• $d\in\mathbb N$
  • $\Lambda\subseteq\mathbb R^d$ 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?

  • $\begingroup$ 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

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 $\lambda_n=\mu_n^{-1}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.