Scalar unitisation of L^1 of an LCA group: characters, spectrum and identity criterion

($A^{+}$ is a unital commutative Banach $\ast$-algebra.) Bilinearity, associativity, commutativity and the involution laws of the product $(z,f)(w,g)=(zw,zg+wf+f\ast g)$ follow from the corresponding laws of convolution in $A$ and the fact that $(z,f)=z(1,0)+(0,f)$ with $(1,0)$ a unit; the norm is submultiplicative because $\mathrm{\parallel }(z,f)(w,g){\mathrm{\parallel }}_1\le \mathrm{\mid }z\mathrm{\mid }\mathrm{\mid }w\mathrm{\mid }+\mathrm{\mid }z\mathrm{\mid }\mathrm{\parallel }g{\mathrm{\parallel }}_1+\mathrm{\mid }w\mathrm{\mid }\mathrm{\parallel }f{\mathrm{\parallel }}_1+\mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{\parallel }g{\mathrm{\parallel }}_1=(\mathrm{\mid }z\mathrm{\mid }+\mathrm{\parallel }f{\mathrm{\parallel }}_1)(\mathrm{\mid }w\mathrm{\mid }+\mathrm{\parallel }g{\mathrm{\parallel }}_1)$, and $A^{+}=\mathbb{C}\oplus A$ with the ${\mathrm{\ell }}^1$-norm is complete because $\mathbb{C}$ and $A$ are. Moreover $\mathrm{\parallel }(z,f{)}^{\ast }\mathrm{\parallel }=\mathrm{\mid }\bar{z}\mathrm{\mid }+\mathrm{\parallel }{f}^{\ast }{\mathrm{\parallel }}_1=\mathrm{\mid }z\mathrm{\mid }+\mathrm{\parallel }f{\mathrm{\parallel }}_1=\mathrm{\parallel }(z,f)\mathrm{\parallel }$ by the isometry of the involution.

(Characters of $A^{+}$.) Let $\chi$ be a character of $A^{+}$. Since $\chi$ is unital, $\chi (z,0)=z$. Let $h:=\chi {\mathrm{\mid }}_A$ be the restriction to the ideal $\{0\}\oplus A$, identified with $A$. If $h=0$ then $\chi (z,f)=z=q(z,f)$. If $h\mathrm{\ne }0$, then $h$ is a nonzero multiplicative linear functional on $A$, so by [F2] there is a unique $\gamma \in \widehat{G}$ with $h(f)=\widehat{f}(\gamma )$ for all $f$; hence $\chi (z,f)=z+\widehat{f}(\gamma )=h_{\gamma }^{+}(z,f)$. Conversely $q$ is the character of the quotient $A^{+}\mathrm{/}(A)$ and each $h_{\gamma }^{+}$ is a character: $h_{\gamma }^{+}((z,f)(w,g))=zw+\widehat{zg+wf+f\ast g}(\gamma )=zw+z\widehat{g}(\gamma )+w\widehat{f}(\gamma )+\widehat{f}(\gamma )\widehat{g}(\gamma )$ by linearity and the convolution identity of [F8], which is the product of $h_{\gamma }^{+}(z,f)$ and $h_{\gamma }^{+}(w,g)$.

(The identity criterion, discrete case.) If $G$ is discrete then $\{0\}$ is open and nonempty, so $c:=m_G(\{0\})>0$ by [F6]; the class $u:=c^{-1}{\mathbf{1}}_{\{0\}}$ lies in $A$ and for every $f\in A$, using $m_G(\{x\})=c$ and the definition of convolution, $(f\ast u)(x)=c^{-1}{\int }_{G}f(y){\mathbf{1}}_{\{0\}}(x-y)d{m}_G(y)=c^{-1}f(x)m_G(\{x\})=f(x)$ for a.e. $x$; hence $u$ is an identity of $A$.

(The spectrum.) By [F3] applied to $A^{+}$ and the character list of step 1.2, ${\sigma }_{A^{+}}(0,f)=\{\chi (0,f):\chi \in \mathrm{\Delta }(A^{+})\}=\{0\}\cup \{\widehat{f}(\gamma ):\gamma \in \widehat{G}\}$.

(The identity criterion, nondiscrete case.) Suppose $G$ is nondiscrete and $A$ has an identity $u$. By [F6], $m_G(\{0\})=0$, so [F7] provides an open identity neighbourhood $V_0$ with ${\int }_{V_0}\mathrm{\mid }u\mathrm{\mid }d{m}_G<1$; replacing it by $V_0\cap (-V_0)$ we may take $V$ symmetric. Choose a symmetric open $W$ with compact closure and $W-W\subseteq V$ (continuity of addition and local compactness). Then $W$ has $m_G(W)>0$ and ${\mathbf{1}}_W\in A$, so $1_W=u\ast 1_W$ a.e.; but for every $x\in W$ the defining integral $(u\ast 1_W)(x)={\int }_{x-W}u(y)d{m}_G(y)$ satisfies $x-W=x+W\subseteq W+W\subseteq V$ (symmetry of $W$), hence $\mathrm{\mid }(u\ast 1_W)(x)\mathrm{\mid }\le {\int }_V\mathrm{\mid }u\mathrm{\mid }<1$, contradicting $(u\ast 1_W)(x)=1$ on a set of positive measure. Therefore $A$ has no identity.

(The character space of $A^{+}$.) By [F4] the space $\mathrm{\Delta }(A^{+})$ is compact Hausdorff and by step 1.2 the map sending $\gamma$ to $h_{\gamma }^{+}$ and the point $q$ to the restriction character is a bijection onto $\mathrm{\Delta }(A^{+})$. The topology is the topology of pointwise convergence on $A^{+}$: a net $h_{{\gamma }_i}^{+}\to h_{{\gamma }_0}^{+}$ iff $\widehat{f}({\gamma }_i)\to \widehat{f}({\gamma }_0)$ for all $f\in A$, which by [F2] is exactly the compact-open convergence ${\gamma }_i\to {\gamma }_0$; hence the restriction of the homeomorphism to $\widehat{G}$ identifies $\widehat{G}$ with the open subspace $\mathrm{\Delta }(A^{+})\setminus \{q\}$.

(Semisimplicity of $A$.) For $f\in L^1(G)$ and $g\in L^2(G)$ choose $\sigma$-compact essential supports $S_f,S_g$ using [F9], and set $S=S_f+S_g$, also $\sigma$-compact and of $\sigma$-finite Haar measure. The measurable function $H(t,x):=\mathrm{\mid }f(x)\mathrm{\mid }\mathrm{\mid }g(t-x)\mathrm{\mid }$ on $S\times S_f$ has, by Minkowski [F10] and translation isometry [F9], ${\parallel {\int }_{S_f}\mathrm{\mid }f(x)\mathrm{\mid }\mathrm{\mid }g(\cdot -x)\mathrm{\mid }d{m}_G(x)\parallel }_{L^2(S)}\le {\int }_{S_f}\mathrm{\mid }f(x)\mathrm{\mid }\mathrm{\parallel }g(\cdot -x){\mathrm{\parallel }}_{L^2(S)}d{m}_G(x)=\mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{\parallel }g{\mathrm{\parallel }}_2$. The integral on the left is finite a.e.; therefore the defining convolution integral $(f\ast g)(t):={\int }_{G}f(x)g(t-x)d{m}_G(x)$ converges absolutely for a.e. $t$ and determines an $L^2$ class with $\mathrm{\parallel }f\ast g{\mathrm{\parallel }}_2\le \mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{\parallel }g{\mathrm{\parallel }}_2$. Thus $\lambda (f)g:=f\ast g$ is a well-defined bounded linear operator and $\mathrm{\parallel }\lambda (f)\mathrm{\parallel }\le \mathrm{\parallel }f{\mathrm{\parallel }}_1$. For $f_1,f_2\in L^1$ and $g\in C_c(G)\subset L^1\cap L^2$, the L${}^1$ convolution associativity supplier gives $(f_1\ast f_2)\ast g=f_1\ast (f_2\ast g)$ a.e.; the L${}^1$ and L${}^2$ definitions here use the same a.e.-defined convolution integrals, and both sides are in $L^2$ by the bound just proved. Since $C_c(G)$ is dense in $L^2$ [F9] and all three convolution operators are bounded, this identity extends to every $g\in L^2$. Hence $\lambda (f_1\ast f_2)=\lambda (f_1)\lambda (f_2)$, and ${\lambda }^{+}:A^{+}\to \mathcal{B}(L^2)$, ${\lambda }^{+}(z,f)=zI+\lambda (f)$, is a unital algebra homomorphism. For $g,h\in L^2(G)$, the integral of $\mathrm{\mid }f(x)g(y)h(x+y)\mathrm{\mid }$ over $S_f\times S_g$ is bounded by $\mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{\parallel }g{\mathrm{\parallel }}_2\mathrm{\parallel }h{\mathrm{\parallel }}_2$: for each $x$, Cauchy--Schwarz and translation isometry give ${\int }_G\mathrm{\mid }g(y)h(x+y)\mathrm{\mid }d{m}_G(y)\le \mathrm{\parallel }g{\mathrm{\parallel }}_2\mathrm{\parallel }h{\mathrm{\parallel }}_2$, and then integrate against $\mathrm{\mid }f(x)\mathrm{\mid }$. Thus Fubini [F11] and the substitution $z=x-y$ yield $⟨\lambda (f)g,h⟩={\int }_G{\int }_{G}f(y)g(z)\overline{h(y+z)}d{m}_G(z)d{m}_G(y)=⟨g,\lambda (f^{\ast })h⟩$. Here the last equality follows from $(f^{\ast }\ast h)(z)={\int }_G\overline{f(-w)}h(z-w)d{m}_G(w)={\int }_G\overline{f(y)}h(z+y)d{m}_G(y)$ by the inversion-invariant Haar substitution $y=-w$. Therefore $\lambda (f{)}^{\ast }=\lambda (f^{\ast })$; in particular, $f=f^{\ast }$ makes $\lambda (f)$ self-adjoint and hence normal. The representation is faithful: if $\lambda (f)=0$, then $f\ast u_U=\lambda (f)u_U=0$ for every approximate-identity function $u_U\in C_c(G)\subset L^2(G)$, while $f\ast u_U\to f$ in $A$ by [F8], so $f=0$. For self-adjoint $f=f^{\ast }$, the unital homomorphism ${\lambda }^{+}$ gives spectral inclusion ${\sigma }_{\mathcal{B}(L^2)}(\lambda (f))\subseteq {\sigma }_{A^{+}}(0,f)=\{0\}\cup \widehat{f}(\widehat{G})$ by step 2.1; as $\lambda (f)$ is normal, [F5] yields $\mathrm{\parallel }\lambda (f)\mathrm{\parallel }=r(\lambda (f))\le {\sup }_{\gamma \in \widehat{G}}\mathrm{\mid }\widehat{f}(\gamma )\mathrm{\mid }$. Thus $\widehat{f}\equiv 0$ implies $\lambda (f)=0$ and then $f=0$ for every self-adjoint $f$. For arbitrary $f$, $f^{\ast }\ast f$ is self-adjoint and $\widehat{f^{\ast }\ast f}=\bar{\widehat{f}}\widehat{f}=\mathrm{\mid }\widehat{f}{\mathrm{\mid }}^2$ by [F8]; if $\widehat{f}\equiv 0$, the preceding self-adjoint case gives $f^{\ast }\ast f=0$, and the adjoint identity gives $\lambda (f{)}^{\ast }\lambda (f)=\lambda (f^{\ast }\ast f)=0$, whence $\lambda (f)=0$ and faithfulness gives $f=0$. Therefore $A$ is semisimple and (5) holds.

(Isolation of $q$ and the one-point compactification.) If $G$ is discrete then by step 1.3 $A$ has an identity $u$ with $\widehat{u}\equiv 1$; the continuous function $\chi \mapsto \chi (1,-u)$ on $\mathrm{\Delta }(A^{+})$ equals $1$ at $q$ and $0$ at every $h_{\gamma }^{+}$, so $\{q\}$ is open and $q$ is isolated. If $G$ is nondiscrete, suppose $q$ were isolated; then $\widehat{G}\cong \mathrm{\Delta }(A^{+})\setminus \{q\}$ would be compact, and we choose finitely many $f_1,\dots ,f_m\in A$ with ${\bigcup }_j\{{\widehat{f}}_j\mathrm{\ne }0\}=\widehat{G}$ (a finite subcover of the cover by the open sets $\{\widehat{f}\mathrm{\ne }0\}$). Put $b:={\sum }_j{f}_j\ast f_j^{\ast }$, so $\widehat{b}={\sum }_j\mathrm{\mid }{\widehat{f}}_j{\mathrm{\mid }}^2>0$ on $\widehat{G}$ and, $\widehat{G}$ being compact, $\widehat{b}\ge c>0$ there; by step 2.1, ${\sigma }_{A^{+}}(0,b)=\{0\}\cup \widehat{b}(\widehat{G})\subseteq \{0\}\cup [c,\mathrm{\infty })$. By [F5] applied to the function that is $0$ near $0$ and $1$ near $[c,\mathrm{\infty })$, the element $e:=\phi (0,b)$ is an idempotent of $A^{+}$ with $h_{\gamma }^{+}(e)=\phi (\widehat{b}(\gamma ))=1$ for every $\gamma$ and $q(e)=\phi (0)=0$; thus $e\in A$ and $\widehat{e\ast f}=\widehat{e}\widehat{f}=\widehat{f}$ for every $f\in A$. Semisimplicity (step 3.1) gives $e\ast f=f$ for all $f$, so $e$ is an identity of $A$, contradicting step 2.2. Hence $q$ is not isolated, $\mathrm{\Delta }(A^{+})\setminus \{q\}$ is dense, and $\mathrm{\Delta }(A^{+})$ is the one-point compactification of its open dense subspace $\widehat{G}$: neighbourhoods of $q$ are exactly the complements of compact subsets of $\widehat{G}$ (The one-point (Alexandroff) compactification $X^{\ast }=X\cup \{\mathrm{\infty }\}$, whose open sets are the open sets of $X$ together with the complements in $X^{\ast }$ of the closed compact subsets of $X$).

Steps 1.1 and 1.2 prove (1), step 2.1 proves (2), steps 1.3 and 2.2 prove (3), steps 2.3 and 4.1 prove (4), and step 3.1 proves (5).