Definition 6.4.9 (Functional Calculus).label Using the same notation as Proposition 6.4.8. Let $f\in C(\spec{A}{x})$ then
\begin{align*}f(x)\define\p(f)\end{align*}
is well-defined and we have the following properties:
$(fg)(x)=f(x)g(x)$
$\cl{f}(x)=f(x)^{*}$
$\spec{A}{f(x)}=\curl{f(\lam):\lam\in\spec{A}{x}}$
$\norm{f(x)}{}=\SUP{\lam\in\spec{A}{x}}\abs{f(\lam)}$
If $g\in C(f(\spec{A}{x}))$ then
\begin{align*}(g\circ f)(x)=g(f(x))\end{align*}
Proof. If $\lam\nin\spec{A}{x}$ then $x-\lam$ is invertible in $B$ because $\lam-\indi{\spec{A}{x}}$ is invertible in $C(\spec{A}{x})$. Let $F$ be a holomorphic function on a neighborhood of $\spec{A}{x}$ and $F|_{\spec{A}{x}}=f$ then for any $\om\in\OM(B)$
\begin{align*}\inner{F(x)}{\om}&=\om(F(x))\\&=\frac{1}{2\pi i}\integral{C}{}\inner{(\lam-x)\inv}{\om}F(\lam)d\lam\\&=\frac{1}{2\pi i}\integral{C}{}\parens{\lam-\om(x)}\inv F(\lam)d\lam\\&=F(\om(x))=f(\om(x))=f(\psi(\om))\end{align*}
where $C$ is a smooth simple closed curve enclosing $\spec{A}{x}$. Observe that
\begin{align*}\scr{F}(\p(f))=f\circ \psi\implies\scr{F}(\p(f))(\om)=f\circ \psi(\om)=f(\om(x))\end{align*}
So we have for all $\om\in\OM(B)$
\begin{align*}\om(F(x))=\om(\p(f))\end{align*}
hence $F(x)=\p(f)$ and since $\p$ is a $*$-isomorphism we extend by density to $C(\spec{A}{x})$.$\square$