Putnam 2001, Problem B-5

Let $a$ and $b$ be real numbers in the interval $(0, 1/2)$, and let $g$ be a continuous real-valued function such that $g(g(x)) = a g(x) + b x$ for all real $x$. Prove that $g(x) = c x$ for some constant $c$.

Solution:
Note that $g(x) = g(y)$ implies that $g(g(x)) = g(g(y))$ and hence $x = y$ from the given equation. That is, $g$ is injective. Since $g$ is also continuous, $g$ is either strictly increasing or strictly decreasing. Moreover, $g$ cannot tend to a finite limit $L$ as $x \to +\infty$, or else we'd have $g(g(x)) - ag(x) = bx$, with the left side bounded and the right side unbounded. Similarly, $g$ cannot tend to a finite limit as $x \to -\infty$. Together with monotonicity, this yields that $g$ is also surjective.

Pick $x_0$ arbitrary, and define $x_n$ for all $n \in \mathbb{Z}$ recursively by $x_{n+1} = g(x_n)$ for $n > 0$, and $x_{n-1} = g^{-1}(x_n)$ for $n < 0$. Let $r_1 = (a + \sqrt{a^2 + 4b})/2$ and $r_2 = (a - \sqrt{a^2 + 4b})/2$ and $r_2$ be the roots of $x^2 - ax - b = 0$, so that $r_1 > 0 > r_2$ and $1 > |r_1| > |r_2|$. Then there exist $c_1, c_2 \in \mathbb{R}$ such that $x_n = c_1 r_1^n + c_2 r_2^n$ for all $n \in \mathbb{Z}$.

Suppose $g$ is strictly increasing. If $c_2 \neq 0$ for some choice of $x_0$, then $x_n$ is dominated by $r_2^n$ for $n$ sufficiently negative. But taking $x_n$ and $x_{n+2}$ for $n$ sufficiently negative of the right parity, we get $0 < x_n < x_{n+2}$ but $g(x_n) > g(x_{n+2})$, contradiction. Thus $c_2 = 0$; since $x_0 = c_1$ and $x_1 = c_1 r_1$, we have $g(x) = r_1 x$ for all $x$.

Analogously, if $g$ is strictly decreasing, then $c_2 = 0$ or else $x_n$ is dominated by $r_1^n$ for $n$ sufficiently positive. But taking $x_n$ and $x_{n+2}$ for $n$ sufficiently positive of the right parity, we get $0 < x_n < x_{n+2}$ but $g(x_n) < g(x_{n+2})$, contradiction. Thus in that case, $g(x) = r_2 x$ for all $x$.