My question concerns some nomenclature on page 4 of this paper.

I've always viewed the Fourier transform as a continuous analogue of Fourier coefficients: the Fourier coefficients usually defined by a sum, and the Fourier transform by an integral. However, a paper I'm reading doesn't seem to reflect that idea, and I'm not sure why.

Consider a lattice $\Gamma := (2 \pi \mathbb{Z})^d$ as a subset of $\mathbb{R}^d$. Then the lattice analytically dual to $\Gamma$ is given by $\Gamma^{\ast} = \mathbb{Z}^d.$ We take the fundamental domains of the lattices $\Gamma$ and $\Gamma^{\ast}$ to be $\mathcal{O} := [0, 2\pi)^d$ and $\mathcal{O}^{\ast} := [0, 1)^d$, respectively. Let $\mathrm{det}(\Gamma)$ denote the volume of the lattice $\Gamma$. The paper then says this (on page 4):

For any $u \in L^2(\mathcal{O})$ and $f \in L^2(\mathbb{R}^d)$, define the Fourier coefficients and Fourier transform respectively:

$$\displaystyle\hat{u}(\mathbf{\theta}) = \frac{1}{\sqrt{\mathrm{det}(\Gamma)}} \int_{\mathcal{O}}e^{-i \langle\mathbf{\theta}, \mathbf{x} \rangle}u(\mathbf{x})d\mathbf{x}, \ \ \mathbf{\theta} \in \mathcal{O}^{\ast},$$

$$\displaystyle (\mathcal{F}f)(\mathbf{\xi}) = \frac{1}{(2\pi)^{\frac{d}{2}}} \int_{\mathbb{R}^d}e^{-i \langle \xi, \mathbf{x}\rangle} f(\mathbf{x})d\mathbf{x}, \ \ \mathbf{\xi} \in \mathbb{R}^d.$$

Could someone explain the reasoning behind these differing definitions? I can understand the definition of the Fourier transform here, but I don't understand why the Fourier coefficients of $u$ have been defined in this way. Aren't they both Fourier transforms?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.