Weak approximative compactness of hyperplane and Asplund property in Musielak-Orlicz-Bochner function spaces

In this paper, some criteria for weakly approximative compactness and approximative compactness of weak \begin{document}$ ^{*} $\end{document} hyperplane for Musielak-Orlicz-Bochner function spaces are given. Moreover, we also prove that, in Musielak-Orlicz-Bochner function spaces generated by strongly smooth Banach space, \begin{document}$ L_{M}^{0}(X) $\end{document} (resp \begin{document}$ L_{M}(X) $\end{document} ) is an Asplund space if and only if \begin{document}$ M $\end{document} and \begin{document}$ N $\end{document} satisfy condition \begin{document}$ \Delta $\end{document} . As a corollary, we obtain that \begin{document}$ L_{M}^{0}(R) $\end{document} (resp \begin{document}$ L_{M}(R) $\end{document} ) is an Asplund space if and only if \begin{document}$ M $\end{document} and \begin{document}$ N $\end{document} satisfy condition \begin{document}$ \Delta $\end{document} .


Introduction and preliminaries
The study of Orlicz function space originated in the last century. Orlicz function space is an important class of Banach spaces. Orlicz function space has important applications in the field of partial differential equations. However, with the development of differential equation theory, Orlicz function space can no longer satisfy the development of theory of differential equation (see [1], [6]- [11] and [15]- [21]). Mathematicians began to pay attention to the extended form of Orlicz function space. Musielak-Orlicz-Bochner function space is an important extension of Orlicz function space. The development of theory of Musielak-Orlicz-Bochner function space provides theoretical basis for the development of differential equations. In this paper, some criteria for weakly approximative compactness and approximative compactness of weak * hyperplane for Musielak-Orlicz-Bochner function spaces are given. Moreover, we also prove that, in Musielak-Orlicz-Bochner function spaces generated by strongly smooth Banach space, L 0 M (X) (resp L M (X)) is an Asplund space if and only if M ∈ ∆ and N ∈ ∆. As a corollary, we obtain that L 0 M (R) (resp L M (R)) is an Asplund space if and only if M ∈ ∆ and N ∈ ∆.
Let (X, · ) be a real Banach space, S(X) and B(X) denote the unit sphere and unit ball of X, respectively. By X * we denote the dual space of X. Let N, R and R + denote the sets natural numbers, reals and nonnegative reals, respectively. Let us take a point x ∈ S(X) and let H x = {x * ∈ X * : x * (x) = 1}. Let C ⊂ X be a nonempty subset of X. Then the set-valued mapping P C : X → C P C (x) = z ∈ C : x − z = dist (x, C) = inf y∈C x − y is called the metric projection operator from X onto C. First let us recall some definitions and results that will be used in the further part of the paper. Definition 1.1. A nonempty subset C of X is said to be approximatively compact (weakly approximatively compact) if for any sequence {y n } ∞ n=1 ⊂ C and any x ∈ X satisfying x − y n → inf y∈C x − y as n → ∞, there exists a subsequence converging (weakly) to an element of C. Definition 1.2. A point x ∈ S(X) is called a smooth point if it has a unique supporting functional f x ∈ S(X * ). If every x ∈ S(X) is a smooth point, then X is called a smooth space.
Consider a convex subset A of a Banach space X. A point x ∈ 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 strictly convex. Moreover, it is well known that if X * is strictly convex, then X is smooth. Definition 1.3. A point x ∈ S(X) is said to be a strongly smooth point of X if there exist {x * n } ∞ n=1 ⊂ S(X * ) and x * 0 ∈ S(X * ) such that x * n → x * 0 whenever x * n (x) → 1. A Banach space X is said to be strongly smooth if every point of S(X) is strongly smooth point of X. Definition 1.4. A Banach space X is said to have the Radon-Nikodym property if let (T, Σ, µ) be nonatomic measurable space. v is a measure and v is bounded variation and absolutely continuous with respect to µ, then there exists an integrable function f such that for any A ∈ Σ, we have It is well known that if X is strongly smooth, then X has the Radon-Nikodym property. Moreover, it is easy to see that if X is a strongly smooth space, then X is smooth. Let f be a real continuous convex function on X. Recall that f is said to be Gâteaux differentiable at the point x in X if the limit exists for all y ∈ X. If the difference quotient in ( * ) converges to df (x)(y) uniformly for y in the unit ball B(X), then f is said to be Frechet differentiable at x. X is called an Asplund space if for every continuous convex function on X, there exists a dense G δ subset G of X such that it is Frechet differentiable at each point of G.
It is well known that if X is an Asplund space, then X * is separable if and only if X is separable. Moreover, It is well known that X is an Asplund space if and only if X * has the Radon-Nikodym property. Let (T, Σ, µ) be a complete nonatomic measurable space. Suppose that a function M : T × [0, ∞) → [0, ∞] statisfies the following conditions.
(1) for µ-a.e, t ∈ T , M (t, 0) = 0, lim u→∞ M (t, u) = ∞ and M (t, u ) < ∞ for some Every such a function M is called a Musielak-Orlicz-function. Let p(t, ·) denote the right derivative of M (t, ·) at u ∈ R + (where p(t, u) = ∞ if M (t, u) = ∞) and let q(t, ·) be the generalized inverse function of p(t, ·) defined on R + by Then the function N (t, v) defined by N (t, v) = |v| 0 q(t, s)ds for any v ∈ R and Σ-a.e. t ∈ T is called the complementary function to M in the sense of Young. It is well known that the Young inequality uv ≤ M (t, u) + N (t, v) holds for all u, v, ∈ R and µ-a.e. t ∈ T . Moreover, for any u ∈ R the equality uv = M (t, u) + N (t, v) holds if and only if v ∈ [p − (t, u), p(t, u)]. Let Lemma 1.8. (see [2]) M / ∈ ∆ ⇔ for any ε > 0, there exists u ∈ L M (X) such that ρ M (u) = ε and u(t) < E(t) for almost all t ∈ T .

Strongly smooth point and approximative compactness in
Musielak-Orlicz-Bochner function spaces Theorem 2.1. Suppose that X is a strongly smooth space and v ∈ S(E 0 N (X)). Then the following statements are equivalent: (1) The point v is a strongly smooth point of E 0 N (X); In order to prove the theorem, we first give some lemmas.

Lemma 2.2.
Suppose that X is a Banach space and x ∈ S(X). Then the following statements are equivalent: (1) The hyperplane H x of X * is approximatively (weakly approximatively) compact.
Proof. (2) ⇒ (1). By Theorem 2.1 of [18], we obtain that the hyperplane H x is a proximinal set. We will prove that if For clarity, we will divide the proof into two cases.
Analogous to the proof of Case I, we have Analogous to the proof of Case I, we obtain that {y * n } ∞ n=1 is relatively (weakly) compact. Hence weak * hyperplane H x is approximatively (weakly approximatively) compact.
there exists a natural number n 0 ∈ N such that µH n0 > 0, where where n = 1, 2, 3, ... and i = 1, 2, ..., 2 n . Then we define two function sequences t ∈ H n 2 n and r 0 = 1/n 0 . Moreover, by formula (2.1) and the definition of H, we have Hence the sequence {z n } ∞ n=1 is not relatively weakly compact, a contradiction. This implies that µ {t ∈ T : u(t) ∈ K t } = 0, which completes the proof.
is weakly approximatively compact and X is a strongly smooth space. Then M ∈ ∆.
We next prove that Theorem 2.1.
(4)⇒(1). First we will prove that v is a smooth point of E 0 N (X). In fact, since X is strongly smooth, we obtain that X * has the Radon-Nikodym property. Then (E 0 N (X)) * = L M (X * ). Suppose that u 1 , v = u 2 , v = 1 and u 1 = u 2 = 1. Then u, v = 1, where 2u = u 1 + u 2 . Hence µ {t ∈ T : u(t) ∈ K t } = 0. Then u(t) ∈ L M (R) and v(t) ∈ E 0 N (R). Since M ∈ ∆ and µ{t ∈ T : u(t) ∈ K t } = 0, by Theorem 5.10 of [2], we get that u is an extreme point of L M (R). Moreover, by u 1 = u 2 = 1 and v ∈ S(E 0 N (X)), we have This implies that Moreover, by u, v = 1 and 2u = u 1 + u 2 , we get that u = 1. Hence for almost all t ∈ T , by the smoothness of X and u 1 (t) = u 2 (t) for almost all t ∈ T , we get that u 1 (t) = u 2 (t) for almost all t ∈ T . This implies that u 1 = u 2 . Hence we obtain that v is a smooth point of E 0 N (X). Next we will prove that the point v is a strongly smooth point of E 0 N (X). Let u, v = 1 and u n , v → 1 as n → ∞, where {u n } ∞ n=1 ∈ S(L M (X * )) and u ∈ S(L M (X * )). Since B(L M (R)) is weakly * sequentially compact, by (E 0 N (R)) * = L M (R) and u n ∈ L M (X * ), we may assume without loss of generality that there exists a functional h ∈ S(L M (R)) such that T u n (t) w(t)dt → T h(t)w(t)dt whenever w ∈ E 0 N (R). This implies that T h(t) v(t) dt = 1. Since v is a smooth point of E 0 N (X), it is easy to see v is a smooth point of E 0 N (R). Therefore, by formula we obtain that u(t) = h(t) µ-a.e. on T . Hence we have the following formula whenever w ∈ E 0 N (R). Therefore, by formula (2.3), we obtain that { u n } ∞ n=1 converges weakly * to u . We claim that ρ M (u n χ E ) → ρ M (uχ G ) for all G ⊂ T . In fact, let for any m, n ∈ N . Since v ∈ E 0 N (X), we get that p(uχ E m n ) ∈ E 0 N (R). Therefore, by formula (2.3), we have the following formula This implies that lim inf n→∞ ρ M (u n χ Em ) ≥ ρ M (uχ Em ). Let m → ∞. Then, for any be a set of all the extreme points of linear interval of M (t, u) and let Since µ {t ∈ T : u(t) ∈ K t } = 0, we get that M (t, u(t) ) > 0 whenever t ∈ F . We claim that u n (t) → u(t) in measure on F . Otherwise, we may assume without loss of generality that for each n ∈ N , there exists E n ⊆ F , ε 0 > 0 and σ 0 > 0 such that µE n ≥ ε 0 , where E n = {t ∈ F : | u n (t) − u(t) | ≥ σ 0 }. Let us define the sets Similarly, we have µB ≤ ε 0 /8. Hence, for µ-a.e. t ∈ T , we define the bounded closed sets for any (u, v) ∈ C t and for µ-a.e. t ∈ T . Hence we define the function We claim that δ(t) is Σ-measurable. In fact, pick a dense subset {r i } ∞ i=1 of (0, ∞) and define the function then by the definition of M (t, u), it is easy to see that 1 − δ ri,rj (t) is Σ-measurable and On the other hand, since . Therefore, by the definition of the function 1 − δ(t), for µ-a.e. t ∈ T, ε > 0, there exists Since ε is arbitrary, we have the following formula µ-a.e. on T . Hence 1 − δ(t) = sup 1 − δ ri,rj (t) : |r i − r j | ≥ σ 0 /8 µ-a.e. on T . This implies that δ(t) is Σ-measurable. Since δ(t) > 0, there exists a real number δ 0 ∈ (0, 1) such that µG < ε 0 /8, where G = {t ∈ T : δ(t) ≤ δ 0 }. Moreover, by M (t, u(t) ) > 0, t ∈ F , there exist F 0 ⊂ F and r > 0 such that M (t, u(t) ) > r, t ∈ F \F 0 and µF 0 < ε 0 16 .
whenever t ∈ H n . This implies that Moreover, by formula u n , v → 1 and u, v = 1, we get that u = 1, u n → 1 and u n + u /2 → 1 as n → ∞. Therefore, by M ∈ ∆, we obtain that ρ M (u) = 1, ρ M (u n ) → 1 and ρ M ((u n + u)/2) → 1 as n → ∞. This implies that this is a contradiction. Hence we obtain that u n (t) → u(t) in measure on F . Let {r 1 (i, t)} ∞ i=1 denote the set of all the right extreme points of linear interval of M (t, u). Define the set We will prove that u n (t) → u(t) in measure on G. Otherwise, we may assume without loss of generality that for each n ∈ N , there exists E n ⊆ G, ε 0 > 0 and . For clarity, we will divide the proof into two cases.
Case II. Let lim sup n→∞ µ (G 0 ∩ E n ) = η 0 > 0, where t ∈ G 0 if and only if M (t, u(t) ) is a left extreme point and is a right extreme point of linear interval of M (t, u). Then G 0 ⊂ G and p (t, u(t) ) − p − (t, u(t) ) > 0 whenever t ∈ G 0 .
Hence there exist a set H ⊂ G 0 and a real number h > 0 such that µH < η 0 /16 and p (t, u(t) ) − p − (t, u(t) ) > h whenever t ∈ G 0 \ H. Moreover, there exists m ∈ N such that µ(T \T m ) < η 0 /16. Let F = (G 0 ∩ T m ) \H. Then, by formula lim sup n→∞ µ (G 0 ∩ E n ) = η 0 , we may assume that µ (F ∩ E n ) > η 0 /8. Therefore, by formula (2.3), we get the following inequalities Similarly, we get that u n (t) → u(t) in measure on G. In summary, we have u n (t) → u(t) in measure on T . Therefore, by the Riesz Theorem, there exists a subsequence {n} of {n} such that u n (t) → u(t) µ−a.e. in T . Noticing that we get the following formula Moreover, it is easy to see that This implies that u n (t) · v(t) − (u n (t), v(t)) µ → 0 in measure. Therefore, by the Riesz theorem, there exists a subsequence {n} of {n} such that u n (t) · v(t) − (u n (t), v(t)) → 0 µ−a.e. in T . Since u n (t) → u(t) µ-a.e. on T , we get that (u n (t), v(t)) → u(t) · v(t) µ−a.e. on T . Hence we may assume without loss of generality that Hence we may assume without loss of generality that Since X is a strongly smooth space, we obtain that sequence whenever n > n 1 . This implies that (u n − u 0 )χ H < 2ε whenever n > n 1 . Since whenever n > n 1 . Hence we have u n − u → 0 as n → ∞. This implies that v is a strongly smooth point of E 0 N (X), which completes the proof. Corollary 2.5. Suppose that v ∈ S(E 0 N (R)). Then the following statements are equivalent: (1) The point v is a strongly smooth point of E 0 N (R); Corollary 2.6. Suppose that X is a strongly smooth space. Then the following statements are equivalent: (1) E 0 N (X) is a strongly smooth space; (2) Every weak * hyperplane of L M (X * ) is approximatively compact; (3) Every weak * hyperplane of L M (X * ) is weakly approximatively compact.

Asplund property and Radon-Nikodym property in
Musielak-Orlicz-Bochner function spaces Theorem 3.1. Suppose that X is a strongly smooth space. Then the following statements are equivalent: (1) L 0 M (X) is an Asplund space; (2) M ∈ ∆ and N ∈ ∆.
Next we will prove that θ(·) is continuous. Otherwise, there exist ε 0 > 0, u ∈ L M (X) and {u n } ∞ n=1 ⊂ L M (X) such that u n − u → 0 and |θ(u n ) − θ(u)| ≥ 2ε 0 . Then we may assume that θ(u) ≥ θ(u n ) + 2ε 0 or θ(u n ) ≥ θ(u) + 2ε 0 . Hence, if θ(u) ≥ θ(u n ) + 2ε 0 , then for all n ∈ N . Decompose G(n) into G 1 (n) and G 2 (n) such that G 1 (n) ∪ G 2 (n) = G(n) and ρ N (lvχ G1(n) ) = ρ N (lvχ G2(n) ). Decompose G into G 1 and G 2 such that G 1 ∪ G 2 = G and ρ N (lvχ G1 ) = ρ N (lvχ G2 ). Let Then ρ N (lvχ E ) = ρ N (lvχ F ). This implies that vχ E ≥ 1/l and vχ F ≥ 1/l. Hence there exists a sequence of set {E n } ∞ n=1 such that vχ En ≥ 1/l and E i ∩E j = ∅ whenever i = j. Define the set Then it is easy to see that C is a bounded closed convex set of L N (X * ). We claim that C has no extreme points. In fact, Then there exists a natural number j ∈ N such that |ε j | < 1/4. Define Then u 1 , u 2 ∈ C, u 1 = u 2 and 2u = u 1 + u 2 . This implies that u is not an extreme point. Since u is arbitrary, we have ExtC = ∅. Hence L N (X * ) has not the Krein-Milman property. Then L N (X * ) has not the Radon-Nikodym property. Hence E 0 M (X) is not an Asplund space. However, since L 0 M (X) is an Asplund space, we obtain that E 0 M (X) is an Asplund space, a contradiction. Then N ∈ ∆. (2)⇒(1). By Lemma 1.15 of [2], there exists a function M 1 such that and right derivative of p 1 (t, u) of M 1 (t, u) is continuous with respect to u for almost all t ∈ T . Therefore, by formula 3.1, we obtain that u ∈ L M (X) for any u ∈ L M1 (X). Since M ∈ ∆, we have the following inequalities T M 1 (t, λ u(t) )dt ≤ T 2M (t, λ u(t) )dt < +∞ for any λ > 0. This implies that M 1 ∈ ∆. Moreover, if v ∈ L N1 (X * ), then there exists λ 0 > 0 such that ρ N1 (λ 0 v) < +∞. Since we have the following inequalities This implies that v ∈ L N (X * ). Therefore, by N ∈ ∆, we obtain that v ∈ E N (X * ). Therefore, by formula we have the following inequalities T N 1 (t, λ v(t) ) dt ≤ T N (t, λ v(t) ) dt < +∞ for every λ > 0. This implies that N 1 ∈ ∆. Therefore, by theorem 2.7, we obtain that L 0 M1 (X) is a strongly smooth space. This implies that L 0 M1 (X) is an Asplund space. Moreover, we have for any u ∈ L 0 M (X). This means that L 0 M (X) is an Asplund space, which completes the proof.
By Theorem 3.1, we obtain that Corollary 3.2 and Corollary 3.3.

Corollary 3.2.
Suppose that X is a strongly smooth space. Then the following statements are equivalent: (1) L M (X) is an Asplund space; (2) M ∈ ∆ and N ∈ ∆.  (1) M ∈ ∆ and X is a strongly smooth space; (2) p(t, u) is continuous with respect to u for almost all t ∈ T . Then E 0 N (X) is a strongly smooth space. Proof. By Theorem 2.1, it is easy to see that Theorem 3.4 is true, which completes the proof. Proof. Sufficiency. Let X be a strongly smooth space. Then, by Lemma 1.15 of [2], there exists a function M 1 such that M (t, u) ≤ M 1 (t, u) ≤ 2M (t, u), u ∈ R and right derivative of p 1 (t, u) of M 1 (t, u) is continuous with respect to u for almost all t ∈ T . Moreover, by the proof of Theorem 3.1, we get that M 1 ∈ ∆. Therefore, by Theorem 3.4, we obtain that E 0 N (X) is a strongly smooth space. Hence E 0 N (X) is an Asplund space. This implies that L M (X * ) has the Radon-Nikodym property.
Necessity. Suppose that M / ∈ ∆. Then, by the proof of Theorem 3.1, we obtain that L M (X * ) has not the Krein-Milman property. Hence L M (X * ) has not the Radon-Nikodym property, a contradiction. This implies that M ∈ ∆, which completes the proof.