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 $(4)$, $$\left|\varphi\_{\mu\_n}\right|^2\xrightarrow{n\to\infty}\varphi\tag5$$.



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> 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 $\varphi$.
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> 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.
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> 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?
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> **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](https://en.wikipedia.org/wiki/Convergence_of_measures#Weak_convergence_of_measures) of measures.



Can we show that the characteristic function of an infinitely divisible probability measure has no zeros
$\newcommand\vpi\varphi\newcommand\R{\mathbb R}$


1. The approach involving (4) will not work, because [Lévy's continuity theorem](https://en.wikipedia.org/wiki/L%C3%A9vy%27s_continuity_theorem#:%7E:text=In%20probability%20theory%2C%20L%C3%A9vy%27s%20continuity,convergence%20of%20their%20characteristic%20functions.) 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](https://people.cam.cornell.edu/av395/levy-khintchine.pdf), $\vpi\_{x'\sharp\mu}(t)\ne0$ 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.](https://www.wiley.com/en-us/An+Introduction+to+Probability+Theory+and+Its+Applications%2C+Volume+2%2C+2nd+Edition-p-9780471257097))
Taking now $t=1$ in (1), we get $\vpi\_\mu(x')\ne0$ for all $x'\in E'$, as desired.


