13
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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
    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 $n \in \mathbb{N}$. $\endgroup$ 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. $\endgroup$ Jul 19, 2017 at 12:54

2 Answers 2

3
$\begingroup$

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

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

$\endgroup$
8
  • $\begingroup$ I think you meant $Q=-(\lambda-T)^{-1}$, right? So that $QT=\lambda Q+1$. $\endgroup$ 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. $\endgroup$ 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. $\endgroup$ 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
    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
    Apr 13, 2012 at 10:24
1
$\begingroup$

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 $\lambda$.

$\endgroup$
5
  • 3
    $\begingroup$ Doesn't this reference only concern bounded operators? $\endgroup$ 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. $\endgroup$ 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
    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 :) $\endgroup$ 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
    Apr 13, 2012 at 6:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.