Positive definite functions give positive bounded functionals on the transform core

(Boundedness.) By [F1], $\mathrm{\mid }\varphi (-x)\mathrm{\mid }=\mathrm{\mid }\varphi (x)\mathrm{\mid }\le k$ for every $x$, so the integral defining $L_{\varphi }(f)$ converges absolutely for every $f\in A$ and $\mathrm{\mid }{L}_{\varphi }(f)\mathrm{\mid }\le k\mathrm{\parallel }f{\mathrm{\parallel }}_1$; linearity in $f$ is immediate.

(Integrated positivity.) For $f\in A$, $(f\ast f^{\ast })(u)={\int }_{G}f(y)\overline{f(y-u)}d{m}_G(y)$, so by inversion invariance of $m_G$ the substitution $z=y-u$ in the inner integral (a reflection followed by a translation) gives $L_{\varphi }(f\ast f^{\ast })={\int }_G{\int }_{G}f(y)\overline{f(z)}\varphi (z-y)d{m}_G(z)d{m}_G(y),$ and the integrand is carried by the $\sigma$-finite product of a $\sigma$-compact essential support of $f$ with itself ([F6]). For $f\in C_c(G)$ put $K=\mathrm{supp}f$. The continuous kernel $f(y)\overline{f(z)}\varphi (z-y)$ on $K\times K$ admits finite Borel partitions of $K$ on whose product cells its oscillation is arbitrarily small: use compactness and continuity in the group uniformity to take a finite sufficiently small cover, then disjointify it. Choose a point $y_j$ in each nonempty cell $P_j$. The resulting sums ${\sum }_{j,k}{c}_j\overline{c_k}\varphi (y_k-y_j)$, where $c_j=f(y_j)m_G(P_j)$, converge to the double integral since their error is bounded by the kernel oscillation times $m_G(K{)}^2$. Each sum is nonnegative by positive definiteness applied to the points $-y_j$ and coefficients $c_j$. Hence $L_{\varphi }(f\ast f^{\ast })\ge 0$ for $f\in C_c(G)$; for general $f\in A$ choose $f_n\in C_c(G)$ with $f_n\to f$ in $A$ ([F5]), then $f_n\ast f_n^{\ast }\to f\ast f^{\ast }$ in $A$ by [F2] and $L_{\varphi }(f_n\ast f_n^{\ast })\to L_{\varphi }(f\ast f^{\ast })$ by step 1.1, so the inequality passes to the limit.

(The Cauchy-Schwarz bound.) The form $[g,h]:=L_{\varphi }(g\ast h^{\ast })$ is sesquilinear by [F2] and positive semidefinite by step 2.1; for such a form $\mathrm{\mid }[g,h]{\mathrm{\mid }}^2\le [g,g][h,h]$ (the quadratic $[g+th,g+th]\ge 0$ in $t\in \mathbb{C}$ has nonnegative discriminant). With $h=u_U^{\ast }=u_U$ from [F3] this gives $\mathrm{\mid }{L}_{\varphi }(f\ast u_U){\mathrm{\mid }}^2=\mathrm{\mid }[f,u_U]{\mathrm{\mid }}^2\le L_{\varphi }(f\ast f^{\ast })L_{\varphi }(u_U\ast u_U^{\ast })$. As $U$ shrinks, $f\ast u_U\to f$ in $A$ ([F3]) so $L_{\varphi }(f\ast u_U)\to L_{\varphi }(f)$ by step 1.1; and $L_{\varphi }(u_U\ast u_U^{\ast })\to k$: the functions $u_U\ast u_U^{\ast }$ are nonnegative with integral $1$ and support shrinking to $\{0\}$, so $\mathrm{\mid }{L}_{\varphi }(u_U\ast u_U^{\ast })-k\mathrm{\mid }=\mathrm{\mid }{\int }_G(u_U\ast u_U^{\ast })(x)(\varphi (-x)-k)d{m}_G(x)\mathrm{\mid }\le {\sup }_{x\in \mathrm{supp}(u_U\ast u_U^{\ast })}\mathrm{\mid }\varphi (-x)-k\mathrm{\mid }\to 0$ by continuity of $\varphi$ at $0$ and [F1]. Hence $\mathrm{\mid }{L}_{\varphi }(f){\mathrm{\mid }}^2\le k{L}_{\varphi }(f\ast f^{\ast })$.

(The sup-norm bound.) If $k=0$, [F1] makes $L_{\varphi }=0$ and all bounds hold. Assume $k>0$ and put $a:=f\ast f^{\ast }$, so $a^{\ast }=a$ and $\widehat{a}=\mathrm{\mid }\widehat{f}{\mathrm{\mid }}^2$ by [F2]. Applying step 3.1 to $a,a^2,a^4,\dots$ gives by induction $\mathrm{\mid }{L}_{\varphi }(f)\mathrm{\mid }\le k^{1-2^{-n}}{L}_{\varphi }(a^{2^{n-1}}{)}^{2^{-n}}$ for $n\ge 1$, and the elementary bound of step 1.1 applied to the last factor yields $\mathrm{\mid }{L}_{\varphi }(f)\mathrm{\mid }\le k\mathrm{\parallel }{a}^{2^{n-1}}{\mathrm{\parallel }}_1^{2^{-n}}$. Since $\mathrm{\parallel }{a}^{2^{n-1}}{\mathrm{\parallel }}_1^{2^{-n}}=(\mathrm{\parallel }{a}^{2^{n-1}}{\mathrm{\parallel }}_1^{1\mathrm{/}{2}^{n-1}}{)}^{1\mathrm{/}2}\to r_{A^{+}}(0,a{)}^{1\mathrm{/}2}$ by the spectral radius formula [F4], while $r_{A^{+}}(0,a)={\max }_{\lambda \in {\sigma }_{A^{+}}(0,a)}\mathrm{\mid }\lambda \mathrm{\mid }={\sup }_{\gamma }\mathrm{\mid }\widehat{a}(\gamma )\mathrm{\mid }=\mathrm{\parallel }\widehat{f}{\mathrm{\parallel }}_{\mathrm{\infty }}^2$ by [F4], we obtain $\mathrm{\mid }{L}_{\varphi }(f)\mathrm{\mid }\le k\mathrm{\parallel }\widehat{f}{\mathrm{\parallel }}_{\mathrm{\infty }}$. Applying this bound to $a=f\ast f^{\ast }$ gives $0\le L_{\varphi }(a)\le k\mathrm{\parallel }\widehat{a}{\mathrm{\parallel }}_{\mathrm{\infty }}=k\mathrm{\parallel }\widehat{f}{\mathrm{\parallel }}_{\mathrm{\infty }}^2$, proving the second inequality in the stated chain. The bound holds for every representative, so $L_{\varphi }$ vanishes on $\{f:\widehat{f}=0\}$.

(Descent and positivity.) Since $L_{\varphi }$ vanishes on the kernel of the transform, $F_{\varphi }(\widehat{f}):=L_{\varphi }(f)$ is a well-defined linear functional on the transform core, and step 4.1 gives $\mathrm{\mid }{F}_{\varphi }(h)\mathrm{\mid }\le k\mathrm{\parallel }h{\mathrm{\parallel }}_{\mathrm{\infty }}$. For positivity let $h\ge 0$ belong to the core. The core is a self-adjoint algebra of functions vanishing at infinity ([F5]), so choose real polynomials $p_n$ with $p_n(0)=0$ and $p_n(t{)}^2\to t$ uniformly on $[0,\mathrm{\parallel }h{\mathrm{\parallel }}_{\mathrm{\infty }}]$ ([F5]); then $p_n(h)$ lies in the core with $\mathrm{\mid }{p}_n(h){\mathrm{\mid }}^2=p_n(h{)}^2\to h$ uniformly, and writing $p_n(h)=\widehat{g_n}$ for some $g_n\in A$ we get $\mathrm{\mid }{p}_n(h){\mathrm{\mid }}^2=\widehat{g_n\ast g_n^{\ast }}$, so $F_{\varphi }(\mathrm{\mid }{p}_n(h){\mathrm{\mid }}^2)=L_{\varphi }(g_n\ast g_n^{\ast })\ge 0$ by step 2.1; passing to the uniform limit using the bound of step 4.1 gives $F_{\varphi }(h)\ge 0$.

Steps 1.1, 2.1, 3.1, 4.1 and 5.1 establish every displayed claim: well-definedness and the $L^1$ bound, integrated positivity, the Cauchy-Schwarz bound $\mathrm{\mid }{L}_{\varphi }(f){\mathrm{\mid }}^2\le k{L}_{\varphi }(f\ast f^{\ast })\le k^2\mathrm{\parallel }\widehat{f}{\mathrm{\parallel }}_{\mathrm{\infty }}^2$, the vanishing on $\{f:\widehat{f}=0\}$ and the descent to a positive functional with $\mathrm{\mid }{F}_{\varphi }(h)\mathrm{\mid }\le k\mathrm{\parallel }h{\mathrm{\parallel }}_{\mathrm{\infty }}$.