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(x0) = 0, then φf(x0) = 0.
So take any f and any x0 such that f(x0) = 0. Let L(x) = f(x) for x < x0 and 0 for x ≥ x0, and le R(x) = 0 for x ≤ x0 and f(x) for x > x0. Then f = L + R. Also φL(x) = 0 for any x > 0. But φL is continuous, so φL(x0) = 0. Similarly, φR(x0) = 0. Hence φf(x0) = 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 x0 the function f - f(x0)u is zero at x0 and hence φf(x0) = h(x0)f(x0).
© John Scholes
jscholes@kalva.demon.co.uk
25 Jan 2002