If $f$ is continuous and $f(\frac{x+y}{2})\leq \frac{f(x)+f(y)}2$ for all $x , y \in (a,b)$, prove that $f$ is convex.

Question

I know a proof which starts with proving the convexity for "dyadic rational" numbers. I would like to know if someone has some other ideas.

Answer 1

The continuity assumption is necessary. Observe that any additive function fulfils this inequality (with equality in fact). There are discontinuous additive functions. Of course they are not convex (for instance, ther graph are dense in a strip $\text{domain}\times\Bbb R$; this is not the case for a continuous function).

The dyadic numbers is the simplest way. We could also try to show that differential quotients are nondecreasing which is the necessary and suficient condition of convexity. For the inequality in question we have that

$$ \frac{f(v)-f(x)}{v-x}\le\frac{f(v)-f(y)}{v-y}, $$

whenever $x<y$ and $v=\dfrac{x+y}{2}.$ Under continuity assumption try to extend it to any $v$ between $x$ and $y$, not only the midpoint.