3 Integration of Unbounded Functions with Respect to Projection Valued Measures
Tải bản đầy đủ - 0trang
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)
5.3 Integration of Unbounded Functions with Respect …
109
when n → ∞. Since the claim holds for each f n , for each ϕ ∈ D f the dominated
convergence theorem shows that
L( f )ϕ
2
= lim L( f n )ϕ
n→∞
2
= lim
n→∞ Ω
| f n |2 dEϕ,ϕ =
Ω
| f |2 dEϕ,ϕ .
Proposition 5.8 If f, g : Ω → C are measurable functions, then L( f )L(g) ⊂
L( f g) and D(L( f )L(g)) = Dg ∩ D f g ; in particular, L( f )L(g) = L( f g) if and
only if D f g ⊂ Dg .
Proof Assume first that f is bounded. Then Dg ⊂ D f g . Define gn (ω) = g(ω)
if |g(ω)| ≤ n, and gn (ω) = 0 if |g(ω)| > n, n ∈ N. Suppose that ϕ ∈ Dg .
Now L( f ) is bounded, and since by (5.1) we have L(g)ϕ = limn→∞ L(gn )ϕ and
L( f g)ϕ = limn→∞ L( f gn )ϕ, Proposition 4.13(b) implies
L( f )L(g)ϕ = lim L( f )L(gn )ϕ = lim L( f gn )ϕ = L( f g)ϕ.
n→∞
n→∞
(5.2)
If ψ = L(g)ϕ, we thus get using Lemma 5.2,
Ω
| f |2 dEψ,ψ = L( f )ψ
2
= L( f g)ϕ
2
=
Ω
| f g|2 dEϕ,ϕ .
(5.3)
Let now f be arbitrary (not necessarily bounded). Since
D(L( f )L(g)) = ϕ ∈ Dg L(g)ϕ ∈ D(L( f ))
and since by the monotone convergence theorem (5.3) implies, when applied to
the functions f n used in the proof of Lemma 5.2, that for a vector ϕ ∈ Dg we have
L(g)ϕ ∈ D f if and only if ϕ ∈ D f g , we see that D(L( f )L(g)) = Dg ∩D f g . Suppose
now that ϕ ∈ Dg ∩D f g , ψ = L(g)ϕ, and f n is as in the proof of Lemma 5.2. It follows
from the dominated convergence theorem that f n → f in L 2 (Eψ,ψ ) (for | f − f n |2 ≤
| f |2 ) and f n g → f g in L 2 (Eϕ,ϕ ). Hence because by (5.2) L( f n )L(g)ϕ = L( f n g)ϕ,
we find using (5.1) that L( f )L(g)ϕ = L( f )ψ = limn→∞ L( f n )ψ = L( f n g)ϕ =
L( f g)ϕ.
Proposition 5.9 For any measurable function f : Ω → C we have L( f )∗ = L( f )
and L( f )L( f )∗ = L(| f |2 ) = L( f )∗ L( f ).
Proof Let ϕ ∈ D f and ψ ∈ D f = D f . If f n is as in the proof of Lemma 5.2,
using (5.1) and Proposition 4.12(b) we get ψ | L( f )ϕ = limn→∞ ψ | L( f n )ϕ
= limn→∞ L( f n )ψ | ϕ = L( f )ψ | ϕ . It follows that ψ ∈ D(L( f )∗ ) and
L( f )ψ = L( f )∗ ψ. To prove the equality L( f ) = L( f )∗ it now suffices to show
that D(L( f )∗ ) ⊂ D f . Let ξ ∈ D(L( f )∗ ) and η = L( f )∗ ξ. As f n = f h n where
h n is the characteristic function of the set ω ∈ Ω | f (ω)| ≤ n , by Proposition 5.8
we get L( f n ) = L( f )L(h n ), since Dh n = H. Since L(h n ) is a projection, using
110
5 Operator Integrals and Spectral Representations: The Unbounded Case
Propositions 5.1 and 4.12(b) we find that L(h n )L( f )∗ ⊂ [L( f )L(h n )]∗ = L( f n )∗ =
L( f n ), and so
(5.4)
L(h n )η = L( f n )ξ for all n ∈ N.
Since |h n | ≤ 1, (5.4) and Lemma 5.2 together imply
Ω
| f n |2 dEξ,ξ =
=
Ω
Ω
| f n |2 dEξ,ξ = L( f n )ξ
2
= L(h n )η
2
h n dEη,η ≤ Eη,η (Ω),
and so by the monotone convergence theorem we have ξ ∈ D f . The latter equations
in the claim follow from the above by Proposition 5.8, since D f f ⊂ D f in view of
the fact that
| f | dEϕ,ϕ ≤
2
Ω
2
Ω
1 dEϕ,ϕ
1
2
| f | dEϕϕ
4
Ω
1
2
<∞
if ϕ ∈ D f f .
We still mention some immediate consequences of the preceding discussion.
Corollary 5.1 Let f, g : Ω → C be A-measurable functions.
(a) The operator L( f ) is closed.
(b) If g is bounded, then L(g)(D f ) ⊂ D f .
(c) If g is bounded and vanishes outside the set ω ∈ Ω | f (ω)| ≤ M for some
M ∈ [0, ∞), then L(g)(H) ⊂ D f .
Proof (a) This follows from Proposition 5.9 (applied to f ) since any adjoint is closed
by Proposition 5.2.
(b) Since D f g ⊃ D f and Dg = H, Proposition 5.8 shows that D(L( f )L(g)) ⊃ D f ,
implying the claim.
(c) Now f g is bounded, so that D(L( f )L(g)) = H.
5.4 The Cayley Transform
If z ∈ C\{−i}, we define f (z) = z−i
. In this way we get a homeomorphism
z+i
.
f : C\{−i} → C\{1} whose inverse is given by the formula z = f −1 (w) = i 1+w
1−w
Since
1 − |w|2
1 1+w 1+w
1
+
=
,
Im z = (z − z) =
2i
2 1−w 1−w
|1 − w|2
f defines a homeomorphism between the real axis and the set T\{1} where T =
w ∈ C |w| = 1 is the unit circle. This simple observation provides a blueprint for
5.4 The Cayley Transform
111
the definition of the so-called Cayley transform, which will turn out to be an efficient
tool in the study of a (generally unbounded) selfadjoint operator: such an operator
can be transformed into a unitary operator, for which we already have a spectral
representation. Before defining the Cayley transform we make some preparations.
Recall that an operator T in H is isometric if T ϕ = ϕ for all ϕ ∈ D(T ).
Proposition 5.10 Let S be a symmetric operator in H.
(a) For all ϕ ∈ D(S) we have (S ± i I )ϕ 2 = ϕ 2 + Sϕ 2 .
(b) The map S + i I : D(S) → R(S + i I ) is a bijection and its inverse (S + i I )−1 :
R(S + i I ) → D(S) is continuous. Moreover, R(S + i I ) is closed if S is a closed
operator.
(c) The map V = (S − i I )(S + i I )−1 is isometric, D(V ) = R(S + i I ) and
R(V ) = R(S − i I ). Moreover, R(V ) is closed if S is a closed operator.
(d) The map I − V is injective, D(S) = R(I − V ), and S = i(I + V )(I − V )−1 .
Proof (a) Since S is symmetric we have (S ± i I )ϕ 2 = Sϕ | Sϕ ± Sϕ | iϕ
± iϕ | Sϕ + ϕ | ϕ = Sϕ 2 + ϕ 2 for all ϕ ∈ D(S).
(b) If (S + i I )ϕ = 0, it follows from (a) that ϕ = 0. Thus S + i I is an injection.
Moreover,
ϕ ≤ (S + i I )ϕ ,
(5.5)
and so (S + i I )−1 is continuous. If Sϕn + iϕn → ψ where ϕn ∈ D(S) for all n ∈ N,
by (5.5) (ϕn ) is a Cauchy sequence, so that it converges to some ϕ ∈ H. If S is closed,
it is easily seen that S + i I is also closed, and so ϕ ∈ D(S + i I ) and ψ = Sϕ + iϕ.
Hence ψ ∈ R(S + i I ), so that R(S + i I ) is closed.
(c) Clearly V is defined in R(S + i I ) (since D(S − i I ) = D(S) = D(S + i I ))
and R(V ) = R(S − i I ). Moreover, by (a) we have (S − i I )ϕ = (S + i I )ϕ for
all ϕ ∈ D(S) = R (S + i I )−1 implying that V is isometric. As in (b), we see that
R(V ) = R(S − i I ) is closed if S is a closed operator.
(d) Between the sets R (S +i I )−1 = D(S +i I ) = D(S) and D(V ) = R(S +i I )
there is in view of (b) and (c) a bijective correspondence making ϕ ∈ D(S) correspond
to ψ ∈ D(V ) where
ψ = (S + i I )ϕ, V ψ = (S − i I )ϕ
(5.6)
or, equivalently,
(I − V )ψ = 2iϕ, (I + V )ψ = 2Sϕ.
(5.7)
If (I − V )ψ = 0, then denoting ϕ = (S +i I )−1 ψ we have (I − V )ψ = 2iϕ, so ϕ = 0
and hence ψ = 0. This shows that I − V is an injection and R(I − V ) = D(S). In
addition Sϕ = 21 (I + V )ψ = i(I + V )(I − V )−1 ϕ for all ϕ ∈ D(S) = D(S + i I )
by (5.7).
112
5 Operator Integrals and Spectral Representations: The Unbounded Case
Definition 5.5 The operator V = (S−i I )(S+i I )−1 defined in the above proposition
is called the Cayley transform of the operator S.
A symmetric operator S can be studied with the help of its Cayley transform. It
will turn out that a densely defined symmetric operator is selfajoint if and only if its
Cayley transform is unitary. We prove this with the help of the following result.
Proposition 5.11 For a symmetric densely defined operator S, the following three
conditions are equivalent:
(i) S is selfadjoint;
(ii) S is closed, and zero is the only vector ψ ∈ D(S ∗ ) such that (S ∗ + i I )ψ = 0
or (S ∗ − i I )ψ = 0;
(iii) R(S + i I ) = R(S − i I ) = H.
Proof Assume first (i). According to Proposition 5.2 S is closed. Let now ψ ∈ D(S ∗ )
be such that S ∗ ψ + iψ = 0. Then Sψ + iψ = 0 since S is selfadjoint, and Proposition 5.10 (a) shows that ψ = 0. Similarly, ψ = 0 if (S ∗ ψ − i I )ψ = 0. Thus (ii)
holds.
Next assume (ii). Let ψ ∈ R(S − i I )⊥ . Then
ψ | Sϕ = ψ | (S − i I )ϕ + ψ | iϕ = −iψ | ϕ
for all ϕ ∈ D(S), so ψ ∈ D(S ∗ ) and S ∗ ψ = −iψ, which implies by (ii) that ψ = 0.
Therefore by the projection theorem, Theorem 2.4, R(S − i I ) is dense in H. But
according to Proposition 5.10 (c) R(S − i I ) is closed, so R(S − i I ) = H. It is
similarly seen (using Proposition 5.10 (b)) that R(S + i I ) = H.
Finally assume (iii). Let ϕ ∈ D(S ∗ ). Since R(S − i I ) = H, there is a vector
ψ ∈ D(S) such that (S − i I )ψ = (S ∗ − i I )ϕ. As S ⊂ S ∗ , we have ϕ − ψ ∈ D(S ∗ )
and (S ∗ − i I )(ϕ − ψ) = 0. For any ξ ∈ D(S) we have ϕ − ψ | (S + i I )ξ =
S ∗ (ϕ − ψ) | ξ + −i(ϕ − ψ) | ξ = (S ∗ − i I )(ϕ − ψ) | ξ = 0, so ϕ = ψ ∈
D(S), since R(S + i I ) = H. It follows that D(S ∗ ) = D(S), and so S is selfadjoint.
Theorem 5.1 A symmetric densely defined operator is selfadjoint if and only if its
Cayley transform is unitary.
Proof The Cayley transform is unitary if and only if it is an isometric surjection
defined on all of H. Therefore the claim follows from Propositions 5.10 (c) and 5.11.
5.5 The Spectral Representation of an Unbounded Selfadjoint Operator
113
5.5 The Spectral Representation of an Unbounded
Selfadjoint Operator
We are now in a position to prove the spectral representation theorem for a general,
not necessarily bounded, selfadjoint operator. Now the spectral measure will be
defined on B(R), and in general there is no compact set whose complement would
be mapped to the null operator. The proof depends on the Cayley transform and the
spectral representation of a unitary operator.
Theorem 5.2 Let A be a selfadjoint operator in H. There is a uniquely determined
spectral measure E : B(R) → L(H) such that
ψ | Aϕ =
R
λ dEψ,ϕ (λ)
(5.8)
for all ϕ ∈ D(A) and ψ ∈ H. Moreover, D(A) is the set of those ϕ ∈ H for which
the function λ → λ2 on R is Eϕ,ϕ -integrable, so that
A=
R
λ dE(λ).
Proof We first prove the existence part. Let V be the Cayley transform of A. Since
V is unitary by Theorem 5.1, Corollary 4.3 yields a spectral measure E1 : B(T) →
L(H) satisfying V = T w dE1 (w). Proposition 4.13 shows that
(I − V )E1 ({1}) =
T
(1 − w)χ{1} (w) dE1 (w) = 0,
and since I − V is an injection by Proposition 5.10 (d), we must have E1 ({1}) = 0.
˜
Therefore the formula E(B)
= E1 (B), B ∈ B(T\{1}), defines a spectral measure
E˜ : B(T\{1}) → L(H) satisfying
T\{1}
˜
w d E(w)
= V.
(5.9)
We define g(w) = i(1 + w)(1 − w)−1 for all w ∈ T\{1}. Then g : T\{1} → R is a
homeomorphism and g −1 (x) = (x −i)(x +i)−1 . We now define E(B) = E˜ g −1 (B) ,
B ∈ B(R). Then E : B(R) → L(H) is a spectral measure, and for a vector ϕ ∈ H the
function |g|2 is E˜ ϕ,ϕ -integrable if and only if the function λ → λ2 is Eϕ,ϕ -integrable;
˜ : Dg → H, as
we let Dg denote the set of all such vectors ϕ. We define L(g, E)
usual, to be the operator satisfying
˜
ψ | L(g, E)ϕ
=
T\{1}
g d E˜ ψ,ϕ
(5.10)
114
5 Operator Integrals and Spectral Representations: The Unbounded Case
for all ϕ ∈ Dg , ψ ∈ H (see Proposition 5.8). Since g is real valued, the operator
L(g) is selfadjoint (Proposition 5.9), and since g(w)(1 − w) = i(1 + w), using
Proposition 5.8 we see that
L(g)(I − V ) = i(I + V );
R(I − V ) ⊂ D(L(g)).
(5.11)
D(A) = R(I − V ) ⊂ D(L(g))
(5.12)
On the other hand,
A(I − V ) = i(I + V );
by Proposition 5.10 (d). Thus in view of (5.11) and (5.12) L(g) is a selfadjoint
extension of A, implying by Proposition 5.3 that A = L(g). Since g : T\{1} → R is
a homeomorphism, by (5.10) we get ψ | Aϕ = R x dEψ,ϕ (x) for all ϕ ∈ D(A) =
Dg , ψ ∈ H. Thus (5.8) holds, and also A = R λ dE(λ), since Dg = D(L(id, E)).
We now prove the uniqueness claim. Let Eˆ : B(R) → L(H) be an arbitrary
spectral measure such that ψ | Aϕ = R λ d Eˆ ψ,ϕ for all ϕ ∈ D(A), ψ ∈ H.
Let g : T\{1} → R be the homeomorphism defined above. The formula E2 (X ) =
Eˆ g(X \{1}) defines a spectral measure E2 : B(T) → L(H). We define V1 =
T λ dE2 (λ) (∈ L(H)). Using Proposition 4.14 we see that
V1 =
T\{1}
λ dE2 (λ) =
R
ˆ
(x − i)(x + i)−1 d E(x).
Let ϕ ∈ D(A). If ψ ∈ H, by assumption the function x → x and hence x → |x| is
Eˆ ψ,ϕ -integrable (see Proposition 4.5(c)) so that the dominated convergence theorem,
Theorem 4.1, implies that R x d Eˆ ψ,ϕ (x) = limn→∞ ψ | An ϕ where
An =
R
ˆ
xχ[−n,n] (x) d E(x)
(∈ L(H)).
We also have ψ | I ϕ = ψ | ϕ = limn→∞ ψ In ϕ where In =
by Proposition 4.13(b)
R
ˆ Hence
χ[−n,n] d E.
ψ | V1 (A + i I )ϕ = V1∗ ψ | (A + i I )ϕ = lim V1∗ ψ | (An + i I )ϕ
n→∞
= lim ψ | V1 (An + i In )ϕ
n→∞
t −i
(t + i) d Eˆ ψ,ϕ (x)
n→∞ R
t +i
= ψ | (A − i I )ϕ = ψ | V (A + i I )ϕ
= lim
χ[−n,n] (x)
for all ψ ∈ H, ϕ ∈ D(A) (see Proposition 5.10). From the uniqueness part of
ˆ
Corollary 4.3 it thus follows that E1 = E2 , and this implies that E = E.
5.5 The Spectral Representation of an Unbounded Selfadjoint Operator
115
We call the spectral measure E in the preceding theorem the spectral measure of
the selfadjoint operator A. In the case of a bounded (normal) operator we saw that
the spectral measure is actually carried by the spectrum of the operator. In the next
section we consider the analogue of this for unbounded selfadjoint operators.
5.6 The Support of the Spectral Measure of a Selfadjoint
Operator
We assume in this section that A is a selfadjoint operator in H and E : B(R) → L(H)
is its spectral measure. In accordance with Theorem 5.2, E is characterised by the
condition A = R λ dE(λ).
We recall from Definition 4.6 the notion of the support of a positive operator
measure: Let U denote the union of all open sets V ⊂ R for which E(V ) = 0.
Then E(U ) = 0, and the complement R\U is the support of E, denoted by supp(E).
Clearly supp(E) is the smallest closed subset of R mapped to the identity operator I
by E. In the following we regard R in the usual way as a subset of C, so that supp(E)
is also a (closed) subset of C. We denote B(z, ) = w ∈ C |w − z| < for z ∈ C,
> 0.
Theorem 5.3 For a number z ∈ C the following conditions are equivalent:
(i) E R ∩ B(z, ) = 0 for some > 0;
(ii) z ∈ C\ supp(E);
(iii) there is an operator B ∈ L(H) such that B(z I − A) = I |D(A) and
(z I − A)B = I .
Proof The implication (i) =⇒ (ii), as well as its converse, follows immediately
from the definition.
Assume now (ii). 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 5.8 it follows that (z I − A)B = I and B(z I − A) = I |D(A). Thus
(iii) holds.
Now assume (iii). Suppose that (ii) does not hold. Then z is real and (by the
equivalence of (i) and (ii)) for each n ∈ N E (z − 1/n, z + 1/n) is a nonzero
projection, so there is a ϕn ∈ H such that ϕn = 1 and E (z − 1/n, z + 1/n) ϕn =
ϕn . According to Corollary 5.1 (c) ϕn ∈ D(A). Now
1 = ϕn = B(z I − A)E (z − 1/n, z + 1/n) ϕn
= B
(z−1/n,z+1/n)
(z − λ) dE(λ) ϕn ≤ B
1
ϕn → 0,
n
116
5 Operator Integrals and Spectral Representations: The Unbounded Case
when n → ∞ (see Propositions 5.8, 4.13). This contradiction shows that z ∈
/
supp(E).
The set of those z ∈ C for which the condition (iii) in the above theorem does
not hold, is called the spectrum of A and denoted by σ(A). We have thus proved that
σ(A) = supp(E). In particular, σ(A) ⊂ R for the selfadjoint operator A, and the
spectral representation of A may be written in the form A = σ(A) λ dE(λ). As in
the case of a bounded normal operator, the spectral measure E restricted to the Borel
σ-algebra of σ(A) may also be called the spectral measure of A. The complement
C\σ(A) is called the resolvent set of A and denoted by ρ(A).
It is clear that this definition of σ(A) and ρ(A) is consistent with that which was
used in connection with bounded operators. (See the discussion preceding Theorem 3.4.)
For general (bounded) operators T ∈ L(H) we considered the subdivision of the
spectrum of T into three parts: σ(T ) = σ p (T ) ∪ σc (T ) ∪ σr (T ). We can generalise
this to the case of the not necessarily bounded selfadjoint operator A, but in this case
the residual spectrum σr (A) turns out to be empty. To see this we first prove a lemma.
Lemma 5.3 (a) For the selfadjoint operator A and z ∈ R, denote A z = z I − A. If
A z is an injective operator, then its range is dense in H.
(b) If T : D(T ) → L(H) is a closed operator which is continuous on D(T ), then
D(T ) is closed.
Proof (a) A vector ψ ∈ H is orthogonal to R(A z ) if and only if ψ | A z ϕ = 0 for
all ϕ ∈ D(A z ) = D(A), i.e. if and only if ψ ∈ D((A z )∗ ) and (A z )∗ ψ = 0. Since
clearly A z = (A z )∗ , this is equivalent to ψ ∈ D(A z ) and (A z )ψ = 0. Since A z
is injective, ψ = 0. Thus R(A z ) is dense in H.
(b) Suppose (ϕn ) is a sequence in D(T ) converging to some ϕ ∈ H. Since T is
continuous (T ϕn ) is a Cauchy sequence, and so there is some ψ ∈ H such that
limn→∞ T ϕn = ψ. Since T is closed, (ϕ, ψ) ∈ G(T ), so ϕ ∈ D(T ).
Proposition 5.12 (a) A number z ∈ C belongs to ρ(A) if and only if z I − A is
injective and its inverse (z I − A)−1 : R(z I − A) → D(A) is continuous. If this
is the case, then R(z I − A) = H.
(b) If z ∈ σ(A), then one and only one of the following two conditions holds:
(i) there is some ϕ ∈ D(A)\{0} satisfying Aϕ = zϕ;
(ii) the operator z I − A is injective, its range is dense in H and the inverse
(z I − A)−1 : R(z I − A) → D(A) is discontinuous.
Proof (a) Suppose z ∈ R and z I − A is injective. By Lemma 5.3 (a) D(z I − A)
is dense in H. It is easily seen that z I − A is closed (since A is) and so is its
inverse (since the map (ξ, η) → (η, ξ) is a homeomorphism on H × H). Thus if
(z I − A)−1 is continuous, its domain is all of H by Lemma 5.3 (b), and it follows
that z ∈ ρ(A). The rest of the claim is covered by Theorem 5.3.
(b) This follows immediately from (a).
5.6 The Support of the Spectral Measure of a Selfadjoint Operator
117
The set of those points in σ(A) which satisfy (i) in Proposition 5.12 (b) is called
the point spectrum of A. Its elements are the eigenvalues of A, and a vector ϕ ∈
D(A)\{0} satisfying Aϕ = zϕ is an eigenvector belonging to the eigenvalue z. The
set σc (A) = σ(A)\σ p (A) is called the continuous spectrum of A. Thus σc (A) is
characterised by condition (ii) in Proposition 5.12 (b). In view of Proposition 5.12
this terminology is consistent with the usage of Sect. 3.2. In particular, the residual
spectrum of a selfadjoint operator is empty.
Proposition 5.13 Suppose that x ∈ R. The following conditions are equivalent:
(i) E({x}) = 0;
(ii) Eϕ,ϕ ({x}) = 1 for some ϕ ∈ D(A), ϕ = 1;
(iii) x ∈ σ p (A).
Proof Assume first (i). By Corollary 5.1 (c) E({x}) maps H into D(A), so by choosing
ϕ ∈ E({x})(H), ϕ = 1, we get a vector satisfying (ii).
Next assume (ii). From Lemma 5.2 it follows that
Aϕ − xϕ
2
=
R
|λ − x|2 dEϕ,ϕ (λ)
(5.13)
for all ϕ ∈ D(A). Since Eϕ,ϕ ({x}) = 1 = ϕ 2 = Eϕ,ϕ (R), we have Eϕ,ϕ (R\{x}) =
0, so (5.13) implies that Aϕ = xϕ. Hence x ∈ σ p (A).
Finally assume (iii). Choose ϕ ∈ D(A)\{0} such that Aϕ = xϕ. By (5.13) R |λ−
x|2 dEϕ,ϕ (λ) = 0, so the function λ → λ − x vanishes Eϕ,ϕ -almost everywhere.
Thus Eϕ,ϕ (R\{x}) = 0, so that E({x})ϕ 2 = Eϕ,ϕ ({x}) = Eϕ,ϕ (R) = ϕ 2 > 0.
Thus (i) holds.
Corollary 5.2 If x is an isolated point of σ(A), then x is an eigenvalue of A.
Proof If x is an isolated point of σ(A), there is some > 0 such that σ(A) ∩ (x −
, x + )) = {x}. Then E({x}) = 0 by Theorem 5.3, and the above proposition implies
the claim.
Simple examples show that the converse of the above corollary does not hold.
5.7 Applying a Borel Function to a Selfadjoint Operator
We assume in this section that A is a selfadjoint operator in H and E : B(R) → L(H)
is its spectral measure.
Definition 5.6 If X ∈ B(R) is a set such that E(X ) = I , and f : X → C is a
B(X )-measurable function, we denote f (A) = X f (λ)dE(λ).
118
5 Operator Integrals and Spectral Representations: The Unbounded Case
The notation f (A) is clearly unambiguous in the sense that for the choice of another
Y ∈ B(R) with E(Y ) = I and a B(Y )-measurable function g : Y → C such that f
and g agree on X ∩ Y , we have X f (λ)dE(λ) = Y g(λ)dE(λ). We may also denote
f (A) = L( f, E) consistently with our earlier usage (where f was defined on all of
R).
Example
√ 5.1 (a) If A ≥ 0 and1 f : [0, ∞) → R is the function defined by f (λ) =
λ, we denote f (A) = A 2 ; this is the square root of A. In Example 5.2 it will
1
be shown that A 2 is the only positive operator whose square is A. Thus this
1
definition of A 2 is consistent with the notion of the square root of a bounded
positive operator considered earlier.
(b) If f (λ) = λ2 , noting that |λ2 |2 ≥ |λ|2 when |λ| ≥ 1, we see from Proposition 5.8 that f (A) = A2 . By induction it is seen that f (A) = An if f (λ) = λn ,
n ∈ N.
According to Corollary 5.1 (a) f (A) is always a closed operator, and if f is real
valued, then f (A) is selfadjoint (Proposition 5.9). The following result shows the
connection between the spectral measures of A and f (A) in case f is real valued.
Theorem 5.4 Suppose X ∈ B(R) and let f : X → R be a B(R)-measurable
function. Assume that E(X ) = I . Let E f (A) : B(R) → L(H) be the spectral measure
of the selfadjoint operator f (A). Then
(a) E f (A) (Y ) = E( f −1 (Y )) for all Y ∈ B(R) and
(b) if Z is a Borel set containing the set f (X ) and g : Z → C is a B(Z )-measurable
function, then E f (A) (Z ) = I and g( f (A)) = (g ◦ f )(A).
Proof Let Z ⊂ f (X ) be a Borel set and g : Z → C a B(Z )-measurable function. Let
E f denote the spectral measure defined by E f (Y ) = E( f −1 (Y )) for all Y ∈ B(R).
f
(Y ) = Eϕ,ϕ ( f −1 (Y )), and so using the well-known theorem
If ϕ ∈ H, we have Eϕ,ϕ
on transformations of integrals for positive measures we see that the function | f |2
f
is Eϕ,ϕ
-integrable if and only if the composite function |g ◦ f |2 = |g|2 ◦ f is Eϕ,ϕ integrable. (As E f (Z ) = E( f −1 (Z )) = E(X ) = I , we may consider E f restricted
to B(Z ).) It follows that Dgf = Dg◦ f where the (obvious) notation Dgf refers to the
spectral measure E f . Let now ϕ ∈ Dgf = Dg◦ f . Choose a sequence of bounded
B(Z )-measurable functions gn such that |gn | ≤ |g| and limn→∞ gn = g pointwise.
For every ψ ∈ H we then get
ψ|(
Z
gdE f )ϕ =
= limn→∞
X (gn
f
Z
gdEψ,ϕ = limn→∞
f
◦ f )dEψ,ϕ = ψ | (
X (g
f
Z
gn dEψ,ϕ
◦ f )dE)ϕ
by the dominated convergence theorem and Proposition 4.14, for we have |gn ◦ f | ≤
|g ◦ f | where the function |g ◦ f | is Eψ,ϕ -integrable (see Proposition 5.5). Thus we
have seen that
gdE f =
Z
(g ◦ f )dE.
X
(5.14)