Convolution on a discrete group

The measure and the involution. By [F1], $\mu =c\cdot \text{counting}$ and $\mu$ is right invariant, so $G$ is unimodular; by [F2] the involution of $L^1(G)$ is $f^{\ast }(x)=\overline{f(x^{-1})}$.

The formula on finitely supported functions. For $f,g\in C_c(G)$ and $x\in G$, only finitely many $y$ have $f(y)\mathrm{\ne }0$, and by [F3] and [F1] $(f\ast g)(x)={\int }_{G}f(y)g(y^{-1}x)d\mu (y)=c{\sum }_{y\in G}f(y)g(y^{-1}x)$.

The series map is bilinear and bounded. For $f,g\in L^1(G)$ define $h(x):=c{\sum }_{y\in G}f(y)g(y^{-1}x)$; on this discrete group each $L^1$ class has a unique pointwise representative because every singleton has measure $c>0$. For each $x$, Cauchy--Schwarz gives ${\sum }_{y\in G}\mathrm{\mid }f(y)\mathrm{\mid }\mathrm{\mid }g(y^{-1}x)\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_{{\mathrm{\ell }}^2}\mathrm{\parallel }g{\mathrm{\parallel }}_{{\mathrm{\ell }}^2}\le \mathrm{\parallel }f{\mathrm{\parallel }}_{{\mathrm{\ell }}^1}\mathrm{\parallel }g{\mathrm{\parallel }}_{{\mathrm{\ell }}^1}<\mathrm{\infty },$ where $y\mapsto y^{-1}x$ is a bijection and $\mathrm{\parallel }f{\mathrm{\parallel }}_{{\mathrm{\ell }}^2}\le \mathrm{\parallel }f{\mathrm{\parallel }}_{{\mathrm{\ell }}^1}$ for every summable family; this proves pointwise absolute convergence by [F1, F5]. For the $L^1$ bound, rearrange the nonnegative double sum using the finite-subsum definition of sums in [F5]: ${\sum }_{x\in G}\mathrm{\mid }h(x)\mathrm{\mid }\le c{\sum }_{x\in G}{\sum }_{y\in G}\mathrm{\mid }f(y)\mathrm{\mid }\mathrm{\mid }g(y^{-1}x)\mathrm{\mid }=c{\sum }_{y\in G}\mathrm{\mid }f(y)\mathrm{\mid }{\sum }_{x\in G}\mathrm{\mid }g(y^{-1}x)\mathrm{\mid }=c\mathrm{\parallel }f{\mathrm{\parallel }}_{{\mathrm{\ell }}^1}\mathrm{\parallel }g{\mathrm{\parallel }}_{{\mathrm{\ell }}^1}\mathrm{.}$ The last equality again uses the bijection $x\mapsto y^{-1}x$. Thus $h\in L^1(G)$ and $\mathrm{\parallel }h{\mathrm{\parallel }}_1=c{\sum }_x\mathrm{\mid }h(x)\mathrm{\mid }\le c^2\mathrm{\parallel }f{\mathrm{\parallel }}_{{\mathrm{\ell }}^1}\mathrm{\parallel }g{\mathrm{\parallel }}_{{\mathrm{\ell }}^1}=\mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{\parallel }g{\mathrm{\parallel }}_1\mathrm{.}$ Absolute convergence also gives bilinearity, so $(f,g)\mapsto h$ is a bounded, hence continuous, bilinear map into $L^1(G)$.

The series computes the convolution. On $C_c(G)\times C_c(G)$ the map $(f,g)\mapsto h$ of step 1.3 agrees with the $C_c$ convolution by step 1.2, and both $(f,g)\mapsto h$ and the $L^1$ convolution are continuous bilinear maps on $L^1(G)\times L^1(G)$ by steps 1.3 and [F3]. Since $C_c(G)$ is dense in $L^1(G)$ by [F4], they agree as $L^1$ classes for all $f,g\in L^1(G)$. Every singleton has positive measure $c$, so equality of classes on this discrete group is pointwise equality; hence $(f\ast g)(x)=c{\sum }_{y\in G}f(y)g(y^{-1}x)$ for all $x\in G$.

The convolution unit. Let $u:=c^{-1}{\mathbf{1}}_{\{e\}}$; then ${\int }_G\mathrm{\mid }u\mathrm{\mid }d\mu =c\cdot c^{-1}=1$, so $u\in L^1(G)$ and $\mathrm{\parallel }u{\mathrm{\parallel }}_1=1$. By step 2.1, $(u\ast g)(x)=c{\sum }_{y\in G}u(y)g(y^{-1}x)=c\cdot c^{-1}g(x)=g(x)$ and $(g\ast u)(x)=c{\sum }_{y\in G}g(y)u(y^{-1}x)=c\cdot c^{-1}g(x)=g(x)$ for every $g\in L^1(G)$ and every $x$, the only contributing index being $y=e$, respectively $y=x$. So $u$ is a two-sided identity.

Collecting: on a discrete group with $\mu =c\cdot \text{counting}$ the group is unimodular with $f^{\ast }(x)=\overline{f(x^{-1})}$, convolution is the absolutely convergent series of step 2.1, and $c^{-1}{\mathbf{1}}_{\{e\}}$ is the norm-one convolution identity. ∎