5 Connection of the Ideals mathcalT(mathcalH) and mathcalHS(mathcalH) with the Sequence Spaces ell1 and ell2
Tải bản đầy đủ - 0trang
50
3 Classes of Compact Operators
∞
T =
cn |ψn ϕn |
(3.9)
n=1
where cn ≥ 0, lim cn = 0 and (ϕn ) and (ψn ) are orthonormal sequences in H.
n→∞
(We assume in this and the next section that H is infinite-dimensional; the simpler
finite-dimensional case would only require minor changes in notation.) We denote
as usual
1
= (cn ) ∈ RN
∞
|cn | < ∞ and
2
∞
= (cn ) ∈ RN
|cn |2 < ∞ .
n=1
n=1
The following theorem explains the connections mentioned in the title of this
section.
Theorem 3.10 In the situation of formula (3.9) we have
∞
(a) T ∈ HS(H), if and only if (cn ) ∈
2
, and then T
2
=
cn2
1
2
;
n=1
∞
(b) T ∈ T (H), if and only if (cn ) ∈
1
, and then
T
1
=
cn and tr[T ] =
n=1
∞
cn ϕn |ψn .
n=1
Proof First, suppose that T ∈ HS(H). Choose for H an orthonormal basis K containing the set {ϕn |n ∈ N} (see Theorem 2.21). Then
∞
Tξ
ξ∈K
2
=
∞
cn ψn
2
∞
=
n=1
cn2 , for T ξ =
n=1
cm ϕm |ξ ψm = cn ψn ,
m=1
if ξ = ϕn , and T ξ = 0, if ξ ∈ K \ {ϕn }. This implies (a). Since
1
∞
1
|T | 2 =
cn2 |ϕn ϕn |
n=1
(see Remark 3.1), from the above we see that
the first claim in (b) and the equality T 1 =
∞
ξ∈K ξ| |T |ξ =
n=1 cn , implying
∞
n=1 cn . If T ∈ T (H), we further get
∞
tr T =
ξ|T ξ =
ξ∈K
∞
ϕn |cn ψn =
n=1
cn ϕn |ψn .
n=1
3.5 Connection of the Ideals T (H) and HS (H) with the Sequence Spaces
1
and
2
51
We now list some immediate consequences of the preceding theorem.
Theorem 3.11 (a) If ϕ, ψ ∈ H then |ϕ ψ| ∈ T (H) (⊂ HS(H)) and
|ϕ ψ|
1
=
|ϕ ψ|
2
= ϕ
ψ .
n
n
(b) If T ∈ F(H) then T ∈ T (H). If T =
|ϕi ψi | then tr[T ] =
i=1
ψi |ϕi .
i=1
(c) If P ∈ L(H) is a projection then P ∈ T (H) if and only if P ∈ F(H), and then
tr[P] = dim P(H).
(d) If T ∈ T (H), then the series ∞
n=1 cn |ψn ϕn | in (3.9) converges with respect to
the norm · 1 , and if T ∈ HS(H), then it converges with respect to the norm
· 2.
(e) The set F(H) is dense in T (H) with respect to the norm · 1 and in the set
HS(H) with respect to the norm · 2 .
(f) If T ∈ T (H) then T ≤ T 2 ≤ T 1 .
Proof (a) We may assume that ϕ = 0, ψ = 0. As
|ϕ ψ| = ϕ
−1
ϕ
ψ
ϕ
ψ
−1
ψ,
the claim follows from Theorem 3.10.
(b) If ϕi = 0, ψi = 0, then by Theorem 3.10 (b)
tr |ϕi ψi | = ϕi
ψi
ψi
−1
ϕi
ψi
−1
ϕi = ψi |ϕi ,
and the claim follows form the linearity of the trace.
(c) If P ∈ T (H), then P ∈ C(H) (see Lemma 3.2 (b), Theorem 3.7), so that P ∈
F(H) (Theorem 3.3 (f)). The equality tr P = dim(P(H)) follows from (b) and
Theorem 3.3 (a).
(d) The claims follow from Theorem 3.10, since for example
∞
p
cn |ψn ϕn |
T−
n=1
2
=
cn2
1
2
→ 0, when p → ∞.
n= p+1
(e) This is a direct consequence of (d).
(f) The inequality T
∞
then
n=1
(cn c−1 )2 ≤
∞
n=1
2
≤ T
∞
1 follows from Theorem 3.10, for if 0 = c =
cn c−1 = 1, so that
∞
n=1
cn
n=1
(cn c−1 )2
1
2
≤ 1, and so
52
3 Classes of Compact Operators
∞
cn2
1
2
≤ c.
n=1
The inequality T ≤ T
2
was already seen in Theorem 3.7 (a).
3.6 The Dualities C(H)∗ = T (H) and T (H)∗ = L(H)
The trace class of the Hilbert space H has an important role in operator theory. For
example, equipped with the norm introduced in Definition 3.5, as a normed space
T (H) can be identified with the dual of the space C(H) of compact operators, and
the dual of T (H) in turn with L(H). We prove these results in this section.
Theorem 3.12 Define for each T ∈ T (H) the mapping f T : C(H) → C via the formula f T (S) = tr T S , S ∈ C(H). In this way we obtain a linear isometric bijection
T → f T from the space T (H) (equipped with the norm · 1 ) onto the dual of C(H).
Proof By Theorem 3.8 f T is a linear functional defined on C(H). It follows from Theorems 3.9 and 3.8 that f T is continuous and f T ≤ T 1 . Let now f ∈ C(H)∗ . Since
by Theorem 3.7 HS(H) ⊂ C(H) and S 2 ≥ S for all S ∈ HS(H), the restriction f |HS(H) is a continuous linear functional on the Hilbert space HS(H), so that
by the Fréchet–Riesz theorem there is a T ∈ HS(H) satisfying f (S) = T ∗ |S =
∗
ξ∈K T ξ|Sξ =
ξ∈K ξ|T Sξ = tr T S for all S ∈ HS(H) (here T S ∈ T (H)
by Lemma 3.2 (a)). We prove that T ∈ T (H), f = f T and f T ≥ T 1 . Let
T = V |T | be the polar decomposition of T . Then |T | = V ∗ T and hence by Theorem 3.8
tr S|T | = tr SV ∗ T
= |tr T SV ∗ | = | f (SV ∗ )| ≤ f
S
(3.10)
whenever S ∈ F(H) (implying SV ∗ ∈ F(H) ⊂ T (H)). Let now
∞
|T | =
cn |ϕn ϕn |
n=1
be the representation given by (3.1”) (so that cn ≥ 0). If Pk is the projection onto the
subspace spanned by ϕ1 , . . . , ϕk , then by (3.10) and Theorem 3.10 (b)
k
k
cn = tr
n=1
cn |ϕn ϕn | = tr Pk |T | ≤ f ,
n=1
3.6 The Dualities C (H)∗ = T (H) and T (H)∗ = L(H)
53
so that ∞
k=1 ck < ∞, and by Theorem 3.10 (b) |T | ∈ T (H). Hence T ∈ T (H) and
T 1 = tr |T | ≤ f . The functional f and f T , which are continuous functions on
C(H), agree on the dense subspace F(H), and so f = f T . The mapping T → f T is
clearly linear, and when it is shown to be an injection, the theorem is proved. But if
T ∈ T (H) is such that f T = 0, then ϕ|T ψ = tr T |ψ ϕ| = f T (|ψ ϕ|) = 0 for
all ϕ, ψ ∈ H (see Theorems 3.3 (b), 3.11 (b)), so that T = 0.
Since C(H)∗ is a Banach space, we immediately obtain the following corollary:
Corollary 3.2 The space (T (H), ·
1)
is complete.
Theorem 3.13 For each S ∈ L(H) define the mapping f S : T (H) → C via the formula f S (T ) = tr ST for all T ∈ T (H). In this way we get an isometric linear bijection S → f s from the space L(H) onto the dual of the Banach space (T (H), · 1 ).
Proof From Theorem 3.8 it follows that each f S is defined on T (H) and linear.
By Theorem 3.9 | f S (T )| = |tr ST | ≤ S T 1 for all T ∈ T (H), so that f S is
continuous and f S ≤ S . The mapping S → f S is clearly linear. If ϕ, ψ ∈ H,
then by using Theorems 3.3 (b) and 3.11 (b) we see that
| ϕ|Sψ | = |tr[|Sψ ϕ|]| = | f S (|ψ ϕ|)|
≤ fS
|ψ ϕ| 1 = f S ψ
ϕ
so that
S = sup | ϕ|Sψ |
ϕ ≤ 1, ψ ≤ 1 ≤ f S .
Thus the mapping S → f S is isometric and in particular injective. We still show that it
is a surjection onto the dual of T (H). Let f ∈ T (H)∗ . We define a mapping B : H ×
H → C via the formula B(ϕ, ψ) = f (|ψ ϕ|). Then |B(ϕ, ψ)| ≤ f
|ψ ϕ| 1 =
f ϕ ψ by Theorem 3.11 (a). Since B is also conjugate linear with respect to
the first and linear with respect to the second argument, by Proposition 2.1 there is
S ∈ L(H) such that B(ϕ, ψ) = ϕ|Sψ for all ϕ, ψ ∈ H. We show that f = f S , i.e.
tr ST = f (T ) for all T ∈ T (H).
(3.11)
Both sides of (3.11) are linear functions of T that are continuous on T (H) with
respect to the norm · 1 (see Theorem 3.9). Since F(H) is dense in T (H) (see
Theorem 3.11 (e)), in view of Theorem 3.3 (c) it is enough to show that (3.11) holds
whenever T = |ψ ϕ| for some ϕ, ψ ∈ H. But according to Theorems 3.3 (b) and
3.11 (b) we get
tr S|ψ ϕ| = tr |Sψ ϕ| = ϕ|Sψ = B(ϕ, ψ) = f (|ψ ϕ|).