A self-contained context can be found in the picture below (from the book An Introduction to Local Spectral Theory by Kjeld Bagger Laursen, Michael Neumann). I'm stuck at the authors' claim that "a simple argument shows that there are no others" (bold text). The term "Beurling algebra" is irrelevant here but I'll add the definition at the bottom.
Here's my attempt before being stuck: Suppose there are others, namely $\varphi:\ell^1(\omega) \to \mathbb{C}$ is another nontrivial multiplicative linear functional (complex homomorphism) that cannot be represented in the form $\varphi_\lambda$, then there exists $y \in \ell^1(\omega)$ such that for all $z \in R_\omega$, we have $\varphi(y) \neq \sum\limits_{n = -\infty}^{\infty} y_n z^n$... Which should gives a contradiction but I don't know how.
Is my attempt sensible at all? I'm afraid my assumption is not perfect for a contradiction and maybe there is something wrong with it. If my attempt is OK so far, how could I go further? Is it possible to "solve" the Laurent series for given the complex number $\varphi(y)$ and a sequence $\{y_n\}$? I have never seen that in any complex analysis theory.
Anyways, any hint or solution would be sincerely appreciated!
We now turn to the case of Beurling algebras on the additive group of integers. In this case, the weight function appears as a two-sided sequence $\omega=(\omega_n)_{n\in\mathbb{Z}}$ of positive real numbers for which $\omega_{m+n}\leq\omega_m\omega_n$, for all $m,n\in\mathbb{Z}$, and $L^1(\omega)$ becomes the space $\ell^1(\omega)$ of all two-sided complex sequences $x=(x_n)_{n\in\mathbb{Z}}$ for which $\sum_{n\in\mathbb{Z}}|x_n|\omega_n$ is finite.
For $x,y\in\ell^1(\omega)$, the convolution product $x*y$ is the sequence given by $$(x*y)_n=\sum_{j=-\infty}^\infty x_{n-j}y_j\quad\text{for all}\ n\in\mathbb{Z}.$$ Obviously, $\ell^1(\omega)$ is a commutative Banach algebra with identity, and therefore coincides with its multiplier algebra. We note that the spectrum of the two-sided Beurling sequence algebra $\ell^1(\omega)$ may be identified with the annulus $$R_\omega:=\{\lambda\in\mathbb{C}:\ \rho_\omega^-\leq|\lambda|\leq\rho_\omega^+\},$$ where, due to the submultiplicativity of the weight sequence $\omega$, the two limits $$\rho_\omega^+:=\lim_{n\to\infty}\omega_n^{1/n}=\inf_{n>0}\omega_n^{1/n}\quad\text{and}\quad\rho_\omega^-:=\lim_{n\to-\infty}\omega_n^{1/n}=\sup_{n<0}\omega_n^{1/n}$$ exist and satisfy $0<\rho_\omega^-\leq\rho_\omega^+<\infty$.
In fact, given an arbitrary $\lambda\in R_\omega$, it is immediate that the definition $$\varphi_\lambda(x):=\sum_{n=-\infty}^\infty x_n\lambda^n\quad\text{for all}\ x\in\ell^1(\omega)$$ yields a non-trivial multiplicative linear functional $\varphi_\lambda$ on $\ell^1(\omega)$, and a simple argument shows that there are no others. Thus elementary Gelfand theory allows us to think of the elements of $\ell^1(\omega)$ as absolutely convergent Laurent series on the annulus $R_\omega$.
Let $G$ be a locally compact abelian group endowed with a weight function $\omega$, where $\omega$ is positive and for all $s,t \in G$, $\omega(s+t) \leq \omega(s)\omega(t)$ (i.e., $\omega$ is submultiplicative). Let $L^1(\omega)$ denote the space of all (equivalence classes of) Borel measurable functions $f:G \to \mathbb{C}$ for which $f\omega \in L^1(G)$. Endowed with convolution as multiplication and the weighted $L^1$-norm $\lVert \cdot \rVert_\omega$ given by
$$ \lVert f \rVert_\omega = \int_G |f(t)|\omega(t)dm(t), $$
the space $L^1(\omega)$ turns into a commutative Banach algebra, the Beurling algebra for the weight $\omega$. In this question, $G$ is actually $\mathbb{Z}$.
I found the solution and hope my solution will help some random people.
The key is the generator. Note we by this product rule, we have the unit:
$$ e:e_n = \begin{cases} 1 & n=0, \\ 0 &\text{otherwise}. \end{cases} $$
Also we have the generator
$$ g:g_n = \begin{cases} 1 & n = 1, \\ 0 &\text{otherwise}. \end{cases} $$
and its inverse is given by
$$ g^{-1}:g_n^{-1} = \begin{cases} 1 & n = -1, \\ 0 &\text{otherwise}. \end{cases} $$
Note for $k>0$ and $g^{\ast k} = g \ast \cdots \ast g$ ($k$ times), the $k$th component is $1$ and other components are $0$. Similarly, the $-k$ th component of $g^{\ast -k} = g^{-1} \ast \cdots \ast g^{-1}$ (also $k$ times) is $1$ and other components are $0$. Now for any $x \in \ell^1(\omega)$, we have a unique expansion (in the form of Laurent series) given by
$$ x = \sum_{k=-\infty}^{\infty}x_k \cdot g^{\ast k} $$
where $\cdot$ is merely the scalar multiplication.
For any complex homomorphism $\varphi$, suppose $\varphi(g)=\lambda$, then the correspondence of addition and multiplication gives us
$$ \varphi(x)= \sum_{k=-\infty}^{\infty}x_k\lambda^k. $$
This is exactly what we want. $\lambda$ must be in $R_\omega$, since if not the absolute convergence of $\ell^1(\omega)$ would be violated.