Prove that an attractive point of a map, then its conjugate is also attractive in the conjugate map

Question

I have the following problem:

Consider $I, J$ be intervals of $\mathbb{R}$ and $f : I \rightarrow I$ and $g: J \rightarrow J$ be continuous maps. We say $f$ is conjugate to $g$ if there exists a homeomorphism $h : I \rightarrow J$ such that $$h \circ f = g \circ h$$

Let $x_0$ be a periodic point of period $k$ of $f$ and $f$ be $\mathcal{C}^1$ in a neighbourhood of $x_0$ and $h$ is a local diffeomorphism in a neighbourhood of $x_0$. Assume $f$ is conjugate to $g$.

Prove that if $x_0$ is attracting (not neutral), then $h(x_0)$ is an attracting periodic point of $g$.

We know that $g = h \circ f \circ h^{-1}$, and the rule for a periodic point $x_0$ whose orbit is $x_0,\dots,x_{k-1}$ to be attractive is $$|f'(x_0)f'(x_1)\dots f'(x_{k-1})|<1$$

I tried to use chain rule to write $g'$ in terms of $f'$ and prove $|g'(x_0)g'(x_1)\dots g'(x_{k-1})|<1$, but I couldn't manage to do it.

Any help would be appreciated.

Answer 1

Hint: Consider the fact that $h \circ f = g \circ h \implies h \circ f^{k} = g^{k} \circ h$.

Solution:

Differentiating both sides of the hint we get $$ Dh(f^{k}(x))Df^{k}(x) = Dg^{k}(h(x))Dh(x)$$ Now if $x = x_0$ then $f^{k}(x_0) = x_0$ since $x_0$ is periodic with period $k$ and $Dh(x_0) \neq 0$ since $h$ is a diffeomorphism on some neighborhood of $x_0$. Thus we can manipulate the above equation to get $$\begin{align} Dg^{k}(h(x_0)) &= \frac{Dh(f^{k}(x_0))}{Dh(x_0)}Df^{k}(x_0) \\ &= \frac{Dh(x_0)}{Dh(x_0)}Df^{k}(x_0) \\ &= Df^{k}(x_0)\end{align}$$ Thus $|Dg^{k}(h(x_0))| = |Df^{k}(x_0)| < 1,$ so $h(x_0)$ is an attracting periodic point for $g$. Note that by the chain rule $Df^{k}(x_0) = \prod_{i=0}^{k-1}Df(f^{i}(x_0)) = f'(x_0)f'(x_1) \ldots f'(x_{k-1})$ which matches your definition (similarly for $Dg^{k}(h(x_0))$.