$u_n\rightarrow u$ and $\nabla u_n\rightarrow g$ implies $g=\nabla u$?

Question

Suppose that $u_n\in H^1(\Omega)$ and $u_n\rightarrow u$ in $L^2(\Omega)$ and one shows that $\nabla u_n$ has a limit in $L^2(\Omega)$, let's say $\nabla u_n\rightarrow g$. I wonder if $g=\nabla u$. Any hint to prove it?

Answer 1

Everything follows from $$ \langle u,\partial_k\phi\rangle = \lim_n\langle u_n,\partial_k\phi\rangle = -\lim_n\langle\partial_k u_n,\phi\rangle = -\langle g_k,\phi\rangle $$ for $\phi\in C_c^\infty(\Omega)$.