IMO 1994 SL A4

Origin: MON | Category: Algebra

Problem

Let R denote the set of all real numbers and R+ the subset of all positive ones. Let \alpha and \beta be given elements in R, not necessarily distinct. Find all functions f : R+ \toR such that f(x)f(y) = y\alphaf x

• x\betaf y for all x and y in R+.

Solution

Suppose that \alpha = \beta. The given functional equation for x = y yields f(x/2) = x−\alphaf(x)2/2; hence the functional equation can be written as f(x)f(y) = 1 2x\alphay−\alphaf(y)2 + 1 2y\alphax−\alphaf(x)2, i.e.,  (x/y)\alpha/2f(y) −(y/x)\alpha/2f(x) = 0. Hence f(x)/x\alpha = f(y)/y\alpha for all x, y \inR+, so f(x) = \lambdax\alpha for some \lambda. Substituting into the functional equation we obtain that \lambda = 21−\alpha or \lambda = 0. Thus either f(x) \equiv21−\alphax\alpha or f(x) \equiv0. Now let \alpha ̸= \beta. Interchanging x with y in the given equation and sub- tracting these equalities from each other, we get (x\alpha −x\beta)f(y/2) = (y\alpha − y\beta)f(x/2), so for some constant \lambda \geq0 and all x ̸= 1, f(x/2) = \lambda(x\alpha−x\beta). Substituting this into the given equation, we obtain that only \lambda = 0 is possible, i.e., f(x) \equiv0.