Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 3 down vote accepted

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.

share|cite|improve this answer
nice solution. Thanks. – user 1618033 Jun 3 '12 at 20:38
is it correct $f(f(-s))\leq f(s)$ ? I suppose you meant $f(f(-s))\leq f(f(s))$ – user 1618033 Jun 3 '12 at 20:40
@Chris : i think thats typo, should have been $ff(s)$ – Theorem Jun 3 '12 at 20:43
Right, fixed now. – Davide Giraudo Jun 3 '12 at 20:46
@Davide Giraudo: in the last row we have that $f\circ f$ must be odd or even for getting equality? – user 1618033 Jun 3 '12 at 20:54

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.