Journal of Function Spaces

Journal of Function Spaces / 2019 / Article

Research Article | Open Access

Volume 2019 |Article ID 6852859 | 12 pages | https://doi.org/10.1155/2019/6852859

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

Academic Editor: Gestur Ólafsson
Received15 Jan 2019
Accepted31 Mar 2019
Published02 May 2019

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