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Let $\Omega$ be an open bounded subset of $\mathbb{R}^N$ and $1<p<N$. Let $\{u_n\}$ in $W_0^{1, p}(\Omega)$ be such that $$ u_n \rightharpoonup u \quad \text { in } W_0^{1, p}(\Omega) . $$

Then, it is true that up to subsequence $$ \nabla u_n \rightarrow \nabla u, $$ almost everywhere in $\Omega$ It was on my notes, so I guess it is true.

Could anyone help, please?

Thank you in advance.

kilose2013
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    No, for the same reason that weak convergence in $L^p$ does not imply pointwise convergence. – daw Dec 16 '23 at 11:26
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    I believe that if $1 < p < N$ the convergence a.e holds, because the embedding $W^{1,p}_0(\Omega) \hookrightarrow L^q(\Omega)$ is compact for $1 < p < N$, and $q < \frac{np}{n-p} = p^*$. We can use the fact that $u_n \rightharpoonup u$ in $W_0^{1,p}$ if and only if $u_n \rightharpoonup u$ in $L^p$ and $\nabla u_n \rightharpoonup \nabla u$ in $L^p$. – kilose2013 Dec 17 '23 at 00:39
  • I didn't get the point of cmk in this question: https://math.stackexchange.com/questions/3723208/application-of-weak-convergence-in-sobolev-space-w-01-omega – kilose2013 Dec 17 '23 at 00:44
  • Lp convergence does not imply almost everywhere: https://math.stackexchange.com/questions/138043/does-convergence-in-lp-imply-convergence-almost-everywhere – Thomas Kojar Dec 17 '23 at 00:44
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    @kilose2013 You only have convergence a.e. to $u$ up to some subsequence of $u_n$. – BigbearZzz Dec 17 '23 at 00:45
  • take a look, for example, at this question: https://math.stackexchange.com/questions/4792474/weak-convergence-implies-pointwise-convergence/4792479#4792479 there, you take $u'_n = f_n$ so that your $u_n$ is a sequence of sawtooth functions weakly converging to $0$. – BigbearZzz Dec 17 '23 at 00:48
  • @kilose2013 in the link you provided, $\lim_{n\to +\infty}\int_{\Omega} \nabla (u_n - u) dx =0$ because $\nabla (u_n - u) $ is integrated against the constant function $1$, which is clearly in all $L^p$ when $|\Omega|<\infty$. We don't need a.e. convergence, just weak convergence suffices. – BigbearZzz Dec 17 '23 at 00:52
  • @ BigbearZzz in the link: he sad that $\nabla u_n \rightarrow \nabla u $ in $L^1(\Omega)$ and thus up to subsequence $\nabla u_n \rightarrow \nabla u $ a.e ? – kilose2013 Dec 17 '23 at 01:05
  • @BigbearZzz Alright, get it from https://math.stackexchange.com/questions/4792474/weak-convergence-implies-pointwise-convergence/4792479#4792479, thank you very much for helping – kilose2013 Dec 17 '23 at 01:08

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