Does weak convergence in $H^1_0(\Omega)$ imply weak convergence of the weak derivatives in $L^2(\Omega)$

functional-analysispartial differential equationssobolev-spacesweak-convergence

Let $\Omega\subset\mathbb{R}^N$ be bounded and consider a sequence of functions $(u_n)_n\subset H^1_0(\Omega)$ such that $u_n\rightharpoonup u\in H^1_0(\Omega)$. Can we then say that $\partial_i u_n \rightharpoonup \partial_iu$ in $L^2(\Omega)$ for $i=\overline{1,N}$?

My progress: since the inclusion $H^1_0(\Omega)\subset L^2(\Omega)$ is compact, eventually passing to a subsequence we can assume that $u_n\to u$ in $L^2(\Omega)$. This allows us to assume that $u_n(x)\to u(x)$ for a.e. $x\in \Omega$ by again eventually passing to a subsequence. I wasn't able to use any of these general facts, I only thought that maybe it is enough to write the definition of $u_n \rightharpoonup u$ in $H^1_0(\Omega)$, i.e. $$\int_\Omega \nabla u_n\cdot \nabla v \to \int_\Omega \nabla u \cdot \nabla v$$ for all $v\in H^1_0(\Omega)$. This implies that $$\int_\Omega \partial_i u \cdot \partial_iv\to \int_\Omega \partial_i u \cdot \partial_i v$$ for all $v\in H^1_0(\Omega)$. Now if for any $w\in L^2(\Omega)$ there would be some $v\in H^1_0(\Omega)$ such that $w=\partial_i v$, then we would be done. I am not sure if this is true however and I don't know how to proceed.

Best Answer

The mapping $u \mapsto \partial_i u$ is linear and continuous (bounded) from $H^1_0(\Omega)$ to $L^2(\Omega)$. Both are easy to verify. (If you have any trouble with these claims: please leave a comment.)

Since linear and continuous operators map weakly converging sequences to weakly converging sequences, the claim follows.