Consider a semigroup $\Gamma$ and the space
$$l^1(\Gamma) := \left\{f: \Gamma \to \mathbb{C}: \sum_{x \in \Gamma} |f(x)| < \infty\right\}$$
where the summation is understood as in the following definition:
Let $S$ be any set. Let $f: S \to \mathbb{C}$ be a function. We say
$\sum_{n \in S}f(n)$ converges to $F\in \mathbb{C}$ if the following
condition is satisfied:For all $\epsilon > 0$, there is a finite subset $T_0$ of $S$ such
that if $T\supseteq T_0$ and $T$ is a finite subset of $S$, then$$\left|\sum_{n \in T} f(n)-F\right| < \epsilon$$
I know the basic properties of this summation, i.e. Fubini etc.
Define the convolution $f * g$ by
$$(f*g)(x) = \sum_{\{(y,z)\in \Gamma^2: yz = x\}} f(y)g(z)$$
I'm trying to prove that
$$((f*g)*h)(x)=(f*(g*h))(x)$$
or equivalently
$$\sum_{ab=x}\sum_{st = a}f(s)g(t)h(b) = \sum_{ab=x}\sum_{st=b}f(a)g(s)h(t)$$
but I can't formally justify why these two sums must coincide.
Any help is appreciated!
Best Answer
If I interpret your sums correctly, your equality is just different notation. The outer sum is $ab=x$, so given $x$ sum over all pairs $(a,b)$ such that $ab=x$. The inner sum is all pairs $(s,t)$ such that $st=a$. This is exactly the same as given $x$, sum over all triples $(s,t,b)$ such that $stb=x$. Similarly you can rewrite the right side of the equation as the single sum over all triples $(a,s,t)$ such that $ast=x$. Now rename the variables $(s,t,b) \mapsto (a,s,t)$ and you see that both sums are equivalent.