Strongly extreme points of Orlicz function spaces equipped with Φ-Amemiya norm

Abstract

In this paper, the criterion that points of Orlicz function spaces equipped with Φ-Amemiya norm generated by an Orlicz function are strongly extreme is given. As a corollary, the sufficient and necessary conditions of midpoint local uniform rotundity of Orlicz function spaces equipped with Φ-Amemiya norm are obtained.

1 Introduction

An extreme point plays a crucial role in functional analysis, convex analysis, and optimization. In fact, any compact convex set is the convex hull of its extreme point set, the result is called Krein–Milman theorem. The notion of a dentable subset of a Banach space was introduced by Rieffel in conjunction with a Radon–Nikodym theorem for Banach space-valued measures. Subsequent work by Maynard and Davis and Phelos has shown those Banach spaces in which Rieffel’s Radon–Nikodym theorem is valid and every bounded closed convex set is dentable. This has been a significant breakthrough in studying the nature of Radon–Nikodym as a geometric property. In 1988, Bor-Luh Lin, Pei-Kee Lin, and Troyanski described the characteristic of denting points (see [1, 2]) and obtained that there is a close relationship between denting points and strongly extreme points. It is easy to see that every denting point of Banach space X is a strongly extreme point (see [3]) of X, and it is known that every strongly extreme point of X is a $w^{\ast }$ extreme point of X. Orlicz space is an important class of Banach space, it was introduced by the famous Polish mathematician Wladyslaw Orlicz in 1932. The theory of Orlicz space has been greatly developed because of its important theoretical properties and application value. Up to now, the criterion that an element in the unit sphere of Orlicz spaces equipped with the Orlicz norm, the Luxemburg norm, and the p-Amemiya norm is a strongly extreme point has been obtained (see [4–6]). In this paper, we introduce a new norm, namely Φ-Amemiya norm, whose calculation formula is given as follows: $\Vert x \Vert _{\varPhi ,\varPhi _{1}}=\inf_{k>0} \{ \frac{1}{k} (1+ \varPhi (I_{\varPhi _{1}}(kx)) ) \}$. When we take some special functions, the previous norms are special cases of this new norm. This new norm also has wider applicability than before. We give the criterion that an element in the unit sphere of Orlicz spaces equipped with Φ-Amemiya norm is a strongly extreme point. Incidentally, the sufficient and necessary conditions of midpoint local uniform rotundity of Orlicz function spaces equipped with Φ-Amemiya norm are obtained.

2 Preliminaries

Let $[X, \Vert \cdot \Vert ]$ be a Banach space. $S(X)$ and $B(X)$ denote the unit sphere and the unit ball of X, respectively. $X^{\ast }$ is said to be the dual space of X.

Definition 2.1

A mapping $\varPhi :R\rightarrow [0,\infty )$ is called an Orlicz function: if Φ is even, continuous, convex and $\varPhi (u)=0$ if and only if $u=0$. If Φ also satisfies $\lim_{u\rightarrow 0}\frac{\varPhi (u)}{u}=0$ and $\lim_{u\rightarrow \infty }\frac{\varPhi (u)}{u}=\infty$, then Φ is called an N-function.

Definition 2.2

The function Ψ defined by the formula $\varPsi (u)=\sup \{ \vert u \vert v-\varPhi (v): v\geq 0\}$ is called complementary function of Φ in the sense of Young.

Definition 2.3

Let $(G,\varSigma ,\mu )$ be a nonatomic finite measure space. Let $L^{0}$ denote the whole of the measurable real function on G. We define the modular $I_{\varPhi }$: $L^{0}\rightarrow R^{+}=[0,+\infty ]$ as follows:

$$\begin{aligned} I_{\varPhi }(x)= \int _{G} \varPhi \bigl(x(t)\bigr)\,dt, \end{aligned}$$

it is called the modular (see [7]) of x.

Definition 2.4

The Orlicz function space (see [8]) $L_{\varPhi }$ generated by an Orlicz function is defined by the formula $L_{\varPhi }=\{x\in L^{0}: I_{\varPhi }(kx )<+\infty \text{ for some }k>0\}$.

Those spaces that are equipped with the Orlicz norm (Amemiya norm) (see [9])

$$\begin{aligned} \Vert x \Vert _{\varPhi }^{0}=\inf_{k>0} \frac{1}{k}\bigl(1+I_{\varPhi }(kx)\bigr), \end{aligned}$$

or equipped with the Luxemburg norm

$$\begin{aligned} \Vert x \Vert _{\varPhi }=\inf \biggl\{ k>0:I_{\varPhi } \biggl( \frac{x}{k} \biggr) \leq 1 \biggr\} , \end{aligned}$$

or equipped with the p-Amemiya norm ($1\leq p < +\infty$) (see [10, 11])

$$\begin{aligned} \Vert x \Vert _{\varPhi ,p}=\inf_{k>0}\frac{1}{k} \bigl(1+I_{\varPhi }^{p}(kx)\bigr)^{ \frac{1}{p}}, \end{aligned}$$

are Banach spaces, abbreviated as

$$\begin{aligned} L^{0}_{\varPhi }= \bigl[L_{\varPhi }, \Vert \cdot \Vert _{\varPhi }^{0}\bigr]; \qquad L_{\varPhi }= \bigl[L_{\varPhi }, \Vert \cdot \Vert _{\varPhi }\bigr]; \qquad L_{\varPhi ,p} = \bigl[L_{\varPhi }, \Vert \cdot \Vert _{\varPhi ,p} \bigr]. \end{aligned}$$

Definition 2.5

We say that Orlicz function Φ satisfies the $\Delta _{2}$ condition if there exist $k>2$ and $u_{0}\geq 0$ such that the inequality

$$\begin{aligned} \varPhi (2u) \leq k \varPhi (u) \end{aligned}$$

holds for $\vert u \vert \geq u_{0}$.

Definition 2.6

If, for every $y,z\in R$ and $y\neq z$ with $\frac{y+z}{2}=x$, we have $\varPhi (x)< \frac{\varPhi (y)+\varPhi (z)}{2}$, then x is called a strictly convex point of Φ. The set of all strictly convex points of Φ will be denoted by $S_{\varPhi }$.

For any Orlicz functions Φ and $\varPhi _{1}$, we put $L_{\varPhi ,\varPhi _{1}}=\{x\in L^{0}:\varPhi (I_{\varPhi _{1}}(kx))<+\infty \text{ for some}\ k>0 \}$. The calculation formula

$$\begin{aligned} \Vert x \Vert _{\varPhi ,\varPhi _{1}}=\inf_{k>0}\frac{1}{k} \bigl(1+\varPhi \bigl(I_{ \varPhi _{1}}(kx)\bigr)\bigr) \end{aligned}$$

is called Φ-Amemiya norm.

Remark

• If we take $\varPhi (u)=\max \{0,u-1\}$, then $\Vert x \Vert _{\varPhi ,\varPhi _{1}}$ is the Luxemburg norm $\Vert \cdot \Vert _{\varPhi }$;

• If we take $\varPhi (u)=u$, then $\Vert x \Vert _{\varPhi ,\varPhi _{1}}$ is the Orlicz norm $\Vert \cdot \Vert _{\varPhi }^{0}$;

• If we take $\max \{0,u-1\} \leq \varPhi (u)\leq u$, then $\Vert x \Vert _{\varPhi ,\varPhi _{1}}$ is the s-norm $\Vert \cdot \Vert _{\varPhi }^{s}$;

• If we take $\varPhi (u)\geq \vert u \vert$, then $\Vert x \Vert _{\varPhi ,\varPhi _{1}}\geq \Vert \cdot \Vert _{\varPhi }^{0}$.

An important question is the attainability of the “inf” in $\Vert x \Vert _{\varPhi ,\varPhi _{1}}=\inf_{k>0}\frac{1}{k}(1+ \varPhi (I_{ \varPhi _{1}}(kx)))$. For any $x\in L_{\varPhi ,\varPhi _{1}}$, $x\neq 0$, we define

$$\begin{aligned} K(x)=\biggl\{ k>0: \Vert x \Vert _{\varPhi ,\varPhi _{1}}=\frac{1}{k}\bigl(1+ \varPhi \bigl(I_{\varPhi _{1}}(kx)\bigr)\bigr) \biggr\} . \end{aligned}$$

We will prove that if $\lim_{u\rightarrow \infty }\frac{\varPhi _{1}(u)}{u}=+\infty$, then $K(x)\neq \phi$.

Proof

Put $F(k)=\frac{1}{k}(1+\varPhi (\int _{G}\varPhi _{1}(kx(t))\,dt))$ and $\theta (x)=\inf \{k>0,I_{\varPhi _{1}}(\frac{x}{k})<\infty \}$.

Then there exists $d>0$ such that $\mu (\{t\in G: \vert x(t) \vert >d\})>0$ and $F(k)$ is continuous on $(0,\theta (x))$. So $\lim_{k\rightarrow 0^{+}}F(k)=+\infty$.

Suppose that $\theta (x)=+\infty$. Then

$$\begin{aligned} \lim_{k\rightarrow +\infty }F(k) =&\lim_{k \rightarrow +\infty } \frac{\varPhi (\int _{G}\varPhi _{1}(kx(t))\,dt)}{k} \\ \geq &\lim_{k\rightarrow +\infty }\varPhi \biggl( \frac{\int _{G}\varPhi _{1}(kx(t))\,dt}{k}\biggr) \\ \geq &\lim_{k\rightarrow +\infty }\varPhi \biggl( \frac{\int _{\{t\in G: \vert x(t) \vert >d\}}\varPhi _{1}(kx(t))\,dt}{k}\biggr) \\ \geq &d\lim_{k\rightarrow +\infty }\varPhi \biggl( \frac{\varPhi _{1}(kd)\cdot \mu (\{t\in G: \vert x(t) \vert >d\})}{kd}\biggr) \\ =&+\infty . \end{aligned}$$

Since $F(k)$ is a continuous function, then there exists $k_{0}\in (0,\theta (x))$ such that $F(k)\geq F(k_{0})$.

Suppose that $\theta (x)<+\infty$. If $I_{\varPhi _{1}}(\theta (x)x(t))=+\infty$, we have

$$\begin{aligned} \lim_{k\rightarrow \theta (x)-0}F(k =\frac{1}{\theta (x)}\biggl(1+ \varPhi \biggl( \int _{G}\varPhi _{1}\bigl(\theta (x)x(t)\bigr)\,dt \biggr)\biggr) =+\infty . \end{aligned}$$

If $I_{\varPhi _{1}}(\theta (x)x(t))<+\infty$, we have

$$\begin{aligned} \lim_{k\rightarrow \theta (x)-0}F(k)=\frac{1}{\theta (x)}\biggl(1+ \varPhi \biggl( \int _{G}\varPhi _{1}\bigl(\theta (x)x(t)\bigr)\,dt \biggr)\biggr)< +\infty . \end{aligned}$$

Since $F(k)$ is a continuous function, there exists $k_{0}\in (0,\theta (x)]$ such that $F(k)\geq F(k_{0})$.

Thus $K(x)\neq \phi$. □

Definition 2.7

A point $x\in S(X)$ is said to be an extreme point (see [12]) of $B(X)$ if, for any $y,z\in S(X)$ and $x=\frac{y+z}{2}$, we have $y=z$. The set of all extreme points of the unit ball $B(X)$ will be denoted by $ExtB(X)$. X is said to be strictly convex (see [13]) if and only if $ExtB(X)=S(X)$.

Definition 2.8

A point $x\in S(X)$ is called a strongly extreme point (see [14–16]) of $B(X)$ if, for any $\{x_{n}\}\subseteq X$, $\{y_{n}\}\subseteq X$, $\lim_{n\rightarrow \infty } \Vert x_{n} \Vert =\lim_{n \rightarrow \infty } \Vert y_{n} \Vert =1$, and $\frac{x_{n}+y_{n}}{2} =x$, we have $\lim_{n\rightarrow \infty } \Vert x_{n}-y_{n} \Vert =0$.

Definition 2.9

Banach space X is called middle point local uniform convex (see [17, 18]) if and only if each point on $S(X)$ is a strongly extreme point.

Lemma 2.10

(EropoB theorem)

Let$\{f_{n}\}_{n=1}^{\infty }$be a measurable function and$\vert f_{n}(x) \vert <\infty$a.e. $x\in E$with$m(E)<+\infty$. If$f_{n}(x)\rightarrow f(x)$a.e. $x\in E$, then for any$\delta >0$there exists$E_{0}\subset E$such that$m(E_{0})<\delta$and$f_{n}(x)\rightarrow f(x)$uniformly in$x\in E\setminus E_{0}$.

Lemma 2.11

Assume$\varPhi \in \bigtriangleup _{2}$ (see [19]). Then, for any$L>0$and$\varepsilon >0$, there exists$\delta >0$such that

$$\begin{aligned} \bigl\vert I_{\varPhi }(u+v)-I_{\varPhi }(v) \bigr\vert < \varepsilon, \end{aligned}$$

whenever$I_{\varPhi }(u)\leq L$, $I_{\varPhi }(v)\leq \delta$.

Lemma 2.12

Let$\varPhi \in \bigtriangleup _{2}$. If$I_{\varPhi }(x_{n})\rightarrow I_{\varPhi }(x)$, $x_{n}\xrightarrow{\mu }x$, then$\Vert x_{n}-x \Vert _{\varPhi }\rightarrow 0$ (see [19]).

3 Main results

Theorem 3.1

Let$\varPhi _{1}$be an N-function. Then$x_{0}\in S(L_{\varPhi ,\varPhi _{1}})$is a strongly extreme point of$B(L_{\varPhi ,\varPhi _{1}})$if and only if$\varPhi _{1}\in \Delta _{2}$and$k_{0} x_{0}(t)\in S_{\varPhi _{1}}$, where$k_{0} \in K(x_{0})$.

Proof

Necessity. Suppose that $\mu (\{t\in G:k_{0}x_{0}(t)\notin S_{\varPhi _{1}}\})>0$ for some $k_{0} \in K(x_{0})$. There exists an interval $(a,b)$ such that $\mu (\{t\in G:\frac{a}{k_{0}}+\varepsilon < x_{0}(t)<\frac{b}{k_{0}}- \varepsilon \})>0$ ($\varepsilon >0$) and $\varPhi _{1}$ is affine on $(a,b)$, i.e., $\varPhi _{1}(x)=px+q$. Divide $\{t\in G:\frac{a}{k_{0}}+\varepsilon < x_{0}(t)<\frac{b}{k_{0}}- \varepsilon \}$ into two sets E and F with $E\cap F=\emptyset$ and $\mu (E)=\mu (F)$. Define

$$\begin{aligned}& y(t)= \textstyle\begin{cases} x_{0}(t),&t\in G\setminus (E\cup F), \\ x_{0}(t)-\varepsilon ,&t\in E, \\ x_{0}(t)+\varepsilon ,&t\in F, \end{cases}\displaystyle \\& z(t)= \textstyle\begin{cases} x_{0}(t),&t\in G\setminus (E\cup F), \\ x_{0}(t)+\varepsilon ,&t\in E, \\ x_{0}(t)-\varepsilon ,&t\in F. \end{cases}\displaystyle \end{aligned}$$

Then $x_{0}=\frac{y+z}{2}$, $y\neq z$ and

$$\begin{aligned} I_{\varPhi _{1}}(k_{0}y) =& \int _{E\cup F}\varPhi _{1}\bigl(k_{0}y(t) \bigr)\,dt+ \int _{G \setminus E\cup F}\varPhi _{1}\bigl(k_{0}y(t) \bigr)\,dt \\ =& \int _{E}\bigl(p\bigl(k_{0}\bigl(x_{0}(t)- \varepsilon \bigr)\bigr)+q\bigr)\,dt+ \int _{F}\bigl(p\bigl(k_{0}\bigl(x_{0}(t)+ \varepsilon \bigr)\bigr)+q\bigr)\,dt \\ &{}+ \int _{G\setminus E\cup F}\varPhi _{1}\bigl(k_{0}x_{0}(t) \bigr)\,dt \\ =& \int _{E\cup F}\bigl(pk_{0}x_{0}(t)+q\bigr)\,dt+ \int _{G\setminus E\cup F} \varPhi _{1}\bigl(k_{0}x_{0}(t) \bigr)\,dt \\ =& \int _{E\cup F}\varPhi _{1}\bigl(k_{0}x_{0}(t) \bigr)\,dt+ \int _{G\setminus E\cup F} \varPhi _{1}\bigl(k_{0}x_{0}(t) \bigr)\,dt \\ =&I_{\varPhi _{1}}(k_{0}x_{0}). \end{aligned}$$

Thus $\Vert y \Vert _{\varPhi ,\varPhi _{1}}\leq \frac{1}{k_{0}}(1+\varPhi (I_{\varPhi _{1}}(k_{0}y)))= \frac{1}{k_{0}}(1+\varPhi (I_{\varPhi _{1}}(k_{0}x_{0})))= \Vert x_{0} \Vert _{\varPhi , \varPhi _{1}}=1$. In the same way, we can get $\Vert z \Vert _{\varPhi ,\varPhi _{1}}\leq 1$. This contradicts the fact that $x_{0}$ is an extreme point of $S(L_{\varPhi ,\varPhi _{1}})$.

In order to complete this proof, we need to prove that if $\varPhi _{1} \notin \Delta _{2}$, there is not a strongly extreme point on the unit sphere of $L_{\varPhi ,\varPhi _{1}}$. If $x_{0}\in S(L_{\varPhi ,\varPhi _{1}})$, then there exists $d>0$ such that $\mu (\{t\in G: \vert x_{0}(t) \vert \leq d \}) >0$. Suppose $\varPhi _{1} \notin \Delta _{2}$, then there exists $u_{n}>0$, $u_{n}\uparrow \infty$ such that $\varPhi _{1}(2u_{n})>2^{n} \varPhi _{1}(u_{n})$ ($n=1,2,\ldots$). Without loss of generality, we can assume that $\frac{1}{\varPhi _{1} (u_{1})}< \mu (\{t\in G: \vert x(t) \vert \leq d \})$. Take $\{G_{n}\}\subset \{t\in G: \vert x(t) \vert < d \}$ with $G_{m} \cap G_{n} = \emptyset$ for any $m\neq n$, satisfying

$$\begin{aligned} \mu (G_{n})=\frac{1}{2^{n} \varPhi _{1} (u_{n})} \quad (n=1,2,\ldots ). \end{aligned}$$

Define

$$\begin{aligned}& x_{n}(t)= \textstyle\begin{cases} x_{0}(t), & t\in G \setminus G_{n}, \\ x_{0}(t)+\frac{u_{n}}{k_{0}}, & t\in G_{n} , \end{cases}\displaystyle \\& y_{n}(t)= \textstyle\begin{cases} x_{0}(t), & t\in G \setminus G_{n}, \\ x_{0}(t)-\frac{u_{n}}{k_{0}}, & t\in G_{n}. \end{cases}\displaystyle \end{aligned}$$

Then $x_{0}=\frac{x_{n}+y_{n}}{2}$ for each $n\in N$.

Put

$$\begin{aligned} x_{n}(t) = &x_{n}^{\prime }(t) + x_{n}^{\prime \prime }(t), \end{aligned}$$

where $x_{n}^{\prime }(t) = x_{0} \chi _{G \setminus G_{n}}(t)+ \frac{u_{n}}{k_{0}} \chi _{G_{n}}(t)$, $x_{n}^{\prime \prime }(t)= x_{0} \chi _{G_{n}}(t)$.

Since $\Vert x_{n}^{\prime \prime } \Vert _{\varPhi ,\varPhi _{1}}= \Vert x_{0} \chi _{ G_{n}} \Vert _{\varPhi , \varPhi _{1}} \leq d \Vert \chi _{ G_{n}} \Vert _{\varPhi ,\varPhi _{1}} \rightarrow 0$ ($n\rightarrow \infty$), we have the inequality $\Vert x_{n}^{\prime } \Vert _{\varPhi ,\varPhi _{1}}\geq \Vert x_{0} \chi _{G \setminus G_{n}} \Vert _{\varPhi ,\varPhi _{1}} \geq \Vert x_{0} \Vert _{\varPhi ,\varPhi _{1}}- \Vert x_{0} \chi _{G_{n}} \Vert _{\varPhi ,\varPhi _{1}}$ holds. Therefore $\lim_{\overline{n\rightarrow \infty }} \Vert x_{n}^{\prime } \Vert _{\varPhi , \varPhi _{1}}\geq \Vert x_{0} \Vert _{\varPhi ,\varPhi _{1}} =1$.

By the definition of Φ-Amemiya norm, we deduce that

$$\begin{aligned} \bigl\Vert x_{n}^{\prime } \bigr\Vert _{\varPhi ,\varPhi _{1}} =& \inf_{k>0} \frac{1}{k} \bigl(1+\varPhi \bigl(I_{\varPhi _{1}}\bigl(kx_{n}^{\prime }\bigr) \bigr) \bigr) \\ \leq & \frac{1}{k_{0}} \bigl(1+\varPhi \bigl(I_{\varPhi _{1}} \bigl(k_{0}x_{n}^{\prime }\bigr) \bigr) \bigr) \\ \leq & \frac{1}{k_{0}} + \frac{1}{k_{0}}\varPhi \biggl( \int _{G} \varPhi _{1} \biggl(k_{0} \biggl( x_{0} \chi _{G \setminus G_{n}} (t) + \frac{u_{n}}{k_{0}} \chi _{G_{n}}(t) \biggr) \biggr)\,dt \biggr) \\ \leq & \frac{1}{k_{0}} + \frac{1}{k_{0}}\varPhi \biggl( \int _{G \setminus G_{n}} \varPhi _{1} \bigl(k_{0} x_{0} \chi _{G \setminus G_{n}} (t) \bigr)\,dt + \int _{G_{n}} \varPhi _{1} \bigl(u_{n} \chi _{G_{n}} (t) \bigr) \,dt \biggr) \\ \leq & \frac{1}{k_{0}} + \frac{1}{k_{0}}\varPhi \bigl( I_{\varPhi _{1}}(k_{0}x_{0})+ \varPhi _{1} (u_{n}) \mu (G_{n}) \bigr) \\ =& \frac{1}{k_{0}} \biggl(1+ \varPhi \biggl( I_{\varPhi _{1}}(k_{0}x_{0}) + \frac{1}{2^{n}} \biggr) \biggr). \end{aligned}$$

Then

$$\begin{aligned} \overline{\lim_{n\rightarrow \infty }} \bigl\Vert x_{n}^{\prime } \bigr\Vert _{\varPhi , \varPhi _{1}} \leq \Vert x_{0} \Vert _{\varPhi ,\varPhi _{1}}=1. \end{aligned}$$

Hence

$$\begin{aligned} \lim_{n\rightarrow \infty } \Vert x_{n} \Vert _{\varPhi ,\varPhi _{1}} =1. \end{aligned}$$

In the same way, we have

$$\begin{aligned} \lim_{n\rightarrow \infty } \Vert y_{n} \Vert _{\varPhi ,\varPhi _{1}} =1. \end{aligned}$$

But

$$\begin{aligned} I_{\varPhi _{1}} \bigl(k_{0} (x_{n}-y_{n}) \bigr) =& \int _{G_{n}}\varPhi _{1} \biggl( k_{0} \frac{2u_{n}(t)}{k_{0}} \biggr) \,dt \\ =& \varPhi _{1} (2u_{n}) \mu (G_{n}) \geq 1 \quad (n=1,2,\ldots ). \end{aligned}$$

Therefore

$$\begin{aligned} \Vert x_{n}-y_{n} \Vert _{\varPhi ,\varPhi _{1}} = \frac{1}{k_{0}} \Vert 2u_{n} \chi _{G_{n}} \Vert _{\varPhi ,\varPhi _{1}} \geq \frac{1}{k_{0}} \Vert 2u_{n} \chi _{G_{n}} \Vert _{ \varPhi _{1}} \geq \frac{1}{k_{0}}, \end{aligned}$$

a contradiction.

Sufficiency. Let $\varPhi _{1}\in \bigtriangleup _{2}$ and $x_{0} \in S(L_{\varPhi ,\varPhi _{1}})$ with $k_{0}x_{0}(t) \in S_{\varPhi _{1}}$ for $k_{0} \in K(x_{0})$. For any $x_{n},y_{n} \in L_{\varPhi ,\varPhi _{1}}$ such that

$$\begin{aligned}& \lim_{n\rightarrow \infty } \Vert x_{n} \Vert _{\varPhi ,\varPhi _{1}}=1, \\& \lim_{n\rightarrow \infty } \Vert y_{n} \Vert _{\varPhi ,\varPhi _{1}}=1, \\& x_{n} +y_{n}=2 x_{0} \end{aligned}$$

for each $n\in N$.

Take sequences of positive numbers $\{k_{n}\}$ and $\{h_{n}\}$ such that

$$\begin{aligned}& \begin{gathered} \Vert x_{n} \Vert _{\varPhi ,\varPhi _{1}} \geq \frac{1}{k_{n}} \bigl( 1+ \varPhi \bigl(I_{\varPhi _{1}}(k_{n}x_{n}) \bigr) \bigr) - \frac{1}{n}, \\ \Vert y_{n} \Vert _{\varPhi ,\varPhi _{1}} \geq \frac{1}{h_{n}} \bigl( 1+ \varPhi \bigl(I_{\varPhi _{1}}(h_{n}y_{n}) \bigr) \bigr) - \frac{1}{n}. \end{gathered} \end{aligned}$$

Define

$$\begin{aligned}& \widetilde{x_{n}}(t)= \frac{x_{n}+x_{0}}{2}, \\& \widetilde{y_{n}}(t)= \frac{y_{n}+x_{0}}{2}. \end{aligned}$$

Then

$$\begin{aligned} \widetilde{x_{n}}+\widetilde{y_{n}} =2x_{0} \end{aligned}$$

and

$$\begin{aligned}& \lim_{n\rightarrow \infty } \Vert \widetilde{x_{n}} \Vert _{\varPhi , \varPhi _{1}} \leq 1, \\& \lim_{n\rightarrow \infty } \Vert \widetilde{y_{n}} \Vert _{\varPhi , \varPhi _{1}} \leq 1. \end{aligned}$$

Now, we will prove that $\lim_{n\rightarrow \infty } \Vert \widetilde{x_{n}} \Vert _{\varPhi , \varPhi _{1}}=\lim_{n\rightarrow \infty } \Vert \widetilde{y_{n}} \Vert _{\varPhi ,\varPhi _{1}}=1$. Otherwise, we can assume that $\lim_{n\rightarrow \infty } \Vert \widetilde{x_{n}} \Vert _{\varPhi , \varPhi _{1}}<1$ and there exist $\delta >0$, $n_{0} \in N$ such that

$$\begin{aligned}& \Vert \widetilde{x_{n}} \Vert _{\varPhi ,\varPhi _{1}} \leq 1- \delta , \\& \Vert \widetilde{y_{n}} \Vert _{\varPhi ,\varPhi _{1}} \leq 1+ \frac{\delta }{2} \end{aligned}$$

for all $n \geq n_{0}$. Then

$$\begin{aligned} 1= \Vert x_{0} \Vert _{\varPhi ,\varPhi _{1}} = \biggl\Vert \frac{\widetilde{x_{n}} + \widetilde{y_{n}}}{2} \biggr\Vert _{\varPhi , \varPhi _{1}} \leq \frac{1}{2} \biggl(1- \delta +1 + \frac{\delta }{2} \biggr) < 1, \end{aligned}$$

a contradiction.

Thus

$$\begin{aligned} \lim_{n\rightarrow \infty } \Vert \widetilde{x_{n}} \Vert _{\varPhi , \varPhi _{1}} = \lim_{n\rightarrow \infty } \Vert \widetilde{y_{n}} \Vert _{\varPhi ,\varPhi _{1}} =1. \end{aligned}$$

Since $\Vert \widetilde{x_{n}} - \widetilde{y_{n}} \Vert _{\varPhi ,\varPhi _{1}} \rightarrow 0$ if and only if $\Vert x_{n} - y_{n} \Vert _{\varPhi ,\varPhi _{1}} \rightarrow 0$ ($n\rightarrow \infty$), we will use the sequences $\{\widetilde{x_{n}}\}$ and $\{\widetilde{y_{n}}\}$ instead of $\{x_{n}\}$ and $\{y_{n}\}$, respectively. Put $k_{n}^{\prime }= \frac{2k_{n}k_{0}}{k_{n}+k_{0}}$, $h_{n}^{\prime }= \frac{2h_{n}k_{0}}{h_{n}+k_{0}}$. Then $d=\sup \{k_{n}^{\prime },h_{n}^{\prime }\}<+\infty$.

Taking advantage of the forced convergence theorem and

$$\begin{aligned} \Vert \widetilde{x_{n}} \Vert _{\varPhi ,\varPhi _{1}} \leq & \frac{1}{k_{n}^{\prime }} \bigl(1+ \varPhi \bigl( I_{\varPhi _{1}} \bigl( k_{n}^{\prime } \widetilde{x_{n}}\bigr) \bigr) \bigr) \\ \leq & \frac{k_{n}+k_{0}}{2k_{n}k_{0}} \biggl( 1+ \varPhi \biggl( I_{ \varPhi _{1}} \biggl( \frac{k_{n}k_{0}}{k_{n}+k_{0}} (x_{n}+x_{0}) \biggr) \biggr) \biggr) \\ \leq & \frac{1}{2} \biggl(\frac{1}{k_{0}} + \frac{1}{k_{0}} \varPhi \bigl( I_{\varPhi _{1}} ( k_{0} x_{0}) \bigr) + \frac{1}{k_{n}} + \frac{1}{k_{n}} \varPhi \bigl( I_{\varPhi _{1}} ( k_{n} x_{n}) \bigr) \biggr) \\ \leq & \frac{1}{2} \biggl( \Vert x_{0} \Vert _{\varPhi ,\varPhi _{1}} + \Vert x_{n} \Vert _{ \varPhi ,\varPhi _{1}} + \frac{1}{n} \biggr) \\ \rightarrow & 1 \quad (n\rightarrow \infty ), \end{aligned}$$

we have

$$\begin{aligned} \lim_{n\rightarrow \infty }\frac{1}{k_{n}^{\prime }} \bigl(1+ \varPhi \bigl( I_{\varPhi _{1}} \bigl( k_{n}^{\prime } \widetilde{x_{n}} \bigr) \bigr) \bigr) = 1 . \end{aligned}$$

In the same way, we also have

$$\begin{aligned} \lim_{n\rightarrow \infty }\frac{1}{h_{n}^{\prime }} \bigl(1+ \varPhi \bigl( I_{\varPhi _{1}} \bigl( h_{n}^{\prime } \widetilde{y_{n}} \bigr) \bigr) \bigr) =1 . \end{aligned}$$

Assume $k_{n}^{\prime } \rightarrow k$, $h_{n}^{\prime } \rightarrow h$ ($n\rightarrow \infty$). We will prove that $k, h\geq 1$. Since $\lim_{n\rightarrow \infty }\frac{1}{k_{n}^{\prime }} (1+ \varPhi ( I_{\varPhi _{1}} ( k_{n}^{\prime } \widetilde{x_{n}}) ) ) = 1$, then $\lim_{n\rightarrow \infty }\varPhi ( I_{\varPhi _{1}} ( k_{n}^{\prime } \widetilde{x_{n}}) ) = k-1$. If $k<1$, then $\varPhi ( I_{\varPhi _{1}} ( k_{n}^{\prime } \widetilde{x_{n}}) ) <0$ as $n\rightarrow \infty$, a contradiction. Therefore, $k\geq 1$. Similarly, $h\geq 1$.

Hence

$$\begin{aligned} \frac{k}{k+h}, \frac{h}{k+h} \in \biggl[ \frac{1}{1+d} , \frac{d}{1+d} \biggr]. \end{aligned}$$

In order to finish the proof of the theorem, we divide the left proof of the theorem into three steps.

Step 1: We will show that $k_{0} = \frac{2kh}{k+h}\in K(x_{0})$. In fact

$$\begin{aligned} \Vert x_{0} \Vert _{\varPhi ,\varPhi _{1}} \leq & \frac{k_{n}^{\prime }+h_{n}^{\prime }}{2k_{n}^{\prime }h_{n}^{\prime }} \biggl(1 + \varPhi \biggl( I_{\varPhi _{1}} \biggl( \frac{2k_{n}^{\prime }h_{n}^{\prime }}{k_{n}^{\prime }+h_{n}^{\prime }} x_{0} \biggr) \biggr) \biggr) \\ \leq & \frac{k_{n}^{\prime }+h_{n}^{\prime }}{2k_{n}^{\prime }h_{n}^{\prime }} \biggl(1 + \varPhi \biggl( I_{\varPhi _{1}} \biggl( \frac{k_{n}^{\prime }h_{n}^{\prime }}{k_{n}^{\prime }+h_{n}^{\prime }} (\widetilde{x_{n}} + \widetilde{y_{n}} ) \biggr) \biggr) \biggr) \\ \leq & \frac{k_{n}^{\prime }+h_{n}^{\prime }}{2k_{n}^{\prime }h_{n}^{\prime }} \biggl(1 + \varPhi \biggl( I_{\varPhi _{1}} \biggl( \frac{h_{n}^{\prime }}{k_{n}^{\prime }+h_{n}^{\prime }} k_{n}^{\prime } \widetilde{x_{n}}+ \frac{k_{n}^{\prime }}{k_{n}^{\prime }+h_{n}^{\prime }} h_{n}^{\prime }\widetilde{y_{n}} \biggr) \biggr) \biggr) ) \\ \leq & \frac{1}{2} \biggl( \frac{1}{ k_{n}^{\prime }} \bigl( 1+ \varPhi \bigl( I_{\varPhi _{1}}\bigl(k_{n}^{\prime } \widetilde{x_{n}} \bigr) \bigr) \bigr) + \frac{1}{ h_{n}^{\prime }} \bigl( 1+ \varPhi \bigl( I_{\varPhi _{1}}\bigl(h_{n}^{\prime } \widetilde{y_{n}} \bigr) \bigr) \bigr) \biggr) \\ \rightarrow & 1 \quad (n\rightarrow \infty ). \end{aligned}$$

Since $\Vert x_{0} \Vert _{\varPhi ,\varPhi _{1}}=1$, we get $\frac{2k_{n}^{\prime }h_{n}^{\prime }}{k_{n}^{\prime }+h_{n}^{\prime }} \rightarrow \frac{2kh}{k+h} = k_{0} \in K(x_{0})$.

Step 2: We will prove that $k_{n}^{\prime }\widetilde{x_{n}} - k_{0} x_{0} \xrightarrow{\mu } 0$ ($n\rightarrow \infty$).

Firstly, we will show that

$$\begin{aligned} k \widetilde{x_{n}}-h \widetilde{y_{n}} \xrightarrow{\mu } 0 \quad (n \rightarrow \infty ). \end{aligned}$$

Otherwise, there exist $\sigma _{0}, \varepsilon _{0} >0$ such that

$$\begin{aligned} \mu \bigl(\bigl\{ t\in G : \bigl\vert k \widetilde{x_{n}}(t)-h \widetilde{y_{n}}(t) \bigr\vert \geq \sigma _{0} \bigr\} \bigr) \geq \varepsilon _{0}. \end{aligned}$$

Let

$$\begin{aligned}& D= \varPhi _{1} ^{-1} \biggl( \frac{3}{\varepsilon _{0}} \biggr), \\& D_{1} = 2kD. \end{aligned}$$

Put $G_{n}=\{ t\in G : \vert k\widetilde{x_{n}}(t) \vert \leq D_{1}, \vert h \widetilde{y_{n}}(t) \vert \leq D_{1}, \vert k\widetilde{x_{n}}(t)-h \widetilde{y_{n}}(t) \vert \geq \sigma _{0} \}$. We will show that

$$\begin{aligned} \mu (G_{n})> \frac{\varepsilon _{0}}{3}. \end{aligned}$$

Indeed, since $\lim_{n\rightarrow \infty } \Vert \widetilde{x_{n}} \Vert _{\varPhi , \varPhi _{1}}=1$, we may assume $\Vert \widetilde{x_{n}} \Vert _{\varPhi } \leq \Vert \widetilde{x_{n}} \Vert _{\varPhi ,\varPhi _{1}} \leq 2$. Then

$$\begin{aligned} 1 \geq & I_{\varPhi _{1}} \biggl(\frac{\widetilde{x_{n}}}{2} \biggr) \\ \geq & \int _{ \{ t\in G: \vert \frac{\widetilde{x_{n}}(t)}{2} \vert >D \} }\varPhi _{1} \biggl( \frac{\widetilde{x_{n}}(t)}{2} \biggr)\,dt \\ >& \varPhi _{1}(D)\mu \biggl( \biggl\{ t\in G: \biggl\vert \frac{\widetilde{x_{n}}(t)}{2} \biggr\vert >D \biggr\} \biggr) \\ =& \frac{3}{\varepsilon _{0}}\mu \biggl( \biggl\{ t\in G: \biggl\vert \frac{\widetilde{x_{n}}(t)}{2} \biggr\vert >D \biggr\} \biggr). \end{aligned}$$

Hence

$$\begin{aligned} \mu \biggl( \biggl\{ t\in G: \biggl\vert \frac{\widetilde{x_{n}}(t)}{2} \biggr\vert >D \biggr\} \biggr)< \frac{\varepsilon _{0}}{3}. \end{aligned}$$

Consequently,

$$\begin{aligned} \mu \bigl(\bigl\{ t\in G: \bigl\vert k\widetilde{x_{n}}(t) \bigr\vert >D_{1}\bigr\} \bigr)< \frac{\varepsilon _{0}}{3}. \end{aligned}$$

Therefore,

$$\begin{aligned} \mu (G_{n}) =& \mu \bigl(\bigl\{ t\in G: \bigl\vert k \widetilde{x_{n}}(t)-h \widetilde{y_{n}}(t) \bigr\vert \geq \sigma _{0}\bigr\} \bigr)-\mu \bigl(\bigl\{ t\in G: \bigl\vert k \widetilde{x_{n}}(t) \bigr\vert >D_{1}\bigr\} \bigr) \\ &{}-\mu \bigl(\bigl\{ t\in G: \bigl\vert h\widetilde{y_{n}}(t) \bigr\vert >D_{1}\bigr\} \bigr) \\ >& \varepsilon _{0}-\frac{\varepsilon _{0}}{3}- \frac{\varepsilon _{0}}{3} \\ =&\frac{\varepsilon _{0}}{3}. \end{aligned}$$

Let

$$\begin{aligned} F =& \biggl\{ (x,y): \vert x \vert \leq D_{1}, \vert y \vert \leq D_{1}, \vert x-y \vert \geq \sigma _{0}, \frac{h}{k+h}x+\frac{k}{k+h}y\in S_{\varPhi }\biggr\} . \end{aligned}$$

By virtue of the fact that $S_{\varPhi _{1}}$ is a closed set, we know that F is a bounded closed set and

$$\begin{aligned} f(x,y)= \frac{\varPhi _{1} (\frac{h}{k+h}x+\frac{k}{k+h}y )}{\frac{h}{k+h} \varPhi _{1}(x) +\frac{k}{k+h} \varPhi _{1}(y)} < 1 \end{aligned}$$

for every $(x,y)\in F$.

By $f(x,y)$ is continuous on F, there exists $(x_{0},y_{0})\in F$ such that $f(x,y)\leq f(x_{0},y_{0})$. We next will prove that $f(x_{0},y_{0})<1$. If $f(x_{0},y_{0})=1$, then $\frac{\varPhi _{1}(\frac{h}{k+h}x_{0}+\frac{k}{k+h}y_{0})}{\frac{h}{k+h}\varPhi _{1}(x_{0})+\frac{k}{k+h}\varPhi _{1}(y_{0})}=1$, this contradicts $\frac{h}{k+h}x_{0}+\frac{k}{k+h}y_{0}\in S_{\varPhi _{1}}$. Put $f(x_{0},y_{0})=1-\delta$. For every $(x,y)\in F$, we have

$$\begin{aligned} \varPhi _{1} \biggl(\frac{h}{k+h}x+\frac{k}{k+h}y \biggr) \leq (1-\delta ) \biggl(\frac{h}{k+h} \varPhi _{1}(x) + \frac{k}{k+h} \varPhi _{1}(y) \biggr). \end{aligned}$$

By the definition of $k_{0}$ and $x_{0}$, we derive that

$$\begin{aligned} \frac{h}{k+h}k\widetilde{x_{n}}(t) +\frac{k}{k+h}h \widetilde{y_{n}}(t)= \frac{2kh}{k+h} x_{0}(t) = k_{0}x_{0} (t) \in S_{\varPhi _{1}}. \end{aligned}$$

Since $k_{0}x_{0}(t) \in S_{\varPhi _{1}}$, then $( k\widetilde{x_{n}}(t), h\widetilde{y_{n}}(t) ) \in F$, i.e., for $t\in G_{n}$ and

$$\begin{aligned} \varPhi _{1} \biggl( \frac{h}{k+h}k\widetilde{x_{n}}(t) +\frac{k}{k+h}h \widetilde{y_{n}}(t) \biggr) \leq (1-\delta ) \biggl(\frac{h}{k+h} \varPhi _{1} \bigl(k\widetilde{x_{n}}(t) \bigr) +\frac{k}{k+h} \varPhi _{1} \bigl(h\widetilde{y_{n}}(t) \bigr) \biggr). \end{aligned}$$

Hence

$$\begin{aligned}& \Vert \widetilde{x_{n}} +\widetilde{y_{n}} \Vert _{\varPhi ,\varPhi _{1}} \\ & \quad \leq \frac{k+h}{kh} \biggl( 1+ \varPhi \biggl( I_{\varPhi _{1} } \biggl( \frac{kh}{k+h} (\widetilde{x_{n}} + \widetilde{y_{n}}) \biggr) \biggr) \biggr) \\ & \quad \leq \frac{k+h}{kh}+\frac{k+h}{kh} \varPhi \biggl( \int _{G} \varPhi _{1} \biggl( \frac{kh}{k+h} \bigl(\widetilde{x_{n}}(t) + \widetilde{y_{n}}(t)\bigr) \biggr) \,dt \biggr) \\ & \quad \leq \frac{k+h}{kh}+\frac{k+h}{kh} \varPhi \biggl( (1-\delta ) \int _{G_{n}} \biggl[ \frac{h}{k+h}\varPhi _{1} \bigl(k \widetilde{x_{n}}(t) \bigr) +\frac{k}{k+h} \varPhi _{1} \bigl(h \widetilde{y_{n}}(t) \bigr) \biggr] \,dt \biggr) \\ & \quad \quad {}+ \frac{k+h}{kh} \varPhi \biggl( \int _{G \setminus G_{n}} \biggl[ \frac{h}{k+h}\varPhi _{1} \bigl(k\widetilde{x_{n}}(t) \bigr) + \frac{k}{k+h} \varPhi _{1} \bigl(h\widetilde{y_{n}}(t) \bigr) \biggr] \,dt \biggr) \\ & \quad \leq \frac{k+h}{kh}+\frac{k+h}{kh} \varPhi \biggl( \int _{G} \biggl[ \frac{h}{k+h}\varPhi _{1} \bigl(k\widetilde{x_{n}}(t) \bigr) + \frac{k}{k+h} \varPhi _{1} \bigl(h\widetilde{y_{n}}(t) \bigr) \biggr] \,dt \biggr) \\ & \quad \quad {}- \frac{k+h}{kh} \varPhi \biggl( \delta \int _{G_{n}} \biggl[ \frac{h}{k+h}\varPhi _{1} \bigl(k\widetilde{x_{n}}(t) \bigr) + \frac{k}{k+h} \varPhi _{1} \bigl(h\widetilde{y_{n}}(t) \bigr) \biggr] \,dt \biggr) \\ & \quad \leq \frac{1}{k} ( 1+ \varPhi \bigl( I_{\varPhi _{1}} \bigl(k \widetilde{x_{n}}(t) \bigr) \bigr) + \frac{1}{h} ( 1+ \varPhi \bigl( I_{\varPhi _{1}} \bigl(k\widetilde{y_{n}}(t) \bigr) \bigr) \\ & \quad \quad {}- \frac{k+h}{kh} \varPhi \biggl(\delta \int _{G_{n}} \biggl[ \frac{h}{k+h}\varPhi _{1} \bigl(k\widetilde{x_{n}}(t) \bigr) + \frac{k}{k+h} \varPhi _{1} \bigl(h\widetilde{y_{n}}(t) \bigr) \biggr] \,dt \biggr). \end{aligned}$$

Notice that

$$\begin{aligned} I_{\varPhi _{1}} \bigl(\bigl(k-k_{n}^{\prime }\bigr) \widetilde{x_{n}}\bigr)\leq \bigl\vert k-k_{n}^{\prime } \bigr\vert I_{ \varPhi _{1}} (\widetilde{x_{n}}) \rightarrow 0 \quad (n \rightarrow \infty ). \end{aligned}$$

By Lemma 2.11, we get

$$\begin{aligned} I_{\varPhi _{1}} (k \widetilde{x_{n}}) -I_{\varPhi _{1}} \bigl(k_{n}^{\prime } \widetilde{x_{n}}\bigr)= I_{\varPhi _{1}} \bigl(k_{n}^{\prime } \widetilde{x_{n}} + \bigl(k-k_{n}^{\prime }\bigr) \widetilde{x_{n}}\bigr) -I_{\varPhi _{1}} \bigl(k_{n}^{\prime }\widetilde{x_{n}} \bigr) \rightarrow 0 \quad (n \rightarrow \infty ). \end{aligned}$$

Thus

$$\begin{aligned} 0 \leq & \frac{1}{k} \bigl(1+ \varPhi \bigl(I_{\varPhi _{1}}(k \widetilde{x_{n}})\bigr) \bigr) - \Vert \widetilde{x_{n}} \Vert _{\varPhi ,\varPhi _{1}} \\ =& \frac{1}{k} \bigl(1+ \varPhi \bigl(I_{\varPhi _{1}}(k \widetilde{x_{n}})\bigr) \bigr)-\frac{1}{k_{n}^{\prime }} \bigl(1+ \varPhi \bigl(I_{\varPhi _{1}}\bigl(k_{n}^{\prime } \widetilde{x_{n}} \bigr)\bigr) \bigr) +\frac{1}{n}\rightarrow 0 \quad (n \rightarrow \infty ). \end{aligned}$$

Similarly, $\frac{1}{h}(1+\varPhi (I_{\varPhi _{1}}(h\widetilde{y_{n}})))- \Vert \widetilde{y_{n}} \Vert _{\varPhi ,\varPhi _{1}}\rightarrow 0$ as $n\rightarrow \infty$. Since $u>0$, $\varPhi (u)>0$, and

$$\begin{aligned} \Vert \widetilde{x_{n}}+\widetilde{y_{n}} \Vert _{\varPhi ,\varPhi _{1}}\leq 2- \frac{k+h}{kh} \biggl(\varPhi \biggl( \frac{2\delta }{1+d}\varPhi _{1}\biggl( \frac{\delta _{0}}{2}\biggr) \frac{\varepsilon _{0}}{3} \biggr) \biggr) \quad (n \rightarrow \infty ), \end{aligned}$$

we have $\lim_{n\rightarrow \infty } \Vert \widetilde{x_{n}}+ \widetilde{y_{n}} \Vert _{\varPhi ,\varPhi _{1}}<2$. A contradiction. Hence $k\widetilde{x_{n}}-h\widetilde{y_{n}}\xrightarrow{\mu } 0$.

Thanks to the Φ-Amemiya norm being equivalent with the Luxemburg norm, their weak topology and weak star topology are all equivalent. So $L_{\varPhi ,\varPhi _{1}}$ is $w^{\ast }$ compact. Take $\{x''_{n}\}\subset \{\widetilde{x_{n}}\}$, $\{y''_{n}\}\subset \{\widetilde{y_{n}}\}$ such that $x''_{n}\xrightarrow{w^{\ast }} x'$ and $y''_{n}\xrightarrow{w^{\ast }} y'$. We get $x'+y'=2x_{0}$.

Since $\varPhi _{1}\in \bigtriangleup _{2}$, we have

$$\begin{aligned} \Vert x \Vert _{\varPhi ,\varPhi _{1}}=\sup \biggl\{ \int _{G}x(t)y(t)\,dt:y\in B\bigl(L^{\ast }_{\varPhi ,\varPhi _{1}} \bigr) \biggr\} , \end{aligned}$$

where $L^{\ast }_{\varPhi ,\varPhi _{1}}$ is the dual space of $L_{\varPhi ,\varPhi _{1}}$.

So

$$\begin{aligned} \Vert x \Vert _{\varPhi ,\varPhi _{1}}=\sup \biggl\{ \int _{G}x(t)y(t)\,dt:y\in B\bigl(E^{\ast }_{\varPhi ,\varPhi _{1}} \bigr) \biggr\} . \end{aligned}$$

Since $\Vert 2x_{0} \Vert _{\varPhi ,\varPhi _{1}}=2$, then

$$\begin{aligned} \Vert 2x_{0} \Vert _{\varPhi ,\varPhi _{1}} \leq \bigl\Vert x' \bigr\Vert _{\varPhi ,\varPhi _{1}}+ \bigl\Vert y' \bigr\Vert _{ \varPhi ,\varPhi _{1}}\leq \varliminf _{n\rightarrow \infty } \Vert \widetilde{x_{n}} \Vert _{\varPhi ,\varPhi _{1}}+\varliminf _{n \rightarrow \infty } \Vert \widetilde{y_{n}} \Vert _{\varPhi ,\varPhi _{1}}=2. \end{aligned}$$

This shows

$$\begin{aligned} \bigl\Vert x' \bigr\Vert _{\varPhi ,\varPhi _{1}} = \bigl\Vert y' \bigr\Vert _{\varPhi ,\varPhi _{1}}=1. \end{aligned}$$

Hence, there exist $k, h>1$ such that

$$\begin{aligned}& 1= \bigl\Vert x^{\prime } \bigr\Vert _{\varPhi ,\varPhi _{1}}=\frac{1}{k} \bigl(1+\varPhi \bigl(I_{\varPhi _{1}}\bigl(kx^{\prime }\bigr)\bigr)\bigr), \\& 1= \bigl\Vert y^{\prime } \bigr\Vert _{\varPhi ,\varPhi _{1}}=\frac{1}{h} \bigl(1+\varPhi \bigl(I_{\varPhi _{1}}\bigl(hy^{\prime }\bigr)\bigr)\bigr). \end{aligned}$$

Since $\Vert x^{\prime } \Vert _{\varPhi ,\varPhi _{1}}+ \Vert y^{\prime } \Vert _{\varPhi ,\varPhi _{1}}=2$, then

$$\begin{aligned} \bigl\Vert x^{\prime } \bigr\Vert _{\varPhi ,\varPhi _{1}}+ \bigl\Vert y^{\prime } \bigr\Vert _{\varPhi ,\varPhi _{1}} =&\frac{1}{k} \bigl(1+ \varPhi \bigl(I_{\varPhi _{1}}\bigl(kx^{\prime }\bigr)\bigr)\bigr)+ \frac{1}{h}\bigl(1+\varPhi \bigl(I_{\varPhi _{1}}\bigl(hy^{\prime } \bigr)\bigr)\bigr) \\ =&\frac{k+h}{kh}\biggl[1+\frac{h}{k+h}\varPhi \bigl(I_{\varPhi _{1}} \bigl(kx^{\prime }\bigr)\bigr)+ \frac{k}{k+h}\varPhi \bigl(I_{\varPhi _{1}}\bigl(hy^{\prime }\bigr)\bigr)\biggr] \\ \geq & \frac{k+h}{kh}\biggl[1+\varPhi \biggl(\frac{h}{k+h} \bigl(I_{\varPhi _{1}}\bigl(kx^{\prime }\bigr)\bigr)+ \frac{k}{k+h} \bigl(I_{\varPhi _{1}}\bigl(hy^{\prime }\bigr)\bigr)\biggr)\biggr] \\ \geq &\frac{k+h}{kh}\biggl[1+\varPhi \biggl(I_{\varPhi _{1}}\biggl( \frac{kh}{k+h}x^{\prime }+ \frac{kh}{k+h}y^{\prime }\biggr) \biggr)\biggr] \\ =&2\cdot \frac{1}{k_{0}}\bigl(1+\varPhi \bigl(I_{\varPhi _{1}}(k_{0}x_{0}) \bigr)\bigr) \\ =&2, \end{aligned}$$

and

$$\begin{aligned} \Vert x_{0} \Vert _{\varPhi ,\varPhi _{1}}=1=\frac{1}{k_{0}}\bigl(1+ \varPhi \bigl(I_{\varPhi _{1}}(k_{0}x_{0})\bigr)\bigr). \end{aligned}$$

Hence $x^{\prime }=y^{\prime }=x_{0}$. Combining this with $kx''_{n}-hy''_{n}\xrightarrow{\mu } 0$, we can prove that $kx''_{n}-hy''_{n}\xrightarrow{w^{\ast }} 0$. Since $v(t)\in E_{\varPsi }$, then for any $\varepsilon >0$ there exists $\delta >0$ such that $\Vert v\chi _{G_{0}} \Vert _{\varPhi }<\varepsilon$, whence $\mu (G_{0})<\delta$. Using the EropoB theorem, there exists $G_{0}\subset G$, $\mu (G_{0})<\delta$ such that $kx''_{n}-hy''_{n}\xrightarrow{\mu } 0$ for $t\in G\setminus G_{0}$.

Put $\Vert x_{n}^{\prime \prime } \Vert _{\varPhi }^{o}=1$ and $\Vert y_{n}^{\prime \prime } \Vert _{\varPhi }^{o}=1$. We have

$$\begin{aligned} \int _{G}\bigl(kx''_{n}-hy''_{n} \bigr)v(t)\,dt =& \int _{G\setminus G_{0}}\bigl(kx''_{n}-hy''_{n} \bigr)v(t)\,dt+ \int _{G_{0}}\bigl(kx''_{n}-hy''_{n} \bigr)v(t)\,dt \\ \leq &M\cdot \mu (G\setminus G_{0})\cdot \varepsilon +k \bigl\Vert x''_{n} \bigr\Vert _{ \varPhi }^{o} \bigl\Vert v(t) \bigr\Vert _{\varPhi }+h \bigl\Vert y''_{n} \bigr\Vert _{\varPhi }^{o} \bigl\Vert v(t) \bigr\Vert _{\varPhi } \\ \leq &M\cdot \mu (G\setminus G_{0})\cdot \varepsilon +k\cdot \varepsilon +h\cdot \varepsilon . \end{aligned}$$

By the arbitrariness of ε, we have $\int _{G}(kx^{\prime \prime }_{n}-hy^{\prime \prime }_{n})v(t)\,dt< \varepsilon$. Thus $kx''_{n}-hy''_{n}\xrightarrow{w^{\ast }} 0$. Since $x''_{n}-y''_{n}\xrightarrow{w^{\ast }} 0$, then $k=h$. Thus $\widetilde{x_{n}}-\widetilde{y_{n}}\xrightarrow{\mu } 0$ as $n\rightarrow \infty$. Therefore

$$\begin{aligned} k'_{n}\widetilde{x_{n}}-k_{0}x_{0} \xrightarrow{\mu } 0 \quad (n \rightarrow \infty ). \end{aligned}$$

Step 3: We will prove that $I_{\varPhi _{1}}(k_{n}x_{n})\rightarrow I_{\varPhi _{1}}(k_{0}x_{0})$. In fact

$$\begin{aligned}& \varPhi \bigl(I_{\varPhi _{1}}(k_{0}x_{0}) \bigr)=k_{0}-1, \\& \varPhi \bigl(I_{\varPhi _{1}}\bigl(k'_{n} \widetilde{x_{n}}\bigr)\bigr)\rightarrow k \quad (n \rightarrow \infty ). \end{aligned}$$

We deduce that $\varPhi (I_{\varPhi _{1}}(k'_{n}\widetilde{x_{n}}))\rightarrow \varPhi (I_{ \varPhi _{1}}(k_{0}x_{0}))$ ($n\rightarrow \infty$). Using $u>0$, $\varPhi (u)>0$ and $\varPhi (u)$ is strictly increasing, we get

$$\begin{aligned} I_{\varPhi _{1}}\bigl(k'_{n}\widetilde{x_{n}} \bigr))\rightarrow I_{\varPhi _{1}}(k_{0}x_{0}) \quad (n \rightarrow \infty ). \end{aligned}$$

By Lemma 2.12, we have

$$\begin{aligned} \bigl\Vert k'_{n}\widetilde{x_{n}}-k_{0}x_{0} \bigr\Vert _{\varPhi ,\varPhi _{1}}\rightarrow 0. \end{aligned}$$

 □

Corollary 3.2

LetΦbe an Orlicz function. Then$L_{\varPhi ,\varPhi _{1}}$is midpoint local uniform rotundity if and only if$\varPhi _{1}\in \Delta _{2}$and$L_{\varPhi ,\varPhi _{1}}$is strictly convex.