$-Δ$ with domain $\{u∈H_0^1:Δu∈L^2\}$ admits an orthonormal basis of $L^2$ consisting of eigenfunctions of $-Δ$ with corresp. positive eigenvalues Let


*

*$d\in\mathbb N$

*$\Lambda\subseteq\mathbb R^d$ be bounded and open

*$\mathcal D(A):=\left\{u\in H_0^1(\Lambda):\Delta u\in L^2(\Lambda)\right\}$ and $$Au:=-\Delta u\;\;\;\text{for }u\in\mathcal D(A)$$


It's easy to see that $(\mathcal D(A),A)$ is a densely-defined linear symmetric operator on $L^2(\Lambda)$. Since $\Lambda$ is bounded, $(\mathcal D(A),A)$ is positive, i.e. $$\langle u,Au\rangle_{L^2(\Lambda)}=\left\|\nabla u\right\|_{L^2(\Lambda)}^2>0\;\;\;\text{ for all }u\in\mathcal D(A)\setminus\left\{0\right\}\;,\tag 1$$ and hence invertible, i.e. there is a unique linear operator $(\mathcal R(A),A^{-1})$ with $$\mathcal R(A):=\left\{Au:u\in\mathcal D(A)\right\}$$ and $$Au=v\Leftrightarrow u=A^{-1}v\;\;\;\text{for all }u\in\mathcal D(A)\text{ and }v\in\mathcal R(A)\;.\tag 2$$

I want to show that there is an orthonormal basis $(e_n)_{n\in\mathbb N}\subseteq\mathcal D(A)$ of $L^2(\Lambda)$ with $$Ae_n=\lambda_ne_n\;\;\;\text{for all }n\in\mathbb N\tag 3$$ for some $(\lambda_n)_{n\in\mathbb N}\subseteq(0,\infty)$ with $$\lambda_{n+1}\ge\lambda_n\text{ for all }n\in\mathbb N\;.\tag 4$$

I'm not sure what the easiest way is to obtain the desired result. Maybe we should use the Hilbert-Schmidt theorem and maybe it's easier to apply it to $(\mathcal R(A),A^{-1})$ instead of $(\mathcal D(A),A)$. If that's a good idea, we need to show that the corresponding operator is compact. In that case: How can we do that?
 A: Define $B:D(B)\to L^2$ by $Bu=-\Delta u$, where $D(B)=H_0^1\cap H^2$.
Due to elliptic regularity, $B$ is bijective and thus it has an inverse $B^{-1}: L^2\to D(B)$.
By the Rellich Kondrachov Theorem, $\iota B^{-1}:L^2\to L^2$ is compact where $\iota$ is the inclusion from $D(B)$ to $L^2$.
As $\iota B^{-1}$ is symmetric, it follows from the Spectral Theorem (for self-adjoint compact operators) that $L^2$ has an orthonormal basis $(e_n)_{n\in\mathbb{N}}$ consisting of eigenvectors of $\iota B^{-1}$ whose corresponding eigenvalues $(\mu_n)_{n\in\mathbb N}$ satisfy
$$|\mu_{n+1}|\leq|\mu_n|\neq 0,\quad\forall\ n\in\mathbb N.$$
Note that
$$e_n=\frac{1}{\mu_n}\iota B^{-1}e_n=\frac{1}{\mu_n}B^{-1}e_n\in D(B),\quad\forall\ n\in\mathbb N$$
and
$$\mu_n=\mu_n\langle e_n,e_n\rangle_{L^2}=\langle e_n,B^{-1}e_n\rangle_{L^2}=\langle B^{-1}e_n,BB^{-1}e_n\rangle_{L^2}\geq0,\quad \forall\ n\in\mathbb N.$$
Thus we have proven the following result.

There is an orthonormal basis $(e_n)_{n\in\mathbb N}\subset D(B)$ of $L^2$ with
  $$Be_n=\frac{1}{\mu_n}e_n\;\;\;\text{for all }n\in\mathbb N$$
  for some $(\mu_n)_{n\in\mathbb N}\subset(0,\infty)$ with $$\mu_{n+1}\le\mu_n\text{ for all }n\in\mathbb N.$$

As $D(B)\subset D(A)$ and $A|_{D(B)}=B$, we get the desired result by taking $\lambda_n=\mu_n^{-1}$.
