Timeline for A bounded operator $T$ is compact if and only if $\sigma_{\mathrm{ess}}(T)=\{0\}$ [closed]
Current License: CC BY-SA 4.0
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| Feb 19 at 21:23 | history | closed |
Yemon Choi Daniele Tampieri Mark Wildon Christian Remling Max Horn |
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| Feb 15 at 21:05 | review | Close votes | |||
| Feb 19 at 21:23 | |||||
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| Oct 3, 2020 at 10:35 | history | edited | YCor | CC BY-SA 4.0 |
formatting (the question was bumped anyway)
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| Oct 3, 2020 at 10:08 | history | edited | Glorfindel | CC BY-SA 4.0 |
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| Oct 2, 2020 at 7:03 | history | edited | Andrés Felipe | CC BY-SA 4.0 |
added 4 characters in body; edited title
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| Oct 2, 2020 at 6:51 | answer | added | Andrés Felipe | timeline score: 0 | |
| Oct 2, 2020 at 6:38 | history | edited | YCor | CC BY-SA 4.0 |
romanized abbreviation 'ess'
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| Oct 2, 2020 at 6:34 | comment | added | Andrés Felipe | I mean, $\lambda\in\sigma_{ess}(T)$ if and only if $\lambda$ is accumulation point of the spectrum of $T$ or $\dim\ker(T-\lambda)=\infty$. So $T\equiv 0$ implies that $\sigma(T)=\sigma_{ess}(T)=\{0\}$ because $H$ is infinite dimensional. | |
| Oct 2, 2020 at 6:28 | comment | added | Dieter Kadelka | $T \equiv 0$ is compact and self-adjoint with $\sigma(T) = \sigma_p(T) = \{0\}$. Other counterexamples are operators with finite dimensional range. | |
| Oct 2, 2020 at 6:26 | history | edited | Andrés Felipe | CC BY-SA 4.0 |
added 2 characters in body
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| Oct 2, 2020 at 4:10 | history | edited | Andrés Felipe | CC BY-SA 4.0 |
added 6 characters in body
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| Oct 2, 2020 at 3:50 | history | asked | Andrés Felipe | CC BY-SA 4.0 |