Question 117
Let $0 \lt a \lt b \lt 1$ be real numbers and
$$
g(x) = \begin{cases} x + 1 - a, & \text{if} 0 \lt x \lt a, \\ b - a, & \text{if} x = a, \\ x - a, & \text{if} a \lt x \lt b, \\ 1 - a, & \text{if} x = b, \\ x - a, & \text{if} b \lt x \lt 1. \end{cases}
$$
Assume that for some positive integer $n$ there are $n+1$ real numbers $0 \lt x_0 \lt x_1 \lt \dots \lt x_n \lt 1$ such that $g^n(x_i) = x_i$ for $0 \le i \le n$. Prove that there is a positive integer $N$ such that $g^N(x) = x$ for all $0 \lt x \lt 1$. (Notation: $g^k(x) = \underbrace{g(g(\dots(g(x))\dots))}_{k \text{times}}$).
$$
g(x) = \begin{cases} x + 1 - a, & \text{if} 0 \lt x \lt a, \\ b - a, & \text{if} x = a, \\ x - a, & \text{if} a \lt x \lt b, \\ 1 - a, & \text{if} x = b, \\ x - a, & \text{if} b \lt x \lt 1. \end{cases}
$$
Assume that for some positive integer $n$ there are $n+1$ real numbers $0 \lt x_0 \lt x_1 \lt \dots \lt x_n \lt 1$ such that $g^n(x_i) = x_i$ for $0 \le i \le n$. Prove that there is a positive integer $N$ such that $g^N(x) = x$ for all $0 \lt x \lt 1$. (Notation: $g^k(x) = \underbrace{g(g(\dots(g(x))\dots))}_{k \text{times}}$).