Let us set some notation first:

• $\sigma(T)$ will denote the spectrum of $T$.

• $C(\sigma(T))$ will denote the $C^{*}$-algebra of continuous functions $\sigma(T)\to\mathbb{C}$.

• $B(\sigma(T))$ will denote the $C^{*}$-algebra of bounded Borel functions $\sigma(T)\to\mathbb{C}$, endowed with the sup norm.

The continuous functional calculus for $T$ allows the expression $f(T)$ to make sense for continuous functions $f\in C(\sigma(T))$, by the assignment of a unital *-homomorphism

that sends the identity function to $T$. This unital *-homomorphism is in fact uniquely determined by this property (see the entry on the continuous functional calculus for more details).

The objective is to extend $\pi$ to a unital *-homomorphism $\widetilde{\pi}:B(\sigma(T))\longrightarrow B(H)$.

Since $B(\sigma(T))$ is a much larger $C^{*}$-algebra than $C(\sigma(T))$, there is no reson to presume that there is only one extension of $\pi$. Which extension would be the most natural then? It turns out that there is a unique extension that satisfies a good continuity property.

It is known that *-homomorphisms between $C^{*}$-algebras are continuous (see this entry), so that whenever a net $f_{i}\in B(\sigma(T))$ converges in the sup norm to a function $f\in B(\sigma(T))$ we will have that $f_{i}(T)\to f(T)$ in the operator norm. All extensions of $\pi$ will automaticaly satisfy this continuity property, but this can be improved in a satisfactory manner.

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Notation - Let $X$ be a compact Hausdorff space, $M(X)$ the space of all finite regular Borel measures in $X$ and $B(X)$ the $C^{*}$-algebra of all bounded Borel functions in $X$. The weakest topology in $B(X)$ for which integration against any measure $\nu$ is continuous will be reffered to as the $\mu$-topology. This means that $f_{i}\to f$ in the $\mu$-topology if and only if $\int f_{i}d\nu\to\int fd\nu$ for all $\nu\in M(X)$.

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Notice that we can identify each function $f\in B(X)$ with a bounded linear functional $\omega_{f}$ in $M(X)$, given by

and the $\mu$-topology corresponds exactly to the weak-* topology under this identification.

We will see in the next section that there is an unique extension of $\pi$ that is continuous from the $\mu$-topology to the weak operator topology.

Just like the Stone-Weierstrass theorem allowed the passage from the polynomial functional calculus to the continuous functional calculus, the Riesz representation theorem will allow the passage from the latter to the Borel functional calculus.

The following result is the key for the definition of the Borel functional calculus.

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Theorem 1 - Let $T$ be a normal operator in $B(H)$ and $\pi:C(\sigma(T))\longrightarrow B(H)$ the unital *-homomorphism corresponding to the continuous functional calculus for $T$. Then, $\pi$ extends uniquely to a *-homomorphism $\widetilde{\pi}:B(\sigma(T))\longrightarrow B(H)$ that is continuous from the $\mu$-topology to the weak operator topology. Moreover, each operator $\pi(f)$ lies in strong operator closure of the unital *-algebra generated by $T$.

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Proof: See this attached entry

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We are now able to define the Borel functional calculus:

Definition - Let $T$ be a normal operator in $B(H)$. Let $\widetilde{\pi}:B(\sigma(T))\longrightarrow B(H)$ be the unique *-homomorphism defined in Theorem 1. This *-homomorphism is denoted by

and it is called the Borel functional calculus for $T$.

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Since this functional calculus extends the polynomial functional calculus, we have that for any polynomial $p(z):=\sum c_{{n,m}}z^{n}\overline{z}^{m}$,

Moreover, since $f(T)$ lies in the strong operator closure of the unital *-algebra generated by $T$, for any function $f\in B(\sigma(T))$, we see that $f(T)$ is the strong operator limit of polynomials $\sum c_{{n,m}}T^{n}(T^{*})^{m}$.

The Borel functional calculus is in fact applicable for any normal operator $T$ in any von Neumann algebra $\mathcal{M}$.

That is due to the fact, expressed in Theorem 1, that for every $f\in B(\sigma(T))$ the operator $f(T)$ belongs to the strong operator closure of the unital *-algebra generated by $T$. Being a von Neumann algebra, $\mathcal{M}$ is closed in the strong operator topology, and therefore all operators $f(T)$ belong to $\mathcal{M}$.

Thus, by restriction, we have in fact a *-homomorphism

satisfying the properties of Theorem 1, i.e. we have a Borel functional calculus for normal operators of a von Neumann algebra.