈ 2^{Y} |f^{−1}(A) ∈ G}.

Definition 2.8. Let E ⊂ 2^{X} be a collection of sets, A ⊂ X, i_{A} : A → X be the inclusion map (i_{A}(x) = x for all x ∈ A) and

E_{A} = i_{A}^{−1}(E) = {A ∩ E : E ∈ E} .

2.1 Exercises

Let f : X → Y be a function and {A_{i}}_{i∈I} be an indexed family of subsets of Y, verify the following assertions.

Exercise 2.1. (∩_{i∈I}A_{i})^{c} = ∪_{i∈I}A_{i}^{c}.

Exercise 2.2. Suppose that B ⊂ Y, show that B \ (∪_{i∈I}A_{i}) = ∩_{i∈I}(B \ A_{i}).

Exercise 2.3. f^{−1}(∪_{i∈I}A_{i}) = ∪_{i∈I}f^{−1}(A_{i}).

Exercise 2.4. f^{−1}(∩_{i∈I}A_{i}) = ∩_{i∈I}f^{−1}(A_{i}).

Exercise 2.5. Find a counter example which shows that f(C ∩ D) = f(C) ∩ f(D) need not hold.

A Brief Review of Real and Complex Numbers

Although it is assumed that the reader of this book is familiar with the properties of the real numbers, ℝ, nevertheless I feel it is instructive to define them here and sketch the development of their basic properties. It will most certainly be assumed that the reader is familiar with basic algebraic properties of the natural numbers ℕ and the ordered field of rational numbers,

ℚ = \{\frac{m}{n} : m, n ∈ ℤ : n ≠ 0\} .

As usual, for q ∈ ℚ, we define

|q| = \{ \begin{matrix}q & if & q ≥ 0 \\ −q & if & q ≤ 0. \end{matrix}

Notice that if q ∈ ℚ and |q| ≤ \frac{1}{n} for all n, then q = 0. Indeed q ≠ 0 then |q| = \frac{m}{n} for some m, n ∈ ℕ and hence |q| ≥ \frac{1}{n}. A similar argument shows q ≥ 0 iff q ≥ −\frac{1}{n} for all n ∈ ℕ. These trivial remarks will be used in the future without further reference.

Definition 3.1. A sequence {q_{n}}_{n=1}^{∞} ⊂ ℚ converges to q ∈ ℚ if |q − q_{n}| → 0 as n → ∞, i.e. if for all N ∈ ℕ, |q − q_{n}| ≤ \frac{1}{N} for a.a. n. As usual if \{q_{n}\}_{n=1}^{∞} converges to q we will write q_{n} → q as n → ∞ or q = lim_{n→∞} q_{n}.

Definition 3.2. A sequence {q_{n}}_{n=1}^{∞} ⊂ ℚ is Cauchy if |q_{n} − q_{m}| → 0 as m, n → ∞. More precisely we require for each N ∈ ℕ that |q_{m} − q_{n}| ≤ \frac{1}{N} for a.a. pairs (m, n) .

Exercise 3.1. Show that all convergent sequences {q_{n}}_{n=1}^{∞} ⊂ ℚ are Cauchy and that all Cauchy sequences {q_{n}}_{n=1}^{∞} are bounded – i.e. there exists M ∈ ℕ such that

|q_{n}| ≤ M for all n ∈ ℕ.

Exercise 3.2. Suppose {q_{n}}_{n=1}^{∞} and {r_{n}}_{n=1}^{∞} are Cauchy sequences in ℚ.

1. Show {q_{n} + r_{n}}_{n=1}^{∞} and {q_{n} · r_{n}}_{n=1}^{∞} are Cauchy.

Now assume that {q_{n}}_{n=1}^{∞} and {r_{n}}_{n=1}^{∞} are convergent sequences in ℚ.

2. Show {q_{n} + r_{n}}_{n=1}^{∞} {q_{n} · r_{n}}_{n=1}^{∞} are convergent in ℚ and

lim_{n→∞} (q_{n} + r_{n}) = lim_{n→∞} q_{n} + lim_{n→∞} r_{n} and
lim_{n→∞} (q_{n}r_{n}) = lim_{n→∞} q_{n} · lim_{n→∞} r_{n}.

3. If we further assume q_{n} ≤ r_{n} for all n, show lim_{n→∞} q_{n} ≤ lim_{n→∞} r_{n}. (It suffices to consider the case where q_{n} = 0 for all n.)

The rational numbers ℚ suffer from the defect that they are not complete, i.e. not all Cauchy sequences are convergent. In fact, according to Corollary 3.14 below, “most” Cauchy sequences of rational numbers do not converge to a rational number.

Exercise 3.3. Use the following outline to construct a Cauchy sequence {q_{n}}_{n=1}^{∞} ⊂ ℚ which is not convergent in ℚ.

1. Recall that there is no element q ∈ ℚ such that q^{2} = 2.^{1} To each n ∈ ℕ let m_{n} ∈ ℕ be chosen so that

\frac{m_{n}^{2}}{n^{2}} < 2 < \frac{(m_{n} + 1)^{2}}{n^{2}}  (3.1)

and let q_{n} := \frac{m_{n}}{n} .

2. Verify that q_{n}^{2} → 2 as n → ∞ and that {q_{n}}_{n=1}^{∞} is a Cauchy sequence in ℚ.

3. Show {q_{n}}_{n=1}^{∞} does not have a limit in ℚ.

3.1 The Real Numbers

Let C denote the collection of Cauchy sequences a = {a_{n}}_{n=1}^{∞} ⊂ ℚ and say a, b ∈ C are equivalent (write a ∼ b) iff lim_{n→∞} |a_{n} − b_{n}| = 0. (The reader should check that “ ∼ ” is an equivalence relation.)

Definition 3.3. A real number is an equivalence class, ā := {b ∈ C : b ∼ a} associated to some element a ∈ C. The collection of real numbers will be denoted by ℝ. For q ∈ ℚ, let i (q) = ā where a is the constant sequence a_{n} = q for all n ∈ ℕ. We will simply write 0 for i (0) and 1 for i (1) .

Exercise 3.4. Given ā, b̄ ∈ ℝ show that the definitions

−ā = \overline{(−a)}, ā + b̄:= \overline{(a + b)} and ā · b̄:= \overline{a · b}

1 This fact also shows that the intermediate value theorem, (See Theorem 10.50 below.) fails when working with continuous functions defined over ℚ.

are well defined. Here −a, a + b and a · b denote the sequences {−a_{n}}_{n=1}^{∞} , {a_{n} + b_{n}}_{n=1}^{∞} and {a_{n} · b_{n}}_{n=1}^{∞} respectively. Further verify that with these operations, ℝ becomes a field and the map i : ℚ → ℝ is injective homomorphism of fields. Hint: if ā ≠ 0 show that ā may be represented by a sequence a ∈ C with |a_{n}| ≥ \frac{1}{N} for all n and some N ∈ ℕ. For this representative show the sequence a^{−1} := {a_{n}^{−1}}_{n=1}^{∞} ∈ C. The multiplicative inverse to ā may now be

constructed as: \frac{1}{ā} = ā^{−1} := \overline{\{a_{n}^{−1}\}_{n=1}^{∞}}.

Definition 3.4. Let ā, b̄ ∈ ℝ. Then

1. ā > 0 if there exists an N ∈ ℕ such that a_{n} > \frac{1}{N} for a.a. n.

2. ā ≥ 0 iff either ā > 0 or ā = 0. Equivalently (as the reader should verify), ā ≥ 0 iff for all N ∈ ℕ, a_{n} ≥ −\frac{1}{N} for a.a. n.

3. Write ā > b̄ or b̄ < ā if ā − b̄ > 0

4. Write ā ≥ b̄ or b̄ ≤ ā if ā − b̄ ≥ 0.

Exercise 3.5. Show “ ≥ ” make ℝ into a linearly ordered field and the map i : ℚ → ℝ preserves order. Namely if ā, b̄ ∈ ℝ then

1. exactly one of the following relations hold: ā < b̄ or ā > b̄ or ā = b̄.

2. If ā ≥ 0 and b̄ ≥ 0 then ā + b̄ ≥ 0 and ā · b̄ ≥ 0.

3. If q, r ∈ Q then q ≤ r iff i (q) ≤ i (r) .

The absolute value of a real number ā is defined analogously to that of a rational number by

\left|{ \atop ā}\right| = \{ \begin{matrix}ā & if & ā ≥ 0 \\ −ā & if & ā < 0 \end{matrix} .

Observe this definition is consistent with our previous definition of the absolute value on ℚ, namely i (|q|) = |i (q)| . Also notice that ā = 0 (i.e. a ∼ 0 where 0 denotes the constant sequence of all zeros) iff for all N ∈ ℕ, |a_{n}| ≤ \frac{1}{N} for a.a. n. This is equivalent to saying \left|{ \atop ā}\right| ≤ i (\frac{1}{N} ) for all N ∈ ℕ iff ā = 0.

Exercise 3.6. Given ā, b̄ ∈ ℝ show

|āb̄| = \left|{ \atop ā}\right| |b̄| and |ā + b̄| ≤ \left|{ \atop ā}\right| + |b̄| .

The latter inequality being referred to as the triangle inequality.

By exercise 3.6,

\left|{ \atop ā}\right| = |ā − b̄ + b̄| ≤ |ā − b̄| + |b̄|

and hence

\left|{ \atop ā}\right| − |b̄| ≤ |ā − b̄|

and by reversing the roles of ā and b̄ we also have

− (\left|{ \atop ā}\right| − |b̄|) = |b̄| − \left|{ \atop ā}\right| ≤ |b̄ − ā| = |ā − b̄| .

Therefore |\left|{ \atop ā}\right| − |b̄|| ≤ |ā − b̄| and in particular if \left\{{ \atop ā_{n}}\right\}_{n=1}^{∞} ⊂ ℝ converges to ā ∈ ℝ then

|\left|{ \atop ā_{n}}\right| − \left|{ \atop ā}\right|| ≤ |ā_{n} − ā| → 0 as n → ∞.

Definition 3.5. A sequence \left\{{ \atop ā_{n}}\right\}_{n=1}^{∞} ⊂ ℝ converges to ā ∈ ℝ if |ā − ā_{n}| → 0 as n → ∞, i.e. if for all N ∈ ℕ, |ā − ā_{n}| ≤ i (\frac{1}{N} ) for a.a. n. As before if \left\{{ \atop ā_{n}}\right\}_{n=1}^{∞} converges to ā we will write ā_{n} → ā as n → ∞ or ā = lim_{n→∞} ā_{n}.

Remark 3.6. The field i (ℚ) is dense in ℝ in the sense that if ā ∈ ℝ there exists {q_{n}}_{n=1}^{∞} ⊂ ℚ such that i (q_{n}) → ā as n → ∞. Indeed, simply let q_{n} = a_{n} where a represents ā. Since a is a Cauchy sequence, to any N ∈ ℕ there exits M ∈ ℕ such that

−\frac{1}{N} ≤ a_{m} − a_{n} ≤ \frac{1}{N} for all m, n ≥ M

and therefore

−i (\frac{1}{N} ) ≤ i (a_{m}) − ā ≤ i (\frac{1}{N} ) for all m ≥ M.

This shows

|i (q_{m}) − ā| = |i (a_{m}) − ā| ≤ i (\frac{1}{N} ) for all m ≥ M

and since N is arbitrary that i (q_{m}) → ā as m → ∞.

Definition 3.7. A sequence \left\{{ \atop ā_{n}}\right\}_{n=1}^{∞} ⊂ ℝ is Cauchy if |ā_{n} − ā_{m}| → 0 as m, n → ∞. More precisely we require for each N ∈ ℕ that |ā_{m} − ā_{n}| ≤ i (\frac{1}{N} ) for a.a. pairs (m, n) .

Exercise 3.7. The analogues of the results in Exercises 3.1 and 3.2 hold with ℚ replaced by ℝ. (We now say a subset Λ ⊂ ℝ is bounded if there exists M ∈ ℕ such that |λ| ≤ i (M) for all λ ∈ Λ.)

For the purposes of real analysis the most important property of ℝ is that it is “complete.”

Theorem 3.8. The ordered field ℝ is complete, i.e. all Cauchy sequences in ℝ are convergent.

Proof. Suppose that {ā (m)}_{m=1}^{∞} is a Cauchy sequence in ℝ. By Remark 3.6, we may choose q_{m} ∈ ℚ such that

|ā (m) − i (q_{m})| ≤ i (m^{−1}) for all m ∈ ℕ.

Given N ∈ ℕ, choose M ∈ ℕ such that |ā (m) − ā (n)| ≤ i (N^{−1}) for all m, n ≥ M. Then

|i (q_{m}) − i (q_{n})| ≤ |i (q_{m}) − ā (m)| + |ā (m) − ā (n)| + |ā (n) − i (q_{n})|
≤ i (m^{−1}) + i (n^{−1}) + i (N^{−1})

and therefore

|q_{m} − q_{n}| ≤ m^{−1} + n^{−1} + N^{−1} for all m, n ≥ M.

It now follows that q = {q_{m}}_{m=1}^{∞} ∈ C and therefore q represents a point q̄ ∈ ℝ. Using Remark 3.6 and the triangle inequality,

|ā (m) − q̄| ≤ |ā (m) − i (q_{m})| + |i (q_{m}) − q̄|
≤ i (m^{−1}) + |i (q_{m}) − q̄| → 0 as m → ∞

and therefore lim_{m→∞} ā (m) = q̄.

Definition 3.9. A number M ∈ ℝ is an upper bound for a set Λ ⊂ ℝ if λ ≤ M for all λ ∈ Λ and a number m ∈ ℝ is an lower bound for a set Λ ⊂ ℝ if λ ≥ m for all λ ∈ Λ. Upper and lower bounds need not exist. If Λ has upper (lower) bound, Λ is said to be bounded from above (below).

Theorem 3.10. To each non-empty set Λ ⊂ ℝ which is bounded from above (below) there is a unique least upper bound denoted by sup Λ ∈ ℝ (respectively greatest lower bound denoted by inf Λ ∈ ℝ).

Proof. Suppose Λ is bounded from above and for each n ∈ ℕ, let m_{n} ∈ ℤ be the smallest integer such that i (\frac{m_{n}}{2^{n}} ) is an upper bound for Λ. The sequence q_{n} := \frac{m_{n}}{2^{n}} is Cauchy because q_{m} ∈ [q_{n} − 2^{−n}, q_{n}] ∩ ℚ for all m ≥ n, i.e.

|q_{m} − q_{n}| ≤ 2^{− min(m,n)} → 0 as m, n → ∞.

Passing to the limit, n → ∞, in the inequality i (q_{n}) ≥ λ, which is valid for all λ ∈ Λ implies

q̄ = lim_{n→∞} i (q_{n}) ≥ λ for all λ ∈ Λ.

Thus q̄ is an upper bound for Λ. If there were another upper bound M ∈ ℝ for Λ such that M < q̄, it would follow that M ≤ i (q_{n}) < q̄ for some n. But this is a contradiction because {q_{n}}_{n=1}^{∞} is a decreasing sequence, i (q_{n}) ≥ i (q_{m}) for all m ≥ n and therefore i (q_{n}) ≥ q̄ for all n. Therefore q̄ is the unique least upper bound for Λ. The existence of lower bounds is proved analogously.

Proposition 3.11. If {a_{n}}_{n=1}^{∞} ⊂ ℝ is an increasing (decreasing) sequence which is bounded from above (below), then {a_{n}}_{n=1}^{∞} is convergent and

lim_{n→∞} a_{n} = sup {a_{n} : n ∈ ℕ} ( lim_{n→∞} a_{n} = inf {a_{n} : n ∈ ℕ}).

If Λ ⊂ ℝ is a set bounded from above then there exists {λ_{n}} ⊂ Λ such that λ_{n} ↑ M := sup Λ, as n → ∞, i.e. {λ_{n}} is increasing and lim_{n→∞} λ_{n} = M.

Proof. Let M := sup {a_{n} : n ∈ ℕ} , then for each N ∈ ℕ there must exist m ∈ ℕ such that M − i (N^{−1}) < a_{m} ≤ M. Since a_{n} is increasing, it follows that

M − i (N^{−1}) < a_{n} ≤ M for all n ≥ m.

From this we conclude that lim a_{n} exists and lim a_{n} = M. If M = sup Λ, for each n ∈ ℕ we may choose λ_{n} ∈ Λ such that

M − i (n^{−1}) < λ_{n} ≤ M.  (3.2)

By replacing λ_{n} by max {λ_{1}, …, λ_{n}}^{2} if necessary we may assume that λ_{n} is increasing in n. It now follows easily from Eq. (3.2) that lim_{n→∞} λ_{n} = M.

3.1.1 The Decimal Representation of a Real Number

Let α ∈ ℝ or α ∈ ℚ, m, n ∈ ℤ and S := ∑_{k=n}^{m} α^{k}. If α = 1 then ∑_{k=n}^{m} α^{k} = m − n + 1 while for α ≠ 1,

αS − S = α^{m+1} − α^{n}

and solving for S gives the important geometric summation formula,

∑_{k=n}^{m} α^{k} = \frac{α^{m+1} − α^{n}}{α − 1} if α ≠ 1.  (3.3)

Taking α = 10^{−1} in Eq. (3.3) implies

∑_{k=n}^{m} 10^{−k} = \frac{10^{−(m+1)} − 10^{−n}}{10^{−1} − 1} = \frac{1}{10^{n−1}} \frac{1 − 10^{−(m−n)}}{9}

and in particular, for all M ≥ n,

lim_{m→∞} ∑_{k=n}^{m} 10^{−k} = \frac{1}{9 · 10^{n−1}} ≥ ∑_{k=n}^{M} 10^{−k}.

Let 𝔻 denote those sequences α ∈ {0, 1, 2, …, 9}^{ℤ} with the following properties:

1. there exists N ∈ ℕ such that α_{−n} = 0 for all n ≥ N and

2. α_{n} ≠ 0 for some n ∈ ℤ.

2 The notation, max Λ, denotes sup Λ along with the assertion that sup Λ ∈ Λ. Similarly, min Λ = inf Λ along with the assertion that inf Λ ∈ Λ.

Associated to each α ∈ 𝔻 is the sequence a = a (α) defined by

a_{n} := ∑_{k=−∞}^{n} α_{k}10^{−k}.

Since for m > n,

|a_{m} − a_{n}| = | ∑_{k=n+1}^{m} α_{k}10^{−k}| ≤ 9 ∑_{k=n+1}^{m} 10^{−k} ≤ 9\frac{1}{9 · 10^{n}} = \frac{1}{10^{n}} ,

it follows that

|a_{m} − a_{n}| ≤ \frac{1}{10^{min(m,n)}} → 0 as m, n → ∞.

Therefore a = a (α) ∈ C and we may define a map D : {±1} × 𝔻 → ℝ defined by D (ε, α) = ε\overline{a (α)}. As is customary we will denote D (ε, α) = ε\overline{a (α)} as

ε · α_{m} … α_{0}.α_{1}α_{2} … α_{n} …  (3.4)

where m is the largest integer in ℤ such that α_{k} = 0 for all k < m. If m > 0 the expression in Eq. (3.4) should be interpreted as

ε · 0.0 … 0α_{m}α_{m+1} ….

An element α ∈ 𝔻 has a tail of all 9’s starting at N ∈ ℕ if α_{n} = 9 and for all n ≥ N and α_{N−1} ≠ 9. If α has a tail of 9’s starting at N ∈ ℕ, then for n > N,

a_{n} (α) = ∑_{k=−∞}^{N−1} α_{k}10^{−k} + 9 ∑_{k=N}^{n} 10^{−k}

= ∑_{k=−∞}^{N−1} α_{k}10^{−k} + \frac{9}{10^{N−1}} · \frac{1 − 10^{−(n−N)}}{9}

→ ∑_{k=−∞}^{N−1} α_{k}10^{−k} + 10^{−(N−1)} as n → ∞.

If α′ is the digits in the decimal expansion of ∑_{k=−∞}^{N−1} α_{k}10^{−k} + 10^{−(N−1)}, then

α′ ∈ 𝔻′ := {α ∈ 𝔻 : α does not have a tail of all 9’s} .

and we have just shown that D (ε, α) = D (ε, α′) . In particular this implies

D ({±1} × 𝔻′) = D ({±1} × 𝔻) .  (3.5)

Theorem 3.12 (Decimal Representation). The map

D : {±1} × 𝔻′→ ℝ\ {0}

is a bijection.

Proof. Suppose D (ε, α) = D (δ, β) for some (ε, α) and (δ, β) in {±1} × 𝔻. Since D (ε, α) > 0 if ε = 1 and D (ε, α) < 0 if ε = −1 it follows that ε = δ. Let a = a (α) and b = a (β) be the sequences associated to α and β respectively. Suppose that α ≠ β and let j ∈ ℤ be the position where α and β first disagree, i.e. α_{n} = β_{n} for all n < j while α_{j} ≠ β_{j}. For sake of definiteness suppose β_{j} > α_{j}. Then for n > j we have

b_{n} − a_{n} = (β_{j} − α_{j}) 10^{−j} + ∑_{k=j+1}^{n} (β_{k} − α_{k}) 10^{−k}

≥ 10^{−j} − 9 ∑_{k=j+1}^{n} 10^{−k} ≥ 10^{−j} − 9\frac{1}{9 · 10^{j}} = 0.

Therefore b_{n}−a_{n} ≥ 0 for all n and lim (b_{n} − a_{n}) = 0 iff β_{j} = α_{j} +1 and β_{k} = 9 and α_{k} = 0 for all k > j. In summary, D (ε, α) = D (δ, β) with α ≠ β implies either α or β has an infinite tail of nines which shows that D is injective when restricted to {±1} × 𝔻′. To see that D is surjective it suffices to show any b̄ ∈ ℝ with 0 < b̄ < 1 is in the range of D. For each n ∈ ℕ, let a_{n} = .α_{1} … α_{n} with α_{i} ∈ {0, 1, 2, …, 9} such that

i (a_{n}) < b̄ ≤ i (a_{n}) + i (10^{−n}) .  (3.6)

Since a_{n+1} = a_{n} + α_{n+1}10^{−(n+1)} for some α_{n+1} ∈ {0, 1, 2, …, 9} , we see that a_{n+1} = .α_{1} … α_{n}α_{n+1}, i.e. the first n digits in the decimal expansion of a_{n+1} are the same as in the decimal expansion of a_{n}. Hence this defines α_{n} uniquely for all n ≥ 1. By setting α_{n} = 0 when n ≤ 0, we have constructed from b̄ an element α ∈ 𝔻. Because of Eq. (3.6), D (1, α) = b̄.

Notation 3.13 From now on we will identify ℚ with i (ℚ) ⊂ ℝ and elements in ℝ with their decimal expansions.

To summarize, we have constructed a complete ordered field ℝ “containing” ℚ as a dense subset. Moreover every element in ℝ (modulo those of the form m10^{−n} for some m ∈ ℤ and n ∈ ℕ) has a unique decimal expansion.

Corollary 3.14. The set (0, 1) := {a ∈ ℝ : 0 < a < 1} is uncountable while ℚ ∩ (0, 1) is countable.

Proof. By Theorem 3.12, the set {0, 1, 2 …, 8}^{ℕ} can be mapped injectively into (0, 1) and therefore it follows from Lemma 2.6 that (0, 1) is uncountable. For each m ∈ ℕ, let A_{m} := \{\frac{n}{m} : n ∈ ℕ with n < m\} . Since ℚ ∩ (0, 1) = ∪_{m=1}^{∞}A_{m} and # (A_{m}) < ∞ for all m, another application of Lemma 2.6 shows ℚ ∩ (0, 1) is countable.

3.2 The Complex Numbers

Definition 3.15 (Complex Numbers). Let ℂ = ℝ^{2} equipped with multiplication rule

(a, b)(c, d) := (ac − bd, bc + ad)  (3.7)

and the usual rule for vector addition. As is standard we will write 0 = (0, 0) , 1 = (1, 0) and i = (0, 1) so that every element z of ℂ may be written as z = x1 + yi which in the future will be written simply as z = x + iy. If z = x + iy, let Re z = x and Im z = y.

Writing z = a + ib and w = c + id, the multiplication rule in Eq. (3.7) becomes

(a + ib)(c + id) := (ac − bd) + i(bc + ad)  (3.8)

and in particular 1^{2} = 1 and i^{2} = −1.

Proposition 3.16. The complex numbers ℂ with the above multiplication rule satisfies the usual definitions of a field. For example wz = zw and z (w_{1} + w_{2}) = zw_{1} + zw_{2}, etc. Moreover if z ≠ 0, z has a multiplicative inverse given by

z^{−1} = \frac{a}{a^{2} + b^{2}} − i\frac{b}{a^{2} + b^{2}} .  (3.9)

Proof. The proof is a straightforward verification. Only the last assertion will be verified here. Suppose z = a + ib ≠ 0, we wish to find w = c + id such that zw = 1 and this happens by Eq. (3.8) iff

ac − bd = 1 and  (3.10)
bc + ad = 0.  (3.11)

Solving these equations for c and d gives c = \frac{a}{a^{2}+b^{2}} and d = −\frac{b}{a^{2}+b^{2}} as claimed.

Notation 3.17 (Conjugation and Modulus) If z = a + ib with a, b ∈ ℝ let z̄ = a − ib and

|z| := \sqrt{zz̄} = \sqrt{a^{2} + b^{2}} = \sqrt{|Re z|^{2} + |Im z|^{2}}.

See Exercise 3.8 for the existence of the square root as a positive real number.

Notice that

Re z = \frac{1}{2} (z + z̄) and Im z = \frac{1}{2i} (z − z̄) .  (3.12)

Proposition 3.18. Complex conjugation and the modulus operators satisfy the following properties.

1. \overline{z̄} = z, 2. \overline{zw} = z̄w̄ and z̄ + w̄ = \overline{z + w}.

3. \left|{ \atop z̄}\right| = |z|

4. |zw| = |z| |w| and in particular |z^{n}| = |z|^{n} for all n ∈ ℕ.

5. |Re z| ≤ |z| and |Im z| ≤ |z|

6. |z + w| ≤ |z| + |w| .

7. z = 0 iff |z| = 0.

8. If z ≠ 0 then z^{−1} := \frac{z̄}{|z|^{2}} (also written as \frac{1}{z}) is the inverse of z.

9. |z^{−1}| = |z|^{−1} and more generally |z^{n}| = |z|^{n} for all n ∈ ℤ.

Proof. All of these properties are direct computations except for possibly the triangle inequality in item 6 which is verified by the following computation;

|z + w|^{2} = (z + w) (\overline{z + w}) = |z|^{2} + |w|^{2} + wz̄ + w̄z
= |z|^{2} + |w|^{2} + wz̄ + \overline{wz̄}
= |z|^{2} + |w|^{2} + 2 Re (wz̄) ≤ |z|^{2} + |w|^{2} + 2 |z| |w|
= (|z| + |w|)^{2} .

Definition 3.19. A sequence {z_{n}}_{n=1}^{∞} ⊂ ℂ is Cauchy if |z_{n} − z_{m}| → 0 as m, n → ∞ and is convergent to z ∈ ℂ if |z − z_{n}| → 0 as n → ∞. As usual if {z_{n}}_{n=1}^{∞} converges to z we will write z_{n} → z as n → ∞ or z = lim_{n→∞} z_{n}.

Theorem 3.20. The complex numbers are complete,i.e. all Cauchy sequences are convergent.

Proof. This follows from the completeness of real numbers and the easily proved observations that if z_{n} = a_{n} + ib_{n} ∈ ℂ, then

1. {z_{n}}_{n=1}^{∞} ⊂ ℂ is Cauchy iff {a_{n}}_{n=1}^{∞} ⊂ ℝ and {b_{n}}_{n=1}^{∞} ⊂ ℝ are Cauchy and

2. z_{n} → z = a + ib as n → ∞ iff a_{n} → a and b_{n} → b as n → ∞.

3.3 Exercises

Exercise 3.8. Show to every a ∈ ℝ with a ≥ 0 there exists a unique number b ∈ ℝ such that b ≥ 0 and b^{2} = a. Of course we will call b = \sqrt{a}. Also show that a → \sqrt{a} is an increasing function on [0, ∞). Hint: To construct b = \sqrt{a} for a > 0, to each n ∈ ℕ let m_{n} ∈ ℕ_{0} be chosen so that

\frac{m_{n}^{2}}{n^{2}} < a ≤ \frac{(m_{n} + 1)^{2}}{n^{2}} i.e. i (\frac{m_{n}^{2}}{n^{2}} ) < a ≤ i (\frac{(m_{n} + 1)^{2}}{n^{2}} )

and let q_{n} := \frac{m_{n}}{n} . Then show b = \overline{\{q_{n}\}_{n=1}^{∞}} ∈ ℝ satisfies b > 0 and b^{2} = a.

Limits and Sums

4.1 Limsups, Liminfs and Extended Limits

Notation 4.1 The extended real numbers is the set ℝ̄ := ℝ∪ {±∞} , i.e. it is ℝ with two new points called ∞ and −∞. We use the following conventions, ±∞ · 0 = 0, ±∞ + a = ±∞ for any a ∈ ℝ, ∞ + ∞ = ∞ and −∞ − ∞ = −∞ while ∞ − ∞ is not defined. A sequence a_{n} ∈ ℝ̄ is said to converge to ∞ (−∞) if for all M ∈ ℝ there exists m ∈ ℕ such that a_{n} ≥ M (a_{n} ≤ M) for all n ≥ m.

Lemma 4.2. Suppose {a_{n}}_{n=1}^{∞} and {b_{n}}_{n=1}^{∞} are convergent sequences in ℝ̄, then:

1. If a_{n} ≤ b_{n} for a.a. n then lim_{n→∞} a_{n} ≤ lim_{n→∞} b_{n}.

2. If c ∈ ℝ̄, lim_{n→∞} (ca_{n}) = c lim_{n→∞} a_{n}.

3. If {a_{n} + b_{n}}_{n=1}^{∞} is convergent and

lim_{n→∞} (a_{n} + b_{n}) = lim_{n→∞} a_{n} + lim_{n→∞} b_{n}  (4.1)

provided the right side is not of the form ∞ − ∞.

4. {a_{n}b_{n}}_{n=1}^{∞} is convergent and

lim_{n→∞} (a_{n}b_{n}) = lim_{n→∞} a_{n} · lim_{n→∞} b_{n}  (4.2)

provided the right hand side is not of the for ∞ · 0.

Before going to the proof consider the simple example where a_{n} = n and b_{n} = −αn with α > 0. Then

lim (a_{n} + b_{n}) = \{ \begin{matrix}∞ & if & α < 1 \\ 0 & if & α = 1 \\ −∞ & if & α > 1 \end{matrix}

while

lim_{n→∞} a_{n} + lim_{n→∞} b_{n}“ = ”∞ − ∞.

This shows that the requirement that the right side of Eq. (4.1) is not of form ∞ − ∞ is necessary in Lemma 4.2. Similarly by considering the examples a_{n} = n and b_{n} = n^{−α} with α > 0 shows the necessity for assuming right hand side of Eq. (4.2) is not of the form ∞ · 0.

Proof. The proofs of items 1. and 2. are left to the reader.

Proof of Eq. (4.1). Let a := lim_{n→∞} a_{n} and b = lim_{n→∞} b_{n}. Case 1., suppose b = ∞ in which case we must assume a > −∞. In this case, for every M > 0, there exists N such that b_{n} ≥ M and a_{n} ≥ a − 1 for all n ≥ N and this implies

a_{n} + b_{n} ≥ M + a − 1 for all n ≥ N.

Since M is arbitrary it follows that a_{n} + b_{n} → ∞ as n → b = ∞. The cases where b = −∞ or a = ±∞ are handled similarly. Case 2. If a, b ∈ ℝ, then for every ε > 0 there exists N ∈ ℕ such that

|a − a_{n}| ≤ ε and |b − b_{n}| ≤ ε for all n ≥ N.

Therefore,

|a + b − (a_{n} + b_{n})| = |a − a_{n} + b − b_{n}| ≤ |a − a_{n}| + |b − b_{n}| ≤ 2ε

for all n ≥ N. Since n is arbitrary, it follows that lim_{n→∞} (a_{n} + b_{n}) = a + b.

Proof of Eq. (4.2). It will be left to the reader to prove the case where lim a_{n} and lim b_{n} exist in ℝ. I will only consider the case where a = lim_{n→∞} a_{n} ≠ 0 and lim_{n→∞} b_{n} = ∞ here. Let us also suppose that a > 0 (the case a < 0 is handled similarly) and let α := min (\frac{a}{2}, 1) . Given any M < ∞, there exists N ∈ ℕ such that a_{n} ≥ α and b_{n} ≥ M for all n ≥ N and for this choice of N, a_{n}b_{n} ≥ Mα for all n ≥ N. Since α > 0 is fixed and M is arbitrary it follows that lim_{n→∞} (a_{n}b_{n}) = ∞ as desired.

For any subset Λ ⊂ ℝ̄, let sup Λ and inf Λ denote the least upper bound and greatest lower bound of Λ respectively. The convention being that sup Λ = ∞ if ∞ ∈ Λ or Λ is not bounded from above and inf Λ = −∞ if −∞ ∈ Λ or Λ is not bounded from below. We will also use the conventions that sup ∅ = −∞ and inf ∅ = +∞.

Notation 4.3 Suppose that {x_{n}}_{n=1}^{∞} ⊂ ℝ̄ is a sequence of numbers. Then

lim inf_{n→∞} x_{n} = lim_{n→∞} inf{x_{k} : k ≥ n} and  (4.3)
lim sup_{n→∞} x_{n} = lim_{n→∞} sup{x_{k} : k ≥ n}.  (4.4)

We will also write \underline{lim} for lim inf and \overline{lim} for lim sup .

Remark 4.4. Notice that if a_{k} := inf{x_{k} : k ≥ n} and b_{k} := sup{x_{k} : k ≥ n}, then {a_{k}} is an increasing sequence while {b_{k}} is a decreasing sequence. Therefore the limits in Eq. (4.3) and Eq. (4.4) always exist in ℝ̄ and

lim inf_{n→∞} x_{n} = sup_{n} inf{x_{k} : k ≥ n} and
lim sup_{n→∞} x_{n} = inf_{n} sup{x_{k} : k ≥ n}.

The following proposition contains some basic properties of liminfs and limsups.

Proposition 4.5. Let {a_{n}}_{n=1}^{∞} and {b_{n}}_{n=1}^{∞} be two sequences of real numbers. Then

1. lim inf_{n→∞} a_{n} ≤ lim sup_{n→∞} a_{n} and lim_{n→∞} a_{n} exists in ℝ̄ iff

lim inf_{n→∞} a_{n} = lim sup_{n→∞} a_{n} ∈ ℝ̄.

2. There is a subsequence {a_{n_{k}}}_{k=1}^{∞} of {a_{n}}_{n=1}^{∞} such that lim_{k→∞} a_{n_{k}} = lim sup_{n→∞} a_{n}.

3.

lim sup_{n→∞}(a_{n} + b_{n}) ≤ lim sup_{n→∞} a_{n} + lim sup_{n→∞} b_{n}  (4.5)

whenever the right side of this equation is not of the form ∞ − ∞.

4. If a_{n} ≥ 0 and b_{n} ≥ 0 for all n ∈ ℕ, then

lim sup_{n→∞}(a_{n}b_{n}) ≤ lim sup_{n→∞} a_{n} · lim sup_{n→∞} b_{n},  (4.6)

provided the right hand side of (4.6) is not of the form 0 · ∞ or ∞ · 0.

Proof. Item 1. will be proved here leaving the remaining items as an exercise to the reader. Since

inf{a_{k} : k ≥ n} ≤ sup{a_{k} : k ≥ n} ∀ n,

lim inf_{n→∞} a_{n} ≤ lim sup_{n→∞} a_{n}.

Now suppose that lim inf_{n→∞} a_{n} = lim sup_{n→∞} a_{n} = a ∈ ℝ. Then for all ε > 0, there is an integer N such that

a − ε ≤ inf{a_{k} : k ≥ N} ≤ sup{a_{k} : k ≥ N} ≤ a + ε,

i.e.

a − ε ≤ a_{k} ≤ a + ε for all k ≥ N.

Hence by the definition of the limit, lim_{k→∞} a_{k} = a. If lim inf_{n→∞} a_{n} = ∞, then we know for all M ∈ (0, ∞) there is an integer N such that

M ≤ inf{a_{k} : k ≥ N}

and hence lim_{n→∞} a_{n} = ∞. The case where lim sup_{n→∞} a_{n} = −∞ is handled similarly.

Conversely, suppose that lim_{n→∞} a_{n} = A ∈ ℝ̄ exists. If A ∈ ℝ, then for every ε > 0 there exists N(ε) ∈ ℕ such that |A − a_{n}| ≤ ε for all n ≥ N(ε), i.e.

A − ε ≤ a_{n} ≤ A + ε for all n ≥ N(ε).

From this we learn that

A − ε ≤ lim inf_{n→∞} a_{n} ≤ lim sup_{n→∞} a_{n} ≤ A + ε.

Since ε > 0 is arbitrary, it follows that

A ≤ lim inf_{n→∞} a_{n} ≤ lim sup_{n→∞} a_{n} ≤ A,

i.e. that A = lim inf_{n→∞} a_{n} = lim sup_{n→∞} a_{n}. If A = ∞, then for all M > 0 there exists N(M) such that a_{n} ≥ M for all n ≥ N(M). This show that lim inf_{n→∞} a_{n} ≥ M and since M is arbitrary it follows that

∞ ≤ lim inf_{n→∞} a_{n} ≤ lim sup_{n→∞} a_{n}.

The proof for the case A = −∞ is analogous to the A = ∞ case.

4.2 Sums of positive functions

In this and the next few sections, let X and Y be two sets. We will write α ⊂⊂ X to denote that α is a finite subset of X and write 2_{f}^{X} for those α ⊂⊂ X.

Definition 4.6. Suppose that a : X → [0, ∞] is a function and F ⊂ X is a subset, then

∑_{F} a = ∑_{x∈F} a(x) := sup {∑_{x∈α} a(x) : α ⊂⊂ F } .

Remark 4.7. Suppose that X = ℕ = {1, 2, 3, … } and a : X → [0, ∞], then

∑_{ℕ} a = ∑_{n=1}^{∞} a(n) := lim_{N→∞} ∑_{n=1}^{N} a(n).

Indeed for all N, ∑_{n=1}^{N} a(n) ≤ ∑_{ℕ} a, and thus passing to the limit we learn that

∑_{n=1}^{∞} a(n) ≤ ∑_{ℕ} a.

Conversely, if α ⊂⊂ ℕ, then for all N large enough so that α ⊂ {1, 2, …, N}, we have ∑_{α} a ≤ ∑_{n=1}^{N} a(n) which upon passing to the limit implies that

∑_{α} a ≤ ∑_{n=1}^{∞} a(n).

Taking the supremum over α in the previous equation shows

∑_{ℕ} a ≤ ∑_{n=1}^{∞} a(n).

Remark 4.8. Suppose a : X → [0, ∞] and ∑_{X} a < ∞, then {x ∈ X : a(x) > 0} is at most countable. To see this first notice that for any ε > 0, the set {x : a(x) ≥ ε} must be finite for otherwise ∑_{X} a = ∞. Thus

{x ∈ X : a(x) > 0} = ⋃ _{k=1}^{∞}{x : a(x) ≥ 1/k}

which shows that {x ∈ X : a(x) > 0} is a countable union of finite sets and thus countable by Lemma 2.6.

Lemma 4.9. Suppose that a, b : X → [0, ∞] are two functions, then

∑_{X}(a + b) = ∑_{X} a + ∑_{X} b and
∑_{X} λa = λ ∑_{X} a

for all λ ≥ 0.

I will only prove the first assertion, the second being easy. Let α ⊂⊂ X be a finite set, then

∑_{α}(a + b) = ∑_{α} a + ∑_{α} b ≤ ∑_{X} a + ∑_{X} b

which after taking sups over α shows that

∑_{X}(a + b) ≤ ∑_{X} a + ∑_{X} b.

Similarly, if α, β ⊂⊂ X, then

∑_{α} a + ∑_{β} b ≤ ∑_{α∪β} a + ∑_{α∪β} b = ∑_{α∪β}(a + b) ≤ ∑_{X}(a + b).

Taking sups over α and β then shows that

∑_{X} a + ∑_{X} b ≤ ∑_{X}(a + b).

Lemma 4.10. Let X and Y be sets, R ⊂ X × Y and suppose that a : R → ℝ̄ is a function. Let _{x}R := {y ∈ Y : (x, y) ∈ R} and R_{y} := {x ∈ X : (x, y) ∈ R} . Then

sup_{(x,y)∈R} a(x, y) = sup_{x∈X} sup_{y∈_{x}R} a(x, y) = sup_{y∈Y} sup_{x∈R_{y}} a(x, y) and
inf_{(x,y)∈R} a(x, y) = inf_{x∈X} inf_{y∈_{x}R} a(x, y) = inf_{y∈Y} inf_{x∈R_{y}} a(x, y).

(Recall the conventions: sup ∅ = −∞ and inf ∅ = +∞.)

Proof. Let M = sup_{(x,y)∈R} a(x, y), N_{x} := sup_{y∈_{x}R} a(x, y). Then a(x, y) ≤ M for all (x, y) ∈ R implies N_{x} = sup_{y∈_{x}R} a(x, y) ≤ M and therefore that

sup_{x∈X} sup_{y∈_{x}R} a(x, y) = sup_{x∈X} N_{x} ≤ M.  (4.7)

Similarly for any (x, y) ∈ R,

a(x, y) ≤ N_{x} ≤ sup_{x∈X} N_{x} = sup_{x∈X} sup_{y∈_{x}R} a(x, y)

and therefore

M = sup_{(x,y)∈R} a(x, y) ≤ sup_{x∈X} sup_{y∈_{x}R} a(x, y)  (4.8)

Equations (4.7) and (4.8) show that

sup_{(x,y)∈R} a(x, y) = sup_{x∈X} sup_{y∈_{x}R} a(x, y).

The assertions involving infimums are proved analogously or follow from what we have just proved applied to the function −a.

Fig. 4.1. The x and y – slices of a set R ⊂ X × Y.

Theorem 4.11 (Monotone Convergence Theorem for Sums). Suppose that f_{n} : X → [0, ∞] is an increasing sequence of functions and

f(x) := lim_{n→∞} f_{n}(x) = sup_{n} f_{n}(x).

Then

lim_{n→∞} ∑_{X} f_{n} = ∑_{X} f

Proof. We will give two proves.

First proof. Let

2_{f}^{X} := {A ⊂ X : A ⊂⊂ X}.

Then

lim_{n→∞} ∑_{X} f_{n} = sup_{n} ∑_{X} f_{n} = sup_{n} sup_{α∈2_{f}^{X}} ∑_{α} f_{n} = sup_{α∈2_{f}^{X}} sup_{n} ∑_{α} f_{n}

= sup_{α∈2_{f}^{X}} lim_{n→∞} ∑_{α} f_{n} = sup_{α∈2_{f}^{X}} ∑_{α} lim_{n→∞} f_{n}

= sup_{α∈2_{f}^{X}} ∑_{α} f = ∑_{X} f.

Second Proof. Let S_{n} = ∑_{X} f_{n} and S = ∑_{X} f. Since f_{n} ≤ f_{m} ≤ f for all n ≤ m, it follows that

S_{n} ≤ S_{m} ≤ S

which shows that lim_{n→∞} S_{n} exists and is less that S, i.e.

A := lim_{n→∞} ∑_{X} f_{n} ≤ ∑_{X} f.  (4.9)

Noting that ∑_{α} f_{n} ≤ ∑_{X} f_{n} = S_{n} ≤ A for all α ⊂⊂ X and in particular,

∑_{α} f_{n} ≤ A for all n and α ⊂⊂ X.

Letting n tend to infinity in this equation shows that

∑_{α} f ≤ A for all α ⊂⊂ X

and then taking the sup over all α ⊂⊂ X gives

∑_{X} f ≤ A = lim_{n→∞} ∑_{X} f_{n}  (4.10)

which combined with Eq. (4.9) proves the theorem.

Lemma 4.12 (Fatou’s Lemma for Sums). Suppose that f_{n} : X → [0, ∞] is a sequence of functions, then

∑_{X} lim inf_{n→∞} f_{n} ≤ lim inf_{n→∞} ∑_{X} f_{n}.

Proof. Define g_{k} := inf_{n≥k} f_{n} so that g_{k} ↑ lim inf_{n→∞} f_{n} as k → ∞. Since

g_{k} ≤ f_{n} for all k ≤ n,

∑_{X} g_{k} ≤ ∑_{X} f_{n} for all n ≥ k

and therefore

∑_{X} g_{k} ≤ lim inf_{n→∞} ∑_{X} f_{n} for all k.

We may now use the monotone convergence theorem to let k → ∞ to find

∑_{X} lim inf_{n→∞} f_{n} = ∑_{X} lim_{k→∞} g_{k} =^{MCT} lim_{k→∞} ∑_{X} g_{k} ≤ lim inf_{n→∞} ∑_{X} f_{n}.

Remark 4.13. If A = ∑_{X} a < ∞, then for all ε > 0 there exists α_{ε} ⊂⊂ X such that

A ≥ ∑_{α} a ≥ A − ε

for all α ⊂⊂ X containing α_{ε} or equivalently,

|A − ∑_{α} a| ≤ ε  (4.11)

for all α ⊂⊂ X containing α_{ε}. Indeed, choose α_{ε} so that ∑_{α_{ε}} a ≥ A − ε.

4.3 Sums of complex functions

Definition 4.14. Suppose that a : X → ℂ is a function, we say that

∑_{X} a = ∑_{x∈X} a(x)

exists and is equal to A ∈ ℂ, if for all ε > 0 there is a finite subset α_{ε} ⊂ X such that for all α ⊂⊂ X containing α_{ε} we have

|A − ∑_{α} a| ≤ ε.

The following lemma is left as an exercise to the reader.

Lemma 4.15. Suppose that a, b : X → ℂ are two functions such that ∑_{X} a and ∑_{X} b exist, then ∑_{X}(a + λb) exists for all λ ∈ ℂ and

∑_{X}(a + λb) = ∑_{X} a + λ ∑_{X} b.

Definition 4.16 (Summable). We call a function a : X → ℂ summable if

∑_{X} |a| < ∞.

Proposition 4.17. Let a : X → ℂ be a function, then ∑_{X} a exists iff ∑_{X} |a| < ∞, i.e. iff a is summable. Moreover if a is summable, then

|∑_{X} a| ≤ ∑_{X} |a| .

Proof. If ∑_{X} |a| < ∞, then ∑_{X} (Re a)^{±} < ∞ and ∑_{X} (Im a)^{±} < ∞ and hence by Remark 4.13 these sums exists in the sense of Definition 4.14. Therefore by Lemma 4.15, ∑_{X} a exists and

∑_{X} a = ∑_{X} (Re a)^{+} − ∑_{X} (Re a)^{−} + i (∑_{X} (Im a)^{+} − ∑_{X} (Im a)^{−}) .

Conversely, if ∑_{X} |a| = ∞ then, because |a| ≤ |Re a| + |Im a| , we must have

∑_{X} |Re a| = ∞ or ∑_{X} |Im a| = ∞.

Thus it suffices to consider the case where a : X → ℝ is a real function. Write a = a^{+} − a^{−} where

a^{+}(x) = max(a(x), 0) and a^{−}(x) = max(−a(x), 0).  (4.12)

Then |a| = a^{+} + a^{−} and

∞ = ∑_{X} |a| = ∑_{X} a^{+} + ∑_{X} a^{−}

which shows that either ∑_{X} a^{+} = ∞ or ∑_{X} a^{−} = ∞. Suppose, with out loss of generality, that ∑_{X} a^{+} = ∞. Let X′ := {x ∈ X : a(x) ≥ 0}, then we know that ∑_{X′} a = ∞ which means there are finite subsets α_{n} ⊂ X′ ⊂ X such that ∑_{α_{n}} a ≥ n for all n. Thus if α ⊂⊂ X is any finite set, it follows that lim_{n→∞} ∑_{α_{n}∪α} a = ∞, and therefore ∑_{X} a can not exist as a number in ℝ. Finally if a is summable, write ∑_{X} a = ρe^{iθ} with ρ ≥ 0 and θ ∈ ℝ, then

|∑_{X} a| = ρ = e^{−iθ} ∑_{X} a = ∑_{X} e^{−iθ}a

= ∑_{X} Re [e^{−iθ}a] ≤ ∑_{X} (Re [e^{−iθ}a])^{+}

≤ ∑_{X} |Re [e^{−iθ}a]| ≤ ∑_{X} |e^{−iθ}a| ≤ ∑_{X} |a| .

Alternatively, this may be proved by approximating ∑_{X} a by a finite sum and then using the triangle inequality of |·| .

Remark 4.18. Suppose that X = ℕ and a : ℕ → ℂ is a sequence, then it is not necessarily true that

∑_{n=1}^{∞} a(n) = ∑_{n∈ℕ} a(n).  (4.13)

This is because

∑_{n=1}^{∞} a(n) = lim_{N→∞} ∑_{n=1}^{N} a(n)

depends on the ordering of the sequence a where as ∑_{n∈ℕ} a(n) does not. For example, take a(n) = (−1)^{n}/n then ∑_{n∈ℕ} |a(n)| = ∞ i.e. ∑_{n∈ℕ} a(n) does not exist while ∑_{n=1}^{∞} a(n) does exist. On the other hand, if

∑_{n∈ℕ} |a(n)| = ∑_{n=1}^{∞} |a(n)| < ∞

then Eq. (4.13) is valid.

Theorem 4.19 (Dominated Convergence Theorem for Sums). Suppose that f_{n} : X → ℂ is a sequence of functions on X such that f(x) = lim_{n→∞} f_{n}(x) ∈ ℂ exists for all x ∈ X. Further assume there is a dominating function g : X → [0, ∞) such that

|f_{n}(x)| ≤ g(x) for all x ∈ X and n ∈ ℕ  (4.14)

and that g is summable. Then

lim_{n→∞} ∑_{x∈X} f_{n}(x) = ∑_{x∈X} f(x).  (4.15)

Proof. Notice that |f| = lim |f_{n}| ≤ g so that f is summable. By considering the real and imaginary parts of f separately, it suffices to prove the theorem in the case where f is real. By Fatou’s Lemma,

4.3 Sums of complex functions 31

∑_{X}(g ± f) = ∑_{X} lim inf_{n→∞} (g ± f_{n}) ≤ lim inf_{n→∞} ∑_{X} (g ± f_{n})

= ∑_{X} g + lim inf_{n→∞} (± ∑_{X} f_{n}) .

Since lim inf_{n→∞}(−a_{n}) = − lim sup_{n→∞} a_{n}, we have shown,

∑_{X} g ± ∑_{X} f ≤ ∑_{X} g + \{ \begin{matrix}lim inf_{n→∞} ∑_{X} f_{n} \\ − lim sup_{n→∞} ∑_{X} f_{n} \end{matrix}

and therefore

lim sup_{n→∞} ∑_{X} f_{n} ≤ ∑_{X} f ≤ lim inf_{n→∞} ∑_{X} f_{n}.

This shows that lim_{n→∞} ∑_{X} f_{n}exists and is equal to ∑_{X} f.

Proof. (Second Proof.) Passing to the limit in Eq. (4.14) shows that |f| ≤ g and in particular that f is summable. Given ε > 0, let α ⊂⊂ X such that

∑_{X\α} g ≤ ε.

Then for β ⊂⊂ X such that α ⊂ β,

|∑_{β} f − ∑_{β} f_{n}| = |∑_{β} (f − f_{n})|

≤ ∑_{β} |f − f_{n}| = ∑_{α} |f − f_{n}| + ∑_{β\α} |f − f_{n}|

≤ ∑_{α} |f − f_{n}| + 2 ∑_{β\α} g

≤ ∑_{α} |f − f_{n}| + 2ε.

and hence that

|∑_{β} f − ∑_{β} f_{n}| ≤ ∑_{α} |f − f_{n}| + 2ε.

Since this last equation is true for all such β ⊂⊂ X, we learn that

|∑_{X} f − ∑_{X} f_{n}| ≤ ∑_{α} |f − f_{n}| + 2ε

which then implies that

lim sup_{n→∞} |∑_{X} f − ∑_{X} f_{n}| ≤ lim sup_{n→∞} ∑_{α} |f − f_{n}| + 2ε
= 2ε.

Because ε > 0 is arbitrary we conclude that

lim sup_{n→∞} |∑_{X} f − ∑_{X} f_{n}| = 0.

which is the same as Eq. (4.15).

Remark 4.20. Theorem 4.19 may easily be generalized as follows. Suppose f_{n}, g_{n}, g are summable functions on X such that f_{n} → f and g_{n} → g pointwise, |f_{n}| ≤ g_{n} and ∑_{X} g_{n} → ∑_{X} g as n → ∞. Then f is summable and Eq. (4.15) still holds. For the proof we use Fatou’s Lemma to again conclude

∑_{X}(g ± f) = ∑_{X} lim inf_{n→∞} (g_{n} ± f_{n}) ≤ lim inf_{n→∞} ∑_{X} (g_{n} ± f_{n})

= ∑_{X} g + lim inf_{n→∞} (± ∑_{X} f_{n})

and then proceed exactly as in the first proof of Theorem 4.19.

4.4 Iterated sums and the Fubini and Tonelli Theorems

Let X and Y be two sets. The proof of the following lemma is left to the reader.

Lemma 4.21. Suppose that a : X → ℂ is function and F ⊂ X is a subset such that a(x) = 0 for all x ∉ F. Then ∑_{F} a exists iff ∑_{X} a exists and when the sums exists,

∑_{X} a = ∑_{F} a.

Theorem 4.22 (Tonelli’s Theorem for Sums). Suppose that a : X ×Y → [0, ∞], then

∑_{X×Y} a = ∑_{X} ∑_{Y} a = ∑_{Y} ∑_{X} a.

Proof. It suffices to show, by symmetry, that

∑_{X×Y} a = ∑_{X} ∑_{Y} a

Let Λ ⊂⊂ X × Y. The for any α ⊂⊂ X and β ⊂⊂ Y such that Λ ⊂ α × β, we have

∑_{Λ} a ≤ ∑_{α×β} a = ∑_{α} ∑_{β} a ≤ ∑_{α} ∑_{Y} a ≤ ∑_{X} ∑_{Y} a,

i.e. ∑_{Λ} a ≤ ∑_{X} ∑_{Y} a. Taking the sup over Λ in this last equation shows

∑_{X×Y} a ≤ ∑_{X} ∑_{Y} a.

For the reverse inequality, for each x ∈ X choose β_{n}^{x} ⊂⊂ X such that β_{n}^{x} ↑ as n ↑ and

∑_{y∈Y} a(x, y) = lim_{n→∞} ∑_{y∈β_{n}^{x}} a(x, y).

If α ⊂⊂ X is a given finite subset of X, then

∑_{y∈Y} a(x, y) = lim_{n→∞} ∑_{y∈β_{n}} a(x, y) for all x ∈ α

where β_{n} := ∪_{x∈α}β_{n}^{x} ⊂⊂ X. Hence

∑_{x∈α} ∑_{y∈Y} a(x, y) = ∑_{x∈α} lim_{n→∞} ∑_{y∈β_{n}} a(x, y) = lim_{n→∞} ∑_{x∈α} ∑_{y∈β_{n}} a(x, y)

= lim_{n→∞} ∑_{(x,y)∈α×β_{n}} a(x, y) ≤ ∑_{X×Y} a.

Since α is arbitrary, it follows that

∑_{x∈X} ∑_{y∈Y} a(x, y) = sup_{α⊂⊂X} ∑_{x∈α} ∑_{y∈Y} a(x, y) ≤ ∑_{X×Y} a

which completes the proof.

Theorem 4.23 (Fubini’s Theorem for Sums). Now suppose that a : X × Y → ℂ is a summable function, i.e. by Theorem 4.22 any one of the following equivalent conditions hold:

1. ∑_{X×Y} |a| < ∞,

2. ∑_{X} ∑_{Y} |a| < ∞ or

3. ∑_{Y} ∑_{X} |a| < ∞.

Then

∑_{X×Y} a = ∑_{X} ∑_{Y} a = ∑_{Y} ∑_{X} a.

Proof. If a : X → ℝ is real valued the theorem follows by applying Theorem 4.22 to a^{±} – the positive and negative parts of a. The general result holds for complex valued functions a by applying the real version just proved to the real and imaginary parts of a.

34 4 Limits and Sums

4.5 Exercises

Exercise 4.1. Now suppose for each n ∈ ℕ := {1, 2, …} that f_{n} : X → ℝ is a function. Let

D := {x ∈ X : lim_{n→∞} f_{n}(x) = +∞}

show that

D = ∩_{M=1}^{∞} ∪_{N=1}^{∞} ∩_{n≥N}{x ∈ X : f_{n}(x) ≥ M}.  (4.16)

Exercise 4.2. Let f_{n} : X → ℝ be as in the last problem. Let

C := {x ∈ X : lim_{n→∞} f_{n}(x) exists in ℝ}.

Find an expression for C similar to the expression for D in (4.16). (Hint: use the Cauchy criteria for convergence.)

4.5.1 Limit Problems Exercise 4.3. Show lim inf_{n→∞}(−a_{n}) = − lim sup_{n→∞} a_{n}. Exercise 4.4. Suppose that lim sup_{n→∞} a_{n} = M ∈ ℝ̄, show that there is a subsequence {a_{n_{k}}}_{k=1}^{∞} of {a_{n}}_{n=1}^{∞} such that lim_{k→∞} a_{n_{k}} = M. Exercise 4.5. Show that

lim sup_{n→∞}(a_{n} + b_{n}) ≤ lim sup_{n→∞} a_{n} + lim sup_{n→∞} b_{n}  (4.17)

provided that the right side of Eq. (4.17) is well defined, i.e. no ∞ − ∞ or −∞ + ∞ type expressions. (It is OK to have ∞ + ∞ = ∞ or −∞ − ∞ = −∞, etc.) Exercise 4.6. Suppose that a_{n} ≥ 0 and b_{n} ≥ 0 for all n ∈ ℕ. Show

lim sup_{n→∞}(a_{n}b_{n}) ≤ lim sup_{n→∞} a_{n} · lim sup_{n→∞} b_{n},  (4.18)

provided the right hand side of (4.18) is not of the form 0 · ∞ or ∞ · 0. Exercise 4.7. Prove Lemma 4.15. Exercise 4.8. Prove Lemma 4.21.

Let {a_{n}}_{n=1}^{∞} and {b_{n}}_{n=1}^{∞} be two sequences of real numbers.

4.5.2 Dominated Convergence Theorem Problems

Notation 4.24 For u_{0} ∈ ℝ^{n} and δ > 0, let B_{u_{0}}(δ) := {x ∈ ℝ^{n} : |x − u_{0}| < δ} be the ball in ℝ^{n} centered at u_{0} with radius δ.

Exercise 4.9. Suppose U ⊂ ℝ^{n} is a set and u_{0} ∈ U is a point such that U ∩ (B_{u_{0}}(δ) \ {u_{0}}) ≠ ∅ for all δ > 0. Let G : U \ {u_{0}} → ℂ be a function on U \ {u_{0}}. Show that lim_{u→u_{0}} G(u) exists and is equal to λ ∈ ℂ,^{1} iff for all sequences {u_{n}}_{n=1}^{∞} ⊂ U \ {u_{0}} which converge to u_{0} (i.e. lim_{n→∞} u_{n} = u_{0}) we have lim_{n→∞} G(u_{n}) = λ.

Exercise 4.10. Suppose that Y is a set, U ⊂ ℝ^{n} is a set, and f : U × Y → ℂ is a function satisfying:

1. For each y ∈ Y, the function u ∈ U → f(u, y) is continuous on U.^{2}

2. There is a summable function g : Y → [0, ∞) such that

|f(u, y)| ≤ g(y) for all y ∈ Y and u ∈ U.

Show that

F (u) := ∑_{y∈Y} f(u, y)  (4.19)

is a continuous function for u ∈ U.

Exercise 4.11. Suppose that Y is a set, J = (a, b) ⊂ ℝ is an interval, and f : J × Y → ℂ is a function satisfying:

1. For each y ∈ Y, the function u → f(u, y) is differentiable on J,

2. There is a summable function g : Y → [0, ∞) such that

|\frac{∂}{∂u}f(u, y)| ≤ g(y) for all y ∈ Y and u ∈ J.

3. There is a u_{0} ∈ J such that ∑_{y∈Y} |f(u_{0}, y)| < ∞.

Show:

a) for all u ∈ J that ∑_{y∈Y} |f(u, y)| < ∞.

1 More explicitly, lim_{u→u_{0}} G(u) = λ means for every every ϵ > 0 there exists a δ > 0 such that

|G(u) − λ| < ϵ whenerver u ∈ U ∩ (B_{u_{0}}(δ) \ {u_{0}}) .

2 To say g := f(·, y) is continuous on U means that g : U → ℂ is continuous relative to the metric on ℝ^{n} restricted to U.

b) Let F (u) := ∑_{y∈Y} f(u, y), show F is differentiable on J and that

Ḟ (u) = ∑_{y∈Y} \frac{∂}{∂u}f(u, y).

(Hint: Use the mean value theorem.)

Exercise 4.12 (Differentiation of Power Series). Suppose R > 0 and {a_{n}}_{n=0}^{∞} is a sequence of complex numbers such that ∑_{n=0}^{∞} |a_{n}| r^{n} < ∞ for all r ∈ (0, R). Show, using Exercise 4.11, f(x) := ∑_{n=0}^{∞} a_{n}x^{n} is continuously differentiable for x ∈ (−R, R) and

f′(x) = ∑_{n=0}^{∞} na_{n}x^{n−1} = ∑_{n=1}^{∞} na_{n}x^{n−1}.

Exercise 4.13. Show the functions

e^{x} := ∑_{n=0}^{∞} \frac{x^{n}}{n!} ,  (4.20)

sin x := ∑_{n=0}^{∞} (−1)^{n} \frac{x^{2n+1}}{(2n + 1)!} and  (4.21)

cos x = ∑_{n=0}^{∞} (−1)^{n} \frac{x^{2n}}{(2n)!}  (4.22)

are infinitely differentiable and they satisfy

\frac{d}{dx}e^{x} = e^{x} with e^{0} = 1
\frac{d}{dx} sin x = cos x with sin (0) = 0
\frac{d}{dx} cos x = − sin x with cos (0) = 1.

Exercise 4.14. Continue the notation of Exercise 4.13.

1. Use the product and the chain rule to show,

\frac{d}{dx} [e^{−x}e^{(x+y)}] = 0

and conclude from this, that e^{−x}e^{(x+y)} = e^{y} for all x, y ∈ ℝ. In particular taking y = 0 this implies that e^{−x} = 1/e^{x} and hence that e^{(x+y)} = e^{x}e^{y}. Use this result to show e^{x} ↑ ∞ as x ↑ ∞ and e^{x} ↓ 0 as x ↓ −∞. Remark: since e^{x} ≥ ∑_{n=0}^{N} \frac{x^{n}}{n!} when x ≥ 0, it follows that lim_{x→∞} \frac{x^{n}}{e^{x}} = 0 for any n ∈ ℕ, i.e. e^{x} grows at a rate faster than any polynomial in x as x → ∞.

2. Use the product rule to show

\frac{d}{dx} (cos^{2} x + sin^{2} x) = 0

and use this to conclude that cos^{2} x + sin^{2} x = 1 for all x ∈ ℝ.

Exercise 4.15. Let {a_{n}}_{n=−∞}^{∞} be a summable sequence of complex numbers, i.e. ∑_{n=−∞}^{∞} |a_{n}| < ∞. For t ≥ 0 and x ∈ ℝ, define

F (t, x) = ∑_{n=−∞}^{∞} a_{n}e^{−tn^{2}}e^{inx},

where as usual e^{ix} = cos(x) + i sin(x), this is motivated by replacing x in Eq. (4.20) by ix and comparing the result to Eqs. (4.21) and (4.22).

1. F (t, x) is continuous for (t, x) ∈ [0, ∞)×ℝ. Hint: Let Y = ℤ and u = (t, x) and use Exercise 4.10.

2. ∂F (t, x)/∂t, ∂F (t, x)/∂x and ∂^{2}F (t, x)/∂x^{2} exist for t > 0 and x ∈ ℝ. Hint: Let Y = ℤ and u = t for computing ∂F (t, x)/∂t and u = x for computing ∂F (t, x)/∂x and ∂^{2}F (t, x)/∂x^{2}. See Exercise 4.11.

3. F satisfies the heat equation, namely

∂F (t, x)/∂t = ∂^{2}F (t, x)/∂x^{2} for t > 0 and x ∈ ℝ.

ℓ^{p} – spaces, Minkowski and Holder Inequalities

In this chapter, let μ : X → (0, ∞) be a given function. Let 𝔽 denote either ℝ or ℂ. For p ∈ (0, ∞) and f : X → 𝔽, let

∥f∥_{p} := (∑_{x∈X} |f(x)|^{p} μ(x))^{1/p}

and for p = ∞ let

∥f∥_{∞} = sup {|f(x)| : x ∈ X} .

Also, for p > 0, let

ℓ^{p}(μ) = {f : X → 𝔽 : ∥f∥_{p} < ∞}.

In the case where μ(x) = 1 for all x ∈ X we will simply write ℓ^{p}(X) for ℓ^{p}(μ).

Definition 5.1. A norm on a vector space Z is a function ∥·∥ : Z → [0, ∞) such that

1. (Homogeneity) ∥λf∥ = |λ| ∥f∥ for all λ ∈ 𝔽 and f ∈ Z.

2. (Triangle inequality) ∥f + g∥ ≤ ∥f∥ + ∥g∥ for all f, g ∈ Z.

3. (Positive definite) ∥f∥ = 0 implies f = 0.

A pair (Z, ∥·∥) where Z is a vector space and ∥·∥ is a norm on Z is called a normed vector space.

The rest of this section is devoted to the proof of the following theorem.

Theorem 5.2. For p ∈ [1, ∞], (ℓ^{p}(μ), ∥ · ∥_{p}) is a normed vector space.

Proof. The only difficulty is the proof of the triangle inequality which is the content of Minkowski’s Inequality proved in Theorem 5.8 below.

40 5 ℓ^{p} – spaces, Minkowski and Holder Inequalities

Proposition 5.3. Let f : [0, ∞) → [0, ∞) be a continuous strictly increasing function such that f(0) = 0 (for simplicity) and lim_{s→∞} f(s) = ∞. Let g = f^{−1} and for s, t ≥ 0 let

F (s) = ∫_{0}^{s} f(s′)ds′ and G(t) = ∫_{0}^{t} g(t′)dt′.

Then for all s, t ≥ 0,

st ≤ F (s) + G(t)

and equality holds iff t = f(s).

Proof. Let

A_{s} := {(σ, τ) : 0 ≤ τ ≤ f(σ) for 0 ≤ σ ≤ s} and
B_{t} := {(σ, τ) : 0 ≤ σ ≤ g(τ) for 0 ≤ τ ≤ t}

then as one sees from Figure 5.1, [0, s] × [0, t] ⊂ A_{s} ∪ B_{t}. (In the figure: s = 3, t = 1, A_{3} is the region under t = f(s) for 0 ≤ s ≤ 3 and B_{1} is the region to the left of the curve s = g(t) for 0 ≤ t ≤ 1.) Hence if m denotes the area of a region in the plane, then

st = m ([0, s] × [0, t]) ≤ m(A_{s}) + m(B_{t}) = F (s) + G(t).

As it stands, this proof is a bit on the intuitive side. However, it will become rigorous if one takes m to be Lebesgue measure on the plane which will be introduced later. We can also give a calculus proof of this theorem under the additional assumption that f is C^{1}. (This restricted version of the theorem is all we need in this section.) To do this fix t ≥ 0 and let

h(s) = st − F (s) = ∫_{0}^{s}(t − f(σ))dσ.

If σ > g(t) = f^{−1}(t), then t − f(σ) < 0 and hence if s > g(t), we have

h(s) = ∫_{0}^{s}(t − f(σ))dσ = ∫_{0}^{g(t)}(t − f(σ))dσ + ∫_{g(t)}^{s}(t − f(σ))dσ

≤ ∫_{0}^{g(t)}(t − f(σ))dσ = h(g(t)).

Combining this with h(0) = 0 we see that h(s) takes its maximum at some point s ∈ (0, t] and hence at a point where 0 = h′(s) = t − f(s). The only solution to this equation is s = g(t) and we have thus shown

st − F (s) = h(s) ≤ ∫_{0}^{g(t)}(t − f(σ))dσ = h(g(t))

with equality when s = g(t). To finish the proof we must show ∫_{0}^{g(t)}(t − f(σ))dσ = G(t). This is verified by making the change of variables σ = g(τ) and then integrating by parts as follows:

∫_{0}^{g(t)}(t − f(σ))dσ = ∫_{0}^{t}(t − f(g(τ)))g′(τ)dτ = ∫_{0}^{t}(t − τ)g′(τ)dτ
= ∫_{0}^{t} g(τ)dτ = G(t).

Fig. 5.1. A picture proof of Proposition 5.3.

Definition 5.4. The conjugate exponent q ∈ [1, ∞] to p ∈ [1, ∞] is q := \frac{p}{p−1} with the conventions that q = ∞ if p = 1 and q = 1 if p = ∞. Notice that q is characterized by any of the following identities:

\frac{1}{p} + \frac{1}{q} = 1, 1 + \frac{q}{p} = q, p − \frac{p}{q} = 1 and q(p − 1) = p.  (5.1)

Lemma 5.5. Let p ∈ (1, ∞) and q := \frac{p}{p−1} ∈ (1, ∞) be the conjugate exponent. Then

st ≤ \frac{s^{p}}{p} + \frac{t^{q}}{q} for all s, t ≥ 0

with equality if and only if t^{q} = s^{p}.

Proof. Let F (s) = \frac{s^{p}}{p} for p > 1. Then f(s) = s^{p−1} = t and g(t) = t^{\frac{1}{p−1}} = t^{q−1}, wherein we have used q − 1 = p/ (p − 1) − 1 = 1/ (p − 1) . Therefore G(t) = t^{q}/q and hence by Proposition 5.3,

st ≤ \frac{s^{p}}{p} + \frac{t^{q}}{q}

with equality iff t = s^{p−1}, i.e. t^{q} = s^{q(p−1)} = s^{p}. For those who do not want to use Proposition 5.3, here is a direct calculus proof. Fix t > 0 and let

h (s) := st − \frac{s^{p}}{p} .

Then h (0) = 0, lim_{s→∞} h (s) = −∞ and h′ (s) = t − s^{p−1} which equals zero iff s = t^{\frac{1}{p−1}} . Since

h (t^{\frac{1}{p−1}} ) = t^{\frac{1}{p−1}} t − \frac{t^{\frac{p}{p−1}}}{p} = t^{\frac{p}{p−1}} − \frac{t^{\frac{p}{p−1}}}{p} = t^{q} (1 − \frac{1}{p}) = \frac{t^{q}}{q} ,

it follows from the first derivative test that

max h = max \{h (0) , h (t^{\frac{1}{p−1}} )\} = max \{0, \frac{t^{q}}{q} \} = \frac{t^{q}}{q} .

So we have shown

st − \frac{s^{p}}{p} ≤ \frac{t^{q}}{q} with equality iff t = s^{p−1}.

Theorem 5.6 (Hölder’s inequality). Let p, q ∈ [1, ∞] be conjugate exponents. For all f, g : X → 𝔽,

∥fg∥_{1} ≤ ∥f∥_{p} · ∥g∥_{q}.  (5.2)

If p ∈ (1, ∞) and f and g are not identically zero, then equality holds in Eq. (5.2) iff

(\frac{|f|}{∥f∥_{p}} )^{p} = (\frac{|g|}{∥g∥_{q}} )^{q} .  (5.3)

Proof. The proof of Eq. (5.2) for p ∈ {1, ∞} is easy and will be left to the reader. The cases where ∥f∥_{q} = 0 or ∞ or ∥g∥_{p} = 0 or ∞ are easily dealt with and are also left to the reader. So we will assume that p ∈ (1, ∞) and 0 < ∥f∥_{q}, ∥g∥_{p} < ∞. Letting s = |f (x)| /∥f∥_{p} and t = |g|/∥g∥_{q} in Lemma 5.5 implies

\frac{|f (x) g (x)|}{∥f∥_{p}∥g∥_{q}} ≤ \frac{1}{p} \frac{|f (x)|^{p}}{∥f∥_{p}} + \frac{1}{q} \frac{|g (x)|^{q}}{∥g∥^{q}}

with equality iff

\frac{|f (x)|^{p}}{∥f∥_{p}} = s^{p} = t^{q} = \frac{|g (x)|^{q}}{∥g∥^{q}} .  (5.4)

Multiplying this equation by μ (x) and then summing on x gives

\frac{∥fg∥_{1}}{∥f∥_{p}∥g∥_{q}} ≤ \frac{1}{p} + \frac{1}{q} = 1

with equality iff Eq. (5.4) holds for all x ∈ X, i.e. iff Eq. (5.3) holds.

Definition 5.7. For a complex number λ ∈ ℂ, let

sgn(λ) = \{ \begin{matrix}\frac{λ}{|λ|} & if & λ ≠ 0 \\ 0 & if & λ = 0. \end{matrix}

For λ, μ ∈ ℂ we will write sgn(λ) ≗ sgn(μ) if either λμ = 0 or λμ ≠ 0 and sgn(λ) = sgn(μ).

Theorem 5.8 (Minkowski’s Inequality). If 1 ≤ p ≤ ∞ and f, g ∈ ℓ^{p}(μ) then

∥f + g∥_{p} ≤ ∥f∥_{p} + ∥g∥_{p}.  (5.5)

Moreover, assuming f and g are not identically zero, equality holds in Eq. (5.5) iff

sgn(f) ≗ sgn(g) when p = 1 and
f = cg for some c > 0 when p ∈ (1, ∞).

Proof. For p = 1,

∥f + g∥_{1} = ∑_{X} |f + g|μ ≤ ∑_{X} (|f| μ + |g|μ) = ∑_{X} |f| μ + ∑_{X} |g|μ

with equality iff

|f| + |g| = |f + g| ⟺ sgn(f) ≗ sgn(g).

For p = ∞,

∥f + g∥_{∞} = sup_{X} |f + g| ≤ sup_{X} (|f| + |g|)

≤ sup_{X} |f| + sup_{X} |g| = ∥f∥_{∞} + ∥g∥_{∞}.

Now assume that p ∈ (1, ∞). Since

|f + g|^{p} ≤ (2 max (|f| , |g|))^{p} = 2^{p} max (|f|^{p} , |g|^{p}) ≤ 2^{p} (|f|^{p} + |g|^{p})

it follows that

∥f + g∥_{p}^{p} ≤ 2^{p} (∥f∥_{p}^{p} + ∥g∥_{p}^{p}) < ∞.

Eq. (5.5) is easily verified if ∥f + g∥_{p} = 0, so we may assume ∥f + g∥_{p} > 0. Multiplying the inequality,

|f + g|^{p} = |f + g||f + g|^{p−1} ≤ (|f| + |g|)|f + g|^{p−1}  (5.6)

by μ, then summing on x and applying Holder’s inequality two times gives

∑_{X} |f + g|^{p}μ ≤ ∑_{X} |f| |f + g|^{p−1}μ + ∑_{X} |g| |f + g|^{p−1}μ
≤ (∥f∥_{p} + ∥g∥_{p}) ∥ |f + g|^{p−1} ∥_{q}.  (5.7)

Since q(p − 1) = p, as in Eq. (5.1),

∥|f + g|^{p−1}∥_{q}^{q} = ∑_{X}(|f + g|^{p−1})^{q}μ = ∑_{X} |f + g|^{p}μ = ∥f + g∥_{p}^{p}.  (5.8)

Combining Eqs. (5.7) and (5.8) shows

∥f + g∥_{p}^{p} ≤ (∥f∥_{p} + ∥g∥_{p}) ∥f + g∥_{p}^{p/q}  (5.9)

and solving this equation for ∥f + g∥_{p} (making use of Eq. (5.1)) implies Eq. (5.5). Now suppose that f and g are not identically zero and p ∈ (1, ∞) . Equality holds in Eq. (5.5) iff equality holds in Eq. (5.9) iff equality holds in Eq. (5.7) and Eq. (5.6). The latter happens iff

sgn(f) ≗ sgn(g) and
(\frac{|f|}{∥f∥_{p}} )^{p} = \frac{|f + g|^{p}}{∥f + g∥_{p}^{p}} = (\frac{|g|}{∥g∥_{p}} )^{p} .  (5.10)

wherein we have used

(\frac{|f + g|^{p−1}}{∥|f + g|^{p−1}∥_{q}} )^{q} = \frac{|f + g|^{p}}{∥f + g∥_{p}^{p}} .

Finally Eq. (5.10) is equivalent |f| = c |g| with c = (∥f∥_{p}/∥g∥_{p}) > 0 and this equality along with sgn(f) ≗ sgn(g) implies f = cg.

5.1 Exercises

Exercise 5.1. Generalize Proposition 5.3 as follows. Let a ∈ [−∞, 0] and f : ℝ ∩ [a, ∞) → [0, ∞) be a continuous strictly increasing function such that lim_{s→∞} f(s) = ∞, f(a) = 0 if a > −∞ or lim_{s→−∞} f(s) = 0 if a = −∞. Also let

g = f^{−1}, b = f(0) ≥ 0,

F (s) = ∫_{0}^{s} f(s′)ds′ and G(t) = ∫_{0}^{t} g(t′)dt′.

Then for all s, t ≥ 0,

st ≤ F (s) + G(t ∨ b) ≤ F (s) + G(t)

and equality holds iff t = f(s). In particular, taking f(s) = e^{s}, prove Young’s inequality stating

st ≤ e^{s} + (t ∨ 1) ln (t ∨ 1) − (t ∨ 1) ≤ e^{s} + t ln t − t.

Hint: Refer to Figures 5.2 and 5.3..

Fig. 5.2. Comparing areas when t ≥ b goes the same way as in the text. Fig. 5.3. When t ≤ b, notice that g(t) ≤ 0 but G(t) ≥ 0. Also notice that G(t) is no longer needed to estimate st.

Part II

Metric and Banach Space Basics Metric Spaces

Definition 6.1. A function d : X × X → [0, ∞) is called a metric if

1. (Symmetry) d(x, y) = d(y, x) for all x, y ∈ X

2. (Non-degenerate) d(x, y) = 0 if and only if x = y ∈ X

3. (Triangle inequality) d(x, z) ≤ d(x, y) + d(y, z) for all x, y, z ∈ X.

As primary examples, any normed space (X, ∥·∥) (see Definition 5.1) is a metric space with d(x, y) := ∥x − y∥ . Thus the space ℓ^{p}(μ) (as in Theorem 5.2) is a metric space for all p ∈ [1, ∞]. Also any subset of a metric space is a metric space. For example a surface Σ in ℝ^{3} is a metric space with the distance between two points on Σ being the usual distance in ℝ^{3}.

Definition 6.2. Let (X, d) be a metric space. The open ball B(x, δ) ⊂ X centered at x ∈ X with radius δ > 0 is the set

B(x, δ) := {y ∈ X : d(x, y) < δ}.

We will often also write B(x, δ) as B_{x}(δ). We also define the closed ball centered at x ∈ X with radius δ > 0 as the set C_{x}(δ) := {y ∈ X : d(x, y) ≤ δ}.

Definition 6.3. A sequence {x_{n}}_{n=1}^{∞} in a metric space (X, d) is said to be convergent if there exists a point x ∈ X such that lim_{n→∞} d(x, x_{n}) = 0. In this case we write lim_{n→∞} x_{n} = x of x_{n} → x as n → ∞.

Exercise 6.1. Show that x in Definition 6.3 is necessarily unique.

Definition 6.4. A set E ⊂ X is bounded if E ⊂ B (x, R) for some x ∈ X and R < ∞. A set F ⊂ X is closed iff every convergent sequence {x_{n}}_{n=1}^{∞} which is contained in F has its limit back in F. A set V ⊂ X is open iff V^{c} is closed. We will write F ⊏ X to indicate the F is a closed subset of X and V ⊂_{o} X to indicate the V is an open subset of X. We also let τ_{d} denote the collection of open subsets of X relative to the metric d.

Definition 6.5. A subset A ⊂ X is a neighborhood of x if there exists an open set V ⊂_{o} X such that x ∈ V ⊂ A. We will say that A ⊂ X is an open neighborhood of x if A is open and x ∈ A.

Exercise 6.2. Let F be a collection of closed subsets of X, show ∩F := ∩_{F ∈F}F is closed. Also show that finite unions of closed sets are closed, i.e. if {F_{k}}_{k=1}^{n} are closed sets then ∪_{k=1}^{n}F_{k} is closed. (By taking complements, this shows that the collection of open sets, τ_{d}, is closed under finite intersections and arbitrary unions.)

The following “continuity” facts of the metric d will be used frequently in the remainder of this book.

Lemma 6.6. For any non empty subset A ⊂ X, let d_{A}(x) := inf{d(x, a)|a ∈ A}, then

|d_{A}(x) − d_{A}(y)| ≤ d(x, y) ∀ x, y ∈ X  (6.1)

and in particular if x_{n} → x in X then d_{A} (x_{n}) → d_{A} (x) as n → ∞. Moreover the set F_{ε} := {x ∈ X|d_{A}(x) ≥ ε} is closed in X.

Proof. Let a ∈ A and x, y ∈ X, then

d(x, a) ≤ d(x, y) + d(y, a).

Take the inf over a in the above equation shows that

d_{A}(x) ≤ d(x, y) + d_{A}(y) ∀x, y ∈ X.

Therefore, d_{A}(x)−d_{A}(y) ≤ d(x, y) and by interchanging x and y we also have that d_{A}(y) − d_{A}(x) ≤ d(x, y) which implies Eq. (6.1). If x_{n} → x ∈ X, then by Eq. (6.1),

|d_{A}(x) − d_{A}(x_{n})| ≤ d(x, x_{n}) → 0 as n → ∞

so that lim_{n→∞} d_{A} (x_{n}) = d_{A} (x) . Now suppose that {x_{n}}_{n=1}^{∞} ⊂ F_{ε} and x_{n} → x in X, then

d_{A} (x) = lim_{n→∞} d_{A} (x_{n}) ≥ ε

since d_{A} (x_{n}) ≥ ε for all n. This shows that x ∈ F_{ε} and hence F_{ε} is closed.

Corollary 6.7. The function d satisfies,

|d(x, y) − d(x′, y′)| ≤ d(y, y′) + d(x, x′).

In particular d : X × X → [0, ∞) is “continuous” in the sense that d(x, y) is close to d(x′, y′) if x is close to x′ and y is close to y′. (The notion of continuity will be developed shortly.)

Proof. By Lemma 6.6 for single point sets and the triangle inequality for the absolute value of real numbers,

|d(x, y) − d(x′, y′)| ≤ |d(x, y) − d(x, y′)| + |d(x, y′) − d(x′, y′)|
≤ d(y, y′) + d(x, x′).

Example 6.8. Let x ∈ X and δ > 0, then C_{x} (δ) and B_{x} (δ)^{c} are closed subsets of X. For example if {y_{n}}_{n=1}^{∞} ⊂ C_{x} (δ) and y_{n} → y ∈ X, then d (y_{n}, x) ≤ δ for all n and using Corollary 6.7 it follows d (y, x) ≤ δ, i.e. y ∈ C_{x} (δ) . A similar proof shows B_{x} (δ)^{c} is open, see Exercise 6.3.

Exercise 6.3. Show that V ⊂ X is open iff for every x ∈ V there is a δ > 0 such that B_{x}(δ) ⊂ V. In particular show B_{x}(δ) is open for all x ∈ X and δ > 0. Hint: by definition V is not open iff V^{c} is not closed.

Lemma 6.9 (Approximating open sets from the inside by closed sets). Let A be a closed subset of X and F_{ε} := {x ∈ X|d_{A}(x) ≥ ε} ⊏ X be as in Lemma 6.6. Then F_{ε} ↑ A^{c} as ε ↓ 0.

Proof. It is clear that d_{A}(x) = 0 for x ∈ A so that F_{ε} ⊂ A^{c} for each ε > 0 and hence ∪_{ε>0}F_{ε} ⊂ A^{c}. Now suppose that x ∈ A^{c} ⊂_{o} X. By Exercise 6.3 there exists an ε > 0 such that B_{x}(ε) ⊂ A^{c}, i.e. d(x, y) ≥ ε for all y ∈ A. Hence x ∈ F_{ε} and we have shown that A^{c} ⊂ ∪_{ε>0}F_{ε}. Finally it is clear that F_{ε} ⊂ F_{ε′} whenever ε′ ≤ ε.

Definition 6.10. Given a set A contained a metric space X, let Ā ⊂ X be the closure of A defined by

Ā := {x ∈ X : ∃ {x_{n}} ⊂ A ∋ x = lim_{n→∞} x_{n}}.

That is to say Ā contains all limit points of A. We say A is dense in X if Ā = X, i.e. every element x ∈ X is a limit of a sequence of elements from A.

Exercise 6.4. Given A ⊂ X, show Ā is a closed set and in fact

Ā = ∩{F : A ⊂ F ⊂ X with F closed}.  (6.2)

That is to say Ā is the smallest closed set containing A.

6.1 Continuity

Suppose that (X, ρ) and (Y, d) are two metric spaces and f : X → Y is a function.

Definition 6.11. A function f : X → Y is continuous at x ∈ X if for all ε > 0 there is a δ > 0 such that

d(f(x), f(x′)) < ε provided that ρ(x, x′) < δ.  (6.3)

The function f is said to be continuous if f is continuous at all points x ∈ X.

The following lemma gives two other characterizations of continuity of a function at a point.

Lemma 6.12 (Local Continuity Lemma). Suppose that (X, ρ) and (Y, d) are two metric spaces and f : X → Y is a function defined in a neighborhood of a point x ∈ X. Then the following are equivalent:

1. f is continuous at x ∈ X.

2. For all neighborhoods A ⊂ Y of f(x), f^{−1}(A) is a neighborhood of x ∈ X.

3. For all sequences {x_{n}}_{n=1}^{∞} ⊂ X such that x = lim_{n→∞} x_{n}, {f(x_{n})} is convergent in Y and

lim_{n→∞} f(x_{n}) = f ( lim_{n→∞} x_{n}) .

Proof. 1 ⟹ 2. If A ⊂ Y is a neighborhood of f (x) , there exists ε > 0 such that B_{f(x)} (ε) ⊂ A and because f is continuous there exists a δ > 0 such that Eq. (6.3) holds. Therefore

B_{x} (δ) ⊂ f^{−1} (B_{f(x)} (ε)) ⊂ f^{−1} (A)

showing f^{−1} (A) is a neighborhood of x. 2 ⟹ 3. Suppose that {x_{n}}_{n=1}^{∞} ⊂ X and x = lim_{n→∞} x_{n}. Then for any ε > 0, B_{f(x)} (ε) is a neighborhood of f (x) and so f^{−1} (B_{f(x)} (ε)) is a neighborhood of x which must containing B_{x} (δ) for some δ > 0. Because x_{n} → x, it follows that x_{n} ∈ B_{x} (δ) ⊂ f^{−1} (B_{f(x)} (ε)) for a.a. n and this implies f (x_{n}) ∈ B_{f(x)} (ε) for a.a. n, i.e. d(f(x), f (x_{n})) < ε for a.a. n. Since ε > 0 is arbitrary it follows that lim_{n→∞} f (x_{n}) = f (x) .

3. ⟹ 1. We will show not 1. ⟹ not 3. If f is not continuous at x, there exists an ε > 0 such that for all n ∈ ℕ there exists a point x_{n} ∈ X with ρ (x_{n}, x) < \frac{1}{n} yet d (f (x_{n}) , f (x)) ≥ ε. Hence x_{n} → x as n → ∞ yet f (x_{n}) does not converge to f (x) .

Here is a global version of the previous lemma.

Lemma 6.13 (Global Continuity Lemma). Suppose that (X, ρ) and (Y, d) are two metric spaces and f : X → Y is a function defined on all of X. Then the following are equivalent:

1. f is continuous.

2. f^{−1}(V) ∈ τ_{ρ} for all V ∈ τ_{d}, i.e. f^{−1}(V) is open in X if V is open in Y.

3. f^{−1}(C) is closed in X if C is closed in Y.

4. For all convergent sequences {x_{n}} ⊂ X, {f(x_{n})} is convergent in Y and

lim_{n→∞} f(x_{n}) = f ( lim_{n→∞} x_{n}) .

Proof. Since f^{−1} (A^{c}) = [f^{−1} (A)]^{c} , it is easily seen that 2. and 3. are equivalent. So because of Lemma 6.12 it only remains to show 1. and 2. are equivalent. If f is continuous and V ⊂ Y is open, then for every x ∈ f^{−1} (V) , V is a neighborhood of f (x) and so f^{−1} (V) is a neighborhood of x. Hence f^{−1} (V) is a neighborhood of all of its points and from this and Exercise 6.3 it follows that f^{−1} (V) is open. Conversely if x ∈ X and A ⊂ Y is a neighborhood of f (x) , then there exists V ⊂_{o} X such that f (x) ∈ V ⊂ A. Hence x ∈ f^{−1} (V) ⊂ f^{−1} (A) and by assumption f^{−1} (V) is open showing f^{−1} (A) is a neighborhood of x. Therefore f is continuous at x and since x ∈ X was arbitrary, f is continuous.

Example 6.14. The function d_{A} defined in Lemma 6.6 is continuous for each A ⊂ X. In particular, if A = {x} , it follows that y ∈ X → d(y, x) is continuous for each x ∈ X.

Exercise 6.5. Use Example 6.14 and Lemma 6.13 to recover the results of Example 6.8.

The next result shows that there are lots of continuous functions on a metric space (X, d) .

Lemma 6.15 (Urysohn’s Lemma for Metric Spaces). Let (X, d) be a metric space and suppose that A and B are two disjoint closed subsets of X. Then

f(x) = \frac{d_{B}(x)}{d_{A}(x) + d_{B}(x)} for x ∈ X  (6.4)

defines a continuous function, f : X → [0, 1], such that f(x) = 1 for x ∈ A and f(x) = 0 if x ∈ B.

Proof. By Lemma 6.6, d_{A} and d_{B} are continuous functions on X. Since A and B are closed, d_{A}(x) > 0 if x ∉ A and d_{B}(x) > 0 if x ∉ B. Since A ∩ B = ∅, d_{A}(x) + d_{B}(x) > 0 for all x and (d_{A} + d_{B})^{−1} is continuous as well. The remaining assertions about f are all easy to verify.

Sometimes Urysohn’s lemma will be use in the following form. Suppose F ⊂ V ⊂ X with F being closed and V being open, then there exists f ∈ C (X, [0, 1])) such that f = 1 on F while f = 0 on V^{c}. This of course follows from Lemma 6.15 by taking A = F and B = V^{c}.

6.2 Completeness in Metric Spaces

Definition 6.16 (Cauchy sequences). A sequence {x_{n}}_{n=1}^{∞} in a metric space (X, d) is Cauchy provided that

lim_{m,n→∞} d(x_{n}, x_{m}) = 0.

Exercise 6.6. Show that convergent sequences are always Cauchy sequences. The converse is not always true. For example, let X = ℚ be the set of rational numbers and d(x, y) = |x − y|. Choose a sequence {x_{n}}_{n=1}^{∞} ⊂ ℚ which converges to \sqrt{2} ∈ ℝ, then \{x_{n}\}_{n=1}^{∞} is (ℚ, d) – Cauchy but not (ℚ, d) – convergent. The sequence does converge in ℝ however.

Definition 6.17. A metric space (X, d) is complete if all Cauchy sequences are convergent sequences.

Exercise 6.7. Let (X, d) be a complete metric space. Let A ⊂ X be a subset of X viewed as a metric space using d|_{A×A}. Show that (A, d|_{A×A}) is complete iff A is a closed subset of X.

Example 6.18. Examples 2. – 4. of complete metric spaces will be verified in Chapter 7 below.

1. X = ℝ and d(x, y) = |x − y|, see Theorem 3.8 above.

2. X = ℝ^{n} and d(x, y) = ∥x − y∥_{2} = ∑_{i=1}^{n}(x_{i} − y_{i})^{2}.

3. X = ℓ^{p}(μ) for p ∈ [1, ∞] and any weight function μ : X → (0, ∞).

4. X = C([0, 1], ℝ) – the space of continuous functions from [0, 1] to ℝ and

d(f, g) := max_{t∈[0,1]} |f(t) − g(t)|.

This is a special case of Lemma 7.3 below.

5. Let X = C([0, 1], ℝ) and

d(f, g) := ∫_{0}^{1} |f(t) − g(t)| dt.

You are asked in Exercise 7.10 to verify that (X, d) is a metric space which is not complete.

Exercise 6.8 (Completions of Metric Spaces). Suppose that (X, d) is a (not necessarily complete) metric space. Using the following outline show there exists a complete metric space (X̄, d̄) and an isometric map i : X → X̄ such that i (X) is dense in X̄, see Definition 6.10.

1. Let C denote the collection of Cauchy sequences a = {a_{n}}_{n=1}^{∞} ⊂ X. Given two element a, b ∈ C show

d_{C} (a, b) := lim_{n→∞} d (a_{n}, b_{n}) exists,

d_{C} (a, b) ≥ 0 for all a, b ∈ C and d_{C} satisfies the triangle inequality,

d_{C} (a, c) ≤ d_{C} (a, b) + d_{C} (b, c) for all a, b, c ∈ C.

Thus (C, d_{C}) would be a metric space if it were true that d_{C}(a, b) = 0 iff a = b. This however is false, for example if a_{n} = b_{n} for all n ≥ 100, then d_{C}(a, b) = 0 while a need not equal b.

2. Define two elements a, b ∈ C to be equivalent (write a ∼ b) whenever d_{C}(a, b) = 0. Show “ ∼ ” is an equivalence relation on C and that d_{C} (a′, b′) = d_{C} (a, b) if a ∼ a′ and b ∼ b′. (Hint: see Corollary 6.7.)

3. Given a ∈ C let ā := {b ∈ C : b ∼ a} denote the equivalence class containing a and let X̄ := {ā : a ∈ C} denote the collection of such equivalence classes. Show that d̄(ā, b̄) := d_{C} (a, b) is well defined on X̄ × X̄ and verify (X̄, d̄) is a metric space.

4. For x ∈ X let i (x) = ā where a is the constant sequence, a_{n} = x for all n. Verify that i : X → X̄ is an isometric map and that i (X) is dense in X̄.

5. Verify (X̄, d̄) is complete. Hint: if {ā(m)}_{m=1}^{∞} is a Cauchy sequence in X̄ choose b_{m} ∈ X such that d̄(i (b_{m}) , ā(m)) ≤ 1/m. Then show ā(m) → b̄ where b = {b_{m}}_{m=1}^{∞} .

6.3 Supplementary Remarks

6.3.1 Word of Caution

Example 6.19. Let (X, d) be a metric space. It is always true that \overline{B_{x}(ε)} ⊂ C_{x}(ε) since C_{x}(ε) is a closed set containing B_{x}(ε). However, it is not always true that \overline{B_{x}(ε)} = C_{x}(ε). For example let X = \{1, 2\} and d(1, 2) = 1, then B_{1}(1) = \{1\} , \overline{B_{1}(1)} = \{1\} while C_{1}(1) = X. For another counter example, take

X = {(x, y) ∈ ℝ^{2} : x = 0 or x = 1}

with the usually Euclidean metric coming from the plane. Then

B_{(0,0)}(1) = {(0, y) ∈ ℝ^{2} : |y| < 1} ,
\overline{B_{(0,0)}(1)} = \{(0, y) ∈ ℝ^{2} : |y| ≤ 1\} , while
C_{(0,0)}(1) = \overline{B_{(0,0)}(1)} ∪ \{(0, 1)\} .

In spite of the above examples, Lemmas 6.20 and 6.21 below shows that for certain metric spaces of interest it is true that \overline{B_{x}(ε)} = C_{x}(ε).

Lemma 6.20. Suppose that (X, |·|) is a normed vector space and d is the metric on X defined by d(x, y) = |x − y| . Then

\overline{B_{x}(ε)} = C_{x}(ε) and
bd(B_{x}(ε)) = {y ∈ X : d(x, y) = ε}.

where the boundary operation, bd(·) is defined in Definition 10.29 below.

Proof. We must show that C := C_{x}(ε) ⊂ \overline{B_{x}(ε)} =: B̄. For y ∈ C, let v = y − x, then

|v| = |y − x| = d(x, y) ≤ ε.

Let α_{n} = 1 − 1/n so that α_{n} ↑ 1 as n → ∞. Let y_{n} = x + α_{n}v, then d(x, y_{n}) = α_{n}d(x, y) < ε, so that y_{n} ∈ B_{x}(ε) and d(y, y_{n}) = 1 − α_{n} → 0 as n → ∞. This shows that y_{n} → y as n → ∞ and hence that y ∈ B̄.

Fig. 6.1. An almost length minimizing curve joining x to y.

6.3.2 Riemannian Metrics

This subsection is not completely self contained and may safely be skipped.

Lemma 6.21. Suppose that X is a Riemannian (or sub-Riemannian) manifold and d is the metric on X defined by

d(x, y) = inf {ℓ(σ) : σ(0) = x and σ(1) = y}

where ℓ(σ) is the length of the curve σ. We define ℓ(σ) = ∞ if σ is not piecewise smooth.

Then

\overline{B_{x}(ε)} = C_{x}(ε) and
bd(B_{x}(ε)) = {y ∈ X : d(x, y) = ε}

where the boundary operation, bd(·) is defined in Definition 10.29 below.

Proof. Let C := C_{x}(ε) ⊂ \overline{B_{x}(ε)} =: B̄. We will show that C ⊂ B̄ by showing B̄^{c} ⊂ C^{c}. Suppose that y ∈ B̄^{c} and choose δ > 0 such that B_{y}(δ) ∩ B̄ = ∅. In particular this implies that

B_{y}(δ) ∩ B_{x}(ε) = ∅.

We will finish the proof by showing that d(x, y) ≥ ε + δ > ε and hence that y ∈ C^{c}. This will be accomplished by showing: if d(x, y) < ε + δ then B_{y}(δ) ∩ B_{x}(ε) ≠ ∅. If d(x, y) < max(ε, δ) then either x ∈ B_{y}(δ) or y ∈ B_{x}(ε). In either case B_{y}(δ) ∩ B_{x}(ε) ≠ ∅. Hence we may assume that max(ε, δ) ≤ d(x, y) < ε + δ. Let α > 0 be a number such that

max(ε, δ) ≤ d(x, y) < α < ε + δ

and choose a curve σ from x to y such that ℓ(σ) < α. Also choose 0 < δ′ < δ such that 0 < α−δ′ < ε which can be done since α−δ < ε. Let k(t) = d(y, σ(t)) a continuous function on [0, 1] and therefore k([0, 1]) ⊂ ℝ is a connected set which contains 0 and d(x, y). Therefore there exists t_{0} ∈ [0, 1] such that d(y, σ(t_{0})) = k(t_{0}) = δ′. Let z = σ(t_{0}) ∈ B_{y}(δ) then

d(x, z) ≤ ℓ(σ|_{[0,t_{0}]}) = ℓ(σ) − ℓ(σ|_{[t_{0},1]}) < α − d(z, y) = α − δ′ < ε

and therefore z ∈ B_{x}(ε) ∩ B_{x}(δ) ≠ ∅.

Remark 6.22. Suppose again that X is a Riemannian (or sub-Riemannian) manifold and

d(x, y) = inf {ℓ(σ) : σ(0) = x and σ(1) = y} .

Let σ be a curve from x to y and let ε = ℓ(σ) − d(x, y). Then for all 0 ≤ u < v ≤ 1,

d(σ(u), σ(v)) ≤ ℓ(σ|_{[u,v]}) + ε.

So if σ is within ε of a length minimizing curve from x to y that σ|_{[u,v]} is within ε of a length minimizing curve from σ(u) to σ(v). In particular if d(x, y) = ℓ(σ) then d(σ(u), σ(v)) = ℓ(σ|_{[u,v]}) for all 0 ≤ u < v ≤ 1, i.e. if σ is a length minimizing curve from x to y that σ|_{[u,v]} is a length minimizing curve from σ(u) to σ(v).

To prove these assertions notice that

d(x, y) + ε = ℓ(σ) = ℓ(σ|_{[0,u]}) + ℓ(σ|_{[u,v]}) + ℓ(σ|_{[v,1]})
≥ d(x, σ(u)) + ℓ(σ|_{[u,v]}) + d(σ(v), y)

and therefore

ℓ(σ|_{[u,v]}) ≤ d(x, y) + ε − d(x, σ(u)) − d(σ(v), y)
≤ d(σ(u), σ(v)) + ε.

6.4 Exercises

Exercise 6.9. Let (X, d) be a metric space. Suppose that {x_{n}}_{n=1}^{∞} ⊂ X is a sequence and set ε_{n} := d(x_{n}, x_{n+1}). Show that for m > n that

d(x_{n}, x_{m}) ≤ ∑_{k=n}^{m−1} ε_{k} ≤ ∑_{k=n}^{∞} ε_{k}.

Conclude from this that if

∑_{k=1}^{∞} ε_{k} = ∑_{n=1}^{∞} d(x_{n}, x_{n+1}) < ∞

then {x_{n}}_{n=1}^{∞} is Cauchy. Moreover, show that if {x_{n}}_{n=1}^{∞} is a convergent sequence and x = lim_{n→∞} x_{n} then

d(x, x_{n}) ≤ ∑_{k=n}^{∞} ε_{k}.

Exercise 6.10. Show that (X, d) is a complete metric space iff every sequence {x_{n}}_{n=1}^{∞} ⊂ X such that ∑_{n=1}^{∞} d(x_{n}, x_{n+1}) < ∞ is a convergent sequence in X. You may find it useful to prove the following statements in the course of the proof.

1. If {x_{n}} is Cauchy sequence, then there is a subsequence y_{j} := x_{n_{j}} such that ∑_{j=1}^{∞} d(y_{j+1}, y_{j}) < ∞.

2. If {x_{n}}_{n=1}^{∞} is Cauchy and there exists a subsequence y_{j} := x_{n_{j}} of {x_{n}} such that x = lim_{j→∞} y_{j} exists, then lim_{n→∞} x_{n} also exists and is equal to x.

Exercise 6.11. Suppose that f : [0, ∞) → [0, ∞) is a C^{2} – function such that f(0) = 0, f′ > 0 and f′′ ≤ 0 and (X, ρ) is a metric space. Show that d(x, y) = f(ρ(x, y)) is a metric on X. In particular show that

d(x, y) := \frac{ρ(x, y)}{1 + ρ(x, y)}

is a metric on X. (Hint: use calculus to verify that f(a + b) ≤ f(a) + f(b) for all a, b ∈ [0, ∞).)

Exercise 6.12. Let {(X_{n}, d_{n})}_{n=1}^{∞} be a sequence of metric spaces, X := ∏_{n=1}^{∞} X_{n}, and for x = (x(n))_{n=1}^{∞} and y = (y(n))_{n=1}^{∞} in X let

d(x, y) = ∑_{n=1}^{∞} 2^{−n} \frac{d_{n}(x(n), y(n))}{1 + d_{n}(x(n), y(n))}.

Show:

1. (X, d) is a metric space,

2. a sequence {x_{k}}_{k=1}^{∞} ⊂ X converges to x ∈ X iff x_{k}(n) → x(n) ∈ X_{n} as k → ∞ for each n ∈ ℕ and

3. X is complete if X_{n} is complete for all n.

Exercise 6.13. Suppose (X, ρ) and (Y, d) are metric spaces and A is a dense subset of X.

1. Show that if F : X → Y and G : X → Y are two continuous functions such that F = G on A then F = G on X. Hint: consider the set C := {x ∈ X : F (x) = G (x)} .

2. Suppose f : A → Y is a function which is uniformly continuous, i.e. for every ε > 0 there exists a δ > 0 such that

d (f (a) , f (b)) < ε for all a, b ∈ A with ρ (a, b) < δ.

Show there is a unique continuous function F : X → Y such that F = f on A. Hint: each point x ∈ X is a limit of a sequence consisting of elements from A.

3. Let X = ℝ = Y and A = ℚ ⊂ X, find a function f : ℚ → ℝ which is continuous on ℚ but does not extend to a continuous function on ℝ.

Banach Spaces

Let (X, ∥·∥) be a normed vector space and d (x, y) := ∥x − y∥ be the associated metric on X. We say {x_{n}}_{n=1}^{∞} ⊂ X converges to x ∈ X (and write lim_{n→∞} x_{n} = x or x_{n} → x) if

0 = lim_{n→∞} d (x, x_{n}) = lim_{n→∞} ∥x − x_{n}∥ .

Similarly {x_{n}}_{n=1}^{∞} ⊂ X is said to be a Cauchy sequence if

0 = lim_{m,n→∞} d (x_{m}, x_{n}) = lim_{m,n→∞} ∥x_{m} − x_{n}∥ .

Definition 7.1 (Banach space). A normed vector space (X, ∥·∥) is a Banach space if the associated metric space (X, d) is complete, i.e. all Cauchy sequences are convergent.

Remark 7.2. Since ∥x∥ = d (x, 0) , it follows from Lemma 6.6 that ∥·∥ is a continuous function on X and that

|∥x∥ − ∥y∥| ≤ ∥x − y∥ for all x, y ∈ X.

It is also easily seen that the vector addition and scalar multiplication are continuos on any normed space as the reader is asked to verify in Exercise 7.4. These facts will often be used in the sequel without further mention.

7.1 Examples

Lemma 7.3. Suppose that X is a set then the bounded functions, ℓ^{∞}(X), on X is a Banach space with the norm

∥f∥ = ∥f∥_{∞} = sup_{x∈X} |f(x)| .

Moreover if X is a metric space (more generally a topological space, see Chapter 10) the set BC(X) ⊂ ℓ^{∞}(X) = B(X) is closed subspace of ℓ^{∞}(X) and hence is also a Banach space.

Proof. Let {f_{n}}_{n=1}^{∞} ⊂ ℓ^{∞}(X) be a Cauchy sequence. Since for any x ∈ X, we have

|f_{n}(x) − f_{m}(x)| ≤ ∥f_{n} − f_{m}∥_{∞}  (7.1)

which shows that {f_{n}(x)}_{n=1}^{∞} ⊂ 𝔽 is a Cauchy sequence of numbers. Because 𝔽 (𝔽 = ℝ or ℂ) is complete, f(x) := lim_{n→∞} f_{n}(x) exists for all x ∈ X. Passing to the limit n → ∞ in Eq. (7.1) implies

|f(x) − f_{m}(x)| ≤ lim inf_{n→∞} ∥f_{n} − f_{m}∥_{∞}

and taking the supremum over x ∈ X of this inequality implies

∥f − f_{m}∥_{∞} ≤ lim inf_{n→∞} ∥f_{n} − f_{m}∥_{∞} → 0 as m → ∞

showing f_{m} → f in ℓ^{∞}(X). For the second assertion, suppose that {f_{n}}_{n=1}^{∞} ⊂ BC(X) ⊂ ℓ^{∞}(X) and f_{n} → f ∈ ℓ^{∞}(X). We must show that f ∈ BC(X), i.e. that f is continuous. To this end let x, y ∈ X, then

|f(x) − f(y)| ≤ |f(x) − f_{n}(x)| + |f_{n}(x) − f_{n}(y)| + |f_{n}(y) − f(y)|
≤ 2 ∥f − f_{n}∥_{∞} + |f_{n}(x) − f_{n}(y)| .

Thus if ε > 0, we may choose n large so that 2 ∥f − f_{n}∥_{∞} < ε/2 and then for this n there exists an open neighborhood V_{x} of x ∈ X such that |f_{n}(x) − f_{n}(y)| < ε/2 for y ∈ V_{x}. Thus |f(x) − f(y)| < ε for y ∈ V_{x} showing the limiting function f is continuous.

Here is an application of this theorem.

Theorem 7.4 (Metric Space Tietze Extension Theorem). Let (X, d) be a metric space, D be a closed subset of X, −∞ < a < b < ∞ and f ∈ C(D, [a, b]). (Here we are viewing D as a metric space with metric d_{D} := d_{D×D}.) Then there exists F ∈ C(X, [a, b]) such that F |_{D} = f.

Proof.

1. By scaling and translation (i.e. by replacing f by (b − a)^{−1} (f − a)), it suffices to prove Theorem 7.4 with a = 0 and b = 1.

2. Suppose α ∈ (0, 1] and f : D → [0, α] is continuous function. Let A := f^{−1}([0, \frac{1}{3}α]) and B := f^{−1}([\frac{2}{3}α, α]). By Lemma 6.15 there exists a function g̃ ∈ C(X, [0, α/3]) such that g̃ = 0 on A and g̃ = 1 on B. Letting g := \frac{α}{3} g̃, we have g ∈ C(X, [0, α/3]) such that g = 0 on A and g = α/3 on B. Further notice that

0 ≤ f(x) − g(x) ≤ \frac{2}{3}α for all x ∈ D.

3. Now suppose f : D → [0, 1] is a continuous function as in step 1. Let g_{1} ∈ C(X, [0, 1/3]) be as in step 2, see Figure 7.1. with α = 1 and let f_{1} := f − g_{1}|_{D} ∈ C(D, [0, 2/3]). Apply step 2. with α = 2/3 and f = f_{1} to find g_{2} ∈ C(X, [0, \frac{1}{3} \frac{2}{3}]) such that f_{2} := f − (g_{1} + g_{2}) |_{D} ∈ C(D, [0, (\frac{2}{3})^{2}]). Continue this way inductively to find g_{n} ∈ C(X, [0, \frac{1}{3} (\frac{2}{3})^{n−1}]) such that

f − ∑_{n=1}^{N} g_{n}|_{D} =: f_{N} ∈ C(D, [0, (\frac{2}{3})^{N}]).  (7.2)

4. Define F := ∑_{n=1}^{∞} g_{n}. Since

∑_{n=1}^{∞} ∥g_{n}∥_{∞} ≤ ∑_{n=1}^{∞} \frac{1}{3} (\frac{2}{3})^{n−1} = \frac{1}{3} \frac{1}{1 − \frac{2}{3}} = 1,

the series defining F is uniformly convergent so F ∈ C(X, [0, 1]). Passing to the limit in Eq. (7.2) shows f = F |_{D}.

Fig. 7.1. Reducing f by subtracting off a globally defined function g_{1} ∈ C (X, [0, \frac{1}{3} ]) .

Theorem 7.5 (Completeness of ℓ^{p}(μ)). Let X be a set and μ : X → (0, ∞) be a given function. Then for any p ∈ [1, ∞], (ℓ^{p}(μ), ∥·∥_{p}) is a Banach space.

Proof. We have already proved this for p = ∞ in Lemma 7.3 so we now assume that p ∈ [1, ∞). Let {f_{n}}_{n=1}^{∞} ⊂ ℓ^{p}(μ) be a Cauchy sequence. Since for any x ∈ X,

|f_{n}(x) − f_{m}(x)| ≤ \frac{1}{μ(x)} ∥f_{n} − f_{m}∥_{p} → 0 as m, n → ∞

it follows that {f_{n}(x)}_{n=1}^{∞} is a Cauchy sequence of numbers and f(x) := lim_{n→∞} f_{n}(x) exists for all x ∈ X. By Fatou’s Lemma,

∥f_{n} − f∥_{p}^{p} = ∑_{X} μ · lim_{m→∞} inf |f_{n} − f_{m}|^{p} ≤ lim_{m→∞} inf ∑_{X} μ · |f_{n} − f_{m}|^{p}

= lim_{m→∞} inf ∥f_{n} − f_{m}∥_{p}^{p} → 0 as n → ∞.

This then shows that f = (f − f_{n}) + f_{n} ∈ ℓ^{p}(μ) (being the sum of two ℓ^{p} – functions) and that f_{n} ⟶^{ℓ^{p}} f.

Remark 7.6. Let X be a set, Y be a Banach space and ℓ^{∞}(X, Y) denote the bounded functions f : X → Y equipped with the norm

∥f∥ = ∥f∥_{∞} = sup_{x∈X} ∥f(x)∥_{Y} .

If X is a metric space (or a general topological space, see Chapter 10), let BC(X, Y) denote those f ∈ ℓ^{∞}(X, Y) which are continuous. The same proof used in Lemma 7.3 shows that ℓ^{∞}(X, Y) is a Banach space and that BC(X, Y) is a closed subspace of ℓ^{∞}(X, Y). Similarly, if 1 ≤ p < ∞ we may define

ℓ^{p} (X, Y) = {f : X → Y : ∥f∥_{p} = (∑_{x∈X} ∥f (x)∥_{Y}^{p} )^{1/p} < ∞} .

The same proof as in Theorem 7.5 would then show that (ℓ^{p} (X, Y) , ∥·∥_{p}) is a Banach space.

7.2 Bounded Linear Operators Basics

Definition 7.7. Let X and Y be normed spaces and T : X → Y be a linear map. Then T is said to be bounded provided there exists C < ∞ such that ∥T (x)∥ ≤ C∥x∥_{X} for all x ∈ X. We denote the best constant by ∥T ∥, i.e.

∥T ∥ = sup_{x≠0} \frac{∥T (x)∥}{∥x∥} = sup_{x≠0} \{∥T (x)∥ : ∥x∥ = 1\} .

The number ∥T ∥ is called the operator norm of T.

Proposition 7.8. Suppose that X and Y are normed spaces and T : X → Y is a linear map. The the following are equivalent:

(a) T is continuous.

(b) T is continuous at 0.

(c) T is bounded.

Proof. (a) ⇒ (b) trivial. (b) ⇒ (c) If T continuous at 0 then there exist δ > 0 such that ∥T (x)∥ ≤ 1 if ∥x∥ ≤ δ. Therefore for any x ∈ X, ∥T (δx/∥x∥) ∥ ≤ 1 which implies that ∥T (x)∥ ≤ \frac{1}{δ}∥x∥ and hence ∥T ∥ ≤ \frac{1}{δ} < ∞. (c) ⇒ (a) Let x ∈ X and ε > 0 be given. Then

∥T y − T x∥ = ∥T (y − x)∥ ≤ ∥T ∥ ∥y − x∥ < ε

provided ∥y − x∥ < ε/∥T ∥ := δ.

For the next three exercises, let X = ℝ^{n} and Y = ℝ^{m} and T : X → Y be a linear transformation so that T is given by matrix multiplication by an m × n matrix. Let us identify the linear transformation T with this matrix.

Exercise 7.1. Assume the norms on X and Y are the ℓ^{1} – norms, i.e. for x ∈ ℝ^{n}, ∥x∥ = ∑_{j=1}^{n} |x_{j}| . Then the operator norm of T is given by

∥T ∥ = max_{1≤j≤n} ∑_{i=1}^{m} |T_{ij}| .

Exercise 7.2. Suppose that norms on X and Y are the ℓ^{∞} – norms, i.e. for x ∈ ℝ^{n}, ∥x∥ = max_{1≤j≤n} |x_{j}| . Then the operator norm of T is given by

∥T ∥ = max_{1≤i≤m} ∑_{j=1}^{n} |T_{ij}| .

Exercise 7.3. Assume the norms on X and Y are the ℓ^{2} – norms, i.e. for x ∈ ℝ^{n}, ∥x∥^{2} = ∑_{j=1}^{n} x_{j}^{2}. Show ∥T ∥^{2} is the largest eigenvalue of the matrix T^{tr}T : ℝ^{n} → ℝ^{n}. Hint: Use the spectral theorem for orthogonal matrices.

Notation 7.9 Let L(X, Y) denote the bounded linear operators from X to Y and L (X) = L (X, X) . If Y = 𝔽 we write X^{∗} for L(X, 𝔽) and call X^{∗} the (continuous) dual space to X.

Lemma 7.10. Let X, Y be normed spaces, then the operator norm ∥·∥ on L(X, Y) is a norm. Moreover if Z is another normed space and T : X → Y and S : Y → Z are linear maps, then ∥ST ∥ ≤ ∥S∥∥T ∥, where ST := S ∘ T.

Proof. As usual, the main point in checking the operator norm is a norm is to verify the triangle inequality, the other axioms being easy to check. If A, B ∈ L(X, Y) then the triangle inequality is verified as follows:

∥A + B∥ = sup_{x≠0} \frac{∥Ax + Bx∥}{∥x∥} ≤ sup_{x≠0} \frac{∥Ax∥ + ∥Bx∥}{∥x∥}

≤ sup_{x≠0} \frac{∥Ax∥}{∥x∥} + sup_{x≠0} \frac{∥Bx∥}{∥x∥} = ∥A∥ + ∥B∥ .

For the second assertion, we have for x ∈ X, that

∥ST x∥ ≤ ∥S∥∥T x∥ ≤ ∥S∥∥T ∥∥x∥.

From this inequality and the definition of ∥ST ∥, it follows that ∥ST ∥ ≤ ∥S∥∥T ∥.

The reader is asked to prove the following continuity lemma in Exercise 7.8.

Lemma 7.11. Let X, Y and Z be normed spaces. Then the maps

(S, x) ∈ L(X, Y) × X ⟶ Sx ∈ Y

and

(S, T) ∈ L(X, Y) × L(Y, Z) ⟶ ST ∈ L(X, Z)

are continuous relative to the norms

∥(S, x)∥_{L(X,Y)×X} := ∥S∥_{L(X,Y)} + ∥x∥_{X} and
∥(S, T)∥_{L(X,Y)×L(Y,Z)} := ∥S∥_{L(X,Y)} + ∥T ∥_{L(Y,Z)}

on L(X, Y) × X and L(X, Y) × L(Y, Z) respectively.

Proposition 7.12. Suppose that X is a normed vector space and Y is a Banach space. Then (L(X, Y), ∥ · ∥_{op}) is a Banach space. In particular the dual space X^{∗} is always a Banach space.

Proof. Let {T_{n}}_{n=1}^{∞} be a Cauchy sequence in L(X, Y). Then for each x ∈ X,

∥T_{n}x − T_{m}x∥ ≤ ∥T_{n} − T_{m}∥ ∥x∥ → 0 as m, n → ∞

showing {T_{n}x}_{n=1}^{∞} is Cauchy in Y. Using the completeness of Y, there exists an element T x ∈ Y such that

lim_{n→∞} ∥T_{n}x − T x∥ = 0.

The map T : X → Y is linear map, since for x, x′ ∈ X and λ ∈ 𝔽 we have

T (x + λx′) = lim_{n→∞} T_{n} (x + λx′) = lim_{n→∞} [T_{n}x + λT_{n}x′] = T x + λT x′,

wherein we have used the continuity of the vector space operations in the last equality. Moreover,

∥T x − T_{n}x∥ ≤ ∥T x − T_{m}x∥ + ∥T_{m}x − T_{n}x∥ ≤ ∥T x − T_{m}x∥ + ∥T_{m} − T_{n}∥ ∥x∥

and therefore

∥T x − T_{n}x∥ ≤ lim inf_{m→∞} (∥T x − T_{m}x∥ + ∥T_{m} − T_{n}∥ ∥x∥)

= ∥x∥ · lim inf_{m→∞} ∥T_{m} − T_{n}∥ .

Hence

∥T − T_{n}∥ ≤ lim inf_{m→∞} ∥T_{m} − T_{n}∥ → 0 as n → ∞.

Thus we have shown that T_{n} → T in L(X, Y) as desired.

The following characterization of a Banach space will sometimes be useful in the sequel.

7.2 Bounded Linear Operators Basics 65

Theorem 7.13. A normed space (X, ∥ · ∥) is a Banach space iff for every sequence {x_{n}}_{n=1}^{∞} such that ∑_{n=1}^{∞} ∥x_{n}∥ < ∞ implies lim_{N→∞} ∑_{n=1}^{N} x_{n} = s exists in X (that is to say every absolutely convergent series is a convergent series in X.) As usual we will denote s by ∑_{n=1}^{∞} x_{n}.

Proof. This is very similar to Exercise 6.10. (⇒)If X is complete and ∑_{n=1}^{∞} ∥x_{n}∥ < ∞ then sequence s_{N} := ∑_{n=1}^{N} x_{n} for N ∈ ℕ is Cauchy because (for N > M)

∥s_{N} − s_{M}∥ ≤ ∑_{n=M+1}^{N} ∥x_{n}∥ → 0 as M, N → ∞.

Therefore s = ∑_{n=1}^{∞} x_{n} := lim_{N→∞} ∑_{n=1}^{N} x_{n} exists in X. (⟸) Suppose that {x_{n}}_{n=1}^{∞} is a Cauchy sequence and let {y_{k} = x_{n_{k}}}_{k=1}^{∞} be a subsequence of

{x_{n}}_{n=1}^{∞} such that ∑_{n=1}^{∞} ∥y_{n+1} − y_{n}∥ < ∞. By assumption

y_{N+1} − y_{1} = ∑_{n=1}^{N}(y_{n+1} − y_{n}) → s = ∑_{n=1}^{∞}(y_{n+1} − y_{n}) ∈ X as N → ∞.

This shows that lim_{N→∞} y_{N} exists and is equal to x := y_{1} + s. Since {x_{n}}_{n=1}^{∞} is Cauchy,

∥x − x_{n}∥ ≤ ∥x − y_{k}∥ + ∥y_{k} − x_{n}∥ → 0 as k, n → ∞

showing that lim_{n→∞} x_{n} exists and is equal to x. Example 7.14. Here is another proof of Theorem 7.12 which makes use of Proposition 7.12. Suppose that T_{n} ∈ L(X, Y) is a sequence of operators such

that ∑_{n=1}^{∞} ∥T_{n}∥ < ∞. Then

∑_{n=1}^{∞} ∥T_{n}x∥ ≤ ∑_{n=1}^{∞} ∥T_{n}∥ ∥x∥ < ∞

and therefore by the completeness of Y, Sx := ∑_{n=1}^{∞} T_{n}x = lim_{N→∞} S_{N}x exists

in Y, where S_{N} := ∑_{n=1}^{N} T_{n}. The reader should check that S : X → Y so defined is linear. Since,

∥Sx∥ = lim_{N→∞} ∥S_{N}x∥ ≤ lim_{N→∞} ∑_{n=1}^{N} ∥T_{n}x∥ ≤ ∑_{n=1}^{∞} ∥T_{n}∥ ∥x∥ ,

S is bounded and

∥S∥ ≤ ∑_{n=1}^{∞} ∥T_{n}∥.  (7.3)

Similarly,

∥Sx − S_{M}x∥ = lim_{N→∞} ∥S_{N}x − S_{M}x∥

≤ lim_{N→∞} ∑_{n=M+1}^{N} ∥T_{n}∥ ∥x∥ = ∑_{n=M+1}^{∞} ∥T_{n}∥ ∥x∥

and therefore,

∥S − S_{M}∥ ≤ ∑_{n=M}^{∞} ∥T_{n}∥ → 0 as M → ∞.

7.3 General Sums in Banach Spaces

Definition 7.15. Suppose X is a normed space.

1. Suppose that {x_{n}}_{n=1}^{∞} is a sequence in X, then we say ∑_{n=1}^{∞} x_{n} converges in X and ∑_{n=1}^{∞} x_{n} = s if

lim_{N→∞} ∑_{n=1}^{N} x_{n} = s in X.

2. Suppose that {x_{α} : α ∈ A} is a given collection of vectors in X. We say the sum ∑_{α∈A} x_{α} converges in X and write s = ∑_{α∈A} x_{α} ∈ X if for all ε > 0 there exists a finite set Γ_{ε} ⊂ A such that ∥s − ∑_{α∈Λ} x_{α}∥ < ε for any Λ ⊂⊂ A such that Γ_{ε} ⊂ Λ.

Warning: As usual if ∑_{α∈A} ∥x_{α}∥ < ∞ then ∑_{α∈A} x_{α} exists in X, see Exercise 7.12. However, unlike the case of real valued sums the existence of ∑_{α∈A} x_{α} does not imply ∑_{α∈Λ} ∥x_{α}∥ < ∞. See Proposition 14.19 below, from which one may manufacture counter-examples to this false premise.

Lemma 7.16. Suppose that {x_{α} ∈ X : α ∈ A} is a given collection of vectors in a normed space, X.

1. If s = ∑_{α∈A} x_{α} ∈ X exists and T : X → Y is a bounded linear map between normed spaces, then ∑_{α∈A} T x_{α} exists in Y and

T s = T ∑_{α∈A} x_{α} = ∑_{α∈A} T x_{α}.

2. If s = ∑_{α∈A} x_{α} exists in X then for every ε > 0 there exists Γ_{ε} ⊂⊂ A such that ∥∑_{α∈Λ} x_{α}∥ < ε for all Λ ⊂⊂ A \ Γ_{ε}.

3. If s = ∑_{α∈A} x_{α} exists in X, the set Γ := {α ∈ A : x_{a} ≠ 0} is at most countable. Moreover if Γ is infinite and {α_{n}}_{n=1}^{∞} is an enumeration of Γ, then

s = ∑_{n=1}^{∞} x_{α_{n}} := lim_{N→∞} ∑_{n=1}^{N} x_{α_{n}}.  (7.4)

4. If we further assume that X is a Banach space and suppose for all ε > 0 there exists Γ_{ε} ⊂⊂ A such that ∥∑_{α∈Λ} x_{α}∥ < ε whenever Λ ⊂⊂ A \ Γ_{ε}, then ∑_{α∈A} x_{α} exists in X.

Proof.

1. Let Γ_{ε} be as in Definition 7.15 and Λ ⊂⊂ A such that Γ_{ε} ⊂ Λ. Then

∥T s − ∑_{α∈Λ} T x_{α}∥ ≤ ∥T ∥ ∥s − ∑_{α∈Λ} x_{α}∥ < ∥T ∥ ε

which shows that ∑_{α∈Λ} T x_{α} exists and is equal to T s.

2. Suppose that s = ∑_{α∈A} x_{α} exists and ε > 0. Let Γ_{ε} ⊂⊂ A be as in Definition 7.15. Then for Λ ⊂⊂ A \ Γ_{ε},

∥∑_{α∈Λ} x_{α}∥ = ∥ ∑_{α∈Γ_{ε}∪Λ} x_{α} − ∑_{α∈Γ_{ε}} x_{α}∥

≤ ∥ ∑_{α∈Γ_{ε}∪Λ} x_{α} − s∥ + ∥ ∑_{α∈Γ_{ε}} x_{α} − s∥ < 2ε.

3. If s = ∑_{α∈A} x_{α} exists in X, for each n ∈ ℕ there exists a finite subset Γ_{n} ⊂ A such that ∥∑_{α∈Λ} x_{α}∥ < \frac{1}{n} for all Λ ⊂⊂ A \ Γ_{n}. Without loss of generality we may assume x_{α} ≠ 0 for all α ∈ Γ_{n}. Let Γ_{∞} := ∪_{n=1}^{∞}Γ_{n} – a countable subset of A. Then for any β ∉ Γ_{∞}, we have {β} ∩ Γ_{n} = ∅ and therefore

∥x_{β}∥ = ∥ ∑_{α∈\{β\}} x_{α}∥ ≤ \frac{1}{n} → 0 as n → ∞.

Let {α_{n}}_{n=1}^{∞} be an enumeration of Γ and define γ_{N} := {α_{n} : 1 ≤ n ≤ N} . Since for any M ∈ ℕ, γ_{N} will eventually contain Γ_{M} for N sufficiently large, we have

lim sup_{N→∞} ∥s − ∑_{n=1}^{N} x_{α_{n}} ∥ ≤ \frac{1}{M} → 0 as M → ∞.

Therefore Eq. (7.4) holds.

4. For n ∈ ℕ, let Γ_{n} ⊂⊂ A such that ∥∑_{α∈Λ} x_{α}∥ < \frac{1}{n} for all Λ ⊂⊂ A \ Γ_{n}. Define γ_{n} := ∪_{k=1}^{n}Γ_{k} ⊂ A and s_{n} := ∑_{α∈γ_{n}} x_{α}. Then for m > n,

∥s_{m} − s_{n}∥ = ∥ ∑_{α∈γ_{m}\γ_{n}} x_{α}∥ ≤ 1/n → 0 as m, n → ∞.

Therefore {s_{n}}_{n=1}^{∞} is Cauchy and hence convergent in X, because X is a Banach space. Let s := lim_{n→∞} s_{n}. Then for Λ ⊂⊂ A such that γ_{n} ⊂ Λ, we have

∥s − ∑_{α∈Λ} x_{α}∥ ≤ ∥s − s_{n}∥ + ∥ ∑_{α∈Λ\γ_{n}} x_{α}∥ ≤ ∥s − s_{n}∥ + \frac{1}{n}.

Since the right side of this equation goes to zero as n → ∞, it follows that ∑_{α∈A} x_{α} exists and is equal to s.

7.4 Inverting Elements in L(X)

Definition 7.17. A linear map T : X → Y is an isometry if ∥T x∥_{Y} = ∥x∥_{X} for all x ∈ X. T is said to be invertible if T is a bijection and T^{−1} is bounded.

Notation 7.18 We will write GL(X, Y) for those T ∈ L(X, Y) which are invertible. If X = Y we simply write L(X) and GL(X) for L(X, X) and GL(X, X) respectively.

Proposition 7.19. Suppose X is a Banach space and Λ ∈ L(X) := L(X, X)

satisfies ∑_{n=0}^{∞} ∥Λ^{n}∥ < ∞. Then I − Λ is invertible and

(I − Λ)^{−1} = “\frac{1}{I − Λ}” = ∑_{n=0}^{∞} Λ^{n} and ∥(I − Λ)^{−1}∥ ≤ ∑_{n=0}^{∞} ∥Λ^{n}∥.

In particular if ∥Λ∥ < 1 then the above formula holds and

∥(I − Λ)^{−1}∥ ≤ \frac{1}{1 − ∥Λ∥}.

Proof. Since L(X) is a Banach space and ∑_{n=0}^{∞} ∥Λ^{n}∥ < ∞, it follows from

Theorem 7.13 that

S := lim_{N→∞} S_{N} := lim_{N→∞} ∑_{n=0}^{N} Λ^{n}

7.4 Inverting Elements in L(X) 69

exists in L(X). Moreover, by Lemma 7.11,

(I − Λ) S = (I − Λ) lim_{N→∞} S_{N} = lim_{N→∞} (I − Λ) S_{N}

= lim_{N→∞} (I − Λ) ∑_{n=0}^{N} Λ^{n} = lim_{N→∞}(I − Λ^{N+1}) = I

and similarly S (I − Λ) = I. This shows that (I − Λ)^{−1} exists and is equal to S. Moreover, (I − Λ)^{−1} is bounded because

∥(I − Λ)^{−1}∥ = ∥S∥ ≤ ∑_{n=0}^{∞} ∥Λ^{n}∥.

If we further assume ∥Λ∥ < 1, then ∥Λ^{n}∥ ≤ ∥Λ∥^{n} and

∑_{n=0}^{∞} ∥Λ^{n}∥ ≤ ∑_{n=0}^{∞} ∥Λ∥^{n} = \frac{1}{1 − ∥Λ∥} < ∞.

Corollary 7.20. Let X and Y be Banach spaces. Then GL(X, Y) is an open (possibly empty) subset of L(X, Y). More specifically, if A ∈ GL(X, Y) and B ∈ L(X, Y) satisfies

∥B − A∥ < ∥A^{−1}∥^{−1}  (7.5)

then B ∈ GL(X, Y)

B^{−1} = ∑_{n=0}^{∞} [I_{X} − A^{−1}B]^{n} A^{−1} ∈ L(Y, X),  (7.6)

∥B^{−1}∥ ≤ ∥A^{−1}∥\frac{1}{1 − ∥A^{−1}∥ ∥A − B∥}  (7.7)

and

∥B^{−1} − A^{−1}∥ ≤ \frac{∥A^{−1}∥^{2} ∥A − B∥}{1 − ∥A^{−1}∥ ∥A − B∥}.  (7.8)

In particular the map

A ∈ GL(X, Y) → A^{−1} ∈ GL(Y, X)  (7.9)

is continuous.

Proof. Let A and B be as above, then

B = A − (A − B) = A [I_{X} − A^{−1}(A − B))] = A(I_{X} − Λ)

where Λ : X → X is given by

Λ := A^{−1}(A − B) = I_{X} − A^{−1}B.

Now

∥Λ∥ = ∥A^{−1}(A − B))∥ ≤ ∥A^{−1}∥ ∥A − B∥ < ∥A^{−1}∥∥A^{−1}∥^{−1} = 1.

Therefore I −Λ is invertible and hence so is B (being the product of invertible elements) with

B^{−1} = (I_{X} − Λ)^{−1}A^{−1} = [I_{X} − A^{−1}(A − B))]^{−1} A^{−1}.

Taking norms of the previous equation gives

∥B^{−1}∥ ≤ ∥(I_{X} − Λ)^{−1}∥ ∥A^{−1}∥ ≤ ∥A^{−1}∥\frac{1}{1 − ∥Λ∥}

≤ \frac{∥A^{−1}∥}{1 − ∥A^{−1}∥ ∥A − B∥}

which is the bound in Eq. (7.7). The bound in Eq. (7.8) holds because

∥B^{−1} − A^{−1}∥ = ∥B^{−1} (A − B) A^{−1}∥ ≤ ∥B^{−1}∥ ∥A^{−1}∥ ∥A − B∥

≤ \frac{∥A^{−1}∥^{2} ∥A − B∥}{1 − ∥A^{−1}∥ ∥A − B∥}.

For an application of these results to linear ordinary differential equations, see Section 8.3.

7.5 Exercises

Exercise 7.4. Let (X, ∥·∥) be a normed space over 𝔽 (ℝ or ℂ). Show the map

(λ, x, y) ∈ 𝔽 × X × X → x + λy ∈ X

is continuous relative to the norm on 𝔽 × X × X defined by

∥(λ, x, y)∥_{𝔽×X×X} := |λ| + ∥x∥ + ∥y∥ .

(See Exercise 10.25 for more on the metric associated to this norm.) Also show that ∥·∥ : X → [0, ∞) is continuous.

Exercise 7.5. Let X = ℕ and for p, q ∈ [1, ∞) let ∥·∥_{p} denote the ℓ^{p}(ℕ) – norm. Show ∥·∥_{p} and ∥·∥_{q} are inequivalent norms for p ≠ q by showing

sup_{f≠0} \begin{matrix}∥f∥ & p \\ ∥f∥ & q \end{matrix} = ∞ if p < q.

Exercise 7.6. Suppose that (X, ∥·∥) is a normed space and S ⊂ X is a linear subspace.

1. Show the closure S̄ of S is also a linear subspace.

2. Now suppose that X is a Banach space. Show that S with the inherited norm from X is a Banach space iff S is closed.

Exercise 7.7. Folland Problem 5.9. Showing C^{k}([0, 1]) is a Banach space.

Exercise 7.8. Suppose that X, Y and Z are Banach spaces and Q : X ×Y → Z is a bilinear form, i.e. we are assuming x ∈ X → Q (x, y) ∈ Z is linear for each y ∈ Y and y ∈ Y → Q (x, y) ∈ Z is linear for each x ∈ X. Show Q is continuous relative to the product norm, ∥(x, y)∥_{X×Y} := ∥x∥_{X} + ∥y∥_{Y} , on X × Y iff there is a constant M < ∞ such that

∥Q (x, y)∥_{Z} ≤ M ∥x∥_{X} · ∥y∥_{Y} for all (x, y) ∈ X × Y.  (7.10)

Then apply this result to prove Lemma 7.11.

Exercise 7.9. Let d : C(ℝ) × C(ℝ) → [0, ∞) be defined by

d(f, g) = ∑_{n=1}^{∞} 2^{−n} \frac{∥f − g∥_{n}}{1 + ∥f − g∥_{n}} ,

where ∥f∥_{n} := sup{|f(x)| : |x| ≤ n} = max{|f(x)| : |x| ≤ n}.

1. Show that d is a metric on C(ℝ).

2. Show that a sequence {f_{n}}_{n=1}^{∞} ⊂ C(ℝ) converges to f ∈ C(ℝ) as n → ∞ iff f_{n} converges to f uniformly on bounded subsets of ℝ.

3. Show that (C(ℝ), d) is a complete metric space.

Exercise 7.10. Let X = C([0, 1], ℝ) and for f ∈ X, let

∥f∥_{1} := ∫_{0}^{1} |f(t)| dt.

Show that (X, ∥·∥_{1}) is normed space and show by example that this space is not complete. Hint: For the last assertion find a sequence of {f_{n}}_{n=1}^{∞} ⊂ X which is “trying” to converge to the function f = 1_{[ \frac{1}{2} ,1]} ∉ X.

Exercise 7.11. Let (X, ∥·∥_{1}) be the normed space in Exercise 7.10. Compute the closure of A when

1. A = {f ∈ X : f (1/2) = 0} .

2. A = {f ∈ X : sup_{t∈[0,1]} f (t) ≤ 5} .

3. A = {f ∈ X : ∫_{0}^{1/2} f (t) dt = 0} .

Exercise 7.12. Suppose {x_{α} ∈ X : α ∈ A} is a given collection of vectors in a Banach space X. Show ∑_{α∈A} x_{α} exists in X and

∥∑_{α∈A} x_{α}∥ ≤ ∑_{α∈A} ∥x_{α}∥

if ∑_{α∈A} ∥x_{α}∥ < ∞. That is to say “absolute convergence” implies convergence in a Banach space.

Exercise 7.13. Suppose X is a Banach space and {f_{n} : n ∈ ℕ} is a sequence in X such that lim_{n→∞} f_{n} = f ∈ X. Show s_{N} := \frac{1}{N} ∑_{n=1}^{N} f_{n} for N ∈ ℕ is still a convergent sequence and

lim_{N→∞} \frac{1}{N} ∑_{n=1}^{N} f_{n} = lim_{N→∞} s_{N} = f.

Exercise 7.14 (Dominated Convergence Theorem Again). Let X be a Banach space, A be a set and suppose f_{n} : A → X is a sequence of functions such that f (α) := lim_{n→∞} f_{n} (α) exists for all α ∈ A. Further assume there exists a summable function g : A → [0, ∞) such that ∥f_{n} (α)∥ ≤ g (α) for all α ∈ A. Show ∑_{α∈A} f (α) exists in X and

lim_{n→∞} ∑_{α∈A} f_{n} (α) = ∑_{α∈A} f (α) .

The Riemann Integral

In this Chapter, the Riemann integral for Banach space valued functions is defined and developed. Our exposition will be brief, since the Lebesgue integral and the Bochner Lebesgue integral will subsume the content of this chapter. In Definition 11.1 below, we will give a general notion of a compact subset of a “topological” space. However, by Corollary 11.9 below, when we are working with subsets of ℝ^{d} this definition is equivalent to the following definition.

Definition 8.1. A subset A ⊂ ℝ^{d} is said to be compact if A is closed and bounded.

Theorem 8.2. Suppose that K ⊂ ℝ^{d} is a compact set and f ∈ C (K, X) . Then

1. Every sequence {u_{n}}_{n=1}^{∞} ⊂ K has a convergent subsequence.

2. The function f is uniformly continuous on K, namely for every ε > 0 there exists a δ > 0 only depending on ε such that ∥f (u) − f (v)∥ < ε whenever u, v ∈ K and |u − v| < δ where |·| is the standard Euclidean norm on ℝ^{d}.

Proof.

1. (This is a special case of Theorem 11.7 and Corollary 11.9 below.) Since K is bounded, K ⊂ [−R, R]^{d} for some sufficiently large d. Let t_{n} be the first component of u_{n} so that t_{n} ∈ [−R, R] for all n. Let J_{1} = [0, R] if t_{n} ∈ J_{1} for infinitely many n otherwise let J_{1} = [−R, 0]. Similarly split J_{1} in half and let J_{2} ⊂ J_{1} be one of the halves such that t_{n} ∈ J_{2} for infinitely many n. Continue this way inductively to find a nested sequence of intervals J_{1} ⊃ J_{2} ⊃ J_{3} ⊃ J_{4} ⊃ … such that the length of J_{k} is 2^{−(k−1)}R and for each k, t_{n} ∈ J_{k} for infinitely many n. We may now choose a subsequence, {n_{k}}_{k=1}^{∞} of {n}_{n=1}^{∞} such that τ_{k} := t_{n_{k}} ∈ J_{k} for all k. The sequence {τ_{k}}_{k=1}^{∞} is Cauchy and hence convergent. Thus by replacing {u_{n}}_{n=1}^{∞} by a subsequence if necessary we may assume the first component of {u_{n}}_{n=1}^{∞} is convergent. Repeating this argument for the second, then the third and all the way through the d^{th} – components of {u_{n}}_{n=1}^{∞} , we may, by passing to further subsequences, assume all of the components of u_{n} are convergent. But this implies lim u_{n} = u exists and since K is closed, u ∈ K.

2. (This is a special case of Exercise 11.6 below.) If f were not uniformly continuous on K, there would exists an ε > 0 and sequences {u_{n}}_{n=1}^{∞} and {v_{n}}_{n=1}^{∞} in K such that

∥f (u_{n}) − f (v_{n})∥ ≥ ε while lim_{n→∞} |u_{n} − v_{n}| = 0.

By passing to subsequences if necessary we may assume that lim_{n→∞} u_{n} and lim_{n→∞} v_{n} exists. Since lim_{n→∞} |u_{n} − v_{n}| = 0, we must have

lim_{n→∞} u_{n} = u = lim_{n→∞} v_{n}

for some u ∈ K. Since f is continuous, vector addition is continuous and the norm is continuous, we may now conclude that

ε ≤ lim_{n→∞} ∥f (u_{n}) − f (v_{n})∥ = ∥f (u) − f (u)∥ = 0

which is a contradiction.

For the remainder of the chapter, let [a, b] be a fixed compact interval and X be a Banach space. The collection S = S([a, b], X) of step functions, f : [a, b] → X, consists of those functions f which may be written in the form

f(t) = x_{0}1_{[a,t_{1}]}(t) + ∑_{i=1}^{n−1} x_{i}1_{(t_{i},t_{i+1}]}(t),  (8.1)

where π := {a = t_{0} < t_{1} < ⋯ < t_{n} = b} is a partition of [a, b] and x_{i} ∈ X. For f as in Eq. (8.1), let

I(f) := ∑_{i=0}^{n−1}(t_{i+1} − t_{i})x_{i} ∈ X.  (8.2)

Exercise 8.1. Show that I(f) is well defined, independent of how f is represented as a step function. (Hint: show that adding a point to a partition π of [a, b] does not change the right side of Eq. (8.2).) Also verify that I : S → X is a linear operator.

Notation 8.3 Let S̄ denote the closure of S inside the Banach space, ℓ^{∞}([a, b], X) as defined in Remark 7.6.

The following simple “Bounded Linear Transformation” theorem will often be used in the sequel to define linear transformations.

Theorem 8.4 (B. L. T. Theorem). Suppose that Z is a normed space, X is a Banach space, and S ⊂ Z is a dense linear subspace of Z. If T : S → X is a bounded linear transformation (i.e. there exists C < ∞ such that ∥T z∥ ≤ C ∥z∥ for all z ∈ S), then T has a unique extension to an element T̄ ∈ L(Z, X) and this extension still satisfies

∥T̄z∥ ≤ C ∥z∥ for all z ∈ S̄.

Exercise 8.2. Prove Theorem 8.4.

Proposition 8.5 (Riemann Integral). The linear function I : S → X extends uniquely to a continuous linear operator Ī from S̄ to X and this operator satisfies,

∥Ī(f)∥ ≤ (b − a) ∥f∥_{∞} for all f ∈ S̄.  (8.3)

Furthermore, C([a, b], X) ⊂ S̄ ⊂ ℓ^{∞}([a, b], X) and for f ∈, Ī(f) may be computed as

Ī(f) = lim_{|π|→0} ∑_{i=0}^{n−1} f(c_{i}^{π})(t_{i+1} − t_{i})  (8.4)

where π := {a = t_{0} < t_{1} < ⋯ < t_{n} = b} denotes a partition of [a, b], |π| = max {|t_{i+1} − t_{i}| : i = 0, …, n − 1} is the mesh size of π and c_{i}^{π} may be chosen arbitrarily inside [t_{i}, t_{i+1}]. See Figure 8.1.

Fig. 8.1. The usual picture associated to the Riemann integral.

Proof. Taking the norm of Eq. (8.2) and using the triangle inequality shows,

∥I(f)∥ ≤ ∑_{i=0}^{n−1}(t_{i+1} − t_{i})∥x_{i}∥ ≤ ∑_{i=0}^{n−1}(t_{i+1} − t_{i})∥f∥_{∞} ≤ (b − a)∥f∥_{∞}. (8.5)

The existence of Ī satisfying Eq. (8.3) is a consequence of Theorem 8.4. Given f ∈ C([a, b], X), π := {a = t_{0} < t_{1} < ⋯ < t_{n} = b} a partition of [a, b], and c_{i}^{π} ∈ [t_{i}, t_{i+1}] for i = 0, 1, 2 …, n − 1, let f_{π} ∈ S be defined by

f_{π}(t) := f(c_{0})_{0}1_{[t_{0},t_{1}]}(t) + ∑_{i=1}^{n−1} f(c_{i}^{π})1_{(t_{i},t_{i+1}]}(t).

Then by the uniform continuity of f on [a, b] (Theorem 8.2), lim_{|π|→0} ∥f − f_{π}∥_{∞} = 0 and therefore f ∈ S̄. Moreover,

I (f) = lim_{|π|→0} I(f_{π}) = lim_{|π|→0} ∑_{i=0}^{n−1} f(c_{i}^{π})(t_{i+1} − t_{i})

which proves Eq. (8.4).

If f_{n} ∈ S and f ∈ S̄ such that lim_{n→∞} ∥f − f_{n}∥_{∞} = 0, then for a ≤ α < β ≤ b, then 1_{(α,β]}f_{n} ∈ S and lim_{n→∞} ∥1_{(α,β]}f − 1_{(α,β]}f_{n}∥_{∞} = 0. This shows 1_{(α,β]}f ∈ S̄ whenever f ∈ S̄.

Notation 8.6 For f ∈ S̄ and a ≤ α ≤ β ≤ b we will write denote Ī(1_{(α,β]}f) by ∫_{α}^{β} f(t) dt or ∫_{(α,β]} f(t)dt. Also following the usual convention, if a ≤ β ≤ α ≤ b, we will let

∫_{α}^{β} f(t) dt = −Ī(1_{(β,α]}f) = − ∫_{β}^{α} f(t) dt.

The next Lemma, whose proof is left to the reader contains some of the many familiar properties of the Riemann integral.

Lemma 8.7. For f ∈ S̄([a, b], X) and α, β, γ ∈ [a, b], the Riemann integral satisfies:

1. ∥∫_{α}^{β} f(t) dt∥_{X} ≤ (β − α) sup {∥f(t)∥ : α ≤ t ≤ β} .

2. ∫_{α}^{γ} f(t) dt = ∫_{α}^{β} f(t) dt + ∫_{β}^{γ} f(t) dt.

3. The function G(t) := ∫_{a}^{t} f(τ)dτ is continuous on [a, b].

4. If Y is another Banach space and T ∈ L(X, Y), then T f ∈ S̄([a, b], Y) and

T (∫_{α}^{β} f(t) dt) = ∫_{α}^{β} T f(t) dt.

5. The function t → ∥f(t)∥_{X} is in S̄([a, b], ℝ) and

∥∫_{a}^{b} f(t) dt∥_{X} ≤ ∫_{a}^{b} ∥f(t)∥_{X} dt.

6. If f, g ∈ S̄([a, b], ℝ) and f ≤ g, then

∫_{a}^{b} f(t) dt ≤ ∫_{a}^{b} g(t) dt.

Exercise 8.3. Prove Lemma 8.7.

8.1 The Fundamental Theorem of Calculus

Our next goal is to show that our Riemann integral interacts well with differentiation, namely the fundamental theorem of calculus holds. Before doing this we will need a couple of basic definitions and results of differential calculus, more details and the next few results below will be done in greater detail in Chapter 16.

Definition 8.8. Let (a, b) ⊂ ℝ. A function f : (a, b) → X is differentiable at t ∈ (a, b) iff

L := lim_{h→0} (h^{−1} [f(t + h) − f(t)]) = lim_{h→0} “\frac{f(t + h) − f(t)}{h}”

exists in X. The limit L, if it exists, will be denoted by ḟ(t) or \frac{df}{dt} (t). We also say that f ∈ C^{1}((a, b) → X) if f is differentiable at all points t ∈ (a, b) and ḟ ∈ C((a, b) → X).

As for the case of real valued functions, the derivative operator \frac{d}{dt} is easily seen to be linear. The next two results have proves very similar to their real valued function analogues.

Lemma 8.9 (Product Rules). Suppose that t → U (t) ∈ L (X) , t → V (t) ∈ L (X) and t → x (t) ∈ X are differentiable at t = t_{0}, then

1. \frac{d}{dt}|_{t_{0}} [U (t) x (t)] ∈ X exists and

\frac{d}{dt}|_{t_{0}} [U (t) x (t)] = [U̇ (t_{0}) x (t_{0}) + U (t_{0}) ẋ (t_{0})]

and

2. \frac{d}{dt}|_{t_{0}} [U (t) V (t)] ∈ L (X) exists and

\frac{d}{dt}|_{t_{0}} [U (t) V (t)] = [U̇ (t_{0}) V (t_{0}) + U (t_{0}) V̇ (t_{0})] .

3. If U (t_{0}) is invertible, then t → U (t)^{−1} is differentiable at t = t_{0} and

\frac{d}{dt}|_{t_{0}}U (t)^{−1} = −U (t_{0})^{−1} U̇ (t_{0}) U (t_{0})^{−1} .  (8.6)

Proof. The reader is asked to supply the proof of the first two items in Exercise 8.10. Before proving item 3., let us assume that U (t)^{−1} is differentiable, then using the product rule we would learn

0 = \frac{d}{dt}|_{t_{0}}I = \frac{d}{dt}|_{t_{0}} [U (t)^{−1} U (t)] = [\frac{d}{dt}|_{t_{0}}U (t)^{−1}] U (t_{0}) + U (t_{0})^{−1} U̇ (t_{0}) .

Solving this equation for \frac{d}{dt}|_{t_{0}}U (t)^{−1} gives the formula in Eq. (8.6). The problem with this argument is that we have not yet shown t → U (t)^{−1} is invertible at t_{0}. Here is the formal proof. Since U (t) is differentiable at t_{0}, U (t) → U (t_{0}) as t → t_{0} and by Corollary 7.20, U (t_{0} + h) is invertible for h near 0 and

U (t_{0} + h)^{−1} → U (t_{0})^{−1} as h → 0.

Therefore, using Lemma 7.11, we may let h → 0 in the identity,

\frac{U (t_{0} + h)^{−1} − U (t_{0})^{−1}}{h} = U (t_{0} + h)^{−1} (\frac{U (t_{0}) − U (t_{0} + h)}{h}) U (t_{0})^{−1} ,

to learn

lim_{h→0} \frac{U (t_{0} + h)^{−1} − U (t_{0})^{−1}}{h} = −U (t_{0})^{−1} U̇ (t_{0}) U (t_{0})^{−1} .

Proposition 8.10 (Chain Rule). Suppose s → x (s) ∈ X is differentiable at s = s_{0} and t → T (t) ∈ ℝ is differentiable at t = t_{0} and T (t_{0}) = s_{0}, then t → x (T (t)) is differentiable at t_{0} and

\frac{d}{dt}|_{t_{0}}x (T (t)) = x′ (T (t_{0})) T′ (t_{0}) .

The proof of the chain rule is essentially the same as the real valued function case, see Exercise 8.11.

Proposition 8.11. Suppose that f : [a, b] → X is a continuous function such that ḟ(t) exists and is equal to zero for t ∈ (a, b). Then f is constant.

Proof. Let ε > 0 and α ∈ (a, b) be given. (We will later let ε ↓ 0.) By the definition of the derivative, for all τ ∈ (a, b) there exists δ_{τ} > 0 such that

∥f(t) − f(τ)∥ = ∥f(t) − f(τ) − ḟ(τ)(t − τ)∥ ≤ ε |t − τ| if |t − τ| < δ_{τ} .

(8.7)

Let

A = {t ∈ [α, b] : ∥f(t) − f(α)∥ ≤ ε(t − α)}  (8.8)

and t_{0} be the least upper bound for A. We will now use a standard argument which is referred to as continuous induction to show t_{0} = b. Eq. (8.7) with τ = α shows t_{0} > α and a simple continuity argument shows t_{0} ∈ A, i.e.

∥f(t_{0}) − f(α)∥ ≤ ε(t_{0} − α).  (8.9)

For the sake of contradiction, suppose that t_{0} < b. By Eqs. (8.7) and (8.9),

∥f(t) − f(α)∥ ≤ ∥f(t) − f(t_{0})∥ + ∥f(t_{0}) − f(α)∥
≤ ε(t_{0} − α) + ε(t − t_{0}) = ε(t − α)

for 0 ≤ t − t_{0} < δ_{t_{0}} which violates the definition of t_{0} being an upper bound. Thus we have shown b ∈ A and hence

∥f(b) − f(α)∥ ≤ ε(b − α).

Since ε > 0 was arbitrary we may let ε ↓ 0 in the last equation to conclude f(b) = f (α) . Since α ∈ (a, b) was arbitrary it follows that f(b) = f (α) for all α ∈ (a, b] and then by continuity for all α ∈ [a, b], i.e. f is constant.

Remark 8.12. The usual real variable proof of Proposition 8.11 makes use Rolle’s theorem which in turn uses the extreme value theorem. This latter theorem is not available to vector valued functions. However with the aid of the Hahn Banach Theorem ?? below and Lemma 8.7, it is possible to reduce the proof of Proposition 8.11 and the proof of the Fundamental Theorem of Calculus 8.13 to the real valued case, see Exercise ??.

Theorem 8.13 (Fundamental Theorem of Calculus). Suppose that f ∈ C([a, b], X), Then

1. \frac{d}{dt} ∫_{a}^{t} f(τ) dτ = f(t) for all t ∈ (a, b).

2. Now assume that F ∈ C([a, b], X), F is continuously differentiable on (a, b) (i.e. Ḟ (t) exists and is continuous for t ∈ (a, b)) and Ḟ extends to a continuous function on [a, b] which is still denoted by Ḟ . Then

∫_{a}^{b} Ḟ (t) dt = F (b) − F (a).

Proof. Let h > 0 be a small number and consider

∥∫_{a}^{t+h} f(τ)dτ − ∫_{a}^{t} f(τ)dτ − f(t)h∥ = ∥∫_{t}^{t+h}(f(τ) − f(t)) dτ∥

≤ ∫_{t}^{t+h} ∥(f(τ) − f(t))∥ dτ ≤ hε(h),

where ε(h) := max_{τ∈[t,t+h]} ∥(f(τ) − f(t))∥. Combining this with a similar computation when h < 0 shows, for all h ∈ ℝ sufficiently small, that

∥ ∫_{a}^{t+h} f(τ)dτ − ∫_{a}^{t} f(τ)dτ − f(t)h∥ ≤ |h|ε(h),

where now ε(h) := max_{τ∈[t−|h|,t+|h|]} ∥(f(τ) − f(t))∥. By continuity of f at t, ε(h) → 0 and hence \frac{d}{dt} ∫_{a}^{t} f(τ) dτ exists and is equal to f(t). For the second item, set G(t) := ∫_{a}^{t} Ḟ (τ) dτ − F (t). Then G is continuous by Lemma 8.7 and Ġ(t) = 0 for all t ∈ (a, b) by item 1. An application of Proposition 8.11 shows G is a constant and in particular G(b) = G(a), i.e. ∫_{a}^{b} Ḟ (τ) dτ −F (b) = −F (a).

Corollary 8.14 (Mean Value Inequality). Suppose that f : [a, b] → X is a continuous function such that ḟ(t) exists for t ∈ (a, b) and ḟ extends to a continuous function on [a, b]. Then

∥f(b) − f(a)∥ ≤ ∫_{a}^{b} ∥ḟ(t)∥dt ≤ (b − a) · ∥ḟ∥_{∞} .  (8.10)

Proof. By the fundamental theorem of calculus, f(b) − f(a) = ∫_{a}^{b} ḟ(t)dt and then by Lemma 8.7,

∥f(b) − f(a)∥ = ∥∫_{a}^{b} ḟ(t)dt∥ ≤ ∫_{a}^{b} ∥ḟ(t)∥dt

≤ ∫_{a}^{b} ∥ḟ∥_{∞} dt = (b − a) · ∥ḟ∥_{∞} .

Corollary 8.15 (Change of Variable Formula). Suppose that f ∈ C([a, b], X) and T : [c, d] → (a, b) is a continuous function such that T (s) is continuously differentiable for s ∈ (c, d) and T′ (s) extends to a continuous function on [c, d]. Then

∫_{c}^{d} f (T (s)) T′ (s) ds = ∫_{T (c)}^{T (d)} f (t) dt.

Proof. For s ∈ (a, b) define F (t) := ∫_{T (c)}^{t} f (τ) dτ. Then F ∈ C^{1} ((a, b) , X) and by the fundamental theorem of calculus and the chain rule,

\frac{d}{ds}F (T (s)) = F′ (T (s)) T′ (s) = f (T (s)) T′ (s) .

Integrating this equation on s ∈ [c, d] and using the chain rule again gives

∫_{c}^{d} f (T (s)) T′ (s) ds = F (T (d)) − F (T (c)) = ∫_{T (c)}^{T (d)} f (t) dt.

8.2 Integral Operators as Examples of Bounded Operators

In the examples to follow all integrals are the standard Riemann integrals and we will make use of the following notation.

Notation 8.16 Given an open set U ⊂ ℝ^{d}, let C_{c} (U) denote the collection of real valued continuous functions f on U such that

supp(f) := \overline{\{x ∈ U : f (x) ≠ 0\}}

is a compact subset of U.

Example 8.17. Suppose that K : [0, 1] × [0, 1] → ℂ is a continuous function. For f ∈ C([0, 1]), let

T f(x) = ∫_{0}^{1} K(x, y)f(y)dy.

Since

|T f(x) − T f(z)| ≤ ∫_{0}^{1} |K(x, y) − K(z, y)| |f(y)| dy
≤ ∥f∥_{∞} max_{y} |K(x, y) − K(z, y)|  (8.11)

and the latter expression tends to 0 as x → z by uniform continuity of K. Therefore T f ∈ C([0, 1]) and by the linearity of the Riemann integral, T : C([0, 1]) → C([0, 1]) is a linear map. Moreover,

|T f(x)| ≤ ∫_{0}^{1} |K(x, y)| |f(y)| dy ≤ ∫_{0}^{1} |K(x, y)| dy · ∥f∥_{∞} ≤ A ∥f∥_{∞}

where

A := sup_{x∈[0,1]} ∫_{0}^{1} |K(x, y)| dy < ∞.  (8.12)

This shows ∥T ∥ ≤ A < ∞ and therefore T is bounded. We may in fact show ∥T ∥ = A. To do this let x_{0} ∈ [0, 1] be such that

sup_{x∈[0,1]} ∫_{0}^{1} |K(x, y)| dy = ∫_{0}^{1} |K(x_{0}, y)| dy.

Such an x_{0} can be found since, using a similar argument to that in Eq. (8.11), x → ∫_{0}^{1} |K(x, y)| dy is continuous. Given ε > 0, let

f_{ε}(y) := \frac{\overline{K(x_{0}, y)}}{\sqrt{ε + |K(x_{0}, y)|^{2}}}

and notice that lim_{ε↓0} ∥f_{ε}∥_{∞} = 1 and

∥T f_{ε}∥_{∞} ≥ |T f_{ε}(x_{0})| = T f_{ε}(x_{0}) = ∫_{0}^{1} \begin{matrix}|K(x_{0}, y)| & 2 \\ √ & \overline{ε + |K(x_{0}, y)|^{2}} \end{matrix} dy.

Therefore,

∥T ∥ ≥ lim_{ε↓0} \frac{1}{∥f_{ε}∥_{∞}} ∫_{0}^{1} \begin{matrix}|K(x_{0}, y)| & 2 \\ √ & \overline{ε + |K(x_{0}, y)|^{2}} \end{matrix} dy

= lim_{ε↓0} ∫_{0}^{1} \begin{matrix}|K(x_{0}, y)| & 2 \\ √ & \overline{ε + |K(x_{0}, y)|^{2}} \end{matrix} dy = A

since

0 ≤ |K(x_{0}, y)| − \begin{matrix}|K(x_{0}, y)| & 2 \\ √ & \overline{ε + |K(x_{0}, y)|^{2}} \end{matrix}

= \frac{|K(x_{0}, y)|}{\sqrt{ε + |K(x_{0}, y)|^{2}}} [\sqrt{ε + |K(x_{0}, y)|^{2}} − |K(x_{0}, y)|]

≤ \sqrt{ε + |K(x_{0}, y)|^{2}} − |K(x_{0}, y)|

and the latter expression tends to zero uniformly in y as ε ↓ 0.

We may also consider other norms on C([0, 1]). Let (for now) L^{1} ([0, 1]) denote C([0, 1]) with the norm

∥f∥_{1} = ∫_{0}^{1} |f(x)| dx,

then T : L^{1} ([0, 1], dm) → C([0, 1]) is bounded as well. Indeed, let M = sup {|K(x, y)| : x, y ∈ [0, 1]} , then

|(T f)(x)| ≤ ∫_{0}^{1} |K(x, y)f(y)| dy ≤ M ∥f∥_{1}

which shows ∥T f∥_{∞} ≤ M ∥f∥_{1} and hence,

∥T ∥_{L^{1}→C} ≤ max {|K(x, y)| : x, y ∈ [0, 1]} < ∞.

We can in fact show that ∥T ∥ = M as follows. Let (x_{0}, y_{0}) ∈ [0, 1]^{2} satisfying |K(x_{0}, y_{0})| = M. Then given ε > 0, there exists a neighborhood U = I × J of (x_{0}, y_{0}) such that |K(x, y) − K(x_{0}, y_{0})| < ε for all (x, y) ∈ U. Let f ∈ C_{c}(I, [0, ∞)) such that ∫_{0}^{1} f(x)dx = 1. Choose α ∈ ℂ such that |α| = 1 and αK(x_{0}, y_{0}) = M, then

|(T αf)(x_{0})| = |∫_{0}^{1} K(x_{0}, y)αf(y)dy| = |∫_{I} K(x_{0}, y)αf(y)dy|
≥ Re ∫_{I} αK(x_{0}, y)f(y)dy
≥ ∫_{I} (M − ε) f(y)dy = (M − ε) ∥αf∥_{L^{1}}

and hence

∥T αf∥_{C} ≥ (M − ε) ∥αf∥_{L^{1}}

showing that ∥T ∥ ≥ M − ε. Since ε > 0 is arbitrary, we learn that ∥T ∥ ≥ M and hence ∥T ∥ = M.

One may also view T as a map from T : C([0, 1]) → L^{1}([0, 1]) in which case one may show

∥T ∥_{L^{1}→C} ≤ ∫_{0}^{1} max_{y} |K(x, y)| dx < ∞.

8.3 Linear Ordinary Differential Equations

Let X be a Banach space, J = (a, b) ⊂ ℝ be an open interval with 0 ∈ J, h ∈ C(J → X) and A ∈ C(J → L(X)). In this section we are going to consider the ordinary differential equation,

ẏ(t) = A(t)y(t) + h (t) where y(0) = x ∈ X,  (8.13)

where y is an unknown function in C^{1}(J → X). This equation may be written in its equivalent (as the reader should verify) integral form, namely we are looking for y ∈ C(J, X) such that

y(t) = x + ∫_{0}^{t} h (τ) dτ + ∫_{0}^{t} A(τ)y(τ)dτ.  (8.14)

In what follows, we will abuse notation and use ∥·∥ to denote the operator norm on L (X) associated to then norm, ∥·∥ , on X and let ∥ϕ∥_{∞} := max_{t∈J} ∥ϕ(t)∥ for ϕ ∈ BC(J, X) or BC(J, L (X)).

Notation 8.18 For t ∈ ℝ and n ∈ ℕ, let

Δ_{n}(t) = \{ \begin{matrix}\{(τ_{1}, …, τ_{n}) ∈ ℝ^{n} : 0 ≤ τ_{1} ≤ ⋯ ≤ τ_{n} ≤ t\} & if & t ≥ 0 \\ \{(τ_{1}, …, τ_{n}) ∈ ℝ^{n} : t ≤ τ_{n} ≤ ⋯ ≤ τ_{1} ≤ 0\} & if & t ≤ 0 \end{matrix}

and also write dτ = dτ_{1} … dτ_{n} and

∫_{Δ_{n}(t)} f(τ_{1}, … τ_{n})dτ : = (−1)^{n·1_{t<0}} ∫_{0}^{t} dτ_{n} ∫_{0}^{τ_{n}} dτ_{n−1} … ∫_{0}^{τ_{2}} dτ_{1}f(τ_{1}, … τ_{n}).

Lemma 8.19. Suppose that ψ ∈ C (ℝ, ℝ) , then

(−1)^{n·1_{t<0}} ∫_{Δ_{n}(t)} ψ(τ_{1}) … ψ(τ_{n})dτ = \frac{1}{n!} (∫_{0}^{t} ψ(τ)dτ)^{n} .  (8.15)

Proof. Let Ψ(t) := ∫_{0}^{t} ψ(τ)dτ. The proof will go by induction on n. The case n = 1 is easily verified since

(−1)^{1·1_{t<0}} ∫_{Δ_{1}(t)} ψ(τ_{1})dτ_{1} = ∫_{0}^{t} ψ(τ)dτ = Ψ(t).

Now assume the truth of Eq. (8.15) for n − 1 for some n ≥ 2, then

(−1)^{n·1_{t<0}} ∫_{Δ_{n}(t)} ψ(τ_{1}) … ψ(τ_{n})dτ

= ∫_{0}^{t} dτ_{n} ∫_{0}^{τ_{n}} dτ_{n−1} … ∫_{0}^{τ_{2}} dτ_{1}ψ(τ_{1}) … ψ(τ_{n})

= ∫_{0}^{t} dτ_{n} \frac{Ψ^{n−1}(τ_{n})}{(n − 1)!}ψ(τ_{n}) = ∫_{0}^{t} dτ_{n} \frac{Ψ^{n−1}(τ_{n})}{(n − 1)!}Ψ̇(τ_{n})

= ∫_{0}^{Ψ(t)} \frac{u^{n−1}}{(n − 1)!}du = \frac{Ψ^{n}(t)}{n!},

wherein we made the change of variables, u = Ψ(τ_{n}), in the second to last equality.

Remark 8.20. Eq. (8.15) is equivalent to

∫_{Δ_{n}(t)} ψ(τ_{1}) … ψ(τ_{n})dτ = \frac{1}{n!} (∫_{Δ_{1}(t)} ψ(τ)dτ)^{n}

and another way to understand this equality is to view ∫_{Δ_{n}(t)} ψ(τ_{1}) … ψ(τ_{n})dτ as a multiple integral (see Chapter 20 below) rather than an iterated integral. Indeed, taking t > 0 for simplicity and letting S_{n} be the permutation group on {1, 2, …, n} we have

[0, t]^{n} = ∪_{σ∈S_{n}}{(τ_{1}, …, τ_{n}) ∈ ℝ^{n} : 0 ≤ τ_{σ1} ≤ ⋯ ≤ τ_{σn} ≤ t}

with the union being “essentially” disjoint. Therefore, making a change of variables and using the fact that ψ(τ_{1}) … ψ(τ_{n}) is invariant under permutations, we find

(∫_{0}^{t} ψ(τ)dτ)^{n} = ∫_{[0,t]^{n}} ψ(τ_{1}) … ψ(τ_{n})dτ

= ∑_{σ∈S_{n}} ∫_{{(τ_{1},...,τ_{n})∈ℝ^{n}:0≤τ_{σ1}≤···≤τ_{σn}≤t}} ψ(τ_{1}) … ψ(τ_{n})dτ

= ∑_{σ∈S_{n}} ∫_{{(s_{1},...,s_{n})∈ℝ^{n}:0≤s_{1}≤···≤s_{n}≤t}} ψ(s_{σ^{−1}1}) … ψ(s_{σ^{−1}n})ds

= ∑_{σ∈S_{n}} ∫_{{(s_{1},...,s_{n})∈ℝ^{n}:0≤s_{1}≤···≤s_{n}≤t}} ψ(s_{1}) … ψ(s_{n})ds

= n! ∫_{Δ_{n}(t)} ψ(τ_{1}) … ψ(τ_{n})dτ.

Theorem 8.21. Let ϕ ∈ BC(J, X), then the integral equation

y(t) = ϕ(t) + ∫_{0}^{t} A(τ)y(τ)dτ  (8.16)

has a unique solution given by

y(t) = ϕ(t) + ∑_{n=1}^{∞} (−1)^{n·1_{t<0}} ∫_{Δ_{n}(t)} A(τ_{n}) … A(τ_{1})ϕ(τ_{1})dτ  (8.17)

and this solution satisfies the bound

∥y∥_{∞} ≤ ∥ϕ∥_{∞} e^{∫_{J} ∥A(τ)∥dτ} .

Proof. Define Λ : BC(J, X) → BC(J, X) by

(Λy)(t) = ∫_{0}^{t} A(τ)y(τ)dτ.

Then y solves Eq. (8.14) iff y = ϕ + Λy or equivalently iff (I − Λ)y = ϕ. An induction argument shows

(Λ^{n}ϕ)(t) = ∫_{0}^{t} dτ_{n}A(τ_{n})(Λ^{n−1}ϕ)(τ_{n})

= ∫_{0}^{t} dτ_{n} ∫_{0}^{τ_{n}} dτ_{n−1}A(τ_{n})A(τ_{n−1})(Λ^{n−2}ϕ)(τ_{n−1})

⋮

= ∫_{0}^{t} dτ_{n} ∫_{0}^{τ_{n}} dτ_{n−1} … ∫_{0}^{τ_{2}} dτ_{1}A(τ_{n}) … A(τ_{1})ϕ(τ_{1})

= (−1)^{n·1_{t<0}} ∫_{Δ_{n}(t)} A(τ_{n}) … A(τ_{1})ϕ(τ_{1})dτ.

Taking norms of this equation and using the triangle inequality along with Lemma 8.19 gives,

∥(Λ^{n}ϕ)(t)∥ ≤ ∥ϕ∥_{∞} · ∫_{Δ_{n}(t)} ∥A(τ_{n})∥ … ∥A(τ_{1})∥dτ

≤∥ϕ∥_{∞} · \frac{1}{n!} (∫_{Δ_{1}(t)} ∥A(τ)∥dτ)^{n}

≤∥ϕ∥_{∞} · \frac{1}{n!} (∫_{J} ∥A(τ)∥dτ)^{n} .

Therefore,

∥Λ^{n}∥_{op} ≤ \frac{1}{n!} (∫_{J} ∥A(τ)∥dτ)^{n}  (8.18)

and

∑_{n=0}^{∞} ∥Λ^{n}∥_{op} ≤ e^{∫_{J} ∥A(τ)∥dτ} < ∞

where ∥·∥_{op} denotes the operator norm on L (BC(J, X)) . An application of

Proposition 7.19 now shows (I − Λ)^{−1} = ∑_{n=0}^{∞} Λ^{n} exists and

∥(I − Λ)^{−1}∥_{op} ≤ e^{∫_{J} ∥A(τ)∥dτ} .

It is now only a matter of working through the notation to see that these assertions prove the theorem.

Corollary 8.22. Suppose h ∈ C(J → X) and x ∈ X, then there exits a unique solution, y ∈ C^{1} (J, X) , to the linear ordinary differential Eq. (8.13).

Proof. Let

ϕ (t) = x + ∫_{0}^{t} h (τ) dτ.

By applying Theorem 8.21 with and J replaced by any open interval J_{0} such that 0 ∈ J_{0} and J̄_{0} is a compact subinterval^{1} of J, there exists a unique solution y_{J_{0}} to Eq. (8.13) which is valid for t ∈ J_{0}. By uniqueness of solutions, if J_{1} is a subinterval of J such that J_{0} ⊂ J_{1} and J̄_{1} is a compact subinterval of J, we have y_{J_{1}} = y_{J_{0}} on J_{0}. Because of this observation, we may construct a solution y to Eq. (8.13) which is defined on the full interval J by setting y (t) = y_{J_{0}} (t) for any J_{0} as above which also contains t ∈ J.

Corollary 8.23. Suppose that A ∈ L(X) is independent of time, then the solution to

ẏ(t) = Ay(t) with y(0) = x

1 We do this so that ϕ|_{J_{0}} will be bounded.

is given by y(t) = e^{tA}x where

e^{tA} = ∑_{n=0}^{∞} \frac{t^{n}}{n!} A^{n}.  (8.19)

Moreover,

e^{(t+s)A} = e^{tA}e^{sA} for all s, t ∈ ℝ.  (8.20)

Proof. The first assertion is a simple consequence of Eq. 8.17 and Lemma 8.19 with ψ = 1. The assertion in Eq. (8.20) may be proved by explicit computation but the following proof is more instructive. Given x ∈ X, let y (t) := e^{(t+s)A}x. By the chain rule,

\frac{d}{dt}y (t) = \frac{d}{dτ} |_{τ=t+s}e^{τA}x = Ae^{τA}x|_{τ=t+s}
= Ae^{(t+s)A}x = Ay (t) with y (0) = e^{sA}x.

The unique solution to this equation is given by

y (t) = e^{tA}x (0) = e^{tA}e^{sA}x.

This completes the proof since, by definition, y (t) = e^{(t+s)A}x.

We also have the following converse to this corollary whose proof is outlined in Exercise 8.21 below.

Theorem 8.24. Suppose that T_{t} ∈ L(X) for t ≥ 0 satisfies

1. (Semi-group property.) T_{0} = Id_{X} and T_{t}T_{s} = T_{t+s} for all s, t ≥ 0.

2. (Norm Continuity) t → T_{t} is continuous at 0, i.e. ∥T_{t} − I∥_{L(X)} → 0 as t ↓ 0.

Then there exists A ∈ L(X) such that T_{t} = e^{tA} where e^{tA} is defined in Eq. (8.19).

8.4 Classical Weierstrass Approximation Theorem

Definition 8.25 (Support). Let f : X → Z be a function from a metric space (X, ρ) to a vector space Z. The support of f is the closed subset, supp(f), of X defined by

supp(f) := \overline{\{x ∈ X : f(x) ≠ 0\}}.

Example 8.26. For example if f : ℝ → ℝ is defined by f(x) = sin(x)1_{[0,4π]}(x) ∈ ℝ, then

{f ≠ 0} = (0, 4π) \ {π, 2π, 3π}

and therefore supp(f) = [0, 4π].

For the remainder of this section, Z will be used to denote a Banach space.

Definition 8.27 (Convolution). For f, g ∈ C (ℝ) with either f or g having compact support, we define the convolution of f and g by

f ∗ g(x) = ∫_{ℝ} f(x − y)g(y)dy = ∫_{ℝ} f(y)g(x − y)dy.

We will also use this definition when one of the functions, either f or g, takes values in a Banach space Z.

Lemma 8.28 (Approximate δ – sequences). Suppose that {q_{n}}_{n=1}^{∞} is a sequence non-negative continuous real valued functions on ℝ with compact support that satisfy

∫_{ℝ} q_{n}(x) dx = 1 and  (8.21)
lim_{n→∞} ∫_{|x|≥ε} q_{n}(x)dx = 0 for all ε > 0.  (8.22)

If f ∈ BC(ℝ, Z), then

q_{n} ∗ f (x) := ∫_{ℝ} q_{n}(y)f(x − y)dy

converges to f uniformly on compact subsets of ℝ×W ⊂ ℝ^{d+1}.

Proof. Let x ∈ ℝ, then because of Eq. (8.21),

∥q_{n} ∗ f(x) − f(x)∥ = ∥∫_{ℝ} q_{n}(y) (f(x − y) − f(x)) dy∥
≤ ∫_{ℝ} q_{n}(y) ∥f(x − y) − f(x)∥ dy.

Let M = sup {∥f(x)∥ : x ∈ ℝ} . Then for any ε > 0, using Eq. (8.21),

∥q_{n} ∗ f(x) − f(x)∥ ≤ ∫_{|y|≤ε} q_{n}(y) ∥f(x − y) − f(x)∥ dy

+ ∫_{|y|>ε} q_{n}(y) ∥f(x − y) − f(x)∥ dy

≤ sup_{|w|≤ε} ∥f(x + w) − f(x)∥ + 2M ∫_{|y|>ε} q_{n}(y)dy.

So if K is a compact subset of ℝ (for example a large interval) we have

sup_{(x)∈K} ∥q_{n} ∗ f(x) − f(x)∥

≤ sup_{|w|≤ε, x∈K} ∥f(x + w) − f(x)∥ + 2M ∫_{∥y∥>ε} q_{n}(y)dy

and hence by Eq. (8.22),

lim sup_{n→∞} sup_{x∈K} ∥q_{n} ∗ f(x) − f(x)∥

≤ sup_{|w|≤ε, x∈K} ∥f(x + w) − f(x)∥ .

This finishes the proof since the right member of this equation tends to 0 as ε ↓ 0 by uniform continuity of f on compact subsets of ℝ.

Let q_{n} : ℝ →[0, ∞) be defined by

q_{n}(x) := \frac{1}{c_{n}} (1 − x^{2})^{n}1_{|x|≤1} where c_{n} := ∫_{−1}^{1}(1 − x^{2})^{n}dx.  (8.23)

Figure 8.2 displays the key features of the functions q_{n}.

Fig. 8.2. A plot of q_{1}, q_{50}, and q_{100}. The most peaked curve is q_{100} and the least is q_{1}. The total area under each of these curves is one.

Lemma 8.29. The sequence {q_{n}}_{n=1}^{∞} is an approximate δ – sequence, i.e. they satisfy Eqs. (8.21) and (8.22).

Proof. By construction, q_{n} ∈ C_{c} (ℝ, [0, ∞)) for each n and Eq. 8.21 holds. Since

∫_{|x|≥ε} q_{n}(x)dx = \frac{2 ∫_{ε}^{1}(1 − x^{2})^{n}dx}{2 ∫_{0}^{ε}(1 − x^{2})^{n}dx + 2 ∫_{ε}^{1}(1 − x^{2})^{n}dx}

≤ \frac{∫_{ε}^{1} \frac{x}{ε}(1 − x^{2})^{n}dx}{∫_{0}^{ε} \frac{x}{ε}(1 − x^{2})^{n}dx} = \frac{(1 − x^{2})^{n+1}|_{ε}^{1}}{(1 − x^{2})^{n+1}|_{0}^{ε}}

= \frac{(1 − ε^{2})^{n+1}}{1 − (1 − ε^{2})^{n+1}} → 0 as n → ∞,

the proof is complete.

Notation 8.30 Let ℤ_{+} := ℕ ∪ {0} and for x ∈ ℝ^{d} and α ∈ ℤ_{+}^{d} let x^{α} = ∏_{i=1}^{d} x_{i}^{α_{i}} and |α| = ∑_{i=1}^{d} α_{i}. A polynomial on ℝ^{d} with values in Z is a function p : ℝ^{d} → Z of the form

p(x) = ∑_{α:|α|≤N} p_{α}x^{α} with p_{α} ∈ Z and N ∈ ℤ_{+}.

If p_{α} ≠ 0 for some α such that |α| = N, then we define deg(p) := N to be the degree of p. If Z is a complex Banach space, the function p has a natural extension to z ∈ ℂ^{d}, namely p(z) = ∑_{α:|α|≤N} p_{α}z^{α} where z^{α} = ∏_{i=1}^{d} z_{i}^{α_{i}}.

Given a compact subset K ⊂ ℝ^{d} and f ∈ C (K, ℂ)^{2}, we are going to show, in the Weierstrass approximation Theorem 8.34 below, that f may be uniformly approximated by polynomial functions on K. The next theorem addresses this question when K is a compact subinterval of ℝ.

Theorem 8.31 (Weierstrass Approximation Theorem). Suppose −∞ < a < b < ∞, J = [a, b] and f ∈ C(J, Z). Then there exists polynomials p_{n} on ℝ such that p_{n} → f uniformly on J.

Proof. By replacing f by F where

F (t) := f (a + t (b − a)) − [f (a) + t (f (b) − f (a))] for t ∈ [0, 1] ,

it suffices to assume a = 0, b = 1 and f (0) = f (1) = 0. Furthermore we may now extend f to a continuous function on all ℝ by setting f ≡ 0 on ℝ \ [0, 1] .

With q_{n} defined as in Eq. (8.23), let f_{n}(x) := (q_{n} ∗ f)(x) and recall from Lemma 8.28 that f_{n} (x) → f (x) as n → ∞ with the convergence being uniform in x ∈ [0, 1]. This completes the proof since f_{n} is equal to a polynomial function on [0, 1] . Indeed, there are polynomials, a_{k} (y) , such that

(1 − (x − y)^{2})^{n} = ∑_{k=0}^{2n} a_{k} (y) x^{k},

and therefore, for x ∈ [0, 1] ,

2 Note that f is automatically bounded because if not there would exist u_{n} ∈ K such that lim_{n→∞} |f (u_{n})| = ∞. Using Theorem 8.2 we may, by passing to a subsequence if necessary, assume u_{n} → u ∈ K as n → ∞ . Now the continuity of f would then imply

∞ = lim_{n→∞} |f (u_{n})| = |f (u)|

which is absurd since f takes values in ℂ.

f_{n}(x) = ∫_{ℝ} q_{n}(x − y)f(y)dy

= \frac{1}{c_{n}} ∫_{[0,1]} f(y) [(1 − (x − y)^{2})^{n}1_{|x−y|≤1}] dy

= \frac{1}{c_{n}} ∫_{[0,1]} f(y)(1 − (x − y)^{2})^{n}dy

= \frac{1}{c_{n}} ∫_{[0,1]} f(y) ∑_{k=0}^{2n} a_{k} (y) x^{k}dy = ∑_{k=0}^{2n} A_{k}x^{k}

where

A_{k} = ∫_{[0,1]} f (y) a_{k} (y) dy.

Lemma 8.32. Suppose J = [a, b] is a compact subinterval of ℝ and K is a compact subset of ℝ^{d−1}, then the linear mapping R : C (J × K, Z) → C (J, C (K, Z)) defined by (Rf) (t) = f (t, ·) ∈ C (K, Z) for t ∈ J is an isometric isomorphism of Banach spaces.

Proof. By uniform continuity of f on J × K (see Theorem 8.2),

∥(Rf) (t) − (Rf) (s)∥_{C(K,Z)} = max_{y∈K} ∥f (t, y) − f (s, y)∥_{Z} → 0 as s → t

which shows that Rf is indeed in C (J → C (K, Z)) . Moreover,

∥Rf∥_{C(J→C(K,Z))} = max_{t∈J} ∥(Rf) (t)∥_{C(K,Z)}

= max_{t∈J} max_{y∈K} ∥f (t, y)∥_{Z} = ∥f∥_{C(J×K,Z)} ,

showing R is isometric and therefore injective.

To see that R is surjective, let F ∈ C (J → C (K, Z)) and define f (t, y) := F (t) (y) . Since

∥f (t, y) − f (s, y′)∥_{Z} ≤ ∥f (t, y) − f (s, y)∥_{Z} + ∥f (s, y) − f (s, y′)∥_{Z}
≤ ∥F (t) − F (s)∥_{C(K,Z)} + ∥F (s) (y) − F (s) (y′)∥_{Z} ,

it follows by the continuity of t → F (t) and y → F (s) (y) that

∥f (t, y) − f (s, y′)∥_{Z} → 0 as (t, y) → (s, y′) .

This shows f ∈ C (J × K, Z) and thus completes the proof because Rf = F by construction.

Corollary 8.33 (Weierstrass Approximation Theorem). Let d ∈ ℕ, J_{i} = [a_{i}, b_{i}] be compact subintervals of ℝ for i = 1, 2, …, d, J := J_{1} × ⋯ × J_{d} and f ∈ C(J, Z). Then there exists polynomials p_{n} on ℝ^{d} such that p_{n} → f uniformly on J.

Proof. The proof will be by induction on d with the case d = 1 being the content of Theorem 8.31. Now suppose that d > 1 and the theorem holds with d replaced by d−1. Let K := J_{2}×⋯×J_{d}, Z_{0} = C (K, Z) , R : C (J_{1} × K, Z) → C (J_{1}, Z_{0}) be as in Lemma 8.32 and F := Rf. By Theorem 8.31, for any ε > 0 there exists a polynomial function

p (t) = ∑_{k=0}^{n} c_{k}t^{k}

with c_{k} ∈ Z_{0} = C (K, Z) such that ∥F − p∥_{C(J_{1},Z_{0})} < ε. By the induction hypothesis, there exists polynomial functions q_{k} : K → Z such that

∥c_{k} − q_{k}∥_{Z_{0}} < \frac{ε}{n (|a| + |b|)^{k}} .

It is now easily verified (you check) that the polynomial function,

ρ (x) := ∑_{k=0}^{n} x_{1}^{k}q_{k} (x_{2}, …, x_{d}) for x ∈ J

satisfies ∥f − ρ∥_{C(J,Z)} < 2ε and this completes the induction argument and hence the proof.

The reader is referred to Chapter 20 for a two more alternative proofs of this corollary.

Theorem 8.34 (Weierstrass Approximation Theorem). Suppose that K ⊂ ℝ^{d} is a compact subset and f ∈ C(K, ℂ). Then there exists polynomials p_{n} on ℝ^{d} such that p_{n} → f uniformly on K.

Proof. Choose λ > 0 and b ∈ ℝ^{d} such that

K_{0} := λK − b := {λx − b : x ∈ K} ⊂ B_{d}

where B_{d} := (0, 1)^{d} . The function F (y) := f (λ^{−1} (y + b)) for y ∈ K_{0} is in C (K_{0}, ℂ) and if p̂_{n} (y) are polynomials on ℝ^{d} such that p̂_{n} → F uniformly on K_{0} then p_{n} (x) := p̂_{n} (λx − b) are polynomials on ℝ^{d} such that p_{n} → f uniformly on K. Hence we may now assume that K is a compact subset of B_{d}. Let g ∈ C (K ∪ B_{d}^{c}) be defined by

g (x) = \{ \begin{matrix}f (x) & if & x ∈ K \\ 0 & if & x ∈ B_{d}^{c} \end{matrix}

and then use the Tietze extension Theorem 7.4 (applied to the real and imaginary parts of F) to find a continuous function F ∈ C(ℝ^{d}, ℂ) such that F = g|_{K∪B_{d}^{c}}. If p_{n} are polynomials on ℝ^{d} such that p_{n} → F uniformly on [0, 1]^{d} then p_{n} also converges to f uniformly on K. Hence, by replacing f by F, we may now assume that f ∈ C(ℝ^{d}, ℂ), K = B̄_{d} = [0, 1]^{d} , and f ≡ 0 on B_{d}^{c}. The result now follows by an application of Corollary 8.33 with Z = ℂ.

Remark 8.35. The mapping (x, y) ∈ ℝ^{d} × ℝ^{d} → z = x + iy ∈ ℂ^{d} is an isomorphism of vector spaces. Letting z̄ = x − iy as usual, we have x = \frac{z+z̄}{2} and y = \frac{z−z̄}{2i}. Therefore under this identification any polynomial p(x, y) on ℝ^{d} × ℝ^{d} may be written as a polynomial q in (z, z̄), namely

q(z, z̄) = p(\frac{z + z̄}{2}, \frac{z − z̄}{2i}).

Conversely a polynomial q in (z, z̄) may be thought of as a polynomial p in (x, y), namely p(x, y) = q(x + iy, x − iy).

Corollary 8.36 (Complex Weierstrass Approximation Theorem). Suppose that K ⊂ ℂ^{d} is a compact set and f ∈ C(K, ℂ). Then there exists polynomials p_{n}(z, z̄) for z ∈ ℂ^{d} such that sup_{z∈K} |p_{n}(z, z̄) − f(z)| → 0 as n → ∞.

Proof. This is an immediate consequence of Theorem 8.34 and Remark 8.35.

Example 8.37. Let K = S^{1} = {z ∈ ℂ : |z| = 1} and A be the set of polynomials in (z, z̄) restricted to S^{1}. Then A is dense in C(S^{1}).^{3} Since z̄ = z^{−1} on S^{1}, we have shown polynomials in z and z^{−1} are dense in C(S^{1}). This example generalizes in an obvious way to K = (S^{1})^{d} ⊂ ℂ^{d}.

Exercise 8.4. Suppose −∞ < a < b < ∞ and f ∈ C ([a, b], ℂ) satisfies

∫_{a}^{b} f (t) t^{n}dt = 0 for n = 0, 1, 2 ….

Show f ≡ 0.

Exercise 8.5. Suppose f ∈ C (ℝ, ℂ) is a 2π – periodic function (i.e. f (x + 2π) = f (x) for all x ∈ ℝ) and

∫_{0}^{2π} f (x) e^{inx}dx = 0 for all n ∈ ℤ,

show again that f ≡ 0. Hint: Use Example 8.37 to show that any 2π – periodic continuous function g on ℝ is the uniform limit of trigonometric polynomials of the form

p (x) = ∑_{k=−n}^{n} p_{k}e^{ikx} with p_{k} ∈ ℂ for all k.

3 Note that it is easy to extend f ∈ C(S^{1}) to a function F ∈ C(ℂ) by setting F (z) = zf( \frac{z}{|z|}) for z ≠ 0 and F (0) = 0. So this special case does not require the Tietze extension theorem.

8.5 Iterated Integrals

Theorem 8.38 (Baby Fubini Theorem). Let a_{i}, b_{i} ∈ ℝ with a_{i} ≠ b_{i} for i = 1, 2 …, n, f(t_{1}, t_{2}, …, t_{n}) ∈ Z be a continuous function of (t_{1}, t_{2}, …, t_{n}) where t_{i} between a_{i} and b_{i} for each i and for any given permutation, σ, of {1, 2 …, n} let

I_{σ} (f) := ∫_{a_{σ_{1}}}^{b_{σ_{1}}} dt_{σ_{1}} … ∫_{a_{σn}}^{b_{σn}} dt_{σ_{n}}f(t_{1}, t_{2}, …, t_{n}).  (8.24)

Then I_{σ} (f) is well defined and independent of σ, i.e. the order of iterated integrals is irrelevant under these hypothesis.

Proof. Let J_{i} := [min (a_{i}, b_{i}) , max (a_{i}, b_{i})] , J := J_{1} × ⋯ × J_{n} and |J_{i}| := max (a_{i}, b_{i}) − min (a_{i}, b_{i}) . Using the uniform continuity of f (Theorem 8.2) and the continuity of the Riemann integral, it is easy to prove (compare with the proof of Lemma 8.32) that the map

(t_{1}, … t̂_{σn}, …, t_{n}) ∈ (J_{1} × ⋯ × Ĵ_{σn} × ⋯ × J_{n}) → ∫_{a_{σn}}^{b_{σn}} dt_{σ_{n}}f(t_{1}, t_{2}, …, t_{n})

is continuous, where the hat is used to denote a missing element from a list. From this remark, it follows that each of the integrals in Eq. (8.24) are well defined and hence so is I_{σ} (f) . Moreover by an induction argument using Lemma 8.32 and the boundedness of the Riemann integral, we have the estimate,

∥I_{σ} (f)∥_{Z} ≤ (∏_{i=1}^{n} |J_{i}|) ∥f∥_{C(J,Z)} .  (8.25)

Now suppose τ is another permutation. Because of Eq. (8.25), I_{σ} and I_{τ} are bounded operators on C (J, Z) and so to shows I_{σ} = I_{τ} is suffices to shows there are equal on the dense set of polynomial functions (see Corollary 8.33) in C (J, Z) . Moreover by linearity, it suffices to show I_{σ} (f) = I_{τ} (f) when f has the form

f(t_{1}, t_{2}, …, t_{n}) = t_{1}^{k_{1}} … t_{n}^{k_{n}}z

for some k_{i} ∈ ℕ_{0} and z ∈ Z. However for this function, explicit computations show

I_{σ} (f) = I_{τ} (f) = (∏_{i=1}^{n} \frac{b_{i}^{k_{i}+1} − a_{i}^{k_{i}+1}}{k_{i} + 1} ) · z.

Proposition 8.39 (Equality of Mixed Partial Derivatives). Let Q = (a, b) × (c, d) be an open rectangle in ℝ^{2} and f ∈ C(Q, Z). Assume that \frac{∂}{∂t}f(s, t), \frac{∂}{∂s}f(s, t) and \frac{∂}{∂t} \frac{∂}{∂s}f(s, t) exists and are continuous for (s, t) ∈ Q, then \frac{∂}{∂s} \frac{∂}{∂t}f(s, t) exists for (s, t) ∈ Q and

\frac{∂}{∂s} \frac{∂}{∂t}f(s, t) = \frac{∂}{∂t} \frac{∂}{∂s}f(s, t) for (s, t) ∈ Q.  (8.26)

Proof. Fix (s_{0}, t_{0}) ∈ Q. By two applications of Theorem 8.13,

f(s, t) = f(s_{t_{0}}, t) + ∫_{s_{0}}^{s} \frac{∂}{∂σ} f(σ, t)dσ
= f(s_{0}, t) + ∫_{s_{0}}^{s} \frac{∂}{∂σ} f(σ, t_{0})dσ + ∫_{s_{0}}^{s} dσ ∫_{t_{0}}^{t} dτ \frac{∂}{∂τ} \frac{∂}{∂σ} f(σ, τ) (8.27)

and then by Fubini’s Theorem 8.38 we learn

f(s, t) = f(s_{0}, t) + ∫_{s_{0}}^{s} \frac{∂}{∂σ} f(σ, t_{0})dσ + ∫_{t_{0}}^{t} dτ ∫_{s_{0}}^{s} dσ \frac{∂}{∂τ} \frac{∂}{∂σ} f(σ, τ).

Differentiating this equation in t and then in s (again using two more applications of Theorem 8.13) shows Eq. (8.26) holds.

8.6 Exercises

Throughout these problems, (X, ∥·∥) is a Banach space.

Exercise 8.6. Show f = (f_{1}, …, f_{n}) ∈ S̄([a, b], ℝ^{n}) iff f_{i} ∈ S̄([a, b], ℝ) for i = 1, 2, …, n and

∫_{a}^{b} f(t)dt = (∫_{a}^{b} f_{1}(t)dt, …, ∫_{a}^{b} f_{n}(t)dt) .

Here ℝ^{n} is to be equipped with the usual Euclidean norm. Hint: Use Lemma 8.7 to prove the forward implication.

Exercise 8.7. Prove Theorem 8.38 using the following strategy.

1. Use the results from the proof in the text of Theorem 8.38 that

s → ∫_{c}^{d} f(s, t)dt and t → ∫_{a}^{b} f(s, t)ds

are continuous maps.

2. For the moment take X = ℝ and prove Eq. (8.24) holds by first proving it holds when f (s, t) = s^{m}t^{n} with m, n ∈ ℕ_{0}. Then use this result along with Theorem 8.34 to show Eq. (8.24) holds for all f ∈ C ([a, b] × [c, d], ℝ) .

3. For the general case, use the special case proved in item 2. along with Hahn - Banach theorem.

Exercise 8.8. Give another proof of Proposition 8.39 which does not use Fubini’s Theorem 8.38 as follows.

1. By a simple translation argument we may assume (0, 0) ∈ Q and we are trying to prove Eq. (8.26) holds at (s, t) = (0, 0).

2. Let h(s, t) := \frac{∂}{∂t} \frac{∂}{∂s}f(s, t) and

G(s, t) := ∫_{0}^{s} dσ ∫_{0}^{t} dτh(σ, τ)

so that Eq. (8.27) states

f(s, t) = f(0, t) + ∫_{0}^{s} \frac{∂}{∂σ} f(σ, t_{0})dσ + G(s, t)

and differentiating this equation at t = 0 shows

\frac{∂}{∂t}f(s, 0) = \frac{∂}{∂t}f(0, 0) + \frac{∂}{∂t}G(s, 0).  (8.28)

Now show using the definition of the derivative that

\frac{∂}{∂t}G(s, 0) = ∫_{0}^{s} dσh(σ, 0).  (8.29)

Hint: Consider

G(s, t) − t ∫_{0}^{s} dσh(σ, 0) = ∫_{0}^{s} dσ ∫_{0}^{t} dτ [h(σ, τ) − h(σ, 0)] .

3. Now differentiate Eq. (8.28) in s using Theorem 8.13 to finish the proof.

Exercise 8.9. Give another proof of Eq. (8.24) in Theorem 8.38 based on Proposition 8.39. To do this let t_{0} ∈ (c, d) and s_{0} ∈ (a, b) and define

G(s, t) := ∫_{t_{0}}^{t} dτ ∫_{s_{0}}^{s} dσf(σ, τ)

Show G satisfies the hypothesis of Proposition 8.39 which combined with two applications of the fundamental theorem of calculus implies

\frac{∂}{∂t} \frac{∂}{∂s}G(s, t) = \frac{∂}{∂s} \frac{∂}{∂t}G(s, t) = f(s, t).

Use two more applications of the fundamental theorem of calculus along with the observation that G = 0 if t = t_{0} or s = s_{0} to conclude

G(s, t) = ∫_{s_{0}}^{s} dσ ∫_{t_{0}}^{t} dτ \frac{∂}{∂τ} \frac{∂}{∂σ} G(σ, τ) = ∫_{s_{0}}^{s} dσ ∫_{t_{0}}^{t} dτ \frac{∂}{∂τ} f(σ, τ).  (8.30)

Finally let s = b and t = d in Eq. (8.30) and then let s_{0} ↓ a and t_{0} ↓ c to prove Eq. (8.24).

Exercise 8.10 (Product Rule). Prove items 1. and 2. of Lemma 8.9. This can be modeled on the standard proof for real valued functions.

Exercise 8.11 (Chain Rule). Prove the chain rule in Proposition 8.10. Again this may be modeled on the on the standard proof for real valued functions.

Exercise 8.12. To each A ∈ L (X) , we may define L_{A}, R_{A} : L (X) → L (X) by

L_{A}B = AB and R_{A}B = BA for all B ∈ L (X) .

Show L_{A}, R_{A} ∈ L (L (X)) and that

∥L_{A}∥_{L(L(X))} = ∥A∥_{L(X)} = ∥R_{A}∥_{L(L(X))} .

Exercise 8.13. Suppose that A : ℝ → L(X) is a continuous function and U, V : ℝ → L(X) are the unique solution to the linear differential equations

V̇ (t) = A(t)V (t) with V (0) = I  (8.31)

and

U̇(t) = −U(t)A(t) with U(0) = I.  (8.32)

Prove that V (t) is invertible and that V^{−1}(t) = U(t)^{4}, where by abuse of notation I am writing V^{−1} (t) for [V (t)]^{−1} . Hints: 1) show \frac{d}{dt} [U(t)V (t)] = 0 (which is sufficient if dim(X) < ∞) and 2) show compute y(t) := V (t)U(t) solves a linear differential ordinary differential equation that has y ≡ Id as an obvious solution. (The results of Exercise 8.12 may be useful here.) Then use the uniqueness of solutions to linear ODEs.

Exercise 8.14. Suppose that (X, ∥·∥) is a Banach space, J = (a, b) with −∞ ≤ a < b ≤ ∞ and f_{n} : ℝ → X are continuously differentiable functions such that there exists a summable sequence {a_{n}}_{n=1}^{∞} satisfying

∥f_{n} (t)∥ + ∥ḟ_{n} (t)∥ ≤ a_{n} for all t ∈ J and n ∈ ℕ.

Show:

1. sup \{∥\frac{f_{n}(t+h)−f_{n}(t)}{h} ∥ : (t, h) ∈ J × ℝ ∋ t + h ∈ J and h ≠ 0\} ≤ a_{n}.

2. The function F : ℝ → X defined by

F (t) := ∑_{n=1}^{∞} f_{n} (t) for all t ∈ J

is differentiable and for t ∈ J,

Ḟ (t) = ∑_{n=1}^{∞} ḟ_{n} (t) .

4 The fact that U(t) must be defined as in Eq. (8.32) follows from Lemma 8.9.

Exercise 8.15. Suppose that A ∈ L(X). Show directly that:

1. e^{tA} define in Eq. (8.19) is convergent in L(X) when equipped with the operator norm.

2. e^{tA} is differentiable in t and that \frac{d}{dt}e^{tA} = Ae^{tA}.

Exercise 8.16. Suppose that A ∈ L(X) and v ∈ X is an eigenvector of A with eigenvalue λ, i.e. that Av = λv. Show e^{tA}v = e^{tλ}v. Also show that if X = ℝ^{n} and A is a diagonalizable n × n matrix with

A = SDS^{−1} with D = diag(λ_{1}, …, λ_{n})

then e^{tA} = Se^{tD}S^{−1} where e^{tD} = diag(e^{tλ_{1}}, …, e^{tλ_{n}}). Here diag(λ_{1}, …, λ_{n}) denotes the diagonal matrix Λ such that Λ_{ii} = λ_{i} for i = 1, 2, …, n.

Exercise 8.17. Suppose that A, B ∈ L(X) and [A, B] := AB − BA = 0. Show that e^{(A+B)} = e^{A}e^{B}.

Exercise 8.18. Suppose A ∈ C(ℝ, L(X)) satisfies [A(t), A(s)] = 0 for all s, t ∈ ℝ. Show

y(t) := e^{(∫_{0}^{t} A(τ)dτ)}x

is the unique solution to ẏ(t) = A(t)y(t) with y(0) = x.

Exercise 8.19. Compute e^{tA} when

A = ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} )

and use the result to prove the formula

cos(s + t) = cos s cos t − sin s sin t.

Hint: Sum the series and use e^{tA}e^{sA} = e^{(t+s)A}.

Exercise 8.20. Compute e^{tA} when

A = ( \begin{matrix}0 & a & b \\ 0 & 0 & c \\ 0 & 0 & 0 \end{matrix} )

with a, b, c ∈ ℝ. Use your result to compute e^{t(λI+A)} where λ ∈ ℝ and I is the 3 × 3 identity matrix. Hint: Sum the series.

Exercise 8.21. Prove Theorem 8.24 using the following outline.

1. Using the right continuity at 0 and the semi-group property for T_{t}, show there are constants M and C such that ∥T_{t}∥_{L(X)} ≤ MC^{t} for all t > 0.

2. Show t ∈ [0, ∞) → T_{t} ∈ L(X) is continuous.

3. For ε > 0, let S_{ε} := \frac{1}{ε} ∫_{0}^{ε} T_{τ} dτ ∈ L(X). Show S_{ε} → I as ε ↓ 0 and conclude from this that S_{ε} is invertible when ε > 0 is sufficiently small. For the remainder of the proof fix such a small ε > 0.

4. Show

T_{t}S_{ε} = \frac{1}{ε} ∫_{t}^{t+ε} T_{τ} dτ

and conclude from this that

lim_{t↓0} (\frac{T_{t} − I}{t} ) S_{ε} = \frac{1}{ε} (T_{ε} − Id_{X}) .

5. Using the fact that S_{ε} is invertible, conclude A = lim_{t↓0} t^{−1} (T_{t} − I) exists in L(X) and that

A = \frac{1}{ε} (T_{ε} − I) S_{ε}^{−1}.

6. Now show, using the semigroup property and step 4., that \frac{d}{dt}T_{t} = AT_{t} for all t > 0.

7. Using step 5, show \frac{d}{dt}e^{−tA}T_{t} = 0 for all t > 0 and therefore e^{−tA}T_{t} = e^{−0A}T_{0} = I.

Exercise 8.22 (Duhamel’ s Principle I). Suppose that A : ℝ → L(X) is a continuous function and V : ℝ → L(X) is the unique solution to the linear differential equation in Eq. (8.31). Let x ∈ X and h ∈ C(ℝ, X) be given. Show that the unique solution to the differential equation:

ẏ(t) = A(t)y(t) + h(t) with y(0) = x  (8.33)

is given by

y(t) = V (t)x + V (t) ∫_{0}^{t} V (τ)^{−1}h(τ) dτ.  (8.34)

Hint: compute \frac{d}{dt}[V^{−1}(t)y(t)] (see Exercise 8.13) when y solves Eq. (8.33).

Exercise 8.23 (Duhamel’ s Principle II). Suppose that A : ℝ → L(X) is a continuous function and V : ℝ → L(X) is the unique solution to the linear differential equation in Eq. (8.31). Let W_{0} ∈ L(X) and H ∈ C(ℝ, L(X)) be given. Show that the unique solution to the differential equation:

Ẇ (t) = A(t)W (t) + H(t) with W (0) = W_{0}  (8.35)

is given by

W (t) = V (t)W_{0} + V (t) ∫_{0}^{t} V (τ)^{−1}H(τ) dτ.  (8.36)

Hölder Spaces as Banach Spaces

Notation 9.1 Let Ω be an open subset of ℝ^{d}, BC(Ω) and BC(Ω̄) be the bounded continuous functions on Ω and Ω̄ respectively. By identifying f ∈ BC(Ω̄) with f|_{Ω} ∈ BC(Ω), we will consider BC(Ω̄) as a subset of BC(Ω). For u ∈ BC(Ω) and 0 < β ≤ 1 let

∥u∥_{u} := sup_{x∈Ω} |u(x)| and [u]_{β} := sup_{x,y∈Ω_{x≠y}} \{\frac{|u(x) − u(y)|}{|x − y|^{β}}\} .

If [u]_{β} < ∞, then u is Hölder continuous with holder exponent^{1} β. The collection of β – Hölder continuous function on Ω will be denoted by

C^{0,β}(Ω) := {u ∈ BC(Ω) : [u]_{β} < ∞}

and for u ∈ C^{0,β}(Ω) let

∥u∥_{C^{0,β}(Ω)} := ∥u∥_{u} + [u]_{β}.  (9.1)

Remark 9.2. If u : Ω → ℂ and [u]_{β} < ∞ for some β > 1, then u is constant on each connected component of Ω. Indeed, if x ∈ Ω and h ∈ ℝ^{d} then

|\frac{u(x + th) − u(x)}{t}| ≤ [u]_{β}t^{β}/t → 0 as t → 0

which shows ∂_{h}u(x) = 0 for all x ∈ Ω. If y ∈ Ω is in the same connected component as x, then by Exercise 22.8 below there exists a smooth curve σ : [0, 1] → Ω such that σ(0) = x and σ(1) = y. So by the fundamental theorem of calculus and the chain rule,

u(y) − u(x) = ∫_{0}^{1} \frac{d}{dt}u(σ(t))dt = ∫_{0}^{1} 0 dt = 0.

This is why we do not talk about Hölder spaces with Hölder exponents larger than 1.

1 If β = 1, u is is said to be Lipschitz continuous.

Lemma 9.3. Suppose u ∈ C^{1}(Ω) ∩ BC(Ω) and ∂_{i}u ∈ BC(Ω) for i = 1, 2, …, d, then u ∈ C^{0,1}(Ω), i.e. [u]_{1} < ∞.

The proof of this lemma is left to the reader as Exercise 9.1.

Theorem 9.4. Let Ω be an open subset of ℝ^{d}. Then

1. Under the identification of u ∈ BC \binom{}{Ω̄} with u|_{Ω} ∈ BC (Ω) , BC(Ω̄) is a closed subspace of BC(Ω).

2. Every element u ∈ C^{0,β}(Ω) has a unique extension to a continuous function (still denoted by u) on Ω̄. Therefore we may identify C^{0,β}(Ω) with C^{0,β}(Ω̄) ⊂ BC(Ω̄). (In particular we may consider C^{0,β}(Ω) and C^{0,β}(Ω̄) to be the same when β > 0.)

3. The function u ∈ C^{0,β}(Ω) → ∥u∥_{C^{0,β}(Ω)} ∈ [0, ∞) is a norm on C^{0,β}(Ω) which make C^{0,β}(Ω) into a Banach space.

Proof. 1. The first item is trivial since for u ∈ BC(Ω̄), the sup-norm of u on Ω̄ agrees with the sup-norm on Ω and BC(Ω̄) is complete in this norm.

2. Suppose that [u]_{β} < ∞ and x_{0} ∈ bd(Ω). Let {x_{n}}_{n=1}^{∞} ⊂ Ω be a sequence such that x_{0} = lim_{n→∞} x_{n}. Then

|u(x_{n}) − u(x_{m})| ≤ [u]_{β} |x_{n} − x_{m}|^{β} → 0 as m, n → ∞

showing {u(x_{n})}_{n=1}^{∞} is Cauchy so that ū(x_{0}) := lim_{n→∞} u(x_{n}) exists. If {y_{n}}_{n=1}^{∞} ⊂ Ω is another sequence converging to x_{0}, then

|u(x_{n}) − u(y_{n})| ≤ [u]_{β} |x_{n} − y_{n}|^{β} → 0 as n → ∞,

showing ū(x_{0}) is well defined. In this way we define ū(x) for all x ∈ bd(Ω) and let ū(x) = u(x) for x ∈ Ω. Since a similar limiting argument shows

|ū(x) − ū(y)| ≤ [u]_{β} |x − y|^{β} for all x, y ∈ Ω̄

it follows that ū is still continuous and [ū]_{β} = [u]_{β}. In the sequel we will abuse notation and simply denote ū by u. 3. For u, v ∈ C^{0,β}(Ω),

[v + u]_{β} = sup_{x,y∈Ω_{x≠y}} \{\frac{|v(y) + u(y) − v(x) − u(x)|}{|x − y|^{β}}\}

≤ sup_{x,y∈Ω_{x≠y}} \{\frac{|v(y) − v(x)| + |u(y) − u(x)|}{|x − y|^{β}}\} ≤ [v]_{β} + [u]_{β}

and for λ ∈ ℂ it is easily seen that [λu]_{β} = |λ| [u]_{β}. This shows [·]_{β} is a seminorm on C^{0,β}(Ω) and therefore ∥ · ∥_{C^{0,β}(Ω)} defined in Eq. (9.1) is a norm. To see that C^{0,β}(Ω) is complete, let {u_{n}}_{n=1}^{∞} be a C^{0,β}(Ω)–Cauchy sequence. Since BC(Ω̄) is complete, there exists u ∈ BC(Ω̄) such that ∥u − u_{n}∥_{∞} → 0 as n → ∞. For x, y ∈ Ω with x ≠ y,

\frac{|u(x) − u(y)|}{|x − y|^{β}} = lim_{n→∞} \frac{|u_{n}(x) − u_{n}(y)|}{|x − y|^{β}}

≤ lim sup_{n→∞}[u_{n}]_{β} ≤ lim_{n→∞} ∥u_{n}∥_{C^{0,β}(Ω)} < ∞,

and so we see that u ∈ C^{0,β}(Ω). Similarly,

\frac{|u(x) − u_{n}(x) − (u(y) − u_{n}(y))|}{|x − y|^{β}} = lim_{m→∞} \frac{|(u_{m} − u_{n})(x) − (u_{m} − u_{n})(y)|}{|x − y|^{β}}

≤ lim sup_{m→∞}[u_{m} − u_{n}]_{β} → 0 as n → ∞,

showing [u − u_{n}]_{β} → 0 as n → ∞ and therefore lim_{n→∞} ∥u − u_{n}∥_{C^{0,β}(Ω)} = 0.

Notation 9.5 Since Ω and Ω̄ are locally compact Hausdorff spaces, we may define C_{0}(Ω) and C_{0}(Ω̄) as in Definition 12.22. We will also let

C_{0}^{0,β}(Ω) := C^{0,β}(Ω) ∩ C_{0}(Ω) and C_{0}^{0,β}(Ω̄) := C^{0,β}(Ω) ∩ C_{0}(Ω̄).

It has already been shown in Proposition 12.23 that C_{0}(Ω) and C_{0}(Ω̄) are closed subspaces of BC(Ω) and BC(Ω̄) respectively. The next proposition describes the relation between C_{0}(Ω) and C_{0}(Ω̄).

Proposition 9.6. Each u ∈ C_{0}(Ω) has a unique extension to a continuous function on Ω̄ given by ū = u on Ω and ū = 0 on bd(Ω) and the extension ū is in C_{0}(Ω̄). Conversely if u ∈ C_{0}(Ω̄) and u|_{bd(Ω)} = 0, then u|_{Ω} ∈ C_{0}(Ω). In this way we may identify C_{0}(Ω) with those u ∈ C_{0}(Ω̄) such that u|_{bd(Ω)} = 0.

Proof. Any extension u ∈ C_{0}(Ω) to an element ū ∈ C(Ω̄) is necessarily unique, since Ω is dense inside Ω̄. So define ū = u on Ω and ū = 0 on bd(Ω). We must show ū is continuous on Ω̄ and ū ∈ C_{0}(Ω̄). For the continuity assertion it is enough to show ū is continuous at all points in bd(Ω). For any ε > 0, by assumption, the set K_{ε} := {x ∈ Ω : |u(x)| ≥ ε} is a compact subset of Ω. Since bd(Ω) = Ω̄ \ Ω, bd(Ω) ∩ K_{ε} = ∅ and therefore the distance, δ := d(K_{ε}, bd(Ω)), between K_{ε} and bd(Ω) is positive. So if x ∈ bd(Ω) and y ∈ Ω̄ and |y − x| < δ, then |ū(x) − ū(y)| = |u(y)| < ε which shows ū : Ω̄ → ℂ is continuous. This also shows \{\left|{ \atop ū}\right| ≥ ε\} = \{|u| ≥ ε\} = K_{ε} is compact in Ω and hence also in Ω̄. Since ε > 0 was arbitrary, this shows ū ∈ C_{0}(Ω̄). Conversely if u ∈ C_{0}(Ω̄) such that u|_{bd(Ω)} = 0 and ε > 0, then K_{ε} := {x ∈ Ω̄ : |u(x)| ≥ ε} is a compact subset of Ω̄ which is contained in Ω since bd(Ω) ∩ K_{ε} = ∅. Therefore K_{ε} is a compact subset of Ω showing u|_{Ω} ∈ C_{0}(Ω̄).

Definition 9.7. Let Ω be an open subset of ℝ^{d}, k ∈ ℕ∪ {0} and β ∈ (0, 1]. Let BC^{k}(Ω) (BC^{k}(Ω̄)) denote the set of k – times continuously differentiable functions u on Ω such that ∂^{α}u ∈ BC(Ω) (∂^{α}u ∈ BC(Ω̄))^{2} for all |α| ≤ k.

2 To say ∂^{α}u ∈ BC(Ω̄) means that ∂^{α}u ∈ BC(Ω) and ∂^{α}u extends to a continuous function on Ω̄.

Similarly, let BC^{k,β}(Ω) denote those u ∈ BC^{k}(Ω) such that [∂^{α}u]_{β} < ∞ for all |α| = k. For u ∈ BC^{k}(Ω) let

∥u∥_{C^{k}(Ω)} = ∑_{|α|≤k} ∥∂^{α}u∥_{u} and
∥u∥_{C^{k,β}(\overline{Ω})} = ∑_{|α|≤k} ∥∂^{α}u∥_{u} + ∑_{|α|=k}[∂^{α}u]_{β}.

Theorem 9.8. The spaces BC^{k}(Ω) and BC^{k,β}(Ω) equipped with ∥ · ∥_{C^{k}(Ω)} and ∥·∥_{C^{k,β}(\overline{Ω})} respectively are Banach spaces and BC^{k}(Ω̄) is a closed subspace of BC^{k}(Ω) and BC^{k,β}(Ω) ⊂ BC^{k}(Ω̄). Also

C_{0}^{k,β}(Ω) = C_{0}^{k,β}(Ω̄) = {u ∈ BC^{k,β}(Ω) : ∂^{α}u ∈ C_{0}(Ω) ∀ |α| ≤ k}

is a closed subspace of BC^{k,β}(Ω).

Proof. Suppose that {u_{n}}_{n=1}^{∞} ⊂ BC^{k}(Ω) is a Cauchy sequence, then {∂^{α}u_{n}}_{n=1}^{∞} is a Cauchy sequence in BC(Ω) for |α| ≤ k. Since BC(Ω) is complete, there exists g_{α} ∈ BC(Ω) such that lim_{n→∞} ∥∂^{α}u_{n} − g_{α}∥_{∞} = 0 for all |α| ≤ k. Letting u := g_{0}, we must show u ∈ C^{k}(Ω) and ∂^{α}u = g_{α} for all |α| ≤ k. This will be done by induction on |α| . If |α| = 0 there is nothing to prove. Suppose that we have verified u ∈ C^{l}(Ω) and ∂^{α}u = g_{α} for all |α| ≤ l for some l < k. Then for x ∈ Ω, i ∈ {1, 2, …, d} and t ∈ ℝ sufficiently small,

∂^{a}u_{n}(x + te_{i}) = ∂^{a}u_{n}(x) + ∫_{0}^{t} ∂_{i}∂^{a}u_{n}(x + τe_{i})dτ.

Letting n → ∞ in this equation gives

∂^{a}u(x + te_{i}) = ∂^{a}u(x) + ∫_{0}^{t} g_{α+e_{i}}(x + τe_{i})dτ

from which it follows that ∂_{i}∂^{α}u(x) exists for all x ∈ Ω and ∂_{i}∂^{α}u = g_{α+e_{i}}. This completes the induction argument and also the proof that BC^{k}(Ω) is complete. It is easy to check that BC^{k}(Ω̄) is a closed subspace of BC^{k}(Ω) and by using Exercise 9.1 and Theorem 9.4 that that BC^{k,β}(Ω) is a subspace of BC^{k}(Ω̄). The fact that C_{0}^{k,β}(Ω) is a closed subspace of BC^{k,β}(Ω) is a consequence of Proposition 12.23. To prove BC^{k,β}(Ω) is complete, let {u_{n}}_{n=1}^{∞} ⊂ BC^{k,β}(Ω) be a ∥·∥_{C^{k,β}(\overline{Ω})} – Cauchy sequence. By the completeness of BC^{k}(Ω) just proved, there exists u ∈ BC^{k}(Ω) such that lim_{n→∞} ∥u − u_{n}∥_{C^{k}(Ω)} = 0. An application of Theorem 9.4 then shows lim_{n→∞} ∥∂^{α}u_{n} − ∂^{α}u∥_{C^{0,β}(Ω)} = 0 for |α| = k and therefore lim_{n→∞} ∥u − u_{n}∥_{C^{k,β}(\overline{Ω})} = 0.

The reader is asked to supply the proof of the following lemma.

Lemma 9.9. The following inclusions hold. For any β ∈ [0, 1]

BC^{k+1,0}(Ω) ⊂ BC^{k,1}(Ω) ⊂ BC^{k,β}(Ω)
BC^{k+1,0}(Ω̄) ⊂ BC^{k,1}(Ω̄) ⊂ BC^{k,β}(Ω).

9.1 Exercises 105

9.1 Exercises

Exercise 9.1. Prove Lemma 9.3.

Part III

Topological Spaces Topological Space Basics

Using the metric space results above as motivation we will axiomatize the notion of being an open set to more general settings.

Definition 10.1. A collection of subsets τ of X is a topology if

1. ∅, X ∈ τ

2. τ is closed under arbitrary unions, i.e. if V_{α} ∈ τ, for α ∈ I then ⋃_{α∈I} V_{α} ∈ τ.

3. τ is closed under finite intersections, i.e. if V_{1}, …, V_{n} ∈ τ then V_{1} ∩ ⋯ ∩ V_{n} ∈ τ. A pair (X, τ) where τ is a topology on X will be called a topological space.

Notation 10.2 Let (X, τ) be a topological space.

1. The elements, V ∈ τ, are called open sets. We will often write V ⊂_{o} X to indicate V is an open subset of X.

2. A subset F ⊂ X is closed if F^{c} is open and we will write F ⊏ X if F is a closed subset of X.

3. An open neighborhood of a point x ∈ X is an open set V ⊂ X such that x ∈ V. Let τ_{x} = {V ∈ τ : x ∈ V } denote the collection of open neighborhoods of x.

4. A subset W ⊂ X is a neighborhood of x if there exists V ∈ τ_{x} such that V ⊂ W.

5. A collection η ⊂ τ_{x} is called a neighborhood base at x ∈ X if for all V ∈ τ_{x} there exists W ∈ η such that W ⊂ V.

The notation τ_{x} should not be confused with

τ_{{x}} := i_{{x}}^{−1}(τ) = {{x} ∩ V : V ∈ τ} = {∅, {x}} .

Example 10.3. 1. Let (X, d) be a metric space, we write τ_{d} for the collection of d – open sets in X. We have already seen that τ_{d} is a topology, see Exercise 6.2. The collection of sets η = {B_{x}(ε) : ε ∈ 𝔻} where 𝔻 is any dense subset of (0, 1] is a neighborhood base at x.

2. Let X be any set, then τ = 2^{X} is a topology. In this topology all subsets of X are both open and closed. At the opposite extreme we have the trivial topology, τ = {∅, X} . In this topology only the empty set and X are open (closed).

3. Let X = {1, 2, 3}, then τ = {∅, X, {2, 3}} is a topology on X which does not come from a metric.

4. Again let X = {1, 2, 3}. Then τ = {{1}, {2, 3}, ∅, X}. is a topology, and the sets X, {1}, {2, 3}, ∅ are open and closed. The sets {1, 2} and {1, 3} are neither open nor closed.

Fig. 10.1. A topology.

Definition 10.4. Let (X, τ_{X}) and (Y, τ_{Y}) be topological spaces. A function f : X → Y is continuous if

f^{−1}(τ_{Y}) := {f^{−1} (V) : V ∈ τ_{Y} } ⊂ τ_{X}.

We will also say that f is τ_{X}/τ_{Y} –continuous or (τ_{X}, τ_{Y}) – continuous. Let C(X, Y) denote the set of continuous functions from X to Y.

Exercise 10.1. Show f : X → Y is continuous iff f^{−1}(C) is closed in X for all closed subsets C of Y.

Definition 10.5. A map f : X → Y between topological spaces is called a homeomorphism provided that f is bijective, f is continuous and f^{−1} : Y → X is continuous. If there exists f : X → Y which is a homeomorphism, we say that X and Y are homeomorphic. (As topological spaces X and Y are essentially the same.)

10.1 Constructing Topologies and Checking Continuity

Proposition 10.6. Let E be any collection of subsets of X. Then there exists a unique smallest topology τ(E) which contains E.

Proof. Since 2^{X} is a topology and E ⊂ 2^{X}, E is always a subset of a topology. It is now easily seen that

τ(E) := ⋂{τ : τ is a topology and E ⊂ τ}

is a topology which is clearly the smallest possible topology containing E. The following proposition gives an explicit descriptions of τ(E).

Proposition 10.7. Let X be a set and E ⊂ 2^{X}. For simplicity of notation, assume that X, ∅ ∈ E. (If this is not the case simply replace E by E ∪ {X, ∅} .) Then

τ (E) := {arbitrary unions of finite intersections of elements from E}. (10.1)

Proof. Let τ be given as in the right side of Eq. (10.1). From the definition of a topology any topology containing E must contain τ and hence E ⊂ τ ⊂ τ(E). The proof will be completed by showing τ is a topology. The validation of τ being a topology is routine except for showing that τ is closed under taking finite intersections. Let V, W ∈ τ which by definition may be expressed as

V = ∪_{α∈A}V_{α} and W = ∪_{β∈B}W_{β},

where V_{α} and W_{β} are sets which are finite intersection of elements from E. Then

V ∩ W = (∪_{α∈A}V_{α}) ∩ (∪_{β∈B}W_{β}) = ⋃_{(α,β)∈A×B} V_{α} ∩ W_{β}.

Since for each (α, β) ∈ A × B, V_{α} ∩ W_{β} is still a finite intersection of elements from E, V ∩ W ∈ τ showing τ is closed under taking finite intersections.

Definition 10.8. Let (X, τ) be a topological space. We say that S ⊂ τ is a sub-base for the topology τ iff τ = τ(S) and X = ∪S := ∪_{V ∈S}V. We say V ⊂ τ is a base for the topology τ iff V is a sub-base with the property that every element V ∈ τ may be written as

V = ∪{B ∈ V : B ⊂ V }.

Exercise 10.2. Suppose that S is a sub-base for a topology τ on a set X.

1. Show V := S_{f} (S_{f} is the collection of finite intersections of elements from S) is a base for τ.

2. Show S is itself a base for τ iff

V_{1} ∩ V_{2} = ∪{S ∈ S : S ⊂ V_{1} ∩ V_{2}}.

for every pair of sets V_{1}, V_{2} ∈ S.

Fig. 10.2. Fitting balls in the intersection.

Remark 10.9. Let (X, d) be a metric space, then E = {B_{x}(δ) : x ∈ X and δ > 0} is a base for τ_{d} – the topology associated to the metric d. This is the content of Exercise 6.3.

Let us check directly that E is a base for a topology. Suppose that x, y ∈ X and ε, δ > 0. If z ∈ B(x, δ) ∩ B(y, ε), then

B(z, α) ⊂ B(x, δ) ∩ B(y, ε)  (10.2)

where α = min{δ − d(x, z), ε − d(y, z)}, see Figure 10.2. This is a formal consequence of the triangle inequality. For example let us show that B(z, α) ⊂ B(x, δ). By the definition of α, we have that α ≤ δ − d(x, z) or that d(x, z) ≤ δ − α. Hence if w ∈ B(z, α), then

d(x, w) ≤ d(x, z) + d(z, w) ≤ δ − α + d(z, w) < δ − α + α = δ

which shows that w ∈ B(x, δ). Similarly we show that w ∈ B(y, ε) as well.

Owing to Exercise 10.2, this shows E is a base for a topology. We do not need to use Exercise 10.2 here since in fact Equation (10.2) may be generalized to finite intersection of balls. Namely if x_{i} ∈ X, δ_{i} > 0 and z ∈ ∩_{i=1}^{n}B(x_{i}, δ_{i}), then

B(z, α) ⊂ ∩_{i=1}^{n}B(x_{i}, δ_{i})  (10.3)

where now α := min {δ_{i} − d(x_{i}, z) : i = 1, 2, …, n} . By Eq. (10.3) it follows that any finite intersection of open balls may be written as a union of open balls.

Exercise 10.3. Suppose f : X → Y is a function and τ_{X} and τ_{Y} are topologies on X and Y respectively. Show

f^{−1}τ_{Y} := {f^{−1} (V) ⊂ X : V ∈ τ_{Y} } and f_{∗}τ_{X} := {V ⊂ Y : f^{−1} (V) ∈ τ_{X}}

(as in Notation 2.7) are also topologies on X and Y respectively.

Remark 10.10. Let f : X → Y be a function. Given a topology τ_{Y} ⊂ 2^{Y} , the topology τ_{X} := f^{−1}(τ_{Y}) is the smallest topology on X such that f is (τ_{X}, τ_{Y})

- continuous. Similarly, if τ_{X} is a topology on X then τ_{Y} = f_{∗}τ_{X} is the largest topology on Y such that f is (τ_{X}, τ_{Y}) - continuous.

Definition 10.11. Let (X, τ) be a topological space and A subset of X. The relative topology or induced topology on A is the collection of sets

τ_{A} = i_{A}^{−1}(τ) = {A ∩ V : V ∈ τ} ,

where i_{A} : A → X be the inclusion map as in Definition 2.8.

Lemma 10.12. The relative topology, τ_{A}, is a topology on A. Moreover a subset B ⊂ A is τ_{A} – closed iff there is a τ – closed subset, C, of X such that B = C ∩ A.

Proof. The first assertion is a consequence of Exercise 10.3. For the second, B ⊂ A is τ_{A} – closed iff A \ B = A ∩ V for some V ∈ τ which is equivalent to B = A \ (A ∩ V) = A ∩ V^{c} for some V ∈ τ.

Exercise 10.4. Show if (X, d) is a metric space and τ = τ_{d} is the topology coming from d, then (τ_{d})_{A} is the topology induced by making A into a metric space using the metric d|_{A×A}.

Lemma 10.13. Suppose that (X, τ_{X}), (Y, τ_{Y}) and (Z, τ_{Z}) are topological spaces. If f : (X, τ_{X}) → (Y, τ_{Y}) and g : (Y, τ_{Y}) → (Z, τ_{Z}) are continuous functions then g ∘ f : (X, τ_{X}) → (Z, τ_{Z}) is continuous as well.

Proof. This is easy since by assumption g^{−1}(τ_{Z}) ⊂ τ_{Y} and f^{−1} (τ_{Y}) ⊂ τ_{X} so that

(g ∘ f)^{−1} (τ_{Z}) = f^{−1} (g^{−1} (τ_{Z})) ⊂ f^{−1} (τ_{Y}) ⊂ τ_{X}.

The following elementary lemma turns out to be extremely useful because it may be used to greatly simplify the verification that a given function is continuous.

Lemma 10.14. Suppose that f : X → Y is a function, E ⊂ 2^{Y} and A ⊂ Y, then

τ (f^{−1}(E)) = f^{−1}(τ(E)) and  (10.4)
τ (E_{A}) = (τ(E))_{A} .  (10.5)

Moreover, if τ_{Y} = τ (E) and τ_{X} is a topology on X, then f is (τ_{X}, τ_{Y}) – continuous iff f^{−1}(E) ⊂ τ_{X}.

Proof. We will give two proof of Eq. (10.4). The first proof is more constructive than the second, but the second proof will work in the context of σ – algebras to be developed later. First Proof. There is no harm (as the reader should verify) in replacing E by E ∪ {∅, Y } if necessary so that we may assume that ∅, Y ∈ E. By Proposition 10.7, the general element V of τ(E) is an arbitrary unions of finite intersections of elements from E. Since f^{−1} preserves all of the set operations, it follows that f^{−1}τ(E) consists of sets which are arbitrary unions of finite intersections of elements from f^{−1}E, which is precisely τ (f^{−1}(E)) by another application of Proposition 10.7. Second Proof. By Exercise 10.3, f^{−1}(τ(E)) is a topology and since E ⊂ τ (E) , f^{−1}(E) ⊂ f^{−1}(τ(E)). It now follows that τ(f^{−1}(E)) ⊂ f^{−1}(τ(E)). For the reverse inclusion notice that

f_{∗}τ (f^{−1}(E)) = {B ⊂ Y : f^{−1}(B) ∈ τ (f^{−1}(E))}

is a topology which contains E and thus τ(E) ⊂ f_{∗}τ (f^{−1}(E)) . Hence if B ∈ τ(E) we know that f^{−1}(B) ∈ τ (f^{−1}(E)) , i.e. f^{−1}(τ(E)) ⊂ τ (f^{−1}(E)) and Eq. (10.4) has been proved. Applying Eq. (10.4) with X = A and f = i_{A} being the inclusion map implies

(τ(E))_{A} = i_{A}^{−1}(τ(E)) = τ(i_{A}^{−1}(E)) = τ(E_{A}).

Lastly if f^{−1}E ⊂ τ_{X}, then f^{−1}τ (E) = τ (f^{−1}E) ⊂ τ_{X} which shows f is (τ_{X}, τ_{Y}) – continuous.

Corollary 10.15. If (X, τ) is a topological space and f : X → ℝ is a function then the following are equivalent:

1. f is (τ, τ_{ℝ}) - continuous,

2. f^{−1}((a, b)) ∈ τ for all −∞ < a < b < ∞,

3. f^{−1}((a, ∞)) ∈ τ and f^{−1}((−∞, b)) ∈ τ for all a, b ∈ ℚ.

(We are using τ_{ℝ} to denote the standard topology on ℝ induced by the metric d(x, y) = |x − y|.)

Proof. Apply Lemma 10.14 with appropriate choices of E.

Definition 10.16. Let (X, τ_{X}) and (Y, τ_{Y}) be topological spaces. A function f : X → Y is continuous at a point x ∈ X if for every open neighborhood V of f(x) there is an open neighborhood U of x such that U ⊂ f^{−1}(V). See Figure 10.3.

Exercise 10.5. Show f : X → Y is continuous (Definition 10.16) iff f is continuous at all points x ∈ X.

Definition 10.17. Given topological spaces (X, τ) and (Y, τ′) and a subset A ⊂ X. We say a function f : A → Y is continuous iff f is τ_{A}/τ′ – continuous.

Fig. 10.3. Checking that a function is continuous at x ∈ X.

Definition 10.18. Let (X, τ) be a topological space and A ⊂ X. A collection of subsets U ⊂ τ is an open cover of A if A ⊂ ⋃ U := ⋃_{U∈U} U.

Proposition 10.19 (Localizing Continuity). Let (X, τ) and (Y, τ′) be topological spaces and f : X → Y be a function.

1. If f is continuous and A ⊂ X then f|_{A} : A → Y is continuous.

2. Suppose there exist an open cover, U ⊂ τ, of X such that f|_{A} is continuous for all A ∈ U, then f is continuous.

Proof. 1. If f : X → Y is a continuous, f^{−1}(V) ∈ τ for all V ∈ τ′ and therefore

f|_{A}^{−1} (V) = A ∩ f^{−1}(V) ∈ τ_{A} for all V ∈ τ′.

2. Let V ∈ τ′, then

f^{−1}(V) = ∪_{A∈U} (f^{−1}(V) ∩ A) = ∪_{A∈U}f|_{A}^{−1}(V).  (10.6)

Since each A ∈ U is open, τ_{A} ⊂ τ and by assumption, f|_{A}^{−1}(V) ∈ τ_{A} ⊂ τ. Hence Eq. (10.6) shows f^{−1} (V) is a union of τ – open sets and hence is also τ – open.

Exercise 10.6 (A Baby Extension Theorem). Suppose V ∈ τ and f : V → ℂ is a continuous function. Further assume there is a closed subset C such that {x ∈ V : f (x) ≠ 0} ⊂ C ⊂ V, then F : X → ℂ defined by

F (x) = \{ \begin{matrix}f(x) & if & x ∈ V \\ 0 & if & x ∉ V \end{matrix}

is continuous.

Exercise 10.7 (Building Continuous Functions). Prove the following variant of item 2. of Proposition 10.19. Namely, suppose there exists a finite collection F of closed subsets of X such that X = ∪_{A∈F}A and f|_{A} is continuous for all A ∈ F, then f is continuous. Given an example showing that the assumption that F is finite can not be eliminated. Hint: consider f^{−1} (C) where C is a closed subset of Y.

10.2 Product Spaces I

Definition 10.20. Let X be a set and suppose there is a collection of topological spaces {(Y_{α}, τ_{α}) : α ∈ A} and functions f_{α} : X → Y_{α} for all α ∈ A. Let τ(f_{α} : α ∈ A) denote the smallest topology on X such that each f_{α} is continuous, i.e.

τ(f_{α} : α ∈ A) = τ(∪_{α}f_{α}^{−1}(τ_{α})).

Proposition 10.21 (Topologies Generated by Functions). Assuming the notation in Definition 10.20 and additionally let (Z, τ_{Z}) be a topological space and g : Z → X be a function. Then g is (τ_{Z}, τ(f_{α} : α ∈ A)) – continuous iff f_{α} ∘ g is (τ_{Z}, τ_{α})–continuous for all α ∈ A.

Proof. (⇒) If g is (τ_{Z}, τ(f_{α} : α ∈ A)) – continuous, then the composition f_{α} ∘ g is (τ_{Z}, τ_{α}) – continuous by Lemma 10.13. (⇐) Let

τ_{X} = τ(f_{α} : α ∈ A) = τ (∪_{α∈A}f_{α}^{−1}(τ_{α})) .

If f_{α} ∘ g is (τ_{Z}, τ_{α}) – continuous for all α, then

g^{−1}f_{α}^{−1}(τ_{α}) ⊂ τ_{Z} ∀ α ∈ A

and therefore

g^{−1} (∪_{α∈A}f_{α}^{−1}(τ_{α})) = ∪_{α∈A}g^{−1}f_{α}^{−1}(τ_{α}) ⊂ τ_{Z}

Hence

g^{−1} (τ_{X}) = g^{−1} (τ (∪_{α∈A}f_{α}^{−1}(τ_{α}))) = τ(g^{−1} (∪_{α∈A}f_{α}^{−1}(τ_{α})) ⊂ τ_{Z}

which shows that g is (τ_{Z}, τ_{X}) – continuous.

Let {(X_{α}, τ_{α})}_{α∈A} be a collection of topological spaces, X = X_{A} = ∏_{α∈A} X_{α}

and π_{α} : X_{A} → X_{α} be the canonical projection map as in Notation 2.2.

Definition 10.22. The product topology τ = ⊗_{α∈A}τ_{α} is the smallest topology on X_{A} such that each projection π_{α} is continuous. Explicitly, τ is the topology generated by the collection of sets,

E = {π_{α}^{−1}(V_{α}) : α ∈ A, V_{α} ∈ τ_{α}} = ∪_{α∈A}π^{−1}τ_{α}.  (10.7)

Applying Proposition 10.21 in this setting implies the following proposition.

Proposition 10.23. Suppose Y is a topological space and f : Y → X_{A} is a map. Then f is continuous iff π_{α} ∘ f : Y → X_{α} is continuous for all α ∈ A. In particular if A = {1, 2, …, n} so that X_{A} = X_{1} × X_{2} × ⋯ × X_{n} and f(y) = (f_{1}(y), f_{2}(y), …, f_{n}(y)) ∈ X_{1} × X_{2} × ⋯ × X_{n}, then f : Y → X_{A} is continuous iff f_{i} : Y → X_{i} is continuous for all i.

Proposition 10.24. Suppose that (X, τ) is a topological space and {f_{n}} ⊂ X^{A} (see Notation 2.2) is a sequence. Then f_{n} → f in the product topology of X^{A} iff f_{n}(α) → f(α) for all α ∈ A.

Proof. Since π_{α} is continuous, if f_{n} → f then f_{n}(α) = π_{α}(f_{n}) → π_{α}(f) = f(α) for all α ∈ A. Conversely, f_{n}(α) → f(α) for all α ∈ A iff π_{α}(f_{n}) → π_{α}(f) for all α ∈ A. Therefore if V = π_{α}^{−1}(V_{α}) ∈ E (with E as in Eq. (10.7)) and f ∈ V, then π_{α}(f) ∈ V_{α} and π_{α}(f_{n}) ∈ V_{α} for a.a. n and hence f_{n} ∈ V for a.a. n. This shows that f_{n} → f as n → ∞.

Proposition 10.25. Suppose that (X_{α}, τ_{α})_{α∈A} is a collection of topological spaces and ⊗_{α∈A}τ_{α} is the product topology on X := ∏_{α∈A} X_{α}.

1. If E_{α} ⊂ τ_{α} generates τ_{α} for each α ∈ A, then

⊗_{α∈A} τ_{α} = τ (∪_{α∈A}π_{α}^{−1}(E_{α}))  (10.8)

2. If B_{α} ⊂ τ_{α} is a base for τ_{α} for each α, then the collection of sets, V, of the form

V = ∩_{α∈Λ}π_{α}^{−1}V_{α} = ∏_{α∈Λ} V_{α} × ∏_{α∉Λ} X_{α} =: V_{Λ} × X_{A\Λ},  (10.9)

where Λ ⊂⊂ A and V_{α} ∈ B_{α} for all α ∈ Λ is base for ⊗_{α∈A}τ_{α}.

Proof. 1. Since

∪_{α}π_{α}^{−1}E_{α} ⊂ ∪_{α}π_{α}^{−1}τ_{α} = ∪_{α}π_{α}^{−1}(τ(E_{α}))
= ∪_{α}τ(π_{α}^{−1}E_{α}) ⊂ τ (∪_{α}π_{α}^{−1}E_{α}) ,

it follows that

τ (∪_{α}π_{α}^{−1}E_{α}) ⊂ ⊗_{α}τ_{α} ⊂ τ (∪_{α}π_{α}^{−1}E_{α}) .

2. Now let U = [∪_{α}π_{α}^{−1}τ_{α}]_{f} denote the collection of sets consisting of finite intersections of elements from ∪_{α}π_{α}^{−1}τ_{α}. Notice that U may be described as those sets in Eq. (10.9) where V_{α} ∈ τ_{α} for all α ∈ Λ. By Exercise 10.2, U is a base for the product topology, ⊗_{α∈A}τ_{α}. Hence for W ∈ ⊗_{α∈A}τ_{α} and x ∈ W, there exists a V ∈ U of the form in Eq. (10.9) such that x ∈ V ⊂ W. Since B_{α} is a base for τ_{α}, there exists U_{α} ∈ B_{α} such that x_{α} ∈ U_{α} ⊂ V_{α} for each α ∈ Λ. With this notation, the set U_{Λ} × X_{A\Λ} ∈ V and x ∈ U_{Λ} × X_{A\Λ} ⊂ V ⊂ W. This shows that every open set in X may be written as a union of elements from V, i.e. V is a base for the product topology.

Notation 10.26 Let E_{i} ⊂ 2^{X_{i}} be a collection of subsets of a set X_{i} for each i = 1, 2, …, n. We will write, by abuse of notation, E_{1} × E_{2} × ⋯ × E_{n} for the collection of subsets of X_{1}×⋯×X_{n} of the form A_{1}×A_{2}×⋯×A_{n} with A_{i} ∈ E_{i} for all i. That is we are identifying (A_{1}, A_{2}, …, A_{n}) with A_{1} × A_{2} × ⋯ × A_{n}.

Corollary 10.27. Suppose A = {1, 2, …, n} so X = X_{1} × X_{2} × ⋯ × X_{n}.

1. If E_{i} ⊂ 2^{X_{i}}, τ_{i} = τ (E_{i}) and X_{i} ∈ E_{i} for each i, then

τ_{1} ⊗ τ_{2} ⊗ ⋯ ⊗ τ_{n} = τ(E_{1} × E_{2} × ⋯ × E_{n})  (10.10)

and in particular

τ_{1} ⊗ τ_{2} ⊗ ⋯ ⊗ τ_{n} = τ(τ_{1} × ⋯ × τ_{n}).  (10.11)

2. Furthermore if B_{i} ⊂ τ_{i} is a base for the topology τ_{i} for each i, then B_{1} × ⋯ × B_{n} is a base for the product topology, τ_{1} ⊗ τ_{2} ⊗ ⋯ ⊗ τ_{n}.

Proof. (The proof is a minor variation on the proof of Proposition 10.25.)

1. Let [∪_{i∈A}π_{i}^{−1}(E_{i})]_{f} denotes the collection of sets which are finite intersections from ∪_{i∈A}π_{i}^{−1}(E_{i}), then, using X_{i} ∈ E_{i} for all i,

∪_{i∈A}π_{i}^{−1}(E_{i}) ⊂ E_{1} × E_{2} × ⋯ × E_{n} ⊂ [∪_{i∈A}π_{i}^{−1}(E_{i})]_{f} .

Therefore

τ = τ (∪_{i∈A}π_{i}^{−1}(E_{i})) ⊂ τ (E_{1} × E_{2} × ⋯ × E_{n}) ⊂ τ ([∪_{i∈A}π_{i}^{−1}(E_{i})]_{f}) = τ.

2. Observe that τ_{1} × ⋯ × τ_{n} is closed under finite intersections and generates τ_{1} ⊗ τ_{2} ⊗ ⋯ ⊗ τ_{n}, therefore τ_{1} × ⋯ × τ_{n} is a base for the product topology. The proof that B_{1} × ⋯ × B_{n} is also a base for τ_{1} ⊗ τ_{2} ⊗ ⋯ ⊗ τ_{n} follows the same method used to prove item 2. in Proposition 10.25.

Lemma 10.28. Let (X_{i}, d_{i}) for i = 1, …, n be metric spaces, X := X_{1} ×⋯× X_{n} and for x = (x_{1}, x_{2}, …, x_{n}) and y = (y_{1}, y_{2}, …, y_{n}) in X let

d(x, y) = ∑_{i=1}^{n} d_{i}(x_{i}, y_{i}).  (10.12)

Then the topology, τ_{d}, associated to the metric d is the product topology on X, i.e.

τ_{d} = τ_{d_{1}} ⊗ τ_{d_{2}} ⊗ ⋯ ⊗ τ_{d_{n}}.

Proof. Let ρ(x, y) = max{d_{i}(x_{i}, y_{i}) : i = 1, 2, …, n}. Then ρ is equivalent to d and hence τ_{ρ} = τ_{d}. Moreover if ε > 0 and x = (x_{1}, x_{2}, …, x_{n}) ∈ X, then

B_{x}^{ρ}(ε) = B_{x_{1}}^{d_{1}}(ε) × ⋯ × B_{x_{n}}^{d_{n}}(ε).

By Remark 10.9,

E := {B_{x}^{ρ}(ε) : x ∈ X and ε > 0}

is a base for τ_{ρ} and by Proposition 10.25 E is also a base for τ_{d_{1}} ⊗τ_{d_{2}} ⊗⋯⊗τ_{d_{n}}. Therefore,

τ_{d_{1}} ⊗ τ_{d_{2}} ⊗ ⋯ ⊗ τ_{d_{n}} = τ(E) = τ_{ρ} = τ_{d}.

10.3 Closure operations

Definition 10.29. Let (X, τ) be a topological space and A be a subset of X.

1. The closure of A is the smallest closed set Ā containing A, i.e.

Ā := ∩ {F : A ⊂ F ⊏ X} .

(Because of Exercise 6.4 this is consistent with Definition 6.10 for the closure of a set in a metric space.)

2. The interior of A is the largest open set A^{o} contained in A, i.e.

A^{o} = ∪ {V ∈ τ : V ⊂ A} .

(With this notation the definition of a neighborhood of x ∈ X may be stated as: A ⊂ X is a neighborhood of a point x ∈ X if x ∈ A^{o}.)

3. The accumulation points of A is the set

acc(A) = {x ∈ X : V ∩ A \ {x} ≠ ∅ for all V ∈ τ_{x}}.

4. The boundary of A is the set bd(A) := Ā \ A^{o}.

Remark 10.30. The relationships between the interior and the closure of a set are:

(A^{o})^{c} = ⋂\{V^{c} : V ∈ τ and V ⊂ A\} = ⋂\{C : C is closed C ⊃ A^{c}\} = \overline{A^{c}}

and similarly, (Ā)^{c} = (A^{c})^{o}. Hence the boundary of A may be written as

bd(A) := Ā \ A^{o} = Ā ∩ (A^{o})^{c} = Ā ∩ \overline{A^{c}},  (10.13)

which is to say bd(A) consists of the points in both the closure of A and A^{c}.

Proposition 10.31. Let A ⊂ X and x ∈ X.

1. If V ⊂_{o} X and A ∩ V = ∅ then Ā ∩ V = ∅.

2. x ∈ Ā iff V ∩ A ≠ ∅ for all V ∈ τ_{x}.

3. x ∈ bd(A) iff V ∩ A ≠ ∅ and V ∩ A^{c} ≠ ∅ for all V ∈ τ_{x}.

4. Ā = A ∪ acc(A).

Proof. 1. Since A ∩ V = ∅, A ⊂ V^{c} and since V^{c} is closed, Ā ⊂ V^{c}. That is to say Ā∩ V = ∅. 2. By Remark 10.30^{1}, Ā = ((A^{c})^{o})^{c} so x ∈ Ā iff x ∉ (A^{c})^{o} which happens iff V ⊈ A^{c} for all V ∈ τ_{x}, i.e. iff V ∩ A ≠ ∅ for all V ∈ τ_{x}. 3. This assertion easily follows from the Item 2. and Eq. (10.13). 4. Item 4. is an easy consequence of the definition of acc(A) and item 2.

1 Here is another direct proof of item 2. which goes by showing x ∉ Ā iff there exists

V ∈ τ_{x} such that V ∩A = ∅. If x ∉ Ā then V = \binom{}{Ā}^{c} ∈ τ_{x} and V ∩A ⊂ V ∩Ā = ∅.

Conversely if there exists V ∈ τ_{x} such that A ∩ V = ∅ then by Item 1. Ā ∩ V = ∅.

Lemma 10.32. Let A ⊂ Y ⊂ X, Ā^{Y} denote the closure of A in Y with its relative topology and Ā = Ā^{X} be the closure of A in X, then Ā^{Y} = Ā^{X} ∩ Y.

Proof. Using Lemma 10.12,

Ā^{Y} = ∩ {B ⊏ Y : A ⊂ B} = ∩ {C ∩ Y : A ⊂ C ⊏ X}
= Y ∩ (∩ {C : A ⊂ C ⊏ X}) = Y ∩ Ā^{X}.

Alternative proof. Let x ∈ Y then x ∈ Ā^{Y} iff V ∩ A ≠ ∅ for all V ∈ τ_{Y} such that x ∈ V. This happens iff for all U ∈ τ_{x}, U ∩ Y ∩ A = U ∩ A ≠ ∅ which happens iff x ∈ Ā^{X}. That is to say Ā^{Y} = Ā^{X} ∩ Y.

The support of a function may now be defined as in Definition 8.25 above.

Definition 10.33 (Support). Let f : X → Y be a function from a topological space (X, τ_{X}) to a vector space Y. Then we define the support of f by

supp(f) := \overline{\{x ∈ X : f(x) ≠ 0\}},

a closed subset of X.

The next result is included for completeness but will not be used in the sequel so may be omitted.

Lemma 10.34. Suppose that f : X → Y is a map between topological spaces. Then the following are equivalent:

1. f is continuous.

2. f(Ā) ⊂ \overline{f(A)} for all A ⊂ X

3. \overline{f^{−1}(B)} ⊂ f^{−1}(B̄) for all B ⊂ X.

Proof. If f is continuous, then f^{−1} (\overline{f(A)}) is closed and since A ⊂

f^{−1} (f(A)) ⊂ f^{−1} (\overline{f(A)}) it follows that Ā ⊂ f^{−1} (\overline{f(A)}) . From this equa-

tion we learn that f(Ā) ⊂ \overline{f(A)} so that (1) implies (2) Now assume (2), then for B ⊂ Y (taking A = f^{−1}(B̄)) we have

f(\overline{f^{−1}(B)}) ⊂ f(\overline{f^{−1}(B̄)}) ⊂ \overline{f(f^{−1}(B̄))} ⊂ B̄

and therefore

\overline{f^{−1}(B)} ⊂ f^{−1}(B̄).  (10.14)

This shows that (2) implies (3) Finally if Eq. (10.14) holds for all B, then when B is closed this shows that

\overline{f^{−1}(B)} ⊂ f^{−1}(B̄) = f^{−1}(B) ⊂ \overline{f^{−1}(B)}

which shows that

f^{−1}(B) = \overline{f^{−1}(B)}.

Therefore f^{−1}(B) is closed whenever B is closed which implies that f is continuous.

10.4 Countability Axioms

Definition 10.35. Let (X, τ) be a topological space. A sequence {x_{n}}_{n=1}^{∞} ⊂ X converges to a point x ∈ X if for all V ∈ τ_{x}, x_{n} ∈ V almost always (abbreviated a.a.), i.e. # ({n : x_{n} ∉ V }) < ∞. We will write x_{n} → x as n → ∞ or lim_{n→∞} x_{n} = x when x_{n} converges to x.

Example 10.36. Let X = {1, 2, 3} and τ = {X, ∅, {1, 2}, {2, 3}, {2}} and x_{n} = 2 for all n. Then x_{n} → x for every x ∈ X. So limits need not be unique!

Definition 10.37 (First Countable). A topological space, (X, τ), is first countable iff every point x ∈ X has a countable neighborhood base as defined in Notation 10.2

Example 10.38. All metric spaces, (X, d) , are first countable. Indeed, if x ∈ X then \{B (x, \frac{1}{n}) : n ∈ ℕ\} is a countable neigborhood base at x ∈ X.

Exercise 10.8. Suppose X is an uncountable set and let V ∈ τ iff V^{c} is finite or countable of V = ∅. Show τ is a topology on X which is closed under countable intersections and that (X, τ) is not first countable.

Exercise 10.9. Let {0, 1} be equipped with the discrete topology and X = {0, 1}^{ℝ} be equipped with the product topology, τ. Show (X, τ) is not first countable.

The spaces described in Exercises 10.8 and 10.9 are examples of topological spaces which are not metrizable, i.e. the topology is not induced by any metric on X. Like for metric spaces, when τ is first countable, we may formulate many topological notions in terms of sequences.

Proposition 10.39. If f : X → Y is continuous at x ∈ X and lim_{n→∞} x_{n} = x ∈ X, then lim_{n→∞} f(x_{n}) = f(x) ∈ Y. Moreover, if there exists a countable neighborhood base η of x ∈ X, then f is continuous at x iff lim_{n→∞} f(x_{n}) = f(x)

for all sequences {x_{n}}_{n=1}^{∞} ⊂ X such that x_{n} → x as n → ∞.

Proof. If f : X → Y is continuous and W ∈ τ_{Y} is a neighborhood of f(x) ∈ Y, then there exists a neighborhood V of x ∈ X such that f(V) ⊂ W. Since x_{n} → x, x_{n} ∈ V a.a. and therefore f(x_{n}) ∈ f(V) ⊂ W a.a., i.e. f(x_{n}) → f(x) as n → ∞. Conversely suppose that η := {W_{n}}_{n=1}^{∞} is a countable neighborhood base at x and lim_{n→∞} f(x_{n}) = f(x) for all sequences

{x_{n}}_{n=1}^{∞} ⊂ X such that x_{n} → x. By replacing W_{n} by W_{1} ∩ ⋯ ∩ W_{n} if necessary, we may assume that {W_{n}}_{n=1}^{∞} is a decreasing sequence of sets. If f were not continuous at x then there exists V ∈ τ_{f(x)} such that x ∉ [f^{−1}(V)]^{o} . Therefore, W_{n} is not a subset of f^{−1}(V) for all n. Hence for each n, we may choose x_{n} ∈ W_{n} \ f^{−1}(V). This sequence then has the property that x_{n} → x as n → ∞ while f(x_{n}) ∉ V for all n and hence lim_{n→∞} f(x_{n}) ≠ f(x).

Lemma 10.40. Suppose there exists {x_{n}}_{n=1}^{∞} ⊂ A such that x_{n} → x, then x ∈ Ā. Conversely if (X, τ) is a first countable space (like a metric space) then if x ∈ Ā there exists {x_{n}}_{n=1}^{∞} ⊂ A such that x_{n} → x.

Proof. Suppose {x_{n}}_{n=1}^{∞} ⊂ A and x_{n} → x ∈ X. Since Ā^{c} is an open set, if x ∈ Ā^{c} then x_{n} ∈ Ā^{c} ⊂ A^{c} a.a. contradicting the assumption that {x_{n}}_{n=1}^{∞} ⊂ A. Hence x ∈ Ā. For the converse we now assume that (X, τ) is first countable and that {V_{n}}_{n=1}^{∞} is a countable neighborhood base at x such that V_{1} ⊃ V_{2} ⊃ V_{3} ⊃ …. By Proposition 10.31, x ∈ Ā iff V ∩ A ≠ ∅ for all V ∈ τ_{x}. Hence x ∈ Ā implies there exists x_{n} ∈ V_{n} ∩ A for all n. It is now easily seen that x_{n} → x as n → ∞.

Definition 10.41. A topological space, (X, τ), is second countable if there exists a countable base V for τ, i.e. V ⊂ τ is a countable set such that for every W ∈ τ,

W = ∪{V : V ∈ V ∋ V ⊂ W }.

Definition 10.42. A subset D of a topological space X is dense if D̄ = X. A topological space is said to be separable if it contains a countable dense subset, D.

Example 10.43. The following are examples of countable dense sets.

1. The rational number ℚ are dense in ℝ equipped with the usual topology.

2. More generally, ℚ^{d} is a countable dense subset of ℝ^{d} for any d ∈ ℕ.

3. Even more generally, for any function μ : ℕ → (0, ∞), ℓ^{p}(μ) is separable for all 1 ≤ p < ∞. For example, let Γ ⊂ 𝔽 be a countable dense set, then

D := {x ∈ ℓ^{p}(μ) : x_{i} ∈ Γ for all i and #{j : x_{j} ≠ 0} < ∞}.

The set Γ can be taken to be ℚ if 𝔽 = ℝ or ℚ + iℚ if 𝔽 = ℂ.

4. If (X, d) is a metric space which is separable then every subset Y ⊂ X is also separable in the induced topology.

To prove 4. above, let A = {x_{n}}_{n=1}^{∞} ⊂ X be a countable dense subset of X. Let d_{Y} (x) = inf{d(x, y) : y ∈ Y } be the distance from x to Y and recall that d_{Y} : X → [0, ∞) is continuous. Let ε_{n} = max \{d_{Y} (x_{n}), \frac{1}{n}\} ≥ 0 and for each n let y_{n} ∈ B_{x_{n}}(2ε_{n}). Then if y ∈ Y and ε > 0 we may choose n ∈ ℕ such that d(y, x_{n}) ≤ ε_{n} < ε/3. Then d(y_{n}, x_{n}) ≤ 2ε_{n} < 2ε/3 and therefore

d(y, y_{n}) ≤ d(y, x_{n}) + d(x_{n}, y_{n}) < ε.

This shows that B := {y_{n}}_{n=1}^{∞} is a countable dense subset of Y.

Exercise 10.10. Show ℓ^{∞} (ℕ) is not separable.

Exercise 10.11. Show every second countable topological space (X, τ) is separable. Show the converse is not true by showing X := ℝ with τ = {∅} ∪ {V ⊂ ℝ : 0 ∈ V } is a separable, first countable but not a second countable topological space.

Exercise 10.12. Every separable metric space, (X, d) is second countable.

Exercise 10.13. Suppose E ⊂ 2^{X} is a countable collection of subsets of X, then τ = τ(E) is a second countable topology on X.

10.5 Connectedness

Definition 10.44. (X, τ) is disconnected if there exists non-empty open sets U and V of X such that U ∩ V = ∅ and X = U ∪ V. We say {U, V } is a disconnection of X. The topological space (X, τ) is called connected if it is not disconnected, i.e. if there is no disconnection of X. If A ⊂ X we say A is connected iff (A, τ_{A}) is connected where τ_{A} is the relative topology on A. Explicitly, A is disconnected in (X, τ) iff there exists U, V ∈ τ such that U ∩ A ≠ ∅, U ∩ A ≠ ∅, A ∩ U ∩ V = ∅ and A ⊂ U ∪ V.

The reader should check that the following statement is an equivalent definition of connectivity. A topological space (X, τ) is connected iff the only sets A ⊂ X which are both open and closed are the sets X and ∅. This version of the definition is often used in practice.

Remark 10.45. Let A ⊂ Y ⊂ X. Then A is connected in X iff A is connected in Y.

Proof. Since

τ_{A} := {V ∩ A : V ⊂ X} = {V ∩ A ∩ Y : V ⊂ X} = {U ∩ A : U ⊂_{o} Y },

the relative topology on A inherited from X is the same as the relative topology on A inherited from Y. Since connectivity is a statement about the relative topologies on A, A is connected in X iff A is connected in Y.

The following elementary but important lemma is left as an exercise to the reader.

Lemma 10.46. Suppose that f : X → Y is a continuous map between topological spaces. Then f(X) ⊂ Y is connected if X is connected.

Here is a typical way these connectedness ideas are used.

Example 10.47. Suppose that f : X → Y is a continuous map between two topological spaces, the space X is connected and the space Y is “T_{1},” i.e. {y} is a closed set for all y ∈ Y as in Definition 12.35 below. Further assume f is locally constant, i.e. for all x ∈ X there exists an open neighborhood V of x in X such that f|_{V} is constant. Then f is constant, i.e. f(X) = {y_{0}} for some y_{0} ∈ Y. To prove this, let y_{0} ∈ f(X) and let W := f^{−1}({y_{0}}). Since {y_{0}} ⊂ Y is a closed set and since f is continuous W ⊂ X is also closed. Since f is locally constant, W is open as well and since X is connected it follows that W = X, i.e. f(X) = {y_{0}} .

As a concrete application of this result, suppose that X is a connected open subset of ℝ^{d} and f : X → ℝ is a C^{1} – function such that ∇f ≡ 0. If x ∈ X and ε > 0 such that B (x, ε) ⊂ X, we have, for any |v| < ε and t ∈ [−1, 1] , that

\frac{d}{dt}f (x + tv) = ∇f (x + tv) · v = 0.

Therefore f (x + v) = f (x) for all |v| < ε and this shows f is locally constant. Hence, by what we have just proved, f is constant on X.

Theorem 10.48 (Properties of Connected Sets). Let (X, τ) be a topological space.

1. If B ⊂ X is a connected set and X is the disjoint union of two open sets U and V, then either B ⊂ U or B ⊂ V.

2. If A ⊂ X is connected, a) then Ā is connected.

b) More generally, if A is connected and B ⊂ acc(A), then A ∪ B is connected as well. (Recall that acc(A) – the set of accumulation points of A was defined in Definition 10.29 above.)

3. If {E_{α}}_{α∈A} is a collection of connected sets such that ⋂_{α∈A} E_{α} ≠ ∅, then Y := ⋃_{α∈A} E_{α} is connected as well.

4. Suppose A, B ⊂ X are non-empty connected subsets of X such that Ā ∩ B ≠ ∅, then A ∪ B is connected in X.

5. Every point x ∈ X is contained in a unique maximal connected subset C_{x} of X and this subset is closed. The set C_{x} is called the connected component of x.

Proof.

1. Since B is the disjoint union of the relatively open sets B ∩ U and B ∩ V, we must have B ∩ U = B or B ∩ V = B for otherwise {B ∩ U, B ∩ V } would be a disconnection of B.

2. a. Let Y = Ā be equipped with the relative topology from X. Suppose that U, V ⊂_{o} Y form a disconnection of Y = Ā. Then by 1. either A ⊂ U or A ⊂ V. Say that A ⊂ U. Since U is both open an closed in Y, it follows that Y = Ā ⊂ U. Therefore V = ∅ and we have a contradiction to the assumption that {U, V } is a disconnection of Y = Ā. Hence we must conclude that Y = Ā is connected as well. b. Now let Y = A ∪ B with B ⊂ acc(A), then

Ā^{Y} = Ā ∩ Y = (A ∪ acc(A)) ∩ Y = A ∪ B.

Because A is connected in Y, by (2a) Y = A ∪ B = Ā^{Y} is also connected.

3. Let Y := ⋃_{α∈A} E_{α}. By Remark 10.45, we know that E_{α} is connected in Y for each α ∈ A. If {U, V } were a disconnection of Y, by item (1), either E_{α} ⊂ U or E_{α} ⊂ V for all α. Let Λ = {α ∈ A : E_{α} ⊂ U} then U = ∪_{α∈Λ}E_{α} and V = ∪_{α∈A\Λ}E_{α}. (Notice that neither Λ or A \ Λ can be empty since U and V are not empty.) Since

∅ = U ∩ V = ⋃_{α∈Λ,β∈Λ^{c}} (E_{α} ∩ E_{β}) ⊃ ⋂_{α∈A} E_{α} ≠ ∅.

we have reached a contradiction and hence no such disconnection exists.

4. (A good example to keep in mind here is X = ℝ, A = (0, 1) and B = [1, 2).) For sake of contradiction suppose that {U, V } were a disconnection of Y = A ∪ B. By item (1) either A ⊂ U or A ⊂ V, say A ⊂ U in which case B ⊂ V. Since Y = A ∪ B we must have A = U and B = V and so we may conclude: A and B are disjoint subsets of Y which are both open and closed. This implies

A = Ā^{Y} = Ā ∩ Y = Ā ∩ (A ∪ B) = A ∪ (Ā ∩ B)

and therefore

∅ = A ∩ B = [A ∪ (Ā ∩ B)] ∩ B = Ā ∩ B ≠ ∅

which gives us the desired contradiction.

5. Let C denote the collection of connected subsets C ⊂ X such that x ∈ C. Then by item 3., the set C_{x} := ∪C is also a connected subset of X which contains x and clearly this is the unique maximal connected set containing x. Since C̄_{x} is also connected by item (2) and C_{x} is maximal, C_{x} = C̄_{x}, i.e. C_{x} is closed.

Theorem 10.49 (The Connected Subsets of ℝ). The connected subsets of ℝ are intervals.

Proof. Suppose that A ⊂ ℝ is a connected subset and that a, b ∈ A with a < b. If there exists c ∈ (a, b) such that c ∉ A, then U := (−∞, c) ∩ A and V := (c, ∞) ∩ A would form a disconnection of A. Hence (a, b) ⊂ A. Let α := inf(A) and β := sup(A) and choose α_{n}, β_{n} ∈ A such that α_{n} < β_{n} and α_{n} ↓ α and β_{n} ↑ β as n → ∞. By what we have just shown, (α_{n}, β_{n}) ⊂ A for all n and hence (α, β) = ∪_{n=1}^{∞}(α_{n}, β_{n}) ⊂ A. From this it follows that A = (α, β), [α, β), (α, β] or [α, β], i.e. A is an interval.

Conversely suppose that A is an interval, and for sake of contradiction, suppose that {U, V } is a disconnection of A with a ∈ U, b ∈ V. After relabelling U and V if necessary we may assume that a < b. Since A is an interval [a, b] ⊂ A. Let p = sup ([a, b] ∩ U) , then because U and V are open, a < p < b. Now p can not be in U for otherwise sup ([a, b] ∩ U) > p and p can not be in V for otherwise p < sup ([a, b] ∩ U) . From this it follows that p ∉ U ∪ V and hence A ≠ U ∪V contradicting the assumption that {U, V } is a disconnection.

Theorem 10.50 (Intermediate Value Theorem). Suppose that (X, τ) is a connected topological space and f : X → ℝ is a continuous map. Then f satisfies the intermediate value property. Namely, for every pair x, y ∈ X such that f (x) < f(y) and c ∈ (f (x) , f(y)), there exits z ∈ X such that f(z) = c.

Proof. By Lemma 10.46, f (X) is connected subset of ℝ. So by Theorem 10.49, f (X) is a subinterval of ℝ and this completes the proof.

Definition 10.51. A topological space X is path connected if to every pair of points {x_{0}, x_{1}} ⊂ X there exists a continuous path, σ ∈ C([0, 1], X), such that σ(0) = x_{0} and σ(1) = x_{1}. The space X is said to be locally path connected if for each x ∈ X, there is an open neighborhood V ⊂ X of x which is path connected.

Proposition 10.52. Let X be a topological space.

1. If X is path connected then X is connected.

2. If X is connected and locally path connected, then X is path connected.

3. If X is any connected open subset of ℝ^{n}, then X is path connected.

Proof. The reader is asked to prove this proposition in Exercises 10.20 – 10.22 below.

Proposition 10.53 (Stability of Connectedness Under Products). Let (X_{α}, τ_{α}) be connected topological spaces. Then the product space X_{A} = ∏_{α∈A} X_{α} equipped with the product topology is connected.

Proof. Let us begin with the case of two factors, namely assume that X and Y are connected topological spaces, then we will show that X × Y is connected as well. Given x ∈ X, let f_{x} : Y → X ×Y be the map f_{x}(y) = (x, y) and notice that f_{x} is continuous since π_{X} ∘ f_{x}(y) = x and π_{Y} ∘ f_{x}(y) = y are continuous maps. From this we conclude that {x} × Y = f_{x}(Y) is connected by Lemma 10.46. A similar argument shows X ×{y} is connected for all y ∈ Y.

Let p = (x_{0}, y_{0}) ∈ X × Y and C_{p} denote the connected component of p. Since {x_{0}} × Y is connected and p ∈ {x_{0}} × Y it follows that {x_{0}} × Y ⊂ C_{p} and hence C_{p} is also the connected component (x_{0}, y) for all y ∈ Y. Similarly, X ×{y} ⊂ C_{(x_{0},y)} = C_{p} is connected, and therefore X ×{y} ⊂ C_{p}. So we have shown (x, y) ∈ C_{p} for all x ∈ X and y ∈ Y, see Figure 10.4. By induction the theorem holds whenever A is a finite set, i.e. for products of a finite number of connected spaces.

For the general case, again choose a point p ∈ X_{A} = X^{A} and again let C = C_{p} be the connected component of p. Recall that C_{p} is closed and therefore if C_{p} is a proper subset of X_{A}, then X_{A} \ C_{p} is a non-empty open set. By the definition of the product topology, this would imply that X_{A} \ C_{p} contains an open set of the form

V := ∩_{α∈Λ}π_{α}^{−1}(V_{α}) = V_{Λ} × X_{A\Λ}

Fig. 10.4. This picture illustrates why the connected component of p in X × Y must contain all points of X × Y.

where Λ ⊂⊂ A and V_{α} ∈ τ_{α} for all α ∈ Λ. We will now show that no such V can exist and hence X_{A} = C_{p}, i.e. X_{A} is connected.

Define ϕ : X_{Λ} → X_{A} by ϕ(y) = x where

x_{α} = \{ \begin{matrix}y_{α} & if & α ∈ Λ \\ p_{α} & if & α ∉ Λ. \end{matrix}

If α ∈ Λ, π_{α}∘ϕ(y) = y_{α} = π_{α}(y) and if α ∈ A\Λ then π_{α}∘ϕ(y) = p_{α} so that in every case π_{α}∘ϕ : X_{Λ} → X_{α} is continuous and therefore ϕ is continuous. Since X_{Λ} is a product of a finite number of connected spaces and so is connected and thus so is the continuous image, ϕ(X_{Λ}) = X_{Λ} × {p_{α}}_{α∈A\Λ} ⊂ X_{Λ}. Now p ∈ ϕ(X_{Λ}) and ϕ(X_{Λ}) is connected implies that ϕ(X_{Λ}) ⊂ C. On the other hand one easily sees that

∅ ≠ V ∩ ϕ(X_{Λ}) ⊂ V ∩ C

contradicting the assumption that V ⊂ C^{c}.

10.6 Exercises

10.6.1 General Topological Space Problems

Exercise 10.14. Let V be an open subset of ℝ. Show V may be written as a disjoint union of open intervals J_{n} = (a_{n}, b_{n}), where a_{n}, b_{n} ∈ ℝ∪ {±∞} for n = 1, 2, ⋯ < N with N = ∞ possible.

Exercise 10.15. Let (X, τ) and (Y, τ′) be a topological spaces, f : X → Y be a function, U be an open cover of X and {F_{j}}_{j=1}^{n} be a finite cover of X by closed sets.

1. If A ⊂ X is any set and f : X → Y is (τ, τ′) – continuous then f|_{A} : A → Y is (τ_{A}, τ′) – continuous.

2. Show f : X → Y is (τ, τ′) – continuous iff f|_{U} : U → Y is (τ_{U}, τ′) – continuous for all U ∈ U.

3. Show f : X → Y is (τ, τ′) – continuous iff f|_{F_{j}} : F_{j} → Y is (τ_{F_{j}}, τ′) – continuous for all j = 1, 2, …, n.

Exercise 10.16. Suppose that X is a set, {(Y_{α}, τ_{α}) : α ∈ A} is a family of topological spaces and f_{α} : X → Y_{α} is a given function for all α ∈ A. Assuming that S_{α} ⊂ τ_{α} is a sub-base for the topology τ_{α} for each α ∈ A, show S := ∪_{α∈A}f_{α}^{−1}(S_{α}) is a sub-base for the topology τ := τ(f_{α} : α ∈ A).

10.6.2 Connectedness Problems

Exercise 10.17. Show any non-trivial interval in ℚ is disconnected.

Exercise 10.18. Suppose a < b and f : (a, b) → ℝ is a non-decreasing function. Show if f satisfies the intermediate value property (see Theorem 10.50), then f is continuous.

Exercise 10.19. Suppose −∞ < a < b ≤ ∞ and f : [a, b) → ℝ is a strictly increasing continuous function. By Lemma 10.46, f ([a, b)) is an interval and since f is strictly increasing it must of the form [c, d) for some c ∈ ℝ and d ∈ ℝ̄ with c < d. Show the inverse function f^{−1} : [c, d) → [a, b) is continuous and is strictly increasing. In particular if n ∈ ℕ, apply this result to f (x) = x^{n} for x ∈ [0, ∞) to construct the positive n^{th} – root of a real number. Compare with Exercise 3.8

Exercise 10.20. Prove item 1. of Proposition 10.52. Hint: show X is not connected implies X is not path connected.

Exercise 10.21. Prove item 2. of Proposition 10.52. Hint: fix x_{0} ∈ X and let W denote the set of x ∈ X such that there exists σ ∈ C([0, 1], X) satisfying σ(0) = x_{0} and σ(1) = x. Then show W is both open and closed.

Exercise 10.22. Prove item 3. of Proposition 10.52.

Exercise 10.23. Let

X := {(x, y) ∈ ℝ^{2} : y = sin(x^{−1})} ∪ {(0, 0)}

equipped with the relative topology induced from the standard topology on ℝ^{2}. Show X is connected but not path connected.

10.6.3 Metric Spaces as Topological Spaces

Definition 10.54. Two metrics d and ρ on a set X are said to be equivalent if there exists a constant c ∈ (0, ∞) such that c^{−1}ρ ≤ d ≤ cρ.

Exercise 10.24. Suppose that d and ρ are two metrics on X.

1. Show τ_{d} = τ_{ρ} if d and ρ are equivalent.

2. Show by example that it is possible for τ_{d} = τ_{ρ} even thought d and ρ are inequivalent.

Exercise 10.25. Let (X_{i}, d_{i}) for i = 1, …, n be a finite collection of metric spaces and for 1 ≤ p ≤ ∞ and x = (x_{1}, x_{2}, …, x_{n}) and y = (y_{1}, …, y_{n}) in X := ∏_{i=1}^{n} X_{i}, let

ρ_{p}(x, y) = \{ \begin{matrix}(∑_{i=1}^{n} [d_{i}(x_{i}, y_{i})]^{p})^{1/p} & if & p ≠ ∞ \\ max_{i} d_{i}(x_{i}, y_{i}) & if & p = ∞ \end{matrix} .

1. Show (X, ρ_{p}) is a metric space for p ∈ [1, ∞]. Hint: Minkowski’s inequality.

2. Show for any p, q ∈ [1, ∞], the metrics ρ_{p} and ρ_{q} are equivalent. Hint: This can be done with explicit estimates or you could use Theorem 11.12 below.

Notation 10.55 Let X be a set and p := {p_{n}}_{n=0}^{∞} be a family of semi-metrics on X, i.e. p_{n} : X × X → [0, ∞) are functions satisfying the assumptions of metric except for the assertion that p_{n}(x, y) = 0 implies x = y. Further assume that p_{n}(x, y) ≤ p_{n+1}(x, y) for all n and if p_{n}(x, y) = 0 for all n ∈ ℕ then x = y. Given n ∈ ℕ and x ∈ X let

B_{n}(x, ε) := {y ∈ X : p_{n}(x, y) < ε} .

We will write τ(p) form the smallest topology on X such that p_{n}(x, ·) : X → [0, ∞) is continuous for all n ∈ ℕ and x ∈ X, i.e. τ(p) := τ(p_{n}(x·) : n ∈ ℕ and x ∈ X).

Exercise 10.26. Using Notation 10.55, show that collection of balls,

B := {B_{n}(x, ε) : n ∈ ℕ, x ∈ X and ε > 0} ,

forms a base for the topology τ(p). Hint: Use Exercise 10.16 to show B is a sub-base for the topology τ(p) and then use Exercise 10.2 to show B is in fact a base for the topology τ(p).

Exercise 10.27 (A minor variant of Exercise 6.12). Let p_{n} be as in Notation 10.55 and

d(x, y) := ∑_{n=0}^{∞} 2^{−n} \frac{p_{n}(x, y)}{1 + p_{n}(x, y)}.

Show d is a metric on X and τ_{d} = τ(p). Conclude that a sequence {x_{k}}_{k=1}^{∞} ⊂ X converges to x ∈ X iff

lim_{k→∞} p_{n}(x_{k}, x) = 0 for all n ∈ ℕ.

Exercise 10.28. Let {(X_{n}, d_{n})}_{n=1}^{∞} be a sequence of metric spaces, X := ∏_{n=1}^{∞} X_{n}, and for x = (x(n))_{n=1}^{∞} and y = (y(n))_{n=1}^{∞} in X let

d(x, y) = ∑_{n=1}^{∞} 2^{−n} \frac{d_{n}(x(n), y(n))}{1 + d_{n}(x(n), y(n))}.

(See Exercise 6.12.) Moreover, let π_{n} : X → X_{n} be the projection maps, show

τ_{d} = ⊗_{n=1}^{∞}τ_{d_{n}} := τ({π_{n} : n ∈ ℕ}).

That is show the d – metric topology is the same as the product topology on X. Suggestions: 1) show π_{n} is τ_{d} continuous for each n and 2) show for each x ∈ X that d (x, ·) is ⊗_{n=1}^{∞}τ_{d_{n}} – continuous. For the second assertion notice that d (x, ·) = ∑_{n=1}^{∞} f_{n} where f_{n} = 2^{−n} (\frac{d_{n}(x(n),·)}{1+d_{n}(x(n),·)}) ∘ π_{n}.

Compactness

Definition 11.1. The subset A of a topological space (X τ) is said to be compact if every open cover (Definition 10.18) of A has finite a sub-cover, i.e. if U is an open cover of A there exists U_{0} ⊂⊂ U such that U_{0} is a cover of A. (We will write A ⊏⊏ X to denote that A ⊂ X and A is compact.) A subset A ⊂ X is precompact if Ā is compact.

Proposition 11.2. Suppose that K ⊂ X is a compact set and F ⊂ K is a closed subset. Then F is compact. If {K_{i}}_{i=1}^{n} is a finite collections of compact subsets of X then K = ∪_{i=1}^{n}K_{i} is also a compact subset of X.

Proof. Let U ⊂ τ be an open cover of F, then U∪ {F^{c}} is an open cover of K. The cover U∪ {F^{c}} of K has a finite subcover which we denote by U_{0}∪ {F^{c}} where U_{0} ⊂⊂ U. Since F ∩ F^{c} = ∅, it follows that U_{0} is the desired subcover of F. For the second assertion suppose U ⊂ τ is an open cover of K. Then U covers each compact set K_{i} and therefore there exists a finite subset U_{i} ⊂⊂ U for each i such that K_{i} ⊂ ∪U_{i}. Then U_{0} := ∪_{i=1}^{n}U_{i} is a finite cover of K.

Exercise 11.1 (Suggested by Michael Gurvich). Show by example that the intersection of two compact sets need not be compact. (This pathology disappears if one assumes the topology is Hausdorff, see Definition 12.2 below.)

Exercise 11.2. Suppose f : X → Y is continuous and K ⊂ X is compact, then f(K) is a compact subset of Y. Give an example of continuous map, f : X → Y, and a compact subset K of Y such that f^{−1}(K) is not compact.

Exercise 11.3 (Dini’s Theorem). Let X be a compact topological space and f_{n} : X → [0, ∞) be a sequence of continuous functions such that f_{n}(x) ↓ 0 as n → ∞ for each x ∈ X. Show that in fact f_{n} ↓ 0 uniformly in x, i.e. sup_{x∈X} f_{n}(x) ↓ 0 as n → ∞. Hint: Given ε > 0, consider the open sets V_{n} := {x ∈ X : f_{n}(x) < ε}.

Definition 11.3. A collection F of closed subsets of a topological space (X, τ) has the finite intersection property if ∩F_{0} ≠ ∅ for all F_{0} ⊂⊂ F.

The notion of compactness may be expressed in terms of closed sets as follows.

Proposition 11.4. A topological space X is compact iff every family of closed sets F ⊂ 2^{X} having the finite intersection property satisfies ⋂ F ≠ ∅.

Proof. (⇒) Suppose that X is compact and F ⊂ 2^{X} is a collection of closed sets such that ⋂ F = ∅. Let

U = F^{c} := {C^{c} : C ∈ F} ⊂ τ,

then U is a cover of X and hence has a finite subcover, U_{0}. Let F_{0} = U_{0}^{c} ⊂⊂ F, then ∩F_{0} = ∅ so that F does not have the finite intersection property. (⇐) If X is not compact, there exists an open cover U of X with no finite subcover. Let

F = U^{c} := {U^{c} : U ∈ U} ,

then F is a collection of closed sets with the finite intersection property while ⋂ F = ∅.

Exercise 11.4. Let (X, τ) be a topological space. Show that A ⊂ X is compact iff (A, τ_{A}) is a compact topological space.

11.1 Metric Space Compactness Criteria

Let (X, d) be a metric space and for x ∈ X and ε > 0 let B′_{x}(ε) = B_{x}(ε)\{x} – the deleted ball centered at x of radius ε > 0. Recall from Definition 10.29 that a point x ∈ X is an accumulation point of a subset E ⊂ X if ∅ ≠ E ∩ V \ {x} for all open neighborhoods, V, of x. The proof of the following elementary lemma is left to the reader.

Lemma 11.5. Let E ⊂ X be a subset of a metric space (X, d) . Then the following are equivalent:

1. x ∈ X is an accumulation point of E.

2. B′_{x}(ε) ∩ E ≠ ∅ for all ε > 0.

3. B_{x}(ε) ∩ E is an infinite set for all ε > 0.

4. There exists {x_{n}}_{n=1}^{∞} ⊂ E \ {x} with lim_{n→∞} x_{n} = x.

Definition 11.6. A metric space (X, d) is ε – bounded (ε > 0) if there exists a finite cover of X by balls of radius ε and it is totally bounded if it is ε – bounded for all ε > 0.

Theorem 11.7. Let (X, d) be a metric space. The following are equivalent.

(a) X is compact.

(b) Every infinite subset of X has an accumulation point.

(c) Every sequence {x_{n}}_{n=1}^{∞} ⊂ X has a convergent subsequence.

(d) X is totally bounded and complete.

Proof. The proof will consist of showing that a ⇒ b ⇒ c ⇒ d ⇒ a.

(a ⇒ b) We will show that not b ⇒ not a. Suppose there exists an infinite subset E ⊂ X which has no accumulation points. Then for all x ∈ X there exists δ_{x} > 0 such that V_{x} := B_{x}(δ_{x}) satisfies (V_{x} \ {x}) ∩ E = ∅. Clearly V = {V_{x}}_{x∈X} is a cover of X, yet V has no finite sub cover. Indeed, for each x ∈ X, V_{x} ∩ E ⊂ {x} and hence if Λ ⊂⊂ X, ∪_{x∈Λ}V_{x} can only contain a finite number of points from E (namely Λ∩E). Thus for any Λ ⊂⊂ X, E ⫅̸ ∪_{x∈Λ}V_{x} and in particular X ≠ ∪_{x∈Λ}V_{x}. (See Figure 11.1.)

Fig. 11.1. The construction of an open cover with no finite sub-cover.

(b ⇒ c) Let {x_{n}}_{n=1}^{∞} ⊂ X be a sequence and E := {x_{n} : n ∈ ℕ} . If #(E) < ∞, then {x_{n}}_{n=1}^{∞} has a subsequence {x_{n_{k}}}_{k=1}^{∞} which is constant and hence convergent. On the other hand if #(E) = ∞ then by assumption E has an accumulation point and hence by Lemma 11.5, {x_{n}}_{n=1}^{∞} has a convergent subsequence.

(c ⇒ d) Suppose {x_{n}}_{n=1}^{∞} ⊂ X is a Cauchy sequence. By assumption there exists a subsequence {x_{n_{k}}}_{k=1}^{∞} which is convergent to some point x ∈ X. Since {x_{n}}_{n=1}^{∞} is Cauchy it follows that x_{n} → x as n → ∞ showing X is complete. We now show that X is totally bounded. Let ε > 0 be given and choose an arbitrary point x_{1} ∈ X. If possible choose x_{2} ∈ X such that d(x_{2}, x_{1}) ≥ ε, then if possible choose x_{3} ∈ X such that d_{{x_{1},x_{2}}}(x_{3}) ≥ ε and continue inductively choosing points {x_{j}}_{j=1}^{n} ⊂ X such that d_{{x_{1},...,x_{n−1}}}(x_{n}) ≥ ε. (See Figure 11.2.) This process must terminate, for otherwise we would produce a sequence {x_{n}}_{n=1}^{∞} ⊂ X which can have no convergent subsequences. Indeed, the x_{n} have been chosen so that d (x_{n}, x_{m}) ≥ ε > 0 for every m ≠ n and hence no subsequence of {x_{n}}_{n=1}^{∞} can be Cauchy.

(d ⇒ a) For sake of contradiction, assume there exists an open cover V = {V_{α}}_{α∈A} of X with no finite subcover. Since X is totally bounded for each n ∈ ℕ there exists Λ_{n} ⊂⊂ X such that

X = ⋃_{x∈Λ_{n}} B_{x}(1/n) ⊂ ⋃_{x∈Λ_{n}} C_{x}(1/n).

Fig. 11.2. Constructing a set with out an accumulation point.

Choose x_{1} ∈ Λ_{1} such that no finite subset of V covers K_{1} := C_{x_{1}}(1). Since K_{1} = ∪_{x∈Λ_{2}}K_{1} ∩C_{x}(1/2), there exists x_{2} ∈ Λ_{2} such that K_{2} := K_{1} ∩C_{x_{2}}(1/2) can not be covered by a finite subset of V, see Figure 11.3. Continuing this way inductively, we construct sets K_{n} = K_{n−1} ∩ C_{x_{n}}(1/n) with x_{n} ∈ Λ_{n} such that no K_{n} can be covered by a finite subset of V. Now choose y_{n} ∈ K_{n} for each n. Since {K_{n}}_{n=1}^{∞} is a decreasing sequence of closed sets such that diam(K_{n}) ≤ 2/n, it follows that {y_{n}} is a Cauchy and hence convergent with

y = lim_{n→∞} y_{n} ∈ ∩_{m=1}^{∞}K_{m}.

Since V is a cover of X, there exists V ∈ V such that y ∈ V. Since K_{n} ↓ {y} and diam(K_{n}) → 0, it now follows that K_{n} ⊂ V for some n large. But this violates the assertion that K_{n} can not be covered by a finite subset of V.

Fig. 11.3. Nested Sequence of cubes.

Corollary 11.8. Any compact metric space (X, d) is second countable and hence also separable by Exercise 10.11. (See Example 12.25 below for an example of a compact topological space which is not separable.)

Proof. To each integer n, there exists Λ_{n} ⊂⊂ X such that X = ∪_{x∈Λ_{n}}B(x, 1/n). The collection of open balls,

V := ∪_{n∈ℕ} ∪_{x∈Λ_{n}} {B(x, 1/n)}

forms a countable basis for the metric topology on X. To check this, suppose that x_{0} ∈ X and ε > 0 are given and choose n ∈ ℕ such that 1/n < ε/2 and x ∈ Λ_{n} such that d (x_{0}, x) < 1/n. Then B(x, 1/n) ⊂ B (x_{0}, ε) because for y ∈ B(x, 1/n),

d (y, x_{0}) ≤ d (y, x) + d (x, x_{0}) < 2/n < ε.

Corollary 11.9. The compact subsets of ℝ^{n} are the closed and bounded sets.

Proof. This is a consequence of Theorem 8.2 and Theorem 11.7. Here is another proof. If K is closed and bounded then K is complete (being the closed subset of a complete space) and K is contained in [−M, M]^{n} for some positive integer M. For δ > 0, let

Λ_{δ} = δℤ^{n} ∩ [−M, M]^{n} := {δx : x ∈ ℤ^{n} and δ|x_{i}| ≤ M for i = 1, 2, …, n}.

We will show, by choosing δ > 0 sufficiently small, that

K ⊂ [−M, M]^{n} ⊂ ∪_{x∈Λ_{δ}}B(x, ε)  (11.1)

which shows that K is totally bounded. Hence by Theorem 11.7, K is compact. Suppose that y ∈ [−M, M]^{n}, then there exists x ∈ Λ_{δ} such that |y_{i} − x_{i}| ≤ δ for i = 1, 2, …, n. Hence

d^{2}(x, y) = ∑_{i=1}^{n} (y_{i} − x_{i})^{2} ≤ nδ^{2}

which shows that d(x, y) ≤ \sqrt{n}δ. Hence if choose δ < ε/\sqrt{n} we have shows that d(x, y) < ε, i.e. Eq. (11.1) holds.

Example 11.10. Let X = ℓ^{p}(ℕ) with p ∈ [1, ∞) and μ ∈ ℓ^{p}(ℕ) such that μ(k) ≥ 0 for all k ∈ ℕ. The set

K := {x ∈ X : |x(k)| ≤ μ(k) for all k ∈ ℕ}

is compact. To prove this, let {x_{n}}_{n=1}^{∞} ⊂ K be a sequence. By compactness of closed bounded sets in ℂ, for each k ∈ ℕ there is a subsequence of {x_{n}(k)}_{n=1}^{∞} ⊂ ℂ which is convergent. By Cantor’s diagonalization trick, we may choose a subsequence {y_{n}}_{n=1}^{∞} of {x_{n}}_{n=1}^{∞} such that y(k) := lim_{n→∞} y_{n}(k) exists for all k ∈ ℕ.^{1} Since |y_{n}(k)| ≤ μ(k) for all n it follows that |y(k)| ≤ μ(k), i.e. y ∈ K. Finally

lim_{n→∞} ∥y − y_{n}∥_{p}^{p} = lim_{n→∞} ∑_{k=1}^{∞} |y(k) − y_{n}(k)|^{p} = ∑_{k=1}^{∞} lim_{n→∞} |y(k) − y_{n}(k)|^{p} = 0

wherein we have used the Dominated convergence theorem. (Note |y(k) − y_{n}(k)|^{p} ≤ 2^{p}μ^{p}(k) and μ^{p} is summable.) Therefore y_{n} → y and we are done.

Alternatively, we can prove K is compact by showing that K is closed and totally bounded. It is simple to show K is closed, for if {x_{n}}_{n=1}^{∞} ⊂ K is a convergent sequence in X, x := lim_{n→∞} x_{n}, then

|x(k)| ≤ lim_{n→∞} |x_{n}(k)| ≤ μ(k) ∀ k ∈ ℕ.

This shows that x ∈ K and hence K is closed. To see that K is totally bounded, let ε > 0 and choose N such that (∑_{k=N+1}^{∞} |μ(k)|^{p})^{1/p} < ε. Since ∏_{k=1}^{N} C_{μ(k)}(0) ⊂ ℂ^{N} is closed and bounded, it is compact. Therefore there exists a finite subset Λ ⊂ ∏_{k=1}^{N} C_{μ(k)}(0) such that

∏_{k=1}^{N} C_{μ(k)}(0) ⊂ ∪_{z∈Λ}B_{z}^{N}(ε)

where B_{z}^{N}(ε) is the open ball centered at z ∈ ℂ^{N} relative to the ℓ^{p}({1, 2, 3, …, N}) – norm. For each z ∈ Λ, let z̃ ∈ X be defined by z̃(k) = z(k) if k ≤ N and z̃(k) = 0 for k ≥ N + 1. I now claim that

K ⊂ ∪_{z∈Λ}Bz̃(2ε)  (11.2)

which, when verified, shows K is totally bounded. To verify Eq. (11.2), let x ∈ K and write x = u + v where u(k) = x(k) for k ≤ N and u(k) = 0 for k < N. Then by construction u ∈ Bz̃(ε) for some z̃ ∈ Λ and

∥v∥_{p} ≤ ( ∑_{k=N+1}^{∞} |μ(k)|^{p})^{1/p} < ε.

So we have

∥x − z̃∥_{p} = ∥u + v − z̃∥_{p} ≤ ∥u − z̃∥_{p} + ∥v∥_{p} < 2ε.

1 The argument is as follows. Let {n_{j}^{1}}_{j=1}^{∞} be a subsequence of ℕ ={n}_{n=1}^{∞} such that lim_{j→∞} x_{n_{j}^{1}}(1) exists. Now choose a subsequence {n_{j}^{2}}_{j=1}^{∞} of {n_{j}^{1}}_{j=1}^{∞} such that lim_{j→∞} x_{n_{j}^{2}}(2) exists and similarly {n_{j}^{3}}_{j=1}^{∞} of {n_{j}^{2}}_{j=1}^{∞} such that lim_{j→∞} x_{n_{j}^{3}}(3) exists. Continue on this way inductively to get

{n}_{n=1}^{∞} ⊃ {n_{j}^{1}}_{j=1}^{∞} ⊃ {n_{j}^{2}}_{j=1}^{∞} ⊃ {n_{j}^{3}}_{j=1}^{∞} ⊃ …

such that lim_{j→∞} x_{n_{j}^{k}}(k) exists for all k ∈ ℕ. Let m_{j} := n_{j}^{j} so that eventually {m_{j}}_{j=1}^{∞} is a subsequence of {n_{j}^{k}}_{j=1}^{∞} for all k. Therefore, we may take y_{j} := x_{m_{j}} .

Exercise 11.5 (Extreme value theorem). Let (X, τ) be a compact topological space and f : X → ℝ be a continuous function. Show −∞ < inf f ≤ sup f < ∞ and there exists a, b ∈ X such that f(a) = inf f and f(b) = sup f^{2}. Hint: use Exercise 11.2 and Corollary 11.9.

Exercise 11.6 (Uniform Continuity). Let (X, d) be a compact metric space, (Y, ρ) be a metric space and f : X → Y be a continuous function. Show that f is uniformly continuous, i.e. if ε > 0 there exists δ > 0 such that ρ(f(y), f(x)) < ε if x, y ∈ X with d(x, y) < δ. Hint: you could follow the argument in the proof of Theorem 8.2.

Definition 11.11. Let L be a vector space. We say that two norms, |·| and ∥·∥ , on L are equivalent if there exists constants α, β ∈ (0, ∞) such that

∥f∥ ≤ α |f| and |f| ≤ β ∥f∥ for all f ∈ L.

Theorem 11.12. Let L be a finite dimensional vector space. Then any two norms |·| and ∥·∥ on L are equivalent. (This is typically not true for norms on infinite dimensional spaces, see for example Exercise 7.5.)

Proof. Let {f_{i}}_{i=1}^{n} be a basis for L and define a new norm on L by

∥∑_{i=1}^{n} a_{i}f_{i}∥_{2} := \sqrt{∑_{i=1}^{n} |a_{i}|^{2}} for a_{i} ∈ 𝔽.

By the triangle inequality for the norm |·| , we find

|∑_{i=1}^{n} a_{i}f_{i}| ≤ ∑_{i=1}^{n} |a_{i}| |f_{i}| ≤ \sqrt{∑_{i=1}^{n} |f_{i}|^{2}}\sqrt{∑_{i=1}^{n} |a_{i}|^{2}} ≤ M ∥∑_{i=1}^{n} a_{i}f_{i}∥_{2}

where M = \sqrt{∑_{i=1}^{n} |f_{i}|^{2}}. Thus we have

|f| ≤ M ∥f∥_{2}

for all f ∈ L and this inequality shows that |·| is continuous relative to ∥·∥_{2} . Since the normed space (L, ∥·∥_{2}) is homeomorphic and isomorphic to 𝔽^{n} with the standard euclidean norm, the closed bounded set, S := {f ∈ L : ∥f∥_{2} = 1} ⊂ L, is a compact subset of L relative to ∥·∥_{2} . Therefore by Exercise 11.5 there exists f_{0} ∈ S such that

m = inf {|f| : f ∈ S} = |f_{0}| > 0.

2 Here is a proof if X is a metric space. Let {x_{n}}_{n=1}^{∞} ⊂ X be a sequence such that f(x_{n}) ↑ sup f. By compactness of X we may assume, by passing to a subsequence if necessary that x_{n} → b ∈ X as n → ∞. By continuity of f, f(b) = sup f.

Hence given 0 ≠ f ∈ L, then \frac{f}{∥f∥_{2}} ∈ S so that

m ≤ |\frac{f}{∥f∥_{2}} | = |f| \frac{1}{∥f∥_{2}}

or equivalently

∥f∥_{2} ≤ \frac{1}{m} |f| .

This shows that |·| and ∥·∥_{2} are equivalent norms. Similarly one shows that ∥·∥ and ∥·∥_{2} are equivalent and hence so are |·| and ∥·∥ .

Corollary 11.13. If (L, ∥·∥) is a finite dimensional normed space, then A ⊂ L is compact iff A is closed and bounded relative to the given norm, ∥·∥ .

Corollary 11.14. Every finite dimensional normed vector space (L, ∥·∥) is complete. In particular any finite dimensional subspace of a normed vector space is automatically closed.

Proof. If {f_{n}}_{n=1}^{∞} ⊂ L is a Cauchy sequence, then {f_{n}}_{n=1}^{∞} is bounded and hence has a convergent subsequence, g_{k} = f_{n_{k}}, by Corollary 11.13. It is now routine to show lim_{n→∞} f_{n} = f := lim_{k→∞} g_{k}.

Theorem 11.15. Suppose that (X, ∥·∥) is a normed vector in which the unit ball, V := B_{0} (1) , is precompact. Then dim X < ∞.

Proof. Since V̄ is compact, we may choose Λ ⊂⊂ X such that

V̄ ⊂ ∪_{x∈Λ} (x + \frac{1}{2}V )  (11.3)

where, for any δ > 0,

δV := {δx : x ∈ V } = B_{0} (δ) .

Let Y := span(Λ), then Eq. (11.3) implies,

V ⊂ V̄ ⊂ Y + \frac{1}{2}V.

Multiplying this equation by \frac{1}{2} then shows

\frac{1}{2}V ⊂ \frac{1}{2}Y + \frac{1}{4}V = Y + \frac{1}{4}V

and hence

V ⊂ Y + \frac{1}{2}V ⊂ Y + Y + \frac{1}{4}V = Y + \frac{1}{4}V.

Continuing this way inductively then shows that

V ⊂ Y + \frac{1}{2^{n}} V for all n ∈ ℕ.  (11.4)

Indeed, if Eq. (11.4) holds, then

V ⊂ Y + \frac{1}{2}V ⊂ Y + \frac{1}{2} (Y + \frac{1}{2^{n}} V ) = Y + \frac{1}{2^{n+1}} V.

Hence if x ∈ V, there exists y_{n} ∈ Y and z_{n} ∈ B_{0} (2^{−n}) such that y_{n} + z_{n} → x. Since lim_{n→∞} z_{n} = 0, it follows that x = lim_{n→∞} y_{n} ∈ Ȳ . Since dim Y ≤ # (Λ) < ∞, Corollary 11.14 implies Y = Ȳ and so we have shown that V ⊂ Y. Since for any x ∈ X, \frac{1}{2∥x∥}x ∈ V ⊂ Y, we have x ∈ Y for all x ∈ X, i.e. X = Y.

Exercise 11.7. Suppose (Y, ∥·∥_{Y}) is a normed space and (X, ∥·∥_{X}) is a finite dimensional normed space. Show every linear transformation T : X → Y is necessarily bounded.

11.2 Compact Operators

Definition 11.16. Let A : X → Y be a bounded operator between two Banach spaces. Then A is compact if A [B_{X}(0, 1)] is precompact in Y or equivalently for any {x_{n}}_{n=1}^{∞} ⊂ X such that ∥x_{n}∥ ≤ 1 for all n the sequence y_{n} := Ax_{n} ∈ Y has a convergent subsequence.

Example 11.17. Let X = ℓ^{2} = Y and λ_{n} ∈ ℂ such that lim_{n→∞} λ_{n} = 0, then A : X → Y defined by (Ax)(n) = λ_{n}x(n) is compact.

Proof. Suppose {x_{j}}_{j=1}^{∞} ⊂ ℓ^{2} such that ∥x_{j}∥^{2} = ∑ |x_{j}(n)|^{2} ≤ 1 for all j. By Cantor’s Diagonalization argument, there exists {j_{k}} ⊂ {j} such that, for each n, x̃_{k}(n) = x_{j_{k}}(n) converges to some x̃(n) ∈ ℂ as k → ∞. By Fatou’s Lemma 4.12,

∑_{n=1}^{∞} |x̃(n)|^{2} = ∑_{n=1}^{∞} lim inf_{k→∞} |x̃_{k}(n)|^{2} ≤ lim inf_{k→∞} ∑_{n=1}^{∞} |x̃_{k}(n)|^{2} ≤ 1,

which shows x̃ ∈ ℓ^{2}.

Let λ_{M}^{∗} = max_{n≥M} |λ_{n}|. Then

∥Ax̃_{k} − Ax̃∥^{2} = ∑_{n=1}^{∞} |λ_{n}|^{2} |x̃_{k}(n) − x̃(n)|^{2}

≤ ∑_{n=1}^{M} |λ_{n}|^{2}|x̃_{k}(n) − x̃(n)|^{2} + |λ_{M}^{∗}|^{2} ∑_{M+1}^{∞} |x̃_{k}(n) − x̃(n)|^{2}

≤ ∑_{n=1}^{M} |λ_{n}|^{2}|x̃_{k}(n) − x̃(n)|^{2} + |λ_{M}^{∗}|^{2} ∥x̃_{k} − x̃∥^{2}

≤ ∑_{n=1}^{M} |λ_{n}|^{2}|x̃_{k}(n) − x̃(n)|^{2} + 4|λ_{M}^{∗}|^{2}.

Passing to the limit in this inequality then implies

lim sup_{k→∞} ∥Ax̃_{k} − Ax̃∥^{2} ≤ 4|λ_{M}^{∗}|^{2} → 0 as M → ∞

and this completes the proof the A is a compact operator.

Lemma 11.18. If X ⟶^{A} Y ⟶^{B} Z are bounded operators such the either A or B is compact then the composition BA : X → Z is also compact.

Proof. Let B_{X}(0, 1) be the open unit ball in X. If A is compact and B is bounded, then BA(B_{X}(0, 1)) ⊂ B(\overline{AB_{X}(0, 1)}) which is compact since the image of compact sets under continuous maps are compact. Hence we conclude that \overline{BA(B_{X}(0, 1))} is compact, being the closed subset of the compact set B(\overline{AB_{X}(0, 1)}). If A is continuous and B is compact, then A(B_{X}(0, 1)) is a bounded set and so by the compactness of B, BA(B_{X}(0, 1)) is a precompact subset of Z, i.e. BA is compact.

11.3 Local and σ – Compactness

Notation 11.19 If X is a topological spaces and Y is a normed space, let

BC(X, Y) := {f ∈ C(X, Y) : sup_{x∈X} ∥f(x)∥_{Y} < ∞}

and

C_{c}(X, Y) := {f ∈ C(X, Y) : supp(f) is compact}.

If Y = ℝ or ℂ we will simply write C(X), BC(X) and C_{c}(X) for C(X, Y), BC(X, Y) and C_{c}(X, Y) respectively.

Remark 11.20. Let X be a topological space and Y be a Banach space. By combining Exercise 11.2 and Theorem 11.7 it follows that C_{c}(X, Y) ⊂ BC(X, Y).

11.3 Local and σ – Compactness 141

Definition 11.21 (Local and σ – compactness). Let (X, τ) be a topological space.

1. (X, τ) is locally compact if for all x ∈ X there exists an open neighborhood V ⊂ X of x such that V̄ is compact. (Alternatively, in light of Definition 10.29 (also see Definition 6.5), this is equivalent to requiring that to each x ∈ X there exists a compact neighborhood N_{x} of x.)

2. (X, τ) is σ – compact if there exists compact sets K_{n} ⊂ X such that X = ∪_{n=1}^{∞}K_{n}. (Notice that we may assume, by replacing K_{n} by K_{1} ∪K_{2} ∪ ⋯ ∪ K_{n} if necessary, that K_{n} ↑ X.)

Example 11.22. Any open subset of U ⊂ ℝ^{n} is a locally compact and σ – compact metric space. The proof of local compactness is easy and is left to the reader. To see that U is σ – compact, for k ∈ ℕ, let

K_{k} := {x ∈ U : |x| ≤ k and d_{U^{c}}(x) ≥ 1/k} .

Then K_{k} is a closed and bounded subset of ℝ^{n} and hence compact. Moreover K_{k}^{o} ↑ U as k → ∞ since^{3}

K_{k}^{o} ⊃ {x ∈ U : |x| < k and d_{U^{c}}(x) > 1/k} ↑ U as k → ∞.

Exercise 11.8. If (X, τ) is locally compact and second countable, then there is a countable basis B_{0} for the topology consisting of precompact open sets. Use this to show (X, τ) is σ - compact.

Exercise 11.9. Every separable locally compact metric space is σ – compact.

Exercise 11.10. Every σ – compact metric space is second countable (or equivalently separable), see Corollary 11.8.

Exercise 11.11. Suppose that (X, d) is a metric space and U ⊂ X is an open subset.

1. If X is locally compact then (U, d) is locally compact.

2. If X is σ – compact then (U, d) is σ – compact. Hint: Mimic Example 11.22, replacing {x ∈ ℝ^{n} : |x| ≤ k} by compact sets X_{k} ⊏⊏ X such that X_{k} ↑ X.

Lemma 11.23. Let (X, τ) be locally and σ – compact. Then there exists compact sets K_{n} ↑ X such that K_{n} ⊂ K_{n+1}^{o} ⊂ K_{n+1} for all n.

Proof. Suppose that C ⊂ X is a compact set. For each x ∈ C let V_{x} ⊂_{o} X be an open neighborhood of x such that V̄_{x} is compact. Then C ⊂ ∪_{x∈C}V_{x} so there exists Λ ⊂⊂ C such that

C ⊂ ∪_{x∈Λ}V_{x} ⊂ ∪_{x∈Λ}V̄_{x} =: K.

3 In fact this is an equality, but we will not need this here.

Then K is a compact set, being a finite union of compact subsets of X, and C ⊂ ∪_{x∈Λ}V_{x} ⊂ K^{o}. Now let C_{n} ⊂ X be compact sets such that C_{n} ↑ X as n → ∞. Let K_{1} = C_{1} and then choose a compact set K_{2} such that C_{2} ⊂ K_{2}^{o}. Similarly, choose a compact set K_{3} such that K_{2} ∪ C_{3} ⊂ K_{3}^{o} and continue inductively to find compact sets K_{n} such that K_{n} ∪ C_{n+1} ⊂ K_{n+1}^{o} for all n. Then {K_{n}}_{n=1}^{∞} is the desired sequence.

Remark 11.24. Lemma 11.23 may also be stated as saying there exists precompact open sets {G_{n}}_{n=1}^{∞} such that G_{n} ⊂ Ḡ_{n} ⊂ G_{n+1} for all n and G_{n} ↑ X as n → ∞. Indeed if {G_{n}}_{n=1}^{∞} are as above, let K_{n} := Ḡ_{n} and if {K_{n}}_{n=1}^{∞} are as in Lemma 11.23, let G_{n} := K_{n}^{o}.

Proposition 11.25. Suppose X is a locally compact metric space and U ⊂_{o} X and K ⊏⊏ U. Then there exists V ⊂_{o} X such that K ⊂ V ⊂ \overline{V} ⊂ U ⊂ X and V̄ is compact.

Proof. (This is done more generally in Proposition 12.7 below.) By local compactness or X, for each x ∈ K there exists ε_{x} > 0 such that \overline{B_{x}(ε_{x})} is compact and by shrinking ε_{x} if necessary we may assume,

\overline{B_{x}(ε_{x})} ⊂ C_{x}(ε_{x}) ⊂ B_{x}(2ε_{x}) ⊂ U

for each x ∈ K. By compactness of K, there exists Λ ⊂⊂ K such that K ⊂ ∪_{x∈Λ}B_{x}(ε_{x}) =: V. Notice that V̄ ⊂ ∪_{x∈Λ}\overline{B_{x}(ε_{x})} ⊂ U and V̄ is a closed subset of the compact set ∪_{x∈Λ}\overline{B_{x}(ε_{x})} and hence compact as well.

Definition 11.26. Let U be an open subset of a topological space (X, τ). We will write f ≺ U to mean a function f ∈ C_{c}(X, [0, 1]) such that supp(f) := \overline{\{f ≠ 0\}} ⊂ U.

Lemma 11.27 (Urysohn’s Lemma for Metric Spaces). Let X be a locally compact metric space and K ⊏⊏ U ⊂_{o} X. Then there exists f ≺ U such that f = 1 on K. In particular, if K is compact and C is closed in X such that K ∩ C = ∅, there exists f ∈ C_{c}(X, [0, 1]) such that f = 1 on K and f = 0 on C.

Proof. Let V be as in Proposition 11.25 and then use Lemma 6.15 to find a function f ∈ C(X, [0, 1]) such that f = 1 on K and f = 0 on V^{c}. Then supp(f) ⊂ V̄ ⊂ U and hence f ≺ U.

11.4 Function Space Compactness Criteria

In this section, let (X, τ) be a topological space.

Definition 11.28. Let F ⊂ C(X).

1. F is equicontinuous at x ∈ X iff for all ε > 0 there exists U ∈ τ_{x} such that |f(y) − f(x)| < ε for all y ∈ U and f ∈ F.

2. F is equicontinuous if F is equicontinuous at all points x ∈ X.

3. F is pointwise bounded if sup{|f(x)| : |f ∈ F} < ∞ for all x ∈ X.

Theorem 11.29 (Ascoli-Arzela Theorem). Let (X, τ) be a compact topological space and F ⊂ C(X). Then F is precompact in C(X) iff F is equicontinuous and point-wise bounded.

Proof. (⇐) Since C(X) ⊂ ℓ^{∞}(X) is a complete metric space, we must show F is totally bounded. Let ε > 0 be given. By equicontinuity, for all x ∈ X, there exists V_{x} ∈ τ_{x} such that |f(y) − f(x)| < ε/2 if y ∈ V_{x} and f ∈ F. Since X is compact we may choose Λ ⊂⊂ X such that X = ∪_{x∈Λ}V_{x}. We have now decomposed X into “blocks” {V_{x}}_{x∈Λ} such that each f ∈ F is constant to within ε on V_{x}. Since sup {|f(x)| : x ∈ Λ and f ∈ F} < ∞, it is now evident that

M = sup {|f(x)| : x ∈ X and f ∈ F}
≤ sup {|f(x)| : x ∈ Λ and f ∈ F} + ε < ∞.

Let 𝔻 := {kε/2 : k ∈ ℤ} ∩ [−M, M]. If f ∈ F and ϕ ∈ 𝔻^{Λ} (i.e. ϕ : Λ → 𝔻 is a function) is chosen so that |ϕ(x) − f(x)| ≤ ε/2 for all x ∈ Λ, then

|f(y) − ϕ(x)| ≤ |f(y) − f(x)| + |f(x) − ϕ(x)| < ε ∀ x ∈ Λ and y ∈ V_{x}.

From this it follows that F = ⋃ {F_{ϕ} : ϕ ∈ 𝔻^{Λ}} where, for ϕ ∈ 𝔻^{Λ},

F_{ϕ} := {f ∈ F : |f(y) − ϕ(x)| < ε for y ∈ V_{x} and x ∈ Λ}.

Let Γ := {ϕ ∈ 𝔻^{Λ} : F_{ϕ} ≠ ∅} and for each ϕ ∈ Γ choose f_{ϕ} ∈ F_{ϕ} ∩ F. For f ∈ F_{ϕ}, x ∈ Λ and y ∈ V_{x} we have

|f(y) − f_{ϕ}(y)| ≤ |f(y) − ϕ(x))| + |ϕ(x) − f_{ϕ}(y)| < 2ε.

So ∥f − f_{ϕ}∥_{∞} < 2ε for all f ∈ F_{ϕ} showing that F_{ϕ} ⊂ B_{f_{ϕ}}(2ε). Therefore,

F = ∪_{ϕ∈Γ} F_{ϕ} ⊂ ∪_{ϕ∈Γ} B_{f_{ϕ}}(2ε)

and because ε > 0 was arbitrary we have shown that F is totally bounded.

(⇒) (*The rest of this proof may safely be skipped.) Since ∥·∥_{∞} : C(X) → [0, ∞) is a continuous function on C(X) it is bounded on any compact subset F ⊂ C(X). This shows that sup {∥f∥_{∞} : f ∈ F} < ∞ which clearly implies that F is pointwise bounded.^{4} Suppose F were not equicontinuous at some

4 One could also prove that F is pointwise bounded by considering the continuous evaluation maps e_{x} : C(X) → ℝ given by e_{x}(f) = f(x) for all x ∈ X.

point x ∈ X that is to say there exists ε > 0 such that for all V ∈ τ_{x}, sup_{y∈V} sup_{f∈F} |f(y) − f(x)| > ε.^{5} Equivalently said, to each V ∈ τ_{x} we may choose

f_{V} ∈ F and x_{V} ∈ V ∋ |f_{V} (x) − f_{V} (x_{V})| ≥ ε.  (11.5)

Set C_{V} = \overline{\{f_{W} : W ∈ τ_{x} and W ⊂ V \}}^{∥·∥_{∞}} ⊂ F and notice for any V ⊂⊂ τ_{x} that

∩_{V ∈V}C_{V} ⊇ C_{∩V} ≠ ∅,

so that {C_{V} }_{V} ∈ τ_{x} ⊂ F has the finite intersection property.^{6} Since F is compact, it follows that there exists some

f ∈ ⋂_{V ∈τ_{x}} C_{V} ≠ ∅.

Since f is continuous, there exists V ∈ τ_{x} such that |f(x) − f(y)| < ε/3 for all y ∈ V. Because f ∈ C_{V} , there exists W ⊂ V such that ∥f − f_{W} ∥ < ε/3. We now arrive at a contradiction;

ε ≤ |f_{W} (x) − f_{W} (x_{W})|
≤ |f_{W} (x) − f(x)| + |f(x) − f(x_{W})| + |f(x_{W}) − f_{W} (x_{W})|
< ε/3 + ε/3 + ε/3 = ε.

Exercise 11.12. Give an alternative proof of the implication, (⇐) , in Theorem 11.29 by showing every subsequence {f_{n} : n ∈ ℕ} ⊂ F has a convergence sub-sequence.

5 If X is first countable we could finish the proof with the following argument. Let {V_{n}}_{n=1}^{∞} be a neighborhood base at x such that V_{1} ⊃ V_{2} ⊃ V_{3} ⊃ …. By the assumption that F is not equicontinuous at x, there exist f_{n} ∈ F and x_{n} ∈ V_{n} such that |f_{n}(x) − f_{n}(x_{n})| ≥ ϵ ∀ n. Since F is a compact metric space by passing to a subsequence if necessary we may assume that f_{n} converges uniformly to some f ∈ F. Because x_{n} → x as n → ∞ we learn that

ϵ ≤ |f_{n}(x) − f_{n}(x_{n})| ≤ |f_{n}(x) − f(x)| + |f(x) − f(x_{n})| + |f(x_{n}) − f_{n}(x_{n})|
≤ 2∥f_{n} − f∥ + |f(x) − f(x_{n})| → 0 as n → ∞

which is a contradiction. 6 If we are willing to use Net’s described in Appendix C below we could finish

the proof as follows. Since F is compact, the net {f_{V} }_{V ∈τ_{x}} ⊂ F has a cluster point f ∈ F ⊂ C(X). Choose a subnet {g_{α}}_{α∈A} of {f_{V} }_{V ∈τ_{X}} such that g_{α} → f uniformly. Then, since x_{V} → x implies x_{V_{α}} → x, we may conclude from Eq. (11.5) that

ϵ ≤ |g_{α}(x) − g_{α}(x_{V_{α}})| → |g(x) − g(x)| = 0

which is a contradiction.

Exercise 11.13. Suppose k ∈ C ([0, 1]^{2} , ℝ) and for f ∈ C ([0, 1] , ℝ) , let

Kf (x) := ∫_{0}^{1} k (x, y) f (y) dy for all x ∈ [0, 1] .

Show K is a compact operator on (C ([0, 1] , ℝ) , ∥·∥_{∞}) .

The following result is a corollary of Lemma 11.23 and Theorem 11.29.

Corollary 11.30 (Locally Compact Ascoli-Arzela Theorem). Let (X, τ) be a locally compact and σ – compact topological space and {f_{m}} ⊂ C(X) be a pointwise bounded sequence of functions such that {f_{m}|_{K}} is equicontinuous for any compact subset K ⊂ X. Then there exists a subsequence {m_{n}} ⊂ {m} such that {g_{n} := f_{m_{n}}}_{n=1}^{∞} ⊂ C(X) is a sequence which is uniformly convergent on compact subsets of X.

Proof. Let {K_{n}}_{n=1}^{∞} be the compact subsets of X constructed in Lemma 11.23. We may now apply Theorem 11.29 repeatedly to find a nested family of subsequences

{f_{m}} ⊃ {g_{m}^{1}} ⊃ {g_{m}^{2}} ⊃ {g_{m}^{3}} ⊃ …

such that the sequence {g_{m}^{n}}_{m=1}^{∞} ⊂ C(X) is uniformly convergent on K_{n}. Using Cantor’s trick, define the subsequence {h_{n}} of {f_{m}} by h_{n} := g_{n}^{n}. Then {h_{n}} is uniformly convergent on K_{l} for each l ∈ ℕ. Now if K ⊂ X is an arbitrary compact set, there exists l < ∞ such that K ⊂ K_{l}^{o} ⊂ K_{l} and therefore {h_{n}} is uniformly convergent on K as well.

Proposition 11.31. Let Ω ⊂_{o} ℝ^{d} such that Ω̄ is compact and 0 ≤ α < β ≤ 1. Then the inclusion map i : C^{β}(\overline{Ω}) ↪ C^{α}(\overline{Ω}) is a compact operator. See Chapter 9 and Lemma 9.9 for the notation being used here.

Let \{u_{n}\}_{n=1}^{∞} ⊂ C^{β}(\overline{Ω}) such that ∥u_{n}∥_{C^{β}} ≤ 1, i.e. ∥u_{n}∥_{∞} ≤ 1 and

|u_{n}(x) − u_{n}(y)| ≤ |x − y|^{β} for all x, y ∈ \overline{Ω}.

By the Arzela-Ascoli Theorem 11.29, there exists a subsequence of {ũ_{n}}_{n=1}^{∞} of {u_{n}}_{n=1}^{∞} and u ∈ C^{o}(Ω̄) such that ũ_{n} → u in C^{0}. Since

|u(x) − u(y)| = lim_{n→∞} |ũ_{n}(x) − ũ_{n}(y)| ≤ |x − y|^{β},

u ∈ C^{β} as well. Define g_{n} := u − ũ_{n} ∈ C^{β}, then

[g_{n}]_{β} + ∥g_{n}∥_{C^{0}} = ∥g_{n}∥_{C^{β}} ≤ 2

and g_{n} → 0 in C^{0}. To finish the proof we must show that g_{n} → 0 in C^{α}. Given δ > 0,

[g_{n}]_{α} = sup_{x≠y} \frac{|g_{n}(x) − g_{n}(y)|}{|x − y|^{α}} ≤ A_{n} + B_{n}

where

A_{n} = sup \{\frac{|g_{n}(x) − g_{n}(y)|}{|x − y|^{α}} : x ≠ y and |x − y| ≤ δ\}

= sup \{\frac{|g_{n}(x) − g_{n}(y)|}{|x − y|^{β}} · |x − y|^{β−α} : x ≠ y and |x − y| ≤ δ\}

≤ δ^{β−α} · [g_{n}]_{β} ≤ 2δ^{β−α}

and

B_{n} = sup \{\frac{|g_{n}(x) − g_{n}(y)|}{|x − y|^{α}} : |x − y| > δ\} ≤ 2δ^{−α} ∥g_{n}∥_{C^{0}} → 0 as n → ∞.

Therefore,

lim sup_{n→∞}[g_{n}]_{α} ≤ lim sup_{n→∞} A_{n} + lim sup_{n→∞} B_{n} ≤ 2δ^{β−α} + 0 → 0 as δ ↓ 0.

This proposition generalizes to the following theorem which the reader is asked to prove in Exercise 11.20 below.

Theorem 11.32. Let Ω be a precompact open subset of ℝ^{d}, α, β ∈ [0, 1] and k, j ∈ ℕ_{0}. If j +β > k +α, then C^{j,β} \binom{}{Ω̄} is compactly contained in C^{k,α} \binom{}{Ω̄} .

11.5 Tychonoff’s Theorem

The goal of this section is to show that arbitrary products of compact spaces is still compact. Before going to the general case of an arbitrary number of factors let us start with only two factors.

Proposition 11.33. Suppose that X and Y are non-empty compact topological spaces, then X × Y is compact in the product topology.

Proof. Let U be an open cover of X × Y. Then for each (x, y) ∈ X × Y there exist U ∈ U such that (x, y) ∈ U. By definition of the product topology, there also exist V_{x} ∈ τ_{x}^{X} and W_{y} ∈ τ_{y}^{Y} such that V_{x} × W_{y} ⊂ U. Therefore V := {V_{x} × W_{y} : (x, y) ∈ X × Y } is also an open cover of X × Y. We will now show that V has a finite sub-cover, say V_{0} ⊂⊂ V. Assuming this is proved for the moment, this implies that U also has a finite subcover because each V ∈ V_{0} is contained in some U_{V} ∈ U. So to complete the proof it suffices to show every cover V of the form V = {V_{α} × W_{α} : α ∈ A} where V_{α} ⊂_{o} X and W_{α} ⊂_{o} Y has a finite subcover. Given x ∈ X, let f_{x} : Y → X × Y be the map f_{x}(y) = (x, y) and notice that f_{x} is continuous since π_{X} ∘ f_{x}(y) = x and π_{Y} ∘ f_{x}(y) = y are continuous maps. From this we conclude that {x} × Y = f_{x}(Y) is compact. Similarly, it follows that X × {y} is compact for all y ∈ Y. Since V is a cover of {x} × Y, there exist Γ_{x} ⊂⊂ A such that {x} × Y ⊂ ⋃_{α∈Γ_{x}}(V_{α} × W_{α}) without loss of generality we may assume that Γ_{x} is chosen so that x ∈ V_{α} for all

α ∈ Γ_{x}. Let U_{x} := ⋂_{α∈Γ_{x}} V_{α} ⊂_{o} X, and notice that

⋃_{α∈Γ_{x}}(V_{α} × W_{α}) ⊃ ⋃_{α∈Γ_{x}}(U_{x} × W_{α}) = U_{x} × Y,  (11.6)

see Figure 11.4 below. Since {U_{x}}_{x∈X} is now an open cover of X and X is

Fig. 11.4. Constructing the open set U_{x}.

compact, there exists Λ ⊂⊂ X such that X = ∪_{x∈Λ}U_{x}. The finite subcollection, V_{0} := {V_{α} × W_{α} : α ∈ ∪_{x∈Λ}Γ_{x}}, of V is the desired finite subcover. Indeed using Eq. (11.6),

∪V_{0} = ∪_{x∈Λ} ∪_{α∈Γ_{x}} (V_{α} × W_{α}) ⊃ ∪_{x∈Λ} (U_{x} × Y) = X × Y.

The results of Exercises 11.21 and 10.28 prove Tychonoff’s Theorem for a countable product of compact metric spaces. We now state the general version of the theorem.

Theorem 11.34 (Tychonoff’s Theorem). Let {X_{α}}_{α∈A} be a collection of non-empty compact spaces. Then X := X_{A} = ∏_{α∈A} X_{α} is compact in the prod-

uct space topology. (Compare with Exercise 11.21 which covers the special case of a countable product of compact metric spaces.)

Proof. (The proof is taken from Loomis [1] which followed Bourbaki. Remark 11.35 below should help the reader understand the strategy of the proof to follow.) The proof requires a form of “induction” known as Zorn’s lemma which is equivalent to the axiom of choice, see Theorem B.7 of Appendix B below.

For α ∈ A let π_{α} denote the projection map from X to X_{α}. Suppose that F is a family of closed subsets of X which has the finite intersection property, see Definition 11.3. By Proposition 11.4 the proof will be complete if we can show ∩F ≠ ∅.

The first step is to apply Zorn’s lemma to construct a maximal collection, F_{0}, of (not necessarily closed) subsets of X with the finite intersection property such that F ⊂ F_{0}. To do this, let Γ := {G ⊂ 2^{X} : F ⊂ G} equipped with the partial order, G_{1} < G_{2} if G_{1} ⊂ G_{2}. If Φ is a linearly ordered subset of Γ, then G:= ∪Φ is an upper bound for Γ which still has the finite intersection property as the reader should check. So by Zorn’s lemma, Γ has a maximal element F_{0}. The maximal F_{0} has the following properties.

1. F_{0} is closed under finite intersections. Indeed, if we let (F_{0})_{f} denote the collection of all finite intersections of elements from F_{0}, then (F_{0})_{f} has the finite intersection property and contains F_{0}. Since F_{0} is maximal, this implies (F_{0})_{f} = F_{0}.

2. If B ⊂ X and B ∩ F ≠ ∅ for all F ∈ F_{0} then B ∈ F_{0}. For if not F_{0} ∪ {B} would still satisfy the finite intersection property and would properly contain F_{0} and this would violate the maximallity of F_{0}.

3. For each α ∈ A,

π_{a}(F_{0}) := {π_{α}(F) ⊂ X_{α} : F ∈ F_{0}}

has the finite intersection property. Indeed, if {F_{i}}_{i=1}^{n} ⊂ F_{0}, then ∩_{i=1}^{n}π_{α}(F_{i}) ⊃ π_{α} (∩_{i=1}^{n}F_{i}) ≠ ∅.

Since X_{α} is compact, property 3. above along with Proposition 11.4 implies ∩_{F ∈F_{0}} \overline{π_{α}(F)} ≠ ∅. Since this true for each α ∈ A, using the axiom of choice, there exists p ∈ X such that p_{α} = π_{α}(p) ∈ ∩_{F ∈F_{0}} \overline{π_{α}(F)} for all α ∈ A. The proof will be completed by showing ∩F ≠ ∅ by showing p ∈ ∩F.

Since C := ∩ {F̄ : F ∈ F_{0}} ⊂ ∩F, it suffices to show p ∈ C. Let U be an open neighborhood of p in X. By the definition of the product topology (or item 2. of Proposition 10.25), there exists Λ ⊂⊂ A and open sets U_{α} ⊂ X_{α} for all α ∈ Λ such that p ∈ ∩_{α∈Λ}π_{α}^{−1}(U_{α}) ⊂ U. Since p_{α} ∈ ∩_{F ∈F_{0}} \overline{π_{α}(F)} and p_{α} ∈ U_{α} for all α ∈ Λ, it follows that U_{α} ∩ π_{α}(F) ≠ ∅ for all F ∈ F_{0} and all α ∈ Λ. This then implies π_{α}^{−1} (U_{α}) ∩ F ≠ ∅ for all F ∈ F_{0} and all α ∈ Λ. By property 2.^{7} above we concluded that π_{α}^{−1} (U_{α}) ∈ F_{0} for all α ∈ Λ and then by property 1. that ∩_{α∈Λ}π_{α}^{−1} (U_{α}) ∈ F_{0}. In particular

∅ ≠ F ∩ (∩_{α∈Λ}π_{α}^{−1} (U_{α})) ⊂ F ∩ U for all F ∈ F_{0}

which shows p ∈ F̄ for each F ∈ F_{0}, i.e. p ∈ C.

Remark 11.35. Consider the following simple example where X = [−1, 1] × [−1, 1] and F = {F_{1}, F_{2}} as in Figure 11.5. Notice that π_{i}(F_{1}) ∩ π_{i} (F_{2}) =

7 Here is where we use that F_{0} is maximal among the collection of all, not just closed, sets having the finite intersection property and containing F.

[−1, 1] for each i and so gives no help in trying to find the i^{th} – coordinate of one of the two points in F_{1} ∩ F_{2}. This is why it is necessary to introduce the collection F_{0} in the proof of Theorem 11.34. In this case one might take F_{0} to be the collection of all subsets F ⊂ X such that p ∈ F. We then have ∩_{F ∈F_{0}}π_{i} (F) = {p_{i}} , so the i^{th} – coordinate of p may now be determined by observing the sets, {π_{i} (F) : F ∈ F_{0}} .

Fig. 11.5. Here F = {F_{1}, F_{2}} where F_{1} and F_{2} are the two parabolic arcs and F_{1} ∩ F_{2} = {p, q}.

11.6 Exercises

Exercise 11.14. Prove Lemma 11.5.

Exercise 11.15. Let C be a closed proper subset of ℝ^{n} and x ∈ ℝ^{n} \C. Show there exists a y ∈ C such that d(x, y) = d_{C}(x).

Exercise 11.16. Let 𝔽 = ℝ in this problem and A ⊂ ℓ^{2}(ℕ) be defined by

A = {x ∈ ℓ^{2}(ℕ) : x(n) ≥ 1 + 1/n for some n ∈ ℕ}
= ∪_{n=1}^{∞}{x ∈ ℓ^{2}(ℕ) : x(n) ≥ 1 + 1/n}.

Show A is a closed subset of ℓ^{2}(ℕ) with the property that d_{A}(0) = 1 while there is no y ∈ A such that d(0, y) = 1. (Remember that in general an infinite union of closed sets need not be closed.)

Exercise 11.17. Let p ∈ [1, ∞] and X be an infinite set. Show directly, without using Theorem 11.15, the closed unit ball in ℓ^{p}(X) is not compact.

11.6.1 Ascoli-Arzela Theorem Problems

Exercise 11.18. Let T ∈ (0, ∞) and F ⊂ C([0, T]) be a family of functions such that:

1. ḟ(t) exists for all t ∈ (0, T) and f ∈ F.

2. sup_{f∈F} |f(0)| < ∞ and

3. M := sup_{f∈F} sup_{t∈(0,T)} |ḟ(t)| < ∞.

Show F is precompact in the Banach space C([0, T]) equipped with the norm ∥f∥_{∞} = sup_{t∈[0,T]} |f(t)| .

Exercise 11.19 (Peano’s Existence Theorem). Suppose Z : ℝ × ℝ^{d} → ℝ^{d} is a bounded continuous function. Then for each T < ∞^{8} there exists a solution to the differential equation

ẋ(t) = Z(t, x(t)) for − T < t < T with x(0) = x_{0}.  (11.7)

Do this by filling in the following outline for the proof.

1. Given ε > 0, show there exists a unique function x_{ε} ∈ C([−ε, ∞) → ℝ^{d}) such that x_{ε}(t) := x_{0} for −ε ≤ t ≤ 0 and

x_{ε}(t) = x_{0} + ∫_{0}^{t} Z(τ, x_{ε}(τ − ε))dτ for all t ≥ 0.  (11.8)

Here

∫_{0}^{t} Z(τ, x_{ε}(τ−ε))dτ = (∫_{0}^{t} Z_{1}(τ, x_{ε}(τ − ε))dτ, …, ∫_{0}^{t} Z_{d}(τ, x_{ε}(τ − ε))dτ)

where Z = (Z_{1}, …, Z_{d}) and the integrals are either the Lebesgue or the Riemann integral since they are equal on continuous functions. Hint: For t ∈ [0, ε], it follows from Eq. (11.8) that

x_{ε}(t) = x_{0} + ∫_{0}^{t} Z(τ, x_{0})dτ.

Now that x_{ε}(t) is known for t ∈ [−ε, ε] it can be found by integration for t ∈ [−ε, 2ε]. The process can be repeated.

2. Then use Exercise 11.18 to show there exists {ε_{k}}_{k=1}^{∞} ⊂ (0, ∞) such that lim_{k→∞} ε_{k} = 0 and x_{ε_{k}} converges to some x ∈ C([0, T]) with respect to the sup-norm: ∥x∥_{∞} = sup_{t∈[0,T]} |x(t)|). Also show for this sequence that

lim_{k→∞} sup_{ε_{k}≤τ≤T} |x_{ε_{k}}(τ − ε_{k}) − x (τ)| = 0.

8 Using Corollary 11.30, we may in fact allow T = ∞.

3. Pass to the limit (with justification) in Eq. (11.8) with ε replaced by ε_{k} to show x satisfies

x(t) = x_{0} + ∫_{0}^{t} Z(τ, x(τ))dτ ∀ t ∈ [0, T].

4. Conclude from this that ẋ(t) exists for t ∈ (0, T) and that x solves Eq. (11.7).

5. Apply what you have just proved to the ODE,

ẏ(t) = −Z(−t, y(t)) for 0 ≤ t < T with y(0) = x_{0}.

Then extend x(t) above to (−T, T) by setting x(t) = y(−t) if t ∈ (−T, 0]. Show x so defined solves Eq. (11.7) for t ∈ (−T, T).

Exercise 11.20. Prove Theorem 11.32. Hint: First prove C^{j,β} \binom{}{Ω̄} ⊏⊏ C^{j,α} \binom{}{Ω̄} is compact if 0 ≤ α < β ≤ 1. Then use Lemma 11.18 repeatedly to handle all of the other cases.

11.6.2 Tychonoff’s Theorem Problem

Exercise 11.21 (Tychonoff’s Theorem for Compact Metric Spaces). Let us continue the Notation used in Exercise 6.12. Further assume that the spaces X_{n} are compact for all n. Show, without using Theorem 11.34, (X, d) is compact. Hint: Either use Cantor’s method to show every sequence {x_{m}}_{m=1}^{∞} ⊂ X has a convergent subsequence or alternatively show (X, d) is complete and totally bounded. (Compare with Example 11.10.)

*** Beginning of WORK material. ***

Exercise 11.22. Let X := {0, 1}^{ℝ} and τ be the product topology on X where {0, 1} is equipped with the discrete topology. By Tychonoff’s Theorem 11.34, (X, τ) is a compact space. Show (X, τ) is not separable and hence not metrizable.

Solution to Exercise (11.22). I don’t know how to do this but here is a start. Suppose D := {f_{n}}_{n=1}^{∞} ⊂ X were a dense set and g ∈ X. We begin by observing that a basic open neighborhood of g is of the form

V_{Λ} := {f ∈ X : f = g on Λ}

where Λ ⊂⊂ ℝ. Therefore to see that X is not separable, we must find a finite set Λ ⊂ ℝ and a function g : ℝ → {0, 1} such that f_{n} ≠ g on Λ for any n ∈ ℕ. It is clear, to be able to do this we must assume that # (Λ) > 1 for if # (Λ) = 1, and f_{1} ≡ 0 and f_{2} ≡ 1, then g = f_{1} or g = f_{2} on Λ.

So the idea now is to show, there must exist some s < t in ℝ such that

(0, 1) ∉ {(f_{n} (s) , f_{n} (t)) : n ∈ ℕ} .

If this can be done, then we may choose g : ℝ → {0, 1} such that g (s) = 0 and g (t) = 1 and we will have shown that g ∉ D̄.

*** End of WORK material. ***

Locally Compact Hausdorff Spaces

In this section X will always be a topological space with topology τ. We are now interested in restrictions on τ in order to insure there are “plenty” of continuous functions. One such restriction is to assume τ = τ_{d} – is the topology induced from a metric on X. For example the results in Lemma 6.15 and Theorem 7.4 above shows that metric spaces have lots of continuous functions.

The main thrust of this section is to study locally compact (and σ – compact) “Hausdorff” spaces as defined in Definitions 12.2 and 11.21. We will see again that this class of topological spaces have an ample supply of continuous functions. We will start out with the notion of a Hausdorff topology. The following example shows a pathology which occurs when there are not enough open sets in a topology.

Example 12.1. As in Example 10.36, let

X := {1, 2, 3} with τ := {X, ∅, {1, 2}, {2, 3}, {2}}.

Example 10.36 shows limits need not be unique in this space and moreover it is easy to verify that the only continuous functions, f : Y → ℝ, are necessarily constant.

Definition 12.2 (Hausdorff Topology). A topological space, (X, τ), is Hausdorff if for each pair of distinct points, x, y ∈ X, there exists disjoint open neighborhoods, U and V of x and y respectively. (Metric spaces are typical examples of Hausdorff spaces.)

Remark 12.3. When τ is Hausdorff the “pathologies” appearing in Example 12.1 do not occur. Indeed if x_{n} → x ∈ X and y ∈ X \ {x} we may choose V ∈ τ_{x} and W ∈ τ_{y} such that V ∩ W = ∅. Then x_{n} ∈ V a.a. implies x_{n} ∉ W for all but a finite number of n and hence x_{n} ↛ y, so limits are unique.

Proposition 12.4. Let (X_{α}, τ_{α}) be Hausdorff topological spaces. Then the product space X_{A} = ∏_{α∈A} X_{α} equipped with the product topology is Hausdorff.

Proof. Let x, y ∈ X_{A} be distinct points. Then there exists α ∈ A such that π_{α}(x) = x_{α} ≠ y_{α} = π_{α}(y). Since X_{α} is Hausdorff, there exists disjoint open sets U, V ⊂ X_{α} such π_{α}(x) ∈ U and π_{α}(y) ∈ V. Then π_{α}^{−1}(U) and π_{α}^{−1}(V) are disjoint open sets in X_{A} containing x and y respectively.

Proposition 12.5. Suppose that (X, τ) is a Hausdorff space, K ⊏⊏ X and x ∈ K^{c}. Then there exists U, V ∈ τ such that U ∩ V = ∅, x ∈ U and K ⊂ V. In particular K is closed. (So compact subsets of Hausdorff topological spaces are closed.) More generally if K and F are two disjoint compact subsets of X, there exist disjoint open sets U, V ∈ τ such that K ⊂ V and F ⊂ U.

Proof. Because X is Hausdorff, for all y ∈ K there exists V_{y} ∈ τ_{y} and U_{y} ∈ τ_{x} such that V_{y} ∩ U_{y} = ∅. The cover {V_{y}}_{y∈K} of K has a finite subcover, {V_{y}}_{y∈Λ} for some Λ ⊂⊂ K. Let V = ∪_{y∈Λ}V_{y} and U = ∩_{y∈Λ}U_{y}, then U, V ∈ τ satisfy x ∈ U, K ⊂ V and U ∩ V = ∅. This shows that K^{c} is open and hence that K is closed. Suppose that K and F are two disjoint compact subsets of X. For each x ∈ F there exists disjoint open sets U_{x} and V_{x} such that K ⊂ V_{x} and x ∈ U_{x}. Since {U_{x}}_{x∈F} is an open cover of F, there exists a finite subset Λ of F such that F ⊂ U := ∪_{x∈Λ}U_{x}. The proof is completed by defining V := ∩_{x∈Λ}V_{x}.

Exercise 12.1. Show any finite set X admits exactly one Hausdorff topology τ.

Exercise 12.2. Let (X, τ) and (Y, τ_{Y}) be topological spaces.

1. Show τ is Hausdorff iff Δ := {(x, x) : x ∈ X} is a closed in X×X equipped with the product topology τ ⊗ τ.

2. Suppose τ is Hausdorff and f, g : Y → X are continuous maps. If \overline{\{f = g\}}^{Y} = Y then f = g. Hint: make use of the map f ×g : Y → X ×X defined by (f × g) (y) = (f(y), g(y)).

Exercise 12.3. Given an example of a topological space which has a nonclosed compact subset.

Proposition 12.6. Suppose that X is a compact topological space, Y is a Hausdorff topological space, and f : X → Y is a continuous bijection then f is a homeomorphism, i.e. f^{−1} : Y → X is continuous as well.

Proof. Since closed subsets of compact sets are compact, continuous images of compact subsets are compact and compact subsets of Hausdorff spaces are closed, it follows that (f^{−1})^{−1} (C) = f(C) is closed in X for all closed subsets C of X. Thus f^{−1} is continuous.

The next two results shows that locally compact Hausdorff spaces have plenty of open sets and plenty of continuous functions.

Proposition 12.7. Suppose X is a locally compact Hausdorff space and U ⊂_{o} X and K ⊏⊏ U. Then there exists V ⊂_{o} X such that K ⊂ V ⊂ \overline{V} ⊂ U ⊂ X and V̄ is compact. (Compare with Proposition 11.25 above.)

Proof. By local compactness, for all x ∈ K, there exists U_{x} ∈ τ_{x} such that Ū_{x} is compact. Since K is compact, there exists Λ ⊂⊂ K such that {U_{x}}_{x∈Λ} is a cover of K. The set O = U ∩ (∪_{x∈Λ}U_{x}) is an open set such that K ⊂ O ⊂ U and O is precompact since Ō is a closed subset of the compact set ∪_{x∈Λ}Ū_{x}. (∪_{x∈Λ}Ū_{x}. is compact because it is a finite union of compact sets.) So by replacing U by O if necessary, we may assume that Ū is compact. Since Ū is compact and bd(U) = Ū ∩ U^{c} is a closed subset of Ū, bd(U) is compact. Because bd(U) ⊂ U^{c}, it follows that bd(U) ∩ K = ∅, so by Proposition 12.5, there exists disjoint open sets V and W such that K ⊂ V and bd(U) ⊂ W. By replacing V by V ∩U if necessary we may further assume that K ⊂ V ⊂ U, see Figure 12.1. Because Ū ∩W^{c} is a closed set containing V and bd(U)∩W^{c} = ∅,

Fig. 12.1. The construction of V.

V̄ ⊂ Ū ∩ W^{c} = (U ∪ bd(U)) ∩ W^{c} = U ∩ W^{c} ⊂ U ⊂ Ū.

Since Ū is compact it follows that V̄ is compact and the proof is complete. The following Lemma is analogous to Lemma 11.27.

Lemma 12.8 (Urysohn’s Lemma for LCH Spaces). Let X be a locally compact Hausdorff space and K ⊏⊏ U ⊂_{o} X. Then there exists f ≺ U (see Definition 11.26) such that f = 1 on K. In particular, if K is compact and C is closed in X such that K ∩ C = ∅, there exists f ∈ C_{c}(X, [0, 1]) such that f = 1 on K and f = 0 on C.

Proof. For notational ease later it is more convenient to construct g := 1 − f rather than f. To motivate the proof, suppose g ∈ C(X, [0, 1]) such that g = 0 on K and g = 1 on U^{c}. For r > 0, let U_{r} = {g < r} . Then for 0 < r < s ≤ 1, U_{r} ⊂ {g ≤ r} ⊂ U_{s} and since {g ≤ r} is closed this implies

K ⊂ U_{r} ⊂ Ū_{r} ⊂ {g ≤ r} ⊂ U_{s} ⊂ U.

Therefore associated to the function g is the collection open sets {U_{r}}_{r>0} ⊂ τ with the property that K ⊂ U_{r} ⊂ Ū_{r} ⊂ U_{s} ⊂ U for all 0 < r < s ≤ 1 and U_{r} = X if r > 1. Finally let us notice that we may recover the function g from the sequence {U_{r}}_{r>0} by the formula

g(x) = inf{r > 0 : x ∈ U_{r}}.  (12.1)

The idea of the proof to follow is to turn these remarks around and define g by Eq. (12.1).

Step 1. (Construction of the U_{r}.) Let

𝔻 := {k2^{−n} : k = 1, 2, …, 2^{−n}, n = 1, 2, … }

be the dyadic rationals in (0, 1]. Use Proposition 12.7 to find a precompact open set U_{1} such that K ⊂ U_{1} ⊂ Ū_{1} ⊂ U. Apply Proposition 12.7 again to construct an open set U_{1/2} such that

K ⊂ U_{1/2} ⊂ Ū_{1/2} ⊂ U_{1}

and similarly use Proposition 12.7 to find open sets U_{1/2}, U_{3/4} ⊂_{o} X such that

K ⊂ U_{1/4} ⊂ Ū_{1/4} ⊂ U_{1/2} ⊂ Ū_{1/2} ⊂ U_{3/4} ⊂ Ū_{3/4} ⊂ U_{1}.

Likewise there exists open set U_{1/8}, U_{3/8}, U_{5/8}, U_{7/8} such that

K ⊂ U_{1/8} ⊂ Ū_{1/8} ⊂ U_{1/4} ⊂ Ū_{1/4} ⊂ U_{3/8} ⊂ Ū_{3/8} ⊂ U_{1/2}
⊂ Ū_{1/2} ⊂ U_{5/8} ⊂ Ū_{5/8} ⊂ U_{3/4} ⊂ Ū_{3/4} ⊂ U_{7/8} ⊂ Ū_{7/8} ⊂ U_{1}.

Continuing this way inductively, one shows there exists precompact open sets {U_{r}}_{r∈𝔻} ⊂ τ such that

K ⊂ U_{r} ⊂ \overline{U}_{r} ⊂ U_{s} ⊂ U_{1} ⊂ Ū_{1} ⊂ U

for all r, s ∈ 𝔻 with 0 < r < s ≤ 1.

Step 2. Let U_{r} := X if r > 1 and define

g(x) = inf{r ∈ 𝔻 ∪ (1, 2) : x ∈ U_{r}},

see Figure 12.2. Then g(x) ∈ [0, 1] for all x ∈ X, g(x) = 0 for x ∈ K since x ∈ K ⊂ U_{r} for all r ∈ 𝔻. If x ∈ U_{1}^{c}, then x ∉ U_{r} for all r ∈ 𝔻 and hence g(x) = 1. Therefore f := 1 − g is a function such that f = 1 on K and \{f ≠ 0\} = \{g ≠ 1\} ⊂ U_{1} ⊂ Ū_{1} ⊂ U so that supp(f) = \overline{\{f ≠ 0\}} ⊂ Ū_{1} ⊂ U is

Fig. 12.2. Determining g from {U_{r}} .

a compact subset of U. Thus it only remains to show f, or equivalently g, is continuous.

Since E = {(α, ∞), (−∞, α) : α ∈ ℝ} generates the standard topology on ℝ, to prove g is continuous it suffices to show {g < α} and {g > α} are open sets for all α ∈ ℝ. But g(x) < α iff there exists r ∈ 𝔻 ∪ (1, ∞) with r < α such that x ∈ U_{r}. Therefore

{g < α} = ⋃ {U_{r} : r ∈ 𝔻 ∪ (1, ∞) ∋ r < α}

which is open in X. If α ≥ 1, {g > α} = ∅ and if α < 0, {g > α} = X. If α ∈ (0, 1), then g(x) > α iff there exists r ∈ 𝔻 such that r > α and x ∉ U_{r}. Now if r > α and x ∉ U_{r} then for s ∈ 𝔻 ∩ (α, r), x ∉ Ū_{s} ⊂ U_{r}. Thus we have shown that

\{g > α\} = ⋃ \{(\overline{U}_{s})^{c} : s ∈ 𝔻 ∋ s > α\}

which is again an open subset of X.

Theorem 12.9 (Locally Compact Tietz Extension Theorem). Let (X, τ) be a locally compact Hausdorff space, K ⊏⊏ U ⊂_{o} X, f ∈ C(K, ℝ), a = min f(K) and b = max f(K). Then there exists F ∈ C(X, [a, b]) such that F |_{K} = f. Moreover given c ∈ [a, b], F can be chosen so that supp(F − c) = \overline{\{F ≠ c\}} ⊂ U.

The proof of this theorem is similar to Theorem 7.4 and will be left to the reader, see Exercise 12.5.

12.1 Locally compact form of Urysohn’s Metrization Theorem

Notation 12.10 Let Q := [0, 1]^{ℕ} denote the (infinite dimensional) unit cube in ℝ^{ℕ}. For a, b ∈ Q let

d(a, b) := ∑_{n=1}^{∞} \frac{1}{2^{n}} |a_{n} − b_{n}| .  (12.2)

The metric introduced in Exercise 11.21 would be defined, in this context, as d̃(a, b) := ∑_{n=1}^{∞} \frac{1}{2^{n}} \frac{|a_{n}−b_{n}|}{1+|a_{n}−b_{n}|}. Since 1 ≤ 1+|a_{n} − b_{n}| ≤ 2, it follows that d̃≤ d ≤ 2d. So the metrics d and d̃ are equivalent and in particular the topologies induced by d and d̃ are the same. By Exercises 10.28, the d – topology on Q is the same as the product topology and by Tychonoff’s Theorem 11.34 or by Exercise 11.21, (Q, d) is a compact metric space.

Theorem 12.11. To every separable metric space (X, ρ), there exists a continuous injective map G : X → Q such that G : X → G(X) ⊂ Q is a homeomorphism. In short, any separable metrizable space X is homeomorphic to a subset of (Q, d).

Remark 12.12. Notice that if we let ρ′(x, y) := d(G(x), G(y)), then ρ′ induces the same topology on X as ρ and G : (X, ρ′) → (Q, d) is isometric.

Proof. Let D = {x_{n}}_{n=1}^{∞} be a countable dense subset of X,

ϕ(t) := \{ \begin{matrix}1 & if & t ≤ 0 \\ 1 − t & if & 0 ≤ t ≤ 1 \\ 0 & if & t ≥ 1, \end{matrix}

(see Figure 12.3) and for m, n ∈ ℕ let

f_{m,n}(x) := 1 − ϕ (mρ(x_{n}, x)).

Then f_{m,n} = 0 if ρ(x, x_{n}) < 1/m and f_{m,n} = 1 if ρ(x, x_{n}) > 2/m. Let {g_{k}}_{k=1}^{∞} be an enumeration of {f_{m,n} : m, n ∈ ℕ} and define G : X → Q by

G(x) = (g_{1}(x), g_{2}(x), …) ∈ Q.

We will now show G : X → G(X) ⊂ Q is a homeomorphism. To show G is injective suppose x, y ∈ X and ρ(x, y) = δ ≥ 1/m. In this case we may find x_{n} ∈ X such that ρ(x, x_{n}) ≤ \frac{1}{2m}, ρ(y, x_{n}) ≥ δ − \frac{1}{2m} ≥ \frac{1}{2m} and hence f_{4m,n}(y) = 1 while f_{4m,n}(y) = 0. From this it follows that G(x) ≠ G(y) if x ≠ y and hence G is injective. The continuity of G is a consequence of the continuity of each of the components g_{i} of G. So it only remains to show G^{−1} : G(X) → X is continuous. Given a = G(x) ∈ G(X) ⊂ Q and ε > 0, choose m ∈ ℕ and x_{n} ∈ X such that ρ(x_{n}, x) < \frac{1}{2m} < \frac{ε}{2}. Then f_{m,n}(x) = 0

Fig. 12.3. The graph of the function ϕ.

and for y ∉ B(x_{n}, \frac{2}{m}), f_{m,n}(y) = 1. So if k is chosen so that g_{k} = f_{m,n}, we have shown that for

d(G(y), G(x)) ≥ 2^{−k} for y ∉ B(x_{n}, 2/m)

or equivalently put, if

d(G(y), G(x)) < 2^{−k} then y ∈ B(x_{n}, 2/m) ⊂ B(x, 1/m) ⊂ B(x, ε).

This shows that if G(y) is sufficiently close to G(x) then ρ(y, x) < ε, i.e. G^{−1} is continuous at a = G(x).

Theorem 12.13 (Urysohn Metrization Theorem for LCH’s). Every second countable locally compact Hausdorff space, (X, τ) , is metrizable, i.e. there is a metric ρ on X such that τ = τ_{ρ}. Moreover, ρ may be chosen so that X is isometric to a subset Q_{0} ⊂ Q equipped with the metric d in Eq. (12.2). In this metric X is totally bounded and hence the completion of X (which is isometric to Q̄_{0} ⊂ Q) is compact. (Also see Theorem 12.43.)

Proof. Let B be a countable base for τ and set

Γ := {(U, V) ∈ B × B | Ū ⊂ V and Ū is compact}.

To each O ∈ τ and x ∈ O there exist (U, V) ∈ Γ such that x ∈ U ⊂ V ⊂ O. Indeed, since B is a base for τ, there exists V ∈ B such that x ∈ V ⊂ O. Now apply Proposition 12.7 to find U′ ⊂_{o} X such that x ∈ U′ ⊂ Ū′ ⊂ V with Ū′ being compact. Since B is a base for τ, there exists U ∈ B such that x ∈ U ⊂ U′ and since Ū ⊂ Ū′, Ū is compact so (U, V) ∈ Γ. In particular this shows that B′ := {U ∈ B : (U, V) ∈ Γ for some V ∈ B} is still a base for τ. If Γ is a finite, then B′ is finite and τ only has a finite number of elements as well. Since (X, τ) is Hausdorff, it follows that X is a finite set. Letting {x_{n}}_{n=1}^{N} be an enumeration of X, define T : X → Q by T (x_{n}) = e_{n} for n = 1, 2, …, N where e_{n} = (0, 0, …, 0, 1, 0, …), with the 1 occurring in the n^{th} spot. Then ρ(x, y) := d(T (x), T (y)) for x, y ∈ X is the desired metric.

So we may now assume that Γ is an infinite set and let {(U_{n}, V_{n})}_{n=1}^{∞} be an enumeration of Γ. By Urysohn’s Lemma 12.8 there exists f_{U,V} ∈ C(X, [0, 1]) such that f_{U,V} = 0 on Ū and f_{U,V} = 1 on V^{c}. Let F := {f_{U,V} | (U, V) ∈ Γ } and set f_{n} := f_{U_{n},V_{n}} – an enumeration of F. We will now show that

ρ(x, y) := ∑_{n=1}^{∞} \frac{1}{2^{n}} |f_{n}(x) − f_{n}(y)|

is the desired metric on X. The proof will involve a number of steps.

1. (ρ is a metric on X.) It is routine to show ρ satisfies the triangle inequality and ρ is symmetric. If x, y ∈ X are distinct points then there exists (U_{n_{0}}, V_{n_{0}}) ∈ Γ such that x ∈ U_{n_{0}} and V_{n_{0}} ⊂ O := {y}^{c} . Since f_{n_{0}}(x) = 0 and f_{n_{0}}(y) = 1, it follows that ρ(x, y) ≥ 2^{−n_{0}} > 0.

2. (Let τ_{0} = τ (f_{n} : n ∈ ℕ) , then τ = τ_{0} = τ_{ρ}.) As usual we have τ_{0} ⊂ τ. Since, for each x ∈ X, y → ρ(x, y) is τ_{0} – continuous (being the uniformly convergent sum of continuous functions), it follows that B_{x}(ε) := {y ∈ X : ρ(x, y) < ε} ∈ τ_{0} for all x ∈ X and ε > 0. Thus τ_{ρ} ⊂ τ_{0} ⊂ τ. Suppose that O ∈ τ and x ∈ O. Let (U_{n_{0}}, V_{n_{0}}) ∈ Γ be such that x ∈ U_{n_{0}} and V_{n_{0}} ⊂ O. Then f_{n_{0}}(x) = 0 and f_{n_{0}} = 1 on O^{c}. Therefore if y ∈ X and f_{n_{0}}(y) < 1, then y ∈ O so x ∈ {f_{n_{0}} < 1} ⊂ O. This shows that O may be written as a union of elements from τ_{0} and therefore O ∈ τ_{0}. So τ ⊂ τ_{0} and hence τ = τ_{0}. Moreover, if y ∈ B_{x}(2^{−n_{0}}) then 2^{−n_{0}} > ρ(x, y) ≥ 2^{−n_{0}}f_{n_{0}}(y) and therefore x ∈ B_{x}(2^{−n_{0}}) ⊂ {f_{n_{0}} < 1} ⊂ O. This shows O is ρ – open and hence τ_{ρ} ⊂ τ_{0} ⊂ τ ⊂ τ_{ρ}.

3. (X is isometric to some Q_{0} ⊂ Q.) Let T : X → Q be defined by T (x) = (f_{1}(x), f_{2}(x), …, f_{n}(x), …). Then T is an isometry by the very definitions of d and ρ and therefore X is isometric to Q_{0} := T (X). Since Q_{0} is a subset of the compact metric space (Q, d), Q_{0} is totally bounded and therefore X is totally bounded.

BRUCE: Add Stone Chech Compactification results.

12.2 Partitions of Unity

Definition 12.14. Let (X, τ) be a topological space and X_{0} ⊂ X be a set. A collection of sets {B_{α}}_{α∈A} ⊂ 2^{X} is locally finite on X_{0} if for all x ∈ X_{0}, there is an open neighborhood N_{x} ∈ τ of x such that #{α ∈ A : B_{α} ∩ N_{x} ≠ ∅} < ∞.

Definition 12.15. Suppose that U is an open cover of X_{0} ⊂ X. A collection {ϕ_{α}}_{α∈A} ⊂ C(X, [0, 1]) (N = ∞ is allowed here) is a partition of unity on X_{0} subordinate to the cover U if:

1. for all α there is a U ∈ U such that supp(ϕ_{α}) ⊂ U, 2. the collection of sets, {supp(ϕ_{α})}_{α∈A}, is locally finite on X, and

3. ∑_{α∈A} ϕ_{α} = 1 on X_{0}.

Notice by item 2. that, for each x ∈ X, there is a neighborhood N_{x} such that

Λ := {α ∈ A : supp(ϕ_{α}) ∩ N_{x} ≠ ∅}

is a finite set. Therefore, ∑_{α∈A} ϕ_{α}|_{N_{x}} = ∑_{α∈Λ} ϕ_{α}|_{N_{x}} which shows the sum ∑_{α∈A} ϕ_{α} is well defined and defines a continuous function on N_{x} and therefore on X since continuity is a local property. We will summarize these last comments by saying the sum, ∑_{α∈A} ϕ_{α}, is locally finite.

Proposition 12.16 (Partitions of Unity: The Compact Case). Suppose that X is a locally compact Hausdorff space, K ⊂ X is a compact set and U = {U_{j}}_{j=1}^{n} is an open cover of K. Then there exists a partition of unity {h_{j}}_{j=1}^{n} of K such that h_{j} ≺ U_{j} for all j = 1, 2, …, n.

Proof. For all x ∈ K choose a precompact open neighborhood, V_{x}, of x such that \overline{V}_{x} ⊂ U_{j}. Since K is compact, there exists a finite subset, Λ, of K such that K ⊂ ⋃_{x∈Λ} V_{x}. Let

F_{j} = ∪ \{V̄_{x} : x ∈ Λ and \overline{V}_{x} ⊂ U_{j}\} .

Then F_{j} is compact, F_{j} ⊂ U_{j} for all j, and K ⊂ ∪_{j=1}^{n}F_{j}. By Urysohn’s Lemma 12.8 there exists f_{j} ≺ U_{j} such that f_{j} = 1 on F_{j} for j = 1, 2, …, n and by convention let f_{n+1} ≡ 1. We will now give two methods to finish the proof.

Method 1. Let h_{1} = f_{1}, h_{2} = f_{2}(1 − h_{1}) = f_{2}(1 − f_{1}),

h_{3} = f_{3}(1 − h_{1} − h_{2}) = f_{3}(1 − f_{1} − (1 − f_{1})f_{2}) = f_{3}(1 − f_{1})(1 − f_{2})

and continue on inductively to define

h_{k} = (1 − h_{1} − ⋯ − h_{k−1})f_{k} = f_{k} · ∏_{j=1}^{k−1}(1 − f_{j}) ∀ k = 2, 3, …, n (12.3)

and to show

h_{n+1} = (1 − h_{1} − ⋯ − h_{n}) · 1 = 1 · ∏_{j=1}^{n}(1 − f_{j}).  (12.4)

From these equations it clearly follows that h_{j} ∈ C_{c}(X, [0, 1]) and that supp(h_{j}) ⊂ supp(f_{j}) ⊂ U_{j}, i.e. h_{j} ≺ U_{j}. Since ∏_{j=1}^{n}(1 − f_{j}) = 0 on K, ∑_{j=1}^{n} h_{j} = 1 on K and {h_{j}}_{j=1}^{n} is the desired partition of unity.

Method 2. Let g := ∑_{j=1}^{n} f_{j} ∈ C_{c}(X). Then g ≥ 1 on K and hence

K ⊂ \{g > \frac{1}{2}\}. Choose ϕ ∈ C_{c}(X, [0, 1]) such that ϕ = 1 on K and supp(ϕ) ⊂ \{g > \frac{1}{2}\} and define f_{0} := 1 − ϕ. Then f_{0} = 0 on K, f_{0} = 1 if g ≤ \frac{1}{2} and therefore,

f_{0} + f_{1} + ⋯ + f_{n} = f_{0} + g > 0

on X. The desired partition of unity may be constructed as

h_{j}(x) = \frac{f_{j}(x)}{f_{0}(x) + ⋯ + f_{n}(x)}.

Indeed supp (h_{j}) = supp (f_{j}) ⊂ U_{j}, h_{j} ∈ C_{c}(X, [0, 1]) and on K,

h_{1} + ⋯ + h_{n} = \frac{f_{1} + ⋯ + f_{n}}{f_{0} + f_{1} + ⋯ + f_{n}} = \frac{f_{1} + ⋯ + f_{n}}{f_{1} + ⋯ + f_{n}} = 1.

Proposition 12.17. Let (X, τ) be a locally compact and σ – compact Hausdorff space. Suppose that U ⊂ τ is an open cover of X. Then we may construct two locally finite open covers V = {V_{i}}_{i=1}^{N} and W = {W_{i}}_{i=1}^{N} of X (N = ∞ is allowed here) such that:

1. W_{i} ⊂ W̄_{i} ⊂ V_{i} ⊂ V̄_{i} and V̄_{i} is compact for all i.

2. For each i there exist U ∈ U such that V̄_{i} ⊂ U.

Proof. By Remark 11.24, there exists an open cover of G = {G_{n}}_{n=1}^{∞} of X such that G_{n} ⊂ Ḡ_{n} ⊂ G_{n+1}. Then X = ∪_{k=1}^{∞}(Ḡ_{k} \ Ḡ_{k−1}), where by convention G_{−1} = G_{0} = ∅. For the moment fix k ≥ 1. For each x ∈ Ḡ_{k} \G_{k−1}, let U_{x} ∈ U be chosen so that x ∈ U_{x} and by Proposition 12.7 choose an open neighborhood N_{x} of x such that N̄_{x} ⊂ U_{x} ∩ (G_{k+1} \ Ḡ_{k−2}), see Figure 12.4 below. Since {N_{x}}_{x∈Ḡ_{k}\G_{k−1}} is an open cover of the compact set Ḡ_{k} \ G_{k−1}, there exist a finite subset Γ_{k} ⊂ {N_{x}}_{x∈Ḡ_{k}\G_{k−1}} which also covers Ḡ_{k} \ G_{k−1}.

By construction, for each W ∈ Γ_{k}, there is a U ∈ U such that W̄ ⊂ U ∩ (G_{k+1} \ Ḡ_{k−2}) and by another application of Proposition 12.7, there exists an open set V_{W} such that W̄ ⊂ V_{W} ⊂ V̄_{W} ⊂ U ∩ (G_{k+1} \ Ḡ_{k−2}). We now choose and enumeration {W_{i}}_{i=1}^{N} of the countable open cover, ∪_{k=1}^{∞}Γ_{k}, of X and define V_{i} = V_{W_{i}}. Then the collection {W_{i}}_{i=1}^{N} and {V_{i}}_{i=1}^{N} are easily checked to satisfy all the conclusions of the proposition. In particular notice that for each k; V_{i} ∩ G_{k} ≠ ∅ for only a finite number of i’s.

Theorem 12.18 (Partitions of Unity for σ – Compact LCH Spaces). Let (X, τ) be locally compact, σ – compact and Hausdorff and let U ⊂ τ be an open cover of X. Then there exists a partition of unity of {h_{i}}_{i=1}^{N} (N = ∞ is allowed here) subordinate to the cover U such that supp(h_{i}) is compact for all i.

Proof. Let V = {V_{i}}_{i=1}^{N} and W = {W_{i}}_{i=1}^{N} be open covers of X with the properties described in Proposition 12.17. By Urysohn’s Lemma 12.8, there

Fig. 12.4. Constructing the {W_{i}}_{i=1}^{N} .

exists f_{i} ≺ V_{i} such that f_{i} = 1 on W̄_{i} for each i. As in the proof of Proposition 12.16 there are two methods to finish the proof.

Method 1. Define h_{1} = f_{1}, h_{j} by Eq. (12.3) for all other j. Then as in Eq. (12.4), for all n < N + 1,

1 − ∑_{j=1}^{∞} h_{j} = lim_{n→∞} (f_{n} ∏_{j=1}^{n}(1 − f_{j})) = 0

since for x ∈ X, f_{j}(x) = 1 for some j. As in the proof of Proposition 12.16, it is easily checked that {h_{i}}_{i=1}^{N} is the desired partition of unity.

Method 2. Let f := ∑_{i=1}^{N} f_{i}, a locally finite sum, so that f ∈ C(X). Since {W_{i}}_{i=1}^{∞} is a cover of X, f ≥ 1 on X so that 1/f ∈ C (X)) as well. The functions h_{i} := f_{i}/f for i = 1, 2, …, N give the desired partition of unity.

Lemma 12.19. Let (X, τ) be a locally compact Hausdorff space.

1. A subset E ⊂ X is closed iff E ∩ K is closed for all K ⊏⊏ X.

2. Let {C_{α}}_{α∈A} be a locally finite collection of closed subsets of X, then C = ∪_{α∈A}C_{α} is closed in X. (Recall that in general closed sets are only closed under finite unions.)

Proof. 1. Since compact subsets of Hausdorff spaces are closed, E ∩ K is closed if E is closed and K is compact. Now suppose that E ∩ K is closed for all compact subsets K ⊂ X and let x ∈ E^{c}. Since X is locally compact, there exists a precompact open neighborhood, V, of x.^{1} By assumption E ∩ V̄

1 If X were a metric space we could finish the proof as follows. If there does not exist an open neighborhood of x which is disjoint from E, then there would exists

is closed so x ∈ (E ∩ V̄ )^{c} – an open subset of X. By Proposition 12.7 there exists an open set U such that x ∈ U ⊂ Ū ⊂ (E ∩ V̄ )^{c} , see Figure 12.5. Let

Fig. 12.5. Showing E^{c} is open.

W := U ∩ V. Since

W ∩ E = U ∩ V ∩ E ⊂ U ∩ V̄ ∩ E = ∅,

and W is an open neighborhood of x and x ∈ E^{c} was arbitrary, we have shown E^{c} is open hence E is closed.

2. Let K be a compact subset of X and for each x ∈ K let N_{x} be an open neighborhood of x such that #{α ∈ A : C_{α} ∩ N_{x} ≠ ∅} < ∞. Since K is compact, there exists a finite subset Λ ⊂ K such that K ⊂ ∪_{x∈Λ}N_{x}. Letting Λ_{0} := {α ∈ A : C_{α} ∩ K ≠ ∅}, then

# (Λ_{0}) ≤ ∑_{x∈Λ} #{α ∈ A : C_{α} ∩ N_{x} ≠ ∅} < ∞

and hence K ∩ (∪_{α∈A}C_{α}) = K ∩ (∪_{α∈Λ_{0}}C_{α}) . The set (∪_{α∈Λ_{0}}C_{α}) is a finite union of closed sets and hence closed. Therefore, K ∩ (∪_{α∈A}C_{α}) is closed and by item 1. it follows that ∪_{α∈A}C_{α} is closed as well.

Corollary 12.20. Let (X, τ) be a locally compact and σ – compact Hausdorff space and U = {U_{α}}_{α∈A} ⊂ τ be an open cover of X. Then there exists a partition of unity of {h_{α}}_{α∈A} subordinate to the cover U such that supp(h_{α}) ⊂ U_{α} for all α ∈ A. (Notice that we do not assert that h_{α} has compact support. However if Ū_{α} is compact then supp(h_{α}) will be compact.)

x_{n} ∈ E such that x_{n} → x. Since E ∩ V̄ is closed and x_{n} ∈ E ∩ V̄ for all large n, it follows (see Exercise 6.4) that x ∈ E ∩ V̄ and in particular that x ∈ E. But we chose x ∈ E^{c}.

12.3 C_{0}(X) and the Alexanderov Compactification 165

Proof. By the σ – compactness of X, we may choose a countable subset, {α_{i}}_{i=1}^{N} (N = ∞ allowed here), of A such that {U_{i} := U_{α_{i}}}_{i=1}^{N} is still an open cover of X. Let {g_{j}}_{j=1}^{∞} be a partition of unity^{2} subordinate to the cover {U_{i}}_{i=1}^{N} as in Theorem 12.18. Define Γ̃_{k} := {j : supp(g_{j}) ⊂ U_{k}} and Γ_{k} := Γ̃_{k} \ (∪_{j=1}^{k−1}Γ̃_{k}), where by convention Γ̃_{0} = ∅. Then

ℕ = ⋃_{k=1}^{∞} Γ̃_{k} = ∐_{k=1}^{∞} Γ_{k}.

If Γ_{k} = ∅ let h_{k} := 0 otherwise let h_{k} := ∑_{j∈Γ_{k}} g_{j}, a locally finite sum. Then

∑_{k=1}^{N} h_{k} = ∑_{j=1}^{∞} g_{j} = 1.

By Item 2. of Lemma 12.19, ∪_{j∈Γ_{k}}supp(g_{j}) is closed and therefore,

supp(h_{k}) = \overline{\{h_{k} ≠ 0\}} = \overline{∪_{j∈Γ_{k}} \{g_{j} ≠ 0\}} ⊂ ∪_{j∈Γ_{k}}supp(g_{j}) ⊂ U_{k}

and hence h_{k} ≺ U_{k} and the sum ∑_{k=1}^{N} h_{k} is still locally finite. (Why?) The desired partition of unity is now formed by letting h_{α_{k}} := h_{k} for k < N + 1 and h_{α} ≡ 0 if α ∉ {α_{i}}_{i=1}^{N}.

Corollary 12.21. Let (X, τ) be a locally compact and σ – compact Hausdorff space and A, B be disjoint closed subsets of X. Then there exists f ∈ C(X, [0, 1]) such that f = 1 on A and f = 0 on B. In fact f can be chosen so that supp(f) ⊂ B^{c}.

Proof. Let U_{1} = A^{c} and U_{2} = B^{c}, then {U_{1}, U_{2}} is an open cover of X. By Corollary 12.20 there exists h_{1}, h_{2} ∈ C(X, [0, 1]) such that supp(h_{i}) ⊂ U_{i} for i = 1, 2 and h_{1} + h_{2} = 1 on X. The function f = h_{2} satisfies the desired properties.

12.3 C_{0}(X) and the Alexanderov Compactification

Definition 12.22. Let (X, τ) be a topological space. A continuous function f : X → ℂ is said to vanish at infinity if {|f| ≥ ε} is compact in X for all ε > 0. The functions, f ∈ C(X), vanishing at infinity will be denoted by C_{0}(X). (Notice that C_{0} (X) = C (X) whenever X is compact.)

Proposition 12.23. Let X be a topological space, BC(X) be the space of bounded continuous functions on X with the supremum norm topology. Then

2 So as to simplify the indexing we assume there countable number of g_{j}’s. This can always be arranged by taking g_{k} ≡ 0 for large k if necessary.

1. C_{0}(X) is a closed subspace of BC(X).

2. If we further assume that X is a locally compact Hausdorff space, then C_{0}(X) = \overline{C_{c}(X)}.

Proof.

1. If f ∈ C_{0}(X), K_{1} := {|f| ≥ 1} is a compact subset of X and therefore f(K_{1}) is a compact and hence bounded subset of ℂ and so M := sup_{x∈K_{1}} |f(x)| < ∞. Therefore ∥f∥_{∞} ≤ M ∨ 1 < ∞ showing f ∈ BC(X). Now suppose f_{n} ∈ C_{0}(X) and f_{n} → f in BC(X). Let ε > 0 be given and choose n sufficiently large so that ∥f − f_{n}∥_{∞} ≤ ε/2. Since

|f| ≤ |f_{n}| + |f − f_{n}| ≤ |f_{n}| + ∥f − f_{n}∥_{∞} ≤ |f_{n}| + ε/2,
{|f| ≥ ε} ⊂ {|f_{n}| + ε/2 ≥ ε} = {|f_{n}| ≥ ε/2} .

Because {|f| ≥ ε} is a closed subset of the compact set {|f_{n}| ≥ ε/2} , {|f| ≥ ε} is compact and we have shown f ∈ C_{0}(X).

2. Since C_{0}(X) is a closed subspace of BC(X) and C_{c}(X) ⊂ C_{0}(X), we always have \overline{C_{c}(X)} ⊂ C_{0}(X). Now suppose that f ∈ C_{0}(X) and let K_{n} := \{|f| ≥ \frac{1}{n}\} ⊏⊏ X. By Lemma 12.8 we may choose ϕ_{n} ∈ C_{c}(X, [0, 1]) such that ϕ_{n} ≡ 1 on K_{n}. Define f_{n} := ϕ_{n}f ∈ C_{c}(X). Then

∥f − f_{n}∥_{u} = ∥(1 − ϕ_{n})f∥_{∞} ≤ \frac{1}{n} → 0 as n → ∞.

This shows that f ∈ \overline{C_{c}(X)}.

Proposition 12.24 (Alexanderov Compactification). Suppose that (X, τ) is a non-compact locally compact Hausdorff space. Let X^{∗} = X ∪ {∞} , where {∞} is a new symbol not in X. The collection of sets,

τ^{∗} = τ ∪ {X^{∗} \ K : K ⊏⊏ X} ⊂ 2^{X^{∗}},

is a topology on X^{∗} and (X^{∗}, τ^{∗}) is a compact Hausdorff space. Moreover f ∈ C(X) extends continuously to X^{∗} iff f = g + c with g ∈ C_{0}(X) and c ∈ ℂ in which case the extension is given by f(∞) = c.

Proof. 1. (τ^{∗} is a topology.) Let F := {F ⊂ X^{∗} : X^{∗} \ F ∈ τ^{∗}}, i.e. F ∈ F iff F is a compact subset of X or F = F_{0} ∪ {∞} with F_{0} being a closed subset of X. Since the finite union of compact (closed) subsets is compact (closed), it is easily seen that F is closed under finite unions. Because arbitrary intersections of closed subsets of X are closed and closed subsets of compact subsets of X are compact, it is also easily checked that F is closed under arbitrary intersections. Therefore F satisfies the axioms of the closed subsets associated to a topology and hence τ^{∗} is a topology.

2. ((X^{∗}, τ^{∗}) is a Hausdorff space.) It suffices to show any point x ∈ X can be separated from ∞. To do this use Proposition 12.7 to find an open precompact neighborhood, U, of x. Then U and V := X^{∗} \Ū are disjoint open subsets of X^{∗} such that x ∈ U and ∞ ∈ V.

3. ((X^{∗}, τ^{∗}) is compact.) Suppose that U ⊂ τ^{∗} is an open cover of X^{∗}. Since U covers ∞, there exists a compact set K ⊂ X such that X^{∗} \ K ∈ U. Clearly X is covered by U_{0} := {V \ {∞} : V ∈ U} and by the definition of τ^{∗} (or using (X^{∗}, τ^{∗}) is Hausdorff), U_{0} is an open cover of X. In particular U_{0} is an open cover of K and since K is compact there exists Λ ⊂⊂ U such that K ⊂ ∪ {V \ {∞} : V ∈ Λ} . It is now easily checked that Λ ∪ {X^{∗} \ K} ⊂ U is a finite subcover of X^{∗}.

4. (Continuous functions on C(X^{∗}) statements.) Let i : X → X^{∗} be the inclusion map. Then i is continuous and open, i.e. i(V) is open in X^{∗} for all V open in X. If f ∈ C(X^{∗}), then g = f|_{X} −f(∞) = f ∘i−f(∞) is continuous on X. Moreover, for all ε > 0 there exists an open neighborhood V ∈ τ^{∗} of ∞ such that

|g(x)| = |f(x) − f(∞)| < ε for all x ∈ V.

Since V is an open neighborhood of ∞, there exists a compact subset, K ⊂ X, such that V = X^{∗} \ K. By the previous equation we see that {x ∈ X : |g(x)| ≥ ε} ⊂ K, so {|g| ≥ ε} is compact and we have shown g vanishes at ∞.

Conversely if g ∈ C_{0}(X), extend g to X^{∗} by setting g(∞) = 0. Given ε > 0, the set K = {|g| ≥ ε} is compact, hence X^{∗} \ K is open in X^{∗}. Since g(X^{∗} \ K) ⊂ (−ε, ε) we have shown that g is continuous at ∞. Since g is also continuous at all points in X it follows that g is continuous on X^{∗}. Now it f = g + c with c ∈ ℂ and g ∈ C_{0}(X), it follows by what we just proved that defining f(∞) = c extends f to a continuous function on X^{∗}.

Example 12.25. Let X be an uncountable set and τ be the discrete topology on X. Let (X^{∗} = X ∪ {∞} , τ^{∗}) be the one point compactification of X. The smallest dense subset of X^{∗} is the uncountable set X. Hence X^{∗} is a compact but non-separable and hence non-metrizable space.

The next proposition gathers a number of results involving countability assumptions which have appeared in the exercises.

Proposition 12.26 (Summary). Let (X, τ) be a topological space.

1. If (X, τ) is second countable, then (X, τ) is separable; see Exercise 10.11.

2. If (X, τ) is separable and metrizable then (X, τ) is second countable; see Exercise 10.12.

3. If (X, τ) is locally compact and metrizable then (X, τ) is σ – compact iff (X, τ) is separable; see Exercises 11.10 and 11.11.

4. If (X, τ) is locally compact and second countable, then (X, τ) is σ - compact, see Exercise 11.8.

5. If (X, τ) is locally compact and metrizable, then (X, τ) is σ – compact iff (X, τ) is separable, see Exercises 11.9 and 11.10.

12.4 Stone-Weierstrass Theorem

We now wish to generalize Theorem 8.34 to more general topological spaces. We will first need some definitions.

Definition 12.27. Let X be a topological space and A ⊂ C(X) = C(X, ℝ) or C(X, ℂ) be a collection of functions. Then

1. A is said to separate points if for all distinct points x, y ∈ X there exists f ∈ A such that f(x) ≠ f(y).

2. A is an algebra if A is a vector subspace of C(X) which is closed under pointwise multiplication. (Note well: we do not assume 1 ∈ A.)

3. A ⊂ C(X, ℝ) is called a lattice if f ∨ g := max(f, g) and f ∧ g = min(f, g) ∈ A for all f, g ∈ A.

4. A ⊂ C(X, ℂ) is closed under conjugation if f̄ ∈ A whenever f ∈ A.

Remark 12.28. If X is a topological space such that C(X, ℝ) separates points then X is Hausdorff. Indeed if x, y ∈ X and f ∈ C(X, ℝ) such that f(x) ≠ f(y), then f^{−1}(J) and f^{−1}(I) are disjoint open sets containing x and y respectively when I and J are disjoint intervals containing f(x) and f(y) respectively.

Lemma 12.29. If A is a closed sub-algebra of BC(X, ℝ) then |f| ∈ A for all f ∈ A and A is a lattice.

Proof. Let f ∈ A and let M = sup_{x∈X} |f(x)| . Using Theorem 8.34 or Exercise

12.12, there are polynomials p_{n}(t) such that

lim_{n→∞} sup_{|t|≤M} ||t| − p_{n}(t)| = 0.

By replacing p_{n} by p_{n} − p_{n}(0) if necessary we may assume that p_{n}(0) = 0. Since A is an algebra, it follows that f_{n} = p_{n}(f) ∈ A and |f| ∈ A, because |f| is the uniform limit of the f_{n}’s. Since

f ∨ g = \frac{1}{2} (f + g + |f − g|) and
f ∧ g = \frac{1}{2} (f + g − |f − g|),

we have shown A is a lattice.

Lemma 12.30. Let A ⊂ C(X, ℝ) be an algebra which separates points and suppose x and y are distinct points of X. If there exits such that f, g ∈ A such that

f(x) ≠ 0 and g(y) ≠ 0,  (12.5)

then

V := {(f(x), f(y)) : f ∈ A}= ℝ^{2}.  (12.6)

Proof. It is clear that V is a non-zero subspace of ℝ^{2.} If dim(V) = 1, then V = span(a, b) for some (a, b) ∈ ℝ^{2} which, necessarily by Eq. (12.5), satisfy a ≠ 0 ≠ b. Since (a, b) = (f(x), f(y)) for some f ∈ A and f^{2} ∈ A, it follows that (a^{2}, b^{2}) = (f^{2}(x), f^{2}(y)) ∈ V as well. Since dim V = 1, (a, b) and (a^{2}, b^{2}) are linearly dependent and therefore

0 = det ( \begin{matrix}a & b \\ a^{2} & b^{2} \end{matrix} ) = ab^{2} − a^{2}b = ab(b − a)

which implies that a = b. But this the implies that f(x) = f(y) for all f ∈ A, violating the assumption that A separates points. Therefore we conclude that dim(V) = 2, i.e. V = ℝ^{2}.

Theorem 12.31 (Stone-Weierstrass Theorem). Suppose X is a locally compact Hausdorff space and A ⊂ C_{0}(X, ℝ) is a closed subalgebra which separates points. For x ∈ X let

A_{x} := {f(x) : f ∈ A} and
I_{x} = {f ∈ C_{0}(X, ℝ) : f(x) = 0}.

Then either one of the following two cases hold.

1. A = C_{0}(X, ℝ) or

2. there exists a unique point x_{0} ∈ X such that A = I_{x_{0}}.

Moreover, case 1. holds iff A_{x} = ℝ for all x ∈ X and case 2. holds iff there exists a point x_{0} ∈ X such that A_{x_{0}} = {0} .

Proof. If there exists x_{0} such that A_{x_{0}} = {0} (x_{0} is unique since A separates points) then A ⊂ I_{x_{0}}. If such an x_{0} exists let C = I_{x_{0}} and if A_{x} = ℝ for all x, set C = C_{0}(X, ℝ). Let f ∈ C be given. By Lemma 12.30, for all x, y ∈ X such that x ≠ y, there exists g_{xy} ∈ A such that f = g_{xy} on {x, y}.^{3} When X is compact the basic idea of the proof is contained in the following identity,

f(z) = inf_{x∈X} sup_{y∈X} g_{xy}(z) for all z ∈ X.  (12.7)

To prove this identity, let g_{x} := sup_{y∈X} g_{xy} and notice that g_{x} ≥ f since g_{xy}(y) = f(y) for all y ∈ X. Moreover, g_{x}(x) = f(x) for all x ∈ X since g_{xy}(x) = f(x) for all x. Therefore,

inf_{x∈X} sup_{y∈X} g_{xy} = inf_{x∈X} g_{x} = f.

The rest of the proof is devoted to replacing the inf and the sup above by min and max over finite sets at the expense of Eq. (12.7) becoming only an approximate identity. We also have to modify Eq. (12.7) slightly to take care of the non-compact case.

3 If A_{x_{0}} = {0} and x = x_{0} or y = x_{0}, then g_{xy} exists merely by the fact that A separates points.

Claim. Given ε > 0 and x ∈ X there exists g_{x} ∈ A such that g_{x}(x) = f(x) and f < g_{x} + ε on X.

To prove this, let V_{y} be an open neighborhood of y such that |f − g_{xy}| < ε on V_{y}; in particular f < ε + g_{xy} on V_{y}. Also let g_{x,∞} be any fixed element in A such that g_{x,∞} (x) = f (x) and let

K = \{|f| ≥ \frac{ε}{2}\} ∪ \{|g_{x,∞}| ≥ \frac{ε}{2}\} .  (12.8)

Since K is compact, there exists Λ ⊂⊂ K such that K ⊂ ⋃_{y∈Λ} V_{y}. Define

g_{x}(z) = max{g_{xy} : y ∈ Λ ∪ {∞}}.

Since

f < ε + g_{xy} < ε + g_{x} on V_{y},

for any y ∈ Λ, and

f < \frac{ε}{2} < ε + g_{x,∞} ≤ g_{x} + ε on K^{c},

f < ε + g_{x} on X and by construction f(x) = g_{x}(x), see Figure ??. This completes the proof of the claim.

Fig. 12.6. Constructing the “dominating approximates,” g_{x} for each x ∈ X.

To complete the proof of the theorem, let g_{∞} be a fixed element of A such that f < g_{∞} + ε on X; for example let g_{∞} = g_{x_{0}} ∈ A for some fixed x_{0} ∈ X. For each x ∈ X, let U_{x} be a neighborhood of x such that |f − g_{x}| < ε on U_{x}. Choose

Γ ⊂⊂ F := \{|f| ≥ \frac{ε}{2}\} ∪ \{|g_{∞}| ≥ \frac{ε}{2}\}

such that F ⊂ ⋃_{x∈Γ} U_{x} (Γ exists since F is compact) and define

g = min{g_{x} : x ∈ Γ ∪ {∞}} ∈ A.

Then, for x ∈ F, g_{x} < f + ε on U_{x} and hence g < f + ε on ⋃_{x∈Γ} U_{x} ⊃ F.

Likewise,

g ≤ g_{∞} < ε/2 < f + ε on F^{c}.

Therefore we have now shown,

f < g + ε and g < f + ε on X,

i.e. |f − g| < ε on X. Since ε > 0 is arbitrary it follows that f ∈ Ā = A and so A = C.

Corollary 12.32 (Complex Stone-Weierstrass Theorem). Let X be a locally compact Hausdorff space. Suppose A ⊂ C_{0}(X, ℂ) is closed in the uniform topology, separates points, and is closed under complex conjugation. Then either A = C_{0}(X, ℂ) or

A = I_{x_{0}}^{ℂ} := {f ∈ C_{0}(X, ℂ) : f(x_{0}) = 0}

for some x_{0} ∈ X.

Proof. Since

Re f = \frac{f + f̄}{2} and Im f = \frac{f − f̄}{2i},

Re f and Im f are both in A. Therefore

A_{ℝ} = {Re f, Im f : f ∈ A}

is a real sub-algebra of C_{0}(X, ℝ) which separates points. Therefore either A_{ℝ} = C_{0}(X, ℝ) or A_{ℝ} = I_{x_{0}} ∩ C_{0}(X, ℝ) for some x_{0} and hence A = C_{0}(X, ℂ) or I_{x_{0}}^{ℂ} respectively.

As an easy application, Theorem 12.31 and Corollary 12.32 imply Theorem 8.34 and Corollary 8.36 respectively. Here are a few more applications.

Example 12.33. Let f ∈ C([a, b]) be a positive function which is injective. Then functions of the form ∑_{k=1}^{N} a_{k}f^{k} with a_{k} ∈ ℂ and N ∈ ℕ are dense in C([a, b]). For example if a = 1 and b = 2, then one may take f(x) = x^{α} for any α ≠ 0, or f(x) = e^{x}, etc.

Exercise 12.4. Let (X, d) be a separable compact metric space. Show that C(X) is also separable. Hint: Let E ⊂ X be a countable dense set and then consider the algebra, A ⊂ C(X), generated by {d(x, ·)}_{x∈E} .

Example 12.34. Let X = [0, ∞), λ > 0 be fixed, A be the real algebra generated by t → e^{−λt}. So the general element f ∈ A is of the form f(t) = p(e^{−λt}), where p(x) is a polynomial function in x with real coefficients. Since A ⊂ C_{0}(X, ℝ) separates points and e^{−λt} ∈ A is pointwise positive, Ā = C_{0}(X, ℝ).

As an application of Example 12.34, suppose that g ∈ C_{c} (X, ℝ) satisfies,

∫_{0}^{∞} g (t) e^{−λt}dt = 0 for all λ > 0.  (12.9)

(Note well that the integral in Eq. (12.9) is really over a finite interval since g is compactly supported.) Equation (12.9) along with linearity of the Riemann integral implies

∫_{0}^{∞} g (t) f (t) dt = 0 for all f ∈ A.

We may now choose f_{n} ∈ A such that f_{n} → g uniformly and therefore, using the continuity of the Riemann integral under uniform convergence (see Proposition 8.5),

0 = lim_{n→∞} ∫_{0}^{∞} g (t) f_{n} (t) dt = ∫_{0}^{∞} g^{2} (t) dt.

From this last equation it is easily deduced, using the continuity of g, that g ≡ 0. See Theorem 22.12 below, where this is done in greater generality.

12.5 *More on Separation Axioms: Normal Spaces

(This section may safely be omitted on the first reading.)

Definition 12.35 (T_{0} – T_{2} Separation Axioms). Let (X, τ) be a topological space. The topology τ is said to be:

1. T_{0} if for x ≠ y in X there exists V ∈ τ such that x ∈ V and y ∉ V or V such that y ∈ V but x ∉ V.

2. T_{1} if for every x, y ∈ X with x ≠ y there exists V ∈ τ such that x ∈ V and y ∉ V. Equivalently, τ is T_{1} iff all one point subsets of X are closed.^{4}

3. T_{2} if it is Hausdorff.

Note T_{2} implies T_{1} which implies T_{0}. The topology in Example 12.1 is T_{0} but not T_{1}. If X is a finite set and τ is a T_{1} – topology on X then τ = 2^{X}. To prove this let x ∈ X be fixed. Then for every y ≠ x in X there exists V_{y} ∈ τ such that x ∈ V_{y} while y ∉ V_{y}. Thus {x} = ∩_{y≠x}V_{y} ∈ τ showing τ contains all one point subsets of X and therefore all subsets of X. So we have to look to infinite sets for an example of T_{1} topology which is not T_{2}.

4 If one point subsets are closed and x ≠ y in X then V := {x}^{c} is an open set containing y but not x. Conversely if τ is T_{1} and x ∈ X there exists V_{y} ∈ τ such that y ∈ V_{y} and x ∉ V_{y} for all y ≠ x. Therefore, {x}^{c} = ∪_{y≠x}V_{y} ∈ τ.

Example 12.36. Let X be any infinite set and let τ = {A ⊂ X : #(A^{c}) < ∞}∪ {∅} – the so called cofinite topology. This topology is T_{1} because if x ≠ y in X, then V = {x}^{c} ∈ τ with x ∉ V while y ∈ V. This topology however is not T_{2}. Indeed if U, V ∈ τ are open sets such that x ∈ U, y ∈ V and U ∩ V = ∅ then U ⊂ V^{c}. But this implies #(U) < ∞ which is impossible unless U = ∅ which is impossible since x ∈ U.

The uniqueness of limits of sequences which occurs for Hausdorff topologies (see Remark 12.3) need not occur for T_{1} – spaces. For example, let X = ℕ and τ be the cofinite topology on X as in Example 12.36. Then x_{n} = n is a sequence in X such that x_{n} → x as n → ∞ for all x ∈ ℕ. For the most part we will avoid these pathologies in the future by only considering Hausdorff topologies.

Definition 12.37 (Normal Spaces: T_{4} – Separation Axiom). A topological space (X, τ) is said to be normal or T_{4} if:

1. X is Hausdorff and

2. if for any two closed disjoint subsets A, B ⊂ X there exists disjoint open sets V, W ⊂ X such that A ⊂ V and B ⊂ W.

Example 12.38. By Lemma 6.15 and Corollary 12.21 it follows that metric spaces and topological spaces which are locally compact, σ – compact and Hausdorff (in particular compact Hausdorff spaces) are normal. Indeed, in each case if A, B are disjoint closed subsets of X, there exists f ∈ C(X, [0, 1]) such that f = 1 on A and f = 0 on B. Now let U = \{f > \frac{1}{2}\} and V = \{f < \frac{1}{2}\}.

Remark 12.39. A topological space, (X, τ), is normal iff for any C ⊂ W ⊂ X with C being closed and W being open there exists an open set U ⊂_{o} X such that

C ⊂ U ⊂ Ū ⊂ W.

To prove this first suppose X is normal. Since W^{c} is closed and C ∩ W^{c} = ∅, there exists disjoint open sets U and V such that C ⊂ U and W^{c} ⊂ V. Therefore C ⊂ U ⊂ V^{c} ⊂ W and since V^{c} is closed, C ⊂ U ⊂ Ū ⊂ V^{c} ⊂ W.

For the converse direction suppose A and B are disjoint closed subsets of X. Then A ⊂ B^{c} and B^{c} is open, and so by assumption there exists U ⊂_{o} X such that A ⊂ U ⊂ Ū ⊂ B^{c} and by the same token there exists W ⊂_{o} X such that Ū ⊂ W ⊂ W̄ ⊂ B^{c}. Taking complements of the last expression implies

B ⊂ W̄^{c} ⊂ W^{c} ⊂ Ū^{c}.

Let V = W̄^{c}. Then A ⊂ U ⊂_{o} X, B ⊂ V ⊂_{o} X and U ∩ V ⊂ U ∩ W^{c} = ∅.

Theorem 12.40 (Urysohn’s Lemma for Normal Spaces). Let X be a normal space. Assume A, B are disjoint closed subsets of X. Then there exists f ∈ C(X, [0, 1]) such that f = 0 on A and f = 1 on B.

Proof. To make the notation match Lemma 12.8, let U = A^{c} and K = B. Then K ⊂ U and it suffices to produce a function f ∈ C(X, [0, 1]) such that f = 1 on K and supp(f) ⊂ U. The proof is now identical to that for Lemma 12.8 except we now use Remark 12.39 in place of Proposition 12.7.

Theorem 12.41 (Tietze Extension Theorem). Let (X, τ) be a normal space, D be a closed subset of X, −∞ < a < b < ∞ and f ∈ C(D, [a, b]). Then there exists F ∈ C(X, [a, b]) such that F |_{D} = f.

Proof. The proof is identical to that of Theorem 7.4 except we now use Theorem 12.40 in place of Lemma 6.15.

Corollary 12.42. Suppose that X is a normal topological space, D ⊂ X is closed, F ∈ C(D, ℝ). Then there exists F ∈ C(X) such that F |_{D} = f.

Proof. Let g = arctan(f) ∈ C(D, (−\frac{π}{2} , \frac{π}{2} )). Then by the Tietze extension theorem, there exists G ∈ C(X, [−\frac{π}{2} , \frac{π}{2} ]) such that G|_{D} = g. Let B := G^{−1}(\{−\frac{π}{2} , \frac{π}{2} \}) ⊏ X, then B ∩ D = ∅. By Urysohn’s lemma (Theorem 12.40) there exists h ∈ C(X, [0, 1]) such that h ≡ 1 on D and h = 0 on B and in particular hG ∈ C(D, (−\frac{π}{2} , \frac{π}{2} )) and (hG) |_{D} = g. The function F := tan(hG) ∈ C(X) is an extension of f.

Theorem 12.43 (Urysohn Metrization Theorem for Normal Spaces). Every second countable normal space, (X, τ) , is metrizable, i.e. there is a metric ρ on X such that τ = τ_{ρ}. Moreover, ρ may be chosen so that X is isometric to a subset Q_{0} ⊂ Q (Q is as in Notation 12.10) equipped with the metric d in Eq. (12.2). In this metric X is totally bounded and hence the completion of X (which is isometric to Q̄_{0} ⊂ Q) is compact.

Proof. (The proof here will be very similar to the proof of Theorem 12.13.) Let B be a countable base for τ and set

Γ := {(U, V) ∈ B × B | Ū ⊂ V }.

To each O ∈ τ and x ∈ O there exist (U, V) ∈ Γ such that x ∈ U ⊂ V ⊂ O. Indeed, since B is a base for τ, there exists V ∈ B such that x ∈ V ⊂ O. Because {x}∩V^{c} = ∅, there exists disjoint open sets Ũ and W such that x ∈ Ũ, V^{c} ⊂ W and Ũ ∩ W = ∅. Choose U ∈ B such that x ∈ U ⊂ Ũ. Since U ⊂ Ũ ⊂ W^{c}, \overline{U} ⊂ W^{c} ⊂ V and hence (U, V) ∈ Γ. See Figure 12.7 below. In particular this shows that

B_{0} := {U ∈ B : (U, V) ∈ Γ for some V ∈ B}

is still a base for τ.

If Γ is a finite set, the previous comment shows that τ only has a finite number of elements as well. Since (X, τ) is Hausdorff, it follows that X is a finite set. Letting {x_{n}}_{n=1}^{N} be an enumeration of X, define T : X → Q by

Fig. 12.7. Constructing (U, V) ∈ Γ.

T (x_{n}) = e_{n} for n = 1, 2, …, N where e_{n} = (0, 0, …, 0, 1, 0, …), with the 1 occurring in the n^{th} spot. Then ρ(x, y) := d(T (x), T (y)) for x, y ∈ X is the desired metric.

So we may now assume that Γ is an infinite set and let {(U_{n}, V_{n})}_{n=1}^{∞} be an enumeration of Γ. By Urysohn’s Lemma for normal spaces (Theorem 12.40) there exists f_{U,V} ∈ C(X, [0, 1]) such that f_{U,V} = 0 on Ū and f_{U,V} = 1 on V^{c}. Let F := {f_{U,V} | (U, V) ∈ Γ } and set f_{n} := f_{U_{n},V_{n}} – an enumeration of F. The proof that

ρ(x, y) := ∑_{n=1}^{∞} \frac{1}{2^{n}} |f_{n}(x) − f_{n}(y)|

is the desired metric on X now follows exactly as the corresponding argument in the proof of Theorem 12.13.

12.6 Exercises

Exercise 12.5. Prove Theorem 12.9. Hints:

1. By Proposition 12.7, there exists a precompact open set V such that K ⊂ V ⊂ V̄ ⊂ U. Now suppose that f : K → [0, α] is continuous with α ∈ (0, 1] and let A := f^{−1}([0, \frac{1}{3}α]) and B := f^{−1}([\frac{2}{3}α, 1]). Appeal to Lemma 12.8 to find a function g ∈ C(X, [0, α/3]) such that g = α/3 on B and supp(g) ⊂ V \ A.

2. Now follow the argument in the proof of Theorem 7.4 to construct F ∈ C(X, [a, b]) such that F |_{K} = f.

3. For c ∈ [a, b], choose ϕ ≺ U such that ϕ = 1 on K and replace F by F_{c} := ϕF + (1 − ϕ)c.

Exercise 12.6 (Sterographic Projection). Let X = ℝ^{n}, X^{∗} := X ∪ {∞} be the one point compactification of X, S^{n} := {y ∈ ℝ^{n+1} : |y| = 1} be the unit sphere in ℝ^{n+1} and N = (0, …, 0, 1) ∈ ℝ^{n+1}. Define f : S^{n} → X^{∗} by f(N) = ∞, and for y ∈ S^{n} \ {N} let f(y) = b ∈ ℝ^{n} be the unique point such that (b, 0) is on the line containing N and y, see Figure 12.8 below. Find a formula for f and show f : S^{n} → X^{∗} is a homeomorphism. (So the one point compactification of ℝ^{n} is homeomorphic to the n sphere.)

Fig. 12.8. Sterographic projection and the one point compactification of ℝ^{n}.

Exercise 12.7. Let (X, τ) be a locally compact Hausdorff space. Show (X, τ) is separable iff (X^{∗}, τ^{∗}) is separable.

Exercise 12.8. Show by example that there exists a locally compact metric space (X, d) such that the one point compactification, (X^{∗} := X ∪ {∞} , τ^{∗}) , is not metrizable. Hint: use exercise 12.7.

Exercise 12.9. Suppose (X, d) is a locally compact and σ – compact metric space. Show the one point compactification, (X^{∗} := X ∪ {∞} , τ^{∗}) , is metrizable.

Exercise 12.10. In this problem, suppose Theorem 12.31 has only been proved when X is compact. Show that it is possible to prove Theorem 12.31 by using Proposition 12.24 to reduce the non-compact case to the compact case.

Hints.

Exercise 12.11. 1. If A_{x} = ℝ for all x ∈ X let X^{∗} = X ∪ {∞} be the one

point compactification of X.

2. If A_{x_{0}} = {0} for some x_{0} ∈ X, let Y := X \ {x_{0}} and Y^{∗} = Y ∪ {∞} be the one point compactification of Y. For f ∈ A define f (∞) = 0. In this way A may be considered to be a

sub-algebra of C(X^{∗}, ℝ) in case 1. or a sub-algebra of C(Y^{∗}, ℝ) in case 2.

Exercise 12.12. Let M < ∞, show there are polynomials p_{n}(t) such that

lim_{n→∞} sup_{|t|≤M} ||t| − p_{n}(t)| = 0

using the following outline.

1. Let f(x) = \sqrt{1 − x} for |x| ≤ 1 and use Taylor’s theorem with integral remainder (see Eq. A.15 of Appendix A), or analytic function theory if you know it, to show there are constants^{5} c_{n} > 0 for n ∈ ℕ such that

\sqrt{1 − x} = 1 − ∑_{n=1}^{∞} c_{n}x^{n} for all |x| < 1.  (12.10)

2. Let q_{m}(x) := 1 − ∑_{n=1}^{m} c_{n}x^{n}. Use (12.10) to show ∑_{n=1}^{∞} c_{n} = 1 and conclude from this that

lim_{m→∞} sup_{|x|≤1} |\sqrt{1 − x} − q_{m}(x)| = 0.  (12.11)

3. Let 1 − x = t^{2}/M^{2}, i.e. x = 1 − t^{2}/M^{2}, then

lim_{m→∞} sup_{|t|≤M} |\frac{|t|}{M} − q_{m}(1 − t^{2}/M^{2})| = 0

so that p_{m}(t) := Mq_{m}(1 − t^{2}/M^{2}) are the desired polynomials.

Exercise 12.13. Given a continuous function f : ℝ → ℂ which is 2π - periodic and ε > 0. Show there exists a trigonometric polynomial, p(θ) =

∑_{n=−N}^{n} α_{n}e^{inθ}, such that |f(θ) − P (θ)| < ε for all θ ∈ ℝ. Hint: show that

there exists a unique function F ∈ C(S^{1}) such that f(θ) = F (e^{iθ}) for all θ ∈ ℝ.

Remark 12.44. Exercise 12.13 generalizes to 2π – periodic functions on ℝ^{d}, i.e. functions such that f(θ + 2πe_{i}) = f(θ) for all i = 1, 2, …, d where {e_{i}}_{i=1}^{d} is the standard basis for ℝ^{d}. A trigonometric polynomial p(θ) is a function of θ ∈ ℝ^{d} of the form

p(θ) = ∑_{n∈Γ} α_{n}e^{in·θ}

where Γ is a finite subset of ℤ^{d}. The assertion is again that these trigonometric polynomials are dense in the 2π – periodic functions relative to the supremum norm.

5 In fact c_{n} := \frac{(2n−3)!!}{2^{n}n!} , but this is not needed.

Baire Category Theorem

Definition 13.1. Let (X, τ) be a topological space. A set E ⊂ X is said to be nowhere dense if \binom{}{Ē}^{o} = ∅ i.e. Ē has empty interior.

Notice that E is nowhere dense is equivalent to

X = (\binom{}{Ē}^{o})^{c} = \overline{\binom{}{Ē}^{c}} = \overline{(E^{c})^{o}}.

That is to say E is nowhere dense iff E^{c} has dense interior.

13.1 Metric Space Baire Category Theorem

Theorem 13.2 (Baire Category Theorem). Let (X, ρ) be a complete metric space.

1. If {V_{n}}_{n=1}^{∞} is a sequence of dense open sets, then G := ⋂_{n=1}^{∞} V_{n} is dense in

X.

2. If {E_{n}}_{n=1}^{∞} is a sequence of nowhere dense sets, then ⋃_{n=1}^{∞} E_{n} ⊂ ⋃_{n=1}^{∞} Ē_{n} ⫋ X and in particular X ≠ ⋃_{n=1}^{∞} E_{n}.

Proof. 1) We must shows that Ḡ = X which is equivalent to showing that W ∩ G ≠ ∅ for all non-empty open sets W ⊂ X. Since V_{1} is dense, W ∩ V_{1} ≠ ∅ and hence there exists x_{1} ∈ X and ε_{1} > 0 such that

\overline{B(x_{1}, ε_{1})} ⊂ W ∩ V_{1}.

Since V_{2} is dense, B(x_{1}, ε_{1})∩V_{2} ≠ ∅ and hence there exists x_{2} ∈ X and ε_{2} > 0 such that

\overline{B(x_{2}, ε_{2})} ⊂ B(x_{1}, ε_{1}) ∩ V_{2}.

Continuing this way inductively, we may choose {x_{n} ∈ X and ε_{n} > 0}_{n=1}^{∞} such that

\overline{B(x_{n}, ε_{n})} ⊂ B(x_{n−1}, ε_{n−1}) ∩ V_{n} ∀n.

Furthermore we can clearly do this construction in such a way that ε_{n} ↓ 0 as n ↑ ∞. Hence {x_{n}}_{n=1}^{∞} is Cauchy sequence and x = lim_{n→∞} x_{n} exists in X

since X is complete. Since \overline{B(x_{n}, ε_{n})} is closed, x ∈ \overline{B(x_{n}, ε_{n})} ⊂ V_{n} so that x ∈ V_{n} for all n and hence x ∈ G. Moreover, x ∈ \overline{B(x_{1}, ε_{1})} ⊂ W ∩ V_{1} implies x ∈ W and hence x ∈ W ∩ G showing W ∩ G ≠ ∅. 2) The second assertion is equivalently to showing

∅ ≠ ( ⋃_{n=1}^{∞} Ē_{n})^{c} = ⋂_{n=1}^{∞} \binom{}{Ē_{n}}^{c} = ⋂_{n=1}^{∞} (E_{n}^{c})^{o} .

As we have observed, E_{n} is nowhere dense is equivalent to (E_{n}^{c})^{o} being a dense open set, hence by part 1), ⋂_{n=1}^{∞} (E_{n}^{c})^{o} is dense in X and hence not empty.

13.2 Locally Compact Hausdorff Space Baire Category Theorem

Here is another version of the Baire Category theorem when X is a locally compact Hausdorff space.

Proposition 13.3. Let X be a locally compact Hausdorff space.

1. If {V_{n}}_{n=1}^{∞} is a sequence of dense open sets, then G := ⋂_{n=1}^{∞} V_{n} is dense in

X.

2. If {E_{n}}_{n=1}^{∞} is a sequence of nowhere dense sets, then X ≠ ⋃_{n=1}^{∞} E_{n}.

Proof. As in the previous proof, the second assertion is a consequence of the first. To finish the proof, if suffices to show G ∩ W ≠ ∅ for all open sets W ⊂ X. Since V_{1} is dense, there exists x_{1} ∈ V_{1} ∩ W and by Proposition 12.7 there exists U_{1} ⊂_{o} X such that x_{1} ∈ U_{1} ⊂ Ū_{1} ⊂ V_{1}∩W with Ū_{1} being compact. Similarly, there exists a non-empty open set U_{2} such that U_{2} ⊂ Ū_{2} ⊂ U_{1} ∩ V_{2}. Working inductively, we may find non-empty open sets {U_{k}}_{k=1}^{∞} such that U_{k} ⊂ Ū_{k} ⊂ U_{k−1} ∩ V_{k}. Since ∩_{k=1}^{n}Ū_{k} = Ū_{n} ≠ ∅ for all n, the finite intersection characterization of Ū_{1} being compact implies that

∅ ≠ ∩_{k=1}^{∞}Ū_{k} ⊂ G ∩ W.

Definition 13.4. A subset E ⊂ X is meager or of the first category if

E = ⋃_{n=1}^{∞} E_{n} where each E_{n} is nowhere dense. And a set R ⊂ X is called

residual if R^{c} is meager.

Remarks 13.5 For those readers that already know some measure theory may want to think of meager as being the topological analogue of sets of measure 0 and residual as being the topological analogue of sets of full measure. (This analogy should not be taken too seriously, see Exercise 19.19.)

1. R is residual iff R contains a countable intersection of dense open sets. Indeed if R is a residual set, then there exists nowhere dense sets {E_{n}} such that

R^{c} = ∪_{n=1}^{∞}E_{n} ⊂ ∪_{n=1}^{∞}Ē_{n}.

Taking complements of this equation shows that

∩_{n=1}^{∞}Ē_{n}^{c} ⊂ R,

i.e. R contains a set of the form ∩_{n=1}^{∞}V_{n} with each V_{n} (= Ē_{n}^{c}) being an open dense subset of X. Conversely, if ∩_{n=1}^{∞}V_{n} ⊂ R with each V_{n} being an open dense subset of X, then R^{c} ⊂ ∪_{n=1}^{∞}V_{n}^{c} and hence R^{c} = ∪_{n=1}^{∞}E_{n} where each E_{n} = R^{c} ∩ V_{n}^{c}, is a nowhere dense subset of X.

2. A countable union of meager sets is meager and any subset of a meager set is meager.

3. A countable intersection of residual sets is residual.

Remarks 13.6 The Baire Category Theorems may now be stated as follows. If X is a complete metric space or X is a locally compact Hausdorff space, then

1. all residual sets are dense in X and

2. X is not meager.

It should also be remarked that incomplete metric spaces may be meager. For example, let X ⊂ C([0, 1]) be the subspace of polynomial functions on [0, 1] equipped with the supremum norm. Then X = ∪_{n=1}^{∞}E_{n} where E_{n} ⊂ X denotes the subspace of polynomials of degree less than or equal to n. You are asked to show in Exercise 13.1 below that E_{n} is nowhere dense for all n. Hence X is meager and the empty set is residual in X.

Here is an application of Theorem 13.2.

Theorem 13.7. Let N ⊂ C([0, 1], ℝ) be the set of nowhere differentiable functions. (Here a function f is said to be differentiable at 0 if f′(0) := lim_{t↓0} \frac{f(t)−f(0)}{t} exists and at 1 if f′(1) := lim_{t↑0} \frac{f(1)−f(t)}{1−t} exists.) Then N is a residual set so the “generic” continuous functions is nowhere differentiable.

Proof. If f ∉ N , then f′(x_{0}) exists for some x_{0} ∈ [0, 1] and by the definition of the derivative and compactness of [0, 1], there exists n ∈ ℕ such that |f(x) − f(x_{0})| ≤ n|x − x_{0}| ∀ x ∈ [0, 1]. Thus if we define

E_{n} := {f ∈ C([0, 1]) : ∃ x_{0} ∈ [0, 1] ∋ |f(x) − f(x_{0})| ≤ n|x − x_{0}| ∀ x ∈ [0, 1]} , then we have just shown N^{c} ⊂ E := ∪_{n=1}^{∞}E_{n}. So to finish the proof it suffices to show (for each n) E_{n} is a closed subset of C([0, 1], ℝ) with empty interior.

1) To prove E_{n} is closed, let {f_{m}}_{m=1}^{∞} ⊂ E_{n} be a sequence of functions such that there exists f ∈ C([0, 1], ℝ) such that ∥f − f_{m}∥_{∞} → 0 as m → ∞. Since f_{m} ∈ E_{n}, there exists x_{m} ∈ [0, 1] such that

|f_{m}(x) − f_{m}(x_{m})| ≤ n|x − x_{m}| ∀ x ∈ [0, 1].  (13.1)

Since [0, 1] is a compact metric space, by passing to a subsequence if necessary, we may assume x_{0} = lim_{m→∞} x_{m} ∈ [0, 1] exists. Passing to the limit in Eq. (13.1), making use of the uniform convergence of f_{n} → f to show lim_{m→∞} f_{m}(x_{m}) = f(x_{0}), implies

|f(x) − f(x_{0})| ≤ n|x − x_{0}| ∀ x ∈ [0, 1]

and therefore that f ∈ E_{n}. This shows E_{n} is a closed subset of C([0, 1], ℝ).

2) To finish the proof, we will show E_{n}^{0} = ∅ by showing for each f ∈ E_{n} and ε > 0 given, there exists g ∈ C([0, 1], ℝ)\E_{n} such that ∥f − g∥_{∞} < ε. We now construct g. Since [0, 1] is compact and f is continuous there exists N ∈ ℕ such that |f(x) − f(y)| < ε/2 whenever |y − x| < 1/N. Let k denote the piecewise linear function on [0, 1] such that k(\frac{m}{N}) = f(\frac{m}{N}) for m = 0, 1, …, N and k′′(x) = 0 for x ∉ π_{N} := {m/N : m = 0, 1, …, N} . Then it is easily seen that ∥f − k∥_{u} < ε/2 and for x ∈ (\frac{m}{N}, \frac{m+1}{N}) that

|k′(x)| = \frac{|f(\frac{m+1}{N}) − f(\frac{m}{N})|}{\frac{1}{N}} < Nε/2.

We now make k “rougher” by adding a small wiggly function h which we define as follows. Let M ∈ ℕ be chosen so that 4εM > 2n and define h uniquely by h(\frac{m}{M}) = (−1)^{m}ε/2 for m = 0, 1, …, M and h′′(x) = 0 for x ∉ π_{M}. Then ∥h∥_{∞} < ε and |h′(x)| = 4εM > 2n for x ∉ π_{M}. See Figure 13.1 below. Finally define g := k + h. Then

∥f − g∥_{∞} ≤ ∥f − k∥_{∞} + ∥h∥_{∞} < ε/2 + ε/2 = ε

and

|g′(x)| ≥ |h′(x)| − |k′ (x)| > 2n − n = n ∀x ∉ π_{M} ∪ π_{N}.

It now follows from this last equation and the mean value theorem that for any x_{0} ∈ [0, 1],

|\frac{g(x) − g(x_{0})}{x − x_{0}} | > n

for all x ∈ [0, 1] sufficiently close to x_{0}. This shows g ∉ E_{n} and so the proof is complete.

Here is an application of the Baire Category Theorem in Proposition 13.3.

Fig. 13.1. Constgructing a rough approximation, g, to a continuous function f.

Proposition 13.8. Suppose that f : ℝ → ℝ is a function such that f′(x) exists for all x ∈ ℝ. Let

U := ∪_{ε>0} {x ∈ ℝ : sup_{|y|<ε} |f′(x + y)| < ∞} .

Then U is a dense open set. (It is not true that U = ℝ in general, see Example ?? below.)

Proof. It is easily seen from the definition of U that U is open. Let W ⊂_{o} ℝ be an open subset of ℝ. For k ∈ ℕ, let

E_{k} := \{x ∈ W : |f(y) − f(x)| ≤ k |y − x| when |y − x| ≤ \frac{1}{k} \}

= ⋂_{z:|z|≤k^{−1}} {x ∈ W : |f(x + z) − f(x)| ≤ k |z|} ,

which is a closed subset of ℝ since f is continuous. Moreover, if x ∈ W and M = |f′(x)| , then

|f(y) − f(x)| = |f′(x) (y − x) + o (y − x)|
≤ (M + 1) |y − x|

for y close to x. (Here o(y − x) denotes a function such that lim_{y→x} o(y − x)/(y − x) = 0.) In particular, this shows that x ∈ E_{k} for all k sufficiently large. Therefore W = ∪_{k=1}^{∞}E_{k} and since W is not meager by the Baire category Theorem in Proposition 13.3, some E_{k} has non-empty interior. That is there exists x_{0} ∈ E_{k} ⊂ W and ε > 0 such that

J := (x_{0} − ε, x_{0} + ε) ⊂ E_{k} ⊂ W.

For x ∈ J, we have |f(x + z) − f(x)| ≤ k |z| provided that |z| ≤ k^{−1} and therefore that |f′(x)| ≤ k for x ∈ J. Therefore x_{0} ∈ U ∩ W showing U is dense.

Remark 13.9. This proposition generalizes to functions f : ℝ^{n} → ℝ^{m} in an obvious way.

For our next application of Theorem 13.2, let X := BC^{∞} ((−1, 1)) denote the set of smooth functions f on (−1, 1) such that f and all of its derivatives are bounded. In the metric

ρ(f, g) := ∑_{k=0}^{∞} 2^{−k} \frac{∥f^{(k)} − g^{(k)}∥_{∞}}{1 + ∥f^{(k)} − g^{(k)}∥_{∞}} for f, g ∈ X,

X becomes a complete metric space.

Theorem 13.10. Given an increasing sequence of positive numbers {M_{n}}_{n=1}^{∞} , the set

F := \{f ∈ X : lim sup_{n→∞} |\frac{f^{(n)}(0)}{M_{n}} | ≥ 1\}

is dense in X. In particular, there is a dense set of f ∈ X such that the power series expansion of f at 0 has zero radius of convergence.

Proof. Step 1. Let n ∈ ℕ. Choose g ∈ C_{c}^{∞}((−1, 1)) such that ∥g∥_{∞} < 2^{−n} while g′(0) = 2M_{n} and define

f_{n}(x) := ∫_{0}^{x} dt_{n−1} ∫_{0}^{t_{n−1}} dt_{n−2} … ∫_{0}^{t_{2}} dt_{1}g(t_{1}).

Then for k < n,

f_{n}^{(k)}(x) = ∫_{0}^{x} dt_{n−k−1} ∫_{0}^{t_{n−k−1}} dt_{n−k−2} … ∫_{0}^{t_{2}} dt_{1}g(t_{1}),

f^{(n)}(x) = g′(x), f_{n}^{(n)}(0) = 2M_{n} and f_{n}^{(k)} satisfies

∥f_{n}^{(k)}∥_{∞} ≤ \frac{2^{−n}}{(n − 1 − k)!} ≤ 2^{−n} for k < n.

Consequently,

ρ(f_{n}, 0) = ∑_{k=0}^{∞} 2^{−k} \frac{∥f_{n}^{(k)}∥_{∞}}{1 + ∥f_{n}^{(k)}∥_{∞}}

≤ ∑_{k=0}^{n−1} 2^{−k}2^{−n} + ∑_{k=n}^{∞} 2^{−k} · 1 ≤ 2 (2^{−n} + 2^{−n}) = 4 · 2^{−n}.

Thus we have constructed f_{n} ∈ X such that lim_{n→∞} ρ(f_{n}, 0) = 0 while f_{n}^{(n)}(0) = 2M_{n} for all n. Step 2. The set

G_{n} := ∪_{m≥n} {f ∈ X : |f^{(m)}(0)| > M_{m}}

is a dense open subset of X. The fact that G_{n} is open is clear. To see that G_{n} is dense, let g ∈ X be given and define g_{m} := g + ε_{m}f_{m} where ε_{m} := sgn(g^{(m)}(0)). Then

|g_{m}^{(m)}(0)| = |g^{(m)}(0)| + |f_{m}^{(m)}(0)| ≥ 2M_{m} > M_{m} for all m.

Therefore, g_{m} ∈ G_{n} for all m ≥ n and since

ρ(g_{m}, g) = ρ(f_{m}, 0) → 0 as m → ∞

it follows that g ∈ Ḡ_{n}. Step 3. By the Baire Category theorem, ∩G_{n} is a dense subset of X. This completes the proof of the first assertion since

F = \{f ∈ X : lim sup_{n→∞} |\frac{f^{(n)}(0)}{M_{n}} | ≥ 1\}

= ∩_{n=1}^{∞} \{f ∈ X : |\frac{f^{(n)}(0)}{M_{n}} | ≥ 1 for some n ≥ m\} ⊃ ∩_{n=1}^{∞}G_{n}.

Step 4. Take M_{n} = (n!)^{2} and recall that the power series expansion for f near 0 is given by ∑_{n=0}^{∞} \frac{f_{n}(0)}{n!}x^{n}. This series can not converge for any f ∈ F and any x ≠ 0 because

lim sup_{n→∞} |\frac{f_{n}(0)}{n!}x^{n}| = lim sup_{n→∞} |\frac{f_{n}(0)}{(n!)^{2}}n!x^{n}|

= lim sup_{n→∞} |\frac{f_{n}(0)}{(n!)^{2}} | · lim_{n→∞} n! |x^{n}| = ∞

where we have used lim_{n→∞} n! |x^{n}| = ∞ and lim sup_{n→∞} |\frac{f_{n}(0)}{(n!)^{2}} | ≥ 1.

Remark 13.11. Given a sequence of real number {a_{n}}_{n=0}^{∞} there always exists f ∈ X such that f^{(n)}(0) = a_{n}. To construct such a function f, let ϕ ∈ C_{c}^{∞}(−1, 1) be a function such that ϕ = 1 in a neighborhood of 0 and ε_{n} ∈ (0, 1) be chosen so that ε_{n} ↓ 0 as n → ∞ and ∑_{n=0}^{∞} |a_{n}| ε_{n}^{n} < ∞. The desired function f can then be defined by

f(x) = ∑_{n=0}^{∞} \frac{a_{n}}{n!} x^{n}ϕ(x/ε_{n}) =: ∑_{n=0}^{∞} g_{n}(x).  (13.2)

The fact that f is well defined and continuous follows from the estimate:

|g_{n}(x)| = |\frac{a_{n}}{n!} x^{n}ϕ(x/ε_{n})| ≤ \frac{∥ϕ∥_{∞}}{n!} |a_{n}| ε_{n}^{n}

and the assumption that ∑_{n=0}^{∞} |a_{n}| ε_{n}^{n} < ∞. The estimate

|g′_{n}(x)| = |\frac{a_{n}}{(n − 1)!}x^{n−1}ϕ(x/ε_{n}) + \frac{a_{n}}{n!ε_{n}} x^{n}ϕ′(x/ε_{n})|

≤ \frac{∥ϕ∥_{∞}}{(n − 1)!} |a_{n}| ε_{n}^{n−1} + \frac{∥ϕ′∥_{∞}}{n!} |a_{n}| ε_{n}^{n}

≤ (∥ϕ∥_{∞} + ∥ϕ′∥_{∞}) |a_{n}| ε_{n}^{n}

and the assumption that ∑_{n=0}^{∞} |a_{n}| ε_{n}^{n} < ∞ shows f ∈ C^{1}(−1, 1) and f′(x) = ∑_{n=0}^{∞} g′_{n}(x). Similar arguments show f ∈ C_{c}^{k}(−1, 1) and f^{(k)}(x) = ∑_{n=0}^{∞} g_{n}^{(k)}(x) for all x and k ∈ ℕ. This completes the proof since, using ϕ(x/ε_{n}) = 1 for x in a neighborhood of 0, g_{n}^{(k)}(0) = δ_{k,n}a_{k} and hence

f^{(k)}(0) = ∑_{n=0}^{∞} g_{n}^{(k)}(0) = a_{k}.

13.3 Exercises

Exercise 13.1. Let (X, ∥·∥) be an infinite dimensional normed space and E ⊂ X be a finite dimensional subspace. Show that E ⊂ X is nowhere dense.

Exercise 13.2. Now suppose that (X, ∥·∥) is an infinite dimensional Banach space. Show that X can not have a countable algebraic basis. More explicitly, there is no countable subset S ⊂ X such that every element x ∈ X may be written as a finite linear combination of elements from S. Hint: make use of Exercise 13.1 and the Baire category theorem.

Hilbert Space Basics

(BRUCE: Perhaps this should be move to between Chapters 7 & 8?)

Definition 14.1. Let H be a complex vector space. An inner product on H is a function, ⟨·|·⟩ : H × H → ℂ, such that

1. ⟨ax + by|z⟩ = a⟨x|z⟩ + b⟨y|z⟩ i.e. x → ⟨x|z⟩ is linear.

2. \overline{⟨x|y⟩} = ⟨y|x⟩.

3. ∥x∥^{2} := ⟨x|x⟩ ≥ 0 with equality ∥x∥^{2} = 0 iff x = 0.

Notice that combining properties (1) and (2) that x → ⟨z|x⟩ is anti-linear for fixed z ∈ H, i.e.

⟨z|ax + by⟩ = ā⟨z|x⟩ + b̄⟨z|y⟩.

The following identity will be used frequently in the sequel without further mention,

∥x + y∥^{2} = ⟨x + y|x + y⟩ = ∥x∥^{2} + ∥y∥^{2} + ⟨x|y⟩ + ⟨y|x⟩
= ∥x∥^{2} + ∥y∥^{2} + 2Re⟨x|y⟩.  (14.1)

Theorem 14.2 (Schwarz Inequality). Let (H, ⟨·|·⟩) be an inner product space, then for all x, y ∈ H

|⟨x|y⟩| ≤ ∥x∥∥y∥

and equality holds iff x and y are linearly dependent.

Proof. If y = 0, the result holds trivially. So assume that y ≠ 0 and observe; if x = αy for some α ∈ ℂ, then ⟨x|y⟩ = α ∥y∥^{2} and hence

|⟨x|y⟩| = |α| ∥y∥^{2} = ∥x∥∥y∥.

Now suppose that x ∈ H is arbitrary, let z := x − ∥y∥^{−2}⟨x|y⟩y. (So z is the “orthogonal projection” of x onto y, see Figure 14.1.) Then

Fig. 14.1. The picture behind the proof of the Schwarz inequality.

0 ≤ ∥z∥^{2} = ∥x − \frac{⟨x|y⟩}{∥y∥^{2}}y∥^{2} = ∥x∥^{2} + \frac{|⟨x|y⟩|^{2}}{∥y∥^{4}} ∥y∥^{2} − 2Re⟨x|\frac{⟨x|y⟩}{∥y∥^{2}}y⟩

= ∥x∥^{2} − \frac{|⟨x|y⟩|^{2}}{∥y∥^{2}}

from which it follows that 0 ≤ ∥y∥^{2}∥x∥^{2} − |⟨x|y⟩|^{2} with equality iff z = 0 or equivalently iff x = ∥y∥^{−2}⟨x|y⟩y.

Corollary 14.3. Let (H, ⟨·|·⟩) be an inner product space and ∥x∥ := \sqrt{⟨x|x⟩}. Then the Hilbertian norm, ∥·∥, is a norm on H. Moreover ⟨·|·⟩ is continuous on H × H, where H is viewed as the normed space (H, ∥·∥).

Proof. If x, y ∈ H, then, using the Schwarz’s inequality,

∥x + y∥^{2} = ∥x∥^{2} + ∥y∥^{2} + 2Re⟨x|y⟩
≤ ∥x∥^{2} + ∥y∥^{2} + 2∥x∥∥y∥ = (∥x∥ + ∥y∥)^{2}.

Taking the square root of this inequality shows ∥·∥ satisfies the triangle inequality.

Checking that ∥·∥ satisfies the remaining axioms of a norm is now routine and will be left to the reader. If x, x′y, y′ ∈ H, then

|⟨x|y⟩ − ⟨x′|y′⟩| = |⟨x − x′|y⟩ + ⟨x′|y − y′⟩|
≤ ∥y∥∥x − x′∥ + ∥x′∥∥y − y′∥
≤ ∥y∥∥x − x′∥ + (∥x∥ + ∥x − x′∥) ∥y − y′∥
= ∥y∥∥x − x′∥ + ∥x∥∥y − y′∥ + ∥x − x′∥∥y − y′∥

from which it follows that ⟨·|·⟩ is continuous.

Definition 14.4. Let (H, ⟨·|·⟩) be an inner product space, we say x, y ∈ H are orthogonal and write x ⊥ y iff ⟨x|y⟩ = 0. More generally if A ⊂ H is a set, x ∈ H is orthogonal to A (write x ⊥ A) iff ⟨x|y⟩ = 0 for all y ∈ A. Let A^{⊥} = {x ∈ H : x ⊥ A} be the set of vectors orthogonal to A. A subset S ⊂ H is an orthogonal set if x ⊥ y for all distinct elements x, y ∈ S. If S further satisfies, ∥x∥ = 1 for all x ∈ S, then S is said to be orthonormal set.

Proposition 14.5. Let (H, ⟨·|·⟩) be an inner product space then

1. (Parallelogram Law)

∥x + y∥^{2} + ∥x − y∥^{2} = 2∥x∥^{2} + 2∥y∥^{2}  (14.2)

for all x, y ∈ H.

2. (Pythagorean Theorem) If S ⊂ H is a finite orthogonal set, then

∥∑_{x∈S} x∥^{2} = ∑_{x∈S} ∥x∥^{2}.  (14.3)

3. If A ⊂ H is a set, then A^{⊥} is a closed linear subspace of H.

Remark 14.6. See Proposition 14.54 for the “converse” of the parallelogram law.

Proof. I will assume that H is a complex Hilbert space, the real case being easier. Items 1. and 2. are proved by the following elementary computations;

∥x + y∥^{2} + ∥x − y∥^{2}
= ∥x∥^{2} + ∥y∥^{2} + 2Re⟨x|y⟩ + ∥x∥^{2} + ∥y∥^{2} − 2Re⟨x|y⟩
= 2∥x∥^{2} + 2∥y∥^{2},

and

∥∑_{x∈S} x∥^{2} = ⟨∑_{x∈S} x| ∑_{y∈S} y⟩ = ∑_{x,y∈S}⟨x|y⟩

= ∑_{x∈S}⟨x|x⟩ = ∑_{x∈S} ∥x∥^{2}.

Item 3. is a consequence of the continuity of ⟨·|·⟩ and the fact that

A^{⊥} = ∩_{x∈A} Nul(⟨·|x⟩)

where Nul(⟨·|x⟩) = {y ∈ H : ⟨y|x⟩ = 0} – a closed subspace of H.

Definition 14.7. A Hilbert space is an inner product space (H, ⟨·|·⟩) such that the induced Hilbertian norm is complete.

Example 14.8. Suppose X is a set and μ : X → (0, ∞) , then H := ℓ^{2} (μ) is a Hilbert space when equipped with the inner product,

⟨f|g⟩ := ∑_{x∈X} f (x) ḡ (x) μ (x) .

In Exercise 14.7 you will show every Hilbert space H is “equivalent” to a Hilbert space of this form with μ ≡ 1.

More example of Hilbert spaces will be given later after we develop the Lebesgue integral, see Example 23.1 below.

Definition 14.9. A subset C of a vector space X is said to be convex if for all x, y ∈ C the line segment [x, y] := {tx + (1 − t)y : 0 ≤ t ≤ 1} joining x to y is contained in C as well. (Notice that any vector subspace of X is convex.)

Theorem 14.10 (Best Approximation Theorem). Suppose that H is a Hilbert space and M ⊂ H is a closed convex subset of H. Then for any x ∈ H there exists a unique y ∈ M such that

∥x − y∥ = d(x, M) = inf_{z∈M} ∥x − z∥.

Moreover, if M is a vector subspace of H, then the point y may also be characterized as the unique point in M such that (x − y) ⊥ M.

Proof. Uniqueness. By replacing M by M − x := {m − x : m ∈ M} we may assume x = 0. Let δ := d(0, M) = inf_{m∈M} ∥m∥ and y, z ∈ M, see Figure 14.2.

Fig. 14.2. The geometry of convex sets.

By the parallelogram law and the convexity of M,

2∥y∥^{2} + 2∥z∥^{2} = ∥y + z∥^{2} + ∥y − z∥^{2}
= 4 ∥\frac{y + z}{2} ∥^{2} + ∥y − z∥^{2} ≥ 4δ^{2} + ∥y − z∥^{2}.  (14.4)

Hence if ∥y∥ = ∥z∥ = δ, then 2δ^{2} + 2δ^{2} ≥ 4δ^{2} + ∥y − z∥^{2}, so that ∥y − z∥^{2} = 0. Therefore, if a minimizer for d(0, ·)|_{M} exists, it is unique.

Existence. Let y_{n} ∈ M be chosen such that ∥y_{n}∥ = δ_{n} → δ ≡ d(0, M). Taking y = y_{m} and z = y_{n} in Eq. (14.4) shows

2δ_{m}^{2} + 2δ_{n}^{2} ≥ 4δ^{2} + ∥y_{n} − y_{m}∥^{2}.

Passing to the limit m, n → ∞ in this equation implies,

2δ^{2} + 2δ^{2} ≥ 4δ^{2} + lim sup_{m,n→∞} ∥y_{n} − y_{m}∥^{2},

i.e. lim sup_{m,n→∞} ∥y_{n} − y_{m}∥^{2} = 0. Therefore, by completeness of H, {y_{n}}_{n=1}^{∞} is convergent. Because M is closed, y := lim_{n→∞} y_{n} ∈ M and because the norm is continuous,

∥y∥ = lim_{n→∞} ∥y_{n}∥ = δ = d(0, M).

So y is the desired point in M which is closest to 0.

Now suppose M is a closed subspace of H and x ∈ H. Let y ∈ M be the closest point in M to x. Then for w ∈ M, the function

g(t) := ∥x − (y + tw)∥^{2} = ∥x − y∥^{2} − 2tRe⟨x − y|w⟩ + t^{2}∥w∥^{2}

has a minimum at t = 0 and therefore 0 = g′(0) = −2Re⟨x − y|w⟩. Since w ∈ M is arbitrary, this implies that (x − y) ⊥ M.

Finally suppose y ∈ M is any point such that (x − y) ⊥ M. Then for z ∈ M, by Pythagorean’s theorem,

∥x − z∥^{2} = ∥x − y + y − z∥^{2} = ∥x − y∥^{2} + ∥y − z∥^{2} ≥ ∥x − y∥^{2}

which shows d(x, M)^{2} ≥ ∥x − y∥^{2}. That is to say y is the point in M closest to x.

Definition 14.11. Suppose that A : H → H is a bounded operator. The adjoint of A, denote A^{∗}, is the unique operator A^{∗} : H → H such that ⟨Ax|y⟩ = ⟨x|A^{∗}y⟩. (The proof that A^{∗} exists and is unique will be given in Proposition 14.16 below.) A bounded operator A : H → H is self - adjoint or Hermitian if A = A^{∗}.

Definition 14.12. Let H be a Hilbert space and M ⊂ H be a closed subspace. The orthogonal projection of H onto M is the function P_{M} : H → H such that for x ∈ H, P_{M}(x) is the unique element in M such that (x − P_{M}(x)) ⊥ M.

Theorem 14.13 (Projection Theorem). Let H be a Hilbert space and M ⊂ H be a closed subspace. The orthogonal projection P_{M} satisfies:

1. P_{M} is linear and hence we will write P_{M}x rather than P_{M}(x).

2. P_{M}^{2} = P_{M} (P_{M} is a projection).

3. P_{M}^{∗} = P_{M}, (P_{M} is self-adjoint).

4. Ran(P_{M}) = M and Nul(P_{M}) = M^{⊥}.

Proof.

1. Let x_{1}, x_{2} ∈ H and α ∈ 𝔽, then P_{M}x_{1} + αP_{M}x_{2} ∈ M and

P_{M}x_{1} + αP_{M}x_{2} − (x_{1} + αx_{2}) = [P_{M}x_{1} − x_{1} + α(P_{M}x_{2} − x_{2})] ∈ M^{⊥}

showing P_{M}x_{1} + αP_{M}x_{2} = P_{M}(x_{1} + αx_{2}), i.e. P_{M} is linear.

2. Obviously Ran(P_{M}) = M and P_{M}x = x for all x ∈ M. Therefore P_{M}^{2} = P_{M}.

3. Let x, y ∈ H, then since (x − P_{M}x) and (y − P_{M}y) are in M^{⊥},

⟨P_{M}x|y⟩ = ⟨P_{M}x|P_{M}y + y − P_{M}y⟩ = ⟨P_{M}x|P_{M}y⟩
= ⟨P_{M}x + (x − P_{M}x)|P_{M}y⟩ = ⟨x|P_{M}y⟩.

4. We have already seen, Ran(P_{M}) = M and P_{M}x = 0 iff x = x − 0 ∈ M^{⊥}, i.e. Nul(P_{M}) = M^{⊥}.

Corollary 14.14. If M ⊂ H is a proper closed subspace of a Hilbert space H, then H = M ⊕ M^{⊥}.

Proof. Given x ∈ H, let y = P_{M}x so that x − y ∈ M^{⊥}. Then x = y + (x − y) ∈ M + M^{⊥}. If x ∈ M ∩ M^{⊥}, then x ⊥ x, i.e. ∥x∥^{2} = ⟨x|x⟩ = 0. So M ∩ M^{⊥} = {0} .

Exercise 14.1. Suppose M is a subset of H, then M^{⊥⊥} = \overline{span(M)}.

Theorem 14.15 (Riesz Theorem). Let H^{∗} be the dual space of H (Notation 7.9). The map

z ∈ H ⟶^{j} ⟨·|z⟩ ∈ H^{∗}  (14.5)

is a conjugate linear^{1} isometric isomorphism.

Proof. The map j is conjugate linear by the axioms of the inner products. Moreover, for x, z ∈ H,

|⟨x|z⟩| ≤ ∥x∥ ∥z∥ for all x ∈ H

with equality when x = z. This implies that ∥jz∥_{H^{∗}} = ∥⟨·|z⟩∥_{H^{∗}} = ∥z∥ . Therefore j is isometric and this implies j is injective. To finish the proof we

1 Recall that j is conjugate linear if

j (z_{1} + αz_{2}) = jz_{1} + ᾱjz_{2}

for all z_{1}, z_{2} ∈ H and α ∈ ℂ.

must show that j is surjective. So let f ∈ H^{∗} which we assume, with out loss of generality, is non-zero. Then M =Nul(f) – a closed proper subspace of H. Since, by Corollary 14.14, H = M ⊕ M^{⊥}, f : H/M ≅ M^{⊥} → 𝔽 is a linear isomorphism. This shows that dim(M^{⊥}) = 1 and hence H = M ⊕ 𝔽x_{0} where x_{0} ∈ M^{⊥} \ {0} .^{2} Choose z = λx_{0} ∈ M^{⊥} such that f(x_{0}) = ⟨x_{0}|z⟩, i.e. λ = f̄(x_{0})/ ∥x_{0}∥^{2} . Then for x = m + λx_{0} with m ∈ M and λ ∈ 𝔽,

f(x) = λf(x_{0}) = λ⟨x_{0}|z⟩ = ⟨λx_{0}|z⟩ = ⟨m + λx_{0}|z⟩ = ⟨x|z⟩

which shows that f = jz.

Proposition 14.16 (Adjoints). Let H and K be Hilbert spaces and A : H → K be a bounded operator. Then there exists a unique bounded operator A^{∗} : K → H such that

⟨Ax|y⟩_{K} = ⟨x|A^{∗}y⟩_{H} for all x ∈ H and y ∈ K.  (14.6)

Moreover, for all A, B ∈ L(H, K) and λ ∈ ℂ,

1. (A + λB)^{∗} = A^{∗} + λ̄B^{∗},

2. A^{∗∗} := (A^{∗})^{∗} = A,

3. ∥A^{∗}∥ = ∥A∥ and

4. ∥A^{∗}A∥ = ∥A∥^{2} .

5. If K = H, then (AB)^{∗} = B^{∗}A^{∗}. In particular A ∈ L (H) has a bounded inverse iff A^{∗} has a bounded inverse and (A^{∗})^{−1} = (A^{−1})^{∗} .

Proof. For each y ∈ K, the map x → ⟨Ax|y⟩_{K} is in H^{∗} and therefore there exists, by Theorem 14.15, a unique vector z ∈ H such that

⟨Ax|y⟩_{K} = ⟨x|z⟩_{H} for all x ∈ H.

This shows there is a unique map A^{∗} : K → H such that ⟨Ax|y⟩_{K} = ⟨x|A^{∗}(y)⟩_{H} for all x ∈ H and y ∈ K.

To see A^{∗} is linear, let y_{1}, y_{2} ∈ K and λ ∈ ℂ, then for any x ∈ H,

⟨Ax|y_{1} + λy_{2}⟩_{K} = ⟨Ax|y_{1}⟩_{K} + λ̄⟨Ax|y_{2}⟩_{K}
= ⟨x|A^{∗}(y_{1})⟩_{K} + λ̄⟨x|A^{∗}(y_{2})⟩_{K}
= ⟨x|A^{∗}(y_{1}) + λA^{∗}(y_{2})⟩_{K}

and by the uniqueness of A^{∗}(y_{1} + λy_{2}) we find

A^{∗}(y_{1} + λy_{2}) = A^{∗}(y_{1}) + λA^{∗}(y_{2}).

This shows A^{∗} is linear and so we will now write A^{∗}y instead of A^{∗}(y).

2 Alternatively, choose x_{0} ∈ M^{⊥} \ {0} such that f(x_{0}) = 1. For x ∈ M^{⊥} we have f(x − λx_{0}) = 0 provided that λ := f(x). Therefore x − λx_{0} ∈ M ∩ M^{⊥} = {0} , i.e. x = λx_{0}. This again shows that M^{⊥} is spanned by x_{0}.

Since

⟨A^{∗}y|x⟩_{H} = \overline{⟨x|A^{∗}y⟩_{H}} = \overline{⟨Ax|y⟩_{K}} = ⟨y|Ax⟩_{K}

it follows that A^{∗∗} = A. The assertion that (A + λB)^{∗} = A^{∗} +λ̄B^{∗} is Exercise 14.2.

Items 3. and 4. Making use of the Schwarz inequality (Theorem 14.2), we have

∥A^{∗}∥ = sup_{k∈K:∥k∥=1} ∥A^{∗}k∥

= sup_{k∈K:∥k∥=1} sup_{h∈H:∥h∥=1} |⟨A^{∗}k|h⟩|

= sup_{h∈H:∥h∥=1} sup_{k∈K:∥k∥=1} |⟨k|Ah⟩| = sup_{h∈H:∥h∥=1} ∥Ah∥ = ∥A∥

so that ∥A^{∗}∥ = ∥A∥ . Since

∥A^{∗}A∥ ≤ ∥A^{∗}∥ ∥A∥ = ∥A∥^{2}

and

∥A∥^{2} = sup_{h∈H:∥h∥=1} ∥Ah∥^{2} = sup_{h∈H:∥h∥=1} |⟨Ah|Ah⟩|
= sup_{h∈H:∥h∥=1} |⟨h|A^{∗}Ah⟩| ≤ sup_{h∈H:∥h∥=1} ∥A^{∗}Ah∥ = ∥A^{∗}A∥  (14.7)

we also have ∥A^{∗}A∥ ≤ ∥A∥^{2} ≤ ∥A^{∗}A∥ which shows ∥A∥^{2} = ∥A^{∗}A∥ . Alternatively, from Eq. (14.7),

∥A∥^{2} ≤ ∥A^{∗}A∥ ≤ ∥A∥ ∥A^{∗}∥  (14.8)

which then implies ∥A∥ ≤ ∥A^{∗}∥ . Replacing A by A^{∗} in this last inequality shows ∥A^{∗}∥ ≤ ∥A∥ and hence that ∥A^{∗}∥ = ∥A∥ . Using this identity back in Eq. (14.8) proves ∥A∥^{2} = ∥A^{∗}A∥ .

Now suppose that K = H. Then

⟨ABh|k⟩ = ⟨Bh|A^{∗}k⟩ = ⟨h|B^{∗}A^{∗}k⟩

which shows (AB)^{∗} = B^{∗}A^{∗}. If A^{−1} exists then

(A^{−1})^{∗} A^{∗} = (AA^{−1})^{∗} = I^{∗} = I and
A^{∗} (A^{−1})^{∗} = (A^{−1}A)^{∗} = I^{∗} = I.

This shows that A^{∗} is invertible and (A^{∗})^{−1} = (A^{−1})^{∗} . Similarly if A^{∗} is invertible then so is A = A^{∗∗}.

Exercise 14.2. Let H, K, M be Hilbert spaces, A, B ∈ L(H, K), C ∈ L(K, M) and λ ∈ ℂ. Show (A + λB)^{∗} = A^{∗} + λ̄B^{∗} and (CA)^{∗} = A^{∗}C^{∗} ∈ L(M, H).

Exercise 14.3. Let H = ℂ^{n} and K = ℂ^{m} equipped with the usual inner products, i.e. ⟨z|w⟩_{H} = z·w̄ for z, w ∈ H. Let A be an m×n matrix thought of as a linear operator from H to K. Show the matrix associated to A^{∗} : K → H is the conjugate transpose of A.

Lemma 14.17. Suppose A : H → K is a bounded operator, then:

1. Nul(A^{∗}) = Ran(A)^{⊥}.

2. \overline{Ran(A)} = Nul(A^{∗})^{⊥}.

3. if K = H and V ⊂ H is an A – invariant subspace (i.e. A(V) ⊂ V), then V^{⊥} is A^{∗} – invariant.

Proof. An element y ∈ K is in Nul(A^{∗}) iff 0 = ⟨A^{∗}y|x⟩ = ⟨y|Ax⟩ for all x ∈ H which happens iff y ∈ Ran(A)^{⊥}. Because, by Exercise 14.1, \overline{Ran(A)} = Ran(A)^{⊥⊥}, and so by the first item, \overline{Ran(A)} = Nul(A^{∗})^{⊥}. Now suppose A(V) ⊂ V and y ∈ V^{⊥}, then

⟨A^{∗}y|x⟩ = ⟨y|Ax⟩ = 0 for all x ∈ V

which shows A^{∗}y ∈ V^{⊥}.

14.1 Hilbert Space Basis

Proposition 14.18 (Bessel’s Inequality). Let T be an orthonormal set, then for any x ∈ H,

∑_{v∈T} |⟨x|v⟩|^{2} ≤ ∥x∥^{2} for all x ∈ H.  (14.9)

In particular the set T_{x} := {v ∈ T : ⟨x|v⟩ ≠ 0} is at most countable for all x ∈ H.

Proof. Let Γ ⊂ T be any finite set. Then

0 ≤ ∥x − ∑_{v∈Γ}⟨x|v⟩v∥^{2} = ∥x∥^{2} − 2Re ∑_{v∈Γ}⟨x|v⟩ ⟨v|x⟩ + ∑_{v∈Γ} |⟨x|v⟩|^{2}

= ∥x∥^{2} − ∑_{v∈Γ} |⟨x|v⟩|^{2}

showing that ∑_{v∈Γ} |⟨x|v⟩|^{2} ≤ ∥x∥^{2}. Taking the supremum of this inequality over

Γ ⊂⊂ T then proves Eq. (14.9).

Proposition 14.19. Suppose T ⊂ H is an orthogonal set. Then s = ∑_{v∈T} v exists in H (see Definition 7.15) iff ∑_{v∈T} ∥v∥^{2} < ∞. (In particular T must be at most a countable set.) Moreover, if ∑_{v∈T} ∥v∥^{2} < ∞, then 1. ∥s∥^{2} = ∑_{v∈T} ∥v∥^{2} and

2. ⟨s|x⟩ = ∑_{v∈T} ⟨v|x⟩ for all x ∈ H.

Similarly if {v_{n}}_{n=1}^{∞} is an orthogonal set, then s = ∑_{n=1}^{∞} v_{n} exists in H

iff ∑_{n=1}^{∞} ∥v_{n}∥^{2} < ∞. In particular if ∑_{n=1}^{∞} v_{n} exists, then it is independent of

rearrangements of {v_{n}}_{n=1}^{∞}.

Proof. Suppose s = ∑_{v∈T} v exists. Then there exists Γ ⊂⊂ T such that

∑_{v∈Λ} ∥v∥^{2} = ∥∑_{v∈Λ} v∥^{2} ≤ 1

for all Λ ⊂⊂ T \ Γ, wherein the first inequality we have used Pythagorean’s theorem. Taking the supremum over such Λ shows that ∑_{v∈T \Γ} ∥v∥^{2} ≤ 1 and therefore

∑_{v∈T} ∥v∥^{2} ≤ 1 + ∑_{v∈Γ} ∥v∥^{2} < ∞.

Conversely, suppose that ∑_{v∈T} ∥v∥^{2} < ∞. Then for all ε > 0 there exists Γ_{ε} ⊂⊂ T such that if Λ ⊂⊂ T \ Γ_{ε},

∥∑_{v∈Λ} v∥^{2} = ∑_{v∈Λ} ∥v∥^{2} < ε^{2}.  (14.10)

Hence by Lemma 7.16, ∑_{v∈T} v exists.

For item 1, let Γ_{ε} be as above and set s_{ε} := ∑_{v∈Γ_{ε}} v. Then

|∥s∥ − ∥s_{ε}∥| ≤ ∥s − s_{ε}∥ < ε

and by Eq. (14.10),

0 ≤ ∑_{v∈T} ∥v∥^{2} − ∥s_{ε}∥^{2} = ∑_{v∉Γ_{ε}} ∥v∥^{2} ≤ ε^{2}.

Letting ε ↓ 0 we deduce from the previous two equations that ∥s_{ε}∥ → ∥s∥ and ∥s_{ε}∥^{2} → ∑_{v∈T} ∥v∥^{2} as ε ↓ 0 and therefore ∥s∥^{2} = ∑_{v∈T} ∥v∥^{2}.

Item 2. is a special case of Lemma 7.16. For the final assertion, let

s_{N} := ∑_{n=1}^{N} v_{n} and suppose that lim_{N→∞} s_{N} = s exists in H and in partic-

ular {s_{N}}_{N=1}^{∞} is Cauchy. So for N > M.

∑_{n=M+1}^{N} ∥v_{n}∥^{2} = ∥s_{N} − s_{M}∥^{2} → 0 as M, N → ∞

which shows that ∑_{n=1}^{∞} ∥v_{n}∥^{2} is convergent, i.e. ∑_{n=1}^{∞} ∥v_{n}∥^{2} < ∞.

Alternative proof of item 1. We could use the last result to prove Item 1. Indeed, if ∑_{v∈T} ∥v∥^{2} < ∞, then T is countable and so we may write T = {v_{n}}_{n=1}^{∞} . Then s = lim_{N→∞} s_{N} with s_{N} as above. Since the norm, ∥·∥ , is continuous on H,

∥s∥^{2} = lim_{N→∞} ∥s_{N}∥^{2} = lim_{N→∞} ∥∑_{n=1}^{N} v_{n}∥^{2} = lim_{N→∞} ∑_{n=1}^{N} ∥v_{n}∥^{2}

= ∑_{n=1}^{∞} ∥v_{n}∥^{2} = ∑_{v∈T} ∥v∥^{2}.

Corollary 14.20. Suppose H is a Hilbert space, β ⊂ H is an orthonormal set and M = \overline{span β}. Then

P_{M}x = ∑_{u∈β}⟨x|u⟩u,  (14.11)
∑_{u∈β} |⟨x|u⟩|^{2} = ∥P_{M}x∥^{2} and  (14.12)
∑_{u∈β}⟨x|u⟩⟨u|y⟩ = ⟨P_{M}x|y⟩  (14.13)

for all x, y ∈ H.

Proof. By Bessel’s inequality, ∑_{u∈β} |⟨x|u⟩|^{2} ≤ ∥x∥^{2} for all x ∈ H and hence by Proposition 14.18, P x := ∑_{u∈β}⟨x|u⟩u exists in H and for all x, y ∈ H,

⟨P x|y⟩ = ∑_{u∈β}⟨⟨x|u⟩u|y⟩ = ∑_{u∈β}⟨x|u⟩⟨u|y⟩.  (14.14)

Taking y ∈ β in Eq. (14.14) gives ⟨P x|y⟩ = ⟨x|y⟩, i.e. that ⟨x − P x|y⟩ = 0 for all y ∈ β. So (x − P x) ⊥ span β and by continuity we also have (x − P x) ⊥ M = \overline{span β.} Since P x is also in M, it follows from the definition of P_{M} that P x = P_{M}x proving Eq. (14.11). Equations (14.12) and (14.13) now follow from (14.14), Proposition 14.19 and the fact that ⟨P_{M}x|y⟩ = ⟨P_{M}^{2}x|y⟩ = ⟨P_{M}x|P_{M}y⟩ for all x, y ∈ H.

Exercise 14.4. Let (H, ⟨·|·⟩) be a Hilbert space and suppose that {P_{n}}_{n=1}^{∞} is a sequence of orthogonal projection operators on H such that P_{n}(H) ⊂ P_{n+1}(H) for all n. Let M := ∪_{n=1}^{∞}P_{n}(H) (a subspace of H) and let P denote orthonormal projection onto M̄. Show lim_{n→∞} P_{n}x = P x for all x ∈ H. Hint: first prove the result for x ∈ M^{⊥}, then for x ∈ M and then for x ∈ M̄.

Definition 14.21 (Basis). Let H be a Hilbert space. A basis β of H is a maximal orthonormal subset β ⊂ H.

Proposition 14.22. Every Hilbert space has an orthonormal basis.

Proof. Let F be the collection of all orthonormal subsets of H ordered by inclusion. If Φ ⊂ F is linearly ordered then ∪Φ is an upper bound. By Zorn’s Lemma (see Theorem B.7) there exists a maximal element β ∈ F.

An orthonormal set β ⊂ H is said to be complete if β^{⊥} = {0} . That is to say if ⟨x|u⟩ = 0 for all u ∈ β then x = 0.

Lemma 14.23. Let β be an orthonormal subset of H then the following are equivalent:

1. β is a basis,

2. β is complete and

3. \overline{span β} = H.

Proof. (1. ⟺ 2.) If β is not complete, then there exists a unit vector x ∈ β^{⊥} \ {0} . The set β ∪ {x} is an orthonormal set properly containing β, so β is not maximal. Conversely, if β is not maximal, there exists an orthonormal set β_{1} ⊂ H such that β ⫋ β_{1}. Then if x ∈ β_{1} \ β, we have ⟨x|u⟩ = 0 for all u ∈ β showing β is not complete.

(2. ⟺ 3.) If β is not complete and x ∈ β^{⊥} \ \{0\} , then \overline{span β} ⊂ x^{⊥} which is a proper subspace of H. Conversely if \overline{span β} is a proper subspace of H, β^{⊥} = \overline{span β}^{⊥} is a non-trivial subspace by Corollary 14.14 and β is not complete.

Theorem 14.24. Let β ⊂ H be an orthonormal set. Then the following are equivalent:

1. β is complete, i.e. β is an orthonormal basis for H.

2. x = ∑_{u∈β}⟨x|u⟩u for all x ∈ H.

3. ⟨x|y⟩ = ∑_{u∈β}⟨x|u⟩ ⟨u|y⟩ for all x, y ∈ H.

4. ∥x∥^{2} = ∑_{u∈β} |⟨x|u⟩|^{2} for all x ∈ H.

Proof. Let M = \overline{span β} and P = P_{M}.

(1) ⇒ (2) By Corollary 14.20, ∑_{u∈β}⟨x|u⟩u = P_{M}x. Therefore

x − ∑_{u∈β}⟨x|u⟩u = x − P_{M}x ∈ M^{⊥} = β^{⊥} = {0} .

(2) ⇒ (3) is a consequence of Proposition 14.19.

(3) ⇒ (4) is obvious, just take y = x.

(4) ⇒ (1) If x ∈ β^{⊥}, then by 4), ∥x∥ = 0, i.e. x = 0. This shows that β is complete.

Suppose Γ := {u_{n}}_{n=1}^{∞} is a collection of vectors in an inner product space (H, ⟨·|·⟩) . The standard Gram-Schmidt process produces from Γ an orthonormal subset, β = {v_{n}}_{n=1}^{∞}, such that every element u_{n} ∈ Γ is a finite linear combination of elements from β. Recall the procedure is to define v_{n} inductively by setting

ṽ_{n+1} := v_{n+1} − ∑_{j=1}^{n}⟨u_{n+1}|v_{j}⟩v_{j} = v_{n+1} − P_{n}v_{n+1}

where P_{n} is orthogonal projection onto M_{n} := span({v_{k}}_{k=1}^{n}). If v_{n+1} := 0, let ṽ_{n+1} = 0, otherwise set v_{n+1} := \left∥{ \atop ṽ_{n+1}}\right∥^{−1} ṽ_{n+1}. Finally re-index the resulting sequence so as to throw out those v_{n} with v_{n} = 0. The result is an orthonormal subset, β ⊂ H, with the desired properties.

Definition 14.25. As subset, Γ, of a normed space X is said to be total if span(Γ) is a dense in X.

Remark 14.26. Suppose that {u_{n}}_{n=1}^{∞} is a total subset of H. Let {v_{n}}_{n=1}^{∞} be the vectors found by performing Gram-Schmidt on the set {u_{n}}_{n=1}^{∞}. Then β := {v_{n}}_{n=1}^{∞} is an orthonormal basis for H. Indeed, if h ∈ H is orthogonal to β then h is orthogonal to \{u_{n}\}_{n=1}^{∞} and hence also \overline{span \{u_{n}\}_{n=1}^{∞}} = H. In particular h is orthogonal to itself and so h = 0.

Proposition 14.27. A Hilbert space H is separable iff H has a countable orthonormal basis β ⊂ H. Moreover, if H is separable, all orthonormal bases of H are countable. (See Proposition 4.14 in Conway’s, “A Course in Functional Analysis,” for a more general version of this proposition.)

Proof. Let 𝔻 ⊂ H be a countable dense set 𝔻 = {u_{n}}_{n=1}^{∞}. By Gram-Schmidt process there exists β = {v_{n}}_{n=1}^{∞} an orthonormal set such that span{v_{n} : n = 1, 2 …, N} ⊇ span{u_{n} : n = 1, 2 …, N}. So if ⟨x|v_{n}⟩ = 0 for all n then ⟨x|u_{n}⟩ = 0 for all n. Since 𝔻 ⊂ H is dense we may choose {w_{k}} ⊂ 𝔻 such that x = lim_{k→∞} w_{k} and therefore ⟨x|x⟩ = lim_{k→∞}⟨x|w_{k}⟩ = 0. That is to say x = 0 and β is complete. Conversely if β ⊂ H is a countable orthonormal basis, then the countable set

𝔻 = {∑_{u∈β} a_{u}u : a_{u} ∈ ℚ + iℚ : #{u : a_{u} ≠ 0} < ∞}

is dense in H. Finally let β = {u_{n}}_{n=1}^{∞} be an orthonormal basis and β_{1} ⊂ H be another orthonormal basis. Then the sets

B_{n} = {v ∈ β_{1} : ⟨v|u_{n}⟩ ≠ 0}

are countable for each n ∈ ℕ and hence B := ⋃_{n=1}^{∞} B_{n} is a countable subset

of β_{1}. Suppose there exists v ∈ β_{1} \ B, then ⟨v|u_{n}⟩ = 0 for all n and since β = {u_{n}}_{n=1}^{∞} is an orthonormal basis, this implies v = 0 which is impossible since ∥v∥ = 1. Therefore β_{1} \ B = ∅ and hence β_{1} = B is countable.

Proposition 14.28. Suppose X and Y are sets and μ : X → (0, ∞) and ν : Y → (0, ∞) are give weight functions. For functions f : X → ℂ and g : Y → ℂ let f ⊗ g : X × Y → ℂ be defined by f ⊗ g (x, y) := f (x) g (y) . If β ⊂ ℓ^{2} (μ) and γ ⊂ ℓ^{2} (ν) are orthonormal bases, then

β ⊗ γ := {f ⊗ g : f ∈ β and g ∈ γ}

is an orthonormal basis for ℓ^{2} (μ ⊗ ν) .

Proof. Let f, f′ ∈ ℓ^{2} (μ) and g, g′ ∈ ℓ^{2} (ν) , then by the Tonelli’s Theorem 4.22 for sums and Hölder’s inequality,

∑_{X×Y} |f ⊗ g · f′ ⊗ g′| μ ⊗ ν = ∑_{X} |ff′| μ · ∑_{Y} |gg′| ν

≤ ∥f∥_{ℓ^{2}(μ)} ∥f′∥_{ℓ^{2}(μ)} ∥g∥_{ℓ^{2}(ν)} ∥g′∥_{ℓ^{2}(ν)} = 1 < ∞.

So by Fubini’s Theorem 4.23 for sums,

⟨f ⊗ g|f′ ⊗ g′⟩_{ℓ^{2}(μ⊗ν)} = ∑_{X} ff′μ · ∑_{Y} gḡ′ν
= ⟨f|f′⟩_{ℓ^{2}(μ)}⟨g|g′⟩_{ℓ^{2}(ν)} = δ_{f,f′}δ_{g,g′}.

Therefore, β ⊗ γ is an orthonormal subset of ℓ^{2}(μ ⊗ ν). So it only remains to show β ⊗γ is complete. We will give two proofs of this fact. Let F ∈ ℓ^{2}(μ⊗ν). In the first proof we will verify item 4. of Theorem 14.24 while in the second we will verify item 1 of Theorem 14.24.

First Proof. By Tonelli’s Theorem,

∑_{x∈X} μ (x) ∑_{y∈Y} ν (y) |F (x, y)|^{2} = ∥F ∥_{ℓ^{2}(μ⊗ν)}^{2} < ∞

and since μ > 0, it follows that

∑_{y∈Y} |F (x, y)|^{2} ν (y) < ∞ for all x ∈ X,

i.e. F (x, ·) ∈ ℓ^{2}(ν) for all x ∈ X. By the completeness of γ,

∑_{Y} |F (x, y)|^{2} ν (y) = ⟨F (x, ·) |F (x, ·)⟩_{ℓ^{2}(ν)} = ∑_{g∈γ} |⟨F (x, ·) |g⟩_{ℓ^{2}(ν)}|^{2}

∥F ∥_{ℓ^{2}(μ⊗ν)}^{2} = ∑_{x∈X} μ (x) ∑_{y∈Y} ν (y) |F (x, y)|^{2}
= ∑_{x∈X} ∑_{g∈γ} |⟨F (x, ·) |g⟩_{ℓ^{2}(ν)}|^{2} μ (x) .  (14.15)

and in particular, x → ⟨F (x, ·) |g⟩_{ℓ^{2}(ν)} is in ℓ^{2} (μ) . So by the completeness of β and the Fubini and Tonelli theorems, we find

∑_{X} |⟨F (x, ·) |g⟩_{ℓ^{2}(ν)}|^{2} μ (x) = ∑_{f∈β} |∑_{X}⟨F (x, ·) |g⟩_{ℓ^{2}(ν)}f (x) μ (x)|^{2}

= ∑_{f∈β} |∑_{X} (∑_{Y} F (x, y) g (y) ν (y)) f (x) μ (x)|^{2}

= ∑_{f∈β} | ∑_{X×Y} F (x, y) \overline{f ⊗ g} (x, y) μ ⊗ ν (x, y)|^{2}

= ∑_{f∈β} |⟨F |f ⊗ g⟩_{ℓ^{2}(μ⊗ν)}|^{2} .

Combining this result with Eq. (14.15) shows

∥F ∥_{ℓ^{2}(μ⊗ν)}^{2} = ∑_{f∈β, g∈γ} |⟨F |f ⊗ g⟩_{ℓ^{2}(μ⊗ν)}|^{2}

as desired.

Second Proof. Suppose, for all f ∈ β and g ∈ γ that ⟨F |f ⊗ g⟩ = 0, i.e.

0 = ⟨F |f ⊗ g⟩_{ℓ^{2}(μ⊗ν)} = ∑_{x∈X} μ (x) ∑_{y∈Y} ν (y) F (x, y)f̄(x)g(y)
= ∑_{x∈X} μ (x) ⟨F (x, ·)|g⟩_{ℓ^{2}(ν)}f̄(x).  (14.16)

Since

∑_{x∈X} |⟨F (x, ·)|g⟩_{ℓ^{2}(ν)}|^{2} μ (x) ≤ ∑_{x∈X} μ (x) ∑_{y∈Y} |F (x, y)|^{2} ν (y) < ∞, (14.17)

it follows from Eq. (14.16) and the completeness of β that ⟨F (x, ·)|g⟩_{ℓ^{2}(ν)} = 0 for all x ∈ X. By the completeness of γ we conclude that F (x, y) = 0 for all (x, y) ∈ X × Y.

Definition 14.29. A linear map U : H → K is an isometry if ∥Ux∥_{K} = ∥x∥_{H} for all x ∈ H and U is unitary if U is also surjective.

Exercise 14.5. Let U : H → K be a linear map, show the following are equivalent:

1. U : H → K is an isometry,

2. ⟨Ux|Ux′⟩_{K} = ⟨x|x′⟩_{H} for all x, x′ ∈ H, (see Eq. (14.33) below)

3. U^{∗}U = id_{H}.

Exercise 14.6. Let U : H → K be a linear map, show the following are equivalent:

1. U : H → K is unitary

2. U^{∗}U = id_{H} and UU^{∗} = id_{K}.

3. U is invertible and U^{−1} = U^{∗}.

Exercise 14.7. Let H be a Hilbert space. Use Theorem 14.24 to show there exists a set X and a unitary map U : H → ℓ^{2}(X). Moreover, if H is separable and dim(H) = ∞, then X can be taken to be ℕ so that H is unitarily equivalent to ℓ^{2} = ℓ^{2}(ℕ).

14.2 Some Spectral Theory

For this section let H and K be two Hilbert space over ℂ.

Exercise 14.8. Suppose A : H → H is a bounded self-adjoint operator. Show:

1. If λ is an eigenvalue of A, i.e. Ax = λx for some x ∈ H \ {0} , then λ ∈ ℝ.

2. If λ and μ are two distinct eigenvalues of A with eigenvectors x and y respectively, then x ⊥ y.

Unlike in finite dimensions, it is possible that an operator on a complex Hilbert space may have no eigenvalues, see Example 14.35 and Lemma 14.36 below for a couple of examples. For this reason it is useful to generalize the notion of an eigenvalue as follows.

Definition 14.30. Suppose X is a Banach space over 𝔽 (𝔽 = ℝ or ℂ) and A ∈ L (X) . We say λ ∈ 𝔽 is in the spectrum of A if A − λI does not have a bounded^{3} inverse. The spectrum will be denoted by σ (A) ⊂ 𝔽. The resolvent set for A is ρ (A) := 𝔽\σ (A) .

Remark 14.31. If λ is an eigenvalue of A, then A − λI is not injective and hence not invertible. Therefore any eigenvalue of A is in the spectrum of A. If H is a Hilbert space ant A ∈ L (H) , it follows from item 5. of Proposition 14.16 that λ ∈ σ (A) iff λ̄ ∈ σ (A^{∗}) , i.e.

σ (A^{∗}) = {λ̄ : λ ∈ σ (A)} .

3 It will follow by the open mapping Theorem ?? or the closed graph Theorem ?? that the word bounded may be omitted from this definition.

Exercise 14.9. Suppose X is a complex Banach space and A ∈ GL (X) . Show

σ (A^{−1}) = σ (A)^{−1} := {λ^{−1} : λ ∈ σ (A)} .

If we further assume A is both invertible and isometric, i.e. ∥Ax∥ = ∥x∥ for all x ∈ X, then show

σ (A) ⊂ S^{1} := {z ∈ ℂ : |z| = 1} .

Hint: working formally,

(A^{−1} − λ^{−1})^{−1} = \frac{1}{\frac{1}{A} − \frac{1}{λ}} = \frac{1}{\frac{λ−A}{Aλ}} = \frac{Aλ}{λ − A}

from which you might expect that (A^{−1} − λ^{−1})^{−1} = −λA (A − λ)^{−1} if λ ∈ ρ (A) .

Exercise 14.10. Suppose X is a Banach space and A ∈ L (X) . Use Corollary 7.20 to show σ (A) is a closed subset of {λ ∈ 𝔽 : |λ| ≤ ∥A∥ := ∥A∥_{L(X)}} .

Lemma 14.32. Suppose that A ∈ L(H) is a normal operator, i.e. [A, A^{∗}] = 0. Then λ ∈ σ(A) iff

inf_{∥ψ∥=1} ∥(A − λ1)ψ∥ = 0.  (14.18)

In other words, λ ∈ σ (A) iff there is an “approximate sequence of eigenvectors” for (A, λ) , i.e. there exists ψ_{n} ∈ H such that ∥ψ_{n}∥ = 1 and Aψ_{n} − λψ_{n} → 0 as n → ∞.

Proof. By replacing A by A − λI we may assume that λ = 0. If 0 ∉ σ(A), then

inf_{∥ψ∥=1} ∥Aψ∥ = inf \frac{∥Aψ∥}{∥ψ∥} = inf \frac{∥ψ∥}{∥A^{−1}ψ∥} = 1/ ∥A^{−1}∥ > 0.

Now suppose that inf_{∥ψ∥=1} ∥Aψ∥ = ε > 0 or equivalently we have

∥Aψ∥ ≥ ε ∥ψ∥

for all ψ ∈ H. Because A is normal,

∥Aψ∥^{2} = ⟨A^{∗}Aψ|ψ⟩ = ⟨AA^{∗}ψ|ψ⟩ = ⟨A^{∗}ψ|A^{∗}ψ⟩ = ∥A^{∗}ψ∥^{2} .

Therefore we also have

∥A^{∗}ψ∥ = ∥Aψ∥ ≥ ε ∥ψ∥ ∀ ψ ∈ H.  (14.19)

This shows in particular that A and A^{∗} are injective, Ran(A) is closed and hence by Lemma 14.17

Ran(A) = \overline{Ran(A)} = Nul(A^{∗})^{⊥} = \{0\}^{⊥} = H.

Therefore A is algebraically invertible and the inverse is bounded by Eq. (14.19).

Lemma 14.33. Suppose that A ∈ L(H) is self-adjoint (i.e. A = A^{∗}) then

σ(A) ⊂ [− ∥A∥_{op} , ∥A∥_{op}] ⊂ ℝ.

Proof. Writting λ = α + iβ with α, β ∈ ℝ, then

∥(A + α + iβ) ψ∥^{2} = ∥(A + α)ψ∥^{2} + |β|^{2} ∥ψ∥^{2} + 2 Re((A + α) ψ, iβψ)
= ∥(A + α)ψ∥^{2} + |β|^{2} ∥ψ∥^{2}  (14.20)

wherein we have used

Re [iβ((A + α) ψ, ψ)] = β Im((A + α) ψ, ψ) = 0

since

((A + α) ψ, ψ) = (ψ, (A + α) ψ) = \overline{((A + α) ψ, ψ)}.

Eq. (14.20) along with Lemma 14.32 shows that λ ∉ σ(A) if β ≠ 0, i.e. σ(A) ⊂ ℝ. The fact that σ (A) is now contained in [− ∥A∥_{op} , ∥A∥_{op}] is a consequence of Exercise 14.9.

Remark 14.34. It is not true that σ(A) ⊂ ℝ implies A = A^{∗}. For example let

A = ( \begin{matrix}0 & 1 \\ 0 & 0 \end{matrix} ) on H = ℂ^{2}, then σ(A) = \{0\} yet A ≠ A^{∗}.

Example 14.35. Let S ∈ L(H) be a (not necessarily) normal operator. The proof of Lemma 14.32 gives λ ∈ σ(S) if Eq. (14.18) holds. However the converse is not always valid unless S is normal. For example, let S : ℓ^{2} → ℓ^{2} be the shift, S(ω_{1}, ω_{2}, …) = (0, ω_{1}, ω_{2}, …). Then for any λ ∈ D := {z ∈ ℂ : |z| < 1} ,

∥(S − λ) ψ∥ = ∥Sψ − λψ∥ ≥ |∥Sψ∥ − |λ| ∥ψ∥| = (1 − |λ|) ∥ψ∥

and so there does not exists an approximate sequence of eigenvectors for (S, λ) . However, as we will now show, σ (S) = D̄.

To prove this it suffices to show by Remark 14.31 and Exercise 14.9 that D ⊂ σ (S^{∗}) . For if this is the cae then D̄ ⊂ σ (S^{∗}) ⊂ D̄ and hence σ (S) = D̄ since D̄ is invariant under complex conjugation.

A simple computation shows,

S^{∗}(ω_{1}, ω_{2}, …) = (ω_{2}, ω_{3}, …)

and ω = (ω_{1}, ω_{2}, …) is an eigenvector for S^{∗} with eigenvalue λ ∈ ℂ iff

0 = (S^{∗} − λI) (ω_{1}, ω_{2}, …) = (ω_{2} − λω_{1}, ω_{3} − λω_{2}, …).

Solving these equation shows

ω_{2} = λω_{1}, ω_{3} = λω_{2} = λ^{2}ω_{1} , …, ω_{n} = λ^{n−1}ω_{1}.

Hence if λ ∈ D, we may let ω_{1} = 1 above to find

S^{∗}(1, λ, λ^{2}, …) = λ(1, λ, λ^{2}, …)

where (1, λ, λ^{2}, …) ∈ ℓ^{2}. Thus we have shown λ is an eigenvalue for S^{∗} for all λ ∈ D and hence D ⊂ σ(S^{∗}).

Lemma 14.36. Let H = ℓ^{2} (ℤ) and let A : H → H be defined by

Af (k) = i (f (k + 1) − f (k − 1)) for all k ∈ ℤ.

Then:

1. A is a bounded self-adjoint operator.

2. A has no eigenvalues.

3. σ (A) = [−2, 2] .

Proof. For another (simpler) proof of this lemma, see Exercise 23.8 below.

1. Since

∥Af∥_{2} ≤ ∥f (· + 1)∥_{2} + ∥f (· − 1)∥_{2} = 2 ∥f∥_{2} ,

∥A∥_{op} ≤ 2 < ∞. Moreover, for f, g ∈ ℓ^{2} (ℤ) ,

⟨Af|g⟩ = ∑_{k} i (f (k + 1) − f (k − 1)) ḡ (k)

= ∑_{k} if (k) ḡ (k − 1) − ∑_{k} if (k) ḡ (k + 1)

= ∑_{k} f (k) \overline{Ag (k)} = ⟨f|Ag⟩,

which shows A = A^{∗}.

2. From Lemma 14.33, we know that σ (A) ⊂ [−2, 2] . If λ ∈ [−2, 2] and f ∈ H satisfies Af = λf, then

f (k + 1) = −iλf (k) + f (k − 1) for all k ∈ ℤ.  (14.21)

This is a second order difference equations which can be solved analogously to second order ordinary differential equations. The idea is to start by looking for a solution of the form f (k) = α^{k}. Then Eq. (14.21) beocmes, α^{k+1} = −iλα^{k} + α^{k−1} or equivalently that

α^{2} + iλα − 1 = 0.

So we will have a solution if α ∈ {α_{±}} where

α_{±} = \frac{−iλ ± \sqrt{4 − λ^{2}}}{2} .

For |λ| ≠ 2, there are two distinct roots and the general solution to Eq. (14.21) is of the form

f (k) = c_{+}α_{+}^{k} + c_{−}α_{−}^{k}  (14.22)

for some constants c_{±} ∈ ℂ and |λ| = 2, the general solution has the form

f (k) = cα_{+}^{k} + dkα_{+}^{k}  (14.23)

Since in all cases, |α_{±}| = \frac{1}{4} (λ^{2} + 4 − λ^{2}) = 1, it follows that neither of these functions, f, will be in ℓ^{2} (ℤ) unless they are identically zero. This shows that A has no eigenvalues.

3. The above argument suggest a method for constructing approximate eigenfucntions. Namely, let λ ∈ [−2, 2] and define f_{n} (k) := 1_{|k|≤n}α^{k} where α = α_{+}. Then a simple computation shows

lim_{n→∞} \begin{matrix}∥(A − λI) f_{n}∥ & 2 \\ ∥f_{n}∥ & 2 \end{matrix} = 0  (14.24)

and therefore λ ∈ σ (A) .

Exercise 14.11. Verify Eq. (14.24). Also show by explicit computations that

lim_{n→∞} \begin{matrix}∥(A − λI) f_{n}∥ & 2 \\ ∥f_{n}∥ & 2 \end{matrix} ≠ 0

if λ ∉ [−2, 2] .

The next couple of results will be needed for the next section.

Theorem 14.37 (Rayleigh quotient). Suppose T ∈ L(H) := L(H, H) is a bounded self-adjoint operator, then

∥T ∥ = sup_{f≠0} \frac{|⟨f|T f⟩|}{∥f∥^{2}}.

Moreover if there exists a non-zero element g ∈ H such that

\frac{|⟨T g|g⟩|}{∥g∥^{2}} = ∥T ∥,

then g is an eigenvector of T with T g = λg and λ ∈ {±∥T ∥}.

Proof. Let

M := sup_{f≠0} \frac{|⟨f|T f⟩|}{∥f∥^{2}}.

We wish to show M = ∥T ∥. Since

|⟨f|T f⟩| ≤ ∥f∥∥T f∥ ≤ ∥T ∥∥f∥^{2},

we see M ≤ ∥T ∥. Conversely let f, g ∈ H and compute

⟨f + g|T (f + g)⟩ − (f − g|T (f − g)⟩
= ⟨f|T g⟩ + ⟨g|T f⟩ + ⟨f|T g⟩ + ⟨g|T f⟩
= 2[⟨f|T g⟩ + ⟨T g|f⟩] = 2[⟨f|T g⟩ + \overline{⟨f|T g⟩}]
= 4Re⟨f|T g⟩.

Therefore, if ∥f∥ = ∥g∥ = 1, it follows that

|Re⟨f|T g⟩| ≤ \frac{M}{4} \{∥f + g∥^{2} + ∥f − g∥^{2}\} = \frac{M}{4} \{2∥f∥^{2} + 2∥g∥^{2}\} = M.

By replacing f be e^{iθ}f where θ is chosen so that e^{iθ}⟨f|T g⟩ is real, we find

|⟨f|T g⟩| ≤ M for all ∥f∥ = ∥g∥ = 1.

Hence

∥T ∥ = sup_{∥f∥=∥g∥=1} |⟨f|T g⟩| ≤ M.

If g ∈ H \ {0} and ∥T ∥ = |⟨T g|g⟩|/∥g∥^{2} then, using the Cauchy Schwarz inequality,

∥T ∥ = \frac{|⟨T g|g⟩|}{∥g∥^{2}} ≤ \frac{∥T g∥}{∥g∥} ≤ ∥T ∥.  (14.25)

This implies |⟨T g|g⟩| = ∥T g∥∥g∥ and forces equality in the Cauchy Schwarz inequality. So by Theorem 14.2, T g and g are linearly dependent, i.e. T g = λg for some λ ∈ ℂ. Substituting this into (14.25) shows that |λ| = ∥T ∥. Since T is self-adjoint,

λ∥g∥^{2} = ⟨λg|g⟩ = ⟨T g|g⟩ = ⟨g|T g⟩ = ⟨g|λg⟩ = λ̄⟨g|g⟩,

which implies that λ ∈ ℝ and therefore, λ ∈ {±∥T ∥}.

Lemma 14.38 (Invariant subspaces). Let T : H → H be a self-adjoint operator and M be a T – invariant subspace of H, i.e. T (M) ⊂ M. Then M^{⊥} is also a T – invariant subspace, i.e. T (M^{⊥}) ⊂ M^{⊥}.

Proof. Let x ∈ M and y ∈ M^{⊥}, then T x ∈ M and hence

0 = ⟨T x|y⟩ = ⟨x|T y⟩ for all x ∈ M.

Thus T y ∈ M^{⊥}.

14.3 Compact Operators on a Hilbert Space

In this section let H and B be Hilbert spaces and U := {x ∈ H : ∥x∥ < 1} be the unit ball in H. Recall from Definition 11.16 that a bounded operator, K : H → B, is compact iff \overline{K(U)} is compact in B. Equivalently, for all bounded sequences {x_{n}}_{n=1}^{∞} ⊂ H, the sequence {Kx_{n}}_{n=1}^{∞} has a convergent subsequence in B. Because of Theorem 11.15, if dim(H) = ∞ and K : H → B is invertible, then K is not compact.

Definition 14.39. K : H → B is said to have finite rank if Ran(K) ⊂ B is finite dimensional.

The following result is a simple consequence of Corollaries 11.13 and 11.14.

Corollary 14.40. If K : H → B is a finite rank operator, then K is compact. In particular if either dim(H) < ∞ or dim(B) < ∞ then any bounded operator K : H → B is finite rank and hence compact.

Lemma 14.41. Let K := K(H, B) denote the compact operators from H to B. Then K(H, B) is a norm closed subspace of L(H, B).

Proof. The fact that K is a vector subspace of L(H, B) will be left to the reader. To finish the proof, we must show that K ∈ L(H, B) is compact if there exists K_{n} ∈ K(H, B) such that lim_{n→∞} ∥K_{n} − K∥_{op} = 0.

First Proof. Given ε > 0, choose N = N(ε) such that ∥K_{N} − K∥ < ε. Using the fact that K_{N}U is precompact, choose a finite subset Λ ⊂ U such that min_{x∈Λ} ∥y − K_{N}x∥ < ε for all y ∈ K_{N} (U) . Then for z = Kx_{0} ∈ K(U) and x ∈ Λ,

∥z − Kx∥ = ∥(K − K_{N})x_{0} + K_{N}(x_{0} − x) + (K_{N} − K)x∥
≤ 2ε + ∥K_{N}x_{0} − K_{N}x∥.

Therefore min_{x∈Λ} ∥z − K_{N}x∥ < 3ε, which shows K(U) is 3ε bounded for all ε > 0, K(U) is totally bounded and hence precompact.

Second Proof. Suppose {x_{n}}_{n=1}^{∞} is a bounded sequence in H. By compactness, there is a subsequence {x_{n}^{1}}_{n=1}^{∞} of {x_{n}}_{n=1}^{∞} such that {K_{1}x_{n}^{1}}_{n=1}^{∞} is convergent in B. Working inductively, we may construct subsequences

{x_{n}}_{n=1}^{∞} ⊃ {x_{n}^{1}}_{n=1}^{∞} ⊃ {x_{n}^{2}}_{n=1}^{∞} ⋯ ⊃ {x_{n}^{m}}_{n=1}^{∞} ⊃ …

such that {K_{m}x_{n}^{m}}_{n=1}^{∞} is convergent in B for each m. By the usual Cantor’s diagonalization procedure, let y_{n} := x_{n}^{n}, then {y_{n}}_{n=1}^{∞} is a subsequence of {x_{n}}_{n=1}^{∞} such that {K_{m}y_{n}}_{n=1}^{∞} is convergent for all m. Since

∥Ky_{n} − Ky_{l}∥ ≤ ∥(K − K_{m}) y_{n}∥ + ∥K_{m}(y_{n} − y_{l})∥ + ∥(K_{m} − K) y_{l})∥
≤ 2 ∥K − K_{m}∥ + ∥K_{m}(y_{n} − y_{l})∥ ,

lim sup_{n,l→∞} ∥Ky_{n} − Ky_{l}∥ ≤ 2 ∥K − K_{m}∥ → 0 as m → ∞,

which shows {Ky_{n}}_{n=1}^{∞} is Cauchy and hence convergent.

Proposition 14.42. A bounded operator K : H → B is compact iff there exists finite rank operators, K_{n} : H → B, such that ∥K − K_{n}∥ → 0 as n → ∞.

Proof. Since \overline{K(U)} is compact it contains a countable dense subset and from this it follows that \overline{K (H)} is a separable subspace of B. Let \{ϕ_{n}\} be an orthonormal basis for \overline{K (H)} ⊂ B and

P_{N}y = ∑_{n=1}^{N}⟨y|ϕ_{n}⟩ϕ_{n}

be the orthogonal projection of y onto span{ϕ_{n}}_{n=1}^{N}. Then lim_{N→∞} ∥P_{N}y − y∥ = 0 for all y ∈ K(H). Define K_{n} := P_{n}K – a finite rank operator on H. For sake of contradiction suppose that

lim sup_{n→∞} ∥K − K_{n}∥ = ε > 0,

in which case there exists x_{n_{k}} ∈ U such that ∥(K − K_{n_{k}})x_{n_{k}}∥ ≥ ε for all n_{k}. Since K is compact, by passing to a subsequence if necessary, we may assume {Kx_{n_{k}}}_{n_{k}=1}^{∞} is convergent in B. Letting y := lim_{k→∞} Kx_{n_{k}},

∥(K − K_{n_{k}})x_{n_{k}}∥ = ∥(1 − P_{n_{k}})Kx_{n_{k}}∥
≤ ∥(1 − P_{n_{k}})(Kx_{n_{k}} − y)∥ + ∥(1 − P_{n_{k}})y∥
≤ ∥Kx_{n_{k}} − y∥ + ∥(1 − P_{n_{k}})y∥ → 0 as k → ∞.

But this contradicts the assumption that ε is positive and hence we must have lim_{n→∞} ∥K − K_{n}∥ = 0, i.e. K is an operator norm limit of finite rank operators. The converse direction follows from Corollary 14.40 and Lemma 14.41.

Corollary 14.43. If K is compact then so is K^{∗}.

Proof. First Proof. Let K_{n} = P_{n}K be as in the proof of Proposition 14.42, then K_{n}^{∗} = K^{∗}P_{n} is still finite rank. Furthermore, using Proposition 14.16,

∥K^{∗} − K_{n}^{∗}∥ = ∥K − K_{n}∥ → 0 as n → ∞

showing K^{∗} is a limit of finite rank operators and hence compact. Second Proof. Let {x_{n}}_{n=1}^{∞} be a bounded sequence in B, then

∥K^{∗}x_{n} − K^{∗}x_{m}∥^{2} = (x_{n} − x_{m}, KK^{∗} (x_{n} − x_{m})) ≤ 2C ∥KK^{∗} (x_{n} − x_{m})∥

(14.26)

where C is a bound on the norms of the x_{n}. Since {K^{∗}x_{n}}_{n=1}^{∞} is also a bounded sequence, by the compactness of K there is a subsequence {x′_{n}} of the {x_{n}} such that KK^{∗}x′_{n} is convergent and hence by Eq. (14.26), so is the sequence {K^{∗}x′_{n}} .

14.3.1 The Spectral Theorem for Self Adjoint Compact Operators

For the rest of this section, K ∈ K(H) := K(H, H) will be a self-adjoint compact operator or S.A.C.O. for short. Because of Proposition 14.42, we might expect compact operators to behave very much like finite dimensional matrices. This is typically the case as we will see below.

Example 14.44 (Model S.A.C.O.). Let H = ℓ_{2} and K be the diagonal matrix

K = ( \begin{matrix}λ_{1} & 0 & 0 & ⋯ \\ 0 & λ_{2} & 0 & ⋯ \\ 0 & 0 & λ_{3} & ⋯ \\ ⋮ & ⋮ & ... & ... \end{matrix} ) ,

where lim_{n→∞} |λ_{n}| = 0 and λ_{n} ∈ ℝ. Then K is a self-adjoint compact operator. This assertion was proved in Example 11.17 above.

The main theorem (Theorem 14.46) of this subsection states that up to unitary equivalence, Example 14.44 is essentially the most general example of an S.A.C.O.

Theorem 14.45. Let K be a S.A.C.O., then either λ = ∥K∥ or λ = − ∥K∥ is an eigenvalue of K.

Proof. Without loss of generality we may assume that K is non-zero since otherwise the result is trivial. By Theorem 14.37, there exists f_{n} ∈ H such that ∥f_{n}∥ = 1 and

\frac{|⟨f_{n}|Kf_{n}⟩|}{∥f_{n}∥^{2}} = |⟨f_{n}|Kf_{n}⟩| ⟶ ∥K∥ as n → ∞.  (14.27)

By passing to a subsequence if necessary, we may assume that λ := lim_{n→∞}⟨f_{n}|Kf_{n}⟩ exists and λ ∈ {±∥K∥}. By passing to a further subsequence if necessary, we may assume, using the compactness of K, that Kf_{n} is convergent as well. We now compute:

0 ≤ ∥Kf_{n} − λf_{n}∥^{2} = ∥Kf_{n}∥^{2} − 2λ⟨Kf_{n}|f_{n}⟩ + λ^{2}
≤ λ^{2} − 2λ⟨Kf_{n}|f_{n}⟩ + λ^{2}
→ λ^{2} − 2λ^{2} + λ^{2} = 0 as n → ∞.

Hence

Kf_{n} − λf_{n} → 0 as n → ∞  (14.28)

and therefore

f := lim_{n→∞} f_{n} = \frac{1}{λ} lim_{n→∞} Kf_{n}

exists. By the continuity of the inner product, ∥f∥ = 1 ≠ 0. By passing to the limit in Eq. (14.28) we find that Kf = λf.

Theorem 14.46 (Compact Operator Spectral Theorem). Suppose that K : H → H is a non-zero S.A.C.O., then

1. there exists at least one eigenvalue λ ∈ {±∥K∥}.

2. There are at most countable many non-zero eigenvalues, {λ_{n}}_{n=1}^{N}, where N = ∞ is allowed. (Unless K is finite rank (i.e. dim Ran (K) < ∞), N will be infinite.)

3. The λ_{n}’s (including multiplicities) may be arranged so that |λ_{n}| ≥ |λ_{n+1}| for all n. If N = ∞ then lim_{n→∞} |λ_{n}| = 0. (In particular any eigenspace for K with non-zero eigenvalue is finite dimensional.)

4. The eigenvectors {ϕ_{n}}_{n=1}^{N} can be chosen to be an O.N. set such that H = \overline{span\{ϕ_{n}\}} ⊕ Nul(K).

5. Using the {ϕ_{n}}_{n=1}^{N} above,

Kψ = ∑_{n=1}^{N} λ_{n}⟨ψ|ϕ_{n}⟩ϕ_{n} for all ψ ∈ H.

6. The spectrum of K is, σ(K) = {0} ∪ {λ_{n} : n < N + 1} .

Proof. We will find λ_{n}’s and ϕ_{n}’s recursively. Let λ_{1} ∈ {±∥K∥} and ϕ_{1} ∈ H such that Kϕ_{1} = λ_{1}ϕ_{1} as in Theorem 14.45. Take M_{1} = span(ϕ_{1}) so K(M_{1}) ⊂ M_{1}. By Lemma 14.38, KM_{1}^{⊥} ⊂ M_{1}^{⊥}. Define K_{1} : M_{1}^{⊥} → M_{1}^{⊥} via K_{1} = K|_{M_{1}^{⊥}}. Then K_{1} is again a compact operator. If K_{1} = 0, we are done. If K_{1} ≠ 0, by Theorem 14.45 there exists λ_{2} ∈ {±∥K∥_{1}} and ϕ_{2} ∈ M_{1}^{⊥} such that ∥ϕ_{2}∥ = 1 and K_{1}ϕ_{2} = Kϕ_{2} = λ_{2}ϕ_{2}. Let M_{2} := \overline{span(ϕ_{1}, ϕ_{2}).} Again K(M_{2}) ⊂ M_{2} and hence K_{2} := K|_{M_{2}^{⊥}} : M_{2}^{⊥} → M_{2}^{⊥} is compact. Again if K_{2} = 0 we are done. If K_{2} ≠ 0. Then by Theorem 14.45 there exists λ_{3} ∈ {±∥K∥_{2}} and ϕ_{3} ∈ M_{2}^{⊥} such that ∥ϕ_{3}∥ = 1 and K_{2}ϕ_{3} = Kϕ_{3} = λ_{3}ϕ_{3}. Continuing this way indefinitely or until we reach a point where K_{n} = 0, we construct a sequence {λ_{n}}_{n=1}^{N} of eigenvalues and orthonormal eigenvectors {ϕ_{n}}_{n=1}^{N} such that |λ_{i}| ≥ |λ_{i+1}| with the further property that

|λ_{i}| = sup_{ϕ⊥\{ϕ_{1},ϕ_{2},...ϕ_{i−1}\}} \frac{∥Kϕ∥}{∥ϕ∥}  (14.29)

If N = ∞ then lim_{i→∞} |λ_{i}| = 0 for if not there would exist ε > 0 such that |λ_{i}| ≥ ε > 0 for all i. In this case {ϕ_{i}/λ_{i}}_{i=1}^{∞} is sequence in H bounded by ε^{−1}. By compactness of K, there exists a subsequence i_{k} such that ϕ_{i_{k}} = Kϕ_{i_{k}}/λ_{i_{k}} is convergent. But this is impossible since {ϕ_{i_{k}}} is an orthonormal set. Hence we must have that ε = 0. Let M := span{ϕ_{i}}_{i=1}^{N} with N = ∞ possible. Then K(M) ⊂ M and hence K(M^{⊥}) ⊂ M^{⊥}. Using Eq. (14.29),

∥K|_{M^{⊥}}∥ ≤ ∥K|_{M_{n}^{⊥}}∥ = |λ_{n}| ⟶ 0 as n → ∞

showing K|M^{⊥} ≡ 0. Define P_{0} to be orthogonal projection onto M^{⊥}. Then for ψ ∈ H,

ψ = P_{0}ψ + (1 − P_{0})ψ = P_{0}ψ + ∑_{i=1}^{N}⟨ψ|ϕ_{i}⟩ϕ_{i}

and

Kψ = KP_{0}ψ + K ∑_{i=1}^{N}⟨ψ|ϕ_{i}⟩ϕ_{i} = ∑_{i=1}^{N} λ_{i}⟨ψ|ϕ_{i}⟩ϕ_{i}.

Since {λ_{n}} ⊂ σ(K) and σ(K) is closed, it follows that 0 ∈ σ(K) and hence {λ_{n}}_{n=1}^{∞} ∪ {0} ⊂ σ(K). Suppose that z ∉ {λ_{n}}_{n=1}^{∞} ∪ {0} and let d be the distance between z and {λ_{n}}_{n=1}^{∞}∪{0}. Notice that d > 0 because lim_{n→∞} λ_{n} = 0. A few simple computations show that:

(K − zI)ψ = ∑_{i=1}^{N}⟨ψ|ϕ_{i}⟩(λ_{i} − z)ϕ_{i} − zP_{0}ψ,

(K − z)^{−1} exists,

(K − zI)^{−1}ψ = ∑_{i=1}^{N}⟨ψ|ϕ_{i}⟩(λ_{i} − z)^{−1}ϕ_{i} − z^{−1}P_{0}ψ,

and

∥(K − zI)^{−1}ψ∥^{2} = ∑_{i=1}^{N} |⟨ψ|ϕ_{i}⟩|^{2} \frac{1}{|λ_{i} − z|^{2}} + \frac{1}{|z|^{2}} ∥P_{0}ψ∥^{2}

≤ (\frac{1}{d})^{2} (∑_{i=1}^{N} |⟨ψ|ϕ_{i}⟩|^{2} + ∥P_{0}ψ∥^{2}) = \frac{1}{d^{2}} ∥ψ∥^{2}.

We have thus shown that (K − zI)^{−1} exists, ∥(K − zI)^{−1}∥ ≤ d^{−1} < ∞ and hence z ∉ σ(K).

Theorem 14.47 (Structure of Compact Operators). Let K : H → B be a compact operator. Then there exists N ∈ ℕ∪ {∞} , orthonormal subsets {ϕ_{n}}_{n=1}^{N} ⊂ H and {ψ_{n}}_{n=1}^{N} ⊂ B and a sequences {α_{n}}_{n=1}^{N} ⊂ ℝ_{+} such that λ_{1} ≥ λ_{2} ≥ …, lim_{n→∞} α_{n} = 0 if N = ∞, ∥ψ_{n}∥ ≤ 1 for all n and

Kf = ∑_{n=1}^{N} α_{n}⟨f|ϕ_{n}⟩ψ_{n} for all f ∈ H.  (14.30)

Proof. Since K^{∗}K is a selfadjoint compact operator, Theorem 14.46 implies there exists an orthonormal set {ϕ_{n}}_{n=1}^{N} ⊂ H and positive numbers {λ_{n}}_{n=1}^{N} such that

K^{∗}Kψ = ∑_{n=1}^{N} λ_{n}⟨ψ|ϕ_{n}⟩ϕ_{n} for all ψ ∈ H.

Let A be the positive square root of K^{∗}K defined by

Aψ := ∑_{n=1}^{N} \sqrt{λ_{n}}⟨ψ|ϕ_{n}⟩ϕ_{n} for all ψ ∈ H.

A simple computation shows, A^{2} = K^{∗}K, and therefore,

∥Aψ∥^{2} = ⟨Aψ|Aψ⟩ = ⟨ψ|A^{2}ψ⟩
= ⟨ψ|K^{∗}Kψ⟩ = ⟨Kψ|Kψ⟩ = ∥Kψ∥^{2}

for all ψ ∈ H. Hence we may define a unitary operator, u : \overline{Ran(A)} → \overline{Ran(K)} by the formula

uAψ = Kψ for all ψ ∈ H.

We then have

Kψ = uAψ = ∑_{n=1}^{N} \sqrt{λ_{n}}⟨ψ|ϕ_{n}⟩uϕ_{n}  (14.31)

which proves the result with ψ_{n} := uϕ_{n} and α_{n} = \sqrt{λ_{n}}.

It is instructive to find ψ_{n} explicitly and to verify Eq. (14.31) by bruit force. Since ϕ_{n} = λ_{n}^{−1/2}Aϕ_{n},

ψ_{n} = λ_{n}^{−1/2}uAϕ_{n} = λ_{n}^{−1/2}uAϕ_{n} = λ_{n}^{−1/2}Kϕ_{n}

and

⟨Kϕ_{n}|Kϕ_{m}⟩ = ⟨ϕ_{n}|K^{∗}Kϕ_{m}⟩ = λ_{n}δ_{mn}.

This verifies that {ψ_{n}}_{n=1}^{N} is an orthonormal set. Moreover,

∑_{n=1}^{N} \sqrt{λ_{n}}⟨ψ|ϕ_{n}⟩ψ_{n} = ∑_{n=1}^{N} \sqrt{λ_{n}}⟨ψ|ϕ_{n}⟩λ_{n}^{−1/2}Kϕ_{n}

= K ∑_{n=1}^{N}⟨ψ|ϕ_{n}⟩ϕ_{n} = Kψ

since ∑_{n=1}^{N}⟨ψ|ϕ_{n}⟩ϕ_{n} = P ψ where P is orthogonal projection onto Nul(K)^{⊥}.

Second Proof. Let K = u |K| be the polar decomposition of K. Then |K| is self-adjoint and compact, by Corollary ??, and hence by Theorem 14.46 there exists an orthonormal basis {ϕ_{n}}_{n=1}^{N} for Nul(|K|)^{⊥} = Nul(K)^{⊥} such that |K| ϕ_{n} = λ_{n}ϕ_{n}, λ_{1} ≥ λ_{2} ≥ … and lim_{n→∞} λ_{n} = 0 if N = ∞. For f ∈ H,

Kf = u |K| ∑_{n=1}^{N}⟨f|ϕ_{n}⟩ϕ_{n} = ∑_{n=1}^{N}⟨f|ϕ_{n}⟩u |K| ϕ_{n} = ∑_{n=1}^{N} λ_{n}⟨f|ϕ_{n}⟩uϕ_{n}

which is Eq. (14.30) with ψ_{n} := uϕ_{n}.

14.4 Weak Convergence

Suppose H is an infinite dimensional Hilbert space and {x_{n}}_{n=1}^{∞} is an orthonormal subset of H. Then, by Eq. (14.1), ∥x_{n} − x_{m}∥^{2} = 2 for all m ≠ n and in particular, {x_{n}}_{n=1}^{∞} has no convergent subsequences. From this we conclude that C := {x ∈ H : ∥x∥ ≤ 1} , the closed unit ball in H, is not compact. To overcome this problems it is sometimes useful to introduce a weaker topology on X having the property that C is compact.

Definition 14.48. Let (X, ∥·∥) be a Banach space and X^{∗} be its continuous dual. The weak topology, τ_{w}, on X is the topology generated by X^{∗}. If {x_{n}}_{n=1}^{∞} ⊂ X is a sequence we will write x_{n} →^{w} x as n → ∞ to mean that x_{n} → x in the weak topology.

Because τ_{w} = τ(X^{∗}) ⊂ τ_{∥·∥} := τ({∥x − ·∥ : x ∈ X} , it is harder for a function f : X → 𝔽 to be continuous in the τ_{w} – topology than in the norm topology, τ_{∥·∥}. In particular if ϕ : X → 𝔽 is a linear functional which is τ_{w} – continuous, then ϕ is τ_{∥·∥} – continuous and hence ϕ ∈ X^{∗}.

Exercise 14.12. Show the vector space operations of X are continuous in the weak topology, i.e. show:

1. (x, y) ∈ X × X → x + y ∈ X is (τ_{w} ⊗ τ_{w}, τ_{w}) – continuous and

2. (λ, x) ∈ 𝔽 × X → λx ∈ X is (τ_{𝔽} ⊗ τ_{w}, τ_{w}) – continuous.

Proposition 14.49. Let {x_{n}}_{n=1}^{∞} ⊂ X be a sequence, then x_{n} →^{w} x ∈ X as n → ∞ iff ϕ(x) = lim_{n→∞} ϕ(x_{n}) for all ϕ ∈ X^{∗}.

Proof. By definition of τ_{w}, we have x_{n} →^{w} x ∈ X iff for all Γ ⊂⊂ X^{∗} and ε > 0 there exists an N ∈ ℕ such that |ϕ(x) − ϕ(x_{n})| < ε for all n ≥ N and ϕ ∈ Γ. This later condition is easily seen to be equivalent to ϕ(x) = lim_{n→∞} ϕ(x_{n}) for all ϕ ∈ X^{∗}.

The topological space (X, τ_{w}) is still Hausdorff as follows from the Hahn Banach Theorem, see Theorem ??. For the moment we will concentrate on the special case where X = H is a Hilbert space in which case H^{∗} = {ϕ_{z} := ⟨·|z⟩ : z ∈ H} , see Theorem 14.15. If x, y ∈ H and z := y − x ≠ 0, then

0 < ε := ∥z∥^{2} = ϕ_{z}(z) = ϕ_{z}(y) − ϕ_{z}(x).

Thus

V_{x} := {w ∈ H : |ϕ_{z}(x) − ϕ_{z}(w)| < ε/2} and
V_{y} := {w ∈ H : |ϕ_{z}(y) − ϕ_{z}(w)| < ε/2}

are disjoint sets from τ_{w} which contain x and y respectively. This shows that (H, τ_{w}) is a Hausdorff space. In particular, this shows that weak limits are unique if they exist.

Remark 14.50. Suppose that H is an infinite dimensional Hilbert space {x_{n}}_{n=1}^{∞} is an orthonormal subset of H. Then Bessel’s inequality (Proposition 14.18) implies x_{n} →^{w} 0 ∈ H as n → ∞. This points out the fact that if x_{n} →^{w} x ∈ H as n → ∞, it is no longer necessarily true that ∥x∥ = lim_{n→∞} ∥x_{n}∥ . However we do always have ∥x∥ ≤ lim inf_{n→∞} ∥x_{n}∥ because,

∥x∥^{2} = lim_{n→∞}⟨x_{n}|x⟩ ≤ lim inf_{n→∞} [∥x_{n}∥ ∥x∥] = ∥x∥ lim inf_{n→∞} ∥x_{n}∥ .

Proposition 14.51. Let H be a Hilbert space, β ⊂ H be an orthonormal basis for H and {x_{n}}_{n=1}^{∞} ⊂ H be a bounded sequence, then the following are equivalent:

1. x_{n} →^{w} x ∈ H as n → ∞.

2. ⟨x|y⟩ = lim_{n→∞}⟨x_{n}|y⟩ for all y ∈ H.

3. ⟨x|y⟩ = lim_{n→∞}⟨x_{n}|y⟩ for all y ∈ β.

Moreover, if c_{y} := lim_{n→∞}⟨x_{n}|y⟩ exists for all y ∈ β, then ∑_{y∈β} |c_{y}|^{2} < ∞

and x_{n} →^{w} x := ∑_{y∈β} c_{y}y ∈ H as n → ∞.

Proof. 1. ⟹ 2. This is a consequence of Theorem 14.15 and Proposition 14.49. 2. ⟹ 3. is trivial. 3. ⟹ 1. Let M := sup_{n} ∥x_{n}∥ and H_{0} denote the algebraic span of β. Then for y ∈ H and z ∈ H_{0},

|⟨x − x_{n}|y⟩| ≤ |⟨x − x_{n}|z⟩| + |⟨x − x_{n}|y − z⟩| ≤ |⟨x − x_{n}|z⟩| + 2M ∥y − z∥ .

Passing to the limit in this equation implies lim sup_{n→∞} |⟨x − x_{n}|y⟩| ≤ 2M ∥y − z∥ which shows lim sup_{n→∞} |⟨x − x_{n}|y⟩| = 0 since H_{0} is dense in H. To prove the last assertion, let Γ ⊂⊂ β. Then by Bessel’s inequality (Proposition 14.18),

∑_{y∈Γ} |c_{y}|^{2} = lim_{n→∞} ∑_{y∈Γ} |⟨x_{n}|y⟩|^{2} ≤ lim inf_{n→∞} ∥x_{n}∥^{2} ≤ M^{2}.

Since Γ ⊂⊂ β was arbitrary, we conclude that ∑_{y∈β} |c_{y}|^{2} ≤ M < ∞ and hence we may define x := ∑_{y∈β} c_{y}y. By construction we have

⟨x|y⟩ = c_{y} = lim_{n→∞}⟨x_{n}|y⟩ for all y ∈ β

and hence x_{n} →^{w} x ∈ H as n → ∞ by what we have just proved.

Theorem 14.52. Suppose {x_{n}}_{n=1}^{∞} is a bounded sequence in a Hilbert space, H. Then there exists a subsequence y_{k} := x_{n_{k}} of {x_{n}}_{n=1}^{∞} and x ∈ X such

that y_{k} →^{w} x as k → ∞.

Proof. This is a consequence of Proposition 14.51 and a Cantor’s diagonalization argument which is left to the reader, see Exercise 14.13.

Theorem 14.53 (Alaoglu’s Theorem for Hilbert Spaces). Suppose that H is a separable Hilbert space, C := {x ∈ H : ∥x∥ ≤ 1} is the closed unit ball in H and {e_{n}}_{n=1}^{∞} is an orthonormal basis for H. Then

ρ(x, y) := ∑_{n=1}^{∞} \frac{1}{2^{n}} |⟨x − y|e_{n}⟩|  (14.32)

defines a metric on C which is compatible with the weak topology on C, τ_{C} := (τ_{w})_{C} = {V ∩ C : V ∈ τ_{w}} . Moreover (C, ρ) is a compact metric space. (This theorem will be extended to Banach spaces, see Theorems ?? and ?? below.)

Proof. The routine check that ρ is a metric is left to the reader. Let τ_{ρ} be the topology on C induced by ρ. For any y ∈ H and n ∈ ℕ, the map x ∈ H → ⟨x − y|e_{n}⟩ = ⟨x|e_{n}⟩ − ⟨y|e_{n}⟩ is τ_{w} continuous and since the sum in Eq. (14.32) is uniformly convergent for x, y ∈ C, it follows that x → ρ(x, y) is τ_{C} – continuous. This implies the open balls relative to ρ are contained in τ_{C} and therefore τ_{ρ} ⊂ τ_{C}. For the converse inclusion, let z ∈ H, x → ϕ_{z}(x) = ⟨x|z⟩ be an element of H^{∗}, and for N ∈ ℕ let z_{N} := ∑_{n=1}^{N}⟨z|e_{n}⟩e_{n}. Then ϕ_{z_{N}} = ∑_{n=1}^{N} \overline{⟨z|e_{n}⟩}ϕ_{e_{n}} is ρ continuous, being a finite linear combination of the ϕ_{e_{n}} which are easily seen to be ρ – continuous. Because z_{N} → z as N → ∞ it follows that

sup_{x∈C} |ϕ_{z}(x) − ϕ_{z_{N}} (x)| = ∥z − z_{N}∥ → 0 as N → ∞.

Therefore ϕ_{z}|_{C} is ρ – continuous as well and hence τ_{C} = τ(ϕ_{z}|_{C} : z ∈ H) ⊂ τ_{ρ}. The last assertion follows directly from Theorem 14.52 and the fact that sequential compactness is equivalent to compactness for metric spaces.

14.5 Supplement 1: Converse of the Parallelogram Law

Proposition 14.54 (Parallelogram Law Converse). If (X, ∥·∥) is a normed space such that Eq. (14.2) holds for all x, y ∈ X, then there exists a unique inner product on ⟨·|·⟩ such that ∥x∥ := \sqrt{⟨x|x⟩} for all x ∈ X. In this case we say that ∥·∥ is a Hilbertian norm.

Proof. If ∥·∥ is going to come from an inner product ⟨·|·⟩, it follows from Eq. (14.1) that

2Re⟨x|y⟩ = ∥x + y∥^{2} − ∥x∥^{2} − ∥y∥^{2}

and

−2Re⟨x|y⟩ = ∥x − y∥^{2} − ∥x∥^{2} − ∥y∥^{2}.

Subtracting these two equations gives the “polarization identity,”

4Re⟨x|y⟩ = ∥x + y∥^{2} − ∥x − y∥^{2}.

Replacing y by iy in this equation then implies that

4Im⟨x|y⟩ = ∥x + iy∥^{2} − ∥x − iy∥^{2}

from which we find

⟨x|y⟩ = \frac{1}{4} ∑_{ε∈G} ε∥x + εy∥^{2}  (14.33)

where G = {±1, ±i} – a cyclic subgroup of S^{1} ⊂ ℂ. Hence if ⟨·|·⟩ is going to exists we must define it by Eq. (14.33). Notice that

⟨x|x⟩ = \frac{1}{4} ∑_{ε∈G} ε∥x + εx∥^{2} = ∥x∥^{2} + i∥x + ix∥^{2} − i∥x − ix∥^{2}
= ∥x∥^{2} + i |1 + i|^{2}| ∥x∥^{2} − i |1 − i|^{2}| ∥x∥^{2} = ∥x∥^{2} .

So to finish the proof of (4) we must show that ⟨x|y⟩ in Eq. (14.33) is an inner product. Since

4⟨y|x⟩ = ∑_{ε∈G} ε∥y + εx∥^{2} = ∑_{ε∈G} ε∥ε (y + εx) ∥^{2}

= ∑_{ε∈G} ε∥εy + ε^{2}x∥^{2}

= ∥y + x∥^{2} + ∥ − y + x∥^{2} + i∥iy − x∥^{2} − i∥ − iy − x∥^{2}

= ∥x + y∥^{2} + ∥x − y∥^{2} + i∥x − iy∥^{2} − i∥x + iy∥^{2}

= 4\overline{⟨x|y⟩}

it suffices to show x → ⟨x|y⟩ is linear for all y ∈ H. (The rest of this proof may safely be skipped by the reader.) For this we will need to derive an identity from Eq. (14.2). To do this we make use of Eq. (14.2) three times to find

∥x + y + z∥^{2} = −∥x + y − z∥^{2} + 2∥x + y∥^{2} + 2∥z∥^{2}

= ∥x − y − z∥^{2} − 2∥x − z∥^{2} − 2∥y∥^{2} + 2∥x + y∥^{2} + 2∥z∥^{2}

= ∥y + z − x∥^{2} − 2∥x − z∥^{2} − 2∥y∥^{2} + 2∥x + y∥^{2} + 2∥z∥^{2}

= −∥y + z + x∥^{2} + 2∥y + z∥^{2} + 2∥x∥^{2}
− 2∥x − z∥^{2} − 2∥y∥^{2} + 2∥x + y∥^{2} + 2∥z∥^{2}.

Solving this equation for ∥x + y + z∥^{2} gives

∥x + y + z∥^{2} = ∥y + z∥^{2} + ∥x + y∥^{2} − ∥x − z∥^{2} + ∥x∥^{2} + ∥z∥^{2} − ∥y∥^{2}. (14.34)

Using Eq. (14.34), for x, y, z ∈ H,

4 Re⟨x + z|y⟩ = ∥x + z + y∥^{2} − ∥x + z − y∥^{2}
= ∥y + z∥^{2} + ∥x + y∥^{2} − ∥x − z∥^{2} + ∥x∥^{2} + ∥z∥^{2} − ∥y∥^{2}

− (∥z − y∥^{2} + ∥x − y∥^{2} − ∥x − z∥^{2} + ∥x∥^{2} + ∥z∥^{2} − ∥y∥^{2})

= ∥z + y∥^{2} − ∥z − y∥^{2} + ∥x + y∥^{2} − ∥x − y∥^{2}
= 4 Re⟨x|y⟩ + 4 Re⟨z|y⟩.  (14.35)

Now suppose that δ ∈ G, then since |δ| = 1,

4⟨δx|y⟩ = \frac{1}{4} ∑_{ε∈G} ε∥δx + εy∥^{2} = \frac{1}{4} ∑_{ε∈G} ε∥x + δ^{−1}εy∥^{2}

= \frac{1}{4} ∑_{ε∈G} εδ∥x + δεy∥^{2} = 4δ⟨x|y⟩  (14.36)

where in the third inequality, the substitution ε → εδ was made in the sum. So Eq. (14.36) says ⟨±ix|y⟩ = ±i⟨ix|y⟩ and ⟨−x|y⟩ = −⟨x|y⟩. Therefore

Im⟨x|y⟩ = Re (−i⟨x|y⟩) = Re⟨−ix|y⟩

which combined with Eq. (14.35) shows

Im⟨x + z|y⟩ = Re⟨−ix − iz|y⟩ = Re⟨−ix|y⟩ + Re⟨−iz|y⟩
= Im⟨x|y⟩ + Im⟨z|y⟩

and therefore (again in combination with Eq. (14.35)),

⟨x + z|y⟩ = ⟨x|y⟩ + ⟨z|y⟩ for all x, y ∈ H.

Because of this equation and Eq. (14.36) to finish the proof that x → ⟨x|y⟩ is linear, it suffices to show ⟨λx|y⟩ = λ⟨x|y⟩ for all λ > 0. Now if λ = m ∈ ℕ, then

⟨mx|y⟩ = ⟨x + (m − 1)x|y⟩ = ⟨x|y⟩ + ⟨(m − 1)x|y⟩

so that by induction ⟨mx|y⟩ = m⟨x|y⟩. Replacing x by x/m then shows that ⟨x|y⟩ = m⟨m^{−1}x|y⟩ so that ⟨m^{−1}x|y⟩ = m^{−1}⟨x|y⟩ and so if m, n ∈ ℕ, we find

⟨\frac{n}{m}x|y⟩ = n⟨\frac{1}{m}x|y⟩ = \frac{n}{m}⟨x|y⟩

so that ⟨λx|y⟩ = λ⟨x|y⟩ for all λ > 0 and λ ∈ ℚ. By continuity, it now follows that ⟨λx|y⟩ = λ⟨x|y⟩ for all λ > 0.

14.6 Supplement 2. Non-complete inner product spaces

Part of Theorem 14.24 goes through when H is a not necessarily complete inner product space. We have the following proposition.

Proposition 14.55. Let (H, ⟨·|·⟩) be a not necessarily complete inner product space and β ⊂ H be an orthonormal set. Then the following two conditions are equivalent:

1. x = ∑_{u∈β}⟨x|u⟩u for all x ∈ H.

2. ∥x∥^{2} = ∑_{u∈β} |⟨x|u⟩|^{2} for all x ∈ H.

Moreover, either of these two conditions implies that β ⊂ H is a maximal orthonormal set. However β ⊂ H being a maximal orthonormal set is not sufficient to conditions for 1) and 2) hold!

Proof. As in the proof of Theorem 14.24, 1) implies 2). For 2) implies 1) let Λ ⊂⊂ β and consider

∥x − ∑_{u∈Λ}⟨x|u⟩u∥^{2} = ∥x∥^{2} − 2 ∑_{u∈Λ} |⟨x|u⟩|^{2} + ∑_{u∈Λ} |⟨x|u⟩|^{2}

= ∥x∥^{2} − ∑_{u∈Λ} |⟨x|u⟩|^{2} .

Since ∥x∥^{2} = ∑_{u∈β} |⟨x|u⟩|^{2}, it follows that for every ε > 0 there exists Λ_{ε} ⊂⊂ β

such that for all Λ ⊂⊂ β such that Λ_{ε} ⊂ Λ,

∥x − ∑_{u∈Λ}⟨x|u⟩u∥^{2} = ∥x∥^{2} − ∑_{u∈Λ} |⟨x|u⟩|^{2} < ε

showing that x = ∑_{u∈β}⟨x|u⟩u. Suppose x = (x_{1}, x_{2}, …, x_{n}, …) ∈ β^{⊥}. If 2)

is valid then ∥x∥^{2} = 0, i.e. x = 0. So β is maximal. Let us now construct a counter example to prove the last assertion. Take H = Span{e_{i}}_{i=1}^{∞} ⊂ ℓ^{2} and let ũ_{n} = e_{1} − (n + 1)e_{n+1} for n = 1, 2 …. Applying Gramn-Schmidt to \left\{{ \atop ũ_{n}}\right\}_{n=1}^{∞} we construct an orthonormal set β = \{u_{n}\}_{n=1}^{∞} ⊂ H. I now claim that β ⊂ H is maximal. Indeed if x = (x_{1}, x_{2}, …, x_{n}, …) ∈ β^{⊥} then x ⊥ u_{n} for all n, i.e.

0 = ⟨x|ũ_{n}⟩ = x_{1} − (n + 1)x_{n+1}.

Therefore x_{n+1} = (n + 1)^{−1} x_{1} for all n. Since x ∈ Span{e_{i}}_{i=1}^{∞}, x_{N} = 0 for some N sufficiently large and therefore x_{1} = 0 which in turn implies that x_{n} = 0 for all n. So x = 0 and hence β is maximal in H. On the other hand, β is not maximal in ℓ^{2}. In fact the above argument shows that β^{⊥} in ℓ^{2} is given by the span of v = (1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, …). Let P be the orthogonal projection of ℓ^{2} onto the Span(β) = v^{⊥}. Then

∑_{i=1}^{∞}⟨x|u_{n}⟩u_{n} = P x = x − \frac{⟨x|v⟩}{∥v∥^{2}}v,

so that ∑_{i=1}^{∞}⟨x|u_{n}⟩u_{n} = x iff x ∈ Span(β) = v^{⊥} ⊂ ℓ^{2}. For example if x =

(1, 0, 0, …) ∈ H (or more generally for x = e_{i} for any i), x ∉ v^{⊥} and hence

∑_{i=1}^{∞}⟨x|u_{n}⟩u_{n} ≠ x.

14.7 Exercises

Exercise 14.13. Prove Theorem 14.52. Hint: Let H_{0} := \overline{span \{x_{n} : n ∈ ℕ\}} – a separable Hilbert subspace of H. Let {λ_{m}}_{m=1}^{∞} ⊂ H_{0} be an orthonormal basis and use Cantor’s diagonalization argument to find a subsequence y_{k} := x_{n_{k}} such that c_{m} := lim_{k→∞}⟨y_{k}|λ_{m}⟩ exists for all m ∈ ℕ. Finish the proof by appealing to Proposition 14.51.

Exercise 14.14. Suppose that {x_{n}}_{n=1}^{∞} ⊂ H and x_{n} →^{w} x ∈ H as n → ∞. Show x_{n} → x as n → ∞ (i.e. lim_{n→∞} ∥x − x_{n}∥ = 0) iff lim_{n→∞} ∥x_{n}∥ = ∥x∥ .

Exercise 14.15 (Banach-Saks). Suppose that {x_{n}}_{n=1}^{∞} ⊂ H, x_{n} →^{w} x ∈ H as n → ∞, and c := sup_{n} ∥x_{n}∥ < ∞.^{4} Show there exists a subsequence, y_{k} = x_{n_{k}} such that

lim_{N→∞} ∥x − \frac{1}{N} ∑_{k=1}^{N} y_{k}∥ = 0,

i.e. \frac{1}{N} ∑_{k=1}^{N} y_{k} → x as N → ∞. Hints: 1. show it suffices to assume x = 0 and then choose {y_{k}}_{k=1}^{∞} so that |⟨y_{k}|y_{l}⟩| ≤ l^{−1} (or even smaller if you like) for all k ≤ l.

Exercise 14.16 (The Mean Ergodic Theorem). Let U : H → H be a unitary operator on a Hilbert space H, M = Nul(U −I), P = P_{M} be orthogonal projection onto M, and S_{n} = \frac{1}{n} ∑_{k=0}^{n−1} U^{k}. Show S_{n} → P_{M} strongly, i.e. lim_{n→∞} S_{n}x = P_{M}x for all x ∈ H. Hints: 1. verify the result for x ∈ Nul(U − I) and x ∈ Ran(U − I), 2. show Nul(U^{∗} − I) = Nul(U − I), 3. finish the result with a limiting argument making use of items 1. and 2. and Lemma 14.17.

4 The assumption that c < ∞ is superfluous because of the “uniform boundedness principle,” see Theorem ?? below.

Part IV

Calculus and Ordinary Differential Equations in Banach Spaces Ordinary Differential Equations in a Banach Space

Let X be a Banach space, U ⊂_{o} X, J = (a, b) ∋ 0 and Z ∈ C (J × U, X) – Z is to be interpreted as a time dependent vector-field on U ⊂ X. In this section we will consider the ordinary differential equation (ODE for short)

ẏ(t) = Z(t, y(t)) with y(0) = x ∈ U.  (15.1)

The reader should check that any solution y ∈ C^{1}(J, U) to Eq. (15.1) gives a solution y ∈ C(J, U) to the integral equation:

y(t) = x + ∫_{0}^{t} Z(τ, y(τ))dτ  (15.2)

and conversely if y ∈ C(J, U) solves Eq. (15.2) then y ∈ C^{1}(J, U) and y solves Eq. (15.1).

Remark 15.1. For notational simplicity we have assumed that the initial condition for the ODE in Eq. (15.1) is taken at t = 0. There is no loss in generality in doing this since if ỹ solves

\frac{dỹ}{dt}(t) = Z̃(t, ỹ(t)) with ỹ(t_{0}) = x ∈ U

iff y(t) := ỹ(t + t_{0}) solves Eq. (15.1) with Z(t, x) = Z̃(t + t_{0}, x).

15.1 Examples

Let X = ℝ, Z(x) = x^{n} with n ∈ ℕ and consider the ordinary differential equation

ẏ(t) = Z(y(t)) = y^{n}(t) with y(0) = x ∈ ℝ.  (15.3)

If y solves Eq. (15.3) with x ≠ 0, then y(t) is not zero for t near 0. Therefore up to the first time y possibly hits 0, we must have

t = ∫_{0}^{t} \frac{ẏ(τ)}{y(τ)^{n}} dτ = ∫_{0}^{y(t)} u^{−n}du = \{ \begin{matrix}\frac{[y(t)]^{1−n}−x^{1−n}}{1−n} & if & n > 1 \\ ln |\frac{y(t)}{x} | & if & n = 1 \end{matrix}

and solving these equations for y(t) implies

y(t) = y(t, x) = \{ \begin{matrix}\frac{x}{\sqrt[n−1]{1−(n−1)tx^{n−1}}} & if & n > 1 \\ e^{t}x & if & n = 1. \end{matrix}  (15.4)

The reader should verify by direct calculation that y(t, x) defined above does indeed solve Eq. (15.3). The above argument shows that these are the only possible solutions to the Equations in (15.3).

Notice that when n = 1, the solution exists for all time while for n > 1, we must require

1 − (n − 1)tx^{n−1} > 0

or equivalently that

t < \frac{1}{(1 − n)x^{n−1}} if x^{n−1} > 0 and
t > −\frac{1}{(1 − n) |x|^{n−1}} if x^{n−1} < 0.

Moreover for n > 1, y(t, x) blows up as t approaches the value for which 1 − (n − 1)tx^{n−1} = 0. The reader should also observe that, at least for s and t close to 0,

y(t, y(s, x)) = y(t + s, x)  (15.5)

for each of the solutions above. Indeed, if n = 1 Eq. (15.5) is equivalent to the well know identity, e^{t}e^{s} = e^{t+s} and for n > 1,

y(t, y(s, x)) = \frac{y(s, x)}{\sqrt[n−1]{1 − (n − 1)ty(s, x)^{n−1}}}

= \frac{\frac{x}{\sqrt[n−1]{1−(n−1)sx^{n−1}}}}{\sqrt[n−1]{1 − (n − 1)t [ \frac{x}{\sqrt[n−1]{1−(n−1)sx^{n−1}}} ]^{n−1}}}

= \frac{\frac{x}{\sqrt[n−1]{1−(n−1)sx^{n−1}}}}{\sqrt[n−1]{1 − (n − 1)t\frac{x^{n−1}}{1−(n−1)sx^{n−1}}}}

= \frac{x}{\sqrt[n−1]{1 − (n − 1)sx^{n−1} − (n − 1)tx^{n−1}}}

= \frac{x}{\sqrt[n−1]{1 − (n − 1)(s + t)x^{n−1}}} = y(t + s, x).

Now suppose Z(x) = |x|^{α} with 0 < α < 1 and we now consider the ordinary differential equation

ẏ(t) = Z(y(t)) = |y(t)|^{α} with y(0) = x ∈ ℝ.  (15.6)

Working as above we find, if x ≠ 0 that

t = ∫_{0}^{t} \frac{ẏ(τ)}{|y(t)|^{α}} dτ = ∫_{0}^{y(t)} |u|^{−α} du = \frac{[y(t)]^{1−α} − x^{1−α}}{1 − α} ,

where u^{1−α} := |u|^{1−α} sgn(u). Since sgn(y(t)) = sgn(x) the previous equation implies

sgn(x)(1 − α)t = sgn(x) [sgn(y(t)) |y(t)|^{1−α} − sgn(x) |x|^{1−α}]
= |y(t)|^{1−α} − |x|^{1−α}

and therefore,

y(t, x) = sgn(x) (|x|^{1−α} + sgn(x)(1 − α)t)^{\frac{1}{1−α}}  (15.7)

is uniquely determined by this formula until the first time t where |x|^{1−α} + sgn(x)(1 − α)t = 0. As before y(t) = 0 is a solution to Eq. (15.6), however it is far from being the unique solution. For example letting x ↓ 0 in Eq. (15.7) gives a function

y(t, 0+) = ((1 − α)t)^{\frac{1}{1−α}}

which solves Eq. (15.6) for t > 0. Moreover if we define

y(t) := \{ \begin{matrix}((1 − α)t)^{\frac{1}{1−α}} & if & t > 0 \\ 0 & if & t ≤ 0 \end{matrix} ,

(for example if α = 1/2 then y(t) = \frac{1}{4}t^{2}1_{t≥0}) then the reader may easily check y also solve Eq. (15.6). Furthermore, y_{a}(t) := y(t − a) also solves Eq. (15.6) for all a ≥ 0, see Figure 15.1 below.

With these examples in mind, let us now go to the general theory. The case of linear ODE’s has already been studied in Section 8.3 above.

15.2 Uniqueness Theorem and Continuous Dependence on Initial Data

Lemma 15.2. Gronwall’s Lemma. Suppose that f, ε, and k are nonnegative functions of a real variable t such that

f(t) ≤ ε(t) + |∫_{0}^{t} k(τ)f(τ)dτ| .  (15.8)

Fig. 15.1. Three different solutions to the ODE ẏ(t) = |y(t)|^{1/2} with y(0) = 0.

f(t) ≤ ε(t) + |∫_{0}^{t} k(τ)ε(τ)e^{|∫_{τ}^{t} k(s)ds|}dτ| ,  (15.9)

and in particular if ε and k are constants we find that

f(t) ≤ εe^{k|t|}.  (15.10)

Proof. I will only prove the case t ≥ 0. The case t ≤ 0 can be derived by applying the t ≥ 0 to f̃(t) = f(−t), k̃(t) = k(−t) and ε(t) = ε(−t). Set F (t) = ∫_{0}^{t} k(τ)f(τ)dτ. Then by (15.8),

Ḟ = kf ≤ kε + kF.

Hence,

\frac{d}{dt}(e^{− ∫_{0}^{t} k(s)ds}F) = e^{− ∫_{0}^{t} k(s)ds}(Ḟ − kF) ≤ kεe^{− ∫_{0}^{t} k(s)ds}.

Integrating this last inequality from 0 to t and then solving for F yields:

F (t) ≤ e^{∫_{0}^{t} k(s)ds} · ∫_{0}^{t} dτk(τ)ε(τ)e^{− ∫_{0}^{τ} k(s)ds} = ∫_{0}^{t} dτk(τ)ε(τ)e^{∫_{τ}^{t} k(s)ds}.

But by the definition of F we have that

f ≤ ε + F,

and hence the last two displayed equations imply (15.9). Equation (15.10) follows from (15.9) by a simple integration.

Corollary 15.3 (Continuous Dependence on Initial Data). Let U ⊂_{o} X, 0 ∈ (a, b) and Z : (a, b) × U → X be a continuous function which is K– Lipschitz function on U, i.e. ∥Z(t, x) − Z(t, x′)∥ ≤ K∥x − x′∥ for all x and x′ in U. Suppose y_{1}, y_{2} : (a, b) → U solve

\frac{dy_{i}(t)}{dt} = Z(t, y_{i}(t)) with y_{i}(0) = x_{i} for i = 1, 2.  (15.11)

Then

∥y_{2}(t) − y_{1}(t)∥ ≤ ∥x_{2} − x_{1}∥e^{K|t|} for t ∈ (a, b)  (15.12)

and in particular, there is at most one solution to Eq. (15.1) under the above Lipschitz assumption on Z.

Proof. Let f(t) := ∥y_{2}(t) − y_{1}(t)∥. Then by the fundamental theorem of calculus,

f(t) = ∥y_{2}(0) − y_{1}(0) + ∫_{0}^{t}(ẏ_{2}(τ) − ẏ_{1}(τ)) dτ∥
≤ f(0) + |∫_{0}^{t} ∥Z(τ, y_{2}(τ)) − Z(τ, y_{1}(τ))∥ dτ|
= ∥x_{2} − x_{1}∥ + K |∫_{0}^{t} f(τ) dτ| .

Therefore by Gronwall’s inequality we have,

∥y_{2}(t) − y_{1}(t)∥ = f(t) ≤ ∥x_{2} − x_{1}∥e^{K|t|}.

15.3 Local Existence (Non-Linear ODE)

We now show that Eq. (15.1) under a Lipschitz condition on Z. Another existence theorem was given in Exercise 11.19.

Theorem 15.4 (Local Existence). Let T > 0, J = (−T, T), x_{0} ∈ X, r > 0 and

C(x_{0}, r) := {x ∈ X : ∥x − x_{0}∥ ≤ r}

be the closed r – ball centered at x_{0} ∈ X. Assume

M = sup {∥Z(t, x)∥ : (t, x) ∈ J × C(x_{0}, r)} < ∞  (15.13)

and there exists K < ∞ such that

∥Z(t, x) − Z(t, y)∥ ≤ K ∥x − y∥ for all x, y ∈ C(x_{0}, r) and t ∈ J. (15.14)

Let T_{0} < min {r/M, T } and J_{0} := (−T_{0}, T_{0}), then for each x ∈ B(x_{0}, r−MT_{0}) there exists a unique solution y(t) = y(t, x) to Eq. (15.2) in C (J_{0}, C(x_{0}, r)) . Moreover y(t, x) is jointly continuous in (t, x), y(t, x) is differentiable in t, ẏ(t, x) is jointly continuous for all (t, x) ∈ J_{0} × B(x_{0}, r − MT_{0}) and satisfies Eq. (15.1).

Proof. The uniqueness assertion has already been proved in Corollary 15.3. To prove existence, let C_{r} := C(x_{0}, r), Y := C (J_{0}, C(x_{0}, r)) and

S_{x}(y)(t) := x + ∫_{0}^{t} Z(τ, y(τ))dτ.  (15.15)

With this notation, Eq. (15.2) becomes y = S_{x}(y), i.e. we are looking for a fixed point of S_{x}. If y ∈ Y, then

∥S_{x}(y)(t) − x_{0}∥ ≤ ∥x − x_{0}∥ + |∫_{0}^{t} ∥Z(τ, y(τ))∥ dτ| ≤ ∥x − x_{0}∥ + M |t|
≤ ∥x − x_{0}∥ + MT_{0} ≤ r − MT_{0} + MT_{0} = r,

showing S_{x} (Y) ⊂ Y for all x ∈ B(x_{0}, r − MT_{0}). Moreover if y, z ∈ Y,

∥S_{x}(y)(t) − S_{x}(z)(t)∥ = ∥∫_{0}^{t} [Z(τ, y(τ)) − Z(τ, z(τ))] dτ∥
≤ |∫_{0}^{t} ∥Z(τ, y(τ)) − Z(τ, z(τ))∥ dτ|
≤ K |∫_{0}^{t} ∥y(τ) − z(τ)∥ dτ| .  (15.16)

Let y_{0}(t, x) = x and y_{n}(·, x) ∈ Y defined inductively by

y_{n}(·, x) := S_{x}(y_{n−1}(·, x)) = x + ∫_{0}^{t} Z(τ, y_{n−1}(τ, x))dτ.  (15.17)

Using the estimate in Eq. (15.16) repeatedly we find

|| y_{n+1}(t) − y_{n}(t) ||
≤ K |∫_{0}^{t} ∥y_{n}(τ) − y_{n−1}(τ)∥ dτ|
≤ K^{2} |∫_{0}^{t} dt_{1} |∫_{0}^{t_{1}} dt_{2} ∥y_{n−1}(t_{2}) − y_{n−2}(t_{2})∥||
⋮
≤ K^{n} |∫_{0}^{t} dt_{1} |∫_{0}^{t_{1}} dt_{2} … |∫_{0}^{t_{n−1}} dt_{n} ∥y_{1}(t_{n}) − y_{0}(t_{n})∥| … ||
≤ K^{n} ∥y_{1}(·, x) − y_{0}(·, x)∥_{∞} ∫_{Δ_{n}(t)} dτ
= \frac{K^{n} |t|^{n}}{n!} ∥y_{1}(·, x) − y_{0}(·, x)∥_{∞} ≤ 2r \frac{K^{n} |t|^{n}}{n!}  (15.18)

wherein we have also made use of Lemma 8.19. Combining this estimate with

∥y_{1}(t, x) − y_{0}(t, x)∥ = ∥∫_{0}^{t} Z(τ, x)dτ∥ ≤ |∫_{0}^{t} ∥Z(τ, x)∥ dτ| ≤ M_{0},

where

M_{0} = T_{0} max {∫_{0}^{T_{0}} ∥Z(τ, x)∥ dτ, ∫_{−T_{0}}^{0} ∥Z(τ, x)∥ dτ} ≤ MT_{0},

shows

∥y_{n+1}(t, x) − y_{n}(t, x)∥ ≤ M_{0} \frac{K^{n} |t|^{n}}{n!} ≤ M_{0} \frac{K^{n}T_{0}^{n}}{n!}

and this implies

∑_{n=0}^{∞} sup{ ∥y_{n+1}(·, x) − y_{n}(·, x)∥_{∞,J_{0}} : t ∈ J_{0}}

≤ ∑_{n=0}^{∞} M_{0} \frac{K^{n}T_{0}^{n}}{n!} = M_{0}e^{KT_{0}} < ∞

where

∥y_{n+1}(·, x) − y_{n}(·, x)∥_{∞,J_{0}} := sup {∥y_{n+1}(t, x) − y_{n}(t, x)∥ : t ∈ J_{0}} .

So y(t, x) := lim_{n→∞} y_{n}(t, x) exists uniformly for t ∈ J and using Eq. (15.14) we also have

sup{ ∥Z(t, y(t)) − Z(t, y_{n−1}(t))∥ : t ∈ J_{0}}
≤ K ∥y(·, x) − y_{n−1}(·, x)∥_{∞,J_{0}} → 0 as n → ∞.

Now passing to the limit in Eq. (15.17) shows y solves Eq. (15.2). From this equation it follows that y(t, x) is differentiable in t and y satisfies Eq. (15.1). The continuity of y(t, x) follows from Corollary 15.3 and mean value inequality (Corollary 8.14):

∥y(t, x) − y(t′, x′)∥ ≤ ∥y(t, x) − y(t, x′)∥ + ∥y(t, x′) − y(t′, x′)∥
= ∥y(t, x) − y(t, x′)∥ + ∥∫_{t′}^{t} Z(τ, y(τ, x′))dτ∥
≤ ∥y(t, x) − y(t, x′)∥ + |∫_{t′}^{t} ∥Z(τ, y(τ, x′))∥ dτ|
≤ ∥x − x′∥e^{KT} + |∫_{t′}^{t} ∥Z(τ, y(τ, x′))∥ dτ|  (15.19)
≤ ∥x − x′∥e^{KT} + M |t − t′| .

The continuity of ẏ(t, x) is now a consequence Eq. (15.1) and the continuity of y and Z.

Corollary 15.5. Let J = (a, b) ∋ 0 and suppose Z ∈ C(J × X, X) satisfies

∥Z(t, x) − Z(t, y)∥ ≤ K ∥x − y∥ for all x, y ∈ X and t ∈ J.  (15.20)

Then for all x ∈ X, there is a unique solution y(t, x) (for t ∈ J) to Eq. (15.1). Moreover y(t, x) and ẏ(t, x) are jointly continuous in (t, x).

Proof. Let J_{0} = (a_{0}, b_{0}) ∋ 0 be a precompact subinterval of J and Y := BC (J_{0}, X) . By compactness, M := sup_{t∈J̄_{0}} ∥Z(t, 0)∥ < ∞ which combined with Eq. (15.20) implies

sup_{t∈J̄_{0}} ∥Z(t, x)∥ ≤ M + K ∥x∥ for all x ∈ X.

Using this estimate and Lemma 8.7 one easily shows S_{x}(Y) ⊂ Y for all x ∈ X. The proof of Theorem 15.4 now goes through without any further change.

15.4 Global Properties

Definition 15.6 (Local Lipschitz Functions). Let U ⊂_{o} X, J be an open interval and Z ∈ C(J ×U, X). The function Z is said to be locally Lipschitz in x if for all x ∈ U and all compact intervals I ⊂ J there exists K = K(x, I) < ∞ and ε = ε(x, I) > 0 such that B(x, ε(x, I)) ⊂ U and

∥Z(t, x_{1}) − Z(t, x_{0})∥ ≤ K(x, I)∥x_{1} − x_{0}∥ ∀ x_{0}, x_{1} ∈ B(x, ε(x, I)) & t ∈ I. (15.21)

For the rest of this section, we will assume J is an open interval containing 0, U is an open subset of X and Z ∈ C(J×U, X) is a locally Lipschitz function.

Lemma 15.7. Let Z ∈ C(J × U, X) be a locally Lipschitz function in X and E be a compact subset of U and I be a compact subset of J. Then there exists ε > 0 such that Z(t, x) is bounded for (t, x) ∈ I × E_{ε} and and Z(t, x) is K – Lipschitz on E_{ε} for all t ∈ I, where

E_{ε} := {x ∈ U : dist(x, E) < ε} .

Proof. Let ε(x, I) and K(x, I) be as in Definition 15.6 above. Since E is compact, there exists a finite subset Λ ⊂ E such that E ⊂ V := ∪_{x∈Λ}B(x, ε(x, I)/2). If y ∈ V, there exists x ∈ Λ such that ∥y − x∥ < ε(x, I)/2 and therefore

∥Z(t, y)∥ ≤ ∥Z(t, x)∥ + K(x, I) ∥y − x∥ ≤ ∥Z(t, x)∥ + K(x, I)ε(x, I)/2

≤ sup_{x∈Λ,t∈I} {∥Z(t, x)∥ + K(x, I)ε(x, I)/2} =: M < ∞.

This shows Z is bounded on I × V. Let

ε := d(E, V^{c}) ≤ \frac{1}{2} min_{x∈Λ} ε(x, I)

and notice that ε > 0 since E is compact, V^{c} is closed and E ∩ V^{c} = ∅. If y, z ∈ E_{ε} and ∥y − z∥ < ε, then as before there exists x ∈ Λ such that ∥y − x∥ < ε(x, I)/2. Therefore

∥z − x∥ ≤ ∥z − y∥ + ∥y − x∥ < ε + ε(x, I)/2 ≤ ε(x, I)

and since y, z ∈ B(x, ε(x, I)), it follows that

∥Z(t, y) − Z(t, z)∥ ≤ K(x, I)∥y − z∥ ≤ K_{0}∥y − z∥

where K_{0} := max_{x∈Λ} K(x, I) < ∞. On the other hand if y, z ∈ E_{ε} and ∥y − z∥ ≥ ε, then

∥Z(t, y) − Z(t, z)∥ ≤ 2M ≤ \frac{2M}{ε} ∥y − z∥ .

Thus if we let K := max {2M/ε, K_{0}} , we have shown

∥Z(t, y) − Z(t, z)∥ ≤ K∥y − z∥ for all y, z ∈ E_{ε} and t ∈ I.

Proposition 15.8 (Maximal Solutions). Let Z ∈ C(J × U, X) be a locally Lipschitz function in x and let x ∈ U be fixed. Then there is an interval J_{x} = (a(x), b(x)) with a ∈ [−∞, 0) and b ∈ (0, ∞] and a C^{1}–function y : J → U with the following properties:

1. y solves ODE in Eq. (15.1).

2. If ỹ : J̃ = (ã, b̃) → U is another solution of Eq. (15.1) (we assume that 0 ∈ J̃) then J̃ ⊂ J and ỹ = y|_{J̃}.

The function y : J → U is called the maximal solution to Eq. (15.1).

Proof. Suppose that y_{i} : J_{i} = (a_{i}, b_{i}) → U, i = 1, 2, are two solutions to Eq. (15.1). We will start by showing the y_{1} = y_{2} on J_{1} ∩ J_{2}. To do this^{1} let

1 Here is an alternate proof of the uniqueness. Let

T ≡ sup{t ∈ [0, min{b_{1}, b_{2}}) : y_{1} = y_{2} on [0, t]}.

(T is the first positive time after which y_{1} and y_{2} disagree.

Suppose, for sake of contradiction, that T < min{b_{1}, b_{2}}. Notice that y_{1}(T) = y_{2}(T) =: x′. Applying the local uniqueness theorem to y_{1}(· − T) and y_{2}(· − T) thought as function from (−δ, δ) → B(x′, ϵ(x′)) for some δ sufficiently small, we learn that y_{1}(·−T) = y_{2}(·−T) on (−δ, δ). But this shows that y_{1} = y_{2} on [0, T +δ) which contradicts the definition of T. Hence we must have the T = min{b_{1}, b_{2}}, i.e. y_{1} = y_{2} on J_{1} ∩ J_{2} ∩ [0, ∞). A similar argument shows that y_{1} = y_{2} on J_{1} ∩ J_{2} ∩ (−∞, 0] as well.

J_{0} = (a_{0}, b_{0}) be chosen so that 0 ∈ J_{0} ⊂ J_{1} ∩J_{2}, and let E := y_{1}(J_{0})∪y_{2}(J_{0}) – a compact subset of X. Choose ε > 0 as in Lemma 15.7 so that Z is Lipschitz on E_{ε}. Then y_{1}|_{J_{0}}, y_{2}|_{J_{0}} : J_{0} → E_{ε} both solve Eq. (15.1) and therefore are equal by Corollary 15.3. Since J_{0} = (a_{0}, b_{0}) was chosen arbitrarily so that [a, b] ⊂ J_{1} ∩ J_{2}, we may conclude that y_{1} = y_{2} on J_{1} ∩ J_{2}. Let (y_{α}, J_{α} = (a_{α}, b_{α}))_{α∈A} denote the possible solutions to (15.1) such that 0 ∈ J_{α}. Define J_{x} = ∪J_{α} and set y = y_{α} on J_{α}. We have just checked that y is well defined and the reader may easily check that this function y : J_{x} → U satisfies all the conclusions of the theorem.

Notation 15.9 For each x ∈ U, let J_{x} = (a(x), b(x)) be the maximal interval on which Eq. (15.1) may be solved, see Proposition 15.8. Set D(Z) := ∪_{x∈U}(J_{x} × {x}) ⊂ J × U and let ϕ : D(Z) → U be defined by ϕ(t, x) = y(t) where y is the maximal solution to Eq. (15.1). (So for each x ∈ U, ϕ(·, x) is the maximal solution to Eq. (15.1).)

Proposition 15.10. Let Z ∈ C(J × U, X) be a locally Lipschitz function in x and y : J_{x} = (a(x), b(x)) → U be the maximal solution to Eq. (15.1). If b(x) < b, then either lim sup_{t↑b(x)} ∥Z(t, y(t))∥ = ∞ or y(b(x)−) := lim_{t↑b(x)} y(t) exists and y(b(x)−) ∉ U. Similarly, if a > a(x), then either lim sup_{t↓a(x)} ∥y(t)∥ = ∞ or y(a(x)+) := lim_{t↓a} y(t) exists and y(a(x)+) ∉ U.

Proof. Suppose that b < b(x) and M := lim sup_{t↑b(x)} ∥Z(t, y(t))∥ < ∞. Then there is a b_{0} ∈ (0, b(x)) such that ∥Z(t, y(t))∥ ≤ 2M for all t ∈ (b_{0}, b(x)). Thus, by the usual fundamental theorem of calculus argument,

∥y(t) − y(t′)∥ ≤ |∫_{t}^{t′} ∥Z(t, y(τ))∥ dτ| ≤ 2M|t − t′|

for all t, t′ ∈ (b_{0}, b(x)). From this it is easy to conclude that y(b(x)−) = lim_{t↑b(x)} y(t) exists. If y(b(x)−) ∈ U, by the local existence Theorem 15.4, there exists δ > 0 and w ∈ C^{1} ((b(x) − δ, b(x) + δ), U) such that

ẇ(t) = Z(t, w(t)) and w(b(x)) = y(b(x)−).

Now define ỹ: (a, b(x) + δ) → U by

ỹ(t) = \{ \begin{matrix}y(t) & if t ∈ J_{x} \\ w(t) & if t ∈ [b(x), b(x) + δ) \end{matrix} .

The reader may now easily show ỹ solves the integral Eq. (15.2) and hence also solves Eq. 15.1 for t ∈ (a(x), b(x) + δ).^{2} But this violates the maximality of y and hence we must have that y(b(x)−) ∉ U. The assertions for t near a(x) are proved similarly.

2 See the argument in Proposition 15.13 for a slightly different method of extending y which avoids the use of the integral equation (15.2).

Example 15.11. Let X = ℝ^{2}, J = ℝ, U = {(x, y) ∈ ℝ^{2} : 0 < r < 1} where r^{2} = x^{2} + y^{2} and

Z(x, y) = \frac{1}{r}(x, y) + \frac{1}{1 − r^{2}} (−y, x).

The the unique solution (x(t), y(t)) to

\frac{d}{dt}(x(t), y(t)) = Z(x(t), y(t)) with (x(0), y(0)) = (\frac{1}{2}, 0)

is given by

(x(t), y(t)) = (t + \frac{1}{2}) (cos (\frac{1}{1/2 − t}) , sin (\frac{1}{1/2 − t}))

for t ∈ J_{(1/2,0)} = (−∞, 1/2) . Notice that ∥Z(x(t), y(t))∥ → ∞ as t ↑ 1/2 and dist((x(t), y(t)), U^{c}) → 0 as t ↑ 1/2.

Example 15.12. (Not worked out completely.) Let X = U = ℓ^{2}, ψ ∈ C^{∞}(ℝ^{2}) be a smooth function such that ψ = 1 in a neighborhood of the line segment joining (1, 0) to (0, 1) and being supported within the 1/10 – neighborhood of this segment. Choose a_{n} ↑ ∞ and b_{n} ↑ ∞ and define

Z(x) = ∑_{n=1}^{∞} a_{n}ψ(b_{n}(x_{n}, x_{n+1}))(e_{n+1} − e_{n}).  (15.22)

For any x ∈ ℓ^{2}, only a finite number of terms are non-zero in the above some in a neighborhood of x. Therefor Z : ℓ^{2} → ℓ^{2} is a smooth and hence locally Lipshcitz vector field. Let (y(t), J = (a, b)) denote the maximal solution to

ẏ(t) = Z(y(t)) with y(0) = e_{1}.

Then if the a_{n} and b_{n} are chosen appropriately, then b < ∞ and there will exist t_{n} ↑ b such that y(t_{n}) is approximately e_{n} for all n. So again y(t_{n}) does not have a limit yet sup_{t∈[0,b)} ∥y(t)∥ < ∞. The idea is that Z is constructed to blow the particle form e_{1} to e_{2} to e_{3} to e_{4} etc. etc. with the time it takes to travel from e_{n} to e_{n+1} being on order 1/2^{n}. The vector field in Eq. (15.22) is a first approximation at such a vector field, it may have to be adjusted a little more to provide an honest example. In this example, we are having problems because y(t) is “going off in dimensions.”

Here is another version of Proposition 15.10 which is more useful when dim(X) < ∞.

Proposition 15.13. Let Z ∈ C(J × U, X) be a locally Lipschitz function in x and y : J_{x} = (a(x), b(x)) → U be the maximal solution to Eq. (15.1).

1. If b(x) < b, then for every compact subset K ⊂ U there exists T_{K} < b(x) such that y(t) ∉ K for all t ∈ [T_{K}, b(x)).

2. When dim(X) < ∞, we may write this condition as: if b(x) < b, then

either

lim sup_{t↑b(x)} ∥y(t)∥ = ∞ or lim inf_{t↑b(x)} dist(y(t), U^{c}) = 0.

Proof. 1) Suppose that b(x) < b and, for sake of contradiction, there exists a compact set K ⊂ U and t_{n} ↑ b(x) such that y(t_{n}) ∈ K for all n. Since K is compact, by passing to a subsequence if necessary, we may assume y_{∞} := lim_{n→∞} y(t_{n}) exists in K ⊂ U. By the local existence Theorem 15.4, there exists T_{0} > 0 and δ > 0 such that for each x′ ∈ B (y_{∞}, δ) there exists a unique solution w(·, x′) ∈ C^{1}((−T_{0}, T_{0}), U) solving

w(t, x′) = Z(t, w(t, x′)) and w(0, x′) = x′.

Now choose n sufficiently large so that t_{n} ∈ (b(x) − T_{0}/2, b(x)) and y(t_{n}) ∈ B (y_{∞}, δ) . Define ỹ: (a(x), b(x) + T_{0}/2) → U by

ỹ(t) = \{ \begin{matrix}y(t) & if t ∈ J_{x} \\ w(t − t_{n}, y(t_{n})) & if t ∈ (t_{n} − T_{0}, b(x) + T_{0}/2). \end{matrix}

wherein we have used (t_{n} −T_{0}, b(x)+T_{0}/2) ⊂ (t_{n} −T_{0}, t_{n} +T_{0}). By uniqueness of solutions to ODE’s ỹ is well defined, ỹ ∈ C^{1}((a(x), b(x) + T_{0}/2) , X) and ỹ solves the ODE in Eq. 15.1. But this violates the maximality of y. 2) For each n ∈ ℕ let

K_{n} := {x ∈ U : ∥x∥ ≤ n and dist(x, U^{c}) ≥ 1/n} .

Then K_{n} ↑ U and each K_{n} is a closed bounded set and hence compact if dim(X) < ∞. Therefore if b(x) < b, by item 1., there exists T_{n} ∈ [0, b(x)) such that y(t) ∉ K_{n} for all t ∈ [T_{n}, b(x)) or equivalently ∥y(t)∥ > n or dist(y(t), U^{c}) < 1/n for all t ∈ [T_{n}, b(x)).

Remark 15.14. In general it is not true that the functions a and b are continuous. For example, let U be the region in ℝ^{2} described in polar coordinates by r > 0 and 0 < θ < 3π/4 and Z(x, y) = (0, −1) as in Figure 15.2 below. Then b(x, y) = y for all x, y > 0 while b(x, y) = ∞ for all x < 0 and y ∈ ℝ which shows b is discontinuous. On the other hand notice that

{b > t} = {x < 0} ∪ {(x, y) : x ≥ 0, y > t}

is an open set for all t > 0. An example of a vector field for which b(x) is discontinuous is given in the top left hand corner of Figure 15.2. The map ψ would allow the reader to find an example on ℝ^{2} if so desired. Some calculations shows that Z transferred to ℝ^{2} by the map ψ is given by the new vector

Z̃(x, y) = −e^{−x} (sin (\frac{3π}{8} + \frac{3}{4} tan^{−1} (y)) , cos (\frac{3π}{8} + \frac{3}{4} tan^{−1} (y))) .

Fig. 15.2. Manufacturing vector fields where b(x) is discontinuous.

Theorem 15.15 (Global Continuity). Let Z ∈ C(J × U, X) be a locally Lipschitz function in x. Then D(Z) is an open subset of J × U and the functions ϕ : D(Z) → U and ϕ̇ : D(Z) → U are continuous. More precisely, for all x_{0} ∈ U and all open intervals J_{0} such that 0 ∈ J_{0} ⊏⊏ J_{x_{0}} there exists δ = δ(x_{0}, J_{0}, Z) > 0 and C = C(x_{0}, J_{0}, Z) < ∞ such that for all x ∈ B(x_{0}, δ), J_{0} ⊂ J_{x} and

∥ϕ(·, x) − ϕ(·, x_{0})∥_{BC(J_{0},U)} ≤ C ∥x − x_{0}∥ .  (15.23)

Proof. Let |J_{0}| = b_{0} − a_{0}, I = J̄_{0} and E := y(J̄_{0}) – a compact subset of U and let ε > 0 and K < ∞ be given as in Lemma 15.7, i.e. K is the Lipschitz constant for Z on E_{ε}. Also recall the notation: Δ_{1}(t) = [0, t] if t > 0 and Δ_{1}(t) = [t, 0] if t < 0. Suppose that x ∈ E_{ε}, then by Corollary 15.3,

∥ϕ(t, x) − ϕ(t, x_{0})∥ ≤ ∥x − x_{0}∥e^{K|t|} ≤ ∥x − x_{0}∥e^{K|J_{0}|}  (15.24)

for all t ∈ J_{0} ∩ J_{x} such that such that ϕ (Δ_{1}(t), x) ⊂ E_{ε}. Letting δ := εe^{−K|J_{0}|}/2, and assuming x ∈ B(x_{0}, δ), the previous equation implies

∥ϕ(t, x) − ϕ(t, x_{0})∥ ≤ ε/2 < ε ∀ t ∈ J_{0} ∩ J_{x} ∋ ϕ (Δ_{1}(t), x) ⊂ E_{ε}.

This estimate further shows that ϕ(t, x) remains bounded and strictly away from the boundary of U for all such t. Therefore, it follows from Proposition 15.8 and “continuous induction^{3}” that J_{0} ⊂ J_{x} and Eq. (15.24) is valid for all

3 See the argument in the proof of Proposition 8.11.

t ∈ J_{0}. This proves Eq. (15.23) with C := e^{K|J_{0}|}. Suppose that (t_{0}, x_{0}) ∈ D(Z) and let 0 ∈ J_{0} ⊏⊏ J_{x_{0}} such that t_{0} ∈ J_{0} and δ be as above. Then we have just shown J_{0} × B(x_{0}, δ) ⊂ D(Z) which proves D(Z) is open. Furthermore, since the evaluation map

(t_{0}, y) ∈ J_{0} × BC(J_{0}, U) →^{e} y(t_{0}) ∈ X

is continuous (as the reader should check) it follows that ϕ = e∘(x → ϕ(·, x)) : J_{0} × B(x_{0}, δ) → U is also continuous; being the composition of continuous maps. The continuity of ϕ̇(t_{0}, x) is a consequences of the continuity of ϕ and the differential equation 15.1 Alternatively using Eq. (15.2),

∥ϕ(t_{0}, x) − ϕ(t, x_{0})∥ ≤ ∥ϕ(t_{0}, x) − ϕ(t_{0}, x_{0})∥ + ∥ϕ(t_{0}, x_{0}) − ϕ(t, x_{0})∥
≤ C ∥x − x_{0}∥ + |∫_{t}^{t_{0}} ∥Z(τ, ϕ(τ, x_{0}))∥ dτ|
≤ C ∥x − x_{0}∥ + M |t_{0} − t|

where C is the constant in Eq. (15.23) and M = sup_{τ∈J_{0}} ∥Z(τ, ϕ(τ, x_{0}))∥ < ∞. This clearly shows ϕ is continuous.

15.5 Semi-Group Properties of time independent flows

To end this chapter we investigate the semi-group property of the flow associated to the vector-field Z. It will be convenient to introduce the following suggestive notation. For (t, x) ∈ D(Z), set e^{tZ}(x) = ϕ(t, x). So the path t → e^{tZ}(x) is the maximal solution to

\frac{d}{dt}e^{tZ}(x) = Z(e^{tZ}(x)) with e^{0Z}(x) = x.

This exponential notation will be justified shortly. It is convenient to have the following conventions.

Notation 15.16 We write f : X → X to mean a function defined on some open subset D(f) ⊂ X. The open set D(f) will be called the domain of f. Given two functions f : X → X and g : X → X with domains D(f) and D(g) respectively, we define the composite function f ∘ g : X → X to be the function with domain

D(f ∘ g) = {x ∈ X : x ∈ D(g) and g(x) ∈ D(f)} = g^{−1}(D(f))

given by the rule f ∘ g(x) = f(g(x)) for all x ∈ D(f ∘ g). We now write f = g iff D(f) = D(g) and f(x) = g(x) for all x ∈ D(f) = D(g). We will also write f ⊂ g iff D(f) ⊂ D(g) and g|_{D(f)} = f.

Theorem 15.17. For fixed t ∈ ℝ we consider e^{tZ} as a function from X to X with domain D(e^{tZ}) = {x ∈ U : (t, x) ∈ D(Z)}, where D(ϕ) = D(Z) ⊂ ℝ × U, D(Z) and ϕ are defined in Notation 15.9. Conclusions:

1. If t, s ∈ ℝ and t · s ≥ 0, then e^{tZ} ∘ e^{sZ} = e^{(t+s)Z}.

2. If t ∈ ℝ, then e^{tZ} ∘ e^{−tZ} = Id_{D(e^{−tZ})}.

3. For arbitrary t, s ∈ ℝ, e^{tZ} ∘ e^{sZ} ⊂ e^{(t+s)Z}.

Proof. Item 1. For simplicity assume that t, s ≥ 0. The case t, s ≤ 0 is left to the reader. Suppose that x ∈ D(e^{tZ} ∘e^{sZ}). Then by assumption x ∈ D(e^{sZ}) and e^{sZ}(x) ∈ D(e^{tZ}). Define the path y(τ) via:

y(τ) = \{ \begin{matrix}e^{τZ}(x) & if & 0 ≤ τ ≤ s \\ e^{(τ−s)Z}(x) & if & s ≤ τ ≤ t + s \end{matrix} .

It is easy to check that y solves ẏ(τ) = Z(y(τ)) with y(0) = x. But since, e^{τZ}(x) is the maximal solution we must have that x ∈ D(e^{(t+s)Z}) and y(t + s) = e^{(t+s)Z}(x). That is e^{(t+s)Z}(x) = e^{tZ} ∘ e^{sZ}(x). Hence we have shown that e^{tZ} ∘ e^{sZ} ⊂ e^{(t+s)Z}. To finish the proof of item 1. it suffices to show that D(e^{(t+s)Z}) ⊂ D(e^{tZ} ∘ e^{sZ}). Take x ∈ D(e^{(t+s)Z}), then clearly x ∈ D(e^{sZ}). Set y(τ) = e^{(τ+s)Z}(x) defined for 0 ≤ τ ≤ t. Then y solves

ẏ(τ) = Z(y(τ)) with y(0) = e^{sZ}(x).

But since τ → e^{τZ}(e^{sZ}(x)) is the maximal solution to the above initial valued problem we must have that y(τ) = e^{τZ}(e^{sZ}(x)), and in particular at τ = t, e^{(t+s)Z}(x) = e^{tZ}(e^{sZ}(x)). This shows that x ∈ D(e^{tZ} ∘ e^{sZ}) and in fact e^{(t+s)Z} ⊂ e^{tZ} ∘ e^{sZ}.

Item 2. Let x ∈ D(e^{−tZ}) – again assume for simplicity that t ≥ 0. Set y(τ) = e^{(τ−t)Z}(x) defined for 0 ≤ τ ≤ t. Notice that y(0) = e^{−tZ}(x) and ẏ(τ) = Z(y(τ)). This shows that y(τ) = e^{τZ}(e^{−tZ}(x)) and in particular that x ∈ D(e^{tZ} ∘ e^{−tZ}) and e^{tZ} ∘ e^{−tZ}(x) = x. This proves item 2.

Item 3. I will only consider the case that s < 0 and t + s ≥ 0, the other cases are handled similarly. Write u for t + s, so that t = −s + u. We know that e^{tZ} = e^{uZ} ∘ e^{−sZ} by item 1. Therefore

e^{tZ} ∘ e^{sZ} = (e^{uZ} ∘ e^{−sZ}) ∘ e^{sZ}.

Notice in general, one has (f ∘ g) ∘ h = f ∘ (g ∘ h) (you prove). Hence, the above displayed equation and item 2. imply that

e^{tZ} ∘ e^{sZ} = e^{uZ} ∘ (e^{−sZ} ∘ e^{sZ}) = e^{(t+s)Z} ∘ I_{D(e^{sZ})} ⊂ e^{(t+s)Z}.

The following result is trivial but conceptually illuminating partial converse to Theorem 15.17.

Proposition 15.18 (Flows and Complete Vector Fields). Suppose U ⊂_{o} X, ϕ ∈ C(ℝ × U, U) and ϕ_{t}(x) = ϕ(t, x). Suppose ϕ satisfies:

1. ϕ_{0} = I_{U},

2. ϕ_{t} ∘ ϕ_{s} = ϕ_{t+s} for all t, s ∈ ℝ, and

3. Z(x) := ϕ̇(0, x) exists for all x ∈ U and Z ∈ C(U, X) is locally Lipschitz.

Then ϕ_{t} = e^{tZ}.

Proof. Let x ∈ U and y(t) := ϕ_{t}(x). Then using Item 2.,

ẏ(t) = \frac{d}{ds}|_{0}y(t + s) = \frac{d}{ds}|_{0}ϕ_{(t+s)}(x) = \frac{d}{ds}|_{0}ϕ_{s} ∘ ϕ_{t}(x) = Z(y(t)).

Since y(0) = x by Item 1. and Z is locally Lipschitz by Item 3., we know by uniqueness of solutions to ODE’s (Corollary 15.3) that ϕ_{t}(x) = y(t) = e^{tZ}(x).

15.6 Exercises

Exercise 15.1. Find a vector field Z such that e^{(t+s)Z} is not contained in e^{tZ} ∘ e^{sZ}.

Definition 15.19. A locally Lipschitz function Z : U ⊂_{o} X → X is said to be a complete vector field if D(Z) = ℝ × U. That is for any x ∈ U, t → e^{tZ}(x) is defined for all t ∈ ℝ.

Exercise 15.2. Suppose that Z : X → X is a locally Lipschitz function. Assume there is a constant C > 0 such that

∥Z(x)∥ ≤ C(1 + ∥x∥) for all x ∈ X.

Then Z is complete. Hint: use Gronwall’s Lemma 15.2 and Proposition 15.10.

Exercise 15.3. Suppose y is a solution to ẏ(t) = |y(t)|^{1/2} with y(0) = 0. Show there exists a, b ∈ [0, ∞] such that

y(t) = \{ \begin{matrix}\frac{1}{4}(t − b)^{2} & if & t ≥ b \\ 0 & if & −a < t < b \\ −\frac{1}{4}(t + a)^{2} & if & t ≤ −a. \end{matrix}

Exercise 15.4. Using the fact that the solutions to Eq. (15.3) are never 0 if x ≠ 0, show that y(t) = 0 is the only solution to Eq. (15.3) with y(0) = 0.

Exercise 15.5 (Higher Order ODE). Let X be a Banach space, , U ⊂_{o} X^{n} and f ∈ C (J × U, X) be a Locally Lipschitz function in x = (x_{1}, …, x_{n}). Show the n^{th} ordinary differential equation,

y^{(n)}(t) = f(t, y(t), ẏ(t), … y^{(n−1)}(t)) with y^{(k)}(0) = y_{0}^{k} for k < n (15.25)

where (y_{0}^{0}, …, y_{0}^{n−1}) is given in U, has a unique solution for small t ∈ J. Hint: let y(t) = (y(t), ẏ(t), … y^{(n−1)}(t)) and rewrite Eq. (15.25) as a first order ODE of the form

ẏ(t) = Z(t, y(t)) with y(0) = (y_{0}^{0}, …, y_{0}^{n−1}).

Exercise 15.6. Use the results of Exercises 8.20 and 15.5 to solve

ÿ(t) − 2ẏ(t) + y(t) = 0 with y(0) = a and ẏ(0) = b.

Hint: The 2 × 2 matrix associated to this system, A, has only one eigenvalue 1 and may be written as A = I + B where B^{2} = 0.

Exercise 15.7 (Non-Homogeneous ODE). Suppose that U ⊂_{o} X is open and Z : ℝ×U → X is a continuous function. Let J = (a, b) be an interval and t_{0} ∈ J. Suppose that y ∈ C^{1}(J, U) is a solution to the “non-homogeneous” differential equation:

ẏ(t) = Z(t, y(t)) with y(t_{o}) = x ∈ U.  (15.26)

Define Y ∈ C^{1}(J − t_{0}, ℝ × U) by Y (t) := (t + t_{0}, y(t + t_{0})). Show that Y solves the “homogeneous” differential equation

Ẏ (t) = Z̃(Y (t)) with Y (0) = (t_{0}, y_{0}),  (15.27)

where Z̃(t, x) := (1, Z(x)). Conversely, suppose that Y ∈ C^{1}(J − t_{0}, ℝ × U) is a solution to Eq. (15.27). Show that Y (t) = (t + t_{0}, y(t + t_{0})) for some y ∈ C^{1}(J, U) satisfying Eq. (15.26). (In this way the theory of non-homogeneous ode’s may be reduced to the theory of homogeneous ode’s.)

Exercise 15.8 (Differential Equations with Parameters). Let W be another Banach space, U × V ⊂_{o} X × W and Z ∈ C(U × V, X) be a locally Lipschitz function on U × V. For each (x, w) ∈ U × V, let t ∈ J_{x,w} → ϕ(t, x, w) denote the maximal solution to the ODE

ẏ(t) = Z(y(t), w) with y(0) = x.  (15.28)

Prove

D := {(t, x, w) ∈ ℝ × U × V : t ∈ J_{x,w}}  (15.29)

is open in ℝ × U × V and ϕ and ϕ̇ are continuous functions on D.

Hint: If y(t) solves the differential equation in (15.28), then v(t) := (y(t), w) solves the differential equation,

v̇(t) = Z̃(v(t)) with v(0) = (x, w),  (15.30)

where Z̃(x, w) := (Z(x, w), 0) ∈ X ×W and let ψ(t, (x, w)) := v(t). Now apply the Theorem 15.15 to the differential equation (15.30).

Exercise 15.9 (Abstract Wave Equation). For A ∈ L(X) and t ∈ ℝ, let

cos(tA) := ∑_{n=0}^{∞} \frac{(−1)^{n}}{(2n)!} t^{2n}A^{2n} and

\frac{sin(tA)}{A} := ∑_{n=0}^{∞} \frac{(−1)^{n}}{(2n + 1)!}t^{2n+1}A^{2n}.

Show that the unique solution y ∈ C^{2} (ℝ, X) to

ÿ(t) + A^{2}y(t) = 0 with y(0) = y_{0} and ẏ(0) = ẏ_{0} ∈ X  (15.31)

is given by

y(t) = cos(tA)y_{0} + \frac{sin(tA)}{A}ẏ_{0}.

Remark 15.20. Exercise 15.9 can be done by direct verification. Alternatively and more instructively, rewrite Eq. (15.31) as a first order ODE using Exercise 15.5. In doing so you will be lead to compute e^{tB} where B ∈ L(X × X) is given by

B = ( \begin{matrix}0 & I \\ −A^{2} & 0 \end{matrix} ) ,

where we are writing elements of X ×X as column vectors, ( \begin{matrix}x_{1} \\ x_{2} \end{matrix} ) . You should

then show

e^{tB} = ( \begin{matrix}cos(tA) & \frac{sin(tA)}{A} \\ −A sin(tA) & cos(tA) \end{matrix} )

where

A sin(tA) := ∑_{n=0}^{∞} \frac{(−1)^{n}}{(2n + 1)!}t^{2n+1}A^{2(n+1)}.

Exercise 15.10 (Duhamel’s Principle for the Abstract Wave Equation). Continue the notation in Exercise 15.9, but now consider the ODE,

ÿ(t) + A^{2}y(t) = f(t) with y(0) = y_{0} and ẏ(0) = ẏ_{0} ∈ X  (15.32)

where f ∈ C(ℝ, X). Show the unique solution to Eq. (15.32) is given by

y(t) = cos(tA)y_{0} + \frac{sin(tA)}{A}ẏ_{0} + ∫_{0}^{t} \frac{sin((t − τ) A)}{A}f(τ)dτ  (15.33)

Hint: Again this could be proved by direct calculation. However it is more instructive to deduce Eq. (15.33) from Exercise 8.22 and the comments in Remark 15.20.

Banach Space Calculus

In this section, X and Y will be Banach space and U will be an open subset of X.

Notation 16.1 (ε, O, and o notation) Let 0 ∈ U ⊂_{o} X, and f : U → Y be a function. We will write:

1. f(x) = ε(x) if lim_{x→0} ∥f(x)∥ = 0.

2. f(x) = O(x) if there are constants C < ∞ and r > 0 such that ∥f(x)∥ ≤ C∥x∥ for all x ∈ B(0, r). This is equivalent to the condition that lim sup_{x→0} (∥x∥^{−1}∥f(x)∥) < ∞, where

lim sup_{x→0} \frac{∥f(x)∥}{∥x∥} := lim_{r↓0} sup\{∥f(x)∥ : 0 < ∥x∥ ≤ r\}.

3. f(x) = o(x) if f(x) = ε(x)O(x), i.e. lim_{x→0} ∥f(x)∥/∥x∥ = 0.

Example 16.2. Here are some examples of properties of these symbols.

1. A function f : U ⊂_{o} X → Y is continuous at x_{0} ∈ U if f(x_{0} + h) = f(x_{0}) + ε(h).

2. If f(x) = ε(x) and g(x) = ε(x) then f(x) + g(x) = ε(x).

Now let g : Y → Z be another function where Z is another Banach space.

3. If f(x) = O(x) and g(y) = o(y) then g ∘ f(x) = o(x).

4. If f(x) = ε(x) and g(y) = ε(y) then g ∘ f(x) = ε(x).

16.1 The Differential

Definition 16.3. A function f : U ⊂_{o} X → Y is differentiable at x_{0} ∈ U if there exists a linear transformation Λ ∈ L(X, Y) such that

f(x_{0} + h) − f(x_{0}) − Λh = o(h).  (16.1)

We denote Λ by f′(x_{0}) or Df(x_{0}) if it exists. As with continuity, f is differentiable on U if f is differentiable at all points in U.

Remark 16.4. The linear transformation Λ in Definition 16.3 is necessarily unique. Indeed if Λ_{1} is another linear transformation such that Eq. (16.1) holds with Λ replaced by Λ_{1}, then

(Λ − Λ_{1})h = o(h),

i.e.

lim sup_{h→0} \frac{∥(Λ − Λ_{1})h∥}{∥h∥} = 0.

On the other hand, by definition of the operator norm,

lim sup_{h→0} \frac{∥(Λ − Λ_{1})h∥}{∥h∥} = ∥Λ − Λ_{1}∥.

The last two equations show that Λ = Λ_{1}.

Exercise 16.1. Show that a function f : (a, b) → X is a differentiable at t ∈ (a, b) in the sense of Definition 8.8 iff it is differentiable in the sense of Definition 16.3. Also show Df(t)v = vḟ(t) for all v ∈ ℝ.

Example 16.5. If T ∈ L (X, Y) and x, h ∈ X, then

T (x + h) − T (x) − T h = 0

which shows T′ (x) = T for all x ∈ X.

Example 16.6. Assume that GL(X, Y) is non-empty. Then by Corollary 7.20, GL(X, Y) is an open subset of L(X, Y) and the inverse map f : GL(X, Y) → GL(Y, X), defined by f(A) := A^{−1}, is continuous. We will now show that f is differentiable and

f′(A)B = −A^{−1}BA^{−1} for all B ∈ L(X, Y).

This is a consequence of the identity,

f(A + H) − f(A) = (A + H)^{−1} (A − (A + H)) A^{−1} = −(A + H)^{−1}HA^{−1}

which may be used to find the estimate,

∥f(A + H) − f(A) + A^{−1}HA^{−1}∥ = ∥[A^{−1} − (A + H)^{−1}] HA^{−1}∥

≤ ∥A^{−1} − (A + H)^{−1}∥ ∥H∥ ∥A^{−1}∥

≤ \frac{∥A^{−1}∥^{3} ∥H∥^{2}}{1 − ∥A^{−1}∥ ∥H∥} = O (∥H∥^{2})

wherein we have used the bound in Eq. (7.8) of Corollary 7.20 for the last inequality.

16.2 Product and Chain Rules

The following theorem summarizes some basic properties of the differential.

Theorem 16.7. The differential D has the following properties:

1. Linearity: D is linear, i.e. D(f + λg) = Df + λDg.

2. Product Rule: If f : U ⊂_{o} X → Y and A : U ⊂_{o} X → L(X, Z) are differentiable at x_{0} then so is x → (Af)(x) := A(x)f(x) and

D(Af)(x_{0})h = (DA(x_{0})h)f(x_{0}) + A(x_{0})Df(x_{0})h.

3. Chain Rule: If f : U ⊂_{o} X → V ⊂_{o} Y is differentiable at x_{0} ∈ U, and g : V ⊂_{o} Y → Z is differentiable at y_{0} := f(x_{0}), then g ∘f is differentiable at x_{0} and (g ∘ f)′(x_{0}) = g′(y_{0})f′(x_{0}).

4. Converse Chain Rule: Suppose that f : U ⊂_{o} X → V ⊂_{o} Y is continuous at x_{0} ∈ U, g : V ⊂_{o} Y → Z is differentiable y_{0} := f(h_{o}), g′(y_{0}) is invertible, and g ∘ f is differentiable at x_{0}, then f is differentiable at x_{0} and

f′(x_{0}) := [g′(x_{0})]^{−1}(g ∘ f)′(x_{0}).  (16.2)

Proof. Linearity. Let f, g : U ⊂_{o} X → Y be two functions which are differentiable at x_{0} ∈ U and λ ∈ ℝ, then

(f + λg)(x_{0} + h)
= f(x_{0}) + Df(x_{0})h + o(h) + λ(g(x_{0}) + Dg(x_{0})h + o(h)
= (f + λg)(x_{0}) + (Df(x_{0}) + λDg(x_{0}))h + o(h),

which implies that (f + λg) is differentiable at x_{0} and that

D(f + λg)(x_{0}) = Df(x_{0}) + λDg(x_{0}).

Product Rule. The computation,

A(x_{0} + h)f(x_{0} + h)
= (A(x_{0}) + DA(x_{0})h + o(h))(f(x_{0}) + f′(x_{0})h + o(h))
= A(x_{0})f(x_{0}) + A(x_{0})f′(x_{0})h + [DA(x_{0})h]f(x_{0}) + o(h),

verifies the product rule holds. This may also be considered as a special case of Proposition 16.9. Chain Rule. Using f(x_{0} + h) − f(x_{0}) = O(h) (see Eq. (16.1)) and o(O(h)) = o(h),

(g∘f)(x_{0} + h)

= g(f(x_{0})) + g′(f(x_{0}))(f(x_{0} + h) − f(x_{0})) + o(f(x_{0} + h) − f(x_{0}))

= g(f(x_{0})) + g′(f(x_{0}))(Df(x_{0})x_{0} + o(h)) + o(f(x_{0} + h) − f(x_{0})
= g(f(x_{0})) + g′(f(x_{0}))Df(x_{0})h + o(h).

Converse Chain Rule. Since g is differentiable at y_{0} = f(x_{0}) and g′ (y_{0}) is invertible,

g(f(x_{0} + h)) − g(f(x_{0}))
= g′(f(x_{0}))(f(x_{0} + h) − f(x_{0})) + o(f(x_{0} + h) − f(x_{0}))
= g′(f(x_{0})) [f(x_{0} + h) − f(x_{0}) + o(f(x_{0} + h) − f(x_{0}))] .

And since g ∘ f is differentiable at x_{0},

(g ∘ f)(x_{0} + h) − g(f(x_{0})) = (g ∘ f)′(x_{0})h + o(h).

Comparing these two equations shows that

f(x_{0} + h) − f(x_{0}) + o(f(x_{0} + h) − f(x_{0}))
= g′(f(x_{0}))^{−1} [(g ∘ f)′(x_{0})h + o(h)]

which is equivalent to

f(x_{0} + h) − f(x_{0}) + o(f(x_{0} + h) − f(x_{0}))
= g′(f(x_{0}))^{−1} [(g ∘ f)′(x_{0})h + o(h)]
= g′(f(x_{0}))^{−1}{(g ∘ f)′(x_{0})h + o(h) − o(f(x_{0} + h) − f(x_{0}))}
= g′(f(x_{0}))^{−1}(g ∘ f)′(x_{0})h + o(h) + o(f(x_{0} + h) − f(x_{0})). (16.3)

Using the continuity of f, f(x_{0} + h) − f(x_{0}) is close to 0 if h is close to zero, and hence

∥o(f(x_{0} + h) − f(x_{0}))∥ ≤ \frac{1}{2}∥f(x_{0} + h) − f(x_{0})∥  (16.4)

for all h sufficiently close to 0. (We may replace \frac{1}{2} by any number α > 0 above.) Taking the norm of both sides of Eq. (16.3) and making use of Eq. (16.4) shows, for h close to 0, that

∥f(x_{0} + h) − f(x_{0})∥

≤ ∥g′(f(x_{0}))^{−1}(g ∘ f)′(x_{0})∥∥h∥ + o(∥h∥) + \frac{1}{2}∥f(x_{0} + h) − f(x_{0})∥.

Solving for ∥f(x_{0} + h) − f(x_{0})∥ in this last equation shows that

f(x_{0} + h) − f(x_{0}) = O(h).  (16.5)

(This is an improvement, since the continuity of f only guaranteed that f(x_{0}+ h) − f(x_{0}) = ε(h).) Because of Eq. (16.5), we now know that o(f(x_{0} + h) − f(x_{0})) = o(h), which combined with Eq. (16.3) shows that

f(x_{0} + h) − f(x_{0}) = g′(f(x_{0}))^{−1}(g ∘ f)′(x_{0})h + o(h),

i.e. f is differentiable at x_{0} and f′(x_{0}) = g′(f(x_{0}))^{−1}(g ∘ f)′(x_{0}).

Corollary 16.8 (Chain Rule). Suppose that σ : (a, b) → U ⊂_{o} X is differentiable at t ∈ (a, b) and f : U ⊂_{o} X → Y is differentiable at σ(t) ∈ U. Then f ∘ σ is differentiable at t and

d(f ∘ σ)(t)/dt = f′(σ(t))σ̇(t).

Proposition 16.9 (Product Rule II). Suppose that X := X_{1} × ⋯ × X_{n} with each X_{i} being a Banach space and T : X_{1} ×⋯×X_{n} → Y is a multilinear map, i.e.

x_{i} ∈ X_{i} → T (x_{1}, …, x_{i−1}, x_{i}, x_{i+1}, …, x_{n}) ∈ Y

is linear when x_{1}, …, x_{i−1}, x_{i+1}, …, x_{n} are held fixed. Then the following are equivalent:

1. T is continuous.

2. T is continuous at 0 ∈ X.

3. There exists a constant C < ∞ such that

∥T (x)∥_{Y} ≤ C ∏_{i=1}^{n} ∥x_{i}∥_{X_{i}}  (16.6)

for all x = (x_{1}, …, x_{n}) ∈ X.

4. T is differentiable at all x ∈ X_{1} × ⋯ × X_{n}.

Moreover if T the differential of T is given by

T′ (x) h = ∑_{i=1}^{n} T (x_{1}, …, x_{i−1}, h_{i}, x_{i+1}, …, x_{n})  (16.7)

where h = (h_{1}, …, h_{n}) ∈ X.

Proof. Let us equip X with the norm

∥x∥_{X} := max {∥x_{i}∥_{X_{i}}} .

If T is continuous then T is continuous at 0. If T is continuous at 0, using T (0) = 0, there exists a δ > 0 such that ∥T (x)∥_{Y} ≤ 1 whenever ∥x∥_{X} ≤ δ. Now if x ∈ X is arbitrary, let x′ := δ (∥x_{1}∥_{X_{1}}^{−1} x_{1}, …, ∥x_{n}∥_{X_{n}}^{−1} x_{n}) . Then ∥x′∥_{X} ≤ δ and hence

∥(δ^{n} ∏_{i=1}^{n} ∥x_{i}∥_{X_{i}}^{−1}) T (x_{1}, …, x_{n})∥_{Y} = ∥T (x′)∥ ≤ 1

from which Eq. (16.6) follows with C = δ^{−n}.

Now suppose that Eq. (16.6) holds. For x, h ∈ X and ε ∈ {0, 1}^{n} let |ε| = ∑_{i=1}^{n} ε_{i} and

x^{ε} (h) := ((1 − ε_{1}) x_{1} + ε_{1}h_{1}, …, (1 − ε_{n}) x_{n} + ε_{n}h_{n}) ∈ X.

By the multi-linearity of T,

T (x + h) = T (x_{1} + h_{1}, …, x_{n} + h_{n}) = ∑_{ε∈{0,1}^{n}} T (x^{ε} (h))

= T (x) + ∑_{i=1}^{n} T (x_{1}, …, x_{i−1}, h_{i}, x_{i+1}, …, x_{n})

+ ∑_{ε∈{0,1}^{n}:|ε|≥2} T (x^{ε} (h)) .  (16.8)

From Eq. (16.6),

∥ ∑_{ε∈{0,1}^{n}:|ε|≥2} T (x^{ε} (h))∥ = O (∥h∥^{2}) ,

and so it follows from Eq. (16.8) that T′ (x) exists and is given by Eq. (16.7). This completes the proof since it is trivial to check that T being differentiable at x ∈ X implies continuity of T at x ∈ X.

Exercise 16.2. Let det : L (ℝ^{n}) → ℝ be the determinant function on n × n matrices and for A ∈ L ( ℝ̸^{n}) we will let A_{i} denote the i^{th} – column of A and write A = (A_{1}|A_{2}| … |A_{n}) .

1. Show det′ (A) exists for all A ∈ L ( ℝ̸^{n}) and

det^{′} (A) H = ∑_{i=1}^{n} det (A_{1}| … |A_{i−1}|H_{i}|A_{i+1}| … |A_{n})  (16.9)

for all H ∈ L (ℝ^{n}) . Hint: recall that det (A) is a multilinear function of its columns.

2. Use Eq. (16.9) along with basic properties of the determinant to show det′ (I) H = tr(H).

3. Suppose now that A ∈ GL (ℝ^{n}) , show

det^{′} (A) H = det (A) tr(A^{−1}H).

Hint: Notice that det (A + H) = det (A) det (I + A^{−1}H) .

4. If A ∈ L (ℝ^{n}) , show det (e^{A}) = e^{tr(A)}. Hint: use the previous item and Corollary 16.8 to show

\frac{d}{dt} det (e^{tA}) = det (e^{tA}) tr(A).

Definition 16.10. Let X and Y be Banach spaces and let L^{1}(X, Y) := L(X, Y) and for k ≥ 2 let L^{k}(X, Y) be defined inductively by L^{k+1}(X, Y) = L(X, L^{k}(X, Y)). For example L^{2}(X, Y) = L(X, L(X, Y)) and L^{3}(X, Y) = L (X, L(X, L(X, Y))) .

Suppose f : U ⊂_{o} X → Y is a function. If f is differentiable on U, then it makes sense to ask if f′ = Df : U → L(X, Y) = L^{1}(X, Y) is differentiable. If Df is differentiable on U then f′′ = D^{2}f := DDf : U → L^{2}(X, Y). Similarly we define f^{(n)} = D^{n}f : U → L^{n}(X, Y) inductively.

Definition 16.11. Given k ∈ ℕ, let C^{k} (U, Y) denote those functions f : U → Y such that f^{(j)} := D^{j}f : U → L^{j} (X, Y) exists and is continuous for j = 1, 2, …, k.

Example 16.12. Let us continue on with Example 16.6 but now let X = Y to simplify the notation. So f : GL(X) → GL(X) is the map f(A) = A^{−1} and

f′(A) = −L_{A^{−1}}R_{A^{−1}}, i.e. f′ = −L_{f}R_{f}.

where L_{A}B = AB and R_{A}B = AB for all A, B ∈ L(X). As the reader may easily check, the maps

A ∈ L(X) → L_{A}, R_{A} ∈ L(L(X))

are linear and bounded. So by the chain and the product rule we find f′′(A) exists for all A ∈ L(X) and

f′′(A)B = −L_{f′(A)B}R_{f} − L_{f}R_{f′(A)B}.

More explicitly

[f′′(A)B] C = A^{−1}BA^{−1}CA^{−1} + A^{−1}CA^{−1}BA^{−1}.  (16.10)

Working inductively one shows f : GL(X) → GL(X) defined by f(A) := A^{−1} is C^{∞}.

16.3 Partial Derivatives

Definition 16.13 (Partial or Directional Derivative). Let f : U ⊂_{o} X → Y be a function, x_{0} ∈ U, and v ∈ X. We say that f is differentiable at x_{0} in the direction v iff \frac{d}{dt}|_{0}(f(x_{0} + tv)) =: (∂_{v}f)(x_{0}) exists. We call (∂_{v}f)(x_{0}) the directional or partial derivative of f at x_{0} in the direction v.

Notice that if f is differentiable at x_{0}, then ∂_{v}f(x_{0}) exists and is equal to f′(x_{0})v, see Corollary 16.8.

Proposition 16.14. Let f : U ⊂_{o} X → Y be a continuous function and D ⊂ X be a dense subspace of X. Assume ∂_{v}f(x) exists for all x ∈ U and v ∈ D, and there exists a continuous function A : U → L(X, Y) such that ∂_{v}f(x) = A(x)v for all v ∈ D and x ∈ U ∩ D. Then f ∈ C^{1}(U, Y) and Df = A.

Proof. Let x_{0} ∈ U, ε > 0 such that B(x_{0}, 2ε) ⊂ U and M := sup{∥A(x)∥ : x ∈ B(x_{0}, 2ε)} < ∞^{1}. For x ∈ B(x_{0}, ε) ∩ D and v ∈ D ∩ B(0, ε), by the fundamental theorem of calculus,

f(x + v) − f(x) = ∫_{0}^{1} \frac{df(x + tv)}{dt} dt
= ∫_{0}^{1}(∂_{v}f)(x + tv) dt = ∫_{0}^{1} A(x + tv) v dt.  (16.11)

For general x ∈ B(x_{0}, ε) and v ∈ B(0, ε), choose x_{n} ∈ B(x_{0}, ε) ∩ D and v_{n} ∈ D ∩ B(0, ε) such that x_{n} → x and v_{n} → v. Then

f(x_{n} + v_{n}) − f(x_{n}) = ∫_{0}^{1} A(x_{n} + tv_{n}) v_{n} dt  (16.12)

holds for all n. The left side of this last equation tends to f(x + v) − f(x) by the continuity of f. For the right side of Eq. (16.12) we have

∥ ∫_{0}^{1} A(x + tv) v dt − ∫_{0}^{1} A(x_{n} + tv_{n}) v_{n} dt∥
≤ ∫_{0}^{1} ∥A(x + tv) − A(x_{n} + tv_{n}) ∥∥v∥ dt + M∥v − v_{n}∥.

It now follows by the continuity of A, the fact that ∥A(x+tv)−A(x_{n}+tv_{n}) ∥ ≤ M, and the dominated convergence theorem that right side of Eq. (16.12) converges to ∫_{0}^{1} A(x + tv) v dt. Hence Eq. (16.11) is valid for all x ∈ B(x_{0}, ε) and v ∈ B(0, ε). We also see that

f(x + v) − f(x) − A(x)v = ε(v)v,  (16.13)

where ε(v) := ∫_{0}^{1}[A(x + tv) − A(x)] dt. Now

1 It should be noted well, unlike in finite dimensions closed and bounded sets need not be compact, so it is not sufficient to choose ϵ sufficiently small so that \overline{B(x_{0}, 2ϵ)} ⊂ U. Here is a counter example. Let X ≡ H be a Hilbert space, \{e_{n}\}_{n=1}^{∞} be an orthonormal set. Define f(x) ≡ ∑_{n=1}^{∞} nϕ(∥x − e_{n}∥), where ϕ is any continuous function on ℝ such that ϕ(0) = 1 and ϕ is supported in (−1, 1). Notice that ∥e_{n} − e_{m}∥^{2} = 2 for all m ≠ n, so that ∥e_{n} − e_{m}∥ = \sqrt{2}. Using this fact it is rather easy to check that for any x_{0} ∈ H, there is an ϵ > 0 such that for all x ∈ B(x_{0}, ϵ), only one term in the sum defining f is non-zero. Hence, f is continuous. However, f(e_{n}) = n → ∞ as n → ∞.

∥ε(v)∥ ≤ ∫_{0}^{1} ∥A(x + tv) − A(x)∥ dt

≤ max_{t∈[0,1]} ∥A(x + tv) − A(x)∥ → 0 as v → 0,

by the continuity of A. Thus, we have shown that f is differentiable and that Df(x) = A(x).

Corollary 16.15. Suppose now that X = ℝ^{d}, f : U ⊂_{o} X → Y be a continuous function such that ∂_{i}f(x) := ∂_{e_{i}}f (x) exists and is continuous on U for i = 1, 2, …, d, where {e_{i}}_{i=1}^{d} is the standard basis for ℝ^{d}. Then f ∈ C^{1}(U, Y) and Df (x) e_{i} = ∂_{i}f (x) for all i.

Proof. For x ∈ U, let A (x) : ℝ^{d} → Y be the unique linear map such that A (x) e_{i} = ∂_{i}f (x) for i = 1, 2, …, d. Then A : U → L(ℝ^{d}, Y) is a continuous map. Now let v ∈ ℝ^{d} and v^{(i)} := (v_{1}, v_{2}, …, v_{i}, 0, …, 0) for i = 1, 2, …, d and v^{(0)} := 0. Then for t ∈ ℝ near 0, using the fundamental theorem of calculus and the definition of ∂_{i}f (x) ,

f (x + tv) − f (x) = ∑_{i=1}^{d} [f (x + tv^{(i)}) − f (x + tv^{(i−1)})]

= ∑_{i=1}^{d} ∫_{0}^{1} \frac{d}{ds}f (x + tv^{(i−1)} + stv_{i}e_{i}) ds

= ∑_{i=1}^{d} tv_{i} ∫_{0}^{1} ∂_{i}f (x + tv^{(i−1)} + stv_{i}e_{i}) ds

= ∑_{i=1}^{d} tv_{i} ∫_{0}^{1} A (x + tv^{(i−1)} + stv_{i}e_{i}) e_{i}ds.

Using the continuity of A, it now follows that

lim_{t→0} \frac{f (x + tv) − f (x)}{t} = ∑_{i=1}^{d} v_{i} lim_{t→0} ∫_{0}^{1} A (x + tv^{(i−1)} + stv_{i}e_{i}) e_{i}ds

= ∑_{i=1}^{d} v_{i} ∫_{0}^{1} A (x) e_{i}ds = A (x) v

which shows ∂_{v}f (x) exists and ∂_{v}f (x) = A (x) v. The result now follows from an application of Proposition 16.14.

16.4 Higher Order Derivatives

It is somewhat inconvenient to work with the Banach spaces L^{k}(X, Y) in Definition 16.10. For this reason we will introduce an isomorphic Banach space, M_{k}(X, Y).

Definition 16.16. For k ∈ {1, 2, 3, …}, let M_{k}(X, Y) denote the set of functions f : X^{k} → Y such that

1. For i ∈ {1, 2, …, k}, v ∈ X → f⟨v_{1}, v_{2}, …, v_{i−1}, v, v_{i+1}, …, v_{k}⟩ ∈ Y is linear ^{2} for all {v_{i}}_{i=1}^{n} ⊂ X.

2. The norm ∥f∥_{M_{k}(X,Y)} should be finite, where

∥f∥_{M_{k}(X,Y)} := sup\{\frac{∥f⟨v_{1}, v_{2}, …, v_{k}⟩∥_{Y}}{∥v_{1}∥∥v_{2}∥ ⋯ ∥v_{k}∥} : \{v_{i}\}_{i=1}^{k} ⊂ X \ \{0\}\}.

Lemma 16.17. There are linear operators j_{k} : L^{k}(X, Y) → M_{k}(X, Y) defined inductively as follows: j_{1} = Id_{L(X,Y)} (notice that M_{1}(X, Y) = L^{1}(X, Y) = L(X, Y)) and

(j_{k+1}A)⟨v_{0}, v_{1}, …, v_{k}⟩ = (j_{k}(Av_{0}))⟨v_{1}, v_{2}, …, v_{k}⟩ ∀v_{i} ∈ X.

(Notice that Av_{0} ∈ L^{k}(X, Y).) Moreover, the maps j_{k} are isometric isomorphisms.

Proof. To get a feeling for what j_{k} is let us write out j_{2} and j_{3} explicitly. If A ∈ L^{2}(X, Y) = L(X, L(X, Y)), then (j_{2}A)⟨v_{1}, v_{2}⟩ = (Av_{1})v_{2} and if A ∈ L^{3}(X, Y) = L(X, L(X, L(X, Y))), (j_{3}A)⟨v_{1}, v_{2}, v_{3}⟩ = ((Av_{1})v_{2})v_{3} for all v_{i} ∈ X. It is easily checked that j_{k} is linear for all k. We will now show by induction that j_{k} is an isometry and in particular that j_{k} is injective. Clearly this is true if k = 1 since j_{1} is the identity map. For A ∈ L^{k+1}(X, Y),

∥j_{k+1}A∥_{M_{k+1}(X,Y)}

:= sup\{\frac{∥(j_{k}(Av_{0}))⟨v_{1}, v_{2}, …, v_{k}⟩∥_{Y}}{∥v_{0}∥∥v_{1}∥∥v_{2}∥ ⋯ ∥v_{k}∥} : \{v_{i}\}_{i=0}^{k} ⊂ X \ \{0\}\}

= sup\{\frac{∥(j_{k}(Av_{0}))∥_{M_{k}(X,Y)}}{∥v_{0}∥} : v_{0} ∈ X \ \{0\}\}

= sup\{\frac{∥Av_{0}∥_{L^{k}(X,Y)}}{∥v_{0}∥} : v_{0} ∈ X \ \{0\}\}

= ∥A∥_{L(X,L^{k}(X,Y))} := ∥A∥_{L^{k+1}(X,Y)},

wherein the second to last inequality we have used the induction hypothesis. This shows that j_{k+1} is an isometry provided j_{k} is an isometry. To finish the proof it suffices to show that j_{k} is surjective for all k. Again this is true for k = 1. Suppose that j_{k} is invertible for some k ≥ 1. Given f ∈ M_{k+1}(X, Y) we must produce A ∈ L^{k+1}(X, Y) = L(X, L^{k}(X, Y)) such that j_{k+1}A = f. If such an equation is to hold, then for v_{0} ∈ X, we would have j_{k}(Av_{0}) = f⟨v_{0}, ⋯ ⟩. That is Av_{0} = j_{k}^{−1}(f⟨v_{0}, ⋯ ⟩). It is easily checked that A so defined is linear, bounded, and j_{k+1}A = f.

From now on we will identify L^{k} with M_{k} without further mention. In particular, we will view D^{k}f as function on U with values in M_{k}(X, Y).

2 I will routinely write f⟨v_{1}, v_{2}, …, v_{k}⟩ rather than f(v_{1}, v_{2}, …, v_{k}) when the function f depends on each of variables linearly, i.e. f is a multi-linear function.

Theorem 16.18 (Differentiability). Suppose k ∈ {1, 2, …} and D is a dense subspace of X, f : U ⊂_{o} X → Y is a function such that (∂_{v_{1}}∂_{v_{2}} ⋯ ∂_{v_{l}}f)(x) exists for all x ∈ D ∩ U, {v_{i}}_{i=1}^{l} ⊂ D, and l = 1, 2, … k. Further assume there exists continuous functions A_{l} : U ⊂_{o} X → M_{l}(X, Y) such that such that (∂_{v_{1}}∂_{v_{2}} ⋯ ∂_{v_{l}}f)(x) = A_{l}(x)⟨v_{1}, v_{2}, …, v_{l}⟩ for all x ∈ D ∩ U, {v_{i}}_{i=1}^{l} ⊂ D, and l = 1, 2, … k. Then D^{l}f(x) exists and is equal to A_{l}(x) for all x ∈ U and l = 1, 2, …, k.

Proof. We will prove the theorem by induction on k. We have already proved the theorem when k = 1, see Proposition 16.14. Now suppose that k > 1 and that the statement of the theorem holds when k is replaced by k−1. Hence we know that D^{l}f(x) = A_{l}(x) for all x ∈ U and l = 1, 2, …, k − 1. We are also given that

(∂_{v_{1}}∂_{v_{2}} ⋯ ∂_{v_{k}}f)(x) = A_{k}(x)⟨v_{1}, v_{2}, …, v_{k}⟩ ∀x ∈ U ∩ D, {v_{i}} ⊂ D. (16.14)

Now we may write (∂_{v_{2}} ⋯ ∂_{v_{k}}f)(x) as (D^{k−1}f)(x)⟨v_{2}, v_{3}, …, v_{k}⟩ so that Eq. (16.14) may be written as

∂_{v_{1}}(D^{k−1}f)(x)⟨v_{2}, v_{3}, …, v_{k}⟩)
= A_{k}(x)⟨v_{1}, v_{2}, …, v_{k}⟩ ∀x ∈ U ∩ D, {v_{i}} ⊂ D.  (16.15)

So by the fundamental theorem of calculus, we have that

((D^{k−1}f)(x + v_{1}) − (D^{k−1}f)(x))⟨v_{2}, v_{3}, …, v_{k}⟩
= ∫_{0}^{1} A_{k}(x + tv_{1})⟨v_{1}, v_{2}, …, v_{k}⟩ dt  (16.16)

for all x ∈ U ∩ D and {v_{i}} ⊂ D with v_{1} sufficiently small. By the same argument given in the proof of Proposition 16.14, Eq. (16.16) remains valid for all x ∈ U and {v_{i}} ⊂ X with v_{1} sufficiently small. We may write this last equation alternatively as,

(D^{k−1}f)(x + v_{1}) − (D^{k−1}f)(x) = ∫_{0}^{1} A_{k}(x + tv_{1})⟨v_{1}, ⋯ ⟩ dt.  (16.17)

Hence

(D^{k−1}f)(x + v_{1}) − (D^{k−1}f)(x) − A_{k}(x)⟨v_{1}, ⋯ ⟩
= ∫_{0}^{1}[A_{k}(x + tv_{1}) − A_{k}(x)]⟨v_{1}, ⋯ ⟩ dt

from which we get the estimate,

∥(D^{k−1}f)(x + v_{1}) − (D^{k−1}f)(x) − A_{k}(x)⟨v_{1}, ⋯ ⟩∥ ≤ ε(v_{1})∥v_{1}∥ (16.18)

where ε(v_{1}) := ∫_{0}^{1} ∥A_{k}(x + tv_{1}) − A_{k}(x)∥ dt. Notice by the continuity of A_{k} that ε(v_{1}) → 0 as v_{1} → 0. Thus it follow from Eq. (16.18) that D^{k−1}f is differentiable and that (D^{k}f)(x) = A_{k}(x).

Example 16.19. Let f : GL(X, Y) → GL(Y, X) be defined by f(A) := A^{−1}. We assume that GL(X, Y) is not empty. Then f is infinitely differentiable and

(D^{k}f)(A)⟨V_{1}, V_{2}, …, V_{k}⟩
= (−1)^{k} ∑_{σ}{B^{−1}V_{σ(1)}B^{−1}V_{σ(2)}B^{−1} ⋯ B^{−1}V_{σ(k)}B^{−1}}, (16.19)

where sum is over all permutations of σ of {1, 2, …, k}.

Let me check Eq. (16.19) in the case that k = 2. Notice that we have already shown that (∂_{V_{1}}f)(B) = Df(B)V_{1} = −B^{−1}V_{1}B^{−1}. Using the product rule we find that

(∂_{V_{2}}∂_{V_{1}}f)(B) = B^{−1}V_{2}B^{−1}V_{1}B^{−1} + B^{−1}V_{1}B^{−1}V_{2}B^{−1} =: A_{2}(B)⟨V_{1}, V_{2}⟩.

Notice that ∥A_{2}(B)⟨V_{1}, V_{2}⟩∥ ≤ 2∥B^{−1}∥^{3}∥V_{1}∥ · ∥V_{2}∥, so that ∥A_{2}(B)∥ ≤ 2∥B^{−1}∥^{3} < ∞. Hence A_{2} : GL(X, Y) → M_{2}(L(X, Y), L(Y, X)). Also

∥(A_{2}(B) − A_{2}(C))⟨V_{1}, V_{2}⟩∥ ≤ 2∥B^{−1}V_{2}B^{−1}V_{1}B^{−1} − C^{−1}V_{2}C^{−1}V_{1}C^{−1}∥
≤ 2∥B^{−1}V_{2}B^{−1}V_{1}B^{−1} − B^{−1}V_{2}B^{−1}V_{1}C^{−1}∥
+ 2∥B^{−1}V_{2}B^{−1}V_{1}C^{−1} − B^{−1}V_{2}C^{−1}V_{1}C^{−1}∥
+ 2∥B^{−1}V_{2}C^{−1}V_{1}C^{−1} − C^{−1}V_{2}C^{−1}V_{1}C^{−1}∥
≤ 2∥B^{−1}∥^{2}∥V_{2}∥∥V_{1}∥∥B^{−1} − C^{−1}∥
+ 2∥B^{−1}∥∥C^{−1}∥∥V_{2}∥∥V_{1}∥∥B^{−1} − C^{−1}∥
+ 2∥C^{−1}∥^{2}∥V_{2}∥∥V_{1}∥∥B^{−1} − C^{−1}∥.

This shows that

∥A_{2}(B) − A_{2}(C)∥ ≤ 2∥B^{−1} − C^{−1}∥{∥B^{−1}∥^{2} + ∥B^{−1}∥∥C^{−1}∥ + ∥C^{−1}∥^{2}}.

Since B → B^{−1} is differentiable and hence continuous, it follows that A_{2}(B) is also continuous in B. Hence by Theorem 16.18 D^{2}f(A) exists and is given as in Eq. (16.19)

Example 16.20. Suppose that f : ℝ → ℝ is a C^{∞}– function and F (x) := ∫_{0}^{1} f(x(t)) dt for x ∈ X := C([0, 1], ℝ) equipped with the norm ∥x∥ := max_{t∈[0,1]} |x(t)|. Then F : X → ℝ is also infinitely differentiable and

(D^{k}F)(x)⟨v_{1}, v_{2}, …, v_{k}⟩ = ∫_{0}^{1} f^{(k)}(x(t))v_{1}(t) ⋯ v_{k}(t) dt,  (16.20)

for all x ∈ X and {v_{i}} ⊂ X.

To verify this example, notice that

(∂_{v}F)(x) := \frac{d}{ds}|_{0}F (x + sv) = \frac{d}{ds}|_{0} ∫_{0}^{1} f(x(t) + sv(t)) dt

= ∫_{0}^{1} \frac{d}{ds}|_{0}f(x(t) + sv(t)) dt = ∫_{0}^{1} f′(x(t))v(t) dt.

Similar computations show that

(∂_{v_{1}}∂_{v_{2}} ⋯ ∂_{v_{k}}f)(x) = ∫_{0}^{1} f^{(k)}(x(t))v_{1}(t) ⋯ v_{k}(t) dt =: A_{k}(x)⟨v_{1}, v_{2}, …, v_{k}⟩.

Now for x, y ∈ X,

|A_{k}(x)⟨v_{1}, v_{2}, …, v_{k}⟩ − A_{k}(y)⟨v_{1}, v_{2}, …, v_{k}⟩|

≤ ∫_{0}^{1} |f^{(k)}(x(t)) − f^{(k)}(y(t))| · |v_{1}(t) ⋯ v_{k}(t) |dt

≤ ∏_{i=1}^{k} ∥v_{i}∥ ∫_{0}^{1} |f^{(k)}(x(t)) − f^{(k)}(y(t))|dt,

which shows that

∥A_{k}(x) − A_{k}(y)∥ ≤ ∫_{0}^{1} |f^{(k)}(x(t)) − f^{(k)}(y(t))|dt.

This last expression is easily seen to go to zero as y → x in X. Hence A_{k} is continuous. Thus we may apply Theorem 16.18 to conclude that Eq. (16.20) is valid.

16.5 Inverse and Implicit Function Theorems

In this section, let X be a Banach space, R > 0, U = B = B(0, R) ⊂ X and ε : U → X be a continuous function such that ε (0) = 0. Our immediate goal is to give a sufficient condition on ε so that F (x) := x + ε(x) is a homeomorphism from U to F (U) with F (U) being an open subset of X. Let’s start by looking at the one dimensional case first. So for the moment assume that X = ℝ, U = (−1, 1), and ε : U → ℝ is C^{1}. Then F will be injective iff F is either strictly increasing or decreasing. Since we are thinking that F is a “small” perturbation of the identity function we will assume that F is strictly increasing, i.e. F′ = 1 + ε′ > 0. This positivity condition is not so easily interpreted for operators on a Banach space. However the condition that |ε′| ≤ α < 1 is easily interpreted in the Banach space setting and it implies 1 + ε′ > 0.

Lemma 16.21. Suppose that U = B = B(0, R) (R > 0) is a ball in X and ε : B → X is a C^{1} function such that ∥Dε∥ ≤ α < ∞ on U. Then

∥ε(x) − ε(y)∥ ≤ α∥x − y∥ for all x, y ∈ U.  (16.21)

Proof. By the fundamental theorem of calculus and the chain rule:

ε(y) − ε(x) = ∫_{0}^{1} \frac{d}{dt}ε(x + t(y − x))dt
= ∫_{0}^{1}[Dε(x + t(y − x))](y − x)dt.

Therefore, by the triangle inequality and the assumption that ∥Dε(x)∥ ≤ α on B,

∥ε(y) − ε(x)∥ ≤ ∫_{0}^{1} ∥Dε(x + t(y − x))∥dt · ∥(y − x)∥ ≤ α∥(y − x)∥.

Remark 16.22. It is easily checked that if ε : U = B(0, R) → X is C^{1} and satisfies (16.21) then ∥Dε∥ ≤ α on U.

Using the above remark and the analogy to the one dimensional example, one is lead to the following proposition.

Proposition 16.23. Suppose α ∈ (0, 1), R > 0, U = B(0, R) ⊂_{o} X and ε : U → X is a continuous function such that ε (0) = 0 and

∥ε(x) − ε(y)∥ ≤ α∥x − y∥ ∀ x, y ∈ U.  (16.22)

Then F : U → X defined by F (x) := x + ε(x) for x ∈ U satisfies:

1. F is an injective map and G = F^{−1} : V := F (U) → U is continuous.

2. If x_{0} ∈ U, z_{0} = F (x_{0}) and r > 0 such the B(x_{0}, r) ⊂ U, then

B(z_{0}, (1 − α)r) ⊂ F (B(x_{0}, r)) ⊂ B(z_{0}, (1 + α)r).  (16.23)

In particular, for all r ≤ R,

B(0, (1 − α) r) ⊂ F (B(0, r)) ⊂ B(0, (1 + α) r),  (16.24)

see Figure 16.1 below.

3. V := F (U) is open subset of X and F : U → V is a homeomorphism.

Proof.

1. Using the definition of F and the estimate in Eq. (16.22),

∥x − y∥ = ∥(F (x) − F (y)) − (ε(x) − ε(y))∥
≤ ∥F (x) − F (y)∥ + ∥ε(x) − ε(y)∥
≤ ∥F (x) − F (y)∥ + α∥(x − y)∥

for all x, y ∈ U. This implies

Fig. 16.1. Nesting of F (B(x_{0}, r)) between B(z_{0}, (1 − α)r) and B(z_{0}, (1 + α)r).

∥x − y∥ ≤ (1 − α)^{−1}∥F (x) − F (y)∥  (16.25)

which shows F is injective on U and hence shows the inverse function G = F^{−1} : V := F (U) → U is well defined. Moreover, replacing x, y in Eq. (16.25) by G (x) and G (y) respectively with x, y ∈ V shows

∥G (x) − G (y) ∥ ≤ (1 − α)^{−1}∥x − y∥ for all x, y ∈ V.  (16.26)

Hence G is Lipschitz on V and hence continuous.

2. Let x_{0} ∈ U, r > 0 and z_{0} = F (x_{0}) = x_{0} + ε(x_{0}) be as in item 2. The second inclusion in Eq. (16.23) follows from the simple computation:

∥F (x_{0} + h) − z_{0}∥ = ∥h + ε (x_{0} + h) − ε (x_{0})∥
≤ ∥h∥ + ∥ε (x_{0} + h) − ε (x_{0})∥
≤ (1 + α) ∥h∥ < (1 + α) r

for all h ∈ B (0, r) . To prove the first inclusion in Eq. (16.23) we must find, for every z ∈ B(z_{0}, (1−α)r), an h ∈ B (0, r) such that z = F (x_{0} + h) or equivalently an h ∈ B (0, r) solving

z − z_{0} = F (x_{0} + h) − F (x_{0}) = h + ε(x_{0} + h) − ε(x_{0}).

Let k := z − z_{0} and for h ∈ B (0, r) , let δ (h) := ε(x_{0} + h) − ε(x_{0}). With this notation it suffices to show for each k ∈ B(z_{0}, (1 − α)r) there exists h ∈ B (0, r) such that k = h + δ (h) . Notice that δ (0) = 0 and

∥δ (h_{1}) − δ (h_{2})∥ = ∥ε(x_{0} + h_{1}) − ε(x_{0} + h_{2})∥ ≤ α ∥h_{1} − h_{2}∥ (16.27)

for all h_{1}, h_{2} ∈ B (0, r) . We are now going to solve the equation k = h + δ (h) for h by the method of successive approximations starting with h_{0} = 0 and then defining h_{n} inductively by

h_{n+1} = k − δ (h_{n}) .  (16.28)

A simple induction argument using Eq. (16.27) shows that

∥h_{n+1} − h_{n}∥ ≤ α^{n} ∥k∥ for all n ∈ ℕ_{0}

and in particular that

∥h_{N}∥ = ∥ ∑_{n=0}^{N−1} (h_{n+1} − h_{n})∥ ≤ ∑_{n=0}^{N−1} ∥h_{n+1} − h_{n}∥

≤ ∑_{n=0}^{N−1} α^{n} ∥k∥ = \frac{1 − α^{N}}{1 − α} ∥k∥ .  (16.29)

Since ∥k∥ < (1 − α) r, this implies that ∥h_{N}∥ < r for all N showing the approximation procedure is well defined. Let

h := lim_{N→∞} h_{n} = ∑_{n=0}^{∞} (h_{n+1} − h_{n}) ∈ X

which exists since the sum in the previous equation is absolutely convergent. Passing to the limit in Eqs. (16.29) and (16.28) shows that ∥h∥ ≤ (1 − α)^{−1} ∥k∥ < r and h = k − δ (h) , i.e. h ∈ B (0, r) solves k = h + δ (h) as desired.

3. Given x_{0} ∈ U, the first inclusion in Eq. (16.23) shows that z_{0} = F (x_{0}) is in the interior of F (U) . Since z_{0} ∈ F (U) was arbitrary, it follows that V = F (U) is open. The continuity of the inverse function has already been proved in item 1.

For the remainder of this section let X and Y be two Banach spaces, U ⊂_{o} X, k ≥ 1, and f ∈ C^{k}(U, Y).

Lemma 16.24. Suppose x_{0} ∈ U, R > 0 is such that B^{X}(x_{0}, R) ⊂ U and T : B^{X}(x_{0}, R) → Y is a C^{1} – function such that T′(x_{0}) is invertible. Let

α (R) := sup_{x∈B^{X}(x_{0},R)} ∥T′(x_{0})^{−1}T′(x) − I∥_{L(X)}  (16.30)

and ε ∈ C^{1} (B^{X}(0, R), X) be defined by

ε (h) = T′(x_{0})^{−1} [T (x_{0} + h) − T (x_{0})] − h  (16.31)

so that

T (x_{0} + h) = T (x_{0}) + T′(x_{0}) (h + ε(h)) .  (16.32)

Then ε(h) = o(h) as h → 0 and

∥ε(h′) − ε(h)∥ ≤ α (R) ∥h′ − h∥ for all h, h′ ∈ B^{X}(0, R).  (16.33)

If α (R) < 1 (which may be achieved by shrinking R if necessary), then T′(x) is invertible for all x ∈ B^{X}(x_{0}, R) and

sup_{x∈B^{X}(x_{0},R)} ∥T′(x)^{−1}∥_{L(Y,X)} ≤ \frac{1}{1 − α (R)} ∥T′(x_{0})^{−1}∥_{L(Y,X)} .  (16.34)

Proof. By definition of T′ (x_{0}) and using T′ (x_{0})^{−1} exists,

T (x_{0} + h) − T (x_{0}) = T′(x_{0})h + o(h)

from which it follows that ε(h) = o(h). In fact by the fundamental theorem of calculus,

ε(h) = ∫_{0}^{1} (T′(x_{0})^{−1}T′(x_{0} + th) − I) hdt

but we will not use this here. Let h, h′ ∈ B^{X}(0, R) and apply the fundamental theorem of calculus to t → T (x_{0} + t(h′ − h)) to conclude

ε(h′) − ε(h) = T′(x_{0})^{−1} [T (x_{0} + h′) − T (x_{0} + h)] − (h′ − h)
= [∫_{0}^{1} (T′(x_{0})^{−1}T′(x_{0} + t(h′ − h)) − I) dt] (h′ − h).

Taking norms of this equation gives

∥ε(h′) − ε(h)∥ ≤ [∫_{0}^{1} ∥T′(x_{0})^{−1}T′(x_{0} + t(h′ − h)) − I∥ dt] ∥h′ − h∥
≤ α (R) ∥h′ − h∥

It only remains to prove Eq. (16.34), so suppose now that α (R) < 1. Then by Proposition 7.19, T′(x_{0})^{−1}T′(x) = I − (I − T′(x_{0})^{−1}T′(x)) is invertible and

∥[T′(x_{0})^{−1}T′(x)]^{−1}∥ ≤ \frac{1}{1 − α (R)} for all x ∈ B^{X}(x_{0}, R).

Since T′(x) = T′(x_{0}) [T′(x_{0})^{−1}T′(x)] this implies T′(x) is invertible and

∥T′(x)^{−1}∥ = ∥[T′(x_{0})^{−1}T′(x)]^{−1} T′(x_{0})^{−1}∥ ≤ \frac{1}{1 − α (R)} ∥T′(x_{0})^{−1}∥

for all x ∈ B^{X}(x_{0}, R).

Theorem 16.25 (Inverse Function Theorem). Suppose U ⊂_{o} X, k ≥ 1 and T ∈ C^{k}(U, Y) such that T′(x) is invertible for all x ∈ U. Further assume x_{0} ∈ U and R > 0 such that B^{X}(x_{0}, R) ⊂ U.

1. For all r ≤ R,

T (B^{X}(x_{0}, r)) ⊂ T (x_{0}) + T′ (x_{0}) B^{X} (0, (1 + α (r))r) .  (16.35)

2. If we further assume that

α (R) := sup_{x∈B^{X}(x_{0},R)} ∥T′(x_{0})^{−1}T′(x) − I∥ < 1,

which may always be achieved by taking R sufficiently small, then

T (x_{0}) + T′ (x_{0}) B^{X} (0, (1 − α (r))r) ⊂ T (B^{X}(x_{0}, r))  (16.36)

for all r ≤ R, see Figure 16.2.

3. T : U → Y is an open mapping, in particular V := T (U) ⊂_{o} Y.

4. Again if R is sufficiently small so that α (R) < 1, then T |_{B^{X}(x_{0},R)} : B^{X}(x_{0}, R) → T (B^{X}(x_{0}, R)) is invertible and T |_{B^{X}(x_{0},R)}^{−1} : T (B^{X}(x_{0}, R)) → B^{X}(x_{0}, R) is a C^{k} – map.

5. If T is injective, then T^{−1} : V → U is also a C^{k} – map and

(T^{−1})′ (y) = [T′(T^{−1}(y))]^{−1} for all y ∈ V.

Fig. 16.2. The nesting of T (B^{X}(x_{0}, r)) between T (x_{0})+T′ (x_{0}) B^{X} (0, (1 − α (r))r) andT (x_{0}) + T′ (x_{0}) B^{X} (0, (1 + α (r))r) .

Proof. Let ε ∈ C^{1} (B^{X}(0, R), X) be as defined in Eq. (16.31).

1. Using Eqs. (16.32) and (16.24),

T (B^{X}(x_{0}, r)) = T (x_{0}) + T′ (x_{0}) (I + ε) (B^{X} (0, r))  (16.37)
⊂ T (x_{0}) + T′ (x_{0}) B^{X} (0, (1 + α (r)) r)

which proves Eq. (16.35).

2. Now assume α (R) < 1, then by Eqs. (16.37) and (16.24),

T (x_{0}) + T′ (x_{0}) B^{X} (0, (1 − α (r)) r)
⊂ T (x_{0}) + T′ (x_{0}) (I + ε) (B^{X} (0, r)) = T (B^{X} (x_{0}, r))

which proves Eq. (16.36).

3. Notice that h ∈ X → T (x_{0}) + T′ (x_{0}) h ∈ Y is a homeomorphism. The fact that T is an open map follows easily from Eq. (16.36) which shows that T (x_{0}) is interior of T (W) for any W ⊂_{o} X with x_{0} ∈ W.

4. The fact that T |_{B^{X}(x_{0},R)} : B^{X}(x_{0}, R) → T (B^{X}(x_{0}, R)) is invertible with a continuous inverse follows from Eq. (16.32) and Proposition 16.23. It now follows from the converse to the chain rule, Theorem 16.7, that g := T |_{B^{X}(x_{0},R)}^{−1} : T (B^{X}(x_{0}, R)) → B^{X}(x_{0}, R) is differentiable and

g′ (y) = [T′ (g (y))]^{−1} for all y ∈ T (B^{X}(x_{0}, R)) .

This equation shows g is C^{1}. Now suppose that k ≥ 2. Since T′ ∈ C^{k−1}(B, L(X)) and i(A) := A^{−1} is a smooth map by Example 16.19, g′ = i ∘ T′ ∘ g is C^{1}, i.e. g is C^{2}. If k ≥ 2, we may use the same argument to now show g is C^{3}. Continuing this way inductively, we learn g is C^{k}.

5. Since differentiability and smoothness is local, the assertion in item 5. follows directly from what has already been proved.

Theorem 16.26 (Implicit Function Theorem). Suppose that X, Y, and W are three Banach spaces, k ≥ 1, A ⊂ X × Y is an open set, (x_{0}, y_{0}) is a point in A, and f : A → W is a C^{k} – map such f(x_{0}, y_{0}) = 0. Assume that D_{2}f(x_{0}, y_{0}) := D(f(x_{0}, ·))(y_{0}) : Y → W is a bounded invertible linear transformation. Then there is an open neighborhood U_{0} of x_{0} in X such that for all connected open neighborhoods U of x_{0} contained in U_{0}, there is a unique continuous function u : U → Y such that u(x_{0}) = y_{o}, (x, u(x)) ∈ A and f(x, u(x)) = 0 for all x ∈ U. Moreover u is necessarily C^{k} and

Du(x) = −D_{2}f(x, u(x))^{−1}D_{1}f(x, u(x)) for all x ∈ U.  (16.38)

Proof. By replacing f by (x, y) → D_{2}f(x_{0}, y_{0})^{−1}f(x, y) if necessary, we may assume with out loss of generality that W = Y and D_{2}f(x_{0}, y_{0}) = I_{Y} . Define F : A → X ×Y by F (x, y) := (x, f(x, y)) for all (x, y) ∈ A. Notice that

DF (x, y) = [ \begin{matrix}I & D_{1}f(x, y) \\ 0 & D_{2}f(x, y) \end{matrix} ]

which is invertible iff D_{2}f(x, y) is invertible and if D_{2}f(x, y) is invertible then

DF (x, y)^{−1} = [ \begin{matrix}I & −D_{1}f(x, y)D_{2}f(x, y)^{−1} \\ 0 & D_{2}f(x, y)^{−1} \end{matrix} ] .

Since D_{2}f(x_{0}, y_{0}) = I is invertible, the inverse function theorem guarantees that there exists a neighborhood U_{0} of x_{0} and V_{0} of y_{0} such that U_{0} ×V_{0} ⊂ A, F (U_{0} × V_{0}) is open in X × Y, F |_{(U_{0}×V_{0})} has a C^{k}–inverse which we call F^{−1}. Let π_{2}(x, y) := y for all (x, y) ∈ X × Y and define C^{k} – function u_{0} on U_{0} by u_{0}(x) := π_{2} ∘ F^{−1}(x, 0). Since F^{−1}(x, 0) = (x̃, u_{0}(x)) iff

(x, 0) = F (x̃, u_{0}(x)) = (x̃, f(x̃, u_{0}(x))),

it follows that x = x̃ and f(x, u_{0}(x)) = 0. Thus

(x, u_{0}(x)) = F^{−1}(x, 0) ∈ U_{0} × V_{0} ⊂ A

and f(x, u_{0}(x)) = 0 for all x ∈ U_{0}. Moreover, u_{0} is C^{k} being the composition of the C^{k}– functions, x → (x, 0), F^{−1}, and π_{2}. So if U ⊂ U_{0} is a connected set containing x_{0}, we may define u := u_{0}|_{U} to show the existence of the functions u as described in the statement of the theorem. The only statement left to prove is the uniqueness of such a function u. Suppose that u_{1} : U → Y is another continuous function such that u_{1}(x_{0}) = y_{0}, and (x, u_{1}(x)) ∈ A and f(x, u_{1}(x)) = 0 for all x ∈ U. Let

O := {x ∈ U|u(x) = u_{1}(x)} = {x ∈ U|u_{0}(x) = u_{1}(x)}.

Clearly O is a (relatively) closed subset of U which is not empty since x_{0} ∈ O. Because U is connected, if we show that O is also an open set we will have shown that O = U or equivalently that u_{1} = u_{0} on U. So suppose that x ∈ O, i.e. u_{0}(x) = u_{1}(x). For x̃ near x ∈ U,

0 = 0 − 0 = f(x̃, u_{0}(x̃)) − f(x̃, u_{1}(x̃)) = R(x̃)(u_{1}(x̃) − u_{0}(x̃))  (16.39)

where

R(x̃) := ∫_{0}^{1} D_{2}f((x̃, u_{0}(x̃) + t(u_{1}(x̃) − u_{0}(x̃)))dt.  (16.40)

From Eq. (16.40) and the continuity of u_{0} and u_{1}, lim_{x̃→x} R(x̃) = D_{2}f(x, u_{0}(x)) which is invertible.^{3} Thus R(x̃) is invertible for all x̃ sufficiently close to x which combined with Eq. (16.39) implies that u_{1}(x̃) = u_{0}(x̃) for all x̃ sufficiently close to x. Since x ∈ O was arbitrary, we have shown that O is open.

16.6 Smooth Dependence of ODE’s on Initial Conditions*

In this subsection, let X be a Banach space, U ⊂_{o} X and J be an open interval with 0 ∈ J.

Lemma 16.27. If Z ∈ C(J × U, X) such that D_{x}Z(t, x) exists for all (t, x) ∈ J × U and D_{x}Z(t, x) ∈ C(J × U, X) then Z is locally Lipschitz in x, see Definition 15.6.

Proof. Suppose I ⊏⊏ J and x ∈ U. By the continuity of DZ, for every t ∈ I there an open neighborhood N_{t} of t ∈ I and ε_{t} > 0 such that B(x, ε_{t}) ⊂ U and

sup {∥D_{x}Z(t′, x′)∥ : (t′, x′) ∈ N_{t} × B(x, ε_{t})} < ∞.

By the compactness of I, there exists a finite subset Λ ⊂ I such that I ⊂ ∪_{t∈I}N_{t}. Let ε(x, I) := min {ε_{t} : t ∈ Λ} and

3 Notice that DF (x, u_{0}(x)) is invertible for all x ∈ U_{0} since F |_{U_{0}×V_{0}} has a C^{1} inverse. Therefore D_{2}f(x, u_{0}(x)) is also invertible for all x ∈ U_{0}.

K(x, I) := sup {∥DZ(t, x′)∥(t, x′) ∈ I × B(x, ε(x, I))} < ∞.

Then by the fundamental theorem of calculus and the triangle inequality,

∥Z(t, x_{1}) − Z(t, x_{0})∥ ≤ (∫_{0}^{1} ∥D_{x}Z(t, x_{0} + s(x_{1} − x_{0})∥ ds) ∥x_{1} − x_{0}∥
≤ K(x, I)∥x_{1} − x_{0}∥

for all x_{0}, x_{1} ∈ B(x, ε(x, I)) and t ∈ I.

Theorem 16.28 (Smooth Dependence of ODE’s on Initial Conditions). Let X be a Banach space, U ⊂_{o} X, Z ∈ C(ℝ × U, X) such that D_{x}Z ∈ C(ℝ×U, X) and ϕ : D(Z) ⊂ ℝ×X → X denote the maximal solution operator to the ordinary differential equation

ẏ(t) = Z(t, y(t)) with y(0) = x ∈ U,  (16.41)

see Notation 15.9 and Theorem 15.15. Then ϕ ∈ C^{1}(D(Z), U), ∂_{t}D_{x}ϕ(t, x) exists and is continuous for (t, x) ∈ D(Z) and D_{x}ϕ(t, x) satisfies the linear differential equation,

\frac{d}{dt}D_{x}ϕ(t, x) = [(D_{x}Z) (t, ϕ(t, x))]D_{x}ϕ(t, x) with D_{x}ϕ(0, x) = I_{X} (16.42)

for t ∈ J_{x}.

Proof. Let x_{0} ∈ U and J be an open interval such that 0 ∈ J ⊂ J̄ ⊏⊏ J_{x_{0}}, y_{0} := y(·, x_{0})|_{J} and

O_{ε} := {y ∈ BC(J, U) : ∥y − y_{0}∥_{∞} < ε} ⊂_{o} BC(J, X).

By Lemma 16.27, Z is locally Lipschitz and therefore Theorem 15.15 is applicable. By Eq. (15.23) of Theorem 15.15, there exists ε > 0 and δ > 0 such that G : B(x_{0}, δ) → O_{ε} defined by G(x) := ϕ(·, x)|_{J} is continuous. By Lemma 16.29 below, for ε > 0 sufficiently small the function F : O_{ε} → BC(J, X) defined by

F (y) := y − ∫_{0}^{·} Z(t, y(t))dt.  (16.43)

is C^{1} and

DF (y)v = v − ∫_{0}^{·} D_{y}Z(t, y(t))v(t)dt.  (16.44)

By the existence and uniqueness Theorem 8.21 for linear ordinary differential equations, DF (y) is invertible for any y ∈ BC(J, U). By the definition of ϕ, F (G(x)) = h(x) for all x ∈ B(x_{0}, δ) where h : X → BC(J, X) is defined by h(x)(t) = x for all t ∈ J, i.e. h(x) is the constant path at x. Since h is a bounded linear map, h is smooth and Dh(x) = h for all x ∈ X.

We may now apply the converse to the chain rule in Theorem 16.7 to conclude G ∈ C^{1} (B(x_{0}, δ), O) and DG(x) = [DF (G(x))]^{−1}Dh(x) or equivalently, DF (G(x))DG(x) = h which in turn is equivalent to

D_{x}ϕ(t, x) − ∫_{0}^{t}[DZ(ϕ(τ, x)]D_{x}ϕ(τ, x) dτ = I_{X}.

As usual this equation implies D_{x}ϕ(t, x) is differentiable in t, D_{x}ϕ(t, x) is continuous in (t, x) and D_{x}ϕ(t, x) satisfies Eq. (16.42).

Lemma 16.29. Continuing the notation used in the proof of Theorem 16.28 and further let

f(y) := ∫_{0}^{·} Z(τ, y(τ)) dτ for y ∈ O_{ε}.

Then f ∈ C^{1}(O_{ε}, Y) and for all y ∈ O_{ε},

f′(y)h = ∫_{0}^{·} D_{x}Z(τ, y(τ))h(τ) dτ =: Λ_{y}h.

Proof. Let h ∈ Y be sufficiently small and τ ∈ J, then by fundamental theorem of calculus,

Z(τ,y(τ) + h(τ)) − Z(τ, y(τ))
= ∫_{0}^{1}[D_{x}Z(τ, y(τ) + rh(τ)) − D_{x}Z(τ, y(τ))]dr

and therefore,

f(y + h) − f(y) − Λ_{y}h(t)
= ∫_{0}^{t}[Z(τ, y(τ) + h(τ)) − Z(τ, y(τ)) − D_{x}Z(τ, y(τ))h(τ) ] dτ
= ∫_{0}^{t} dτ ∫_{0}^{1} dr[D_{x}Z(τ, y(τ) + rh(τ)) − D_{x}Z(τ, y(τ))]h(τ).

Therefore,

∥(f(y + h) − f(y) − Λ_{y}h)∥_{∞} ≤ ∥h∥_{∞}δ(h)  (16.45)

where

δ(h) := ∫_{J} dτ ∫_{0}^{1} dr ∥D_{x}Z(τ, y(τ) + rh(τ)) − D_{x}Z(τ, y(τ))∥ .

With the aide of Lemmas 16.27 and Lemma 15.7,

(r, τ, h) ∈ [0, 1] × J × Y → ∥D_{x}Z(τ, y(τ) + rh(τ))∥

is bounded for small h provided ε > 0 is sufficiently small. Thus it follows from the dominated convergence theorem that δ(h) → 0 as h → 0 and hence Eq. (16.45) implies f′(y) exists and is given by Λ_{y}. Similarly,

||f′(y + h) − f′(y)||_{op}
≤ ∫_{J} ∥D_{x}Z(τ, y(τ) + h(τ)) − D_{x}Z(τ, y(τ))∥ dτ → 0 as h → 0

showing f′ is continuous.

Remark 16.30. If Z ∈ C^{k}(U, X), then an inductive argument shows that ϕ ∈ C^{k}(D(Z), X). For example if Z ∈ C^{2}(U, X) then (y(t), u(t)) := (ϕ(t, x), D_{x}ϕ(t, x)) solves the ODE,

\frac{d}{dt}(y(t), u(t)) = Z̃ ((y(t), u(t))) with (y(0), u(0)) = (x, Id_{X})

where Z̃ is the C^{1} – vector field defined by

Z̃ (x, u) = (Z(x), D_{x}Z(x)u) .

Therefore Theorem 16.28 may be applied to this equation to deduce: D_{x}^{2}ϕ(t, x) and D_{x}^{2}ϕ̇(t, x) exist and are continuous. We may now differentiate Eq. (16.42) to find D_{x}^{2}ϕ(t, x) satisfies the ODE,

\frac{d}{dt}D_{x}^{2}ϕ(t, x) = [(∂_{D_{x}ϕ(t,x)}D_{x}Z) (t, ϕ(t, x))]D_{x}ϕ(t, x)
+ [(D_{x}Z) (t, ϕ(t, x))]D_{x}^{2}ϕ(t, x)

with D_{x}^{2}ϕ(0, x) = 0.

16.7 Existence of Periodic Solutions

A detailed discussion of the inverse function theorem on Banach and Frechét spaces may be found in Richard Hamilton’s, “The Inverse Function Theorem of Nash and Moser.” The applications in this section are taken from this paper. In what follows we say f ∈ C_{2π}^{k}(ℝ, (c, d)) if f ∈ C_{2π}^{k}(ℝ, (c, d)) and f is 2π – periodic, i.e. f (x + 2π) = f (x) for all x ∈ ℝ.

Theorem 16.31 (Taken from Hamilton, p. 110.). Let p : U := (a, b) → V := (c, d) be a smooth function with p′ > 0 on (a, b). For every g ∈ C_{2π}^{∞}(ℝ, (c, d)) there exists a unique function y ∈ C_{2π}^{∞}(ℝ, (a, b)) such that

ẏ(t) + p(y(t)) = g(t).

Proof. Let Ṽ := C_{2π}^{0}(ℝ, (c, d)) ⊂_{o} C_{2π}^{0}(ℝ, ℝ) and Ũ ⊂_{o} C_{2π}^{1}(ℝ, (a, b)) be given by

Ũ := {y ∈ C_{2π}^{1}(ℝ, ℝ) : a < y(t) < b & c < ẏ(t) + p(y(t)) < d ∀ t} .

The proof will be completed by showing P : Ũ → Ṽ defined by

P (y)(t) = ẏ(t) + p(y(t)) for y ∈ Ũ and t ∈ ℝ

is bijective. Note that if P (y) is smooth then so is y.

Step 1. The differential of P is given by P′(y)h = ḣ + p′(y)h, see Exercise 16.8. We will now show that the linear mapping P′(y) is invertible. Indeed let f = p′(y) > 0, then the general solution to the Eq. ḣ + fh = k is given by

h(t) = e^{− ∫_{0}^{t} f(τ)dτ} h_{0} + ∫_{0}^{t} e^{− ∫_{τ}^{t} f(s)ds}k(τ)dτ

where h_{0} is a constant. We wish to choose h_{0} so that h(2π) = h_{0}, i.e. so that

h_{0} (1 − e^{−c(f)}) = ∫_{0}^{2π} e^{− ∫_{τ}^{t} f(s)ds}k(τ)dτ

where

c(f) = ∫_{0}^{2π} f(τ)dτ = ∫_{0}^{2π} p′(y(τ))dτ > 0.

The unique solution h ∈ C_{2π}^{1}(ℝ, ℝ) to P′(y)h = k is given by

h(t) = (1 − e^{−c(f)})^{−1} e^{− ∫_{0}^{t} f(τ)dτ} ∫_{0}^{2π} e^{− ∫_{τ}^{t} f(s)ds}k(τ)dτ + ∫_{0}^{t} e^{− ∫_{τ}^{t} f(s)ds}k(τ)dτ

= (1 − e^{−c(f)})^{−1} e^{− ∫_{0}^{t} f(s)ds} ∫_{0}^{2π} e^{− ∫_{τ}^{t} f(s)ds}k(τ)dτ + ∫_{0}^{t} e^{− ∫_{τ}^{t} f(s)ds}k(τ)dτ.

Therefore P′(y) is invertible for all y. Hence by the inverse function Theorem 16.25, P : Ũ → Ṽ is an open mapping which is locally invertible.

Step 2. Let us now prove P : Ũ → Ṽ is injective. For this suppose y_{1}, y_{2} ∈ Ũ such that P (y_{1}) = g = P (y_{2}) and let z = y_{2} − y_{1}. Since

ż(t) + p(y_{2}(t)) − p(y_{1}(t)) = g(t) − g(t) = 0,

if t_{m} ∈ ℝ is point where z(t_{m}) takes on its maximum, then ż(t_{m}) = 0 and hence

p(y_{2}(t_{m})) − p(y_{1}(t_{m})) = 0.

Since p is increasing this implies y_{2}(t_{m}) = y_{1}(t_{m}) and hence z(t_{m}) = 0. This shows z(t) ≤ 0 for all t and a similar argument using a minimizer of z shows z(t) ≥ 0 for all t. So we conclude y_{1} = y_{2}.

Step 3. Let W := P (Ũ), we wish to show W = Ṽ . By step 1., we know W is an open subset of Ṽ and since Ṽ is connected, to finish the proof it suffices to show W is relatively closed in Ṽ . So suppose y_{j} ∈ Ũ such that g_{j} := P (y_{j}) → g ∈ Ṽ . We must now show g ∈ W, i.e. g = P (y) for some y ∈ W. If t_{m} is a maximizer of y_{j}, then ẏ_{j}(t_{m}) = 0 and hence g_{j}(t_{m}) = p(y_{j}(t_{m})) < d and therefore y_{j}(t_{m}) < b because p is increasing. A similar argument works for the minimizers then allows us to conclude Ran p ∘ y_{j}) ⊂ Ran g_{j}) ⊏⊏ (c, d) for all j. Since g_{j} is converging uniformly to g, there exists c < γ < δ < d such that Ran(p ∘ y_{j}) ⊂ Ran(g_{j}) ⊂ [γ, δ] for all j. Again since p′ > 0,

Ran(y_{j}) ⊂ p^{−1} ([γ, δ]) = [α, β] ⊏⊏ (a, b) for all j.

In particular sup {|ẏ_{j}(t)| : t ∈ ℝ and j} < ∞ since

ẏ_{j}(t) = g_{j}(t) − p(y_{j}(t)) ⊂ [γ, δ] − [γ, δ]  (16.46)

which is a compact subset of ℝ. The Ascoli-Arzela Theorem 11.29 now allows us to assume, by passing to a subsequence if necessary, that y_{j} is converging uniformly to y ∈ C_{2π}^{0}(ℝ, [α, β]). It now follows that

ẏ_{j}(t) = g_{j}(t) − p(y_{j}(t)) → g − p(y)

uniformly in t. Hence we concluded that y ∈ C_{2π}^{1}(ℝ, ℝ)∩C_{2π}^{0}(ℝ, [α, β]), ẏ_{j} → y and P (y) = g. This has proved that g ∈ W and hence that W is relatively closed in Ṽ .

16.8 Contraction Mapping Principle

Some of the arguments uses in this chapter and in Chapter 15 may be abstracted to a general principle of finding fixed points on a complete metric space. This is the content of this chapter.

Theorem 16.32. Suppose that (X, ρ) is a complete metric space and S : X → X is a contraction, i.e. there exists α ∈ (0, 1) such that ρ(S(x), S(y)) ≤ αρ(x, y) for all x, y ∈ X. Then S has a unique fixed point in X, i.e. there exists a unique point x ∈ X such that S(x) = x.

Proof. For uniqueness suppose that x and x′ are two fixed points of S, then

ρ(x, x′) = ρ(S(x), S(x′)) ≤ αρ(x, x′).

Therefore (1 − α)ρ(x, x′) ≤ 0 which implies that ρ(x, x′) = 0 since 1 − α > 0. Thus x = x′. For existence, let x_{0} ∈ X be any point in X and define x_{n} ∈ X inductively by x_{n+1} = S(x_{n}) for n ≥ 0. We will show that x := lim_{n→∞} x_{n} exists in X and because S is continuous this will imply,

x = lim_{n→∞} x_{n+1} = lim_{n→∞} S(x_{n}) = S( lim_{n→∞} x_{n}) = S(x),

showing x is a fixed point of S. So to finish the proof, because X is complete, it suffices to show {x_{n}}_{n=1}^{∞} is a Cauchy sequence in X. An easy inductive computation shows, for n ≥ 0, that

ρ(x_{n+1}, x_{n}) = ρ(S(x_{n}), S(x_{n−1})) ≤ αρ(x_{n}, x_{n−1}) ≤ ⋯ ≤ α^{n}ρ(x_{1}, x_{0}).

Another inductive argument using the triangle inequality shows, for m > n, that,

ρ(x_{m}, x_{n}) ≤ ρ(x_{m}, x_{m−1}) + ρ(x_{m−1}, x_{n}) ≤ ⋯ ≤ ∑_{k=n}^{m−1} ρ(x_{k+1}, x_{k}).

Combining the last two inequalities gives (using again that α ∈ (0, 1)),

ρ(x_{m}, x_{n}) ≤ ∑_{k=n}^{m−1} α^{k}ρ(x_{1}, x_{0}) ≤ ρ(x_{1}, x_{0})α^{n} ∑_{l=0}^{∞} α^{l} = ρ(x_{1}, x_{0})\frac{α^{n}}{1 − α}.

This last equation shows that ρ(x_{m}, x_{n}) → 0 as m, n → ∞, i.e. {x_{n}}_{n=0}^{∞} is a Cauchy sequence.

Corollary 16.33 (Contraction Mapping Principle II). Suppose that (X, ρ) is a complete metric space and S : X → X is a continuous map such that S^{(n)} is a contraction for some n ∈ ℕ. Here

S^{(n)} := \overbrace{S ∘ S ∘ … ∘ S}^{n times}

and we are assuming there exists α ∈ (0, 1) such that ρ(S^{(n)}(x), S^{(n)}(y)) ≤ αρ(x, y) for all x, y ∈ X. Then S has a unique fixed point in X.

Proof. Let T := S^{(n)}, then T : X → X is a contraction and hence T has a unique fixed point x ∈ X. Since any fixed point of S is also a fixed point of T, we see if S has a fixed point then it must be x. Now

T (S(x)) = S^{(n)}(S(x)) = S(S^{(n)}(x)) = S(T (x)) = S(x),

which shows that S(x) is also a fixed point of T. Since T has only one fixed point, we must have that S(x) = x. So we have shown that x is a fixed point of S and this fixed point is unique.

Lemma 16.34. Suppose that (X, ρ) is a complete metric space, n ∈ ℕ, Z is a topological space, and α ∈ (0, 1). Suppose for each z ∈ Z there is a map S_{z} : X → X with the following properties:

Contraction property ρ(S_{z}^{(n)}(x), S_{z}^{(n)}(y)) ≤ αρ(x, y) for all x, y ∈ X and z ∈

Z. Continuity in z For each x ∈ X the map z ∈ Z → S_{z}(x) ∈ X is continuous.

By Corollary 16.33 above, for each z ∈ Z there is a unique fixed point G(z) ∈ X of S_{z}.

Conclusion: The map G : Z → X is continuous.

Proof. Let T_{z} := S_{z}^{(n)}. If z, w ∈ Z, then

ρ(G(z), G(w)) = ρ(T_{z}(G(z)), T_{w}(G(w)))
≤ ρ(T_{z}(G(z)), T_{w}(G(z))) + ρ(T_{w}(G(z)), T_{w}(G(w)))
≤ ρ(T_{z}(G(z)), T_{w}(G(z))) + αρ(G(z), G(w)).

Solving this inequality for ρ(G(z), G(w)) gives

ρ(G(z), G(w)) ≤ \frac{1}{1 − α}ρ(T_{z}(G(z)), T_{w}(G(z))).

Since w → T_{w}(G(z)) is continuous it follows from the above equation that G(w) → G(z) as w → z, i.e. G is continuous.

16.9 Exercises

Exercise 16.3. Suppose that A : ℝ → L(X) is a continuous function and V : ℝ → L(X) is the unique solution to the linear differential equation

V̇ (t) = A(t)V (t) with V (0) = I.  (16.47)

Assuming that V (t) is invertible for all t ∈ ℝ, show that V^{−1}(t) := [V (t)]^{−1} must solve the differential equation

\frac{d}{dt}V^{−1}(t) = −V^{−1}(t)A(t) with V^{−1}(0) = I.  (16.48)

See Exercise 8.13 as well.

Exercise 16.4 (Differential Equations with Parameters). Let W be another Banach space, U × V ⊂_{o} X × W and Z ∈ C^{1}(U × V, X). For each (x, w) ∈ U × V, let t ∈ J_{x,w} → ϕ(t, x, w) denote the maximal solution to the ODE

ẏ(t) = Z(y(t), w) with y(0) = x  (16.49)

and

D := {(t, x, w) ∈ ℝ × U × V : t ∈ J_{x,w}}

as in Exercise 15.8.

1. Prove that ϕ is C^{1} and that D_{w}ϕ(t, x, w) solves the differential equation:

\frac{d}{dt}D_{w}ϕ(t, x, w) = (D_{x}Z)(ϕ(t, x, w), w)D_{w}ϕ(t, x, w)+(D_{w}Z)(ϕ(t, x, w), w)

with D_{w}ϕ(0, x, w) = 0 ∈ L(W, X). Hint: See the hint for Exercise 15.8 with the reference to Theorem 15.15 being replace by Theorem 16.28.

2. Also show with the aid of Duhamel’s principle (Exercise 8.23) and Theorem 16.28 that

D_{w}ϕ(t, x, w) = D_{x}ϕ(t, x, w) ∫_{0}^{t} D_{x}ϕ(τ, x, w)^{−1}(D_{w}Z)(ϕ(τ, x, w), w)dτ

Exercise 16.5. (Differential of e^{A}) Let f : L(X) → GL(X) be the exponential function f(A) = e^{A}. Prove that f is differentiable and that

Df(A)B = ∫_{0}^{1} e^{(1−t)A}Be^{tA} dt.  (16.50)

Hint: Let B ∈ L(X) and define w(t, s) = e^{t(A+sB)} for all t, s ∈ ℝ. Notice that

dw(t, s)/dt = (A + sB)w(t, s) with w(0, s) = I ∈ L(X).  (16.51)

Use Exercise 16.4 to conclude that w is C^{1} and that w′(t, 0) := dw(t, s)/ds|_{s=0} satisfies the differential equation,

\frac{d}{dt}w′(t, 0) = Aw′(t, 0) + Be^{tA} with w(0, 0) = 0 ∈ L(X).  (16.52)

Solve this equation by Duhamel’s principle (Exercise 8.23) and then apply Proposition 16.14 to conclude that f is differentiable with differential given by Eq. (16.50).

Exercise 16.6 (Local ODE Existence). Let S_{x} be defined as in Eq. (15.15) from the proof of Theorem 15.4. Verify that S_{x} satisfies the hypothesis of Corollary 16.33. In particular we could have used Corollary 16.33 to prove Theorem 15.4.

Exercise 16.7 (Local ODE Existence Again). Let J = (−1, 1) , Z ∈ C^{1}(X, X), Y := BC(J, X) and for y ∈ Y and s ∈ J let y_{s} ∈ Y be defined by y_{s}(t) := y(st). Use the following outline to prove the ODE

ẏ(t) = Z(y(t)) with y(0) = x  (16.53)

has a unique solution for small t and this solution is C^{1} in x.

1. If y solves Eq. (16.53) then y_{s} solves

ẏ_{s}(t) = sZ(y_{s}(t)) with y_{s}(0) = x

or equivalently

y_{s}(t) = x + s ∫_{0}^{t} Z(y_{s}(τ))dτ.  (16.54)

Notice that when s = 0, the unique solution to this equation is y_{0}(t) = x.

2. Let F : J × Y → J × Y be defined by

F (s, y) := (s, y(t) − s ∫_{0}^{t} Z(y(τ))dτ).

Show the differential of F is given by

F′(s, y)(a, v) = (a, t → v(t) − s ∫_{0}^{t} Z′(y(τ))v(τ)dτ − a ∫_{0}^{·} Z(y(τ))dτ) .

3. Verify F′(0, y) : ℝ × Y → ℝ × Y is invertible for all y ∈ Y and notice that F (0, y) = (0, y).

4. For x ∈ X, let C_{x} ∈ Y be the constant path at x, i.e. C_{x}(t) = x for all t ∈ J. Use the inverse function Theorem 16.25 to conclude there exists ε > 0 and a C^{1} map ϕ : (−ε, ε) × B(x_{0}, ε) → Y such that

F (s, ϕ(s, x)) = (s, C_{x}) for all (s, x) ∈ (−ε, ε) × B(x_{0}, ε).

5. Show, for s ≤ ε that y_{s}(t) := ϕ(s, x)(t) satisfies Eq. (16.54). Now define y(t, x) = ϕ(ε/2, x)(2t/ε) and show y(t, x) solve Eq. (16.53) for |t| < ε/2 and x ∈ B(x_{0}, ε).

Exercise 16.8. Show P defined in Theorem 16.31 is continuously differentiable and P′(y)h = ḣ + p′(y)h.

Exercise 16.9. Embedded sub-manifold problems.

Exercise 16.10. Lagrange Multiplier problems.

16.9.1 Alternate construction of g. To be made into an exercise.

Suppose U ⊂_{o} X and f : U → Y is a C^{2} – function. Then we are looking for a function g(y) such that f(g(y)) = y. Fix an x_{0} ∈ U and y_{0} = f(x_{0}) ∈ Y. Suppose such a g exists and let x(t) = g(y_{0} + th) for some h ∈ Y. Then differentiating f(x(t)) = y_{0} + th implies

\frac{d}{dt}f(x(t)) = f′(x(t))ẋ(t) = h

or equivalently that

ẋ(t) = [f′(x(t))]^{−1} h = Z(h, x(t)) with x(0) = x_{0}  (16.55)

where Z(h, x) = [f′(x(t))]^{−1} h. Conversely if x solves Eq. (16.55) we have \frac{d}{dt}f(x(t)) = h and hence that

f(x(1)) = y_{0} + h.

Thus if we define

g(y_{0} + h) := e^{Z(h,·)}(x_{0}),

then f(g(y_{0} + h)) = y_{0} + h for all h sufficiently small. This shows f is an open mapping.

Part V

Lebesgue Integration Theory Introduction: What are measures and why “measurable” sets

Definition 17.1 (Preliminary). A measure μ “on” a set X is a function μ : 2^{X} → [0, ∞] such that

1. μ(∅) = 0

2. If {A_{i}}_{i=1}^{N} is a finite (N < ∞) or countable (N = ∞) collection of subsets of X which are pair-wise disjoint (i.e. A_{i} ∩ A_{j} = ∅ if i ≠ j) then

μ(∪_{i=1}^{N}A_{i}) = ∑_{i=1}^{N} μ(A_{i}).

Example 17.2. Suppose that X is any set and x ∈ X is a point. For A ⊂ X, let

δ_{x}(A) = \{ \begin{matrix}1 & if & x ∈ A \\ 0 & if & x ∉ A. \end{matrix}

Then μ = δ_{x} is a measure on X called the Dirac delta measure at x.

Example 17.3. Suppose that μ is a measure on X and λ > 0, then λ · μ is also a measure on X. Moreover, if {μ_{α}}_{α∈J} are all measures on X, then μ = ∑_{α∈J} μ_{α}, i.e.

μ(A) = ∑_{α∈J} μ_{α}(A) for all A ⊂ X

is a measure on X. (See Section 2 for the meaning of this sum.) To prove this we must show that μ is countably additive. Suppose that {A_{i}}_{i=1}^{∞} is a collection of pair-wise disjoint subsets of X, then

μ(∪_{i=1}^{∞}A_{i}) = ∑_{i=1}^{∞} μ(A_{i}) = ∑_{i=1}^{∞} ∑_{α∈J} μ_{α}(A_{i})

= ∑_{α∈J} ∑_{i=1}^{∞} μ_{α}(A_{i}) = ∑_{α∈J} μ_{α}(∪_{i=1}^{∞}A_{i})

= μ(∪_{i=1}^{∞}A_{i}) wherein the third equality we used Theorem 4.22 and in the fourth we used that fact that μ_{α} is a measure.

Example 17.4. Suppose that X is a set λ : X → [0, ∞] is a function. Then

μ := ∑_{x∈X} λ(x)δ_{x}

is a measure, explicitly

μ(A) = ∑_{x∈A} λ(x)

for all A ⊂ X.

17.1 The problem with Lebesgue “measure”

So far all of the examples of measures given above are “counting” type measures, i.e. a weighted count of the number of points in a set. We certainly are going to want other types of measures too. In particular, it will be of great interest to have a measure on ℝ (called Lebesgue measure) which measures the “length” of a subset of ℝ. Unfortunately as the next theorem shows, there is no such reasonable measure of length if we insist on measuring all subsets of ℝ.

Theorem 17.5. There is no measure μ : 2^{ℝ}→[0, ∞] such that

1. μ([a, b)) = (b − a) for all a < b and

2. is translation invariant, i.e. μ(A + x) = μ(A) for all x ∈ ℝ and A ∈ 2^{ℝ},

where

A + x := {y + x : y ∈ A} ⊂ ℝ.

In fact the theorem is still true even if (1) is replaced by the weaker condition that 0 < μ((0, 1]) < ∞.

The counting measure μ (A) = # (A) is translation invariant. However μ((0, 1]) = ∞ in this case and so μ does not satisfy condition 1.

Proof. First proof. Let us identify [0, 1) with the unit circle S^{1} := {z ∈ ℂ : |z| = 1} by the map

ϕ(t) = e^{i2πt} = (cos 2πt + i sin 2πt) ∈ S^{1}

for t ∈ [0, 1). Using this identification we may use μ to define a function ν on 2^{S^{1}} by ν(ϕ(A)) = μ(A) for all A ⊂ [0, 1). This new function is a measure on S^{1} with the property that 0 < ν((0, 1]) < ∞. For z ∈ S^{1} and N ⊂ S^{1} let

zN := {zn ∈ S^{1} : n ∈ N},  (17.1)

that is to say e^{iθ}N is N rotated counter clockwise by angle θ. We now claim that ν is invariant under these rotations, i.e.

ν(zN) = ν(N)  (17.2)

for all z ∈ S^{1} and N ⊂ S^{1}. To verify this, write N = ϕ(A) and z = ϕ(t) for some t ∈ [0, 1) and A ⊂ [0, 1). Then

ϕ(t)ϕ(A) = ϕ(t + A mod 1)

where for A ⊂ [0, 1) and α ∈ [0, 1),

t + A mod 1 := {a + t mod 1 ∈ [0, 1) : a ∈ N}
= (a + A ∩ {a < 1 − t}) ∪ ((t − 1) + A ∩ {a ≥ 1 − t}) .

Thus

ν(ϕ(t)ϕ(A)) = μ(t + A mod 1)
= μ ((a + A ∩ {a < 1 − t}) ∪ ((t − 1) + A ∩ {a ≥ 1 − t}))
= μ ((a + A ∩ {a < 1 − t})) + μ (((t − 1) + A ∩ {a ≥ 1 − t}))
= μ (A ∩ {a < 1 − t}) + μ (A ∩ {a ≥ 1 − t})
= μ ((A ∩ {a < 1 − t}) ∪ (A ∩ {a ≥ 1 − t}))
= μ(A) = ν(ϕ(A)).

Therefore it suffices to prove that no finite non-trivial measure ν on S^{1} such that Eq. (17.2) holds. To do this we will “construct” a non-measurable set N = ϕ(A) for some A ⊂ [0, 1). Let

R := {z = e^{i2πt} : t ∈ ℚ} = {z = e^{i2πt} : t ∈ [0, 1) ∩ ℚ}

– a countable subgroup of S^{1}. As above R acts on S^{1} by rotations and divides S^{1} up into equivalence classes, where z, w ∈ S^{1} are equivalent if z = rw for some r ∈ R. Choose (using the axiom of choice) one representative point n from each of these equivalence classes and let N ⊂ S^{1} be the set of these representative points. Then every point z ∈ S^{1} may be uniquely written as z = nr with n ∈ N and r ∈ R. That is to say

S^{1} = ∐_{r∈R} (rN)  (17.3)

where ∐_{α} A_{α} is used to denote the union of pair-wise disjoint sets {A_{α}} . By Eqs. (17.2) and (17.3),

ν(S^{1}) = ∑_{r∈R} ν(rN) = ∑_{r∈R} ν(N).

The right member from this equation is either 0 or ∞, 0 if ν(N) = 0 and ∞ if ν(N) > 0. In either case it is not equal ν(S^{1}) ∈ (0, 1). Thus we have reached the desired contradiction.

Proof. Second proof of Theorem 17.5. For N ⊂ [0, 1) and α ∈ [0, 1), let

N^{α} = N + α mod 1
= {a + α mod 1 ∈ [0, 1) : a ∈ N}
= (α + N ∩ {a < 1 − α}) ∪ ((α − 1) + N ∩ {a ≥ 1 − α}) .

Then

μ (N^{α}) = μ (α + N ∩ {a < 1 − α}) + μ ((α − 1) + N ∩ {a ≥ 1 − α})
= μ (N ∩ {a < 1 − α}) + μ (N ∩ {a ≥ 1 − α})
= μ (N ∩ {a < 1 − α} ∪ (N ∩ {a ≥ 1 − α}))
= μ(N).  (17.4)

We will now construct a bad set N which coupled with Eq. (17.4) will lead to a contradiction. Set

Q_{x} := {x + r ∈ ℝ : r∈ ℚ} =x + ℚ.

Notice that Q_{x} ∩ Q_{y} ≠ ∅ implies that Q_{x} = Q_{y}. Let O = {Q_{x} : x ∈ ℝ} – the orbit space of the ℚ action. For all A ∈ O choose f(A) ∈ [0, 1/3) ∩ A^{1} and define N = f(O). Then observe:

1. f(A) = f(B) implies that A ∩ B ≠ ∅ which implies that A = B so that f is injective.

2. O = {Q_{n} : n ∈ N}.

Let R be the countable set,

R := ℚ ∩ [0, 1).

We now claim that

N^{r} ∩ N^{s} = ∅ if r ≠ s and  (17.5)
[0, 1) = ∪_{r∈R}N^{r}.  (17.6)

Indeed, if x ∈ N^{r} ∩ N^{s} ≠ ∅ then x = r + n mod 1 and x = s + n′ mod 1, then n−n′ ∈ ℚ, i.e. Q_{n} = Q_{n′}. That is to say, n = f(Q_{n}) = f(Q_{n′}) = n′ and hence that s = r mod 1, but s, r ∈ [0, 1) implies that s = r. Furthermore, if x ∈ [0, 1) and n := f(Q_{x}), then x − n = r ∈ ℚ and x ∈ N^{r mod 1}. Now that we have constructed N, we are ready for the contradiction. By Equations (17.4–17.6) we find

1 We have used the Axiom of choice here, i.e. ∏_{A∈F}(A ∩ [0, 1/3]) ≠ ∅

1 = μ([0, 1)) = ∑_{r∈R} μ(N^{r}) = ∑_{r∈R} μ(N)

= \{ \begin{matrix}∞ & if & μ(N) > 0 \\ 0 & if & μ(N) = 0 \end{matrix} .

which is certainly inconsistent. Incidentally we have just produced an example of so called “non – measurable” set.

Because of Theorem 17.5, it is necessary to modify Definition 17.1. Theorem 17.5 points out that we will have to give up the idea of trying to measure all subsets of ℝ but only measure some sub-collections of “measurable” sets. This leads us to the notion of σ – algebra discussed in the next chapter. Our revised notion of a measure will appear in Definition 19.1 of Chapter 19 below.

Measurability

18.1 Algebras and σ – Algebras

Definition 18.1. A collection of subsets A of a set X is an algebra if

1. ∅, X ∈ A

2. A ∈ A implies that A^{c} ∈ A

3. A is closed under finite unions, i.e. if A_{1}, …, A_{n} ∈ A then A_{1} ∪⋯∪A_{n} ∈ A. In view of conditions 1. and 2., 3. is equivalent to

3′. A is closed under finite intersections.

Definition 18.2. A collection of subsets M of X is a σ – algebra (or sometimes called a σ – field) if M is an algebra which also closed under countable unions, i.e. if {A_{i}}_{i=1}^{∞} ⊂ M, then ∪_{i=1}^{∞}A_{i} ∈ M. (Notice that since M is also closed under taking complements, M is also closed under taking countable intersections.) A pair (X, M), where X is a set and M is a σ – algebra on X, is called a measurable space.

The reader should compare these definitions with that of a topology in Definition 10.1. Recall that the elements of a topology are called open sets. Analogously, elements of and algebra A or a σ – algebra M will be called measurable sets.

Example 18.3. Here are some examples of algebras.

1. M = 2^{X}, then M is a topology, an algebra and a σ – algebra.

2. Let X = {1, 2, 3}, then τ = {∅, X, {2, 3}} is a topology on X which is not an algebra.

3. τ = A = {{1}, {2, 3}, ∅, X} is a topology, an algebra, and a σ – algebra on X. The sets X, {1}, {2, 3}, ∅ are open and closed. The sets {1, 2} and {1, 3} are neither open nor closed and are not measurable.

The reader should compare this example with Example 10.3.

Proposition 18.4. Let E be any collection of subsets of X. Then there exists a unique smallest algebra A(E) and σ – algebra σ(E) which contains E.

Proof. The proof is the same as the analogous Proposition 10.6 for topologies, i.e.

A(E) := ⋂{A : A is an algebra such that E ⊂ A}

and

σ(E) := ⋂{M : M is a σ – algebra such that E ⊂ M}.

Example 18.5. Suppose X = {1, 2, 3} and E = {∅, X, {1, 2}, {1, 3}}, see Figure 18.1.

Fig. 18.1. A collection of subsets.

Then

τ(E) = {∅, X, {1}, {1, 2}, {1, 3}}
A(E) = σ(E) = 2^{X}.

The next proposition is the analogue to Proposition 10.7 for topologies and enables us to give and explicit descriptions of A(E). On the other hand it should be noted that σ(E) typically does not admit a simple concrete description.

Proposition 18.6. Let X be a set and E ⊂ 2^{X}. Let E^{c} := {A^{c} : A ∈ E} and E_{c} := E ∪ {X, ∅} ∪ E^{c} Then

A(E) := {finite unions of finite intersections of elements from E_{c}}. (18.1)

18.1 Algebras and σ – Algebras 281

Proof. Let A denote the right member of Eq. (18.1). From the definition of an algebra, it is clear that E ⊂ A ⊂ A(E). Hence to finish that proof it suffices to show A is an algebra. The proof of these assertions are routine except for possibly showing that A is closed under complementation. To check A is closed under complementation, let Z ∈ A be expressed as

Z = ⋃_{i=1}^{N} ⋂_{j=1}^{K} A_{ij}

where A_{ij} ∈ E_{c}. Therefore, writing B_{ij} = A_{ij}^{c} ∈ E_{c}, we find that

Z^{c} = ⋂_{i=1}^{N} ⋃_{j=1}^{K} B_{ij} = ⋃_{j_{1},...,j_{N} =1}^{K} (B_{1j_{1}} ∩ B_{2j_{2}} ∩ ⋯ ∩ B_{Nj_{N}}) ∈ A

wherein we have used the fact that B_{1j_{1}} ∩ B_{2j_{2}} ∩ ⋯ ∩ B_{Nj_{N}} is a finite intersection of sets from E_{c}.

Remark 18.7. One might think that in general σ(E) may be described as the countable unions of countable intersections of sets in E^{c}. However this is in general false, since if

Z = ⋃_{i=1}^{∞} ⋂_{j=1}^{∞} A_{ij}

with A_{ij} ∈ E_{c}, then

Z^{c} = ⋃_{j_{1}=1,j_{2}=1,...j_{N} =1,...}^{∞} (⋂_{ℓ=1}^{∞} A_{ℓ,j_{ℓ}}^{c})

which is now an uncountable union. Thus the above description is not correct. In general it is complicated to explicitly describe σ(E), see Proposition 1.23 on page 39 of Folland for details. Also see Proposition 18.13 below.

Exercise 18.1. Let τ be a topology on a set X and A = A(τ) be the algebra generated by τ. Show A is the collection of subsets of X which may be written as finite union of sets of the form F ∩ V where F is closed and V is open.

The following notion will be useful in the sequel and plays an analogous role for algebras as a base (Definition 10.8) does for a topology.

Definition 18.8. A set E ⊂ 2^{X} is said to be an elementary family or elementary class provided that

• ∅ ∈ E

• E is closed under finite intersections

• if E ∈ E, then E^{c} is a finite disjoint union of sets from E. (In particular X = ∅^{c} is a finite disjoint union of elements from E.)

Example 18.9. Let X = ℝ, then

E := {(a, b] ∩ ℝ : a, b ∈ ℝ̄}
= {(a, b] : a ∈ [−∞, ∞) and a < b < ∞} ∪ {∅, ℝ}

is an elementary family.

Exercise 18.2. Let A ⊂ 2^{X} and B ⊂ 2^{Y} be elementary families. Show the collection

E = A × B = {A × B : A ∈ A and B ∈ B}

is also an elementary family.

Proposition 18.10. Suppose E ⊂ 2^{X} is an elementary family, then A = A(E) consists of sets which may be written as finite disjoint unions of sets from E.

Proof. This could be proved making use of Proposition 18.6. However it is easier to give a direct proof. Let A denote the collection of sets which may be written as finite disjoint unions of sets from E. Clearly E ⊂ A ⊂ A(E) so it suffices to show A is an algebra since A(E) is the smallest algebra containing E. By the properties of E, we know that ∅, X ∈ A. Now suppose that A_{i} = ∐_{F ∈Λ_{i}} F ∈ A where, for i = 1, 2, …, n, Λ_{i} is a finite collection of disjoint sets from E. Then

⋂_{i=1}^{n} A_{i} = ⋂_{i=1}^{n} ( ∐_{F ∈Λ_{i}} F ) = ⋃_{(F_{1},,...,F_{n})∈Λ_{1}×···×Λ_{n}}(F_{1} ∩ F_{2} ∩ ⋯ ∩ F_{n})

and this is a disjoint (you check) union of elements from E. Therefore A is closed under finite intersections. Similarly, if A = ∐_{F ∈Λ} F with Λ being a finite collection of disjoint sets from E, then A^{c} = ⋂_{F ∈Λ} F^{c}. Since by assumption F^{c} ∈ A for F ∈ Λ ⊂ E and A is closed under finite intersections, it follows that A^{c} ∈ A.

Definition 18.11. Let X be a set. We say that a family of sets F ⊂ 2^{X} is a partition of X if distinct members of F are disjoint and if X is the union of the sets in F.

Example 18.12. Let X be a set and E = {A_{1}, …, A_{n}} where A_{1}, …, A_{n} is a partition of X. In this case

A(E) = σ(E) = τ(E) = {∪_{i∈Λ}A_{i} : Λ ⊂ {1, 2, …, n}}

where ∪_{i∈Λ}A_{i} := ∅ when Λ = ∅. Notice that

# (A(E)) = #(2^{{1,2,...,n}}) = 2^{n}.

Proposition 18.13. Suppose that M ⊂ 2^{X} is a σ – algebra and M is at most a countable set. Then there exists a unique finite partition F of X such that F ⊂ M and every element B ∈ M is of the form

B = ∪ {A ∈ F : A ⊂ B} .  (18.2)

In particular M is actually a finite set and # (M) = 2^{n} for some n ∈ ℕ.

Proof. For each x ∈ X let

A_{x} = ∩ {A ∈ M : x ∈ A} ∈ M,

wherein we have used M is a countable σ – algebra to insure A_{x} ∈ M. Hence A_{x} is the smallest set in M which contains x. Let C = A_{x} ∩ A_{y}. If x ∉ C then A_{x} \C ⊂ A_{x} is an element of M which contains x and since A_{x} is the smallest member of M containing x, we must have that C = ∅. Similarly if y ∉ C then C = ∅. Therefore if C ≠ ∅, then x, y ∈ A_{x} ∩ A_{y} ∈ M and A_{x} ∩ A_{y} ⊂ A_{x} and A_{x} ∩A_{y} ⊂ A_{y} from which it follows that A_{x} = A_{x} ∩A_{y} = A_{y}. This shows that F = {A_{x} : x ∈ X} ⊂ M is a (necessarily countable) partition of X for which Eq. (18.2) holds for all B ∈ M. Enumerate the elements of F as F = {P_{n}}_{n=1}^{N} where N ∈ ℕ or N = ∞. If N = ∞, then the correspondence

a ∈ {0, 1}^{ℕ} → A_{a} = ∪{P_{n} : a_{n} = 1} ∈ M

is bijective and therefore, by Lemma 2.6, M is uncountable. Thus any countable σ – algebra is necessarily finite. This finishes the proof modulo the uniqueness assertion which is left as an exercise to the reader.

Example 18.14. Let X = ℝ and

E = {(a, ∞) : a ∈ ℝ} ∪ {ℝ, ∅} = {(a, ∞) ∩ ℝ : a ∈ ℝ̄} ⊂ 2^{ℝ}.

Notice that E_{f} = E and that E is closed under unions, which shows that τ(E) = E, i.e. E is already a topology. Since (a, ∞)^{c} = (−∞, a] we find that E_{c} = {(a, ∞), (−∞, a], −∞ ≤ a < ∞} ∪ {ℝ, ∅}. Noting that

(a, ∞) ∩ (−∞, b] = (a, b]

it follows that A(E) = A(Ẽ) where

Ẽ:= {(a, b] ∩ ℝ : a, b ∈ ℝ̄} .

Since Ẽ is an elementary family of subsets of ℝ, Proposition 18.10 implies A(E) may be described as being those sets which are finite disjoint unions of sets from Ẽ. The σ – algebra, σ(E), generated by E is very complicated. Here are some sets in σ(E) – most of which are not in A(E).

(a) (a, b) = ⋃_{n=1}^{∞}(a, b − \frac{1}{n}] ∈ σ(E).

(b) All of the standard open subsets of ℝ are in σ(E).

(c) \{x\} = ⋂_{n} (x − \frac{1}{n}, x] ∈ σ(E)

(d) [a, b] = {a} ∪ (a, b] ∈ σ(E)

(e) Any countable subset of ℝ is in σ(E).

Remark 18.15. In the above example, one may replace E by E = {(a, ∞) : a ∈ ℚ} ∪ {ℝ, ∅}, in which case A(E) may be described as being those sets which are finite disjoint unions of sets from the following list

{(a, ∞), (−∞, a], (a, b] : a, b ∈ ℚ} ∪ {∅, ℝ} .

This shows that A(E) is a countable set – a useful fact which will be needed later.

Notation 18.16 For a general topological space (X, τ), the Borel σ – algebra is the σ – algebra B_{X} := σ(τ) on X. In particular if X = ℝ^{n}, B_{ℝ^{n}} will be used to denote the Borel σ – algebra on ℝ^{n} when ℝ^{n} is equipped with its standard Euclidean topology.

Exercise 18.3. Verify the σ – algebra, B_{ℝ}, is generated by any of the following collection of sets:

1. {(a, ∞) : a ∈ ℝ} , 2. {(a, ∞) : a ∈ ℚ} or 3. {[a, ∞) : a ∈ ℚ} .

Proposition 18.17. If τ is a second countable topology on X and E is a countable collection of subsets of X such that τ = τ(E), then B_{X} := σ(τ) = σ(E), i.e. σ(τ(E)) = σ(E).

Proof. Let E_{f} denote the collection of subsets of X which are finite intersection of elements from E along with X and ∅. Notice that E_{f} is still countable (you prove). A set Z is in τ(E) iff Z is an arbitrary union of sets from E_{f}. Therefore Z = ⋃_{A∈F} A for some subset F ⊂ E_{f} which is necessarily count-

able. Since E_{f} ⊂ σ(E) and σ(E) is closed under countable unions it follows that Z ∈ σ(E) and hence that τ(E) ⊂ σ(E). Lastly, since E ⊂ τ(E) ⊂ σ(E), σ(E) ⊂ σ(τ(E)) ⊂ σ(E).

18.2 Measurable Functions

Our notion of a “measurable” function will be analogous to that for a continuous function. For motivational purposes, suppose (X, M, μ) is a measure space and f : X → ℝ_{+}. Roughly speaking, in the next Chapter we are going to define ∫_{X} fdμ as a certain limit of sums of the form,

∑_{0<a_{1}<a_{2}<a_{3}<...}^{∞} a_{i}μ(f^{−1}(a_{i}, a_{i+1}]).

For this to make sense we will need to require f^{−1}((a, b]) ∈ M for all a < b. Because of Lemma 18.22 below, this last condition is equivalent to the condition f^{−1}(B_{ℝ}) ⊂ M.

Definition 18.18. Let (X, M) and (Y, F) be measurable spaces. A function f : X → Y is measurable if f^{−1}(F) ⊂ M. We will also say that f is M/F – measurable or (M, F) – measurable.

Example 18.19 (Characteristic Functions). Let (X, M) be a measurable space and A ⊂ X. We define the characteristic function 1_{A} : X → ℝ by

1_{A}(x) = \{ \begin{matrix}1 & if & x ∈ A \\ 0 & if & x ∉ A. \end{matrix}

If A ∈ M, then 1_{A} is (M, 2^{ℝ}) – measurable because 1_{A}^{−1}(W) is either ∅, X, A or A^{c} for any W ⊂ ℝ. Conversely, if F is any σ – algebra on ℝ containing a set W ⊂ ℝ such that 1 ∈ W and 0 ∈ W^{c}, then A ∈ M if 1_{A} is (M, F) – measurable. This is because A = 1_{A}^{−1}(W) ∈ M.

Exercise 18.4. Suppose f : X → Y is a function, F ⊂ 2^{Y} and M ⊂ 2^{X}. Show f^{−1}F and f_{∗}M (see Notation 2.7) are algebras (σ – algebras) provided F and M are algebras (σ – algebras).

Remark 18.20. Let f : X → Y be a function. Given a σ – algebra F ⊂ 2^{Y} , the σ – algebra M := f^{−1}(F) is the smallest σ – algebra on X such that f is (M, F) - measurable . Similarly, if M is a σ - algebra on X then F = f_{∗}M is the largest σ – algebra on Y such that f is (M, F) - measurable .

Recall from Definition 2.8 that for E ⊂ 2^{X} and A ⊂ X that

E_{A} = i_{A}^{−1}(E) = {A ∩ E : E ∈ E}

where i_{A} : A → X is the inclusion map. Because of Exercise 10.3, when E = M is an algebra (σ – algebra), M_{A} is an algebra (σ – algebra) on A and we call M_{A} the relative or induced algebra (σ – algebra) on A.

The next two Lemmas are direct analogues of their topological counter parts in Lemmas 10.13 and 10.14. For completeness, the proofs will be given even though they are same as those for Lemmas 10.13 and 10.14.

Lemma 18.21. Suppose that (X, M), (Y, F) and (Z, G) are measurable spaces. If f : (X, M) → (Y, F) and g : (Y, F) → (Z, G) are measurable functions then g ∘ f : (X, M) → (Z, G) is measurable as well.

Proof. By assumption g^{−1}(G) ⊂ F and f^{−1} (F) ⊂ M so that

(g ∘ f)^{−1} (G) = f^{−1} (g^{−1} (G)) ⊂ f^{−1} (F) ⊂ M.

Lemma 18.22. Suppose that f : X → Y is a function and E ⊂ 2^{Y} and A ⊂ Y then

σ (f^{−1}(E)) = f^{−1}(σ(E)) and  (18.3)
(σ(E))_{A} = σ(E_{A} ).  (18.4)

(Similar assertion hold with σ (·) being replaced by A (·) .) Moreover, if F = σ(E) and M is a σ – algebra on X, then f is (M, F) – measurable iff f^{−1}(E) ⊂ M.

Proof. By Exercise 18.4, f^{−1}(σ(E)) is a σ – algebra and since E ⊂ F, f^{−1}(E) ⊂ f^{−1}(σ(E)). It now follows that σ (f^{−1}(E)) ⊂ f^{−1}(σ (E)). For the reverse inclusion, notice that

f_{∗}σ (f^{−1}(E)) = {B ⊂ Y : f^{−1}(B) ∈ σ (f^{−1}(E))}

is a σ – algebra which contains E and thus σ(E) ⊂ f_{∗}σ (f^{−1}(E)) . Hence if B ∈ σ(E) we know that f^{−1}(B) ∈ σ (f^{−1}(E)) , i.e. f^{−1}(σ(E)) ⊂ σ (f^{−1}(E)) and Eq. (18.3) has been proved. Applying Eq. (18.3) with X = A and f = i_{A} being the inclusion map implies

(σ(E))_{A} = i_{A}^{−1}(σ(E)) = σ(i_{A}^{−1}(E)) = σ(E_{A}).

Lastly if f^{−1}E ⊂ M, then f^{−1}σ (E) = σ (f^{−1}E) ⊂ M which shows f is (M, F) – measurable.

Corollary 18.23. Suppose that (X, M) is a measurable space. Then the following conditions on a function f : X → ℝ are equivalent:

1. f is (M, B_{ℝ}) – measurable,

2. f^{−1}((a, ∞)) ∈ M for all a ∈ ℝ,

3. f^{−1}((a, ∞)) ∈ M for all a ∈ ℚ,

4. f^{−1}((−∞, a]) ∈ M for all a ∈ ℝ.

Proof. An exercise in using Lemma 18.22 and is the content of Exercise 18.8.

Here is yet another way to generate σ – algebras. (Compare with the analogous topological Definition 10.20.)

Definition 18.24 (σ – Algebras Generated by Functions). Let X be a set and suppose there is a collection of measurable spaces {(Y_{α}, F_{α}) : α ∈ A} and functions f_{α} : X → Y_{α} for all α ∈ A. Let σ(f_{α} : α ∈ A) denote the smallest σ – algebra on X such that each f_{α} is measurable, i.e.

σ(f_{α} : α ∈ A) = σ(∪_{α}f_{α}^{−1}(F_{α})).

Proposition 18.25. Assuming the notation in Definition 18.24 and additionally let (Z, M) be a measurable space and g : Z → X be a function. Then g is (M, σ(f_{α} : α ∈ A)) – measurable iff f_{α} ∘ g is (M, F_{α})–measurable for all α ∈ A.

Proof. This proof is essentially the same as the proof of the topological analogue in Proposition 10.21. (⇒) If g is (M, σ(f_{α} : α ∈ A)) – measurable, then the composition f_{α} ∘ g is (M, F_{α}) – measurable by Lemma 18.21. (⇐) Let

G = σ(f_{α} : α ∈ A) = σ (∪_{α∈A}f_{α}^{−1}(F_{α})) .

If f_{α} ∘ g is (M, F_{α}) – measurable for all α, then

g^{−1}f_{α}^{−1}(F_{α}) ⊂ M ∀ α ∈ A

and therefore

g^{−1} (∪_{α∈A}f_{α}^{−1}(F_{α})) = ∪_{α∈A}g^{−1}f_{α}^{−1}(F_{α}) ⊂ M.

Hence

g^{−1} (G) = g^{−1} (σ (∪_{α∈A}f_{α}^{−1}(F_{α}))) = σ(g^{−1} (∪_{α∈A}f_{α}^{−1}(F_{α})) ⊂ M

which shows that g is (M, G) – measurable.

Definition 18.26. A function f : X → Y between two topological spaces is Borel measurable if f^{−1}(B_{Y}) ⊂ B_{X}.

Proposition 18.27. Let X and Y be two topological spaces and f : X → Y be a continuous function. Then f is Borel measurable.

Proof. Using Lemma 18.22 and B_{Y} = σ(τ_{Y}),

f^{−1}(B_{Y}) = f^{−1}(σ(τ_{Y})) = σ(f^{−1}(τ_{Y})) ⊂ σ(τ_{X}) = B_{X}.

Definition 18.28. Given measurable spaces (X, M) and (Y, F) and a subset A ⊂ X. We say a function f : A → Y is measurable iff f is M_{A}/F – measurable.

Proposition 18.29 (Localizing Measurability). Let (X, M) and (Y, F) be measurable spaces and f : X → Y be a function.

1. If f is measurable and A ⊂ X then f|_{A} : A → Y is measurable.

2. Suppose there exist A_{n} ∈ M such that X = ∪_{n=1}^{∞}A_{n} and f|A_{n} is M_{A_{n}} measurable for all n, then f is M – measurable.

Proof. As the reader will notice, the proof given below is essentially identical to the proof of Proposition 10.19 which is the topological analogue of this proposition. 1. If f : X → Y is measurable, f^{−1}(B) ∈ M for all B ∈ F and therefore

f|_{A}^{−1} (B) = A ∩ f^{−1}(B) ∈ M_{A} for all B ∈ F.

2. If B ∈ F, then

f^{−1}(B) = ∪_{n=1}^{∞} (f^{−1}(B) ∩ A_{n}) = ∪_{n=1}^{∞}f|_{A_{n}}^{−1}(B).

Since each A_{n} ∈ M, M_{A_{n}} ⊂ M and so the previous displayed equation shows f^{−1}(B) ∈ M.

Proposition 18.30. If (X, M) is a measurable space, then

f = (f_{1}, f_{2}, …, f_{n}) : X → ℝ^{n}

is (M, B_{ℝ^{n}}) – measurable iff f_{i} : X → ℝ is (M, B_{ℝ}) – measurable for each i. In particular, a function f : X → ℂ is (M, B_{ℂ}) – measurable iff Re f and Im f are (M, B_{ℝ}) – measurable.

Proof. This is formally a consequence of Corollary 18.65 and Proposition 18.60 below. Nevertheless it is instructive to give a direct proof now. Let τ = τ_{ℝ^{n}} denote the usual topology on ℝ^{n} and π_{i} : ℝ^{n} → ℝ be projection onto the i^{th} – factor. Since π_{i} is continuous, π_{i} is B_{ℝ^{n}}/B_{ℝ} – measurable and therefore if f : X → ℝ^{n} is measurable then so is f_{i} = π_{i} ∘ f. Now suppose f_{i} : X → ℝ is measurable for all i = 1, 2, …, n. Let

E := {(a, b) : a, b ∈ ℚ^{n} ∋ a < b} ,

where, for a, b ∈ ℝ^{n}, we write a < b iff a_{i} < b_{i} for i = 1, 2, …, n and let

(a, b) = (a_{1}, b_{1}) × ⋯ × (a_{n}, b_{n}) .

Since E ⊂ τ and every element V ∈ τ may be written as a (necessarily) countable union of elements from E, we have σ (E) ⊂ B_{ℝ^{n}} = σ (τ) ⊂ σ (E) , i.e. σ (E) = B_{ℝ^{n}}. (This part of the proof is essentially a direct proof of Corollary 18.65 below.) Because

f^{−1} ((a, b)) = f_{1}^{−1} ((a_{1}, b_{1})) ∩ f_{2}^{−1} ((a_{2}, b_{2})) ∩ ⋯ ∩ f_{n}^{−1} ((a_{n}, b_{n})) ∈ M

for all a, b ∈ ℚ with a < b, it follows that f^{−1}E ⊂ M and therefore

f^{−1}B_{ℝ^{n}} = f^{−1}σ (E) = σ (f^{−1}E) ⊂ M.

Corollary 18.31. Let (X, M) be a measurable space and f, g : X → ℂ be (M, B_{ℂ}) – measurable functions. Then f ± g and f · g are also (M, B_{ℂ}) – measurable.

Proof. Define F : X → ℂ × ℂ, A_{±} : ℂ × ℂ → ℂ and M : ℂ × ℂ ⟶ ℂ by F (x) = (f(x), g(x)), A_{±}(w, z) = w ± z and M(w, z) = wz. Then A_{±} and M are continuous and hence (B_{ℂ^{2}}, B_{ℂ}) – measurable. Also F is (M, B_{ℂ} ⊗ B_{ℂ}) = (M, B_{ℂ^{2}}) – measurable since π_{1} ∘ F = f and π_{2} ∘ F = g are (M, B_{ℂ}) – measurable. Therefore A_{±}∘F = f ±g and M ∘F = f ·g, being the composition of measurable functions, are also measurable.

Lemma 18.32. Let α ∈ ℂ, (X, M) be a measurable space and f : X → ℂ be a (M, B_{ℂ}) – measurable function. Then

F (x) := \{ \begin{matrix}\frac{1}{f(x)} & if f(x) ≠ 0 \\ α & if f(x) = 0 \end{matrix}

is measurable.

Proof. Define i : ℂ → ℂ by

i(z) = \{ \begin{matrix}\frac{1}{z} & if z ≠ 0 \\ 0 & if z = 0. \end{matrix}

For any open set V ⊂ ℂ we have

i^{−1}(V) = i^{−1}(V \ {0}) ∪ i^{−1}(V ∩ {0})

Because i is continuous except at z = 0, i^{−1}(V \ {0}) is an open set and hence in B_{ℂ}. Moreover, i^{−1}(V ∩ {0}) ∈ B_{ℂ} since i^{−1}(V ∩ {0}) is either the empty set or the one point set {0} . Therefore i^{−1}(τ_{ℂ}) ⊂ B_{ℂ} and hence i^{−1}(B_{ℂ}) = i^{−1}(σ(τ_{ℂ})) = σ(i^{−1}(τ_{ℂ})) ⊂ B_{ℂ} which shows that i is Borel measurable. Since F = i ∘ f is the composition of measurable functions, F is also measurable.

We will often deal with functions f : X → ℝ̄ = ℝ∪ {±∞} . When talking about measurability in this context we will refer to the σ – algebra on ℝ̄ defined by

Bℝ̄ := σ ({[a, ∞] : a ∈ ℝ}) .  (18.5)

Proposition 18.33 (The Structure of Bℝ̄). Let B_{ℝ} and Bℝ̄ be as above, then

Bℝ̄ = {A ⊂ ℝ̄ : A ∩ ℝ ∈B_{ℝ}}.  (18.6)

In particular {∞} , {−∞} ∈ Bℝ̄ and B_{ℝ} ⊂ Bℝ̄.

Proof. Let us first observe that

{−∞} = ∩_{n=1}^{∞}[−∞, −n) = ∩_{n=1}^{∞}[−n, ∞]^{c} ∈ Bℝ̄,
{∞} = ∩_{n=1}^{∞}[n, ∞] ∈ Bℝ̄ and ℝ = ℝ̄\ {±∞} ∈ Bℝ̄.

Letting i : ℝ → ℝ̄ be the inclusion map,

i^{−1} (Bℝ̄) = σ (i^{−1} ({[a, ∞] : a ∈ ℝ̄})) = σ ({i^{−1} ([a, ∞]) : a ∈ ℝ̄})
= σ ({[a, ∞] ∩ ℝ : a ∈ ℝ̄}) = σ ({[a, ∞) : a ∈ ℝ}) = B_{ℝ}.

Thus we have shown

B_{ℝ} = i^{−1} (Bℝ̄) = {A ∩ ℝ : A ∈ Bℝ̄}.

This implies:

1. A ∈ Bℝ̄ ⟹ A ∩ ℝ ∈B_{ℝ} and

2. if A ⊂ ℝ̄ is such that A ∩ ℝ ∈B_{ℝ} there exists B ∈ Bℝ̄ such that A ∩ ℝ = B ∩ ℝ. Because AΔB ⊂ {±∞} and {∞} , {−∞} ∈ Bℝ̄ we may conclude that A ∈ Bℝ̄ as well.

This proves Eq. (18.6).

The proofs of the next two corollaries are left to the reader, see Exercises 18.5 and 18.6.

Corollary 18.34. Let (X, M) be a measurable space and f : X → ℝ̄ be a function. Then the following are equivalent

1. f is (M, Bℝ̄) - measurable,

2. f^{−1}((a, ∞]) ∈ M for all a ∈ ℝ,

3. f^{−1}((−∞, a]) ∈ M for all a ∈ ℝ,

4. f^{−1}({−∞}) ∈ M, f^{−1}({∞}) ∈ M and f^{0} : X → ℝ defined by

f^{0} (x) := 1_{ℝ} (f (x)) = \{ \begin{matrix}f (x) & if & f (x) ∈ ℝ \\ 0 & if & f (x) ∈ \{±∞\} \end{matrix}

is measurable.

Corollary 18.35. Let (X, M) be a measurable space, f, g : X → ℝ̄ be functions and define f · g : X → ℝ̄ and (f + g) : X → ℝ̄ using the conventions, 0 · ∞ = 0 and (f + g) (x) = 0 if f (x) = ∞ and g (x) = −∞ or f (x) = −∞ and g (x) = ∞. Then f · g and f + g are measurable functions on X if both f and g are measurable.

Exercise 18.5. Prove Corollary 18.34 noting that the equivalence of items 1.

– 3. is a direct analogue of Corollary 18.23. Use Proposition 18.33 to handle item 4.

Exercise 18.6. Prove Corollary 18.35.

Proposition 18.36 (Closure under sups, infs and limits). Suppose that (X, M) is a measurable space and f_{j} : (X, M) → \overline{ℝ} for j ∈ ℕ is a sequence of M/B\overline{ℝ} – measurable functions. Then

sup_{j}f_{j}, inf_{j}f_{j}, lim sup_{j→∞} f_{j} and lim inf_{j→∞} f_{j}

are all M/B\overline{ℝ} – measurable functions. (Note that this result is in generally false when (X, M) is a topological space and measurable is replaced by continuous in the statement.)

Proof. Define g_{+}(x) := sup_{j}f_{j}(x), then

{x : g_{+}(x) ≤ a} = {x : f_{j}(x) ≤ a ∀ j}
= ∩_{j}{x : f_{j}(x) ≤ a} ∈ M

so that g_{+} is measurable. Similarly if g_{−}(x) = inf_{j} f_{j}(x) then

{x : g_{−}(x) ≥ a} = ∩_{j}{x : f_{j}(x) ≥ a} ∈ M.

Since

lim sup_{j→∞} f_{j} = inf_{n} sup {f_{j} : j ≥ n} and
lim inf_{j→∞} f_{j} = sup_{n} inf {f_{j} : j ≥ n}

we are done by what we have already proved.

Definition 18.37. Given a function f : X → ℝ̄ let f_{+}(x) := max {f(x), 0} and f_{−} (x) := max (−f(x), 0) = − min (f(x), 0) . Notice that f = f_{+} − f_{−}.

Corollary 18.38. Suppose (X, M) is a measurable space and f : X → ℝ̄ is a function. Then f is measurable iff f_{±} are measurable.

Proof. If f is measurable, then Proposition 18.36 implies f_{±} are measurable. Conversely if f_{±} are measurable then so is f = f_{+} − f_{−}.

18.2.1 More general pointwise limits

Lemma 18.39. Suppose that (X, M) is a measurable space, (Y, d) is a metric space and f_{j} : X → Y is (M, B_{Y}) – measurable for all j. Also assume that for each x ∈ X, f(x) = lim_{n→∞} f_{n}(x) exists. Then f : X → Y is also (M, B_{Y}) – measurable.

Proof. Let V ∈ τ_{d} and W_{m} := {y ∈ Y : d_{V^{c}}(y) > 1/m} for m = 1, 2, …. Then W_{m} ∈ τ_{d},

W_{m} ⊂ W̄_{m} ⊂ {y ∈ Y : d_{V^{c}}(y) ≥ 1/m} ⊂ V

for all m and W_{m} ↑ V as m → ∞. The proof will be completed by verifying the identity,

f^{−1}(V) = ∪_{m=1}^{∞} ∪_{N=1}^{∞} ∩_{n≥N}f_{n}^{−1}(W_{m}) ∈ M.

If x ∈ f^{−1}(V) then f(x) ∈ V and hence f(x) ∈ W_{m} for some m. Since f_{n}(x) → f(x), f_{n}(x) ∈ W_{m} for almost all n. That is x ∈ ∪_{m=1}^{∞} ∪_{N=1}^{∞} ∩_{n≥N}f_{n}^{−1}(W_{m}). Conversely when x ∈ ∪_{m=1}^{∞} ∪_{N=1}^{∞} ∩_{n≥N}f_{n}^{−1}(W_{m}) there exists an m such that f_{n}(x) ∈ W_{m} ⊂ W̄_{m} for almost all n. Since f_{n}(x) → f(x) ∈ W̄_{m} ⊂ V, it follows that x ∈ f^{−1}(V).

Remark 18.40. In the previous Lemma 18.39 it is possible to let (Y, τ) be any topological space which has the “regularity” property that if V ∈ τ there exists W_{m} ∈ τ such that W_{m} ⊂ W̄_{m} ⊂ V and V = ∪_{m=1}^{∞}W_{m}. Moreover, some extra condition is necessary on the topology τ in order for Lemma 18.39 to be correct. For example if Y = {1, 2, 3} and τ = {Y, ∅, {1, 2}, {2, 3}, {2}} as in Example 10.36 and X = {a, b} with the trivial σ – algebra. Let f_{j}(a) = f_{j}(b) = 2 for all j, then f_{j} is constant and hence measurable. Let f(a) = 1 and f(b) = 2, then f_{j} → f as j → ∞ with f being non-measurable. Notice that the Borel σ – algebra on Y is 2^{Y} .

18.3 σ – Function Algebras

In this subsection, we are going to relate σ – algebras of subsets of a set X to certain algebras of functions on X. We will begin this endeavor after proving the simple but very useful approximation Theorem 18.42 below.

Definition 18.41. Let (X, M) be a measurable space. A function ϕ : X → 𝔽 (𝔽 denotes either ℝ, ℂ or [0, ∞] ⊂ ℝ̄) is a simple function if ϕ is M – B_{𝔽} measurable and ϕ(X) contains only finitely many elements.

Any such simple functions can be written as

ϕ = ∑_{i=1}^{n} λ_{i}1_{A_{i}} with A_{i} ∈ M and λ_{i} ∈ 𝔽.  (18.7)

Indeed, take λ_{1}, λ_{2}, …, λ_{n} to be an enumeration of the range of ϕ and A_{i} = ϕ^{−1}({λ_{i}}). Note that this argument shows that any simple function may be written intrinsically as

ϕ = ∑_{y∈𝔽} y1_{ϕ^{−1}({y})}.  (18.8)

The next theorem shows that simple functions are “pointwise dense” in the space of measurable functions.

Theorem 18.42 (Approximation Theorem). Let f : X → [0, ∞] be measurable and define, see Figure 18.2,

ϕ_{n}(x) := ∑_{k=0}^{2^{2n}−1} \frac{k}{2^{n}} 1_{f^{−1}(( \frac{k}{2^{n}} , \frac{k+1}{2^{n}}])}(x) + 2^{n}1_{f^{−1}((2^{n},∞])}(x)

= ∑_{k=0}^{2^{2n}−1} \frac{k}{2^{n}} 1_{\{\frac{k}{2^{n}} <f≤ \frac{k+1}{2^{n}}\}}(x) + 2^{n}1_{\{f>2^{n}\}}(x)

then ϕ_{n} ≤ f for all n, ϕ_{n}(x) ↑ f(x) for all x ∈ X and ϕ_{n} ↑ f uniformly on the sets X_{M} := {x ∈ X : f(x) ≤ M} with M < ∞. Moreover, if f : X → ℂ is a measurable function, then there exists simple functions ϕ_{n} such that lim_{n→∞} ϕ_{n}(x) = f(x) for all x and |ϕ_{n}| ↑ |f| as n → ∞.

Proof. Since

(\frac{k}{2^{n}} , \frac{k + 1}{2^{n}}] = (\frac{2k}{2^{n+1}} , \frac{2k + 1}{2^{n+1}}] ∪ (\frac{2k + 1}{2^{n+1}}, \frac{2k + 2}{2^{n+1}}],

if x ∈ f^{−1} ((\frac{2k}{2^{n+1}} , \frac{2k+1}{2^{n+1}}]) then ϕ_{n}(x) = ϕ_{n+1}(x) = \frac{2k}{2^{n+1}} and if x ∈ f^{−1} ((\frac{2k+1}{2^{n+1}}, \frac{2k+2}{2^{n+1}}]) then ϕ_{n}(x) = \frac{2k}{2^{n+1}} < \frac{2k+1}{2^{n+1}} = ϕ_{n+1}(x). Similarly

(2^{n}, ∞] = (2^{n}, 2^{n+1}] ∪ (2^{n+1}, ∞],

18.3 σ – Function Algebras 293

Fig. 18.2. Constructing simple functions approximating a function, f : X → [0, ∞].

and so for x ∈ f^{−1}((2^{n+1}, ∞]), ϕ_{n}(x) = 2^{n} < 2^{n+1} = ϕ_{n+1}(x) and for x ∈ f^{−1}((2^{n}, 2^{n+1}]), ϕ_{n+1}(x) ≥ 2^{n} = ϕ_{n}(x). Therefore ϕ_{n} ≤ ϕ_{n+1} for all n. It is clear by construction that ϕ_{n}(x) ≤ f(x) for all x and that 0 ≤ f(x)−ϕ_{n}(x) ≤ 2^{−n} if x ∈ X_{2^{n}}. Hence we have shown that ϕ_{n}(x) ↑ f(x) for all x ∈ X and ϕ_{n} ↑ f uniformly on bounded sets. For the second assertion, first assume that f : X → ℝ is a measurable function and choose ϕ_{n}^{±} to be simple functions such that ϕ_{n}^{±} ↑ f_{±} as n → ∞ and define ϕ_{n} = ϕ_{n}^{+} − ϕ_{n}^{−}. Then

|ϕ_{n}| = ϕ_{n}^{+} + ϕ_{n}^{−} ≤ ϕ_{n+1}^{+} + ϕ_{n+1}^{−} = |ϕ_{n+1}|

and clearly |ϕ_{n}| = ϕ_{n}^{+} +ϕ_{n}^{−} ↑ f_{+} +f_{−} = |f| and ϕ_{n} = ϕ_{n}^{+} −ϕ_{n}^{−} → f_{+} −f_{−} = f as n → ∞. Now suppose that f : X → ℂ is measurable. We may now choose simple function u_{n} and v_{n} such that |u_{n}| ↑ |Re f| , |v_{n}| ↑ |Im f| , u_{n} → Re f and v_{n} → Im f as n → ∞. Let ϕ_{n} = u_{n} + iv_{n}, then

|ϕ_{n}|^{2} = u_{n}^{2} + v_{n}^{2} ↑ |Re f|^{2} + |Im f|^{2} = |f|^{2}

and ϕ_{n} = u_{n} + iv_{n} → Re f + i Im f = f as n → ∞.

For the rest of this section let X be a given set.

Definition 18.43 (Bounded Convergence). We say that a sequence of functions f_{n} from X to ℝ or ℂ converges boundedly to a function f if lim_{n→∞} f_{n}(x) = f(x) for all x ∈ X and

sup{|f_{n}(x)| : x ∈ X and n = 1, 2, …} < ∞.

Definition 18.44. A function algebra H on X is a linear subspace of ℓ^{∞} (X, ℝ) which contains 1 and is closed under pointwise multiplication, i.e. H is a subalgebra of ℓ^{∞} (X, ℝ) which contains 1. If H is further closed under bounded convergence then H is said to be a σ – function algebra.

Example 18.45. Suppose M is a σ – algebra on X, then

ℓ^{∞} (M, ℝ) := {f ∈ ℓ^{∞} (X, ℝ) : f is M/B_{ℝ} – measurable}  (18.9)

is a σ – function algebra. The next theorem will show that these are the only example of σ – function algebras. (See Exercise 18.7 below for examples of function algebras on X.)

Notation 18.46 If H ⊂ ℓ^{∞} (X, ℝ) be a function algebra, let

M (H) := {A ⊂ X : 1_{A} ∈ H} .  (18.10)

Theorem 18.47. Let H be a σ – function algebra on a set X. Then

1. M (H) is a σ – algebra on X.

2. H = ℓ^{∞} (M (H) , ℝ) .

3. The map

M ∈ {σ – algebras on X} → ℓ^{∞} (M, ℝ) ∈ {σ – function algebras on X}

(18.11)

is bijective with inverse given by H → M (H) .

Proof. Let M := M (H) .

1. Since 0, 1 ∈ H, ∅, X ∈ M. If A ∈ M then, since H is a linear subspace of ℓ^{∞} (X, ℝ) , 1_{A^{c}} = 1 − 1_{A} ∈ H which shows A^{c} ∈ M. If {A_{n}}_{n=1}^{∞} ⊂ M, then since H is an algebra,

1_{∩_{n=1}^{N}A_{n}} = ∏_{n=1}^{N} 1_{A_{n}} =: f_{N} ∈ H

for all N ∈ ℕ. Because H is closed under bounded convergence it follows that

1_{∩_{n=1}^{∞}A_{n}} = lim_{N→∞} f_{N} ∈ H

and this implies ∩_{n=1}^{∞}A_{n} ∈ M. Hence we have shown M is a σ – algebra.

2. Since H is an algebra, p (f) ∈ H for any f ∈ H and any polynomial p on ℝ. The Weierstrass approximation Theorem 8.34, asserts that polynomials on ℝ are uniformly dense in the space of continuos functions on any compact subinterval of ℝ. Hence if f ∈ H and ϕ ∈ C (ℝ) , there exists polynomials p_{n} on ℝ such that p_{n} ∘ f (x) converges to ϕ ∘ f (x) uniformly (and hence boundedly) in x ∈ X as n → ∞. Therefore ϕ ∘ f ∈ H for all f ∈ H and ϕ ∈ C (ℝ) and in particular |f| ∈ H and f_{±} := \frac{|f|±f}{2} ∈ H if f ∈ H. Fix an α ∈ ℝ and for n ∈ ℕ let ϕ_{n} (t) := (t − α)_{+}^{1/n} , where (t − α)_{+} := max {t − α, 0} . Then ϕ_{n} ∈ C (ℝ) and ϕ_{n} (t) → 1_{t>α} as n → ∞ and the convergence is bounded when t is restricted to any compact subset of ℝ. Hence if f ∈ H it follows that 1_{f>α} = lim_{n→∞} ϕ_{n} (f) ∈ H for all α ∈ ℝ, i.e. {f > α} ∈ M for all α ∈ ℝ. Therefore if f ∈ H then f ∈ ℓ^{∞} (M, ℝ) and we have shown H ⊂ ℓ^{∞} (M, ℝ) . Conversely if f ∈ ℓ^{∞} (M, ℝ) , then for any α < β, {α < f ≤ β} ∈ M = M (H) and so by assumption 1_{{α<f≤β}} ∈ H. Combining this remark with the approximation Theorem 18.42 and the fact that H is closed under bounded convergence shows that f ∈ H. Hence we have shown ℓ^{∞} (M, ℝ) ⊂ H which combined with H ⊂ ℓ^{∞} (M, ℝ) already proved shows H = ℓ^{∞} (M (H) , ℝ) .

3. Items 1. and 2. shows the map in Eq. (18.11) is surjective. To see the map is injective suppose M and F are two σ – algebras on X such that ℓ^{∞} (M, ℝ) = ℓ^{∞} (F, ℝ) , then

M = {A ⊂ X : 1_{A} ∈ ℓ^{∞} (M, ℝ)}
= {A ⊂ X : 1_{A} ∈ ℓ^{∞} (F, ℝ)} = F.

Notation 18.48 Suppose M is a subset of ℓ^{∞} (X, ℝ) .

1. Let H (M) denote the smallest subspace of ℓ^{∞} (X, ℝ) which contains M and the constant functions and is closed under bounded convergence.

2. Let H_{σ} (M) denote the smallest σ – function algebra containing M.

Theorem 18.49. Suppose M is a subset of ℓ^{∞} (X, ℝ) , then H_{σ} (M) = ℓ^{∞} (σ (M) , ℝ) or in other words the following diagram commutes:

M  ⟶  σ (M)
M {Multiplicative Subsets} ⟶  {σ – algebras}  M
↓  ↓  ↓  ↓
H_{σ} (M) {σ– function algebras} = {σ– function algebras} ℓ^{∞} (M, ℝ) .

Proof. Since ℓ^{∞} (σ (M) , ℝ) is σ – function algebra which contains M it follows that

H_{σ} (M) ⊂ ℓ^{∞} (σ (M) , ℝ) .

For the opposite inclusion, let

M = M (H_{σ} (M)) := {A ⊂ X : 1_{A} ∈ H_{σ} (M)} .

By Theorem 18.47, M ⊂ H_{σ} (M) = ℓ^{∞} (M, ℝ) which implies that every f ∈ M is M – measurable. This then implies σ (M) ⊂ M and therefore

ℓ^{∞} (σ (M) , ℝ) ⊂ ℓ^{∞} (M, ℝ) = H_{σ} (M) .

Definition 18.50 (Multiplicative System). A collection of bounded real or complex valued functions, M, on a set X is called a multiplicative system if f · g ∈ M whenever f and g are in M.

Theorem 18.51 (Dynkin’s Multiplicative System Theorem). Suppose M ⊂ ℓ^{∞} (X, ℝ) is a multiplicative system, then

H (M) = H_{σ} (M) = ℓ^{∞} (σ (M) , ℝ) .  (18.12)

In words, the smallest subspace of bounded real valued functions on X which contains M that is closed under bounded convergence is the same as the space of bounded real valued σ (M) – measurable functions on X.

Proof. We begin by proving H := H (M) is a σ – function algebra. To do this, for any f ∈ H let

H_{f} := {g ∈ H : fg ∈ H} ⊂ H

and notice that H_{f} is a linear subspace of ℓ^{∞} (X, ℝ) which is closed under bounded convergence. Moreover if f ∈ M, M ⊂ H_{f} since M is multiplicative. Therefore H_{f} = H and we have shown that fg ∈ H whenever f ∈ M and g ∈ H. Given this it now follows that M ⊂ H_{f} for any f ∈ H and by the same reasoning just used, H_{f} = H. Since f ∈ H is arbitrary, we have shown fg ∈ H for all f, g ∈ H, i.e. H is an algebra. Since it is harder to be an algebra of functions containing M (see Exercise 18.13) than it is to be a subspace of functions containing M it follows that H (M) ⊂ H_{σ} (M) . But as we have just seen H (M) is a σ – function algebra which contains M so we must have H_{σ} (M) ⊂ H (M) because H_{σ} (M) is by definition the smallest such σ – function algebra. Hence H_{σ} (M) = H (M) . The assertion that H_{σ} (M) = ℓ^{∞} (σ (M) , ℝ) has already been proved in Theorem 18.49.

Theorem 18.52 (Complex Multiplicative System Theorem). Suppose H is a complex linear subspace of ℓ^{∞}(X, ℂ) such that: 1 ∈ H, H is closed under complex conjugation, and H is closed under bounded convergence. If M ⊂ H is multiplicative system which is closed under conjugation, then H contains all bounded complex valued σ(M)-measurable functions, i.e. ℓ^{∞} (σ (M) , ℂ) ⊂ H.

Proof. Let M_{0} = span_{ℂ}(M ∪ {1}) be the complex span of M. As the reader should verify, M_{0} is an algebra, M_{0} ⊂ H, M_{0} is closed under complex conjugation and that σ (M_{0}) = σ (M) . Let H^{ℝ} := H ∩ ℓ^{∞}(X, ℝ) and M_{0}^{ℝ} = M ∩ℓ^{∞}(X, ℝ). Then (you verify) M_{0}^{ℝ} is a multiplicative system, M_{0}^{ℝ} ⊂ H^{ℝ} and H^{ℝ} is a linear space containing 1 which is closed under bounded convergence. Therefore by Theorem 18.51, ℓ^{∞} (σ (M_{0}^{ℝ}) , ℝ) ⊂ H^{ℝ}. Since H and M_{0} are complex linear spaces closed under complex conjugation, for any f ∈ H or f ∈ M_{0}, the functions Re f = \frac{1}{2} (f + f̄) and Im f = \frac{1}{2i} (f − f̄) are in H (M_{0}) or M_{0} respectively. Therefore H = H^{ℝ} + iH^{ℝ}, M_{0} = M_{0}^{ℝ} + iM_{0}^{ℝ}, σ (M_{0}^{ℝ}) = σ (M_{0}) = σ (M) and

ℓ^{∞} (σ (M) , ℂ) = ℓ^{∞} (σ (M_{0}^{ℝ}) , ℝ) + iℓ^{∞} (σ (M_{0}^{ℝ}) , ℝ)
⊂ H^{ℝ} + iH^{ℝ} = H.

Exercise 18.7 (Algebra analogue of Theorem 18.47). Call a function algebra H ⊂ ℓ^{∞} (X, ℝ) a simple function algebra if the range of each function f ∈ H is a finite subset of ℝ. Prove there is a one to one correspondence between algebras A on a set X and simple function algebras H on X.

Definition 18.53. A collection of subsets, C, of X is a multiplicative class(or a π – class) if C is closed under finite intersections.

Corollary 18.54. Suppose H is a subspace of ℓ^{∞}(X, ℝ) which is closed under bounded convergence and 1 ∈ H. If C ⊂ 2^{X} is a multiplicative class such that 1_{A} ∈ H for all A ∈ C, then H contains all bounded σ(C) – measurable functions.

Proof. Let M = {1} ∪ {1_{A} : A ∈ C} . Then M ⊂ H is a multiplicative system and the proof is completed with an application of Theorem 18.51.

Corollary 18.55. Suppose that (X, d) is a metric space and B_{X} = σ(τ_{d}) is the Borel σ – algebra on X and H is a subspace of ℓ^{∞}(X, ℝ) such that BC(X, ℝ) ⊂ H and H is closed under bounded convergence^{1}. Then H contains all bounded B_{X} – measurable real valued functions on X. (This may be stated as follows: the smallest vector space of bounded functions which is closed under bounded convergence and contains BC(X, ℝ) is the space of bounded B_{X} – measurable real valued functions on X.)

Proof. Let V ∈ τ_{d} be an open subset of X and for n ∈ ℕ let

f_{n}(x) := min(n · d_{V^{c}}(x), 1) for all x ∈ X.

Notice that f_{n} = ϕ_{n} ∘ d_{V^{c}} where ϕ_{n}(t) = min(nt, 1) (see Figure 18.3) which is continuous and hence f_{n} ∈ BC(X, ℝ) for all n. Furthermore, f_{n} converges boundedly to 1_{d_{V ^{c}} >0} = 1_{V} as n → ∞ and therefore 1_{V} ∈ H for all V ∈ τ. Since τ is a π – class, the result now follows by an application of Corollary 18.54.

Plots of ϕ_{1}, ϕ_{2} and ϕ_{3}.

1 Recall that BC(X, ℝ) are the bounded continuous functions on X.

Here are some more variants of Corollary 18.55.

Proposition 18.56. Let (X, d) be a metric space, B_{X} = σ(τ_{d}) be the Borel σ – algebra and assume there exists compact sets K_{k} ⊂ X such that K_{k}^{o} ↑ X. Suppose that H is a subspace of ℓ^{∞}(X, ℝ) such that C_{c}(X, ℝ) ⊂ H (C_{c}(X, ℝ) is the space of continuous functions with compact support) and H is closed under bounded convergence. Then H contains all bounded B_{X} – measurable real valued functions on X.

Proof. Let k and n be positive integers and set ψ_{n,k}(x) = min(1, n · d_{(K_{k}^{o})}^{c}(x)). Then ψ_{n,k} ∈ C_{c}(X, ℝ) and {ψ_{n,k} ≠ 0} ⊂ K_{k}^{o}. Let H_{n,k} denote those bounded B_{X} – measurable functions, f : X → ℝ, such that ψ_{n,k}f ∈ H. It is easily seen that H_{n,k} is closed under bounded convergence and that H_{n,k} contains BC(X, ℝ) and therefore by Corollary 18.55, ψ_{n,k}f ∈ H for all bounded measurable functions f : X → ℝ. Since ψ_{n,k}f → 1_{K_{k}^{o}}f boundedly as n → ∞, 1_{K_{k}^{o}}f ∈ H for all k and similarly 1_{K_{k}^{o}}f → f boundedly as k → ∞ and therefore f ∈ H.

Lemma 18.57. Suppose that (X, τ) is a locally compact second countable Hausdorff space.^{2} Then:

1. every open subset U ⊂ X is σ – compact. In fact U is still a locally compact second countable Hausdorff space.

2. If F ⊂ X is a closed set, there exist open sets V_{n} ⊂ X such that V_{n} ↓ F as n → ∞.

3. To each open set U ⊂ X there exists f_{n} ≺ U (i.e. f_{n} ∈ C_{c} (U, [0, 1])) such that lim_{n→∞} f_{n} = 1_{U}.

4. B_{X} = σ(C_{c}(X, ℝ)), i.e. the σ – algebra generated by C_{c}(X) is the Borel σ – algebra on X.

Proof.

1. Let U be an open subset of X, V be a countable base for τ and

V^{U} := {W ∈ V : W̄ ⊂ U and W̄ is compact}.

For each x ∈ U, by Proposition 12.7, there exists an open neighborhood V of x such that V̄ ⊂ U and V̄ is compact. Since V is a base for the topology τ, there exists W ∈ V such that x ∈ W ⊂ V. Because W̄ ⊂ V̄ , it follows that W̄ is compact and hence W ∈ V^{U}. As x ∈ U was arbitrary, U = ∪V^{U}. This shows V^{U} is a countable basis for the topology on U and that U is still locally compact. Let {W_{n}}_{n=1}^{∞} be an enumeration of V^{U} and set K_{n} := ∪_{k=1}^{n}W̄_{k}. Then K_{n} ↑ U as n → ∞ and K_{n} is compact for each n. This shows U is σ – compact. (See Exercise 11.7.)

2 For example any separable locally compact metric space and in particular any open subset of ℝ^{n}.

18.4 Product σ – Algebras 299

2. Let {K_{n}}_{n=1}^{∞} be compact subsets of F^{c} such that K_{n} ↑ F^{c} as n → ∞ and set V_{n} := K_{n}^{c} = X \ K_{n}. Then V_{n} ↓ F and by Proposition 12.5, V_{n} is open for each n.

3. Let U ⊂ X be an open set and {K_{n}}_{n=1}^{∞} be compact subsets of U such that K_{n} ↑ U. By Urysohn’s Lemma 12.8, there exist f_{n} ≺ U such that f_{n} = 1 on K_{n}. These functions satisfy, 1_{U} = lim_{n→∞} f_{n}.

4. By item 3., 1_{U} is σ(C_{c}(X, ℝ)) – measurable for all U ∈ τ and hence τ ⊂ σ(C_{c}(X, ℝ)). Therefore B_{X} = σ(τ) ⊂ σ(C_{c}(X, ℝ)). The converse inclusion holds because continuous functions are always Borel measurable.

Here is a variant of Corollary 18.55.

Corollary 18.58. Suppose that (X, τ) is a second countable locally compact Hausdorff space and B_{X} = σ(τ) is the Borel σ – algebra on X. If H is a subspace of ℓ^{∞}(X, ℝ) which is closed under bounded convergence and contains C_{c}(X, ℝ), then H contains all bounded B_{X} – measurable real valued functions on X.

Proof. By Item 3. of Lemma 18.57, for every U ∈ τ the characteristic function, 1_{U}, may be written as a bounded pointwise limit of functions from C_{c} (X, ℝ) . Therefore 1_{U} ∈ H for all U ∈ τ. Since τ is a π – class, the proof is finished with an application of Corollary 18.54

18.4 Product σ – Algebras

Let {(X_{α}, M_{α})}_{α∈A} be a collection of measurable spaces X = X_{A} = ∏_{α∈A} X_{α}

and π_{α} : X_{A} → X_{α} be the canonical projection map as in Notation 2.2.

Definition 18.59 (Product σ – Algebra). The product σ – algebra, ⊗_{α∈A}M_{α}, is the smallest σ – algebra on X such that each π_{α} for α ∈ A is measurable, i.e.

⊗_{α∈A}M_{α} := σ(π_{α} : α ∈ A) = σ (∪_{α}π_{α}^{−1}(M_{α})) .

Applying Proposition 18.25 in this setting implies the following proposition.

Proposition 18.60. Suppose Y is a measurable space and f : Y → X = X_{A} is a map. Then f is measurable iff π_{α} ∘ f : Y → X_{α} is measurable for all α ∈ A. In particular if A = {1, 2, …, n} so that X = X_{1} × X_{2} × ⋯ × X_{n} and f(y) = (f_{1}(y), f_{2}(y), …, f_{n}(y)) ∈ X_{1} × X_{2} × ⋯ × X_{n}, then f : Y → X_{A} is measurable iff f_{i} : Y → X_{i} is measurable for all i.

Proposition 18.61. Suppose that (X_{α}, M_{α})_{α∈A} is a collection of measurable spaces and E_{α} ⊂ M_{α} generates M_{α} for each α ∈ A, then

⊗_{α∈A} M_{α} = σ (∪_{α∈A}π_{α}^{−1}(E_{α}))  (18.13)

Moreover, suppose that A is either finite or countably infinite, X_{α} ∈ E_{α} for each α ∈ A, and M_{α} = σ(E_{α}) for each α ∈ A. Then the product σ – algebra satisfies

⊗_{α∈A} M_{α} = σ ({ ∏_{α∈A} E_{α} : E_{α} ∈ E_{α} for all α ∈ A}) .  (18.14)

In particular if A = {1, 2, …, n} , then X = X_{1} × X_{2} × ⋯ × X_{n} and

M_{1} ⊗ M_{2} ⊗ ⋯ ⊗ M_{n} = σ(M_{1} × M_{2} × ⋯ × M_{n}),

where M_{1} × M_{2} × ⋯ × M_{n} is as defined in Notation 10.26.

Proof. Since ∪_{α}π_{α}^{−1}(E_{α}) ⊂ ∪_{α}π_{α}^{−1}(M_{α}), it follows that

F := σ (∪_{α}π_{α}^{−1}(E_{α})) ⊂ σ (∪_{α}π_{α}^{−1}(M_{α})) = ⊗_{α∈A}M_{α}.

Conversely,

F ⊃ σ(π_{α}^{−1}(E_{α})) = π_{α}^{−1}(σ(E_{α})) = π_{α}^{−1}(M_{α})

holds for all α implies that

∪_{α}π_{α}^{−1}(M_{α}) ⊂ F

and hence that ⊗_{α∈A}M_{α} ⊂ F. We now prove Eq. (18.14). Since we are assuming that X_{α} ∈ E_{α} for each α ∈ A, we see that

∪_{α}π_{α}^{−1}(E_{α}) ⊂ { ∏_{α∈A} E_{α} : E_{α} ∈ E_{α} for all α ∈ A}

and therefore by Eq. (18.13)

⊗_{α∈A}M_{α} = σ (∪_{α}π_{α}^{−1}(E_{α})) ⊂ σ ({ ∏_{α∈A} E_{α} : E_{α} ∈ E_{α} for all α ∈ A}) .

This last statement is true independent as to whether A is countable or not. For the reverse inclusion it suffices to notice that since A is countable,

∏_{α∈A} E_{α} = ∩_{α∈A}π_{α}^{−1}(E_{α}) ∈ ⊗_{α∈A}M_{α}

and hence

σ ({ ∏_{α∈A} E_{α} : E_{α} ∈ E_{α} for all α ∈ A}) ⊂ ⊗_{α∈A}M_{α}.

Remark 18.62. One can not relax the assumption that X_{α} ∈ E_{α} in the second part of Proposition 18.61. For example, if X_{1} = X_{2} = {1, 2} and E_{1} = E_{2} = {{1}} , then σ(E_{1} × E_{2}) = {∅, X_{1} × X_{2}, {(1, 1)}} while σ(σ(E_{1}) × σ(E_{2})) = 2^{X_{1}×X_{2}}.

Theorem 18.63. Let {X_{α}}_{α∈A} be a sequence of sets where A is at most countable. Suppose for each α ∈ A we are given a countable set E_{α} ⊂ 2^{X_{α}}. Let τ_{α} = τ(E_{α}) be the topology on X_{α} generated by E_{α} and X be the product space ∏_{α∈A} X_{α} with equipped with the product topology τ := ⊗_{α∈A}τ(E_{α}). Then the Borel σ – algebra B_{X} = σ(τ) is the same as the product σ – algebra:

B_{X} = ⊗_{α∈A}B_{X_{α}},

where B_{X_{α}} = σ(τ(E_{α})) = σ(E_{α}) for all α ∈ A.

In particular if A = {1, 2, …, n} and each (X_{i}, τ_{i}) is a second countable topological space, then

B_{X} := σ(τ_{1} ⊗ τ_{2} ⊗ ⋯ ⊗ τ_{n}) = σ(B_{X_{1}} × ⋯ × B_{X_{n}}) =: B_{X_{1}} ⊗ ⋯ ⊗ B_{X_{n}}.

Proof. By Proposition 10.25, the topology τ may be described as the smallest topology containing E = ∪_{α∈A}π_{α}^{−1}(E_{α}). Now E is the countable union of countable sets so is still countable. Therefore by Proposition 18.17 and Proposition 18.61,

B_{X} = σ(τ) = σ(τ(E)) = σ(E) = ⊗_{α∈A}σ(E_{α})
= ⊗_{α∈A}σ(τ_{α}) = ⊗_{α∈A}B_{X_{α}}.

Corollary 18.64. If (X_{i}, d_{i}) are separable metric spaces for i = 1, …, n, then

B_{X_{1}} ⊗ ⋯ ⊗ B_{X_{n}} = B_{(X_{1}×···×X_{n})}

where B_{X_{i}} is the Borel σ – algebra on X_{i} and B_{(X_{1}×···×X_{n})} is the Borel σ – algebra on X_{1} × ⋯ × X_{n} equipped with the metric topology associated to the metric d(x, y) = ∑_{i=1}^{n} d_{i}(x_{i}, y_{i}) where x = (x_{1}, x_{2}, …, x_{n}) and y = (y_{1}, y_{2}, …, y_{n}).

Proof. This is a combination of the results in Lemma 10.28, Exercise 10.12 and Theorem 18.63.

Because all norms on finite dimensional spaces are equivalent, the usual Euclidean norm on ℝ^{m} × ℝ^{n} is equivalent to the “product” norm defined by

∥(x, y)∥_{ℝ^{m}×ℝ^{n}} = ∥x∥_{ℝ^{m}} + ∥y∥_{ℝ^{n}} .

Hence by Lemma 10.28, the Euclidean topology on ℝ^{m+n} is the same as the product topology on ℝ^{m+n} ≅ ℝ^{m} ×ℝ^{n}. Here we are identifying ℝ^{m} ×ℝ^{n} with ℝ^{m+n} by the map

(x, y) ∈ ℝ^{m} × ℝ^{n} → (x_{1}, …, x_{m}, y_{1}, …, y_{n}) ∈ ℝ^{m+n}.

These comments along with Corollary 18.64 proves the following result.

Corollary 18.65. After identifying ℝ^{m} × ℝ^{n} with ℝ^{m+n} as above and letting B_{ℝ^{n}} denote the Borel σ –algebra on ℝ^{n}, we have

B_{ℝ^{m+n}} = B_{ℝ^{n}} ⊗ B_{ℝ^{m}} and B_{ℝ^{n}} = \overbrace{B_{ℝ} ⊗ ⋯ ⊗ B_{ℝ}}^{n–times}.

18.4.1 Factoring of Measurable Maps

Lemma 18.66. Suppose that (Y, F) is a measurable space and F : X → Y is a map. Then to every (σ(F), Bℝ̄) – measurable function, H : X → ℝ̄, there is a (F, Bℝ̄) – measurable function h : Y → ℝ̄ such that H = h ∘ F.

Proof. First suppose that H = 1_{A} where A ∈ σ(F) = F^{−1}(F). Let B ∈ F such that A = F^{−1}(B) then 1_{A} = 1_{F^{−1}(B)} = 1_{B} ∘ F and hence the Lemma is valid in this case with h = 1_{B}. More generally if H = ∑ a_{i}1_{A_{i}} is a simple function, then there exists B_{i} ∈ F such that 1_{A_{i}} = 1_{B_{i}} ∘ F and hence H = h ∘ F with h := ∑ a_{i}1_{B_{i}} – a simple function on ℝ̄. For general (σ(F), F) – measurable function, H, from X → ℝ̄, choose simple functions H_{n} converging to H. Let h_{n} be simple functions on ℝ̄ such that H_{n} = h_{n} ∘ F. Then it follows that

H = lim_{n→∞} H_{n} = lim sup_{n→∞} H_{n} = lim sup_{n→∞} h_{n} ∘ F = h ∘ F

where h := lim sup_{n→∞} h_{n} – a measurable function from Y to ℝ̄.

The following is an immediate corollary of Proposition 18.25 and Lemma 18.66.

Corollary 18.67. Let X and A be sets, and suppose for α ∈ A we are give a measurable space (Y_{α}, F_{α}) and a function f_{α} : X → Y_{α}. Let Y := ∏_{α∈A} Y_{α}, F := ⊗_{α∈A}F_{α} be the product σ – algebra on Y and M := σ(f_{α} : α ∈ A) be the smallest σ – algebra on X such that each f_{α} is measurable. Then the function F : X → Y defined by [F (x)]_{α} := f_{α}(x) for each α ∈ A is (M, F)

– measurable and a function H : X → ℝ̄ is (M, Bℝ̄) – measurable iff there exists a (F, Bℝ̄) – measurable function h from Y to ℝ̄ such that H = h ∘ F.

18.5 Exercises

Exercise 18.8. Prove Corollary 18.23. Hint: See Exercise 18.3.

Exercise 18.9. If M is the σ – algebra generated by E ⊂ 2^{X}, then M is the union of the σ – algebras generated by countable subsets F ⊂ E.

Exercise 18.10. Let (X, M) be a measure space and f_{n} : X → 𝔽 be a sequence of measurable functions on X. Show that {x : lim_{n→∞} f_{n}(x) exists in 𝔽} ∈ M.

Exercise 18.11. Show that every monotone function f : ℝ → ℝ is (B_{ℝ}, B_{ℝ})

– measurable.

Exercise 18.12. Show by example that the supremum of an uncountable family of measurable functions need not be measurable. (Folland problem 2.6 on p. 48.)

Exercise 18.13. Let X = {1, 2, 3, 4} , A = {1, 2} , B = {2, 3} and M := {1_{A}, 1_{B}} . Show H_{σ} (M) ≠ H (M) in this case.

Measures and Integration

Definition 19.1. A measure μ on a measurable space (X, M) is a function μ : M → [0, ∞] such that

1. μ(∅) = 0 and

2. (Finite Additivity) If {A_{i}}_{i=1}^{n} ⊂ M are pairwise disjoint, i.e. A_{i} ∩ A_{j} = ∅ when i ≠ j, then

μ(⋃_{i=1}^{n} A_{i}) = ∑_{i=1}^{n} μ(A_{i}).

3. (Continuity) If A_{n} ∈ M and A_{n} ↑ A, then μ(A_{n}) ↑ μ(A).

We call a triple (X, M, μ), where (X, M) is a measurable space and μ : M → [0, ∞] is a measure, a measure space.

Remark 19.2. Properties 2) and 3) in Definition 19.1 are equivalent to the following condition. If {A_{i}}_{i=1}^{∞} ⊂ M are pairwise disjoint then

μ(⋃_{i=1}^{∞} A_{i}) = ∑_{i=1}^{∞} μ(A_{i}).  (19.1)

To prove this assume that Properties 2) and 3) in Definition 19.1 hold and

{A_{i}}_{i=1}^{∞} ⊂ M are pairwise disjoint. Letting B_{n} := ⋃_{i=1}^{n} A_{i} ↑ B := ⋃_{i=1}^{∞} A_{i}, we

have

μ(⋃_{i=1}^{∞} A_{i}) = μ(B) =^{(3)} lim_{n→∞} μ(B_{n}) =^{(2)} lim_{n→∞} ∑_{i=1}^{n} μ(A_{i}) = ∑_{i=1}^{∞} μ(A_{i}).

Conversely, if Eq. (19.1) holds we may take A_{j} = ∅ for all j > n to see that Property 2) of Definition 19.1 holds. Also if A_{n} ↑ A, let B_{n} := A_{n} \A_{n−1} with A_{0} := ∅. Then {B_{n}}_{n=1}^{∞} are pairwise disjoint, A_{n} = ∪_{j=1}^{n}B_{j} and A = ∪_{j=1}^{∞}B_{j}. So if Eq. (19.1) holds we have

μ(A) = μ (∪_{j=1}^{∞}B_{j}) = ∑_{j=1}^{∞} μ(B_{j})

= lim_{n→∞} ∑_{j=1}^{n} μ(B_{j}) = lim_{n→∞} μ(∪_{j=1}^{n}B_{j}) = lim_{n→∞} μ(A_{n}).

Proposition 19.3 (Basic properties of measures). Suppose that (X, M, μ) is a measure space and E, F ∈ M and {E_{j}}_{j=1}^{∞} ⊂ M, then :

1. μ(E) ≤ μ(F) if E ⊂ F.

2. μ(∪E_{j}) ≤ ∑ μ(E_{j}).

3. If μ(E_{1}) < ∞ and E_{j} ↓ E, i.e. E_{1} ⊃ E_{2} ⊃ E_{3} ⊃ … and E = ∩_{j}E_{j}, then μ(E_{j}) ↓ μ(E) as j → ∞.

Proof.

1. Since F = E ∪ (F \ E),

μ(F) = μ(E) + μ(F \ E) ≥ μ(E).

2. Let Ẽ_{j} = E_{j} \ (E_{1} ∪ ⋯ ∪ E_{j−1}) so that the Ẽ_{j} ’s are pair-wise disjoint and E = ∪Ẽ_{j}. Since Ẽ_{j} ⊂ E_{j} it follows from Remark 19.2 and part (1), that

μ(E) = ∑ μ(Ẽ_{j}) ≤ ∑ μ(E_{j}).

3. Define D_{i} := E_{1} \ E_{i} then D_{i} ↑ E_{1} \ E which implies that

μ(E_{1}) − μ(E) = lim_{i→∞} μ(D_{i}) = μ(E_{1}) − lim_{i→∞} μ(E_{i})

which shows that lim_{i→∞} μ(E_{i}) = μ(E).

Definition 19.4. A set E ⊂ X is a null set if E ∈ M and μ(E) = 0. If P is some “property” which is either true or false for each x ∈ X, we will use the terminology P a.e. (to be read P almost everywhere) to mean

E := {x ∈ X : P is false for x}

is a null set. For example if f and g are two measurable functions on (X, M, μ), f = g a.e. means that μ(f ≠ g) = 0.

Definition 19.5. A measure space (X, M, μ) is complete if every subset of a null set is in M, i.e. for all F ⊂ X such that F ⊂ E ∈ M with μ(E) = 0 implies that F ∈ M.

Proposition 19.6 (Completion of a Measure). Let (X, M, μ) be a measure space. Set

N = N^{μ} := {N ⊂ X : ∃ F ∈ M ∋ N ⊂ F and μ(F) = 0} ,
M̄ = M̄^{μ} := {A ∪ N : A ∈ M and N ∈ N } and
μ̄(A ∪ N) := μ(A) for A ∈ M and N ∈ N ,

see Fig. 19.1. Then M̄ is a σ – algebra, μ̄ is a well defined measure on M̄, μ̄ is the unique measure on M̄ which extends μ on M, and (X, M̄, μ̄) is complete measure space. The σ-algebra, M̄, is called the completion of M relative to μ and μ̄, is called the completion of μ.

Proof. Clearly X, ∅ ∈ M̄. Let A ∈ M and N ∈ N and choose F ∈ M

Fig. 19.1. Completing a σ – algebra.

such that N ⊂ F and μ(F) = 0. Since N^{c} = (F \ N) ∪ F^{c},

(A ∪ N)^{c} = A^{c} ∩ N^{c} = A^{c} ∩ (F \ N ∪ F^{c})
= [A^{c} ∩ (F \ N)] ∪ [A^{c} ∩ F^{c}]

where [A^{c} ∩ (F \ N)] ∈ N and [A^{c} ∩ F^{c}] ∈ M. Thus M̄ is closed under complements. If A_{i} ∈ M and N_{i} ⊂ F_{i} ∈ M such that μ(F_{i}) = 0 then ∪(A_{i}∪N_{i}) = (∪A_{i})∪(∪N_{i}) ∈ M̄ since ∪A_{i} ∈ M and ∪N_{i} ⊂ ∪F_{i} and μ(∪F_{i}) ≤ ∑ μ(F_{i}) = 0. Therefore, M̄ is a σ – algebra. Suppose A ∪ N_{1} = B ∪ N_{2} with A, B ∈ M and N_{1}, N_{2}, ∈ N. Then A ⊂ A ∪ N_{1} ⊂ A ∪ N_{1} ∪ F_{2} = B ∪ F_{2} which shows that

μ(A) ≤ μ(B) + μ(F_{2}) = μ(B).

Similarly, we show that μ(B) ≤ μ(A) so that μ(A) = μ(B) and hence μ̄(A ∪ N) := μ(A) is well defined. It is left as an exercise to show μ̄ is a measure, i.e. that it is countable additive.

Many theorems in the sequel will require some control on the size of a measure μ. The relevant notion for our purposes (and most purposes) is that of a σ – finite measure defined next.

Definition 19.7. Suppose X is a set, E ⊂ M ⊂ 2^{X} and μ : M → [0, ∞] is a function. The function μ is σ – finite on E if there exists E_{n} ∈ E such that μ(E_{n}) < ∞ and X = ∪_{n=1}^{∞}E_{n}. If M is a σ – algebra and μ is a measure on M which is σ – finite on M we will say (X, M, μ) is a σ – finite measure space.

The reader should check that if μ is a finitely additive measure on an algebra, M, then μ is σ – finite on M iff there exists X_{n} ∈ M such that X_{n} ↑ X and μ(X_{n}) < ∞.

19.1 Example of Measures

Most σ – algebras and σ -additive measures are somewhat difficult to describe and define. However, one special case is fairly easy to understand. Namely suppose that F ⊂ 2^{X} is a countable or finite partition of X and M ⊂ 2^{X} is the σ – algebra which consists of the collection of sets A ⊂ X such that

A = ∪ {α ∈ F : α ⊂ A} .  (19.2)

It is easily seen that M is a σ – algebra.

Any measure μ : M → [0, ∞] is determined uniquely by its values on F. Conversely, if we are given any function λ : F → [0, ∞] we may define, for A ∈ M,

μ(A) = ∑_{α∈F∋α⊂A} λ(α) = ∑_{α∈F} λ(α)1_{α⊂A}

where 1_{α⊂A} is one if α ⊂ A and zero otherwise. We may check that μ is a measure on M. Indeed, if A = ∐_{i=1}^{∞} A_{i} and α ∈ F, then α ⊂ A iff α ⊂ A_{i} for one and hence exactly one A_{i}. Therefore 1_{α⊂A} = ∑_{i=1}^{∞} 1_{α⊂A_{i}} and hence

μ(A) = ∑_{α∈F} λ(α)1_{α⊂A} = ∑_{α∈F} λ(α) ∑_{i=1}^{∞} 1_{α⊂A_{i}}

= ∑_{i=1}^{∞} ∑_{α∈F} λ(α)1_{α⊂A_{i}} = ∑_{i=1}^{∞} μ(A_{i})

as desired. Thus we have shown that there is a one to one correspondence between measures μ on M and functions λ : F → [0, ∞].

The construction of measures will be covered in Chapters ?? – ?? below. However, let us record here the existence of an interesting class of measures.

Theorem 19.8. To every right continuous non-decreasing function F : ℝ → ℝ there exists a unique measure μ_{F} on B_{ℝ} such that

μ_{F} ((a, b]) = F (b) − F (a) ∀ − ∞ < a ≤ b < ∞  (19.3)

Moreover, if A ∈ B_{ℝ} then

μ_{F} (A) = inf {∑_{i=1}^{∞}(F (b_{i}) − F (a_{i})) : A ⊂ ∪_{i=1}^{∞}(a_{i}, b_{i}]}  (19.4)

= inf {∑_{i=1}^{∞}(F (b_{i}) − F (a_{i})) : A ⊂ ∐_{i=1}^{∞}(a_{i}, b_{i}]} .  (19.5)

In fact the map F → μ_{F} is a one to one correspondence between right continuous functions F with F (0) = 0 on one hand and measures μ on B_{ℝ} such that μ(J) < ∞ on any bounded set J ∈ B_{ℝ} on the other.

Proof. See Section ?? below or Theorem ?? below.

Example 19.9. The most important special case of Theorem 19.8 is when F (x) = x, in which case we write m for μ_{F} . The measure m is called Lebesgue measure.

Theorem 19.10. Lebesgue measure m is invariant under translations, i.e. for B ∈ B_{ℝ} and x ∈ ℝ,

m(x + B) = m(B).  (19.6)

Moreover, m is the unique measure on B_{ℝ} such that m((0, 1]) = 1 and Eq. (19.6) holds for B ∈ B_{ℝ} and x ∈ ℝ. Moreover, m has the scaling property

m(λB) = |λ| m(B)  (19.7)

where λ ∈ ℝ, B ∈ B_{ℝ} and λB := {λx : x ∈ B}.

Proof. Let m_{x}(B) := m(x + B), then one easily shows that m_{x} is a measure on B_{ℝ} such that m_{x}((a, b]) = b − a for all a < b. Therefore, m_{x} = m by the uniqueness assertion in Theorem 19.8. For the converse, suppose that m is translation invariant and m((0, 1]) = 1. Given n ∈ ℕ, we have

(0, 1] = ∪_{k=1}^{n}(\frac{k − 1}{n}, \frac{k}{n}] = ∪_{k=1}^{n} (\frac{k − 1}{n} + (0, \frac{1}{n}]) .

Therefore,

1 = m((0, 1]) = ∑_{k=1}^{n} m (\frac{k − 1}{n} + (0, \frac{1}{n}])

= ∑_{k=1}^{n} m((0, \frac{1}{n}]) = n · m((0, \frac{1}{n}]).

That is to say

m((0, \frac{1}{n}]) = 1/n.

Similarly, m((0, \frac{l}{n}]) = l/n for all l, n ∈ ℕ and therefore by the translation invariance of m,

m((a, b]) = b − a for all a, b ∈ ℚ with a < b.

Finally for a, b ∈ ℝ such that a < b, choose a_{n}, b_{n} ∈ ℚ such that b_{n} ↓ b and a_{n} ↑ a, then (a_{n}, b_{n}] ↓ (a, b] and thus

m((a, b]) = lim_{n→∞} m((a_{n}, b_{n}]) = lim_{n→∞} (b_{n} − a_{n}) = b − a,

i.e. m is Lebesgue measure. To prove Eq. (19.7) we may assume that λ ≠ 0 since this case is trivial to prove. Now let m_{λ}(B) := |λ|^{−1} m(λB). It is easily checked that m_{λ} is again a measure on B_{ℝ} which satisfies

m_{λ}((a, b]) = λ^{−1}m ((λa, λb]) = λ^{−1}(λb − λa) = b − a

if λ > 0 and

m_{λ}((a, b]) = |λ|^{−1} m ([λb, λa)) = − |λ|^{−1} (λb − λa) = b − a

if λ < 0. Hence m_{λ} = m.

We are now going to develop integration theory relative to a measure. The integral defined in the case for Lebesgue measure, m, will be an extension of the standard Riemann integral on ℝ.

19.1.1 ADD: Examples of Measures

BRUCE: ADD details.

1. Product measure for the flipping of a coin.

2. Haar Measure

3. Measure on embedded submanifolds, i.e. Hausdorff measure.

4. Wiener measure.

5. Gibbs states.

6. Measure associated to self-adjoint operators and classifying them.

19.2 Integrals of Simple functions

Let (X, M, μ) be a fixed measure space in this section.

Definition 19.11. Let 𝔽 = ℂ or [0, ∞) and suppose that ϕ : X → 𝔽 is a simple function as in Definition 18.41. If 𝔽 = ℂ assume further that μ(ϕ^{−1}({y})) < ∞ for all y ≠ 0 in ℂ. For such functions ϕ, define I_{μ}(ϕ) by

I_{μ}(ϕ) = ∑_{y∈𝔽} yμ(ϕ^{−1}({y})).

Proposition 19.12. Let λ ∈ 𝔽 and ϕ and ψ be two simple functions, then I_{μ} satisfies:

1.

I_{μ}(λϕ) = λI_{μ}(ϕ).  (19.8)

2.

I_{μ}(ϕ + ψ) = I_{μ}(ψ) + I_{μ}(ϕ).

3. If ϕ and ψ are non-negative simple functions such that ϕ ≤ ψ then

I_{μ}(ϕ) ≤ I_{μ}(ψ).

Proof. Let us write {ϕ = y} for the set ϕ^{−1}({y}) ⊂ X and μ(ϕ = y) for μ({ϕ = y}) = μ(ϕ^{−1} ({y})) so that

I_{μ}(ϕ) = ∑_{y∈𝔽} yμ(ϕ = y).

We will also write {ϕ = a, ψ = b} for ϕ^{−1}({a}) ∩ ψ^{−1}({b}). This notation is more intuitive for the purposes of this proof. Suppose that λ ∈ 𝔽 then

I_{μ}(λϕ) = ∑_{y∈𝔽} y μ(λϕ = y) = ∑_{y∈𝔽} y μ(ϕ = y/λ)

= ∑_{z∈𝔽} λz μ(ϕ = z) = λI_{μ}(ϕ)

provided that λ ≠ 0. The case λ = 0 is clear, so we have proved 1. Suppose that ϕ and ψ are two simple functions, then

I_{μ}(ϕ + ψ) = ∑_{z∈𝔽} z μ(ϕ + ψ = z)

= ∑_{z∈𝔽} z μ (∪_{w∈𝔽} {ϕ = w, ψ = z − w})

= ∑_{z∈𝔽} z ∑_{w∈𝔽} μ(ϕ = w, ψ = z − w)

= ∑_{z,w∈𝔽}(z + w)μ(ϕ = w, ψ = z)

= ∑_{z∈𝔽} z μ(ψ = z) + ∑_{w∈𝔽} w μ(ϕ = w)

= I_{μ}(ψ) + I_{μ}(ϕ).

which proves 2. For 3. if ϕ and ψ are non-negative simple functions such that ϕ ≤ ψ

I_{μ}(ϕ) = ∑_{a≥0} aμ(ϕ = a) = ∑_{a,b≥0} aμ(ϕ = a, ψ = b)

≤ ∑_{a,b≥0} bμ(ϕ = a, ψ = b) = ∑_{b≥0} bμ(ψ = b) = I_{μ}(ψ),

wherein the third inequality we have used {ϕ = a, ψ = b} = ∅ if a > b.

19.3 Integrals of positive functions

Definition 19.13. Let L^{+} = L^{+} (M) = {f : X → [0, ∞] : f is measurable}. Define

∫_{X} f (x) dμ (x) = ∫_{X} fdμ := sup {I_{μ}(ϕ) : ϕ is simple and ϕ ≤ f} .

We say the f ∈ L^{+} is integrable if ∫_{X} fdμ < ∞. If A ∈ M, let

∫_{A} f (x) dμ (x) = ∫_{A} fdμ := ∫_{X} 1_{A}f dμ.

Remark 19.14. Because of item 3. of Proposition 19.12, if ϕ is a non-negative simple function, ∫_{X} ϕdμ = I_{μ}(ϕ) so that ∫_{X} is an extension of I_{μ}. This extension still has the monotonicity property if I_{μ} : namely if 0 ≤ f ≤ g then

∫_{X} fdμ = sup {I_{μ}(ϕ) : ϕ is simple and ϕ ≤ f}
≤ sup {I_{μ}(ϕ) : ϕ is simple and ϕ ≤ g} ≤ ∫_{X} gdμ.

Similarly if c > 0,

∫_{X} cfdμ = c ∫_{X} fdμ.

Also notice that if f is integrable, then μ ({f = ∞}) = 0.

Lemma 19.15 (Sums as Integrals). Let X be a set and ρ : X → [0, ∞] be a function, let μ = ∑_{x∈X} ρ(x)δ_{x} on M = 2^{X}, i.e.

μ(A) = ∑_{x∈A} ρ(x).

If f : X → [0, ∞] is a function (which is necessarily measurable), then

∫_{X} fdμ = ∑_{X} fρ.

Proof. Suppose that ϕ : X → [0, ∞) is a simple function, then ϕ = ∑_{z∈[0,∞)} z1_{{ϕ=z}} and

∑_{X} ϕρ = ∑_{x∈X} ρ(x) ∑_{z∈[0,∞)} z1_{{ϕ=z}}(x) = ∑_{z∈[0,∞)} z ∑_{x∈X} ρ(x)1_{{ϕ=z}}(x)

= ∑_{z∈[0,∞)} zμ({ϕ = z}) = ∫_{X} ϕdμ.

So if ϕ : X → [0, ∞) is a simple function such that ϕ ≤ f, then

∫_{X} ϕdμ = ∑_{X} ϕρ ≤ ∑_{X} fρ.

Taking the sup over ϕ in this last equation then shows that

∫_{X} fdμ ≤ ∑_{X} fρ.

For the reverse inequality, let Λ ⊂⊂ X be a finite set and N ∈ (0, ∞). Set f^{N}(x) = min {N, f(x)} and let ϕ_{N,Λ} be the simple function given by ϕ_{N,Λ}(x) := 1_{Λ}(x)f^{N}(x). Because ϕ_{N,Λ}(x) ≤ f(x),

∑_{Λ} f^{N}ρ = ∑_{X} ϕ_{N,Λ}ρ = ∫_{X} ϕ_{N,Λ}dμ ≤ ∫_{X} fdμ.

Since f^{N} ↑ f as N → ∞, we may let N → ∞ in this last equation to concluded

∑_{Λ} fρ ≤ ∫_{X} fdμ.

Since Λ is arbitrary, this implies

∑_{X} fρ ≤ ∫_{X} fdμ.

Theorem 19.16 (Monotone Convergence Theorem). Suppose f_{n} ∈ L^{+} is a sequence of functions such that f_{n} ↑ f (f is necessarily in L^{+}) then

∫ f_{n} ↑ ∫ f as n → ∞.

Proof. Since f_{n} ≤ f_{m} ≤ f, for all n ≤ m < ∞,

∫ f_{n} ≤ ∫ f_{m} ≤ ∫ f

from which if follows ∫ f_{n} is increasing in n and

lim_{n→∞} ∫ f_{n} ≤ ∫ f.  (19.9)

For the opposite inequality, let ϕ : X → [0, ∞) be a simple function such that 0 ≤ ϕ ≤ f, α ∈ (0, 1) and X_{n} := {f_{n} ≥ αϕ} . Notice that X_{n} ↑ X and f_{n} ≥ α1_{X_{n}}ϕ and so by definition of ∫ f_{n},

∫ f_{n} ≥ ∫ α1_{X_{n}}ϕ = α ∫ 1_{X_{n}}ϕ.  (19.10)

Then using the continuity property of μ,

lim_{n→∞} ∫ 1_{X_{n}}ϕ = lim_{n→∞} ∫ 1_{X_{n}} ∑_{y>0} y1_{{ϕ=y}}

= lim_{n→∞} ∑_{y>0} yμ(X_{n} ∩ {ϕ = y}) = ∑_{y>0} y lim_{n→∞} μ(X_{n} ∩ {ϕ = y})

= ∑_{y>0} y lim_{n→∞} μ({ϕ = y}) = ∫ ϕ.

This identity allows us to let n → ∞ in Eq. (19.10) to conclude

∫_{X} ϕ ≤ \frac{1}{α} lim_{n→∞} ∫ f_{n}.

Since this is true for all non-negative simple functions ϕ with ϕ ≤ f;

∫ f = sup \{∫_{X} ϕ : ϕ is simple and ϕ ≤ f\} ≤ \frac{1}{α} lim_{n→∞} ∫ f_{n}.

Because α ∈ (0, 1) was arbitrary, it follows that ∫ f ≤ lim_{n→∞} ∫ f_{n} which combined with Eq. (19.9) proves the theorem.

The following simple lemma will be use often in the sequel.

Lemma 19.17 (Chebyshev’s Inequality). Suppose that f ≥ 0 is a measurable function, then for any ε > 0,

μ(f ≥ ε) ≤ \frac{1}{ε} ∫_{X} fdμ.  (19.11)

In particular if ∫_{X} fdμ < ∞ then μ(f = ∞) = 0 (i.e. f < ∞ a.e.) and the set {f > 0} is σ – finite.

Proof. Since 1_{\{f≥ε\}} ≤ 1_{\{f≥ε\}} \frac{1}{ε}f ≤ \frac{1}{ε}f,

μ(f ≥ ε) = ∫_{X} 1_{\{f≥ε\}}dμ ≤ ∫_{X} 1_{\{f≥ε\}} \frac{1}{ε}fdμ ≤ \frac{1}{ε} ∫_{X} fdμ.

If M := ∫_{X} fdμ < ∞, then

μ(f = ∞) ≤ μ(f ≥ n) ≤ \frac{M}{n} → 0 as n → ∞

and {f ≥ 1/n} ↑ {f > 0} with μ(f ≥ 1/n) ≤ nM < ∞ for all n.

Corollary 19.18. If f_{n} ∈ L^{+} is a sequence of functions then

∫ ∑_{n=1}^{∞} f_{n} = ∑_{n=1}^{∞} ∫ f_{n}.

In particular, if ∑_{n=1}^{∞} ∫ f_{n} < ∞ then ∑_{n=1}^{∞} f_{n} < ∞ a.e.

Proof. First off we show that

∫ (f_{1} + f_{2}) = ∫ f_{1} + ∫ f_{2}

by choosing non-negative simple function ϕ_{n} and ψ_{n} such that ϕ_{n} ↑ f_{1} and ψ_{n} ↑ f_{2}. Then (ϕ_{n} + ψ_{n}) is simple as well and (ϕ_{n} + ψ_{n}) ↑ (f_{1} + f_{2}) so by the monotone convergence theorem,

∫ (f_{1} + f_{2}) = lim_{n→∞} ∫ (ϕ_{n} + ψ_{n}) = lim_{n→∞} (∫ ϕ_{n} + ∫ ψ_{n})

= lim_{n→∞} ∫ ϕ_{n} + lim_{n→∞} ∫ ψ_{n} = ∫ f_{1} + ∫ f_{2}.

Now to the general case. Let g_{N} := ∑_{n=1}^{N} f_{n} and g = ∑_{1}^{∞} f_{n}, then g_{N} ↑ g and so

again by monotone convergence theorem and the additivity just proved,

∑_{n=1}^{∞} ∫ f_{n} := lim_{N→∞} ∑_{n=1}^{N} ∫ f_{n} = lim_{N→∞} ∫ ∑_{n=1}^{N} f_{n}

= lim_{N→∞} ∫ g_{N} = ∫ g =: ∫ ∑_{n=1}^{∞} f_{n}.

Remark 19.19. It is in the proof of this corollary (i.e. the linearity of the integral) that we really make use of the assumption that all of our functions are measurable. In fact the definition ∫ fdμ makes sense for all functions f : X → [0, ∞] not just measurable functions. Moreover the monotone convergence theorem holds in this generality with no change in the proof. However, in the proof of Corollary 19.18, we use the approximation Theorem 18.42 which relies heavily on the measurability of the functions to be approximated.

The following Lemma and the next Corollary are simple applications of Corollary 19.18.

Lemma 19.20 (The First Borell – Carntelli Lemma). Let (X, M, μ) be a measure space, A_{n} ∈ M, and set

{A_{n} i.o.} = {x ∈ X : x ∈ A_{n} for infinitely many n’s} = ⋂_{N=1}^{∞} ⋃_{n≥N} A_{n}.

If ∑_{n=1}^{∞} μ(A_{n}) < ∞ then μ({A_{n} i.o.}) = 0.

Proof. (First Proof.) Let us first observe that

{A_{n} i.o.} = {x ∈ X : ∑_{n=1}^{∞} 1_{A_{n}}(x) = ∞} .

Hence if ∑_{n=1}^{∞} μ(A_{n}) < ∞ then

∞ > ∑_{n=1}^{∞} μ(A_{n}) = ∑_{n=1}^{∞} ∫_{X} 1_{A_{n}} dμ = ∫_{X} ∑_{n=1}^{∞} 1_{A_{n}} dμ

implies that ∑_{n=1}^{∞} 1_{A_{n}}(x) < ∞ for μ - a.e. x. That is to say μ({A_{n} i.o.}) = 0.

(Second Proof.) Of course we may give a strictly measure theoretic proof of this fact:

μ(A_{n} i.o.) = lim_{N→∞} μ ( ⋃_{n≥N} A_{n})

≤ lim_{N→∞} ∑_{n≥N} μ(A_{n})

and the last limit is zero since ∑_{n=1}^{∞} μ(A_{n}) < ∞.

Corollary 19.21. Suppose that (X, M, μ) is a measure space and {A_{n}}_{n=1}^{∞} ⊂ M is a collection of sets such that μ(A_{i} ∩ A_{j}) = 0 for all i ≠ j, then

μ (∪_{n=1}^{∞}A_{n}) = ∑_{n=1}^{∞} μ(A_{n}).

Proof. Since

μ (∪_{n=1}^{∞}A_{n}) = ∫_{X} 1_{∪_{n=1}^{∞}A_{n}}dμ and
∑_{n=1}^{∞} μ(A_{n}) = ∫_{X} ∑_{n=1}^{∞} 1_{A_{n}}dμ

it suffices to show

∑_{n=1}^{∞} 1_{A_{n}} = 1_{∪_{n=1}^{∞}A_{n}} μ – a.e.  (19.12)

Now ∑_{n=1}^{∞} 1_{A_{n}} ≥ 1_{∪_{n=1}^{∞}A_{n}} and ∑_{n=1}^{∞} 1_{A_{n}}(x) ≠ 1_{∪_{n=1}^{∞}A_{n}}(x) iff x ∈ A_{i} ∩ A_{j} for some i ≠ j, that is

{x : ∑_{n=1}^{∞} 1_{A_{n}}(x) ≠ 1_{∪_{n=1}^{∞}A_{n}}(x)} = ∪_{i<j}A_{i} ∩ A_{j}

and the latter set has measure 0 being the countable union of sets of measure zero. This proves Eq. (19.12) and hence the corollary.

Notation 19.22 If m is Lebesgue measure on B_{ℝ}, f is a non-negative Borel measurable function and a < b with a, b ∈ ℝ̄, we will often write ∫_{a}^{b} f (x) dx or ∫_{a}^{b} fdm for ∫_{(a,b]∩ℝ} fdm.

Example 19.23. Suppose −∞ < a < b < ∞, f ∈ C([a, b], [0, ∞)) and m be Lebesgue measure on ℝ. Also let π_{k} = {a = a_{0}^{k} < a_{1}^{k} < ⋯ < a_{n_{k}}^{k} = b} be a sequence of refining partitions (i.e. π_{k} ⊂ π_{k+1} for all k) such that

mesh(π_{k}) := max{|a_{j}^{k} − a_{j−1}^{k+1}| : j = 1, …, n_{k}} → 0 as k → ∞.

For each k, let

f_{k}(x) = f(a)1_{{a}} + ∑_{l=0}^{n_{k}−1} min {f(x) : a_{l}^{k} ≤ x ≤ a_{l+1}^{k}} 1_{(a_{l}^{k},a_{l+1}^{k}]}(x)

then f_{k} ↑ f as k → ∞ and so by the monotone convergence theorem,

∫_{a}^{b} fdm := ∫_{[a,b]} fdm = lim_{k→∞} ∫_{a}^{b} f_{k} dm

= lim_{k→∞} ∑_{l=0}^{n_{k}−1} min {f(x) : a_{l}^{k} ≤ x ≤ a_{l+1}^{k}} m ((a_{l}^{k}, a_{l+1}^{k}])

= ∫_{a}^{b} f(x)dx.

The latter integral being the Riemann integral.

We can use the above result to integrate some non-Riemann integrable functions:

Example 19.24. For all λ > 0,

∫_{0}^{∞} e^{−λx}dm(x) = λ^{−1} and ∫_{ℝ} \frac{1}{1 + x^{2}} dm(x) = π.

The proof of these identities are similar. By the monotone convergence theorem, Example 19.23 and the fundamental theorem of calculus for Riemann integrals (or see Theorem 8.13 above or Theorem 19.40 below),

∫_{0}^{∞} e^{−λx}dm(x) = lim_{N→∞} ∫_{0}^{N} e^{−λx}dm(x) = lim_{N→∞} ∫_{0}^{N} e^{−λx}dx

= − lim_{N→∞} \frac{1}{λ}e^{−λx}|_{0}^{N} = λ^{−1}

and

∫_{ℝ} \frac{1}{1 + x^{2}} dm(x) = lim_{N→∞} ∫_{−N}^{N} \frac{1}{1 + x^{2}} dm(x) = lim_{N→∞} ∫_{−N}^{N} \frac{1}{1 + x^{2}} dx

= lim_{N→∞} [tan^{−1}(N) − tan^{−1}(−N)] = π.

Let us also consider the functions x^{−p},

∫_{(0,1]} \frac{1}{x^{p}} dm(x) = lim_{n→∞} ∫_{0}^{1} 1_{( \frac{1}{n} ,1]}(x)\frac{1}{x^{p}} dm(x)

= lim_{n→∞} ∫ 1_{\frac{1}{n}} \frac{1}{x^{p}} dx = lim_{n→∞} \frac{x^{−p+1}}{1 − p} |_{1/n}^{1}

= \{ \begin{matrix}\frac{1}{1−p} & if & p < 1 \\ ∞ & if & p > 1 \end{matrix}

If p = 1 we find

∫_{(0,1]} \frac{1}{x^{p}} dm(x) = lim_{n→∞} ∫ 1_{\frac{1}{n}} \frac{1}{x}dx = lim_{n→∞} ln(x)|_{1/n}^{1} = ∞.

Example 19.25. Let {r_{n}}_{n=1}^{∞} be an enumeration of the points in ℚ ∩ [0, 1] and define

f(x) = ∑_{n=1}^{∞} 2^{−n} \frac{1}{\sqrt{|x − r_{n}|}}

with the convention that

\frac{1}{\sqrt{|x − r_{n}|}} = 5 if x = r_{n}.

Since, By Theorem 19.40,

∫_{0}^{1} \frac{1}{\sqrt{|x − r_{n}|}} dx = ∫_{r_{n}}^{1} \frac{1}{\sqrt{x − r_{n}}} dx + ∫_{0}^{r_{n}} \frac{1}{\sqrt{r_{n} − x}}dx
= 2\sqrt{x − r_{n}}|_{r_{n}}^{1} − 2\sqrt{r_{n} − x}|_{0}^{r_{n}} = 2 (\sqrt{1 − r_{n}} − \sqrt{r_{n}})
≤ 4,

we find

∫_{[0,1]} f(x)dm(x) = ∑_{n=1}^{∞} 2^{−n} ∫_{[0,1]} \frac{1}{\sqrt{|x − r_{n}|}}dx ≤ ∑_{n=1}^{∞} 2^{−n}4 = 4 < ∞.

In particular, m(f = ∞) = 0, i.e. that f < ∞ for almost every x ∈ [0, 1] and this implies that

∑_{n=1}^{∞} 2^{−n} \frac{1}{\sqrt{|x − r_{n}|}} < ∞ for a.e. x ∈ [0, 1].

This result is somewhat surprising since the singularities of the summands form a dense subset of [0, 1].

Proposition 19.26. Suppose that f ≥ 0 is a measurable function. Then ∫_{X} fdμ = 0 iff f = 0 a.e. Also if f, g ≥ 0 are measurable functions such that f ≤ g a.e. then ∫ fdμ ≤ ∫ gdμ. In particular if f = g a.e. then ∫ fdμ = ∫ gdμ.

Proof. If f = 0 a.e. and ϕ ≤ f is a simple function then ϕ = 0 a.e. This implies that μ(ϕ^{−1}({y})) = 0 for all y > 0 and hence ∫_{X} ϕdμ = 0 and therefore ∫_{X} fdμ = 0. Conversely, if ∫ fdμ = 0, then by (Lemma 19.17),

μ(f ≥ 1/n) ≤ n ∫ fdμ = 0 for all n.

Therefore, μ(f > 0) ≤ ∑_{n=1}^{∞} μ(f ≥ 1/n) = 0, i.e. f = 0 a.e. For the second assertion let E be the exceptional set where f > g, i.e. E := {x ∈ X : f(x) > g(x)}. By assumption E is a null set and 1_{E^{c}}f ≤ 1_{E^{c}}g everywhere. Because g = 1_{E^{c}}g + 1_{E}g and 1_{E}g = 0 a.e.,

∫ gdμ = ∫ 1_{E^{c}}gdμ + ∫ 1_{E}gdμ = ∫ 1_{E^{c}}gdμ

and similarly ∫ fdμ = ∫ 1_{E^{c}}fdμ. Since 1_{E^{c}}f ≤ 1_{E^{c}}g everywhere,

∫ fdμ = ∫ 1_{E^{c}}fdμ ≤ ∫ 1_{E^{c}}gdμ = ∫ gdμ.

Corollary 19.27. Suppose that {f_{n}} is a sequence of non-negative measurable functions and f is a measurable function such that f_{n} ↑ f off a null set, then

∫ f_{n} ↑ ∫ f as n → ∞.

Proof. Let E ⊂ X be a null set such that f_{n}1_{E^{c}} ↑ f1_{E^{c}} as n → ∞. Then by the monotone convergence theorem and Proposition 19.26,

∫ f_{n} = ∫ f_{n}1_{E^{c}} ↑ ∫ f1_{E^{c}} = ∫ f as n → ∞.

Lemma 19.28 (Fatou’s Lemma). If f_{n} : X → [0, ∞] is a sequence of measurable functions then

∫ lim inf_{n→∞} f_{n} ≤ lim inf_{n→∞} ∫ f_{n}

Proof. Define g_{k} := inf_{n≥k} f_{n} so that g_{k} ↑ lim inf_{n→∞} f_{n} as k → ∞. Since

g_{k} ≤ f_{n} for all k ≤ n,

∫ g_{k} ≤ ∫ f_{n} for all n ≥ k

and therefore

∫ g_{k} ≤ lim inf_{n→∞} ∫ f_{n} for all k.

We may now use the monotone convergence theorem to let k → ∞ to find

∫ lim inf_{n→∞} f_{n} = ∫ lim_{k→∞} g_{k} =^{MCT} lim_{k→∞} ∫ g_{k} ≤ lim inf_{n→∞} ∫ f_{n}.

19.4 Integrals of Complex Valued Functions

Definition 19.29. A measurable function f : X → ℝ̄ is integrable if f_{+} := f1_{{f≥0}} and f_{−} = −f 1_{{f≤0}} are integrable. We write L^{1} (μ; ℝ) for the space of real valued integrable functions. For f ∈ L^{1} (μ; ℝ) , let

∫ fdμ = ∫ f_{+}dμ − ∫ f_{−}dμ

Convention: If f, g : X → ℝ̄ are two measurable functions, let f + g denote the collection of measurable functions h : X → ℝ̄ such that h(x) = f(x)+g(x) whenever f(x)+g(x) is well defined, i.e. is not of the form ∞−∞ or −∞ + ∞. We use a similar convention for f − g. Notice that if f, g ∈ L^{1} (μ; ℝ) and h_{1}, h_{2} ∈ f + g, then h_{1} = h_{2} a.e. because |f| < ∞ and |g| < ∞ a.e.

Notation 19.30 (Abuse of notation) We will sometimes denote the integral ∫_{X} fdμ by μ (f) . With this notation we have μ (A) = μ (1_{A}) for all A ∈ M.

Remark 19.31. Since

f_{±} ≤ |f| ≤ f_{+} + f_{−},

a measurable function f is integrable iff ∫ |f| dμ < ∞. Hence

L^{1} (μ; ℝ) := {f : X → ℝ̄ : f is measurable and ∫_{X} |f| dμ < ∞} .

If f, g ∈ L^{1} (μ; ℝ) and f = g a.e. then f_{±} = g_{±} a.e. and so it follows from Proposition 19.26 that ∫ fdμ = ∫ gdμ. In particular if f, g ∈ L^{1} (μ; ℝ) we may define

∫_{X} (f + g) dμ = ∫_{X} hdμ

where h is any element of f + g.

Proposition 19.32. The map

f ∈ L^{1} (μ; ℝ) → ∫_{X} fdμ ∈ ℝ

is linear and has the monotonicity property: ∫ fdμ ≤ ∫ gdμ for all f, g ∈ L^{1} (μ; ℝ) such that f ≤ g a.e.

Proof. Let f, g ∈ L^{1} (μ; ℝ) and a, b ∈ ℝ. By modifying f and g on a null set, we may assume that f, g are real valued functions. We have af + bg ∈ L^{1} (μ; ℝ) because

|af + bg| ≤ |a| |f| + |b| |g| ∈ L^{1} (μ; ℝ) .

If a < 0, then

(af)_{+} = −af_{−} and (af)_{−} = −af_{+}

so that

∫ af = −a ∫ f_{−} + a ∫ f_{+} = a(∫ f_{+} − ∫ f_{−}) = a ∫ f.

A similar calculation works for a > 0 and the case a = 0 is trivial so we have shown that

∫ af = a ∫ f.

Now set h = f + g. Since h = h_{+} − h_{−},

h_{+} − h_{−} = f_{+} − f_{−} + g_{+} − g_{−}

or

h_{+} + f_{−} + g_{−} = h_{−} + f_{+} + g_{+}.

Therefore,

∫ h_{+} + ∫ f_{−} + ∫ g_{−} = ∫ h_{−} + ∫ f_{+} + ∫ g_{+}

and hence

∫ h = ∫ h_{+} − ∫ h_{−} = ∫ f_{+} + ∫ g_{+} − ∫ f_{−} − ∫ g_{−} = ∫ f + ∫ g.

Finally if f_{+} − f_{−} = f ≤ g = g_{+} − g_{−} then f_{+} + g_{−} ≤ g_{+} + f_{−} which implies that

∫ f_{+} + ∫ g_{−} ≤ ∫ g_{+} + ∫ f_{−}

or equivalently that

∫ f = ∫ f_{+} − ∫ f_{−} ≤ ∫ g_{+} − ∫ g_{−} = ∫ g.

The monotonicity property is also a consequence of the linearity of the integral, the fact that f ≤ g a.e. implies 0 ≤ g − f a.e. and Proposition 19.26.

Definition 19.33. A measurable function f : X → ℂ is integrable if ∫_{X} |f| dμ < ∞. Analogously to the real case, let

L^{1} (μ; ℂ) := {f : X → ℂ : f is measurable and ∫_{X} |f| dμ < ∞} .

denote the complex valued integrable functions. Because, max (|Re f| , |Im f|) ≤ |f| ≤ \sqrt{2} max (|Re f| , |Im f|) , ∫ |f| dμ < ∞ iff

∫ |Re f| dμ + ∫ |Im f| dμ < ∞.

For f ∈ L^{1} (μ; ℂ) define

∫ f dμ = ∫ Re f dμ + i ∫ Im f dμ.

It is routine to show the integral is still linear on L^{1} (μ; ℂ) (prove!). In the remainder of this section, let L^{1} (μ) be either L^{1} (μ; ℂ) or L^{1} (μ; ℝ) . If A ∈ M and f ∈ L^{1} (μ; ℂ) or f : X → [0, ∞] is a measurable function, let

∫_{A} fdμ := ∫_{X} 1_{A}fdμ.

Proposition 19.34. Suppose that f ∈ L^{1} (μ; ℂ) , then

|∫_{X} fdμ| ≤ ∫_{X} |f| dμ.  (19.13)

Proof. Start by writing ∫_{X} f dμ = Re^{iθ} with R ≥ 0. We may assume that R = |∫_{X} fdμ| > 0 since otherwise there is nothing to prove. Since

R = e^{−iθ} ∫_{X} f dμ = ∫_{X} e^{−iθ}f dμ = ∫_{X} Re (e^{−iθ}f) dμ + i ∫_{X} Im (e^{−iθ}f) dμ,

it must be that ∫_{X} Im [e^{−iθ}f] dμ = 0. Using the monotonicity in Proposition 19.26,

|∫_{X} fdμ| = ∫_{X} Re (e^{−iθ}f) dμ ≤ ∫_{X} |Re (e^{−iθ}f)| dμ ≤ ∫_{X} |f| dμ.

Proposition 19.35. Let f, g ∈ L^{1} (μ) , then

1. The set \{f ≠ 0\} is σ – finite, in fact \{|f| ≥ \frac{1}{n}\} ↑ \{f ≠ 0\} and μ(|f| ≥ \frac{1}{n}) < ∞ for all n.

2. The following are equivalent a) ∫_{E} f = ∫_{E} g for all E ∈ M

b) ∫_{X} |f − g| = 0

c) f = g a.e.

Proof. 1. By Chebyshev’s inequality, Lemma 19.17,

μ(|f| ≥ \frac{1}{n}) ≤ n ∫_{X} |f| dμ < ∞

for all n. 2. (a) ⟹ (c) Notice that

∫_{E} f = ∫_{E} g ⇔ ∫_{E}(f − g) = 0

for all E ∈ M. Taking E = {Re(f − g) > 0} and using 1_{E} Re(f − g) ≥ 0, we learn that

0 = Re ∫_{E}(f − g)dμ = ∫ 1_{E} Re(f − g) ⟹ 1_{E} Re(f − g) = 0 a.e.

This implies that 1_{E} = 0 a.e. which happens iff

μ ({Re(f − g) > 0}) = μ(E) = 0.

Similar μ(Re(f −g) < 0) = 0 so that Re(f −g) = 0 a.e. Similarly, Im(f −g) = 0 a.e and hence f − g = 0 a.e., i.e. f = g a.e. (c) ⟹ (b) is clear and so is (b) ⟹ (a) since

|∫_{E} f − ∫_{E} g| ≤ ∫ |f − g| = 0.

Definition 19.36. Let (X, M, μ) be a measure space and L^{1}(μ) = L^{1}(X, M, μ) denote the set of L^{1} (μ) functions modulo the equivalence relation; f ∼ g iff f = g a.e. We make this into a normed space using the norm

∥f − g∥_{L^{1}} = ∫ |f − g| dμ

and into a metric space using ρ_{1}(f, g) = ∥f − g∥_{L^{1}} .

Warning: in the future we will often not make much of a distinction between L^{1}(μ) and L^{1} (μ) . On occasion this can be dangerous and this danger will be pointed out when necessary.

Remark 19.37. More generally we may define L^{p}(μ) = L^{p}(X, M, μ) for p ∈ [1, ∞) as the set of measurable functions f such that

∫_{X} |f|^{p} dμ < ∞

modulo the equivalence relation; f ∼ g iff f = g a.e.

We will see in Chapter 21 that

∥f∥_{L^{p}} = (∫ |f|^{p} dμ)^{1/p} for f ∈ L^{p}(μ)

is a norm and (L^{p}(μ), ∥·∥_{L^{p}}) is a Banach space in this norm.

Theorem 19.38 (Dominated Convergence Theorem). Suppose f_{n}, g_{n}, g ∈ L^{1} (μ) , f_{n} → f a.e., |f_{n}| ≤ g_{n} ∈ L^{1} (μ) , g_{n} → g a.e. and ∫_{X} g_{n}dμ → ∫_{X} gdμ. Then f ∈ L^{1} (μ) and

∫_{X} fdμ = lim_{h→∞} ∫_{X} f_{n}dμ.

(In most typical applications of this theorem g_{n} = g ∈ L^{1} (μ) for all n.)

Proof. Notice that |f| = lim_{n→∞} |f_{n}| ≤ lim_{n→∞} |g_{n}| ≤ g a.e. so that f ∈ L^{1} (μ) . By considering the real and imaginary parts of f separately, it suffices to prove the theorem in the case where f is real. By Fatou’s Lemma,

∫_{X}(g ± f)dμ = ∫_{X} lim inf_{n→∞} (g_{n} ± f_{n}) dμ ≤ lim inf_{n→∞} ∫_{X} (g_{n} ± f_{n}) dμ

= lim_{n→∞} ∫_{X} g_{n}dμ + lim inf_{n→∞} (± ∫_{X} f_{n}dμ)

= ∫_{X} gdμ + lim inf_{n→∞} (± ∫_{X} f_{n}dμ)

Since lim inf_{n→∞}(−a_{n}) = − lim sup_{n→∞} a_{n}, we have shown,

∫_{X} gdμ ± ∫_{X} fdμ ≤ ∫_{X} gdμ + \{ \begin{matrix}lim inf_{n→∞} ∫_{X} f_{n}dμ \\ − lim sup_{n→∞} ∫_{X} f_{n}dμ \end{matrix}

and therefore

lim sup_{n→∞} ∫_{X} f_{n}dμ ≤ ∫_{X} fdμ ≤ lim inf_{n→∞} ∫_{X} f_{n}dμ.

This shows that lim_{n→∞} ∫_{X} f_{n}dμ exists and is equal to ∫_{X} fdμ.

Exercise 19.1. Give another proof of Proposition 19.34 by first proving Eq. (19.13) with f being a cylinder function in which case the triangle inequality for complex numbers will do the trick. Then use the approximation Theorem 18.42 along with the dominated convergence Theorem 19.38 to handle the general case.

Corollary 19.39. Let {f_{n}}_{n=1}^{∞} ⊂ L^{1} (μ) be a sequence such that ∑_{n=1}^{∞} ∥f_{n}∥_{L^{1}(μ)} < ∞, then ∑_{n=1}^{∞} f_{n} is convergent a.e. and

∫_{X} (∑_{n=1}^{∞} f_{n}) dμ = ∑_{n=1}^{∞} ∫_{X} f_{n}dμ.

Proof. The condition ∑_{n=1}^{∞} ∥f_{n}∥_{L^{1}(μ)} < ∞ is equivalent to ∑_{n=1}^{∞} |f_{n}| ∈ L^{1} (μ) . Hence ∑_{n=1}^{∞} f_{n} is almost everywhere convergent and if S_{N} := ∑_{n=1}^{N} f_{n}, then

|S_{N}| ≤ ∑_{n=1}^{N} |f_{n}| ≤ ∑_{n=1}^{∞} |f_{n}| ∈ L^{1} (μ) .

So by the dominated convergence theorem,

∫_{X} (∑_{n=1}^{∞} f_{n}) dμ = ∫_{X} lim_{N→∞} S_{N}dμ = lim_{N→∞} ∫_{X} S_{N}dμ

= lim_{N→∞} ∑_{n=1}^{N} ∫_{X} f_{n}dμ = ∑_{n=1}^{∞} ∫_{X} f_{n}dμ.

Theorem 19.40 (The Fundamental Theorem of Calculus). Suppose −∞ < a < b < ∞, f ∈ C((a, b), ℝ)∩L^{1}((a, b), m) and F (x) := ∫_{a}^{x} f(y)dm(y). Then

1. F ∈ C([a, b], ℝ) ∩ C^{1}((a, b), ℝ).

2. F′(x) = f(x) for all x ∈ (a, b).

3. If G ∈ C([a, b], ℝ) ∩ C^{1}((a, b), ℝ) is an anti-derivative of f on (a, b) (i.e. f = G′|_{(a,b)}) then

∫_{a}^{b} f(x)dm(x) = G(b) − G(a).

Proof. Since F (x) := ∫_{ℝ} 1_{(a,x)}(y)f(y)dm(y), lim_{x→z} 1_{(a,x)}(y) = 1_{(a,z)}(y) for m – a.e. y and |1_{(a,x)}(y)f(y)| ≤ 1_{(a,b)}(y) |f(y)| is an L^{1} – function, it follows from the dominated convergence Theorem 19.38 that F is continuous on [a, b]. Simple manipulations show,

|\frac{F (x + h) − F (x)}{h} − f(x)| = \frac{1}{|h|} \{ \begin{matrix}|∫_{x}^{x+h} [f(y) − f(x)] dm(y)| & if & h > 0 \\ |∫_{x+h}^{x} [f(y) − f(x)] dm(y)| & if & h < 0 \end{matrix}

≤ \frac{1}{|h|} \{ \begin{matrix}∫_{x}^{x+h} |f(y) − f(x)| dm(y) & if & h > 0 \\ ∫_{x+h}^{x} |f(y) − f(x)| dm(y) & if & h < 0 \end{matrix}

≤ sup {|f(y) − f(x)| : y ∈ [x − |h| , x + |h|]} and the latter expression, by the continuity of f, goes to zero as h → 0 . This shows F′ = f on (a, b). For the converse direction, we have by assumption that G′(x) = F′(x) for x ∈ (a, b). Therefore by the mean value theorem, F − G = C for some constant C. Hence

∫_{a}^{b} f(x)dm(x) = F (b) = F (b) − F (a)
= (G(b) + C) − (G(a) + C) = G(b) − G(a).

Example 19.41. The following limit holds,

lim_{n→∞} ∫_{0}^{n}(1 − \frac{x}{n})^{n}dm(x) = 1.

Let f_{n}(x) = (1 − \frac{x}{n})^{n}1_{[0,n]}(x) and notice that lim_{n→∞} f_{n}(x) = e^{−x}. We will now show

0 ≤ f_{n}(x) ≤ e^{−x} for all x ≥ 0.

It suffices to consider x ∈ [0, n]. Let g(x) = e^{x}f_{n}(x), then for x ∈ (0, n),

\frac{d}{dx} ln g(x) = 1 + n\frac{1}{(1 − \frac{x}{n})}(−\frac{1}{n}) = 1 − \frac{1}{(1 − \frac{x}{n})} ≤ 0

which shows that ln g(x) and hence g(x) is decreasing on [0, n]. Therefore g(x) ≤ g(0) = 1, i.e.

0 ≤ f_{n}(x) ≤ e^{−x}.

From Example 19.24, we know

∫_{0}^{∞} e^{−x}dm(x) = 1 < ∞,

so that e^{−x} is an integrable function on [0, ∞). Hence by the dominated convergence theorem,

lim_{n→∞} ∫_{0}^{n}(1 − \frac{x}{n})^{n}dm(x) = lim_{n→∞} ∫_{0}^{∞} f_{n}(x)dm(x)

= ∫_{0}^{∞} lim_{n→∞} f_{n}(x)dm(x) = ∫_{0}^{∞} e^{−x}dm(x) = 1.

Example 19.42 (Integration of Power Series). Suppose R > 0 and {a_{n}}_{n=0}^{∞} is a sequence of complex numbers such that ∑_{n=0}^{∞} |a_{n}| r^{n} < ∞ for all r ∈ (0, R). Then

∫_{α}^{β} (∑_{n=0}^{∞} a_{n}x^{n}) dm(x) = ∑_{n=0}^{∞} a_{n} ∫_{α}^{β} x^{n}dm(x) = ∑_{n=0}^{∞} a_{n} \frac{β^{n+1} − α^{n+1}}{n + 1}

for all −R < α < β < R. Indeed this follows from Corollary 19.39 since

∑_{n=0}^{∞} ∫_{α}^{β} |a_{n}| |x|^{n} dm(x) ≤ ∑_{n=0}^{∞} (∫_{0}^{|β|} |a_{n}| |x|^{n} dm(x) + ∫_{0}^{|α|} |a_{n}| |x|^{n} dm(x))

≤ ∑_{n=0}^{∞} |a_{n}| \frac{|β|^{n+1} + |α|^{n+1}}{n + 1} ≤ 2r ∑_{n=0}^{∞} |a_{n}| r^{n} < ∞

where r = max(|β| , |α|).

Corollary 19.43 (Differentiation Under the Integral). Suppose that J ⊂ ℝ is an open interval and f : J × X → ℂ is a function such that

1. x → f(t, x) is measurable for each t ∈ J.

2. f(t_{0}, ·) ∈ L^{1}(μ) for some t_{0} ∈ J.

3. \frac{∂f}{∂t} (t, x) exists for all (t, x).

4. There is a function g ∈ L^{1} (μ) such that |\frac{∂f}{∂t} (t, ·)| ≤ g ∈ L^{1} (μ) for each t ∈ J. Then f(t, ·) ∈ L^{1} (μ) for all t ∈ J (i.e. ∫_{X} |f(t, x)| dμ(x) < ∞), t → ∫_{X} f(t, x)dμ(x) is a differentiable function on J and

\frac{d}{dt} ∫_{X} f(t, x)dμ(x) = ∫_{X} \frac{∂f}{∂t} (t, x)dμ(x).

Proof. (The proof is essentially the same as for sums.) By considering the real and imaginary parts of f separately, we may assume that f is real. Also notice that

\frac{∂f}{∂t} (t, x) = lim_{n→∞} n(f(t + n^{−1}, x) − f(t, x))

and therefore, for x → \frac{∂f}{∂t} (t, x) is a sequential limit of measurable functions and hence is measurable for all t ∈ J. By the mean value theorem,

|f(t, x) − f(t_{0}, x)| ≤ g(x) |t − t_{0}| for all t ∈ J  (19.14)

and hence

|f(t, x)| ≤ |f(t, x) − f(t_{0}, x)| + |f(t_{0}, x)| ≤ g(x) |t − t_{0}| + |f(t_{0}, x)| .

This shows f(t, ·) ∈ L^{1} (μ) for all t ∈ J. Let G(t) := ∫_{X} f(t, x)dμ(x), then

\frac{G(t) − G(t_{0})}{t − t_{0}} = ∫_{X} \frac{f(t, x) − f(t_{0}, x)}{t − t_{0}}dμ(x).

By assumption,

lim_{t→t_{0}} \frac{f(t, x) − f(t_{0}, x)}{t − t_{0}} = \frac{∂f}{∂t} (t, x) for all x ∈ X

328 19 Measures and Integration

and by Eq. (19.14),

|\frac{f(t, x) − f(t_{0}, x)}{t − t_{0}} | ≤ g(x) for all t ∈ J and x ∈ X.

Therefore, we may apply the dominated convergence theorem to conclude

lim_{n→∞} \frac{G(t_{n}) − G(t_{0})}{t_{n} − t_{0}} = lim_{n→∞} ∫_{X} \frac{f(t_{n}, x) − f(t_{0}, x)}{t_{n} − t_{0}}dμ(x)

= ∫_{X} lim_{n→∞} \frac{f(t_{n}, x) − f(t_{0}, x)}{t_{n} − t_{0}}dμ(x)

= ∫_{X} \frac{∂f}{∂t} (t_{0}, x)dμ(x)

for all sequences t_{n} ∈ J \ {t_{0}} such that t_{n} → t_{0}. Therefore, Ġ(t_{0}) = lim_{t→t_{0}} \frac{G(t)−G(t_{0})}{t−t_{0}} exists and

Ġ(t_{0}) = ∫_{X} \frac{∂f}{∂t} (t_{0}, x)dμ(x).

Example 19.44. Recall from Example 19.24 that λ^{−1} = ∫_{[0,∞)} e^{−λx}dm(x) for all λ > 0.

Let ε > 0. For λ ≥ 2ε > 0 and n ∈ ℕ there exists C_{n}(ε) < ∞ such that

0 ≤ (−\frac{d}{dλ})^{n} e^{−λx} = x^{n}e^{−λx} ≤ C(ε)e^{−εx}.

Using this fact, Corollary 19.43 and induction gives

n!λ^{−n−1} = (−\frac{d}{dλ})^{n} λ^{−1} = ∫_{[0,∞)} (−\frac{d}{dλ})^{n} e^{−λx}dm(x)
= ∫_{[0,∞)} x^{n}e^{−λx}dm(x).

That is n! = λ^{n} ∫_{[0,∞)} x^{n}e^{−λx}dm(x). Recall that

Γ (t) := ∫_{[0,∞)} x^{t−1}e^{−x}dx for t > 0.

(The reader should check that Γ (t) < ∞ for all t > 0.) We have just shown that Γ (n + 1) = n! for all n ∈ ℕ.

Remark 19.45. Corollary 19.43 may be generalized by allowing the hypothesis to hold for x ∈ X \ E where E ∈ M is a fixed null set, i.e. E must be independent of t. Consider what happens if we formally apply Corollary 19.43 to g(t) := ∫_{0}^{∞} 1_{x≤t}dm(x),

ġ(t) = \frac{d}{dt} ∫_{0}^{∞} 1_{x≤t}dm(x) =^{?} ∫_{0}^{∞} \frac{∂}{∂t}1_{x≤t}dm(x).

The last integral is zero since \frac{∂}{∂t}1_{x≤t} = 0 unless t = x in which case it is not defined. On the other hand g(t) = t so that ġ(t) = 1. (The reader should decide which hypothesis of Corollary 19.43 has been violated in this example.)

19.5 Measurability on Complete Measure Spaces

In this subsection we will discuss a couple of measurability results concerning completions of measure spaces.

Proposition 19.46. Suppose that (X, M, μ) is a complete measure space^{1} and f : X → ℝ is measurable.

1. If g : X → ℝ is a function such that f(x) = g(x) for μ – a.e. x, then g is measurable.

2. If f_{n} : X → ℝ are measurable and f : X → ℝ is a function such that lim_{n→∞} f_{n} = f, μ - a.e., then f is measurable as well.

Proof. 1. Let E = {x : f(x) ≠ g(x)} which is assumed to be in M and μ(E) = 0. Then g = 1_{E^{c}}f + 1_{E}g since f = g on E^{c}. Now 1_{E^{c}}f is measurable so g will be measurable if we show 1_{E}g is measurable. For this consider,

(1_{E}g)^{−1}(A) = \{ \begin{matrix}E^{c} ∪ (1_{E}g)^{−1}(A \ \{0\}) & if 0 ∈ A \\ (1_{E}g)^{−1}(A) & if 0 ∉ A \end{matrix}  (19.15)

Since (1_{E}g)^{−1}(B) ⊂ E if 0 ∉ B and μ(E) = 0, it follow by completeness of M that (1_{E}g)^{−1}(B) ∈ M if 0 ∉ B. Therefore Eq. (19.15) shows that 1_{E}g is measurable. 2. Let E = {x : lim_{n→∞} f_{n}(x) ≠ f(x)} by assumption E ∈ M and μ(E) = 0. Since g := 1_{E}f = lim_{n→∞} 1_{E^{c}}f_{n}, g is measurable. Because f = g on E^{c} and μ(E) = 0, f = g a.e. so by part 1. f is also measurable.

The above results are in general false if (X, M, μ) is not complete. For example, let X = {0, 1, 2}, M = {{0}, {1, 2}, X, ϕ} and μ = δ_{0}. Take g(0) = 0, g(1) = 1, g(2) = 2, then g = 0 a.e. yet g is not measurable.

1 Recall this means that if N ⊂ X is a set such that N ⊂ A ∈ M and μ(A) = 0, then N ∈ M as well.

Lemma 19.47. Suppose that (X, M, μ) is a measure space and M̄ is the completion of M relative to μ and μ̄ is the extension of μ to M̄. Then a function f : X → ℝ is (M̄, B = B_{ℝ}) – measurable iff there exists a function g : X → ℝ that is (M, B) – measurable such E = {x : f(x) ≠ g(x)} ∈ M̄ and μ̄ (E) = 0, i.e. f(x) = g(x) for μ̄ – a.e. x. Moreover for such a pair f and g, f ∈ L^{1}(μ̄) iff g ∈ L^{1}(μ) and in which case

∫_{X} fdμ̄ = ∫_{X} gdμ.

Proof. Suppose first that such a function g exists so that μ̄(E) = 0. Since g is also (M̄, B) – measurable, we see from Proposition 19.46 that f is (M̄, B)

– measurable. Conversely if f is (M̄, B) – measurable, by considering f_{±} we may assume that f ≥ 0. Choose (M̄, B) – measurable simple function ϕ_{n} ≥ 0 such that ϕ_{n} ↑ f as n → ∞. Writing

ϕ_{n} = ∑ a_{k}1_{A_{k}}

with A_{k} ∈ M̄, we may choose B_{k} ∈ M such that B_{k} ⊂ A_{k} and μ̄(A_{k}\B_{k}) = 0. Letting

ϕ̃_{n} := ∑ a_{k}1_{B_{k}}

we have produced a (M, B) – measurable simple function ϕ̃_{n} ≥ 0 such that E_{n} := {ϕ_{n} ≠ ϕ̃_{n}} has zero μ̄ – measure. Since μ̄ (∪_{n}E_{n}) ≤ ∑_{n} μ̄ (E_{n}) , there exists F ∈ M such that ∪_{n}E_{n} ⊂ F and μ(F) = 0. It now follows that

1_{F} ϕ̃_{n} = 1_{F} ϕ_{n} ↑ g := 1_{F} f as n → ∞.

This shows that g = 1_{F} f is (M, B) – measurable and that {f ≠ g} ⊂ F has μ̄ – measure zero. Since f = g, μ̄ – a.e., ∫_{X} fdμ̄ = ∫_{X} gdμ̄ so to prove Eq. (19.16) it suffices to prove

∫_{X} gdμ̄ = ∫_{X} gdμ.  (19.16)

Because μ̄ = μ on M, Eq. (19.16) is easily verified for non-negative M – measurable simple functions. Then by the monotone convergence theorem and the approximation Theorem 18.42 it holds for all M – measurable functions g : X → [0, ∞]. The rest of the assertions follow in the standard way by considering (Re g)_{±} and (Im g)_{±} .

19.6 Comparison of the Lebesgue and the Riemann Integral

For the rest of this chapter, let −∞ < a < b < ∞ and f : [a, b] → ℝ be a bounded function. A partition of [a, b] is a finite subset π ⊂ [a, b] containing {a, b}. To each partition

π = {a = t_{0} < t_{1} < ⋯ < t_{n} = b}  (19.17)

of [a, b] let

mesh(π) := max{|t_{j} − t_{j−1}| : j = 1, …, n},
M_{j} = sup{f(x) : t_{j} ≤ x ≤ t_{j−1}}, m_{j} = inf{f(x) : t_{j} ≤ x ≤ t_{j−1}}
G_{π} = f(a)1_{{a}} + ∑_{1}^{n} M_{j}1_{(t_{j−1},t_{j}]}, g_{π} = f(a)1_{{a}} + ∑_{1}^{n} m_{j}1_{(t_{j−1},t_{j}]} and

S_{π}f = ∑ M_{j}(t_{j} − t_{j−1}) and s_{π}f = ∑ m_{j}(t_{j} − t_{j−1}).

Notice that

S_{π}f = ∫_{a}^{b} G_{π}dm and s_{π}f = ∫_{a}^{b} g_{π}dm.

The upper and lower Riemann integrals are defined respectively by

\overline{∫{b \atop a}}f(x)dx = inf_{π} S_{π}f and \underline{∫} _{b}^{a}f(x)dx = sup_{π} s_{π}f.

Definition 19.48. The function f is Riemann integrable iff \overline{∫{b \atop a}}f = \underline{∫} _{a}^{b}f ∈

ℝ and which case the Riemann integral ∫_{a}^{b} f is defined to be the common value:

∫_{a}^{b} f(x)dx = \overline{∫{b \atop a}}f(x)dx = \underline{∫{b \atop a}}f(x)dx.

The proof of the following Lemma is left to the reader as Exercise 19.20.

Lemma 19.49. If π′ and π are two partitions of [a, b] and π ⊂ π′ then

G_{π} ≥ G_{π′} ≥ f ≥ g_{π′} ≥ g_{π} and
S_{π}f ≥ S_{π′}f ≥ s_{π′}f ≥ s_{π}f.

There exists an increasing sequence of partitions {π_{k}}_{k=1}^{∞} such that mesh(π_{k}) ↓ 0 and

S_{π_{k}}f ↓ \overline{∫{b \atop a}}f and s_{π_{k}}f ↑ \underline{∫} _{a}^{b}f as k → ∞.

If we let

G := lim_{k→∞} G_{π_{k}} and g := lim_{k→∞} g_{π_{k}}  (19.18)

then by the dominated convergence theorem,

∫_{[a,b]} gdm = lim_{k→∞} ∫_{[a,b]} g_{π_{k}} = lim_{k→∞} s_{π_{k}}f = \underline{∫{b \atop a}}f(x)dx  (19.19)

and

∫_{[a,b]} Gdm = lim_{k→∞} ∫_{[a,b]} G_{π_{k}} = lim_{k→∞} S_{π_{k}}f = \overline{∫{b \atop a}}f(x)dx.  (19.20)

Notation 19.50 For x ∈ [a, b], let

H(x) = lim sup_{y→x} f(y) := lim_{ε↓0} sup{f(y) : |y − x| ≤ ε, y ∈ [a, b]} and
h(x) = lim inf_{y→x} f(y) := lim_{ε↓0} inf {f(y) : |y − x| ≤ ε, y ∈ [a, b]}.

Lemma 19.51. The functions H, h : [a, b] → ℝ satisfy:

1. h(x) ≤ f(x) ≤ H(x) for all x ∈ [a, b] and h(x) = H(x) iff f is continuous at x.

2. If {π_{k}}_{k=1}^{∞} is any increasing sequence of partitions such that mesh(π_{k}) ↓ 0 and G and g are defined as in Eq. (19.18), then

G(x) = H(x) ≥ f(x) ≥ h(x) = g(x) ∀ x ∉ π := ∪_{k=1}^{∞}π_{k}. (19.21)

(Note π is a countable set.)

3. H and h are Borel measurable.

Proof. Let G_{k} := G_{π_{k}} ↓ G and g_{k} := g_{π_{k}} ↑ g.

1. It is clear that h(x) ≤ f(x) ≤ H(x) for all x and H(x) = h(x) iff lim_{y→x} f(y)

exists and is equal to f(x). That is H(x) = h(x) iff f is continuous at x.

2. For x ∉ π,

G_{k}(x) ≥ H(x) ≥ f(x) ≥ h(x) ≥ g_{k}(x) ∀ k

and letting k → ∞ in this equation implies

G(x) ≥ H(x) ≥ f(x) ≥ h(x) ≥ g(x) ∀ x ∉ π.  (19.22)

Moreover, given ε > 0 and x ∉ π,

sup{f(y) : |y − x| ≤ ε, y ∈ [a, b]} ≥ G_{k}(x)

for all k large enough, since eventually G_{k}(x) is the supremum of f(y) over some interval contained in [x − ε, x + ε]. Again letting k → ∞ implies sup_{|y−x|≤ε} f(y) ≥ G(x) and therefore, that

H(x) = lim sup_{y→x} f(y) ≥ G(x)

for all x ∉ π. Combining this equation with Eq. (19.22) then implies H(x) = G(x) if x ∉ π. A similar argument shows that h(x) = g(x) if x ∉ π and hence Eq. (19.21) is proved.

3. The functions G and g are limits of measurable functions and hence measurable. Since H = G and h = g except possibly on the countable set π, both H and h are also Borel measurable. (You justify this statement.)

Theorem 19.52. Let f : [a, b] → ℝ be a bounded function. Then

\overline{∫{b \atop a}}f = ∫_{[a,b]} Hdm and \underline{∫{b \atop a}}f = ∫_{[a,b]} hdm  (19.23)

and the following statements are equivalent:

1. H(x) = h(x) for m -a.e. x,

2. the set

E := {x ∈ [a, b] : f is discontinuous at x}

is an m̄ – null set.

3. f is Riemann integrable.

If f is Riemann integrable then f is Lebesgue measurable^{2}, i.e. f is L/B – measurable where L is the Lebesgue σ – algebra and B is the Borel σ – algebra on [a, b]. Moreover if we let m̄ denote the completion of m, then

∫_{[a,b]} Hdm = ∫_{a}^{b} f(x)dx = ∫_{[a,b]} fdm̄ = ∫_{[a,b]} hdm.  (19.24)

Proof. Let {π_{k}}_{k=1}^{∞} be an increasing sequence of partitions of [a, b] as described in Lemma 19.49 and let G and g be defined as in Lemma 19.51. Since m(π) = 0, H = G a.e., Eq. (19.23) is a consequence of Eqs. (19.19) and (19.20). From Eq. (19.23), f is Riemann integrable iff

∫_{[a,b]} Hdm = ∫_{[a,b]} hdm

and because h ≤ f ≤ H this happens iff h(x) = H(x) for m - a.e. x. Since E = {x : H(x) ≠ h(x)}, this last condition is equivalent to E being a m – null set. In light of these results and Eq. (19.21), the remaining assertions including Eq. (19.24) are now consequences of Lemma 19.47.

Notation 19.53 In view of this theorem we will often write ∫_{a}^{b} f(x)dx for ∫_{a}^{b} fdm.

19.7 Determining Classes of Measures

Definition 19.54 (σ – finite). Let X be a set and E ⊂ F ⊂ 2^{X}. We say that a function μ : F → [0, ∞] is σ – finite on E if there exist X_{n} ∈ E such that X_{n} ↑ X and μ(X_{n}) < ∞ for all n.

2 f need not be Borel measurable.

Theorem 19.55 (Uniqueness). Suppose that C ⊂ 2^{X} is a π – class (see Definition 18.53), M = σ(C) and μ and ν are two measure on M. If μ and ν are σ – finite on C and μ = ν on C, then μ = ν on M.

Proof. We begin first with the special case where μ(X) < ∞ and therefore also

ν(X) = lim_{n→∞} ν (X_{n}) = lim_{n→∞} μ (X_{n}) = μ(X) < ∞.

Let

H := {f ∈ ℓ^{∞} (M, ℝ) : μ (f) = ν (f)} .

Then H is a linear subspace which is closed under bounded convergence (by the dominated convergence theorem), contains 1 and contains the multiplicative system, M := {1_{C} : C ∈ C} . Therefore, by Theorem 18.51 or Corollary 18.54, H = ℓ^{∞} (M, ℝ) and hence μ = ν. For the general case, let X_{n}^{1}, X_{n}^{2} ∈ C be chosen so that X_{n}^{1} ↑ X and X_{n}^{2} ↑ X as n → ∞ and μ (X_{n}^{1}) + ν (X_{n}^{2}) < ∞ for all n. Then X_{n} := X_{n}^{1} ∩ X_{n}^{2} ∈ C increases to X and ν (X_{n}) = μ (X_{n}) < ∞ for all n. For each n ∈ ℕ, define two measures μ_{n} and ν_{n} on M by

μ_{n}(A) := μ(A ∩ X_{n}) and ν_{n}(A) = ν(A ∩ X_{n}).

Then, as the reader should verify, μ_{n} and ν_{n} are finite measure on M such that μ_{n} = ν_{n} on C. Therefore, by the special case just proved, μ_{n} = ν_{n} on M. Finally, using the continuity properties of measures,

μ(A) = lim_{n→∞} μ(A ∩ X_{n}) = lim_{n→∞} ν(A ∩ X_{n}) = ν(A)

for all A ∈ M.

As an immediate consequence we have the following corollaries.

Corollary 19.56. Suppose that (X, τ) is a topological space, B_{X} = σ(τ) is the Borel σ – algebra on X and μ and ν are two measures on B_{X} which are σ – finite on τ. If μ = ν on τ then μ = ν on B_{X}, i.e. μ ≡ ν.

Corollary 19.57. Suppose that μ and ν are two measures on B_{ℝ^{n}} which are finite on bounded sets and such that μ(A) = ν(A) for all sets A of the form

A = (a, b] = (a_{1}, b_{1}] × ⋯ × (a_{n}, b_{n}]

with a, b ∈ ℝ^{n} and a < b, i.e. a_{i} < b_{i} for all i. Then μ = ν on B_{ℝ^{n}}.

Proposition 19.58. Suppose that (X, d) is a metric space, μ and ν are two measures on B_{X} := σ(τ_{d}) which are finite on bounded measurable subsets of X and

∫_{X} fdμ = ∫_{X} fdν  (19.25)

for all f ∈ BC_{b}(X, ℝ) where

BC_{b}(X, ℝ) = {f ∈ BC(X, ℝ) : supp(f) is bounded}.  (19.26)

Then μ ≡ ν.

Proof. To prove this fix a o ∈ X and let

ψ_{R}(x) = ([R + 1 − d(x, o)] ∧ 1) ∨ 0  (19.27)

so that ψ_{R} ∈ BC_{b}(X, [0, 1]), supp(ψ_{R}) ⊂ B(o, R + 2) and ψ_{R} ↑ 1 as R → ∞. Let H_{R} denote the space of bounded real valued B_{X} – measurable functions f such that

∫_{X} ψ_{R}fdμ = ∫_{X} ψ_{R}fdν.  (19.28)

Then H_{R} is closed under bounded convergence and because of Eq. (19.25) contains BC(X, ℝ). Therefore by Corollary 18.55, H_{R} contains all bounded measurable functions on X. Take f = 1_{A} in Eq. (19.28) with A ∈ B_{X}, and then use the monotone convergence theorem to let R → ∞. The result is μ(A) = ν(A) for all A ∈ B_{X}.

Here is another version of Proposition 19.58.

Proposition 19.59. Suppose that (X, d) is a metric space, μ and ν are two measures on B_{X} = σ(τ_{d}) which are both finite on compact sets. Further assume there exists compact sets K_{k} ⊂ X such that K_{k}^{o} ↑ X. If

∫_{X} fdμ = ∫_{X} fdν  (19.29)

for all f ∈ C_{c}(X, ℝ) then μ ≡ ν.

Proof. Let ψ_{n,k} be defined as in the proof of Proposition 18.56 and let H_{n,k} denote those bounded B_{X} – measurable functions, f : X → ℝ such that

∫_{X} fψ_{n,k}dμ = ∫_{X} fψ_{n,k}dν.

By assumption BC(X, ℝ) ⊂ H_{n,k} and one easily checks that H_{n,k} is closed under bounded convergence. Therefore, by Corollary 18.55, H_{n,k} contains all bounded measurable function. In particular for A ∈ B_{X},

∫_{X} 1_{A}ψ_{n,k}dμ = ∫_{X} 1_{A}ψ_{n,k}dν.

Letting n → ∞ in this equation, using the dominated convergence theorem, one shows

∫_{X} 1_{A}1_{K_{k}^{o}}dμ = ∫_{X} 1_{A}1_{K_{k}^{o}}dν

holds for k. Finally using the monotone convergence theorem we may let k → ∞ to conclude

μ(A) = ∫_{X} 1_{A}dμ = ∫_{X} 1_{A}dν = ν(A)

for all A ∈ B_{X}.

19.8 Exercises

Exercise 19.2. Let μ be a measure on an algebra A ⊂ 2^{X}, then μ(A) + μ(B) = μ(A ∪ B) + μ(A ∩ B) for all A, B ∈ A.

Exercise 19.3 (From problem 12 on p. 27 of Folland.). Let (X, M, μ) be a finite measure space and for A, B ∈ M let ρ(A, B) = μ(AΔB) where AΔB = (A \ B) ∪ (B \ A) . It is clear that ρ (A, B) = ρ (B, A) . Show:

1. ρ satisfies the triangle inequality:

ρ (A, C) ≤ ρ (A, B) + ρ (B, C) for all A, B, C ∈ M.

2. Define A ∼ B iff μ(AΔB) = 0 and notice that ρ (A, B) = 0 iff A ∼ B. Show “∼ ” is an equivalence relation.

3. Let M/ ∼ denote M modulo the equivalence relation, ∼, and let [A] := {B ∈ M : B ∼ A} . Show that ρ̄([A] , [B]) := ρ (A, B) is gives a well defined metric on M/ ∼ .

4. Similarly show μ̃ ([A]) = μ (A) is a well defined function on M/ ∼ and show μ̃ : (M/ ∼) → ℝ_{+} is ρ̄ – continuous.

Exercise 19.4. Suppose that μ_{n} : M → [0, ∞] are measures on M for n ∈ ℕ. Also suppose that μ_{n}(A) is increasing in n for all A ∈ M. Prove that μ : M → [0, ∞] defined by μ(A) := lim_{n→∞} μ_{n}(A) is also a measure.

Exercise 19.5. Now suppose that Λ is some index set and for each λ ∈ Λ, μ_{λ} : M → [0, ∞] is a measure on M. Define μ : M → [0, ∞] by μ(A) = ∑_{λ∈Λ} μ_{λ}(A) for each A ∈ M. Show that μ is also a measure.

Exercise 19.6. Let (X, M, μ) be a measure space and ρ : X → [0, ∞] be a measurable function. For A ∈ M, set ν(A) := ∫_{A} ρdμ.

1. Show ν : M → [0, ∞] is a measure.

2. Let f : X → [0, ∞] be a measurable function, show

∫_{X} fdν = ∫_{X} fρdμ.  (19.30)

Hint: first prove the relationship for characteristic functions, then for simple functions, and then for general positive measurable functions.

3. Show that a measurable function f : X → ℂ is in L^{1}(ν) iff |f| ρ ∈ L^{1}(μ) and if f ∈ L^{1}(ν) then Eq. (19.30) still holds.

Notation 19.60 It is customary to informally describe ν defined in Exercise 19.6 by writing dν = ρdμ.

Exercise 19.7. Let (X, M, μ) be a measure space, (Y, F) be a measurable space and f : X → Y be a measurable map. Define a function ν : F → [0, ∞] by ν(A) := μ(f^{−1}(A)) for all A ∈ F.

1. Show ν is a measure. (We will write ν = f_{∗}μ or ν = μ ∘ f^{−1}.)

2. Show

∫_{Y} gdν = ∫_{X} (g ∘ f) dμ  (19.31)

for all measurable functions g : Y → [0, ∞]. Hint: see the hint from Exercise 19.6.

3. Show a measurable function g : Y → ℂ is in L^{1}(ν) iff g ∘ f ∈ L^{1}(μ) and that Eq. (19.31) holds for all g ∈ L^{1}(ν).

Exercise 19.8. Let F : ℝ → ℝ be a C^{1}-function such that F′(x) > 0 for all x ∈ ℝ and lim_{x→±∞} F (x) = ±∞. (Notice that F is strictly increasing so that F^{−1} : ℝ → ℝ exists and moreover, by the inverse function theorem that F^{−1} is a C^{1} – function.) Let m be Lebesgue measure on B_{ℝ} and

ν(A) = m(F (A)) = m((F^{−1})^{−1} (A)) = (F_{∗}^{−1}m) (A)

for all A ∈ B_{ℝ}. Show dν = F′dm. Use this result to prove the change of variable formula,

∫_{ℝ} h ∘ F · F′dm = ∫_{ℝ} hdm  (19.32)

which is valid for all Borel measurable functions h : ℝ → [0, ∞].

Hint: Start by showing dν = F′dm on sets of the form A = (a, b] with a, b ∈ ℝ and a < b. Then use the uniqueness assertions in Theorem 19.8 (or see Corollary 19.57) to conclude dν = F′dm on all of B_{ℝ}. To prove Eq. (19.32) apply Exercise 19.7 with g = h ∘ F and f = F^{−1}.

Exercise 19.9. Let (X, M, μ) be a measure space and {A_{n}}_{n=1}^{∞} ⊂ M, show

μ({A_{n} a.a.}) ≤ lim inf_{n→∞} μ (A_{n})

and if μ (∪_{m≥n}A_{m}) < ∞ for some n, then

μ({A_{n} i.o.}) ≥ lim sup_{n→∞} μ (A_{n}) .

Exercise 19.10. BRUCE: Delete this exercise which is contained in Lemma 19.17. Suppose (X, M, μ) be a measure space and f : X → [0 ∞] be a measurable function such that ∫_{X} fdμ < ∞. Show μ ({f = ∞}) = 0 and the set {f > 0} is σ – finite.

Exercise 19.11. Folland 2.13 on p. 52. Hint: “Fatou times two.”

Exercise 19.12. Folland 2.14 on p. 52. BRUCE: delete this exercise

Exercise 19.13. Give examples of measurable functions {f_{n}} on ℝ such that f_{n} decreases to 0 uniformly yet ∫ f_{n}dm = ∞ for all n. Also give an example of a sequence of measurable functions {g_{n}} on [0, 1] such that g_{n} → 0 while ∫ g_{n}dm = 1 for all n.

Exercise 19.14. Folland 2.19 on p. 59. (This problem is essentially covered in the previous exercise.)

Exercise 19.15. Suppose {a_{n}}_{n=−∞}^{∞} ⊂ ℂ is a summable sequence (i.e. ∑_{n=−∞}^{∞} |a_{n}| < ∞), then f(θ) := ∑_{n=−∞}^{∞} a_{n}e^{inθ} is a continuous function for θ ∈ ℝ and

a_{n} = \frac{1}{2π} ∫_{−π}^{π} f(θ)e^{−inθ}dθ.

Exercise 19.16. For any function f ∈ L^{1} (m) , show x ∈ ℝ → ∫_{(−∞,x]} f (t) dm (t) is continuous in x. Also find a finite measure, μ, on B_{ℝ} such that x → ∫_{(−∞,x]} f (t) dμ (t) is not continuous.

Exercise 19.17. Folland 2.28 on p. 60.

Exercise 19.18. Folland 2.31b and 2.31e on p. 60. (The answer in 2.13b is wrong by a factor of −1 and the sum is on k = 1 to ∞. In part e, s should be taken to be a. You may also freely use the Taylor series expansion

(1 − z)^{−1/2} = ∑_{n=0}^{∞} \frac{(2n − 1)!!}{2^{n}n!}z^{n} = ∑_{n=0}^{∞} \frac{(2n)!}{4^{n} (n!)^{2}} z^{n} for |z| < 1.

Exercise 19.19. There exists a meager (see Definition 13.4 and Proposition 13.3) subsets of ℝ which have full Lebesgue measure, i.e. whose complement is a Lebesgue null set. (This is Folland 5.27. Hint: Consider the generalized Cantor sets discussed on p. 39 of Folland.)

Exercise 19.20. Prove Lemma 19.49.

Multiple Integrals

In this chapter we will introduce iterated integrals and product measures. We are particularly interested in when it is permissible to interchange the order of integration in multiple integrals.

Example 20.1. As an example let X = [1, ∞) and Y = [0, 1] equipped with their Borel σ - algebras and let μ = ν = m, where m is Lebesgue measure. The iterated integrals of the function f (x, y) := e^{−xy} − 2e^{−2xy} satisfy,

∫_{0}^{1} [∫_{1}^{∞}(e^{−xy} − 2e^{−2xy})dx] dy = ∫_{0}^{1} e^{−y} (\frac{1 − e^{−y}}{y} ) dy ∈ (0, ∞)

and

∫_{1}^{∞} [∫_{0}^{1}(e^{−xy} − 2e^{−2xy})dy] dx = − ∫_{1}^{∞} e^{−x} [\frac{1 − e^{−x}}{x} ] dx ∈ (−∞, 0)

and therefore are not equal. Hence it is not always true that order of integration is irrelevant.

Lemma 20.2. Let 𝔽 be either [0, ∞), ℝ or ℂ. Suppose (X, M) and (Y, N) are two measurable spaces and f : X × Y → 𝔽 is a (M ⊗ N , B_{𝔽}) – measurable function, then for each y ∈ Y,

x → f(x, y) is (M, B_{𝔽}) measurable,  (20.1)

for each x ∈ X,

y → f(x, y) is (N , B_{𝔽}) measurable.  (20.2)

Proof. Suppose that E = A × B ∈ E := M × N and f = 1_{E}. Then

f(x, y) = 1_{A×B}(x, y) = 1_{A}(x)1_{B}(y)

from which it follows that Eqs. (20.1) and (20.2) for this function. Let H be the collection of all bounded (M ⊗ N , B_{𝔽}) – measurable functions on X × Y such that Eqs. (20.1) and (20.2) hold, here we assume 𝔽 = ℝ or ℂ. Because measurable functions are closed under taking linear combinations and pointwise limits, H is linear subspace of ℓ^{∞} (M ⊗ N , 𝔽) which is closed under bounded convergence and contain 1_{E} ∈ H for all E in the π – class, E. Therefore by by Corollary 18.54, that H = ℓ^{∞} (M ⊗ N , 𝔽) .

For the general (M ⊗ N , B_{ℝ}) – measurable functions f : X × Y → 𝔽 and M ∈ ℕ, let f_{M} := 1_{|f|≤M}f ∈ ℓ^{∞} (M ⊗ N , 𝔽) . Then Eqs. (20.1) and (20.2) hold with f replaced by f_{M} and hence for f as well by letting M → ∞.

Notation 20.3 (Iterated Integrals) If (X, M, μ) and (Y, N , ν) are two measure spaces and f : X × Y → ℂ is a M ⊗ N – measurable function, the iterated integrals of f (when they make sense) are:

∫_{X} dμ(x) ∫_{Y} dν(y)f(x, y) := ∫_{X} [∫_{Y} f(x, y)dν(y)] dμ(x)

and

∫_{Y} dν(y) ∫_{X} dμ(x)f(x, y) := ∫_{Y} [∫_{X} f(x, y)dμ(x)] dν(y).

Notation 20.4 Suppose that f : X → ℂ and g : Y → ℂ are functions, let f ⊗ g denote the function on X × Y given by

f ⊗ g(x, y) = f(x)g(y).

Notice that if f, g are measurable, then f ⊗g is (M ⊗ N , B_{ℂ}) – measurable. To prove this let F (x, y) = f(x) and G(x, y) = g(y) so that f ⊗ g = F · G will be measurable provided that F and G are measurable. Now F = f ∘ π_{1} where π_{1} : X × Y → X is the projection map. This shows that F is the composition of measurable functions and hence measurable. Similarly one shows that G is measurable.

20.1 Fubini-Tonelli’s Theorem and Product Measure

Theorem 20.5. Suppose (X, M, μ) and (Y, N , ν) are σ-finite measure spaces and f is a nonnegative (M ⊗ N , B_{ℝ}) – measurable function, then for each y ∈ Y,

x → f(x, y) is M – B_{[0,∞]} measurable,  (20.3)

for each x ∈ X,

y → f(x, y) is N – B_{[0,∞]} measurable,  (20.4)

x → ∫_{Y} f(x, y)dν(y) is M – B_{[0,∞]} measurable,  (20.5)
y → ∫_{X} f(x, y)dμ(x) is N – B_{[0,∞]} measurable,  (20.6)

and

∫_{X} dμ(x) ∫_{Y} dν(y)f(x, y) = ∫_{Y} dν(y) ∫_{X} dμ(x)f(x, y).  (20.7)

Proof. Suppose that E = A × B ∈ E := M × N and f = 1_{E}. Then

f(x, y) = 1_{A×B}(x, y) = 1_{A}(x)1_{B}(y)

and one sees that Eqs. (20.3) and (20.4) hold. Moreover

∫_{Y} f(x, y)dν(y) = ∫_{Y} 1_{A}(x)1_{B}(y)dν(y) = 1_{A}(x)ν(B),

so that Eq. (20.5) holds and we have

∫_{X} dμ(x) ∫_{Y} dν(y)f(x, y) = ν(B)μ(A).  (20.8)

Similarly,

∫_{X} f(x, y)dμ(x) = μ(A)1_{B}(y) and
∫_{Y} dν(y) ∫_{X} dμ(x)f(x, y) = ν(B)μ(A)

from which it follows that Eqs. (20.6) and (20.7) hold in this case as well. For the moment let us further assume that μ(X) < ∞ and ν(Y) < ∞ and let H be the collection of all bounded (M ⊗ N , B_{ℝ}) – measurable functions on X × Y such that Eqs. (20.3) – (20.7) hold. Using the fact that measurable functions are closed under pointwise limits and the dominated convergence theorem (the dominating function always being a constant), one easily shows that H closed under bounded convergence. Since we have just verified that 1_{E} ∈ H for all E in the π – class, E, it follows by Corollary 18.54 that H is the space of all bounded (M ⊗ N , B_{ℝ}) – measurable functions on X ×Y. Finally if f : X × Y → [0, ∞] is a (M ⊗ N , Bℝ̄) – measurable function, let f_{M} = M ∧ f so that f_{M} ↑ f as M → ∞ and Eqs. (20.3) – (20.7) hold with f replaced by f_{M} for all M ∈ ℕ. Repeated use of the monotone convergence theorem allows us to pass to the limit M → ∞ in these equations to deduce the theorem in the case μ and ν are finite measures. For the σ – finite case, choose X_{n} ∈ M, Y_{n} ∈ N such that X_{n} ↑ X, Y_{n} ↑ Y, μ(X_{n}) < ∞ and ν(Y_{n}) < ∞ for all m, n ∈ ℕ. Then define μ_{m}(A) = μ(X_{m} ∩ A) and ν_{n}(B) = ν(Y_{n} ∩ B) for all A ∈ M and B ∈ N or equivalently dμ_{m} = 1_{X_{m}}dμ and dν_{n} = 1_{Y_{n}}dν. By what we have just proved Eqs. (20.3) – (20.7) with μ replaced by μ_{m} and ν by ν_{n} for all (M ⊗ N , Bℝ̄) – measurable functions, f : X × Y → [0, ∞]. The validity of Eqs. (20.3) – (20.7) then follows by passing to the limits m → ∞ and then n → ∞ making use of the monotone convergence theorem in the form,

∫_{X} udμ_{m} = ∫_{X} u1_{X_{m}}dμ ↑ ∫_{X} udμ as m → ∞

and

∫_{Y} vdμ_{n} = ∫_{Y} v1_{Y_{n}}dμ ↑ ∫_{Y} vdμ as n → ∞

for all u ∈ L^{+}(X, M) and v ∈ L^{+}(Y, N).

Corollary 20.6. Suppose (X, M, μ) and (Y, N , ν) are σ – finite measure spaces. Then there exists a unique measure π on M⊗N such that π(A×B) = μ(A)ν(B) for all A ∈ M and B ∈ N . Moreover π is given by

π(E) = ∫_{X} dμ(x) ∫_{Y} dν(y)1_{E}(x, y) = ∫_{Y} dν(y) ∫_{X} dμ(x)1_{E}(x, y) (20.9)

for all E ∈ M ⊗ N and π is σ – finite.

Proof. Notice that any measure π such that π(A × B) = μ(A)ν(B) for all A ∈ M and B ∈ N is necessarily σ – finite. Indeed, let X_{n} ∈ M and Y_{n} ∈ N be chosen so that μ(X_{n}) < ∞, ν(Y_{n}) < ∞, X_{n} ↑ X and Y_{n} ↑ Y, then X_{n} × Y_{n} ∈ M ⊗ N , X_{n} × Y_{n} ↑ X × Y and π(X_{n} × Y_{n}) < ∞ for all n. The uniqueness assertion is a consequence of Theorem 19.55 or see Theorem ?? below with E = M × N . For the existence, it suffices to observe, using the monotone convergence theorem, that π defined in Eq. (20.9) is a measure on M ⊗ N . Moreover this measure satisfies π(A × B) = μ(A)ν(B) for all A ∈ M and B ∈ N from Eq. (20.8

Notation 20.7 The measure π is called the product measure of μ and ν and will be denoted by μ ⊗ ν.

Theorem 20.8 (Tonelli’s Theorem). Suppose (X, M, μ) and (Y, N , ν) are σ – finite measure spaces and π = μ ⊗ ν is the product measure on M ⊗ N . If f ∈ L^{+}(X × Y, M ⊗ N), then f(·, y) ∈ L^{+}(X, M) for all y ∈ Y, f(x, ·) ∈ L^{+}(Y, N) for all x ∈ X,

∫_{Y} f(·, y)dν(y) ∈ L^{+}(X, M), ∫_{X} f(x, ·)dμ(x) ∈ L^{+}(Y, N)

and

∫_{X×Y} f dπ = ∫_{X} dμ(x) ∫_{Y} dν(y)f(x, y)  (20.10)
= ∫_{Y} dν(y) ∫_{X} dμ(x)f(x, y).  (20.11)

Proof. By Theorem 20.5 and Corollary 20.6, the theorem holds when f = 1_{E} with E ∈ M ⊗ N . Using the linearity of all of the statements, the theorem is also true for non-negative simple functions. Then using the monotone convergence theorem repeatedly along with the approximation Theorem 18.42, one deduces the theorem for general f ∈ L^{+}(X × Y, M ⊗ N).

The following convention will be in force for the rest of this chapter.

Convention: If (X, M, μ) is a measure space and f : X → ℂ is a measurable but non-integrable function, i.e. ∫_{X} |f| dμ = ∞, by convention we will define ∫_{X} fdμ := 0. However if f is a non-negative function (i.e. f : X → [0, ∞]) is a non-integrable function we will still write ∫_{X} fdμ = ∞.

Theorem 20.9 (Fubini’s Theorem). Suppose (X, M, μ) and (Y, N , ν) are σ – finite measure spaces, π = μ ⊗ ν is the product measure on M ⊗ N and f : X × Y → ℂ is a M ⊗ N – measurable function. Then the following three conditions are equivalent:

∫_{X×Y} |f| dπ < ∞, i.e. f ∈ L^{1}(π),  (20.12)
∫_{X} (∫_{Y} |f(x, y)| dν(y)) dμ(x) < ∞ and  (20.13)
∫_{Y} (∫_{X} |f(x, y)| dμ(x)) dν(y) < ∞.  (20.14)

If any one (and hence all) of these condition hold, then f(x, ·) ∈ L^{1}(ν) for μ a.e. x, f(·, y) ∈ L^{1}(μ) for ν a.e. y, ∫_{Y} f(·, y)dv(y) ∈ L^{1}(μ), ∫_{X} f(x, ·)dμ(x) ∈ L^{1}(ν) and Eqs. (20.10) and (20.11) are still valid.

Proof. The equivalence of Eqs. (20.12) – (20.14) is a direct consequence of Tonelli’s Theorem 20.8. Now suppose f ∈ L^{1}(π) is a real valued function and let

E := {x ∈ X : ∫_{Y} |f (x, y)| dν (y) = ∞} .  (20.15)

Then by Tonelli’s theorem, x → ∫_{Y} |f (x, y)| dν (y) is measurable and hence E ∈ M. Moreover Tonelli’s theorem implies

∫_{X} [∫_{Y} |f (x, y)| dν (y)] dμ (x) = ∫_{X×Y} |f| dπ < ∞

which implies that μ (E) = 0. Let f_{±} be the positive and negative parts of f, then using the above convention we have

∫_{Y} f (x, y) dν (y) = ∫_{Y} 1_{E} (x) f (x, y) dν (y)
= ∫_{Y} 1_{E} (x) [f_{+} (x, y) − f_{−} (x, y)] dν (y)

= ∫_{Y} 1_{E} (x) f_{+} (x, y) dν (y) − ∫_{Y} 1_{E} (x) f_{−} (x, y) dν (y) . (20.16)

Noting that 1_{E} (x) f_{±} (x, y) = (1_{E} ⊗ 1_{Y} · f_{±}) (x, y) is a positive M ⊗ N – measurable function, it follows from another application of Tonelli’s theorem that x → ∫_{Y} f (x, y) dν (y) is M – measurable, being the difference of two measurable functions. Moreover

∫_{X} |∫_{Y} f (x, y) dν (y)| dμ (x) ≤ ∫_{X} [∫_{Y} |f (x, y)| dν (y)] dμ (x) < ∞,

which shows ∫_{Y} f(·, y)dv(y) ∈ L^{1}(μ). Integrating Eq. (20.16) on x and using Tonelli’s theorem repeatedly implies,

∫X [∫_{Y} f (x, y) dν (y)] dμ (x)

= ∫_{X} dμ (x) ∫_{Y} dν (y) 1_{E} (x) f_{+} (x, y) − ∫_{X} dμ (x) ∫_{Y} dν (y) 1_{E} (x) f_{−} (x, y)

= ∫_{Y} dν (y) ∫_{X} dμ (x) 1_{E} (x) f_{+} (x, y) − ∫_{Y} dν (y) ∫_{X} dμ (x) 1_{E} (x) f_{−} (x, y)

= ∫_{Y} dν (y) ∫_{X} dμ (x) f_{+} (x, y) − ∫_{Y} dν (y) ∫_{X} dμ (x) f_{−} (x, y)
= ∫_{X×Y} f_{+}dπ − ∫_{X×Y} f_{−}dπ = ∫_{X×Y} (f_{+} − f_{−}) dπ = ∫_{X×Y} fdπ (20.17)

which proves Eq. (20.10) holds.

Now suppose that f = u + iv is complex valued and again let E be as in Eq. (20.15). Just as above we still have E ∈ M and μ (E) < ∞. By our convention,

∫_{Y} f (x, y) dν (y) = ∫_{Y} 1_{E} (x) f (x, y) dν (y) = ∫_{Y} 1_{E} (x) [u (x, y) + iv (x, y)] dν (y)
= ∫_{Y} 1_{E} (x) u (x, y) dν (y) + i ∫_{Y} 1_{E} (x) v (x, y) dν (y)

which is measurable in x by what we have just proved. Similarly one shows ∫_{Y} f (·, y) dν (y) ∈ L^{1} (μ) and Eq. (20.10) still holds by a computation similar to that done in Eq. (20.17). The assertions pertaining to Eq. (20.11) may be proved in the same way.

Notation 20.10 Given E ⊂ X × Y and x ∈ X, let

xE := {y ∈ Y : (x, y) ∈ E}.

Similarly if y ∈ Y is given let

E_{y} := {x ∈ X : (x, y) ∈ E}.

If f : X × Y → ℂ is a function let f_{x} = f(x, ·) and f^{y} := f(·, y) so that f_{x} : Y → ℂ and f^{y} : X → ℂ.

Theorem 20.11. Suppose (X, M, μ) and (Y, N , ν) are complete σ – finite measure spaces. Let (X × Y, L, λ) be the completion of (X × Y, M ⊗ N , μ ⊗ ν). If f is L – measurable and (a) f ≥ 0 or (b) f ∈ L^{1}(λ) then f_{x} is N – measurable for μ a.e. x and f^{y} is M – measurable for ν a.e. y and in case (b) f_{x} ∈ L^{1}(ν) and f^{y} ∈ L^{1}(μ) for μ a.e. x and ν a.e. y respectively. Moreover,

(x → ∫_{Y} f_{x}dν) ∈ L^{1} (μ) and (y → ∫_{X} f^{y}dμ) ∈ L^{1} (ν)

and

∫_{X×Y} fdλ = ∫_{Y} dν ∫_{X} dμ f = ∫_{X} dμ ∫_{Y} dν f.

Proof. If E ∈ M ⊗ N is a μ ⊗ ν null set (i.e. (μ ⊗ ν)(E) = 0), then

0 = (μ ⊗ ν)(E) = ∫_{X} ν(_{x}E)dμ(x) = ∫_{X} μ(E_{y})dν(y).

This shows that

μ({x : ν(_{x}E) ≠ 0}) = 0 and ν({y : μ(E_{y}) ≠ 0}) = 0,

i.e. ν(_{x}E) = 0 for μ a.e. x and μ(E_{y}) = 0 for ν a.e. y. If h is L measurable and h = 0 for λ – a.e., then there exists E ∈ M ⊗ N such that {(x, y) : h(x, y) ≠ 0} ⊂ E and (μ⊗ν)(E) = 0. Therefore |h(x, y)| ≤ 1_{E}(x, y) and (μ⊗ν)(E) = 0. Since

{h_{x} ≠ 0} = {y ∈ Y : h(x, y) ≠ 0} ⊂ _{x}E and
{h_{y} ≠ 0} = {x ∈ X : h(x, y) ≠ 0} ⊂ E_{y}

we learn that for μ a.e. x and ν a.e. y that {h_{x} ≠ 0} ∈ M, {h_{y} ≠ 0} ∈ N , ν({h_{x} ≠ 0}) = 0 and a.e. and μ({h_{y} ≠ 0}) = 0. This implies ∫_{Y} h(x, y)dν(y) exists and equals 0 for μ a.e. x and similarly that ∫_{X} h(x, y)dμ(x) exists and equals 0 for ν a.e. y. Therefore

0 = ∫_{X×Y} hdλ = ∫_{Y} (∫_{X} hdμ) dν = ∫_{X} (∫_{Y} hdν) dμ.

For general f ∈ L^{1}(λ), we may choose g ∈ L^{1}(M ⊗ N , μ ⊗ ν) such that f(x, y) = g(x, y) for λ− a.e. (x, y). Define h := f − g. Then h = 0, λ− a.e. Hence by what we have just proved and Theorem 20.8 f = g + h has the following properties:

1. For μ a.e. x, y → f(x, y) = g(x, y) + h(x, y) is in L^{1}(ν) and

∫_{Y} f(x, y)dν(y) = ∫_{Y} g(x, y)dν(y).

2. For ν a.e. y, x → f(x, y) = g(x, y) + h(x, y) is in L^{1}(μ) and

∫_{X} f(x, y)dμ(x) = ∫_{X} g(x, y)dμ(x).

From these assertions and Theorem 20.8, it follows that

∫_{X} dμ(x) ∫_{Y} dν(y)f(x, y) = ∫_{X} dμ(x) ∫_{Y} dν(y)g(x, y)
= ∫_{Y} dν(y) ∫_{Y} dν(x)g(x, y)
= ∫_{X×Y} g(x, y)d(μ ⊗ ν)(x, y)
= ∫_{X×Y} f(x, y)dλ(x, y).

Similarly it is shown that

∫_{Y} dν(y) ∫_{X} dμ(x)f(x, y) = ∫_{X×Y} f(x, y)dλ(x, y).

The previous theorems have obvious generalizations to products of any finite number of σ – finite measure spaces. For example the following theorem holds.

Theorem 20.12. Suppose {(X_{i}, M_{i}, μ_{i})}_{i=1}^{n} are σ – finite measure spaces and X := X_{1} × ⋯ × X_{n}. Then there exists a unique measure, π, on (X, M_{1} ⊗ ⋯ ⊗ M_{n}) such that π(A_{1} × ⋯ × A_{n}) = μ_{1}(A_{1}) … μ_{n}(A_{n}) for all A_{i} ∈ M_{i}. (This measure and its completion will be denote by μ_{1} ⊗ ⋯ ⊗ μ_{n}.) If f : X → [0, ∞] is a M_{1} ⊗ ⋯ ⊗ M_{n} – measurable function then

∫_{X} fdπ = ∫_{X_{σ(1)}} dμ_{σ(1)}(x_{σ(1)}) … ∫_{X_{σ(n)}} dμ_{σ(n)}(x_{σ(n)}) f(x_{1}, …, x_{n}) (20.18)

where σ is any permutation of {1, 2, …, n}. This equation also holds for any f ∈ L^{1}(π) and moreover, f ∈ L^{1}(π) iff

∫_{X_{σ(1)}} dμ_{σ(1)}(x_{σ(1)}) … ∫_{X_{σ(n)}} dμ_{σ(n)}(x_{σ(n)}) |f(x_{1}, …, x_{n})| < ∞

for some (and hence all) permutations, σ.

This theorem can be proved by the same methods as in the two factor case, see Exercise 20.5. Alternatively, one can use the theorems already proved and induction on n, see Exercise 20.6 in this regard.

20.1 Fubini-Tonelli’s Theorem and Product Measure 347

Example 20.13. In this example we will show

lim_{M→∞} ∫_{0}^{M} \frac{sin x}{x}dx = π/2.  (20.19)

To see this write \frac{1}{x} = ∫_{0}^{∞} e^{−tx}dt and use Fubini-Tonelli to conclude that

∫_{0}^{M} \frac{sin x}{x}dx = ∫_{0}^{M} [∫_{0}^{∞} e^{−tx} sin x dt] dx

= ∫_{0}^{∞} [∫_{0}^{M} e^{−tx} sin x dx] dt

= ∫_{0}^{∞} \frac{1}{1 + t^{2}} (1 − te^{−Mt} sin M − e^{−Mt} cos M) dt

→ ∫_{0}^{∞} \frac{1}{1 + t^{2}} dt = \frac{π}{2} as M → ∞,

wherein we have used the dominated convergence theorem to pass to the limit.

The next example is a refinement of this result. Example 20.14. We have

∫_{0}^{∞} \frac{sin x}{x}e^{−Λx}dx = \frac{1}{2}π − arctan Λ for all Λ > 0  (20.20)

and forΛ, M ∈ [0, ∞),

|∫_{0}^{M} \frac{sin x}{x}e^{−Λx}dx − \frac{1}{2}π + arctan Λ| ≤ C \frac{e^{−MΛ}}{M}  (20.21)

where C = max_{x≥0} \frac{1+x}{1+x^{2}} = \frac{1}{2\sqrt{2}−2} ≅ 1.2. In particular Eq. (20.19) is valid.

To verify these assertions, first notice that by the fundamental theorem of calculus,

|sin x| = |∫_{0}^{x} cos ydy| ≤ |∫_{0}^{x} |cos y| dy| ≤ |∫_{0}^{x} 1dy| = |x|

so |\frac{sin x}{x} | ≤ 1 for all x ≠ 0. Making use of the identity

∫_{0}^{∞} e^{−tx}dt = 1/x

and Fubini’s theorem,

∫_{0}^{M} \frac{sin x}{x}e^{−Λx}dx = ∫_{0}^{M} dx sin x e^{−Λx} ∫_{0}^{∞} e^{−tx}dt

= ∫_{0}^{∞} dt ∫_{0}^{M} dx sin x e^{−(Λ+t)x}

= ∫_{0}^{∞} \frac{1 − (cos M + (Λ + t) sin M) e^{−M(Λ+t)}}{(Λ + t)^{2} + 1} dt

= ∫_{0}^{∞} \frac{1}{(Λ + t)^{2} + 1}dt − ∫_{0}^{∞} \frac{cos M + (Λ + t) sin M}{(Λ + t)^{2} + 1} e^{−M(Λ+t)}dt

= \frac{1}{2}π − arctan Λ − ε(M, Λ)  (20.22)

where

ε(M, Λ) = ∫_{0}^{∞} \frac{cos M + (Λ + t) sin M}{(Λ + t)^{2} + 1} e^{−M(Λ+t)}dt.

Since

|\frac{cos M + (Λ + t) sin M}{(Λ + t)^{2} + 1} | ≤ \frac{1 + (Λ + t)}{(Λ + t)^{2} + 1} ≤ C,

|ε(M, Λ)| ≤ ∫_{0}^{∞} e^{−M(Λ+t)}dt = C \frac{e^{−MΛ}}{M} .

This estimate along with Eq. (20.22) proves Eq. (20.21) from which Eq. (20.19) follows by taking Λ → ∞ and Eq. (20.20) follows (using the dominated convergence theorem again) by letting M → ∞.

20.2 Lebesgue Measure on ℝ^{d} and the Change of Variables Theorem

Notation 20.15 Let

m^{d} := \overbrace{m ⊗ ⋯ ⊗ m}^{d times} on B_{ℝ^{d}} = \overbrace{B_{ℝ} ⊗ ⋯ ⊗ B_{ℝ}}^{d times}

be the d – fold product of Lebesgue measure m on B_{ℝ}. We will also use m^{d} to denote its completion and let L_{d} be the completion of B_{ℝ^{d}} relative to m^{d}. A subset A ∈ L_{d} is called a Lebesgue measurable set and m^{d} is called d – dimensional Lebesgue measure, or just Lebesgue measure for short.

Definition 20.16. A function f : ℝ^{d} → ℝ is Lebesgue measurable if f^{−1}(B_{ℝ}) ⊂ L_{d}.

Notation 20.17 I will often be sloppy in the sequel and write m for m^{d} and dx for dm(x) = dm^{d}(x), i.e.

20.2 Lebesgue Measure on ℝ^{d} and the Change of Variables Theorem 349

∫_{ℝ^{d}} f (x) dx = ∫_{ℝ^{d}} fdm = ∫_{ℝ^{d}} fdm^{d}.

Hopefully the reader will understand the meaning from the context.

Theorem 20.18. Lebesgue measure m^{d} is translation invariant. Moreover m^{d} is the unique translation invariant measure on B_{ℝ^{d}} such that m^{d}((0, 1]^{d}) = 1.

Proof. Let A = J_{1} × ⋯ × J_{d} with J_{i} ∈ B_{ℝ} and x ∈ ℝ^{d}. Then

x + A = (x_{1} + J_{1}) × (x_{2} + J_{2}) × ⋯ × (x_{d} + J_{d})

and therefore by translation invariance of m on B_{ℝ} we find that

m^{d}(x + A) = m(x_{1} + J_{1}) … m(x_{d} + J_{d}) = m(J_{1}) … m(J_{d}) = m^{d}(A)

and hence m^{d}(x + A) = m^{d}(A) for all A ∈ B_{ℝ^{d}} by Corollary 19.57. From this fact we see that the measure m^{d}(x + ·) and m^{d}(·) have the same null sets. Using this it is easily seen that m(x + A) = m(A) for all A ∈ L_{d}. The proof of the second assertion is Exercise 20.7.

Exercise 20.1. In this problem you are asked to show there is no reasonable notion of Lebesgue measure on an infinite dimensional Hilbert space. To be more precise, suppose H is an infinite dimensional Hilbert space and m is a countably additive measure on B_{H} which is invariant under translations and satisfies, m(B_{0}(ε)) > 0 for all ε > 0. Show m(V) = ∞ for all non-empty open subsets V ⊂ H.

Theorem 20.19 (Change of Variables Theorem). Let Ω ⊂_{o} ℝ^{d} be an open set and T : Ω → T (Ω) ⊂_{o} ℝ^{d} be a C^{1} – diffeomorphism,^{1} see Figure 20.1. Then for any Borel measurable function, f : T (Ω) → [0, ∞],

∫_{Ω} f (T (x)) | det T′ (x) |dx = ∫_{T (Ω)} f (y) dy,  (20.23)

where T′(x) is the linear transformation on ℝ^{d} defined by T′(x)v := \frac{d}{dt}|_{0}T (x+ tv). More explicitly, viewing vectors in ℝ^{d} as columns, T′ (x) may be represented by the matrix

T′ (x) = [ \begin{matrix}∂_{1}T_{1} (x) & … & ∂_{d}T_{1} (x) \\ ⋮ & ... & ⋮ \\ ∂_{1}T_{d} (x) & … & ∂_{d}T_{d} (x) \end{matrix} ] ,  (20.24)

i.e. the i - j – matrix entry of T′(x) is given by T′(x)_{ij} = ∂_{i}T_{j}(x) where T (x) = (T_{1}(x), …, T_{d}(x))^{tr} and ∂_{i} = ∂/∂x_{i}.

1 That is T : Ω → T (Ω) ⊂_{o} ℝ^{d} is a continuously differentiable bijection and the inverse map T^{−1} : T (Ω) → Ω is also continuously differentiable.

Fig. 20.1. The geometric setup of Theorem 20.19.

Remark 20.20. Theorem 20.19 is best remembered as the statement: if we make the change of variables y = T (x) , then dy = | det T′ (x) |dx. As usual, you must also change the limits of integration appropriately, i.e. if x ranges through Ω then y must range through T (Ω) .

Proof. The proof will be by induction on d. The case d = 1 was essentially done in Exercise 19.8. Nevertheless, for the sake of completeness let us give a proof here. Suppose d = 1, a < α < β < b such that [a, b] is a compact subinterval of Ω. Then | det T′| = |T′| and

∫_{[a,b]} 1_{T ((α,β])} (T (x)) |T′ (x)| dx = ∫_{[a,b]} 1_{(α,β]} (x) |T′ (x)| dx = ∫_{α}^{β} |T′ (x)| dx.

If T′ (x) > 0 on [a, b] , then

∫_{α}^{β} |T′ (x)| dx = ∫_{α}^{β} T′ (x) dx = T (β) − T (α)
= m (T ((α, β])) = ∫_{T ([a,b])} 1_{T ((α,β])} (y) dy

while if T′ (x) < 0 on [a, b] , then

∫_{α}^{β} |T′ (x)| dx = − ∫_{α}^{β} T′ (x) dx = T (α) − T (β)
= m (T ((α, β])) = ∫_{T ([a,b])} 1_{T ((α,β])} (y) dy.

Combining the previous three equations shows

∫_{[a,b]} f (T (x)) |T′ (x)| dx = ∫_{T ([a,b])} f (y) dy  (20.25)

whenever f is of the form f = 1_{T ((α,β])} with a < α < β < b. An application of Dynkin’s multiplicative system Theorem 18.51 then implies that Eq. (20.25) holds for every bounded measurable function f : T ([a, b]) → ℝ. (Observe that |T′ (x)| is continuous and hence bounded for x in the compact interval, [a, b] . From Exercise 10.14, Ω = ∐_{n=1}^{N} (a_{n}, b_{n}) where a_{n}, b_{n} ∈ ℝ∪ {±∞} for n = 1, 2, ⋯ < N with N = ∞ possible. Hence if f : T (Ω) → ℝ _{+} is a Borel measurable function and a_{n} < α_{k} < β_{k} < b_{n} with α_{k} ↓ a_{n} and β_{k} ↑ b_{n}, then by what we have already proved and the monotone convergence theorem

∫_{Ω} 1_{(a_{n},b_{n})} · (f ∘ T) · |T′|dm = ∫_{Ω} (1_{T ((a_{n},b_{n}))} · f) ∘ T · |T′|dm

= lim_{k→∞} ∫_{Ω} (1_{T ([α_{k},β_{k}])} · f) ∘ T · |T′| dm

= lim_{k→∞} ∫_{T (Ω)} 1_{T ([α_{k},β_{k}])} · f dm

= ∫_{T (Ω)} 1_{T ((a_{n},b_{n}))} · f dm.

Summing this equality on n, then shows Eq. (20.23) holds.

To carry out the induction step, we now suppose d > 1 and suppose the theorem is valid with d being replaced by d−1. For notational compactness, let us write vectors in ℝ^{d} as row vectors rather than column vectors. Nevertheless, the matrix associated to the differential, T′ (x) , will always be taken to be given as in Eq. (20.24).

Case 1. Suppose T (x) has the form

T (x) = (x_{i}, T_{2} (x) , …, T_{d} (x))  (20.26)

or

T (x) = (T_{1} (x) , …, T_{d−1} (x) , x_{i})  (20.27)

for some i ∈ {1, …, d} . For definiteness we will assume T is as in Eq. (20.26), the case of T in Eq. (20.27) may be handled similarly. For t ∈ ℝ, let i_{t} : ℝ^{d−1} → ℝ^{d} be the inclusion map defined by

i_{t} (w) := w_{t} := (w_{1}, …, w_{i−1}, t, w_{i+1}, …, w_{d−1}) ,

Ω_{t} be the (possibly empty) open subset of ℝ^{d−1} defined by

Ω_{t} := {w ∈ ℝ^{d−1} : (w_{1}, …, w_{i−1}, t, w_{i+1}, …, w_{d−1}) ∈ Ω}

and T_{t} : Ω_{t} → ℝ^{d−1} be defined by

T_{t} (w) = (T_{2} (w_{t}) , …, T_{d} (w_{t})) ,

see Figure 20.2. Expanding det T′ (w_{t}) along the first row of the matrix T′ (w_{t})

Fig. 20.2. In this picture d = i = 3 and Ω is an egg-shaped region with an eggshaped hole. The picture indicates the geometry associated with the map T and slicing the set Ω along planes where x_{3} = t.

shows

|det T′ (w_{t})| = |det T′_{t} (w)| .

Now by the Fubini-Tonelli Theorem and the induction hypothesis,

∫_{Ω} f ∘ T | det T′|dm = ∫_{ℝ^{d}} 1_{Ω} · f ∘ T | det T′|dm

= ∫_{ℝ^{d}} 1_{Ω} (w_{t}) (f ∘ T) (w_{t}) | det T′ (w_{t}) |dwdt

= ∫_{ℝ} [∫_{Ω_{t}} (f ∘ T) (w_{t}) | det T′ (w_{t}) |dw] dt

= ∫_{ℝ} [∫_{Ω_{t}} f (t, T_{t} (w)) | det T′_{t} (w) |dw] dt

= ∫_{ℝ} [ ∫_{T_{t}(Ω_{t})} f (t, z) dz] dt = ∫_{ℝ} [ ∫_{ℝ^{d−1}} 1_{T (Ω)} (t, z) f (t, z) dz] dt

= ∫_{T (Ω)} f (y) dy

wherein the last two equalities we have used Fubini-Tonelli along with the identity;

T (Ω) = ∐_{t∈ℝ} T (i_{t} (Ω)) = ∐_{t∈ℝ} {(t, z) : z ∈ T_{t} (Ω_{t})} .

Case 2. (Eq. (20.23) is true locally.) Suppose that T : Ω → ℝ^{d} is a general map as in the statement of the theorem and x_{0} ∈ Ω is an arbitrary point. We will now show there exists an open neighborhood W ⊂ Ω of x_{0} such that

∫_{W} f ∘ T | det T′|dm = ∫_{T (W)} fdm

holds for all Borel measurable function, f : T (W) → [0, ∞]. Let M_{i} be the 1-i minor of T′ (x_{0}) , i.e. the determinant of T′ (x_{0}) with the first row and i^{th} – column removed. Since

0 ≠ det T′ (x_{0}) = ∑_{i=1}^{d} (−1)^{i+1} ∂_{i}T_{j} (x_{0}) · M_{i},

there must be some i such that M_{i} ≠ 0. Fix an i such that M_{i} ≠ 0 and let,

S (x) := (x_{i}, T_{2} (x) , …, T_{d} (x)) .  (20.28)

Observe that |det S′ (x_{0})| = |M_{i}| ≠ 0. Hence by the inverse function Theorem 16.25, there exist an open neighborhood W of x_{0} such that W ⊂_{o} Ω and S (W) ⊂_{o} ℝ^{d} and S : W → S (W) is a C^{1} – diffeomorphism. Let R : S (W) → T (W) ⊂_{o} ℝ^{d} to be the C^{1} – diffeomorphism defined by

R (z) := T ∘ S^{−1} (z) for all z ∈ S (W) .

Because

(T_{1} (x) , …, T_{d} (x)) = T (x) = R (S (x)) = R ((x_{i}, T_{2} (x) , …, T_{d} (x)))

for all x ∈ W, if

(z_{1}, z_{2}, …, z_{d}) = S (x) = (x_{i}, T_{2} (x) , …, T_{d} (x))

then

R (z) = (T_{1} (S^{−1} (z)) , z_{2}, …, z_{d}) .  (20.29)

Observe that S is a map of the form in Eq. (20.26), R is a map of the form in Eq. (20.27), T′ (x) = R′ (S (x)) S′ (x) (by the chain rule) and (by the multiplicative property of the determinant)

|det T′ (x)| = | det R′ (S (x)) | |det S′ (x)| ∀ x ∈ W.

So if f : T (W) → [0, ∞] is a Borel measurable function, two applications of the results in Case 1. shows,

∫_{W} f ∘ T · | det T′|dm = ∫_{W} (f ∘ R · | det R′|) ∘ S · |det S′| dm

= ∫_{S(W)} f ∘ R · | det R′|dm = ∫_{R(S(W))} fdm

= ∫_{T (W)} fdm

and Case 2. is proved.

Case 3. (General Case.) Let f : Ω → [0, ∞] be a general non-negative Borel measurable function and let

K_{n} := {x ∈ Ω : dist(x, Ω^{c}) ≥ 1/n and |x| ≤ n} .

Then each K_{n} is a compact subset of Ω and K_{n} ↑ Ω as n → ∞. Using the compactness of K_{n} and case 2, for each n ∈ ℕ, there is a finite open cover W_{n} of K_{n} such that W ⊂ Ω and Eq. (20.23) holds with Ω replaced by W for each W ∈ W_{n}. Let {W_{i}}_{i=1}^{∞} be an enumeration of ∪_{n=1}^{∞}W_{n} and set W̃_{1} = W_{1} and W̃_{i} := W_{i} \ (W_{1} ∪ ⋯ ∪ W_{i−1}) for all i ≥ 2. Then Ω = ∐_{i=1}^{∞} W̃_{i} and by repeated use of case 2.,

∫_{Ω} f ∘ T | det T′|dm = ∑_{i=1}^{∞} ∫_{Ω} 1W̃_{i} · (f ∘ T) · | det T′|dm

= ∑_{i=1}^{∞} ∫_{W_{i}} [(1_{T \binom{}{W̃_{i}}}f) ∘ T ] · | det T′|dm

= ∑_{i=1}^{∞} ∫_{T (W_{i})} 1_{T \binom{}{W̃_{i}}} · f dm = ∑_{i=1}^{n} ∫_{T (Ω)} 1_{T \binom{}{W̃_{i}}} · f dm

= ∫_{T (Ω)} fdm.

Remark 20.21. When d = 1, one often learns the change of variables formula as

∫_{a}^{b} f (T (x)) T′ (x) dx = ∫_{T (a)}^{T (b)} f (y) dy  (20.30)

where f : [a, b] → ℝ is a continuous function and T is C^{1} – function defined in a neighborhood of [a, b] . If T′ > 0 on (a, b) then T ((a, b)) = (T (a) , T (b)) and Eq. (20.30) is implies Eq. (20.23) with Ω = (a, b) . On the other hand if T′ < 0 on (a, b) then T ((a, b)) = (T (b) , T (a)) and Eq. (20.30) is equivalent to

∫_{(a,b)} f (T (x)) (− |T′ (x)|) dx = − ∫_{T (b)}^{T (a)} f (y) dy = − ∫_{T ((a,b))} f (y) dy

which is again implies Eq. (20.23). On the other hand Eq. Eq. (20.30) is more general than Eq. (20.23) since it does not require T to be injective. The standard proof of Eq. (20.30) is as follows. For z ∈ T ([a, b]) , let

F (z) := ∫_{T (a)}^{z} f (y) dy.

Then by the chain rule and the fundamental theorem of calculus,

∫_{a}^{b} f (T (x)) T′ (x) dx = ∫_{a}^{b} F′ (T (x)) T′ (x) dx = ∫_{a}^{b} \frac{d}{dx} [F (T (x))] dx
= F (T (x)) |_{a}^{b} = ∫_{T (a)}^{T (b)} f (y) dy.

An application of Dynkin’s multiplicative systems theorem (in the form of Corollary 18.55) now shows that Eq. (20.30) holds for all bounded measurable functions f on (a, b) . Then by the usual truncation argument, it also holds for all positive measurable functions on (a, b) .

Example 20.22. Continuing the setup in Theorem 20.19, if A ∈ B_{Ω}, then

m (T (A)) = ∫_{ℝ^{d}} 1_{T (A)} (y) dy = ∫_{ℝ^{d}} 1_{T (A)} (T x) |det T′ (x)| dx
= ∫_{ℝ^{d}} 1_{A} (x) |det T′ (x)| dx

wherein the second equality we have made the change of variables, y = T (x) . Hence we have shown

d (m ∘ T) = |det T′ (·)| dm.

In particular if T ∈ GL(d, ℝ) = GL(ℝ^{d}) – the space of d×d invertible matrices, then m ∘ T = |det T | m, i.e.

m (T (A)) = |det T | m (A) for allA ∈ B_{ℝ^{d}}.  (20.31)

This equation also shows that m ∘ T and m have the same null sets and hence the equality in Eq. (20.31) is valid for any A ∈ L_{d}.

Exercise 20.2. Show that f ∈ L^{1} (T (Ω) , m^{d}) iff

∫_{Ω} |f ∘ T | | det T′|dm < ∞

and if f ∈ L^{1} (T (Ω) , m^{d}) , then Eq. (20.23) holds.

Example 20.23 (Polar Coordinates). Suppose T : (0, ∞) × (0, 2π) → ℝ^{2} is defined by

x = T (r, θ) = (r cos θ, r sin θ) ,

i.e. we are making the change of variable,

x_{1} = r cos θ and x_{2} = r sin θ for 0 < r < ∞ and 0 < θ < 2π.

In this case

T′(r, θ) = ( \begin{matrix}cos θ & − r sin θ \\ sin θ & r cos θ \end{matrix} )

and therefore

dx = |det T′(r, θ)| drdθ = rdrdθ.

Observing that

ℝ^{2} \ T ((0, ∞) × (0, 2π)) = ℓ := {(x, 0) : x ≥ 0}

has m^{2} – measure zero, it follows from the change of variables Theorem 20.19 that

∫_{ℝ^{2}} f(x)dx = ∫_{0}^{2π} dθ ∫_{0}^{∞} dr r · f(r (cos θ, sin θ))  (20.32)

for any Borel measurable function f : ℝ^{2} → [0, ∞].

Example 20.24 (Holomorphic Change of Variables). Suppose that f : Ω ⊂_{o} ℂ ≅ ℝ^{2}→ ℂ is an injective holomorphic function such that f′ (z) ≠ 0 for all z ∈ Ω. We may express f as

f (x + iy) = U (x, y) + iV (x, y)

for all z = x + iy ∈ Ω. Hence if we make the change of variables,

w = u + iv = f (x + iy) = U (x, y) + iV (x, y)

then

dudv = |det [ \begin{matrix}U_{x} & U_{y} \\ V_{x} & V_{y} \end{matrix} ]| dxdy = |U_{x}V_{y} − U_{y}V_{x}| dxdy.

Recalling that U and V satisfy the Cauchy Riemann equations, U_{x} = V_{y} and U_{y} = −V_{x} with f′ = U_{x} + iV_{x}, we learn

U_{x}V_{y} − U_{y}V_{x} = U_{x}^{2} + V_{x}^{2} = |f′|^{2} .

Therefore

dudv = |f′ (x + iy)|^{2} dxdy.

Example 20.25. In this example we will evaluate the integral

I := ∫∫_{Ω} (x^{4} − y^{4}) dxdy

where

Ω = {(x, y) : 1 < x^{2} − y^{2} < 2, 0 < xy < 1} ,

see Figure 20.3 We are going to do this by making the change of variables,

Fig. 20.3. The region Ω consists of the two curved rectangular regions shown.

(u, v) := T (x, y) = (x^{2} − y^{2}, xy) ,

in which case

dudv = |det [ \begin{matrix}2x & −2y \\ y & x \end{matrix} ]| dxdy = 2 (x^{2} + y^{2}) dxdy

Notice that

(x^{4} − y^{4}) = (x^{2} − y^{2}) (x^{2} + y^{2}) = u (x^{2} + y^{2}) = \frac{1}{2}ududv.

The function T is not injective on Ω but it is injective on each of its connected components. Let D be the connected component in the first quadrant so that Ω = −D ∪ D and T (±D) = (1, 2) × (0, 1) The change of variables theorem then implies

I_{±} := ∫∫_{±D} (x^{4} − y^{4}) dxdy = \frac{1}{2} ∫∫_{(1,2)×(0,1)} ududv = \frac{1}{2} \frac{u^{2}}{2} |_{1}^{2} · 1 = \frac{3}{4}

and therefore I = I_{+} + I_{−} = 2 · (3/4) = 3/2.

Exercise 20.3 (Spherical Coordinates). Let T : (0, ∞)×(0, π)×(0, 2π) → ℝ^{3} be defined by

T (r, ϕ, θ) = (r sin ϕ cos θ, r sin ϕ sin θ, r cos ϕ)
= r (sin ϕ cos θ, sin ϕ sin θ, cos ϕ) ,

see Figure 20.4. By making the change of variables x = T (r, ϕ, θ) , show

Fig. 20.4. The relation of x to (r, ϕ, θ) in spherical coordinates.

∫_{ℝ^{3}} f(x)dx = ∫_{0}^{π} dϕ ∫_{0}^{2π} dθ ∫_{0}^{∞} dr r^{2} sin ϕ · f(T (r, ϕ, θ))

for any Borel measurable function, f : ℝ^{3} → [0, ∞].

Lemma 20.26. Let a > 0 and

I_{d}(a) := ∫_{ℝ^{d}} e^{−a|x|^{2}}dm(x).

Then I_{d}(a) = (π/a)^{d/2}.

Proof. By Tonelli’s theorem and induction,

I_{d}(a) = ∫_{ℝ^{d−1}×ℝ} e^{−a|y|^{2}}e^{−at^{2}}m_{d−1}(dy) dt
= I_{d−1}(a)I_{1}(a) = I_{1}^{d}(a).  (20.33)

So it suffices to compute:

I_{2}(a) = ∫_{ℝ^{2}} e^{−a|x|^{2}}dm(x) = ∫_{ℝ^{2}\{0}} e^{−a(x_{1}^{2}+x_{2}^{2})}dx_{1}dx_{2}.

Using polar coordinates, see Eq. (20.32), we find,

I_{2}(a) = ∫_{0}^{∞} dr r ∫_{0}^{2π} dθ e^{−ar^{2}} = 2π ∫_{0}^{∞} re^{−ar^{2}}dr

= 2π lim_{M→∞} ∫_{0}^{M} re^{−ar^{2}}dr = 2π lim_{M→∞} \frac{e^{−ar^{2}}}{−2a} ∫_{0}^{M} = \frac{2π}{2a} = π/a.

This shows that I_{2}(a) = π/a and the result now follows from Eq. (20.33).

20.3 The Polar Decomposition of Lebesgue Measure

Let

S^{d−1} = {x ∈ ℝ^{d} : |x|^{2} := ∑_{i=1}^{d} x_{i}^{2} = 1}

be the unit sphere in ℝ^{d} equipped with its Borel σ – algebra, B_{S^{d−1}} and Φ : ℝ^{d} \ {0} → (0, ∞) × S^{d−1} be defined by Φ(x) := (|x| , |x|^{−1} x). The inverse map, Φ^{−1} : (0, ∞) × S^{d−1} → ℝ^{d} \ {0} , is given by Φ^{−1}(r, ω) = rω. Since Φ and Φ^{−1} are continuous, they are both Borel measurable. For E ∈ B_{S^{d−1}} and a > 0, let

E_{a} := {rω : r ∈ (0, a] and ω ∈ E} = Φ^{−1}((0, a] × E) ∈ B_{ℝ^{d}}.

Definition 20.27. For E ∈ B_{S^{d−1}}, let σ(E) := d · m(E_{1}). We call σ the surface measure on S^{d−1}.

It is easy to check that σ is a measure. Indeed if E ∈ B_{S^{d−1}}, then E_{1} = Φ^{−1} ((0, 1] × E) ∈ B_{ℝ^{d}} so that m(E_{1}) is well defined. Moreover if E = ∐_{i=1}^{∞} E_{i}, then E_{1} = ∐_{i=1}^{∞} (E_{i})_{1} and

σ(E) = d · m(E_{1}) = ∑_{i=1}^{∞} m ((E_{i})_{1}) = ∑_{i=1}^{∞} σ(E_{i}).

The intuition behind this definition is as follows. If E ⊂ S^{d−1} is a set and ε > 0 is a small number, then the volume of

(1, 1 + ε] · E = {rω : r ∈ (1, 1 + ε] and ω ∈ E}

should be approximately given by m ((1, 1 + ε] · E) ≅ σ(E)ε, see Figure 20.5 below. On the other hand

Fig. 20.5. Motivating the definition of surface measure for a sphere.

m ((1, 1 + ε]E) = m (E_{1+ε} \ E_{1}) = {(1 + ε)^{d} − 1} m(E_{1}).

Therefore we expect the area of E should be given by

σ(E) = lim_{ε↓0} \frac{\{(1 + ε)^{d} − 1\} m(E_{1})}{ε} = d · m(E_{1}).

The following theorem is motivated by Example 20.23 and Exercise 20.3.

Theorem 20.28 (Polar Coordinates). If f : ℝ^{d} → [0, ∞] is a (B_{R^{d}}, B)– measurable function then

∫_{ℝ^{d}} f(x)dm(x) =  ∫_{(0,∞)×S^{d−1}} f(rω)r^{d−1} drdσ(ω).  (20.34)

Proof. By Exercise 19.7,

∫_{ℝ^{d}} fdm = ∫_{ℝ^{d}\{0}} (f ∘ Φ^{−1}) ∘ Φ dm =  ∫_{(0,∞)×S^{d−1}} (f ∘ Φ^{−1}) d (Φ_{∗}m) (20.35)

and therefore to prove Eq. (20.34) we must work out the measure Φ_{∗}m on B_{(0,∞)} ⊗ B_{S^{d−1}} defined by

Φ_{∗}m(A) := m (Φ^{−1}(A)) ∀ A ∈ B_{(0,∞)} ⊗ B_{S^{d−1}}.  (20.36)

If A = (a, b] × E with 0 < a < b and E ∈ B_{S^{d−1}}, then

Φ^{−1}(A) = {rω : r ∈ (a, b] and ω ∈ E} = bE_{1} \ aE_{1}

wherein we have used E_{a} = aE_{1} in the last equality. Therefore by the basic scaling properties of m and the fundamental theorem of calculus,

(Φ_{∗}m) ((a, b] × E) = m (bE_{1} \ aE_{1}) = m(bE_{1}) − m(aE_{1})
= b^{d}m(E_{1}) − a^{d}m(E_{1}) = d · m(E_{1}) ∫_{a}^{b} r^{d−1}dr. (20.37)

Letting dρ(r) = r^{d−1}dr, i.e.

ρ(J) = ∫_{J} r^{d−1}dr ∀ J ∈ B_{(0,∞)},  (20.38)

Eq. (20.37) may be written as

(Φ_{∗}m) ((a, b] × E) = ρ((a, b]) · σ(E) = (ρ ⊗ σ) ((a, b] × E) .  (20.39)

Since

E = {(a, b] × E : 0 < a < b and E ∈ B_{S^{d−1}}} ,

is a π class (in fact it is an elementary class) such that σ(E) = B_{(0,∞)} ⊗B_{S^{d−1}}, it follows from Theorem 19.55 and Eq. (20.39) that Φ_{∗}m = ρ ⊗ σ. Using this result in Eq. (20.35) gives

∫_{ℝ^{d}} fdm = ∫_{(0,∞)×S^{d−1}} (f ∘ Φ^{−1}) d (ρ ⊗ σ)

which combined with Tonelli’s Theorem 20.8 proves Eq. (20.35).

Corollary 20.29. The surface area σ(S^{d−1}) of the unit sphere S^{d−1} ⊂ ℝ^{d} is

σ(S^{d−1}) = \frac{2π^{d/2}}{Γ (d/2)}  (20.40)

where Γ is the gamma function given by

Γ (x) := ∫_{0}^{∞} u^{x−1}e^{−u}dr  (20.41)

Moreover, Γ (1/2) = \sqrt{π}, Γ (1) = 1 and Γ (x + 1) = xΓ (x) for x > 0.

Proof. Using Theorem 20.28 we find

I_{d}(1) = ∫_{0}^{∞} dr r^{d−1}e^{−r^{2}} ∫_{S^{d−1}} dσ = σ(S^{d−1}) ∫_{0}^{∞} r^{d−1}e^{−r^{2}}dr.

We simplify this last integral by making the change of variables u = r^{2} so that r = u^{1/2} and dr = \frac{1}{2}u^{−1/2}du. The result is

∫_{0}^{∞} r^{d−1}e^{−r^{2}}dr = ∫_{0}^{∞} u^{\frac{d−1}{2}} e^{−u} \frac{1}{2}u^{−1/2}du
= \frac{1}{2} ∫_{0}^{∞} u^{\frac{d}{2} −1}e^{−u}du = \frac{1}{2}Γ (d/2).  (20.42)

Combing the the last two equations with Lemma 20.26 which states that I_{d}(1) = π^{d/2}, we conclude that

π^{d/2} = I_{d}(1) = \frac{1}{2}σ(S^{d−1})Γ (d/2)

which proves Eq. (20.40). Example 19.24 implies Γ (1) = 1 and from Eq. (20.42),

Γ (1/2) = 2 ∫_{0}^{∞} e^{−r^{2}}dr = ∫_{−∞}^{∞} e^{−r^{2}}dr
= I_{1}(1) = \sqrt{π}.

The relation, Γ (x+1) = xΓ (x) is the consequence of the following integration by parts argument:

Γ (x + 1) = ∫_{0}^{∞} e^{−u} u^{x+1} \frac{du}{u} = ∫_{0}^{∞} u^{x} (−\frac{d}{du} e^{−u}) du
= x ∫_{0}^{∞} u^{x−1} e^{−u} du = x Γ (x).

BRUCE: add Morrey’s Inequality ?? here.

20.4 More proofs of the classical Weierstrass approximation Theorem 8.34

In each of these proofs we will use the reduction explained the previous proof of Theorem 8.34 to reduce to the case where f ∈ C([0, 1]^{d}). The first proof we will give here is based on the “weak law” of large numbers. The second will be another approximate δ – function argument.

Proof. of Theorem 8.34. Let 0 : = (0, 0, …, 0), 1 : = (1, 1, …, 1) and [0, 1] := [0, 1]^{d}. By considering the real and imaginary parts of f separately, it suffices to assume f ∈ C([0, 1], ℝ). For x ∈ [0, 1], let ν_{x} be the measure on {0, 1} such that ν_{x} ({0}) = 1 − x and ν_{x} ({1}) = x. Then

∫_{{0,1}} ydν_{x}(y) = 0 · (1 − x) + 1 · x = x and  (20.43)
∫_{{0,1}}(y − x)^{2}dν_{x}(y) = x^{2}(1 − x) + (1 − x)^{2} · x = x(1 − x).  (20.44)

For x ∈ [0, 1] let μ_{x} = ν_{x_{1}} ⊗ ⋯ ⊗ ν_{x_{d}} be the product of ν_{x_{1}}, …, ν_{x_{d}} on Ω := {0, 1}^{d} . Alternatively the measure μ_{x} may be described by

μ_{x} ({ε}) = ∏_{i=1}^{d} (1 − x_{i})^{1−ε_{i}} x_{i}^{ε_{i}}  (20.45)

for ε ∈ Ω. Notice that μ_{x} ({ε}) is a degree d polynomial in x for each ε ∈ Ω. For n ∈ ℕ and x ∈ [0, 1], let μ_{x}^{n} denote the n – fold product of μ_{x} with itself on Ω^{n}, X_{i}(ω) = ω_{i} ∈ Ω ⊂ ℝ^{d} for ω ∈ Ω^{n} and let

S_{n} = (S_{n}^{1}, …, S_{n}^{d}) := (X_{1} + X_{2} + ⋯ + X_{n})/n,

so S_{n} : Ω^{n} → ℝ^{d}. The reader is asked to verify (Exercise 20.4) that

∫_{Ω^{n}} S_{n}dμ_{x}^{n} := (∫_{Ω^{n}} S_{n}^{1}dμ_{x}^{n}, …, ∫_{Ω^{n}} S_{n}^{d}dμ_{x}^{n}) = (x_{1}, …, x_{d}) = x (20.46)

and

∫_{Ω^{n}} |S_{n} − x|^{2} dμ_{x}^{n} = \frac{1}{n} ∑_{i=1}^{d} x_{i}(1 − x_{i}) ≤ \frac{d}{n}.  (20.47)

From these equations it follows that S_{n} is concentrating near x as n → ∞, a manifestation of the law of large numbers. Therefore it is reasonable to expect

p_{n}(x) := ∫_{Ω^{n}} f(S_{n})dμ_{x}^{n}  (20.48)

should approach f(x) as n → ∞. Let ε > 0 be given, M = sup {|f(x)| : x ∈ [0, 1]} and

δ_{ε} = sup {|f(y) − f(x)| : x, y ∈ [0, 1] and |y − x| ≤ ε} .

By uniform continuity of f on [0, 1], lim_{ε↓0} δ_{ε} = 0. Using these definitions and the fact that μ_{x}^{n}(Ω^{n}) = 1,

|f(x) − p_{n}(x)| = |∫_{Ω^{n}} (f(x) − f(S_{n})) dμ_{x}^{n}| ≤ ∫_{Ω^{n}} |f(x) − f(S_{n})| dμ_{x}^{n}

≤ ∫_{{|S_{n}−x|>ε}} |f(x) − f(S_{n})| dμ_{x}^{n} + ∫_{{|S_{n}−x|≤ε}} |f(x) − f(S_{n})| dμ_{x}^{n}

≤ 2Mμ_{x}^{n} (|S_{n} − x| > ε) + δ_{ε}.  (20.49)

By Chebyshev’s inequality,

μ_{x}^{n} (|S_{n} − x| > ε) ≤ \frac{1}{ε^{2}} ∫_{Ω^{n}}(S_{n} − x)^{2}dμ_{x}^{n} = \frac{d}{nε^{2}} ,

and therefore, Eq. (20.49) yields the estimate

∥f − p_{n}∥_{∞} ≤ \frac{2dM}{nε^{2}} + δ_{ε}

and hence

lim sup_{n→∞} ∥f − p_{n}∥_{∞} ≤ δ_{ε} → 0 as ε ↓ 0.

This completes the proof since, using Eq. (20.45),

p_{n}(x) = ∑_{ω∈Ω^{n}} f(S_{n}(ω))μ_{x}^{n}({ω}) = ∑_{ω∈Ω^{n}} f(S_{n}(ω)) ∏_{i=1}^{n} μ_{x}({ω_{i}}),

is an nd – degree polynomial in x ∈ ℝ^{d}).

Exercise 20.4. Verify Eqs. (20.46) and (20.47). This is most easily done using Eqs. (20.43) and (20.44) and Fubini’s theorem repeatedly. (Of course Fubini’s theorem here is over kill since these are only finite sums after all. Nevertheless it is convenient to use this formulation.)

The second proof requires the next two lemmas.

Lemma 20.30 (Approximate δ – sequences). Suppose that {Q_{n}}_{n=1}^{∞} is a sequence of positive functions on ℝ^{d} such that

∫_{ℝ^{d}} Q_{n}(x) dx = 1 and  (20.50)
lim_{n→∞} ∫_{|x|≥ε} Q_{n}(x)dx = 0 for all ε > 0.  (20.51)

For f ∈ BC(ℝ^{d}), Q_{n} ∗ f converges to f uniformly on compact subsets of ℝ^{d}.

Proof. The proof is exactly the same as the proof of Lemma 8.28, it is only necessary to replace ℝ by ℝ^{d} everywhere in the proof.

Define

Q_{n} : ℝ^{n} → [0, ∞) by Q_{n}(x) = q_{n}(x_{1}) … q_{n}(x_{d}).  (20.52)

where q_{n} is defined in Eq. (8.23).

Lemma 20.31. The sequence {Q_{n}}_{n=1}^{∞} is an approximate δ – sequence, i.e. they satisfy Eqs. (20.50) and (20.51).

Proof. The fact that Q_{n} integrates to one is an easy consequence of Tonelli’s theorem and the fact that q_{n} integrates to one. Since all norms on ℝ^{d} are equivalent, we may assume that |x| = max {|x_{i}| : i = 1, 2, …, d} when proving Eq. (20.51). With this norm

{x ∈ ℝ^{d} : |x| ≥ ε} = ∪_{i=1}^{d} {x ∈ ℝ^{d} : |x_{i}| ≥ ε}

and therefore by Tonelli’s theorem,

∫_{{|x|≥ε}} Q_{n}(x)dx ≤ ∑_{i=1}^{d} ∫_{{|x_{i}|≥ε}} Q_{n}(x)dx = d ∫_{{x∈ℝ|x|≥ε}} q_{n}(t)dt

which tends to zero as n → ∞ by Lemma 8.29.

Proof. Proof of Theorem 8.34. Again we assume f ∈ C (ℝ^{d}, ℂ) and f ≡ 0 on Q_{d}^{c} where Q_{d} := (0, 1)^{d} . Let Q_{n}(x) be defined as in Eq. (20.52). Then by Lemma 20.31 and 20.30, p_{n}(x) := (Q_{n} ∗F)(x) → F (x) uniformly for x ∈ [0, 1] as n → ∞. So to finish the proof it only remains to show p_{n}(x) is a polynomial when x ∈ [0, 1]. For x ∈ [0, 1],

p_{n}(x) = ∫_{ℝ^{d}} Q_{n}(x − y)f(y)dy

= \frac{1}{c_{n}} ∫_{[0,1]} f(y) ∏_{i=1}^{d} [c_{n}^{−1}(1 − (x_{i} − y_{i})^{2})^{n}1_{|x_{i}−y_{i}|≤1}] dy

= \frac{1}{c_{n}} ∫_{[0,1]} f(y) ∏_{i=1}^{d} [c_{n}^{−1}(1 − (x_{i} − y_{i})^{2})^{n}] dy.

Since the product in the above integrand is a polynomial if (x, y) ∈ ℝ^{d} × ℝ^{d}, it follows easily that p_{n}(x) is polynomial in x.

20.5 More Spherical Coordinates

In this section we will define spherical coordinates in all dimensions. Along the way we will develop an explicit method for computing surface integrals on spheres. As usual when n = 2 define spherical coordinates (r, θ) ∈ (0, ∞)× [0, 2π) so that

( \begin{matrix}x_{1} \\ x_{2} \end{matrix} ) = ( \begin{matrix}r cos θ \\ r sin θ \end{matrix} ) = T_{2}(θ, r).

For n = 3 we let x_{3} = r cos ϕ_{1} and then

( \begin{matrix}x_{1} \\ x_{2} \end{matrix} ) = T_{2}(θ, r sin ϕ_{1}),

as can be seen from Figure 20.6, so that

Fig. 20.6. Setting up polar coordinates in two and three dimensions.

( \begin{matrix}x_{1} \\ x_{2} \\ x_{3} \end{matrix} ) = ( \begin{matrix}T_{2}(θ, r sin ϕ_{1}) \\ r cos ϕ_{1} \end{matrix} ) = ( \begin{matrix}r sin ϕ_{1} cos θ \\ r sin ϕ_{1} sin θ \\ r cos ϕ_{1} \end{matrix} ) =: T_{3}(θ, ϕ_{1}, r,).

We continue to work inductively this way to define

( x_{1} ; ⋮ ; x_{n} ; x_{n+1} ) = ( \begin{matrix}T_{n}(θ, ϕ_{1}, …, ϕ_{n−2}, r sin ϕ_{n−1},) \\ r cos ϕ_{n−1} \end{matrix} ) = T_{n+1}(θ, ϕ_{1}, …, ϕ_{n−2}, ϕ_{n−1}, r).

So for example,

x_{1} = r sin ϕ_{2} sin ϕ_{1} cos θ
x_{2} = r sin ϕ_{2} sin ϕ_{1} sin θ
x_{3} = r sin ϕ_{2} cos ϕ_{1}
x_{4} = r cos ϕ_{2}

and more generally,

x_{1} = r sin ϕ_{n−2} … sin ϕ_{2} sin ϕ_{1} cos θ
x_{2} = r sin ϕ_{n−2} … sin ϕ_{2} sin ϕ_{1} sin θ
x_{3} = r sin ϕ_{n−2} … sin ϕ_{2} cos ϕ_{1}
⋮
x_{n−2} = r sin ϕ_{n−2} sin ϕ_{n−3} cos ϕ_{n−4}
x_{n−1} = r sin ϕ_{n−2} cos ϕ_{n−3}
x_{n} = r cos ϕ_{n−2}.  (20.53)

By the change of variables formula, ∫_{ℝ^{n}} f(x)dm(x)

= ∫_{0}^{∞} dr ∫_{0≤ϕ_{i}≤π,0≤θ≤2π} dϕ_{1} … dϕ_{n−2}dθΔ_{n}(θ, ϕ_{1}, …, ϕ_{n−2}, r)f(T_{n}(θ, ϕ_{1}, …, ϕ_{n−2}, r)) (20.54)

where

Δ_{n}(θ, ϕ_{1}, …, ϕ_{n−2}, r) := |det T′_{n}(θ, ϕ_{1}, …, ϕ_{n−2}, r)| .

Proposition 20.32. The Jacobian, Δ_{n} is given by

Δ_{n}(θ, ϕ_{1}, …, ϕ_{n−2}, r) = r^{n−1} sin^{n−2} ϕ_{n−2} … sin^{2} ϕ_{2} sin ϕ_{1}.  (20.55)

If f is a function on rS^{n−1} – the sphere of radius r centered at 0 inside of ℝ^{n}, then

∫_{rS^{n−1}} f(x)dσ(x) = r^{n−1} ∫_{S^{n−1}} f(rω)dσ(ω)

= ∫_{0≤ϕ_{i}≤π,0≤θ≤2π} f(T_{n}(θ, ϕ_{1}, …, ϕ_{n−2}, r))Δ_{n}(θ, ϕ_{1}, …, ϕ_{n−2}, r)dϕ_{1} … dϕ_{n−2}dθ (20.56)

Proof. We are going to compute Δ_{n} inductively. Letting ρ := r sin ϕ_{n−1} and writing \frac{∂T_{n}}{∂ξ} for \frac{∂T_{n}}{∂ξ} (θ, ϕ_{1}, …, ϕ_{n−2}, ρ) we have

Δ_{n+1}(θ,ϕ_{1}, …, ϕ_{n−2}, ϕ_{n−1}, r)

= |[ 0 0^{\frac{∂T_{n}}{∂θ} \frac{∂T_{n}}{∂ϕ_{1}}} \begin{matrix}… & \frac{∂T_{n}}{∂ϕ_{n−2}} \\ … & 0 \end{matrix} \begin{matrix}\frac{∂T_{n}}{∂ρ} r cos ϕ_{n−1} \\ −r sin ϕ_{n−1} \end{matrix} \begin{matrix}\frac{∂T_{n}}{∂ρ} sin ϕ_{n−1} \\ cos ϕ_{n−1} \end{matrix} ]|

= r (cos^{2} ϕ_{n−1} + sin^{2} ϕ_{n−1}) Δ_{n}(, θ, ϕ_{1}, …, ϕ_{n−2}, ρ) = rΔ_{n}(θ, ϕ_{1}, …, ϕ_{n−2}, r sin ϕ_{n−1}),

i.e.

Δ_{n+1}(θ, ϕ_{1}, …, ϕ_{n−2}, ϕ_{n−1}, r) = rΔ_{n}(θ, ϕ_{1}, …, ϕ_{n−2}, r sin ϕ_{n−1}). (20.57)

To arrive at this result we have expanded the determinant along the bottom row. Staring with Δ_{2}(θ, r) = r already derived in Example 20.23, Eq. (20.57) implies,

Δ_{3}(θ, ϕ_{1}, r) = rΔ_{2}(θ, r sin ϕ_{1}) = r^{2} sin ϕ_{1}
Δ_{4}(θ, ϕ_{1}, ϕ_{2}, r) = rΔ_{3}(θ, ϕ_{1}, r sin ϕ_{2}) = r^{3} sin^{2} ϕ_{2} sin ϕ_{1}
⋮
Δ_{n}(θ, ϕ_{1}, …, ϕ_{n−2}, r) = r^{n−1} sin^{n−2} ϕ_{n−2} … sin^{2} ϕ_{2} sin ϕ_{1}

which proves Eq. (20.55). Eq. (20.56) now follows from Eqs. (??), (20.54) and (20.55).

As a simple application, Eq. (20.56) implies

σ(S^{n−1}) = ∫_{0≤ϕ_{i}≤π,0≤θ≤2π} sin^{n−2} ϕ_{n−2} … sin^{2} ϕ_{2} sin ϕ_{1}dϕ_{1} … dϕ_{n−2}dθ

= 2π ∏_{k=1}^{n−2} γ_{k} = σ(S^{n−2})γ_{n−2}  (20.58)

where γ_{k} := ∫_{0}^{π} sin^{k} ϕdϕ. If k ≥ 1, we have by integration by parts that,

γ_{k} = ∫_{0}^{π} sin^{k} ϕdϕ = − ∫_{0}^{π} sin^{k−1} ϕ d cos ϕ = 2δ_{k,1} + (k − 1) ∫_{0}^{π} sin^{k−2} ϕ cos^{2} ϕdϕ

= 2δ_{k,1} + (k − 1) ∫_{0}^{π} sin^{k−2} ϕ (1 − sin^{2} ϕ) dϕ = 2δ_{k,1} + (k − 1) [γ_{k−2} − γ_{k}]

and hence γ_{k} satisfies γ_{0} = π, γ_{1} = 2 and the recursion relation

γ_{k} = \frac{k − 1}{k}γ_{k−2} for k ≥ 2.

Hence we may conclude

γ_{0} = π, γ_{1} = 2, γ_{2} = \frac{1}{2}π, γ_{3} = \frac{2}{3}2, γ_{4} = \frac{3}{4} \frac{1}{2}π, γ_{5} = \frac{4}{5} \frac{2}{3}2, γ_{6} = \frac{5}{6} \frac{3}{4} \frac{1}{2}π

and more generally by induction that

γ_{2k} = π \frac{(2k − 1)!!}{(2k)!!} and γ_{2k+1} = 2\frac{(2k)!!}{(2k + 1)!!}.

Indeed,

γ_{2(k+1)+1} = \frac{2k + 2}{2k + 3}γ_{2k+1} = \frac{2k + 2}{2k + 3}2\frac{(2k)!!}{(2k + 1)!!} = 2\frac{[2(k + 1)]!!}{(2(k + 1) + 1)!!}

and

γ_{2(k+1)} = \frac{2k + 1}{2k + 1}γ_{2k} = \frac{2k + 1}{2k + 2}π \frac{(2k − 1)!!}{(2k)!!} = π \frac{(2k + 1)!!}{(2k + 2)!!}.

The recursion relation in Eq. (20.58) may be written as

σ(S^{n}) = σ (S^{n−1}) γ_{n−1}  (20.59)

which combined with σ (S^{1}) = 2π implies

σ (S^{1}) = 2π,
σ(S^{2}) = 2π · γ_{1} = 2π · 2,
σ(S^{3}) = 2π · 2 · γ_{2} = 2π · 2 · \frac{1}{2}π = \frac{2^{2}π^{2}}{2!!} ,
σ(S^{4}) = \frac{2^{2}π^{2}}{2!!} · γ_{3} = \frac{2^{2}π^{2}}{2!!} · 2\frac{2}{3} = \frac{2^{3}π^{2}}{3!!}
σ(S^{5}) = 2π · 2 · \frac{1}{2}π · \frac{2}{3}2 · \frac{3}{4} \frac{1}{2}π = \frac{2^{3}π^{3}}{4!!} ,
σ(S^{6}) = 2π · 2 · \frac{1}{2}π · \frac{2}{3}2 · \frac{3}{4} \frac{1}{2}π · \frac{4}{5} \frac{2}{3}2 = \frac{2^{4}π^{3}}{5!!}

and more generally that

σ(S^{2n}) = \frac{2 (2π)^{n}}{(2n − 1)!!} and σ(S^{2n+1}) = \frac{(2π)^{n+1}}{(2n)!!}  (20.60)

which is verified inductively using Eq. (20.59). Indeed,

σ(S^{2n+1}) = σ(S^{2n})γ_{2n} = \frac{2 (2π)^{n}}{(2n − 1)!!}π \frac{(2n − 1)!!}{(2n)!!} = \frac{(2π)^{n+1}}{(2n)!!}

and

σ(S^{(n+1)}) = σ(S^{2n+2}) = σ(S^{2n+1})γ_{2n+1} = \frac{(2π)^{n+1}}{(2n)!!} 2\frac{(2n)!!}{(2n + 1)!!} = \frac{2 (2π)^{n+1}}{(2n + 1)!!}.

Using

(2n)!! = 2n (2(n − 1)) … (2 · 1) = 2^{n}n!

we may write σ(S^{2n+1}) = \frac{2π^{n+1}}{n!} which shows that Eqs. (??) and (20.60 in agreement. We may also write the formula in Eq. (20.60) as

σ(S^{n}) = \{ \begin{matrix}\frac{2(2π)^{n/2}}{(n−1)!!} & for & n even \\ \frac{(2π)^{\frac{n+1}{2}}}{(n−1)!!} & for & n odd. \end{matrix}

20.6 Sard’s Theorem

See p. 538 of Taylor and references. Also see Milnor’s topology book. Add in the Brower’s Fixed point theorem here as well. Also Spivak’s calculus on manifolds.

Theorem 20.33. Let U ⊂_{o} ℝ^{m}, f ∈ C^{∞}(U, ℝ^{d}) and C := {x ∈ U : Ran k(f′(x)) < n} be the set of critical points of f. Then the critical values, f(C), is a Borel measurable subset of ℝ^{d} of Lebesgue measure 0.

Remark 20.34. This result clearly extends to manifolds.

For simplicity in the proof given below it will be convenient to use the norm, |x| := max_{i} |x_{i}| . Recall that if f ∈ C^{1}(U, ℝ^{d}) and p ∈ U, then

f(p+x) = f(p)+∫_{0}^{1} f′(p+tx)xdt = f(p)+f′(p)x+∫_{0}^{1} [f′(p + tx) − f′(p)] xdt

so that if

R(p, x) := f(p + x) − f(p) − f′(p)x = ∫_{0}^{1} [f′(p + tx) − f′(p)] xdt

we have

|R(p, x)| ≤ |x| ∫_{0}^{1} |f′(p + tx) − f′(p)| dt = |x| ε(p, x).

By uniform continuity, it follows for any compact subset K ⊂ U that

sup {|ε(p, x)| : p ∈ K and |x| ≤ δ} → 0 as δ ↓ 0.

Proof. Notice that if x ∈ U \ C, then f′(x) : ℝ^{m} → ℝ^{n} is surjective, which is an open condition, so that U \ C is an open subset of U. This shows C is relatively closed in U, i.e. there exists C̃ ⊏ ℝ^{m} such that C = C̃ ∩ U. Let K_{n} ⊂ U be compact subsets of U such that K_{n} ↑ U, then K_{n} ∩ C ↑ C and K_{n} ∩ C = K_{n} ∩ C̃ is compact for each n. Therefore, f(K_{n} ∩ C) ↑ f(C) i.e. f(C) = ∪_{n}f(K_{n} ∩ C) is a countable union of compact sets and therefore is Borel measurable. Moreover, since m(f(C)) = lim_{n→∞} m(f(K_{n} ∩ C)), it suffices to show m(f(K)) = 0 for all compact subsets K ⊂ C. Case 1. (n ≤ m) Let K = [a, a + γ] be a cube contained in U and by scaling the domain we may assume γ = (1, 1, 1, …, 1). For N ∈ ℕ and j ∈ S_{N} := {0, 1, …, N − 1}^{n} let K_{j} = j/N + [a, a + γ/N] so that K = ∪_{j∈S_{N}} K_{j} with K_{j}^{o} ∩ K_{j′}^{o} = ∅ if j ≠ j′. Let {Q_{j} : j = 1 …, M} be the collection of those {K_{j} : j ∈ S_{N}} which intersect C. For each j, let p_{j} ∈ Q_{j} ∩ C and for x ∈ Q_{j} − p_{j} we have

f(p_{j} + x) = f(p_{j}) + f′(p_{j})x + R_{j}(x)

where |R_{j}(x)| ≤ ε_{j}(N)/N and ε(N) := max_{j} ε_{j}(N) → 0 as N → ∞. Now

m (f(Q_{j})) = m (f(p_{j}) + (f′(p_{j}) + R_{j}) (Q_{j} − p_{j}))
= m ((f′(p_{j}) + R_{j}) (Q_{j} − p_{j}))
= m (O_{j} (f′(p_{j}) + R_{j}) (Q_{j} − p_{j}))  (20.61)

where O_{j} ∈ SO(n) is chosen so that O_{j}f′(p_{j})ℝ^{n} ⊂ ℝ^{m−1} × {0} . Now O_{j}f′(p_{j})(Q_{j} − p_{j}) is contained in Γ × {0} where Γ ⊂ ℝ^{m−1} is a cube centered at 0 ∈ ℝ^{m−1} with side length at most 2 |f′(p_{j})| /N ≤ 2M/N where M = max_{p∈K} |f′(p)| . It now follows that O_{j} (f′(p_{j}) + R_{j}) (Q_{j} − p_{j}) is contained the set of all points within ε(N)/N of Γ × {0} and in particular

O_{j} (f′(p_{j}) + R_{j}) (Q_{j} − p_{j}) ⊂ (1 + ε(N)/N) Γ × [ε(N)/N, ε(N)/N].

From this inclusion and Eq. (20.61) it follows that

m (f(Q_{j})) ≤ [2\frac{M}{N} (1 + ε(N)/N)]^{m−1} 2ε(N)/N

= 2^{m}M^{m−1} [(1 + ε(N)/N)]^{m−1} ε(N)\frac{1}{N^{m}}

and therefore,

m (f(C ∩ K)) ≤ ∑_{j} m (f(Q_{j})) ≤ N^{n}2^{m}M^{m−1} [(1 + ε(N)/N)]^{m−1} ε(N)\frac{1}{N^{m}}

= 2^{n}M^{n−1} [(1 + ε(N)/N)]^{n−1} ε(N)\frac{1}{N^{m−n}} → 0 as N → ∞

since m ≥ n. This proves the easy case since we may write U as a countable union of cubes K as above. Remark. The case (m < n) also follows from the case m = n as follows. When m < n, C = U and we must show m(f(U)) = 0. Letting F : U×ℝ^{n−m} → ℝ^{n} be the map F (x, y) = f(x). Then F′(x, y)(v, w) = f′(x)v, and hence C_{F} := U × ℝ^{n−m}. So if the assertion holds for m = n we have

m(f(U)) = m(F (U × ℝ^{n−m})) = 0.

Case 2. (m > n) This is the hard case and the case we will need in the co-area formula to be proved later. Here I will follow the proof in Milnor. Let

C_{i} := {x ∈ U : ∂^{α}f(x) = 0 when |α| ≤ i}

so that C ⊃ C_{1} ⊃ C_{2} ⊃ C_{3} ⊃ …. The proof is by induction on n and goes by the following steps:

1. m(f(C \ C_{1})) = 0.

2. m(f(C_{i} \ C_{i+1})) = 0 for all i ≥ 1.

3. m(f(C_{i})) = 0 for all i sufficiently large.

Step 1. If m = 1, there is nothing to prove since C = C_{1} so we may assume m ≥ 2. Suppose that x ∈ C \ C_{1}, then f′(p) ≠ 0 and so by reordering the components of x and f(p) if necessary we may assume that ∂_{1}f_{1} (p) ≠ 0 where we are writing ∂f(p)/∂x_{i} as ∂_{i}f (p) . The map h(x) := (f_{1}(x), x_{2}, …, x_{n}) has differential

h′(p) = [ \begin{matrix}∂_{1}f_{1} (p) & ∂_{2}f_{1} (p) & … & ∂_{n}f_{1} (p) \\ 0 & 1 & 0 & 0 \\ ⋮ & ⋮ & ... & ⋮ \\ 0 & 0 & 0 & 1 \end{matrix} ]

which is not singular. So by the implicit function theorem, there exists there exists V ∈ τ_{p} such that h : V → h(V) ∈ τ_{h(p)} is a diffeomorphism and in particular ∂f_{1}(x)/∂x_{1} ≠ 0 for x ∈ V and hence V ⊂ U \ C_{1}. Consider the map g := f ∘ h^{−1} : V′ := h(V) → ℝ^{m}, which satisfies

(f_{1}(x), f_{2}(x), …, f_{m}(x)) = f(x) = g(h(x)) = g((f_{1}(x), x_{2}, …, x_{n}))

which implies g(t, y) = (t, u(t, y)) for (t, y) ∈ V′ := h(V) ∈ τ_{h(p)}, see Figure 20.7 below where p = x̄ and m = p. Since

Fig. 20.7. Making a change of variable so as to apply induction.

g′(t, y) = [ \begin{matrix}1 & 0 \\ ∂_{t}u(t, y) & ∂_{y}u(t, y) \end{matrix} ]

it follows that (t, y) is a critical point of g iff y ∈ C′_{t} – the set of critical points of y → u(t, y). Since h is a diffeomorphism we have C′ := h(C ∩ V) are the critical points of g in V′ and

f(C ∩ V) = g(C′) = ∪_{t} [{t} × u_{t}(C′_{t})] .

By the induction hypothesis, m_{m−1}(u_{t}(C′_{t})) = 0 for all t, and therefore by Fubini’s theorem,

m(f(C ∩ V)) = ∫_{ℝ} m_{m−1}(u_{t}(C′_{t}))1_{V′_{t} ≠∅}dt = 0.

Since C \C_{1} may be covered by a countable collection of open sets V as above, it follows that m(f(C \ C_{1})) = 0. Step 2. Suppose that p ∈ C_{k} \ C_{k+1}, then there is an α such that |α| = k + 1 such that ∂^{α}f(p) = 0 while ∂^{β}f(p) = 0 for all |β| ≤ k. Again by permuting coordinates we may assume that α_{1} ≠ 0 and ∂^{α}f_{1}(p) ≠ 0. Let w(x) := ∂^{α−e_{1}}f_{1}(x), then w(p) = 0 while ∂_{1}w(p) ≠ 0.

So again the implicit function theorem there exists V ∈ τ_{p} such that h(x) := (w (x) , x_{2}, …, x_{n}) maps V → V′ := h(V) ∈ τ_{h(p)} in a diffeomorphic way and in particular ∂_{1}w(x) ≠ 0 on V so that V ⊂ U \C_{k+1}. As before, let g := f ∘h^{−1} and notice that C′_{k} := h(C_{k} ∩ V) ⊂ {0} × ℝ^{n−1} and

f(C_{k} ∩ V) = g(C′_{k}) = ḡ (C′_{k})

where ḡ := g|_{({0}×ℝ^{n−1})∩V′}. Clearly C′_{k} is contained in the critical points of ḡ, and therefore, by induction

0 = m(ḡ(C′_{k})) = m(f(C_{k} ∩ V)).

Since C_{k}\C_{k+1} is covered by a countable collection of such open sets, it follows that

m(f(C_{k} \ C_{k+1})) = 0 for all k ≥ 1.

Step 3. Suppose that Q is a closed cube with edge length δ contained in U and k > n/m − 1. We will show m(f(Q ∩ C_{k})) = 0 and since Q is arbitrary it will follows that m(f(C_{k})) = 0 as desired. By Taylor’s theorem with (integral) remainder, it follows for x ∈ Q ∩ C_{k} and h such that x + h ∈ Q that

f(x + h) = f(x) + R(x, h)

where

|R(x, h)| ≤ c ∥h∥^{k+1}

where c = c(Q, k). Now subdivide Q into r^{n} cubes of edge size δ/r and let Q′ be one of the cubes in this subdivision such that Q′ ∩ C_{k} ≠ ∅ and let x ∈ Q′ ∩ C_{k}. It then follows that f(Q′) is contained in a cube centered at f(x) ∈ ℝ^{m} with side length at most 2c (δ/r)^{k+1} and hence volume at most (2c)^{m} (δ/r)^{m(k+1)} . Therefore, f(Q ∩ C_{k}) is contained in the union of at most r^{n} cubes of volume (2c)^{m} (δ/r)^{m(k+1)} and hence meach

m (f(Q ∩ C_{k})) ≤ (2c)^{m} (δ/r)^{m(k+1)} r^{n} = (2c)^{m} δ^{m(k+1)}r^{n−m(k+1)} → 0 as r ↑ ∞

provided that n − m(k + 1) < 0, i.e. provided k > n/m − 1.

20.7 Exercises

Exercise 20.5. Prove Theorem 20.12. Suggestion, to get started define

π (A) := ∫_{X_{1}} dμ (x_{1}) … ∫_{X_{n}} dμ (x_{n}) 1_{A} (x_{1}, …, x_{n})

and then show Eq. (20.18) holds. Use the case of two factors as the model of your proof.

Exercise 20.6. Let (X_{j}, M_{j}, μ_{j}) for j = 1, 2, 3 be σ – finite measure spaces. Let F : (X_{1} × X_{2}) × X_{3} → X_{1} × X_{2} × X_{3} be defined by

F ((x_{1}, x_{2}), x_{3}) = (x_{1}, x_{2}, x_{3}).

1. Show F is ((M_{1} ⊗ M_{2}) ⊗ M_{3}, M_{1} ⊗ M_{2} ⊗ M_{3}) – measurable and F^{−1} is (M_{1} ⊗ M_{2} ⊗ M_{3}, (M_{1} ⊗ M_{2}) ⊗ M_{3}) – measurable. That is

F : ((X_{1} × X_{2})×X_{3}, (M_{1} ⊗ M_{2})⊗M_{3}) → (X_{1}×X_{2}×X_{3}, M_{1}⊗M_{2}⊗M_{3})

is a “measure theoretic isomorphism.”

2. Let π := F_{∗} [(μ_{1} ⊗ μ_{2}) ⊗ μ_{3}] , i.e. π(A) = [(μ_{1} ⊗ μ_{2}) ⊗ μ_{3}] (F^{−1}(A)) for all A ∈ M_{1} ⊗ M_{2} ⊗ M_{3}. Then π is the unique measure on M_{1} ⊗ M_{2} ⊗ M_{3} such that

π(A_{1} × A_{2} × A_{3}) = μ_{1}(A_{1})μ_{2}(A_{2})μ_{3}(A_{3})

for all A_{i} ∈ M_{i}. We will write π := μ_{1} ⊗ μ_{2} ⊗ μ_{3}.

3. Let f : X_{1} × X_{2} × X_{3} → [0, ∞] be a (M_{1} ⊗ M_{2} ⊗ M_{3}, Bℝ̄) – measurable function. Verify the identity,

∫_{X_{1}×X_{2}×X_{3}} fdπ = ∫_{X_{3}} dμ_{3}(x_{3}) ∫_{X_{2}} dμ_{2}(x_{2}) ∫_{X_{1}} dμ_{1}(x_{1})f(x_{1}, x_{2}, x_{3}),

makes sense and is correct.

4. (Optional.) Also show the above identity holds for any one of the six possible orderings of the iterated integrals.

Exercise 20.7. Prove the second assertion of Theorem 20.18. That is show m^{d} is the unique translation invariant measure on B_{ℝ^{d}} such that m^{d}((0, 1]^{d}) = 1. Hint: Look at the proof of Theorem 19.10.

Exercise 20.8. (Part of Folland Problem 2.46 on p. 69.) Let X = [0, 1], M = B_{[0,1]} be the Borel σ – field on X, m be Lebesgue measure on [0, 1] and ν be counting measure, ν(A) = #(A). Finally let D = {(x, x) ∈ X^{2} : x ∈ X} be the diagonal in X^{2}. Show

∫_{X} [∫_{X} 1_{D}(x, y)dν(y)] dm(x) ≠ ∫_{X} [∫_{X} 1_{D}(x, y)dm(x)] dν(y)

by explicitly computing both sides of this equation.

Exercise 20.9. Folland Problem 2.48 on p. 69. (Counter example related to Fubini Theorem involving counting measures.)

Exercise 20.10. Folland Problem 2.50 on p. 69 pertaining to area under a curve. (Note the M × B_{ℝ} should be M ⊗ Bℝ̄ in this problem.)

Exercise 20.11. Folland Problem 2.55 on p. 77. (Explicit integrations.)

20.7 Exercises 375

Exercise 20.12. Folland Problem 2.56 on p. 77. Let f ∈ L^{1}((0, a), dm), g(x) = ∫_{x}^{a} \frac{f(t)}{t}dt for x ∈ (0, a), show g ∈ L^{1}((0, a), dm) and

∫_{0}^{a} g(x)dx = ∫_{0}^{a} f(t)dt.

Exercise 20.13. Show ∫_{0}^{∞} |\frac{sin x}{x} | dm(x) = ∞. So \frac{sin x}{x} ∉ L^{1}([0, ∞), m) and ∫_{0}^{∞} \frac{sin x}{x}dm(x) is not defined as a Lebesgue integral. Exercise 20.14. Folland Problem 2.57 on p. 77. Exercise 20.15. Folland Problem 2.58 on p. 77. Exercise 20.16. Folland Problem 2.60 on p. 77. Properties of the Γ – function. Exercise 20.17. Folland Problem 2.61 on p. 77. Fractional integration. Exercise 20.18. Folland Problem 2.62 on p. 80. Rotation invariance of surface measure on S^{n−1}. Exercise 20.19. Folland Problem 2.64 on p. 80. On the integrability of |x|^{a} |log |x||^{b} for x near 0 and x near ∞ in ℝ^{n}. Exercise 20.20. Show, using Problem 20.18 that

∫_{S^{d−1}} ω_{i}ω_{j}dσ (ω) = \frac{1}{d}δ_{ij}σ (S^{d−1}) .

Hint: show ∫_{S^{d−1}} ω_{i}^{2}dσ (ω) is independent of i and therefore

∫_{S^{d−1}} ω_{i}^{2}dσ (ω) = \frac{1}{d} ∑_{j=1}^{d} ∫_{S^{d−1}} ω_{j}^{2}dσ (ω) .

L^{p}-spaces

Let (X, M, μ) be a measure space and for 0 < p < ∞ and a measurable function f : X → ℂ let

∥f∥_{p} := (∫_{X} |f|^{p} dμ)^{1/p} .  (21.1)

When p = ∞, let

∥f∥_{∞} = inf {a ≥ 0 : μ(|f| > a) = 0}  (21.2)

For 0 < p ≤ ∞, let

L^{p}(X, M, μ) = {f : X → ℂ : f is measurable and ∥f∥_{p} < ∞}/ ∼

where f ∼ g iff f = g a.e. Notice that ∥f − g∥_{p} = 0 iff f ∼ g and if f ∼ g then ∥f∥_{p} = ∥g∥_{p}. In general we will (by abuse of notation) use f to denote both the function f and the equivalence class containing f.

Remark 21.1. Suppose that ∥f∥_{∞} ≤ M, then for all a > M, μ(|f| > a) = 0 and therefore μ(|f| > M) = lim_{n→∞} μ(|f| > M + 1/n) = 0, i.e. |f(x)| ≤ M for μ - a.e. x. Conversely, if |f| ≤ M a.e. and a > M then μ(|f| > a) = 0 and hence ∥f∥_{∞} ≤ M. This leads to the identity:

∥f∥_{∞} = inf {a ≥ 0 : |f(x)| ≤ a for μ – a.e. x} .

The next theorem is a generalization Theorem 5.6 to general integrals and the proof is essentially identical to the proof of Theorem 5.6.

Theorem 21.2 (Hölder’s inequality). Suppose that 1 ≤ p ≤ ∞ and q := \frac{p}{p−1}, or equivalently p^{−1} + q^{−1} = 1. If f and g are measurable functions then

∥fg∥_{1} ≤ ∥f∥_{p} · ∥g∥_{q}.  (21.3)

Assuming p ∈ (1, ∞) and ∥f∥_{p} · ∥g∥_{q} < ∞, equality holds in Eq. (21.3) iff |f|^{p} and |g|^{q} are linearly dependent as elements of L^{1} which happens iff

|g|^{q}∥f∥_{p}^{p} = ∥g∥_{q}^{q} |f|^{p} a.e.  (21.4)

378 21 L^{p}-spaces

Proof. The cases where ∥f∥_{q} = 0 or ∞ or ∥g∥_{p} = 0 or ∞ are easy to deal with and are left to the reader. So we will now assume that 0 < ∥f∥_{q}, ∥g∥_{p} < ∞. Let s = |f| /∥f∥_{p} and t = |g|/∥g∥_{q} then Lemma 5.5 implies

\frac{|fg|}{∥f∥_{p}∥g∥_{q}} ≤ \frac{1}{p} \frac{|f|^{p}}{∥f∥_{p}} + \frac{1}{q} \frac{|g|^{q}}{∥g∥^{q}}  (21.5)

with equality iff |g/∥g∥_{q}| = |f|^{p−1} /∥f∥_{p}^{(p−1)} = |f|^{p/q} /∥f∥_{p}^{p/q}, i.e. |g|^{q}∥f∥_{p}^{p} = ∥g∥_{q}^{q} |f|^{p} . Integrating Eq. (21.5) implies

\frac{∥fg∥_{1}}{∥f∥_{p}∥g∥_{q}} ≤ \frac{1}{p} + \frac{1}{q} = 1

with equality iff Eq. (21.4) holds. The proof is finished since it is easily checked that equality holds in Eq. (21.3) when |f|^{p} = c |g|^{q} of |g|^{q} = c |f|^{p} for some constant c.

The following corollary is an easy extension of Hölder’s inequality.

Corollary 21.3. Suppose that f_{i} : X → ℂ are measurable functions for i = 1, …, n and p_{1}, …, p_{n} and r are positive numbers such that ∑_{i=1}^{n} p_{i}^{−1} = r^{−1}, then

∥∏_{i=1}^{n} f_{i}∥_{r} ≤ ∏_{i=1}^{n} ∥f_{i}∥_{p_{i}} where ∑_{i=1}^{n} p_{i}^{−1} = r^{−1}.

Proof. To prove this inequality, start with n = 2, then for any p ∈ [1, ∞],

∥fg∥_{r}^{r} = ∫_{X} |f|^{r} |g|^{r} dμ ≤ ∥f^{r}∥_{p} ∥g^{r}∥_{p^{∗}}

where p^{∗} = \frac{p}{p−1} is the conjugate exponent. Let p_{1} = pr and p_{2} = p^{∗}r so that p_{1}^{−1} + p_{2}^{−1} = r^{−1} as desired. Then the previous equation states that

∥fg∥_{r} ≤ ∥f∥_{p_{1}} ∥g∥_{p_{2}}

as desired. The general case is now proved by induction. Indeed,

∥∏_{i=1}^{n+1} f_{i}∥_{r} = ∥∏_{i=1}^{n} f_{i} · f_{n+1}∥_{r} ≤ ∥∏_{i=1}^{n} f_{i}∥_{q} ∥f_{n+1}∥_{p_{n+1}}

where q^{−1}+p_{n+1}^{−1} = r^{−1}. Since ∑_{i=1}^{n} p_{i}^{−1} = q^{−1}, we may now use the induction hypothesis to conclude

∥∏_{i=1}^{n} f_{i}∥_{q} ≤ ∏_{i=1}^{n} ∥f_{i}∥_{p_{i}} ,

which combined with the previous displayed equation proves the generalized form of Holder’s inequality.

Theorem 21.4 (Minkowski’s Inequality). If 1 ≤ p ≤ ∞ and f, g ∈ L^{p} then

∥f + g∥_{p} ≤ ∥f∥_{p} + ∥g∥_{p}.  (21.6)

Moreover, assuming f and g are not identically zero, equality holds in Eq. (21.6) iff sgn(f) ≗ sgn(g) a.e. (see the notation in Definition 5.7) when p = 1 and f = cg a.e. for some c > 0 for p ∈ (1, ∞).

Proof. When p = ∞, |f| ≤ ∥f∥_{∞} a.e. and |g| ≤ ∥g∥_{∞} a.e. so that |f + g| ≤ |f| + |g| ≤ ∥f∥_{∞} + ∥g∥_{∞} a.e. and therefore

∥f + g∥_{∞} ≤ ∥f∥_{∞} + ∥g∥_{∞} .

When p < ∞,

|f + g|^{p} ≤ (2 max (|f| , |g|))^{p} = 2^{p} max (|f|^{p} , |g|^{p}) ≤ 2^{p} (|f|^{p} + |g|^{p}) ,
∥f + g∥_{p}^{p} ≤ 2^{p} (∥f∥_{p}^{p} + ∥g∥_{p}^{p}) < ∞.

In case p = 1,

∥f + g∥_{1} = ∫_{X} |f + g|dμ ≤ ∫_{X} |f| dμ + ∫_{X} |g|dμ

with equality iff |f| + |g| = |f + g| a.e. which happens iff sgn(f) ≗ sgn(g) a.e. In case p ∈ (1, ∞), we may assume ∥f + g∥_{p}, ∥f∥_{p} and ∥g∥_{p} are all positive since otherwise the theorem is easily verified. Now

|f + g|^{p} = |f + g||f + g|^{p−1} ≤ (|f| + |g|)|f + g|^{p−1}

with equality iff sgn(f) ≗ sgn(g). Integrating this equation and applying Holder’s inequality with q = p/(p − 1) gives

∫_{X} |f + g|^{p}dμ ≤ ∫_{X} |f| |f + g|^{p−1}dμ + ∫_{X} |g| |f + g|^{p−1}dμ
≤ (∥f∥_{p} + ∥g∥_{p}) ∥ |f + g|^{p−1} ∥_{q}  (21.7)

with equality iff

sgn(f) ≗ sgn(g) and
(\frac{|f|}{∥f∥_{p}} )^{p} = \frac{|f + g|^{p}}{∥f + g∥_{p}^{p}} = (\frac{|g|}{∥g∥_{p}} )^{p} a.e.  (21.8)

Therefore

∥|f + g|^{p−1}∥_{q}^{q} = ∫_{X}(|f + g|^{p−1})^{q}dμ = ∫_{X} |f + g|^{p}dμ.  (21.9)

Combining Eqs. (21.7) and (21.9) implies

∥f + g∥_{p}^{p} ≤ ∥f∥_{p}∥f + g∥_{p}^{p/q} + ∥g∥_{p}∥f + g∥_{p}^{p/q}  (21.10)

with equality iff Eq. (21.8) holds which happens iff f = cg a.e. with c > 0. Solving for ∥f + g∥_{p} in Eq. (21.10) gives Eq. (21.6).

The next theorem gives another example of using Hölder’s inequality

Theorem 21.5. Suppose that (X, M, μ) and (Y, N , ν) be σ – finite measure spaces, p ∈ [1, ∞], q = p/(p−1) and k : X ×Y → ℂ be a M⊗N – measurable function. Assume there exist finite constants C_{1} and C_{2} such that

∫_{X} |k(x, y)| dμ(x) ≤ C_{1} for ν a.e. y and
∫_{Y} |k(x, y)| dν(y) ≤ C_{2} for μ a.e. x.

If f ∈ L^{p}(ν), then

∫_{Y} |k(x, y)f(y)| dν(y) < ∞ for μ – a.e. x,

x → Kf(x) := ∫_{Y} k(x, y)f(y)dν(y) ∈ L^{p}(μ) and

∥Kf∥_{L^{p}(μ)} ≤ C_{1}^{1/p}C_{2}^{1/q} ∥f∥_{L^{p}(ν)}  (21.11)

Proof. Suppose p ∈ (1, ∞) to begin with and let q = p/(p − 1), then by Hölder’s inequality,

∫_{Y} |k(x, y)f(y)| dν(y) = ∫_{Y} |k(x, y)|^{1/q} |k(x, y)|^{1/p} |f(y)| dν(y)

≤ [∫_{Y} |k(x, y)| dν(y)]^{1/q} [∫_{Y} |k(x, y)| |f(y)|^{p} dν(y)]^{1/p}

≤ C_{2}^{1/q} [∫_{Y} |k(x, y)| |f(y)|^{p} dν(y)]^{1/p} .

Therefore,

∥∫_{Y} |k(·, y)f(y)| dν(y)∥_{L^{p}(μ)}^{p} = ∫_{X} dμ(x) [∫_{Y} |k(x, y)f(y)| dν(y)]^{p}
≤ C_{2}^{p/q} ∫_{X} dμ(x) ∫_{Y} dν(y) |k(x, y)| |f(y)|^{p}
= C_{2}^{p/q} ∫_{Y} dν(y) |f(y)|^{p} ∫_{X} dμ(x) |k(x, y)|
≤ C_{2}^{p/q}C_{1} ∫_{Y} dν(y) |f(y)|^{p} = C_{2}^{p/q}C_{1} ∥f∥_{L^{p}(ν)}^{p} ,

wherein we used Tonelli’s theorem in third line. From this it follows that ∫_{Y} |k(x, y)f(y)| dν(y) < ∞ for μ - a.e. x,

x → Kf(x) := ∫_{Y} k(x, y)f(y)dν(y) ∈ L^{p}(μ)

and that Eq. (21.11) holds.

Similarly if p = ∞,

∫_{Y} |k(x, y)f(y)| dν(y) ≤ ∥f∥_{L^{∞}(ν)}·∫_{Y} |k(x, y)| dν(y) ≤ C_{2} ∥f∥_{L^{∞}(ν)} for μ – a.e. x.

so that ∥Kf∥_{L^{∞}(μ)} ≤ C_{2} ∥f∥_{L^{∞}(ν)} . If p = 1, then

∫_{X} dμ(x) ∫_{Y} dν(y) |k(x, y)f(y)| = ∫_{Y} dν(y) |f(y)| ∫_{X} dμ(x) |k(x, y)|
≤ C_{1} ∫_{Y} dν(y) |f(y)|

which shows ∥Kf∥_{L^{1}(μ)} ≤ C_{1} ∥f∥_{L^{1}(ν)} .

21.1 Jensen’s Inequality

Definition 21.6. A function ϕ : (a, b) → ℝ is convex if for all a < x_{0} < x_{1} < b and t ∈ [0, 1] ϕ(x_{t}) ≤ tϕ(x_{1}) + (1 − t)ϕ(x_{0}) where x_{t} = tx_{1} + (1 − t)x_{0}.

Example 21.7. The functions exp(x) and − log(x) are convex and x^{p} is convex iff p ≥ 1 as follows from Corollary 21.9 below which in part states that any ϕ ∈ C^{2} ((a, b) , ℝ) such that ϕ′′ ≥ 0 is convex.

The following Proposition is clearly motivated by Figure 21.1.

Proposition 21.8. Suppose ϕ : (a, b) → ℝ is a convex function, then

1. For all u, v, w, z ∈ (a, b) such that u < z, w ∈ [u, z) and v ∈ (u, z],

\frac{ϕ(v) − ϕ(u)}{v − u} ≤ \frac{ϕ(z) − ϕ(w)}{z − w}.  (21.12)

2. For each c ∈ (a, b), the right and left sided derivatives ϕ′_{±}(c) exists in ℝ and if a < u < v < b, then ϕ′_{+}(u) ≤ ϕ′_{−}(v) ≤ ϕ′_{+}(v).

3. The function ϕ is continuous.

4. For all t ∈ (a, b) and β ∈ [ϕ′_{−}(t), ϕ′_{+}(t)], ϕ(x) ≥ ϕ(t) + β(x − t) for all x ∈ (a, b). In particular,

ϕ(x) ≥ ϕ(t) + ϕ′_{−}(t)(x − t) for all x, t ∈ (a, b).

Fig. 21.1. A convex function along with two cords corresponding to x_{0} = −2 and x_{1} = 4 and x_{0} = −5 and x_{1} = −2.

Proof. 1a) Suppose first that u < v = w < z, in which case Eq. (21.12) is equivalent to

(ϕ(v) − ϕ(u)) (z − v) ≤ (ϕ(z) − ϕ(v)) (v − u)

which after solving for ϕ(v) is equivalent to the following equations holding:

ϕ(v) ≤ ϕ(z)\frac{v − u}{z − u} + ϕ(u)\frac{z − v}{z − u}.

But this last equation states that ϕ(v) ≤ ϕ(z)t + ϕ(u) (1 − t) where t = \frac{v−u}{z−u} and v = tz + (1 − t)u and hence is valid by the definition of ϕ being convex. 1b) Now assume u = w < v < z, in which case Eq. (21.12) is equivalent to

(ϕ(v) − ϕ(u)) (z − u) ≤ (ϕ(z) − ϕ(u)) (v − u)

which after solving for ϕ(v) is equivalent to

ϕ(v) (z − u) ≤ ϕ(z) (v − u) + ϕ(u) (z − v)

which is equivalent to

ϕ(v) ≤ ϕ(z)\frac{v − u}{z − u} + ϕ(u)\frac{z − v}{z − u}.

Again this equation is valid by the convexity of ϕ. 1c) u < w < v = z, in which case Eq. (21.12) is equivalent to

(ϕ(z) − ϕ(u)) (z − w) ≤ (ϕ(z) − ϕ(w)) (z − u)

and this is equivalent to the inequality,

ϕ(w) ≤ ϕ(z)\frac{w − u}{z − u} + ϕ(u)\frac{z − w}{z − u}

which again is true by the convexity of ϕ. 1) General case. If u < w < v < z, then by 1a-1c)

\frac{ϕ(z) − ϕ(w)}{z − w} ≥ \frac{ϕ(v) − ϕ(w)}{v − w} ≥ \frac{ϕ(v) − ϕ(u)}{v − u}

and if u < v < w < z

\frac{ϕ(z) − ϕ(w)}{z − w} ≥ \frac{ϕ(w) − ϕ(v)}{w − v} ≥ \frac{ϕ(w) − ϕ(u)}{w − u}.

We have now taken care of all possible cases. 2) On the set a < w < z < b, Eq. (21.12) shows that (ϕ(z) − ϕ(w)) / (z − w) is a decreasing function in w and an increasing function in z and therefore ϕ′_{±}(x) exists for all x ∈ (a, b). Also from Eq. (21.12) we learn that

ϕ′_{+}(u) ≤ \frac{ϕ(z) − ϕ(w)}{z − w} for all a < u < w < z < b,  (21.13)
\frac{ϕ(v) − ϕ(u)}{v − u} ≤ ϕ′_{−}(z) for all a < u < v < z < b,  (21.14)

and letting w ↑ z in the first equation also implies that

ϕ′_{+}(u) ≤ ϕ′_{−}(z) for all a < u < z < b.

The inequality, ϕ′_{−}(z) ≤ ϕ′_{+}(z), is also an easy consequence of Eq. (21.12). 3) Since ϕ(x) has both left and right finite derivatives, it follows that ϕ is continuous. (For an alternative proof, see Rudin.) 4) Given t, let β ∈ [ϕ′_{−}(t), ϕ′_{+}(t)], then by Eqs. (21.13) and (21.14),

\frac{ϕ(t) − ϕ(u)}{t − u} ≤ ϕ′_{−}(t) ≤ β ≤ ϕ′_{+}(t) ≤ \frac{ϕ(z) − ϕ(t)}{z − t}

for all a < u < t < z < b. Item 4. now follows.

Corollary 21.9. Suppose ϕ : (a, b) → ℝ is differential then ϕ is convex iff ϕ′ is non decreasing. In particular if ϕ ∈ C^{2}(a, b) then ϕ is convex iff ϕ′′ ≥ 0.

Proof. By Proposition 21.8, if ϕ is convex then ϕ′ is non-decreasing. Conversely if ϕ′ is increasing then by the mean value theorem,

\frac{ϕ(x_{1}) − ϕ(c)}{x_{1} − c} = ϕ′(ξ_{1}) for some ξ_{1} ∈ (c, x_{1})

and

\frac{ϕ(c) − ϕ(x_{0})}{c − x_{0}} = ϕ′(ξ_{2}) for some ξ_{2} ∈ (x_{0}, c).

Hence

\frac{ϕ(x_{1}) − ϕ(c)}{x_{1} − c} ≥ \frac{ϕ(c) − ϕ(x_{0})}{c − x_{0}}

for all x_{0} < c < x_{1}. Solving this inequality for ϕ(c) gives

ϕ(c) ≤ \frac{c − x_{0}}{x_{1} − x_{0}} ϕ(x_{1}) + \frac{x_{1} − c}{x_{1} − x_{0}} ϕ(x_{0})

showing ϕ is convex.

Theorem 21.10 (Jensen’s Inequality). Suppose that (X, M, μ) is a probability space, i.e. μ is a positive measure and μ(X) = 1. Also suppose that f ∈ L^{1}(μ), f : X → (a, b), and ϕ : (a, b) → ℝ is a convex function. Then

ϕ (∫_{X} fdμ) ≤ ∫_{X} ϕ(f)dμ

where if ϕ ∘ f ∉ L^{1}(μ), then ϕ ∘ f is integrable in the extended sense and ∫_{X} ϕ(f)dμ = ∞.

Proof. Let t = ∫_{X} fdμ ∈ (a, b) and let β ∈ ℝ be such that ϕ(s) − ϕ(t) ≥ β(s−t) for all s ∈ (a, b). Then integrating the inequality, ϕ(f)−ϕ(t) ≥ β(f −t), implies that

0 ≤ ∫_{X} ϕ(f)dμ − ϕ(t) = ∫_{X} ϕ(f)dμ − ϕ(∫_{X} fdμ).

Moreover, if ϕ(f) is not integrable, then ϕ(f) ≥ ϕ(t) + β(f − t) which shows that negative part of ϕ(f) is integrable. Therefore, ∫_{X} ϕ(f)dμ = ∞ in this case.

Example 21.11. The convex functions in Example 21.7 lead to the following inequalities,

exp (∫_{X} fdμ) ≤ ∫_{X} e^{f}dμ,  (21.15)
∫_{X} log(|f|)dμ ≤ log (∫_{X} |f| dμ)

and for p ≥ 1,

|∫_{X} fdμ|^{p} ≤ (∫_{X} |f| dμ)^{p} ≤ ∫_{X} |f|^{p} dμ.

The last equation may also easily be derived using Hölder’s inequality. As a special case of the first equation, we get another proof of Lemma 5.5. Indeed, more generally, suppose p_{i}, s_{i} > 0 for i = 1, 2, …, n and ∑_{i=1}^{n} \frac{1}{p_{i}} = 1, then

s_{1} … s_{n} = e^{∑_{i=1}^{n} ln s_{i}} = e^{∑_{i=1}^{n} \frac{1}{p_{i}} ln s_{i}^{p_{i}}} ≤ ∑_{i=1}^{n} \frac{1}{p_{i}} e^{ln s_{i}^{p_{i}}} = ∑_{i=1}^{n} \frac{s_{i}^{p_{i}}}{p_{i}}  (21.16)

where the inequality follows from Eq. (21.15) with X = {1, 2, …, n} , μ = ∑_{i=1}^{n} \frac{1}{p_{i}} δ_{i} and f (i) := ln s_{i}^{p_{i}}. Of course Eq. (21.16) may be proved directly using the convexity of the exponential function.

21.2 Modes of Convergence

As usual let (X, M, μ) be a fixed measure space, assume 1 ≤ p ≤ ∞ and let {f_{n}}_{n=1}^{∞} ∪ {f} be a collection of complex valued measurable functions on X. We have the following notions of convergence and Cauchy sequences.

Definition 21.12. 1. f_{n} → f a.e. if there is a set E ∈ M such that μ(E) =

0 and lim_{n→∞} 1_{E^{c}}f_{n} = 1_{E^{c}}f.

2. f_{n} → f in μ – measure if lim_{n→∞} μ(|f_{n} − f| > ε) = 0 for all ε > 0. We will abbreviate this by saying f_{n} → f in L^{0} or by f_{n} →^{μ} f.

3. f_{n} → f in L^{p} iff f ∈ L^{p} and f_{n} ∈ L^{p} for all n, and lim_{n→∞} ∥f_{n} − f∥_{p} = 0.

Definition 21.13. 1. {f_{n}} is a.e. Cauchy if there is a set E ∈ M such that

μ(E) = 0 and{1_{E^{c}} f_{n}} is a pointwise Cauchy sequences.

2. {f_{n}} is Cauchy in μ – measure (or L^{0} – Cauchy) if lim_{m,n→∞} μ(|f_{n} − f_{m}| > ε) = 0 for all ε > 0.

3. {f_{n}} is Cauchy in L^{p} if lim_{m,n→∞} ∥f_{n} − f_{m}∥_{p} = 0.

Lemma 21.14 (Chebyshev’s inequality again). Let p ∈ [1, ∞) and f ∈ L^{p}, then

μ (|f| ≥ ε) ≤ \frac{1}{ε^{p}} ∥f∥_{p}^{p} for all ε > 0.

In particular if {f_{n}} ⊂ L^{p} is L^{p} – convergent (Cauchy) then {f_{n}} is also convergent (Cauchy) in measure.

Proof. By Chebyshev’s inequality (19.11),

μ (|f| ≥ ε) = μ (|f|^{p} ≥ ε^{p}) ≤ \frac{1}{ε^{p}} ∫_{X} |f|^{p} dμ = \frac{1}{ε^{p}} ∥f∥_{p}^{p}

and therefore if {f_{n}} is L^{p} – Cauchy, then

μ (|f_{n} − f_{m}| ≥ ε) ≤ \frac{1}{ε^{p}} ∥f_{n} − f_{m}∥_{p}^{p} → 0 as m, n → ∞

showing {f_{n}} is L^{0} – Cauchy. A similar argument holds for the L^{p} – convergent case.

Here is a sequence of functions where f_{n} → 0 a.e., f_{n} ↛ 0 in L^{1}, f_{n} →^{m} 0.

Above is a sequence of functions where f_{n} → 0 a.e., yet f_{n} ↛ 0 in L^{1}. or in measure.

Here is a sequence of functions where f_{n} → 0 a.e., f_{n} →^{m} 0 but f_{n} ↛ 0 in L^{1}.

Above is a sequence of functions where f_{n} → 0 in L^{1}, f_{n} ↛ 0 a.e., and

f_{n} →^{m} 0.

Lemma 21.15. Suppose a_{n} ∈ ℂ and |a_{n+1} − a_{n}| ≤ ε_{n} and ∑_{n=1}^{∞} ε_{n} < ∞. Then

lim_{n→∞} a_{n} = a ∈ ℂ exists and |a − a_{n}| ≤ δ_{n} := ∑_{k=n}^{∞} ε_{k}.

Proof. (This is a special case of Exercise 6.9.) Let m > n then

|a_{m} − a_{n}| = | ∑_{k=n}^{m−1}(a_{k+1} − a_{k})| ≤ ∑_{k=n}^{m−1} |a_{k+1} − a_{k}| ≤ ∑_{k=n}^{∞} ε_{k} := δ_{n} . (21.17)

So |a_{m} − a_{n}| ≤ δ_{min(m,n)} → 0 as , m, n → ∞, i.e. {a_{n}} is Cauchy. Let m → ∞ in (21.17) to find |a − a_{n}| ≤ δ_{n}.

Theorem 21.16. Suppose {f_{n}} is L^{0}-Cauchy. Then there exists a subsequence g_{j} = f_{n_{j}} of {f_{n}} such that lim g_{j} := f exists a.e. and f_{n} →^{μ} f as n → ∞. Moreover if g is a measurable function such that f_{n} →^{μ} g as n → ∞, then f = g a.e.

Proof. Let ε_{n} > 0 such that ∑_{n=1}^{∞} ε_{n} < ∞ (ε_{n} = 2^{−n} would do) and set

δ_{n} = ∑_{k=n}^{∞} ε_{k}. Choose g_{j} = f_{n_{j}} such that {n_{j}} is a subsequence of ℕ and

μ({|g_{j+1} − g_{j}| > ε_{j}}) ≤ ε_{j}.

Let E_{j} = {|g_{j+1} − g_{j}| > ε_{j}} ,

F_{N} = ⋃_{j=N}^{∞} E_{j} = ⋃_{j=N}^{∞} {|g_{j+1} − g_{j}| > ε_{j}}

and

E := ⋂_{N=1}^{∞} F_{N} = ⋂_{N=1}^{∞} ⋃_{j=N}^{∞} E_{j} = {|g_{j+1} − g_{j}| > ε_{j} i.o.}.

Then μ(E) = 0 by Lemma 19.20 or the computation

μ(E) ≤ ∑_{j=N}^{∞} μ(E_{j}) ≤ ∑_{j=N}^{∞} ε_{j} = δ_{N} → 0 as N → ∞.

If x ∉ F_{N}, i.e. |g_{j+1}(x) − g_{j}(x)| ≤ ε_{j} for all j ≥ N, then by Lemma 21.15, f(x) = lim_{j→∞} g_{j}(x) exists and |f(x) − g_{j}(x)| ≤ δ_{j} for all j ≥ N. Therefore,

since E^{c} = ⋃_{N=1}^{∞} F_{N}^{c}, lim_{j→∞} g_{j}(x) = f(x) exists for all x ∉ E. Moreover, {x :

|f(x) − g_{j}(x)| > δ_{j}} ⊂ F_{j} for all j ≥ N and hence

μ(|f − g_{j}| > δ_{j}) ≤ μ(F_{j}) ≤ δ_{j} → 0 as j → ∞.

Therefore g_{j} →^{μ} f as j → ∞. Since

{|f_{n} − f| > ε} = {|f − g_{j} + g_{j} − f_{n}| > ε}
⊂ {|f − g_{j}| > ε/2} ∪ {|g_{j} − f_{n}| > ε/2},

μ({|f_{n} − f| > ε}) ≤ μ({|f − g_{j}| > ε/2}) + μ(|g_{j} − f_{n}| > ε/2)

and

μ({|f_{n} − f| > ε}) ≤ lim_{j→∞} sup μ(|g_{j} − f_{n}| > ε/2) → 0 as n → ∞.

If there is another function g such that f_{n} →^{μ} g as n → ∞, then arguing as above

μ(|f − g| > ε) ≤ μ({|f − f_{n}| > ε/2}) + μ(|g − f_{n}| > ε/2) → 0 as n → ∞.

Hence

μ(|f − g| > 0) = μ(∪_{n=1}^{∞}\{|f − g| > \frac{1}{n}\}) ≤ ∑_{n=1}^{∞} μ(|f − g| > \frac{1}{n}) = 0,

i.e. f = g a.e.

Corollary 21.17 (Dominated Convergence Theorem). Suppose {f_{n}} , {g_{n}} , and g are in L^{1} and f ∈ L^{0} are functions such that

|f_{n}| ≤ g_{n} a.e., f_{n} ⟶^{μ} f, g_{n} ⟶^{μ} g, and ∫ g_{n} → ∫ g as n → ∞.

Then f ∈ L^{1} and lim_{n→∞} ∥f − f_{n}∥_{1} = 0, i.e. f_{n} → f in L^{1}. In particular lim_{n→∞} ∫ f_{n} = ∫ f.

Proof. First notice that |f| ≤ g a.e. and hence f ∈ L^{1} since g ∈ L^{1}. To see that |f| ≤ g, use Theorem 21.16 to find subsequences {f_{n_{k}}} and {g_{n_{k}}} of {f_{n}} and {g_{n}} respectively which are almost everywhere convergent. Then

|f| = lim_{k→∞} |f_{n_{k}}| ≤ lim_{k→∞} g_{n_{k}} = g a.e.

If (for sake of contradiction) lim_{n→∞} ∥f − f_{n}∥_{1} ≠ 0 there exists ε > 0 and a subsequence {f_{n_{k}}} of {f_{n}} such that

∫ |f − f_{n_{k}}| ≥ ε for all k.  (21.18)

Using Theorem 21.16 again, we may assume (by passing to a further subsequences if necessary) that f_{n_{k}} → f and g_{n_{k}} → g almost everywhere. Noting, |f − f_{n_{k}}| ≤ g + g_{n_{k}} → 2g and ∫ (g + g_{n_{k}}) → ∫ 2g, an application of the dominated convergence Theorem 19.38 implies lim_{k→∞} ∫ |f − f_{n_{k}}| = 0 which contradicts Eq. (21.18).

Exercise 21.1 (Fatou’s Lemma). If f_{n} ≥ 0 and f_{n} → f in measure, then ∫ f ≤ lim inf_{n→∞} ∫ f_{n}.

21.3 Completeness of L^{p} – spaces 389

Theorem 21.18 (Egoroff’s Theorem). Suppose μ(X) < ∞ and f_{n} → f a.e. Then for all ε > 0 there exists E ∈ M such that μ(E) < ε and f_{n} → f uniformly on E^{c}. In particular f_{n} ⟶^{μ} f as n → ∞.

Proof. Let f_{n} → f a.e. Then μ(\{|f_{n} − f| > \frac{1}{k} i.o. n\}) = 0 for all k > 0, i.e.

lim_{N→∞} μ ( ⋃_{n≥N}\{|f_{n} − f| > \frac{1}{k} \}) = μ ( ⋂_{N=1}^{∞} ⋃_{n≥N}\{|f_{n} − f| > \frac{1}{k} \}) = 0.

Let E_{k} := ⋃_{n≥N_{k}}\{|f_{n} − f| > \frac{1}{k} \} and choose an increasing sequence \{N_{k}\}_{k=1}^{∞}

such that μ(E_{k}) < ε2^{−k} for all k. Setting E := ∪E_{k}, μ(E) < ∑_{k} ε2^{−k} = ε and if x ∉ E, then |f_{n} − f| ≤ \frac{1}{k} for all n ≥ N_{k} and all k. That is f_{n} → f uniformly on E^{c}.

Exercise 21.2. Show that Egoroff’s Theorem remains valid when the assumption μ(X) < ∞ is replaced by the assumption that |f_{n}| ≤ g ∈ L^{1} for all n. Hint: make use of Theorem 21.18 applied to f_{n}|_{X_{k}} where X_{k} := {|g| ≥ k^{−1}} .

21.3 Completeness of L^{p} – spaces

Theorem 21.19. Let ∥·∥_{∞} be as defined in Eq. (21.2), then (L^{∞}(X, M, μ), ∥·∥_{∞}) is a Banach space. A sequence {f_{n}}_{n=1}^{∞} ⊂ L^{∞} converges to f ∈ L^{∞} iff there exists E ∈ M such that μ(E) = 0 and f_{n} → f uniformly on E^{c}. Moreover, bounded simple functions are dense in L^{∞}.

Proof. By Minkowski’s Theorem 21.4, ∥·∥_{∞} satisfies the triangle inequality. The reader may easily check the remaining conditions that ensure ∥·∥_{∞} is a norm. Suppose that {f_{n}}_{n=1}^{∞} ⊂ L^{∞} is a sequence such f_{n} → f ∈ L^{∞}, i.e. ∥f − f_{n}∥_{∞} → 0 as n → ∞. Then for all k ∈ ℕ, there exists N_{k} < ∞ such that

μ (|f − f_{n}| > k^{−1}) = 0 for all n ≥ N_{k}.

Let

E = ∪_{k=1}^{∞} ∪_{n≥N_{k}} {|f − f_{n}| > k^{−1}} .

Then μ(E) = 0 and for x ∈ E^{c}, |f(x) − f_{n}(x)| ≤ k^{−1} for all n ≥ N_{k}. This shows that f_{n} → f uniformly on E^{c}. Conversely, if there exists E ∈ M such that μ(E) = 0 and f_{n} → f uniformly on E^{c}, then for any ε > 0,

μ (|f − f_{n}| ≥ ε) = μ ({|f − f_{n}| ≥ ε} ∩ E^{c}) = 0

for all n sufficiently large. That is to say lim sup_{n→∞} ∥f − f_{n}∥_{∞} ≤ ε for all ε > 0. The density of simple functions follows from the approximation Theorem 18.42. So the last item to prove is the completeness of L^{∞} for which we will use Theorem 7.13.

Suppose that {f_{n}}_{n=1}^{∞} ⊂ L^{∞} is a sequence such that ∑_{n=1}^{∞} ∥f_{n}∥_{∞} < ∞. Let M_{n} := ∥f_{n}∥_{∞} , E_{n} := {|f_{n}| > M_{n}} , and E := ∪_{n=1}^{∞}E_{n} so that μ(E) = 0. Then

∑_{n=1}^{∞} sup_{x∈E^{c}} |f_{n}(x)| ≤ ∑_{n=1}^{∞} M_{n} < ∞

which shows that S_{N}(x) = ∑_{n=1}^{N} f_{n}(x) converges uniformly to S(x) := ∑_{n=1}^{∞} f_{n}(x) on E^{c}, i.e. lim_{n→∞} ∥S − S_{n}∥_{∞} = 0.

Alternatively, suppose ε_{m,n} := ∥f_{m} − f_{n}∥_{∞} → 0 as m, n → ∞. Let E_{m,n} = {|f_{n} − f_{m}| > ε_{m,n}} and E := ∪E_{m,n}, then μ(E) = 0 and

sup_{x∈E^{c}} |f_{m} (x) − f_{n} (x)| ≤ ε_{m,n} → 0 as m, n → ∞.

Therefore, f := lim_{n→∞} f_{n} exists on E^{c} and the limit is uniform on E^{c}. Letting f = lim_{n→∞} 1_{E^{c}}f_{n}, it then follows that lim_{n→∞} ∥f_{n} − f∥_{∞} = 0.

Theorem 21.20 (Completeness of L^{p}(μ)). For 1 ≤ p ≤ ∞, L^{p}(μ) equipped with the L^{p} – norm, ∥·∥_{p} (see Eq. (21.1)), is a Banach space.

Proof. By Minkowski’s Theorem 21.4, ∥·∥_{p} satisfies the triangle inequality. As above the reader may easily check the remaining conditions that ensure ∥·∥_{p} is a norm. So we are left to prove the completeness of L^{p}(μ) for 1 ≤ p < ∞, the case p = ∞ being done in Theorem 21.19.

Let {f_{n}}_{n=1}^{∞} ⊂ L^{p}(μ) be a Cauchy sequence. By Chebyshev’s inequality (Lemma 21.14), {f_{n}} is L^{0}-Cauchy (i.e. Cauchy in measure) and by Theorem 21.16 there exists a subsequence {g_{j}} of {f_{n}} such that g_{j} → f a.e. By Fatou’s Lemma,

∥g_{j} − f∥_{p}^{p} = ∫ lim_{k→∞} inf |g_{j} − g_{k}|^{p}dμ ≤ lim_{k→∞} inf ∫ |g_{j} − g_{k}|^{p}dμ

= lim_{k→∞} inf ∥g_{j} − g_{k}∥_{p}^{p} → 0 as j → ∞.

In particular, ∥f∥_{p} ≤ ∥g_{j} − f∥_{p} + ∥g_{j}∥_{p} < ∞ so the f ∈ L^{p} and g_{j} ⟶^{L^{p}} f. The

proof is finished because,

∥f_{n} − f∥_{p} ≤ ∥f_{n} − g_{j}∥_{p} + ∥g_{j} − f∥_{p} → 0 as j, n → ∞.

The L^{p}(μ) – norm controls two types of behaviors of f, namely the “behavior at infinity” and the behavior of “local singularities.” So in particular, if f is blows up at a point x_{0} ∈ X, then locally near x_{0} it is harder for f to be in L^{p}(μ) as p increases. On the other hand a function f ∈ L^{p}(μ) is allowed to decay at “infinity” slower and slower as p increases. With these insights in mind, we should not in general expect L^{p}(μ) ⊂ L^{q}(μ) or L^{q}(μ) ⊂ L^{p}(μ). However, there are two notable exceptions. (1) If μ(X) < ∞, then there is no behavior at infinity to worry about and L^{q}(μ) ⊂ L^{p}(μ) for all q ≤ p as is shown in Corollary 21.21 below. (2) If μ is counting measure, i.e. μ(A) = #(A), then all functions in L^{p}(μ) for any p can not blow up on a set of positive measure, so there are no local singularities. In this case L^{p}(μ) ⊂ L^{q}(μ) for all q ≤ p, see Corollary 21.25 below.

Corollary 21.21. If μ(X) < ∞ and 0 < p < q ≤ ∞, then L^{q}(μ) ⊂ L^{p}(μ), the inclusion map is bounded and in fact

∥f∥_{p} ≤ [μ(X)]^{(\frac{1}{p} − \frac{1}{q})} ∥f∥_{q} .

Proof. Take a ∈ [1, ∞] such that

\frac{1}{p} = \frac{1}{a} + \frac{1}{q}, i.e. a = \frac{pq}{q − p}.

Then by Corollary 21.3,

∥f∥_{p} = ∥f · 1∥_{p} ≤ ∥f∥_{q} · ∥1∥_{a} = μ(X)^{1/a}∥f∥_{q} = μ(X)^{( \frac{1}{p} − \frac{1}{q})}∥f∥_{q}.

The reader may easily check this final formula is correct even when q = ∞ provided we interpret 1/p − 1/∞ to be 1/p.

Proposition 21.22. Suppose that 0 < p_{0} < p_{1} ≤ ∞, λ ∈ (0, 1) and p_{λ} ∈ (p_{0}, p_{1}) be defined by

\frac{1}{p_{λ}} = \frac{1 − λ}{p_{0}} + \frac{λ}{p_{1}}  (21.19)

with the interpretation that λ/p_{1} = 0 if p_{1} = ∞.^{1} Then L^{p_{λ}} ⊂ L^{p_{0}} + L^{p_{1}}, i.e. every function f ∈ L^{p_{λ}} may be written as f = g +h with g ∈ L^{p_{0}} and h ∈ L^{p_{1}}. For 1 ≤ p_{0} < p_{1} ≤ ∞ and f ∈ L^{p_{0}} + L^{p_{1}} let

∥f∥ := inf {∥g∥_{p_{0}} + ∥h∥_{p_{1}} : f = g + h} .

Then (L^{p_{0}} + L^{p_{1}}, ∥·∥) is a Banach space and the inclusion map from L^{p_{λ}} to L^{p_{0}} + L^{p_{1}} is bounded; in fact ∥f∥ ≤ 2 ∥f∥_{p_{λ}} for all f ∈ L^{p_{λ}}.

Proof. Let M > 0, then the local singularities of f are contained in the set E := {|f| > M} and the behavior of f at “infinity” is solely determined by f on E^{c}. Hence let g = f1_{E} and h = f1_{E^{c}} so that f = g +h. By our earlier discussion we expect that g ∈ L^{p_{0}} and h ∈ L^{p_{1}} and this is the case since,

1 A little algebra shows that λ may be computed in terms of p_{0}, p_{λ} and p_{1} by

λ = \frac{p_{0}}{p_{λ}} · \frac{p_{1} − p_{λ}}{p_{1} − p_{0}} .

∥g∥_{p_{0}}^{p_{0}} = ∫ |f|^{p_{0}} 1_{|f|>M} = M^{p_{0}} ∫ |\frac{f}{M} |^{p_{0}} 1_{|f|>M}

≤ M^{p_{0}} ∫ |\frac{f}{M} |^{p_{λ}} 1_{|f|>M} ≤ M^{p_{0}−p_{λ}} ∥f∥_{p_{λ}}^{p_{λ}} < ∞

and

∥h∥_{p_{1}}^{p_{1}} = ∥f1_{|f|≤M} ∥_{p_{1}}^{p_{1}} = ∫ |f|^{p_{1}} 1_{|f|≤M} = M^{p_{1}} ∫ |\frac{f}{M} |^{p_{1}} 1_{|f|≤M}

≤ M^{p_{1}} ∫ |\frac{f}{M} |^{p_{λ}} 1_{|f|≤M} ≤ M^{p_{1}−p_{λ}} ∥f∥_{p_{λ}}^{p_{λ}} < ∞.

Moreover this shows

∥f∥ ≤ M^{1−p_{λ}/p_{0}} ∥f∥_{p_{λ}}^{p_{λ}/p_{0}} + M^{1−p_{λ}/p_{1}} ∥f∥_{p_{λ}}^{p_{λ}/p_{1}} .

Taking M = λ ∥f∥_{p_{λ}} then gives

∥f∥ ≤ (λ^{1−p_{λ}/p_{0}} + λ^{1−p_{λ}/p_{1}}) ∥f∥_{p_{λ}}

and then taking λ = 1 shows ∥f∥ ≤ 2 ∥f∥_{p_{λ}} . The the proof that (L^{p_{0}} + L^{p_{1}}, ∥·∥) is a Banach space is left as Exercise 21.7 to the reader.

Corollary 21.23 (Interpolation of L^{p} – norms). Suppose that 0 < p_{0} < p_{1} ≤ ∞, λ ∈ (0, 1) and p_{λ} ∈ (p_{0}, p_{1}) be defined as in Eq. (21.19), then L^{p_{0}} ∩ L^{p_{1}} ⊂ L^{p_{λ}} and

∥f∥_{p_{λ}} ≤ ∥f∥_{p_{0}}^{λ} ∥f∥_{p_{1}}^{1−λ} .  (21.20)

Further assume 1 ≤ p_{0} < p_{λ} < p_{1} ≤ ∞, and for f ∈ L^{p_{0}} ∩ L^{p_{1}} let

∥f∥ := ∥f∥_{p_{0}} + ∥f∥_{p_{1}} .

Then (L^{p_{0}} ∩ L^{p_{1}}, ∥·∥) is a Banach space and the inclusion map of L^{p_{0}} ∩ L^{p_{1}} into L^{p_{λ}} is bounded, in fact

∥f∥_{p_{λ}} ≤ max (λ^{−1}, (1 − λ)^{−1}) (∥f∥_{p_{0}} + ∥f∥_{p_{1}}) .  (21.21)

The heuristic explanation of this corollary is that if f ∈ L^{p_{0}} ∩ L^{p_{1}}, then f has local singularities no worse than an L^{p_{1}} function and behavior at infinity no worse than an L^{p_{0}} function. Hence f ∈ L^{p_{λ}} for any p_{λ} between p_{0} and p_{1}.

Proof. Let λ be determined as above, a = p_{0}/λ and b = p_{1}/(1 − λ), then by Corollary 21.3,

∥f∥_{p_{λ}} = ∥|f|^{λ} |f|^{1−λ}∥_{p_{λ}} ≤ ∥|f|^{λ}∥_{a} ∥|f|^{1−λ}∥_{b} = ∥f∥_{p_{0}}^{λ} ∥f∥_{p_{1}}^{1−λ} .

It is easily checked that ∥·∥ is a norm on L^{p_{0}} ∩ L^{p_{1}}. To show this space is complete, suppose that {f_{n}} ⊂ L^{p_{0}} ∩ L^{p_{1}} is a ∥·∥ – Cauchy sequence. Then {f_{n}} is both L^{p_{0}} and L^{p_{1}} – Cauchy. Hence there exist f ∈ L^{p_{0}} and g ∈ L^{p_{1}} such that lim_{n→∞} ∥f − f_{n}∥_{p_{0}} = 0 and lim_{n→∞} ∥g − f_{n}∥_{p_{λ}} = 0. By Chebyshev’s inequality (Lemma 21.14) f_{n} → f and f_{n} → g in measure and therefore by Theorem 21.16, f = g a.e. It now is clear that lim_{n→∞} ∥f − f_{n}∥ = 0. The estimate in Eq. (21.21) is left as Exercise 21.6 to the reader.

Remark 21.24. Combining Proposition 21.22 and Corollary 21.23 gives

L^{p_{0}} ∩ L^{p_{1}} ⊂ L^{p_{λ}} ⊂ L^{p_{0}} + L^{p_{1}}

for 0 < p_{0} < p_{1} ≤ ∞, λ ∈ (0, 1) and p_{λ} ∈ (p_{0}, p_{1}) as in Eq. (21.19).

Corollary 21.25. Suppose now that μ is counting measure on X. Then L^{p}(μ) ⊂ L^{q}(μ) for all 0 < p < q ≤ ∞ and ∥f∥_{q} ≤ ∥f∥_{p} .

Proof. Suppose that 0 < p < q = ∞, then

∥f∥_{∞}^{p} = sup {|f(x)|^{p} : x ∈ X} ≤ ∑_{x∈X} |f(x)|^{p} = ∥f∥_{p}^{p} ,

i.e. ∥f∥_{∞} ≤ ∥f∥_{p} for all 0 < p < ∞. For 0 < p ≤ q ≤ ∞, apply Corollary 21.23 with p_{0} = p and p_{1} = ∞ to find

∥f∥_{q} ≤ ∥f∥_{p}^{p/q} ∥f∥_{∞}^{1−p/q} ≤ ∥f∥_{p}^{p/q} ∥f∥_{p}^{1−p/q} = ∥f∥_{p} .

21.3.1 Summary:

1. Since μ(|f| > ε) ≤ ε^{−p} ∥f∥_{p}^{p} , L^{p} – convergence implies L^{0} – convergence.

2. L^{0} – convergence implies almost everywhere convergence for some subsequence.

3. If μ(X) < ∞ then almost everywhere convergence implies uniform convergence off certain sets of small measure and in particular we have L^{0} – convergence.

4. If μ(X) < ∞, then L^{q} ⊂ L^{p} for all p ≤ q and L^{q} – convergence implies L^{p} – convergence.

5. L^{p_{0}} ∩ L^{p_{1}} ⊂ L^{q} ⊂ L^{p_{0}} + L^{p_{1}} for any q ∈ (p_{0}, p_{1}).

6. If p ≤ q, then ℓ^{p} ⊂ ℓ^{q} and∥f∥_{q} ≤ ∥f∥_{p} .

21.4 Converse of Hölder’s Inequality

Throughout this section we assume (X, M, μ) is a σ – finite measure space, q ∈ [1, ∞] and p ∈ [1, ∞] are conjugate exponents, i.e. p^{−1} + q^{−1} = 1. For g ∈ L^{q}, let ϕ_{g} ∈ (L^{p})^{∗} be given by

ϕ_{g}(f) = ∫ gf dμ =: ⟨g, f⟩.  (21.22)

By Hölder’s inequality

|ϕ_{g}(f)| ≤ ∫ |gf|dμ ≤ ∥g∥_{q}∥f∥_{p}  (21.23)

which implies that

∥ϕ_{g}∥_{(L^{p})^{∗}} := sup{|ϕ_{g}(f)| : ∥f∥_{p} = 1} ≤ ∥g∥_{q}.  (21.24)

Proposition 21.26 (Converse of Hölder’s Inequality). Let (X, M, μ) be a σ – finite measure space and 1 ≤ p ≤ ∞ as above. For all g ∈ L^{q},

∥g∥_{q} = ∥ϕ_{g}∥_{(L^{p})^{∗}} := sup {|ϕ_{g}(f)| : ∥f∥_{p} = 1}  (21.25)

and for any measurable function g : X → ℂ,

∥g∥_{q} = sup {∫_{X} |g| fdμ : ∥f∥_{p} = 1 and f ≥ 0} .  (21.26)

Proof. We begin by proving Eq. (21.25). Assume first that q < ∞ so p > 1. Then

|ϕ_{g}(f)| = |∫ gf dμ| ≤ ∫ |gf| dμ ≤ ∥g∥_{q}∥f∥_{p}

and equality occurs in the first inequality when sgn(gf) is constant a.e. while equality in the second occurs, by Theorem 21.2, when |f|^{p} = c|g|^{q} for some constant c > 0. So let f := \overline{sgn(g)}|g|^{q/p} which for p = ∞ is to be interpreted as f = \overline{sgn(g)}, i.e. |g|^{q/∞} ≡ 1. When p = ∞,

|ϕ_{g}(f)| = ∫_{X} g \overline{sgn(g)}dμ = ∥g∥_{L^{1}(μ)} = ∥g∥_{1} ∥f∥_{∞}

which shows that ∥ϕ_{g}∥_{(L^{∞})^{∗}} ≥ ∥g∥_{1}. If p < ∞, then

∥f∥_{p}^{p} = ∫ |f|^{p} = ∫ |g|^{q} = ∥g∥_{q}^{q}

while

ϕ_{g}(f) = ∫ gfdμ = ∫ |g||g|^{q/p}dμ = ∫ |g|^{q}dμ = ∥g∥_{q}^{q}.

Hence

\frac{|ϕ_{g}(f)|}{∥f∥_{p}} = \frac{∥g∥_{q}^{q}}{∥g∥_{q}^{q/p}} = ∥g∥_{q}^{q(1− \frac{1}{p})} = ∥g∥_{q}.

This shows that ||ϕ_{g}∥ ≥ ∥g∥_{q} which combined with Eq. (21.24) implies Eq. (21.25).

The last case to consider is p = 1 and q = ∞. Let M := ∥g∥_{∞} and choose X_{n} ∈ M such that X_{n} ↑ X as n → ∞ and μ(X_{n}) < ∞ for all n. For any ε > 0, μ(|g| ≥ M −ε) > 0 and X_{n} ∩{|g| ≥ M −ε} ↑ {|g| ≥ M −ε}. Therefore, μ(X_{n} ∩ {|g| ≥ M − ε}) > 0 for n sufficiently large. Let

f = \overline{sgn(g)}1_{X_{n}∩\{|g|≥M−ε\}},

then

∥f∥_{1} = μ(X_{n} ∩ {|g| ≥ M − ε}) ∈ (0, ∞)

and

|ϕ_{g}(f)| = ∫_{X_{n}∩\{|g|≥M−ε\}} \overline{sgn(g)}gdμ = ∫_{X_{n}∩\{|g|≥M−ε\}} |g|dμ
≥ (M − ε)μ(X_{n} ∩ {|g| ≥ M − ε}) = (M − ε)∥f∥_{1}.

Since ε > 0 is arbitrary, it follows from this equation that ∥ϕ_{g}∥_{(L^{1})}^{∗} ≥ M = ∥g∥_{∞}.

Now for the proof of Eq. (21.26). The key new point is that we no longer are assuming that g ∈ L^{q}. Let M(g) denote the right member in Eq. (21.26) and set g_{n} := 1_{X_{n}∩{|g|≤n}}g. Then |g_{n}| ↑ |g| as n → ∞ and it is clear that M(g_{n}) is increasing in n. Therefore using Lemma 4.10 and the monotone convergence theorem,

lim_{n→∞} M(g_{n}) = sup_{n} M(g_{n}) = sup_{n} sup {∫_{X} |g_{n}| fdμ : ∥f∥_{p} = 1 and f ≥ 0}

= sup {sup_{n} ∫_{X} |g_{n}| fdμ : ∥f∥_{p} = 1 and f ≥ 0}

= sup { lim_{n→∞} ∫_{X} |g_{n}| fdμ : ∥f∥_{p} = 1 and f ≥ 0}

= sup {∫_{X} |g| fdμ : ∥f∥_{p} = 1 and f ≥ 0} = M(g).

Since g_{n} ∈ L^{q} for all n and M(g_{n}) = ∥ϕ_{g_{n}}∥_{(L^{p})}_{∗} (as you should verify), it follows from Eq. (21.25) that M(g_{n}) = ∥g_{n}∥_{q} . When q < ∞ (by the monotone convergence theorem) and when q = ∞ (directly from the definitions) one learns that lim_{n→∞} ∥g_{n}∥_{q} = ∥g∥_{q} . Combining this fact with lim_{n→∞} M(g_{n}) = M(g) just proved shows M(g) = ∥g∥_{q} .

As an application we can derive a sweeping generalization of Minkowski’s inequality. (See Reed and Simon, Vol II. Appendix IX.4 for a more thorough discussion of complex interpolation theory.)

Theorem 21.27 (Minkowski’s Inequality for Integrals). Let (X, M, μ) and (Y, N , ν) be σ – finite measure spaces and 1 ≤ p ≤ ∞. If f is a M ⊗ N measurable function, then y → ∥f(·, y)∥_{L^{p}(μ)} is measurable and 1. if f is a positive M ⊗ N measurable function, then

∥ ∫_{Y} f(·, y)dν(y)∥_{L^{p}(μ)} ≤ ∫_{Y} ∥f(·, y)∥_{L^{p}(μ)}dν(y).  (21.27)

2. If f : X×Y → ℂ is a M⊗N measurable function and ∫_{Y} ∥f(·, y)∥_{L^{p}(μ)}dν(y) < ∞ then

a) for μ – a.e. x, f(x, ·) ∈ L^{1}(ν),

b) the μ –a.e. defined function, x → ∫_{Y} f(x, y)dν(y), is in L^{p}(μ) and

c) the bound in Eq. (21.27) holds.

Proof. For p ∈ [1, ∞], let F_{p}(y) := ∥f(·, y)∥_{L^{p}(μ)}. If p ∈ [1, ∞)

F_{p}(y) = ∥f(·, y)∥_{L^{p}(μ)} = (∫_{X} |f(x, y)|^{p} dμ(x))^{1/p}

is a measurable function on Y by Fubini’s theorem. To see that F_{∞} is measurable, let X_{n} ∈ M such that X_{n} ↑ X and μ(X_{n}) < ∞ for all n. Then by Exercise 21.5,

F_{∞}(y) = lim_{n→∞} lim_{p→∞} ∥f(·, y)1_{X_{n}}∥_{L^{p}(μ)}

which shows that F_{∞} is (Y, N) – measurable as well. This shows that integral on the right side of Eq. (21.27) is well defined.

Now suppose that f ≥ 0, q = p/(p − 1)and g ∈ L^{q}(μ) such that g ≥ 0 and ∥g∥_{L^{q}(μ)} = 1. Then by Tonelli’s theorem and Hölder’s inequality,

∫_{X} [∫_{Y} f(x, y)dν(y)] g(x)dμ(x) = ∫_{Y} dν(y) ∫_{X} dμ(x)f(x, y)g(x)
≤ ∥g∥_{L^{q}(μ)} ∫_{Y} ∥f(·, y)∥_{L^{p}(μ)}dν(y)
= ∫_{Y} ∥f(·, y)∥_{L^{p}(μ)}dν(y).

Therefore by the converse to Hölder’s inequality (Proposition 21.26),

∥ ∫_{Y} f(·, y)dν(y)∥_{L^{p}(μ)}
= sup {∫_{X} [∫_{Y} f(x, y)dν(y)] g(x)dμ(x) : ∥g∥_{L^{q}(μ)} = 1 and g ≥ 0}
≤ ∫_{Y} ∥f(·, y)∥_{L^{p}(μ)}dν(y)

proving Eq. (21.27) in this case.

Now let f : X × Y → ℂ be as in item 2) of the theorem. Applying the first part of the theorem to |f| shows

∫_{Y} |f(x, y)| dν(y) < ∞ for μ– a.e. x,

i.e. f(x, ·) ∈ L^{1}(ν) for the μ –a.e. x. Since |∫_{Y} f(x, y)dν(y)| ≤ ∫_{Y} |f(x, y)| dν(y) it follows by item 1) that

∥∫_{Y} f(·, y)dν(y)∥_{L^{p}(μ)} ≤ ∥∫_{Y} |f(·, y)| dν(y)∥_{L^{p}(μ)} ≤ ∫_{Y} ∥f(·, y)∥_{L^{p}(μ)} dν(y).

Hence the function, x ∈ X → ∫_{Y} f(x, y)dν(y), is in L^{p}(μ) and the bound in Eq. (21.27) holds.

Here is an application of Minkowski’s inequality for integrals. In this theorem we will be using the convention that x^{−1/∞} := 1.

Theorem 21.28 (Theorem 6.20 in Folland). Suppose that k : (0, ∞) × (0, ∞) → ℂ is a measurable function such that k is homogenous of degree −1, i.e. k(λx, λy) = λ^{−1}k(x, y) for all λ > 0. If, for some p ∈ [1, ∞],

C_{p} := ∫_{0}^{∞} |k(x, 1)| x^{−1/p}dx < ∞

then for f ∈ L^{p}((0, ∞), m), k(x, ·)f(·) ∈ L^{1}((0, ∞), m) for m – a.e. x. Moreover, the m – a.e. defined function

(Kf)(x) = ∫_{0}^{∞} k(x, y)f(y)dy  (21.28)

is in L^{p}((0, ∞), m) and

∥Kf∥_{L^{p}((0,∞),m)} ≤ C_{p}∥f∥_{L^{p}((0,∞),m)}.

Proof. By the homogeneity of k, k(x, y) = x^{−1}k(1, \frac{y}{x}). Using this relation and making the change of variables, y = zx, gives

∫_{0}^{∞} |k(x, y)f(y)| dy = ∫_{0}^{∞} x^{−1} |k(1, \frac{y}{x})f(y)| dy
= ∫_{0}^{∞} x^{−1} |k(1, z)f(xz)| xdz = ∫_{0}^{∞} |k(1, z)f(xz)| dz.

Since

∥f(· z)∥_{L^{p}((0,∞),m)}^{p} = ∫_{0}^{∞} |f(yz)|^{p}dy = ∫_{0}^{∞} |f(x)|^{p} \frac{dx}{z},

∥f(· z)∥_{L^{p}((0,∞),m)} = z^{−1/p}∥f∥_{L^{p}((0,∞),m)}.

Using Minkowski’s inequality for integrals then shows

∥∫_{0}^{∞} |k(·, y)f(y)| dy∥_{L^{p}((0,∞),m)} ≤ ∫_{0}^{∞} |k(1, z)| ∥f(·z)∥_{L^{p}((0,∞),m)} dz
= ∥f∥_{L^{p}((0,∞),m)} ∫_{0}^{∞} |k(1, z)| z^{−1/p}dz
= C_{p}∥f∥_{L^{p}((0,∞),m)} < ∞.

This shows that Kf in Eq. (21.28) is well defined from m – a.e. x. The proof is finished by observing

∥Kf∥_{L^{p}((0,∞),m)} ≤ ∥∫_{0}^{∞} |k(·, y)f(y)| dy∥_{L^{p}((0,∞),m)} ≤ C_{p}∥f∥_{L^{p}((0,∞),m)}

for all f ∈ L^{p}((0, ∞), m).

The following theorem is a strengthening of Proposition 21.26. It may be skipped on the first reading.

Theorem 21.29 (Converse of Hölder’s Inequality II). Assume that (X, M, μ) is a σ – finite measure space, q, p ∈ [1, ∞] are conjugate exponents and let 𝕊_{f} denote the set of simple functions ϕ on X such that μ (ϕ ≠ 0) < ∞. Let g : X → ℂ be a measurable function such that ϕg ∈ L^{1} (μ) for all ϕ ∈ 𝕊_{f},^{2} and define

M_{q}(g) := sup {|∫_{X} ϕgdμ| : ϕ ∈ 𝕊_{f} with ∥ϕ∥_{p} = 1} .  (21.29)

If M_{q}(g) < ∞ then g ∈ L^{q} (μ) and M_{q}(g) = ∥g∥_{q} .

Proof. Let X_{n} ∈ M be sets such that μ(X_{n}) < ∞ and X_{n} ↑ X as n ↑ ∞. Suppose that q = 1 and hence p = ∞. Choose simple functions ϕ_{n} on X such that |ϕ_{n}| ≤ 1 and \overline{sgn(g)} = lim_{n→∞} ϕ_{n} in the pointwise sense. Then 1_{X_{m}}ϕ_{n} ∈ 𝕊_{f} and therefore

|∫_{X} 1_{X_{m}}ϕ_{n}gdμ| ≤ M_{q}(g)

for all m, n. By assumption 1_{X_{m}}g ∈ L^{1}(μ) and therefore by the dominated convergence theorem we may let n → ∞ in this equation to find

∫_{X} 1_{X_{m}} |g| dμ ≤ M_{q}(g)

for all m. The monotone convergence theorem then implies that

∫_{X} |g| dμ = lim_{m→∞} ∫_{X} 1_{X_{m}} |g| dμ ≤ M_{q}(g)

2 This is equivalent to requiring 1_{A}g ∈ L^{1}(μ) for all A ∈ M such that μ(A) < ∞.

showing g ∈ L^{1}(μ) and ∥g∥_{1} ≤ M_{q}(g). Since Holder’s inequality implies that M_{q}(g) ≤ ∥g∥_{1} , we have proved the theorem in case q = 1. For q > 1, we will begin by assuming that g ∈ L^{q}(μ). Since p ∈ [1, ∞) we know that 𝕊_{f} is a dense subspace of L^{p}(μ) and therefore, using ϕ_{g} is continuous on L^{p}(μ),

M_{q}(g) = sup {|∫_{X} ϕgdμ| : ϕ ∈ L^{p}(μ) with ∥ϕ∥_{p} = 1} = ∥g∥_{q}

where the last equality follows by Proposition 21.26. So it remains to show that if ϕg ∈ L^{1} for all ϕ ∈ 𝕊_{f} and M_{q}(g) < ∞ then g ∈ L^{q}(μ). For n ∈ ℕ, let g_{n} := 1_{X_{n}}1_{|g|≤n}g. Then g_{n} ∈ L^{q}(μ), in fact ∥g_{n}∥_{q} ≤ nμ(X_{n})^{1/q} < ∞. So by the previous paragraph, ∥g_{n}∥_{q} = M_{q}(g_{n}) and hence

∥g_{n}∥_{q} = sup {|∫_{X} ϕ1_{X_{n}}1_{|g|≤n}gdμ| : ϕ ∈ L^{p}(μ) with ∥ϕ∥_{p} = 1}
≤ M_{q}(g) ∥ϕ1_{X_{n}}1_{|g|≤n}∥_{p} ≤ M_{q}(g) · 1 = M_{q}(g)

wherein the second to last inequality we have made use of the definition of M_{q}(g) and the fact that ϕ1_{X_{n}}1_{|g|≤n} ∈ 𝕊_{f}. If q ∈ (1, ∞), an application of the monotone convergence theorem (or Fatou’s Lemma) along with the continuity of the norm, ∥·∥_{p} , implies

∥g∥_{q} = lim_{n→∞} ∥g_{n}∥_{q} ≤ M_{q}(g) < ∞.

If q = ∞, then ∥g_{n}∥_{∞} ≤ M_{q}(g) < ∞ for all n implies |g_{n}| ≤ M_{q}(g) a.e. which then implies that |g| ≤ M_{q}(g) a.e. since |g| = lim_{n→∞} |g_{n}| . That is g ∈ L^{∞}(μ) and ∥g∥_{∞} ≤ M_{∞}(g).

21.5 Uniform Integrability

This section will address the question as to what extra conditions are needed in order that an L^{0} – convergent sequence is L^{p} – convergent.

Notation 21.30 For f ∈ L^{1}(μ) and E ∈ M, let

μ(f : E) := ∫_{E} fdμ.

and more generally if A, B ∈ M let

μ(f : A, B) := ∫_{A∩B} fdμ.

Lemma 21.31. Suppose g ∈ L^{1}(μ), then for any ε > 0 there exist a δ > 0 such that μ(|g| : E) < ε whenever μ(E) < δ.

Proof. If the Lemma is false, there would exist ε > 0 and sets E_{n} such that μ(E_{n}) → 0 while μ(|g| : E_{n}) ≥ ε for all n. Since |1_{E_{n}}g| ≤ |g| ∈ L^{1} and for any δ ∈ (0, 1), μ(1_{E_{n}} |g| > δ) ≤ μ(E_{n}) → 0 as n → ∞, the dominated convergence theorem of Corollary 21.17 implies lim_{n→∞} μ(|g| : E_{n}) = 0. This contradicts μ(|g| : E_{n}) ≥ ε for all n and the proof is complete.

Suppose that {f_{n}}_{n=1}^{∞} is a sequence of measurable functions which converge in L^{1}(μ) to a function f. Then for E ∈ M and n ∈ ℕ,

|μ(f_{n} : E)| ≤ |μ(f − f_{n} : E)| + |μ(f : E)| ≤ ∥f − f_{n}∥_{1} + |μ(f : E)| .

Let ε_{N} := sup_{n>N} ∥f − f_{n}∥_{1} , then ε_{N} ↓ 0 as N ↑ ∞ and

sup_{n} |μ(f_{n} : E)| ≤ sup_{n≤N} |μ(f_{n} : E)| ∨ (ε_{N} + |μ(f : E)|) ≤ ε_{N} + μ (g_{N} : E) ,

(21.30)

where g_{N} = |f| + ∑_{n=1}^{N} |f_{n}| ∈ L^{1}. From Lemma 21.31 and Eq. (21.30) one easily concludes,

∀ ε > 0 ∃ δ > 0 ∋ sup_{n} |μ(f_{n} : E)| < ε when μ(E) < δ.  (21.31)

Definition 21.32. Functions {f_{n}}_{n=1}^{∞} ⊂ L^{1}(μ) satisfying Eq. (21.31) are said to be uniformly integrable.

Remark 21.33. Let {f_{n}} be real functions satisfying Eq. (21.31), E be a set where μ(E) < δ and E_{n} = E ∩ {f_{n} ≥ 0} . Then μ(E_{n}) < δ so that μ(f_{n}^{+} : E) = μ(f_{n} : E_{n}) < ε and similarly μ(f_{n}^{−} : E) < ε. Therefore if Eq. (21.31) holds then

sup_{n} μ(|f_{n}| : E) < 2ε when μ(E) < δ.  (21.32)

Similar arguments work for the complex case by looking at the real and imaginary parts of f_{n}. Therefore {f_{n}}_{n=1}^{∞} ⊂ L^{1}(μ) is uniformly integrable iff

∀ ε > 0 ∃ δ > 0 ∋ sup_{n} μ(|f_{n}| : E) < ε when μ(E) < δ.  (21.33)

Lemma 21.34. Assume that μ(X) < ∞, then {f_{n}} is uniformly bounded in L^{1}(μ) (i.e. K = sup_{n} ∥f_{n}∥_{1} < ∞) and {f_{n}} is uniformly integrable iff

lim_{M→∞} sup_{n} μ(|f_{n}| : |f_{n}| ≥ M) = 0.  (21.34)

Proof. Since {f_{n}} is uniformly bounded in L^{1}(μ), μ(|f_{n}| ≥ M) ≤ K/M. So if (21.33) holds and ε > 0 is given, we may choose M sufficiently large so that μ(|f_{n}| ≥ M) < δ(ε) for all n and therefore,

sup_{n} μ(|f_{n}| : |f_{n}| ≥ M) ≤ ε.

Since ε is arbitrary, we concluded that Eq. (21.34) must hold. Conversely, suppose that Eq. (21.34) holds, then automatically K = sup_{n} μ(|f_{n}|) < ∞ because

μ(|f_{n}|) = μ(|f_{n}| : |f_{n}| ≥ M) + μ(|f_{n}| : |f_{n}| < M)

≤ sup_{n} μ(|f_{n}| : |f_{n}| ≥ M) + Mμ(X) < ∞.

Moreover,

μ(|f_{n}| : E) = μ(|f_{n}| : |f_{n}| ≥ M, E) + μ(|f_{n}| : |f_{n}| < M, E)

≤ sup_{n} μ(|f_{n}| : |f_{n}| ≥ M) + Mμ(E).

So given ε > 0 choose M so large that sup_{n} μ(|f_{n}| : |f_{n}| ≥ M) < ε/2 and then take δ = ε/ (2M) .

Remark 21.35. It is not in general true that if {f_{n}} ⊂ L^{1}(μ) is uniformly integrable then sup_{n} μ(|f_{n}|) < ∞. For example take X = {∗} and μ({∗}) = 1. Let f_{n}(∗) = n. Since for δ < 1 a set E ⊂ X such that μ(E) < δ is in fact the empty set, we see that Eq. (21.32) holds in this example. However, for finite measure spaces with out “atoms”, for every δ > 0 we may find a finite partition of X by sets {E_{ℓ}}_{ℓ=1}^{k} with μ(E_{ℓ}) < δ. Then if Eq. (21.32) holds with 2ε = 1, then

μ(|f_{n}|) = ∑_{ℓ=1}^{k} μ(|f_{n}| : E_{ℓ}) ≤ k

showing that μ(|f_{n}|) ≤ k for all n.

The following Lemmas gives a concrete necessary and sufficient conditions for verifying a sequence of functions is uniformly bounded and uniformly integrable.

Lemma 21.36. Suppose that μ(X) < ∞, and Λ ⊂ L^{0}(X) is a collection of functions.

1. If there exists a non decreasing function ϕ : ℝ_{+} → ℝ_{+} such that lim_{x→∞} ϕ(x)/x = ∞ and

K := sup_{f∈Λ} μ(ϕ(|f|)) < ∞  (21.35)

then

lim_{M→∞} sup_{f∈Λ} μ (|f| 1_{|f|≥M}) = 0.  (21.36)

2. Conversely if Eq. (21.36) holds, there exists a non-decreasing continuous function ϕ : ℝ_{+} → ℝ_{+} such that ϕ(0) = 0, lim_{x→∞} ϕ(x)/x = ∞ and Eq. (21.35) is valid.

Proof. 1. Let ϕ be as in item 1. above and set ε_{M} := sup_{x≥M} \frac{x}{ϕ(x)} → 0 as M → ∞ by assumption. Then for f ∈ Λ

μ(|f| : |f| ≥ M) = μ(\frac{|f|}{ϕ (|f|)}ϕ (|f|) : |f| ≥ M) ≤ ε_{M}μ(ϕ (|f|) : |f| ≥ M)
≤ ε_{M}μ(ϕ (|f|)) ≤ Kε_{M}

and hence

lim_{M→∞} sup_{f∈Λ} μ (|f| 1_{|f|≥M}) ≤ lim_{M→∞} Kε_{M} = 0.

2. By assumption, ε_{M} := sup_{f∈Λ} μ (|f| 1_{|f|≥M}) → 0 as M → ∞. Therefore we may choose M_{n} ↑ ∞ such that

∑_{n=0}^{∞} (n + 1) ε_{M_{n}} < ∞

where by convention M_{0} := 0. Now define ϕ so that ϕ(0) = 0 and

ϕ′(x) = ∑_{n=0}^{∞} (n + 1) 1_{(M_{n},M_{n+1}]}(x),

i.e.

ϕ(x) = ∫_{0}^{x} ϕ′(y)dy = ∑_{n=0}^{∞} (n + 1) (x ∧ M_{n+1} − x ∧ M_{n}) .

By construction ϕ is continuous, ϕ(0) = 0, ϕ′(x) is increasing (so ϕ is convex) and ϕ′(x) ≥ (n + 1) for x ≥ M_{n}. In particular

\frac{ϕ(x)}{x} ≥ \frac{ϕ(M_{n}) + (n + 1)x}{x} ≥ n + 1 for x ≥ M_{n}

from which we conclude lim_{x→∞} ϕ(x)/x = ∞. We also have ϕ′(x) ≤ (n + 1) on [0, M_{n+1}] and therefore

ϕ(x) ≤ (n + 1)x for x ≤ M_{n+1}.

So for f ∈ Λ,

μ (ϕ(|f|)) = ∑_{n=0}^{∞} μ (ϕ(|f|)1_{(M_{n},M_{n+1}]}(|f|))

≤ ∑_{n=0}^{∞} (n + 1) μ (|f| 1_{(M_{n},M_{n+1}]}(|f|))

≤ ∑_{n=0}^{∞} (n + 1) μ (|f| 1_{|f|≥M_{n}}) ≤ ∑_{n=0}^{∞} (n + 1) ε_{M_{n}}

and hence

sup_{f∈Λ} μ (ϕ(|f|)) ≤ ∑_{n=0}^{∞} (n + 1) ε_{M_{n}} < ∞.

Theorem 21.37 (Vitali Convergence Theorem). (Folland 6.15) Suppose that 1 ≤ p < ∞. A sequence {f_{n}} ⊂ L^{p} is Cauchy iff

1. {f_{n}} is L^{0} – Cauchy,

2. {|f_{n}|^{p}} – is uniformly integrable.

3. For all ε > 0, there exists a set E ∈ M such that μ(E) < ∞ and ∫_{E^{c}} |f_{n}|^{p} dμ < ε for all n. (This condition is vacuous when μ(X) < ∞.)

Proof. (⟹) Suppose {f_{n}} ⊂ L^{p} is Cauchy. Then (1) {f_{n}} is L^{0} – Cauchy by Lemma 21.14. (2) By completeness of L^{p}, there exists f ∈ L^{p} such that ∥f_{n} − f∥_{p} → 0 as n → ∞. By the mean value theorem,

||f|^{p} − |f_{n}|^{p}| ≤ p(max(|f| , |f_{n}|))^{p−1} ||f| − |f_{n}|| ≤ p(|f| + |f_{n}|)^{p−1} ||f| − |f_{n}||

and therefore by Hölder’s inequality,

∫ ||f|^{p} − |f_{n}|^{p}| dμ ≤ p ∫ (|f| + |f_{n}|)^{p−1} ||f| − |f_{n}|| dμ ≤ p ∫ (|f| + |f_{n}|)^{p−1}|f − f_{n}|dμ

≤ p∥f − f_{n}∥_{p}∥(|f| + |f_{n}|)^{p−1}∥_{q} = p∥ |f| + |f_{n}|∥_{p}^{p/q}∥f − f_{n}∥_{p}

≤ p(∥f∥_{p} + ∥f_{n}∥_{p})^{p/q}∥f − f_{n}∥_{p}

where q := p/(p − 1). This shows that ∫ ||f|^{p} − |f_{n}|^{p}| dμ → 0 as n → ∞.^{3} By the remarks prior to Definition 21.32, {|f_{n}|^{p}} is uniformly integrable. To verify (3), for M > 0 and n ∈ ℕ let E_{M} = {|f| ≥ M} and E_{M}(n) = {|f_{n}| ≥ M}. Then μ(E_{M}) ≤ \frac{1}{M^{p}} ∥f||_{p}^{p} < ∞ and by the dominated convergence theorem,

∫_{E_{M}^{c}} |f|^{p} dμ = ∫ |f|^{p} 1_{|f|<M}dμ → 0 as M → 0.

Moreover,

∥f_{n}1_{E_{M}^{c}} ∥_{p} ≤ ∥f1_{E_{M}^{c}} ∥_{p} + ∥(f_{n} − f)1_{E_{M}^{c}} ∥_{p} ≤ ∥f1_{E_{M}^{c}} ∥_{p} + ∥f_{n} − f∥_{p} . (21.37)

So given ε > 0, choose N sufficiently large such that for all n ≥ N, ∥f − f_{n}∥_{p}^{p} < ε. Then choose M sufficiently small such that ∫_{E_{M}^{c}} |f|^{p} dμ < ε and ∫_{E_{M}^{c} (n)} |f|^{p} dμ < ε for all n = 1, 2, …, N − 1. Letting E := E_{M} ∪ E_{M}(1) ∪ ⋯ ∪ E_{M}(N − 1), we have

μ(E) < ∞, ∫_{E^{c}} |f_{n}|^{p} dμ < ε for n ≤ N − 1

and by Eq. (21.37)

3 Here is an alternative proof. Let h_{n} ≡ ||f_{n}|^{p} − |f|^{p}| ≤ |f_{n}|^{p} + |f|^{p} =: g_{n} ∈ L^{1} and g ≡ 2|f|^{p}. Then g_{n} →^{μ} g, h_{n} →^{μ} 0 and ∫ g_{n} → ∫ g. Therefore by the dominated convergence theorem in Corollary 21.17, lim_{n→∞} ∫ h_{n} dμ = 0.

∫_{E^{c}} |f_{n}|^{p} dμ < (ε^{1/p} + ε^{1/p})^{p} ≤ 2^{p}ε for n ≥ N.

Therefore we have found E ∈ M such that μ(E) < ∞ and

sup_{n} ∫_{E^{c}} |f_{n}|^{p} dμ ≤ 2^{p}ε

which verifies (3) since ε > 0 was arbitrary. (⟸) Now suppose{f_{n}} ⊂ L^{p} satisfies conditions (1) - (3). Let ε > 0, E be as in (3) and

A_{mn} := {x ∈ E|f_{m}(x) − f_{n}(x)| ≥ ε}.

Then

∥(f_{n} − f_{m}) 1_{E^{c}}∥_{p} ≤ ∥f_{n}1_{E^{c}}∥_{p} + ∥f_{m} 1_{E^{c}}∥_{p} < 2ε^{1/p}

and

∥f_{n} − f_{m}∥_{p} = ∥(f_{n} − f_{m})1_{E^{c}}∥_{p} + ∥(f_{n} − f_{m})1_{E\A_{mn}}∥_{p}
+ ∥(f_{n} − f_{m})1_{A_{mn}}∥_{p}
≤ ∥(f_{n} − f_{m})1_{E\A_{mn}}∥_{p} + ∥(f_{n} − f_{m})1_{A_{mn}}∥_{p} + 2ε^{1/p}. (21.38)

Using properties (1) and (3) and 1_{E∩{|f_{m}−f_{n}|<ε}}|f_{m} − f_{n}|^{p} ≤ ε^{p}1_{E} ∈ L^{1}, the dominated convergence theorem in Corollary 21.17 implies

∥(f_{n} − f_{m}) 1_{E\A_{mn}}∥_{p}^{p} = ∫ 1_{E∩{|f_{m}−f_{n}|<ε}} |f_{m} − f_{n}|^{p} ⟶_{m,n→∞} 0.

which combined with Eq. (21.38) implies

lim sup_{m,n→∞} ∥f_{n} − f_{m}∥_{p} ≤ lim sup_{m,n→∞} ∥(f_{n} − f_{m})1_{A_{mn}}∥_{p} + 2ε^{1/p}.

Finally

∥(f_{n} − f_{m})1_{A_{mn}}∥_{p} ≤ ∥f_{n}1_{A_{mn}}∥_{p} + ∥f_{m} 1_{A_{mn}}∥_{p} ≤ 2δ(ε)

where

δ(ε) := sup_{n} sup{ ∥f_{n} 1_{E}∥_{p} : E ∈ M ∋ μ(E) ≤ ε}

By property (2), δ(ε) → 0 as ε → 0. Therefore

lim sup_{m,n→∞} ∥f_{n} − f_{m}∥_{p} ≤ 2ε^{1/p} + 0 + 2δ(ε) → 0 as ε ↓ 0

and therefore {f_{n}} is L^{p}-Cauchy.

Here is another version of Vitali’s Convergence Theorem.

Theorem 21.38 (Vitali Convergence Theorem). (This is problem 9 on p. 133 in Rudin.) Assume that μ(X) < ∞, {f_{n}} is uniformly integrable, f_{n} → f a.e. and |f| < ∞ a.e., then f ∈ L^{1}(μ) and f_{n} → f in L^{1}(μ).

Proof. Let ε > 0 be given and choose δ > 0 as in the Eq. (21.32). Now use Egoroff’s Theorem 21.18 to choose a set E^{c} where {f_{n}} converges uniformly on E^{c} and μ(E) < δ. By uniform convergence on E^{c}, there is an integer N < ∞ such that |f_{n} − f_{m}| ≤ 1 on E^{c} for all m, n ≥ N. Letting m → ∞, we learn that

|f_{N} − f| ≤ 1 on E^{c}.

Therefore |f| ≤ |f_{N}| + 1 on E^{c} and hence

μ(|f|) = μ(|f| : E^{c}) + μ(|f| : E)
≤ μ(|f_{N}|) + μ(X) + μ(|f| : E).

Now by Fatou’s lemma,

μ(|f| : E) ≤ lim inf_{n→∞} μ(|f_{n}| : E) ≤ 2ε < ∞

by Eq. (21.32). This shows that f ∈ L^{1}. Finally

μ(|f − f_{n}|) = μ(|f − f_{n}| : E^{c}) + μ(|f − f_{n}| : E)
≤ μ(|f − f_{n}| : E^{c}) + μ(|f| + |f_{n}| : E)
≤ μ(|f − f_{n}| : E^{c}) + 4ε

and so by the Dominated convergence theorem we learn that

lim sup_{n→∞} μ(|f − f_{n}|) ≤ 4ε.

Since ε > 0 was arbitrary this completes the proof.

Theorem 21.39 (Vitali again). Suppose that f_{n} → f in μ measure and Eq. (21.34) holds, then f_{n} → f in L^{1}.

Proof. This could of course be proved using 21.38 after passing to subsequences to get {f_{n}} to converge a.s. However I wish to give another proof. First off, by Fatou’s lemma, f ∈ L^{1}(μ). Now let

ϕ_{K}(x) = x1_{|x|≤K} + K1_{|x|>K}.

then ϕ_{K}(f_{n}) →^{μ} ϕ_{K}(f) because |ϕ_{K}(f) − ϕ_{K}(f_{n})| ≤ |f − f_{n}| and since

|f − f_{n}| ≤ |f − ϕ_{K}(f)| + |ϕ_{K}(f) − ϕ_{K}(f_{n})| + |ϕ_{K}(f_{n}) − f_{n}|

we have that

μ|f − f_{n}| ≤ μ |f − ϕ_{K}(f)| + μ|ϕ_{K}(f) − ϕ_{K}(f_{n})| + μ |ϕ_{K}(f_{n}) − f_{n}|

= μ(|f| : |f| ≥ K) + μ|ϕ_{K}(f) − ϕ_{K}(f_{n})| + μ(|f_{n}| : |f_{n}| ≥ K).

Therefore by the dominated convergence theorem

lim sup_{n→∞} μ|f − f_{n}| ≤ μ(|f| : |f| ≥ K) + lim sup_{n→∞} μ(|f_{n}| : |f_{n}| ≥ K).

This last expression goes to zero as K → ∞ by uniform integrability.

21.6 Exercises

Definition 21.40. The essential range of f, essran(f), consists of those λ ∈ ℂ such that μ(|f − λ| < ε) > 0 for all ε > 0.

Definition 21.41. Let (X, τ) be a topological space and ν be a measure on B_{X} = σ(τ). The support of ν, supp(ν), consists of those x ∈ X such that ν(V) > 0 for all open neighborhoods, V, of x.

Exercise 21.3. Let (X, τ) be a second countable topological space and ν be a measure on B_{X} – the Borel σ – algebra on X. Show

1. supp(ν) is a closed set. (This is actually true on all topological spaces.)

2. ν(X \ supp(ν)) = 0 and use this to conclude that W := X \ supp(ν) is the largest open set in X such that ν(W) = 0. Hint: let U ⊂ τ be a countable base for the topology τ. Show that W may be written as a union of elements from V ∈ V with the property that μ(V) = 0.

Exercise 21.4. Prove the following facts about essran(f).

1. Let ν = f_{∗}μ := μ∘f^{−1} – a Borel measure on ℂ. Show essran(f) = supp(ν).

2. essran(f) is a closed set and f(x) ∈ essran(f) for almost every x, i.e. μ(f ∉ essran(f)) = 0.

3. If F ⊂ ℂ is a closed set such that f(x) ∈ F for almost every x then essran(f) ⊂ F. So essran(f) is the smallest closed set F such that f(x) ∈ F for almost every x.

4. ∥f∥_{∞} = sup {|λ| : λ ∈ essran(f)} .

Exercise 21.5. Let f ∈ L^{p} ∩ L^{∞} for some p < ∞. Show ∥f∥_{∞} = lim_{q→∞} ∥f∥_{q} . If we further assume μ(X) < ∞, show ∥f∥_{∞} = lim_{q→∞} ∥f∥_{q} for all measurable functions f : X → ℂ. In particular, f ∈ L^{∞} iff lim_{q→∞} ∥f∥_{q} < ∞. Hints: Use Corollary 21.23 to show lim sup_{q→∞} ∥f∥_{q} ≤ ∥f∥_{∞} and to show lim inf_{q→∞} ∥f∥_{q} ≥ ∥f∥_{∞} , let M < ∥f∥_{∞} and make use of Chebyshev’s inequality.

Exercise 21.6. Prove Eq. (21.21) in Corollary 21.23. (Part of Folland 6.3 on p. 186.) Hint: Use the inequality, with a, b ≥ 1 with a^{−1} + b^{−1} = 1 chosen appropriately,

st ≤ \frac{s^{a}}{a} + \frac{t^{b}}{b} ,

(see Lemma 5.5 for Eq. (21.16)) applied to the right side of Eq. (21.20).

Exercise 21.7. Complete the proof of Proposition 21.22 by showing (L^{p} + L^{r}, ∥·∥) is a Banach space. Hint: you may find using Theorem 7.13 is helpful here.

Exercise 21.8. Folland 6.5 on p. 186.

Exercise 21.9. By making the change of variables, u = ln x, prove the following facts:

∫_{0}^{1/2} x^{−a} |ln x|^{b} dx < ∞ ⟺ a < 1 or a = 1 and b < −1
∫_{2}^{∞} x^{−a} |ln x|^{b} dx < ∞ ⟺ a > 1 or a = 1 and b < −1
∫_{0}^{1} x^{−a} |ln x|^{b} dx < ∞ ⟺ a < 1 and b > −1
∫_{1}^{∞} x^{−a} |ln x|^{b} dx < ∞ ⟺ a > 1 and b > −1.

Suppose 0 < p_{0} < p_{1} ≤ ∞ and m is Lebesgue measure on (0, ∞) . Use the above results to manufacture a function f on (0, ∞) such that f ∈ L^{p} ((0, ∞) , m) iff (a) p ∈ (p_{0}, p_{1}) , (b) p ∈ [p_{0}, p_{1}] and (c) p = p_{0}.

Exercise 21.10. Folland 6.9 on p. 186.

Exercise 21.11. Folland 6.10 on p. 186. Use the strong form of Theorem 19.38.

Exercise 21.12. Let (X, M, μ) and (Y, N , ν) be σ – finite measure spaces, f ∈ L^{2}(ν) and k ∈ L^{2}(μ ⊗ ν). Show

∫ |k(x, y)f(y)| dν(y) < ∞ for μ – a.e. x.

Let Kf(x) := ∫_{Y} k(x, y)f(y)dν(y) when the integral is defined. Show Kf ∈ L^{2}(μ) and K : L^{2}(ν) → L^{2}(μ) is a bounded operator with ∥K∥_{op} ≤ ∥k∥_{L^{2}(μ⊗ν)} .

Exercise 21.13. Folland 6.27 on p. 196. Hint: Theorem 21.28.

Exercise 21.14. Folland 2.32 on p. 63.

Exercise 21.15. Folland 2.38 on p. 63.

Approximation Theorems and Convolutions

22.1 Density Theorems

In this section, (X, M, μ) will be a measure space A will be a subalgebra of M.

Notation 22.1 Suppose (X, M, μ) is a measure space and A ⊂ M is a subalgebra of M. Let 𝕊(A) denote those simple functions ϕ : X → ℂ such that ϕ^{−1}({λ}) ∈ A for all λ ∈ ℂ and let 𝕊_{f}(A, μ) denote those ϕ ∈ 𝕊(A) such that μ(ϕ ≠ 0) < ∞.

Remark 22.2. For ϕ ∈ 𝕊_{f}(A, μ) and p ∈ [1, ∞), |ϕ|^{p} = ∑_{z≠0} |z|^{p}1_{{ϕ=z}} and hence

∫ |ϕ|^{p} dμ = ∑_{z≠0} |z|^{p}μ(ϕ = z) < ∞  (22.1)

so that 𝕊_{f}(A, μ) ⊂ L^{p}(μ). Conversely if ϕ ∈ 𝕊(A)∩L^{p}(μ), then from Eq. (22.1) it follows that μ (ϕ = z) < ∞ for all z ≠ 0 and therefore μ (ϕ ≠ 0) < ∞. Hence we have shown, for any 1 ≤ p < ∞,

𝕊_{f}(A, μ) = 𝕊(A) ∩ L^{p}(μ).

Lemma 22.3 (Simple Functions are Dense). The simple functions, 𝕊_{f}(M, μ), form a dense subspace of L^{p}(μ) for all 1 ≤ p < ∞.

Proof. Let {ϕ_{n}}_{n=1}^{∞} be the simple functions in the approximation Theorem 18.42. Since |ϕ_{n}| ≤ |f| for all n, ϕ_{n} ∈ 𝕊_{f}(M, μ) and

|f − ϕ_{n}|^{p} ≤ (|f| + |ϕ_{n}|)^{p} ≤ 2^{p} |f|^{p} ∈ L^{1} (μ) .

Therefore, by the dominated convergence theorem,

lim_{n→∞} ∫ |f − ϕ_{n}|^{p}dμ = ∫ lim_{n→∞} |f − ϕ_{n}|^{p}dμ = 0.

The goal of this section is to find a number of other dense subspaces of L^{p} (μ) for p ∈ [1, ∞). The next theorem is the key result of this section.

Theorem 22.4 (Density Theorem). Let p ∈ [1, ∞), (X, M, μ) be a measure space and M be an algebra of bounded 𝔽 – valued (𝔽 = ℝ or 𝔽 = ℂ) measurable functions such that

1. M ⊂ L^{p} (μ, 𝔽) and σ (M) = M.

2. There exists ψ_{k} ∈ M such that ψ_{k} → 1 boundedly.

3. If 𝔽 = ℂ we further assume that M is closed under complex conjugation.

Then to every function f ∈ L^{p} (μ, 𝔽) , there exists ϕ_{n} ∈ M such that lim_{n→∞} ∥f − ϕ_{n}∥_{L^{p}(μ)} = 0, i.e. M is dense in L^{p} (μ, 𝔽) .

Proof. Fix k ∈ ℕ for the moment and let H denote those bounded M – measurable functions, f : X → 𝔽, for which there exists {ϕ_{n}}_{n=1}^{∞} ⊂ M such that lim_{n→∞} ∥ψ_{k}f − ϕ_{n}∥_{L^{p}(μ)} = 0. A routine check shows H is a subspace of ℓ^{∞} (M, 𝔽) such that 1 ∈ H, M ⊂ H and H is closed under complex conjugation if 𝔽 = ℂ. Moreover, H is closed under bounded convergence. To see this suppose f_{n} ∈ H and f_{n} → f boundedly. Then, by the dominated convergence theorem, lim_{n→∞} ∥ψ_{k} (f − f_{n})∥_{L^{p}(μ)} = 0.^{1} (Take the dominating function to be g = [2C |ψ_{k}|]^{p} where C is a constant bounding all of the \{|f_{n}|\}_{n=1}^{∞} .) We may now choose ϕ_{n} ∈ M such that ∥ϕ_{n} − ψ_{k}f_{n}∥_{L^{p}(μ)} ≤ \frac{1}{n} then

lim sup_{n→∞} ∥ψ_{k}f − ϕ_{n}∥_{L^{p}(μ)} ≤ lim sup_{n→∞} ∥ψ_{k} (f − f_{n})∥_{L^{p}(μ)}
+ lim sup_{n→∞} ∥ψ_{k}f_{n} − ϕ_{n}∥_{L^{p}(μ)} = 0 (22.2)

which implies f ∈ H. An application of Dynkin’s Multiplicative System Theorem 18.51 if 𝔽 = ℝ or Theorem 18.52 if 𝔽 = ℂ now shows H contains all bounded measurable functions on X.

Let f ∈ L^{p} (μ) be given. The dominated convergence theorem implies lim_{k→∞} ∥ψ_{k}1_{{|f|≤k}}f − f∥_{L^{p}(μ)} = 0. (Take the dominating function to be g = [2C |f|]^{p} where C is a bound on all of the |ψ_{k}| .) Using this and what we have just proved, there exists ϕ_{k} ∈ M such that

∥ψ_{k}1_{\{|f|≤k\}}f − ϕ_{k}∥_{L^{p}(μ)} ≤ \frac{1}{k} .

The same line of reasoning used in Eq. (22.2) now implies lim_{k→∞} ∥f − ϕ_{k}∥_{L^{p}(μ)} = 0.

1 It is at this point that the proof would break down if p = ∞.

Definition 22.5. Let (X, τ) be a topological space and μ be a measure on B_{X} = σ (τ) . A locally integrable function is a Borel measurable function f : X → ℂ such that ∫_{K} |f| dμ < ∞ for all compact subsets K ⊂ X. We will write L_{loc}^{1}(μ) for the space of locally integrable functions. More generally we say f ∈ L_{loc}^{p} (μ) iff ∥1_{K}f∥_{L^{p}(μ)} < ∞ for all compact subsets K ⊂ X.

Definition 22.6. Let (X, τ) be a topological space. A K-finite measure on X is Borel measure μ such that μ (K) < ∞ for all compact subsets K ⊂ X.

Lebesgue measure on ℝ is an example of a K-finite measure while counting measure on ℝ is not a K-finite measure.

Example 22.7. Suppose that μ is a K-finite measure on B_{ℝ^{d}}. An application of Theorem 22.4 shows C_{c} (ℝ, ℂ) is dense in L^{p}(ℝ^{d}, B_{ℝ^{d}}, μ; ℂ). To apply Theorem 22.4, let M := C_{c} (ℝ^{d}, ℂ) and ψ_{k} (x) := ψ (x/k) where ψ ∈ C_{c} (ℝ^{d}, ℂ) with ψ (x) = 1 in a neighborhood of 0. The proof is completed by showing σ (M) = σ (C_{c} (ℝ^{d}, ℂ)) = B_{ℝ^{d}}, which follows directly from Lemma 18.57.

We may also give a more down to earth proof as follows. Let x_{0} ∈ ℝ^{d}, R > 0, A := B (x_{0}, R)^{c} and f_{n} (x) := d_{A}^{1/n} (x) . Then f_{n} ∈ M and f_{n} → 1_{B(x_{0},R)} as n → ∞ which shows 1_{B(x_{0},R)} is σ (M)-measurable, i.e. B (x_{0}, R) ∈ σ (M) . Since x_{0} ∈ ℝ^{d} and R > 0 were arbitrary, σ (M) = B_{ℝ^{d}}.

More generally we have the following result.

Theorem 22.8. Let (X, τ) be a second countable locally compact Hausdorff space and μ : B_{X} → [0, ∞] be a K-finite measure. Then C_{c}(X) (the space of continuous functions with compact support) is dense in L^{p}(μ) for all p ∈ [1, ∞). (See also Proposition ?? below.)

Proof. Let M := C_{c}(X) and use Item 3. of Lemma 18.57 to find functions ψ_{k} ∈ M such that ψ_{k} → 1 to boundedly as k → ∞. The result now follows from an application of Theorem 22.4 along with the aid of item 4. of Lemma 18.57.

Exercise 22.1. Show that BC (ℝ, ℂ) is not dense in L^{∞}(ℝ, B_{ℝ}, m; ℂ). Hence the hypothesis that p < ∞ in Theorem 22.4 can not be removed.

Corollary 22.9. Suppose X ⊂ ℝ^{n} is an open set, B_{X} is the Borel σ – algebra on X and μ be a K-finite measure on (X, B_{X}) . Then C_{c}(X) is dense in L^{p}(μ) for all p ∈ [1, ∞).

Corollary 22.10. Suppose that X is a compact subset of ℝ^{n} and μ is a finite measure on (X, B_{X}), then polynomials are dense in L^{p}(X, μ) for all 1 ≤ p < ∞.

Proof. Consider X to be a metric space with usual metric induced from ℝ^{n}. Then X is a locally compact separable metric space and therefore C_{c}(X, ℂ) = C(X, ℂ) is dense in L^{p}(μ) for all p ∈ [1, ∞). Since, by the dominated convergence theorem, uniform convergence implies L^{p}(μ) – convergence, it follows from the Weierstrass approximation theorem (see Theorem 8.34 and Corollary 8.36 or Theorem 12.31 and Corollary 12.32) that polynomials are also dense in L^{p}(μ).

Lemma 22.11. Let (X, τ) be a second countable locally compact Hausdorff space and μ : B_{X} → [0, ∞] be a K-finite measure on X. If h ∈ L_{loc}^{1}(μ) is a function such that

∫_{X} fhdμ = 0 for all f ∈ C_{c}(X)  (22.3)

then h(x) = 0 for μ – a.e. x. (See also Corollary ?? below.)

Proof. Let dν(x) = |h(x)| dx, then ν is a K-finite measure on X and hence C_{c}(X) is dense in L^{1}(ν) by Theorem 22.8. Notice that

∫_{X} f · sgn(h)dν = ∫_{X} fhdμ = 0 for all f ∈ C_{c}(X).  (22.4)

Let {K_{k}}_{k=1}^{∞} be a sequence of compact sets such that K_{k} ↑ X as in Lemma 11.23. Then 1_{K_{k}} \overline{sgn(h)} ∈ L^{1}(ν) and therefore there exists f_{m} ∈ C_{c}(X) such that f_{m} → 1_{K_{k}} \overline{sgn(h)} in L^{1}(ν). So by Eq. (22.4),

ν(K_{k}) = ∫_{X} 1_{K_{k}}dν = lim_{m→∞} ∫_{X} f_{m}sgn(h)dν = 0.

Since K_{k} ↑ X as k → ∞, 0 = ν(X) = ∫_{X} |h| dμ, i.e. h(x) = 0 for μ – a.e. x.

As an application of Lemma 22.11 and Example 12.34, we will show that the Laplace transform is injective.

Theorem 22.12 (Injectivity of the Laplace Transform). For f ∈ L^{1}([0, ∞), dx), the Laplace transform of f is defined by

Lf(λ) := ∫_{0}^{∞} e^{−λx}f(x)dx for all λ > 0.

If Lf(λ) := 0 then f(x) = 0 for m -a.e. x.

Proof. Suppose that f ∈ L^{1}([0, ∞), dx) such that Lf(λ) ≡ 0. Let g ∈ C_{0}([0, ∞), ℝ) and ε > 0 be given. By Example 12.34 we may choose {a_{λ}}_{λ>0} such that # ({λ > 0 : a_{λ} ≠ 0}) < ∞ and

|g(x) − ∑_{λ>0} a_{λ}e^{−λx}| < ε for all x ≥ 0.

|∫_{0}^{∞} g(x)f(x)dx| = |∫_{0}^{∞} (g(x) − ∑_{λ>0} a_{λ}e^{−λx}) f(x)dx|

≤ ∫_{0}^{∞} |g(x) − ∑_{λ>0} a_{λ}e^{−λx}| |f(x)| dx ≤ ε∥f∥_{1}.

Since ε > 0 is arbitrary, it follows that ∫_{0}^{∞} g(x)f(x)dx = 0 for all g ∈ C_{0}([0, ∞), ℝ). The proof is finished by an application of Lemma 22.11.

Here is another variant of Theorem 22.8.

Theorem 22.13. Let (X, d) be a metric space, τ_{d} be the topology on X generated by d and B_{X} = σ(τ_{d}) be the Borel σ – algebra. Suppose μ : B_{X} → [0, ∞] is a measure which is σ – finite on τ_{d} and let BC_{f}(X) denote the bounded continuous functions on X such that μ(f ≠ 0) < ∞. Then BC_{f}(X) is a dense subspace of L^{p}(μ) for any p ∈ [1, ∞).

Proof. Let X_{k} ∈ τ_{d} be open sets such that X_{k} ↑ X and μ(X_{k}) < ∞ and let

ψ_{k}(x) = min(1, k · d_{X_{k}^{c}}(x)) = ϕ_{k}(d_{X_{k}^{c}}(x)),

see Figure 22.1 below. It is easily verified that M := BC_{f}(X) is an algebra,

Fig. 22.1. The plot of ϕ_{n} for n = 1, 2, and 4. Notice that ϕ_{n} → 1_{(0,∞)}.

ψ_{k} ∈ M for all k and ψ_{k} → 1 boundedly as k → ∞. Given V ∈ τ and k, n ∈ ℕ,let

f_{k,n} (x) := min(1, n · d_{(V ∩X_{k})}^{c}(x)).

Then {f_{k,n} ≠ 0} = V ∩ X_{k} so f_{k,n} ∈ BC_{f}(X). Moreover

lim_{k→∞} lim_{n→∞} f_{k,n} = lim_{k→∞} 1_{V ∩X_{k}} = 1_{V}

which shows V ∈ σ (M) and hence σ (M) = B_{X}. The proof is now completed by an application of Theorem 22.4.

Exercise 22.2. (BRUCE: Should drop this exercise.) Suppose that (X, d) is a metric space, μ is a measure on B_{X} := σ(τ_{d}) which is finite on bounded measurable subsets of X. Show BC_{b}(X, ℝ), defined in Eq. (19.26), is dense in L^{p} (μ) . Hints: let ψ_{k} be as defined in Eq. (19.27) which incidentally may be used to show σ (BC_{b}(X, ℝ)) = σ (BC(X, ℝ)) . Then use the argument in the proof of Corollary 18.55 to show σ (BC(X, ℝ)) = B_{X}.

Theorem 22.14. Suppose p ∈ [1, ∞), A ⊂ M is an algebra such that σ(A) = M and μ is σ – finite on A. Then 𝕊_{f}(A, μ) is dense in L^{p}(μ). (See also Remark ?? below.)

Proof. Let M := 𝕊_{f}(A, μ). By assumption there exits X_{k} ∈ A such that μ(X_{k}) < ∞ and X_{k} ↑ X as k → ∞. If A ∈ A, then X_{k} ∩ A ∈ A and μ (X_{k} ∩ A) < ∞ so that 1_{X_{k}∩A} ∈ M. Therefore 1_{A} = lim_{k→∞} 1_{X_{k}∩A} is σ (M)

– measurable for every A ∈ A. So we have shown that A ⊂ σ (M) ⊂ M and therefore M = σ (A) ⊂ σ (M) ⊂ M, i.e. σ (M) = M. The theorem now follows from Theorem 22.4 after observing ψ_{k} := 1_{X_{k}} ∈ M and ψ_{k} → 1 boundedly.

Theorem 22.15 (Separability of L^{p} – Spaces). Suppose, p ∈ [1, ∞), A ⊂ M is a countable algebra such that σ(A) = M and μ is σ – finite on A. Then L^{p}(μ) is separable and

𝔻 = {∑ a_{j}1_{A_{j}} : a_{j} ∈ ℚ + iℚ, A_{j} ∈ A with μ(A_{j}) < ∞}

is a countable dense subset.

Proof. It is left to reader to check 𝔻 is dense in 𝕊_{f}(A, μ) relative to the L^{p}(μ) – norm. The proof is then complete since 𝕊_{f}(A, μ) is a dense subspace of L^{p} (μ) by Theorem 22.14.

Example 22.16. The collection of functions of the form ϕ = ∑_{k=1}^{n} c_{k}1_{(a_{k},b_{k}]} with a_{k}, b_{k} ∈ ℚ and a_{k} < b_{k} are dense in L^{p}(ℝ, B_{ℝ}, m; ℂ) and L^{p}(ℝ, B_{ℝ}, m; ℂ) is separable for any p ∈ [1, ∞). To prove this simply apply Theorem 22.14 with A being the algebra on ℝ generated by the half open intervals (a, b] ∩ ℝ with a < b and a, b ∈ ℚ∪ {±∞} , i.e. A consists of sets of the form ∐_{k=1}^{n}(a_{k}, b_{k}]∩ℝ, where a_{k}, b_{k} ∈ ℚ∪ {±∞} .

Exercise 22.3. Show L^{∞}([0, 1] , B_{ℝ}, m; ℂ) is not separable. Hint: Suppose Γ is a dense subset of L^{∞}([0, 1] , B_{ℝ}, m; ℂ) and for λ ∈ (0, 1) , let f_{λ} (x) := 1_{[0,λ]} (x) . For each λ ∈ (0, 1) , choose g_{λ} ∈ Γ such that ∥f_{λ} − g_{λ}∥_{∞} < 1/2 and then show the map λ ∈ (0, 1) → g_{λ} ∈ Γ is injective. Use this to conclude that Γ must be uncountable.

Corollary 22.17 (Riemann Lebesgue Lemma). Suppose that f ∈ L^{1}(ℝ, m), then

lim_{λ→±∞} ∫_{ℝ} f(x)e^{iλx}dm(x) = 0.

Proof. By Example 22.16, given ε > 0 there exists ϕ = ∑_{k=1}^{n} c_{k}1_{(a_{k},b_{k}]} with a_{k}, b_{k} ∈ ℝ such that

∫_{ℝ} |f − ϕ|dm < ε.

Notice that

∫_{ℝ} ϕ(x)e^{iλx}dm(x) = ∫_{ℝ} ∑_{k=1}^{n} c_{k}1_{(a_{k},b_{k}]}(x)e^{iλx}dm(x)

= ∑_{k=1}^{n} c_{k} ∫_{a_{k}}^{b_{k}} e^{iλx}dm(x) = ∑_{k=1}^{n} c_{k}λ^{−1}e^{iλx}|_{a_{k}}^{b_{k}}

= λ^{−1} ∑_{k=1}^{n} c_{k} (e^{iλb_{k}} − e^{iλa_{k}}) → 0 as |λ| → ∞.

Combining these two equations with

|∫_{ℝ} f(x)e^{iλx}dm(x)| ≤ |∫_{ℝ} (f(x) − ϕ(x)) e^{iλx}dm(x)| + |∫_{ℝ} ϕ(x)e^{iλx}dm(x)|
≤ ∫_{ℝ} |f − ϕ|dm + |∫_{ℝ} ϕ(x)e^{iλx}dm(x)|
≤ ε + |∫_{ℝ} ϕ(x)e^{iλx}dm(x)|

we learn that

lim sup_{|λ|→∞} |∫_{ℝ} f(x)e^{iλx}dm(x)| ≤ ε + lim sup_{|λ|→∞} |∫_{ℝ} ϕ(x)e^{iλx}dm(x)| = ε.

Since ε > 0 is arbitrary, this completes the proof of the Riemann Lebesgue lemma.

Corollary 22.18. Suppose A ⊂ M is an algebra such that σ(A) = M and μ is σ – finite on A. Then for every B ∈ M such that μ(B) < ∞ and ε > 0 there exists D ∈ A such that μ(B△D) < ε. (See also Remark ?? below.)

Proof. By Theorem 22.14, there exists a collection, {A_{i}}_{i=1}^{n} , of pairwise disjoint subsets of A and λ_{i} ∈ ℝ such that ∫_{X} |1_{B} − f| dμ < ε where f = ∑_{i=1}^{n} λ_{i}1_{A_{i}}. Let A_{0} := X \ ∪_{i=1}^{n}A_{i} ∈ A then

∫X |1_{B} − f| dμ = ∑_{i=0}^{n} ∫_{A_{i}} |1_{B} − f| dμ

= μ (A_{0} ∩ B) + ∑_{i=1}^{n} [∫_{A_{i}∩B} |1_{B} − λ_{i}| dμ + ∫_{A_{i}\B} |1_{B} − λ_{i}| dμ]

= μ (A_{0} ∩ B) + ∑_{i=1}^{n} [|1 − λ_{i}| μ (B ∩ A_{i}) + |λ_{i}| μ (A_{i} \ B)]  (22.5)

≥ μ (A_{0} ∩ B) + ∑_{i=1}^{n} min {μ (B ∩ A_{i}) , μ (A_{i} \ B)}  (22.6)

where the last equality is a consequence of the fact that 1 ≤ |λ_{i}| + |1 − λ_{i}| . Let

α_{i} = \{ \begin{matrix}0 & if & μ (B ∩ A_{i}) < μ (A_{i} \ B) \\ 1 & if & μ (B ∩ A_{i}) ≥ μ (A_{i} \ B) \end{matrix}

and g = ∑_{i=1}^{n} α_{i}1_{A_{i}} = 1_{D} where

D := ∪ {A_{i} : i > 0 & α_{i} = 1} ∈ A.

Equation (22.5) with λ_{i} replaced by α_{i} and f by g implies

∫_{X} |1_{B} − 1_{D}| dμ = μ (A_{0} ∩ B) + ∑_{i=1}^{n} min {μ (B ∩ A_{i}) , μ (A_{i} \ B)} .

The latter expression, by Eq. (22.6), is bounded by ∫_{X} |1_{B} − f| dμ < ε and therefore,

μ(B△D) = ∫_{X} |1_{B} − 1_{D}| dμ < ε.

Remark 22.19. We have to assume that μ(B) < ∞ as the following example shows. Let X = ℝ, M = B, μ = m, A be the algebra generated by half open intervals of the form (a, b], and B = ∪_{n=1}^{∞}(2n, 2n + 1]. It is easily checked that for every D ∈ A, that m(BΔD) = ∞.

22.2 Convolution and Young’s Inequalities

Throughout this section we will be solely concerned with d – dimensional Lebesgue measure, m, and we will simply write L^{p} for L^{p} (ℝ^{d}, m) .

Definition 22.20 (Convolution). Let f, g : ℝ^{d} → ℂ be measurable functions. We define

f ∗ g(x) = ∫_{ℝ^{d}} f(x − y)g(y)dy  (22.7)

whenever the integral is defined, i.e. either f (x − ·) g (·) ∈ L^{1}(ℝ^{d}, m) or f (x − ·) g (·) ≥ 0. Notice that the condition that f (x − ·) g (·) ∈ L^{1}(ℝ^{d}, m) is equivalent to writing |f| ∗ |g| (x) < ∞. By convention, if the integral in Eq. (22.7) is not defined, let f ∗ g(x) := 0.

Notation 22.21 Given a multi-index α ∈ ℤ_{+}^{d}, let |α| = α_{1} + ⋯ + α_{d},

x^{α} := ∏_{j=1}^{d} x_{j}^{α_{j}}, and ∂_{x}^{α} = (\frac{∂}{∂x})^{α} := ∏_{j=1}^{d} (\frac{∂}{∂x_{j}} )^{α_{j}} .

For z ∈ ℝ^{d} and f : ℝ^{d} → ℂ, let τ_{z}f : ℝ^{d} → ℂ be defined by τ_{z}f(x) = f(x−z).

Remark 22.22 (The Significance of Convolution).

1. Suppose that f, g ∈ L^{1} (m) are positive functions and let μ be the measure on (ℝ^{d})^{2} defined by

dμ (x, y) := f (x) g (y) dm (x) dm (y) .

Then if h : ℝ → [0, ∞] is a measurable function we have

∫_{(ℝ^{d})_{2}} h (x + y) dμ (x, y) = ∫_{(ℝ^{d})_{2}} h (x + y) f (x) g (y) dm (x) dm (y)

= ∫_{(ℝ^{d})_{2}} h (x) f (x − y) g (y) dm (x) dm (y)

= ∫_{ℝ^{d}} h (x) f ∗ g (x) dm (x) .

In other words, this shows the measure (f ∗ g) m is the same as S_{∗}μ where S (x, y) := x+y. In probability lingo, the distribution of a sum of two “independent” (i.e. product measure) random variables is the the convolution of the individual distributions.

2. Suppose that L = ∑_{|α|≤k} a_{α}∂^{α} is a constant coefficient differential operator and suppose that we can solve (uniquely) the equation Lu = g in the form

u(x) = Kg(x) := ∫_{ℝ^{d}} k(x, y)g(y)dy

where k(x, y) is an “integral kernel.” (This is a natural sort of assumption since, in view of the fundamental theorem of calculus, integration is the inverse operation to differentiation.) Since τ_{z}L = Lτ_{z} for all z ∈ ℝ^{d}, (this is another way to characterize constant coefficient differential operators) and L^{−1} = K we should have τ_{z}K = Kτ_{z}. Writing out this equation then says

∫_{ℝ^{d}} k(x − z, y)g(y)dy = (Kg) (x − z) = τ_{z}Kg(x) = (Kτ_{z}g) (x)
= ∫_{ℝ^{d}} k(x, y)g(y − z)dy = ∫_{ℝ^{d}} k(x, y + z)g(y)dy.

Since g is arbitrary we conclude that k(x − z, y) = k(x, y + z). Taking y = 0 then gives

k(x, z) = k(x − z, 0) =: ρ(x − z).

We thus find that Kg = ρ ∗ g. Hence we expect the convolution operation to appear naturally when solving constant coefficient partial differential equations. More about this point later.

Proposition 22.23. Suppose p ∈ [1, ∞], f ∈ L^{1} and g ∈ L^{p}, then f ∗ g(x) exists for almost every x, f ∗ g ∈ L^{p} and

∥f ∗ g∥_{p} ≤ ∥f∥_{1} ∥g∥_{p} .

Proof. This follows directly from Minkowski’s inequality for integrals, Theorem 21.27.

Proposition 22.24. Suppose that p ∈ [1, ∞), then τ_{z} : L^{p} → L^{p} is an isometric isomorphism and for f ∈ L^{p}, z ∈ ℝ^{d} → τ_{z}f ∈ L^{p} is continuous.

Proof. The assertion that τ_{z} : L^{p} → L^{p} is an isometric isomorphism follows from translation invariance of Lebesgue measure and the fact that τ_{−z} ∘ τ_{z} = id. For the continuity assertion, observe that

∥τ_{z}f − τ_{y}f∥_{p} = ∥τ_{−y} (τ_{z}f − τ_{y}f)∥_{p} = ∥τ_{z−y}f − f∥_{p}

from which it follows that it is enough to show τ_{z}f → f in L^{p} as z → 0 ∈ ℝ^{d}. When f ∈ C_{c}(ℝ^{d}), τ_{z}f → f uniformly and since the K := ∪_{|z|≤1}supp(τ_{z}f) is compact, it follows by the dominated convergence theorem that τ_{z}f → f in L^{p} as z → 0 ∈ ℝ^{d}. For general g ∈ L^{p} and f ∈ C_{c}(ℝ^{d}),

∥τ_{z}g − g∥_{p} ≤ ∥τ_{z}g − τ_{z}f∥_{p} + ∥τ_{z}f − f∥_{p} + ∥f − g∥_{p}
= ∥τ_{z}f − f∥_{p} + 2 ∥f − g∥_{p}

and thus

lim sup_{z→0} ∥τ_{z}g − g∥_{p} ≤ lim sup_{z→0} ∥τ_{z}f − f∥_{p} + 2 ∥f − g∥_{p} = 2 ∥f − g∥_{p} .

Because C_{c}(ℝ^{d}) is dense in L^{p}, the term ∥f − g∥_{p} may be made as small as we please.

Exercise 22.4. Let p ∈ [1, ∞] and ∥τ_{z} − I∥_{L(L^{p}(m))} be the operator norm τ_{z} − I. Show ∥τ_{z} − I∥_{L(L^{p}(m))} = 2 for all z ∈ ℝ^{d} \ {0} and conclude from this that z ∈ ℝ^{d} → τ_{z} ∈ L (L^{p} (m)) is not continuous. Hints: 1) Show ∥τ_{z} − I∥_{L(L^{p}(m))} = ∥τ_{|z|e_{1}} − I∥_{L(L^{p}(m))} . 2) Let z = te_{1} with t > 0 and look for f ∈ L^{p} (m) such that τ_{z}f is approximately equal to −f. (In fact, if p = ∞, you can find f ∈ L^{∞} (m) such that τ_{z}f = −f.) (BRUCE: add on a problem somewhere showing that σ (τ_{z}) = S^{1} ⊂ ℂ. This is very simple to prove if p = 2 by using the Fourier transform.)

Definition 22.25. Suppose that (X, τ) is a topological space and μ is a measure on B_{X} = σ(τ). For a measurable function f : X → ℂ we define the essential support of f by

supp_{μ}(f) = {x ∈ X : μ({y ∈ V : f(y) ≠ 0}}) > 0 ∀ neighborhoods V of x}.

(22.8)

Equivalently, x ∉ supp_{μ}(f) iff there exists an open neighborhood V of x such that 1_{V} f = 0 a.e.

It is not hard to show that if supp(μ) = X (see Definition 21.41) and f ∈ C(X) then supp_{μ}(f) = supp(f) := \overline{\{f ≠ 0\} ,} see Exercise 22.7.

Lemma 22.26. Suppose (X, τ) is second countable and f : X → ℂ is a measurable function and μ is a measure on B_{X}. Then X := U \ supp_{μ}(f) may be described as the largest open set W such that f1_{W} (x) = 0 for μ – a.e. x. Equivalently put, C := supp_{μ}(f) is the smallest closed subset of X such that f = f1_{C} a.e.

Proof. To verify that the two descriptions of supp_{μ}(f) are equivalent, suppose supp_{μ}(f) is defined as in Eq. (22.8) and W := X \ supp_{μ}(f). Then

W = {x ∈ X : ∃ τ ∋ V ∋ x such that μ({y ∈ V : f(y) ≠ 0}}) = 0}
= ∪ {V ⊂_{o} X : μ (f1_{V} ≠ 0) = 0}
= ∪ {V ⊂_{o} X : f1_{V} = 0 for μ – a.e.} .

So to finish the argument it suffices to show μ (f1_{W} ≠ 0) = 0. To to this let U be a countable base for τ and set

U_{f} := {V ∈ U : f1_{V} = 0 a.e.}.

Then it is easily seen that W = ∪U_{f} and since U_{f} is countable

μ (f1_{W} ≠ 0) ≤ ∑_{V ∈U_{f}} μ (f1_{V} ≠ 0) = 0.

Lemma 22.27. Suppose f, g, h : ℝ^{d} → ℂ are measurable functions and assume that x is a point in ℝ^{d} such that |f|∗|g| (x) < ∞ and |f|∗(|g| ∗ |h|) (x) < ∞, then

1. f ∗ g(x) = g ∗ f(x)

2. f ∗ (g ∗ h)(x) = (f ∗ g) ∗ h(x)

3. If z ∈ ℝ^{d} and τ_{z}(|f| ∗ |g|)(x) = |f| ∗ |g| (x − z) < ∞, then

τ_{z}(f ∗ g)(x) = τ_{z}f ∗ g(x) = f ∗ τ_{z}g(x)

4. If x ∉ supp_{m}(f) + supp_{m}(g) then f ∗ g(x) = 0 and in particular,

supp_{m}(f ∗ g) ⊂ \overline{supp_{m}(f) + supp_{m}(g)}

where in defining supp_{m}(f ∗g) we will use the convention that “f ∗g(x) ≠ 0” when |f| ∗ |g| (x) = ∞.

Proof. For item 1.,

|f| ∗ |g| (x) = ∫_{ℝ^{d}} |f| (x − y) |g| (y)dy = ∫_{ℝ^{d}} |f| (y) |g| (y − x)dy = |g| ∗ |f| (x)

where in the second equality we made use of the fact that Lebesgue measure invariant under the transformation y → x − y. Similar computations prove all of the remaining assertions of the first three items of the lemma. Item 4. Since f ∗ g(x) = f̃ ∗ g̃(x) if f = f̃ and g = g̃ a.e. we may, by replacing f by f1_{supp_{m}(f)} and g by g1_{supp_{m}(g)} if necessary, assume that {f ≠ 0} ⊂ supp_{m}(f) and {g ≠ 0} ⊂ supp_{m}(g). So if x ∉ (supp_{m}(f) + supp_{m}(g)) then x ∉ ({f ≠ 0} + {g ≠ 0}) and for all y ∈ ℝ^{d}, either x − y ∉ {f ≠ 0} or y ∉ {g ≠ 0} . That is to say either x − y ∈ {f = 0} or y ∈ {g = 0} and hence f(x−y)g(y) = 0 for all y and therefore f ∗g(x) = 0. This shows that f ∗g = 0 on ℝ^{d} \ (\overline{supp_{m}(f) + supp_{m}(g)}) and therefore

ℝ^{d} \ (\overline{supp_{m}(f) + supp_{m}(g)}) ⊂ ℝ^{d} \ supp_{m}(f ∗ g),

i.e. supp_{m}(f ∗ g) ⊂ supp_{m}(f) + supp_{m}(g).

Remark 22.28. Let A, B be closed sets of ℝ^{d}, it is not necessarily true that A + B is still closed. For example, take

A = {(x, y) : x > 0 and y ≥ 1/x} and B = {(x, y) : x < 0 and y ≥ 1/|x|} ,

then every point of A + B has a positive y - component and hence is not zero. On the other hand, for x > 0 we have (x, 1/x) + (−x, 1/x) = (0, 2/x) ∈ A + B for all x and hence 0 ∈ \overline{A + B} showing A + B is not closed. Nevertheless if one of the sets A or B is compact, then A + B is closed again. Indeed, if A is compact and x_{n} = a_{n} + b_{n} ∈ A + B and x_{n} → x ∈ ℝ^{d}, then by passing to a subsequence if necessary we may assume lim_{n→∞} a_{n} = a ∈ A exists. In this case

lim_{n→∞} b_{n} = lim_{n→∞} (x_{n} − a_{n}) = x − a ∈ B

exists as well, showing x = a + b ∈ A + B.

Proposition 22.29. Suppose that p, q ∈ [1, ∞] and p and q are conjugate exponents, f ∈ L^{p} and g ∈ L^{q}, then f ∗ g ∈ BC(ℝ^{d}), ∥f ∗ g∥_{∞} ≤ ∥f∥_{p} ∥g∥_{q} and if p, q ∈ (1, ∞) then f ∗ g ∈ C_{0}(ℝ^{d}).

Proof. The existence of f∗g(x) and the estimate |f ∗ g| (x) ≤ ∥f∥_{p} ∥g∥_{q} for all x ∈ ℝ^{d} is a simple consequence of Holders inequality and the translation invariance of Lebesgue measure. In particular this shows ∥f ∗ g∥_{∞} ≤ ∥f∥_{p} ∥g∥_{q} . By relabeling p and q if necessary we may assume that p ∈ [1, ∞). Since

∥τ_{z} (f ∗ g) − f ∗ g∥_{u} = ∥τ_{z}f ∗ g − f ∗ g∥_{u}
≤ ∥τ_{z}f − f∥_{p} ∥g∥_{q} → 0 as z → 0

it follows that f ∗ g is uniformly continuous. Finally if p, q ∈ (1, ∞), we learn from Lemma 22.27 and what we have just proved that f_{m} ∗ g_{m} ∈ C_{c}(ℝ^{d}) where f_{m} = f1_{|f|≤m} and g_{m} = g1_{|g|≤m}. Moreover,

∥f ∗ g − f_{m} ∗ g_{m}∥_{∞} ≤ ∥f ∗ g − f_{m} ∗ g∥_{∞} + ∥f_{m} ∗ g − f_{m} ∗ g_{m}∥_{∞}
≤ ∥f − f_{m}∥_{p} ∥g∥_{q} + ∥f_{m}∥_{p} ∥g − g_{m}∥_{q}

≤ ∥f − f_{m}∥_{p} ∥g∥_{q} + ∥f∥_{p} ∥g − g_{m}∥_{q} → 0 as m → ∞

showing, with the aid of Proposition 12.23, f ∗ g ∈ C_{0}(ℝ^{d}).

Theorem 22.30 (Young’s Inequality). Let p, q, r ∈ [1, ∞] satisfy

\frac{1}{p} + \frac{1}{q} = 1 + \frac{1}{r}.  (22.9)

If f ∈ L^{p} and g ∈ L^{q} then |f| ∗ |g| (x) < ∞ for m – a.e. x and

∥f ∗ g∥_{r} ≤ ∥f∥_{p} ∥g∥_{q} .  (22.10)

In particular L^{1} is closed under convolution. (The space (L^{1}, ∗) is an example of a “Banach algebra” without unit.)

Remark 22.31. Before going to the formal proof, let us first understand Eq. (22.9) by the following scaling argument. For λ > 0, let f_{λ}(x) := f(λx), then after a few simple change of variables we find

∥f_{λ}∥_{p} = λ^{−d/p} ∥f∥ and (f ∗ g)_{λ} = λ^{d}f_{λ} ∗ g_{λ}.

Therefore if Eq. (22.10) holds for some p, q, r ∈ [1, ∞], we would also have

∥f ∗ g∥_{r} = λ^{d/r} ∥(f ∗ g)_{λ}∥_{r} ≤ λ^{d/r}λ^{d} ∥f_{λ}∥_{p} ∥g_{λ}∥_{q} = λ^{(d+d/r−d/p−d/q)} ∥f∥_{p} ∥g∥_{q}

for all λ > 0. This is only possible if Eq. (22.9) holds.

Proof. By the usual sorts of arguments, we may assume f and g are positive functions. Let α, β ∈ [0, 1] and p_{1}, p_{2} ∈ (0, ∞] satisfy p_{1}^{−1}+p_{2}^{−1}+r^{−1} = 1. Then by Hölder’s inequality, Corollary 21.3,

f ∗ g(x) = ∫_{ℝ^{d}} [f(x − y)^{(1−α)}g(y)^{(1−β)}] f(x − y)^{α}g(y)^{β}dy

≤ (∫_{ℝ^{d}} f(x − y)^{(1−α)r}g(y)^{(1−β)r}dy)^{1/r} (∫_{ℝ^{d}} f(x − y)^{αp_{1}}dy)^{1/p_{1}} ×

× (∫_{ℝ^{d}} g(y)^{βp_{2}}dy)^{1/p_{2}}

= (∫_{ℝ^{d}} f(x − y)^{(1−α)r}g(y)^{(1−β)r}dy)^{1/r} ∥f∥_{αp_{1}}^{α} ∥g∥_{βp_{2}}^{β} .

Taking the r^{th} power of this equation and integrating on x gives

∥f ∗ g∥_{r}^{r} ≤ ∫_{ℝ^{d}} (∫_{ℝ^{d}} f(x − y)^{(1−α)r}g(y)^{(1−β)r}dy) dx · ∥f∥_{αp_{1}}^{α} ∥g∥_{βp_{2}}^{β}
= ∥f∥_{(1−α)r}^{(1−α)r} ∥g∥_{(1−β)r}^{(1−β)r} ∥f∥_{αp_{1}}^{αr} ∥g∥_{βp_{2}}^{βr} .  (22.11)

Let us now suppose, (1 − α)r = αp_{1} and (1 − β)r = βp_{2}, in which case Eq. (22.11) becomes,

∥f ∗ g∥_{r}^{r} ≤ ∥f∥_{αp_{1}}^{r} ∥g∥_{βp_{2}}^{r}

which is Eq. (22.10) with

p := (1 − α)r = αp_{1} and q := (1 − β)r = βp_{2}.  (22.12)

So to finish the proof, it suffices to show p and q are arbitrary indices in [1, ∞] satisfying p^{−1} +q^{−1} = 1+r^{−1}. If α, β, p_{1}, p_{2} satisfy the relations above, then

α = \frac{r}{r + p_{1}} and β = \frac{r}{r + p_{2}}

and

\frac{1}{p} + \frac{1}{q} = \frac{1}{αp_{1}} + \frac{1}{αp_{2}} = \frac{1}{p_{1}} \frac{r + p_{1}}{r} + \frac{1}{p_{2}} \frac{r + p_{2}}{r}

= \frac{1}{p_{1}} + \frac{1}{p_{2}} + \frac{2}{r} = 1 + \frac{1}{r}.

Conversely, if p, q, r satisfy Eq. (22.9), then let α and β satisfy p = (1 − α)r and q = (1 − β)r, i.e.

α := \frac{r − p}{r} = 1 − \frac{p}{r} ≤ 1 and β = \frac{r − q}{r} = 1 − \frac{q}{r} ≤ 1.

Using Eq. (22.9) we may also express α and β as

α = p(1 − \frac{1}{q}) ≥ 0 and β = q(1 − \frac{1}{p}) ≥ 0

and in particular we have shown α, β ∈ [0, 1]. If we now define p_{1} := p/α ∈ (0, ∞] and p_{2} := q/β ∈ (0, ∞], then

\frac{1}{p_{1}} + \frac{1}{p_{2}} + \frac{1}{r} = β \frac{1}{q} + α\frac{1}{p} + \frac{1}{r}

= (1 − \frac{1}{q}) + (1 − \frac{1}{p}) + \frac{1}{r}

= 2 − (1 + \frac{1}{r}) + \frac{1}{r} = 1

as desired.

Theorem 22.32 (Approximate δ – functions). Let p ∈ [1, ∞], ϕ ∈ L^{1}(ℝ^{d}), a := ∫_{ℝ^{d}} ϕ(x)dx, and for t > 0 let ϕ_{t}(x) = t^{−d}ϕ(x/t). Then

1. If f ∈ L^{p} with p < ∞ then ϕ_{t} ∗ f → af in L^{p} as t ↓ 0.

2. If f ∈ BC(ℝ^{d}) and f is uniformly continuous then ∥ϕ_{t} ∗ f − af∥_{∞} → 0 as t ↓ 0.

3. If f ∈ L^{∞} and f is continuous on U ⊂_{o} ℝ^{d} then ϕ_{t} ∗ f → af uniformly on compact subsets of U as t ↓ 0.

Proof. Making the change of variables y = tz implies

ϕ_{t} ∗ f(x) = ∫_{ℝ^{d}} f(x − y)ϕ_{t}(y)dy = ∫_{ℝ^{d}} f(x − tz)ϕ(z)dz

so that

ϕ_{t} ∗ f(x) − af(x) = ∫_{ℝ^{d}} [f(x − tz) − f(x)] ϕ(z)dz
= ∫_{ℝ^{d}} [τ_{tz}f(x) − f(x)] ϕ(z)dz.  (22.13)

Hence by Minkowski’s inequality for integrals (Theorem 21.27), Proposition 22.24 and the dominated convergence theorem,

∥ϕ_{t} ∗ f − af∥_{p} ≤ ∫_{ℝ^{d}} ∥τ_{tz}f − f∥_{p} |ϕ(z)| dz → 0 as t ↓ 0.

Item 2. is proved similarly. Indeed, form Eq. (22.13)

∥ϕ_{t} ∗ f − af∥_{∞} ≤ ∫_{ℝ^{d}} ∥τ_{tz}f − f∥_{∞} |ϕ(z)| dz

which again tends to zero by the dominated convergence theorem because lim_{t↓0} ∥τ_{tz}f − f∥_{∞} = 0 uniformly in z by the uniform continuity of f.

Item 3. Let B_{R} = B(0, R) be a large ball in ℝ^{d} and K ⊏⊏ U, then

sup_{x∈K} |ϕ_{t} ∗ f(x) − af(x)|

≤ |∫_{B_{R}} [f(x − tz) − f(x)] ϕ(z)dz| + |∫_{B_{R}^{c}} [f(x − tz) − f(x)] ϕ(z)dz|

≤ ∫_{B_{R}} |ϕ(z)| dz · sup_{x∈K,z∈B_{R}} |f(x − tz) − f(x)| + 2 ∥f∥_{∞} ∫_{B_{R}^{c}} |ϕ(z)| dz

≤ ∥ϕ∥_{1} · sup_{x∈K,z∈B_{R}} |f(x − tz) − f(x)| + 2 ∥f∥_{∞} ∫_{|z|>R} |ϕ(z)| dz

so that using the uniform continuity of f on compact subsets of U,

lim sup_{t↓0} sup_{x∈K} |ϕ_{t} ∗ f(x) − af(x)| ≤ 2 ∥f∥_{∞} ∫_{|z|>R} |ϕ(z)| dz → 0 as R → ∞.

See Theorem 8.15 of Folland for a statement about almost everywhere convergence.

Exercise 22.5. Let

f(t) = \{ \begin{matrix}e^{−1/t} & if & t > 0 \\ 0 & if & t ≤ 0. \end{matrix}

Show f ∈ C^{∞}(ℝ, [0, 1]).

Lemma 22.33. There exists ϕ ∈ C_{c}^{∞}(ℝ^{d}, [0, ∞)) such that ϕ(0) > 0, supp(ϕ) ⊂ B̄(0, 1) and ∫_{ℝ^{d}} ϕ(x)dx = 1.

Proof. Define h(t) = f(1 − t)f(t + 1) where f is as in Exercise 22.5. Then h ∈ C_{c}^{∞}(ℝ, [0, 1]), supp(h) ⊂ [−1, 1] and h(0) = e^{−2} > 0. Define c = ∫_{ℝ^{d}} h(|x|^{2})dx. Then ϕ(x) = c^{−1}h(|x|^{2}) is the desired function.

The reader asked to prove the following proposition in Exercise 22.9 below.

Proposition 22.34. Suppose that f ∈ L_{loc}^{1}(ℝ^{d}, m) and ϕ ∈ C_{c}^{1}(ℝ^{d}), then f ∗ ϕ ∈ C^{1}(ℝ^{d}) and ∂_{i}(f ∗ ϕ) = f ∗ ∂_{i}ϕ. Moreover if ϕ ∈ C_{c}^{∞}(ℝ^{d}) then f ∗ ϕ ∈ C^{∞}(ℝ^{d}).

Corollary 22.35 (C^{∞} – Uryhson’s Lemma). Given K ⊏⊏ U ⊂_{o} ℝ^{d}, there exists f ∈ C_{c}^{∞}(ℝ^{d}, [0, 1]) such that supp(f) ⊂ U and f = 1 on K.

Proof. Let ϕ be as in Lemma 22.33, ϕ_{t}(x) = t^{−d}ϕ(x/t) be as in Theorem 22.32, d be the standard metric on ℝ^{d} and ε = d(K, U^{c}). Since K is compact and U^{c} is closed, ε > 0. Let V_{δ} = {x ∈ ℝ^{d} : d(x, K) < δ} and f = ϕ_{ε/3} ∗1_{V_{ε/3}}, then

supp(f) ⊂ \overline{supp(ϕ_{ε/3}) + V_{ε/3}} ⊂ V̄_{2ε/3} ⊂ U.

Since V̄_{2ε/3} is closed and bounded, f ∈ C_{c}^{∞}(U) and for x ∈ K,

f(x) = ∫_{ℝ^{d}} 1_{d(y,K)<ε/3} · ϕ_{ε/3}(x − y)dy = ∫_{ℝ^{d}} ϕ_{ε/3}(x − y)dy = 1.

The proof will be finished after the reader (easily) verifies 0 ≤ f ≤ 1.

Here is an application of this corollary whose proof is left to the reader, Exercise 22.10.

Lemma 22.36 (Integration by Parts). Suppose f and g are measurable functions on ℝ^{d} such that t → f(x_{1}, …, x_{i−1}, t, x_{i+1}, …, x_{d}) and t → g(x_{1}, …, x_{i−1}, t, x_{i+1}, …, x_{d}) are continuously differentiable functions on ℝ for each fixed x = (x_{1}, …, x_{d}) ∈ ℝ^{d}. Moreover assume f · g, \frac{∂f}{∂x_{i}} · g and f · \frac{∂g}{∂x_{i}} are in L^{1}(ℝ^{d}, m). Then

∫_{ℝ^{d}} \frac{∂f}{∂x_{i}} · gdm = − ∫_{ℝ^{d}} f · \frac{∂g}{∂x_{i}} dm.

With this result we may give another proof of the Riemann Lebesgue Lemma.

Lemma 22.37 (Riemann Lebesgue Lemma). For f ∈ L^{1}(ℝ^{d}, m) let

f̂(ξ) := (2π)^{−d/2} ∫_{ℝ^{d}} f(x)e^{−iξ·x}dm(x)

be the Fourier transform of f. Then f̂ ∈ C_{0}(ℝ^{d}) and ∥f̂∥_{∞} ≤ (2π)^{−d/2} ∥f∥_{1} .

(The choice of the normalization factor, (2π)^{−d/2}, in f̂ is for later convenience.)

Proof. The fact that f̂ is continuous is a simple application of the dominated convergence theorem. Moreover,

|f̂(ξ)| ≤ ∫_{ℝ^{d}} |f(x)| dm(x) ≤ (2π)^{−d/2} ∥f∥_{1}

so it only remains to see that f̂(ξ) → 0 as |ξ| → ∞. First suppose that f ∈ C_{c}^{∞}(ℝ^{d}) and let Δ = ∑_{j=1}^{d} \frac{∂^{2}}{∂x_{j}^{2}} be the Laplacian on ℝ^{d}. Notice that

\frac{∂}{∂x_{j}} e^{−iξ·x} = −iξ_{j}e^{−iξ·x} and Δe^{−iξ·x} = − |ξ|^{2} e^{−iξ·x}. Using Lemma 22.36 repeatedly,

∫_{ℝ^{d}} Δ^{k}f(x)e^{−iξ·x}dm(x) = ∫_{ℝ^{d}} f(x)Δ_{x}^{k}e^{−iξ·x}dm(x) = − |ξ|^{2k} ∫_{ℝ^{d}} f(x)e^{−iξ·x}dm(x)
= −(2π)^{d/2} |ξ|^{2k} f̂(ξ)

for any k ∈ ℕ. Hence

(2π)^{d/2} |f̂(ξ)| ≤ |ξ|^{−2k} ∥Δ^{k}f∥_{1} → 0

as |ξ| → ∞ and f̂ ∈ C_{0}(ℝ^{d}). Suppose that f ∈ L^{1}(m) and f_{k} ∈ C_{c}^{∞}(ℝ^{d}) is a sequence such that lim_{k→∞} ∥f − f_{k}∥_{1} = 0, then lim_{k→∞} ∥f̂− f̂_{k}∥_{∞} = 0.

Hence f̂ ∈ C_{0}(ℝ^{d}) by an application of Proposition 12.23.

Corollary 22.38. Let X ⊂ ℝ^{d} be an open set and μ be a K-finite measure on B_{X}.

1. Then C_{c}^{∞}(X) is dense in L^{p}(μ) for all 1 ≤ p < ∞.

2. If h ∈ L_{loc}^{1}(μ) satisfies

∫_{X} fhdμ = 0 for all f ∈ C_{c}^{∞}(X)  (22.14)

then h(x) = 0 for μ – a.e. x.

Proof. Let f ∈ C_{c}(X), ϕ be as in Lemma 22.33, ϕ_{t} be as in Theorem 22.32 and set ψ_{t} := ϕ_{t} ∗ (f1_{X}) . Then by Proposition 22.34 ψ_{t} ∈ C^{∞}(X) and by Lemma 22.27 there exists a compact set K ⊂ X such that supp(ψ_{t}) ⊂ K for all t sufficiently small. By Theorem 22.32, ψ_{t} → f uniformly on X as t ↓ 0

1. The dominated convergence theorem (with dominating function being ∥f∥_{∞} 1_{K}), shows ψ_{t} → f in L^{p}(μ) as t ↓ 0. This proves Item 1., since Theorem 22.8 guarantees that C_{c}(X) is dense in L^{p}(μ).

2. Keeping the same notation as above, the dominated convergence theorem (with dominating function being ∥f∥_{∞} |h| 1_{K}) implies

0 = lim_{t↓0} ∫_{X} ψ_{t}hdμ = ∫_{X} lim_{t↓0} ψ_{t}hdμ = ∫_{X} fhdμ.

The proof is now finished by an application of Lemma 22.11.

22.2.1 Smooth Partitions of Unity

We have the following smooth variants of Proposition 12.16, Theorem 12.18 and Corollary 12.20. The proofs of these results are the same as their continuous counterparts. One simply uses the smooth version of Urysohn’s Lemma of Corollary 22.35 in place of Lemma 12.8.

Proposition 22.39 (Smooth Partitions of Unity for Compacts). Suppose that X is an open subset of ℝ^{d}, K ⊂ X is a compact set and U = {U_{j}}_{j=1}^{n} is an open cover of K. Then there exists a smooth (i.e. h_{j} ∈ C^{∞}(X, [0, 1])) partition of unity {h_{j}}_{j=1}^{n} of K such that h_{j} ≺ U_{j} for all j = 1, 2, …, n.

Theorem 22.40 (Locally Compact Partitions of Unity). Suppose that X is an open subset of ℝ^{d} and U is an open cover of X. Then there exists a smooth partition of unity of {h_{i}}_{i=1}^{N} (N = ∞ is allowed here) subordinate to the cover U such that supp(h_{i}) is compact for all i.

Corollary 22.41. Suppose that X is an open subset of ℝ^{d} and U = {U_{α}}_{α∈A} ⊂ τ is an open cover of X. Then there exists a smooth partition of unity of {h_{α}}_{α∈A} subordinate to the cover U such that supp(h_{α}) ⊂ U_{α} for all α ∈ A. Moreover if Ū_{α} is compact for each α ∈ A we may choose h_{α} so that h_{α} ≺ U_{α}.

22.3 Exercises

Exercise 22.6. Let (X, τ) be a topological space, μ a measure on B_{X} = σ(τ) and f : X → ℂ be a measurable function. Letting ν be the measure, dν = |f| dμ, show supp(ν) = supp_{μ}(f), where supp(ν) is defined in Definition 21.41).

Exercise 22.7. Let (X, τ) be a topological space, μ a measure on B_{X} = σ(τ) such that supp(μ) = X (see Definition 21.41). Show supp_{μ}(f) = supp(f) = \overline{\{f ≠ 0\}} for all f ∈ C(X).

Exercise 22.8. Prove the following strong version of item 3. of Proposition 10.52, namely to every pair of points, x_{0}, x_{1}, in a connected open subset V of ℝ^{d} there exists σ ∈ C^{∞}(ℝ, V) such that σ(0) = x_{0} and σ(1) = x_{1}. Hint: First choose a continuous path γ : [0, 1] → V such that γ (t) = x_{0} for t near 0 and γ (t) = x_{1} for t near 1 and then use a convolution argument to smooth γ.

Exercise 22.9. Prove Proposition 22.34 by appealing to Corollary 19.43.

Exercise 22.10 (Integration by Parts). Suppose that (x, y) ∈ ℝ × ℝ^{d−1} → f(x, y) ∈ ℂ and (x, y) ∈ ℝ × ℝ^{d−1} → g(x, y) ∈ ℂ are measurable functions such that for each fixed y ∈ ℝ^{d}, x → f(x, y) and x → g(x, y) are continuously differentiable. Also assume f · g, ∂_{x}f · g and f · ∂_{x}g are integrable relative to Lebesgue measure on ℝ × ℝ^{d−1}, where ∂_{x}f(x, y) := \frac{d}{dt}f(x + t, y)|_{t=0}. Show

∫_{ℝ×ℝ_{d−1}} ∂_{x}f(x, y) · g(x, y)dxdy = − ∫_{ℝ×ℝ^{d−1}} f(x, y) · ∂_{x}g(x, y)dxdy. (22.15)

(Note: this result and Fubini’s theorem proves Lemma 22.36.)

Hints: Let ψ ∈ C_{c}^{∞}(ℝ) be a function which is 1 in a neighborhood of 0 ∈ ℝ and set ψ_{ε}(x) = ψ(εx). First verify Eq. (22.15) with f(x, y) replaced by ψ_{ε}(x)f(x, y) by doing the x – integral first. Then use the dominated convergence theorem to prove Eq. (22.15) by passing to the limit, ε ↓ 0.

Exercise 22.11. Let μ be a finite measure on B_{ℝ^{d}}, then 𝔻 := span{e^{iλ·x} : λ ∈ ℝ^{d}} is a dense subspace of L^{p}(μ) for all 1 ≤ p < ∞. Hints: By Theorem 22.8, C_{c}(ℝ^{d}) is a dense subspace of L^{p}(μ). For f ∈ C_{c}(ℝ^{d}) and N ∈ ℕ, let

f_{N}(x) := ∑_{n∈ℤ^{d}} f(x + 2πNn).

Show f_{N} ∈ BC(ℝ^{d}) and x → f_{N}(Nx) is 2π – periodic, so by Exercise 12.13, x → f_{N}(Nx) can be approximated uniformly by trigonometric polynomials. Use this fact to conclude that f_{N} ∈ 𝔻̄^{L^{p}(μ)}. After this show f_{N} → f in L^{p}(μ).

Exercise 22.12. Suppose that μ and ν are two finite measures on ℝ^{d} such that

∫_{ℝ^{d}} e^{iλ·x}dμ(x) = ∫_{ℝ^{d}} e^{iλ·x}dν(x)  (22.16)

for all λ ∈ ℝ^{d}. Show μ = ν.

Hint: Perhaps the easiest way to do this is to use Exercise 22.11 with the measure μ being replaced by μ + ν. Alternatively, use the method of proof of Exercise 22.11 to show Eq. (22.16) implies ∫_{ℝ^{d}} fdμ(x) = ∫_{ℝ^{d}} fdν(x) for all f ∈ C_{c}(ℝ^{d}) and then apply Corollary 18.58.

Exercise 22.13. Again let μ be a finite measure on B_{ℝ^{d}}. Further assume that C_{M} := ∫_{ℝ^{d}} e^{M|x|}dμ(x) < ∞ for all M ∈ (0, ∞). Let P(ℝ^{d}) be the space of polynomials, ρ(x) = ∑_{|α|≤N} ρ_{α}x^{α} with ρ_{α} ∈ ℂ, on ℝ^{d}. (Notice that |ρ(x)|^{p} ≤ Ce^{M|x|} for some constant C = C(ρ, p, M), so that P(ℝ^{d}) ⊂ L^{p}(μ) for all 1 ≤ p < ∞.) Show P(ℝ^{d}) is dense in L^{p}(μ) for all 1 ≤ p < ∞. Here is a possible outline.

Outline: Fix a λ ∈ ℝ^{d} and let f_{n}(x) = (λ · x)^{n} /n! for all n ∈ ℕ.

1. Use calculus to verify sup_{t≥0} t^{α}e^{−Mt} = (α/M)^{α} e^{−α} for all α ≥ 0 where (0/M)^{0} := 1. Use this estimate along with the identity

|λ · x|^{pn} ≤ |λ|^{pn} |x|^{pn} = (|x|^{pn} e^{−M|x|}) |λ|^{pn} e^{M|x|}

to find an estimate on ∥f_{n}∥_{p} .

2. Use your estimate on ∥f_{n}∥_{p} to show ∑_{n=0}^{∞} ∥f_{n}∥_{p} < ∞ and conclude

lim_{N→∞} ∥e^{iλ·(·)} − ∑_{n=0}^{N} i^{n}f_{n}∥_{p} = 0.

3. Now finish by appealing to Exercise 22.11.

Exercise 22.14. Again let μ be a finite measure on B_{ℝ^{d}} but now assume there exists an ε > 0 such that C := ∫_{ℝ^{d}} e^{ε|x|}dμ(x) < ∞. Also let q > 1 and h ∈ L^{q}(μ) be a function such that ∫_{ℝ^{d}} h(x)x^{α}dμ(x) = 0 for all α ∈ ℕ_{0}^{d}. (As mentioned in Exercise 22.14, P(ℝ^{d}) ⊂ L^{p}(μ) for all 1 ≤ p < ∞, so x → h(x)x^{α} is in L^{1}(μ).) Show h(x) = 0 for μ– a.e. x using the following outline.

Outline: Fix a λ ∈ ℝ^{d}, let f_{n}(x) = (λ · x)^{n} /n! for all n ∈ ℕ, and let p = q/(q − 1) be the conjugate exponent to q.

1. Use calculus to verify sup_{t≥0} t^{α}e^{−εt} = (α/ε)^{α} e^{−α} for all α ≥ 0 where (0/ε)^{0} := 1. Use this estimate along with the identity

|λ · x|^{pn} ≤ |λ|^{pn} |x|^{pn} = (|x|^{pn} e^{−ε|x|}) |λ|^{pn} e^{ε|x|}

to find an estimate on ∥f_{n}∥_{p} .

2. Use your estimate on ∥f_{n}∥_{p} to show there exists δ > 0 such that ∑_{n=0}^{∞} ∥f_{n}∥_{p} < ∞ when |λ| ≤ δ and conclude for |λ| ≤ δ that e^{iλ·x} = L^{p}(μ)-∑_{n=0}^{∞} i^{n}f_{n}(x). Conclude from this that

∫_{ℝ^{d}} h(x)e^{iλ·x}dμ(x) = 0 when |λ| ≤ δ.

3. Let λ ∈ ℝ^{d} (|λ| not necessarily small) and set g(t) := ∫_{ℝ^{d}} e^{itλ·x}h(x)dμ(x) for t ∈ ℝ. Show g ∈ C^{∞}(ℝ) and

g^{(n)}(t) = ∫_{ℝ^{d}}(iλ · x)^{n}e^{itλ·x}h(x)dμ(x) for all n ∈ ℕ.

4. Let T = sup{τ ≥ 0 : g|_{[0,τ]} ≡ 0}. By Step 2., T ≥ δ. If T < ∞, then

0 = g^{(n)}(T) = ∫_{ℝ^{d}}(iλ · x)^{n}e^{iT λ·x}h(x)dμ(x) for all n ∈ ℕ.

Use Step 3. with h replaced by e^{iT λ·x}h(x) to conclude

g(T + t) = ∫_{ℝ^{d}} e^{i(T +t)λ·x}h(x)dμ(x) = 0 for all t ≤ δ/ |λ| .

This violates the definition of T and therefore T = ∞ and in particular we may take T = 1 to learn

∫_{ℝ^{d}} h(x)e^{iλ·x}dμ(x) = 0 for all λ ∈ ℝ^{d}.

5. Use Exercise 22.11 to conclude that

∫_{ℝ^{d}} h(x)g(x)dμ(x) = 0

for all g ∈ L^{p}(μ). Now choose g judiciously to finish the proof.

L^{2} - Hilbert Spaces Techniques and Fourier Series

This section is concerned with Hilbert spaces presented as in the following example.

Example 23.1. Let (X, M, μ) be a measure space. Then H := L^{2}(X, M, μ) with inner product

⟨f|g⟩ = ∫_{X} f · ḡdμ

is a Hilbert space.

It will be convenient to define

⟨f, g⟩ := ∫_{X} f (x) g (x) dμ (x)  (23.1)

for all measurable functions f, g on X such that fg ∈ L^{1} (μ) . So with this notation we have ⟨f|g⟩ = ⟨f, ḡ⟩ for all f, g ∈ H.

Exercise 23.1. Let K : L^{2}(ν) → L^{2}(μ) be the operator defined in Exercise 21.12. Show K^{∗} : L^{2}(μ) → L^{2}(ν) is the operator given by

K^{∗}g(y) = ∫_{X} k̄(x, y)g(x)dμ(x).

23.1 L^{2}-Orthonoramal Basis

Example 23.2. 1. Let H = L^{2}([−1, 1], dm), A := {1, x, x^{2}, x^{3} … } and β ⊂ H be the result of doing the Gram-Schmidt procedure on A. By the Stone-Weierstrass theorem or by Exercise 22.13 directly, A is total in H. Hence by Remark 14.26, β is an orthonormal basis for H. The basis, β, consists of polynomials which up to normalization are the so called “Legendre polynomials.”

432 23 L^{2} - Hilbert Spaces Techniques and Fourier Series

2. Let H = L^{2}(ℝ, e^{− \frac{1}{2} x^{2}}dx) and A := \{1, x, x^{2}, x^{3} … \}. Again by Exercise 22.13, A is total in H and hence the Gram-Schmidt procedure applied to A produces an orthonormal basis, β, of polynomial functions for H. This basis consists, up to normalizations, of the so called “Hermite polynomials” on ℝ.

Remark 23.3 (An Interesting Phenomena). Let H = L^{2}([−1, 1], dm) and B := {1, x^{3}, x^{6}, x^{9}, … }. Then again A is total in H by the same argument as in item 2. Example 23.2. This is true even though B is a proper subset of A. Notice that A is an algebraic basis for the polynomials on [−1, 1] while B is not! The following computations may help relieve some of the reader’s anxiety. Let f ∈ L^{2}([−1, 1], dm), then, making the change of variables x = y^{1/3}, shows that

∫_{−1}^{1} |f(x)|^{2} dx = ∫_{−1}^{1} |f(y^{1/3})|^{2} \frac{1}{3}y^{−2/3}dy = ∫_{−1}^{1} |f(y^{1/3})|^{2} dμ(y) (23.2)

where dμ(y) = \frac{1}{3}y^{−2/3}dy. Since μ([−1, 1]) = m([−1, 1]) = 2, μ is a finite measure on [−1, 1] and hence by Exercise 22.13 A := {1, x, x^{2}, x^{3} … } is a total (see Definition 14.25) in L^{2}([−1, 1], dμ). In particular for any ε > 0 there exists a polynomial p(y) such that

∫_{−1}^{1} |f(y^{1/3}) − p(y)|^{2} dμ(y) < ε^{2}.

However, by Eq. (23.2) we have

ε^{2} > ∫_{−1}^{1} |f(y^{1/3}) − p(y)|^{2} dμ(y) = ∫_{−1}^{1} |f(x) − p(x^{3})|^{2} dx.

Alternatively, if f ∈ C([−1, 1]), then g(y) = f(y^{1/3}) is back in C([−1, 1]). Therefore for any ε > 0, there exists a polynomial p(y) such that

ε > ∥g − p∥_{∞} = sup {|g(y) − p(y)| : y ∈ [−1, 1]}
= sup {|g(x^{3}) − p(x^{3})| : x ∈ [−1, 1]}
= sup {|f(x) − p(x^{3})| : x ∈ [−1, 1]} .

This gives another proof the polynomials in x^{3} are dense in C([−1, 1]) and hence in L^{2}([−1, 1]).

Exercise 23.2. Suppose (X, M, μ) and (Y, N , ν) are σ-finite measure spaces such that L^{2} (μ) and L^{2} (ν) are separable. If {f_{n}}_{n=1}^{∞} and {g_{m}}_{m=1}^{∞} are orthonormal bases for L^{2} (μ) and L^{2} (ν) respectively, then β := {f_{n} ⊗ g_{m} : m, n ∈ ℕ} is an orthonormal basis for L^{2} (μ ⊗ ν) . (Recall that f ⊗ g (x, y) := f (x) g (y) , see Notation 20.4.) Hint: model your proof of the proof of Proposition 14.28.

Exercise 23.3. Suppose H is a Hilbert space and {H_{n} : n ∈ ℕ} are closed subspaces of H such that H_{n} ⊥ H_{m} for all m ≠ n and if f ∈ H with f ⊥ H_{n} for all n ∈ ℕ, then f = 0. Show:

1. If f_{n} ∈ H_{n} for all n ∈ ℕ satisfy ∑_{n=1}^{∞} ∥f_{n}∥^{2} < ∞ then ∑_{n=1}^{∞} f_{n} exists in H.

2. Every element f ∈ H may be uniquely written as f = ∑_{n=1}^{∞} f_{n} with f_{n} ∈ H as in item 1. (For this reason we will write H = ⊕_{n=1}^{∞}H_{n} under the hypothesis of this exercise.)

Exercise 23.4. Suppose (X, M, μ) is a measure space and X = ∐_{n=1}^{∞} X_{n} with X_{n} ∈ M and μ (X_{n}) > 0 for all n. Then U : L^{2} (X, μ) → ⊕_{n=1}^{∞}L^{2} (X_{n}, μ) defined by (Uf)_{n} := f1_{X_{n}} is unitary.

23.2 Hilbert Schmidt Operators

In this section H and B will be Hilbert spaces.

Proposition 23.4. Let H and B be a separable Hilbert spaces, K : H → B be a bounded linear operator, {e_{n}}_{n=1}^{∞} and {u_{m}}_{m=1}^{∞} be orthonormal basis for H and B respectively. Then:

1. ∑_{n=1}^{∞} ∥Ke_{n}∥^{2} = ∑_{m=1}^{∞} ∥K^{∗}u_{m}∥^{2} allowing for the possibility that the sums are infinite. In particular the Hilbert Schmidt norm of K,

∥K∥_{HS}^{2} := ∑_{n=1}^{∞} ∥Ke_{n}∥^{2} ,

is well defined independent of the choice of orthonormal basis {e_{n}}_{n=1}^{∞}. We say K : H → B is a Hilbert Schmidt operator if ∥K∥_{HS} < ∞ and let HS(H, B) denote the space of Hilbert Schmidt operators from H to B.

2. For all K ∈ L(H, B), ∥K∥_{HS} = ∥K^{∗}∥_{HS} and

∥K∥_{HS} ≥ ∥K∥_{op} := sup {∥Kh∥ : h ∈ H ∋ ∥h∥ = 1} .

3. The set HS(H, B) is a subspace of L (H, B) (the bounded operators from H → B), ∥·∥_{HS} is a norm on HS(H, B) for which (HS(H, B), ∥·∥_{HS}) is a Hilbert space, and the corresponding inner product is given by

⟨K_{1}|K_{2}⟩_{HS} = ∑_{n=1}^{∞} ⟨K_{1}e_{n}|K_{2}e_{n}⟩ .  (23.3)

4. If K : H → B is a bounded finite rank operator, then K is Hilbert Schmidt.

5. Let P_{N}x := ∑_{n=1}^{N} ⟨x|e_{n}⟩ e_{n} be orthogonal projection onto span{e_{n} : n ≤ N} ⊂ H and for K ∈ HS(H, B), let K_{N} := KP_{N}. Then

∥K − K_{N}∥_{op}^{2} ≤ ∥K − K_{N}∥_{HS}^{2} → 0 as N → ∞,

which shows that finite rank operators are dense in (HS(H, B), ∥·∥_{HS}) . In particular of HS(H, B) ⊂ K(H, B) – the space of compact operators from H → B.

6. If Y is another Hilbert space and A : Y → H and C : B → Y are bounded operators, then

∥KA∥_{HS} ≤ ∥K∥_{HS} ∥A∥_{op} and ∥CK∥_{HS} ≤ ∥K∥_{HS} ∥C∥_{op} ,

in particular HS(H, H) is an ideal in L (H) .

Proof. Items 1. and 2. By Parseval’s equality and Fubini’s theorem for sums,

∑_{n=1}^{∞} ∥Ke_{n}∥^{2} = ∑_{n=1}^{∞} ∑_{m=1}^{∞} |⟨Ke_{n}|u_{m}⟩|^{2}

= ∑_{m=1}^{∞} ∑_{n=1}^{∞} |⟨e|K^{∗}u_{m}⟩|^{2} = ∑_{m=1}^{∞} ∥K^{∗}u_{m}∥^{2} .

This proves ∥K∥_{HS} is well defined independent of basis and that ∥K∥_{HS} = ∥K^{∗}∥_{HS} . For x ∈ H \ {0} , x/ ∥x∥ may be taken to be the first element in an orthonormal basis for H and hence

∥K \frac{x}{∥x∥}∥ ≤ ∥K∥_{HS} .

Multiplying this inequality by ∥x∥ shows ∥Kx∥ ≤ ∥K∥_{HS} ∥x∥ and hence ∥K∥_{op} ≤ ∥K∥_{HS} .

Item 3. For K_{1}, K_{2} ∈ L(H, B),

∥K_{1} + K_{2}∥_{HS} = \sqrt{∑_{n=1}^{∞} ∥K_{1}e_{n} + K_{2}e_{n}∥^{2}}

≤ \sqrt{∑_{n=1}^{∞} [∥K_{1}e_{n}∥ + ∥K_{2}e_{n}∥]^{2}}

= ∥{∥K_{1}e_{n}∥ + ∥K_{2}e_{n}∥}_{n=1}^{∞}∥_{ℓ_{2}}

≤ ∥{∥K_{1}e_{n}∥}_{n=1}^{∞}∥_{ℓ_{2}} + ∥{∥K_{2}e_{n}∥}_{n=1}^{∞}∥_{ℓ_{2}}

= ∥K_{1}∥_{HS} + ∥K_{2}∥_{HS} .

From this triangle inequality and the homogeneity properties of ∥·∥_{HS} , we now easily see that HS(H, B) is a subspace of L(H, B) and ∥·∥_{HS} is a norm on HS(H, B). Since

∑_{n=1}^{∞} |⟨K_{1}e_{n}|K_{2}e_{n}⟩| ≤ ∑_{n=1}^{∞} ∥K_{1}e_{n}∥ ∥K_{2}e_{n}∥

≤ \sqrt{∑_{n=1}^{∞} ∥K_{1}e_{n}∥^{2}}\sqrt{∑_{n=1}^{∞} ∥K_{2}e_{n}∥^{2}} = ∥K_{1}∥_{HS} ∥K_{2}∥_{HS} ,

the sum in Eq. (23.3) is well defined and is easily checked to define an inner product on HS(H, B) such that ∥K∥_{HS}^{2} = ⟨K|K⟩_{HS} .

To see that HS(H, B) is complete in this inner product suppose {K_{m}}_{m=1}^{∞} is a ∥·∥_{HS} – Cauchy sequence in HS(H, B). Because L(H, B) is complete, there exists K ∈ L(H, B) such that ∥K − K_{m}∥_{op} → 0 as m → ∞. Thus, making use of Fatou’s Lemma 4.12,

∥K − K_{m}∥_{HS}^{2} = ∑_{n=1}^{∞} ∥(K − K_{m}) e_{n}∥^{2}

= ∑_{n=1}^{∞} lim inf_{l→∞} ∥(K_{l} − K_{m}) e_{n}∥^{2}

≤ lim inf_{l→∞} ∑_{n=1}^{∞} ∥(K_{l} − K_{m}) e_{n}∥^{2}

= lim inf_{l→∞} ∥K_{l} − K_{m}∥_{HS}^{2} → 0 as m → ∞.

Hence K ∈ HS(H, B) and lim_{m→∞} ∥K − K_{m}∥_{HS}^{2} = 0.

Item 4. Let N := dim K (H) and {v_{n}}_{n=1}^{N} be an orthonormal basis for Ran(K) = K (H) . Then, for all h ∈ H,

∥Kh∥_{B}^{2} = ∑_{n=1}^{N} |⟨Kh|v_{n}⟩|^{2} = ∑_{n=1}^{N} |⟨h|K^{∗}v_{n}⟩|^{2} .

Summing this equation on h ∈ β where an β is an orthonormal basis for H shows

∥K∥_{HS}^{2} = ∑_{h∈β} ∥Kh∥_{B}^{2} = ∑_{n=1}^{N} ∥K^{∗}v_{n}∥_{H}^{2} < ∞.

Item 5. Simply observe,

∥K − K_{N}∥_{op}^{2} ≤ ∥K − K_{N}∥_{HS}^{2} = ∑_{n>N} ∥Ke_{n}∥^{2} → 0 as N → ∞.

Item 6. For C ∈ L(B, Y) and K ∈ L(H, B) then

∥CK∥_{HS}^{2} = ∑_{n=1}^{∞} ∥CKe_{n}∥^{2} ≤ ∥C∥_{op}^{2} ∑_{n=1}^{∞} ∥Ke_{n}∥^{2} = ∥C∥_{op}^{2} ∥K∥_{HS}^{2}

and for A ∈ L (Y, H) ,

∥KA∥_{HS} = ∥A^{∗}K^{∗}∥_{HS} ≤ ∥A^{∗}∥_{op} ∥K^{∗}∥_{HS} = ∥A∥_{op} ∥K∥_{HS} .

Remark 23.5. The separability assumptions made in Proposition 23.4 are unnecessary. In general, we define

∥K∥_{HS}^{2} = ∑_{e∈β} ∥Ke∥^{2}

where β ⊂ H is an orthonormal basis. The same proof of Item 1. of Proposition 23.4 shows ∥K∥_{HS} is well defined and ∥K∥_{HS} = ∥K^{∗}∥_{HS} . If ∥K∥_{HS}^{2} < ∞, then there exists a countable subset β_{0} ⊂ β such that Ke = 0 if e ∈ β \β_{0}. Let H_{0} := \overline{span(β_{0})} and B_{0} := \overline{K(H_{0})}. Then K (H) ⊂ B_{0}, K|_{H_{0}^{⊥}} = 0 and hence by applying the results of Proposition 23.4 to K|_{H_{0}} : H_{0} → B_{0} one easily sees that the separability of H and B are unnecessary in Proposition 23.4.

Example 23.6. Let (X, μ) be a measure space, H = L^{2}(X, μ) and

k(x, y) := ∑_{i=1}^{n} f_{i}(x)g_{i}(y)

where

f_{i}, g_{i} ∈ L^{2}(X, μ) for i = 1, …, n.

Define (Kf)(x) = ∫_{X} k(x, y)f(y)dμ(y), then K : L^{2}(X, μ) → L^{2}(X, μ) is a finite rank operator and hence Hilbert Schmidt.

Exercise 23.5. Suppose that (X, μ) is a σ–finite measure space such that H = L^{2}(X, μ) is separable and k : X × X → ℝ is a measurable function, such that

∥k∥_{L^{2}(X×X,μ⊗μ)}^{2} := ∫_{X×X} |k(x, y)|^{2}dμ(x)dμ(y) < ∞.

Define, for f ∈ H,

Kf(x) = ∫_{X} k(x, y)f(y)dμ(y),

when the integral makes sense. Show:

1. Kf(x) is defined for μ–a.e. x in X.

2. The resulting function Kf is in H and K : H → H is linear.

3. ∥K∥_{HS} = ∥k∥_{L^{2}(X×X,μ⊗μ)} < ∞. (This implies K ∈ HS(H, H).)

Example 23.7. Suppose that Ω ⊂ ℝ^{n} is a bounded set, α < n, then the operator K : L^{2}(Ω, m) → L^{2}(Ω, m) defined by

Kf(x) := ∫_{Ω} \frac{1}{|x − y|^{α}} f(y)dy

is compact.

Proof. For ε ≥ 0, let

K_{ε}f(x) := ∫_{Ω} \frac{1}{|x − y|^{α} + ε}f(y)dy = [g_{ε} ∗ (1_{Ω}f)] (x)

where g_{ε}(x) = \frac{1}{|x|^{α}+ε}1_{C}(x) with C ⊂ ℝ^{n} a sufficiently large ball such that Ω − Ω ⊂ C. Since α < n, it follows that

g_{ε} ≤ g_{0} = |·|^{−α} 1_{C} ∈ L^{1}(ℝ^{n}, m).

Hence it follows by Proposition 22.23 that

∥(K − K_{ε}) f∥_{L^{2}(Ω)} ≤ ∥(g_{0} − g_{ε}) ∗ (1_{Ω}f)∥_{L^{2}(ℝ^{n})}
≤ ∥(g_{0} − g_{ε})∥_{L^{1}(ℝ^{n})} ∥1_{Ω}f∥_{L^{2}(ℝ^{n})}
= ∥(g_{0} − g_{ε})∥_{L^{1}(ℝ^{n})} ∥f∥_{L^{2}(Ω)}

which implies

∥K − K_{ε}∥_{B(L^{2}(Ω))} ≤ ∥g_{0} − g_{ε}∥_{L^{1}(ℝ^{n})}
= ∫_{C} |\frac{1}{|x|^{α} + ε} − \frac{1}{|x|^{α}} | dx → 0 as ε ↓ 0  (23.4)

by the dominated convergence theorem. For any ε > 0,

∫_{Ω×Ω} [\frac{1}{|x − y|^{α} + ε}]^{2} dxdy < ∞,

and hence K_{ε} is Hilbert Schmidt and hence compact. By Eq. (23.4), K_{ε} → K as ε ↓ 0 and hence it follows that K is compact as well.

Exercise 23.6. Let H := L^{2} ([0, 1] , m) , k (x, y) := min (x, y) for x, y ∈ [0, 1] and define K : H → H by

Kf (x) = ∫_{0}^{1} k (x, y) f (y) dy.

By Exercise 23.5, K is a Hilbert Schmidt operator and it is easily seen that K is self-adjoint. Show:

1. Show ⟨Kf|g′′⟩ = −⟨f|g⟩ for all g ∈ C_{c}^{∞} ((0, 1)) and use this to conclude that Nul(K) = {0} .

2. Now suppose that f ∈ H is an eigenvector of K with eigenvalue λ ≠ 0. Show that there is a version of f in C ([0, 1])∩C^{2} ((0, 1)) and this version, still denoted by f, solves

λf′′ = −f with f (0) = f′ (1) = 0.  (23.5)

where f′ (1) := lim_{x↑1} f′ (x) .

3. Use Eq. (23.5) to find all the eigenvalues and eigenfunctions of K.

4. Use the results above along with the spectral Theorem 14.45, to show

\{\sqrt{2} sin (n\frac{π}{2} x) : n ∈ ℕ\}

is an orthonormal basis for L^{2} ([0, 1] , m) .

Exercise 23.7. Let (X, M, μ) be a σ – finite measure space, a ∈ L^{∞}(μ) and let A be the bounded operator on H := L^{2}(μ) defined by Af (x) = a (x) f (x) for all f ∈ H. (We will denote A by M_{a} in the future.) Show:

1. ∥A∥_{op} = ∥a∥_{L^{∞}(μ)} .

2. A^{∗} = Mā.

3. σ (A) = essran(a) where σ (A) is the spectrum of A and essran(a) is the essential range of a, see Definitions 14.30 and 21.40 respectively.

4. Show λ is an eigenvalue for A = M_{a} iff μ ({a = λ}) > 0, i.e. iff a has a “flat spot of height λ.”

23.3 Fourier Series Considerations

Throughout this section we will let dθ, dx, dα, etc. denote Lebesgue measure on ℝ^{d} normalized so that the cube, Q := (−π, π]^{d}, has measure one, i.e. dθ = (2π)^{−d}dm(θ) where m is standard Lebesgue measure on ℝ^{d}. As usual, for α ∈ ℕ_{0}^{d}, let

D_{θ}^{α} = (\frac{1}{i})^{|α|} \frac{∂^{|α|}}{∂θ_{1}^{α_{1}} … ∂θ_{d}^{α_{d}}} .

Notation 23.8 Let C_{per}^{k}(ℝ^{d}) denote the 2π – periodic functions in C^{k}(ℝ^{d}), that is f ∈ C_{per}^{k}(ℝ^{d}) iff f ∈ C^{k}(ℝ^{d}) and f(θ +2πe_{i}) = f(θ) for all θ ∈ ℝ^{d} and i = 1, 2, …, d. Further let ⟨·|·⟩ denote the inner product on the Hilbert space, H := L^{2}([−π, π]^{d}), given by

⟨f|g⟩ := ∫_{Q} f(θ)ḡ(θ)dθ = (\frac{1}{2π} )^{d} ∫_{Q} f(θ)ḡ(θ)dm (θ)

and define e_{k}(θ) := e^{ik·θ} for all k ∈ ℤ^{d}. For f ∈ L^{1}(Q), we will write f̃(k) for the Fourier coefficient,

f̃(k) := ⟨f|e_{k}⟩ = ∫_{Q} f(θ)e^{−ik·θ}dθ.  (23.6)

Since any 2π – periodic functions on ℝ^{d} may be identified with function on the d - dimensional torus, 𝕋^{d} ≅ ℝ^{d}/ (2πℤ)^{d} ≅ (S^{1})^{d} , I may also write C^{k}(𝕋^{d}) for C_{per}^{k}(ℝ^{d}) and L^{p} (𝕋^{d}) for L^{p} (Q) where elements in f ∈ L^{p} (Q) are to be thought of as there extensions to 2π – periodic functions on ℝ^{d}.

Theorem 23.9 (Fourier Series). The functions β := {e_{k} : k ∈ ℤ^{d}} form an orthonormal basis for H, i.e. if f ∈ H then

f = ∑_{k∈ℤ^{d}}⟨f|e_{k}⟩e_{k} = ∑_{k∈ℤ^{d}} f̃(k)e_{k}  (23.7)

where the convergence takes place in L^{2}([−π, π]^{d}).

Proof. Simple computations show β := {e_{k} : k ∈ ℤ^{d}} is an orthonormal set. We now claim that β is an orthonormal basis. To see this recall that C_{c}((−π, π)^{d}) is dense in L^{2}((−π, π)^{d}, dm). Any f ∈ C_{c}((−π, π)) may be extended to be a continuous 2π – periodic function on ℝ and hence by Exercise 12.13 and Remark 12.44, f may uniformly (and hence in L^{2}) be approximated by a trigonometric polynomial. Therefore β is a total orthonormal set, i.e. β is an orthonormal basis.

This may also be proved by first proving the case d = 1 as above and then using Exercise 23.2 inductively to get the result for any d.

Exercise 23.8. Let A be the operator defined in Lemma 14.36 and for g ∈ L^{2} (𝕋) , let Ug (k) := g̃ (k) so that U : L^{2} (𝕋) → ℓ^{2} (ℤ) is unitary. Show U^{−1}AU = M_{a} where a ∈ C_{per}^{∞} (ℝ) is a function to be found. Use this representation and the results in Exercise 23.7 to give a simple proof of the results in Lemma 14.36.

23.3.1 Dirichlet, Fejér and Kernels

Although the sum in Eq. (23.7) is guaranteed to converge relative to the Hilbertian norm on H it certainly need not converge pointwise even if f ∈ C_{per} (ℝ^{d}) as will be proved in Section ?? below. Nevertheless, if f is sufficiently regular, then the sum in Eq. (23.7) will converge pointwise as we will now show. In the process we will give a direct and constructive proof of the result in Exercise 12.13, see Theorem 23.11 below.

Let us restrict our attention to d = 1 here. Consider

f_{n} (θ) = ∑_{|k|≤n} f̃(k)e_{k} (θ) = ∑_{|k|≤n} \frac{1}{2π} [∫_{[−π,π]} f(x)e^{−ik·x}dx] e_{k} (θ)

= \frac{1}{2π} ∫_{[−π,π]} f(x) ∑_{|k|≤n} e^{ik·(θ−x)}dx = \frac{1}{2π} ∫_{[−π,π]} f(x)D_{n}(θ − x)dx (23.8)

where

D_{n}(θ) := ∑_{k=−n}^{n} e^{ikθ}

is called the Dirichlet kernel. Letting α = e^{iθ/2}, we have

D_{n}(θ) = ∑_{k=−n}^{n} α^{2k} = \frac{α^{2(n+1)} − α^{−2n}}{α^{2} − 1} = \frac{α^{2n+1} − α^{−(2n+1)}}{α − α^{−1}}

= \frac{2i sin(n + \frac{1}{2})θ}{2i sin \frac{1}{2}θ} = \frac{sin(n + \frac{1}{2})θ}{sin \frac{1}{2}θ} .

and therefore

D_{n}(θ) := ∑_{k=−n}^{n} e^{ikθ} = \frac{sin(n + \frac{1}{2})θ}{sin \frac{1}{2}θ} ,  (23.9)

see Figure 23.3.1.

This is a plot D_{1} and D_{10}.

with the understanding that the right side of this equation is 2n + 1 whenever θ ∈ 2πℤ.

Theorem 23.10. Suppose f ∈ L^{1} ([−π, π] , dm) and f is differentiable at some θ ∈ [−π, π] , then lim_{n→∞} f_{n} (θ) = f (θ) where f_{n} is as in Eq. (23.8).

Proof. Observe that

\frac{1}{2π} ∫_{[−π,π]} D_{n}(θ − x)dx = \frac{1}{2π} ∫_{[−π,π]} ∑_{|k|≤n} e^{ik·(θ−x)}dx = 1

f_{n} (θ) − f (θ) = \frac{1}{2π} ∫_{[−π,π]} [f(x) − f (θ)] D_{n}(θ − x)dx

= \frac{1}{2π} ∫_{[−π,π]} [f(x) − f (θ − x)] D_{n}(x)dx

= \frac{1}{2π} ∫_{[−π,π]} [\frac{f(θ − x) − f (θ)}{sin \frac{1}{2}x}] sin(n + \frac{1}{2})x dx. (23.10)

If f is differentiable at θ, the last expression in Eq. (23.10) tends to 0 as n → ∞ by the Riemann Lebesgue Lemma (Corollary 22.17 or Lemma 22.37) and the fact that 1_{[−π,π]} (x) \frac{f(θ−x)−f(θ)}{sin \frac{1}{2} x} ∈ L^{1} (dx) .

Despite the Dirichlet kernel not being positive, it still satisfies the approximate δ – sequence property, \frac{1}{2π} D_{n} → δ_{0} as n → ∞, when acting on C^{1} – periodic functions in θ. In order to improve the convergence properties it is reasonable to try to replace {f_{n} : n ∈ ℕ_{0}} by the sequence of averages (see Exercise 7.13),

F_{N} (θ) = \frac{1}{N + 1} ∑_{n=0}^{N} f_{n} (θ) = \frac{1}{N + 1} ∑_{n=0}^{N} \frac{1}{2π} ∫_{[−π,π]} f(x) ∑_{|k|≤n} e^{ik·(θ−x)}dx

= \frac{1}{2π} ∫_{[−π,π]} K_{N}(θ − x)f(x)dx

where

K_{N}(θ) := \frac{1}{N + 1} ∑_{n=0}^{N} ∑_{|k|≤n} e^{ik·θ}  (23.11)

is the Fejér kernel.

Theorem 23.11. The Fejér kernel K_{N} in Eq. (23.11) satisfies:

1.

K_{N}(θ) = ∑_{n=−N}^{N} [1 − \frac{|n|}{N + 1}] e^{inθ}  (23.12)

= \frac{1}{N + 1} \begin{matrix}sin & 2 & (\frac{N+1}{2}θ) \\ sin & 2 & (\frac{θ}{2}) \end{matrix}.  (23.13)

2. K_{N}(θ) ≥ 0.

3. \frac{1}{2π} ∫_{−π}^{π} K_{N}(θ)dθ = 1

4. sup_{ε≤|θ|≤π} K_{N}(θ) → 0 as N → ∞ for all ε > 0, see Figure 23.1.

5. For any continuous 2π – periodic function f on ℝ, K_{N} ∗ f(θ) → f(θ) uniformly in θ as N → ∞, where

K_{N} ∗ f(θ) = \frac{1}{2π} ∫_{−π}^{π} K_{N}(θ − α)f(α)dα

= ∑_{n=−N}^{N} [1 − \frac{|n|}{N + 1}] f̃(n) e^{inθ}.  (23.14)

Fig. 23.1. Plots of K_{N}(θ) for N = 2, 7 and 13.

Proof. 1. Equation (23.12) is a consequence of the identity,

∑_{n=0}^{N} ∑_{|k|≤n} e^{ik·θ} = ∑_{|k|≤n≤N} e^{ik·θ} = ∑_{|k|≤N} (N + 1 − |k|) e^{ik·θ}.

Moreover, letting α = e^{iθ/2} and using Eq. (3.3) shows

K_{N}(θ) = \frac{1}{N + 1} ∑_{n=0}^{N} ∑_{|k|≤n} α^{2k} = \frac{1}{N + 1} ∑_{n=0}^{N} \frac{α^{2n+2} − α^{−2n}}{α^{2} − 1}

= \frac{1}{(N + 1) (α − α^{−1})} ∑_{n=0}^{N} [α^{2n+1} − α^{−2n−1}]

= \frac{1}{(N + 1) (α − α^{−1})} ∑_{n=0}^{N} [αα^{2n} − α^{−1}α^{−2n}]

= \frac{1}{(N + 1) (α − α^{−1})} [α\frac{α^{2N+2} − 1}{α^{2} − 1} − α^{−1} \frac{α^{−2N−2} − 1}{α^{−2} − 1}]

= \frac{1}{(N + 1) (α − α^{−1})^{2}} [α^{2(N+1)} − 1 + α^{−2(N+1)} − 1]

= \frac{1}{(N + 1) (α − α^{−1})^{2}} [α^{(N+1)} − α^{−(N+1)}]^{2}

= \frac{1}{N + 1} \begin{matrix}sin & 2 & ((N + 1) θ/2) \\ sin & 2 & (θ/2) \end{matrix}.

Items 2. and 3. follow easily from Eqs. (23.13) and (23.12) respectively. Item 4. is a consequence of the elementary estimate;

sup_{ε≤|θ|≤π} K_{N}(θ) ≤ \frac{1}{N + 1} \frac{1}{sin^{2} (\frac{ε}{2})}

and is clearly indicated in Figure 23.1. Item 5. now follows by the standard approximate δ – function arguments, namely,

|K_{N} ∗ f(θ) − f (θ)| = \frac{1}{2π} |∫_{−π}^{π} K_{N}(θ − α) [f(α) − f (θ)] dα|

≤ \frac{1}{2π} ∫_{−π}^{π} K_{N}(α) |f(θ − α) − f (θ)| dα

≤ \frac{1}{π} \frac{1}{N + 1} \frac{1}{sin^{2} (\frac{ε}{2})} ∥f∥_{∞} + \frac{1}{2π} ∫_{|α|≤ε} K_{N}(α) |f(θ − α) − f (θ)| dα

≤ \frac{1}{π} \frac{1}{N + 1} \frac{1}{sin^{2} (\frac{ε}{2})} ∥f∥_{∞} + sup_{|α|≤ε} |f(θ − α) − f (θ)| .

Therefore,

lim sup_{N→∞} ∥K_{N} ∗ f − f∥_{∞} ≤ sup_{θ} sup_{|α|≤ε} |f(θ − α) − f (θ)| → 0 as ε ↓ 0.

23.3.2 The Dirichlet Problems on D and the Poisson Kernel

Let D := {z ∈ ℂ : |z| < 1} be the open unit disk in ℂ ≅ ℝ^{2}, write z ∈ ℂ as z = x + iy or z = re^{iθ}, and let Δ = \frac{∂^{2}}{∂x^{2}} + \frac{∂^{2}}{∂y^{2}} be the Laplacian acting on C^{2} (D) .

Theorem 23.12 (Dirichlet problem for D). To every continuous function g ∈ C (bd(D)) there exists a unique function u ∈ C(D̄) ∩ C^{2}(D) solving

Δu(z) = 0 for z ∈ D and u|_{∂D} = g.  (23.15)

Moreover for r < 1, u is given by,

u(re^{iθ}) = \frac{1}{2π} ∫_{−π}^{π} P_{r}(θ − α)u(e^{iα})dα =: P_{r} ∗ u(e^{iθ})  (23.16)
= \frac{1}{2π} Re ∫_{−π}^{π} \frac{1 + re^{i(θ−α)}}{1 − re^{i(θ−α)}} u(e^{iα})dα  (23.17)

where P_{r} is the Poisson kernel defined by

P_{r}(δ) := \frac{1 − r^{2}}{1 − 2r cos δ + r^{2}} .

(The problem posed in Eq. (23.15) is called the Dirichlet problem for D.)

Proof. In this proof, we are going to be identifying S^{1} = bd(D) := {z ∈ D̄ : |z| = 1} with [−π, π]/ (π ∼ −π) by the map θ ∈ [−π, π] → e^{iθ} ∈ S^{1}. Also recall that the Laplacian Δ may be expressed in polar coordinates as,

Δu = r^{−1}∂_{r} (r^{−1}∂_{r}u) + \frac{1}{r^{2}} ∂_{θ}^{2}u,

where

(∂_{r}u) (re^{iθ}) = \frac{∂}{∂r} u (re^{iθ}) and (∂_{θ}u) (re^{iθ}) = \frac{∂}{∂θ} u (re^{iθ}) .

Uniqueness. Suppose u is a solution to Eq. (23.15) and let

g̃(k) := \frac{1}{2π} ∫_{−π}^{π} g(e^{ikθ})e^{−ikθ}dθ

and

ũ(r, k) := \frac{1}{2π} ∫_{−π}^{π} u(re^{iθ})e^{−ikθ}dθ  (23.18)

be the Fourier coefficients of g (θ) and θ → u (re^{iθ}) respectively. Then for r ∈ (0, 1) ,

r^{−1}∂_{r} (r∂_{r}ũ(r, k)) = \frac{1}{2π} ∫_{−π}^{π} r^{−1}∂_{r} (r^{−1}∂_{r}u) (re^{iθ})e^{−ikθ}dθ

= −\frac{1}{2π} ∫_{−π}^{π} \frac{1}{r^{2}} ∂_{θ}^{2}u(re^{iθ})e^{−ikθ}dθ

= −\frac{1}{r^{2}} \frac{1}{2π} ∫_{−π}^{π} u(re^{iθ})∂_{θ}^{2}e^{−ikθ}dθ

= \frac{1}{r^{2}} k^{2}ũ(r, k)

or equivalently

r∂_{r} (r∂_{r}ũ(r, k)) = k^{2}ũ(r, k).  (23.19)

Recall the general solution to

r∂_{r} (r∂_{r}y(r)) = k^{2}y(r)  (23.20)

may be found by trying solutions of the form y(r) = r^{α} which then implies α^{2} = k^{2} or α = ±k. From this one sees that ũ(r, k) solving Eq. (23.19) may be written as ũ(r, k) = A_{k}r^{|k|} + B_{k}r^{−|k|} for some constants A_{k} and B_{k} when k ≠ 0. If k = 0, the solution to Eq. (23.20) is gotten by simple integration and the result is ũ(r, 0) = A_{0} + B_{0} ln r. Since ũ(r, k) is bounded near the origin for each k it must be that B_{k} = 0 for all k ∈ ℤ. Hence we have shown there exists A_{k} ∈ ℂ such that, for all r ∈ (0, 1),

A_{k}r^{|k|} = ũ(r, k) = \frac{1}{2π} ∫_{−π}^{π} u(re^{iθ})e^{−ikθ}dθ.  (23.21)

Since all terms of this equation are continuous for r ∈ [0, 1], Eq. (23.21) remains valid for all r ∈ [0, 1] and in particular we have, at r = 1, that

A_{k} = \frac{1}{2π} ∫_{−π}^{π} u(e^{iθ})e^{−ikθ}dθ = g̃(k).

Hence if u is a solution to Eq. (23.15) then u must be given by

u(re^{iθ}) = ∑_{k∈ℤ} g̃(k)r^{|k|}e^{ikθ} for r < 1.  (23.22)

or equivalently,

u(z) = ∑_{k∈ℕ_{0}} g̃(k)z^{k} + ∑_{k∈ℕ} g̃(−k)z̄^{k}.

Notice that the theory of the Fourier series implies Eq. (23.22) is valid in the L^{2} (dθ) - sense. However more is true, since for r < 1, the series in Eq. (23.22) is absolutely convergent and in fact defines a C^{∞} – function (see Exercise 4.11 or Corollary 19.43) which must agree with the continuous function, θ → u (re^{iθ}) , for almost every θ and hence for all θ. This completes the proof of uniqueness.

Existence. Given g ∈ C (bd(D)) , let u be defined as in Eq. (23.22). Then, again by Exercise 4.11 or Corollary 19.43, u ∈ C^{∞} (D) . So to finish the proof it suffices to show lim_{x→y} u (x) = g (y) for all y ∈ bd(D). Inserting the formula for g̃(k) into Eq. (23.22) gives

u(re^{iθ}) = \frac{1}{2π} ∫_{−π}^{π} P_{r} (θ − α) u(e^{iα})dα for all r < 1

where

P_{r} (δ) = ∑_{k∈ℤ} r^{|k|}e^{ikδ} = ∑_{k=0}^{∞} r^{k}e^{ikδ} + ∑_{k=0}^{∞} r^{k}e^{−ikδ} − 1 =

= Re [2\frac{1}{1 − re^{iδ}} − 1] = Re [\frac{1 + re^{iδ}}{1 − re^{iδ}} ]

= Re [\begin{matrix}(1 + re^{iδ}) & (1 − re^{−iδ}) \\ |1 − re^{iδ}| & 2 \end{matrix}] = Re [\frac{1 − r^{2} + 2ir sin δ}{1 − 2r cos δ + r^{2}} ] (23.23)

= \frac{1 − r^{2}}{1 − 2r cos δ + r^{2}} .

The Poisson kernel again solves the usual approximate δ – function properties (see Figure 2), namely:

1. P_{r} (δ) > 0 and

\frac{1}{2π} ∫_{−π}^{π} P_{r} (θ − α) dα = \frac{1}{2π} ∫_{−π}^{π} ∑_{k∈ℤ} r^{|k|}e^{ik(θ−α)}dα

= \frac{1}{2π} ∑_{k∈ℤ} r^{|k|} ∫_{−π}^{π} e^{ik(θ−α)}dα = 1

and 2.

sup_{ε≤|θ|≤π} P_{r}(θ) ≤ \frac{1 − r^{2}}{1 − 2r cos ε + r^{2}} → 0 as r ↑ 1.

A plot of P_{r}(δ) for r = 0.2, 0.5 and 0.7.

23.4 Weak L^{2}-Derivatives 447

Therefore by the same argument used in the proof of Theorem 23.11,

lim_{r↑1} sup_{θ} |u (re^{iθ}) − g (e^{iθ})| = lim_{r↑1} sup_{θ} |(P_{r} ∗ g) (e^{iθ}) − g (e^{iθ})| = 0

which certainly implies lim_{x→y} u (x) = g (y) for all y ∈ bd(D).

Remark 23.13 (Harmonic Conjugate). Writing z = re^{iθ}, Eq. (23.17) may be rewritten as

u(z) = \frac{1}{2π} Re ∫_{−π}^{π} \frac{1 + ze^{−iα}}{1 − ze^{−iα}} u(e^{iα})dα

which shows u = Re F where

F (z) := \frac{1}{2π} ∫_{−π}^{π} \frac{1 + ze^{−iα}}{1 − ze^{−iα}} u(e^{iα})dα.

Moreover it follows from Eq. (23.23) that

Im F (re^{iθ}) = \frac{1}{π} Im ∫_{−π}^{π} \frac{r sin(θ − α)}{1 − 2r cos(θ − α) + r^{2}} g(e^{iα})dα
=: (Q_{r} ∗ u) (e^{iθ})

where

Q_{r}(δ) := \frac{r sin(δ)}{1 − 2r cos(δ) + r^{2}} .

From these remarks it follows that v =: (Q_{r} ∗ g) (e^{iθ}) is the harmonic conjugate of u and P̃_{r} = Q_{r}. For more on this point see Section ?? below.

23.4 Weak L^{2}-Derivatives

Theorem 23.14 (Weak and Strong Differentiability). Suppose that f ∈ L^{2}(ℝ^{n}) and v ∈ ℝ^{n} \ {0} . Then the following are equivalent:

1. There exists {t_{n}}_{n=1}^{∞} ⊂ ℝ\ {0} such that lim_{n→∞} t_{n} = 0 and

sup_{n} ∥\frac{f(· + t_{n}v) − f(·)}{t_{n}} ∥_{2} < ∞.

2. There exists g ∈ L^{2}(ℝ^{n}) such that ⟨f, ∂_{v}ϕ⟩ = −⟨g, ϕ⟩ for all ϕ ∈ C_{c}^{∞}(ℝ^{n}).

3. There exists g ∈ L^{2}(ℝ^{n}) and f_{n} ∈ C_{c}^{∞}(ℝ^{n}) such that f_{n} →^{L^{2}} f and ∂_{v}f_{n} →^{L^{2}} g as n → ∞.

4. There exists g ∈ L^{2} such that

\frac{f(· + tv) − f(·)}{t} →^{L^{2}} g as t → 0.

(See Theorem ?? for the L^{p} generalization of this theorem.)

Proof. 1. ⟹ 2. We may assume, using Theorem 14.52 and passing to a subsequence if necessary, that \frac{f(·+t_{n}v)−f(·)}{t_{n}} →^{w} g for some g ∈ L^{2}(ℝ^{n}). Now for ϕ ∈ C_{c}^{∞}(ℝ^{n}),

⟨g|ϕ⟩ = lim_{n→∞} ⟨\frac{f(· + t_{n}v) − f(·)}{t_{n}}, ϕ⟩ = lim_{n→∞} ⟨f, \frac{ϕ(· − t_{n}v) − ϕ(·)}{t_{n}}⟩

= ⟨f, lim_{n→∞} \frac{ϕ(· − t_{n}v) − ϕ(·)}{t_{n}}⟩ = −⟨f, ∂_{v}ϕ⟩,

wherein we have used the translation invariance of Lebesgue measure and the dominated convergence theorem. 2. ⟹ 3. Let ϕ ∈ C_{c}^{∞}(ℝ^{n}, ℝ) such that ∫_{ℝ^{n}} ϕ(x)dx = 1 and let ϕ_{m}(x) = m^{n}ϕ(mx), then by Proposition 22.34, h_{m} := ϕ_{m} ∗ f ∈ C^{∞}(ℝ^{n}) for all m and

∂_{v}h_{m}(x) = ∂_{v}ϕ_{m} ∗ f(x) = ∫_{ℝ^{n}} ∂_{v}ϕ_{m}(x − y)f(y)dy = ⟨f, −∂_{v} [ϕ_{m} (x − ·)]⟩
= ⟨g, ϕ_{m} (x − ·)⟩ = ϕ_{m} ∗ g(x).

By Theorem 22.32, h_{m} → f ∈ L^{2}(ℝ^{n}) and ∂_{v}h_{m} = ϕ_{m} ∗ g → g in L^{2}(ℝ^{n}) as m → ∞. This shows 3. holds except for the fact that h_{m} need not have compact support. To fix this let ψ ∈ C_{c}^{∞}(ℝ^{n}, [0, 1]) such that ψ = 1 in a neighborhood of 0 and let ψ_{ε}(x) = ψ(εx) and (∂_{v}ψ)_{ε} (x) := (∂_{v}ψ) (εx). Then

∂_{v} (ψ_{ε}h_{m}) = ∂_{v}ψ_{ε}h_{m} + ψ_{ε}∂_{v}h_{m} = ε (∂_{v}ψ)_{ε} h_{m} + ψ_{ε}∂_{v}h_{m}

so that ψ_{ε}h_{m} → h_{m} in L^{2} and ∂_{v} (ψ_{ε}h_{m}) → ∂_{v}h_{m} in L^{2} as ε ↓ 0. Let f_{m} = ψ_{ε_{m}}h_{m} where ε_{m} is chosen to be greater than zero but small enough so that

∥ψ_{ε_{m}}h_{m} − h_{m}∥_{2} + ∥∂_{v} (ψ_{ε_{m}}h_{m}) → ∂_{v}h_{m}∥_{2} < 1/m.

Then f_{m} ∈ C_{c}^{∞}(ℝ^{n}), f_{m} → f and ∂_{v}f_{m} → g in L^{2} as m → ∞. 3. ⟹ 4. By the fundamental theorem of calculus

\frac{τ_{−tv}f_{m}(x) − f_{m}(x)}{t} = \frac{f_{m}(x + tv) − f_{m}(x)}{t}

= \frac{1}{t} ∫_{0}^{1} \frac{d}{ds}f_{m}(x + stv)ds = ∫_{0}^{1} (∂_{v}f_{m}) (x + stv)ds. (23.24)

Let

G_{t}(x) := ∫_{0}^{1} τ_{−stv}g(x)ds = ∫_{0}^{1} g(x + stv)ds

which is defined for almost every x and is in L^{2}(ℝ^{n}) by Minkowski’s inequality for integrals, Theorem 21.27. Therefore

\frac{τ_{−tv}f_{m}(x) − f_{m}(x)}{t} − G_{t}(x) = ∫_{0}^{1} [(∂_{v}f_{m}) (x + stv) − g(x + stv)] ds

and hence again by Minkowski’s inequality for integrals,

∥\frac{τ_{−tv}f_{m} − f_{m}}{t} − G_{t}∥_{2} ≤ ∫_{0}^{1} ∥τ_{−stv} (∂_{v}f_{m}) − τ_{−stv}g∥_{2} ds
= ∫_{0}^{1} ∥∂_{v}f_{m} − g∥_{2} ds.

Letting m → ∞ in this equation implies (τ_{−tv}f − f) /t = G_{t} a.e. Finally one more application of Minkowski’s inequality for integrals implies,

∥\frac{τ_{−tv}f − f}{t} − g∥_{2} = ∥G_{t} − g∥_{2} = ∥∫_{0}^{1} (τ_{−stv}g − g) ds∥_{2}
≤ ∫_{0}^{1} ∥τ_{−stv}g − g∥_{2} ds.

By the dominated convergence theorem and Proposition 22.24, the latter term tends to 0 as t → 0 and this proves 4. The proof is now complete since 4. ⟹ 1. is trivial.

23.5 *Conditional Expectation

In this section let (Ω, F, P) be a probability space, i.e. (Ω, F, P) is a measure space and P (Ω) = 1. Let G ⊂ F be a sub – sigma algebra of F and write f ∈ G_{b} if f : Ω → ℂ is bounded and f is (G, B_{ℂ}) – measurable. In this section we will write

Ef := ∫_{Ω} fdP.

Definition 23.15 (Conditional Expectation). Let E_{G} : L^{2}(Ω, F, P) → L^{2}(Ω, G, P) denote orthogonal projection of L^{2}(Ω, F, P) onto the closed subspace L^{2}(Ω, G, P). For f ∈ L^{2}(Ω, G, P), we say that E_{G}f ∈ L^{2}(Ω, F, P) is the conditional expectation of f.

Theorem 23.16. Let (Ω, F, P) and G ⊂ F be as above and f, g ∈ L^{2}(Ω, F, P).

1. If f ≥ 0, P – a.e. then E_{G}f ≥ 0, P – a.e.

2. If f ≥ g, P – a.e. there E_{G}f ≥ E_{G}g, P – a.e.

3. |E_{G}f| ≤ E_{G} |f| , P – a.e.

4. ∥E_{G}f∥_{L^{1}} ≤ ∥f∥_{L^{1}} for all f ∈ L^{2}. So by the B.L.T. Theorem 8.4, E_{G} extends uniquely to a bounded linear map from L^{1}(Ω, F, P) to L^{1}(Ω, G, P) which we will still denote by E_{G}.

5. If f ∈ L^{1}(Ω, F, P) then F = E_{G}f ∈ L^{1}(Ω, G, P) iff

E(F h) = E(fh) for all h ∈ G_{b}.

6. If g ∈ G_{b} and f ∈ L^{1}(Ω, F, P), then E_{G}(gf) = g · E_{G}f, P – a.e.

Proof. By the definition of orthogonal projection for h ∈ G_{b},

E(fh) = E(f · E_{G}h) = E(E_{G}f · h).

So if f, h ≥ 0 then 0 ≤ E(fh) ≤ E(E_{G}f · h) and since this holds for all h ≥ 0 in G_{b}, E_{G}f ≥ 0, P – a.e. This proves (1). Item (2) follows by applying item (1). to f − g. If f is real, ±f ≤ |f| and so by Item (2), ±E_{G}f ≤ E_{G} |f| , i.e. |E_{G}f| ≤ E_{G} |f| , P – a.e. For complex f, let h ≥ 0 be a bounded and G – measurable function. Then

E [|E_{G}f| h] = E [E_{G}f · \overline{sgn (E_{G}f)}h] = E [f · \overline{sgn (E_{G}f)}h]
≤ E [|f| h] = E [E_{G} |f| · h] .

Since h is arbitrary, it follows that |E_{G}f| ≤ E_{G} |f| , P – a.e. Integrating this inequality implies

∥E_{G}f∥_{L^{1}} ≤ E |E_{G}f| ≤ E [E_{G} |f| · 1] = E [|f|] = ∥f∥_{L^{1}}.

Item (5). Suppose f ∈ L^{1}(Ω, F, P) and h ∈ G_{b}. Let f_{n} ∈ L^{2}(Ω, F, P) be a sequence of functions such that f_{n} → f in L^{1}(Ω, F, P). Then

E(E_{G}f · h) = E( lim_{n→∞} E_{G}f_{n} · h) = lim_{n→∞} E(E_{G}f_{n} · h)
= lim_{n→∞} E(f_{n} · h) = E(f · h).  (23.25)

This equation uniquely determines E_{G}, for if F ∈ L^{1}(Ω, G, P) also satisfies E(F · h) = E(f · h) for all h ∈ G_{b}, then taking h = \overline{sgn (F − E_{G}f)} in Eq. (23.25) gives

0 = E((F − E_{G}f) h) = E(|F − E_{G}f|).

This shows F = E_{G}f, P – a.e. Item (6) is now an easy consequence of this characterization, since if h ∈ G_{b},

E [(gE_{G}f) h] = E [E_{G}f · hg] = E [f · hg] = E [gf · h] = E [E_{G} (gf) · h] .

Thus E_{G} (gf) = g · E_{G}f, P – a.e.

Proposition 23.17. If G_{0} ⊂ G_{1} ⊂ F. Then

E_{G_{0}}E_{G_{1}} = E_{G_{1}}E_{G_{0}} = E_{G_{0}}.  (23.26)

Proof. Equation (23.26) holds on L^{2}(Ω, F, P) by the basic properties of orthogonal projections. It then hold on L^{1}(Ω, F, P) by continuity and the density of L^{2}(Ω, F, P) in L^{1}(Ω, F, P).

Example 23.18. Suppose that (X, M, μ) and (Y, N , ν) are two σ – finite measure spaces. Let Ω = X × Y, F = M ⊗ N and P (dx, dy) = ρ(x, y)μ(dx)ν(dy) where ρ ∈ L^{1}(Ω, F, μ⊗ν) is a positive function such that ∫_{X×Y} ρd (μ ⊗ ν) = 1. Let π_{X} : Ω → X be the projection map, π_{X}(x, y) = x, and

G := σ(π_{X}) = π_{X}^{−1}(M) = {A × Y : A ∈ M} .

Then f : Ω → ℝ is G – measurable iff f = F ∘π_{X} for some function F : X → ℝ which is N – measurable, see Lemma 18.66. For f ∈ L^{1}(Ω, F, P), we will now show E_{G}f = F ∘ π_{X} where

F (x) = \frac{1}{ρ̄(x)}1_{(0,∞)}(ρ̄(x)) · ∫_{Y} f(x, y)ρ(x, y)ν(dy),

ρ̄(x) := ∫_{Y} ρ(x, y)ν(dy). (By convention, ∫_{Y} f(x, y)ρ(x, y)ν(dy) := 0 if ∫_{Y} |f(x, y)| ρ(x, y)ν(dy) = ∞.)

By Tonelli’s theorem, the set

E := {x ∈ X : ρ̄(x) = ∞} ∪ {x ∈ X : ∫_{Y} |f(x, y)| ρ(x, y)ν(dy) = ∞}

is a μ – null set. Since

E [|F ∘ π_{X}|] = ∫_{X} dμ(x) ∫_{Y} dν(y) |F (x)| ρ(x, y) = ∫_{X} dμ(x) |F (x)| ρ̄(x)
= ∫_{X} dμ(x) |∫_{Y} ν(dy)f(x, y)ρ(x, y)|
≤ ∫_{X} dμ(x) ∫_{Y} ν(dy) |f(x, y)| ρ(x, y) < ∞,

F ∘ π_{X} ∈ L^{1}(Ω, G, P). Let h = H ∘ π_{X} be a bounded G – measurable function, then

E [F ∘ π_{X} · h] = ∫_{X} dμ(x) ∫_{Y} dν(y)F (x)H(x)ρ(x, y)
= ∫_{X} dμ(x)F (x)H(x)ρ̄(x)
= ∫_{X} dμ(x)H(x) ∫_{Y} ν(dy)f(x, y)ρ(x, y)
= E [hf]

and hence E_{G}f = F ∘ π_{X} as claimed.

This example shows that conditional expectation is a generalization of the notion of performing integration over a partial subset of the variables in the integrand. Whereas to compute the expectation, one should integrate over all of the variables. See also Exercise 23.25 to gain more intuition about conditional expectations.

Theorem 23.19 (Jensen’s inequality). Let (Ω, F, P) be a probability space and φ : ℝ → ℝ be a convex function. Assume f ∈ L^{1}(Ω, F, P; ℝ) is a function such that (for simplicity) φ(f) ∈ L^{1}(Ω, F, P; ℝ), then φ(E_{G}f) ≤ E_{G} [φ(f)] , P – a.e.

Proof. Let us first assume that ϕ is C^{1} and f is bounded. In this case

φ(x) − φ(x_{0}) ≥ φ′(x_{0})(x − x_{0}) for all x_{0}, x ∈ ℝ.  (23.27)

Taking x_{0} = E_{G}f and x = f in this inequality implies

φ(f) − φ(E_{G}f) ≥ φ′(E_{G}f)(f − E_{G}f)

and then applying E_{G} to this inequality gives

E_{G} [φ(f)] − φ(E_{G}f) = E_{G}[φ(f) − φ(E_{G}f)]
≥ φ′(E_{G}f)(E_{G}f − E_{G}E_{G}f) = 0

The same proof works for general ϕ, one need only use Proposition 21.8 to replace Eq. (23.27) by

φ(x) − φ(x_{0}) ≥ φ′_{−}(x_{0})(x − x_{0}) for all x_{0}, x ∈ ℝ

where φ′_{−}(x_{0}) is the left hand derivative of ϕ at x_{0}. If f is not bounded, apply what we have just proved to f^{M} = f1_{|f|≤M}, to find

E_{G} [φ(f^{M})] ≥ φ(E_{G}f^{M}).  (23.28)

Since E_{G} : L^{1}(Ω, F, P; ℝ) → L^{1}(Ω, F, P; ℝ) is a bounded operator and f^{M} → f and φ(f^{M}) → ϕ(f) in L^{1}(Ω, F, P; ℝ) as M → ∞, there exists {M_{k}}_{k=1}^{∞} such that M_{k} ↑ ∞ and f^{M_{k}} → f and φ(f^{M_{k}}) → ϕ(f), P – a.e. So passing to the limit in Eq. (23.28) shows E_{G} [φ(f)] ≥ φ(E_{G}f), P – a.e.

23.6 Exercises

Exercise 23.9. Let (X, M, μ) be a measure space and H := L^{2}(X, M, μ). Given f ∈ L^{∞}(μ) let M_{f} : H → H be the multiplication operator defined by M_{f}g = fg. Show M_{f}^{2} = M_{f} iff there exists A ∈ M such that f = 1_{A} a.e.

Exercise 23.10 (Haar Basis). In this problem, let L^{2} denote L^{2}([0, 1], m) with the standard inner product,

ψ(x) = 1_{[0,1/2)}(x) − 1_{[1/2,1)}(x)

and for k, j ∈ ℕ_{0} := ℕ∪{0} with 0 ≤ j < 2^{k} let

ψ_{kj}(x) = 2^{k/2}ψ(2^{k}x − j)
= 2^{k/2} (1_{2^{−k}[j,j+1/2)}(x) − 1_{2^{−k}[j+1/2,j+1)}(x)) .

The following pictures shows the graphs of ψ_{00}, ψ_{1,0}, ψ_{1,1}, ψ_{2,1}, ψ_{2,2} and ψ_{2,3} respectively.

Plot of ψ_{0}, 0.

Plot of ψ_{1}0.  Plot of ψ_{1}1.

Plot of ψ_{2}0.  Plot of ψ_{2}1.

Plot of ψ_{2}2.  Plot of ψ_{2}3.

1. For n ∈ ℕ, let M_{0} = span({1}) and M_{n} := span ({1} ∪ {ψ_{kj} : 0 ≤ k < n and 0 ≤ j < 2^{k}}) for n ∈ ℕ, where 1 denotes the constant function 1. Show

M_{n} = span ({1_{[j2^{−n},(j+1)2^{−n})} : and 0 ≤ j < 2^{n}) .

2. Show β := {1} ∪ {ψ_{kj} : 0 ≤ k and 0 ≤ j < 2^{k}} is an orthonormal set. Hint: show ψ_{k+1,j} ∈ M_{k}^{⊥} for all 0 ≤ j < 2^{k+1} and show {ψ_{kj} : 0 ≤ j < 2^{k}} is an orthonormal set for fixed k.

3. Show ∪_{n=1}^{∞}M_{n} is a dense subspace of L^{2} and therefore β is an orthonormal basis for L^{2}. Hint: see Theorem 22.15.

4. For f ∈ L^{2}, let

H_{n}f := ⟨f|1⟩1 + ∑_{k=0}^{n−1} ∑_{j=0}^{2^{k}−1}⟨f|ψ_{kj}⟩ψ_{kj}.

Show (compare with Exercise 23.25)

H_{n}f = ∑_{j=0}^{2^{n}−1} (2^{n} ∫_{j2^{−n}}^{(j+1)2^{−n}} f(x)dx) 1_{[j2^{−n},(j+1)2^{−n})}

and use this to show ∥f − H_{n}f∥_{∞} → 0 as n → ∞ for all f ∈ C([0, 1]). Hint: Compute orthogonal projection onto M_{n} using a judiciously chosen basis for M_{n}.

Exercise 23.11. Let O(n) be the orthogonal groups consisting of n × n real orthogonal matrices O, i.e. O^{tr}O = I. For O ∈ O(n) and f ∈ L^{2}(ℝ^{n}) let U_{O}f(x) = f(O^{−1}x). Show

1. U_{O}f is well defined, namely if f = g a.e. then U_{O}f = U_{O}g a.e.

2. U_{O} : L^{2}(ℝ^{n}) → L^{2}(ℝ^{n}) is unitary and satisfies U_{O_{1}}U_{O_{2}} = U_{O_{1}O_{2}} for all O_{1}, O_{2} ∈ O(n). That is to say the map O ∈ O(n) → U(L^{2}(ℝ^{n})) – the unitary operators on L^{2}(ℝ^{n}) is a group homomorphism, i.e. a “unitary representation” of O(n).

3. For each f ∈ L^{2}(ℝ^{n}), the map O ∈ O(n) → U_{O}f ∈ L^{2}(ℝ^{n}) is continuous. Take the topology on O(n) to be that inherited from the Euclidean topology on the vector space of all n × n matrices. Hint: see the proof of Proposition 22.24.

Exercise 23.12. Euclidean group representation and its infinitesimal generators including momentum and angular momentum operators.

Exercise 23.13. Spherical Harmonics.

Exercise 23.14. The gradient and the Laplacian in spherical coordinates.

Exercise 23.15. Legendre polynomials.

23.7 Fourier Series Exercises

Exercise 23.16. Show ∑_{k=1}^{∞} k^{−2} = π^{2}/6, by taking f(x) = x on [−π, π] and computing ∥f∥_{2}^{2} directly and then in terms of the Fourier Coefficients f̃ of f.

Exercise 23.17 (Riemann Lebesgue Lemma for Fourier Series). Show for f ∈ L^{1}([−π, π]^{d}) that f̃ ∈ c_{0}(ℤ^{d}), i.e. f̃ : ℤ^{d} → ℂ and lim_{k→∞} f̃(k) = 0. Hint: If f ∈ H, this follows form Bessel’s inequality. Now use a density argument.

Exercise 23.18. Suppose f ∈ L^{1}([−π, π]^{d}) is a function such that f̃ ∈ ℓ^{1}(ℤ^{d}) and set

g(x) := ∑_{k∈ℤ^{d}} f̃(k)e^{ik·x} (pointwise).

1. Show g ∈ C_{per}(ℝ^{d}).

2. Show g(x) = f(x) for m – a.e. x in [−π, π]^{d}. Hint: Show g̃(k) = f̃(k) and then use approximation arguments to show

∫_{[−π,π]^{d}} f(x)h(x)dx = ∫_{[−π,π]^{d}} g(x)h(x)dx ∀ h ∈ C([−π, π]^{d}).

3. Conclude that f ∈ L^{1}([−π, π]^{d}) ∩ L^{∞}([−π, π]^{d}) and in particular f ∈ L^{p}([−π, π]^{d}) for all p ∈ [1, ∞].

Exercise 23.19. Suppose m ∈ ℕ_{0}, α is a multi-index such that |α| ≤ 2m and f ∈ C_{per}^{2m}(ℝ^{d})^{1}.

1. Using integration by parts, show (using Notation 22.21) that

(ik)^{α}f̃(k) = ⟨∂^{α}f|e_{k}⟩ for all k ∈ ℤ^{d}.

Note: This equality implies

|f̃(k)| ≤ \frac{1}{k^{α}} ∥∂^{α}f∥_{H} ≤ \frac{1}{k^{α}} ∥∂^{α}f∥_{∞} .

2. Now let Δf = ∑_{i=1}^{d} ∂^{2}f/∂x_{i}^{2}, Working as in part 1) show

⟨(1 − Δ)^{m}f|e_{k}⟩ = (1 + |k|^{2})^{m}f̃(k).  (23.29)

Remark 23.20. Suppose that m is an even integer, α is a multi-index and f ∈ C_{per}^{m+|α|}(ℝ^{d}), then

1 We view C_{per}(ℝ) as a subspace of H = L^{2} ([−π, π]) by identifying f ∈ C_{per}(ℝ) with f|_{[−π,π]} ∈ H.

( ∑_{k∈ℤ^{d}} |k^{α}| |f̃(k)|)^{2} = ( ∑_{k∈ℤ^{d}} |⟨∂^{α}f|e_{k}⟩| (1 + |k|^{2})^{m/2}(1 + |k|^{2})^{−m/2})^{2}

= ( ∑_{k∈ℤ^{d}} |⟨(1 − Δ)^{m/2}∂^{α}f|e_{k}⟩| (1 + |k|^{2})^{−m/2})^{2}

≤ ∑_{k∈ℤ^{d}} |⟨(1 − Δ)^{m/2}∂^{α}f|e_{k}⟩|^{2} · ∑_{k∈ℤ^{d}}(1 + |k|^{2})^{−m}

= C_{m} ∥(1 − Δ)^{m/2}∂^{α}f∥_{H}^{2}

where C_{m} := ∑_{k∈ℤ^{d}}(1 + |k|^{2})^{−m} < ∞ iff m > d/2. So the smoother f is the faster f̃ decays at infinity. The next problem is the converse of this assertion and hence smoothness of f corresponds to decay of f̃at infinity and visa-versa.

Exercise 23.20 (A Sobolev Imbedding Theorem). Suppose s ∈ ℝ and {c_{k} ∈ ℂ : k ∈ ℤ^{d}} are coefficients such that

∑_{k∈ℤ^{d}} |c_{k}|^{2} (1 + |k|^{2})^{s} < ∞.

Show if s > \frac{d}{2} + m, the function f defined by

f(x) = ∑_{k∈ℤ^{d}} c_{k}e^{ik·x}

is in C_{per}^{m}(ℝ^{d}). Hint: Work as in the above remark to show

∑_{k∈ℤ^{d}} |c_{k}| |k^{α}| < ∞ for all |α| ≤ m.

Exercise 23.21 (Poisson Summation Formula). Let F ∈ L^{1}(ℝ^{d}),

E := {x ∈ ℝ^{d} : ∑_{k∈ℤ^{d}} |F (x + 2πk)| = ∞}

and set

F̂(k) := (2π)^{−d/2} ∫_{ℝ^{d}} F (x)e^{−ik·x}dx.

Further assume F̂ ∈ ℓ^{1}(ℤ^{d}).

1. Show m(E) = 0 and E + 2πk = E for all k ∈ ℤ^{d}. Hint: Compute ∫_{[−π,π]^{d}} ∑_{k∈ℤ^{d}} |F (x + 2πk)| dx.

2. Let

f(x) := \{ \begin{matrix}∑_{k∈ℤ^{d}} F (x + 2πk) & for & x ∉ E \\ 0 & if & x ∈ E. \end{matrix}

Show f ∈ L^{1}([−π, π]^{d}) and f̃(k) = (2π)^{−d/2} F̂(k).

3. Using item 2) and the assumptions on F, show f ∈ L^{1}([−π, π]^{d}) ∩ L^{∞}([−π, π]^{d}) and

f(x) = ∑_{k∈ℤ^{d}} f̃(k)e^{ik·x} = ∑_{k∈ℤ^{d}} (2π)^{−d/2} F̂(k)e^{ik·x} for m – a.e. x,

i.e.

∑_{k∈ℤ^{d}} F (x + 2πk) = (2π)^{−d/2} ∑_{k∈ℤ^{d}} F̂(k)e^{ik·x} for m – a.e. x. (23.30)

4. Suppose we now assume that F ∈ C(ℝ^{d}) and F satisfies 1) |F (x)| ≤ C(1 + |x|)^{−s} for some s > d and C < ∞ and 2) F̂ ∈ ℓ^{1}(ℤ^{d}), then show Eq. (23.30) holds for all x ∈ ℝ^{d} and in particular

∑_{k∈ℤ^{d}} F (2πk) = (2π)^{−d/2} ∑_{k∈ℤ^{d}} F̂(k).

For notational simplicity, in the remaining problems we will assume that d = 1.

Exercise 23.22 (Heat Equation 1.). Let (t, x) ∈ [0, ∞) × ℝ → u(t, x) be a continuous function such that u(t, ·) ∈ C_{per}(ℝ) for all t ≥ 0, u̇ := u_{t}, u_{x}, and u_{xx} exists and are continuous when t > 0. Further assume that u satisfies the heat equation u̇ = \frac{1}{2}u_{xx}. Let ũ(t, k) := ⟨u(t, ·)|e_{k}⟩ for k ∈ ℤ. Show for t > 0 and k ∈ ℤ that ũ(t, k) is differentiable in t and \frac{d}{dt} ũ(t, k) = −k^{2}ũ(t, k)/2. Use this result to show

u(t, x) = ∑_{k∈ℤ} e^{− \frac{t}{2} k^{2}} f̃(k)e^{ikx}  (23.31)

where f(x) := u(0, x) and as above

f̃(k) = ⟨f|e_{k}⟩ = ∫_{−π}^{π} f(y)e^{−iky}dy = \frac{1}{2π} ∫_{−π}^{π} f(y)e^{−iky}dm (y) .

Notice from Eq. (23.31) that (t, x) → u(t, x) is C^{∞} for t > 0.

Exercise 23.23 (Heat Equation 2.). Let q_{t}(x) := \frac{1}{2π} ∑_{k∈ℤ} e^{− \frac{t}{2} k^{2}}e^{ikx}. Show that Eq. (23.31) may be rewritten as

u(t, x) = ∫_{−π}^{π} q_{t}(x − y)f(y)dy

and

q_{t}(x) = ∑_{k∈ℤ} p_{t}(x + k2π)

where p_{t}(x) := \frac{1}{\sqrt{2πt}}e^{− \frac{1}{2t} x^{2}}. Also show u(t, x) may be written as

u(t, x) = p_{t} ∗ f(x) := ∫_{ℝ^{d}} p_{t}(x − y)f(y)dy.

Hint: To show q_{t}(x) = ∑_{k∈ℤ} p_{t}(x+k2π), use the Poisson summation formula and the Gaussian integration identity,

p̂_{t}(ω) = \frac{1}{\sqrt{2π}} ∫_{ℝ} p_{t}(x)e^{iωx}dx = \frac{1}{\sqrt{2π}} e^{− \frac{t}{2} ω^{2}}.  (23.32)

Equation (23.32) will be discussed in Example ?? below.

Exercise 23.24 (Wave Equation). Let u ∈ C^{2}(ℝ×ℝ) be such that u(t, ·) ∈ C_{per}(ℝ) for all t ∈ ℝ. Further assume that u solves the wave equation, u_{tt} = u_{xx}. Let f(x) := u(0, x) and g(x) = u̇(0, x). Show ũ(t, k) := ⟨u(t, ·), e_{k}⟩ for k ∈ ℤ is twice continuously differentiable in t and \frac{d^{2}}{dt^{2}} ũ(t, k) = −k^{2}ũ(t, k). Use this result to show

u(t, x) = ∑_{k∈ℤ} (f̃(k) cos(kt) + g̃(k)\frac{sin kt}{k}) e^{ikx}  (23.33)

with the sum converging absolutely. Also show that u(t, x) may be written as

u(t, x) = \frac{1}{2} [f(x + t) + f(x − t)] + \frac{1}{2} ∫_{−t}^{t} g(x + τ)dτ.  (23.34)

Hint: To show Eq. (23.33) implies (23.34) use

cos kt = \frac{e^{ikt} + e^{−ikt}}{2} ,
sin kt = \frac{e^{ikt} − e^{−ikt}}{2i} , and
\frac{e^{ik(x+t)} − e^{ik(x−t)}}{ik} = ∫_{−t}^{t} e^{ik(x+τ)}dτ.

23.8 Conditional Expectation Exercises

Exercise 23.25. Suppose (Ω, F, P) is a probability space and A := {A_{i}}_{i=1}^{∞} ⊂ F is a partition of Ω. (Recall this means Ω = ∐_{i=1}^{∞} A_{i}.) Let G be the σ – algebra generated by A. Show:

1. B ∈ G iff B = ∪_{i∈Λ}A_{i} for some Λ ⊂ ℕ.

2. g : Ω → ℝ is G – measurable iff g = ∑_{i=1}^{∞} λ_{i}1_{A_{i}} for some λ_{i} ∈ ℝ.

3. For f ∈ L^{1}(Ω, F, P), let E(f|A_{i}) := E [1_{A_{i}}f] /P (A_{i}) if P (A_{i}) ≠ 0 and E(f|A_{i}) = 0 otherwise. Show

E_{G}f = ∑_{i=1}^{∞} E(f|A_{i})1_{A_{i}}.

Part VI Appendices A

Multinomial Theorems and Calculus Results

Given a multi-index α ∈ ℤ_{+}^{n}, let |α| = α_{1} + ⋯ + α_{n}, α! := α_{1}! ⋯ α_{n}!,

x^{α} := ∏_{j=1}^{n} x_{j}^{α_{j}} and ∂_{x}^{α} = (\frac{∂}{∂x})^{α} := ∏_{j=1}^{n} (\frac{∂}{∂x_{j}} )^{α_{j}} .

We also write

∂_{v}f(x) := \frac{d}{dt}f(x + tv)|_{t=0}.

A.1 Multinomial Theorems and Product Rules

For a = (a_{1}, a_{2}, …, a_{n}) ∈ ℂ^{n}, m ∈ ℕ and (i_{1}, …, i_{m}) ∈ {1, 2, …, n}^{m} let α̂_{j} (i_{1}, …, i_{m}) = # {k : i_{k} = j} . Then

(∑_{i=1}^{n} a_{i})^{m} = ∑_{i_{1},...,i_{m}=1}^{n} a_{i_{1}} … a_{i_{m}} = ∑_{|α|=m} C(α)a^{α}

where

C(α) = # {(i_{1}, …, i_{m}) : α̂_{j} (i_{1}, …, i_{m}) = α_{j} for j = 1, 2, …, n}

I claim that C(α) = \frac{m!}{α!}. Indeed, one possibility for such a sequence (a_{1}, …, a_{i_{m}}) for a given α is gotten by choosing

(\overbrace{a_{1}, …, a_{1}}^{α_{1}}, \overbrace{a_{2}, …, a_{2}}^{α_{2}}, …, \overbrace{a_{n}, …, a_{n}}^{α_{n}}).

Now there are m! permutations of this list. However, only those permutations leading to a distinct list are to be counted. So for each of these m! permutations we must divide by the number of permutation which just rearrange the groups of a_{i}’s among themselves for each i. There are α! := α_{1}! ⋯ α_{n}! such permutations. Therefore, C(α) = m!/α! as advertised. So we have proved

(∑_{i=1}^{n} a_{i})^{m} = ∑_{|α|=m} \frac{m!}{α!}a^{α}.  (A.1)

Now suppose that a, b ∈ ℝ^{n} and α is a multi-index, we have

(a + b)^{α} = ∑_{β≤α} \frac{α!}{β!(α − β)!}a^{β}b^{α−β} = ∑_{β+δ=α} \frac{α!}{β!δ!}a^{β}b^{δ}  (A.2)

Indeed, by the standard Binomial formula,

(a_{i} + b_{i})^{α_{i}} = ∑_{β_{i}≤α_{i}} \frac{α_{i}!}{β_{i}!(α_{i} − β_{i})!}a^{β_{i}}b^{α_{i}−β_{i}}

from which Eq. (A.2) follows. Eq. (A.2) generalizes in the obvious way to

(a_{1} + ⋯ + a_{k})^{α} = ∑_{β_{1}+···+β_{k}=α} \frac{α!}{β_{1}! ⋯ β_{k}!}a_{1}^{β_{1}} … a_{k}^{β_{k}}  (A.3)

where a_{1}, a_{2}, …, a_{k} ∈ ℝ^{n} and α ∈ ℤ_{+}^{n}.

Now let us consider the product rule for derivatives. Let us begin with the one variable case (write d^{n}f for f^{(n)} = \frac{d^{n}}{dx^{n}} f) where we will show by induction that

d^{n}(fg) = ∑_{k=0}^{n} \binom{n}{k}d^{k}f · d^{n−k}g.  (A.4)

Indeed assuming Eq. (A.4) we find

d^{n+1}(fg) = ∑_{k=0}^{n} \binom{n}{k}d^{k+1}f · d^{n−k}g + ∑_{k=0}^{n} \binom{n}{k}d^{k}f · d^{n−k+1}g

= ∑_{k=1}^{n+1} \binom{n}{k − 1}d^{k}f · d^{n−k+1}g + ∑_{k=0}^{n} \binom{n}{k}d^{k}f · d^{n−k+1}g

= ∑_{k=1}^{n+1} [\binom{n}{k − 1} + \binom{n}{k}] d^{k}f · d^{n−k+1}g + d^{n+1}f · g + f · d^{n+1}g.

Since

\binom{n}{k − 1} + \binom{n}{k} = \frac{n!}{(n − k + 1)!(k − 1)!} + \frac{n!}{(n − k)!k!}

= \frac{n!}{(k − 1)! (n − k)!} [\frac{1}{(n − k + 1)} + \frac{1}{k} ]

= \frac{n!}{(k − 1)! (n − k)!} \frac{n + 1}{(n − k + 1) k} = \binom{n + 1}{k}

the result follows.

Now consider the multi-variable case

∂^{α}(fg) = (∏_{i=1}^{n} ∂_{i}^{α_{i}}) (fg) = ∏_{i=1}^{n} [ ∑_{k_{i}=0}^{α_{i}} \binom{α_{i}}{k_{i}}∂_{i}^{k_{i}}f · ∂_{i}^{α_{i}−k_{i}}g]

= ∑_{k_{1}=0}^{α_{1}} ⋯ ∑_{k_{n}=0}^{α_{n}} ∏_{i=1}^{n} \binom{α_{i}}{k_{i}}∂^{k}f · ∂^{α−k}g = ∑_{k≤α} \binom{α}{k}∂^{k}f · ∂^{α−k}g

where k = (k_{1}, k_{2}, …, k_{n}) and

\binom{α}{k} := ∏_{i=1}^{n} \binom{α_{i}}{k_{i}} = \frac{α!}{k!(α − k)!}.

So we have proved

∂^{α}(fg) = ∑_{β≤α} \binom{α}{β}∂^{β}f · ∂^{α−β}g.  (A.5)

A.2 Taylor’s Theorem

Theorem A.1. Suppose X ⊂ ℝ^{n} is an open set, x : [0, 1] → X is a C^{1} – path, and f ∈ C^{N}(X, ℂ). Let v_{s} := x(1) − x(s) and v = v_{1} = x(1) − x(0), then

f(x(1)) = ∑_{m=0}^{N−1} \frac{1}{m!} (∂_{v}^{m}f) (x(0)) + R_{N}  (A.6)

where

R_{N} = \frac{1}{(N − 1)!} ∫_{0}^{1} (∂_{ẋ(s)}∂_{v_{s}}^{N−1}f) (x(s))ds = \frac{1}{N!} ∫_{0}^{1} (−\frac{d}{ds}∂_{v_{s}}^{N}f) (x(s))ds.

(A.7)

and 0! := 1.

Proof. By the fundamental theorem of calculus and the chain rule,

f(x(t)) = f(x(0)) + ∫_{0}^{t} \frac{d}{ds}f(x(s))ds = f(x(0)) + ∫_{0}^{t} (∂_{ẋ(s)}f) (x(s))ds (A.8)

and in particular,

f(x(1)) = f(x(0)) + ∫_{0}^{1} (∂_{ẋ(s)}f) (x(s))ds.

This proves Eq. (A.6) when N = 1. We will now complete the proof using induction on N. Applying Eq. (A.8) with f replaced by \frac{1}{(N−1)!} (∂_{ẋ(s)}∂_{v_{s}}^{N−1}f) gives

\frac{1}{(N − 1)!} (∂_{ẋ(s)}∂_{v_{s}}^{N−1}f) (x(s)) = \frac{1}{(N − 1)!} (∂_{ẋ(s)}∂_{v_{s}}^{N−1}f) (x(0))

+ \frac{1}{(N − 1)!} ∫_{0}^{s} (∂_{ẋ(s)}∂_{v_{s}}^{N−1}∂_{ẋ(t)}f) (x(t))dt

= −\frac{1}{N!} (\frac{d}{ds}∂_{v_{s}}^{N}f) (x(0)) − \frac{1}{N!} ∫_{0}^{s} (\frac{d}{ds}∂_{v_{s}}^{N}∂_{ẋ(t)}f) (x(t))dt

wherein we have used the fact that mixed partial derivatives commute to show \frac{d}{ds}∂_{v_{s}}^{N}f = N∂_{ẋ(s)}∂_{v_{s}}^{N−1}f. Integrating this equation on s ∈ [0, 1] shows, using the fundamental theorem of calculus,

R_{N} = \frac{1}{N!} (∂_{v}^{N}f) (x(0)) − \frac{1}{N!} ∫_{0≤t≤s≤1} (\frac{d}{ds}∂_{v_{s}}^{N}∂_{ẋ(t)}f) (x(t))dsdt

= \frac{1}{N!} (∂_{v}^{N}f) (x(0)) + \frac{1}{(N + 1)!} ∫_{0≤t≤1} (∂_{w_{t}}^{N}∂_{ẋ(t)}f) (x(t))dt

= \frac{1}{N!} (∂_{v}^{N}f) (x(0)) + R_{N+1}

which completes the inductive proof.

Remark A.2. Using Eq. (A.1) with a_{i} replaced by v_{i}∂_{i} (although {v_{i}∂_{i}}_{i=1}^{n} are not complex numbers they are commuting symbols), we find

∂_{v}^{m}f = (∑_{i=1}^{n} v_{i}∂_{i})^{m} f = ∑_{|α|=m} \frac{m!}{α!}v^{α}∂^{α}.

Using this fact we may write Eqs. (A.6) and (A.7) as

f(x(1)) = ∑_{|α|≤N−1} \frac{1}{α!}v^{α}∂^{α}f(x(0)) + R_{N}

and

R_{N} = ∑_{|α|=N} \frac{1}{α!} ∫_{0}^{1} (−\frac{d}{ds}v_{s}^{α}∂^{α}f) (x(s))ds.

Corollary A.3. Suppose X ⊂ ℝ^{n} is an open set which contains x(s) = (1 − s)x_{0} + sx_{1} for 0 ≤ s ≤ 1 and f ∈ C^{N}(X, ℂ). Then

f(x_{1}) = ∑_{m=0}^{N−1} \frac{1}{m!} (∂_{v}^{m}f) (x_{0}) + \frac{1}{N!} ∫_{0}^{1} (∂_{v}^{N}f) (x(s))dν_{N}(s)  (A.9)

= ∑_{|α|<N} \frac{1}{α!}∂^{α}f(x(0))(x_{1} − x_{0})^{α} + ∑_{α:|α|=N} \frac{1}{α!} [∫_{0}^{1} ∂^{α}f(x(s))dν_{N}(s)] (x_{1} − x_{0})^{α} (A.10)

where v := x_{1} − x_{0} and dν_{N} is the probability measure on [0, 1] given by

dν_{N}(s) := N(1 − s)^{N−1}ds.  (A.11)

If we let x = x_{0} and y = x_{1} − x_{0} (so x + y = x_{1}) Eq. (A.10) may be written as

f(x + y) = ∑_{|α|<N} \frac{∂_{x}^{α}f(x)}{α!}y^{α} + ∑_{α:|α|=N} \frac{1}{α!} (∫_{0}^{1} ∂_{x}^{α}f(x + sy)dν_{N}(s)) y^{α}.

(A.12)

Proof. This is a special case of Theorem A.1. Notice that

v_{s} = x(1) − x(s) = (1 − s)(x_{1} − x_{0}) = (1 − s)v

and hence

R_{N} = \frac{1}{N!} ∫_{0}^{1} (−\frac{d}{ds}(1 − s)^{N}∂_{v}^{N}f) (x(s))ds = \frac{1}{N!} ∫_{0}^{1} (∂_{v}^{N}f) (x(s))N(1−s)^{N−1}ds.

Example A.4. Let X = (−1, 1) ⊂ ℝ, β ∈ ℝ and f(x) = (1 − x)^{β}. The reader should verify

f^{(m)}(x) = (−1)^{m}β(β − 1) … (β − m + 1)(1 − x)^{β−m}

and therefore by Taylor’s theorem (Eq. (??) with x = 0 and y = x)

(1 − x)^{β} = 1 + ∑_{m=1}^{N−1} \frac{1}{m!}(−1)^{m}β(β − 1) … (β − m + 1)x^{m} + R_{N}(x) (A.13)

where

R_{N}(x) = \frac{x^{N}}{N!} ∫_{0}^{1}(−1)^{N}β(β − 1) … (β − N + 1)(1 − sx)^{β−N}dν_{N}(s)

= \frac{x^{N}}{N!} (−1)^{N}β(β − 1) … (β − N + 1) ∫_{0}^{1} \frac{N(1 − s)^{N−1}}{(1 − sx)^{N−β}} ds.

Now for x ∈ (−1, 1) and N > β,

0 ≤ ∫_{0}^{1} \frac{N(1 − s)^{N−1}}{(1 − sx)^{N−β}} ds ≤ ∫_{0}^{1} \frac{N(1 − s)^{N−1}}{(1 − s)^{N−β}} ds = ∫_{0}^{1} N(1 − s)^{β−1}ds = \frac{N}{β}

and therefore,

|R_{N}(x)| ≤ \frac{|x|^{N}}{(N − 1)!} |(β − 1) … (β − N + 1)| =: ρ_{N}.

Since

lim sup_{N→∞} \frac{ρ_{N+1}}{ρ_{N}} = |x| · lim sup_{N→∞} \frac{N − β}{N} = |x| < 1

and so by the Ratio test, |R_{N}(x)| ≤ ρ_{N} → 0 (exponentially fast) as N → ∞. Therefore by passing to the limit in Eq. (A.13) we have proved

(1 − x)^{β} = 1 + ∑_{m=1}^{∞} \frac{(−1)^{m}}{m!} β(β − 1) … (β − m + 1)x^{m}  (A.14)

which is valid for |x| < 1 and β ∈ ℝ. An important special cases is β = −1 in which case, Eq. (A.14) becomes \frac{1}{1−x} = ∑_{m=0}^{∞} x^{m}, the standard geometric series formula. Another another useful special case is β = 1/2 in which case Eq. (A.14) becomes

\sqrt{1 − x} = 1 + ∑_{m=1}^{∞} \frac{(−1)^{m}}{m!} \frac{1}{2}(\frac{1}{2} − 1) … (\frac{1}{2} − m + 1)x^{m}

= 1 − ∑_{m=1}^{∞} \frac{(2m − 3)!!}{2^{m}m!}x^{m} for all |x| < 1.  (A.15)

B

Zorn’s Lemma and the Hausdorff Maximal Principle

Definition B.1. A partial order ≤ on X is a relation with following properties

(i) If x ≤ y and y ≤ z then x ≤ z. (ii)If x ≤ y and y ≤ x then x = y. (iii)x ≤ x for all x ∈ X.

Example B.2. Let Y be a set and X = 2^{Y} . There are two natural partial orders on X.

1. Ordered by inclusion, A ≤ B is A ⊂ B and

2. Ordered by reverse inclusion, A ≤ B if B ⊂ A.

Definition B.3. Let (X, ≤) be a partially ordered set we say X is linearly a totally ordered if for all x, y ∈ X either x ≤ y or y ≤ x. The real numbers ℝ with the usual order ≤ is a typical example.

Definition B.4. Let (X, ≤) be a partial ordered set. We say x ∈ X is a maximal element if for all y ∈ X such that y ≥ x implies y = x, i.e. there is no element larger than x. An upper bound for a subset E of X is an element x ∈ X such that x ≥ y for all y ∈ E.

Example B.5. Let

X = { a = {1} b = {1, 2} c = {3} d = {2, 4} e = {2} }

ordered by set inclusion. Then b and d are maximal elements despite that fact that b ≰ a and a ≰ b. We also have

• If E = {a, e, c}, then E has no upper bound.

Definition B.6. • If E = {a, e}, then b is an upper bound.

• E = {e}, then b and d are upper bounds.

Theorem B.7. The following are equivalent.

1. The axiom of choice: to each collection, {X_{α}}_{α∈A} , of non-empty sets there exists a “choice function,” x : A → ∐_{α∈A} X_{α} such that x(α) ∈ X_{α} for

all α ∈ A, i.e. ∏_{α∈A} X_{α} ≠ ∅.

2. The Hausdorff Maximal Principle: Every partially ordered set has a maximal (relative to the inclusion order) linearly ordered subset.

3. Zorn’s Lemma: If X is partially ordered set such that every linearly ordered subset of X has an upper bound, then X has a maximal element.^{1}

Proof. (2 ⇒ 3) Let X be a partially ordered subset as in 3 and let F = {E ⊂ X : E is linearly ordered} which we equip with the inclusion partial ordering. By 2. there exist a maximal element E ∈ F. By assumption, the linearly ordered set E has an upper bound x ∈ X. The element x is maximal, for if y ∈ Y and y ≥ x, then E ∪ {y} is still an linearly ordered set containing E. So by maximality of E, E = E∪{y} , i.e. y ∈ E and therefore y ≤ x showing which combined with y ≥ x implies that y = x.^{2} (3 ⇒ 1) Let {X_{α}}_{α∈A} be a collection of non-empty sets, we must show ∏_{α∈A} X_{α} is not empty. Let G denote the collection of functions g : D(g) → ∐_{α∈A} X_{α} such that D(g) is a subset of A, and for all α ∈ D(g), g(α) ∈ X_{α}. Notice that G is not empty, for we may let α_{0} ∈ A and x_{0} ∈ X_{α} and then set D(g) = {α_{0}} and g(α_{0}) = x_{0} to construct an element of G. We now put a partial order on G as follows. We say that f ≤ g for f, g ∈ G provided that D(f) ⊂ D(g) and f = g|_{D(f)}. If Φ ⊂ G is a linearly ordered set, let D(h) = ∪_{g∈Φ}D(g) and for α ∈ D(g) let h(α) = g(α). Then h ∈ G is an upper bound for Φ. So by Zorn’s Lemma there exists a maximal element h ∈ G. To finish the proof we need only show that D(h) = A. If this were not the case, then let α_{0} ∈ A \ D(h) and x_{0} ∈ X_{α_{0}}. We may now define D(h̃) = D(h) ∪ {α_{0}} and

h̃(α) = \{ \begin{matrix}h(α) & if & α ∈ D(h) \\ x_{0} & if & α = α_{0}. \end{matrix}

1 If X is a countable set we may prove Zorn’s Lemma by induction. Let {x_{n}}_{n=1}^{∞} be an enumeration of X, and define E_{n} ⊂ X inductively as follows. For n = 1 let E_{1} = {x_{1}}, and if E_{n} have been chosen, let E_{n+1} = E_{n} ∪ {x_{n+1}} if x_{n+1} is an upper bound for E_{n} otherwise let E_{n+1} = E_{n}. The set E = ∪_{n=1}^{∞}E_{n} is a linearly ordered (you check) subset of X and hence by assumption E has an upper bound, x ∈ X. I claim that his element is maximal, for if there exists y = x_{m} ∈ X such that y ≥ x, then x_{m} would be an upper bound for E_{m−1} and therefore y = x_{m} ∈ E_{m} ⊂ E. That is to say if y ≥ x, then y ∈ E and hence y ≤ x, so y = x. (Hence we may view Zorn’s lemma as a “ jazzed” up version of induction.)

2 Similarly one may show that 3 ⇒ 2. Let F = {E ⊂ X : E is linearly ordered} and order F by inclusion. If M ⊂ F is linearly ordered, let E = ∪M = ⋃_{A∈M} A.

If x, y ∈ E then x ∈ A and y ∈ B for some A, B ⊂ M. Now M is linearly ordered by set inclusion so A ⊂ B or B ⊂ A i.e. x, y ∈ A or x, y ∈ B. Since A and B are linearly order we must have either x ≤ y or y ≤ x, that is to say E is linearly ordered. Hence by 3. there exists a maximal element E ∈ F which is the assertion in 2.

Then h ≤ h̃ while h ≠ h̃ violating the fact that h was a maximal element. (1 ⇒ 2) Let (X, ≤) be a partially ordered set. Let F be the collection of linearly ordered subsets of X which we order by set inclusion. Given x_{0} ∈ X, {x_{0}} ∈ F is linearly ordered set so that F ≠ ∅. Fix an element P_{0} ∈ F. If P_{0} is not maximal there exists P_{1} ∈ F such that P_{0} ⊊ P_{1}. In particular we may choose x ∉ P_{0} such that P_{0} ∪ {x} ∈ F. The idea now is to keep repeating this process of adding points x ∈ X until we construct a maximal element P of F. We now have to take care of some details. We may assume with out loss of generality that F̃ = {P ∈ F : P is not maximal} is a non-empty set. For P ∈ F̃, let P^{∗} = {x ∈ X : P ∪ {x} ∈ F} . As the above argument shows, P^{∗} ≠ ∅ for all P ∈ F̃. Using the axiom of choice, there exists f ∈ ∏_{P ∈F̃} P^{∗}. We now define g : F → F by

g(P) = \{ \begin{matrix}P & if & P is maximal \\ P ∪ \{f(x)\} & if & P is not maximal. \end{matrix}  (B.1)

The proof is completed by Lemma B.8 below which shows that g must have a fixed point P ∈ F. This fixed point is maximal by construction of g.

Lemma B.8. The function g : F → F defined in Eq. (B.1) has a fixed point.^{3}

Proof. The idea of the proof is as follows. Let P_{0} ∈ F be chosen arbitrarily. Notice that Φ = {g^{(n)}(P_{0})}_{n=0}^{∞} ⊂ F is a linearly ordered set and it

is therefore easily verified that P_{1} = ⋃_{n=0}^{∞} g^{(n)}(P_{0}) ∈ F. Similarly we may repeat

the process to construct P_{2} = ⋃_{n=0}^{∞} g^{(n)}(P_{1}) ∈ F and P_{3} = ⋃_{n=0}^{∞} g^{(n)}(P_{2}) ∈ F,

etc. etc. Then take P_{∞} = ∪_{n=0}^{∞}P_{n} and start again with P_{0} replaced by P_{∞}. Then keep going this way until eventually the sets stop increasing in size, in which case we have found our fixed point. The problem with this strategy is that we may never win. (This is very reminiscent of constructing measurable sets and the way out is to use measure theoretic like arguments.) Let us now start the formal proof. Again let P_{0} ∈ F and let F_{1} = {P ∈ F : P_{0} ⊂ P }. Notice that F_{1} has the following properties:

1. P_{0} ∈ F_{1}.

2. If Φ ⊂ F_{1} is a totally ordered (by set inclusion) subset then ∪Φ ∈ F_{1}.

3. If P ∈ F_{1} then g(P) ∈ F_{1}.

Let us call a general subset F′ ⊂ F satisfying these three conditions a tower and let

F_{0} = ∩ {F′ : F′ is a tower} .

3 Here is an easy proof if the elements of F happened to all be finite sets and there existed a set P ∈ F with a maximal number of elements. In this case the condition that P ⊂ g(P) would imply that P = g(P), otherwise g(P) would have more elements than P.

Standard arguments show that F_{0} is still a tower and clearly is the smallest tower containing P_{0}. (Morally speaking F_{0} consists of all of the sets we were trying to constructed in the “idea section” of the proof.) We now claim that F_{0} is a linearly ordered subset of F. To prove this let Γ ⊂ F_{0} be the linearly ordered set

Γ = {C ∈ F_{0} : for all A ∈ F_{0} either A ⊂ C or C ⊂ A} .

Shortly we will show that Γ ⊂ F_{0} is a tower and hence that F_{0} = Γ. That is to say F_{0} is linearly ordered. Assuming this for the moment let us finish the proof. Let P ≡ ∪F_{0} which is in F_{0} by property 2 and is clearly the largest element in F_{0}. By 3. it now follows that P ⊂ g(P) ∈ F_{0} and by maximality of P, we have g(P) = P, the desired fixed point. So to finish the proof, we must show that Γ is a tower. First off it is clear that P_{0} ∈ Γ so in particular Γ is not empty. For each C ∈ Γ let

Φ_{C} := {A ∈ F_{0} : either A ⊂ C or g(C) ⊂ A} .

We will begin by showing that Φ_{C} ⊂ F_{0} is a tower and therefore that Φ_{C} = F_{0}.

1. P_{0} ∈ Φ_{C} since P_{0} ⊂ C for all C ∈ Γ ⊂ F_{0}. 2. If Φ ⊂ Φ_{C} ⊂ F_{0} is totally ordered by set inclusion, then A_{Φ} := ∪Φ ∈ F_{0}. We must show A_{Φ} ∈ Φ_{C}, that is that A_{Φ} ⊂ C or C ⊂ A_{Φ}. Now if A ⊂ C for all A ∈ Φ, then A_{Φ} ⊂ C and hence A_{Φ} ∈ Φ_{C}. On the other hand if there is some A ∈ Φ such that g(C) ⊂ A then clearly g(C) ⊂ A_{Φ} and again A_{Φ} ∈ Φ_{C}. 3. Given A ∈ Φ_{C} we must show g(A) ∈ Φ_{C}, i.e. that

g(A) ⊂ C or g(C) ⊂ g(A).  (B.2)

There are three cases to consider: either A ⊊ C, A = C, or g(C) ⊂ A. In the case A = C, g(C) = g(A) ⊂ g(A) and if g(C) ⊂ A then g(C) ⊂ A ⊂ g(A) and Eq. (B.2) holds in either of these cases. So assume that A ⊊ C. Since C ∈ Γ, either g(A) ⊂ C (in which case we are done) or C ⊂ g(A). Hence we may assume that

A ⊊ C ⊂ g(A).

Now if C were a proper subset of g(A) it would then follow that g(A)\A would consist of at least two points which contradicts the definition of g. Hence we must have g(A) = C ⊂ C and again Eq. (B.2) holds, so Φ_{C} is a tower. It is now easy to show Γ is a tower. It is again clear that P_{0} ∈ Γ and Property 2. may be checked for Γ in the same way as it was done for Φ_{C} above. For Property 3., if C ∈ Γ we may use Φ_{C} = F_{0} to conclude for all A ∈ F_{0}, either A ⊂ C ⊂ g(C) or g(C) ⊂ A, i.e. g(C) ∈ Γ. Thus Γ is a tower and we are done.

Nets

In this section (which may be skipped) we develop the notion of nets. Nets are generalization of sequences. Here is an example which shows that for general topological spaces, sequences are not always adequate.

Example C.1. Equip ℂ^{ℝ} with the topology of pointwise convergence, i.e. the product topology and consider C(ℝ, ℂ) ⊂ ℂ^{ℝ}. If {f_{n}} ⊂ C(ℝ, ℂ) is a sequence which converges such that f_{n} → f ∈ ℂ^{ℝ} pointwise then f is a Borel measurable function. Hence the sequential limits of elements in C(ℝ, ℂ) is necessarily contained in the Borel measurable functions which is properly contained in ℂ^{ℝ}. In short the sequential closure of C(ℝ, ℂ) is a proper subset of ℂ^{ℝ}. On the other hand we have \overline{C(ℝ, ℂ)} = ℂ^{ℝ}. Indeed a typical open neighborhood of f ∈ ℂ^{ℝ} is of the form

N = {g ∈ ℂ^{ℝ} : |g(x) − f(x)| < ε for x ∈ Λ},

where ε > 0 and Λ is a finite subset of ℝ. Since N ∩ C(ℝ, ℂ) ≠ ∅ it follows that f ∈ \overline{C(ℝ, ℂ)}.

Definition C.2. A directed set (A, ≤) is a set with a relation “≤” such that

1. α ≤ α for all α ∈ A.

2. If α ≤ β and β ≤ γ then α ≤ γ.

3. A is cofinite, i.e. α, β ∈ A there exists γ ∈ A such that α ≤ γ and β ≤ γ.

A net is function x : A → X where A is a directed set. We will often denote a net x by {x_{α}}_{α∈A}.

Example C.3 (Directed sets).

1. A = 2^{X} ordered by inclusion, i.e. α ≤ β if α ⊂ β. If α ≤ β and β ≤ γ then α ⊂ β ⊂ γ and hence α ≤ γ. Similalry if α, β ∈ 2^{X} then α, β ≤ α∪β =: γ.

2. A = 2^{X} ordered by reverse inclusion, i.e. α ≤ β if β ⊂ α. If α ≤ β and β ≤ γ then α ⊇ β ⊇ γ and so α ≤ γ and if α, β ∈ A then α, β ≤ α ∩ β.

3. Let A = ℕ equipped with the usual ordering on ℕ. In this case nets are simply sequences.

Definition C.4. Let {x_{α}}_{α∈A} ⊂ X be a net then:

1. x_{α} converges to x ∈ X (written x_{α} → x) iff for all V ∈ τ_{x}, x_{α} ∈ V eventually, i.e. there exists β = β_{V} ∈ A such that x_{α} ∈ V for all α ≥ β.

2. x is a cluster point of {x_{α}}_{α∈A} if for all V ∈ τ_{x}, x_{α} ∈ V frequently, i.e. for all β ∈ A there exists α ≥ β such that x_{α} ∈ V.

Proposition C.5. Let X be a topological space and E ⊂ X. Then

1. x is an accumulation point of E (see Definition 10.29) iff there exists net {x_{α}} ⊂ E \ {x} such that x_{α} → x.

2. x ∈ Ē iff there exists {x_{α}} ⊂ E such that x_{α} → x.

Proof.

1. Suppose x is an accumulation point of E and let A = τ_{x} be ordered by reverse set inclusion. To each α ∈ A = τ_{x} choose x_{α} ∈ (α \ {x}) ∩ E which is possible sine x is an accumulation point of E. Then given V ∈ τ_{x} for all α ≥ V (i.e. and α ⊂ V), x_{α} ∈ V and hence x_{α} → x. Conversely if {x_{α}}_{α∈A} ⊂ E \ {x} and x_{α} → x then for all V ∈ τ_{x} there exists β ∈ A such that x_{α} ∈ V for all α ≥ β. In particular x_{α} ∈ (E \ {x}) ∩ V ≠ ∅ and so x ∈ acc(E) – the accumulation points of E.

2. If {x_{α}} ⊂ E such that x_{α} → x then for all V ∈ τ_{x} there exists β ∈ A such that x_{α} ∈ V ∩ E for all α ≥ β. In particular V ∩ E ≠ ∅ for all V ∈ τ_{x} and this implies x ∈ Ē. For the converse recall Proposition 10.31 implies \overline{E} = E ∪ acc(E). If x ∈ acc(E) there exists a net \{x_{α}\} ⊂ E such that x_{α} → x by item 1. If x ∈ E we may simply take x_{n} = x for all n ∈ A := ℕ.

Proposition C.6. Let X and Y be two topological spaces and f : X → Y be a function. Then f is continuous at x ∈ X iff f(x_{α}) → f(x) for all nets x_{α} → x.

Proof. If f is continuous at x and x_{α} → x then for any V ∈ τ_{f(x)} there exists W ∈ τ_{x} such that f(W) ⊂ V. Since x_{α} ∈ W eventually, f(x_{α}) ∈ V eventually and we have shown f(x_{α}) → f(x). Conversely, if f is not continuous at x then there exists W ∈ τ_{f(x)} such that f(V) ⊈ W for all V ∈ τ_{x}. Let A = τ_{x} be ordered by reverse set inclusion and for V ∈ τ_{x} choose (axiom of choice) x_{V} ∈ V such that f(x_{V}) ∉ W. Then x_{V} → x since for any U ∈ τ_{x}, x_{V} ∈ U if V ≥ U (i.e. V ⊂ U). On the over hand f(x_{V}) ∉ W for all V ∈ τ_{x} showing f(x_{V}) ↛ f(x).

Definition C.7 ( Subnet). A net ⟨y_{β}⟩_{β∈B} is a subnet of a net ⟨x_{α}⟩_{α∈A} if there exists a map β ∈ B → α_{β} ∈ A such that 1. y_{β} = x_{α_{β}} for all β ∈ B and

2. for all α_{0} ∈ A there exists β_{0} ∈ B such that α_{β} ≥ α_{0} whenever β ≥ β_{0}, i.e. α_{β} ≥ α_{0} eventually.

Proposition C.8. A point x ∈ X is a cluster point of a net ⟨x_{α}⟩_{α∈A} iff there exists a subnet ⟨y_{β}⟩_{β∈B} such that y_{β} → x.

Proof. Suppose ⟨y_{β}⟩_{β∈B} is a subnet of ⟨x_{α}⟩_{α∈A} such that y_{β} = x_{α_{β}} → x. Then for W ∈ τ_{x} and α_{0} ∈ A there exists β_{0} ∈ B such that y_{β} = x_{α_{β}} ∈ W for all β ≥ β_{0}. Choose β_{1} ∈ B such that α_{β} ≥ α_{0} for all β ≥ β_{1} then choose β_{3} ∈ B such that β_{3} ≥ β_{1} and β_{3} ≥ β_{2} then α_{β} ≥ α_{0} and x_{α_{β}} ∈ W for all β ≥ β_{3} which implies x_{α} ∈ W frequently. Conversely assume x is a cluster point of a net ⟨x_{α}⟩_{α∈A}. We mak B := τ_{x} × A into a directed set by defining (U, α) ≤ (U′, α′) iff α ≤ α′ and U ⊇ U′. For all (U, γ) ∈ B = τ_{x} × A, choose α_{(U,γ)} ≥ γ in A such that y_{(U,γ)} = x_{α_{(U,γ)}} ∈ U. Then if α_{0} ∈ A for all (U′, γ′) ≥ (U, α_{0}), i.e. γ′ ≥ α_{0} and U′ ⊂ U, α_{(U′,γ′)} ≥ γ′ ≥ α_{0}. Now if W ∈ τ_{x} is given, then y_{(U,γ)} ∈ U ⊂ W for all U ⊂ W. Hence fixing α ∈ A we see if (U, γ) ≥ (W, α) then y_{(U,γ)} = x_{α_{(U,γ)}} ∈ U ⊂ W showing that y_{(U,γ)} → x.

Exercise C.1 (Folland #34, p. 121). Let ⟨x_{α}⟩_{α∈A} be a net in a topological space and for each α ∈ A let E_{α} ≡ {x_{β} : β ≥ α}. Then x is a cluster point of

⟨x_{α}⟩ iff x ∈ ⋂_{α∈A} \overline{E}_{α}.

Solution to Exercise (C.1). If x is a cluster point, then given W ∈ τ_{x} we know E_{α} ∩ W ≠ ∅ for all α ∈ E since x_{β} ∈ W frequently thus x ∈ \overline{E}_{α} for all α, i.e. x ∈ ⋂_{α∈A} \overline{E}_{α}. Conversely if x is not a cluster point of ⟨x_{α}⟩ then there

exists W ∈ τ_{x} and α ∈ A such that x_{β} ∉ W for all β ≥ α, i.e. W ∩ E_{α} = ∅. But this shows x ∉ \overline{E}_{α} and hence x ∉ ⋂_{α∈A} \overline{E}_{α}.

Theorem C.9. A topological space X is compact iff every net has a cluster point iff every net has a convergent subnet.

Proof. Suppose X is compact, ⟨x_{α}⟩_{α∈A} ⊂ X is a net and let F_{α} := \overline{\{x_{β} : β ≥ α\}}. Then F_{α} is closed for all α ∈ A, F_{α} ⊂ F_{α′} if α ≥ α′ and F_{α_{1}} ∩⋯∩F_{α_{n}} ⊇ F_{γ} whenever γ ≥ α_{i} for i = 1, …, n. (Such a γ always exists since A is a directed set.) Therefore F_{α_{1}} ∩ ⋯ ∩ F_{α_{n}} ≠ ∅ i.e. {F_{α}}_{α∈A} has the finite intersection property and since X is compact this implies there exists x ∈ ⋂_{α∈a} F_{α} By Exercise C.1, it follows that x is a cluster point of ⟨x_{α}⟩_{α∈A}.

Conversely, if X is not compact let {U_{j}}_{j∈J} be an infinite cover with no finite subcover. Let A be the directed set A = {α ⊂ J : # (α) < ∞} with α ≤ β iff α ⊂ β. Define a net ⟨x_{α}⟩_{α∈A} in X by choosing

x_{α} ∈ X \ ( ⋃_{j∈α} U_{j}) ≠ ∅ for all α ∈ A.

This net has no cluster point. To see this suppose x ∈ X and j ∈ J is chosen

so that x ∈ U_{j}. Then for all α ≥ {j} (i.e. j ∈ α), x_{α} ∉ ⋃_{γ∈α} U_{α} ⊇ U_{j} and

in particular x_{α} ∉ U_{j}. This shows x_{α} ∉ U_{j} frequently and hence x is not a cluster point.

Study Guides

D.1 Study Guide For Math 240A: Fall 2003

D.1.1 Basic things you should know about numbers and limits

1. I am taking for granted that you know the basic properties of ℝ and ℂ and that they are complete.

2. Should know how to compute lim a_{n}, lim sup a_{n} and lim inf a_{n} and their basic properties. See Lemma 4.2 and Proposition 4.5 for example.

D.1.2 Basic things you should know about topological and measurable spaces:

1. You should know the basic definitions, Definition 10.1 and Definition 18.1.

2. It would be good to understand the notion of generating a topology or a σ – algebra by either a collection of sets or functions. This is key to understanding product topologies and product σ – algebras. See Propositions 10.7, 10.21 and 18.4 and Definition 18.24 and Proposition 18.25.

3. You should be able to check whether a given function is continuous or measurable. Hints:

a) If possible avoid going back to the definition of continuity or measurability. Do this by using the stability properties of continuous (measurable) functions. For example continuous (measurable) functions are stable under compositions and algebraic operations, under uniform (pointwise) limits and sums. Measurable functions are also stable under taking sup, inf lim inf and lim sup of a sequence of measruable functions, see Proposition 18.36. Also recall if we are using the Borel σ – algebras, then continuous functions are automatically measurable.

b) It is also possible to check continuity and measurability by splitting the space up and checking continuity and measurability on the individual pieces. See Proposition 10.19 and Exercise 10.7 and Proposition 18.29. c) If you must go back to first principles, then the fact that σ (f^{−1}(E)) =

f^{−1}(σ(E)) and τ (f^{−1}(E)) = f^{−1}(τ(E)) is key, see Lemma 18.22 and 10.14 respectively.

4. Dynkin’s multiplicative system Theorems 18.51 and 18.52 are extremely useful for understanding the structure of measurable functions. They are also very useful for proving general theorems which are to hold for all bounded measurable functions. See the examples following Theorem 18.52 and the examples in Section 19.7.

D.1.3 Basic things you should know about Metric Spaces

1. The associated topology, see Example 10.3.

2. How to find the closure of a set. I typically would use the sequential definition of closure here.

3. Continuity is equivalent to the sequential notion of continuity, see Section 6.1.

4. The continuity properties of the metric, see Lemma 6.6.

5. The notions of Cauchy sequences and completeness.

D.1.4 Basic things you should know about Banach spaces

1. They are complete normed spaces.

2. ℓ^{p} (μ) – spaces are Banach spaces, see Theorems 5.6, 5.8, and 7.5. Later we will see that all of these theorem hold for more general L^{p} (μ) – spaces as well.

3. BC (X) is a closed subset of the Banach space ℓ^{∞} (X) and hence is a Banach space, see Lemma 7.3.

4. The space of operators L (X, Y) between two Banach spaces is a Banach space. In particular the dual space X^{∗} is a Banach space, see Proposition 7.12.

5. How to find the norm of an operator and the basic properties of the operator norm, Lemma 7.10.

6. Boundedness of an operator is equivalent to continuity, Proposition 7.8.

7. Small perturbations of an invertible operator is still invertible, see Proposition 7.19 and Corollary 7.20.

D.1.5 The Riemann integral

The material on Riemann integral in Chapter 8 served as an illustration of much of the general Banach space theory described above. We also saw interesting applications to linear ODE.

However the most important result from Chapter 8 is the Weierstrass Approximation Theorem 8.34 and its complex version in Corollary 8.36.

D.1.6 Basic things you should know about Lebesgue integration theory and infinite sums

Recall that the Lebesgue integral relative to a counting type measure corresponds to an infinite sum, see Lemma 19.15. As a rule one does not need to go back to the definitions of integrals to work with them. The key points to working with integrals (and hence sums as well) are the following facts.

1. The integral is linear and satisfies the monotonicity properties: ∫ f ≤ ∫ g if f ≤ g a.e. and |∫ f| ≤ ∫ |f| .

2. The monotone convergence Theorem 19.16 and its Corollary 19.18 about interchanging sums and integrals.

3. The dominated convergence Theorem 19.38 and its Corollary 19.39 about interchanging sums and integrals.

4. Fatou’s Lemma 19.28 is used to a lesser extent.

5. Fubini and Tonelli theorems for computing multiple integrals. We have not done this yet for integrals, but the result for sums is in Theorems 4.22 and 4.23.

6. To compute integrals involving Lebesgue measure you will need to know the basic properties of Lebesgue measure, Theorem 19.10 and the fundamental theorem of calculus, Theorem 19.40.

7. You should understand when it is permissible to differentiate past the integral, see Corollary 19.43.

Remark D.1. Again let me stress that the above properties are typically all that are needed to work with integrals (sums). In particular to understand ∫_{X} fdμ for a general measurable f it suffices to understand:

1. If A ∈ M, then ∫_{X} 1_{A}dμ = μ (A) . By linearity of the integral this determines ∫_{X} fdμ on simple functions f.

2. Using either the monotone or dominated convergence theorem along with the approximation Theorem 18.42, ∫_{X} fdμ may be written as a limit of integrals of simple functions.

D.2 Study Guide For Math 240B: Winter 2004

D.2.1 Basic things you should know about Multiple Integrals:

1. Product measures, Fubini and Tonelli theorems for computing multiple integrals, see Theorems 20.8 and 20.9. Keep in mind Driver’s “rule;” if you see a multiple integral you should probably try to change the order of integration.

2. Lebesgue Measure on ℝ^{d} and the change of variables formula, see Theorem 20.19. Also how to work in “abstract polar” coordinates, see Theorem 20.28.

D.2.2 Basic things you should know about L^{p} – spaces

1. L^{p} – spaces are Banach spaces, Theorems 21.19 and 21.20.

2. Key inequalities:

a) Holder inequality, Theorem 21.2.

b) Minkowski’s Inequality, Theorem 21.4.

c) Jensen’s Inequality, Theorem 21.10.

d) Chebyshev’s inequality, Lemma 21.14.

e) Minkowski’s Inequality for Integrals, Theorem 21.27. You should be able to use these inequalities in basic situations.

3. Recall that the L^{p}(μ) – norm controls two types of behaviors of f, namely the “behavior at infinity” and the behavior of “local singularities.” See the comments after Theorem 21.20.

4. You should have some feeling for the different modes of convergence, see Section 21.2.

D.2.3 Additional Basic things you should know about topological spaces:

1. The operations of closure, boundary and interior and in particular the interaction of closure with relative topologies. See Proposition 10.31 and Lemma 10.32.

2. The basic definitions of first countability, second countability, separability, density, etc., see Section 10.4.

3. The basic properties of connected sets, Theorems 10.48, 10.49, 10.50 and Proposition 10.53.

4. Compactness:

a) The continuous image of compact sets are compact, Exercise 11.2.

b) Dini’s Theorem, Exercise 11.3.

c) Equivalent characterizations of compactness in metric spaces, Theorem 11.7. Also see Corollary 11.9. You should be able to check compactness of a set in basic situations. d) Extreme value theorem (Exercise 11.5), uniform continuity (Exercise

11.6).

e) The consequences for normed vector spaces, see Theorem 11.12, Corollary 11.13, Corollary 11.14 and Theorem 11.15.

f) Ascoli-Arzela Theorem 11.29 for checking function space compactness.

g) The definition of a compact operator, Definition 11.16.

h) The notions of locally and σ - compact spaces, Section 11.3.

i) Tychonoff’s Theorem 11.34, i.e. the product of compact sets is still compact.

D.2.4 Things you should know about Locally Compact Hausdorff Spaces:

1. Know the definition.

2. They have lots of open sets and lots of continuous functions, see Propositions 12.5 and 12.7 and Urysohn’s Lemma 12.8 for LCH Spaces and the Locally Compact Tietz Extension Theorem12.9.

3. Basic knowledge of partitions of unity, Section 12.2.

4. Alexanderov Compactification, Proposition 12.24. (Probably will not appear on any test.)

5. The Stone-Weierstrass Theorem, see Theorem 12.31 and Corollary 12.32.

D.2.5 Approximation Theorems and Convolutions

1. The density of C_{c}(X) in L^{p}(μ) for all p ∈ [1, ∞) when (X, τ) is a second countable locally compact Hausdorff space and μ : B_{X} → [0, ∞] be a K-finite measure, see Theorem 22.8. See the important special cases in Corollaries 22.9 and 22.10. Also see the closely related Lemma 22.11.

2. Density of smaller spaces of functions by using the results in item 1. with the Stone Weierstrass theorem, see Exercises 22.11 – 22.14.

3. The density of 𝕊_{f}(A, μ) in L^{p}(μ) when μ is σ - finite on A and M = σ (A) , see Theorem 22.14. Also see Theorem 22.15 on the separability of L^{p} - spaces and Example 22.16.

4. Convolution a) Know the Definition 22.20

b) Know ∥f ∗ g∥_{p} ≤ ∥f∥_{1} ∥g∥_{p} , Proposition 22.23.

c) Understand the basic properties of convolution in Lemma 22.27.

d) Understand Theorem 22.32 about approximate δ – functions.

e) Know that f ∗ g is smooth if g ∈ C_{c}^{∞} (ℝ^{d}), see Proposition 22.34. Coupling this with Theorem 22.32 shows (for example): 1) continuous functions may be locally approximated by C^{∞} - functions, 2) C_{c}^{∞} (ℝ^{d}) is dense in L^{p} (ℝ^{d}, μ) where p ∈ [1, ∞) and μ is any K – finite measure on B_{ℝ^{d}} (see Corollary 22.38 more generally), 3) there are C^{∞} versions of Urysohn’s Lemma (Corollary 22.35) and smooth versions of partitions of unity, see Section 22.2.1.

f) The integration by parts Lemma 22.36 is also often very useful.

D.2.6 Things you should know about Hilbert Spaces

1. The definition and the fact that L^{2} (μ) is an example.

2. The Schwarz Inequality Theorem 14.2 and the fact that the Hilbert norm is a norm, Corollary 14.3.

3. The notions of orthogonality, see Proposition 14.5.

4. The Best Approximation Theorem 14.10 and the Projection Theorem 14.13, see also Corollary 14.14.

5. The very important Riesz Theorem 14.15.

6. The notion of the adjoint of operators and their properties in Proposition 14.16 and Lemma 14.17.

7. The notions of orthonormal bases on Hilbert Spaces and their basic properties, see Section 14.1. Basically the results of this section, show you may manipulate with orthonormal bases on Hilbert spaces as you would in finite dimensional inner product spaces. Understand the examples in Example 23.2 and the important Fourier Series example in Theorem 23.9.

8. Many of the basic properties about Hilbert spaces can easily be deduced from your knowledge about ℓ^{2} (X) and the fact that every Hilbert space is unitarily equivalent (see Definition 14.29) to such a Hilbert space, see Exercise 14.7.

9. The notion of the spectrum of an operator, Definition 14.30.

References

1. Lynn H. Loomis, An introduction to abstract harmonic analysis, D. Van Nostrand Company, Inc., Toronto-New York-London, 1953. MR 14,883c

Index

Summable  Integrable, 29