I want a proof of this proposition:
"The spectrum of an operator with compact resolvent is discrete"???
Thank you very much
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1@Eric Auld: I have no idea – mario Jul 27 '13 at 14:34
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To start with you should know the following property of compact operators. – Mhenni Benghorbal Jul 27 '13 at 15:24
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1@Mhenni Benghorbal: what property? – mario Jul 27 '13 at 15:36
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@mario: The spectrum of a compact operator in the shaded area. – Mhenni Benghorbal Jul 27 '13 at 15:49
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1@Mhenni Benghorbal: I don't understand you – mario Jul 27 '13 at 15:55
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@mario: The propert is "What kind of spectrum does a compact operator have?" I gave you the link to see this property. – Mhenni Benghorbal Jul 28 '13 at 10:03
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Is $T$ is a self adjoint operator with compact resolvent then:
(a) either $\sigma(T)= \Phi$ (b) or $\sigma(T)={\lambda_n , n\ge{1}}$ such that $limit |\lambda_n |= +\infty$ Furthermore, there exists $\lambda_0 \in\rho(T)$ such that $|\lambda_1 - \lambda_0| \le |\lambda_2 - \lambda_0| \le ... $
$ker(T-\lambda_n)$ is of finite dimension and we notice that the elements of the spectrum of $T$ is a sequence so all the points are isolated, thus $\sigma(T)=\sigma_d(T)$
Math1995
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This is not a proof, which is what the question asks for. – Mariano Suárez-Álvarez Jan 19 '18 at 17:11