%<PRE>
\input amstex
\documentstyle{amsppt}
\magnification\magstep1
\redefine\gg{\gamma}
\define\hh{\theta}
\predefine\abscont{\ll}
\redefine\ll{\lambda}
\redefine\ss{\sigma}
\define\ve{\varepsilon}
\define\DD{\Delta}
\define\GG{\Gamma}
\define\LL{\Lambda}
\define\a{\alpha}
\predefine\barunder{\b}
\redefine\b{\beta}
\define\F{{\Cal F}}
\define\Ft{{\Cal F}_t}
\define\Int{\DOTSI\int_0^{\infty}}
\define\PR{{\bold P}}  
\define\Px{{\bold P}^x}
\define\Py{{\bold P}^y}
\define\R{{\bold R}}
\define\Rp{{\bold R}^+}
\define\W{\Omega}
\define\w{\omega}
\define\({\left(}
\define\){\DOTSX\right)}
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\def\litem"#1"{\par\leftskip3\parindent\noindent\hbox
to0pt{\kern-3\parindent{\rm #1}\hss}\ignorespaces}%addition to roster_item

\def\conv#1{\overset{d^{#1}}\to\implies}
\def\eqd#1{\overset{d^{#1}}\to=}
\def\em#1{e^{-#1}}
\def\intt{\int_0^t }
\def\Rpp{\,]0,\infty[\,}
\def\vf{\varphi}
\def\Pz{\PR^0}
\def\notequiv{\not\equiv}
\def\sgn{\operatorname{sgn}}
\def\var{\operatorname{Var}}
\hfuzz2pt
\topmatter
\affil University of California, San Diego\endaffil
\address Department of Mathematics---0112,
9500 Gilman Drive,
La Jolla CA 92093-0112\endaddress
\email rgetoor\@ucsd.edu, mjsharpe\@ucsd.edu\endemail
\thanks Research supported in part by NSF Grant DMS
91-01675\endthanks
\author R\. K\. Getoor and M\. J\. Sharpe\endauthor
\title Arc-sine laws for Levy processes\endtitle
\subjclass 60J30\endsubjclass
\keywords arc-sine law, L\'evy process, stable
process\endkeywords
\abstract
Let $X$ be a L\'evy process on the real line, and
let $F_c$ denote the generalized arc-sine law on $[0,1]$ with
parameter $c$. Then $t^{-1}\int_0^t \Pz(X_s>0)\,ds\to
c$ as $t\to\infty$ is a necessary and sufficient condition for
$t^{-1}\int_0^t 1_{\{X_s>0\}}\,ds$ to converge in $\Pz$ law to
$F_c$. Moreover, if $X_t$ has a diffuse
distribution for all $t>0$, $\Pz(X_t>0)=c$ for all $t>0$ is a
necessary and sufficient condition for $t^{-1}\int_0^t
1_{\{X_s>0\}}\,ds$ under $\Pz$ to have  law $F_c$ for all
$t>0$. We show how to derive Spitzer's theorem for random walks
in a simple way from the L\'evy process version. We also use the
main theorem to derive distributional limit theorems for
functionals of the form $t^{-1}\int_0^t f(X_s)\,ds$. 
\endabstract \endtopmatter
\document
\outer\def\beginsection#1\par{\vskip0pt
plus.2\vsize\penalty-250\vskip0pt
plus-.2\vsize\bigskip\vskip\parskip
\message{#1}\leftline{\bf#1}\nobreak\smallskip\noindent}

 \beginsection 1. Introduction

In 1939, L\'evy \cite{L1},\cite{L2} proved his celebrated
arc-sine laws for occupation time for
Brownian motion and coin tossing. More explicitly, he
showed that if $X=(X_t)$ is standard Brownian motion on
$\R$ with $X_0=0$, then for each $t>0$, $\frac1t\intt
1_{\{X_s>0\}}\,ds$ has an arc-sine distribution, and if
$(S_n)$ is the random walk with $S_0=0$ and
$\PR(S_1=1)=\PR(S_1=-1)=1/2$, the distribution of
$\frac1n\sum_{k=1}^n1_{\{S_k>0\}}$ tends to an arc-sine law as
$n\to\infty$. These results have been the starting point for
much research during the ensuing years.
 In the case of random walks, the
theory culminated with Spitzer's theorem \cite{S, 7.1} which
showed that if $(S_n)$ is a random walk starting
at 0, then  $n^{-1}\sum_{k=1}^n\PR(S_k>0)\to c$ as $n\to\infty$
is necessary and sufficient in order that the distribution of
$n^{-1}\sum_{k=1}^n1_{\{S_k>0\}}$ converge to the {\it
generalized arc-sine law with parameter\/} $c$; {\it viz.\/},
the distribution $F_c$ on $[0,1]$ defined as follows. For
$0<c<1$, let 
$$f_c(x):=\frac{\sin
c\pi}{\pi}x^{c-1}(1-x)^{-c}1_{]0,1[}(x)=
\frac{x^{c-1}(1-x)^{-c}}{\GG(c)\GG(1-c)}1_{]0,1[}(x),\tag1.1$$
 and then, for $0\le
c\le 1$, define probability measures $F_c$ on $[0,1]$ by
$$F_c(dx):=\cases f_c(x)\,dx,&\text{if $0<c<1$}\\
\varepsilon_0(dx),&\text{if $c=0$}\\
\varepsilon_1(dx),&\text{if $c=1$,}\endcases\tag1.2$$
 where
$\varepsilon_a$ denotes unit mass at location $a$.  (The case
$c=1/2$ gives the standard arc-sine distribution with
density $\frac{1}{\pi\sqrt{x(1-x)}}1_{]0,1[}(x)$ referred
to above.) 

The counterpart of a random walk in continuous time is a
L\'evy process; that is, a process with stationary independent
increments. In this direction, in 1951, Kac \cite{K} proved
that if $X$ is a symmetric stable process, then $\frac1t\intt
1_{\{X_s>0\}}\,ds$ has distribution $F_{1/2}$, as in the
Brownian case. His proof works just as well for any symmetric
L\'evy process $X$ such that $X_1$ has a bounded (necessarily
continuous) density. The result for symmetric L\'evy
processes such that $\Pz(X_1=0)=0$ is well-known, and may be
obtained by observing that $s\to1_{\{X_s>0\}}$ has a
discontinuity set of measure zero, and so it is Riemann
integrable, hence the ``discrete'' arc-sine law for random
walks \cite{F, XII--8} can be used to provide approximating
Riemann sums which converge in distribution to $F_{1/2}$.
Moreover, using (8.12) and (8.13) in \cite{F, Ch\. XII}, it
follows similarly that if $X=(X_t)$ is strictly stable then
$\frac1t\intt 1_{\{X_s>0\}}\,ds$ has distribution $F_c$ with
$c:=\PR(X_t>0)$, which does not depend on $t$. See also
\cite{D}.

The main result of this paper is the exact analog for L\'evy
processes of Spitzer's result.
If $X=(X_t)$ is a 
L\'evy process on $\R$, the condition
$$\lim_{T\to\infty}\frac1T\int_0^T\Pz(X_t>0)\,dt= c\tag1.3$$
 is necessary and sufficient in order that the law,
under~$\Pz$, of $\frac1T\int_0^T1_{\{X_t>0\}}\,dt$ converge to
$F_c$ as $T\to\infty$. Aside from the cases mentioned above, the
only result of this nature of which we are aware is for a
spectrally negative, mean zero, L\'evy process in the domain of
attraction of a spectrally negative stable process of index
$\a$, $1<\a<2$, due to Bingham and Hawkes \cite{BH}. 

The main theorem is proved in \S2, and at the
same time, we show that if $X_t$ has a diffuse distribution
for all $t>0$, the stronger condition
$$\frac1T\int_0^T\Pz(X_t>0)\,dt= c\qquad\text{for all
$T>0$}\tag1.4$$  
is necessary and sufficient in order that  
 $\frac1T\int_0^T1_{\{X_t>0\}}\,dt$ have law $F_c$
for all $T>0$ under~$\Pz$. In particular, this is the case
when $X$ is a strictly stable process, and in \S4 we evaluate
the constant $c$ in terms of the parameters defining the stable
process.

In
\S3 we show that  Spitzer's theorem for random walks can be
deduced rapidly from the compound Poisson case of the main
theorem, thus giving a unified treatment of arc-sine laws for
occupation times. In \S4 we show that (1.3) implies that for
every starting point~$x$, the law of
$\frac1T\int_0^T1_{\{X_t>0\}}\,dt$ under $\Px$ converges to
$F_c$ as $T\to\infty$. We also evaluate the constant $c$ in
(1.3) and (1.4) in a number of special cases.
Finally in \S5 we consider some limit theorems for
functionals of the form $\frac1T\int_0^T
f(X_t)\,dt$ for a bounded Borel function
$f$ which is not integrable. Under various conditions on $f$,
we obtain  results which generalize results of Davydov
\cite{D} for stable processes. 

Our proof of the main result is very simple and direct. Other
than standard facts about L\'evy processes, the only thing we
use is Sparre Anderson's formula \cite{A} relating the
generating functions of $p_n:=\PR(S_1>0,\dots,S_n>0)$ and
$r_n:=\PR(S_n>0)$ for a random walk. See \cite{F, XII--7, Th\.
4}. The observation by P\. J\. Fitzsimmons that using this
formula for the random walk with $\PR(S_1\in\,\cdot\,)$ given by
$q\Int\em{qt}\Pz(X_t\in\,\cdot\,)\,dt$ leads quickly to the
arc-sine law for a symmetric process was the starting point of
our investigation. We wish to thank him for sharing his
insights  with us and also for numerous enlightening
discussions during the course of this work.


\beginsection 2. Main theorem for L\'evy processes

Throughout this paper, $X=(\W,\F,\Ft,X_t,\hh_t,\Px)$ denotes
the canonical realization of a L\'evy process. Thus $X$ is a
Hunt process with stationary independent increments specified
by 
$$\gather
\Pz\(e^{i\xi X_t}\)=\em{t\psi(\xi)},\tag2.1\\
\psi(\xi)=ia\xi+\ss^2\xi^2/2+\int[1-e^{i\xi
x}+i\xi x1_{\{|x|<1\}}]\, \nu(dx).\tag2.2
\endgather$$
Here $a\in\R$, $\ss^2\ge0$, and $\nu$ is a measure
on $\R$ such that $\int
x^2(1+x^2)^{-1}\,\nu(dx)<\infty$ and  not charging zero. The
measure $\nu$ is called the L\'evy measure of $X$ and
$\psi$ is called the L\'evy exponent of $X$. We
write $(\mu_t)$ for the associated convolution semigroup of
probability measures on $\R$ so that the transition function
of $X$ is given by 
$$P_t(x,dy)=\mu_t(dy-x),$$
and we write $U^q(dx)$ for resolvent measure given by 
$$U^q(B):=\Int\em{qt}\mu_t(B)\,dt,\qquad q\ge0.\tag2.3$$

The main theorem, (2.7) below, is trivially true when $X_t$ is
a\.s\. constant (equivalently, $\psi\equiv0$, or
$\mu_t\equiv\varepsilon_0$ for  all $t\ge0$), and we rule out
this case in all subsequent arguments.  {\it Thus in the
remainder of the paper, when we say\/} $X$ {\it is a L\'evy
process, we mean a non-constant L\'evy process.\/} For future
reference, we record the well-known fact  that $\mu_t$ is
diffuse for some (equivalently, all) $t>0$ if and only if at
least one of the following two conditions holds:
\roster
\litem"(2.4)" (i) $\ss^2>0$;
\litem"" (ii) $\nu(\R)=\infty$.
\endroster
(If $X$ is symmetric, this is the case if and only if 
$\mu_t(\{0\})=0$ for some $t>0$.)


 We begin with the following lemmas which are
undoubtedly well-known. 
\proclaim{(2.5) Lemma} Let $X$ be a
L\'evy process. If
$K\subset\R$ is compact, then $\mu_t(K)\to0$ as
$t\to\infty$. \endproclaim \demo{Proof} For every $t>0$,
$$\int_{-\infty}^\infty e^{-x^2/2}\,
\mu_t(dx)=\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty
\int_{-\infty}^\infty
e^{i\xi
x}e^{-\xi^2/2}\,d\xi\,\mu_t(dx)=\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty
e^{-t\psi(\xi)}e^{-\xi^2/2}\,d\xi.$$ 
 Suppose
first that $|e^{-\psi(\xi)}|\equiv1$, which happens only when
$\mu_t=\varepsilon_{kt}$ for some constant $k$, non-zero by the
hypothesis $\psi\notequiv0$. In this case, the result is
clearly true. If $|e^{- \psi(\xi)}|\notequiv1$, then
$\{\xi:|e^{-\psi(\xi)}|=1\}$ has the form
$\{nh:n=0,\pm1,\pm2,\dots\}$ for some  $h\ge0$, and so in this
case $e^{-t\psi(\xi)}\to0$ for a.a. $\xi$ as $t\to\infty$.
Consequently $\int_{-\infty}^\infty e^{-x^2/2}\,\mu_t(dx)\to0$
as $t\to\infty$, clearly yielding the conclusion of the lemma.
\enddemo

\proclaim{(2.6) Lemma} $t\to \mu_t([0,\infty[\,)$ is upper
semicontinuous in all cases, and there is a continuous
function $t\to \rho(t)\le\mu_t(\,]0,\infty[\,)\le 
\mu_t(\,[0,\infty[\,)$ such that
$\mu_t([0,\infty[\,)-\rho(t)\to0$ as $t\to\infty$. (Continuity
of $t\to \mu_t([0,\infty[\,)$ may fail only if $(X_t)$ is a
compound Poisson process with non-zero drift.)\endproclaim
\demo{Proof} If $f$ is
bounded and continuous, then
$t\to\mu_t(f)=\PR^0 f(X_t)$ is continuous since $X$ is a Hunt
process. It follows that $t\to\mu_t(\LL)$ is 
upper semicontinuous if $\LL$ is  closed, proving the first
assertion. If $(X_t)$ is not a
compound Poisson process  (with or without drift), then by
(2.4),  $\mu_t$ is diffuse for all $t>0$, and as
$t\to\mu_t(\Rpp)$ is  lower semicontinuous, continuity of 
$t\to \mu_t([0,\infty[\,)$ follows. If $X$ is compound
Poisson without drift, say $X_t=S_{N(t)}$
where $(S_n)$ is a random walk generated by
$\nu$ and $N(t)$ is an independent Poisson
process with rate $\ll$, then
$\Pz(X_t=0)=\em{\ll t}\sum_{k=0}^\infty (\ll
t)^k/k!\PR(S_k=0)$ is clearly continuous in $t$, so
that
$t\to\mu_t([0,\infty[\,)=1-\mu_t(\,]-\infty,0]\,)+\mu_t(\{0\})$
is both upper and lower semicontinuous, hence continuous.
Finally, let $\rho(t):=\Pz \phi(X_t)$, where $\phi(x):=1\land
x1_{\{x\ge 0\}}$, a continuous function bounded above by
$1_{\,]0,\infty[}(x)$. Then $\rho(t)$ is continuous and
$\rho(t)\le\Pz(X_t>0)$, while  $\Pz(X_t\ge0)-\rho(t)\le
\Pz(0\le X_t\le 1)\to0$ by (2.5). \enddemo Let $Z_c$ denote a
random variable with distribution $F_c$. In the sequel, the
symbol $\conv x$ denotes convergence in law relative to $\Px$,
and $\eqd x$ denotes equality in law relative to $\Px$.
 \proclaim{(2.7)
Theorem} Let $X$ be a L\'evy process and let
$A_t:=\int_0^t1_{\{X_s>0\}}\,ds$. Then the following are
equivalent. \roster
\item"(i)"$\lim_{t\to\infty}\frac1t\intt\mu_s(\Rpp)\,ds=c$.
\vskip6pt
\item"(ii)"$\lim_{q\to0}qU^q(\Rpp)=c$.
\vskip6pt
\item"(iii)"$\frac{A_{\b t}}{\b}\conv0 tZ_c$ as
$\b\to\infty$ for each $t>0$.
\endroster
In addition, if $\mu_t$ is diffuse for all $t>0$, the following
are equivalent to one another, but not to (i)--(iii).
\roster
\item"(iv)"$\frac1t\intt\mu_s(\Rpp)\,ds=c$ for each $t>0$.
 \vskip6pt
\item"(v)"$qU^q(\Rpp)=c$ for each $q>0$.
\vskip6pt
\item"(vi)"${A_{t}}\overset{d^0}\to{=} tZ_c$
for each $t>0$.
\endroster
\endproclaim
\noindent{\bf (2.8) Remarks:} (a) If $X$ is symmetric and
$\mu_t(\{0\})=0$  for some (and hence all) $t>0$, then (iv)
holds with $c=1/2$, and so for each fixed  $t>0$, the law of
$A_{t}$ under $\PR^0$ is the same as that of $tZ_{1/2}$. 

(b) If $X$ is symmetric but $\mu_t(\{0\})>0$, then by (2.5), (i)
holds with $c=1/2$, and so  the law of
$A_{t}/t$ under $\PR^0$ converges to that of $Z_{1/2}$. 

(c) If $X$ is strictly stable with index $\a$, $0<\a\le2$,
there is  a constant $c$
which may be explicitly
determined in terms of the L\'evy exponent function
such that (iv) holds (see (4.8)
and (4.9)), and then
$A_t\overset{d^0}\to{=} tZ_c$ for each $t>0$. 

(d) Condition (iii) for one fixed $t>0$ is enough to imply (i)
and (ii), and hence (iii) for all $t>0$.

(e) If any of (i), (ii), (iii) is satisfied, then in fact
convergence in (iii) takes place in $\Px$ distribution for all
$x$.  See \S4.

(f) The assertion ${A_{\b t}}/{\b}\conv0 tZ_c$ as $\b\to\infty$ for
each $t>0$ is equivalent to the
assertion ${A_{t}}/{t}\conv0 Z_c$ as
$t\to\infty$, as one may see by writing the second assertion in
the form $\Pz\exp(-\ll A_t/t)\to\PR(-\ll Z_c)$ as
$t\to\infty$, and making a simple change of variables.

(g) If $t_n^{-1}\int_0^{t_n}1_{\{X_s>0\}}\,ds$ has a limiting
distribution $F$ as $n\to\infty$ for some sequence $t_n$ such
that $t_{n+1}/t_n\to1$, then (i) holds with $c:=\int_0^1
x\,F(dx)$ and so $F=F_c$.

(h) By (2.6), if $\mu_t$ is diffuse, $t\to\mu_t(\Rpp)$ is
continuous, and so (iv) is equivalent to the condition
$\mu_t(\Rpp)=c$ for each $t>0$.

\demo{Proof} Since $t\to H(t):=\intt \mu_s(\Rpp)\,ds$ is a
continuous increasing function with $H(0)=0$ and 
$$U^q(\Rpp)=\Int\em{qt}\,dH(t),$$
the equivalence of (i) and (ii) is an immediate consequence of
the Karamata Tauberian Theorem \cite{BGT, Th\. 1.7.1}. The
equivalence of (iv) and (v) amounts to the uniqueness of
Laplace transforms. Now suppose (iii) holds for a fixed $t>0$.
Since $\b^{-1}A_{\b t}\le t$, it follows that as $\b\to\infty$,
$$(\b t)^{-1}\PR^0(A_{\b t})\to t^{-1}t\int_0^1 xF_c(dx)=c,$$
so (i) holds. Likewise, if (vi) holds, $\Pz A_t=t\PR Z_c=tc$
shows that (iv) holds. It remains only to prove that (i) implies
(iii) and (iv) implies (vi), the most significant assertions of
the theorem. We assume condition (i), and note
parenthetically the modifications required if (iv) is assumed
instead.

Before proceeding with the main line of the argument, we
shall eliminate some trivial cases. By (2.6),
$t\to\mu_t([0,\infty[\,)$ is  upper semicontinuous.
Let $\LL_t$ denote the closed support of $\mu_t$, so that
$\LL_{s+t}$ is the closure of $\LL_s+\LL_t$. If
$\mu_{t_0}([0,\infty[\,)=1$ for some $t_0>0$, then
$\LL_{t_0}\subset[0,\infty[\,$, and it follows that
$\LL_t\subset[0,\infty[$ for all $t$ of the form $k2^{-n}t_0$,
and hence $\mu_t([0,\infty[\,)=1$ for all $t>0$ by upper
semicontinuity. Since $\mu_t(\{0\})\to 0$, this shows
$\Pz|A_{\b t}/\b-t|=\frac1{\b}\int_0^{\b t} \mu_s(\{0\})\,ds\to
0$ as $\b\to\infty$, so (iii)  holds with $c=1$. (If, in
addition, $\mu_t(\{0\})=0$ for all $t>0$, we could conclude that
$A_t=t$ a.s. $\PR^0$ for all $t>0$, and so (vi) holds with
$c=1$.) The case $\mu_t(\,]-\infty,0])=1$ for some (and hence
all) $t>0$ is similar, but with $c=0$. 
We may therefore assume from now on that
$0<\mu_t(]0,\infty[)\le\mu_t([0,\infty[)<1$ for all $t>0$.
Hence for each $q>0$, $qU^q$ is a probability measure on $\R$
with $0<qU^q(\Rpp)\le qU^q([0,\infty[)<1$.

Fix $q>0$ for the moment and let $(S_n)$ denote a random walk
on $\R$ with one-step transition probability $qU^q$ and
$S_0=0$. Let $p_0(q):=1$, and define for $n\ge1$
$$\align
p_n(q):\,&=\PR\(S_1>0,S_2>0,\dots,S_n>0\),\tag2.9\\
&=
\underset{x_1>0,\dots,x_n>0}\to{\int\dots\int}
qU^q(dx_1)qU^q(dx_2-x_1)\dots qU^q(dx_n-x_{n-1})\\
r_n(q):\,&=P\(S_n>0\)=(qU^q)^{n*}(\Rpp),\tag2.10\endalign$$
where, for a finite measure $\nu$ on $\R$,
 $\nu^{n*}$ denotes the $n^{\text{th}}$ convolution power of
$\nu$. Then (7.3), (7.18) and (7.20) of \cite{F, XII} combine
to give $$\sum_{n=0}^\infty
p_n(q)s^n=\exp\(\sum_{n=1}^\infty
r_n(q)\frac{s^n}{n}\),\qquad|s|<1.\tag2.11$$ 
Note that by a simple induction
argument, for every Borel set $J\subset\R$, one has
$$(qU^q)^{n*}(J)=q^n\Int\!\dots\!\Int\em{q(t_1+\dots+t_n)}\mu_{t_1+\dots+t_n}(J)\,dt_1\dots
dt_n.$$
Hence if $h(t):=\mu_t(\Rpp)$ and $s_k:=t_1+\dots+t_k$,
$$\align
r_n(q)&=q^n\Int\!\dots\!\Int\em{q(t_1+\dots+t_n)}
h({t_1+\dots+t_n})\,dt_1\dots dt_n\\
&=q^n\Int
ds_1\int_{s_1}^\infty\!\dots\!ds_{n-1}\int_{s_{n-1}}^\infty
ds_n\em{qs_n}h(s_n)\\
&=q^n\Int
ds_n\em{qs_n}h(s_n)\int_0^{s_n}ds_{n-1}\dots
\int_0^{s_2}ds_1\\
&=\frac{q^n}{(n-1)!}\Int t^{n-1}\em{qt} h(t)\,dt.\endalign$$
As before, let $H(t):=\intt h(s)\,ds$. Then, integrating by
parts and using $H(0)=0$, $H(t)\le t$, one finds
$$\align
r_n(q)&=\frac{q^n}{(n-1)!}\(q\Int
t^{n-1}\em{qt}H(t)\,dt-(n-1)\Int t^{n-2}\em{qt}H(t)\,dt\)\\
&=\frac{1}{(n-1)!}\Int \frac qs H(\frac sq)\em
s\(s^{n}-(n-1)s^{n-1}\)\,ds.
 \endalign
$$
By (i), $\frac qs H(\frac sq)\to c$ as $q\to 0$, boundedly.
Therefore
$$\lim_{q\to 0}
r_n(q)=\frac{c}{(n-1)!}(\GG(n+1)-(n-1)\GG(n))=c.\tag2.12$$
(Under (iv), we would have $H(t)=ct$ for all $t>0$, and so
$r_n(q)=c$ for all $q>0$.)
In view
of (2.11),  (2.12) yields 
$$\lim_{q\to 0}\sum_{n=0}^\infty
p_n(q)s^n=\exp\(-c\,\log(1-s)\)=(1-s)^{-c},\quad |s|<1, \quad0\le
c\le 1.\tag2.13$$
(Under (iv), (2.13) holds with the leading $\lim_{q\to0}$
excised.)  

We are now ready to prove (iii). A straightforward
calculation using the Markov property leads for $n\ge 1$ to
$$\align
\Int\em{qt}\PR^0(A_t^n)\,dt&=\frac{n!}{q}
\underset{x_1>0,\dots,x_n>0}\to{\int\dots\int}
U^q(dx_1)U^q(dx_2-x_1)\dots U^q(dx_n-x_{n-1})\\
&=\frac{n!}{q^{n+1}} p_n(q).\tag2.14\endalign$$
Since $p_0(q)=1$, (2.14) is also valid for $n=0$. Therefore,
if $\ll>0$, 
$$\align
\PR^0\Int\em{qt}\exp(-\ll\b^{-1}A_{\b t})\,dt
&=\b^{-1}\PR^0\Int\em{qt/\b}\exp(-\ll\b^{-1}A_t)\,dt\tag2.15\\
&=\b^{-1}\sum_{n=0}^\infty\frac{(-1)^n}{n!}\(\frac{\ll}{\b}\)^n
\Int\em{qt/\b}\PR^0(A_t^n)\,dt\\
&=\b^{-1}\sum_{n=0}^\infty(-1)^n\(\frac{\ll}{\b}\)^n\(\frac{\b}{q}\)^{n+1}p_n(q/\b)\\
&=q^{-1}\sum_{n=0}^\infty p_n\({q}/{\b}\)\(-\frac{\ll}{q}\)^n.\endalign$$
Letting $\b\to\infty$ and using (2.13) yields, for  $0<\ll<q$,
$$\align
\lim_{\b\to\infty}\PR^0\Int\em{qt}\exp(-\ll\b^{-1}A_{\b t})\,dt&=
q^{-1}(1+\ll/q)^{-c}
=q^{c-1}(\ll+q)^{-c}\tag2.16\\
(&=\Int\em{qt}\,dt\intt \em{\ll
u}\frac{u^{c-1}(t-u)^{-c}}{\GG(c)\GG(1-c)}\,du\quad\text{,
$0<c<1$})\\
 &=\Int\em{qt}\,dt\,\PR\exp(-\ll tZ_c)\quad\text{ in all
cases}. \endalign$$

 To prove (iii), it suffices to prove that
for $\ll>0$ and $t>0$, $\Pz \exp(-\ll\b^{-1}A_{\b t})
\to\PR\exp(-\ll tZ_c)$ as $\b\to\infty$. 
Fix $\ll>0$, and define increasing functions
$G_{\b}(t)$, $K(t)$ by $G_{\b}(t):=1-\Pz \exp(-\ll\b^{-1}A_{\b
t})$, $K(t):=1-\PR\exp(-\ll tZ_c)$. Clearly $G_{\b}(0)=K(0)=0$.
By (2.16), for $q>\ll$,
$$\lim_{\b\to\infty}\Int\em{qt}G_{\b}(t)\,dt =\Int\em{qt}
K(t)\,dt.\tag2.17$$ 
Integration by parts shows
$$\lim_{\b\to\infty}q^{-1}\Int\em{qt}\,d_tG_{\b}(t)
=q^{-1}\Int\em{qt} \,d_tK(t).\tag2.18$$ 
We see from the
extended continuity theorem for
Laplace transforms \cite{F, Th\. 2a, p\. 433} that the measures
$d_tG_{\b}(t)$ converge in the weak${}^*$
sense to $d_tK(t)$.
As the limit measure is continuous, it follows that
$G_{\b}(t)
\to K(t)$ as $\b\to\infty$, which proves (iii), and completes
the proof of the first set of equivalences. (Under (iv), note
that (2.16), (2.17) and (2.18) hold with the leading
$\lim_{\b\to\infty}$ erased, and then, in the part of the
argument following (2.16), each ``$\to$'' may be replaced with
``$=$''. The  extended continuity theorem is not
used---uniqueness of Laplace transform suffices.)
\enddemo


\beginsection 3. Applications to random walk

We show how Spitzer's result may be derived in a very simple
manner from (2.7). Let $S_0:=0$, $S_n:=X_1+\dots+ X_n$, where
$(X_k)$ is an i\.i\.d\. sequence with $\PR(X_1\neq0)>0$. We
assume
$$\frac 1n\sum_{k=1}^n\PR(S_k>0)\to c\qquad\text{ as
$n\to\infty$}.\tag3.1$$
Let $N(t)$ denote a Poisson process of rate 1 independent of
$(S_n)$ so that
$Y_t:=S_{N(t)}$
is a L\'evy process---more precisely, a compound Poisson
process. Let $T_0:=0$ and $T_1$, $T_2$,\dots denote the
successive arrival times for $N(t)$, and for $k\ge0$, let
$U_k:=T_{k+1}-T_k$ denote the interarrival times, so that the
$U_k$ are i\.i\.d\. with $\PR U_k=1$ and $\PR(U_k-1)^2=1$ and
independent of the sequence $(S_k)$. Then
$$\frac1n\sum_{k=1}^n1_{\{S_k>0\}}U_k=\frac1n\int_0^{T_{n+1}}
1_{\{Y_t>0\}}\,dt=\frac1n\int_0^n 
1_{\{Y_t>0\}}\,dt+V_n,\tag3.2$$ where $|V_n|\le
\frac1n|n-T_{n+1}|$. By the strong law of large numbers,
$T_{n+1}/n\to1$ a\.s\. and in $L^1$ as $n\to\infty$, so
$V_n\to0$ a\.s\. and in $L^1$. Therefore, taking expectations
in (3.2) and letting $n\to\infty$, we see that (3.1) yields
$\lim_{n\to\infty} n^{-1}\int_0^n\PR(Y_t>0)\,dt=c$. This
clearly implies that (2.7--i) holds for $Y$, and so Theorem
(2.7) asserts that $$\frac1n \int_0^n 1_{\{Y_t>0\}}\,dt\conv0
Z_c.\tag3.3$$ Combining this with (3.2) gives
$$\frac1n\sum_{k=1}^n 1_{\{S_k>0\}} U_k\conv0 Z_c\quad\text{ as
$n\to\infty$}.\tag3.4$$
 But a trivial calculation using
independence gives 
$$\PR(\frac1n\sum_{k=1}^n 1_{\{S_k>0\}}
(U_k-1))^2=n^{-2}\sum_{k=1}^n \PR1_{\{S_k>0\}}
\PR(U_k-1)^2\le n^{-1},$$
 and so $\frac1n\sum_{k=1}^n
1_{\{S_k>0\}} (U_k-1)\to0$ in probability as $n\to\infty$.
Hence, by (3.4),
$$\frac1n\sum_{k=1}^n
1_{\{S_k>0\}}\conv0 Z_c\quad\text{ as $n\to\infty$},\tag3.5$$
proving the arc-sine law for the random walk $(S_n)$. That
(3.5) implies (3.1) is clear. 

\beginsection 4. Complements to the main theorem

We shall show that (2.7iii) implies that for every $x$,
$\b^{-1}A_{\b t}\conv x tZ_c$ as $\b\to\infty$.
 In fact, the convergence will turn out to be uniform in
$(x,t)$ on compacts.

The next lemma is almost surely well-known.
\proclaim{(4.1) Lemma} For each compact $K\subset\R$,
$\frac1{\b}\int_0^{\b t}1_K(X_s)\,ds\to0$
 in $L^2(\Px)$ 
as $\b\to\infty$, for every $x\in\R$.\endproclaim
\demo{Proof}Let $C_t:=\intt 1_K(X_s)\,ds$. As in (2.14),
$$\Int\em{qt}\Px(C^2_t)\,dt=\frac{2}{q^{3}}b_2(q),\quad\text{where
} b_2(q):=\int_K\!\int_K qU^q(dx_1-x) qU^q(dx_2-x_1).$$ 
If $L\subset\R$ is compact,
$\mu_t(L)\to0$ as $t\to\infty$ implies that $qU^q(L)\to0$ as
$q\to0$. Therefore, if $y\in K$, $qU^q(K-y) \le qU^q(K-K)\to0$,
and it follows readily from this that $b_2(q)\to0$ as
$q\to0$. 
Therefore Karamata's Tauberian theorem and the monotone
density theorem \cite{BGT; 1.7.1, 1.7.2} imply that $t^{-2}\Px
C_t^2\to0$ as $t\to\infty$. Consequently $t^{-1}C_t\to0$ in
$L^2(\Px)$, and this yields (4.1).\enddemo

\noindent{\bf (4.2) Remark:} Actually, using the ergodic
theorem it is  not difficult to see that the convergence in
(4.1) takes place
 a\.s\. $\PR^x$ for every $x$, at least if $\mu_t$ is absolutely
continuous for some $t>0$. 
\proclaim{(4.3) Theorem} Each of the
conditions (i)--(iii) in (2.7) is equivalent to $\b^{-1} A_{\b
t}\conv x tZ_c$ as $\b\to\infty$, and the convergence is
uniform on compacts in the sense that if $g$ is bounded
continuous function, then $\Px g(\b^{-1} A_{\b t})\to\PR
g(tZ_c)$ as $\b\to\infty$, uniformly for $(x,t)\in K\times L$,
where $K\subset\R$ and $L\subset[0,\infty[$ are
compact.\endproclaim  
\demo{Proof} Let
$A^x_t:=\intt 1_{\{X_s>x\}}\,ds$ so $A_t=A^0_t$. Clearly
$A=(A_t)_{t\ge 0}$ under $\Px$ has the same law as
$A^{-x}=(A^{-x}_t)_{t\ge0}$ under $\PR^0$. Suppose $x>0$ for
definiteness. Then 
$$A^{-x}_t=\intt 1_{]-x,0]}(X_s)\,ds+A_t.$$
Now, for $x\le a$ and $t\le T$,
$$\b^{-1}\int_0^{\b t} 1_{]-x,0]}(X_s)\,ds\le
\b^{-1}\int_0^{\b T}1_{]-a,0]}(X_s)\,ds\to0\text{  as }
\b\to\infty \tag4.4$$
in $\Pz$ probability by (4.1). Let $g$ be a bounded continuous
function. Then 
$$\Px g(\b^{-1}A_{\b t})=\PR^0 g(\b^{-1}A_{\b t})
 +\PR^0[ g(\b^{-1}A_{\b t}+\b^{-1}\int_0^{\b t}1_{]-x,0]}(X_s)\,ds)-
 g(\b^{-1}A_{\b t})].$$
Since the arguments of $g$ in the formula above lie in the interval
$[0,T]$ for $t\le T$, we may suppose that $g$ is uniformly continuous.
Combining this with (4.4) it is routine to see that (2.7--iii)
implies that $$\lim_{\b\to\infty} \Px g(\b^{-1}A_{\b t})=\PR
g(tZ_c)$$ uniformly for $0\le x\le a$ and $t\le T$.
The case $x<0$ is handled similarly, proving (4.3).
\enddemo

Next we are going to evaluate the constant $c$ appearing in (2.7)
under additional hypotheses on $X$.
Of course, if $X$ is symmetric, then $c=1/2$.
We shall need the following lemma, which is due to
Croft.  See \cite{BGT, Th\. 1.9.1}.
\proclaim{(4.5) Lemma} Let $f\!\!:\,]0,\infty[\to\R$ be
continuous and suppose there exists $c$ such that
$\lim_{n\to\infty} f(nh)=c$ for each $h>0$. Then
$\lim_{x\to\infty} f(x)=c$.\endproclaim It is trivial to see
that (4.5) also applies when it is known that $f$ is the sum of
a continuous function and one with limit 0 at infinity.
 \proclaim{(4.6) Proposition} Suppose $X_1$
has finite variance $\ss^2$ and mean $m$ (under $\PR^0$). Then
$c=0$, $1/2$ or $1$ according as $m<0$, $m=0$ or
$m>0$.\endproclaim 
\demo{Proof} If $h>0$, we have
$\PR^0(X_h)=mh$ and $\var(X_h)=\ss^2 h$. Since
$X_{nh}=\sum_{k=1}^n (X_{kh}-X_{(k-1)h})$ under $\PR^0$, one has
$$\frac{X_{nh}-nmh}{\ss\sqrt{nh}}\conv0 N(0,1).$$ But
$$\PR^0(X_{nh}>0)=
\PR^0\(\frac{X_{nh}-nmh}{\ss\sqrt{nh}}>-\sqrt{n}\frac{m\sqrt{h}}{\ss}\)$$
and so $\mu_{nh}(\Rpp)\to0$, $1/2$ or $1$ according as $m<0$,
$m=0$ or $m>0$. By (2.6), $t\to \mu_t(\Rpp)$ is the sum of a
continuous function and a function having zero limit at
infinity, and so the desired conclusion follows from (4.5).
\enddemo

Suppose next that $X=(X_t)$ is a strictly stable process of index $\a$,
$0<\a\le 2$, $\a\neq1$. Then the exponent function $\psi$ may be
written in the form 
$$\psi(\xi)= k|\xi|^{\a}\exp(i\vf\sgn
\xi)\tag4.7$$
where $k>0$,  $|\vf|\le\pi\a/2$ if
$0<\a<1$ and $|\vf|\le (2-\a)\pi/2$ if
$1<\a\le 2$. See \cite{F, p\. 581}. The density $p(t,x)$  of
$X_t$ scales according to the rule
$p(t,x)=t^{-1/\a}p(1,t^{-1/\a}x)$, and so 
$$\mu_t(\Rpp)=\Int
p(t,x)\,dx=\Int p(1,x)\,dx.\tag4.8$$
 Thus $c=\Int p(1,x)\,dx$,
and this integral is clearly independent of the value of $k$ in
(4.7). 
This integral has been calculated \cite{Z, \S2.6} in the form of
a Mellin transform of a strictly stable density, and more
directly in
 \cite{DI}, to the effect that  
 $$c=c(\a,\vf):=\Int
p(1,x)\,dx=1/2-\frac{\vf}{\pi\a}.\tag4.9$$
  Note that if
$0<\a<1$, then $c(\a,\vf)$ ranges over $[0,1]$, while if
$1<\a<2$, $c(\a,\vf)$ ranges over $[1-\a^{-1},\a^{-1}]$. 

Now suppose that $X=(X_t)$ is a L\'evy process with $X_1$ in
the domain of attraction of a strictly stable law with
characteristic function $e^{-\psi}$, where $\psi$ is given by (4.7).
We suppose in addition that $\PR^0(X_1)=0$ if $\a>1$.
Then there exist positive constants $a_n\to\infty$ such that
$X_n/a_n\conv0 Y_1$, where $Y=(Y_t)$ is the stable process with
L\'evy exponent $\psi$. See \cite{F, Th\. 3, p\. 580}.
It follows that $X_{nh}/a_n\conv0 Y_h$ for $h>0$.
Therefore, using the notation of (4.8) and (4.9),
$$\PR^0(X_{nh}>0)=\Pz(X_{nh}/a_n>0)\to\Int p(h,x)\,dx=c(\a,\vf).$$
Another appeal to (2.6) and (4.5) establishes the following
result. 
\proclaim{(4.10) Proposition} For a process $X$
satisfying the conditions of the preceding paragraph, one has
$\mu_t(\Rpp)\to1/2-\frac{\vf}{\pi\a}$ as $t\to\infty$.\endproclaim

Next suppose $X=(X_t)$ is a stable process of index $\a=1$. 
Then $\psi$ has the form
$$\psi(\xi)=-i\gg
\xi+k|\xi|[1+i\frac{2\b}{\pi}\sgn\xi\log|\xi|]$$ 
where $\gg\in\R$, $k>0$ and
$-1\le\b\le1$. 
%Again, we may set $k=1$ without loss of generality. 
Since $t\psi(\xi)=\psi(t\xi)-i \frac{2\b}{\pi} k\xi
t\log t$ for $t>0$, $X_t$ has the same law as $tX_1+
 \frac{2\b}{\pi} kt\log t$ under $\Pz$.
Therefore $\Pz(X_t>0)=\Pz(X_1>-\frac{2\b}{\pi} k\log t)$, and
if $\b\neq0$, this approaches $1$ or $0$ according as $\b>0$ or
$\b<0$. If $\b=0$ and $\gg=0$, $X$ is a symmetric Cauchy
process, and this is covered by (2.8a). If,
however, $\b=0$ and $\gg\neq0$, since $X_1+\gg$ is Cauchy,
$\Pz(X_t>0)=\Pz(X_1>0)=\Pz(X_1+\gg>\gg)=1/2-\arctan\gg$.
Thus the Cauchy process with drift can lead to an arc-sine law
with any preassigned parameter $c\in]0,1[$ by suitable choice of
the drift parameter $\gg$.

If $X_1$ is in the domain of {\it normal\/} attraction of a stable law $Y$
of index $\a>1$, then the norming constants $a_n$ satisfy $a_n\sim bn^{1/\a}$ as $n\to\infty$,
where $b>0$. In this situation, as in the finite variance case (4.6), it
follows that if $\PR(Y)\neq0$, then $\mu_t(\Rpp)\to0$ or $1$  as $t\to\infty$ according as
$\PR(Y)<0$ or $\PR(Y)>0$.

Perhaps it is worthwhile to re-write (4.9) when $\psi$ is written in a more familiar form
($\a\neq 1$)
$$\psi(\xi)=k|\xi|^{\a}(1+i\b\sgn(\xi)\tan\frac{\pi\a}{2})$$
where $|\b|\le 1$. Then $\vf=\arctan(\b\tan(\pi\a/2))$ if $0<\a<1$, and
$\vf=\arctan(-\b\tan(\pi-\pi\a/2))$ if $1<\a\le 2$. In particular, letting
$c^*(\a,\b):=c(\a,\vf)$, one has $c^*(\a,1)=0$, $c^*(\a,-1)=1$ when
$0<\a<1$, and $c^*(\a,1)=1/\a$, $c^*(\a,-1)=1-1/\a$ when $1<\a<2$. Since
$\b=1$ corresponds to a spectrally negative process, this is in accordance
with the results in \cite{BH}.

In all of the above situations, $\mu_t(\Rpp)$ itself
approaches a limit. On the other hand, examples (g) and (h) of
\cite{F, XVII.9} are easily modified to provide an example of a
L\'evy process for which the set of accumulation points of
$t\to\mu_t(\Rpp)$ at infinity is the entire unit interval.
Although we do not have an example for which (2.7--i) fails, it
appears extremely unlikely that it is always true.

There is another interesting class of examples in which $c$ may
be evaluated. Let $X$ be a L\'evy process for which
$\mu_t(\Rpp)\to c$ as $t\to\infty$. Let $T(t)$ be a
subordinator (i.e., a L\'evy process with non-decreasing
sample paths) which is independent of $X$. Then
$Y_t:=X_{T(t)}$ is a L\'evy process. If $(\eta_t)$ denotes the
semigroup of $Y$ and $(\pi_t)$ the semigroup of $T$, then
$\eta_t=\Int \mu_s\,\pi_t(ds)$, and in particular,
$$\eta_t(\Rpp)=\Int\mu_s(\Rpp)\,\pi_t(ds).$$
If $\pi_t\notequiv\varepsilon_0$, then by (2.5),
$\pi_t([0,a])\to 0$ as $t\to\infty$ for each $a<\infty$. Since
each $\pi_t$ is a probability on $[0,\infty[$, it follows
readily that $\eta_t(\Rpp)\to c$ as $t\to\infty$. Note the
agreement with (4.10) if $T$ is a stable subordinator of index
$\gg$, $0<\gg<1$, and $X_1$ is in the domain of attraction of
a stable law of index $\a$.

\beginsection 5. Related limit distributions

Let $X$ be a L\'evy process satisfying
condition (2.7--i) so that the results of the preceding
sections apply. We study in this section the distribution
limits as $t\to\infty$ of  functionals 
having the form $t^{-1}\intt f(X_s)\,ds$, with $f$ a bounded,
not necessarily integrable, function on~$\R$. The special case
in which $X$ is a stable process has been treated in \cite{D}.
(In the case of
integrable $f$, one  could expect the
Darling-Kac theory to apply, but with normalizations
other than $t^{-1}$. See \cite{DK} or \cite{BGT, 8.11.1}.)

 To begin with, let $f$ be a bounded Borel function on $\R$
having finite limits $f(\pm\infty):=\lim_{x\to\pm\infty} f(x)$
at $+\infty$ and $-\infty$. Define 
$$g(x):=f(x)-f(+\infty)1_{\Rpp}(x)-f(-\infty)1_{\,]-\infty,0[\,}(x).$$
Then $g$ is a bounded Borel function with $g(\pm\infty)=0$.
Hence $g_n:=1_{[-n,n]}g$ is bounded with compact support and
$g_n\to g$ uniformly as $n\to\infty$.  Let $M:=\|g\|_{\infty}$,
so that (4.1) implies that for each $x$, $$\align
| \b^{-1}\int_0^{\b t} g_n(X_s)\,ds|
&\le M\b^{-1}\int_0^{\b t}1_{[-n,n]}(X_s)\,ds\\
&\to0\quad\text{ in $\Px$ probability as
$\b\to\infty$.}\endalign
$$
But $g_n\to g$ uniformly as $n\to\infty$, and consequently
$\b^{-1}\int_0^{\b t} g(X_s)\,ds\to 0$ in $\Px$ probability as
$\b\to\infty$. Now, 
$$\int_0^{\b t}f(X_s)\,ds=\int_0^{\b t} g(X_s)\,ds
+[f(+\infty)-f(-\infty)]\int_0^{\b
t}1_{\{X_s>0\}}\,ds+f(-\infty)\b t.$$
Divide by $\b$ and let $\b\to\infty$. Then, combining the
above remarks and (4.3) establishes the following result, in
which $Z_c$ denotes, as in the preceding sections, a generalized
arc-sine variable. 
\proclaim{(5.1) Theorem} Let $X$ be a L\'evy
process satisfying (2.7--i). Let $f$ be a bounded
Borel function such that $f(\pm\infty)$ exists. Then, for each
$x$, $$\b^{-1}\int_0^{\b t}f(X_s)\,ds\conv
x[f(+\infty)-f(-\infty)]tZ_c+f(-\infty)t=f(+\infty)tZ_c+
f(-\infty)t(1-Z_c).\tag5.2$$
\endproclaim
When $X$ is a strictly stable process of index $\a$, $0<\a\le
2$, one may weaken the hypothesis on $f$. This will depend on
the following simple lemma.
\proclaim{(5.3) Lemma} Let $f$ be a bounded Borel function on
$\Rp:=[0,\infty[$ such that  as
$T\to\infty$, $\frac1T\int_0^T f(x)\,dx\to c$. If $g\in
L^1(\Rp)$, then $\Int f(Tx)g(x)\,dx\to c\Int g(x)\,dx$ as
$T\to\infty$.\endproclaim 
\demo{Proof} If $0\le a<b$, then
$\frac1T\int_{aT}^{bT}f(x)\,dx\to c(b-a)$. Let $h$ be a step
function of the form
$h:=\sum_{k=1}^n\a_k1_{[a_k,b_k[}$, $0\le a_k<b_k$. Then
$$\Int f(Tx)h(x)\,dx=\frac1T\Int
f(x)h(x/T)\,dx=\frac1T\sum_{k=1}^n\a_k\int_{a_kT}^{b_kT}
f(x)\,dx\to c\Int h(x)\,dx.$$
But any $L^1$ function $g$ is the limit in $L^1$ of a sequence
$(h_n)$ of such step functions, and since $f$ is bounded,
$\Int f(Tx)h_n(x)\,dx\to\Int f(Tx)g(x)\,dx$ uniformly in $T$
as $n\to\infty$. Combining these remarks, we obtain the
conclusion of the lemma.
\enddemo
Now suppose $X$ is a strictly stable process of index $\a>1$.
Then $X$ has a jointly continuous local time $(L^x_t)$ which
we normalize so that for every positive Borel function $f$ on
$\R$,
 $$\intt f(X_s)\,ds=\int f(x)L^x_t\,dx.\tag5.4$$
Moreover, $L$ has the following scaling property. For each
$\b>0$, the following equality in law holds between
two-parameter processes:
$$(\b^{1-1/\a}L_t^{x\b^{-1/\a}};x\in\R;t\ge0:\PR^y)\eqd{}
(L^x_{\b t};x\in\R;t\ge0:\PR^{y\b^{1/\a}}).\tag5.5$$
See for example \cite{FG}.
\proclaim{(5.6) Theorem} Let $f$ be a bounded Borel function
such that $f^+:=\lim_{T\to\infty}\frac1T\int_0^T f(x)\,dx$ and 
 $f^-:=\lim_{T\to\infty}\frac1T\int_{-T}^0 f(x)\,dx$ both
exist. Let $X$ be a strictly stable process of index $\a$,
$1<\a\le 2$. Then for each $x$ as $\b\to\infty$,
$$\align
\b^{-1}\int_0^{\b t} f(X_s)\,ds&\conv x f^+\intt
1_{\{X_s>0\}}\,ds+f^-\intt1_{\{X_s<0\}}\,ds\tag5.7\\
&=(f^+-f^-)\intt1_{\{X_s>0\}}\,ds+f^-t\\
&\eqd{}(f^+-f^-)tZ_c+f^-t.\endalign$$
\endproclaim
\demo{Proof} Using (5.4), (5.5) and (5.3),
$$\align
\b^{-1}\int_0^{\b t}f(X_s)\,ds&=\int_{-\infty}^\infty
f(x)L^x_{\b t}\,dx\tag5.8\\
&\eqd0\b^{-1/\a}\int_{-\infty}^\infty
f(x)L^{x\b^{-1/\a}}_t\,dx\\
&=\Int f(\b^{1/\a}x)L^x_t\,dx+\int_{-\infty}^0
f(\b^{1/\a}x)L^x_t\,dx\\
&\to f^+\Int L^x_t\,dx+f^-\int_{-\infty}^0 L^x_t\,dx\\
&=f^+\intt 1_{\{X_s>0\}}\,ds+f^-\intt 1_{\{X_s<0\}}\,ds,
\endalign$$
almost surely $\Pz$ since, a\.s\. for each $t$, $x\to L^x_t$ is
a continuous function with compact support; in particular, it is
in $L^1$ (this is clear form (5.4)). This establishes (5.7)
when $x=0$. The law of $(X_t)_{t\ge 0}$ under $\Px$ is the
same as the law of $(x+X_t)_{t\ge 0}$ under $\Pz$. If
$f_x(y):=f(x+y)$, then $f^{\pm}_x=f^{\pm}$, and so (5.7) for
all $x$ follows from the case $x=0$.\enddemo

\noindent{\bf Remarks:} If $X$ is strictly stable of index $\a$,
$0<\a\le 1$, then no local time exists. However, one may still
use the scaling and a straightforward moment calculation to
see that Theorem 5.6 is valid for any strictly stable process.
See, for example, \cite{D}. We conjecture that the existence
of the Ces\`aro limits $f^\pm$ suffices for the validity of
Theorem 5.1 provided $\mu_t$ is absolutely continuous for all
sufficiently large $t$.

\def\wk#1{\overset{{\Cal L}^{#1}}\to\longrightarrow}
For strictly stable processes $X$ with index $\a>1$, Theorem
5.1 may be strengthened to a weak convergence statement. Let
$f$ satisfy the conditions of Theorem 5.1 and let $X$ be as in
Theorem 5.6. Then the calculation (5.8) is valid with $f^\pm$
replaced by $f(\pm\infty)$. But, if $0\le t\le T$,
$$\Int|f(\b^{1/\a}x)-f(+\infty)|L^x_t\,dx\le
\Int|f(\b^{1/\a}x)-f(+\infty)|L^x_T\,dx\to0$$
as $\b\to\infty$. It follows that the convergence in (5.8) is
uniform on compact $t$ sets, and consequently
$$\b^{-1}\int_0^{\b t}f(X_s)\,ds\wk0 f(+\infty)\intt
1_{\{X_s>0\}}\,ds+f(-\infty)\intt 1_{\{X_s<0\}}\,ds.\tag5.9$$
Here, $\wk0$ means weak convergence of the laws that the
indicated processes under $\Pz$ induce on the space
$C([0,\infty[,\R)$ equipped with the topology of uniform
convergence on compacts. As before, the identity in law of
$(X_t)_{t\ge 0}$ under $\Px$ and $(x+X_t)_{t\ge 0}$ under $\Pz$
shows that $\wk0$ may be replaced by $\wk x$ in (5.9).
\Refs
\define\ky#1 {\key {\bf #1}\ }
\define\TPA{\jour Theory Probab\. Appl\.}

\widestnumber\key{\bf BGT}

\ref \ky A \by E. Sparre Anderson \paper On the
fluctuation of sums of independent random variables II\jour
Math. Scand. \yr 1954\pages195--223\vol 2\endref

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\endRefs \enddocument
