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.

Theorem - Let $T$ be a normal operator in $B\mathit{}\mathrm{(}H\mathrm{)}$ and $\pi \mathrm{:}C\mathit{}\mathrm{(}\sigma \mathit{}\mathrm{(}T\mathrm{)}\mathrm{)}\mathrm{⟶}B\mathit{}\mathrm{(}H\mathrm{)}$ the unital *-homomorphism corresponding to the continuous functional calculus for $T$. Then, $\pi$ extends uniquely to a *-homomorphism $\stackrel{\mathrm{~}}{\pi }\mathrm{:}B\mathit{}\mathrm{(}\sigma \mathit{}\mathrm{(}T\mathrm{)}\mathrm{)}\mathrm{⟶}B\mathit{}\mathrm{(}H\mathrm{)}$ that is continuous from the $\mu$-topology to the weak operator topology. Moreover, each operator $\pi \mathit{}\mathrm{(}f\mathrm{)}$ 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) $\tilde{\pi }$ 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 $\tilde{\pi }$. 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 ${\varphi }_{\xi ,\eta }:C(\sigma (T))\to ℂ$ given by

${\varphi }_{\xi ,\eta }(f):=⟨\pi (f)\xi ,\eta ⟩$

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

$\parallel {\varphi }_{\xi ,\eta }(f)\parallel \le \parallel \pi \parallel \parallel f\parallel \parallel \xi \parallel \parallel \eta \parallel =\parallel f\parallel \parallel \xi \parallel \parallel \eta \parallel$

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

$⟨\pi (f)\xi ,\eta ⟩={\displaystyle {\int }_{\sigma (T)}}f𝑑{\mu }_{\xi ,\eta },\forall f\in C(\sigma (T))$ (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 ${\lambda }_1,{\lambda }_2\in ℂ$.

• 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 ${\lambda }_1,{\lambda }_2\in ℂ$.

• 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

$[\xi ,\eta ]:={\displaystyle {\int }_{\sigma (T)}}f𝑑{\mu }_{\xi ,\eta }$

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

$⟨\tilde{\pi }(f)\xi ,\eta ⟩:={\displaystyle {\int }_{\sigma (T)}}f𝑑{\mu }_{\xi ,\eta },\xi ,\eta \in H$

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

First, it is clear that $\tilde{\pi }$ is linear. Also clear is the fact that $\tilde{\pi }$ 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

$⟨\tilde{\pi }{(f)}^{*}\xi ,\eta ⟩$ $=$ $\overline{⟨\tilde{\pi }(f)\eta ,\xi ⟩}$

$=$ $\overline{{\displaystyle {\int }_{\sigma (T)}}f𝑑{\mu }_{\eta ,\xi }}$

$=$ ${\displaystyle {\int }_{\sigma (T)}}f𝑑{\mu }_{\xi ,\eta }$

$=$ $⟨\pi (f)\xi ,\eta ⟩$

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

We now show that $\tilde{\pi }$ is multiplicative, i.e. $\tilde{\pi }(fg)=\tilde{\pi }(f)\tilde{\pi }(g)$ 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) $f\cdot {\mu }_{\xi ,\eta }={\mu }_{\xi ,\tilde{\pi }{(f)}^{*}\eta }$, for all $f\in B(\sigma (T))$.

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

$⟨\tilde{\pi }(fg)\xi ,\eta ⟩$ $=$ ${\displaystyle {\int }_{\sigma (T)}}fg𝑑{\mu }_{\xi ,\eta }={\displaystyle {\int }_{{\sigma }_{(}T)}}g𝑑{\mu }_{\xi ,\tilde{\pi }(f)*\eta }$

$=$ $⟨\tilde{\pi }(g)\xi ,\tilde{\pi }{(f)}^{*}\eta ⟩=⟨\tilde{\pi }(f)\tilde{\pi }(g)\xi ,\eta ⟩$

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

0.0.2 Continuity Property

We now prove that the above defined $\tilde{\pi }$ 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 $\int f_i𝑑\nu \to \int f𝑑\nu$.

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

$|⟨\tilde{\pi }(f_i-f)\xi ,\eta ⟩|=\left|{\displaystyle {\int }_{\sigma (T)}}{f}_i-fd{\mu }_{\xi ,\eta }\right|\to 0$

Hence, $\tilde{\pi }(f_i)$ converges to $\tilde{\pi }(f)$ 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

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

We give this set the partial order $\le$ such that $(U_1,K_1)\le (U_2,K_2)$ 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 $ϵ>0$ a pair $(U_0,K_0)\in W$ such that . Of course, for all pairs $(U,K)$ such that $(U_0,K_0)\le (U,K)$ we also have . Hence, we have

$\left|{\displaystyle {\int }_{\sigma (T)}}{f}_{U,K}-{\chi }_{S}d\nu \right|\le {\displaystyle {\int }_{U\setminus K}}|f_{U,K}-{\chi }_S|d|\nu |\le ϵ$

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

$⟨{\pi }^{\prime }(f_{U,V})\xi ,\eta ⟩⟶⟨{\pi }^{\prime }({\chi }_S)\xi ,\eta ⟩$

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

$⟨{\pi }^{\prime }(f_{U,V})\xi ,\eta ⟩=⟨\tilde{\pi }(f_{U,V})\xi ,\eta ⟩⟶⟨\tilde{\pi }({\chi }_S)\xi ,\eta ⟩$

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

0.0.4 Image of $\tilde{\pi }$

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

Recall from the continuous functional calculus that $\pi (f)$ is in the norm closure of $𝒜$, and hence in ${𝒜}^{\prime \prime }$, 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 $\tilde{\pi }(f_{U,K})\to \tilde{\pi }({\chi }_S)$ in the weak operator topology. Given an element $R$ in the commutant of $𝒜$ we have

$⟨\tilde{\pi }(f_{U,K})R\xi ,\eta ⟩=⟨R\tilde{\pi }(f_{U,K})\xi ,\eta ⟩,\forall \xi ,\eta \in H$

The first term converges to $⟨\tilde{\pi }({\chi }_S)R\xi ,\eta ⟩$, whereas the second to $⟨R\tilde{\pi }({\chi }_S)\xi ,\eta ⟩$. Thus, $\tilde{\pi }({\chi }_S)R=R\tilde{\pi }({\chi }_S)$, and therefore $\tilde{\pi }({\chi }_S)\in {𝒜}^{\prime \prime }$.

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

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 $⟨\pi (f)({\lambda }_1{\xi }_1+{\lambda }_2{\xi }_2),\eta ⟩={\lambda }_1⟨\pi (f){\xi }_1,\eta ⟩+{\lambda }_2⟨\pi (f){\xi }_2,\eta ⟩$. Hence,

  ${\displaystyle {\int }_{\sigma (T)}}f𝑑{\mu }_{{\lambda }_1{\xi }_1+{\lambda }_2{\xi }_2,\eta }={\lambda }_1{\displaystyle {\int }_{\sigma (T)}}f𝑑{\mu }_{{\xi }_1,\eta }+{\lambda }_2{\displaystyle {\int }_{\sigma (T)}}f𝑑{\mu }_{{\xi }_2,\eta }$

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

  ${\mu }_{{\lambda }_1{\xi }_1+{\lambda }_2{\xi }_2,\eta }={\lambda }_1{\mu }_{{\xi }_1,\eta }+{\lambda }_2{\mu }_{{\xi }_2,\eta }$

• b) The proof is similar to a).

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

  ${\displaystyle {\int }_{\sigma (T)}}f𝑑{\mu }_{\xi ,\eta }$ $=$ $⟨\pi (f)\xi ,\eta ⟩=\overline{⟨\pi (\overline{f})\eta ,\xi ⟩}$

  $=$ $\overline{{\displaystyle {\int }_{\sigma (T)}}\overline{f}𝑑{\mu }_{\eta ,\xi }}={\displaystyle {\int }_{\sigma (T)}}f𝑑\overline{{\mu }_{\eta ,\xi }}$

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

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

  ${\displaystyle {\int }_{\sigma (T)}}f𝑑g\cdot {\mu }_{\xi ,\eta }$ $=$ ${\displaystyle {\int }_{\sigma (T)}}fg𝑑{\mu }_{\xi ,\eta }=⟨\pi (fg)\xi ,\eta ⟩$

  $=$ $⟨\pi (f)\pi (g)\xi ,\eta ⟩={\displaystyle {\int }_{\sigma (T)}}f𝑑{\mu }_{\pi (g)\xi ,\eta }$

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

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

  ${\displaystyle {\int }_{\sigma (T)}}h𝑑f\cdot {\mu }_{\xi ,\eta }$ $=$ ${\displaystyle {\int }_{\sigma (T)}}hf𝑑{\mu }_{\xi ,\eta }={\displaystyle {\int }_{\sigma (T)}}f𝑑h\cdot {\mu }_{\xi ,\eta }$

  $=$ ${\displaystyle {\int }_{\sigma (T)}}f𝑑{\mu }_{\pi (f)\xi ,\eta }=⟨\tilde{\pi }(f)\pi (g)\xi ,\eta ⟩$

  $=$ $⟨\pi (h)\xi ,\tilde{\pi }{(f)}^{*}\eta ⟩={\displaystyle {\int }_{\sigma (T)}}h𝑑{\mu }_{\xi ,\tilde{\pi }{(f)}^{*}\eta }$

  Hence, $f\cdot {\mu }_{\xi ,\eta }={\mu }_{\xi ,\tilde{\pi }{(f)}^{*}\eta }$.

Title	proof of Borel functional calculus
Canonical name	ProofOfBorelFunctionalCalculus
Date of creation	2013-03-22 18:50:26
Last modified on	2013-03-22 18:50:26
Owner	asteroid (17536)
Last modified by	asteroid (17536)
Numerical id	8
Author	asteroid (17536)
Entry type	Proof
Classification	msc 47A60
Classification	msc 46L10
Classification	msc 46H30