Characters of the L1 algebra of an abelian group

Put $\tilde{A}=\mathbb{C}\oplus L^1(N)$ with $(z,f)(w,g)=(zw,zg+wf+f\ast g)$ and $\mathrm{\parallel }(z,f)\mathrm{\parallel }=\mathrm{\mid }z\mathrm{\mid }+\mathrm{\parallel }f{\mathrm{\parallel }}_1$. The product is bilinear and associative, $\mathrm{\parallel }(z,f)(w,g)\mathrm{\parallel }\le (\mathrm{\mid }z\mathrm{\mid }+\mathrm{\parallel }f{\mathrm{\parallel }}_1)(\mathrm{\mid }w\mathrm{\mid }+\mathrm{\parallel }g{\mathrm{\parallel }}_1)$, and $\tilde{A}$ is complete, so it is a nonzero unital complex Banach algebra with unit $(1,0)$; the map $\tilde{\lambda }(z,f)=z+\lambda (f)$ is complex-linear, multiplicative because $\lambda$ is multiplicative, and $\tilde{\lambda }(1,0)=1$, so it is a character. By [F4], $\mathrm{\mid }\lambda (f)\mathrm{\mid }=\mathrm{\mid }\tilde{\lambda }(0,f)\mathrm{\mid }\le \mathrm{\parallel }(0,f)\mathrm{\parallel }=\mathrm{\parallel }f{\mathrm{\parallel }}_1$ for every $f$.

Since $\lambda \mathrm{\ne }0$ there is $k\in L^1(N)$ with $\lambda (k)\mathrm{\ne }0$; fix such a $k$ and set $\chi (n)=\lambda (L_{n}k)\mathrm{/}\lambda (k)$ for $n\in N$.

For all $f,k\in L^1(N)$ and all $n\in N$ one has $(L_{n}f)\ast k=f\ast (L_{n}k)$: for $f,k\in C_c(N)$ both sides are continuous functions computed by the pointwise convolution formula, and substituting $y=nz$ in ${\int }_{N}f(n^{-1}y)k(y^{-1}x)dy$ uses left invariance of $dy$ to give ${\int }_{N}f(z)k(z^{-1}{n}^{-1}x)dz=(f\ast L_{n}k)(x)$; both sides are bounded bilinear in $(f,k)$ by [F1] and [F3], and $C_c(N)\times C_c(N)$ is dense in $L^1(N)\times L^1(N)$ by [F2], so the identity extends to all $f,k\in L^1(N)$.

For all $f,k\in L^1(N)$, $f\ast k={\int }_{N}f(n)L_{n}kdn$ as a Bochner integral. Approximate $f$ in $L^1$ by $u_m\in C_c(N)$ and pass to a subsequence with $u_m\to f$ a.e.; each $n\mapsto u_m(n)L_{n}k$ is continuous and compactly supported hence strongly measurable, and a diagonal selection of their defining simple approximants shows that the a.e. limit $n\mapsto f(n)L_{n}k$ is strongly measurable; since ${\int }_N\mathrm{\parallel }f(n)L_{n}k{\mathrm{\parallel }}_{1}dn=\mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{\parallel }k{\mathrm{\parallel }}_1<\mathrm{\infty }$, it is Bochner integrable by [F5]. The assignment $f\mapsto {\int }_{N}f(n)L_{n}kdn$ is bounded linear, and for $f,k\in C_c(N)$ pairing with any $\phi \in L^{\mathrm{\infty }}(N)$ and commuting the bounded functional through the Bochner integral reduces the identity to ${\int }_N{\int }_{N}f(n)k(n^{-1}x)\phi (x)dxdn={\int }_N(f\ast k)(x)\phi (x)dx$, which follows from [F6] because the kernel is compactly supported; the pairing with all of $L^{\mathrm{\infty }}(N)$ separates points of $L^1(N)$, and both sides are bounded linear in $f$ with $C_c(N)$ dense by [F2], so the identity holds for all $f\in L^1(N)$; repeating the same density argument in the second variable gives it for all $k$ as well.

Conversely, for a continuous character $\chi$ define ${\lambda }_{\chi }(f)={\int }_{N}f\chi dn$. Then ${\lambda }_{\chi }$ is complex-linear with $\mathrm{\mid }{\lambda }_{\chi }(f)\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_1$, and it is multiplicative: for $f,g\in C_c(N)$ the double integral ${\int }_N{\int }_{N}f(y)g(y^{-1}x)\chi (x)dydx$ equals by [F6] the iterated integral ${\int }_{N}f(y){\int }_{N}g(z)\chi (yz)dzdy={\lambda }_{\chi }(f){\lambda }_{\chi }(g)$ after the substitution $z=y^{-1}x$ and using $\chi (yz)=\chi (y)\chi (z)$; both ${\lambda }_{\chi }(f\ast g)$ and ${\lambda }_{\chi }(f){\lambda }_{\chi }(g)$ are bounded bilinear in $(f,g)$, so density of $C_c(N)$ ([F2]) extends multiplicativity to all $f,g\in L^1(N)$. And ${\lambda }_{\chi }\mathrm{\ne }0$: by continuity of $\chi$ at $e$ there is a nonempty open set $U$ with $\mathrm{Re}\chi >1\mathrm{/}2$ on $U$, and by [F7] there is $c\in C_c(N)$ with $c\ge 0$, $c\mathrm{\ne }0$, supported in $U$; then $\mathrm{Re}{\lambda }_{\chi }(c)={\int }_{N}c\mathrm{Re}\chi dn>0$, so ${\lambda }_{\chi }(c)\mathrm{\ne }0$.

Multiplicativity of $\lambda$ applied to [step 1.3] with this $k$ gives $\lambda (L_{n}f)\lambda (k)=\lambda (f)\lambda (L_{n}k)$, hence $\lambda (L_{n}f)=\chi (n)\lambda (f)$ for every $f\in L^1(N)$ and every $n\in N$.

If ${\chi }_i\to \chi$ in the compact-open topology, then ${\lambda }_{{\chi }_i}(f)\to {\lambda }_{\chi }(f)$ for every $f\in L^1(N)$: given $\epsilon >0$ choose $u\in C_c(N)$ with $\mathrm{\parallel }f-u{\mathrm{\parallel }}_1<\epsilon \mathrm{/}4$ ([F2]); then $\mathrm{\mid }{\lambda }_{{\chi }_i}(f)-{\lambda }_{\chi }(f)\mathrm{\mid }\le 2\mathrm{\parallel }f-u{\mathrm{\parallel }}_1+\mathrm{\parallel }u{\mathrm{\parallel }}_{\mathrm{\infty }}{\int }_{\mathrm{supp}u}\mathrm{\mid }{\chi }_i-\chi \mathrm{\mid }dn$, and ${\chi }_i\to \chi$ uniformly on the compact set $\mathrm{supp}u$ directly from the compact-open subbasis, while the Haar measure of $\mathrm{supp}u$ is finite by [F7]. Thus the map $\chi \mapsto {\lambda }_{\chi }$ is continuous for the two stated topologies.

$\chi$ is multiplicative: since $L_n{L}_m=L_{nm}$ and $\lambda (k)\mathrm{\ne }0$, applying [step 2.1] to $L_{m}k$ gives $\chi (nm)\lambda (k)=\lambda (L_n{L}_{m}k)=\chi (n)\lambda (L_{m}k)=\chi (n)\chi (m)\lambda (k)$, so $\chi (nm)=\chi (n)\chi (m)$; in particular $\chi (e)=1$ and $\chi (n^{-1})=\chi (n{)}^{-1}$.

$\chi$ is continuous: for $n\to n_0$ in $N$ one has $\mathrm{\mid }\chi (n)-\chi (n_0)\mathrm{\mid }=\mathrm{\mid }\lambda (L_{n}k-L_{n_0}k)\mathrm{\mid }\mathrm{/}\mathrm{\mid }\lambda (k)\mathrm{\mid }\le \mathrm{\parallel }{L}_{n}k-L_{n_0}k{\mathrm{\parallel }}_1\mathrm{/}\mathrm{\mid }\lambda (k)\mathrm{\mid }\to 0$ by [step 1.1] and [F3].

Apply the bounded functional $\lambda$ to [step 1.4] and commute it through the Bochner integral: $\lambda (f)\lambda (k)=\lambda (f\ast k)={\int }_{N}f(n)\lambda (L_{n}k)dn=\lambda (k){\int }_{N}f(n)\chi (n)dn$ by [step 2.1]; since $\lambda (k)\mathrm{\ne }0$, dividing gives the classification formula $\lambda (f)={\int }_{N}f(n)\chi (n)dn$ for every $f\in L^1(N)$.

Conversely, suppose ${\lambda }_i\to \lambda$ in the pointwise-evaluation topology of the character space. Fix the $k$ of [step 1.2] and a compact $C\subseteq N$. The set $\{L_{n}k:n\in C\}$ is norm compact in $L^1(N)$ as the continuous image of $C$ under [F3], so for each $\epsilon >0$ it has a finite $\epsilon \mathrm{/}3$-net $L_{n_1}k,\dots ,L_{n_r}k$. For all sufficiently large $i$ one has $\mathrm{\mid }{\lambda }_i(L_{n_j}k)-\lambda (L_{n_j}k)\mathrm{\mid }<\epsilon \mathrm{/}3$ for every $j$ and $\mathrm{\mid }{\lambda }_i(k)-\lambda (k)\mathrm{\mid }<\min \{\epsilon ,\mathrm{\mid }\lambda (k)\mathrm{\mid }\mathrm{/}2\}$, using [step 1.1] for the bounds $\mathrm{\parallel }{\lambda }_i\mathrm{\parallel }\le 1$ and $\mathrm{\parallel }\lambda \mathrm{\parallel }\le 1$; then for every $n\in C$ and the corresponding $j$ one gets $\mathrm{\mid }{\lambda }_i(L_{n}k)-\lambda (L_{n}k)\mathrm{\mid }<\epsilon$, and division by the eventually nonvanishing ${\lambda }_i(k)$ gives $\mathrm{\mid }{\chi }_i(n)-\chi (n)\mathrm{\mid }\le M\epsilon$ uniformly on $C$ for a constant $M$ depending only on $\lambda (k)$ and $\mathrm{\parallel }k{\mathrm{\parallel }}_1$. Hence ${\lambda }_i\to \lambda$ pointwise implies ${\chi }_i\to \chi$ uniformly on compacta, that is, the inverse map is continuous.

$\mathrm{\mid }\chi (n)\mathrm{\mid }=1$ for every $n$: [step 1.1] gives $\mathrm{\mid }\chi (n)\mathrm{\mid }\le \mathrm{\parallel }k{\mathrm{\parallel }}_1\mathrm{/}\mathrm{\mid }\lambda (k)\mathrm{\mid }$, and applying [step 3.1] to the powers $n^j$ gives $\mathrm{\mid }\chi (n){\mathrm{\mid }}^j\le \mathrm{\parallel }k{\mathrm{\parallel }}_1\mathrm{/}\mathrm{\mid }\lambda (k)\mathrm{\mid }$ for all $j\ge 1$, whence $\mathrm{\mid }\chi (n)\mathrm{\mid }\le 1$; replacing $n$ by $n^{-1}$ and using $\chi (n^{-1})=\chi (n{)}^{-1}$ gives $\mathrm{\mid }\chi (n)\mathrm{\mid }\ge 1$ as well. Thus $\chi :N\to \mathbb{T}$ is a continuous character.

If ${\lambda }_{{\chi }_1}={\lambda }_{{\chi }_2}=\lambda$, choose $k$ with $\lambda (k)\mathrm{\ne }0$. Substitution $x=ny$ in the defining integral gives ${\lambda }_{{\chi }_i}(L_{n}k)={\chi }_i(n){\lambda }_{{\chi }_i}(k)$ for $i=1,2$. Hence ${\chi }_i(n)=\lambda (L_{n}k)\mathrm{/}\lambda (k)$ for every $n$, so ${\chi }_1={\chi }_2$. Together with step 3.3 this proves the bijection.

Steps [2.2] and [3.4] show that $\chi \mapsto {\lambda }_{\chi }$ is a homeomorphism from $\widehat{N}$ with the compact-open topology onto the character space with the pointwise-evaluation topology, and [step 3.3] with [step 5.1] shows every nonzero complex-linear multiplicative functional is uniquely of the form ${\lambda }_{\chi }$.