Timeline for Semifinite measure and spectral theorem
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 31, 2018 at 15:29 | comment | added | Student | @NikWeaver You misunderstand me. I look at you references but i don't find what i look for. | |
| May 31, 2018 at 15:02 | comment | added | Nik Weaver | @Student: I told you where to look. I understand you want me to do the work for you, but I'm not going to. | |
| May 31, 2018 at 12:39 | comment | added | Student | @NikWeaver In all references which I find $H$ is assumed to be separable. I hope that if you find a reference which contains the proof in a general cae ( $H$ is not separable) the tell me in order to cite it. Thank you very much for your help. | |
| May 29, 2018 at 23:56 | comment | added | Nik Weaver | @Student: I would just say "it is standard". The argument I gave is very standard stuff. | |
| May 29, 2018 at 21:58 | comment | added | Student | @NikWeaver thank you. I can say in my paper that the measure can be taken semifinite without citing a reference? | |
| May 29, 2018 at 21:42 | comment | added | Nik Weaver | @Student: I would just say "it is standard". If you really want a reference, you're going to have to go to the library and search through some books until you find it. Could be in Dunford & Schwarz, or in Reed & Simon, or Kadison-Ringrose, or Takesaki. | |
| May 29, 2018 at 19:34 | comment | added | Student | @NikWeaver Could you please give me a reference to cite it in a paper? In the paper the Hilbert space is not necessarely separable. I want to use the spectral theorem and I want to say that the measure $\mu$ can be taken semifinite but I don't find any references to cite it. Thanks a lot for your help. | |
| Mar 9, 2018 at 17:03 | comment | added | David C. Ullrich | @NateEldredge I was about to say that - didn't want to be pedantic... | |
| Mar 9, 2018 at 16:08 | comment | added | Nate Eldredge | To be pedantic, I guess we should say that $\mu$ can be taken localizable. There is not a unique $\mu$ satisfying the theorem, and there will certainly exist such $\mu$ that are not semifinite. | |
| Mar 8, 2018 at 20:50 | comment | added | Nik Weaver | No, for example counting measure on any set is localizable. It is a generalization of $\sigma$-finiteness to the nonseparable setting. Here is a little primer on localizablility for you (from a book of mine which is in press). | |
| Mar 8, 2018 at 20:16 | comment | added | Student | Thank you. I think that a localizable measure is not in general $\sigma$-finite. Do you agree with me? | |
| Mar 8, 2018 at 19:45 | vote | accept | Student | ||
| Mar 8, 2018 at 16:16 | comment | added | Nik Weaver | Well, these are exercises. $E_0$ is the smallest subspace which is closed in norm and closed under application of $A_1$, $A_1^*$, $A_2$, and $A_2^*$. | |
| Mar 8, 2018 at 15:59 | comment | added | Student | Thank you for your answer but I don't understand why $A_1E_0 \subseteq E_0$ and $A_1E_0^\perp \subseteq E_0^\perp$? | |
| Mar 7, 2018 at 15:35 | history | answered | Nik Weaver | CC BY-SA 3.0 |