Inequality for $\int_Ω (F(f_1(x))-F(f_2(x)))(f_1(x)-f_2(x))\,\mathrm dx$ for an increasing function $F$

Let $$F\colon \mathbb{R}\to\mathbb{R}$$ be of class $$C^1$$ and increasing and let $$f_1,f_2\colon\Omega\to\mathbb{R}$$ be bounded, $$\Omega\subset\mathbb{R}^n$$ and $$\int_{\Omega}1 dx=1$$.

I am trying to prove or disprove the following inequality: $$\begin{gather*}\int_\Omega \Big(F(f_1(x))-F(f_2(x)\Big)(f_1(x)-f_2(x))\,\mathrm dx\\ \geq \left( \int_\Omega (F(f_1(x))-F(f_2(x))\,\mathrm dx \right)\left( \int_\Omega (f_1(x)-f_2(x))\,\mathrm dx \right)\end{gather*}$$

If $$f_2\equiv 0$$, then it is the known fact, written here in the probabilistic setting.

I was trying to use it to estimate $$\displaystyle \int_\Omega F(f_1(x))f_1(x)\,\mathrm dx$$ and $$\displaystyle\int_\Omega F(f_2(x))f_2(x)\,\mathrm dx$$, but I couldn't cope with mixed terms $$\displaystyle\int_\Omega F(f_1(x))f_2(x)\,\mathrm dx$$ and $$\displaystyle\int_\Omega F(f_2(x))f_1(x)\,\mathrm dx$$.

I tried to find a counterexample for $$f_1$$, $$f_2$$ piecewise constant or linear. I also played with the trigonometric functions and in these cases the inequality was satisfied. For the function $$F$$ I took $$F(f)=|f|f$$, as this is really function I am working with, although I'm curious if the inequality holds in the more general case.

Is this inequality true? I would really appreciate any clues, or suggestions what could work as a counterexample.

• The link does not work for me May 17 at 10:47
• @Thomas Sorry about that, now it should be ok
– M_S
May 17 at 10:52

$$\def\d{\mathrm{d}}\def\Ω{{\mit Ω}}$$Counterexamples: Let $$\Ω = [0, 1]$$ and $$F(y) = y|y|$$. Take any nonconstant $$g: [0, 1] → [a, 1]$$ where $$0 < a < 1$$ and let $$f_1 = \dfrac{1}{g^2} + g$$, $$f_2 = \dfrac{1}{g^2} - g$$.
Now, since $$f_1 - f_2 = 2g$$ and $$F(f_1) - F(f_2) = f_1^2 - f_2^2 = \dfrac{4}{g}$$, then$$\begin{gather*} \int_0^1 (F(f_1(x)) - F(f_2(x))) (f_1(x) - f_2(x)) \,\d x = 8 \int_0^1 \d x = 8,\\ \int_0^1 (F(f_1(x)) - F(f_2(x))) \,\d x = 4 \int_0^1 \frac{\d x}{g(x)},\\ \int_0^1 (f_1(x) - f_2(x)) \,\d x = 2 \int_0^1 g(x) \,\d x. \end{gather*}$$ The Cauchy-Schwartz inequality implies that$$\left( \int_0^1 \frac{\d x}{g(x)} \right)\left( \int_0^1 g(x) \,\d x \right) \geqslant \left( \int_0^1 \d x \right)^2 = 1,$$ and the equality cannot hold since $$g$$ is not constant.
• Thank you very much for your answer. I assume that you meant $f_2=-g+\frac{1}{g^2}$?