finished FT

1 files changed, 9 insertions, 3 deletions

diff --git a/ClassJune28.page b/ClassJune28.page
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+++ b/ClassJune28.page
@@ -217,11 +217,17 @@ It is important to note at this point that we have simply expressed the periodic
 
 $$\sum_{n=-\infty}^\infty a_n e^{inx}$$ converges for all $x$. Then 
 
-$$f(x) = \sum_{n=-\infty}^\infty a_n e^{inx}.$$
+$$f(x) = \frac{1}{\sqrt{2\pi}}\sum_{n=-\infty}^\infty a_n e^{inx}.$$
 *Proof.* 
-Consider the difference $g:=f-\sum_{n=-\infty}^\infty a_n e^{inx}$.
-****inner product with delta functions
+Consider the difference $g:=f-\sum_{n=-\infty}^\infty a_n e^{inx}$. We will show that $g=0$. Since inner products are linear in each variable, 
 
+$$(g,f_m)=(f,f_m)-\sum{a_n (f_n,f_m)}=a_n-a_n=0.$$
+
+In other words, $g$ is orthogonal to all of the complex exponentials $e^{imx}$. In a previous section, we showed that the function $\delta_n(x):=\cos^{2n}(x)+\cos^{2n+1}(x)$ has a peak at 0, and is very small everywhere else. Moreover, since sums and products and shifts of complex exponentials, and thus sinusoids, are complex exponentials, we can write $\delta_n(x-x_0)$ as a sum of complex exponentials, for any real shift $x_0$. This gives a function with a peak wherever we want it. 
+
+Since the inner product of $g$ with every complex exponential vanishes, its inner product with all the $\delta_n$'s also vanishes. This implies that $g$ is everywhere $0$! For suppose otherwise. Since $g$ is continuous, there is some point $x_0$ on the interval where $|g|$ attains its maximum $M>0$. We have
+$$0=(g,\delta_n(x-x_0))=\int g \delta_n(x-x_0) dx.$$
+but if $n$ is sufficiently small, we can prove that almost all of the mass of $\delta_n$ is contained in a narrow window (width $O(1/\sqrt{n}) of the peak. Since the peak is centered at the maximum of $g$, and $g$ is continuous, we can choose $n$ large enough so that more than 80% of the mass of $\delta_n$ is contained in a window around $x_0$ where $g$ still attains 90% of its own peak value. Thus the integral of their inner product is at least, say, $M/2$ times the total mass of $\delta_n$, which is, to wit, nonzero. Contradiction.
   
 #<b>Why the Fourier decomposition is useful? </b>
 Applications will be covered on Monday July 5, 2010. 
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