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Which functions $f:\mathbb{R} \to \mathbb{R}$ do satisfy Jensen's functional equation $$f\left(\frac{x+y}{2}\right) = \frac{f(x)+f(y)}{2}$$ for all $x,y \in \mathbb{R}$?

I think the only ones are of type $f(x) = c$ for some constant $c\in \mathbb{R}$ and the solutions of the Cauchy functional equation $f(x+y) = f(x)+f(y)$ and the sums and constant multiples of these functions. Are there other functions which are both midpoint convex and concave?

Mohsen Shahriari
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Andrei Kh
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1 Answers1

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Without loss, translate so that $f(0) = 0$. Then we have

$$f(x) = f\left(\frac{2x + 0}{2}\right) = \frac{f(2x)}{2}$$

so that $f(2x) = 2 f(x)$.

Now suppose that $f$ is midpoint convex and concave. We show it satisfies the Cauchy equation:

$$f(x + y) = f\left(\frac{2x + 2y}{2}\right) = \frac{f(2x) + f(2y)}{2} = f(x) + f(y)$$ as claimed. Now just remember that multiples of solutions to the Cauchy equation are still solutions to the Cauchy equation - hence, functions with your property are exactly translates of functions which solve the Cauchy equation.

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    "functions with your property are exactly functions which solve the Cauchy equation" - That is false: $f(x) = 1$ is a solution to the equation above but not a solution of Cauchy equation. Have I missed something? – Andrei Kh Jul 05 '16 at 19:46
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    @AndreiKh Yes, you're quite right. That's a translate of the zero function which does solve the Cauchy equation. –  Jul 05 '16 at 19:52
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    Ah - I didn't see "translates of functions". Thanks! – Andrei Kh Jul 05 '16 at 20:33
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    @AndreiKh Actually, what I had written was wrong until your correction - thanks. –  Jul 05 '16 at 20:36
  • @Andrei, do you mean continuous solutions? I suppose then, that we are looking at functions that are traditionally called aff-ine (convex and concave) $F(\sigma x+ (1-\sigma) y)=\sigma F( x)+ (1-\sigma) F( y)$ where$ \sigma, (1-\sigma) \in [0,1]$ , note the last weakening, where one cannot resolve $F(0)=0$ where otherwise this restriction is not kept when linear or cauchy and continuous. Generally of the form, $F(x)=Ax+B$, where if $F(0)=0$ and $F:[0,1] \to [0,1]$ is cauchy. I am not sure if there counter-examples when not defined on the unit interval. – William Balthes May 10 '17 at 04:18
  • As 'affinity' given this restriction, even with $F(0)=0$, seems less expressive due to the restriction(by affinity i mean continuous convex and concave) with the $sigma$ terms having to sum to one – William Balthes May 10 '17 at 04:18
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    Nice answer. I will add for other users that more on Cauchy functional equation can be found in this post: Overview of basic facts about Cauchy functional equation. – Martin Sleziak Aug 08 '17 at 05:56
  • @MartinSleziak Thanks, that's a nice FAQ I wasn't aware of. –  Aug 08 '17 at 05:58