Problem 11, Chapter 18 of Calculus by Spivak 
The problem:
Let $f$ be a nondecreasing function on $[1;\infty)$. Define $F(x)$ as follows:
$$F(x) = \int_{1}^{x}\frac{f(t)}{t}dt.$$
Prove that $f$ is bounded on $[1;\infty)$ if only if $\frac{F}{\log}$ is bounded on $[1;\infty)$.

The left-to-right can be easily proved. Let's just consider the converse.
Here's what i did:
Since $f$ is nondecreasing, $\forall x\in[1;\infty]$: We have: $\forall t\in[1;x]: f(t) \leq f(x)$.
So, $$F(x)=\int_{1}^{x}\frac{f(t)}{t}dt \leq \int_{1}^{x}\frac{f(x)}{t}dt = f(x).log(x) (*)$$
(In Spivak's Calculus, he defines $log(x) = \int_{1}^{x}\frac{1}{t}dt$)
If $f<0$ on $[1;+\infty)$ then: $$F(x) = \int_{1}^{x}\frac{f(t)}{t}dt < \int_{1}^{x}0dt = 0$$. And: $(*)$ becomes:
$$-|F(x)| \leq -|f(x)|.logx \iff |F(x)| \geq f(x).logx \iff |\frac{F(x)}{logx}|\geq|f(x)|$$. Since $F/log$ bounded, so is $f$
If $\exists x\in[1;+\infty): f(x) \geq 0$ then the set $A=\{x\geq 1:f<0\ \text{on}\ [1;x]\}$ must be bounded above.
Consider $\alpha = \text{sup}A$.
So, $f<0$ on $[1;\alpha]$ but $\geq 0$ on $(\alpha;+\infty)$. Similarly, we can prove $f$ is bounded on $[1;\alpha]$. But i'm not quite sure how to handle the $(\alpha;+\infty)$. On $(\alpha;+\infty)$, what we have is: $$F(x)=(\int_{1}^{\alpha}+\int_{\alpha}^{x})\frac{f(t)}{t}dt \leq c_0 + f(x)(logx-log(\alpha)) \iff |\frac{F(x)}{logx - log\alpha}| - \frac{c_0}{logx - log\alpha} \leq |f(x)|$$. We cannot conclude $f$ is bounded at all. And the relation "$\leq$" seems to be the only way to take out $f(x)$. Any ideas on how to proceed ?

Why not prove it by contradiction?
If $f$ is not bounded, then for any large $M > 0$ there exists $X$ such that $f(x) \geq M$ for all $x\geq X$. This means that $F(x)\geq \int_1^X \frac{f(t)}{t} dt + M\ln x - M\ln X$. Thus
$$
\frac{F(x)}{\ln x} \geq \frac{C}{\ln x} + M
$$
where $C$ is a constant depending on $M$ and $X$. Letting $x\to\infty$, we see that $F(x)/\ln x \geq M/2$ for all large $x$. This result holds for any $M$, and thus $F(x)/\ln x$ cannot be bounded.
