how to prove that $u\in W^{1,p}(\Omega)$ and $u_k\to u$ weakly in $W^{1,p}(\Omega)$?

Question

Let $\Omega \subset \mathbb{R}^n$, $p>1$. Let $(u_k)_{k\in\mathbb{N}}\subset W^{1,p}(\Omega)$, $u\in L^p(\Omega)$ such that

(i)if $p<\infty$: $u_k\to u$ weakly in $L^p(\Omega)$,

(ii)if $p=\infty$ $u_k\to u$ weak* in $L^{\infty}(\Omega)$.

And let $$\ \sup_k \| \nabla u_k\|_{L^p}<\infty$$

Prove that $u\in W^{1,p}(\Omega)$ and

1)if $p<\infty$: $u_k\to u$ weakly in $W^{1,p}(\Omega)$

2)if $p=\infty$: $u_k\to u$ weak* in $W^{1,\infty}(\Omega)$.

Sorry, I don't know how to prove it. Especially, I don't know how to conclude from (i) that $u$ has a weak derivative in $L^p$, thus I appreciate any hints how to do it.

Answer 1

Without the uniform bound on the gradient it's obviously false, even if you have strong convergence in $L^p(\Omega)$ (at least for $p<\infty$): The space $C_c^\infty(\Omega)$ is dense in $L^p(\Omega)$.

With the gradient bound we can appeal to the reflexivity of $W^{1,p}(\Omega)$ (the case $p=\infty$ appeals to Banach-Alaoglu instead) to get the existence of $v\in W^{1,p}(\Omega)$ such that for some subsequence (which we call the same) we have $u_k\to v$ weakly in $W^{1,p}(\Omega)$. Now to show that $u=v$ notice that the inclusion $W^{1,p}(\Omega)\hookrightarrow L^p(\Omega)$ is a continuous linear operator, and as such preserves weak convergence. Therefore $u=v$ in $L^p(\Omega)$.