Proof. Let $r(x)=\limitinf{n\to\infty}\norm{x^n}{}^{1/n}$ then the power series $-\sums{\infty}{n=0}\frac{x^{n}}{\lam^{n+1}}$ converges for $\abs{\lam}>r(x)$ and this is the inverse $(x-\lam)\inv$. Hence

Let $f(\lam)=(x-\lam)\inv$ for $\lam\nin\spec{A}{x}$. By Proposition 6.1.17 with $x\bij x-\lam$ and $x_{0}\bij x-\lam_{0}$ then if $\abs{(x-\lam)-(x-\lam_0)}<\norm{f(\lam_0)}{}\inv$ for $\lam_{0}\nin\spec{A}{x}$ and $\lam\in\com$ then $\lam\in B\parens{\lam_0,\norm{f(\lam_0)}{}\inv}\suf G(A)$ thus $\lam\nin \spec{A}{x}$ and

Hence $f(\lam)$ is a holomorphic function on the resolvent of $x$. The power series $\sums{\infty}{n=0}\frac{x^{n}}{\lam^{n+1}}$ is the Laurent expansion of $f(\lam)$ at $\infty$. Let $\p$ be a bounded linear functional of $A$ then

is a holomorphic function on the resolvent of $x$, where $\inner{a}{\p}$ denotes the value of $\p$ at $a\in A$. The function $f_{\p}(\lam)$ is holomorphic for $\abs{\lam}>\norm{x}{sp};$ hence its Laurent expansion $\sums{\infty}{n=0}\frac{\inner{x^n}{\p}}{\lam^{n+1}}$ converges for $\abs{\lam}>\norm{x}{sp}$. Hence we have

By uniform boundedness theorem, the sequence $\curl{\frac{x^{n}}{\lam^{n+1}}}$ is bounded for $\abs{\lam}>\norm{x}{sp}$ so that for any $\abs{\lam}>\norm{x}{sp}$, there exists $\al>0$ such that $\norm{x^n}{}\leq\al\abs{\lam}^{n+1}$, $n=1,2,...$. Therefore, we have

Thus

$\square$