4
$\begingroup$

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!

$\endgroup$
1
$\begingroup$

$$ \sum_{ab=x}\sum_{st=b}f(s)g(t)h(b)=\sum_{ast=x}f(s)g(t)h(b)=\sum_{s't'=x}\sum_{kl=t'}f(s')g(k)h(l)= $$ $$ =\sum_{ab=x}\sum_{s_1t_1=b}f(a)g(s_1)h(t_1)=\sum_{ab=x}\sum_{st=b}f(a)g(s)h(t). $$ QED

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I'm sorry. $x$ is an element of the semigroup $\Gamma$. Sorry if that was unclear! $\endgroup$ – user745578 Mar 13 at 15:17
  • $\begingroup$ Is this look ok? $\endgroup$ – Nikos Bagis Mar 13 at 15:42
  • $\begingroup$ Looks better now! I will go through it in detail. $\endgroup$ – user745578 Mar 13 at 16:49
2
+100
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.