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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.

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  • $\begingroup$ How did you obtain $\frac{1}{4(1-\varepsilon)}$? $\endgroup$ – Uskebasi Apr 12 at 14:26
  • $\begingroup$ @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 $\frac{1}{4(1-\epsilon)}$. $\endgroup$ – xskxzr Apr 12 at 15:16
  • $\begingroup$ @xskxzr Where does this nice problem come from? $\endgroup$ – Robert Z Apr 16 at 11:32
  • $\begingroup$ @RobertZ It comes from a research, and I formalize it as this mathematical problem by myself. $\endgroup$ – xskxzr Apr 16 at 14:29
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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 $\epsilon \to 0^+$ $$\frac{\sup_{x\in [0,1]}(1-x)f(x)}{\int_0^1f(t)\,dt }=\frac{\epsilon(1-\epsilon)}{\frac{\epsilon^2}{2}-\epsilon\log(\epsilon)}=\frac{1-\epsilon}{\frac{\epsilon}{2}+\log(1/\epsilon)}\to 0^+.$$

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