IMC 2024 · Day 2 · P6

We can replace $f(x)$ by the function $g(x) = f(1 - x)$, so without loss of generality we can assume $f(0) \le f(1)$.

If $f(1) \ge f(2)$ then we can choose $(a, b, c) = (0, 1, 2)$. Otherwise we have $f(0) \le f(1) < f(2)$. If there is some $x \in (1, 2)$ such that $f(x) \ge f(2)$ then we can chose $(a, b, c) = (1, x, 2)$; similarly, if there is some $x \in (1, 2)$ with $f(x) \le f(1)$ then choose $(a, b, c) = (0, 1, x)$. Hence, in the remaining cases we have $f(1) \le f(x) \le f(2)$ for all $x \in (1, 2)$.

Now $f$ is bonded on the interval $[1, 2]$, so it has only finitely many values on this interval. Since there are infintely many rational numbers in $[0, 1]$, there is a value $y$ that is attained infinitely many times. The we can choose $1 \le a < b < c \le 2$ such that $f(a) = f(b) = f(c) = y$.

Assume towards a contradiction that there is a function $f$ which does not satisfy the claim: for all rationals $a, b, c$ with $a < b < c$ we have $f(b) < f(a)$ or $f(b) < f(c)$.

Let $x$ and $y$ be arbitrary rationals with $x < y$. Let $I(x, y) = [x, y] \cap \mathbb{Q}$. We first observe that $\inf f(I(x, y)) = -\infty$. Indeed, if the infimum was finite, then, as the set $f(I(x, y))$ is bounded ($\sup f(I(x, y)) = \max\{f(x), f(y)\}$) and thus finite, there are three points having the same value under $f$, which leads to a contradiction regarding our assumption on $f$.

So, going back to the question at hand, let $x$, $b$, $y$ be arbitrary rationals with $x < b < y$. Applying the above observation to the set $I(x, b)$, there exists a point $a \in I(x, b)$ such that $f(a) < f(b)$. Similarly, there exists a point $c \in I(b, y)$ such that $f(c) < f(b)$. Hence we have the points $a$, $b$, $c$ with $a < b < c$ and $f(b) > \max\{f(a), f(c)\}$, which contradicts our assumption on $f$.