How to find an everywhere discontinuous real function with $F((a+b)/2)<(F(a)+F(b))/2$? In here I posted a non-constructive everywhere discontinuous real function with
$$F((a+b)/2)=(F(a)+F(b))/2$$
based on the using of Hamel basis. And Conifold answered there that there is no explicit way to construct an every where discontinuous function with $F((a+b)/2)\leq(F(a)+F(b))/2$. 
So here another question is, how to give a non-constructive everywhere discontinuous real function with a strict inequality
$$F((a+b)/2)<(F(a)+F(b))/2?$$

By non-constructive I mean the using of axiom of choice, but maybe someone more familiar with set theory have different ideas and welcome edits.

Note that your strict inequality is for $a\not =b$. With this restriction, the function $f(x)=x^2$ is such that your strict inequality is satisfied. Now if you take $G(x)$ an everywhere discontinuous function such that $G(\frac{a+b}{2})=\frac{G(a)+G(b)}{2}$ for all $a,b$, I think that $F(x)=G(x)+x^2$ is a solution.
