Proposition 6

Let A be a commutative Banach algebra and let $x\in A$.

a) The union of the set of values of $\mathscr{G}(x)$ and of $\{0\}$ is equal to Sp$'_A(x)$ ;

b) If A has a unit element, the set of values of $\mathscr{G}(x)$ is Sp$_A(x)$. In particular, in order that $x$ be invertible, it is necessary and sufficient that $\mathscr{G}(x)$ should not vanish.

Suppose that A admits a unit element. It is known that, for every $\chi \in \mathsf{X}(A)$, one has $\chi (x)\in$ Sp$_A(x)$. Conversely, let $\lambda \in$ Sp$_A(x)$. Then $x-\lambda$ is not invertible, hence belongs to a maximal ideal of A. There then exists $\chi \in \mathsf{X}(A)$ such that $\chi (x-\lambda ) = 0$ (th. 2 of I, p. 30), whence b).

Let us pass to the general case. Let $\widetilde{A}$ be the Banach algebra obtained from A by adjoining a unit element; it is commutative. The set Sp$'_A(x)$ is equal to Sp$_{\widetilde{A}}(x)$, that is to say to the set of values of $\mathscr{G}_{\widetilde{A}}(x)$ on $\mathsf{X}(\widetilde{A}) =\mathsf{X}'(A)$. Whence a).