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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.