# Gateaux Differentiability of Convex Functions and Weak Dentable Set in Nonseparable Banach Spaces

Authors: Shaoqiang Shang; Yunan Cui
Journal: Journal of Function Spaces
 (2019)

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## Abstract

In this paper, we prove that ifC⁎⁎ is a ε-separable bounded subset of X⁎⁎, then every convex function g≤σC is Ga^teaux differentiable at a dense Gδ subset G of X⁎ if and only if every subset of ∂σC(0)∩X is weakly dentable. Moreover, we also prove that if C is a closed convex set, then dσC(x⁎)=x if and only if x is a weakly exposed point of C exposed by x⁎. Finally, we prove that X is an Asplund space if and only if, for every bounded closed convex set C⁎ of X⁎, there exists a dense subset G of X⁎⁎ such that σC⁎ is Ga^teaux differentiable on G and dσC⁎(G)⊂C⁎. We also prove that X is an Asplund space if and only if, for every w⁎-lower semicontinuous convex function f, there exists a dense subset G of X⁎⁎ such that f is Ga^teaux differentiable on G and df(G)⊂X⁎.


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## Body

## 1. Introduction and Preliminaries

Let(X,·) denote a real Banach space. B(X) and S(X) denote the unit ball and unit sphere of X, respectively. Let X∗ denote the dual space of X. Let N,R, and R+ denote the sets of natural number, reals, and nonnegative reals, respectively. Let B(x,r) denote the closed ball centered at x and of radius r>0. Let xn→wx denote that {xn}n=1∞ is weakly convergent to x.LetD be a nonempty open convex subset of X and f a continuous convex function on D. We called that f is said to be Ga^teaux differentiable at the point x in D if the limit∗dfxy=limt→0fx+ty-fxtexists for all y∈X. Moreover, 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.Definition 1 (see [1]).
X is called a weak Asplund space [Asplund space] if, for every f and D as above, there exists a dense Gδ subset G of D such that f is Ga^teaux [Frechet] differentiable at each point of G.It is well known thatl1 is weak Asplund space, but not Asplund space. Moreover, it is well known that X is an Asplund space if and only if X∗ has the Radon-Nikodym property (see [1]). In 1933, Mazur proved that separable Banach spaces have the weak Asplund property (see [1]).Definition 2 (see [1]).
A Banach space is said to be a Ga^teaux differentiable space if every convex continuous function on it is Ga^teaux differentiable at the points of a dense set.In 2006, Waren B. Moors and Sivajah Somasundaram proved that there exists a Ga^teaux differentiable space that is not a weak Asplund space (see [2]). In 1979, D.G. Larman and R.R.Phelps proved that if X∗ is a strictly convex space, then X is a weak Asplund space (see [3]). In 1997, Cheng proved that if f is a continuous convex function on a Banach space X, then every proper convex function g on X with g≤f is generically Frechet differentiable if and only if the image of the subdifferential map ∂f has the Radon-Nikodym property (see [4]).Definition 3 (see [1]).
A pointx0∗∈C∗ is said to be weak∗ exposed point of C∗ if there exists x∈S(X) such that x0∗(x)>x∗(x) whenever x∗∈C∗\{x0∗}.Definition 4 (see [1]).
Suppose thatf is a convex function on X, then the set-valued mapping ∂f(x)={x∗∈X∗:(x∗,y-x)≤f(y)-f(x)forally∈X} is said to be subdifferential mapping.Remark 5.
It is well known that iff is a continuous convex function, then the set-valued mapping ∂f is norm-weak∗ upper semicontinuous (see [1]). Moreover, it is well known that ∂f is a sington at x if and if f is Ga^teaux differentiable at x (see [1]).LetC be a bounded subset of X. Let C∗=x∗∈X∗:x∗(x)≤1,x∈C and C∗∗=x∗∗∈X∗∗:x∗∗(x∗)≤1,x∗∈C∗. Then it is easy to see that C∗∗ is a weak∗ closed convex set. Define the sublinear functional(1)σCx∗=supx∗x:x∈C=supx∗x∗∗:x∗∗∈C∗∗.Then σC is a continuous sublinear functional. It is well known that ∂σC(0)=C∗∗, cow∗¯C=C∗∗, and C∗=x∗∈X∗:σC(x∗)≤1. Let p be a continuous Minkowski functional on X and C=x∈X:px≤1. Then p(x)=inf⁡{λ≥0:λ-1x∈C} whenever x∈X. In this case, we called that p is generated by C. Let C∗=x∗∈X∗:x∗(x)≤1,x∈C. Then(2)px=σC∗x=supx∗x:x∗∈C∗.Moreover, it is well known that(3)1∂p0=C∗;2x∗∈∂px⇔x∗∈C∗andx∗x=pxIt is well known that if C is a bounded subset of X, then σco¯C∪(-C)(x∗)=infλ:λ-1x∗∈C∗. If f is a convex function, then the set(4)epif=x,r∈X×R:fx≤r.denotes epigraph of f. It is well known that epif is closed if and only if f is lower semicontinuous.Lemma 6 (see [4]).
Letf be a continuous convex function on X and f(0)=-1. Let p:E×R→R be the Minkowski functional generated by epif. Then x∗∈∂f(x) if and only if (y∗,r)∈∂p(x,f(x)) with y∗=[x∗(x)-f(x)]-1x∗ and r=-[x∗(x)-f(x)]-1.Definition 7.
A setD⊂X is said to be weak dentable set if for any weak neighborhood U of origin, there exists z∈D such that z∉co¯D\(z+U).Definition 8 (see [5]).
A setD⊂X is said to be dentable set if, for any ε>0, there exists z∈D such that z∉co¯D\B(z,ε).Definition 9 (see [5]).
A Banach spaceX is said to have the Radon-Nikodym property (see [1]) 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 a bounded variation, then there exists f∈L1(X) such that, for any A∈Σ,(5)vA=∫Aftdt.
It is well known that a Banach spaceX has the Radon-Nikodym property if and only if every bounded subset of X is dentable. By Definitions 8 and 9, it is easy to see that if C is dentable, then C is weak dentable. Moreover, there exists a weak dentable set such that it is not dentable. We will give two examples in Sections 3 and 4.Proposition 10.
The weak neighborhoodA={x∈X:⋂i=1kxi∗,x≤r} is not a weak dentable set, where x1∗,x2∗,…,xk∗⊂S(X∗) and r∈(0,1).Proof.
LetA={x∈X:⋂i=1kxi∗,x≤r}, where x1∗,x2∗,…,xk∗⊂S(X∗) and r∈(0,1). Pick x0∈S(X) and(6)x0∈⋂i=1kx∈X:xi∗,x=0.Then, by the Hahn-Banach theorem, there exists x0∗∈S(X∗) such that x0∗(x0)=1. Define a weak neighborhood(7)V=x∈X:x0∗,x<14of origin. Then, for any x∈A, we have(8)x0∗,x+x0-x=x0∗,x0=1andx0∗,x-x0-x=x0∗,-x0=1.Therefore, by formula (7), we have(9)x+x0∉x+Vandx-x0∉x+V.Moreover, for any i∈{1,2,…,k}, we have(10)xi∗,x+x0=xi∗,x≤randxi∗,x-x0=xi∗,x≤r.Therefore, by formula (6), we have(11)x+x0∈A\x+Vandx-x0∈A\x+V.Hence we obtain that(12)x=12x+x0+12x-x0∈coA\x+V⊂co¯A\x+V.This implies that A is not weak dentable, which finishes the proof.Definition 11.
A setA⊂X∗ is said to be ε-separable if there exists a sequence {xn}n=1∞⊂X such that supn≥1x∗(xn)>0 for any x∗∈A\{0}.It is well known that ifX is a separable space, then every subset of X∗ is ε-separable.Proposition 12.
Suppose thatA⊂X∗ is separable and A≠{0}. Then A is ε-separable.Proof.
SinceA is a separable subset of X∗, there exists a sequence {xn∗}n=1∞⊂A such that {xn∗}n=1∞¯=A. Then we may assume without loss of generality that xn∗≠0 for any n∈N. Hence there exists a sequence {xn}n=1∞⊂S(X) such that xn∗,xn≥xn∗/2.
Picky∗∈A\{0}. We will prove that supn≥1y∗(xn)>0. In fact, suppose that supn≥1y∗(xn)≤0. Then xn,y∗≤0 for all n∈N. Since {xn∗}n=1∞¯=A, there exists a natural number n0 such that xn0∗-y∗<(1/8)y∗. Then y∗-xn0∗<(1/8)y∗. Hence we obtain that y∗<(8/7)xn0∗. Therefore, by xn,y∗≤0 for any n∈N, we have (13)xn0,xn0∗=xn0,xn0∗-y∗+xn0,y∗≤xn0,xn0∗-y∗+0≤xn0∗-y∗<18y∗<18·87xn0∗=17xn0∗,which contradicts xn∗,xn≥xn∗/2 for every n∈N. This implies that supn≥1y∗(xn)>0. Hence A is ε-separable, which finishes the proof.Example 13.
LetX=c0. Then X∗=l1 is separable and X∗∗=l∞ is not a separable space. Let Y be a Banach space and Y be not a separable space. Define C=B(X)×{0}⊂X×Y. Then(14)C∗=BX∗×Y∗⊂X∗×Y∗andC∗∗=BX∗∗×0⊂X∗∗×Y∗∗.Let (x∗∗,0)∈C∗∗\{(0,0)}. Since X∗ is separable, there exists {xn∗}n=1∞⊂S(X∗) such that {xn∗}n=1∞¯=S(X∗). Then supn≥1x∗∗(xn∗)>0. Hence(15)supn≥1x∗∗,0xn∗,0=supn≥1x∗∗,xn∗>0.This implies that C∗∗ is ε-separable and bounded. Moreover, it is easy to see that C∗ and C∗∗ are not separable and C is not dentable.The paper is organized as follows. In Section1 some necessary definitions and notations are collected. In Section 2 we prove that if C∗∗ is a ε-separable bounded subset of X∗∗, then every convex function g≤σC is Ga^teaux differentiable at a dense Gδ subset G of X∗ if and only if every subset of ∂σC(0)∩X is weakly dentable. In Section 3 we prove that if C is a closed convex set, then dσC(x∗)=x if and only if x is a weakly exposed point of C exposed by x∗. Moreover, we also prove that X is an Asplund space if and only if for every bounded closed convex set C∗ of X∗, there exists a dense subset G of X∗∗ such that σC∗ is Ga^teaux differentiable on G and dσC∗(G)⊂C∗. We also prove that X is an Asplund space if and only if for every w∗-lower semicontinuous convex function f, there exists a dense subset G of X∗∗ such that f is Ga^teaux differentiable on G and df(G)⊂X∗. In Section 4 we prove that there exists an exposed point such that it is not a weak exposed point in Orlicz function spaces. The topic of this paper is related to the topic of [5–12].


## 2. Gateaux Differentiability, Weakly Dentable Set, andε-Separable Set

Theorem 14.
Suppose thatC∗∗ is a ε-separable bounded subset of X∗∗. Then the following statements are equivalent.
(1) Everyw∗-lower semicontinuous convex function g≤σC is Ga^teaux differentiable at a dense Gδ subset G of X∗.
(2) Every convex subset of∂σC(0)∩X is a weakly dentable set of X.
(3) Everyweak∗ closed convex subset of ∂σC(0) is the weak∗ closed convex hull of its weak∗ exposed points.In order to prove the theorem, we give some lemmas.Lemma 15.
Suppose that
(1)C∗∗ is a ε-separable bounded subset of X∗∗ and C is a closed convex set;
(2)f is a continuous convex function and ∂f(X∗)⊂C∗∗;
(3) for anyD⊂C and weak neighborhood U of origin, there exists a slice S(x∗,C,α) such that S(x∗,D,α)-S(x∗,D,α)⊂U.
Thenf is Ga^teaux differentiable on a dense Gδ subset G of X∗.Proof.
We claim that ifC∗∗ is ε-separable, then the set C∗∗-C∗∗ is ε-separable. In fact, since C∗∗ is ε-separable, there exists a sequence {xn∗}n=1∞⊂X∗ such that supn≥1x∗∗(xn∗)>0 for any x∗∗∈C∗∗\{0}. Let x0∗∗∈C∗∗, y0∗∗∈C∗∗, and x0∗∗-y0∗∗≠0. Then, for any x∗∈C∗, we have(16)12x0∗∗-12y0∗∗,x∗≤12x0∗∗,x∗+12y0∗∗,x∗≤12+12=1.This implies that (x0∗∗-y0∗∗)/2∈C∗∗. Then supn≥1(x0∗∗-y0∗∗,xn∗)>0. Hence we obtain that C∗∗-C∗∗ is ε-separable.
SinceC∗∗-C∗∗ is a ε-separable bounded subset of X∗∗, there exists a sequence {xn∗}n=1∞⊂X∗ such that supn≥1x∗∗(xn∗)>0 for any x∗∗∈(C∗∗-C∗∗)\{0}. Hence, for every natural number n∈N, we define a weak∗ neighborhood(17)Wn=x∗∗∈X∗∗:⋂i=1nxi∗,x∗∗≤1nof origin in X∗∗. Moreover, for every natural number n∈N, we define a weak neighborhood(18)Un=x∈X:⋂i=1nxi∗,x≤1nof origin in X. Hence, if y∗∗∈⋂n=1∞Wn and y∗∗∈C∗∗-C∗∗, then xn∗,y∗∗=0 for all n∈N. Since supn≥1x∗∗(xn∗)>0 for every x∗∗∈C∗∗\{0}, we have y∗∗=0. Hence, for each n>1, let Gn be the set of all x∗∈X∗ for which there exists a norm neighborhood Vn of x∗ such that ∂f(Vn)-∂f(Vn)⊂Wn. Let x∗∈⋂n=1∞Gn. Pick x∗∗∈∂f(x∗) and y∗∗∈∂f(x∗). Then x∗∗∈∂f(Vn) and y∗∗∈∂f(Vn) for every n∈N. Hence we obtain that(19)x∗∗-y∗∗⊂∂fVn-∂fVn⊂Wnfor every n∈N. This implies that x∗∗-y∗∗,x∗=0 for every x∗∈{xn∗}n=1∞. Since ∂f(X∗)⊂C∗∗, we have x∗∗-y∗∗∈C∗∗-C∗∗. Therefore, by the previous proof, we have x∗∗=y∗∗. This implies that(20)⋂n=1∞Gn=x∗∈X∗:∂fx∗isasingleton. Hence we obtain that f is Ga^teaux differentiable at each point of G=⋂n=1∞Gn.
SinceX∗ is a Baire space, we next will prove that, for any n∈N, the set Gn is open and dense in X∗. It is easy to see that Gn is an open set. We next will prove that Gn is dense in X∗. Let x∗∈X∗ and let U be a neighborhood of x∗ in X∗. We claim that ∂f(U)∩X≠∅. In fact, since f is a w∗-lower semicontinuous function on X∗, we obtain that the set epif is a weak∗ closed set of X∗. Moreover, we may assume without loss of generality that f(0)=-1. Let C∗=epif and μC∗(x∗,r)=inf{λ∈R+:λ-1(x∗,r)∈C∗}. Since 0,0∈intC∗, we obtain that μC∗ is continuous. Pick (x∗,f(x∗)-1)∉epif. Since epif is weak∗ closed, by the separation theorem, there exists (x,r)∈X×R such that(21)x∗x+rfx∗-r≥sup⁡z∗x+rh:z∗,h∈epif. Hence we may assume without loss of generality that sup{z∗(x)+rh:z∗,h∈epif}=1. This implies that the set(22)C=x,r∈X×R:x∗x+rh≤1,x∗,h∈epif is a nonempty bounded closed convex subset of X×R. Therefore, by the Bishop-Phelps Theorem, we obtain that(23)x∗,r∈X∗×R:x∗x+rh=sup⁡x∗z+rl:z,l∈C,x,h∈C is a dense set of X∗×R. Hence(24)x∗,r∈X∗×R:∂μC∗x∗,r∩X×R≠∅ is a dense set of X∗×R. Therefore, by Lemma 6, it is easy to see that ∂f(U)∩X≠∅. Therefore, by formulas ∂f(U)⊂C∗∗ and C∗∗∩X=C, we obtain that ∂f(U)∩X⊂C. Pick n∈N. Then, by hypothesis, there exist a slice(25)Sz∗,∂fU∩X,α=x∈∂σCU∩X:z∗,x>σ∂fU∩Xz∗-α and x0∈S(z∗,∂f(U)∩X,α) such that S(z∗,∂f(U)∩X,α)⊂x0+Un. Moreover, if x∈S(z∗,∂f(U)∩X,α), then x∈∂f(x1∗)∩X for some point x1∗∈U and x0∗=x1∗+rz∗ is in U for sufficiently small r>0. We claim that(26)∂fx0∗⊂x∗∗∈X∗∗:z∗,x∗∗>σ∂fU∩Xz∗-α. Indeed, if y∗∗∈∂f(x0∗), then we have(27)0≤y∗∗-x,x0∗-x1∗=ry∗∗-x,z∗. This implies that(28)y∗∗∈x∗∗∈X∗∗:z∗,x∗∗>σ∂fU∩Xz∗-α. Since the set {x∗∗∈X∗∗:(z∗,x∗∗)>σ∂f(U)∩X(z∗)-α} is a weak∗ open set in X∗∗ and since ∂f is norm-to-weak∗ upper semicontinuous, there exists δ>0 such that B(x0∗,δ)⊂U and(29)∂fy∗⊂x∗∗∈X∗∗:z∗,x∗∗>σ∂fU∩Xz∗-α for any point y∗∈B(x0∗,δ). Moreover, since ∂f(y∗)⊂∂f(U), we obtain that(30)∂fy∗⊂x∗∗∈∂fU:z∗,x∗∗>σ∂fU∩Xz∗-α. Pick(31)z0∗∗∈x∗∗∈∂fU:z∗,x∗∗>σ∂fU∩Xz∗-α. Since Cw∗¯=C∗∗, there exists a net {zβ}β∈Δ⊂C such that zβ→w∗z0∗∗. Therefore, by formula (31), we obtain that (z∗,z0∗∗)>σ∂f(U)∩X(z∗)-α. Hence we may assume that (z∗,zβ)≥σ∂f(U)∩X(z∗)-α. Moreover, by formula (31), there exists z0∗∈U such that(32)z0∗∗,x∗-z0∗≤fx∗-fz0∗ for any x∗∈X∗. Therefore, by zβ→w∗z0∗∗, we obtain that(33)zβ,x∗-z0∗≤fx∗-fz0∗ for any x∗∈X∗. This implies that zβ∈∂f(U)∩X. Therefore, by formula (z∗,zβ)≥σ∂f(U)∩X(z∗)-α, we have (34)zβ∈x∈∂fU∩X:z∗,x≥σ∂fU∩Xz∗-α=Sz∗,∂fU∩X,α. Therefore, by the previous proof, we obtain that(35)zβ∈Sz∗,∂fU∩X,α⊂x0∗+Un. We claim that Unw∗¯⊂Wn for all n∈N. In fact, let z∗∗∈Unw∗¯. Then there exists a net {zα}α∈Δ⊂Un such that zα→w∗z∗∗. Hence, for any i∈{1,2,…,k}, we obtain that xi∗(zα)→xi∗(z∗∗). Since xi∗(zα)≤1/n, we have xi∗(z∗∗)≤1/n. This implies that z∗∗∈Wn. Hence Unw∗¯⊂Wn. Since zα→w∗z∗∗, by formulas (31) and (35), we have(36)z0∗∗∈x0∗+Unw∗¯=x0∗+Unw∗¯⊂x0∗+Wn. Since z0∗∗ is arbitrary, we have ∂f(y∗)⊂x0∗+Wn. It follows that ∂f(B(x0∗,δ))⊂x0∗+Wn. This implies that x0∗∈Gn∩U. Hence Gn is a dense open subset, which finishes the proof.Lemma 16.
Suppose thatX is a Banach space and C is a bounded convex subset of X. Then (1) ⇒(2) is true, where
(1) for any continuous convex functionf on X∗, if ∂f(X∗)⊂C¯w∗, then f has the Ga^teaux differentiable points on X∗;
(2) for any weak neighborhoodU of origin and D⊂C, there exist a slice S(x∗,D,α) and x∈S(x∗,D,α) such that S(x∗,D,α)⊂x+U.Proof.
Suppose that there existD⊂C and a weak neighborhood U of origin such that, for any weak slice S(x∗,D,α) and x∈S(x∗,D,α), we have S(x∗,D,α)⊂x+U. Since(37)σDx∗=sup⁡x,x∗:x∈D=sup⁡x∗∗,x∗:x∈cow∗¯D, by formula (3) and convexity of C, we have(38)∂σDX∗=∂σD0⊂cow∗¯D⊂C¯w∗. Hence the sublinear functional σD has the Ga^teaux differentiable points on X∗. Since C is a bounded subset of X, we obtain that D is a bounded subset of X. Hence there exists M>0 such that x<M whenever x∈D. This implies that σD(x∗)≤Mx∗ for every x∈D. Hence σD is a continuous sublinear functional. Moreover, since U is a weak neighborhood of origin, there exist ε>0 and {x1∗,…,xk∗}⊂X∗ such that(39)x∈X:⋂i=1kxi∗,x<ε⊂U. We will show that the function σD is nowhere Ga^teaux differentiable. Indeed, given any x∗∈X∗, for each slices S(x∗,D,ε/3n), there exists xn∈S(x∗,D,ε/3n) such that S(x∗,D,ε/3n)⊂xn+U. Hence there exist yn∈S(x∗,D,ε/3n) and i∈{1,…,k} such that |(xi∗,xn-yn)|≥ε. Otherwise, for any y∈S(x∗,D,ε/3n) and i∈{1,…,k}, we have |(xi∗,xn-yn)|<ε. Hence we have(40)Sx∗,D,13nε⊂x∈X:⋂i=1kxi∗,x-xn<ε⊂xn+U, a contradiction. Hence we may assume without loss of generality that there exists a subsequence {n} denoted again by {n}, such that |(x1∗,xn-yn)|≥ε. Moreover, we may assume without loss of generality that there exists a subsequence {n} denoted again by {n}, such that (x1∗,xn-yn)≥ε. Therefore, by xn∈S(x∗,D,ε/3n) and yn∈S(x∗,D,ε/3n), we obtain that(41)x∗,xn>σDx∗-13nεandx∗,xn>σDx∗-13nε. This implies that(42)2σDx∗<x∗,xn+13nε+x∗,yn+13nε=x∗,xn+yn+23nε. Therefore, by formula (42), we have (43)1nσDx∗+1nx1∗+σDx∗-1nx1∗-2σDx∗≥1nx∗+1nx1∗,xn+x∗-1nx1∗,yn-x∗,xn+yn-23nε≥1n1nx1∗,xn-yn-23nε≥1nεn-23nε=13ε. This implies that the sublinear functional σD is nowhere Ga^teaux differentiable, a contradiction, which finishes the proof.Lemma 17.
LetC be a bounded subset of X. Then the following statements are equivalent.
(1) For any weak neighborhoodU of origin, there exist a slice S(x∗,α,C) and x0∈S(x∗,α,C) such that S(x∗,α,C)⊂x0+U.
(2) For any weak neighborhoodU of origin, there exists a point x0∈C such that x0∉co¯(C\(x0+U)).
(3) For any weak neighborhoodU of origin, there exists a slice S(x∗,α,C) such that x1-x2∈U for any x1,x2∈S(x∗,α,C).Proof.
(3)⇒(2). Let σC(x∗)=sup{(x∗,y):y∈C}. Then, by condition (3), it is easy to see that, for any weak neighborhood U of origin, there exists an open slice S(x∗,α,C)={x∈C:x∗(x)>σC(x∗)-α} such that x1-x2∈U for any x1∈S(x∗,α,C) and x2∈S(x∗,α,C). Hence, for any weak neighborhood U of origin, there exists x0∈S(x∗,α,C) such that S(x∗,α,C)⊂x0+U, it follows that x∗(x0)>σC(x∗)-α. Since(44)C\x0+U⊂C\Sx∗,α,C⊂x∈C:x∗x≤σCx∗-αand(45)x∈C:x∗x≤σCx∗-α=C∩x∈X:x∗x≤σCx∗-α, we obtain that C\(x0+U)⊂{x∈C:x∗(x)≤σC(x∗)-α}. This implies that (46)σCx∗-α≥sup⁡x∗x:x∈C\x0+U=sup⁡x∗x:x∈coC\x0+U=sup⁡x∗x:x∈co¯C\x0+U. Therefore, by x∗(x0)>σC(x∗)-α, we obtain that x0∉co¯C\(x0+U).
(2)⇒(1). For any weak neighborhood U of origin, there exists x0∈C such that x0∉co¯(C\(x0+U)). Therefore, by the separation theorem, there exist x∗∈X∗ and r>0 such that(47)x∗x0-r>sup⁡x∗x:x∗∈co¯C\x0+U. Let α=σC(x∗)-x∗(x0)+r. Then x∗(x0)=σC(x∗)-α+r>σC(x∗)-α. This implies that x0∈S(x∗,C∗,α). Hence, for any y∈S(x∗,C,α), we obtain that(48)x∗y>σCx∗-α=σCx∗-σCx∗+x∗x0-r=x∗x0-r. Therefore, by x∗(y)>x∗(x0)-r and formula (47), we obtain that y∈x0+U.
(1)⇒(3). For any weak neighborhood U of origin, there exists a weak neighborhood V of origin such that V-V⊂U and there exists a slice S(x∗,C,α) and x0∈S(x∗,C,α) such that S(x∗,C,α)⊂x0+V. Hence, if x1,x2∈S(x∗,C,α), then x1∈x0+V and x2∈x0+V. This implies that(49)x1-x2=x0+V-x0+V=V-V⊂U.which finishes the proof.Lemma 18.
Suppose thatC∗∗ is a ε-separable bounded subset of X∗∗. Then the following statements are equivalent.
(1) Everyw∗-lower convex function g≤σC is Ga^teaux differentiable at a dense Gδ subset G of X∗.
(2) For any weak neighborhoodU of origin and D⊂∂σC(0)∩X, there exist a slice S(x∗,D,α) and x∈S(x∗,D,α) such that S(x∗,D,α)⊂x+U.Proof.
(1)⇒(2). Let p=σC. Suppose that there exist a set D⊂∂p(0)∩X and a weak neighborhood U of origin such that for any weak slice S(x∗,D,α) and x∈S(x∗,D,α), we obtain that S(x∗,D,α)⊂x+U. Since p=σC, we obtain that C∗={x∗∈X∗:p(x∗)≤1}. Then(50)∂p0=x∗∗∈X∗∗:x∗∗x∗≤1,x∗∈C∗and∂p0∩X¯w∗=∂p0. Hence(51)∂p0∩X=x∈X:x∗x≤1,x∗∈C∗andpx∗=σ∂p0x∗. Moreover, by formula(52)x∈X:x∗x≤1,x∗∈C∗¯w∗=x∗∗∈X∗∗:x∗∗x∗≤1,x∗∈C∗, we obtain that(53)px∗=σ∂p0x∗=σ∂p0∩Xx∗=sup⁡x∗x:x∈∂p0∩X. Since D⊂∂p(0)∩X, we obtain that(54)σ∂p0∩Xx∗=sup⁡x∗x:x∈∂p0∩X≥sup⁡x∗x:x∈D=σDx∗. Therefore, by formula σC=σ∂p(0)∩X≥σD, we obtain that σD is Ga^teaux differentiable at a dense Gδ subset G of X∗. However, by the proof of Lemma 16, we obtain that σD is nowhere Ga^teaux differentiable, a contradiction.
(2)⇒(1). Let g be a w∗-lower semicontinuous convex function on X∗ and g≤p. Then g is a continuous function. We claim that ∂g(X∗)⊂∂p(0). In fact, suppose that there exists a point x∗∗∈∂g(x∗) such that x∗∗∉∂p(0). Then, by the separation theorem, there exists a point x∗∈X∗ and a real number r>0 such that(55)x∗∗z∗>r+sup⁡z∗z∗∗:z∗∗∈∂p0≥r+pz∗. Since x∗∗∈∂g(x∗), we obtain that gy∗-gx∗≥x∗∗(y∗-x∗) for all y∗∈X∗. Let y∗=kz∗. Then(56)gkz∗-gx∗≥x∗∗kz∗-x∗∗x∗>12kr+kpz∗-x∗∗x∗. This implies that gkz∗>pkz∗ for all sufficiently large k>0, which contradicts formula g≤p. Therefore, by Lemma 16, we obtain that g is Ga^teaux differentiable at a dense Gδ subset G of X∗, which finishes the proof.Proof of Theorem14.
By Lemmas15–18, we obtain that (1)⇔(2) is true. (1)⇒(3). Let p=σC and D be a weak∗ closed convex subset of ∂p(0). Then p≥σD. This implies that σD′ is Ga^teaux differentiable at a dense Gδ subset for any D′⊂D. Therefore, by Theorem 2 of [3], we obtain that D is the weak∗ closed convex hull of its weak∗ exposed points.
( 3 ) ⇒ ( 2 ). Let D be a closed convex subset of ∂σC(0)∩X and V⊂X be a weak neighborhood of origin. Since C∗∗ is bounded, by the definition of C∗∗, we obtain that C is bounded and C∗∗ is a weak∗ bounded closed subset of X∗∗. Hence D is bounded and D∗∗ is a weak∗ bounded closed subset of X∗∗. Since V is a weak neighborhood of origin, we may assume that there exist {x1∗,x2∗,…,xk∗}⊂S(X∗) and ε>0 such that(57)V=⋂i=1kx∈X:xi∗,x<12ε.Let(58)V0=⋂i=1kx∗∗∈X∗∗:xi∗,x∗∗<12ε. Then V0 is a weak∗ neighborhood of origin in X∗∗. Since D∗∗ is a weak∗ bounded closed subset of X∗∗ and D∗∗⊂∂σC(0), there exists z0∗∗∈D∗∗ such that z0∗∗ is a weak∗ exposed point of D∗∗. Hence there exists x0∗∈S(X∗) such that(59)z0∗∗x0∗=sup⁡x∗∗x0∗:x∗∗∈D∗∗andy∗∗x0∗<sup⁡x∗∗x0∗:x∗∗∈D∗∗ for each y∗∗∈D∗∗\{z0∗∗}. We claim that for any weak∗ neighborhood U of origin, there exists k>0 such that(60)z0∗∗x0∗-4k≥sup⁡x∗∗x0∗:x∗∗∈D∗∗\z0∗∗+U. In fact, suppose that there exists a sequence {xn∗∗}n=1∞⊂D∗∗\(z0∗∗+U) such that xn∗∗(x0∗)→sup{x∗∗(x0∗):x∗∗∈D∗∗} as n→∞. Then we may assume without loss of generality that xn∗∗≠xm∗∗ for any m≠n. Since D∗∗ is a weak∗ bounded closed convex set of X∗∗, we obtain that D∗∗ is weak∗ compact. Then there exists a point x0∗∗∈D∗∗ such that x0∗∗ is a weak∗ accumulation point of {xn∗∗}n=1∞. Put(61)Δ=Ux0∗∗:Ux0∗∗isweak∗neighborhoodofx0∗∗. Hence we define an order by the containing relations, i.e., Ux0∗∗⊃Vx0∗∗ if and only if Vx0∗∗>Ux0∗∗. This implies that Δ is an order set. Hence(62)Ω=Ux0∗∗∩xn∗∗n=1∞:Ux0∗∗isweak∗neighbourhoodofx0∗∗ is an order set whenever Vx0∗∗∩{xn∗∗}n=1∞>Ux0∗∗∩{xn∗∗}n=1∞ if and only if Ux0∗∗∩{xn∗∗}n=1∞⊃Vx0∗∗∩{xn∗∗}n=1∞. Therefore, by the Zermelo Lemma, we obtain that there exists a mapping f on Ω such that f(Ux0∗∗∩{xn∗∗}n=1∞)∈Ux0∗∗∩{xn∗∗}n=1∞. Put xα∗∗=f(Ux0∗∗∩{xn∗∗}n=1∞). Hence we define a net {xα∗∗}α∈Δ⊂{xn∗∗}n=1∞. Therefore, by the definition of net {xα∗∗}α∈Δ, we have(63)x0∗∗x0∗=sup⁡x∗∗x0∗:x∗∗∈D∗∗andxα∗∗→w∗x0∗∗,which contradicts {xα∗∗}α∈Δ⊂{xn∗∗}n=1∞⊂D∗∗\(z0∗∗+U). Moreover, since V0 is a weak∗ neighborhood of origin in X∗∗, there exists weak∗ neighborhood W of origin, such that W+W⊂V0. Since there exists k>0 such that(64)z0∗∗x0∗-4k≥sup⁡x∗∗x0∗:x∗∗∈D∗∗\z0∗∗+W, by formula D¯w∗,X∗∗=D∗∗⊂X∗∗, there exists x∈D such that z0∗∗∈x+W and |z0∗∗(x0∗)-x(x0∗)|<k. Then(65)xx0∗-3k≥supx∗∗x0∗:x∗∗∈D∗∗\z0∗∗+W. Therefore, by z0∗∗∈x+W, we have z0∗∗+W⊂x+W+W⊂x+V0. Moreover, it is easy to see that D\(x+V)⊂D∗∗\(x+V0). Then(66)xx0∗-3k≥sup⁡x∗∗x0∗:x∗∗∈D∗∗\z0∗∗+W≥sup⁡x∗∗x0∗:x∗∗∈D∗∗\x+V0≥sup⁡zx0∗:z∈D\x+V. This implies that (67)xx0∗-3k≥sup⁡zx0∗:z∈D\x+V=sup⁡zx0∗:z∈coD\y+V=sup⁡zx0∗:z∈co¯D\x+V. Hence we obtain that x∉co¯(D\(x+V)). This implies that D is weak dentable, which finishes the proof.Theorem 19.
Suppose thatX∗ is a Ga^teaux differentiable space. Then every bounded subset of X is weak dentable.Proof.
By the proof of Theorem14, we obtain that every closed convex subset of X is weak dentable. Let C be a bounded subset of X. Suppose that C is not weak dentable. Then, by Lemma 17, there exists a weak neighborhood U of origin such that S(x∗,α,C)-S(x∗,α,C)⊂U for any x∗∈X∗ and α>0. Since σC(x∗)=σco¯(C)(x∗) for any x∗∈X∗, we have S(x∗,α,C)⊂S(x∗,α,co¯(C)). This implies that(68)Sx∗,α,co¯C-Sx∗,α,co¯C⊃Sx∗,α,C-Sx∗,α,C⊃U. Therefore, by Lemma 17, we obtain that co¯(C) is not weak dentable, a contradiction, which finishes the proof.Example 20.
LetX=c0. Then X∗=l1. Since l1 is separable, by Theorem 19, we obtain that every bounded subset of c0 is weak dentable. Moreover, it is well known that c0 has not the Radon-Nikodym property. Hence there exists a bounded subset D of c0 such that D is not dentable.Example 21.
LetX=c0. Then X∗=l1 is separable and X∗∗=l∞ is not a separable space. Let Y be a Banach space and Y be not a separable space. Define C=B(X)×{0}⊂X×Y. Then(69)C∗=BX∗×Y∗⊂X∗×Y∗andC∗∗=BX∗∗×0⊂X∗∗×Y∗∗. By Theorem 19, we obtain that C is weak dentable. By Example 13, we obtain that C∗∗ is ε-separable and bounded. Therefore, by Theorem 14, we obtain that every convex function g≤σC is Ga^teaux differentiable at a dense Gδ subset G of X∗.


## 3. Gateaux Differentiability and Weakly Exposed Point

Definition 22.
A pointx0∈C is said to be weakly exposed point of C if there exist x∗∈S(X∗) and {xn}n=1∞⊂C such that x∗(xn)→σC(x∗); then xn→wx0.Definition 23.
A pointx0∈C is said to be exposed point of C if there exists x∗∈S(X∗) such that x∗(x0)>x∗(x) whenever x∈C\{x0}.Definition 24.
A pointx0∈C is said to be strongly exposed point of C if there exist x∗∈S(X∗) and {xn}n=1∞⊂C such that x∗(xn)→σC(x∗); then xn→x0.Definition 25.
A pointx∈A is said to be an extreme point of A if 2x=y+z and y,z∈A imply y=z. The set of all extreme points of A is denoted by ExtA. If ExtB(X)=S(X), then X is said to be a strictly convex space.
It is easy to see that ifx is a strongly exposed point of C, then x is a weakly exposed point of C and if x is a weakly exposed point of C, then x is a exposed point of C. Moreover, weakly exposed point, exposed point, and strongly exposed point are different. We will give two examples in Sections 3 and 4. A Banach space X is said to have the Krein-Milman property if every bounded closed convex subset of X is the closed convex hull of its extreme points. It is well known that if X has the Radon-Nikodym property, then X has the Krein-Milman property. Moreover, we know that X∗ has the Krein-Milman property if and only if X∗ has the Radon-Nikodym property (see [12]).Theorem 26.
Suppose thatC is a bounded closed convex set. Then dσC(x∗)=x if and only if x is a weakly exposed point of C exposed by x∗.Proof.
Necessity. Letp(x∗)=σC(x∗) and dσC(x∗)=x. Then we have x∈C∗∗=∂p(0). Therefore, by Lemma 1 of [1], we obtain that x∗ exposes ∂p(0) at x; i.e., x∗(x)=σC(x∗). Let {xn}n=1∞⊂C and x∗(xn)→σC(x∗) as n→∞. We next will prove that xn→wx as n→∞. In fact, we may assume without loss of generality that xn≠xm for any m≠n. Since C is a bounded closed convex set, we obtain that C∗∗ is a bounded set. Hence we obtain that ∂p(0)=C∗∗ is a bounded set. This implies that ∂p(0) is a weak∗ compact set in X∗∗. Hence there exists x0∗∗∈∂p(0) such that x0∗∗ is a weak∗ accumulation point of {xn}n=1∞. Hence there exists a net {xα}α∈Δ⊂{xn}n=1∞ such that(70)x0∗∗x∗=sup⁡x∗∗x∗:x∗∗∈∂p0andxα→w∗x0∗∗. This implies that x0∗∗∈∂p(x∗). Since dp(x∗)=x, we have x0∗∗=x. Moreover, by xα→w∗x0∗∗, we obtain that xα→wx.
Suppose that{xn}n=1∞ does not converge weakly to x. Then there exist a weak neighbourhood V of x and a subsequence {nk} of {n} such that xnk∉V. Repeat the previous proof; there exists a net {xβ}β∈Δ⊂{xnk}k=1∞ such that xβ→wx, a contradiction. Hence we have xn→wx as n→∞. This implies that x is a weakly exposed point of C.
Sufficiency. Suppose that there existsx∗∗∈C∗∗ such that x∗∗(x∗)=x(x∗). Since x is a weakly exposed point of C and Cw∗¯=C∗∗, we have(71)x∗∗x∗=xx∗=sup⁡yx∗:y∈C=sup⁡y∗∗y∗:y∗∗∈C∗∗. We next will prove that x∗∗=x. Suppose that x∗∗≠x. Then there exists a weak∗ neighborhood V of origin such that x∗∗+V∩x+V=∅. Let(72)Un=y∗∗∈X∗∗:y∗∗x∗-x∗∗x∗<1n. Then, by x∗∗+V∩x+V=∅, we obtain that x∗∗+V∩Un∩x+V=∅. Moreover, by x∗∗∈C∗∗ and Cw∗¯=C∗∗, we obtain that there exists xn∈x∗∗+(V∩Un) such that xn∈C. Hence we have x∗(xn)→x∗(x)=supx∗(y):y∈C. Since x is a weakly exposed point of C and exposed by x∗, by formula(73)limn→∞x∗xn=x∗x=sup⁡x∗y:y∈C, we obtain that xn→wx, which contradicts x∗∗+V∩Un∩x+V=∅. Hence x∗∗=x. This implies that x∈C∗∗ is a weak∗ exposed point of C∗∗ exposed by x∗. Therefore, by Lemma 1 of [3], we obtain that dσC(x∗)=x, which completes the proof.Lemma 27 (see [1]).
Suppose thatx∗∈S(X∗), y∗∈S(X∗), and ε>0. If y∗(x)≤1 whenever x∈X satisfies(74)x∗x=0andx≤2ε-1, then either x∗-y∗≤ε or x∗+y∗≤ε.Theorem 28.
The following statements are equivalent:
(1)X is an Asplund space.
(2)X∗ has the Radon-Nikodym property.
(3) Every bounded closed convex subset ofX∗ is the closed convex hull of its weakly exposed points.
(4) For every bounded closed convex setC∗ of X∗, there exists x∗∗∈X∗∗ such that σC∗ is Ga^teaux differentiable at x∗∗ and dσC∗(x∗∗)∈C∗.
(5) For every bounded closed convex setC∗ of X∗, there exists a dense subset G of X∗∗ such that σC∗ is Ga^teaux differentiable on G and dσC∗(G)⊂C∗.Proof.
It is well known that (1)⇔(2) and (2)⇒(3) are true. Moreover, by Theorem 26, it is easy to see that (3)⇒(4) is true. (3)⇒(2). Suppose that every bounded closed convex subset of X∗ is the closed convex hull of its weakly exposed points. Then every bounded closed convex subset of X∗ is the closed convex hull of its extreme points. Hence X∗ has the Krein-Milman property. This implies that X∗ has the Radon-Nikodym property.
(4)⇒(2). Let C∗ be a bounded closed convex set of X∗ and J be the closed convex hull of the extreme points of C∗. Suppose that J≠C∗. Then, by the separation theorem and the Bishop-Phelps Theorem, there exist x0∗∗∈X∗∗ and x0∗∈X∗ such that(75)sup⁡x0∗∗x∗:x∗∈C∗=x0∗∗x0∗>sup⁡x0∗∗x∗:x∗∈J. Then(76)D=x∈X:x0∗∗x∗=sup⁡x0∗∗y∗:y∗∈C∗ is a nonempty bounded closed convex set; by Theorem 26, there exists x1∗∈D∗ such that x1∗ is a weakly exposed point of D∗. Then x1∗ is an extreme point of D∗. Let 2x1∗=y1∗+z1∗, y1∗∈C∗, and z1∗∈C∗. Then(77)x0∗∗2x1∗=x0∗∗y1∗+z1∗=x0∗∗y1∗+x0∗∗z1∗=2sup⁡x0∗∗y∗:y∈C∗. This implies that(78)x0∗∗y1∗=x0∗∗z1∗=sup⁡x0∗∗y∗∗:y∗∗∈C∗∗. Hence y1∗∈D∗ and z1∗∈D∗. Since x1∗ is an extreme point of D∗, we have y1∗=z1∗. This implies that x1∗ is an extreme point of C∗, which contradicts J≠C∗. Hence every bounded closed convex subset of X∗ is the closed convex hull of its extreme points. This implies that X∗ has the Krein-Milman property. Hence X∗ has the Radon-Nikodym property.
It is easy to see that (5)⇒(4) is true. We next will prove that (4)⇒(5) is true. Let C∗ be a bounded closed convex subset of X∗ and C∗ be not a singleton. Then we may assume without loss of generality that C∗⊂B(X∗) and 0∈C∗. Let x∗∗=1 and ε∈(0,1/8). Then we define the bounded closed convex set C1∗=co¯wC∗∪N, where(79)N=x∗∈X∗:x∗,x∗∗=0,x∗≤2ε-1. Then, by C∗⊂B(X∗) and ε∈(0,1/8), we have C1∗⊂B(X∗). Moreover, we may assume without loss of generality that σC1∗(x∗∗)-ε>0. Let α=σC1∗(x∗∗)-ε. By hypothesis, we obtain that C1∗ is the closed convex hull of its weak exposed points, so any slice of C1∗ contains a weakly exposed point of C∗. Hence there exist y∗∈S(x∗∗,C1∗,α) and y∗∗∈S(X∗∗) such that y∗ is a weakly exposed point of C1∗ and y∗∗ is an exposed functional.
We claim thaty∗∈C∗∪N. In fact, suppose that y∗∉C∗∪N. Since C∗ and N are weak closed, we obtain that C∗∪N is weak closed. Therefore, by y∗∉C∗∪N, there exists a weak neighborhood U of origin such that (y∗+U)∩C∗∪N=∅. Moreover, by 0∈C∗, we may assume without loss of generality that y∗∗(y∗)>0. Therefore, by C1∗=co¯wC∗∪N, there exists a sequence {yn∗}n=1∞⊂C∗∪N such that y∗∗(yn∗)→σC1∗(y∗∗) as n→∞. Since y∗ is a weakly exposed point of C1∗ and y∗∗ is an exposed functional, we have yn∗→wy∗ as n→∞, which contradicts (y∗+U)∩C∗∪N=∅. Hence we have y∗∈C∗∪N. Since y∗∈S(x∗∗,C1∗,α) and α=σC1∗(x∗∗)-ε, we have(80)x∗∗y∗≥σC1∗x∗∗-α=σC1∗x∗∗-σC1∗x∗∗-ε=ε. This implies that y∗∉N. Therefore, by formula y∗∈C∗∪N, we have y∗∈C∗. Let x∗∈N⊂C1∗. Then, by C1∗⊂B(X∗), we have(81)x∗,y∗∗≤σC1∗y∗∗=y∗∗,y∗≤y∗∗y∗≤1. Therefore, by symmetry of N, we obtain that x∗,y∗∗≤1 for all x∗⊂N⊂C1∗. Therefore, by Lemma 27, we obtain that either x∗∗-y∗∗≤ε or x∗∗+y∗∗≤ε. Since y∗≤1, we have(82)x∗∗+y∗∗≥y∗,x∗∗+y∗∗>y∗,x∗∗≥ε. This implies that x∗∗-y∗∗≤ε. Therefore, by Theorem 26, there exists a dense subset G of X∗∗ such that σC∗ is Ga^teaux differentiable on G and dσC∗(G)⊂C∗, which finishes the proof.Moreover, by the proof of Theorems26 and 28, it is easy to see the following.Corollary 29.
A Banach spaceX is an Asplund space if and only if for every bounded closed convex set C∗ of X∗, there exists a dense subset G of X∗∗ such that σC∗ is Ga^teaux differentiable on G and dσC∗(G)⊂X∗.Theorem 30.
A Banach spaceX is an Asplund space if and only if, for every w∗-lower semicontinuous convex function f, there exists a dense subset G of X∗∗ such that f is Ga^teaux differentiable on G and df(G)⊂X∗.Proof.
Noticing the proof of Theorem 2.1 of [9] and Corollary 29, it is easy to see that Theorem 30 is true, which finishes the proof.Theorem 31.
LetX∗ be a weak Asplund space and continuous convex function f be above bounded in weak∗ neighborhood U. Then f is Ga^teaux differentiable at a dense Gδ subset G of X∗ and df(G)⊂X.In order to prove the theorem, we give a lemma.Lemma 32.
Suppose that graph of convex functionf has an interior point in (X∗×R,w∗). Then for any x∗∈X∗, we have ∂f(x∗)⊂X.Proof.
Since continuous convex functionf has an interior point in (X∗×R,w∗), we obtain that w∗-int(epif)≠∅. Pick x∗∈X∗. Then it is easy to see that (x∗,f(x∗))∉w∗-int(epif). Therefore, by the separation theorem, there exists (x,r)∈(X×R)\(0,0) such that(83)x∗x+rfx∗≥sup⁡z∗x+rξ:z∗,ξ∈w∗-intepif. Hence, for any y∗∈X∗, we have x∗(x)+rf(x∗)≥y∗(x)+rf(y∗) and x∗(x)+rf(x∗)≥y∗(x)+r(f(y∗)+1). Let y∗=x∗. Then rf(x∗)≤r(f(x∗)+1). This implies that r≤0. Suppose that r=0. Then, for any y∗∈X∗, we obtain that x∗(x)≥y∗(x). This implies that x=0. Hence we have (x,r)=(0,0), a contradiction. This implies that r<0. Hence we may assume without loss of generality that r=-1. This implies that x∗(x)-f(x∗)≥y∗(x)-f(y∗). Hence x∈∂f(x∗). This implies that ∂f(x∗)≠∅ for any x∗∈X∗.
Letx∗∗∈∂f(x∗). We next will prove that x∗∗∈X. Since x∗∗∈∂f(x∗), we obtain that x∗∗,y∗-x∗≤f(y∗)-f(x∗). This implies that x∗∗(y∗)-f(y∗)≤x∗∗(x∗)-f(x∗). Hence, for any (y∗,t)∈w∗-int(epif), we have(84)x∗∗y∗-t≤x∗∗y∗-fy∗≤x∗∗x∗-fx∗. This implies that(85)x∗∗x∗-fx∗≥sup⁡x∗∗y∗-t:y∗,t∈w∗-intepif. We next will prove that the functional x∗∗ is a continuous functional of (X∗,w∗). Suppose that x∗∗ is not continuous at origin. Then there exist a net {xα∗}α∈Δ⊂X∗ and r>0 such that xα∗→w∗0 and x∗∗(xα∗)>r. Pick x0∗∈{xα∗}α∈Δ. Then,(86)xα∗x∗∗xα∗-x0∗x∗∗x0∗→w∗-x0∗x∗∗x0∗,x∗∗,-x0∗x∗∗x0∗=-1 and(87)x∗∗,xα∗x∗∗xα∗-x0∗x∗∗x0∗=0. This implies that the hyperplane h∗∈X∗:x∗∗(h∗)=0 is not a weak∗ closed set. Pick(88)z∗∈h∗∈X∗:x∗∗h∗=0¯w∗\h∗∈X∗:x∗∗h∗=0. Then(89)λz∗:λ∈R⊂h∗∈X∗:x∗∗h∗=0¯w∗. Hence we have (90)X∗=λz∗:λ∈R⊕h∗∈X∗:x∗∗h∗=0⊂h∗∈X∗:x∗∗h∗=0¯w∗. This implies that(91)X∗=h∗∈X∗:x∗∗h∗=0¯w∗. Moreover, there exist a weak∗ open set U of X∗ and an open interval (a,b) such that U×(a,b)⊂w∗-int(epif). Pick h0∗∈h∗∈X∗:x∗∗(h∗)=0. Then (92)X∗=h0∗+X∗=h0∗+h∗∈X∗:x∗∗h∗=0¯w∗=h0∗+h∗∈X∗:x∗∗h∗=0¯w∗=h∗∈X∗:x∗∗h∗=x∗∗x∗-fx∗+b¯w∗. Therefore, by formula (92), there exists x0∗∈U such that x∗∗(x0∗)=x∗∗(x∗)-f(x∗)+b. Pick r∈(a,b). Then (x0∗,r)∈U×(a,b)⊂w∗-int(epif). Therefore, by formula (85), we have(93)x∗∗x∗-fx∗≥x∗∗x0∗-r=x∗∗x∗-fx∗+b-r. This implies that r≥b, a contradiction. Hence we obtain that x∗∗ is continuous at origin. This implies that x∗∗ is a continuous functional of (X∗,w∗). Since (X∗,w∗)∗=X, we have x∗∗∈X, which finishes the proof.Proof of Theorem31.
LetU be a weak∗ neighbourhood and f is above bounded on U. Then we may assume without loss of generality that f(x∗)<0 whenever x∗∈U. Then U×(0,1)⊂epif. Therefore, by Lemma 32, we have ∂f(x∗)⊂X for any x∗∈X∗. Since X∗ is a weak Asplund space, we obtain that f is Ga^teaux differentiable at a dense Gδ subset G of X∗. Hence convex function f is Ga^teaux differentiable at a dense Gδ subset G of X∗ and df(G)⊂X, which finishes the proof.Definition 33.
A pointx∈S(X) is called a smooth point if it has an unique supporting functional fx. If every x∈S(X) is a smooth point, then X is called a smooth space.Definition 34.
A Banach spaceX is said to have the H-property if {xn}n=1∞⊂S(X), x∈S(X), and xn→wx; then xn→x as n→∞.Example 35.
It is well known that there exists a Banach spaceX such that X is reflexive and strictly convex and does not have the H-property. Then it is easy to see that there exists x∈B(X) such that x is a weakly exposed point of B(X) and not a strongly exposed point of B(X).


## 4. Some Examples in Orlicz Function Spaces

Definition 36.
M : R → R is called a N-function if it has the following properties:
(1)M is even, continuous, convex and M(0)=0.
(2)M(u)>0 for all u≠0.
(3)limu→0M(u)/u=0 and limu→∞M(u)/u=∞.LetM be a N-function and (G,Σ,μ) be a finite nonatomic measure space. Let p(u) denote the right derivative of M(u) and q(v) be the generalized inverse function of p(u) by(94)qv=supu≥0u≥0:pu≤v. Then the function N(v) defined by N(v)=∫0|v|q(s)ds for any v∈R is called the complementary function to M in the sense of Young. We define the modular of x by(95)ρMx=∫GMxtdt. Let us define the Orlicz function space LM by(96)LM=xt:ρMλx<∞forsomeλ>0EM=xt:ρMλx<∞forallλ>0. It is well known that LM and EM are Banach spaces when it is equipped with the Luxemburg norm(97)x=inf⁡λ>0:ρMxλ≤1 or equipped with the Orlicz norm(98)x0=infk>01k1+ρMkx.LM,EM denote Orlicz spaces equipped with the Luxemburg norm. LM0,EM0 denote Orlicz spaces equipped with the Orlicz norm.Definition 37 (see [12]).
We say that anN-function M satisfies condition Δ2 if there exist K>2 and u0≥0 such that M(2u)≤KM(u) whenever u≥u0.It is well known that(EN)∗=LM0 and (EN0)∗=LM (see [12]). Moreover, it is well known that EM=LM if and only if M∈Δ2.Theorem 38 (see [12]).
Orlicz spaceEM is smooth if and only if p is continuous.Theorem 39 (see [12]).
Orlicz spaceLM is strictly convex if and only if M is strictly convex and M∈Δ2.Theorem 40 (see [12]).
Orlicz spaceEM(EM0) has the Radon-Nikodym property if and only if M∈Δ2.Example 41.
There exist a bounded setC⊂X and x∈C such that x is a exposed point of C and is not a weakly exposed point. Let X=LN0, where q is continuous, N is strictly convex, N∈Δ2, and M∉Δ2. Since N is strictly convex and N∈Δ2, we obtain that LN0 is strictly convex. Since q is continuous and N∈Δ2, we obtain that LN0 is smooth. Since q is continuous and N is strictly convex, we obtain that p is continuous. Since M∉Δ2, we obtain that LM\EM≠∅. Pick r∈(0,1/8) and u1∈LM\EM such that u1<r. Pick u2∈EM such that u2=8. Then it is easy to see that u1+u2∈LM\EM, u1+u2>8-r and(99)distu1+u2,EM≤u1+u2-u2=u1<r. Let(100)u′=u1+u2u1+u2. Then, by Theorem 1.44 of [12], there exists a∈(0,1) such that (101)θu′=inf⁡λ>0:ρMu′λ<∞=distu′,EM=distu1+u2,EMu1+u2<r8-r=8a<1-4a<1. Moreover, by the Bishop-Phelp theorem, there exist u∈S(LM0) and v∈S(LN) such that u,v=∫Gu(t),v(t)dt=1 and u-u′<a. Since dist(u′,EM)=8a, we have dist(u,EM)>4a. This implies that u∈LM\EM. Therefore, by formula (101), we have(102)θu=inf⁡λ>0:ρMuλ<∞=distu,EM≤10a<1. Since p is continuous, by formula (102) and Theorem 2.49 of [12], we obtain that u is a smooth point. Let(103)En=t∈G:ut≤nandun=uχEn. Then, by holder inequality, we have(104)∫GuχEnt-ut,vtdt≤uχEn-u0vχG\En≤vχG\En→0. This implies that uχEn-u,v→1 as n→∞. Moreover, by the Hahn-Banach theorem, there exists ϕ∈LM0∗ such that ϕ(v)>0 and ϕ(EM0)=0. Hence(105)ϕuχEn-u=ϕuχEn-ϕu=-ϕu<0. Since u is a smooth point and u,v=1, by formulas (104) and (105), we obtain that u is not a weakly exposed point of B(LM0). Since q is continuous, we obtain that M is strictly convex. Therefore, by Theorem 2.4 of [12], we obtain that LM0 is strictly convex. This implies that u is a exposed point of B(LM0).Example 42.
LetX=EN0 and M∈Δ2 and N∉Δ2. Then X∗=LM. Since M∈Δ2, we obtain that LM is separable. Therefore, by Theorem 19 and (EN0)∗=LM, we obtain that every bounded subset of EN0 is weak dentable. Moreover, by Theorem 40, we obtain that EN0 has not the Radon-Nikodym property. Hence there exists a bounded subset D of EN0 such that D is not dentable.



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*Source: 6852859-2019-05-02.xml*