I am currently reading through the text "Complete Normed Algebras" by F.F. Bonsall and J. Duncan. I am very new to this topic area so please bare with me!
Example 18 on page 6 says that if $ G $ is a group and $ \ell^1(G) $ denotes the set of all functions $ f : G \to \mathbb{C} $ such that:
$$ \sum_{s \in G} |f(x)| < \infty $$
Then $\ell^1(G) $ with pointwise addition, scalar multiplication, and convolution as product given by:
$$ (f * g)(s) = \sum_{t \in G} f(t)g(t^{-1}s) $$
And with norm:
$$ \| f \| = \sum_{s \in G} |f(s)| $$
is a Banach algebra and is called the discrete group algebra of $ G $.
I am currently trying to verify this example. Clearly $\ell^1(G) $ with addition and scalar multiplication is normed linear space. I believe that all I need to verify is that $\ell^1(G) $ with convolution as the product makes $\ell^1(G) $ an algebra, and that $ \| \cdot \| $ is furthermore an algebra norm on $\ell^1(G) $.
First, to show that $\ell^1(G) $ is an algebra I need to show that convolution is associative, distributive with addition, and that $ (\alpha \beta)(f * g) = (\alpha f) * (\beta g) $ for all $ \alpha, \beta \in \mathbf{F}$ (= $\mathbb{R}$ or $\mathbb{C} $) and all $ f, g \in\ell^1(G) $. I think I am good at verifying all of these EXCEPT for verifying associativity.
For associativity, if $ f, g, h \in\ell^1(G) $ and $ x \in G $ then:
\begin{align*} [(f * g) * h](x) &= \sum_{t \in G} (f * g)(t) h(t^{-1}x) \\ &= \sum_{t \in G} \left [ \sum_{s \in G} f(s)g(s^{-1}t) \right ] h(t^{-1}x) \\ &= \sum_{t \in G} \sum_{s \in G} f(s)g(s^{-1}t)h(t^{-1}x) \end{align*}
From this point on I have read that I should be using Fubini's theorem to swap the order of summation. It has been a LONG time since I have looked at Fubini's theorem so I am not entirely sure why I can use it here (any comments would be greatly appreciated), but then:
\begin{align*} [(f * g) * h](x) &= \sum_{s \in G} \sum_{t \in G} f(s)g(s^{-1}t)h(t^{-1}x) \\ &= \sum_{s \in G} f(s) \left [ \sum_{t \in G} g(s^{-1}t)h(t^{-1}x) \right ] \\ \end{align*}
I am now supposed to replace the $ t $ in the inner sum with $ st $ to get:
\begin{align*} \sum_{t \in G} g(s^{-1}t)h(t^{-1}x) &= \sum_{t \in G} g(s^{-1}(st)) h((st)^{-1}x) \\ &= \sum_{t \in G} g(t) h(t^{-1}s^{-1}x) \end{align*}
(I have not entirely convinced myself as to why the above sums are equal. I am a bit rusty here, so any help would be appreciated).
And so:
\begin{align*} [(f * g) * h](x) &= \sum_{s \in G} \left [ \sum_{t \in G} g(t) h(t^{-1}s^{-1}x) \right ] \\ &= \sum_{s \in G} f(s) (g * h)(s^{-1}x) \\ &= [f * (g * h)](x) \end{align*}
So $ (f * g) * h = f * (g * h) $. Is this proof correct and could anyone help fill in my gaps? Thanks!
Also, why is this space a Banach algebra?
To apply Fubini, you just need to verify that either of the sums $$\sum_{t \in G} \sum_{s \in G} |f(s)g(s^{-1}t)h(t^{-1}x)|,\qquad\sum_{s \in G} \sum_{t \in G} |f(s)g(s^{-1}t)h(t^{-1}x)|$$ is finite.
To see why $\sum_{t \in G} g(s^{-1}t)h(t^{-1}x) = \sum_{t \in G} g(s^{-1}(st)) h((st)^{-1}x)$, note that for any $s\in G$ the map $G\to G$, $t\mapsto st$ is a bijection. So you're just permuting the index of the sum, and since we're in $L^1$ that's fine.
Showing that $\ell^1(G)$ is a Banach space is essentially the same proof as showing $\ell^1(\mathbb N)$ is a Banach space. To show the norm is submultiplicative, just observe that $$\|f*g\|_1=\sum_{s\in G}\left|\sum_{t\in G}f(t)g(t^{-1}s)\right|\leq\sum_{s\in G}\sum_{t\in G}|f(t)||g(t^{-1}s)|$$ then switch the order of summation (Fubini allows this), pull a $|f(t)|$ out of the first sum, then reindex the inner sum.