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Jul 3, 2015 at 13:11 comment added D. Thomine @chaut: no, this is false. Let $T$ be the $\times 2$ map, and let $(u_n)$ be an enumeration of the dyadic rationals. Take $x = (0.u_1 v_1 u_2 v_2...)$, where the $v_n$ are finite sequences of $0$. Then, if the length of the $v_n$ grow fast enough, the orbit of $x$ spends most of its time close to $0$, so $\mu_x = \delta_0$. The orbit of $x$ will still be dense, as it goes very close to each dyadic rational.
May 11, 2015 at 14:13 comment added chaut Excuse me,@IanMorris but I have another question: Is it truth that if x has a dense positive orbit and $x$ generates an ergodic measure $\mu_x=\lim \frac{1}{n}\sum \delta_{T^jx}$ then its support is total?
Apr 27, 2015 at 18:52 comment added Pietro Majer ah, ok thanks -I assumed T was a diffeo :)
Apr 27, 2015 at 18:38 comment added Ian Morris @PietroMajer: let $T$ have degree two or higher, for example $T \colon \mathbb{R}/\mathbb{Z} \to \mathbb{R}/\mathbb{Z}$ defined by $T(x):=2x \mod \mathbb{Z}$.
Apr 11, 2015 at 22:02 comment added Asaf Even for the $\times p$ map, there are very wild sets who are the support for an invariant and ergodic measure, basically you can get Cantor sets of every possible dimension.
Apr 11, 2015 at 22:02 comment added Pietro Majer I don't understand... how can $T:\mathbb{S}^1\to \mathbb{S}^1$ have $T'(x)>1$ for all $x$ ?
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Apr 11, 2015 at 21:41 history asked chaut CC BY-SA 3.0