# What is the difference between Fourier coefficients and Fourier transforms in this context?

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?