Let be a compact Hausdorff space and a Hilbert Space. A *spectral measure *relative to is a map from the -algebra of all Borel sets of to the set of projections in such that

- , ;
- for all Borel sets of ;
- for all , the function , is a regular Borel complex measure on .

A Borel measure is a measure defined on Borel sets. If every Borel set in is both outer and inner regular, then is called regular. A measurable is inner and outer regular if

, and

Denote by the Banach space of all regular Borel complex measures on , and by the C*-algebra of all bounded Borel-measurable complex-valued functions on (I assume with respect to the Borel -algebra on ).

# Example 2.5.1

Let be a compact Hausdorff space and let be a positive regular Borel measure on . Define by (I’m pretty sure we need to say that here. Also is a product rather than a composition)

That is bounded is given by

which implies that . The operator is called a *multiplicative operator*. The map

, ,

is a *-homomorphism of C*-algebras. In particular, the adjoint of is , and is normal (routine).

If is a Borel set of , then is a projection in , so is a projection in . The map is a spectral measure relative to the pair .

Since the multiplication operators are a very important class we linger with this example a little longer to show that if , then . Suppose this is false. Then such that and therefore, there is a Borel set such that and for all . Since is regular,

,

so we may suppose that is compact. Then , again by regularity of (?). However

and therefore we get , a contradiction. Hence as claimed.

This result means that the map is in fact an isometric *-isomorphism of onto a C*-subalgebra of . We therefore have .

# Lemma 2.5.1

*Let be a compact Hausdorff space, let be a Hilbert space, and suppose that for all . Suppose also that for each Borel set the function*

*, ,*

*is a sesquilinear form. Then for each the function*

*, *

*is a sesquilinear form.*

## Proof

Suppose is simple, so we can write (finite sum) , where are pairwise disjoint Borel Sets and . Then

.

The set of sesquilinear forms on is a vector space with the point-wise defined operations, and we have just shown that is a linear combination of the sesquilinear forms .

Now let . Then is the uniform limit of a sequence . Hence

, so

for each .

It immediately follows that is a sesquilinear form on (if we prove that the vector space of bounded sesquilinear forms is complete – this isn’t too difficult using Th. 2.3.6 – to all bounded sesquilinear forms there exists a bounded operator such that . Using the completeness of finishes the proof)

# Theorem 2.5.2

*Let be a compact Hausdorff space, a Hilbert space, and a spectral measure relative to . Then for each the function*

*, *

*is a bounded sesquilinear form on , and .*

## Proof

That is a sesquilinear form follows from the preceding lemma, we need only show . Suppose that disjointly, where . Then (here we use the fact that the are projections)

Hence, . Therefore,

,

so

# Theorem 2.5.3

*Let be a compact Hausdorff space, a Hilbert space, and a spectral measure relative to . Then for each there is a unique bounded operator such that*

*, *

We write for and call it the *integral of with respect to . *Note that for each Borel set .

# Theorem 2.5.4

*With the same assumptions on and as in the last theorem, the map*

*, *

*is a unital *-homomorphism.*

## Proof

Linearity is routine and boundedness follows from Theorems 2.5.2 and 2.3.6. To show that and , we need only show these results when , because . Hence, by way of linearity, we may suppose that and . Then

Also

# Theorem 2.5.5

*Let be a compact Hausdorff space and a Hilbert space, and suppose that is a unital *-homomorphism. Then there is a unique spectral measure relative to such that*

*, *

*Moreover, if , then commutes with for all iff commutes with for all Borel sets .*

## Proof

If , then the function

, ,

is linear and . By the Riesz-Kakutani Theorem (to each continuous linear functional on we may isometrically associate a measure on ), there is a unique measure such that for all . Also (here the “norm” on is total variation). Since the function

,

is sesquilinear, the maps from given by

and

are, respectively, linear and conjugate-linear. Hence for each the function

,

is a sesquilinear form (Th.2.5.1: if is sesquilinear so is where ). Also

so this sesquilinear form is bounded with norm bounded by . By Theorem 2.3.6 there is a unique operator, say, such that

, .

Moreover, .

Now suppose that . Then

, ,

so .

It is straightforward to check that the map

,

is linear and we already know that it is norm-decreasing. We now show that is a *-homomorphism.

If and , then is hermitian (it is a *-homomorphism), so is a real number. Thus the measure is real (if we can use continuous functions to approximate to for any Borel set – perhaps via Urysohn’s Lemma – then for all Borel sets which is real and we are done), that is , and therefore if , then is real. Since is arbitrary this shows that is hermitian. Therefore, preserves involutions.

Let and .

**Claim**:

If the equation

(*)

holds for all , then it also holds for all .

#### Proof

Observe that (*) is equivalent to

(**)

To prove the assertion, note that if (*) holds for all , then the regular measures and are equal because (**) holds for all . Hence (**) holds for all ; that is (*) holds for all such , as claimed

Since is a *-homomorphism, (*) holds for all . Hence, by the assertion, (1_ holds if and . Replacing with their conjugates, we get

.

Taking conjugates of both sides and using the fact that preserves involutions, we get

(***)

for all and . Using the claim again, with roles of and reversed, we get that (***) holds for all . Hence , so is a homomorphism.

If , define . is a projection as required, and the map from the -algebra to is a spectral measure relative to (just a few calculations – Let . . Hence, on at least, ) .We have that , since .

If , then

,

so . In particular, for all .

To see uniqueness of , suppose that is another spectral measure relative to such that for all . Then

.

Hence and therefore so .

Now suppose is an operator on commuting with all of the elements of the range of . Then if ,

.

Hence , so for all ( and take the adjoints on the right).

Conversely now suppose that commutes with all of the projections . Then

so . Hence for every ,

;

that is so

# Spectral Theorem

*Let be a normal operator on a Hilbert space . Then there is a unique spectral measure relative to such that , where is the inclusion map of in .*

## Proof

Let be the functional calculus at . By the preceding theorem, there exists a unique spectral measure relative to such that for all . In particular, . If is another spectral measure such that , then for all , since and generate . Therefore

The spectral measure constructed here is called the *resolution of the identity for *. Since for all , we can define for all . The unital *-homomorphism

,

is called *the Borel functional calculus at .*

If commutes with both and , then commutes with for all . For in this case commutes with all polynomials in and , and since and generate by the Stone-Weierstrass theorem, commutes with for all . By Th.2.5.5 ( commutes with for all iff commutes with for all ), therefore, commutes with for all . It follows that for all . Hence if ,

.

Therefore .

Incidentally if , then .

# Theorem 2.5.7

*Let be a normal operator on a Hilbert space , and suppose that is a continuous function. Then *

*for all .*

## Proof

The result is easily seen (have to do this yet!) by first showing it for a polynomial in and , and then observing that an arbitrary continuous function is a uniform limit of such polynomials on the compact disc , using the Stone-Weierstrass theorem applied to

# Theorem 2.5.8

*Let be a unitary operator in , where is a Hilbert space. Then there exists a hermitian operator such that and .*

## Proof

The function

,

is a continuous bijection with Borel measurable inverse . Since , we can set . The operator is self-adjoint because is real-valued. Moreover, . By Th 2.5.7, . But for all , so . Therefore,

## 2 comments

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June 29, 2011 at 12:56 pm

Von Neumann Algebras: The Double Commutant Theorem « J.P. McCarthy: Math Page[…] space, and suppose that is a finite positive regular Borel measure on . We saw in Example 2.5.1 (http://irishjip.wordpress.com/2011/01/19/c-algebras-and-operator-theory-2-5-the-spectral-theorem/) that the […]

July 7, 2011 at 10:35 am

Von Neumann Algebras: The Weak and Ultraweak Topologies « J.P. McCarthy: Math Page[…] set of , then the spectral projection belongs to by Theorem 4.1.11. From this and the proof of Theorem 2.5.8, it follows that if is a unitary, then for some hermitian […]