Proving $l^{1}(\mathbb{Z})$ with $\ast$ is a commutative Banach algebra Define the $\ast$ on $l^{1}(\mathbb{Z})$ by $$(x\ast y)(n)=\sum_{k=-\infty}^{\infty}x(n-k)y(k),\qquad\text{with } n\in\mathbb{Z}\text{ and } x,y\in l^{1}(\mathbb{Z})$$
I want to show that this defines a commutative Banach algebra.
Firstly I must show that for $x,y,z\in l^{1}(\mathbb{Z})$,
$$(x\ast(y\ast z))(n)=((x\ast y)\ast z)(n)$$ However, I am struggling to derive the expression for $(x\ast(y\ast z))(n)$. I came up with something like $$(x\ast(y\ast z))(n)=\sum_{k=-\infty}^{\infty}x(n-k)(y(n-k)z(k))$$ But I am sure that is wrong.
 A: Convolution really comes out of gathering like powers of things such as powers of exponentials or of powers of a complex variable.
\begin{align}
                \sum_{n=-\infty}^{\infty}a_n e^{in\theta}\sum_{n=-\infty}^{\infty}b_ne^{in\theta}&=\sum_{n=-\infty}^{\infty}\left(\sum_{j+k=n}a_j b_k\right)e^{in\theta} \\
       & = \sum_{n=-\infty}^{\infty}\left(\sum_{k=-\infty}^{\infty}a_{k}b_{n-k}\right)e^{in\theta}.
\end{align}
The above is rigorously justified for $\{ a_n \},\{ b_n \} \in \ell^1(\mathbb{Z})$ for all $\theta\in \mathbb{R}$ because the series are absolutely convergent. And that's another way you can show that the convolution identities apply. The convolution problems are reduced to function multiplication of periodic functions on $[0,2\pi]$, for example.
$$
             (\sum_n a_n e^{in\theta}\sum_n b_n e^{in\theta})\sum_n c_ne^{in\theta} \\ =\sum_n a_n e^{in\theta}(\sum_n b_n e^{in\theta}\sum_n a_n e^{in\theta}).
$$
By uniqueness of coefficients, you find convolution is commutative and associative.
