Timeline for System with invariant measure, but no ergodic measure.
Current License: CC BY-SA 3.0
14 events
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| Oct 4, 2011 at 19:14 | comment | added | Asaf | @Andre, thanks for the correction.About the proof of the Banach-Alouglu theorem (and the Riesz theorem),I think you can get them both (at-least for seperable metric cases) from Hilbert's cube. Embed X inside H,using the Titze extesnion theorem (or Urysohn's theorem),extend every function in C_{c}(X) to C_{c}(H).Given a measure on X, take its push-forward to H. Now using the Banach-Alouglu theorem for C_{c}(H)*,you are getting a weak-* compact set. There you have ergodic measures by Krein-Milman (or ergodic decomposition by Choquet) and you take the pre-images. Notice that mass can escape! | |
| Oct 4, 2011 at 17:24 | comment | added | André Caldas | @Asaf: Subspaces of locally-compact spaces might not be locally-compact. The rationals (or irrationals) are not locally-compact. As for the Krein-Milman theorem, what one needs is the topology to be generated by a separating family of linear functionals, and the set in question to be compact in this topology. | |
| Oct 4, 2011 at 10:20 | comment | added | Asaf | @Andre, By general metrization theorems, if your space is separable and metric, it can be embededd into the Hilbert's cube, hence locally compact. I think that losing so much topological assumptions might break down the Riesz Rep. theorem. Notice that the proof of Riesz's theorem uses Urysohn's functions, which are basically what you need in order to prove convergence of measurs in the non-compact case (up to some escape of mass). I must also mention that the Krein-Milman theorem (or Choquet's theorem) works in a very general settings (probably a locally convex topological vector space). | |
| Oct 4, 2011 at 2:46 | comment | added | André Caldas | @Theo: Thank you very much for your comment. You really made me think! Metrizability is for the Riesz Representation Theorem, so we can identify the finite signed measures and the functional linears on $C(X)$. How do you prove the $M(X)$ is weak$^*$ closed? I am more interested in dropping locally-compactness then metrizability. | |
| Oct 4, 2011 at 2:26 | comment | added | Theo Buehler | I don't understand the problem here. Metrizability of $X$ doesn't enter in any of your arguments as long as you have local compactness. By its definition the set of positive measures is a weak$^∗$-closed cone in $M(X)$, and thus it cuts out a compact set out of the unit ball, so as soon as you have invariant measures you have invariant ergodic measures by Krein-Milman. | |
| Oct 4, 2011 at 2:11 | history | edited | André Caldas | CC BY-SA 3.0 |
X was changed to a topological space and T to a continuous transformation.
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| Oct 4, 2011 at 0:53 | answer | added | Daniel Mansfield | timeline score: 1 | |
| Oct 3, 2011 at 15:58 | comment | added | André Caldas | @Jesse, The Krein-Milman needs a compact (in some kind of weak topology) convex set. For the compact metrizable case, the set of probability measures is compact int the weak* topology. For the locally-compact metrizable case, the set of (positive) measures $\mu$ with $\mu(X) \leq 1$ is weak* compact. | |
| Oct 3, 2011 at 15:49 | comment | added | Jesse Peterson | What's wrong with using the Krein-Milman theorem in the general situation? | |
| Oct 3, 2011 at 15:46 | comment | added | André Caldas | @Asaf, I am interested on probability measures. The entropy is the Kolmogorov-Sinai entropy. | |
| Oct 3, 2011 at 15:13 | comment | added | Asaf | Are you interested in finite measures or infinite measures? What is the notion of entropy you are referring to in the infinite case? Anyways, usually at any question in ergodic theory (especially in entropy theory), one usually deals with "standard" probability spaces (and maybe even Lebesgue spaces). | |
| Oct 3, 2011 at 14:40 | history | edited | André Caldas | CC BY-SA 3.0 |
Specify the topology
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| Oct 3, 2011 at 11:44 | comment | added | André Caldas | This post is related to mathoverflow.net/questions/76908/… | |
| Oct 3, 2011 at 11:30 | history | asked | André Caldas | CC BY-SA 3.0 |