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?

1$\begingroup$ If $A$ is normal, then you need $A$ to have purely discrete spectrum and $\lambda_n\to\infty$. $\endgroup$– Christian RemlingOct 7, 2014 at 15:09

$\begingroup$ 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 IsraelOct 7, 2014 at 15:23

$\begingroup$ @Christian: Great, what is a good (basic) reference for this. $\endgroup$– Juan CorridaOct 7, 2014 at 15:59

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 ReedSimon 1. $\endgroup$– Christian RemlingOct 7, 2014 at 16:54
1 Answer
It is enough that just one among the operators $(A\lambda)^{1}$ be compact. Because of the formula $$(A\mu)^{1}=(A\lambda)^{1}+(\mu\lambda)(A\lambda)^{1}(A\mu)^{1}.$$