Earlier HR and scalar Plancherel proof bodies supply the complete stated analytic premises.

Reduced Fourier norms, biduality and the returned measure

The H1 Fourier completion identifies the character group and its topology. We use the scalar Fourier unitary to identify the reduced norm, prove topological biduality, and determine the returned Haar measure exactly. The earlier H0: compact topology and Hilbert tensor and HR-02, HR-03, HR-05, HR-06, HR-07, HR-08, HR-09 supply the Haar and Radon measure facts used below.

The earlier dual Haar construction, scalar Plancherel unitary and dense inverse-integral proof supply the negative-character transform for every LCA group, including its dual. These proofs use the earlier CF/SC, topology, HR, L24 and H1 bodies. The later deductions below use those exact results.

H2. The reduced norm and inverse integral

Write $\Gamma=\widehat G$. Apply the earlier scalar Plancherel theorem to obtain the onto unitary $F_G:L^2(G,m)\to L^2(\Gamma,\widehat m)$. For $h\in L^1(G)$, integration of the unitary translations defines $\lambda(h)$, with norm at most $\|h\|_1$, by L24. On integrable square-integrable $f$, qualified Fubini gives $F_G\lambda(h)f=\widehat h\,F_Gf. \tag{H2.1}$ One may first prove this for $h,f\in C_c$ and extend in $L^1$ and $L^2$, using the common operator bounds. A multiplier by continuous $k\in C_0(\Gamma)$ has norm $\|k\|_\infty$: the upper bound is pointwise; for any smaller bound, the open set where its modulus exceeds it contains a nonzero compact bump, and Haar full support makes that bump an $L^2$ norm test. Thus $\|\lambda(h)\|=\|\widehat h\|_\infty=\|h\|_u. \tag{H2.2}$ The group quotient is an isometric onto isomorphism by L24 Section 11. This statement applies separately to $G$ and to every other LCA group, including $\Gamma$.

An integrable nonnegative density $w$ against Radon Haar gives a finite Radon measure $w\,dm$. Here are the regularity details used below. L24's finite-exponent comparison gives a Borel $L^1$ representative on a sigma compact carrier. SC's simple approximation gives nonnegative finite sums $s=\sum_jc_j1_{E_j}$, with all $E_j$ of finite Haar measure, and $\|w-s\|_1$ arbitrarily small. For any Borel $E$, finite-set inner regularity supplies a compact subset of $E\cap E_j$ of nearly that measure. It also supplies a compact subset $C\subset E_j\setminus E$ of nearly its measure; the open set $X\setminus C$ contains $E$, and its excess measure for the restriction $1_{E_j}dm$ is arbitrarily small. Thus each finite restricted measure is inner and outer regular, and so is their finite weighted sum. The bound $\left|\int_B(w-s)\,dm\right|\le\|w-s\|_1 \quad\text{for every Borel }B \tag{H2.3a}$ transfers these compact and open approximations to $w\,dm$. This proves the assertion, including compact tail approximation for its whole finite mass. The same estimate applies to the variation density of an integrable complex function.

We will need the inverse integral with its sign fixed. For $k\in C_c(\Gamma)$, set $v(x)=\int_\Gamma k(\chi)\chi(x)\,d\widehat m(\chi)$. Compact-tail uniform estimates make $v$ continuous. Testing $f\in C_c(G)$, the earlier finite-measure interchange and Plancherel inverse-domain theorem give $\int_G f(x)\overline{v(x)}\,dm(x) =\langle F_Gf,k\rangle =\langle f,F_G^*k\rangle. \tag{H2.3}$ This identifies the continuous function $v$ locally with the $L^2$ class $F_G^*k$. The identification needs no global separability assumption: on each sigma compact open coset of L24 Section 2, both functions are locally square integrable; compact cutoff approximation of finite-measure indicators shows that a locally integrable difference with zero integrals against every compact continuous test is zero almost everywhere. On a countable compact exhaustion of that coset this gives the asserted local equality. Only countably many cosets of $F_G^*k$ have a nonzero class. On every other coset the continuous $v$ is zero everywhere, by Haar full support. Hence $v$ itself has a sigma compact carrier, belongs to $L^2(G)$, and equals $F_G^*k$ there.

The same formula holds for $k\in L^1(\Gamma)\cap L^2(\Gamma)$: approximate simultaneously in both norms by $C_c(\Gamma)$. Such approximation follows by truncation to finite-measure bounded functions and the cutoff proof of L24 Lemma 3.1. The inverse integrals converge uniformly by the $L^1$ bound, while the inverse transforms converge in $L^2$. On each sigma compact coset a common subsequence converges almost everywhere, so the two limits agree locally, and the preceding carrier argument gives global $L^2$ equality. The integral has the positive character sign; $F_\Gamma$ below has the negative sign.

H3. Biduality from the scalar unitary and compact tests

Let $P=\widehat\Gamma$, which is LCH by H1 applied to $\Gamma$, and let $j(x)(\chi)=\chi(x)$. Evaluation $G\times\Gamma\to\mathbb T$ is jointly continuous: choose a compact neighbourhood of a given $x$, restrict uniform character error on that neighbourhood, and use continuity of the fixed character in the $x$ variable. Finite covers of any compact $K\subset\Gamma$ consequently show $\sup_{\chi\in K}|\chi(x)-1|\longrightarrow0\quad(x\to0). \tag{H3.1}$ Thus $j:G\to P$ is a continuous homomorphism.

Given an identity neighbourhood $U\subset G$, choose relatively compact $V$ with $V-V\subset U$, and $0\ne h\in C_c(G)$ supported in $V$, using H0. The normalized autocorrelation $c=h*\widetilde h/\|h\|_2^2$, with $\widetilde h(x)=\overline{h(-x)}$, has support in $U$, satisfies $c(0)=1$, and has transform $w=|\widehat h|^2/\|h\|_2^2\in L^1(\Gamma)\cap L^2(\Gamma)$. The last membership follows from $\widehat h\in L^2\cap L^\infty$. H2's inverse integral and continuity on both sides give $c(x)=\int_\Gamma w(\chi)\chi(x)\,d\widehat m(\chi), \qquad\int_\Gamma w=1. \tag{H3.2}$ Choose compact $K$ with $\int_{\Gamma\setminus K}w<1/8$. If $\sup_K|\chi(x)-1|<1/4$, then $|c(x)-1|<\tfrac14+2\cdot\tfrac18=\tfrac12. \tag{H3.3}$ Hence $x\in U$. This proves that $j^{-1}$ is continuous on its image and that its kernel is zero: a kernel element lies in every identity neighbourhood. Therefore $j$ is a topological embedding.

Every locally compact subgroup $B\subset P$ is closed. Choose an identity neighbourhood in $B$ with compact closure $C\subset B$, and an ambient open $O\subset P$ whose intersection with $B$ is inside that neighbourhood. Every point of $O\cap\overline B$ is in the ambient closure of $O\cap B$, hence in $C\subset B$. Thus $B$ is open in $\overline B$, and its cosets make its complement open there. A dense closed subgroup of its closure is the closure itself. Apply this to $B=j(G)$.

We prove the finite-density uniqueness needed for surjectivity directly. If $a\in L^1(\Gamma)$ and $\int a(\chi)\chi(x)\,d\widehat m=0$ for every $x\in G$, then $a=0$. Put $d\omega=|a|\,d\widehat m$, a finite Radon measure. The evaluation polynomials $\sum_l c_l\chi(x_l)$ separate points of $\Gamma$ and are a unital self-adjoint algebra. On compact $K\subset\Gamma$, CF Section 5 approximates every continuous $q:K\to\mathbb C$, $|q|\le1$, by such polynomials with a global modulus bound one. Here is the bound: first choose a polynomial $p_0$ close on $K$, and let $M\ge1$ bound its modulus globally. Approximate $T(z)=z/\max(1,|z|)$ on $|z|\le M$ by a polynomial $Q(z,\bar z)$, with error $\delta$, using CF Section 5 again. Then $p=Q(p_0,\bar p_0)/(1+\delta)$ is still an evaluation polynomial, has global modulus at most one, and has error at most $2\|p_0-q\|_K+2\delta$ on $K$. The inequality used is $|T(z)-q|\le2|z-q|$ for $|q|\le1$.

If $m=\|a\|_1>0$, approximate the measurable phase $\bar a/|a|$, defined as zero at zeros, in $L^1(\omega)$ by $q_0\in C_c(\Gamma)$, $|q_0|\le1$, to error $m/8$. Finite-measure simple approximation, finite Radon regularity and H0 cutoffs prove this density; clipping by $T$ preserves an arbitrarily small error. Choose compact $K$ with $\omega(\Gamma\setminus K)<m/8$ and an evaluation polynomial $p$, $|p|\le1$ globally, with $|p-q_0|_K<1/8$. The hypothesis makes $\int ap=0$, whereas $\left|m-\int ap\right| \le\left\|\bar a/|a|-q_0\right\|_{L^1(\omega)} +\tfrac18\omega(K)+2\omega(\Gamma\setminus K)<\tfrac12m, \tag{H3.4}$ a contradiction. This uniqueness argument uses one finite measure and never an uncountable union of null exceptions.

If the closed subgroup $B=j(G)$ were proper, choose nonzero nonnegative $\phi,\psi\in C_c(P)$ with support product contained in an open set disjoint from $B$. Small neighbourhoods at a point of that open set and at the identity, followed by H0, supply them. Their convolution is nonzero: its integral is the positive product of their integrals. Take $u=F_\Gamma^{-1}\phi$, $v=F_\Gamma^{-1}\psi$; they are in $L^2(\Gamma)$, so $a=uv\in L^1(\Gamma)$. Parseval gives $\widehat a=\phi*\psi\quad\text{pointwise on }P. \tag{H3.5}$ To verify the formula, for each $\tau\in P$ write $\widehat{uv}(\tau)=\langle u,\tau\bar v\rangle$. The conjugation and modulation identities, checked by the integral on $C_c(\Gamma)$ and extended by unitary density, give $F_\Gamma(\tau\bar v)(\eta)=\overline{F_\Gamma v(\eta^{-1}\tau)}$. Parseval now gives the convolution integral. Cauchy–Schwarz proves its absolute convergence at each $\tau$; compact continuous approximations show it is continuous. This justifies (H3.5) for the stated arbitrary $L^2$ vectors.

Since $\widehat a(j(-x))=0$ for every $x$, the just-proved finite-density uniqueness forces $a=0$, contrary to the nonzero convolution. Therefore $j:G\xrightarrow{\cong}\widehat{\widehat G} \tag{H3.6}$ is a topological group isomorphism. This proof uses the earlier scalar Plancherel theorem on $G$ and on $\Gamma$, but imports no separate Bochner or Pontryagin theorem.

H4. The Haar measure returns exactly

Let $m_P$ be the Haar measure on $P$ paired with $\widehat m$ by the dual Haar construction and scalar Plancherel. Haar uniqueness and H3 give $m_P=cj_*m$ for some $c>0$. For $0\ne\phi\in C_c(\Gamma)$, H2's inverse integral gives $F_\Gamma\phi(j(x)) =\int_\Gamma\phi(\chi)\overline{\chi(x)}\,d\widehat m(\chi) =F_G^{-1}\phi(-x). \tag{H4.1}$ Reflection preserves abelian Haar measure, by the inversion formula in L24 and unimodularity. Plancherel on both groups yields $\|\phi\|_2^2=\|F_\Gamma\phi\|_{L^2(P,m_P)}^2 =c\|F_G^{-1}\phi\|_{L^2(G,m)}^2=c\|\phi\|_2^2. \tag{H4.2}$ Its norm is positive, so $c=1$. Density and unitarity extend (H4.1) to all $L^2(\Gamma)$. Equivalently $F_\Gamma F_Gf(x)=f(-x)$ as an $L^2$ equality. No arbitrary scalar is left in the returned Haar measure.

H5. Further reading and prerequisites

Dana Williams, author draft v3.1, Proposition 3.1, printed page 82 (PDF page 94), motivates the group Fourier completion; its representation-theoretic inputs are replaced by H1's complete CF/L24 deduction. Siegfried Echterhoff, arXiv:1006.4975v4, Remark 3.4(8), pages 9–10, records the same identification with positive characters. H1 uses negative characters by indexing that representation with $\chi^{-1}$. H0 and H2–H4 are independently expressed deductions from the precise earlier inputs above. Neither source's statement of duality is used as a premise for H3.

The harmonic analytic premises are the earlier finite-measure interchange, positive-coefficient representation, dual Haar normalization, onto scalar Fourier unitary and inverse-integral domain proofs, at arbitrary LCA generality, together with HR-01, HR-02, HR-03, HR-04, HR-05, HR-06, HR-07, HR-08, HR-09.

The prerequisites are H0, H1, H2, H3, H4. Their Haar and Radon measure arguments are proved in the earlier HR lessons. The scalar Plancherel theorem is the earlier complete proof, with paired dual Haar normalization and inverse domain.

H6. Fourier inversion for every integrable transform

Let $G$ be an arbitrary locally compact Hausdorff abelian group, $\Gamma=\widehat G$, with the paired Haar measures of P2. Keep the negative forward transform $\widehat a(\chi)=\int_G a(t)\overline{\chi(t)}\,dm(t).$ For $a\in L^1(G)$, square integrability of this actual transform already forces $a\in L^2(G)$. More precisely, $\widehat a\in L^2(\Gamma) \quad\Longrightarrow\quad a\in L^1(G)\cap L^2(G),\qquad \mathcal F_2a=\widehat a,\qquad \|a\|_2=\|\widehat a\|_2 , \tag{H6.1}$ where $\mathcal F_2$ is the onto unitary of P3. The equality identifies the original $L^1$ class, not just another function with a formal inverse transform.

Choose the nonnegative mass-one $b_V\in C_c(G)$ of L24, Theorem 5.2, directed by shrinking identity neighbourhoods. Set $a_V=a*b_V$. P3's full common-domain convolution proof applies to $a\in L^1(G)$ and $b_V\in L^1(G)\cap L^2(G)$; hence $\begin{aligned} a_V&\in L^1(G)\cap L^2(G),& \|a_V\|_2&\leq\|a\|_1\|b_V\|_2,\\ \|a_V-a\|_1&\longrightarrow0,& \mathcal F_2a_V&=\widehat a\,\widehat b_V . \end{aligned} \tag{H6.2}$ The last equality combines the $L^1$ convolution identity with P3's proved compatibility on the common domain.

Write $f=\widehat a$. Positivity and normalization of $b_V$ give $|\widehat b_V|\leq1$. For every compact $K\subset\Gamma$, the joint character continuity proved in H3 gives $\delta_V(K):=\sup_{\chi\in K}|\widehat b_V(\chi)-1| \leq\sup_{\substack{t\in\operatorname{supp}b_V\\\chi\in K}} |\chi(t)-1| \longrightarrow0 . \tag{H6.3}$ This yields the estimate on the entire dual group $\|f\widehat b_V-f\|_2^2 \leq \delta_V(K)^2\|f\|_2^2 +4\int_{\Gamma\setminus K}|f(\chi)|^2\,d\widehat m(\chi). \tag{H6.4}$ The finite-density regularity of HR3 makes the second term arbitrarily small by choosing $K$ compact. With that $K$ fixed, (H6.3) controls the first term for all sufficiently small $V$. Thus $f\widehat b_V\to f$ in $L^2(\Gamma)$. This is a proof for the neighbourhood net; dominated convergence for arbitrary nets is not being assumed.

Let $v=\mathcal F_2^{-1}f\in L^2(G)$. Unitarity and (H6.2) now give $a_V\to v$ in $L^2(G)$, while $a_V\to a$ in $L^1(G)$. To identify the two limits without a countability assumption on $G$, choose a sequence of indices $V_n$ far enough in these two norm-convergent nets that $\|a_{V_n}-a\|_1<2^{-n},\qquad \|a_{V_n}-v\|_2^2<2^{-n}. \tag{H6.5}$ No countable neighbourhood base is claimed. These indices are chosen only to control two real-valued norm errors.

Use Borel representatives on the sigma compact carriers supplied by L24's finite-exponent convention. The countably many representatives involved have one sigma compact union of carriers. SC4 gives $\int_G\sum_{n\geq1}|a_{V_n}-a|\,dm<\infty,\qquad \int_G\sum_{n\geq1}|a_{V_n}-v|^2\,dm<\infty. \tag{H6.6}$ Both sums are finite outside one common null set. Their terms tend to zero there, so the same sequence $a_{V_n}(t)$ tends both to $a(t)$ and to $v(t)$. Consequently $a=v$ almost everywhere. This proves (H6.1), including the exact transform and norm identities.

Now suppose merely that $a\in L^1(G)$ and $\widehat a\in L^1(\Gamma)$. The integral bound $|\widehat a|\leq\|a\|_1$ gives $\|\widehat a\|_2^2 \leq \|a\|_1\|\widehat a\|_1<\infty. \tag{H6.7}$ Apply (H6.1), then H2's inverse integral to this $L^1\cap L^2$ transform. The original class $a$ has the continuous representative $a^\circ(t)=\int_\Gamma\widehat a(\chi)\chi(t)\, d\widehat m(\chi), \qquad a=a^\circ\ \text{almost everywhere},\qquad \|a^\circ\|_\infty\leq\|\widehat a\|_1 . \tag{H6.8}$ The integral converges absolutely for every $t$, with the positive inverse character and the precise dual Haar normalization.

In fact this representative is uniformly continuous. For any compact $K\subset\Gamma$ and $s,t\in G$, $|a^\circ(t+s)-a^\circ(t)| \leq \|\widehat a\|_1\sup_{\chi\in K}|\chi(s)-1| +2\int_{\Gamma\setminus K}|\widehat a|\,d\widehat m . \tag{H6.9}$ Choose $K$ by finite-density regularity, then $s$ by H3's compact character estimate. The bound is independent of $t$, as required. If a chosen representative of $a$ is continuous at $t_0$, it agrees with $a^\circ$ at $t_0$: otherwise their difference would stay bounded away from zero on a neighbourhood of $t_0$, which has positive Haar measure, contradicting their almost-everywhere equality. Uniqueness among continuous representatives follows by the same argument.

Equivalently, if $f\in A(\Gamma)\cap L^1(\Gamma)$, where $A(\Gamma)=\{\widehat a:a\in L^1(G)\}$, then its uniquely determined inverse $L^1$ class is represented by $\check f(t)=\int_\Gamma f(\chi)\chi(t)\,d\widehat m(\chi)$. No compact-support assumption on $f$, or sigma compactness or separability assumption on either group, occurs in this conclusion. This is the inverse formula used in Takesaki, Theory of Operator Algebras II, XI §1, printed p.312.