R is the reals. f **:** R → R is continuous and satisfies f(r) = f(x) f(y) for all x, y, where r = √(x^{2} + y^{2}). Show that f(x) = f(1) to the power of x^{2}.

**Solution**

Induction on n shows that f(x √n) = f(x)^{n}, and hence f(n x) = f(x) to the power of n^{2}. In particular, taking x = 1/n, f(1) = f(1/n) to the power of n^{2}. Hence, provided f(1) is non-zero, f(1/n) = f(1) to the power of 1/n^{2}. Hence f(m/n) = f(1) to the power of (m/n)^{2}. So we have established that f(x) = f(1) to the power of x^{2} for all rational x. But f is continuous, so the relation holds for all x.

If f(1) = 0, then the same reasoning establishes that f(x) = 0 for all x.

© John Scholes

jscholes@kalva.demon.co.uk

5 Mar 2002