30th Swedish 1990 Problem 5 Find all monotonic positive functions f(x) defined on the positive reals such that f(xy) f( f(y)/x) = 1 for all x, y.
f(x) = 1/x or (if f is not required to be strictly monotonic) f(x) = 1 for all x. Solution Putting x = f(y) gives f(yf(y)) f(1) = 1 (*). Let f(1) = a. Then putting y = 1 in (*) gives f(a) = 1/a. Putting y = a in (*) gives a 2 = 1. But a is positive, so a = 1. Hence f(yf(y)) = 1. If monotonic means strictly monotonic (so that x ≠ y implies f(x) ≠ f(y)), then yf(y) = 1, so f(y) = 1/y for all y. Suppose it does not mean strictly monotonic. So suppose we have b ≠ 1, such that f(b) = 1. Putting y = 1 in the original equation we have f(x) f(1/x) = 1, so f(1/b) = 1. So wlog b > 1. Putting x = y = b i the original equation, f(b 2 ) = f(1/b) = 1, and by an easy induction f(b n ) = 1 for all positive integers n. Hence we have f(k) = 1 for arbitrarily large numbers k. But f(1) = 1 and f is monotonic, so f(x) = 1 for all x ≥ 1. But f(1/x) = 1/f(x), so f(x) = 1 for all x ≤ 1 also. Thanks to Suat Namli 30th Swedish 1990 © John Scholes jscholes@kalva.demon.co.uk 16 January 2004 Last corrected/updated 16 Jan 04