Timeline for Convergence of Radon Nikodym derivatives
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 9, 2017 at 18:05 | vote | accept | Eduardo | ||
| Mar 7, 2017 at 6:43 | comment | added | Julian Newman | This is a very nicely formulated question: although it initially appears as though some of your assumptions should be redundant, I have had to use every single one for my proof below (except the continuity of $h_n$, and the added statement about the measures having compact support). | |
| Mar 7, 2017 at 6:40 | answer | added | Julian Newman | timeline score: 3 | |
| Mar 6, 2017 at 23:38 | history | edited | Eduardo | CC BY-SA 3.0 |
added 64 characters in body
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| Mar 6, 2017 at 20:44 | comment | added | Eduardo | @Nate Eldredge the convergence is locally compact | |
| S Mar 6, 2017 at 20:11 | history | suggested | J.J. Green | CC BY-SA 3.0 |
minor typographic fixes
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| Mar 6, 2017 at 20:05 | review | Suggested edits | |||
| S Mar 6, 2017 at 20:11 | |||||
| Mar 6, 2017 at 19:35 | comment | added | Christian Remling | This will work as soon as you can guarantee that $h_n\, d\nu_n$ is close to $h\, d\nu_n$, and this will be the case under mild additional assumptions (for example, $\| h_n -h\|_{L^1(d\nu_n)}\to 0$ would give you that these measures are in fact close in norm). | |
| Mar 6, 2017 at 19:27 | comment | added | Nate Eldredge | In what sense does $h_n \to h$? Pointwise convergence is not enough. | |
| Mar 6, 2017 at 19:24 | history | edited | Nate Eldredge | CC BY-SA 3.0 |
spelling and grammar corrections
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| Mar 6, 2017 at 19:17 | history | asked | Eduardo | CC BY-SA 3.0 |