Consider the map $f(x)=3^x$ mod 1. Using the the iterated function system $T_{0}x=\log_{3}(x+1), T_{1}x=\log_{3}(x+2)$ we see that $f$ is dynamical conjugated to a full shift on two symbols. Moreover i think it follows from Lasota and York (Trans AMS, 1973) that there is an absolutely continuous ergodic measure for $f$. What is the density of this measure? Is this measure Bernoulli or Markov (an image of a Bernoulli or Markov measure under the conjugation)? What is the entropy of this measure?
1 Answer
As you say, it's an expanding map (min derivative at 0 is $\log 3$), so Lasota-Yorke and a bunch of other papers give that it has an absolutely continuous invariant measure. It's too much to hope that this measure is conjugate to a one-sided Bernoulli shift or a Markov chain. Instead, you can consider the natural extension, a canonically-define invertible map $\hat f$ that factors onto $f$. It is known that $\hat f$ is isomorphic to a Bernoulli shift. This is part of the industry dating back to the 1970's of studying expanding maps. Quite a good summary of everything is contained in Góra's 1994 paper in Ergodic Theory and Dynamical Systems.
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$\begingroup$ Dear Anthony, thanks for Your answer. Is there the possibility to calculate the entropy of the a.c. measure using the techniques You refer to? Best 9i $\endgroup$ Commented Oct 1, 2013 at 16:00
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2$\begingroup$ Estimate? Yes, easily; Calculate? Probably not. You can get a good estimate of the entropy using the fact that for absolutely continuous invariant measures, the entropy is the same as the Lyapunov exponent. Just pick a point arbitrarily, and compute $(1/n)\log (f^n)'(x)$. This is easy because you're just averaging the log of the derivative along the orbit. $\endgroup$ Commented Oct 1, 2013 at 21:26