IMC 2020 · Day 2 · P5

We shall show that only positive constant functions satisfy the condition.

Let $g(x) = \dfrac{1}{f(x)}$. Notice that $$g'' = \left( \frac1f \right)'' = \left( \frac{-f'}{f^2} \right)' = \left( \frac{2 (f')^2 - f'' f}{f^3} \right) \le 0,$$ so the positive function $g(x)$ is concave. We show that $g$ must be constant.

Take two arbitrary real numbers $a < b$. By the concavity of $g$, for all $u < a$ and $v > b$ we have $$\frac{g(a) - g(u)}{a - u} \ge \frac{g(b) - g(a)}{b - a} \ge \frac{g(v) - g(b)}{v - b}.$$ Combining this with $g(u), g(v) > 0$ we get $$\frac{g(a)}{a - u} > \frac{g(b) - g(a)}{b - a} > \frac{-g(b)}{v - b}$$ Now by taking limits $u \to -\infty$ and $v \to \infty$ we obtain $$0 \ge \frac{g(b) - g(a)}{b - a} \ge 0,$$ so $g(a) = g(b)$. This holds for any pair $(a, b)$, so $g(x)$ is constant and $f(x) = 1/g(x)$ also is constant.

If $f$ is constant then $f' = f'' = 0$, so the condition is satisfied.

Remark. Instead of the function $1/f(x)$, the same idea works with $\arctan f(x)$: $$(\arctan f(x))'' = \frac{f'' (1 + f^2) - 2 (f')^2}{(1 + f^2)^2} = \frac{f'' (1 + f^2) - 2 (f')^2 (1 + f^2)}{(1 + f^2)^2} = \frac{f'' - 2 (f')^2}{1 + f^2} \ge 0.$$ As can be seen, $\arctan f(x)$ is a bounded convex function, therefore it must be constant.