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Schrodinger operators, operators on manifolds, general differential operators, numerical studies, integral operators, discrete models, resonances, non-self-adjoint operators, random operators/matrices
2
votes
1
answer
91
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On the dimension of the range of the resolution of the identity
I want to prove the following: Let $A,B$ be bounded self-adjoint operators in a complex-Hilbert space and $E_A(\lambda)$, $E_B(\lambda)$ its corresponding spectral resolutions, i.e.,
$$A=\int_{[m_A,M_ …
0
votes
A bounded operator $T$ is compact if and only if $\sigma_{\mathrm{ess}}(T)=\{0\}$
Suppose that $\sigma_{\mathrm{ess}}(T)=\{0\}$. So, zero is the only one possible accumulation point.
Thus if $\sigma_d(T)$ is finite then $E(\{\lambda\in\mathbb{R}:|\lambda|>\varepsilon\})$ has finite …
1
vote
1
answer
752
views
A bounded operator $T$ is compact if and only if $\sigma_{\mathrm{ess}}(T)=\{0\}$ [closed]
Theorem: Let $T$ be a bounded self-adjoint operator on a complex infinite-dimensional Hilbert space $H$. Then $T$ is compact if and only if $\sigma_{\mathrm{ess}}(T)=\{0\}$.
Proof: If $T$ is compact t …