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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 $f:\mathbb R\to \mathbb R$?

J. Netzing
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    How did you come across it? – LSpice May 08 '20 at 20:14
  • Taking $y = 0$ gives $f(x) = 2f(x/2)$, so that $f(0) = 0$ and $f : x \mapsto \lim_{n \to \infty} 2^n f(x/2^n)$ is entirely determined by its germ at $0$. Taking $y = -x$ gives $f(x^2) + f(-x^2) = 2f(0) = 0$, so $f$ is odd. – LSpice May 08 '20 at 20:20
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    Taking $y=1$ gives $f(x)+f(1)=2f((x+1)/2)$, so combining with $f(x)=2f(x/2)$ gives $f(x'+1/2)=f(x')+f(1)/2$, where $x'=x/2$. – Wojowu May 08 '20 at 20:25
  • There is the trivial $f(z)=az+b$. – DSM May 09 '20 at 01:32
  • $f(xy) = axy+b$, $-2f((x+y)/2) = -2(a((x+y)/2)+b) = -(a(x+y)+2b)$, $f(x+y-xy) = a(x+y-xy)+b$. They add up to zero. – DSM May 09 '20 at 01:59
  • (I should have said that taking $y = 0$ gives $f(x) + f(0) = 2f(x/2)$, as @MichaelRenardy correctly points out in their answer below. Oops!) – LSpice May 09 '20 at 02:55

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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.

Michael Renardy
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