Isolated points of the spectrum are always eigenvalues?

Question

Let $A=A^*:D(A)\subset H \rightarrow H$ be a linear operator, where $H$ is a seperable Hilbert space. The discrete spectrum of $A$ is defined to be $\sigma_{disc}(A):=\left\lbrace \lambda \in \sigma(A)\Big|\exists\varepsilon>0 :B_\varepsilon(\lambda)\cap \sigma(A)=\lbrace \lambda\rbrace\text{ }\& \text{ }\mathrm{dim}PH<\infty\right\rbrace.$

Where $P:=\oint\limits_{|z-\lambda|=\varepsilon}R(z,A)dz$ is the corresponding Riesz-projector of $A$.

$R(z,A):=(z-A)^{-1}$ is the resolvent of $A$.

It is probably known among you analysts, that the spectral measure of an isolated point $\lambda \in \sigma(A)$, can be presented as an Riesz-projector i.e.

$E_{\lbrace \lambda \rbrace}=\oint\limits_{|z-\lambda|=\varepsilon}R(z,A)dz$

Where $E$ is the spectral measure of $A$. At the same time we have $E_{\lbrace \lambda\rbrace}=E_{(\lambda-\varepsilon,\lambda+\varepsilon)}\neq 0$ for some $\varepsilon>0$, due to the properties of the spectral measure. This implies $\lambda$ is an eigenvalue.

I suppose that means if any point is isolated in the spectrum, then it is also an eigenvalue if $A=A^*$. Hence we could also define the discrete spectrum by

$\sigma_{disc}(A):=\left\lbrace \lambda \in \sigma_p(A)\Big|\exists\varepsilon>0 :B_\varepsilon(\lambda)\cap \sigma(A)=\lbrace \lambda\rbrace\text{ }\& \text{ }\mathrm{dim}PH<\infty\right\rbrace.$

Question: How is it in case that $A$ is not selfadjoint? Are isolated points in the spectrum also eigenvalues?

Answer 1

A counterexample in the non-selfadjoint case is any compact operator on an infinite-dimensional complex Banach space with empty point spectrum, like the operator $$T : u \in L^2(0,1) \mapsto \left(x \mapsto \int_0^x u(t)\mathrm{d}t\right)$$ For a compact operator in an infinite-dimensional setting, and when the scalars are the complex numbers, $0$ is always in the spectrum, and the spectrum outside of $0$ is exactly the point spectrum outside of $0$, thus with the operator mentioned above $0$ is an isolated point of the spectrum which isn't an eigenvalue.

Showing that the example I mentioned is compact but has empty point spectrum is a classic exercise, but if you need a proof there is one here: Volterra Operator is compact but has no eigenvalue.