h and k are reals. Find all real-valued functions f defined on the positive reals such that f(x) f(y) = y^{h} f(x/2) + x^{k} f(y/2) for all x, y.

**Solution**

Answer: f(x) = 0 is always a solution. If h = k, then there is an additional solution f(x) = 2(x/2)^{h}.

Consider first the case h = k. Putting x = y, f(x)^{2} = 2x^{h}f(x/2), so f(x/2) = x^{-h}f(x)^{2}/2. Hence f(x) f(y) = (y/x)^{h} f(x)^{2}/2 + (x/y)^{h} f(y)^{2} or ( (y/x)^{h/2}f(x) - (x/y)^{h/2}f(y) )^{2} = 0. So f(x)/x^{h} = f(y)/y^{h}. This is true for all x, y, so we have f(x) = A x^{h} for some constant A. Substituting back into the original equation gives A^{2} = A/2^{h} + A/2^{h}, so A = 0 or A = 2^{1-h}. Checking, it is clear that A = 0, corresponding to f(x) = 0, is a solution. If A = 2^{1-h}, then f(x) = 2 (x/2)^{h}. The lhs of the original equation then becomes 4 (x/2)^{h}(y/2)^{h}, and the rhs becomes y^{h} 2 (x/4)^{h} + x^{h} 2 (y/4)^{h} = 4 (x/2)^{h}(y/2)^{h} = lhs. So it is indeed a solution.

Now consider h and k unequal. We have f(x) f(y) = y^{h} f(x/2) + x^{k} f(y/2) and f(y) f(x) = x^{h} f(y/2) + y^{k} f(x/2). Subtracting, 0 = (y^{h} - y^{k}) f(x/2) - (x^{h} - x^{k}) f(y/2). So f(x/2) = A (x^{h} - x^{k}) for some constant A for all x except possibly x = 1.

Substituting back in the original equation, we get A^{2}( (2x)^{h} - (2x)^{k})( (2y)^{h} - (2y)^{k}) = A(x^{h} - x^{k}) y^{h} + A(y^{h} - y^{k}) x^{k}, except possibly for x or y = 1/2 or 1. Comparing coefficients, 4^{h}A^{2} = A, 4^{k}A^{2} = -A. So A = 0, and hence f(x) = 0 for all x except possibly 1/2 and 1. So, in particular f(2) = 0. Putting x = y = 2 in the original equation gives: 0 = (2^{h} + 2^{k}) f(1), so f(1) = 0 also. Similarly, putting x = y = 1 in the original equation gives f(1/2) = 0. So f(x) = 0 for all x.

© John Scholes

jscholes@kalva.demon.co.uk

18 Sep 2002