Find all real-valued functions f(x) on the reals such that f(f(x) + y) = 2x + f(f(y) - x) for all x, y.
Answer
f(x) = x + c
Solution
It is easy to check that f(x) = x + c is a solution. We have to show that it is the only solution.
Given any x, put x' = (f(0) - x)/2, then x = f(0) - 2x'. Put y = -f(x'), then f(f(x') + y) = f(0) = 2x' + f(f(y) - x'), so if z = f(y) - x', then f(z) = x. Thus f is surjective. In particular, f(a) = 0 for some a. So for any y, f(y) = f(f(a) + y) = 2a + f(f(y) - a). Also, given any x, we can find y such that x = f(y) - a, then x = f(y) - a = (2a + f(f(y) - a) ) - a = f(x) + a, or f(x) = x - a.
(C) John Scholes
jscholes@kalva.demon.co.uk
8 Aug 2003
Last corrected/updated 8 Aug 2003