INVARIANT SUBSPACES OF POSITIVE STRICTLY SINGULAR OPERATORS ON BANACH LATTICES

JULIO FLORES, PEDRO TRADACETE, AND VLADIMIR G. TROITSKY

Abstract. It is shown that every positive strictly singular operator T on a Banach lattice satisfying certain conditions is AM-compact and has invariant subspaces. Moreover, every positive operator commuting with T has an invariant subspace. It is also proved that on such spaces the product of a disjointly strictly singular and a regular AM-compact operator is strictly singular. Finally, we prove that on these spaces the known invariant subspace results for compact-friendly operators can be extended to strictly singular-friendly operators.

introduction

Read [Read91] presented an example of a strictly singular operator with no (closed non-zero proper) invariant subspaces. It remains an open question whether every positive strictly singular operator on a Banach lattice has an invariant subspace. The present paper contains several results in this direction.

Recall that a bounded operator T on a Banach space X is said to be strictly singular if its restriction to any infinite-dimensional subspace is not an isomorphism. We say that T is ℓ_{2}-singular if the preceding definition is true for every subspace isomorphic to ℓ_{2}. Furthermore, if X is a Banach lattice, we say that T is disjointly strictly singular if the preceding is true for every subspace spanned by a pairwise disjoint infinite sequence. Disjointly strictly singular operators were introduced in [HR89] as a lattice version of strictly singular operators. Unlike strictly singular operators, they do not form an operator ideal. Clearly, every strictly singular operator is ℓ_{2}-singular and disjointly strictly singular. An operator between Banach spaces is called Dunford-Pettis if it takes weakly null sequences to norm null sequences. An operator from a Banach lattice to a Banach space is AM-compact if it takes order intervals into

J. Math. Anal. and Appl., 343 (2008), 743–751.

1 2 J. FLORES, P. TRADACETE, AND V. G. TROITSKY

2000 Mathematics Subject Classification. Primary: 47B65. Secondary: 47B07, 47A15.

Key words and phrases. Strictly singular operator, disjointly strictly singular operator, positive operator, AM-compact operator, Dunford-Pettis operator, Banach lattice, invariant subspace, invariant ideal.

The first author was partially supported by grants MTM2005-00082 and CS-PR27/05-14045. The second author was partially supported by grants PR27/05-14045, MTM2005-00082, and AP-2004-4841.

The third author was supported by NSERC.

relatively compact sets. If E is a Banach lattice and T ∈ L(E), we say that T is regular if it can be written as the difference of two positive operators. We write L(E) for the space of all (linear bounded) operators on E, and L_{r}(E) and L(E)_{+} for the collections of all regular or positive operators on E.

Throughout this paper, E will be a fixed order continuous Banach lattice with a weak order unit. We can assume by [LT79, Proposition 1.b.14] that there is a probability space (Ω, Σ, μ) such that E is an order ideal of L_{1}(μ) such that L_{∞}(μ) ⊆ E and ∥x∥_{1} ⩽ ∥x∥_{E} ⩽ 2∥x∥_{∞}. We fix (Ω, Σ, μ) throughout the paper. We will also make use of the following fact due to Weis [Weis82, Theorem 2.2]: every regular operator T: E → E can be extended to a bounded operator T̃: L_{1}(μ) → L_{1}(μ).

Consider a sequence (f_{n}) in E which is bounded in ∥·∥_{∞} and which, viewed as a sequence in L_{1}(μ), is equivalent to the unit vector basis of ℓ_{2}. We say that E satisfies the R-condition if every such sequence has a subsequence which remains equivalent to the unit vector basis of ℓ_{2} when viewed as a sequence in E. We show in Section 1 that every p-concave Banach lattice with 1 ⩽ p < ∞ satisfies the R-condition. In particular, if E contains L_{p}(μ) then E satisfies the R-condition.

In Section 2 we establish certain connections between various special classes of operators on Banach lattices. We show that a regular operator T on E is AM-compact iff its extension to L_{1}(μ) is a Dunford-Pettis operator. We prove that the product of a disjointly strictly singular and a regular AM-compact operators is strictly singular. We also show that if E satisfies the R-condition then every regular ℓ_{2}-singular operator is AM-compact. In Section 3 we use the results of Section 2 to show that certain strictly singular operators as well as many other operators related to them have invariant subspaces.

In [AAB94] the authors introduced the concept of a compact-friendly operator, and proved the existence of invariant subspaces for these operators under certain additional assumptions. In Section 4 we define strictly singular-friendly operators in a similar fashion, and show that under the same assumptions plus the R-condition strictly singularfriendly operators have invariant subspaces.

We will also use the following standard notations. We say that E contains a copy of ℓ_{1} if there is a subspace Z of E such that Z is isomorphic to ℓ_{1}. If, in addition, Z is a sublattice of E, we say that Z is a lattice copy of ℓ_{1} in E. It is well known that E contains a copy of ℓ_{1} if and only if it contains a lattice copy of ℓ_{1} if and only if E^{∗} is not order continuous [AB85, pp. 190,238]. We say that an operator T ∈ L(E) preserves a (lattice) copy of ℓ_{1} if E contains a (lattice) copy of ℓ_{1} such that the restriction of T to this subspace is an isomorphism. For more information on Banach lattices we refer the reader to [AA02, AB85, LT79, MN91].

1. R-condition

We will say that E satisfies the R-condition if every bounded sequence (f_{n}) in L_{∞}(μ) which is equivalent in L_{1}(μ) to the unit basis of ℓ_{2}, has a subsequence which is equivalent in E to the unit basis of ℓ_{2}. In this section we show that many Banach lattices enjoy the R-condition.

Lemma 1.1. Suppose that the inclusion i: L_{∞}(μ) ↪ E factorizes through L_{p}(ν) for some probability measure ν and 1 ⩽ p < ∞ with positive factors. Then E satisfies the R-condition.

Proof. Let (f_{n}) be a bounded sequence in L_{∞}(μ) which is equivalent in L_{1}(μ) to the unit vector basis of ℓ_{2}. By hypothesis, we have the following factorization

\xymatrix{L_{∞}(μ) \ar@{^{(}->}[rr]^{i} \ar[dr]_{T_{1}} &  & E \\  & L_{p}(ν) \ar[ur]_{T_{2}}}

Since (f_{n}) viewed as a sequence in L_{1}(μ) is equivalent to the unit vector basis of ℓ_{2}, it is weakly null in L_{1}(μ). Since (f_{n}) is order bounded in E, (f_{n}) is weakly null in E by Amemiya’s Theorem [MN91, 2.4.8].

The sequence (T_{1}f_{n}) has a subsequence which converges weakly to some g ∈ L_{p}(ν); therefore T_{2}g = 0. Consider the sequence y_{n} = T_{1}f_{n}−g; it has a weakly null subsequence. It cannot be null in L_{p}(ν) because (f_{n}) is not null in E.

Therefore, by passing to a subsequence, we may assume that (y_{n}) is weakly null, seminormalized, and (y_{n}) ⊂ [−y, y] for some y in L_{p}(ν). Since L_{p}(ν) has an unconditional basis, we can extract a subsequence (y_{n_{k}}) which is unconditional with constant K. For every n ∈ ℕ, let r_{n}: [0, 1] → [−1, 1] be the n-th Rademacher function r_{n}(t) = sign sin 2^{n}πt. By [LT79, Theorem 1.d.6], there exists C > 0 such that

∥∑_{k=1}^{m} a_{k}y_{n_{k}} ∥_{p} ⩽ K ∫_{0}^{1}∥∑_{k=1}^{m} r_{k}(s)a_{k}y_{n_{k}} ∥_{p}ds ⩽ KC∥(∑_{k=1}^{m}|a_{k}y_{n_{k}}|^{2})^{\frac{1}{2}} ∥_{p} ⩽ KC∥y∥_{p}∥(a_{i})∥_{ℓ_{2}^{m}}

for all (a_{k})_{k=1}^{m}. Then, for some c > 0 we have

c∥(a_{i})∥_{ℓ_{2}^{m}} ⩽ ∥∑_{k=1}^{m} a_{k}f_{n_{k}} ∥_{1} ⩽ ∥∑_{k=1}^{m} a_{k}f_{n_{k}} ∥_{E}

= ∥∑_{k=1}^{m} a_{k}T_{2}y_{n_{k}} ∥_{E} ⩽ ∥T_{2}∥∥∑_{k=1}^{m} a_{k}y_{n_{k}} ∥_{p} ⩽ KC∥T_{2}∥∥(a_{i})∥_{ℓ_{2}^{m}}

Therefore (f_{n_{k}}) is equivalent in E to the unit vector basis of ℓ_{2}. □

We claim that if L_{p}(μ) ⊆ E for some 1 ⩽ p < ∞ then E satisfies R-condition. This is a special case of a more general fact: if E is p-concave for some 1 ⩽ p < ∞ then E has R-condition. This follows from Lemma 1.1 together with Krivine’s factorization theorem [LT79, Theorem 1.d.11] since the inclusion map i: L_{∞}(μ) → E is p-convex (an inspection of the proof of [LT79, Theorem 1.d.11] reveals that in our setting the factors are positive).

The last statement can be extended as follows: we will show that if E has property (U_{2}) then it satisfies R-condition. Property (U_{2}) was introduced by Räbiger in [Rab91]: a Banach lattice F has property (U_{2}) if for every seminormalized weakly null order bounded sequence (x_{n}) in F there is a subsequence (x_{n_{i}}) and a constant C > 0 such that

∥∑_{i=1}^{m} a_{i}x_{n_{i}} ∥ ⩽ C∥(a_{i})∥_{ℓ_{2}^{m}}

for any coefficients a_{1}, …, a_{m}. It was proved in [Rab91] that every Banach lattice which is p-concave for some 1 ⩽ p < ∞ has property (U_{2}). However, the following example shows that the converse is false.

Example. The Banach lattice ℓ_{2}(ℓ_{∞}^{2^{n}}) has property (U_{2}) but it is not p-concave for any 1 ⩽ p < ∞.

Proof. Let us write every element x ∈ ℓ_{2}(ℓ_{∞}^{2^{n}}) as a sequence (x_{i})_{i=1}^{∞}, with x_{1} ∈ ℓ_{∞}^{1}, (x_{2}, x_{3}) ∈ ℓ_{∞}^{2}, (x_{4}, x_{5}, x_{6}, x_{7}) ∈ ℓ_{∞}^{4}, and so on. Therefore,

∥x∥ = ( ∑_{n=0}^{∞} max_{2^{n}⩽i<2^{n+1}}|x_{i}|^{2})^{\frac{1}{2}} .

For x ∈ ℓ_{2}(ℓ_{∞}^{2^{n}}) we consider \overline{x} = (\overline{x}_{i})_{i=1}^{∞}, defined by \overline{x}_{i} = max\{|x_{j}| : 2^{n} ≤ j < 2^{n+1}\} whenever 2^{n} ⩽ i < 2^{n+1}. Clearly, ∥\overline{x}∥ = ∥x∥, and \overline{x} belongs to the closed linear span of (e^{n}), where

e_{k}^{n} = \begin{cases}1 & if 2^{n} ⩽ k < 2^{n+1}, \\ 0 & otherwise. \end{cases}

Note that (e^{n}) is equivalent to the unit vector basis of ℓ_{2}.

POSITIVE STRICTLY SINGULAR OPERATORS 5

Let (x^{(n)}) be a seminormalized order bounded weakly null sequence in ℓ_{2}(ℓ_{∞}^{2^{n}}). For each n, put y^{(n)} = \overline{x^{(n)}}. We can write y^{(n)} = ∑_{i=1}^{∞} α_{n,i}e_{i}. Since (x^{(n)}) is weakly null, both (x^{(n)}) and (y^{(n)}) converge to zero coordinate-wise, hence lim_{n} α_{n,i} = 0 for every i. It follows from [LT77, Proposition 1.a.12] that there is a subsequence (y^{(n_{k})}) equivalent to the unit vector basis of ℓ_{2}. Therefore, we have:

∥∑_{k=1}^{∞} a_{k}x^{n_{k}} ∥ ⩽ ∥∑_{k=1}^{∞}|a_{k}|y^{(n_{k})}∥ ⩽ C(∑_{k=1}^{∞}|a_{k}|^{2})^{\frac{1}{2}}

for some constant C > 0 independent of the sequence (a_{k})_{k=1}^{∞}. Therefore, ℓ_{2}(ℓ_{∞}^{2^{n}}) has property (U_{2}). On the other hand, this space contains ℓ_{∞}^{n}’s uniformly, so that it is not p-concave for any 1 ⩽ p < ∞. □

Proposition 1.2. If E has property (U_{2}) then it satisfies the R-condition.

Proof. Let (x_{n}) be a bounded sequence in L_{∞}(μ) equivalent in L_{1}(μ) to the unit vector basis of ℓ_{2}. Again, (x_{n}) is weakly null in E. By property (U_{2}), there exists a constant C > 0 such that, after passing to a subsequence, we have

∥∑_{i=1}^{m} a_{i}x_{n_{i}} ∥_{E} ⩽ C∥(a_{i})∥_{ℓ_{2}^{m}}

for any (a_{i})_{i=1}^{m}. On the other hand, by our choice of (x_{n}) there exists another constant c such that

c∥(a_{i})∥_{ℓ_{2}^{m}} ⩽ ∥∑_{i=1}^{m} a_{i}x_{n_{i}} ∥_{1} ⩽ ∥∑_{i=1}^{m} a_{i}x_{n_{i}} ∥_{E}

for any coefficients a_{1}, …, a_{m}. Hence (x_{n_{i}}) is also equivalent to the unit vector basis of ℓ_{2} in E. □

2. Strictly singular and AM-compact operators

We start by showing that a regular operator T on E is AM-compact iff its extension T̃ to L_{1}(μ) is Dunford-Pettis. This result is related to Theorem 19.18 of [AB85] which asserts that a regular operator from E to L_{1}(μ) is Dunford-Pettis iff it maps order intervals onto norm compact sets. We will use the following observation of Uhl ([Ros76], [Bour80]).

Theorem 2.1 (Uhl). An operator T: L_{1}(μ) → L_{1}(μ) is Dunford-Pettis iff its restriction to L_{∞}(μ) is compact as an operator from L_{∞}(μ) to L_{1}(μ).

Theorem 2.2. Let T be a regular operator on E. Then T is AM-compact iff T̃: L_{1}(μ) → L_{1}(μ) is Dunford-Pettis.

6 J. FLORES, P. TRADACETE, AND V. G. TROITSKY

Proof. Suppose that T̃ is Dunford-Pettis. It suffices to show that if (x_{n}) is a sequence in [0, x] for some x ∈ E_{+} then (T x_{n}) has a convergent subsequence. Without loss of generality we can take (x_{n}) normalized. Since order intervals in L_{1}(μ) are weakly compact, there exists a subsequence (x_{n_{k}}) which converges weakly to some g in L_{1}(μ). We then have T x_{n_{k}} = T̃x_{n_{k}} ⟶^{∥·∥_{1}} T̃g because T̃ is Dunford-Pettis. Since T is regular then (T x_{n}) is contained in [−|T |x, |T |x]. Also, T̃g ∈ [−|T |x, |T |x] since order intervals are closed. By Amemiya’s Theorem [MN91, 2.4.8] it follows that T̃g ∈ E and T x_{n_{k}} → T̃g in E.

Conversely, suppose that T is AM-compact. Then T is AM-compact as an operator from L_{∞}(μ) to L_{1}(μ), because L_{∞}(μ) is an ideal in E and the inclusion E ↪ L_{1}(μ) is continuous. Since the unit ball in L_{∞}(μ) is an order interval, it follows that this operator is in fact compact. Now Theorem 2.1 implies that T̃ is Dunford-Pettis. □

Remark 2.3. It was shown in [Ros76] that an operator S: L_{1}(μ) → L_{1}(μ) is Dunford-Pettis iff it is ℓ_{2}-strictly singular. Moreover, if S is not Dunford-Pettis then one can find a sequence (f_{n}) bounded in L_{∞}(μ) such that (f_{n}) viewed as a sequence in L_{1}(μ) is weakly null and equivalent to the unit vector basis of ℓ_{2}, and the restriction of S to the span of (f_{n}) in L_{1}(μ) is an isomorphism.

Proposition 2.4. Suppose that E satisfies the R-condition and T ∈ L_{r}(E) is ℓ_{2}-strictly singular. Then T is AM-compact.

Proof. In view of Theorem 2.2 it suffices to show that T̃: L_{1}(μ) → L_{1}(μ) is Dunford-Pettis. Suppose it is not. Let (f_{n}) be as in Remark 2.3 for S = T̃. Since E satisfies the R-condition, after passing to a subsequence we have the following chain of inequalities with appropriate constants:

∥T (∑_{n=1}^{∞} α_{n}f_{n})∥_{E} ⩾ ∥T (∑_{n=1}^{∞} α_{n}f_{n})∥_{1} ⩾ C_{1}∥∑_{n=1}^{∞} α_{n}f_{n}∥_{1} ⩾ C_{2}∥(α_{n})∥_{2} ⩾ C_{3}∥∑_{n=1}^{∞} α_{n}f_{n}∥_{E}

for any ∑_{n=1}^{∞} α_{n}f_{n} in E. This contradicts T being ℓ_{2}-singular. □

Corollary 2.5. Suppose that X is an arbitrary Banach lattice, T: X → X is strictly singular and factors with positive factors through E, and E satisfies the R-condition. Then T^{3} is AM-compact.

Proof. Suppose that we can factor T = RS where X ⟶^{S} E ⟶^{R} X and S, R ⩾ 0. Then ST R: E → E is positive and strictly singular, hence AM-compact by Proposition 2.4. Since AM-compact operators form an algebraic ideal among regular operators, it follows from T^{3} = R(ST R)S that T^{3} is AM-compact □ Theorem 2.6. Suppose that S, T ∈ L(E) such that S is disjointly strictly singular and T is regular and AM-compact. Then ST is strictly singular.

The proof of this theorem will be based on the following two facts. The first one is an observation which follows easily from the results of [Chen99].

Theorem 2.7. Let T be an operator on an order continuous Banach lattice. If T preserves a copy of ℓ_{1}, then T preserves a lattice copy of ℓ_{1}.

Proof. If T preserves a copy of ℓ_{1} then there is a normalized sequence (x_{n}) such that (x_{n}) and (T x_{n}) are both equivalent to the unit vector basis of ℓ_{1}. Therefore, (T x_{n}) has no weakly Cauchy subsequence. Thus, T is not weakly sequentially precompact (see [Chen99, Section 1]) and the required conclusion follows from [Chen99, Theorem 1.1]. □

The second fact needed for the proof of Theorem 2.6 is related to the well-known Kadec-Pełczyński sets. Recall that, given ε > 0, the Kadec-Pełczyński set M(ε) is defined as follows:

M(ε) = {x ∈ E : μ(σ(x, ε)) ⩾ ε}

where σ(x, ε) = {t ∈ Ω : |x(t)| ⩾ ε∥x∥_{E}}. It is known ([LT79, Proposition 1.c.8]) that ∥x∥_{1} ⩾ ε^{2}∥x∥_{E} for all x ∈ M(ε); hence the norms ∥·∥_{E} and ∥·∥_{1} are equivalent on every subspace of E contained in M(ε) for some ε > 0. On the other hand, if a normalized sequence (x_{n}) in E is not contained in any M(ε), then there is a subsequence (x_{n_{k}}) and a normalized disjoint (unconditional basic) sequence (y_{k}) in E equivalent to (x_{n_{k}}) with ∥x_{n_{k}} − y_{k}∥_{E} → 0.

Lemma 2.8. Suppose that T ∈ L_{r}(E) is AM-compact and (x_{n}) is a normalized weakly null sequence in E such that (T x_{n}) is not null. Then there is a subsequence (x_{n_{k}}) and a disjoint seminormalized basic sequence (y_{k}) in E equivalent to (T x_{n_{k}}) and ∥T x_{n_{k}} − y_{k}∥_{E} → 0.

Proof. By Theorem 2.2, T̃ is Dunford-Pettis. Clearly, (x_{n}) is still weakly null viewed as a sequence in L_{1}(μ), so that ∥T x_{n}∥_{1} → 0. It follows that (T x_{n}) can not be entirely contained in some M(ε) as this would imply ∥T x_{n}∥_{E} → 0. The conclusion follows from the preceding remark. □

Proof of Theorem 2.6. Let Y be an infinite-dimensional subspace of E such that ST is an isomorphism on Y. Suppose first that Y contains no isomorphic copy of ℓ_{1}. Applying Rosenthal’s theorem [Ros74] to any bounded sequence in Y with no convergent subsequences, we conclude that Y contains a normalized weakly null sequence (x_{n}). Since

8 J. FLORES, P. TRADACETE, AND V. G. TROITSKY

T is an isomorphism on Y, we may assume by Lemma 2.8 that (T x_{n}) is equivalent to a disjoint seminormalized basic sequence (y_{n}) and ∥T x_{n} − y_{n}∥_{E} → 0. Since S is disjointly strictly singular, we can choose a normalized block sequence (w_{n}) of (y_{n}) such that ∥Sw_{n}∥_{E} → 0. If (v_{n}) is the corresponding block sequence of (T x_{n}) with the same coefficients, then (v_{n}) is seminormalized and ∥Sv_{n}∥_{E} → 0. This contradicts ST being an isomorphism on Y.

Now suppose that Y contains a copy of ℓ_{1}. It follows that S preserves a copy of ℓ_{1}. Theorem 2.7 yields that S preserves a lattice copy of ℓ_{1}. This contradicts S being disjointly strictly singular. □

Corollary 2.9. If T ∈ L_{r}(E) is disjointly strictly singular and AM-compact, then T^{2} is strictly singular.

3. Invariant subspaces of positive strictly singular operators

In this section we apply the results of the preceding sections to the Invariant Subspace Problem. Invariant subspaces are always assumed to be non-zero and proper. A subspace is said to be hyperinvariant under an operator T if it is invariant under every operator commuting with T. Recall that T is said to be quasinilpotent if its spectrum is {0} or, equivalently, if lim_{n} \sqrt[n]{∥T^{n}∥} = 0. We use the following standard lemma.

Lemma 3.1. Suppose that T is an operator on a Banach space. If T is not quasinilpotent and some power of T is strictly singular then T has a finite-dimensional hyperinvariant subspace.

Proof. Suppose that T is an operator on a Banach space X such that T is not quasinilpotent and T^{n} is strictly singular for some n. Clearly, T^{n} is not quasinilpotent.

Suppose first that X is a Banach space over ℂ. Then T^{n} has non-trivial eigenspaces by [AA02, Theorem 7.11]. Let Z be a non-trivial eigenspace of T^{n}. Since T^{n} is strictly singular, we have dim Z < ∞. It is easy to see that Z is hyperinvariant under T.

Now suppose that X is a Banach space over ℝ. The complexification T_{ℂ}^{n} of T^{n} is still strictly singular by [AA02, p. 177]. Again, let Z be a non-trivial eigenspace of T_{ℂ}^{n} in X_{ℂ}. Then dim Z < ∞ and Z is hyperinvariant under T_{ℂ}. Let (x_{1} + iy_{1}), …, (x_{m} + iy_{m}) be a basis of Z, put M = span{x_{1}, …, x_{m}, y_{1}, …, y_{m}} in X. Clearly, 0 < dim M ⩽ 2m. We claim that M is hyperinvariant under T. Indeed, suppose that S is an operator such that ST = T S. Then S_{ℂ}T_{ℂ} = T_{ℂ}S_{ℂ}. Since Z in hyperinvariant under T_{ℂ}, for every k ⩽ m we have S_{ℂ}(x_{k} + iy_{k}) ∈ Z, so that Sx_{k} and Sy_{k} are both in M. Hence, S(M) ⊆ M. □

POSITIVE STRICTLY SINGULAR OPERATORS 9

We make use of the following statement, which is a special case of Theorem 10.26 of [AA02].

Theorem 3.2 ([AA02]). Every quasinilpotent AM-compact positive operator on a Banach lattice has an invariant subspace.

Combining Theorem 3.2 with Lemma 3.1 we obtain the following result.

Proposition 3.3. Suppose that T is a positive AM-compact operator on a Banach lattice. If T^{n} is strictly singular for some n then T has an invariant subspace.

Together with Theorem 2.4, this yields the following.

Corollary 3.4. If E satisfies the R-condition and T ∈ L(E)_{+} is strictly singular then T is AM-compact and has an invariant subspace.

Corollary 3.5. Suppose that E satisfies the R-condition and S, T ∈ L(E) are such that 0 ⩽ S ⩽ T. If T is strictly singular then S has an invariant subspace.

Proof. It was shown in [FHT] that every operator on a Banach lattice with order continuous norm which is dominated by a strictly singular operator has strictly singular square; hence S^{2} is strictly singular. Theorem 2.4 yields that T is AM-compact, hence 0 ⩽ S ⩽ T implies that S is AM-compact. The conclusion now follows from Proposition 3.3. □

Proposition 3.6. Every positive disjointly strictly singular AM-compact operator on E has an invariant subspace.

Proof. Follows from Corollary 2.9 and Proposition 3.3. □

Propositions 2.4 and 3.6 immediately yield the following result.

Theorem 3.7. If E satisfies the R-condition and T ∈ L(E)_{+} is ℓ_{2}-singular and disjointly strictly singular then T is AM-compact and has an invariant subspace.

Recall that if T is a positive operator on a Banach lattice then its left and right semi-commutants are defined as follows:

⟨T] = {S ⩾ 0 : ST ⩽ T S} and [T ⟩ = {S ⩾ 0 : ST ⩾ T S}.

Theorem 3.8 ([AT05, GT]). Suppose that T is a positive quasinilpotent operator on a Banach lattice X. Suppose that there exists a closed ball B(x_{0}, r) in X centered at some x_{0} ⩾ 0, of positive radius r < ∥x_{0}∥ such that for every sequence (x_{n}) in B(x_{0}, r) ∩ [0, x_{0}] there exists a subsequence (x_{n_{i}}) and a sequence of operators (K_{i}) such that 0 ⩽ K_{i} ⩽ T for each i and (K_{i}x_{n_{i}}) converges to a non-zero vector. Then ⟨T] has a (common) invariant closed order ideal. In particular, T has an invariant subspace.

Theorem 3.9. Suppose that E satisfies R-condition and T is a positive quasinilpotent strictly singular operator on E. Then ⟨T] has an invariant closed ideal.

Proof. Choose x_{0} ∈ X_{+} and r > 0 so that B(x_{0}, r) ∩ ker T = ∅. Suppose that (x_{n}) is a sequence in B(x_{0}, r) ∩ [0, x_{0}]. We claim that there is a subsequence (x_{n_{i}}) such that (T x_{n_{i}}) converges in norm to a non-zero vector; then the result will follow from Theorem 3.8. We may assume without loss of generality that (x_{n}) has no norm convergent subsequences. Since order intervals are weakly compact in E, we may assume by passing to a subsequence that (x_{n}) converges weakly to some x. Since B(x_{0}, r) is weakly closed and B(x_{0}, r)∩ker T = ∅, we have x ≠ 0 and T x ≠ 0. Notice that ∥T x_{n}−T x∥_{1} → 0 since the extension T̃: L_{1}(μ) → L_{1}(μ) is Dunford-Pettis by Theorem 2.2 and Corollary 3.4. Again, since the sequence (T x_{n}) is order bounded (T is positive) we obtain ∥T x_{n} − T x∥_{E} → 0 by Amemiya’s theorem. □

To produce a similar result for [T ⟩ we use the following version of a theorem of Drnovs̆ek [AA02, Theorem 10.50]. We start by recalling a few definitions. Let X be a Banach lattice. An operator T on X is locally quasinilpotent at a point x if lim_{n} \sqrt[n]{∥T^{n}x∥} = 0. A point w in X is called quasi-interior if E_{w} is norm dense in X, where

E_{w} = {x ∈ X : |x| ⩽ λ|w| for some λ > 0},

the principal ideal generated by w. Suppose that S and T are two operators on X, we say that T dominates S if |Sx| ⩽ T |x| for every x ∈ X. Please see Section 10.4 of [AA02] for the other terminology used in the proof.

Theorem 3.10. Suppose that T is a positive operator on a Banach lattice X with a quasi-interior point w such that

(i) T is locally quasinilpotent at some x_{0} > 0, and (ii) there is S ∈ [T ⟩ such that S dominates a non-zero AM-compact operator K.

Then [T ⟩ has an invariant closed ideal.

Proof. Since the null ideal N_{T} of T is [T ⟩-invariant, we may assume that N_{T} = {0}. Let z ∈ X such that Kz ≠ 0. We may assume that |Kz| ⩽ w as otherwise we can replace w with w ∨ |Kz|. By Lemma 4.16(1) of [AA02] there exists an operator M dominated by the identity operator such that MKz > 0. Put K_{1} = MK. It follows from N_{T} = 0 that

POSITIVE STRICTLY SINGULAR OPERATORS 11

T K_{1}z ≠ 0, hence T K_{1} ≠ 0. It is easy to see that T K_{1} is AM-compact and is dominated by T S.

Let J be the algebraic ideal in [T ⟩ generated by T S, i.e., J = {AT SB : A, B ∈ [T ⟩}. It can be verified directly that J is finitely quasinilpotent at x_{0}. Since T S ∈ J and T S dominates a non-zero AM-compact operator, J has an invariant closed ideal by Theorem 10.44 of [AA02]. Now Theorem 10.49 of [AA02] yields that [T ⟩ has an invariant closed ideal. □

Corollary 3.11. Suppose that E has R-condition and T is a positive quasinilpotent strictly singular operator on E. Then [T ⟩ has an invariant closed ideal.

Proof. By Corollay 3.4, T is AM-compact. Now apply Theorem 3.10. □

Corollary 3.12. Suppose that E has R-condition and T is a positive strictly singular operator on E. Then every positive operator commuting with T has an invariant subspace.

Proof. If T is quasinilpotent then the conclusion follows from Theorem 3.9 or Corollary 3.11. If T is not quasinilpotent, then the result follows from Lemma 3.1. □

Proposition 3.13. Suppose that X is an arbitrary Banach lattice, T: X → X is strictly singular and factors with positive factors through E, and E satisfies the R-condition. If T is locally quasinilpotent at a positive vector then [T ⟩ has an invariant closed ideal.

Proof. Corollary 2.5 yields that T^{3} is AM-compact. The result now follows from Theorem 3.10. □

4. Invariant subspaces of SS-friendly operators

It is well known that compact operators enjoy good properties concerning the Invariant Subspace Problem. The compactness properties were relaxed in [AAB94], where the authors introduced the class of compact-friendly operators and showed that these operators also have invariant subspaces. We present here the analogous concept for strictly singular operators.

We call an operator B ∈ L(E)_{+} strictly singular-friendly if there is a positive operator that commutes with B and dominates a non-zero operator which is dominated by a strictly singular positive operator. In particular, every operator dominating a positive strictly singular operator is strictly singular-friendly.

12 J. FLORES, P. TRADACETE, AND V. G. TROITSKY

Theorem 4.1. Suppose that E satisfies the R-condition. If B ∈ L(E)_{+} is a non-zero strictly singular-friendly operator which is locally quasinilpotent at some x_{0} > 0, then B has a non-trivial closed invariant ideal. Moreover, if (T_{n}) is a sequence in [B⟩, then there exist a non-trivial closed ideal that is invariant under B and each T_{n}.

Proof. Our argument is similar to the one in [AAB94]. We can suppose without loss of generality that ∥B∥ < 1. Pick small enough scalars α_{n} > 0 such that the positive operator T = ∑_{n=1}^{∞} α_{n}T_{n} exists and ∥B + T ∥ < 1. It is clear that T ∈ [B⟩, and (B + T)^{n} ∈ [B⟩ for every n, so that the positive operator A = ∑_{n=0}^{∞}(B + T)^{n} also belongs to [B⟩.

For each x > 0, let J_{x} be the principal ideal generated by Ax, that is,

J_{x} = {y ∈ E : |y| ⩽ λAx for some λ > 0}.

Since x ⩽ Ax, we have that x ∈ J_{x} so this is a non-zero ideal.

Note that J_{x} is (B + T)-invariant. Indeed, if y ∈ J_{x}, then |y| ⩽ λAx for some λ > 0 so

|(B + T)y| ⩽ (B + T)|y| ⩽ λ(B + T) ∑_{n=0}^{∞}(B + T)^{n}x = λ ∑_{n=1}^{∞}(B + T)^{n}x ⩽ λAx.

Clearly J_{x} is also invariant under B and T, since 0 ⩽ B, T ⩽ B + T, so it is also T_{n}-invariant for each n.

Therefore, for our purposes, it suffices to prove that there exists a positive x ∈ E such that the ideal J_{x} is not norm dense in E. Suppose the contrary, that is, Ax is a quasi-interior point in E for each x > 0. By assumption, there exist operators R, S, and C in L(E) such that R and S are positive, S is strictly singular, C ≠ 0, RB = BR, and C is dominated by both R and S.

Since C ≠ 0, there exists some x_{1} > 0 such that Cx_{1} ≠ 0. Then A|Cx_{1}| is a quasiinterior point satisfying A|Cx_{1}| ⩾ |Cx_{1}|. By [AA02, Lemma 4.16]) there exists an operator M_{1} ∈ L(E) dominated by the identity operator such that x_{2} = M_{1}Cx_{1} > 0. Let U_{1} = M_{1}C. Note that U_{1} is dominated by S and by R.

Now we have \overline{J_{x_{2}}} = E. Therefore, since C ≠ 0, there exists 0 < y < Ax_{2} such that Cy ≠ 0. Because A|Cy| is a quasi-interior point and |Cy| ⩽ A|Cy|, then, as before, there exists M_{2} ∈ L(E) dominated by the identity operator such that x_{3} = M_{2}Cy > 0. Since |y| ⩽ Ax_{2} and Ax_{2} is a quasi-interior point, it follows that there is an operator M ∈ L(E) dominated by the identity such that MAx_{2} = y. So x_{3} = M_{2}Cy = M_{2}CMAx_{2}. And the operator U_{2} = M_{2}CMA is dominated by SA and by RA.

Consider the operator U_{2}U_{1}. From U_{2}U_{1}x_{1} = x_{3} > 0, we see that U_{2}U_{1} is a non-zero operator. Since both U_{1} and U_{2} are dominated by strictly singular positive operators, U_{2}U_{1} is strictly singular by [FHT]. Moreover

|U_{2}U_{1}x| = |M_{2}CMAM_{1}Cx| ⩽ RAR|x|

for each x ∈ E.

Let V = RAR. Since A and R belong to [B⟩ then V also belongs to [B⟩. Observe that V dominates U_{2}U_{1} which is strictly singular, therefore AM-compact. The result now follows from Theorem 3.10. □

References

[AA02] Abramovich, Y.A. and Aliprantis, C.D. An invitation to operator theory, Graduate Studies in Mathematics, vol. 50, American Mathematical Society, Providence, RI, 2002.

[AAB94] Abramovich, Y.A., Aliprantis, C.D. and Burkinshaw, O., Invariant subspaces for positive operators. J. Funct. Anal., 124, 95–111 (1994).

[AB85] Aliprantis, C.D. and Burkinshaw, O. Positive operators, Academic Press Inc., Orlando, Fla., 1985.

[AT05] Anisca, R. and Troitsky, V.G., Minimal vectors of positive operators, Indiana Univ. Math. J., 54 (2005), no. 3, 861–872.

[Bour80] Bourgain, J., Dunford-Pettis operators on L^{1} and the Radon-Nikodym property, Israel J. Math., 37 (1980), 34–47.

[Chen99] Chen, Z., Weak sequential precompactness in Banach lattices, Chinese Ann. Math. Ser. A 20, (1999), no. 5, 567–574; translation in Chinese J. Contemp. Math. 20, (1999), no. 4, 477–486 (2000).

[FHT] Flores, J., Hernández, F.L. and Tradacete, P., Powers of operators dominated by strictly singular operators, to appear in Q. J. Math.

[GT] Gessesse, H. and Troitsky, V.G., Invariant subspaces of positive quasinilpotent operators on Banach spaces with a closed generating cone, to appear in Positivity.

[HR89] Hernández, F.L. and Rodríguez-Salinas, B., On l^{p}-complemented copies in Orlicz spaces. II. Israel J. Math., 68 (1989), no. 1, 27–55.

[LT77] Lindenstrauss, J. and Tzafriri, L. Classical Banach spaces. I, Springer-Verlag, Berlin, 1977, Sequence spaces, Ergebnisse der Mathematik und ihrer Grenzgebiete, Vol. 92.

[LT79] , Classical Banach spaces. II, Springer-Verlag, Berlin, 1979, Function spaces.

[Lom73] Lomonosov, V.I., Invariant subspaces of the family of operators that commute with a completely continuous operator, Funkcional. Anal. i Priložen., 7 (1973), no. 3, 55–56.

[MN91] Meyer-Nieberg, P., Banach lattices, Springer-Verlag, Berlin, 1991.

[Rab91] Räbiger, F., Lower and upper 2-estimates for order bounded sequences and Dunford-Pettis operators between certain class of Banach lattices, Functional analysis (Austin, TX, 1987/1989), 159–170, Lecture Notes in Math., 1470, Springer, Berlin, 1991.

[Read91] Read, C.J., Strictly singular operators and the invariant subspace problem, Studia Math., 132 No. 3, (1991), 203–226.

[Ros74] Rosenthal, H.P., A characterization of Banach spaces containing l^{1}, Proc. Nat. Acad. Sci. U.S.A., 71 (1974), 2411–2413.

[Ros76] Rosenthal, H.P., Convolution by a biased coin, The Altgeld Book, University of Illinois, Functional Analysis Seminar (1975/76).

[Weis82] Weis, L., Integral operators and changes of density, Indiana Univ. Math. J., 31 (1982), no. 1, 83–96.

Department of Applied Mathematics, Escet, Universidad Rey Juan Carlos, 28933, Móstoles, Madrid, Spain.

E-mail address: julio.flores@urjc.es

Department of Mathematical Analysis, Universidad Complutense de Madrid, 28040, Madrid, Spain.

E-mail address: tradacete@mat.ucm.es

Department of Mathematical and Statistical Sciences, University of Alberta, Edmonton, AB, T6G 2G1, Canada.

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