13 The Spectral Representations of Unitary and Other Normal Operators
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4.13 The Spectral Representations of Unitary and Other Normal Operators
95
We now prove the uniqueness statement. Assume that E is as in the claim. Define
E˜ 1 (X) = E(X × [a2 , b2 ]) for all X ∈ B [a1 , b1 ] . Then E˜ 1 is a spectral measure and
by Proposition 4.14
[a1 ,b1 ]×[a2 ,b2 ]
Re λ dE(λ) =
[a1 ,b1 ]
x d E˜ 1 (x).
Since
A1 =
1
(T + T ∗ ) =
2
[a1 ,b1 ]×[a2 ,b2 ]
1
(λ + λ) dE(λ) =
2
[a1 ,b1 ]
x d E˜ 1 (x)
it follows from Theorem 4.7 that E˜ 1 = E1 . A similar argument shows that E2 (Y ) =
E([a1 , b2 ] × Y ) for all Y ∈ B [a2 , b2 ] . Hence
E(X × Y ) = E (X × [a2 , b2 ]) ∩ ([a1 , b1 ] × Y ) = E((X × [a2 , b2 ])E([a1 , b1 ] × Y ))
= E1 (X)E2 (Y )
for all X ∈ B [a1 , b1 ] , Y ∈ B [a2 , b2 ] . But by Example 4.1 there is only one spectral
measure on B [a1 , b1 ] × [a2 , b2 ] with this property.
We next prove the spectral representation of a normal operator in a form involving
the spectrum.
Theorem 4.10 There is a unique spectral measure E : B(σ(T )) → L(H) such that
T = σ(T ) λ dE(λ). Morever, σ(T ) is the support of E.
Proof Since σ(T ) is compact, in the situation described before Theorem 4.9 we may
assume that σ(T ) ⊂ [a1 , b1 ] × [a2 , b2 ]. Let E : B [a1 , b1 ] × [a2 , b2 ] → L(H) be
a spectral measure such that T = [a1 ,b1 ]×[a2 ,b2 ] λ dE(λ). We show that E([a1 , b1 ] ×
[a2 , b2 ] \ σ(T )) = 0. Since [a1 , b1 ] × [a2 , b2 ] \ σ(T ) is the union of a countable
number of compact sets, using the σ-subadditivity of the measures Eϕ,ϕ we see that
it suffices to show that E(K) = 0 for any compact set K ⊂ [a1 , b1 ] × [a2 , b2 ] \ σ(T ).
By compactness and subadditivity, it is enough to show that for each z ∈ K there is a
δ > 0 such that E(B(z, δ)) = 0 where B(z, δ) = {w ∈ [a1 , b1 ] × [a2 , b2 ] | w − z
< δ}. Suppose that for some z ∈ [a1 , b1 ] × [a2 , b2 ] \ σ(T ) we have E(B(z, n1 )) = 0
whenever n ∈ N. Choose ϕn ∈ E(B(z, n1 ))(H) with ϕn = 1 for each n ∈ N. Then
by Proposition 4.13
96
4 Operator Integrals and Spectral Representations: The Bounded Case
1 = ϕn = (zI − T )−1 (zI − T )E B(z, 1/n) ϕn
= (zI − T )−1
(z − λ) dE(λ)ϕn
B(z,1/n)
≤ (zI − T )−1
(z − λ) dE(λ)
B(z,1/n)
≤ (T − zI)−1
1
→ 0,
n
when n → ∞. This contradiction proves our claim, and thus the existence of E as
a consequence of Theorem 4.9. The uniqueness part also follows from Theorem 4.9,
since any spectral measure on B(σ(T )) may be regarded as a spectral measure on
B([a1 , b1 ] × [a2 , b2 ]) vanishing on [a1 , b1 ] × [a2 , b2 ] \ σ(T ). To prove that σ(T ) =
supp(E) we now only need to show that σ(T ) ⊂ supp(E). Suppose that z ∈ C \
supp(E). Then the function λ → (z − λ)−1 is defined, continuous and bounded on
the closed set supp(E). Define
(z − λ)−1 dE(λ) (∈ L(H)).
B=
supp(E)
From Proposition 4.13(b) it follows that (zI − T )B = B(zI − T ) = I, and so z ∈
/
σ(T ).
The spectral measure E : B(σ(T )) → L(H) in the above theorem is called the
spectral measure of the normal operator T . For convenience we may use the same
expression for the spectral measure obtained by extending E, as zero outside σ(T ), to
the Borel σ-algebra of a suitable set containing σ(T ). Thus in the following corollary
dealing with the spectral representation of a unitary operator, E could be called the
spectral measure of U.
Since the spectrum of a unitary operator U ∈ L(H) is contained in T = {z ∈
C | |z| ≤ 1} by Corollary 4.2, the above result yields the following corollary. (The
uniqueness part is seen by modifying an argument at the end of the proof of
Theorem 4.10.)
Corollary 4.3 If U ∈ L(H) is a unitary operator, there is a unique spectral measure
E : B(T) → L(H) such that U = T λ dE(λ).
4.14 Exercises
Unless otherwise stated, H is an arbitrary Hilbert space.
1. Let Ω be a set and F the set of the singletons {x}, x ∈ Ω. Describe
(a) the ring generated by F,
(b) the σ-algebra generated by F.
4.14 Exercises
97
2. Let Ω be a set. The counting measure μ : 2Ω → [0, ∞] is defined as follows:
μ(A) is the number of the elements in the set A ⊂ Ω, if A is a finite set, and
μ(A) = ∞ otherwise. Show that μ is a measure.
3. In the situation of the preceding exercise, show that a function f : Ω → C is
μ-integrable if and only if (f (x))x∈Ω is a summable family, and if this is the case,
then
f dμ =
f (x).
Ω
x∈Ω
4. In the situation of exercise 2, let f : Ω → C be a μ-integrable function. For each
X ⊂ Ω, define ν(X) = X f (x)dμ(x) (= x∈X f (x)). Show that ν : 2Ω → C is
a complex measure.
5. Describe the total variation measure |ν| of the complex measure ν defined in the
previous exercise.
6. Let Ω be a set and A ⊂ 2Ω an algebra, i.e. a ring having Ω as one of its
elements. Let bfa(Ω, A) denote the set of the bounded (finitely) additive set
functions μ : A → C. (A function μ : A → C is (finitely) additive if μ(X ∪
Y ) = μ(X) + μ(Y ) for any two disjoint sets X, Y ∈ A.) Define the total variation
|ν| : A → [0, ∞] as in the case of a complex measure. Show that bfa(Ω, A) is a
Banach space (with the obvious vector space structure) with respect to the norm
μ → μ = |μ|(Ω). (In particular, μ < ∞.)
7. Let Ω be a set and A ⊂ 2Ω a σ-algebra. Show that ca(Ω, A) is a closed subspace
of the Banach space bfa(Ω, A) defined in the previous exercise.
8. Let Ω be a set and A ⊂ 2Ω a σ-algebra. Show that the total variation |μ| of μ is
the smallest among the measures ν : A → [0, ∞] satisfying |μ(A)| ≤ ν(A) for
all A ∈ A.
9. Let g : N → [0, ∞) be a bounded function. For each X ⊂ N and f ∈ 2 , denote
Eg (X)f = gχX f . Show that the map X → Eg (X) on 2N is a positive operator
measure.
10. Give a necessary and sufficient condition on the function g in the previous exercise for Eg to be
(a) projection valued;
(b) a semispectral measure;
(c) a spectral measure.
11. Give a necessary and sufficient condition on the function g in Exercise 9 for
Eg to be countably additive with respect to the norm topology of L(H). (First
state the obvious definition of being countably additive with respect to the norm
topology of L(H).)
12. Is it possible for the positive operator measure in exercise 9 to be both countably
additive with respect to the norm topology of L(H) and
(a) projection-valued;
(b) a semispectral measure;
(c) a spectral measure?
98
4 Operator Integrals and Spectral Representations: The Bounded Case
13. Explain in what sense the spectral representation of a compact selfadjoint operator on H may be expressed in the form
A=
Ω
λ dE(λ)
where E : A → L(H) is a spectral measure for a suitable set Ω and a σ-algebra
A ⊂ 2Ω .
14. Prove the following statement in the proof of Proposition 4.13: “It easily follows
from (ii) in view of Proposition 4.6(c) that (i) holds for simple functions”.
15. Let (Ω, A, μ) be a measure space, so that μ is a not necessarily a finite (positive)
measure. Suppose that h : Ω → [−∞, ∞] = R is an A-measurable function.
Denote
B = α ∈ R μ(h−1 ((α, ∞])) = 0 .
16.
17.
18.
19.
20.
21.
22.
23.
Show that inf B (∈ R) is the smallest λ ∈ R having the property that h(x) ≤ λ
for μ-almost every x ∈ Ω. We call inf B the essential supremum of h and denote
it by ess sup h or ess supx∈Ω h(x).
Let (Ω, A, μ) be a measure space where μ(Ω) > 0 and consider the Hilbert
space L 2 (μ) = L 2 (Ω, A, μ). Let g : Ω → C be an A-measurable function and
assume that g is essentially bounded, i.e. ess supx∈Ω |g(x)| < ∞. We then denote
ess supx∈Ω |g(x)| = g ∞ . Show that via the formula Tg f = gf we get a welldefined bounded linear map Tg : L 2 (μ) → L 2 (μ) such that Tg ≤ g ∞ .
(a) Show by an example that in the previous exercise the strict inequality Tg <
g ∞ is possible.
(b) Show that the equality Tg = g ∞ holds in the case of the Lebesgue
measure μ on Ω = R.
(c) Can you abstract some general feature from (b) to guarantee the equality
Tg = g ∞ ?
Let A ⊂ 2Ω be a σ-algebra and μ ∈ ca(Ω, A). Let f : Ω → C be a μ-integrable
function, and denote ν(A) = A f dμ for all A ∈ A. Show that ν is a complex
measure and prove the formula |ν|(A) = A |f | d|μ| for all A ∈ A.
Let A ⊂ 2Ω be a σ-algebra and μ ∈ ca(Ω, A). Show that there is an Ameasurable |μ|-integrable function g : Ω → C such that |g(x)| = 1 for all x ∈ Ω
and μ(A) = A g(x) d|μ|(x) for all x ∈ Ω. (Hint: Consider the Hilbert space
L 2 (Ω, A, |μ|) and the bounded (why?) linear functional ϕ → Ω ϕ(x) dμ(x) on
it.)
Let A ⊂ 2Ω be a σ-algebra and μ : A → [0, ∞) a (finite positive) measure.
Show that there is a Hilbert space H with a spectral measure E : A → L(H)
and a vector ϕ ∈ H such that μ(A) = ϕ | E(A)ϕ for all A ∈ A.
Let A ⊂ 2Ω be a σ-algebra and μ : A → C a complex measure. Show that there
is a Hilbert space H with a spectral measure E : A → L(H) and two vectors
ϕ, ψ ∈ H such that μ(A) = ψ | E(A)ϕ for all A ∈ A.
Complete the details in the proof of Lemma 4.2.
Complete the details in Remark 4.8.
4.14 Exercises
99
24. Prove the commutation claim left as an exercise in Example 4.1.
25. Let Ω be a closed or open subset of Rn and E : B(Ω) → L(H) a positive operator
measure. Show that for any X ∈ B(Ω), E(X) is the infimum of the set of the
operators E(U) where each U is a relatively open subset of Ω.
26. Let Ω be a closed or open subset of Rn and E : B(Ω) → L(H) a positive operator
measure. Show that for any X ∈ B(Ω), E(X) is the supremum of the set of the
operators E(K) where each K is a compact subset of Ω.
27. Let Ω be a locally compact Hausdorff space and E : B(Ω) → L(H) a positive
operator measure. Show that the complex measure ξ | E(·)η is regular for all
ξ, η ∈ H if the measure ξ | E(·)ξ is regular for all ξ ∈ H.
28. Complete the details in the proof of Theorem 4.4(c).
29. Complete the details in the proof of Lemma 4.10.
30. Let g : N → C be a bounded function and Tg : 2 → 2 the bounded linear
operator defined by the formula Tg f = gf . Show that the spectrum of Tg is the
closure of the set g(n) n ∈ N .
31. The operator Tg in the previous exercise is clearly normal, since its adjoint is Tg .
Describe the spectral measure of Tg .
32. Give an example of an isometric operator whose spectrum is not contained in
the set T = z ∈ C |z| = 1 .
33. (a) Let K be a compact subset of C. Show that there is a normal operator T on
2
such that σ(T ) = K.
(b) Let K ⊂ T = z ∈ C |z| = 1 be a compact set. Show that there is a unitary
operator U on 2 such that σ(T ) = K.
(c) State and prove the analogue of (b) for (bounded) selfadjoint operators.
34. (a) Let U ∈ L(H) be a normal operator whose spectrum is a subset of T = z ∈
C |z| = 1 . Show that T is unitary.
(b) State and prove the analogue of (a) for (bounded) selfadjoint operators.
35. Let T ∈ L(H) be a normal operator and T = σ(T ) λ dE(λ) its spectral represen1 n
tation. Show that the series I + ∞
n=1 n! T converges in norm, and that if exp(T )
denotes its sum, then exp(T ) = σ(T ) exp(λ) dE(λ).
36. Let T ∈ L(H) be a normal operator. Suppose that there is some w ∈ C, w = 0,
such that λw λ ≥ 0 ∩ σ(T ) = ∅. Show that there is a normal operator S ∈
L(H) such that T = exp(S).
37. Show that Theorem 3.5 remains valid if the selfadjoint compact operator A is
replaced by a normal compact operator. (Then the eigenvalues λn are of course
not necessarily real.)
References
1. Dunford, N., Schwartz, J.T.: Linear Operators. Part I. Wiley Classics Library. Wiley, New York
(1988). Reprint of the 1958 original, A Wiley-Interscience Publication
2. Friedman, A.: Foundations of Modern Analysis. Holt Rinehart and Winston Inc, New York
(1970)
100
4 Operator Integrals and Spectral Representations: The Bounded Case
3. Halmos, P.R.: Measure Theory, 4th edn. Springer, Heidelberg (1988). Reprint of the 1950 edition
4. Berg, C., Christensen, J.P.R., Ressel, P.: Harmonic Analysis on Semigroups. Graduate Texts in
Mathematics. Springer, New York (1984)
5. Ylinen, K.: Positive operator bimeasures and a noncommutative generalization. Studia Math.
118(2), 157–168 (1996)
Chapter 5
Operator Integrals and Spectral
Representations: The Unbounded Case
In the traditional Hilbert space quantum mechanics, (generally unbounded) selfadjoint operators are taken as physical observables. In the present work a more comprehensive approach is used: observables are represented as normalised positive operator
measures. Selfadjoint operators, however, have independent interest, and their spectral representation in terms of spectral measures shows the connection to the approach
to quantum mechanics with positive operator measures. The spectral representation
theory of selfadjoint operators utilises the integration of unbounded functions with
respect to spectral measures. We develop this theory in the more general setting of
positive operator measures, paving the way for important physical applications in
the sequel. As the final highlight of the chapter we prove the Stone representation
theorem for strongly continuous one-parameter groups of unitary operators.
5.1 Elementary Notes on Unbounded Operators
In many important applications of Hilbert space theory, e.g. in quantum mechanics,
one is naturally led to study linear maps that are neither continuous nor defined on
the whole of H. In this section we make some basic observations related to such
situations.
From now on, we use the term operator (in H) for any linear map T whose domain
D(T ) is a linear subspace of H and which satisfies R(T ) = T ϕ ϕ ∈ D(T ) ⊂ H.
The graph of an operator T in H is the subset G(T ) = (ϕ, T ϕ) ϕ ∈ D(T ) of the
Cartesian product H × H. For operators S and T , we write S ⊂ T if G(S) ⊂ G(T ).
(From a set theoretical point of view, of course T is the same as its graph; the
introduction of the notion of graph is a concession to historical usage.) An operator
T is said to be densely defined if its domain is dense in H, i.e. D(T ) = H. We now
generalise the concept of the adjoint of an element of L(H).
© Springer International Publishing Switzerland 2016
P. Busch et al., Quantum Measurement, Theoretical and Mathematical Physics,
DOI 10.1007/978-3-319-43389-9_5
101
102
5 Operator Integrals and Spectral Representations: The Unbounded Case
Definition 5.1 Let T be a densely defined operator in H. The adjoint of T is the
operator T ∗ defined as follows:
(i) The domain D(T ∗ ) is the set of those ψ ∈ H for which the linear functional
ϕ → ψ | T ϕ is continuous on D(T ), and
(ii) whenever ψ ∈ D(T ∗ ), T ∗ ψ is the vector in H satisfying T ∗ ψ | ϕ = ψ | T ϕ
for all ϕ ∈ D(T ).
The vector T ∗ ψ mentioned in the definition exists (and is unique) by the
Fréchet–Riesz theorem, for if ψ ∈ D(T ), the map ϕ → ψ | T ϕ can be uniquely
extended to a continuous linear functional on H.
It is immediately seen that T ∗ is an operator, i.e. D(T ∗ ) is a vector subspace of
H and T ∗ is linear. However, in general T ∗ need not be densely defined.
Definition 5.2 If S and T are operators in H, then the operators S + T and ST are
defined as follows: D(S + T ) = D(S) ∩ D(T ) and (S + T )ϕ = Sϕ + T ϕ for all
ϕ ∈ D(S + T ); D(ST ) = ϕ ∈ D(T ) T ϕ ∈ D(S) and (ST )ϕ = S(T ϕ) for all
ϕ ∈ D(ST ).
Proposition 5.1 If S, T , and ST are densely defined operators in H, then T ∗ S ∗ ⊂
(ST )∗ .
Proof Suppose ψ ∈ D(T ∗ S ∗ ). Then S ∗ ψ ∈ D(T ∗ ). Thus the map
ϕ → ψ | ST ϕ = S ∗ ψ | T ϕ
on D(ST ) is continuous, so that ψ ∈ D((ST )∗ ). Moreover,
T ∗ S ∗ ψ | ϕ = S ∗ ψ | T ϕ = ψ | ST ϕ = (ST )∗ ψ | ϕ
for all ϕ in the dense subspace D(ST ) of H, so that T ∗ S ∗ ψ = (ST )∗ ψ.
Definition 5.3 Let T be an operator in H. We say that T is symmetric if ψ | T ϕ =
T ψ | ϕ for all ϕ, ψ ∈ D(T ). If T is densely defined and T = T ∗ , then T is said
to be selfadjoint.
Thus in our usage a symmetric operator need not be densely defined. It is clear that
a densely defined operator T is symmetric if and only if T ⊂ T ∗ .
Next we study an operator and its adjoint in terms of their graphs. An operator T
is said to be a closed operator if its graph G(T ) is a closed subspace of the Cartesian
product H × H (regarded as the direct sum Hilbert space H ⊕ H in the usual way).
Clearly T is closed if and only if ϕ ∈ D(T ) and ψ = T ϕ, whenever (ϕn ) and (ψn )
are convergent sequences in H such that ϕn ∈ D(T ) and ψn = T ϕn for all n ∈ N
and ϕ = limn→∞ ϕn , ψ = limn→∞ ψn .
If the closure (in H × H) of the graph of an operator T has the property that it
can contain the vector pairs (ϕ, ξ) and (ϕ, η) only if ξ = η, then an easy argument
shows that this closure is the graph of a linear map defined on a vector subspace of
5.1 Elementary Notes on Unbounded Operators
103
H. We then denote this linear map by T and call it the closure of the operator T . In
this situation T is said to be closable. Clearly T is closable if and only if there is
some closed operator S such that T ⊂ S.
Proposition 5.2 Let T be a densely defined operator in H.
(a) The graph of T ∗ is the orthogonal complement (in the Hilbert space H × H) of
the set (−T ξ, ξ) ∈ H × H ξ ∈ D(T ) .
(b) The adjoint T ∗ is a closed operator.
(c) The operator T is closable if and only if D(T ∗ ) is dense in which case T = T ∗∗ .
(d) If T is symmetric, then it is closable and thus T = T ∗∗ .
(e) If T is selfadjoint, then T is closed.
Proof The following statements are equivalent for a pair (ϕ, ψ) of vectors in H:
(ϕ, ψ) | (−T ξ, ξ) = 0 for all ξ ∈ D(T ); ϕ | T ξ = ψ | ξ for all ξ ∈ D(T );
ϕ ∈ D(T ∗ ) and ψ = T ∗ ϕ. Thus (a) holds. Since any orthogonal complement is
closed, (b) and (e) follow from (a).
To prove (c), suppose first that D(T ∗ ) is dense. We denote by U the unitary
operator on H × H defined by U (ϕ, ψ) = (−ψ, ϕ). In the proof of (a) we saw
that G(T ∗ ) = (U (G(T )))⊥ , and so by using this also for T ∗ in place of T and
noting that U −1 = −U , we get G(T ) = (−U G(T ∗ ))⊥ = (U G(T ∗ ))⊥ = G(T ∗∗ ).
Conversely, assume that D(T ∗ ) is not dense and choose ψ ∈ D(T ∗ )⊥ \{0}. Then
(ψ, 0) ∈ G(T ∗ )⊥ which shows that (U (G(T ∗ )))⊥ contains (0, ψ) and hence is not
the graph of an operator. Since G(T ) = (U (G(T ∗ )))⊥ , we see that T is not closable.
Part (d) follows from (c) as T ⊂ T ∗ .
We show next that a selfadjoint operator is maximally symmetric.
Proposition 5.3 If S is a selfadjoint operator and T a symmetric operator such that
S ⊂ T , then S = T .
Proof Since also T is densely defined, T ∗ exists. If ϕ ∈ D(S) and ψ ∈ D(T ∗ ), then
T ∗ ψ | ϕ = ψ | T ϕ = ψ | Sϕ , so that also ψ ∈ D(S ∗ ) and S ∗ ψ = T ∗ ψ. We
thus have T ⊂ T ∗ ⊂ S ∗ = S, and so S = T .
A densely defined symmetric operator T (which is closable by Proposition 5.2
(d)) is called essentially selfadjoint if its closure T = T ∗∗ is selfadjoint. If T is
essentially selfadjoint, then by Proposition 5.3 it has only one selfadjoint extension,
since any selfadjoint extension is also an extension of T . In typical applications such
a T appears as the sum of two selfadjoint operators (or the restriction of the sum to
a dense subspace).
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5 Operator Integrals and Spectral Representations: The Unbounded Case
5.2 Integration of Unbounded Functions with Respect
to Positive Operator Measures
In this section we extend the definition of the operator integral, which we previously
only studied for bounded functions, to the case where the integrand is not necessarily
bounded. Then the integral L( f, E) is not necessarily a bounded operator.
Throughout this section, unless specified otherwise, we assume that A ⊂ 2Ω is
a σ-algebra and E : A → L(H) is a positive operator measure. Let f : Ω → C
be an A-measurable function. We let D( f, E), or just D( f ) if there is no ambiguity
about E, denote the set of those ϕ ∈ H for which f is integrable with respect to the
complex measure Eψ,ϕ whenever ψ ∈ H.
Lemma 5.1 Suppose that ϕ ∈ D( f, E). Then there is a unique vector L( f, E)ϕ ∈ H
such that
ψ | L( f, E)ϕ =
f dEψ,ϕ
for all ψ ∈ H.
Proof Choose a sequence ( f n ) of simple functions converging pointwise to f , with
| f n | ≤ | f | for all n ∈ N. For each n ∈ N, we let αn denote the integral (with the
obvious definition) of f n with respect to the additive set function X → E(X )ϕ.
Consider the bounded linear functionals h n : H → C defined by h(ψ) = ψ | αn .
It follows from the dominated convergence theorem that the sequence (h n (ψ)) converges to the complex conjugate of f dEψ,ϕ for every ψ ∈ H. The principle of
uniform boundedness can then be used to show that the function ψ → Ω f dEψ,ϕ is
a bounded linear functional on H, and so the claim follows from the Fréchet–Riesz
theorem.
Definition 5.4 In the situation considered above, we call L( f, E) (i.e. the map ϕ →
L( f, E)ϕ) the integral of f with respect to E. We may also denote L( f, E) by E[ f ],
or by L( f ) if E is clear from the context.
Proposition 5.4 The set D( f, E) is a vector subspace of H, and the map L( f, E) :
D( f, E) → H defined in Lemma 5.1 is linear.
Proof Suppose now that ξ, η ∈ D( f, E) and ψ ∈ H. From the definition of the total
variation it follows that |Eψ,ξ+η |(X ) ≤ |Eψ,ξ |(X ) + |Eψ,η |(X ) for any X ∈ A. Thus
from the monotone convergence theorem (applied to a suitable increasing sequence
of nonnegative simple functions) it follows that a nonnegative measurable function
that is integrable with respect to |Eψ,ξ | and |Eψ,η | is also |Eψ,ξ+η |-integrable. In
particular | f |, and hence f , is |Eψ,ξ+η |-integrable. The equality Ω f dEψ,ξ+η =
Ω f dEψ,ξ + Ω f dEψ,η then follows from the dominated convergence theorem as
is obvious for f n in place of f . The scalar multiplication part is obvious.
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