Given: $f\in\mathcal C$ ($f$ is continuous), the range of $f$ is $(0,1)$.
Solve: $f\in\arg\inf_{f\in\mathcal C}(\max_{x\in[0,1])}x(1-f(x)))$
Here is my attempt:
Since $f(x)\in (0,1)$, $1-f(x)\in(0,1)$. So we first try to solve a simpler problem $g\in\arg\inf_{g\in\mathcal C}(\max_{x\in(0,1)}xg(x))$.
Take first order condition: $(xg(x))'=g(x)+xg'(x)$. I don't know how to proceed...
However, by thinking about the intuition behind this problem, it is sure that $f(0)\to 0$, $f(\epsilon)\to 0$ and $f(x)\to 1$ whenever $x>\epsilon$. The solution of the optimization problem would be: $x=\epsilon$ and the min is $\epsilon$.