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Let $E$ be a $\mathbb R$-Banach space, $\mathcal M_1(E)$ (resp. $\mathcal M_1^\infty(E)$) denote the set of probability measures (resp. infinitely divisible probability measures) on $E$, $\varphi_\mu$ denote the characteristic function of $\mu\in\mathcal M_1(E)$, $$\mathcal C_1(E):=\left\{\varphi_\mu:\mu\in\mathcal M_1(E)\right\}$$ and $$\mathcal C_1^\infty(E):=\left\{\varphi_\mu:\mu\in\mathcal M_1^\infty(E)\right\}.$$

Remember that $\mathcal M_1(E)$ is infinitely divisible, i.e. $\mu\in\mathcal M_1^\infty(E)$, if and only if $$\forall n\in\mathbb N:\exists\nu\in\mathcal M_1(E):\mu=\nu^{\ast k}\tag1$$ or, equivalently, $$\forall n\in\mathbb N:\exists\psi\in\mathcal C_1(E):\varphi_\mu=\psi^n\tag2.$$

The first question is whether $\nu$ (resp. $\psi$) in $(1)$ (resp. $(2)$) are unique (if they exist). The second question is whether, given $\mu\in\mathcal M_1^\infty(E)$, there is a unique continuous convolution semigroup $(\mu_t)_{t\ge0}$ on $E$ with $\mu_1=\mu$.

For both questions, I'm only able to give a positive result when $E=\mathbb R^d$ for some $d\in\mathbb N$ and I would really like to know whether there are generalizations.

Assuming $E=\mathbb R^d$ for some $d\in\mathbb N$, we are able to show that for every $\varphi\in C^0(\mathbb R^d,\mathbb C\setminus\{0\})$ with $\varphi(0)=1$, there is a unique $f\in C^0(\mathbb R^d,\mathbb C)$ with $f(0)=0$ and $\varphi=e^f$. Moreover, for every $k\in\mathbb N$, there is a unique $g\in\mathbb C^0(\mathbb R^d,\mathbb C)$ with $\varphi=g^k$; in fact, $g=e^{f/k}$. This can be applied to every $\varphi\in C_1(\mathbb R^d)$ and hence we not only obtain uniqueness of $\nu$ (resp. $\psi$) in $(1)$ (resp. $(2)$), but even that $\nu$ (resp. $\psi$) are infinitely divisible as well.

By this result it's easy to see that, for every $m,n\in\mathbb N$, there is a unique $\mu_{m/n}\in\mathcal M_1(\mathbb R^d)$ with $\mu^{\ast m}=\mu_{m/n}^{\ast n}$; in fact, $\mu_{m/n}=\mu_{1/n}^{\ast m}$. If $t\ge0$, there is a $(t_n)_{n\in\mathbb N}\subseteq[0,\infty)\cap\mathbb Q$ with $t_n\xrightarrow{n\to\infty}t$ and hence $$\varphi_{\mu_{t_n}}=e^{t_nf}\xrightarrow{n\to\infty}e^{tf}\tag3,$$ where $f$ is as above, but corresponding to the choice $\varphi=\varphi_\mu$ for our given $\mu$. And here it seems like that we again need that $E=\mathbb R^d$, since only then Lèvy's continuity theorem is applicable and yields the existence of a unique $\mu_t\in\mathcal M_1(\mathbb R^d)$ with $\varphi_{\mu_t}=e^{tf}$. Once again, it's easy to see that $\mu_t$ is infinitely divisible as well.

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A quick Google search on "infinitely divisible" and "Banach space" leads to Linde's Probability in Banach Spaces: Stable and Infinitely Divisible Distributions (John Wiley & Sons, 1986). There we find:

  • Proposition 5.1.1: If $\mu$ is infinitely divisible on $E$, then $\hat\mu(a) \ne 0$ for every $a \in E'$.

  • Corollary 5.1.3: For each infinitely divisible $\mu$ on $E$ there exists a unique continuous function $\operatorname{Log} \hat\mu$ with domain $E'$ and range $\mathbb C$ such that $\hat\mu(a) = \exp(\operatorname{Log} \hat\mu(a))$ and $\operatorname{Log} \hat\mu(0) = 0$.

  • Proposition 5.1.4: If $\mu$ is infinitely divisible on $E$, then the measures $\mu_n$ with $(\mu_n)^{\star n} = \mu$ are uniquely determined. Moreover, their characteristic functions are given by $\hat\mu_n(a) = \exp(\operatorname{Log} \hat\mu(a) / n)$ for $a \in E'$.

  • Proposition 5.1.5: If $\mu$ is infinitely divisible on $E$, then the measures $\mu_n$ defined above converge weakly to $\delta_0$ as $n \to \infty$.

  • Corollary 5.1.8: For each infinitely divisible $\mu$ on $E$ and each $\alpha \geqslant 0$ there exists a measure $\mu^{\star\alpha}$ with characteristic function $\hat\mu^{\star\alpha}(a) = \exp(\alpha \operatorname{Log} \hat\mu(a))$ for $a \in E'$. Moreover, $\mu^{\star\alpha} \star \mu^{\star\beta} = \mu^{\star\alpha + \beta}$, and $\mu^{\star 0} = \delta_0$.

I believe this answers your present questions, and you will find much more in that book, including Lévy measures and Lévy–Khintchine representation.

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  • $\begingroup$ Thank you. This looks exactly like the reference, I was looking for. Meanwhile, I've found a claim in the book of Kallenberg, which confused me: i.sstatic.net/6Ykkz.png. If $μ_t$ is the distribution of $X_t$ and $X$ is a Lévy process, then $(μ_t)_{t\ge0}$ is a continuous convolution semigroup. The characteristic function $φ_{μ_t}$ of $μ_t$ is equal to $e^{tψ}$, where $ψ$ is the characteristic exponent of $μ_1$. So shouldn't we have $μ_t^{*1/t}=μ_1$ (since $φ_{μ_t^{\ast1/t}}=e^f=φ_{μ_1}$)? So, it's weird that he's claiming that $μ_t^{*1/t}$ weakly converges to $μ_1$ as $t\to0$. $\endgroup$
    – 0xbadf00d
    Commented Nov 16, 2020 at 5:40
  • $\begingroup$ I'm asking all these questions, cause I'm searching for an alternative proof for the characterization of the generator. They either use Fourier transforms on $\mathbb R^d$ or consider the decomposition of the process, which I've described in my answer to the other question. I really would like to know whether there is a more semigroup-theoretic proof or (maybe) a proof in terms of martingale problems. Maybe one can use that a process $M:=X-\int_0^{\;\cdot}Y_s\:{\rm d}s$ is a martingale iff $N_t:=e^{-\lambda t}X_t+\int_0^te^{-\lambda s}(\lambda X_s-Y_s)\:{\rm d}s$ is a martingale. $\endgroup$
    – 0xbadf00d
    Commented Nov 16, 2020 at 11:58
  • $\begingroup$ I've asked for that here: mathoverflow.net/q/376585/91890. $\endgroup$
    – 0xbadf00d
    Commented Nov 16, 2020 at 11:58
  • $\begingroup$ @0xbadf00d: I always had troubles understanding people that insist on "alternative proofs", so I guess I am not the right person to ask for advice. $\endgroup$ Commented Nov 16, 2020 at 20:35
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    $\begingroup$ @0xbadf00d: I fail to understand what are you trying to prove. An expression for the generator in what context? Banach-space-valued Lévy processes? What class of functions? What for? This looks like a typical example of the XY problem. $\endgroup$ Commented Nov 17, 2020 at 11:27

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