Find all functions f defined on the set of positive reals which take positive real values and satisfy:
f(x(f(y)) = yf(x) for all x, y; and f(x) → 0 as x → ∞.
Solution
If f(k) = 1, then f(x) = f(xf(k)) = kf(x), so k =1. Let y = 1/f(x) and set k = xf(y), then f(k) = f(xf(y)) = yf(x) = 1. Hence f(1) = 1 and f(1/f(x)) = 1/x. Also f(f(y)) = f(1f(y)) = y. Hence f(1/x) = 1/f(x). Finally, let z = f(y), so that f(z) = y. Then f(xy) = f(xf(z)) = zf(x) = f(x)f(y).
Now notice that f(xf(x)) = xf(x). Let k = xf(x). We show that k = 1. f(k2) = f(k)f(k) = k2 and by a simple induction f(kn) = kn, so we cannot have k > 1, or f(x) would not tend to 0 as x tends to infinity. But f(1/k) = 1/k and the same argument shows that we cannot have 1/k > 1. Hence k = 1.
So the only such function f is f(x) = 1/x.
Solutions are also available in Murray S Klamkin, International Mathematical Olympiads 1978-1985, MAA 1986, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
14 Oct 1998