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 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$^1$ $$\mu=\mu_n^{\ast n}\tag1$$ and hence $$\varphi_\mu=\varphi_{\mu_n}^n\tag2$$ for all $n\in\mathbb N$. We can easily show that$^2$ $$\left|\varphi_\mu\right|^{\frac2n}=\left|\varphi_{\mu_n}\right|^2=\varphi_{|\mu_n|^2}\;\;\;\text{for all }n\in\mathbb N.\tag3$$
How can we show that $$\varphi(x'):=\lim_{n\to\infty}\left|\varphi_{\mu_n}\right|^2=\left.\begin{cases}1&\text{, if }\varphi_\mu(x')\ne0\\0&\text{, otherwise}\end{cases}\right\}\tag4$$ for all $x'\in E'$?
Note that the claim is made (in the case $E=\mathbb R^d$, $d\in\mathbb N$) in the proof of Lemma 7.5 in Sato's book Lévy Processes and Infinitely Divisible Distributions.
Remark: If you can provide an answer for the case $E=\mathbb R^d$, $d\in\mathbb N$, I would be interested in it as well.
$^1$ 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 $\nu_1^{\ast k}:=\nu_1\ast\cdots\ast\nu_k$.
$^2$ If $\nu$ is a finite measure on $\mathcal B(E)$, then $$\nu^-(B):=\nu(-B)\;\;\;\text{for }B\in\mathcal B(E)$$ and $$|\nu|^2:=\nu\ast\nu^-.$$