Let $f:\mathbb{R} \rightarrow \mathbb{R}$ twice differentiable. Suppose $f \geqslant 0$ and $f''\leqslant 0$. Show that $\lim_{x\to\infty} \frac{f(x)}{x}$ exist.
My attempt was, as $f$ is concave, to pass by next inequalities:
Let $a<b<x$ so we have
$\frac{f(b)-f(a)}{b-a}>\frac{f(x)-f(a)}{x-a}>\frac{f(x)-f(b)}{x-b}$
and then pass to limit and observe that $\frac{f(x)}{x}$ is bounded as $x$ goes to $\infty$. But, then i recognised that it doesn't hold…
So i need some help please. Thanks in advance
Best Answer
As you noticed, $f$ is concave, so that for $0 < x < y$ $$ f(x) \ge \frac{y-x}{y-0} f(0) + \frac{x-0}{y-0} f(y) \ge \frac x y f(y) \\ \implies \frac{f(x)}{x} \ge \frac{f(y)}{y} \, . $$ So $f(x)/x$ is decreasing and bounded below (by zero) and therefore convergent.
Note that the conclusion holds for all non-negative concave functions, even if they are not differentiable.