0

Suppose f:R↦R is a continuous bijection. Show that the system x_n+1=f(xn) cannot have periodic points of prime period greater than 2. Hint: Use Sharkovskii's Theorem to reduce the problem to the case of periodic points of prime period 4, then use the Intermediate Value Theorem to prove the result by contraction.

I am not sure how to start disproving a periodic point of prime period 4 using IVT.

user147527
  • 1

1 Answers1

0

Assume $x_1$ has period $4$, so $f(x_1)=x_2$, $f(x_2)=x_3$, $f(x_3)=x_4$, $f(x_4)=x_1$. Wlog. $x_1=\min\{x_1,x_2,x_3,x_4\}$. Also we may replace $f$ with $f^{-1}$ and thus swap $x_2$ vs. $x_4$ if we like. Therefore wlog. $x_2<x_4$.

  • Either $x_3>x_2$; then $x_1<x_2<x_4$ and $f(x_1)=x_2<f(x_2)=x_3>f(x_4)=x_1$, hence IVT allows us to find $\xi_1\in(x_1,x_2)$ and $\xi_2\in(x_2,x_4]$ with $f(\xi_1)=f(\xi_2)=\frac{x_2+x_3}{2}$.
  • Or $x_3<x_2$; then $x_1<x_3<x_2$ and $f(x_1)<f(x_3)>f(x_2)$, hence the same IVT argument applies.
Hagen von Eitzen
  • 374,180