%
\magnification\magstephalf \input amstex \documentstyle{amsppt} \hsize5.5truein \hoffset.5in \voffset.25in \hfuzz3pt\define\Supp{\operatorname{Supp}} \define\R{\bold R} \define\abscont{\ll} \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} \define\G{{\frak G}} \define\PR{{\bold P}} \define\da{\downarrow} \define\ua{\uparrow} \define\a{\alpha} \define\rb{\roman b} \define\rp{\roman p} \define\ip#1#2{\langle#1,#2\rangle} \define\Sgp{\operatorname{Sgp}} \def\litem"#1"{\par\leftskip3\parindent\noindent\hbox to0pt{\kern-3\parindent{\rm #1}\hss}\ignorespaces}%addition to roster_item \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 $00$ for all $u$ close to $t$. Therefore, there exist $00$, $H_0+S_t=S_t$.\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 $s$, $00$ 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 $k0}$ 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. \Refs \widestnumber\key{WW82} \ref\key BG68 \by \BG\book Markov Processes and Potential Theory\publ Academic Press\yr 1968\publaddr San Diego\endref \comment \ref\key Br77 \by P. L. Brockett \paper Supports of infinitely divisible measures on Hilbert space\jour Ann. Prob.\vol 5\pages 1012--1017\yr 1977\endref \endcomment \ref\key BH80 \by P. L. Brockett and W. N. Hudson \paper Zeros of the densities of infinitely divisible measures\jour Ann. 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