On generalized García-Falset coefficient in Musielak-Orlicz sequence spaces

Abstract

We introduce a new geometric coefficient which is related García-Falset coefficient and weak star fixed point property. The García-Falset coefficient that was introduced by Benavides in (Houst. J. Math. 22:835-849, 1996) is calculated in this paper for Musielak-Orlicz sequence spaces equipped with the Luxemburg norm. Specifically, in reflexive Banach spaces, the new geometric coefficient and the García-Falset coefficient are the same.

1 Introduction and preliminaries

Throughout this paper X is a Banach space which is assumed not to have the Schur property, i.e., X has a weakly convergent sequence that is not norm convergent. $S(X)$ and $B(X)$ denote the unit sphere and the unit ball of X, respectively and $l^{0}$ denotes the set of all real sequences.

A Banach space X is said to have the fixed point property (FPP, for short) if every nonexpansive mapping $T :C\rightarrow C$, i.e., the mapping satisfying

$$\Vert Tx-Ty\Vert \leq \Vert x-y\Vert ,\quad \forall x,y\in C, $$

and acting on a nonempty bounded closed and convex subset C of X has a fixed point in C. A natural generalization of FPP is the weak fixed point property (WFPP, for short). A Banach space X is said to have the WFPP whenever it satisfies the above condition from the definition of FPP with ‘weakly compact’ in place of ‘bounded closed’. In 1965 Kirk [2] proved that any reflexive Banach space with normal structure has the FPP. In 1989 Prus [3] introduced a property of a Banach space X, called nearly uniformly smoothness. In 1992 Prus [4] also proved that a weakly nearly uniformly smooth Banach space X with the weak Opial property has the FPP. To obtain the weak fixed point property in Banach spaces, García-Falset introduced in [5] the following coefficient:

$$ R(X)=\sup \Bigl\{ \liminf_{n\rightarrow\infty}\Vert x_{n}-x\Vert:\{ x_{n}\} \subset B(X), {x_{n}} \stackrel{w}{\rightarrow} 0, x\in B(X) \Bigr\} . $$

He has proved that a Banach space X with $R(X)<2$ has the weak fixed point property, i.e., every nonexpansive mapping T from a weakly compact nonempty and convex set $A\subset X$ into itself has a fixed point in A (see [6] and [7]).

A Banach space X is said to be NUS provided that for every $\varepsilon>0$ there is $\eta>0$ such that if $t\in(0,\eta)$ and $\{ x_{n}\}$ is a basic sequence in $B(X)$, then there exists $k>1$ so that $\Vert x_{1}+tx_{k}\Vert \leq1+t\varepsilon$ (see [3]).

A natural generalization of this notion is said to be WNUS. A Banach space X is WNUS whenever it satisfies the above condition with ‘for some $\varepsilon\in(0,1)$’ in place of ‘for every $\varepsilon>0$’ (see [6]).

It is well known that a Banach space X is WNUS if and only if X is reflexive and $R(X)<2$ (see [6]).

The coefficient $R(X,a)$ of a Banach space X was defined in 1996 by Benavides [1], as a generalization of the coefficient $R(X)$, which also plays an important role in the fixed point theory for nonexpansive mappings. Benavides defined the coefficient $R(X,a)$ for a Banach space X as follows.

Definition 1.1

For a given $a\geq0$,

$$ R(X,a)=\sup \Bigl\{ \liminf_{n\rightarrow\infty} \Vert x_{n}-x \Vert :x_{n}\in B(X), x_{n} \stackrel{w}{\rightarrow} 0,D\bigl[(x_{n})\bigr]\leq1,\Vert x\Vert \leq a \Bigr\} , $$

where $D[(x_{n})]={ \liminf_{n\rightarrow\infty} } \{ \Vert x_{i}-x_{j}\Vert :i\neq j,i,j\geq n \}$.

He also defined the following coefficient:

$$M(X)=\sup \biggl\{ \frac{1+a}{R(X,a)}:a\geq0 \biggr\} . $$

Moreover, the coefficient $R(X,a)$ remains unaltered if in the definition we replace lim inf by lim sup. He obtained the following: for a given $a\geq0$, if $R(X,a)<1+a$, then the Banach space X has the fixed point property; the result means that the condition $M(X)>1$ implies that X has the weak fixed point property for nonexpansive mappings [1].

In this paper, we introduce a new geometric coefficient that is a related García-Falset coefficient and a weak^∗ fixed point property, written $R^{\ast} ( X^{\ast},a )$. For a given $a\geq0$, let

$${R^{* }\bigl(X^{* },a\bigr)= \sup\Bigl\{ \lim _{n\rightarrow\infty}\inf\Vert x_{n}-x\Vert:x_{n}\in B \bigl(X^{* }\bigr), x_{n} \stackrel{w^{\ast}}{ \rightarrow} 0, D\bigl[(x_{n})\bigr]\leq1,\Vert x\Vert \leq a, x\in X\Bigr\} } , $$

where $D[(x_{n})]= \lim_{n\rightarrow\infty}\inf \{ \Vert x_{i}-x_{j}\Vert :i\neq j,i,j\geq n \}$.

Similarly to [1], we can prove that if $R^{\ast} ( X^{\ast},a ) <1+a$ for some $a\in(0,1]$ then $T:C\rightarrow C$ has a fixed points $a\in C$, if C is a weak^∗-compact and weak^∗ sequentially complete subset of $X^{*}$ and T is a nonexpansive mapping. It is clear that $R(X,a)=R^{\ast} ( X^{\ast},a )$ if the Banach space X is reflexive.

A Banach space X is called a Köthe sequence space if it is a subspace of $l^{0}$ and for every $x\in l^{0}$ and $y\in X$ satisfying $\vert x(i)\vert \leq \vert y(i)\vert$ for all $i\in \mathcal{N}$, we have $x\in X$ and $\Vert x\Vert \leq \Vert y\Vert$ and if there is a $x=(x(i))\in X$ with $x(i)>0$ for all $i\in\mathcal{N}$ (see [8, 9] and [10]).

Let

$$ X_{a}= \Bigl\{ x\in X:\lim_{n\rightarrow\infty}\bigl\Vert \bigl(0,0, \ldots ,0,x(n+1),x(n+2), \ldots \bigr)\bigr\Vert =0 \Bigr\} . $$

A Köthe sequence space X is said to have an absolutely continuous norm if $X_{a}=X$. A Köthe sequence space X is said to have the semi-Fatou property if for every sequence $\{x_{n}\}\subset X$ and $x\in X$ satisfying $\vert x_{n}(i)\vert \uparrow \vert x(i)\vert$ for all $i\in N$ we have $\Vert x_{n}\Vert \rightarrow \Vert x\Vert$.

A mapping $\Phi:\mathcal{R}\rightarrow\mathcal{R}_{+}$ is said to be an Orlicz function if Φ vanishes only at 0, Φ is even and convex on the whole line R. For every Orlicz function Φ we define its complementary function $\Psi:\mathcal{R}\rightarrow{}[0,\infty)$ by the formula

$$\Psi ( v ) = {\sup_{{u>0}}} \bigl\{ u\vert v\vert -\Phi ( u ) \bigr\} $$

for every $v\in\mathcal{R}$.

Denote by $\Phi= \{ \Phi_{i} \} _{i=1}^{\infty}$ a sequence Orlicz function. Such a sequence is called a Musielak-Orlicz function on $N\times R$.

We say that Φ satisfies the $\delta_{2}$-condition ($\Phi\in \delta_{2}$ for short) if there exist $k>0$, $u_{0}>0$,and $c_{i}\geq 0$, with ${{\sum_{i\geq1}c_{i}}}<\infty$ such that we have the inequality

$$\Phi_{i} ( 2u ) \leq k\Phi_{i} ( u ) +c_{i} \quad \bigl( i\geq1,\Phi_{i} ( u ) \leq u_{0} \bigr). $$

We now introduce a new definition, namely $\overline{\delta}_{2}(k)$. We say that Φ satisfies the $\overline{\delta}_{2}(k)$-condition ($\Phi \in\overline{\delta}_{2}(k)$ for short) if there exist $\varepsilon \in ( 0,1 )$ and $i_{\varepsilon}\in\mathcal{N}$ such that

$$I_{\Phi} \biggl( \frac{x}{2} \biggr) \leq\frac{1-\varepsilon}{2}I_{\Phi } ( x ) $$

whenever $I_{\Phi} ( x ) =k$ and $N(x)\geq i_{\varepsilon}$, where $N(x)=\{i\in\mathcal{N}:x(i)\neq0\}$ and $N(x)\geq i_{\varepsilon}$, which means that $\min \{ i:i\in N(x) \} \geq i_{\varepsilon}$.

Proposition 1.2

The following are equivalent (see [11]):

• (1) $\Phi\in\delta_{2}$;

• (2) for any $\varepsilon>0$, there exist $k_{\varepsilon }>0$, $u_{\varepsilon}>0$, and $c_{i}\geq0$ ($i\geq1$), ${\sum_{i\geq1}}c_{i}<\infty$ such that

  $$\Phi_{i} \biggl( \frac{u}{\varepsilon} \biggr) \leq k_{\varepsilon}\Phi _{i} ( u ) +c_{i}\quad \bigl( i\geq1,\Phi_{i} ( u ) \leq u_{\varepsilon} \bigr) ; $$

• (3) there exist $\varepsilon\in ( 0,1 )$, $i_{\varepsilon}\in M$, $c_{i}\geq0$ ($i\geq1$), ${\sum_{i\geq1}c_{i}}<\infty$ and $v_{\varepsilon}>0$ such that

  $$\Psi_{i} \biggl( \frac{v}{2} \biggr) \leq\frac{1-\varepsilon}{2}\Psi _{i} ( v ) +c_{i} \quad \bigl( i\geq1,\Psi_{i} ( v ) \leq v_{\varepsilon} \bigr) . $$

The Musielak-Orlicz sequence space $l_{\Phi}$ is defined to be the set $\{x\in l_{0}:I_{\Phi}(\lambda x)=\sum_{i=1}^{\infty}\Phi _{i}(\lambda x(i))<\infty\mbox{ for some }\lambda>0\}$ and its subspace $h_{\Phi}$ is defined to be the set $\{x\in l_{0}:I_{\Phi}(\lambda x)=\sum_{i=1}^{\infty}\Phi _{i}(\lambda x(i))<\infty\mbox{ for any }\lambda>0\}$ both equipped with the Luxemburg norm

$$\Vert x\Vert =\inf \biggl\{ k>0:I_{\Phi} \biggl( \frac {x}{k} \biggr) \leq1 \biggr\} . $$

To simplify notations, we put $l_{\Phi}=(l_{\Phi}, \Vert \, \Vert _{\Phi})$ and $h_{\Phi}=(h_{\Phi}, \Vert \, \Vert _{\Phi})$.

We say that a Musielak-Orlicz function Φ satisfies the $\overline{\delta}_{2}$-condition if its complementary function Ψ satisfies the $\delta_{2}$-condition.

The basic information on Musielak-Orlicz spaces can be found in [11–13], and [14].

2 Results

The idea of Theorem 2.1 is similar to Corollary 8.2 in [15]. In order to keep the consistency of this paper, we accept it in the following.

Theorem 2.1

Let X be a Köthe sequence space with the Fatou property. If X has no absolute continuous norm, then $R(X,a)=1+a$.

Proof

Suppose that X does not have an absolutely continuous norm. Then there exists $\varepsilon_{0}>0$ and $x_{0}\in S(X)$ such that

$$\lim_{n\rightarrow\infty}\Biggl\Vert \sum_{i=n+1}^{\infty }x_{0}(i)e_{i} \Biggr\Vert =\varepsilon_{0}, $$

where $e_{i}=(0,0, \ldots , \stackrel{i\mathrm{th}}{1},0, \ldots )$.

Take a sequence of positive numbers $\{\varepsilon_{n}\}$ such that $\varepsilon_{n}\downarrow0$. By $\lim_{n\rightarrow\infty} \Vert \sum_{i=n+1}^{\infty}x_{0}(i)e_{i}\Vert =\varepsilon _{0}$, there exists $n_{1}\in N$ such that

$$\Biggl\Vert \sum_{i=n_{1}}^{\infty}x_{0}(i)e_{i} \Biggr\Vert \leq (1+\varepsilon_{1})\varepsilon_{0}. $$

Notice that

$${\lim_{m\rightarrow\infty} }\Biggl\Vert \sum_{i=n_{1}+1}^{m}x_{0}(i)e_{i} \Biggr\Vert =\varepsilon_{0}, $$

so there exists $n_{2}>n_{1}$ such that

$$(1-\varepsilon_{1})\varepsilon_{0}\leq\Biggl\Vert \sum _{i=n_{1}+1}^{n_{2}}x_{0}(i)e_{i} \Biggr\Vert \leq(1+\varepsilon _{1})\varepsilon_{0}. $$

In this way, we get by induction a sequence $\{n_{i}\}$ of natural numbers such that

$$(1-\varepsilon_{i})\varepsilon_{0}\leq\Biggl\Vert \sum _{j=n_{i}+1}^{n_{i+1}}x_{0}(i)e_{i} \Biggr\Vert \leq (1+\varepsilon_{i})\varepsilon_{0}, \quad i=1,2, \ldots . $$

Put $x_{i}=\sum_{j=n_{i}+1}^{n_{i+1}}x_{0}(i)e_{i}$ and $v_{k}=\sum_{j=n_{k}+1}^{\infty}x_{0}(i)e_{i}$. Then:

(a) $\Vert x_{i}\Vert \rightarrow\varepsilon_{0}$ as $i\rightarrow \infty$.

(b) $x_{i} \stackrel{w}{\rightarrow} 0$ as $i\rightarrow\infty$. It is well known that for any Köthe space X we have

$$X^{*}=X'\oplus S, $$

where S is the space of all singular functionals over X, i.e., functionals which vanish on the subspace $X_{a}=\{x\in X: x \mbox{ has absolutely continuous norm}\}$ and $X'=\{y\in l^{0}:\sum_{i=1}^{\infty}x(i)y(i)<\infty\mbox{ for all }x\in X\}$ (see [11]). This means that every $f\in X^{*}$ is uniquely represented in the form

$$f=T_{y}+\varphi, $$

where $\varphi\in S$ and for $y\in X'$ the function $T_{y}$ is defined by

$$T_{y}(x)= \sum_{i=1}^{\infty}x(i)y(i) $$

for all $x\in X$.

Taking any $y\in X$, we have

$${\lim_{i\rightarrow\infty} }\sum_{j=1}^{\infty}x_{n}(j)y(j)=\lim_{i\rightarrow\infty} \sum _{j=n_{i}+1}^{n_{i+1}}x_{i}(j)y(j)=0. $$

(c) Put $z_{i}=\frac{x_{i}}{\Vert x_{i}\Vert }$ and $w_{k}=\frac{v_{k}}{\Vert v_{k}\Vert }$ for all $i,k\in\mathcal{N}$. It is easy to check $D(z_{k}) \leq1$. Then

$$\begin{aligned}& \liminf_{i\rightarrow\infty} \Vert z_{i}+aw_{k}\Vert \\& \quad = \liminf_{i\rightarrow\infty} \biggl\Vert \frac{x_{i}}{\Vert x_{i}\Vert }+a \frac{v_{k}}{\Vert v_{k}\Vert }\biggr\Vert \\& \quad = \liminf_{i\rightarrow\infty}\frac{1}{\Vert x_{i}\Vert \Vert v_{k}\Vert }\bigl\Vert \Vert v_{k}\Vert x_{i}+a\Vert x_{i}\Vert v_{k}\bigr\Vert \\& \quad \geq\frac{1}{\varepsilon_{0}(1+\varepsilon_{k})\varepsilon_{0}} \liminf_{i\rightarrow\infty} \bigl\Vert \Vert v_{k}\Vert x_{i}+a\Vert x_{i}\Vert v_{k}\bigr\Vert \\& \quad =\frac{1}{\varepsilon_{0}(1+\varepsilon_{kj})\varepsilon_{0}} \liminf_{i\rightarrow\infty}\bigl\Vert \bigl( \Vert v_{k}\Vert +a\Vert x_{i}\Vert \bigr) x_{i}+a\Vert x_{i}\Vert (v_{k}-x_{i}) \bigr\Vert \\& \quad \geq\frac{1}{\varepsilon_{0}(1+\varepsilon_{k})\varepsilon_{0}} \Bigl( \liminf_{i\rightarrow\infty}\bigl( \varepsilon_{0}+a(1-\varepsilon _{i})\varepsilon_{0} \bigr)\Vert x_{i}\Vert - \limsup _{i\rightarrow\infty} \bigl( \Vert v_{k}\Vert -\Vert x_{i}\Vert \bigr) \Vert x_{i}\Vert \Bigr) \\& \quad \geq\frac{1}{\varepsilon_{0}(1+\varepsilon_{k})\varepsilon_{0}} \Bigl( \liminf_{i\rightarrow\infty}\bigl( \varepsilon_{0}+a(1-\varepsilon _{i})\varepsilon_{0} \bigr) (1-\varepsilon_{i})\varepsilon_{0}- \limsup _{i\rightarrow\infty} \bigl( \Vert v_{k}\Vert -\Vert x_{i} \Vert \bigr) (1+\varepsilon_{i})\varepsilon _{0} \Bigr) \\& \quad \geq\frac{1}{\varepsilon_{0}(1+\varepsilon_{k})\varepsilon_{0}} \Bigl( \liminf_{i\rightarrow\infty}\bigl( \varepsilon_{0}+a(1-\varepsilon _{i})\varepsilon_{0} \bigr) (1-\varepsilon_{i})\varepsilon_{0} \\& \qquad {}- \limsup _{i\rightarrow\infty} \bigl( (1+\varepsilon _{k})\varepsilon _{0}-(1-\varepsilon_{i})\varepsilon_{0} \bigr) \varepsilon_{0} \Bigr) \\& \quad =\frac{1}{\varepsilon_{0}(1+\varepsilon_{k})\varepsilon_{0}} \bigl( (\varepsilon_{0}+a \varepsilon_{0})\varepsilon _{0}- \bigl((1+\varepsilon _{k})\varepsilon_{0}-\varepsilon_{0} \bigr) \varepsilon_{0} \bigr) \\& \quad =\frac{1}{\varepsilon_{0}(1+\varepsilon_{k})} \bigl( (\varepsilon _{0}+a\varepsilon _{0})-\varepsilon_{k}\varepsilon_{0} \bigr) \\& \quad =\frac{1}{(1+\varepsilon_{k})} \bigl( (1+a)-\varepsilon_{k} \bigr). \end{aligned}$$

By the arbitrariness of k and $\lim_{{k\rightarrow\infty}}\varepsilon _{k}=0$, we get $R(X,a)\geq1+a$. It is clear that $R(X,a)\leq1+a$. Therefore $R(X,a)=1+a$. □

Corollary 2.2

If $\Phi\notin{\delta}_{2}$ then $R(l_{\Phi },a)=R^{\ast} ( l_{\Phi},a ) =1+a$ for any $0< a\leq1$.

Proof

Since $\Phi\notin\overline{\delta}_{2}$, we see that $l_{\Phi}$ has no absolutely continuous norm. So we have $R(l_{\Phi},a)=1+a$. Since $h_{\Psi}$ is separable and $( h_{\Psi} ) ^{\ast}=l_{\Phi}$, we have $R^{\ast} ( l_{\Phi},a ) =1+a$. □

For any $x\in\ell_{\Phi}$ with $\Vert x\Vert =a$ and $N(x)= \{ i\in N:x(i)\neq0 \}$ being finite, we define $c_{x}$ by the formula

$$ c_{x}={\lim_{n\rightarrow\infty} }\sup\biggl\{ c_{x,y}>0:I_{\Phi } \biggl( \frac{x}{c_{x,y}} \biggr) +I_{\Phi} \biggl( \frac {y}{c_{x,y}} \biggr) =1: { y}\in \ell_{\Phi}, I_{\Phi} ( y ) \leq \frac{1}{2},n\leq N(y)< \infty\biggr\} . $$

Theorem 2.3

Suppose that $\Phi\in{\delta}_{2}$. Then for the Musielak-Orlicz sequence space $\ell_{\Phi}$ we have

$$ R^{\ast} ( \ell_{\Phi},a ) =\sup \bigl\{ c_{x}:x\in \ell _{\Phi}\textit{ with }\Vert x\Vert =a \textit{ and } N(x) \textit{ being finite} \bigr\} . $$

Proof

Let

$$d_{\Phi}=\sup \bigl\{ c_{x}:x\in\ell_{\Phi}\mbox{ with }\Vert x\Vert =a\mbox{ and }N(x)\mbox{ being finite} \bigr\} . $$

Then for any $\varepsilon\in ( 0,d_{\Phi} )$, there exists $\Vert x\Vert =a$ with finite $N(x)$ such that

$$c_{x}>d_{\Phi}-\varepsilon. $$

By the definition of $c_{x}$ there exists $n_{1}\in\mathcal{N}$ such that

$$\sup\biggl\{ c_{x,y}>0:I_{\Phi} \biggl( \frac{x}{c_{x,y}} \biggr) +I_{\Phi } \biggl( \frac{y}{c_{x,y}} \biggr) =1\mbox{ for }I_{\Phi} ( y ) \leq \frac{1}{2}\mbox{ and }N(y)\geq n_{1}\biggr\} > d_{\Phi}-\varepsilon, $$

whenever $n\geq n_{1}$. By the definition of the supremum, there exists $y_{1}\in S(\ell_{\Phi})$ with $N(y_{1})>n_{1}$ such that $c_{x,y_{1}}>d_{\Phi}-\varepsilon$, i.e., $I_{\Phi} ( \frac{x}{d_{\Phi }-\varepsilon} ) +I_{\Phi} ( \frac{y_{1}}{d_{\Phi }-\varepsilon} ) >1$. Hence there exists $n_{2}>n_{1}$ such that $I_{\Phi} ( \frac{x}{d_{\Phi}-\varepsilon} ) +\sum_{i=n_{1}+1}^{n_{2}}\Phi _{i} ( \frac{y_{1} ( i ) }{d_{\Phi}-\varepsilon} ) >1$. Since $n_{2}>n_{1}$, we also have

$$\sup\biggl\{ c_{x,y}>0:I_{\Phi} \biggl( \frac{x}{c_{x,y}} \biggr) +I_{\Phi} \biggl( \frac{y}{c_{x,y}} \biggr) =1\mbox{ for }I_{\Phi} ( y ) \leq\frac{1}{2}\mbox{ and }N(y)\geq n_{2}\biggr\} > d_{\Phi }-\varepsilon. $$

There exists $y_{2}\in\ell_{\Phi}$ with $I_{\Phi} ( y ) \leq \frac{1}{2}$ and $N(y_{2})>n_{1}$ such that $c_{x,y_{2}}>d_{\Phi }-\varepsilon$, i.e., $I_{\Phi} ( \frac{x}{d_{\Phi }-\varepsilon} ) +I_{\Phi} ( \frac{y_{2}}{d_{\Phi}-\varepsilon} ) >1$. Hence there exists $n_{2}>n_{1}$ such that $I_{\Phi} ( \frac {x}{d_{\Phi }-\varepsilon} ) +\sum_{i=n_{1}+1}^{n_{2}}\Phi_{i} ( \frac{y_{2} ( i ) }{d_{\Phi}-\varepsilon} ) >1$. Furthermore, there exists $n_{3}>n_{2}$ such that $I_{\Phi} ( \frac{x}{d_{\Phi }-\varepsilon} ) +\sum_{i=n_{2}+1}^{n_{3}}\Phi_{i} ( \frac{y_{2} ( i ) }{d_{\Phi}-\varepsilon} ) >1$. In such a way, we can prove by induction that there exist a sequence $\{ y_{k } \} _{k=1}^{\infty }\subset\ell_{\Phi}$ with $I_{\Phi} ( y_{k} ) \leq\frac {1}{2}$ for any natural k and a sequence of natural numbers $n_{1}< n_{2 }< n_{3}<\cdots$ such that $I_{\Phi} ( \frac{x}{d_{\Phi}-\varepsilon} ) + \sum_{i=n_{k}+1}^{n_{k+1}}\Phi_{i} ( \frac{y_{k} ( i ) }{d_{\Phi}-\varepsilon} ) >1$ for all $k\in\mathcal{N}$. It is clear that $y_{k}$ is weakly star convergent to 0. Since the supports of $y_{k}$ are pairwise disjoint, we have $I_{\Phi} ( y_{i}-y_{i} ) =I_{\Phi} ( y_{i} ) +I_{\Phi} ( y_{i} ) \leq\frac {1}{2}+\frac{1}{2}=1$ for $i,j\in N$ with $i\neq j$. Therefore, $D[(y_{k})]\leq1$.

For any $k>i_{0}$, we have

$$I_{\Phi}\biggl(\frac{y_{k}-x}{d_{\Phi}-\varepsilon}\biggr)=\sum _{i=n_{k}+1}^{n_{k+1}} \Phi_{i} \biggl( \frac{y_{k} ( i ) }{d_{\Phi}-\varepsilon} \biggr) +I_{\Phi} \biggl( \frac{x ( i ) }{d_{\Phi}-\varepsilon} \biggr) >1, $$

i.e., $\Vert y_{k}-x\Vert >d_{\Phi}-\varepsilon$. Therefore, $R^{\ast} ( \ell_{\Phi},a ) \geq d_{\Phi}-\varepsilon$ and by the arbitrariness of $\varepsilon>0$, we have $R^{\ast} ( \ell _{\Phi},a ) \geq d_{\Phi}$.

Now, we will prove that $R^{\ast} ( \ell_{\Phi},a ) \leq d_{\Phi}$. By the definition of $d_{\Phi}$, we always have

$${\lim_{n\rightarrow\infty} }\sup \biggl\{ c_{x,y}>0:I_{\Phi} \biggl( \frac{x}{c_{x,y}} \biggr) +I_{\Phi} \biggl( \frac{y}{c_{x,y}} \biggr) =1\mbox{ for }y\in S(\ell_{\Phi})\mbox{ and }N(y)\geq n \biggr\} \leq d_{\Phi} $$

for any $\Vert x\Vert =a$ with finite $N(x)$.

First of all, we want to prove that for any weak star null sequence $\{x_{n} \} \subset l_{\Phi}$ and $\varepsilon>0$ there exists a subsequence $\{ x_{n_{i}} \} \subset \{ x_{n} \}$ such that $I_{\Phi} ( x_{n_{i}} ) \leq\frac{1}{2}+\varepsilon$ for each $i\in N$. Otherwise there exists $\varepsilon_{0}>0$. Without loss of generality, we may assume that $I_{\Phi} ( x_{n} ) >\frac {1}{2}+\varepsilon_{0}$ for all $n\in\dot{N}$. By $\Phi\in\delta_{2}$ there exists a $\delta_{1}>0$ such that $\Vert x\Vert >1+5\delta_{1}$ whenever $I_{\Phi} ( x ) >1+\frac{2\varepsilon_{0}}{3}$. Using $\Phi\in\delta_{2}$again, there exists a $\delta_{2}>0$ such that $I_{\Phi} ( x ) <\delta_{2}$ whenever $\Vert x \Vert <\frac{\varepsilon_{0}}{3}$. Set $\delta_{0}=\min \{ \delta_{1}, \delta_{2} \}$.

Put $n_{1}=1$. Then there exists a $i_{1}>1$ such that $\Vert \sum_{i=i_{1}+1}^{\infty}x_{n_{1}}(i)e_{i}\Vert <\delta _{0}\leq\delta_{2}$. Since the sequence $\{ x_{n} \}$ is a weakly null sequence, there exists $n_{2}>n_{1}$ such that

$$\Biggl\Vert \sum_{i=1}^{i_{1}}x_{n}(i)e_{i} \Biggr\Vert < \delta_{0}\leq \delta_{2}\quad \mbox{whenever }n\geq n_{2}. $$

Using $\Phi\in\delta_{2}$ again, we see that there exists a $i_{2}>i_{1}$ such that $\Vert \sum_{i=i_{2}+1}^{\infty}x_{n_{2}}(i)e_{i}\Vert <\delta_{0}\leq\delta_{2}$. Hence

$$I_{\Phi} \Biggl( \sum_{i=1}^{i_{1}}x_{n_{2}}(i)e_{i} \Biggr) < \frac{\varepsilon_{0}}{3}\quad \mbox{and}\quad I_{\Phi} \Biggl( \sum _{i=i_{2}+1}^{\infty }x_{n_{2}}(i)e_{i} \Biggr) < \frac{\varepsilon_{0}}{3}. $$

Therefore $I_{\Phi} ( \sum_{i=i_{1}+1} ^{i_{2}}x_{n_{2}}(i)e_{i} ) >\frac{1}{2}+\frac{\varepsilon_{0}}{3}$.

In such a way, we get a subsequence $\{ x_{n_{j}} \} \subset \{ x_{n} \}$ such that

$$I_{\Phi} \Biggl( \sum_{i=1}^{i_{j-1}}x_{n_{j}}(i)e_{i} \Biggr) < \frac{\varepsilon_{0}}{3},\qquad I_{\Phi} \Biggl( \sum _{i=i_{j}+1}^{\infty }x_{n_{2}}(i)e_{i} \Biggr) < \frac{\varepsilon_{0}}{3} $$

and

$$I_{\Phi } \Biggl( \sum_{i=i_{j-1}+1}^{i_{j}}x_{n_{2}}(i)e_{i} \Biggr) >\frac{1}{2}+ \frac{\varepsilon_{0}}{3} $$

for all $j\in N$.

So

$$\begin{aligned} \Vert x_{n_{k}}-x_{n_{j}}\Vert = & \Biggl\Vert \sum _{i=1}^{i_{k-1}}x_{n_{k}}(i)e_{i}+\sum_{i=i_{k-1}+1}^{i_{k}}x_{n_{k}}(i)e_{i}+ \sum_{i=i_{k}+1}^{\infty }x_{n_{k}}(i)e_{i} \\ &{}- \sum_{i=1}^{i_{j-1}}x_{n_{j}}(i)e_{i}-\sum_{i=i_{j-1}+1}^{i_{j}}x_{n_{j}}(i)e_{i}- \sum_{i=i_{j}+1}^{\infty }x_{n_{j}}(i)e_{i} \Biggr\Vert \\ \geq& \Biggl\Vert \sum_{i=i_{k-1}+1}^{i_{k}}x_{n_{k}}(i)e_{i}-\sum_{i=i_{j-1}+1}^{i_{j}}x_{n_{j}}(i)e_{i} \Biggr\Vert -4\delta_{0}. \end{aligned}$$

By

$$\begin{aligned}& I_{\Phi} \Biggl( \sum_{i=i_{k-1}+1}^{i_{k}}x_{n_{k}}(i)e_{i}-\sum_{i=i_{j-1}+1}^{i_{j}}x_{n_{j}}(i)e_{i} \Biggr) \\& \quad =I_{\Phi} \Biggl( \sum_{i=i_{k-1}+1}^{i_{k}}x_{n_{k}}(i)e_{i} \Biggr) +I_{\Phi} \Biggl( \sum_{i=i_{j-1}+1}^{i_{j}}x_{n_{j}}(i)e_{i} \Biggr) \\& \quad \geq\frac{1}{2}+\frac{\varepsilon_{0}}{3}+\frac{1}{2}+ \frac{\varepsilon _{0}}{3}=1+\frac{2\varepsilon_{0}}{3}, \end{aligned}$$

we have

$$\Biggl\Vert \Biggl( \sum_{i=i_{k-1}+1}^{i_{k}}x_{n_{k}}(i)e_{i}-\sum_{i=i_{j-1}+1}^{i_{j}}x_{n_{j}}(i)e_{i} \Biggr) \Biggr\Vert \geq 1+5\delta_{1}\geq1+5\delta_{0}. $$

Hence

$$\Vert x_{n_{k}}-x_{n_{j}}\Vert \geq\Biggl\Vert \sum _{i=i_{k-1}+1}^{i_{k}}x_{n_{k}}(i)e_{i}-\sum_{i=i_{j-1}+1}^{i_{j}}x_{n_{j}}(i)e_{i} \Biggr\Vert -4\delta_{0}\geq 1+\delta_{0}. $$

This contradicts with the inequality $D[ ( x_{n} ) ]\leq1$.

For convenience, we may assume that $I_{\Phi} ( x_{n} ) \leq \frac{1}{2}+\varepsilon$ for all $n\in N$. Hence

$$I_{\Phi} \biggl( \frac{1-\varepsilon}{1+\varepsilon}x_{n} \biggr) \leq \frac{1-\varepsilon}{1+\varepsilon}I_{\Phi} ( x_{n} ) \leq\frac{1-\varepsilon}{1+\varepsilon} \biggl(\frac{1}{2}+\varepsilon\biggr)=\frac{1}{2} $$

for all $n\in N$.

Let us take an element $x\in l_{\Phi}$ with $\Vert x\Vert =a$. By $\Phi\in\delta_{2}$, for any $\varepsilon>0$ there exists $i_{0}>0$ such that $\Vert \sum_{i=i_{0}+1}^{\infty}x(i)e_{i}\Vert <\varepsilon$. Put $x_{0}=a\frac{\sum_{i=i_{0}+1}^{\infty }x(i)e_{i}}{\Vert \sum_{i=i_{0}+1}^{\infty}x(i)e_{i}\Vert }$. Since ${x_{n}}\stackrel{w^{\ast}}{\rightarrow} 0$, there is a $n_{0}\in N$ such that $\Vert \sum_{i=1}^{i_{0}}x_{n} ( i ) e_{i}\Vert <\varepsilon$ when $n\geq n_{0}$.

Hence

$$\begin{aligned} \Vert x_{n}-x\Vert &\leq\Biggl\Vert \sum _{i=1}^{i_{0}}\bigl(x_{n}(i)-x(i) \bigr)e_{i}+\sum_{i=i_{0}+1}^{\infty} \bigl(x_{n}(i)-x(i)\bigr)e_{i}\Biggr\Vert \\ &\leq\Biggl\Vert \sum_{i=1}^{i_{0}}x(i)e_{i}+ \sum_{i=i_{0}+1}^{\infty }x_{n}(i)e_{i} \Biggr\Vert +2\varepsilon. \end{aligned}$$

We next estimate the $\Vert \sum_{i=1}^{i_{0}}x(i)e_{i}+\sum_{i=i_{0}+1}^{\infty }x_{n}(i)e_{i}\Vert$. Put $z_{n}=\frac{1-\varepsilon }{1+\varepsilon}\sum_{i=i_{0}+1}^{\infty}x_{n}(i)e_{i}$ for $n\geq n_{0}$. We have

$$\begin{aligned}& I_{\Phi} \biggl( \frac{a\frac{\sum_{i=1}^{i_{0}}x(i)e_{i}}{\Vert \sum_{i=1}^{i_{0}}x(i)e_{i}\Vert }-\frac{1-\varepsilon }{1+\varepsilon }\sum_{i=i_{0}+1}^{\infty}x_{n}(i)e_{i}}{d_{\Phi}+\varepsilon} \biggr) \\& \quad =I_{\Phi} \biggl( \frac{x_{0}}{d_{\Phi}+\varepsilon} \biggr) +I_{\Phi } \biggl( \frac{z_{n}}{d_{\Phi}+\varepsilon} \biggr) \\& \quad \leq I_{\Phi} \biggl( \frac{x_{0}}{c_{x_{0},z_{n}}} \biggr) +I_{\Phi } \biggl( \frac{z_{n}}{c_{x_{0},z_{n}}} \biggr) \\& \quad =1, \end{aligned}$$

whence

$$\Biggl\Vert a\frac{\sum_{i=1}^{i_{0}}x(i)e_{i}}{\Vert \sum_{i=1}^{i_{0}}x(i)e_{i}\Vert }-\frac{1-\varepsilon }{1+\varepsilon }\sum _{i=i_{0}+1}^{\infty}x_{n}(i)e_{i}\Biggr\Vert \leq d_{\Phi }+\varepsilon $$

for $n\geq n_{0}$. Therefore, we obtain the inequalities

$$\begin{aligned}& \Biggl\Vert \sum_{i=1}^{i_{0}}x(i)e_{i}- \sum_{i=i_{0}+1}^{\infty }x_{n}(i)e_{i} \Biggr\Vert \\& \quad \leq\Biggl\Vert x_{0}\frac{\Vert \sum_{i=1}^{i_{0}}x(i)e_{i}\Vert }{a}-\frac{1-\varepsilon}{1+\varepsilon}\sum _{i=i_{0}+1}^{\infty}x_{n}(i)e_{i}- \frac {2\varepsilon}{1+\varepsilon}\sum_{i=i_{0}+1}^{\infty}x_{n}(i)e_{i} \Biggr\Vert \\& \quad \leq a \biggl( \frac{\Vert \sum_{i=1}^{i_{0}}x(i)e_{i}\Vert }{a}-1 \biggr) +\biggl\Vert a \frac{\sum_{i=1}^{i_{0}}x(i)e_{i}}{\Vert \sum_{i=1}^{i_{0}}x(i)e_{i}\Vert }-z_{n}\biggr\Vert +\frac{2\varepsilon}{1+\varepsilon}\Biggl\Vert \sum_{i=i_{0}+1}^{\infty }x_{n}(i)e_{i} \Biggr\Vert \\& \quad \leq a \biggl( \frac{a+\varepsilon}{a}-1 \biggr) +\biggl\Vert a \frac{\sum_{i=1}^{i_{0}}x(i)e_{i}}{\Vert \sum_{i=1}^{i_{0}}x(i)e_{i}\Vert }-z_{n}\biggr\Vert +\frac{2\varepsilon}{1+\varepsilon} \\& \quad \leq a\varepsilon+d_{\Phi}+\varepsilon+\frac{2\varepsilon }{1+\varepsilon }=d_{\Phi}+ \biggl( \frac{2+(1+a)(1+\varepsilon)}{1+\varepsilon} \biggr) \varepsilon. \end{aligned}$$

Therefore, we have $\Vert x_{n}-x_{0}\Vert \leq d_{\Phi }+ ( \frac{2+(1+a)(1+\varepsilon)}{1+\varepsilon} ) \varepsilon +2\varepsilon$. By the arbitrariness of $\varepsilon>0$, we get the inequality $R ( \ell _{\Phi} ) \leq d_{\Phi}$.

Summing up, we see that the equality $R ( \ell_{\Phi} ) =d_{\Phi}$ holds. □

For any $x\in\ell_{\Phi}$ with $\Vert x\Vert = 1$ and $N(x)= \{ i\in N:x(i)\neq0 \}$ being finite we define $c_{x}$ as follows:

$$\begin{aligned} \tilde{c}_{x} =&{\lim_{n\rightarrow\infty} }\sup\biggl\{ c_{x,y}>0:I_{\Phi } \biggl( \frac{x}{c_{x,y}} \biggr) +I_{\Phi} \biggl( \frac {y}{c_{x,y}} \biggr) =1\mbox{ for } y\in \ell_{\Phi} \mbox{ with } I_{\Phi} ( y ) \leq 1, \\ & n\leq N(y)< \infty\biggr\} . \end{aligned}$$

Corollary 2.4

If $\Phi\in\delta_{2}$ then $R(l_{\Phi})= \sup \{ \tilde{c}_{x}:x\in\ell_{\Phi}\textit { with }\Vert x\Vert =1 \textit{ and } N(x) \textit{ being finite} \}$.

Proof

The proof is similar to the proof of Theorem 2.3. □

Corollary 2.5

If $\Phi\in\delta_{2}$ and $\Phi\in \overline{\delta}_{2}$ then $R(l_{\Phi},a)=R^{\ast} ( l_{\Phi},a )$ for any $0< a\leq1$.

Theorem 2.6

$R^{\ast}(\ell_{\Phi},1)<2$ if and only if $\ell_{\Phi}\in\delta_{2}$ and $\Phi\in\overline{\delta}_{2}(1)$.

Proof

Necessity. We only need to prove the necessity of $\Phi \in\overline{\delta}_{2}(1)$. Suppose that $\Phi\notin\overline{\delta}_{2}(1)$. Then for any natural number k there exists $x_{k}\in S ( \ell_{\Phi} )$ with $N(x_{k})\geq k$ such that

$$I_{\Phi} \biggl( \frac{x_{k}}{2} \biggr) >\frac{1-\frac{1}{k}}{2}I_{\Phi } ( x_{k} ) . $$

For any fixed $k\in N$ and for any $\varepsilon>0$, there exists $i_{k}\in N$ such that $\sum_{i=k}^{i_{k}}\Phi_{i}(x_{k}(i))>1-\varepsilon$. Put $\overline{x}_{k}=\frac{\sum_{i=k}^{i_{k}}x_{k}(i)e_{i}}{\Vert \sum_{i=k}^{i_{k}}x_{k}(i)e_{i}\Vert }$. Then

$$\begin{aligned} I_{\Phi} \biggl( \frac{\overline{x}_{k}}{2} \biggr) +I_{\Phi} \biggl( \frac {x_{i_{k}}}{2} \biggr) &=I_{\Phi} \biggl( \frac{\sum _{i=k}^{i_{k}}x_{k}(i)e_{i}}{2\Vert \sum_{i=k}^{i_{k}}x_{k}(i)e_{i}\Vert } \biggr) +I_{\Phi} \biggl( \frac{x_{i_{k}}}{2} \biggr) \\ &\geq I_{\Phi} \biggl( \frac{\sum_{i=k}^{i_{k}}x_{k}(i)e_{i}}{2} \biggr) +I_{\Phi} \biggl( \frac{x_{i_{k}}}{2} \biggr) +I_{\Phi} \biggl( \frac{\sum_{i=i_{k}+1}^{\infty}x_{k}(i)e_{i}}{2} \biggr) -\varepsilon \\ &=I_{\Phi} \biggl( \frac{x_{k}}{2} \biggr) +I_{\Phi} \biggl( \frac {x_{i_{k}}}{2} \biggr) -\varepsilon\geq\frac{1-\frac{1}{k}}{2}+ \frac{1-\frac {1}{i_{k}}}{2}-\varepsilon \\ &=1-\frac{1}{k}-\varepsilon, \end{aligned}$$

which shows that $\Vert \overline{x}_{k}+x_{i_{k}}\Vert \geq 2(1-\frac{1}{k}-\varepsilon)$. Hence $c_{\overline{x}_{k},x_{i_{k}}}\geq2(1-\frac {1}{k}-\varepsilon)$. By the arbitrariness of $\varepsilon>0$ and $k\in N$, we see that $R(\ell_{\Phi})=2$.

Sufficiency. Since $\Phi\in\overline{\delta}_{2}(1)$, there exist $\varepsilon_{0}\in ( 0,1 )$ and $i_{\varepsilon}\in \mathcal{N}$ such that

$$I_{\Phi} \biggl( \frac{x}{2} \biggr) \leq\frac{1-\varepsilon _{0}}{2}I_{\Phi } ( x ), $$

whenever $I_{\Phi} ( x ) =1$ and $N(x)\geq i_{\varepsilon}$. For any $x\in S(\ell_{\Phi})$ with finite $N(x)$ and $y\in S(\ell_{\Phi})$ with $N(y)\geq n$, we may assume without loss of generality that $\max \{ i:i\in N(x) \} < n$ and $n\geq i_{\varepsilon}$. Hence

$$I_{\Phi} \biggl( \frac{x}{2} \biggr) +I_{\Phi} \biggl( \frac{y}{2} \biggr) \leq \frac{1}{2}+\frac{1-\varepsilon}{2}= \frac{2-\varepsilon}{2}. $$

Since $\Phi\in\delta_{2}$, there exists $0<\alpha<1$ such that

$$\Vert z\Vert \leq\alpha\quad \mbox{whenever }I_{\Phi} ( z ) \leq \frac{2-\varepsilon}{2}. $$

Therefore $\Vert \frac{x+y}{2}\Vert \leq\alpha$, i.e., $\Vert x+y\Vert \leq2\alpha$. Note that $c_{x,y}=\Vert x+y\Vert$. Hence $R(\ell_{\Phi})\leq2\alpha<2$. □

Theorem 2.7

$R^{\ast}(\ell_{\Phi},a)<1+a$ for $0< a<1$ if and only if $\Phi \in\delta_{2}$.

Proof

We only need to prove the sufficiency. For any $0<\varepsilon<\frac{1}{2}$, by $\Phi\in\delta_{2}$, there exists a $d_{0}>0$ such that $\Vert x\Vert \leq1-d$ whenever $I_{\Phi} ( x ) \leq\frac {1}{2}+\varepsilon$. Hence $\Vert x_{n}\Vert \leq1-d$ if n large enough for any weakly star null sequence $\{ x_{n} \} \subset B ( l_{\Phi} )$ with $D[ ( x_{n} ) ]\leq1$. Hence

$$\liminf_{n\rightarrow\infty} \Vert x_{n}-x\Vert \leq1-d+a< 1+a, $$

that is, $R^{\ast} ( l_{\Phi},a ) <1+a$. □

Example 2.8

Let $\Phi_{n}(u)=\bigl \{\scriptsize{ \begin{array}{l@{\quad}l} u^{2}& \mbox{if }u\leq\frac{1}{n}, \\ a_{n}u+b_{n}& \mbox{if }\frac{1}{n}\leq u\leq\infty, \end{array}} \bigr.$ where $a_{n}=\frac{2}{n}$, $b_{n}=-\frac{1}{n^{2}}$. Then $\Phi _{n}$ is an Orlicz function for each $n\in N$.

If $u\leq\frac{1}{n}$ and $2u>\frac{1}{n}$, then $\frac{1}{2n}< u\leq \frac{1}{n}$. Hence

$$\Phi_{n} ( 2u ) =\frac{2}{n}2u-\frac{1}{n^{2}}\leq \frac {4}{n^{2}}-\frac{1}{n^{2}}=\frac{3}{n^{2}}\leq12 \frac{1}{4n^{2}}\leq 12u^{2}=12\Phi _{n} ( u ) . $$

If $0< u\leq\frac{1}{2n}$ then $\Phi_{n} ( 2u ) =4\Phi _{n} ( u )$. If $\frac{1}{n}< u\leq1$ then $\Phi_{n} ( 2u ) \leq 2\Phi_{n} ( u )$. If we put $K=24$ and $u_{0}=1$, then

$$\Phi_{n}(2u)\leq K\Phi_{n} ( u )\quad \mbox{for all }n\in N, $$

that is, $\Phi\in\delta_{2}$.

If $0< a <1$ then $R^{\ast} ( l_{\Phi},a ) <1+a$.

Let us take $x_{n}= ( 0,0,\ldots,0, \overbrace{\tfrac{n}{2}+\tfrac{1}{2n}}^{n\mathrm{th}},0,\ldots )$ for any $n\in N$. Then $I_{\Phi} ( x_{n} ) =1$ and

$$I_{\Phi} \biggl( \frac{x_{n}}{2} \biggr) =\frac{2}{n} \biggl( \frac {n}{4}+\frac{1}{4n} \biggr) -\frac{1}{n^{2}}= \frac{1}{2}-\frac{1}{2n^{2}}, $$

whence ${\lim_{n\rightarrow\infty} }I_{\Phi} ( \frac{x_{n}}{2} ) =\frac{1}{2}$. Therefore $\Phi\notin\delta_{2} ( 1 )$, which implies that $R^{\ast} ( l_{\Phi},1 ) =2$.

Let $\{ p_{k} \} _{k=1}^{\infty}$ be a sequence of real increasing numbers with $1< p_{1}$ and ${\lim_{n\rightarrow\infty} }p_{n}=p<\infty$. Then we have the following.

Theorem 2.9

Let $l^{(p_{i})}$ be a Nakano sequence space equipped with the Luxemburg norm. Then $R(l^{(p_{i})})=2^{\frac{1}{p}}$ and $R(l_{\Phi},a)= ( \frac{1}{2}+a^{p} ) ^{\frac{1}{p}}$, $0< a\leq1$.

Proof

Since $1<\inf \{ p_{i} \} \leq\sup \{ p_{i} \} =p<\infty$, the Nakano space equipped with the Luxemburg norm is reflexive. For any $x,y\in S ( l^{(p_{i})} )$ with $N(x)$, $N(y)$ being finite. We now consider the following equation:

$$I_{\Phi} \biggl( \frac{x}{c} \biggr) +I_{\Phi} \biggl( \frac{y}{c} \biggr) =1, $$

that is,

$$\sum_{i=1}^{\infty}\biggl\vert \frac{x(i)}{c}\biggr\vert ^{p_{i}}+\sum_{i=1}^{\infty} \biggl\vert \frac{y ( i ) }{c}\biggr\vert ^{p_{i}}=1. $$

Then

$$\frac{1}{c^{p}}\sum_{i=1}^{\infty}\bigl\vert x(i)\bigr\vert ^{p_{i}}+\frac{1}{c^{p}}\sum _{i=1}^{\infty}\bigl\vert x_{n}(i)\bigr\vert ^{p_{i}}=\frac{2}{c^{p}}\leq1, $$

i.e., $c\leq2^{\frac{1}{p}}$. This shows the inequality $R(l^{(p_{i})},1)\leq2^{\frac{1}{p}}$. Take the classical basic sequence $\{ e_{n} \} \subset S(l^{(p_{i})})$. If $c_{n,m}$ is a solution of the equation

$$I_{\Phi} \biggl( \frac{e_{n}}{c} \biggr) +I_{\Phi} \biggl( \frac{e_{m}}{c} \biggr) =1, $$

and assuming without loss of generality that we may take $n>m$, we have the inequality

$$\frac{2}{c^{p_{n}}}\geq1. $$

Hence $R(l^{(p_{i})})\geq2^{\frac{1}{p}}$. Together with the opposite inequality proved already, we have $R(l^{(p_{i})})=2^{\frac{1}{p}}$.

If $0< a<1$, for $x\in B ( l^{(p_{i})} )$ with finite $N(x)$ and $I_{\Phi} ( x ) =\frac{1}{2}$ and $y\in B ( l^{(p_{i})} )$ with finite $N(y)$ and $\Vert x\Vert =a$, we consider the following equation:

$$I_{\Phi} \biggl( \frac{x}{c_{x,y}} \biggr) +I_{\Phi} \biggl( \frac {y}{c_{x,y}} \biggr) =1, $$

that is, the equation

$$\sum_{i=1}^{\infty}\biggl\vert \frac{x(i)}{c_{x,y}}\biggr\vert ^{p_{i}}+\sum _{i=1}^{\infty}\biggl\vert \frac{y ( i ) }{c_{x,y}}\biggr\vert ^{p_{i}}=1. $$

Hence

$$\frac{1}{c_{x,y}^{p}}\sum_{i=1}^{\infty}\bigl\vert x(i)\bigr\vert ^{p_{i}}+\frac{a^{p_{\max \{ i:i\in N(y) \} }}}{c_{x,y}^{p}}\sum_{i=1}^{\infty}\biggl\vert \frac{y(i)}{a}\biggr\vert ^{p_{i}}=\frac {1}{2c_{x,y}^{p}}+ \frac{a^{p_{\max \{ i:i\in N(y) \} }}}{c_{x,y}^{p}}\geq1, $$

where $c_{x,y}^{p}\leq\frac{1}{2}+a^{p_{\max \{ i:i\in N(y) \} }}$. Therefore $R(l_{\Phi},a)\leq ( \frac{1}{2}+a^{p} ) ^{\frac{1}{p}}$.

Taking $x_{n}=(0,\ldots,0, ( \frac{1}{2} ) ^{\frac{1}{p_{n}}},0,\ldots)$ and $x_{m}=(0,\ldots,0, \stackrel{m\mathrm{th}}{a} ,0,\ldots)$, we get $I_{\Phi} ( x_{n} ) =\frac{1}{2}$ and $I_{\Phi} ( \frac{x_{m}}{a} ) =1$, which implies the equality $\Vert x_{m}\Vert =a$.

For any $n\neq m$, if $c>0$ is such that

$$I_{\Phi} \biggl( \frac{x_{n}+x_{m}}{c} \biggr) =1, $$

then

$$\biggl( \frac{1}{2}+a^{p_{m}} \biggr) ^{\frac{1}{\max \{ p_{n},p_{m} \} }}\leq c. $$

Letting $n,m\rightarrow\infty$, we get $R(l_{\Phi},a)\leq ( \frac {1}{2}+a^{p} ) ^{\frac{1}{p}}$, that is, $R(l_{\Phi},a)= ( \frac {1}{2}+a^{p} ) ^{\frac{1}{p}}$. □