Find all real-valued functions f on the reals such that (f(x) + f(y)) (f(u) + f(v)) = f(xu - yv) + f(xv + yu) for all x, y, u, v.

**Solution**

Answer: there are three possible functions: (1) f(x) = 0 for all x; (2) f(x) = 1/2 for all x; or (3) f(x) = x^{2}.

Put x = y = 0, u = v, then 4 f(0) f(u) = 2 f(0). So either f(u) = 1/2 for all u, or f(0) = 0. f(u) = 1/2 for all u is certainly a solution. So assume f(0) = 0.

Putting y = v = 0, f(x) f(u) = f(xu) (*). In particular, taking x = u = 1, f(1)^{2} = f(1). So f(1) = 0 or 1. Suppose f(1) = 0. Putting x = y = 1, v = 0, we get 0 = 2f(u), so f(x) = 0 or all x. That is certainly a solution. So assume f(1) = 1.

Putting x = 0, u = v = 1 we get 2 f(y) = f(y) + f(-y), so f(-y) = f(y). So we need only consider f(x) for x positive. We show next that f(r) = r^{2} for r rational. The first step is to show that f(n) = n^{2} for n an integer. We use induction on n. It is true for n = 0 and 1. Suppose it is true for n-1 and n. Then putting x = n, y = u = v = 1, we get 2f(n) + 2 = f(n-1) + f(n+1), so f(n+1) = 2n^{2} + 2 - (n-1)^{2} = (n+1)^{2} and it is true for n+1. Now (*) implies that f(n) f(m/n) = f(m), so f(m/n) = m^{2}/n^{2} for integers m, n. So we have established f(r) = r^{2} for all rational r.

From (*) above, we have f(x^{2}) = f(x)^{2} ≥ 0, so f(x) is always non-negative for positive x and hence for all x. Putting u = y, v = x, we get ( f(x) + f(y) )^{2} = f(x^{2} + y^{2}), so f(x^{2} + y^{2}) = f(x)^{2} + 2f(x)f(y) + f(y)^{2} ≥ f(x)^{2} = f(x^{2}). For any u > v > 0, we may put u = x^{2} + y^{2}, v = x^{2} and hence f(u) ≥ f(v). In other words, f is an increasing function.

So for any x we may take a sequence of rationals r_{n} all less than x we converge to x and another sequence of rationals s_{n} all greater than x which converge to x. Then r_{n}^{2} = f(r_{n}) ≤ f(x) ≤ f(s_{n}) = s_{n}^{2} for all x and hence f(x) = x^{2}.

© John Scholes

jscholes@kalva.demon.co.uk

28 Jul 2002

Last corrected/updated 28 Jul 2002