# An explicit construction for an everywhere discontinuous real function with $F((a+b)/2)\leq(F(a)+F(b))/2$?

I would like to know an explicit method on constructing an everywhere discontinuous real function $F$ with the property: $$F((a+b)/2)\leq(F(a)+F(b))/2.$$

There is a non-constructive example (with the inequality been trivial):

Take a Hamel basis $S$ of the $\mathbb{Q}$-linear space $\mathbb{R}$, take in $S$ a countably-infinite subset $X=\{x_1, x_2, ...\}$, then by multiplying a rational number $c_n$ to $x_n$ for each $n$ we can produce a set $Y=\{y_1, y_2, ...\}$ with $0< y_n\leq1/n$. Now we replace the $X$ in $S$ with $Y$ and obtain a new Hamel basis $T$. Take a $t_0\in T$; let $F(t_0)=0$ and let $F(t)=1$ for any other $t\in T$, then $F$ extends to a function on $\mathbb{R}$ linearly, and it is clear this is a required function.

By the answer of Conifold below, such an explicit method does not exist. But it would still be nice to know how to give such a non-constructive function with the inequality been strict.

• If you have an answer to your question it;s better that answer it separately. – Hoseyn Heydari Jul 23 '14 at 23:56
• @HoseynHeydari No, I don't have an answer (the example given in the question is non-constructive since it uses axiom of choice). – user148212 Jul 24 '14 at 0:07
• Can you clarify what you mean by "constructive proof"? – Asaf Karagila Jul 24 '14 at 0:56
• @AsafKaragila I don't have a precise idea for this, but I hope one can drop the axiom of choice (by the answer of Conifold, this is not possible, so I will edit to add one more question). – user148212 Jul 24 '14 at 8:40
• – Martin Sleziak Jun 19 '16 at 11:59