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.

  • $\begingroup$ The link does not work for me $\endgroup$
    – Thomas
    May 17 at 10:47
  • $\begingroup$ @Thomas Sorry about that, now it should be ok $\endgroup$
    – 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.

  • $\begingroup$ Thank you very much for your answer. I assume that you meant $f_2=-g+\frac{1}{g^2}$? $\endgroup$
    – M_S
    May 20 at 10:49
  • $\begingroup$ @M_S Yes, it's corrected now. $\endgroup$
    – Saad
    May 20 at 11:00

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