Can this integral diverge?

Question

Let $f:\mathbb{R}_+\to\mathbb{R}_+$ be a continuous non-increasing bounded function, and suppose $$\limsup_{x\to +\infty}\frac{f(x)}{f(2x)} = +\infty. $$ Can $\int_0^{+\infty} f(x)dx$ diverges?

I suspect that no, since the naivest example of such function that I can imagine is $e^{-x}$. However, it seems like it's possible to construct a counterexample by taking $x^{-2}$ and making arbitrarily large intervals where it remains constant, and then "continuously gluing" these. Anyway, I couldn't formalize this idea.

Any tips are welcome!

Answer 1

Yes, I think that integral can diverge. Hint: For $n= 1,2,\dots, $ let $I_n = [e^{n^2},e^{(n+1)^2}-1].$ Consider the function

$$f(x) = e^{-n^2}, x \in I_n.$$

This isn't quite what was asked for, but we're getting there.

Answer 2

If we don't restrict $f$ to be non-increasing and bounded, the answer is: Yes, the integral can diverge.

Let $f(x)=x\sin^2x$. Then $f(2x)=2x(2\sin x \cos x)^2=8x(\sin^2 x)(\cos^2 x)$. So we have $$\limsup_{x\to +\infty}\frac{f(x)}{f(2x)}= \limsup_{x\to +\infty}\frac{1}{8(\cos^2 x)}= +\infty. $$

On the other hand, it easy to show that $\int_0^{+\infty} f(x)dx=+\infty$.