Do there exist real-valued functions f and g on the reals such that f( g(x) ) = x2 and g( f(x) ) = x3? Do there exist real-valued functions f and g on the reals such that f( g(x) ) = x2 and g( f(x) ) = x4?
Answer
No, yes.
Solution
Suppose such functions exist. Note that f must be injective, for if f(a) = f(b), then a3 = g(f(a)) = g(f(b)) = b3, so a = b. But we also have f(x)2 = f(g(f(x))) = f(x3). Substituting x = 0, ±1 we see that f(x) = f(x)2 is satisfied by x = 0, x = 1, x = -1. There are only two solutions to y = y2, namely y = 0, 1, so two of f(0), f(1), f(-1) must be equal. Contradiction.
Put p(x) = 2√x, q(x) = x2 for x ≥ 0. Then p(q(x)) = 2x, q(p(x)) = 4x. We can extend these to negative x, whilst preserving these relations by putting p(-x) = -p(x), q(-x) = -q(x).
Now put f(2t) = 2p(t), g(2t) = 2q(t). It is easy to check that f(g(2t) = f(2q(t)) = 2p(q(t)) = 22t = (2t)2 and similarly g(f(2t)) = (2t)4. So f and g satisfy the required relations for positive reals. But we can now extend them by taking f(0) = g(0) = 0 and f(-x) = f(x), g(-x) = g(x).
Thanks to Kamil Duszenko
© John Scholes
jscholes@kalva.demon.co.uk
2 Dec 2003
Last corrected/updated 2 Dec 2003