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Romanian Mathematical Olympiad

Algebra Difficulty 5.4 AIME, harder Prove it Romania

Determine the continuous functions f:R→Rf : \mathbb{R} \to \mathbb{R} having the property that
f(x+1n)≤f(x)+1n, for all x∈R and n∈Z∗. f\left(x + \frac{1}{n}\right) \le f(x) + \frac{1}{n}, \text{ for all } x \in \mathbb{R} \text{ and } n \in \mathbb{Z}^*.

Solution

Inductively we obtain f(x+r)≤f(x)+rf(x+r) \le f(x)+r, for any x∈Rx \in \mathbb{R} and r∈Qr \in \mathbb{Q}.

The continuity of ff and the density of Q\mathbb{Q} in R\mathbb{R} give f(x+y)≤f(x)+yf(x+y) \le f(x)+y, for all x∈Rx \in \mathbb{R} and y∈Ry \in \mathbb{R}. We get thus the functions defined by fa(x)=x+af_a(x) = x + a, for x∈Rx \in \mathbb{R} where the parameter aa runs over R\mathbb{R}. It is obvious that all these functions verify the hypothesis.

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