Proposition 5

Let A be a regular commutative Banach algebra without radical, satisfying Ditkin's condition. Let I be a closed ideal of A such that the boundary F of V(I) contains no nonempty perfect set. Then I = Υ(V(I)), that is to say that I is the set of $x\in A$ such that $\mathscr{G}(x)$ vanishes on V(I). In particular, if V(I) reduces to a point $\chi$, one has I = Ker($\chi$ ).

One has $I\subset \Upsilon (V(I))$. Let now $x\in \Upsilon (V(I))$. Let G be the set of characters $\chi \in \mathsf{X}'(A)$ such that $x$ belongs to I in a neighbourhood of $\chi$. It is open and its complement K is perfect (Lemma 1). Since $\mathscr{G}'(x)$ is zero on V(I), the set G contains the interior of $V(I)\cup \{0\}$ (Remark 1). It also contains $\mathsf{X}(A)$- V(I) by Remark 5. Therefore $K =\mathsf{X}'(A)$ - G is contained in the boundary $F_0$ of $V(I)\cup \{0\}$. One has $F_0\subset F\cup \{0\}$. The assumption therefore implies that $F_0$ contains no nonempty perfect set (Lemma 2). It therefore follows from Lemma 1 that the perfect set K is empty. Therefore $x$ belongs to I in a neighbourhood of every $\chi \in \mathsf{X}'(A)$, which means that $x\in I$ (Remark 4). Thus $\Upsilon (V(I))\subset$ I, which concludes the proof.