We have to prove that at least one of the integrals $$\int^\infty_af(x)g(x)dx, \int^\infty_a\frac{f(x)}{g(x)}dx$$ is divergent, knowing that:

  • $f,g:[a,\infty)\to\mathbb{R}$
  • $\forall x\in[a,\infty)$ $f(x)\geq 0$, $g(x)>0$
  • $\int^\infty_a f(x)dx$ is divergent

What I have so far.

If I knew that $x\in[a,\infty)$ $g(x)\geq 1$, then I would know that $0\leq f(x)\leq f(x)g(x)$ and as $\int^\infty_a f(x)dx$ is divergent $\int^\infty_af(x)g(x)dx$ is divergent. When $x\in[a,\infty)$ $1>g(x)>0$ $0\leq f(x)\leq \frac{f(x)}{g(x)}$ and as $\int^\infty_a f(x)dx$ is divergent $\int^\infty_a\frac{f(x)}{g(x)}dx$ is divergent. But I think I can't make such assumptions for $g$.

A similar question can be found here, however the solutions proposed to that question are above my knowledge. So I was wondering (how)can it be done with more basic methods (like integral evaluation)?


Since $f,g \ge 0$, by C-S, $\int_a^\infty f(x)dx = \int_a^\infty \sqrt{f(x)g(x)}\sqrt{\frac{f(x)}{g(x)}} \overset{*}{\le} (\int_a^\infty f(x)g(x))^{1/2}(\int_a^\infty \frac{f(x)}{g(x)})^{1/2}$

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  • $\begingroup$ Can you explain why $*$ holds? $\endgroup$ – pls_halp Nov 27 '17 at 22:40
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    $\begingroup$ OH ITS CAUCHY SCHWARZ $\endgroup$ – mathworker21 Nov 27 '17 at 22:41
  • $\begingroup$ I edited your proposed answer. $\endgroup$ – pls_halp Nov 27 '17 at 22:41
  • $\begingroup$ Do you happen to know where I can find a proof for this case of C.-S.? $\endgroup$ – pls_halp Nov 27 '17 at 22:53
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    $\begingroup$ Not specifically. The point is you can make this stuff into a Hilbert space. And Cauchy Schwarz holds in any Hilbert space. Here the inner product would be $\langle f,g \rangle = \int_a^\infty f(x)\overline{g(x)}dx$. $\endgroup$ – mathworker21 Nov 28 '17 at 6:46

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