# How to show convolution is associative?

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!

$$\sum_{ab=x}\sum_{st=b}f(s)g(t)h(b)=\sum_{ast=x}f(s)g(t)h(b)=\sum_{s't'=x}\sum_{kl=t'}f(s')g(k)h(l)=$$ $$=\sum_{ab=x}\sum_{s_1t_1=b}f(a)g(s_1)h(t_1)=\sum_{ab=x}\sum_{st=b}f(a)g(s)h(t).$$ QED
• I'm sorry. $x$ is an element of the semigroup $\Gamma$. Sorry if that was unclear! – user745578 Mar 13 at 15:17
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.