Gelfand transform of ell one of Z

All integer-index sums are transported by the fixed bijection $b:\mathbb{N}\to \mathbb{Z}$; double-index sums use $b\times b$. For nonnegative families the sum is the supremum of finite subsums, so a bijective reindexing preserves it. For absolutely summable complex families apply [L3] to real and imaginary parts. Thus the $\mathbb{N}$-indexed Tonelli and Fubini statements apply to the displayed integer-index sums. Convolution is well defined and $\mathrm{\parallel }a\ast b{\mathrm{\parallel }}_1\le \mathrm{\parallel }a{\mathrm{\parallel }}_1\mathrm{\parallel }b{\mathrm{\parallel }}_1$: for each $n$ the family $(\mathrm{\mid }{a}_j{b}_{n-j}\mathrm{\mid }{)}_j$ has finite sum at most $\mathrm{\parallel }a{\mathrm{\parallel }}_1\mathrm{\parallel }b{\mathrm{\parallel }}_1$ whenever ${\sum }_j\mathrm{\mid }{a}_j\mathrm{\mid }<\mathrm{\infty }$ and $\mathrm{\mid }{b}_{n-j}\mathrm{\mid }$ is bounded along $j$; summing over $n$ and interchanging the order of summation by Tonelli's theorem [L3] gives ${\sum }_n{\sum }_j\mathrm{\mid }{a}_j{b}_{n-j}\mathrm{\mid }=\mathrm{\parallel }a{\mathrm{\parallel }}_1\mathrm{\parallel }b{\mathrm{\parallel }}_1$, so every convolution coordinate is absolutely convergent and the estimate follows.

Convolution is commutative and associative and ${\delta }_0$ is the identity: commutativity is the change of variable $j\mapsto n-j$; associativity is the regrouping of the absolutely summable triple family $(a_i{b}_j{c}_{n-i-j}{)}_{i,j}$ along the two possible bracketing orders, licensed by Tonelli and Fubini for the real and imaginary parts [L3]; and $(a\ast {\delta }_0{)}_n=a_n=({\delta }_0\ast a{)}_n$.

${\mathrm{\ell }}^1(\mathbb{Z})$ is complete: if $(a^{(k)})$ is Cauchy in $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_1$, then each coordinate sequence $(a_n^{(k)}{)}_k$ is Cauchy in $\mathbb{C}$ and converges by [L4] to some $a_n$; for every finite set $F\subseteq \mathbb{Z}$ one has ${\sum }_{n\in F}\mathrm{\mid }{a}_n\mathrm{\mid }={\lim }_k{\sum }_{n\in F}\mathrm{\mid }{a}_n^{(k)}\mathrm{\mid }\le {\sup }_k\mathrm{\parallel }{a}^{(k)}{\mathrm{\parallel }}_1<\mathrm{\infty }$, so $a\in {\mathrm{\ell }}^1(\mathbb{Z})$ with $\mathrm{\parallel }a{\mathrm{\parallel }}_1\le {\sup }_k\mathrm{\parallel }{a}^{(k)}{\mathrm{\parallel }}_1$, and the same finite-subset estimate applied to $a-a^{(k)}$ gives $\mathrm{\parallel }a-a^{(k)}{\mathrm{\parallel }}_1\le {\sup }_{l\ge k}\mathrm{\parallel }{a}^{(l)}-a^{(k)}{\mathrm{\parallel }}_1\to 0$.

${\delta }_1$ is invertible with inverse ${\delta }_{-1}$, since ${\delta }_1\ast {\delta }_{-1}={\delta }_0$; if $\chi$ is a character and $z:=\chi ({\delta }_1)$, then $1=\chi ({\delta }_0)=\chi ({\delta }_1)\chi ({\delta }_{-1})$ so $\mathrm{\mid }z\mathrm{\mid }\ge 1$ by [L1], while $\mathrm{\mid }z\mathrm{\mid }\le \mathrm{\parallel }{\delta }_1{\mathrm{\parallel }}_1=1$; hence $\mathrm{\mid }z\mathrm{\mid }=1$, and multiplicativity gives $\chi ({\delta }_n)=z^n$ for all $n\in \mathbb{Z}$.

For $a\in {\mathrm{\ell }}^1(\mathbb{Z})$ define $h_a(z)={\sum }_n{a}_n{z}^n$ on $\mathbb{T}$. This series converges absolutely by [L3]. Given $ϵ>0$, choose a finite initial segment of the fixed enumeration with tail sum less than $ϵ\mathrm{/}4$, and choose $N$ so that $[-N,N]$ contains that segment. Then ${\sum }_{\mathrm{\mid }n\mathrm{\mid }>N}\mathrm{\mid }{a}_n\mathrm{\mid }<ϵ\mathrm{/}4$. Put $M={\sum }_{\mathrm{\mid }n\mathrm{\mid }\le N}\mathrm{\mid }n\mathrm{\mid }\mathrm{\mid }{a}_n\mathrm{\mid }$. For $z,w\in \mathbb{T}$, telescoping positive powers and the identity $\mathrm{\mid }{z}^{-1}-w^{-1}\mathrm{\mid }=\mathrm{\mid }z-w\mathrm{\mid }$ give $\mathrm{\mid }{z}^n-w^n\mathrm{\mid }\le \mathrm{\mid }n\mathrm{\mid }\mathrm{\mid }z-w\mathrm{\mid }$ for every integer $n$. Thus $\mathrm{\mid }{h}_a(z)-h_a(w)\mathrm{\mid }\le M\mathrm{\mid }z-w\mathrm{\mid }+ϵ\mathrm{/}2$. Taking $\mathrm{\mid }z-w\mathrm{\mid }<ϵ\mathrm{/}(2(M+1))$ proves continuity, including $M=0$. Once these maps are shown to be characters, this proves continuity into the evaluation topology [L1]; evaluation at ${\delta }_1$ is continuous in the reverse direction.

For $\mathrm{\mid }z\mathrm{\mid }=1$ the map ${\chi }_z(a):={\sum }_n{a}_n{z}^n$ is a character: it is complex-linear, nonzero (${\chi }_z({\delta }_0)=1$) and multiplicative, because expanding ${\chi }_z(a){\chi }_z(b)={\sum }_j{\sum }_k{a}_j{b}_k{z}^{j+k}$ and regrouping along $n=j+k$ (Tonelli and Fubini on the absolutely summable family $(a_j{b}_k{z}^{j+k})$, [L3]) gives ${\sum }_n(a\ast b{)}_n{z}^n={\chi }_z(a\ast b)$.

For every $a\in {\mathrm{\ell }}^1(\mathbb{Z})$ and every character $\chi$ with $z=\chi ({\delta }_1)$ one has $\chi (a)={\sum }_n{a}_n{z}^n$: let $U(a{)}_k=a_{b(k)}$. The definition of the norm gives $\mathrm{\parallel }Ua{\mathrm{\parallel }}_1=\mathrm{\parallel }a{\mathrm{\parallel }}_1$, and $U$ has inverse $(U^{-1}v{)}_n=v_{b^{-1}(n)}$. Define $Q_N=U^{-1}{P}_{N}U$ using precisely the $\mathbb{N}$-indexed $P_N$ of [L2]. Then $\mathrm{\parallel }a-Q_{N}a{\mathrm{\parallel }}_1=\mathrm{\parallel }Ua-P_{N}Ua{\mathrm{\parallel }}_1\to 0$; $Q_{N}a$ is the finite sum ${\sum }_{k=0}^N{a}_{b(k)}{\delta }_{b(k)}$. Approximate $a$ by these $Q_{N}a$, apply linearity and [step 1.4] to each truncation, and pass to the limit using continuity of $\chi$ from [L1]; the series converges absolutely because $\mathrm{\mid }{a}_n{z}^n\mathrm{\mid }=\mathrm{\mid }{a}_n\mathrm{\mid }$.

The map $z\mapsto {\chi }_z$ is a bijection $\mathbb{T}\to \mathrm{\Delta }({\mathrm{\ell }}^1(\mathbb{Z}))$: it is injective because ${\chi }_z={\chi }_{z^{\mathrm{\prime }}}$ forces $z={\chi }_z({\delta }_1)={\chi }_{z^{\mathrm{\prime }}}({\delta }_1)=z^{\mathrm{\prime }}$, and it is surjective by [step 2.1] applied to the character $\chi$ and its value $z=\chi ({\delta }_1)\in \mathbb{T}$ from [step 1.4].

By [step 1.5] and [step 3.1] the assignment is a continuous bijection with continuous inverse between the compact Hausdorff space $\mathbb{T}$ and the Hausdorff character space $\mathrm{\Delta }({\mathrm{\ell }}^1(\mathbb{Z}))$, hence a homeomorphism; and by [step 2.1] the Gelfand transform of $a$ is the Laurent series $\widehat{a}({\chi }_z)={\sum }_n{a}_n{z}^n$.