2
$\begingroup$

I'm working on a homework problem which seems obvious, but I am having a hard time proving/completing. The problem can be stated as follows:

Let $f,g:$ $\mathbb R$ $\rightarrow$ $\mathbb R$ be $\mathcal C^1$ functions and suppose they are topologically conjugated by a diffeomorphism $h$. Let $x_0$ be a periodic point of period $n$ of $f$. Prove that $y_0=h(x_0)$ is a periodic point of period $n$ of $g$, and that $(f^n)'(x_0)=(g^n)'(h(x_0))$

The first part seems pretty obvious from the definition of topological conjugacy - it preserves all the dynamics, so the periodic point is preserved as well. Anything beyond this seems to be lost on me, and I can't figure out how to start the second part.

$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

Since you have $h(f^n) = g^n(h)$, just differentiate and use the chain rule: $$(h(f^n))' = (g^n(h))'$$ $$h'(f^n)\cdot(f^n)' = (g^n)'(h)\cdot h'$$ Evaluating at $x_0$, and recalling that $f^n(x_0)=x_0$ and $g^n(h(x_0))=h(x_0)$, $$h'(\;\underbrace{f^n(x_0)}_{x_0}\;)\cdot(f^n)'(x_0) = (g^n)'(h(x_0))\cdot h'(x_0)$$ $$\require{cancel}\cancel{h'(x_0)}\cdot(f^n)'(x_0) = (g^n)'(h(x_0))\cdot \cancel{h'(x_0)}$$ Cancelling $h'(x_0)$ (since it isn't zero), $$\boxed{(f^n)'(x_0) = (g^n)'(h(x_0))}$$ as desired.

$\endgroup$
3
  • $\begingroup$ Ah. I didn't know you could make that first claim about conjugates. I thought the conjugacy only applied to the functions, not their iterates. Thanks for clearing that up. $\endgroup$
    – A. Thomas Yerger
    Mar 18, 2014 at 21:16
  • 1
    $\begingroup$ Well, you have to prove to yourself that it's true. You can write (abbreviating composition) $f = h^{-1}gh$, so just note that $f^n = (h^{-1}gh)(h^{-1}gh)\cdots(h^{-1}gh)$ and all of the "interior" $h^{-1}h$ are identity, leaving $f^n = h^{-1}g^nh$. This gives what you want, right? $\endgroup$
    – MPW
    Mar 18, 2014 at 21:29
  • $\begingroup$ No of course this is true. As soon as you wrote it, it hit me. $\endgroup$
    – A. Thomas Yerger
    Mar 19, 2014 at 2:26

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .