Timeline for Does the Krylov-Bogolyubov construction preserve "ergodic statistics"?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Aug 6 at 23:20 | vote | accept | Julian Newman | ||
| Aug 6 at 21:51 | comment | added | Terry Tao | You're right; I've updated the answer. What the analysis shows is that the answer to the question is positive when $\mu$ is ergodic and negative otherwise. | |
| Aug 6 at 21:50 | history | edited | Terry Tao | CC BY-SA 4.0 |
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| Aug 6 at 19:49 | comment | added | Julian Newman | In your last sentence (before "The claim follows"), are you not assuming that $\mu$ is ergodic, as opposed to just invariant? [Incidentally, I think the dominated convergence theorem should imply that the whole sequence $\left(\frac{1}{N} \sum_{i=0}^{N-1} f^i_\ast\nu\right)_{N \geq 1}$ - without having to take a subsequence - will converge weakly to $\mu$ as $N \to \infty$, and so $\mathbb{P}$ is just equal to $\mu$.] | |
| Aug 6 at 19:31 | comment | added | Julian Newman | Thanks. I'm a bit confused - I think I've just realised that any example of the Bowen-Mañé phenomenon (physical measures that are not ergodic) will immediately be a counterexample to my claim, and yet you have proved that the claim is true. | |
| Aug 6 at 16:09 | history | edited | Terry Tao | CC BY-SA 4.0 |
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| Aug 6 at 16:03 | history | answered | Terry Tao | CC BY-SA 4.0 |