Timeline for Do invariant measures maximize the integral?
Current License: CC BY-SA 3.0
21 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 20, 2015 at 15:14 | history | edited | YCor | CC BY-SA 3.0 |
Removed non-mathematical part of the title
|
| Nov 21, 2011 at 14:00 | history | edited | Valerio Capraro | CC BY-SA 3.0 |
deleted 1163 characters in body; edited title
|
| Nov 18, 2011 at 21:28 | comment | added | Kim Morrison | There is a meta discussion, specifically discussing the bounty, at tea.mathoverflow.net/discussion/1212/…. My suggestion would be to explicitly spell out that the bounty is entirely independent of the usual mathematical conventions regarding acknowledgement and priority. | |
| Nov 18, 2011 at 18:58 | comment | added | Gerhard Paseman | Indeed he did Alain. And how many of those were phrased in a fashion appropriate for MathOverflow? I am not opposing the occasional practice; I just think the presentation should be improved. Gerhard "It Might Be Others Too" Paseman, 2011.11.18 | |
| Nov 18, 2011 at 18:47 | comment | added | Alain Valette | @ Gerhard: After all, the late Paul Erdös also offered cash prizes for questions... (:-) | |
| Nov 18, 2011 at 17:27 | comment | added | Gerhard Paseman | I think the title should be "Do invariant measures maximize the integral? (Bounty offered)" . That phrasing is slightly less crass, and stirs more curiosity; people will read through the post to find out what the bounty is and how much; I think the present phrasing of the title is in the grey area of acceptability on MathOverflow. Gerhard "It May Be Just Me" Paseman, 2011.11.18 | |
| Nov 18, 2011 at 16:27 | answer | added | Matthew Daws | timeline score: 23 | |
| Nov 18, 2011 at 13:25 | history | edited | Valerio Capraro | CC BY-SA 3.0 |
added 462 characters in body; edited title
|
| Nov 14, 2011 at 14:47 | history | edited | Valerio Capraro | CC BY-SA 3.0 |
added 112 characters in body
|
| Nov 13, 2011 at 22:28 | history | edited | Valerio Capraro | CC BY-SA 3.0 |
added 587 characters in body
|
| Oct 23, 2011 at 16:22 | history | bounty ended | Valerio Capraro | ||
| Oct 16, 2011 at 16:08 | history | bounty started | Valerio Capraro | ||
| Sep 25, 2011 at 1:45 | comment | added | Valerio Capraro | By the way, I do not remember quite well, but maybe that functional cannot be continuous, since otherwise Fubini's theorem would hold - and this is false. | |
| Sep 25, 2011 at 1:42 | comment | added | Valerio Capraro | one of the reason why I cancelled my tentative of answer is that I used the continuity of the functional that is absolutely not clear. The reason why I believe that the result is true is the following: a positive answer to the question would imply that a certain game has a Nash equilibrium. Now, it turns that the players of this game have no particular information, so one expects that the solution is some kind of casual choice, i.e. a uniform distribution.. Now the question is What is a uniform distribution on the integers? My intuitive answer is that it is an invariant mean.. | |
| Sep 24, 2011 at 21:03 | comment | added | Pietro Majer | Valerio, your question seems to give for granted the existence of a maximizer. How do you see it? Actually the domain $\mathcal{M}(\mathbb{Z})$ is a nice convex $w^*$ compact subset of $(\ell_\infty)^*$, but is that functional $w^*$ continuous? | |
| Sep 3, 2011 at 11:35 | history | edited | Valerio Capraro | CC BY-SA 3.0 |
edited title
|
| Sep 3, 2011 at 8:51 | comment | added | Tapio Rajala | Yes, you are right. Silly, as advertised. | |
| Sep 3, 2011 at 8:44 | comment | added | Valerio Capraro | Unfortunately Fubini's theorem does not hold. For intance, let $\phi=\chi_{\mathbb N}, $\mu$ a invariant measure such that $\int\phi d\mu=1$ e $\vu$ an invariant measure such that $\int\phi d\vu=0$. The lackness of Fubini's theorem is, at the end, the point. | |
| Sep 3, 2011 at 8:34 | comment | added | Tapio Rajala | A simple observation which might be silly because I don't know if Fubini's theorem holds for finitely additive probability measures: If we are in a situation where we can use Fubini's theorem the mapping in the question is constant since $$\int\int\phi(x+y)d\nu(x)d\mu(y) = \int\int\phi(x+y)d\mu(y)d\nu(x)$$ $$= \int\int\phi(y)d\mu(y)d\nu(x) = \int\phi(y)d\mu(y).$$ | |
| Sep 3, 2011 at 1:15 | history | edited | Valerio Capraro | CC BY-SA 3.0 |
deleted 3 characters in body; edited title; added 3 characters in body; added 12 characters in body
|
| Sep 2, 2011 at 18:46 | history | asked | Valerio Capraro | CC BY-SA 3.0 |