COMPACT INTERPOLATION ON AX = Y IN ALG𝓛

Abstract

In this paper the following is proved: Let $\mathcal{L}$ be a subspace lattice on a Hilbert space $\mathcal{H}$ and X and Y be operators acting on $\mathcal{H}$. Then there exists a compact operator A in $Alg\mathcal{L}$ such that AX = Y if and only if ${\sup}\{\frac{{\parallel}E^{\perp}Yf{\parallel}}{{\parallel}E^{\perp}Xf{\parallel}}\;:\;f{\in}\mathcal{H},\;E{\in}\mathcal{L}\}$ = K < ${\infty}$ and Y is compact. Moreover, if the necessary condition holds, then we may choose an operator A such that AX = Y and ${\parallel}A{\parallel}=K$.

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 AlgN 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 AlgN that maps x to y, whereupon Hopenwasser [4] again extended to AlgL. In [1], authors studied the problem of finding A so that Ax = y and A is required to lie in certain ideals contained in AlgL (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(ballH) has compact closure in H, where ballH = {h ∈ H : ∥h∥ ≤ 1}. We denote B0(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 AlgL 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 X1, X2,... and Y1, Y2,...

Theorem 1.1 ([2]). Let X and Y be bounded operators acting on a Hilbert space H. Then the following statements are equivalent:

Moreover, if (1), (2), and (3) are valid, then there exists a unique operator A so that

Lemma 1.2. Let A and X be bounded operators acting on a Hilbert space H.

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.

Theorem 1.3. Let X and Y be bounded operators acting on a Hilbert space H. Then the following are equivalent:

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)Xƒ = ⟨Xƒ, x⟩y. So Y is a rank-one operator.

Conversely, suppose rangeY∗ ⊆ rangeX∗ and Y = y1⊗x1. Then there exists an operator B in B(H) such that BX = Y. Then for each h ∈ H,

So B(rangeX) = rangeY = sp{y1}. Define by Ah = Bh if h ∈ rangeX and Ah = 0 if h ∈ B(H). Since rangeA ⊂ sp{y1}. Since rangeA is a linear subspace containing y1, 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 where y1, ..., yn are linearly independent. Since rangeY∗ ⊆ rangeX∗, there exists an operator B in B(H) such that BX = Y. For each h ∈ H,

So B(rangeX) = sp{y1, ..., yn} Define by Ah = Bh if h ∈ rangeX and Ah = 0 if Then A ∈ B(H). And 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:

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

Proof. Assume that sup Then there exists an operator A in AlgL 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 for each n ∈ ℕ, 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 X1, ..., Xn and Y1, ..., 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:

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

Proof. If sup and for given k in {1, 2, ,,, n}, Yk is compact, then by Theorem 3.2 [5], there exists an operator A in AlgL 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 for each m ∈ ℕ. Therefore for each m ∈ ℕ. By Theorem 2.1, for each m ∈ ℕ, there is a finite-rank operator Akm such that AkmXk = 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 X1, ..., Xn and Y1, ..., 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:

By the Theorem 2.3, we can get the following Theorem.

Theorem 2.4. Let L be a subspace lattice on H and let X1, ..., Xn and Y1, ..., 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:

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:

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: