Periodic point of dynamical system

Question

Hi please help me someone with the proof:

We have a function $f:\mathbb{R}\longrightarrow\mathbb{R}$ continous and invertible, discrete dynamical system is given by $x_{n+1}=f(x_n)$

(a): prove that any periodic point of $f$ must have minimal period $p\le2$

(b): prove that $f$ has no eventually periodic point

[Eventually periodic point means: point $x$ is an eventually periodic of period $p$ if $x$ is not periodic but there exists $m>0$ s.t. $x_m$ is a periodic point of period $p$]

Answer 1

Hint: Notice that $f$ is either strictly increasing or strictly decreasing. Show that:

• If $f$ is strictly increasing, then the only periodic points are fixed points.

• If $f$ is strictly decreasing, then the only periodic points are fixed points or have period $2$.

You can proceed by contradiction. For example, in the first case say that $x$ is a periodic point that is not a fixed point. Either $x<f(x)$ or $x>f(x)$. In the first case, you have $f(x)<f^2(x)$, because $f$ is increasing. You can continue iterating and you find that $x<f(x)<f^2(x)<\cdots$, which shows that $x$ is not a periodic point. Contradiction!