# Weak convergence and convergence almost everywhere

If a bounded sequence $(u_n)$ converge weakly to $u$ in $W^{1,p}(\Omega)$ (where $\Omega$ is an open bounded subset of $\mathbb{R}^N$ with $N>p$),

is it true that $u_n(x)$ converges to $u(x)$ for almost every $x\in \Omega$?

thank you

• If a subsequence $u_{n_k}(x)$ that converges to $u(x)$ a.e in $\Omega$ works for you- there is such. Commented Jul 29, 2015 at 12:20
• What i need is when i have that $(u_n)$ is bounded in $W^{1,p}_0$ then because $W^{1,p}_0$ is reflexive $(u_n)$ has a sub sequence which converge weakly in $W^{1,p}_0$ then can i deduce that this sub-sequence converge almost everywhere ? Commented Jul 29, 2015 at 16:04
• Yes, but for a furthter subsequence, because $W_0^{1,p}$ is compactly embedded in $L_p$. Now, because the embedding operator is compact, it maps weakly convergent sequences in strongly convergent, i.e this weakly convergent subsequence in $W_0^{1,p}$ is strongly convergent in $L_p$. Finally, you can extract a further subsequence, which is pointwise a.e convergent. Commented Jul 29, 2015 at 16:22
• Why strong convergence in L^p imply convergence almost everywhere ? please @Svetoslav write it as an answer to obtein +50 Commented Jul 29, 2015 at 16:27
• Strong convergence in $L_p$ implies almost everywhere convergence of a subsequence. Commented Jul 29, 2015 at 16:31

Because $W_0^{1,p}$ is compactly embedded in $L_p$, the embedding operator is obviously compact so it maps weakly convergent sequences in strongly convergent, i.e a weakly convergent subsequence in $W_0^{1,p}$ is strongly convergent in $L_p$. Finally, you can extract a further subsequence, which is pointwise a.e convergent: see Does convergence in $L^{p}$ implies convergence almost everywhere?
• Is there an easier proof of that convergence $L^p$ implies almost everywhere ? Commented Jul 30, 2015 at 7:58
• As it was mentioned in one of the comments in the link that I posted, The extraction of a pointwise a.e convergent sequence is the first step in proving that $L_1$ is a complete space. So look on this proof. For $L_p$ and for finite measure space (like $\Omega\subset \mathbb R^n$ - bounded ), then you have $L_p (\Omega)\subset L_1(\Omega)$ for $p\in[1,\infty]$ Commented Jul 30, 2015 at 9:16