Proposition 11

Let A be a unital Banach algebra, $a\in A$ and U an open neighbourhood of Sp$_A(a)$. Let $\delta$ denote the distance from Sp$_A(a)$ to $\mathbf{C}-$ U. Let $f\in \mathscr{O}(U)$.

a) For every real number $\eta$ such that $0< \eta < \delta$, there exists a real number $C\geqslant 0$ such that $\|f^{(n)}(a)\|\leqslant Cn$!$\eta^{-n}$ for every integer $n\in \mathbf{N}$;

b) If $b\in A$ commutes with $a$ and if $\varrho (b)< \delta$, one has Sp$_A(a+b)\subset U$, and

$\infty (n)$

$f(a+b) =\sum fn$!$(a)b^n$,

$n=0$

where the series converges absolutely.

Let $\eta$ be a real number such that $0< \eta < \delta$. Put $\varepsilon =\delta -\eta >0$. Let K be the compact neighbourhood of Sp$_A(a)$ formed by the points of $\mathbf{C}$ whose distance from Sp$_A(a)$ is $\leqslant \varepsilon /2$. Since $f$ is holomorphic in every open disk of radius $\eta +\varepsilon /2$ whose center belongs to K, there exists, by Cauchy's inequalities (VAR, R1, p. 29, 3.3.4), a real number $C\geqslant 0$ such that

sup$_{z\in K}|f^(\overset{n)}{n}$!$(z)|\leqslant \frac{C}{\eta^n}$

for every integer $n\geqslant 0$. Then assertion a) results from Prop. 1 of I, p. 61 applied to $f^{(n)}$ and to a piece V contained in K.

Let $b$ be an element of A commuting with $a$ such that $\varrho (b)< \delta$. Replacing A by the full closed subalgebra B generated by $a$ and $b$, which satisfies Sp$_A(a) =$ Sp$_B(a)$ and Sp$_A(a+b) =$ Sp$_B(a+b)$, one is reduced, in order to prove b), to the case where A is commutative.

Since $\varrho (b)< \delta$, one can choose $\eta$ such that $\varrho (b)< \eta < \delta$. Let $V_1$ be the set of points of $\mathbf{C}$ whose distance from Sp$_A(a)$ is $< \delta -\eta$, and $V_2$ the open disk with center 0 and radius $\eta$ in $\mathbf{C}$. Let $g$ be the mapping $(z_1, z_2)\mapsto z_1+z_2$ of $V_1\times V_2$ into U. Then $h=f\circ g$ is the mapping $(z_1, z_2)\mapsto f(z_1+z_2)$ of $V_1\times V_2$ into $\mathbf{C}$. One has Sp$^2_A(a, b)\subset V_1\times V_2$, hence Sp$_A(a+b)\subset U$ (cf. cor. 2 of I, p. 67), and moreover $f(a+b) =h(a, b)$ by th. 4 of I, p. 72. Now, in the space $\mathscr{O}(V_1\times V_2)$, one has

$h(z_1, z_2) =\sum_{n\geqslant 0}f^{(n}\overset{)}{n}($!$z_1)z_2^n$,

(VAR, R1, p. 29, 3.3.4) hence the series

$\sum f^(\overset{n)}{n}$!$(a)b^n$

$n\geqslant 0$ converges in A and its sum is $h(a, b) =f(a+b)$. Moreover, this series is absolutely convergent by assertion a).