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 $2f(0)=f(1)=\pm 1$, i.e., $a=b=\pm \frac{1}{2}$, we get $ab=\frac{1}{4}$. Therefore, the maximum of $ab$ is $\frac{1}{4}$. The answer is $\frac{1}{4}$.