Fourier coefficients and trigonometric polynomials on the torus

Assume the Axiom of Countable Choice (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)) and let $\mathbb{T}=\mathbb{R}\mathrm{/}\mathbb{Z}$ with its normalized Haar integral $d{m}_{\mathbb{T}}$ (The one-dimensional torus and its normalized Haar integral).

Characters. For $k\in \mathbb{Z}$ define $e_k:\mathbb{T}\to \mathbb{C}$ by $e_k(x):=\exp (2\pi ik\tilde{x})$ for $x=[\tilde{x}]\in \mathbb{T}$, where $\exp$ is the complex exponential (The complex exponential by its power series). The definition is independent of the representative $\tilde{x}$: replacing $\tilde{x}$ by $\tilde{x}+n$ with $n\in \mathbb{Z}$ adds the period $2\pi kn$ to the argument of sine and cosine. Each $e_k$ is continuous, hence Borel, and satisfies

$e_k(x)e_l(x)=e_{k+l}(x),\mathrm{\mid }{e}_k(x)\mathrm{\mid }=1,\overline{e_k(x)}=e_{-k}(x),$

the last two by the cartesian form $\exp (i\theta )=\cos \theta +i\sin \theta$ of $\exp (x+iy)=e^x(\cos y+i\sin y)$, $\mathrm{\mid }\exp (x+iy)\mathrm{\mid }=e^x$, and $e^{i\pi }+1=0$ and Conjugation is an involutive real-field automorphism, $z\bar{z}=\mathrm{\mid }z{\mathrm{\mid }}^2$, and modulus is definite, multiplicative, and subadditive. In particular $e_0=1$ and every $e_k$ is bounded and nonzero everywhere.

Fourier coefficients. Let $f\in L^1(\mathbb{T};\mathbb{C})$ (Complex Lp classes and Euclidean test-function conventions), that is, a class of Borel functions with ${\int }_{\mathbb{T}}\mathrm{\mid }f\mathrm{\mid }d{m}_{\mathbb{T}}<+\mathrm{\infty }$. Define the Fourier coefficient of $f$ at $k\in \mathbb{Z}$ by

$\widehat{f}(k):={\int }_{\mathbb{T}}f(x)e_{-k}(x)d{m}_{\mathbb{T}}={\int }_{\mathbb{T}}f(x)\exp (-2\pi ikx)d{m}_{\mathbb{T}}\mathrm{.}$

This is well defined: $\mathrm{\mid }f{e}_{-k}\mathrm{\mid }=\mathrm{\mid }f\mathrm{\mid }$ because $\mathrm{\mid }{e}_{-k}\mathrm{\mid }=1$, so $f{e}_{-k}\in L^1(\mathbb{T};\mathbb{C})$ and its integral is finite; and the integral depends only on the class of $f$, because it is unchanged when $f$ is modified on a null set (Two integrable functions are equal almost everywhere exactly when all of their indefinite integrals agree). The map $f\mapsto \widehat{f}(k)$ is complex-linear for each $k$ (The Lebesgue integral is linear on $L^1(\mu )$).

Trigonometric polynomials. A trigonometric polynomial on $\mathbb{T}$ is a finite complex linear combination of characters, $p={\sum }_{k\in F}{c}_k{e}_k$, for a finite $F\subseteq \mathbb{Z}$ and scalars $c_k\in \mathbb{C}$. The set of all trigonometric polynomials is the span of $\{e_k:k\in \mathbb{Z}\}$; it is closed under addition, scalar multiplication, multiplication and complex conjugation (by the identities for $e_k{e}_l$ and $\overline{e_k}$), and it contains $e_0=1$. Nothing is claimed here about uniqueness of the coefficients in an expansion, nor about the size of $\widehat{p}$; those are properties of the characters proved below.

The finite torus. On ${\mathbb{T}}^n$, $n\ge 1$, the characters are $e_k(x):=\exp (2\pi ik\cdot x)$ for $k\in {\mathbb{Z}}^n$, where $k\cdot x$ is the Euclidean dot product of a representative tuple; trigonometric polynomials are finite complex linear combinations of these, and Fourier coefficients of $f\in L^1({\mathbb{T}}^n;\mathbb{C})$ are $\widehat{f}(k)={\int }_{{\mathbb{T}}^n}f{e}_{-k}d{m}_{{\mathbb{T}}^n}$. Fubini computes integrals of products of characters on the product measure (Fubini's theorem for L^1 functions on a sigma-finite product); for $f\in L^2({\mathbb{T}}^n;\mathbb{C})$ Hölder's inequality makes $f{e}_{-k}$ integrable (Complex Holder, Minkowski, and the quotient norm, Finite-measure $L^r$ includes into $L^p$ for $p<r$).