Properties about the Laplace Operator

Question

Let $\Omega \subset \mathbb{R}^n$ open and bounded. Consider the operator $$A:=-\Delta: H^2(\Omega) \cap H_0^1(\Omega) \longrightarrow L^2(\Omega)$$ where $\Delta$ is the Laplace operator in $\Omega$ with the Dirichlet boundary conditions. I know that $A$ is linear, self-adjoint and a positive operator densely defined on $L^2(\Omega)$.

Question 1. The operator $A$ is invertible?

I think so, because of the Lax-Milgram Lemma. In addition, I think that the inverse is a compact operator. (Right?)

Question 2. If $ A $ is invertible, it is possible explicit the inverse $ A ^ {- 1} $ of $A$? Moreover, it's possible also explicit $A^{-1/2}$?

I confess that for me it is not clear the real meaning of $A^{-1/2}$.

Answer 1

Question 1. Yes, it is invertible by Lax-Milgram theorem, and compact by Rellich theorem.

Question 2. Yes, you can prove the existence of a hilbertian basis of $L^2(\Omega)$ constructed from eigenfunctions of $A^{-1}$. For $B=A^{-1}$, you can define the square root of $B$ (generally the fractional powers) by: $$D(B^{1/2}):=\{f\in L^2(\Omega) : \sum_{n=1}^{\infty} \lambda_n (f|e_n)^2_{L^2} <\infty\},$$ $$B^{1/2} f = \sum_{n=1}^{\infty} \lambda_n^{1/2} (f|e_n)_{L^2}e_n,$$ where $(e_n)$ is a basis of eigenfunctions of $B$ associated to eigenvalues $(\lambda_n)$.

Answer 2

I only address Q1:

First of all, Lax-Milgram cannot be used, since this theorem only applies to operators from $H$ to $H^*$, where $H$ is a Hilbert space. (Or, equivalently, to bilinear forms on a Hilbert space)

The invertibility of $A$ depends on the regularity of the domain $\Omega$. E.g. if the boundary is smooth or if $\Omega \subset \mathbb R^2$ is a convex polyhedron, then $A$ is invertible.

On other domains, $A$ might not be invertible, see, e.g., https://mathoverflow.net/a/38803.

Finally, if $A$ is invertible, the inverse $A^{-1}$ can never be compact as an operator from $L^2$ to $H^2 \cap H_0^1$. (Compactness of an invertible operator implies that the spaces are finite dimensional). However, if you embed the image of $A^{-1}$ into $L^2$ (or $H_0^1$), you get an operator $B \colon L^2 \to L^2$ and this operator is compact since it is the concatenation of a bounded operator $A^{-1}$ and a compact operator (the embedding).