REFLEXIVITY AND HYPERREFLEXIVITY OF THE SPACE OF LOCALLY INTERTWINING OPERATORS

J. BRAČIČ, V. MÜLLER, AND M. ZAJAC

Abstract. We characterize the spaces of all local intertwiners I(A, B; e) that are reflexive (hyperreflexive). We show that if e is not an eigenvector of A, then the reflexivity (hyperreflexivity) of I(A, B; e) depends only on B and is independent of A and e. This has consequences concerning the reflexivity of the space of intertwiners I(A, B) and of the commutant of an operator.

1. Introduction

For complex Banach spaces 𝒳 and 𝒴, let B(𝒳, 𝒴) be the Banach space of all bounded linear operators from 𝒳 to 𝒴; similarly, let B(𝒳) be the Banach algebra of all bounded linear operators on 𝒳. The topological dual of 𝒳 is denoted by 𝒳^{∗}.

Let A ∈ B(𝒳), B ∈ B(𝒴), and e ∈ 𝒳 be given. An operator S ∈ B(𝒳, 𝒴) intertwines A and B at e, if SAe = BSe. The set of all operators that intertwine A and B at e is denoted by I(A, B; e). In particular, if 𝒳 = 𝒴 and A = B, then C(A, e) := I(A, A; e) is the local commutant of A at e. Local commutants were introduced and studied by Larson [8], see also [3].

It is obvious that I(A, B; e) is a linear space of operators and it is not hard to see that I(A, B; e) is closed in the strong operator topology, which means, by convexity, that it is closed in the weak operator topology as well.

For a linear subspace 𝒮 ⊆ B(𝒳, 𝒴), the reflexive closure of 𝒮 is given by

Ref 𝒮 = {T ∈ B(𝒳, 𝒴); T x ∈ [𝒮x] for all x ∈ 𝒳},

where 𝒮x = {Sx; S ∈ 𝒮} is the orbit of 𝒮 at x and [𝒮x] is its closure. It is obvious that Ref 𝒮 ⊇ 𝒮. If Ref 𝒮 = 𝒮, then the space 𝒮 is said to be reflexive.

In Section 2 we give a complete description of subspaces I(A, B; e) that are reflexive. It is easy to see that this space is reflexive if e is an eigenvector of A. If e and Ae are linearly independent then the space I(A, B; e) is reflexive if and only if ⋂_{λ∈ℂ}[im (B − λ)] = {0}. It is interesting that this condition depends only on B and is independent of A and e. This has applications for the reflexivity of the space of intertwiners between A and B.

Section 3 is devoted to the hyperreflexivity (for the definition see that section). It is wellknown that any hyperreflexive subspace of operators is reflexive and that the converse does not hold, see [7], Theorem 6. We shall show that spaces of locally intertwining operators provide natural examples of spaces of operators that are reflexive but not hyperreflexive.

In the last section we discuss the k-reflexivity and k-hyperreflexivity of spaces of local intertwiners.

Key words and phrases. Commutant, local commutant, reflexivity, hyperreflexivity.

2. Reflexivity of the space of locally intertwining operators

In this section we shall characterize those spaces I(A, B; e) that are reflexive. The following proposition describes the orbits of spaces of local intertwiners.

Proposition 2.1. Let A ∈ B(𝒳), B ∈ B(𝒴), and e, x ∈ 𝒳 \ {0} be arbitrary.

(i) If x is not in the linear span of the vectors e and Ae, i.e. x ∉ ∨{e, Ae}, then I(A, B; e)x = 𝒴.

(ii) If Ae = λe for some λ ∈ ℂ and x is a scalar multiple of e, then I(A, B; e)x = ker (B − λ). (iii) If e and Ae are linearly independent, α, β ∈ ℂ and x = αAe + βe, then I(A, B; e)x =

im (αB + β).

Proof. (i) Since x ∉ ∨{e, Ae} there exists ξ ∈ 𝒳^{∗} that annihilates ∨{e, Ae}, that is ξ ∈ (∨{e, Ae})^{⊥} such that ⟨x, ξ⟩ = 1. Let y ∈ 𝒴 be artitrary. The operator y ⊗ ξ, which is given by (y ⊗ξ)z = ⟨z, ξ⟩y (z ∈ 𝒳), maps x to y and it is in I(A, B; e) because (y ⊗ξ)Ae = 0 = B(y ⊗ξ)e.

(ii) Let μ ∈ ℂ\{0} be such that x = μe. If S ∈ I(A, B; e), then (B −λ)Sx = μS(Ae−λe) = 0. Thus, I(A, B; e)x ⊆ ker (B − λ). For the opposite inclusion, let y ∈ ker (B − λ) be arbitrary. Then there exists S ∈ B(𝒳, 𝒴) such that Sx = y. Since (B − λ)Se = μ^{−1}(B − λ)y = 0 we have BSe = λSe = SAe and S ∈ I(A, B; e).

(iii) If S ∈ I(A, B; e), then Sx = S(αAe + βe) = (αB + β)Se, which shows that I(A, B; e)x ⊆ im (αB + β). On the other hand, let y = (αB + β)w, where w ∈ 𝒴, be an arbitrary vector in the range im (αB + β). Since e and Ae are linearly independent there exist ξ, η ∈ 𝒳^{∗} such that ⟨e, ξ⟩ = 1 = ⟨Ae, η⟩ and ⟨Ae, ξ⟩ = 0 = ⟨e, η⟩. Set S := w ⊗ ξ + Bw ⊗ η. Then it is easily seen that S ∈ I(A, B; e) and Sx = y. □

Let σ_{p}(T) be the point spectrum (the set of eigenvalues) of a given linear operator T ∈ B(𝒳). It is well-known that a number λ is in σ_{p}(T^{∗}) if and only if the range im (T − λ) is not dense in 𝒳. Recall that a nonempty set 𝒮 ⊆ B(𝒳) is transitive if, for any x ≠ 0, the orbit 𝒮x is dense in 𝒳.

Corollary 2.2. Let A, B ∈ B(𝒳) and e ∈ 𝒳. Assume that e and Ae are linearly independent. Then it is an immediate consequence of Proposition 2.1 that I(A, B; e) is transitive if and only if the point spectrum of B^{∗} is empty. In particular, the local commutant C(A, e) is transitive if and only if σ_{p}(A^{∗}) = ∅.

Now we describe the reflexive closure of the space of local intertwiners.

Proposition 2.3. Let A ∈ B(𝒳), B ∈ B(𝒴), and e ∈ 𝒳 be arbitrary. If e and Ae are linearly independent, then

Ref I(A, B; e) = {T ∈ B(𝒳, 𝒴); T (A − λ)e ∈ [im (B − λ)] for all λ ∈ ℂ}.

Proof. Let T ∈ Ref I(A, B; e) be arbitrary. Choose λ ∈ ℂ and set x_{λ} = Ae − λe. By Proposition 2.1 (iii), I(A, B; e)x_{λ} ∈ im (B−λ). Since T x ∈ [I(A, B; e)x] for any x ∈ 𝒳 we conclude T (A−λ)e = T x_{λ} ∈ [I(A, B; e)x_{λ}] = [im (B − λ)].

Now, assume that T ∈ B(𝒳, 𝒴) satisfies T (A − λ)e ∈ [im (B − λ)] for all λ ∈ ℂ. Let x ∈ 𝒳 be arbitrary. It is obvious that T x ∈ [I(A, B; e)x] for x = 0. Suppose therefore that x ≠ 0. If x ∉ [{e, Ae}], then, by Proposition 2.1 (i), I(A, B; e)x = 𝒴 which gives T x ∈ [I(A, B; e)x]. If x is a scalar multiple of e, say x = βe for some β ≠ 0, then I(A, B; e)x = im (βI) = 𝒴, by Proposition 2.1 (iii), and again T x ∈ [I(A, B; e)x]. Finally, assume that x = αAe + βe with α ≠ 0. Then T x = αT (A + β/α)e ∈ [im (B + β/α)]. Since, by Proposition 2.1 (iii), im (B + β/α) = I(A, B; e)(A + β/α)e we conclude that T x ∈ [I(A, B; e)x]. □

Corollary 2.4. If e and Ae are linearly independent, then Ref I(A, B; e) = B(𝒳, 𝒴) if and only if σ_{p}(B^{∗}) = ∅.

Proof. If σ_{p}(B^{∗}) = ∅, then [im (B − λ)] = 𝒴 for all λ ∈ ℂ. Thus, every T ∈ B(𝒳, 𝒴) satisfies the condition T (A−λ)e ∈ [im (B−λ)] (λ ∈ ℂ), which means, by Proposition 2.3, that T ∈ I(A, B; e).

On the other hand, if there exists λ ∈ σ_{p}(B^{∗}), then [im (B − λ)] ≠ 𝒴. Since (A − λ)e is a nonzero vector there exists T ∈ B(𝒳, 𝒴) such that T (A − λ)e ∉ [im (B − λ)]. □

It follows from Proposition 2.1 that I(A, B; e) is reflexive whenever e is an eigenvector of A.

Proposition 2.5. Let A ∈ B(𝒳) and B ∈ B(𝒴). If e ∈ 𝒳 is an eigenvector of A, then I(A, B; e) is reflexive.

Proof. Let Ae = λe and assume that T ∈ Ref I(A, B; e). Then, by Proposition 2.1, we have T e ∈ ker (B − λ). It follows that BT e = λT e = T Ae, i.e., T ∈ I(A, B; e). □

For an operator T ∈ B(𝒳) such that σ_{p}(T^{∗}) ≠ ∅, let Eig(T^{∗}) be the weak-∗ closure of the subspace of 𝒳^{∗} that is spanned by the eigenvectors of T^{∗}. If σ_{p}(T^{∗}) is empty, then we set Eig(T^{∗}) = {0}.

Theorem 2.6. Let A ∈ B(𝒳), B ∈ B(𝒴), and e ∈ 𝒳 be arbitrary. If e and Ae are linearly independent, then the following is equivalent:

(i) I(A, B; e) is reflexive;

(ii) Eig(B^{∗}) = 𝒴^{∗};

(iii) ⋂_{λ∈ℂ}[im (B − λ)] = {0}.

Proof. First we shall prove the equivalence of (i) and (ii). If Eig(B^{∗}) is a proper subspace of 𝒴^{∗}, then there exists a non-zero vector y ∈ Eig(B^{∗})_{⊥}. Let ξ ∈ 𝒳^{∗} be such that ⟨e, ξ⟩ = 0 and ⟨Ae, ξ⟩ = 1. Then T := y ⊗ ξ is not in I(A, B; e), since T Ae = y ≠ 0 = BT e. However, for an arbitrary number λ_{0}, we have

T (A − λ_{0})e = y ∈ Eig(B^{∗})_{⊥} = ⋂_{λ∈ℂ}[im (B − λ)] ⊆ [im (B − λ_{0})],

which gives T ∈ Ref I(A, B; e), by Proposition 2.3.

For the opposite implication, assume that Eig(B^{∗}) = 𝒴^{∗}. Let T ∈ Ref I(A, B; e) be arbitary. By Proposition 2.3, we have T (A−λ)e ∈ [im (B −λ)] for all λ ∈ ℂ. Choose and fix λ_{0} ∈ σ_{p}(B^{∗}). Then ⟨T (A − λ_{0})e, η⟩ = 0 for each η ∈ ker (B^{∗} − λ_{0}). It follows that

⟨T Ae, η⟩ = λ_{0}⟨T e, η⟩ = ⟨T e, B^{∗}η⟩ = ⟨BT e, η⟩.

Thus, ⟨(BT − T A)e, η⟩ = 0 for all η ∈ ker (B^{∗} − λ_{0}). Since λ_{0} ∈ σ_{p}(B^{∗}) is arbitrary and since Eig(B^{∗}) = 𝒴^{∗} we conclude that (BT − T A)e = 0, i.e. operator T is in I(A, B; e).

Now about the equivalence of (ii) and (iii). It is well known that [im (B −λ)] = ker (B^{∗} −λ)_{⊥}. Thus, if x ∈ [im (B − λ)], for all λ ∈ ℂ, then ⟨ξ, x⟩ = 0, for any eigenvector ξ of B^{∗}. It follows that x ∈ Eig(B^{∗})_{⊥}. On the other hand, if x ∈ 𝒳 is not in the intersection ∩_{λ∈ℂ}[im (B − λ)], then there exists a number λ_{0} such that x ∉ [im (B − λ_{0})] = ker (B^{∗} − λ_{0})_{⊥}. Thus, there exists an eigenvector ξ of B^{∗} such that ⟨ξ, x⟩ ≠ 0, which means x ∉ Eig(B^{∗})_{⊥}. □

Note that conditions (ii) and (iii) do not depend on vector e. Thus, the following assertion holds.

Corollary 2.7. If I(A, B; e) is reflexive for e ∈ 𝒳 \ {0} that is not an eigenvector for A, then I(A, B; f) is reflexive for any f ∈ 𝒳. □

Clearly

⋂_{e∈𝒳} I(A, B; e) = I(A, B) := {S ∈ B(𝒳, 𝒴); SA = BS}.

Since an arbitary intersection of reflexive spaces is a reflexive space we have the following corollary, which extends Lemma 1 [9].

Corollary 2.8. Let A ∈ B(𝒳) and B ∈ B(𝒴). If Eig(B^{∗}) = 𝒴^{∗}, then I(A, B) is reflexive. □

Note however that the condition Eig(B^{∗}) = 𝒴^{∗} is not necessary for reflexivity of I(A, B). For instance, let N be a normal operator without eigenvalues on a complex Hilbert space ℋ. Then, of course, Eig(N^{∗}) = {0}. On the other hand, the commutant {N}′ is reflexive since it is a von Neumann algebra ([2], Proposition 56.6).

Corollary 2.9. Let A ∈ B(𝒳) be an arbitrary operator and let B ∈ B(𝒴) be a non-zero nilpotent operator. If I(A, B; e) is reflexive for some non-zero e ∈ 𝒳, then e is an eigenvector of A.

Proof. Since B is non-zero nilpotent the adjoint operator B^{∗} is a non-zero nilpotent as well. It follows that Eig(B^{∗}) ≠ 𝒴^{∗}. By Theorem 2.6, I(A, B; e) cannot be reflexive if e is not an eigenvector of A. □

Proposition 2.10. Let T ∈ B(𝒳) and S ∈ B(𝒴) be operators such that there exists an injective operator V ∈ I(T, S). If S satisfies condition (iii) of Theorem 2.6, then T satisfies this condition as well.

Proof. Assume that T does not satisfy the conditions. Then there exists a non-zero vector x ∈ ⋂_{λ∈ℂ}[im (T − λ)]. The intertwiner V is injective, therefore V x ∈ 𝒴 is also a non-zero vector. Let λ ∈ ℂ be an arbitary number. Since x ∈ [im (T − λ)], there exists a sequence (x_{n}) in 𝒳 such that lim_{n→∞} ∥(T − λ)x_{n} − x∥ = 0. It follows lim_{n→∞} ∥(S − λ)V x_{n} − V x∥ ≤ ∥V ∥ lim_{n→∞} ∥(T − λ)x_{n} − x∥ = 0, which gives V x ∈ [im (S − λ)]. We conclude that S does not satisfy condition (iii) of Theorem 2.6. □

Note that the condition in Proposition 2.10 is satisfied if T is a quasi-affine transform of S. In particular, it is weaker than the quasi-similarity of operators T and S.

Now we shall give a description of operators that satisfy the equivalent conditions (ii) and (iii) of Theorem 2.6. Our description is based on the idea presented in [5], Solution 69.

Let Ω be a non-empty set and let X(Ω) be a Banach space of complex-valued functions on Ω satisfying the following two conditions:

for each ω ∈ Ω, there exists f ∈ X(Ω) such that f(ω) ≠ 0;

|f(ω)| ≤ ∥f∥, for f ∈ X(Ω) and ω ∈ Ω. (1)

An operator M ∈ B(X(Ω)) is a multiplication operator if there exists a complex-valued function φ on Ω such that (Mf)(ω) = φ(ω)f(ω) for all ω ∈ Ω. If M is a multiplication operator, then the corresponding function φ is uniquely determined. In the sequel we shall write M_{φ} to indicate the connection between a multiplication operator and the corresponding function.

For each ω ∈ Ω, define the point evaluation ξ_{ω} on X(Ω) by ⟨f, ξ_{ω}⟩ = f(ω) (f ∈ X(Ω)). Since

|⟨f, ξ_{ω}⟩| = |f(ω)| ≤ ∥f∥ (f ∈ X(Ω))

each ξ_{ω} is a linear functional with norm at most 1. By the first condition in (1), each ξ_{ω} is non-zero and it is not hard to see that the linear span of {ξ_{ω}; ω ∈ Ω} is weak-∗ dense in X(Ω)^{∗}. Let M_{φ} ∈ B(X(Ω)) be an arbitrary multiplication operator. Then

⟨f, (M_{φ})^{∗}ξ_{ω}⟩ = ⟨M_{φ}f, ξ_{ω}⟩ = φ(ω)f(ω) = ⟨f, φ(ω)ξ_{ω}⟩ (f ∈ X(Ω))

holds for any ω ∈ Ω. Thus, each ξ_{ω} is an eigenvector for (M_{φ})^{∗} (with φ(ω) as the corresponding eigenvalue) and consequently Eig((M_{φ})^{∗}) = X(Ω)^{∗}.

Now, let 𝒳 be a Banach space that is isometrically isomorphic to X(Ω), i.e. there exists a (bijective) linear isometry U : 𝒳 → X(Ω). Assume that T ∈ B(𝒳) is equivalent to a multiplication operator M_{φ} ∈ B(X(Ω)), which means T = U^{−1}M_{φ}U. It is easily seen that the linear span of {U^{∗}ξ_{ω}; ω ∈ Ω} is weak-∗ dense in 𝒳^{∗} and that T^{∗}U^{∗}ξ_{ω} = φ(ω)U^{∗}ξ_{ω} (ω ∈ Ω). Thus, Eig(T^{∗}) = 𝒳^{∗}. We have proved one implication in the following theorem.

Theorem 2.11. Let 𝒳 be a Banach space. An operator T ∈ B(𝒳) satisfies Eig(T^{∗}) = 𝒳^{∗} if and only if T is equivalent to a multiplication operator M_{φ} on a Banach space X(Ω) satisfying (1).

Proof. Let Ω be the set of all eigenvectors of T^{∗} of norm 1. For each x ∈ 𝒳, let Ux be the complex function on Ω defined by (Ux)(ω) = ⟨x, ω⟩. Of course X(Ω) := {Ux; x ∈ 𝒳} is a linear space of complex-valued functions on Ω and U : x ↦ Ux is a linear surjection from 𝒳 to X(Ω). The map U is also injective since the weak-∗ closed linear span of Ω is Eig(T^{∗}) = 𝒳^{∗}. If we equip X(Ω) with the norm ∥Ux∥ := ∥x∥ (x ∈ 𝒳), then X(Ω) becomes a Banach space satisfying (1) and U becomes an isometry, which means that 𝒳 and X(Ω) are isometrically isomorphic Banach spaces. Define φ : Ω → ℂ through T^{∗}ω = φ(ω)ω and let M_{φ} : X(Ω) → X(Ω) be given by (M_{φ}Ux)(ω) = φ(ω)(Ux)(ω). Then

(M_{φ}Ux)(ω) = φ(ω)⟨x, ω⟩ = ⟨x, T^{∗}ω⟩ = (UT x)(ω),

which gives M_{φ} = UT U^{−1}. Thus, M_{φ} is bounded and it is a multiplication operator equivalent to T. □

Corollary 2.12. Let A ∈ B(𝒳), B ∈ B(𝒴), and e ∈ 𝒳 be arbitrary. If e and Ae are linearly independent, then I(A, B; e) is reflexive if and only if B is equivalent to a multiplication operator M_{φ} on a Banach space X(Ω) satisfying (1).

Assume that a multiplication operator M_{φ} on X(Ω) (satisfying (1)) is also an algebraic operator. Let m(z) = (z − λ_{1})^{r_{1}} ⋯ (z − λ_{k})^{r_{k}} be the minimal polynomial. It is easily seen that the condition m(M_{φ}) = 0 is equivalent to the condition

(φ(ω) − λ_{1})^{r_{1}} ⋯ (φ(ω) − λ_{k})^{r_{k}} = 0 for all ω ∈ Ω.

However, (φ(ω) − λ_{1})^{r_{1}} ⋯ (φ(ω) − λ_{k})^{r_{k}} = 0 if and only if (φ(ω) − λ_{1}) ⋯ (φ(ω) − λ_{k}) = 0. Thus, if M_{φ} is an algebraic operator, then each zero of its minimal polynomial is simple. On the other hand, if φ(Ω) = {λ_{1}, …, λ_{k}}, then M_{φ} is an algebraic multiplication operator with the minimal polynomial m(z) = (z − λ_{1}) ⋯ (z − λ_{k}).

Corollary 2.13 (Cf. [1], Section 3). If B ∈ B(𝒴) is an algebraic operator such that its minimal polynomial has only simple zeroes, then I(A, B; e) is reflexive for any A ∈ B(𝒳) and any e ∈ 𝒳. On the other hand, if B is algebraic and I(A, B; e) is reflexive for an operator A ∈ B(𝒳) and a vector e ∈ 𝒳 that is not an eigenvector for A, then the minimal polynomial of B has only simple zeroes.

Proof. Let m(z) = (z − λ_{1}) ⋯ (z − λ_{k}) be the minimal polynomial of B (thus, λ_{i} ≠ λ_{j} if i ≠ j). For each 1 ≤ i ≤ k, let q_{i}(z) := m(z)/(z − λ_{i}). Since m(B) = 0 we have [im (B − λ_{i})] ⊆ ker q_{i}(B) and consequently

⋂_{λ∈ℂ}[im (B − λ)] ⊆ ⋂_{i=1}^{k}[im (B − λ_{i})] ⊆ ⋂_{i=1}^{k} ker q_{i}(B).

However, the intersection ⋂_{i=1}^{k} ker q_{i}(B) is trivial since the greatest common divisor of the polynomials q_{i} is equal to 1.

Conversely, suppose that I(A, B; e) is reflexive for some A and e, such that e is not an eigenvector of A. Then B is equivalent to a multiplication operator M_{φ}, by Theorems 2.6 and 2.11. Of course, M_{φ} is an algebraic operator with the same minimal polynomial as B. By the observation above, we conclude that the minimal polynomial has only simple zeroes. □

Example 2.14. (1) An operator B ∈ B(𝒴) will be called semi-shift if it is bounded below and ⋂_{n=1}^{∞} im B^{n} = {0}. Any semi-shift satisfies the equivalent conditions of Theorem 2.6. Indeed, there is an open neighbourhood U of 0 such that B − z is bounded below for z ∈ U. Then ⋂_{z∈U} im (B − z) = ⋂_{n=1}^{∞} im B^{n} = {0}. Hence the spaces of intertwiners I(A, B; e) are reflexive for all A ∈ B(𝒳) and e ∈ 𝒳, which gives the reflexivity of I(A, B) for any A ∈ B(𝒳).

(2) In particular, let B ∈ B(ℋ) be a unilateral weighted shift on a Hilbert space ℋ. Thus, Be_{i} = w_{i}e_{i+1} (i = 0, 1, …), where e_{0}, e_{1}, … is an orthonormal basis for ℋ and w_{i} ∈ ℂ form a bounded sequence. Suppose that im B is closed, i.e. inf_{i} |w_{i}| > 0. Then B is a semi-shift and therefore it satisfies the conditions of Theorem 2.6.

Assumption that im B is closed is necessary. For example, let B be the weighted shift with weights w_{i} = \frac{1}{i+1}. Then ∥B^{n}∥ = \frac{1}{n!} and so B is quasinilpotent. Hence ⋂_{z∈ℂ}[im (B − z)] = [im B] = ∨{e_{i}; i ≥ 1} and B does not satisfy the conditions of Theorem 2.6.

(3) Let V be an isometry acting in a Hilbert space ℋ. Let V = U ⊕ S be the Wold decomposition of V, where U is unitary and S is a unilateral shift (of some multiplicity). Clearly, the commutant {U}′ is reflexive since it is a von Neumann algebra and {S}′ is reflexive by (1). However, in general {V }′ is not reflexive. For example, let U be the bilateral shift and S the unilateral shift. Then V = U ⊕ S may be represented as the operator of multiplication by z in L^{2} ⊕ H^{2}, where L^{2} is considered with respect to Lebesgue measure on the unit circle and H^{2} is the Hardy space. For f_{1}, f_{2} ∈ L^{∞}, f_{3} ∈ H^{∞} the operator of multiplication by the matrix

[\begin{matrix}f_{1} & f_{2} \\ 0 & f_{3} \end{matrix}] belongs to \{V \}′. For g ∈ H^{2}, g ≠ 0, we have \{V \}′(0 ⊕ g) ⊃ \overline{gL^{∞}} ⊕ 0 = L^{2} ⊕ 0. Hence

for any X ∈ B(H^{2}, L^{2}) the operator [\begin{matrix}0 & X \\ 0 & 0 \end{matrix}] ∈ Ref\{V \}′ and \{V \}′ is not reflexive.

3. Hyperreflexivity of the space of locally intertwining operators

Let 𝒮 ⊆ B(𝒳, 𝒴) be a closed subspace. For an operator T ∈ B(𝒳, 𝒴), define

α(T, 𝒮) = sup{dist(T x, 𝒮x); x ∈ 𝒳, ∥x∥ = 1}.

The space 𝒮 is said to be hyperreflexive if there is a constant c > 0 such that the inequality dist(T, 𝒮) ≤ c α(T, 𝒮) holds for all T ∈ B(𝒳, 𝒴). It is well known that the hyperreflexivity is stronger condition than reflexivity, that is, each hyperreflexive space is reflexive. In this section we shall show that some spaces of local intertwiners can serve as natural examples of spaces that are reflexive but not hyperreflexive.

First we give a characterisation of hyperreflexive spaces of local intertwiners.

Proposition 3.1. Let A ∈ B(𝒳) and B ∈ B(𝒴) be arbitary operators and assume that Ae = λe for some λ ∈ ℂ. Then I(A, B; e) is hyperreflexive.

Proof. Without loss of generality we may assume that ∥e∥ = 1. Let S ∈ B(𝒳, 𝒴). By Proposition 2.1, we have α(S, I(A, B; e)) = dist(Se, ker (B − λ)).

We shall prove that dist(S, I(A, B; e)) = dist(Se, ker (B−λ)). Let ε > 0 and let y ∈ ker (B−λ) satisfy ∥Se − y∥ < dist(Se, ker (B − λ)) + ε. Let y^{∗} ∈ 𝒴^{∗} satisfy ⟨e, y^{∗}⟩ = 1 = ∥y^{∗}∥. Define S_{0} ∈ B(𝒳, 𝒴) by S_{0}e = Se−y and S_{0}|_{ker y^{∗}} = 0. Then S−S_{0} ∈ I(A, B; e) and dist(S, I(A, B; e)) ≤ ∥S_{0}∥. Let x ∈ 𝒳 have norm 1. Write x = αe + x_{0} with α ∈ ℂ and x_{0} ∈ ker y^{∗}. Then

\begin{aligned}∥(S_{0})x∥ & = ∥α(S_{0})e∥ = |⟨x, y^{∗}⟩| · ∥Se − y∥ \\ & ≤ ∥Se − y∥ ≤ dist(Se, ker (B − λ)) + ε. \end{aligned}

Hence dist(S, I(A, B; e)) ≤ dist(Se, ker (B − λ)). □

Lemma 3.2. Let A ∈ B(𝒳) and B ∈ B(𝒴) be arbitary operators. Let e ∈ 𝒳 and Ae be linearly independent. Then there exists a constant k > 0 such that for any S ∈ B(𝒳, 𝒴) it is possible to find S_{0} ∈ B(𝒳, 𝒴) with the properties

S_{0}e = 0,  S − S_{0} ∈ I(A, B; e)  and  ∥S_{0}∥ ≤ k∥SAe − BSe∥.

Consequently, dist(S, I(A, B; e)) ≤ k∥SAe − BSe∥.

Proof. Since e and Ae are linearly independent there exists k > 0 such that |β| ≤ \frac{k}{2}∥αe + βAe∥ for arbitrary α, β ∈ ℂ. Choose and fix a projection P ∈ B(𝒳) whose image is ∨{e, Ae} and ∥P ∥ ≤ 2. Let S ∈ B(𝒳, 𝒴) be arbitrary. Now let S_{0} ∈ B(𝒳, 𝒴) be defined by conditions

S_{0}e = 0,  S_{0}Ae = SAe − BSe  and S_{0}|_{ker P} = 0.

Since (S − S_{0})Ae = SAe − SAe + BSe = B(S − S_{0})e, the operator S − S_{0} is in I(A, B; e). Let x ∈ 𝒳 be an arbitrary vector of norm 1 and let x = αe + βAe + x_{0} with x_{0} ∈ ker P. Then

\begin{aligned}∥S_{0}x∥ & = ∥βS_{0}Ae∥ = |β| · ∥SAe − BSe∥ ≤ \frac{k}{2} ∥αe + βAe∥ · ∥SAe − BSe∥ \\ & = \frac{k}{2} ∥P x∥ · ∥SAe − BSe∥ ≤ k∥SAe − BSe∥. \end{aligned}

It follows now that dist(S, I(A, B; e)) ≤ ∥S_{0}∥ ≤ k∥SAe − BSe∥. □

Theorem 3.3. Let A ∈ B(𝒳) and B ∈ B(𝒴) be arbitary operators and assume that e ∈ 𝒳 and Ae are linearly independent. Then I(A, B; e) is hyperreflexive if and only if there exists a number ϵ > 0 such that sup{dist(y, im (B − λ)); λ ∈ ℂ} > ϵ, for all y ∈ 𝒴, ∥y∥ = 1.

Proof. Withot loss of generality we assume that ∥e∥ = 1, ∥A∥ ≤ 1, and ∥B∥ ≤ 1.

Suppose that for any ϵ > 0 there exists a vector y_{ϵ} ∈ 𝒴 of norm one such that

(2)  sup{dist(y_{ϵ}, im (B − λ)); λ ∈ ℂ} < ϵ.

Since e and Ae are linearly independent, there exists ξ ∈ 𝒳^{∗} such that ⟨ξ, e⟩ = 0 and ⟨ξ, Ae⟩ = 1. Let F_{ϵ} := y_{ϵ} ⊗ ξ. Thus F_{ϵ} is a rank-one operator that maps e to 0 and Ae to y_{ϵ}. We show that dist(F_{ϵ}, I(A, B; e)) ≥ 1/2. Towards contradiction suppose that there exists an operator S ∈ I(A, B; e) such that ∥F_{ϵ} − S∥ < 1/2. Then ∥Se∥ = ∥F_{ϵ}e − Se∥ ≤ ∥F_{ϵ} − S∥ < 1/2 and therefore ∥SAe∥ = ∥BSe∥ ≤ ∥B∥ · ∥Se∥ < 1/2. It follows that

∥(F_{ϵ} − S)Ae∥ = ∥y_{ϵ} − SAe∥ ≥ ∥y_{ϵ}∥ − ∥SAe∥ > 1 − 1/2 = 1/2.

Since ∥Ae∥ ≤ 1 we conclude that ∥F_{ϵ} − S∥ > 1/2, which contradicts to the assumption.

We have seen that for any ϵ > 0 there exists a rank-one operator F_{ϵ} such that dist(F_{ϵ}, I(A, B; e)) ≥ 1/2. Now we shall estimate α(F_{ϵ}, I(A, B; e)).

If a vector x ∈ 𝒳 is not in [{e, Ae}], then I(A, B; e)x = 𝒴, by Proposition 2.1. Thus, dist(F_{ϵ}x, I(A, B; e)x) = 0 in this case.

Assume that x = αAe + βe, for some α, β ∈ ℂ, and ∥x∥ = 1. Of course, there is a number M > 0 such that M ≥ |α| for all α ∈ ℂ that satisfy condition ∥αAe + βe∥ = 1 for some β ∈ ℂ. Note that M does not depend on ϵ. By Proposition 2.1, if x = αAe + βe, then I(A, B; e)x = im (αB + β). Thus, dist(F_{ϵ}x, I(A, B; e)x) = dist(αy_{ϵ}, im (αB + β)) ≤ Mdist(y_{ϵ}, im (αB + β)) and therefore, by (2), dist(F_{ϵ}x, I(A, B; e)x) < Mϵ. We conclude that α(F_{ϵ}, I(A, B; e)) < Mϵ. Now, since lim_{ϵ→0} α(F_{ϵ}, I(A, B; e)) = 0 and dist(F_{ϵ}, I(A, B; e)) ≥ 1/2 for any ϵ > 0, the space I(A, B; e) is not hyperreflexive.

For the opposite implication, let S ∈ B(𝒳, 𝒴) be arbitrary and let S_{0} ∈ B(𝒳, 𝒴) be an operator that satisfies the conditions from Lemma 3.2, so dist(S, I(A, B; e)) ≤ ∥S_{0}∥ ≤ k∥SAe − BSe∥. Since S − S_{0} ∈ I(A, B; e) we have α(S, I(A, B; e)) = α(S_{0}, I(A, B; e)). By the assumption, there exists λ ∈ ℂ such that dist(S_{0}Ae, im (B − λ)) ≥ ε∥S_{0}Ae∥. Clearly λ ∈ σ(B), and so |λ| ≤ ∥B∥. Note also that I(A, B; e)(Ae − λe) = im (B − λ), by Proposition 2.1. So we have

α(S, I(A, B; e)) = α(S_{0}, I(A, B; e)) ≥ ∥Ae − λe∥^{−1}dist(S_{0}(Ae − λe), I(A, B; e)(Ae − λe))

≥ \frac{dist(S_{0}Ae, im (B − λ))}{(∥A∥ + ∥B∥)∥e∥} ≥ \frac{ε∥S_{0}Ae∥}{(∥A∥ + ∥B∥)∥e∥}.

Recall that S_{0}Ae = SAe−BSe (see proof of the claim) and so α(S, I(A, B; e)) ≥ c∥SAe−BSe∥, where c = \frac{ε}{(∥A∥+∥B∥)∥e∥}. □

Example 3.4. Let 𝒴 = ℓ^{2} and let B ∈ B(ℓ^{2}) be given by

B : (x_{1}, x_{2}, x_{3}, …) ↦ (x_{1}, \frac{1}{2}x_{2}, \frac{1}{3}x_{3}, …).

It is easily seen that im (B − \frac{1}{n}) = \{(x_{i}) ∈ ℓ^{2}; x_{n} = 0\}, for any n ∈ ℕ, and that im (B −λ) = ℓ^{2} if λ ≠ \frac{1}{n} (∀n ∈ ℕ). Thus B satisfies condition (iii) of Theorem 2.6 and we conclude that I(A, B; e) is reflexive for any Banach space 𝒳 and arbitary A ∈ B(𝒳) and e ∈ 𝒳. On the other hand, these spaces are hyperreflexive if and only if e is eigenvector of A or e = 0. Namely, we shall see that B does not satisfy condition (ii) of Theorem 3.3.

For a positive integer k, let f^{(k)} = (f_{j}^{(k)}) ∈ ℓ^{2} be given by

f_{j}^{(k)} = \{ \begin{matrix}\frac{1}{k}; & 1 ≤ j ≤ k^{2} \\ 0; & k^{2} < j. \end{matrix}

Then ∥f^{(k)}∥ = 1 and f^{(k)} ∈ im (B − λ) if λ ∉ \{1, \frac{1}{2}, …, \frac{1}{k^{2}} \}. Thus dist(f^{(k)}, im (B − λ)) = 0 if λ ∉ \{1, \frac{1}{2}, …, \frac{1}{k^{2}} \}. For 1 ≤ n ≤ k^{2} we have

dist(f^{(k)}, im (B − \frac{1}{n}) = min\{∥f^{(k)} − (x_{j})∥; x_{n} = 0\} = \frac{1}{k} .

We conclude that

sup\{dist(f^{(k)}, im (B − λ)); λ ∈ ℂ\} = \frac{1}{k} ,

which means that condition (ii) of Theorem 3.3 is not fulfilled.

4. k-reflexivity and k-hyperreflexivity of the space of locally intertwining operators

Let 𝒳 and 𝒴 be complex Banach spaces and let F (𝒴, 𝒳) be the space of all operators of finite rank from 𝒴 to 𝒳, that is the linear span of all operators of finite rank. Thus, an operator F ∈ B(𝒴, 𝒳) is of finite rank if and only if there exist a positive integer n and x_{1}, …, x_{n} ∈ 𝒳, η_{1}, …, η_{n} ∈ 𝒴^{∗} such that F = x_{1} ⊗ η_{1} + ⋯ + x_{n} ⊗ η_{n}. The pair (B(𝒳, 𝒴), F (𝒴, 𝒳)) is a dual pair via the pairing

⟨T, F ⟩ = ⟨T x_{1}, η_{1}⟩ + ⋯ + ⟨T x_{n}, η_{n}⟩,

where T ∈ B(𝒳, 𝒴) and F = x_{1} ⊗ η_{1} + ⋯ + x_{n} ⊗ η_{n} ∈ F (𝒴, 𝒳) are arbitrary. If U ⊆ B(𝒳, 𝒴), then let U^{⊥} := {F ∈ F (𝒴, 𝒳); ⟨S, F ⟩ = 0 for all S ∈ U} and, similarly, for W ⊆ F (𝒴, 𝒳), let W_{⊥} := {S ∈ B(𝒳, 𝒴); ⟨S, F ⟩ = 0 for all F ∈ W}.

For a positive integer k, let F_{k}(𝒴, 𝒳) ⊆ F (𝒴, 𝒳) be the subset of all operators from 𝒴 to 𝒳 whose rank is at most k. Since F_{k}(𝒴, 𝒳)_{⊥} = {0} and F_{k}(𝒴, 𝒳) is closed under multiplication by the scalars, (B(𝒳, 𝒴), F (𝒴, 𝒳), F_{k}(𝒴, 𝒳)) satisfies the conditions of a reflexive triple (over ℂ) in the sense of [4]. Thus, for a linear subspace 𝒮 ⊆ B(𝒳, 𝒴) we define the k-reflexive cover of 𝒮 as Ref_{k}𝒮 := (𝒮^{⊥} ∩ F_{k}(𝒴, 𝒳))_{⊥}. The sets Ref_{k}𝒮 are linear subspaces of B(𝒳, 𝒴) closed in the weak operator topology. Of course, 𝒮 ⊆ Ref_{k}𝒮 and 𝒮 is said to be k-reflexive if 𝒮 = Ref_{k}𝒮. Clearly, the 1-reflexivity coincides with the notion of reflexivity. The reader is referred to [4] for details; especially for the relation to the classical notion of a reflexive algebra.

Let 𝒮 ⊆ B(𝒳, 𝒴) be a weakly closed subspace such that 𝒮 = W_{⊥} with W ⊆ F_{k}(𝒴, 𝒳). Then 𝒮^{⊥} ∩ F (𝒴, 𝒳) = (W_{⊥})^{⊥} ∩ F (𝒴, 𝒳) ⊇ W and consequently Ref_{k}𝒮 = (𝒮^{⊥} ∩ F (𝒴, 𝒳))_{⊥} ⊆ W_{⊥} = 𝒮. It follows that 𝒮 is k-reflexive. On the other hand, if 𝒮 is k-reflexive, then 𝒮 = W_{⊥} with W = 𝒮^{⊥}∩F_{k}(𝒴, 𝒳) ⊆ F_{k}(𝒴, 𝒳). Thus, 𝒮 is k-reflexive if and only if there is a subset W ⊆ F_{k}(𝒴, 𝒳) such that 𝒮 = W_{⊥}.

Proposition 4.1. For arbitrary A ∈ B(𝒳), B ∈ B(𝒴), and e ∈ 𝒳, the subspace I(A, B; e) ⊆ B(𝒳, 𝒴) is 2-reflexive.

Proof. It is obvious that an operator S ∈ B(𝒳, 𝒴) satisfies SAe = BSe if and only if ⟨S, Ae ⊗ η − e ⊗ B^{∗}η⟩ = 0 holds for all η ∈ 𝒴^{∗}. Thus, I(A, B; e) = G(A, B; e)_{⊥}, where G(A, B; e) := {Ae ⊗ η − e ⊗ B^{∗}η; η ∈ 𝒴^{∗}} ⊆ F_{2}(𝒴, 𝒳). □

Let 𝒮 ⊆ B(𝒳, 𝒴) be a subspace and T ∈ B(𝒳, 𝒴). For a positive integer k, define

α_{k}(T, 𝒮) = sup{ inf_{A∈𝒮} ∑_{i=1}^{k} ∥T x_{i} − Ax_{i}∥; x_{1}, …, x_{k} ∈ 𝒳, ∥x_{1}∥ + ⋯ + ∥x_{k}∥ = 1}.

In particular, for k = 1, we have α_{1}(T, 𝒮) = α(T, 𝒮). The space 𝒮 is said to be k-hyperreflexive if the seminorms dist(·, 𝒮) and α_{k}(·, 𝒮) are equivalent.

Again, the notion of 1-hyperreflexivity coincides with that of hyperreflexivity.

Denote by dist_{1} the distance in the space 𝒴^{k} (the ℓ_{1}-direct sum of k copies of 𝒴). We have

α_{k}(T, 𝒮) = sup_{{x_{1},...,x_{k}∈𝒳 \atop ∥x_{1}∥+···+∥x_{k}∥=1}} dist_{1}((T x_{1}, …, T x_{k}), \{(Ax_{1}, …, Ax_{k}); A ∈ 𝒮\})

= sup_{{x_{1},...,x_{k}∈𝒳 \atop ∥x_{1}∥+···+∥x_{k}∥=1}} sup_{{y^{∗}_{1},...,y^{∗}_{k}∈𝒴^{∗} \atop ∥y^{∗}_{1}∥≤1,...,∥y^{∗}_{k}∥≤1}} \{|∑_{i=1}^{k}⟨T x_{i}, y_{i}^{∗}⟩|; ∑_{i=1}^{k}⟨Ax_{i}, y_{i}^{∗}⟩ = 0 for all A ∈ 𝒮\}

= sup_{{F ∈F_{k}(𝒴,𝒳) \atop ∥F ∥_{1}≤1}} |⟨T, F ⟩|.

Thus, this definition agrees with that given by Kliś and Ptak in [6] for Hilbert spaces.

Theorem 4.2. For arbitrary A ∈ B(𝒳), B ∈ B(𝒴), and e ∈ 𝒳, the subspace I(A, B; e) ⊆ B(𝒳, 𝒴) is 2-hyperreflexive.

Proof. If e is an eigenvector of A, then the space I(A, B; e) is even hyperreflexive, by Proposition 3.1.

Assume that the vectors e and Ae are linearly independent and let T ∈ B(𝒳, 𝒴) be arbitrary. By Lemma 3.2, there is a constant k > 0 such that dist(T, I(A, B; e)) ≤ k∥T Ae − BT e∥. On the other hand, let y^{∗} ∈ 𝒴^{∗} satisfy ∥y^{∗}∥ = 1 and ⟨T Ae − BT e, y^{∗}⟩ = ∥T Ae − BT e∥. We have

\begin{aligned}α_{2}(T, I(A, B, e)) & ≥ ∥Ae ⊗ y^{∗} − e ⊗ B^{∗}y^{∗}∥_{1}^{−1}|⟨T, Ae ⊗ y^{∗} − e ⊗ B^{∗}y^{∗}⟩| \\ & ≥ ((∥A∥ + ∥B∥)∥e∥)^{−1}|⟨T Ae − BT e, y^{∗}⟩| = ((∥A∥ + ∥B∥)∥e∥)^{−1}∥T Ae − BT e∥. \end{aligned}

Hence I(A, B; e) is 2-hyperreflexive. □

References

[1] J. Bračič, Reflexivity of the commutant and local commutants of an algebraic operator, to apear in Linear Alg. Appl.

[2] J. B. Conway, A Course in Operator Theory, AMS, Graduate Studies in Mathematics, Vol. 21, Providence, RI, (2000).

[3] X. Dai and D. R. Larson, Wandering Vectors for Unitary Systems and Orthogonal Wavelets, Mem. Amer. Math. Soc., 134 No. 640 (1998). MR1432142 (98m:47067).

[4] D. Hadwin, A general view of reflexivity, Trans. Amer. Math. Soc., 344 (1994), 325-360.

[5] P. R. Halmos, A Hilbert Space Problem Book, Springer, 1974.

[6] K. Kliś and M. Ptak, k-hyperreflexive subspaces, Houston Journal of Mathematics, 32 (2006), 299-313.

[7] J. Kraus and D. R. Larson, Some applications of a technique for constructing reflexive operator algebras, J. Operator Theory, 13 (1985), 227-236.

[8] D. R. Larson, Von Neumann algebras and wavelets, Operator algebras and applications (Samos, 1996), 267– 312, NATO Adv. Sci. Inst. Ser. C Math. Phys. Sci., 495, Kluwer Acad. Publ., Dordrecht, 1997. MR1462685 (98g:46091)

[9] M. Zajac, Operators whose eigenvectors span the space, Tatra Mountains Math. Publ. 3 (1993), 245-249.

University of Ljubljana, IMFM, Jadranska ul. 19, SI-1000 Ljubljana, Slovenia

E-mail address: janko.bracic@fmf.uni-lj.si

Mathematical Institute, Czech Academy of Sciences, Žitná 25, 115 67 Prague 1, Czech Republic E-mail address: muller@math.cas.cz

Department of Mathematics, Slovak University of Technology, SK-812 19 Bratislava, Slovakia

E-mail address: michal.zajac@stuba.sk