Proposition 1

Let $f\mapsto \|f\|_B$ be a norm making B into a Banach algebra.

a) The mapping $j$ has norm $\leqslant 1$ in $\mathscr{L}(B;\mathscr{C}(X))$ ;

b) The radical of the algebra B is zero ;

c) The mapping ev is continuous from X into $\mathsf{X}(B)$ ;

d) If B separates the points of X, then the mapping ev is a homeomorphism of X onto a closed subset of $\mathsf{X}(B)$.

One can identify X with $\mathsf{X}(\mathscr{C}(X))$ and $\mathscr{G}_{\mathscr{C}(X)}$ with the identity mapping (example 1 of I, p. 36). The mapping ev is then identified with $\mathsf{X}(j)$, which proves c).

For every function $f\in B$ and every $x\in X$, one has $\mathscr{G}_B(f$)(ev$_x$) $=f(x)$, whence

$\|f\|_{\mathscr{C}(X)}=$ sup$_{x\in X}|\mathscr{G}_B(f$)(ev$_x$)$|\leqslant \varrho_B(f)\leqslant \|f\|_B$,

(cf. prop. 7 of I, p. 38, a)) which entails a). Moreover, this shows that the Gelfand transform of B is injective, whence b) (prop. 8 of I, p. 38). If B separates the points of X, then the mapping ev is injective, whence d) since X is compact.

We now consider the surjectivity of the mapping ev.