For a function $f:\mathbf{R}_+\rightarrow\mathbf{R}_+,x\mapsto f(x)$, assume that :
- $f(0^+)=0$
- $\forall\ x,y>0:f(x)+f(y)<f(x+y)$
- $\forall\ 0<x<y: f(x)<f(y)$
For example, $f(x)=2^x-1$.
Must $f$ be convex? If not, is there any counterexample?
I tried to calculate the second derivative of $f$ and prove that it's always positive, but I failed. Moreover, even if it's proved, it's not enough to solve the problem ($f$ may be non-differentiable).
Obviously, it's impossible to prove $f(x)+f(y)>2f(\frac{x+y}2)$ or other similar conclusions, so the Cauchy trick is hard to be available.