Let $X$ be a Banach space and $T \in \mathcal L(X)$.

The authors Engel and Nagel introduce in their book "One-Parameter Semigroups for Linear Evolution Equations" on p. 248 the concept of the Fredholm domain of $T$ defined by $$\rho_F(T) := \{\lambda \in \mathbb C: \lambda - T \text{ is a Fredholm operator} \}.$$ On the next page the following is stated:

"Here, we only recall that the poles of $R(\cdot, T)$ with finite algebraic multiplicity belong to $\rho_F(T)$. Conversely, an element of the unbounded connected component of $\rho_F(T)$ either belongs to $\rho(T)$ or is a pole of finite algebraic multiplicity."

I can prove the first statement very elementary just by using properties of spectral projections and some very basic functional calculus. But the second statement seems to be quite difficult to prove. In the cited literature I found a proof of the stament (cf. the proof Corollary XI.8.5 in "Classes of Linear Operators Vol. I" by Gohberg, Goldberg and Kaashoek). But it seems to rely on quite some theorems about Fredholm operator valued functions.

So my question is whether there is a more elementary way to see that the statement holds, maybe just by using some basic facts on spectral projections? I thought quite some time about it but couldn't prove it. So is there maybe a reference for the statement which uses more elementary arguments? Or does someone know another way how to prove it? I am looking forward to your answers.

  • 1
    $\begingroup$ I'm not sure whether this helps (or whether it is sufficiently elementary), but it might be worthwhile to note that $\lambda - T$ has Fredholm index $0$ for every $\lambda \in \rho_F(T)$; this follows from the fact that $\rho_F(T)$ intersects the resolvent set of $T$. $\endgroup$ – Jochen Glueck Apr 16 '19 at 17:03
  • $\begingroup$ @JochenGlueck It helps indeed as it yields that each element $\rho_F^\infty(T) \cap \sigma(T)$ is isolated. $\endgroup$ – Adriano Apr 18 '19 at 8:01

A maybe different proof can be obtained using techniques related with the single-valued extension property (SVEP for short) of an operator. A good reference for this topic is the book [Aiena]: Pietro Aiena. Fredholm and local spectral theory, with applications to multipliers. Kluwer, 2004.

Let us denote by $\rho_F^\infty(T)$ the unbounded connected component of $\rho_F(T)$. First you have to show that $\rho_F^\infty(T)\setminus\rho(T)$ consists of isolated points of the spectrum of $T$. This fact can be proved as follows:

(1) The index $\textrm{ind}(zI-T):= \dim \ker(zI-T) - \dim X/R(zI-T)$ is constant for $z$ in each connected component of $\rho_F(T)$. Hence $\textrm{ind}(zI-T)=0$ for $z\in\rho_F^\infty(T)$.

Proof: This is not trivial, but it is a standard result.

(2) The set $\rho_F^\infty(T)\setminus\rho(T)$ does not have accumulation points in the spectrum of $T$.

Proof: If $T$ is a Fredholm operator with $\textrm{ind}(T)=0$, then so is the conjugate operator $T^*:X^*\to X^*$. Therefore you can find closed subspaces $M$ of $X$ and $N$ of $X^*$ such that $X=\ker(T)\oplus M$, $X^*=\ker(T^*)\oplus M$ and there exists $c>0$ such that $\|Tm\|\geq c \|m\|$ for each $m\in M$ ($T$ is bounded below on $M$) and $\|T^*n\|\geq c \|n\|$ for each $n\in N$. As a consequence, if $0<|z|<c$ then $zI-T$ and $zI-T^*$ are bounded below, hence $zI-T$ and $zI-T^*$ are bijective for $0<|z|<c$.

Now it remains to show that for each $z_0\in \rho_F^\infty(T)\setminus\rho(T)$, the spectral projection associated to the spectral set $\{z_0\}$ has finite dimensional rank. Here is where we apply the SVEP:

It is clear that $T$ and $T^*$ have the SVEP at $z_0$ [Aiena, Definition 2.3]. Thus [Aiena, Theorem 3.76] and part (iii) of [Aiena, Theorem 3.79] give you the required result.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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