LCA Fourier transforms form a dense algebra in C0 of the dual

The Fourier transform is well defined and bounded: $\mathrm{\mid }\widehat{f}(\chi )\mathrm{\mid }\le {\int }_N\mathrm{\mid }f\mathrm{\mid }dn=\mathrm{\parallel }f{\mathrm{\parallel }}_1$, and $\widehat{f}:\widehat{N}\to \mathbb{C}$ is continuous, since compact-open convergence ${\chi }_i\to \chi$ gives uniform convergence on a compact set carrying all but $\epsilon$ of $\mathrm{\mid }f\mathrm{\mid }dn$ after choosing a compactly supported $L^1$-approximant of $f$.

Let $\tilde{A}=\mathbb{C}\oplus L^1(N)$ be the sum-norm unitization and $K=\mathrm{\Delta }(\tilde{A})$ its character space, a compact Hausdorff space by [F2]. Every $\psi \in K$ is unital; if $\psi$ does not vanish on $L^1(N)$ its restriction is a character of $L^1(N)$, hence equal to ${\lambda }_{\chi }$ for exactly one $\chi \in \widehat{N}$ by [F1], and then $\psi (z,f)=z+{\lambda }_{\chi }(f)$; otherwise $\psi (1,0)=1$ and $\psi (0,f)=0$ for all $f$, so $\psi =q$ with $q(z,f)=z$. Thus $K=\{{\tilde{\lambda }}_{\chi }:\chi \in \widehat{N}\}\cup \{q\}$, the map $\chi \mapsto {\tilde{\lambda }}_{\chi }$ is a homeomorphism onto the open subset $K\setminus \{q\}$ (openness because ${\tilde{\lambda }}_{\chi }\mapsto {\tilde{\lambda }}_{\chi }(0,f)=\widehat{f}(\chi )$), and $K$ is the one-point compactification of $\widehat{N}$ in the sense of [F4].

Each $\widehat{f}$ lies in $C_0(\widehat{N})$: the evaluation function $\psi \mapsto \psi (0,f)$ is continuous on $K$ by its pointwise-evaluation topology, equals $\widehat{f}$ on $\widehat{N}$, and is zero at $q$. Hence its closed superlevel set $\{\psi :\mathrm{\mid }\psi (0,f)\mathrm{\mid }\ge \epsilon \}$ is compact and misses $q$, for every $\epsilon >0$. This is a compact superlevel set of $\widehat{f}$ in $\widehat{N}$, proving the required vanishing at infinity.

The algebra $\mathcal{A}=\{\widehat{f}+c:f\in L^1(N),c\in \mathbb{C}\}$ of continuous functions on $K$ contains the constants, is self-adjoint and separates points: $\widehat{f}\widehat{g}+c$ corresponds to the $L^1$-function $f\ast g$ up to constants, $\widehat{f^{\ast }}=\bar{\widehat{f}}$ by [F5], distinct points of $\widehat{N}$ are separated by some $\widehat{f}$ by [F1], and $q$ is separated from any ${\tilde{\lambda }}_{\chi }$ by a $\widehat{f}$ with $\widehat{f}(\chi )\mathrm{\ne }0$, which exists by [F6]. By [F3] its uniform closure is $C(K)$.

The transforms alone have uniform closure $C_0(\widehat{N})$: if $g\in C_0(\widehat{N})$, regard it as an element of $C(K)$ with $g(q)=0$ by [F4] and choose ${\widehat{f}}_n+c_n\in \mathcal{A}$ with $\mathrm{\parallel }{\widehat{f}}_n+c_n-g{\mathrm{\parallel }}_{\mathrm{\infty }}<\epsilon$; evaluating at $q$ gives $\mathrm{\mid }{c}_n\mathrm{\mid }<\epsilon$ because ${\widehat{f}}_n(q)=0$ and $g(q)=0$, so $\mathrm{\parallel }{\widehat{f}}_n-g{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }{\widehat{f}}_n+c_n-g{\mathrm{\parallel }}_{\mathrm{\infty }}+\mathrm{\mid }{c}_n\mathrm{\mid }<2\epsilon$; hence $g$ is a uniform limit of Fourier transforms.

By [step 4.1] the Fourier transforms are uniformly dense in $C_0(\widehat{N})$; by [step 2.1] each lies in $C_0(\widehat{N})$; by [F5] the family is a self-adjoint algebra; it separates points by the separation used in [step 3.1]; and it vanishes nowhere by the bump construction of [F6], which at each $\chi$ supplies $c$ with $\widehat{c}(\chi )\mathrm{\ne }0$. No injectivity of the transform and no inversion formula were used.