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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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    There is a demonstration there of this result. Did you understood it?2012-12-29

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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.