Incidentally, I came across the following functional equation


$$f(xy) - 2 f(\frac{x+y}{2}) + f(x+y- x\cdot y) = 0$$


that is to hold for all $x,y\in \mathbb R$. Is there a neat way to find all solutions $f:\mathbb R\to \mathbb R$?



The functional equation $f(xy) - 2 f(\frac{x+y}{2}) + f(x+y- x\cdot y) = 0$
Adding a constant to f does not change the property, so we may assume f(0)=0. Under this assumption, I claim that the property holds if and only if f is additive. The if part is obvious. For the only if part, we first set y=0 to obtain f(x)=2f(x/2)-f(0)=2f(x/2). Next, we set x+y=a, xy=b, and we obtain f(b)+f(a-b)=2f(a/2)=f(a), at least for b<0 (in which case there is always a corresponding x and y which is real). To deal with the opposite sign of b, simply reverse the roles of a and a-b. This proves f is additive. If f is also continuous, it must be linear. If you are pro-choice, you can characterize discontinuous solutions after you pick a basis for the real numbers as a vector space over the rationals.


