Timeline for Essential spectrum of multiplication operator
Current License: CC BY-SA 4.0
19 events
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| Jan 24, 2021 at 18:06 | answer | added | Martin Väth | timeline score: 1 | |
| Jan 23, 2021 at 19:42 | comment | added | potionowner | @JochenGlueck Thanks for the detailed answers, comments, and discussion here. I guess the self-adjoint case is well-addressed. I still wonder whether there are some nice results for the non-self-adjoint case. Maybe my question could be translated as "Given the spectrum of a non-self-adjoint operator $a$, denoted as $\sigma(a)$, how can we identify $\sigma_{\mathrm{ess}}(a)"$ and $\sigma(a)\setminus \sigma_{\mathrm{ess}}(a)$"? If $\lambda\in \sigma (a)$ is isolated, then that's automatically a boundary point of $\sigma(a)$, isn't it? | |
| Jan 23, 2021 at 3:57 | history | became hot network question | |||
| Jan 22, 2021 at 21:37 | comment | added | Jochen Glueck | @potionowner: Please note that the last sentence of your previous comment is only true under additional assumptions: we can only conclude that a value $\lambda \in \sigma(a) \setminus \sigma_{ess}(a)$ is automatically isolated with finite multiplicity if $\lambda$ is a priori known to be a boundary point of the spectrum. (Of course, this is automatically true for self-adjoint operators - which are considered in the note you link). That's the main reason why I asked in the beginning whether you are interested in self-adjoint operators only. | |
| Jan 22, 2021 at 21:29 | answer | added | Jochen Glueck | timeline score: 2 | |
| Jan 22, 2021 at 21:09 | answer | added | Christian Remling | timeline score: 8 | |
| Jan 22, 2021 at 20:57 | comment | added | potionowner | @JochenGlueck (i) That's a very good catch. I mean $L^2([0, 1], \mathbb{R}^n)$ so that $a$ is an operator looks like an $\mathbb{R}^{n\times n}$-valued function over $[0, 1]$. (ii) From the note I attach in the OP (see Remark 1.4), the 'discrete spectrum' is defined $\sigma(a) \setminus \sigma_{\mathrm{ess}}(a)$. This concept is especially interesting since any $\lambda$ in the discrete spectrum must be isolated eigenvalues with finite multiplicity. | |
| Jan 22, 2021 at 20:53 | history | edited | potionowner | CC BY-SA 4.0 |
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| Jan 22, 2021 at 20:51 | comment | added | Jochen Glueck | I'm about to expand my comment into an answer including more details, but I'm still confused by two things: (i) In the case $n \not= 1$, do you really want to consider the space $L^2([0,1]; \mathbb{R}^{n \times n})$ - or are you rather interested in the space $L^2([0,1]; \mathbb{R}^{n})$, which means that the symbol of your multiplication operator takes values in $\mathbb{R}^{n \times n}$? (ii) Could you please specifiy what you mean by the notion discrete spectrum? | |
| Jan 22, 2021 at 20:48 | comment | added | potionowner | @JochenGlueck Would you mind giving me a reference on this result? Also, does it mean that the discrete spectrum of a self-adjoint operator is always empty? | |
| Jan 22, 2021 at 20:41 | comment | added | Jochen Glueck | @potionowner: Thanks for your reply! In the self-adjoint case the essential spectrum of $a$ coincides with the spectrum. This follows from a spectral projection argument and from the fact that $[0,1]$ does not contain any atoms. I'd guess that it is true for the non-self-adjoint case, as well, but I'm not quite sure right now. | |
| Jan 22, 2021 at 20:38 | comment | added | potionowner | @JochenGlueck But any result on the self-adjoint case would definitely be a good start for me! | |
| Jan 22, 2021 at 20:37 | comment | added | potionowner | @JochenGlueck Thanks for pointing this out! I'm interested in the case where $a$ is real. But this does not necessarily mean that $a$ is self-adjoint when $a$ is not $1$-dimensional. | |
| Jan 22, 2021 at 20:34 | history | edited | potionowner | CC BY-SA 4.0 |
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| Jan 22, 2021 at 20:29 | comment | added | Jochen Glueck | The notes that you linked for the definition of the essential spectrum only refer to self-adjoint case. So are you mainly interested in the case where $a$ is self-adjoint (i.e., where the symbol of $a$ is real)? | |
| Jan 22, 2021 at 20:14 | comment | added | Willie Wong | Ah, sorry, I didn't read carefully. | |
| Jan 22, 2021 at 20:11 | history | edited | potionowner | CC BY-SA 4.0 |
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| Jan 22, 2021 at 20:10 | comment | added | potionowner | @WillieWong Thanks for the comments. I know of this note. But I'm asking to compute the 'essential spectrum' of a multiplication operator, not the spectrum decomposition of `point, continuous and residual spectra'. | |
| Jan 22, 2021 at 19:56 | history | asked | potionowner | CC BY-SA 4.0 |