Ball-Covering Property in Uniformly Non-𝑙₃⁽¹⁾ Banach Spaces and Application

Abstract

This paper shows the following. (1) 𝑋 is a uniformly non-𝑙₃⁽¹⁾ space if and only if there exist two constants 𝛼,𝛽>0 such that, for every 3-dimensional subspace 𝑌 of 𝑋, there exists a ball-covering 𝔅 of 𝑌 with 𝑐(𝔅)=4 or 5 which is 𝛼-off the origin and 𝑟(𝔅)≤𝛽. (2) If a separable space 𝑋 has the Radon-Nikodym property, then 𝑋^∗ 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 (𝑋,‖⋅‖) be a real Banach space. 𝑆(𝑋) and 𝐵(𝑋) denote the unit sphere and unit ball, respectively. 𝑋^∗ denote the dual space of 𝑋. Let 𝐵(𝑥,𝑟) denote the closed ball centered at 𝑥 and of radius 𝑟>0. Let 𝑁, 𝑅, and 𝑅⁺ 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 𝑋 is a locally uniformly convex space and 𝐵(𝑋^∗) is 𝑤^∗-separable, then 𝑋 has the ball-covering property. In [3], Cheng proved that by constructing the equivalent norms on 𝑙^∞, there exists a Banach space (𝑙^∞,‖⋅‖⁰) such that (𝑙^∞,‖⋅‖⁰) has not ball-covering property. In [4], it was established that for every 𝜀>0 every Banach space with a 𝑤^∗-separable dual has an 1+𝜀-equivalent norm with the ball-covering property. For a ball-covering 𝔅={𝐵(𝑥_𝑖;𝑟_𝑖)}_𝑖∈𝐼 of 𝑋, we denote by 𝑐(𝔅) its cardinality and by 𝑟(𝔅) the least upper bound of the radius set {𝑟_𝑖}_𝑖∈𝐼, and we call it the radius of 𝔅. 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 𝔅𝛼-off the origin if inf{‖𝑥‖∶𝑥∈∪𝔅}≥𝛼. Let 𝔅_min=𝔅_min(𝑋) be any minimal ball covering of 𝑋. Cheng [1] showed the following results.

Proposition 2. Suppose that 𝑋 is an 𝑛-dimensional Banach space. Then (1)𝑛+1≤𝑐(𝔅_min)≤2𝑛;(2)if 𝑋 is smooth, then 𝑐(𝔅_min)=𝑛+1;(3)𝑐(𝔅_min)=2𝑛 if and only if 𝑋 is isometric to (𝑅^𝑛,‖⋅‖_∞).

It is easy to see that (𝑅^𝑛,‖⋅‖₁) is isometric to (𝑅^𝑛,‖⋅‖_∞). Moreover, Cheng [5, 6] showed the following results.

Proposition 3. Suppose that 𝑋 is a Banach space. Then 𝑋 is a uniformly nonsquare space if and only if there exist two constants 𝛼,𝛽>0 such that, for every 2-dimensional subspace 𝑌 of 𝑋, there exists a ball-covering 𝔅 of 𝑌 with 𝑐(𝔅)=3 which is 𝛼-off the origin and 𝑟(𝔅)≤𝛽.

Definition 4. A Banach space 𝑋 is said to be non-𝑙_𝑛⁽¹⁾ space, if, for all 𝑥₁,𝑥₂,…,𝑥_𝑛∈𝑆(𝑋), min𝜉₁𝑥₁+𝜉₂𝑥₂+⋯+𝜉_𝑛𝑥_𝑛∶𝜉_𝑖=±1,𝑖∈{1,2,…,𝑛}<𝑛.(1)

Definition 5. A Banach space 𝑋 is said to be uniformly non-𝑙_𝑛⁽¹⁾ space, if there exists 𝛿>0 such that, for all 𝑥₁,𝑥₂,…,𝑥_𝑛∈𝑆(𝑋), min𝜉₁𝑥₁+𝜉₂𝑥₂+⋯+𝜉_𝑛𝑥_𝑛∶𝜉_𝑖=±1,𝑖∈{1,2,…,𝑛}<𝑛−𝛿.(2)

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-𝑙_𝑛⁽¹⁾ with some positive integer 𝑛≥2 are reflexive. James [8] settled the question affirmatively for 𝑛=2 and gave a partial result for 𝑛=3. Afterwards, the same author presented in [9] an example of a nonreflexive uniformly non-𝑙₃⁽¹⁾ Banach space.

Definition 6. A Banach space 𝑋 is said to have the Radon-Nikodym property whenever if (𝑇,Σ,𝜇) is a nonatomic measure space and 𝑣 is a vector measure on Σ with values in 𝑋 which is absolutely continuous with respect to 𝜇 and has bounded variation; then there exists 𝑓∈𝐿₁(𝑋) such that, for any 𝐴∈Σ, 𝑣(𝐴)=_𝐴𝑓(𝑡)𝑑𝑡.(3)

Let us recall some geometrical notions that will be used in the further part of this paper. A point 𝑥∈𝐶 is said to be a strongly exposed point of 𝐶 if there exists 𝑥^∗∈𝑋^∗ such that 𝑥_𝑛→𝑥 whenever 𝑥^∗(𝑥_𝑛)→𝑥^∗(𝑥)=sup{𝑥^∗(𝑥)∶𝑥∈𝐶}. It is well known that Banach spaces have the Radon-Nikodym property if and only if every bounded closed convex subset of 𝑋 is the closed convex hull of its strongly exposed points. A point 𝑥∈𝑆(𝑋) is said to be a smooth point if it has a unique supporting functional 𝑓_𝑥. If every 𝑥∈𝑆(𝑋) is a smooth point, then 𝑋 is called smooth. Let 𝐷 be a nonempty open convex subset of 𝑋 and let 𝑓 be a real-valued continuous convex function on 𝐷. Recall that 𝑓 is said to be Gateaux differentiable at the point 𝑥 in 𝐷 if the limit 𝑓^′(𝑥)𝑦=lim_𝑡→0𝑓𝑥+𝑡𝑦−𝑓(𝑥)𝑡(∗) exists for all 𝑦∈𝑋. When this is the case, the limit is a continuous linear function of 𝑦, denoted by 𝑓^′(𝑥).

In this paper, firstly, we prove that 𝑋 is a uniformly non-𝑙₃⁽¹⁾ nonsquare if and only if there exist two constants 𝛼,𝛽>0 such that, for every 3-dimensional subspace 𝑌 of 𝑋, there exists a ball-covering 𝔅 of 𝑌 with 𝑐(𝔅)=4 or 5 which is 𝛼-off the origin and 𝑟(𝔅)≤𝛽. Secondly, we will also prove that if a separable Banach space 𝑋 has the Radon-Nikodym property, then 𝑋^∗ 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 [1–6, 10–12].

2. Main Results

Theorem 7. Suppose that 𝑋 is a Banach space. Then, 𝑋 is a uniformly non-𝑙₃⁽¹⁾ space if and only if there exist two constants 𝛼,𝛽>0 such that, for every 3-dimensional subspace 𝑌 of 𝑋, there exists a ball-covering 𝔅 of 𝑌 with 𝑐(𝔅)=4 or 5 which is 𝛼-off the origin and 𝑟(𝔅)≤𝛽.

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

Lemma 8. Let {𝑥_𝑛}^∞_𝑛=1 and {𝑦_𝑛}^∞_𝑛=1 be sequences in 𝑋. If 𝛼_𝑛≥0, 𝛽_𝑛≥0, and lim_𝑛→∞‖𝑥_𝑛+𝑦_𝑛‖=lim_𝑛→∞‖2𝑥_𝑛‖=lim_𝑛→∞‖2𝑦_𝑛‖=2, then lim_𝑛→∞‖𝛼_𝑛𝑥_𝑛+𝛽_𝑛𝑦_𝑛‖=1 if and only if lim_𝑛→∞(𝛼_𝑛+𝛽_𝑛)=1.

Proof 
Sufficiency. Let lim_𝑛→∞(𝛼_𝑛+𝛽_𝑛)=1. By lim_𝑛→∞‖𝑥_𝑛‖=lim_𝑛→∞‖𝑦_𝑛‖=1, we obtain that limsup_𝑛→∞𝛼_𝑛𝑥_𝑛+𝛽_𝑛𝑦_𝑛≤limsup_𝑛→∞𝛼_𝑛𝑥_𝑛+𝛽_𝑛𝑦_𝑛=limsup_𝑛→∞𝛼_𝑛+𝛽_𝑛=1.(4) Moreover, we may assume without loss of generality that 𝛼_𝑛≥𝛽_𝑛. Noticing that lim_𝑛→∞‖𝑥_𝑛+𝑦_𝑛‖=lim_𝑛→∞‖2𝑥_𝑛‖=lim_𝑛→∞‖2𝑦_𝑛‖=2, we have liminf_𝑛→∞𝛼_𝑛𝑥_𝑛+𝛽_𝑛𝑦_𝑛=liminf_𝑛→∞𝛼_𝑛𝑥_𝑛+𝑦_𝑛−𝛼_𝑛−𝛽_𝑛𝑦_𝑛≥liminf_𝑛→∞𝛼_𝑛𝑥_𝑛+𝑦_𝑛−𝛼_𝑛−𝛽_𝑛𝑦_𝑛=liminf_𝑛→∞𝛼_𝑛+𝛽_𝑛=1.(5) Hence, we have lim_𝑛→∞‖𝛼_𝑛𝑥_𝑛+𝛽_𝑛𝑦_𝑛‖=1.
Necessity. Let lim_𝑛→∞‖𝛼_𝑛𝑥_𝑛+𝛽_𝑛𝑦_𝑛‖=1. By (𝛼_𝑛+𝛽_𝑛)⁻¹(𝛼_𝑛+𝛽_𝑛)=1 for any 𝑛∈𝑁 and the sufficiently part of the proof that has been just finished, we obtain that lim_𝑛→∞𝛼_𝑛𝛼_𝑛+𝛽_𝑛𝑥_𝑛+𝛽_𝑛𝛼_𝑛+𝛽_𝑛𝑦_𝑛=1.(6) This implies that 1=lim_𝑛→∞𝛼_𝑛𝑥_𝑛+𝛽_𝑛𝑦_𝑛=lim_𝑛→∞𝛼_𝑛+𝛽_𝑛𝛼_𝑛𝛼_𝑛+𝛽_𝑛𝑥_𝑛+𝛽_𝑛𝛼_𝑛+𝛽_𝑛𝑦_𝑛=lim_𝑛→∞𝛼_𝑛+𝛽_𝑛,(7) which completes the proof.

Lemma 9. If {𝑥_𝑛}^∞_𝑛=1, {𝑦_𝑛}^∞_𝑛=1, and {𝑧_𝑛}^∞_𝑛=1 are three sequences and lim_𝑛→∞min{‖𝜉₁𝑥_𝑛  +  𝜉₂𝑦_𝑛+𝜉₃𝑧_𝑛‖∶𝜉_𝑖=±1}=lim_𝑛→∞‖3𝑥_𝑛‖=lim_𝑛→∞‖3𝑦_𝑛‖=lim_𝑛→∞‖3𝑧_𝑛‖=3, then lim_𝑛→∞𝑧_𝑛+𝛼_𝑛𝛼_𝑛+𝛽_𝑛𝑥_𝑛+𝛽_𝑛𝛼_𝑛+𝛽_𝑛𝑦_𝑛=2.(8)

Proof. Since lim_𝑛→∞min{‖𝜉₁𝑥_𝑛+𝜉₂𝑦_𝑛+𝜉₃𝑧_𝑛‖∶𝜉_𝑖=±1}=lim_𝑛→∞‖3𝑥_𝑛‖=lim_𝑛→∞‖3𝑦_𝑛‖=lim_𝑛→∞‖3𝑧_𝑛‖=3, we obtain that liminf_𝑛→∞‖𝑥_𝑛+𝑦_𝑛+𝑧_𝑛‖=3 and liminf_𝑛→∞‖𝑥_𝑛−𝑦_𝑛+𝑧_𝑛‖=3. Therefore, by lim_𝑛→∞‖𝑥_𝑛‖=1 and lim_𝑛→∞‖𝑦_𝑛‖=1, we have 2=lim_𝑛→∞𝑥_𝑛+lim_𝑛→∞𝑦_𝑛≥liminf_𝑛→∞𝑥_𝑛+𝑦_𝑛≥liminf_𝑛→∞𝑥_𝑛+𝑦_𝑛+𝑧_𝑛−𝑧_𝑛≥liminf_𝑛→∞𝑥_𝑛+𝑦_𝑛+𝑧_𝑛−limsup_𝑛→∞𝑧_𝑛=2.(9) Noticing that limsup_𝑛→∞‖𝑥_𝑛+𝑦_𝑛‖≤2, we have lim_𝑛→∞‖𝑥_𝑛+𝑦_𝑛‖=2. Similarly, we have lim_𝑛→∞‖𝑥_𝑛−𝑦_𝑛‖=2. This implies that lim_𝑛→∞min{‖𝑥_𝑛−𝑦_𝑛‖,‖𝑥_𝑛+𝑦_𝑛‖}=2. Moreover, we may assume without loss of generality that 𝛼_𝑛≥𝛽_𝑛. Thenliminf_𝑛→∞𝑧_𝑛+𝛼_𝑛𝛼_𝑛+𝛽_𝑛𝑥_𝑛+𝛽_𝑛𝛼_𝑛+𝛽_𝑛𝑦_𝑛=liminf_𝑛→∞𝑥_𝑛+𝑦_𝑛+𝑧_𝑛−𝛽_𝑛𝛼_𝑛+𝛽_𝑛𝑥_𝑛−𝑦_𝑛−𝛽_𝑛−𝛼_𝑛𝛼_𝑛+𝛽_𝑛𝑦_𝑛≥liminf_𝑛→∞𝑥_𝑛+𝑦_𝑛+𝑧_𝑛−𝛽_𝑛𝛼_𝑛+𝛽_𝑛𝑥_𝑛−𝑦_𝑛−𝛼_𝑛−𝛽_𝑛𝛼_𝑛+𝛽_𝑛𝑦_𝑛=liminf_𝑛→∞3−2𝛽_𝑛𝛼_𝑛+𝛽_𝑛−𝛼_𝑛−𝛽_𝑛𝛼_𝑛+𝛽_𝑛=2.(10)Moreover, it is easy to see that limsup_𝑛→∞𝑧_𝑛+𝛼_𝑛𝛼_𝑛+𝛽_𝑛𝑥_𝑛+𝛽_𝑛𝛼_𝑛+𝛽_𝑛𝑦_𝑛≤2.(11) This implies that lim_𝑛→∞𝑧_𝑛+𝛼_𝑛𝛼_𝑛+𝛽_𝑛𝑥_𝑛+𝛽_𝑛𝛼_𝑛+𝛽_𝑛𝑦_𝑛=2,(12) which completes the proof.

Lemma 10. Suppose that 𝑋 is a Banach space. Then, 𝑋 is a uniformly non-𝑙₃⁽¹⁾ space if and only if there exists 𝜀>0 such that, for every 3-dimensional subspace 𝑋₃ of 𝑋, if 𝑇∶𝑋₃→(𝑅³,‖⋅‖₁) is a linear isomorphism, then ‖𝑇‖⋅‖𝑇⁻¹‖≥1+𝜀.

Proof 
Necessity. Suppose that, for any natural number 𝑘, there exist a 3-dimensional subspace 𝑋_3,𝑘 of 𝑋 and a linear operator 𝑇_𝑘 such that 𝑇_𝑘∶𝑋_3,𝑘→(𝑅³,‖⋅‖₁) is a linear isomorphism and ‖𝑇_𝑘‖⋅‖𝑇_𝑘⁻¹‖<1+1/𝑘. We may assume without loss of generality that ‖𝑇_𝑘‖=1. Moreover, it is easy to see that there exist 𝑦₁,𝑦₂,𝑦₃∈𝑆((𝑅³,‖⋅‖₁)) such that min𝜉₁𝑦₁+𝜉₂𝑦₂+𝜉_𝑛𝑦₃∶𝜉_𝑖=±1=3.(13) By ‖𝑇_𝑘‖⋅‖𝑇_𝑘⁻¹‖<1+1/𝑘 and ‖𝑇_𝑘‖=1, we have 1≤𝑇_𝑘⁻¹𝑦₁1≤1+𝑘,1≤𝑇_𝑘⁻¹𝑦₂1≤1+𝑘,1≤𝑇_𝑘⁻¹𝑦₃1≤1+𝑘.(14) Let 𝑥_1,𝑘=𝑇_𝑘⁻¹𝑦₁𝑇_𝑘⁻¹𝑦₁,𝑥_2,𝑘=𝑇_𝑘⁻¹𝑦₂𝑇_𝑘⁻¹𝑦₂,𝑥_3,𝑘=𝑇_𝑘⁻¹𝑦₃𝑇_𝑘⁻¹𝑦₃.(15) Then 3≥min𝜉₁𝑥_1,𝑘+𝜉₂𝑥_2,𝑘+𝜉₃𝑥_3,𝑘∶𝜉_𝑖𝜉=±1=min₁𝑇_𝑘⁻¹𝑦₁𝑇_𝑘⁻¹𝑦₁+𝜉₂𝑇_𝑘⁻¹𝑦₂𝑇_𝑘⁻¹𝑦₂+𝜉₃𝑇_𝑘⁻¹𝑦₃𝑇_𝑘⁻¹𝑦₃∶𝜉_𝑖=𝑇=±1_𝑘min𝜉₁𝑇_𝑘⁻¹𝑦₁𝑇_𝑘⁻¹𝑦₁+𝜉₂𝑇_𝑘⁻¹𝑦₂𝑇_𝑘⁻¹𝑦₂+𝜉₃𝑇_𝑘⁻¹𝑦₃𝑇_𝑘⁻¹𝑦₃∶𝜉_𝑖𝜉=±1≥min₁𝑇_𝑘𝑇_𝑘⁻¹𝑦₁𝑇_𝑘⁻¹𝑦₁+𝜉₂𝑇_𝑘𝑇_𝑘⁻¹𝑦₂𝑇_𝑘⁻¹𝑦₂+𝜉₃𝑇_𝑘𝑇_𝑘⁻¹𝑦₃𝑇_𝑘⁻¹𝑦₃∶𝜉_𝑖𝜉=±1=min₁𝑦₁𝑇_𝑘⁻¹𝑦₁+𝜉₂𝑦₂𝑇_𝑘⁻¹𝑦₂+𝜉₃𝑦₃𝑇_𝑘⁻¹𝑦₃∶𝜉_𝑖=±1.(16) Therefore, by (13) and (14), we have lim_𝑘→∞min𝜉₁𝑥_1,𝑘+𝜉₂𝑥_2,𝑘+𝜉₃𝑥_3,𝑘∶𝜉_𝑖=±1=3,(17) a contradiction. This implies that if there exists 𝜀>0 such that, for every 3-dimensional subspace 𝑋₃ of 𝑋, if 𝑇∶𝑋₃→(𝑅³,‖⋅‖₁) is a linear isomorphism, then ‖𝑇‖⋅‖𝑇⁻¹‖≥1+𝜀.
Sufficiency. Suppose that 𝑋 is not a uniformly non-𝑙₃⁽¹⁾ space. Then, for any natural number 𝑘, there exist 𝑥_1,𝑘,𝑥_2,𝑘,𝑥_3,𝑘∈𝑆(𝑋) such that lim_𝑘→∞min𝜉₁𝑥_1,𝑘+𝜉₂𝑥_2,𝑘+𝜉₃𝑥_3,𝑘∶𝜉_𝑖=±1=3.(18) We define the subspace 𝑌_𝑘={𝑡₁𝑥_1,𝑘+𝑡₂𝑥_2,𝑘+𝑡₃𝑥_3,𝑘∶{𝑡_𝑖}³_𝑖=1⊂𝑅} of 𝑋. We claim that dim𝑌_𝑘=3. In fact, suppose that dim𝑌_𝑘<3. By (18), we obtain that lim_𝑘→∞min{‖𝑥_1,𝑘+𝑥_2,𝑘‖,‖𝑥_1,𝑘−𝑥_2,𝑘‖}=2. Hence, for any natural number 𝑘, we may assume without loss of generality that 𝑥_1,𝑘 and 𝑥_2,𝑘 are linearly independent. Notice that dim𝑌_𝑘<3, so we obtain that dim𝑌_𝑘=2. This implies that there exist 𝑡_𝑘∈𝑅 and ℎ_𝑘∈𝑅 such that 𝑥_3,𝑘=𝑡_𝑘𝑥_1,𝑘+ℎ_𝑘𝑥_2,𝑘. By (18) and lim_𝑘→∞min{‖𝑥_1,𝑘+𝑥_2,𝑘‖,‖𝑥_1,𝑘−𝑥_2,𝑘‖}=2, we may assume without loss of generality that 𝑡_𝑘≥ℎ_𝑘≥0. By 1=‖𝑥_3,𝑘‖=‖𝑡_𝑘𝑥_1,𝑘+ℎ_𝑘𝑥_2,𝑘‖ and Lemma 8, we have lim_𝑘→∞(𝑡_𝑘+ℎ_𝑘)=1. Hence, we have 3=lim_𝑘→∞𝑥_1,𝑘+𝑥_2,𝑘−𝑥_3,𝑘=lim_𝑘→∞𝑥_1,𝑘+𝑥_2,𝑘−𝑡_𝑘𝑥_1,𝑘+ℎ_𝑘𝑥_2,𝑘=lim_𝑘→∞1−𝑡_𝑘𝑥_1,𝑘+1−ℎ_𝑘𝑥_2,𝑘≤limsup_𝑘→∞1−𝑡_𝑘𝑥_1,𝑘+limsup_𝑘→∞1−ℎ_𝑘𝑥_2,𝑘≤2,(19) a contradiction. This implies that 𝑥_1,𝑘, 𝑥_2,𝑘, and 𝑥_3,𝑘 are linearly independent. Then, dim𝑌_𝑘=3. We define the linear operator 𝑇_𝑘∶𝑌_𝑘→(𝑅³,‖⋅‖₁) by the formula 𝑇_𝑘𝑡_1,𝑘𝑥_1,𝑘+𝑡_2,𝑘𝑥_2,𝑘+𝑡_3,𝑘𝑥_3,𝑘=𝑡_1,𝑘100+𝑡_2,𝑘010+𝑡_3,𝑘001.(20) It is easy to see that 𝑇_𝑘 is one-one mapping.
Next, we will prove that lim_𝑘→∞‖𝑇_𝑘‖=1. In fact, it is easy to see that, for any natural number 𝑘, we have ‖𝑇_𝑘‖≥1. Suppose that lim_𝑘→∞‖𝑇_𝑘‖≠1. Then, we may assume without loss of generality that there exists 𝑟>0 such that lim_𝑘→∞‖𝑇_𝑘‖>1+𝑟. This implies that there exists a sequence {𝑡_1,𝑘𝑥_1,𝑘+𝑡_2,𝑘𝑥_2,𝑘+𝑡_3,𝑘𝑥_3,𝑘}^∞_𝑘=1⊂𝑆(𝑌_𝑘) such that lim_𝑘→∞𝑇_𝑘𝑡_1,𝑘𝑥_1,𝑘+𝑡_2,𝑘𝑥_2,𝑘+𝑡_3,𝑘𝑥_3,𝑘1>1+2𝑟.(21) By (18), we may assume without loss of generality that 𝑡_1,𝑘≥0, 𝑡_2,𝑘≥0, and 𝑡_3,𝑘≥0. By Lemma 8, we have lim_𝑘→∞𝑡_2,𝑘𝑡_2,𝑘+𝑡_3,𝑘𝑥_2,𝑘+𝑡_3,𝑘𝑡_2,𝑘+𝑡_3,𝑘𝑥_3,𝑘=1.(22) Moreover, by Lemma 9, we have lim_𝑘→∞𝑥_1,𝑘+𝑡_2,𝑘𝑡_2,𝑘+𝑡_3,𝑘𝑥_2,𝑘+𝑡_3,𝑘𝑡_2,𝑘+𝑡_3,𝑘𝑥_3,𝑘=2.(23) By ‖𝑡_1,𝑘𝑥_1,𝑘+𝑡_2,𝑘𝑥_2,𝑘+𝑡_3,𝑘𝑥_3,𝑘‖=1, we obtain 1=𝑡_1,𝑘𝑥_1,𝑘+𝑡_2,𝑘𝑥_2,𝑘+𝑡_3,𝑘𝑥_3,𝑘=𝑡_1,𝑘𝑥_1,𝑘+𝑡_2,𝑘+𝑡_3,𝑘𝑡_2,𝑘𝑡_2,𝑘+𝑡_3,𝑘𝑥_2,𝑘+𝑡_3,𝑘𝑡_2,𝑘+𝑡_3,𝑘𝑥_3,𝑘.(24) Therefore, by (22)–(24) and Lemma 8, we obtain lim_𝑘→∞(𝑡_1,𝑘+𝑡_2,𝑘+𝑡_3,𝑘)=1. Noticing that 𝑇_𝑘(𝑡_1,𝑘𝑥_1,𝑘+𝑡_2,𝑘𝑥_2,𝑘+𝑡_3,𝑘𝑥_3,𝑘)=(𝑡_1,𝑘,𝑡_2,𝑘,𝑡_3,𝑘), we have lim_𝑘→∞𝑇_𝑘𝑡_1,𝑘𝑥_1,𝑘+𝑡_2,𝑘𝑥_2,𝑘+𝑡_3,𝑘𝑥_3,𝑘=lim_𝑘→∞𝑡_1,𝑘+𝑡_2,𝑘+𝑡_3,𝑘=1,(25) a contradiction. This implies that lim_𝑘→∞‖𝑇_𝑘‖=1.
Moreover, we claim that lim_𝑘→∞‖𝑇_𝑘⁻¹‖=1. In fact, it is easy to see that, for any natural number 𝑘, we have ‖𝑇_𝑘⁻¹‖≥1. Suppose that lim_𝑘→∞‖𝑇_𝑘⁻¹‖≠1. Then, we may assume without loss of generality that there exists 𝑑>0 such that lim_𝑘→∞‖𝑇_𝑘⁻¹‖>1+𝑑. This implies that there exists a sequence {(𝑡_1,𝑘,𝑡_2,𝑘,𝑡_3,𝑘)}^∞_𝑘=1⊂𝑆((𝑅³,‖⋅‖₁)) such that lim_𝑘→∞𝑇_𝑘⁻¹𝑡_1,𝑘,𝑡_2,𝑘,𝑡_3,𝑘=lim_𝑘→∞𝑡_1,𝑘𝑥_1,𝑘+𝑡_2,𝑘𝑥_2,𝑘+𝑡_3,𝑘𝑥_3,𝑘1>1+2𝑑.(26) By (18), we may assume without loss of generality that 𝑡_1,𝑘≥0, 𝑡_2,𝑘≥0, and 𝑡_3,𝑘≥0. Noticing that {(𝑡_1,𝑘,𝑡_2,𝑘,𝑡_3,𝑘)}^∞_𝑘=1⊂𝑆((𝑅³,‖⋅‖₁)), we have 𝑡_1,𝑘+𝑡_2,𝑘+𝑡_3,𝑘=1. Therefore, by (22)–(24) and Lemma 8, we have 1=lim_𝑘→∞𝑡_1,𝑘𝑥_1,𝑘+𝑡_2,𝑘+𝑡_3,𝑘𝑡_2,𝑘𝑡_2,𝑘+𝑡_3,𝑘𝑥_2,𝑘+𝑡_3,𝑘𝑡_2,𝑘+𝑡_3,𝑘𝑥_3,𝑘=lim_𝑘→∞𝑡_1,𝑘𝑥_1,𝑘+𝑡_2,𝑘𝑥_2,𝑘+𝑡_3,𝑘𝑥_3,𝑘,(27) a contradiction. This implies that lim_𝑘→∞‖𝑇_𝑘⁻¹‖=1. Thus lim_𝑘→∞‖𝑇_𝑘‖⋅‖𝑇_𝑘⁻¹‖=1, a contradiction. Hence, we obtain that 𝑋 is a uniformly non-𝑙₃⁽¹⁾ space, which completes the proof.

Lemma 11. Suppose that (1) there exists a ball-covering 𝔅_𝔫 of (𝑅³,‖⋅‖_𝑛) and 𝑐(𝔅_𝔫)=5, (2) ‖⋅‖_𝑛 is uniformly convergent to ‖⋅‖_∞ in 𝐵(𝑅³,‖⋅‖_∞) and (3) 𝔅_𝔫 is 𝛼-off the origin for any 𝑛. Then, 𝑟(𝔅_𝔫)→∞ as 𝑛→∞.

Proof. Let 𝑆(𝑅²,‖⋅‖_𝑛)⊂∪⁵_𝑖=1𝐵(𝑦^𝑖_𝑛,𝑟^𝑖_𝑛), ‖𝑦^𝑖_𝑛‖≥𝑟^𝑖_𝑛>0, and inf{‖𝑦^𝑖_𝑛‖−𝑟^𝑖_𝑛∶1≤𝑖≤5,𝑛∈𝑁}≥𝛼. Suppose that there exists 𝛿>0 such that max_1≤𝑖≤5{𝑟^𝑖_𝑛}=𝑟(𝔅_𝔫)≤𝛿. Then, {𝑦^𝑖_𝑛}^∞_𝑛=1 and {𝑟^𝑖_𝑛}^∞_𝑛=1 are bounded sequences. Hence, we may assume without loss of generality that 𝑦^𝑖_𝑛→𝑦^𝑖 and 𝑟^𝑖_𝑛→𝑟^𝑖 in ‖⋅‖_∞ for any {1,…,5}. Then, 𝛼≤‖𝑦^𝑖_𝑛‖−𝑟^𝑛_𝑖→‖𝑦^𝑖‖−𝑟 as 𝑛→∞. We claim that 𝑆𝑅³,‖⋅‖_∞⊂⁵_𝑖=1𝐵𝑦^𝑖,𝑟^𝑖.(28) In fact, for any 𝐴,𝐵⊂𝑅³ and 𝑛∈𝑁∪+∞, let 𝑑_𝑛=maxsup_𝑎∈𝐴inf_𝑏∈𝐵‖𝑎−𝑏‖_𝑛,sup_𝑏∈𝐵inf_𝑎∈𝐴‖𝑎−𝑏‖_𝑛.(29) Since ‖⋅‖_𝑛 is uniformly convergent to ‖⋅‖_∞ in 𝐵(𝑅³,‖⋅‖_∞), we obtain that, for any bounded set, ‖⋅‖_𝑛 is uniformly convergent to ‖⋅‖_∞. Moreover, it is easy to see that lim_𝑛→∞𝑑_∞𝐵𝑅³,‖⋅‖_𝑛,𝐵𝑅³,‖⋅‖_∞=0,lim_𝑛→∞𝑑_∞𝐵𝑦^𝑖_𝑛,𝑟^𝑖_𝑛,𝐵𝑦^𝑖,𝑟^𝑖=0.(30) This implies that lim_𝑛→∞𝑑_∞𝑆𝑅³,‖⋅‖_𝑛,𝑆𝑅³,‖⋅‖_∞=0,lim_𝑛→∞𝑑_∞⁵_𝑖=1𝐵𝑦^𝑖_𝑛,𝑟^𝑖_𝑛,⁵_𝑖=1𝐵𝑦^𝑖,𝑟^𝑖=0.(31)
Since lim_𝑛→∞𝑑_∞(𝑆(𝑅³,‖⋅‖_𝑛),𝑆(𝑅³,‖⋅‖_∞))=0, then, for any 𝑥∈𝑆(𝑅³,‖⋅‖_∞), there exists a sequence {𝑥_𝑛}^∞_𝑛=1⊂𝑆(𝑅³,‖⋅‖_𝑛) such that 𝑥_𝑛→𝑥 as 𝑛→∞. Since 𝑆(𝑅³,‖⋅‖_𝑛)⊂∪⁵_𝑖=1𝐵(𝑦^𝑖_𝑛,𝑟^𝑖_𝑛), then there exists 𝑖∈{1,…,5} such that 𝑥_𝑛∈𝐵(𝑦^𝑖_𝑛,𝑟^𝑖_𝑛) for any 𝑛∈𝑁. Noticing that lim_𝑛→∞𝑑_∞𝐵𝑦^𝑖_𝑛,𝑟^𝑖_𝑛,𝐵𝑦^𝑖,𝑟^𝑖=0,(32)
we have 𝑥∈𝐵(𝑦^𝑖,𝑟^𝑖). This implies that 𝑆(𝑅³,‖⋅‖_∞)⊂∪⁵_𝑖=1𝐵(𝑦^𝑖,𝑟^𝑖). By Proposition 2, we have 𝑐(𝔅_min(𝑅³,‖⋅‖_∞))=6, a contradiction. This implies that 𝑟(𝔅_𝔫)→∞ as 𝑛→∞, which completes the proof.

Proof of Theorem 7 
Sufficiency. It is easy to see that there exist two constants 𝛼,𝛽>0 such that, for every 3-dimensional subspace 𝑌 of 𝑋, there exists a ball-covering 𝔅 of 𝑌 with 𝑐(𝔅)=5 which is 𝛼-off the origin and 𝑟(𝔅)≤𝛽. Suppose that 𝑋 is not a uniformly non-𝑙₃⁽¹⁾ space. By Lemma 10, for any natural number 𝑘, there exist a 3-dimensional subspace 𝑋_3,𝑘 of 𝑋 and a linear operator 𝑇_𝑘 such that 𝑇_𝑘∶𝑋_3,𝑘→(𝑅³,‖⋅‖₁) is a linear isomorphism and ‖𝑇_𝑘‖⋅‖𝑇_𝑘⁻¹‖<1+1/𝑘. Since (𝑅³,‖⋅‖₁) and (𝑅³,‖⋅‖_∞) are isomorphism, there exists a linear operator 𝐺_𝑘 such that 𝐺_𝑘∶𝑋_3,𝑘→(𝑅³,‖⋅‖_∞) is a linear isomorphism and 𝐺_𝑘⋅𝐺_𝑘⁻¹1<1+𝑘.(33) Moreover, we may assume without loss of generality that ‖𝐺_𝑘‖=1. Let ‖𝑥‖_𝑘=𝐺_𝑘⁻¹𝑥,∀𝑥∈𝑅³,(34) and let 𝔅_𝑘 be a ball covering of 𝑋_3,𝑘, where 𝑐(𝔅_𝑘)=5 which is 𝛼-off the origin and 𝑟(𝔅_𝑘)≤𝛽. It is easy to see that ‖⋅‖_𝑘 is uniformly convergent to ‖⋅‖_∞ in 𝐵(𝑅³,‖⋅‖_∞). By Lemma 11, we have that 𝑟(𝔅_𝑘)→∞ as 𝑘→∞, a contradiction. Hence, we obtain that 𝑋 is a uniformly non-𝑙₃⁽¹⁾ space.
Necessity. By Lemma 10, we have 𝑐(𝔅)=4 or 5. Using the method of Theorem  3.5 in [6] and Lemma 10, similarly, we obtain that 𝔅 is 𝛼-off the origin and 𝑟(𝔅)≤𝛽, which completes the proof.

Theorem 12. Suppose that 𝑋 is a uniformly non-𝑙₃⁽¹⁾ space and smooth space. Then, there exist two constants 𝛼,𝛽>0 such that for every 3-dimensional subspace 𝑌 of 𝑋, there exists a minimal ball-covering 𝔅 of 𝑌 with 𝑐(𝔅)=4 which is 𝛼-off the origin and 𝑟(𝔅)≤𝛽.

Proof. By Proposition 2, we obtain that there exist two constants 𝛼,𝛽>0 such that, for every 3-dimensional subspace 𝑌 of 𝑋, there exists a minimal ball-covering 𝔅 of 𝑌 with 𝑐(𝔅)=4. Using the method of Theorem  3.5 in [6] and Lemma 10, similarly, we obtain that 𝔅 is 𝛼-off the origin and 𝑟(𝔅)≤𝛽, which completes the proof.

Theorem 13. Suppose that there exist two constants 𝛼,𝛽>0 such that, for every 3-dimensional subspace 𝑌 of 𝑋, there exists a minimal ball-covering 𝔅 of 𝑌 with 𝑐(𝔅)=5 which is 𝛼-off the origin and 𝑟(𝔅)≤𝛽. Then, 𝑋 is a uniformly non-𝑙₃⁽¹⁾ space and not smooth space.

Proof. By Theorem 7, we obtain that 𝑋 is a uniformly non-𝑙₃⁽¹⁾ space. Suppose that 𝑋 is a smooth space. Then, every 3-dimensional subspace 𝑌 of 𝑋 is smooth. By Proposition 2, we obtain that 𝑐(𝔅_min(𝑌))=4. However, there exist two constants 𝛼,𝛽>0 such that, for every 3-dimensional subspace 𝑌 of 𝑋, there exists a minimal ball-covering 𝔅 of 𝑌 with 𝑐(𝔅)=5 which is 𝛼-off the origin and 𝑟(𝔅)≤𝛽, a contradiction. Hence, we obtain that 𝑋 is not a smooth space, which completes the proof.

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

Lemma 14 (see [5]). Suppose that 𝑝 is a Minkowski functional defined on the space 𝑋. Then, 𝑝 is Gateaux differentiable at 𝑥 and with the Gateaux derivative 𝑥^∗ if and only if 𝑥^∗ is a 𝑤^∗-exposed point of 𝐶^∗ and exposed by 𝑥, where 𝐶^∗ is the polar of the level set 𝐶={𝑦∈𝑋∶𝑝(𝑦)≤1}.

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

Proof. (a) First we will prove that there exists a sequence {𝑥_𝑛}^∞_𝑛=1 of 𝑤^∗-exposed points of 𝐵(𝑋^∗∗) such that sup_𝑛𝑥^∗𝑥_𝑛=𝑥^∗,∀𝑥^∗∈𝑋^∗.(35) Since 𝑋 has the Radon-Nikodym property, then the closed convex hull co(𝐸) of 𝐸 is the whole 𝐵(𝑋), where 𝐸 denotes strongly exposed points of 𝐵(𝑋).
Pick 𝑦∈𝐸. Since 𝑦∈𝐸 is a strongly exposed point of 𝐵(𝑋), there exists 𝑦^∗∈𝑆(𝑋^∗) such that 𝑥_𝑛→𝑦 whenever 𝑦^∗(𝑥_𝑛)→𝑦^∗(𝑦)=sup{𝑦^∗(𝑥)∶𝑥∈𝐵(𝑋)}=1. Next we will prove that 𝑦∈𝑆(𝑋^∗∗) is 𝑤^∗-exposed point of 𝐵(𝑋^∗∗) and exposing by 𝑦^∗. In fact, suppose that there exists 𝑦^∗∗∈𝑆(𝑋^∗∗) such that 𝑦^∗(𝑦^∗∗)=1. Since weak^∗ topology is a Hausdorff topology, there exist a weak^∗ neighbourhood 𝑈_𝑦 of 𝑦 and a weak^∗ neighbourhood 𝑈_𝑦∗∗ of 𝑦^∗∗ such that 𝑈_𝑦∩𝑈_𝑦∗∗=∅. Define the weak^∗ neighbourhood as follows: 𝑈_𝑛=𝑥^∗∗∈𝑋^∗∗∶𝑦^∗𝑥^∗∗−𝑦^∗𝑦^∗∗<1𝑛𝑈_𝑦∗∗.(36) By the Goldstine theorem, there exists 𝑦_𝑛∈𝐵(𝑋^∗∗) such that 𝑦_𝑛∈𝑈_𝑛. Hence, we have that 𝑦^∗(𝑦_𝑛)→1 as 𝑛→∞. Since 𝑦∈𝐸 is a strongly exposed point of 𝐵(𝑋), we obtain that 𝑦_𝑛→𝑦 as 𝑛→∞. This implies that 𝑦_𝑛∈𝑈_𝑦, when 𝑛 is large enough. This contradicts the fact that 𝑦_𝑛∈𝑈_𝑛. Hence, we obtain that 𝑦∈𝑆(𝑋^∗∗) is a 𝑤^∗-exposed point of 𝐵(𝑋^∗∗).
Since 𝑋 is a separable space, then there exists a sequence {𝑥_𝑛}^∞_𝑛=1 such that {𝑥_𝑛}^∞_𝑛=1 is a dense sequence in 𝐸. Noticing that co(𝐸)=𝐵(𝑋), we have 𝑥^∗≥sup_𝑛𝑥^∗𝑥_𝑛=sup_𝑥∈𝐸𝑥^∗(𝑥)=sup_{𝑥∈co(𝐸)}𝑥^∗(𝑥)=sup_{𝑥∈𝐵(𝑋)}𝑥^∗(𝑥)=𝑥^∗.(37)
(b) Next we will prove that 𝑋^∗ has the ball-covering property. By Lemma 14, for any 𝑥_𝑖∈{𝑥_𝑛}^∞_𝑛=1⊂𝑆(𝑋^∗∗), there exist 𝑥^∗_𝑖∈𝑆(𝑋^∗) such that ‖⋅‖ is Gateaux differentiable at 𝑥^∗_𝑖 and with the Gateaux derivative 𝑥_𝑖. For each fixed 1<𝑖<∞, let 𝐵_𝑖,𝑚 be the balls defined by 𝐵_𝑖,𝑚=𝐵𝑚𝑥^∗_𝑖1,𝑚−𝑚,𝑖=1,2,….(38) Clearly, every 𝐵_𝑖,𝑚 has the distance 1/𝑚 from the origin. We claim that 𝑆𝑋^∗⊂𝐵_𝑖,𝑚∶𝑖=1,2,…𝑚=1,2,….(39) In fact, pick 𝛼∈(0,1). Noticing that sup_𝑛𝑥^∗𝑥_𝑛=𝑥^∗,∀𝑥^∗∈𝑋^∗,(40) we obtain that, for 𝑦^∗∈𝑆(𝑋^∗), there exists 𝑥∈{𝑥_𝑛}^∞_𝑛=1 such that 𝑦^∗(𝑥)≥𝛼𝑦^∗=𝛼>0.(41) We can assume that 𝑥=𝑥_𝑗 for some 1≤𝑗<∞. Thus, there exist 𝛽≥𝛼 and ℎ^∗_𝑗∈𝐻^∗_𝑗={𝑥^∗∈𝑋^∗∶𝑥^∗(𝑥_𝑗)=0} such that 𝑦^∗=𝛽𝑥^∗+ℎ^∗_𝑗.(42) We want to show that 𝑦^∗∈∪^∞_𝑚=1𝐵_𝑗,𝑚. Otherwise, for every 𝑚∈𝑁, 1𝑚−𝑚≤𝑚𝑥^∗_𝑗−𝑦^∗=𝑚−𝛽𝑥^∗_𝑗−ℎ^∗_𝑗.(43) Thus, −1𝑚≤𝑚−𝛽𝑥^∗_𝑗−ℎ^∗_𝑗=𝑥−𝑚𝑚−𝛽^∗_𝑗−ℎ^∗_𝑗−𝑚𝑥^∗_𝑗=𝑚−𝛽𝑥^∗_𝑗−1ℎ𝑚−𝛽^∗𝑗−𝑥^∗_𝑗=𝑥−𝛽_∗𝑗−𝑡ℎ^∗_𝑗−𝑥^∗_𝑗𝑡−𝛽,(44) where 𝑡=1/(𝑚−𝛽). Letting 𝑚→∞, we observe that 0≤𝑥^∗_𝑗^′ℎ^∗_𝑗−𝛽≤ℎ^∗_𝑗𝑥_𝑗−𝛽=−𝛽<0,(45) which is a contradiction. Therefore, 𝑆𝑋^∗⊂𝐵_𝑖,𝑚∶𝑖=1,2,…𝑚=1,2,….(46) Hence, 𝑋^∗ has the ball-covering property, which completes the proof.

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

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

3. Applications to Orlicz Function Spaces

It is easy to see that if 𝑋 is separable, then 𝑋 has the ball-covering property. Cheng [1] proved that the sequence space 𝑙^∞ 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. 𝑀∶𝑅→𝑅 is called an 𝑁-function if it has the following properties: (1)𝑀 is even, convex and 𝑀(0)=0;(2)𝑀(𝑢)>0 for all 𝑢≠0;(3)lim_𝑢→0𝑀(𝑢)/𝑢=0 and lim_𝑢→∞𝑀(𝑢)/𝑢=∞.

Let (𝐺,Σ,𝜇) be a finite nonatomic and complete measure space. Denote by 𝑝 and 𝑞 the right derivative of 𝑀 and 𝑁, respectively. We define 𝜌_𝑀(𝑥)=_𝐺𝑀(𝑥(𝑡𝐿))𝑑𝑡,_𝑀=𝑥(𝑡)∶𝜌_𝑀(𝜆𝑥)<∞,forsome,𝐸𝜆>0_𝑀=𝑥(𝑡)∶𝜌_𝑀(𝜆𝑥)<∞,∀𝜆>0.(47) It is well known that the Orlicz function space 𝐿_𝑀 is a Banach space when it is equipped with the Luxemburg norm ‖𝑥‖=inf𝜆>0∶𝜌_𝑀𝑥𝜆≤1(48) or equipped with the Amemiya-Orlicz norm ‖𝑥‖⁰=inf_𝑘>01𝑘1+𝜌_𝑀(𝑘𝑥).(49)

Let 𝑝(𝑢) denote the right derivative of 𝑀(𝑢) at 𝑢∈𝑅⁺ and let 𝑞(𝑣) be the generalized inverse function of 𝑝(𝑢) defined on 𝑅⁺ by 𝑞(𝑣)=sup_𝑢≥0𝑢≥0∶𝑝(𝑢)≤𝑣.(50) Then, we call 𝑁(𝑣)=∫₀^|𝑣|𝑞(𝑠)𝑑𝑠 the complementary function of 𝑀. It is well known that there holds the Young inequality 𝑢𝑣≤𝑀(𝑢)+𝑁(𝑣) and 𝑢𝑣=𝑀(𝑢)+𝑁(𝑣)⇔𝑢=|𝑞(𝑣)|sign𝑣 or 𝑣=|𝑝(𝑢)|sign𝑢. Moreover, it is well known that 𝑀 and 𝑁 are complementary to each other.

We say that an 𝑁-function 𝑀∈Δ₂(𝑁∈∇₂) if there exist 𝐾>2 and 𝑢₀≥0 such that 𝑀(2𝑢)≤𝐾𝑀(𝑢)𝑢≥𝑢₀.(51) By [13], we know that 𝐿_𝑀(𝐿⁰_𝑀) is separable, ⇔𝐿_𝑀(𝐿⁰_𝑀) has the Radon-Nikodym property ⇔𝑀∈Δ₂, and 𝑀∈Δ₂⇔𝐿_𝑀=𝐸_𝑀(𝐿⁰_𝑀=𝐸⁰_𝑀)⇔𝐿_𝑀(𝐿⁰_𝑀) is separable. Moreover, by [13], we know that (𝐸_𝑀)^∗=𝐿⁰_𝑀 and (𝐸⁰_𝑀)^∗=𝐿_𝑀.

Theorem 18. If 𝑀∈∇₂ or 𝑀∈Δ₂, then 𝐿_𝑀(𝐿⁰_𝑀) has the ball-covering property.

Proof. By [13], we know that (𝐸_𝑀)^∗=𝐿⁰_𝑀,(𝐸⁰_𝑀)^∗=𝐿_𝑀. Using Theorem 15, we obtain that if 𝑀∈∇₂, then 𝐿_𝑀(𝐿⁰_𝑀) has the ball-covering property. Moreover, by 𝑀∈Δ₂, we obtain that 𝐿_𝑀(𝐿⁰_𝑀) is separable. Hence, 𝐿_𝑀(𝐿⁰_𝑀) has the ball-covering property, which completes the proof.

Remark 19. It is well known that there exists an 𝑁-function 𝑀 such that 𝑀∈∇₂ and 𝑀∉Δ₂. This means that 𝐿_𝑀(𝐿⁰_𝑀) is not a separable space. However, by Theorem 18, we obtain that 𝐿_𝑀(𝐿⁰_𝑀) 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.