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 was asking. Does Shub and Sullivan's theorem work for $C^{1+\epsilon}?

  • 3
    $\begingroup$ 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. $\endgroup$ Feb 23, 2019 at 1:45
  • $\begingroup$ @AnthonyQuas :Thanks $\endgroup$
    – Adam
    Feb 23, 2019 at 21:37


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