The Plancherel transform range is dense in L^2 of the dual

Given: A locally compact Hausdorff abelian group $G$ with Haar measure $m_G$, dual $\widehat{G}$, compatible dual Haar measure $m_{\widehat{G}}$, and the isometric extension $\mathcal{F}$ of the Fourier transform.

$\mathcal{F}$ is a linear isometry $L^2(G)\to L^2(\widehat{G})$ extending the transform ${\mathcal{F}}_{0}f=\widehat{f}$ on the dense subspace $L^1(G)\cap L^2(G)$; the transform of $f\in L^1(G)$ is $\widehat{f}(\chi )={\int }_{G}f(x)\overline{\chi (x)}d{m}_G(x)$. (Plancherel isometric extension on LCA groups, The Fourier transform on an LCA group, The space $L^p(\mu )$ as the quotient by null functions)

For $f\in L^1(G)$ and $x\in G$ the translation $T_{x}f=f(\cdot -x)$ lies in $L^1(G)$ and $\widehat{T_{x}f}(\chi )=\overline{\chi (x)}\widehat{f}(\chi )$; translations preserve $L^1\cap L^2$ and are isometries of $L^2$. (Fourier transform intertwines translation, modulation and convolution, Translation continuity and normalised local approximate identities on an LCA group)

If $u,v\in L^2$ then $uv\in L^1$ and $\mathrm{\parallel }uv{\mathrm{\parallel }}_1\le \mathrm{\parallel }u{\mathrm{\parallel }}_2\mathrm{\parallel }v{\mathrm{\parallel }}_2$. (Cauchy-Schwarz inequality for $L^2$)

A finite regular complex Borel measure on $\widehat{G}$ whose inverse transform $x\mapsto {\int }_{\widehat{G}}\chi (x)d\mu (\chi )$ vanishes for every $x\in G$ is zero. (Fourier-Stieltjes transforms determine finite Radon measures, Regular complex Borel measures)

For $h\in L^1(\widehat{G})$ the measure $h{m}_{\widehat{G}}$ has total variation $\mathrm{\mid }h\mathrm{\mid }{m}_{\widehat{G}}$, finite because $h\in L^1$. Haar measure on the locally compact space $\widehat{G}$ is Radon: finite on compact sets, outer regular on Borel sets and inner regular on open sets. For a bounded density $h$ supported in a compact set $K$, put $M=\mathrm{\parallel }h{\mathrm{\parallel }}_{\mathrm{\infty }}$. Given a Borel set $E$ and $ϵ>0$, outer regularity supplies open $O_1\supseteq E\cap K$ and $O_2\supseteq K\setminus E$ with $m(O_1\setminus (E\cap K)),m(O_2\setminus (K\setminus E))<ϵ\mathrm{/}(1+M)$. Then $V=O_1\cup (\widehat{G}\setminus K)$ is open and contains $E$, while $F=K\setminus O_2$ is compact and contained in $E$; both ${\int }_{V\setminus E}\mathrm{\mid }h\mathrm{\mid }dm$ and ${\int }_{E\setminus F}\mathrm{\mid }h\mathrm{\mid }dm$ are less than $ϵ$. Thus $hm$ is a finite regular complex measure, using compact closedness and closed-subset compactness (In a Hausdorff space a point and a disjoint compact set, and two disjoint compact sets, have disjoint open neighbourhoods; hence every compact subset is closed, and in a compact Hausdorff space the compact subsets are exactly the closed ones, A closed subspace of a compact space is compact, and a finite union of compact subspaces is compact). (A complex L^1 density defines a complex measure whose total variation is |h| dmu, Radon measure on an LCH space, Haar measure is positive on nonempty open sets and finite on compact sets)

In a locally compact Hausdorff space, every open neighbourhood of a point contains an open neighbourhood with compact closure (In a locally compact Hausdorff space every open set containing a point contains an open set containing it whose closure is compact and still inside; such a space is regular). If $f\in L^1(G)$ then $\widehat{f}\in C_0(\widehat{G})$, so the set where $\widehat{f}\mathrm{\ne }0$ is open. A nonempty open subset of $G$ has strictly positive Haar measure, and $C_c\subseteq L^1\cap L^2$. (Riemann-Lebesgue lemma on LCA groups, Haar measure is positive on nonempty open sets and finite on compact sets, Compact support, $C_c(X)$, and $C_0(X)$, C_c(X) is dense in L^p(mu) for a Radon measure)

A closed linear subspace of a Hilbert space whose orthogonal complement is trivial is the whole space. (A closed L2 subspace with trivial orthogonal complement fills L2, Riesz-Fischer completeness of $L^p$ for $1\le p\le \mathrm{\infty }$)

The support of an $L^2(\widehat{G})$ class can be restricted, up to a null set, to a $\sigma$-compact set: for a measurable representative $q$, each level set $E_n=\{\mathrm{\mid }q\mathrm{\mid }>1\mathrm{/}n\}$ has finite measure since $n^{-2}m(E_n)\le \mathrm{\parallel }q{\mathrm{\parallel }}_2^2$; outer regularity puts $E_n$ inside an open set $U_n$ of finite measure, and inner regularity exhausts $U_n$ up to a null set by countably many compact subsets $K_{n,j}$. Their countable union contains $\{q\mathrm{\ne }0\}$ up to a null set. Countable choices are licensed by DC, and every compact subset admits finite subcovers from covers by ambient open sets (The space $L^p(\mu )$ as the quotient by null functions, Left Haar integral and left Haar measure, Radon measure on an LCH space, The axiom of dependent choice: a relation in which every element is related to something admits an $\mathbb{N}$-indexed chain, Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right, A subspace is compact exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it).

With Dependent Choice, $C_c(\widehat{G})$ is dense in $L^1(\widehat{G})$ for the Radon Haar measure. (C_c(X) is dense in L^p(mu) for a Radon measure)