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'''Proof''' | '''Proof''' | ||
In progress | In progress | ||
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| + | 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. | ||
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| + | 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>. | 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>. | ||
Revision as of 06:15, 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
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
