On some classes of Dunford–Pettis-like operators

Abstract

In this paper we give characterizations of weak Dunford–Pettis, $\hbox {weak}^*$ Dunford–Pettis, weak p-convergent, and $\hbox {weak}^*$p-convergent operators.

1 Introduction

In this paper Dunford–Pettis sets and limited sets are used to characterize the classes of weak Dunford–Pettis and $\hbox {weak}^*$ Dunford–Pettis operators. The classes of weak p-convergent and $\hbox {weak}^*$p-convergent operators are also studied.

Our major results are Theorems 2, 6, 14, and 18. As consequences, we obtain equivalent characterizations of Banach spaces with the Dunford–Pettis property, $DP^*$-property, Dunford–Pettis property of order p, and $DP^*$-property of order p. We generalize some results in [1, 5, 11, 15, 18].

2 Definitions and notation

Throughout this paper, X and Y will denote real Banach spaces. The unit ball of X will be denoted by $B_X$ and $X^*$ will denote the continuous linear dual of X. An operator $T: X \rightarrow Y$ will be a continuous and linear function. The space of all operators from X to Y will be denoted by L(X, Y).

The operator T is completely continuous (or Dunford–Pettis) if T maps weakly convergent sequences to norm convergent sequences.

A Banach space X has the Dunford–Pettis property (DPP) if every weakly compact operator T with domain X is completely continuous. Equivalently, X has the DPP if and only if $x_n^*(x_n)\rightarrow 0$ for all weakly null sequences $(x_n)$ in X and $(x_n^*)$ in $X^*$ [8, Theorem 1]. Schur spaces, C(K) spaces, and $L_1(\mu )$ spaces have the DPP. The reader can check [7,8,9], and [2] for a guide to the extensive classical literature dealing with the DPP, equivalent formulations of the preceding definitions, and undefined notation and terminology.

A subset A of X is called a Dunford–Pettis (DP) subset (resp. limited subset) of X if each weakly null (resp. $w^*$ -null) sequence $(x_n^*)$ in $X^*$ tends to 0 uniformly on A [2] (resp. [3, 24]); i.e.

$$\begin{aligned} \sup _{x \in A} |x_n^* (x) | \rightarrow 0. \end{aligned}$$

If A is a limited subset of X, then T(A) is relatively compact for any operator $T:X\rightarrow c_0$ [3, p. 56], [24, p. 23]. The subset A of X is a DP subset of X if and only if T(A) is relatively compact whenever $T: X \rightarrow Y$ is a weakly compact operator [2] if and only if T(A) is relatively compact whenever $T: X \rightarrow Y$ is an operator with weakly precompact adjoint [20].

A bounded subset S of X is said to be weakly precompact provided that every sequence from S has a weakly Cauchy subsequence. Every DP subset of X is weakly precompact [2]. Since any limited set is a DP set, any limited set is weakly precompact. An operator $T:X\rightarrow Y$ is called weakly precompact (or almost weakly compact) if $T(B_X)$ is weakly precompact.

A Banach space X has the $DP^*$-property ($DP^*P$) if all weakly compact sets in X are limited [4, 5, 21]. The space X has the $DP^*P$ if and only if $x_n^*(x_n)\rightarrow 0$ for all weakly null sequences $(x_n)$ in X and $w^*$-null sequences $(x_n^*)$ in $X^*$ [16]. If X has the $DP^*P$, then it has the DPP. If X is a Schur space or if X has the DPP and the Grothendieck property (weak and $\hbox {weak}^*$ convergence of sequences in $X^*$ coincide), then X has the $DP^*P$.

3 Weak Dunford–Pettis operators and $\hbox {weak}^*$ Dunford–Pettis operators

An operator $T:X\rightarrow Y$ is called weak Dunford–Pettis [1, p. 349] if $\langle T(x_n), y_n^*\rangle \rightarrow 0$, whenever $(x_n)$ is a weakly null sequence in X and $(y_n^*)$ is a weakly null sequence in $Y^*$. An operator $T:X\rightarrow Y$ is called $\hbox {weak}^*$Dunford–Pettis [11] if $\langle T(x_n), y_n^*\rangle \rightarrow 0$, whenever $(x_n)$ is a weakly null sequence in X and $(y_n^*)$ is a $w^*$-null sequence in $Y^*$.

In this section we give some characterizations of weak Dunford–Pettis and $\hbox {weak}^*$ Dunford–Pettis operators.

Observation 1 If $T:X\rightarrow Y$ is an operator, then $T(B_X)$ is a DP (resp. limited) subset of Y if and only if $T^* : Y^* \rightarrow X^*$ is completely continuous (resp. $T^*$ is $w^*$-norm sequentially continuous).

To see this, note that $T(B_X)$ is a DP (resp. limited) subset of Y if and only if

$$\begin{aligned} 0=\lim _n \sup \{|\langle y_n^*, T(x)\rangle |: x\in B_X\} = \lim _n \sup \{|\langle T^*(y_n^*), x\rangle |: x\in B_X\} =\lim _n \Vert T^*(y_n^*)\Vert \end{aligned}$$

for each weakly null (resp. $w^*$-null) sequence $(y_n^*)$ in $Y^*$; that is, $T^* : Y^* \rightarrow X^*$ is completely continuous (resp. $T^*$ is $w^*$-norm sequentially continuous).

Theorem 1

[1, Theorem 5.99, p. 351] Let $T:X\rightarrow Y$ be an operator. The following statements are equivalent:

• (1) T is a weak Dunford–Pettis operator.

• (2) T carries weakly compact subsets of X to Dunford–Pettis subsets of Y.

• (3) If $S:Y\rightarrow Z$ is a weakly compact operator, then $ST:X\rightarrow Z$ is completely continuous, for any Banach space Z.

We are now giving our first major result. It gives characterizations of weak Dunford–Pettis operators and generalizes [1, Theorem 5.99, p. 351].

Theorem 2

Let X and Y be Banach spaces, and let $T:X\rightarrow Y$ be an operator. The following statements are equivalent.

• (1) T is a weak Dunford–Pettis operator.

• (2) T carries weakly precompact subsets of X to Dunford–Pettis subsets of Y.

• (3) For all Banach spaces Z, if $S:Y\rightarrow Z$ has a weakly precompact adjoint, then $ST:X\rightarrow Z$ is completely continuous.

• (4) If $S:Y\rightarrow c_0$ has a weakly precompact adjoint, then $ST:X\rightarrow c_0$ is completely continuous.

• (5) If $(x_n)$ is a weakly null sequence in X and $(y_n^*)$ is a weakly Cauchy sequence in $Y^*$, then $\langle y_n^*, T(x_n) \rangle \rightarrow 0$.

Proof

$(1) \Rightarrow (2)$ Let A be a weakly precompact subset of X. Suppose by contradiction that T(A) is not a Dunford–Pettis subset of Y. Suppose that $(y_n^*)$ is a weakly null sequence in $Y^*$, $(x_n)$ is a sequence in A, and $\epsilon >0$ such that $|\langle y_n^*, T(x_n) \rangle |>\epsilon$ for all n. Without loss of generality assume that $(x_n)$ is weakly Cauchy.

Let $n_1=1$ and choose $n_2>n_1$ so that $|\langle y_{n_2}^*, T(x_{n_1})\rangle |<\epsilon /2$. We can do this since $(y_n^*)$ is $w^*$-null. Continue inductively. Choose $n_k>n_{k-1}$ so that $|\langle y_{n_k}^*, T(x_{n_{k-1}}) \rangle |<\epsilon /2$. Since T is a weak Dunford–Pettis operator, $\langle y^*_{n_k}, T(x_{n_k}-x_{n_{k-1}})\rangle \rightarrow 0$. However,

$$\begin{aligned} |\langle y_{n_k}^*, T(x_{n_k}-x_{n_{k-1}})\rangle |\ge |\langle y_{n_k}^*, T(x_{n_k})\rangle |-|\langle y_{n_k}^*, T(x_{n_{k-1}})\rangle |> \epsilon /2, \end{aligned}$$

a contradiction.

$(2) \Rightarrow (3)$ Let $S:Y\rightarrow Z$ be an operator such that $S^*:Z^*\rightarrow Y^*$ is weakly precompact. Suppose $(x_n)$ is a weakly null sequence in X. Since $\{T(x_n):n\in {\mathbb {N}}\}$ is a Dunford–Pettis set in Y and $S^*$ is weakly precompact, $\{ST(x_n):n\in {\mathbb {N}}\}$ is relatively compact [20, Corollary 4]. Then $\Vert ST(x_n)\Vert \rightarrow 0$, and thus $ST:X\rightarrow Z$ is completely continuous.

$(3) \Rightarrow (4)$ and $(5) \Rightarrow (1)$ are obvious.

$(4) \Rightarrow (1)$ Suppose $(x_n)$ is weakly null in X and $(y_n^*)$ is weakly null in $Y^*$. Define $S:Y\rightarrow c_0$ by $S(y)=(y_i^*(y))$. Then $S^*:\ell _1\rightarrow Y^*$, $S^*(b)=\sum b_i y_i^*$. Note that $S^*$ maps $B_{\ell _1}$ into the closed and absolutely convex hull of $\{y_i^*:i\in {\mathbb {N}}\}$, which is relatively weakly compact [9, p. 51]. Then $S^*$ is weakly compact. Hence $ST:X\rightarrow c_0$ is completely continuous. Therefore $\langle T(x_n), y_n^*\rangle \le \Vert ST(x_n)\Vert =\sup _i |\langle y_i^*, T(x_n)\rangle |\rightarrow 0$, and T is weak Dunford–Pettis.

$(1) \Rightarrow (5)$ Suppose that $(x_n)$ is a weakly null sequence in X, $(y_n^*)$ is a weakly Cauchy sequence in $Y^*$, and $\langle y_n^*, T(x_n) \rangle \not \rightarrow 0$. Without loss of generality suppose that $|\langle y_n^*, T(x_n) \rangle |> \epsilon$ for each $n\in {\mathbb {N}}$, for some $\epsilon > 0$.

Let $n_1=1$ and choose $n_2>n_1$ so that $|\langle y_{n_1}^*, T(x_{n_2})\rangle |<\epsilon /2$. We can do this since $(T(x_n))$ is weakly null. Continue inductively. Choose $n_{k+1}>n_{k}$ so that $|\langle y_{n_k}^*, T(x_{n_{k+1}})\rangle |<\epsilon /2$. Since T is weak Dunford–Pettis, $\langle y^*_{n_{k+1}}-y_{n_k}^*, T(x_{n_{k+1}}) \rangle \rightarrow 0$. Since

$$\begin{aligned} |\langle y^*_{n_{k+1}}-y_{n_k}^*, T(x_{n_{k+1})}\rangle |\ge |\langle y_{n_{k+1}}^*, T(x_{n_{k+1}})\rangle | -|\langle y_{n_k}^*, T(x_{n_{k+1}})\rangle |>\epsilon /2, \end{aligned}$$

we have a contradiction. $\square$

Corollary 3

Let X and Y be Banach spaces, and let $T:X\rightarrow Y$ be an operator. The following statements are equivalent.

• (i) T is a weak Dunford–Pettis operator.

• (ii) For all Banach spaces Z, if $S:Z\rightarrow X$ is a weakly precompact operator, then $TS:Z\rightarrow Y$ has a completely continuous adjoint.

• (iii) If $S:\ell _1\rightarrow X$ is a weakly precompact operator, then $TS:\ell _1\rightarrow Y$ has a completely continuous adjoint.

• (iv) If $(x_n)$ is a weakly Cauchy sequence in X and $(y_n^*)$ is a weakly null sequence in $Y^*$, then $\langle y_n^*, T(x_n) \rangle \rightarrow 0$.

Proof

$\mathrm{(i)} \Rightarrow \mathrm{(ii)}$ Let $S:Z\rightarrow X$ be a weakly precompact operator. Then $TS(B_{Z})$ is a Dunford–Pettis set. Thus $TS:Z\rightarrow Y$ has a completely continuous adjoint.

$\mathrm{(iii)} \Rightarrow \mathrm{(iv)}$ Suppose $(x_n)$ is a weakly Cauchy sequence in X and $(y_n^*)$ is a weakly null sequence in $Y^*$. Define $S:\ell _1\rightarrow X$ by

$$\begin{aligned} S(b)=\sum b_n x_n, \end{aligned}$$

where $b=(b_n)\in \ell _1$. Since $S(B_{\ell _1} )$ is contained in the closed and absolutely convex hull of $\{ x_n : n \in {\mathbb {N}}\}$, which is weakly precompact [24, p. 27], S is weakly precompact.

By assumption, $(TS)^*=S^*T^*$ is completely continuous. Note that $S^*(x^*)=(\langle x^*, x_i\rangle )_i$, $x^*\in X^*$, and $S^*T^*(y_n^*)=(\langle T^*(y_n^*), x_i\rangle )_i$. Hence

$$\begin{aligned} \langle y_n^*, T(x_n) \rangle =\langle T^*(y_n^*), x_n \rangle \le \Vert S^*T^*(y_n^*)\Vert = \sup _i |\langle T^*(y_n^*), x_i\rangle | \rightarrow 0. \end{aligned}$$

$\mathrm{(ii)} \Rightarrow \mathrm{(iii)}$ and $\mathrm{(iv)} \Rightarrow \mathrm{(i)}$ are obvious. $\square$

The following two corollaries provide equivalent characterizations of spaces with the Dunford–Pettis property.

Corollary 4

Let X be a Banach space. Then the following statements are equivalent:

• (i) X has the DPP.

• (ii) The identity operator $i:X\rightarrow X$ is a weak Dunford–Pettis operator; that is, every weakly precompact subset of X is a Dunford–Pettis set.

• (iii) [8] Every operator $S:X\rightarrow Z$ with weakly precompact adjoint is completely continuous, for any Banach space Z.

• (iv) [8] Every operator $S:X\rightarrow c_0$ with weakly precompact adjoint is completely continuous.

• (v) [8] If $(x_n)$ is a weakly null sequence in X and $(x_n^*)$ is a weakly Cauchy sequence in $X^*$, then $x_n^*(x_n) \rightarrow 0$.

Proof

Apply Theorem 2 to the identity operator $i:X\rightarrow X$. $\square$

We note that X has the DPP if and only if weakly precompact sets and DP sets coincide (since every DP set is weakly precompact [2]).

Corollary 5

Let X be a Banach space. Then the following statements are equivalent:

• (i) X has the DPP.

• (ii) [8] For all Banach spaces Z, every weakly precompact operator $S:Z\rightarrow X$ has a completely continuous adjoint.

• (iii) Every weakly precompact operator $S:\ell _1 \rightarrow X$ has a completely continuous adjoint.

• (iv) [8] If $(x_n)$ is a weakly Cauchy sequence in X and $(x_n^*)$ is a weakly null sequence in $X^*$, then $x_n^*(x_n) \rightarrow 0$.

Proof

Apply Corollary 3 to the identity operator $i:X\rightarrow X$. $\square$

An operator $T:X \rightarrow Y$ is called limited if $T(B_X)$ is limited in Y. The operator $T:X \rightarrow Y$ is limited if and only if $T^*:Y^* \rightarrow X^*$ is $w^*$-norm sequentially continuous (by Observation 1).

We are now giving our second major result. It gives a characterization of $\hbox {weak}^*$ Dunford–Pettis operators and generalizes [11, Theorem 3.2].

Theorem 6

Let X and Y be Banach spaces and let $T:X\rightarrow Y$ be an operator. The following statements are equivalent.

• (1) T is a $weak^*$ Dunford–Pettis operator.

• (2) T carries weakly precompact subsets of X to limited subsets of Y.

• (3) If $S:Z\rightarrow X$ is a weakly precompact operator, then $TS:Z\rightarrow Y$ is limited, for any Banach space Z.

• (4) If $S:\ell _1\rightarrow X$ is a weakly precompact operator, then $TS:\ell _1\rightarrow Y$ is limited.

• (5) If $(x_n)$ is a weakly null sequence in X and $(y_n^*)$ is a $w^*$-Cauchy sequence in $Y^*$, then $\langle y_n^*, T(x_n) \rangle \rightarrow 0$.

Proof

$(1) \Rightarrow (2)$ is similar to the proof of $(1) \Rightarrow (2)$ in Theorem 2.

$(2) \Rightarrow (3)$ Suppose $S:Z\rightarrow X$ is weakly precompact. Then $TS(B_Z)$ is limited, and thus TS is limited.

$(3) \Rightarrow (4)$ and $(5) \Rightarrow (1)$ are obvious.

$(4) \Rightarrow (1)$ Let $(x_n)$ be a weakly null sequence in X and $(y_n^*)$ be a $w^*$-null sequence in $Y^*$. Define $S:\ell _1\rightarrow X$ by

$$\begin{aligned} S(b)=\sum b_n x_n, \end{aligned}$$

where $b=(b_n)\in \ell _1$. Since $S(B_{\ell _1} )$ is contained in the closed and absolutely convex hull of $\{ x_n : n \in {\mathbb {N}}\}$, which is relatively weakly compact [9, p. 51], S is weakly compact. By assumption, TS is limited. Suppose $(e_n^*)$ denotes the unit vector basis of $\ell _1$. Then

$$\begin{aligned} \langle y_n^*, T(x_n) \rangle =\langle y_n^*, TS(e_n^*)\rangle \rightarrow 0. \end{aligned}$$

$(1) \Rightarrow (5)$ is similar to the proof of $(1) \Rightarrow (5)$ in Theorem 2. $\square$

Corollary 7

Let X and Y be Banach spaces and let $T:X\rightarrow Y$ be an operator. The following statements are equivalent.

• (i) T is a $weak^*$ Dunford–Pettis operator.

• (ii) If $(x_n)$ is a weakly Cauchy sequence in X and $(y_n^*)$ is a $w^*$-null sequence in $Y^*$, then $\langle y_n^*, T(x_n) \rangle \rightarrow 0$.

• (iii) If $S:Y\rightarrow Z$ is an operator such that $S^*(B_{Z^*})$ is $w^*$-sequentially compact, then $ST:X\rightarrow Z$ is completely continuous.

• (iv) If $S:Y\rightarrow c_0$ is an operator, then $ST:X\rightarrow c_0$ is completely continuous.

Proof

$\mathrm{(i)} \Rightarrow \mathrm{(ii)}$ Suppose that $(x_n)$ is a weakly Cauchy sequence in X and $(y_n^*)$ is a $w^*$-null sequence in $Y^*$. Since $(T(x_n))$ is limited in Y, $\langle y_n^*, T(x_n)\rangle \rightarrow 0$.

$\mathrm{(ii)} \Rightarrow \mathrm{(iii)}$ Let $S:Y\rightarrow Z$ be an operator such that $S^*(B_{Z^*})$ is $w^*$-sequentially compact, but $ST:X\rightarrow Z$ is not completely continuous. Let $(x_n)$ be weakly null in X so that $\Vert ST(x_n)\Vert >\epsilon$, for some $\epsilon >0$. Choose $(z_n^*)$ in $B_{Z^*}$ so that $\langle z_n^*, ST(x_n)\rangle > \epsilon$. Without loss of generality $(S^*(z_n^*))$ is $w^*$-convergent. Hence $\langle S^*(z_n^*), T(x_n) \rangle = \langle z_n^*, ST(x_n) \rangle \rightarrow 0$, a contradiction.

$\mathrm{(iii)} \Rightarrow \mathrm{(iv)}$ Let $S:Y\rightarrow c_0$ be an operator. Note that $B_{\ell _1}$, and thus $S^*(B_{\ell _1})$ is $w^*$-sequentially compact (since $c_0$ is separable). Then $ST:X\rightarrow c_0$ is completely continuous.

$\mathrm{(iv)} \Rightarrow \mathrm{(i)}$ Suppose $(x_n)$ is a weakly null sequence in X and $(y_n^*)$ is a $w^*$-null sequence in $Y^*$. Define $S:Y\rightarrow c_0$ by $S(y)=(y_i^*(y))$. Since $ST:X\rightarrow c_0$ is completely continuous, $\langle y_n^*, T(x_n)\rangle \le \Vert ST(x_n)\Vert \rightarrow 0$. $\square$

The following corollary provides a characterization of spaces with the $DP^*P$ and generalizes [11, Corollary 3.3].

Corollary 8

Let X be a Banach space. Then the following statements are equivalent:

• (i) X has the $DP^*P$.

• (ii) [16] The identity operator $i:X\rightarrow X$ is a $weak^*$ Dunford–Pettis operator ; that is, every weakly precompact subset of X is a limited set.

• (iii) [16] Every weakly precompact operator $S:Z\rightarrow X$ is limited, for any Banach space Z.

• (iv) Every weakly precompact operator $S:\ell _1\rightarrow X$ is limited.

• (v) [16] If $(x_n)$ is a weakly null sequence in X and $(x_n^*)$ is a $w^*$-Cauchy sequence in $X^*$, then $x_n^*(x_n) \rightarrow 0$.

Proof

Apply Theorem 6 to the identity operator $i:X\rightarrow X$. $\square$

We note that X has the $DP^*P$ if and only if weakly precompact sets and limited sets coincide (since every limited set is weakly precompact [3]).

Corollary 9

Let X be a Banach space. Then the following statements are equivalent:

• (i) X has the $DP^*P$.

• (ii) [16] If $(x_n)$ is a weakly Cauchy sequence in X and $(x_n^*)$ is a $w^*$-null sequence in $Y^*$, then $x_n^*(x_n) \rightarrow 0$.

• (iii) [16] If $S:X\rightarrow Z$ is an operator such that $S^*(B_{Z^*})$ is $w^*$-sequentially compact, then S is completely continuous.

• (iv) [5] Every operator $S:X\rightarrow c_0$ is completely continuous.

Proof

Apply Corollary 7 to the identity operator $i:X\rightarrow X$. $\square$

Corollary 10

• (i) If $Y^*$ does not contain a copy of $\ell _1$, then every weak Dunford–Pettis operator $T:X\rightarrow Y$ is completely continuous.

• (ii) If $B_{Y^*}$ is $w^*$-sequentially compact (in particular if Y is separable), then every $\hbox {weak}^*$ Dunford–Pettis operator $T:X\rightarrow Y$ is completely continuous.

• (iii) If X or Y has the DPP, then every operator $T:X\rightarrow Y$ is weak Dunford–Pettis.

• (iv) If X or Y has the $DP^*P$, then every operator $T:X\rightarrow Y$ is $\hbox {weak}^*$ Dunford–Pettis.

Proof

• (i) Let $i:Y\rightarrow Y$ be the identity operator on Y. Suppose that $T:X\rightarrow Y$ is a weak Dunford–Pettis operator. Since $Y^*$ does not contain a copy of $\ell _1$, $i^*$ is weakly precompact (by Rosenthal’s $\ell _1$ theorem). Then $T=iT$ is completely continuous by Theorem 2.

• (ii) Let $i:Y\rightarrow Y$ be the identity operator on Y. Suppose $T:X\rightarrow Y$ is a $\hbox {weak}^*$ Dunford–Pettis operator. Since $i^*(B_{Y^*})$ is $w^*$-sequentially compact, $T=iT$ is completely continuous by Corollary 7.

• (iii) Let $T:X\rightarrow Y$ be an operator. If Y has the DPP, then the identity operator $i:Y\rightarrow Y$ is weak Dunford–Pettis. Hence $T=iT$ is weak Dunford–Pettis. If X has the DPP, then the identity operator $i:X\rightarrow X$ is weak Dunford–Pettis. Hence $T=Ti$ is weak Dunford–Pettis.

• (iv) The proof is similar to that of (iii).

$\square$

Clearly each completely continuous operator $T:X\rightarrow Y$ is $\hbox {weak}^*$ Dunford–Pettis and each $\hbox {weak}^*$ Dunford–Pettis operator is weak Dunford–Pettis. By Corollary 10, we obtain the following result.

Corollary 11

If $Y^*$ does not contain a copy of $\ell _1$, then the families of completely continuous operators, $weak^*$ Dunford–Pettis operators, and weak Dunford–Pettis operators $T:X\rightarrow Y$ coincide.

Examples (a) Note that $\ell _\infty$ has the $DP^*P$ (since it has the DPP and the Grothendieck property [5]). Then the identity operator $i:\ell _\infty \rightarrow \ell _\infty$ is $\hbox {weak}^*$ Dunford–Pettis and is not completely continuous.

(b) A space X has the $DP^*P$ if and only if every operator $T:X\rightarrow c_0$ is completetely continuous [5]. Since the identity operator $i:c_0 \rightarrow c_0$ is not completetely continuous, $c_0$ does not have the $DP^*P$. Thus $i:c_0 \rightarrow c_0$ is weak Dunford–Pettis (since $c_0$ has the DPP) and not $\hbox {weak}^*$ Dunford–Pettis.

Corollary 12

• (i) Suppose that Y has the DPP. If $T:X\rightarrow Y$ is an operator such that $T^*$ is not completely continuous, then T fixes a copy of $\ell _1$.

• (ii) Suppose that Y has the $DP^*P$. If $T:X\rightarrow Y$ is a non-limited operator, then T fixes a copy of $\ell _1$.

Proof

• (i) Suppose that $T^*$ is not completely continuous. Let $(y_n^*)$ be weakly null in $Y^*$ so that $\Vert T^*(y_n^*)\Vert \not \rightarrow 0$. Suppose $(x_n)$ is a sequence in $B_X$ such that $|\langle y_n^*, T(x_n)\rangle | >\epsilon$ for some $\epsilon >0$. We claim that $(x_n)$ has no weakly Cauchy subsequence. If the claim is false, suppose without loss of generality that $(x_n)$ is weakly Cauchy. Since Y has the DPP, T is weak Dunford–Pettis. Then $\langle y_n^*, T(x_n)\rangle \rightarrow 0$ by Corollary 3. This contradiction shows that $(x_n)$ has no weakly Cauchy subsequence. By Rosenthal’s $\ell _1$ theorem, $(x_n)$ has a subsequence equivalent to the $\ell _1$ basis. Suppose without loss of generality that $(x_n)$ is equivalent to $(e_n^*)$, where $(e_n^*)$ denotes the basis of $\ell _1$.

  Now, since $|\langle y_n^*, T(x_n)\rangle | >\epsilon$ and Y has the DPP, $(T(x_n))$ has no weakly Cauchy subsequence (by Corollary 5). By Rosenthal’s $\ell _1$ theorem, $(T(x_n))$ has a subsequence equivalent to $(e_n^*)$. Suppose without loss of generality that $(T(x_n))$ is equivalent to $(e_n^*)$. Hence T fixes a copy of $\ell _1$.

• (ii) The proof is similar to that of (i).

$\square$

Corollary 12 (ii) generalizes [5, Theorem 2.3] (which states that if X and Y have the the $DP^*P$ and $T:X\rightarrow Y$ is a non-limited operator, then T fixes a copy of $\ell _1$).

A Banach space X has the Dunford–Pettis relatively compact property (DPrcP) if every DP subset of X is relatively compact [13]. Schur spaces have the DPrcP. The space X does not contain a copy of $\ell _1$ if and only if $X^*$ has the DPrcP [12, 13]. We note that if $X^*$ does not contain a copy of $\ell _1$, then $X^{**}$, thus X, has the DPrcP [12, 13].

The space X has the Gelfand–Phillips (GP) property (or X is a Gelfand–Phillips space) if every limited subset of X is relatively compact. Schur spaces and separable spaces have the Gelfand–Phillips property [3].

An operator $T:X\rightarrow Y$ is called Dunford–Pettis completely continuous (DPcc) if T carries weakly null and DP sequences to norm null ones [22]. An operator $T:X\rightarrow Y$ is called limited completely continuous (lcc) if T maps weakly null limited sequences to norm null sequences [23].

The sets of all limited completely continuous, Dunford–Pettis completely continuous operators, weak Dunford Pettis, and $\hbox {weak}^*$ Dunford Pettis operators from X to Y will be respectively denoted by LCC(X, Y), DPCC(X, Y), WDP(X, Y), and $W^*DP(X,Y)$.

In the following result, we characterize Banach spaces X on which every weak (resp. $\hbox {weak}^*$) Dunford–Pettis operator is a DPcc (resp. lcc) operator.

Corollary 13

• (i) A Banach space X has the DPrcP if and only if $DPCC(X, \ell _\infty )=WDP(X,\ell _\infty )$.

• (ii) A Banach space X has the GP property if and only if $LCC(X, \ell _\infty )=W^*DP(X,\ell _\infty )$.

Proof

• (i) A Banach space X has the DPrcP if and only if $DPCC(X, \ell _\infty )=L(X,\ell _\infty )$ [22]. Since $\ell _\infty$ has the DPP, $L(X,\ell _\infty )=WDP(X,\ell _\infty )$.

• (ii) A Banach space X has the GP property if and only if $LCC(X, \ell _\infty )=L(X,\ell _\infty )$ [23]. Since $\ell _\infty$ has the $DP^*P$, $L(X,\ell _\infty )=W^*DP(X,\ell _\infty )$.$\square$

If X has the DPrcP, then X has the GP property (since any limited set is a DP set). Thus, if X has the DPrcP, then $L(X,\ell _\infty )=LCC(X,\ell _\infty )=DPCC(X,\ell _\infty )= WDP(X,\ell _\infty )=W^*DP(X,\ell _\infty )$.

Example We note that the identity operator $i:\ell _\infty \rightarrow \ell _\infty$ is $\hbox {weak}^*$ Dunford–Pettis and not lcc (since $\ell _\infty$ does not have the GP property). Further, $i:\ell _\infty \rightarrow \ell _\infty$ is weak Dunford–Pettis (since $\ell _\infty$ has the DPP) and not DPcc (since $\ell _\infty$ does not have the DPrcP).

4 Weak p-convergent operators and $\hbox {weak}^*$p-convergent operators

For $1\le p<\infty$, $p^*$ denotes the conjugate of p. If $p=1$, we take $c_0$ instead of $\ell _{p^*}$. The unit vector basis of $\ell _p$ will be denoted by $(e_n)$.

Let $1\le p < \infty$. A sequence $(x_n)$ in X is called weakly p-summable if $(x^*(x_n))\in \ell _p$ for each $x^*\in X^*$ [10, p. 32]. Let $\ell _p^w(X)$ denote the set of all weakly p-summable sequences in X. The space $\ell _p^w(X)$ is a Banach space with the norm

$$\begin{aligned} \Vert (x_n)\Vert _p^w=\sup \left\{ \left( \sum _{n=1}^\infty |\langle x^*, x_n \rangle |^p \right) ^{1/p}: x^*\in B_{X^*}\right\} \end{aligned}$$

We recall the following isometries: $L(\ell _{p^*},X) \simeq \ell _p^w(X)$ for $1<p<\infty$; $L(c_0,X)\simeq \ell _p^w(X)$ for $p=1$; that are obtained via the isometry $T \rightarrow (T(e_n))$ [10, Proposition 2.2, p. 36].

A series $\sum x_n$ in X is said to be weakly unconditionally convergent (wuc) if for every $x^*\in X^*$, the series $\sum |x^*(x_n)|$ is convergent. An operator $T:X\rightarrow Y$ is unconditionally converging if it maps weakly unconditionally convergent series to unconditionally convergent ones.

Let $1\le p\le \infty$. An operator $T:X\rightarrow Y$ is called p-convergent if T maps weakly p-summable sequences into norm null sequences. The set of all p-convergent operators from X to Y is denoted by $C_p(X,Y)$ [6].

The 1-convergent operators are precisely the unconditionally converging operators and the $\infty$-convergent operators are precisely the completely continuous operators. If $p<q$, then $C_q(X,Y)\subseteq C_p(X,Y)$.

A sequence $(x_n)$ in X is called weakly p-convergent to $x\in X$ if the sequence $(x_{n}-x)$ is weakly p-summable [6]. Weakly $\infty$-convergent sequences are precisely the weakly convergent sequences.

Let $1\le p\le \infty$. A bounded subset K of X is relatively weakly p-compact if every sequence in K has a weakly p-convergent subsequence. An operator $T:X\rightarrow Y$ is weakly p-compact if $T(B_X)$ is relatively weakly p-compact [6].

The set of weakly p-compact operators $T:X\rightarrow Y$ is denoted by $W_p(X,Y)$. If $p<q$, then $W_p(X,Y)\subseteq W_q(X,Y)$. A Banach space $X\in C_p$ (resp. $X\in W_p$) if $id(X)\in C_p(X,X)$ (resp. $id(X)\in W_p(X,X)$) [6], where id(X) is the identity map on X.

A sequence $(x_n)$ in X is called weakly p-Cauchy if $(x_{n_k}-x_{m_k})$ is weakly p-summable for any increasing sequences $(n_k)$ and $(m_k)$ of positive integers.

Every weakly p-convergent sequence is weakly p-Cauchy, and the weakly $\infty$-Cauchy sequences are precisely the weakly Cauchy sequences.

Let $1\le p\le \infty$. A subset S of X is called weakly p-precompact if every sequence from S has a weakly p-Cauchy subsequence [18]. An operator $T:X\rightarrow Y$ is called weakly p-precompact if $T(B_X)$ is weakly p-precompact.

Let $1\le p\le \infty$. A Banach space X has the Dunford–Pettis property of order p $(DPP_p)$ ($1\le p\le \infty$) if every weakly compact operator $T:X \rightarrow Y$ is p-convergent, for any Banach space Y [6]. Equivalently, X has the $DPP_p$ if and only if $x_n^*(x_n)\rightarrow 0$ whenever $(x_n)$ is weakly p-summable in X and $(x_n^*)$ is weakly null in $X^*$ [6, Proposition 3.2].

If X has the $DPP_p$, then it has the $DPP_q$, if $q<p$. Also, the $DPP_\infty$ is precisely the DPP, and every Banach space has the $DPP_1$. C(K) spaces and $L_1$ have the DPP, and thus the $DPP_p$ for all p. If $1<r<\infty$, then $\ell _r$ has the $DPP_p$ for $p<r^*$. If $1<r<\infty$, then $L_r(\mu )$ has the $DPP_p$ for $p<min(2,r^*)$. Tsirelson’s space T has the $DPP_p$ for all $p<\infty$. Since T is reflexive, it does not have the DPP. Tsirelson’s dual space $T^*$ does not have the $DPP_p$, if $p>1$ [6].

Let $1\le p\le \infty$. A Banach space X has the $DP^*$-property of order p ($DP^*P_p$) if all weakly p-compact sets in X are limited [14]. Equivalently, X has the $DP^*P_p$ if and only if $x_n^*(x_n)\rightarrow 0$ whenever $(x_n)$ is weakly p-summable in X and $(x_n^*)$ is weakly null in $X^*$ [14].

If X has the $DP^*P_q$, then it has the $DP^*P_p$, if $q>p$. Further, the $DP^*P_\infty$ is precisely the $DP^*P$ and every Banach space has the $DP^*P_1$. If X has the $DP^*P$, then X has the $DP^*P_p$, $1 \le p \le \infty$. If X has the $DP^*P_p$, then X has the $DPP_p$.

Let $1\le p<\infty$. An operator $T:X\rightarrow Y$ is called weak p-convergent if $\langle y_n^*,T(x_n) \rangle \rightarrow 0$ whenever $(x_n)$ is weakly p-summable in X and $(y_n^*)$ is weakly null in $Y^*$ [15]. An operator $T:X\rightarrow Y$ is called $weak^*$p-convergent if $\langle y_n^*,T(x_n) \rangle \rightarrow 0$ whenever $(x_n)$ is weakly p-summable in X and $(y_n^*)$ is $w^*$-null in $Y^*$ [15].

In the following we study weak p-convergent and $\hbox {weak}^*$p-convergent operators. The following result generalizes [18, Theorem 8].

Theorem 14

Let X and Y be Banach spaces, and let $1< p<\infty$. The following statements are equivalent about an operator $T:X\rightarrow Y$.

• (1) T is weak p-convergent.

• (2) T takes weakly p-precompact subsets of X to DP subsets of Y.

• (3) For any Banach space Z, if $S:Y\rightarrow Z$ has a weakly precompact adjoint, then $ST:X\rightarrow Z$ is p-convergent.

• (4) If $S:Y\rightarrow c_0$ has a weakly precompact adjoint, then $ST:X\rightarrow c_0$ is p-convergent.

• (5) If $(x_n)$ is a weakly p-summable sequence in X and $(y_n^*)$ is a weakly Cauchy sequence in $Y^*$, then $\langle y_n^*, T(x_n)\rangle \rightarrow 0$.

Proof

$(1)\Rightarrow (2)$ Let A be a weakly p-precompact subset of X. Suppose by contradiction that T(A) is not a Dunford–Pettis subset of Y. Let $(y_n^*)$ be a weakly null sequence in $Y^*$, and let $(x_n)$ be a sequence in A such that $|\langle y_n^*, T(x_n) \rangle |>\epsilon$ for all n, for some $\epsilon >0$. By passing to a subsequence, we can assume that $(x_n)$ is weakly p-Cauchy.

Let $n_1=1$ and choose $n_2>n_1$ so that $|\langle y_{n_2}^*, T(x_{n_1})\rangle |<\epsilon /2$. We can do this since $(y_n^*)$ is $w^*$-null. Continue inductively. Choose $n_k>n_{k-1}$ so that $|\langle y_{n_k}^*, T(x_{n_{k-1}}) \rangle |<\epsilon /2$. Since T is weak p-convergent, $\langle y^*_{n_k}, T(x_{n_k}-x_{n_{k-1}})\rangle \rightarrow 0$. However,

$$\begin{aligned} |\langle y_{n_k}^*, T(x_{n_k}-x_{n_{k-1}})\rangle |\ge |\langle y_{n_k}^*, T(x_{n_k})\rangle |-|\langle y_{n_k}^*, T(x_{n_{k-1}})\rangle |> \epsilon /2, \end{aligned}$$

a contradiction.

$(2)\Rightarrow (3)$ Suppose $S:Y\rightarrow Z$ is an operator with weakly precompact adjoint. Let $(x_n)$ be a weakly p-summable sequence in X. By (2), $(T(x_n))$ is a DP subset of Y. Therefore $(ST(x_n))$ is relatively compact [20, Corollary 4]. Hence $\Vert ST(x_n)\Vert \rightarrow 0$, and thus ST is p-convergent.

$(3) \Rightarrow (4)$ and $(5) \Rightarrow (1)$ are obvious.

$(4) \Rightarrow (1)$ Let $(x_n)$ be a weakly p-summable sequence in X and $(y_n^*)$ be a weakly null sequence in $Y^*$. Define $S:Y\rightarrow c_0$ by $S(y)=(y_i^*(y))$. Then $S^*:\ell _1\rightarrow Y^*$, $S^*(b)=\sum b_i y_i^*$. Note that $S^*$ maps $B_{\ell _1}$ into the closed and absolutely convex hull of $\{y_i^*:i\in {\mathbb {N}}\}$, which is relatively weakly compact [9, p. 51]. Then $S^*$ is weakly compact. Hence $ST:X\rightarrow c_0$ is p-convergent. Therefore $\langle T(x_n), y_n^*\rangle \le \Vert ST(x_n)\Vert =\sup _i |\langle y_i^*, T(x_n)\rangle |\rightarrow 0$, and T is weak p-convergent.

$(1) \Rightarrow (5)$ Let $(x_n)$ be a weakly p-summable sequence in X and $(y_n^*)$ be a weakly Cauchy sequence in $Y^*$. Suppose $\langle y_n^*, T(x_n) \rangle \not \rightarrow 0$. Without loss of generality suppose that $|\langle y_n^*, T(x_n) \rangle |> \epsilon$ for each $n\in {\mathbb {N}}$, for some $\epsilon > 0$.

Let $n_1=1$ and choose $n_2>n_1$ so that $|\langle y_{n_1}^*, T(x_{n_2})\rangle |<\epsilon /2$. We can do this since $(T(x_n))$ is weakly null. Continue inductively. Choose $n_{k+1}>n_{k}$ so that $|\langle y_{n_k}^*, T(x_{n_{k+1}})\rangle |<\epsilon /2$. Since T is a weak p-convergent operator, $|\langle y^*_{n_{k+1}}-y_{n_k}^*, T(x_{n_{k+1}}) \rangle |\rightarrow 0$. However,

$$\begin{aligned} |\langle y^*_{n_{k+1}}-y_{n_k}^*, T(x_{n_{k+1})}\rangle |\ge |\langle y_{n_{k+1}}^*, T(x_{n_{k+1}})\rangle |-|\langle y_{n_k}^*, T(x_{n_{k+1}})\rangle |>\epsilon /2, \end{aligned}$$

and we have a contradiction. $\square$

Corollary 15

Let X and Y be Banach spaces, and let $1< p<\infty$. The following statements are equivalent about an operator $T:X\rightarrow Y$.

• (i) T is weak p-convergent.

• (ii) For every Banach space Z, if $S:Z\rightarrow X$ is a weakly p-precompact operator, then $TS:Z\rightarrow Y$ has a completely continuous adjoint.

• (iii) [18] If $S:\ell _{p^*}\rightarrow X$ is an operator, then $TS:\ell _{p^*} \rightarrow Y$ has a completely continuous adjoint.

• (iv) If $(x_n)$ is a weakly p-Cauchy sequence in X and $(y_n^*)$ is a weakly null sequence in $Y^*$, then $\langle y_n^*, T(x_n)\rangle \rightarrow 0$.

Proof

$\mathrm{(i)}\Rightarrow \mathrm{(ii)}$ Let $S:Z\rightarrow X$ be a weakly p-precompact operator. Then $TS(B_Z)$ is a DP set in Y. Hence $(TS)^*$ is completely continuous.

$\mathrm{(ii)}\Rightarrow \mathrm{(iii)}$ Let $S:\ell _{p^*}\rightarrow X$ be an operator. Since $1< p<\infty$, $\ell _{p^*} \in W_p$ [6]. Hence S is weakly p-compact, and thus $(TS)^*$ is completely continuous.

$\mathrm{(iii)}\Rightarrow \mathrm{(i)}$ Let $(x_n)$ be weakly p-summable in X and let $(y_n^*)$ be weakly null in $Y^*$. Define $S:\ell _{p^*}\rightarrow X$ by $S(b)=\sum b_i x_i$ [10, Proposition 2.2, p. 36]. Since $TS(B_{\ell _{p^*}})$ is a DP set in Y, $\langle y_n^*, TS(e_n)\rangle = \langle y_n^*, T(x_n)\rangle \rightarrow 0$.

$\mathrm{(i)}\Rightarrow \mathrm{(iv)}$ Let $(x_n)$ be weakly p-Cauchy in X and let $(y_n^*)$ be weakly null in $Y^*$. Since $(T(x_n))$ is a DP set in Y, $\langle y_n^*, T(x_n)\rangle \rightarrow 0$.

$\mathrm{(iv)} \Rightarrow \mathrm{(i)}$ is obvious. $\square$

As a consequence of the previous two results we obtain the following characterizations of Banach spaces with the $DPP_p$.

Corollary 16

[19, Theorem 1] Let $1< p<\infty$. The following statements are equivalent about a Banach space X.

• (1) X has the $DPP_p$.

• (2) The identity operator $i:X\rightarrow X$ is weak p-convergent; that is, every weakly p-precompact subset of X is a Dunford–Pettis set.

• (3) Every operator $S:X\rightarrow Z$ with weakly precompact adjoint is p-convergent, for any Banach space Z.

• (4) Every operator $S:X\rightarrow c_0$ with weakly precompact adjoint is p-convergent.

• (5) If $(x_n)$ is a weakly p-summable sequence in X and $(x_n^*)$ is a weakly Cauchy sequence in $X^*$, then $x_n^*(x_n) \rightarrow 0$.

Proof

Apply Theorem 14 to the identity operator $i:X\rightarrow X$. $\square$

Corollary 17

[19, Theorem 1] Let $1< p<\infty$. The following statements are equivalent about a Banach space X.

• (i) X has the $DPP_p$.

• (ii) For all Banach spaces Z, every weakly p-precompact operator $S:Z\rightarrow X$ has a completely continuous adjoint.

• (iii) Every operator $S:\ell _{p^*}\rightarrow X$ has a completely continuous adjoint.

• (iv) If $(x_n)$ is a weakly p-Cauchy sequence in X and $(x_n^*)$ is a weakly null sequence in $X^*$, then $x_n^*(x_n) \rightarrow 0$.

Proof

Apply Corollary 15 to the identity map $i:X\rightarrow X$. $\square$

The following result generalizes [15, Theorem 2.11].

Theorem 18

Let X and Y be Banach spaces and $T:X\rightarrow Y$ be an operator. Let $1< p<\infty$. The following statements are equivalent.

• (1) T is $weak^*$p-convergent.

• (2) T carries weakly p-precompact subsets of X to limited subsets of Y.

• (3) If $S:Z\rightarrow X$ is a weakly p-precompact operator, then $TS:Z\rightarrow Y$ is limited, for any Banach space Z.

• (4) If $S:\ell _{p^*}\rightarrow X$ is an operator, then $TS:\ell _{p^*}\rightarrow Y$ is limited.

• (5) If $(x_n)$ is a weakly p-summable sequence in X and $(y_n^*)$ is a $w^*$-Cauchy sequence in $Y^*$, then $\langle y_n^*, T(x_n) \rangle \rightarrow 0$.

Proof

$(1)\Rightarrow (2)$ is similar to the proof of $(1) \Rightarrow (2)$ in Theorem 14.

$(2)\Rightarrow (3)$ Let $S:Z\rightarrow X$ be a weakly p-precompact operator. Then $TS(B_Z)$ is limited, and thus $TS:Z\rightarrow Y$ is limited.

$(3)\Rightarrow (4)$ Let $S:\ell _{p^*}\rightarrow X$ be an operator. Since $1< p<\infty$, $\ell _{p^*} \in W_p$ [6]. Hence S is weakly p-compact, and thus TS limited.

$(4)\Rightarrow (1)$ Suppose $(x_n)$ is weakly p-summable in X and $(y_n^*)$ is $w^*$-null in $Y^*$. Define $S:\ell _{p^*}\rightarrow X$ by $S(b)=\sum b_i x_i$ [10, Proposition 2.2, p. 36]. Since $TS(B_{\ell _{p^*}})$ is a limited set in Y, $\langle y_n^*, TS(e_n)\rangle = \langle y_n^*, T(x_n)\rangle \rightarrow 0$.

$(1) \Rightarrow (5)$ is similar to the proof of $(1) \Rightarrow (5)$ in Theorem 14. $\square$

Corollary 19

Let X and Y be Banach spaces and $T:X\rightarrow Y$ be an operator. Let $1< p<\infty$. The following statements are equivalent.

• (i) T is $weak^*$p-convergent.

• (ii) If $(x_n)$ is a weakly p-Cauchy sequence in X and $(y_n^*)$ is a $w^*$-null sequence in $Y^*$, then $\langle y_n^*, T(x_n) \rangle \rightarrow 0$.

• (iii) If $S:Y\rightarrow Z$ is an operator such that $S^*(B_{Z^*})$ is $w^*$-sequentially compact, then $ST:X\rightarrow Z$ is p-convergent.

• (iv) If $S:Y\rightarrow c_0$ is an operator, then $ST:X\rightarrow c_0$ is p-convergent.

Proof

$\mathrm{(i)} \Rightarrow \mathrm{(ii)}$ Suppose that $(x_n)$ is a weakly p-Cauchy sequence in X and $(y_n^*)$ is a $w^*$-null sequence in $Y^*$. Since $(T(x_n))$ is limited in Y, $\langle y_n^*, T(x_n)\rangle \rightarrow 0$.

$\mathrm{(ii)} \Rightarrow \mathrm{(iii)}$ Let $S:Y\rightarrow Z$ be an operator such that $S^*(B_{Z^*})$ is $w^*$-sequentially compact, but $ST:X\rightarrow Z$ is not p-convergent. Let $(x_n)$ be weakly p-summable in X so that $\Vert ST(x_n)\Vert >\epsilon$, for some $\epsilon >0$. Choose $(z_n^*)$ in $B_{Z^*}$ so that $\langle z_n^*, ST(x_n)\rangle > \epsilon$. Without loss of generality, $(S^*(z_n^*))$ is $w^*$-convergent. Then $\langle S^*(z_n^*), T(x_n) \rangle = \langle z_n^*, ST(x_n) \rangle \rightarrow 0$, a contradiction.

$\mathrm{(iii)} \Rightarrow \mathrm{(iv)}$ Let $S:Y\rightarrow c_0$ be an operator. Note that $B_{\ell _1}$, and thus $S^*(B_{\ell _1})$ is $w^*$-sequentially compact. Then $ST:X\rightarrow c_0$ is p-convergent.

$\mathrm{(iv)} \Rightarrow \mathrm{(i)}$ Let $(x_n)$ be a weakly p-summable sequence in X and let $(y_n^*)$ be a $w^*$-null sequence in $Y^*$. Define $S:Y\rightarrow c_0$ by $S(y)=(y_i^*(y))$. Since ST is p-convergent, $\langle y_n^*, T(x_n)\rangle \le \Vert ST(x_n)\Vert \rightarrow 0$. $\square$

The following two corollaries provide equivalent characterizations of spaces with the $DP^*P_p$.

Corollary 20

Let $1< p<\infty$. The following statements are equivalent about a Banach space X.

• (i) X has the $DP^*P_p$.

• (ii) [15] The identity operator $i:X\rightarrow X$ is $weak^*$p-convergent; that is, every weakly p-precompact subset of X is a limited set.

• (iii) [18] Every weakly p-precompact operator $S:Z\rightarrow X$ is limited, for any Banach space Z.

• (iv) [15] Every operator $S:\ell _{p^*}\rightarrow X$ is limited.

• (v) [18] If $(x_n)$ is a weakly p-summable sequence in X and $(x_n^*)$ is a $w^*$-Cauchy sequence in $X^*$, then $x_n^*(x_n) \rightarrow 0$.

Proof

Apply Theorem 18 to the identity operator $i:X\rightarrow X$. $\square$

Corollary 21

Let $1< p<\infty$. The following statements are equivalent about a Banach space X.

• (i) X has the $DP^*P_p$.

• (ii) [18] If $(x_n)$ is a weakly p-Cauchy sequence in X and $(x_n^*)$ is a $w^*$-null sequence in $X^*$, then $x_n^*(x_n) \rightarrow 0$.

• (iii) [18] If $S:X\rightarrow Z$ is an operator such that $S^*(B_{Z^*})$ is $w^*$-sequentially compact, then S is p-convergent.

• (iv) [15] Every operator $S:X \rightarrow c_0$ is p-convergent.

Proof

Apply Corollary 19 to the identity operator $i:X\rightarrow X$. $\square$

We note that an operator $T:X\rightarrow Y$ is p-convergent if and only if T takes weakly p-precompact subsets of X into norm compact subsets of Y.

Corollary 22

Let $1< p< \infty$.

• (i) Suppose $S:X\rightarrow Y$ is weakly p-precompact and $T:Y\rightarrow Z$ is an operator with weakly precompact adjoint. If Y has the $DPP_p$, then TS is compact.

• (ii) Suppose $S:X\rightarrow Y$ is weakly p-precompact and $T:Y\rightarrow Z$ is an operator such that $T^*(B_{Z^*})$ is $w^*$-sequentially compact. If Y has the $DP^*P_p$, then TS is compact.

Proof

• (i) Suppose $S:X\rightarrow Y$ is weakly p-precompact and $T:Y\rightarrow Z$ is an operator such that $T^*$ is weakly precompact. Since Y has the $DPP_p$, T is p-convergent by Corollary 16. Then $TS(B_X)$ is relatively compact, and thus TS is compact.

• (ii) The proof is similar to that of (i).

$\square$

Corollary 23

Let $1< p<\infty$.

• (i) If $Y^*$ does not contain a copy of $\ell _1$, then every weak p-convergent operator $T:X\rightarrow Y$ is p-convergent.

• (ii) If $B_{Y^*}$ is $w^*$-sequentially compact (in particular if Y is separable), then every $weak^*$p-convergent operator $T:X\rightarrow Y$ is p-convergent.

• (iii) If X or Y has the $DPP_p$, then every operator $T:X\rightarrow Y$ is weak p-convergent.

• (iv) If X or Y has the $DP^*P_p$, then every operator $T:X\rightarrow Y$ is $weak^*$p-convergent.

Proof

• (i) Let $i:Y\rightarrow Y$ be the identity operator on Y. Suppose $T:X\rightarrow Y$ is a weak p-convergent operator. By Rosenthal’s $\ell _1$ theorem, $i^*$ is weakly precompact. Then $T=iT$ is p-convergent by Theorem 14.

• (ii) The proof is similar to that of (ii).

• (iii) Let $T:X\rightarrow Y$ be an operator. If Y has the $DPP_p$, then the identity operator $i:Y\rightarrow Y$ is weak p-convergent. Hence $T=iT$ is weak p-convergent. If X has the $DPP_p$, then the identity operator $i: X\rightarrow X$ is weak p-convergent. Hence $T=Ti$ is weak p-convergent.

• (iv) The proof is similar to that of (iii).

$\square$

Clearly each p-convergent operator $T:X\rightarrow Y$ is $\hbox {weak}^*$p-convergent and each $\hbox {weak}^*$p-convergent operator is weak p-convergent. By Corollay 23, we obtain the following result. It generalizes [15, Proposition 2.5].

Corollary 24

If $Y^*$ does not contain a copy of $\ell _1$, then the families of p-convergent operators, $weak^*$p-convergent operators, and weak p-convergent operators $T:X\rightarrow Y$ coincide.

Let $1\le p<\infty$. A Banach space X has the p-Gelfand–Phillips (p-GP) property (or is a p-Gelfand–Phillips space) if every limited weakly p-summable sequence in X is norm null [15].

If X has the GP property, then X has the p-GP property for any $1\le p<\infty$. The space $\ell _\infty$ does not have the p-GP property for any $1\le p<\infty$ [15].

Let $1\le p<\infty$. A space X has the p-Dunford Pettis relatively compact property (p-DPrcP) if every DP weakly p-summable sequence $(x_n)$ in X is norm null [17].

If X has the DPrcP property, then X has the p-DPrcP property for any $1\le p<\infty$.

Corollary 25

Let $1\le p<\infty$. If X has the p-GP (resp. the p-DPrcP) property, then the following are equivalent.

• (i) X has the $DP^*P_p$ (resp. the $DPP_p$).

• (ii) $X\in C_p$.

Proof

$(i) \Rightarrow (ii)$ We only prove the result when X has the the p-GP and the $DP^*P_p$. The other case is similar.

Let $(x_n)$ be weakly p-summable in X. Then $(x_n)$ is limited by Corollary 20. Therefore $\Vert x_n\Vert \rightarrow 0$, and thus $X\in C_p$. $\square$

Let $1\le p< \infty$. An operator $T:X\rightarrow Y$ is called limited p-convergent if it carries limited weakly p-summable sequences in X to norm null ones in Y [15]. An operator $T:X\rightarrow Y$ is called DP p-convergent if it takes DP weakly p-summable sequences to norm null sequences [17].

The sets of all limited p-convergent, DP p-convergent, weak p-convergent, and $\hbox {weak}^*$p-convergent operators from X to Y will be respectively denoted by $LC_p(X,Y)$, $DPC_p(X,Y)$, $WC_p(X,Y)$, and $W^*C_p(X,Y)$.

Corollary 26

Let $1\le p<\infty$. Let X be a Banach space. The following statements hold.

• (i) X has the p-DPrcP if and only if $WC_p(X,\ell _\infty )=DPC_p(X,\ell _\infty )$.

• (ii) X has the p-GP property if and only if $W^*C_p(X,\ell _\infty )=LC_p(X,\ell _\infty )$.

Proof

• (i) A Banach space X has the p-DPrcP if and only if $DPC_p(X, \ell _\infty )=L(X,\ell _\infty )$ [17]. Since $\ell _\infty$ has the $DPP_p$, $L(X,\ell _\infty )=WC_p(X,\ell _\infty )$.

• (ii) A Banach space X has the p-GP if and only if $LC_p(X, \ell _\infty )=L(X,\ell _\infty )$ [17]. Since $\ell _\infty$ has the $DP^*P_p$, $L(X,\ell _\infty )=W^*C_p(X,\ell _\infty )$.

$\square$

Since any limited set is a DP set, any limited weakly p-summable sequence is also DP weakly p-summable. Hence if X has the p-DPrcP, then X has the p-GP property. Thus, if X has the p-DPrcP, then $L(X,\ell _\infty )=LC_p(X,\ell _\infty )=DPC_p(X,\ell _\infty )= WC_p(X, \ell _\infty )= W^*C_p(X, \ell _\infty )$.