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Let $\Omega \subset \mathbb{R}^N$ be a smooth domain with bounded complement (e.g. $\Omega = \mathbb{R}^N \setminus \overline{B(0;1)}$), and $(u_n)$ be a bounded sequence in $H^1_0(\Omega) = W_0^{1,2}(\Omega)$.

Since $H^1_0(\Omega)$ is a reflexive Banach space, weak compactness implies the existence of a subsequence $(u_{n_k})$ and a function $u_0 \in H_0^1(\Omega)$ such that $u_{n_k} \rightharpoonup u_0$ weakly in $H^1_0(\Omega)$. May I assume that (going if necessary to a subsequence, which I still denote by $(u_{n_k})$) that $u_{n_k} \rightarrow u_0$ almost everywhere on $\Omega$?

According to the Rellich theorem, for $\omega \Subset \Omega$ (compactly embedded), the embedding $$H_0^1(\omega) \subset L^p(\omega)$$ is compact for all $1 \leq p < 2^*$. Hence, we may assume that $u_n \rightarrow u_0$ in $L^p(\omega)$, with $1 \leq p < 2^*$. Thus, by some theorem, there exists a subsequence (still denoted $(u_n)$) such that $u_n \rightarrow u_0$ almost everywhere on $\Omega$. Can we (as proposed here) exhaust $\Omega$ by balls of increasing radius and conclude that $u_n \rightarrow u_0$ almost everywhere on $\Omega$?

Thanks in advance.

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  • $\begingroup$ Could you post your version of Rellich's theorem? There is a version that states that bounded sequences in $H^1_0(\Omega)$ converge strongly in $L^2(\Omega)$. Thus up to a subsequence pointwise a.e. $\endgroup$ – Quickbeam2k1 Nov 8 '14 at 19:38
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    $\begingroup$ Rellich theorem: If $\vert \Omega \vert < \infty$, the embeddings $H_0^1(\Omega) \subset L^p(\Omega)$, $1 \leq p < 2^*$, are compact. Here, $\Omega$ is a smooth domain with bounded complement, meaning $\vert \Omega \vert = \infty$. Hence the Rellich theorem does not apply directly. $\endgroup$ – Gatz' Nov 8 '14 at 19:50
  • $\begingroup$ What about the assumptions of bounded complement? I think in your example domain one could use the reflection $x/|x|$ (or $x/|x|^2$). Have you tried this? $\endgroup$ – Quickbeam2k1 Nov 9 '14 at 6:56
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Let's consider the following:

Fix the sequence $u_n\to u$ weakly in $H_0^1(\Omega)$ as you suggest in your post. Define $\Omega_k:=\Omega\cap B(0,k)$. Then each $\Omega_k$ is compact and hence we could consider to use Rellich theorem.

But before we do that, let me point out that the boundary $\partial \Omega_k$ may not be smooth, actually it can even be pretty bad, even if $\partial\Omega$ is smooth... So the Rellich theorem won't work here directly.

But now, let me wave my hand and assume that $\partial \Omega_k$ is good enough so that we could use Rellich on it. Take $k=1$, then by Rellich, from the sequence $u_n$ we could extract a future subsequence, call it $u_{n_{(1)}}$, such that $u_{n_{(1)}}\to u$ at least in $L^1(\Omega_1)$ strongly. Hence you know that $u_{n_{(1)}}\to u$ a.e. (subject to a subsequence of course) in $\Omega_1$. Continuous on this method, by diagonalization argument, you could obtain a subsequence, still denote as $u_n$, such that $u_n\to u$ weakly and $u_n\to u$ a.e. on $\Omega$. But be careful that you can not conclude $u_n\to u$ strongly in $L^1(\Omega)$, you could only say that $u_n\to u$ in $L^1_{\text{loc}}(\Omega)$

Having said that, now let's consider what if $\partial \Omega_k$ is not smooth, or, not a extension domain. Here is one thing can probably rescue us. On Leoni's book, exercise 11.17, he gives the following statement:

The space $W^{1,p}(\Omega)$ is compact embedded in $L^q(\Omega)$ when $q<p$ and $\Omega$ is finite.

In our argument, of course $\Omega_k$ is finite for each $k$ and hence we at least can have $H_0^1(\Omega_k)$ is compact embedded in $L^1(\Omega_K)$ and hence we are good.

However, here I must warn you that I only see this statement in Leoni's book but no-where-else. (On Adam's book, it has the inverse statement, i.e., compact embedding implies that $\Omega$ is finite.) I also receive concern from my friend that this exercise is wrong but they are not sure... Anyhow, at least I have a prove of this exercise by myself but maybe you could work out a proof yourself too...

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  • $\begingroup$ In the book of Alt (Funktionalanalysis, german) no boundary regularity is needed to obtain the compact embedding $H^1_0(\Omega)\to L^2(\Omega)$ (as long as $\Omega$ is bounded). Without the subscript zero Lipschitz regularity is needed. However, there $H^1_0(\Omega)$ is defined as the closure of $C_c^\infty(\Omega)$ with respect to the $H^1$ norm. $\endgroup$ – Quickbeam2k1 Nov 9 '14 at 6:26
  • $\begingroup$ What about the assumption of bounded complement? This is not used in your approach. Any idea on how to use this? If the doamin is really $\mathbb{R}^n\setminus \overline B_1(0)$. A reflection might help $\endgroup$ – Quickbeam2k1 Nov 9 '14 at 6:53
  • $\begingroup$ I confirm: If $N > 2$, the embeddings $H^1(\Omega) \hookrightarrow L^p(\Omega)$ for $1 \leq p < 2^*$ are compact provided that $\Omega \subset \mathbb{R}^N$ is a bounded domain with Lipschitz boundary. The embeddings are still compact for $H_0^1(\Omega)$-spaces on arbitrary bounded domains $\Omega$. This can be found in Adams' book. The geometry of $\Omega$ should not matter. Hence we have no guarantee that the $\Omega_k$ will have Lipschitz boundary. Anyway, thanks to both of you for your help. $\endgroup$ – Gatz' Nov 9 '14 at 11:52
  • $\begingroup$ We could extend the bounded sequence $(u_n) \subset H_0^1(\Omega)$ by zero on the bounded complement of $\Omega$. Let $(\widetilde{u}_n) \subset H^1(\mathbb{R}^N)$ be such an extension. How can we be sure that if $u_n \rightharpoonup u_0$ in $H^1_0(\Omega)$, then $\widetilde{u}_n \rightarrow \widetilde{u}_0$ almost everywhere (hence $u_n \rightarrow u_0$ a.e. on $\Omega$), where $\widetilde{u}_0$ is the extension by zero of $u_0$ ? $\endgroup$ – Gatz' Nov 9 '14 at 12:17
  • $\begingroup$ Sure, $W_0^{1,p}(\Omega)$ will be compact embedded in $L^{p^*}(\Omega)$ as long as $\Omega$ is bounded, but no boundary condition is needed whatsoever. However, in your setting that $u\in H_0^1(\Omega)$ only, so for $\Omega_k\subset \Omega$, we never know $u\in H_0^1(\Omega_k)$ or not, it probably in $H^1(\Omega_k)$. Hence, boundary of $\Omega_k$ is still important. But, as I said in my answer, we only need $L^1$ convergence to obtain a.e. convergence, so my prove should work. $\endgroup$ – spatially Nov 9 '14 at 15:36

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