User john griesmer - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T06:13:51Z http://mathoverflow.net/feeds/user/10457 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/4411/quantitative-versions-of-ergodic-theorem/44385#44385 Answer by John Griesmer for Quantitative versions of ergodic theorem John Griesmer 2010-10-31T22:20:07Z 2010-10-31T22:20:07Z <p>A. Leibman <a href="http://www.math.osu.edu/~leibman/preprints/ErgEst.pdf" rel="nofollow">proved a quantitative lower bound</a> 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.</p> http://mathoverflow.net/questions/44234/furstenbergs-conjecture-on-2-3-invariant-continuous-probability-measures-on-the/44297#44297 Answer by John Griesmer for Furstenberg's Conjecture on 2-3-invariant continuous probability measures on the circle John Griesmer 2010-10-31T01:44:13Z 2010-10-31T01:44:13Z <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/44183/dual-of-bounded-uniformly-continuous-functions/44508#44508 Comment by John Griesmer John Griesmer 2010-11-02T19:32:36Z 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.