Is there a closed form for $\inf_{f\in\mathcal{F}}\sup_{0\le x\le 1}(1-x)f(x)/\int_0^1f(t)\,\mathrm{d}t$?

Let $$\mathcal{F}$$ be the set of continuous and strictly increasing functions from $$[0,1]$$ to $$[0,1]$$ with $$f(0)=0$$. Is there a closed form for $$\inf_{f\in\mathcal{F}}\sup_{0\le x\le 1}\frac{(1-x)f(x)}{\int_0^1f(t)\,\mathrm{d}t}?$$

That is, the ratio of the following dark area to the integral area.

Note: this question is similar to this one, but this time I want to minimize the ratio when $$f$$ varies.

• How did you obtain $\frac{1}{4(1-\varepsilon)}$? – Uskebasi Apr 12 at 14:26
• @Uskebasi By symmetry we can assume $x\le 1-\epsilon$, then $f(x)=\frac{x\epsilon}{1-\epsilon}$, so we have $\frac{(1-x)f(x)}{\int_0^1f(t)\,\mathrm{d}t}=\frac{(1-x)x\epsilon}{(1-\epsilon)\epsilon}$, which has a maximum of $\frac{1}{4(1-\epsilon)}$. – xskxzr Apr 12 at 15:16
• @xskxzr Where does this nice problem come from? – Robert Z Apr 16 at 11:32
• @RobertZ It comes from a research, and I formalize it as this mathematical problem by myself. – xskxzr Apr 16 at 14:29

The desired infimum is zero. For $$\epsilon\in (0,1/2)$$, consider the strictly increasing continuous function $$f(x)=\begin{cases} \frac{\epsilon x}{1-x}& \text{if x\in [0,1-\epsilon]},\\ x& \text{if x\in [1-\epsilon,1].} \end{cases}$$ Then $$\int_0^1f(t)\,dt =\frac{\epsilon^2}{2}-\epsilon\log(\epsilon)$$ and $$\sup_{x\in [0,1]}(1-x)f(x)=\epsilon(1-\epsilon)$$. Hence, as $$\epsilon \to 0^+$$ $$\frac{\sup_{x\in [0,1]}(1-x)f(x)}{\int_0^1f(t)\,dt }=\frac{\epsilon(1-\epsilon)}{\frac{\epsilon^2}{2}-\epsilon\log(\epsilon)}=\frac{1-\epsilon}{\frac{\epsilon}{2}+\log(1/\epsilon)}\to 0^+.$$