Timeline for If $(\exp(\mu_n))_{n\in\mathbb N}$ is weakly convergent, is the normalized sequence convergent as well?
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16 events
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| Jan 1, 2021 at 17:06 | comment | added | 0xbadf00d | Please take note of my answer: mathoverflow.net/a/380160/91890. | |
| Jan 1, 2021 at 15:41 | comment | added | 0xbadf00d | Didn't I already proven the equivalence? One direction is given in my comment above. And the other direction is given in this post. Is there any step in these proofs which is not clear to you or which you think is not justified? If so, I'll ask a separate question. Please let me know. | |
| Jan 1, 2021 at 15:19 | comment | added | Iosif Pinelis | @0xbadf00d : I am not sure about this, but my guess would be that that would not suffice. However, it would be much harder to construct a counterexample with that additional condition. I suggest you post that as a separate question. | |
| Jan 1, 2021 at 15:05 | comment | added | 0xbadf00d | Ah, I've mistakenly looked at $\left\|\nu_n\right\|=1$. So, that convergence is what we need to assume. Then the equivalence should hold. Do you agree? | |
| Jan 1, 2021 at 14:44 | comment | added | Iosif Pinelis | @0xbadf00d : In my example, $\|\mu_n\|$ does not converge. But, of course, you are (and everyone is) welcome to find a mistake there. | |
| Jan 1, 2021 at 5:51 | comment | added | 0xbadf00d | Am I missing something? Since the $\nu_n$ in your counterexample is a probability measure, something would be wrong in your post ... | |
| Jan 1, 2021 at 5:51 | comment | added | 0xbadf00d | Thank you for the confirmation. Thinking about that, don't we simply need to add the assumption $c:=\lim_{n\to\to\infty}\left\|\mu_n\right\|\ne0$ in order to conclude the implication I'm asking for in the question as well? Indeed, if $c\ne0$ and $\mu_n\xrightarrow{n\to\infty}\mu$ weakly, then it obviously holds $$\frac{\mu_n}{\left\|\mu_n\right\|}f\to\frac\mu cf$$ for all $f\in C_b(E)$, i.e. $\frac{\mu_n}{\left\|\mu_n\right\|}\to\frac\mu c$ weakly. In conclusion, as long as $c\ne0$, weak convergence of $\mu_n$ and $\mu_n/\left\|\mu_n\right\|$ are equivalent. Am I missing something? | |
| Dec 31, 2020 at 20:49 | comment | added | Iosif Pinelis | @0xbadf00d : That seems fine. | |
| Dec 31, 2020 at 18:44 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Dec 31, 2020 at 18:30 | comment | added | 0xbadf00d | (b) Yes, sorry, that was a typo. I've meant "converse implication". I've provided an answer for this implication in this post. Please let me know whether you agree or not. | |
| Dec 31, 2020 at 18:29 | vote | accept | 0xbadf00d | ||
| Dec 31, 2020 at 16:50 | comment | added | Iosif Pinelis | @0xbadf00d : (a) Yes, my $\exp^*(\mu)$ (showing that the exponentiation is convolutional) is your $\exp(\mu)$. (b) What do you mean by "converge implication"? If by this you mean "converse implication", I think it is better to ask this additional question in a separate post. Generally, I think multiple questions in one post should be avoided. | |
| Dec 31, 2020 at 16:46 | comment | added | 0xbadf00d | Regarding (b): This implication should generally hold. If $\mu_t$ is a net of finite signed measures such that $\mu_t/\left\|\mu_t\right\|$ weakly converges to $\mu$ and we additionally assume that $\left\|\mu_t\right\|\to a\ne 0$, then $\mu_tf=\left\|\mu_t\right\|\frac{\mu_t}{\left\|\mu_t\right\|}\to a\mu f$ for all $f\in C_b(E)$. Thus $\mu_t\to a\mu$. | |
| Dec 31, 2020 at 16:39 | comment | added | 0xbadf00d | Thank you for your answer. (a) Your "$\exp^\ast$" is simply my "$\exp$" or is there a difference? (b) Does the converge implication fail to hold as well? | |
| Dec 31, 2020 at 16:38 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Dec 31, 2020 at 16:16 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |