# $C^{1+\epsilon}$ conjugacy of expanding map on circle

A continuously differentiable map $$f:S^{1}\rightarrow S^{1}$$ is called expanding if $$|f^{'}(x)|>1$$ for all $$x\in S^{1}$$.

We can define the degree of f, def(f) to be number of preimage $$f^{-1}(x)$$, for any $$x\in S^{1}.$$

Theorem Any expanding map $$f:S^{1}\rightarrow S^{1}$$ of degree $$d\geq2$$ is topologically conjugate to linear of degree d.

QuestionLet f be $$C^{1+\epsilon}$$. Does the theorem work for $$C^{1+\epsilon}$$ conjugacy?if No,Does anyone know under what assumption we have $$C^{1+\epsilon}$$conjugacy above theorem?

Remark We have following theorem: Let $$2\leq r\leq t$$. If two orientation preserving expanding $$C^r$$ endomorphisms f and g of $$S^{l}$$ are absolutely continuously conjugate, then they are conjugate by a $$C^r$$ diffeomorphism. But, i do not assume $$C^{2}$$. In fact, i asked. Does Shub and Sullivan's theorem work for $C^{1+\epsilon}? • Certainly not: any$C^1$(or smoother) conjugacy preserves multipliers of fixed points. For the linear map of degree$d$, each periodic point of period$n$has multiplier$d^n$, so a necessary condition for being conjugate to linear is that each point of period$n$has multiplied$d^n\$. This is probably sufficient also by some version of Livsic’s theorem/Shub and Sullivan. – Anthony Quas Feb 23 at 1:45
• @AnthonyQuas :Thanks – A M Feb 23 at 21:37