Let $\Omega\subset \mathbb R^N$ be open bounded, smooth boundary. Assume $u\in L^\infty(\Omega)$. We know a sequence $u_n\in L^\infty(\Omega)$ such that $$ \sup_{n}\|u_n\|_{L^\infty}<+\infty $$ and $u_n\to u$ in $L^2$ weakly.
Now, assume that for any $\Omega'\subset\subset \Omega$, i.e., compactly contained in, we have $u_n\to u$ strongly in $L^2(\Omega')$.
My question: can we deduce that $u_n\to u$ strongly in $L^2(\Omega)$ as well? (up to a subsequence of course)
Let $C = \sup_{n\in \mathbb N} \|u_n\|_\infty$. Let $\epsilon >0$ and $\Omega' \subset\subset \Omega$ so that $L^n(\Omega\setminus \Omega') <\epsilon$. Then
$$\|u_n\|^2_{L^2(\Omega)} \le \|u_n\|^2_{L^2(\Omega')} + C^2\epsilon.$$
Given that $u_n \to u$ strongly in $\Omega'$, we have
$$ \begin{split} \limsup_{n\to \infty} \|u_n\|^2_{L^2(\Omega)} &\le \|u\|^2_{L^2(\Omega')} + C^2\epsilon \\ &\le \|u\|^2_{L^2 (\Omega)} + C^2 \epsilon. \end{split}$$
Let $\epsilon \to 0$, we have
$$\limsup_{n\to \infty} \|u_n\|_{L^2(\Omega)} \le \|u\|_{L^2(\Omega)}.$$
Together with
$$\liminf_{n\to \infty} \|u\|_{L^2(\Omega)} \ge \|u\|_{L^2(\Omega)}$$
coming from the weak convergence $u_n\to u$, then
$$\lim_{n\to \infty}\|u\|_{L^2(\Omega)} = \|u\|_{L^2(\Omega)}$$
and so the convergence $u_n \to u$ is a strong convergence in $\Omega$.