Proposition 10

Let A be a unital involutive algebra. Let U be a relatively compact open subset of $\mathbf{C}$. Let $\Omega_n$ denote the set of normal elements of A such that Sp$_A(a)$ is contained in U. The mapping $(f, a)\mapsto f(a)$ of $\mathscr{C}(U)\times \Omega_n$ into A is continuous.

Let $q$ denote the mapping $(f, a)\mapsto f(a)$ of $\mathscr{C}(U)\times \Omega_n$ into A. The set of functional-calculus mappings $f\mapsto f(a)$ for $a\in \Omega_n$ is equicontinuous in $\mathscr{L}(\mathscr{C}(U); A)$, since each is a continuous linear mapping of norm $\leqslant 1$. To prove the assertion, it is therefore enough to verify that, for every $f\in \mathscr{C}$ (U), the mapping of $\Omega_n$ into A defined by $a\mapsto f(a)$ is continuous (TG, X, p. 13, cor. 3).

Let $\mathscr{A}$ be the set of $f\in \mathscr{C}(U)$ such that the mapping $a\mapsto f(a)$ of $\Omega_n$ into A is continuous; it remains to prove that $\mathscr{A}=\mathscr{C}(U)$.

The set $\mathscr{A}$ is a unital involutive subalgebra of $\mathscr{C}(U)$. It contains the identity function of U, and therefore separates points. Consequently, it is dense in $\mathscr{C}(U)$ (TG, X, p. 39, prop. 7). Let us prove that it is closed.

Let $(f_n)$ be a sequence in $\mathscr{A}$ converging to $f\in \mathscr{C}(U)$. Let $\varepsilon >0$ and choose $n\in \mathbf{N}$ such that $\|f-f_n\|_{\infty}\leqslant \varepsilon /4$. For every $(a_1, a_2)\in \Omega^2_n$ we have

2

Since $f_n$ belongs to $\mathscr{A}$, there exists a neighbourhood V of $a_1$ in $\Omega_n$ such that $\|f_n(a_1)-f_n(a_2)\|\leqslant \varepsilon /2$ for every $a_2\in$ V. Hence $\|f(a_1)-f(a_2)\|\leqslant \varepsilon$ for every $a_2\in V$. The mapping $a\mapsto f(a)$ is therefore continuous at $a_1$; the assertion is proved.