f is a real valued function on the reals satisfying (1) f(0) = 1/2, (2) for some real a we have f(x+y) = f(x) f(a-y) + f(y) f(a-x) for all x, y. Prove that f is constant.
Solution
Put x = y = 0. We get f(0) = 2 f(0) f(a), so f(a) = 1/2. Put y = 0, we get f(x) = f(x) f(a) + f(0) f(a-x), so f(x) = f(a-x). Put y = a-x, we get f(0) = f(x)2 + f(a-x)2, so f(x) = 1/2 or -1/2.
Now take any x. We have f(x/2) = 1/2 or -1/2 and f(a - x/2) = f(x/2). Hence f(x) = f(x/2 + x/2) = 2 f(x) f(a - x/2) = 1/2.
© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002