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Let $X$ be some metric measure space on which Sobolev spaces can be defined in a reasonable way. In many cases, $H^1(X)$ is compactly embedded in $L^2(X)$ (e.g., if $X=\Omega$ is a bounded open set of $\mathbb R^d$), and in that case, $H^1_0(X)$ is of course compactly embedded in $L^2(X)$, too. In many cases, on the other hand, $H^1_0(X)$ is not compactly embedded in $L^2(X)$ (e.g., $X=\Omega=\mathbb R^d_+$), let alone $H^1(X)$.

My question is now, whether structures $X$ are known such that the embedding of $H^1_0(X)$ in $L^2(X)$ is compact but that of $H^1(X)$ is not.

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    $\begingroup$ How confident are you that $H^1(\Omega)$ embeds compactly into $L^2(\Omega)$ when $\Omega \subset \mathbf{R}^n$ is bounded? It looks fishy to me when $\Omega$ is not Lipschitz-regular. In fact a sufficiently irregular bounded domain would be my first guess. $\endgroup$
    – Leo Moos
    Commented Jun 17, 2021 at 15:08
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    $\begingroup$ According to Adams-Fournier, domains satisfying the cone condition are good enough. But you are right, I was implicitly assuming some regularity when talking about bounded sets. A counterexample in the class of rough domains would be sufficient for my purposes. $\endgroup$
    – Delio Mugnolo
    Commented Jun 17, 2021 at 15:15
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    $\begingroup$ This seems like a delicate question, since (for a bounded domain) as soon as you have $H^1(\Omega) \subset L^p(\Omega)$ continuously, the embedding $H^1(\Omega) \subset L^2(\Omega)$ will be compact. So my guess would have been a domain with an exponential cusp as in Adams/Fournier 4.47/4.48. $\endgroup$
    – Hannes
    Commented Jun 17, 2021 at 15:30
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    $\begingroup$ In fact, digging in Adams/Fournier, there is an example in section 6.35 for the desired situation: $\Omega$ is the countable union of disjoint balls $B_j$ and $u_j(x) = \chi_{\overline{B_j}}(x) |B_j|^{-1/2}$ is claimed to be bounded in $H^1(\Omega)$ but not precompact in $L^2(\Omega)$. But if the radii $r_j$ go to zero, then $H^1_0(\Omega) \subset L^2(\Omega)$ continuously. $\endgroup$
    – Hannes
    Commented Jun 17, 2021 at 15:33
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    $\begingroup$ (Sorry, I forgot to add the crucial "$p>2$" in my first comment and cannot edit any more..) $\endgroup$
    – Hannes
    Commented Jun 17, 2021 at 16:13

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