Differentiability and ball-covering property in Banach spaces

J. Math. Anal. Appl. 434 (2016) 182–190

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Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa

Differentiability and ball-covering property in Banach spaces ✩ Shaoqiang Shang Department of Mathematics, Northeast Forestry University, Harbin 150040, PR China

a r t i c l e

i n f o

Article history: Received 12 May 2015 Available online 9 September 2015 Submitted by Richard M. Aron Keywords: Convex function Gâteaux differentiable space Ball-covering property

a b s t r a c t In this paper, author proves that if X1 and X2 are Gâteaux differentiable space, then X1 and X2 have the ball-covering property if and only if (X1 × X2 ,  · p ) and 1 (X1 ×X2 ,  ·∞ ) have the ball-covering property, where (x, y)p = (xp1 + yp2 ) p , p ∈ [1, +∞) and (x, y)∞ = max{x , y}. © 2015 Elsevier Inc. All rights reserved.

1. Introduction w

Let (X,  · ) be a real Banach space. By xn → x we denote that {xn }∞ n=1 is weakly convergent to x. w C(C ) denotes closed hull of C (weak closed hull) and dist(x, C) denotes the distance of x and C. Let N, R and R+ denote the set natural number, reals and nonnegative reals, respectively. Let D be a nonempty open convex subset of X and f 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 df (x)(y) = lim

t→0

f (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 df (x). If the difference quotient in (∗) converges to df (x)(y) uniformly for y in the unit ball, then f is said to be Frechet differentiable at x. X is called a weak Asplund space [Asplund space] or said to have the weak Asplund property if for every f and D as above, f is “generically” Gâteaux [Frechet] differentiable, that is, there exists a dense Gδ subset G of D such that f is Gâteaux [Frechet] differentiable at each point of G. X is called a Gâteaux differentiability space if every convex continuous function on it is Gâteaux differentiable at the points of a dense set. In 1933, Mazur proved that separable Banach spaces have the weak Asplund property (see [15]). Moreover, it is well known that if X is an Asplund space if and only if X ∗ has the ✩

Supported by China Natural Science Fund under grant 11401084. E-mail address: [email protected].

http://dx.doi.org/10.1016/j.jmaa.2015.09.009 0022-247X/© 2015 Elsevier Inc. All rights reserved.

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Radon–Nikodym property, X is a Gâteaux differentiable space if and only if every weak∗ compact convex subset of X ∗ is the weak∗ closed convex hull of its weak∗ exposed points. It is easy to see that Asplund space ⇒ weak Asplund space ⇒ Gâteaux differentiable space. It is well known there exists a weak Asplund space, but not Asplund space, for example, l1 is a weak Asplund space, but not Asplund space. In 2006, Waren B. Moors and Sivajah Somasundaram proved that there exists a Gâteaux differentiable space that is not a weak Asplund space (see [14]). In 2002, L. Cheng and M. Fabian proved that the product of a Gâteaux space and a separable space is a Gâteaux differentiable spaces (see [4]). The study of geometric and topological properties of unit balls of Banach spaces plays a central rule in the geometry of Banach spaces. Almost all properties of Banach spaces, such as convexity, smoothness, reflexivity and the Radon–Nikodym property, can be viewed as properties of the unit ball. We should mention here that there are many topics studying the behaviour of collections of balls. For example, the Mazur intersection property, the packing sphere problem of unit balls, the measures of non-compactness, and the ball topology have all received a great deal of attention by 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 open balls that do not contain the origin. In this case, we also say that the norm has the ball-covering property. In [2], Cheng proved that if X is a locally uniformly convex space and B(X ∗ ) is w∗ -separable, then X has the ball-covering property. In [7], 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. Clearly, every separable space has ball-covering property, but the converse version is not true. For example, ∞ is not a separable space, but ∞ has the ball-covering property (see [1]). In [18], Shang and Cui proved that if a separable space X has the Radon–Nikodym property, then X ∗ has the ball-covering property. As a corollary, Shang and Cui proved that there exists a non-separable Orlicz function space LM such that LM has the ball-covering property. In [3], Cheng and Liu proved that by constructing the equivalent norms on ∞, there exists a Banach space (∞ ,  · 0 ) such that (∞ ,  · 0 ) does not possess the ball covering property. In [19], Shang and Cui proved that if X is separable, X is locally 2-uniformly convex and X is uniformly nonsquare, then ∞ there exists a sequence {xn }∞ n=1 of strongly extreme points such that ∪n=1 B(xn , rn ) is a ball-covering of X, where supn≥1 {rn } < 1. The paper is organized as follows. In Section 1, some necessary definitions and notations are collected. In Section 2, author proves that if X1 and X2 are Gâteaux differentiable space, then X1 and X2 have the ball-covering property if and only if (X1 × X2 ,  · p ) and (X1 × X2 ,  · ∞ ) have p p 1 the ball-covering property, where (x, y)p = (x1 + y2 ) p , p ∈ [1, +∞) and (x, y)∞ = max{x , y}. The topic of this paper is related to the topic of [5,6,8–13,16,17,20,21]. First let us recall some definitions that will be used in the further part of this paper. Definition 2. A point x ∈ S(X) is called a smooth point if it has an unique supporting functional fx . The set of all smooth points of X is denoted by SmoX. If every x ∈ S(X) is a smooth point, then X is called smooth. It is well known that if x0 ∈ S(X) is a smooth point, then convex function f (x) = x is Gâteaux differentiable at x0 . Definition 3. A point x∗0 ∈ C ∗ is said to be weak∗ exposed point of C ∗ if there exists x ∈ S(X) such that x∗0 (x) > x∗ (x) whenever x∗ ∈ C ∗ \ {x∗0 }.

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2. Main results Theorem 1. Suppose that X1 and X2 are Gâteaux differentiable space. Then the following statements are equivalent: (1) X1 and X2 have the ball-covering property. (2) Product spaces (X1 × X2 ,  · p ) and (X1 × X2 ,  · ∞ ) have the ball-covering property, where (x, y)p = p

p

1

(x1 + y2 ) p , p ∈ [1, +∞) and (x, y)∞ = max{x , y}. In order to prove the theorem, we give some lemmas. Lemma 1. (See [15].) Suppose that p is a Minkowski functional defined on the space X. Then p is Gateaux differentiable at x and with the Gâteaux 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}. Lemma 2. Suppose that the smooth points of X are dense in S(X). Then the following statements are equivalent: (1) X has the ball-covering property. ∗ ∗ ∗ (2) There exists a sequence {x∗n }∞ n=1 of w -exposed points of B(X ) such that supn∈N (xn , x) > 0 for any x ∈ S(X). ∗ ∗ ∗ Proof. (2)⇒(1). Let {x∗n }∞ n=1 be a sequence of w -exposed points of B(X ) and supn∈N (xn , x) > 0 for any ∗ x ∈ X. Then, by Lemma 1, for any x∗i ∈ {x∗n }∞ n=1 ⊂ S(X ), there exists xi ∈ S(X) such that  ·  is Gâteaux ∗ differentiable at xi and with the Gâteaux derivative xi . For each fixed 1 < i < ∞, let Bi,m be the balls defined by

  1 , Bi,m = B mxi , m − m

i = 1, 2, . . .

It is easy to see that every Bi,m has the distance 1/m from the origin. Hence, for any y ∈ S(X), pick ∗ αy ∈ (0, supn∈N (x∗n , y)), there exists x∗j ∈ {x∗n }∞ n=1 such that xj (y) > αy . Hence there exist βy ≥ αy and hj ∈ Hj = {x ∈ X : x∗j (x) = 0} such that y = βy xj + hj . We claim that y ∈ ∪∞ m=1 Bj,m . Otherwise, for every m ∈ N , we have m−

1 ≤ mxj − y = (m − βy )xj − hj  . m

Thus −

1 m

≤ (m − βy )xj − hj  − m = (m − βy )xj − hj  − m xj       1   = (m − βy ) xj − hj − xj  − βy m − βy  =

xj − thj  − xj  − βy , t

where t = 1/(m − β). Letting m → ∞, we observe that

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0 ≤ x∗j (hj ) − βy = −βy < 0, a contradiction. Therefore, S(X) ⊂

 {Bi,m : i = 1, 2, . . . m = 1, 2, . . .}.

This implies that X has the ball-covering property. (1)⇒(2). From the proof of Theorem 0.2 of [2], it is easy to see that (1)⇒(2) is true, which completes the proof. 2 Lemma 3. Suppose that X is a Gâteaux differentiable space. Then smooth points of X are dense in S(X). Proof. Let s : X/{0} → S(X) be defined by s(x) := x/x. Then clearly s is a continuous surjection onto S(X). Let W be a nonempty open subset of S(X). Then s−1 (W ) is a nonempty open subset of X/{0}. Hence there exists a point x0 ∈ s−1 (W ) such that  ·  is Gateaux differentiable at x0 . It now follows that  ·  is Gateaux differentiable at s(x0 ) ∈ W , which completes the proof. 2 Proof of Theorem 2. (1)⇒(2). (a) First we will prove that smooth points of X1 ×X2 are dense in S(X1 ×X2 ). For clarity, we will divide the proof into two cases. Case I. Let (x1 , x2 )∞ = max{x1 , x2 } for any (x1 , x2 ) ∈ X1 × X2 . Since X1 and X2 are Gâteaux differentiable space, for any ε > 0, there exist y1 ∈ X1 and y2 ∈ X2 such that (1) y1 is a Gâteaux differentiable point of function f (x) = x1 ; (2) y2 is a Gâteaux differentiable point of function g(x) = x2 ; (3) x1 − y1  < ε/2 and x2 − y2  < ε/2. Therefore, by condition (3), we obtain that (x1 , x2 ) − (y1 , y2 ) = max{x1 − y1  , x2 − y2 } <

1 1 ε + ε = ε. 2 2

Moreover, we may assume without loss of generality that y1  > y2 . Hence there exists t0 (x, y) > 0 such that y1 + tx > y2 + ty and y1 − tx > y2 − ty whenever |t| < t0 (x, y). Then

= = =

1 [(y1 , y2 ) + t(x, y) + (y1 , y2 ) − t(x, y) − 2 (y1 , y2 )] t 1 [(y1 + tx, y2 + ty) + (y1 − tx, y2 − ty) − 2 (y1 , y2 )] t 1 [max{y1 + tx , y2 + ty} + max{y1 − tx , y2 − ty} − 2 (y1 , y2 )] t 1 [y1 + tx + y1 − tx − 2 y1 ] t

whenever |t| < t0 (x, y). This implies that 1 lim [(y1 , y2 ) + t(x, y) + (y1 , y2 ) − t(x, y) − 2 (y1 , y2 )] t→0 t 1 = lim [y1 + tx + y1 − tx − 2 y1 ] = 0. t→0 t Hence (y1 , y2 ) is a Gâteaux differentiable point of function f (x, y) = (x, y). Therefore, by the proof of Lemma 3, we obtain that smooth points of X1 × X2 are dense in S(X1 × X2 ).

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Case II. Let 1 ≤ p < +∞. We claim that the convex function f (x, y) = x is Gâteaux differentiable on a dense subset of X1 × X2 . In fact, it is differentiable on a set of the form D1 × X2 where D1 = {x ∈ X1 : ·1 is Gâteaux differentiable at x}. Similarly, for the mapping f (x, y) = y, we get a dense subset D2 of X2 such that f (x, y) = y is Gâteaux differentiable at each point of X1 × D2 . Therefore, by chain-rule for differentiation, we obtain that p p 1 (x, y)p = (x1 + y2 ) p is Gâteaux differentiable at each point of D1 × D2 , which is dense X1 × X2 . (b) Next we will prove that (X1 × X2 ,  · p ) and (X1 × X2 ,  · ∞ ) have the ball-covering property. For clarity, we will divide the proof into three cases. Case I. Let (x1 , x2 )∞ = max{x1 , x2 } for any (x1 , x2 ) ∈ X1 × X2 . Since Xi has the ball∗ ∗ covering property, by Lemma 2, there exists a sequence {x∗i,n }∞ n=1 of w -exposed points of B(Xi ) such ∗ ∗ that supn∈N (xi,n , x) > 0 for any x ∈ S(Xi ), where i ∈ {1, 2}. Let xi,n (xi,n ) = 1 and xi,n be a smooth point of Xi . Then x∗i,n is a w∗ -exposed point of B(Xi∗ ) and exposed by xi,n . Hence  ∗ (x∗1,n , 0) ∈ B (X1 × X2 ) and (x∗1,n , 0)(x1,n , 0) = x∗1,n (x1,n ) = 1.  ∗ Moreover, for any (z1∗ , z2∗ ) ∈ B (X1 × X2 ) and (z1∗ , z2∗ ) = (x∗1,n , 0), we obtain that z1∗ = x∗1,n . Since x∗1,n is a w∗ -exposed points of B(X1∗ ) and exposed by x1,n , we obtain that (z1∗ , z2∗ )(x1,n , 0) = z1∗ (x1,n ) < x∗1,n (x1,n ) = 1. This implies that (x∗1,n , 0) is a w∗ -exposed points of B((X1 × X2 )∗ ) for all n ∈ N . Similarly, we obtain that (0, x∗2,n ) is a w∗ -exposed points of B((X1 × X2 )∗ ) for all n ∈ N . Put ∗ ∗ ∗ ∞ ∗ ∞ {(z1,n , z2,n )}∞ n=1 = {(x1,n , 0)}n=1 ∪ {(0, x2,n )}n=1 ∗ ∞ ∪ {(−x∗1,n , 0)}∞ n=1 ∪ {(0, −x2,n )}n=1 .

Moreover, it is easy to see that if (x1 , x2 ) ∈ S(X1 × X2 ), then max{x1  , x2 } = 1. Hence ∗ ∗ ∗ ∗ sup (z1,n , z2,n )(x1 , x2 ) = sup [z1,n (x1 ) + z1,n (x2 )] > 0.

n∈N

n∈N

Therefore, by Lemma 2, we obtain that if (x1 , x2 )∞ = max{x1  , x2 } for any (x1 , x2 ) ∈ X1 × X2 , then (2) is true. Case II. Let 1 < p < +∞. Since Xi has the ball-covering property, by Lemma 2, there exists a sequence ∗ ∗ ∗ {x∗i,n }∞ n=1 of w -exposed points of B(Xi ) such that supn∈N (xi,n , x) > 0 for any x ∈ S(Xi ), where i ∈ {1, 2}. 1

1

Let x∗i,n (xi,n ) = 1 and xi,n be a smooth point of Xi . Then (1/2 q x∗1,n , 1/2 q x∗2,n ) ∈ S((X1 × X2 )∗ ) and 1 1 q

= = =

x∗1,n (x1,n ) +

1 1 q

x∗2,n (x2,n )

2 2     1  ∗ x1,n  · x1,n  + x∗2,n  · x2,n  1 2q    1 1 1  x∗1,n q + x∗2,n q q (x1,n p + x2,n p ) p 1 2q 1 1 q 1 · 2 (x1,n , x2,n )p = (x1,n , x2,n )p , q 2 1

1

where (1/p) + (1/q) = 1. Moreover, for any (x∗1 , x∗2 ) ∈ B((X1 × X2 )∗ ) and (x∗1 , x∗2 ) = (1/2 q x∗1,n , 1/2 q x∗2,n ), we have

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(x∗1 , x∗2 )(x1,n , x2,n ) = x∗1 (x1,n ) + x∗2 (x2,n ) < x∗1  x1,n  + x∗2  x2,n  q  q 1  p p 1 ≤ x∗1,n  + x∗2,n  q (x1,n  + x2,n  ) p = (x1,n , x2,n )p . 1

1

∗ ∗ This implies that the sequence {(x∗1,n /2 q , x∗2,n /2 q )}∞ n=1 is a sequence of w -exposed points of B((X1 × X2 ) ). 1

1

1

1

1

1

∗ ∗ ∞ ∗ ∗ ∞ q q q q Hence {(x∗1,n /2 q , −x∗2,n /2 q )}∞ n=1 , {(−x1,n /2 , x2,n /2 )}n=1 and {(−x1,n /2 , −x2,n /2 )}n=1 are sequences ∗ ∗ of w -exposed points of B((X1 × X2 ) ). Put

∗ ∗ {(z1,n , z2,n )}∞ n=1 =

 −

1

∗ 1 x1,n , −

1

∗ 1 x2,n

∞   1

∗ 1 x1,n , −

1

∗ 1 x2,n

∞

2q 2q 2q 2q n=1 n=1    ∞ ∞   1 1 1 ∗ 1 ∗ − 1 x∗1,n , 1 x∗2,n . 1 x1,n , 1 x2,n q q 2q 2q 2 2 n=1 n=1 1

It is easy to see that if (x1 , x2 ) ∈ S(X1 × X2 ) then max{x1  , x2 } ≥ 1/2 p . Hence ∗ ∗ ∗ ∗ sup (z1,n , z2,n )(x1 , x2 ) = sup [z1,n (x1 ) + z1,n (x2 )] > 0.

n∈N

n∈N

Therefore, by Lemma 2, we obtain that if 1 < p < +∞ then (2) is true. Case III. Let p = 1. Since Xi has the ball-covering property, by Lemma 2, there exists a sequence {x∗i,n }∞ n=1 of w∗ -exposed points of B(Xi∗ ) such that supn∈N (x∗i,n , x) > 0 for any x ∈ S(Xi∗ ), where i ∈ {1, 2}. Let x∗i,n (xi,n ) = 1 and xi,n ∈ S(Xi ) be a smooth point of Xi , where i ∈ {1, 2}. Then (x∗1,n , x∗2,n ) ∈ S((X1 × X2 )∗ ) and (x∗1,n , x∗2,n )(x1,n , x2,n ) = x∗1,n (x1,n ) + x∗2,n (x2,n ) = x1,n  + x2,n  = (x1,n , x2,n )p . Moreover, for any (x∗1 , x∗2 ) ∈ S((X1 × X2 )∗ ) and (x∗1 , x∗2 ) = (x∗1,n , x∗2,n ), we obtain that (x∗1 , x∗2 )(x1,n , x2,n ) = x∗1 (x1,n ) + x∗2 (x2,n ) < x1,n  + x2,n  = (x1,n , x2,n )p . ∗ ∗ This implies that sequence {(x∗1,n , x∗2,n )}∞ n=1 is a sequence of w -exposed points of B((X1 × X2 ) ). Hence ∗ ∗ ∞ ∗ ∗ ∞ ∗ ∗ ∞ ∗ we obtain that {(x1,n , −x2,n )}n=1 , {(−x1,n , x2,n )}n=1 and {(−x1,n , −x2,n )}n=1 are sequences of w -exposed points of B((X1 × X2 )∗ ). Put ∗ ∗ ∗ ∗ ∞ ∗ ∗ ∞ {(z1,n , z2,n )}∞ n=1 = {(−x1,n , −x2,n )}n=1 ∪ {(x1,n , −x2,n )}n=1 ∗ ∗ ∞ ∪ {(−x∗1,n , x∗2,n )}∞ n=1 ∪ {(x1,n , x2,n )}n=1 .

It is easy to see that if (x1 , x2 ) ∈ S(X1 × X2 ) then max{x1  , x2 } ≥ 1/2. Hence ∗ ∗ ∗ ∗ sup (z1,n , z2,n )(x1 , x2 ) = sup [z1,n (x1 ) + z1,n (x2 )] > 0.

n∈N

n∈N

Therefore, by Lemma 2, we obtain that if p = 1 then (2) is true. (2)⇒(1). Since (X1 × X2 ,  · p ) and (X1 × X2 ,  · ∞ ) have the ball-covering property, by Lemma 2, ∗ ∗ we obtain that there exists a sequence {(x∗1n , x∗2n )}∞ n=1 of w -exposed points of B((X1 × X2 ) ) such that sup (x∗1n , x∗2n )(x1 , x2 ) = sup [x∗1n (x1 ) + x∗2n (x2 )] > 0

n∈N

n∈N

for any (x1 , x2 ) ∈ S(X1 × X2 ). For clarity, we will divide the proof into three cases.

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Case I. Let (x1 , x2 )∞ = max{x1 , x2 } for any (x1 , x2 ) ∈ X1 × X2 . Then there exists (x1n , x2n ) ∈ S(X1 × X2 ) such that (x∗1n , x∗2n ) is exposed by (x1n , x2n ). Hence (x1n , x2n ) is a smooth point of X1 × X2 for all n ∈ N . Therefore, by max{x1,n  , x2,n } = 1, we obtain that x1,n  = x2,n . Otherwise, it is easy to see that (x1n , x2n ) is not a smooth point of X1 × X2 . Since     max{x1,n  , x2,n } = 1 and x∗1,n  + x∗2,n  = 1, we obtain that     x∗1,n , x∗2,n (x1,n , x2,n ) ≤ x∗1,n  x1,n  + x∗2,n  x2,n      ≤ x∗1,n  + x∗2,n  = 1.

1 =



Then all inequalities are equalities in fact.  ∗ Therefore,   ∗  by max{x1,n  , x2,n } = 1 and x1,n  = x2,n , we  obtain that x1,n  · x2,n  = 0. Then x1,n  · x2,n  = 0. Hence ∗ ∞ ∗ ∞ {(x∗1,n , x∗2,n )}∞ n=1 = {(z1,n , 0)}n=1 ∪ {(0, z2,n )}n=1 . ∗ ∗ It is easy to see that the sequence {z1,n }∞ n=1 is a sequence of w -exposed point of X1 . Hence, for any x1 ∈ S(X1 ), we have ∗ ∗ sup z1,n (x1 ) = sup (z1,n , 0)(x1 , 0) = sup (x∗1,n , x∗2,n )(x1 , 0) > 0.

n∈N

n∈N

n∈N

Therefore, by Lemma 2, we obtain that if (x1 , x2 )∞ = max{x1 , x2 } for any (x1 , x2 ) ∈ X1 × X2 , then X1 has the ball-covering property. Similarly, we obtain that X2 has the ball-covering property. Case II. Let p ∈ (1, +∞). Then there exists (x1n , x2n ) ∈ S(X1 × X2 ) such that (x∗1n , x∗2n ) is exposed by (x1n , x2n ). Since 1 = (x∗1n , x∗2n )(x1n , x2n ) = x∗1n (x1n ) + x∗2n (x2n ) ≤ x∗1n  x1n  + x∗2n  x2n  ≤ (x1n  + x2n  ) p (x∗1n  + x∗2n  ) q = 1, p

p

1

q

1

q

∗ ∗ we obtain that all inequalities are equalities in fact. Hence, if x∗1n = 0 and y1n (x1n ) = y1n  x1n  = ∗ x1n  x1n , then ∗ ∗ ∗ (y1n , x∗2n )(x1n , x2n ) = y1n (x1n ) + x∗2n (x2n ) = y1n  x1n  + x∗2n  x2n  ∗ = (x1n  + x2n  ) p (y1n  + x∗2n  ) p = 1. p

p

1

p

p

1

∗ Moreover, since (x∗1n , x∗2n ) is a w∗ -exposed point of B((X1 × X2 )∗ ), we obtain that (x∗1n , x∗2n ) = (y1n , x∗2n ). ∗ Hence x∗1n = y1n . This implies that if x∗1n = 0 then x∗1n / x∗1n  is a w∗ -exposed point of B(X1∗ ). Similarly, we obtain that if x∗2n = 0 then x∗2n / x∗2n  is a w∗ -exposed point of B(X2∗ ). Moreover, it is easy to see ∗ that there exists a subsequence {nk } of {n} such that x∗1nk / x∗1nk  is a w∗ -exposed point of B(X 1 ) and   x∗1nh = 0, where nh ∈ {n}\{nk }. Similarly, there exists a subsequence {ni } of {n} such that x∗2ni / x∗2ni  is a w∗ -exposed point of B(X2∗ ) and x∗2nh = 0, where nh ∈ {n}\{ni }. Hence

sup x∗1nk (x1 ) = sup (x∗1nk , x∗2nk )(x1 , 0) = sup (x∗1n , x∗2n )(x1 , 0) > 0

k∈N

k∈N

n∈N

  for any x1 ∈ S(X1 ). Hence we obtain that supk∈N (x∗1nk / x∗1nk )(x1 ) > 0 for any x1 ∈ S(X1 ). Similarly,   we have supi∈N (x∗2ni / x∗2ni )(x2 ) > 0 for any x2 ∈ S(X2 ). Therefore, by Lemma 2, we obtain that if p ∈ (1, +∞), then X1 and X2 have the ball-covering property.

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Case III. Let p = 1. Since (x∗1n , x∗2n ) ∈ S((X1 × X2 )∗ ) is a w∗ -exposed points of B((X1 × X2 )∗ ), there exists (x1n , x2n ) ∈ S(X1 × X2 ) such that (x∗1n , x∗2n ) is exposed by (x1n , x2n ). Since 1 = (x∗1n , x∗2n )(x1n , x2n ) = x∗1n (x1n ) + x∗2n (x2n ) ≤ x∗1n  x1n  + x∗2n  x2n  ≤ x1n  + x2n  = 1, we obtain that all inequalities are equalities in fact and x∗1n  = x∗2n  = 1. Hence, if y1∗ (x1n ) = 1 and y1∗ ∈ S(X1∗ ), then 1 = (y1∗ , x∗2n )(x1n , x2n ) = y1∗ (x1n ) + x∗2n (x2n ) = y1∗  x1n  + x∗2n  x2n  ≤ x1n  + x2n  = 1. Since (x∗1n , x∗2n ) ∈ S((X1 × X2 )∗ ) is a w∗ -exposed points of B((X1 × X2 )∗ ), we obtain that (x∗1n , x∗2n ) = (y1∗ , x∗2n ). Hence x∗1n = y1∗ . Since x∗1n (x1n ) = 1, we obtain that x∗1n is a w∗ -exposed points of B(X1∗ ). Hence sup x∗1n (x1 ) = sup (x∗1n , 0)(x1 , 0) = sup (x∗1n , x∗2n )(x1 , 0) > 0

n∈N

n∈N

n∈N

for any x1 ∈ S(X1 ). Similarly, we obtain that x∗2n is a w∗ -exposed points of B(X2∗ ) and supn∈N x∗2n (x2 ) > 0 for any x2 ∈ S(X2 ). Therefore, by Lemma 2, we obtain that if p = 1, then X1 and X2 have the ball-covering property, which completes the proof. 2 Acknowledgment The authors would like to thank the anonymous referee for some suggestions to improve the manuscript. References [1] L. Cheng, Ball-covering property of Banach spaces, Israel J. Math. 156 (1) (2006) 111–123. [2] L. Cheng, Erratum to: “Ball-covering property of Banach spaces”, Israel J. Math. 184 (1) (2011) 505–507. [3] L. Cheng, Q. Cheng, X. Liu, Ball-covering property of Banach spaces is not preserved under linear isomorphisms, Sci. China Ser. A 51 (1) (2008) 143–147. [4] L. Cheng, M. Fabian, The product of a Gateaux differentiability space and a separable space is a Gateaux differentiability space, Proc. Amer. Math. Soc. 129 (2001) 3539–3541. [5] L. Cheng, Y. Ruan, Y. Teng, Approximation of convex functions on the dual of Banach spaces, J. Approx. Theory 116 (2002) 126–140. [6] L. Cheng, S. Shi, B. Wang, E.S. Lee, Generic Frechet differentiability of convex functions dominated by a lower semicontinuous convex function, J. Math. Anal. Appl. 225 (1998) 389–400. [7] L. Cheng, H. Shi, W. Zhang, Every Banach spaces with a w∗ -separable dual has an 1 + ε-equivalent norm with the ball covering property, Sci. China Ser. A 52 (9) (2009) 1869–1874. [8] L. Cheng, B. Wang, W. Zhang, Y. Zhou, Some geometric and topological properties of Banach spaces via ball coverings, J. Math. Anal. Appl. 377 (2011) 874–880. [9] M. Contreras, R. Paya, On upper semicontinuity of duality mappings, Proc. Amer. Math. Soc. 121 (2) (1994) 451–459. [10] J.R. Giles, D.A. Gregory, B. Sims, Geometrical implications of upper semi-continuity of the duality mapping on a Banach space, Pacific J. Math. 79 (1) (1978) 99–109. [11] J.R. Giles, S. Sciffer, Separable determination of Frechet differentiability of convex functions, Bull. Aust. Math. Soc. 52 (1995) 161–167. [12] W.B. Moors, Some more recent results concerning weak Asplund spaces, Abstr. Appl. Anal. 2005 (2005) 307–318. [13] W.B. Moors, S. Somasundaram, Some recent results concerning weak Asplund spaces, Acta Univ. Carolin. Math. Phys. 43 (2002) 67–86. [14] Warren B. Moors, Sivajah Somasundaram, A Gateaux differentiability space that is not weak Asplund, Proc. Amer. Math. Soc. 134 (2006) 2745–2754. [15] R.R. Phelps, Convex Functions, Monotone Operators and Differentiability, Lecture Notes in Math., vol. 1364, SpringerVerlag, New York, 1989. [16] D. Preiss, Differentiability of Lipschitz functions on Banach spaces, J. Funct. Anal. 91 (1990) 312–345. [17] D. Preiss, R. Phelps, I. Namioka, Smooth Banach spaces, weak Asplund spaces and monotone operators or usco mappings, Israel J. Math. 72 (1990) 257–279.

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