Why does harmonic analysis give a series on the circle, but an integral over the line?

Question

Why this dichotomy between series and integrals? How would one view a circle and think, ah, harmonic analysis here must involve series?

Answer 1

Hint: Because the dual group of the torus is the integers and the dual group of the reals is itself.

Here's a Reference https://en.m.wikipedia.org/wiki/Pontryagin_duality

Answer 2

One of the earliest approaches for understanding the problems on the disk and on the half plane comes from Fourier. If you use separation of variables to solve the Laplace equation on the unit disk, you try solutions $R(r)\Theta(\theta)$ and obtain the equation \begin{align} 0 & =\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right)R(r)\Theta(\theta) \\ & = \left(\frac{1}{r}\frac{\partial}{\partial r}r\frac{\partial}{\partial r}+\frac{1}{r^2}\frac{\partial^2}{\partial\theta^2}\right)R(r)\Theta(\theta) \\ & = R''(r)\Theta(\theta)+\frac{1}{r}R'(r)\Theta(\theta)+\frac{1}{r^2}R(r)\Theta'(\theta). \end{align} This is a separable equation. Dividing by $R\Theta$ and multiplying by $r^2$ gives $$ r^2\frac{R''}{R}+r\frac{R'}{R}=-\frac{\Theta''(\theta)}{\Theta(\theta)}. $$ Therefore, there is a constant $\lambda$ such that $$ r^2\frac{R''}{R}+r\frac{R'}{R}=\lambda,\;\;\;\lambda = -\frac{\Theta''}{\Theta}, $$ which leads to two equations $$ r^{2}R''+rR'-\lambda R = 0,\;\; R \mbox{ bounded near } 0. \\ \Theta''+\lambda\Theta = 0,\;\;\;\Theta(0)=\Theta(2\pi). $$ The periodicity condition forces $\Theta(\theta)=Ce^{in\theta}$, $n=0,\pm 1,\pm 2,\cdots$ and corresponding $\lambda_n = n^2$. So the only acceptable solutions on the full disk (i.e., non-singular at the origin) are $$ \{ r^{|n|}e^{in\theta} \}_{n=-\infty}^{\infty} $$ And this is a spanning set. The discrete nature of the eigenfunctions is determined by the periodicity condition. General solutions have the the form $$ u(r,\theta)=\sum_{n=-\infty}^{\infty}A_n r^{|n|}e^{in\theta}, $$ where the $A_n$ are Fourier coefficients of the boundary function $u(1,\theta)$ on the unit circle.

By contrast, if you're looking at the upper half plane, there are no conditions to impose that lead to a discrete set of parameters. For $\Pi^+ = \{ (x,y) : x,y\in\mathbb{R},\; y \ge 0 \}$, boundedness can be imposed, which leads to solutions $$ e^{isx}e^{-|s|y},\;\;\; -\infty < s < \infty. $$ The separation parameter $\lambda$ is $s^2$ in this case, where $s$ can be any real number. You end up using integrals to form general linear combinations $$ u(x,y)=\int_{-\infty}^{\infty}A(s)e^{isx}e^{-|s|y}ds, $$ where the coefficient function $A(s)$ is determined as a Fourier transform of the boundary function $u(x,0)$ on $y=0$. Without endpoint conditions on an interval, the eigenvalue parameter is allowed to assume a continuum of values, and the sums give way to integrals.

Answer 3

I agree with Jacky Chong that Pontryagin Duality is important here. However, the fact that, on $\mathbb{R}$, the Fourier Transform of the basic periodic functions, in the sense of distributions, are discrete is also important. For example, the Fourier Transform of $e^{i\alpha x}=\cos(\alpha x)+i\sin(\alpha x)$ is $$ \begin{align} &\lim_{\lambda\to0}\int_{\mathbb{R}} e^{-\lambda x^2}e^{i\alpha x}e^{-2\pi ix\xi}\mathrm{d}x\\ &=\lim_{\lambda\to0}e^{-\left.\left(\pi\xi-\frac\alpha2\right)^2\right/\lambda}\int_{\mathbb{R}}e^{-\left.\lambda\left(x+i\left(\pi\xi-\frac\alpha2\right)\right/\lambda\right)^2}\mathrm{d}x\\ &=\lim_{\lambda\to0}\sqrt{\frac\pi\lambda}\,e^{-\left.\left(\pi\xi-\frac\alpha2\right)^2\right/\lambda}\\ &=\delta\!\left(\xi-\frac\alpha{2\pi}\right)\tag{1} \end{align} $$ When looking at functions on $[0,1]$, we can extend them periodically to $\mathbb{R}$ and expand them as a sum $$ f(x)=\sum_{k\in\mathbb{Z}}a_ke^{2\pi ikx}\tag{2} $$ Applying $(1)$ to $(2)$ gives $$ \hat{f}(\xi)=\sum_{k\in\mathbb{Z}}a_k\delta(\xi-k)\tag{3} $$ which is essentially the Fourier Series for a function on $[0,1]$.