$f$ is bounded $\iff$ $F/\log$ where $F(x)= \int_{[1,x]}f(t)/t \,dt$ Hi everyone I'm stuck with one exercise. This says the following: 

Let $F(x)= \int_{[1,x]}f(t)/t \,dt$ where $f$ is a non-decreasing function. Show that $f$ is bounded $\iff$ $F/\log$ is also bounded on $[1,\infty)$. 

($\Rightarrow$)  Let $M$ be a number such that $|f|\le M$,  we have 
$$\left| \int_{1}^x\frac{f(t)}{t}dt\right|\le \int_{1}^x\frac{|f(t)|}{t}dt\le M\log x$$
Then $|F/\log|\le M$ for $x\ge 1$. 
($\Leftarrow$) This direction is really the difficult part to me. The first thing that I've tried to do is assuming the continuity of $f$ and using the Cauchy Mean Value Thm.
By continuity of $f$, clearly $F$ is differentiable and for $\,y<x \in [1,\infty)$ we have 
$$\log'(c) [F(x)-F(y)]=[\log(x)-\log(y)]F'(c)$$
$$ [F(x)-F(y)]=[\log(x)-\log(y)]f(c)$$
and since $\log(x)-\log(y)\not=0$ we must  have 
$$f(c)=\frac{F(x)-F(y)}{\log(x)-\log(y)} \tag{1}$$
and I've tried to compare (1) with $F/\log$ and see what happens. The problem is how to make the correct estimate of this two values and that is not as general as I wish, because I'm assuming that $f$ is continuous. Can somebody give me a hint, please?
Thanks in advance. 

Show the contrapositive. If $f$ is not bounded, then for every $K \in (0,\infty)$, there is an $x\in [1,\infty)$ with $F(x) > K\log x$.
Given $K$, since $f$ is nondecreasing and unbounded, there is an $x_0 \in (1,\infty)$ such that $f(x) > 2K$ for all $x \geqslant x_0$. Now, for $x > x_0$, we have
$$F(x) = \int_1^x \frac{f(t)}{t}\,dt = \int_{x_0}^x \frac{f(t)}{t}\,dt + \int_1^{x_0} \frac{f(t)}{t}\,dt.$$
Use that to deduce that $F(x) > K\log x$ for all sufficiently large $x$.
