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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.

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  • $\begingroup$ You mean $\lambda I - A$ is bijective from $((1 - P_{\lambda})E_0) \cap E_1$ to $(1 - P_{\lambda})E_0$ , or ? $\endgroup$
    – jjcale
    Commented Apr 14, 2012 at 8:18
  • $\begingroup$ @jjcale - Yes. It should be a bijection when considering it's restriction to the residual space, where we need to be careful to only consider elements in $((1 - P_{\lambda})E_0) \cap E_1$, as you mentioned. This care is not necessary on $P_{\lambda}E_0$ however, since one can show that $P_{\lambda}E_0 \subset D(A^n)$ for any $n \in \mathbb{N}$. $\endgroup$
    – Jeremy LeCrone
    Commented Apr 15, 2012 at 3:14
  • $\begingroup$ Let $E_0=E_1$ be a Hilbert space, let $i: E_0 \to E_0$ be compact embedding with dense image. Let $B: E_1 \to E_0$ be the identity map. Define an unbounded operator $A:E_0 \to E_0$ with domain $i(E_0)$ by setting $A= B \circ i^{-1}$. Then $A^{-1}= i \circ B^{-1}= i$ is compact and $B-1 \cdot B$ is not Fredholm. $\endgroup$
    – Chris Judge
    Commented Jul 19, 2017 at 12:54

2 Answers 2

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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

$$ QT-1=\lambda Q =\mbox{compact}. $$

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  • $\begingroup$ I think you meant $Q=-(\lambda-T)^{-1}$, right? So that $QT=\lambda Q+1$. $\endgroup$
    – Renato G. Bettiol
    Commented Apr 12, 2012 at 22:02
  • $\begingroup$ @Liviu- I had originally considered exactly this argument, but there seems to be a flaw in it. By assumption I only know that $Q = (\lambda - T)^{-1}$ is compact from $E_0$ into itself. However, when you are considering it in this context, you wind up with $Q$ as an operator from $E_0$ into $E_1$ (This is the only way that the equations $1 - QT$ and $1 - TQ$ make sense...) We, however, do not have compactness of $Q$ between the two spaces. I would love if I am wrong in this, but I think that this consideration hinders your argument. $\endgroup$
    – Jeremy LeCrone
    Commented Apr 13, 2012 at 1:10
  • $\begingroup$ @Renato Thanks for pointing up that error. @Jeremy: Note that $E_0=E_1$ or else the operator $\lambda-T$ is meaningless. $\endgroup$
    – Liviu Nicolaescu
    Commented Apr 13, 2012 at 9:11
  • $\begingroup$ @Liviu : $E_1$ is a subset of $E_0$, so it makes sense. I was wondering if your proof goes well indeed in the unbounded case (thanks for all your notes btw). $\endgroup$
    – Amin
    Commented Apr 13, 2012 at 10:21
  • $\begingroup$ @Liviu : I remember that you showed that elliptic operators were unbounded Fredholm in other notes, and I don't know if the parallel can be done just based on the compact resolvent hypothesis (haven't checked at all). $\endgroup$
    – Amin
    Commented Apr 13, 2012 at 10:24
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It is essentially stated in Theorem 4.3.7 in E.B. Davies, Linear Operators and their Spectra. The theorem says that the essential spectrum of an operator A is the spectrum of A as an element in the Calkin algebra.

An operator with compact resolvent has a discrete set of eigenvalues, each of finite multiplicity, so its essential spectrum is empty. By the above theorem, $A-\lambda$ is invertible in the Calkin algebra, hence Fredholm, for all $\lambda$.

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    $\begingroup$ Doesn't this reference only concern bounded operators? $\endgroup$
    – Aaron Tikuisis
    Commented Apr 12, 2012 at 11:58
  • $\begingroup$ Thank you Heiko, I will look into the reference in more detail, though at first glance it does appear to specifically concern bounded operators. $\endgroup$
    – Jeremy LeCrone
    Commented Apr 12, 2012 at 15:25
  • $\begingroup$ I may sound completely fool, but in your question, it's written bounded on $E_1$; is it then not directly extendable to a bounded op on $E_0$ ? Or it's a misspell and you meant unbounded? $\endgroup$
    – Amin
    Commented Apr 12, 2012 at 20:55
  • $\begingroup$ @Amin- It is bounded FROM $E_1$ TO $E_0$, not ON $E_1$. I.e. For every $x \in E_1$ we know that $\| Ax \|_0 \leq M \| x \|_1$, where $\| \cdot \|_i$ is the norm on $E_i$. This does not mean that we can extend $A$ to all of $E_0$. Specifically, you are likely thinking, let $x_0 \in E_0$ and take a sequence $(x_n) \subset E_1$, then define $A x_0$ by a limiting argument. BUT $\| Ax_0 - Ax_n \|_0 \leq M \|x_0 - x_n \|_1$ is the inequality you get and you only know that $\| x_0 - x_n \|_0 \rightarrow 0$. So, no it is not a misspell :) $\endgroup$
    – Jeremy LeCrone
    Commented Apr 13, 2012 at 1:03
  • $\begingroup$ Oh right right, sorry about that, I recalled this morning about the usual case of elliptic operators with Sobolev as $E_1$ and figured out I was mistaken ;). $\endgroup$
    – Amin
    Commented Apr 13, 2012 at 6:53

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