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 $n\in\mathbb N$.
Moreover, let $$\nu^-(B):=\nu(-B)\;\;\;\text{for }B\in\mathcal B(E)$$ and $$|\nu|^2:=\nu\ast\nu^-$$ for any finite measure $\nu$ on $\mathcal B(E)$.
Are we able to show $|\mu|^2$ is infinitely divisible as well? And does the same hold true for $\nu^-$?
I suspect that the latter is true, but we should be able to show the former. However, applying the definitions directly seems to be complicated. So, how can we show this?
$^1$ 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 $\nu_1^{\ast k}:=\nu_1\ast\cdots\ast\nu_k$.
Best Answer
It's actually quite easy: Let $n\in\mathbb N$ and $\nu:=\mu^{\ast\frac1n}$. Then, $$\varphi_{|\mu|^2}=|\varphi_\mu|^2=|\varphi_{\nu^{\ast n}}|^2=|\varphi_\nu^n|^2=|\varphi|^{2n}=\varphi_{|\nu|^2}^n$$ and hence $$|\mu|^2=(|\nu|^2)^{\ast n}.$$