China Girls' Mathematical Olympiad

Question

A function f: (0, +) R satisfies the following conditions: a. f(a) = 1 for a positive real number a, b. f(x)f(y) + f(ax)f(ay) = 2f(xy), for any positive real number x, y. Prove that f(x) is constant.

Accepted Answer

Setting x = y = 1 in (1) gives f^2(1) + f^2(a) = 2f(1), (f(1) - 1)^2 = 0, so f(1) = 1. Setting y = 1 in (1) yields f(x)f(1) + f(ax)f(a) = 2f(x), f(x) = f(ax), x > 0. (2) Setting y = ax in (1) yields f(x)f(ax) + f(ax)f(x) = 2f(a), f(x)f(ax) = 1. (3) Combining (2) and (3) gives f^2(x) = 1, x > 0. Setting x = y = t in (1) gives f^2(t) + f^2(at) = 2f(t), f(t) > 0. So f(x) = 1, x > 0, as desired.