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'''Proof'''
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'''PROOF'''
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In progress
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Suppose not.  Then for some <math>x_0 \in I</math>, then <math>|f(x_0)| \neq 0</math>.  Choose <math>x_0</math> such that <math>|f|</math> attains its maximum; let's call the maximum value of <math>|f|</math> on I <math>\varepsilon</math>.  WLOG we assume <math>f(x_0) > 0</math>, for if not, then <math>f < 0</math> on <math>I</math>, and carry out the following argument, replacing <math>f</math> with <math>-f</math>.
 
  
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By continuity of <math>f</math>, <math>\exist \delta > 0</math> such that on <math>U = (x_0 - \delta, x_0 + \delta)</math>, <math>f > \frac{\varepsilon}{2}</math>.  Now, we have 2 cases:
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So, since <math>|f|</math> is integrable, we can use Weierstrauss' Approximation theorem as follows.  Since
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<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
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<math>\int_{I} (f(x))^2 dx \leq 0</math>, hence <math>f^2</math> is 0 almost everywhere, hence so too is f.
  
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<math>x_0 \neq 0</math>:
 
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<math>\int_{I} |f| x^n dx \geq \int_{U} |f| x^n dx > \int_{U} \frac{\varepsilon}{2} x^n = \frac{\varepsilon}{2} \left(\frac{(x_0 + \delta)^{n+1}}{n+1} - \frac{(x_0 - \delta)^{n+1}}{n+1} \right) \neq 0</math>.  (Note we can only say it's non-zero, because if n is even and x_0 is positive, then the integral is positive, but if n is odd and x_0 is negative, then the integral is negative.  If <math>x_0 \neq 0</math> then <math>\frac{(x_0 + \delta)^{n+1}}{n+1} \neq \frac{(x_0 - \delta)^{n+1}}{n+1}</math>
 
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<math>x_0 = 0</math>:
 
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sub-case: <math>n</math> is odd.  Then <math>\frac{(x_0 + \delta)^{n+1}}{n+1} = -\frac{(x_0 - \delta)^{n+1}}{n+1}</math>, and the conclusion remains unchanged.
 
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sub-case: <math>n</math> is even.  Then <math>\frac{(x_0 + \delta)^{n+1}}{n+1} = \frac{(x_0 - \delta)^{n+1}}{n+1}</math>, but we can alter our argument as follows:
 
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<math>\int_{I} |f| x^n dx \geq \int_{U} |f| x^n dx > \int_{(0, \delta)} \frac{\varepsilon}{2} x^n dx = \frac{\varepsilon \delta^{n+1}}{n+1} \neq 0</math>
 
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In either case, we get a contradiction, since we only assumed that <math>|f| > 0</math> at a point <math>x_0 \in I</math>, and so we see that <math>|f| = 0</math> on <math>I</math>, hence <math>f = 0</math> on <math>I</math>
 
  
 
Written by Nicholas Stull
 
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

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