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

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 [5–12].

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 15–18, 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 such that is a exposed point of . Hence there exists such that for each . We claim that for any neighborhood of origin, there exists such that In fact, suppose that there exists a sequence such that as . Then we may assume without loss of generality that for any . Since is a bounded closed convex set of , we obtain that is compact. Then there exists a point such that is a accumulation point of . Put Hence we define an order by the containing relations, i.e., if and only if . This implies that is an order set. Hence is an order set whenever if and only if . Therefore, by the Zermelo Lemma, we obtain that there exists a mapping on such that Put . Hence we define a net . Therefore, by the definition of net , we havewhich contradicts . Moreover, since is a neighborhood of origin in , there exists neighborhood of origin, such that . Since there exists such that by formula , there exists such that and . Then Therefore, by , we have . Moreover, it is easy to see that . Then This implies that Hence we obtain that . This implies that is weak dentable, which finishes the proof.

Theorem 19. Suppose that is a Gteaux differentiable space. Then every bounded subset of is weak dentable.

Proof. By the proof of Theorem 14, we obtain that every closed convex subset of is weak dentable. Let be a bounded subset of . Suppose that is not weak dentable. Then, by Lemma 17, there exists a weak neighborhood of origin such that for any and . Since for any , we have . This implies that Therefore, by Lemma 17, we obtain that is not weak dentable, a contradiction, which finishes the proof.

Example 20. Let . Then . Since is separable, by Theorem 19, we obtain that every bounded subset of is weak dentable. Moreover, it is well known that has not the Radon-Nikodym property. Hence there exists a bounded subset of such that is not dentable.

Example 21. Let . Then is separable and is not a separable space. Let be a Banach space and be not a separable space. Define . Then By Theorem 19, we obtain that is weak dentable. By Example 13, we obtain that is -separable and bounded. Therefore, by Theorem 14, we obtain that every convex function is Gteaux differentiable at a dense subset of .

3. Gateaux Differentiability and Weakly Exposed Point

Definition 22. A point is said to be weakly exposed point of if there exist and such that ; then .

Definition 23. A point is said to be exposed point of if there exists such that whenever .

Definition 24. A point is said to be strongly exposed point of if there exist and such that ; then .

Definition 25. A point is said to be an extreme point of if and imply . The set of all extreme points of is denoted by . If , then is said to be a strictly convex space.
It is easy to see that if is a strongly exposed point of , then is a weakly exposed point of and if is a weakly exposed point of , then is a exposed point of . 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 is said to have the Krein-Milman property if every bounded closed convex subset of is the closed convex hull of its extreme points. It is well known that if has the Radon-Nikodym property, then has the Krein-Milman property. Moreover, we know that has the Krein-Milman property if and only if has the Radon-Nikodym property (see [12]).

Theorem 26. Suppose that is a bounded closed convex set. Then if and only if is a weakly exposed point of exposed by .

Proof. Necessity. Let and . Then we have . Therefore, by Lemma 1 of [1], we obtain that exposes at ; i.e., . Let and as . We next will prove that as . In fact, we may assume without loss of generality that for any . Since is a bounded closed convex set, we obtain that is a bounded set. Hence we obtain that is a bounded set. This implies that is a compact set in . Hence there exists such that is a accumulation point of . Hence there exists a net such that This implies that . Since , we have . Moreover, by , we obtain that .
Suppose that does not converge weakly to . Then there exist a weak neighbourhood of and a subsequence of such that . Repeat the previous proof; there exists a net such that , a contradiction. Hence we have as . This implies that is a weakly exposed point of .
Sufficiency. Suppose that there exists such that . Since is a weakly exposed point of and , we have We next will prove that . Suppose that . Then there exists a neighborhood of origin such that . Let Then, by , we obtain that . Moreover, by and , we obtain that there exists such that . Hence we have . Since is a weakly exposed point of and exposed by , by formula we obtain that , which contradicts . Hence . This implies that is a exposed point of exposed by . Therefore, by Lemma 1 of [3], we obtain that , which completes the proof.

Lemma 27 (see [1]). Suppose that , , and . If whenever satisfies then either or .

Theorem 28. The following statements are equivalent:
(1) is an Asplund space.
(2) has the Radon-Nikodym property.
(3) Every bounded closed convex subset of is the closed convex hull of its weakly exposed points.
(4) For every bounded closed convex set of , there exists such that is Gteaux differentiable at and .
(5) For every bounded closed convex set of , there exists a dense subset of such that is Gteaux differentiable on and .

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 is the closed convex hull of its weakly exposed points. Then every bounded closed convex subset of is the closed convex hull of its extreme points. Hence has the Krein-Milman property. This implies that has the Radon-Nikodym property.
(4)(2). Let be a bounded closed convex set of and be the closed convex hull of the extreme points of . Suppose that . Then, by the separation theorem and the Bishop-Phelps Theorem, there exist and such that Then is a nonempty bounded closed convex set; by Theorem 26, there exists such that is a weakly exposed point of . Then is an extreme point of . Let , , and . Then This implies that Hence and . Since is an extreme point of , we have . This implies that is an extreme point of , which contradicts . Hence every bounded closed convex subset of is the closed convex hull of its extreme points. This implies that has the Krein-Milman property. Hence 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 be a bounded closed convex subset of and be not a singleton. Then we may assume without loss of generality that and . Let and . Then we define the bounded closed convex set , where Then, by and , we have . Moreover, we may assume without loss of generality that . Let . By hypothesis, we obtain that is the closed convex hull of its weak exposed points, so any slice of contains a weakly exposed point of . Hence there exist and such that is a weakly exposed point of and is an exposed functional.
We claim that . In fact, suppose that . Since and are weak closed, we obtain that is weak closed. Therefore, by , there exists a weak neighborhood of origin such that . Moreover, by , we may assume without loss of generality that . Therefore, by , there exists a sequence such that as . Since is a weakly exposed point of and is an exposed functional, we have as , which contradicts . Hence we have . Since and , we have This implies that . Therefore, by formula , we have . Let . Then, by , we have Therefore, by symmetry of , we obtain that for all . Therefore, by Lemma 27, we obtain that either or . Since , we have This implies that . Therefore, by Theorem 26, there exists a dense subset of such that is Gteaux differentiable on and , which finishes the proof.

Moreover, by the proof of Theorems 26 and 28, it is easy to see the following.

Corollary 29. A Banach space 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 .

Theorem 30. A Banach space 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 .

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. Let be a weak Asplund space and continuous convex function be above bounded in neighborhood . Then is Gteaux differentiable at a dense subset of and .

In order to prove the theorem, we give a lemma.

Lemma 32. Suppose that graph of convex function has an interior point in . Then for any , we have .

Proof. Since continuous convex function has an interior point in , we obtain that . Pick . Then it is easy to see that . Therefore, by the separation theorem, there exists such that Hence, for any , we have and . Let . Then . This implies that . Suppose that . Then, for any , we obtain that . This implies that . Hence we have , a contradiction. This implies that . Hence we may assume without loss of generality that . This implies that . Hence . This implies that for any .
Let . We next will prove that . Since , we obtain that . This implies that . Hence, for any , we have This implies that We next will prove that the functional is a continuous functional of . Suppose that is not continuous at origin. Then there exist a net and such that and . Pick . Then, and This implies that the hyperplane is not a closed set. Pick Then Hence we have This implies that Moreover, there exist a open set of and an open interval such that . Pick . Then Therefore, by formula (92), there exists such that . Pick . Then . Therefore, by formula (85), we have This implies that , a contradiction. Hence we obtain that is continuous at origin. This implies that is a continuous functional of . Since , we have , which finishes the proof.

Proof of Theorem 31. Let be a neighbourhood and is above bounded on . Then we may assume without loss of generality that whenever . Then . Therefore, by Lemma 32, we have for any . Since is a weak Asplund space, we obtain that is Gteaux differentiable at a dense subset of . Hence convex function is Gteaux differentiable at a dense subset of and , which finishes the proof.

Definition 33. A point is called a smooth point if it has an unique supporting functional . If every is a smooth point, then is called a smooth space.

Definition 34. A Banach space is said to have the -property if , , and ; then as .

Example 35. It is well known that there exists a Banach space such that is reflexive and strictly convex and does not have the -property. Then it is easy to see that there exists such that is a weakly exposed point of and not a strongly exposed point of .

4. Some Examples in Orlicz Function Spaces

Definition 36. is called a -function if it has the following properties:
(1) is even, continuous, convex and .
(2) for all .
(3) and .

Let be a function and be a finite nonatomic measure space. Let denote the right derivative of and be the generalized inverse function of by Then the function defined by for any is called the complementary function to in the sense of Young. We define the modular of by Let us define the Orlicz function space by It is well known that and are Banach spaces when it is equipped with the Luxemburg norm or equipped with the Orlicz norm, denote Orlicz spaces equipped with the Luxemburg norm. , denote Orlicz spaces equipped with the Orlicz norm.

Definition 37 (see [12]). We say that an -function satisfies condition if there exist and such that whenever .

It is well known that and (see [12]). Moreover, it is well known that if and only if .

Theorem 38 (see [12]). Orlicz space is smooth if and only if is continuous.

Theorem 39 (see [12]). Orlicz space is strictly convex if and only if is strictly convex and .

Theorem 40 (see [12]). Orlicz space has the Radon-Nikodym property if and only if .

Example 41. There exist a bounded set and such that is a exposed point of and is not a weakly exposed point. Let , where is continuous, is strictly convex, , and . Since is strictly convex and , we obtain that is strictly convex. Since is continuous and , we obtain that is smooth. Since is continuous and is strictly convex, we obtain that is continuous. Since , we obtain that . Pick and such that . Pick such that . Then it is easy to see that , and Let Then, by Theorem 1.44 of [12], there exists such that Moreover, by the Bishop-Phelp theorem, there exist and such that and . Since , we have . This implies that . Therefore, by formula (101), we have Since is continuous, by formula (102) and Theorem 2.49 of [12], we obtain that is a smooth point. Let Then, by holder inequality, we have This implies that as . Moreover, by the Hahn-Banach theorem, there exists such that and . Hence Since is a smooth point and , by formulas (104) and (105), we obtain that is not a weakly exposed point of . Since is continuous, we obtain that is strictly convex. Therefore, by Theorem 2.4 of [12], we obtain that is strictly convex. This implies that is a exposed point of .

Example 42. Let and and . Then . Since , we obtain that is separable. Therefore, by Theorem 19 and , we obtain that every bounded subset of is weak dentable. Moreover, by Theorem 40, we obtain that has not the Radon-Nikodym property. Hence there exists a bounded subset of such that is not dentable.

Data Availability

The data used to support the findings of this study are available from the corresponding author upon request.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

Acknowledgments

This research is supported by “China Natural Science Fund under Grant 11871181” and “China Natural Science Fund under Grant 11561053”.