Fix mistake in notes on Fourer on a flat surface

1 files changed, 28 insertions, 20 deletions

diff --git a/notes/FourierOnS2.tex b/notes/FourierOnS2.tex
index 213a9d6..aeae7ef 100644
--- a/notes/FourierOnS2.tex
+++ b/notes/FourierOnS2.tex
@@ -72,7 +72,7 @@ where \(m, n \in \mathbb{Z}\), be our basis functions in the space of ``nice'' f
 And finally by the Fourier theorem we can reconstruct the original function using an unoriginally named Fourier series:
 \[
   f(\mu, \nu) =
-    \sum_{m = -\infty}^{\infty} \sum_{n = -\infty}^{\infty} c_{m, n} B_{m, n}(\mu, \nu).
+    \sum_{m\in\mathbb{Z}} \sum_{n\in\mathbb{Z}} c_{m, n} B_{m, n}(\mu, \nu).
 \]
 
 % \begin{definition}[\(L^2\) norm]
@@ -124,48 +124,56 @@ The solutions to this elementary ODE are of course complex exponentials, the sam
 is easy to compute. This is shown in the next lemma.
 
 \begin{lemma}
-  Let \(f\) be a ``nice'' function. Then
+  Let \(f \in C(\mathbb{R}^2/\mathbb{Z}^2)\), then
   \[
     \langle \nabla^2 f, B_{m, n} \rangle
-    = \left( \frac{1}{m^2} + \frac{1}{n^2} \right) \langle f, B_{m, n} \rangle.
+    = (2\pi i)^2 \left( m^2 + n^2 \right) \langle f, B_{m, n} \rangle.
   \]
 \end{lemma}
 \begin{proof}
-To show this, we first expand the left side of the statement:
+To begin this proof, we first expand the left side of the statement:
 \begin{gather} 
   \nonumber
   \langle \nabla^2 f, B_{m, n} \rangle
-    = \iint_{\mathbb{R}^2} \nabla^2 f B_{m,n} d\mu d\nu \\
-    = \iint_{\mathbb{R}^2} \left(
+    = \iint_{[0, 1]^2} \nabla^2 f B_{m,n} d\mu d\nu \\
+    = \iint_{[0, 1]^2} \left(
         \frac{\partial^2 f}{\partial \mu^2} 
         + \frac{\partial^2 f}{\partial \nu^2} 
-      \right) e^{-im\mu} e^{-in\nu} d\mu d\nu.
+      \right) e^{-i2\pi m\mu} e^{-i2\pi n\nu} d\mu d\nu.
     \label{eqn:laplacian-coeffs-expanded}
 \end{gather}
 Since the integrand is a sum of partial derivatives, we now have 2 integrals. Notice that inside each integral we have an expressions of the form:
 \begin{equation} \label{eqn:inner-by-parts}
-    \int_{\mathbb{R}} \frac{\partial^2 f}{\partial \xi^2} e^{-ix\xi} d\xi,
+  \int_{[0, 1]} \frac{\partial^2 f}{\partial \xi^2} e^{-i2\pi x\xi} d\xi,
 \end{equation}
 once with \(x = m, \xi = \mu\) and the second time with \(x = n, \xi = \nu\).
 The integral \eqref{eqn:inner-by-parts} can be integrated by parts twice resulting in this ugly expression:
 \[
-  e^{-x\xi} \left(
-    \frac{\partial f}{\partial \xi} + \frac{f}{ix}
-  \right)\Bigg|_{-\infty}^{+\infty}
-  - \frac{1}{(ix)^2} \int_\mathbb{R} f e^{-jx\xi} d\xi.
+  e^{-i2\pi x\xi} \left(
+    \frac{\partial f}{\partial \xi} - i2\pi f
+  \right)\Bigg|_0^1
+  + (i2\pi x)^2 \int_{[0, 1]} f e^{-i2\pi x\xi} d\xi.
 \]
-However, because \(f\) is ``nice'', both \(f\) and its derivative vanish at infinity, leaving only the integral. By substituting back this result into \eqref{eqn:laplacian-coeffs-expanded} we get:
+However, actually this is not too bad. That is because once we substitute the bounds two things happen: the exponential in the front always equals 1 and what is inside of the parenthesis can be rewritten as
+\[
+  \frac{\partial f}{\partial\xi}(1) - \frac{\partial f}{\partial\xi}(0)
+  + i2\pi \left[ f(1) - f(0) \right],
+\]
+which equals zero, since \(f\) and its derivative are continuous and periodic. Hence, we are left with two integrals, that when substituted back into \eqref{eqn:laplacian-coeffs-expanded} give:
 \begin{align*}
-  \langle \nabla^2 f, B_{m, n} \rangle
-    &= \frac{1}{m^2} \iint f e^{-im\mu} e^{-in\nu} d\mu d\nu \\
-    &\qquad + \frac{1}{n^2} \iint f e^{-im\mu} e^{-in\nu} d\mu d\nu \\
-    &= \frac{1}{m^2} \langle f, B_{m, n} \rangle
-       + \frac{1}{n^2} \langle f, B_{m, n} \rangle
-       \qedhere
+  \langle \nabla^2 & f, B_{m, n} \rangle = \\
+    &(i2\pi m)^2 \iint_{[0, 1]^2} f e^{-i2\pi m\mu} e^{-i2\pi n\nu} d\mu d\nu \\
+    &\quad + (i2\pi n)^2 \iint_{[0, 1]^2} f e^{-i2\pi m\mu} e^{-i2\pi n\nu} d\mu d\nu.
 \end{align*}
-
+Finally, to complete the proof we rewrite the right side using the compact notation:
+\[
+  (i2\pi m)^2 \langle f, B_{m, n} \rangle
+  + (i2\pi n)^2 \langle f, B_{m, n} \rangle.
+  \qedhere
+\]
 \end{proof}
 
+% TODO: closing words: this is why fourier is useful
 
 \section{Fourier on the Sphere}