Timeline for Properties of the total variation norm on space of totally finite measure (from Bogachev)
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Jun 28, 2020 at 14:22 | vote | accept | Léo D | ||
| Jun 28, 2020 at 14:04 | answer | added | Nate Eldredge | timeline score: 3 | |
| Jun 28, 2020 at 13:56 | comment | added | Dieter Kadelka | I have difficulties to understand the problem. Let $\|\mu\| := \sup_{A \in \cal{B}} |\mu(A)$. Then as noted in (2) $\|,\|$ and $\|,\|_{TV}$ are equivalent norms. Further $\|\mu\| = \max\{\mu^+(X),\mu^-(X)\}$. Where is the problem? In (3) there is no problem, both sides define equivalent metrics. | |
| Jun 28, 2020 at 13:52 | vote | accept | Léo D | ||
| Jun 28, 2020 at 14:22 | |||||
| Jun 28, 2020 at 11:38 | comment | added | Léo D | Thanks, if I understand: for $P$ and $Q$ probability measures, $d_{TV}(P,Q) \neq ||P-Q||_{TV}$ where $d_{TV}$ is defined in \eqref{3} and $||\cdot||_{TV}$ is defined in \eqref{0}. This is where my mistake comes from I think. | |
| Jun 28, 2020 at 11:34 | history | edited | Léo D | CC BY-SA 4.0 |
added 34 characters in body
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| Jun 28, 2020 at 10:41 | comment | added | leo monsaingeon | There's no contradiction, since for probability measures $\mu^-=0$. Also, $\|P\|_{TV}=d_{TV}(P,0)$. The "problem" arises when BOTH $\mu^\pm$ are nonzero, as clearly illustrated by Stefan's counterexample. | |
| Jun 28, 2020 at 10:25 | answer | added | Stefan Waldmann | timeline score: 2 | |
| S Jun 28, 2020 at 9:48 | history | suggested | Daniele Tampieri | CC BY-SA 4.0 |
Added hyperlinks to formulas + minor Math Jaxing ($\|$ instead of $||$) + typo correction
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| Jun 28, 2020 at 9:46 | review | Suggested edits | |||
| S Jun 28, 2020 at 9:48 | |||||
| Jun 28, 2020 at 9:36 | review | First posts | |||
| Jun 28, 2020 at 10:21 | |||||
| Jun 28, 2020 at 9:32 | history | asked | Léo D | CC BY-SA 4.0 |