# Prove that at least one the following is integrals is divergent.

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, $\int_a^\infty f(x)dx = \int_a^\infty \sqrt{f(x)g(x)}\sqrt{\frac{f(x)}{g(x)}} \overset{*}{\le} (\int_a^\infty f(x)g(x))^{1/2}(\int_a^\infty \frac{f(x)}{g(x)})^{1/2}$
• Can you explain why $*$ holds? – pls_halp Nov 27 '17 at 22:40
• 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 $\langle f,g \rangle = \int_a^\infty f(x)\overline{g(x)}dx$. – mathworker21 Nov 28 '17 at 6:46