Sunday 2 June 2019

real analysis - Function that is both midpoint convex and concave



Which functions $f:\mathbb{R} \to \mathbb{R}$ do satisfy $$f\left(\frac{x+y}{2}\right) = \frac{f(x)+f(y)}{2}$$ 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?


Answer



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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