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\hfuzz3pt\define\Supp{\operatorname{Supp}}
\define\R{\bold R}
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\define\ky#1 {\key {\bf #1}\ }
\define\TPA{\jour Theory Probab\. Appl\.}
\define\ZfW{\jour Z\. Wahrs\-chein\-lich\-keits\-theorie
 verw\. Ge\-b\-iete}
\define\BG{R\. M\. Blumenthal and R\. K\. Getoor}
\define\Spr{\publ Spr\-ing\-er\publaddr Berlin Heidelberg
 New York} 
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\topmatter
\title Supports of convolution semigroups and
densities\endtitle
\author Michael J. Sharpe\endauthor
\affil University of California, San Diego\endaffil
\address Math\. Dept\.-0112, UCSD, 9500 Gilman Dr\.,
La Jolla, CA 92093-0112\endaddress
\email msharpe\@ucsd.edu\endemail
\thanks
Research supported in part by NSF Grant DMS
91-01675\endthanks
\subjclass 60J30\endsubjclass
\keywords convolution semigroup, L\'evy process,
support, density, zero\endkeywords
\abstract Let $\mu_t$ be a weakly continuous
convolution semigroup of probability measures on a
LCA group $\G$.
 We give a simpler proof of Tortrat's
characterization of the support
$S_t$ of
$\mu_t$ in the case $\G=\R^d$. In addition, in case
$\mu_t$ is absolutely continuous for all $t>0$ and
the densities $p_t$ are chosen so as to satisfy the
Chapman-Kolmogorov equation identically, we
derive a number of regularity properties of the
$p_t$, including the fact that
$p_t$ is strictly positive on the interior of $S_t$. 
\endabstract
 \endtopmatter
\outer\def\beginsection#1\par{\vskip0pt plus .1\vsize\penalty-250
 \vskip0pt plus-.1\vsize\bigskip\vskip\parskip
 \leftline{\bf#1}\nobreak\smallskip\noindent}
\document
\beginsection 1. Introduction

 Let 
$(\mu_t)_{t>0}$ be a weakly continuous convolution
semigroup  with L\'evy measure
$\nu$ on the LCA group $\G$, and let $S_t$ denote the
closed support of
$\mu_t$. 
Let $H$ denote the closure in $[0,\infty[\times\G$
of  $\{(t,x):t>0,x\in S_t\}$. Then
$H_0:=\{x\in\G:(0,x)\in H\}$ is called the
invariant semigroup for $(\mu_t)$. A theorem of
Tortrat
\cite{To88}, when specialized to the case $\G=\R^d$,
shows that there is a vector
$b$ and a closed additive semigroup
$\Sigma\subset\R^d$ such that
$S_t=tb+\Sigma$. From this, one concludes that
$H_0=\Sigma$. Tortrat gave a version of his theorem
for a general LCA group, but the
hard part of his proof was for
$\R^d$. We provide a simpler, self-contained proof
of Tortrat's theorem in $\R^d$ using the
properties of  the invariant semigroup $H_0$.

 We
further show that if, for every
$t>0$,
$\mu_t\abscont m$ (Haar measure on $\G$) and
if (as we may, at least in the separable metric case)
we select the densities
$p_t(x)$ so as to be jointly measurable in $(t,x)$
and to satisfy the convolution equation
$$p_{t+s}(x)=\int p_t(x-y)p_s(y)\,dy\qquad
\forall t>0, \forall s>0, \forall x\in
\G,\tag1.1$$ 
then the set
$G_t:=\{x\in\G:p_t(x)>0\}$ is equal to the interior
of
$S_t$. In particular, if
$S_t=\G$, then $p_t$ never vanishes. 

 In the
one-dimensional case $\G=\R$, assuming
$\nu(\R)=\infty$,
$H_0$ must be the entire real line or one of the closed half
lines
$\R^+$,
$\R^-$, and the results therefore include as a
special case the results of
\cite{Sh69}, the hypotheses of which
were much more restrictive. See
 \cite{BH80} and \cite{HS78}
for related results. Information about the zeroes of
the densities
$p_t$ is important in connection with the bridge
processes constructed over the original L\'evy
process. See for example
\cite{GS82}.

In case $\G$ is separable and
metrizable and $\mu_t\abscont m$ for all
$t>0$, the fact that we can choose jointly
measurable densities satisfying (1.1)
 was proved in \cite{Ha79}.
Alternatively, a simple modification of the proof of
\cite{GS82, (3.7)} (which provides more detail
than the proof in \cite{Ha79}) may be used to the same
effect.

The author wishes to thank the referee for
rectifying an error in the original version of
the proof of (2.8). 

\beginsection 2. Characterization of the closed
supports

Throughout this section, we assume $\G=\R^d$.
Let $\langle
y,x\rangle$ denote the inner product on $\R^d$, let
$B_{x,r}$ denote the open ball of radius $r$ about
$x$, and let $T$ denote the unit sphere in $\R^d$.  
We assume that
$(\mu_t)$ is a weakly continuous convolution
semigroup with L\'evy measure $\nu$ on $\R^d$, and
$X_t$ is a corresponding L\'evy process. Because
$S_{t+r}$ is the closure of the algebraic sum
$S_t+S_r$, the set $\{(t,x):x\in S_t\}$ is an
additive  semigroup in
$[0,\infty[\times\R^d$, and so therefore is its
closure $H$. It follows that $H_0$ is a closed
(additive) semigroup in $\R^d$.

Given a measure $\lambda$ on $\R^d$, let
$\Sgp({\lambda})$ denote the smallest closed
(additive) semigroup containing 0 and carrying
$\lambda$. The set
$\Sgp(\lambda)$ may be constructed explicitly as
follows. Let $J_0:=\{0\}$, $J_1:=\Supp(\lambda)$ and
 for $n\ge2$, let
$J_n$ be the algebraic sum of $J_1$ with itself $n$
times. Then $\Sgp(\lambda):=(\cup_{n\ge0} J_n)^-$.

We shall say that $X$ is of type I in case
$\int_{B_{0,1}}\|x\|\,\nu(dx)<\infty$, and otherwise
$X$ will be said to be of type II. 
\proclaim{(2.1) Theorem (Tortrat)} There is a vector
$b$ such that for every $t>0$, 
$S_t=tb+H_0$. Let $V$ be the
subspace carrying the centered Gaussian component
of $X$. (i) If
$X$ is of type I, then $H_0=\Sgp(\nu)+V$. (ii) 
Suppose $X$ is of type II. Define the subspace
$L\subset\R^d$ by
$$L:=\{y\in\R^d:\int_{B_{0,1}}|\ip
xy|\,\nu(dx)<\infty\}\tag2.2$$
and let $\Pi_L$ denote the orthogonal projection
map of $\R^d$ onto $L$. Let $\nu_L$ denote the
restriction of $\Pi_L\nu$ to $\{0\}^c$, so that
$\nu_L$  is the L\'evy measure of $\Pi_L X$. Then
$\Pi_L X$ is of type I and 
$$H_0=\Pi_L^{-1} \Sgp(\nu_L)+V.\tag2.3$$
\endproclaim
\noindent{\bf (2.4) Remarks:} It is easy to see that
$\nu$ is of type I if and only if
$\int_{B_{0,1}}|\ip xy|\,\nu(dx)$ is finite for every
$y\in\R^d$, which is the same as the condition
$L=\R^d$. Thus the type~II characterization (2.3) is
consistent with the type I case. Tortrat gave a
description of $L^\perp$ as the ``complete
asymptotic direction of $1_{B_{0,r}^c}\nu$ as
$r\to 0$'', but it is not easy to disengage a
precise definition of this object from his proof.
\demo{Proof} We shall assume
that the Gaussian component vanishes, and that $\nu$
is carried by $B_{0,1}$. The general case follows by
simple considerations of the effects of adding 
an independent Gaussian component and an
independent compound Poisson process to
$X$. 
Define type I L\'evy measures
$\nu_n$ and vectors $a_n$ by
$$\align
\nu_n(dx)&:=(\|x\|1_{B_{0,1/n}}+1_{B^c_{0,1/n}})(x)\nu(dx).\tag2.5\\
a_n&:=\int x1_{B_{0,1}}(x)\,
\nu_n(dx).\tag2.6\endalign$$
Then $\Supp(\nu_n)=\Supp(\nu)$ for all $n$. Let
 $X^n$ denote a pure jump L\'evy process with
L\'evy measure $\nu_n$, and let $Y^n$ denote a
L\'evy process with L\'evy measure $\nu-\nu_n$,
independent of $X^n$, such that $(X^n_t-ta_n)+Y^n_t$
has the same law as $X_t$. The theory of L\'evy
processes tells us that there is a vector $b$
such that 
$X^n_t-ta_n$ converges to $X_t-tb$ as
$n\to\infty$, and therefore 
$$S_t-tb=\cap_n(\cup_{k\ge n}
\Supp(X^k_t-ta_k))^-=\cap_n(\cup_{k\ge n}
(\Sgp(\nu)-ta_k))^-.$$ 
(This seems to have been noticed first in
\cite{YL76}, using the L\'evy-Khintchine formula
rather than L\'evy processes.) If
$X$ is of type I, the
$a_k$ are bounded, and we may assume $a_k\to a$, hence
$S_t-tb=\Sgp(\nu)-ta$, completing the proof in the
type I case.
\comment It does not seem to be true,
as claimed in
\cite{Br77, Theorem 2.1}, that this implies
$S_t-tb=(\Sgp(\nu)-A)^-$, where $A:=\{a_n:n\ge1\}$,
but it is true that
$$S_t=tb+\{x\in\R^d:\exists
k_n\to\infty,x_n\in\Sgp(\nu)\text{ with
}x_n-ta_{k_n}\to x\}.
$$
\endcomment
The general case of the theorem will
follow from lemmas (2.7)--(2.13)
below.\enddemo
\proclaim{(2.7) Lemma} Suppose
$k_n\to\infty$,
$t_n\da\da0$ and $x_n\in\Sgp(\nu)$ have the property
$x_n-t_n a_{k_n}\to x$ as $n\to\infty$. Then $x\in
H_0$.
\endproclaim
\demo{Proof} Let $Z^n_t:=X^n_t-ta_n$. Clearly,
$y_n:=x_n-t_n a_{k_n}\in\Supp(Z^{k_n}_{t_n})$. Then
$Z^{k_n}_{t_n}+Y^{k_n}_{t_n}$ has the same
distribution as $X_{t_n}$, the summands are
independent and $Y^{k_n}_{t_n}\to0$ in distribution as
$n\to\infty$. By hypothesis, for every $r>0$, 
$\PR\{Z^{k_n}_{t_n}\in B_{x,r}\}>0$ for all
sufficiently large
$n$. In addition, $\PR\{Y^{k_n}_{t_n}\in B_{0,r}\}>0$ 
for all sufficiently large $n$. It follows that 
$\PR\{X_{t_n}\in
B_{x,2r}\}\ge\PR\{Z^{k_n}_{t_n}\in
B_{x,r}\}\PR\{Y^{k_n}_{t_n}\in B_{0,r}\}>0$ for all
sufficiently large $n$. As $r>0$ is arbitrary,
$x\in H_0$.\enddemo
\comment
\proclaim{(2.8) Lemma} For every
$t>0$,
 $H_0+S_t=S_t$.\endproclaim
\demo{Proof} If not, there would exist $t>0$ and
points $x_0\in S_{2t}$,
$z\in H_0$ such that 
$x_1=x_0+z\notin S_{2t}$. We show that this
supposition leads to a contradiction. Since
$x_1\notin S_{2t}$, there exists
$r>0$ such that
$\mu_{t}(B_{x_1,2r})=0$. Note that, since $B_{x,r}$ is
open, 
$y\to\mu_t(B_{x,r}-y)$ is lower semicontinuous.
Since
$\mu_s(B_{0,r})\to 1$ as
$s\da0$,
$\mu_s(B_{0,r})>0$ for all sufficiently small $s$,
say $0<s<\epsilon$. As
$0=\mu_t(B_{x_1,2r})=
\int\mu_s(dy)\mu_{t-s}(B_{x_1,2r}-y)$, lower
semicontinuity of the integrand shows that for
$0<s<\epsilon$, $\mu_{t-s}(B_{x_1,2r}-y)=0$ for all
$y\in B_{0,r}\cap S_s\neq\emptyset$.  This shows in
particular that $H$ does not contain the set
$]t-\epsilon,t[\times B_{x_1+z,r}$. On the other hand,
lower semicontinuity of
$u\to \mu_u(B_{x_0,r})$ and the fact that
$x_0\in S_{t}$ imply
$\mu_u(B_{x_0,r})>0$ for all $u$ close to $t$.
Therefore, there exist $0<s<t$ and $y\in B_{x_0,r}$,
with $(t-s,y)\in H$.  However, $(t-s,y+z)\notin H$,
which violates the semigroup property of $H$. This
contradiction establishes the lemma.
\enddemo
\endcomment
\proclaim{(2.8) Lemma} For every
$t>0$,
 $H_0+S_t=S_t$.\endproclaim
\demo{Proof} Since $0\in H_0$, it is enough to show
that $H_0+S_t\subset S_t$. Suppose that $z\in H_0$
and $x_0\in S_t$. Since $(0,z)\in H$, by
definition of $H$, for every
$r>0$ and $\epsilon>0$ there exists $s$,
$0<s<\epsilon$, such that $B_{z,r/2}\cap
S_s\neq\emptyset$. On the other hand,
lower semicontinuity of
$u\to \mu_u(B_{x_0,r/2})$ and the fact that
$x_0\in S_{t}$ imply
$\mu_u(B_{x_0,r/2})>0$ for all $u$ close to $t$, so we
may also assume that
$\mu_{t-s}(B_{x_0,r/2})>0$. Fix an arbitrary
$z_0\in B_{z,r/2}\cap
S_s$. Note that
$B_{z_0,r}+B_{x_0+(z-z_0),r}\subset B_{x_0+z,2r}$,
while $B_{x_0,r/2}\subset B_{x_0+(z-z_0),r}$ so
$B_{x_0+z,r}-y\supset B_{x_0,r/2}$
for all
$y\in B_{z_0,r}$. Thus we have
$$\mu_t(B_{x_0+z,2r})\ge\int_{B_{x_0,r}}
\mu_{t-s}(B_{x_0+z,2r}
-y)\,\mu(dy)\ge\mu_s(B_{x_0,r})\mu_{t-s}(B_{x_0,r/2})
>0$$
which proves that $x_0+z\in S_t$.
\enddemo
A
cone $K$ in $\R^d$ will be called open in case
$K\cap T$ is relatively open in $T$.  Let
$\hat K_\infty$ denote the set of
$y\in\R^d$ with the property that, for every open cone
$K$ containing $\{ty:t\ge0\}$, $\int_{K\cap
B_{0,1}}\|x\|\,\nu(dx)=\infty\}$, and let
$K_\infty$ denote the closed cone in $\R^d$ generated
by the points in
$\hat K_\infty$. 
(Alternatively, let
$\eta$ denote the radial projection of
$\|x\|\nu(dx)$ on $T$. Then, for $y\in T$,  $y\in\hat
K_\infty$ if and only $\eta(G)=\infty$ for every
neighbourhood $G$ of $y$ in $T$.) 
\comment
 It will turn out that $K_\infty$
generates $L^\perp$.
\endcomment

The objects $K_\infty$ and 
$\hat K_\infty$ are related to the ``asymptotic
cones'' that were the basis of the proof in
\cite{To88}.

\proclaim{(2.9) Lemma}
 $K_\infty\subset L^\perp$, and for
every open cone
$K\supset K_\infty$, 
$\int_{K^c}\|x\|\,\nu(dx)<\infty$. In
particular, $K_\infty\neq\{0\}$ if
$\int_{B_{0,1}} \|x\|\,\nu(dx)=\infty$.
\endproclaim
\demo{Proof} 
 For any $y\in L$ and $x_0\in\hat
K_\infty$, if $|\ip y{x_0}|>0$, then there exists an
open cone
$K_0$ about $x_0$ and a constant $\beta>0$ such that
$|\ip yx|\ge\beta\|x\|$ for all $x\in K_0$, and
therefore $\int|\ip yx|\,\nu(dx)=\infty$, contrary
to the hypothesis $y\in L$. Thus $K_\infty\subset
L^\perp$. Now let $K$ be an arbitrary open cone
containing $K_\infty$. Let
$W:=\{x\in T:x\notin K\}$, so that
$W$ is compact. For every
$\theta\in W$, there exists
$\epsilon>0$ such that
 the circular cone $K'$ with angle $\epsilon$
about $\theta$ satisfies 
$\int_{K'}\|x\|\,\nu(dx)<\infty$. By compactness of
$W$, a finite number of such
$K'$ would cover $\R^d\setminus K$, implying
$\int_{K^c}\|x\|\,\nu(dx)<\infty$. The
last assertion follows upon setting $K:=\{0\}$.
\enddemo
\proclaim{(2.10) Lemma} $H_0$ contains 0 and
$\Sgp(\nu)$, and $K_\infty\subset
H_0$.\endproclaim
\demo{Proof} The first assertion is an elementary
consequence of the fact that, as
$t\to0$, $\mu_t\to\delta_0$  and
$\mu_t/t\to\nu$ outside any neighbourhood of 0. If
$x_0\in\hat K_\infty$ and $\|x_0\|=1$, for every
$\epsilon>0$, the circular cone $K^\epsilon_0$ of
angle
$\epsilon$ and axis $\{tx_0:t\ge0\}$ has the property
$\int_{K^\epsilon_0} \|x\|\,\nu(dx)=\infty$, and
therefore, for every $r>0$, $\nu$ charges
$K^\epsilon_0\cap B_{0,r}$. Given $n\ge1$, let
$\epsilon:=1/n$ and $r:=1/n$, and choose $x_n\in
K^{1/n}_0\cap B_{0,1/n}\cap S_{\nu}$, $x_n\neq0$.
Then ${\Bbb N}x_n:=\{kx_n:k=1,2\dots\}\subset
K^{1/n}_0$ consists of points separated by a distance
no more than $1/n$, so there must exist $k_n\ge1$
such that
$\|k_nx_n-x_0\|<2/n$. (If not, the region in
$K^{1/n}_0$ given by $\{y\in
K^{1/n}_0:1-1/n<\|y\|<1+1/n\}$ would contain no
points of ${\Bbb N}x_n$, violating the separation
property.) As
$k_nx_n\in
\Sgp(\nu)\subset H_0$ and $H_0$ is closed, 
 it follows that $x_0\in H_0$. This proves $\hat
K_\infty\subset H_0$. It suffices now to note that
$K_\infty$ is the smallest closed semigroup
containing $\hat K_\infty$, hence
$K_\infty\subset H_0$.
\enddemo

\proclaim{(2.11) Lemma} Define  $a_n$ as in
(2.6), and suppose
$\|a_n\|\to\infty$ as
$n\to\infty$. Then, for every open cone
$K$ containing $K_\infty$,  $a_n\in K$ for all
sufficiently large
$n$.
\endproclaim
\demo{Proof} Choose another closed cone $\tilde
K\subset K$ such that the interior (in $T$) of $\tilde
K\cap T$ contains
$K_\infty\cap T$. Let
$\nu=\nu'+\nu''$, where
$\nu'$ is the restriction of $\nu$ to $\tilde K$ and
let
$a_n=a_n'+a_n''$ with $a_n':=\int
x1_{B_{0,1}}(x)\,\nu'_n(dx)$, where $\nu_n'$ is
defined relative to $\nu'$ as in (2.5). Then
$a_n''$ is uniformly bounded by (2.9).  Since
$\Supp(\nu')\subset\tilde K$, $a_n'\in \tilde K$ for
all
$n$ and $\|a_n'\|\to\infty$. It is then clear that
$a_n\in K$ for all sufficiently large $n$.
\enddemo
\proclaim{(2.12) Lemma} If
$X$ is type II, then $H_0$
contains a subspace $[\xi]$,
with $0\neq\xi\in K_\infty$.\endproclaim
\demo{Proof} If $K_\infty$ contains $[\xi]$ for some
non-zero 
$\xi\in K_\infty$, the proof is finished because
$K_\infty\subset H_0$. We assume from now on
that $K_\infty$ contains no non-trivial subspace.
It is therefore equal to the intersection of the
open cones containing it. Let
$\nu_n$ and
$a_n$ be defined as in (2.5) and (2.6). The argument
varies according as
$\sup_n\|a_n\|$ is finite or infinite. Suppose first
that 
$\sup_n\|a_n\|=\infty$. We may suppose, passing to a
subsequence if necessary, that
$\|a_n\|\to\infty$ and
$a_n/\|a_n\|$ converges to
a unit vector
$\xi$.  For any
open cone $K$ containing $K_\infty$, 
$a_n\in K$ for sufficiently large $n$ by (2.11), so
$\xi\in\bar K$. As $K$ is an arbitrary open
cone containing $K_\infty$, it follows that $\xi\in
K_\infty$. Let
$t_n:=1/\|a_n\|$, and apply (2.7) to conclude that
$-\xi=\lim_n 0-t_n a_n\in H_0$. Consequently $H_0$
contains $-\xi+r\xi$ for all $r\ge0$, and therefore
$H_0$ contains the subspace
$[\xi]$. 

We next suppose $\sup_n\|a_n\|<\infty$. If $K_\infty$ 
contains no subspace $[\xi]$, the cones $K_\infty$
and $-K_\infty$ meet only at the origin, and
according to the theory of convex cones, there is
then a linear functional $\lambda$ separating
$K_\infty$ and $-K_\infty$, in the sense that
$\lambda>0$ on $K_\infty\setminus\{0\}$ and
therefore $\lambda<0$ on
$-K_\infty\setminus\{0\}$.
Then
$\lambda(a_n)=\int\lambda(x)\,\nu_n(dx)$
remains bounded since $a_n$ is bounded. On the
other hand, $K_\infty\subset\{\lambda>0\}$ implies
by (2.9) that
$\int_{\{\lambda\le0\}}|\lambda(x)|\,\nu(dx)<\infty$.
Since
$$\lambda(a_n)=\int_{\{\lambda>0\}}\lambda(x)\,\nu_n(dx)
+\int_{\{\lambda\le 0\}}\lambda(x)\,\nu_n(dx)$$
it follows that
$\int_{\{\lambda>0\}}\lambda(x)\,\nu(dx)=\lim_n
\int_{\{\lambda>0\}}\lambda(x)\,\nu_n(dx)<\infty$,
and so by (2.9), $K_\infty=\{0\}$, violating the
hypothesis
$\int\|x\|\,\nu(dx)=\infty$.
\enddemo
\proclaim{(2.13) Lemma} $H_0=\Pi_L^{-1}
\Sgp(\nu_L)$.\endproclaim
\demo{Proof} If $L=\R^d$, $X$ is type I by (2.4),
a case already proved. Suppose from now on that
$L\neq\R^d$, so that 
$X$ is of type II. By (2.12), there exists $\xi\in
K_\infty$ with
$\|\xi\|=1$ and 
$[\xi]\subset H_0$. Let $\Pi$ denote the orthogonal
projection from $\R^d$ onto $[\xi]^\perp$, let
$X':=\Pi X$, $\nu':=\Pi\nu$ restricted to $\{0\}^c$, 
$L':=\{y\in
[\xi]^\perp:\int|\ip yx|\,\nu'(dx)<\infty$, and let
$H_0'$ denote the invariant semigroup
for $X'$. 
\comment
$H_0=\Pi^{-1} H_0'$. We make a series of
reductions based on (2.9). Let
$k:=\dim L$. If $k=d$, 
$L^\perp=\{0\}$ and there is nothing to prove, so
suppose $k<d$. Let $V^1:=\R^d$, $\mu_t^1:=\mu_t$,
$H_0^1:=H_0$, $L^1:=L$, $K_\infty^1:=K_\infty$ and
$\nu^1:=\nu$. Since $L^1\neq V^1$, 
$\int_{B_{0,1}}\|x\|\,\nu^1(dx)=\infty$ by (2.13), so
$K_\infty\neq\{0\}$ by (2.10). By (2.12), we may
choose a unit vector $\xi_1\in K^1_\infty$ with
$[\xi_1]\subset H_0$. Let $V^2:=[\xi_1]^\perp$ so that
$L\subset V^2$ by (2.10). Let $\Pi^2$ denote the
orthogonal projection of $V^1$ onto $V^2$ and let
$\nu^2$ denote the restriction of $\Pi^2\nu^1$ to
$\{0\}^c$, so that $\nu^2$ is the L\'evy measure for
$\mu_t^2:=\Pi^2\mu^1_t$. If $k=d-1$, then $L=V^2$
since $L\subset V^2$ and both have dimension $k$, and
so
$L^\perp=[\xi_1]$, and the theorem is proved.
We may therefore suppose $k<d-1$. Let
$K_\infty^2$ denote the cone $K_\infty$
defined relative to
$\mu_t^2$,  let
$H_0^2$ be the corresponding invariant semigroup for
$\mu_t^2$, and let $L^2:=\{y\in
V^2:\int_{B_{0,1}\cap V^2}|\ip
yx|\,\nu^2(dx)<\infty\}$. I claim
\endcomment
It is easy to see from properties of the
projection that $L'=L$.
I claim next that $H_0=\Pi^{-1} H_0'$.
\comment 
For (2.15), if $y\in L^1$, then, because
$\Pi^2 y=y$,  
$$\align
\int_{B_{0,1}\cap V^2}|\ip{
y}x|\,\nu^2(dx)&=\int_{(\Pi^2)^{-1}(B_{0,1}\cap
V^2)}|\ip{y}{\Pi^2 x}|\,\nu^1(dx)\\
&=
\int_{(\Pi^2)^{-1}(B_{0,1}\cap
V^2)}|\ip{\Pi^2 y}{x}|\,\nu^1(dx)\\
&=
\int_{(\Pi^2)^{-1}(B_{0,1}\cap
V^2)}|\ip{ y}{x}|\,\nu^1(dx).\endalign
$$
Because the last integrand is uniformly bounded
and $\nu^1$ is finite away from the origin, the
last integral is finite if and only if
$\int_{B_{0,1}}|\ip yx|\,\nu^1(dx)<\infty$. This
proves
$L^2= L^1$. 
\endcomment
For, if $x\in
H_0$, say
$t_n\to 0$ and
$x_n\to x$ with $x_n\in S_{t_n}$ for all $n$, then
because every $S_t$ is invariant under
translation in the direction $[\xi]$, we have
$\Pi x_n\in S_{t_n}'$ and $\Pi x_n\to \Pi x$,
proving $\Pi H_0\subset H_0'$. On the other
hand, if $x'\in H_0'$, there exists $t_n\to 0$ and
$x_n'\in S'_{t_n}$ with $x_n'\to x'$ in
$[\xi]^\perp$. Then, because
$S_t=\Pi^{-1}(S'_t)$, $x'_n\in
S_{t_n}$, hence $x'=\Pi x'\in H_0$, proving
the claim. It follows that if $k:=\dim (L)$, then
after applying the previous projection $d-k$ times,
we obtain orthonormal vectors $\xi_1,\dots,\xi_{d-k}$ 
generating $L^\perp$, and after the 
$d-k^{\text{th}}$ projection, we have 
$H_0=\Pi_L^{-1} H_0^L$, where $H_0^L$ is the
invariant semigroup for $\Pi_L X$. By construction,
$\Pi_L X$ is of type~I, and therefore
$H_0^L=\Sgp(\nu_L)$.
\enddemo

\beginsection 3. Properties of the densities

We suppose throughout this section that
$(\mu_t)_{t>0}$ is a weakly continuous
convolution semigroup of probability measures on a
locally compact Abelian group $\G$ written additively
and with Haar measure $m$, such that for every $t>0$,
$\mu_t(dx)=p_t(x)\,m(dx)$, where
$p_t(x)$ is jointly measurable in $(t,x)$
and satisfies (1.1). (If
$\G$ is separable and metrizable, it is enough by the
discussion at the end of the first section to require
$\mu_t\abscont m$ for every $t>0$.) 

Because the convolution of two
probability density functions is lower
semicontinuous (approximate from below by bounded,
positive integrable functions), 
for
every
$t>0$, $p_t$ is lower semicontinuous and
the set
$\{x:p_t(x)>0\}$ is a non-empty open 
set. The same sort of idea can be extended to yield
the following stronger result.
\proclaim{(3.1) Proposition} The function
$(t,x)\to p_t(x)$ is lower semicontinuous. For each  
$t>0$, $\liminf_{s\to t}p_s(x)=p_t(x)$ for $m$
a\.a\. $x$.
\endproclaim  
\demo{Proof} Formula (1.1) implies 
$$p_t(x)=\frac1t \int_0^t ds\int_{\G}m(dy)\,
p_{t-s}(x-y)p_s(y),$$
which may be interpreted as a convolution
identity on $\R^+\times\G$. Lower semicontinuity then
follows as before, approximating
$(r,x)\to p_r(x)$ monotonically from below by
bounded, positive, integrable functions on
$\R^+\times\G$. The last assertion is a consequence
of Fatou's lemma.
\enddemo
\proclaim{(3.2) Lemma} For every $t>0$, 
$H_0+G_t\subset G_t$.
\endproclaim
\demo{Proof}  If the conclusion of the lemma were
false, there would exist $t>0$ and points $x_0\in
G_{2t}$,
$z\in H_0$ such that 
$x_1:=x_0+z\notin G_{2t}$. We show that this
supposition leads to a contradiction. Since
$p_{2t}(x_1)=0$, $p_t(x_1-y)p_t(y)=0$ a\.e\.,
and hence everywhere by lower semicontinuity. On the
other hand, $p_{2t}(x_0)>0$ implies
$p_t(x_0-y)p_t(y)>0$ on a non-empty open subset $B$
of $G_t$. Let $x_2\in B$ and let $D:=B-x_2$, a
neighborhood of 0. For every
$y\in D$, $y+x_2\in B$, hence $p_t(y+x_2)>0$ and
therefore
$p_t(x_1-x_2-y)=0$. Let $x:=x_1-x_2$, so that 
$p_t(x)>0$
 while $p_t$ vanishes on
$x+z-D$. This proves that
$x+z\notin S_t$, but
$x\in S_t$, violating the fact that
$S_t=tb+H_0$ is closed under shifts in $H_0$.\enddemo

\proclaim{(3.3) Theorem} 
$p_t>0$ everywhere on the interior,
$S_t^o=tb+H_0^o$,  of
$S_t$. 
\endproclaim
\demo{Proof}  We have $S_t=tb+H_0$, and it suffices
to assume
$b=0$, as translation of
$S_t$ and
$G_t$ by
$-bt$ will not affect the conclusion of the theorem.
Thus $S_t=H_0$ for all $t>0$.  As
$H_0=S_t$ is the closure of the open set $G_t$, it
follows that
$H_0^o\supset G_t$ and therefore $H_0$
is the closure of its interior. Now let $x\in H_0^o$,
say
$x\in x-D\subset H_0$ with $D$ an open neighborhood
of 0. If
$p_t(x)=0$, (3.2) would imply $p_t(x-z)=0$ for all
$z\in H_0$, hence for all $z$ of the form $z=x-y$
with $y\in D$. But then
$p_t(y)=0$ for every $y\in D$, showing that
$H_0=S_t$ omits some neighborhood of~0, which is
clearly impossible.
\enddemo
\noindent
{\bf Note added in proof:} After this paper had been
refereed, Steve Evans brought to my attention an
unpublished manuscript of John Hawkes entitled {\it
Transition and resolvent densities for L\'evy
Processes\/}, containing a number of interesting
results about infinitely divisible densities on
$\R^d$,
 including the fact that
$p_t>0$ on the interior of
$S_t$. His proof uses a Baire category
argument.  

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\enddocument

