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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?

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  • $\begingroup$ It is compact because it sends bounded sets in relatively compact sets. In fact it is bounded and there is a Theorem saying that the closed unit ball in $H_0^1$ is compact in $L^2$.The proof uses Ascoli-Arzela compactness. $\endgroup$ – Maffred Jan 6 '17 at 16:34
  • $\begingroup$ In your domain, you have $\Lambda$ instead of $\Delta$. $\endgroup$ – DisintegratingByParts Jan 6 '17 at 16:41
  • $\begingroup$ I think this is what states the Rellich–Kondrachov Theorem. Maybe you can start from here www4.ncsu.edu/~aalexan3/articles/rellich.pdf $\endgroup$ – Maffred Jan 6 '17 at 16:45
  • $\begingroup$ If you can show that a symmetric operator $A : \mathcal{D}(A)\subseteq H\rightarrow H$ is semibounded below and has the property that there exists $f \in H$ such that $\frac{\langle Af,f\rangle}{\langle f,f\rangle}=\inf_{f\ne 0,f\in\mathcal{D}(A)}\frac{\langle Af,f\rangle}{\langle f,f\rangle}$ then $Af=\lambda f$ where $\lambda$ is that minimum. $\endgroup$ – DisintegratingByParts Jan 6 '17 at 16:47
  • $\begingroup$ @TrialAndError Thank you for mentioning the typo. I've corrected it. $\endgroup$ – 0xbadf00d Jan 6 '17 at 19:39
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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 $\lambda_n=\mu_n^{-1}$.

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