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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)?

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    $\begingroup$ There is some related discussion at math.stackexchange.com/questions/876410/… which in particular mentions a model of set theory where such functions are convex. $\endgroup$
    – Suvrit
    Commented Nov 29, 2015 at 20:14
  • $\begingroup$ (+1) Nice question, Mohammad! I doubt if one can find a proof for this theorem without using (any degree of) Axiom of Choice. By the way, can you add some explanations about the motivation for building a model of $ZF+\neg AC$ in which all (continuous or discontinuous) functions which satisfy the condition (1) are convex? Does it have any application in your work or any particular interpretation, say in physics, analysis or statistics? Or this is just a question out of curiosity? $\endgroup$
    – user82740
    Commented Nov 30, 2015 at 4:50
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    $\begingroup$ somehow related question: Are there any non-linear solutions of Cauchy's equation (f(x+y)=f(x)+f(y)) without assuming the Axiom of Choice?. $\endgroup$
    – Mohammad Golshani
    Commented Nov 30, 2015 at 11:52
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    $\begingroup$ @Amir, I'm working on some problems in convex analysis right now, but this question was more out of curiosity than for any particular application. $\endgroup$
    – Mohammad Safdari
    Commented Nov 30, 2015 at 15:31
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    $\begingroup$ @YemonChoi to be fair to the original author on the answer on m.SE, I've requested him to post his answer on MO. If he does not post, then I'll post an answer. Thanks! $\endgroup$
    – Suvrit
    Commented Nov 30, 2015 at 19:18

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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.

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  • $\begingroup$ Can you give a reference to Sierpinski's paper, or another source where the result and its proof can be found? $\endgroup$
    – Nate Eldredge
    Commented Jun 8, 2016 at 22:06
  • $\begingroup$ @Nate, The Wikipedia article on convex functions gives this reference: books.google.com/books?id=P30Y7daiGvQC&pg=PA12 $\endgroup$
    – Mohammad Safdari
    Commented Jun 9, 2016 at 1:36
  • $\begingroup$ @NateEldredge Some further references are mentioned in my answer here. (Including the references to the papers by Sierpinski and Blumberg.) $\endgroup$
    – Martin Sleziak
    Commented Jun 9, 2016 at 10:25

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