R is the reals. f : R → R is continuous and satisfies f(r) = f(x) f(y) for all x, y, where r = √(x2 + y2). Show that f(x) = f(1) to the power of x2.
Solution
Induction on n shows that f(x √n) = f(x)n, and hence f(n x) = f(x) to the power of n2. In particular, taking x = 1/n, f(1) = f(1/n) to the power of n2. Hence, provided f(1) is non-zero, f(1/n) = f(1) to the power of 1/n2. 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 x2 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