Does there exist a real-valued function f on the reals such that f(x) is bounded, f(1) = 1 and f(x + 1/x2) = f(x) + f(1/x)2 for all non-zero x?
Solution
Answer: no.
Suppose there is such a function. Let c be the least upper bound of the set of values f(x). We have f(2) = f(1 + 1/12) = f(1) + f(1/1)2 = 2. So c >= 2. But definition we can find y such that f(y) > c - 1/4. So c ≥ f(y + 1/y2) = f(y) + f(1/y)2 > c - 1/4 + f(1/y)2. So f(1/y)2 < 1/4 and hence f(1/y) > -1/2.
We also have c ≥ f(1/y + y2) = f(1/y) + f(y)2 > -1/2 + (c - 1/4)2 = c2 - c/2 - 7/16. So c2 - 3c/2 - 7/16 < 0, or (c - 3/4)2 < 1. But c ≥ 2, so that is false. Contradiction. So there cannot be any such function.
© John Scholes
jscholes@kalva.demon.co.uk
4 Sep 2002