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

Authors: Shaoqiang Shang; Yunan Cui
Journal: Abstract and Applied Analysis
 (2013)

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## 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.


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## Body

## 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(xi;ri)}i∈I of X, we denote by c(𝔅) its cardinality and by r(𝔅) the least upper bound of the radius set {ri}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 𝔅min=𝔅min(X) be any minimal ball covering of X. Cheng [1] showed the following results.Proposition 2.
Suppose thatX is an n-dimensional Banach space. Then (1)
n+1≤c(𝔅min)≤2n;(2)
ifX is smooth, then c(𝔅min)=n+1;(3)
c(𝔅min)=2n if and only if X is isometric to (Rn,∥·∥∞).It is easy to see that(Rn,∥·∥1) is isometric to (Rn,∥·∥∞). Moreover, Cheng [5, 6] showed the following results.Proposition 3.
Suppose thatX 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 spaceX is said to be non-ln(1) space, if, for all x1,x2,…,xn∈S(X),
(1)min{∥ξ1x1+ξ2x2+⋯+ξnxn∥:ξi=±1,i∈{1,2,…,n}}<n.Definition 5.
A Banach spaceX is said to be uniformly non-ln(1) space, if there exists δ>0 such that, for all x1,x2,…,xn∈S(X),
(2)min{∥ξ1x1+ξ2x2+⋯+ξnxn∥:ξi=±1,i∈{1,2,…,n}}<n-δ.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-ln(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 spaceX 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,
(3)v(A)=∫Af(t)dt.Let us recall some geometrical notions that will be used in the further part of this paper. A pointx∈C is said to be a strongly exposed point of C if there exists x*∈X* such that xn→x whenever x*(xn)→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 fx. 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)=limt→0f(x+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 thatX 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 thatX 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{xn}n=1∞ and {yn}n=1∞ be sequences in X. If αn≥0, βn≥0, and limn→∞∥xn+yn∥=limn→∞∥2xn∥=limn→∞∥2yn∥=2, then limn→∞∥αnxn+βnyn∥=1 if and only if limn→∞(αn+βn)=1.Proof
Sufficiency. Let limn→∞(αn+βn)=1. By limn→∞∥xn∥=limn→∞∥yn∥=1, we obtain that
(4)limsupn→∞∥αnxn+βnyn∥≤limsupn→∞(αn∥xn∥+βn∥yn∥)=limsupn→∞(αn+βn)=1.
Moreover, we may assume without loss of generality that αn≥βn. Noticing that limn→∞∥xn+yn∥=limn→∞∥2xn∥=limn→∞∥2yn∥=2, we have
(5)liminfn→∞∥αnxn+βnyn∥=liminfn→∞∥αn(xn+yn)-(αn-βn)yn∥≥liminfn→∞(αn∥xn+yn∥-(αn-βn)∥yn∥)=liminfn→∞(αn+βn)=1.
Hence, we have limn→∞∥αnxn+βnyn∥=1.
Necessity. Let limn→∞∥αnxn+βnyn∥=1. By (αn+βn)-1(αn+βn)=1 for any n∈N and the sufficiently part of the proof that has been just finished, we obtain that
(6)limn→∞∥αnαn+βnxn+βnαn+βnyn∥=1.
This implies that
(7)1=limn→∞∥αnxn+βnyn∥=limn→∞(αn+βn)∥αnαn+βnxn+βnαn+βnyn∥=limn→∞(αn+βn),
which completes the proof.Lemma 9.
If{xn}n=1∞, {yn}n=1∞, and {zn}n=1∞ are three sequences and limn→∞min{∥ξ1xn  +  ξ2yn+ξ3zn∥:ξi=±1}=limn→∞∥3xn∥=limn→∞∥3yn∥=limn→∞∥3zn∥=3, then
(8)limn→∞∥zn+αnαn+βnxn+βnαn+βnyn∥=2.Proof.
Sincelimn→∞min{∥ξ1xn+ξ2yn+ξ3zn∥:ξi=±1}=limn→∞∥3xn∥=limn→∞∥3yn∥=limn→∞∥3zn∥=3, we obtain that liminfn→∞∥xn+yn+zn∥=3 and liminfn→∞∥xn-yn+zn∥=3. Therefore, by limn→∞∥xn∥=1 and limn→∞∥yn∥=1, we have
(9)2=limn→∞∥xn∥+limn→∞∥yn∥≥liminfn→∞∥xn+yn∥≥liminfn→∞(∥xn+yn+zn∥-∥zn∥)≥liminfn→∞∥xn+yn+zn∥-limsupn→∞∥zn∥=2.
Noticing that limsupn→∞∥xn+yn∥≤2, we have limn→∞∥xn+yn∥=2. Similarly, we have limn→∞∥xn-yn∥=2. This implies that limn→∞min{∥xn-yn∥,∥xn+yn∥}=2. Moreover, we may assume without loss of generality that αn≥βn. Then(10)liminfn→∞∥zn+αnαn+βnxn+βnαn+βnyn∥=liminfn→∞∥(xn+yn+zn)-βnαn+βn(xn-yn)-βn-αnαn+βnyn∥≥liminfn→∞(∥xn+yn+zn∥-βnαn+βn∥xn-yn∥-αn-βnαn+βn∥yn∥)=liminfn→∞(3-2βnαn+βn-αn-βnαn+βn)=2.Moreover, it is easy to see that
(11)limsupn→∞∥zn+αnαn+βnxn+βnαn+βnyn∥≤2.
This implies that
(12)limn→∞∥zn+αnαn+βnxn+βnαn+βnyn∥=2,
which completes the proof.Lemma 10.
Suppose thatX 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 X3,k of X and a linear operator Tk such that Tk:X3,k→(R3,∥·∥1) is a linear isomorphism and ∥Tk∥·∥Tk-1∥<1+1/k. We may assume without loss of generality that ∥Tk∥=1. Moreover, it is easy to see that there exist y1,y2,y3∈S((R3,∥·∥1)) such that
(13)min{∥ξ1y1+ξ2y2+ξny3∥:ξi=±1}=3.
By ∥Tk∥·∥Tk-1∥<1+1/k and ∥Tk∥=1, we have
(14)1≤∥Tk-1y1∥≤1+1k,1≤∥Tk-1y2∥≤1+1k,1≤∥Tk-1y3∥≤1+1k.
Let
(15)x1,k=Tk-1y1∥Tk-1y1∥,x2,k=Tk-1y2∥Tk-1y2∥,x3,k=Tk-1y3∥Tk-1y3∥.
Then
(16)3≥min{∥ξ1x1,k+ξ2x2,k+ξ3x3,k∥:ξi=±1}=min{∥ξ1Tk-1y1∥Tk-1y1∥+ξ2Tk-1y2∥Tk-1y2∥+ξ3Tk-1y3∥Tk-1y3∥∥:ξi=±1}=∥Tk∥min{∥ξ1Tk-1y1∥Tk-1y1∥+ξ2Tk-1y2∥Tk-1y2∥+ξ3Tk-1y3∥Tk-1y3∥∥:ξi=±1}≥min{∥ξ1TkTk-1y1∥Tk-1y1∥+ξ2TkTk-1y2∥Tk-1y2∥+ξ3TkTk-1y3∥Tk-1y3∥∥:ξi=±1}=min{∥ξ1y1∥Tk-1y1∥+ξ2y2∥Tk-1y2∥+ξ3y3∥Tk-1y3∥∥:ξi=±1}.
Therefore, by (13) and (14), we have
(17)limk→∞min{∥ξ1x1,k+ξ2x2,k+ξ3x3,k∥:ξi=±1}=3,
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 x1,k,x2,k,x3,k∈S(X) such that
(18)limk→∞min{∥ξ1x1,k+ξ2x2,k+ξ3x3,k∥:ξi=±1}=3.
We define the subspace Yk={t1x1,k+t2x2,k+t3x3,k:{ti}i=13⊂R} of X. We claim that dimYk=3. In fact, suppose that dimYk<3. By (18), we obtain that limk→∞min{∥x1,k+x2,k∥,∥x1,k-x2,k∥}=2. Hence, for any natural number k, we may assume without loss of generality that x1,k and x2,k are linearly independent. Notice that dimYk<3, so we obtain that dimYk=2. This implies that there exist tk∈R and hk∈R such that x3,k=tkx1,k+hkx2,k. By (18) and limk→∞min{∥x1,k+x2,k∥,∥x1,k-x2,k∥}=2, we may assume without loss of generality that tk≥hk≥0. By 1=∥x3,k∥=∥tkx1,k+hkx2,k∥ and Lemma 8, we have limk→∞(tk+hk)=1. Hence, we have
(19)3=limk→∞∥x1,k+x2,k-x3,k∥=limk→∞∥x1,k+x2,k-(tkx1,k+hkx2,k)∥=limk→∞∥(1-tk)x1,k+(1-hk)x2,k∥≤limsupk→∞(1-tk)∥x1,k∥+limsupk→∞(1-hk)∥x2,k∥≤2,
a contradiction. This implies that x1,k, x2,k, and x3,k are linearly independent. Then, dimYk=3. We define the linear operator Tk:Yk→(R3,∥·∥1) by the formula
(20)Tk(t1,kx1,k+t2,kx2,k+t3,kx3,k)=t1,k(100)+t2,k(010)+t3,k(001).
It is easy to see that Tk is one-one mapping.
Next, we will prove thatlimk→∞∥Tk∥=1. In fact, it is easy to see that, for any natural number k, we have ∥Tk∥≥1. Suppose that limk→∞∥Tk∥≠1. Then, we may assume without loss of generality that there exists r>0 such that limk→∞∥Tk∥>1+r. This implies that there exists a sequence {t1,kx1,k+t2,kx2,k+t3,kx3,k}k=1∞⊂S(Yk) such that
(21)limk→∞∥Tk(t1,kx1,k+t2,kx2,k+t3,kx3,k)∥>1+12r.
By (18), we may assume without loss of generality that t1,k≥0, t2,k≥0, and t3,k≥0. By Lemma 8, we have
(22)limk→∞∥t2,kt2,k+t3,kx2,k+t3,kt2,k+t3,kx3,k∥=1.
Moreover, by Lemma 9, we have
(23)limk→∞∥x1,k+t2,kt2,k+t3,kx2,k+t3,kt2,k+t3,kx3,k∥=2.
By ∥t1,kx1,k+t2,kx2,k+t3,kx3,k∥=1, we obtain
(24)1=∥t1,kx1,k+t2,kx2,k+t3,kx3,k∥=∥(t2,kt2,k+t3,kx2,k+t3,kt2,k+t3,kx3,k)t1,kx1,k+(t2,k+t3,k)(t2,kt2,k+t3,kx2,k+t3,kt2,k+t3,kx3,k)∥.
Therefore, by (22)–(24) and Lemma 8, we obtain limk→∞(t1,k+t2,k+t3,k)=1. Noticing that Tk(t1,kx1,k+t2,kx2,k+t3,kx3,k)=(t1,k,t2,k,t3,k), we have
(25)limk→∞∥Tk(t1,kx1,k+t2,kx2,k+t3,kx3,k)∥=limk→∞(t1,k+t2,k+t3,k)=1,
a contradiction. This implies that limk→∞∥Tk∥=1.
Moreover, we claim thatlimk→∞∥Tk-1∥=1. In fact, it is easy to see that, for any natural number k, we have ∥Tk-1∥≥1. Suppose that limk→∞∥Tk-1∥≠1. Then, we may assume without loss of generality that there exists d>0 such that limk→∞∥Tk-1∥>1+d. This implies that there exists a sequence {(t1,k,t2,k,t3,k)}k=1∞⊂S((R3,∥·∥1)) such that
(26)limk→∞∥Tk-1(t1,k,t2,k,t3,k)∥=limk→∞∥t1,kx1,k+t2,kx2,k+t3,kx3,k∥>1+12d.
By (18), we may assume without loss of generality that t1,k≥0, t2,k≥0, and t3,k≥0. Noticing that {(t1,k,t2,k,t3,k)}k=1∞⊂S((R3,∥·∥1)), we have t1,k+t2,k+t3,k=1. Therefore, by (22)–(24) and Lemma 8, we have
(27)1=limk→∞∥(t2,kt2,k+t3,kx2,k+t3,kt2,k+t3,kx3,k)t1,kx1,k+(t2,k+t3,k)(t2,kt2,k+t3,kx2,k+t3,kt2,k+t3,kx3,k)∥=limk→∞∥t1,kx1,k+t2,kx2,k+t3,kx3,k∥,
a contradiction. This implies that limk→∞∥Tk-1∥=1. Thus limk→∞∥Tk∥·∥Tk-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-covering𝔅𝔫 of (R3,∥·∥n) and c(𝔅𝔫)=5, (2) ∥·∥n is uniformly convergent to ∥·∥∞ in B(R3,∥·∥∞) and (3) 𝔅𝔫 is α-off the origin for any n. Then, r(𝔅𝔫)→∞ as n→∞.Proof.
LetS(R2,∥·∥n)⊂∪i=15B(yni,rni), ∥yni∥≥rni>0, and inf{∥yni∥-rni:1≤i≤5,n∈N}≥α. Suppose that there exists δ>0 such that max1≤i≤5{rni}=r(𝔅𝔫)≤δ. Then, {yni}n=1∞ and {rni}n=1∞ are bounded sequences. Hence, we may assume without loss of generality that yni→yi and rni→ri in ∥·∥∞ for any {1,…,5}. Then, α≤∥yni∥-rin→∥yi∥-r as n→∞. We claim that
(28)S(R3,∥·∥∞)⊂⋃i=15B(yi,ri).
In fact, for any A,B⊂R3 and n∈N∪+∞, let
(29)dn=max{supa∈A{infb∈B∥a-b∥n},supb∈B{infa∈A∥a-b∥n}}.
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
(30)limn→∞d∞(B(R3,∥·∥n),B(R3,∥·∥∞))=0,limn→∞d∞(B(yni,rni),B(yi,ri))=0.
This implies that
(31)limn→∞d∞(S(R3,∥·∥n),S(R3,∥·∥∞))=0,limn→∞d∞(⋃i=15B(yni,rni),⋃i=15B(yi,ri))=0.
Sincelimn→∞d∞(S(R3,∥·∥n),S(R3,∥·∥∞))=0, then, for any x∈S(R3,∥·∥∞), there exists a sequence {xn}n=1∞⊂S(R3,∥·∥n) such that xn→x as n→∞. Since S(R3,∥·∥n)⊂∪i=15B(yni,rni), then there exists i∈{1,…,5} such that xn∈B(yni,rni) for any n∈N. Noticing that
(32)limn→∞d∞(B(yni,rni),B(yi,ri))=0,
we havex∈B(yi,ri). This implies that S(R3,∥·∥∞)⊂∪i=15B(yi,ri). By Proposition 2, we have c(𝔅min(R3,∥·∥∞))=6, a contradiction. This implies that r(𝔅𝔫)→∞ as n→∞, which completes the proof.Proof of Theorem7
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 X3,k of X and a linear operator Tk such that Tk:X3,k→(R3,∥·∥1) is a linear isomorphism and ∥Tk∥·∥Tk-1∥<1+1/k. Since (R3,∥·∥1) and (R3,∥·∥∞) are isomorphism, there exists a linear operator Gk such that Gk:X3,k→(R3,∥·∥∞) is a linear isomorphism and
(33)∥Gk∥·∥Gk-1∥<1+1k.
Moreover, we may assume without loss of generality that ∥Gk∥=1. Let
(34)∥x∥k=∥Gk-1x∥,∀x∈R3,
and let 𝔅k be a ball covering of X3,k, where c(𝔅k)=5 which is α-off the origin and r(𝔅k)≤β. It is easy to see that ∥·∥k is uniformly convergent to ∥·∥∞ in B(R3,∥·∥∞). By Lemma 11, we have that r(𝔅k)→∞ 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 thatX 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 Proposition2, 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 constantsα,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 Theorem7, 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(𝔅min(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 (Theorem15) 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 thatp 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 spaceX has the Radon-Nikodym property. Then, X* has the ball-covering property.Proof.
(a) First we will prove that there exists a sequence{xn}n=1∞ of w*-exposed points of B(X**) such that
(35)supnx*(xn)=∥x*∥,∀x*∈X*.
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).
Picky∈E. Since y∈E is a strongly exposed point of B(X), there exists y*∈S(X*) such that xn→y whenever y*(xn)→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 Uy of y and a weak* neighbourhood Uy** of y** such that Uy∩Uy**=∅. Define the weak* neighbourhood as follows:
(36)Un={x**∈X**:|y*(x**)-y*(y**)|<1n}⋂Uy**.
By the Goldstine theorem, there exists yn∈B(X**) such that yn∈Un. Hence, we have that y*(yn)→1 as n→∞. Since y∈E is a strongly exposed point of B(X), we obtain that yn→y as n→∞. This implies that yn∈Uy, when n is large enough. This contradicts the fact that yn∈Un. Hence, we obtain that y∈S(X**) is a w*-exposed point of B(X**).
SinceX is a separable space, then there exists a sequence {xn}n=1∞ such that {xn}n=1∞ is a dense sequence in E. Noticing that co¯(E)=B(X), we have
(37)∥x*∥≥supnx*(xn)=supx∈Ex*(x)=supx∈co(E)x*(x)=supx∈B(X)x*(x)=∥x*∥.
(b) Next we will prove thatX* has the ball-covering property. By Lemma 14, for any xi∈{xn}n=1∞⊂S(X**), there exist xi*∈S(X*) such that ∥·∥ is Gateaux differentiable at xi* and with the Gateaux derivative xi. For each fixed 1<i<∞, let Bi,m be the balls defined by
(38)Bi,m=B(mxi*,m-1m),i=1,2,….
Clearly, every Bi,m has the distance 1/m from the origin. We claim that
(39)S(X*)⊂⋃‍{Bi,m:i=1,2,…m=1,2,…}.
In fact, pick α∈(0,1). Noticing that
(40)supnx*(xn)=∥x*∥,∀x*∈X*,
we obtain that, for y*∈S(X*), there exists x∈{xn}n=1∞ such that
(41)y*(x)≥α∥y*∥=α>0.
We can assume that x=xj for some 1≤j<∞. Thus, there exist β≥α and hj*∈Hj*={x*∈X*:x*(xj)=0} such that
(42)y*=βx*+hj*.
We want to show that y*∈∪m=1∞Bj,m. Otherwise, for every m∈N,
(43)m-1m≤∥mxj*-y*∥=∥(m-β)xj*-hj*∥.
Thus,
(44)-1m≤∥(m-β)xj*-hj*∥-m=∥(m-β)xj*-hj*∥-m∥xj*∥=(m-β){∥xj*-1m-βhj*∥-∥xj*∥}-β=∥xj*-thj*∥-∥xj*∥t-β,
where t=1/(m-β). Letting m→∞, we observe that
(45)0≤∥xj*∥′(hj*)-β≤hj*(xj)-β=-β<0,
which is a contradiction. Therefore,
(46)S(X*)⊂⋃‍{Bi,m:i=1,2,…m=1,2,…}.
Hence, X* has the ball-covering property, which completes the proof.Corollary 16.
IfX* is a separable space, then X** has the ball-covering property.Proof.
IfX* 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 ifX 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)
limu→0M(u)/u=0 and limu→∞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
(47)ρM(x)=∫GM(x(t))dt,LM={x(t):ρM(λx)<∞,forsomeλ>0},EM={x(t):ρM(λx)<∞,∀λ>0}.
It is well known that the Orlicz function space LM is a Banach space when it is equipped with the Luxemburg norm
(48)∥x∥=inf{λ>0:ρM(xλ)≤1}
or equipped with the Amemiya-Orlicz norm
(49)∥x∥0=infk>01k(1+ρM(kx)).Letp(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
(50)q(v)=supu≥0{u≥0:p(u)≤v}.
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 anN-function M∈Δ2(N∈∇2) if there exist K>2 and u0≥0 such that
(51)M(2u)≤KM(u)(u≥u0).
By [13], we know that LM(LM0) is separable, ⇔LM(LM0) has the Radon-Nikodym property ⇔M∈Δ2, and M∈Δ2⇔LM=EM(LM0=EM0)⇔LM(LM0) is separable. Moreover, by [13], we know that (EM)*=LM0 and (EM0)*=LM.Theorem 18.
IfM∈∇2 or M∈Δ2, then LM(LM0) has the ball-covering property.Proof.
By [13], we know that (EM)*=LM0,(EM0)*=LM. Using Theorem 15, we obtain that if M∈∇2, then LM(LM0) has the ball-covering property. Moreover, by M∈Δ2, we obtain that LM(LM0) is separable. Hence, LM(LM0) has the ball-covering property, which completes the proof.Remark 19.
It is well known that there exists anN-function M such that M∈∇2 and M∉Δ2. This means that LM(LM0) is not a separable space. However, by Theorem 18, we obtain that LM(LM0) has ball-covering property.



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*Source: 873943-2013-12-08.xml*