The eigenfunctions of $\Delta \colon H_0^1(\Omega)\cap H^2(\Omega) \to L^2(\Omega)$ form an orthonormal basis? Let $\Omega$ be a bounded open subset of $\mathbb{R}^n$ with smooth boundary. Consider the Dirichlet Laplace operator
$$\begin{cases}
D(A)=H^1_0(\Omega)\cap H^2(\Omega)\\
Au=\Delta u.
\end{cases}
$$
 I usually see in some references that the eigenfunctions of this operator are an orthonormal basis for the Hilbert space $L^2(\Omega)$. It is stated in Wikipedia that this result follows from the spectral theorem on compact self-adjoint operators, applied to the inverse of the Laplacian (which is compact, by the Poincaré inequality and Rellich–Kondrachov theorem).
I have some questions about this:
1) first why the operator $A$ is invertible
2) why the inverse is a bounded operator. (my guess is that the boundedness maybe follows from the closed graph theorem, $A$ closed operator implies $A^{-1}$ is also closed? I am I right?) however I didn't prove that $A$ is a closed operator, I just have a feeling it is closed.
3) I don't see how the compactness of the operator $A^{-1}$ follows from the Poincaré inequality and Rellich–Kondrachov.
 A: We can prove the existence and boundedness of the inverse Laplacian using the Riesz representation theorem for Hilbert spaces. First, let us define the bilinear form $B[ \ , \ ]$ on $H_0^1(\Omega)$ as follows:
$$ B[u, v] = \int_{\Omega} \sum_i \partial_{x_i} u \partial_{x_i}v.$$
It is possible to prove a couple of inequalities:


*

*$|B[u,u]| \leq || u ||^2_{H_0^1(\Omega)} $

*$|| u ||_{H_0^1(\Omega)}^2 \leq c B[u, u]$ for a suitably chosen value of $c$.
The first inequality is obvious. The second isn't much harder - it just requires some fiddling around with the Poincare inequality.
These inequalities tell us that the norm associated to the inner product $B[ \ , \ ]$ is equivalent to the original Sobolev norm $|| . ||_{H_0(\Omega)}^1$. Therefore, since $H_0^1(\Omega)$ is complete with respect to the Sobolev norm, it must also be complete with respect to $B[\ , \ ].\ $ As a consequence, we can legitimately apply the Riesz representation theorem in $H_0^1(\Omega)$ using $B[ \ , \ ]$ instead of $( \ , \ )_{H_0^1(\Omega)}$ as our inner product.

Now let's use the Riesz representation theorem in this way to deduce that the Laplacian operator has a bounded inverse. To be more precise, we want to show that, for every $g \in L^2(\Omega)$, there exists a unique $u_g \in H_0^1(\Omega)$ such that
$$ B[u_g, v ] = \int_{\Omega} g v  \ \ \ \ \ \ \ \ \ {\rm for \ all \ } v \in H_0^1(\Omega) $$
and moreover, the mapping
$$ g \mapsto u_g $$
is a bounded linear map from $L^2(\Omega)$ to $H_0^1(\Omega)$.
This conclusion does indeed follow from the Riesz representation theorem, and the way to apply the Riesz representation theorem here is to think of $v \mapsto \int_\Omega g v $ as a linear functional on $H_0^1(\Omega)$ whose norm is no greater than $|| g ||_{L^2(\Omega)}$.
Notice that the $u_g$ that we have constructed is a solution to the equation $ - \nabla^2 u = g$ in the weak sense. So if we are content to use the notation $\mathcal L$ for the $ - \nabla^2$ operator, then we may as well use the notation $\mathcal L^{-1}$ for our newly constructed bounded operator $L^2(\Omega) \to H_0^1(\Omega)$ sending $g \mapsto u_g$.

Having defined our bounded inverse operator $\mathcal L^{-1} : L^2(\Omega) \to H_0^1(\Omega)$, I'll now discuss eigenfunctions. This is where we get to apply the Rellich-Kondrachov theorem and the spectral theorem for compact operators.
Let us define a weak eigenfunction of the Laplacian $\mathcal L$ (corresponding to the eigenvalue $k$) to be a $u \in H_0^1(\Omega)$ such that
$$ B[u, v] = k  \int_\Omega u v \ \ \ \ \ \ \ \ \ {\rm for \ all \ } v \in H_0^1(\Omega)  $$
We can immediately rephrase this definition in terms of our inverse operator $\mathcal L^{-1}$: A function $u \in H_0^1(\Omega)$ is a weak eigenfunction of $\mathcal L$ with eigenvalue $k$ if and only if
$$ u = k \left( \mathcal L^{-1} (u)\right),$$
Notice that the $u$ on the right-hand side of this equation is thought of as an element of $L^2(\Omega)$ whereas the $u$ on the left-hand side is thought of as an element of $H_0^1(\Omega)$. This makes sense, because $H_0^1(\Omega) \subset L^2(\Omega)$
I'm now going to massage this definition into a form that we can apply the spectral theorem to. Let's use the symbol $\iota$ to denote the inclusion $H_0^1(\Omega) \hookrightarrow L^2(\Omega)$. If you think about it, the previous paragraph can be written like this: A function $u \in L^2(\Omega)$ is a weak eigenfunction of $\mathcal L$ (and, in particular, is contained within the subspace  $H_0^1(\Omega) \subset L^2(\Omega)$) iff it satisfies
$$ u = k \left( (\iota \circ \mathcal L^{-1}) (u)\right).$$
But $\iota : H_0^1(\Omega) \hookrightarrow L^2(\Omega)$ is a compact operator by Rellich, and $\mathcal L^{-1} : L^2(\Omega) \to H_0^1(\Omega)$ is a bounded operator, so the composition $$\iota \circ \mathcal L^{-1} : L^2(\Omega) \to L^2(\Omega)$$ is compact. The composition $\iota \circ \mathcal L^{-1}$ is also self-adjoint (and to check this, it suffices to verify self-adjointness on $C_{c}^\infty(\Omega)$, which is dense in $L^2(\Omega)$ and is contained inside $H_0^1(\Omega)$).
We can therefore legitimately apply the spectral theorem to $\iota \circ \mathcal L^{-1}$. This tells us that the weak eigenfunctions of $\mathcal L$ form a complete (countable) orthogonal basis for the orthogonal complement of the kernel of $\iota \circ \mathcal L^{-1}$ within $L^2(\Omega)$. But the kernel of $\iota \circ \mathcal L^{-1}$ is zero (since any $u$ in this kernel obeys $\int_\Omega uv = 0$ for every $v \in H_0^1(\Omega)$, and in particular, for every $v \in C_c^\infty(\Omega)$). The conclusion then is that the weak eigenfunctions of $\mathcal L$  form a complete (countable) orthogonal basis for $L^2(\Omega)$.

At the moment, these eigenfunctions are only weak eigenfunctions, living in $L^2(\Omega)$, and obeying only the weak condition $B[u, v] = k [u , v]$ for $v \in H_0^1(\Omega)$. It would be nice if we could show that these eigenfunctions are genuine smooth functions in $C^\infty(\Omega)$ obeying $\mathcal L u = k u$! 
We can prove this as follows: By a regularity theorem in Evans Chapter 6.3, any weak solution $u$ to $\mathcal L u = f$, where $f$ is an element of $H^m(\Omega)$, must automatically be in $H_{\rm loc}^{m + 2}(\Omega)$. In our case, $u$ is a weak solution to $\mathcal L u = k u$. So the fact that $u$ is in $L^2(\Omega)$ implies that $u$ is in $H_{\rm loc}^2(\Omega)$, which in turn implies that $u$ is in $H_{\rm loc}^4(\Omega)$, which in turn implies that $u$ is in $H_{\rm loc}^6(\Omega)$, etc. Thus, $u$ is in $H_{\rm loc}^m(\Omega)$ for all $m$. But then, by the Sobolev inequalities, $u $ must also be in $C^\infty(\Omega)$, and we are done.
With further conditions on the smoothness of $\mathcal \Omega$, I believe it it is possible to prove that $u$ is also in $C^\infty_c(\bar \Omega)$, where $\bar \Omega$ is the closure of $\Omega$. (See Evans 6.3.) It then makes sense to evaluate $u$ on the boundary $\partial \Omega$, and the fact that $u$ is in $ H_0^1(\Omega)$ rather than just $H^1(\Omega)$ ensures that $u|_{\partial \Omega} = 0$ (see Evans 5.5), which is to say that $u$ satisfies the Dirichlet boundary condition. 
A: *

*Notice that $A$ is the Riesz isomorphism between $H_0^1$ and $H^{-1}$ as it is defined via 
$$\langle Au,v \rangle_{H^{-1} \times H_0^1} = (u,v)_{H_0^1}$$
and therefore its inverse exists and is continuous (hence bounded).

*We have $$L^2 \xrightarrow{\,J\,} H^{-1} \xrightarrow{A^{-1}} H_0^1 \xrightarrow{\text{Id}} L^2$$
where $J:L^2 \to H^{-1}$ such that $\langle Jh,v\rangle_{H^{-1} \times H_0^1}=(h,v)_{L^2}.$ And thus we consider the mapping $$\text{Id} \circ A^{-1} \circ J:L^2 \to L^2$$ which is mostly denoted just by $A^{-1}$. Notice that $J$ is the composition of the Riesz isomorphism of $L^2$ and the restriction mapping from $(L^2)' \to H^{-1}$. The mapping $\text{Id} \circ A^{-1} \circ J$ is compact as it is the composition of a compact operator ($\text{Id}$ by Rellich) and a continuous operator ($A^{-1}J$). Self-adjointness follows by Lax-Milgram's lemma since for all $f \in H^{-1}$ there is a unique $u \in H_0^1$ such that for all $v \in H_0^1$
$$(u,v)_{H_0^1}=(f,v)_{L^2}=\langle Jf,v \rangle_{H^{-1} \times H_0^1}=\langle AA^{-1}Jf,v \rangle_{H^{-1} \times H_0^1}=(A^{-1}Jf,v)_{H_0^1}$$
and taking the test function $v=A^{-1}Jg$ for a $g \in L^2$ yields
$$(f,A^{-1} Jg)_{L^2}=(A^{-1}Jf,A^{-1} Jg)_{H_0^1}=\langle AA^{-1} Jg,A^{-1}Jf \rangle_{H^{-1} \times H_0^1}=(g,A^{-1}Jf)_{L^2}.$$
Finally, the spectral theorem yields an orthonormal basis in $L^2$. 
