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Given a bounded operator $A$ on a Banach space $X$, one may find the spectrum $\sigma(A)\subset{\bf C}$.

Here are my questions:

  • Given some set in the complex plane, say, $S\subset{\bf C}$, can one find an operator $A$ such that $\sigma(A)=S$?
  • Is there a "big picture" for this kind of questions?
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Certainly the operator would not be unique. –  Alex Becker Apr 26 '11 at 22:08
The result of Argyros and Haydon cited by Jonas Meyer below is fascinating. But, as a deleted comment formerly asked, is there a simple example of a compact $K \subset \mathbb{C}$ and a Banach space where no operator has spectrum $K$? –  Nate Eldredge Apr 27 '11 at 3:55
@Nate: I'd very much like to know that, too. Please let me know if you can think of an example. –  t.b. Apr 27 '11 at 5:40
@Nate: I've now asked a question about that –  t.b. Apr 28 '11 at 14:38
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3 Answers

up vote 14 down vote accepted

Let $A$ be a bounded operator on the Banach space $X$.

The spectrum of $A$ must be closed. The set of invertible operators on a Banach space is open, and $\lambda\mapsto A-\lambda I$ is continuous. The resolvent set of $A$ (complement of the spectrum) is the inverse image of the invertible operators under this map.

The spectrum of $A$ must be bounded. If $|\lambda|>\|A\|$, then $\|\frac{1}{\lambda}A\|=\|I-(I-\frac{1}{\lambda}A)\|<1$. This implies that $I-\frac{1}{\lambda}A$ is invertible, which in turn implies that $A-\lambda I$ is invertible.

The spectrum is nonempty. The function $\lambda\mapsto (A-\lambda I)^{-1}$ is holomorphic on the resolvent set and goes to $0$ at infinity. If it were defined on the whole complex plane, it would be identically $0$ by Liouville's theorem (you can apply Hahn-Banach and the scalar-valued version of Liouville). But this is absurd, because $(A-\lambda I)^{-1}$ is invertible whenever it exists.

So to have any hope, $S$ should be compact and nonempty. If you are allowing $X$ to vary, then this is sufficient, and it is enough to consider Hilbert space as Rasmus already mentioned. For example, you could let $\mu$ be a regular Borel measure with support $S$, and then let $A$ be the operator on $L^2(\mu)$ defined by $(Af)(x)=xf(x)$. (Or you could consider diagonal operators on spaces with chosen bases.)

If you mean that $X$ is fixed, then the answer depends on $X$, and I don't know what can be said in general. Of course if $X$ is finite dimensional, then the possible spectra are the sets with cardinality no greater than the dimension of $X$. There are also infinite dimensional spaces for which not every nonempty compact set can be the spectrum of an operator. As mentioned in a comment on Rasmus's answer, Argyros and Haydon showed that there are infinite dimensional Banach spaces on which every operator has the form $\lambda I +K$ with $K$ compact. Since compact operators have countable spectrum with $0$ as the only possible limit point, $\lambda I+K$ has countable spectrum with $\lambda$ as the only possible limit point.

Some searching inspired by Theo Buehler's question (which in turn was inspired by this question and Nate Eldredge's comment above) has turned up the fact that hereditarily indecomposable Banach spaces also have the property that all operators have countable spectrum with at most one limit point. Every operator on such a space is scalar plus strictly singular, and there were known examples well before Argyros and Haydon's breakthrough, as mentioned on Gowers's blog (and constructed by Gowers himself as well as Maurey). It is also known that there are hereditarily indecomposable spaces such that not every operator is scalar plus compact. Maurey's chapter in the Handbook of the geometry of Banach spaces, Volume 2, titled "Banach spaces with few operators," gives an introduction to these and much more. (I cannot really add anything better than a pointer to this wonderful reference, due to my ignorance.)

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The following only holds if you are working over a Hilbert space:

A subset of the complex numbers is the spectrum of a bounded operator if and only if it is non-empty and compact.

You can find this in every decent book on functional analysis.

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Silly question: Is this true for any Banach space (for Hilbert spaces this is clear)? But in general I wouldn't see why. –  t.b. Apr 26 '11 at 22:25
@Theo: No. Argyros and Haydon showed in 2009 that there are infinite dimensional separable Banach spaces on which every bounded linear operator is scalar plus compact. Such an operator has countable spectrum with at most one limit point. (So if the present question is about operators on a fixed Banach space $X$, then the answer depends on $X$.) –  Jonas Meyer Apr 26 '11 at 22:38
Whoops, thank you for pointing this out. @Theo –  Rasmus Apr 26 '11 at 22:40
@Jonas: Ah, so this is finally solved, then? Great, didn't know that. Thanks! –  t.b. Apr 26 '11 at 22:50
@Theo: I first learned about it from Gowers's Weblog: gowers.wordpress.com/2009/02/07/… –  Jonas Meyer Apr 26 '11 at 23:18
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In the case of Hilbert spaces, there is something like a big picture concerning related questions. Briefly this goes as follows:

let $X$ be a compact subset of the complex plane $\mathbb{C}$. Denote by $\mathrm{Ext}(X)$ the set of equivalence classes of essentially normal operators with spectrum equal to $X$. Then this set $\mathrm{Ext}(X)$ is an abelian group and moreover one model for the $K$-Homology of $X$, the homology theory dual to Atiyah-Hirzebruch $K$-Theory. All of this is part of the socalled Brown Douglas Filmore Theorem which was one of the starting points of $K$-Homology.

So classifying certain operators (essentially normal ones up to equivalence) with fixed spectrum $X \subset \mathbb{C}$ has a connection to classical algebraic topology

A great reference for this is the book "Analytic K-Homology" by Nigel Higson and John Roe.

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