R is the reals. f : [a, b] → R is continuous and ∫ab xnf(x) dx = 0 for all non-negative integers n. Show that f(x) = 0 for all x.
Solution
We deduce immediately that ∫ab xnf(x) p(x) dx = 0 for any polynomial p(x). We now need the Weierstrass approximation theorem: we can uniformly approximate any continuous function on a compact set by polynomials. In other words, we can find a polynomial p(x) such that |f(x) - p(x)| < ε for all x in [a, b]. Also, since [a, b] is compact, f must be bounded, say by k, so |f(x)| < k for all x in [a, b]. Now we have that |∫abf(x)2 dx| = |∫ f(x) (f(x) - p(x)) dx| ≤ ∫ |f(x)| |f(x) - p(x)| dx ≤ (b - a) k ε, which can be made arbitrarily small by choosing ε sufficiently small. Hence ∫abf(x)2 dx = 0, so f(x) = 0 throughout the interval.
© John Scholes
jscholes@kalva.demon.co.uk
25 Feb 2002