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Let $\Omega \subset \mathbb{R}^d$ be a domain of finite Lebesgue measure, not assumed to be smooth or bounded. Is it true that the embedding of, say, $W^{1,p}_0(\Omega)$ (Sobolev functions with zero boundary value) into $L^q(\Omega)$ is compact for $1/q > 1/p - 1/d$?

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There are necessary and sufficient conditions on $\Omega$ for the compactness of such embedding. Check Mazy'a's book Sobolev Spaces, Springer Verlag, 2011, Section 5.5.2. –  Liviu Nicolaescu Sep 28 '12 at 20:57
    
Since you want this only for functions with compact support, the embedding theorem follows directly from the one for functions on $\mathbb{R}^n$. –  Deane Yang Mar 3 '13 at 23:04
    
@Deane, I am not sure I follow your argument here. There is no compact embedding for functions on $\mathbb{R}^n$. Could you axpand what do you mean here, I am sure I misunderstand something. –  András Bátkai Mar 4 '13 at 9:01
    
@Andras There is no compact embedding on $\mathbb{R}^N$ but since the functions have zero boundary value they can be extended to a large ball where the embedding will hold, and therefore the restriction to $\Omega$ also embeds into $L^p$ compactly for $p<p^*$, as holds for smooth domains. –  Daniel Spector Mar 4 '13 at 11:40
    
@Daniel, what disturbs me in this argument is only that the ball depends on the functions. –  András Bátkai Mar 4 '13 at 16:56

1 Answer 1

You can find a rather general answer in the book of Adams and Fournier, Theorem 6.16. You do have the embedding you wish, under a mild measure theoretical assumption.

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