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Let $G = \{0,1\}^{\mathbb{N}} = \mathbb{Z}_{2}^{\mathbb{N}}$ be the Bernoulli space of two symbols, let $\sigma$ be the shift map and $M(G)$ the set of $\sigma$-invariant probabilities. Let $\bar{d}$ be the distance defined here: joining or coupling. I put a bar now. We already know that, in this context, we have $$\bar{d}(\eta,\mu)= \lambda([0]\times[1]) + \lambda([1]\times[0]),$$ where $\lambda$ is the joining which achieves the infimum in joining or coupling. We already know too that the entropy function $h:M(G) \rightarrow [0,\log2]$ is $\bar{d}$-continuous.

My question is now the converse: Is $\bar{d}$ h continuous (in the point $(\frac{1}{2},\frac{1}{2})$ the Bernoulli uniform probability?

More than that, would we have some good estimative like: if $h(\eta) > (\log2 - \varepsilon)$ so $\bar{d}(\eta,(\frac{1}{2},\frac{1}{2}))<\varepsilon$?

Thanks a lot for your attention

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I don't know bounds, but you can get the continuity that you want. The key point is that the Bernoulli $(\frac 12,\frac 12)$ measure, $\mu$, is finitely determined. This means that if another measure $\nu$ satisfies (1) $\nu$ is weak$^*$-close to $\mu$ (that is $\mu$ and $\nu$ are close on cylinder sets of length $n$) and (2) $h(\nu)>h(\mu)-\epsilon$, then $\nu$ is $\bar d$-close to $\mu$. The condition that $h(\nu)>\log 2-\epsilon$ ensures that on $n$-cylinders, $\nu$ is close to uniform; and then the result you want follows from the finitely determined property.

See http://www.scholarpedia.org/article/Ornstein_theory#Finitely_determined_.28FD.29_systems for more details.

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  • $\begingroup$ Mnay thanks for your answer, professor @Anthony Quas. I guess the measure $\nu$ close to $\mu$ has to be ergodic... I mean, the definition of finitely determined only works with ergodic process... but good to know this property... $\endgroup$ – Bruno Brogni Uggioni Oct 24 '15 at 19:37
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    $\begingroup$ I think for non-ergodic measures, you get what you want from ergodic decomposition and the fact that entropy is an affine function. $\endgroup$ – Anthony Quas Oct 24 '15 at 20:59

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