Can we show that the characteristic function of an infinitely divisible probability measure has no zeros

Question

Let $E$ be a normed $\mathbb R$-vector space, $\mu$ be a probability measure on $\mathcal B(E)$ and $\varphi_\mu$ denote the characteristic function$^1$ of $\mu$.

Assume $\mu$ is infinitely divisible, i.e. there is a sequence $(\mu_n)_{n\in\mathbb N}$ of probability measures on $\mathcal B(E)$ such that$^2$ $$\mu=\mu_n^{\ast n}\tag1$$ and hence $$\varphi_\mu=\varphi_{\mu_n}^n\tag2$$ for all $n\in\mathbb N$. We can easily show that$^3$ $$\left|\varphi_\mu\right|^{\frac2n}=\left|\varphi_{\mu_n}\right|^2=\varphi_{|\mu_n|^2}\;\;\;\text{for all }n\in\mathbb N.\tag3$$

Let $$\varphi(x'):=\left.\begin{cases}1&\text{, if }\varphi_\mu(x')\ne0\\0&\text{, otherwise}\end{cases}\right\}\tag4\;\;\;\text{for }x'\in E'.$$ By $(3)$ and $(4)$, $$\left|\varphi_{\mu_n}\right|^2\xrightarrow{n\to\infty}\varphi\tag5$$.

By $(3)$ and $(5)$, $\left(\left|\varphi_{\mu_n}\right|^2\right)_{n\in\mathbb N}$ is a sequence of characteristic functions convering pointwise to $\varphi$.

If $E=\mathbb R^d$ for some $d\in\mathbb N$, we can immediately conclude that $\varphi$ is the characteristic function of a probability measure on $\mathcal B(E)$ by Lèvy’s continuity theorem.

If $E$ is a general normed $\mathbb R$-vector space, are we able to deduce the same result using the Itō-Nisio theorem$^4$? If not, can we come up with a different approach which yields this result at least in the particular case of a separable $\mathbb R$-Banach space?

Remark: Note that my actual goal is to deduce that an infinitely divisible probabilty measures $\mu$ on $\mathcal B(E)$ satisfies $\varphi_\mu(x')\ne0$ for all $x'\in E'$. The conclusion I've asked for above is obviously sufficient to obtain this result; but it clearly is not necessary, so there might be a better approach for this.

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$^1$ i.e. $$\varphi_\mu(x'):=\int\mu({\rm d}x)e^{{\rm i}\langle x,\:x'\rangle}\;\;\;\text{for }x'\in E'.$$

$^2$ If $\nu_1,\ldots,\nu_k$ are measures on $\mathcal B(E)$ and $$\theta_k:E^k\to E\;,\;\;\;x\mapsto x_1+\cdots+x_k,$$ then the convolution of $\nu_1,\ldots,\nu_k$ is defined to be the pushforward measure $$\nu_1\ast\cdots\ast\nu_k:=\theta_k(\nu_1\otimes\cdots\otimes\nu_k)$$ of the product measure $\nu_1\otimes\cdots\otimes\nu_k$ with respect to $\theta_k$. If $\nu_1=\cdots=\nu_k$, we simply write $\nu_1^{\ast k}:=\nu_1\ast\cdots\ast\nu_k$.

$^3$ If $\nu$ is a finite measure on $\mathcal B(E)$, then $$\nu^-(B):=\nu(-B)\;\;\;\text{for }B\in\mathcal B(E)$$ and $$|\nu|^2:=\nu\ast\nu^-.$$

$^4$ Itō-Nisio theorem: Let $\nu_n,\nu$ be tight (hence Radon) probability measures on $\mathcal B(E)$ such that $\nu_n$ is symmetric (i.e. $\nu_n=\nu_n^-$), $\nu_n\prec\nu_{n+1}$ (i.e. there is a probability measure $\sigma_n$ such that $\nu_{n+1}=\nu_n\ast\sigma$) and $\varphi_{\nu_n}\xrightarrow{n\to\infty}\varphi_\nu$, then $(\nu_n)_{n\in\mathbb N}\to\nu$ in the topology of weak convergence of measures.

Answer 1

$\newcommand\vpi\varphi\newcommand\R{\mathbb R}$

1. The approach involving (4) will not work, because Lévy's continuity theorem will guarantee that the pointwise limit of characteristic functions (c.f.'s) is a c.f. only when the limit function function is continuous (everywhere or, equivalently, at the origin).

2. Nonetheless, your c.f. $\vpi_\mu$ is nonzero everywhere. Indeed, for any $x'\in E'$ and any real $t$, $$\vpi_\mu(tx')=\int_E\mu(dx)\,e^{itx'(x)}=\vpi_{x'\sharp\mu}(t),\tag{1}$$ where $\vpi_{x'\sharp\mu}$ is the c.f of the probability measure $x'\sharp\mu$ over $\R$ that is the pushforward of $\mu$ under the map $x'$. The probability measure $x'\sharp\mu$ over $\R$ is infinitely divisible, since $\mu$ is infinitely divisible: if $\mu=\mu_n^{*n}$, then $x'\sharp\mu=(x'\sharp\mu_n)^{*n}$. So, by the Lévy--Khintchine formula, $\vpi_{x'\sharp\mu}(t)\ne0$ for all real $t$. (Instead of the Lévy--Khintchine formula, it suffices to use e.g. Theorem 1 in Section 1 of Chapter XVII of Feller, Vol. II, Second Ed.) Taking now $t=1$ in (1), we get $\vpi_\mu(x')\ne0$ for all $x'\in E'$, as desired.