The Banach-Saks Properties in Orlicz-Lorentz Spaces

Kamińska, Anna; Lee, Han Ju

doi:https://doi.org/10.1155/2014/423198

Abstract and Applied Analysis

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Research Article | Open Access

Volume 2014 | Article ID 423198 | https://doi.org/10.1155/2014/423198

The Banach-Saks Properties in Orlicz-Lorentz Spaces

Anna Kamińska¹and Han Ju Lee²

Academic Editor: Mieczysław Mastyło

Received30 Nov 2013

Accepted07 Feb 2014

Published13 Mar 2014

Abstract

The Banach-Saks index of an Orlicz-Lorentz space for both function and sequence case, is computed with respect to its Matuszewska-Orlicz indices of . It is also shown that an Orlicz-Lorentz function space has weak Banach-Saks (resp., Banach-Saks) property if and only if it is separable (resp., reflexive).

1. Introduction

Let be a real Banach space. A bounded sequence is called a weak Banach-Saks sequence whenever there exists a subsequence such that its Cesàro means converge in norm to zero; that is, The space is said to satisfy the weak Banach-Saks property, in short , if every weakly null sequence in is a weak Banach-Saks sequence. Recall that a Banach space satisfies the Banach-Saks property if for every bounded sequence in , there is a subsequence such that its Cesàro means converge; that is, the sequence is convergent in norm. It is well known that a Banach space has the -property if and only if it is reflexive and it has the -property [1].

Schreier constructed a separable Banach space without the weak Banach-Saks property [2] (which is called a Schreier space) and later it was shown by Baernstein [3] that there is a reflexive separable Banach space without the weak Banach-Saks property. Hence the space does not have the weak Banach-Saks property, since every separable space is embeddable in .

Following Johnson [4], given , a bounded sequence is called a sequence if there exists a subsequence such that Here for all . We then say that has the -Banach-Saks property, shortly , if each weakly null sequence contains a subsequence. It is clear that every Banach space has a property. The Banach-Saks index is the number . The set is always an interval or . The Bessaga-Pełczyński method [5, Proposition ] shows that and the Schur property of shows that . The Banach-Saks index of Nakano sequence space was computed in [6].

The stronger property () was introduced by Knaust and Odell in [7]. It is said that has property , , if every weakly null sequence in has a subsequence so that there is a constant such that, for all and all real sequences , where is the norm.

It is clear that for all . The Elton -theorem [8, Theorem ] states that and a Banach space has if and only if every normalized weakly null sequence contains a subsequence which is equivalent to the unit vector basis of , which is equivalent to the case in which every subspace of has the Dunford-Pettis property (for details see [9]). In general, the two properties and are not equivalent if [10]; however, Rakov [11, Theorem 3] showed that if , then implies . Various weak Banach-Saks properties and their relations were shown in [6] for Musielak-Orlicz sequence spaces.

In this paper we compute the Banach-Saks index of Orlicz-Lorentz space and show that Orlicz-Lorentz function space has the weak Banach-Saks property (resp., Banach-Saks property) if and only if it is separable (resp., reflexive). We start with several preliminary results about Orlicz-Lorentz spaces.

2. Preliminaries

Let or with , equipped with the counting measure if and the Lebesgue measure if . If is a measurable set, we denote by its measure. By we denote the collection of all real valued measurable functions on . In the case when the elements are sequences , and in the other cases they are real valued Lebesgue measurable functions .

For a measurable function , the decreasing rearrangement of will be denoted by , defined as , , where , . For two measurable functions and , the relation means that, for all , we have

For a sequence , will denote the decreasing rearrangement of . That is, letting , we have for all . For sequences and , means that for all , we have

Let be an Orlicz function; that is, is convex, strictly increasing, and . Let the weight function be nonnegative and nonincreasing. For , , the Orlicz-Lorentz space is the collection of all functions such that where . Analogously in case when , the Orlicz-Lorentz sequence space contains all real sequences such that where . It is well known that the space is a rearrangement invariant (r.i.) Banach function space satisfying the Fatou property [12–15]. The space (resp., ) is the order continuous part of (resp., ). In the sequence case is the closure of the linear span of the unit vectors in . If and , then and [16]. If , then is an Orlicz function space, and is an Orlicz sequence space. In this case also and are subspaces of order continuous elements in Orlicz spaces.

We may assume that . In fact, the space , where with , is isometric to ; we have .

Definition 1. Given an Orlicz function , one defines the growth conditions connected to its lower and upper Matuszewska-Orlicz indices in three different categories, at zero, at infinity, and on .
We say that, for , , , and , whenever respectively. Analogously we say that, given , , , and , whenever respectively.
The Matuszewska-Orlicz indices, lower , , and and upper , , and , are then defined as follows:
It is clear that and . We say that satisfies conditions , , and , whenever respectively. Let and be Orlicz functions. We call them equivalent at zero, equivalent at infinity, or just equivalent, if there is such that , for all , , or for all , respectively. It is well known and easy to prove that Matuszewska-Orlicz indices of Orlicz functions are preserved by the corresponding equivalence relations (e.g., see [17]); that is, if and are equivalent (resp., equivalent at and equivalent at ), then (resp., and ). Similar equalities hold for upper indices. Recall also that any upper index of is finite if and only if satisfies the corresponding condition [17, 18].
We also have that if the Orlicz functions and are equivalent (resp., equivalent at infinity and equivalent at zero), then (resp., , ; ) with equivalent norms.
The space (resp., (); ) is separable if and only if and (resp., ; and ), if (resp., ; ). The separability is also equivalent to the case in which the space does not contain the isomorphic copy of [12, 15, 19].

Lemma 2. Let ;(1)([20, Lemma 6], [21, Lemma 3], and [22]) An Orlicz function (resp., ; ) if and only if there exists an Orlicz function equivalent (resp., equivalent at ; equivalent at ) to such that is convex;(2)([20, Lemma 5], [21, Lemma 2], and [22]) An Orlicz function (resp., ; ) if and only if there exists an Orlicz function equivalent (resp., equivalent at and equivalent at ) to such that is concave.

We will use the following functions: Recall that the weight is regular whenever there exists such that for all . Since always , the weight is regular if the functions and are equivalent on . Notice that for any regular weight on or it holds that .

A Banach lattice is said to be -convex (resp., -concave) for some if there is a constant such that for every choice of vectors in . A Banach lattice is said to satisfy upper -estimate (resp., lower -estimate) if the definition of -convexity (resp., -concavity) holds true for any choice of disjointly supported elements in . It is known that, given , if is -convex (resp., -concave), then is -convex (resp., -concave) for (resp., ) [23, 24].

Theorem 3 (see [25]). The space (resp., (); ) has finite concavity if and only if satisfies the (resp., ; ) condition and the weight is regular.

Recall that a Banach lattice has the finite concavity if and only if it is of finite cotype [26].

Theorem 4 (see [19]). Let ;(1)[19, Theorem 7.18] For , the following conditions are equivalent:(i) (replaced by when ) is order isomorphic to a sublattice of ,(ii)either or, for some , ;(2)[19, Theorem 6.13] For , the following assertions are equivalent:(i) (replaced by when ) is order isomorphic to a sublattice of ,(ii),(iii) (replaced by when ) is order isomorphic to a sublattice of ;(3)[19, Theorem 7.8] For the following statements are equivalent (where is meant in place of if ):(i) is order isomorphic to a sublattice of ,(ii) is isomorphic to a subspace of ,(iii),(iv) is isomorphic to a subspace of .

3. Main Results

Theorem 5. If and (resp., and ), then (resp., ; ) is -convex. If (resp., ; ) is -convex for some , then (resp., ; ).

Proof. (1) By Lemma 2 we assume that is convex. Then we have the equality , where is a -convexification of . Indeed,
Since is a convex Orlicz function, the space is a Banach space, so its -convexification is -convex [26].
We only show the second part in the case of . If , then, by Theorem 4(1), there is such that is an order copy in , and thus the space cannot be -convex.

Example 6. In general, the parallel statement on concavity in Theorem 5 does not hold. As shown by Pisier in [26, Example ], the Lorentz space , where and , is not 2-concave even though the function satisfies condition and is regular.

Proposition 7. Let . If and if (resp., and in the case when ) and is regular, then , , has type .

Proof. By Theorem 5 the space is -convex, and thus in view of Theorem 3 and by applying the well-known fact [26, Proposition ], the space has type .

Rakov [27, Theorem 1] showed that if a Banach space has type , then it has -Banach-Saks property. Hence we get the following.

Theorem 8. If a Banach space has type for some , then . Consequently .

Notice that for every separable r.i. space on , it is shown in [28] that . Because every r.i. function space on contains the r.i. space isometrically, we get the following.

Proposition 9 (see [29]). If is a separable r.i. space on , , then .

Recall that a Banach space is said to be -smooth () if its modulus of smoothness is majorized by the function . That is, there is a constant such that, for all ,

Rakov [27, Theorem 1] showed that if is a -smooth space, then it has the property. It is well known [26] that a Banach lattice is isomorphic to a -smooth Banach space for some if and only if it has the type and satisfies a lower -estimate for some . Notice also that the finite concavity is equivalent to the lower -estimate for some [26]. Then we obtain immediately the following corollary as a consequence of definition of lower indices and by Propositions 7 and 9 and Theorem 8.

Corollary 10. Let . If and if (resp., and in the case when ) and is regular, then , , has the -property.
Consequently, if when (resp., in the case when ) and is regular, then

The next corollary is a result of the facts that and that if does not satisfy appropriate condition , then contains an isomorphic copy of .

Corollary 11. If when , in the case when , and when , then

Corollary 12. An Orlicz-Lorentz space (resp., , ; ) does not contain an isomorphic copy of if and only if (resp., ; ).

Proof. This hypothesis for sequence spaces is an immediate consequence of Theorem 4(3).
Let now . From [19, Corollary 3.5(2), (3)], the space contains an isomorphic copy of if and only if either is an order isomorphic copy, that is, in view of Theorem 4(2), , or the sequence of functions is bounded in . However implies that there exists such that for all and some . Hence for any constant and large enough , Thus whenever the sequence is not bounded, and the proof is done in case of finite interval.
Assume that . Again applying Corollary 3.5(2) in [19], we get dichotomy that is an isomorphic copy in if and only if either is an order isomorphic copy in or the sequence is bounded in . Assuming now that , we can show similarly as above that the second alternative cannot occur. Recall then that, by Theorem 4(1), is an order isomorphic copy in the space if and only if either or, for some , . We will exclude again the second possibility. In fact if , then for some and and for all and , . Therefore for every and some such that , we have and this completes the proof.

Corollary 13. Assume that , , does not contain an isomorphic copy of , equivalently that in case and in case . Then

Proof. By Proposition 9 the Banach-Saks index of any r.i. space over the interval cannot be bigger than . On the other hand, recall that for and . The space does not have property, so . Hence Theorem 4 shows that where when and if .

Finally by Corollaries 10 and 11 we get the result on Banach-Saks index in the Orlicz-Lorentz spaces over the interval , .

Theorem 14. Assume that , , does not contain an isomorphic copy of and equivalently that in case and in case . Assume also that when , when , and is regular. Then

Using the Bessaga-Pełczynski method, we get the following.

Theorem 15. Let . If an order continuous sequence space is -convex, then has the property. Consequently .

Proof. Since is order continuous, the standard unit vectors form a basis of . Suppose that is a normalized weakly null sequence in . By the Bessaga-Pełczyński method [5, Proposition ], we may assume that for all , where is a block basic sequence of and for all . Then for real numbers with , the -convexity of and disjointness of the supports of ’s show that where is the constant from the -convexity of and is the conjugate of satisfying . The proof is done.

As a consequence of the above Theorems 15 and 5 about -convexity and Theorem 4 about -copies, we get the Banach-Saks index in sequence spaces as follows.

Theorem 16. Assume that does not contain an isomorphic copy of and equivalently that . Assume also that and . Then

Proof. By Theorems 5 and 15, and Since (), Theorem 4 completes the proof.

We finish with a result on the weak Banach-Saks property.

Theorem 17. A separable Orlicz-Lorentz space , , has the weak Banach-Saks property.

Proof. Let be weakly null and assume without loss of generality that , . Then by [29, Proposition 3.2] there exists a subsequence splitted into three subsequences such that are bounded and(a), , and weakly;(b) for , , in measure, and weakly;(c).We need to show that there is a subsequence of whose Cesàro means approach zero in norm. It is obvious that we can do it for the sequence satisfying (c). We will show this now for the sequence . By [29, Proposition 4.3] there exists a subsequence , , such that in for all further subsequences (we can assume that ). Hence (Recall that if in any r.i. space , then in measure.) Since weakly in , we have, for any subsequence of , Applying now [29, Lemma 4.4] we get that for all subsequences . Let now be an arbitrary subsequence of . Set By and by subadditivity of the operator , , we get, for , that is, Since in measure and for every , by [29, Proposition 4.2], in , which finishes the case (a).
In the case (b), by [19, Theorem 5.5], there exists a subsequence such that any further subsequence of is equivalent to the unit vector basis in some Orlicz sequence space . Thus weakly in , and by -property of (see, e.g., [27]), there is a subsequence such that in and thus in . This completes case (b) and proves the theorem.

A Banach space is said to have the (resp., weakly) alternative signs Banach-Saks property (ABS) (resp., (wABS)) if, for every bounded (resp., weakly null) sequence , there is a subsequence and a sequence of signs such that the Cesàro averages are norm convergent to zero. That is, It is shown by Rosenthal [30, 31] that if has the (ABS), then has the weak Banach-Saks property. In case that does not contain , then (ABS) is equivalent to the weak Banach-Saks property [31]. It is clear that does not have the (ABS) and has the weak Banach-Saks property. Hence a Banach space has the (ABS) if and only if has the weak Banach-Saks property and does not contain . The notion of (wABS) was introduced in [32] and it was shown that (wABS) is equivalent to (wBS).

Corollary 18. Let for some . Then the following are equivalent: (1)Orlicz-Lorentz space has the Banach-Saks property (BS);(2)Orlicz-Lorentz space is reflexive;(3)Orlicz-Lorentz space does not contain isomorphic copies of or ;(4)Orlicz-Lorentz space is separable and does not contain an isomorphic copy of ;(5)Orlicz-Lorentz space has the weak Banach-Saks property (wBS) and does not contain an isomorphic copy of ;(6)Orlicz-Lorentz space has the alternative signs Banach-Saks property (ABS).

Proof. Recall that a Banach lattice is reflexive if and only if it does not contain isomorphic copies of or [26, Theorem ]. Hence we get . Recall also that a Banach lattice is a KB space if and only if does not contain an isomorphic copy of [33, Theorem 14.12]. That is, if is a Köthe function space, then it is equivalent to the case in which is order continuous and satisfies the Fatou property. Since an Orlicz-Lorentz space satisfies the Fatou property and the order continuity is equivalent to the separability of the Orlicz-Lorentz space, follows.
Since a nonseparable Orlicz-Lorentz space contains and it does not have the weak Banach-Saks property, Theorem 17 shows that . The preceding argument of the corollary shows that . Since (6) implies both (5) and (2), (6) implies (1). The proof is done.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgment

Han Ju Lee was supported by Basic Science Research Program through the National Research Foundation of Korea(NRF) funded by the Ministry of Education, Science and Technology (NRF-2012R1A1A1006869).

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Copyright © 2014 Anna Kamińska and Han Ju Lee. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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