A ball in $H^m(\Omega)$ is closed under $L^2$-convergence

Question

I read it in a paper in which the author aims to claim that if a sequence $\{u_n\}$ is bounded in $H^m(\Omega)$ (here $\Omega$ is a bounded domain), and $u_n$ has a limit $u$ in $L^2$ norm, then $u$ is in $H^m(\Omega)$. I don't know why it stands. please help me understand it.

Answer 1

Since $\{ u_n \}$ is bounded in $H^m(\Omega)$, it has a subsequence $\{ u_{\varphi(n)} \}$ which converges weakly in $H^m(\Omega)$ towards some $v \in H^m(\Omega)$. Hence, $\{ u_{\varphi(n)} \}$ converges strongly towards $v$ in $L^2(\Omega)$. By uniqueness of the limit, you have $u = v$. Hence $u \in H^m(\Omega)$.

However, of course, this does not imply that $\{ u_n \}$ converges to $u$ for the $H^m(\Omega)$ topology.