Timeline for Complex Borel measures: does $\mu_n \to \mu$ weakly imply $|\mu|(\Theta) \le \liminf_n |\mu_n|(\Theta)$ for every open subset $\Theta$?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| S Jan 15, 2023 at 19:04 | history | suggested | Oliver Díaz | CC BY-SA 4.0 |
Emphasize the question in the posting
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| Jan 15, 2023 at 17:57 | review | Suggested edits | |||
| S Jan 15, 2023 at 19:04 | |||||
| Nov 8, 2022 at 0:51 | vote | accept | Analyst | ||
| Nov 7, 2022 at 20:02 | history | edited | Iosif Pinelis |
edited tags
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| Nov 7, 2022 at 19:22 | comment | added | Oliver Díaz | @Analyst: Notice that if $\mu(dx)=e^{i\theta(x)}\,|\mu|(dx)$, then since $|e^{i\theta(x)}|=1$, $|\mu|(dx)=e^{-i \theta}\mu(dx)$ | |
| Nov 7, 2022 at 18:45 | comment | added | Analyst | @GiorgioMetafune From this Wikipedia page, the polar form is $\mathrm d \mu=e^{i \theta} \mathrm d|\mu|$, whereas yours is $\mathrm d |\mu| = e^{i \theta} \mathrm d\mu$. Could you elaborate on this difference? | |
| Nov 7, 2022 at 16:25 | answer | added | Iosif Pinelis | timeline score: 4 | |
| Nov 7, 2022 at 15:04 | comment | added | Giorgio Metafune | Write $|\mu(U)|=e^{i\theta}\mu(U)=\nu (U)$, where $\nu$ is the real measure $Re\, ( e^{i\theta}\mu)$. Then take $|f| \leq \chi_U$ real such that $$|\nu(U)| \leq \epsilon \int fd\nu=\epsilon +Re \int f e^{i\theta} d\mu.$$ Then it should go on as in your post. | |
| Nov 6, 2022 at 21:01 | comment | added | Analyst | @IosifPinelis I have tried but could not prove that $$ \liminf _{n \rightarrow \infty} \big (\left|\mu^1_n\right|(\Theta) + \left|\mu^2_n\right|(\Theta) \big ) \le \liminf_n \left|\mu_n\right|(\Theta) . $$ Please see my update. | |
| Nov 6, 2022 at 21:00 | history | edited | Analyst | CC BY-SA 4.0 |
added 2040 characters in body
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| Nov 6, 2022 at 11:14 | history | asked | Analyst | CC BY-SA 4.0 |