Proposition 1

Let A be a commutative Banach algebra. The following conditions are equivalent:

(i) The weak topology and the Jacobson topology on $\mathsf{X}(A)$ coincide ;

(ii) For every $\chi \in \mathsf{X}(A)$ and every weakly closed subset F of $\mathsf{X}(A)$ such that $\chi \notin F$, there exists an $x\in A$ such that $\mathscr{G}(x)$ is equal to 1 at $\chi$ and to 0 on F;

(iii) For every weakly compact subset K and every weakly closed subset F of $\mathsf{X}(A)$ such that $K\cap F =\emptyset$, there exists an element $x\in A$ such that $\mathscr{G}(x)$ is equal to 1 on K and to 0 on F.

Let $M\subset \mathsf{X}(A)$. To say that M is closed for the Jacobson topology means that, for every $\chi \in \mathsf{X}(A)$ - M, there exists an $x\in A$ such that $\mathscr{G}(x)$ vanishes on M but not at $\chi$ (lemma 2 of I, p. 39). Condition (ii) therefore means that every weakly closed subset of $\mathsf{X}(A)$ is closed for the Jacobson topology, which shows that (ii) $=\Rightarrow$ (i). Moreover (iii) $=\Rightarrow$ (ii) since the subset $\{\chi \}$ is weakly compact in $\mathsf{X}(A)$. Finally (i) $=\Rightarrow$ (iii) according to the cor. of prop. 15 of I, p. 81.