Nonzero multiplicative functionals on L^1 of an LCA group are Fourier evaluations

(Boundedness of $h$.) In the unitisation $A^{+}$ of [F1] the functional $h^{+}(z,f)=z+h(f)$ is multiplicative and unital: $h^{+}((z,f)(w,g))=zw+h(zg+wf+f\ast g)=zw+zh(g)+wh(f)+h(f)h(g)=h^{+}(z,f)h^{+}(w,g)$ by linearity and multiplicativity of $h$. By [F2] applied to the character $h^{+}$, $\mathrm{\mid }h(f)\mathrm{\mid }=\mathrm{\mid }{h}^{+}(0,f)\mathrm{\mid }\le \mathrm{\parallel }(0,f)\mathrm{\parallel }=\mathrm{\parallel }f{\mathrm{\parallel }}_1$ for every $f\in A$; in particular $h$ is bounded with $\mathrm{\parallel }h\mathrm{\parallel }\le 1$.

(The character ratio is independent of the reference function.) For $x\in G$ and $f,g\in A$ one has $T_x(f\ast g)=(T_{x}f)\ast g=f\ast (T_{x}g)$: evaluating at $z$ and substituting $y\mapsto y-x$ in the defining integral gives $(T_x(f\ast g))(z)=(f\ast g)(z-x)={\int }_{G}f(y)g(z-x-y)d{m}_G(y)$ and likewise for the other two expressions. Choose $f_0\in A$ with $h(f_0)\mathrm{\ne }0$ and put $a(x):=h(T_x{f}_0)\mathrm{/}h(f_0)$. For every $g\in A$, multiplicativity gives $h(T_{x}g)h(f_0)=h(T_x(g\ast f_0))=h(g)h(T_x{f}_0)=h(g)a(x)h(f_0)$. Dividing by the fixed nonzero $h(f_0)$ proves $h(T_{x}g)=a(x)h(g)$, including when $h(g)=0$.

($a$ is a continuous character of $G$.) From $T_0=\mathrm{i}\mathrm{d}$ we get $a(0)=1$, and from $T_{x+y}=T_x{T}_y$ and step 2.1 applied twice, $a(x+y)=\frac{h(T_x{T}_y{f}_0)}{h(f_0)}=\frac{a(x)h(T_y{f}_0)}{h(f_0)}=a(x)a(y)\mathrm{.}$ Moreover $a$ is continuous: by step 1.1 $\mathrm{\mid }a(x)-a(x_0)\mathrm{\mid }=\mathrm{\mid }h(T_x{f}_0-T_{x_0}{f}_0)\mathrm{\mid }\mathrm{/}\mathrm{\mid }h(f_0)\mathrm{\mid }\le \mathrm{\parallel }{T}_x{f}_0-T_{x_0}{f}_0{\mathrm{\parallel }}_1\mathrm{/}\mathrm{\mid }h(f_0)\mathrm{\mid }\to 0$ as $x\to x_0$, by the norm continuity of translations [F4]. Finally $a$ is bounded, $\mathrm{\mid }a(x)\mathrm{\mid }\le \mathrm{\parallel }{f}_0{\mathrm{\parallel }}_1\mathrm{/}\mathrm{\mid }h(f_0)\mathrm{\mid }=:M$, and multiplicativity with $a(0)=1$ gives $a(nx)=a(x{)}^n$ for every $n\in \mathbb{Z}$; since $\mathrm{\mid }a(x){\mathrm{\mid }}^n=\mathrm{\mid }a(nx)\mathrm{\mid }\le M$ for every $n\ge 0$, necessarily $\mathrm{\mid }a(x)\mathrm{\mid }\le 1$, and applying this bound to $-x$, where $a(-x)=a(x{)}^{-1}$, gives $\mathrm{\mid }a(x)\mathrm{\mid }\ge 1$. Hence $\mathrm{\mid }a(x)\mathrm{\mid }=1$ and $\gamma :=\bar{a}$ takes values in the unit circle $\mathbb{T}$.

(The integral identity.) Let $f\in A$ and $u\in C_c(G)$. By the $\sigma$-compact essential-support reduction in the convolution-algebra supplier, represent $f$ as zero outside a countable union of compact sets $S$. The continuous orbit $x\mapsto T_{x}u$ has compact metric image on each of those compact sets, hence separable image there; Dependent Choice makes their countable union separable. Thus $F(x):=f(x)T_{x}u$, zero outside $S$, is strongly measurable by measurable scalar multiplication and countable simple approximations in this separable range. Its norm has integral $\mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{\parallel }u{\mathrm{\parallel }}_1$, so [F6] makes it Bochner integrable. Its integral equals $f\ast u$: pairing against any $\psi \in C_c(G)$ commutes with the Bochner integral and the $\sigma$-finite Fubini calculation gives the same pairing as $f\ast u$. The uniqueness of $L^1$ densities from their $C_c$ pairings, proved in the convolution-algebra supplier, identifies the two elements of $A$. Since $h$ is bounded linear, bounded linear maps commute with Bochner integration, so $h(f\ast u)={\int }_{G}f(x)h(T_{x}u)d{m}_G(x)$, and by step 2.1 $h(T_{x}u)=a(x)h(u)$; hence $h(f\ast u)=h(u){\int }_{G}f(x)a(x)d{m}_G(x)\mathrm{.}$

($\gamma$ is a character.) The conjugate $\gamma =\bar{a}$ of the continuous homomorphism $a$ is a continuous homomorphism $G\to \mathbb{T}$, hence $\gamma \in \widehat{G}$.

(Identification of $h$.) Apply step 3.2 with $u=u_i$ from the all-admissible-pair net [F3] and let $i$ tend along that directed set. Since $f\ast u_i\to f$ in $A$ and $h$ is continuous, $h(f\ast u_i)\to h(f)$; since $f_0\ast u_i\to f_0$ and $h(u_i)h(f_0)=h(f_0\ast u_i)\to h(f_0)\mathrm{\ne }0$, we get $h(u_i)\to 1$. Therefore $h(f)={\int }_{G}f(x)a(x)d{m}_G(x)={\int }_{G}f(x)\overline{\gamma (x)}d{m}_G(x)=\widehat{f}(\gamma )$ for every $f\in A$.

(Uniqueness and norm.) The Fourier evaluations separate points of $\widehat{G}$: if $\widehat{f}({\gamma }_1)=\widehat{f}({\gamma }_2)$ for all $f\in A$, put $d:=\overline{{\gamma }_1}-\overline{{\gamma }_2}$. If $d(x_0)\mathrm{\ne }0$, continuity gives a relatively compact open neighbourhood where $\mathrm{\mid }d\mathrm{\mid }$ is bounded below; a nonnegative cutoff $v\in C_c$ with $v(x_0)=1$ yields $f=\bar{d}v\in A$ and ${\int }_{G}fdd{m}_G={\int }_G\mathrm{\mid }d{\mathrm{\mid }}^{2}vd{m}_G>0$ by [F5], a contradiction. Hence $d=0$ and ${\gamma }_1={\gamma }_2$. Finally $\mathrm{\parallel }h\mathrm{\parallel }=1$: step 1.1 gives $\mathrm{\parallel }h\mathrm{\parallel }\le 1$, while $h(u_i)\to 1$ and $\mathrm{\parallel }{u}_i{\mathrm{\parallel }}_1=1$ give $\mathrm{\parallel }h\mathrm{\parallel }\ge 1$. The same argument applies to $h_{\gamma }(f):=\widehat{f}(\gamma )$: it is multiplicative by [F4] and bounded of norm at most $1$, and continuity of $\gamma$ at $0$ gives $\mathrm{\mid }{h}_{\gamma }(u_i)-1\mathrm{\mid }\le {\sup }_{x\in U_i}\mathrm{\mid }\gamma (x)-1\mathrm{\mid }\to 0$ by the mass-one and support properties of [F3]. Thus it is nonzero and its norm is $1$; hence the converse holds for every $\gamma \in \widehat{G}$.

Steps 1.1, 2.1, 3.1, 3.2, 4.1 and 5.1 produce the unique $\gamma \in \widehat{G}$ with $h=h_{\gamma }$ and the bound $\mathrm{\mid }h(f)\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_1$, step 6.1 proves uniqueness and that every such functional has norm $1$, and the converse is included in step 6.1.