Strongly Exposed Points of Orlicz Sequence Spaces Equipped with the p-Amemiya Norm

Abstract

Using some new techniques, criteria for strongly exposed points of Orlicz sequence spaces generated by arbitrary Orlicz function and equipped with the p-Amemiya $(1 < p < \infty )$ norm are obtained. Next, strongly exposed property of these spaces is deduced. The obtained results broaden the knowledge about these notions in Orlicz sequence spaces equipped with the p-Amemiya norm.

1. Introduction

Let X be a real Banach space, and let $B\left(X\right)$ and $S\left(X\right)$ be the closed unit ball and the unit sphere of X, respectively. By $X^{*}$ we denote the dual space of X. For any $u\in S\left(X\right)$ we denote by $Grad\left(u\right)$ the set of all supporting functionals at x, i.e.,

$$Grad\left(u\right)=\{f\in S\left(X^{*}\right):f\left(u\right)=\parallel u\parallel \}.$$

Exposed points were first defined by Straszewicz in 1935 in the case of finite-dimensional spaces (see [1]). The concept of strongly exposed points were introduced by Lindenstrauss in 1963 (see [2]). Both exposed points and strongly exposed points are basic concepts in the geometric theory of Banach spaces. There is a connection of strongly exposed points and the Radon-Nikodym property. Phelps showed in 1974 that a Banach space X has a Radon-Nikodym property if every non-empty closed, bounded convex subset is contained in a closed convex hull of its strongly exposed points (see [3]). Criteria for exposed points in classical Orlicz spaces equipped the Orlicz norm, Luxemburg nom and p-Amemiya norms were presented in [4,5,6,7,8]. Criteria for strongly exposed points in classical Orlicz spaces and Musielak-Orlicz spaces equipped the Orlicz norm and Luxemburg nom were given in [9,10,11,12,13]. The p-Amemiya norms are generalization of the Orlicz norm and Luxemburg norm. Up to now, no characterization of strong exposed points and strongly exposed property of Orlicz sequence spaces equipped with the p-Amemiya norm was known. The main aim of the paper is to complete this gap.

 Definition 1 

([1]). A point $u\in S\left(X\right)$ is said to be an exposed point if there exists $f\in Grad\left(u\right)$, such that for all $v\in B\left(X\right)\setminus \left\{u\right\},f\left(v\right)<f\left(u\right)=1$. Then f is called an exposed functional of u.

 Definition 2 

([2]). A point $u\in S\left(X\right)$ is called a strongly exposed point if there exists $f\in Grad\left(u\right)$, such that for any ${\left\{u_n\right\}}_{n=1}^{\infty }\subset B\left(X\right),f\left(u_n\right)\to f\left(u\right)=1$ implies $\parallel u_n-u\parallel \to 0$ as $n\to \infty$. Then f is called a strongly exposed functional of u.

Obviously, a strongly exposed point is an exposed point. If every $x\in S\left(X\right)$ is an (strongly) exposed point, then X has the (strongly) exposed property. In a strictly convex Banach space all points of its unit sphere are exposed, while in a locally uniformly convex Banach space all points of its unit sphere are strongly exposed (see [5,9]).

In the sequel, $\mathbb{N}$ and $\mathbb{R}$ denote the set of natural numbers and the set of real numbers, respectively.

The triple $(\mathbb{N},2^{\mathbb{N}},\mu )$ stands for the counting measure space and $l^0\left(\mu \right)=l^0$ denotes the space of all real sequences $u=\left(u\right(i\left)\right)$. A Banach space X is called a Köthe sequence space if it is a subspace of $l^0$ such that

(i) if $u\in l^0,v\in X$ and $\left|u\right(i\left)\right|\le \left|v\right(i\left)\right|$ for $i\in \mathbb{N},$ then $u\in X$ and $\parallel u\parallel \le \parallel v\parallel$,

(ii) there exists $u\in X$ such that $suppu=\mathbb{N}$, where $suppu=\{i\in \mathbb{N}:u(i)\ne 0\}$.

For any map $\Phi :\mathbb{R}\to [0,\infty ]$, define

$$a_{\Phi }=sup\{x\ge 0:\Phi \left(x\right)=0\},b_{\Phi }=sup\{x>0:\Phi \left(x\right)<\infty \}.$$

Note that if $\Phi$ is even on $\mathbb{R}$, $a_{\Phi }=0$ means that $\Phi$ vanishes only at zero while $b_{\Phi }=\infty$ means that $\Phi$ takes only finite values.

A map $\Phi :\mathbb{R}\to [0,\infty ]$ is said to be an Orlicz function if $\Phi \left(0\right)=0$, $\Phi$ is not identically equal to zero, $\Phi$ is even and convex on the interval $(-b_{\Phi },b_{\Phi })$ and $\Phi$ is left continuous at $b_{\Phi }$, i.e., $\underset{x\to b_{\Phi }^{-}}{lim}\Phi \left(x\right)=\Phi \left(b_{\Phi }\right)$.

For every Orlicz function $\Phi$, we define the complementary function, in the sense of Young, $\Psi :\mathbb{R}\to [0,\infty ]$ by the formula

$$\Psi \left(y\right)=sup\left\{x\right|y|-\Phi (x):x\ge 0\}.$$

Let ${\Phi }_{+}^{\prime }\left(x\right)$ and ${\Phi }_{-}^{\prime }\left(x\right)$ (${\Psi }_{+}^{\prime }\left(y\right)$ and ${\Psi }_{-}^{\prime }\left(y\right)$) stand for the right and left derivatives of $\Phi (\Psi )$ at $x\in \mathbb{R}$ with $0\le x<b_{\Phi }$ (at $y\in \mathbb{R}$ with $0\le y<b_{\Psi }$), respectively. Here we define ${\Phi }_{+}^{\prime }\left(b_{\Phi }\right)=\infty$ and ${\Phi }_{-}^{\prime }\left(x\right)={\Phi }_{+}^{\prime }\left(x\right)=\infty$ for all $x>b_{\Phi }$, ${\Psi }_{+}^{\prime }\left(b_{\Psi }\right)=\infty$ and ${\Psi }_{-}^{\prime }\left(y\right)={\Psi }_{+}^{\prime }\left(y\right)=\infty$ for all $y>b_{\Psi }$.

For every $x,y\in \mathbb{R}$, we have the following Young Inequality:

$$\left|xy\right|\le \Phi \left(x\right)+\Psi \left(y\right)$$

(1)

which reduces to an equality when $y\in \partial \Phi \left(x\right)$ if x is given, or when $x\in \partial \Psi \left(y\right)$ if y is given, and

$$\partial \Phi \left(x\right)=\left\{\begin{array}{cc}[{\Phi }_{-}^{\prime }\left(x\right),{\Phi }_{+}^{\prime }\left(x\right)],\hfill & \mathrm{for}-b_{\Phi }<x<b_{\Phi },\hfill \\ [{\Phi }_{-}^{\prime }\left(x\right),+\infty ),\hfill & \mathrm{for}x=b_{\Phi }\mathrm{and}{\Phi }_{-}^{\prime }\left(b_{\Phi }\right)<\infty ,\hfill \\ (-\infty ,{\Phi }_{+}^{\prime }\left(x\right)],\hfill & \mathrm{for}x=-b_{\Phi }\mathrm{and}{\Phi }_{+}^{\prime }(-b_{\Phi })>-\infty .\hfill \end{array}\right.$$

Given any Orlicz function $\Phi$, we define on $l^0$ the convex modular of u by

$$I_{\Phi }\left(u\right)=\sum _{i=1}^{\infty }\Phi \left(u\left(i\right)\right),\mathrm{for} \mathrm{any}u=\left(u\left(i\right)\right)\in l^0.$$

The Orlicz sequence space $l_{\Phi }$ generated by an Orlicz function $\Phi$ is a linear space of real sequences defined by

$$l_{\Phi }=\{u\in l^0:I_{\Phi }\left(cu\right)<\infty \mathrm{for} \mathrm{some}c>0\mathrm{depending} \mathrm{on} u\}.$$

The space $h_{\Phi }$ is the subspace of all order continuous elements of $l_{\Phi }$ defined by

$$h_{\Phi }=\{u\in l^0:I_{\Phi }\left(cu\right)<\infty \mathrm{for} \mathrm{any}c>0\}.$$

The Orlicz sequence space $l_{\Phi }$ is a Köthe sequence space when it is equipped with the Orlicz norm

$${\parallel u\parallel }^o=\underset{I_{\Psi }\left(v\right)\le 1}{sup}\sum _{i=1}^{\infty }u\left(i\right)v\left(i\right)$$

or the Luxemburg norm

$${\parallel u\parallel }_{\Phi }=inf\left\{\lambda >0:I_{\Phi }\left({\displaystyle \frac{u}{\lambda }}\right)\le 1\right\}.$$

At first sight the Orlicz and Luxemburg norms seem far from similar. In fact, in many cases the geometric properties of Orlicz spaces under each of these norms differ from each other. But in the fifties, I. Amemiya (see [14]) considered a norm defined by the formula

$${\parallel u\parallel }^A=\underset{k>0}{inf}{\displaystyle \frac{1}{k}}(1+I_{\Phi }\left(ku\right)).$$

In 2000, Hudzik and Maligranda proved the Amemiya norm and the Orlicz norm coincide for any Orlicz function $\Phi$, i.e., ${\parallel ·\parallel }^o={\parallel ·\parallel }^A$ (see [15]). Moreover, it is not difficult to verify that the Luxemburg norm can be expressed by an Amemiya-like formula (see [16,17]), namely

$${\parallel u\parallel }_{\Phi }=\underset{k>0}{inf}{\displaystyle \frac{1}{k}}max\{1,I_{\Phi }\left(ku\right)\}.$$

The only difference between the two Amemyia formulas is the function under the infimum operation (we will call it the outer function): for all $t\ge 0$, $s\left(t\right)=1+t$ (for the Orlicz norm) and $s\left(t\right)=max\{1,t\}$ (for the Luxemburg norm). In [15], Hudzik and Maligranda proposed to investigate the Amemiya formula-norms generated by the outer functions of the type

$$s_p\left(t\right)=\left\{\begin{array}{cc}{(1+t^p)}^{{\displaystyle \frac{1}{p}}},\hfill & \mathrm{for}1\le p<\infty ,\hfill \\ max\{1,t\},\hfill & \mathrm{for}p=\infty ,\hfill \end{array}\right.$$

and for any $u\in l_{\Phi }$,

$${\parallel u\parallel }_{\Phi ,p}=\underset{k>0}{inf}{\displaystyle \frac{1}{k}}{s}_p\left(I_{\Phi }\left(ku\right)\right)(1\le p\le \infty ).$$

We obtain a family of topologically equivalent norms (called the p-Amemiya norm), indexed by $1\le p\le \infty$ and satisfying the inequalities

$${\parallel u\parallel }_{\Phi }={\parallel u\parallel }_{\Phi ,\infty }\le {\parallel u\parallel }_{\Phi ,p}\le {\parallel u\parallel }_{\Phi ,p^{\prime }}\le {\parallel u\parallel }_{\Phi ,1}={\parallel u\parallel }^o\le 2^{{\displaystyle \frac{1}{p}}}{\parallel u\parallel }_{\Phi ,p},$$

for all $1\le p^{\prime }\le p\le \infty$.

The first paper constituting some basic and crutial results allowing further research and containing the complete characterization of rotundity and extreme points in Orlicz spaces equipped with the p-Amemiya norms ($1\le p\le \infty$) was written by Cui et al. (see [17]). After that, an intensive development of research connected with Orlicz spaces equipped with the p-Amemiya norm have taken place. Necessary and sufficient conditions for non-squareness properties, smoothness, midpoint local uniform rotundity, local uniform rotundity, etc., were presented in [18,19,20,21,22,23]. These results broaden the knowledge about the geometry of these spaces.

In order to simplify notation, the Orlicz sequence spaces equipped with p-Amemiya norms are denoted by $l_{\Phi ,p}=(l_{\Phi }{,\parallel ·\parallel }_{\Phi ,p})$, $h_{\Phi ,p}=(h_{\Phi }{,\parallel ·\parallel }_{\Phi ,p})$.

An interval $[a,b]$ is called a structural affine interval of $\Phi$, or simply a SAI of $\Phi$, provided that $\Phi$ is affine on $[a,b]$ and it is not affine on either $[a-\epsilon ,b]$ or $[a,b+\epsilon ]$ for any $\epsilon >0$. Let ${\left([a_i,b_i]\right)}_i$ be all SAIs of $\Phi$. We call $S{C}_{\Phi }=\mathbb{R}\setminus \left({\cup }_{i=1}^{\infty }(a_i,b_i)\right)$ the set of strictly convex points of $\Phi$. Denote

$$A^{\prime }=\bigcup \{a_i:SAI(a_i,b_i)\mathrm{with}{\Phi }_{-}^{\prime }\left(a_i\right)={\Phi }_{+}^{\prime }\left(a_i\right)\},$$

$$B^{\prime }=\bigcup \{b_i:SAI(a_i,b_i)\mathrm{with}{\Phi }_{-}^{\prime }\left(b_i\right)={\Phi }_{+}^{\prime }\left(b_i\right)\},$$

$$A=\bigcup \{a_i:SAI(a_i,b_i)\mathrm{with}{\Phi }_{-}^{\prime }\left(a_i\right)<{\Phi }_{+}^{\prime }\left(a_i\right)\},$$

$$B=\bigcup \{b_i:SAI(a_i,b_i)\mathrm{with}{\Phi }_{-}^{\prime }\left(b_i\right)<{\Phi }_{+}^{\prime }\left(b_i\right)\}.$$

We say an Orlicz function $\Phi$ satisfies the ${\Delta }_2$-condition at zero ($\Phi \in {\Delta }_2\left(0\right)$, for short) if there are positive constants k and $0<x_0<\infty$ with $0<\Phi \left(x_0\right)<\infty$ such that $\Phi \left(2x\right)\le k\Phi \left(x\right)$ for $\left|x\right|\le x_0$.

For any Orlicz function $\Phi$, $\Phi \in {\Delta }_2\left(0\right)$, we also write that its complementary function $\Psi$ satisfies the ${\nabla }_2$-condition at zero, i.e.,$\Psi \in {\nabla }_2\left(0\right)$. If $\Phi \in {\Delta }_2\left(0\right)$ and $\Psi \in {\Delta }_2\left(0\right)$, then we write $\Phi \in {\Delta }_2\left(0\right)\cap {\nabla }_2\left(0\right)$. It is not difficult to prove that $l_{\Phi }=h_{\Phi }\iff \Phi \in {\Delta }_2\left(0\right)\cap {\nabla }_2\left(0\right)$ (see [24]).

By $sgnx$ we will denote the sign function defined on $\mathbb{R}$. This function $sgnx$ is equal to 1 for $x>0$, −1 for $x<0$ and 0 for $x=0$.

Further details about Orlicz spaces, we refer to [24,25,26,27]. Some fundamental topological and geometrical properties of Orlicz spaces equipped with p-Amemiya norms have been investigated in [17,22].

2. Auxiliary Results

In this section we will present a few auxiliary definitions and results that will be used in the main part of the paper.

Let $1\le p\le \infty$, for any $u\in l_{\Phi ,p},i<j$, we denote

$$\begin{array}{ccc}& & {\left[u\right]}_j^i=(0,\cdots ,0,u(i+1),\cdots ,u\left(j\right),0,\cdots ),\hfill \\ & & {\left[u\right]}_j=(u\left(1\right),u\left(2\right),\cdots ,u\left(j\right),0,\cdots ).\hfill \end{array}$$

By Theorem 1.43 in [24] and equivalence of p-Amemiya $(1\le p\le \infty )$ norms, we have

$$\underset{n\to \infty }{lim}{\parallel u-{\left[u\right]}_n\parallel }_{\Phi ,p}={\theta }_{\Phi }\left(u\right)=inf\left\{\lambda >0,I_{\Phi }\left({\displaystyle \frac{u}{\lambda }}\right)<\infty \right\}.$$

For any $u\in l_{\Phi ,p}$, define

$$\begin{array}{ccc}& & k_p^{*}\left(u\right)=inf\{k\ge 0:I_{\Phi }^{p-1}\left(ku\right)·I_{\Psi }\left({\Phi }_{+}^{\prime }\left(ku\right)\right)\ge 1\}(\mathrm{with}inf\varnothing =\infty ),\hfill \\ & & k_p^{**}\left(u\right)=sup\{k\ge 0:I_{\Phi }^{p-1}\left(ku\right)·I_{\Psi }\left({\Phi }_{+}^{\prime }\left(ku\right)\right)\le 1\}.\hfill \end{array}$$

It is evident that $k_p^{*}\left(u\right)\le k_p^{**}\left(u\right)$ for every $u\in l_{\Phi ,p}\setminus \left\{0\right\}$. Set $K_p\left(u\right)=\{0<k<\infty :k_p^{*}\left(u\right)\le k\le k_p^{**}\left(u\right)\}$.

 Lemma 1 

([17]). For any $1\le p\le \infty$ the set of those k’s at which the p-Amemiya norm of an element $u\in l_{\Phi ,p}\setminus \left\{0\right\}$ is attainded is equal to the interval $K_p\left(u\right)$.

 Lemma 2 

([17]). Let $1<p<\infty$, the interval of norm attainable of all $u\in l_{\Phi ,p}\setminus \left\{0\right\}$ is nonempty, closed and bounded (in fact, equal to $[k_p^{*}\left(u\right),k_p^{**}\left(u\right)]$) if and only if Φ is not linear on $[0,\infty )$.

 Lemma 3 

([22]). Let $1\le p<\infty$. For every $u\in l_{\Phi ,p}\setminus \left\{0\right\}$, if $K_p\left(u\right)=\varnothing$, then ${\parallel u\parallel }_{\Phi ,p}=b_{\Psi }{\displaystyle \sum _{i=1}^{\infty }}\left|u\left(i\right)\right|$ where $b_{\Psi }=\underset{x\to \infty }{lim}{\Phi }_{+}^{\prime }\left(x\right)=\underset{x\to \infty }{lim}{\displaystyle \frac{\Phi \left(x\right)}{x}}$.

 Lemma 4 

([19]). Let $1\le p\le \infty$. If $u\in B\left(l_{\Phi ,p}\right)\setminus \left\{0\right\}$, then $I_{\Phi }\left(u\right)\le {\parallel u\parallel }_{\Phi ,p}$.

We say that $\phi \in {\left(l_{\Phi ,p}\right)}^{*}$ is a singular functional, or simply, $\phi \in F$, if $\phi \left(h_{\Phi }\right)=\left\{0\right\}$.

 Lemma 5 

([22]). Let $\Phi ,\Psi$ be Orlicz functions complementary to each other in the sense of Young that take finite values only and let $1<p, q<\infty$ with ${\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1$. If the p-Amemiya norm is $k^{*}$-finite, then $f\in {\left(l_{\Phi ,p}\right)}^{*}\setminus \left\{0\right\}$ has the unique decomposition $f=v+\phi$ where $v\in l_{\Psi ,q}$, $\phi \in F$, and

$$\parallel f\parallel ={\parallel f\parallel }_{\Psi ,q}=\underset{l>0}{inf}{\displaystyle \frac{1}{l}}{s}_q(I_{\Psi }\left(lv\right)+l\parallel \phi \parallel ).$$

 Lemma 6 

([22]). Let $1<p, q<\infty$, $u\in l_{\Phi ,p}\setminus \left\{0\right\}$ and $K_p\left(u\right)\ne \varnothing$. $v\in S\left(l_{\Psi ,q}\right)$ is a supporting functional of u if and only if:

(i) $v\left(i\right)={\displaystyle \frac{w\left(i\right)}{{\parallel w\parallel }_{\Psi ,q}}}·sgnu\left(i\right)$, for some w satisfying ${\Phi }_{-}^{\prime }\left(k\right|u\left(i\right)\left|\right)\le w\left(i\right)\le {\Phi }_{+}^{\prime }\left(k\right|u\left(i\right)\left|\right),$ $i\in \mathbb{N},k\in K_p\left(u\right)$,

(ii) $I_{\Phi }^{p-1}\left(ku\right)·I_{\Psi }\left(w\right)=1$.

 Lemma 7 

([20]). Let $1<p<\infty$, $l_{\Phi ,p}$ is strictly convex if and only if $a_{\Phi }=0$ and Φ is strictly convex on the interval $[0,{\pi }_{\Phi ,p}\left({\displaystyle \frac{1}{2}}\right)]$ where

$${\pi }_{\Phi ,p}\left({\displaystyle \frac{1}{2}}\right)=sup\left\{t>0:{(\Phi \left(t\right))}^{p-1}·\Psi \left({\Phi }_{+}^{\prime }\left(t\right)\right)\le {\displaystyle \frac{1}{2^p}}\right\}.$$

 Lemma 8 

([23]). Let $1<p<\infty$, $l_{\Phi ,p}$ is locally uniformly convex if and only if

(i) $\Phi \in {\Delta }_2\left(0\right)$ and $\Psi \in {\Delta }_2\left(0\right)$ (i.e., $\Phi \in {\Delta }_2\left(0\right)\bigcap {\nabla }_2\left(0\right)$).

(ii) Φ is strictly convex on the interval $\left[0,{\pi }_{\Phi ,p}\left({\displaystyle \frac{1}{2}}\right)\right]$.

 Lemma 9 

([22]). Let $1<p<\infty$, $\Phi \in {\Delta }_2\left(0\right)$ and $u\in l_{\Phi ,p}\setminus \left\{0\right\}$. For any $i\in \mathbb{N}$, if $u_n\left(i\right)\to u\left(i\right)$ (as $n\to \infty$) and

$$\underset{i_0\to \infty }{lim}\underset{n}{sup}\sum _{i>i_0}\Phi \left(u_n\left(i\right)\right)=0,$$

then $\parallel u_n{-u\parallel }_{\Phi ,p}\to 0$ as $n\to \infty$.

 Lemma 10 

([28]). If $s_p\left(x\right)={(1+x^p)}^{{\displaystyle \frac{1}{p}}},s_q\left(y\right)={(1+y^q)}^{{\displaystyle \frac{1}{q}}}$, for any $1<p,q<\infty$ with ${\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1$, and the Hölder equality $x+y=s_p\left(x\right)s_q\left(y\right)={(1+x^p)}^{{\displaystyle \frac{1}{p}}}{(1+y^q)}^{{\displaystyle \frac{1}{q}}}$ for all $0<x, y<\infty$ holds true, it is necessary and sufficient that $x^{{\displaystyle \frac{1}{q}}}{y}^{{\displaystyle \frac{1}{p}}}=1$$(i.e.,x^{p-1}y=1orx{y}^{q-1}=1)$.

 Lemma 11 

([22]). (Minkowski Inequality) Let $1\le q<\infty$. For any sequences $\left\{{\xi }_k\right\},\left\{{\eta }_k\right\}\subset \mathbb{R}$, we have

(i) ${\left({\sum }_k{|{\xi }_k+{\eta }_k|}^q\right)}^{{\displaystyle \frac{1}{q}}}\le ({\sum }_k|{\xi }_k{|}^q{)}^{{\displaystyle \frac{1}{q}}}+({\sum }_k|{\eta }_k{{|}^q)}^{{\displaystyle \frac{1}{q}}}$.

(ii) ${(1+{(x+y)}^q)}^{{\displaystyle \frac{1}{q}}}\le {(1+x^q)}^{{\displaystyle \frac{1}{q}}}+y$ for all $x,y\ge 0$.

 Lemma 12 

([8]). Let $1<p<\infty$, $u\in S\left(l_{\Phi ,p}\right)$ and $K_p\left(u\right)=\varnothing$. u is an exposed point if and only if $card\left(supp\right(u\left)\right)=1$.

 Lemma 13 

([8]). Let $1<p<\infty$, $u\in S\left(l_{\Phi ,p}\right)$ and $K_p\left(u\right)\ne \varnothing$. u is an exposed point of $B\left(l_{\Phi ,p}\right)$ if and only if:

(i) $K_p\left(u\right)$ is a singleton set, i.e., $K_p\left(u\right)=\left\{k_u\right\}$.

(ii) $card\left(supp\right(u\left)\right)=1$ or $card\{i\in \mathbb{N}:k_{u}u\left(i\right)\in (\mathbb{R}\setminus S{C}_{\Phi })\}\le 1$.

(iii) $card\{i\in \mathbb{N}:k_{u}u\left(i\right)\in A^{\prime }\}·card\{i\in \mathbb{N}:k_{u}u\left(i\right)\in B^{\prime }\}=0.$

(iv) If $I_{\Phi }^{p-1}\left(k_{u}u\right)·I_{\Psi }\left({\Phi }_{-}^{\prime }\left(k_{u}u\right)\right)=1$, then

$$card\{i\in \mathbb{N}:k_{u}u\left(i\right)\in A^{\prime }\}·card\{i\in \mathbb{N}:k_{u}u\left(i\right)\in B\}=0.$$

If $I_{\Phi }^{p-1}\left(k_{u}u\right)·I_{\Psi }\left({\Phi }_{+}^{\prime }\left(k_{u}u\right)\right)=1,$ and ${\theta }_{\Phi }\left(k_{u}u\right)<1$, then

$$card\{i\in \mathbb{N}:k_{u}u\left(i\right)\in A\}·card\{i\in \mathbb{N}:k_{u}u\left(i\right)\in B^{\prime }\}=0.$$

3. Main Results

 Theorem 1. 

Let $1<p<\infty$, $u\in S\left(l_{\Phi ,p}\right)$ and $K_p\left(u\right)=\varnothing$. u is a strongly exposed point if and only if $card\left(supp\right(u\left)\right)=1$.

Proof. 

Necessity. Since a strongly exposed point is an exposed point, by Lemma 12, $card\left(supp\right(u\left)\right)=1$ is necessary.

Sufficiency. Without loss of generality, we assume $u=\left(u\right(1),0,\cdots )$ and $u\left(1\right)>0$. By $K_p\left(u\right)=\varnothing$ and Lemma 3, ${\parallel u\parallel }_{\Phi ,p}=b_{\Psi }u\left(1\right)$ where $b_{\Psi }=\underset{x\to \infty }{lim}{\displaystyle \frac{\Phi \left(x\right)}{x}}$. Thus $v=(b_{\Psi },0,\cdots )$ is the supporting functional of u. If there is ${\left\{u_n\right\}}_{n=1}^{\infty }\subset B\left(l_{\Phi ,p}\right)$ that satisfies

$$<v,u_n>=b_{\Psi }{u}_n\left(1\right)\to b_{\Psi }u\left(1\right),$$

then $u_n\left(1\right)\to u\left(1\right)$ as $n\to \infty .$

We will consider the following two cases separately.

(I) There are infinite numbers of n such that $K_p\left(u_n\right)=\varnothing$. Passing to a subsequence of $\left\{u_n\right\}$, if necessary, we may assume that for any $n\in \mathbb{N}$, $K_p\left(u_n\right)=\varnothing$. By Lemma 3, we have

$$\begin{array}{ccc}\hfill 1\ge \parallel u_n{\parallel }_{\Phi ,p}& =& b_{\Psi }\sum _{i=1}^{\infty }\left|u_n\left(i\right)\right|\hfill \\ & =& b_{\Psi }|u_n\left(1\right)|+b_{\Psi }\sum _{i=2}^{\infty }|u_n\left(i\right)|\hfill \\ & \to & 1+b_{\Psi }\sum _{i=2}^{\infty }\left|u_n\left(i\right)\right|\hfill \end{array}$$

as $n\to \infty$. Therefore, ${\sum }_{i=2}^{\infty }\left|u_n\left(i\right)\right|\to 0$ as $n\to \infty$. Hence,

$$\begin{array}{ccc}\hfill \parallel u-u_n{\parallel }_{\Phi ,p}& =& \underset{k>0}{inf}{\displaystyle \frac{1}{k}}{(1+I_{\Phi }^p\left(k(u-u_n)\right))}^{{\displaystyle \frac{1}{p}}}\hfill \\ & \le & \underset{k\to \infty }{lim}{\displaystyle \frac{1}{k}}{(1+I_{\Phi }^p\left(k(u-u_n)\right))}^{{\displaystyle \frac{1}{p}}}\hfill \\ & =& \underset{k\to \infty }{lim}\left({\displaystyle \frac{\Phi \left(k(u\left(1\right)-u_n\left(1\right))\right)}{k}}+\sum _{i=2}^{\infty }{\displaystyle \frac{\Phi \left(k{u}_n\left(i\right)\right)}{k}}\right)\hfill \\ & =& \underset{k\to \infty }{lim}\left({\displaystyle \frac{\Phi \left(k(u\left(1\right)-u_n\left(1\right))\right)}{k|u\left(1\right)-u_n\left(1\right)|}}|u\left(1\right)-u_n\left(1\right)|+\sum _{i=2}^{\infty }{\displaystyle \frac{\Phi \left(k{u}_n\left(i\right)\right)}{k|u_n\left(i\right)|}}\left|u_n\left(i\right)\right|\right)\hfill \\ & =& b_{\Psi }|u\left(1\right)-u_n\left(1\right)|+b_{\Psi }\sum _{i=2}^{\infty }\left|u_n\left(i\right)\right|\to 0\hfill \end{array}$$

as $n\to \infty .$ Thus, u is a strongly exposed point of $S\left(l_{\Phi ,p}\right)$.

(II) There are infinite numbers of n such that $K_p\left(u_n\right)\ne \varnothing$. Passing to a subsequence of $\left\{u_n\right\}$, if necessary, we may assume that for any $n\in \mathbb{N}$, $K_p\left(u_n\right)\ne \varnothing$.

Take $k_n\in K_p\left(u_n\right)$. We will show that $k_n\to \infty$ as $n\to \infty$. Otherwise, without loss of generality, we may assume that $k_n\to k_0$ as $n\to \infty$. By the definition of ${\parallel u\parallel }_{\Phi ,p}$, we get

$$\begin{array}{ccc}\hfill {\parallel u\parallel }_{\Phi ,p}& <& {\displaystyle \frac{1}{k_0}}{(1+I_{\Phi }^p\left(k_{0}u\right))}^{{\displaystyle \frac{1}{p}}}\hfill \\ & =& {\displaystyle \frac{1}{k_0}}{(1+{(\Phi \left(k_{0}u\left(1\right)\right))}^p)}^{{\displaystyle \frac{1}{p}}}\hfill \\ & =& \underset{n\to \infty }{lim}{\displaystyle \frac{1}{k_n}}{\left(1+{\left(\Phi \left(k_n{u}_n\left(1\right)\right)\right)}^p\right)}^{{\displaystyle \frac{1}{p}}}\hfill \\ & \le & \underset{n\to \infty }{lim}{\displaystyle \frac{1}{k_n}}{(1+I_{\Phi }^p\left(k_n{u}_n\right))}^{{\displaystyle \frac{1}{p}}}\hfill \\ & =& \underset{n\to \infty }{lim}\parallel u_n{\parallel }_{\Phi ,p}\le 1={\parallel u\parallel }_{\Phi ,p}\hfill \end{array}$$

a contradiction.

Since,

$$\begin{array}{ccc}\hfill 1& \ge & \parallel u_n{\parallel }_{\Phi ,p}={\displaystyle \frac{1}{k_n}}{(1+I_{\Phi }^p\left(k_n{u}_n\right))}^{{\displaystyle \frac{1}{p}}}\hfill \\ & =& {\displaystyle \frac{1}{k_n}}{\left(1+{\left(\Phi \left(k_n{u}_n\left(1\right)\right)+\sum _{i\ge 2}\Phi \left(k_n{u}_n\left(i\right)\right)\right)}^p\right)}^{{\displaystyle \frac{1}{p}}}\hfill \\ & \ge & {\displaystyle \frac{1}{k_n}}{(1+{(\Phi \left(k_n{u}_n\left(1\right)\right))}^p)}^{{\displaystyle \frac{1}{p}}}\ge \parallel {\left[u_n\right]}_1{\parallel }_{\Phi ,p}\to {\parallel u\parallel }_{\Phi ,p}=1.\hfill \end{array}$$

Therefore, ${\sum }_{i=2}^{\infty }\Phi \left(k_n{u}_n\left(i\right)\right)\to 0$ as $n\to \infty$. Thus, we obtain

$$\parallel u_n-{\left[u_n\right]}_1{\parallel }_{\Phi ,p}\le {\displaystyle \frac{1}{k_n}}{\left(1+{\left(\sum _{i>1}\Phi \left(k_n{u}_n\left(i\right)\right)\right)}^p\right)}^{{\displaystyle \frac{1}{p}}}\to 0$$

as $n\to \infty$. Hence,

$$\parallel u-u_n{\parallel }_{\Phi ,p}\le \parallel u-{\left[u_n\right]}_1{\parallel }_{\Phi ,p}+{\parallel u_n-{\left[u_n\right]}_1\parallel }_{\Phi ,p}\to 0$$

as $n\to \infty$. Thus, u is a strongly exposed point of $S\left(l_{\Phi ,p}\right)$. □

 Theorem 2. 

Let $1<q<\infty$, $v\in S\left(l_{\Psi ,q}\right)$ and ${\theta }_{\Psi }\left(v\right)<1$. If there is ${\left\{u_n\right\}}_{n=1}^{\infty }\subset B\left(l_{\Phi ,p}\right)$ such that $<v,u_n>\to 1$ as $n\to \infty$, then

$$\underset{i_0\to \infty }{lim}\underset{n}{sup}\sum _{i>i_0}\Phi \left(u_n\left(i\right)\right)=0.$$

(2)

Proof. 

By $v\in S\left(l_{\Psi ,q}\right)$, ${\left\{u_n\right\}}_{n=1}^{\infty }\subset B\left(l_{\Phi ,p}\right)$ and $<v,u_n>\to 1$ as $n\to \infty$, we get $\parallel u_n{\parallel }_{\Phi ,p}\to 1$ as $n\to \infty$. If there is $i_0\nearrow \infty$, such that ${\sum }_{i>i_0}\Phi \left(u_n\left(i\right)\right)\ge {\delta }_0$. We will consider the following two cases separately.

(I) $K_p\left(u_n\right)\ne \varnothing (n=1,2,\cdots )$.

Take $k_n\in K_p\left(u_n\right)$, $l\in K_q\left(v\right)$. By ${\left\{u_n\right\}}_{n=1}^{\infty }\subset B\left(l_{\Phi ,p}\right)$, $v\in S\left(l_{\Psi ,q}\right)$, we get $k_n\ge 1,$ $l\ge 1$. Since ${\theta }_{\Psi }\left(v\right)<1$, there exists $\alpha >0$ such that $I_{\Psi }\left(l(1+\alpha )v\right)<\infty$. For $0<\epsilon <{\displaystyle \frac{\alpha {\delta }_0}{2}}$, take $j_0\in \mathbb{N}$ large enough satisfying ${\sum }_{i>j_0}\Psi \left(l(1+\alpha )v\left(i\right)\right)<\epsilon$. Thus,

$$\begin{array}{ccc}\hfill 1& \leftarrow & {\displaystyle \frac{1}{k_{n}l}}<k_n{u}_n,lv>\hfill \\ & =& {\displaystyle \frac{1}{k_{n}l}}\left(\sum _{i=1}^{j_0}{k}_n{u}_n\left(i\right)lv\left(i\right)+{\displaystyle \frac{1}{1+\alpha }}\sum _{i>j_0}{k}_n{u}_n\left(i\right)l(1+\alpha )v\left(i\right)\right)\hfill \\ & \le & {\displaystyle \frac{1}{k_{n}l}}(\sum _{i=1}^{j_0}\Phi \left(k_n{u}_n\left(i\right)\right)+\sum _{i=1}^{j_0}\Psi \left(lv\left(i\right)\right)\hfill \\ & & +{\displaystyle \frac{1}{1+\alpha }}\left(\sum _{i>j_0}\Phi \left(k_n{u}_n\left(i\right)\right)+\sum _{i>j_0}\Psi \left(l(1+\alpha )v\left(i\right)\right)\right))\hfill \\ & \le & {\displaystyle \frac{1}{k_{n}l}}\left(I_{\Phi }\left(k_n{u}_n\right)+I_{\Psi }\left(lv\right)-{\displaystyle \frac{\alpha }{1+\alpha }}\sum _{i>j_0}\Phi \left(k_n{u}_n\left(i\right)\right)+{\displaystyle \frac{1}{1+\alpha }}\sum _{i>j_0}\Psi \left(l(1+\alpha )v\left(i\right)\right)\right)\hfill \\ & \le & {\displaystyle \frac{1}{k_n}}{s}_p\left(I_{\Phi }\left(k_n{u}_n\right)\right){\displaystyle \frac{1}{l}}{s}_q\left(I_{\Psi }\left(lv\right)\right)-{\displaystyle \frac{1}{k_{n}l}}{\displaystyle \frac{\alpha }{1+\alpha }}{k}_n{\delta }_0+{\displaystyle \frac{1}{k_{n}l}}{\displaystyle \frac{\epsilon }{1+\alpha }}\hfill \\ & \to & 1-{\displaystyle \frac{1}{k_{n}l}}{\displaystyle \frac{\alpha }{1+\alpha }}{k}_n{\delta }_0+{\displaystyle \frac{1}{k_{n}l}}{\displaystyle \frac{\epsilon }{1+\alpha }}\hfill \\ & <& 1-{\displaystyle \frac{1}{l}}{\displaystyle \frac{\alpha {\delta }_0}{1+\alpha }}\left(1-{\displaystyle \frac{1}{2k_n}}\right)<1-{\displaystyle \frac{\alpha {\delta }_0}{2l(1+\alpha )}}\hfill \end{array}$$

as $n\to \infty$, this is a contradiction.

(II) $K_p\left(u_n\right)=\varnothing (n=1,2,\cdots )$.

By Lemma 3, We have $\parallel u_n{\parallel }_{\Phi ,p}=b_{\Psi }{\sum }_{i=1}^{\infty }\left|u_n\left(i\right)\right|$. By ${\theta }_{\Psi }\left(v\right)<1$, there is $\lambda >0$ and $i_0\in \mathbb{N}$ such that

$$\sum _{i>i_0}^{\infty }\Psi \left({\displaystyle \frac{v\left(i\right)}{1-\lambda }}\right)<\infty .$$

So, for any $i>i_0$, ${\displaystyle \frac{v\left(i\right)}{1-\lambda }}<b_{\Psi }$. From

$$\Phi \left(u_n\left(i\right)\right)={\int }_0^{|u_n\left(i\right)|}{\Phi }_{+}^{\prime }\left(t\right)dt\le {\int }_0^{|u_n\left(i\right)|}{b}_{\Psi }dt=b_{\Psi }\left|u_n\left(i\right)\right|.$$

We obtain ${\sum }_{i>i_n}{b}_{\Psi }\left|u_n\left(i\right)\right|\ge {\sum }_{i>i_n}\Phi \left(u_n\left(i\right)\right)\ge {\delta }_0$ for $i_n>i_0$. Thus,

$$\begin{array}{ccc}\hfill 1\leftarrow <v,u_n>& =& \sum _{i=1}^{i_n}v\left(i\right)u_n\left(i\right)+(1-\lambda )\sum _{i\ge i_n}{\displaystyle \frac{v\left(i\right)}{1-\lambda }}{u}_n\left(i\right)\hfill \\ & \le & b_{\Psi }\sum _{i=1}^{i_n}|u_n\left(i\right)|+(1-\lambda )\sum _{i\ge i_n}{b}_{\Psi }\left|u_n\left(i\right)\right|\hfill \\ & =& b_{\Psi }\sum _{i=1}^{\infty }|u_n\left(i\right)|-\lambda \sum _{i\ge i_n}{b}_{\Psi }\left|u_n\left(i\right)\right|\hfill \\ & \le & \parallel u_n{\parallel }_{\Phi ,p}-\lambda {\delta }_0\to 1-\lambda {\delta }_0\hfill \end{array}$$

as $n\to \infty$, this is a contradiction. □

 Theorem 3. 

Let $1<p<\infty$, if $\Phi \notin {\Delta }_2\left(0\right)$ and the p-Amemiya norm ${\parallel ·\parallel }_{\Phi ,p}$ is $k_p^{*}$-finite, then $S\left(l_{\Phi ,p}\right)$ has no strongly exposed point.

Proof. 

Let $u\in S\left(l_{\Phi ,p}\right)$. Since $k_p^{*}\left(u\right)<\infty$, take $k\in K_p\left(u\right)$, then ${\theta }_{\Phi }\left(ku\right)\le 1$. We will consider the following two cases separately.

(I) ${\theta }_{\Phi }\left(ku\right)=1$.

For $\epsilon >0$, $j\in \mathbb{N}$, by the definition of ${\theta }_{\Phi }\left(ku\right)$, we have ${\sum }_{i=j}^{\infty }\Phi \left({\displaystyle \frac{ku\left(i\right)}{1-\epsilon }}\right)=\infty$. Take $0=n_0<n_1<n_2<\cdots$, such that

$$I_{\Phi }\left({\displaystyle \frac{k{\left[u\right]}_{n_j}^{n_{j-1}}}{1-\frac{1}{j}}}\right)=\sum _{i=n_{j-1}+1}^{n_j}\Phi \left({\displaystyle \frac{ku\left(i\right)}{1-\frac{1}{j}}}\right)>1(j=2,3,\cdots ).$$

By the definition of ${\parallel ·\parallel }_{\Phi ,\infty }$, we obtain

$${∥{\left[u\right]}_{n_j}^{n_{j-1}}∥}_{\Phi ,\infty }\ge {\displaystyle \frac{1}{k}}\left(1-{\displaystyle \frac{1}{j}}\right).$$

Set $u_j=u-{\left[u\right]}_{n_j}^{n_{j-1}}$, then $u_j\in B\left(l_{\Phi ,p}\right)$. Since $\Phi \notin {\Delta }_2\left(0\right)$, the supporting functional of u must be in $l_{\Psi ,q}\oplus F$. Suppose $f=(v+\phi )\in Grad\left(u\right)$ where $v\in l_{\Psi ,q}$, $\phi \in F$. We have

$$\begin{array}{ccc}\hfill 1\ge f\left(u_j\right)& =& 〈v,u-{\left[u\right]}_{n_j}^{n_{j-1}}〉+\phi \left(u-{\left[u\right]}_{n_j}^{n_{j-1}}\right)\hfill \\ & \ge & \sum _{i=1}^{n_{j-1}}v\left(i\right)u\left(i\right)+\phi \left(u\right)\hfill \\ & \to & <v,u>+\phi \left(u\right)=f\left(u\right)=1\hfill \end{array}$$

as $j\to \infty$. However,

$$\parallel u-u_j{\parallel }_{\Phi ,p}={∥{\left[u\right]}_{n_j}^{n_{j-1}}∥}_{\Phi ,p}\ge {∥{\left[u\right]}_{n_j}^{n_{j-1}}∥}_{\Phi ,\infty }\ge {\displaystyle \frac{1}{k}}\left(1-{\displaystyle \frac{1}{j}}\right)\to {\displaystyle \frac{1}{k}}$$

as $j\to \infty$. This shows that u is not a strongly exposed point of $S\left(l_{\Phi ,p}\right)$.

(II) ${\theta }_{\Phi }\left(ku\right)<1$.

Take $w\in l_{\Phi ,p}$ such that $I_{\Phi }\left(w\right)<\infty$ and ${\theta }_{\Phi }(ku-w)\ne 0$ (indeed, if ${\theta }_{\Phi }\left(ku\right)=0$, we take $w\in S\left(l_{\Phi ,p}\right)$ with ${\theta }_{\Phi }\left(w\right)\ne 0$, if ${\theta }_{\Phi }\left(ku\right)\ne 0$, we take $w=0$). Since $\Phi \notin {\Delta }_2\left(0\right)$, there exists ${\phi }_0\in S\left(F\right)$ such that ${\phi }_0(ku-w)\ne 0$. Set

$$u_n=\left(u\left(1\right),\cdots ,u\left(n\right),{\displaystyle \frac{1}{k}}w(n+1),{\displaystyle \frac{1}{k}}w(n+2),\cdots \right).$$

By Minkowski Inequality, we have

$$\begin{array}{ccc}\hfill \parallel u_n{\parallel }_{\Phi ,p}& \le & {\displaystyle \frac{1}{k}}{(1+I_{\Phi }^p\left(k{u}_n\right))}^{{\displaystyle \frac{1}{p}}}\hfill \\ & =& {\displaystyle \frac{1}{k}}{\left(1+{\left(\sum _{i=1}^n\Phi \left(ku\left(i\right)\right)+\sum _{i=n+1}^{\infty }\Phi \left(w\left(i\right)\right)\right)}^p\right)}^{{\displaystyle \frac{1}{p}}}\hfill \\ & \le & {\displaystyle \frac{1}{k}}{\left(1+{\left(\sum _{i=1}^n\Phi \left(ku\left(i\right)\right)\right)}^p\right)}^{{\displaystyle \frac{1}{p}}}+{\displaystyle \frac{1}{k}}\sum _{i=n+1}^{\infty }\Phi \left(w\left(i\right)\right)\hfill \\ & \le & {\displaystyle \frac{1}{k}}{(1+I_{\Phi }^p\left(ku\right))}^{{\displaystyle \frac{1}{p}}}+{\displaystyle \frac{1}{k}}\sum _{i=n+1}^{\infty }\Phi \left(w\left(i\right)\right)\hfill \\ & \to & {\parallel u\parallel }_{\Phi ,p}\hfill \end{array}$$

as $n\to \infty$ where ${\sum }_{i=n+1}^{\infty }\Phi \left(w\left(i\right)\right)\to 0$ is deduced from $I_{\Phi }\left(w\right)<\infty$. By ${\theta }_{\Phi }\left(ku\right)<1$ and Lemma 8 in [22], we have $Grad\left(u\right)\subset l_{\Psi ,q}$, take $f=v\in Grad\left(u\right)$, then

$$\begin{array}{ccc}\hfill 1\ge f\left(u_n\right)& =& <v,u_n>\hfill \\ & =& \sum _{i=1}^{n}u\left(i\right)v\left(i\right)+{\displaystyle \frac{1}{k}}\sum _{i=n+1}^{\infty }w\left(i\right)v\left(i\right)\hfill \\ & \to & <v,u>=f\left(u\right)=1\hfill \end{array}$$

as $n\to \infty$. But

$$\parallel u-u_n{\parallel }_{\Phi ,p}\ge \left|{\phi }_0(u-u_n)\right|=\left|{\phi }_0\left(u-{\displaystyle \frac{1}{k}}w\right)\right|>0.$$

So, u is not a strongly exposed point of $S\left(l_{\Phi ,p}\right)$. □

 Theorem 4. 

Let $1<p<\infty$, $u\in S\left(l_{\Phi ,p}\right)$ and $K_p\left(u\right)\ne \varnothing$. u is a strongly exposed point of $B\left(l_{\Phi ,p}\right)$ if and only if

(i) u is an exposed point.

(ii) $\Phi \in {\Delta }_2\left(0\right)$.

(iii) u has a supporting functional $v\in l_{\Psi ,q}$ with ${\theta }_{\Psi }\left(v\right)<1$.

Proof. 

Necessity. Since a strongly exposed point is an exposed point, (i) is trivial. By Theorem 3, $\Phi \in {\Delta }_2\left(0\right)$ is necessary. We only need to prove (iii). If (iii) does not hold, then for any supporting functional $v\in l_{\Psi ,q}$, ${\theta }_{\Psi }\left(v\right)=1$, i.e., for and any $\tau >0$, $I_{\Psi }\left((1+\tau )v\right)=\infty$. By Theorem 1.43 in [24] and equivalence of p-Amemiya norms, we have

$$\underset{n\to \infty }{lim}{\parallel v-{\left[v\right]}_n\parallel }_{\Psi ,q}={\theta }_{\Psi }\left(v\right)=1.$$

Since ${∥v-{\left[v\right]}_n∥}_{\Psi ,q}={sup}_{{\parallel u\parallel }_{\Phi ,p}=1}〈u,v-{\left[v\right]}_n〉,$ there exists ${\left\{u_n\right\}}_{n=1}^{\infty }\subset S\left(l_{\Phi ,p}\right)$ satisfying

$$〈v-{\left[v\right]}_n,u_n〉=\sum _{i=n+1}^{\infty }v\left(i\right)u_n\left(i\right)\ge {\parallel v-{\left[v\right]}_n\parallel }_{\Psi ,q}-{\displaystyle \frac{1}{n}}.$$

Without loss of generality, let

$$u_n=(0,\cdots ,0,u_n(n+1),u_n(n+2),\cdots ).$$

Put

$$z_n={\displaystyle \frac{1}{2}}(u\left(1\right),\cdots ,u\left(n\right),u_n(n+1),u_n(n+2),\cdots )={\displaystyle \frac{{\left[u\right]}_n+u_n}{2}}.$$

Then

$$\begin{array}{ccc}\hfill 1& \ge & {\displaystyle \frac{1}{2}}\left({\parallel \left[u\right]}_n{\parallel }_{\Phi ,p}+{\parallel u_n\parallel }_{\Phi ,p}\right)\hfill \\ & \ge & \parallel z_n{\parallel }_{\Phi ,p}\ge \sum _{i=1}^{\infty }{z}_n\left(i\right)v\left(i\right)\hfill \\ & =& {\displaystyle \frac{1}{2}}\left(\sum _{i=1}^{n}u\left(i\right)v\left(i\right)+\sum _{i=n+1}^{\infty }{u}_n\left(i\right)v\left(i\right)\right)\hfill \\ & \ge & {\displaystyle \frac{1}{2}}\left(\sum _{i=1}^{n}u\left(i\right)v\left(i\right)+{\parallel v-{\left[v\right]}_n\parallel }_{\Psi ,q}-{\displaystyle \frac{1}{n}}\right)\to 1\hfill \end{array}$$

as $n\to \infty$. But $\parallel z_n{-u\parallel }_{\Phi ,p}\ge {\displaystyle \frac{1}{2}}{\parallel {\left[u\right]}_n\parallel }_{\Phi ,p}\to {\displaystyle \frac{1}{2}}>0$ as $n\to \infty$. We obtain that u is not a strongly exposed point of $S\left(l_{\Phi ,p}\right)$.

Sufficiency. Now, we prove the sufficiency for $card\left(supp\right(u\left)\right)=1$.

Assume $u=\left(u\right(1),0,\cdots )$ and $u\left(1\right)>0$. Since $u\in h_{\Phi ,p}$, we have $Grad\left(u\right)\subset l_{\Psi ,q}$. Suppose $v=\left(v\right(1),0,\cdots )\in Grad\left(u\right)$, then ${\parallel u\parallel }_{\Phi ,p}=u\left(1\right)v\left(1\right)=1$. If there exists ${\left\{u_n\right\}}_{n=1}^{\infty }\subset B\left(l_{\Phi ,p}\right)$, satisfying $<u_n,v>\to <u,v>$, then $u_n\left(1\right)\to u\left(1\right)$ as $n\to \infty$. Since $K_p\left(u\right)\ne \varnothing$, by Lemma 2, $K_p\left(u_n\right)\ne \varnothing$, take $k_n\in K_p\left(u_n\right)$, we have

$$\begin{array}{ccc}\hfill \parallel u_n{\parallel }_{\Phi ,p}& =& {\displaystyle \frac{1}{k_n}}{(1+I_{\Phi }^p\left(k_n{u}_n\right))}^{{\displaystyle \frac{1}{p}}}\hfill \\ & \ge & {\displaystyle \frac{1}{k_n}}{(1+{(\Phi \left(k_n{u}_n\left(1\right)\right))}^p)}^{{\displaystyle \frac{1}{p}}}\hfill \\ & \ge & \parallel {\left[u_n\right]}_1{\parallel }_{\Phi ,p}\to {\parallel u\parallel }_{\Phi ,p}=1\hfill \end{array}$$

as $n\to \infty$. Hence, ${\sum }_{i=2}^{\infty }\Phi \left(k_n{u}_n\left(i\right)\right)\to 0$ as $n\to \infty$, which implies ${\sum }_{i=2}^{\infty }\Phi \left(u_n\left(i\right)\right)\to 0$ as $n\to \infty$. By $\Phi \in {\Delta }_2\left(0\right)$ and Theorem 1.39 in [24], we have

$$\parallel u_n-{\left[u_n\right]}_1{\parallel }_{\Phi ,\infty }\to 0$$

as $n\to \infty$. By equivalence of p-Amemiya norms,

$$\parallel u_n-{\left[u_n\right]}_1{\parallel }_{\Phi ,p}\to 0$$

as $n\to \infty$. So,

$$\parallel u-u_n{\parallel }_{\Phi ,p} \le \parallel u-{\left[u_n\right]}_1{\parallel }_{\Phi ,p}+{\parallel u_n-{\left[u_n\right]}_1\parallel }_{\Phi ,p}\to 0$$

as $n\to \infty$. Thus, u is a strongly exposed point of $S\left(l_{\Phi ,p}\right)$.

Next, we will prove the sufficiency for $card\left(supp\right(u\left)\right)>1$.

Suppose ${\left\{u_n\right\}}_{n=1}^{\infty }\subset B\left(l_{\Phi ,p}\right)$, $<u_n,v>\to <u,v>=1$. By Theorem 2,

$$\underset{i_0\to \infty }{lim}\underset{n}{sup}\sum _{i>i_0}\Phi \left(u_n\left(i\right)\right)=0$$

(3)

holds. Combine $\Phi \in {\Delta }_2\left(0\right)$ and Lemma 9, to prove $\parallel u_n{-u\parallel }_{\Phi ,p}\to 0$ as $n\to \infty$, we only need to prove, for any $i\in \mathbb{N}$, $u_n\left(i\right)\to u\left(i\right)$ as $n\to \infty .$ Take $k_n\in K_p\left(u_n\right)$. We can prove, for any $i\in \mathbb{N}$, $k_n{u}_n\left(i\right)\to k_{u}u\left(i\right)$ as $n\to \infty$ where $k_u\in K_p\left(u\right)$.

(I) $I_{\Phi }^{p-1}\left(k_{u}u\right)·I_{\Psi }\left({\Phi }_{-}^{\prime }\left(k_{u}u\right)\right)=1$.

By Lemma 6, $v={\displaystyle \frac{{\Phi }_{-}^{\prime }\left(k_{u}u\right)}{\parallel {\Phi }_{-}^{\prime }\left(k_u\left(u\right)\right){\parallel }_{\Psi ,q}}}$ is the unique supporting functional of u. Since $I_{\Phi }^{p-1}\left(k_{u}u\right)·I_{\Psi }\left({\Phi }_{-}^{\prime }\left(k_{u}u\right)\right)=1$, we get $1\in K_q\left({\Phi }_{-}^{\prime }\left(k_{u}u\right)\right)$. By the Hölder Inequality ([28]) and the Young Inequality, we have

$$\begin{array}{ccc}\hfill 0& \leftarrow & {\displaystyle \frac{1}{k_n}}{(1+I_{\Phi }^p\left(k_n{u}_n\right))}^{{\displaystyle \frac{1}{p}}}-〈{\displaystyle \frac{{\Phi }_{-}^{\prime }\left(k_{u}u\right)}{\parallel {\Phi }_{-}^{\prime }\left(k_{u}u\right){\parallel }_{\Psi ,q}}},u_n〉\hfill \\ & =& {\displaystyle \frac{{(1+I_{\Phi }^p\left(k_n{u}_n\right))}^{\frac{1}{p}}·{\parallel {\Phi }_{-}^{\prime }\left(k_{u}u\right)\parallel }_{\Psi ,q}-\langle {\Phi }_{-}^{\prime }\left(k_{u}u\right),k_n{u}_n\rangle }{k_n{\parallel {\Phi }_{-}^{\prime }\left(k_{u}u\right)\parallel }_{\Psi ,q}}}\hfill \\ & =& {\displaystyle \frac{{(1+I_{\Phi }^p\left(k_n{u}_n\right))}^{\frac{1}{p}}·{(1+I_{\Psi }^q\left({\Phi }_{-}^{\prime }\left(k_{u}u\right)\right))}^{\frac{1}{q}}-{\sum }_{i=1}^{\infty }{k}_n{u}_n\left(i\right){\Phi }_{-}^{\prime }\left(k_{u}u\left(i\right)\right)}{k_n{\parallel {\Phi }_{-}^{\prime }\left(k_{u}u\right)\parallel }_{\Psi ,q}}}\hfill \\ & \ge & {\displaystyle \frac{I_{\Phi }\left(k_n{u}_n\right)+I_{\Psi }\left({\Phi }_{-}^{\prime }\left(k_{u}u\right)\right)-{\sum }_{i=1}^{\infty }{k}_n{u}_n\left(i\right){\Phi }_{-}^{\prime }\left(k_{u}u\left(i\right)\right)}{k_n{\parallel {\Phi }_{-}^{\prime }\left(k_{u}u\right)\parallel }_{\Psi ,q}}}\hfill \\ & \ge & {\displaystyle \frac{I_{\Phi }\left(k_n{u}_n\right)+I_{\Psi }\left({\Phi }_{-}^{\prime }\left(k_{u}u\right)\right)-I_{\Phi }\left(k_n{u}_n\right)-I_{\Psi }\left({\Phi }_{-}^{\prime }\left(k_{u}u\right)\right)}{k_n{\parallel {\Phi }_{-}^{\prime }\left(k_{u}u\right)\parallel }_{\Psi ,q}}}\hfill \\ & =& 0\hfill \end{array}$$

as $n\to \infty$. Thus, for any $i\in \mathbb{N}$,

$$\Phi \left(k_n{u}_n\left(i\right)\right)+\Psi \left({\Phi }_{-}^{\prime }\left(k_{u}u\left(i\right)\right)\right)-k_n{u}_n\left(i\right)·{\Phi }_{-}^{\prime }\left(k_{u}u\left(i\right)\right)\to 0$$

(4)

as $n\to \infty$. And

$${(1+I_{\Phi }^p\left(k_n{u}_n\right))}^{{\displaystyle \frac{1}{p}}}·{(1+I_{\Psi }^q\left({\Phi }_{-}^{\prime }\left(k_{u}u\right)\right))}^{{\displaystyle \frac{1}{q}}}-I_{\Phi }\left(k_n{u}_n\right)-I_{\Psi }\left({\Phi }_{-}^{\prime }\left(k_{u}u\right)\right)\to 0$$

as $n\to \infty$, which means

$$I_{\Phi }^{p-1}\left(k_n{u}_n\right)·I_{\Psi }\left({\Phi }_{-}^{\prime }\left(k_{u}u\right)\right)\to 1=I_{\Phi }^{p-1}\left(k_{u}u\right)·I_{\Psi }\left({\Phi }_{-}^{\prime }\left(k_{u}u\right)\right)$$

as $n\to \infty$, i.e.,

$$I_{\Phi }\left(k_n{u}_n\right)\to I_{\Phi }\left(k_{u}u\right)$$

(5)

as $n\to \infty$.

We will discuss the following cases.

Case-1 If $k_{u}u\left(i\right)\in S{C}_{\Phi }$, then $k_n{u}_n\left(i\right)\le k_{u}u\left(i\right)-\delta$ for any $\delta >0$ implies

$$\begin{array}{ccc}& & \Phi \left(k_n{u}_n\left(i\right)\right)+\Psi \left({\Phi }_{-}^{\prime }\left(k_{u}u\left(i\right)\right)\right)-k_n{u}_n\left(i\right)·{\Phi }_{-}^{\prime }\left(k_{u}u\left(i\right)\right)\hfill \\ & \ge & \mathrm{the} \mathrm{area} \mathrm{of} \mathrm{figure} \mathrm{ABC} > 0.\hfill \end{array}$$

(see Figure 1). This contradicts to (4). So $k_n{u}_n\left(i\right)>k_{u}u\left(i\right)-\delta$ for n large enough and any $\delta >0$.

Moreover, $k_n{u}_n\left(i\right)\ge k_{u}u\left(i\right)+\delta$ for any $\delta >0$ implies

$$\begin{array}{ccc}& & \Phi \left(k_n{u}_n\left(i\right)\right)+\Psi \left({\Phi }_{-}^{\prime }\left(k_{u}u\left(i\right)\right)\right)-k_n{u}_n\left(i\right)·{\Phi }_{-}^{\prime }\left(k_{u}u\left(i\right)\right)\hfill \\ & \ge & \mathrm{the} \mathrm{area} \mathrm{of} \mathrm{figure} \mathrm{ABC} > 0.\hfill \end{array}$$

(see Figure 1). This contradicts to (4). So $k_n{u}_n\left(i\right)<k_{u}u\left(i\right)+\delta$ for n large enough and any $\delta >0$. Thus, $k_n{u}_n\left(i\right)\to k_{u}u\left(i\right)$ as $n\to \infty$.

Case-2 If $k_{u}u\left(i\right)=a\in A$, then $k_n{u}_n\left(i\right)\le a-\delta$ for any $\delta >0$ implies

$$\begin{array}{ccc}& & \Phi \left(k_n{u}_n\left(i\right)\right)+\Psi \left({\Phi }_{-}^{\prime }\left(k_{u}u\left(i\right)\right)\right)-k_n{u}_n\left(i\right)·{\Phi }_{-}^{\prime }\left(k_{u}u\left(i\right)\right)\hfill \\ & =& \Phi \left(k_n{u}_n\left(i\right)\right)+\Psi \left({\Phi }_{-}^{\prime }\left(a\right)\right)-k_n{u}_n\left(i\right)·{\Phi }_{-}^{\prime }\left(a\right)\hfill \\ & \ge & \mathrm{the} \mathrm{area} \mathrm{of} \mathrm{figure} \mathrm{ABC} > 0.\hfill \end{array}$$

(see Figure 2). This contradicts to (4). So $k_n{u}_n\left(i\right)>a-\delta$ for n large enough and any $\delta >0$.

Moreover, $k_n{u}_n\left(i\right)\ge a+\delta$ for any $\delta >0$ implies

$$\begin{array}{ccc}& & \Phi \left(k_n{u}_n\left(i\right)\right)+\Psi \left({\Phi }_{-}^{\prime }\left(a\right)\right)-k_n{u}_n\left(i\right)·{\Phi }_{-}^{\prime }\left(a\right)\hfill \\ & \ge & \mathrm{the} \mathrm{area} \mathrm{of} \mathrm{the} rectangle \mathrm{ADEF} \hfill \\ & \ge & ({\Phi }_{+}^{\prime }\left(a\right)-{\Phi }_{-}^{\prime }\left(a\right))\delta >0.\hfill \end{array}$$

(see Figure 2). This contradicts to (4). So $k_n{u}_n\left(i\right)<a+\delta$ for n large enough and any $\delta >0$. Thus, $k_n{u}_n\left(i\right)\to a$ as $n\to \infty$.

Case-3 If $k_{u}u\left(i\right)=b\in B$, then $k_n{u}_n\left(i\right)\ge b+\delta$ for any $\delta >0$ implies

$$\begin{array}{ccc}& & \Phi \left(k_n{u}_n\left(i\right)\right)+\Psi \left({\Phi }_{-}^{\prime }\left(k_{u}u\left(i\right)\right)\right)-k_n{u}_n\left(i\right)·{\Phi }_{-}^{\prime }\left(k_{u}u\left(i\right)\right)\hfill \\ & =& \Phi \left(k_n{u}_n\left(i\right)\right)+\Psi \left({\Phi }_{-}^{\prime }\left(b\right)\right)-k_n{u}_n\left(i\right)·{\Phi }_{-}^{\prime }\left(b\right)\hfill \\ & \ge & \mathrm{the} \mathrm{area} \mathrm{of} \mathrm{the} figure \mathrm{ABCD} \hfill \\ & \ge & ({\Phi }_{+}^{\prime }\left(b\right)-{\Phi }_{-}^{\prime }\left(b\right))\delta >0.\hfill \end{array}$$

(see Figure 3). This contradicts to (4). So $k_n{u}_n\left(i\right)<b+\delta$ for n large enough and any $\delta >0$.

Moreover, $k_n{u}_n\left(i\right)\le b-\delta$ for any $\delta \in (0,b-a)$ or $(0,b-a^{\prime })$ implies

$$\Phi \left(k_n{u}_n\left(i\right)\right)+\Psi \left({\Phi }_{-}^{\prime }\left(b\right)\right)-k_n{u}_n\left(i\right)·{\Phi }_{-}^{\prime }\left(b\right)=0.$$

(see Figure 3). Thus, $k_n{u}_n\left(i\right)\le b$ as $n\to \infty$.

Case-4 If $k_{u}u\left(i\right)=a^{\prime }\in A^{\prime }$, then $k_n{u}_n\left(i\right)\le a^{\prime }-\delta$ for any $\delta >0$ implies

$$\begin{array}{ccc}& & \Phi \left(k_n{u}_n\left(i\right)\right)+\Psi \left({\Phi }_{-}^{\prime }\left(k_{u}u\left(i\right)\right)\right)-k_n{u}_n\left(i\right)·{\Phi }_{-}^{\prime }\left(k_{u}u\left(i\right)\right)\hfill \\ & =& \Phi \left(k_n{u}_n\left(i\right)\right)+\Psi \left({\Phi }_{-}^{\prime }\left(a^{\prime }\right)\right)-k_n{u}_n\left(i\right)·{\Phi }_{-}^{\prime }\left(a^{\prime }\right)\hfill \\ & \ge & \mathrm{the} \mathrm{area} \mathrm{of} \mathrm{the} figure \mathrm{ABC} >0.\hfill \end{array}$$

(see Figure 4). This contradicts to (4). So $k_n{u}_n\left(i\right)>a^{\prime }-\delta$ for n large enough and any $\delta >0$.

Moreover, $k_n{u}_n\left(i\right)\ge a^{\prime }+\delta$ for any $\delta \in (0,b^{\prime }-a^{\prime })$ or $(0,b-a^{\prime })$ implies

$$\Phi \left(k_n{u}_n\left(i\right)\right)+\Psi \left({\Phi }_{-}^{\prime }\left(a^{\prime }\right)\right)-k_n{u}_n\left(i\right)·{\Phi }_{-}^{\prime }\left(a^{\prime }\right)=0.$$

(see Figure 4). Thus, $k_n{u}_n\left(i\right)\ge a^{\prime }$ as $n\to \infty$.

Case-5 If $k_{u}u\left(i\right)=b^{\prime }\in B^{\prime }$, then $k_n{u}_n\left(i\right)\ge b^{\prime }+\delta$ for any $\delta >0$ implies

$$\begin{array}{ccc}& & \Phi \left(k_n{u}_n\left(i\right)\right)+\Psi \left({\Phi }_{-}^{\prime }\left(k_{u}u\left(i\right)\right)\right)-k_n{u}_n\left(i\right)·{\Phi }_{-}^{\prime }\left(k_{u}u\left(i\right)\right)\hfill \\ & =& \Phi \left(k_n{u}_n\left(i\right)\right)+\Psi \left({\Phi }_{-}^{\prime }\left(b^{\prime }\right)\right)-k_n{u}_n\left(i\right)·{\Phi }_{-}^{\prime }\left(b^{\prime }\right)\hfill \\ & \ge & \mathrm{the} \mathrm{area} \mathrm{of} \mathrm{the} figure \mathrm{ABC} >0.\hfill \end{array}$$

(see Figure 5). This contradicts to (4). So $k_n{u}_n\left(i\right)<b^{\prime }+\delta$ for n large enough and any $\delta >0$.

Moreover, $k_n{u}_n\left(i\right)\le b^{\prime }-\delta$ for any $\delta \in (0,b^{\prime }-a^{\prime })$ or $(0,b^{\prime }-a)$ implies

$$\Phi \left(k_n{u}_n\left(i\right)\right)+\Psi \left({\Phi }_{-}^{\prime }\left(b^{\prime }\right)\right)-k_n{u}_n\left(i\right)·{\Phi }_{-}^{\prime }\left(b^{\prime }\right)=0.$$

(see Figure 5). Thus, $k_n{u}_n\left(i\right)\le b^{\prime }$ as $n\to \infty$.

If $k_{u}u\left(i\right)\in A^{\prime }$, by u is an exposed point and Lemma 13, we get

$$\{i\in \mathbb{N}:k_{u}u\left(i\right)\in B^{\prime }\}=\{i\in \mathbb{N}:k_{u}u\left(i\right)\in B\}=\varnothing .$$

Set $I_0=\{i\in \mathbb{N}:k_{u}u\left(i\right)\in A^{\prime }\}$. By case-1, case-2 and case-4, we have

$$\begin{array}{ccc}& & k_n{u}_n\left(i\right)\to k_{u}u\left(i\right)(i\in \mathbb{N}\setminus I_0)\mathrm{as}n\to \infty ,\hfill \\ & & k_n{u}_n\left(i\right)\ge k_{u}u\left(i\right)(i\in I_0)\mathrm{as}n\to \infty .\hfill \end{array}$$

In fact, for any $i\in I_0$, we can claim that $k_n{u}_n\left(i\right)\to k_{u}u\left(i\right)$ as $n\to \infty .$ Otherwise, there exists $i_0\in I_0$, ${\epsilon }_0>0,n_0\in \mathbb{N}$ such that $k_n{u}_n\left(i_0\right)>k_{u}u\left(i_0\right)+{\epsilon }_0$ for all $n>n_0$. Since $\Phi$ is affine on $[a^{\prime },b]$ or $[a^{\prime },b^{\prime }]$: $\Phi \left(u\right)=cu+d$ where $c,d\in \mathbb{R}$ for $u\in [a^{\prime },b]$ or $[a^{\prime },b^{\prime }]$. For all $n>n_0$, we have

$$\begin{array}{ccc}\hfill \Phi \left(k_n{u}_n\left(i_0\right)\right)& =& c{k}_n{u}_n\left(i_0\right)+d\hfill \\ & >& c(k_{u}u\left(i_0\right)+{\epsilon }_0)+d\hfill \\ & =& c{k}_{u}u\left(i_0\right)+d+c{\epsilon }_0\hfill \\ & =& \Phi \left(k_{u}u\left(i_0\right)\right)+c{\epsilon }_0.\hfill \end{array}$$

Then,

$$\begin{array}{ccc}\hfill I_{\Phi }\left(k_n{u}_n\right)& =& \sum _{i\ne i_0}\Phi \left(k_n{u}_n\left(i\right)\right)+\Phi \left(k_n{u}_n\left(i_0\right)\right)\hfill \\ & >& \sum _{i\ne i_0}\Phi \left(k_n{u}_n\left(i\right)\right)+(\Phi \left(k_{u}u\left(i_0\right)\right)+c{\epsilon }_0)\hfill \\ & \ge & I_{\Phi }\left(k_{u}u\right)+c{\epsilon }_0\hfill \end{array}$$

as $n\to \infty$, this contradicts to (5). Hence, for any $i\in \mathbb{N}$, $k_n{u}_n\left(i\right)\to k_{u}u\left(i\right)$ as $n\to \infty .$

If $k_{u}u\left(i\right)\in B\cup B^{\prime }$, by u is an exposed point and Lemma 13, we get

$$\{i\in \mathbb{N}:k_{u}u\left(i\right)\in A^{\prime }\}=\varnothing .$$

Set $I_1=\{i\in \mathbb{N}:k_{u}u\left(i\right)\in B\cup B^{\prime }\}$. By case-1, case-2, case-3 and case-5, we have

$$\begin{array}{ccc}& & k_n{u}_n\left(i\right)\to k_{u}u\left(i\right)(i\in \mathbb{N}\setminus I_1)\mathrm{as}n\to \infty ,\hfill \\ & & k_n{u}_n\left(i\right)\le k_{u}u\left(i\right)(i\in I_1)\mathrm{as}n\to \infty .\hfill \end{array}$$

Use the analogous way as above, we obtain, for any $i\in \mathbb{N}$, $k_n{u}_n\left(i\right)\to k_{u}u\left(i\right)$ as $n\to \infty .$

Thus, for any $i\in \mathbb{N}$, we have $k_n{u}_n\left(i\right)\to k_{u}u\left(i\right)$ as $n\to \infty .$ By $\parallel u_n{\parallel }_{\Phi ,p}\to 1$, ${\parallel u\parallel }_{\Phi ,p}=1$, we have ${(1+I_{\Phi }^p\left(k_n{u}_n\right))}^{{\displaystyle \frac{1}{p}}}\to k_n$ and ${(1+I_{\Phi }^p\left(k_{u}u\right))}^{{\displaystyle \frac{1}{p}}}=k_u$. From (5), we get $k_n\to k_u$ as $n\to \infty$. Hence, for any $i\in \mathbb{N}$, $u_n\left(i\right)\to u\left(i\right)$ as $n\to \infty$. By Lemma 9, u is a strongly exposed point of $S\left(l_{\Phi ,p}\right)$.

(II) $I_{\Phi }^{p-1}\left(k_{u}u\right)I_{\Psi }\left({\Phi }_{+}^{\prime }\left(k_{u}u\right)\right)=1$.

By Lemma 6, $v={\displaystyle \frac{{\Phi }_{+}^{\prime }\left(k_{u}u\right)}{\parallel {\Phi }_{+}^{\prime }\left(k_{u}u\right){\parallel }_{\Psi ,q}}}$ is the unique supporting functional of u. Use the analogous way as case (I), we have, for any $i\in \mathbb{N}$, $u_n\left(i\right)\to u\left(i\right)$ as $n\to \infty$, and u is a strongly exposed point of $S\left(l_{\Phi ,p}\right)$.

(III) $I_{\Phi }^{p-1}\left(k_{u}u\right)I_{\Psi }\left({\Phi }_{-}^{\prime }\left(k_{u}u\right)\right)<1<I_{\Phi }^{p-1}\left(k_{u}u\right)I_{\Psi }\left({\Phi }_{+}^{\prime }\left(k_{u}u\right)\right)$.

By $\Phi \in {\Delta }_2\left(0\right)$, we have ${\left(l_{\Phi ,p}\right)}^{*}={\left(h_{\Phi ,p}\right)}^{*}=l_{\Psi ,q}$. Thus, there exists $v\in l_{\Psi ,q}$ such that $<u,v>={\parallel u\parallel }_{\Phi ,p}·{\parallel v\parallel }_{\Psi ,q}$, i.e., ${\displaystyle \frac{v}{{\parallel v\parallel }_{\Psi ,q}}}\in Grad\left(u\right)$, and $\theta \left({\displaystyle \frac{v}{{\parallel v\parallel }_{\Psi ,q}}}\right)<1$. By Lemma 6, ${\Phi }_{-}^{\prime }\left(k_{u}u\left(i\right)\right)\le v\left(i\right)\le {\Phi }_{+}^{\prime }\left(k_{u}u\left(i\right)\right)$ for any $i\in \mathbb{N}$, and $I_{\Phi }^{p-1}\left(k_{u}u\right)·I_{\Psi }\left(v\right)=1$.

Set $J_0=\{i\in \mathbb{N}:{\Phi }_{-}^{\prime }\left(k_{u}u\left(i\right)\right)<{\Phi }_{+}^{\prime }\left(k_{u}u\left(i\right)\right)\}$. Then

$$v\left(i\right)=\left\{\begin{array}{cc}{\Phi }_{-}^{\prime }\left(k_{u}u\left(i\right)\right),\hfill & i\notin J_0,\hfill \\ {\Phi }_{-}^{\prime }\left(k_{u}u\left(i\right)\right)+{\epsilon }_i,\hfill & i\in J_0,\hfill \end{array}\right.$$

(6)

where for any $i\in J_0$, ${\Phi }_{-}^{\prime }\left(k_{u}u\left(i\right)\right)+{\epsilon }_i\le {\Phi }_{+}^{\prime }\left(k_{u}u\left(i\right)\right)$. By $I_{\Phi }^{p-1}\left(k_{u}u\right)·I_{\Psi }\left(v\right)=1$, we obtain $1\in K_q\left(v\right)$. For any ${\left\{u_n\right\}}_{n=1}^{\infty }\subset B\left(l_{\Phi ,p}\right)$ with $\langle {\displaystyle \frac{v}{{\parallel v\parallel }_{\Psi ,q}}},u_n\rangle \to 1$ as $n\to \infty$. Take $k_n\in K_p\left(u_n\right)$. By the Young Inequality and the Hölder Inequality, we have

$$\begin{array}{ccc}\hfill 〈{\displaystyle \frac{v}{{\parallel v\parallel }_{\Psi ,q}}},u_n〉& =& {\displaystyle \frac{1}{k_n{\parallel v\parallel }_{\Psi ,q}}}\langle v,k_n{u}_n\rangle \le {\displaystyle \frac{1}{k_n{\parallel v\parallel }_{\Psi ,q}}}(I_{\Psi }\left(v\right)+I_{\Phi }\left(k_n{u}_n\right))\hfill \\ & \le & {\displaystyle \frac{1}{k_n}}{s}_p\left(I_{\Phi }\left(k_n{u}_n\right)\right)·{\displaystyle \frac{1}{{\parallel v\parallel }_{\Psi ,q}}}{s}_q\left(I_{\Psi }\left(v\right)\right)\to 1.\hfill \end{array}$$

as $n\to \infty .$ Thus, for any $i\in \mathbb{N}$,

$$\Phi \left(k_n{u}_n\left(i\right)\right)+\Psi \left(v\left(i\right)\right)-k_n{u}_n\left(i\right)·v\left(i\right)\to 0,$$

(7)

as $n\to \infty$. And

$${(1+I_{\Phi }^p\left(k_n{u}_n\right))}^{{\displaystyle \frac{1}{p}}}·{(1+I_{\Psi }^q\left(v\right))}^{{\displaystyle \frac{1}{q}}}-I_{\Phi }\left(k_n{u}_n\right)-I_{\Psi }\left(v\right)\to 0,$$

as $n\to \infty$, which means

$$I_{\Phi }^{p-1}\left(k_n{u}_n\right)·I_{\Psi }\left(v\right)\to 1=I_{\Phi }^{p-1}\left(k_{u}u\right)·I_{\Psi }\left(v\right),$$

as $n\to \infty$, i.e.,

$$I_{\Phi }\left(k_n{u}_n\right)\to I_{\Phi }\left(k_{u}u\right),$$

(8)

as $n\to \infty$.

If $i\in J_0$, then $v\left(i\right)={\Phi }_{-}^{\prime }\left(k_{u}u\left(i\right)\right)+{\epsilon }_i\le {\Phi }_{+}^{\prime }\left(k_{u}u\left(i\right)\right)$. For any $\delta >0$, $k_n{u}_n\left(i\right)\le k_{u}u\left(i\right)-\delta$ implies

$$\begin{array}{ccc}& & \Phi \left(k_n{u}_n\left(i\right)\right)+\Psi \left(v\left(i\right)\right)-k_n{u}_n\left(i\right)·v\left(i\right)\hfill \\ & \ge & \mathrm{the} \mathrm{area} \mathrm{of} \mathrm{the} figure \mathrm{ABCD} \hfill \\ & \ge & {\epsilon }_i\delta >0\hfill \end{array}$$

(see Figure 6). This contradicts to (7). So $k_n{u}_n\left(i\right)\le k_{u}u\left(i\right)-\delta$ for n large enough and any $\delta >0$.

For any $\delta >0$, $k_n{u}_n\left(i\right)\ge k_{u}u\left(i\right)+\delta$ implies

$$\begin{array}{ccc}& & \Phi \left(k_n{u}_n\left(i\right)\right)+\Psi \left(v\left(i\right)\right)-k_n{u}_n\left(i\right)·v\left(i\right)\hfill \\ & \ge & \mathrm{the} \mathrm{area} \mathrm{of} \mathrm{the} figure \mathrm{AEFG} \hfill \\ & >& ({\Phi }_{+}^{\prime }\left(k_{u}u\left(i\right)\right)-v\left(i\right))\delta >0\hfill \end{array}$$

(see Figure 6). This contradicts to (7). So $k_n{u}_n\left(i\right)\le k_{u}u\left(i\right)+\delta$ for n large enough and any $\delta >0$. Thus, for any $i\in J_0$, $k_n{u}_n\left(i\right)\to k_{u}u\left(i\right)$ as $n\to \infty$. If $i\in \mathbb{N}\setminus J_0$, then $v\left(i\right)={\Phi }_{-}\left(k_{u}u\left(i\right)\right)={\Phi }_{+}\left(k_{u}u\left(i\right)\right)$. We have $k_n{u}_n\left(i\right)\to k_{u}u\left(i\right)$ as $n\to \infty$. Similarly, by (8), we can prove $k_n\to k_u$ as $n\to \infty$. Hence, for any $i\in \mathbb{N}$, $u_n\left(i\right)\to u\left(i\right)$ as $n\to \infty$. By Lemma 9, u is a strongly exposed point of $S\left(l_{\Phi ,p}\right)$. □

 Theorem 5. 

Let $1<p<\infty$, $l_{\Phi ,p}$ has the strongly exposed property if and only if

(i) $\Phi \in {\Delta }_2\left(0\right)$ and $\Psi \in {\Delta }_2\left(0\right)$ (i.e., $\Phi \in {\Delta }_2\left(0\right)\bigcap {\nabla }_2\left(0\right)$).

(ii) Φ is strictly convex on the interval $\left[0,{\pi }_{\Phi ,p}\left({\displaystyle \frac{1}{2}}\right)\right]$.

Proof. 

Sufficiency. By Lemma 8, $l_{\Phi ,p}$ is a locally uniform rotundity space. So $l_{\Phi ,p}$ has the strongly exposed property.

Necessity. $\Phi \in {\Delta }_2\left(0\right)$ follows from Theorem 4. If condition (ii) fails, by Lemma 7, $l_{\Phi ,p}$ is not strictly convex, and therefore it does not have the strongly exposed property either. So (ii) is necessary. If $\Psi \notin {\Delta }_2\left(0\right)$, for any $n>2$, there exist $a_i\searrow 0$ such that $a_i{\Psi }_{+}^{\prime }\left(a_i\right)<{\displaystyle \frac{1}{n^i}}$ and ${\Psi }_{+}^{\prime }\left(\left(1+{\displaystyle \frac{1}{i}}\right)a_i\right)>n^i{\Psi }_{+}^{\prime }\left(a_i\right)$. Choose natural numbers $m_i$ satisfying

$${\displaystyle \frac{1}{n^{i+1}}}\le m_i{a}_i{\Psi }_{+}^{\prime }\left(a_i\right)<{\displaystyle \frac{1}{n^i}}(i=1,2\cdots ),$$

and define

$$y=\left(\stackrel{m_1}{\overbrace{a_1,\cdots ,a_1}},\stackrel{m_2}{\overbrace{a_2,\cdots ,a_2}},\stackrel{m_3}{\overbrace{a_3,\cdots ,a_3}},\cdots \right).$$

Then we have

$$<y,{\Psi }_{+}^{\prime }\left(y\right)>=\sum _{i=1}^{\infty }{m}_i{a}_i{\Psi }_{+}^{\prime }\left(a_i\right)<\sum _{i=1}^{\infty }{\displaystyle \frac{1}{n^i}}={\displaystyle \frac{1}{n-1}}.$$

Whence by $<y,{\Psi }_{+}^{\prime }\left(y\right)>=I_{\Psi }\left(y\right)+I_{\Phi }\left({\Psi }_{+}^{\prime }\left(y\right)\right)$, we get $I_{\Psi }\left(y\right)<{\displaystyle \frac{1}{n-1}}$ and $I_{\Phi }\left({\Psi }_{+}^{\prime }\left(y\right)\right)<{\displaystyle \frac{1}{n-1}}$. Thus, $I_{\Psi }^{q-1}\left(y\right)I_{\Phi }\left({\Psi }_{+}^{\prime }\left(y\right)\right)<1$.

Notice that for any $\lambda >0,$

$$\begin{array}{ccc}& & I_{\Phi }\left({\Psi }_{+}^{\prime }\left((1+\lambda )y\right)\right)+I_{\Psi }\left(y\right)\ge <y,{\Psi }_{+}^{\prime }\left((1+\lambda )y\right)>\hfill \\ & =& \sum _{i=1}^{\infty }{m}_i{a}_i{\Psi }_{+}^{\prime }\left((1+\lambda )a_i\right)\ge \sum _{i\ge {\displaystyle \frac{1}{\lambda }}}{m}_i{a}_i{\Psi }_{+}^{\prime }\left(\left(1+{\displaystyle \frac{1}{i}}\right)a_i\right)\hfill \\ & \ge & \sum _{i\ge {\displaystyle \frac{1}{\lambda }}}{m}_i{a}_i{n}^i{\Psi }_{+}^{\prime }\left(a_i\right)\ge \sum _{i\ge {\displaystyle \frac{1}{\lambda }}}{\displaystyle \frac{1}{n}}=\infty .\hfill \end{array}$$

So we derive that $I_{\Phi }\left({\Psi }_{+}^{\prime }\left((1+\lambda )y\right)\right)=\infty$. We also have that for any $\lambda >0$,

$$\begin{array}{ccc}\hfill I_{\Psi }\left((1+2\lambda )y\right)& =& \sum _{i=1}^{\infty }{m}_i\Phi \left((1+2\lambda )a_i\right)\hfill \\ & \ge & \sum _{i=1}^{\infty }{m}_i{\int }_{(1+\lambda )a_i}^{(1+2\lambda )a_i}{\Psi }_{+}^{\prime }\left(s\right)ds\hfill \\ & \ge & \sum _{i=1}^{\infty }{m}_i\lambda a_i{\Psi }_{+}^{\prime }\left((1+\lambda )a_i\right)\hfill \\ & =& {\displaystyle \frac{\lambda }{1+\lambda }}\sum _{i=1}^{\infty }{m}_i(1+\lambda )a_i{\Psi }_{+}^{\prime }\left((1+\lambda )a_i\right)\hfill \\ & =& {\displaystyle \frac{\lambda }{1+\lambda }}<(1+\lambda )y,{\Psi }_{+}^{\prime }\left((1+\lambda )y\right)>\hfill \\ & \ge & {\displaystyle \frac{\lambda }{1+\lambda }}{I}_{\Phi }\left({\Psi }_{+}^{\prime }\left((1+\lambda )y\right)\right)=\infty .\hfill \end{array}$$

So $I_{\Psi }^{q-1}\left((1+\lambda )y\right)·I_{\Phi }\left({\Psi }_{+}^{\prime }\left((1+\lambda )y\right)\right)=\infty$, which shows $K_q\left(y\right)=\left\{1\right\}$, i.e., ${\parallel y\parallel }_{\Psi ,q}=1+I_{\Psi }\left(y\right)$. Set $x={\Psi }_{+}^{\prime }\left(y\right)$, by Lemma 6, ${\displaystyle \frac{y}{{\parallel y\parallel }_{\Psi ,q}}}$ is the unique supporting functional of x. For any $\epsilon >0$, there exists $n_0\in \mathbb{N}$ such that ${\displaystyle \frac{2}{n-1}}<\epsilon$ for all $n>n_0$, then

$$I_{\Psi }\left((1+\epsilon ){\displaystyle \frac{y}{{\parallel y\parallel }_{\Psi ,q}}}\right)>I_{\Psi }\left(\left(1+{\displaystyle \frac{2}{n-1}}\right){\displaystyle \frac{y}{1+\frac{1}{n-1}}}\right)=I_{\Psi }\left(\left(1+{\displaystyle \frac{1}{n}}\right)y\right)=\infty .$$

This is contradicts to condition (iii) of Theorem 4. So $\Psi \in {\Delta }_2\left(0\right)$ is necessary. □

4. Conclusions

The classical Orlicz and Luxemburg norms generated by an Orlicz function $\Phi$ can be defined with the use of the Amemiya formula: ${\parallel u\parallel }_{\Phi }^o={inf}_{k>0}{\displaystyle \frac{1}{k}}(1+I_{\Phi }\left(ku\right))$ and ${\parallel u\parallel }_{\Phi }={inf}_{k>0}{\displaystyle \frac{1}{k}}\{1,I_{\Phi }\left(ku\right)\}$ respectively (see [15]). Furthermore, the family of p-Amemiya norms ${\parallel u\parallel }_{\Phi ,p}={inf}_{k>0}{\displaystyle \frac{1}{k}}{(1+I_{\Phi }^p\left(ku\right))}^{{\displaystyle \frac{1}{p}}}(1\le p\le \infty )$ (under the convention: ${(1+x^{\infty })}^{{\displaystyle \frac{1}{\infty }}}=\underset{p\to \infty }{lim}{(1+x^p)}^{{\displaystyle \frac{1}{p}}}=max\{1,x\}$ for all $x\ge 0$) were introduced in the same paper. In last few years, a number of papers about the geometrical properties of Orlicz spaces equipped with the p-Amemiya norms have been published. The criteria for the strongly exposed points and strongly exposed property of Orlicz spaces equipped with the classical norms were presented in [9,10,11]. However, the criteria for the strongly exposed points and strongly exposed property of Orlicz sequence spaces equipped with the p-Amemiya norms have not been found. In this paper, Theorems 1, 4 and 5 resolves these issues. The obtained results broaden the knowledge about strongly exposed points and strongly exposed property in Orlicz spaces equipped with the p-Amemiya norms.