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Aug 10 at 10:05 comment added terceira The situation where the operator has no classical eigenvectors (i.e., not in the original Hilbert space) but does have so-called generalised ones (in this case exponential functions or, in the the FT mirror, delta distributions) was studied in great detail 50 years ago (key words: generalised eigenvectors, Gelfand triples, distribution theory). The literature is vast but you could start with the monographs of K. Maurin, easy to find with a search machine.
Aug 10 at 1:32 comment added Nate Eldredge The Laplacian $\Delta$ indeed doesn't have any eigenfunctions in $L^2(\mathbb{R}^n)$. One way to to see this is to apply the Fourier transform (an isometric isomorphism), under which $\Delta$ becomes multiplication by $|x|^2$ which clearly doesn't have any eigenfunctions.
Aug 10 at 0:30 comment added Nate Eldredge You can use LaTeX style formatting on this site, which would be more readable than what you have. See math.stackexchange.com/help/notation for info.
Aug 9 at 10:55 history edited YCor CC BY-SA 4.0
fixed typos
Aug 9 at 10:50 history asked Alucard-o Ming CC BY-SA 4.0