On some geometric properties of sequence spaces of generalized arithmetic divisor sum function

Abstract

Recently, some new sequence spaces $\ell _{p}(\mathfrak{A}^{\alpha })$ $(0< p<\infty )$, $c_{0}(\mathfrak{A}^{\alpha })$, $c(\mathfrak{A}^{\alpha })$, and $\ell _{\infty }(\mathfrak{A}^{\alpha })$ have been studied by Yaying et al. (Forum Math., 2024, https://doi.org/10.1515/forum-2023-0138) as matrix domains of $\mathfrak{A}^{\alpha }=(a_{n,v}^{\alpha })$, where

$$ a_{\mathfrak{m},v}^{\alpha }=\left \{ \textstyle\begin{array}{c@{\quad}c@{\quad}c} \dfrac{v^{\alpha }}{\rho ^{(\alpha )}(\mathfrak{m})} & , & v\mid \mathfrak{m}, \\ 0 & , & v\nmid \mathfrak{m},\end{array}\displaystyle \right . $$

and $\rho ^{(\alpha )}(\mathfrak{m}):=$ sum of the $\alpha ^{\text{th}}$ power of the positive divisors of $\mathfrak{m}\in \mathbb{N}$. They obtained their duals, matrix transformations and associated compact matrix operators for these matrix classes.

This article deals with some geometric properties of these sequence spaces.

1 Introduction

We recall some known arithmetic functions [1, 16]:

$$\begin{aligned} d(\mathfrak{m}) =&\sum _{\mathfrak{v}\mid \mathfrak{m}}1,\text{(Divisor function)} \\ \rho (\mathfrak{m}) =&\sum _{\mathfrak{v}\mid \mathfrak{m}} \mathfrak{v},\text{(Divisor sum function)} \\ \rho ^{(\alpha )}(\mathfrak{m}) =&\sum _{\mathfrak{v}\mid \mathfrak{m}}v^{\alpha },\text{(Divisor sum function of order $\alpha $)} \\ \mu (\mathfrak{m}) =&\left \{ \textstyle\begin{array}{c@{\quad}c@{\quad}c} 1 & , & \mathfrak{m}=1, \\ (-1)^{\mathfrak{v}} & , & \mathfrak{m}=p_{1}p_{2}\cdots p_{v}, \\ 0 & , & p^{2}|\mathfrak{m,}\text{ for any prime }p,\end{array}\displaystyle \right . \text{ (M\"{o}bius function)} \\ \varphi (\mathfrak{m}) =&\mathfrak{m}\sum _{\mathfrak{v}\mid \mathfrak{m}}\dfrac{\mu (\mathfrak{v})}{\mathfrak{v}}, \text{ (Euler's totient function)} \\ J_{r}(\mathfrak{m}) =&\mathfrak{m}^{r}\sum _{\mathfrak{v}\mid \mathfrak{m}}\dfrac{\mu (\mathfrak{v})}{\mathfrak{v}^{r}},\text{ (Jordan's totient function),} \end{aligned}$$

where $\mathfrak{m\in }\mathbb{N}$ and $p_{\mathfrak{v}}$ denote successive prime numbers.

Lemma 1.1

[16] For any $\mathfrak{m}\in N$, $f(\mathfrak{m})=\sum _{v\mid \mathfrak{m}}g(v)$ iff $g(\mathfrak{m})=\sum _{\mathfrak{v}\mid \mathfrak{m}}\mu (v)g\left ( \frac{\mathfrak{m}}{\mathfrak{v}}\right ) = \sum _{\mathfrak{v}\mid \mathfrak{m}}\mu \left ( \frac{\mathfrak{m}}{\mathfrak{v}}\right ) g(\mathfrak{v})$.

We highlight some of the interesting properties of $\rho ^{(\alpha )}(\mathfrak{m})$ (see [1]):

• (a) $\rho ^{(\alpha )}(\mathfrak{m}\mathfrak{n})=\rho ^{(\alpha )}(\mathfrak{m})\rho ^{(\alpha )}(\mathfrak{n})$.

• (b) By Lemma 1.1,

  $$ \rho ^{(\alpha )}(\mathfrak{m})=\sum _{\mathfrak{v}\mid m} \mathfrak{v}\text{ iff }\mathfrak{m}^{\alpha }=\sum _{v\mid m}\mu \left ( \frac{\mathfrak{m}}{\mathfrak{v}}\right ) \rho ^{(\alpha )}(\mathfrak{v}). $$

  (1.1)

• (c) For any prime p,

  $$ \rho ^{(\alpha )}(p^{\mathfrak{v}})=\left \{ \textstyle\begin{array}{c@{\quad}c@{\quad}c} \dfrac{p^{\alpha (\mathfrak{v}+1)}}{p^{\alpha }-1} & , & \alpha \neq 0, \\ \mathfrak{v}+1 & , & \alpha =0.\end{array}\displaystyle \right . $$

  In general, if $\mathfrak{m}=p_{1}^{k_{1}}p_{2}^{k_{2}}\cdots p_{\mathfrak{v}}^{k_{v}}$, then

  $$ \rho ^{(\alpha )}(\mathfrak{m})= \dfrac{p^{\alpha (k_{1}+1)}}{p^{\alpha }-1}\cdot \dfrac{p^{\alpha (k_{2}+1)}}{p^{\alpha }-1}\cdots \dfrac{p^{\alpha (k_{v}+1)}}{p^{\alpha }-1}. $$

For $\alpha =0$, $\rho ^{(\alpha )}(\mathfrak{m})=\rho ^{(0)}(\mathfrak{m})=d(\mathfrak{m})$. For $\alpha =1$, $\rho ^{(\alpha )}(\mathfrak{m})=\rho ^{(1)}(\mathfrak{m})=\rho ( \mathfrak{m})$.

We write ω for the set of all real or complex valued sequeces. We further denote by $\ell _{p}~(1\leq p<\infty )$ the set of all p-absolutely summable sequences, $\ell _{\infty}$ for all bounded sequences, $c_{0}$ for all convergent to zero sequences), and c for all convergent sequences [14].

Let $\mathsf{A}=(\mathsf{a}_{rs})$ be an infinite matrix and $\mathsf{A}_{r}$ denotes its $r^{\text{th}}$ row. Then, we term the sequence $\mathsf{A}x=\{(\mathsf{A}x)_{r}\}=\left \{ \sum _{s=0}^{r}\mathsf{a}_{rs}x_{s}\right \}$ as the $\mathsf{A}$-transform of the sequence $x=(x_{s})$. Let X and Y be any two sequence spaces. We say that $\mathsf{A}$ defines a matrix mapping from X to Y if for each $x\in X$, $\mathsf{A}x\in Y$. We use the notation $(X,Y)$ to denote the family of all matrix mappings such that $X\rightarrow Y$. Further, for any sequence space X, the set $X_{A}$ that contains all the sequences whose $\mathsf{A}$-transforms belong to X is called as the domain of $\mathsf{A}$ in X, i.e., $X_{\mathsf{A}}=\{x\in \omega :\mathsf{A}x\in X\}$. For different matrix domains in classical sequence spaces, one can refer to [2, 8–11, 15].

Recently, Yaying et al. [22] defined the following sequence spaces via $\rho ^{(\alpha )}(\mathfrak{n})$:

$$\begin{aligned} \ell _{p}(\mathfrak{A}^{\alpha }):= &\left \{ \mathfrak{x}=( \mathfrak{x}_{v})\in \omega :\mathfrak{A}^{\alpha }\mathfrak{x}\in \ell _{p} \right \} , \\ c_{0}(\mathfrak{A}^{\alpha }):= &\left \{ \mathfrak{x}=(\mathfrak{x}_{v}) \in \omega :\mathfrak{A}^{\alpha }\mathfrak{x}\in c_{0}\right \} , \\ c(\mathfrak{A}^{\alpha }):= &\left \{ \mathfrak{x}=(\mathfrak{x}_{v}) \in \omega :\mathfrak{A}^{\alpha }\mathfrak{x}\in c\right \} , \\ \ell _{\infty }(\mathfrak{A}^{\alpha }):= &\left \{ \mathfrak{x}=( \mathfrak{x}_{v})\in \omega :\mathfrak{A}^{\alpha }\mathfrak{x}\in \ell _{ \infty }\right \} , \end{aligned}$$

where the matrix $\mathfrak{A}^{\alpha }=(a_{\mathfrak{n},\mathfrak{v}}^{\alpha })_{\mathfrak{n},\mathfrak{v}\in \mathbb{N}}$ is

$$ a_{\mathfrak{n},v}^{\alpha }=\left \{ \textstyle\begin{array}{c@{\quad}c@{\quad}c} \dfrac{\mathfrak{v}^{\alpha }}{\rho ^{(\alpha )}(\mathfrak{n})} & , & \mathfrak{v}\mid \mathfrak{n}, \\ 0 & , & \mathfrak{v}\nmid \mathfrak{n}.\end{array}\displaystyle \right . $$

That is

$$ \mathfrak{A}^{\alpha }= \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & \cdots & \\ \dfrac{1^{\alpha }}{1^{\alpha }+2^{\alpha }} & \dfrac{2^{\alpha }}{1^{\alpha }+2^{\alpha }} & 0 & 0 & 0 & \cdots & \\ \dfrac{1^{\alpha }}{1^{\alpha }+3^{\alpha }} & 0 & \dfrac{3^{\alpha }}{1^{\alpha }+3^{\alpha }} & 0 & 0 & \cdots & \\ \dfrac{1^{\alpha }}{1^{\alpha }+2^{\alpha }+4^{\alpha }} & \dfrac{2^{\alpha }}{1^{\alpha }+2^{\alpha }+4^{\alpha }} & 0 & \dfrac{4^{\alpha }}{1^{\alpha }+2^{\alpha }+4^{\alpha }} & 0 & \cdots & \\ \dfrac{1^{\alpha }}{1^{\alpha }+5^{\alpha }} & 0 & 0 & 0 & \dfrac{5^{\alpha }}{1^{\alpha }+5^{\alpha }} & \cdots & \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \end{bmatrix}. $$

Since $\mathfrak{A}^{\alpha }$ is a triangle, its unique inverse by (1.1) is $\left ( \mathfrak{A}^{\alpha }\right ) ^{-1}=(a_{\mathfrak{n},\mathfrak{v}}^{-\alpha })$, where

$$ a_{\mathfrak{n},\mathfrak{v}}^{-\alpha }=\left \{ \textstyle\begin{array}{c@{\quad}c@{\quad}c} \dfrac{\mu \big(\frac{\mathfrak{n}}{\mathfrak{v}}\big)\rho ^{(\alpha )}(\mathfrak{v})}{\mathfrak{n}^{\alpha }} & , & \mathfrak{v}\mid \mathfrak{n}, \\ 0 & , & \mathfrak{v}\nmid \mathfrak{n}.\end{array}\displaystyle \right . $$

$\mathfrak{A}^{\alpha }$-transform of a sequence $\mathfrak{x}=(\mathfrak{x}_{\mathfrak{v}})$ is given by $\eta =(\eta _{\mathfrak{n}})$

$$ \eta _{\mathfrak{n}}=(\mathfrak{A}^{\alpha }\mathfrak{x})_{ \mathfrak{n}}=\sum _{\mathfrak{v}\mid \mathfrak{n}} \dfrac{\mathfrak{v}^{\alpha }}{\rho ^{(\alpha )}(\mathfrak{n})} \mathfrak{x}_{v}~(\mathfrak{n}\in \mathbb{N}). $$

(1.2)

The relation (1.2) is represented by

$$ \mathfrak{x}_{\mathfrak{n}}=\left ( (\mathfrak{A}^{\alpha })^{-1} \eta \right ) _{\mathfrak{n}}=\sum _{\mathfrak{v}\mid \mathfrak{n}} \frac{\mu \big(\frac{\mathfrak{n}}{\mathfrak{v}}\big)\rho ^{(\alpha )}(\mathfrak{v})}{\mathfrak{n}^{\alpha }}\eta _{\mathfrak{v}}~(\mathfrak{n}\in \mathbb{N}). $$

The readers are suggested to consult the papers [18–20] for more insights into sequence spaces that are constructed by using arithmetic functions. Clearly $X(\mathfrak{A}^{\alpha })=X_{\mathfrak{A}^{\alpha }}$, where $X=\ell _{p},c_{0},c$, or $\ell _{\infty }$.

Remark 1.2

For $\alpha =1$, $\ell _{p}(\mathfrak{A}^{\alpha })$, $c_{0}(\mathfrak{A}^{\alpha })$, $c(\mathfrak{A}^{\alpha })$ and $\ell _{\infty }(\mathfrak{A}^{\alpha })$ reduce to the spaces defined in [21].

Theorem 1.3

We have

• (1) $c_{0}(\mathfrak{A}^{\alpha })$, $c(\mathfrak{A}^{\alpha })$, $\ell _{ \infty }(\mathfrak{A}^{\alpha })$ are BK-spaces with the norm

  $$ \left \Vert \mathfrak{x}\right \Vert _{\ell _{\infty }(\mathfrak{A}^{ \alpha })}=\left \Vert \mathfrak{A}^{\alpha }\mathfrak{x}\right \Vert _{\ell _{\infty }}=\sup _{\mathfrak{n}\in \mathbb{N}}\left \vert \sum _{\mathfrak{v}\mid \mathfrak{n}} \frac{\mathfrak{v}^{\alpha }}{\rho ^{(\alpha )}(\mathfrak{n})}\mathfrak{x}_{\mathfrak{v}}\right \vert . $$

• (2) $\ell _{p}(\mathfrak{A}^{\alpha })(1\leq p<\infty )$ is a BK-space with the norm

  $$ \Vert \mathfrak{x}\Vert _{\ell _{p}(\mathfrak{A}^{\alpha })}=\left \Vert \mathfrak{A}^{\alpha }\mathfrak{x}\right \Vert _{\ell _{p}}= \left [ \sum _{\mathfrak{n}=0}^{\infty }\left \vert \sum _{\mathfrak{v}\mid \mathfrak{n}}\dfrac{v^{\alpha }}{\rho ^{(\alpha )}(\mathfrak{n})}\mathfrak{x}_{ \mathfrak{v}}\right \vert ^{p}\right ] ^{1/p}< \infty . $$

In this paper, we study some geometric properties of these sequence spaces.

2 Geometric properties

We recall some geometric properties to study in our case. For Banach spaces λ and μ, let $L:\lambda \rightarrow \mu$ be a linear operator. We denote $B(\lambda ,\mu )$ and $C(\lambda ,\mu )$ for the spaces of bounded linear operators and compact linear operators, respectively.

L is weakly compact [13, Definition 3.5.1] if $L(Q)$ is a relatively weakly compact subset of μ whenever Q is a bounded subset of λ.

Approximation property [13, Definition 3.4.26] is possesed by λ if the set of finite rank members of $B(\mu ,\lambda )$ is dense in $C(\mu ,\lambda )$ for any μ.

A Banach space is said to have the approximation property (AP), if every compact operator is a limit of finite-rank operators.

The approximation property is satisfied by the space $\ell _{p}$ $(1\leq p<\infty )$ (see [13]).

The Dunford–Pettis property (in short, D-P property) is possessed by λ if every continuous weakly compact operator $L:\lambda \rightarrow \mu$ transforms weakly compact sets in λ into a compact sets in μ (such operators are called completely continuous).

Theorem 2.1

[17] Let $L_{0}\in B(\nu ,\ell _{\infty })$. Then, the operator $L_{0}$ may be extended to $L\in B(\lambda ,\ell _{\infty })$ with $\left \Vert L_{0}\right \Vert =\left \Vert L\right \Vert$, where ν is a linear subspace of λ. In this case, $\ell _{\infty }$ is said to have Hahn–Banach extension property.

Let

$$ S_{\lambda}=\left \lbrace s\in \lambda : \left \|s \right \|=1 \right \rbrace . $$

A normed space λ is said to be rotund (or strictly convex) [13, Definition 5.1.1] if for any $s_{1},s_{2}\in S_{\lambda }$ $(s_{1}\neq s_{2})$ and $0<\alpha <1$,

$$ \left \Vert \alpha s_{1}+(1-\alpha )s_{2}\right \Vert < 1. $$

A normed space λ is rotund [13] iff

$$ \left \Vert \frac{s_{1}+s_{2}}{2}\right \Vert < 1 $$

for any $s_{1},s_{2}\in S_{\lambda }$ $(s_{1}\neq s_{2})$.

Proposition 2.2

[13, Proposition 5.1.9] Any normed space that is isometrically isomorphic to a rotund space is also rotund.

Let X be a Banach space.

If every bounded sequence $(\xi _{r})$ in X has a subsequence $(\chi _{r})$ such that the sequence $\{t_{k}(\chi )\}$ converges in the norm, then X has the Banach–Saks property [15], where

$$ \{t_{i}(\chi )\}=\frac{1}{i+1}(\chi _{0}+\chi _{1}+\cdots )~(i\in \mathbb{N}). $$

If any weakly null sequence $(\xi _{r})$ in X has a subsequence $(\chi _{r})$ such that $\{t_{i}(\chi )\}$ is strongly convergent to zero, then X has the weak Banach–Saks property.

The following coefficient is provided by Garcia-Falset [4],

$$ R(X)=\sup \{\lim _{r\rightarrow \infty }\inf \Vert \xi _{r}-\xi \Vert :(\xi _{r})\subset D(X),\xi _{r}\rightarrow \xi (w),~\xi \in D(X) \}, $$

where $D(X)$ represents $X^{\prime }s$ unit ball.

Remark 2.3

X has weak fixed point characteristics when $R(X)<2$ [5].

For $1< p<\infty$, the property $(BS)_{p}$, also known as Banach–Saks type p, is that if a subsequence $(\xi _{k_{l}})$ of every weakly null sequence $(\xi _{k})$ satisfies

$$ \bigg\| \sum _{l=0}^{u}\xi _{k_{l}}\bigg\| < Q.(u+1)^{\frac{1}{p}} $$

for each $Q>0$ and for all $u\in \mathbb{N}$ ([12]).

The Gurarii’s modulus of convexity (see [6, 7]) is defined by

$$ \beta _{X}{(\epsilon )}=\inf _{{}}{\left \{ 1-\inf _{0\leq \delta \leq 1}{||\delta x+(1-\delta )y||;x,y\in }S_{X},||x-y||=\epsilon \right \} }, $$

where $0\leq \epsilon \leq 2$, and $S_{X}$ denotes the unit sphere in X.

Most recently such properties are studied in [3].

3 Main results

Here we study such geometric properties for our sequence spaces.

Theorem 3.1

The approximation property is possessed by the space $\ell _{p}(\mathfrak{A}^{\alpha })$ for $1\leq p<\infty$.

Proof

Let $L\in C(\lambda ,\ell _{p}(\mathfrak{A}^{\alpha }))$ for any Banach space λ. It follows that for each bounded sequence $s=(s_{n})\in \lambda$, the sequence $\left ( Ls_{n}\right )$ has a convergent sub-sequence $\left ( Ls_{n_{v}}\right )$ in $\ell _{p}(\mathfrak{A}^{\alpha })$, i.e.,

$$ \left \Vert Ls_{n_{u}}-Ls_{n_{v}}\right \Vert _{\ell _{p}( \mathfrak{A}^{\alpha })}^{p}=\left \Vert L\left ( s_{n_{u}}-s_{n_{v}} \right ) \right \Vert _{\ell _{p}(\mathfrak{A}^{\alpha })}^{p}=\left \Vert \left ( \mathfrak{A}^{\alpha }L\right ) \left ( s_{n_{u}}-s_{n_{v}} \right ) \right \Vert _{\ell _{p}}^{p}\rightarrow 0 $$

as $u,v\rightarrow \infty$. Then, $\mathfrak{A}^{\alpha }L\in C(\lambda ,\ell _{p})$. Since $\ell _{p}$ possesses the approximation property, there exists a sequence $T_{n}\in B(\lambda ,\ell _{p})$ of finite rank operators such that

$$ \left \Vert \mathfrak{A}^{\alpha }L-T_{n}\right \Vert \rightarrow 0. $$

Consequently, the sequence $\left ( {(\mathfrak{A}^{\alpha })}^{-1}T_{n}\right ) \in B(\lambda ,\ell _{p}({\mathfrak{A}^{\alpha }}))$ is the required sequence of finite rank. Also

$$\begin{aligned} \left \Vert L-{(\mathfrak{A}^{\alpha })}^{-1}T_{n}\right \Vert =& \sup _{\left \Vert s\right \Vert =1}\left \Vert \left ( L-{( \mathfrak{A}^{\alpha })}^{-1}T_{n}\right ) s\right \Vert _{\ell _{p}(\mathfrak{A}^{ \alpha })}^{p} \\ =&\sup _{\left \Vert s\right \Vert =1}\left \Vert Ls-\left ( {( \mathfrak{A}^{\alpha })}^{-1}T_{n}\right ) s\right \Vert _{\ell _{p}(\mathfrak{A}^{ \alpha })}^{p} \\ =&\sup _{\left \Vert s\right \Vert =1}\left \Vert \mathfrak{A}^{ \alpha }Ls-T_{n}s\right \Vert _{\ell _{p}}^{p} \\ =&\sup _{\left \Vert s\right \Vert =1}\left \Vert \left ( \mathfrak{A}^{\alpha }L-T_{n}\right ) s\right \Vert _{\ell _{p}}^{p} \\ \rightarrow &0\text{ as $n\rightarrow \infty $}. \end{aligned}$$

This completes the proof. □

Theorem 3.2

The D-P property is possessed by the space $\ell _{1}(\mathfrak{A}^{\alpha })$.

Proof

Suppose that L: $\ell _{1}(\mathfrak{A}^{\alpha })\rightarrow \lambda$ is a weakly compact operator. Then, $L\{\mathfrak{A}^{\alpha }\}^{-1}:\ell _{1}\rightarrow \lambda$ is a bounded linear operator. Let $B\subset \ell _{1}$ be bounded. Then, it follows that $\{\mathfrak{A}^{\alpha }\}^{-1}B\subset \ell _{1}(\mathfrak{A}^{ \alpha })$ is bounded. It follows that the set

$$ L\left ( \{\mathfrak{A}^{\alpha }\}^{-1}B\right ) =\left ( L\{ \mathfrak{A}^{\alpha }\}^{-1}\right ) B $$

is relatively weakly compact in λ, since L is weakly compact. Therefore, $L\{\mathfrak{A}^{\alpha }\}^{-1}$: $\ell _{1}\rightarrow \lambda$ is a weakly compact operator. Now, the operator $L\{\mathfrak{A}^{\alpha }\}^{-1}$ is completely continuous, since the space $\ell _{1}$ has the D-P property. Suppose that Q is a weakly compact subset of $\ell _{1}(\mathfrak{A}^{\alpha })$. Then, $\mathfrak{A}^{\alpha }Q$ is a weakly compact subset of $\ell _{1}$. Therefore, $L\{\mathfrak{A}^{\alpha }\}^{-1}({\mathfrak{A}^{\alpha }})(Q)=L(Q)$ is a compact set in μ, since $L\{\mathfrak{A}^{\alpha }\}^{-1}$ is completely continuous. Hence, L is completely continuous as required. □

Theorem 3.3

The space $\ell _{\infty }(\mathfrak{A}^{\alpha })$ has the Hahn–Banach extension property.

Proof

Let ν be a linear subspace of a Banach space λ and $L_{0}\in B(\nu ,\ell _{\infty }(\mathfrak{A}^{\alpha }))$. Then, $\mathfrak{A}^{\alpha }L_{0}\in B(\nu ,\ell _{\infty })$. Then the operator $\mathfrak{A}^{\alpha }L_{0}$ can be extended to $T\in B(\lambda ,\ell _{\infty })$ with $\left \Vert \mathfrak{A}^{\alpha }L_{0}\right \Vert =\left \Vert T \right \Vert$, since by Theorem 2.1$\ell _{\infty }$ has the Hahn–Banach extension property. Choose the operator $L=\{\mathfrak{A}^{\alpha }\}^{-1}T$. Then, $L\in B(\lambda ,\ell _{\infty }(\mathfrak{A}^{\alpha }))$. Also, we observe that

$$ Ls=\left ( \{\mathfrak{A}^{\alpha }\}^{-1}T\right ) s=\{\mathfrak{A}^{ \alpha }\}^{-1}(Ts)=\{\mathfrak{A}^{\alpha }\}^{-1}\left ( ( \mathfrak{A}^{\alpha }L_{0})s\right ) =L_{0}s. $$

for any $s\in \nu$. Additionally

$$ \left \Vert L\right \Vert =\left \Vert \{\mathfrak{A}^{\alpha }\}^{-1}T \right \Vert =\left \Vert \{\mathfrak{A}^{\alpha }\}^{-1}\left ( \mathfrak{A}^{\alpha }L_{0}\right ) \right \Vert =\left \Vert L_{0} \right \Vert , $$

as desired. □

Theorem 3.4

The space $\ell _{p}(\mathfrak{A}^{\alpha })$ $(1< p<\infty )$ is rotund.

Proof

Since $\ell _{p}$ $(1< p<\infty )$ is a rotund, using Proposition 2.2 we get the result. □

Theorem 3.5

The spaces $\ell _{1}(\mathfrak{A}^{\alpha })$ and $\ell _{\infty }(\mathfrak{A}^{\alpha })$ are not rotund.

Proof

Choose $a_{v},b_{v}\in \ell _{1}(\mathfrak{A}^{\alpha })$ given by

$$\begin{aligned} a_{v}&=\left \{ \textstyle\begin{array}{c@{\quad}c@{\quad}c} \dfrac{\mu (v)+(1^{\alpha }+2^{\alpha })\mu \left ( \frac{v}{2}\right ) }{v^{\alpha }} & , & v\text{ is even} \\ \dfrac{\mu (v)}{v^{\alpha }} & , & v\text{ is odd},\end{array}\displaystyle \right . \quad \text{and} \\ b_{v}&=\left \{ \textstyle\begin{array}{c@{\quad}c@{\quad}c} \dfrac{\mu (v)-(1^{\alpha }+2^{\alpha })\mu \left ( \frac{v}{2}\right ) }{v^{\alpha }} & , & v\text{ is even} \\ \dfrac{\mu (v)}{v^{\alpha }} & , & v\text{ is odd},\end{array}\displaystyle \right . \end{aligned}$$

for all $v\in \mathbb{N}$. Then, $\mathfrak{A}^{\alpha }a=(1,1,0,0,\ldots )\in \ell _{p}$ and $\mathfrak{A}^{\alpha }b=(1,-1,0,0,\ldots )\in \ell _{p}$. It follows that $\left \Vert a\right \Vert _{\ell _{1}(\mathfrak{A}^{\alpha })}=1$ and $\left \Vert b\right \Vert _{\ell _{1}(\mathfrak{A}^{\alpha })}=1$. That is $a,b\in S_{\ell _{1}(\mathfrak{A}^{\alpha })}$.

Let $s=\frac{a+b}{2}$. Then, $\mathfrak{A}^{\alpha }s=\{\dfrac{\mu (v)}{v^{\alpha }}\}$. Thus,

$$ \left \Vert s\right \Vert _{\ell _{1}(\mathfrak{A}^{\alpha })}=\left \Vert \mathfrak{A}^{\alpha }s\right \Vert _{\ell _{1}}=1. $$

Hence, we see that

$$ \left \Vert s\right \Vert _{\ell _{1}(\mathfrak{A}^{\alpha })}\nless 1. $$

Therefore, the space $\ell _{1}(\mathfrak{A}^{\alpha })$ is not rotund. Similarly, non-rotundness of $\ell _{\infty }(\mathfrak{A}^{\alpha })$ can be proved. □

Theorem 3.6

The space $\ell _{p}(\mathfrak{A}^{\alpha })$ $(1< p<\infty )$ has the property $(BS)_{p}$.

Proof

For a positive number sequence $(\epsilon _{r})$ such that $\displaystyle \sum _{r=1}^{\infty }\epsilon _{r}\leq \frac{1}{2}$ and a weakly null sequence $(\xi _{r})\in B(\ell _{p}(\mathfrak{A}^{\alpha }))$. Put $\chi _{0}=\xi _{0}=0$ and $\chi _{1}=\xi _{r_{1}}=\xi _{1}$. Therefore, there exists $v_{1}\in \mathbb{N}$ such that

$$ \bigg\| \sum _{k=v_{1}+1}^{\infty }\chi _{1}(k)e^{(k)}\bigg\| _{\ell _{p}(\mathfrak{A}^{\alpha })}< \epsilon _{1}. $$

There is an $r_{2}\in \mathbb{N}$ such that

$$ \bigg\| \sum _{k=0}^{v_{1}}\xi _{r}(k)e^{(k)}\bigg\| _{\ell _{p}( \mathfrak{A}^{\alpha })}< \epsilon _{1}, $$

when $r\geq r_{2}$, since $(\xi _{r})$ is a weakly null sequence, then $\xi _{r}\rightarrow 0$ coordinatewise. Set $\chi _{2}=\xi _{r_{2}}$. Therefore there exists an $r_{2}>r_{1}$ such that

$$ \bigg\| \sum _{k=v_{2}+1}^{\infty }\chi _{2}(k)e^{(k)}\bigg\| _{\ell _{p}(\mathfrak{A}^{\alpha })}< \epsilon _{2}. $$

By using $\xi _{r}\rightarrow 0$ coordinatewise, there exists $r_{3}>r_{2}$ such that

$$ \bigg\| \sum _{k=0}^{v_{2}}\xi _{r}(k)e^{(k)}\bigg\| _{\ell _{p}( \mathfrak{A}^{\alpha })}< \epsilon _{2}, $$

when $r\geq r_{3}$.

By following this procedure, two increasing subsequences $(v_{k})$ and $(r_{k})$ can be obtained such that

$$ \bigg\| \sum _{k=0}^{v_{\alpha }}\xi _{r}(k)e^{(k)}\bigg\| _{\ell _{p}(\mathfrak{A}^{\alpha })}< \epsilon _{\alpha }, $$

for each $r\geq r_{\alpha +1}$ and

$$ \bigg\| \sum _{k=v_{\alpha }+1}^{\infty }\chi _{j}(k)e^{(k)}\bigg\| _{ \ell _{p}(\mathfrak{A}^{\alpha })}< \epsilon _{\alpha }, $$

where $\chi _{\alpha }=\xi _{r_{\alpha }}$. Thus

$$\begin{aligned} &\displaystyle \bigg\| \sum _{\alpha =0}^{r}\chi _{\alpha }\bigg\| _{ \ell _{p}(\mathfrak{A}^{\alpha })} \\ &\quad =\bigg\| \sum _{\alpha =0}^{r}\left ( \sum _{k=0}^{v_{\alpha -1}} \chi _{\alpha }(k)e^{(k)}+\sum _{k=v_{\alpha -1}+1}^{v_{j}}\chi _{ \alpha }(k)e^{(k)}+\sum _{k=v_{\alpha }+1}^{\infty }\chi _{\alpha }(k)e^{(k)} \right ) \bigg\| _{\ell _{p}(\mathfrak{A}^{\alpha })} \\ &\quad \leq \bigg\| \sum _{\alpha =0}^{r}\left ( \sum _{k=v_{\alpha -1}+1}^{v_{ \alpha }}\chi _{\alpha }(k)e^{(k)}\right ) \bigg\| _{\ell _{p}( \mathfrak{A}^{\alpha })}+2\sum _{\alpha =0}^{r}\epsilon _{\alpha }. \end{aligned}$$

However, we see that $\Vert \xi \Vert _{\ell _{p}(\mathfrak{A}^{\alpha })}\leq 1$. Thus, we have

$$ \bigg\| \sum _{\alpha =0}^{r}\left ( \sum _{k=v_{\alpha -1}+1}^{v_{ \alpha }}\chi _{\alpha }(k)e^{(k)}\right ) \bigg\| _{\ell _{p}( \mathfrak{A}^{\alpha })}^{p}\leq (r+1). $$

So, we have

$$ \bigg\| \sum _{\alpha =0}^{r}\sum _{k=v_{\alpha -1}+1}^{v_{\alpha }} \chi _{\alpha }(k)e^{(k)}\bigg\| _{\ell _{p}(\mathfrak{A}^{\alpha })}^{p} \leq (r+1)^{\frac{1}{p}}. $$

By using $1\leq (r+1)^{\frac{1}{p}}$ for all $r\in \mathbb{N}$ and $1< p<\infty$, we have

$$ \bigg\| \sum _{\alpha =0}^{r}\chi _{\alpha }\bigg\| _{\ell _{p}( \mathfrak{A}^{\alpha })}\leq (r+1)^{\frac{1}{p}}+1\leq 2(r+1)^{\frac{1}{p}}. $$

Therefore, $\ell _{p}(\mathfrak{A}^{\alpha })$ has Banach–Saks type p. □

Remark 3.7

The space $\ell _{p}(\mathfrak{A}^{\alpha })$ is linearly isomorphic to $\ell _{p}$ and $R(\ell _{p}(\mathfrak{A}^{\alpha }))=R(\ell _{p})=2^{\frac{1}{p}}$.

Theorem 3.8

The space $\ell _{p}(\mathfrak{A}^{\alpha })$ $(1< p<\infty )$ has weak fixed-point property.

Proof

The proof is straightforward and follows from Remark 2.3 and 3.7. □

Theorem 3.9

The Gurarii’s modulus of convexity for $\ell _{p}(\mathfrak{A}^{\alpha })$ $(p\geq 1)$ is

$$ \beta _{\ell _{p}(\mathfrak{A}^{\alpha })}{(\delta )}\leq 1-\left ( 1- \left ( \frac{{\delta }}{2}\right ) ^{p}\right ) ^{1/p}, $$

where $0\leq {\delta }\leq 2$.

Proof

Let $\mathfrak{x\in }\ell _{p}(\mathfrak{A}^{\alpha })$. Then

$$ \Vert \mathfrak{x}\Vert _{\ell _{p}(\mathfrak{A}^{\alpha })}=\left \Vert \mathfrak{A}^{\alpha }\mathfrak{x}\right \Vert _{\ell _{p}}= \left [ \sum _{\mathfrak{n}=0}^{\infty }\left \vert \sum _{\mathfrak{v}\mid \mathfrak{n}}\dfrac{v^{\alpha }}{\rho ^{(\alpha )}(\mathfrak{n})}\mathfrak{x}_{ \mathfrak{v}}\right \vert ^{p}\right ] ^{1/p}. $$

For $0\leq {\delta }\leq 2$, define

$$ x=\left ( \{\mathfrak{A}^{\alpha }\}^{-1}\left ( 1-\left ( \frac{{\delta }}{2}\right ) ^{p}\right ) ^{1/p},\{\mathfrak{A}^{\alpha }\}^{-1}\left ( \frac{{\delta }}{2}\right ) ,0,0,0,\ldots\right ) $$

and

$$ y=\left ( \{\mathfrak{A}^{\alpha }\}^{-1}\left ( 1-\left ( \frac{{\delta }}{2}\right ) ^{p}\right ) ^{1/p},\{\mathfrak{A}^{\alpha }\}^{-1}\left ( - \frac{{\delta }}{2}\right ) ,0,0,0,\ldots\right ) . $$

Then, $\left \Vert \mathfrak{A}^{\alpha }x\right \Vert _{\ell _{p}}=\Vert x \Vert _{\ell _{p}(\mathfrak{A}^{\alpha })}=1$ and $\left \Vert \mathfrak{A}^{\alpha }y\right \Vert _{\ell _{p}}=\Vert y\Vert _{\ell _{p}( \mathfrak{A}^{\alpha })}=1$. That is, $x,y\in S(\ell _{p}(\mathfrak{A}^{\alpha }))$ and $\left \Vert \mathfrak{A}^{\alpha }x-\mathfrak{A}^{\alpha }y\right \Vert _{\ell _{p}}=\Vert x-y\Vert _{\ell _{p}(\mathfrak{A}^{\alpha })}={ \delta }$. For $0\leq {\delta }\leq 1$,

$$\begin{aligned}& \Vert \alpha x+(1-\alpha )y\Vert _{\ell _{p}(\mathfrak{A}^{\alpha })}^{p}= \Vert \alpha \mathfrak{A}^{\alpha }x+(1-\alpha )\mathfrak{A}^{\alpha }y \Vert _{\ell _{p}}^{p}\\& =1-\left ( \frac{{\delta }}{2}\right ) ^{p}+[2\alpha -1]\left ( \frac{{\delta }}{2}\right ) ^{p}. \end{aligned}$$

Hence

$$ \inf _{0\leq {\delta }\leq 1}\Vert \alpha x+(1-\alpha )y\Vert _{\ell _{p}(\mathfrak{A}^{\alpha })}^{p}=1-\left ( \frac{{\delta }}{2}\right ) ^{p}. $$

That is, for $p\geq 1$,

$$ \beta _{\ell _{p}(\mathfrak{A}^{\alpha })}{(\delta )}\leq 1-\left ( 1- \left ( \frac{{\delta }}{2}\right ) ^{p}\right ) ^{1/p}. $$

Hence proved. □

Corollary 3.10

(i) If ${\delta }=2$, then $\beta _{\ell _{p}(\mathfrak{A}^{\alpha })}{(\delta )=1}$. So, $\ell _{p}(\mathfrak{A}^{\alpha })$ is strictly convex. (ii) If $0<{\delta }\leq 2$, then $0<\beta _{\ell _{p}(\mathfrak{A}^{\alpha })}{(\delta )\leq 1.}$ So, $\ell _{p}(\mathfrak{A}^{\alpha })$ is uniformly convex.