For $A$ an unbounded (densely defined) operator on a separable Hilbert space, what conditions on its eigenvalues will show that, for $\lambda \notin $spec$(A)$, we have that $(A-\lambda)^{-1}$ is a compact operator?
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1$\begingroup$ If $A$ is normal, then you need $A$ to have purely discrete spectrum and $|\lambda_n|\to\infty$. $\endgroup$– Christian RemlingCommented Oct 7, 2014 at 15:09
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$\begingroup$ On the other hand, if $A$ is not normal, the spectrum could even be empty and you wouldn't be able to conclude anything about $(A - \lambda)^{-1}$ being compact. $\endgroup$– Robert IsraelCommented Oct 7, 2014 at 15:23
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$\begingroup$ @Christian: Great, what is a good (basic) reference for this. $\endgroup$– Juan CorridaCommented Oct 7, 2014 at 15:59
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1$\begingroup$ @User1298: This just follows from the definitions; pretty much any book on spectral theory on Hilbert spaces should have the needed background material. For example Weidmann or Reed-Simon 1. $\endgroup$– Christian RemlingCommented Oct 7, 2014 at 16:54
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1 Answer
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It is enough that just one among the operators $(A-\lambda)^{-1}$ be compact. Because of the formula $$(A-\mu)^{-1}=(A-\lambda)^{-1}+(\mu-\lambda)(A-\lambda)^{-1}(A-\mu)^{-1}.$$
answered Oct 7, 2014 at 15:14