Compatible dual Haar normalisation

(Generator measures.) Let $p=g\ast \tilde{g}$ for some $g\in C_c(G;\mathbb{C})$. By [F1], $p\in C_c(G;\mathbb{C})$ is continuous and positive definite with $p(0)={\int }_G\mathrm{\mid }g{\mathrm{\mid }}^{2}d{m}_G\ge 0$ and $\widehat{p}=\mathrm{\mid }\widehat{g}{\mathrm{\mid }}^2\ge 0$; by [F2] there is a unique finite positive Radon measure ${\mu }_p$ on $\widehat{G}$ with $p(x)={\int }_{\widehat{G}}\gamma (x)d{\mu }_p(\gamma )$ for all $x$ and ${\mu }_p(\widehat{G})=p(0)$. For a finite sum $r=p_1+\cdots +p_N$ of such generators, the measure ${\mu }_r:={\mu }_{p_1}+\cdots +{\mu }_{p_N}$ represents the continuous positive definite function $r$; uniqueness in [F2] gives $r(x)={\int }_{\widehat{G}}\gamma (x)d{\mu }_r(\gamma )$ and ${\mu }_r(\widehat{G})=r(0)$, and ${\mu }_{r+s}={\mu }_r+{\mu }_s$ for finite sums of generators.

(The sets $\{\widehat{p}>0\}$ cover $\widehat{G}$.) Put $U_p:=\{\gamma \in \widehat{G}:\widehat{p}(\gamma )>0\}$ for a generator $p$. By [F3], for every ${\gamma }_0$ there is $g$ with $\widehat{g}({\gamma }_0)\mathrm{\ne }0$; for $p=g\ast \tilde{g}$ we have $\widehat{p}({\gamma }_0)=\mathrm{\mid }\widehat{g}({\gamma }_0){\mathrm{\mid }}^2>0$, and $\widehat{p}$ is continuous, so ${\gamma }_0\in U_p$. Hence the family $\{U_p\}$ over all generators covers $\widehat{G}$.

(Density of the transform algebra.) The functions $\chi \mapsto \chi (0,f)$, $f\in L^1(G,m_G)$, form a self-adjoint complex subalgebra of $C(\mathrm{\Delta }(A^{+}))$ by [F4]. Distinct characters of $A^{+}$ are separated by some element of $A^{+}$ because the Gelfand topology is Hausdorff, and since all characters agree on the constants the separating element may be taken as $(0,f)$; thus the algebra separates points, and its only common zero is $q$. By the vanishing-at-one-point case of Stone-Weierstrass in [F4], its uniform closure is $\{G\in C(\mathrm{\Delta }(A^{+})):G(q)=0\}$, which is $C_0(\widehat{G})$ under $\mathrm{\Delta }(A^{+})\cong \widehat{G}\cup \{q\}$. Hence $\{\widehat{f}:f\in L^1(G,m_G)\}$ is uniformly dense in $C_0(\widehat{G})$.

(The consistency identity.) Let $p=g\ast \tilde{g}$ and $q=k\ast \tilde{k}$ be generators. For $x\in L^1(G,m_G)$, absolute integrability over $\mathrm{supp}x\times \mathrm{supp}p\times \widehat{G}$ and [F5] permit Fubini in ${\int }_{\widehat{G}}\widehat{x}\widehat{p}d{\mu }_q={\int }_{\widehat{G}}{\int }_G{\int }_{G}x(y)p(z)\overline{\gamma (y)}\overline{\gamma (z)}d{m}_G(y)d{m}_G(z)d{\mu }_q(\gamma )={\int }_G{\int }_{G}x(y)p(z)q(-y-z)d{m}_G(y)d{m}_G(z),$ because ${\int }_{\widehat{G}}\overline{\gamma (y+z)}d{\mu }_q(\gamma )=q(-y-z)$ by [F2]. The same computation with $p$ and $q$ interchanged gives ${\int }_{\widehat{G}}\widehat{x}\widehat{q}d{\mu }_p={\int }_G{\int }_{G}x(y)q(z)p(-y-z)d{m}_G(y)d{m}_G(z)$, and the substitution $z\mapsto -y-z$ in the first double integral shows it equals the second. Therefore ${\int }_{\widehat{G}}\widehat{x}d(\widehat{p}{\mu }_q)={\int }_{\widehat{G}}\widehat{x}d(\widehat{q}{\mu }_p)$ for every $x\in L^1(G,m_G)$. Both $\widehat{p}{\mu }_q$ and $\widehat{q}{\mu }_p$ are finite measures (dominated by $\mathrm{\parallel }\widehat{p}{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\mid }{\mu }_q\mathrm{\mid }$ and $\mathrm{\parallel }\widehat{q}{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\mid }{\mu }_p\mathrm{\mid }$), and step 1.3 makes the functions $\widehat{x}$ uniformly dense in $C_0(\widehat{G})$, so $\widehat{p}{\mu }_q=\widehat{q}{\mu }_p\mathrm{.}$ By step 1.1 the same identity holds for finite sums $r,s$ of generators: $\widehat{r}{\mu }_s=\widehat{s}{\mu }_r$.

(Gluing the local measures.) For $f\in C_c(\widehat{G};\mathbb{C})$, step 1.2 and compactness of $\mathrm{supp}f$ give finitely many generators $p_1,\dots ,p_N$ with $\mathrm{supp}f\subseteq {\bigcup }_j{U}_{p_j}$; put $p:=p_1+\cdots +p_N$, so $\widehat{p}>0$ on $\mathrm{supp}f$. Define $m(f):={\int }_{\widehat{G}}\frac{f}{\widehat{p}}d{\mu }_p,$ where $f\mathrm{/}\widehat{p}$ is set to $0$ off $\mathrm{supp}f$; the integral converges because $f$ is bounded and $\widehat{p}$ is bounded below on $\mathrm{supp}f$. If $r$ is another finite sum of generators with $\widehat{r}>0$ on $\mathrm{supp}f$, then by step 2.1, $\widehat{r}{\mu }_p=\widehat{p}{\mu }_r$, so ${\int }_{\widehat{G}}\frac{f}{\widehat{p}}d{\mu }_p={\int }_{\widehat{G}}\frac{f}{\widehat{p}\widehat{r}}\widehat{r}d{\mu }_p={\int }_{\widehat{G}}\frac{f}{\widehat{p}\widehat{r}}\widehat{p}d{\mu }_r={\int }_{\widehat{G}}\frac{f}{\widehat{r}}d{\mu }_r;$ hence $m(f)$ is well defined. The map $m:C_c(\widehat{G};\mathbb{C})\to \mathbb{C}$ is linear and positive: for $f\ge 0$ one has $f\mathrm{/}\widehat{p}\ge 0$ on $\mathrm{supp}f$.

(The measure $m$.) By step 3.1 the functional $m$ is a positive linear functional on $C_c(\widehat{G};\mathbb{R})$; by [F6] there is a Radon measure, again written $m$, on $\widehat{G}$ with $m(f)={\int }_{\widehat{G}}fdm$ for every $f\in C_c(\widehat{G};\mathbb{C})$, and we identify $m$ with this measure.

(${\mu }_p=\widehat{p}m$ for every generator.) Fix a generator $p$. On the open set $U_p=\{\widehat{p}>0\}$, step 3.1 applied to $f\in C_c(U_p)$ with the single generator $p$ gives $m(f)={\int }_{\widehat{G}}f\mathrm{/}\widehat{p}d{\mu }_p$, that is, ${\mu }_p\mathrm{\mid }{U}_p=\widehat{p}m\mathrm{\mid }{U}_p$. On the closed set $Z:=\{\widehat{p}=0\}$, let $K\subseteq Z$ be compact; by step 1.2 choose finitely many generators $q_1,\dots ,q_N$ with ${\sum }_j{\widehat{q}}_j>0$ on $K$, put $s:={\sum }_j{q}_j$. By step 2.1, $\widehat{s}{\mu }_p=\widehat{p}{\mu }_s$, so ${\int }_K\widehat{s}d{\mu }_p={\int }_K\widehat{p}d{\mu }_s=0,$ and $\widehat{s}>0$ on $K$ gives ${\mu }_p(K)=0$. To pass from compact subsets to the whole closed set, fix $\epsilon >0$ and use outer regularity [F6] to choose an open $U\supseteq Z$ with ${\mu }_p(U)<{\mu }_p(Z)+\epsilon$. By inner regularity on the open set $U$ [F6], choose compact $L\subseteq U$ with ${\mu }_p(L)>{\mu }_p(U)-\epsilon$. Then $L\cap Z$ is compact and $0={\mu }_p(L\cap Z)\ge {\mu }_p(L)-{\mu }_p(U\setminus Z)>{\mu }_p(Z)-2\epsilon ,$ because ${\mu }_p(U\setminus Z)={\mu }_p(U)-{\mu }_p(Z)<\epsilon$. Letting $\epsilon \downarrow 0$ gives ${\mu }_p(Z)=0$. Hence ${\mu }_p=\widehat{p}m$ as measures on $\widehat{G}$.

(Inversion for the core.) Let $h\in E$, say $h={\sum }_j{c}_j{p}_j$ with generators $p_j$ and $c_j\in \mathbb{C}$. By step 5.1, $p_j(x)={\int }_{\widehat{G}}\gamma (x){\widehat{p}}_j(\gamma )dm(\gamma )$ for all $x$, and by step 1.1, $\widehat{h}={\sum }_j{c}_j{\widehat{p}}_j$ and ${\int }_{\widehat{G}}\mathrm{\mid }\widehat{h}\mathrm{\mid }dm\le {\sum }_j\mathrm{\mid }{c}_j\mathrm{\mid }{\int }_{\widehat{G}}{\widehat{p}}_{j}dm={\sum }_j\mathrm{\mid }{c}_j\mathrm{\mid }{p}_j(0)<+\mathrm{\infty }$. Therefore $h(x)={\sum }_j{c}_j{p}_j(x)={\int }_{\widehat{G}}\gamma (x){\sum }_j{c}_j{\widehat{p}}_j(\gamma )dm(\gamma )={\int }_{\widehat{G}}\widehat{h}(\gamma )\gamma (x)dm(\gamma )$ for every $x\in G$, and in particular for $m_G$-almost every $x$.

($m$ is a Haar measure.) First $m\mathrm{\ne }0$: a nonzero generator $p$ has $p(0)>0$, and step 5.1 gives ${\int }_{\widehat{G}}\widehat{p}dm={\mu }_p(\widehat{G})=p(0)>0$. Next, $m$ is translation invariant. Fix ${\gamma }_0\in \widehat{G}$ and a generator $p=g\ast \tilde{g}$. The modulation ${\gamma }_{0}g$ lies in $C_c(G;\mathbb{C})$ and $({\gamma }_{0}g)\ast \widetilde{({\gamma }_{0}g)}(x)={\int }_G{\gamma }_0(y)g(y)\overline{{\gamma }_0(y-x)g(y-x)}d{m}_G(y)={\gamma }_0(x){\int }_{G}g(y)\overline{g(y-x)}d{m}_G(y)={\gamma }_0(x)p(x),$ so ${\gamma }_{0}p=({\gamma }_{0}g)\ast \widetilde{({\gamma }_{0}g)}$ is again a generator, and its transform is $\widehat{{\gamma }_{0}p}(\gamma )=\widehat{p}({\gamma }_0^{-1}\gamma )$ by the modulation identity of [F1]. Applying step 6.1 to $p$ and to ${\gamma }_{0}p$ gives, for every $x\in G$, $p(x)={\int }_{\widehat{G}}\gamma (x)\widehat{p}(\gamma )dm(\gamma ),{\gamma }_0(x)p(x)={\int }_{\widehat{G}}\gamma (x)\widehat{p}({\gamma }_0^{-1}\gamma )dm(\gamma )\mathrm{.}$ In the second integral substitute $\eta ={\gamma }_0^{-1}\gamma$: it becomes ${\int }_{\widehat{G}}{\gamma }_0(x)\eta (x)\widehat{p}(\eta )d((T_{{\gamma }_0^{-1}}{)}_{\ast }m)(\eta )$, where $T_{\delta }(\eta )=\delta \eta$ and $(T_{{\gamma }_0^{-1}}{)}_{\ast }m$ is the pushforward of $m$ under $T_{{\gamma }_0^{-1}}$. Comparing with ${\gamma }_0(x)$ times the first integral yields ${\int }_{\widehat{G}}\eta (x)\widehat{p}(\eta )d((T_{{\gamma }_0^{-1}}{)}_{\ast }m)(\eta )={\int }_{\widehat{G}}\eta (x)\widehat{p}(\eta )dm(\eta )$ for every $x\in G$. Thus the finite measures $\widehat{p}(T_{{\gamma }_0^{-1}}{)}_{\ast }m$ and $\widehat{p}m$ have the same inverse transform, so they are equal by [F7]. Since ${\gamma }_0^{-1}$ ranges over all of $\widehat{G}$, this gives $\widehat{p}(T_{\delta }{)}_{\ast }m=\widehat{p}m$ for every $\delta \in \widehat{G}$. For a compact $K\subseteq \widehat{G}$, step 1.2 provides finitely many generators $p_1,\dots ,p_N$ with $\widehat{P}:={\sum }_j{\widehat{p}}_j>0$ on $K$; summing the identities ${\widehat{p}}_j(T_{\delta }{)}_{\ast }m={\widehat{p}}_{j}m$ gives $\widehat{P}(T_{\delta }{)}_{\ast }m=\widehat{P}m$, and dividing by the strictly positive continuous function $\widehat{P}$ on $K$ gives $(T_{\delta }{)}_{\ast }m\mathrm{\mid }K=m\mathrm{\mid }K$. Since $K$ and $\delta$ are arbitrary, $m$ is translation invariant. Finally, $m$ is positive on every nonempty open set: if $m(V)=0$ for a nonempty open $V$, then by invariance $m(\gamma +V)=0$ for every $\gamma$, and any compact $K$ is covered by finitely many translates of $V$, so $m(K)=0$; inner regularity of the Radon measure $m$ then gives $m=0$, contradicting $m\mathrm{\ne }0$. Hence $m$ is a Haar measure on $\widehat{G}$.

(Uniqueness of the scale and reciprocal scaling.) Let $m^{\mathrm{\prime }}$ be any Haar measure on $\widehat{G}$ for which inversion holds for every $h\in E$. By [F8], $m^{\mathrm{\prime }}=\lambda m$ for some $\lambda >0$. For a nonzero generator $p$, the inversion property gives $p(x)={\int }_{\widehat{G}}\widehat{p}(\gamma )\gamma (x)d{m}^{\mathrm{\prime }}(\gamma )$ almost everywhere, with $\widehat{p}\in L^1(\widehat{G},m^{\mathrm{\prime }})$. By [F9], this inverse transform is continuous; since $p$ is continuous and Haar measure is positive on every nonempty open set, the almost-everywhere identity is everywhere. Evaluating at $x=0$, and using step 6.1 for $m$, gives $p(0)={\int }_{\widehat{G}}\widehat{p}d{m}^{\mathrm{\prime }}=\lambda {\int }_{\widehat{G}}\widehat{p}dm=\lambda p(0)$. Since $p(0)>0$, $\lambda =1$ and $m^{\mathrm{\prime }}=m$. Thus the inversion property determines $m$ uniquely. If $m_G$ is replaced by $c{m}_G$ with $c>0$, then for every $h\in E$ the transform of the same function $h$ becomes $c\widehat{h}$; inversion under a Haar measure $m^{\mathrm{\prime }\mathrm{\prime }}$ reads $h(x)={\int }_{\widehat{G}}c\widehat{h}(\gamma )\gamma (x)d{m}^{\mathrm{\prime }\mathrm{\prime }}(\gamma )$, which holds exactly when $c{m}^{\mathrm{\prime }\mathrm{\prime }}$ satisfies the original inversion identity; by uniqueness this means $c{m}^{\mathrm{\prime }\mathrm{\prime }}=m$, that is, $m^{\mathrm{\prime }\mathrm{\prime }}=c^{-1}m$.

Steps 4.1 and 7.1 construct a Haar measure $m_{\widehat{G}}:=m$ on $\widehat{G}$; step 6.1 gives $\widehat{h}\in L^1(\widehat{G},m)$ and the inversion identity for every $h\in E$ and every $x$; step 8.1 proves uniqueness of the scale and the reciprocal scaling law.