Exercise 3, § 8

Let X be a compact topological space and, for every $x\in X$, let $A_x$ be a commutative unital Banach algebra, whose unit element is denoted by $e_x$. Suppose that $\|e_x\|= 1$. Let B be the product Banach algebra of the $A_x$ for $x\in X$. Let C be a Banach subalgebra of B possessing the following properties:

(i) $e= (e_x)_{x\in X}\in C$;

(ii) If $a= (a_x)\in C$ and $f\in \mathscr{C}(X)$, the function $f a:x\mapsto f(x)a_x$ is an element of C;

(iii) For every $a= (a_x)\in C$, the function $x\mapsto \|a_x\|$ is upper semi-continuous.

Let $\widehat{X}$ be the sum set of the $\mathsf{X}(A_x)$. For every $x_0\in X$ and $\chi \in \mathsf{X}(A_{x_0})$, let $\varphi (\chi )$ be the character of C defined by $\varphi (\chi )((a_x)_{x\in X}) =\chi (a_{x_0})$. Prove that $\varphi$ defines a bijection of $\widehat{X}$ onto $\mathsf{X}(C)$. (Let $\xi \in \mathsf{X}(C)$. Since $\mathscr{C}(X)$ can be identified with a closed subalgebra of C$,\xi$ defines a character of $\mathscr{C}(X)$, hence an $x_0\in X$. Let $a= (a_x)\in C$ and $b= (b_x)\in C$, with $a_{x_0}=b_{x_0}$; let $\varepsilon >0$; one has $\|a_x-b_x\|< \varepsilon$ in a neighbourhood U of $x_0$; let $f\in \mathscr{C}(X)$ such that $0\leqslant f\leqslant 1$, equal to 1 at $x_0$ and to 0 outside U; one has $\|f a-f b\|\leqslant \varepsilon$, whence $|\xi (f a-f b)|\leqslant \varepsilon$, whence $|\xi (a)-\xi (b)|\leqslant \varepsilon$, whence $\xi (a) =\xi (b)$. Deduce that there exists a $\chi \in \mathsf{X}(A_{x_0})$ such that $\xi = (\varphi (\chi ))$.)