TITLE: Is $f(x)x$ convex for increasing function $f$?
QUESTION [5 upvotes]: Suppose that $f:(0,\infty)\to[0,1]$ is strictly increasing and infinitely differentiable. I have an intuition that 
$$
g(x)=f(x)x
$$ 
should be convex in $x$ (i.e. increasing at accelerating rate) because $g(x)$ can be interpreted as a proportion of a number both of which go up when the number goes up, but I can't prove it nor come up with a counterexample.
I tried differentiating:
$$
g'(x)=xf'(x)+f(x)\implies g''(x)=xf''(x)+2\underbrace{f'(x)}_{>0}\overset{?}{>}0.
$$
Thank you for your help.

REPLY [4 votes]: Let $f(x) = 1- e^{-x}$.
Then for $x > 2$
$$g''(x) = e^{-x}(2-x) < 0$$
and $g(x) = xf(x)$ is not convex.