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Research Article | Open Access

Volume 2013 |Article ID 873943 | https://doi.org/10.1155/2013/873943

Shaoqiang Shang, Yunan Cui, "Ball-Covering Property in Uniformly Non- Banach Spaces and Application", Abstract and Applied Analysis, vol. 2013, Article ID 873943, 7 pages, 2013. https://doi.org/10.1155/2013/873943

# Ball-Covering Property in Uniformly Non- Banach Spaces and Application

Revised09 Nov 2013
Accepted11 Nov 2013
Published08 Dec 2013

#### Abstract

This paper shows the following. (1) is a uniformly non- space if and only if there exist two constants such that, for every 3-dimensional subspace of , there exists a ball-covering of with or which is -off the origin and . (2) If a separable space has the Radon-Nikodym property, then has the ball-covering property. Using this general result, we find sufficient conditions in order that an Orlicz function space has the ball-covering property.

#### 1. Introduction

Let be a real Banach space. and denote the unit sphere and unit ball, respectively. denote the dual space of . Let denote the closed ball centered at and of radius . Let , , and denote the set of natural numbers, reals, and nonnegative reals, respectively.

It is no doubt that the study of geometric and topological properties of unit balls of normed spaces has played an important role in geometry of Banach spaces. Almost all properties of Banach spaces, such as convexity, smoothness, reflexivity, and the Radon-Nikodym property, can be viewed as the corresponding properties of their unit balls. We should mention here that there are many topics studying behavior of ball collections. For example, the Mazur intersection property, the packing sphere problem of unit balls, the measure of noncompactness with respect to topological degree, and the ball topology have also brought great attention of many mathematicians.

Starting with a different viewpoint, a notion of ball-covering property is introduced by Cheng [1].

Definition 1. A Banach space is said to have the ball-covering property if its unit sphere can be contained in the union of countably many balls off the origin. In this case, we also say that the norm has ball-covering property.

In [2], it was established that if is a locally uniformly convex space and is -separable, then has the ball-covering property. In [3], Cheng proved that by constructing the equivalent norms on , there exists a Banach space such that has not ball-covering property. In [4], it was established that for every every Banach space with a -separable dual has an -equivalent norm with the ball-covering property. For a ball-covering of , we denote by its cardinality and by the least upper bound of the radius set , and we call it the radius of . We say that a ball covering is minimal if its cardinality is the smallest of all cardinalities of ball coverings. We call a given ball covering -off the origin if . Let be any minimal ball covering of . Cheng [1] showed the following results.

Proposition 2. Suppose that is an -dimensional Banach space. Then (1);(2)if is smooth, then ;(3) if and only if is isometric to .

It is easy to see that is isometric to . Moreover, Cheng [5, 6] showed the following results.

Proposition 3. Suppose that is a Banach space. Then is a uniformly nonsquare space if and only if there exist two constants such that, for every 2-dimensional subspace of , there exists a ball-covering of with which is -off the origin and .

Definition 4. A Banach space is said to be non- space, if, for all ,

Definition 5. A Banach space is said to be uniformly non- space, if there exists such that, for all ,

Relationships between various kinds of convexity of Banach spaces and reflexivity have been developed by many authors. Giesy [7] and James [8] raised the question whether Banach spaces which are uniformly non- with some positive integer are reflexive. James [8] settled the question affirmatively for and gave a partial result for . Afterwards, the same author presented in [9] an example of a nonreflexive uniformly non- Banach space.

Definition 6. A Banach space is said to have the Radon-Nikodym property whenever if is a nonatomic measure space and is a vector measure on with values in which is absolutely continuous with respect to and has bounded variation; then there exists such that, for any ,

Let us recall some geometrical notions that will be used in the further part of this paper. A point is said to be a strongly exposed point of if there exists such that whenever . It is well known that Banach spaces have the Radon-Nikodym property if and only if every bounded closed convex subset of is the closed convex hull of its strongly exposed points. A point is said to be a smooth point if it has a unique supporting functional . If every is a smooth point, then is called smooth. Let be a nonempty open convex subset of and let be a real-valued continuous convex function on . Recall that is said to be Gateaux differentiable at the point in if the limit exists for all . When this is the case, the limit is a continuous linear function of , denoted by .

In this paper, firstly, we prove that is a uniformly non- nonsquare if and only if there exist two constants such that, for every -dimensional subspace of , there exists a ball-covering of with or 5 which is -off the origin and . Secondly, we will also prove that if a separable Banach space has the Radon-Nikodym property, then has the ball-covering property. Using this general result, we find sufficient conditions for an Orlicz function space to have ball-covering property. The topic of this paper is related to the topic of [1â€“6, 10â€“12].

#### 2. Main Results

Theorem 7. Suppose that is a Banach space. Then, is a uniformly non- space if and only if there exist two constants such that, for every -dimensional subspace of , there exists a ball-covering of with or which is -off the origin and .

In order to prove the theorem, we give some lemmas.

Lemma 8. Let and be sequences in . If , , and , then if and only if .

Proofâ€‰
Sufficiency. Let . By , we obtain that Moreover, we may assume without loss of generality that . Noticing that , we have Hence, we have .
Necessity. Let . By for any and the sufficiently part of the proof that has been just finished, we obtain that This implies that which completes the proof.

Lemma 9. If , , and are three sequences and â€‰â€‰+â€‰â€‰, then

Proof. Since , we obtain that and . Therefore, by and , we have Noticing that , we have . Similarly, we have . This implies that . Moreover, we may assume without loss of generality that . ThenMoreover, it is easy to see that This implies that which completes the proof.

Lemma 10. Suppose that is a Banach space. Then, is a uniformly non- space if and only if there exists such that, for every -dimensional subspace of , if is a linear isomorphism, then .

Proofâ€‰
Necessity. Suppose that, for any natural number , there exist a 3-dimensional subspace of and a linear operator such that is a linear isomorphism and . We may assume without loss of generality that . Moreover, it is easy to see that there exist such that By and , we have Let Then Therefore, by (13) and (14), we have a contradiction. This implies that if there exists such that, for every 3-dimensional subspace of , if is a linear isomorphism, then .
Sufficiency. Suppose that is not a uniformly non- space. Then, for any natural number , there exist such that We define the subspace of . We claim that . In fact, suppose that . By (18), we obtain that . Hence, for any natural number , we may assume without loss of generality that and are linearly independent. Notice that , so we obtain that . This implies that there exist and such that . By (18) and , we may assume without loss of generality that . By and Lemma 8, we have . Hence, we have a contradiction. This implies that , , and are linearly independent. Then, . We define the linear operator by the formula It is easy to see that is one-one mapping.
Next, we will prove that . In fact, it is easy to see that, for any natural number , we have . Suppose that . Then, we may assume without loss of generality that there exists such that . This implies that there exists a sequence such that By (18), we may assume without loss of generality that , , and . By Lemma 8, we have Moreover, by Lemma 9, we have By , we obtain Therefore, by (22)â€“(24) and Lemma 8, we obtain . Noticing that , we have a contradiction. This implies that .
Moreover, we claim that . In fact, it is easy to see that, for any natural number , we have . Suppose that . Then, we may assume without loss of generality that there exists such that . This implies that there exists a sequence such that By (18), we may assume without loss of generality that , , and . Noticing that , we have . Therefore, by (22)â€“(24) and Lemma 8, we have a contradiction. This implies that . Thus , a contradiction. Hence, we obtain that is a uniformly non- space, which completes the proof.

Lemma 11. Suppose that (1) there exists a ball-covering of and , (2) is uniformly convergent to in and (3) is -off the origin for any . Then, as .

Proof. Let , , and . Suppose that there exists such that . Then, and are bounded sequences. Hence, we may assume without loss of generality that and in for any . Then, as . We claim that In fact, for any and , let Since is uniformly convergent to in , we obtain that, for any bounded set, is uniformly convergent to . Moreover, it is easy to see that This implies that
Since , then, for any , there exists a sequence such that as . Since , then there exists such that for any . Noticing that
we have . This implies that . By Proposition 2, we have , a contradiction. This implies that as , which completes the proof.

Proof of Theorem 7â€‰
Sufficiency. It is easy to see that there exist two constants such that, for every -dimensional subspace of , there exists a ball-covering of with which is -off the origin and . Suppose that is not a uniformly non- space. By Lemma 10, for any natural number , there exist a 3-dimensional subspace of and a linear operator such that is a linear isomorphism and . Since and are isomorphism, there exists a linear operator such that is a linear isomorphism and Moreover, we may assume without loss of generality that . Let and let be a ball covering of , where which is -off the origin and . It is easy to see that is uniformly convergent to in . By Lemma 11, we have that as , a contradiction. Hence, we obtain that is a uniformly non- space.
Necessity. By Lemma 10, we have or . Using the method of Theoremâ€‰â€‰3.5 in [6] and Lemma 10, similarly, we obtain that is -off the origin and , which completes the proof.

Theorem 12. Suppose that is a uniformly non- space and smooth space. Then, there exist two constants such that for every 3-dimensional subspace of , there exists a minimal ball-covering of with which is -off the origin and .

Proof. By Proposition 2, we obtain that there exist two constants such that, for every 3-dimensional subspace of , there exists a minimal ball-covering of with . Using the method of Theoremâ€‰â€‰3.5 in [6] and Lemma 10, similarly, we obtain that is -off the origin and , which completes the proof.

Theorem 13. Suppose that there exist two constants such that, for every 3-dimensional subspace of , there exists a minimal ball-covering of with which is -off the origin and . Then, is a uniformly non- space and not smooth space.

Proof. By Theorem 7, we obtain that is a uniformly non- space. Suppose that is a smooth space. Then, every -dimensional subspace of is smooth. By Proposition 2, we obtain that . However, there exist two constants such that, for every -dimensional subspace of , there exists a minimal ball-covering of with which is -off the origin and , a contradiction. Hence, we obtain that is not a smooth space, which completes the proof.

The following theorem (Theorem 15) shows that if a separable space has the Radon-Nikodym property, then has the ball-covering property. We first need a lemma.

Lemma 14 (see [5]). Suppose that is a Minkowski functional defined on the space . Then, is Gateaux differentiable at and with the Gateaux derivative if and only if is a -exposed point of and exposed by , where is the polar of the level set .

Theorem 15. Suppose that separable space has the Radon-Nikodym property. Then, has the ball-covering property.

Proof. (a) First we will prove that there exists a sequence of -exposed points of such that Since has the Radon-Nikodym property, then the closed convex hull of is the whole , where denotes strongly exposed points of .
Pick . Since is a strongly exposed point of , there exists such that whenever . Next we will prove that is -exposed point of and exposing by . In fact, suppose that there exists such that . Since topology is a Hausdorff topology, there exist a neighbourhood of and a neighbourhood of such that . Define the neighbourhood as follows: By the Goldstine theorem, there exists such that . Hence, we have that as . Since is a strongly exposed point of , we obtain that as . This implies that , when is large enough. This contradicts the fact that . Hence, we obtain that is a -exposed point of .
Since is a separable space, then there exists a sequence such that is a dense sequence in . Noticing that , we have
(b) Next we will prove that has the ball-covering property. By Lemma 14, for any , there exist such that is Gateaux differentiable at and with the Gateaux derivative . For each fixed , let be the balls defined by Clearly, every has the distance from the origin. We claim that In fact, pick . Noticing that we obtain that, for , there exists such that We can assume that for some . Thus, there exist and such that We want to show that . Otherwise, for every , Thus, where . Letting , we observe that which is a contradiction. Therefore, Hence, has the ball-covering property, which completes the proof.

Corollary 16. If is a separable space, then has the ball-covering property.

Proof. If is separable, then has the Radon-Nikodym property. By Theorem 13, we obtain that has the ball-covering property, which completes the proof.

#### 3. Applications to Orlicz Function Spaces

It is easy to see that if is separable, then has the ball-covering property. Cheng [1] proved that the sequence space which is not separable has the ball-covering property. In this section, we obtain that there exists a nonseparable function space such that it has the ball-covering property.

Definition 17. is called an -function if it has the following properties: (1) is even, convex and ;(2) for all ;(3) and .

Let be a finite nonatomic and complete measure space. Denote by and the right derivative of and , respectively. We define It is well known that the Orlicz function space is a Banach space when it is equipped with the Luxemburg norm or equipped with the Amemiya-Orlicz norm

Let denote the right derivative of at and let be the generalized inverse function of defined on by Then, we call the complementary function of . It is well known that there holds the Young inequality and or . Moreover, it is well known that and are complementary to each other.

We say that an -function if there exist and such that By [13], we know that is separable, has the Radon-Nikodym property , and is separable. Moreover, by [13], we know that and .

Theorem 18. If or , then has the ball-covering property.

Proof. By [13], we know that . Using Theorem 15, we obtain that if , then has the ball-covering property. Moreover, by , we obtain that is separable. Hence, has the ball-covering property, which completes the proof.

Remark 19. It is well known that there exists an -function such that and . This means that is not a separable space. However, by Theorem 18, we obtain that has ball-covering property.

#### Acknowledgment

This research is supported by The Fundamental Research Funds for the Central Universities, DL12BB36 and by Heilongjiang Provincial Department of Education Funds 12521070.

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Copyright © 2013 Shaoqiang Shang and Yunan Cui. 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.