# proof of Borel functional calculus

In this entry we give a proof of the main result about the Borel functional calculus ( 1 on the parent entry). We will restate here the result for convenience. Please, check the parent entry for the details on notation.

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 (http://planetmath.org/OperatorTopologies) closure of the unital *-algebra generated by $T$.

$\,$

Proof : First we shall prove the existence of the extension (http://planetmath.org/ExtensionOfAFunction) $\widetilde{\pi}$ with the described continuity property, then we shall prove its uniqueness and for last we shall prove the last assertion of the theorem about the image of $\widetilde{\pi}$. For simplicity the proofs of some auxiliary results are given at the end of the entry

## 0.0.1 Existence

For each pair of vectors $\xi,\eta\in H$ consider the linear functional $\phi_{\xi,\eta}:C(\sigma(T))\to\mathbb{C}$ given by

 $\displaystyle\phi_{\xi,\eta}(f):=\langle\pi(f)\xi,\eta\rangle$

This linear functional is bounded with norm at most $\|\xi\|\|\eta\|$ because

 $\displaystyle\|\phi_{\xi,\eta}(f)\|\leq\|\pi\|\,\|f\|\,\|\xi\|\|\eta\|=\|f\|\,% \|\xi\|\|\eta\|$

where the last equality comes from the fact that $\pi$ is a *-isomorphism between $C^{*}$-algebras (http://planetmath.org/CAlgebras), therefore having norm 1 (see this entry (http://planetmath.org/HomomorphismsOfCAlgebrasAreContinuous)).

By the Riesz representation theorem (http://planetmath.org/RieszRepresentationTheoremOfLinearFunctionalsOnFunctionSpaces) there is a unique complex Radon measure $\mu_{\xi,\eta}$ in $\sigma(T)$ such that

 $\displaystyle\langle\pi(f)\xi,\eta\rangle=\int_{\sigma(T)}f\,d\mu_{\xi,\eta}\,% ,\qquad\qquad\forall f\in C(\sigma(T))$ (1)

Consider now the mapping $(\xi,\eta)\mapsto\mu_{\xi,\eta}$. This mapping has the following properties, whoose proofs are given at the end of the entry (section Auxiliary Results):

• a) $\mu_{\lambda_{1}\xi_{1}+\lambda_{2}\xi_{2},\eta}=\lambda_{1}\mu_{\xi_{1},\eta}% +\lambda_{2}\mu_{\xi_{2},\eta}$, for all $\lambda_{1},\lambda_{2}\in\mathbb{C}$.

• b) $\mu_{\xi,\lambda_{1}\eta_{1}+\lambda_{2}\eta_{2}}=\overline{\lambda_{1}}\mu_{% \xi,\eta_{1}}+\overline{\lambda_{2}}\mu_{\xi,\eta_{2}}$, for all $\lambda_{1},\lambda_{2}\in\mathbb{C}$.

• c) $\mu_{\xi,\eta}=\overline{\mu_{\eta,\xi}}$.

• d) $g\cdot\mu_{\xi,\eta}=\mu_{\pi(g)\xi,\eta}$, for all $g\in C(\sigma(T)$.

Therefore, for each function $f\in B(\sigma(T))$ we have a sesquilinear form in $H$ given by

 $\displaystyle[\xi,\eta]:=\int_{\sigma(T)}f\,d\mu_{\xi,\eta}$

Moreover, this sesquilinear form is bounded (http://planetmath.org/BoundedSesquilinearForm) with norm at most $\|f\|\,\|\xi\|\|\eta\|$. By the Riesz lemma on bounded sesquilinear forms, there is a unique operator $\widetilde{\pi}(f)\in B(H)$ such that

 $\displaystyle\langle\widetilde{\pi}(f)\xi,\eta\rangle:=\int_{\sigma(T)}f\,d\mu% _{\xi,\eta}\,,\qquad\qquad\xi,\eta\in H$

We will now see that the mapping $\widetilde{\pi}:B(\sigma(T)\to B(H)$, such that $f\mapsto\widetilde{\pi}(f)$, has the desired properties stated in the theorem.

First, it is clear that $\widetilde{\pi}$ is linear. Also clear is the fact that $\widetilde{\pi}$ coincides with $\pi$ in $C(\sigma(T))$, because of equality (1) and the uniqueness part of Riesz lemma on sesquilinear forms. Now, for any real valued function $f\in B(\sigma(T))$ we have that

 $\displaystyle\langle\widetilde{\pi}(f)^{*}\xi,\eta\rangle$ $\displaystyle=$ $\displaystyle\overline{\langle\widetilde{\pi}(f)\eta,\xi\rangle}$ $\displaystyle=$ $\displaystyle\overline{\int_{\sigma(T)}f\,d\mu_{\eta,\xi}}$ $\displaystyle=$ $\displaystyle\int_{\sigma(T)}f\,d\mu_{\xi,\eta}$ $\displaystyle=$ $\displaystyle\langle\pi(f)\xi,\eta\rangle$

which means that $\widetilde{\pi}(f)^{*}=\widetilde{\pi}(f)$, i.e. $\widetilde{\pi}(f)$ is self-adjoint. Decomposing an arbitrary function $f\in B(\sigma(T))$ in its real and imaginary parts we see that $\widetilde{\pi}(f)^{*}=\widetilde{\pi}(\overline{f})$.

We now show that $\widetilde{\pi}$ is multiplicative, i.e. $\widetilde{\pi}(fg)=\widetilde{\pi}(f)\widetilde{\pi}(g)$ for all $f,g\in B(\sigma(T))$. For that we need an additional property of the measures $\mu_{\xi,\eta}$, whoose proof is also at the end of the entry:

• e) $f\cdot\mu_{\xi,\eta}=\mu_{\xi,\widetilde{\pi}(f)^{*}\eta}$, for all $f\in B(\sigma(T))$.

Given $f,g\in B(\sigma(T))$ we have, for every $\xi,\eta\in H$,

 $\displaystyle\langle\widetilde{\pi}(fg)\xi,\eta\rangle$ $\displaystyle=$ $\displaystyle\int_{\sigma(T)}fg\,d\mu_{\xi,\eta}=\int_{\sigma_{(}T)}g\,d\mu_{% \xi,\widetilde{\pi}(f)*\eta}$ $\displaystyle=$ $\displaystyle\langle\widetilde{\pi}(g)\xi,\widetilde{\pi}(f)^{*}\eta\rangle=% \langle\widetilde{\pi}(f)\widetilde{\pi}(g)\xi,\eta\rangle$

and therefore $\widetilde{\pi}(fg)=\widetilde{\pi}(f)\widetilde{\pi}(g)$. Thus, $\widetilde{\pi}:B(\sigma(T))\to B(H)$ is a *-homomorphism that extends $\pi$.

## 0.0.2 Continuity Property

We now prove that the above defined $\widetilde{\pi}$ is continuous from the $\mu$-topology to the weak operator topology.

Let $\{f_{i}\}$ be a net of functions in $B(\sigma(T))$ that converge in the $\mu$-topology to a function $f\in B(\sigma(T))$. This means that for all Radon measures $\nu$ in $\sigma(T)$ we have $\int f_{i}\,d\nu\to\int f\,d\nu$.

Now for all $\xi,\eta\in H$ we have

 $\displaystyle|\langle\widetilde{\pi}(f_{i}-f)\xi,\eta\rangle|=\big{|}\int_{% \sigma(T)}f_{i}-f\,d\mu_{\xi,\eta}\big{|}\to 0$

Hence, $\widetilde{\pi}(f_{i})$ converges to $\widetilde{\pi}(f)$ in the weak operator topology.

## 0.0.3 Uniqueness

Let $\pi^{\prime}:B(\sigma(T))\to B(H)$ be another *-homomorphism that extends $\pi$ and is continuous from the $\mu$-topology to the weak operator topology. For any measurable subset $S\subset\sigma(T)$ consider the set

 $\displaystyle W:=\{(U,K):U\supset S\text{is open and}K\subset S\text{is % compact}\}$

We give this set the partial order $\leq$ such that $(U_{1},K_{1})\leq(U_{2},K_{2})$ whenever $U_{2}\subset U_{1}$ and $K_{1}\subset K_{2}$. For any pair $(U,K)\in W$ there is a continuous function $f_{U,K}\in C(\sigma(T))$ such that $f$ takes values on the interval $[0,1]$, $f|_{K}=1$ and $\mathrm{supp}f\subset U$ (see this entry (http://planetmath.org/ApplicationsOfUrysohnsLemmaToLocallyCompactHausdorffSpaces)).

We claim that $f_{U,K}$ converges to $\chi_{S}$ in the $\mu$-topology. In fact, given a complex Radon measure $\nu$ in $\sigma(T)$, there is for every $\epsilon>0$ a pair $(U_{0},K_{0})\in W$ such that $|\nu|(U_{0}\setminus K_{0})<\epsilon$. Of course, for all pairs $(U,K)$ such that $(U_{0},K_{0})\leq(U,K)$ we also have $|\nu|(U\setminus K)<\epsilon$. Hence, we have

 $\displaystyle\big{|}\int_{\sigma(T)}f_{U,K}-\chi_{S}\,d\nu\big{|}\leq\int_{U% \setminus K}|f_{U,K}-\chi_{S}|\,d|\nu|\leq\epsilon$

We conclude that $f_{U,K}$ converges to $\chi_{S}$ in the $\mu$-topology.

Since $\pi^{\prime}$ is continuous from the $\mu$-topology to the weak operator topology we must have

 $\displaystyle\langle\pi^{\prime}(f_{U,V})\xi,\eta\rangle\longrightarrow\langle% \pi^{\prime}(\chi_{S})\xi,\eta\rangle$

But since $\pi^{\prime}$ and $\widetilde{\pi}$ coincide with $\pi$ on $C(\sigma(T))$ we also have

 $\displaystyle\langle\pi^{\prime}(f_{U,V})\xi,\eta\rangle=\langle\widetilde{\pi% }(f_{U,V})\xi,\eta\rangle\longrightarrow\langle\widetilde{\pi}(\chi_{S})\xi,\eta\rangle$

Hence, for any characteristic function $\chi_{S}$ we have $\pi^{\prime}(\chi_{S})=\widetilde{\pi}(\chi_{S})$. Since any function $f\in B(\sigma(T))$ can be uniformly approximated by simple functions it follows that $\pi^{\prime}(f)=\widetilde{\pi}(f)$, and we have proved the uniqueness of $\widetilde{\pi}$.

## 0.0.4 Image of $\widetilde{\pi}$

Let $\mathcal{A}$ be the unital *-algebra generated by $T$. We now prove that for any $f\in B(\sigma(T))$, the operator $\widetilde{\pi}(f)$ lies in the strong operator closure of $\mathcal{A}$, i.e. lies in the von Neumann algebra generated by $T$. For that it is enough to prove that $\widetilde{\pi}(f)$ is in the double commutant $\mathcal{A}^{\prime\prime}$ of $\mathcal{A}$.

Recall from the continuous functional calculus that $\pi(f)$ is in the norm closure of $\mathcal{A}$, and hence in $\mathcal{A}^{\prime\prime}$, for every $f\in C(\sigma(T))$.

We have seen above that for each characteristic function $\chi_{S}$ there is a net $f_{U,K}$ of functions in $C(\sigma(T))$ such that $\widetilde{\pi}(f_{U,K})\to\widetilde{\pi}(\chi_{S})$ in the weak operator topology. Given an element $R$ in the commutant of $\mathcal{A}$ we have

 $\displaystyle\langle\widetilde{\pi}(f_{U,K})R\;\xi,\eta\rangle=\langle R\,% \widetilde{\pi}(f_{U,K})\;\xi,\eta\rangle\,,\qquad\qquad\forall\xi,\eta\in H$

The first term converges to $\langle\widetilde{\pi}(\chi_{S})R\;\xi,\eta\rangle$, whereas the second to $\langle R\,\widetilde{\pi}(\chi_{S})\;\xi,\eta\rangle$. Thus, $\widetilde{\pi}(\chi_{S})R=R\,\widetilde{\pi}(\chi_{S})$, and therefore $\widetilde{\pi}(\chi_{S})\in\mathcal{A}^{\prime\prime}$.

Since every function $f\in B(\sigma(T))$ can be uniformly approximated by simple functions, it follows that $\widetilde{\pi}(f)\in\mathcal{A}^{\prime\prime}$.

## 0.0.5 Auxiliary Results

In this section we prove the properties of the measures $\mu_{\xi,\eta}$ stated and used above.

• a) For all functions $f\in C(\sigma(T))$ we have $\langle\pi(f)(\lambda_{1}\xi_{1}+\lambda_{2}\xi_{2}),\eta\rangle=\lambda_{1}% \langle\pi(f)\xi_{1},\eta\rangle+\lambda_{2}\langle\pi(f)\xi_{2},\eta\rangle$. Hence,

 $\displaystyle\int_{\sigma(T)}f\,d\mu_{\lambda_{1}\xi_{1}+\lambda_{2}\xi_{2},% \eta}=\lambda_{1}\int_{\sigma(T)}f\,d\mu_{\xi_{1},\eta}+\lambda_{2}\int_{% \sigma(T)}f\,d\mu_{\xi_{2},\eta}$

Since this holds for every $f\in C(\sigma(T))$, the uniqueness part of tells us that

 $\displaystyle\mu_{\lambda_{1}\xi_{1}+\lambda_{2}\xi_{2},\eta}=\lambda_{1}\mu_{% \xi_{1},\eta}+\lambda_{2}\mu_{\xi_{2},\eta}$
• b) The proof is similar to a).

• c) For every $f\in C(\sigma(T))$ we have

 $\displaystyle\int_{\sigma(T)}f\,d\mu_{\xi,\eta}$ $\displaystyle=$ $\displaystyle\langle\pi(f)\xi,\eta\rangle=\overline{\langle\pi(\overline{f})% \eta,\xi\rangle}$ $\displaystyle=$ $\displaystyle\overline{\int_{\sigma(T)}\overline{f}\,d\mu_{\eta,\xi}}=\int_{% \sigma(T)}f\,d\overline{\mu_{\eta,\xi}}$

Hence we conclude that $\mu_{xi,\eta}=\overline{\mu_{\eta,\xi}}$.

• d) For every $f\in C(\sigma(T))$ we have

 $\displaystyle\int_{\sigma(T)}f\,dg\cdot\mu_{\xi,\eta}$ $\displaystyle=$ $\displaystyle\int_{\sigma(T)}fg\,d\mu_{\xi,\eta}=\langle\pi(fg)\xi,\eta\rangle$ $\displaystyle=$ $\displaystyle\langle\pi(f)\pi(g)\xi,\eta\rangle=\int_{\sigma(T)}f\,d\mu_{\pi(g% )\xi,\eta}$

Hence, $g\cdot\mu_{\xi,\eta}=\mu_{\pi(g)\xi,\eta}$.

• e) For all $h\in C(\sigma(T))$ we have

 $\displaystyle\int_{\sigma(T)}h\,df\cdot\mu_{\xi,\eta}$ $\displaystyle=$ $\displaystyle\int_{\sigma(T)}hf\,d\mu_{\xi,\eta}=\int_{\sigma(T)}f\,dh\cdot\mu% _{\xi,\eta}$ $\displaystyle=$ $\displaystyle\int_{\sigma(T)}f\,d\mu_{\pi(f)\xi,\eta}=\langle\widetilde{\pi}(f% )\pi(g)\xi,\eta\rangle$ $\displaystyle=$ $\displaystyle\langle\pi(h)\xi,\widetilde{\pi}(f)^{*}\eta\rangle=\int_{\sigma(T% )}h\,d\mu_{\xi,\widetilde{\pi}(f)^{*}\eta}$

Hence, $f\cdot\mu_{\xi,\eta}=\mu_{\xi,\widetilde{\pi}(f)^{*}\eta}$.

Title proof of Borel functional calculus ProofOfBorelFunctionalCalculus 2013-03-22 18:50:26 2013-03-22 18:50:26 asteroid (17536) asteroid (17536) 8 asteroid (17536) Proof msc 47A60 msc 46L10 msc 46H30