HYPERREFLEXIVITY OF FINITE-DIMENSIONAL SUBSPACES

VLADIMÍR MÜLLER AND MAREK PTAK

Abstract. We show that each reflexive finite-dimensional subspace of operators is hyperreflexive. This gives a positive answer to a problem of Kraus and Larson. We also show that each n– dimensional subspace of Hilbert space operators is [\sqrt{2n}]–hyperreflexive.

1. Introduction

Let X be a complex Banach space and let B(X) be the algebra of all bounded linear operators on X. For an algebra W ⊂ B(X) with identity, let Alg LatW denote the set of all operators which leave invariant all (closed) subspaces of X, which are invariant for all operators from W. The algebra W is called reflexive if W = Alg LatW.

The definition was introduced for the first time in [16] and further studied by a number of authors. The concept of reflexivity is interesting even if the underlying space is finite dimensional. For example, the

algebra \{[\begin{matrix}a & b \\ 0 & a \end{matrix}]⊕[a] : a, b ∈ ℂ\} is reflexive, but the algebra \{[\begin{matrix}a & b \\ 0 & a \end{matrix}] :

a, b ∈ ℂ} is not reflexive (the former example will be used later).

The definition of reflexivity was extended to subspaces of operators in [13]. Let X, Y be Banach spaces and let M be a norm-closed subspace of B(X, Y) — the space of all bounded linear operators from X into Y. Write

refM = \{T ∈ B(X, Y): T x ∈ \overline{Mx} for allx ∈ X\},

where Mx = {Sx : S ∈ M}. The subspace M is called reflexive if M = refM. For algebras with identity both definition coincide.

2000 Mathematics Subject Classification. Primary: 47D15; Secondary: 47D30.

Key words and phrases. Reflexive subspaces, hyperreflexive subspace, hyperreflexive constant, k–hyperreflexive subspaces.

The research of the first author was supported by grant No. 201/03/0041 of GA ČR.

A stronger concept of hyperreflexivity was introduced for algebras in [1] and extended for subspaces of operators in [10]. Denote by dist(·, ·) the usual distance in Y; we use also the same notation for the distance in B(X, Y). Let M ⊂ B(X, Y) be a norm-closed subspace and T ∈ B(X, Y). Write

(1)  α(T, M) = sup{dist(T x, Mx): x ∈ X, ∥x∥ = 1}.

We always have α(T, M) ⩽ dist(T, M). The subspace M is called hyperreflexive if there is a constant C > 0 such that for all T ∈ B(X, Y), we have

(2)  dist(T, M) ⩽ C α(T, M).

The smallest constant C fulfilling (1) is called the hyperreflexive constant and denoted by κ_{M}.

Let us observe that if M is reflexive and T ∈ refM, then α(T, M) = 0. Hence dist(T, M) = 0 and, since M is norm closed, we have T ∈ M. Thus each hyperreflexive subspace is also reflexive. On the other hand there are reflexive non-hyperreflexive subspaces (see [9]). However, if both spaces X and Y are finite dimensional then each reflexive subspace is also hyperreflexive. Namely, as we have observed above the reflexivity of a norm-closed subspace M is equivalent to the condition:

α(T, M) = 0 ⟺ dist(T, M) = 0.

Thus, for the whole conclusion, it is enough to note that all norms on the finite dimensional space B(X, Y)/M are equivalent.

In [10, Problem 3.9], Kraus and Larson posed the question whether the concepts of reflexivity and hyperreflexivity are equivalent for finitedimensional subspaces of operators on infinite dimensional spaces. The problem was considered also in [6].

In [10] it was shown that each one-dimensional subspace is hyperreflexive. By [14], the hyperreflexive constant is equal to 1.

The aim of this paper is to give a positive answer to the problem of Kraus and Larson. The main result of the paper is

Main Theorem. Let M ⊂ B(X, Y) with dim M < ∞. If M is reflexive, then M is hyperreflexive.

In [12] it was shown that each n-dimensional subspace of Hilbert space operators is [\sqrt{2n}]-reflexive, where [\sqrt{2n}] is the integer part of \sqrt{2n}. Using our main result we show in Section 3 that each n-dimensio-nal subspace is even [\sqrt{2n}]-hyperreflexive (for definitions see Section 3).

Remark. Many authors (including [10]) considered the reflexivity and hyperreflexivity only for subspaces of operators on a Hilbert space.

They use a different definition of the distance α(T, M):

α(T, M) = sup{∥QT P ∥ : P, Q are projections andQMP = 0}.

To see the equivalence of both definitions of the distance α(·, ·), note that (see [3, Proposition 58.1]) both distances are equal to (3) α(T, M) = sup{|(T x, y)|: ∥x∥ = ∥y∥ = 1, (Sx, y) = 0 forall S ∈ M}.

It is easy to see that the definitions of reflexivity and hyperreflexivity used in this paper also agree with the more general definitions introduced in [5].

2. Main theorem

Let X, Y be Banach spaces. Denote by F (X, Y) the set of all finiterank operators from X to Y and by F_{k}(X, Y) the set of all operators in B(X, Y) of rank smaller or equal to k. Denote by S_{X} = {x ∈ X : ∥x∥ = 1} the unit sphere in X.

Let n ≥ 1 and let A_{1}, …, A_{n} ∈ B(X, Y). Denote by span{A_{i} : i = 1, …, n} the closed linear space generated by A_{1}, …, A_{n}. Write

s_{0}(A_{1}, …, A_{n}) = inf{∥∑_{i=1}^{n} λ_{i}A_{i}∥ : λ_{1}, …, λ_{n} ∈ ℂ, max |λ_{i}| = 1}.

More generally, for k ∈ ℕ set

s_{k}(A_{1}, …, A_{n}) = inf{s_{0}(A_{1}|_{M}, … A_{n}|_{M}) : M ⊂ X, codimM ⩽ k}.

The following lemma summarizes the properties of the quantities s_{k}.

Lemma 2.1. Let A_{1}, …, A_{n} ∈ B(X, Y). Then:

(1) s_{0}(A_{1}, …, A_{n}) = inf{∥∑_{i=1}^{n} λ_{i}A_{i}∥ : max |λ_{i}| ⩾ 1};

(2) s_{0}(A_{1}) = ∥A_{1}∥;

(3) s_{0}(A_{1}, …, A_{n}) ⩽ min{∥A_{i}∥ : i = 1, …, n};

(4) s_{0}(A_{1}, …, A_{n}) > 0 if and only if the operators A_{1}, …, A_{n} are linearly independent;

(5) s_{k}(A_{1}, …, A_{n}) ⩾ s_{l}(A_{1}, …, A_{n}) for k ⩽ l;

(6) s_{k}(A_{1}, …, Â_{j}, …, A_{n}) ⩾ s_{k}(A_{1}, …, A_{n}) for j = 1, …, n, where the hat denotes the omitted term;

(7) if M is a subspace of X and codimM ⩽ k then for any l we have s_{l}(A_{1}|_{M}, …, A_{n}|_{M}) ⩾ s_{l+k}(A_{1},…,A_{n});

(8) dist(A_{j}, span{A_{i} : i ≠ j}) ⩾ s_{0}(A_{1}, …, A_{n}) for j = 1, …, n;

(9) if k ∈ ℕ and no non-trivial linear combination of A_{1}, …, A_{n} belongs to F_{k}(X, Y), then s_{k}(A_{1}, …, A_{n}) > 0.

Proof. The statements (1)–(7) are trivial. To see (8), fix j and observe that

dist(A_{j}, span{A_{i} : i ≠ j}) = inf{∥∑_{i=1}^{n} λ_{i}A_{i}∥ : |λ_{j}| = 1}

⩾ inf{∥∑_{i=1}^{n} λ_{i}A_{i}∥ : max |λ_{i}| ⩾ 1} = s_{0}(A_{1}, …, A_{n}).

To see (9), let us fix k ⩾ 0. Let Z = {∑_{i=1}^{n} λ_{i}A_{i} : max |λ_{i}| = 1}. Since

Z is compact and F_{k}(X, Y) closed, we have dist(Z, F_{k}(X, Y)) > 0.

Let M ⊂ X, codimM ⩽ k. Let P ∈ B(X) be a projection onto M such that ∥P ∥ ⩽ k + 2, see [4, Exercise 5.24]. Let λ_{1}, …, λ_{n} ∈ ℂ, max |λ_{i}| = 1. Then

dist(∑_{i=1}^{n} λ_{i}A_{i}, F_{k}(X, Y)) ⩽ ∥∑_{i=1}^{n} λ_{i}A_{i} − ∑_{i=1}^{n} λ_{i}A_{i}(I − P)∥

∥∑_{i=1}^{n} λ_{i}A_{i}P ∥ ⩽ ∥∑_{i=1}^{n} λ_{i}A_{i}|_{M} ∥ · ∥P ∥ ⩽ (k + 2)∥∑_{i=1}^{n} λ_{i}A_{i}|_{M} ∥.

Thus

∥∑_{i=1}^{n} λ_{i}A_{i}|_{M} ∥ ⩾ \frac{1}{k + 2} dist(∑_{i=1}^{n} λ_{i}A_{i}, F_{k}(X, Y)),

and so

s_{k}(A_{1}, …, A_{n}) ⩾ \frac{1}{k + 2} dist(Z, F_{k}(X, Y)) > 0.

□

The following lemma is a quantitative version of [15, Lemma 1]. Note that for Hilbert spaces it is possible to take M = F^{⊥}.

Lemma 2.2. Let F ⊂ X, dim F = n < ∞, let ε > 0. Then there exists a subspace M ⊂ X such that codimM ⩽ (4nε^{−1} + 3)^{2n} and

∥f + m∥ ⩾ (1 − ε) max{∥f∥, ∥m∥/2}

for all m ∈ M, f ∈ F.

In particular, there is a subspace M_{0} ⊂ X with codimM_{0} ⩽ (12n + 3)^{2n} such that

∥f + m∥ ⩾ \frac{1}{3} max\{∥f∥, ∥m∥\}

for all f ∈ F and m ∈ M_{0}.

Proof. By the Auerbach lemma there are vectors x_{1}, …, x_{n} ∈ F and x_{1}^{∗}, …, x_{n}^{∗} ∈ F^{∗} of norm one such that ⟨x_{j}, x_{k}^{∗}⟩ = δ_{j,k} (the Kronecker symbol) for all j, k. Thus for all γ_{1}, …, γ_{n} ∈ ℂ we have

∥∑_{j=1}^{n} γ_{j}x_{j}∥ ⩾ max_{k}|⟨∑_{j=1}^{n} γ_{j}x_{j}, x_{k}^{∗}⟩| = max_{k} |γ_{k}|.

In particular, the vectors x_{1}, …, x_{n} are linearly independent and therefore form a basis of F. Let

Z = \{∑_{j=1}^{n}(\frac{k_{j}ε}{2n} + i\frac{l_{j}ε}{2n})x_{j} : k_{j}, l_{j} integers , |k_{j}|, |l_{j}| ⩽ 2nε^{−1} + 1\}.

Then cardZ ⩽ (4nε^{−1} + 3)^{2n}.

Let u ∈ F, ∥u∥ = 1. Write u = ∑_{j=1}^{n}(t_{j} + is_{j})x_{j} for real t_{j}, s_{j}. Clearly

|t_{j}|, |s_{j}| ⩽ 1 and we can find integers k_{j}, l_{j} such that |\frac{k_{j}ε}{2n} − t_{j}| ⩽ \frac{ε}{4n} and |\frac{l_{j}ε}{2n} − s_{j}| ⩽ \frac{ε}{4n}. Thus

∥u − ∑_{j=1}^{n}(\frac{k_{j}ε}{2n} + i\frac{l_{j}ε}{2n})x_{j}∥ ⩽ ∑_{j=1}^{n}(|\frac{k_{j}ε}{2n} − t_{j}| + |\frac{l_{j}ε}{2n} − s_{j}|) ⩽ \frac{ε}{2}.

So dist(u, Z) ⩽ \frac{ε}{2}. For z ∈ Z choose z^{∗} ∈ X^{∗} such that ∥z^{∗}∥ = 1 and ⟨z, z^{∗}⟩ = ∥z∥. Let M = ⋂_{z∈Z} ker z^{∗}. Clearly codimM ⩽ cardZ ⩽

(4nε^{−1} + 3)^{2n}.

Let f ∈ F, ∥f∥ = 1 and m ∈ M. Then there exists z ∈ Z such that ∥z − f∥ ⩽ \frac{ε}{2}. Thus ∥z∥ ⩾ 1 − \frac{ε}{2}. Let z^{∗} ∈ X^{∗} be the functional considered above. Then we have

∥f + m∥ ⩾ |⟨f + m, z^{∗}⟩| = |⟨f, z^{∗}⟩| ⩾ |⟨z, z^{∗}⟩| − |⟨f − z, z^{∗}⟩| ⩾ ∥z∥ − \frac{ε}{2} ⩾ 1 − ε.

Hence ∥f + m∥ ⩾ (1 − ε)∥f∥ for all f ∈ F, m ∈ M.

Furthermore,

∥f + m∥ ⩾ \frac{1}{2}(1 − ε)\frac{2−ε}{1−ε}∥f + m∥ = \frac{1}{2}(1 − ε)(∥f + m∥ + \frac{1}{1−ε}∥f + m∥)

⩾ \frac{1}{2}(1 − ε)(∥m∥ − ∥f∥ + ∥f∥) = \frac{1}{2}(1 − ε)∥m∥.

In particular, for ε = \frac{1}{3} we get that there exists a subspace M_{0} ⊂ X with codimM_{0} ⩽ (12n + 3)^{2n} such that

∥f + m∥ ⩾ \frac{1}{3} max\{∥f∥, ∥m∥\}

for all f ∈ F and m ∈ M_{0}. □

For simplicity we write r(n) = (12n + 3)^{2n} for the codimension of the space M_{0} in the previous lemma.

Theorem 2.3. There are increasing sequences of nonnegative integers h(n), g(n) and sequences of positive numbers c_{n}, c′_{n} with the following properties:

(a) if A_{1}, …, A_{n} ∈ B(X, Y) satisfy ∥A_{j}∥ ⩽ 1 for j = 1, …, n and no non-trivial linear combination of A_{1}, …, A_{n} belongs to F (X, Y), then there exists a unit vector u ∈ X such that

∥∑_{i=1}^{n} λ_{i}A_{i}u∥ ⩾ c_{n} s_{h(n)}^{n}(A_{1}, …, A_{n}) · max{|λ_{i}| : i = 1, …, n}

for all λ_{1}, …, λ_{n} ∈ ℂ;

(b) if T, A_{1}, …, A_{n} ∈ B(X, Y) satisfy ∥A_{1}∥ ⩽ 1 for j = 1, …, n and no non-trivial linear combination of A_{1}, …, A_{n} belongs to F (X, Y), then

α(T, span{A_{1}, …, A_{n}}) ⩾ c′_{n} s_{g(n)}^{n}(A_{1}, …, A_{n}) · dist(T, span{A_{1}, …, A_{n}}).

Proof. We prove both statements by induction on n.

Let n = 1 and let A_{1} ∈ B(X, Y) satisfy ∥A_{1}∥ ⩽ 1. Set c_{1} = \frac{1}{2} and h(1) = 0. There is a vector u ∈ X such that ∥u∥ = 1 and ∥A_{1}u∥ ⩾ \frac{1}{2}∥A_{1}∥. Thus ∥λ_{1}A_{1}u∥ ⩾ \frac{1}{2}|λ_{1}| · ∥A_{1}∥ = \frac{1}{2}|λ_{1}| s_{0}(A_{1}) for all λ_{1} ∈ ℂ. This proves statement (a) for n = 1.

(a)_{n} ⇒ (b)_{n} : Let g(n) = h(n) + 2n + 2 + (n + 1) r((2n + 2)(n + 1))

and c′_{n} = (\frac{12n}{c_{n}} + 6)^{−1}.

Let T ∈ B(X, Y). Write for short ε = α(T, span{A_{1}, …, A_{n}}) and ε′ = \frac{ε}{c′_{n}s_{g(n)}^{n}(A_{1},...,A_{n})}. For x ∈ X with ∥x∥ = 1 and δ > 0 set

D_{x,δ} = {(λ_{1}, …, λ_{n}) ∈ ℂ^{n} : ∥T x − ∑_{j=1}^{n} λ_{j}A_{j}x∥ ⩽ δ}.

Clearly D_{x,δ} is a closed convex set. By the definition of the distance α, D_{x,ε} ≠ ∅ for all x ∈ X, ∥x∥ = 1.

To show property (b)_{n}, we must prove that

(4)  ⋂_{x∈S_{X}} D_{x,ε′} ≠ ∅.

Indeed, for (γ_{1}, …, γ_{n}) ∈ ⋂_{x∈S_{X}} D_{x,ε′} we have ∥T x − ∑_{j=1}^{n} γ_{j}A_{j}x∥ ⩽ ε′

for all x ∈ X, ∥x∥ = 1, and so ∥T − ∑_{j=1}^{n} γ_{j}A_{j}∥ ⩽ ε′. Therefore

dist(T, span{A_{1},…, A_{n}}) ⩽ ε′, and so statement (b) for n is fulfilled.

By (a)_{n} and Lemma 2.1(9), there exists a vector x_{0} ∈ X with ∥x_{0}∥ = 1 and a constant c > 0 such that ∥∑_{i=1}^{n} λ_{i}A_{i}x_{0}∥ ≥ c max |λ_{i}| for all (λ_{1}, …, λ_{n}) ∈ ℂ^{n}. Therefore the set D_{x_{0},ε′} is bounded. Thus (4) is equivalent to

(5)  ⋂_{x∈S_{X}}(D_{x,ε′} ∩ D_{x_{0},ε′}) ≠ ∅,

where the sets D_{x,ε′} ∩ D_{x_{0},ε′} are convex compact subsets of ℂ^{n} ∼ ℝ^{2n}. By the classical Helly theorem (see [7]), it is sufficient to show that

⋂_{i=0}^{2n+1} D_{x_{i},ε′} ≠ ∅

for all (2n + 1)-tuples of unit vectors x_{1}, …, x_{2n+1} ∈ X.

Fix x_{1}, …, x_{2n+1} ∈ X of norm one. Let F_{1} = span{x_{i} : i = 0, …, 2n + 1} and let M_{1} ⊂ X be a subspace such that X = F_{1} ⊕ M_{1}. Then codimM_{1} ≤ 2n + 2 and F_{1} ∩ M_{1} = ∅. Let

F_{2} = span{T x_{i}, A_{j}x_{i} : i = 0, …, 2n + 1, j = 1, …, n}.

Then dim F_{2} ⩽ (2n + 2)(n + 2). By Lemma 2.2, there is a subspace M_{2} ⊂ Y with codimM_{2} ⩽ r((2n + 2)(n + 1)) such that ∥f + m∥ ⩾ \frac{1}{3} max\{∥f∥, ∥m∥\} for all f ∈ F_{2}, m ∈ M_{2}.

Let M = M_{1} ∩ T^{−1}M_{2} ∩ ⋂_{j=1}^{n} A_{j}^{−1}M_{2}. Then codimM ⩽ codimM_{1} +

(n + 1) codimM_{2}, and so h(n) + codimM ⩽ g(n). By the induction assumption (a)_{n} and by Lemma 2.1(7), (5), there exists a vector u ∈ M, ∥u∥ = 1 such that

(6) ∥∑_{i=j}^{n} λ_{j}A_{j}u∥ ⩾ c_{n} s_{h(n)}^{n}(A_{1}|_{M}, …, A_{n}|_{M}) · max_{j} |λ_{j}|

⩾ c_{n} s_{g(n)}^{n}(A_{1}, …, A_{n}) · max_{j} |λ_{j}|

for all λ_{1}, …, λ_{n} ∈ ℂ.

Claim 1. D_{x_{i},6ε} ∩ D_{u,6ε} ≠ ∅ for i = 0, 1, …, 2n + 1.

Proof. Fix i ∈ {0, 1, …, 2n + 1}. Note that x_{i} ∈ F_{1}, u ∈ M ⊂ M_{1}, and so x_{i} + u ≠ 0. Set v = \frac{x_{i}+u}{∥x_{i}+u∥}. Suppose on the contrary that D_{x_{i},6ε} ∩ D_{u,6ε} = ∅. For λ_{1}, …, λ_{n} ∈ ℂ we have

∥T v − ∑_{j=1}^{n} λ_{j}A_{j}v∥ = \frac{1}{∥x_{i} + u∥}∥T x_{i} − ∑_{j=1}^{n} λ_{j}A_{j}x_{i} + T u − ∑_{j=1}^{n} λ_{j}A_{j}u∥

⩾ \frac{1}{6} max\{∥T x_{i} − ∑_{j=1}^{n} λ_{j}A_{j}x_{i}∥, ∥T u − ∑_{j=1}^{n} λ_{j}A_{j}u∥\},

since T x_{i} − ∑_{j=1}^{n} λ_{j}A_{j}x_{i} ∈ F_{2} and T u − ∑_{j=1}^{n} λ_{j}A_{j}u ∈ M_{2}.

Since either (λ_{1}, …, λ_{n}) ∉ D_{x_{i},6ε} or (λ_{1}, …, λ_{n}) ∉ D_{u,6ε}, at least one of the two terms is greater than 6ε. Thus

∥T v − ∑_{j=1}^{n} λ_{j}A_{j}v∥ > ε

for all λ_{1}, …, λ_{n} ∈ ℂ. Hence D_{v,ε} = ∅, a contradiction. Claim 2. Let (λ_{1}, …, λ_{n}), (μ_{1}, …, μ_{n}) ∈ D_{u,6ε}. Then

max_{j} |λ_{j} − μ_{j}| ⩽ \frac{12ε}{c_{n}s_{g(n)}^{n}(A_{1}, …, A_{n})}.

Proof. Let (λ_{1}, …, λ_{n}), (μ_{1}, …, μ_{n}) ∈ D_{u,6ε}. Then

∥T u − ∑_{j=1}^{n} λ_{j}A_{j}u∥ ⩽ 6ε and ∥T u − ∑_{j=1}^{n} μ_{j}A_{j}u∥ ⩽ 6ε.

Hence

∥∑_{j=1}^{n}(λ_{j} − μ_{j})A_{j}u∥ ⩽ 12ε,

and so, by (6) we have

max_{j} |λ_{j} − μ_{j}| ⩽ \frac{∥∑_{j=1}^{n}(λ_{j} − μ_{j})A_{j}u∥}{c_{n} s_{g(n)}^{n}(A_{1}, …, A_{n})} ⩽ \frac{12ε}{c_{n} s_{g(n)}^{n}(A_{1}, …, A_{n})}.

Claim 3. Let i ∈ {0, 1, …, 2n + 1}. Then D_{x_{i},ε′} ⊃ D_{u,6ε}.

Proof. Let (λ_{1}, …, λ_{n}) ∈ D_{u,6ε} ∩ D_{x_{i},6ε}. Let (μ_{1}, …, μ_{n}) ∈ D_{u,6ε} be

arbitrary. Then max_{j} |λ_{j} − μ_{j}| ⩽ \frac{12ε}{c_{n} s_{g(n)}^{n}(A_{1},...,A_{n})} and

∥T x_{i} − ∑_{j=1}^{n} μ_{j}A_{j}x_{i}∥ ⩽ ∥T x_{i} − ∑_{j=1}^{n} λ_{j}A_{j}x_{i}∥ + ∥∑_{j=1}^{n}(λ_{j} − μ_{j})A_{j}x_{i}∥

⩽ 6ε + \frac{12εn}{c_{n}s_{g(n)}^{n}(A_{1}, …, A_{n})} ⩽ \frac{ε}{s_{g(n)}^{n}(A_{1}, …, A_{n})}(6 + \frac{12n}{c_{n}})ε′.

Thus (μ_{1}, …, μ_{n}) ∈ D_{x_{i},ε′}.

Hence ⋂_{i=0}^{2n+1} D_{x_{i},ε′} ⊃ D_{u,6ε} ≠ ∅, and (4) is fulfilled. This proves state-

ment (b) for n.

(b)_{n−1} ⇒ (a)_{n}: Let n ⩾ 2 and suppose that property (b) holds for n − 1. Set c_{n} = \frac{c′_{n−1}}{18n} and h(n) = g(n − 1) + n^{2} · r(n(n − 1)).

We construct inductively vectors u_{1}, …, u_{n} ∈ X of norm one in the following way. Let k ∈ {1, …, n} and suppose that the vectors u_{j}, j = 1, …, k − 1 have already been constructed. Let

F_{k} = span{A_{i}u_{j} : i = 1, …, n, j = 1, …, k − 1}.

Then dim F_{k} ⩽ n(k − 1) ⩽ n(n − 1). By Lemma 2.2, there is a subspace M_{k} ⊂ Y such that codimM_{k} ⩽ r(n(n − 1)) and ∥f + m∥ ⩾

\frac{1}{3} max\{∥f∥, ∥m∥\} for all f ∈ F_{k}, m ∈ M_{k}. Let M′_{k} = ⋂_{j=1}^{k} ⋂_{i=1}^{n} A_{i}^{−1}M_{j}.

Then codimM′_{k} ⩽ n^{2} · codimM_{k}, and so g(n − 1) + codimM′_{k} ⩽ h(n). By property (b)_{n−1}, there is a vector u_{k} ∈ M′_{k} of norm one such that

dist(A_{k}u_{k}, span{A_{i}u_{k} : i ≠ k})

⩾ \frac{1}{2} c′_{n−1}s_{g(n−1)}^{n−1}(A_{1}|_{M′_{k}}, …, { \atop A_{k}|_{M′_{k}}}, …, A_{n}|_{M′_{k}})

· dist(A_{k}|_{M′_{k}}, span{A_{i}|_{M′_{k}} : i ≠ k})

⩾ \frac{1}{2} c′_{n−1}s_{h(n)}^{n−1}(A_{1}, …, A_{n}) · s_{0}(A_{1}|_{M′_{k}}, …, A_{n}|_{M′_{k}})

⩾ \frac{1}{2} c′_{n−1}s_{h(n)}^{n}(A_{1}, …, A_{n}),

where the hat denotes the omitted term; in the estimates we used Lemma 2.1(6),(8) and (5).

Let u_{1}, …, u_{n} ∈ S_{X} be constructed in the above described way. Set

v = ∑_{j=1}^{n} u_{j}. For λ_{1}, …, λ_{n} ∈ ℂ and k ∈ {1, …, n}, since ∑_{j=1}^{k} ∑_{i=1}^{n} λ_{i}A_{i}u_{j} ∈

F_{k+1}, ∑_{j=k+1}^{n} ∑_{i=1}^{n} λ_{i}A_{i}u_{j} ∈ M_{k+1}, ∑_{j=1}^{k−1} ∑_{i=1}^{n} λ_{i}A_{i}u_{j} ∈ F_{k} and ∑_{i=1}^{n} λ_{i}A_{i}u_{k} ∈ M_{k},

we have

∥∑_{i=1}^{n} λ_{i}A_{i}v∥ = ∥∑_{j=1}^{n} ∑_{i=1}^{n} λ_{i}A_{i}u_{j}∥ ⩾ \frac{1}{3} ∥∑_{j=1}^{k} ∑_{i=1}^{n} λ_{i}A_{i}u_{j}∥

⩾ \frac{1}{9} ∥∑_{i=1}^{n} λ_{i}A_{i}u_{k}∥ ⩾ \frac{1}{9} |λ_{k}| · dist(A_{k}u_{k}, span\{A_{i}u_{k} : i ≠ k\})

⩾ \frac{1}{18} c′_{n−1} s_{g(n)}^{n}(A_{1}, …, A_{n}) · |λ_{k}|,

(if k = n then the first inequality is trivial). In particular, v ≠ 0, by Lemma 2.1(9). Hence the vector u = \frac{v}{∥v∥} satisfies ∥u∥ = 1 and

∥∑_{i=1}^{n} λ_{i}A_{i}u∥ ⩾ \frac{1}{18 ∥v∥} c′_{n−1} s_{g(n)}^{n}(A_{1}, …, A_{n}) · max_{k} |λ_{k}|

⩾ c_{n} s_{g(n)}^{n}(A_{1}, …, A_{n}) · max_{k} |λ_{k}|.

This finishes the proof. □

Corollary 2.4. Let M ⊂ B(X, Y) be a finite-dimensional subspace which contains no non-zero finite rank operators. Then M is hyperreflexive.

Proof. Choose a basis A_{1}, …, A_{n} of M. The proof follows from the previous theorem, property (b). □

Now we are ready to prove the main theorem.

Theorem 2.5. Let M ⊂ B(X, Y), dim M < ∞. Then M is hyperreflexive if and only if M is reflexive.

Proof. If M is hyperreflexive then M is clearly reflexive. Conversely, let M be reflexive. Let M_{1} = M ∩ F (X, Y) and let M_{2} be any subspace of M such that M = M_{1} ⊕ M_{2}. Choose a basis A_{1}, …, A_{k} of M_{1} and a basis B_{1}, …, B_{l} of M_{2}.

Let M = ⋂_{i=1}^{k} ker A_{i}. Then codimM < ∞. By the previous result for

the operators B_{i}|_{M}, there is a constant d_{1} > 0 such that (7) dist(T |_{M}, span{B_{1}|_{M}, …, B_{l}|_{M}}) ⩽ d_{1}·α(T |_{M}, span{B_{1}|_{M}, …, B_{l}|_{M}}).

Let P ∈ B(X) be a projection onto M and F = ker P. Let F′ = span{Sf : S ∈ M, f ∈ F }. Clearly dim F′ < ∞. By Lemma 2.2, there is a subspace M′ ⊂ Y such that codimM′ < ∞ and ∥f′ + m′∥ ⩾ \frac{1}{3} max\{∥f′∥, ∥m′∥\} for all f′ ∈ F′ and m′ ∈ M′. Set M′′ = M ∩

⋂_{i=1}^{l} B_{i}^{−1}M′. Clearly codimM′′ < ∞.

Let u ∈ M′′ be a ”separating vector” for the operators B_{i}|_{M′′}, i.e.,

∥u∥ = 1 and there is a constant d_{2} > 0 such that ∥∑_{i=1}^{l} γ_{i}B_{i}u∥ ⩾

d_{2} max |γ_{i}| for all γ_{1}, …, γ_{l} ∈ ℂ. Such a vector exists by Theorem 2.3 and Lemma 2.1(9).

It follows from [11, Corollary 2.8] that since M is reflexive, M_{1} is also reflexive. For the sake of completeness we include the proof of this here. Since M_{1} ⊂ M, we have refM_{1} ⊂ refM. By reflexivity of M, we have refM_{1} ⊂ M. To show the reflexivity of M_{1} = M∩F (X, Y), it is enough to show that refM_{1} ⊂ F (X, Y). Let B ∈ refM_{1}. Then, for all u ∈ X, Bu ∈ span{A_{i} x : i = 1, …, k, x ∈ X}. Hence rankB < ∞ and B ∈ F (X, Y).

Now, for i = 1, …, k consider the operators Ã_{i}: F → span{A_{1}x, …, A_{k}x : x ∈ F } induced by A_{i}. Since the operators A_{1}, …, A_{k} are equal to zero on M, it is easy to see that M̃_{1} = span{Ã_{1}, …, Ã_{k}} is reflexive. As it was observed in the introduction, M̃_{1} is hyperreflexive. Thus there exists a constant d_{3} > 0 with the following property: if ε > 0 and T: F → span{A_{1}x, …, A_{k}x : x ∈ F } satisfies dist(T x, span{A_{1}x, …, A_{k}x}) ⩽ ε for all x ∈ F, ∥x∥ = 1 then there

exist numbers γ_{1}, …, γ_{k} ∈ ℂ such that ∥T − ∑_{i=1}^{n} γ_{i}A_{i}|_{F} ∥ ⩽ d_{3}ε.

We show now that M is hyperreflexive. Let ε > 0, T ∈ B(X, Y) and let

dist(T x, Mx) ⩽ ε

for all x ∈ X, ∥x∥ = 1. By (7) there exist numbers β_{1}, … β_{l} ∈ ℂ such that

(8)  ∥T |_{M} − ∑_{j=1}^{l} β_{j}B_{j}|_{M} ∥ ⩽ d_{1} ε.

Set S = T −∑_{j=1}^{l} β_{j}B_{j}. Thus ∥S|_{M}∥ ⩽ d_{1}ε and S satisfies dist(Sx, Mx) ⩽

ε for all x ∈ X, ∥x∥ = 1.

Let x ∈ F, ∥x∥ = 1. Then there are numbers λ_{1}, …, λ_{k}, μ_{1}, …, μ_{l} ∈ ℂ such that

∥Sx − ∑_{i=1}^{k} λ_{i}A_{i}x − ∑_{j=1}^{l} μ_{j}B_{j}x∥ ⩽ ε.

12 VLADIMÍR MÜLLER AND MAREK PTAK

Similarly, there are numbers λ′_{1}, …, λ′_{k}, μ′_{1}, …, μ′_{l} ∈ ℂ such that

∥S(x + u) − ∑_{i=1}^{k} λ′_{i}A_{i}(x + u) − ∑_{j=1}^{l} μ′_{j}B_{j}(x + u)∥ ⩽ ε∥x + u∥ ⩽ 2ε.

By subtracting we have

∥Su + ∑_{i=1}^{k}(λ_{i} − λ′_{i})A_{i}x − ∑_{j=1}^{l}(μ_{j} − μ′_{j})B_{j}x − ∑_{j=1}^{l} μ′_{j}B_{j}u∥ ⩽ 3ε,

since A_{i} u = 0 for all i. By the definitions of M′′ and F′ and by (8), we have

∥∑_{j=1}^{l} μ′_{j}B_{j}u∥ ⩽ 3∥∑_{i=1}^{k}(λ_{i} − λ′_{i})A_{i}x + ∑_{j=1}^{l}(μ_{j} − μ′_{j})B_{j}x − ∑_{j=1}^{l} μ′_{j}B_{j}u∥

⩽ 3(3ε + ∥Su∥) ⩽ 3ε(3 + d_{1}).

Since ∥∑_{j=1}^{l} μ′_{j}B_{j}u∥ ⩾ d_{2} max |μ′_{j}|, we have max |μ′_{j}| ⩽ 3ε\frac{d_{1}+3}{d_{2}}. Thus we

have

∥Sx − ∑_{i=1}^{n} λ′_{i}A_{i}x∥ ⩽ ∥Sx − S(x + u)∥

+ ∥S(x + u) − ∑_{i=1}^{n} λ′_{i}A_{i}x − ∑_{j=1}^{l} μ′_{j}B_{j}(x + u)∥ + ∥∑_{j=1}^{l} μ′_{j}B_{j}(x + u)∥

⩽ ∥Su∥ + 2ε + ∑_{j=1}^{l} |μ′_{j}| · ∥B_{j}∥ · 2 ⩽ d_{4} ε,

where d_{4} = d_{1} + 2 + \frac{3d_{1}+9}{d_{2}} · 2 ∑_{j=1}^{l} ∥B_{j}∥. Thus there exist numbers

γ_{1}, …, γ_{k} ∈ ℂ such that ∥S|_{F} − ∑_{i=1}^{k} γ_{i}A_{i}|_{F} ∥ ⩽ d_{3} d_{4} ε.

Let f ∈ F, m ∈ M and ∥f + m∥ = 1. Then ∥m∥ = ∥P (f + m)∥ ⩽ ∥P ∥ and ∥f∥ ⩽ ∥f + m∥ + ∥m∥ ⩽ 1 + ∥P ∥. Since A|_{M} = 0, we have

∥T (f + m) − ∑_{i=1}^{k} γ_{i}A_{i}(f + m) − ∑_{j=1}^{l} β_{j}B_{j}(f + m)∥

∥S(f + m) − ∑_{i=1}^{k} γ_{i}A_{i}f∥ ⩽ ∥Sm∥ + ∥Sf − ∑_{i=1}^{k} γ_{i}A_{i}f∥

⩽ d_{1} ε ∥m∥ + d_{3} d_{4} ε ∥f∥.

Thus ∥T − ∑_{i=1}^{k} γ_{i}A_{i} − ∑_{j=1}^{l} β_{j}B_{j}∥ ⩽ ε(d_{1} ∥P ∥ + d_{3} d_{4} (∥P ∥ + 1)), and so

M is hyperreflexive. □

3. Examples and Corollaries

The example from [9], mentioned in the introduction shows also that there is no constant in the condition (2) for the hyperreflexivity of a finite-dimensional subspace depending only on the dimension of the subspace. Bellow we give another example of this kind.

Example 3.1. Let H = ℂ^{3} with the Hilbert norm. For ε > 0 consider

the operators A_{1,ε} = [\begin{matrix}1 & 0 \\ 0 & 1 \end{matrix}] ⊕ [ε] and A_{2,ε} = [\begin{matrix}0 & 1 \\ 0 & 0 \end{matrix}] ⊕ [0]. Let M_{ε} =

span{A_{1,ε}, A_{2,ε}}. Clearly dim M_{ε} = 2. It is easy to verify that M_{ε} is reflexive for all ε.

Let T [\begin{matrix}1 & 0 \\ 0 & 0 \end{matrix}] ⊕ [0]. For β, γ ∈ ℂ we have

∥βA_{1,ε} + γA_{2,ε} − T ∥ = ∥[\begin{matrix}β − 1 & γ \\ 0 & β \end{matrix}] ⊕ [εβ]∥ ⩾ max\{|β − 1|, |β|\} ⩾ \frac{1}{2}.

Thus dist(T, M_{ε}) ⩾ \frac{1}{2} for all ε > 0.

Let x = [\begin{matrix}a \\ b \end{matrix}] ⊕ [c] ∈ H, ∥x∥ = 1. If b ≠ 0 then dist(T x, M_{ε}x) ≤

∥ab^{−1}A_{2,ε}x − T x∥ = 0. If b = 0 then dist(T x, M_{ε}x) ≤ ∥A_{1,ε}x − T x∥ =

∥[\begin{matrix}0 \\ 0 \end{matrix}] ⊕ [εc]∥ ≤ ε. Thus α(T, M_{ε}) ⩽ ε and there is no constant C > 0

such that

dist(T, M_{ε}) ⩽ C · α(T, M_{ε})

for all ε > 0.

Now we consider finite dimensional subspaces of B(H), where H is a Hilbert space. It is well known that B(H) is the dual of the trace class operators. If M is a w^{∗}-closed subspace of B(H), in particular if dim M < ∞, then M is reflexive if and only if M_{⊥} ∩ F_{1}(H) is total in M_{⊥} (see for example [2]). According to [2], a subspace M is called kreflexive if M_{⊥} ∩ F_{k}(H) is total in M_{⊥}. In [12] it was shown that each n dimensional subspace is [\sqrt{2n}]–reflexive ([·] denotes the integer part). For any subspace M ⊂ B(H) and T ∈ B(H), as it was suggested in [8], we can consider

α_{k}(T, M) = sup{|⟨T, t⟩| : t ∈ M_{⊥}, ∥t∥ ⩽ 1, rankt ⩽ k}

(compare with (3)). As in [8] we can call the subspace M ⊂ B(H) k-hyperreflexive if there is a constant C such that

dist(T, M) ⩽ Cα_{k}(T, M)

for each operator T ∈ B(H). We will show the following

Corollary 3.2. Let M ⊂ B(H) and dim M = n. Then M is [\sqrt{2n}]- hyperreflexive.

Proof. Let k = [\sqrt{2n}]. By M^{(k)} we denote the k-th amplification of M

M^{(k)} = \{\underbrace{S ⊕ ⋯ ⊕ S}_{k} : S ∈ M\} ⊂ B (H^{(k)}) ,

where H^{(k)} is the direct sum of k-copies of H, H^{(k)} = \underbrace{H ⊕ ⋯ ⊕ H}_{k}.

Since dim M = n, M is k-reflexive by [12, Theorem 12]. By [2], M^{(k)} is reflexive. Since dim M^{(k)} = n, it is also hyperreflexive. Hence [8, Theorem 3.5] implies that M is k-hyperreflexive. □

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Vladimír Müller, Institute of Mathematics, Academy of Sciences of Czech Republic, Žitná 25, 115 67 Praha 1, Czech Republic

E-mail address: muller@math.cas.cz

Marek Ptak, Institute of Mathematics, University of Agriculture, Al.Mickiewicza 24/28, 30-059 Kraków, Poland

E-mail address: rmptak@cyf-kr.edu.pl