IMC 2018 · Day 1 · P4

First we show that $f$ is infinitely many times differentiable. By substituting $a = \frac12 t$ and $b = 2t$ in (2), $$\tag{3} f'(t) = \frac{f(2t) - f(\frac12 t)}{\frac32 t}.$$ Inductively, if $f$ is $k$ times differentiable then the right-hand side of (3) is $k$ times differentiable, so the $f'(t)$ on the left-hand-side is $k$ times differentiable as well; hence $f$ is $k + 1$ times differentiable.

Now substitute $b = e^h t$ and $a = e^{-h} t$ in (2), differentiate three times with respect to $h$ then take limits with $h \to 0$: $$\begin{gather*} f(e^h t) - f(e^{-h} t) - (e^h t - e^{-h} t) f(t) = 0 \\$$

\left( \frac{\partial}{\partial h} \right)^3 \Bigl( f(e^h t) - f(e^{-h} t) - (e^h t - e^{-h} t) f(t) \Bigr) = 0 \\ e^{3h} t^3 f'''(e^h t) + 3 e^{2h} t^2 f''(e^h t) + e^h t f'(e^h t) + e^{-3h} t^3 f'''(e^{-h} t) + 3 e^{-2h} t^2 f''(e^{-h} t) + e^{-h} t f'(e^{-h} t) - {} \\ {} - (e^h t + e^{-h} t) f'(t) = 0 \\ 2 t^3 f'''(t) + 6 t^2 f''(t) = 0 \\ t f'''(t) + 3 f''(t) = 0 \\ (t f(t))''' = 0. \end{gather*} Consequently, $t f(t)$ is an at most quadratic polynomial of $t$, and therefore $$\tag{4} f(t) = C_1 t + \frac{C_2}{t} + C_3$$ with some constants $C_1$, $C_2$ and $C_3$.

It is easy to verify that all functions of the form (4) satisfy the equation (1).