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Let $\Omega\subset\mathbb{R}^n$ a open bounded set. The Dirichlet laplacian can be defined via it's closed semi-bounded form on $H^1_0(\Omega)$. The fact that it's spectrum is discrete is as far as I can tell proven by that fact that the embedding $H_0^1(\Omega)\rightarrow L^2(\Omega)$ is compact and that the spectrum is discrete if and only if the embedding $H_0^1(\Omega)=(D(q),\lVert\cdot\lVert_q)\rightarrow L^2(\Omega)$ is compact. Where $q$ is the associated form and $D(q)$ is the form domain.

I search for quite some time but didn't find a proof for the second claim. I'd appreciate hints on the proof itself and references very much!

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Where did you find this equivalence: spectrum is discrete if and only if the embedding $H_0^1(\Omega)=(D(q),\lVert\cdot\lVert_q)\rightarrow L^2(\Omega)$ is compact –  Tomás Dec 29 '12 at 1:00
    
I never saw the second claim phrased as "if and only if". The embedding of $H_0^1(\Omega)$ into $L^2(\Omega)$ is compact for any bounded domain. An accessible proof that the spectrum of the Dirichlet Laplacian is discrete (on any bounded domain) can be found in the book Spectral theory and differential operators by E. B. Davies. –  user53153 Dec 29 '12 at 5:43
    
the if and only if is claimed in "Schmüdgen, Unbounded Operators on Hilbert Space, Chapter 10, Prop. 10.6" –  Julian Dec 29 '12 at 7:43
    
There is a demonstration there of this result. Did you understood it? –  Tomás Dec 29 '12 at 12:23

1 Answer 1

up vote 1 down vote accepted

I know how to prove one part of the assertion. Let $f\in L^2$ and consider the problem $$ \left\{ \begin{array}{rl} -\Delta u=f &\mbox{in $\Omega$} \\ u=0 &\mbox{in $\partial\Omega$ } \end{array} \right. $$

We know that for each $f$ there exist a unique weak solution $u\in H_0^1$ of the previous problem, i.e. $$\int_\Omega \nabla u\nabla v=\int_\Omega fv,\ \forall\ v\in H_0^1$$

Define $T:L^2\rightarrow H_0^1$ by $Tf=u$, where $u$ is the weak solution. Moreover, you can conclude from the characterization of weak solution that $\|u\|_{H_0^1}\leq C\|f\|_2$ and $T$ is self-adjoint. Because $H_0^1$ is compactly embedded in $L^2$, you have that $T:L^2\rightarrow L^2$ is a self-adjoint compact operator. Now you can conclude.

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