# Entropy, Convergence

imagine you have a sequence $\eta_{n}$ of (shift) invariant measures in the Bernoulli space $\{0,1\}^{\mathbb{N}}$ that satisfy the following: there are a $0<\delta <1$ and an $N$ such that $$n > N \Rightarrow \sum_{i=1}^{2^{m}}|\eta_{n}(C_{i}^{m}) - \frac{1}{2^{m}}|^{2}<\delta^{2n},$$ for all $m$ where

$C_{1}^{1} = , C_{2}^{1} = $,

$C_{1}^{2} =[0,0],...,C_{4}^{2}=[1,1]$,

$C_{i}^{m}$ is the $i^{th}$ cylinder of length $m$, with $i \in \{1,2,...,2^{m}\}$.

So, of course you have much more than $\eta_{n} \rightarrow \lambda$ in the weak* topology, where $\lambda$ is the Bernoulli uniform probability $(\frac{1}{2},\frac{1}{2})$. For instance, you have that all your cylinders are converging (in measure) to $\frac{1}{2^{m}}$ at the same time!

My question is: $h(\eta_{n})\rightarrow \log2$? Are the entropy of the measures growing to $\log2$?

I think you can make the measures have entropy as small as you like (even zero if you want, but this is slightly more complicated): regard $\{0,1\}^{\mathbb N}$ as a group (with coordinatewise addition modulo 2). Now you can define convolution of measures: $\int f\,d(\mu*\nu):=\int f(x+y)\,d\mu(x)\,d\nu(y)$. Let $P_n$ denote the collection of points fixed by the $n$th power of the shift and let $\nu_n$ be the uniform measure on $P_n$ so that $\nu_n$ is invariant. Finally pick your favourite $\delta<1$, let $\mu$ be any Bernoulli measure and let $\eta_n=\mu*\nu_{k_n}$ where $(k_n)$ is chosen below. Now $\eta_n(C^k_i)=2^{-k}$ for all $i$ and all $k\le k_n$. Now to estimate the sum: the sum is 0 for $m\le k_n$. For $m>k_n$, the sum is bounded above by $2\sum_i \eta_n(C^m_i)^2 + 2\sum_i (2^{-m})^2\le 2^{-m+1} + 2\sum_i \eta_n(C^m_i)^2\le 2^{-k_n}+\max_i \eta_n(C^m_i)$. Now just choose the $k_n$ to make sure this is at most $\delta^{2n}$.
• You are right professor... more than that: you don't need to pickup the measure $\mu$, those measures $\nu_{k_{n}}$ are the example yet... thanks for your answer... – Bruno Brogni Uggioni Jul 17 '15 at 21:15