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Written by Nicholas Stull
 
Written by Nicholas Stull
 
Jacob caught an error in my previous proof...So thanks to him for pointing that out
 

Latest revision as of 06:54, 6 July 2009

4.7

Let $ f $ be a continuous function on $ I = [-1, 1] $ with the property that $ \int_{I} x^n f(x) \ dx = 0 $ for $ n = 0, 1, ... $. Show that $ f $ is identically 0.


PROOF


So, since $ |f| $ is integrable, we can use Weierstrauss' Approximation theorem as follows. Since $ \int_{I} x^n f(x) \ dx = 0 $ for any non-negative n, then for any polynomial P, $ \int_{I} P(x) f(x) \ dx = 0 $. By Weierstrauss' Approximation theorem, we know that we may approach f with polynomials, hence taking a sequence of polynomials $ P_n $ approaching f, then $ \int_{I} \lim_{n} P_n(x) f(x) dx \leq \lim\inf \int_{I} P_n(x) f(x) dx $ by Fatou's Lemma, hence $ \int_{I} (f(x))^2 dx \leq 0 $, hence $ f^2 $ is 0 almost everywhere, hence so too is f.


Written by Nicholas Stull

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang