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.
No comments:
Post a Comment