Domination by positive disjointly strictly singular operators

PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 129, Number 7, Pages 1979–1986 S 0002-9939(00)05948-7 Article electronically published on November 22, 2000 DOMINATION BY POSITIVE DISJOINTLY STRICTLY SINGULAR OPERATORS JULIO FLORES AND FRANCISCO L. HERNÁNDEZ (Communicated by Joseph A. Ball) Abstract. We prove that each positive operator from a Banach lattice E to a Banach lattice F with a disjointly strictly singular majorant is itself disjointly strictly singular provided the norm on F is order continuous. We prove as well that if S : E → E is dominated by a disjointly strictly singular operator, then S 2 is disjointly strictly singular. 1. Introduction The classical problem of domination for positive compact operators on Banach lattices was solved by Dodds and Fremlin ([5]) for a pair of positive operators 0 ≤ S ≤ T defined on a Banach lattice E with order continuous dual norm and taking values in a Banach lattice F with order continuous norm: we can guarantee that S is compact if T is so. A full answer to this problem was given by Aliprantis and Burkinshaw in [2], namely if E = F and either the norm on E or the norm on E ′ is order continuous, then the compactness of T is inherited by the operator S 2 . They also show that for an arbitrary Banach lattice, T compact always implies S 3 compact. More recently Wickstead has given in [14] not only sufficient but necessary conditions for the problem of domination for positive compact operators to have a solution. The problem of domination for weakly compact operators was first considered by Abramovich in [1], giving a positive solution for a Banach lattice E and a KBspace F . Later on, a general result was obtained by Wickstead in [13] where it was shown that the problem has a positive answer if and only if either the norm on E ′ or F is order continuous. Again Aliprantis and Burkinshaw settled the question by considering the case E = F and showing that T weakly compact implies S 2 weakly compact ([3]). The aim of this paper is to study the problem of domination for positive disjointly strictly singular operators. We recall that an operator T between a Banach lattice E and a Banach space Y is said to be disjointly strictly singular (DSS) if there is no disjoint sequence of non-null vectors (xn )n in E such that the restriction of T to the subspace [xn ] spanned by the vectors (xn )n is an isomorphism. DSS operators were introduced by Rodrı́guez-Salinas and the second author in [9]. This class of operators, a generalization of the class of strictly singular (or Kato) operators, is a Received by the editors October 14, 1999. 2000 Mathematics Subject Classification. Primary 47B65, 46B42. This work was partially supported by DGES PB97-0240. c 2000 American Mathematical Society 1979 License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use 1980 JULIO FLORES AND FRANCISCO L. HERNÁNDEZ useful tool to compare the lattice structure of Banach function lattices ([7]). Recall that an operator T between two Banach spaces X and Y is said to be strictly singular if the restriction of T to any infinite dimensional closed subspace is not an isomorphism. Every strictly singular operator is DSS but the converse is not true (f.i. take the natural inclusion i : Lp [0, 1] → Lq [0, 1] with p > q ≥ 1). However if E is a Banach lattice with a Schauder basis of mutually disjoint vectors or a C(K)-space, then every DSS operator from E to Y is strictly singular. The set of all DSS operators between E and Y is a vector space which is stable under the composition by the left but not by the right ([8, Prop. 1]). The main results of the paper are presented now. Theorem 1.1. Let E and F be Banach lattices and 0 ≤ S ≤ T : E → F two positive operators. If the norm on F is order continuous and T is DSS, then S is also DSS. Moreover, if F is σ-Dedekind complete and every positive operator from E to F dominated by a DSS operator is DSS, then either the norm on F or the norm on E ′ is order continuous. In the case E = F we obtain the following: Theorem 1.2. Let 0 ≤ S ≤ T be two operators on a Banach lattice E. If T is DSS, then S 2 is DSS. For any unexplained terms from Banach lattices and regular operators theory we refer to [4], [12] or [15]. 2. Proofs Let us start by recalling a couple of well-known facts. Lemma 2.1. Let E be an L-space. Then every weakly-null sequence of positive vectors is convergent to zero. Lemma 2.2. Let (Ω, Σ, µ) be a finite measure space and (fn )n be a weakly convergent sequence in L1 (µ). If (fn )n converges to zero in µ-measure, then (fn )n converges to zero in norm. Proof. We can assume w.l.o.g. that µ(Ω) = 1. The sequence (fn )n is uniformly absolutely continuous since it is weakly convergent (cf. [6, Cor. IV.8.11]). Hence for every ε > 0 there is δ > 0 such that kχB fn k1 < ε/2 for every integer n and every B ∈ Σ with µ(B) < δ. Consider Bn = {t ∈ Ω : |fn (t)| > ε/2}. By assumption there exists an integer n0 such that µ(Bn ) < δ for n ≥ n0 . Thus, for n ≥ n0 we have Z Z |fn | + |fn | ≤ kχBn fn k1 + ε/2 < ε. kfn k1 = Bn c Bn The following result will be used in the sequel (cf. [12, Cor. 3.4.14 and Thm. 3.4.17]). Proposition 2.3. Let E and F be two Banach lattices such that the norm on F is order continuous. If T : E → F is a positive operator, then T preserves an isomorphic copy of l1 if and only if T preserves a lattice isomorphic copy of l1 . License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use POSITIVE DISJOINTLY STRICTLY SINGULAR OPERATORS 1981 The proposition given next is essential in the proof of Theorem 1.1. Notice that in the definition of a DSS operator the disjoint vectors are not in general positive. Proposition 2.4. Let T be a positive operator defined on a Banach lattice E and with values in a Banach lattice F with order continuous norm. Then T is DSS if and only if there is no disjoint sequence of non-null positive vectors (yn )n in E such that the restriction of T to the span [yn ] is an isomorphism. Proof. We just need to prove the non-trivial implication. Suppose that T is not DSS; then there exists a (normalized) sequence in E of pairwise disjoint elements (xn )n such that the restriction of T to the span [xn ] is an isomorphism, that is, X  ∞ ∞ X T an xn ≥α an xn for some α > 0 n=1 n=1 (note that (xn )n is an unconditional basic sequence, being disjoint). Since the span of the sequence (T xn )n is a separable subspace of the Banach lattice F , we can find a closed order ideal J of F with a weak unit which contains [T xn ] (cf. [11, Prop. 1.a.9]); furthermore J is complemented in F by a positive projection P (cf. [11, Prop. 1.a.11]). Consider the operator P T : E → J. Clearly the restriction of P T to the span [xn ] is not DSS. On the other hand the assumption on T shows that the operator P T is not invertible on the span of any disjoint sequence of positive vectors. Therefore there is no loss of generality in assuming that F itself has a weak unit. In such a case we can represent F as an (in general not closed) order ideal of L1 (Ω, Σ, µ) for some probability space (Ω, Σ, µ), which is continuously included in L1 (Ω, Σ, µ) (cf. [11, Thm. 1.b.14]). (∗) Let us consider the sublattice [ |xn | ]; we claim that [ |xn | ] contains no lattice isomorphic copy of c0 . To prove this we will assume the contrary and get a contradiction by showing that T is invertible on a sublattice of [|xn |]. Suppose then that there exists a normalized sequence (zk )k ⊂ [ |xn | ] of mutually disjoint positive vectors equivalent ∞ P akj |xj | with akj ≥ 0 for all to the unit basis {ek }k of c0 . Let us write zk = j=1  ∞ P kl aj xj j. If k T (zkl )k → 0 for some subsequence (kl )l , then T → 0 and l hence kzkl k = ∞ P j=1 j=1 akj l xj l → 0 since the restriction of T to the span [xn ] is an l isomorphism; however this is a contradiction with kzk k = 1 for all k. Thus we may assume that inf k kT zk k ≥ δ > 0 for some δ > 0. The sequence (zk )k is σ([zk ], [zk ]′ )null being equivalent to the unit basis of c0 . Hence (T zk )k is weakly null in F being T bounded. Note that (T zk )k is also convergent to zero in the weak topology of L1 (µ) since F is continuously included in L1 (µ). In fact (T zk )k converges to zero in L1 (µ) since T zk ≥ 0 for all k (cf. Lemma 2.1). Let us consider the Kadec-Pelczynski set M (ε) = {y ∈ F : µ(σ(y, ε)) ≥ ε}, where ε > 0 and σ(y, ε) = {t ∈ Ω : |y(t)| ≥ εkyk}. If (T zk )k ⊂ M (ε) for some ε > 0, then k T zk k1 ≥ ε2 kT zk k for all k; hence kT zk k → 0, which is a contradiction with k inf k kT zk k ≥ δ > 0. Thus we may assume that (T zk )k * M (ε) for every ε > 0; then, by Kadec-Pelczynsky’s disjointification process (cf. [11, Prop. 1.c.8]) we may choose a subsequence (T zkj )j equivalent to a disjoint sequence in F ; it follows that (T zkj )j is an unconditional basic sequence with unconditional constant, say β > 0. License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use 1982 JULIO FLORES AND FRANCISCO L. HERNÁNDEZ For every integer j we have ∞ ∞ X X aj T zkj ≥ β −1 |aj | T zkj ≥ β −1 |aj | k T zkj k ≥ β −1 |aj |δ j=1 j=1 (note that in the previous inequalities we use that T zkj ≥ 0 for all j). Thus X   _ ∞ ∞ ∞ X ≥ β −1 δ aj zkj T |aj | ≥ K aj zkj , j=1 j=1 j=1 where K is a positive constant. Hence the operator T preserves a lattice copy of c0 , which is a contradiction. Now if we apply Rosenthal’s dichotomy theorem (cf. [10, Thm. 2.e.5]) to the sequence (|xn |)n , we obtain a subsequence (|xnj |)j satisfying either (1) (|xnj |)j is equivalent to the unit basis of l1 or (2) (|xnj |)j is a weakly Cauchy sequence. Suppose first that (1) holds. Then X X X |aj | < ∞; aj |xnj | < ∞ ⇔ aj xnj < ∞ ⇔ j j j hence T preserves an isomorphic copy of l1 or, equivalently by Proposition 2.3, T preserves a lattice copy of l1 . Contradiction. Finally let us show that case (2) also leads to contradiction. Indeed, once the statement (∗) has been proved we may assume that the Banach lattice [ |xn | ] is weakly sequentially complete or equivalently a KB-space (cf. [4, Thm. 14.12]); hence the subsequence (|xnj |)j must be weakly convergent. Thus the separable lattice [ |xn | ] has an order continuous norm and a weak unit, and hence it can be considered as a continuously-included order ideal in L1 (Ω′ , Σ′ , µ′ ) for some probability space (Ω′ , Σ′ , µ′ ) (cf. [11, Thm. 1.b.14]). It follows that (|xnj |)j is convergent in the weak topology of L1 (µ′ ) to a function f . In fact f = 0 since the sequence (|xnj |)j converges to zero in µ′ -measure being pairwise disjoint (cf. Lemma 2.2). Since T is bounded, the sequence (T |xnj |)j converges to zero in the weak topologies of F and L1 (µ); hence k T |xnj | k1 → 0 by Lemma 2.1. j We apply again the Kadec-Pelczynski method: if T (|xnj |)j ⊂ M (ε) for some ε > 0, then k T |xnj | k1 → 0 implies k T |xnj | k → 0 and kxnj k → 0 follows; this is j j j a contradiction with the initial choice of (xn )n . Thus we may assume (T |xnj |)j * M (ε) for all ε > 0; in this case we may choose a subsequence, still denoted by (T |xnj |)j , which is equivalent to a disjoint sequence in F . It follows that (T |xnj |)j is an unconditional basic sequence with unconditional constant, say K > 0. And ∞ ∞ ∞ X X X aj T xnj aj xnj ≤ aj |xnj | = α α j=1 j=1 ≤ ∞ X j=1 j=1 |aj | T |xnj | ≤ K ∞ X aj T |xnj | = K T j=1 X ∞ j=1 aj |xnj |  , that is, T is invertible on the span [ |xnj | ]. This contradiction concludes the proof. Remark 2.5. If the dual norm on E ′ and the norm on F are simultaneously order continuous, then the previous proof becomes shorter by noticing that the sequence (|xn |)n is weakly null (cf. [12, Thm. 2.4.14]). License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use POSITIVE DISJOINTLY STRICTLY SINGULAR OPERATORS 1983 We recall that an operator T between a Banach lattice E and a Banach space Y is said to be order-weakly compact if T [−x, x] is relatively weakly compact for every x ∈ E+ . It is known that T : E → Y is order-weakly compact if and only if T does not preserve a sublattice isomorphic to c0 whose unit ball is order bounded in E (cf. [12, Cor. 3.4.5]). Consequently every DSS operator is order-weakly compact. The following characterization will be used in the sequel (cf. [12, Prop. 3.4.9]). Proposition 2.6. Let E be a Banach lattice, F be a Banach space and T : E → F a bounded operator. Then T is order-weakly compact if and only if T transforms every order bounded and weakly null sequence of positive vectors in E in a sequence convergent to zero. We pass now to prove the main result of the paper. Proof of Theorem 1.1. We prove first that if the norm on F as well as the norm on E ′ are order continuous, then T DSS implies S DSS. Assume that this is not the case. By Proposition 2.4 there exists a disjoint (normalized) sequence (xn )n of positive vectors in E and α > 0 such that kSxk ≥ αkxk for all x ∈ [xn ]. As in the proof of Proposition 2.4 the Banach lattice F can be considered to have a weak unit. Hence there exist a probability space (Ω, Σ, µ), an order ideal I of L1 (Ω, Σ, µ), a lattice norm k kI on I and an order isometry ψ between F and (I, k kI ), such that the canonical inclusion from I in L1 (µ) is continuous with kf k1 ≤ kf kI (cf. [11, Thm. 1.b.14]). Note that ψT : E → I is DSS and that 0 ≤ ψS ≤ ψT . Note too that if ψS were DSS, then S would also be DSS. This observation allows us to reduce the proof to the case that F is an order ideal in L1 (Ω, Σ, µ). We claim that (T xn )n * M (ε) for all ε > 0 where M (ε) denotes a KadecPelczynski set as above. Indeed, the norm-bounded disjoint sequence (xn )n is weakly null since the norm on E ′ is order continuous (cf. [12, Thm. 2.4.14]). Hence (T xn )n is a weakly null sequence in L1 (µ); in fact kT xn k1 → 0 by Lemma 2.1. If (T xn )n ⊆ M (ε) for some ε > 0, then kT xn k1 ≥ ε2 kT xn k; hence (T xn )n converges to zero in F . The inequalities 0 ≤ Sxn ≤ T xn for all n show that (Sxn )n converges to zero in F , and hence kxn k → 0 since kSxn k ≥ αkxn k for all n; n however this is a contradiction with the choice of (xn )n . Now, by Kadec-Pelczynski’s disjointification process (cf. [11, Prop. 1.c.8] ), we may choose a subsequence, still denoted by (T xn )n , equivalent to a pairwise disjoint sequence in F ; hence (T xn )n is an unconditional basic sequence with unconditional constant, say K > 0. We have X  ∞ ∞ ∞ ∞ X X X an xn T = an T xn ≥ K −1 |an |T xn ≥ K −1 |an |Sxn n=1 n=1 ≥ αK −1 n=1 ∞ X |an |xn = αK −1 n=1 n=1 ∞ X an xn , n=1 for all x ∈ [xn ]. However this is impossible since T is a DSS operator. We can now prove the first part of Theorem 1.1 in the general case. Assume the opposite, that is, there is a disjoint sequence (xn )n of positive vectors in E such that the restriction of S to the sublattice [xn ] is an isomorphism while T is DSS. By the lines above we can assume that the dual norm on the sublattice [xn ]′ is not order continuous. Then [xn ] contains a lattice copy of l1 (cf. [12, Thm. 2.4.14]) which is License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use 1984 JULIO FLORES AND FRANCISCO L. HERNÁNDEZ preserved by S. It follows that the adjoint operator S ′ is not order-weakly compact (cf. [12, Cor. 3.4.14]) and so is T ′ by Proposition 2.6 (note that 0 ≤ S ′ ≤ T ′ ); hence T preserves a copy of l1 (cf. [12, Cor. 3.4.14]) or equivalently T preserves a lattice copy of l1 (cf. Proposition 2.3). Contradiction. To prove the second part of Theorem 1.1 assume that the norms on E ′ and F are simultaneously not order continuous. Since the dual norm on E ′ is not order continuous, we can find in E a lattice copy of l1 complemented by a positive projection (cf. [12, Thm. 2.4.14 and Prop. 2.3.11]); let H1 be the sublattice of E lattice isomorphic to l1 , φ1 the lattice isomorphism between H1 and l1 , and P1 the positive projection from E onto H1 . On the other hand, since the norm on F is not order continuous and F is σ-Dedekind complete, there exist a sublattice H2 in F and a lattice isomorphism φ2 between H2 and l∞ (cf. [12, Cor. 2.4.3]). Consider the operators defined from l1 into l∞ by X  X  ∞  ∞ ∞ an (1, 1, · · · ) and S(a = (an )) = xk,n an T (a = (an )) = , n=1 n=1 k=1 where S ≡ (xk,n ) is the infinite matrix with {0, 1, −1}-entries defined as follows:   1 0 0 0 0 0 ... −1 0 0 0 0 0 . . .   1 1 0 0 0 0 . . .    S ≡ (xk,n ) =   1 −1 0 0 0 0 . . . . −1 1 0 0 0 0 . . .   −1 −1 0 0 0 0 . . . .......................... Note that for a given a ∈ l1 there exists an integer k with 2 + 22 + · · · + 2n−1 < k ≤ 2 + 22 + · · · + 2n satisfying n X |am | = m=1 n X xk,j aj . j=1 Clearly S is a linear isometry from l1 into l∞ . Indeed,  X  X ∞ ∞ xk,n an ≤ |an | = kak1 . kS(a)k∞ = sup k n=1 n=1 Pn On the other hand, for a given ε > 0 there exists an integer n such that m=1 |am | ≥ n n P P xk,j aj = |am | ≥ (kak1 − ε). (kak1 − ε); hence there is an integer k satisfying j=1 m=1 The relations kSak∞ ≥ kak1 − ε and kSak∞ = kak1 follow. Consider now the operators S + and S − defined by the sequences (x′k,n )n,k and ′′ (xk,n )n,k where ( ( xk,n if xk,n > 0, −xk,n if xk,n < 0, ′ ′′ xk,n = xk,n = 0 if xk,n ≤ 0, 0 if xk,n ≥ 0. P ∞ 1 ∞ e The operator S(a) =( ∞ clearly factorizes through j=1 |xk,j |aj )k=1 from l into l the space c of all convergent sequences (which is isomorphic to c0 ); thus Se is strictly singular. It follows from the equalities Se = S + + S − and S = 2S + − Se that S + is not strictly singular. Hence S + is neither DSS as an operator from l1 into l∞ (in License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use POSITIVE DISJOINTLY STRICTLY SINGULAR OPERATORS 1985 fact the inequality kS + ak ≥ 1/2kak1 holds for all a ∈ l1 ). Finally it is clear that S + ≤ T and that T is DSS being a rank-one operator. Consider the operators S ′ = φ2 S + φ1 P1 and T ′ = φ2 T φ1 P1 defined on E and with values in F . Clearly 0 ≤ S ′ ≤ T ′ ; moreover φ2 S + φ1 is not strictly singular since φ1 and φ2 are isomorphisms and S + is not strictly singular. The inequalities mkhkE ≤ kS + φ1 (h)kl∞ ≤ kφ2 S + φ1 (h)kF ≤ M khkH show that S ′ is invertible on H, or equivalently that S ′ is not DSS. Remark 2.7. The proof actually shows that if the norms on E ′ and F are simultaneously not order continuous and F is σ-Dedekind complete, then the problem of domination for strictly singular operators has in general a negative answer. This problem requires its own study which will be carried out elsewhere. We consider next the problem of domination in the case E = F . To this end we recall the following factorization result due to Aliprantis and Burkinshaw (cf. [4, Thm. 18.7] or [12, Thm. 3.4.6]): Let E and Y be a Banach lattice and a Banach space respectively and T : E → Y an order-weakly compact operator. Then there exist a Banach lattice F with order continuous norm, a lattice homomorphism Q from E into F and a bounded operator S from F into Y such that T = SQ. Moreover, if Y is a Banach lattice and 0 ≤ T1 ≤ T , then 0 ≤ S1 ≤ S. Proposition 2.8. Let Ei , i = 1, 2, 3, be Banach lattices and 0 ≤ Si ≤ Ti operators defined on Ei and taking values in Ei+1 for i = 1, 2. If T1 is DSS and T2 is order-weakly compact, then S2 S1 is DSS. Proof. Given 0 ≤ S2 ≤ T2 : E2 → E3 , we may find, by the above result, a Banach lattice G with order continuous norm, a lattice homomorphism Q from E2 into f2 from G into E3 such that T2 = T f2 Q and f2 ≤ T G and two positive operators S f S2 = S2 Q. Consider the operators S = QS1 and T = QT1 from E1 into G. Note that T is DSS since T1 is so and we are composing by the left; hence S is DSS by f2 S concludes that S2 S1 f2 S. Finally the equality S2 S1 = S Theorem 1.1 and so is S is DSS. Now Theorem 1.2 is a direct consequence of Proposition 2.8. Remark 2.9. Theorem 1.2 is the best possible. Indeed, consider E = l1 ⊕ l∞ and the operators 0 ≤ Se ≤ Te on E defined via the matrices Se =  0 S+  0 , 0 Te =  0 T  0 0 where S + and T are the operators defined in the proof of Theorem 1.1. Clearly Te is DSS and Se is not. Recall that an orthomorphism on a Banach lattice E is a band preserving operator which is also order bounded. An easy consequence of Theorem 1.2, which is obtained by reasoning as in [4, Thm. 16.21], is Corollary 2.10. Let S and T be two positive operators on a Dedekind complete Banach lattice E such that 0 ≤ S ≤ T holds. If T is DSS and S is an orthomorphism, then S is DSS. License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use 1986 JULIO FLORES AND FRANCISCO L. HERNÁNDEZ The authors wish to thank Y. Abramovich, J. Mendoza and Y. Raynaud for some helpful discussions. They also appreciate the suggestions and remarks made by the referee. References 1. Y. A. 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MR 86b:46001 Departamento de Análisis Matemático, Facultad de Matemáticas, Universidad Complutense, 28040 Madrid, Spain E-mail address: [email protected] Departamento de Análisis Matemático, Facultad de Matemáticas, Universidad Complutense, 28040 Madrid, Spain E-mail address: [email protected] License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use