f and g are real-valued functions defined on the real line. For all x and y, f(x + y) + f(x - y) = 2f(x)g(y). f is not identically zero and |f(x)| ≤ 1 for all x. Prove that |g(x)| ≤ 1 for all x.
Solution
Let k be the least upper bound for |f(x)|. Suppose |g(y)| > 1. Take any x with |f(x)| > 0, then 2k ≥ |f(x+y)| + |f(x-y)| ≥ |f(x+y) + f(x-y)| = 2|g(y)||f(x)|, so |f(x)| < k/|g(y)|. In other words, k/|g(y)| is an upper bound for |f(x)| which is less than k. Contradiction.
Solutions are also available in: Samuel L Greitzer, International Mathematical Olympiads 1959-1977, MAA 1978, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
10 Oct 1998
Last updated/corrected 14 Mar 03