f is an integrable real-valued function on the closed interval [0, 1] such that ∫01 xmf(x) dx = 0 for m = 0, 1, 2, ... , n - 1, and 1 for m = n. Show that |f(x)| ≥ 2n(n + 1) on a set of positive measure.
Solution
The trick is to look at (x - 1/2)n f(x). Using the relations given in the question, we have immediately that ∫01 (x - 1/2)n f(x) dx = 1. But we can also show that it must be small unless f(x) is large. For if |f(x)| < 2n(n + 1) except possibly on a set of zero measure, then | ∫ | < 2n(n + 1) ∫01 |x - 1/2|n dx = 2n+1(n + 1) ∫01/2 xn dx = 2n+1 xn+1|01/2 = 1. Contradiction.
© John Scholes
jscholes@kalva.demon.co.uk
27 Jan 2001