Let $l^1(\mathbb{Z})$ be the collection of all complex valued functions $f$ in $\mathbb{Z}$, where $$\Vert f \Vert_1 = \sum_{n=-\infty}^\infty |f(n)| < \infty$$
Define multiplication in $l^2(\mathbb{Z})$ as $$f*g(m) = \sum_{n=-\infty}^\infty f(m-n)g(n)$$
Then under the $\Vert \Vert_1$ norm this becomes a unital Banach algebra with unit, $$ \chi_0(m) = \begin{cases} 1, & \text{if $m=0$} \\ 0, & \text{otherwise} \end{cases}$$
In general, let $$ \chi_i(m) = \begin{cases} 1, & \text{if $m=i$} \\ 0, & \text{otherwise} \end{cases} $$
Then we get that $$\chi_i*\chi_j = \chi_{i+j}$$ Therefore we can write $$\chi_n = \underbrace{\chi_1*\chi_1* \cdots *\chi_1}_{n \text{ times}} = (\chi_1)^n$$
Now if $f \in l^1(\mathbb{Z})$, \begin{equation}f = \sum_{n=-\infty}^\infty f(n) \chi_n = \sum_{n=-\infty}^\infty f(n) (\chi_1)^n \end{equation}
Let $\Omega(l^1(\mathbb{Z}))$ be the character space of $l^1(\mathbb{Z})$. What I have to show is the existence of a bijection from the unit circle, $\mathbb{T}$ of $\mathbb{C}$ and $\Omega(l^1(\mathbb{Z}))$. I have proved the map from $\mathbb{T} \to \Omega(l^1(\mathbb{Z}))$ defined as $$z \to \sum_{n=-\infty}^\infty f(n)z^n$$
is injective.
But I am confused about proving the surjectivity of the map. From above I see that if $\tau \in \Omega(l^1(\mathbb{Z}))$, $\tau$ is completely characterized by its image of $\chi_1$. So, it is enough to show that if $\tau \in \Omega(l^1(\mathbb{Z})$, then $\tau(\chi_1) \in \mathbb{T}$.