The Bochner functional extends and has a Radon representing measure

(Density of the transform core in $C_0(\widehat{G})$.) The functions $\chi \mapsto \chi (0,f)$, $f\in A$, form a self-adjoint complex subalgebra $B$ of $C(\mathrm{\Delta }(A^{+}))$ by [F2]. It is point-separating: distinct characters of $A^{+}$ are separated by some $a\in A^{+}$ because the Gelfand topology is Hausdorff, and since all characters agree on the constants this element may be taken as $(0,f)$, so $\widehat{f}$ separates them. Its only common zero is $q$: $q(0,f)=0$ for all $f$, while $h_{\gamma }^{+}\mathrm{\ne }q$ gives some $f$ with $\widehat{f}(\gamma )\mathrm{\ne }0$. By the vanishing-at-one-point case of [F3], the uniform closure of $B$ is $\{G\in C(\mathrm{\Delta }(A^{+})):G(q)=0\}$, which under $\mathrm{\Delta }(A^{+})\cong \widehat{G}\cup \{q\}$ is exactly $C_0(\widehat{G})$; hence the transform core is uniformly dense in $C_0(\widehat{G})$.

(The approximate identity.) For the approximate identity of [F5], $L_{\varphi }(u_U)={\int }_G{u}_U(x)\varphi (-x)d{m}_G(x)\to \varphi (0)=k$, because $u_U\ge 0$, ${\int }_G{u}_U=1$, $\mathrm{supp}{u}_U\subseteq U$ and $\varphi$ is continuous at $0$. Also $\widehat{u_U}\in C_0(\widehat{G})$, $\mathrm{\mid }\widehat{u_U}\mathrm{\mid }\le 1$ and $\widehat{u_U}\to 1$ uniformly on compact subsets of $\widehat{G}$.

(Continuity of inverse transforms.) Let $\sigma$ be a finite regular complex Borel measure on $\widehat{G}$ and $\mathrm{\Psi }(x):={\int }_{\widehat{G}}\gamma (x)d\sigma (\gamma )$. For a net $x_i\to x_0$ and $\epsilon >0$, inner regularity of $\mathrm{\mid }\sigma \mathrm{\mid }$ gives a compact $K\subseteq \widehat{G}$ with $\mathrm{\mid }\sigma \mathrm{\mid }(\widehat{G}\setminus K)<\epsilon \mathrm{/}4$; by joint continuity of the pairing and compactness of $K$ one has ${\sup }_{\gamma \in K}\mathrm{\mid }\gamma (x_i)-\gamma (x_0)\mathrm{\mid }<\epsilon \mathrm{/}(2\mathrm{\mid }\sigma \mathrm{\mid }(\widehat{G})+2)$ eventually, so $\mathrm{\mid }\mathrm{\Psi }(x_i)-\mathrm{\Psi }(x_0)\mathrm{\mid }\le {\int }_K\mathrm{\mid }\gamma (x_i)-\gamma (x_0)\mathrm{\mid }d\mathrm{\mid }\sigma \mathrm{\mid }+2\mathrm{\mid }\sigma \mathrm{\mid }(\widehat{G}\setminus K)<\epsilon \mathrm{.}$ Hence $\mathrm{\Psi }$ is continuous.

(Extension of $F_{\varphi }$.) By step 1.1 the transform core is dense in $C_0(\widehat{G})$, and by [F1] $F_{\varphi }$ is linear on it with norm at most $k$. By [F7] it has a unique bounded linear extension $L:C_0(\widehat{G})\to \mathbb{C}$ with $\mathrm{\parallel }L\mathrm{\parallel }=\mathrm{\parallel }{F}_{\varphi }\mathrm{\parallel }\le k$.

(Positivity of $L$.) Let $H\in C_0(\widehat{G})$ with $H\ge 0$. Since $\sqrt{H}\in C_0(\widehat{G})$, step 1.1 gives $g_n\in \{\widehat{f}:f\in A\}$ with $\mathrm{\parallel }{g}_n-\sqrt{H}{\mathrm{\parallel }}_{\mathrm{\infty }}\to 0$; then $\mathrm{\mid }{g}_n{\mathrm{\mid }}^2\in \{\widehat{f}:f\in A\}$ by the self-adjoint algebra property [F2] and $\mathrm{\mid }{g}_n{\mathrm{\mid }}^2\to H$ uniformly, because $\mathrm{\parallel }\mathrm{\mid }{g}_n{\mathrm{\mid }}^2-H{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }{g}_n-\sqrt{H}{\mathrm{\parallel }}_{\mathrm{\infty }}(\mathrm{\parallel }{g}_n{\mathrm{\parallel }}_{\mathrm{\infty }}+\mathrm{\parallel }\sqrt{H}{\mathrm{\parallel }}_{\mathrm{\infty }})$ is eventually bounded by a constant times $\mathrm{\parallel }{g}_n-\sqrt{H}{\mathrm{\parallel }}_{\mathrm{\infty }}$. Each $\mathrm{\mid }{g}_n{\mathrm{\mid }}^2=\widehat{f_n\ast f_n^{\ast }}$ for some $f_n\in A$, so $F_{\varphi }(\mathrm{\mid }{g}_n{\mathrm{\mid }}^2)=L_{\varphi }(f_n\ast f_n^{\ast })\ge 0$ by [F1]. Passing to the limit along step 2.1 gives $L(H)={\lim }_n{F}_{\varphi }(\mathrm{\mid }{g}_n{\mathrm{\mid }}^2)\ge 0$.

(Riesz-Markov representation.) The real part of $L$ is a bounded positive linear functional on $C_0(\widehat{G};\mathbb{R})$, so by [F4] there is a unique finite regular Borel measure ${\mu }_{\varphi }\ge 0$ on $\widehat{G}$ with $L(H)={\int }_{\widehat{G}}Hd{\mu }_{\varphi }$ for all $H\in C_0(\widehat{G})$, and ${\mu }_{\varphi }(\widehat{G})=\mathrm{\parallel }L\mathrm{\parallel }$.

(Mass ${\mu }_{\varphi }(\widehat{G})=k$.) By step 4.1 and step 2.1, ${\int }_{\widehat{G}}\widehat{u_U}d{\mu }_{\varphi }=L(\widehat{u_U})=F_{\varphi }(\widehat{u_U})=L_{\varphi }(u_U)$ for every identity neighbourhood $U$. By step 1.2, $L_{\varphi }(u_U)\to k$. By step 1.2 and step 4.1, ${\int }_{\widehat{G}}\widehat{u_U}d{\mu }_{\varphi }\to {\int }_{\widehat{G}}1d{\mu }_{\varphi }={\mu }_{\varphi }(\widehat{G})$: for $\epsilon >0$ choose compact $K$ with ${\mu }_{\varphi }(\widehat{G}\setminus K)<\epsilon \mathrm{/}4$, then use $\mathrm{\mid }\widehat{u_U}\mathrm{\mid }\le 1$ and eventual ${\sup }_K\mathrm{\mid }1-\widehat{u_U}\mathrm{\mid }<\epsilon \mathrm{/}(2(1+{\mu }_{\varphi }(\widehat{G})))$. Hence ${\mu }_{\varphi }(\widehat{G})=k=\mathrm{\parallel }L\mathrm{\parallel }$, and $L$ has norm exactly $k$.

(Uniqueness of ${\mu }_{\varphi }$.) Let $\nu$ be another finite positive Radon measure with $L(H)={\int }_{\widehat{G}}Hd\nu$ for all $H\in C_0(\widehat{G})$; then $\sigma :={\mu }_{\varphi }-\nu$ is a finite regular complex Borel measure with ${\int }_{\widehat{G}}\widehat{f}d\sigma =0$ for every $f\in C_c(G)$. By Fubini [F6], $0={\int }_{\widehat{G}}\widehat{f}d\sigma ={\int }_{G}f(x)({\int }_{\widehat{G}}\gamma (-x)d\sigma (\gamma ))d{m}_G(x)={\int }_{G}f(x)\mathrm{\Psi }(-x)d{m}_G(x)$ with $\mathrm{\Psi }(x):={\int }_{\widehat{G}}\gamma (x)d\sigma (\gamma )$ continuous by step 1.3. If $\mathrm{\Psi }(y_0)\mathrm{\ne }0$, choose $c\in \mathbb{C}$ with $\mathrm{\mid }c\mathrm{\mid }=1$ and $\mathrm{Re}(c\mathrm{\Psi }(y_0))>0$; by continuity, $\mathrm{Re}(c\mathrm{\Psi })>0$ on a nonempty open neighbourhood $V$ of $y_0$. Choose a nonzero nonnegative $f\in C_c(G)$ supported in $-V$, as provided by [F6]. Then $\mathrm{Re}(c\mathrm{\Psi }(-x))>0$ on $\mathrm{supp}f$, so ${\int }_{G}f(x)\mathrm{Re}(c\mathrm{\Psi }(-x))d{m}_G(x)>0$ by positivity of Haar measure on nonempty open sets, contradicting $c{\int }_{G}f(x)\mathrm{\Psi }(-x)d{m}_G(x)=0$. Hence $\mathrm{\Psi }\equiv 0$. The Fourier-Stieltjes uniqueness theorem [F6] now gives $\sigma =0$, that is, $\nu ={\mu }_{\varphi }$.

Steps 2.1 and 5.1 exhibit the unique bounded linear extension $L$ of $F_{\varphi }$ with $\mathrm{\parallel }L\mathrm{\parallel }=k$, step 3.1 proves it positive, step 4.1 represents it by the finite positive Radon measure ${\mu }_{\varphi }$ of mass ${\mu }_{\varphi }(\widehat{G})=k$, and step 6.1 proves that this representing measure is unique.