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Suppose $f(x)$ is continuous on $[0,+\infty)$. Show that $\int_0^\infty f(x)dx=a, 0<f(x)<1$ implies that $\int_0^\infty xf(x)dx>\frac{a^2}{2}$.

My try: $\int_0^\infty (x-a/2)f(x)dx\geq \int_t^\infty (x-a/2)f(x)dx...$ how to use $0<f<1$?

xrfxlp
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xldd
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1 Answers1

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Hint: $x=\int_0^x 1\,\mathrm{d}y$ and $a^2=\int_0^\infty\int_0^\infty f(x)f(y)\,\mathrm{d}x\mathrm{d}y$.

Spoiler:

\begin{align} \small \int_0^\infty f(x)x\,\mathrm{d}x=\int_0^\infty\int_0^x f(x)1\,\mathrm{d}y\,\mathrm{d}x>\int_0^\infty\int_0^x f(x)f(y)\,\mathrm{d}y\,\mathrm{d}x =\frac{1}{2}\int_0^{\infty}\int_0^\infty f(x)f(y) \,\mathrm{d}y\,\mathrm{d}x \end{align}

anon
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    Can you tell me how I can use that hint? – xrfxlp Jul 09 '19 at 07:38
  • @AjayMishra Substitute these two things into the desired inequality, note both sides are double integrals, the integrands are $f(x)$ vs $f(x)f(y)$ and one can use $1>f(y)$ on the latter, and $f(x)f(y)$ is symmetric across $y=x$ while the domain on the one double integral is only $y<x$... – anon Jul 09 '19 at 07:45
  • I have still no idea, leave it. – xrfxlp Jul 09 '19 at 07:46
  • I'd say you have an "idea," as I just sketched the whole proof. But putting the pieces together does take thought. I added a spoiler for you though @AjayMishra – anon Jul 09 '19 at 07:55