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COMPACT INTERPOLATION ON AX = Y IN ALGL†
COMPACT INTERPOLATION ON AX = Y IN ALGL†
Journal of Applied Mathematics & Informatics. 2014. Sep, 32(3_4): 441-446
Copyright © 2014, Korean Society of Computational and Applied Mathematics
  • Received : November 07, 2013
  • Accepted : December 24, 2013
  • Published : September 28, 2014
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JOO HO KANG

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 H . Then there exists a compact operator A in Alg L such that AX = Y if and only if sup and Y is compact. Moreover, if the necessary condition holds, then we may choose an operator A such that AX = Y and ∥ A ∥ = K . AMS Mathematics Subject Classification : 47L35.
Keywords
1. Introduction
Let C be a collection of operators acting on a Hilbert space H . An interpo-lation question for C asks for which operators X and Y on H when is there a bounded linear operator A (usually satisfying some other conditions) such that AX = Y ? The "other conditions" that have been of interest to us involve re-stricting A to lie in the algebra associated with a subspace lattice. 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 Douglas [2] . Another interpolation question for a given subal-gebra ⌋ of B ( H ) asks for which vectors x and y in H is there a bounded operator A C that maps x to y . Lance [6] initiated the discussion by considering a nest N and asking what conditions on x and y will guarantee the existence of an operator A in Alg N such that Ax = y . Hopenwasser [3] extended Lance's result to the case where the nest N is replaced by an arbitrary commutative subspace lattice L . Munch [7] considered the problem of finding a Hilbert-Schmidt oper-ator A in Alg N that maps x to y , whereupon Hopenwasser [4] again extended to Alg L . In [1] , authors studied the problem of finding A so that Ax = y and A is required to lie in certain ideals contained in Alg L (for a nest L ).
Roughly speaking, when an operator maps one thing to another, we think of the operator as the interpolating operator and the equation representing the mapping as the interpolation equation. The equations Ax = y and AX = Y are indistinguishable if spoken aloud, but we mean the change to capital letters to indicate that we intend to look at fixed operators X and Y , and ask under what conditions there will exist an operator A satisfying the equation AX = Y .
Let x and y be vectors in a Hilbert space. Then ⟨ x , y ⟩ means the inner product of vectors x and y . Note that the "vector interpolation" problem is a special case of the "operator interpolation" problem. Indeed, if we denote by x u the rank-one operator defined by the equation x u ( w ) = ⟨ w , u x , and if we set X = x u , and Y = y u , then the equations AX = Y and Ax = y represent the same restriction on A .
Let H be a Hilbert space and ℕ = {1, 2, ...}. A bounded operator A on H has finite-rank if rangeA is finite dimensional. A bounded operator A on H is called compact if A (ball H ) has compact closure in H , where ball H = { h H : ∥ h ∥ ≤ 1}. We denote B 0 ( H ) the set of all compact operators on H .
We will study finite-rank operator interpolation problems on L and find a compact operator A in Alg L such that AX = Y for given X and Y in B ( H ) as convergence of finite-rank operators. Also, we will study this problem for given countable operators X 1 , X 2 ,... and Y 1 , Y 2 ,...
Theorem 1.1 ( [2] ). Let X and Y be bounded operators acting on a Hilbert space H. Then the following statements are equivalent:
  • (1) rangeY∗⊆range X∗
  • (2) Y∗Y≤λ2X∗Xfor 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)
Lemma 1.2. Let A and X be bounded operators acting on a Hilbert space H.
  • (a) If X=y⊗x is a rank-one operator and Ay≠ 0,then AX is a rank-one operator.
  • (b) If X= (y1⊗x1)+(y2⊗x2)is a rank-two operator and Ay1and Ay2are linearly independent, then AX is a rank-two operator.
  • (c) Ifis a rank-n operator and Ay1, ...,Aynare linearly independent, then AX is a rank-n operator.
  • (d) If X is a rank-n operator, then AX is a rank-m operator for m≤n.
Proof . Let ƒ H . Then (a) AXƒ = A ( y x ) ƒ = A ƒ, x y = ⟨ ƒ, x Ay . So AX is a rank-one operator.
  • (b)AXƒ=A((y1⊗x1)+(y2⊗x2))ƒ=A⟨ƒ,x1⟩y1+A⟨ƒ,x2⟩y2= ⟨ƒ,x1⟩Ay1+⟨ƒ,x2⟩Ay2for allƒ∈H.
  • (c)for allƒ∈H.
  • (d) can be easily obtained by using (a), (b) and (c).
Theorem 1.3. Let X and Y be bounded operators acting on a Hilbert space H. Then the following are equivalent:
  • (1) There is a rank-one operator A=y⊗x such that AX=Y and range X⊄ {x}⊥.
  • (2) rangeY∗⊆rangeX∗and Y is a rank-one operator.
Proof . If there exists a rank-one operator A = y x such that AX = Y . Then rangeY rangeX . Since rangeX ⊄ { x } , there exists a vector h in H such that ⟨ Xh, x ⟩ ≠ 0. For any ƒ in H , Y ƒ = ( y x ) = ⟨ , x y . So Y is a rank-one operator.
Conversely, suppose rangeY rangeX and Y = y 1 x 1 . Then there exists an operator B in B ( H ) such that BX = Y . Then for each h H ,
PPT Slide
Lager Image
So B ( rangeX ) = rangeY = sp { y 1 }. Define
PPT Slide
Lager Image
by Ah = Bh if h rangeX and Ah = 0 if h B ( H ). Since
PPT Slide
Lager Image
rangeA sp { y 1 }. Since rangeA is a linear subspace containing y 1 ,
PPT Slide
Lager Image
Hence A is a rank-one operator and AX = Y .
Theorem 1.4 . Let X and Y be bounded operators acting on a Hilbert space H. If rangeY rangeX. and Y is a finite-rank operator, then there exists a finite-rank operator A such that AX = Y .
Proof . If Y is a finite-rank operator, then
PPT Slide
Lager Image
where y 1 , ..., yn are linearly independent. Since rangeY rangeX , there exists an operator B in B ( H ) such that BX = Y . For each h H ,
PPT Slide
Lager Image
So B ( rangeX ) = sp { y 1 , ..., yn } Define
PPT Slide
Lager Image
by Ah = Bh if h rangeX and Ah = 0 if
PPT Slide
Lager Image
Then A B ( H ). And
PPT Slide
Lager Image
Therefore dim ( rangeA ) = n and AX = Y .
2. Results
Theorem 2.1. Let L be a subspace lattice on H and let X and Y be operators in B ( H ). Assume that the range of X is dense in H. Then the following are equivalent:
  • (1) There exists a compact operator A in AlgL such that AX=Y.
  • (2)supand Y is compact.
Moreover, if condition (2) holds, we may choose an operator A such that A ∥ = K .
Proof . Assume that sup
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Lager Image
Then there exists an operator A in Alg L such that AX = Y by Theorem 3.1 [5] . Since Y is compact, there is a sequence { Yn } of fInite-rank operators that converges to Y in the norm topology on B ( H ). From the construction of Yn , since
PPT Slide
Lager Image
for each n ∈ ℕ,
PPT Slide
Lager Image
for each n ∈ ℕ. By Theorems 1.4, there is a FInite-rank operator An such that AnX = Yn for each n ∈ ℕ. Since Yn Y in the norm topology on B ( H ), ∥ An A ∥ → 0. Hence A is compact. The proof of the converse is obvious.
Theorem 2.2. Let L be a subspace lattice on H and let X 1 , ..., Xn and Y 1 , ..., Yn be bounded operators acting on H. Let k be a fixed natural number in {1, 2, ..., n } and assume that Xk has dense range. Then the following are equivalent:
  • (1) There exists a compact operator A in AlgL such that AXi=Yifor each i= 1, 2, ,,,n.
  • (2)supand Ykis compact.
Moreover, if condition (2) holds, we may choose an operator A such that A ∥ = K .
Proof . If sup
PPT Slide
Lager Image
and for given k in {1, 2, ,,, n }, Yk is compact, then by Theorem 3.2 [5] , there exists an operator A in Alg L such that AXi = Yi for i = 1, 2, ,,, n . Since Yk is compact, there is a sequence { Ykm }of finite-rank operators that converges to Yk in the norm topology on B ( H ). From the construction of Ykm , we know that
PPT Slide
Lager Image
for each m ∈ ℕ. Therefore
PPT Slide
Lager Image
for each m ∈ ℕ. By Theorem 2.1, for each m ∈ ℕ, there is a finite-rank operator Akm such that Akm Xk = Ykm . Since Ykm Yk in the norm topology on B ( H ), ∥ Akm A ∥ → 0. Hence A is compact. We omit the proof of the converse since it can be proved easily.
Theorem 2.3 ( [5] ). Let L be a subspace lattice on H and let X 1 , ..., Xn and Y 1 , ..., Yn be bounded operators acting on H. Let k be a fixed natural number in {1, 2, ..., n } and assume that Xk has dense range and Re E Xif , E Xjg ⟩ ≥ 0 for each E in L, i < j and all f, g in H. Then the following are equivalent:
  • (1)for each E in L and i= 1, 2, ...,n.
  • (2) There exists an operator A in AlgL such that AXi=Yifor i= 1, 2, ...,n.
  • (3)sup
By the Theorem 2.3, we can get the following Theorem.
Theorem 2.4. Let L be a subspace lattice on H and let X 1 , ..., Xn and Y 1 , ..., Yn be bounded operators acting on H. Let k be a fixed natural number in {1, 2, ..., n } and assume that Xk has dense range. If for i < j in {1, 2, ..., n } and all f, g in H, Re E Xif , E Xjg ⟩ ≥ 0, then the following are equivalent:
  • (1)for each E in L and i= 1, 2, ...,n, and Ykis compact.
  • (2) There exists a compact operator A in AlgL such that AXi=Yifor i= 1, 2, ...,n.
If we observe the proof of the above theorems, we can generalize Theorem 2.2 to the countable case easily.
Theorem 2.5. Let L be a subspace lattice on H and let Xi and Yi be bounded operators acting on H for all i = 1, 2, ..., n. Let k be a fixed natural number in {1, 2, ..., n } and assume that Xk has dense range. Then the following are equivalent:
  • (1) There exists a compact operator A in AlgL such that AXi=Yifor i= 1, 2, ...,n.
  • (2)supand Ykis com-pact.
Moreover, if condition (2) holds, we may choose an operator an operator A such that A ∥ = K .
Theorem 2.6. Let L be a subspace lattice on H and let Xi and Yi be bounded operators acting on H for all i = 1, 2, ..., n . Let k be a fixed natural number in {1, 2, ..., n } and assume that Xk has dense range and Re E Yif , E Yjg ⟩ ≤ Re E Xif , E Xjg for each E in L, i < j and allf, g in H, then the following are equivalent:
  • (1) There exists M≥ 0such thatsupfor each i∈ ℕand Ykis compact.
  • (2) There is a compact operator A in AlgL such that AXi=Yifor i= 1, 2, ...,n.
BIO
Joo Ho Kang received her Ph.D. at the University of Alabama under the 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, Kyungpook, Korea.
e-mail: jhkang@daegu.ac.kr
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