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Regarding the Dirichlet Laplacian operator $L^{(D)}f=-\frac{d^2}{dx^2}f$ on the set $D(L^{(D)})=\{f\in C^2([a,b]):f(a)=f(b)=0 \}$ on the Hilbert space $L^2(a,b)$ one can show that it is essentially self-adjoint by showing that the family of vectors $e_n(x)=\left(\frac{2}{b-a}\right)^{\frac{1}{2}}\sin(\frac{\pi n (x-a)}{b-a})$ is an orthonormal basis consisting of eigenvectors for $L^{(D)}$. The orthonormality and the eigenvector property is not difficult to show but how do I show that $(e_n)_{n\in\mathbb{N}}$ is a basis for $D(L^{(D)})$? I could assume that there is another function, say $e_\infty\in D(L^{(D)})$ that is orthonormal to all the other $e_n$ and show that a contradiction arises. But how do I tackle such a problem?

EDIT: $(e_n)$ shall even be a basis for $L^2(a,b)$

mathemagician99
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    One viewpoint would be that you need to know that the resolvent of the Laplacian (restricted to your space...) is compact... – paul garrett Jun 13 '20 at 20:15
  • @paulgarrett And then applying some spectral theorem? – mathemagician99 Jun 13 '20 at 20:21
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    Oh, yes, exactly, apply the spectral theorem for compact self-adjoint operators. – paul garrett Jun 13 '20 at 20:40
  • @paulgarrett The problem I see there is that I would have to use it on $D(L^{(D)})$ since $L^{(D)}$ is only defined there. The spectral theorem would yield that $D(L^{(D)})=\text{ker}(L^{(D)})\oplus_2 {e_n:n\in\mathbb{N}}$ and I'd have to deal with $\text{ker}(L^{(D)})$ and also I don't know if $e_n$ are all of the eigenvalues. There might be missing some. Furthermore the spectral theorem also yields that the limit of the eigenvalues tends to $0$ which is not the case here. – mathemagician99 Jun 13 '20 at 20:54
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    Ah, not applied to the unbounded operator, but to its resolvent. Its resolvent is a continuous operator, defined everywhere, and so on... – paul garrett Jun 13 '20 at 20:55
  • @paulgarrett So you mean, applying the spectral theorem to $L^{(D)}-\lambda Id$? But for which eigenvalues? – mathemagician99 Jun 13 '20 at 20:57
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    If/when $(L-\lambda)^{-1}$ exists and is compact, then it also exists and is compact for $\lambda\in\mathbb C$ off a discrete subset. Yes, this is a little theorem. This is pretty standard, e.g., in Kato's "Perturbation Theory", or substantial books on general functional analysis and operator theory. – paul garrett Jun 13 '20 at 20:59
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    Must you use those sines as your basis? Note that $\left<L^{(D)}\phi,\psi\right> = \left<\phi,L^{(D)}\psi\right> - \left.\phi'(x)\overline{\psi}(x) - \psi(x)\overline{\psi}'(x)\right|_{a}^{b}$, so one could choose $\phi(a) = \phi(b)$ and $\phi'(a) = \phi'(b)$ as BCs. Those BCs permit using the complex exponentials $\exp(2\pi i n(x - a)/(b-a))$ as eigenfunctions, and classical results ensure that they form a complete set for $L^2(a,b)$. – Joe Mack Jun 13 '20 at 23:35
  • @JoeMac Yes, I would like to use the functions I've given. However those exponentials are very closely linked to the sines, aren't they? – mathemagician99 Jun 13 '20 at 23:39
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    @mathemagician99 There is a relation, but not a simple one. Note the $2\pi n$ in my exponents and $n\pi$ in your sine arguments. From the point-of-view of complex exponentials or the alternative sines-and-cosines basis ($\sin(2\pi n(x-a)/(b-a), \cos(2\pi n(x-a)/(b-a))$), you have no cosines and twice the usual number of sines. Requiring both endpoint values to be 0 seems strict, but I am not sure that it is. – Joe Mack Jun 13 '20 at 23:43

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Prepare for a reference to established theorems instead of a nitty-gritty proof.

Your equation is just about the simplest possible Sturm-Liouville problem with separated boundary conditions. S-L theory ensures that the normalized eigenfunctions form an orthonormal basis for $L^2(a,b)$
Joe Mack
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