A Generalization of Uniformly Extremely Convex Banach Spaces

Wulede, Suyalatu; Bai, Wurichaihu; Bao, Wurina

doi:https://doi.org/10.1155/2016/9161252

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Function Spaces, Approximation Theory, and Their Applications

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Volume 2016 | Article ID 9161252 | https://doi.org/10.1155/2016/9161252

A Generalization of Uniformly Extremely Convex Banach Spaces

Suyalatu Wulede,¹Wurichaihu Bai,¹and Wurina Bao²

Academic Editor: Gianluca Vinti

Received19 Jan 2016

Accepted16 Mar 2016

Published10 Apr 2016

Abstract

We discuss a new class of Banach spaces which are the generalization of uniformly extremely convex spaces introduced by Wulede and Ha. We prove that the new class of Banach spaces lies strictly between either the classes of -uniformly rotund spaces and -strongly convex spaces or classes of fully -convex spaces and -strongly convex spaces and has no inclusive relation with the class of locally -uniformly convex spaces. We obtain in addition some characterizations and properties of this new class of Banach spaces. In particular, our results contain the main results of Wulede and Ha.

1. Introduction

Different uniformly convex spaces have been defined between the uniformly convex spaces [1] and the reflexivity of the Banach spaces [2–6]. In the previous paper [7] we introduce a new class of this type, namely, uniformly extremely convex spaces. This new class of Banach spaces lies strictly between either the classes of uniformly convex spaces and strongly convex spaces or the classes of fully -convex spaces and strongly convex spaces.

Here we consider another new class of this type, namely, -uniformly extremely convex spaces, as a generalization of uniformly extremely convex spaces and discuss its relation to the drop property, the -uniformly rotund spaces, the full -convex spaces, the -strongly convex spaces, the nearly uniformly convex spaces, and -nearly uniformly convex spaces. We also give some characterizations of -uniformly extremely convex spaces and find that this new class of Banach spaces has the following features:(1)-uniformly extremely convex spaces (indeed lower case) coincide with uniformly extremely convex spaces;(2)-uniformly extremely convex spaces possess the drop property;(3)-uniformly extremely convex spaces are -uniformly extremely convex spaces, but the converse implication is not true.

Throughout this paper denotes an infinite-dimensional real Banach space with the norm . The symbol denotes the dual of the space . and denote the closed unit ball and the unit sphere of , respectively. The symbol denotes the unit sphere of . The symbol denotes the weak topology of .

Let be norm-1 elements in Banach spaces . The -dimensional volume enclosed by is given by Here, and throughout the sequel, the symbol denotes the determinant.

Sullivan [6] has introduced the -uniformly rotund (UR) spaces and locally -uniformly rotund (LUR) spaces. Fan and Glicksberg [2] have introduced the fully -convex (R) Banach spaces. It is well known that UR and R spaces imply reflexivity. About UR and R spaces, we have the following chain of implications [2, 6, 8]:

A Banach space is said to be a UR space [6] if, for any , there exists a such that, for all norm-1 elements and , we have

A Banach space is said to be a R space [2] if, for any sequence in such that , then is a Cauchy sequence in .

A point is said to be a denting point of [8] if for all , where

Huff [3] has introduced the nearly uniformly convex (NUC) spaces as a generalization of uniformly convex Banach spaces and showed that the NUC spaces are equivalent to reflexive spaces possessing the uniform Kadec-Klee property. The local version of NUC spaces, namely, locally nearly uniformly convex (LNUC), was studied by Kutzarova and Lin [9]. Kutzarova [4] introduced the -nearly uniformly convex (NUC) spaces as a generalization of nearly uniformly convex Banach spaces. In [4, 9], it is pointed out that NUC LNUC and NUC NUC for every .

A Banach space is said to be a NUC [3] space if, for any , there exists a such that, for any sequence , , we have , where and means the convex hull of .

A Banach space is said to be a LNUC [9] space if, for any norm-1 element and , there exists a such that, for any sequence , , we have , where means the convex hull of and .

A Banach space is said to be a NUC [4] space, if for any there exists a such that, for any sequence , , there are indices and scalars , , with so that

Singer [10] has introduced the -strictly convex spaces. It is well known that -strictly convex spaces are -strictly convex spaces; 1-strictly convex spaces (indeed lower case) coincide with strictly convex spaces; R spaces are -strictly convex spaces and have the drop property. Wu and Li [11] have introduced the strongly convex spaces. Wulede and Wu [12] introduced the -strongly convex spaces as a generalization of strongly convex Banach spaces and gave an equivalent definition of -strongly convex spaces (see Theorem 5 in [13]). It is well known that -strongly convex spaces are -strictly convex spaces; 1-strongly convex spaces (indeed lower case) coincide with strongly convex spaces; -strongly convex spaces are -strongly convex spaces, but the converse implication is not true.

A Banach space is said to be a -strictly convex space [10] if, for all norm-1 elements such that , then are linearly dependent.

A Banach space is said to be a strongly convex space [11] if, for any , and for a certain functional such that , then , where , .

A Banach space is said to be a -strongly convex space [12] if, for any norm-1 element , and for any functional , there is a such that, for all norm-1 elements and , we have .

Rolewicz [14] has defined the norm to have the drop property, if for every closed set disjoint from there exists such that , where the set , the convex hull of and , is called the drop generated by .

Lemma 1 (Kadec-Klee property). If any , such that , , and , , then , , where , , means that , , for all .

Lemma 2 (Montesinos [15]). Let be a Banach space. Then has the drop property if and only if is reflexive and has the Kadec-Klee property.

Lemma 3 (Nan and Wang [16]). is -strictly convex space if and only if, for any , one has , where ,

Lemma 4 (Wulede and Wu [12], Zhang and Fang [17]). Let be a Banach space.(i)If is -strongly convex, then is -strictly convex and has the Kadec-Klee property.(ii)If is reflexive, -strictly convex and has the Kadec-Klee property, then is -strongly convex.(iii)If is -strongly convex, , , and , , then , .

Lemma 5 (Zhang and Fang [17]). is -strongly convex if and only if, for any and , , there exist and a compact set with such that , , where the set , is the slice generated by and .

2. -Uniformly Extremely Convex Spaces and Drop Property

Definition 6 (see [7]). A Banach space is said to be a uniformly extremely convex space if, for any sequences consisting of elements of norm-1 and for a certain functional of norm-1, holds; then

On the base of uniformly extremely convex spaces, now we introduce the notion of -uniformly extremely convex spaces as a generalization of uniformly extremely convex spaces.

Definition 7. A Banach space is said to be a -uniformly extremely convex space if, for any sequences consisting of elements of norm-1 and for a certain functional of norm-1, holds; then .

We give first a simple result which shows that the notion of -uniformly extremely convex space is “coherent.”

Theorem 8. If is -uniformly extremely convex space, then is -uniformly extremely convex space.

Proof. If, for any sequences consisting of elements of norm-1 and for a certain functional of norm-1, holds, then, for all , we have by the assumption that is -uniformly extremely convex space. Furthermore, by the properties of determinant we havethis shows that is -uniformly extremely convex space.

Now we give a simple but useful lemma. By using this lemma we can prove that any -uniformly extremely convex space has the drop property. And the fact that -uniformly extremely convex spaces include -strongly convex spaces can be easily found. We also show that -uniformly extremely convex spaces coincide with uniformly extremely convex spaces by using this lemma.

Lemma 9. is -uniformly extremely convex if and only if, for any , there exists a such that, for all norm-1 elements and , one has .

Proof.
Necessity. Suppose the contrary. Then there exist and such that, for any , we have , but . Take ; then and . It follows that . On the other hand, by the definition of the -uniformly extremely convex space, we have ; this contradicts the statement that .
Sufficiency. If, for any sequences consisting of elements of norm-1 and for a certain functional of norm-1, holds, then . Therefore, for any , there exists an integer such that, for all , inequality holds. For any , by the conditions given in Lemma 9, we have ; this means that .

Remark 10. -uniformly extremely convex space (indeed lower case) coincides with uniformly extremely convex space.
In fact, by Lemma 9 we know that is -uniformly extremely convex space if and only if, for any , , there exists a such that, for any norm-1 elements and , we have and also if and only if is uniformly extremely convex space.

Theorem 11. is -uniformly extremely convex space if and only if is -strictly convex space and has the drop property.

Proof.
Necessity. Suppose that is -uniformly extremely convex space; by the definition of -strongly convex space and a condition which characterizes -uniformly extremely convex space in Lemma 9, it is easy to see that is -strongly convex space. From Lemma 4(i), we know that is -strictly convex space and has the Kadec-Klee property.
Now we are going to prove that has the drop property. In fact, from Lemma 2, it is sufficient to prove that is reflexive. Suppose that is not reflexive. Using the well-known James’ theorem, for each , we can choose so that and , and , so that Here is the function required in Lemma 9.
Now we have that On the other hand it is easy to check that which gives the required contradiction.
Sufficiency. From the assumption that is -strictly convex space and has the drop property, we can deduce, by Lemma 4(ii), that is -strongly convex space and reflexive. Observing the definition of -strongly convex space and a condition which characterizes -uniformly extremely convex space in Lemma 9, by the reflexivity of , it is easy to see that is -uniformly extremely convex space.

Corollary 12 (see [7]). is uniformly extremely convex space if and only if is strictly convex space and has the drop property.

Noticing the procedure of proving Theorem 11 we can deduce the following.

Corollary 13. If is -uniformly extremely convex space, then is -strongly convex space.

Now we are going to show that the converse to Corollary 13 is not true. In [12], it is proved that LUR spaces are -strongly convex spaces. In general, LUR spaces need not be reflexive since LUR is just the usual definition of LUR space [18, 19]. It follows that there exists a -strongly convex space which is not reflexive. Hence is not a -uniformly extremely convex space since is not reflexive.

Corollary 14. is -uniformly extremely convex space if and only if is reflexive and, for any and , , there exist and a compact set with such that , where the set is the slice generated by and .

Proof. It is immediate from Corollary 13, Theorem 11, and Lemmas 2, 4, and 5.

Theorem 15. is -uniformly extremely convex space if and only if is reflexive and, for any , one has , , where the set , which includes set , is arbitrary open set with regard to norm topology .

Proof.
Necessity. Suppose that is -uniformly extremely convex space; then by Theorem 11 we know that is -strictly convex space and reflexive. For any , by the reflexivity of , there exists such that ; hence . Combining the fact that is -strictly convex space with Lemma 3 we can deduce that .
Now we are going to prove the equality .
Firstly, we will prove that, for any and every open set (where ) with regard to norm topology , there exists a scalar such that dist.
Noticing that is compact set in , for any , we can deduce that there exists such that . Now we claim that there exists minimum value of denoted by , such that for any . In fact, if does not have minimum value, then is impossible to be minimum value for any integer . Hence, there exist and such that . Since is compact, the above sequence has the convergent subsequence; without loss of generality and letting the convergent subsequence be itself, then . Noticing that , we can deduce that .
On the other hand, combining the fact that is closed set with , we can deduce that This contradicts .
Secondly, we will prove that for there exists a scalar such that the inequality holds for all and .
If the above inequality is not true, then there exists such that . By the condition given in Theorem 15, Corollary 13, and Lemma 4(iii), we have . On the other hand, by , we can deduce that ; this contradicts the statement that . Hence we have this shows that . By the arbitrary of , we can deduce that .
Sufficiency. By Lemmas 2 and 3, Theorem 11, and the condition given in Theorem 15, only we need to prove that has the Kadec-Klee property. Let , , and . By the well-known James’ theorem, there exists such that ; it follows that .
Case 1. If , then is relatively compact. Otherwise, every point of is not accumulation point of . Hence, for any there exists such that does not contain any point of . We construct an open set with regard to norm topology ; then and . Since is bounded closed convex set with regard to norm topology , . Noticing that is bounded set with regard to weak topology , we know that is compact set with regard to weak topology . Hence there is a function which separates and ; that is, there is a scalar such that . Evidently, ; it follows that . This contradicts the assumption that .
Case 2. If , then . According to Case 1 we know that is relatively compact set. Hence is compact set since is bounded closed convex set in certain finite dimensional subspace of . On the other hand, it is obvious that ; hence . This shows that is relatively compact.
Consequently, in Cases 1 and 2, we always conclude that is relatively compact. Furthermore, by the assumption that , we can deduce that . This completes the proof that has the Kadec-Klee property.

In particular, considering the special case of Theorem 15 when , we obtained Theorem 2.5 in [7] as a corollary.

Corollary 16. is uniformly extremely convex space if and only if is reflexive and for any , one has , , where the set , which includes set , is arbitrary open set with regard to norm topology . In other words, is uniformly extremely convex space if and only if is reflexive and every point of is a denting point of .

To show that the converse to Theorem 8 is not true, we consider the following example.

Example 17. There exists a -uniformly extremely convex space which is not a -uniformly extremely convex space.

Let be an integer, and let . For each , define From [20] we know that is a UR space. It is easy to see that UR space is -uniformly extremely convex space from the definition of UR space and a condition which characterizes -uniformly extremely convex space in Lemma 9. Hence is a -uniformly extremely convex space. It follows from Theorem 11 that is a -strictly convex space but is not a -strictly convex space that follows from [16]. Hence is not a -uniformly extremely convex space.

3. The Relations between -Uniformly Extremely Convex Space and Various Other Types of Convex Space

Now we give a list of examples to distinguish -uniformly extremely convex spaces from R, UR, NUC, and NUC spaces.(i)We are ready now to distinguish -uniformly extremely convex and R spaces.

Since R spaces are -strictly convex spaces and have the drop property, it follows from Theorem 11 that R spaces are -uniformly extremely convex, but the converse is not true.

Example 18. There exists a -uniformly extremely convex space which is not a R space for every .

Let be an integer, and let . For each , define and let For , let , . It follows from [20] that is a UR space but is not a R space. Hence is a -uniformly extremely convex space since is a UR space. (ii)We are ready now to distinguish -uniformly extremely convex and UR spaces.

Example 19. For all , there exists a -uniformly extremely convex space which is not a UR space.

Let ; for , define It follows from [2] that is a 2R space; furthermore, is a -uniformly extremely convex space but is not a UR space [20]. (iii)We are ready now to distinguish -uniformly extremely convex and LUR spaces.

We consider a nonreflexive LUR space . Then is not a -uniformly extremely convex space since is not reflexive. On the other hand, we consider Example 19; then is a 2R space and it follows that is a -uniformly extremely convex space for all . But is not a LUR space that follows from [21].(iv)We are ready now to distinguish -uniformly extremely convex spaces and NUC or NUC spaces.

Example 20. For all , there exists a -uniformly extremely convex space which is neither a NUC nor a NUC space for all .

Let be the -sum of ; then is a 2R space with normalized basis . Define, , By Theorem 4 in [9], we know that is a 2R space but is not a LNUC space. It follows that is a -uniformly extremely convex space for all but is neither a NUC nor a NUC space for all .

Remark 21. (i) The class of -uniformly extremely convex spaces lies strictly between the classes of UR spaces and the -strongly convex spaces.
(ii) The class of -uniformly extremely convex spaces lies strictly between the classes of R spaces and the class of -strongly convex spaces.
(iii) The class of -uniformly extremely convex spaces has no inclusive relation with the class of LUR spaces.

In particular, considering the special case of Remark 21 when , we obtained the main conclusions of [7], that is, Remarks 3.5 and 3.7 in [7].

Competing Interests

The authors declare that they have no competing interests.

Acknowledgments

This work was supported by the National Natural Science Foundation of China (Grant no. 11561053).

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Copyright © 2016 Suyalatu Wulede et al. 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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