#### Abstract

It is well known that the modulus of nearly uniform smoothness related with the fixed point property is important in Banach spaces. In this paper, we prove that the modulus of nearly uniform smoothness in Köthe sequence spaces without absolutely continuous norm is . Meanwhile, the formula of the modulus of nearly uniform smoothness in Orlicz sequence spaces equipped with the Luxemburg norm is given. As a corollary, we get a criterion for nearly uniform smoothness of Orlicz sequence spaces equipped with the Luxemburg norm. Finally, the equivalent conditions of and are given.

#### 1. Introduction

In last century, the fixed point property has been studied by many scholars. A Banach space is said to have the fixed point property (FPP, for short) if every nonexpansive mapping acting on a nonempty bounded closed and convex subset of has a fixed point. A natural generalization of FPP is the weak fixed point property (WFPP, for short). A Banach space is said to have the WFPP, whenever it satisfies the above condition from the definition of FPP with “weakly compact” in place of “bounded closed”. In 1965 Kirk  proved that any reflexive Banach space with normal structure has the FPP and he also asserts that a Banach space with weak normal structure has WFPP. In this connection, see also the paper [2, 3]. In 1989 Prus  introduced a property of a Banach space , called nearly uniform smoothness and he got that a Banach space is nearly uniform convex if and only if its dual space is nearly uniformly smooth. In paper , it is proved that weak nearly uniformly smooth Banach spaces have the FPP, which strengthens the result by S. Prus . In , Domínguez-Benavides proved an important result on the existence of fixed points for nonexpansive mappings. In order to do this, he defined, for a Banach space and a nonnegative real number , the parameter where the supremum is taken over all with and all weakly null sequences in such that . He also defined the coefficient The result that a Banach space has the WFPP, whenever was obtained in  using an embedding into .

In this paper we prove that in Kthe sequence spaces without absolutely continuous norm. Furthermore, the formula of the modulus of nearly uniform smoothness in Orlicz sequence spaces equipped with the Luxemburg norm is given. We find the criteria for nearly uniform smoothness of Orlicz sequence spaces equipped with the Luxemburg. Finally, we prove that the condition of is equivalent to . and if and only if exist and such that or when .

#### 2. Preliminaries

Throughout this paper, is a Banach space which is assumed not to have the Schur property, in which has a weakly convergent sequence that is not norm convergent. By and we denote, respectively, the unit sphere and the unit ball of , and by we denote the set of all real sequences.

Inspiring on the underlying ideas in the definition of nearly uniformly smooth spaces given by Prus  as well as on the modulus of uniform smoothness, in  Domínguez-Benavides defined the modulus of nearly uniform smoothness.

Definition 1. The modulus of nearly uniform smoothness is defined by

where the supremum is taken over all basic sequences in .

Definition 2. For each , we define

In , they have the important result that if a Banach space is reflexive and for some then that the Banach space has the FPP.

They use the above result to give a direct proof that the uniformly nonsequare Banach space has the FPP.

In paper  it is shown that a Banach space is nearly uniformly smooth if and only if is reflexive and and where the supremum is taken over all weak null sequences in .

The following theorem was proved in .

Theorem 3. If is a reflexive Banach space and , has the FPP.

Definition 4. A Banach space is called a Kthe sequence space if for every and satisfying for all , we have and and if there is a with for all (see [11, 12]). Let

A Köthe sequence space is said to be norm absolutely continuous if .

Definition 5. A Köthe sequence space is said to have the semi-Fatou property if for every sequence and satisfying for all there holds .

Definition 6. is called an Orlicz function if it has the following properties:
(1) is even, continuous, convex and ;
(2) for all .

Definition 7. For every Orlicz function we define the complementary function by the formula for every .

Definition 8. We say that satisfies the -condition ( for short) if there exist , such that the inequality whenever .

We say that satisfies the -condition ( for short) if its complementary function satisfies the -condition.

The Orlicz sequence space is defined to be the set for some equipped with the Luxemburg norm To simplify notations, we put .

The basic informations on Orlicz spaces can be found in .

#### 3. Main Results

Theorem 9. Let be a Kthe sequence space with the Semi-Fatou property. If the norm of is not absolutely continuous, .

Proof. Suppose that the norm of is not absolutely continuous. Then there exist and such that where .
Take any . By , there exists such that Notice that so there exists such that In this way, we get a sequence of natural numbers such that Put and . Then
(a) as ;
(b) as . It is well known that, for any Köthe space , we have where is the space of all singular functionals over . converges for all . This means that every is uniquely represented in the form where and for the function is defined by for all .
Consider that the series converges. We have (c) Put and for all . Then for the following holds: By the arbitrariness of , we get . It is clear that . Therefore .

Theorem 9 also can be derived from the result of Hudzik and Mastyło .

Corollary 10. If , for any .

Proof. Since , we have that the norm of is not absolutely continuous. Hence for any .

Theorem 11. Let be a reflexive Kthe sequence space and the norm of be absolutely continuous. Then where the supremum is taken over all basic sequences in .

Proof. It is clear that if is a basic sequence in , is also a basic sequence in . We only prove that . For any , by the norm of which is absolutely continuous, there is a such that and since is a basic sequence, we have that the sequence converge to 0 in coordinate, and there exists a such that whenever . So Noting that is a Köthe space, we have Therefore whenever , that is, .

In order to get the criterion that an Orlicz sequence space is nearly uniformly smooth, we may assume that and and that the space is reflexive. Put . For any with finite and , we define as follows: It is clear that is decreasing. Hence .

Theorem 12. Supposed that and . For an Orlicz sequence space , there holds

Proof. Let Then for any , there exists with finite such that . By the definition of , there exists such that whenever . By the definition of , there exists with such that , that is to say, Hence there exists such that Since , we also have and there exists with such that , that is to say, Hence there also exists with such that In such a way, there exist a sequence and a sequence of natural numbers such that for all . It is clear that .
For any , we have that is to say, Therefore and by the arbitrariness of , we have .
Now we prove that . By the definition of , we always have for any with finite .
Taking any weak null sequence and , by there exists such that . Since , there is a such that when . Hence We next estimate the , when . Put and for . There exists such that if , then for . that is, when .
When . It follows that By the arbitrariness of , we get the inequality . Summing up, we obtain the equality .

Lemma 13. Let . Then for any and , there exists such that whenever and .

Proof. By , for any , there exists such that whenever .
For any , there is a such that . Put . Then whenever . Hence whenever and .

Corollary 14. An Orlicz sequence space is nearly uniformly smooth if and only if and .

Proof. Since and imply that the space is reflexive, we only need to prove the sufficiency. For any with finite that is , and with . For any , by Lemma 13, there exists such that when and . Hence if and This shows that , that is, when . Therefore .

Corollary 15. Let be a Lebesgue sequence space. Then .

Proof. It is well known that the Lebesgue sequence space is an Orlicz sequence space generated by . For any , we now consider the following: That is, Hence that is to say, . So for .

It is also well known that if is an Orlicz function, . If , has Schur property .

Theorem 16. Suppose that and . if and only if exist and such that or when .

Proof. Necessity. If there exists a sequence such that and for each , take such that Putting , then . So as . Therefore Take such that It is clear that for large enough. Let , Then and as .
Since that is, .
By we have as .
Therefore that is to say .
Sufficiency. Suppose that there exist and such that or when . That means there exist and such that or when . It is well known that if is an Orlicz function, . If , has Schur property. That is to say, the norm convergence of sequence in coincides with the weak convergence. In order to discuss the coefficient we may assume . Put . By , we have . Since the function is continuous in , there exists a such that for each . Suppose . Then and is affine function in , where . This contradicts with . Therefore . This implies when . It is shown that there exist and such that or when . Using there exists a such that when .
Put and Let and .
We consider the following two cases.
Case I (). In this case, from we get For any with , the following inequality holds: Therefore