Let $\Omega \subset \mathbb { R } ^ n $be a bounded domain, and suppose that in the space $W^{ 1,q}(\Omega) $ $ (1 \leq q < \infty)$, the sequence $\{ u_j \} $ converges weakly in $W^{ 1,q}(\Omega) $, and $\{u_j \} $converges almost everywhere to a function $u_0$. My question is whether or not that $\{ u_ j \} $ has a subsequence that converges in $L^q (\Omega)$? If so, converge to what? Is $u_ 0 $ still in $W ^ { 1, q} (\Omega)$?
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2$\begingroup$ Is this a question on the regularity of $\Omega$? With some regularity the embedding is compact. $\endgroup$– Giorgio MetafuneCommented May 6, 2023 at 11:38
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$\begingroup$ It follows from Rellich-Kondrachov compactness Theorem, only when $\partial\Omega$ is $C^1$, $1\leq q<n$, $1\leq p<q^*$ with $q^*=\frac{qn}{n-q}$, the embedding $W^{1,q}$ to $L^p$ is compact. Are the assumptions sharp? $\endgroup$– Davidi ConeCommented May 7, 2023 at 3:22
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$\begingroup$ If $\Omega $ is $C^1$, the emebedding of $W^{1,q}$ into $L^q$ is compact for every $q$. See Theorem IX.16 in the book of Brezis. $\endgroup$– Giorgio MetafuneCommented May 7, 2023 at 7:33
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