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

Question

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.

Answer 1

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.

Answer 2

Maybe one way to go is the following. Take some $w \in L^2(\Omega)$. Then $\partial_i w \in \mathcal{D}'(\Omega)$ is actually in $H^{-1}(\Omega)$ (the topological dual of $H^1_0(\Omega)$), where the pairing is defined as $$ \langle \partial_i w, u \rangle_{H^{-1},H^1_0} = - \int_\Omega w \partial_i u. $$ By the Riesz representation theorem, since $\partial_i w \in H^{-1}(\Omega)$ is a linear form on $H^1_0(\Omega)$, you can represent it as a scalar product with some $v \in H^1_0(\Omega)$. Hence $$ \langle \partial_i w, u \rangle_{H^{-1},H^1_0} = \int_\Omega \nabla v \nabla u. $$ Thus, $$ - \int_\Omega w \partial_i u_n = \int_\Omega \nabla v \nabla u_n \to \int_\Omega \nabla v \nabla u = - \int_\Omega w \partial_i u. $$