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| − | So, since <math>|f|</math> is integrable | + | So, since <math>|f|</math> is integrable, we can use Weierstrauss' Approximation theorem as follows. Since |
<math>\int_{I} x^n f(x) \ dx = 0</math> for any non-negative n, then for any polynomial P, <math>\int_{I} P(x) f(x) \ dx = 0</math>. By Weierstrauss' Approximation theorem, we know that we may approach f with polynomials, hence taking a sequence of polynomials <math>P_n</math> approaching f, then <math>\int_{I} \lim_{n} P_n(x) f(x) dx \leq \lim\inf \int_{I} P_n(x) f(x) dx</math> by Fatou's Lemma, hence | <math>\int_{I} x^n f(x) \ dx = 0</math> for any non-negative n, then for any polynomial P, <math>\int_{I} P(x) f(x) \ dx = 0</math>. By Weierstrauss' Approximation theorem, we know that we may approach f with polynomials, hence taking a sequence of polynomials <math>P_n</math> approaching f, then <math>\int_{I} \lim_{n} P_n(x) f(x) dx \leq \lim\inf \int_{I} P_n(x) f(x) dx</math> by Fatou's Lemma, hence | ||
<math>\int_{I} (f(x))^2 dx \leq 0</math>, hence <math>f^2</math> is 0 almost everywhere, hence so too is f. | <math>\int_{I} (f(x))^2 dx \leq 0</math>, hence <math>f^2</math> is 0 almost everywhere, hence so too is f. | ||
Revision as of 06:53, 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
