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ON OPERATOR INTERPOLATION PROBLEMS

Young Soo Jo , Joo Ho Kang and Ki Sook Kim

Abstract. In this paper we obtained the following : Let H be a Hilbert space and L be a subspace lattice on H. Let X and Y be operators acting on H. If the range of X is dense in H, then the following are equivalent:

(1) there exists an operator A in AlgL such that AX = Y , (2) sup

 E

Y f

E

Xf : f ∈ H, E ∈ L



= K < ∞.

Moreover, if condition (2) holds, we may choose the operator A such that A = K.

1. Introduction

On the process of solving operator equation AX = Y for two given X and Y in the algebra B(H), the class of all bounded operators acting on a Hilbert space H, many mathematicians investigated the problem on their field of studies. What is the condition for an operator A satisfying AX = Y to be a member of A which is a specified subalgebra of B(H)?

Douglas [2] used the range inclusion property of operators to show necessary and sufficient conditions for the existence of an operator A satisfying AX = Y . In [7], authors investigated the problem on nest algebra. Hopenwasser [3] studied on CSL and Moore [9] showed the condition is a necessary and sufficient condition for the existence of an interpolating operator in Alg L when L is a CSL and X has dense range.

In [5], authors has found the other necessary and sufficient condition for interpolating operator when L is a subspace lattice and X has dense range. In this paper, we are going to show that the condition given in [9] holds for a subspace lattice L.

Received March 3, 2003.

2000 Mathematics Subject Classification: 47L35.

Key words and phrases: interpolation problem, subspace lattice, Alg L, CSL.

This work was supported by Korea Research Foundation Grant(KRF-99-015-

DP0016).

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Let H be a Hilbert space. A subspace lattice L is a strongly closed lat- tice of orthogonal projections in B(H) containing the trivial projections 0 and I. The symbol Alg L denotes the algebra of bounded operators on H that leave invariant every projection in L; AlgL is a weakly closed subalgebra of B(H). A lattice L is a commutative subspace lattice, or CSL, if the projections in L all commute; in this case, AlgL is called a CSL algebra.

We want to introduce the simplest case of the operator interpolation problem that relaxes all restrictions on A, requiring it simply to be a bounded operator. In this case, the existence of A is nicely characterized by the well-known factorization theorem of Douglas.

Theorem A. ([2], [4]) Let X and Y be bounded operators acting on a Hilbert space H. Then the following statements are equivalent:

(1) range Y ⊆ range X ,

(2) Y Y ≤ λ

2

X X for some λ ≥ 0,

(3) there exists a bounded operator A on H so that AX = Y . Moreover, if (1), (2), and (3) are valid, then there exists a unique oper- ator A so that

(a) A

2

= inf {µ : Y Y ≤ µX X}, (b) ker Y = ker A and

(c) range A ⊆ range X .

2. Operator interpolation problems in B(H)

Let H be a Hilbert space and M a subset of H. Then M means the closure of M and M is the orthogonal complement of M.

We use the convention

00

= 0, when necessary.

We will introduce theorems that are obtained easily.

Theorem 2.1. Let X and Y be bounded operators acting on H.

Then the following are equivalent:

(1) there exists λ( ≥ 0) such that Y Y ≤ λ

2

X X,

(2) there exists λ( ≥ 0) such that Y f ≤ λXf for every f ∈ H, (3) sup

 Y f

Xf : f ∈ H



<∞.

We can get the following theorem by Theorems A and 2.1.

Theorem 2.2. Let X and Y be bounded operators acting on H.

Then the following are equivalent:

(1) range Y ⊆ range X ,

(3)

(2) there exists λ( ≥ 0) such that Y Y ≤ λ

2

X X,

(3) there exists an operator A in B(H) such that AX = Y ,

(4) there exists λ( ≥ 0) such that Y f ≤ λXf for every f ∈ H, (5) sup

 Y f

Xf : f ∈ H



< ∞.

Theorem 2.3. Let X

1

, X

2

, . . . , X n and Y

1

, Y

2

, . . . , Y n be bounded operators acting on H. Then the following are equivalent:

(1) there exists an operator A in B(H) such that AX i = Y i for i = 1, 2, . . . , n,

(2) sup

   n

i=1 Y i f i 

  n

i=1 X i f i  : f i ∈ H



< ∞.

Proof. If sup

   n

i=1 Y i f i 

  n

i=1 X i f i  : f i ∈ H



< ∞, then we may assume that sup

   n

i=1 Y i f i 

  n

i=1 X i f i  : f i ∈ H



= 1. Then



 n i=1

Y i f i  ≤ 

 n i=1

X i f i 

for all f i in H. Let M = {  n

i=1 X i f i : f i ∈ H }. Then M is a linear manifold. We define a function A : M −→ H by A(  n

i=1 X i f i ) =

 n

i=1 Y i f i for f i in H. Then A is well-defined.

For, if  n

i=1 f i f i =  n

i=1 X i g i , then  n

i=1 X i f i − (  n

i=1 X i g i ) = 0.

So,   n

i=1 X i f i − (  n

i=1 X i g i )  =   n

i=1 X i (f i − g i )  = 0. Since

  n

i=1 Y i (f i − g i )  ≤   n

i=1 X i (f i − g i )  = 0,  n

i=1 X i f i =  n

i=1 X i g i . And A is continuous. Extend A to M by continuity and define Ah = 0 for h ∈ M . Then, A is a bounded operator on H clearly. Moreover, AX i = Y i for i = 1, 2, . . . , n.

The converse is proved easily. 

We can extend the above fact for infinitely many operators easily.

Theorem 2.4. Let X i and Y i be bounded operators acting on H for each i in N. Then the following are equivalent:

(1) there exists an operator A in B(H) such that AX i = Y i for i = 1, 2, . . . ,

(2) sup

   n

i=1 Y i f i 

  n

i=1 X i f i  : f i ∈ H and n ∈ N



< ∞.

(4)

Theorems 2.3 and 2.4 show that if we get necessary and sufficient conditions which are equivalent to the second case of the above Theo- rems, then it gives conditions for the existence of an operator A such that AX i = Y i for any bounded operators X i and Y i . So we want to find conditions that are equivalent to the second case.

Theorem 2.5. Let X

1

, . . . , X n and Y

1

, . . . , Y n be bounded operators acting on H. Assume that Re < X i f, X j g >≥ 0 for i < j and all f, g in H. If there exists λ(≥ 0) such that Y i Y i ≤ λ

2

X i X i for each i = 1, 2, . . . , n, then sup

   n

i=1 Y i f i 

  n

i=1 X i f i  : f i ∈H



< ∞.

Proof. If Y i Y i ≤ λ

2

X i X i for each i = 1, 2, . . . , n, then Y i f

X i f ≤ λ for each i = 1, 2, . . . , n. Since Re <X i f, X j g>≥ 0 for i < j, X k f k  ≤

  n

i=1 X i f i  for each k = 1, 2, . . . , n. For,



 n i=1

X i f i 

2

= < X k f k , X k f k > + < 

i=k

X i f i , X k f k >

+ < X k f k , 

i=k

X i f i > + < 

i=k

X i f i , 

i=k

X i f i >

= X k f k 

2

+  

i=k

X i f i 

2

+ 

i<j

2Re < X i f i , X j f j >≥ X k f k 

2

.

Hence nλ ≥

 n i=1

Y i f i 

X i f i 

 n i=1

Y i f i 

  n

i=1 X i f i  =

 n

i=1 Y i f i 

  n

i=1 X i f i    n

i=1 Y i f i 

  n

i=1 X i f i  . So, sup

   n

i=1 Y i f i 

  n

i=1 X i f i  : f i ∈ H



< ∞. 

From Theorem 2.2, we have the following corollary.

Corollary 2.6. Let X

1

, . . . , X n and Y

1

, . . . , Y n be bounded oper- ators acting on H. Assume that Re < X i f, X j g >≥ 0 for i < j and all f, g in H. If range Y i ⊆ range X i for each i = 1, 2, . . . , n, then sup

   n

i=1 Y i f i 

  n

i=1 X i f i  : f i ∈ H



< ∞.

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Theorem 2.7. Let X

1

, . . . , X n and Y

1

,. . . ,Y n be bounded operators acting on H. If sup

   n

i=1 Y i f i 

  n

i=1 X i f i  : f i ∈ H



< ∞, then there exists λ(≥ 0) such that Y i Y i ≤ λ

2

X i X i for each i = 1, 2, . . . , n.

Proof. If sup

   n

i=1 Y i f i 

  n

i=1 X i f i  : f i ∈ H



< ∞, then

sup

 Y i f

X i f : f ∈ H



< ∞

for each i = 1, 2, . . . , n. So, by the Theorem 2.2, there exists λ i ( ≥ 0) such that Y i Y i ≤λ

2

i X i X i for each i = 1, 2, . . . , n. Hence we can take λ = max{λ

1

, λ

2

,. . . ,λ n } such that Y i Y i ≤ λ

2

X i X i for each i = 1, 2, . . . , n. 

From the above theorems, we have the following.

Theorem 2.8. Let X

1

, . . . , X n and Y

1

,. . . ,Y n be bounded operators acting on H. If Re < X i f, X j g >≥ 0 for i < j and all f, g in H, then the following are equivalent:

(1) range Y i ⊆ range X i for each i = 1, 2, . . . , n,

(2) there exists λ( ≥ 0) such that Y i Y i ≤ λ

2

X i X i for each i = 1, 2, . . . , n,

(3) there exists an operator A in B(H) such that AX i = Y i for i = 1, 2, . . . , n,

(4) there exists λ( ≥ 0) such that Y i f ≤ λX i f for i = 1, 2, . . . , n, (5) sup

   n

i=1 Y i f i 

  n

i=1 X i f i  : f i ∈ H



< ∞.

If we look through the proof of the above theorems, then we know that it is not extended for the countable bounded operators by the same type. But we have a sufficient condition as the following theorem and corollary.

Theorem 2.9. Let X i and Y i be bounded operators acting on H for i = 1, 2, . . . . If Re < X i f, X j g >≥ 0 for i < j and f, g in H and there exists a sequence n } of non-negative numbers such that 

i=1 λ i < ∞ and Y i Y i ≤ λ

2

i X i X i for i ∈ N, then

sup

   n

i=1 Y i f i 

  n

i=1 X i f i  : f i ∈ H and n ∈ N



< ∞.

(6)

Proof. If Re < X i f, X j g >≥ 0 for i < j and f, g in H, then for each n ∈ N, X k f k  ≤ 

 n i=1

X i f i  for every k ≤ n with the same proof as Theorem 2.5. Hence for each n ∈ N,

∞ > 

i=1

λ i

 n i=1

Y i f i 

X i f i 

 n i=1

Y i f i 

  n

i=1 X i f i 

 n

i=1 Y i f i 

  n

i=1 X i f i 

  n

i=1 Y i f i 

  n

i=1 X i f i  . So, sup

   n

i=1 Y i f i 

  n

i=1 X i f i  : f i ∈ H and n ∈ N



< ∞. 

Corollary 2.10. Let X i and Y i be bounded operators acting on H for i = 1, 2, . . . . Assume that Re < X i f, X j g >≥ 0 for i < j and f, g in H. If there exists a sequence {λ n } of non-negative numbers such that 

i=1 λ i < ∞ and Y i Y i ≤ λ

2

i X i X i for i ∈ N, then there exists a bounded operator A in B(H) such that AX i = Y i for any i ∈ N.

3. Interpolation problems in Alg L

In [9], the authors showed a necessary and sufficient condition that there exists an operator A in Alg L such that AX = Y when L is a commutative subspace lattice. We state the theorem.

Theorem B [9]. Let L be a commutative subspace lattice in B(H) and let X be an operator with dense range, and let Y be a bounded operator acting on H. The following conditions are equivalent:

(1) there exists an operator A in Alg L such that AX = Y , (2) ρ(X, Y ; L) = def sup

 E Y f

E Xf : E ∈ L and f ∈ H



< ∞ .

Furthermore, if condition (2) holds, then there is an operator A ∈AlgL such that AX = Y and whose norm is equal to ρ(X, Y ; L).

In the following theorem, we obtain the above result for non-commuta-

tive subspace lattice case.

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Theorem 3.1. Let L be a subspace lattice on H and let X and Y be in B(H). If the range of X is dense in H, then the following are equivalent:

(1) there exists an operator A in Alg L such that AX = Y , (2) sup

 E Y f

E Xf : f ∈ H, E ∈ L



= K < ∞.

Moreover, if condition (2) holds, we may choose the operator A such that A = K.

Proof. Assume that sup

 E Y f

E Xf : f ∈H, E ∈L



= K < ∞. Then for each E in L, there exists an operator A E in B(H) such that A E (E X) = E Y and A E  ≤ K by Theorem 2.2. In particular, if E = 0, then we have an operator A

0

in B(H) such that A

0

X = Y and A

0

 ≤ K. And for each E ∈ L, A E (E X) = E Y = E (A

0

X). Since the range of X is dense in H, A E E = E A

0

for each E in L. Hence for each E in L,

E A

0

E = A E E E = A E E = E A

0 .

So A

0

is in Alg L. Since A

0

X = Y and E A

0

E = E A

0

,

E Y f = E A

0

Xf = E A

0

E Xf ≤ A

0

E Xf .

So, K ≤ A

0

. Therefore, A

0

 = K.

The proof of the converse is obvious. 

Theorem 3.2. Let X

1

, . . . , X n and Y

1

, . . . , Y n be bounded operators acting on H. If the range of X k is dense in H for some k in {1, 2, . . . , n}, then the following are equivalent:

(1) there exists an operator A in Alg L such that AX i = Y i for i = 1, 2, . . . , n,

(2) sup

 E (  n

i=1 Y i f i ) 

E (  n

i=1 X i f i )  : f i ∈ H, E ∈ L



= K < ∞.

Proof. If sup

 E (  n

i=1 Y i f i ) 

E (  n

i=1 X i f i )  : f i ∈ H, E ∈ L



= K, then

E (  n

i=1 Y i f i ) 

E (  n

i=1 X i f i )  ≤ K

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for each E ∈ L and all f i ∈ H. So for each E ∈ L, there exists an oper- ator A E in B(H) such that A E (E X i ) = E Y i for each i = 1, 2, . . . , n by Theorem 2.3. If E = 0, then A

0

X i = Y i for i = 1, 2, . . . , n. Hence

A E (E X i ) = E Y i = E A

0

X i for each i = 1, 2, . . . , n.

Since range X k = H for some k in {1, 2, . . . , n}, A E E = E A

0

for each E ∈ L. Also,

E A

0

E = A E E E = A E E = E A

0 .

So A

0

is in Alg L.

We omit the proof of the converse since it can be proved easily.  We can generalize Theorem 3.2 to the countable case easily.

Theorem 3.3. Let X i and Y i be bounded operators acting on H. If the range of X k is dense in H for some k ∈ N, then the following are equivalent:

(1) there exists an operator A in Alg L such that AX i = Y i for i = 1, 2, . . . ,

(2) sup

 E (  n

i=1 Y i f i ) 

E (  n

i=1 X i f i )  : f i ∈ H, E ∈ L, n ∈ N



= K < ∞.

Proof of the following theorem can be obtained by the same way as the proof of Theorem 2.5.

Theorem 3.4. Let X

1

, . . . , X n and Y

1

, . . . , Y n be bounded operators acting on a Hilbert space H. Assume that Re <E X i f, E X j g >≥ 0 for each E ∈ L, i < j and all f, g in H. If there exists λ(≥ 0) such that (E Y i ) (E Y i ) ≤ λ

2

(E X i ) (E X i ) for each E ∈ L and i = 1, 2, . . . , n, then sup

 E (  n

i=1 Y i f i ) 

E (  n

i=1 X i f i )  : f i ∈ H, E ∈ L



< ∞.

Theorem 3.5. Let X

1

, . . . , X n and Y

1

, . . . , Y n be bounded operators acting on a Hilbert space H. If

sup

 E (  n

i=1 Y i f i ) 

E (  n

i=1 X i f i )  : f i ∈ H, E ∈ L



< ∞, then there exists λ( ≥ 0) such that

(E Y i ) (E Y i ) ≤λ

2

(E X i ) (E X i )

for each E ∈ L and i = 1, 2, . . . , n.

(9)

Proof. If sup

 E (  n

i=1 Y i f i ) 

E (  n

i=1 X i f i )  : f i ∈ H, E ∈ L



= λ < ∞ for some nonnegative real number λ, then

E (

 n i=1

Y i f i )  ≤ λE (

 n i=1

X i f i ) 

for each E ∈ L and all f i in H. If f i = 0 for i = k, then E Y k f k  ≤ λE X k f k  for k = 1, 2, . . . , n. So

(E Y i ) (E Y i ) ≤ λ

2

(E X i ) (E X i )

for each E ∈ L and i = 1, 2, . . . , n. 

We have the following theorem from the above theorems.

Theorem 3.6. Let X

1

, . . . , X n and Y

1

, . . . , Y n be bounded operators acting on a Hilbert space H. If Re < E X i f, E X j g >≥ 0 for each E ∈ L, i < j and all f, g in H and if the range of X k is dense in H for some k, then the following are equivalent:

(1) range (E Y i ) ⊆range (E X i ) for each E ∈L and i = 1, 2, . . . , n, (2) there exists λ( ≥ 0) such that (E Y i ) (E Y i ) ≤λ

2

(E X i ) (E X i ) for each E ∈ L and i = 1, 2, . . . , n,

(3) there exists λ( ≥ 0) such that E Y i f ≤ λE X i f for each E ∈ L and i = 1, 2, . . . , n,

(4) sup

 E (  n

i=1 Y i f i ) 

E (  n

i=1 X i f i )  : f i ∈ H, E ∈ L



< ∞,

(5) there exists an operator A in B(H) such that AE X i = E Y i for each E ∈ L and i = 1, 2, . . . , n,

(6) there exists an operator A in Alg L such that AX i = Y i for each i = 1, 2, . . . , n.

From Theorem 2.9, we can get the following theorem.

Theorem 3.7. Let X i and Y i be bounded operators on a Hilbert space H for i = 1, 2, . . . . If Re <E X i f, E X j g >≥ 0 for each E ∈ L, i < j and all f, g in H and if there exists a sequence {λ n } of non-negative numbers such that 

i=1 λ i < ∞ and (E Y i ) (E Y i ) λ

2

i (E X i ) (E X i ) for each E ∈ L and i ∈ N, then

sup

 E (  n

i=1 Y i f i ) 

E (  n

i=1 X i f i )  : f i ∈ H, E ∈ L, n ∈ N



< ∞.

(10)

Proof. Let Re <E X i f, E X j g >≥ 0 for each E ∈ L, i < j and all f, g in H and let n ∈ N. Then E X k f k  ≤ E (

 n i=1

X i f i )  for every k ≤ n. Indeed,

E (

 n i=1

X i f i ) 

2

= < E X k f k , E X k f k > + < E ( 

i=k

X i f i ), E X k f k >

+ < E X k f k , E ( 

i=k

X i f i ) > + < E ( 

i=k

X i f i ), E ( 

i=k

X i f i ) >

= E X k f k 

2

+ E ( 

i=k

X i f i ) 

2

+ 

i<j

2Re < E X i f i , E X j f j >

≥ E X k f k 

2

for i, j = 1, 2, . . . , n . Hence for each n ∈ N,

∞ >

 i=1

λ i

 n i=1

E Y i f i 

E X i f i 

 n i=1

E Y i f i 

E (  n

i=1 X i f i ) 

 n

i=1 E Y i f i 

E (  n

i=1 X i f i ) 

E (  n

i=1 Y i f i ) 

E (  n

i=1 X i f i )  . So, sup

 E (  n

i=1 Y i f i ) 

E (  n

i=1 X i f i )  : f i ∈ H, E ∈ L, n ∈ N



< ∞. 

References

[1] M. Anoussis, E. Katsoulis, R. L. Moore and T. T. Trent, Interpolation problems for ideals in nest algebras, Math. Proc. Camb. Phil. Soc. 111 (1992), 151–160.

[2] R. G. Douglas, On majorization, factorization, and range inclusion of operators on Hilbert space, Proc. Amer. Math. Soc. 17 (1966), 413–415.

[3] A. Hopenwasser, The equation T x = y in a reflexive operator algebra, Indiana Univ. Math. J. 29 (1980), 121–126.

[4] , Hilbert-Schmidt Interpolation in CSL-Algebras, Illinois J. Math. 33

(1989), no. 4, 657–672.

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[5] Y. S. Jo and J. H. Kang, Interpolation problems in Alg L, to appear.

[6] R. Kadison, Irreducible Operator Algebras, Pro. Nat. Acad. Sci. U.S.A. (1957), 273–276.

[7] E. Katsoulis, R. L. Moore and T. T. Trent, Interpolation in nest algebras and applications to operator Corona Theorems, J. Operator Theory 29 (1993), 115–

123.

[8] E. C. Lance, Some properties of nest algebras, Proc. London Math. Soc. 19 (1969), no. 3, 45–68.

[9] R. Moore and T. T. Trent, Linear equations in subspaces of operators, Proc.

Amer. Math. Soc. 128 (2000), no. 3, 781–788.

[10] N. Munch, Compact causal data interpolation, J. Math. Anal. Appl. 140 (1989), 407–418.

Young Soo Jo

Department of Mathematics Keimyung University Daegu 704-701, Korea E-mail: [email protected]

Joo Ho Kang and Ki Sook Kim Department of Mathematics Daegu University

Daegu 712-714, Korea

E-mail: [email protected]

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