Borel functional calculus for a bounded normal operator

Assume AC. Let $T$ be a bounded normal operator on a nonzero complex Hilbert space $H$ and let $E$ be its spectral projection valued measure on the Borel $\sigma$-algebra of $\sigma (T)$, the unique regular projection valued measure with $\int zdE(z)=T$ (Spectral theorem for bounded normal operators pvm form). For a bounded Borel function $f:\sigma (T)\to \mathbb{C}$, that is a bounded $\mathrm{\Sigma }$-measurable function on the Borel $\sigma$-algebra of $\sigma (T)$ (A measurable function between measurable spaces), define the Borel functional calculus of $T$ at $f$ by

$f(T):={\mathrm{\Phi }}_E(f)={\int }_{\sigma (T)}fdE,$

the bounded Borel integral of $f$ against $E$ from Bounded borel pvm integral. The map $f\mapsto f(T)$ from bounded Borel functions on $\sigma (T)$ to $\mathcal{B}(H)$ is the bounded Borel functional calculus of $T$.

Well-definedness and consistency. The operator $f(T)$ is well defined because the spectral projection valued measure $E$ of $T$ is unique: if $E^{\mathrm{\prime }}$ were another regular projection valued measure with $\int zd{E}^{\mathrm{\prime }}=T$, then $E^{\mathrm{\prime }}=E$. The construction agrees with the continuous calculus on continuous functions, ${\mathrm{\Phi }}_E(f)=f(T)$ for $f\in C(\sigma (T))$, by the displayed clause of the spectral theorem. It inherits the algebraic behaviour of the projection valued measure integral: $f(T)$ is complex-linear in $f$, unital with $\mathbf{1}(T)=I$, multiplicative, star-preserving with $\bar{f}(T)=f(T{)}^{\ast }$, norm bounded by $\mathrm{\parallel }f(T)\mathrm{\parallel }\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$, strongly continuous for bounded Borel $f_n,f$ when $f_n\to f$ pointwise $E$-almost everywhere and ${\sup }_n\mathrm{\parallel }{f}_n{\mathrm{\parallel }}_{\mathrm{\infty }}<\mathrm{\infty }$, as supplied by Pvm integral is a star homomorphism. In particular ${\mathbf{1}}_B(T)=E(B)$ for every Borel set $B\subseteq \sigma (T)$, and the norm of $f(T)$ is the $E$-essential supremum of $f$ (Bounded borel pvm integral).

Commutation. Every bounded $S$ commuting with $T$ and $T^{\ast }$ commutes with every $f(T)$. Write $\pi (g)=g(T)$ for the continuous calculus. The construction in Continuous functional calculus produces a regular PVM provides finite regular complex measures ${\mu }_{x,y}$ with ${\mu }_{x,y}(B)=⟨E(B)x,y⟩$ and $\int gd{\mu }_{x,y}=⟨\pi (g)x,y⟩$ for continuous $g$; its PVM is the present $E$ by uniqueness. The continuous commutant property (Continuous functional calculus properties) and the defining adjoint identity (The Hilbert-space adjoint of a bounded operator) give $\int gd{\mu }_{Sx,y}=⟨\pi (g)Sx,y⟩=⟨S\pi (g)x,y⟩=⟨\pi (g)x,S^{\ast }y⟩=\int gd{\mu }_{x,S^{\ast }y}\mathrm{.}$ Uniqueness of the finite regular complex representing measure on the compact spectrum, where $C_0(\sigma (T))=C(\sigma (T))$, gives ${\mu }_{Sx,y}={\mu }_{x,S^{\ast }y}$ (The bounded complex dual of C_0(X) is regular complex measures). For bounded Borel $f$, the bounded-integral pairing identity therefore yields $⟨f(T)Sx,y⟩=\int fd{\mu }_{Sx,y}=\int fd{\mu }_{x,S^{\ast }y}=⟨Sf(T)x,y⟩\mathrm{.}$ Testing the difference against itself proves $f(T)S=Sf(T)$. These properties are collected in Borel functional calculus for bounded normal operators.