Prove the discrete spectrum of $A$ equals to the set of those complex $\lambda$ such that $\lambda I-A$ is Fredholm.

Question

There is considerable divergence in the literature concerning the definition of the essential spectrum of a densely defined closed operator $A$ on a Banach space $X$. I want to find a direct definition of the essential spectrum such that it is the complement of the discrete spectrum and is stable under compact perturbations. It seems to me that I have succeeded, but I still have one problem that I really what to know the proof.

It seems that we always define the discrete spectrum $\sigma_{\text{d}}(A)$ of $A$ by $$\sigma_{\text{d}}(A):=\{\lambda\in\mathbb C: \lambda \text{ is an isolated eigenvalue of }\ A \text{ and has finite algebraic multiplicity}\}.\tag 1$$ Here the algebraic multiplicity of an isolated eigenvalue $\lambda$ is defined as the dimension of the range of the Riesz projection $P_\lambda$.

This is a relatively long post. I wrote almost all things that I have done relating to this topic. In the end of the post, I will write a self-contained statement of my problem. If you want to save time, you can skip the heavy body of the post.

For a self-adjiont operator $A$ on a Hilbert space $X$, we can define the essential spectrum $\sigma_{\text{ess}}(A)$ by $\sigma_{\text{ess}}(A):=\sigma(A)\setminus \sigma_{\text{d}}(A)$. Then we can use Weyl's criterion to prove that the essential spectrum of self-adjiont operators is invariant under symmetric compact perturbations.

However, when it comes to general densely defined closed operators on Banach spaces, things become complicated. Kato defined in Chapter 4 of his book Perturbation of Linear Operators that $$\sigma_{\text{ess}}(A):=\sigma(A)\setminus \{\lambda\in \mathbb C: \lambda I-A \text{ is semi-Fredholm}\}.$$ Then Theorem 5.35 in Chapter 4 proves the stability of $\sigma_{\text{ess}}(A)$ under compact perturbations. But as Therorem 5.33 implies, the relation $\sigma_{\text{ess}}(A)=\sigma(A)\setminus \sigma_{\text{d}}(A)$ is not always right.

I searched for some materials and finally find out that the definition given by F.Wolf should be satisfying. He wrote: The spectrum $\sigma(A)$ of an operator $A$ can be divided into two parts: The essential spectrum $\sigma_e(A)$ consisting of the points $\lambda$ at which $\Re(\lambda I-A)$, the range of $\lambda I-A$ is not closed and of eigenvalues of infinite multipilicity. The second part may be called the Fredholm part of $\sigma(A)$ consists beside others of $\rho(A)$, the resolvent set of $A$ and of isolated eigenvalues of finite multiplicity. But I also find that in another paper the author rewrote Wolf's definition of essential spectrum as the subset of $\sigma(A)$ consisting those $\lambda$ such that $\lambda I-A$ is not Fredholm. These two definitions seem different at the first glance, and I wonder why they are the same.

My Problem. Let $A$ be a densely defined closed operator $A$ on a Banach space $X$ and define $\sigma_\text{d}(A)$ as in (1). Prove that $$\sigma_\text{d}(A)=\{\lambda\in\sigma(A): \lambda I-A \text{ is Fredholm}\}.$$ That is to say, $\sigma_\text{d}(A)$ consists of those complex $\lambda\in\sigma(A)$ for which

• $\Re(\lambda I-A)$, the range of $\lambda I-A$, is closed,
• $\text{dim}\ N(\lambda I-A)$, the geometric multiplicity of $\lambda$, is finite,
• $\text{dim}\ \left(X/\Re(\lambda I-A)\right)$, the codimension of $\Re(\lambda I-A)$, is finite.

Any hints or useful references are welcome!

Answer 1

That is not true. The reason is that for non-normal (bounded or unbounded, that doen's matter here) operators, all the nice equivalent characterisations of the essential spectra of normal/self-adjoint operators are no longer equivalent. For this reason, you cannot define 'the' essential spectrum of a general operator and there is downright 'zoo' of various essential spectra.

For instance, the set $\sigma_d(T)$ you defined in $(1)$ is the correct definition of the discrete spectrum. But its complement in the spectrum $$ \sigma_{e5}(T):=\sigma(T)\setminus\sigma_{d}(T), $$ is, first of all, not equal to $\sigma_{ess}(T)$ as you define it above and, secondly, only invariant under commuting compact perturbations but, in general, not invariant under arbitrary compact perturbations.

The largest essential spectrum that is always stable under any compact perturbations is the set $\sigma_{e4}(T) :=\mathbb{C}\setminus \rho_{e4}(T)$, where $$ \rho_{e4}(T):=\{\lambda \in \mathbb{C} \mid T-\lambda\ \text{is Fredholm with index }\ 0\}. $$ However, in general only the inclusion $\sigma_{e4}(T)\subset \sigma_{e5}(T)$ holds which can be strict. An exmple which shows that this can be strict and that $\sigma_{e5}(T)$ is not invariant under arbitrary comapct perturbation can be constructed using the bilateral shift operator.

The essential spectrum defined in Kato is again differtent and can be defined as $$ \sigma_{e1}(T):=\{\lambda \in \mathbb{C} \mid T-\lambda\ \text{is not semi-Fredholm }\}. $$ It is even smaller and satisfies $$ \sigma_{e1}(T) \subset \sigma_{e4}(T) \subset \sigma_{e5}(T), $$ again, strict inclusions being possbible. As my enumeration suggests, there are also two essential spectra in between (and even much more exists in literatur). Among all the essential spectra that can be defined with Fredholm properties, $\sigma_{e1}(T)$ is the smallest one and $\sigma_{e5}(T)$ is the largest one and they all coincide in the normal case (however, I've also seen an essential spectrum that is even smaller than $\sigma_{e1}(T)$, but isn't defined in terms of Fredhol properties).

If you would like or need to know more about the various essential spectra, I would recoomend having a look into Edmunds, D. E., and Evans, W. D. Spectral Theory and Differential Operators. There they introduce the most common essential spectra and consider them in relative detail. The couterexample I mentioned eralier is also contained therein.