Fourier Transform on the circle

Question

I've been studying the Fourier Transform on Riemannian Manifolds. I got the general idea (I think) and I started by trying to define a Fourier Transform on the circle, $S^1$. I have found that this result is already known: $$F(n)=\frac{1}{2\pi}\int_0^{2\pi}f(\theta)e^{-in\theta}d\theta,$$ the same result that I got. But then I got confused... Would the inverse Fourier transform be: $$f(\theta)=\sum_{n\in\mathbb Z}^\infty F(n) e^{in\theta}$$?

If so, if the circle doesn't have unitary radius but some value $R\in \mathbb R$ I imagine that in the definition of the Fourier transform I would have to change $d\theta \to Rd\theta$...is this correct? And what about the inverse Fourier transform?

Answer 1

One way to think about the FT on the circle is that it is really the FT of a periodic function. Let's think about the FT of a function $f$ that has period $p$. We may write $f$ as

$$f(x) = \sum_{n=-\infty}^{\infty} f_0(x+n p)$$

where

$$f_0(x) = \begin{cases} \\ f(x) & |x|<p/2 \\ 0 & |x| > p/2\end{cases}$$

We may rewrite this expression for $f$ as a convolution:

$$f(x) = f_0(x) \otimes \sum_{n=-\infty}^{\infty} \delta(x+n p) $$

where $\delta$ is the Dirac delta function and $\otimes$ denotes convolution. The series of delta functions on the right is known as a comb function. We may then find the FT of $f$, $\hat{f}$:

$$\hat{f}(k) = \int_{-\infty}^{\infty} dx \: f(x) \,e^{i k x}$$

by the convolution theorem. Note that the FT of the comb function is another comb function; the result is

$$\hat{f}(k) = 2 \pi \hat{f_0}(k) \sum_{n=-\infty}^{\infty} \delta(p k - n) = \frac{2 \pi}{p} \sum_{n=-\infty}^{\infty} \hat{f_0}\left(\frac{n}{p}\right) \delta\left(k-\frac{n}{p}\right) $$

That last step results from the sampling property of delta functions. Taking the inverse FT gives us a new representation for $f$:

$$f(x) = \sum_{n=-\infty}^{\infty} \frac{1}{p} \hat{f_0}\left(\frac{n}{p}\right) \exp{\left(i \frac{n x}{p}\right)}$$

and, from above,

$$\frac{1}{p} \hat{f_0}\left(\frac{n}{p}\right) = \frac{1}{p} \int_{-p/2}^{p/2} dx \: f_0(x) \exp{\left(i \frac{n x}{p}\right)}$$

Now, finally, to answer your question: the period $p$ is indicative of a map from a circle of circumference $p$ to the interval $x \in [-p/2,p/2)$. From the above, the coefficients of a Fourier series are simply the FT of a single period of the periodic function. Thus, the inverse transforms for the circle and the real line are one and the same.

Answer 2

There is a general theorem of Pontryagin for locally compact abelian groups.

The dual of $\mathbb R$, meaning the group of all continuous maps $\mathbb R \to \mathbb U$ is $\widehat{\mathbb R} \simeq \mathbb R$.

For the circle $\widehat{\mathbb U} = \mathbb Z$.

And here goes the general definition of the Fourier transform. Let $f \in \mathrm L^2(\mathrm G)$ with $\mathrm G$ a locally compact abelian group, for $\chi \in \widehat{\mathrm G}$, $$ \widehat{f}(\chi) = \int_{g \in \mathrm G} f(g) \overline{\chi(g)} d\mu$$ where $d\mu$ is a Haar measure on $\mathrm G$.

With this, you get back all the theorems known for the Fourier transform on $\mathbb R$. Especially the Fourier Inverse is $$ \hat{h}(g) = \int_{\chi \in \widehat{\mathrm G}} h(g) \chi(g) d\mu$$.

This explains you why in one case you have an integral on $\mathbb R$ (for the Fourier Transform on $\mathbb R$) and on the other case you have a series (for the Fourier Transform on the circle).