A function $f: X \rightarrow \mathbb{R}$ is said to be mid-point convex if for all $x, y \in X$, we have $$f(\frac{x + y}{2}) \leq \frac{f(x) + f(y)}{2}. $$ Can you please give an example of a function which is mid-point convex but not convex?
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1Such a function would have to be non-measurable and so might be a little difficult to just "write down". – Prahlad Vaidyanathan Apr 21 '14 at 13:52
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@PrahladVaidyanathan why non measurable ? – yago Apr 21 '14 at 13:55
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It is a theorem of someone (who I don't remember - but I suspect it is rather deep) that a measurable mid-point convex function is forced to be continuous - and hence convex (A midpoint convex function is always rationally convex) – Prahlad Vaidyanathan Apr 21 '14 at 13:56
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I am really thankful to you all for the responses. – Summation Apr 21 '14 at 19:08
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4This seems to be the same question as this one. I will also add link to this related post. Other questions linked there might be of interest, too. – Martin Sleziak Dec 08 '15 at 15:09
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That is a theorem on convex analysis but it is stated for continuous function. For making counterexample you can remove the continuity from the condition such as $f(x)=x^2$ for $x\in \mathbb{Q}$ and $0$ otherwise. I think that, it is a counter example.
Ali
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In fact, it is known that any counterexample cannot be measurable and that existence of such function cannot be shown in ZF. (See the linked questions.) Your example does not work. Just consider $x=2+\sqrt2$ and $y=-\sqrt2$. (Or any other two irrational numbers such that their average is a non-zero rational number.) – Martin Sleziak Dec 08 '15 at 15:16
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If a Midpoint Convex function F, strictly monotonic, increasing on a real interval $[0,1]\to[0,1]$, with $F(0)=0$ and $F(1)=1$ , $F(0.5)=0.5$ will it be measurable and convex, aside from the possible counter-examples. Generally strict quasi-convexity is sufficient under mild constraints – William Balthes May 05 '17 at 15:56