Timeline for Convergence of probability measures with respect to sequentially weak continuous functions
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 12, 2017 at 16:06 | comment | added | Michael Greinecker | @JorgeE.Cardona Yes, that is what I meant. | |
| May 12, 2017 at 15:13 | history | edited | Jorge E. Cardona | CC BY-SA 3.0 |
I have crossed-out a paragraph referencing an incomplete result, and now I am referencing a different source with the correct result.
|
| May 12, 2017 at 14:48 | comment | added | Jorge E. Cardona | @MichaelGreinecker Do you mean: "Let $\langle x_n \rangle$ be a sequence without a convergent subsequence"? | |
| May 12, 2017 at 14:47 | comment | added | Jorge E. Cardona | @MichaelGreinecker Thank you, I will refer to Bogachev. Since $X$ is also separable I can use the second part of Theorem 8.6.7. Let me check if I get the problem: From a sequence in the unit ball in $C(K_m)'$ we can only extract a subnet that may not be a sequence, right? To allow for a subsequence we need the unit ball in $C(K_m)'$ to be e.g. metrizable, this requires $C(K_m)$ to be separable. If $K_m$ is a compact metric space that works, right? I really appreciate your comment pointing this out with the right reference to Bogachev. | |
| May 12, 2017 at 7:59 | comment | added | Michael Greinecker | Kallianpur uses the compactness of the unit ball of the dual of $C(K)$ in the weak* topology, but to show that this unit ball is metrizable one typically uses the separability of $C(K)$. But $C(K)$ is only separable if $K$ is metrizable. | |
| May 12, 2017 at 7:59 | comment | added | Michael Greinecker | The Eberlein–Šmulian seems not to apply because "weak convergence of measures" is not "weak convergence" but "weak* convergence" from the functional analytic point of view. It is also stated in the Bogachev result mentioned that if the compact approximating sets used for tightness can be taken to be metrizable, then we actually get that every sequence has a converging subsequence. But weakly compact sets are generally not metrizable. | |
| May 12, 2017 at 7:36 | comment | added | Michael Greinecker | @JorgeE.Cardona I looked at the proof, and I don't really see how the subsequence is constructed. I don't think his proof works. Take a compact but not sequentially compact Hausdorff space. The standard example is $[01]^{[0,1]}$ with the product topology. Let $\langle x_n\rangle$ be a convergent sequence without a convergent subsequence. Then the family of Dirac measures $\{\delta_{x_n}:n\in\mathbb{N}\}$ is trivially tight, but not sequentially compact. The result as stated should hold at least for completely regular spaces though, this is Theorem 8.6.7. in Bogachev's book. | |
| May 11, 2017 at 22:29 | comment | added | Jorge E. Cardona | @MichaelGreinecker thanks, I guess I should have been more careful in citing the result in Kallianpur notes. From the Theorem 2.2.1., we have that a tight sequence has a weakly convergent subsequence. The limit is explicitly constructed in the proof. Can I apply Eberlein–Šmulian theorem to say that this sequential relative compactness actually means relative compactness? | |
| May 11, 2017 at 21:45 | comment | added | Michael Greinecker | Related to the remark by @NateEldredge: That a set of probability measures is relatively compact does not imply that the set is sequentially relatively compact. So there might be a sequence of probability measures that has no convergent subsequence but onlya convergent subnet, and a subnet of a sequence need not be representable by a subsequence. Maybe you can rule that out, but there should be some argument then. | |
| May 11, 2017 at 18:50 | history | edited | Jorge E. Cardona | CC BY-SA 3.0 |
added 95 characters in body
|
| May 11, 2017 at 18:48 | comment | added | Jorge E. Cardona | Hi, I'm basing this from the notes of Kallianpur in projecteuclid.org/euclid.lnms/1215451870 Theorem 2.2.1, the direction tight $\implies$ relatively weakly compact holds for Hausdorff spaces. The full Prokhorov requires Polish. Also, $X$ is reflexive as well. | |
| May 11, 2017 at 18:28 | comment | added | Nate Eldredge | Also, given that $\mu^n$ is tight, how do you conclude it has a weakly convergent subsequence? The usual version of Prohorov's theorem requires you are working on a separable metric space, but the weak topology isn't necessarily metrizable, not even on balls. Unless you also want to assume $X$ is reflexive? | |
| May 11, 2017 at 18:23 | history | edited | Jorge E. Cardona | CC BY-SA 3.0 |
added 10 characters in body
|
| May 11, 2017 at 18:23 | comment | added | Jorge E. Cardona | Yes, $X$ is separable, I will add that. Thanks. | |
| May 11, 2017 at 18:17 | comment | added | Nate Eldredge | Is $X$ assumed separable? I am not sure that the weak and strong topologies induce the same Borel $\sigma$-algebra otherwise. | |
| May 11, 2017 at 17:44 | history | asked | Jorge E. Cardona | CC BY-SA 3.0 |