Spectral theorem for bounded normal operators pvm form

Existence: $\sigma (T)$ is a nonempty compact subset of $\mathbb{C}$ and $\pi (f):=f(T)$ is a unital star-homomorphism $C(\sigma (T);\mathbb{C})\to \mathcal{B}(H)$; applying the construction lemma to $K=\sigma (T)$ gives a regular PVM $E$ on the Borel $\sigma$-algebra of $\sigma (T)$ with ${\mathrm{\Phi }}_E(f)=\pi (f)=f(T)$ for every continuous $f$; taking $f$ to be the coordinate function $z$ gives $\int zdE=\pi (z)=T$.

Converse, boundedness and normality: for a regular PVM $E$ on a nonempty compact $\mathrm{\Lambda }$ the coordinate function is bounded by ${\max }_{z\in \mathrm{\Lambda }}\mathrm{\mid }z\mathrm{\mid }$ and measurable, so $T_E={\mathrm{\Phi }}_E(z)$ is a bounded operator with $\mathrm{\parallel }{T}_E\mathrm{\parallel }\le {\max }_{z\in \mathrm{\Lambda }}\mathrm{\mid }z\mathrm{\mid }$, its adjoint is $T_E^{\ast }={\mathrm{\Phi }}_E(\bar{z})$ by conjugation preservation, and $T_E{T}_E^{\ast }={\mathrm{\Phi }}_E(z\bar{z})={\mathrm{\Phi }}_E(\bar{z}z)=T_E^{\ast }{T}_E$ by multiplicativity, so $T_E$ is normal.

Converse, spectrum: if $\lambda \in \mathbb{C}\setminus \mathrm{\Lambda }$ then $\mathrm{\Lambda }$ is closed and $g(z):=(\lambda -z{)}^{-1}$ is continuous on $\mathrm{\Lambda }$ with $(\lambda -z)g(z)=1$; hence $(\lambda I-T_E){\mathrm{\Phi }}_E(g)={\mathrm{\Phi }}_E(\lambda \mathbf{1}-z){\mathrm{\Phi }}_E(g)={\mathrm{\Phi }}_E((\lambda -z)g)={\mathrm{\Phi }}_E(\mathbf{1})=I$ and likewise ${\mathrm{\Phi }}_E(g)(\lambda I-T_E)=I$, so $\lambda I-T_E$ has a two-sided bounded inverse and $\lambda \mathrm{\notin }\sigma (T_E)$; therefore $\sigma (T_E)\subseteq \mathrm{\Lambda }$.

Uniqueness for the direct statement: if $E^{\mathrm{\prime }}$ is a regular PVM on the Borel $\sigma$-algebra of $\sigma (T)$ with $\int zd{E}^{\mathrm{\prime }}=T$, then for every $\ast$-polynomial $p(z,\bar{z})$ one has ${\mathrm{\Phi }}_{E^{\mathrm{\prime }}}(p)=p(T_E^{\mathrm{\prime }},T_E^{\mathrm{\prime }\ast })=p(T,T^{\ast })=p(T)={\mathrm{\Phi }}_E(p)$, using multiplicativity and conjugation preservation of ${\mathrm{\Phi }}_{E^{\mathrm{\prime }}}$ together with $T_E^{\mathrm{\prime }}=T$ and the identification of $T^{\ast }$ with the calculus value of $\bar{z}$; since $\ast$-polynomials are uniformly dense in $C(\sigma (T))$ and both ${\mathrm{\Phi }}_{E^{\mathrm{\prime }}}$ and ${\mathrm{\Phi }}_E$ are bounded linear maps agreeing there, they agree on every continuous function, so $E^{\mathrm{\prime }}=E$ by the uniqueness clause of the construction lemma.

The spectral PVM $E$ of $T$ therefore exists, is unique among regular PVMs on $\sigma (T)$ whose coordinate integral is $T$, and satisfies ${\mathrm{\Phi }}_E(f)=f(T)$ for all continuous $f$; conversely every coordinate integral of a regular PVM on a compact set is a bounded normal operator with spectrum inside that set.