Suppose $T: H\rightarrow H$ is a self-adjoint bounded linear operator that is not compact and $H$ is infinite dimensional real separable Hilbert space. Can $T$ have a discrete spectrum?
Discrete Spectrum
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functional-analysis
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0Thanks you Pavel, I feel like a dumb with your example, haha. – 2012-12-29
1 Answers
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Any compact set in can be a spectrum of some operator.
If operator is self adjoint then its spectrum lies on the real line
Operator is compact iff its spectrum is discrete and the only possible limit point of the spectrum is $0$.
This facts are enough to produce a lot of desired operators.
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1@Tomás I guess Norbert was posting quickly. The spectrum of identity is $\{1\}$ but it it not a compact operator. We should assume that the operator is normal, so that the spectral theorem applies. Also, that the eigenspace for each non-zero eigenvalue is finite-dimensional. Then the compactness follows like this: for any given \epsilon>0, you can remove finitely many dimensions (i.e., subtract a finite-rank operator) so that the remainder has norm less than $\epsilon$. (Recall that an operator is compact iff it is the norm limit of finite-rank operators.) – 2012-12-29