Inequality with definite integrals

Question

This problem has been bugging me for days. A function $f:[0,\,1]\to[0,\,1]$ with $f(0)=0$ and $f(1)=1$ is strictly increasing and differentiable, with $f'$ also strictly increasing. (So $f$ is a convex function.) I want to show that $2\left(\int_0^1f(x)\,\mathrm dx\right)^2 \, \geqslant \, f\left(\int_0^1f(x)\,\mathrm dx\right)$.

I’ve managed so far to show that $f(x)\leqslant x$ for all $0\leqslant x\leqslant1$ and $\int_0^1f(x)\,\mathrm dx\leqslant\frac12$. I’m stuck after this.

I’ve also shown (if it helps) that $f'(0)\leqslant1\leqslant f'(1)$. Any help would be appreciated. Thanks.

Answer 1

Step 1. $f(x)\le x$

By the mean value theorem, there exist $t\in(0,x)$ and $s\in(x,1)$ such that $$\frac{f(x)-f(0)}{x-0}=f'(t)$$ $$\frac{f(1)-f(x)}{1-x}=f'(s)$$ From $t<s$, $$\frac{f(x)}{x}=f'(t)<f'(s)=\frac{1-f(x)}{1-x}$$ so we have $f(x)\le x$.

Let $$I=\int_0^1 f(x)\,\mathrm{d}x$$ By step 1, we have $I\le\frac{1}{2}$

Step 2. $f(x)\ge f(I)+(x-I)f'(I)$

If $x>I$, there exist some $s\in(I,1)$ that $$\frac{f(x)-f(I)}{x-I}=f'(s)>f'(I)$$ and the result follows.

Same goes for the case $x<I$, since there exist some $t\in(0,I)$ that $$\frac{f(I)-f(x)}{I-x}=f'(t)<f'(I)$$

Step 3. $2I^2\ge f(I)$

By step 2, \begin{align} I&=\int_0^1 f(x)\,\mathrm{d}x\\ &\ge\int_0^1 f(I)+(x-I)f'(I)\,\mathrm{d}x\\ &=f(I)+(\frac{1}{2}-I)f'(I)\\ &\ge f(I)+(\frac{1}{2}-I)\frac{f(I)}{I}\\ &= \frac{f(I)}{2I} \end{align} Thus, $$2I^2\ge f(I)$$