SOLVING OPERATOR EQUATIONS Ax = Y AND Ax = y IN ALGL

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 ${\in}$ L, then there exists an operator A in AlgL such that AX = Y if and only if sup $\left{\frac{\parallel{XEf}\parallel}{\parallel{YEf}\parallel}\;:\;f{\in}H,\;E{\in}L\right}$ = K < $\infty$ and YE=EYE. Let x and y be non-zero vectors in H. Let P_x be the orthogonal pro-jection on sp(x). If EP_x = P_xE for each E $\in$ 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 ${\in}$ L and f ${\in}$ H.

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 AlgL satisfying AX = Y for operators X and Y acting on a Hilbert space H and there exists an operator B in AlgL 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) Y∗Y ≤ λ2X∗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 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 is 0 from the proof of (3) by (2).

 

2. The Equation AX = Y in 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 AlgL 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 commute; in this case, AlgL is called a CSL algebra. Let x1,⋯,xn be vectors of H. Then sp({x1,⋯,xn}) = {α1x1 + α2x2 + ⋯ +αnxn : α1;α2, ⋯ αn ∈ ℂ }. Let M be a subset of H. Then means the closure of M and the orthogonal complement of . 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

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 and YE = EY E for each E in L.

Proof. Assume that sup 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 . Let h be in . Since EX = XE for each E in L, < Eh,X f>=< h, EXf>=< h, XEf >= 0. So Eh ∈ . By the definition of AI , (AIE)h = 0 = (EAIE)h. Hence AIE = EAIE on . So AI is an operator in AlgL. □

Assume that X1,⋯,Xn and Y1,⋯,Yn are operators in B(H) and A is an operator in AlgL 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

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

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 and YiE = EYiE for each i = 1,⋯,n and E in L.

Proof. Assume that sup and YiE = EYiE for each i = 1,⋯,n and E ∈ L. Let E be in L and

Define AE : ME → H by . Then AE is welldefined and bounded linear. Extend AE on continuously. Define AEf = 0 for each f ∈ . Then AE : H → H is a bounded linear and AEEXi = 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 = AIEXi and AEXiE = EAIXiE = EAIEXi. Hence AIE = EAIE on . Let h be in . Then since EXi = XiE for each i = 1,⋯,n, < Eh,Xif >==< h,XiEf >= 0 for each f ∈ H. So

By the definition of AI , AIEh = 0 = EAIEh for each E ∈ L. Hence AIE = EAIE on . So AI is an operator in AlgL □

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 and YiE = EYiE for each i = 1,⋯ and E ∈ L.

Proof. Assume that sup and YiE = EYiE for each i = 1,⋯. Let E be in L and

Define AE : NE → H by . Then AE is welldefined and bounded linear. Extend AE on continuously. Define AEf = 0 for each f ∈ . Then AE : H → H is a bounded linear and AEEXi = 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 = AIEXi and AEXiE = EAIXiE = EAIEXi. Hence AIE = EAIE on . Let h be in . Then since EXi = XiE for each i = 1,⋯ , < Eh,Xif>=< h,EXif >=< h,XiEf >= 0 for each f ∈ H. So

for each n ∈ ℕ. By the definition of AI , AIEh = 0 = EAIEh for each E ∈ L. Hence AIE = EAIE on . So AI is an operator in AlgL. □

 

3. The Equation Ax = y in 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,

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,

is

Hence sup , 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

Hence 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 x1,⋯,xn and y1,⋯, 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 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 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 , then there exists an operator A in AlgL such that Ax = y.

Proof. Assume that sup . 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 AlgL. □

Theorem 3.5. Let L be a subspace lattice on a Hilbert space H and x1,⋯,xn and y1,⋯,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

then there exists an operator A in AlgL such that Axi = yi for i = 1,⋯,n.

Proof. Assume that sup . Let E be in L. Define AE : sp({Ex1,⋯,Exn}) → H by . Then AE is well-defined and bounded linear. Define AEf = 0 for each f ∈ sp({Ex1,⋯,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({x1,⋯,xn}). Let h be in sp({x1,⋯,xn})⊥ . Since < Eh,Exi>=< h,Exi >= 0, < Eh, . So Eh ∈ sp({Ex1,⋯,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 AlgL. □

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

then there exists an operator A in AlgL such that Axi = yi for i = 1, 2,⋯.