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SOLVING OPERATOR EQUATIONS AX = Y AND Ax = y IN ALGL†
SOLVING OPERATOR EQUATIONS AX = Y AND Ax = y IN ALGL†
Journal of Applied Mathematics & Informatics. 2015. May, 33(3_4): 417-424
Copyright © 2015, Korean Society of Computational and Applied Mathematics
  • Received : December 20, 2014
  • Accepted : March 09, 2015
  • Published : May 30, 2015
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SANG KI LEE
JOO HO KANG

Abstract
Abstract. In this paper the following is proved: Let L be a subspace lattice on a Hilbert space H and X and Y be operators acting on a Hilbert space H . If XE = EX for each E L , then there exists an operator A in Alg L such that AX = Y if and only if sup and YE = EYE . Let x and y be non-zero vectors in H . Let Px be the orthogonal pro-jection on sp ( x ). If EPx = PxE for each E L , then the following are equivalent. (1) There exists an operator A in Alg L such that Ax = y . (2) < f , Ey > y =< f , Ey > Ey for each E L and f H . AMS Mathematics Subject Classification : 47L35.
Keywords
1. Introduction
Interpolation problems have been developed by many mathematicians since Douglas considered a problem to find a bounded operator A satisfying AX = Y for two operators X and Y acting on a Hilbert space H in 1966 [1 , 2 , 3 , 4 , 5 , 6] . Douglas used the range inclusion property of operators to show necessary and sufficient conditions for the existence of an operator A such that AX = Y . A condition for the operator A to be a member of A which is a specified subalgebra of B ( H ) can be given. In this paper, authors investigated to find sufficient and necessary conditions that there exists an operator A in Alg L satisfying AX = Y for operators X and Y acting on a Hilbert space H and there exists an operator B in Alg L satisfying Bx = y for two vectors x and y in H . And authors investigated the above interpolation problems for finitely or countably many operators and vectors.
The simplest case of the operator interpolation problem 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 1.1 (R.G. Douglas [1] ). Let X and Y be bounded operators acting on a Hilbert space H. Then the following statements are equivalent:
(1) rangeY range X
(2) YY λ 2 XX 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 operator A so that
(a) A 2 = inf{ μ : Y Y μX X }
(b) kerY = kerA and
(c) rangeA rangeX.
We need to look at the proof of Theorem A carefully. Then we know that the image of A on
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is 0 from the proof of (3) by (2).
2. The EquationAX=Yin AlgL
Let H be a Hilbert space. A subspace lattice L is a strongly closed lattice of orthogonal projections on 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 ; Alg L is a weakly closed subalgebra of B ( H ). A lattice L is a commutative subspace lattice , or CSL, if the projections in L commute; in this case, Alg L is called a CSL algebra . Let x 1 ,⋯, xn be vectors of H . Then sp({ x 1 ,⋯, xn }) = { α 1 x 1 + α 2 x 2 + ⋯ + αnxn : α 1 ; α 2 , ⋯ αn ∈ ℂ }. Let M be a subset of H . Then
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means the closure of M and
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the orthogonal complement of
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. Let ℕ be the set of natural numbers and ℂ be the set of complex numbers.
Let L be a subspace lattice and A , X and Y be operators acting on a Hilbert space H such that AX = Y . If XE = EX , then ║ YEf ║ = ║ AXEf ║ = ║ AEXf ║ ≤ ║ A ║║ XEf ║ for all E L and for all f in H . If we adopt the convention that a fraction whose numerator and denominator are both zero is equal to zero, then the inequality above may be stated in the form
PPT Slide
Lager Image
Theorem 2.1. Let L be a subspace lattice on a Hilbert space H and X and Y be operators acting on the Hilbert space H. If XE = EX for each E in L, then the following are equivalent.
(1) There exists an operator A in AlgL such that AX = Y.
(2) sup
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and YE = EY E for each E in L.
Proof . Assume that sup
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and Y E = EY E for each E in L . Then for each E in L , there exists an operator AE in B ( H ) such that AE ( XE ) = Y E = EY E by Theorem A. In particular, if E = I , then we have an operator AI in B ( H ) such that AIX = Y . So AE ( XE ) = AIXE = EAIXE for each E in L . Since EX = XE for each E L , AIXE = EAIEX . Hence AIE = EAIE on
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. Let h be in
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. Since EX = XE for each E in L , < Eh , X f >=< h , EXf >=< h , XEf >= 0. So Eh
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. By the definition of AI , ( AIE ) h = 0 = ( EAIE ) h . Hence AIE = EAIE on
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. So AI is an operator in Alg L . □
Assume that X 1 ,⋯, Xn and Y 1 ,⋯, Yn are operators in B ( H ) and A is an operator in Alg L such that AXi = Yi for each i = 1,⋯, n . Then YiEfi = AXiEfi for each i = 1,⋯, n , E L and each fi in H . Hence
PPT Slide
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for all E L and all fi in H . If, for convenience, we adopt the convention that a fraction whose numerator and denominator are both zero is equal to zero, then the inequality above may be stated in the form
PPT Slide
Lager Image
Theorem 2.2. Let X1,⋯,Xn and Y1,⋯,Yn be bounded operators acting on H. If XiE = EXi for each E in L and i {1, 2,⋯, n }, then the following are equivalent .
(1) There exists an operator A in AlgL such that AXi = Yi for i = 1; 2,⋯,n .
(2) sup
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and YiE = EYiE for each i = 1,⋯, n and E in L .
Proof . Assume that sup
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and YiE = EYiE for each i = 1,⋯, n and E L . Let E be in L and
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Define AE : ME H by
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. Then AE is welldefined and bounded linear. Extend AE on
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continuously. Define AEf = 0 for each f
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. Then AE : H H is a bounded linear and AE EXi = YiE for i = 1,⋯, n . If E = I , then AIXi = Yi for i = 1,⋯, n . Since EXi = XiE and YiE = EYiE for each i = 1,⋯, n , AEXiE = AIXiE = AI EXi and AEXiE = EAIXiE = EAIEXi . Hence AIE = EAIE on
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. Let h be in
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. Then since EXi = XiE for each i = 1,⋯, n , < Eh , Xif >=< h , EXif >=< h , XiEf >= 0 for each f H . So
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By the definition of AI , AIEh = 0 = EAIEh for each E L . Hence AIE = EAIE on
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. So AI is an operator in Alg L
We can generalize the above Theorem to the countable case easily.
Theorem 2.3. Let Xi and Yi be bounded operators acting on H for all i = 1, 2,⋯. If XiE = EXi for each E in L and i in ℕ, then the following are equivalent .
(1) There exists an operator A in AlgL such that AXi = Yi for i = 1, 2,⋯.
(2) sup
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and YiE = EYiE for each i = 1,⋯ and E L .
Proof . Assume that sup
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and YiE = EYiE for each i = 1,⋯. Let E be in L and
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Define AE : NE H by
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. Then AE is welldefined and bounded linear. Extend AE on
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continuously. Define AEf = 0 for each f
PPT Slide
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. Then AE : H H is a bounded linear and AE EXi = YiE for i = 1,⋯. If E = I , then AIXi = Yi for i = 1,⋯. Since EXi = XiE and YiE = EYiE for each i = 1,⋯ , AEXiE = AIXiE = AI EXi and AEXiE = EAIXiE = EAIEXi . Hence AIE = EAIE on
PPT Slide
Lager Image
. Let h be in
PPT Slide
Lager Image
. Then since EXi = XiE for each i = 1,⋯ , < Eh , Xif >=< h , EXif >=< h , XiEf >= 0 for each f H . So
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Lager Image
for each n ∈ ℕ. By the definition of AI , AIEh = 0 = EAIEh for each E L . Hence AIE = EAIE on
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. So AI is an operator in Alg L . □
3. The EquationAx=yin AlgL
Let x and y be non-zero vectors in a Hilbert space H . Let X = x y and Y = y y . Then for f in H and E L ,
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If for convenience, we adopt the convention that a fraction whose numerator and denominator are both zero is equal to zero, then for f in H and E L ,
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is
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Hence sup
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, Y Ef =< f , Ey > y and EY Ef =< f , Ey > Ey for each f in H and each E L .
We can obtain the following theorem by Theorem 2.1.
Theorem 3.1. Let L be a subspace lattice on H and let x and y be non-zero vectors in H. Let Px be the orthogonal projection on sp ( x ). If EPx = PxE for each E L, then the following are equivalent .
(1) There exists an operator A in AlgL such that Ax = y.
(2) < f , Ey > y =< f , Ey > Ey for each E L and f H .
Let xi , yi ( i = 1,⋯, n ) be non-zero vectors in H . Let Xi = xi yi and Yi = yi yi . Then for fi in H and E L
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Lager Image
Hence
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and YiEf =< f , Eyi > yi and < EYiEf >=< f , Eyi > Eyi for each E L , f H and i = 1,⋯, n .
We can obtain the following theorem by Theorem 2.2.
Theorem 3.2. Let L be a subspace lattice on H and let x 1 ,⋯, xn and y 1 ,⋯, yn be vectors in H. Let Pxi be the orthogonal projection on sp ( xi ). If EPxi = PxiE for each E L and i = 1,⋯, n, then the following are equivalent .
(1) There exists an operator A in AlgL such that Axi = yi for i = 1; 2,⋯, n .
(2) sup
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Lager Image
and < f, Eyi > yi =< f, Eyi > Eyi for each E L, f H and i = 1,⋯, n .
We can extend Theorem 3.2 to countably infinite vectors and get the following theorem from Theorem 2.3.
Theorem 3.3. Let L be a subspace lattice on H and let { xi } and { yi } be vectors in H for i ∈ ℕ. Let Pxi be the orthogonal projection on sp ( xi ). If EPxi = PxiE for each E L and = 1, 2,⋯ , then the following are equivalent .
(1) There exists an operator A in AlgL such that Axi = yi for i = 1; 2,⋯.
(2) sup
PPT Slide
Lager Image
and < f , Eyi > yi =< f , Eyi > Eyi for each E ∈ L, f H and i = 1; 2,⋯.
Theorem 3.4. Let L be a subspace lattice on a Hilbert space H and x and y be vectors in H. Let Px be the orthogonal projection on sp ( x ). If EPx = PxE for each E L and sup
PPT Slide
Lager Image
, then there exists an operator A in AlgL such that Ax = y .
Proof . Assume that sup
PPT Slide
Lager Image
. Then for each E in L , there exists an operator AE in B ( H ) such that AEEx = Ey by Theorem 1.1. In particular, if E = I , then we have an operator AI in B ( H ) such that AIx = y . Let’s put AI = A . So AEEx = Ey = EAx for each E L . Hence AEE = EA on sp ( x ). Let h be in sp ( x ) . Since EPx = PxE for each E L , < Eh , Ex >=< h , Ex >=< h , EPxx >=< h , PxEx >= 0. Hence Eh sp ( Ex ) . By the definition of AE and A , AEEh = 0 = EAh for each E in L . Hence AEE = EA on H for each E in L . So A = EA . Therefore A is in Alg L . □
Theorem 3.5. Let L be a subspace lattice on a Hilbert space H and x 1 ,⋯, xn and y 1 ,⋯, yn be vectors in H. Let Pxi be the orthogonal projection on sp ( xi ). If EPxi = PxiE for each E L and = 1,⋯, n and
PPT Slide
Lager Image
then there exists an operator A in AlgL such that Axi = yi for i = 1,⋯, n .
Proof . Assume that sup
PPT Slide
Lager Image
. Let E be in L . Define AE : sp ({ Ex 1 ,⋯, Exn }) → H by
PPT Slide
Lager Image
. Then AE is well-defined and bounded linear. Define AEf = 0 for each f sp ({ Ex 1 ,⋯, Exn }) . Then AE : H H is bounded linear and AEExi = Eyi for i = 1,⋯, n . If E = I , then AIxi = yi for i = 1,⋯, n . Let’s put AI = A . So AEExi = Eyi = EAxi for each E L . Hence AEE = EA on sp ({ x 1 ,⋯, xn }). Let h be in sp ({ x 1 ,⋯, xn }) . Since < Eh , Exi >=< h , Exi >= 0, < Eh ,
PPT Slide
Lager Image
. So Eh sp ({ Ex 1 ,⋯, Exn }) . By the definition of AE and A , AEEh = 0 = EAh for each E in L . Hence AEE = EA on H for each E in L . So A = EA . Therefore A is in Alg L . □
We ca generalize the above theorem for countable case.
Theorem 3.6. Let L be a subspace lattice on a Hilbert space H and { xi } and { yi } be vectors in H. Let Pxi be the orthogonal projection on sp ( xi ) for each i = 1; 2,⋯. If EPxi = PxiE for each E L and i = 1; 2,⋯ and
PPT Slide
Lager Image
then there exists an operator A in AlgL such that Axi = yi for i = 1, 2,⋯.
BIO
Sang Ki Lee received his Ph.D at Keimyung University under direction of Chul Soon Han. He has been a professor of Daegu University since 1982. His research interest is topology and mathematics education.
Department of Mathematics Education, Daegu University, Daegu, Korea.
e-mail: sangklee@daegu.ac.kr
Joo Ho Kang received her Ph.D at the University of Alabama under direction of Tavan T. Trent. She has been a professor of Daegu University since 1977. Her research interest is an operator theory.
Department of Mathematics, Daegu University, Daegu, Korea.
e-mail: jhkang@daegu.ac.kr
References
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Kang J.H. (2014) Compact interpolation on AX = Y in AlgL J. Appl. Math. & Informatics 32 441 - 446    DOI : 10.14317/jami.2014.441
Lance E.C. (1969) Some properties of nest algebras Proc. London Math. Soc. 19 (III) 45 - 68    DOI : 10.1112/plms/s3-19.1.45