Difference between revisions of "4.7 MA598R in progress" - Rhea

Revision as of 06:52, 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, and hence measurable, 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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 '''PROOF'''
-Dammit, Jacob.
 So, since <math>|f|</math> is integrable, and hence measurable, we can use Weierstrauss' Approximation theorem as follows.  Since
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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.
-'''Proof'''
-In progress
-First, let's show that <math>f</math> is bounded, and hence integrable (I originally didn't think I would need this, but I realized this morning that this fact, albeit easy to see, is essential to this proof).  Well, since <math>f</math> is continuous on a compact set, it attains its minimum <math>m</math>, and its maximum <math>M</math>.  Hence <math>f</math> is bounded, and <math>|f| \leq K = \max{\left(|m|, |M|\right)} < \infty</math>.  Hence <math>f</math> is bounded and continuous, hence so is <math>|f|</math>, so f is integrable.
-Now to prove the statement given.
-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>.
-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:
-<math>x_0 \neq 0</math>:
-<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 <math>x_0</math> is positive, then the integral is positive, but if n is odd and <math>x_0</math> 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>, so the above is non-zero.)
-<math>x_0 = 0</math>:
-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.
-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:
-<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>
-In either case, this is a contradiction of our original assumption.  And, since we only assumed that <math>|f| > 0</math> at a point <math>x_0 \in I</math>, we see that <math>|f| = 0</math> on <math>I</math>, hence <math>f = 0</math> on <math>I</math>
 Written by Nicholas Stull

Difference between revisions of "4.7 MA598R in progress" - Rhea

Revision as of 06:52, 6 July 2009

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