Let $\ch$ be a [[Hilbert Space|Hilbert space]], and $T \in L(\ch, \ch)$ be a [[Normal Operator|normal operator]]. Let $\alg \subset_T L(\ch, \ch)$ 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 $\sigma(\alg_t) = \sigma(T)$, then there exists a unique $*$-homomorphism $f \mapsto f(T)$ from the [[Space of Bounded Borel Functions|bounded Borel functions]] $B(\sigma(T))$ to $L(\ch, \ch)$ such that 1. $f(T) = T$ when $f(\lambda) = \lambda$. 2. $f \mapsto f(T)$ is continuous with respect to the weak topology on $B(\sigma(T))$ and the [[Operator Topologies|strong operator topology]] on $L(\ch, \ch)$. 3. If $\alg \subset L(\ch, \ch)$ is a commutative $C^*$-subalgebra containing $T$, and $P_{\alg'}$ is the associated [[Spectral Theorem|projection-valued measure]], then $f(T) = \angles{f \circ \Gamma_{\alg}, P_\alg}$. 4. If $(\Omega, \cm, \mu)$ is the measure space associated with $\alg_T$, and $T$ has representation $\phi_T \in L^\infty(\mu)$, then $f(T)$ has representation $f \circ \phi$. 5. If $S \in L(\ch, \ch)$ commutes with $T$ and $T^*$, then $S$ commutes with $f(T)$ for all $f \in B(\sigma(T))$. *Proof*.