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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} |