4
$\begingroup$

Maybe a trivial question. I think it is true, but I do not succeed in proving.

Suppose $h(x)$ is such that for all $a\geq 0$ $ \int_a^\infty h(x) dx\geq0 $ Prove for all $a\geq 0$ $ \int_a^\infty x h(x) dx\geq0 $

If needed, one may assume $\lim_{x\rightarrow\infty}h(x)=0$ and/or continuity of $h$.

  • 1
    I think I solved it. $ \int_{x=a}^\infty x h(x) dx = \int_{x=a}^\infty \int_{y=0}^x \rm{d}\it y h(x) \rm{d}\it x = \int_{y=0}^\infty\int_{x=\min(y,a)}^\infty h(x) \rm{d} \it x \rm{d} \it y \geq 0 $ where we used that by assumption for all $y\geq 0$ $ \int_{x=\min(y,a)}^\infty h(x) \rm{d} \it x \geq 0 $2012-10-25

1 Answers 1

2

Introduce $H(s)=\displaystyle\int_s^\infty h(x) \mathrm dx$ for every $s$. Then $\displaystyle\int_a^\infty xh(x) \mathrm dx=aH(a)+\int_a^\infty H(s)\mathrm ds$ hence $\displaystyle\int_a^\infty xh(x) \mathrm dx\geqslant0$ for every $a$ as soon as $H(s)\geqslant0$ for every $s$.

More generally, $\displaystyle\int_a^\infty u(x)h(x) \mathrm dx=u(a)H(a)+\int_a^\infty u'(s)H(s)\mathrm ds\geqslant0$ for every nonnegative nondecreasing function $u$ as soon as $H(s)\geqslant0$ for every $s$.