Show that if $\int^{\infty}_0f(x)dx$ conditionally converges then $\int^{\infty}_0f_+(x)dx$, $\int^{\infty}_0f_-(x)dx$ both diverge

I've been battling this one question for quite some time now, and I can't seem to get the hang of it.

Let $f(x)$ be a continuous function in $[0,\infty)$, and also define

$$f_+(x)=\max\{f(x),0\}, f_-(x)=\max\{-f(x),0\}$$

Note:

$$|f(x)|=f_+(x)+f_-(x)$$

I have managed to prove that

$\int^{\infty}_0|f(x)|dx<\infty\iff\int^{\infty}_0f_+(x)dx<\infty\land \int^{\infty}_0f_-(x)dx<\infty$

Fairly easily, according to the definition of these max functions.

However, I've been asked to prove that in the case where $f$ is actually conditionally convergent, both $f_+$ and $f_-$ must satisfy that their improper integrals in $[0,\infty)$ diverge.

From the aforementioned ($\iff$) proof I concluded that at least one of those improper integrals must be non-convergent, but I couldn't figure out how to infer that they both fail to converge altogether.

How would you go about solving this problem? Convergence tests for non-negative functions do not apply to $f$, and "opening $f_{+,-}$ up" by definition has not worked for me.

Thank you in advance!

Solution 1:

We have that

$$L=\int f=-\int f_-+\int f_+$$

Now, if either of the integrals on the right converge (say the integral over $f_-$) then we have

$$L+\int f_-=\int f_+$$

which implies the integral over $f_+$ converges. Since you have already shown that at least one must diverge, this is enough to prove the other one diverges.