Fourier uniqueness for continuous functions on the Euclidean torus

Realize $T$ as the subset of ${\mathbb{R}}^{2n}$ whose $n$ coordinate pairs have squared norm one. It is closed and bounded, hence compact by [F2], and its Euclidean metric is Hausdorff. The continuous map $q:[0,1{]}^n\to T$, $q(x)=(e^{2\pi i{x}_j}{)}_j$, is onto by [F3], [F4]. To determine its fibres without using the affected injectivity assertion, suppose one coordinate has equal sine-cosine pairs at $s=2\pi x_j$ and $t=2\pi y_j$. The subtraction formulas give $\sin (s-t)=0$ and $\cos (s-t)=1$. Hence $s-t=m\pi$ for an integer $m$; the integer-shift formula in [F3] gives $1=(-1{)}^m$, so $m$ is even and $s-t\in 2\pi \mathbb{Z}$. The converse is periodicity. Thus $q(x)=q(y)$ exactly when every $x_j-y_j$ is an integer, which on the cube means equality or the endpoint identification $0\sim 1$ in each coordinate. Periodicity therefore defines a unique function $\tilde{f}$ on $T$ with $f=\tilde{f}\circ q$ on the cube. For any closed $C\subseteq \mathbb{C}$, the set $K=f^{-1}(C)\cap [0,1{]}^n$ is closed bounded and compact by [F2]. Then ${\tilde{f}}^{-1}(C)=q(K)$ is compact by [F5] and closed in the Euclidean ambient space by [F2], hence closed in $T$. This proves continuity of $\tilde{f}$.

The finite linear combinations of characters $z\mapsto {\prod }_j{z}_j^{k_j}$, $k\in {\mathbb{Z}}^n$, form a complex algebra on $T$: character products add indices, the zero index gives one, and conjugation negates indices by [F4]. Coordinate characters separate distinct points of $T$. Therefore [F1] applies. For any $\epsilon >0$, it gives a character polynomial $p$ with $\mathrm{\parallel }p-\overline{\stackrel{~}{f}}{\mathrm{\parallel }}_{\mathrm{\infty }}<\epsilon$. Pullback by $q$ is a finite sum of $e^{2\pi ik\cdot x}$; every integral of $f$ times such a term is zero by the hypothesis with index $-k$. For the integral manipulations, augment any finite disjoint nonnegative-simple display by its complement with coefficient $0$. Intersections of two augmented displays partition the cube and carry equal coefficients on nonempty cells, so finite additivity and $0\cdot (+\mathrm{\infty })=0$ prove representation independence. Common refinements give addition and monotonicity; scalar zero is direct and positive scalars are termwise. Supremum over simple minorants and increasing simple approximation give nonnegative additivity and hence finite complex $L^1$ linearity. Consequently $0\le {\int }_{[0,1{]}^n}\mathrm{\mid }f{\mathrm{\mid }}^2=\int f(\bar{f}-p\circ q)\le \epsilon \int \mathrm{\mid }f\mathrm{\mid }$, where the last bound means the modulus of the integral. Letting $\epsilon \downarrow 0$ gives zero square integral. Directly, $(1\mathrm{/}r){\mathbf{1}}_{\{\mathrm{\mid }f{\mathrm{\mid }}^2\ge 1\mathrm{/}r\}}\le \mathrm{\mid }f{\mathrm{\mid }}^2$ shows that each displayed level set is null; their countable union is $\{\mathrm{\mid }f\mathrm{\mid }>0\}$. Thus $f=0$ a.e. on the cube, proving the needed branch of [F6] locally.

If $f$ were nonzero at a cube point, continuity would give a neighbourhood on which $\mathrm{\mid }f\mathrm{\mid }$ is bounded below by a positive constant. Its intersection with the cube contains a nondegenerate box, even when the point lies on a face or corner, so it has positive measure, contradicting step 2.1. Thus $f=0$ throughout the cube. Every point of ${\mathbb{R}}^n$ differs from a cube point by an integer vector, so periodicity gives the global conclusion. Only the stated Euclidean integral interfaces need countable choice; the compact-space approximation uses one approximant at a time.