TITLE: subadditive implies concave
QUESTION [19 upvotes]: Let $f:R_+\to R_+$ be smooth on $(0,\infty)$, increasing, $f(0)=0$ and 
$\lim_{x\to\infty}=\infty$. Assume also that $f$ is subadditive:
$f(x+y)\le f(x)+f(y)$ for all $x,y\ge 0$. Must $f$ be concave? The converse is obvious.

REPLY [3 votes]: Subadditivity + $f(0)=0$ do not imply concavity, neither for regular functions.
Counterexamples are in Bruckner "Some relationships between locally superadditive functions and convex functions", Proc. Amer. Math. Soc., 15, 1964, 61-65, and in Bruckner-Ostrow "Some function classes related to the class of convex functions", Pacific J. Math. 12, 1962, 1203--1215.