Let $\Omega$ be a bounded regular open set. Let $\Delta^{-1}:L^2(\Omega)\to H^1_0(\Omega)\cap H^2(\Omega)$ be the (weak) inverse Laplace operator. By the Lax-Milgram theorem, we know that $\Delta^{-1}$ is an isomorphism, and by regularity results we also know that it is continuous. Now if we look at the following operator:
$$T:L^2(\Omega)\overset{\Delta^{-1}}{\longrightarrow} H^1_0(\Omega)\overset{i}{\hookrightarrow} L^2(\Omega), $$
then $T$ is continuous and compact since the Sobolev embedding $i$ is continuous, and it is compact by Rellich-Kondrachov. How can we deduce from this $\Delta^{-1}$ is compact? Or is there another way to show that $\Delta^{-1}$ is compact? It seems I'm missing something obvious here!
Compactness of the inverse Laplacian.
elliptic-equationsfunctional-analysissobolev-spaces
Best Answer
Since $L^2(\Omega)$ is infinite-dimensional, the isomorphism $\Delta^{-1} \colon L^2(\Omega) \to H_0^1(\Omega) \cap H^2(\Omega)$ can never be compact. Indeed, if $\Delta^{-1}$ would be compact, the identity $\operatorname{id} = \Delta \colon \Delta^{-1}$ on $L^2(\Omega)$ would be compact.