Expanding maps of the circle and their coding
A continuous smooth ($C^{10}$-smooth, for example) map $f:S^1\to S^1$ of the circle is called expanding if $\inf_{x\in S^1} f'(x) > 1 $. Here $S^1 = [0,1]/\sim$, the segment with identified endpoints. It is well known that the dynamics of linear examples $f(x) = mx \pmod 1$ for natural $m\geq 2$ is conjugated (from the measure point of view) to the (full) shift on the space of one-sided sequences of digits $\{ 0, \ldots, m-1 \}$.
Question: is it true that any (non necessary linear) expanding map of the circle is conjugated to the one-sided full shift (with the number of letters equal to the topological degree of the map)?
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$\begingroup$I think I know the answer now. Yes, the conjugation exists but the measure that is generated on the full shift by the coding is not necessary Bernoulli for nonlinear maps. The reason is simple -- the coding map is the same both for nonlinear map $f$ and for its linearisation (when branches are made linear conserving partitions of injectivity), the ergodic measure on the circle is no longer exactly Lebesgue measure (but something equivalent to it via a positive density).
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