strong convergence in $W^{1,p}$ if $L^p$ convergence of function and all derivatives

Question

Let $\{u_n\}_{n \in \mathbb{N}} \subset W^{1,p}(\Omega)$ for $p \in (1, \infty)$ and $\Omega \subset \mathbb{R^n}$ open, bounded.

If $u_n \to u^* \in L^p(\Omega)$ and $\nabla u_n \to 0 \in L^p(\Omega)$. Is it true that $u^* \in W^{1,p}(\Omega)$ and $\nabla u^* = 0$?

If the answer is yes, why does this hold? In particular, why does $u^*$ need to be weakly differentiable and why is the weak derivative given by the limit of $\nabla u_n$?

Answer 1

If $\nabla u^*$ exists in a weak sense and is equal to $0$ then $\nabla u^*\in L^p$ and so $u^*\in W^{1,p}(\Omega)$. So it suffices to prove that $$\int_{\Omega}u^*\partial_i\varphi=0\qquad \forall \varphi \in C_c^{\infty}(\Omega) $$ We know that $$ \int_{\Omega}u_n\partial_i\varphi=-\int_{\Omega}\varphi\partial_iu_n$$ so it suffices to show the following convergences (can you?) $$\int_{\Omega}u_n\partial_i\varphi \to \int_{\Omega}u^*\partial_i\varphi,\qquad \int_{\Omega}\varphi \partial_iu_n\to 0 $$ and then the thesis follows.