Let $X_n$ be a sequence of random variables converging in mean square to $X$, that is, \begin{equation}\tag{1} \lim_{n\to\infty}\mathbb{E}\Big(\big|X_n-X\big|^2\Big)=0 \end{equation} or, in short-hand notation, $X_n\overset{\mathcal{L}^2}{\longrightarrow}X$
Wish: does $(1)$ imply $X_n^2\overset{\mathcal{L}^2}{\longrightarrow}X^2$?
In the accepted answer to this post it is written that $(1)$ implies $\mathbb{E}(X_n^2)\to\mathbb{E}(X^2)$, but I don't know if it's enough to guarantee the convergence of the squares.
No, this would require $\lim_{n \rightarrow \infty} \mathbb{E}[|X_n^2-X^2|^2] = 0$, but there is no guarantee that $\mathbb{E}[|X_n^2-X^2|^2]$ is even finite.