f is an integrable real-valued function on the closed interval [0, 1] such that ∫_{0}^{1} x^{m}f(x) dx = 0 for m = 0, 1, 2, ... , n - 1, and 1 for m = n. Show that |f(x)| ≥ 2^{n}(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 ∫_{0}^{1} (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)| < 2^{n}(n + 1) except possibly on a set of zero measure, then | ∫ | < 2^{n}(n + 1) ∫_{0}^{1} |x - 1/2|^{n} dx = 2^{n+1}(n + 1) ∫_{0}^{1/2} x^{n} dx = 2^{n+1} x^{n+1}|_{0}^{1/2} = 1. Contradiction.

© John Scholes

jscholes@kalva.demon.co.uk

27 Jan 2001