Timeline for Is the weak derivative a Radon-Nikodym derivative?
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| Aug 8, 2024 at 12:38 | vote | accept | rod | ||
| S Apr 12, 2024 at 19:39 | history | rollback | rod |
Rollback to Revision 2 - Edit approval overridden by post owner or moderator
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| Apr 11, 2024 at 13:50 | history | suggested | postylem | CC BY-SA 4.0 |
spelling fix
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| Apr 11, 2024 at 13:19 | review | Suggested edits | |||
| S Apr 12, 2024 at 19:39 | |||||
| Sep 28, 2023 at 15:09 | vote | accept | rod | ||
| Aug 8, 2024 at 12:38 | |||||
| Sep 26, 2023 at 9:15 | history | removed from network questions | Asaf Karagila♦ | ||
| Sep 25, 2023 at 17:09 | history | became hot network question | |||
| Sep 25, 2023 at 9:39 | answer | added | humanStampedist | timeline score: 3 | |
| Sep 25, 2023 at 9:33 | comment | added | rod | @geetha290krm Also, could you provide an example of a function $f$ such that $LS_f$ is not a.c. w.r.t. the Lebesgue measure, but it has a weak derivative? I don't know any. | |
| Sep 25, 2023 at 9:21 | comment | added | rod | @GiuseppeNegro I had already edited the monotonicity hypothesis, thanks. I think there is indeed some connection. The Fundamental Theorem of Calculus for weak derivatives seems to say something very similar: $f(x)-f(a) = \int_a^x f'(t)dt$, i.e. you're writing $LS_f = f' dx$, or in other words: $f' = \frac{dLS_f}{dx}$ | |
| Sep 25, 2023 at 9:17 | comment | added | rod | @geetha290krm Thanks for the useful comment. I tried to edit the question (I added the request that we only consider the absolutely continuous part of the measure when we take the R-N derivative). | |
| Sep 25, 2023 at 9:16 | comment | added | Giuseppe Negro | You must require $f$ to be monotone increasing also. Otherwise $LS_f$ makes no sense. In any case I seriously doubt there is any connection here. We say "Radon-Nikodym derivative" to hint at the mnemonic $\nu=f\,d\mu\Rightarrow f=\frac{d\mu}{d\nu}$. That is not really a derivative. More of a mnemonic aid than anything else. | |
| Sep 25, 2023 at 9:15 | history | edited | rod | CC BY-SA 4.0 |
The conjecture could not hold true. Added details thaks to a useful comment by geetha290krm
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| Sep 25, 2023 at 9:11 | comment | added | Kavi Rama Murthy | $LS_f$ need not be absolutely continuous w.r.t. Lebesgue measure. | |
| Sep 25, 2023 at 9:05 | history | asked | rod | CC BY-SA 4.0 |