Timeline for Explicit example of a certain weak-* limit
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| Jun 16, 2021 at 10:50 | comment | added | user95282 | @MateuszWasilewski That is correct, the space of measures on the real line is w* sequentially complete; see e.g. Bogachev, Measure Theory (2007), section 8.7. So the set $\cal S$ as defined in the question is empty. | |
| Jun 14, 2021 at 19:19 | history | became hot network question | |||
| Jun 14, 2021 at 15:03 | vote | accept | Nate River | ||
| Jun 14, 2021 at 14:52 | answer | added | Nik Weaver | timeline score: 9 | |
| Jun 14, 2021 at 14:46 | comment | added | Nate River | @Diego Martinez ah you’re right - I had the result mixed up. | |
| Jun 14, 2021 at 14:34 | comment | added | Mateusz Wasilewski | If I remember correctly, the set of probability measures on a separable metric space is sequentially closed in the weak* topology of $C_b^{\ast}$, so $\mathcal{S}$ would be empty, if you didn't allow nets. I don't have a reference, though, so take it with a grain a salt. | |
| Jun 14, 2021 at 14:31 | comment | added | Diego Martinez | @NateRiver that's not the usual use of $C^0$, which is usually the bounded functions that vanish at infinity. Now those are usually denoted $C_0(X)$, which is indeed separable when $X$ is separable. However, the $C_b$ in your question is non-separable, as it contains $\ell^\infty$. | |
| Jun 14, 2021 at 13:53 | comment | added | Nate River | Ah, so I use $C^0$ to mean the set of continuous functions. $C_b$ being an open subset of $C^0$ should then be separable a well. | |
| Jun 14, 2021 at 13:44 | comment | added | Matthew Daws | Ah, okay, I guess I misread "...and the corresponding probability measures". But where in the question does $C^0(X)$ occur? | |
| Jun 14, 2021 at 13:25 | comment | added | Nate River | Also I believe $C^0 (X)$ is separable for $X$ separable. | |
| Jun 14, 2021 at 13:14 | comment | added | Nate River | I didn’t specify that the element of $\mathcal S$ needed to be induced by a probability measure though. | |
| Jun 14, 2021 at 12:07 | comment | added | Matthew Daws | Also, as $C_b$ is non-separable, I am not completely sure (but I could be wrong) whether you really can get the sequence $(\mu_n)$ to converge, or whether you will need to look as a sub-net of $(\mu_n)$, equivalently, as limit-point of $(\mu_n)$. | |
| Jun 14, 2021 at 12:01 | comment | added | Matthew Daws | The question doesn't quite make sense: every probability measure on $[0,\infty)$ induces a member of $C_b^*$, but the converse is very far from being true. So $\mathcal S$ need not actually contain any functional induced by a probability measure. | |
| Jun 14, 2021 at 11:57 | history | edited | Nate River | CC BY-SA 4.0 |
added 89 characters in body
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| Jun 14, 2021 at 11:56 | comment | added | Nate River | Ah yes, sorry. Edited. | |
| Jun 14, 2021 at 11:15 | history | asked | Nate River | CC BY-SA 4.0 |