Timeline for Semifinite measure and spectral theorem
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 27, 2018 at 16:02 | answer | added | Student | timeline score: 1 | |
| Mar 18, 2018 at 14:10 | comment | added | Nate Eldredge | Yes, that is the issue. | |
| Mar 18, 2018 at 9:31 | comment | added | Student | @NateEldredge If I understand very well your comment, the problem is how to show that $$L^2(\mu)"=" L^2(\mu_1)?$$ | |
| Mar 11, 2018 at 14:33 | comment | added | Nate Eldredge | Let $\mu_0$ be the semifinite part of $\mu$. Consider the "identity map" from $L^2(\mu)$ to $L^2(\mu_0)$. I think I can show that it's well defined and is an isometry. What's not so clear is how to show that it is surjective. A function in $L^2(\mu_0)$ may be nonzero on some set which is an infinite atom for $\mu$, hence not in $L^2(\mu)$. One needs to show it is possible to modify it on a $\mu_0$-null set so as to fix this, and I can't see how to do that. | |
| Mar 11, 2018 at 11:17 | comment | added | Student | @NateEldredge Thank you for your details. If we replace $\mu$ by its semifinite part, where is the problem in order to finish the proof? Thank you. | |
| Mar 9, 2018 at 16:12 | comment | added | Nate Eldredge | Every localizable measure is semifinite. But Nik doesn't mean that $\mu$ is localizable, rather that it can be taken localizable, i.e. there always exists a localizable measure satisfying the theorem. (The same applies to your last sentence in your question: it should say that if $H$ is separable then $\mu$ can be taken $\sigma$-finite.) | |
| Mar 9, 2018 at 16:10 | comment | added | Student | @NateEldredge Yes I understand and thank. The answer of Nik Weaver says that $\mu$ is localizable but I don't understand very well the difference between localizable and semifinite. | |
| Mar 9, 2018 at 16:06 | comment | added | Nate Eldredge | What I am asserting is that there always exists a semifinite measure satisfying the given properties. Of course, there will always exist non-semifinite ones as well (take any such measure and if it's semifinite then consider a space with one additional point that has measure infinity). My suspicion is that you can prove there's a semifinite one by saying "let $\mu$ be a measure satisfying the conclusion of the spectral theorem; if it is not semifinite, let $\mu_0$ be its semifinite part and then $\mu_0$ satisfies." But I can't quite finish the proof of this. | |
| Mar 9, 2018 at 5:06 | comment | added | Student | @NateEldredge When we remplace $\mu$ by its semifinite part that doen't mean that $\mu$ is semifinite. Do you agree with me? | |
| Mar 8, 2018 at 19:45 | vote | accept | Student | ||
| Mar 8, 2018 at 14:45 | history | edited | Student | CC BY-SA 3.0 |
improved formatting
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| Mar 8, 2018 at 12:13 | history | edited | Student | CC BY-SA 3.0 |
improved formatting
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| Mar 7, 2018 at 15:39 | comment | added | Nate Eldredge | Sorry, wrong link: I meant math.stackexchange.com/questions/393985/… | |
| Mar 7, 2018 at 15:35 | answer | added | Nik Weaver | timeline score: 2 | |
| Mar 7, 2018 at 15:30 | comment | added | Nate Eldredge | If $\mu$ has an infinite atom, then every $L^2$ function must vanish on that set, so it might as well not be there at all. This suggests that you should be able to replace $\mu$ by its semifinite part and get the same $L^2$ space, though I have not checked the details. | |
| Mar 7, 2018 at 15:15 | history | asked | Student | CC BY-SA 3.0 |