> [!definition] > > Let $\ch$ be a [[Hilbert Space|Hilbert space]], and $T \in L(\ch, \ch)$ be a [[Normal Operator|normal operator]]. Let $\alg_T \subset L(\ch, \ch)$ be the subalgebra generated by $I$, $T$, and $T^*$, then the [[Gelfand Transform|Gelfand transform]] $\Gamma_{\alg_T} T: \sigma(\alg_T) \to \sigma(T)$ is a [[Homeomorphism|homeomorphism]]. Identify $\sigma(\alg_T)$ with $\sigma(T)$, then $\Gamma T$ is the identity map. > > Let $P_T$ be the [[Projection-Valued Measure|projection-valued measure]] [[Spectral Theorem|corresponding to]] $\alg_T$, then > $ > T = \angles{\Gamma T, P_T} = \angles{\text{Id}, P_T} > $ > Let $p = \sum_{j, k}^n c_{jk}\lambda^j\ol{\lambda}^k$ be a [[Ring of Polynomials|polynomial]] in $\lambda$ and $\ol \lambda$, then since $\angles{\cdot, P_T}$ is a $*$-homomorphism, > $ > \angles{p, P_T} = \sum_{j, k = 1}^n c_{jk} T^{j}T^{*k} > $ > given that $\angles{\cdot, P_T}$ is [[Continuity|continuous]] with respect to the $M(\sigma(T))$-weak topology on $B(\sigma(T))$ and the [[Operator Topologies|strong operator topology]], for each $f \in B(\sigma(T))$, define > $ > f(T) = \angles{f, P_T} > $ > as the **Borel functional calculus** for $T$. > [!theorem] > > There exists a unique $*$-homomorphism $P: B(\sigma(T)) \to L(\ch, \ch)$ defined by $f \to f(T)$ such that > 1. $\angles{\text{Id}, P} = T$. > 2. $P$ is continuous with respect to the $M(\sigma(T))$-weak topology on $B(\sigma(T))$ and the strong operator topology on $L(\ch, \ch)$. > > *Proof*. Integration against the projection-valued measure yields the desired homomorphism. On the other hand, let $P: B(\sigma(T)) \to L(\ch, \ch)$ be another such homomorphism. Since $P = P_T$ on the polynomials in $\lambda$ and $\ol \lambda$ by $(1)$ and being a $*$-homomorphism, $P = P_T$ on $B(\sigma(T))$ because the polynomials are dense in $B(\sigma(T))$ with respect to the $M(\sigma(T))$-weak topology. > [!theorem] > > Let $\alg$ be a commutative $C^*$-algebra containing $T$, then > $ > \angles{f, P_T} = \angles{f \circ \Gamma_\alg T, P_\alg} > $ > where $P_\alg$ is the projection-valued measure on $\sigma(\alg)$. > > *Proof*. If $f = \text{Id}$, then > $ > \angles{\text{Id} \circ \Gamma_\alg T, P_\alg} = \angles{\Gamma_\alg T, P_\alg} = T > $ > If $\seq{f_n}$ is uniformly bounded and converges pointwise to $f$, then > $ > \angles{f_n \circ \Gamma_{\alg}T, P_\alg} \to \angles{f \circ \Gamma_\alg T, P_\alg} > $ > so $\angles{\cdot, P_\alg} = \angles{\cdot, P_T}$ by the uniqueness part of the previous theorem. > [!theorem] > > Let $(\Omega, \cm, \mu)$ be the same as in the setting of [[Spectral Theorem|Spectral Theorem (II)]]. If $T \in L(L^2, L^2)$ is a linear operator with representation $\phi_T \in L^\infty$, with $\phi(\Omega) \subset \sigma(T)$, then $\angles{f, P_T}$ has representation $f \circ \phi$. > > *Proof*. Let $\mathcal U: B(\sigma(T)) \to L(\ch, \ch)$ be defined by > $ > f \mapsto U^{-1}(f \circ \phi)U > $ > where $(f \circ \phi)$ is identified with the operator it represents. Since $\mathcal U$ is linear and the representation map is a $*$-homomorphism, $\mathcal U$ is also a $*$-homomorphism $(1)$. If $f = \text{Id}$, then $\mathcal Uf = T$. Moreover, since pointwise and bounded convergence of $f$ implies the strong operator convergence of $f \circ \phi$, $\mathcal U$ satisfies $(2)$. By the uniqueness part of the previous theorem, $\mathcal U = \angles{\cdot, P_T}$.