User john griesmer - MathOverflow

Answer by John Griesmer for Quantitative versions of ergodic theorem

2010-10-31T22:20:07Z

A. Leibman proved a quantitative lower bound for the averages $\frac{1}{N}\sum_{n=0}^{N-1} \mu(A\cap T^{-n}A)$ in terms of $\mu(A)$ (note: the sum begins at $n=0$). The bound is $$ \frac{1}{N}\sum_{n=0}^{N-1} \mu(A\cap T^{-n}A) \geq \sqrt{\mu(A)^2+(1-\mu(A))^2} + \mu(A)-1 $$ for all $N\geq 1$ when $T$ is a measure preserving transformation of a probability space, and this is the best possible such bound.

Answer by John Griesmer for Furstenberg's Conjecture on 2-3-invariant continuous probability measures on the circle

2010-10-31T01:44:13Z

Manfred Einsiedler and Alexander Fish have a paper (arxiv.org/abs/0804.3586) showing that a multiplicative subsemigroup of $\mathbb N$ which is not too sparse satisfies the desired measure classification. The semigroups they consider are still somewhat large, and in particular not contained in finitely generated semigroups.

One can check that the set of perfect squares $E$ satisfies measure classification as follows: since $\frac{1}{N}\sum_{n=1}^N \exp(2\pi i n^2 \alpha)\to 0$ for every irrational $\alpha,$ one can conclude as in Wiener's lemma that an atomless measure $\mu$ on $S^1$ satisfies $\lim_{N\to \infty} \frac{1}{N}\sum_{n=1}^N|\widehat{\mu}(n^2)|=0.$ As in the original post, it follows that an atomless $E$-invariant measure is Lebesgue.

Comment by John Griesmer

2010-11-02T19:32:36Z

I think this description of $C_u(\mathbb R)$ says the extreme points of the unit ball of $C_u(\mathbb R)^*$ are finitely additive $\{0,1\}$-valued measures supported on cosets of $\mathbb Z.$ This makes a bit of sense, since an atomless $\{0,1\}$-valued finitely additive measure supported on $\{n+\frac{1}{n}:n\in \mathbb N\}$ induces an element of $C_u(\mathbb R)^*$ which comes from such a measure supported on $\mathbb N,$ thanks to uniform continuity.