Timeline for Essential spectrum of multiplication operator
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Jan 22, 2021 at 22:43 | comment | added | Christian Remling | @JochenGlueck: Yes, $\cup$, thanks. And I indeed had in mind the self-adjoint case exclusively. | |
| Jan 22, 2021 at 22:35 | comment | added | Jochen Glueck | Thanks for your reply! Yes, for sure your statement is correct, just a bit more complicated than necessary. By the way, the set $P \cup A$ (which you probably mean, rather than $P \cap A$), is equal to the essential spectrum in the self-adjoint case, but it can be larger than the essential spectrum in the non-self adjoint case. (If we use that definition that $\lambda$ is in the essential spectrum of $T$ iff $\lambda - T$ is not Fredholm). | |
| Jan 22, 2021 at 22:27 | comment | added | Christian Remling | @JochenGlueck: That is a good point. Still, my statement is correct (though unnecessary here); I was simply following the identity (definition?) $\sigma_{ess} = P\cap A$, with $P =$ eigenvalues of infinite multiplicity, $A=$ accumulation points of $\sigma$. | |
| Jan 22, 2021 at 21:21 | comment | added | Jochen Glueck | I'm not sure I understand the sentence "Too find all of $\sigma_{ess}$, add the accumulation points of $\sigma$ to this set." As pointed out in a comment above, the essential spectrum is simply the spectrum (and hence equal to the essential range of $a$). | |
| Jan 22, 2021 at 21:15 | history | edited | Christian Remling | CC BY-SA 4.0 |
added 15 characters in body
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| Jan 22, 2021 at 21:09 | history | answered | Christian Remling | CC BY-SA 4.0 |