most recent 30 from http://mathoverflow.net 2013-05-18T16:19:57Z http://mathoverflow.net/feeds/user/30119 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/116713/support-of-an-infinitely-divisible-measure/116876#116876 Gabriel 2012-12-20T15:55:58Z 2012-12-20T15:55:58Z

<p>Actually Martin the support in your exemple would be the closure of ${\exp(ix), x\in \mathbb{N}}$, which is equal to $U(1)$, which is a group. </p> <p>In a finite group the proof would be just that if $x$ is in the support of $\mu$, looking at $\mu$ like more or less the measure of $X_1$ with $X$ a Levy process going from $e$, then it is obvious that we can "speed" up the jumps and so $x$ is in $X_{\frac{1}{2}}$, and so as the support of $X_1$ is equal to the closure of the product of the support of $X_{\frac{1}{2}}$ with himself then $\text{Support}(X_1)$ is stable by multiplication. Thus it is a group as a subset of a finite group which is stable by multiplication is a group. But in a Lie group can we "speed up" the jumps ? If $X$ is a Levy process beginning from $e$ having at any time $e$ in his support, is it true that : $Support(X_{1}) \subset Support(X_{\frac{1}{2}})$ ? </p>