The fixed point property of Orlicz sequence spaces equipped with thep-Amemiya norm

Abstract

In this paper, the Opial modulus and the weakly convergent sequence coefficient ofOrlicz space $l_{\varphi ,p}$endowed with the p-Amemiya norm are calculated, the criteria for the uniformOpial property as well as for weakly uniform normal structure of$l_{\varphi ,p}$are presented. It is shown that the Orlicz sequence space equipped with thep-Amemiya norm has the fixed point property if and only if it isreflexive.

MSC: 47H10, 46E30, 46B20.

1 Introduction and preliminaries

The aim of this paper is to present criteria for some important geometric propertiesrelated to the metric fixed point theory in Orlicz sequence spaces.

The Opial property originates from the fixed point theorem proved by Opial in[1]. The uniform Opial property withrespect to the weak topology was defined in [2]and Opial modulus was introduced in [3]. It iswell known that the Opial property and normal structure of a Banach space Xplay an important role in metric fixed point theory for nonexpansive mappings, as wellas in the theory of differential and integral equations (see [1, 4–6]). The Opial property also plays an important role in the study ofweak convergence of iterates, random products of nonexpansive mappings and theasymptotic behavior of nonlinear semigroups [1, 7–9].Moreover, it can be introduced to the open unit ball of a complex Hilbert space,equipped with the hyperbolic metric, where it is useful in proving the existence offixed points of holomorphic self-mappings of X[8].

The coefficient $WCS(X)$ was introduced by Bynum [10], who established their relations with normal structure andcalculated the value of $WCS(l^p)$. A reflexive Banach space X with$WCS(X)>1$ has normalstructure and consequently it has the weakly fixed point property. This is probably oneof the Banach space constants which has been most widely studied, although withconsiderable confusion because there exist many equivalent definitions. Severaldifferent formulae for $WCS(X)$ were found (see [4]), also see the work by Sims and Smyth [11].

The notion of p-Amemiya norm was introduced by Cui and Hudzik in [12], where they showed that the p-Amemiya norm${\parallel \cdot \parallel }_{\mathrm{\Phi },p}$is equivalent to the Orlicz norm ${\parallel \cdot \parallel }_{\mathrm{\Phi }}^{\circ }$ aswell as to the Luxemburg norm ${\parallel \cdot \parallel }_{\mathrm{\Phi }}$. They also illustrated thedescription of extreme points and strongly extreme points in Orlicz spaces equipped withthe p-Amemiya norm [12, 13]. In 2012, they presented the criteria for non-squareness,uniform non-squareness, and locally uniform non-squareness of these spaces[14]. Chen and Cui (see [15, 16]) gave the criteria forcomplex extreme points and complex strict convexity in Orlicz function spaces equippedwith the p-Amemiya norm, and for complex mid-point locally uniform rotundityand complex rotundity of Orlicz sequence spaces equipped with the p-Amemiyanorm.

The rest of the paper is organized as follows. In the first section, some basic notions,terminology and original results are reviewed, which will be used throughout the paper.In Section 2, the Opial modulus of Orlicz space $l_{\varphi ,p}$endowed with the p-Amemiya norm is calculated, and the criteria for the uniformOpial property of $l_{\varphi ,p}$are presented. The weakly convergent sequence coefficient is calculated inSection 3. Finally, the necessary and sufficient condition for fixed pointproperties to exist in $l_{\varphi ,p}$are given.

Let X be a Banach space. We denote by $B(X)$ the unit ball ofX, by $S(X)$ the unit sphere of X. Now we recall somenotions from fixed point theory.

A mapping $T:C\to X$ defined on a subsetC of a Banach space X is said to be non-expansive if$\parallel Tx-Ty\parallel \le \parallel x-y\parallel$ for all$x,y\in C$.

We say that a Banach space X has the fixed point property if for every weaklycompact convex subset $C\subset X$ and forevery nonexpansive $T:C\to C$, T has a fixedpoint of C.

It is known that $L^1[0,1]$ does not have the fixed point property.

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

$$a_{\mathrm{\Phi }}=max\{u\ge 0:\mathrm{\Phi }(u)=0\},b_{\mathrm{\Phi }}=max\{u\ge 0:\mathrm{\Phi }(u)<\mathrm{\infty }\}.$$

A map Φ is said to be an Orlicz function if $\mathrm{\Phi }(0)=0$, Φ is notidentically equal to zero, it is even and convex on the interval $(-b_{\mathrm{\Phi }},b_{\mathrm{\Phi }})$ and left-continuous at $b_{\mathrm{\Phi }}$.

For every Orlicz function Φ, we define its complementary function$\mathrm{\Psi }:\mathbb{R}\to [0,\mathrm{\infty }]$ by theformula

$$\mathrm{\Psi }(v)=sup\{u|v|-\mathrm{\Phi }(u):u\ge 0\}.$$

And the convex modular by $I_{\mathrm{\Phi }}(x)={\sum }_{i=1}^{\mathrm{\infty }}\mathrm{\Phi }(x(i))$ for any$x=(x(i))$.

Definition 1.1[17–19]

The Orlicz sequence space is defined as the set

$$l_{\mathrm{\Phi }}=\{x=(x(i)):I_{\mathrm{\Phi }}(\lambda x)<\mathrm{\infty }\text{ for some }\lambda >0\}.$$

The Luxemburg norm and the Orlicz norm are expressed as

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

and

$${\parallel x\parallel }_{\mathrm{\Phi }}^{\circ }=\underset{k>0}{inf}\frac{1}{k}(1+I_{\mathrm{\Phi }}(kx)),$$

respectively. The Orlicz space equipped with the Luxemburg norm and the Orlicz norm isdenoted by $l_{\mathrm{\Phi }}$ and$l_{\mathrm{\Phi }}^{\circ }$,respectively.

For any $1\le p\le \mathrm{\infty }$ and$u\ge 0$,define

$$s_p(u)=\{\begin{array}{ll}{(1+u^p)}^{\frac{1}{p}}& \text{for }1\le p<\mathrm{\infty },\\ max\{1,u\}& \text{for }p=\mathrm{\infty }\end{array}$$

and define $s_{\mathrm{\Phi },p}(x)=s_p\circ I_{\mathrm{\Phi }}(x)$ for all $1\le p\le \mathrm{\infty }$. Note that thefunctions $s_p$ and$s_{\mathrm{\Phi },p}$are convex. Moreover, the function $s_p$ isincreasing on ${\mathbb{R}}_{+}$ for$1\le p<\mathrm{\infty }$, but the function$s_{\mathrm{\infty }}$ isincreasing on the interval $[1,\mathrm{\infty })$ only.

Definition 1.2[12, 13]

Let $1\le p\le \mathrm{\infty }$. For any$x=(x(i))$, define the p-Amemiya norm by theformula

$${\parallel x\parallel }_{\mathrm{\Phi },p}=\underset{k>0}{inf}\frac{1}{k}{s}_{\mathrm{\Phi },p}(kx).$$

The Orlicz space equipped with the p-Amemiya norm will be denoted by$l_{\mathrm{\Phi },p}$.

It is known that ${\parallel x\parallel }_{\mathrm{\Phi },1}={\parallel x\parallel }_{\mathrm{\Phi }}^{\circ }$ and${\parallel x\parallel }_{\mathrm{\Phi },\mathrm{\infty }}={\parallel x\parallel }_{\mathrm{\Phi }}$. If $1<p<\mathrm{\infty }$ and$x\ne 0$,

$$\frac{1}{2}{\parallel x\parallel }_{\mathrm{\Phi }}^{\circ }\le {\parallel x\parallel }_{\mathrm{\Phi }}\le {\parallel x\parallel }_{\mathrm{\Phi },p}\le 2^{\frac{1}{p}}{\parallel x\parallel }_{\mathrm{\Phi }}<2^{\frac{1}{p}}{\parallel x\parallel }_{\mathrm{\Phi }}^{\circ }$$

(see [12]).

Let $p_{+}$ bethe right-hand side derivative of Φ on $[0,b_{\mathrm{\Phi }})$ and put $p_{+}(b_{\mathrm{\Phi }})={lim}_{u\to b_{\mathrm{\Phi }}^{-}}{p}_{+}(u)$. Define the function ${\alpha }_p:l_{\mathrm{\Phi },p}\to [-1,\mathrm{\infty }]$ by

$${\alpha }_p(x)=\{\begin{array}{ll}{I}_{\mathrm{\Phi }}^{p-1}(x)I_{\mathrm{\Psi }}(p_{+}(|x|))-1,& 1\le p<\mathrm{\infty },\\ -1,& p=\mathrm{\infty },I_{\mathrm{\Phi }}(x)\le 1,\\ I_{\mathrm{\Psi }}(p_{+}(|x|)),& p=\mathrm{\infty },I_{\mathrm{\Phi }}(x)>1\end{array}$$

and the functions $k_p^{\ast }:l_{\mathrm{\Phi },p}\to [0,\mathrm{\infty })$,$k_p^{\ast \ast }:l_{\mathrm{\Phi },p}\to [0,\mathrm{\infty })$ by

$$\begin{array}{c}{k}_p^{\ast }(x)=inf\{k\ge 0:{\alpha }_p(kx)\ge 0\}(\text{with }inf\varphi =\mathrm{\infty }),\hfill \\ k_p^{\ast \ast }(x)=inf\{k\ge 0:{\alpha }_p(kx)\le 0\}.\hfill \end{array}$$

It is obvious that $k_p^{\ast }(x)\le k_p^{\ast \ast }(x)$ for every $1\le p\le \mathrm{\infty }$ and$x\in l_{\mathrm{\Phi },p}$.

Set $K_p(x)=\{0<k<\mathrm{\infty }:k_p^{\ast }(x)\le k\le k_p^{\ast \ast }(x)\}$.

Definition 1.3[17]

We say that an Orlicz function Φ satisfies the ${\mathrm{\Delta }}_2(0)$-condition ($\mathrm{\Phi }\in {\mathrm{\Delta }}_2(0)$, for short) if there exist constants$K\ge 2$ and$u_0>0$such that

$$\mathrm{\Phi }(2u)\le K\mathrm{\Phi }(u)\text{for every }|u|\le u_0.$$

For more details about Orlicz spaces, we refer to [13, 14, 18, 19].

Lemma 1.1[14]

Let$1\le p\le \mathrm{\infty }$.Then$I_{\mathrm{\Phi }}(x)\le {\parallel x\parallel }_{\mathrm{\Phi },p}$forall$x\in L_{\mathrm{\Phi },p}$with${\parallel x\parallel }_{\mathrm{\Phi },p}\le 1$.

Lemma 1.2[17]

If the Orlicz function Φ vanishes only at zeroand$\mathrm{\Phi }\in {\mathrm{\Delta }}_2(0)$, then the norm convergence and the modularconvergence are equivalent.

Lemma 1.3[12]

For every$1\le p\le \mathrm{\infty }$andeach$x\in l_{\mathrm{\Phi },p}\mathrm{\setminus }\{0\}$, the following conditions hold.

• 1. If$k_p^{\ast }(x)=k_p^{\ast \ast }(x)=\mathrm{\infty }$, $K_p(x)=\varphi$, then

  $${\parallel x\parallel }_{\mathrm{\Phi },p}=\underset{k\to \mathrm{\infty }}{lim}\frac{1}{k}{(1+I_{\mathrm{\Phi }}^p(kx))}^{\frac{1}{p}}.$$

• 2. If$k_p^{\ast }(x)<k_p^{\ast \ast }(x)=\mathrm{\infty }$, then the p-Amemiya norm${\parallel x\parallel }_{\mathrm{\Phi },p}$is attained at every$k\in [k_p^{\ast }(x),\mathrm{\infty })$.

• 3. If$k_P^{\ast \ast }(x)<\mathrm{\infty }$, then the p-Amemiya norm${\parallel x\parallel }_{\mathrm{\Phi },p}$is attained at every$k\in [k_p^{\ast }(x),k_p^{\ast \ast }(x)]$.

Lemma 1.4 If the Orlicz function Φ vanishes only at zero,then$l_{\mathrm{\Phi },p}$isorder continuous if and only if$\mathrm{\Phi }\in {\mathrm{\Delta }}_2(0)$.

Lemma 1.5 Assume$\mathrm{\Phi }\in {\mathrm{\Delta }}_2(0)$, $1\le p<\mathrm{\infty }$. Then,for any$L>0$and$\epsilon >0$, thereexists$\delta >0$suchthat

$$I_{\mathrm{\Phi }}(x)\le L,I_{\mathrm{\Phi }}(y)\le \delta \Rightarrow |I_{\mathrm{\Phi }}^p(x+y)-I_{\mathrm{\Phi }}^p(x)|<\epsilon (x,y\in l_{\mathrm{\Phi },p}).$$

Proof Let

$$h=sup\{I_{\mathrm{\Phi }}(2x+2y):I_{\mathrm{\Phi }}(x)\le L,I_{\mathrm{\Phi }}(y)\le 1\}.$$

Then $L<h<\mathrm{\infty }$, since$\mathrm{\Phi }\in {\mathrm{\Delta }}_2(0)$. Without loss of generality, we assume$L>1$ and$\epsilon <1$.Set $\beta =\frac{\epsilon }{h^p}$.Since the modular convergence implies the norm convergence, so we can find$\delta >0$ suchthat $I_{\mathrm{\Phi }}(y)\le \delta$implies ${\parallel y\parallel }_{\mathrm{\Phi },p}\le min\{\frac{\beta }{2},\frac{{\epsilon }^{\frac{1}{p}}{\beta }^{1-\frac{1}{p}}}{2}\}$. Thus, applying theconvexity of Φ and Lemma 1.1, we have

$$\begin{array}{rcl}{I}_{\mathrm{\Phi }}^p(x+y)& =& I_{\mathrm{\Phi }}^p((1-\beta )x+\beta (x+{\beta }^{-1}y))\\ \le & (1-\beta )I_{\mathrm{\Phi }}^p(x)+\beta I_{\mathrm{\Phi }}^p(x+{\beta }^{-1}y)\\ \le & (1-\beta )I_{\mathrm{\Phi }}^p(x)+\frac{\beta }{2}{I}_{\mathrm{\Phi }}^p(2x)+\frac{\beta }{2}{I}_{\mathrm{\Phi }}^p\left(2{\beta }^{-1}y\right)\\ \le & (1-\beta )I_{\mathrm{\Phi }}^p(x)+\frac{\beta }{2}{I}_{\mathrm{\Phi }}^p(2x)+\frac{\beta }{2}{\parallel 2{\beta }^{-1}y\parallel }_{\mathrm{\Phi },p}^p\\ \le & I_{\mathrm{\Phi }}^p(x)+\frac{\beta h^p}{2}+\frac{2^{p-1}}{{\beta }^{p-1}}{\parallel y\parallel }_{\mathrm{\Phi },p}^p\\ \le & I_{\mathrm{\Phi }}^p(x)+\epsilon .\end{array}$$

Replacing x, y by $x+y$,−y, respectively, in the above inequalities, we have

$$I_{\mathrm{\Phi }}^p(x)\le I_{\mathrm{\Phi }}^p(x+y)+\epsilon .$$

 □

Lemma 1.6[20]

$l_{\mathrm{\Phi },p}$isreflexive if and only if$\mathrm{\Phi }\in {\mathrm{\Delta }}_2(0)$and$\mathrm{\Psi }\in {\mathrm{\Delta }}_2(0)$.

Lemma 1.7 Let$1\le p<\mathrm{\infty }$.$l_{\mathrm{\Phi },p}$hasa subspace isomorphic to$l^1$if andonly if$A>0$,where$A={lim}_{u\to 0}\frac{\mathrm{\Phi }(u)}{u}$.

Proof Since the function $\frac{\mathrm{\Phi }(u)}{u}$ isnondecreasing, ${lim}_{u\to 0}\frac{\mathrm{\Phi }(u)}{u}$exists.

If $A>0$, by thecontinuity of $\frac{\mathrm{\Phi }(u)}{u}$, thereexists $A_1>0$such that

$$A|u|\le \mathrm{\Phi }(u)\le A_1|u|(|u|\in [0,{\mathrm{\Phi }}^{-1}(1)]).$$

Then $\mathrm{\Phi }\in {\mathrm{\Delta }}_2(0)$.

For any $x\in l_{\mathrm{\Phi },p}$,we have $I_{\mathrm{\Phi }}(\frac{x}{{\parallel x\parallel }_{\mathrm{\Phi }}})=1$, then$\frac{|x(i)|}{{\parallel x\parallel }_{\mathrm{\Phi }}}\in [0,{\mathrm{\Phi }}^{-1}(1)]$ ($i=1,2,\dots$),therefore,

$$A\frac{|x(i)|}{{\parallel x\parallel }_{\mathrm{\Phi }}}\le \mathrm{\Phi }\left(\frac{|x(i)|}{{\parallel x\parallel }_{\mathrm{\Phi }}}\right)\le A_1\frac{|x(i)|}{{\parallel x\parallel }_{\mathrm{\Phi }}},$$

we have

$$A\sum _{i=1}^{\mathrm{\infty }}\frac{|x(i)|}{{\parallel x\parallel }_{\mathrm{\Phi }}}\le \sum _{i=1}^{\mathrm{\infty }}\mathrm{\Phi }\left(\frac{|x(i)|}{{\parallel x\parallel }_{\mathrm{\Phi }}}\right)\le A_1\sum _{i=1}^{\mathrm{\infty }}\frac{|x(i)|}{{\parallel x\parallel }_{\mathrm{\Phi }}},$$

then $A{\parallel x\parallel }_{l^1}\le {\parallel x\parallel }_{\mathrm{\Phi }}\le A_1{\parallel x\parallel }_{l^1}$.This yields that the $l^1$ norm and theLuxemburg norm are equivalent. Since the p-Amemiya norm and the Luxemburg normare equivalent, $l_{\mathrm{\Phi },p}$has a subspace isomorphic to $l^1$. □

Therefore, if $A>0$, thespace $l_{\mathrm{\Phi },p}$has the Schur property, and it has the Opial property trivially because there is noweakly null sequence $(x_n)$ in $S(l_{\mathrm{\Phi }})$. The case when $A>0$ is notinteresting if we consider the Opial modulus and weakly convergent sequence coefficient.For this reason we will assume that $A=0$ in thefollowing whenever the Opial modulus and the weakly convergent sequence coefficient areconsidered.

2 Opial modulus for Orlicz sequence spaces

In this section we present some results on the Opial modulus. The obtained resultsextend the existing ones, which were presented by a number of papers studying thegeometry of Orlicz spaces endowed with the Luxemburg norm and the Orlicz norm,respectively. A formula for calculating the Opial modulus in Orlicz andMusielak-Orlicz spaces equipped with the Luxemburg or the Orlicz norm is found in[21–26].

Definition 2.1[1]

We say that a Banach space X has the Opial property if for any weakly nullsequence $(x_n)$ in X and any$x\in X\mathrm{\setminus }\{0\}$ there holds

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_n\parallel <\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x+x_n\parallel .$$

Opial proved in [1] that $l^p$($1<p<\mathrm{\infty }$) has thisproperty, but $L^p[0,2\pi ]$ does not have it if$p\in (1,\mathrm{\infty })$,$p\ne 2$.

Definition 2.2[6]

We say that X has the uniform Opial property if for any $\epsilon >0$ thereexists $r>0$ such thatfor any $x\in X$with $\parallel x\parallel \ge \epsilon$and any weakly null sequence $(x_n)$ in the unit sphere $S(X)$ of X there holds

$$1+r\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_n+x\parallel .$$

It is obvious that the uniform Opial property implies the Opial property.

Definition 2.3[2]

The Opial modulus of X is denoted by ${\delta }_o$ and itis defined for $\epsilon \in (0;1]$ by the formula

$${\delta }_o(\epsilon )=inf\{\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_n+x\parallel :(x_n)\subset S(X),x_n\to 0\text{ weakly},\parallel x\parallel =\epsilon \}.$$

It is easy to see that X has the uniform Opial property if and only if${\delta }_o(\epsilon )>1$ for any$\epsilon \in (0,1]$.

Theorem 2.1 If Φ is an Orlicz function,$1\le p<\mathrm{\infty }$,$a_{\mathrm{\Phi }}>0$,then$l_{\mathrm{\Phi },p}$doesnot have the Opial property.

Proof Divide ℕ into a sequence $(N_n)$ of pairwise disjoint and infinite subsetsof ℕ such that ${inf}_n{N}_n\to \mathrm{\infty }$ as $n\to \mathrm{\infty }$ and define

$$x_n=\sum _{i\in N_{n+1}}{a}_{\mathrm{\Phi }}{e}_i,x=\sum _{i\in N_1}{a}_{\mathrm{\Phi }}{e}_i.$$

Then the sequence $(x_n)$ is weakly convergent to zero. For any$k>1$, we have

$$I_{\mathrm{\Phi }}(k{x}_n)=I_{\mathrm{\Phi }}(kx)=I_{\mathrm{\Phi }}(k(x_n+x))=\mathrm{\infty }$$

and for any $k\in (0,1]$,

$$\frac{1}{k}{(1+I_{\mathrm{\Phi }}^p(k{x}_n))}^{\frac{1}{p}}=\frac{1}{k}{(1+I_{\mathrm{\Phi }}^p(kx))}^{\frac{1}{p}}=\frac{1}{k}{(1+I_{\mathrm{\Phi }}^p(k(x_n+x)))}^{\frac{1}{p}}=\frac{1}{k}.$$

So ${\parallel x_n\parallel }_{\mathrm{\Phi },p}={\parallel x\parallel }_{\mathrm{\Phi },p}={\parallel x_n+x\parallel }_{\mathrm{\Phi },p}=1$.Therefore, $l_{\mathrm{\Phi },p}$does not have the Opial property. □

In the following we may assume that $a_{\mathrm{\Phi }}=0$.

Theorem 2.2 Let Φ be an Orlicz functionsatisfying${\mathrm{\Delta }}_2(0)$, $1\le p<\mathrm{\infty }$. Then forany$\epsilon \in (0,1]$, we have

$$\begin{array}{rcl}{\delta }_o(\epsilon )& =& inf\{c_{xyk}>0:I_{\mathrm{\Phi }}^p\left(\frac{kx}{c_{xyk}}\right)+I_{\mathrm{\Phi }}^p\left(\frac{ky}{c_{xyk}}\right)=k^p-1,\\ k>1,{\parallel x\parallel }_{\mathrm{\Phi },p}=1,{\parallel y\parallel }_{\mathrm{\Phi },p}=\epsilon ,x,y\in l_{\mathrm{\Phi },p}\}.\end{array}$$

Proof Set $d(\epsilon )=inf\{c_{xyk}>0:I_{\mathrm{\Phi }}^p(\frac{kx}{c_{xyk}})+I_{\mathrm{\Phi }}^p(\frac{ky}{c_{xyk}})=k^p-1,k>1,{\parallel x\parallel }_{\mathrm{\Phi },p}=1,{\parallel y\parallel }_{\mathrm{\Phi },p}=\epsilon ,x,y\in l_{\mathrm{\Phi },p}\}$.

(1) For fixed $k>1$, consider the function

$$F(c)=I_{\mathrm{\Phi }}^p\left(\frac{kx}{c}\right)+I_{\mathrm{\Phi }}^p\left(\frac{ky}{c}\right).$$

We have F is continuous on ${\mathbb{R}}_{+}$. Since

$$1={\parallel x\parallel }_{\mathrm{\Phi },p}\le \frac{1}{k}{(1+I_{\mathrm{\Phi }}^p(kx))}^{\frac{1}{p}},$$

we have $I_{\mathrm{\Phi }}^p(kx)\ge k^p-1$for any $k>0$, thus

$$F(1)=I_{\mathrm{\Phi }}^p(kx)+I_{\mathrm{\Phi }}^p(ky)>k^p-1.$$

Moreover, ${lim}_{c\to +\mathrm{\infty }}F(c)=0$, then thereexists unique $c_{xyk}\in (1,+\mathrm{\infty })$ such that

$$F(c_{xyk})=I_{\mathrm{\Phi }}^p\left(\frac{kx}{c_{xyk}}\right)+I_{\mathrm{\Phi }}^p\left(\frac{ky}{c_{xyk}}\right)=k^p-1,$$

which shows that the function $d(\epsilon )$ is well defined.

(2) Now, we will show that ${\delta }_o(\epsilon )\le d(\epsilon )$ for any $\epsilon \in (0,1]$. For any $\theta >0$, there exist $x_0,y_0\in X$ with $\parallel x_0\parallel =1$, $\parallel y_0\parallel =\epsilon$ and $k_0>1$ such that $d(\epsilon )+\theta >c_{x_0{y}_0{k}_0}$. Put

$$\begin{array}{c}x=(y_0(1),0,y_0(2),0,y_0(3),0,y_0(4),0,\dots ),\hfill \\ x_n=\sum _{i=0}^{\mathrm{\infty }}{x}_0(i+1)e_{2n+2^{n}i}(n=1,2,\dots ).\hfill \end{array}$$

Then ${\parallel x_n\parallel }_{\mathrm{\Phi },p}={\parallel x_0\parallel }_{\mathrm{\Phi },p}=1$,${\parallel x\parallel }_{\mathrm{\Phi },p}={\parallel y_0\parallel }_{\mathrm{\Phi },p}=\epsilon$,and $x_n(i)\to 0$ as$n\to \mathrm{\infty }$ for any $i\in \mathbb{N}$.

Since ${\sum }_{i=1}^{\mathrm{\infty }}\mathrm{\Phi }(x(i))<\mathrm{\infty }$, there exists $i_0\in \mathbb{N}$ such that${\sum }_{i=i_0+1}^{\mathrm{\infty }}\mathrm{\Phi }(x(i))<\theta$.

For fixed $i_0$,due to ${lim}_{u\to 0}\frac{\mathrm{\Phi }(u)}{u}=0$,we have

$$\begin{array}{rcl}\underset{l\to 0}{lim}\frac{1}{l}\sum _{i=1}^{i_0}\mathrm{\Phi }(lx(i))& =& \underset{l\to 0}{lim}\sum _{i=1}^{i_0}\frac{\mathrm{\Phi }(lx(i))}{l|x(i)|}|x(i)|\\ =& \sum _{i=1}^{i_0}\underset{l\to 0}{lim}\frac{\mathrm{\Phi }(lx(i))}{l|x(i)|}|x(i)|=0.\end{array}$$

So, for any $n\in \mathbb{N}$ and $l\in (0,1)$ which is small enough, we get

$$\begin{array}{rcl}\frac{I_{\mathrm{\Phi }}(l{x}_n)}{l}& =& \frac{I_{\mathrm{\Phi }}(lx)}{l}=\frac{1}{l}(\sum _{i=1}^{i_0}\mathrm{\Phi }(lx(i))+\sum _{i=i_0+1}^{\mathrm{\infty }}\mathrm{\Phi }(lx(i)))\\ \le & \frac{1}{l}\sum _{i=1}^{i_0}\mathrm{\Phi }(lx(i))+\sum _{i=i_0+1}^{\mathrm{\infty }}\mathrm{\Phi }(x(i))<2\theta .\end{array}$$

Then ${lim}_{l\to 0}\frac{I_{\mathrm{\Phi }}(l{x}_n)}{l}=0$.

Take any $f\in l_{\mathrm{\Psi },q}$,there exists $\lambda >0$satisfying $I_{\mathrm{\Psi }}(\lambda f)<\mathrm{\infty }$. Since

$$inf(supp{x}_n)\to \mathrm{\infty }\text{as }n\to \mathrm{\infty },$$

as well as the Young inequality, we have

$$|f(x_n)|=|\sum _{i=1}^{\mathrm{\infty }}{x}_n(i)f(i)|\le \frac{1}{l\lambda }(I_{\mathrm{\Phi }}(l{x}_n)+I_{\mathrm{\Psi }}(\lambda f{\chi }_{supp{x}_n}))\to 0$$

(as $n\to \mathrm{\infty }$). Then we have proved that$x_n\to 0$ weakly.

Since x and $x_n$ havedisjoint supports, we get

$$\begin{array}{rcl}\frac{{\parallel x_n+x\parallel }_{\mathrm{\Phi },p}}{d(\epsilon )+\theta }& \le & \frac{1}{k_0}{(1+I_{\mathrm{\Phi }}^p\left(\frac{k_0(x_n+x)}{d(\epsilon )+\theta }\right))}^{\frac{1}{p}}\\ =& \frac{1}{k_0}{(1+I_{\mathrm{\Phi }}^p\left(\frac{k_0{x}_n}{d(\epsilon )+\theta }\right)+I_{\mathrm{\Phi }}^p\left(\frac{k_{0}x}{d(\epsilon )+\theta }\right))}^{\frac{1}{p}}\\ \le & \frac{1}{k_0}{(1+I_{\mathrm{\Phi }}^p\left(\frac{k_0{x}_n}{c_{x_0{y}_0{k}_0}}\right)+I_{\mathrm{\Phi }}^p\left(\frac{k_{0}x}{c_{x_0{y}_0{k}_0}}\right))}^{\frac{1}{p}}=1,\end{array}$$

hence ${\parallel x_n+x\parallel }_{\mathrm{\Phi },p}\le d(\epsilon )+\theta$. By thearbitrariness of $\theta >0$, wehave ${\delta }_o(\epsilon )\le d(\epsilon )$.

(3) Assume that ${\delta }_o(\epsilon )<d(\epsilon )$ for some $\epsilon \in (0,1]$. Then there exists ${\epsilon }_0>0$ such that ${\delta }_o(\epsilon )\le d(\epsilon )-2{\epsilon }_0$, so there exist $x\in l_{\mathrm{\Phi },p}$ and $(x_n)$ in $S(l_{\mathrm{\Phi },p})$ such that ${\parallel x\parallel }_{\mathrm{\Phi },p}=\epsilon$, $x_n\to 0$ weakly and

$$\underset{n\to \mathrm{\infty }}{lim}{\parallel x_n+x\parallel }_{\mathrm{\Phi },p}<d(\epsilon )-2{\epsilon }_0.$$

For fixed $\frac{{\epsilon }_0}{4}$,there exists $i_0$such that

$${\parallel \sum _{i=i_0+1}^{\mathrm{\infty }}x(i)e_i\parallel }_{\mathrm{\Phi },p}<\frac{{\epsilon }_0}{4},{\parallel \sum _{i=1}^{i_0}{x}_n(i)e_i\parallel }_{\mathrm{\Phi },p}<\frac{{\epsilon }_0}{4}$$

for $n\in \mathbb{N}$ large enough. Set

$${\parallel \sum _{i=1}^{i_0}x(i)e_i\parallel }_{\mathrm{\Phi },p}=a,{\parallel \sum _{i=i_0+1}^{\mathrm{\infty }}{x}_n(i)e_i\parallel }_{\mathrm{\Phi },p}=b_n.$$

And define

$$\begin{array}{c}y=(\frac{\epsilon x(1)}{a},\frac{\epsilon x(2)}{a},\dots ,\frac{\epsilon x(i_0)}{a},0,0,\dots ),\hfill \\ y_n=(0,0,\dots 0,\frac{x_n(i_0+1)}{b_n},\frac{x_n(i_0+2)}{b_n},\dots ,\frac{x_n(i_0+m)}{b_n},\dots ),\hfill \end{array}$$

then ${\parallel y_n\parallel }_{\mathrm{\Phi },p}=1$(for all $n\in \mathbb{N}$), ${\parallel y\parallel }_{\mathrm{\Phi },p}=\epsilon$,$y_n\stackrel{w}{\to }0$.For n large enough, we have

$$\begin{array}{rcl}{\parallel y_n+y\parallel }_{\mathrm{\Phi },p}& \le & {\parallel \sum _{i=i_0+1}^{\mathrm{\infty }}\frac{x_n(i)}{b_n}{e}_i-x_n\parallel }_{\mathrm{\Phi },p}+{\parallel \sum _{i=1}^{i_0}\frac{\epsilon }{a}x(i)e_i-x\parallel }_{\mathrm{\Phi },p}+{\parallel x_n+x\parallel }_{\mathrm{\Phi },p}\\ \le & {\parallel \sum _{i=1}^{i_0}{x}_n(i)e_i\parallel }_{\mathrm{\Phi },p}+|1-b_n|+{\parallel \sum _{i=i_0+1}^{\mathrm{\infty }}x(i)e_i\parallel }_{\mathrm{\Phi },p}\\ +|\epsilon -a|+d(\epsilon )-2{\epsilon }_0\\ \le & 2{\parallel \sum _{i=1}^{i_0}{x}_n(i)e_i\parallel }_{\mathrm{\Phi },p}+2{\parallel \sum _{i=i_0+1}^{\mathrm{\infty }}x(i)e_i\parallel }_{\mathrm{\Phi },p}+d(\epsilon )-2{\epsilon }_0\\ <& d(\epsilon )-{\epsilon }_0.\end{array}$$

Case 1. $A=\{n\in \mathbb{N}:K_{\mathrm{\Phi }}(\frac{y_n+y}{d(\epsilon )})\ne \varphi \}$ is infinite. For any $n\in A$, we take$k_n\in K_{\mathrm{\Phi }}(\frac{y_n+y}{d(\epsilon )})$. Then

$$\begin{array}{rcl}1-\frac{{\epsilon }_0}{d(\epsilon )}& \ge & \frac{{\parallel y_n+y\parallel }_{\mathrm{\Phi },p}}{d(\epsilon )}\\ =& \frac{1}{k_n}{(1+I_{\mathrm{\Phi }}^p\left(\frac{k_n(y_n+y)}{d(\epsilon )}\right))}^{\frac{1}{p}}\\ =& \frac{1}{k_n}{(1+I_{\mathrm{\Phi }}^p\left(\frac{k_n{y}_n}{d(\epsilon )}\right)+I_{\mathrm{\Phi }}^p\left(\frac{k_{n}y}{d(\epsilon )}\right))}^{\frac{1}{p}}\\ \ge & \frac{1}{k_n}{(1+I_{\mathrm{\Phi }}^p\left(\frac{k_n{y}_n}{c_{y_{n}y{k}_n}}\right)+I_{\mathrm{\Phi }}^p\left(\frac{k_{n}y}{c_{y_{n}y{k}_n}}\right))}^{\frac{1}{p}}=1.\end{array}$$

This is a contradiction.

Case 2. $B=\{n\in \mathbb{N}:K_{\mathrm{\Phi }}(\frac{y_n+y}{d(\epsilon )})=\varphi \}$ is infinite. For any $n\in B$,

$$\begin{array}{rcl}1-\frac{{\epsilon }_0}{d(\epsilon )}& \ge & \frac{{\parallel y_n+y\parallel }_{\mathrm{\Phi },p}}{d(\epsilon )}\\ =& \underset{l\to \mathrm{\infty }}{lim}\frac{1}{l}{(1+I_{\mathrm{\Phi }}^p\left(\frac{l(y_n+y)}{d(\epsilon )}\right))}^{\frac{1}{p}}\\ =& \underset{l\to \mathrm{\infty }}{lim}\frac{1}{l}{(1+I_{\mathrm{\Phi }}^p\left(\frac{l{y}_n}{d(\epsilon )}\right)+I_{\mathrm{\Phi }}^p\left(\frac{ly}{d(\epsilon )}\right))}^{\frac{1}{p}}\\ \ge & \underset{l\to \mathrm{\infty }}{lim}\frac{1}{l}{(1+I_{\mathrm{\Phi }}^p\left(\frac{l{y}_n}{c_{y_{n}yl}}\right)+I_{\mathrm{\Phi }}^p\left(\frac{ly}{c_{y_{n}yl}}\right))}^{\frac{1}{p}}=1,\end{array}$$

this is also a contradiction. The two cases have shown that ${\delta }_o(\epsilon )=d(\epsilon )$. □

Remark The main result presented in this paper generalizes the existing result tothe p-Amemiya norm. In the case that $p=1$, the situationdegrades to the case of the classical Orlicz norm.

If $\mathrm{\Phi }\in {\mathrm{\Delta }}_2(0)$, then

$$\begin{array}{rcl}{\delta }_o(\epsilon )& =& inf\{c_{xyk}>0:I_{\mathrm{\Phi }}\left(\frac{kx}{c_{xyk}}\right)+I_{\mathrm{\Phi }}\left(\frac{ky}{c_{xyk}}\right)=k-1,\\ k>1,{\parallel x\parallel }_{\mathrm{\Phi }}^{\circ }=1,{\parallel y\parallel }_{\mathrm{\Phi }}^{\circ }=\epsilon ,x,y\in l_{\mathrm{\Phi }}^{\circ }\}.\end{array}$$

In the following we will consider the uniform Opial property for the Orlicz sequenceequipped with the p-Amemiya norm.

Lemma 2.3[25]

Let X be a Köthe sequence space with the semi-Fatouproperty and without order continuity of the norm.Then X does not have the uniform Opial property. Infact, we even have that for any$\epsilon \in (0,1)$, ${\delta }_o(\epsilon )=1$.

Theorem 2.4 Let$1\le p<\mathrm{\infty }$,$l_{\mathrm{\Phi },p}$hasthe uniform Opial property if and only if$\mathrm{\Phi }\in {\mathrm{\Delta }}_2(0)$.

Proof If $\mathrm{\Phi }\in {\mathrm{\Delta }}_2(0)$ and $l_{\mathrm{\Phi },p}$does not have the uniform Opial property, then there exists some ${\epsilon }_0\in (0,1]$ such that${\delta }_o({\epsilon }_0)=1$. Therefore, for anysequence $(c_n)$ such that $c_n\searrow 1$,there are two sequences $(x_n)$ and $(y_n)$ in $l_{\mathrm{\Phi },p}$with $\parallel x_n\parallel =1$,$\parallel y_n\parallel ={\epsilon }_0$and $k_n>1$such that

$$I_{\mathrm{\Phi }}^p\left(\frac{k_n{x}_n}{c_n}\right)+I_{\mathrm{\Phi }}^p\left(\frac{k_n{y}_n}{c_n}\right)=k_n^p-1.$$

Since

$$\begin{array}{c}\frac{1}{c_n}=\frac{{\parallel x_n\parallel }_{\mathrm{\Phi },p}}{c_n}\le \frac{1}{k_n}{(1+I_{\mathrm{\Phi }}^p\left(\frac{k_n{x}_n}{c_n}\right))}^{\frac{1}{p}},\hfill \\ \frac{{\epsilon }_0}{c_n}=\frac{{\parallel y_n\parallel }_{\mathrm{\Phi },p}}{c_n}\le \frac{1}{k_n}{(1+I_{\mathrm{\Phi }}^p\left(\frac{k_n{y}_n}{c_n}\right))}^{\frac{1}{p}},\hfill \end{array}$$

then

$$I_{\mathrm{\Phi }}^p\left(\frac{k_n{x}_n}{c_n}\right)\ge {\left(\frac{k_n}{c_n}\right)}^p-1,I_{\mathrm{\Phi }}^p\left(\frac{k_n{y}_n}{c_n}\right)\ge {\left(\frac{{\epsilon }_0{k}_n}{c_n}\right)}^p-1.$$

Thus we obtain ${(\frac{k_n}{c_n})}^p-1+{(\frac{{\epsilon }_0{k}_n}{c_n})}^p-1\le k_n^p-1$,then $k_n\le \frac{c_n^p}{1+{\epsilon }_0^p}$,which means that the sequence $(k_n)$ is bounded. Then

$$I_{\mathrm{\Phi }}\left(\frac{k_n{y}_n}{c_n}\right)=k_n^p-1-I_{\mathrm{\Phi }}^p\left(\frac{k_n{x}_n}{c_n}\right)\le k_n^p-1-{\left(\frac{k_n}{c_n}\right)}^p+1\to 0.$$

But ${\parallel \frac{k_n{y}_n}{c_n}\parallel }_{\mathrm{\Phi },p}=\frac{{\epsilon }_0{k}_n}{c_n}>\frac{{\epsilon }_0}{c_n}\to {\epsilon }_0$,according to Lemma 1.2 and $\mathrm{\Phi }\in {\mathrm{\Delta }}_2(0)$, this is a contradiction.

If $\mathrm{\Phi }\notin {\mathrm{\Delta }}_2(0)$, then $l_{\mathrm{\Phi },p}$does not have the order continuity property, by Lemma 2.3, the proof isfinished. □

3 Weakly convergent sequence coefficient for Orlicz sequence spaces

In this section, our main aim is to calculate the weakly convergent sequence coefficientfor an Orlicz sequence space and further discuss the fixed point property of thisspace.

Let X denote a reflexive infinite dimensional Banach space, without Schurproperty automatically. For each sequence $(x_n)$ in X, we define the asymptoticdiameter and asymptotic radius respectively by

$$\begin{array}{c}{dian}_a(x_n)=\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\{\parallel x_n-x_m\parallel :n,m\ge k\},\hfill \\ r_a(x_n)=inf\{\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_n-y\parallel :y\in \overline{conv}(x_n)\}.\hfill \end{array}$$

The weakly convergent sequence coefficient concerned with normal structure is animportant geometric parameter. It was introduced by Bynum [10] as follows.

$WCS(X)$ is the supremum of the set of all numbersM with the property that for each weakly convergent sequence$(x_n)$ with asymptotic diameter A, thereis some y in the closed convex hull of the sequence such that

$$M\underset{n}{lim\hspace{0.17em}sup}\parallel x_n-y\parallel \le A.$$

A sequence $(x_n)$ in X is said to be asymptoticequidistant if

$${dian}_a(x_n)=\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\{\parallel x_i-x_j\parallel :i\ne j,i,j\ge n\}.$$

In this paper, we use the following equivalent definition of $WCS(X)$. This definition was introduced in [27], where it was proved that

$$\begin{array}{rcl}WCS(X)& =& inf\{{dian}_a(x_n):(x_n)\text{ is an asymptotic equidistant sequence in}\\ (X)\text{ and }{x}_n\to 0\text{ weakly}\}.\end{array}$$

It is obvious that $1\le WCS(X)\le 2$ (see[10]). A Banach space X is saidto have weakly uniform normal structure provided $WCS(X)>1$. See[11] for further information about thiscoefficient.

For $p\ge 1$,$WSC(l^p)=2^{\frac{1}{p}}$and $WCS(L^p(\mathrm{\Omega }))=min\{2^{\frac{1}{p}},2^{1-\frac{1}{p}}\}$. A formula for calculatingthe weakly convergent sequences of reflexive Orlicz and Musielak-Orlicz sequence spacesequipped with the Luxemburg or Amemiya norm is found in [21, 23], respectively.

Theorem 3.1 If$\mathrm{\Phi }\in {\mathrm{\Delta }}_2(0)$, $1\le p<\mathrm{\infty }$,then

$$WCS(l_{\mathrm{\Phi },p})=inf\{\underset{k>1}{inf}\{c_{xk}:I_{\mathrm{\Phi }}^p\left(\frac{kx}{c_{xk}}\right)=\frac{k^p-1}{2}\}:x\in S(l_{\mathrm{\Phi },p})\}.$$

Proof Let $d=inf\{{inf}_{k>1}\{c_{xk}:I_{\mathrm{\Phi }}^p(\frac{kx}{c_{xk}})=\frac{k^p-1}{2}\}:x\in S(l_{\mathrm{\Phi },p})\}$. For any $\epsilon >0$, thereexists $x\in S(l_{\mathrm{\Phi },p})$ such that

$$inf\{\underset{k>1}{inf}\{c_{xk}:I_{\mathrm{\Phi }}^p\left(\frac{kx}{c_{xk}}\right)=\frac{k^p-1}{2}\}:x\in S(l_{\mathrm{\Phi },p})\}<d+\epsilon .$$

Then there exist $k>1$ and$c_{xk}<d+\epsilon$such that $I_{\mathrm{\Phi }}^p(\frac{kx}{c_{xk}})=\frac{k^p-1}{2}$.Define

$$\begin{array}{c}{x}_1=(x(1),0,x(2),0,x(3),0,x(4),0,x(5),0,x(6),0,\dots ),\hfill \\ x_2=(0,x(1),0,0,0,x(2),0,0,0,0,0,0,0,x(3),0,0,0,\dots ),\hfill \\ x_3=(0,0,0,x(1),0,0,0,0,0,0,0,x(2),0,0,0,0,0,0,0,\dots ),\hfill \\ \dots ,\hfill \end{array}$$

where $(x_n)$ have pairwise disjoint supports. Then${\parallel x_n\parallel }_{\mathrm{\Phi },p}={\parallel x\parallel }_{\mathrm{\Phi },p}=1$($n\in \mathbb{N}$) for all $n\ne m$ and all$k>1$.According to the same method as in the proof of Theorem 2.2, we have$x_n\to 0$ weakly,

$$\begin{array}{rcl}{\parallel \frac{x_n-x_m}{d+\epsilon }\parallel }_{\mathrm{\Phi },p}& \le & \frac{1}{k}{(1+I_{\mathrm{\Phi }}^p\left(k\frac{x_n-x_m}{d-\epsilon }\right))}^{\frac{1}{p}}\\ =& \frac{1}{k}{(1+2I_{\mathrm{\Phi }}^p\left(\frac{kx}{d+\epsilon }\right))}^{\frac{1}{p}}\\ <& \frac{1}{k}{(1+2I_{\mathrm{\Phi }}^p\left(\frac{kx}{c_{xk}}\right))}^{\frac{1}{p}}\\ =& \frac{1}{k}{(1+k^p-1)}^{\frac{1}{p}}=1.\end{array}$$

Then ${\parallel x^n-x^m\parallel }_{\mathrm{\Phi },p}\le d+\epsilon$,so ${dian}_a(x_n)\le d+\epsilon$,since ε is arbitrary, we have $WCS(l_{\mathrm{\Phi },p})\le d$.

On the other hand, let $\{x_n\}$ in $S(l_{\mathrm{\Phi },p})$ be an arbitrary asymptoticequidistant sequence such that $x_n\to 0$weakly.

Since $\mathrm{\Phi }\in {\mathrm{\Delta }}_2(0)$, then by Lemma 1.5 for$\epsilon >0$,there exists $0<\delta <\epsilon$such that

$$I_{\mathrm{\Phi }}(x)\le 1,k\le \frac{1}{\epsilon }\text{and}{I}_{\mathrm{\Phi }}(y)\le \delta \Rightarrow |I_{\mathrm{\Phi }}^p\left(\frac{k(x+y)}{d}\right)-I_{\mathrm{\Phi }}^p\left(\frac{kx}{d}\right)|<\epsilon .$$

Let $n_1=1$and pick $m_1$such that ${\sum }_{j>m_1}\mathrm{\Phi }(\frac{x_{n_1}(j)}{\epsilon d})<\delta$ and choose$n_2>n_1$such that ${\sum }_{j\le m_1}\mathrm{\Phi }(\frac{x_{n_2}(j)}{\epsilon d})<\delta$. By$x_n(i)\to 0$ as$n\to \mathrm{\infty }$ for $i=1,2,\dots$ ,we can find $m_2$with $m_2>m_1$such that ${\sum }_{j>m_2}\mathrm{\Phi }(\frac{x_{n_2}(j)}{\epsilon d})<\delta$. And so on,by induction, we find the sequence $\{n_i\}$ and $\{m_i\}$ of natural numbers with$n_1<n_2<\cdots$ ,$m_1<m_2<\cdots$satisfying

$$\sum _{j>m_i}\mathrm{\Phi }\left(\frac{x_{n_i}(j)}{\epsilon d}\right)<\delta ,\sum _{j\le m_i}\mathrm{\Phi }\left(\frac{x_{n_i}(j)}{\epsilon d}\right)<\delta .$$

Take $k_{ij}\in K_{\mathrm{\Phi }}(\frac{x_{n_i}-x_{n_j}}{d})$.

• (1) If $k_{ij}\le 1$, then ${\parallel \frac{x_{n_i}-x_{n_j}}{d}\parallel }_{\mathrm{\Phi },p}\ge \frac{1}{k_{ij}}\ge 1$, so $\parallel x_{n_i}-x_{n_j}\parallel \ge d$.

• (2) If $1<k_{ij}\le \frac{1}{\epsilon }$, set $n<m$, then

  $$\begin{array}{c}{\parallel \frac{x_{n_i}-x_{n_j}}{d}\parallel }_{\mathrm{\Phi },p}^p\hfill \\ \ge \frac{1}{k_{ij}^p}(1+{\left(\sum _{l\le m_i}\mathrm{\Phi }\left(\frac{k_{ij}{x}_{n_i}(l)}{d}\right)\right)}^p-\epsilon +{\left(\sum _{l>m_i}\mathrm{\Phi }\left(\frac{k_{ij}{x}_{n_i}(l)}{d}\right)\right)}^p-\epsilon )\hfill \\ \ge \frac{1}{k_{ij}^p}(1+I_{\mathrm{\Phi }}^p\left(\frac{k_{ij}{x}_{n_i}(l)}{d}\right)-\delta -\epsilon +I_{\mathrm{\Phi }}^p\left(\frac{k_{ij}{x}_{n_i}(l)}{d}\right)-\delta -\epsilon )\hfill \\ \ge \frac{1}{k_{ij}^p}(1+I_{\mathrm{\Phi }}^p\left(\frac{k_{ij}{x}_{n_i}(l)}{c_{x_{n_i}{k}_{ij}}}\right)+I_{\mathrm{\Phi }}^p\left(\frac{k_{ij}{x}_{n_i}(l)}{c_{x_{n_i}{k}_{ij}}}\right)-4\epsilon )\hfill \\ =\frac{1}{k_{ij}^p}(1+2\cdot \frac{k_{ij}^p-1}{2}-4\epsilon )>1-4\epsilon .\hfill \end{array}$$

Therefore ${\parallel x_{n_i}-x_{n_j}\parallel }_{\mathrm{\Phi },p}\ge d$.

(3) If $k_{ij}\ge \frac{1}{\epsilon }$, then

$$\begin{array}{rcl}{\parallel \frac{x_{n_i}-x_{n_j}}{d}\parallel }_{\mathrm{\Phi },p}^p& >& \frac{1}{k_{ij}^p}{I}_{\mathrm{\Phi }}^p\left(k_{ij}\frac{x_{n_i}-x_{n_j}}{d}\right)\\ \ge & {\epsilon }^p{I}_{\mathrm{\Phi }}^p\left(\frac{x_{n_i}-x_{n_j}}{\epsilon d}\right)\\ \ge & {\epsilon }^p(I_{\mathrm{\Phi }}^p\left(\frac{x_{n_i}}{\epsilon d}\right)+I_{\mathrm{\Phi }}^p\left(\frac{x_{n_j}}{\epsilon d}\right)-4\epsilon )\\ \ge & {\epsilon }^p(I_{\mathrm{\Phi }}^p\left(\frac{x_{n_i}}{\epsilon c_{x_{n_i}\frac{1}{\epsilon }}}\right)+I_{\mathrm{\Phi }}^p\left(\frac{x_{n_j}}{\epsilon c_{x_{n_j}\frac{1}{\epsilon }}}\right)-4\epsilon )\\ =& {\epsilon }^p({\epsilon }^{-p}-1-4\epsilon )=1-{\epsilon }^p-4{\epsilon }^{p+1},\end{array}$$

hence we get ${\parallel x_{n_i}-x_{n_j}\parallel }_{\mathrm{\Phi },p}\ge d$again. Consequently ${dian}_a(x_n)\ge d$. By thearbitrariness of $\{x_n\}$ in $S(l_{\mathrm{\Phi },p})$, it follows that$WCS(l_{\mathrm{\Phi },p})\ge d$. □

Theorem 3.2 Let$1\le p<\mathrm{\infty }$and$a_{\mathrm{\Phi }}=0$.$l_{\mathrm{\Phi },p}$hasweakly uniform normal structure if and only if$\mathrm{\Phi }\in {\mathrm{\Delta }}_2(0)$.

Proof Necessity. If $\mathrm{\Phi }\in {\mathrm{\Delta }}_2(0)$, by Theorem 3.1,

$$WCS(l_{\mathrm{\Phi },p})=inf\{\underset{k>1}{inf}\{c_{xk}:I_{\mathrm{\Phi }}^p\left(\frac{kx}{c_{xk}}\right)=\frac{k^p-1}{2}\}:x\in S(l_{\mathrm{\Phi },p})\}.$$

Assume $WCS(l_{\mathrm{\Phi },p})=1$, then for any$0<\epsilon <1$there exist $x\in S(l_{\mathrm{\Phi },p})$ and $k>1$ satisfying$c_{xk}<1+\epsilon$.By ${\parallel x\parallel }_{\mathrm{\Phi },p}=1$and $\mathrm{\Phi }\in {\mathrm{\Delta }}_2(0)$, there exists $\delta >0$ such that$I_{\mathrm{\Phi }}(\frac{x}{2})\ge \delta$. Hence,

$$\begin{array}{rcl}1& =& \frac{1}{k^p}(2I_{\mathrm{\Phi }}^p\left(\frac{kx}{c_{xk}}\right)+1)\\ \ge & \frac{1}{k^p}(1+I_{\mathrm{\Phi }}^p\left(\frac{kx}{1+\epsilon }\right))+\frac{1}{k^p}{I}_{\mathrm{\Phi }}^p\left(\frac{kx}{1+\epsilon }\right)\\ \ge & {\parallel \frac{x}{1+\epsilon }\parallel }_{\mathrm{\Phi },p}^p+I_{\mathrm{\Phi }}^p\left(\frac{x}{1+\epsilon }\right)\\ \ge & {\left(\frac{1}{1+\epsilon }\right)}^p+I_{\mathrm{\Phi }}^p\left(\frac{x}{2}\right)\\ \ge & {\left(\frac{1}{1+\epsilon }\right)}^p+{\delta }^p\to 1+{\delta }^p\text{as }\epsilon \to 0.\end{array}$$

This is a contradiction.

Sufficiency. If not, $\mathrm{\Phi }\notin {\mathrm{\Delta }}_2(0)$, then for any $\epsilon >0$ thereexists $0<u<\epsilon$such that

$$\mathrm{\Phi }((1+\epsilon )u)>\frac{1}{\epsilon }\mathrm{\Phi }(u).$$

We can find $m\in \mathbb{N}$ such that

$$1-\mathrm{\Phi }(2\epsilon )<m\mathrm{\Phi }((1+\epsilon )u)\le 1.$$

Take $c>0$satisfying $m\mathrm{\Phi }((1+\epsilon )u)+\mathrm{\Phi }(c)=1$. Then$\mathrm{\Phi }(c)<\mathrm{\Phi }(2\epsilon )$. Set

Then $\frac{x_n}{{\parallel x_n\parallel }_{\mathrm{\Phi },p}}\in S(l_{\mathrm{\Phi },p})$ and $\{\frac{x_n}{{\parallel x_n\parallel }_{\mathrm{\Phi },p}}\}$ is an asymptoticequidistant sequence. And because ${lim\hspace{0.17em}sup}_{l\to 0}\frac{I_{\mathrm{\Phi }}(l{x}_n)}{l}=0$,we have $x_n\to 0$ weakly. Finally,${\parallel x_n\parallel }_{\mathrm{\Phi },p}={\parallel x_1\parallel }_{\mathrm{\Phi },p}\ge {\parallel x_1\parallel }_{\mathrm{\Phi },\mathrm{\infty }}=1$,then

$$\begin{array}{c}{\parallel \frac{1}{1+\epsilon }(\frac{x_n}{\parallel x_n\parallel }-\frac{x_m}{\parallel x_m\parallel })\parallel }_{\mathrm{\Phi },p}^p\hfill \\ \le {\parallel \frac{x_n-x_m}{1+\epsilon }\parallel }_{\mathrm{\Phi },p}^p\le 1+I_{\mathrm{\Phi }}^p\left(\frac{x_n-x_m}{1+\epsilon }\right)\hfill \\ =1+{(2m\mathrm{\Phi }(u)+2\mathrm{\Phi }\left(\frac{c}{1+\epsilon }\right))}^p\hfill \\ \le 1+{(2m\epsilon \mathrm{\Phi }((1+\epsilon )u)+2\mathrm{\Phi }(c))}^p\hfill \\ \le 1+{(2\epsilon +2\mathrm{\Phi }(2\epsilon ))}^p,\hfill \end{array}$$

which implies that

$${dian}_a(x_n)\le (1+\epsilon )(1+{(2\epsilon +2\mathrm{\Phi }(2\epsilon ))}^p),$$

so $WCS(l_{\mathrm{\Phi },p})=1$. □

According to the above proof, we have the following.

Corollary 3.3 Let$1\le p<\mathrm{\infty }$,$a_{\mathrm{\Phi }}=0$and$\mathrm{\Phi }\notin {\mathrm{\Delta }}_2(0)$, then$WCS(l_{\mathrm{\Phi },p})=1$.

Remark In the case that $p=1$, the situationdegrades to the case of classical Orlicz norm.

• 1. If $\mathrm{\Phi }\notin {\mathrm{\Delta }}_2(0)$, then $WCS(l_{\mathrm{\Phi }}^{\circ })=1$.

• 2. If $\mathrm{\Phi }\in {\mathrm{\Delta }}_2(0)$, then

  $$WCS\left(l_{\mathrm{\Phi }}^{\circ }\right)=inf\{\underset{k>1}{inf}\{c_{xk}:I_{\mathrm{\Phi }}\left(\frac{kx}{c_{xk}}\right)=\frac{k-1}{2}\}:x\in S\left(l_{\mathrm{\Phi }}^{\circ }\right)\}.$$

Corollary 3.4

$$WCS\left(l^p\right)=\{\begin{array}{ll}\frac{1}{2^p},& 1\le p<\mathrm{\infty },\\ 1,& p=\mathrm{\infty }.\end{array}$$

Next, we discuss the fixed point property of $l_{\mathrm{\Phi },p}$.

Theorem 3.5$\mathrm{\Phi }\notin {\mathrm{\Delta }}_2(0)$and$a_{\mathrm{\Phi }}=0$,$1\le p\le \mathrm{\infty }$,then$l_{\mathrm{\Phi },p}$containsan asymptotically isometric copy of$c_0$.

Proof If $\mathrm{\Phi }\notin {\mathrm{\Delta }}_2(0)$, then there exist the sequence$\{u_k\}\downarrow 0$ and$\{n_k\}\subset \mathbb{N}$ such that

$$\begin{array}{c}\mathrm{\Phi }(u_k)\le \frac{1}{2^{k+\frac{1}{p}+1}},\frac{1}{2^{k+\frac{1}{p}+1}}\le n_k\mathrm{\Phi }(u_k)\le \frac{1}{2^{k+\frac{1}{p}}},\hfill \\ \mathrm{\Phi }\left((1+\frac{1}{k})u_k\right)>2^{k+\frac{1}{p}+1}\mathrm{\Phi }(u_k)\text{for all }k\in \mathbb{N}.\hfill \end{array}$$

Set

Define $P:c_0\to l_{\mathrm{\Phi },p}$by

$$Pt=\sum _{n=1}^{\mathrm{\infty }}{t}_n{x}_n$$

for $t=(t_1,t_2,\dots )\in c_0$.It is obvious that P is linear.

For any $\lambda >0$,since $t_n\to 0$, then there exists$j_0\in \mathbb{N}$ such that$\lambda |t_i|<1$ for all$i\ge j_0$,hence

$$\begin{array}{rcl}{I}_{\mathrm{\Phi }}(\lambda Pt)& =& \sum _{i=1}^{\mathrm{\infty }}{n}_i\mathrm{\Phi }(\lambda t_i{u}_i)\\ =& \sum _{i=1}^{j_0}{n}_i\mathrm{\Phi }(\lambda t_i{u}_i)+\sum _{i=j_0+1}^{\mathrm{\infty }}{n}_i\mathrm{\Phi }(\lambda t_i{u}_i)\\ \le & \sum _{i=1}^{j_0}{n}_i\mathrm{\Phi }(\lambda t_i{u}_i)+\sum _{i=j_0+1}^{\mathrm{\infty }}\frac{1}{2^{\frac{1}{p}+i}}\\ <& \sum _{i=1}^{j_0}{n}_i\mathrm{\Phi }(\lambda t_i{u}_i)+1<\mathrm{\infty },\end{array}$$

which implies $Pt\in l_{\mathrm{\Phi },p}$.Moreover, for any $t=(t_1,t_2,\dots )\in c_0$,we have

$$I_{\mathrm{\Phi }}\left(\frac{Pt}{{\parallel t\parallel }_{\mathrm{\infty }}}\right)=\sum _{i=1}^{\mathrm{\infty }}{n}_i\mathrm{\Phi }\left(\frac{t_i}{{\parallel t\parallel }_{\mathrm{\infty }}}{u}_i\right)\le \sum _{i=1}^{\mathrm{\infty }}\frac{1}{2^{\frac{1}{p}+i}}=2^{-\frac{1}{p}},$$

then ${\parallel Pt\parallel }_{\mathrm{\Phi },p}\le 2^{\frac{1}{p}}{\parallel Pt\parallel }_{\mathrm{\Phi },\mathrm{\infty }}\le {\parallel t\parallel }_{\mathrm{\infty }}$.

On the other hand, for any $\lambda \in (0,1)$, there exists $j_1\in \mathbb{N}$ such that

$$\frac{(1-{\epsilon }_{j_1})|t_{j_1}|}{\lambda sup\{(1-{\epsilon }_i)|t_i|:i\in \mathbb{N}\}}>1,$$

where ${\epsilon }_i=\frac{1}{i+1}$($i\in \mathbb{N}$), then

$$\frac{|t_{j_1}|}{\lambda sup\{(1-{\epsilon }_i)|t_i|:i\in \mathbb{N}\}}>\frac{1}{1-{\epsilon }_{j_1}}=1+\frac{1}{j_1}$$

and

$$I_{\mathrm{\Phi }}\left(\frac{Pt}{\lambda sup\{(1-{\epsilon }_i)|t_i|:i\in \mathbb{N}\}}\right)>n_{j_1}\mathrm{\Phi }\left((1+\frac{1}{j_1})u_{j_1}\right)>2^{j_1+\frac{1}{p}+1}{n}_{j_1}\mathrm{\Phi }(u_{j_1})>1.$$

Therefore,

$${\parallel Pt\parallel }_{\mathrm{\Phi },p}\ge {\parallel Pt\parallel }_{\mathrm{\Phi },\mathrm{\infty }}\ge sup\{(1-{\epsilon }_i)|t_i|:i\in \mathbb{N}\},$$

which implies that $l_{\mathrm{\Phi },p}$contains an asymptotically isometric copy of $c_0$. □

Theorem 3.6 Let$1\le p\le \mathrm{\infty }$,$a_{\mathrm{\Phi }}=0$,then$l_{\mathrm{\Phi },p}$hasthe fixed point property if and only if it is reflexive.

Proof Since a reflexive Banach space X with $WCS(X)>1$ has the fixedpoint property, we only need to prove the necessity.

Suppose $\mathrm{\Phi }\notin {\mathrm{\Delta }}_2(0)$, then by Theorem 3.5,$l_{\mathrm{\Phi },p}$contains an asymptotically isometric copy of $c_0$. Hence$l_{\mathrm{\Phi },p}$does not have the fixed point property.

Suppose $\mathrm{\Psi }\notin {\mathrm{\Delta }}_2(0)$, then there exists $y\in S(l_{\mathrm{\Psi },q})$ such that$I_{\mathrm{\Psi }}(ly)=\mathrm{\infty }$ for any $l>1$, and for everysequence $\{{\epsilon }_n\}$ decreasing to 0, thereexist $0=i_1<i_2<i_3<\cdots$such that

$${\parallel \sum _{i=i_n+1}^{i_{n+1}}y(i)e_i\parallel }_{\mathrm{\Psi },q}>1-{\epsilon }_n\text{for all }n\in \mathbb{N}.$$

Set $y_n={\sum }_{i=i_n+1}^{i_{n+1}}y(i)e_i$,then there exists $x_n\in S(l_{\mathrm{\Phi },p})$ such that$〈y_n,x_n〉={\parallel y_n\parallel }_{\mathrm{\Psi },q}$.Hence, for any $\alpha =(\alpha (n))\in l^1$,we have

$$\begin{array}{rcl}\sum _{n=1}^{\mathrm{\infty }}|\alpha (n)|& =& \sum _{n=1}^{\mathrm{\infty }}|\alpha (n)|{\parallel x_n\parallel }_{\mathrm{\Phi },p}\ge {\parallel \sum _{n=1}^{\mathrm{\infty }}\alpha (n)x_n\parallel }_{\mathrm{\Phi },p}\\ \ge & 〈\sum _{n=1}^{\mathrm{\infty }}\alpha (n)x_n,\sum _{n=1}^{\mathrm{\infty }}sign(\alpha (n))y_n〉\\ =& \sum _{n=1}^{\mathrm{\infty }}|\alpha (n)|〈x_n,y_n〉\ge (1-{\epsilon }_n)\sum _{n=1}^{\mathrm{\infty }}|\alpha (n)|.\end{array}$$

Hence $l_{\mathrm{\Phi },p}$contains an asymptotically isometric copy of $l^1$. ByTheorem 2 of [28], $l_{\mathrm{\Phi },p}$does not have the fixed point property. □

Theorem 3.7 If$a_{\mathrm{\Phi }}>0$,$1\le p<\mathrm{\infty }$,then$l_{\mathrm{\Phi },p}$doesnot have the fixed point property.

Proof If $a_{\mathrm{\Phi }}>0$, forany $n\in \mathbb{N}$, define , then

$$\begin{array}{c}1+I_{\mathrm{\Phi }}(x_n)=1,\hfill \\ \frac{1}{k}{(1+I_{\mathrm{\Phi }}^p(k{x}_n))}^{\frac{1}{p}}=\frac{1}{k}>1(\mathrm{\forall }k\in (0,1)),\hfill \\ \frac{1}{k}{(1+I_{\mathrm{\Phi }}^p(k{x}_n))}^{\frac{1}{p}}>1(\mathrm{\forall }k>1).\hfill \end{array}$$

Therefore, ${\parallel x_n\parallel }_{\mathrm{\Phi },p}=1$for any $n\in \mathbb{N}$.

Moreover, we can prove that ${\parallel {\sum }_{n=1}^{\mathrm{\infty }}{x}_n\parallel }_{\mathrm{\Phi },p}=1$.Hence, define $P:l^{\mathrm{\infty }}\to l_{\mathrm{\Phi },p}$by

$$Py=\sum _{n=1}^{\mathrm{\infty }}{y}_n{x}_n$$

for all $y=(y_1,\dots ,y_n,\dots )\in l^{\mathrm{\infty }}$. Thenthe operator P is obviously linear, since

$$\begin{array}{rcl}{\parallel y\parallel }_{\mathrm{\infty }}& =& \underset{n}{sup}|y_n|=\underset{n}{sup}|y_n|{\parallel x_n\parallel }_{\mathrm{\Phi },p}\le {\parallel \sum _{n=1}^{\mathrm{\infty }}{y}_n{x}_n\parallel }_{\mathrm{\Phi },p}\\ =& {\parallel Py\parallel }_{\mathrm{\Phi },p}\le \underset{n}{sup}|y_n|{\parallel \sum _{n=1}^{\mathrm{\infty }}{x}_n\parallel }_{\mathrm{\Phi },p}=\underset{n}{sup}|y_n|={\parallel y\parallel }_{\mathrm{\infty }},\end{array}$$

so we have ${\parallel Py\parallel }_{\mathrm{\Phi },p}={\parallel y\parallel }_{\mathrm{\infty }}$, then P is an orderisometry of $l^{\mathrm{\infty }}$ ontoa closed subspace $P(l^{\mathrm{\infty }})$ of $l_{\mathrm{\Phi },p}$. □