Why $-(\Delta)^{-1}$ is compact operator? Assume
$$
A= -(\Delta)^{-1}: L^2(\Omega)\rightarrow L^2(\Omega),
$$
I know $A$ is self-adjiont. Then why the space $H^1_0(\Omega)$ can be compactly imbedded in $L^2(\Omega)$ by means the fact that  $A$ is compact operator ?
From the point of view of notation, by  $H^1_0(\Omega)$ I mean the Sobolev space $\mathring W^{1,2}(\Omega)$.

I suppose that $\Omega$ is a bounded open subset of $\mathbb R^n$ and $\Delta$ is defined as the Friedrichs extension of Laplace operator $\Delta: \mathscr D(\Omega) \rightarrow L_2(\Omega)$. In this case $-\Delta$ is positive densely defined operator with $D_\Delta \subset H^1_0( \Omega)$. Operator $A = \Delta^{-1}$ is defined and is continuous in the following sense: $A: L_2(\Omega) \rightarrow H^1_0(\Omega)$. Now we discuss the compactness of $A:L_2(\Omega) \rightarrow L_2(\Omega)$.
If you assume Rellich theorem ($j:H^1_0(\Omega) \rightarrow L_2(\Omega)$ is a compact embedding), then $A:L_2(\Omega) \rightarrow L_2(\Omega)$ is equal to a composition of continuous operators $L_2(\Omega) \xrightarrow{A} H^1_0(\Omega) \xrightarrow{j} L_2(\Omega)$, where the second operator is compact. Thus $A$ is compact.
If you assume that $A:L_2(\Omega) \rightarrow L_2(\Omega)$ is compact, then $\sqrt{A}:L_2(\Omega) \rightarrow L_2(\Omega)$ is also compact. It is known that $\sqrt{A}$ is an isomorphism betweeen $L_2(\Omega)$ and $H^1_0(\Omega)$. Therefore $(\sqrt{A})^{-1}:H^1_0(\Omega) \rightarrow L_2(\Omega)$ is a continuous operator. Now you can represent the embedding $H^1_0(\Omega) \rightarrow L_2(\Omega)$ as the composition $H^1_0(\Omega) \xrightarrow{(\sqrt{A})^{-1}} L_2(\Omega) \xrightarrow{\sqrt{A}} L_2(\Omega)$, where the second operator is compact. Thus, the embedding is compact.
