0
$\begingroup$

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)?

$\endgroup$
2
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Can you explain why $*$ holds? $\endgroup$ – pls_halp Nov 27 '17 at 22:40
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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