If a real-valued function $f$ over reals satisfies $$ (1) \; \; \; f({x+y\over2})\le {f(x)+f(y)\over2}, $$and it is continuous, then it is not hard to see that $f$ is indeed convex. On the other hand, a discontinuous additive function, which exists by the axiom of choice, satisfies (1) but is not convex.


My question is, can we find real-valued functions over reals satisfying (1) which are not convex, without using the axiom of choice? Or, are there models of set theory in which all functions satisfying (1) are convex (continuous)?



Finding non convex functions satisfying a weak form of convexity, without the axiom of choice
Since no one posted the answer mentioned in the comments, I write it here to make this question more helpful.


In the Solovay model of set theory, where the axiom of choice does not hold, all functions are measurable; and this implies that all functions satisfying (1) are convex by a theorem of Sierpinski.


