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user 32507
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$H_0^1(\Omega, D) \hookrightarrow L^2(D)$ is compact, for $\Omega$ quasi-open in $D$ - Proof...
You have that $H_0^1(\Omega, D)$ is a subspace of $H^1(D)$ (equipped with the same norm). Due to the Lipschitz condition on $D$, $H^1(D)$ is compactly embedded in $L^2(D)$ (standard-Rellich-Kondrachov …
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