Corollary

Suppose that A is commutative. For every $f\in$ $\mathscr{C}$(Sp$_A(x)$), one has $\mathscr{G}_A(f(x)) =f\circ \mathscr{G}_A(x)$.

It is enough to apply Proposition 8 to the Gelfand transform $\mathscr{G}_A$ of A into $\mathscr{C}(\mathsf{X}(A))$ and to remark (example above) that $f(\mathscr{G}_A(x)) =f\circ \mathscr{G}_A(x)$.