Proving that the eigenfunctions of the Laplacian form a basis of $L^2(\Omega)$ (and of $H_0^1(\Omega)$)

Question

I am studying the eigenfunctions and eigenvalues of the Laplacian on an open, bounded domain $\Omega \subset \mathbb{R}^n$ with homogeneous Dirichlet boundary conditions. I have read about the the weak and variational formulation of the problem. I understand the result that the first eigenvalue is given by: $$ \lambda_1 = \inf_{H_0^1(\Omega)} R = \inf_{v \in H_0^1(\Omega)} \frac{\int_\Omega \| \nabla v\|^2 \, \mathrm{d}x}{\int_\Omega v^2 \, \mathrm{d}x}$$ and the associated eigenfunction is $u_1$ such that $R(u_1) = \lambda_1$, as well as the characterization of the nth eigenvalue/eigenfunction. I have proved some of the basic results, such as the fact that the sequence of eigenvalues is unbounded and eigenfunctions associated to different eigenvalues are orthogonal (both in $L^2(\Omega)$ and $H_0^1(\Omega)$).

Now I am trying to prove that the eigenfunctions $u_1,u_2,\ldots$ form a basis of $L^2(\Omega)$. I have seen some proofs of this fact (e.g. Jost's book on PDEs and Mihai Nica's article), but I am trying to use a different approach. I have a sketch of this proof but I need to fill in some details.

The proof argues by contradiction. We define $V$ to be the closure in $H_0^1(\Omega)$ of the set: $$\left\{ u \in H_0^1(\Omega) : \exists \, N \in \mathbb{N}: u = \sum_{n=1}^N\alpha_nu_n\right\},$$ where $\alpha_n \in \mathbb{R}$ and the $u_n$'s are the eigenfunctions. Then $V$ is a closed subspace of $H_0^1(\Omega)$. We assume, for the sake of contradiction, that $V \neq H_0^1(\Omega)$.

1) This should then imply that $V^\perp \neq \{0\}$, but I am not sure why. I know that it is not necessary for a subset of a Hilbert space $H$ to be equal to the whole space $H$ for its orthogonal complement to be trivial. However here the space is closed so it is possible that this may force the orthogonal complement to be trivial, but I am not sure if that is enough.

Then $V^\perp$ is a non-trivial closed subspace of $H_0^1(\Omega)$ and hence we can apply the same methods used to determine the existence of eigenvalues and eigenfunctions to deduce the existence of an eigenfunction $u$ in $V^\perp$. It can then be shown that the eigenvalue associated to this eigenfunction must equal one of the previously determined eigenvalues (I understand how to do this).

2) Then I am not sure why this leads us to a contradiction. My guess is that then this $u\in V^\perp$ should equal some $u_n \in V$ (it is obvious that $V$ contains all the $u_n$'s) and this would imply that $V\cap V^\perp \neq \varnothing$, which is impossible.

3) This would then prove that $V = H_0^1(\Omega)$ but I have some doubts/confusion as to why this shows that the $u_n$'s form a basis of $H_0^1(\Omega)$, if they do.

4) Then I am not sure how to extend this to $L^2(\Omega)$. I suspect that the fact that $H_0^1(\Omega)$ is dense in $L^2(\Omega)$ should play a role.

I would be very grateful for any help (and/or references) on the numbered points.

Answer 1

This post was too long for a comment, so I write it here and I hope it will be useful.

$1)$ You can write $H^1_0(\Omega)=V\oplus V^{\perp}$ because $V$ is closed (it is a general fact for closed sets in Hilbert spaces), so $V=H^1_0(\Omega)$ iff $V^\perp=\{0\}$.

$2)$ Now if $u=u_n$, then $u\in V\cap V^\perp=\{0\}$, but $u_n\neq0$ because it is an eigenvector.

$3)$ Next, ${u_n}$ is an orthogonal basis of $H^1_0(\Omega)$ if $u_m\perp u_n$ when $m\neq n$ and if for all $u\in H^1_0(\Omega)$, $u=\sum \alpha_n u_n$ (strong limit).

$4)$ Finally, every $v\in L^2(\Omega)$ is a (strong) limit of elements $v_n$ is $H^1_0(\Omega)$ (by density), hence : $$\forall\varepsilon>0,\exists N\in\mathbb{N};\forall n\geq N,\|v-v_n\|<\varepsilon.$$ Write $v_n=\sum_k\alpha_{k,n}u_{k,n}$ with $\alpha_{k,n}$ some real numbers and $u_{k,n}$ some eigenvectors to conclude.

Answer 2

I know a different approach, and I think it is more "standard." Basically it is sufficient to prove that if $\Omega \subset \mathbb{R}^n$ is limited open, then $(-\Delta)^{-1}$ is a compact and injective self-adjoint operator on $L^2(\Omega)$ and on $H^1_{0}(\Omega)$ and then after applies Hilbert-Schmidt spectral theorem. Recalling a little proof of the Hilbert-Schmidt spectral theorem, I think that what you ask lies precisely in the proof of this theorem. More precisely there is the following theorem

"If $H$ is a real (it is true also in the complex case) Hilbert spaces and $K:H \longrightarrow H$ is a compact adjoint operator, then exists an orthonormal basis of eigenvectors $\lbrace u_n \rbrace$ of $K$ with eigenvalues $\lbrace \lambda_n \rbrace$ and it has the representation

$\displaystyle Kx = \sum_{n \geq 1} \lambda_n (x, u_n)_H u_n$ $(x \in H)$"

Now, the facts you say apply generally to elliptic operators in divergence form, ie type

$\displaystyle Lu:= - \sum_{i,j=1}^n (a_{ij} u_{x_i})_{x_j} + \sum_{i=1}^n (b_i u)_{x_i} +c u$

where $a_{ij}, b_i, c \in L^\infty(\Omega)$, and it assumes that $L$ is uniformly elliptic. Basically the case of the Laplace operator is a particular case of $L$. After we introduce the weak solutions, assume $b_i=c=0$, there is the following theorem

"If $a_{ij}=a_{ji} \in L^\infty(\Omega)$, and considering $L^{-1} : L^2(\Omega) \longrightarrow H^1_{0}(\Omega) \subset L^2(\Omega)$, then $L^{-1}$ is a compact and injective self-adjoint operator. In addition there is an orthonormal basis $\lbrace \phi_k : k \in \mathbb{N} \rbrace$ of $L^2(\Omega)$ of eigenfunctions associated to the eigenvalues of $L^{-1}$ and

$\displaystyle L^{-1}f =\sum_{k=1}^\infty \lambda_k (f, \phi_k)_{L^2} \phi_k$

where $Lu=f$ wih $f \in L^2(\Omega)$. In particular $\lim_{k \rightarrow \infty} \lambda_k =0$".

This whole theory is very well explained in the book "Lecture Notes on Functional Analysis: With Applications to Linear Partial Differential Equations" by A. Bressan.