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

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.

• 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. Mar 21, 2016 at 18:57
• Thank you. I haven't seen that theorem yet but I will definitely have a look at its proof. Mar 21, 2016 at 19:33

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.

• I like this approach! You must be italian, right ? (because of the "dove") :) Mar 21, 2016 at 21:03
• Lol, yes I am italian Mar 21, 2016 at 21:09
• Thank you very much for this. It looks very interesting even though it uses a bit more functional analysis than wat I currently know. Thanks for the reference as well, I will look into it. P.S.: I too am Italian, and I had spotted the "dove"! :) Mar 21, 2016 at 21:59
• Well, a lot of Italians :) I suggest you study the proof of the Hilbert-Schmidt spectral theorem, because your questions are the proof of this theorem. Mar 21, 2016 at 22:05
• Will do. Thanks again. Mar 21, 2016 at 22:26

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.

• Thank you very much for this. I am happy you posted it as an answer since it is exactly what I was looking for. 1) This is clear, I have used this result before. I don't know why I missed it here. Thanks. 2) Again, very easy and clear. Thanks. 3) Mar 21, 2016 at 21:38
• @DemK. Thank you, I'm glad to see that it helped you, but YOU did the big part of the job in your original post! Mar 21, 2016 at 21:41
• @Nicholas, I'm sorry, I pressed enter by mistake. 3) I still have some doubts here. I know the definition of a basis (I didn't include orthogonalityt since I have already proved that elsewhere). Essentially I don't see why $V = H_0^1(\Omega)$ implies that $u_n$'s are a basis. Since V is the closure of the set in the question,can we write: $V = \{u:u=\sum_{n=1}^\infty\alpha_nu_n\}$ ? I am not convinced... Mar 21, 2016 at 21:49
• 4) Again, here I think I get it. Basically by replacing $v_n$ with its eigenfunction expansion (which we have from part 3) we conclude that every $v$ in $L^2(\Omega)$ is the limit of some eigenfunction expansion. Is it ok then to conclude here? Because we have not determined the coefficients in the expansion (I know what they "should" be though) or shown that they are unique. Mar 21, 2016 at 21:52
• @DemK. For $3)$, if $v_n=\sum_{k=1}^{N(n)} \alpha_{k,n}u_{k,n}$ is a sequence that strongly converges to $v$, then $N(n)$ can stay finite (and in this case, $v$ is an algebraic (=finite) linear combination of eigenvectors) or $N(n)\to+\infty$ (and $v\in V$). In any case, $v$ lies in the (topological) span of the eigenvectors. As for $4)$, as long as you have a topological linear combination $v=\sum_{k=1}^{+\infty}\alpha_k u_k$, the coefficients $\alpha_k$ are unique, as for the algebraic case ; in fact, $\alpha_k=\|u_k\|^{-2}\langle u_k,v\rangle$ for all $k\in\mathbb{N}$. Mar 21, 2016 at 22:22