Is $W_0^{k,\:p}(\Omega)$ weakly compact?

Question

We know that if $\Omega\subseteq\mathbb R^d$ is open, $k\in\mathbb N$ and $p\in(1,\infty)$, then for every bounded $(u_n)_{n\in\mathbb N}\subseteq W^{k,\:p}(\Omega)$, there is a subsequence $\left(u_{n_k}\right)_{k\in\mathbb N}$ and some $u\in W^{k,\:p}(\Omega)$ s.t. $u_{n_k}\xrightarrow{k\to\infty}u$ weakly in $W^{k,\:p}(\Omega)$.

Does the same hold true with $W^{k,\:p}(\Omega)$ replaced by $W_0^{k,\:p}(\Omega)$? Moreover, if we know that $(v_n)_{n\in\mathbb N}\subseteq W_0^{k,\:p}(\Omega)$ is (uniformly) norm-bounded and almost everywhere pointwisely convergent to some $v:\Omega\to\mathbb R$, are we able to conclude that $v_n\xrightarrow{n\to\infty}v$ weakly in $W_0^{k,\:p}(\Omega)$ from the aforementioned fact?

Answer 1

Since $W^{k,p}_0$ is a closed subspace of $W^{k,p}$, all these results carry over to $W^{k,p}_0$.

If $v_n \rightharpoonup v$ in $W^{k,p}$ and $v_n \to u$ pointwise a.e., then $u=v$.

As the result ``weak equals pointwise limit'' is hard to track down, here is a sketch of a proof:

Let $(\Omega,\mathcal A,\mu)$ be a $\sigma$-finite measure space. Assume $v_n \rightharpoonup v$ in $L^p(\Omega)$, $1<p<\infty$, and $v_n \to u$ pointwise $\mu$-a.e. on $\Omega$. Then $v=u$ $\mu$-a.e.

Proof. By Fatou Lemma, $u\in L^p(\Omega)$. Let $B\subset \Omega$ be bounded. Let $\epsilon>0$. By Egorov's theorem, there is a subset $A\subset B$ such that $v_n \to u$ uniformly on $A$ and $\mu(B\setminus A)<\epsilon$. Let $\phi\in L^\infty(\mu)$. Then $\chi_Bv\in L^q(\mu)$ with $\frac1p+\frac1q=1$ and $1<q<\infty$. Then $$ |\int_B(v_n-u)\phi d \mu | = |\int_{A \cup (B\setminus A)} ...| \le \mu(A) \|v_n-u\|_{L^\infty(A)} \|\phi\|_{L^\infty} + \epsilon^{1/q} \|v_n-u\|_{L^p(B)} \|\phi\|_{L^\infty} $$ which implies $$ \limsup |\int_B(v_n-u)\phi d \mu| \le \epsilon^{1/q} 2M\|\phi\|_{L^\infty} $$ where $M$ is such that $\|v_n\|_{L^p}, \|u\|_{L^p}\le M$. Since $\epsilon>0$ was arbitrary, this proves $\int_B(v_n-u)\phi d \mu \to0$. Since $L^\infty(B)$ is dense in $L^q(B)$, this proves $v_n \rightharpoonup u$ in $L^p(B)$. This proves $u=v$ $\mu$-a.e. on $B$. Since $\Omega$ is a countable union of sets of bounded measure, $u=v$ $\mu$-a.e. on $\Omega$ follows. $\square$

I believe, this result is also true for $p=1$ (and $q=\infty)$. Then one has to use the Dunford theorem (see Diestel & Uhl), which is a characterization of weakly convergent sequences in $L^1$.

A proof by Banach-Saks property can be found here:

Weak and pointwise convergence in a $L^2$ space