Discreteness of eigenvalues for certain operators - can this approach be made rigorous?

Question

I was idly thinking about why one might naïvely expect a discrete spectrum of eigenvalues for a linear operator $L$ when I dreamt up the following argument (which I expect isn't new instead - references?). Can it be made rigorous, with any necessary added restrictions, or does it at least offer some genuine informal insight? (Or did I just mess something up?)

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Suppose $L$ is a linear, self-adjoint operator acting on (sufficiently) continuously differentiable functions $\phi:[0,1]\to\mathbb{R}$ with boundary conditions $\phi(0)=\phi(1)=0$. An eigenfunction $\phi_\lambda(x)$ is a non-zero such function such that

$$\left[L \phi_\lambda\right](x) = \lambda \phi_\lambda(x)$$

We expect that such functions occur only for 'discrete' $\lambda$, in particular not for an interval $(a,b)$ of eigenvalues. Clarification: I'm thinking of $L$ being a 'nice' integral/differential operator in my head. What I'm interested in is what further restrictions are necessary to make the spectrum simple.

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Here's a reason to suspect this is the case: suppose $\phi\equiv \phi(x;\lambda)$ is a (sufficiently) continuously differentiable function $[0,1]\times(a,b)\to\mathbb{R}$ such that $L \phi = \lambda\phi$. Suppose further that $L,\phi$ is sufficiently nicely behaved that $L (\partial_\lambda \phi) \equiv \partial_\lambda(L\phi)$. Note that we expect $\partial_\lambda \phi|_{x=0,1}=0$ since all $\phi$ obey the boundary conditions, so $L$ should be self-adjoint on this function too.

Then, with a simple $L^2$ inner product $\left<\cdot,\cdot\right>$ say,

$$\boxed{\lambda \left<\phi,\partial_\lambda \phi\right> = \left<L\phi,\partial_\lambda \phi\right> = \left<\phi,L\partial_\lambda \phi\right> = \left<\phi,\partial_\lambda L\phi\right> = \left<\phi,\partial_\lambda (\lambda\phi)\right> = \left<\phi,\phi\right> + \lambda \left<\phi,\partial_\lambda \phi\right>}$$

and hence $$\left<\phi,\phi\right> = 0$$

Hence there is no continuously differentiable family of solutions which vary the eigenvalue throughout some interval.

Example By contrast, if we dropped our boundary conditions and worked on the fully infinite domain $x\in(-\infty,\infty)$ the operator $Lf=f''(x)$ acquires all real eigenvalues on the smooth families $\cos kx,\exp kx$. But for s finite interval with homogeneous BCs, we only get e.g. $0,\pi,2\pi,\ldots$

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Any interesting thoughts are welcome! I haven't thought about this in great detail, so it may be that this just leads to a standard spectral theory argument when one tries to make it rigorous, but I'm curious as to whether it has an interesting interpretation, and whether you think it would be a good, informal justification to teach someone first coming to Sturm-Liouville theory.

Mainly, I am interested in whether this can be tweaked to a proof without losing all the simple spirit of the boxed equation above.

Answer 1

Your argument is very interesting. I'm not sure how to best adapt it yet.

The standard arguments would say that eigenfunctions in $L^{2}$ with different eigenvalues are orthogonal. Because $L^{2}[0,1]$ is separable (has a countable orthonormal basis), then there can be at, most, a countably-infinite number of eigenvalues.

For non-singular Sturm-Liouville equations on a finite interval, the classical eigenfunctions are in $L^{2}[0,1]$, but do not always satisfy the correct boundary conditions. However, one can find classical eigenfunctions $\phi_{\lambda}$ which satisfy one of the two endpoint conditions $C(\phi_{\lambda})=0$. By proper normalization, $\phi_{\lambda}$ will be analytic in $\lambda$ and not the trivial zero solution; then the true eigenvalues occur when the second condition $d(\lambda)=D(\phi_{\lambda})=0$ is satisfied. You end up with the eigenfunctions being the zeros of an analytic function $d(\lambda)$. So the eigenvalues are isolated for non-singular problems on finite intervals.

The situation changes for singular Sturm-Liouville problems, where the spectrum may be continuous, and there are no actual eigenvalues because the classical eigenfunctions are not in $L^{2}$, even though "wave packets" of them (resembling expressions such as $\int_{\lambda-\epsilon}^{\lambda+\epsilon}\phi_{\alpha}d\alpha$) are "approximate eigenfunctions" in $L^{2}$. So simple arguments break down (as they must) because the classical $\phi_{\lambda}$ for real $\lambda$ are not necessarily in $L^{2}$, and cannot be used as vectors in an inner-product, except as wave packets.

Answer 2

I'm puzzled by the fact that you are taking derivatives with respect to $\lambda$: is that a typo?

Regarding the "discrete eigenvalues" thing: let $V$ be any infinite-dimensional inner-product space. Let $\{e_n\}$ be a countable orthonormal set, and let $\{q_n\}$ be an enumeration of $\mathbb Q\cap[0,1]$. Define an operator $L$ by $$ Le_n=q_ne_n,\ \ \ L=0\ \mbox{on the orthogonal complement of }\{e_n\}. $$ Then the set of eigenvalues of $L$ is $\mathbb Q\cap[0,1]$.