Timeline for Counterexample of non-negative sequence weakly converging in $\mathscr{M}^1$ but not $L^1$
Current License: CC BY-SA 4.0
17 events
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| S Aug 3, 2018 at 17:33 | history | suggested | anomaly | CC BY-SA 4.0 |
Looks like the math was intended to be set rather than verbatim.
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| Aug 3, 2018 at 17:07 | review | Suggested edits | |||
| S Aug 3, 2018 at 17:33 | |||||
| May 27, 2013 at 6:28 | vote | accept | Ayman Moussa | ||
| May 27, 2013 at 6:26 | history | edited | Ayman Moussa | CC BY-SA 3.0 |
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| May 25, 2013 at 3:50 | answer | added | TaQ | timeline score: 0 | |
| May 24, 2013 at 13:40 | answer | added | Gerald Edgar | timeline score: 4 | |
| May 24, 2013 at 11:22 | history | edited | Bill Johnson |
added tag.
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| May 24, 2013 at 4:18 | answer | added | Bill Johnson | timeline score: 7 | |
| May 23, 2013 at 17:27 | comment | added | Davide Giraudo | Maybe in this aim Polya's theorem can be useful. | |
| May 23, 2013 at 13:09 | comment | added | Davide Giraudo | If $f\neq 0$, then we can work with probability measures. We have a sequence of probability measures $\mu_n$ which converges in law to $\mu$, and all these measures are absolutely continuous with respect to Lebesgue measure. By portmanteau theorem, we know that $\mu_n(A)\to \mu(A)$ provided that $\mu(\partial A)=0$. And the question is whether this holds when $A$ is an arbitrary measurable set. | |
| May 23, 2013 at 10:45 | history | edited | Ayman Moussa | CC BY-SA 3.0 |
added 2414 characters in body; added 6 characters in body; added 6 characters in body; added 2 characters in body
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| May 22, 2013 at 15:36 | vote | accept | Ayman Moussa | ||
| May 22, 2013 at 16:07 | |||||
| May 22, 2013 at 13:03 | comment | added | Rabee Tourky | Aliprantis and Border "infinite dimensional analysis" is perhaps the best for this for you as it caters for applied mathematicians from a pure math perspective and treats order structures of $L_1$ seriously because they are important in economics. In particular, the weak compactness of $L_1$ order intervals. You can get it in pdf form somehow for the internet. | |
| May 22, 2013 at 12:55 | comment | added | Rabee Tourky | $f\wedge g$ is is just $\inf{f,g}$, state by stat a.e. I'm using notation from Banach lattice theory. My conjecture is that $f_n$ converges to $f$ weakly because order intervals $\{g: 0\leq g\leq m(f+1)$ are weakly compact in $L-1$ and thus sequentially compact. | |
| May 22, 2013 at 12:43 | comment | added | Ayman Moussa | Thanks for the comment Rabee. Could you detail a bit more (what is the lattice operation ?), or give a reference about order intervals ? Thanks. | |
| May 22, 2013 at 12:29 | comment | added | Rabee Tourky | Notice order intervals are weakly sequentially compact. So take you convergent function $f$. Fix $m$, the sequence $f_n\wedge m(f+1)\in [0,m(f+1)]$ has a subsequence converges weakly in $L_1$ to some $f_m\leq f$ (note that lattice operations are not weakly continuous). Keep on doing this for $m$ and see what you get. | |
| May 22, 2013 at 8:46 | history | asked | Ayman Moussa | CC BY-SA 3.0 |