Let $(X,\mathcal A, E, \mathcal H)$ be a spectral measure space. Then show that the map $J : \varphi \longmapsto J_{\varphi} = \int_X \varphi\ dE$ is an isometric isomorphism of the Banach algebra $L_{\infty} (X,E)$ with unit $1$ and involution $\varphi \mapsto \overline {\varphi}$ onto a commutative subalgebra of $\mathscr B (\mathcal H)$ with unit $I$ and involution $T \mapsto T^*.$
I can able to prove that $J$ is an algebra homomorphism and the range of $J$ is commutative. The part where I am struggling is to show that $J$ is an isometry. For simple functions I know the result to be true i.e. $$\left \|\int_X \varphi\ dE \right \| = \text {ess} \sup |\varphi|.$$ But how do I extend this result for any bounded $E$-measurable function on $X\ $? Any help would be appreciated.
Thanks for your time.
EDIT $:$ We know that simple functions are dense in $L_{\infty} (X,E).$ Now given any $f \in L_{\infty} (X,E)$ we can get hold of a sequence of simple functions $\{s_n\}_{n \geq 1}$ which converges to $f.$ Then by using isometry of $J$ on the space of all simple functions on $X$ we can show that $\{J(s_n)\}_{n \geq 1}$ is Cauchy and since $\mathscr B(\mathcal H)$ is complete it converges to a bounded linear operator on $\mathcal H.$ Moreover given any two sequences of simple functions $\{s_n\}_{n \geq 1}$ and $\{t_n\}_{n \geq 1}$ converging to $f$ we can prove (using linearity and isometry of $J$) that their images under $J$ converges to the same limit and this observations allow us to define $$J(f) : = \lim\limits_{n \to \infty} J(s_n)$$ where $\{s_n\}_{n \geq 1}$ is any sequence of simple functions converging to $f.$ In this case we have $$\|J(f)\| = \lim\limits_{n \to \infty} \|J(s_n)\| = \lim\limits_{n \to \infty} \text {ess} \sup |s_n| = \text {ess} \sup |f|.$$ This completes the proof.
Can anybody please verify my argument above? Thanks again.