Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.1 AIME, harder Prove it Thailand

Let $f: \mathbb{R} \to \mathbb{R}$ be a function satisfying
$|f(x+y) - f(x) - f(y)| < 1 \text{ for all } x, y \in \mathbb{R}.$
Prove that $\left| f\left(\frac{x}{2008}\right) - \frac{f(x)}{2008} \right| < 1$ for all $x \in \mathbb{R}$ .

Solution

$\begin{aligned} \left| f(2008x) - 2008f(x) \right| &= \left| \sum_{k=1}^{2007} \left( f((k+1)x) - f(x) - f(kx) \right) \right| \\ &\le \sum_{k=1}^{2007} \left| f((k+1)x) - f(x) - f(kx) \right| < 2007. \end{aligned}$
Replacing $x$ with $\frac{x}{2008}$ and simplifying, one gets
$\left| \frac{f(x)}{2008} - f\left(\frac{x}{2008}\right) \right| < \frac{2007}{2008} < 1.$

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