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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 $$Au:=-\Delta u\;\;\;\text{for }u\in\mathcal D(A)$$

$(\mathcal D(A),A)$ is a linear symmetric operator on $L^2(\Lambda)$. Since $$\langle u,Au\rangle_{L^2(\Lambda)}=\left\|\nabla u\right\|_{L^2(\Lambda)}^2>0\;\;\;\text{ for all }u\in\mathcal D(A)\setminus\left\{0\right\}\;,\tag1$$ there is a unique linear operator $(\mathcal R(A),A^{-1})$ on $L^2(\Lambda)$ with $$\mathcal R(A):=\left\{Au:u\in\mathcal D(A)\right\}$$ and $$Au=v\Leftrightarrow u=A^{-1}v\;\;\;\text{for all }u\in\mathcal D(A)\text{and }v\in\mathcal R(A)\;.\tag2$$

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 $$A^{-1}e_n=\lambda_ne_n\tag3\;\;\;\text{or all }n\in\mathbb N$$ for some $(\lambda_n)_{n\in\mathbb N}\subseteq(0,\infty)$ with $$\lambda_{n+1}\le\lambda_n\;\;\;\text{for all }n\in\mathbb N\tag4\;.$$ Is it possible to do so by the Hilbert-Schmidt theorem?

We can show that $$\left\|u\right\|_{L^2(\Lambda)}\le C\left\|Au\right\|_{L^2(\Lambda)}\;\;\;\text{for all }u\in\mathcal D(A)\tag5$$ for some $C\ge 0$ and hence $$\left\|A^{-1}v\right\|_{L^2(\Lambda)}\le C\left\|v\right\|_{L^2(\Lambda)}\tag6\;,$$ i.e. $(\mathcal R(A),A^{-1})$ is a bounded operator on $L^2(\Lambda)$. Thus, there is a bounded linear operator $\tilde A^{-1}$ on $L^2(\Lambda)$ with $$\tilde A^{-1}v=A^{-1}v\;\;\;\text{for all }v\in\mathcal R(A)\tag 7\;.$$ Now, by the Rellich–Kondrachov theorem the inclusion $\iota$ from $H_0^1(\Lambda)$ into $L^2(\Lambda)$ is compact and hence $\tilde A^{-1}\iota$ is compact; but that's of no use here, is it?

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    The existence of the inverse puts $0$ in the resolvent set of $A$. That and positive definiteness of $A$ give you selfadjointness of $A$. The resolvent is compact. That gives you a complete orthonormal basis of eigenvectors for the resolvent, which must also be eigenvectors of $A$. – 2017-01-07
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    @TrialAndError (a) My definition of $0$ being in the resolvent set of $A$ requires that $\mathcal R(A)$ is dense in $L^2(\Lambda)$. Is that the case? (b) And that the resolvent $A^{-1}$ is compact is one of the things I'm trying to prove in the question. How can we prove it? – 2017-01-08
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    For your domain, do you mean $H^2\cap H^1_0$? In that case, $A$ is the Friedrichs extension, which is selfadjoint. So the range is dense because it has trivial null space. – 2017-01-08
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    @TrialAndError I don't understand your last comment. Is it a reply to (a) or (b)? The domain of $A$ is as defined in the question, $$\mathcal D(A):=\left\{u\in H_0^1(\Lambda):\Delta u\in L^2(\Lambda)\right\}\;.$$ $H^2(\Lambda)\cap H_0^1(\Lambda)$ is a proper subset of $\mathcal D(A)$, unless some regularity of $\Lambda$ is assumed. – 2017-01-08
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    How are you defining $\nabla^2$ on $H^1$? This is problematic in the context of the Hilbert space, but you may not be only in the Hilbert space $L^2$. The natural selfadjoint extension of $\Delta$ is on $\mathcal{D}=H^2\cap H_0^1$ ($\Delta : \mathcal{D}\subset L^2 \rightarrow L^2$. – 2017-01-08

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