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# Orlicz Sequence Spaces with a Unique Spreading Model

*Journal of Inequalities and Applications*
**volume 2010**, Article number: 962842 (2010)

## Abstract

We study the set of all spreading models generated by weakly null sequences in Orlicz sequence spaces equipped with partial order by domination. A sufficient and necessary condition for the above-mentioned set whose cardinality is equal to one is obtained.

## 1. Introduction

Let be a separable infinite dimensional real Banach space. There are three general types of questions we often ask. In general, not much can be said in regard to this question "what can be said about the structure of itself" and not much more can be said about the question "does embedded into a nice subspace". The source of the research on spreading models was mainly from the question "finding a nice subspace " [1]. The spreading models usually have a simpler and better structure than the class of subspaces of [2, 3]. In this paper, we study the question concerning the set of all spreading models whose cardinality is equal to one.

The notion of a spreading model is one of the application of Ramsey theory. It is a useful tool of digging asymptotic structure of Banach space, and it is a class of asymptotic unconditional basis. In 1974, Brunel and Sucheston [4] introduced the concept of spreading model and gave a result that every normalized weakly null sequence contains an asymptotic unconditional subsequence, they call the subsequence spreading model. It was not until the last ten years that the theory of spreading models was developed, especially in recent five years. In 2005, Androulakis et al. in [2] put forward several questions on spreading models and solved some of them. Afterwards, Sari et al. discussed some problems among them and obtained fruitful results. This paper is mainly motivated by some results obtained by Sari et al. in their papers [3, 5].

## 2. Preliminaries and Observations

An Orlicz function is a real-valued continuous nondecreasing and convex function defined for such that and If for some , is said to be a degenerate function. is said to satisfy the condition if there exist , such that for . We denote the modular of a sequence of numbers by . It is well known that the space

endowed with the Luxemburg norm

is a Banach sequence space which is called Orlicz sequence space. The space

is a closed subspace of . It is easy to verify that the spaces are just Orlicz sequence spaces, and Orlicz sequence spaces are the generalization of the spaces . Furthermore, if is a degenerate Orlicz function, then and [6]. In the context, the Orlicz functions considered are nondegenerate. Let

They are nonvoid norm compact subsets of consisting entirely of Orlicz functions which might be degenerate [6, lemma ].

Definition 2.1.

Let be a separable infinite dimensional Banach space. For every normalized basic sequence in a Banach space and for every , there exist a subsequence and a normalized basic sequence such that for all , and ,

The sequence is called the spreading model of and it is a suppression-1 unconditional basic sequence if is weakly null [4].

The following theorem guarantees the existence of a spreading model of . We shall give a detailed proof.

Theorem 2.2.

Let be a normalized basic sequence in and let . Then there exists a subsequence of so that for all and integers

In order to prove Theorem 2.2, we should have to recall the following definitions and theorem.

For is the set of all subsets of of size . We may take it as the set of subsequences of length , with . denotes all subsequences of . Similar definitions apply to and if .

Definition 2.3 . (see [1]).

Let and be two disjoint intervals. For any , and scalars if

then we call "color" and . Meanwhile, we say has the same "color" as , where is a sequence of a Banach space. We identify the same "color" subsets of , saying they are 1-colored.

Definition 2.4 (see [1]).

The family of is called finitely colored provided that it only contains finite subsets in "color" sense, and each subset is 1-colored.

Theorem 2.5 . (see [1]).

Let and let be finitely colored. Then there exists so that is 1-colored.

Proof of Theorem 2.2.

We accomplish the proof in two steps.

Step 1.

We shall prove that for any , there exists such that for any

Firstly, for fixed by the Definition 2.4, we can prove that the above inequality holds. In fact, we partition into subintervals of length and "color" by if

In the same way, we can also "color" by .

We can take as the unit ball in finite-dimensional space ; then is sequentially compact; moreover, it is totally bounded and complete. Under -metric, take for -net of . For any element of net , repeat the above process, and let . We partition into subintervals of length and "color" by if

Since the length of , we have

Secondly, we shall prove that for any holds. Since is the -net of , there exists such that

Therefore, we have

Hence,

Similarly, we obtain

Thus

Step 2.

We apply diagonal argument to prove that there exists such that for any

By Step 1, in view of , there exists such that for any , for any , we have

Obviously, is also a normalized basic sequence. So in view of , there exists such that for any ,

Repeating the above process, for any , there exists such that for any , we have

Finally, we choose the diagonal subsequence ; for any , we obtain that

Definition 2.6.

Let be a separable infinite-dimensional Banach space. A normalized basic sequence generates a spreading model if for some , for all , and ,

Theme 2.7.

Definition 2.6 is equivalent to Definition 2.1.

Proof.

We can easily conclude Definition 2.1 from Definition 2.6

By the Definition 2.1, we know that is a spreading model generated by . For any fixed , we partition into some subintervals of length and "color" by if

Let and ; then

where . Using the same procedure of Theorem 2.2, we can get that for any

Thus

Letting , then

That is,

Similarly,

Hence, we obtain that

Let be the set of all spreading models generated by weakly null sequences in endowed with order relation by domination, that is, if there exists a constant such that for scalars ; then is a partial order set. If and , we call equivalent to , denoted by . We identify and in if .

Lemma 2.8 (see [5]).

If an Orlicz sequence space does not contain an isomorphic copy of , then the sets and coincide. That is, .

## 3. Orlicz Sequence Spaces with Equivalent Spreading Models

Definition 3.1 (see [7]).

Let be a normalized Schauder basis of a Banach space . is said to be lower (resp., upper) semihomogeneous if every normalized block basic sequence of the basis dominates (resp., is dominated by) .

Lemma 3.2 (see [7]).

Let be an Orlicz function with , and let denote the unit vector basis of the space . The basis is

(a)lower semi-homogeneous if and only if for all and some ,

(b)upper semi-homogeneous if and only if for as above.

Lemma 3.3 (see [6]).

The space , or if , is isomorphic to a subspace of an Orlicz sequence space if and only if , where

Lemma 3.4 . (see [5]).

Let , be an Orlicz sequence space which is not isomorphic to . Suppose that is countable, up to equivalence. Then

(i)the unit vector basis of is the upper bound of ;

(ii)the unit vector basis of is the lower bound of , where .

Theorem 3.5.

Let , and let be the unit basis of the space . If is lower semi-homogeneous, then if and only if is isomorphic to .

Proof.

Sufficiency. Since , is countable, then by Lemma 3.4, is the upper bound of , and is the lower bound of Since is isomorphic to , we get

Necessity.

If , then by Lemma 2.8, that is, all the functions in are equivalent to .

For , we define the function [6] as follows:

where with . Obviously, ; next we shall prove that is equivalent to

Since , , and is nondecreasing convex function, therefore, ; then

Since and , we have

Notice that for any fixed , the right side of the above inequality is a constant; then we obtain

By , we have and ; hence

Since , , and

Moreover,

We obtain that

Since is lower semihomogeneous; then by Lemma 3.2, we have for some

Therefore,

Thus we get

So by (3.4) and (3.7), we can know that is equivalent to By Lemma 3.3 and its proof ([6], Theorem .a.9), we obtain that uniformly converges to on Since is the closed subset of , we have that , is equivalent to , and therefore is isomorphic to

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## Acknowledgment

The first author was supported by the NSF of China (no. 10671048) and by Haiwai Xueren Research Foundation in Heilongjiang Province (no. 1055HZ003).

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### Cite this article

Hao, C., Lü, L. & Yin, H. Orlicz Sequence Spaces with a Unique Spreading Model.
*J Inequal Appl* **2010, **962842 (2010). https://doi.org/10.1155/2010/962842

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### Keywords

- Banach Space
- Basic Sequence
- Spreading Model
- Unconditional Basis
- Orlicz Function