Proposition 5

With the above notation:

a) The restriction mapping $h$ of $R_E(X_E)$ into $R_E(X)$ is an isometric isomorphism;

b) The algebra $R_E(X_E)$ is a full subalgebra of $\mathscr{C}(X_E)$;

c) Every character of $R_E(X_E)$ is of the form $f\mapsto f(w)$, where $w$ is an element of $X_E$;

d) The mapping $\chi \mapsto \chi (z)$ is a homeomorphism of $\mathsf{X}(R_E(X))$ onto $X_E$;

e) If $E'$ is a subset of $\mathbf{C}-$ X, one has $R_E(X) = R_{E'}(X)$ if and only if $X_E= X_{E'}$, which is also equivalent to $I(E) = I(E')$.

The restriction mapping $h$ of $R_E(X_E)$ into $R_E(X)$ is a morphism of Banach algebras such that $\|h(f)\|\leqslant \|f\|$ for every function $f\in R_E(X_E)$. Let $f\in R_E(X_E)$. Let $i\in I(E)$. Since $f$ is holomorphic in a neighbourhood of the bounded open set $O_i$, the maximum principle (VAR, I, p. 29) implies that there exists an element $z_0$ on the boundary of $O_i$ such that $|f(z_0)|=$ sup$_{z\in O_i}|f(z)|$. Since the boundary of $O_i$ is contained in X (lemma 1), it follows that sup$_{z\in O_i}|f(z)|\leqslant \|h(f)\|$. Since this inequality holds for every $i\in I(E)$, it follows that $\|f\|\leqslant \|h(f)\|$. The morphism $h$ is therefore isometric, and in particular injective. Let us now prove that it is surjective. Let $g\in R_E(X)$. There exists a sequence $(f_n)$ of rational functions whose poles belong to E which converges uniformly to $g$ on X. The $f_n|X_E$ are elements of $R_E(X_E)$. Since $h$ is isometric, the sequence $(f_n|X_E)$ converges in $\mathscr{C}(X_E)$. If $f$ is its limit, then $f\in R_E(X_E)$ and $g=f|X =h(f)$. Hence a).

Let us prove assertion d). Apply prop. 3 of I, p. 144 to the algebra $B = R_E(X)$. Assertion b) of loc. cit. implies that the mapping which to $\chi$ associates $\chi (z)$ is a homeomorphism of $\mathsf{X}(R_E(X))$ onto Sp$_{R_E(X)}(z)$. Let $z_E$ be the identical mapping of $X_E$. By loc. cit., a), we have Sp$_{R_E(X)}(z) =$ Sp$_{R_E(X_E)}(z_E)$. It is therefore enough to show that Sp$_{R_E(X_E)}(z_E) = X_E$. The cor. to prop. 6 of I, p. 28 proves that Sp$_{R_E(X_E)}(z_E)$ is the union of Sp$_{\mathscr{C}(X_E)}(z_E) = X_E$ and of certain bounded connected components of the complement of $X_E$. Let $O_i$ be one of the bounded connected components of the complement of $X_E$. The intersection $E\cap O_i$ is therefore nonempty. Let $\lambda \in E\cap O_i$; since $(\lambda -z_E)^{-1}\in R_E(X_E)$, we have $\lambda \notin$ Sp$_{R_E(X_E)}(z_E)$. Thus, $O_i$ is not contained in Sp$_{R_E(X_E)}(z_E)$. This shows that Sp$_{R_E(X_E)}(z_E) = X_E$.

Moreover, this equality implies that Sp$_{R_E(X_E)}(z_E) =$ Sp$_{\mathscr{C}(X_E)}(z_E)$. This establishes condition (iii) of prop. 11 of I, p. 41, applied to the canonical injection of $R_E(X_E)$ into $\mathscr{C}(X_E)$ and to the element $z_E$. Assertions b) and c) are the equivalent conditions (i) and (ii) of loc. cit.

Assertion d) shows that $X_E= X_{E'}$ if $R_E(X) = R_{E'}(X)$. Conversely, suppose that $X_E= X_{E'}$. Replacing $E'$ by $E\cup E'$, we may suppose that $E\subset E'$. By b), the algebra $R_E(X_E)$ is a closed full subalgebra of $\mathscr{C}(X_E)$, and hence also of $R_{E'}(X_E)$. It contains $z_E$, and therefore $R_E(X_E) = R_{E'}(X_E)$. Applying a), we deduce that $R_E(X) = R_{E'}(X)$. Finally, the equivalence of $X_E= X_{E'}$ and $I(E) = I(E')$ is a consequence of the definitions.