Let S be the set of continuous real-valued functions on the reals. φ **:**S → S is a linear map such that if f, g ∈ S and f(x) = g(x) on an open interval (a, b), then φf = φg on (a, b). Prove that for some h ∈ S, (φf)(x) = h(x)f(x) for all f and x.

**Solution**

Obviously if f = 0 on an open interval (a, b) then φf = 0 on (a, b). But we need the stronger result that if f(x_{0}) = 0, then φf(x_{0}) = 0.

So take any f and any x_{0} such that f(x_{0}) = 0. Let L(x) = f(x) for x < x_{0} and 0 for x ≥ x_{0}, and le R(x) = 0 for x ≤ x_{0} and f(x) for x > x_{0}. Then f = L + R. Also φL(x) = 0 for any x > 0. But φL is continuous, so φL(x_{0}) = 0. Similarly, φR(x_{0}) = 0. Hence φf(x_{0}) = 0.

Let u be the function which has value 1 for all x. Let h = φu. Then h is continuous. Also for any f and any x_{0} the function f - f(x_{0})u is zero at x_{0} and hence φf(x_{0}) = h(x_{0})f(x_{0}).

© John Scholes

jscholes@kalva.demon.co.uk

25 Jan 2002