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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.

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  • $\begingroup$ This isn't really a question about the axiom of choice. This is a question about finding a function which satisfies this inequality. The [axiom-of-choice] tag is unnecessary here. $\endgroup$ – Asaf Karagila Jul 25 '14 at 9:04
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    $\begingroup$ @AsafKaragila I initially added it because I think it can be regarded as an application of the axiom of choice... $\endgroup$ – user148212 Jul 25 '14 at 9:05
  • $\begingroup$ Yeah, that is probably an application of the axiom of choice. As are 60% of modern functional analysis, parts of ring theory, group theory, and in fact even freshman year calculus. $\endgroup$ – Asaf Karagila Jul 25 '14 at 9:06
  • $\begingroup$ Again, the axiom of choice tag is not for "Well, this probably needs the axiom of choice". It's for asking specific questions about the axiom of choice. You already asked that part, you were told that the axiom of choice is needed to some degree in constructing a solution. Now you just seem to ask "Well, how can I construct this sort of solution?" and that's not quite what fits to the axiom of choice tag. $\endgroup$ – Asaf Karagila Jul 25 '14 at 9:41
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    $\begingroup$ 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. $\endgroup$ – Kelenner Jul 25 '14 at 9:45
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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.

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