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Let $\Omega$ be a domain in $\mathbb{R}^d$ (not necessarily bounded, no regularity assumption) and $K \subset \Omega$ a compact.

Is it true that the embedding $H^1_0(\Omega) \rightarrow L^2_K(\Omega)$ is compact, $L^2_K(\Omega)$ being the set of $L^2$ functions that are supported in $K$?

Thanks for any response.

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    $\begingroup$ if you mean the map $u\mapsto u_{|K}$ (and $d>2$) yes, as e.g. it factors through the usual Rellich-Kondrachov. $\endgroup$ Aug 2, 2014 at 13:05
  • $\begingroup$ I do mean the restriction map but I don't understand your argument. The rellich theorem says $H^1_0(\Omega) \rightarrow L^2(\Omega)$ is compact, if $\Omega$ is bounded. How do you factorize the mapping ? $\endgroup$
    – incas
    Aug 2, 2014 at 16:02
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    $\begingroup$ Do you mean, I take $\Omega'$ bounded and containing $K$ and associate to any $v \in H^1_0(\Omega)$ some $v'$ in $H^1_0(\Omega')$ that equals $v$ in $K$ by multiplying by a bump function for example, and then consider the embedding $H^1_0(\Omega') \rightarrow L^2(\Omega') \rightarrow L^2(K)$ $\endgroup$
    – incas
    Aug 2, 2014 at 16:10
  • $\begingroup$ Yes, sorry, that was nonsense. What you suggest looks fine, and you don't even need your function to be zero on the boundary (of $\Omega'$). $\endgroup$ Aug 2, 2014 at 17:17
  • $\begingroup$ exactly $\phantom{.} $ ;) $\endgroup$ Aug 2, 2014 at 18:26

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