IMO 1983 SL 12

Origin: GBR

Problem

Find all functions f defined on the positive real numbers and taking positive real values that satisfy the following conditions: (i) f(xf(y)) = yf(x) for all positive real x, y. (ii) f(x) \to0 as x \to+\infty.

Solution

Putting y = x in (1) we see that there exist positive real numbers z such that f(z) = z (this is true for every z = xf(x)). Let a be any of them.

Then f(a2) = f(af(a)) = af(a) = a2, and by induction, f(an) = an. If a > 1, then an \to+\inftyas n \to\infty, and we have a contradiction with (2). Again, a = f(a) = f(1 \cdot a) = af(1), so f(1) = 1. Then, af(a−1) = f(a−1f(a)) = f(1) = 1, and by induction, f(a−n) = a−n. This shows that a ̸< 1. Hence, a = 1. It follows that xf(x) = 1, i.e., f(x) = 1/x for all x. This function satisfies (1) and (2), so f(x) = 1/x is the unique solution.