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.

  • $\begingroup$ If you have an answer to your question it;s better that answer it separately. $\endgroup$ – Hoseyn Heydari Jul 23 '14 at 23:56
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    $\begingroup$ @HoseynHeydari No, I don't have an answer (the example given in the question is non-constructive since it uses axiom of choice). $\endgroup$ – user148212 Jul 24 '14 at 0:07
  • $\begingroup$ Can you clarify what you mean by "constructive proof"? $\endgroup$ – Asaf Karagila Jul 24 '14 at 0:56
  • $\begingroup$ @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). $\endgroup$ – user148212 Jul 24 '14 at 8:40
  • $\begingroup$ See also: math.stackexchange.com/questions/71019/… $\endgroup$ – Martin Sleziak Jun 19 '16 at 11:59

No constructive example exists. Functions satisfying this inequality are called midpoint convex. Lebesgue measurable functions that are midpoint convex will be convex by a Sierpinski Theorem, therefore not just continuous but differentiable at all but countably many points.

There is a model of set theory, called Solovay model, where all axioms are satisfied except for the axiom of choice, but every function is Lebesgue measurable. In this model only convex functions will be midpoint convex. Therefore existence of everywhere discontinuous midpoint convex functions can only be proved with the axiom of choice, i.e. non-constructively.

  • $\begingroup$ Constructive does not mean "without using choice". And if I recall correctly, some constructive systems prove the axiom of choice. $\endgroup$ – Asaf Karagila Jul 24 '14 at 0:53
  • $\begingroup$ True, but "provable only with choice" is definitely non-constructive in the colloquial sense. "One trivial meaning of "constructive", used informally by mathematicians, is "provable in ZF set theory without the axiom of choice."en.wikipedia.org/wiki/… $\endgroup$ – Conifold Jul 24 '14 at 0:56
  • $\begingroup$ Thank you! Do you have an idea how to give a non-constructive example with the inequality been strict? $\endgroup$ – user148212 Jul 24 '14 at 8:41
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    $\begingroup$ I'd have to think about it, but it's better to ask it as a new question, it will go to the top of the queue and more people will see it. Put axiom of choice tag among others, and link to this question for motivation. $\endgroup$ – Conifold Jul 24 '14 at 20:19
  • $\begingroup$ I linked to your answer at a related MO post: mathoverflow.net/questions/224852/… --- rather than copy your answer and post it there, I thought I'd request you to post an answer on MO $\endgroup$ – suvrit Nov 30 '15 at 19:17

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