# Proving that $f$ has points with prime period 3

Exercise: Let $f:\mathbb{R}\longrightarrow\mathbb{R}$ be continuous, $n>3$, and $x_1,x_2,\cdots,x_n$ be points such that $x_1<x_2<\cdots<x_n$. Show that if $f(x_i)=x_{i+1}$ for $i\in\{1,2,...,n-1\}$ and $f(x_n)=x_1$, then $f$ has points with all prime periods.

Proof: $x_{n-3}$ and $x_{n-2}$ exist, because $n>3$. $f^3(x_{n-3})=x_n>x_{n-3}$ and $f^3(x_{n-2})=x_1>x_{n-2}$. If $g(x)=f^3(x)-x$, then $g(x_{n-3})=x_n-x_{n-3}>0$ and $g(x_{n-2})=x_1-x_{n-2}<0$. $f(x)$ is continous, so $f^3(x)$ is continuous. $x$ is also continuous. Therefore $g(x)$ must be continuous. So we can use the intermediate-value theorem to conclude there exists a $c\in(x_{n-3},x_{n-2})$ such that $g(c)=0$. This means $f^3(c)=c$ for this $c$. So $c$ is a period-3 point.

Question: We should now proof that $c$ is a prime-period 3 point. This means we should prove that $f(c)\neq c$. If we can do this, we can use Sarkovskii to complete the proof. But how can we prove this, if it's possible at all??

-
Small typo --- I think you want $f^3(x_{n-2})=x_1\lt x_{n-2}$. –  Gerry Myerson Jun 20 '13 at 9:46