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The Spectrum of a bounded operator on a Banach space $X$ is always a compact subset of $\mathbb{C}$. What about the converse?

Given any compact subset $K \subset \mathbb{C}$ is it always possible to find a Banach space $X$ and a bounded operator $T: X \to X$ such that $\sigma(T) = K$ ? Is there one Banach space which works for any $K$ ?

I think for countable $K = \{\lambda_0, \lambda_1, \dots\}$ this is always possible: The operator $T: \ell^2 \to \ell^2, (x_0, x_1, \dots) \mapsto (\lambda_0 x_0, \lambda_1 x_1, \dots)$ is bounded and has all elements of $K$ as eigenvalues.

Carlos Esparza
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  • Yes, you are right for the example you give at the end. Indeed, from the many similar questions on the site, this seems to be the go to example which indicates that given $K\subseteq\mathbb C$ one can construct an operator $T$ (as you gave) for which $\sigma(T)=K$. – Jeremy Jeffrey James Jan 25 '19 at 21:46
  • Other questions which touch on this include: https://math.stackexchange.com/questions/83656/prove-forall-compact-m-m-subset-c-quad-exists-al-2-rightarrow-l-2-sig and https://math.stackexchange.com/questions/740904/show-that-any-compact-set-in-mathbbc-is-the-spectrum-of-an-operator – Jeremy Jeffrey James Jan 25 '19 at 21:48
  • A very late comment. I think the simplest example is to consider the Banach space $C(K)$ which is the space of contiuous functions over $K$, and let the operator $T$ be multiplication with function $z$. – Zhang Yuhan Jan 15 '24 at 13:52

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