Hausdorff dimension = entropy/Lyapunov exponent for the baker's map?

Let $\Sigma=\{0,1\}^{\mathbb Z}$ and let $\sigma:\Sigma\to\Sigma$ be the left shift. Then it is well known that $(\Sigma, \sigma)$ is conjugate to the baker's map $B$ of the unit square: $$B(x,y) = \begin{cases} (2x, \frac{y}{2}) &\text{ if } 0 \leq x < \frac{1}{2}, \\ (2x-1, \frac{y+1}{2}) &\text{ if } \frac{1}{2} \leq x < 1. \end{cases}$$ via the conjugating map $\pi:\Sigma\to[0,1]^2$ defined as follows: $$\pi((x_n)_{-\infty}^\infty)=\left(\sum_{n=1}^\infty x_n2^{-n}, \sum_{n=0}^\infty x_{-n}2^{-n-1} \right).$$ Now assume $X$ to be a subshift of $\Sigma$ and let $h$ stand for topological entropy.

Question. Is it true that $$\dim_H(\pi(X))=\frac{2h(X)}{\log 2}?$$ A weaker version (which is still sufficient for my needs) is whether $h(X)=0$ is equivalent to $\dim_H(\pi(X))=0$.

This looks like a standard result; however, the usual settings are one-dimensional or maps on manifolds.

• I think $d_H=2h(X)/\log 2$ would be a better bet: there are roughly $e^{2hn}$ 2-sided $n$-cylinders, mapping to regions of size $2^{-n}$ in the square. Indeed it's not hard to check that $d_H(\pi(X))\le 2h(X)/\log 2$. – Anthony Quas Apr 27 '16 at 19:01
• You are probably right, Anthony. Certainly agrees when $X=\Sigma$. I edited the question. – Nikita Sidorov Apr 27 '16 at 19:05
• I think you can get a lower bound also by taking a measure of maximal entropy, $\mu$, on $X$ and pushing it forward under $\pi$ to a measure on $[0,1]^2$. Now Shannon-MacMillan-Breiman says that for almost every $x$, there is an $n(x)$ such that $\mu([x]_{-k}^k)<\exp(-2k(h-\epsilon))$ for a.e. $x\in X$ and $k\ge n(x)$. Letting $S$ be the collection of $x$'s where $n(x)\le N_0$ for some large $N_0$ (so that $S$ has measure at least $\frac 12$), you should obtain a lower bound on the Haudorff measure of a covering set of $\pi(X)$ – Anthony Quas Apr 27 '16 at 19:41