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I want to show that if $u_{m} \rightharpoonup u$ in $W^{1,\infty}(\Omega)$ then $u_{m} \rightarrow u$ in $L^{\infty}(\Omega)$.

I know that I can't directly use the compactness of Rellich Kondrachov Theorem since I am taking $p = \infty$. From Morrey's Inequality I have $||u||_{C^{0,\alpha}} \leq ||u||_{W^{1,p}}$ where $\alpha = 1 - \frac{n}{p}$. Is it possible to take this further and show that $W^{1,p}(\Omega) \Subset L^{\infty}(\Omega)$ where $\Omega \subset \mathbb{R}^{n}$ is $C^{1}$?

José Carlos Santos
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Here is a direct proof for $L^\infty$, without Hölder spaces or Morrey inequality.

  1. By the Arzelà–Ascoli theorem, every infinite bounded subset of $W^{1,\infty}$ has a limit point in $L^\infty$.
  2. Therefore, the embedding of $W^{1,\infty}$ into $L^\infty$ is compact.
  3. A compact operator maps weakly convergent sequences into convergent sequences.

Old answer: Every $C^1$ domain, and more generally a Lipschitz domain, is a Sobolev extension domain, meaning that Sobolev functions on it can be extended to Sobolev functions on $\mathbb R^n$. In particular, all embedding theorems for Sobolev spaces hold on such domains.

When $p>n$, Morrey's inequality gives a continuous embedding of $W^{1,p}$ into $C^\beta$ with $\beta=1-n/p$. In turn, $C^\beta$ compactly embeds into $C^\alpha$ for $0<\alpha<\beta$. See Is there a reference for compact imbedding of Hölder space? This topic was also discussed in Compact, continuous embeddings of $H^s := W^{s,2} \leftrightarrow C^{(\alpha)}$ (in one dimension).

Under your assumptions, you can get $u_m\to u$ in any space $C^\alpha$ with $\alpha\in (0,1)$. A fortiori, $u_m\to u$ in $L^\infty$.

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user127096
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  • How do I get $u_{m} \rightarrow u$ in $C^{\alpha}$? I showed continuous embedding which it seems would only imply $u_{m} \rightharpoonup u$ in $C^{\alpha}$, since I only assume $u_{m} \rightharpoonup u$ in $W^{1,p}$. –  Mar 11 '14 at 19:41
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    @John The continuous embedding is followed by a compact one. – user127096 Mar 11 '14 at 19:44
  • Okay so basically $W^{1,\infty} \subset C^{\alpha} \Subset L^{\infty}$? –  Mar 11 '14 at 19:50
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    @John Yes, with the second embedding being compact. – user127096 Mar 11 '14 at 19:52
  • When I do the working and use Arzela Ascoli as you state I can show that given the weakly convergent sequence ${ u_{m} }{m}$ in $W^{1,\infty}$ there is a subsequence ${u{m_{j}}}{j}$ which converges uniformly in $C(\Omega)$, does this then imply that $u{m_{j}}$ converges strongly in $L^{\infty}$? –  Mar 12 '14 at 14:41
  • Does this follow since $L^{\infty}$ convergence is uniform convergence a.e.? –  Mar 12 '14 at 15:19
  • @JohnDoe I added a proof to the top of the answer. – user127096 Mar 12 '14 at 16:55
  • Thanks a lot. Would you look at my proof using Morrey's inequality and Holder spaces? I will post as a question. –  Mar 12 '14 at 18:38
  • The post is called 'Sobolev, Holder, Lp spaces continous and compact embeddings proof'. –  Mar 12 '14 at 19:08