Lemma 1

Let A be a commutative regular Banach algebra without radical, satisfying Ditkin's condition. Let I be a closed ideal of A and $x$ an element of Υ(V(I)). Let K be the set of $\chi \in \mathsf{X}'(A)$ such that $x$ does not belong to I in a neighbourhood of $\chi$. Then the set K is a perfect subset of $\mathsf{X}'(A)$.

Let G denote the complement of K in $\mathsf{X}'(A)$. The set G is open in $\mathsf{X}'(A)$ (Remark 3), therefore K is closed.

We proceed by contradiction, and suppose that K has an isolated point $\chi_0$. Let U denote a neighbourhood of $\chi_0$ such that U $-\{\chi_0\} \subset G$. Since $x$ does not belong to I in a neighbourhood of $\chi_0$, Remark 5 shows that $\chi_0\in V(I)$. In particular, one has $\chi_0(x) = 0$ since $x\in \Upsilon (V(I))$.

We shall show that there exists an element $y$ of A which belongs to I in a neighbourhood of every point of $\mathsf{X}'(A)-\{\chi_0\}$, which does not belong to I in a neighbourhood of $\chi_0$, and such that $\chi_0(y) = 0$.

Suppose first that the existence of such an element $y$ has been proved. Since A satisfies Ditkin's condition, there then exists a sequence $(x_n)$ in A such that $x_ny$ tends to $y$ and such that each $\mathscr{G}'(x_n)$ vanishes in a neighbourhood of $\chi_0$. For every $n$, the element $x_ny$ then belongs to I in a neighbourhood of every point of $\mathsf{X}'(A)$ (Remarks 1 and 2) and therefore $x_ny\in I$ (Remark 4). Since I is closed, one deduces that $y\in I$, which contradicts the fact that $y$ does not belong to I in a neighbourhood of $\chi_0$.

It remains to prove the existence of $y$. If $\chi_0\not= 0$, by assertion (iii) of Prop. 1 of I, p. 88, there exists a $u\in A$ such that $\mathscr{G}'(u)$ is equal to 1 in a neighbourhood of $\chi_0$ and equal to 0 in a neighbourhood of $\mathsf{X}'(A)$ - U. Let $y=ux$. Since $x$ belongs to I in a neighbourhood of $\chi$ for every $\chi \in U-\{\chi_0\}$, the same is true of $y$. Moreover, if $\chi \in \mathsf{X}'(A)$ - U, then $\mathscr{G}'(y)$ vanishes in a neighbourhood of $\chi$. Therefore (Remark 5) the element $y=ux$ belongs to I in a neighbourhood of every $\chi \not=\chi_0$. Since $\mathscr{G}'(y)$ coincides with $\mathscr{G}'(x)$ in a neighbourhood of $\chi_0$, the fact that $\chi_0$ belongs to K implies that $y$ does not belong to I in a neighbourhood of $\chi_0$. Finally, one has $\chi_0(y) =\chi_0(u)\chi_0(x) = 0$.

If $\chi_0= 0$, there similarly exists an element $v\in A$ such that $\mathscr{G}'(v)$ is zero in a neighbourhood of $\chi_0$ and equal to 1 in a neighbourhood of $\mathsf{X}'(A)$- U; as above, one deduces that the element $y=x-vx$ belongs to I in a neighbourhood of every $\chi \not=\chi_0$, that it does not belong to I in a neighbourhood of $\chi_0$, and that $\chi_0(y) = 0$.