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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. | ||
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| + | '''PROOF''' | ||
| + | Dammit, Jacob. | ||
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| + | So, since <math>|f|</math> is integrable, and hence measurable, 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. | ||
Revision as of 06:51, 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
Dammit, Jacob.
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.
Proof
In progress
First, let's show that $ f $ 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 $ f $ is continuous on a compact set, it attains its minimum $ m $, and its maximum $ M $. Hence $ f $ is bounded, and $ |f| \leq K = \max{\left(|m|, |M|\right)} < \infty $. Hence $ f $ is bounded and continuous, hence so is $ |f| $, so f is integrable.
Now to prove the statement given.
Suppose not. Then for some $ x_0 \in I $, then $ |f(x_0)| \neq 0 $. Choose $ x_0 $ such that $ |f| $ attains its maximum; let's call the maximum value of $ |f| $ on I $ \varepsilon $. WLOG we assume $ f(x_0) > 0 $, for if not, then $ f < 0 $ on $ I $, and carry out the following argument, replacing $ f $ with $ -f $.
By continuity of $ f $, $ \exist \delta > 0 $ such that on $ U = (x_0 - \delta, x_0 + \delta) $, $ f > \frac{\varepsilon}{2} $. Now, we have 2 cases:
$ x_0 \neq 0 $:
$ \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 $. (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 $ x_0 \neq 0 $ then $ \frac{(x_0 + \delta)^{n+1}}{n+1} \neq \frac{(x_0 - \delta)^{n+1}}{n+1} $, so the above is non-zero.)
$ x_0 = 0 $:
sub-case: $ n $ is odd. Then $ \frac{(x_0 + \delta)^{n+1}}{n+1} = -\frac{(x_0 - \delta)^{n+1}}{n+1} $, and the conclusion remains unchanged.
sub-case: $ n $ is even. Then $ \frac{(x_0 + \delta)^{n+1}}{n+1} = \frac{(x_0 - \delta)^{n+1}}{n+1} $, but we can alter our argument as follows:
$ \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 $
In either case, this is a contradiction of our original assumption. And, since we only assumed that $ |f| > 0 $ at a point $ x_0 \in I $, we see that $ |f| = 0 $ on $ I $, hence $ f = 0 $ on $ I $
Written by Nicholas Stull
