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Let $p\neq \infty$ and let $\Omega\subseteq \mathbb{R}^N$ be a regular domain (a bounded open set with $C^1$ boundary). I want to prove that

If $\{u_n\}$ weakly converges in $W^{1,p}(\Omega)$ then it converges in $L^p(\Omega)$.

My attempt

By definition if $\{u_n\}$ weakly converges in $W^{1,p}(\Omega)$ then it weakly converges in $L^p(\Omega)$ too and this implies (by Banach-Steinhaus theorem) that $\{u_n\}$ is bounded in $L^p(\Omega)$. The same reasoning applies to the gradient of $u$ so $\{u_n\}$ is bounded in $W^{1,p}(\Omega)$.

By Rellich Kondrachov theorem there exists a subsequence of $\{u_n\}$ that converges in $L^p(\Omega)$. So I basically have

  • $u_n\rightharpoonup u$ in $L^p(\Omega)$

  • $u_{n_k}\to u$ in $L^p(\Omega)$

Is this enough to establish the strong convergence? If yes why?

Secondly, I was trying to construct a counterexample if $p=\infty$, but without any success. Could you help me?

Kandinskij
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  • It doesn't seam detailed enough, it basically just says "apply Rellich Kondrachov"... also it doesn't address the counterexample part... – Kandinskij Aug 24 '23 at 20:37
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    Under the regularity assumptions on $\Omega$, the result is true for $p=\infty$, since in this case $W^{1,\infty}(\Omega)$ is exactly the class of bounded Lipschitz functions in $\Omega$, so you can apply the Arzelá-Ascoli theorem in place of Rellich-Kondrachov. – Jose27 Aug 24 '23 at 23:50
  • This is not a duplicate as here the OP asked about the case $p==q$. – daw Aug 25 '23 at 07:41
  • Honestly, I do not understand, what your are asking. At which step do you have doubts? – daw Aug 25 '23 at 07:53
  • @daw I'm basically just asking: is it true that if a sequence ${u_n}$ weakly converges in $L^p(\Omega)$ to $u$ and admits a subsequence that converges in $L^p(\Omega)$ to $u$ then ${u_n}$ converges in $L^p(\Omega)$ to $u$? – Kandinskij Aug 25 '23 at 08:18
  • This is not true. But you know more: $(u_n)$ converges weakly in $W^{1,p}$. This is enough to conclude. – daw Aug 25 '23 at 10:29

2 Answers2

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The proof is false, since the Rellich Kondrachov theorem has two different exponents: $W^{1,p}\subset\subset L^q$ for all $1\leq q<p^*$, assuming $1\leq p<n$ ($p=\infty$ is not allowed! If you want a counterexample, look at the counterexample for the Rellich Kondrachov Theorem)

To complete the proof, one only needs to know that the inclusion map in the Rellich Kondrachov Theorem is compact: $W^{1,p}\subset\subset L^q$. Therefore, weakly $W^{1,p}$ convergent sequences are mapped to strongly $L^q$ convergent sequences. This is what you wanted to prove. If you want to understand this fact about compact operators, look at this proof.

Eric
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There are actually two versions of the Rellich-Kondrachov theorem. The second one on that wiki page is just the answer to the question. The first version is the following: $W^{1,p}(\Omega)$ is compactly embedded into $L^q(\Omega)$ if $q<p^*$ and $p<n$. Since $p^*>p$ this answers the question for $p<n$.

For $p\ge n$, one can use that $W^{1,p}(\Omega)$ embeds (trivially) into $W^{1,n-\epsilon}(\Omega)$. Now the first version of Rellich-Kondrachov is applicable, and yields the compact embedding into $L^p(\Omega)$ provided $\epsilon$ is small enough. This uses that $p^* \to \infty$ for $p\nearrow n$.

This shows that the embedding operator of $W^{1,p}(\Omega)$ into $L^p(\Omega)$ is compact for all $p$, and so sequences converging weakly in $W^{1,p}(\Omega)$ converge strongly in $L^p(\Omega)$.

daw
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