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Nov 11, 2020 at 7:21 comment added 0xbadf00d @MateuszKwaśnicki I'll try to edit the question, but can you say something about the proof without characteristic functions and something about the convolution semigroups?
Nov 11, 2020 at 7:17 comment added Mateusz Kwaśnicki I am afraid I do not know the answer, but the question would be much easier to answer if you stated explicitly what you are after: abelian/non-abelian locally compact groups? Lie groups? Discrete objects? My rough intuition is that what you need is that the dual group is connected, so that characteristic functions of ID measures are non-vanishing.
Nov 11, 2020 at 5:43 comment added 0xbadf00d @MateuszKwaśnicki I see, but can we give a positive result somewhere in-between arbitrary groups and Banach spaces (e.g. for metric spaces)? In any case, as I pointed out in the question, I would really like to know whether we can show the well-defined/uniqueness without the use of the characteristic function (e.g. by using that weak limits of infinitely divisible measures are infinitely divisible); even in the Banach space case. Or at least for continuous convolution semigroups.
Nov 10, 2020 at 21:25 comment added Mateusz Kwaśnicki Another trivial remark: the same construction works for a circle group $\{e^{it} : t \in [0, 2\pi)\}$. In this case $\delta_1$ can be realised as $\mu_1$ for $\mu_t = \delta_{\exp(2\pi i n t)}$ for an arbitrary $n \in \mathbb{Z}$, so again there's no uniqueness here.
S Nov 10, 2020 at 21:13 history suggested RobPratt CC BY-SA 4.0
corrected spelling in title
Nov 10, 2020 at 20:51 comment added Mateusz Kwaśnicki The Dirac measure $\delta_0$ on $\mathbb{Z}_2 = \{0, 1\}$ is clearly infinitely divisible, but since $\delta_0 * \delta_0 = \delta_1 * \delta_1 = \delta_0$, the $2k$-th convolution root of $\delta_0$ is not unique.
Nov 10, 2020 at 20:37 review Suggested edits
S Nov 10, 2020 at 21:13
Nov 10, 2020 at 20:05 history asked 0xbadf00d CC BY-SA 4.0