| (13 intermediate revisions by the same user not shown) | |||
| Line 1: | Line 1: | ||
'''4.7''' | '''4.7''' | ||
| − | Let <math>f</math> be a continuous function on <math>I = [-1, 1]</math> with the property that <math>int_{I} x^n f(x) \ dx = 0</math> for <math>n = 0, 1, ...</math>. Show that <math>f</math> is identically 0. | + | Let <math>f</math> be a continuous function on <math>I = [-1, 1]</math> with the property that <math>\int_{I} x^n f(x) \ dx = 0</math> for <math>n = 0, 1, ...</math>. Show that <math>f</math> is identically 0. |
| − | ''' | + | '''PROOF''' |
| − | + | ||
| + | |||
| + | 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} (f(x))^2 dx \leq 0</math>, hence <math>f^2</math> is 0 almost everywhere, hence so too is f. | ||
| + | |||
| + | |||
| + | Written by Nicholas Stull | ||
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
