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I can't understand why the following fact holds:

I consider a sequence $(u_h)\subset W^{1, \infty}(U, \mathbb{R}^N)$, with $U$ open bounded set in $\mathbb{R}^n$, such that $$u_h \rightharpoonup u$$ $*w-W^{1,\infty}$ (the * weak convergence).

Then it is possible to extract a subsequence such that $u_{h_k}\rightarrow u$ in $L^{\infty}$

Edit

Thanks to @daw 's answer I know how to prove this fact under the hypothesis $U$ bounded and with Lipschitz boundary.

I would like to know if this fact is true also if I don't assume that the boundary of $U$ is Lipschitz.

Thanks for the help!

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This argument works if $U$ is bounded and has Lipschitz boundary.

Since $(u_h)$ converges weak-star in $W^{1,\infty}(U)$, it is bounded in $W^{1,\infty}(U)$ and also bounded in $W^{1,n+1}(U)$ since $U$ is bounded. The space $W^{1,n+1}(U)$ is continuously embeddded into the space of H"older continuous functions $C^{0,1/n}(\bar U)$, Morrey embedding theorem. This space compactly embeds into $C(\bar U)$, thus also into $L^\infty(\Omega)$ by Arzela-Ascoli. So we have this chain of continuous $\hookrightarrow$ and compact $\hookrightarrow\hookrightarrow$ embeddings: $$ W^{1,\infty}(U) \hookrightarrow W^{1,n+1}(U) \hookrightarrow C^{0,1/n}(\bar U)\hookrightarrow\hookrightarrow C(\bar U) \hookrightarrow L^\infty(U). $$

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  • $\begingroup$ thanks a lot for your answer. I have a question about it. The fact that $C^{0,1/n}(U)\hookrightarrow\hookrightarrow C(U)$ is a consequence of Ascoli-Arzelà. But why $C(U) \hookrightarrow L^\infty(U)$? Thank you! $\endgroup$ – Gggl. Jul 21 '16 at 14:34
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    $\begingroup$ It is $C^{0,1/n}(\bar U)$ and $C(\bar u)$. Edited the answer. The space $C(\bar U)$ embeds into $L^\infty(U)$: continuous functions are measurable, functions from $C(\bar U)$ are bounded, hence $L^\infty$. $\endgroup$ – daw Jul 21 '16 at 20:13
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To answer your additional question: Without assumptions on the boundary the statement is false. You can pretty much do the same thing as in the unbounded case. More precisely: look at the bounded part of a spiral (with some thickness) with infinite length. Parameterized along the length of the spiral from the outside in, just let $f_n=0$ on $[0,n]$ and $f_n = sin(\pi x)$ for $x>n$. This converges weak-$\ast$ to $0$ in $W^{1,\infty}$, but clearly not (essentially) uniformly. (The boundary is not Lipschitz in the center of the spiral.)

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