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3 The Hilbert--Schmidt Operator Class mathcalHS(mathcalH)

# 3 The Hilbert--Schmidt Operator Class mathcalHS(mathcalH)

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46

3 Classes of Compact Operators

Proof We first show that HS(H) is a vector subspace. If S, T ∈ HS(H) then clearly

cS ∈ HS(H) for all c ∈ C, and since

(S + T )ξ

2

= Sξ

2

+ Sξ|T ξ + T ξ|Sξ + T ξ

≤ Sξ

2

+ 2 Sξ

Tξ + Tξ

2

2

≤ 2 Sξ

2

+ 2 Tξ

2

for all ξ ∈ K , we have S + T ∈ HS(H). For every finite F ⊂ K we may estimate

using the Cauchy–Schwarz inequality

| Sξ|T ξ | ≤

ξ∈F

2

1

2

ξ∈F

2

1

2

≤ S

2

T

2,

ξ∈F

and it follows that S|T is defined as the sum of a summable family whenever

S, T ∈ HS(H). It is easily seen that in this way we get an inner product for the

space HS(H). For example, if T |T = 0, then Lemma 3.1 (c) shows that T = 0.

As T |T = T 22 is by Lemma 3.1 (b) independent of the choice of K , using the

polarisation identity it is seen that S|T is independent of the choice of K . We still

show the completeness of the space HS(H). Let (Tn ) be a Cauchy sequence in it with

respect to the norm · 2 , so that in view on Lemma 3.1 (c) it is a Cauchy sequence

also with respect to the ordinary operator norm of L(H). Hence there is a T ∈ L(H)

satisfying lim Tn − T = 0. Let ε > 0. Choose a p ∈ N such that Tn − Tm 2 < ε

n→∞

whenever n, m ≥ p. Let {ξ1 , . . . , ξk } be an arbitrary finite subset of K and n ≥ p.

Then

k

k

(Tn − T )ξ j

2

= lim

m→∞

j=1

(Tn − Tm )ξ j

2

≤ ε2 .

j=1

It follows that Tn − T ∈ HS(H), hence T = Tn − (Tn − T ) ∈ HS(H) and Tn −

T 2 ≤ ε. Therefore lim Tn − T 2 = 0.

n→∞

We still prove a theorem showing that HS(H) is a ∗-ideal of L(H).

Theorem 3.7 (a) If T ∈ HS(H) then T ∗ ∈ HS(H) and T ∗ 2 = T

(b) If S ∈ HS(H) and T ∈ L(H)), then ST, T S ∈ HS(H) and

ST

TS

2

2

≤ S

≤ T

2

2

≥ T .

T and

S 2.

(c) A projection P is in HS(H) if and only if P(H) is finite-dimensional, and then

P 22 = dim(P(H)).

(d) F(H) ⊂ HS(H) ⊂ C(H).

Proof Item (a) follows from Lemma 3.1 (b), (c). Moreover,

T Sξ

ξ∈K

2

≤ T

2

ξ∈K

2

= T

2

S 22 ,

3.3 The Hilbert–Schmidt Operator Class HS (H)

47

so that T S ∈ HS(H) and T S 2 ≤ T

S 2 . By (a) we have ST = (T ∗ S ∗ )∗ ∈

∗ ∗

S∗ 2 = T

S 2 . Let now P ∈ L(H)

HS(H) and ST 2 = T S 2 ≤ T

be a projection. Choose for the Hilbert space P(H) an orthonormal basis L 0 and

extend it to an orthonormal basis L of H (see Theorem 2.21). Since P(L\L 0 ) ⊂

{0}, we see that ξ∈L Pξ 2 = ξ∈L 0 ξ 2 immediately implying (c). (d) Every

|ϕ ψ| is in HS(H), since ξ∈K ψ|ξ ϕ 2 ≤ ψ 2 ϕ 2 (Lemma 2.2 (b)). Hence

by Theorems 3.3 (a) and 3.6 we have F(H) ⊂ HS(H). Now let T ∈ HS(H). Since

2

< ∞, we have T ξ = 0 for each ξ ∈ K with the possible exception

ξ∈K T ξ

of a countable set of vectors ξ1 , ξ2 , . . . ∈ K . Let Pn be the projection onto the (by

Theorem 3.3 (a) closed) subspace spanned by {ξ1 , . . . , ξn }. Then by Lemma 3.1 (c)

T − T Pn

2

≤ T − T Pn

2

2

=

T (I − Pn )ξ

ξ∈

K \{ξ1 ,...,ξn }

2

=

T ξj

2

→ 0,

j=n+1

when n → ∞. Since always T Pn ∈ F(H), we have T ∈ C(H) by Theorem 3.2.

3.4 The Trace Class T (H)

In linear algebra the trace of a matrix is defined as the sum of its main diagonal

elements. Even in the case of infinite-dimensional Hilbert spaces the notion of a

trace can be extended to certain operators, those belonging to the so-called trace class

which we now define. As in the previous section, in this section K is an arbitrary

orthonormal basis of H. In the next definition the expression ξ∈K ξ|T ξ is in view

1

of Lemma 3.1 independent of the choice of K , since T = S ∗ S = SS ∗ , where S = T 2 .

Definition 3.3 For each T ∈ Ls (H)+ we denote

tr T =

ξ|T ξ

(≤ ∞).

(3.4)

ξ∈K

We write T (H) = T ∈ L(H)|tr |T | < ∞ and call the set T (H) the trace class.

The trace class has a close connection with the Hilbert–Schmidt operators:

Lemma 3.2 (a) T (H) = {ST |S, T ∈ HS(H)}.

(b) T (H) ⊂ HS(H).

Proof (a) Assume first that S, T ∈ HS(H). Let ST = U |ST | be the polar decomposition of ST . Then

ξ|U ∗ ST ξ =

ξ| |ST |ξ =

ξ∈K

ξ∈K

(U ∗ S)∗ ξ|T ξ ≤ U ∗ S

ξ∈K

2

T

2

<∞

48

3 Classes of Compact Operators

by the Cauchy–Schwarz inequality in HS(H), so that ST ∈ T (H). Next suppose that

1

1

V ∈ T (H) and let V = W |V | be its polar decomposition. Then V = W |V | 2 |V | 2 ,

1

1

where by definition |V | 2 ∈ HS(H), and according to Theorem 3.7 (b) also W |V | 2 ∈

HS(H).

(b) This follows from (a) and Theorem 3.7 (b).

Before defining the trace we prove another auxiliary result.

Lemma 3.3 If T ∈ T (H) and K , L are orthonormal bases in H, then the families

( ξ|T ξ )ξ∈K and ( η|T η )η∈L are summable and

ξ|T ξ =

η|T η .

(3.5)

η∈L

ξ∈K

Proof By Lemma 3.2 T = AB where A, B ∈ HS(H). Since ξ|T ξ = A∗ ξ|Bξ ,

the claim follows from Theorem 3.6, for also A∗ ∈ HS(H) by Theorem 3.7.

Definition 3.4 If T ∈ T (H) the number tr T = ξ∈K ξ|T ξ (which according to

the preceding lemma is independent of the choice of K ) is called the trace of T .

Theorem 3.8 The trace class T (H) is a ∗-ideal of L(H), and the trace tr · is a

linear functional on it. The trace is unitarily invariant, i.e.

tr T = tr U ∗ T U

(3.6)

whenever T ∈ T (H) and U ∈ L(H) is unitary. Moreover,

tr ST = tr T S

(3.7)

whenever S ∈ L(H) and T ∈ T (H).

Proof If T = AB with A, B ∈ HS(H) (see Lemma 3.2), then T ∗ = B ∗ A∗ ∈ T (H)

by Theorem 3.7 (a) and Lemma 3.2. From Lemma 3.2 and Theorem 3.7 (b) it also

follows that T (H) is closed with respect to multiplication by elements of L(H). If

S, T ∈ T (H) and T + S = U |S + T | is the polar decomposition of the sum S + T .

Then by Theorem 2.18 |S + T | = U ∗ (S + T ), and since U ∗ S, U ∗ T ∈ T (H), we

have

ξ| |S + T |ξ ≤

| ξ|U ∗ Sξ | +

| ξ|U ∗ T ξ | < ∞.

ξ∈K

ξ∈K

ξ∈K

Hence T (H) is a ∗-ideal. From Definition 3.4 one sees immediately the linearity of

the trace. If U ∈ L(H) is unitary, then {U ξ | ξ ∈ K } is an orthonormal basis of H,

and so by Lemma 3.3

tr U ∗ T U =

U ξ|T U ξ = tr T

ξ∈K

when T ∈ T (H).

3.4 The Trace Class T (H)

49

Thus (3.6) holds and tr U T = tr U ∗ U T U = tr T U . From this (3.7) follows,

since every S ∈ L(H) is a linear combination of four unitary operators by Proposition 2.6.

Also T (H) turns out to be a Banach space with respect to a suitable norm (which

is in general greater than the ordinary operator norm). We next define this norm and

prove some of its basic properties. The completeness will be proved later.

Definition 3.5 For each T ∈ T (H) denote

T

1

Theorem 3.9 The mapping T → T

|tr ST | ≤ S

1

= tr |T | (=

T

1

|T | 2

2

2 ).

is a norm in T (H), and we have

for all S ∈ L(H), T ∈ T (H).

1

(3.8)

Proof Let T, T0 ∈ T (H). We first prove (3.8). If T = U |T | is the polar decomposition of T and S ∈ L(H), we have

1

1

|tr ST | = |tr SU |T | | = tr SU |T | 2 |T | 2

|T | 2 U ∗ S ∗ ξ| |T | 2 ξ

1

=

1

ξ∈K

1

2

∗ ∗

= |T | U S | |T |

U∗

1

|T | 2

2

1

2

S∗

1

2

∗ ∗

2

= S

|T | U S

1

|T | 2

1

2

|T | 2

T

2

1

by using the Cauchy–Schwarz inequality and Theorem 3.7 (b). If λ ∈ C, then

1

1

λT 1 = |λT | 2 22 = |λ| T 1 . If T 1 = |T | 2 22 = 0, then |T | = 0, so that

T 2 = T ∗ T = 0, i.e. T = 0. Let now T + T0 = U |T + T0 | be the polar decomposition of the sum T + T0 . Then by Theorem 2.18 we get

T + T0

1

= tr[|T + T0 |] = tr[U ∗ T + U ∗ T0 ] = tr[U ∗ T ] + tr[U ∗ T0 ]

≤ |tr[U ∗ T ]| + |tr[U ∗ T0 ]| ≤ U ∗ T 1 + U ∗ T0 1

≤ T

1

+ T0 1 .

Thus the mapping T → T

1

is a norm.

3.5 Connection of the Ideals T (H) and HS(H)

with the Sequence Spaces 1 and 2

We know that T (H) ⊂ HS(H) ⊂ C(H) (Lemma 3.2, Theorem 3.7). In view of

Remark 3.1 the compact operators, on the other hand are precisely those having

the representation

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

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

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