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


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.

  • $\begingroup$ 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
  • 2
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.