Dynamical Systems: Period Orbits

Just want a quick check that my proof holds and I'm not missing something

Question: Consider an orbit with period $m$ whose minimal period is $p$. Prove that $\dfrac{m}{p}$ is an integer.

Proof: Suppose $x_0$ is our fixed period point. Since $p$ is the minimal period, we have that $p \leq m$ and $f^p(x_0) = x_0$ and $f^{p-k}(x_0) \neq x_0$ for $k\in \{1, 2, ..., p-1\}$. Since $m$ is a period of this orbit, then $f^m(x_0) = x_0$.

If $m=p$ then the statement is automatically true.

If $m > p$, then denote $m = rp + s$ where $r \in \mathbb{N}$ and $s \in \{ 0, 1, \ldots, p-1 \}$.

Hence



Hence $f^s(x_0) = x_0$.

But this is a contradiction if $0 < s < p$ where $p$ is the minimal period, hence $s=0$ is the only possibility.

Thus $m=rp \implies \dfrac{m}{p} = r \in \mathbb{N}$ as required.

Looks good to me.

OK so I'm having trouble getting somewhere here. The layout heavily hints at MVT but I'm having trouble incorporating it.

Suppose we have a period-2 orbit for which 2 is the minimal period; $\{ \overline{x_0,x_1} \}$.
So we have that $f(x_0) = x_1$ and $f(x_1) = x_0$, where $x_0 \neq x_1$ and $x_0,x_1 \in (a,b)$

The multiplier here is $\mu = f'(x_0) f'(x_1)$.

Since $|f'(x)| < 1$ for $x \in (a,b)$ we must have that $|\mu| =| f'(x_0) f'(x_1) | < 1$. So the orbit is an attractor (is linearly stable).

By MVT, $\exists c \in (a,b)$ such that $f'(c) = \dfrac{f(b)-f(a)}{b-a}$.

Past this I'm unsure how to reach a contradiction that says $x_0 = x_1$.

Think about

$\dfrac{f(x_{1})-f(x_{0})}{x_{1}-x_{0}}$

Is this what you have in mind?

Yup!