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.
Best Answer
See Theorem 3.4.3, page 93, of these notes for a detailed proof of the fact that $T: H\to H$ is Fredholm if and only if there exists $Q:H\to H$ such that $QT-1$ is compact. If $Q=(T-\lambda)^{-1}$ is compact then
$$Q T= Q(T-\lambda)+\lambda Q=1+\lambda Q$$
so that
$$ QT-1=\lambda Q =\mbox{compact}. $$