IMC 1998 · Day 2 · P9

Let $f_n(x) = \underbrace{f(f(\dots f}_{n}(x)))$. It is easy to see that $f_n(x)$ is a picewise monotone function and its graph contains $2^n$ linear segments; one endpoint is always on $\{(x,y) : 0 \le x \le 1,\ y = 0\}$, the other is on $\{(x,y) : 0 \le x \le 1,\ y = 1\}$. Thus the graph of the identity function intersects each segment once, so the number of points for which $f_n(x) = x$ is $2^n$.

Since for each $n$-periodic points we have $f_n(x) = x$, the number of $n$-periodic points is finite.

A point $x$ is $n$-periodic if $f_n(x) = x$ but $f_k(x) \ne x$ for $k = 1, \dots, n-1$. But as we saw before $f_k(x) = x$ holds only at $2^k$ points, so there are at most $2^1 + 2^2 + \dots + 2^{n-1} = 2^n - 2$ points $x$ for which $f_k(x) = x$ for at least one $k \in \{1, 2, \dots, n-1\}$. Therefore at least two of the $2^n$ points for which $f_n(x) = x$ are $n$-periodic points.