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Consider a continuous function $f:[0,1]\to\mathbb{R}^{+}$. Show that

$$\int_0^1 f(x)dx-\exp\left(\int_0^1 \log(f(x)) dx\right)\le \max_{0\le x,y\le 1}\left(\sqrt{f(x)}-\sqrt{f(y)}\right)^2$$

I tried first to discretize the problem and found that LHS is equal to

$$\int_0^1 f(x)dx-\exp\left(\int_0^1 \log(f(x))dx\right)=\lim_{n\to\infty}\left(\frac{1}{n}\sum_{i=1}^{n}f\left(\frac{i}{n}\right)-\frac{1}{n^n}\prod_{i=1}^{n}f\left(\frac{i}{n}\right)\right)$$ which looks like AM-GM inequality. Now I need to do the same thing for the RHS.

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    $\begingroup$ By setting $a_i = f(i/n)$ we have: $$\int_{0}^{1}f(x)\,dx-\exp\left(\int_{0}^{1}\log(f(x))\,dx\right)=\lim_{n\to +\infty}\left(\frac{1}{n}\sum_{i=1}^{n}a_i-\left(\prod_{i=1}^{n}a_i\right)^{1/n}\right),$$ and your inequality follows from a celebrated result of Cartwright and Field: jstor.org/stable/2042211 $\endgroup$ – Jack D'Aurizio Aug 2 '14 at 19:09
  • $\begingroup$ @JackD'Aurizio I tried to apply the theorem in the link, but I only got an upper bound $\dfrac{1}{2\min f(x) n^2}{n \choose 2} \max(f(x) - f(y))^2$, whose limit is not exactly the RHS. Could you give some more explanations? Thanks $\endgroup$ – Petite Etincelle Sep 24 '14 at 10:35

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