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Take $T$ to be the inverse of a bounded/continuous, self-adjoint operator with eigenvalues (an orthonormal basis) all rationals between $0$ and $1$. Then $T$ has an orthonormal basis of eigenvectors, …

As a qualitative answer: although I do not know any cite-able sources, it is pretty clear that for compact (connected) real Lie groups a notion of Sobolev space parallel to that on products of circles …

The physical idea is that $-\Delta+q$ for a sufficiently-growing "confining potential" $q$ on $\mathbb R^n$ should have compact resolvent, which then proves discreteness of its spectrum. In many expli …

We can find all the tempered distributions $u$ such that $\Delta u=\lambda u$ (thus, including continuous functions going to $0$ at infinity, since these are locally integrable): taking Fourier transf …

R. Rankin's 1939 paper giving a non-trivial estimate on Ramanujan's $\tau$ function used the "real-analytic Eisenstein series" for $SL_2(\mathbb Z)$, at least. Selberg's related paper just-slightly la …

As far as I know, this is a very delicate question. That is, already in two dimensions, there can be only finite discrete spectrum. (See Phillips-Sarnak and Wolpert.) Even on the modular curve $SL(2, …

Too big to fit well as comment: There is a seeming-technicality which is important to not overlook, the question of whether a symmetric operator is "essentially self-adjoint" or not. As I discovered o …

I think it is useful to ask the simpler question, why $f\cdot (1-\Delta)^{-1}$ is compact, on $\mathbb R^n$, when $f$ is a test function. Part of the point is that $\Delta$ itself (nevermind the Dirac …

In addition to Michael Renardy's apt comment about the physical sense of the situation, one can also see the boundary vanishing from the characterization of the Friedrichs extension. Namely, by constr …

For $\lambda$ in the continuous spectrum, while there may not be genuine eigenvectors, there are approximate eigenvectors. That point is related to the fact that (as Bob Israel mentions) normal opera …

It can easily happen that the product of two eigenfunctions has an infinite eigenfunction expansion. Probably this is more typical (in natural problems) than not. For example, the Laplace-Beltrami ope …

I infer that the relevant normalizations of the intertwinings do not make them send the normalized spherical vector to itself, or else the Hilbert integral of these isomorphisms would be the identity …