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

1.   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, $L_{M}^{0}(X)$ (resp $L_{M}(X)$) is an Asplund space if and only if $M\in\Delta$ and $N\in\Delta$. As a corollary, we obtain that $L_{M}^{0}(R)$ (resp $L_{M}(R)$) is an Asplund space if and only if $M\in\Delta$ and $N\in\Delta$.

Let $(X,\|\cdot\|)$ 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 $x\in S(X)$ and let ${H_x} = \left\{ {{x^*} \in {X^*}:{x^*}(x) = 1} \right\}$. Let $C \subset X$ be a nonempty subset of $X$. Then the set-valued mapping $P_C :X \to C$

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 $\{ y_n \} _{n = 1}^\infty \subset C$ and any $x\in X$ satisfying $\left\| {x - y_n } \right\| \to \inf _{y \in C} \left\| {x^{} - y} \right\|$ as $n\rightarrow\infty$, there exists a subsequence converging (weakly) to an element of $C$.

Definition 1.2. A point $x\in S(X)$ is called a smooth point if it has a unique supporting functional $f_{x}\in S(X^{*})$. If every $x\in S(X)$ is a smooth point, then $X$ is called a smooth space.

Consider a convex subset $A$ of a Banach space $X$. A point $x\in A$ is said to be an extreme point of $A$ if $2x = y+z$ and $y,z\in A$ 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 $x\in S(X)$ is said to be a strongly smooth point of $X$ if there exist $\{ x_n^*\} _{n = 1}^\infty \subset S(X^{*})$ and $x_{0}^{*}\in S(X^{*})$ such that ${x_n^* \to x_0^*}$ whenever $x_n^*(x) \to 1$. 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 $(T, \Sigma ,\mu )$ be nonatomic measurable space. $v$ is a measure and $v$ is bounded variation and absolutely continuous with respect to $\mu$, then there exists an integrable function $f$ such that for any $A\in \Sigma$, 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 G$\mathrm{\hat{a}}$teaux differentiable at the point $x$ in $X$ if the limit

exists for all $y \in X$. 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 $G_{\delta}$ 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 $(T,\Sigma,\mu )$ be a complete nonatomic measurable space. Suppose that a function $M:T \times [0,\infty ) \to [0,\infty ]$ statisfies the following conditions.

(1) for $\mu$-a.e, $t\in T$, $M(t,0) = 0$, $\mathop {\lim }\limits_{u \to \infty } M(t,u) = \infty$ and $M(t,u') < \infty$ for some $u'>0$.

(2) for $\mu$-a.e, $t\in T$, $M(t,u)$ is convex in $[0,\infty )$ with respect to $u$.

(3) for each $u \in [0,\infty ),$ $M(t,u)$ is a $\Sigma$-measurable function of $t$ on $T$.

Every such a function $M$ is called a Musielak-Orlicz-function. Let $p(t,\cdot)$ denote the right derivative of $M(t,\cdot)$ at $u\in R^{+}$ (where $p(t,u) = \infty$ if $M(t,u) = \infty$) and let $q(t,\cdot)$ be the generalized inverse function of $p(t,\cdot)$ defined on $R^{+}$ by

Then the function $N(t,v)$ defined by $N(t,v) = \int_0^{|v|} {q(t,s)ds}$ for any $v\in R$ and $\Sigma$-a.e. $t\in T$ is called the complementary function to $M$ in the sense of Young. It is well known that the Young inequality $uv\le M(t,u) + N(t,v)$ holds for all $u,v,\in {R}$ and $\mu$-a.e. $t\in T$. Moreover, for any $u\in {R}$ the equality $uv = M(t,u) + N(t,v)$ holds if and only if $v\in [p_-(t,u),p(t,u)]$. Let

For a fixed $t\in T$ and $v\geq 0$, if there exists $\varepsilon\in (0,1)$ such that

then $v$ is called a point of affinity of $M(t,\cdot)$. The set of all points of affinity of $M(t,\cdot)$ for a fixed $t\in T$ is denoted by $K_{t}$.

Definition 1.5. (see[2]) We say that a Musielak-Orlicz function $M$ satisfies condition $\Delta(M\in \Delta)$ if there exist $K\geq1$ and a measureable nonnegative function $\delta(t)$ on $T$ such that $\int_T {M(t,\delta (t))}$ $dt < \infty$ and $M(t,2u) \le KM(t,u)$ for almost all $t\in T$ and all $u \ge \delta (t)$.

Definition 1.6. (see[2]) A Musielak-Orlicz function $M(t,u)$ is said to be strictly convex with respect to $u$ for $t\in T$ if for almost every $t\in T$ and any $u,v\in R$, $u\neq v$, we have

Given any Banach space $(X,\left\| \cdot \right\|)$, we denote by $X_T$ the set of all strongly $\Sigma$-measurable functions from $T$ to $X$, and for each $u \in X_T$, we define the modular of $u$ by

Let us define the Musielak-Orlicz-Bochner function space $L_M(X)$ by

and its subspace

It is well known that the spaces $L_{M}(X)$ and their subspaces $E_{M}(X)$ are Banach spaces when they are equipped with the Luxemburg norm

or with the Orlicz norm

It is well known that the Luxemburg norm and the Orlicz norm are equivalent. The spaces $(L_M({X}),\|\cdot\|), (L_M({X}),\|\cdot\|^0), (E_M(X),\|\cdot\|_M)$ and $(E_M(X),\|\cdot\|^0)$ are denoted shortly by $L_M(X), L^0_M(X), E_M(X)$ and $E^0_M(X)$, respectively. $L_M(R)$ and $L_M^{0}(R)$ are called Musielak-Orlicz function spaces, and $E_M(R)$ and $E^0_M(R)$ are the subspaces of $L_M(R)$ and $L^0_M(R)$, respectively. Moreover, it is well known that ${\left( {{E_M}(R)} \right)^*} = L_N^0(R)$ and ${\left( {{E_M^{0}}(R)} \right)^*} = L_N(R)$ (see[4], [20]). Moreover, by [2], we know that $E_M(R) = L_M(R)$ if and only if $M\in \Delta$.

Lemma 1.7. (see[2]) $\left\| u \right\| \le 1 \Rightarrow {\rho _M}(u) \le \left\| u \right\|$ and $\left\| u \right\| > 1 \Rightarrow {\rho _M}(u) > \left\| u \right\|$.

Lemma 1.8. (see [2]) $M \notin \Delta \Leftrightarrow$ for any $\varepsilon>0$, there exists $u\in L_{M}(X)$ such that $\rho _M(u) = \varepsilon$ and $\|u(t)\|<E(t)$ for almost all $t\in T$.

2.   Strongly smooth point and approximative compactness in Musielak-Orlicz-Bochner function spaces

Theorem 2.1. Suppose that $X$ is a strongly smooth space and $v\in S(E_{N}^{0}(X))$. Then the following statements are equivalent:

(1) The point $v$ is a strongly smooth point of $E_{N}^{0}(X)$;

(2) The hyperplane $H_{v}$ of $L_M(X^{*})$ is approximatively compact;

(3) The hyperplane $H_{v}$ of $L_M(X^{*})$ is weakly approximatively compact;

(4) $M\in\Delta$ and $\mu \{ t \in T:\| {u(t)} \| \in K_t\} = 0$ whenever $\langle u,v\rangle = \|u\|$.

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

Lemma 2.2. Suppose that $X$ is a Banach space and $x\in S(X)$. Then the following statements are equivalent:

(1) The hyperplane $H_{x}$ of $X^{*}$ is approximatively (weakly approximatively) compact.

(2) If $y_n^* \in S(X^* )$ and $y_n^* (x) \to 1$ as $n \to \infty$, then the sequence $\{ y_n^* \} _{n = 1}^\infty$ is relatively (weakly) compact.

Proof. $(2)\Rightarrow(1)$. By Theorem 2.1 of [18], we obtain that the hyperplane $H_{x}$ is a proximinal set. We will prove that if $\left\| {x^* - y_n^*} \right\| \to \mathrm{dist}\left( {x^* ,{H_x}} \right)$ as $n\to \infty$, then the sequence $\{ y_n^*\} _{n = 1}^\infty$ is relatively (weakly) compact, where $\{ y_n^*\} _{n = 1}^\infty \subset {H_x}$ and $x^* \notin H_x$. Since $\left\| {x^* - y_n^*} \right\| \to \mathrm{dist}\left( {x^* ,{H_x}} \right)$ as $n\to \infty$, we have $\left\| 0-(y_n^*-x^{*}) \right\| \to \mathrm{dist}\left( {0 ,{H_x}-x^{*}} \right)$. Since $H_{x}-x^{*}$ is weakly$^{*}$ closed set, we have $\mathrm{dist}(0,H_{x}-x^{*}) = r > 0$. Pick $y_0^* \in P_{H_{x}-x^{*}} (0)$. Then

and

For clarity, we will divide the proof into two cases.

Case I. $k = \sup \{ x(y^* ):y^* \in H_{x}-x^{*} \}\leq 0$. We claim that

In fact, suppose that there exists $y_{1}^* \in B(0,r)$ such that $x(y_{1}^* )<k$. Then there exists $\lambda\in (0,1)$ such that $x(\lambda y_{1}^*) = k$. It is easy to see that $\lambda y_{1}^*\in B(0,r)$ and $\lambda y_{1}^*\in H_{x}$, a contradiction. Since $y_0^* \in P_{H_{x}-x^{*}} (0) \subset H_{x}-x^{*}$, we have

and

This means that the inequality $x(y_0^* ) \ge x(y_n^* -x^{*})$ holds. Therefore

This implies that $\left\| x^{*} -y_n^*\right\| \to \left\| {y_0^* } \right\|$ and $x(x^{*} - y_n^* ) \to \left\| { y_0^* } \right\|$ as $n \to \infty$. Moreover, we have

Therefore, by $x(x^{*} - y_n^* ) \to \left\| { y_0^* } \right\|$ as $n \to \infty$, we have

Hence the sequence $\{ - {{(x^{*} -y_n^* )}}/{{\left\| {(x^{*} -y_n^*) } \right\|}}\} _{n = 1}^\infty$ is relatively (weakly) compact. Since $\left\| {x^{*} -y_n^* } \right\| \to \left\| {y_0^* } \right\|$ as $n \to \infty$, we obtain that $\{ y_n^* \} _{n = 1}^\infty$ is relatively (weakly) compact. Hence $H_{x}$ is approximatively (weakly approximatively) compact.

Case II. $\sup \{ x(y^* ):y^* \in H_{x}-x^{*} \}> 0$. This implies that $k = \sup \{ -x(y^* ):y^* \in H_{x}-x^{*} \}< 0$. Analogous to the proof of Case I, we have

Analogous to the proof of Case I, we obtain that $\{ y_n^* \} _{n = 1}^\infty$ is relatively (weakly) compact. Hence weak$^{*}$ hyperplane $H_{x}$ is approximatively (weakly approximatively) compact.

$(1)\Rightarrow(2)$. Let $\{y_n^* \}_{n = 1}^{\infty}\subset S(X^* )$ and $x\in S(X)$ satisfy $y_n^* (x) \to 1$ as $n \to \infty$. Since $H_{x}$ is a weak$^{*}$ closed set, by Theorem 2.1 of [18], we obtain that $H_{x}$ is a proximinal set. Hence there exists $z_n^* \in H_{x}$ such that $\left\| {y_n^* - z_n^* } \right\| = \mathrm{dist}(y_n^* ,H_{x})$. Pick $x_{0}^{*}\in H_{x}\cap S(X^{*})$. Since

we have the following formula

Therefore, by formula $\left\| {z_n^* } \right\| \ge 1$, we have $\left\| {0 - z_n^* } \right\| \to \mathrm{dist}\left(0,H_{x}\right)$ as $n \to \infty$. This implies that the sequence $\{ z_n^* \} _{n = 1}^\infty$ is relatively (weakly) compact. Consequently $\{y_n^*\} _{n = 1}^\infty$ is relatively (weakly) compact by $y_n^* = z_n^* + (y_n^* - z_n^* )$, which completes the proof.

Lemma 2.3. Suppose that $u\in S(L_M(X^{*}))$ is norm attainable on $S(E_N^{0}(X))$ and $M\in\Delta$. Then (2)$\Rightarrow$(1) is true, where

(1) $\mu \left\{ t \in T:\left\| {u(t)} \right\| \in {K_t}\right\} = 0$;

(2) If $u = \sum_{i = 1}^\infty {t_i u_i }$, where $u_{i}\in B(L_{M}(X^{*}) ), \; t_i \in (0,1)$ and $\sum_{i = 1}^\infty {t_i = 1}$, then the sequence $\{ u_i\} _{i = 1}^\infty$ is relatively weakly compact.

Proof. (2)$\Rightarrow$(1). Let $u\in L_M(X^{*})$ and $u = \sum_{i = 1}^\infty {t_i u_i }$, where $u_{i}\in B(L_{M}(X^{*}) ), \; t_i \in (0,1)$ and $\sum_{i = 1}^\infty {t_i = 1}$. Suppose that $\mu H'>0$, where

${T_0} = \left\{ {t \in T:\left\| {u(t)} \right\| > 0} \right\}$ and $\left\| {u(t)} \right\| - {\varepsilon _t}>0$. Since $H'\subset \cup _{n = 2}^\infty {H_n}$, where

there exists a natural number $n_{0}\in N$ such that $\mu {H_{{n_0}}}>0$, where

Since $u$ is norm attainable on $S(E_N^0(X))$, there exists a point $v \in S(E_N^0(X))$ such that $\int_{T} \langle u,v\rangle dt = \left\| u\right\| = \left\| v\right\|^{0} = 1$. Let $H = {H_{{n_0}}}$. Then, decompose $H$ into $H_1^1$ and $H_1^2$ such that $H_1^1 \cup H_2^1 = H$, $H_1^1 \cap H_2^1 = \emptyset$ and $\int_{H_1^1} {\langle u,v\rangle dt} = \int_{H_2^1} {\langle u,v\rangle dt}$. Decompose $H_1^1$ into $H_1^2$ and $H_2^2$ such that $H_1^2 \cup H_2^2 = H_1^1$, $H_1^2 \cap H_2^2 = \emptyset$ and $\int_{H_1^2} {\langle u,v\rangle dt} = \int_{H_2^2} {\langle u,v\rangle dt}$. Decompose $H_2^1$ into $H_3^2$ and $H_4^2$ such that $H_3^2 \cup H_4^2 = H_2^1$, $H_3^2 \cap H_4^2 = \emptyset$ and $\int_{H_3^2} {\langle u,v\rangle dt} = \int_{H_4^2} {\langle u,v\rangle dt}$. Generally, decompose $H_i^{n - 1}$ into $H_{2i - 1}^n$ and $H_{2i}^n$ such that

where $n = 1, 2, 3, . . .$ and $i = 1, 2, . . . , 2^n$. Then we define two function sequences $\{ {u_n}\} _{n = 1}^\infty$ and $\{ {u_n^{1}}\} _{n = 1}^\infty$, where

and $r_{0} = 1/n_{0}$. Moreover, by formula (2.1) and the definition of $H$, we have

and

This implies that $\left\| {{u_n}} \right\| \ge 1$. Hence ${\rho _M}({u_n}) \ge 1$. Similarly, we have ${\rho _M}({u_n^{1}}) \ge 1$. Since ${\rho _M}({u_n}) + {\rho _M}(u_n^1) = 2$, we get that ${\rho _M}({u_n}) = {\rho _M}(u_n^1) = 1$. This implies that $\left\| {{u_n}} \right\| = \left\| {{u_n^{1}}} \right\| = 1$. Hence we define a new function sequence $\{ z_n \} _{n = 1}^\infty$ such that

Moreover, it is easy to see that

Since $\int_{T} {(u(t),v(t))dt} = \left\| u\right\| = \|v\|^{0} = 1$ and $M\in\Delta$, by formula $\mu H>0$, we have $(u(t),v(t))>0$ for $\mu$-a.e, $t\in H$. Therefore, by formula $\mu H>0$, we get that $\int_H {(u(t),v(t))} dt > 0$. Moreover, since the sequence $\{ u_n \} _{n = 1}^\infty$ is a subsequence of $\{ z_n \} _{n = 1}^\infty$, we have the following formula

whenever $m\neq n$. Hence the sequence $\{ z_n \} _{n = 1}^\infty$ is not relatively weakly compact, a contradiction. This implies that $\mu \left\{ t \in T:\left\| {u(t)} \right\| \in {K_t}\right\} = 0$, which completes the proof.

Lemma 2.4. Suppose that $v_{0}\in E_{N}^{0}(X)$ and the $w^{*}$-hyperplane $H_{v_{0}}$ of ${L_M}({X^*})$ is weakly approximatively compact and $X$ is a strongly smooth space. Then $M\in\Delta$.

Proof. Since $X$ is a strongly smooth space, we obtain that $X^{*}$ has the Radon-Nikodym property. Then $(E_{N}^{0}(X))^{*} = {L_M}({X^*})$. Hence there exists $u_{0}\in {L_M}({X^*})$ such that $\|u_{0}\| = 1$ and $\langle u_{0},v_{0}\rangle = 1$. Suppose that $M\notin \Delta$. Then, by Lemma 1.8, there exists a point $u \in {L_M}({X^*})$ such that $\|u\| = 1$ and ${\rho _M}(u) < 1$. This implies that

Hence, for any $L>1$, we have $\rho_{M}(Lu) = \infty$. Indeed, suppose that there exists $L_{1}>1$ such that $\rho_{M}(L_{1}u)<\infty$. Moreover, we know that the function $F(k) = \int_T {M(t,k\left\| {u(t)} \right\|)dt}$ is continuous on $[1,L_{1}]$. Then there exists $L_{2}>1$ such that $\rho_{M}(L_{2}u) = 1$. This implies that $\left\| u \right\| \le {1}/{{L_2 }}$, contradicting $\left\| {u } \right\| = 1$. Decompose $T$ into $E_{1}$ and $G_{1}$ such that $\mu E_{1} = \mu G_{1}$. Then for any $L>1$, we obtain that $\int_{E_1 } {M(t,L\left\| {u(t)} \right\|)dt = \infty }$ or $\int_{G_1 } {M(t,L\left\| {u(t)} \right\|)}dt = \infty .$ Hence we may assume without loss of generality that $\int_{E_1 } {M(t,L\left\| {u(t)} \right\|)dt = \infty }$. Decompose $E_{1}$ into $E_{2}$ and $G_{2}$ such that $\mu E_{2} = \mu G_{2}$. Then, for any $L>1$, we have $\int_{E_2 } {M(t,L\left\| {u(t)} \right\|)dt = \infty }$ or $\int_{G_2 } {M(t,L\left\| {u(t)} \right\|)}dt = \infty$. Hence we may assume without loss of generality that $\int_{E_2 } {M(t,L\left\| {u(t)} \right\|)dt = \infty }$. In generical, decompose $E_{n}$ into $E_{n+1}$ and $G_{n+1}$ such that $\mu E_{n+1} = \mu G_{n+1}$. Then, for any $L>1$, we have $\int_{E_{n+1} } {M(t,L\left\| {u(t)} \right\|)dt = \infty }$ or $\int_{G_{n+1}} {M(t,L\left\| {u(t)} \right\|)}dt = \infty$. Hence we may assume without loss of generality that $\int_{E_{n+1}} {M(t,L\left\| {u(t)} \right\|)dt = \infty }$. Hence

This implies that $\left\| {u\chi _{E_i } } \right\| = 1$. Let $u_{n} = u_{0}+u\chi _{T\setminus E_i }$ and $D = \overline {co} \{ {u_n}\} _{n = 1}^\infty$. Then, for every $v = u_{0}+ {\alpha _1}u{\chi _{T\backslash {E_{n(1)}}}} + \cdot \cdot \cdot + {\alpha _k}u{\chi _{T\backslash {E_{n(k)}}}}\in {co} \{ {u_n}\} _{n = 1}^\infty$, there exists a natural number $i(0)>\max\{n(1),...,n(k)\}$ such that

where ${\alpha _1} + {\alpha _2} + \cdot \cdot \cdot + {\alpha _k} = 1$. This implies that dist$(u_{0},D) = 1$. Therefore, by the separable theorem, there exist $f \in {\left( {L_M^0(X)} \right)^*}$ and $r>0$ such that

Moreover, by formula $\langle u_{0},v_{0}\rangle = 1$ and the definition of $u_n$, we have the following inequalities

Hence we have $\langle{u_n},v_{0} \rangle\to 1$ as $n\to\infty$. Since weak$^{*}$ hyperplane $H_{v_{0}}$ of ${L_M}({X^*})$ is weakly approximatively compact, by Lemma 2.2, we get that $\{ {u_n}\} _{n = 1}^\infty$ is relatively weakly compact. Moreover, for any $v \in E_N^0(X)$, we have

This implies that ${u_n}\xrightarrow{w^{*}} u_{0}$ as $n\to\infty$. Since the sequence $\{ {u_n}\} _{n = 1}^\infty$ is relatively weakly compact, we get that ${u_n}\mathop \to \limits^{{w}} u_{0}$ as $n\to\infty$. However, by formula (2.2), we get that ${u_n}\mathop \to \limits^w u_{0}$ is impossible, a contradiction, which completes the proof.

We next prove that Theorem 2.1.

Proof. By Lemma 2.2, it is easy to see that (1)$\Rightarrow$(2) and (2)$\Rightarrow$(3) are true. We next will prove that (3)$\Rightarrow$(4). By Lemma 2.4, we obtain that $M\in\Delta$. Moreover, if $u\in S(L_M(X^{*}))$ is norm attainable at point $v_{0}$ and $u = \sum_{n = 1}^\infty {t_n u_n }$, where $u_{n}\in B(L_{M}(X^{*}) ), \; t_n \in (0,1)$ and $\sum_{n = 1}^\infty {t_n = 1}$, then

This implies that ${\left\langle{{u_n},{v_0}} \right\rangle = 1}$ for all $n\in N$. Therefore, by Lemma 2.2, we obtain that $\{ {u_n}\} _{n = 1}^\infty$ is relatively weakly compact. Therefore, by Lemma 2.3, we get that $\mu \left\{ t \in T:\left\| {u(t)} \right\| \in {K_t}\right\} = 0$.

(4) $\Rightarrow$(1). First we will prove that $v$ is a smooth point of $E_{N}^{0}(X)$. In fact, since $X$ is strongly smooth, we obtain that $X^{*}$ has the Radon-Nikodym property. Then $(E_{N}^{0}(X))^{*} = L_{M}(X^{*})$. Suppose that $\langle u_{1},v\rangle = \langle u_{2},v\rangle = 1$ and $\|u_{1}\| = \|u_{2}\| = 1$. Then $\langle u,v\rangle = 1$, where $2u = u_{1}+u_{2}$. Hence $\mu \left\{ t \in T:\| {u(t)} \| \in K_t\right\} = 0$. Then $\left\| {u(t)} \right\| \in {L_M}(R)$ and $\left\| {v(t)} \right\| \in E_N^0(R)$. Since $M\in\Delta$ and $\mu \{ t \in T:\| {u(t)} \| \in K_t\} = 0$, by Theorem 5.10 of [2], we get that $\left\| {u} \right\|$ is an extreme point of ${L_M}(R)$. Moreover, by $\|u_{1}\| = \|u_{2}\| = 1$ and $v\in S(E_{N}^{0}(X))$, we have

and

This implies that

$\left( {{u_1}(t),v(t)} \right) = \left\| {{u_1}(t)} \right\|{\left\| {v(t)} \right\|}$ and $\left( {{u_2}(t),v(t)} \right) = \left\| {{u_2}(t)} \right\|{\left\| {v(t)} \right\|}$ for almost all $t\in T$. Moreover, by $\langle u,v\rangle = 1$ and $2u = u_{1}+u_{2}$, we get that $\|u\| = 1$. Hence

This implies that $2 \left\| {{u}(t)} \right\| = \left\| {{u_1}(t)} \right\| +\left\| {{u_2}(t)} \right\|$ for almost all $t\in T$. Since $\left\| {u} \right\|$ is an extreme point of ${L_M}(R)$, we have $\left\| {{u_1}(t)} \right\| = \left\| {{u_2}(t)} \right\|$ for almost all $t\in T$. Since $\left( {{u_1}(t),v(t)} \right) = \left\| {{u_1}(t)} \right\|{\left\| {v(t)} \right\|}$ and $\left( {{u_2}(t),v(t)} \right) = \left\| {{u_2}(t)} \right\|{\left\| {v(t)} \right\|}$ for almost all $t\in T$, by the smoothness of $X$ and $\left\| {{u_1}(t)} \right\| = \left\| {{u_2}(t)} \right\|$ for almost all $t\in T$, we get that ${{u_1}(t)} = {{u_2}(t)}$ for almost all $t\in T$. This implies that $u_{1} = u_{2}$. Hence we obtain that $v$ is a smooth point of $E_{N}^{0}(X)$.

Next we will prove that the point $v$ is a strongly smooth point of $E_{N}^{0}(X)$. Let $\langle u,v\rangle = 1$ and $\langle u_{n},v\rangle\to1$ as $n\to\infty$, where $\{u_{n}\}_{n = 1}^{\infty}\in S({L_M}(X^{*}))$ and $u\in S({L_M}(X^{*}))$. Since $B({L_M}(R))$ is weakly$^{*}$ sequentially compact, by ${(E_N^0(R))^*} = {L_M}(R)$ and $u_{n}\in {L_M}(X^{*})$, we may assume without loss of generality that there exists a functional $h\in S({L_M}(R))$ such that $\int_T {\left\| {{u_n}(t)} \right\| {w(t)} dt} \to \int_T {h(t) {w(t)} } dt$ whenever $w\in E_N^0(R)$. This implies that $\int_T {h(t)\left\| {v(t)} \right\|dt} = 1$. Since $v$ is a smooth point of $E_{N}^{0}(X)$, it is easy to see $\|v\|$ is a smooth point of $E_{N}^{0}(R)$. Therefore, by formula

we obtain that $\left\| {u(t)} \right\| = h(t)$ $\mu$-a.e. on $T$. Hence we have the following formula

whenever $w\in E_N^0(R)$. Therefore, by formula (2.3), we obtain that $\{\|u_{n}\|\}_{n = 1}^{\infty}$ converges weakly$^{*}$ to $\|u\|$. We claim that ${\rho _M}({u_n}{\chi _E}) \to {\rho _M}(u{\chi _G})$ for all $G\subset T$. In fact, let

and

for any $m,n\in N$. Since $v\in E_{N}^{0}(X)$, we get that $p(u{\chi _{E_n^m}})\in E_{N}^{0}(R)$. Therefore, by formula (2.3), we have the following formula

This implies that $\lim {\inf _{n \to \infty }}{\rho _M}({u_n}{\chi _{{E_m}}}) \geq {\rho _M}(u{\chi_{E_{m}}})$. Let $m\to\infty$. Then, for any $G\subset T$, we have $\lim {\inf _{n \to \infty }}{\rho _M}({u_n}{\chi _G}) \geq {\rho _M}(u{\chi _G})$. Since ${\rho _M}({u_n}) = {\rho _M}(u) = 1$, we get that ${\rho _M}({u_n}{\chi _G}) \to {\rho _M}(u{\chi _G})$ for any $G\subset T$.

We next will prove that ${\left\| {{u_n}(t)} \right\| \to \left\| {u(t)} \right\|}$ in measure on $T$. Let $\{ r(i,t)\} _{i = 1}^\infty$ be a set of all the extreme points of linear interval of $M(t,u)$ and let

Since $\mu \left\{ t \in T:\left\| {u(t)} \right\| \in {K_t}\right\} = 0$, we get that $M(t,\left\| {u(t)} \right\|)>0$ whenever $t\in F$. We claim that $\left\| {{u_n}(t)} \right\| \to \left\| {u(t)} \right\|$ in measure on $F$. Otherwise, we may assume without loss of generality that for each $n\in N$, there exists $E_{n}\subseteq F$, $\varepsilon _0>0$ and $\sigma _0>0$ such that $\mu E_n \ge \varepsilon _0$, where $E_n = \{ t \in F:\left| {\left\| {{u_n}(t)} \right\| - \left\| {u(t)} \right\|} \right| \ge \sigma _0 \}.$ Let us define the sets

Then

Similarly, we have $\mu B \leq \varepsilon _0 /8$. Hence, for $\mu$-a.e. $t\in T$, we define the bounded closed sets

in the two dimensional space $R^{2}$. Since $C_t$ is compact, there exists $(u_t ,v_t) \in C_t$ such that

for any $(u,v)\in C_{t}$ and for $\mu$-a.e. $t\in T$. Hence we define the function

We claim that $\delta (t)$ is $\Sigma$-measurable. In fact, pick a dense subset $\{ r_i \} _{i = 1}^\infty$ of $(0,\infty)$ and define the function

then by the definition of $M(t,u)$, it is easy to see that $1 - \delta _{r_i ,r_j } (t)$ is $\Sigma$-measurable and

On the other hand, since $\{ r_i \} _{i = 1}^\infty$ is dense in $(0,\infty)$ then $\{ (r_i ,r_j )\} _{i = 1,j = 1}^\infty$ is dense in $(0,\infty ) \times (0,\infty )$. Therefore, by the definition of the function $1 - \delta(t)$, for $\mu$-a.e. $t\in T, \varepsilon>0$, there exists

such that

$\mu$-a.e. on $T$. Since $\varepsilon$ is arbitrary, we have the following formula

$\mu$-a.e. on $T$. Hence $1 - \delta (t) = \sup \left\{ 1 - \delta _{r_i ,r_j } (t):\left| {r_i - r_j } \right| \ge \sigma _0 /8\right\}$ $\mu$-a.e. on $T$. This implies that $\delta (t)$ is $\Sigma$-measurable. Since $\delta (t)>0$, there exists a real number $\delta_{0}\in(0,1)$ such that $\mu G < \varepsilon _0 /8$, where $G = \{ t \in T :\delta (t) \le \delta _0 \}$. Moreover, by $M(t,\left\| {u(t)} \right\|)>0,\; t\in F$, there exist $F_{0}\subset F$ and $r>0$ such that

Let ${H_n} = E_{n}\backslash ({A_n} \cup B \cup G\cup F_{0})$. Then $\mu {H_n}\geq \varepsilon _0 /8$ and

whenever $t\in {H_n}$. This implies that

Moreover, by formula $\langle u_{n},v\rangle\to1$ and $\langle u,v\rangle = 1$, we get that $\|u\| = 1$, $\|u_{n}\|\to 1$ and $\|u_{n}+u\|/2 \to 1$ as $n\to \infty$. Therefore, by $M\in \Delta$, we obtain that ${\rho _M}(u) = 1$, ${\rho _M}({u_n}) \to 1$ and ${\rho _M}\left( {{{({u_n} + u)}}/{2}} \right)\to 1$ as $n\to \infty$. This implies that

this is a contradiction. Hence we obtain that ${\left\| {{u_n}(t)} \right\| \to \left\| {u(t)} \right\|}$ in measure on $F$.

Let $\{ r_{1}(i,t)\} _{i = 1}^\infty$ denote the set of all the right extreme points of linear interval of $M(t,u)$. Define the set

We will prove that ${\left\| {{u_n}(t)} \right\| \to \left\| {u(t)} \right\|}$ in measure on $G$. Otherwise, we may assume without loss of generality that for each $n\in N$, there exists $E_{n}\subseteq G$, $\varepsilon _0>0$ and $\sigma _0>0$ such that $\mu E_n \ge \varepsilon _0$, where $E_n = \{ t \in G:\left| {\left\| {{u_n}(t)} \right\| - \left\| {u(t)} \right\|} \right| \ge \sigma _0 \}$. Hence we may assume that $E_n = \{ t \in T: {\left\| {{u_n}(t)} \right\| - \left\| {u(t)} \right\|} \ge \sigma _0 \}$. Let

and

For clarity, we will divide the proof into two cases.

Case I. Let ${\lim \sup }_{n \to \infty }\mu \left(G_{0}\cap E_{n}\right) = \eta_{0}>0$, where $t\in G_{0}$ if and only if $M(t,\left\| {u(t)} \right\|)$ is a right extreme point and not a left extreme point. Then $G_{0}\subset G$. Therefore, by formula $M(t,u) = \int_0^u {p(t,u)} dt$, we have the following inequalities

whenever $t\in E_{n}$. Hence there exist $H_{n}\subset G_{0}$ and $h>0$ such that $\mu H_{n}<\eta_{0}/16$ and ${h_n}(t)>h$ whenever $t\in E_{n} \setminus H_{n}$. Moreover, there exists a natural number $m\in N$ such that $\mu(T\backslash T_{m})<\eta_{0}/16$, where

Let $G_{n} = \left( {G_{0} \cap {T_m}} \right)\backslash H_{n}$. Then, by ${\lim \sup }_{n \to \infty }\mu (G_{0}\cap E_{n}) = \eta_{0}$ and $\mu(T\backslash T_{m})<\eta_{0}/16$, we may assume that $\mu (G_{n} \cap E_{n})> \eta_{0} /8$. Moreover, by the definition of ${T_m}$, we have $p( t,\| u(t)\| )\chi_{G_{n}}\in E_{N}^{0}(X)$. Therefore, by formula (2.3), we have

for $n$ large enough, this is a contradiction.

Case II. Let ${\lim \sup }_{n \to \infty }\mu \left(G_{0}\cap E_{n}\right) = \eta_{0}>0$, where $t\in G_{0}$ if and only if $M(t,\left\| {u(t)} \right\|)$ is a left extreme point and is a right extreme point of linear interval of $M(t,u)$. Then $G_{0}\subset G$ and $p\left( {t,\left\| {u(t)} \right\|} \right) - {p_ - }\left( {t,\left\| {u(t)} \right\|} \right)>0$ whenever $t\in G_{0}$. Hence there exist a set $H \subset G_{0}$ and a real number $h>0$ such that $\mu H <\eta_{0}/16$ and $p\left( {t,\left\| {u(t)} \right\|} \right) - {p_ - }\left( {t,\left\| {u(t)} \right\|} \right)>h$ whenever $t\in G_{0} \setminus H$. Moreover, there exists $m\in N$ such that $\mu(T\backslash T_{m})<\eta_{0}/16$. Let $F = \left( {G_{0} \cap {T_m}} \right)\backslash H$. Then, by formula ${\lim \sup }_{n \to \infty }\mu \left(G_{0}\cap E_{n}\right) = \eta_{0}$, we may assume that $\mu \left(F\cap E_{n}\right)> \eta_{0}/8$. Therefore, by formula (2.3), we get the following inequalities

for $n$ large enough, this is a contradiction. Hence we get that ${\left\| {{u_n}(t)} \right\| \to \left\| {u(t)} \right\|}$ in measure on $G$. Let $\{ r_{1}(i,t)\} _{i = 1}^\infty$ denote the set of all the left extreme points of linear interval of $M(t,u)$. Define the set

Similarly, we get that ${\left\| {{u_n}(t)} \right\| \to \left\| {u(t)} \right\|}$ in measure on $G$. In summary, we have ${\left\| {{u_n}(t)} \right\| \to \left\| {u(t)} \right\|}$ in measure on $T$. Therefore, by the Riesz Theorem, there exists a subsequence $\{n\}$ of $\{n\}$ such that $\| u_n(t) \| \to \| u(t)\|$ $\mu-$a.e. in $T$. Noticing that

and

we get the following formula

Moreover, it is easy to see that

This implies that $\left\| {u_n (t)} \right\| \cdot \left\| {v(t)} \right\| - (u_n (t),v(t))\mathop \to \limits^\mu 0$ in measure. Therefore, by the Riesz theorem, there exists a subsequence $\{n\}$ of $\{n\}$ such that $\left\| {u_n (t)} \right\| \cdot \left\| {v(t)} \right\| - (u_n (t),v(t))\rightarrow 0$ $\mu-$a.e. in $T$. Since $\| u_n(t) \| \to \| u(t)\|$ $\mu$-a.e. on $T$, we get that $(u_n (t),v(t)) \to \left\| {u(t)} \right\| \cdot \left\| {v(t)} \right\|$ $\mu-$a.e. on $T$. Hence we may assume without loss of generality that

on $\{ t \in T:\left\| {u(t)} \right\|\cdot\left\| {v(t)} \right\| \ne 0\}$. Since $\langle u,v\rangle = 1$, we get that $\mu T_{1} = 0$, where

Hence we may assume without loss of generality that

Since $X$ is a strongly smooth space, we obtain that sequence $\left \{ {{u_n (t)}}/{{\left\| {u(t)} \right\|}}\right \} _{n = 1}^\infty$ is convergent. Hence there exists $x(t)\in S(X)$ such that ${{u_n (t)}}/{{\left\| {u(t)} \right\|}} \to x(t)$ on $\{ t \in T:\left\| {u(t)} \right\| \ne 0\}$. Let

Then it is easy to see that $\|u_{0}\|^{0} = 1$ and $u_n(t) \to u_{0}(t)$ $\mu$-a.e. on $T$. Therefore, by the Fatou Lemma, we obtain the following inequalities

This implies that $\rho _M ((u_{n} - u)/2) \to 0$ as $n \to \infty$. Moreover, from the previous proof, we obtain that ${\rho _M}(2{u_n}{\chi _E}/\varepsilon ) \to {\rho _M}(2{u_0}{\chi _E}/\varepsilon )$ for any $E\subset T$. We next will prove that $\left\| {{u_n} - u} \right\| \to 0$ as $n \to \infty$. Let $\varepsilon\in(0,1/4)$. Since $M\in \Delta$, there exists $\delta>0$ such that ${\rho _M}(2u_{0}{\chi _E}/\varepsilon ) < 1/2$ whenever $\mu E<4\delta$. Moreover, there exists $r>0$ such that $\mu \left\{ {t \in T:0 < \left\| {u(t)} \right\| < r} \right\} < \delta$. Since $u_n(t) \to u_{0}(t)$ $\mu$-a.e. on $T$, by the Egorov theorem, there exist a natural number $n_{0}\in N$ and $F\subset T$ such that $\mu F<\delta$ and ${\left\| {{u_n}(t) - {u_0}(t)} \right\| < \varepsilon^{2}}$, $t\in T\backslash F$ whenever $n>n_{0}$. This implies that $\left\| {({u_n} - {u_0}){\chi _{B\backslash F}}} \right\| < 2\varepsilon$ whenever $n>n_{0}$. Let $B = \{ t \in T:\left\| {{u_0}(t)} \right\|\geq \varepsilon\}$, $D = \left\{ {t \in T:0 < \left\| {u(t)} \right\| < \varepsilon} \right\}$ and $H = \{ t \in T:\left\| {{u_0}(t)} \right\| = 0\} \cup F \cup D$.

Moreover, by ${\rho _M}(2{u_n}{\chi _E}/\varepsilon ) \to {\rho _M}(2{u_0}{\chi _E}/\varepsilon )$ for any $E\subset T$ and $\mu (F \cup D) < 2\delta$, there exists a natural number $n_{1}>n_{0}$ such that

whenever $n>n_{1}$. This implies that $\left\| {({u_n} - {u_0}){\chi _H}} \right\| < 2\varepsilon$ whenever $n>n_{1}$. Since $T = H \cup (B\backslash F)$, we have

whenever $n>n_{1}$. Hence we have $\left\| {{u_n} - u} \right\| \to 0$ as $n \to \infty$. This implies that $v$ is a strongly smooth point of $E_{N}^{0}(X)$, which completes the proof.

Corollary 2.5. Suppose that $v\in S(E_{N}^{0}(R))$. Then the following statements are equivalent:

(1) The point $v$ is a strongly smooth point of $E_{N}^{0}(R)$;

(2) The hyperplane $H_{v}$ of $L_M(R)$ is approximatively compact;

(3) The hyperplane $H_{v}$ of $L_M(R)$ is weakly approximatively compact;

(4) $M\in\Delta$ and $\mu \{ t \in T:\left| {u(t)} \right| \in {K_t}\} = 0$ whenever $\langle u,v\rangle = \|u\|$.

Corollary 2.6. Suppose that $X$ is a strongly smooth space. Then the following statements are equivalent:

(1) $E_{N}^{0}(X)$ 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.

3.   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) $L_{M}^{0}(X)$ is an Asplund space;

(2) $M\in\Delta$ and $N\in\Delta$.

Proof. (1)$\Rightarrow$(2). Suppose that $M\notin\Delta$. Then we define the functional $\theta (u) = \inf \left\{ \lambda \in {R^ + }: {\rho _M}(u/\lambda ) < + \infty \right\}$. Therefore, by the definition of $\theta (\cdot)$, we get that $\theta (ku) = k\theta (u)$ whenever $k>0$. Moreover, by the convexity of $M$, we have

This implies that $\theta ({u_1} + {u_1}) \le \theta ({u_1}) + \theta ({u_2}) + 2\varepsilon$. Since $\varepsilon$ is arbitrary, we have $\theta ({u_1} + {u_1}) \le \theta ({u_1}) + \theta ({u_2})$. It is easy to see that $\theta (u)$ is a convex function. We claim that if $\|u_{n}\|\to 0$ then $\theta ({u_n})\to 0$ as $n\to\infty$. In fact, for any $\varepsilon>0$, we have $\|u_{n}/\varepsilon \|\to 0$ as $n\to\infty$. Hence there exists a natural number $n_{0}$ such that ${\rho _M}\left( {{u_n}/\varepsilon } \right) \le \left\| {{u_n}/\varepsilon } \right\| \le 1$ whenever $n>n_{0}$. This implies that $\theta ({u_n})<\varepsilon$ whenever $n>n_{0}$. Hence $\theta ({u_n})\to 0$ as $n\to\infty$.

Next we will prove that $\theta (\cdot)$ is continuous. Otherwise, there exist $\varepsilon_{0}>0$, $u\in {L_M}(X)$ and $\{ {u_n}\} _{n = 1}^\infty \subset {L_M}(X)$ such that $\|u_{n}-u\|\to 0$ and $\left| {\theta ({u_n}) - \theta (u)} \right| \ge 2{\varepsilon _0}$. Then we may assume that $\theta (u) \ge \theta ({u_n}) + 2{\varepsilon _0}$ or $\theta (u_{n}) \ge \theta ({u}) + 2{\varepsilon _0}$. Hence, if $\theta (u) \ge \theta ({u_n}) + 2{\varepsilon _0}$, then

a contradiction. Moreover, if $\theta (u_{n}) \ge \theta ({u}) + 2{\varepsilon _0}$, then

a contradiction. Hence $\theta (\cdot)$ is a continuous function. Pick $u\in L_{M}^{0}(X)\setminus E_{M}^{0}(X)$. Then $\theta (u)>0$. We next will prove that there exist $E\subset T$ and $F\subset T$ such that $E\cup F = T$, $E\cap F = \emptyset$ and $\theta (u{\chi _E}) = \theta (u{\chi _F}) = \theta (u)$. Define $G(n) = \{ t \in T:n - 1 \le \left\| {u(t)} \right\| < n\}$ and for each $n\in N$, decompose $G(n)$ into $G_{1}(n)$ and $G_{2}(n)$ such that

where $i = 1,2$. We claim that ${u_i} = \sum\nolimits_{n = 1}^\infty {u{\chi _{{G_i}(n)}}}$ satisfy $\theta ({u_i}) = \theta (u)$, where $i = 1,2$. In fact, let $\theta = \theta (u)$. Then for any $\varepsilon \in (0,\theta /2)$, there exists a natural number $m$ such that

Since

for all $n\geq m$ and all $t \in {G_i}(n)$, we have the following inequalities

Therefore, by the definition of $\theta (\cdot)$, we have the following inequalities

This implies that $\theta ({u_i}) = \theta (u)$, where $i = 1,2$. Hence there exist a set $E\subset T$ and $F\subset T$ such that $E\cup F = T$, $E\cap F = \emptyset$ and $\theta (u{\chi _E}) = \theta (u{\chi _F}) = \theta (u)$. Let $v = u{\chi _E} - u{\chi _F}$. Then, if $t>0$ then

This implies that

whenever $t>0$. Moreover, if $t<0$ then

This implies that

whenever $t<0$. Hence, for any $u\in L_{M}^{0}(X)\setminus E_{M}^{0}(X)$, we obtain that $\theta (\cdot)$ is not differentiable at $u$, a contradiction. Then $M\in\Delta$.

Suppose that $N\notin\Delta$. Then there exists a point $v \in L_{N}(X^{*})$ such that $\|v\| = 1$ and ${\rho _N}(v)<1$. Pick a real number $l>1$. Then ${\rho _N}(lv) = \infty$. Define

for all $n\in N$. Decompose $G(n)$ into $G_{1}(n)$ and $G_{2}(n)$ such that $G_{1}(n) \cup G_{2}(n) = G(n)$ and ${\rho _N}(lv{\chi _{{G_1}(n)}}) = {\rho _N}(lv{\chi _{{G_2}(n)}})$. Decompose $G$ into $G_{1}$ and $G_{2}$ such that $G_{1} \cup G_{2} = G$ and ${\rho _N}(lv{\chi _{{G_1}}}) = {\rho _N}(lv{\chi _{{G_2}}})$. Let

Then ${\rho _N}(lv{\chi _E}) = {\rho _N}(lv{\chi _F})$. This implies that $\left\| {v{\chi _E}} \right\| \ge 1/l$ and $\left\| {v{\chi _F}} \right\| \ge 1/l$. Hence there exists a sequence of set $\{ {E_n}\} _{n = 1}^\infty$ such that $\left\| {v{\chi _{{E_n}}}} \right\| \ge 1/l$ and $E_i\cap E_j = \emptyset$ whenever $i\neq j$. 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, Pick $u = \sum_{i = 1}^\infty {{\varepsilon _i}v{\chi _{{E_i}}}} \in C$. Then there exists a natural number $j\in N$ such that $\left| {{\varepsilon _j}} \right| < 1/4$. Define

Then $u_{1}, u_{2}\in C$, $u_{1}\neq u_{2}$ and $2u = u_{1}+ u_{2}$. This implies that $u$ is not an extreme point. Since $u$ is arbitrary, we have $ExtC = \emptyset$. Hence $L_{N}(X^{*})$ has not the Krein-Milman property. Then $L_{N}(X^{*})$ has not the Radon-Nikodym property. Hence $E_{M}^{0}(X)$ is not an Asplund space. However, since $L_{M}^{0}(X)$ is an Asplund space, we obtain that $E_{M}^{0}(X)$ is an Asplund space, a contradiction. Then $N\in\Delta$.

(2) $\Rightarrow$(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 $t \in T$. Therefore, by formula 3.1, we obtain that $u\in L_{M}(X)$ for any $u\in L_{M_{1}}(X)$. Since $M\in \Delta$, we have the following inequalities

for any $\lambda>0$. This implies that $M_{1}\in \Delta$. Moreover, if $v\in L_{N_{1}}(X^{*})$, then there exists $\lambda_{0}>0$ such that ${\rho _{{N_1}}}({\lambda _0}v) < + \infty$. Since

we have the following inequalities

This implies that $v\in L_{N}(X^{*})$. Therefore, by $N\in \Delta$, we obtain that $v\in E_{N}(X^{*})$. Therefore, by formula

we have the following inequalities

for every $\lambda>0$. This implies that $N_{1}\in \Delta$. Therefore, by theorem 2.7, we obtain that $L_{M_{1}}^{0}(X)$ is a strongly smooth space. This implies that $L_{M_{1}}^{0}(X)$ is an Asplund space. Moreover, we have

and

for any $u\in L_{M}^{0}(X)$. This means that $L_{M}^{0}(X)$ 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) $M\in\Delta$ and $N\in\Delta$.

Corollary 3.3. $L_{M}^{0}(R)(L_{M}(R))$ is an Asplund space if and only if $M\in\Delta$ and $N\in\Delta$.

Theorem 3.4. Suppose that

(1) $M\in\Delta$ and $X$ is a strongly smooth space;

(2) $p(t, u)$ is continuous with respect to $u$ for almost all $t \in T$.

Then $E_{N}^{0}(X)$ 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.

Theorem 3.5. Suppose that $X$ is a strongly smooth space. Then $L_{M}(X^{*})$ has the Radon-Nikodym property if and only if $M\in\Delta$.

Proof. Sufficiency. Let $X$ be a strongly smooth space. Then, 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 $t \in T$. Moreover, by the proof of Theorem 3.1, we get that $M_{1}\in\Delta$. Therefore, by Theorem 3.4, we obtain that $E_{N}^{0}(X)$ is a strongly smooth space. Hence $E_{N}^{0}(X)$ is an Asplund space. This implies that $L_{M}(X^{*})$ has the Radon-Nikodym property.

Necessity. Suppose that $M\notin\Delta$. 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 $M\in\Delta$, which completes the proof.

Corollary 3.6. Suppose that $X$ is a strongly smooth space. Then $L_{M}^{0}(X^{*})$ has the Radon-Nikodym property if and only if $M\in\Delta$.

Corollary 3.7. $L_{M}^{0}(R)(L_{M}(R))$ has the Radon-Nikodym property if and only if $M\in\Delta$.

Acknowledgment

This research is supported by "China Natural Science Fund under grant 11871181" and "China Natural Science Fund under grant 11561053".