China Mathematical Competition

It is easy to find that $a = f(1) - f(0)$, $b = f(0)$. Then
$$ab = f(0) \cdot (f(1) - f(0)) = - \left( f(0) - \frac{1}{2} f(1) \right)^2 + \frac{1}{4} (f(1))^2 \le \frac{1}{4} (f(1))^2 \le \frac{1}{4}.$$
When $2 f(0) = f(1) = \pm 1$, i.e., $a = b = \pm \frac{1}{2}$, we get $a b = \frac{1}{4}$.
Therefore, the maximum of $a b$ is $\frac{1}{4}$.
The answer is $\frac{1}{4}$.