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Aug 21, 2014 at 6:17 review Suggested edits
Aug 21, 2014 at 6:28
Aug 19, 2014 at 19:29 vote accept Ali Taghavi
Aug 19, 2014 at 19:29 comment added Ali Taghavi Thank you very much for your beautiful answer and your reference to Putnam theorem. However, regarding the eigenvalue problem, i think that $I+V$ is a counter example where $V$ is the Voltra integral operator. It has a single spectrum $\{1\}$ which is not eigenvalue.
Aug 19, 2014 at 16:37 comment added Mike Jury OK, I no longer trust my claim about eigenvalues; however it is true that if 0 is in the boundary of the spectrum of a Fredholm operator (which is of course the case here), then it must be isolated and the operator must have index 0. This follows from a theorem of Putnam; see Conway's "A Course in Functional Analysis", Theorems XI.6.8 and XI.6.9.
Aug 19, 2014 at 16:26 comment added Mike Jury I thought this should just follow from the Fredholm assumption and the Riesz functional calculus, but perhaps it is more subtle than that. Let me think about it some more.
Aug 19, 2014 at 16:09 comment added Ali Taghavi thank you very much for your interesting answer. I have a question on your statement "Isolated points are eigenvalue" In this statement, is not necessary that we assume T is normal?
Aug 19, 2014 at 15:30 history edited Mike Jury CC BY-SA 3.0
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Aug 19, 2014 at 13:00 history answered Mike Jury CC BY-SA 3.0