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2 Integration of Unbounded Functions with Respect to Positive Operator Measures

2 Integration of Unbounded Functions with Respect to Positive Operator Measures

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5.2 Integration of Unbounded Functions with Respect …



105



We now consider a different, in general more restricted, approach to operator

integrals. For projection valued measures, however, the approaches will in the next

section be shown to coincide. Let D( f, E), or just D( f ), denote the set of those

ϕ ∈ H for which | f |2 is Eϕ,ϕ -integrable.

Proposition 5.5 The set D( f, E) is a vector subspace of H contained in D( f, E).

Moreover, if ϕ ∈ D( f, E), then

| f | d|Eψ,ϕ | ≤



ψ



E(Ω)



for all ψ ∈ H, and the map ψ →



whose norm is at most

E(Ω)



| f |2 dEϕ,ϕ ,



f dEψ,ϕ on H is a bounded linear functional

| f |2 dEϕ,ϕ .



Proof Choose a sequence ( f n ) of simple functions converging pointwise to f , with

p

| f n | ≤ | f | for all n ∈ N, and let f n = k=1 ck χ X k , with X 1 , . . . , X p ∈ A constituting

a partition of Ω. For each k = 1, . . . , p, let Yk1 , . . . , Ykrk ∈ A form a partition of

X k . Then for all ϕ, ψ ∈ H,

|Eψ,ϕ (Yk jk )| = | ψ | E(Yk jk )ϕ | ≤ E(Yk jk )1/2 ψ



E(Yk jk )1/2 ϕ ,



which gives

p



p



rk



|ck |

jk =1



k=1

p



|ck |

p



E(Yk jk )1/2 ψ



|ck |2 E(Yk jk )1/2 ϕ

k=1 jk =1

p



rk



rk



ψ | E(Yk jk )ψ



|ck |2 ϕ | E(Yk jk )ϕ



k=1 jk =1



k=1 jk =1



p



p



ψ | E(X k )ψ



=



|ck |2 ϕ | E(X k )ϕ



k=1



k=1

p



=



ψ | E(Ω)ψ



|ck |2 ϕ | E(X k )ϕ

k=1







E(Ω)



ψ



E(Yk jk )1/2 ϕ



rk



2



k=1 jk =1



=



E(Yk jk )1/2 ψ

jk =1



k=1



rk





p



rk



|Eψ,ϕ (Yk jk )| ≤



| f n |2 dEϕ,ϕ .



2



106



5 Operator Integrals and Spectral Representations: The Unbounded Case



Taking the suprema over all finite partitions of the sets X k one gets

| f n | d|Eψ,ϕ | ≤



ψ



E(Ω)



| f n |2 dEϕ,ϕ ,



where |Eψ,ϕ | denotes the total variation of Eψ,ϕ . If ϕ ∈ D( f ), using the dominated

convergence theorem one obtains limn→∞ | f n |2 dEϕ,ϕ = | f |2 dEϕ,ϕ . Hence, by

Fatou’s lemma, | f | is |Eψ,ϕ |-integrable, and thus Eψ,ϕ -integrable, showing that ϕ ∈

D( f ). This also proves the inequality stated in the proposition, and the last claim is

proved by observing that

|



f dEψ,ϕ | ≤



| f | d|Eψ,ϕ | ≤ lim



n→∞







E(Ω)



ψ lim



=



E(Ω)



ψ



n→∞



| f n |d|Eψ,ϕ |

| f n |2 dEϕ,ϕ



| f |2 dEϕ,ϕ .



Let now ϕ, ψ ∈ H, c, d ∈ C, and X ∈ A. Denoting ξ = cϕ + dψ we have

ξ | E(X )ξ = E(X )1/2 ξ



2



≤ |c| E(X )1/2 ϕ + |d| E(X )1/2 ψ



2



≤ 2|c|2 ϕ | E(X )ϕ + 2|d|2 ψ | E(X )ψ ,

which implies that D( f ) is a linear subspace.

Remark 5.1 If ϕ ∈ D( f ), then the Fréchet–Riesz theorem in conjunction with the

above proposition may be used to give a quick proof of the existence of what we have

denoted by L( f, E)ϕ, independently of the uniform boundedness principle used in

the proof of Lemma 5.1.

Remark 5.2 In general, the inclusion D( f ) ⊂ D( f ) in Proposition 5.5 may be strict.

For example, let μ : A → [0, 1] be a probability measure, and consider the positive operator measure X → E(X ) = μ(X )I . Take any measurable f which is

μ-integrable and such that | f |2 is not μ-integrable. Then {0} = D( f ) = D( f ) = H.

Choosing an f which is not μ-integrable one gets the extreme case of D( f ) being

the null space.

Proposition 5.6 (a) If f is real valued, then L( f, E) is symmetric, that is, for any

ϕ, ψ ∈ D( f, E), ψ | L( f, E)ϕ = L( f, E)ψ | ϕ .

(b) If D( f, E) is a dense subspace of H, then the adjoint L( f, E)∗ is an extension

of L( f , E).

Proof (a) Let ( f n ) be a sequence of real simple functions converging pointwise to

f and satisfying | f n | ≤ | f |. By the dominated convergence theorem



5.2 Integration of Unbounded Functions with Respect …



107



ψ | L( f )ϕ = lim ψ | L( f n )ϕ = lim L( f n )∗ ψ | ϕ

n→∞



n→∞



L( f n )ψ | ϕ = L( f )ψ | ϕ ,



= lim



n→∞



where we have used the obvious fact that for a (simple) bounded real function f n the

operator L( f n ) is selfadjoint.

(b) Since f is measurable, f is also measurable, and the domains of the operators

L( f ) and L( f ) are the same, D( f ) = D( f ). If D( f ) is dense in H, then the adjoint

of L( f ) is defined. For each n ∈ N, let gn (x) = f (x) if | f (x)| ≤ n, and gn (x) = 0

otherwise. Since gn is bounded, L(gn )∗ = L(gn ). By the dominated convergence

theorem we may write for all ψ, ϕ ∈ D( f ) = D( f )

ψ | L( f )ϕ = lim ψ | L(gn )ϕ

n→∞



= lim L(g n )ψ | ϕ = L( f )ψ | ϕ

n→∞



which shows that ψ

L( f ) ⊂ L( f )∗ .







D(L( f )∗ ) and L( f )ψ



=



L( f )∗ ψ, that is,



5.3 Integration of Unbounded Functions with Respect

to Projection Valued Measures

Throughout this section, A ⊂ 2Ω is a σ-algebra and E : A → L(H) is assumed to

be a projection valued measure. We first show that in this case the two approaches

to integration described in the previous section are actually equivalent.

Proposition 5.7 In the case of a projection valued measure E, we have D( f, E) =

D( f, E) for any measurable function f , and this subspace is dense in H.

Proof We already know that D( f, E) ⊂ D( f, E) (see Proposition 5.5). Now suppose

that ϕ ∈ D( f ). Choose a sequence ( f n ) of simple functions converging pointwise to

p

f , with | f n | ≤ | f | for all n ∈ N. For a fixed n ∈ N we write f n = k=1 ck χ X k , where

X 1 , . . . , X p ∈ A are disjoint sets with union Ω. Then a simple calculation based on

the fact that the projections E(X j ) are pairwise orthogonal (by Proposition 4.6 (c))

shows that

p



| f n |2 dEϕ,ϕ =



2



ck E(X k )ϕ



2



=



f n dE(·)ϕ



k=1



with the obvious definition of the integral f n dE(·)ϕ of the simple function f n with

respect to the additive vector valued set function X → E(X )ϕ on A.

For any ψ the dominated convergence theorem shows that the sequence of the

numbers



108



5 Operator Integrals and Spectral Representations: The Unbounded Case



ψ



f n dE(·)ϕ =



f n dEψ,ϕ



tends to the limit f dEψ,ϕ as n → ∞. It thus follows from the uniform boundedness

principle that the sequence of the integrals | f n |2 dEϕ,ϕ is bounded, and so Fatou’s

lemma implies that | f |2 is integrable with respect to Eϕ,ϕ .

We now show that D( f ) is dense in H. Take some ϕ ∈ H. Write ϕ = E(Ω)ϕ + ψ

where ψ ⊥ E(Ω)(H). Since E(Ω)ψ = 0, it is clear that ψ ∈ D( f ), so that it is

enough to show that E(Ω)ϕ can be approximated by vectors from D( f ). For each

n ∈ N denote An = ω ∈ Ω | f (ω)| ≤ n and ϕn = E(An )ϕ. As (An ) is an

increasing sequence, E(Ω)ϕ = limn→∞ ϕn (see Remark 4.4(b)). But ϕn ∈ D( f ),

since Eϕn ,ϕn (Ω\An ) = E(An )ϕ | (E(Ω) − E(An ))E(An )ϕ = 0, and f is bounded

on An .

In the sequel we use the notation D f for the space D( f, E) = D( f, E) for any

measurable function f : Ω → C.

Lemma 5.2 If f : Ω → C is a measurable function, then

L( f )ϕ



2



=



Ω



| f |2 dEϕ,ϕ



for all ϕ ∈ D f .

m

αi χ Ai is a simple function where the sets Ai ∈ A are disjoint

Proof If h = i=1

and their union is Ω, then

m



L(h)ϕ



2



m



=



αi E(Ai )ϕ

i=1

m



m



αi E(Ai )ϕ =

i=1



|αi |2 ϕ | E(Ai )ϕ =



=

i=1



|αi |2 E(Ai )ϕ



2



i=1



Ω



|h|2 dEϕ,ϕ .



The claim follows from this observation in the case when f is bounded, for we can

approximate such an f uniformly by simple functions (see Lemma 4.7) and then

apply Proposition 4.11. In the general case we define for all n ∈ N f n (ω) = f (ω) if

| f (ω)| ≤ n, and f n (ω) = 0 if | f (ω)| > n. Since f n is bounded, we have D f − fn =

D f , and using Proposition 5.5 and the dominated convergence theorem we find that

for all ϕ ∈ D f

L( f )ϕ − L( f n )ϕ = sup | ψ | L( f − f n )ϕ | ≤ sup

ψ ≤1







Ω



| f − f n |2 dEϕ,ϕ



ψ ≤1 Ω



1

2



→0



| f − f n | d|Eψ,ϕ |

(5.1)



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