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Journal of Function Spaces
Volume 2019, Article ID 6852859, 12 pages
https://doi.org/10.1155/2019/6852859
Research Article

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

1Academy of Mathematical Sciences, Harbin Engineering University, Harbin 150001, China
2Department of Mathematics, Harbin University of Science and Technology University, Harbin 150080, China

Correspondence should be addressed to Shaoqiang Shang; moc.361@gnahsqs

Received 15 January 2019; Accepted 31 March 2019; Published 2 May 2019

Academic Editor: Gestur Ólafsson

Copyright © 2019 Shaoqiang Shang and Yunan Cui. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

In this paper, we prove that if is a -separable bounded subset of , then every convex function is Gteaux differentiable at a dense subset of if and only if every subset of is weakly dentable. Moreover, we also prove that if is a closed convex set, then if and only if is a weakly exposed point of exposed by . Finally, we prove that is an Asplund space if and only if, for every bounded closed convex set of , there exists a dense subset of such that is Gteaux differentiable on and . We also prove that is an Asplund space if and only if, for every -lower semicontinuous convex function , there exists a dense subset of such that is Gteaux differentiable on and .

1. Introduction and Preliminaries

Let denote a real Banach space. and denote the unit ball and unit sphere of , respectively. Let denote the dual space of . Let , and denote the sets of natural number, reals, and nonnegative reals, respectively. Let denote the closed ball centered at and of radius . Let denote that is weakly convergent to .

Let be a nonempty open convex subset of and a continuous convex function on . We called that is said to be Gteaux differentiable at the point in if the limitexists for all . Moreover, if the difference quotient in converges to uniformly for in the unit ball, then is said to be Frechet differentiable at .

Definition 1 (see [1]). is called a weak Asplund space [Asplund space] if, for every and as above, there exists a dense subset of such that is Gteaux [Frechet] differentiable at each point of .

It is well known that is weak Asplund space, but not Asplund space. Moreover, it is well known that is an Asplund space if and only if 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 Gteaux differentiable space if every convex continuous function on it is Gteaux differentiable at the points of a dense set.

In 2006, Waren B. Moors and Sivajah Somasundaram proved that there exists a Gteaux differentiable space that is not a weak Asplund space (see [2]). In 1979, D.G. Larman and R.R.Phelps proved that if is a strictly convex space, then is a weak Asplund space (see [3]). In 1997, Cheng proved that if is a continuous convex function on a Banach space , then every proper convex function on with is generically Frechet differentiable if and only if the image of the subdifferential map has the Radon-Nikodym property (see [4]).

Definition 3 (see [1]). A point is said to be exposed point of if there exists such that whenever .

Definition 4 (see [1]). Suppose that is a convex function on , then the set-valued mapping is said to be subdifferential mapping.

Remark 5. It is well known that if is a continuous convex function, then the set-valued mapping is norm- upper semicontinuous (see [1]). Moreover, it is well known that is a sington at if and if is Gteaux differentiable at (see [1]).

Let be a bounded subset of . Let and . Then it is easy to see that is a closed convex set. Define the sublinear functionalThen is a continuous sublinear functional. It is well known that , , and . Let be a continuous Minkowski functional on and . Then whenever . In this case, we called that is generated by . Let . ThenMoreover, it is well known thatIt is well known that if is a bounded subset of , then . If is a convex function, then the setdenotes epigraph of . It is well known that is closed if and only if is lower semicontinuous.

Lemma 6 (see [4]). Let be a continuous convex function on and . Let be the Minkowski functional generated by . Then if and only if with and .

Definition 7. A set is said to be weak dentable set if for any weak neighborhood of origin, there exists such that .

Definition 8 (see [5]). A set is said to be dentable set if, for any , there exists such that .

Definition 9 (see [5]). A Banach space is said to have the Radon-Nikodym property (see [1]) if is a nonatomic measure space and is a vector measure on with values in which is absolutely continuous with respect to and has a bounded variation, then there exists such that, for any ,It is well known that a Banach space has the Radon-Nikodym property if and only if every bounded subset of is dentable. By Definitions 8 and 9, it is easy to see that if is dentable, then 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 neighborhood is not a weak dentable set, where and .

Proof. Let , where and . Pick andThen, by the Hahn-Banach theorem, there exists such that . Define a weak neighborhoodof origin. Then, for any , we haveTherefore, by formula (7), we haveMoreover, for any , we haveTherefore, by formula (6), we haveHence we obtain thatThis implies that is not weak dentable, which finishes the proof.

Definition 11. A set is said to be -separable if there exists a sequence such that for any .

It is well known that if is a separable space, then every subset of is -separable.

Proposition 12. Suppose that is separable and . Then is -separable.

Proof. Since is a separable subset of , there exists a sequence such that . Then we may assume without loss of generality that for any . Hence there exists a sequence such that .
Pick . We will prove that . In fact, suppose that . Then for all . Since , there exists a natural number such that . Then . Hence we obtain that . Therefore, by for any , we have which contradicts for every . This implies that . Hence is -separable, which finishes the proof.

Example 13. Let . Then is separable and is not a separable space. Let be a Banach space and be not a separable space. Define . ThenLet . Since is separable, there exists such that . Then . HenceThis implies that is -separable and bounded. Moreover, it is easy to see that and are not separable and is not dentable.

The paper is organized as follows. In Section 1 some necessary definitions and notations are collected. In Section 2 we prove that if is a -separable bounded subset of , then every convex function is Gteaux differentiable at a dense subset of if and only if every subset of is weakly dentable. In Section 3 we prove that if is a closed convex set, then if and only if is a weakly exposed point of exposed by . Moreover, we also prove that is an Asplund space if and only if for every bounded closed convex set of , there exists a dense subset of such that is Gteaux differentiable on and . We also prove that is an Asplund space if and only if for every -lower semicontinuous convex function , there exists a dense subset of such that is Gteaux differentiable on and . 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 [512].

2. Gateaux Differentiability, Weakly Dentable Set, and -Separable Set

Theorem 14. Suppose that is a -separable bounded subset of . Then the following statements are equivalent.
(1) Every -lower semicontinuous convex function is Gteaux differentiable at a dense subset of .
(2) Every convex subset of is a weakly dentable set of .
(3) Every closed convex subset of is the closed convex hull of its exposed points.

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

Lemma 15. Suppose that
(1) is a -separable bounded subset of and is a closed convex set;
(2) is a continuous convex function and ;
(3) for any and weak neighborhood of origin, there exists a slice such that .
Then is Gteaux differentiable on a dense subset of .

Proof. We claim that if is -separable, then the set is -separable. In fact, since is -separable, there exists a sequence such that for any . Let , , and . Then, for any , we haveThis implies that . Then . Hence we obtain that is -separable.
Since is a -separable bounded subset of , there exists a sequence such that for any . Hence, for every natural number , we define a neighborhoodof origin in . Moreover, for every natural number , we define a weak neighborhoodof origin in . Hence, if and , then for all . Since for every , we have . Hence, for each , let be the set of all for which there exists a norm neighborhood of such that . Let . Pick and . Then and for every . Hence we obtain thatfor every . This implies that for every . Since , we have . Therefore, by the previous proof, we have . This implies that Hence we obtain that is Gteaux differentiable at each point of .
Since is a Baire space, we next will prove that, for any , the set is open and dense in . It is easy to see that is an open set. We next will prove that is dense in . Let and let be a neighborhood of in . We claim that . In fact, since is a -lower semicontinuous function on , we obtain that the set is a closed set of . Moreover, we may assume without loss of generality that . Let and . Since , we obtain that is continuous. Pick . Since is closed, by the separation theorem, there exists such that Hence we may assume without loss of generality that This implies that the set is a nonempty bounded closed convex subset of . Therefore, by the Bishop-Phelps Theorem, we obtain that is a dense set of . Hence is a dense set of . Therefore, by Lemma 6, it is easy to see that . Therefore, by formulas and , we obtain that . Pick . Then, by hypothesis, there exist a slice and such that . Moreover, if , then for some point and is in for sufficiently small . We claim that Indeed, if , then we have This implies that Since the set is a open set in and since is norm-to- upper semicontinuous, there exists such that and for any point . Moreover, since , we obtain that Pick Since , there exists a net such that . Therefore, by formula (31), we obtain that . Hence we may assume that . Moreover, by formula (31), there exists such that for any . Therefore, by , we obtain that for any . This implies that . Therefore, by formula , we have Therefore, by the previous proof, we obtain that We claim that for all . In fact, let . Then there exists a net such that . Hence, for any , we obtain that . Since , we have . This implies that . Hence . Since , by formulas (31) and (35), we have Since is arbitrary, we have . It follows that . This implies that . Hence is a dense open subset, which finishes the proof.

Lemma 16. Suppose that is a Banach space and is a bounded convex subset of . Then (1) is true, where
(1) for any continuous convex function on , if , then has the Gteaux differentiable points on ;
(2) for any weak neighborhood of origin and , there exist a slice and such that .

Proof. Suppose that there exist and a weak neighborhood of origin such that, for any weak slice and , we have . Since by formula (3) and convexity of , we have Hence the sublinear functional has the Gteaux differentiable points on . Since is a bounded subset of , we obtain that is a bounded subset of . Hence there exists such that whenever . This implies that for every . Hence is a continuous sublinear functional. Moreover, since is a weak neighborhood of origin, there exist and such that We will show that the function is nowhere Gteaux differentiable. Indeed, given any , for each slices , there exists such that . Hence there exist and such that . Otherwise, for any and , we have . Hence we have a contradiction. Hence we may assume without loss of generality that there exists a subsequence denoted again by , such that . Moreover, we may assume without loss of generality that there exists a subsequence denoted again by , such that . Therefore, by and , we obtain that This implies that Therefore, by formula (42), we have This implies that the sublinear functional is nowhere Gteaux differentiable, a contradiction, which finishes the proof.

Lemma 17. Let be a bounded subset of . Then the following statements are equivalent.
(1) For any weak neighborhood of origin, there exist a slice and such that .
(2) For any weak neighborhood of origin, there exists a point such that .
(3) For any weak neighborhood of origin, there exists a slice such that for any .

Proof. (3)(2). Let . Then, by condition (3), it is easy to see that, for any weak neighborhood of origin, there exists an open slice such that for any and . Hence, for any weak neighborhood of origin, there exists such that , it follows that . Sinceand we obtain that . This implies that Therefore, by , we obtain that .
(2)(1). For any weak neighborhood of origin, there exists such that . Therefore, by the separation theorem, there exist and such that Let . Then . This implies that . Hence, for any , we obtain that Therefore, by and formula (47), we obtain that .
(1)(3). For any weak neighborhood of origin, there exists a weak neighborhood of origin such that and there exists a slice and such that . Hence, if , then and . This implies thatwhich finishes the proof.

Lemma 18. Suppose that is a -separable bounded subset of . Then the following statements are equivalent.
(1) Every -lower convex function is Gteaux differentiable at a dense subset of .
(2) For any weak neighborhood of origin and , there exist a slice and such that .

Proof. (1)(2). Let . Suppose that there exist a set and a weak neighborhood of origin such that for any weak slice and , we obtain that . Since , we obtain that . Then Hence Moreover, by formula we obtain that Since , we obtain that Therefore, by formula , we obtain that is Gteaux differentiable at a dense subset of . However, by the proof of Lemma 16, we obtain that is nowhere Gteaux differentiable, a contradiction.
(2)(1). Let be a -lower semicontinuous convex function on and . Then is a continuous function. We claim that . In fact, suppose that there exists a point such that . Then, by the separation theorem, there exists a point and a real number such that Since , we obtain that for all . Let . Then This implies that for all sufficiently large , which contradicts formula . Therefore, by Lemma 16, we obtain that is Gteaux differentiable at a dense subset of , which finishes the proof.

Proof of Theorem 14. By Lemmas 1518, we obtain that is true. . Let and be a closed convex subset of . Then . This implies that is Gteaux differentiable at a dense subset for any . Therefore, by Theorem 2 of [3], we obtain that is the closed convex hull of its exposed points.
. Let be a closed convex subset of and be a weak neighborhood of origin. Since is bounded, by the definition of , we obtain that is bounded and is a bounded closed subset of . Hence is bounded and is a bounded closed subset of . Since is a weak neighborhood of origin, we may assume that there exist and such thatLet Then is a neighborhood of origin in . Since is a bounded closed subset of and , there exists </