Some geometric properties of $N(\lambda ,p)$-spaces

Abstract

In this paper, we introduce the sequence spaces $N(\lambda ,p)$ and we show some geometric properties of that spaces. The main purpose of this paper is to show that $N(\lambda ,p)$ is a Banach space and has the rotund property, the Kadec-Klee property, the uniform Opial property, the $(\beta )$-property, the k-NUC property and the Banach-Saks property of type p.

MSC:46A45, 40C05, 46B45, 46A35.

1 Introduction

By ω, we denote the space of all real valued sequences. Any vector subspace of ω is called a sequence space. We write $l_{\mathrm{\infty }}$, c, and $c_0$ for the spaces of all bounded, convergent and null sequences, respectively. Also by bs, cs, $l_1$, and $l_{\mathrm{\infty }}$, we denote the spaces of all bounded, convergent, absolutely convergent and p-absolutely convergent series, respectively; where $1<p<\mathrm{\infty }$. Assume here and after that $(p_k)$ be a bounded sequence of strictly positive real numbers with $sup\{p_k\}=H$ and $M=max\{1,H\}$. Then, the linear space $l(p)$ was defined by Maddox [1] (see also Simons [2] and Nakano [3]) as follows:

$$l(p)=\{x=(x_k)\in \omega :\sum _k{|x_k|}^{p_k}<\mathrm{\infty }\}(0<p_k\le H<\mathrm{\infty })$$

which is complete paranormed space paranormed by

$$g(x)={\left(\sum _k{|x_k|}^{p_k}\right)}^{\frac{1}{M}}.$$

For simplicity in notation, here and in what follows, the summation without limits runs from 1 to ∞.

In [4] was introduced the following numerical sequence $\lambda ={({\lambda }_k)}_{k=0}^{\mathrm{\infty }}$, which is a strictly increasing sequence of positive real numbers tending to infinity, as $k\to \mathrm{\infty }$, that is

$$0<{\lambda }_0<{\lambda }_1<\cdots \text{and}{\lambda }_k\to \mathrm{\infty }\text{as }k\to \mathrm{\infty }.$$

We will introduce the following sequence space:

$$N(\lambda ,p)=\{x=(x_n)\in \omega :\sum _{k=1}^{\mathrm{\infty }}{\left(\frac{1}{{\lambda }_k}\sum _{i=1}^k\right|({\lambda }_i-{\lambda }_{i-1})x_i\left|\right)}^{p_k}<\mathrm{\infty }\}.$$

For ${\lambda }_k=k$, we obtain the Cesaro sequence space $ces(p)$ (see [5]). If ${\lambda }_k=k$ and $p_k=p$, then $N(\lambda ,p)={ces}_p$ (see [6]). In case where $p_k=p$ for all $k\in \mathbb{N}$, then we will denote $N(\lambda ,p)=N_p$. Some results related to the geometric properties of sequence spaces are given in [7–9].

2 Topological properties

Theorem 2.1 The paranorm on $N(\lambda ,p)$ is given by the relation

$$h(x)={\left(\sum _{k=1}^{\mathrm{\infty }}{\left(\frac{1}{{\lambda }_k}\sum _{i=1}^k\right|({\lambda }_i-{\lambda }_{i-1})x_i\left|\right)}^{p_k}\right)}^{\frac{1}{M}},$$

where $M=max\{1,H\}$ and $H=sup{p}_k$.

3 Geometrical properties

In this section we will show some geometric properties of the $N(\lambda ,p)$-spaces, such as the $(\beta )$-property, the k-NUC property, the Banach-Saks property of type p, and the $(H)$-property. It is well known that these properties are most important in Banach spaces (see [10, 11] and [1]).

Definition 3.1 A Banach space X is said to be k-nearly uniformly convex (k-NUC) if for any $ϵ>0$, there exists a $\delta >0$ such that for any sequence $(x_n)\subset B(X)$ with $sep(x_n)\ge ϵ$, there are $n_1,n_2,\dots ,n_k\in \mathbb{N}$ such that

$$\parallel \frac{x_{n_1}+x_{n_2}+\cdots +x_{n_k}}{k}\parallel <1-\delta ,$$

where $sep(x_n)=inf\{\parallel x_n-x_m\parallel :n\ne m\}>ϵ$.

Definition 3.2 A Banach space X has property $(\beta )$ if and only if for each $r>0$ and $ϵ>0$, there exists a $\delta >0$ such that for each element $x\in B(X)$ and each sequence $x_n\in B(X)$ with $sep(x_n)\ge ϵ$, there is an index k for each

$$\parallel \frac{x+x_k}{2}\parallel \le \delta .$$

Definition 3.3 A Banach space X is said to have the Banach-Saks property type p if every weakly null sequence $(x_k)$ has a subsequence $(x_{kl})$ such that for some $C>0$

$$\parallel \sum _{l=0}^n{x}_{kl}\parallel <C{(n+1)}^{\frac{1}{p}}$$

for all $n\in \mathbb{N}$.

Definition 3.4 Let X be a real vector space. A functional $\sigma :X\to [0,\mathrm{\infty })$ is called a modular if

• (1) $\sigma (x)=0$ if and only if $x=\theta$,

• (2) $\sigma (\alpha x)=\sigma (x)$ for all scalars α with $|\alpha |=1$,

• (3) $\sigma (\alpha x+\beta y)\le \sigma (x)+\sigma (y)$ for all $x,y\in X$ and $\alpha ,\beta >0$ with $\alpha +\beta =1$,

• (4) the modular σ is called convex if $\sigma (\alpha x+\beta y)\le \alpha \sigma (x)+\beta \sigma (y)$ for all $x,y\in X$ and $\alpha ,\beta >0$, with $\alpha +\beta =1$.

A modular σ is called:

• (5) right continuous if ${lim}_{\alpha \to 1^{+}}\sigma (\alpha x)=\sigma (x)$ for all $x\in X_{\sigma }$,

• (6) left continuous if ${lim}_{\alpha \to 1^{-}}\sigma (\alpha x)=\sigma (x)$ for all $x\in X_{\sigma }$,

• (7) continuous if it is both right and left continuous,

where $X_{\sigma }=\{x\in X:{lim}_{\alpha \to 0^{+}}\sigma (\alpha x)=0\}$. We define ${\sigma }_p$ on $N(\lambda ,p)$ as follows:

$${\sigma }_p(x)=\left(\sum _{k=1}^{\mathrm{\infty }}{\left(\frac{1}{{\lambda }_k}\sum _{i=0}^k\right|({\lambda }_i-{\lambda }_{i-1})x_i\left|\right)}^{p_k}\right),$$

where ${\lambda }_{-1}=0$.

If $p_k\ge 1$, for all $k\in \mathbb{N}$, by the convexity of the function $t\to {|t|}^{p_k}$, for all $k\in \mathbb{N}$, ${\sigma }_p$ defined above is a modular convex in the $N(\lambda ,p)$.

Definition 3.5 A modular ${\sigma }_p$ is said to satisfy the ${\delta }_2$-conditions if for every $ϵ>0$, there exist constant $M>0$ and $m>0$ such that

$${\sigma }_p(2t)\le M{\sigma }_p(t)+ϵ$$

(3.1)

for all $t\in X_{{\sigma }_p}$ with ${\sigma }_p(t)\le m$.

Lemma 3.6 ([12])

If ${\sigma }_p$ satisfies the ${\delta }_2$-conditions, then for any $A>0$ and $ϵ>0$, there exists $\delta >0$ such that

$$|{\sigma }_p(t+w)-{\sigma }_p(t)|<ϵ$$

(3.2)

whenever $t,w\in X_{{\sigma }_p}$ with ${\sigma }_p(t)\le A$ and ${\sigma }_p(w)\le \delta$.

Theorem 3.7 ([12])

• (1) If ${\sigma }_p$ satisfies the ${\delta }_2$-conditions, then for any $x\in X_{{\sigma }_p}$, $\parallel x\parallel =1$ if and only if ${\sigma }_p(x)=1$.

• (2) If ${\sigma }_p$ satisfies the ${\delta }_2$-conditions, then for any sequence $(x_n)\in X_{{\sigma }_p}$, $\parallel x_n\parallel \to 0$ if and only if ${\sigma }_p(x_n)\to 0$.

Theorem 3.8 If ${\sigma }_p$ satisfies the ${\delta }_2$-conditions, then for any $ϵ\in (0,1)$, there exists $\delta \in (0,1)$ such that ${\sigma }_p(x)\le 1-ϵ$ implies $\parallel x\parallel \le 1-\delta$.

Proof The proof of the theorem follows directly from the above two facts. □

Theorem 3.9 For any $x\in N(\lambda ,p)$ and $ϵ\in (0,1)$, there exists $\delta \in (0,1)$, such that ${\sigma }_p(x)\le 1-ϵ$ implies $\parallel x\parallel \le 1-\delta$.

Proof The proof of the theorem follows directly from Theorem 3.8. □

Proposition 3.10 If $p_k\ge 1$, for all $k\in \mathbb{N}$, then the modular function ${\sigma }_p$, on $N(\lambda ,p)$, satisfies the following conditions:

• (1) If $0<\alpha \le 1$, then ${\alpha }^M{\sigma }_p(\frac{x}{\alpha })\le {\sigma }_p(x)$ and ${\sigma }_p(\alpha x)\le \alpha {\sigma }_p(x)$.

• (2) If $\alpha \ge 1$, then ${\sigma }_p(x)\le {\alpha }^M{\sigma }_p(\frac{x}{\alpha })$.

• (3) If $\alpha \ge 1$, then ${\sigma }_p(x)\le \alpha {\sigma }_p(\frac{x}{\alpha })$.

• (4) The modular function ${\sigma }_p(x)$ is continuous on $N(\lambda ,p)$.

Proof The proof of the proposition is similar to Proposition 2.1 in [13]. □

Now we will define the following two norms (the first is known as the Luxemburg norm and the second as the Amemiya norm) in $N(\lambda ,p)$:

$${\parallel x\parallel }_L=inf\{\alpha >0:{\sigma }_p\left(\frac{x}{\alpha }\right)\le 1\}$$

(3.3)

and

$${\parallel x\parallel }_A=\underset{\alpha >0}{inf}\frac{1}{\alpha }\{1+{\sigma }_p(\alpha \cdot x)\}.$$

(3.4)

Proposition 3.11 Let $x\in N(\lambda ,p)$. Then the following relations are satisfied:

• (1) If ${\parallel x\parallel }_L<1$, then ${\sigma }_p(x)\le {\parallel x\parallel }_L$.

• (2) If ${\parallel x\parallel }_L>1$, then ${\sigma }_p(x)\ge {\parallel x\parallel }_L$.

• (3) ${\parallel x\parallel }_L=1$ if and only if ${\sigma }_p(x)=1$.

• (4) ${\parallel x\parallel }_L<1$ if and only if ${\sigma }_p(x)<1$.

• (5) ${\parallel x\parallel }_L>1$ if and only if ${\sigma }_p(x)>1$.

Proof (1) Let $x\in N(\lambda ,p)$ and ${\parallel x\parallel }_L<1$. Let also $ϵ>0$ such that $0<ϵ<1-{\parallel x\parallel }_L$. On the other hand from the definition of the norm by relation (3.3) we find that there exists a $\alpha >0$ such that ${\parallel x\parallel }_L+ϵ>\alpha$ and ${\sigma }_p(\frac{x}{\alpha })\le 1$. From the above relations and property (1) of Proposition 3.10, we obtain

$$\frac{{\parallel x\parallel }_L+ϵ}{\alpha }>1$$

and

$${\sigma }_p(x)\le \frac{{\parallel x\parallel }_L+ϵ}{\alpha }{\sigma }_p(x)=\frac{{\parallel x\parallel }_L+ϵ}{\alpha }{\sigma }_p(\alpha \cdot \frac{x}{\alpha })\le \frac{{\parallel x\parallel }_L+ϵ}{\alpha }\cdot \alpha \cdot {\sigma }_p\left(\frac{x}{\alpha }\right)\le {\parallel x\parallel }_L+ϵ.$$

The previous statement is valid for every $ϵ>0$, from which it follows that ${\sigma }_p(x)\le {\parallel x\parallel }_L$.

• (2) In this case we will choose $ϵ>0$ such that $0<ϵ<1-\frac{1}{{\parallel x\parallel }_L}$, and we obtain $1<(1-ϵ){\parallel x\parallel }_L<{\parallel x\parallel }_L$. Now using into consideration definition of the norm (3.3) and relation (1) of Proposition 3.10, we get

  $$1<{\sigma }_p\left(\frac{x}{(1-ϵ){\parallel x\parallel }_L}\right)\le \frac{1}{(1-ϵ){\parallel x\parallel }_L}{\sigma }_p(x)\Rightarrow (1-ϵ){\parallel x\parallel }_L\le {\sigma }_p(x)$$

for every $ϵ\in (0,1-\frac{1}{{\parallel x\parallel }_L})$. Finally we have proved that ${\parallel x\parallel }_L\le {\sigma }_p(x)$.

• (3) Since ${\sigma }_p(x)$ is continuous function (see [14]), this property follows immediately.

• (4) Follows from properties (1) and (3).

• (5) Follows from properties (2) and (3). □

Theorem 3.12 $N(\lambda ,p)$ is a Banach space under the Luxemburg and Amemiya norms.

Proof We will prove that $N(\lambda ,p)$ is a Banach space under the Luxemburg norm. In a similar way we can prove that $N(\lambda ,p)$ is a Banach space under the Amemiya norm. In what follows we need to show that every Cauchy sequence in $N(\lambda ,p)$ is convergent according to the Luxemburg norm. Let $\{x_k^n\}$ be any Cauchy sequence in $N(\lambda ,p)$ and $ϵ\in (0,1)$. Thus there exists a natural number $n_0$, such that for any $n,m\ge n_0$ we get ${\parallel x^{(n)}-x^{(m)}\parallel }_L<ϵ$. From Proposition 3.11 we get

$${\sigma }_p(x^{(n)}-x^{(m)})\le {\parallel x^{(n)}-x^{(m)}\parallel }_L<ϵ$$

(3.5)

for all $n,m\ge n_0$. This implies that

$$\sum _{k=1}^{\mathrm{\infty }}{\left(\frac{1}{{\lambda }_k}\sum _{i=1}^k\right|({\lambda }_i-{\lambda }_{i-1})(x_i^{(n)}-x_i^{(m)})\left|\right)}^{p_k}<ϵ.$$

(3.6)

For each fixed k and for all $n,m\ge n_0$,

$$\frac{1}{{\lambda }_k}\sum _{i=1}^k|({\lambda }_i-{\lambda }_{i-1})(x_i^{(n)}-x_i^{(m)})|<ϵ.$$

Hence ${(y_k^{(n)})}_k={(\frac{1}{{\lambda }_k}{\sum }_{i=1}^k|({\lambda }_i-{\lambda }_{i-1})x_i^{(n)}|)}_k$ is a Cauchy sequence in ℝ. Since ℝ is a complete normed space, there exists a ${(y_k)}_k={(\frac{1}{{\lambda }_k}{\sum }_{i=1}^k|({\lambda }_i-{\lambda }_{i-1})x_i|)}_k\in \mathbb{R}$ such that $(y_k^{(n)})\to y_k$ as $n\to \mathrm{\infty }$. Therefore, as $n\to \mathrm{\infty }$, by relation (3.6) we have

$$\sum _{k=1}^{\mathrm{\infty }}{\left(\frac{1}{{\lambda }_k}\sum _{i=1}^k\right|({\lambda }_i-{\lambda }_{i-1})(x_i-x_i^{(m)})\left|\right)}^{p_k}<ϵ$$

for all $m\ge n_0$. In the sequel we will show that $(y_k)$ is a sequence form $N(\lambda ,p)$. From Proposition 3.10 and relation (3.5) we have

$$\underset{n\to \mathrm{\infty }}{lim}{\sigma }_p(x^{(n)}-x^{(m)})={\sigma }_p(x-x^{(m)})\le {\parallel x-x^{(m)}\parallel }_L<ϵ$$

for all $m\ge n_0$. This implies that $(x^{(n)})\to x$ as $m\to \mathrm{\infty }$. So we have $x=x^{(n)}-(x^{(n)}-x)\in N(\lambda ,p)$. This proves that $N(\lambda ,p)$ is a complete normed space under the Luxemburg norm. □

In what follows we will show results related to the Luxemburg norm, and for this reason we will denote it just $\parallel \cdot \parallel$.

Theorem 3.13 The space $N(\lambda ,p)$ is rotund if and only if $p_k>1$ for all $k\in \mathbb{N}$.

Proof Let $N(\lambda ,p)$ be rotund and choose $k\in \mathbb{N}$ such that $p_k=1$. Consider the two sequences given by

$$x=(0,0,\dots ,0,\frac{{\lambda }_k}{2^k\cdot |{\lambda }_k-{\lambda }_{k-1}|},0,0,\dots )$$

and

$$y=(0,0,\dots ,0,\frac{2{\lambda }_k}{3^k\cdot |{\lambda }_k-{\lambda }_{k-1}|},0,0,\dots ).$$

Then obviously $x\ne y$ and

$${\sigma }_p(x)={\sigma }_p(y)={\sigma }_p\left(\frac{x+y}{2}\right)=1.$$

Then from Proposition 3.11, property (3), it follows that $x,y,\frac{x+y}{2}\in S[N(\lambda ,p)]$, which leads to the contradiction that the sequence space $N(\lambda ,p)$ is not rotund. Hence $p_k>1$, for all $k\in \mathbb{N}$.

Conversely, let $x\in S[N(\lambda ,p)]$ and $y,z\in S[N(\lambda ,p)]$ such that $x=\frac{y+z}{2}$. By the convexity of ${\sigma }_p$ and property (3) from Proposition 3.11, we have

$$1={\sigma }_p(x)\le \frac{{\sigma }_p(y)+{\sigma }_p(z)}{2}\le \frac{1}{2}+\frac{1}{2}=1,$$

which gives ${\sigma }_p(y)={\sigma }_p(z)=1$ and

$${\sigma }_p(x)=\frac{{\sigma }_p(y)+{\sigma }_p(z)}{2}.$$

(3.7)

From the previous relation we obtain

$$\begin{array}{c}\sum _{k=1}^{\mathrm{\infty }}{\left(\frac{1}{{\lambda }_k}\sum _{i=1}^k\right|({\lambda }_i-{\lambda }_{i-1})x_i\left|\right)}^{p_k}\hfill \\ =\frac{1}{2}(\sum _{k=1}^{\mathrm{\infty }}{\left(\frac{1}{{\lambda }_k}\sum _{i=1}^k\right|({\lambda }_i-{\lambda }_{i-1})y_i\left|\right)}^{p_k}+\sum _{k=1}^{\mathrm{\infty }}{\left(\frac{1}{{\lambda }_k}\sum _{i=1}^k\right|({\lambda }_i-{\lambda }_{i-1})z_i\left|\right)}^{p_k}).\hfill \end{array}$$

Since $x=\frac{y+z}{2}$, we get

$$\begin{array}{c}\sum _{k=1}^{\mathrm{\infty }}{\left(\frac{1}{{\lambda }_k}\sum _{i=1}^k\right|({\lambda }_i-{\lambda }_{i-1})(y_i+z_i)\left|\right)}^{p_k}\hfill \\ =\frac{1}{2}(\sum _{k=1}^{\mathrm{\infty }}{\left(\frac{1}{{\lambda }_k}\sum _{i=1}^k\right|({\lambda }_i-{\lambda }_{i-1})y_i\left|\right)}^{p_k}+\sum _{k=1}^{\mathrm{\infty }}{\left(\frac{1}{{\lambda }_k}\sum _{i=1}^k\right|({\lambda }_i-{\lambda }_{i-1})z_i\left|\right)}^{p_k}).\hfill \end{array}$$

This implies that

$$\begin{array}{c}{\left(\frac{1}{{\lambda }_k}\sum _{i=1}^k\right|({\lambda }_i-{\lambda }_{i-1})(y_i+z_i)\left|\right)}^{p_k}\hfill \\ =\frac{1}{2}({\left(\frac{1}{{\lambda }_k}\sum _{i=1}^k\right|({\lambda }_i-{\lambda }_{i-1})y_i\left|\right)}^{p_k}+{\left(\frac{1}{{\lambda }_k}\sum _{i=1}^k\right|({\lambda }_i-{\lambda }_{i-1})z_i\left|\right)}^{p_k}).\hfill \end{array}$$

(3.8)

From the previous relation we get $y_i=z_i$ for all $i\in \mathbb{N}$, respectively, $z=y$. That is, the sequence space $N(\lambda ,p)$ is rotund. □

In what follows we will give two facts without proof because their proofs follow directly from Proposition 3.10 and Proposition 3.11.

Theorem 3.14 Let $x\in N(\lambda ,p)$. Then the following statements hold:

• (i) For $0<\alpha <1$ and $\parallel x\parallel >\alpha$ we get ${\sigma }_p(x)>{\alpha }^M$.

• (ii) If $\alpha \ge 1$ and $\parallel x\parallel <\alpha$, then we have ${\sigma }_p(x)<{\alpha }^M$.

Theorem 3.15 Let $(x_n)$ be a sequence in $N(\lambda ,p)$. Then the following statements hold:

• (i) ${lim}_{n\to \mathrm{\infty }}\parallel x_n\parallel =1$ implies ${lim}_{n\to \mathrm{\infty }}{\sigma }_p(x_n)=1$.

• (ii) ${lim}_{n\to \mathrm{\infty }}{\sigma }_p(x_n)=0$ implies ${lim}_{n\to \mathrm{\infty }}\parallel x_n\parallel =0$.

Theorem 3.16 Let $x\in N(\lambda ,p)$ and $(x^{(n)})\subset N(\lambda ,p)$. If ${\sigma }_p(x^{(n)})\to {\sigma }_p(x)$ as $n\to \mathrm{\infty }$ and $x_k^{(n)}\to x_k$ as $n\to \mathrm{\infty }$ for all $k\in \mathbb{N}$, then $x^{(n)}\to x$ as $n\to \mathrm{\infty }$.

Proof The proof of the theorem is similar to Theorem 2.9 in [13]. □

Theorem 3.17 The sequence space $N(\lambda ,p)$ has the Kadec-Klee property.

Proof It is enough to prove that every weakly convergent sequence on the unit sphere is convergent in norm. Let $x\in N(\lambda ,p)$ and $(x^{(n)})\in N(\lambda ,p)$ such that $\parallel x^{(n)}\parallel \to 1$ and $x^{(n)}\stackrel{w}{\to }x$ be given. From the properties of Theorem 3.15 it follows that ${\sigma }_p(x^{(n)})\to 1$ as $n\to \mathrm{\infty }$. On the other hand, from Proposition 3.11, we get ${\sigma }_p(x)=1$. Therefore we have ${\sigma }_p(x^{(n)})\to {\sigma }_p(x)$, as $n\to \mathrm{\infty }$. Since $x^{(n)}\stackrel{w}{\to }x$ and $p_k(x)=x_k$ is a continuous functional, $x_k^{(n)}\to x_k$ as $n\to \mathrm{\infty }$ and for $k\in \mathbb{N}$. Now the proof of the theorem follows from Theorem 3.16. □

Theorem 3.18 For any $1<p<\mathrm{\infty }$, the space $N_p$ has the uniform Opial property.

We omit this proof.

To prove the following theorem we will use the same technique given in [15] and will consider that ${lim}_{n}inf{p}_n>1$.

Theorem 3.19 The Banach space $N(\lambda ,p)$ has the k-NUC property for every $k\ge 2$.

Proof Let $ϵ>0$ and $(x_n)\subset B(N(\lambda ,p))$ with $sep(x_n)\ge ϵ$. For each $m\in \mathbb{N}$, let

$$x_n^m=(\stackrel{m-1}{\overbrace{0,0,\dots ,0}},x_n(m),x(m+1),\dots ).$$

(3.9)

Since for each $i\in \mathbb{N}$, ${(x_n(i))}_{n=1}^{\mathrm{\infty }}$ is bounded, by the diagonal method (see [16]), we find that for each $m\in \mathbb{N}$, we can find a subsequence $(x_{n_j})$ of $(x_n)$ such that $(x_{n_j}(i))$ converges for each $i\in \mathbb{N}$, $1\le i\le m$. Therefore, there exists an increasing sequence of positive integers $(t_m)$ such that $sep({(x_{n_j}^m)}_{j>t_m})\ge ϵ$. Hence, there is a sequence of positive integers ${(r_m)}_{m\in \mathbb{N}}$ with $r_1<r_2<r_3<\cdots$ such that $\parallel x_{r_m}^m\parallel \ge \frac{ϵ}{2}$ for all $m\in \mathbb{N}$. Then by Theorem 3.15, we may assume that there exists $\eta >0$ such that

$${\sigma }_p\left(x_{r_m}^m\right)\ge \eta \text{for all }m\in \mathbb{N}.$$

(3.10)

Let $\alpha >0$ be such that $1<\alpha <{lim\hspace{0.17em}inf}_n{p}_n$. For fixed integer $k\ge 2$, let ${ϵ}_1=(\frac{(k^{\alpha -1}-1)}{(k-1)k^{\alpha }})\cdot \frac{\eta }{2}$. Then by Lemma 3.6, there is a $\delta >0$ such that

$$|{\sigma }_p(u+v)-{\sigma }_p(u)|\le {ϵ}_1,$$

(3.11)

whenever ${\sigma }_p(u)\le 1$ and ${\sigma }_p(v)\le \delta$. Since by Proposition 3.11, property (1), we get ${\sigma }_p(x_n)\le 1$, $\mathrm{\forall }n\in \mathbb{N}$. Then there exist positive integers $m_i$ ($i=1,2,\dots ,k-1$) with $m_1<m_2<\cdots <m_{k-1}$ such that ${\sigma }_p(x_{p_i}^{m_i})\le \delta$ and $\alpha \le p_j$ for all $j\ge m_{k-1}$. Define $m_k=m_{k-1}+1$. By (3.10), we have ${\sigma }_p(x_{r_{m_k}}^{m_k})\ge \eta$. Let $s_i=i$ for $1\le i\le k-1$ and $s_k=r_{m_k}$. From relations (3.10), (3.11), and the convexity of the function $f_i(u)={|u|}^{p_i}$ ($i\in \mathbb{N}$), we have

$$\begin{array}{c}{\sigma }_p\left(\frac{x_{s_1}+x_{s_2}+\cdots +x_{s_k}}{k}\right)\hfill \\ =\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{1}{{\lambda }_n}\sum _{i=1}^n\right|({\lambda }_i-{\lambda }_{i-1})\frac{x_{s_1}(i)+x_{s_2}(i)+\cdots +x_{s_k}(i)}{k}\left|\right)}^{p_n}\hfill \\ =\sum _{n=1}^{m_1}{\left(\frac{1}{{\lambda }_n}\sum _{i=1}^n\right|({\lambda }_i-{\lambda }_{i-1})\frac{x_{s_1}(i)+x_{s_2}(i)+\cdots +x_{s_k}(i)}{k}\left|\right)}^{p_n}\hfill \\ +\sum _{n=m_1+1}^{\mathrm{\infty }}{\left(\frac{1}{{\lambda }_n}\sum _{i=1}^n\right|({\lambda }_i-{\lambda }_{i-1})\frac{x_{s_1}(i)+x_{s_2}(i)+\cdots +x_{s_k}(i)}{k}\left|\right)}^{p_n};\hfill \end{array}$$

from (3.11) we get

$$\begin{array}{c}\sum _{n=1}^{m_1}{\left(\frac{1}{{\lambda }_n}\sum _{i=1}^n\right|({\lambda }_i-{\lambda }_{i-1})\frac{x_{s_1}(i)+x_{s_2}(i)+\cdots +x_{s_k}(i)}{k}\left|\right)}^{p_n}\hfill \\ +\sum _{n=m_1+1}^{\mathrm{\infty }}{\left(\frac{1}{{\lambda }_n}\sum _{i=1}^n\right|({\lambda }_i-{\lambda }_{i-1})\frac{x_{s_2}(i)+\cdots +x_{s_k}(i)}{k}\left|\right)}^{p_n}+{ϵ}_1\le \hfill \end{array}$$

from the convexity of $f_i(u)={|u|}^{p_i}$ ($i\in \mathbb{N}$), it follows that

$$\begin{array}{c}\le \sum _{n=1}^{m_1}\frac{1}{k}\sum _{j=1}^k{\left(\frac{1}{{\lambda }_k}\sum _{i=1}^n\right|({\lambda }_i-{\lambda }_{i-1})x_{s_j}(i)\left|\right)}^{p_n}\hfill \\ +\sum _{n=m_1+1}^{m_2}{\left(\frac{1}{{\lambda }_n}\sum _{i=1}^n\right|({\lambda }_i-{\lambda }_{i-1})\frac{x_{s_2}(i)+x_{s_3}(i)+\cdots +x_{s_k}(i)}{k}\left|\right)}^{p_n}\hfill \\ +\sum _{n=m_2+1}^{\mathrm{\infty }}{\left(\frac{1}{{\lambda }_n}\sum _{i=1}^n\right|({\lambda }_i-{\lambda }_{i-1})\frac{x_{s_2}(i)+x_{s_3}(i)+\cdots +x_{s_k}(i)}{k}\left|\right)}^{p_n}+{ϵ}_1\hfill \\ \le \sum _{n=1}^{m_1}\frac{1}{k}\sum _{j=1}^k{\left(\frac{1}{{\lambda }_k}\sum _{i=1}^n\right|({\lambda }_i-{\lambda }_{i-1})x_{s_j}(i)\left|\right)}^{p_n}\hfill \\ +\sum _{n=m_1+1}^{m_2}{\left(\frac{1}{{\lambda }_n}\sum _{i=1}^n\right|({\lambda }_i-{\lambda }_{i-1})\frac{x_{s_2}(i)+x_{s_3}(i)+\cdots +x_{s_k}(i)}{k}\left|\right)}^{p_n}\hfill \\ +\sum _{n=m_2+1}^{\mathrm{\infty }}{\left(\frac{1}{{\lambda }_n}\sum _{i=1}^n\right|({\lambda }_i-{\lambda }_{i-1})\frac{x_{s_3}(i)+x_{s_4}(i)+\cdots +x_{s_k}(i)}{k}\left|\right)}^{p_n}+2{ϵ}_1\hfill \\ \le \sum _{n=1}^{m_1}\frac{1}{k}\sum _{j=1}^k{\left(\frac{1}{{\lambda }_k}\sum _{i=1}^n\right|({\lambda }_i-{\lambda }_{i-1})x_{s_j}(i)\left|\right)}^{p_n}\hfill \\ +\sum _{n=m_1+1}^{m_2}\frac{1}{k}\sum _{j=2}^k{\left(\frac{1}{{\lambda }_k}\sum _{i=1}^n\right|({\lambda }_i-{\lambda }_{i-1})x_{s_j}(i)\left|\right)}^{p_n}\hfill \\ +\sum _{n=m_2+1}^{m_3}\frac{1}{k}\sum _{j=3}^k{\left(\frac{1}{{\lambda }_k}\sum _{i=1}^n\right|({\lambda }_i-{\lambda }_{i-1})x_{s_j}(i)\left|\right)}^{p_n}\hfill \\ +\sum _{n=m_3+1}^{m_4}\frac{1}{k}\sum _{j=4}^k{\left(\frac{1}{{\lambda }_k}\sum _{i=1}^n\right|({\lambda }_i-{\lambda }_{i-1})x_{s_j}(i)\left|\right)}^{p_n}+\cdots \hfill \\ +\sum _{n=m_{k-1}+1}^{m_k}\frac{1}{k}\sum _{j=k-1}^k{\left(\frac{1}{{\lambda }_k}\sum _{i=1}^n\right|({\lambda }_i-{\lambda }_{i-1})x_{s_j}(i)\left|\right)}^{p_n}\hfill \\ +\sum _{n=m_k+1}^{\mathrm{\infty }}{\left(\frac{1}{{\lambda }_k}\sum _{i=1}^n\right|({\lambda }_i-{\lambda }_{i-1})\frac{x_{s_k}(i)}{k}\left|\right)}^{p_n}+(k-1){ϵ}_1\hfill \\ \le \frac{{\sigma }_p(x_{s_1})+{\sigma }_p(x_{s_2})+\cdots +{\sigma }_p(x_{s_{k-1}})}{k}+\frac{1}{k}\sum _{n=1}^{m_k}{\left(\frac{1}{{\lambda }_k}\sum _{i=1}^n\right|({\lambda }_i-{\lambda }_{i-1})x_{s_k}(i)\left|\right)}^{p_n}\hfill \\ +\sum _{n=m_k+1}^{\mathrm{\infty }}{\left(\frac{1}{{\lambda }_k}\sum _{i=1}^n\right|({\lambda }_i-{\lambda }_{i-1})\frac{x_{s_k}(i)}{k}\left|\right)}^{p_n}+(k-1){ϵ}_1\hfill \\ \le \frac{k-1}{k}+\frac{1}{k}\sum _{n=1}^{m_k}{\left(\frac{1}{{\lambda }_k}\sum _{i=1}^n\right|({\lambda }_i-{\lambda }_{i-1})x_{s_k}(i)\left|\right)}^{p_n}\hfill \\ +\frac{1}{k^{\alpha }}\cdot \sum _{n=m_k+1}^{\mathrm{\infty }}{\left(\frac{1}{{\lambda }_k}\sum _{i=1}^n\right|({\lambda }_i-{\lambda }_{i-1})x_{s_k}(i)\left|\right)}^{p_n}+(k-1){ϵ}_1\hfill \\ \le 1-\frac{1}{k}+\frac{1}{k}[1-\sum _{n=m_k+1}^{\mathrm{\infty }}{\left(\frac{1}{{\lambda }_k}\sum _{i=1}^n\right|({\lambda }_i-{\lambda }_{i-1})\frac{x_{s_k}(i)}{k}\left|\right)}^{p_n}]\hfill \\ +\frac{1}{k^{\alpha }}\cdot \sum _{n=m_k+1}^{\mathrm{\infty }}{\left(\frac{1}{{\lambda }_k}\sum _{i=1}^n\right|({\lambda }_i-{\lambda }_{i-1})\frac{x_{s_k}(i)}{k}\left|\right)}^{p_n}+(k-1){ϵ}_1\hfill \\ \le 1+(k-1){ϵ}_1-\left(\frac{k^{\alpha -1}-1}{k^{\alpha }}\right)\sum _{n=m_k+1}^{\mathrm{\infty }}{\left(\frac{1}{{\lambda }_k}\sum _{i=1}^n\right|({\lambda }_i-{\lambda }_{i-1})x_{s_k}(i)\left|\right)}^{p_n}\hfill \\ \le 1+(k-1){ϵ}_1-\left(\frac{k^{\alpha -1}-1}{k^{\alpha }}\right)\eta \hfill \\ =1-\left(\frac{k^{\alpha -1}-1}{k^{\alpha }}\right)\frac{\eta }{2}.\hfill \end{array}$$

Now from Theorem 3.9, there exists a $\vartheta >0$ such that

$$\parallel \frac{x_{s_1}+x_{s_2}+\cdots +x_{s_k}}{k}\parallel <1-\vartheta .$$

 □

The proof of the following results we omit.

Theorem 3.20 The Banach space $N(\lambda ,p)$ has the $(\beta )$-property.

Theorem 3.21 The Banach space $N(\lambda ,p)$ has the Banach-Saks property of type p.