Continuous functional calculus for bounded normal operators

The algebra $C^{\ast }(I,T)$ is a nonzero unital commutative C*-algebra with the same identity as $\mathcal{B}(H)$, its character space $\mathrm{\Delta }$ is nonempty compact Hausdorff, $\mathrm{\Phi }:\mathrm{\Delta }\to \sigma (T)$ is a homeomorphism, and the pullback map $U(f):=f\circ \mathrm{\Phi }$ is an isometric bijection $C(\sigma (T))\to C(\mathrm{\Delta })$ that preserves pointwise sums, products and conjugation.

The inverse Gelfand transform ${\mathrm{\Gamma }}^{-1}:C(\mathrm{\Delta })\to C^{\ast }(I,T)$ is an isometric unital $\ast$-isomorphism onto $C^{\ast }(I,T)$.

The composite $\mathrm{\Psi }:={\mathrm{\Gamma }}^{-1}\circ U:C(\sigma (T))\to C^{\ast }(I,T)$ is an isometric unital $\ast$-isomorphism onto $C^{\ast }(I,T)$, and $\mathrm{\Psi }(z)={\mathrm{\Gamma }}^{-1}(\widehat{T})=T$ because the Gelfand transform of $T$ is $\widehat{T}(\chi )=\chi (T)=\mathrm{\Phi }(\chi )=z(\mathrm{\Phi }(\chi ))=(z\circ \mathrm{\Phi })(\chi )$.

$\mathrm{\Psi }$ is the only isometric unital $\ast$-isomorphism $C(\sigma (T))\to C^{\ast }(I,T)$ with $z\mapsto T$: if $\mathrm{\Xi }$ is another, then ${\mathrm{\Xi }}^{-1}\circ \mathrm{\Psi }$ is a unital $\ast$-isomorphism of $C(\sigma (T))$ fixing $z$ and $\bar{z}$, hence fixing every $\ast$-polynomial in $z$; these are uniformly dense by Stone–Weierstrass and the map is isometric, so it is the identity on $C(\sigma (T))$ and $\mathrm{\Xi }=\mathrm{\Psi }$.

For self-adjoint $T$ the self-adjoint calculus of [A5] is an isometric unital star-homomorphism $C(\sigma (T))\to \mathcal{B}(H)$ with $z\mapsto T$ and range $C^{\ast }(I,T)$; regarded as a map onto $C^{\ast }(I,T)$ it is an isometric unital $\ast$-isomorphism, so step 3.1 identifies it with $\mathrm{\Psi }$.

The map $f\mapsto f(T):=\mathrm{\Psi }(f)$ is therefore the unique isometric unital star-isomorphism $C(\sigma (T))\to C^{\ast }(I,T)$ sending $z$ to $T$, and it agrees with the self-adjoint calculus when $T$ is self-adjoint.