ALMOST INVARIANT HALF-SPACES OF OPERATORS ON BANACH SPACES

GEORGE ANDROULAKIS, ALEXEY I. POPOV, ADI TCACIUC, AND VLADIMIR G. TROITSKY

Abstract. We introduce and study the following modified version of the Invariant Subspace Problem: whether every operator T on an infinite-dimensional Banach space has an almost invariant half-space, that is, a subspace Y of infinite dimension and infinite codimension such that Y is of finite codimension in T (Y). We solve this problem in the affirmative for a large class of operators which includes quasinilpotent weighted shift operators on ℓ_{p} (1 ⩽ p < ∞) or c_{0}.

1. Introduction

Throughout the paper, X is a Banach space and by L(X) we denote the set of all (bounded linear) operators on X. By a “subspace” of a Banach space we always mean a “closed subspace”. Given a sequence (x_{n}) in X, we write [x_{n}] for the closed linear span of (x_{n}).

Definition 1.1. A subspace Y of a Banach space X is called a half-space if it is both of infinite dimension and of infinite codimension in X.

Definition 1.2. If T ∈ L(X) and Y is a subspace of X, then Y is called almost invariant under T, or T -almost invariant, if there exists a finite dimensional subspace F of X such that T (Y) ⊆ Y + F.

In this work, the following question will be referred to as the almost invariant half-space problem: Does every operator on an infinite-dimensional Banach space have an almost invariant half-space? Observe that every subspace of X that is not a half-space is clearly almost invariant under any operator. Also, note that the almost invariant half-space problem is not weaker than the well known invariant subspace problem, because in the latter the invariant subspaces are not required to be halfspaces.

Integral Equations and Operator Theory

1 2 G. ANDROULAKIS, A.I. POPOV, A. TCACIUC, AND V.G. TROITSKY

Date: May 19, 2009.

The third and the fourth authors were supported by NSERC.

The natural question whether the usual unilateral right shift operator acting on a Hilbert space has almost invariant half-spaces has an affirmative answer. Moreover, it is known that this operator has even invariant half-spaces. Indeed, by [6, Corollary 3.15], this operator has an invariant subspace with infinite-dimensional orthogonal complement (thus the invariant subspace is of infinite codimension). It is not hard to see that the space exhibited in the proof of this statement is in fact infinite dimensional.

It is natural to consider Donoghue operators as candidates for counterexamples to the almost invariant half-space problem, as their invariant subspaces are few and well understood. Recall that a Donoghue operator D ∈ L(ℓ_{2}) is an operator defined by

De_{0} = 0, De_{i} = w_{i}e_{i−1}, i ∈ ℕ,

where (w_{i}) is a sequence of non-zero complex numbers such that (|w_{i}|) is monotone decreasing and in ℓ_{2}. It is known that if D is a Donoghue operator then D has only invariant subspaces of finite dimension and D^{∗} has only invariant subspaces of finite codimension (see [6, Theorem 4.12]). Hence neither D nor D^{∗} have invariant halfspaces. In Section 3 we will employ the tools of Section 2 to show that, nevertheless, every Donoghue operator has almost invariant half-spaces. We do not know whether the operators constructed by Enflo [3] and Read [7] have almost invariant half-spaces.

The following result explains how almost invariant half-spaces of operators are related to invariant subspaces of perturbed operators.

Proposition 1.3. Let T ∈ L(X) and H ⊆ X be a half-space. Then H is almost invariant under T if and only if H is invariant under T + K for some finite rank operator K.

Proof. Suppose that T has an almost invariant half-space H. Let F be a subspace of the smallest dimension satisfying the condition in Definition 1.2. Then we have H ∩ F = {0}. Define P: H + F → F by P (h + f) = f. Since P is a finite rank operator, we can extend it to a finite rank operator on X using Hahn-Banach theorem. That is, there exists P̃: X → F such that P̃|_{H+F} = P. Define K: X → X by K := −P̃T. Clearly K has finite rank and for any h ∈ H we have T h = h′ + f for some h′ ∈ H and f ∈ F, so that

(T + K)(h) = T h − P̃T h = h′ + f − P̃(h′ + f) = h′ + f − f = h′

Therefore, (T + K)H ⊆ H, which shows that T + K has an invariant half-space.

Conversely, from (T +K)(H) ⊆ H it follows immediately that T (H) ⊆ H +K(H),so that H is an almost invariant half space for T. □

ALMOST INVARIANT HALF-SPACES 3

Finally we would like to point out that if an operator has almost invariant halfspaces, then so does its adjoint. For that we will need two simple lemmas. The proof of the first lemma is elementary.

Lemma 1.4. Let X be a Banach space and Y be a subspace of X. Then Y is infinite codimensional if and only if Y^{⊥} is of infinite dimension. Thus Y is a half-space if and only if both Y and Y^{⊥} are of infinite dimension.

Lemma 1.5. A subspace Y of X is a half-space if and only if Y^{⊥} is a half-space in X^{∗}.

Proof. Suppose Y is a half-space. By Lemma 1.4, Y^{⊥} must be infinite-dimensional. Also (Y^{⊥})^{⊥} ⊇ j(Y) where j: X → X^{∗∗} denotes the natural embedding. Thus (Y^{⊥})^{⊥} is infinite dimensional. Now Lemma 1.4 yields that Y^{⊥} is a half-space.

Let’s assume that Y^{⊥} is a half-space. Since Y^{⊥} is infinite codimensional Y must be infinite-dimensional (see, e.g. [1, Theorem 5.110]). On the other hand, since Y^{⊥} is infinite dimensional, by Lemma 1.4 we obtain that Y is of infinite codimension, thus a half-space. □

Remark 1.6. The statement dual to that of Lemma 1.5 is not true in general. That is, if Z is a half-space in X^{∗} then Z_{⊥} need not be a half-space. For example, c_{0} is a half-space in ℓ_{∞} while (c_{0})_{⊥} = {0} ⊆ ℓ_{1} is not.

Proposition 1.7. Let T be an operator on a Banach space X. If T has an almost invariant half-space then so does its adjoint T^{∗}.

Proof. Let Y be a half-space in X such that Y is almost invariant under T, and F be a finite-dimensional subspace of X of smallest dimension such that T Y ⊆ Y + F. Then Y ∩ F = {0}. Thus there exists a subspace W of X such that W + F = X, W ∩ F = {0}, and Y ⊆ W. In particular, W^{⊥} is finite dimensional. Denote Z = (Y + F)^{⊥}. By Lemma 1.5, Z is a half-space in X^{∗}. For every z ∈ Z and y ∈ Y we have ⟨y, T^{∗}z⟩ = ⟨T y, z⟩ = 0 since T y ∈ Y + F. Therefore T^{∗}Z ⊆ Y^{⊥}. To finish the proof, it suffices to show that Y^{⊥} = Z + W^{⊥}.

Indeed, by the definition of Z we have that Z ⊆ Y^{⊥}. Also since Y ⊆ W we have W^{⊥} ⊆ Y^{⊥}. Thus Z + W^{⊥} ⊆ Y^{⊥}. On the other hand, since F is finite dimensional and F ∩ W = {0}, we may choose a basis (f_{i}) of F with biorthogonal functionals (f_{i}^{∗}) such that f_{i}^{∗} ∈ W^{⊥}. Since Y ⊆ W we have that f_{i}^{∗} ∈ Y^{⊥}. Thus, if x^{∗} is an arbitrary element of Y^{⊥} then x^{∗} − ∑_{i} x^{∗}(f_{i})f_{i}^{∗} ∈ (Y + F)^{⊥} = Z, and therefore x^{∗} ∈ Z + W^{⊥}. □

4 G. ANDROULAKIS, A.I. POPOV, A. TCACIUC, AND V.G. TROITSKY

2. Basic tools

All Banach spaces in Sections 2, 3 and 4 are assumed to be complex. For a subset A of ℂ, we will write A^{−1} = \{\frac{1}{λ}: λ ∈ A, λ ≠ 0\}. For a Banach space X and T ∈ L(X), we will use symbols σ(T) for the spectrum of T, r(T) for the spectral radius of T, and ρ(T) for the resolvent set of T. For a nonzero vector e ∈ X and λ ∈ ρ(T)^{−1}, define a vector h(λ, e) in X by

h(λ, e) := (λ^{−1}I − T )^{−1}(e).

Note that if |λ| < \frac{1}{r(T)} then^{1} Neumann’s formula yields

(1)  h(λ, e) = λ ∑_{n=0}^{∞} λ^{n}T^{n}e.

Also, observe that (λ^{−1}I − T )h(λ, e) = e for every λ ∈ ρ(T)^{−1}, so that

(2)  T h(λ, e) = λ^{−1}h(λ, e) − e.

The last identity immediately yields the following result.

Lemma 2.1. Let X be a Banach space, T ∈ L(X), 0 ≠ e ∈ X, and A ⊆ ρ(T)^{−1}. Put

Y = \overline{span}\{h(λ, e): λ ∈ A\}.

Then Y is a T -almost invariant subspace (which is not necessarily a half-space), with T Y ⊆ Y + span{e}.

Remark 2.2. The Replacement procedure. For any nonzero vector e in a Banach space X, we have

h(λ, e) − h(μ, e) = (μ^{−1} − λ^{−1})h(λ, h(μ, e))

whenever λ, μ ∈ ρ(T)^{−1}.

Indeed,

h(λ, e) − h(μ, e) = [(λ^{−1}I − T)^{−1} − (μ^{−1}I − T)^{−1}](e)
= (μ^{−1} − λ^{−1})(λ^{−1}I − T)^{−1}(μ^{−1}I − T)^{−1}(e)
= (μ^{−1} − λ^{−1})(λ^{−1}I − T)^{−1}h(μ, e)
= (μ^{−1} − λ^{−1})h(λ, h(μ, e))

Lemma 2.3. Suppose that T ∈ L(X) has no eigenvectors. Then, for any nonzero vector e ∈ X the set {h(λ, e) : λ ∈ ρ(T)^{−1}} is linearly independent.

1In case r(T) = 0 we take \frac{1}{r(T)} = +∞.

ALMOST INVARIANT HALF-SPACES 5

Proof. We are going to use induction on n to show that for any nonzero vector e ∈ X and any distinct λ_{1}, λ_{2}, …, λ_{n} ∈ ρ(T)^{−1} the set

{h(λ_{1}, e), h(λ_{2}, e), …, h(λ_{n}, e)}

is linearly independent. The statement is clearly true for n = 1; we assume it is true for n − 1 and will prove it for n.

Fix e ∈ X and distinct λ_{1}, λ_{2}, …, λ_{n} ∈ ρ(T)^{−1}. Let a_{1}, a_{2}, …, a_{n} be scalars such that ∑_{k=1}^{n} a_{k}h(λ_{k}, e) = 0. It follows from (2) that

0 = T (∑_{k=1}^{n} a_{k}h(λ_{k}, e)) = ∑_{k=1}^{n} a_{k}λ_{k}^{−1}h(λ_{k}, e) − ∑_{k=1}^{n} a_{k}e.

If ∑_{k=1}^{n} a_{k} ≠ 0 then e ∈ span {h(λ_{k}, e)}_{k=1}^{n}, so that span{h(λ_{k}, e)}_{k=1}^{n} is T -invariant by (2). This subspace is finite-dimensional, so that T has an eigenvalue, which is a contradiction. Therefore ∑_{k=1}^{n} a_{k} = 0, so that a_{1} = − ∑_{k=2}^{n} a_{k}.

Using the Replacement Procedure we obtain

0 = ∑_{k=1}^{n} a_{k}h(λ_{k}, e) = ( − ∑_{k=2}^{n} a_{k})h(λ_{1}, e) + ∑_{k=2}^{n} a_{k}h(λ_{k}, e)
= ∑_{k=2}^{n} a_{k}(h(λ_{k}, e) − h(λ_{1}, e))
= ∑_{k=2}^{n} a_{k}(λ_{1}^{−1} − λ_{k}^{−1})h(λ_{k}, h(λ_{1}, e)).

By the induction hypothesis, the set {h(λ_{k}, h(λ_{1}, e))}_{k=2}^{n} is linearly independent, hence a_{k}(λ_{1}^{−1} − λ_{k}^{−1}) = 0 for any 2 ⩽ k ⩽ n. It follows immediately that a_{k} = 0 for any 1 ⩽ k ⩽ n, and this concludes the proof. □

This gives us a natural way to try to construct almost invariant half-spaces. Indeed, suppose that T has no eigenvectors. Let e ∈ X such that e ≠ 0, and let (λ_{n}) be a sequence of distinct elements of ρ(T)^{−1}. Put Y = [h(λ_{n}, e)]_{n=1}^{∞}. Then Y is almost invariant by Lemma 2.1 and infinite-dimensional by Lemma 2.3. However, the difficult part is to show that Y is infinite codimensional. Even passing to subsequences might not help, as there are sequences whose every subsequence spans a dense subspace (see, e.g., [8, page 58] and also [2]).

3. Weighted shift operators

In this section we give a sufficient condition for a quasinilpotent operator to have almost invariant half-spaces (Theorem 3.2). As an application, we show in Corollary 3.4

6 G. ANDROULAKIS, A.I. POPOV, A. TCACIUC, AND V.G. TROITSKY

that quasinilpotent weighted shifts on ℓ_{p} or c_{0} have invariant half-spaces. In particular, every Donoghue operator has an almost invariant half-space.

Recall that a sequence (x_{i}) in a Banach space is called minimal if x_{k} ∉ [x_{i}]_{i≠k} for every k, (see also [4, section 1.f]). It is easy to see that this is equivalent to saying that for every k, the biorthogonal functional x_{k}^{∗} defined on span{x_{i}} by x_{k}^{∗}(∑_{i=0}^{n} α_{i}x_{i}) = α_{k} is bounded.

We will use the following numerical lemma.

Lemma 3.1. Given a sequence (r_{i}) of positive reals, there exists a sequence (c_{i}) of positive reals such that the series ∑_{i=0}^{∞} c_{i}r_{i+k} converges for every k.

Proof. For every i take c_{i} = \frac{1}{2^{i}} min\{\frac{1}{r_{1}} , …, \frac{1}{r_{2i}} \}. For every i ⩾ k we have k + i ⩽ 2i, so that c_{i}r_{i+k} ⩽ \frac{1}{2^{i}}. It follows that

∑_{i=0}^{∞} c_{i}r_{i+k} ⩽ ∑_{i=0}^{k−1} c_{i}r_{i+k} + ∑_{i=k}^{∞} \frac{1}{2^{i}} < +∞.

□

Theorem 3.2. Let X be a Banach space and T ∈ L(X) satisfying the following conditions:

(i) T has no eigenvalues.

(ii) The unbounded component of ρ(T) contains {z ∈ ℂ : 0 < |z| < ε} for some

ε > 0. (iii) There is a vector whose orbit is a minimal sequence.

Then T has an almost invariant half-space.

Proof. Let e ∈ X be such that (T^{i}e)_{i=0}^{∞} is minimal. For each i put x_{i} = T^{i}e. Then for each k, the biorthogonal functional x_{k}^{∗} defined on span x_{i} by x_{k}^{∗}(∑_{i=0}^{n} α_{i}x_{i}) = α_{k} is bounded. Let r_{k} = ∥x_{k}^{∗}∥. Let (c_{i}) be a sequence of positive real numbers as in Lemma 3.1, so that β_{k} := ∑_{i=0}^{∞} c_{i}r_{i+k} < +∞ for every k. By making c_{i}’s even smaller, if necessary, we may assume that \sqrt[i]{c_{i}} → 0.

Consider a function F : ℂ → ℂ defined by F (z) = ∑_{i=0}^{∞} c_{i}z^{i}. Evidently, F is entire. Observe that we may assume that the set {z ∈ ℂ : F (z) = 0} is infinite. Indeed, by the Picard Theorem there exists a negative real number d such that the set {z ∈ ℂ : F (z) = d} is infinite. Now replace c_{0} with c_{0} − d. This doesn’t affect our other assumptions on the sequence (c_{i}).

ALMOST INVARIANT HALF-SPACES 7

Fix a sequence of distinct complex numbers (λ_{n}) such that F (λ_{n}) = 0 for every n. Since F is non-constant, the sequence (λ_{n}) has no accumulation points. Hence, |λ_{n}| → +∞.

Note that (ii) can be restated as follows: ρ(T)^{−1} has a connected component C such that 0 ∈ \overline{C} and C contains a neighbourhood of ∞. Thus by passing to a subsequence of λ_{n}’s and relabeling, if necessary, we can assume that λ_{n} ∈ C for all n.

Observe that the condition λ_{n} ∈ ρ(T)^{−1} for every n implies that h(λ_{n}, e) is defined for each n. Put Y = [h(λ_{n}, e)]_{n=1}^{∞}. Then Y is almost invariant under T by Lemma 2.1 and dim Y = ∞ by Lemma 2.3. We will prove that Y is actually a half-space by constructing a sequence of linearly independent functionals (f_{n}) such that every f_{n} annihilates Y.

For every k = 0, 1, …, put F_{k}(z) = z^{k}F (z). Let’s write F_{k}(z) in the form of Taylor series, F_{k}(z) = ∑_{i=0}^{∞} c_{i}^{(k)}z^{i}. Then

c_{i}^{(k)} = \begin{cases}0 & if i < k, and \\ c_{i−k} & if i ⩾ k. \end{cases}

Define a functional f_{k} on span{T^{i}e}_{i=0}^{∞} via f_{k}(T^{i}e) = c_{i}^{(k)}. Since T has no eigenvalues, the orbit of T is linearly independent thus f_{k} is well-defined. We will show now that f_{k} is bounded. Let x ∈ span{T^{i}e}_{i=0}^{∞}, then x = ∑_{i=0}^{n} x_{i}^{∗}(x)T^{i}e for some n, so that

|f_{k}(x)| = |f_{k}(∑_{i=0}^{n} x_{i}^{∗}(x)T^{i}e)| ⩽ (∑_{i=0}^{n}∥x_{i}^{∗}∥c_{i}^{(k)})∥x∥

= (∑_{i=k}^{n} r_{i}c_{i−k})∥x∥ ⩽ (∑_{i=k}^{∞} r_{i}c_{i−k})∥x∥ = β_{k}∥x∥,

so that ∥f_{k}∥ ⩽ β_{k}. Hence, f_{k} can be extended by continuity to a bounded functional on [T^{i}e]_{i=1}^{∞}, and then by the Hahn-Banach Theorem to a bounded functional on all of X.

Now we show that each f_{k} annihilates Y. Fix k. Recall that for each λ ∈ ρ(T)^{−1} such that |λ| < \frac{1}{r(T)} we have h(λ, e) = λ ∑_{i=0}^{∞} λ^{i}T^{i}e. Therefore

f_{k}(h(λ, e)) = f_{k}(λ ∑_{i=0}^{∞} λ^{i}T^{i}e) = λ ∑_{i=0}^{∞} λ^{i}c_{i}^{(k)} = λF_{k}(λ) = λ^{k+1}F (λ).

for every λ ∈ C such that |λ| < \frac{1}{r(T)} (recall 0 ∈ \overline{C}). The map λ ↦ h(λ, e) and, therefore, the map λ ↦ f_{k}(h(λ, e)), is analytic on the set ρ(T)^{−1}. Therefore, by the principle of uniqueness of analytic function, the functions f_{k}(h(λ, e)) and λ^{k+1}F (λ) must agree

8 G. ANDROULAKIS, A.I. POPOV, A. TCACIUC, AND V.G. TROITSKY

on C. Since λ_{n} ∈ C for all n, we have f_{k}(h(λ_{n}, e)) = λ_{n}^{k+1}F (λ_{n}) = 0 for all n. Thus, Y is annihilated by every f_{k}.

It is left to prove the linear independence of {f_{k}}_{k=1}^{∞}. Observe that f_{k} ≠ 0 for all k since f_{k}(T^{i}e) ≠ 0 for i ⩾ k. Suppose that f_{N} = ∑_{k=M}^{N−1} a_{k}f_{k} with a_{M} ≠ 0. However f_{N}(T^{M}e) = 0 by definition of f_{N} while ∑_{k=M}^{N−1} a_{k}f_{k}(T^{M}e) = a_{M}c_{0} ≠ 0, contradiction. □

Remark 3.3. Note that condition (ii) of Theorem 3.2 is satisfied by many important classes of operators. For example, it is satisfied if σ(T) is finite (in particular, if T is quasinilpotent) or if 0 belongs in the unbounded component of ρ(T).

Corollary 3.4. Suppose that X = ℓ_{p} (1 ⩽ p < ∞) or c_{0} and T ∈ L(X) is a weighted right shift operator with weights converging to zero but not equal to zero. Then both T and T^{∗} have almost invariant half-spaces.

Proof. It can be easily verified that T is quasinilpotent. Clearly, T has no eigenvalues, and the orbit of e_{1} is evidently a minimal sequence. By Theorem 3.2 and Remark 3.3, T has almost invariant half-spaces. Finally, Proposition 1.7 yields almost invariant half-spaces for T^{∗}. □

The following statement is a special case of Corollary 3.4.

Corollary 3.5. If D is a Donoghue operator then both D and D^{∗} have almost invariant half-spaces.

Recall that a subset D of ℂ is called a cone if D is closed under addition and multiplication by positive scalars.

Remark 3.6. Condition (ii) in Theorem 3.2 can be weakened as follows: instead of requiring that ρ(T) contains a punctured disk centered at zero, we may only require that it contains a non-trivial sector of this disk, i.e., the intersection of the punctured disk with a non-empty open cone. Equivalently, ρ(T)^{−1} has a connected component C such that 0 ∈ \overline{C}, and there exists an open cone D in ℂ and M > 0 such that {z ∈ D : |z| ⩾ M} ⊆ C. Indeed, suppose that such D and M exist. Choose ν in this cone with |ν| = M. Also, suppose that, as in the proof of Theorem 3.2, we have already found a sequence (λ_{m}) of zeros of F. The set \{M \frac{λ_{m}}{|λ_{m}|}\}_{m=0}^{∞} has an accumulation point, say μ. By passing to a subsequence, we may assume that M \frac{λ_{m}}{|λ_{m}|} → μ. Note that the spectrum of \frac{μ}{ν}T is obtained by rotating the spectrum of T. Thus ρ (\frac{μ}{ν}T )^{−1} has a connected component C′ such that 0 ∈ \overline{C′} and there exists an open cone D′ in ℂ

ALMOST INVARIANT HALF-SPACES 9

which contains a neighborhood of μ and {z ∈ D′ : |z| ⩾ M} ⊆ C′. Thus by passing to a subsequence of (λ_{m}) we can assume that λ_{m} ∈ C′ for all m. Replace in the proof T with \frac{μ}{ν}T. Note that this doesn’t affect the assumptions on the operator and the definitions of c_{i}’s, F, and λ_{m}’s. Finally, multiplying an operator by a non-zero number does not affect its almost invariant half-spaces.

Note that every operator T with σ(T) ⊆ ℝ satisfies this weaker version of condition (ii). In particular, it is satisfied by self-adjoint operators on Hilbert spaces.

Corollary 3.7. Suppose that T ∈ L(X) such that T has no eigenvectors, σ(T) ⊆ ℝ, and there is a vector whose orbit is a minimal sequence. Then T has an almost invariant half-space.

4. Non-quasinilpotent operators

In this section we will modify the argument of Theorem 3.2 to extend its statement to another class of operators having non-zero spectral radius. It is a standard fact that if (x_{i}) is a minimal sequence then \frac{1}{∥x_{n}^{∗}∥} = dist(x_{n}, [x_{i}]_{i≠n}) for every n.

Theorem 4.1. Let X be a Banach space and T ∈ L(X) be an operator with r(T) ⩽ 1 having no eigenvectors. Let e ∈ X; put x_{n} = T^{n}e for n ∈ ℕ. If (x_{n}) is a minimal sequence and ∑_{n=1}^{∞} \frac{∥x_{n}^{∗}∥}{n} < ∞ then T has an almost invariant half-space.

Proof. Let D stand for the unit disk in ℂ. For a sequence (λ_{n}) ⊂ D such that

(3)  ∑_{n=1}^{∞}(1 − |λ_{n}|) < ∞.

the corresponding Blaschke product is defined by

(4)  B(z) = ∏_{n=1}^{∞} \frac{|λ_{n}|}{λ_{n}} \frac{λ_{n} − z}{1 − \overline{λ_{n}}z} .

It is well known that B is a bounded analytic function on D with zeros exactly at (λ_{n}). According to [5, Theorem 2] we can choose a sequence (λ_{n}) ⊂ D satisfying (3) such that \frac{B^{(n)}(0)}{n!} = O(\frac{1}{n+1}). Thus B^{(n)}(0) = O(\frac{n!}{n+1}). For m ∈ ℕ set F_{m}(z) = z^{m}B(z). Obviously the functions (F_{m}) are linearly independent. It follows from

F_{m}(z) = z^{m}B(z) = ∑_{n=0}^{∞} \frac{B^{(n)}(0)}{n!}z^{n+m} = ∑_{n=m}^{∞} \frac{B^{(n−m)}(0)}{(n − m)!}z^{n}

that

(5)  F_{m}^{(n)}(0) = \begin{cases}0 & for n < m, and \\ \frac{n!}{(n−m)!}B^{(n−m)}(0) ⩽ C \frac{n!}{n−m+1} & for n ⩾ m. \end{cases}

10 G. ANDROULAKIS, A.I. POPOV, A. TCACIUC, AND V.G. TROITSKY

Put Y = [h(λ_{n}, e)]_{n=1}^{∞}. By Lemma 2.1, Y is almost invariant under T and dim Y = ∞ by Lemma 2.3. As in the proof of Theorem 3.2, we will show that under the conditions of the Theorem 4.1 there is a sequence of linearly independent functionals annihilating Y.

Define a linear functional f_{m} on span\{x_{n}\} by f_{m}(x_{n}) = \frac{F_{m}^{(n)}(0)}{n!}. Since T has no eigenvectors, the orbit of T is linearly independent, so f_{m} is well defined.

Let’s prove that f_{m} is bounded for every m ∈ ℕ. Take any x := ∑ α_{n}x_{n} ∈ span{x_{n}}. Using (5), we obtain

|f_{m}(x)| = |∑ α_{n} \frac{F_{m}^{(n)}(0)}{n!}| ⩽ C ∑_{n⩾m} \frac{|α_{n}|}{n − m + 1}

= C ∑_{n⩾m} \frac{|x_{n}^{∗}(x)|}{n − m + 1} ⩽ C∥x∥ ∑_{n⩾m} \frac{∥x_{n}^{∗}∥}{n − m + 1}.

It suffices to show that ∑_{n⩾m} \frac{∥x_{n}^{∗}∥}{n−m+1} < ∞. Note that

m(n − m + 1) = (m − 1)(n − m) + n ⩾ n

whenever n ⩾ m, so that

∑_{n=m}^{∞} \frac{∥x_{n}^{∗}∥}{n − m + 1} = m ∑_{n=m}^{∞} \frac{∥x_{n}^{∗}∥}{m(n − m + 1)} ⩽ m ∑_{n=m}^{∞} \frac{∥x_{n}^{∗}∥}{n} < ∞

by assumption. Hence, f_{m} is bounded, so that we can extend it to X. Observe that if |λ| < 1 and m ∈ ℕ then (1) yields that

f_{m}(h(λ, e)) = f_{m}(λ ∑_{n=0}^{∞} λ^{n}T^{n}e) = λ ∑_{n=0}^{∞} λ^{n} \frac{F_{m}^{(n)}(0)}{n!} = λF_{m}(λ),

Thus f_{m}(h(λ_{k}, e)) = λ_{k}F_{m}(λ_{k}) = 0 for all m, k ∈ ℕ, hence each f_{m} annihilates Y.

Finally, the set {f_{m}}_{m=1}^{∞} is linearly independent. Indeed if it was linearly dependent and a certain linear non-zero linear combination of them vanished, then by writing the Taylor expansion of each F_{m} on D we see that the same linear combination of F_{m}’s would vanish. This is a contradiction, since the F_{m}’s are linear independent. □

5. Invariant subspaces of operators with many almost invariant half-spaces

Let X be a Banach space and T: X → X be a bounded operator. It is well known that if every subspace of X is invariant under T then T must be a multiple of the identity. In this section we will obtain a result of the same spirit for almost invariant half-spaces.

ALMOST INVARIANT HALF-SPACES 11

Proposition 5.1. Let X be a Banach space and T ∈ L(X). Suppose that every halfspace of X is almost invariant under T. Then T has a non-trivial invariant subspace of finite codimension. Iterating, one can get a chain of such subspaces.

Proof. Let’s assume that T has no non-trivial invariant subspaces of finite codimension. We will now construct by an inductive procedure a half-space that is not almost invariant under T.

Put Y_{0} = X. Fix an arbitrary non-zero z_{1} ∈ X. Choose f_{1} ∈ X^{∗} such that f_{1}(z_{1}) ≠ 0 and put Y_{1} = ker f_{1}.

Since Y_{1} is not invariant under T, there exists z_{2} ∈ Y_{1} such that f_{1}(T z_{2}) ≠ 0. Define g_{2} ∈ Y_{1}^{∗} by g_{2}(y) = f_{1}(T y). Let P_{1} be a projection along span {z_{1}} onto Y_{1}. Define f_{2} = g_{2}∘P_{1} ∈ X^{∗}. Now put Y_{2} = ker g_{2} = ker f_{2}∩Y_{1}. Then we have Y_{1} = Y_{2}⊕span {z_{2}}. Since f_{1}(T y) = g_{2}(y) = 0 for all y ∈ Y_{2}, we have T Y_{2} ⊆ Y_{1}.

Continuing inductively with this procedure, we will build sequences (z_{n}) of vectors, (f_{n}) of functionals, and (Y_{n}) of subspaces such that

(i) z_{n+1} ∈ Y_{n},

(ii) Y_{n+1} = ker f_{n+1} ∩ Y_{n} = ⋂_{k=1}^{n+1} ker f_{k},

(iii) f_{n+1}(y) = f_{n}(T y) for all y ∈ Y_{n}, (iv) Y_{n} = Y_{n+1} ⊕ span {z_{n+1}},

(v) T Y_{n+1} ⊆ Y_{n}, and (vi) f_{n}(z_{i}) = 0 ⇔ i ≠ n,

for all n ∈ ℕ. Indeed, suppose we have defined Y_{i}, z_{i}, and f_{i}, 1 ⩽ i ⩽ n, satisfying (i)– (vi). Define g_{n+1} ∈ Y_{n}^{∗} by g_{n+1}(y) = f_{n}(T y) and put f_{n+1} = g_{n+1} ∘ P_{n} ∈ X^{∗} where P_{n} is a projection along [z_{k}]_{k=1}^{n} onto Y_{n} (take P_{n}(x) = x − ∑_{k=1}^{n} \frac{f_{k}(x)}{f_{k}(z_{k})}z_{k}). Again, there is z_{n+1} ∈ Y_{n} such that f_{n+1}(z_{n+1}) ≠ 0. Put Y_{n+1} = ker f_{n+1} ∩ Y_{n}. Evidently, (i)–(vi) are then satisfied.

It is easily seen that the sequence (z_{k}) is linearly independent. Put Z = [z_{2k}]_{k=1}^{∞}. Clearly dim Z = ∞. It is also easy to see that f_{2k−1}|_{Z} = 0 for all k ∈ ℕ. Thus, Z is actually a half-space.

By assumption of the theorem, there exists F with dim F = m < ∞ such that T Z ⊆ Z+F. For each k ∈ ℕ, pick u_{k} ∈ Z and v_{k} ∈ F such that T z_{2k} = u_{k}+v_{k}. By (iv), we have z_{2k} ∈ Y_{2k−1}. Applying (iii) with n = 2k − 1, we get f_{2k}(z_{2k}) = f_{2k−1}(T z_{2k}). Now (vi) yields f_{2k−1}(T z_{2k}) ≠ 0.

On the other hand, if 1 ⩽ i < k then z_{2k} ∈ Y_{2i−1}, so that analogously f_{2i}(z_{2k}) = f_{2i−1}(T z_{2k}). Therefore f_{2i−1}(T z_{2k}) = 0.

12 G. ANDROULAKIS, A.I. POPOV, A. TCACIUC, AND V.G. TROITSKY

Since f_{2j−1}|_{Z} = 0 for all j ∈ ℕ, we have f_{2j−1}(u_{k}) = 0 for all j and k. Therefore, for all k ∈ ℕ and 1 ⩽ i < k, we have f_{2k−1}(v_{k}) ≠ 0 and f_{2i−1}(v_{k}) = 0. This implies, however, that F is infinite dimensional. □

Using a similar technique, we obtain the following result.

Proposition 5.2. For all T ∈ L(X) and n ∈ ℕ there exists a subspace Y of X with codim Y = n and a vector e_{Y} ∈ X such that T Y ⊆ Y + span{e_{Y} }.

Proof. Proof is by induction on n. For n = 1, any hyperplane satisfies the conclusion of the statement. Suppose that the statement is valid for all k < n.

Suppose that X contains a subspace Y of codimension j ⩽ n that is invariant under T. If j = n then we are done. If j < n then by the induction assumption we can find Z ⊆ Y such that Z has codimension n − j in Y and T Z ⊆ Z + [y] for some y ∈ Y. Indeed, consider the restriction T′ of T to Y. Now we apply the induction assumption to T′ and to n − j and produce a subspace Z ⊆ Y invariant under T of codimension j. But then Z has codimension n in X and still T Z ⊆ Z + [y], so that Z satisfies the conclusion.

Therefore, we can assume that Z has no invariant subspaces of codimension k ⩽ n. Thus we can use the argument of Proposition 5.1 to show that there exist (finite) sequences of vectors (z_{k})_{k=1}^{n+1}, functionals (f_{k})_{k=1}^{n}, and subspaces (Y_{k})_{k=1}^{n} such that the conditions (i)–(vi) are satisfied. In particular, we get:

Y_{n} = ⋂_{k=1}^{n} ker f_{k},

and (f_{k}) are linearly independent, so that codim Y_{n} = n. Finally, by (vi) and (iv), we have T Y_{n} ⊆ Y_{n−1} = Y_{n} + [z_{n}]. □

Acknowledgments. We would like to thank Heydar Radjavi for helpful discussions and suggestions. A part of the work on this paper was done while the first and the fourth authors were attending SUMIRFAS in the summer of 2008; we would like to express our thanks to its organizers.

References

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[2] I. Chalendar, E. Fricain, and J.R. Partington, Overcompleteness of sequences of reproducing kernels in model spaces. Integral Equations Operator Theory, 56 (2006), no. 1, 45–56.

[3] P. Enflo, On the invariant subspace problem for Banach spaces, Acta Math., 158 (34) (1987) 213–313.

[4] J. Lindenstrauss, L. Tzafriri, Classical Banach spaces. I. Sequence spaces. Ergebnisse der Mathematik und ihrer Grenzgebiete, Vol. 92. Springer-Verlag, Berlin-New York, 1977.

[5] D.J. Newman, H.S. Shapiro, The Taylor coefficients of inner functions, Michigan J. Math. 9 (1962) 249–255.

[6] H. Radjavi, P. Rosenthal, Invariant subspaces, Second edition. Dover Publications, Inc., Mineola, NY, 2003.

[7] C.J. Read, A solution to the invariant subspace problem, Bull. London Math. Soc., 16 (4) (1984) 337–401.

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(G.Androulakis) Department of Mathematics, University of South Carolina, Columbia, SC 29208, USA

E-mail address: giorgis@math.sc.edu

(A. I. Popov, and V. G. Troitsky) Department of Mathematical and Statistical Sciences, University of Alberta, Edmonton, AB, T6G 2G1. Canada

E-mail address: apopov@math.ualberta.ca, vtroitsky@math.ualberta.ca

(A. Tcaciuc) Mathematics and Statistics Department, Grant MacEwan College, Edmonton, Alberta, Canada T5J P2P, Canada

E-mail address: tcaciuc@math.ualberta.ca