Proposition 15

Let A be a commutative Banach algebra, let $I_1$ be an ideal of A and $F_1$ the set of $\chi \in \mathsf{X}(A)$ which are zero on $I_1$. Let $F_2$ be a subset of $\mathsf{X}(A)$ disjoint from $F_1$, closed for the Jacobson topology, and compact for the weak topology. Then there exists $u\in I_1$ such that $\mathscr{G}(u) = 1$ on $F_2$.

Let $I_2$ be the intersection of the kernels of the characters belonging to $F_2$. The Banach algebra $A/I_2$ is semisimple (Prop. 8 of I, p. 38). Since $F_2$ is closed for the Jacobson topology, the only elements of $\mathsf{X}(A)$ vanishing on $I_2$ are those of $F_2($cf. I, p. 13). Hence $F_2$, endowed with the topology induced by the weak topology of $\mathsf{X}(A)$, is identified with $\mathsf{X}(A/I_2)$ endowed with the weak topology (I, p. 9, n$^o7$). Since $F_2$ is weakly compact, the algebra $A/I_2$ has a unit element (Prop. 14).

One then has $I_1+ I_2= A$. For otherwise, $(I_1+ I_2)/I_2$ would be a proper ideal, hence contained in the kernel of a nonzero character of $A/I_2($I, p. 30, Th. 2). The latter would define, by composition with the canonical projection $A\rightarrow A/I_2$, a nonzero character $\chi$ of A which would vanish on $I_1$ and $I_2$, and would therefore belong to $F_1\cap F_2$, contrary to the hypothesis.

Since $I_1+ I_2= A$, there exists $u\in I_1$ whose class in $A/I_2$ is a unit element of $A/I_2$. Then $\chi (u) = 1$ for every $\chi \in F_2$, which concludes the proof.