Proposition 19

Let $z$ denote the germ in $\mathscr{O}(S)$ of the identity mapping of $\mathbf{C}$. Then $\mathscr{O}_{\mathbf{R}}(S)$ is the smallest closed full sub-$\mathbf{R}$-algebra of $\mathscr{O}(S)$ containing $z$.

We have $z^*=z$; hence $z$ belong to $\mathscr{O}_{\mathbf{R}}(S)$. Let B be a full closed sub-$\mathbf{R}$-algebra of $\mathscr{O}(S)$ containing $z$. The mapping $f\mapsto f+f^*$ of $\mathscr{O}(S)$ into $\mathscr{O}_{\mathbf{R}}(S)$ is continuous and surjective, and the set of germs of rational functions holomorphic in a neighbourhood of S is dense in $\mathscr{O}(S)$ (Theorem 3 of I, p. 69). To prove that B contains $\mathscr{O}_{\mathbf{R}}(S)$, it therefore suffices to prove that if $f$ is the germ of such a rational function, then $f+f^*\in B$.

There exist polynomials P and Q in $\mathbf{C}[X]$ such that Q vanishes at no point of S and such that one has $f=\frac{P(z)}{Q(z)}$. Let $P^*$ and $Q^*$ denote the polynomials obtained by replacing the coefficients of P and Q by their conjugates. One then has $P(z)^*= P^*(z)$ and $Q(z)^*= Q^*(z)$. Since S is stable under complex conjugation, the polynomial $Q^*$ vanishes at no point of S. The germs $Q^*(z)$ and $(QQ^*)(z)$ are therefore invertible in $\mathscr{O}(S)$, and

$f+f^*=\frac{P(z)}{Q(z)}+\frac{P^*(z)}{Q^*(z)}=\frac{(PQ^*+P^*Q)(z)}{(QQ^*)(z)}$.

Since the polynomials $PQ^*+P^*Q$ and $QQ^*$ have real coefficients and B is a full sub-$\mathbf{R}$-algebra of $\mathscr{O}(S)$ containing $z$, the element $f+f^*$ belong to B. This concludes the proof of the proposition.

Let A be a complete normable unital algebra over $\mathbf{R}$. Let $x$ be an element of A. The spectrum of the element $1\otimes x$ of the algebra $A_{(\mathbf{C})}$ is called the complex spectrum of $x$, and is denoted by Sp$_{A_{(\mathbf{C})}}(x)$. Its intersection with the set $\mathbf{R}$ is none other than the spectrum Sp$_A(x)$ of $x$ relative to A, which is sometimes called the real spectrum of $x$. The complex spectrum Sp$_{A_{(\mathbf{C})}}(x)$ is a compact subset of $\mathbf{C}$, stable under complex conjugation. It is not empty when the algebra A is not reduced to 0.

Let $x$ be an element of A. The spectral radius of $1\otimes x\in A_{(\mathbf{C})}$ is equal to the spectral radius $\varrho (x)$ of $x$. It is the smallest real number $r\geqslant 0$ such that $|\lambda |\leqslant r$ for every $\lambda \in$ Sp$_{A_{(\mathbf{C})}}(x)$. One has

$\varrho (x) =$ lim$_{n\rightarrow+\infty}\|x^n\|^{1/n}=$ inf$_{n>0}\|x^n\|^{1/n}$

for every norm on A which defines the topology of A. In fact, one may suppose that the norm on A is the restriction of a norm on $A_{(\mathbf{C})}$ which defines the topology of $A_{(\mathbf{C})}$ and apply Prop. 1 of I, p. 20.

Let us denote by $u\mapsto \overline{u}$ the endomorphism of the $\mathbf{R}$-algebra $A_{(\mathbf{C})}$ which maps $\lambda \otimes a$ to $\overline{\lambda}\otimes a$. It is continuous.