Timeline for Repartition of 1's in the "Chacon word"
Current License: CC BY-SA 3.0
17 events
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| Mar 17, 2016 at 17:40 | comment | added | user78465 | The Chacon map is uniquely ergodic and therefore by the comment before the product of the Chacon map with an irrational rotation is uniquely ergodic. This implies that for any Riemann integrable function $g$ on $[0,1]\times [0,1]$ the Birkhoff averages (w.r.t. $C\times T$ in your notation) converges at every point. Since the function you are considering is Riemann integrable then (#) holds for all $(x,u)$. | |
| Mar 16, 2016 at 14:43 | comment | added | Stéphane Laurent | @IanMorris How would you conclude with the unique ergodicity ? Does it imply that $(\#)$ hold for every $(x,u)$ ? | |
| Mar 16, 2016 at 14:40 | history | edited | Stéphane Laurent | CC BY-SA 3.0 |
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| Mar 14, 2016 at 11:49 | comment | added | user78465 | @IanMorris. Is the Chacon map itself uniquely ergodic? If that is so then unique ergodicity of the cartesian square of the Chacon map with an irrational rotation will follow from Furstenberg's disjointedness paper since the Chacon map is weakly mixing and an irrational rotation is a Kroenecker (pure point spectrum) transformation. | |
| Mar 10, 2016 at 20:37 | comment | added | Ian Morris | The natural thing to do is try to prove that the product of the Chacon system with an irrational rotation (or a periodic orbit) must be uniquely ergodic. However it is not clear to me how to do this. | |
| Mar 10, 2016 at 0:52 | comment | added | Douglas Zare | The current version is much more reasonable than the original. Can there be any set that satisfies the original? | |
| Mar 9, 2016 at 20:01 | history | edited | Stéphane Laurent | CC BY-SA 3.0 |
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| Mar 9, 2016 at 19:21 | history | edited | Stéphane Laurent | CC BY-SA 3.0 |
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| Mar 9, 2016 at 19:04 | comment | added | Stéphane Laurent | @DouglasZare I did a correction after Ian's comment. Do you still think it is wrong ? | |
| Mar 9, 2016 at 18:56 | comment | added | Douglas Zare | If you mean the standard definition of natural density then your original conjecture is obviously false. | |
| Mar 9, 2016 at 18:53 | history | edited | Stéphane Laurent | CC BY-SA 3.0 |
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| Mar 9, 2016 at 18:51 | comment | added | Stéphane Laurent | @IanMorris Thanks for this remark. Indeed ! I need to do a correction. | |
| Mar 9, 2016 at 18:51 | comment | added | Stéphane Laurent | @DouglasZare en.wikipedia.org/wiki/Natural_density | |
| Mar 9, 2016 at 18:27 | comment | added | Douglas Zare | Do you mean something unusual by "natural density?" Are you only considering sets that are natural in the sense that they are not chosen to make this conjecture fail? | |
| Mar 9, 2016 at 18:12 | history | edited | Ian Morris |
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| Mar 9, 2016 at 18:11 | comment | added | Ian Morris | Perhaps I haven't understood the question properly: what happens if we take $S$ to be the set of all $i$ such that $B_\infty(i)=0$? | |
| Mar 9, 2016 at 15:45 | history | asked | Stéphane Laurent | CC BY-SA 3.0 |