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I have little experience with functional analysis beyond an undergraduate basic course, and I'm dealing with the following problem:

let $V$ be an infinite-dimensional locally convex (but not normed!) vector space, and let $O:V\rightarrow V$ be an invertible continuous operator; how can I decide whether it has eigenvalues or not?

The problem, sits in a larger framework: let $G$ be a topological group of operators as above; is there a common eigenvector for all these operators?

I do not know what are the tools for this type of problems in an infinite-dimensional setting. The only related theorem that I know proves that the spectrum of an element in a Banach algebra is nonempty (the proof that I know is based upon Liouville's theorem from complex analysis and clearly cannot be mimicked here); I do not know of any result regarding the eigenvalues and not the spectrum (except for the classical result in a finite-dimensional setting). The fact that I do not have a norm can only complicate things, I believe. Any suggestion or bibliographical hint would be appreciated. Thank you.

(Somebody already registered please add the tag "eigenvalue", I'm not allowed to.)

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  • $\begingroup$ How are you given $V$ and how are you given $O$? $\endgroup$
    – Qiaochu Yuan
    Commented Aug 30, 2012 at 7:03
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    $\begingroup$ Points of the spectrum need not be eigenvalues, so without much more information specific to your problem, it's hard to see what can be said. (You might have approximate eigenvalues; but there can be points of the spectrum which aren't even approximate eigenvalues. The classic examples is the "right shift" on $\ell^2({\mathbb N})$, aka "multiplication by $z$" as an operator on Hardy sapce $H^2({\mathbb D})$. $\endgroup$
    – Yemon Choi
    Commented Aug 30, 2012 at 8:29
  • $\begingroup$ Every nonzero element of the spectrum of a compact operator on a Banach space is an eigenvalue of this operator. So this is a classical result in the infinite dimensional case. $\endgroup$
    – jjcale
    Commented Aug 30, 2012 at 19:28
  • $\begingroup$ @jjcale Who said anything about compactness? $\endgroup$
    – Yemon Choi
    Commented Aug 30, 2012 at 23:34
  • $\begingroup$ For my previous examples, add $2I$ (with $I$ the identity operator) to get an invertible operator with no eigenvalues $\endgroup$
    – Yemon Choi
    Commented Aug 30, 2012 at 23:35

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