An integral inequality with inverse

Question

Let $f:[0,1]\to [0,1]$ be a non-decreasing concave function, such that $f(0)=0,f(1)=1$. Prove or disprove that : $$ \int_{0}^{1}(f(x)f^{-1}(x))^2\,\mathrm{d}x\ge \frac{1}{12}$$ A friend posed this to me. He hopes to have solved it, but he is not sure. Can someone help? Thanks.

Answer 1

Let $f(x)=x^{\frac{1}{k}}$ for some $k\geq 1$. Then $f^{-1}(x)=x^k$, so we have

$$\int_{0}^{1}(x^{\frac{1}{k}}x^k)^2\,\mathrm{d}x = \int_{0}^{1}(x^{k+\frac{1}{k}})^2\,\mathrm{d}x = \frac{x^{2k+\frac{2}{k}+1}}{2k+\frac{2}{k}+1}\Big|_0^1 = \frac{1}{2k+\frac{2}{k}+1}$$

which can be made arbitrarily small.

Answer 2

The inequality cannot hold since a non-decreasing concave function $f$ satisfing $f(0)=0$ and $f(1)=1$ can be almost everywhere arbitrarily close to one, with its inverse function being almost everywhere arbitrarily close to zero on $[0,1]$. Hence the given integral can be less than $\varepsilon$ for any $\varepsilon>0$.

Examples of such a phenomenon are given by $$ f(x) = \frac{(n+1)x}{n+x}\quad\mbox{for }n\geq 7, $$ $$ f(x) = x^r\quad\mbox{for }r\in(0,1/6),$$ $$ f(x) = x+\min\left(nx,\frac{n}{n+1}(1-x)\right)\quad\mbox{for }n\geq 4.$$

Answer 3

I am not so sure this works. Take $f(x)=nx$ for $x \in [0;\frac{1}{n+1}]$ and $f(x)=\frac{n}{n+1} + (x-\frac{1}{n+1})\frac{1-\frac{n}{n+1}}{1-\frac{1}{n+1}} = \frac{n}{n+1} + \frac{1}{n}(x-\frac{1}{n+1})$ on $]\frac{1}{n+1},1]$.

Then $f^{-1}(y)= \frac{y}{n}$ if $y \in [0,\frac{n}{n+1}]$ and $f^{-1}(y)= n(y-\frac{n}{n+1})+\frac{1}{n+1}$ if $y \in [\frac{n}{n+1},1]$

Then, noticing that$\forall x$ $f(x)\leq 1$ and $f^{-1}(x)\leq 1$, we can rewrite: $\int_0^1 (f f^{-1} )^2\leq \int_0^1 f^{-1}(x )^2dx \leq \int_0^{\frac{n}{n+1}}\frac{x^2}{n^2}dx + \int_{\frac{n}{n+1}}^1 1dx \leq \frac{1}{n ^2} + \frac{1}{n+1}$ which converges towards 0.