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Let be $ f: \mathbb{R} \rightarrow \mathbb{R}$ a continuous, monotone function. Then, if $a>0$ i must prove that the following inequality holds: $$\int_{-a}^{a}xf(f(x)) \geq0$$

I wonder if there is a simple proof for it.

1 Answers 1

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We have \begin{align} \int_{-a}^axf(f(x))dx&=\int_0^axf(f(x))dx+\int_{-a}^0xf(f(x))dx\\ &=\int_0^axf(f(x))dx+\int_{a}^0(-s)f(f(-s))(-ds)\\ &=\int_0^as\left(f(f(s))-f(f(-s))\right)ds \end{align} and we deal with two cases:

  • $f$ is non-decreasing. Since $-s\leq 0\leq s$, we have $f(-s)\leq f(s)$ and $f(f(-s))\leq f(f(s))$.
  • $f$ is non-increasing. We have $f(-s) \geq f(s)$ and $f(f(-s))\leq f(f(s))$.

(in fact $f\circ f$ is non-decreasing in any case)

We can see we have equality if and only if $f\circ f$ is even.

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    nice solution. Thanks.2012-06-03
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    is it correct $f(f(-s))\leq f(s)$ ? I suppose you meant $f(f(-s))\leq f(f(s))$2012-06-03
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    @Chris : i think thats typo, should have been $ff(s)$2012-06-03
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    Right, fixed now.2012-06-03
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    @Davide Giraudo: in the last row we have that $f\circ f$ must be odd or even for getting equality?2012-06-03
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    In fact even, by bad. Thanks!2012-06-03