# Commutative Banach algebra and its Gelfand spectrum

Let $A$ be the set of all functions on $\mathbb{R}^2$ of the form

$$f(t,s):=\sum_{m=-\infty}^{\infty}\sum_{n=-\infty}^{\infty}{a_{mn}e^{i(mt+ns)}},$$

with the following norm:

$$\|f\|:=\sum_{m=-\infty}^{\infty}\sum_{n=-\infty}^{\infty}{|a_{mn}|}.$$

I want to show that $A$ is a commutative Banach algebra. And, I am curious whether it is possible to show that the Gelfand spectrum of $A$ can be identified with the two-dimensional torus, i.e. $\{(e^{it},e^{is})\colon t,s\in\mathbb{R}\}$.

Firstly, it is easy to show that:

1. $A$ is commutative, since for all $f,g\in A$: $(fg)(t,s)=f(t,s)g(t,s)=g(t,s)f(t,s)=(gf)(t,s)$;
2. $A$ is associative, since for all $f,g,h\in A$: $((fg)h)(t,s)=(fg)(t,s)h(t,s)=g(t,s)f(t,s)h(t,s)=f(t,s)(gh)(t,s)=(f(gh))(t,s)$;
3. $A$ is distributive, since for all $f,g,h\in A$: $(f(g+h))(t,s)=f(t,s)(g+h)(t,s)=f(t,s)(g(t,s)+h(t,s))=f(t,s)g(t,s)+f(t,s)h(t,s)=(fg)(x)+ (fh)(t,s)=(fg+fh)(t,s)$, therefore $(g+h)f=gf+hf$;
4. $A$ has the following property: for all $f,g\in A,\alpha\in \mathbb{R}$: $(\alpha(fg))(t,s)=\alpha(fg)(t,s)=\alpha f(t,s)g(t,s)=(\alpha f)(t,s)g(t,s)=((\alpha f)g)(t,s)= f(t,s) \alpha g(t,s)=(f(\alpha g))(t,s)$;

Secondly, I want to show that the norm of the product is less than or equal to the product of the norms, i.e. $$\|fg\|\leq\|f\|\|g\|.$$ So, let, $$f(t,s)=\sum_{m=-\infty}^{\infty}\sum_{n=-\infty}^{\infty}{a_{mn}e^{i(mt+ns)}}$$ and $$g(t,s)=\sum_{m=-\infty}^{\infty}\sum_{n=-\infty}^{\infty}{b_{mn}e^{i(mt+ns)}}$$ We observe that \begin{align} (fg)(t,s)&=f(t,s)g(t,s)\\ &=\left(\sum_{m=-\infty}^{\infty}\sum_{n=-\infty}^{\infty}{a_{mn}e^{i(mt+ns)}}\right)\left(\sum_{m=-\infty}^{\infty}\sum_{n=-\infty}^{\infty}{b_{mn}e^{i(mt+ns)}}\right) \end{align}

Now, I want to conclude somehow that $f(t,s)g(t,s)$ has the following form:

$$\sum_{m=-\infty}^{\infty}\sum_{n=-\infty}^{\infty}{C_{mn}e^{i(mt+ns)}},$$

such that

$$\|fg\|\leq \sum_{m=-\infty}^{\infty}\sum_{n=-\infty}^{\infty}{|a_{mn}|}\sum_{m=-\infty}^{\infty}\sum_{n=-\infty}^{\infty}{|b_{mn}|}.$$

However, I dont know how to finish this proof. How can I conclude that $\|fg\|\leq\|f\|\|g\|.$

Now, let $M$ denote the set of all multiplicative linear functionals on $A$. Then, we can define the Gelfand transform $\hat{x}$ of $x$ as $$\hat{x}(\varphi):=\varphi(x),\qquad\quad\varphi\in M$$

We can easily see that $A(M)$ with the following norm is a Banach space $$\|\hat{x}\|=\sup_{\varphi\in M}{|\hat{x}(\varphi)|}.$$

Now, there exists a topology on $M$ such that $M$ is compact and $\hat{x}\in C(M)$ (Gelfand topology).

Finally, we call set $M$ on $A$ with the Gelfand topology the Gelfand spectrum of $A$.

My question: Is it possible to identify the Gelfand spectrum of $A$ with the two dimensional torus: $T^2:=\{(e^{it},e^{is})\colon t,s\in\mathbb{R}\}$. If so, how?

Any hints are appreciated.