# weak-* versus entropy growth

General question. Let $\eta_{n}$ be a sequence of invariant measures on $\{0,1,2,...,p-1\}^{\mathbb{N}}$ and $B$ the Bernoulli uniform measure. Knowing that $\eta_{n} \rightarrow B$ in the weak-* topology, this, somehow, could help me to prove that $h(\eta_{n})\rightarrow\log p$?

I know it is easy to produce zero-entropy measures that converge in the weak-* topology to $B$. What I would like to know is: having the weak-* topology more some friendly hypothesis on the measures $\eta_{n}$, could I guarantee growth of entropy?

• The finitely determined property of a Bernoulli measure $B$ says if $\eta_n$ converge weakly to $B$ and $h(\eta_n)\to h(B)$, then $\bar d(\eta_n,B)\to 0$. This is a key part of Ornstein theory. – Anthony Quas Jan 29 '16 at 16:09
• It's a bit hard to imagine what kind of property you could be looking for that guarantees $h(\eta_n)\to h(B)$. As you point out, no local property is sufficient. Generally it's not too hard to check that $h(\eta_n)\to h(B)$ if it's true. The conclusion of the finitely determined property is then very strong. – Anthony Quas Jan 29 '16 at 16:13
• @Algernon: as the OP says, it's not hard to produce zero entropy invariant measures that converge to $B$. (the entropy is upper semi-continuous; not lower semi-continuous). – Anthony Quas Jan 29 '16 at 17:55
• @Algernon: I'm not sure what you think I misread. It is not true that weak$^*$-convergence to the uniform Bernoulli measure gives convergence in entropy. For example, here is a construction of a measure: $\eta_n=\frac{1}{np^n}\sum_{w\in\{0,\ldots,p-1\}^n}\sum_{k=0}^{n-1}\delta_{ \sigma ^k(\bar w)}$, where $\bar w$ is the periodic point $(\bar w)_i=w_{i\bmod n}$. Obviously $h(\eta_n)=0$. It's not hard to check that $\eta_n$ converge weakly to $B$. – Anthony Quas Jan 29 '16 at 19:36
• So to prove that the entropy is growing, I would try and understand the distribution of $x_0$ conditional on $x_1,\ldots,x_n$. If this is approximately uniform for most $x$ and for all $n$, then the entropy is close to $\log p$. – Anthony Quas Jan 29 '16 at 21:43