Continuous lattice-periodic functions are determined by their lattice Fourier coefficients

Given: Countable Choice, a continuous $\mathrm{\Lambda }$-periodic function $h:{\mathbb{R}}^n\to \mathbb{C}$ with $\mathrm{\Lambda }=A{\mathbb{Z}}^n$ a full-rank lattice, ${\mathrm{\Lambda }}^{\ast }=A^{-T}{\mathbb{Z}}^n$, $F=A((0,1{]}^n)$ (Full-rank lattices, covolume, and the dual lattice, Fundamental parallelotopes of a lattice tile Euclidean space with covolume volume), and ${\int }_{F}h{e}^{-2\pi i{\lambda }^{\ast }\cdot x}dx=0$ for every ${\lambda }^{\ast }\in {\mathrm{\Lambda }}^{\ast }$.

The pullback $h_A(y):=h(Ay)$ is continuous, as a composite of continuous maps (Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous), and ${\mathbb{Z}}^n$-periodic: $h_A(y+k)=h(Ay+Ak)=h(Ay)$ because $Ak\in \mathrm{\Lambda }$ (Full-rank lattices, covolume, and the dual lattice, Rectangular matrix multiplication and the identity matrix $I_n$, including zero-sized shapes).

Published uniqueness for the unit lattice: a continuous ${\mathbb{Z}}^n$-periodic $g$ with ${\int }_{[0,1{]}^n}g(y)e^{-2\pi ik\cdot y}dy=0$ for every $k\in {\mathbb{Z}}^n$ is zero everywhere; this assumes Countable Choice (Fourier uniqueness for continuous functions on the Euclidean torus).

The cube $[0,1{]}^n$ is compact by Heine-Borel in ${\mathbb{R}}^n$: with the Euclidean metric a subset of ${\mathbb{R}}^n$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line, so the closed parallelotope $\bar{F}=A([0,1{]}^n)$ is compact, and the real and imaginary parts of $h$ are bounded on it by the compact-image and extreme-value clauses of A continuous image of a compact space is compact; a continuous real-valued map on a nonempty compact space attains a maximum and a minimum; and a continuous bijection from a compact space to a Hausdorff space is a homeomorphism. The boundary of the unit cube is a finite union of degenerate boxes, hence null (A box in ${\mathbb{R}}^n$ with parameters $a_i\le b_i$ is Lebesgue measurable of measure ${\prod }_{i<n}(b_i-a_i)$, whichever of its faces are included); its image under $A$ is null by A linear map $T$ of ${\mathbb{R}}^n$ sends Lebesgue measurable sets to Lebesgue measurable sets, with ${\lambda }_n(T[E])=\mathrm{\mid }\det T\mathrm{\mid }{\lambda }_n(E)$ when $T$ is invertible and $T[E]$ Lebesgue null when it is not. Thus the substitution A C^1 diffeomorphism satisfies the change-of-variables formula for L^1 functions on the open box also computes the half-open and closed-domain integrals, since integrals over null sets vanish (A nonnegative integral over a null set vanishes).

Transpose algebra: $k\cdot y=k\cdot A^{-1}x=(A^{-1}{)}^{T}k\cdot x=(A^{-T}k)\cdot x$, and $A^{-T}k\in {\mathrm{\Lambda }}^{\ast }$ for $k\in {\mathbb{Z}}^n$ (Transpose is linear and involutive, and $(AB{)}^{\mathsf{T}}=B^{\mathsf{T}}{A}^{\mathsf{T}}$, Rectangular matrix multiplication and the identity matrix $I_n$, including zero-sized shapes, Full-rank lattices, covolume, and the dual lattice); the exponential is additive, $e^{u+v}=e^u{e}^v$ ($\exp (z+w)=\exp z\exp w$, and the complex exponential extends the real exponential).