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(H) and π:C(σ(T))B(H) the unital *-homomorphismPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath corresponding to the continuous functional calculus for T. Then, π extends uniquely to a *-homomorphism π~:B(σ(T))B(H) that is continuousMathworldPlanetmathPlanetmath from the μ-topologyMathworldPlanetmath to the weak operator topology. Moreover, each operatorMathworldPlanetmath π(f) lies in strong operator ( closureMathworldPlanetmathPlanetmathPlanetmath of the unital *-algebra generated by T.

Proof : First we shall prove the existence of the extensionPlanetmathPlanetmath ( π~ 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 π~. 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 ξ,ηH consider the linear functionalMathworldPlanetmathPlanetmath ϕξ,η:C(σ(T)) given by


This linear functional is boundedPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath with norm at most ξη because


where the last equality comes from the fact that π is a *-isomorphism between C*-algebras (, therefore having norm 1 (see this entry (

By the Riesz representation theoremMathworldPlanetmath ( there is a unique complex Radon measureMathworldPlanetmath μξ,η in σ(T) such that

π(f)ξ,η=σ(T)f𝑑μξ,η,fC(σ(T)) (1)

Consider now the mapping (ξ,η)μξ,η. This mapping has the following properties, whoose proofs are given at the end of the entry (sectionPlanetmathPlanetmath Auxiliary Results):

  • a) μλ1ξ1+λ2ξ2,η=λ1μξ1,η+λ2μξ2,η, for all λ1,λ2.

  • b) μξ,λ1η1+λ2η2=λ1¯μξ,η1+λ2¯μξ,η2, for all λ1,λ2.

  • c) μξ,η=μη,ξ¯.

  • d) gμξ,η=μπ(g)ξ,η, for all gC(σ(T).

Therefore, for each function fB(σ(T)) we have a sesquilinear formPlanetmathPlanetmath in H given by


Moreover, this sesquilinear form is bounded ( with norm at most fξη. By the Riesz lemma on bounded sesquilinear forms, there is a unique operator π~(f)B(H) such that


We will now see that the mapping π~:B(σ(T)B(H), such that fπ~(f), has the desired properties stated in the theorem.

First, it is clear that π~ is linear. Also clear is the fact that π~ coincides with π in C(σ(T)), because of equality (1) and the uniqueness part of Riesz lemma on sesquilinear forms. Now, for any real valued function fB(σ(T)) we have that

π~(f)*ξ,η = π~(f)η,ξ¯
= σ(T)f𝑑μη,ξ¯
= σ(T)f𝑑μξ,η
= π(f)ξ,η

which means that π~(f)*=π~(f), i.e. π~(f) is self-adjointPlanetmathPlanetmathPlanetmath. Decomposing an arbitrary function fB(σ(T)) in its real and imaginary parts we see that π~(f)*=π~(f¯).

We now show that π~ is multiplicative, i.e. π~(fg)=π~(f)π~(g) for all f,gB(σ(T)). For that we need an additional property of the measuresMathworldPlanetmathPlanetmath μξ,η, whoose proof is also at the end of the entry:

  • e) fμξ,η=μξ,π~(f)*η, for all fB(σ(T)).

Given f,gB(σ(T)) we have, for every ξ,ηH,

π~(fg)ξ,η = σ(T)fg𝑑μξ,η=σ(T)g𝑑μξ,π~(f)*η
= π~(g)ξ,π~(f)*η=π~(f)π~(g)ξ,η

and therefore π~(fg)=π~(f)π~(g). Thus, π~:B(σ(T))B(H) is a *-homomorphism that extends π.

0.0.2 Continuity Property

We now prove that the above defined π~ is continuous from the μ-topology to the weak operator topology.

Let {fi} be a net of functions in B(σ(T)) that convergePlanetmathPlanetmath in the μ-topology to a function fB(σ(T)). This means that for all Radon measures ν in σ(T) we have fi𝑑νf𝑑ν.

Now for all ξ,ηH we have


Hence, π~(fi) converges to π~(f) in the weak operator topology.

0.0.3 Uniqueness

Let π:B(σ(T))B(H) be another *-homomorphism that extends π and is continuous from the μ-topology to the weak operator topology. For any measurable subset Sσ(T) consider the set

W:={(U,K):USis open andKSis compact}

We give this set the partial orderMathworldPlanetmath such that (U1,K1)(U2,K2) whenever U2U1 and K1K2. For any pair (U,K)W there is a continuous function fU,KC(σ(T)) such that f takes values on the interval [0,1], f|K=1 and suppfU (see this entry (

We claim that fU,K converges to χS in the μ-topology. In fact, given a complex Radon measure ν in σ(T), there is for every ϵ>0 a pair (U0,K0)W such that |ν|(U0K0)<ϵ. Of course, for all pairs (U,K) such that (U0,K0)(U,K) we also have |ν|(UK)<ϵ. Hence, we have


We conclude that fU,K converges to χS in the μ-topology.

Since π is continuous from the μ-topology to the weak operator topology we must have


But since π and π~ coincide with π on C(σ(T)) we also have


Hence, for any characteristic functionMathworldPlanetmathPlanetmathPlanetmath χS we have π(χS)=π~(χS). Since any function fB(σ(T)) can be uniformly approximated by simple functionsMathworldPlanetmathPlanetmath it follows that π(f)=π~(f), and we have proved the uniqueness of π~.

0.0.4 Image of π~

Let 𝒜 be the unital *-algebra generated by T. We now prove that for any fB(σ(T)), the operator π~(f) lies in the strong operator closure of 𝒜, i.e. lies in the von Neumann algebraMathworldPlanetmath generated by T. For that it is enough to prove that π~(f) is in the double commutant 𝒜′′ of 𝒜.

Recall from the continuous functional calculus that π(f) is in the norm closure of 𝒜, and hence in 𝒜′′, for every fC(σ(T)).

We have seen above that for each characteristic function χS there is a net fU,K of functions in C(σ(T)) such that π~(fU,K)π~(χS) in the weak operator topology. Given an element R in the commutant of 𝒜 we have


The first term converges to π~(χS)Rξ,η, whereas the second to Rπ~(χS)ξ,η. Thus, π~(χS)R=Rπ~(χS), and therefore π~(χS)𝒜′′.

Since every function fB(σ(T)) can be uniformly approximated by simple functions, it follows that π~(f)𝒜′′.

0.0.5 Auxiliary Results

In this section we prove the properties of the measures μξ,η stated and used above.

  • a) For all functions fC(σ(T)) we have π(f)(λ1ξ1+λ2ξ2),η=λ1π(f)ξ1,η+λ2π(f)ξ2,η. Hence,


    Since this holds for every fC(σ(T)), the uniqueness part of tells us that

  • b) The proof is similarPlanetmathPlanetmath to a).

  • c) For every fC(σ(T)) we have

    σ(T)f𝑑μξ,η = π(f)ξ,η=π(f¯)η,ξ¯
    = σ(T)f¯𝑑μη,ξ¯=σ(T)f𝑑μη,ξ¯

    Hence we conclude that μxi,η=μη,ξ¯.

  • d) For every fC(σ(T)) we have

    σ(T)f𝑑gμξ,η = σ(T)fg𝑑μξ,η=π(fg)ξ,η
    = π(f)π(g)ξ,η=σ(T)f𝑑μπ(g)ξ,η

    Hence, gμξ,η=μπ(g)ξ,η.

  • e) For all hC(σ(T)) we have

    σ(T)h𝑑fμξ,η = σ(T)hf𝑑μξ,η=σ(T)f𝑑hμξ,η
    = σ(T)f𝑑μπ(f)ξ,η=π~(f)π(g)ξ,η
    = π(h)ξ,π~(f)*η=σ(T)h𝑑μξ,π~(f)*η

    Hence, fμξ,η=μξ,π~(f)*η.

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