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This question is somewhat ill-posed (due to the word easy) and is triggered by idle curiosity:

Is there an easy example of a (separable, infinite-dimensional) Banach space $X$ and a nonempty compact set $K \subset \mathbb{C}$ such that $K$ is not the spectrum of a bounded linear operator $T: X \to X$?

As I'm not at all knowledgeable beyond the the very first basics in the geometry of Banach spaces I apologize if the following notes completely miss the point. I add them to give some background and in order to clarify what I consider "easy":


  • If $X$ is a Hilbert space (or more generally, if $X$ admits an unconditional basis $\{e_{n}\}$), it is easy to construct a diagonal operator with spectrum $K$ by choosing a countable dense subset $\{\lambda_{n}\} \subset K$ and letting $T$ be the diagonal operator sending $\sum x_n e_n$ to $\sum (\lambda_n x_n)e_n$.

  • Standard examples for spaces without an unconditional basis are $L^1[0,1]$ and $C[0,1]$. I think I convinced myself that in both cases every non-empty compact set of $\mathbb{C}$ arises as the spectrum of an operator, so these obvious candidates don't seem to answer my question. (If this should be wrong, please tell me!)

  • Variants of the Gowers-Maurey space and the Argyros-Haydon space afford examples such that the spectrum $K$ must be countable with at most one accumulation point. See Gowers's blog for background on that. For the Argyros-Haydon space this is easy to see by the very motivation for its construction: It has the remarkable property that a bounded linear operator is of the form $\lambda \cdot \operatorname{id} + C$, where $C$ is compact (thus solving the long-standing scalar-plus-compact problem).

  • I asked a version of this question a few weeks ago on math.SE but with no answers so far. In view of the illuminating comments by Robert Israel and Jonas Meyer I got there I updated it a bit.

  • The present question is related to Pietro Majer's question Banach spaces with few linear operators ? here on MO. I looked at Maurey's chapter Banach spaces with few operators in the Handbook of the Geometry of Banach spaces Vol. 2, Elsevier 2003, (Johnson, Lindenstrauss, eds) but the examples discussed there are way beyond what I would count as easy.

  • It may well be (as I'm rather ignorant on this topic) that the level of difficulty of an example must be comparable to the one of the construction of the Gowers-Maurey space or even the Argyros-Haydon space, so if there's a compelling reason pointing in this direction, please let me know.

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Your operator $Te_n = \sum \lambda_n e_n$ doesn't exists in general, because $\lambda_n$ isn't necessarily in $l^2$ and therefore this series doesn't converge. An operator on a Hilbert space with spectrum $C$ is given gy the multiplication operator $T_m:L^2(\mathbb{C})\to L^2(\mathbb{C})$ with a bounded function $m$ such that $m^{-1}(\lbrace 0\rbrace)=C$ and $|m(x)|\geq\epsilon$ for some $\epsilon>0$ and all $x\notin C$ (for example the indicator function $m=\chi_C$ fullfills these requirements) – Johannes Hahn May 29 '11 at 9:26
@Johannes: Ah, that was a stupid typo. Thanks, I corrected it. Your way works fine in the Hilbert case but it doesn't immediately generalize to other spaces. – Theo Buehler May 29 '11 at 9:48
I am not aware of an example before Gowers-Maurey of a Banach space such that not every compact subset of the plane is the spectrum of an operator on the space. – Bill Johnson May 29 '11 at 11:42
To realize every compact subset of the plane as the spectrum of some operator on a space, it is enough that the space has a complemented subspace with an unconditional basis. Before Gowers-Maurey, there were known to exist separable spaces (such as the Kalton-Peck twisted sum of two Hilbert spaces) such that no complemented subspace has an unconditional basis. – Bill Johnson May 29 '11 at 11:48
Thank you, Bill! I was aware of the first sentence in your second comment, but in order to keep this post as simple as possible, I skipped that easy extension of my remark on unconditional bases. I'll certainly have a look at the Kalton-Peck construction. Your first comment is reassuring in that I really tried very hard to find something towards answering this question and just couldn't locate anything. Thanks a lot, again. – Theo Buehler May 29 '11 at 12:36

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