Weak convergence of symmetries measures Consider a sequence of probability measure $\mu_1, \cdots, \mu_N$ with support on non-negative real numbers, $\mathbb{R}_{\geq 0}$. Now, let for each $1 \leq i \leq N$, $\hat{\mu}_i$ be the symmetrized measure of $\mu_i$, which is defined as:
$$
\hat{\mu} = \mu \, * \, \mu^\#
$$
with $\mu^\#(B) = \mu(-B)$ for each $B \in \mathfrak{B}(\mathbb{R})$.
Are the following two arguments equivalent?
$$
 \mu_i \to \mu \hspace{10 pt} \text{weakly}  \Longleftrightarrow \hat{\mu}_i \to \hat{\mu} \hspace{10 pt} \text{weakly} 
$$
 A: The notation $\hat{\mu}$ is typically reserved for the Fourier transform of $\mu$ (a.k.a. characteristic function of $\mu$) and thus, I would change the notation of the OP just a little.
Presumably, the symmetrization referred by the OP is the usual symmetrization $\tilde{\nu}$  of a measure $\nu$:
$$\tilde{\nu}:=\nu*\nu^\#$$
where $\nu^\#(B)=\nu(-B)$ for all $B\in\mathscr{B}(\mathbb{R})$.
Equivalently, in terms of random variables, if $X\sim \nu$ then the symmetrization $\tilde{\nu}$ of $\nu$ is the distribution of $Z=X-Y$ where $(X,Y)$ is an i.i.d. pair.
Notice that the characteristic function of the symmetrization $\tilde{\nu}$ of $\nu$ is
$$\widehat{\tilde{\nu}}(t)=|\hat{\nu}(t)|^2$$
Now, if $\mu_n\stackrel{n\rightarrow\infty}{\Longrightarrow}\mu$, then
$\hat{\mu}_n(t)\xrightarrow{n\rightarrow\infty}\hat{\mu}(t)$ for all $t$ and so, $|\hat{\mu}_n(t)|^2\xrightarrow{n\rightarrow\infty}|\hat{\mu}(t)|^2$ which means (Lévy-Bochner) that the symmetrization of $\mu_n$ also converges weakly; furthermore,  $\tilde{\mu}_n\stackrel{n\rightarrow\infty}{\Longrightarrow}\tilde{\mu}$.
The converse does not hold in general. Consider any real random variable $X$ (with values on $[0,\infty)$ to satisfy the conditions of the OP)) an define $X_n=X+n$. Let $\mu_n$ denote the law of $X_n$ and $\mu$ the law of $X$. Then
\begin{align}
\hat{\mu_n}(t)&=e^{int}\mu(t)\\
\hat{\tilde{\mu}}_n(t)&= |\mu(t)|^2
\end{align}
It follows that $\tilde{\mu_n}=\tilde{\mu}$ and so the sequence of symmetrizations $\tilde{\mu_n}$ converge weakly (in fact it is a constant sequence) to $\tilde{\mu}$; however, $\mu_n$ (or $X_n$) does not converge weakly.
