A sufficient condition for an unbounded self-adjoint operator to have point spectrum?

Asked Sep 28 '17 at 05:33

Active Sep 28 '17 at 05:33

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For any complex separable Hilbert space $H$, we know that if $A:H\dashrightarrow H$ is an unbounded self-adjoint operator, then there exists some $\sigma$-finite measure space $(X,\mu)$, some unitary map $U:H\to L^2(X,\mu)$, and some $g:X\to\mathbb{R}$ measurable such that $AU=U M_g$, where $M_g:f\mapsto fg$ for $f\in L^2$.

My question is, when can we take $(X,\mu)$ to be some measure space with the counting measure? In the bounded case, it suffices to assume that $A$ is a compact operator, but that's not a clearly-defined notion in the unbounded case.

asked Sep 28 '17 at 05:33

Dominic Wynter

6,279

I think you mean $AU^{-1}=U^{-1}M_g$ or you want $U: L^2(X,\mu)\to H$. Anyway, here is one class of such operators: https://math.stackexchange.com/questions/1778441/spectrum-of-laplace-operator-with-potential-acting-on-l2-mathbb-r-is-discre – s.harp Sep 28 '17 at 05:37
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"An unbounded operator whose resolvent is compact (for some element of the resolvent set) has a discrete spectrum". – Keith McClary Sep 29 '17 at 00:49

A sufficient condition for an unbounded self-adjoint operator to have point spectrum?

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