Proposition 2

Let $f\mapsto \|f\|_B$ be a norm such that $(B,\|\cdot \|_B)$ is a Banach algebra.

a) If the mapping ev is surjective, then the algebra B is a full subalgebra of $\mathscr{C}(X)$ ;

b) Suppose that B is a full subalgebra of $\mathscr{C}(X)$, and that there exists an element $a\in B$ such that the set of elements $f(a)$, where $f$ ranges over the set of rational functions on $\mathbf{C}$ without poles on Sp$_B(a)$, is dense in B. Then the mapping ev is surjective ;

c) If B is a full involutive subalgebra of $\mathscr{C}(X)$, then the mapping ev is surjective.

Assertion a) follows from prop. 10 of I, p. 40, and assertion b) from prop. 11 of I, p. 41, since the full subalgebra of B generated by $a$ is the set of elements $f(a)$, where $f$ ranges over the set of rational functions on $\mathbf{C}$ without poles on Sp$_B(a)$ (lemma I, p. 6 and prop. 6 of I, p. 37, b)).

Let us prove c). Suppose then that B is a full involutive subalgebra of $\mathscr{C}(X)$. To prove that ev is surjective, it is enough to prove that for every $\chi \in \mathsf{X}(B)$, there exists $y\in X$ such that Ker($\chi$ ) $=$ Ker(ev$_y$) (th. 2 of I, p. 30). Let I = Ker($\chi$ ). It is a maximal ideal of B. Let Φ be the set of $x\in X$ such that $f(x) = 0$ for every $f\in I$. Let us prove that Φ is not empty. Otherwise, since X is compact, there would exist an integer $n\geqslant 1$, an open covering $(V_1, . . . ,V_n)$ of X and, for every integer $i$ such that $1\leqslant i\leqslant n$, a function $f_i\in I$ such that $f_i(x)\not= 0$ for every $x\in V_i$. Since the algebra B is an involutive subalgebra of $\mathscr{C}(X)$, the function

would belong to I. Now $f(x)>0$ for every $x\in X$, and therefore $f$ would be invertible in $\mathscr{C}(X)$. Since B is supposed to be a full subalgebra of $\mathscr{C}(X)$, the function $f\in$ I would be invertible in B, which is impossible. Consequently, the set Φ is not empty. Let $y$ be an element of Φ; the kernel of the character ev$_y$ contains I, and is therefore equal to I.