To get a strong convergent subsequence in $L^p_{loc}(\Omega)$ where $\Omega\subset\mathbb{R}^N$ is a smooth bounded domain and $1<,p<\infty.$

Please help me regarding this question: Let $1<p<\infty$ and $\Omega\subset\mathbb{R}^N$ be a smooth bounded domain. Given a uniformly bounded sequence in $W^{1,p}_{loc}(\Omega)$. Then upto a subsequence $u_n\to u\,\,weakly\,\,in\,\,W^{1,p}_{loc}(\Omega)$

$u_n\to u\,\,strongly\,\,in\,\,L^p_{loc}(\Omega)$ and $u_n\to u\,\,a.e.\,\,in\,\,\Omega.$
If this is true , can You give a clear explanation how to prove it?\ Your help is very much appreciated.

If $A\subset\Omega$ is a smooth subset of $\Omega$ with $\overline{A}\subset\Omega$, the injection from $W^{1,p}(A)$ to $L^p(A)$ is compact (by the Rellich-Kondrachov theorem).

You can construct a sequence $(A_j)$ of smooth domains invading $\Omega$. Since $(u_n)$ is bounded in $W^{1,p}(A_j)$, you can extract a subsequence strongly convergent in $L^p(A_j)$ (and also pointwise a.e.). Moreover, up to another subsequence, you have also weak convergence in $W^{1,p}(A_j)$ (since $1<p<\infty$).

Using a diagonal argument, you can construct a subsequence of $(u_n)$ that weakly converges in $W^{1,p}(A_j)$ adn strongly converges in $L^p(A_j)$ for every $j$ (and pointwise a.e. on $\bigcup_j A_j = \Omega$).

• Thank you very much for giving your precious time to explain the answer. I have a small doubt in the answer. To prove the weak convergence or strong convergence in $W^{1,p}_{loc}(\Omega)$, we need to prove that for every relatively compact subset $A$(need not be smooth) of $\Omega$ such that $\overline{A}\Omega$ we have $u_n$ converges to $u$ weakly in $W^{1,p}(A)$. But in the argument it is proved for any such $A$ which is smooth. Can you please explain why it is enough to prove only for smooth $A$. – Mathlover Jan 1 '18 at 13:47
• Thank you very much for giving your precious time to explain the answer. I have a small doubt in the answer. To prove the weak convergence or strong convergence in $W^{1,p}_{loc}(\Omega)$, we need to prove that for every relatively compact subset $A$(need not be smooth) of $\Omega$ such that $\overline{A}\subset\Omega$ we have $u_n$ converges to $u$ weakly in $W^{1,p}(A)$. But in the argument it is proved for any such $A$ which is smooth. Can you please explain why it is enough to prove only for smooth $A$. – Mathlover Jan 1 '18 at 13:53
• Since the sequence $(A_j)$ is invading $\Omega$, i.e. $A_1\subset A_2\subset\cdots$ and $\bigcup_j A_j = \Omega$, then every compact set $K\subset\Omega$ is contained in some $A_j$. – Rigel Jan 1 '18 at 14:32
• Please verify if what I understoodd is correct or not. – Mathlover Jan 1 '18 at 16:28
• Please verify if what I understood is correct or not. For smooth bounded domain $\Omega$ by $u_n$ converges to $u$ weakly in $W^{1,p}_{loc}(\Omega)$, we mean $u_n$ converges to $u$ weakly in $W^{1,p}(A)$ for every $A$ smooth subset of $\Omega$ such that $\overline{A}\subset\Omega$ (due to the exhaustion of $\Omega$ by smooth sets $A_j$ such that $\overline{A_j}\subset\Omega$). I think the same definition is valid for $L^p_{loc}(\Omega)$. And a last question can you tell me how the exhaustion property of smooth $\Omega$ by smooth sets $A_j$ holds. Thanks – Mathlover Jan 1 '18 at 16:37