Sequentially Right-like properties on Banach spaces

TL;DR

This paper introduces and studies the $p$-sequentially Right property and related concepts in Banach spaces, providing new characterizations and relationships between these properties and operator convergence behaviors.

Contribution

It defines the $p$-sequentially Right property, introduces $p$-Right$^{ ext{*}}$ sets, and explores their relationships, extending the understanding of operator convergence in Banach spaces.

Findings

Characterization of $p$-Dunford-Pettis relatively compact property.

Relationships between $p$-Right and $p$-Right$^{ ext{*}}$ sets.

Conditions for Dunford-Pettis $q$-convergent operators to be $p$-convergent.

Abstract

In this paper, we first study the concept of $ p $-sequentially Right property, which is the $ p$-version of the sequentially Right property. Also, we introduce a new class of subsets of Banach spaces which is called $ p$-Right$ ^{\ast} $ set and obtain the relationship between p-Right subsets and p-Right$ ^{\ast} $ subsets of dual spaces. Furthermore, for $ 1\leq p<q\leq\infty, $ we introduce the concepts of properties $ (SR)_{p,q}$ and $ (SR^{\ast})_{p,q}$ in order to find a condition which every Dunford-Pettis $ q $-convergent operator is Dunford-Pettis $p$-convergent. Finally, we apply these concepts and obtain some characterizations of $ p $-Dunford-Pettis relatively compact property of Banach spaces and their dual spaces.

Equations

$$\underset{n\to \infty }{\lim }\underset{x^{\ast }\in K}{\sup }\mid x^{\ast }(x_n)\mid =0.$$

$$\underset{n}{\lim }\underset{x^{\ast }\in K}{\sup }|x^{\ast }(x_n)|=0.$$

$$\underset{n}{\lim }\underset{x\in K}{\sup }|x_n^{\ast }(x)|=0.$$

$$\underset{n}{\lim }\Vert T(x_n)\Vert =\underset{n}{\lim }(\sup \{|⟨y^{\ast },T(x_n)⟩|:y^{\ast }\in B_{Y^{\ast }}\})$$

$$=\underset{n}{\lim }(\sup \{|⟨T^{\ast }(y^{\ast }),x_n⟩|:y^{\ast }\in B_{Y^{\ast }}\})=0$$

$$\underset{n}{\lim }(\sup \{|x^{\ast }(x_n)|:x^{\ast }\in K\})=0$$

$$\underset{n}{\lim }\Vert T^{\ast }(y_n^{\ast })\Vert =\underset{n}{\lim }(\sup \{|⟨x,T^{\ast }(y_n^{\ast })⟩|:x\in B_X\})$$

$$=\underset{n}{\lim }(\sup \{|⟨T(x),y_n^{\ast }⟩|:x\in B_X\})=0$$

$$\underset{n}{\lim }\Vert E(x_n^{\ast })\Vert =\underset{n}{\lim }(\sup \{|x_n^{\ast }(x)|:x\in K\})=0,$$

$$\Vert {\varphi }_x^{\ast }(y_n^{\ast }))\Vert =\sup \{|{\varphi }_x^{\ast }{y}_n^{\ast }(T)|:T\in B_{\mathcal{M}}\}=\sup \{|y_n^{\ast }(T(x))|:T\in B_{\mathcal{M}}\}.$$

$$\vartheta \mapsto tr(T^{\ast \ast }\vartheta )=\sum _{n=1}^{\infty }⟨T^{\ast \ast }{x}_n^{\ast \ast },y_n^{\ast }⟩$$

$$\underset{n\to \infty }{\lim }\Vert T(x_n)\Vert =\underset{n\to \infty }{\lim }\underset{m}{\sup }|x_m^{\ast }(x_n)|=0.$$

$$⟨T_n(x),y_n⟩=⟨T_n,x\otimes y_n⟩.$$

$$|⟨T,x\otimes y_n⟩|=|⟨T(x),y_n⟩|\in {\ell }_p,$$

$$|⟨A_n,x\otimes y_n⟩|=|⟨A_n(x),y_n⟩|\to 0.$$

$$\underset{n\to \infty }{\lim }\sup \{\sum _{k=n}^{\infty }\Vert P_k{x}^{\ast }{\Vert }^{p^{\ast }}:x^{\ast }\in K\}=0.$$

$$P_k{y}_j^{\ast }=\{\begin{array}{ll}{P}_{k_j}{x}_{n_j}^{\ast }& \text{if }{k}_{j-1}+1\le k\le k_j,\\ 0& \text{otherwise }.\end{array}$$

$$\underset{n}{\lim }\underset{i}{\sup }|x_i^{\ast }(x_n)|=0.$$

$$|T(y)(x_n)|=|T^{\ast }(x_n)(y)|\le \Vert T^{\ast }(x_n)\Vert \to 0.$$

Full text

1. Introduction

Peralta et al. [22], proved that for a given Banach space X there is a locally convex topology on $X,$ which is called the“ Right topology ”, such that a linear map $T$ from $X$ into a Banach space $Y$ is weakly compact if and only if it is Right-to-norm sequentially continuous. Later on Kacena [18] by introducing the notion of Right set in $X^{\ast}$(dual of $X$), showed that a Banach space $X$ has the sequentially Right property if and only if every Right subset of $X^{\ast},$ is relatively weakly compact. A bounded subset $K$ of $X^{\ast}$ is a Right set, if every Right-null sequence $(x_{n})_{n}$ in $X$ converges uniformly to zero on $K;$ that is,

[TABLE]

Recall that a bounded subset $K$ of Banach space $X$ is a Dunford-Pettis set, if every weakly null sequence $(x^{\ast}_{n})_{n}$ in $X^{\ast},$ converges uniformly to zero on the set $K$ [2]. A Banach space $X$ has the Dunford-Pettis relatively compact property (in short $X$ has the $(DPrcP)$), if every Dunford-Pettis subset of $X$ is relatively compact [11].

For more information and examples of Banach spaces with Dunford-Pettis relatively compact property and sequentially Right property, we refer to [11, 18, 22].

Recently, Ghenciu [16] introduced the concepts of Dunford-Pettis $p$-convergent operators, $p$-Dunford-Pettis relatively compact property (in short $p$-$(DPrcP)$), $p$-Right sets and $p$-sequentially Right property (in short $p$-$(SR)$) on Banach spaces as follows:

• •

An operator $T:X\rightarrow Y$ is said to be Dunford-Pettis $p$-convergent, if it takes Dunford-Pettis weakly $p$-summable sequences to norm null sequences. The class of Dunford-Pettis $p$-convergent operators from $X$ into $Y$ is denoted by $DPC_{p}(X,Y).$

• •

A Banach space $X$ has the $p$-Dunford-Pettis relatively compact property ($X$ has the $p$-$(DPrcP)$), if every Dunford-Pettis weakly p-summable sequence $(x_{n})_{n}$ in $X$ is norm null.

• •

A bounded subset $K$ of $X^{\ast}$ is called a $p$-Right set, if every Dunford-Pettis weakly $p$-summable sequence $(x_{n})_{n}$ in $X$ converges uniformly to zero on $K,$ that is,

[TABLE]

• •

A Banach space $X$ has the $p$-sequentially Right property ($X$ has the $p$-$(SR)$), if every $p$-Right set in $X^{\ast}$ is relatively weakly compact.

Motivated by the above works, in Section 3, we introduce the concepts of $p$-Right*∗* sets and $p$-sequentially Right*∗* property on Banach spaces.  Then, we obtain the relationship between $p$-Right subsets and $p$-Right*∗* subsets of dual spaces. In addition, the stability of $p$-sequentially Right property for some subspaces of bounded linear operators and projective tensor product between two Banach spaces are investigated.

In the Section 4, for $1\leq p<q\leq\infty$ inspired by the class $\mathcal{P}_{p,q}$ in [23], for those Banach spaces in which relatively $p$-compact sets are relatively $q$-compact, we introduce the concepts of properties $(SR)_{p,q}$ and $(SR^{\ast})_{p,q}$ for those Banach spaces in which Dunford-Pettis $q$-convergent operators are Dunford-Pettis $p$-convergent operators. Finally, by applying these concepts, some characterizations for the $p$-Dunford-Pettis relatively compact property of Banach spaces and their dual spaces are investigated. Note that, the our results are motivated by results in [3, 16, 18, 22, 23].

2. Definitions and Notions

Throughout this paper $1\leq p<\infty,$ $1\leq p<q\leq\infty,$ except for the cases where we consider other assumptions. Also, we suppose $X,Y$ and $Z$ are arbitrary Banach spaces, $p^{\ast}$ is the H$\ddot{\mathrm{o}}$lder conjugate of $p;$ if $p=1,~{}~{}\ell_{p^{\ast}}$ plays the role of $c_{0}.$ The unit coordinate vector in $\ell_{p}$ (resp. $c_{0}$ or $\ell_{\infty}$) is denoted by $e_{n}^{p}$ (resp. $e_{n}$). The space $X$ embeds in $Y,$ if $X$ is isomorphic to a closed subspace of $Y$ (in short we denote $X\hookrightarrow Y$). We denote two isometrically isomorphic spaces $X$ and $Y$ by $X\cong Y.$ The word ‘operator’ will always mean a bounded linear operator. For any Banach space X, the dual space of bounded linear functionals on X will be denoted by $X^{\ast}.$ Also we use $\langle x,x^{\ast}\rangle$ or $x^{\ast}(x)$ for the duality between $x\in X$ and $x^{\ast}\in X^{\ast}.$ We denote the closed unit ball of $X$ by $B_{X}$ and the identity operator on $X$ is denoted by $id_{X}.$ For a bounded linear operator $T:X\rightarrow Y,$ the adjoint of the operator $T$ is denoted by $T^{\ast}.$ The space of all bounded linear operators, weakly compact operators, and compact operators from $X$ to $Y$ will be denoted by $L(X,Y),W(X,Y),$ and $K(X,Y),$ respectively. The projective tensor product of two Banach spaces $X$ and $Y$ will be denoted by $X\widehat{\bigotimes}_{\pi}Y.$

A bounded linear operator $T:X\rightarrow Y$ is called completely continuous, if $T$ maps weakly convergent sequences to norm convergent sequences [1]. The set of all completely continuous operators from $X$ to $Y$ is denoted by $CC(X,Y).$ A bounded linear $T$ from a Banach space $X$ to a Banach space $Y$ is called Dunford-Pettis completely continuous, if it transforms Dunford-Pettis and weakly null sequences to norm null ones. The class of Dunford-Pettis completely continuous operators from $X$ into $Y$ is denoted by $DPcc(X,Y).$

A sequence $(x_{n})_{n}$ in $X$ is called weakly $p$-summable, if $(x^{\ast}(x_{n}))_{n}\in\ell_{p}$ for each $x^{\ast}\in X^{\ast}.$ We denote the set of all weakly $p$-summable sequences in $X$ is denoted by $\ell_{p}^{w}(X)$ [10]. The weakly $\infty$-summable sequences are precisely the weakly null sequences. Note that, a sequence $(x_{n})_{n}$ in $X$ is said to be weakly $p$-convergent to $x\in X$ if $(x_{n}-x)_{n}$ is weakly $p$-summable. A sequence $(x_{n})_{n}$ in a Banach space $X$ is weakly $p$-Cauchy if for each pair of strictly increasing sequences $(k_{n})_{n}$ and $(j_{n})_{n}$ of positive integers, the sequence $(x_{k_{n}}-x_{j_{n}})_{n}$ is weakly $p$-summable in $X$ [6]. Notice that, every weakly $p$-convergent sequence is weakly $p$-Cauchy, and the weakly $\infty$-Cauchy sequences are precisely the weakly Cauchy sequences. A bounded linear operator $T$ between two Banach spaces is called $p$-convergent, if it transforms weakly $p$-summable sequences into norm null sequences [4]. We denote the class of $p$-convergent operators from $X$ into $Y$ by $C_{p}(X,Y).$ A Banach space $X$ has the $p$-Schur property (in short $X\in(C_{p})$), if the identity operator on $X$ is $p$-convergent. A Banach space $X$ has the Dunford-Pettis property of order $p$ ($X$ has the $(DPP_{p})$), if every weakly compact operator on X is p-convergent. Equivalently, $X$ has the $(DPP_{p})$ if and only if for every weakly $p$-summable sequence $(x_{n})_{n}$ in $X$ and weakly-null sequence $(x^{\ast}_{n})_{n}$ in $X^{\ast},$ we have $x^{\ast}_{n}(x_{n})\rightarrow 0$ as $n\rightarrow\infty$ [5].

A subset $K$ of a Banach space $X$ is called relatively weakly $p$-compact, if each sequence in $K$ admits a weakly $p$-convergent subsequence with limit in $X.$ If the “limit point” of each weakly $p$-convergent subsequence lies in $K,$ then we say that $K$ is a weakly p-compact set. Note that, the weakly $\infty$-compact sets are precisely the weakly compact. A bounded linear operator $T:X\rightarrow Y$ is called weakly $p$-compact, if $T(B_{X})$ is a relatively weakly $p$-compact set in $Y.$ The set of all weakly $p$-compact operators $T:X\rightarrow Y$ is denoted by $W_{p}(X,Y).$ We refer the reader for undefined terminologies to the classical references [1, 9].

3. p-sequentially Right and p-sequentially Right*∗* properties

on Banach spaces

The authors in [7, 15] by using Right topology, proved that a sequence $(x_{n})_{n}$ in a Banach space $X$ is Right null if and only if it is Dunford-Pettis and weakly null. Also, they showed that a sequence $(x_{n})_{n}$ in a Banach space $X$ is Right Cauchy if and only if it is Dunford-Pettis and weakly Cauchy.

Inspired by the above works, for convenience, we apply the notions $p$-Right null and $p$-Right Cauchy sequences instead of weakly $p$-summable and weakly $p$-Cauchy sequences which are Dunford-Pettis sets, respectively.

The main aim of this section is to obtain some characterizations of $p$-Right sets that are relatively weakly $q$-compact.

Definition 3.1**.**

$\rm{(i)}$ A bounded subset $K$ of a Banach space $X$ is said to be $p$-Right*∗* set, if for every $p$-Right null sequence $(x^{\ast}_{n})_{n}$ in $X^{\ast}$ it follows:

[TABLE]

$\rm{(ii)}$ We say that $X$ has the $p$-sequentially Right*∗* property (in short $X$ has the $p$-$(SR^{\ast})$ property), if every $p$-Right*∗* set is relatively weakly compact.

It is easy to verify that, $\infty$-Right*∗* sets are precisely Right*∗* sets and the $\infty$-$(SR)^{\ast}$ property is precisely the sequentially Right*∗* property (see, [14]).

Suppose that $K$ is a bounded subset of $X$ and $B(K)$ is the Banach space of all bounded real-valued functions defined on $K$, provided with the superemum norm. The natural evaluation map $E:X^{\ast}\rightarrow B(K)$ defined by $E(x^{\ast})(x)=x^{\ast}(x)$ for every $x\in K,~{}~{}~{}x^{\ast}\in X^{\ast},$ has been used by many authors to study properties of $K.$ Similarly, if $K$ is a bounded subset of $X^{\ast},$ the natural evaluation map $E_{X}:X\rightarrow B(K)$ defined by $E_{X}(x)(x^{\ast})=x^{\ast}(x)$ for every $x\in X,~{}~{}~{}x^{\ast}\in K,$ (see, [3]).

At the first, inspired by Theorem 3.1 of [3], we obtain some characterizations of notions $p$-Right sets and $p$-Right*∗* sets by using evaluation maps which will be used in the sequel.

Lemma 3.2**.**

*The following statements hold:

$\rm{(i)}$ If $T\in L(X,Y),$ then $T$ is Dunford-Pettis $p$-convergent if and only if $T^{\ast}(B_{Y^{\ast}})$ is a $p$-Right subset of $X^{\ast}.$

$\rm{(ii)}$ A bounded subset $K$ of $X^{\ast}$ is a $p$-Right set if and only if $E_{X}:X\rightarrow B(K)$ is Dunford-Pettis $p$-convergent.

$\rm{(iii)}$ If $T\in L(X,Y),$ then $T^{\ast}$ is Dunford-Pettis $p$-convergent if and only if $T(B_{X})$ is a $p$-Right∗ subset of $Y.$

$\rm{(iv)}$ $X^{\ast}$ has the $p$-$(DPrcP)$ if and only if every bounded subset of $X$ is a $p$-Right∗ set.

$\rm{(v)}$ A bounded subset $K$ of $X$ is a $p$-Right∗ set if and only if $E:X^{\ast}\rightarrow B(K)$ is Dunford-Pettis $p$-convergent.

$\rm{(vi)}$ A bounded subset $K$ of $X$ is a $p$-Right∗ set if and only if there is a Banach space $Y$ and a bounded linear operator $T:Y\rightarrow X$ so that $T$ and $T^{\ast}$ are Dunford-Pettis $p$-convergent and $K\subseteq T(B_{Y}).$*

Proof.

$\rm{(i)}$ Suppose that $T:X\rightarrow Y$ is a bounded linear operator. Clearly, $T^{\ast}(B_{Y^{\ast}})$ is a $p$-Right set if and only if

[TABLE]

[TABLE]

for each $p$-Right null sequence $(x_{n})_{n}$ in $X$ if and only if $T$ is Dunford-Pettis $p$-convergent.

$\rm{(ii)}$ Let $K$ be a bounded subset of $X^{\ast}.$ The evaluation map $E_{X}:X\rightarrow B(K)$ is Dunford-Pettis $p$-converging if and only if $\|E_{X}(x_{n})\|\rightarrow 0$ for each $p$-Right null sequence $(x_{n})_{n}$ in $X$ if and only if

[TABLE]

for each $p$-Right null sequence $(x_{n})_{n}$ in $X$ if and only if $K$ is a $p$-Right set.

$\rm{(iii)}$ Suppose that $T:X\rightarrow Y$ is a bounded linear operator. Clearly, $T(B_{X})$ is a $p$-Right*∗* set if and only if

[TABLE]

[TABLE]

for each $p$-Right null sequence $(y^{\ast}_{n})_{n}$ in $X^{\ast}$ if and only if $T^{\ast}$ is Dunford-Pettis $p$-convergent.

$\rm{(iv)}$ is clear.

$\rm{(v)}$ Suppose that $K$ is a bounded subset of $X$ and $E:X^{\ast}\rightarrow B(K)$ is a Dunford-Pettis $p$-convergent operator. Thus $E^{\ast}$ maps the unit ball of $B(K)^{\ast},$ to a $p$-Right set in $X^{\ast\ast}.$ However, if $k\in K$ and $\delta_{k}$ denotes the point mass at $k,$ then $E^{\ast}(\{\delta_{k}:k\in K\})=K,$ and so $K$ is a $p$-Right set in $X^{\ast\ast}.$ Hence $K$ is a $p$-Right*∗* set in $X.$ Conversely, suppose that $K$ is a $p$-Right*∗* set in $X$, and let $E:X^{\ast}\rightarrow B(K)$ be the evaluation map. If $(x^{\ast}_{n})_{n}$ is a $p$-Right null sequence in $X^{\ast},$ then

[TABLE]

and $E$ is a Dunford-Pettis $p$-convergent operator.

$\rm{(vi)}$ Suppose that $K$ is a $p$-Right set and $Y=\ell_{1}(K).$ Define $T:Y\rightarrow X$ by $T(f)=\sum_{k\in K}f(k)k$ for $f\in\ell_{1}(K).$ It is clear that $T$ is a bounded linear operator, and $K\subseteq T(B_{\ell_{1}(K)}).$ Since $\ell_{1}(K)$ has the Schur property, the operator $T$ is completely continuous and so, it is $p$-convergent. Thus, $T$ is a Dunford-Pettis $p$-convergent operator. It is easy to verify that, $T^{\ast}$ is the evaluation map $E:X^{\ast}\rightarrow B(K).$ Hence, $T^{\ast}$ is Dunford-Pettis $p$-convergent by $\rm{(v)}.$ ∎

Recall that, a bounded subset $K$ of $X^{\ast}$ is called an $(L)$ set, if each weakly null sequence $(x_{n})_{n}$ in $X$ tends to [math] uniformly on $K$ [1].

Bator et al. [3] showed that every $(L)$ subset of $X^{\ast}$ is a Dunford-Pettis set in $X^{\ast}$ if and only if $T^{\ast\ast}$ is completely continuous whenever $Y$ is an arbitrary Banach space and $T:X\rightarrow Y$ is a completely continuous operator.

It is easy to verify that, for each $1\leq p\leq\infty,$ every $p$-Right*∗* subset of dual space is a $p$-Right set, while the converse of implication is false. The following theorem continues our study of the relationship between $p$-Right subsets and $p$-Right*∗* subsets of dual spaces.

Theorem 3.3**.**

Every $p$-Right subset of $X^{\ast}$ is a $p$-Right∗ set in $X^{\ast}$ if and only if $T^{\ast\ast}$ is a Dunford-Pettis $p$-convergent operator whenever $Y$ is an arbitrary Banach space and $T:X\rightarrow Y$ is a Dunford-Pettis $p$-convergent operator.

Proof.

Suppose that $T:X\rightarrow Y$ is a Dunford-Pettis $p$-convergent operator. The part (i) of Lemma of 3.2, yields that $T^{\ast}(B_{Y^{\ast}})$ is a $p$-Right set. By the hypothesis $T^{\ast}(B_{Y^{\ast}})$ is a $p$-Right*∗* set. By applying the Lemma 3.2 $\rm{(iii)},$ we see that $T^{\ast\ast}$ is a Dunford-Pettis $p$-convergent operator. Conversely, suppose that $K$ is a $p$-Right subset of $X^{\ast}.$ The part (ii) of Lemma 3.2, implies that $E_{X}$ is Dunford-Pettis $p$-convergent. Therefore, by the hypothesis, $E_{X}^{\ast\ast}$ is Dunford-Pettis $p$-convergent. Hence, if the unit ball of $B(K)^{\ast}$ denoted by $S,$ then $E^{\ast}_{X}(S)$ is a $p$-Right*∗* set. Since $K\subset E_{X}^{\ast}(S),$ $K$ is a $p$-Right*∗* set in $X^{\ast}.$ ∎

Corollary 3.4**.**

Every Right subset of $X^{\ast}$ is a Right∗ set in $X^{\ast}$ if and only if $T^{\ast\ast}$ is Dunford-Pettis completely continuous whenever $Y$ is an arbitrary Banach space and $T:X\rightarrow Y$ is a pseudo weakly compact operator.

Recall from [1], that the space of all finite regular Borel signed measures on the compact space K is denoted by M(K). It is well known that $M(K)=C(K)^{\ast}.$

Corollary 3.5**.**

If $K$ is a compact Hausdorff space, then every $p$-Right subset of $M(K)$ is a $p$-Right∗ set in $M(K).$

Proof.

Suppose that $K$ is a compact Hausdorff space, $Y$ is a Banach space and $T:C(K)\rightarrow Y$ is a Dunford-Pettis $p$-convergent operator. Since $C(K)$ has the $p$-sequentially Right property, $T$ is weakly compact and so, $T^{\ast\ast}$ is weakly compact. Therefore, $T^{\ast\ast}$ is Dunford-Pettis $p$-convergent. Hence, Theorem 3.3 implies that, every $p$-Right subset of $M(K)$ is a $p$-Right*∗* set in $M(K).$ ∎

If $\mathcal{M}$ is a closed subspace of $\mathcal{U}(X,Y),$ then for arbitrary elements $x\in X$ and $y^{\ast}\in Y^{\ast},$ the evaluation operators $\phi_{x}:\mathcal{M}\rightarrow Y$ and $\psi_{y^{\ast}}$ on $\mathcal{M}$ are defined by $\phi_{x}(T)=T(x),$ $\psi_{y^{\ast}}(T)=T^{\ast}(y^{\ast}).$ Also, the point evaluation sets related to $x\in X$ and $y^{\ast}\in Y^{\ast}$ are the images of the closed unit ball $B_{\mathcal{M}}$ of $\mathcal{M},$ under the evaluation operators $\phi_{x}$ and $\psi_{y^{\ast}}$ and they are denoted by $\mathcal{M}_{1}(x)$ and $\widetilde{\mathcal{M}}_{1}(y^{\ast})$ respectively [21]. Note that, if we speak about $\mathcal{U}(X,Y)$ or its linear subspace $\mathcal{M},$ then the related norm is the ideal norm $\mathcal{A}(.)$ while, the operator norm $\|.\|$ is applied when the space is a linear subspace of $L(X,Y).$

Theorem 3.6**.**

Suppose that $1\leq p\leq\infty$ and the dual $\mathcal{M^{\ast}}$ of a closed subspace $\mathcal{M}\subseteq\mathcal{U}(X,Y)$ has the $p$-$(DPrcP).$ Then all of the point evaluations $\mathcal{M}_{1}(x)$ and $\widetilde{\mathcal{M}}_{1}(y^{\ast})$ are $p$-Right sets.

Proof.

Since $\mathcal{M^{\ast}}$ has the $p$-$(DPrcP),$ $\phi_{x}^{\ast}$ is a Dunford-Pettis $p$-convergent operator. Now, suppose that $(y_{n}^{\ast})_{n}$ is a $p$-Right null sequence in $Y^{\ast}.$ It is clear that $\displaystyle\lim_{n\rightarrow\infty}\|\phi_{x}^{\ast}(y_{n}^{\ast}))\|=0,$ for all $x\in X.$ On the other hand,

[TABLE]

This shows that $\mathcal{M}_{1}(x)$ is a $p$-Right set in $Y,$ for all $x\in X.$ A similar proof shows that $\widetilde{\mathcal{M}}_{1}(y^{\ast})$ is a a $p$-Right set in $X^{\ast}.$ ∎

In the following, we obtain some sufficient conditions for which the point evaluations $\mathcal{M}_{1}(x)$ and $\widetilde{\mathcal{M}}_{1}(y^{\ast})$ are relatively weakly compact for all $x\in X$ and all $y^{\ast}\in Y^{\ast}.$

Theorem 3.7**.**

Suppose that $1\leq p\leq\infty$ and $X^{\ast\ast}$ and $Y^{\ast}$ have the $p$-$(SR)$ property. If $\mathcal{M}\subseteq\mathcal{U}(X,Y)$ is a closed subspace so that the natural restriction operator $R:\mathcal{U}(X,Y)^{\ast}\rightarrow\mathcal{M^{\ast}}$ is a Dunford-Pettis $p$-convergent operator, then all of the point evaluations $\mathcal{M}_{1}(x)$ and $\widetilde{\mathcal{M}}_{1}(y^{\ast})$ are relatively weakly compact.

Proof.

It is enough to show that $\phi_{x}$ and $\psi_{y^{\ast}}$ are weakly compact operators. For this purpose suppose that $T\in\mathcal{U}(X,Y).$ Since $\|T\|\leq\mathcal{A}(T),$ it is not difficult to show that, the operator $\psi:X^{\ast\ast}\widehat{\bigotimes}_{\pi}Y^{\ast}\rightarrow\mathcal{U}(X,Y)^{\ast}$ which is defined by

[TABLE]

is linear and continuous, where $\vartheta=\displaystyle\sum_{n=1}^{\infty}x_{n}^{\ast\ast}\bigotimes y_{n}^{\ast}.$ Fix now an arbitrary element $x\in X$ we define $U_{x}:Y^{\ast}\rightarrow X^{\ast\ast}\widehat{\bigotimes}_{\pi}Y^{\ast}$ by $U_{x}(y^{\ast})=x\bigotimes y^{\ast}.$ It is clear that the operator $\phi_{x}^{\ast}=R\circ\psi\circ U_{x}$ is a Dunford-Pettis $p$-convergent. Since $Y^{\ast}$ has the $p$-$(SR)$ property, we conclude that $\phi_{x}^{\ast}$ is a weakly compact operator. Hence, $\phi_{x}$ is weakly compact. Similarly, we can see that $\psi_{y^{\ast}}$ is weakly compact. ∎

Theorem 3.8**.**

*Let $X$ be a Banach space and $1\leq p<q\leq\infty.$ The following statements are equivalent:

$\rm{(i)}$ For every Banach space $Y,$ if $T:X\rightarrow Y$ is a Dunford-Pettis $p$-convergent operator, then $T$ has a weakly $q$-precompact (weakly $q$-compact, $q$-compact) adjoint,

$\rm{(ii)}$ Same as $\rm{(i)}$ with $Y=\ell_{\infty},$

$\rm{(iii)}$ Every $p$-Right subset of $X^{\ast}$ is weakly $q$-precompact (relatively weakly $q$-compact, $q$-compact).*

Proof.

We will show that in the relatively weakly $q$-compact case. The other proof is similar.

(i) $\Rightarrow$ (ii) is obvious.

(ii) $\Rightarrow$ (iii) Let $K$ be a $p$-Right subset of $X^{\ast}$ and let $(x^{\ast}_{n})_{n}$ be a sequence in $K.$ Define $T:X\rightarrow\ell_{\infty}$ by $T(x)=(x^{\ast}_{n}(x)).$ Let $(x_{n})_{n}$ be a $p$-Right null sequence in $X.$ Since $K$ is a $p$-Right set,

[TABLE]

Therefore, $T$ is Dunford-Pettis $p$-convergent. Hence, by the hypothesis, $T^{\ast}$ is weakly $q$-compact and so, $(T^{\ast}(e^{1}_{n}))_{n}=(x^{\ast}_{n})_{n}$ has a weakly $q$-convergent subsequence.

(iii) $\Rightarrow$ (i) Let $T:X\rightarrow Y$ be a Dunford-Pettis $p$-convergent operator. Then $T^{\ast}(B_{Y^{\ast}})$ is a $p$-Right subset of $X.$ Therefore $T^{\ast}(B_{Y^{\ast}})$ is relatively weakly $q$-compact, and thus $T^{\ast}$ is weakly $q$-compact. ∎

Let $A$ and $B$ be nonempty subsets of a Banach space $X,$ we define ordinary distance and non-symmetrized Hausdorff distance respectively, by

$d(A,B)=\inf\{d(a,b):a\in A,b\in B\},\hskip 25.6073pt\hat{d}(A,B)=\sup\{d(a,B):a\in A\}.$

Let $X$ be a Banach space and $K$ be a bounded subset of $X^{\ast}.$ For $1\leq p\leq\infty,$ we set

$\zeta_{p}(K)=\inf\{\hat{d}(A,K):K\subset X^{\ast}$ is a $p$-Right set $\}.$

We can conclude that $\zeta_{p}(K)=0$ if and only if $K\subset X^{\ast}$ is a $p$-Right set. Now, let $K$ be a bounded subset of a Banach space $X.$ The de Blasi measure of weak non-compactness of $K$ is defined by

$\omega(K)=\inf\{\hat{d}(K,A):\emptyset\neq A\subset X$ is weakly-compact $\}.$

It is clear that $\omega(K)=0$ if and only if $K$ is relatively weakly compact. For a bounded linear operator $T:X\rightarrow Y,$ we denote $\zeta_{p}(T(B_{X})),\omega(T(B_{X})$ by $\zeta_{p}(T),\omega(T)$ respectively.

Corollary 3.9**.**

*Let $X$ be a Banach space and $1\leq p\leq\infty.$ The following statements are equivalent:

$\rm{(i)}$ $X$ has the $p$-$(SR)$ property.

$\rm{(ii)}$ For each Banach space $Y,$ adjoint every Dunford-Pettis $p$-convergent $T:X\rightarrow Y$ is weakly compact.

$\rm{(iii)}$ Same as $\rm{(ii)}$ with $Y=\ell_{\infty}.$

$\rm{(iv)}$ $\omega(T^{\ast})\leq\zeta_{p}(T^{\ast})$ for every bounded linear operator $T$ from $X$ into any Banach space $Y.$

$\rm{(v)}$ $\omega(K)\leq\zeta_{p}(K)$ for every bounded subset $K$ of $X^{\ast}.$*

Recently, the notions of $p$-$(V)$ sets and $p$-$(V)$ property as an extension of the notions $(V)$ sets and Pelczyński’s property $(V)$ introduced by Li et al. [19] as follows:

• •

A bounded subset $K$ of $X^{\ast}$ is a $p$-$(V)$ set, if $\displaystyle\lim_{n\rightarrow\infty}\displaystyle\sup_{x^{\ast}\in K}|x^{\ast}(x_{n})|=0,$ for every weakly $p$-summable sequence $(x_{n})_{n}$ in $X.$

• •

A Banach space $X$ has Pelczyński’s property $(V)$ of order $p$ ($p$-$(V)$ property), if every $p$-$(V)$ subset of $X^{\ast}$ is relatively weakly compact.

• •

A bounded subset $K$ of $X$ is a $p$-$(V^{\ast})$ set, if $\displaystyle\lim_{n\rightarrow\infty}\displaystyle\sup_{x\in K}|x^{\ast}_{n}(x)|=0,$ for every weakly $p$-summable sequence $(x^{\ast}_{n})_{n}$ in $X^{\ast}.$

• •

A Banach space $X$ has Pelczyński’s property $(V^{\ast})$ of order $p$ ($p$-$(V^{\ast})$ property), if every $p$-$(V^{\ast})$ subset of $X$ is relatively weakly compact.

The proof of the following proposition is similar to the proof ([16, Corollary 3.19 (ii) ]). Therefore, its proof is omitted

Proposition 3.10**.**

$X^{\ast}$* has the $(DPP_{p})$ if and only if each $p$-Right∗ set in $X$ is a $p$-$(V^{\ast})$ set.*

Proposition 3.11**.**

Let $X$ be a Banach space. The following statements hold:

$\rm{(i)}$ If $X$ has the $p$-$(SR)$ property, then $X^{\ast}$ has the $p$-$(SR^{\ast})$ property.

$\rm{(ii)}$ If $X$ has the $p$-$(SR)$ property, then $X$ has the $p$-$(V)$ property.

$\rm{(iii)}$ If $X^{\ast}$ has the $p$-$(SR)$ property, then $X$ has the $p$-$(SR^{\ast})$ property,

$\rm{(iv)}$ If $X$ has the $p$-$(SR^{\ast})$ property, then $X$ has the $p$-$(V^{\ast})$ property.

$\rm{(v)}$ Let $Y$ be a reflexive subspace of $X.$ If $\frac{X}{Y}$ has the $p$-$(SR^{\ast})$ property, then $X$ has the same property.

Proof.

Since, the proof of $\rm{(i)},$ $\rm{(ii)},$ $\rm{(iii)}$ and $\rm{(iv)}$ are clear, we only prove $\rm{(v)}.$

$\rm{(i)}$ Let $Q:X\rightarrow\frac{X}{Y}$ be the quotient map. Let $K$ be a $p$-Right*∗* set in $X$ and $(x_{n})_{n}$ be an arbitrary sequence in $K.$ Then $(Q(x_{n}))_{n}$ is a $p$-Right*∗* set in $\frac{X}{Y},$ and thus relatively weakly compact. By passing to a subsequence, suppose $(Q(x_{n}))_{n}$ is weakly convergent. By ([17, Theorem 2.7]), $(x_{n})_{n}$ has a weakly convergent subsequence. Thus $X$ has the $p$-$(SR^{\ast})$ property. ∎

Corollary 3.12**.**

If $X$ has the $(DPP_{p}),$ then $X$ has Pelczy$\acute{n}$ski’s property $(V)$ of order $p$ if and only $X$ has the $p$-$(SR)$ property.

Proof.

Suppose that $X$ has Pelczy$\acute{n}$ski’s property $(V)$ of order $p.$ We show that for each Banach space $Y,$ adjoint every Dunford-Pettis $p$-convergent $T:X\rightarrow Y$ is weakly compact. Let $T\in DPC_{p}(X,Y).$ The part $\rm{(i)}$ of Lemma 3.2, implies that $T^{\ast}(B_{Y^{\ast}})$ is a $p$-Right set in $X^{\ast}.$ So, ([16, Corollary 3.19 (ii) ]) implies that $T^{\ast}(B_{Y^{\ast}})$ is a $p$-$(V)$ set in $X^{\ast}.$ Since $X$ has Pelczy$\acute{n}$ski’s property $(V)$ of order $p,$ $T^{\ast}$ is weakly compact. Hence, Corollary 3.9 implies that $X$ has the $p$-$(SR)$ property. Conversely, If $X$ has the $p$-$(SR)$ property, then $X$ has Pelczy$\acute{n}$ski’s property $(V)$ of order $p.$ Since every $p$-$(V)$ set in $X^{\ast}$ is a $p$-Right set. ∎

Suppose that $X$ is a Banach space and $Y$ is a subspace of $X^{\ast}.$ We define ${}^{\bot}Y:=\{x\in X:y^{\ast}(x)=0~{}~{}$ for all $~{}~{}~{}y^{\ast}\in Y^{\ast}\}.$

Corollary 3.13**.**

$\rm{(i)}$* If $X$ is an infinite dimensional non reflexive Banach space with the $p$-Schur property, then $X$ does not have the $p$-$(SR)$ property.

$\rm{(ii)}$ If every separable subspace of $X$ has the $p$-$(SR)$ property, then $X$ has the same property.

$\rm{(iii)}$ Let $Y$ be a reflexive subspace of $X^{\ast}.$ If ${}^{\bot}Y$ has the $p$-$(SR)$ property, then $X$ has the same property.*

Proof.

$\rm{(i)}$ Since $X\in C_{p},$ the identity operator $id_{X}:X\rightarrow X$ is $p$-convergent and so, it is Dunford-Pettis $p$-convergent. It is clear that $id_{X}$ is not weakly compact. Hence, Corollary 3.9 implies that $X$ does not have the $p$-$(SR)$ property.

$\rm{(ii)}$ Let $(x_{n})_{n}$ be a sequence in $B_{X}$ and let $Z=[x_{n}:n\in\mathbb{N}]$ be the closed linear span of $(x_{n})_{n}.$ Since $Z$ is a separable subspace of $X,$ $Z$ has the $p$-$(SR)$ property. Now, let $T:X\rightarrow Y$ be a Dunford-Pettis $p$-convergent operator. It is clear that $T_{|Z}$ is a Dunford-Pettis $p$-convergent operator. Therefore, Corollary 3.9, implies that $T_{|Z}$ is weakly compact. Hence, there is a subsequence $(x_{n_{k}})_{k}$ of $(x_{n})_{n}$ so that $T(x_{n_{k}})$ is weakly convergent. Thus $T$ is weakly compact. Now an appeal to Corollary 3.9 completes the proof.

$\rm{(iii)}$ By ([20, Theorem 1. 10. 6]), there exists a quotient map $Q:X^{\ast}\rightarrow\frac{X^{\ast}}{Y}$ and a surjective isomorphism $i:\frac{X^{\ast}}{Y}\rightarrow~{}(^{\bot}Y)^{\ast}$ such that $i\circ Q:X^{\ast}\rightarrow~{}(^{\bot}Y)^{\ast}$ is $w^{\ast}$-$w^{\ast}$ continuous. Therefore, there is $S:~{}^{\bot}Y\rightarrow X$ with $S^{\ast}=i\circ Q.$ Hence, for any Dunford-Pettis $p$-convergent operator     $T:X\rightarrow Z,$ the operator $T\circ S:~{}^{\bot}Y\rightarrow Z$ is Dunford-Pettis $p$-convergent, that must be weakly compact; hence, $S^{\ast}\circ T^{\ast}=i\circ Q\circ T^{\ast}$ is also weakly compact, this in turn gives that $Q\circ T^{\ast}$ must be weakly compact, since $i$ is a surjective isomorphism. Therefore $T^{\ast}.$  The Corollary 3.9 completes the proof. ∎

Theorem 3.14**.**

$\rm{(i)}$* Suppose that $L_{w^{\ast}}(X^{\ast},Y)=K_{w^{\ast}}(X^{\ast},Y).$ If $X$ and $Y$ have the $p$-$(SR)$ property, then $K_{w^{\ast}}(X^{\ast},Y)$ has the same property.

$\rm{(i)}$ Suppose that $L(X,Y)=K(X,Y).$ If $X^{\ast}$ and $Y$ have the $p$-$(SR)$ property, then $K(X,Y)$ has the same property.*

Proof.

Since the proofs of (i) and (ii) are essentially the same, we only present that of (i).

$\rm{(i)}$ Suppose $X$ and $Y$ have the $p$-$(SR)$ property. Let $H$ be a $p$-Right subset of $K_{w^{\ast}}(X^{\ast},Y).$ For fixed $x^{\ast}$ in $X^{\ast},$ the map $T\mapsto T(x^{\ast})$ is a bounded operator from $K_{w^{\ast}}(X^{\ast},Y)$ into $Y.$ It can easily seen that continuous linear images of $p$-Right sets are $p$-Right sets. Therefore, $H(x^{\ast})$ is a $p$-Right subset of $Y,$ hence it is relatively weakly compact. For fixed $y^{\ast}$ in $Y^{\ast}$ the map $T\mapsto T^{\ast}(y^{\ast})$ is a bounded linear operator from $K_{w^{\ast}}(X^{\ast},Y)$ into $X.$ So, $H^{\ast}(y^{\ast})$ is a $p$-Right subset of $X,$ hence it is relatively weakly compact. Then, ([12, Theorem 4. 8]) implies that $H$ is relatively weakly compact. ∎

Cilia and Emmanuele in [7] investigated whether the projective tensor product of two Banach spaces $X$ and $Y$ has the sequentially Right property when $X$ and $Y$ have the respective property.

In the following, the stability of $p$-sequentially Right property for projective tensor product between Banach spaces is investigated.

Theorem 3.15**.**

Suppose that $X$ has the $p$-$(SR)$ property and $Y$ is a reflexive space. If $L(X,Y^{\ast})=K(X,Y^{\ast}),$ then $X\widehat{\bigotimes}_{\pi}Y$ has the $p$-$(SR)$ property.

Proof.

Let $H$ be a $p$-Right subset of $(X\widehat{\bigotimes}_{\pi}Y)^{\ast}\simeq L(X,Y^{\ast}).$ We claim that $K$ is relatively weakly compact. We show that the conditions (i) and (ii) of ([13, Theorem 4]) are true. Let $(T_{n})$ be an arbitrary sequence in $H.$ If $x\in X,$ it is enough to show that $\{T_{n}(x):n\in\mathbb{N}\}$ is a $p$-Right subset of $Y^{\ast}.$ For this purpose, suppose that $(y_{n})_{n}$ is a $p$-Right null sequence in $Y.$ For each $n\in\mathbb{N},$ we have:

[TABLE]

We claim that $(x\otimes y_{n})_{n}$ is a $p$-Right null sequence in $X\widehat{\bigotimes}_{\pi}Y.$ If $T\in(X\widehat{\bigotimes}_{\pi}Y)^{\ast}\simeq L(X,Y^{\ast}),$ then

[TABLE]

since $(y_{n})_{n}$ is weakly $p$-summable. Thus $(x\otimes y_{n})_{n}$ is weakly $p$-summable in $X\widehat{\bigotimes}_{\pi}Y.$ Let $(A_{n})_{n}$ be a weakly null sequence in $(X\widehat{\bigotimes}_{\pi}Y)^{\ast}\simeq L(X,Y^{\ast}).$ Since the map $\varphi_{x}:L(X,Y^{\ast})\rightarrow Y^{\ast},$ $\varphi_{x}(T)=T(x)$ is linear and bounded, $(A_{n}(x))_{n}$ is weakly null in $Y^{\ast}.$ Since $(y_{n})_{n}$ is a Dunford-Pettis sequence in $Y.$

[TABLE]

Hence, $(x\otimes y_{n})_{n}$ is Dunford-Pettis and so, $(x\otimes y_{n})_{n}$ is $p$-Right null in $X\widehat{\bigotimes}_{\pi}Y.$ Therefore, the equivalence $\rm{(i)}$ and $\rm{(v)}$ in ([16, Theorem 3.26]) implies that $\{T_{n}(x):n\in\mathbb{N}\}$ is a $p$-Right set in $Y^{\ast}.$ Therefore, $\{T_{n}(x):n\in\mathbb{N}\}$ is a relatively weakly compact. Now, let $y\in Y$ and $(x_{n})_{n}$ be a $p$-Right null sequence in $X.$ An argument similar to the above one can see that $(x_{n}\otimes y)_{n}$ is a $p$-Right null sequence in $X\widehat{\bigotimes}_{\pi}Y.$ Therefore, by reapplying ([16, Theorem 3.26]) $\{T^{\ast}_{n}(y):n\in\mathbb{N}\}$ is a $p$-Right subset of $X^{\ast}.$ So, $\{T^{\ast}_{n}(y):n\in\mathbb{N}\}$ is relatively weakly compact. Hence $H$ is relatively weakly compact. ∎

Let $(X_{n})_{n\in\mathbb{N}}$ be a sequence of Banach spaces. If $1\leq r<\infty$ the space of all vector-valued sequences $(\displaystyle\sum_{n=1}^{\infty}\oplus X_{n})_{\ell_{r}}$ is called, the infinite direct sum of $X_{n}$ in the sense of $\ell_{r},$ consisting of all sequences $x=(x_{n})_{n}$ with values in $X_{n}$ such that $\|x\|_{r}=(\displaystyle\sum_{n=1}^{\infty}\|x_{n}\|^{r})^{\frac{1}{r}}<\infty.$ For every $n\in\mathbb{N},$ we denote the canonical projection from $(\displaystyle\sum_{n=1}^{\infty}\oplus X_{n})_{\ell_{r}}$ into $X_{n}$ by $\pi_{n}.$ Also, we denote the canonical projection from $(\displaystyle\sum_{n=1}^{\infty}\oplus X^{\ast}_{n})_{\ell_{r^{\ast}}}$ onto $X^{\ast}_{n}$ by $P_{n}.$

Using the ([16, Corollary 3.19]),and ([19, Theorem 3.1]), we obtain the following result:

Theorem 3.16**.**

*Let $1<p<\infty$ and $(X_{n})_{n}$ be a sequence of Banach spaces with $(DPP_{p})$ and let $X=(\sum_{n=1}^{\infty}\oplus X_{n})_{\ell_{p}}.$ The following are equivalent for a bounded subset $K$ of $X^{\ast}:$

$\rm{(i)}$ $K$ is a $p^{\ast}$-Right set.

$\rm{(ii)}$ $P_{n}(K)$ is a $p^{\ast}$-Right set for each $n\in\mathbb{N}$ and*

[TABLE]

Theorem 3.17**.**

Let $1<p<\infty$ and $(X_{n})_{n}$ be a sequence of Banach spaces. If $X=(\displaystyle\sum_{n=1}^{\infty}\oplus X_{n})_{\ell_{p}}$ and $1\leq q<p^{\ast},$ then a bounded subset $K$ of $X^{\ast}$ is $q$-Right set if and only if each $P_{n}(K)$ is $q$-Right set.

Proof.

It can easily seen that continuous linear images of $q$-Right set is $q$-Right set. Therefore, we only prove the sufficient part. Assume that $K$ is not a $q$-Right set. Therefore, there exist $\varepsilon_{0}>0,$ a $q$-Right null sequence $(x_{n})_{n}$ in $X$ and a sequence $(x^{\ast}_{n})_{n}$ in $K$ such that

$|\langle x^{\ast}_{n},x_{n}\rangle|=|\displaystyle\sum_{k=1}^{\infty}\langle P_{k}x^{\ast}_{n},\pi_{k}x_{n}\rangle|>\varepsilon_{0},~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}n=1,2,3,...~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}$ $~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(\ast)$

By the assumption, we obtain

$\displaystyle\lim_{n\rightarrow\infty}|\langle P_{k}x^{\ast}_{n},\pi_{k}x_{n}\rangle|=0,~{}~{}~{}~{}~{}~{}~{}~{}k=1,2,3,...~{}~{}~{}~{}$ $~{}~{}~{}~{}~{}~{}(\ast\ast)$

By induction on $n$ in $(\ast)$ and $k$ in $(\ast\ast),$ there exist two strictly increasing sequences $(n_{j})_{j}$ and $(k_{j})_{j}$ of positive integers such that

$|\displaystyle\sum_{k=k_{j-1}+1}^{k_{j}}\langle P_{k}x^{\ast}_{n_{j}},\pi_{k}x_{n_{j}}\rangle|>\frac{\varepsilon_{0}}{2},~{}~{}~{}~{}~{}~{}j=1,2,3,...$

For each $j=1,2,...,$ we consider $y_{j}=x_{n_{j}}$ and $y^{\ast}_{j}\in X^{\ast}$ by

[TABLE]

It is clear that $(y_{j})_{j}$ is a $q$-Right null sequence in $X$ such that

$|\langle y^{\ast}_{j},y_{j}\rangle|=|\displaystyle\sum_{k=k_{j-1}+1}^{k_{j}}\langle P_{k}x^{\ast}_{n_{j}},\pi_{k}x_{n_{j}}\rangle|>\frac{\varepsilon_{0}}{2},~{}~{}~{}~{}~{}~{}j=1,2,3,...$

Since the sequence $(y^{\ast}_{j})_{j}$ has pairwise disjoint supports, Proposition 6.4.1 of [1] implies that $(y^{\ast}_{j})_{j}$ is equivalent to the unit vector basis $(e^{p^{\ast}}_{j})_{j}$ of $\ell_{p^{\ast}}.$ Suppose that $R$ is an isomorphic embedding from $\ell_{p^{\ast}}$ into $X^{\ast}$ such that $R(e^{p^{\ast}}_{j})=y^{\ast}_{j}(j=1,2,...).$ Now, let $T$ be a bounded linear operator from $\ell_{q^{\ast}}$ into $X.$ By Pitt’s Theorem [1], the operator $T^{\ast}\circ R$ is compact and so the sequence $(T^{\ast}(y^{\ast}_{j}))_{j}=(T^{\ast}R(e^{\ast}_{j}))_{j}$ is relatively norm compact. Hence, Theorem 2.3 of [19] implies that the sequence $(y^{\ast}_{j})_{j}$ is a $q$-$(V)$ set and so is a $q$-Right set. Since $(y_{j})_{j}$ is $q$-Right null, we have

$|\langle y^{\ast}_{n},y_{n}\rangle|\leq\sup_{j}|\langle y^{\ast}_{j},y_{n}\rangle|\rightarrow 0$ as $~{}~{}~{}~{}~{}~{}n\rightarrow\infty,$

which is a contradiction. ∎

Theorem 3.18**.**

Let $(X_{n})_{n}$ be a sequence of Banach spaces. If $1<r<\infty$ and $1\leq p<\infty,$ then $X=(\displaystyle\sum_{n=1}^{\infty}\oplus X_{n})_{\ell_{r}}$ has the $p$-$(SR)$ property if and only if each $X_{n}$ has the same property.

Proof.

It is clear that if $X$ has the $p$-$(SR)$ property, then each $X_{n}$ has the $p$-$(SR)$ property. Conversely, let $K$ be a $p$-Right subset of $X^{\ast}.$ Since continuous linear images of $p$-Right sets are $p$-Right sets, each $P_{n}(K)$ is also a $p$-Right set. Since $X_{n}$ has the $p$-$(SR)$ property for each $n\in\mathbb{N},$ each $P_{n}(K)$ is relatively weakly compact. It follows from Lemma 3.4 [19] that $K$ is relatively weakly compact. ∎

Proposition 3.19**.**

Let $(X_{n})_{n}$ be a sequence of Banach spaces. If $1<r<\infty$ and $1\leq p<\infty,$ then each $X_{n}$ has the $p$-$(SR^{\ast})$ property if and only if $X=(\displaystyle\sum_{n=1}^{\infty}\oplus X_{n})_{\ell_{r}}$ has the same property.

Proof.

It is clear that if $X=(\displaystyle\sum_{n=1}^{\infty}\oplus X_{n})_{\ell_{r}}$ has the $p$-$(SR^{\ast})$ roperty, then each $X_{n}$ has this property. Conversely, let $K$ be a $p$-Right*∗* subset of $X.$ It is clear that each $\pi_{n}(K)$ is also a $p$-Right set. Since $X_{n}$ has the $p$-$(SR^{\ast})$ property for each $n\in\mathbb{N},$ each $\pi_{n}(K)$ is relatively weakly compact. It follows from Lemma 3.4 [19] that $K$ is relatively weakly compact. ∎

Suppose that $K$ is a bounded subset of Banach space $X.$ For $1\leq p\leq\infty,$ we set

$\vartheta_{p}(K)=\inf\{\hat{d}(A,K):K\subset X^{\ast}$ is a $p$-Right*∗* set $\}.$

We can conclude that $\vartheta_{p}(K)=0$ if and only if $K\subset X$ is a $p$-Right*∗* set. For a bounded linear operator $T:X\rightarrow Y,$ we denote $\vartheta_{p}(T(B_{X}))$ by $\vartheta_{p}(T).$

The proof of the following theorem is similar to the proof of Theorem 3.8, so its proof is omitted.

Theorem 3.20**.**

*Let $X$ be a Banach space and $1\leq p<q\leq\infty.$ The following statements are equivalent:

$\rm{(i)}$ For every Banach space $Y,$ if for every Banach space $Y,$ if $T:Y\rightarrow X$ is an operator such that $T^{\ast}$ is a Dunford-Pettis $p$-convergent operator, then $T$ is a weakly $q$-precompact (weakly $q$-compact, $q$-compact),

$\rm{(ii)}$ Same as $\rm{(i)}$ with $Y=\ell_{\infty},$

$\rm{(iii)}$ Every $p$-Right∗ subset of $X^{\ast}$ is weakly $q$-precompact (relatively weakly $q$-compact, $q$-compact).*

Corollary 3.21**.**

*Let $X$ be a Banach space and $1\leq p<\infty.$ The following statements are equivalent:

$\rm{(i)}$ For every Banach space $Y,$ if $T:Y\rightarrow X$ is an operator such that $T^{\ast}$ is a Dunford-Pettis $p$-convergent operator, then $T$ is weakly compact,

$\rm{(ii)}$ Same as $\rm{(i)}$ with $Y=\ell_{1},$

$\rm{(iii)}$ $X$ has the $p$-$(SR^{\ast})$ property,

$\rm{(iv)}$ $\omega(T^{\ast})\leq\vartheta_{p}(T^{\ast})$ for every operator $T$ from $X$ into any Banach space $Y,$

$\rm{(v)}$ $\omega(K)\leq\vartheta_{p}(K)$ for every bounded subset $K$ of $X.$*

Corollary 3.22**.**

If $X^{\ast}$ has the $p$-$(DPrcP)$ and $Y$ has the $p$-$(SR^{\ast})$ property, then $L(X,Y)=W(X,Y).$

Proof.

It can easily seen that continuous linear image each $p$-Right null sequence is a $p$-Right null sequence. Therefore, if $T\in L(X,Y)$ and $(y^{\ast}_{n})_{n}$ is a $p$-Right null sequence in $Y^{\ast},$ then $(T^{\ast}(y^{\ast}_{n}))_{n}$ is a $p$-Right null sequence in $X^{\ast}.$ Since $X^{\ast}$ has the $p$-$(DPrcP),$ $\|T^{\ast}(y^{\ast}_{n})\|\rightarrow 0$ and so, $T^{\ast}\in DPC_{p}(Y^{\ast},X^{\ast}).$ Hence, Corollary 3.21 implies that $T\in W(X,Y).$ ∎

4. $(p,q)$-sequentially Right property on Banach spaces

In this section, motivated by the class $\mathcal{P}_{p,q}$ in [23] for those Banach spaces in which relatively $p$-compact sets are relatively $q$-compact, we introduce the concepts of properties $(SR)_{p,q}$ and $(SR^{\ast})_{p,q}$ in order to find a condition which every Dunford-Pettis $q$-convergent operator is Dunford-Pettis $p$-convergent.

Definition 4.1**.**

We say that $X$ has the $(p,q)$-sequentially Right property (in short $X$ has the $(SR)_{p,q}$ property), if each $p$-Right set in $X^{\ast}$ is a $q$-Right set in $X^{\ast}.$

Definition 4.2**.**

We say that $X$ has the $(p,q)$-sequentially Right property (in short $X$ has the $(SR^{\ast})_{p,q}$ property), if each $p$-Right*∗* set in $X$ is a $q$-Right*∗* set in $X.$

From Definitions 4.1 and 4.2, we have the following result. Since its proof is obvious, the proof is omitted.

Proposition 4.3**.**

If $X^{\ast}$ has the $(SR)_{p,q}$ property, then $X$ has the $(SR^{\ast})_{p,q}$ property.

Theorem 4.4**.**

Let $1\leq p<q\leq\infty.$ The following statements are equivalent:

1. (i)

$X$* has the $(SR)_{p,q}$ property.* 2. (ii)

$DPC_{p}(X,Y)\subseteq DPC_{q}(X,Y)$, for every Banach space $Y.$ 3. (iii)

Same as $\rm{(ii)}$ for $Y=\ell_{\infty}.$

Proof.

(i) $\Rightarrow$ (ii) If $T\in DPC_{p}(X,Y),$ then the part $\rm{(i)}$ of Lemma 3.2 implies that $T^{\ast}(B_{Y^{\ast}})$ is a $p$-Right set. Since $X$ has the $(SR)_{p,q}$ property, $T^{\ast}(B_{Y^{\ast}})$ is a $q$-Right set. Therefore, the part $\rm{(i)}$ of Lemma 3.2 yields that $T\in DPC_{q}(X,Y).$

(ii) $\Rightarrow$ (iii) is obvious.

(iii) $\Rightarrow$ (i) Suppose that $K$ is a $p$-Right set in $X^{\ast}$ and let $(x_{n}^{\ast})_{n}$ is an arbitrary sequence in $K.$ Assume that $T:\ell_{1}\rightarrow X^{\ast}$ is defined by $T(b_{n})=\sum_{n=1}^{\infty}b_{n}x_{n}^{\ast}.$ It is clear that $T^{\ast}(x)=(x_{i}^{\ast}(x))_{i}$ for all $x\in X.$ Suppose that the sequence $(x_{n})_{n}$ is a $p$-Right null sequence in $X.$ Since, $K$ is a $p$-Right set, we have

[TABLE]

Therefore, $\displaystyle\lim_{n}\|T^{\ast}(x_{n})\|=0.$ Hence, $T^{\ast}_{|_{X}}$ is a Dunford-Pettis $p$-convergent operator and so by the assumption $T^{\ast}_{|_{X}}$ is a Dunford-Pettis $q$-convergent operator. Now, assume that $(x_{n})_{n}$ is a $q$-Right null sequence in $X$ and $y\in B_{\ell_{1}}.$ Hence,

[TABLE]

Therefore, $T(B_{\ell_{1}})$ is a $q$-Right set in $X^{\ast}$ which follows that $(x_{n}^{\ast})_{n}$ is also a $q$-Right set in $X^{\ast}.$ Since $(x_{n}^{\ast})_{n}$ is an arbitrary sequence in $K,$ $K$ is a $q$-Right set. Thus, $X$ has the $(SR)_{p,q}$ property. ∎

Corollary 4.5**.**

If every $p$-Right set in $X^{\ast}$ is relatively compact, then $X$ has the $(SR)_{p,q}$ property.

Corollary 4.6**.**

Let $1\leq p<q\leq\infty.$ The following statements hold.

1. (i)

If $X$ has both properties $(SR)_{p,q}$ and $p$-$(DPrcP)$, then $X$ has the $q$-$(DPrcP).$** 2. (ii)

If $X^{\ast\ast}$ has both properties $(SR)_{p,q}$ and $p$-$(DPrcP)$, then $X$ has the $q$-$(DPrcP).$** 3. (iii)

If $X$ has the $p$-$(SR),$* then $X$ has the $(SR)_{p,q}$ property.*

Proof.

$\rm{(i)}$ Suppose that $T:X\rightarrow Y$ is a bounded linear operator. Since $X$ has the $p$-$(DPrcP)$, then $T\in DPC_{p}(X,Y).$ On the other hand, $X$ has property $(SR)_{p,q},$ thus by Theorem 4.4, $T\in DPC_{q}(X,Y)$. Thus, $X$ has the $q$-$(DPrcP).$

$\rm{(ii)}$ By part (i), $X^{\ast\ast}$ has the $q$-$(DPrcP).$ Hence, $X$ has the $q$-$(DPrcP).$

$\rm{(iii)}$ Let $Y$ be a Banach space and $T\in DPC_{p}(X,Y).$ From part $\rm{(i)}$ of Lemma 3.2, $T^{\ast}(B_{Y^{\ast}})$ is a $p$-Right set. Since $X$ has the ${p}$-$(SR)$ property, $T^{\ast}(B_{Y^{\ast}})$ is relatively weakly compact. It is clear that $T$ is weakly compact and so, $T$ is Dunford-Pettis completely continuous. Thus, $T$ is Dunford-Pettis $q$-convergent. Hence, by Theorem 4.4 $X$ has the $(SR)_{p,q}$ property. ∎

Theorem 4.7**.**

Let $1<p\leq\infty$. The following statements are equivalent.

1. (i)

$X$* has the $p$-*$(DPrcP).$** 2. (ii)

$X$* has the $(SR)_{1,p}$ property and $X$ contains no isomorphic copy of $c_{0}.$*

Proof.

(i) $\Rightarrow$ (ii) is obvious.

(ii) $\Rightarrow$ (i) Since $X$ contains no isomorphic copy of $c_{0},$ $X$ has the $1$-Schur property (see, Theorem 2.4 in [8]) and so has the $1$-$(DPrcP).$ Hence, $B_{X^{\ast}}$ is $1$-Right subset of $X^{\ast}.$ Since $X$ has the $(SR)_{1,p}$ property, $B_{X^{\ast}}$ is a $p$-Right set. It is easy to verify that $X$ has the $p$-$(DPrcP).$ ∎

In the sequel, we characterize property $(SR^{\ast})_{p,q}.$ Since the proof of the following theorem is similar to the proof of Theorem 4.4, its proof is omitted.

Theorem 4.8**.**

Let $1\leq p<q\leq\infty.$ The following statements are equivalent.

1. (i)

$X$* has the $(SR^{\ast})_{p,q}$ property.* 2. (ii)

$DPC_{p}(X^{\ast},Y^{\ast})\subseteq DPC_{q}(X^{\ast},Y^{\ast}),$* for every Banach space $Y.$* 3. (iii)

Same as $\rm{(ii)}$ for $Y=\ell_{1}.$

Corollary 4.9**.**

If $X^{\ast}$ has the $p$-$(SR)$ property, then $X$ has the $(SR^{\ast})_{p,q}$ property.

Proof.

Let $Y$ be an arbitrary Banach space and $T\in L(Y,X)$ such that $T^{\ast}$ be a Dunford-Pettis $p$-convergent operator. Therefore, the part (iii) of Lemma 3.2, $T(B_{Y})$ is a $p$-Right*∗* set in $X.$ Since $X^{\ast}$ has the $p$-$(SR)$ property, the part (iii) of Proposition 3.11 implies that $X$ has the $p$-$(SR^{\ast})$ property. Hence, $T(B_{Y})$ is a relatively weakly compact set in $X$ and so $T$ is weakly compact. Thus, $T^{\ast}$ is weakly compact and so $T^{\ast}$ is Dunford-Pettis completely continuous. So, $T^{\ast}$ is a Dunford-Pettis $q$-convergent operator. Hence, as an immediate consequence of the Theorem 4.8, we can conclude that $X$ has the $(SR^{\ast})_{p,q}$ property. ∎

Theorem 4.10**.**

If $1<p\leq\infty$, then the following statements are equivalent.

1. (i)

$X^{\ast}$* has the $p$-*$(DPrcP).$** 2. (ii)

$X$* has the $(SR^{\ast})_{1,p}$ property and $X^{\ast}$ contains no isomorphic copy of $c_{0}.$*

Proof.

(i) $\Rightarrow$ (ii) Suppose that $X^{\ast}$ has the $p$-$(DPrcP).$ By Theorem 4.10, $X^{\ast}$ has the $(SR)_{1,p}$ property and $X^{\ast}$ contains no isomorphic copy of $c_{0}.$ Thus, Proposition 4.3 implies that $X$ has the $(SR^{\ast})_{1,p}$ property.

(ii) $\Rightarrow$ (i) By the hypothesis $X^{\ast}$ contains no isomorphic copy of $c_{0}.$ Therefore Theorem 2.4 in [8] implies that $X^{\ast}$ has the $1$-Schur property and so $X^{\ast}$ has the $1$-$(DPrcP).$ Therefore, by the part (iii) of Lemma 3.2, $B_{X}$ is a ${1}$-Right*∗* set in $X.$ Since $X$ has the $(SR^{\ast})_{1,p}$ property, $B_{X}$ is a $p$-Right*∗* set. Hence, by reapplying the part (iii) of Lemma 3.2, $X^{\ast}$ has the $p$-$(DPrcP).$ ∎

Finally, we present an example of property $(SR)_{p,q}$ and an example of property $(SR^{\ast})_{p,q}.$

Example*.*

$\rm{(i)}$ If $\Omega$ is a compact Hausdorff space, then $C(\Omega)$ has the $(SR)_{p,q}$ property.

$\rm{(ii)}$ If $(\Omega,\Sigma,\mu)$ is any $\sigma$-finite measure space, then $L_{1}(\mu)$ has the $(SR^{\ast})_{p,q}$ property.