IMC 1999 · Day 1 · P3

Writing (1) with $n-1$ instead of $n$, $$\tag{2} \left| \sum_{k=1}^{n-1} 3^k \bigl( f(x + ky) - f(x - ky) \bigr) \right| \le 1.$$ From the difference of (1) and (2), $$\left| 3^n \bigl( f(x + ny) - f(x - ny) \bigr) \right| \le 2;$$ which means $$\tag{3} \bigl| f(x + ny) - f(x - ny) \bigr| \le \frac{2}{3^n}.$$ For arbitrary $u, v \in \mathbb{R}$ and $n \in \mathbb{N}$ one can choose $x$ and $y$ such that $x - ny = u$ and $x + ny = v$, namely $x = \frac{u+v}{2}$ and $y = \frac{v-u}{2n}$. Thus, (3) yields $$\bigl| f(u) - f(v) \bigr| \le \frac{2}{3^n}$$ for arbitrary positive integer $n$. Because $\frac{2}{3^n}$ can be arbitrary small, this implies $f(u) = f(v)$.