A study on Dunford-Pettis completely continuous like operators

TL;DR

This paper investigates Dunford-Pettis $p$-convergent operators and related compactness properties on Banach spaces, providing conditions under which operators exhibit $p$-convergence and exploring their behavior in spaces of continuous functions.

Contribution

It introduces new conditions for $p$-Dunford-Pettis properties and characterizes $p$-convergence of operators in function spaces, extending existing theory.

Findings

Operators from $C(\

$X$) are $p$-convergent if $X$ has the Dunford-Pettis property of order $p$.

Characterization of $p$-convergent operators via their representing measures and extensions.

Abstract

In this article, the class of all Dunford-Pettis $ p $-convergent operators and $ p $-Dunford-Pettis relatively compact property on Banach spaces are investigated. Moreover, we give some conditions on Banach spaces $ X $ and $ Y $ such that the class of bounded linear operators from $ X$ to $ Y $ and some its subspaces have the $ p $-Dunford-Pettis relatively compact property. In addition, if $ \Omega $ is a compact Hausdorff space, then we prove that dominated operators from the space of all continuous functions from $ K $ to Banach space $ X $ (in short $ C(\Omega,X) $) taking values in a Banach space with the $ p $-$ (DPrcP) $ are $ p $-convergent when $ X $ has the Dunford-Pettis property of order $ p.$\ Furthermore, we show that if $ T:C(\Omega,X)\rightarrow Y $ is a strongly bounded operator with representing measure $ m:\Sigma\rightarrow L(X,Y) $ and $ \hat{T}:B(\Omega,X)\rightarrow Y $ is its extension, then $ T$ is Dunford-Pettis $ p $-convergent if and only if $ \hat{T}$ is Dunford-Pettis $ p $-convergent.

Equations

$$T_x(y)=T(x)T(y)=T(x)(\le n)T(y)+T(x)(>n)T(y).$$

$$M:=\{({\alpha }_i)(\le n):({\alpha }_i)\in c_0\}$$

$$\{T(x)(\le n)T(y):y\in X\}\subseteq \Vert T(x)\Vert B_M,$$

$$\Vert T(x)\Vert B_M\subseteq \bigcup _{i=1}^j(z_i+\epsilon B_{c_0})$$

$$T_x(B_X)=T(x)T(B_X)\subseteq \bigcup _{i=1}^j(z_i+\epsilon B_{c_0})+\epsilon B_{c_0}\subseteq \bigcup _{i=1}^{j+1}(z_i+2\epsilon B_{c_0})$$

$$\underset{n\to \infty }{\lim }\Vert T(x_n){\Vert }^2=\underset{n\to \infty }{\lim }\underset{i}{\sup }(|(T(x_n{)}_i){|}^2)=\underset{n\to \infty }{\lim }\Vert (T(x_n){)}^2\Vert$$

$$=\underset{n\to \infty }{\lim }\Vert (T⨂T)(x_n,x_n)\Vert =0.$$

$$\epsilon <|x_{k_n}^{\ast }(x_{k_n})|\le |(x_{k_n}^{\ast }-x_n^{\ast })(x_{k_n})|+|x_n^{\ast }(x_{k_n})|<\frac{5\epsilon }{6},$$

$$\underset{n\to \infty }{\lim }\sup \{|y_n^{\ast }(T(x))|:T\in B_{\mathcal{M}}\}$$

$$=\underset{n\to \infty }{\lim }\sup \{|{\phi }_x^{\ast }(y_n^{\ast })(T)|:T\in B_{\mathcal{M}}\}=\underset{n\to \infty }{\lim }\Vert {\phi }_x^{\ast }(y_n^{\ast })\Vert =0,$$

$$\Vert T^{\ast }(y_n^{\ast })\Vert =\underset{x\in B_E}{\sup }|T^{\ast }(y_n^{\ast })(x)|=\underset{x\in B_E}{\sup }|y_n^{\ast }(T(x))|=$$

$$\underset{x\in B_E}{\sup }|y_n^{\ast }(S(P(x))|\ge \delta .|y_n^{\ast }(S(e_n^1))|\ge \delta {\epsilon }_0,$$

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

Full text

1. Introduction:

A bounded subset $K$ of a Banach space $X$ is called Dunford-Pettis set, if any weakly null sequence $(x_{n})$ in $X^{\ast}$ converges uniformly on $K$ [2]. It is easy to verify that the class of Dunford-Pettis sets strictly contains the class of relatively compact sets. But in general the converse is not true. The concept of Dunford-Pettis relatively compact property (briefly denoted by $(DPrcP)$) on Banach spaces was introduced by Emmanuele [16]. A Banach space $X$ has the $(DPrcP),$ if every Dunford-Pettis subset of $X$ is relatively compact. It is well known that any dominated operator from $C(\Omega,X)$ spaces taking values in a Banach space with the $(DPrcP)$ is completely continuous when $X$ has the Dunford-Pettis property (see [16]).

Wen and Chen [24], introduced the definition of a Dunford-Pettis completely continuous operator in order to characterize the $(DPrcP)$ on Banach spaces. A bounded linear operator $T$ from a Banach space $X$ to a Banach space $Y$ is called Dunford-Pettis completely continuous, if it transforms weakly null sequences which are Dunford-Pettis sets in $X$ to norm-null sequences in $Y.$

For more details on the rule of the $(DPrcP)$ in different areas of Banach space theory, one can refer to [2, 18, 24]. Let us recall from [15], that a sequence $(x_{n})_{n}$ in $X$ is called weakly $p$-summable, if $(x^{\ast}(x_{n}))_{n}\in\ell_{p}$ for each $x^{\ast}\in X^{\ast}.$ The set of all weakly $p$-summable sequences in $X$ is denoted by $\ell_{p}^{w}(X).$ The weakly $\infty$-summable sequences are precisely the weakly null sequences. A sequence $(x_{n})_{n}$ in $X$ is said to be weakly $p$-convergent to $x\in X$ if $(x_{n}-x)_{n}\in\ell_{p}^{w}(X).$  Recently, the concepts of Dunford-Pettis $p$-convergent operators between Banach spaces and $p$-Dunford-Pettis relatively compact property on Banach spaces was introduced by Ghenciu [20] as follows:

• •

An operator $T:X\rightarrow Y$ is called Dunford-Pettis $p$-convergent, if it takes weakly $p$-summable sequences which are Dunford-Pettis sets into 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 (briefly denoted by $p$-$(DPrcP)$), whenever every Dunford-Pettis and weakly $p$-summable sequence $(x_{n})_{n}$ in $X$ is norm null.

Motivated by the above works, the following interesting questions are posed in this area:

• •

Under which conditions on Banach spaces $X$ and $Y,$ every dominated operator $T:C(\Omega,X)\rightarrow Y$ is $p$-convergent?

• •

Suppose that $T:C(\Omega,X)\rightarrow Y$ is a strongly bounded operator with representing measure $m:\Sigma\rightarrow L(X,Y)$ and $\hat{T}:B(\Omega,X)\rightarrow Y$ is the restriction of $T^{**}$ to $B(\Omega,X),$ then is $T$ Dunford-Pettis $p$-convergent if and only if $\hat{T}$ is Dunford-Pettis $p$-convergent?

These kind of researches have done by many authors for different operators, see [3, 4, 6, 7, 17, 21].

Here, we try answer to the above questions. The article is organized as follows:

Section 2 of this paper provides a wide range of definitions and concepts in Banach spaces. These concepts are mostly well known, but we need them in the sequel.

Section 3 provides the background knowledge of Dunford-Pettis $p$-convergent operators and the $p$-$(DPrcP)$ on Banach spaces. Here, we investigate some characterizations of $p$-$(DPrcP)$ on Banach spaces.  In addition, we prove that if $(X_{n})_{n}$ a sequence of Banach spaces with the $p$-$(DPrcP),$ then the space $(\sum_{n=1}^{\infty}\bigoplus X_{n})_{\ell_{1}}$ has the same property. In the sequel, we show that the space bounded linear operators and some certain subspaces of this space have the $p$-$(DPrcP)$ under suitable conditions. Moreover, if $E$ is a Banach lattice, then we give some sufficient conditions for which each Dunford-Pettis $p$-convergent operator $T:E\rightarrow X$ has an adjoint which is Dunford-Pettis $p$-convergent. Finally, we state that any dominated operator from $C(\Omega,X)$ spaces taking values in a Banach space with the $p$-$(DPrcP)$ is $p$-convergent, when $X$ has the Dunford-Pettis property of order $p.$

In Section 4, we show that if $T:C(\Omega,X)\rightarrow Y$ is strongly bounded, then $T$ is Dunford-Pettis $p$-convergent if and only if $\hat{T}$ is Dunford-Pettis $p$-convergent. Moreover, we prove that $T^{\ast}$ is Dunford-Pettis $p$-convergent if and only if $\hat{T}^{\ast}$ is Dunford-Pettis $p$-convergent. Furthermore, if $\Omega$ is a dispersed compact Hausdorff space and $T:C(\Omega,X)\rightarrow Y$ is strongly bounded, then we show that $T^{\ast}$ is Dunford-Pettis $p$-convergent if and only if $m(A)^{\ast}:Y^{\ast}\rightarrow X^{\ast}$ is Dunford-Pettis $p$-convergent, for each $A\in\Sigma.$ Note that our results in this section are motivated by results in [3, 6, 17, 21].

2. Definitions and Notations:

Throughout this article $X,Y$ and $Z$ are arbitrary Banach spaces, $1\leq p<\infty,$ except for the cases where we consider other assumptions. Also, we suppose $p^{\ast}$ 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. $e_{n}$). We denote the closed unit ball of $X$ by $B_{X}$ and the identity map on $X$ by $id_{X}.$ 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 $X\cong Y.$  Also we use $\langle x,x^{\ast}\rangle$ or $x^{\ast}(x)$ for the duality between $x\in X$ and $x^{\ast}\in X^{\ast}.$  The space of all bounded $\Sigma$-measurable functions on $\Omega$ with separable range in $X$ is denoted by $B(\Omega,X).$ We denote the unit ball of $C(\Omega,X)$ and the unit ball of $B(\Omega,X)$ by $B_{0}$ and $B,$ respectively. 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 (resp. compact operators) from $X$ into $Y$ is denoted by $L(X,Y)$ (resp. $K(X,Y)$) and the topological dual of $X$ is denoted by $X^{\ast}.$ The space of all $w^{\ast}$-$w$ continuous operators from $X^{\ast}$ to $Y$ will denoted by $L_{w^{\ast}}(X^{\ast},Y).$ The projective tensor product of two Banach spaces $X$ and$Y$ will be denoted by $X\widehat{\bigotimes}_{\pi}Y.$

In what follows we introduce some notions which will be used in the sequel. A bounded linear operator $T$ from a Banach space $X$ to a Banach space $Y$ is called completely continuous, if it transforms weakly null sequences in $X$ to norm-null sequences in $Y$ [14]. The class of all completely continuous operators from $X$ to $Y$ is denoted by $CC(X,Y).$ A Banach space $X$ is said to have the Dunford-Pettis property (in short $X$ has the $(DPP)$), if for any Banach space $Y$ every weakly compact operator $T:X\rightarrow Y$ is completely continuous [14]. A bounded subset $K$ of $X$ is relatively weakly $p$-compact (resp. weakly $p$-compact) if every sequence in $K$ has a weakly $p$-convergent subsequence with limit in $X$ (resp. in $K$) [6]. 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 $Y.$  A sequence $(x_{n})_{n}$ in $X$ is called weakly $p$-Cauchy if $(x_{n_{k}}-x_{m_{k}})_{k}$ is weakly $p$-summable for any increasing sequences $(n_{k})_{k}$ and $(m_{k})_{k}$ of positive integers [8]. A bounded linear operator $T:X\rightarrow Y$ is called $p$-convergent, if it takes weakly $p$-summable sequences in $X$ into norm null sequences in $Y$ [6]. 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 ($X\in(C_{p})$), if $id_{X}\in C_{p}(X,X).$ We refer the reader for undefined terminologies to the classical references [1, 13, 14].

3. Dunford-Pettis relatively compact property of order $p$

In this section, we obtain some characterizations of a Banach space with the $p$-$(DPrcP).$ Moreover the stability of $p$-$(DPrcP)$ for some subspaces of bounded linear operators is investigated.

Our aim in this section is to obtain some suitable conditions on $X$ and $Y$ such that any dominated operator $T$ from $C(\Omega,X)$ into $Y$ is $p$-convergent.

Recall that, the authors in [9, 19] by using Right topology, independently, 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.

Proposition 3.1**.**

Let $T:X\rightarrow Y$ be a bounded linear operator. The following statements are equivalent:

$\rm{(i)}$ $T$ maps Dunford-Pettis and weakly $p$-compact subset of $X$ onto relatively norm compact in $Y.$

$\rm{(ii)}$ $T$ maps $p$-Right null sequences onto norm null sequences,

$\rm{(iii)}$ $T$ maps $p$-Right Cauchy sequences onto norm convergent sequences.

Proof.

(i) $\Rightarrow$ (ii) Suppose that $(x_{n})_{n}$ is a $p$-Right null sequence in $X$ and $K:=\{x_{n}:n\in\mathbb{N}\}\cup\{0\}.$ It is clear that $K$ is a Dunfotd-Pettis weakly $p$-compact set in $X.$  By $\rm{(i)},$ $T(K)$ is relatively norm compact in $Y,$ and so $(T(x_{n}))_{n}$ is norm convergent to zero.

(ii) $\Rightarrow$ (iii) Let $(x_{n})_{n}$ is a weakly $p$-Right Cauchy sequence in $X.$ Therefore for any two subsequences $(a_{n})_{n}$ and $(b_{n})_{n}$ of $(x_{n})_{n},$ $(a_{n}-b_{n})_{n}$ is a $p$-Right null sequence in $X.$ So, $\rm{(ii)}$ implies that $(T(a_{n})-T(b_{n}))_{n}$ is a norm null sequence in $Y.$ Hence, $(T(x_{n}))_{n}$ is norm convergent.

(iii) $\Rightarrow$ (i) Suppose that $K$ is a Dunford-Pettis weakly $p$-compact subset of $X.$ Let $(y_{n})_{n}$ be a sequence in $T(K).$ Therefore there is a sequence $(x_{n})_{n}\subseteq K$ such that $y_{n}=T(x_{n}),$ for each $n\in\mathbb{N}.$  Since $K$ is a weakly $p$-compact set, $(x_{n})_{n}$ has a weakly $p$-Cauchy subsequence. Without loss of generality we can assume that $(x_{n})_{n}$ is a $p$-Right Cauchy sequence. Therefore by $\rm{(iii)},$ $(T(x_{n}))_{n}$ is norm-convergent in $Y.$ Hence $T(K)$ is relatively norm compact. ∎

As an immediate consequence of the Proposition 3.1, we can conclude that the following result:

Corollary 3.2**.**

*Suppose that $X$ is a Banach space. The following assertions are equivalent:

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

$\rm{(ii)}$ The identity operator $id_{X}:X\rightarrow X$ is Dunford-Pettis $p$-convergent.

$\rm{(iii)}$ Every $p$-Right Cauchy sequence in $X$ is norm convergent.

$\rm{(iv)}$ Every Dunford-Pettis and weakly $p$-compact subset of $X$ is relatively norm compact.*

Remark 3.3*.*

$\rm{(i)}$ If $X\in C_{p},$ then $X$ has the $p$-$(DPrcP),$ but in general the converse is not true. For example for all $p\geq 2,$ $\ell_{2}$ has the $p$-$(DPrcP),$ while $\ell_{2}\not\in C_{p}.$

$\rm{(ii)}$ It is clear that, if $1\leq r<q\leq\infty$ and $X$ has the $q$-$(DPrcP),$ then $X$ has the $r$-$(DPrcP).$ In particular, for $1\leq p<\infty,$ if $X$ has the $(DPrcP),$ then $X$ has the $p$-$(DPrcP).$ But, the converse is not true. For example, the space $L_{1}[0,1]$ contains no copy of $c_{0}.$ Therefore $L_{1}[0,1]$ has the $1$-Schur property ([11, Corollary 2.9]). Hence, $L_{1}[0,1]$ has the $1$-$(DPrcP),$ while, $L_{1}[0,1]$ does not have the $(DPrcP).$

$\rm{(iii)}$ It is easy to verify that $(e_{n})_{n}$ is a $p$-Right null sequence in $c_{0}$ such that $\|e_{n}\|=1$ for all $n\in\mathbb{N}.$ Hence $c_{0}$ does not have the $p$-$(DPrcP).$

$\rm{(iv)}$ There exists a Banach space $X$ with the $p$-$(DPrcP)$ such that if $Y$ is a closed subspace of it, then the quotient space $\frac{X}{Y}$ does not have this property. For example $c_{0}$ does not have the $p$-$(DPrcP),$ while $\ell_{1}$ has the $p$-$(DPrcP)$ and $c_{0}$ is isometrically isomorphic to a quotient of $\ell_{1}$ ([1, Corollary 2.3.2]).

Theorem 3.4**.**

*Suppose that $X$ is a Banach space. The following assertions are equivalent:

$\rm{(i)}$ $X$ has the $p$-$(DPrcP),$

$\rm{(ii)}$ For each Banach space $Y,~{}$ $DPC_{p}(X,Y)=L(X,Y),$

$\rm{(iii)}$ $X$ is the direct sum of two Banach spaces with the $p$-$(DPrcP).$*

Proof.

The equivalence of $\rm{(i)}$ and $\rm{(ii)}$ are clear. Hence, we only prove the equivalence of $\rm{(i)}$ and $\rm{(iii)}$.

$\rm{(i)}\Rightarrow\rm{(iii)}$ Note that $X=X\bigoplus\{0\}.$

$\rm{(iii)}\Rightarrow\rm{(i)}$ Let $X=Y\bigoplus Z$ such that $Y$ and $Z$ have the $p$-$(DPrcP).$ Consider the projections $P_{1}:X\rightarrow Y$ and $P_{2}:X\rightarrow Z.$  Let $K$ be a Dunford-Pettis and weakly $p$-compact subset of $X.$ Clearly, $P_{1}(K)$ is a Dunford-Pettis and weakly $p$-compact subset of $Y,$ and so it is a norm compact set in $Y.$ Similarly $P_{2}(K)$ is a norm compact set in $Z.$ Also, it is clear that any sequence $(x_{n})_{n}\subseteq K$ can be written as $x_{n}=y_{n}+z_{n},$ where $y_{n}\in P_{1}(K)$ and $z_{n}\in P_{2}(K).$ Thus, there are subsequences $(y_{n_{k}})_{k}$ and $(z_{n_{k}})_{k}$ and $y\in P_{1}(K)$ and $z\in P_{2}(K)$ such that $x_{n_{k}}=y_{n_{k}}+z_{n_{k}}\rightarrow y+z.$ Since, $K$ is a weakly $p$-compact set, $y+z\in K$ and so, $K$ is a norm compact set. ∎

A bilinear operator $\phi:X\times Y\rightarrow Z$ is called separately compact if for each fixed $y\in Y,$ the linear operator $T_{y}:X\rightarrow Z:x\mapsto\phi(x,y)$ and for each fixed $x\in X,$ the linear operator $T_{x}:Y\rightarrow Z:y\mapsto\phi(x,y)$ are compact.

By using the same argument as in the proof of ([27, Lemm 6.3.19]), we obtain the following result.

Theorem 3.5**.**

If every symmetric bilinear separately compact operator $S:X\times X\rightarrow c_{0}$ is Dunford-Pettis $p$-convergent, then $X$ has the $p$-$(DPrcP).$

Proof.

Suppose that $T\in L(X,c_{0})$ is arbitrary. Define $T\bigotimes T:X\times X\rightarrow c_{0}$ by $(T\bigotimes T)(x,y)=T(x)T(y)$ (coordinatewise product of two sequences in $c_{0}$).  It is easy to verify that $T\bigotimes T$ is bilinear and satisfies the following properties:

$\rm{(i)}$ $(T\bigotimes T)$ is symmetric, i.e., $(T\bigotimes T)(x,y)=(T\bigotimes T)(y,x).$

$\rm{(ii)}$ $(T\bigotimes T)$ is separately compact. Since for each fixed $x\in X,$ the operator $T_{x}:X\rightarrow c_{0}$ defined by $T_{x}(y)=T(x)T(y)=(T\bigotimes T)(x,y)$ is bounded linear such that $\|T_{x}\|\leq\|T\|^{2}\|x\|.$ Without loss of generality that $\|T\|=1.$ Since $T(x)$ is a sequence in $c_{0},$ for convenience we agree to denote the $m$-th term of $T(x)$ by $T(x)_{m}.$ Hence, for $\varepsilon>0,$ there exists $n\in\mathbb{N}$such that $|T(x)_{m}|<\varepsilon$ for all $m\geq n.$

For convenience, we denote $(T(x)_{1},T(x)_{2},\cdot\cdot\cdot,T(x)_{n},0,0,\cdot\cdot\cdot)$ by $T(x)(\leq n)$ and $(0,0,\cdot\cdot\cdot,0,T(x)_{n+1},T(x)_{n+2},\cdot\cdot\cdot)$ by $T(x)(>n).$ Therefore

[TABLE]

Obviously, the set $\{T(x)(\leq n)T(y):y\in X\}$ is contained in a finite-dimensional subspace

[TABLE]

of $c_{0}.$ Hence,

[TABLE]

which is compact.

For each $\varepsilon>0$ there exists $z_{1},\cdot\cdot\cdot,z_{j}\in\|T(x)\|B_{M}$ such that

[TABLE]

So,

[TABLE]

with $z_{j+1}=0.$ Therefore, $T_{x}(B_{X})$ is relatively compact. Hence for each fixed $x\in X,$ the operator $T_{x}$ is compact. Hence $T\bigotimes T$ is separately compact. Therefore, by the hypothesis $T\bigotimes T$ is Dunford-Pettis $p$-convergent. So, if $(x_{n})_{n}$ is a $p$-Right null sequence in $X,$ then

[TABLE]

[TABLE]

This shows that $T$ is Dunford-Pettis $p$-convergent. Since $T\in L(X,c_{0})$ was arbitrarily chosen it follows from Theorem of 3.4 that $X$ has $p$-$(DPrcP).$ ∎

Proposition 3.6**.**

*If $X$ has the $p$-$(DPrcP),$ then the following statements hold:

$\rm{(i)}$ $\displaystyle\lim_{n\rightarrow\infty}x_{n}^{\ast}(x_{n})=0,$ for every $p$-Right Cauchy sequence $(x_{n})_{n}$ in $X$ and every weakly null sequence $(x^{\ast}_{n})_{n}$ in $X^{\ast},$

$\rm{(ii)}$ $\displaystyle\lim_{n\rightarrow\infty}x_{n}^{\ast}(x_{n})=0,$ for every $p$-Right null sequence $(x_{n})_{n}$ in $X$ and every weakly null sequence $(x^{\ast}_{n})_{n}$ in $X^{\ast},$

$\rm{(iii)}$ $\displaystyle\lim_{n\rightarrow\infty}x_{n}^{\ast}(x_{n})=0,$ for every $p$-Right null sequence $(x_{n})_{n}$ in $X$ and every weakly Cauchy sequence $(x^{\ast}_{n})_{n}$ in $X^{\ast}.$*

Proof.

$\rm{(i)}$ Let $(x_{n})_{n}$ be a $p$-Right Cauchy sequence $(x_{n})_{n}$ in $X$ and $(x^{\ast}_{n})_{n}$ be a weakly null sequence in $X^{\ast}.$ Define a bounded linear operator $T:X\rightarrow c_{0}$ by $T(x)=(x^{\ast}_{n}(x))_{n}.$ By Theorem 3.4, $T\in DPC_{p}(X,c_{0}).$ Therefore, Proposition 3.1 implies that $(T(x_{n}))_{n}$ converges to some $\alpha=(\alpha_{n})_{n}\in c_{0}$ in norm. Hence, for every $\varepsilon>0$ there exists a positive integer $N_{1}$ such that $\|T(x_{n})-\alpha\|<\frac{\varepsilon}{2}$ for all $n>N_{1}.$  Since $\alpha\in c_{0},$ we choose another positive integer $N_{2}$ such that $|\alpha_{n}|<\frac{\varepsilon}{2}$ for all $n>N_{2}.$ Hence, $|x_{n}^{\ast}(x_{n})|<\varepsilon$ for all $n>max\{N_{1},N_{2}\}.$ Thus, $\displaystyle\lim_{n\rightarrow\infty}x_{n}^{\ast}(x_{n})=0.$

$\rm{(ii)}$ is trivial.

$\rm{(iii)}$ Suppose there exists a $p$-Right null sequence $(x_{n})_{n}$ in $X$ and there exists a weakly Cauchy sequence $(x_{n}^{\ast})_{n}$ in $X^{\ast}$ such that $|x_{n}^{\ast}(x_{n})|>\varepsilon,$ for some $\varepsilon>0$ and all $n\in\mathbb{N}.$  Since $(x_{n})_{n}$ is weakly $p$-summable and in particular weakly null, there exists a subsequence $(x_{k_{n}})_{n}$ of $(x_{n})_{n}$ such that $|x_{n}^{\ast}(x_{k_{n}})|<\frac{\varepsilon}{2}$ for all $n\in\mathbb{N}.$ Since $(x^{\ast}_{n})_{n}$ is weakly Cauchy, we see that $(x^{\ast}_{k_{n}}-x^{\ast}_{n})_{n}$ is weakly null.  Now, by $\rm{(ii)},$ we have $\lim_{n\rightarrow\infty}(x^{\ast}_{k_{n}}-x^{\ast}_{n})(x_{k_{n}})=0.$ This implies that $|(x^{\ast}_{k_{n}}-x^{\ast}_{n})(x_{k_{n}})|<\frac{\varepsilon}{3}$ for $n$ large enough. But for such $n$’s, we have

[TABLE]

which is a contradiction. ∎

A bounded linear operator $T:X\rightarrow Y$ is called strictly singular, if there is no infinite dimensional subspace $Z\subseteq X$ such that $T|_{Z}$ is an isomorphism onto its range [1].

Proposition 3.7**.**

Suppose that $T\in DPC_{p}(X,Y)$ is not strictly singular. Then, $X$ and $Y$ contain simultaneously some infinite dimensional closed subspaces with the $p$-$(DPrcP).$

Proof.

Suppose that $T$ has a bounded inverse on the closed infinite dimensional subspace $Z$ of $X.$ If $(x_{n})_{n}$ is a $p$-Right null sequence in $Z,$ then $(x_{n})_{n}$ is a $p$-Right null sequence in $X.$ By the hypothesis, $\|T(x_{n})\|\rightarrow 0$ and so $\|x_{n}\|\rightarrow 0.$ Hence, $Z$ has the $p$-$(DPrcP).$ Similarly, we can see that $T(Z)$ has the $p$-$(DPrcP).$ ∎

Suppose that $(X_{n})_{n\in\mathbb{N}}$ is a sequence of Banach spaces. The space of all vector-valued sequences $(\displaystyle\sum_{n=1}^{\infty}\oplus X_{n})_{\ell_{p}}$ is called, the infinite direct sum of $X_{n}$ in the sense of $\ell_{p},$ consisting of all sequences $x=(x_{n})_{n}$ with values in $X_{n}$ such that $\|x\|_{p}=(\displaystyle\sum_{n=1}^{\infty}\|x_{n}\|^{p})^{\frac{1}{p}}<\infty.$

Proposition 3.8**.**

Let $(X_{n})_{n\in\mathbb{N}}$ be a family of Banach space. Then, $X_{n}$ has the $p$-$(DPrcP)$ for all $n\in\mathbb{N}$ if and only if $(\displaystyle\sum_{n=1}^{\infty}\oplus X_{n})_{\ell_{1}}$ has the same property.

Proof.

It is clear that if $X=(\displaystyle\sum_{n=1}^{\infty}\oplus X_{n})_{\ell_{1}}$ has the $p$-$(DPrcP),$ then every closed subspace of $X$ has the $p$-$(DPrcP).$ Hence $X_{n}$ has the $p$-$(DPrcP)$ for all $n\in\mathbb{N}.$ Now, suppose that $(x_{n})_{n}$ is a $p$-Right null sequence in $X,$ where $x_{n}=(b_{n,k})_{k\in\mathbb{N}}.$ It is clear that $(b_{n,k})_{k\in\mathbb{N}}$ is a $p$-Right null sequence in $X_{k}$ for all $k\in\mathbb{N}.$ Since $X_{k}$ has the $p$-$(DPrcP),$ $\|b_{n,k}\|_{X_{k}}\rightarrow 0$ as $n\rightarrow\infty$ for all $k\in\mathbb{N}.$ Using the techniques which used in ([10, Lemma, page 31]), we can conclude that the sum $\|x_{n}\|_{1}=\displaystyle\sum_{k=1}^{\infty}\|b_{n,k}\|_{X_{k}}$ is converges uniformly in $n.$ Hence, $\displaystyle\lim_{n\rightarrow\infty}\|x_{n}\|_{1}=\displaystyle\lim_{n\rightarrow\infty}\sum_{k=1}^{\infty}\|b_{n,k}\|_{X_{k}}=\displaystyle\sum_{k=1}^{\infty}\lim_{n\rightarrow\infty}\|b_{n,k}\|_{X_{k}}=0.$ ∎

If $\mathcal{M}$ is a closed subspace of operator ideal $\mathcal{U}(X,Y),$ then for arbitrary elements $x\in X$ and $y^{\ast}\in Y^{\ast},$ the evaluation operators $\varphi_{x}:\mathcal{M}\rightarrow Y$ and $\psi_{y^{\ast}}:\mathcal{M}\rightarrow X^{\ast}$ on $\mathcal{M}$ are defined by $\varphi_{x}(T)=T(x)$ and $\psi_{y^{\ast}}(T)=T^{\ast}(y^{\ast})$ for $T\in\mathcal{M}$ (see [23]).

The following result shows that the Dunford-Pettis $p$-convergent of all evaluation operators is a necessary condition for the $p$-$(DPrcP)$ of closed subspace $\mathcal{M}\subseteq\mathcal{U}(X,Y).$

Corollary 3.9**.**

If $\mathcal{M}$ is a closed subspace of operator ideal $\mathcal{U}(X,Y)$ that has the $p$-$(DPrcP),$ then all of the evaluation operators $\varphi_{x}:\mathcal{M}\rightarrow Y$ and $\psi_{y^{\ast}}:\mathcal{M}\rightarrow X^{\ast}$ are Dunford-Pettis $p$-convergent.

Theorem 3.10**.**

Let $X$ and $Y$ be two Banach spaces such that $Y$ has the Schur property. If $\mathcal{M}$ is a closed subspace of $\mathcal{U}(X,Y)$ such that each evaluation operator $\psi_{y^{\ast}}$ is Dunford-Pettis $p$-convergent on $\mathcal{M},$ then $\mathcal{M}$ has the $p$-$(DPrcP).$

Proof.

Suppose that $\mathcal{M}$ does not have the $p$-$(DPrcP).$ Therefore, there is a $p$-Right null sequence $(T_{n})_{n}$ in $\mathcal{M}$ such that $\|T_{n}\|\geq\varepsilon$ for all positive integer $n$ and some $\varepsilon>0.$ We can choose a sequence $(x_{n})_{n}$ in $B_{X}$ such that $\|T_{n}(x_{n})\|\geq\varepsilon.$ In addition, for each $y^{\ast}\in Y^{\ast},$ the evaluation operator is $\psi_{y^{\ast}}$ is Dunford-Pettis $p$-convergent. So, $|\langle y^{\ast},T_{n}(x_{n})\rangle|\leq\|T^{\ast}_{n}(y^{\ast})\|\|x_{n}\|\rightarrow 0.$ Hence $(T_{n}(x_{n}))_{n}$ is weakly null in $Y,$ and so is norm null, which is a contradiction. ∎

Corollary 3.11**.**

([24, Theorem 2.3]), Let $X$ and $Y$ be two Banach spaces such that $Y$ has the Schur property. If $\mathcal{M}$ is a closed subspace of $\mathcal{U}(X,Y)$ such that each evaluation operator $\psi_{y^{\ast}}$ is Dunford-Pettis completely continuous on $\mathcal{M},$ then $\mathcal{M}$ has the $(DPrcP).$

Corollary 3.12**.**

If $Y$ has the Schur property and $\mathcal{M}$ is a closed subspace of $L(X,Y)$ such that each evaluation operators $\psi_{y^{\ast}}$ is Dunford-Pettis $p$-convergent on $\mathcal{M},$ then $\mathcal{M}$ has the $p$-$(DPrcP).$

By a similar method, we obtain a sufficient condition for the $p$-$(DPrcP)$ of closed subspaces of $L_{w^{\ast}}(X^{\ast},Y).$ Since the proof of the following result is similar to the proof of Theorem 3.10, we omit its proof.

Theorem 3.13**.**

If $X$ has the Schur property and $\mathcal{M}$ is a closed subspace of $L_{w^{\ast}}(X^{\ast},Y)$ such that each evaluation operators $\psi_{y^{\ast}}$ is Dunford-Pettis $p$-convergent on $\mathcal{M},$ then $\mathcal{M}$ has the $p$-$(DPrcP).$

Corollary 3.14**.**

*The following statements hold:

$\rm{(i)}$ If $X^{\ast}$ has the $p$-$(DPrcP)$ and $Y$ has the Schur property, then $L(X,Y)$ has the $p$-$(DPrcP).$

$\rm{(ii)}$ If $X$ has the $p$-$(DPrcP)$ and $Y$ has the Schur property, then $L_{w^{\ast}}(X^{\ast},Y)$ has the $p$-$(DPrcP).$

$\rm{(iii)}$ If $X^{\ast}$ has the $p$-$(DPrcP),$ then $\ell_{1}^{w}(X^{\ast})$ has the same property.*

A Banach space $X$ has the Dunford-Pettis property of order $p$ (in short $(DPP_{p})$), if every weakly compact operator $T:X\rightarrow Y$ is $p$-convergent, for any Banach space $Y$ [6].

Remark 3.15*.*

Every $p$-convergent operator is Dunford-Pettis $p$-convergent, but in general the converse is not true. For example, the identity operator $id_{\ell_{2}}:\ell_{2}\rightarrow\ell_{2}$ is weakly compact and so is Dunford-Pettis $2$-convergent, while it is not $2$-convergent. Ghenciu [18], gave a characterization of those Banach spaces in which the converse of the above assertion holds. In fact, she showed that, a Banach space $X$ has the $(DPP_{p})$ if and only if $C_{p}(X,Y)=DPC_{p}(X,Y),$ for each Banach space $Y.$

Let $\mathcal{M}$ be a bounded subspace of $\mathcal{U}(X,Y).$ 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 $\psi_{y^{\ast}}$ are denoted by $\mathcal{M}_{1}(x)$ and $\widetilde{\mathcal{M}_{1}}(y^{\ast})$ respectively [23]. Let us recall from [22], that 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 $X^{\ast}.$

In the following a necessary condition for the $p$-$(DPrcP)$ of the dual of closed subspace $\mathcal{M}\subseteq\mathcal{U}(X,Y)$ is given.

Theorem 3.16**.**

Suppose that $X^{\ast\ast}$ and $Y^{\ast}$ have the $(DPP_{p}).$ If $\mathcal{M}$ is a closed subspace of $\mathcal{U}(X,Y)$ such that $\mathcal{M}^{\ast}$ has the $p$-$(DPrcP),$ then of all the point evaluations $\mathcal{M}_{1}(x)$ and $\widetilde{\mathcal{M}}_{1}(y^{\ast})$ are $p$-$(V^{\ast})$ sets in $Y$ and $X^{\ast}$ respectively.

Proof.

Since $\mathcal{M}^{\ast}$ has the $p$-$(DPrcP),$ Theorem 3.4, implies that the operators $\varphi^{\ast}_{x}$ and $\psi^{\ast}_{y^{\ast}}$ are Dunford-Pettis $p$-convergent. On the other hand, $X^{\ast\ast}$ and $Y^{\ast}$ have the $(DPP_{p}).$ Hence, $\varphi^{\ast}_{x}$ and $\psi^{\ast}_{y^{\ast}}$ are $p$-convergent (see ([20, Theorem 3.18])). Therefore, if $(y^{\ast}_{n})_{n}$ is a weakly $p$-summable sequence in $Y^{\ast},$ then we have:

[TABLE]

[TABLE]

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

Corollary 3.17**.**

([24, Theorem 2.2])* Suppose that $X^{\ast\ast}$ and $Y^{\ast}$ have the $(DPP).$ If $\mathcal{M}$ is a closed subspace of $\mathcal{U}(X,Y)$ such that $\mathcal{M}^{\ast}$ has the $(DPrcP),$ then of all the point evaluations $\mathcal{M}_{1}(x)$ and $\widetilde{\mathcal{M}}_{1}(y^{\ast})$ are Dunford-Pettis sets in $Y$ and $X^{\ast}$ respectively.*

Let us recall from [14], 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 $K.$

Theorem 3.18**.**

Suppose that $L(X,Y)=K(X,Y).$ If $X^{\ast}$ has the $p$-$(DPrcP)$ and $Y$ has the $(DPrcP),$ then $L(X,Y)$ has the $p$-$(DPrcP).$

Proof.

Suppose that $(T_{n})_{n}$ is a $p$-Right null sequence in $L(X,Y)$ so that $\|T_{n}\|=1$ for each $n\in\mathbb{N}.$ Let $(y^{\ast}_{n})_{n}$ be a sequence in $B_{Y^{\ast}}$ and $(x_{n})_{n}$ be a sequence in $B_{X}$ so that $y^{\ast}_{n}(T_{n}(x_{n}))>\frac{1}{2}$ for each $n\in\mathbb{N}.$ Since $X^{\ast}$ has the $p$-$(DPrcP),$ for each $y^{\ast}\in Y^{\ast},$ the evaluation operator $\psi_{y^{\ast}}:L(X,Y)\rightarrow X^{\ast}$ is Dunford-Pettis $p$-convergent. Therefore,

$|\langle y^{\ast},T_{n}(x_{n})\rangle|\leq\|T^{\ast}_{n}(y^{\ast})\|\|x_{n}\|\rightarrow 0.$

So, $(T_{n}(x_{n}))_{n}$ is weakly null in $Y.$ Now, we claim that $\{T_{n}(x_{n}):n\in\mathbb{N}\}$ is a Dunford-Pettis set in $Y.$ Let $(z^{\ast}_{n})_{n}$ be a weakly null sequence in $Y^{\ast},$ and let $T\in(X\widehat{\bigotimes}_{\pi}Y^{\ast})^{\ast}.$ It is well known that $(X\widehat{\bigotimes}_{\pi}Y^{\ast})^{\ast}\cong L(X,Y^{\ast\ast})$ (see [13], page 230). Therefore, by the hypothesis $T^{\ast}_{|_{Y^{\ast}}}$ is a compact operator. Hence, $\{T^{\ast}(z^{\ast}_{n}):n\in\mathbb{N}\}$ is relatively compact. Thus, $|\langle x_{n}\otimes z_{n}^{\ast},T\rangle|\leq\|T^{\ast}(z^{\ast}_{n})\|\rightarrow 0,$ and so $(x_{n}\otimes z_{n}^{\ast})_{n}$ is weakly null in $X\widehat{\bigotimes}_{\pi}Y^{\ast}.$ Since, $L(X,Y)$ embeds isometrically in $L(X,Y^{\ast\ast}),$ $(T_{n})_{n}$ is a Dunford-Pettis sequence in $L(X,Y^{\ast\ast}).$ But, a Dunford-Pettis subset of a dual space is necessarily an $(L)$-subset of the dual space. Therefore, $z^{\ast}_{n}(T_{n}(x_{n}))\rightarrow 0.$ Thus $(T_{n}(x_{n}))_{n}$ is a Dunford-Pettis and weakly null sequence in $Y.$ Hence, $\|T_{n}(x_{n})\|\rightarrow 0,$ which is a contradiction. ∎

Let us recall from [25] that, a norm $\|.\|$ of a Banach lattice $E$ is order continuous if for each net $(x_{\alpha})$ such that $x_{\alpha}\downarrow 0$ in $E,$ the net $(x_{\alpha})$ is norm converges to $0,$ where the notation $x_{\alpha}\downarrow 0$ means that the sequence $(x_{\alpha})$ is decreasing, its infimum exists and $\inf(x_{\alpha})=0.$

Theorem 3.19**.**

*Let $E$ be a Banach lattice and $T:E\rightarrow X$ be a Dunford-Pettis $p$-convergent operator. If the adjoint of $T$ is Dunford-Pettis $p$-convergent, then one of the following assertions holds.

$\rm{(i)}$ The norm of $E^{\ast}$ is order continuous.

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

Proof.

Assume that norm $E^{\ast}$ is not order continuous and $X^{\ast}$ does not have the $p$-$(DPrcP).$ Since norm $E^{\ast}$ is not order continuous, there exist a sublattice ${M}$ of $E,$ such that it is isomorphic to $\ell_{1}$ and a positive projection $P:E\rightarrow\ell_{1}$ (see Theorem 1 in [25]). Since $X^{\ast}$ does not have the $p$-$(DPrcP),$ there exists a $p$-Right null and normalized sequence $(y^{\ast}_{n})_{n}$ in $X^{\ast}.$ Therefore, there exit a sequence $(y_{n})_{n}$ in $X$ with $\|y_{n}\|\leq 1$ and an $\varepsilon_{0}>0$ such that $|y^{\ast}_{n}(y_{n})|\geq\varepsilon_{0}$ for all $n\in\mathbb{N}.$ Suppose that $T=S\circ P,$ where the operator $S:\ell_{1}\rightarrow X$ is defined by $S(\alpha_{n})=\displaystyle\sum_{n}\alpha_{n}y_{n}.$ Since $\ell_{1}$ is $p$-Schur space, the operator $T$ is Dunford-Pettis $p$-convergent. Now, we claim that $T^{\ast}$ is not a Dunford-Pettis $p$-convergent operator. As the operator $P$ is surjective, there exits $\delta>0$ such that $\delta B_{\ell_{1}}\subset P(B_{E}).$ So, we have

[TABLE]

[TABLE]

where $(e^{1}_{n})_{n}$ is the standard canonical basic of $\ell_{1}.$ Therefore, $T^{\ast}$ is not a Dunford-Pettis $p$-convergent operator, which is a contradiction. ∎

Let us recall from [16], that a bounded linear operator $T:C(\Omega,X)\rightarrow Y$ is dominated, if there exists a positive linear functional $L$ on $C(\Omega)^{\ast}$ such that $\|T(f)\|\leq L(\|f\|),~{}~{}~{}~{}f\in C(\Omega,X).$

By a similar technique we obtain the following result which is the $p$-version of ([16, Theorem 11]).

Theorem 3.20**.**

Suppose that $Y$ has the $p$-$(DPrcP)$ and $K$ is a compact Hausdorff space. If $X$ has the $(DPP_{p}),$ then any dominated operator $T$ from $C(\Omega,X)$ into $Y$ is $p$-convergent.

Proof.

Suppose that $T:C(\Omega,X)\rightarrow Y$ is an arbitrary dominated operator. By Theorem 5 in Chapter $\rm{III}$ of [12], there is a function $G$ from $\Omega$ into $L(X,Y^{\ast\ast})$ such that

$\rm{(i)}$ $\|G(t)\|=1~{}\mu.a.e.$ in $\Omega.$ i.e.; $\mu(\{t\in\Omega:\|G(t)\|\not=1\})=0.$

$\rm{(ii)}$ For each $y^{\ast}\in Y^{\ast}$ and $f\in C(\Omega,X),$ the function $y^{\ast}(G(.)f(.))$ is $\mu$-integrable and moreover

$y^{\ast}(T(f))=\int_{\Omega}y^{\ast}(G(t)f(t))d\mu~{}~{}$ for $~{}~{}f\in C(\Omega,X).$

Where $\mu$ is the least regular Borel measure dominating $T.$ Consider a weakly $p$-summable sequence $(f_{n})_{n}$ in $C(\Omega,X).$ Since continuous linear images of weakly $p$-summable sequences are weakly $p$-summable sequences, $(T(f_{n}))_{n}$ is a weakly $p$-summable sequence in $Y.$ Now, we show that $\{T(f_{n})):n\in\mathbb{N}\}$ is a Dunford-Pettis set in $Y.$ For this purpose, we consider a weak null sequence $(y_{n}^{\ast})_{n}$ in $Y^{\ast}.$ It is not difficult to show that, for each $t\in\Omega,~{}(G^{\ast}(t)y^{\ast}_{n})_{n}$ is weakly null in $X^{\ast}$ and $(f_{n}(t))_{n}$ is a weakly $p$-summable sequence in $X.$ Since $X$ has the $(DPP_{p}),$ we have :

$y_{n}^{\ast}((G(t)f_{n}(t)))=G^{\ast}(t)y_{n}^{\ast}(f_{n}(t))\rightarrow 0.$

Moreover, there exists a constant $M>0$ such that $|y_{n}^{\ast}(G(t)f_{n}(t))|\leq M$ for all $t\in\Omega$ and $n\in\mathbb{N}.$ The Lebesgue dominated convergent Theorem, implies that:

$\displaystyle\lim_{n\rightarrow\infty}y_{n}^{\ast}(T(f_{n}))=\lim_{n\rightarrow\infty}\int_{\Omega}y_{n}^{\ast}(G(t)f_{n}(t))d\mu=0.$

Therefore, $\{T(f_{n}):n\in\mathbb{N}\}$ is a Dunford-Pettis set in $Y$ ([2, Theorem 1]). Hence, $(T(f_{n}))_{n}$ is a $p$-Right null sequence in $Y,$ and so $\|T(f_{n})\|\rightarrow 0.$ Since $Y$ has the $p$-$(DPrcP).$ ∎

4. Characterizations of some classes of operators on $C(\Omega,X)$

Let us recall from [4, 13], every bounded linear operator $T:C(\Omega,X)\rightarrow Y$ may be represented by a vector measure $m:\Sigma\rightarrow L(X,Y^{\ast\ast})$ of finite semi-variation such that

$T(f)=\int_{\Omega}fdm,~{}~{}~{}~{}f\in C(\Omega,X)$ for every $f\in C(\Omega,X).$

This set function $m$ is called the representing measure of $T.$ Also, a bounded linear operator $T:C(\Omega,X)\rightarrow Y$ is called strongly bounded if $m$ is strongly bounded.

Different efforts have been done in order to characterize several types of operators in terms of their representing measure. Suppose that $T:C(\Omega,X)\rightarrow Y$ is a strongly bounded operator with representing measure $m:\Sigma\rightarrow L(X,Y)$ and $\hat{T}:B(\Omega,X)\rightarrow Y$ is its extension. The authors in [3, 4, 5, 6, 7, 17, 21] have found that study of $\hat{T}$ is more suitable than $T.$ Our aim in this section is to obtain some characterizations of Dunford-Pettis $p$-convergent operators in terms of their representing measure.

By using the same argument as in the proof of ([3, Theorem 3]), we obtain the following result.

Theorem 4.1**.**

Suppose that $T:C(\Omega,X)\rightarrow Y$ is a strongly bounded operator. Then, $T$ is Dunford-Pettis $p$-convergent if and only if $\hat{T}$ is Dunford-Pettis $p$-convergent.

Proof.

Suppose that $T$ is strongly bounded such that $T$ is Dunford-Pettis $p$-convergent and $\hat{T}$ is not. Therefore, there exists a $p$-Right null sequence $(y^{\ast}_{n})_{n}$ in $Y^{\ast}$ and a sequence $(f_{n})_{n}$ in $B$ such that $|\langle y^{\ast}_{n},\hat{T}(f_{n})\rangle|=1$ for all $n\in\mathbb{N}.$ Without loss of generality assume $\|y_{n}^{\ast}\|\leq 1$ for all $n.$

By using the existence of a control measure for $m$ and Lusin’s theorem, we can find a compact subset $\Omega_{0}$ of $\Omega$ such that $\tilde{m}(\Omega-\Omega_{0})<\frac{1}{4}$ and $g_{n}=f_{n}|_{\Omega_{0}}$ is continuous for each $n\in\mathbb{N}.$ Let $H=[g_{n}]$ be the closed linear subspace spanned by $(g_{n})_{n}$ in $C(\Omega_{0},X).$ By ([3, Theorem 1]), there is an isometric extension operator $S:H\rightarrow C(\Omega,X)$ such that the sequence $h_{n}=S(g_{n})$ is $p$-Right null in $C(\Omega,X).$ It is easy to verify that $\|T(h_{n})\|\geq\frac{1}{2},$ which is a contradiction with the fact that $T$ is Dunford-Pettis $p$-convergent. ∎

Definition 4.2**.**

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

[TABLE]

Proposition 4.3**.**

$\rm{(i)}$* 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 $Y.$

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

Theorem 4.4**.**

Suppose that $T:C(\Omega,X)\rightarrow Y$ is a strongly bounded operator. Then, $T^{\ast}$ is a Dunford-Pettis $p$-convergent if and only if $\hat{T}^{\ast}$ is Dunford-Pettis $p$-convergent.

Proof.

By Proposition 4.3 (i), it is enough to show that $T(B_{0})$ is a $p$-Right*∗* set if and only if $\hat{T}(B)$ is a $p$-Right*∗* set. Suppose that $T(B_{0})$ is a $p$-Right*∗* set and $\hat{T}(B)$ is not a $p$-Right*∗* set. Therefore there exists $p$-Right null sequence $(y^{\ast}_{n})$ in $Y^{\ast}$ and a sequence $(f_{n})$ in $B$ such that $|\langle y^{\ast}_{n},\hat{T}(f_{n})\rangle|=1$ for each $n\in\mathbb{N}.$ By using the same argument as in the proof of Theorem 4.1, we obtain a sequence $(h_{n})_{n}$ in $B_{0}$ such that $|\langle y^{\ast}_{n},T(h_{n})\rangle|\geq\frac{1}{2},$ which is a contradiction, since $T(B_{0})$ is a $p$-Right*∗* set. Similarly, if $\hat{T}(B)$ is a $p$-Right*∗* set, then $T(B_{0})$ is a $p$-Right*∗* set. ∎

Corollary 4.5**.**

Suppose that $T:C(\Omega,X)\rightarrow Y$ is a strongly bounded operator. If $T^{\ast}$ is Dunford-Pettis $p$-convergent, then for each $A\in\Sigma,$ the operator $m(A)^{\ast}:Y^{\ast}\rightarrow X^{\ast}$ is Dunford-Pettis $p$-convergent.

The proof of the following result is similar to the proof of Theorem 4.4, and will be omitted.

Theorem 4.6**.**

Suppose that $T:C(\Omega,X)\rightarrow Y^{\ast}$ is a strongly bounded operator. Then $T^{\ast}|_{Y}$ is Dunford-Pettis $p$-convergent if and only if $\hat{T}^{\ast}|_{Y}$ is Dunford-Pettis $p$-convergent.

Corollary 4.7**.**

Suppose that $T:C(\Omega,X)\rightarrow Y^{\ast}$ is a strongly bounded operator. If $T^{\ast}|_{Y}$ is Dunford-Pettis $p$-convergent, then for each $A\in\Sigma,$ $m(A)^{\ast}|_{Y}$ is Dunford-Pettis $p$-convergent.

Let us recall from [21], that if $T:C(\Omega,X)\rightarrow Y$ is a bounded linear operator, $\overline{\Omega}$ is a metrizable compact space, and $\pi:\Omega\rightarrow\overline{\Omega}$ a continuous map which is onto, we will call $\overline{\Omega}$ a quotient of $\Omega.$ The map $\overline{\pi}:C(\overline{\Omega})\rightarrow C(\Omega)$ given by $\overline{\pi}(\overline{f})=\overline{f}\circ\pi$ defines an isometric embedding of $C(\overline{\Omega})$ into $C(\Omega).$ Let $\overline{T}:C(\overline{\Omega},X)\rightarrow Y$ be the operator defined by $\overline{T}(\overline{f})=T(\overline{f}\circ\pi)$ where $\overline{f}\in C(\overline{\Omega},X)$ and $\pi:\Omega\rightarrow\overline{\Omega}$ is the canonical mapping.

By using the same argument as in the proof of ([21, Lemma 18]), we obtain the following result.

Theorem 4.8**.**

$\rm{(i)}$* Suppose that $T:C(\Omega,X)\rightarrow Y$ is a bounded linear operator. Then, $T^{\ast}$ is Dunford-Pettis $p$-convergent if and only if for each metrizable quotient $\overline{\Omega}$ of $\Omega,$ $\overline{T}^{\ast}$ is Dunford-Pettis $p$-convergent.

$\rm{(ii)}$ Suppose that $T:C(\Omega,X)\rightarrow Y^{\ast}$ is a bounded linear operator. Then, $T^{\ast}|_{Y}$ is Dunford-Pettis $p$-convergent if and only if for each metrizable quotient $\overline{\Omega}$ of $\Omega,$ $\overline{T}^{\ast}|_{Y}$ is Dunford-Pettis $p$-convergent.*

Proof.

We will only consider the part (i). The proof of (ii) is similar. Suppose that adjoint $T:C(\Omega,X)\rightarrow Y$ is Dunford-Pettis $p$-convergent and $\overline{\Omega}$ is a metrizable quotient space of $\Omega.$ Then $\overline{T}^{\ast}$ is Dunford-Pettis $p$-convergent.

Conversely, let $T:C(\Omega,X)\rightarrow Y$ be a bounded linear operator and let $(f_{n})_{n}$ be a sequence in $B_{0}.$ It is known (see [4]) that there exists a metrizable quotient space $\overline{\Omega}$ of $\Omega$ and a sequence $(\overline{f_{n}})$ in unit ball $C(\overline{\Omega},X)$ such that $(\overline{f_{n}})(\pi(t))=f_{n}(t)$ for all $t\in\Omega$ and $n\in\mathbb{N}.$ Define $\overline{T}:C(\overline{\Omega},X)\rightarrow Y$ by $\overline{T}(\overline{f})=T(\overline{f}\circ\pi),$ where is the $\pi:\Omega\rightarrow\overline{\Omega}$ canonical mapping. By the hypothesis, $\overline{T}^{\ast}$ is Dunford-Pettis $p$-convergent. Therefore $\{T(f_{n}):n\in\mathbb{N}\}=\{\overline{T}(\overline{f_{n}}):n\in\mathbb{N}\}$ is a $p$-Right*∗* set. Hence, the part $\rm{(i)}$ of Proposition 4.3 implies that $T^{\ast}$ is Dunford-Pettis $p$-convergent. ∎

The proofs of the following Lemma is similar to that of ([21, Lemma 21]) and and will be omitted.

Lemma 4.9**.**

Let $K$ be a bounded subset of $X.$ If for each $\varepsilon>0$ there is a $p$-Right∗ subset $K_{\varepsilon}$ of $X$ such that $K\subseteq K_{\varepsilon}+\varepsilon B_{X},$ then $K$ is a $p$-Right∗ set.

A topological space $\Omega$ is called dispersed (or scattered), if every nonempty closed subset of $\Omega$ has an isolated point [26]. By using the same argument as in the proof of ([21, Theorem 22]), we obtain the following result.

Theorem 4.10**.**

$\rm{(i)}$* Suppose that $\Omega$ is a dispersed compact Hausdorff space and $T:C(\Omega,X)\rightarrow Y$ is a strongly bounded operator. Then $T^{\ast}$ is Dunford-Pettis $p$-convergent if and only if $m(A)^{\ast}:Y^{\ast}\rightarrow X^{\ast}$ is Dunford-Pettis $p$-convergent, for each $A\in\Sigma.$

$\rm{(ii)}$ Let $\Omega$ be a dispersed compact Hausdorff space and $T:C(\Omega,X)\rightarrow Y^{\ast}$ be a strongly bounded operator. Then $T^{\ast}|_{Y}:Y\rightarrow C(\Omega,X)^{\ast}$ is Dunford-Pettis $p$-convergent if and only if $m(A)^{\ast}|_{Y}:Y\rightarrow X^{\ast}$ is Dunford-Pettis $p$-convergent, for each $A\in\Sigma,$*

Proof.

We will only consider the part (i). The proof of (ii) is similar. Suppose that $T:C(\Omega,X)\rightarrow Y$ is strongly bounded such that $T^{\ast}$ is Dunford-Pettis $p$-convergent, then for each $A\in\Sigma,$ Corollary 4.5 implies that $m(A)^{\ast}:Y^{\ast}\rightarrow X^{\ast}$ is a Dunford-Pettis $p$-convergent operator.

Conversely, suppose that $m(A)^{\ast}:Y^{\ast}\rightarrow X^{\ast}$ is Dunford-Pettis $p$-convergent for each $A\in\Sigma.$ Since a quotient space of a dispersed space is dispersed (see ([26, 8. 5. 3])), by using Lemmas 18 and 20 in [21], we can suppose without loss of generality that $\Omega$ is metrizable and so, $\Omega$ is countable (see ([26, 8. 5. 5])). Suppose that $\Omega=\{t_{i}:i\in\mathbb{N}\}$ and $(f_{n})$ is a sequence in $B_{0}.$ For each $i\in\mathbb{N}$ the set $\{f_{n}(t_{i}):i\in\mathbb{N}\}$ is bounded in $X.$ Therefore Proposition 4.3 (i) implies that, the set $H_{i}=\{m(\{t_{i}\})(f_{n}(t_{i})):n\in\mathbb{N}\}$ is a $p$-Right*∗* set, for each $i\in\mathbb{N}.$ Now, let $A_{i}=\{t_{j}:j>i\},i\in\mathbb{N}$ and $\varepsilon>0.$ Since $m$ is strongly bounded, there is a $k\in\mathbb{N}$ such that ${m}(A_{k})<\varepsilon.$ Hence, for each $n\in\mathbb{N},$

$T(f_{n})=\int_{\Omega}f_{n}dm=\displaystyle\sum_{i=1}^{k}m(\{t_{i}\})(f_{n}(t_{i}))+\int_{A_{k}}f_{n}dm.$

Further, $\|\int_{A_{k}}f_{n}dm\|<\tilde{m}(A_{k})<\varepsilon.$ Therefore

$T(f_{n})\in H_{1}+H_{2}+\cdot\cdot\cdot+H_{k}+\varepsilon B_{Y}.$

Since $H_{1}+H_{2}+\cdot\cdot\cdot+H_{k}$ is a $p$-Right*∗* set, by Lemma 4.9, $\{T(f_{n}):n\in\mathbb{N}\}$ is a $p$-Right*∗* set. Thus, an application of Proposition 4.3 (i) gives that $T^{\ast}$ is Dunford-Pettis $p$-convergent. ∎