# when convex diverging functions are monotone when divided by $x$

here is a calculus question that someone asked me to help him wuth and I have no answer for him. any help or ideas?

Given $f:(0,\infty) \rightarrow \mathbb{R}$ is a convex function, and $\displaystyle\lim_{x \rightarrow \infty} f(x) = 0$, prove that, $g(x) = \frac{f(x)}{x}$ is an increasing monotone function in $(0,\infty)$.

Thank you! Shir

Under the stated assumptions, $f(x)/x$ is indeed monotone. But it is decreasing rather than increasing.
Step 1 is to prove that $f$ itself is decreasing (if it was not, the limit at infinity would be $\infty$, not $0$).
Step 2. We have $f\ge 0$. If $f(x)=0$ for some $x$, then $f=0$ for all larger $x$, and the claim follows.
Step 3. Suppose $f>0$ everywhere. Fix $x_0\in (0,\infty)$ and let $$g(x) = \frac{f(x_0)}{x_0} x$$ Since $g$ is linear and $f$ is convex, the set $I=\{x: f(x)\le g(x)\}$ is convex; i.e., an interval. We have $x_0\in I$. Also, all large values of $x$ belong to $I$ because $g$ grows at infinity. Thus, $x\in I$ for all $x\ge x_0$. This says precisely that
$$\frac{f(x)}{x} \le \frac{f(x_0)}{x_0},\quad x\ge x_0$$ which was to be proved.