IMC 2006 · Day 2 · P8

The functions $f(x) = x + c$ and $f(x) = -x + c$ with some constant $c$ obviously satisfy the condition of the problem. We will prove now that these are the only functions with the desired property.

Let $f$ be such a function. Then $f$ clearly satisfies $|f(x) - f(y)| \le |x - y|$ for all $x, y$; therefore, $f$ is continuous. Given $x, y$ with $x < y$, let $a, b \in [x, y]$ be such that $f(a)$ is the maximum and $f(b)$ is the minimum of $f$ on $[x, y]$. Then $f([x, y]) = [f(b), f(a)]$; hence $$y - x = f(a) - f(b) \le |a - b| \le y - x$$ This implies $\{a, b\} = \{x, y\}$, and therefore $f$ is a monotone function. Suppose $f$ is increasing. Then $f(x) - f(y) = x - y$ implies $f(x) - x = f(y) - y$, which says that $f(x) = x + c$ for some constant $c$. Similarly, the case of a decreasing function $f$ leads to $f(x) = -x + c$ for some constant $c$.