Corollary 1

Let A be a commutative regular unital Banach algebra, let I be an ideal of A and let $f:\mathsf{X}(A)\rightarrow \mathbf{C}$ be a continuous function. Suppose that, for every $\chi \in \mathsf{X}(A)$, there exists an element $y_{\chi}\in I$ such that $f=\mathscr{G}(y_{\chi})$ in a neighbourhood of $\chi$. Then there exists an element $y\in I$ such that $f=\mathscr{G}(y)$.

Since $\mathsf{X}(A)$ is compact, there exists a finite open covering $(U_1, . . . ,U_n)$ of $\mathsf{X}(A)$, and elements $y_1, . . . , y_n$ of I such that $f=\mathscr{G}(y_i)$ on $U_i$. By prop. 2, there exist elements $x_1, . . . , x_n$ of A of sum 1 such that Supp($\mathscr{G}(x_i)$)$\subset U_i$ for every $i$. Let $y=x_1y_1+\cdots +$ $x_ny_n$. This is an element of I having the required property. Indeed, let $\chi \in \mathsf{X}(A)$. For $1\leqslant i\leqslant n$, one has $\mathscr{G}(x_i)(\chi )\mathscr{G}(y_i)(\chi ) =\mathscr{G}(x_i)(\chi )f(\chi )$ since $\mathscr{G}(y_i)(\chi ) =f(\chi )$ if $\chi \in U_i$, and $\mathscr{G}(x_i)(\chi ) = 0$ if $\chi \notin U_i$. It follows that