Suppose $A \in \mathcal{L}(E_1, E_0)$ is a bounded linear operator between Banach spaces $E_1$ and $E_0$, and we also have that $E_1$ is densely, continuously embedded in $E_0$ (i.e. $A$ can be regarded as a closed, unbounded operator on $E_0$). It is well known that, if the resolvent $R(\lambda, A) := (\lambda I - A)^{-1}$ is a compact operator in $\mathcal{L}(E_0)$ for $\lambda \in \rho(A)$ (which is equivalent to $R(\lambda_0, A)$ is compact for one particular $\lambda_0 \in \rho(A)$) then the operator $\lambda I - A$ is a Fredholm operator for every $\lambda \in \mathbb{C}$, however I cannot seem to find a reference which states this result. It seems that most references regarding unbounded operators with compact resolvents conclude their investigation with a proof that the spectrum is composed of isolated eigenvalues with finite multiplicity and regard Fredholm operators only long enough to discuss the essential spectrum of an operator. I have looked through texts by Dunford and Schwartz, Kato, Engel and Nagel, and Hormander (among others...), without finding the reference which I am hoping to find.

$\bullet$ To be clear, I am looking for a reference which proves that if $R(\lambda, A)$ is compact for $\lambda \in \rho(A)$ then $\lambda I - A$ is Fredholm for $\lambda \in \mathbb{C}$.

A proof might go as follows: In the case that $\lambda \in \rho(A)$, the conditions of a Fredholm operator are trivial. Meanwhile when $\lambda \in \sigma(A)$ I can prove the result using a spectral projection $P_{\lambda}$ and the fact that $E_0$ decomposes into the direct sum of a finite dimensional space $P_{\lambda}E_0$ and a residual space $(1 - P_{\lambda})E_0$ on which $\lambda I - A$ is bijective. Although this proof is not too complicated, it seems unnecessary that I should have to include it, as the result should show up in previous references. This is my last ditch effort before I break down and either include the proof for myself or else pass it off as a "it is well-known" without reference, so any suggestions or opinions would be helpful.

Thank you.