10 A Riesz--Markov--Kakutani Type Representation Theorem for Positive Operator Measures
Tải bản đầy đủ - 0trang
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4 Operator Integrals and Spectral Representations: The Bounded Case
2−n Φ(fn ) ≥ 2−n 4n = 2n for all n ∈ N, which is impossible. Any g ∈ C0 (Ω)
with f ≤ 1 may be written as g = g1 − g2 + ig3 − ig4 with each gk ∈ C0 (Ω)
satisfying 0 ≤ gk ≤ 1, the general boundedness claim follows.
(b) For any ξ, η ∈ H define the bounded linear functional Φξ,η : C0 (Ω) → C by
the formula Φξ,η (f ) = ξ | Φ(f )η . There is a unique regular complex measure μξ,η : B(Ω) → C such that Φξ,η (f ) = Ω f dμξ,η for all f ∈ C0 (Ω). For
X ∈ B(Ω), ξ, η ∈ H, we define SX (ξ, η) = μξ,η (X). The map (ξ, η) → SX (ξ, η)
is easily seen to be sesquilinear since the map (ξ, η) → ξ | Φ(f )η is sesquilinear for all f ∈ C0 (Ω) and the correspondence which maps Φξ,η to μξ,η is injective. Since |SX (ξ, η)| ≤ |μξ,η |(Ω) = Φξ,η ≤ Φ ξ η , SX is a bounded
sesquilinear form, and so using Proposition 2.1 we get a unique operator E(X) ∈
L(H) such that ξ | E(X)η = μξ,η (X) for all ξ, η ∈ H. Since each μξ,ξ is a positive measure, E is a positive operator measure, and the claimed formula holds
by Proposition 4.12(a). By construction, E is regular, and its uniqueness follows
from the uniqueness part of the Riesz–Markov–Kakutani theorem.
(c) Let G ⊂ Ω be an open set and F the set of the functions f ∈ C0 (G) satisfying
0 ≤ f ≤ 1 and having their supports contained in G. By regularity and Urysohn’s
lemma we easily see (exercise) the claimed equality. In fact F is a directed set
with pointwise order and E(G) is the weak limit of the net (Φ(f ))f ∈F .
(d) Assume first the multiplicativity condition Φ(f g) = Φ(f )Φ(g). In the notation
of the proof of (b), using Theorem 2.8(b) we find that E(G)2 = [w-lim Φ(f )]2 =
w-lim Φ(f 2 ) = E(G), since the functions f 2 with f ∈ F are just the elements of F.
Thus E(G)2 = E(G) for any open set G. By regularity, for any X ∈ B(Ω) Φ(X)
is the weak limit of the net (Φ(G)) defined on the set of the open sets containing
X, directed by requiring that G1 ≤ G2 if G2 ⊂ G1 . Using Theorem 2.8(b) again
we get E(X)2 = [w-lim E(G)]2 = w-lim E(G)2 = w-lim E(G) = E(X). Thus E
is projection valued. The converse follows from Proposition 4.13.
In the situation of the above theorem, we call the positive linear map Φ normalised
if the corresponding regular POM E is normalised, i.e. E(Ω) = I.
4.11 The Spectral Representation of Bounded Selfadjoint
Operators
We assume throughout this section that A ∈ L(H) is a selfadjoint operator. We fix
two real numbers a and b such that a < b and aI ≤ A ≤ bI; this is possible since
sup ϕ ≤1 | ϕ | Aϕ | < ∞. As an intermediate step towards proving the spectral representation of a bounded selfadjoint operator, we construct a suitable bounded linear
map from the space C[a, b] of continuous functions (with the supremum norm) on
the interval [a, b] into the operator space L(H). We begin with an auxiliary result.
Lemma 4.8 Let p be a polynomial with real coefficients such that p(t) ≥ 0 for all
t ∈ [a, b]. The polynomial p can be expressed in the form
4.11 The Spectral Representation of Bounded Selfadjoint Operators
m
p(t) =
n
qi (t)2 +
i=1
87
(t − a)rj (t)2 +
j=1
(b − t)sk (t)2
k=1
where qi rj , and sk are polynomials with real coefficients.
Proof We may assume that p = 0 is not constant and write it as a constant α times
a product of first degree polynomials and monic second degree polynomials without
real roots as follows:
n1
n2
n3
s
p(t) = αΠi=1
(t − ci )λi Πj=1
(t − dj )μj Πk=1
(ek − t)νk Πr=1
Qr (t).
Here we assume that ci ≤ a, a < dj < b, and ek ≥ b. Each μj must be even, for
otherwise p could not preserve its sign on the interval (a, b). Clearly α > 0. For each
r = 1, . . . , s there are αr ∈ R and βr > 0 such that Qr (t) = (t − αr )2 + βr . Writing
t − ci = (t − a) + (a − ci ) and ek − t = (ek − b) + (b − t) (where a − ci ≥ 0 and
ek − b ≥ 0) and performing the multiplications we find that p(t) can be expressed
as stated in the lemma, except that in addition there may be terms of the form
(t − a)(b − t)q(t)2 where q is a real polynomial. But the formula
(t − a)(b − t) = (b − a)−1 [(t − a)2 (b − t) + (t − a)(b − t)2 ]
shows that these terms can also be included in a sum of the same form.
The next results contains in an algebraic form the essence of the spectral representation.
Theorem 4.5 There is a uniquely determined continuous linear map Φ : C[a, b] →
L(H) satisfying the following conditions:
(i) Φ(1) = I where 1 is the function with the constant value 1;
(ii) Φ(id) = A where id(t) = t for all t ∈ [a, b];
(iii) Φ(f g) = Φ(f )Φ(g) for all f , g ∈ C[a, b].
Proof If p(t) = a0 + a1 t + · · · + an t n for t ∈ [a, b], the coefficients ak ∈ C are
uniquely determined, and so we may define
n
Φ0 (p) = a0 I +
ak Ak .
k=1
Denoting by P[a, b] the set of the restrictions of the polynomials with complex
coefficients to [a, b], we may in this way define a linear map Φ0 : P[a, b] → L(H)
satisfying the above conditions (i)–(iii). If q ∈ P[a, b] is the restriction of a polynomial with real coefficients, then q ± q(t) ≥ 0. In view of the preceding lemma we
may thus write
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4 Operator Integrals and Spectral Representations: The Bounded Case
m
q I + Φ0 (q) =
i=1
p
n
Φ0 (qi )2 +
(A − aI)Φ0 (rj )2 +
j=1
(bI − A)Φ0 (sk )2
k=1
for certain polynomials qi , rj and sk with real coefficients (restricted to [a, b]). It
follows that q I + Φ0 (q) ≥ 0 since e.g. A − aI ≥ 0 and the operators A − aI and
Φ0 (rj ) commute. Similarly, q I − Φ0 (q) ≥ 0, and it follows from Proposition 2.2
that Φ0 (q) ≤ q . If q is the restriction to [a, b] of an arbitrary polynomial, we
find
Φ0 (q) ≤ Φ0 (Re q) + Φ0 (Im q) ≤ Re q + Im q ≤ 2 q ,
and so Φ0 is a continuous linear map. The Stone–Weierstrass theorem implies
that P[a, b] is dense in C[a, b], and so Φ0 can be extended to a continuous linear map Φ : C[a, b] → L(H). The map Φ also satisfies the conditions (i)–(iii); for
example Φ(f g) = limn→∞ Φ0 (fn gn ) = limn→∞ Φ0 (fn )Φ0 (gn ) = Φ(f )Φ(g), whenever fn , gn ∈ P[a, b] and fn → f , gn → f uniformly. The uniqueness of the map Φ
follows from continuity and the fact that it is by linearity and the conditions (i)–(iii)
completely determined in the dense subspace P[a, b] of C[a, b].
Remark 4.9 If Φ is as in the preceding theorem, then Φ(f ) = Φ(f )∗ for all f ∈
C[a, b] or, equivalently, Φ(f ) is selfadjoint for every real valued function f ∈ C[a, b].
Since Ak is selfadjoint for any k ∈ N, this follows from the fact that the restrictions
of the polynomials with real coefficients are dense in the space of real continuous
functions on [a, b]. Moreover, if f ∈ C[a, b] and f ≥ 0, then Φ(f ) = Φ(f 1/2 )2 ≥ 0,
since ϕ | Φ(f 1/2 )2 ϕ = Φ(f 1/2 )ϕ | Φ(f 1/2 )ϕ ≥ 0.
The map Φ obtained in Theorem 4.5 can be represented by integration with respect
to a suitable spectral measure. Before proving this, we consider a somewhat more
general situation.
Theorem 4.6 Assume that Φ : C[a, b] → L(H) is a linear map satisfying Φ(1) = I
and Φ(f ) ≥ 0 whenever f ≥ 0. There is a uniquely determined semispectral measure
E : B(R) → Ls (H)+ such that Φ(f ) = [a,b] f dE for all f ∈ C[a, b]. If Φ(f g) =
Φ(f )Φ(g) for all f , g ∈ C[a, b], then E is a spectral measure.
Proof (First proof) Since every complex measure on [a, b] is regular by Proposition 4.9, this is a special case of Theorem 4.4.
Since this proof depends on the Riesz–Markov–Kakutani representation theorem
which we have used without proof, we also give a more elementary approach based
on the use of a semispectral function. We retain the assumptions of Theorem 4.6
and define a function F : R → Ls (H)+ in the following way: F(λ) = 0 for λ < a,
F(λ) = I for λ ≥ b, and for any number λ ∈ [a, b), F(λ) is the infimum of the set of
the operators Φ(f ) where f ∈ C[a, b], f ≥ 0 and f (t) = 1 if t ∈ [a, λ]. (According
to Theorem 2.7 this infimum exists.)
4.11 The Spectral Representation of Bounded Selfadjoint Operators
89
In item (a) below, the estimate Φ ≤ 2 could be replaced by Φ = 1, but the
proof would be more complicated and we do not need this result (which of course
also follows from Theorem 4.6).
Lemma 4.9 (a) The linear map Φ is continuous and Φ ≤ 2.
(b) The function F is a semispectral function.
(c) If λ ∈ [a, b), ϕ ∈ H and > o, then there is a δ > 0 such that
| ϕ | Φ(f )ϕ | <
whenever f ∈ C[a, b] is such that f ≤ 1 and f (t) = 0 outside the interval
(λ, λ + δ].
Proof (a) Since f I ± Φ(f ) ≥ 0 when f ∈ C[a, b] is real valued, the corresponding argument in the proof of Theorem 4.5 works here.
(b) We show that F is weakly right continuous; the other conditions in Definition 4.10 are obvious. It is enough to consider a point λ ∈ [a, b). Suppose that
ϕ ∈ H, ϕ = 1, and > 0. We choose a function f ∈ C[a, b], f ≥ 0, such that
f (t) = 1 for all t ∈ [a, λ] and ϕ | Φ(f )ϕ < ϕ | F(λ)ϕ + 2 (see Theorem
2.7). For any number δ ∈ (0, b − λ) we define fδ (t) = f˜ (t − δ) for all t ∈ [a, b],
where f˜ (t) = f (t) when t ∈ [a, b], and f˜ (t) = 1, when t < a. Since f is uniformly continuous, we have f − fδ < 4 , when δ is small enough, so that by
(a) we get
ϕ | F(λ + δ)ϕ ≤ ϕ | Φ(fδ )ϕ ≤ ϕ | Φ(f )ϕ +
2
< ϕ | F(λ)ϕ + .
This implies the claim.
(c) Choose δ > 0 such that ϕ | F(λ + δ)ϕ < ϕ | F(λ)ϕ + (see (b)). If f is
the kind of function mentioned in the claim (c), we can construct a function g ∈ C[a, b], g ≥ 0, such that g(t) = 1 when t ∈ [a, λ], g(t) + |f (t)| ≤ 1
when t ∈ [λ, λ + δ], and g(t) = 0 when t ∈ [λ + δ, b]. (Put e.g. g(t) = (1 −
|f (t)|)(1 − δ −1 (t − λ)) when t ∈ [λ, λ + δ].) If α ∈ C is such that |α| = 1 and
| ϕ | Φ(f )ϕ | = α ϕ | Φ(f )ϕ , then we get
| ϕ | Φ(f )ϕ | = Re ϕ | Φ(αf )ϕ = ϕ | Φ(Re αf )ϕ ≤ ϕ | Φ(|αf |)ϕ
= ϕ | Φ(|f |)ϕ = ϕ | Φ(g + |f |)ϕ − ϕ | Φ(g)ϕ
≤ ϕ | F(λ + δ)ϕ − ϕ | F(λ)ϕ < ,
for g + |f | ≤ h whenever h ∈ C[a, b], h ≥ 0, and h(t) = 1 when a ≤ t ≤
λ + δ.
Proof (Second proof of Theorem 4.6) Consider the semispectral function F studied
in the above lemma and let E : B(R) → Ls (H)+ be the semispectral measure corresponding to it in accordance with Theorem 4.3, i.e. E (−∞, t] = F(t) for all t ∈ R.
Let f ∈ C[a, b]. We are going to show that
90
4 Operator Integrals and Spectral Representations: The Bounded Case
Φ(f ) =
f dE.
(4.1)
[a,b]
Suppose that ϕ ∈ H and > 0. Since f is uniformly continuous, we can find some
points a = t0 < t1 < · · · < tp = b and numbers α1 , . . . , αp ∈ C such that |f (t) −
αi | < whenever ti−1 ≤ t ≤ ti . By Lemma 4.9(b) there is a δ > 0 such that
ϕ | F(ti + δ)ϕ − ϕ | F(ti )ϕ <
(4.2)
p
for i = 1, . . . , p, and that the claim of Lemma 4.9(c) is true at all points λ =
t0 , . . . , tp−1 for the number p in place of . We may assume that δ < ti+1 − ti . We
construct a function g ∈ C[a, b] as follows: g(t) = α1 for t ∈ [a, t1 ], g(t) = αi+1
for t ∈ [ti + δ, ti+1 ], i = 1, . . . , p − 1, and g(t) = αi + δ −1 (αi+1 − αi )(t − ti ) for
t ∈ [ti , ti + δ], i = 1, . . . , p − 1. Since |αi+1 − αi | ≤ |αi+1 − f (ti )| + |f (ti ) − αi | <
2 , we have |g(t) − αi | < 2 when t ∈ [ti−1 , ti ], implying f − g < 3 . Using the
definition of F we may define for each i = 1, . . . , p − 1 a continuous function hi :
[a, b] → [0, 1] such that hi (t) = 1 when t ∈ [a, ti ], hi (t) = 0 when t ∈ [ti + δ, b],
and
ϕ | F(ti )ϕ ≤ ϕ | Φ(hi )ϕ < ϕ | F(ti )ϕ + .
(4.3)
p
By (4.2)
ϕ | F(ti )ϕ ≤
hi dEϕ,ϕ ≤ ϕ | F(ti + δ)ϕ < ϕ | F(ti )ϕ + ,
p
(4.4)
and so
hi dE ϕ <
ϕ | Φ(hi )ϕ − ϕ
We denote h =
(4.5) we see that
ϕ
p
i=1
p
.
(4.5)
αi (hi − hi−1 ) where identically h0 = 0, hp = 1. Then using
h dE ϕ − ϕ | Φ(h) ϕ ≤ max |αi |2 ≤ ( f + )2 .
1≤i≤p
(4.6)
On the other hand, by Lemma 4.9,
| ϕ | Φ(f )ϕ − ϕ | Φ(h)ϕ | ≤ 2 f − g
ϕ
2
+ | ϕ | Φ(g)ϕ − ϕ | Φ(h)ϕ |
p−1
≤6 ϕ
2
+
| ϕ | Φ(gi )ϕ |
i=1
≤6 ϕ
2
+ (2 f + 4 ),
(4.7)
4.11 The Spectral Representation of Bounded Selfadjoint Operators
91
p−1
for we can write h − g = i=1 gi with functions gi ∈ C[a, b] satisfying gi ≤
2 f + 4 and such that each gi vanishes outside the interval [ti , ti + δ) (for
αi hi (t) − hi−1 (t) = |αi | ≤ f + when t ∈ [ti−1 , ti ], and g ≤ f + 3 ). Moreover,
ϕ
f dE ϕ
h dE ϕ − ϕ
p−1
≤ f −g
ϕ
2
+
3( f + ) ϕ | (F(ti + δ) − F(ti ))ϕ
(4.8)
i=1
≤3 ϕ
2
+ 3 ( f + ).
by (4.2). As > 0 is arbitrary, by (4.6)–(4.8) we get
ϕ | Φ(f )ϕ = ϕ
f dE ϕ ,
implying (4.1) by the polarisation identity.
To prove the uniqueness claim, suppose that two semispectral measures E1 , E2
satisfy Φ(f ) = [a,b] f dEj for all f ∈ C[a, b], j = 1, 2. Since Ej ([a, b]) =
Φ(1) = I, we must have Ej (R \ [a, b]) = 0. Thus if f : R → [0, 1] is any continuous
function vanishing outside a compact interval, then
R
f dE1 =
[a,b]
f dE1 = Φ(f |[a, b]) =
[a,b]
f dE2 =
R
f dE2 .
Therefore E1 = E2 by Corollary 4.1.
Assume now that Φ(f g) = Φ(f )Φ(g) for all f , g ∈ C[a, b]. In order to see
that E is a spectral measure, it is by Theorem 4.3 enough to show that F(λ) is
a projection whenever λ ∈ [a, b). Choose a decreasing sequence (fn ) of continuous functions fn : [a, b] → [0, 1] such that fn (t) = 1 when t ∈ [a, λ], and fn (t) → 0
when t ∈ (λ, b]. Then F(λ) = χ[a,λ] dE = w-limn→∞ fn dE according to Proposition 4.13(e). Similarly F(λ) = w-limn→∞ fn2 dE. But w-lim fn2 dE = w-lim
( fn dE)2 = F(λ)2 , so that F(λ)2 = F(λ). Conversely, by Proposition 4.13(b) Φ is
multiplicative if E is a spectral measure.
We are now in a position to prove the spectral representation of a bounded selfadjoint operator.
Theorem 4.7 If A ∈ Ls (H) and aI ≤ A ≤ bI, there is a uniquely determined spectral measure E : B [a, b] → L(H) such that
A=
[a,b]
λ dE(λ).
Proof Let Φ be as in Theorem 4.5 (so that Φ(f ) ≥ 0 whenever f ≥ 0, since
ϕ | Φ(f )ϕ = Φ( f )ϕ | Φ( f )ϕ ≥ 0), and let E : B(R) → L(H) be the
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4 Operator Integrals and Spectral Representations: The Bounded Case
associated spectral measure of Theorem 4.6. Since E(R \ [a, b]) = 0, by restricting E to the σ-algebra B([a, b]) we obtain the spectral measure E of the claim.
Suppose that E is also a spectral measure on B([a, b]) such that A = [a,b] λ dE (λ).
The mappings f → [a,b] f dE and f → [a,b] f dE satisfy the assumptions made on
Φ in Theorem 4.5 (see Proposition 4.13(b)), and so by its uniqueness part we see that
they agree on C[a, b]. The proof of the uniqueness part of Theorem 4.6 then shows
that E = E .
As an application of the spectral representation theorem, we give another proof
for the fact that every positive operator has a unique positive square root.
Proposition 4.15 If A ∈ Ls (H)+ , then there is a uniquely determined B ∈ Ls (H)+
such that B2 = A.
Proof Since 0 ≤ A ≤ A I, by Theorem 4.7 there is a unique spectral measure
√E :
B([0, A ]) → L(H) such that A = [0, A ] λ dE(λ). We define B = [0, A ] λ d
E(λ). From Proposition 4.12 it follows that B ∈ Ls (H)+ , and B2 = A by Proposition 4.13. On the other hand, let B0 ∈ Ls (H)+ be such that B02 = A. For B0 there
is a spectral representation B0 = [0, B0 ] λ dE0 (λ), and if α : [0, B0 ] → [0, A ]
is defined by α(t) = t 2 (note that B0 2 = A by Theorem 2.6 (e)), we may
˜ 0 : B([0, A ]) → L(H) defined
apply Proposition 4.14 to the spectral measure E
−1
˜
by E0 (B) = E0 (α (B)) and find that
[0, A ]
λ d E˜ 0 (λ) =
[0, B0 ]
t 2 dE0 (t) = B02 = A
(see Proposition 4.13(b)). Therefore by the uniqueness part of Theorem 4.7 we have
E˜ 0 = E, and so Proposition 4.14 shows that
B0 =
[0, A ]
√
λ d E˜ 0 (λ) =
[0, A ]
√
λ dE(λ) = B.
We conclude this section with a commutation result.
Proposition 4.16 In the situation of Theorem 4.7 for an operator T ∈ L(H) the
following conditions are equivalent:
(i) TA = AT ;
(ii) T E(X) = E(X)T for all X ∈ B [a, b] .
Proof By Proposition 4.13(a), (ii) implies (i). Assume now (i). We retrace the main
steps in the construction of the spectral measure E. Clearly, T commutes with every
polynomial in A, and it follows from the proof of Theorem 4.5 that T Φ(f ) = Φ(f )T
where Φ is as in Theorem 4.5. The construction of the spectral function F before
Lemma 4.9 involves weakly convergent nets of operators, i.e. infima of certain sets
4.11 The Spectral Representation of Bounded Selfadjoint Operators
93
of positive operators, directed with respect to the order opposite to the usual operator
order (compare Theorem 2.7) and since weak convergence commutes with multiplication from the left or right by T (see Theorem 2.8(a)), it follows that F(t)T = T F(t)
for all t ∈ R. Finally, the construction of the spectral measure E from the spectral
function F in the proof of Theorem 4.3 only involves weakly convergent nets of operators, and so again the commutation condition is inherited by the next generation.
Thus (ii) holds.
4.12 The Spectrum of a Bounded Operator
In this section we assume that T ∈ L(H). In the discussion preceding Theorem 3.4,
the spectrum of an operator was defined, as well as the subdivision of the spectrum
into various parts. It was shown there that the following definition is equivalent to
the definition of the spectrum we used previously.
Definition 4.11 The spectrum σ(T ) of T is the set of those λ ∈ C for which the
operator Tλ = λ − T = λI − T has no inverse belonging to L(H).
We use the following lemma to prove that the spectrum of any bounded linear
operator is a compact subset of the complex plane. We let Inv(L(H)) denote the set
of the operators in L(H) having an inverse in L(H). (The spectrum is known to be
always nonempty, but we do not need or prove this general fact. However, in the case
of a normal operator this will follow from the spectral representation Theorem 4.10.)
Lemma 4.10 (a) If T ∈ L(H) is such that T < 1, then I − T ∈ Inv(L(H)).
(b) The set Inv(L(H)) is open in L(H).
Proof In (a) the inverse of I − T is obtained as the sum of the absolutely convergent
n
series I + ∞
n=1 T , and (b) easily follows from (a) (exercise).
Theorem 4.8 (a) If λ ∈ σ(T ), then |λ| ≤ T .
(b) The spectrum σ(T ) of T is closed.
Proof Part (a) follows at once from Lemma 4.10(a), and (b) is a consequence of
Lemma 4.10(b) and the continuity of the map λ → λI − T .
Corollary 4.2 If U ∈ L(H) is unitary and λ ∈ σ(U), then |λ| = 1.
Proof Since U = 1, |λ| ≤ 1 by Theorem 4.8. On the other hand, if |λ| < 1, then
|λ−1 | > 1, and since U ∗ = 1, the same result shows that λ−1 − U ∗ ∈ Inv(L(H))
implying that λ − U = (−λU)(λ−1 − U ∗ ) is also in Inv(L(H)).
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4 Operator Integrals and Spectral Representations: The Bounded Case
4.13 The Spectral Representations of Unitary
and Other Normal Operators
In this section we assume that T ∈ L(H) is a normal operator. We write T = A1 + iA2
with commuting selfadjoint operators A1 and A2 . We fix real numbers aj , bj such
that aj I ≤ Aj ≤ bj I, j = 1, 2. By Theorem 4.7 there are unique spectral measures
Ej : B [aj , bj ] → L(H) such that
Aj =
[aj ,bj ]
λ dEj (λ),
j = 1, 2. These will be used to construct a spectral measure for T . When convenient,
we may use the complex notation (x, y) = x + iy for elements of R2 (= C).
In the proof of the following theorem we make essential use of the observation
made on commuting spectral measures in Example 4.1, which is applicable by Proposition 4.9(b). In this situation B(Ω1 ) ⊗ B(Ω2 ) = B(Ω1 × Ω2 ) (Proposition 4.10).
Theorem 4.9 There is a unique spectral measure E : B([a1 , b1 ] × [a2 , b2 ]) →
L(H) such that
T=
[a1 ,b1 ]×[a2 ,b2 ]
λ dE(λ).
Proof To prove the existence of a spectral measure E satisfying the formula in the
claim, we note that since the selfadjoint operators A1 and A2 commute, there are
commuting spectral measures Ej : B [aj , bj ] → L(H) such that
Aj =
[aj ,bj ]
λ dEj (λ),
j = 1, 2 (see Theorem 4.7 and Proposition 4.16). By Example 4.1 there is a spectral
measure E : B [a1 , b1 ] × [a2 , b2 ] → L(H) satisfying E(X1 × X2 ) = E1 (X1 )E2 (X2 )
for all Xj ∈ B [aj , bj ] , j = 1, 2. If h : [a1 , b1 ] × [a2 , b2 ] → [a1 , b1 ] is the map
defined by h(x, y) = x, then E h−1 (X) = E1 (X) for all X ∈ B [a1 , b1 ] , and so
[a1 ,b1 ]×[a2 ,b2 ]
Re λ dE(λ) =
[a1 ,b1 ]
x dE1 (x) = A1 .
Similarly,
[a1 ,b1 ]×[a2 ,b2 ]
Im λ dE(λ) = A2 .
It follows that
[a1 ,b2 ]×[a2 ,b2 ]
λ dE(λ) = A1 + iA2 = T .
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