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0 votes
2 answers
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Spectrum equals eigenvalues for unbounded operator

Let $D$ be an unbounded densely defined operator on a separable Hilbert space $H$. If $D$ is diagonalisable with all eigenvalues having finite multiplicity and growing towards infinity, does it follow ...
Bas Winkelman's user avatar
11 votes
2 answers
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How "generalized eigenvalues" combine into producing the spectral measure?

Hi... I am wondering how 'eigenvalues' that don't lie in my Hilbert space combine into producing the spectral measure. I study probability and I am quite ignorant in the field of spectral analysis of ...
Reda's user avatar
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