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# Some results on order bounded almost weak Dunford-Pettis operators

Nabil Machrafi  Université Ibn Tofaïl, Faculté des Sciences, Département de
Mathématiques, B.P. 133, Kénitra 14000, Maroc<EMAIL_ADDRESS>Aziz
Elbour  Université Moulay Ismaïl, Faculté des Sciences et Techniques,
Département de Mathématiques, B.P. 509, Errachidia 52000, Maroc.
<EMAIL_ADDRESS>Mohammed Moussa  Université Ibn Tofaïl, Faculté des
Sciences, Département de Mathématiques, B.P. 133, Kénitra 14000, Maroc.
<EMAIL_ADDRESS>
(Date: 29/09/2015; Revised 25/02/2016)

###### Abstract.

We give some new characterizations of almost weak Dunford-Pettis operators and
we investigate their relationship with weak Dunford-Pettis operators.

###### Key words and phrases:

Almost weak Dunford-Pettis operator, weak Dunford-Pettis operator, almost
Dunford-Pettis set, almost (L)-set, Banach lattice.

###### 2010 Mathematics Subject Classification:

46A40, 46B40, 46B42

##  1\. Introduction and notations

Throughout this paper, $X,$ $Y$ will denote real Banach spaces, and $E,\,F$
will denote real Banach lattices. $B_{X}$ is the closed unit ball of $X$. We
mean by operator between Banach spaces, a bounded linear application.

A real vector space $E$ is said to be a _partially_ _ordered vector space_
whenever it is equipped with a partial order relation $\geq$ (i.e., a
reflexive, antisymmetric, and transitive binary relation on $E$) that is
compatible with the algebraic structure of $E$ in the sense that it satisfies
the following two axioms:

  1. (1)

If $x\geq y$, then $x+z\geq y+z$ holds for all $z\in E$.

  2. (2)

If $x\geq y$, then $\lambda x\geq\lambda y$ holds for all $\lambda\geq 0$.

An alternative notation for $x\geq y$ is $y\leq x$. The _positive cone_ of
$E$, denoted by $E^{+}$, is the set of all positive vectors of $E$, i.e.,
$E^{+}:=\left\\{x\in E:x\geq 0\right\\}$. If furthermore, every set
$\left\\{x,y\right\\}\subset E$ has a supremum $\sup\left\\{x,y\right\\}=x\vee
y$ (or equivalently it has an infimum $\inf\left\\{x,y\right\\}=x\wedge y$)
then $E$ is called a _vector lattice_ (or _Riesz space_). The elements

$x^{+}=x\vee 0\text{, \ \ }x^{-}=\left(-x\right)\vee 0\text{, \ \
}\left|x\right|=x\vee\left(-x\right)$

are called the _positive part_ , _negative part_ , and _modulus_ of the
element $x$, respectively. Note that $x=x^{+}-x^{-}$ and $|x|=x^{+}+x^{-}$. If
$x,y\in E$ and $x\leq y$, then the _order interval_ $\left[x,y\right]$ is
defined by

$\left[x,y\right]=\left\\{z\in E:x\leq z\leq y\right\\}\text{.}$

A subset $A\subset E$ is said to be _order bounded_ if it is contained in some
order interval. It is said to be _solid_ if conditions
$\left|x\right|\leq\left|a\right|$ and $a\in A$ imply $x\in A$. The smallest
solid set containing a set $A$ is called the _solid hull_ of $A$ and denoted
by $sol\left(A\right)$. We have

$sol\left(A\right)=\left\\{x\in E:\left|x\right|\leq\left|a\right|\text{ for
some }a\in A\right\\}.$

The elements $x,y\in E$ are called _disjoint_ if
$\left|x\right|\wedge\left|y\right|=0$. A generalized sequence
$\left(x_{\alpha}\right)$ in a vector lattice $E$ (i.e. a function
$\alpha\rightarrow x_{\alpha}$ from some an upward directed set $I$ to $E$) is
called disjoint, if $\left|x_{\alpha}\right|\wedge\left|x_{\beta}\right|=0$,
$\alpha\neq\beta$. We will use the notation $x_{\alpha}\perp x_{\beta}$ to
mean that the generalized sequence $\left(x_{\alpha}\right)$ is disjoint. A
collection $\left(e_{i}\right)\subset E^{+}\backslash\left\\{0\right\\}$ is
called a _complete disjoint system_ if

$e_{i}\wedge e_{j}=0,\quad i\neq j\quad\text{and}\quad
e_{i}\wedge\left|x\right|=0\quad\text{for all}\ i\ \text{implies}\quad
x=0\text{.}$

A positive non-zero element $x$ of an _Archimedean vector lattice_ $E$ (i.e.,
such vector lattices that $\inf_{n}\left\\{\frac{1}{n}x\right\\}=0$ for every
$x\in E^{+}$) is called _discrete_ , if

$u,v\in\left[0,x\right]\quad\text{and}\quad u\wedge v=0\quad\text{imply}\quad
u=0\quad\text{or}\quad v=0\text{.}$

An Archimedean vector lattice $E$ is called _discrete_ (or _atomic_), if $E$
has a complete disjoint system consisting of discrete elements, or
equivalently, every non-trivial interval $\left[0,x\right]$ contains a
discrete element. A norm $\left\|.\right\|$ on a vector lattice
$\left(E,\leq\right)$ is said to be a _lattice norm_ whenever

$|x|\leq|y|\quad\text{implies}\quad\left\|x\right\|\leq\left\|y\right\|.$

A vector lattice equipped with a lattice norm is known as a _normed vector
lattice_. If a normed vector lattice is also norm complete, then it is
referred to as a _Banach lattice_. Note that If $E$ is a Banach lattice, its
topological dual $E^{\prime}$, endowed with the dual norm, is also a Banach
lattice. A norm $\left\|.\right\|$ of a Banach lattice $E$ is called _order
continuous_ if for each generalized sequence $(x_{\alpha})\subset E$,
$x_{\alpha}\downarrow 0$ implies $\left\|x_{\alpha}\right\|\rightarrow 0$,
where the notation $x_{\alpha}\downarrow 0$ means that the generalized
sequence $(x_{\alpha})$ is decreasing and $\inf\left\\{x_{\alpha}\right\\}=0$.
A Banach lattice $E$ is said to be a _Kantorovich–Banach space_ (briefly _KB-
space_) whenever every increasing norm bounded sequence of $E^{+}$ is norm
convergent. Note that every KB-space has order continuous norm. The lattice
operations in a Banach lattice $E$ are said to be _sequentially weakly_
_continuous_ , if for every weakly null sequence $(x_{n})$ in $E$ we have
$\left|x_{n}\right|\rightarrow 0$ for $\sigma(E,E^{\prime})$. For a sequence
$\left(x_{n}\right)\subset E$ of a Banach lattice, the following fact will be
used throughout this paper (see [2, Theorem 4.34]):

$x_{n}\overset{\sigma(E,E^{\prime})}{\rightarrow}0,\quad x_{n}\perp
x_{m}\quad\text{implies}\quad\left|x_{n}\right|\overset{\sigma(E,E^{\prime})}{\rightarrow}0\text{.}$

Recall that a subset $A$ of a Banach space $X$ is called a _Dunford-Pettis
(DP) set_ , if each weakly null sequence $\left(f_{n}\right)$ in $X^{\prime}$
converges uniformly to zero on $A$. In his paper [9], T.Leavelle considered
the dual version of DP sets, so-called _(L) sets_ , that is, each subset $B$
of the topological dual $X^{\prime}$, on which every weakly null sequence
$(x_{n})$ in $X$ converges uniformly to zero. Recently, the authors of [5] and
[3] considered respectively, the disjoint versions of DP sets and (L)-sets,
which are respectively called _almost Dunford-Pettis_ (almost DP) sets and
_almost (L)-sets_. From [5] (resp. [3]), a norm bounded subset $A\subset E$
(resp. $B\subset E^{\prime}$) is said to be almost DP (resp. almost (L)-) set,
if every disjoint weakly null sequence $(f_{n})\subset E^{\prime}$ (resp.
$(x_{n})\subset E$ ) converges uniformly to zero on $A$ (resp. $B$). Clearly,
every DP set in a Banach lattice (resp. (L)-set in the topological dual of a
Banach lattice) is an almost DP (resp. almost (L)-) set . But the converse is
false in general for the two latter classes of sets.

A Banach lattice $E$ is said to have the (_positive_) _Schur_ _property_ if
every (positive) weakly null sequence $\left(x_{n}\right)\subset E$ is norm
null. Furthermore, A Banach space $X$ is said to have the _Dunford-Pettis_
_(DP)_ _property_ if each relatively weakly compact set in $X$ is a DP set,
alternatively, $f_{n}\left(x_{n}\right)\rightarrow 0$ for every weakly null
sequences $(x_{n})\subset X$ , $(f_{n})\subset X^{\prime}$.

An operator $T:E\rightarrow F$ between two vector lattices is called an _order
bounded operator_ , if it maps order bounded subsets of $E$ into an order
bounded ones of $F$. It is _positive_ if $T(E^{+})\subset F^{+}$. The positive
operators between two vector lattices generate the vector space of all
_regular operators_ , i.e., operators that are written as a difference of two
positive operators. Note that a regular operator between two vector lattices
need not be order bounded (see example of H. P. Lotz [2, Example 1.16]). An
operator $T:X\rightarrow Y$ between two Banach spaces is said to be _Dunford-
Pettis_ , if $T$ carries each weakly null sequence $(x_{n})$ in $X$ to a norm
null one in $Y$, equivalently, $T$ carries relatively weakly compact subsets
of $X$ to a relatively compact ones in $Y$. The following weak versions of
Dunford-Pettis operators are considered in the Banach lattice setting. An
operator $T:E\rightarrow F$ is

  1. -

_weak Dunford-Pettis_ (wDP), if $f_{n}\left(Tx_{n}\right)\rightarrow 0$
whenever $(x_{n})$ converges weakly to $0$ in $E$ and $(f_{n})$ converges
weakly to $0$ in $F^{\prime}$, equivalently, $T$ maps relatively weakly
compact sets in $E$ to a Dunford–Pettis sets in $F$ (see Theorem 5.99 of [2]).

  2. -

_almost Dunford-Pettis_ (almost DP), if $\left\|Tx_{n}\right\|\rightarrow 0$
whenever $\left(x_{n}\right)\subset E$ is a disjoint weakly null sequence,
equivalently, $\left\|Tx_{n}\right\|\rightarrow 0$ for every weakly null
sequence $\left(x_{n}\right)\subset E$ consisting of positive terms [4,
Theorem 2.2].

The class of wDP operators (between two Banach spaces) was introduced by C. D.
Aliprantis and O. Burkinshaw in [1]. It extends the notions of DP operator and
the DP property of a Banach space, in the sense that every DP operator
$T:X\rightarrow Y$ is wDP, and a Banach space $X$ has the DP property iff the
identity operator on $X$ is wDP. Next, J. A. Sanchez [13] introduced the class
of almost DP operators (from a Banach lattice into a Banach space) which also
extends, in the Banach lattice setting, the notions of DP operator and the
positive Schur property, since every DP operator $T:E\rightarrow Y$ is almost
DP, and a Banach lattice $E$ has the positive Schur property iff the identity
operator on $E$ is almost DP. Also, F. Räbiger [12] distinguished a class of
Banach lattices with a weak version of the DP property. A Banach lattice $E$
is said to have the _weak Dunford-Pettis_ _(wDP)_ _property_ if every weakly
compact operator on $E$ is almost DP, equivalently (see [15, Proposition 1]),
for all sequences $(x_{n})\subset E_{+}^{\prime}$, $(f_{n})\subset
E^{\prime}$,

$x_{n}\overset{\sigma(E,E^{\prime})}{\rightarrow}0,\quad x_{n}\perp
x_{m}\quad\text{and}\quad
f_{n}\overset{\sigma(E,E^{\prime})}{\rightarrow}0\quad\text{imply}\quad
f_{n}\left(x_{n}\right)\rightarrow 0.$ (*)

Inspired by the preceding facts, K. Bouras and M. Moussa introduced naturally
in their recent paper [6] the class of _almost weak Dunford-Pettis_ (awDP)
operators, as a class of operators that extends both the notions of wDP
operator, almost DP operator, and the wDP property of a Banach lattice. An
operator $T:E\rightarrow F$ is said to be awDP, if for all sequences
$(x_{n})\subset E$ , $(f_{n})\subset F^{\prime}$,

$x_{n}\overset{\sigma(E,E^{\prime})}{\rightarrow}0,\quad x_{n}\perp
x_{m}\quad\text{and}\quad
f_{n}\overset{\sigma(E,E^{\prime})}{\rightarrow}0,\quad f_{n}\perp
f_{m}\quad\text{imply}\quad f_{n}\left(Tx_{n}\right)\rightarrow 0.$

It follows from [6, Theorem 2.1(3)] combined with (* ‣ 1) that a Banach
lattice $E$ has the wDP property iff the identity operator on $E$ is awDP.
Note that the authors gave in [6, Theorem 2.1] some sequence characterisations
of positive almost weak Dunford-Pettis operators, which allowed them to
establish some new sequence characterisations of the wDP property of a Banach
lattice (see [6, Corollary 2.1]). The class of awDP operators contains
strictly that of wDP operators as well as that of almost DP operators, that
is, every wDP (resp. almost DP) operator is awDP, but there exists an awDP
operator which is not wDP nor almost DP. The example of such operator is the
identity operator on a Banach lattice $\Phi$ with the wDP property but without
the DP property nor the positive Schur property. W. Wnuk gave in [15, p. 231]
an example of such Banach lattice:

###### Example

Let $\omega$ be a positive non-increasing continuous function on
$\left(0,1\right)$ so that

$\lim_{t\rightarrow
0}\omega(t)=\infty\quad\text{and}\quad\int_{0}^{1}\omega(t)\,dt=1.$

The Lorentz function space $E=\wedge(\omega,1)$ is the space of all measurable
functions $f$ on $\left[0,1\right]$ for which

$\left\|f\right\|_{\omega,1}=\int_{0}^{1}f^{\ast}\left(t\right)\omega(t)\,dt<\infty\text{,}$

where $f^{\ast}$ denotes the decreasing rearrangement of $\left|f\right|$ (cf.
[10, p. l17, p. 120]). $E$ is a Banach lattice under the norm
$\left\|.\right\|_{\omega,1}$ and the standard almost everywhere pointwise
order (i.e., $f\leq g$ if $f\left(x\right)\leq g\left(x\right)$ a.e.). Note
that, since $E$ has the Fatou property, then it is a maximal rearrangement
invariant space [10, Definition 2.a.1], and therefore $E$ has a predual, that
is, $E=\Phi^{\prime}$, where $\Phi$ is the closed linear span of the simple
functions in $E_{i}^{\prime}$. Here, $E_{i}^{\prime}$ stands for the linear
subspace of $E^{\prime}$ consisting of all integrals on $E$, i.e., functionals
$\varphi_{g}\in E^{\prime}$ defined by

$\varphi_{g}\left(f\right)=\int_{0}^{1}f\left(t\right)g(t)\,dt,$

where $g$ is any measurable function on $\left[0,1\right]$ so that $gf\in
L_{1}\left(0,1\right)$ for every $f\in E$ (for details about the preceding
facts, see [10, p. 29, p. 118, p. 121]). Now, it follows from [15, p. 231]
that $\Phi$ has the wDP property but not the DP property nor the positive
Schur property.

Moreover, for the class of wDP operators and that of almost DP ones, each one
of the two classes is not included in the other. For instance, as the Lorentz
space $\wedge(\omega,1)$ has the positive Schur property without the DP
property (see [14, Remark 3]), the identity operator on $\wedge(\omega,1)$ is
almost DP but not wDP. Conversely, since $c_{0}$ (the space of real sequences
$\left(x_{n}\right)$ with $\lim x_{n}=0$) has the DP property without the the
positive Schur property, the identity operator on $c_{0}$ is wDP but not
almost DP.

The present paper is devoted to the class of awDP operators. W. Wnuk noted in
[15, Example 4 p. 230] that a positive operator $T:E\rightarrow F$ is almost
DP if and only if it is a DP operator, provided that $F$ is discrete with
order continuous norm. Motivated by this fact, we look at its weak
alternative, that is, when an awDP operator is wDP? As a response, we prove
the following theorem.

###### Theorem A

Let $E$ and $F$ be two Banach lattices. Then, an order bounded operator
$T:E\rightarrow F$ is almost weak Dunford-Pettis if and only if it is weak
Dunford-Pettis, whenever one of the following holds:

  1. $\left(i\right)$

$E$ has sequentially weakly continuous lattice operations.

  2. $\left(ii\right)$

$F^{\prime}$ has sequentially weakly continuous lattice operations.

  3. $\left(iii\right)$

$T$ is positive and $F$ is discrete with order continuous norm.

For that purpose, we present some new characterisations of awDP operators
through almost DP (resp. almost (L)-) sets and some lattice approximations
(Sect. 2). Next, we give the proof of Theorem A and we derive some
consequences (Sect. 3).

We refer the reader to [2, 11] for more details on Banach lattice theory and
positive operators.

##  2\. Characterisation of almost weak Dunford-Pettis operators

We start this paper by the following lemma which is just a particular case of
Theorem 2.4 of [8].

###### Lemma 2.1

Let $E$ be a Banach lattice, and let $\left(f_{n}\right)\subset E^{\prime}$ be
a sequence with $\left|f_{n}\right|\overset{w^{\ast}}{\rightarrow}0$. If
$A\subset E$ is a norm bounded and solid set such that
$f_{n}(x_{n})\rightarrow 0$ for every disjoint sequence
$\left(x_{n}\right)\subset A^{+}:=A\cap E^{+}$, then $\sup_{x\in
A}\left|f_{n}\right|\left(x\right)\rightarrow 0$.

We will use throughout this paper the following lemma.

###### Lemma 2.2

Let $T:E\rightarrow F$ be an order bounded operator between two Banach
lattices, and let $A$ and $B$ be respectively a norm bounded solid subsets of
$E$ and $F^{\prime}$. Then, the following holds:

  1. (1)

If the sequence $\left(f_{n}\right)\subset F^{\prime}$ satisfy
$\left|f_{n}\right|\overset{w^{\ast}}{\rightarrow}0$ and
$f_{n}\left(Tx_{n}\right)\rightarrow 0$ for every disjoint sequence
$\left(x_{n}\right)\subset A^{+}$, then $\left(f_{n}\right)$ converges
uniformly to zero on $T\left(A\right)$.

  2. (2)

If the sequence $\left(x_{n}\right)\subset E$ satisfy
$\left|x_{n}\right|\overset{w}{\rightarrow}0$ and
$f_{n}\left(Tx_{n}\right)\rightarrow 0$ for every disjoint sequence
$\left(f_{n}\right)\subset B^{+}$, then $\left(x_{n}\right)$ converges
uniformly to zero on $T^{\prime}\left(B\right)$.

###### Proof.

$(1)$ We claim that
$\left|T^{\prime}\left(f_{n}\right)\right|\overset{w^{\ast}}{\rightarrow}0$
holds in $E^{\prime}$. Let $x\in E^{+}$ and pick some $y\in F^{+}$ such that
$T\left[-x,x\right]\subseteq\left[-y,y\right]$. Thus

$\left|T^{\prime}\left(f_{n}\right)\right|\left(x\right)=\sup\left\\{\left|T^{\prime}\left(f_{n}\right)\left(u\right)\right|:\left|u\right|\leq
x\right\\}=\sup\left\\{\left|f_{n}\left(T\left(u\right)\right)\right|:\left|u\right|\leq
x\right\\}\leq\left|f_{n}\right|\left(y\right)\text{.}$

Since $\left|f_{n}\right|\overset{w^{\ast}}{\rightarrow}0$, we have
$\left|f_{n}\right|\left(y\right)\rightarrow 0$ and hence
$\left|T^{\prime}\left(f_{n}\right)\right|\left(x\right)\rightarrow 0$.
Therefore
$\left|T^{\prime}\left(f_{n}\right)\right|\overset{w^{\ast}}{\rightarrow}0$.
On the other hand, by hypothesis
$T^{\prime}\left(f_{n}\right)\left(x_{n}\right)=f_{n}\left(Tx_{n}\right)\rightarrow
0$ for every disjoint sequence $\left(x_{n}\right)\subset A^{+}$. Then,
applying Lemma 2.1, we get $\sup_{x\in
A}\left|T^{\prime}\left(f_{n}\right)\right|\left(x\right)\rightarrow 0$. Now,
from the inequality

$\sup_{y\in T\left(A\right)}\left|f_{n}\left(y\right)\right|=\sup_{x\in
A}\left|f_{n}\left(Tx\right)\right|\leq\sup_{x\in
A}\left|T^{\prime}\left(f_{n}\right)\right|\left(x\right),$

we conclude that $\sup_{y\in
T\left(A\right)}\left|f_{n}\left(y\right)\right|\rightarrow 0$ and we are
done.

$(2)$ We claim that $\left|Tx_{n}\right|\overset{w}{\rightarrow}0$ holds in
$F$. Let $f\in\left(F^{\prime}\right)^{+}$. By Theorem 1.73 of [2],
$T^{\prime}:F^{\prime}\rightarrow E^{\prime}$ is order bounded. So there
exists some $g\in\left(E^{\prime}\right)^{+}$ such that
$T^{\prime}\left[-f,f\right]\subseteq\left[-g,g\right]$. For each $n$ pick
$\left|f_{n}\right|\leq f$ with
$f\left(\left|Tx_{n}\right|\right)=f_{n}\left(Tx_{n}\right)=T^{\prime}\left(f_{n}\right)\left(x_{n}\right)$
(see Theorem 1.23 [2]). Thus, for each $n$, we have

$f\left(\left|T\left(x_{n}\right)\right|\right)=f_{n}\left(Tx_{n}\right)\leq\left|T^{\prime}\left(f_{n}\right)\right|\left(\left|x_{n}\right|\right)\leq
g\left(\left|x_{n}\right|\right)\text{.}$

Since $\left|x_{n}\right|\overset{w}{\rightarrow}0$, we have
$g\left(\left|x_{n}\right|\right)\rightarrow 0$ and hence
$f\left(\left|Tx_{n}\right|\right)\rightarrow 0$. Therefore
$\left|Tx_{n}\right|\overset{w}{\rightarrow}0$ holds in $F$. On the other
hand, if $j:F\rightarrow F^{\prime\prime}$ is the natural embedding, then
$\left|j\left(Tx_{n}\right)\right|=j\left(\left|Tx_{n}\right|\right)\overset{w^{\ast}}{\rightarrow}0$
holds in $F^{\prime\prime}$. Also, by hypothesis
$j\left(Tx_{n}\right)\left(f_{n}\right)=f_{n}\left(Tx_{n}\right)\rightarrow 0$
for every disjoint sequence $\left(f_{n}\right)\subset B^{+}$. Then, applying
Lemma 2.1 for the sequence $\left(j\left(Tx_{n}\right)\right)\subset
F^{\prime\prime}$ and the solid subset $B\subset F^{\prime}$, we get
$\sup_{f\in B}\left|j\left(Tx_{n}\right)\right|\left(f\right)\rightarrow 0$.
Now, from

$\displaystyle\sup_{g\in
T^{\prime}\left(B\right)}\left|g\left(x_{n}\right)\right|=\sup_{f\in
B}\left|f\left(Tx_{n}\right)\right|\leq\sup_{f\in
B}f\left(\left|T\left(x_{n}\right)\right|\right)$ $\displaystyle=\sup_{f\in
B}\left(j\left|Tx_{n}\right|\right)\left(f\right)$ $\displaystyle=\sup_{f\in
B}\left|j\left(Tx_{n}\right)\right|\left(f\right)\text{,}$

we conclude that $\sup_{g\in
T^{\prime}\left(B\right)}\left|g\left(x_{n}\right)\right|\rightarrow 0$. This
completes the proof. ∎

The next result gives a new characterisations of order bounded awDP operators
between Banach lattices, through the almost DP (resp. almost (L)-) sets.

###### Theorem 2.3

Let $T:E\rightarrow F$ be an order bounded operator between two Banach
lattices. Then the following assertions are equivalent:

  1. (1)

$T$ is an awDP operator.

  2. (2)

$T$ carries the solid hull of each relatively weakly compact subset of $E$ to
an almost DP set in $F$.

  3. (3)

$T$ carries each relatively weakly compact subset of $E$ to an almost DP set
in $F$.

  4. (4)

$T^{\prime}$ carries the solid hull of each relatively weakly compact subset
of $F^{\prime}$ to an almost (L)-set in $E^{\prime}$.

  5. (5)

$T^{\prime}$ carries each relatively weakly compact subset of $F^{\prime}$ to
an almost (L)-set in $E^{\prime}$.

###### Proof.

$\left(1\right)\Rightarrow\left(2\right)$ Let $A$ be a relatively weakly
compact subset of $E$ and let $\left(f_{n}\right)\subset F^{\prime}$ be a
disjoint weakly null sequence. By Theorem 4.34 of [2] if
$\left(x_{n}\right)\subset\left(\mathrm{sol}\left(A\right)\right)^{+}$ is a
disjoint sequence, then $x_{n}\overset{w}{\rightarrow}0$ and hence by our
hypothesis $f_{n}(Tx_{n})\rightarrow 0$. Now, since
$\left|f_{n}\right|\overset{w}{\rightarrow}0$ by Lemma 2.2 we conclude that
$\sup_{x\in
T\left(\mathrm{sol}\left(A\right)\right)}\left|f_{n}\left(x\right)\right|\rightarrow
0$. Therefore $T\left(\mathrm{sol}\left(A\right)\right)$ is an almost DP set.

$\left(2\right)\Rightarrow\left(3\right)$ Obvious.

$\left(3\right)\Rightarrow\left(1\right)$ Let $\left(x_{n}\right)\subset E$ ,
$\left(f_{n}\right)\subset F^{\prime}$ be two disjoint weakly null sequences.
It follows that the set $\left\\{Tx_{n}:n\in\mathbb{N}\right\\}$ is an almost
DP. Therefore, $\sup_{k}\left|f_{n}\left(Tx_{k}\right)\right|\rightarrow 0$ as
$n\rightarrow\infty$. Now, from the inequality
$\sup_{k}\left|f_{n}\left(Tx_{k}\right)\right|\geq\left|f_{n}\left(Tx_{n}\right)\right|$
we see that $f_{n}\left(Tx_{n}\right)\rightarrow 0$, and hence $T$ is an awDP
operator.

$\left(1\right)\Rightarrow\left(4\right)$ Let $B$ be a relatively weakly
compact subset of $F^{\prime}$ and let $\left(x_{n}\right)\subset E$ be a
disjoint weakly null sequence. Similarly, we have
$\left|x_{n}\right|\overset{w}{\rightarrow}0$ and for each disjoint sequence
$\left(f_{n}\right)\subset\left(\mathrm{sol}\left(B\right)\right)^{+}$,
$f_{n}\overset{w}{\rightarrow}0$ and Thus by hypothesis
$f_{n}(Tx_{n})\rightarrow 0$. We infer by Lemma 2.2 that $\sup_{f\in
T^{\prime}\left(\mathrm{sol}\left(B\right)\right)}\left|f\left(x_{n}\right)\right|\rightarrow
0$ , i.e., $T^{\prime}\left(\mathrm{sol}\left(B\right)\right)$ is an almost
(L)-set.

$\left(4\right)\Rightarrow\left(5\right)$ Obvious.

$\left(5\right)\Rightarrow\left(1\right)$ Let $\left(x_{n}\right)\subset E$ ,
$\left(f_{n}\right)\subset F^{\prime}$ be two disjoint weakly null sequences.
Since the set $\left\\{T^{\prime}\left(f_{n}\right):n\in\mathbb{N}\right\\}$
is an almost (L)-set, it follows by the same justification in
$\left(3\right)\Rightarrow\left(1\right)$ that
$f_{n}\left(Tx_{n}\right)\rightarrow 0$, and hence $T$ is an awDP operator. ∎

The set characterisations in the above theorem enable us to derive the
following result.

###### Corollary 2.4

In the class of all order bounded operators from $E$ into $E$, the order
bounded awDP operators from $E$ into $E$ form a closed two-sided ideal.

###### Proof.

It is easy to see from the characterisation $\left(3\right)$ of Theorem 2.3
that if for two operators $T,S:E\rightarrow E$, $S$ is an awDP operator thus
the product $ST$ is so. Now, if $T$ is an awDP operator, let $B$ be a
relatively weakly compact subset of $E^{\prime}$. Hence,
$S^{\prime}\left(B\right)$ is also a relatively weakly compact subset of
$E^{\prime}$. As $T$ is an awDP operator then by Theorem 2.3$\left(5\right)$,
$T^{\prime}S^{\prime}\left(B\right)$ is an almost (L)-set in $E^{\prime}$.
This shows that $\left(ST\right)^{\prime}=T^{\prime}S^{\prime}$ carries each
relatively weakly compact subset of $E^{\prime}$ to an almost (L)-set in
$E^{\prime}$. Thus, by Theorem 2.3$\left(5\right)$ again, $ST$ is an awDP
operator and we are done. ∎

In our following result, we show that order bounded awDP operators satisfy
some lattice approximations.

###### Theorem 2.5

Let $T:E\rightarrow F$ be an order bounded awDP operator between two Banach
lattices. Then, the following assertions hold:

  1. (1)

For each relatively weakly compact subsets $A\subset E$ , $B\subset
F^{\prime}$, and for every $\varepsilon>0$, there exists some $u\in
E^{+}$satisfying

$\left|f\right|\left(T\left(\left|x\right|-u\right)^{+}\right)\leq\varepsilon$

for all $x\in A$ and all $f\in B$.

  2. (2)

For each relatively weakly compact subsets $A\subset E$ , $B\subset
F^{\prime}$, and for every $\varepsilon>0$, there exists some
$g\in\left(F^{\prime}\right)^{+}$ satisfying

$\left(\left|f\right|-g\right)^{+}\left(T\left|x\right|\right)\leq\varepsilon$

for all $x\in A$ and all $f\in B$.

###### Proof.

Note that the proof is similar for the two assertions, so we present only that
of the first one. To do this, we proceed in two steps:

Step 1: For every disjoint sequence $\left(x_{n}\right)$ in the solid hull of
$A$, the sequence $\left(Tx_{n}\right)$ converges uniformly to zero on the
solid hull of $B$. Indeed, the sequence $\left(x_{n}\right)$ is in this case,
a disjoint weakly null sequence (Theorem 4.34 of [2]). Thus, by Theorem
2.3$\left(4\right)$ we have

$\sup_{f\in\mathrm{sol}\left(B\right)}\left|f\left(Tx_{n}\right)\right|=\sup_{g\in
T^{\prime}\left(\mathrm{sol}\left(B\right)\right)}\left|g\left(x_{n}\right)\right|\rightarrow
0$

as desired.

Step 2: Assume by way of contradiction that there exists a relatively weakly
compact subsets $A\subset E$ , $B\subset F^{\prime}$ and some $\varepsilon>0$
such that for each $u\in E^{+}$ we have

$\left|f\right|\left(T\left(\left|x\right|-u\right)^{+}\right)>\varepsilon$

for at least two elements $x\in A$ and $f\in B$. In particular, an easy
inductive argument shows that there exists a sequences
$\left(x_{n}\right)\subset A$ , $\left(f_{n}\right)\subset B$ such that

$\left|f_{n}\right|\left(T\left(\left|x_{n+1}\right|-4^{n}\sum_{i=1}^{n}\left|x_{i}\right|\right)^{+}\right)>\varepsilon$
(**)

holds for each $n$. Put $y=\sum_{n=1}^{\infty}2^{-n}\left|x_{n}\right|$ ,
$y_{n}=\left(\left|x_{n+1}\right|-4^{n}\sum_{i=1}^{n}\left|x_{i}\right|\right)^{+}$
and
$z_{n}=\left(\left|x_{n+1}\right|-4^{n}\sum_{i=1}^{n}\left|x_{i}\right|-2^{-n}y\right)^{+}$.
From Lemma 4.35 of [2] the sequence $\left(z_{n}\right)$ is disjoint. Also,
since $0\leq z_{n}\leq|x_{n}+1|$ holds, we see that $\left(z_{n}\right)\subset
sol\left(A\right)$, and so by Step 1 $sup_{f\in
sol\left(B\right)}\left|f\left(Tz_{n}\right)\right|\rightarrow 0$. In
particular, $\left|f_{n}\right|\left(Tz_{n}\right)\rightarrow 0$. On the other
hand, we have $0\leq y_{n}-z_{n}\leq 2^{-n}y$ from which we get
$\left\|y_{n}-z_{n}\right\|\leq 2^{-n}\left\|y\right\|$. In particular, we
infer that
$\left|f_{n}\right|\left(T\left(y_{n}-z_{n}\right)\right)\rightarrow 0$.
Therefore, we see that

$\left|f_{n}\right|\left(Ty_{n}\right)=\left|f_{n}\right|\left(T\left(y_{n}-z_{n}\right)\right)+\left|f_{n}\right|\left(Tz_{n}\right)\rightarrow
0\text{,}$

which contradicts (** ‣ 2). This completes the proof. ∎

###### Corollary 2.6

If $T:E\rightarrow F$ is a positive operator between two Banach lattices, then
the following assertions are equivalent:

  1. (1)

$T$ is an awDP operator.

  2. (2)

For each relatively weakly compact subsets $A\subset E$ , $B\subset
F^{\prime}$, and for every $\varepsilon>0$, there exists some $u\in
E^{+}$satisfying

$\left|f\right|\left(T\left(\left|x\right|-u\right)^{+}\right)\leq\varepsilon$

for all $x\in A$ and all $f\in B$.

  3. (3)

For each relatively weakly compact subsets $A\subset E$ , $B\subset
F^{\prime}$, and for every $\varepsilon>0$, there exists some
$g\in\left(F^{\prime}\right)^{+}$ satisfying

$\left(\left|f\right|-g\right)^{+}\left(T\left|x\right|\right)\leq\varepsilon$

for all $x\in A$ and all $f\in B$.

###### Proof.

Note that the proof is similar for the two equivalences $1\Leftrightarrow 2$
and $1\Leftrightarrow 3$. The implication $1\Rightarrow 2$ is exactly Theorem
2.5$(1)$. For the reciprocal one, let $\left(x_{n}\right)\subset E$ ,
$\left(f_{n}\right)\subset F^{\prime}$ be two disjoint weakly null sequences,
and let $\varepsilon>0$. Put $A=\left\\{x_{n}:n\in\mathbb{N}\right\\}$,
$B=\left\\{f_{n}:n\in\mathbb{N}\right\\}$. By hypothesis there exists some
$u\in E^{+}$ so that
$\left|f\right|\left(T\left(\left|x\right|-u\right)^{+}\right)\leq\varepsilon$
holds for all $x\in A$ and all $f\in B$. In particular,
$\left|f_{n}\right|\left(T\left(\left|x_{n}\right|-u\right)^{+}\right)\leq\varepsilon$
for all $n$. Now, as $\left|f_{n}\right|\overset{w}{\rightarrow}0$ choose some
natural number $m$ such that
$\left|f_{n}\right|\left(Tu\right)\leq\varepsilon$ holds for every $n\geq m$.
Thus, for every $n\geq m$ we get

$\displaystyle\left|f_{n}\left(Tx_{n}\right)\right|$ $\displaystyle\leq$
$\displaystyle\left|f_{n}\right|\left(T\left|x_{n}\right|\right)$
$\displaystyle\leq$
$\displaystyle\left|f_{n}\right|\left(T\left(\left|x_{n}\right|-u\right)^{+}\right)+\left|f_{n}\right|\left(Tu\right)\leq
2\varepsilon\text{.}$

This shows that $f_{n}\left(Tx_{n}\right)\rightarrow 0$, and then $T$ is an
awDP operator. ∎

###### Corollary 2.7

For a Banach lattice $E$, the following assertions are equivalent:

  1. (1)

$E$ has the wDP property.

  2. (2)

For each relatively weakly compact subsets $A\subset E$ , $B\subset
F^{\prime}$, and for every $\varepsilon>0$, there exists some $u\in
E^{+}$satisfying

$\left|f\right|\left(\left(\left|x\right|-u\right)^{+}\right)\leq\varepsilon$

for all $x\in A$ and all $f\in B$.

  3. (3)

For each relatively weakly compact subsets $A\subset E$ , $B\subset
F^{\prime}$, and for every $\varepsilon>0$, there exists some
$g\in\left(F^{\prime}\right)^{+}$ satisfying

$\left(\left|f\right|-g\right)^{+}\left(\left|x\right|\right)\leq\varepsilon$

for all $x\in A$ and all $f\in B$.

##  3\. Proof of Theorem A

###  $\left(i\right)$ $E$ has sequentially weakly continuous lattice
operations

Let $\left(x_{n}\right)\subset E$ , $\left(f_{n}\right)\subset F^{\prime}$ be
two weakly null sequences. We shall see that
$f_{n}\left(Tx_{n}\right)\rightarrow 0$. To this end, put
$A=sol\left\\{x_{n}:n\in\mathbb{N}\right\\}$ and
$B=sol\left\\{f_{n}:n\in\mathbb{N}\right\\}$. We proceed in two steps:

Step 1: We claim that $g_{n}\left(Tx_{n}\right)\rightarrow 0$ for every
disjoint sequence $\left(g_{n}\right)\subset B^{+}$. Let
$\left(g_{n}\right)\subset B^{+}$ be such a sequence. Thus, we have
$g_{n}\overset{w}{\rightarrow}0$. As $T$ is an awDP operator, it follows by
Theorem 2.3$\left(2\right)$ that $\left(g_{n}\right)$ converges uniformly to
zero on $T\left(A\right)$, that is, $\sup_{x\in
A}\left|g_{n}\left(Tx\right)\right|\rightarrow 0$. From the inequality
$\left|g_{n}\left(Tx_{n}\right)\right|\leq\sup_{x\in
A}\left|g_{n}\left(Tx\right)\right|$, we conclude that
$g_{n}\left(Tx_{n}\right)\rightarrow 0$.

Step 2: Since the lattice operations in $E$ are sequentially weakly
continuous, we have $\left|x_{n}\right|\overset{w}{\rightarrow}0$. Thus,
taking into account Step 1, we see by Lemma 2.2$\left(2\right)$, that
$\left(x_{n}\right)$ converges uniformly to zero on
$T^{\prime}\left(B\right)$, i.e., $\sup_{f\in
B}\left|f\left(Tx_{n}\right)\right|\rightarrow 0$. From
$\left|f_{n}\left(Tx_{n}\right)\right|\leq\sup_{f\in
B}\left|f\left(Tx_{n}\right)\right|$, we conclude that
$f_{n}\left(Tx_{n}\right)\rightarrow 0$. Therefore $T$ is a wDP operator.

###  $\left(ii\right)$ $F^{\prime}$ has sequentially weakly continuous lattice
operations

Note that the two situations $\left(i\right)$ and $\left(ii\right)$ are
symmetric. Let $\left(x_{n}\right)\subset E$ , $\left(f_{n}\right)\subset
F^{\prime}$ be two weakly null sequences. We shall see that
$f_{n}\left(Tx_{n}\right)\rightarrow 0$. To this end, put
$A=sol\left\\{x_{n}:n\in\mathbb{N}\right\\}$ and
$B=sol\left\\{f_{n}:n\in\mathbb{N}\right\\}$. We proceed as in
$\left(i\right)$:

Step 1: We claim that $f_{n}\left(Ty_{n}\right)\rightarrow 0$ for every
disjoint sequence $\left(y_{n}\right)\subset A^{+}$. Let
$\left(y_{n}\right)\subset A^{+}$ be such a sequence. Thus, we have
$y_{n}\overset{w}{\rightarrow}0$. As $T$ is an awDP operator, it follows by
Theorem 2.3$\left(4\right)$ that $\left(y_{n}\right)$ converges uniformly to
zero on $T^{\prime}\left(B\right)$, that is, $\sup_{g\in
B}\left|\left(T^{\prime}g\right)\left(y_{n}\right)\right|\rightarrow 0$. From
the inequality
$\left|f_{n}\left(Ty_{n}\right)\right|=\left|T^{\prime}\left(f_{n}\right)\left(y_{n}\right)\right|\leq\sup_{g\in
B}\left|\left(T^{\prime}g\right)\left(y_{n}\right)\right|$, we conclude that
$f_{n}\left(Ty_{n}\right)\rightarrow 0$.

Step 2: Since the lattice operations in $F^{\prime}$ are sequentially weakly
continuous, we have $\left|f_{n}\right|\overset{w}{\rightarrow}0$. Thus,
taking into account Step 1, we see by Lemma 2.2$\left(1\right)$, that
$\left(f_{n}\right)$ converges uniformly to zero on $T\left(A\right)$, i.e.,
$\sup_{x\in A}\left|f_{n}\left(Tx\right)\right|\rightarrow 0$. From
$\left|f_{n}\left(Tx_{n}\right)\right|\leq\sup_{x\in
A}\left|f_{n}\left(Tx\right)\right|$, we conclude that
$f_{n}\left(Tx_{n}\right)\rightarrow 0$. Therefore $T$ is a wDP operator.

###  $\left(iii\right)$ $T$ is positive and $F$ is discrete with order
continuous norm

Let us recall that an operator $T:E\rightarrow Y$ is said to be order weakly
compact, if $T$ carries each order bounded subset of $E$ to a relatively
weakly compact one in $Y$, equivalently, $T\left(\left[0,x\right]\right)$ is
relatively weakly compact in $Y$, for every $x\in E^{+}$. We need to prove the
following claim.

###### Claim

The product $ST$ is a wDP operator, for every positive order weakly compact
operator $S:F\rightarrow G$ into an arbitrary Banach lattice.

###### Proof.

Let $\left(x_{n}\right)\subset E$, $\left(f_{n}\right)\subset G^{\prime}$ be a
weakly null sequences. We shall see that
$f_{n}\left(ST\left(x_{n}\right)\right)\rightarrow 0$. To this end, let
$\varepsilon>0$. As $T$ is an awDP operator, then $ST$ is so (Corollary 2.4).
Therefore, by Theorem 2.5, pick some $u\in E^{+}$ such that

$\left|f_{n}\right|\left(ST\left(\left|x_{n}\right|-u\right)^{+}\right)<\varepsilon$

holds for all $n$. Now, from the inequalities

$\displaystyle\left|Tx_{n}\right|-Tu$ $\displaystyle\leq$ $\displaystyle
T\left|x_{n}\right|-T\left(\left|x_{n}\right|\wedge u\right)$
$\displaystyle\leq$
$\displaystyle\left|T\left|x_{n}\right|-T\left(\left|x_{n}\right|\wedge
u\right)\right|=T\left(\left(\left|x_{n}\right|-u\right)^{+}\right)$

we see that $\left(\left|Tx_{n}\right|-Tu\right)^{+}\leq
T\left(\left(\left|x_{n}\right|-u\right)^{+}\right)$ holds for all $n$. Thus,
for every $n$ we have

$\displaystyle\left|f_{n}\left(ST\left(x_{n}\right)\right)\right|$
$\displaystyle\leq$
$\displaystyle\left|f_{n}\right|\left(S\left|T\left(x_{n}\right)\right|\right)$
$\displaystyle\leq$
$\displaystyle\left|f_{n}\right|\left(S\left(\left(\left|T\left(x_{n}\right)\right|-Tu\right)^{+}\right)\right)+\left|f_{n}\right|\left(S\left(\left|T\left(x_{n}\right)\right|\wedge
Tu\right)\right)$ $\displaystyle\leq$
$\displaystyle\left|f_{n}\right|\left(ST\left(\left|x_{n}\right|-u\right)^{+}\right)+\left|f_{n}\right|\left(S\left(\left|T\left(x_{n}\right)\right|\wedge
Tu\right)\right)$ $\displaystyle\leq$
$\displaystyle\varepsilon+\left|f_{n}\right|\left(S\left(\left|T\left(x_{n}\right)\right|\wedge
Tu\right)\right)\text{.}$

Or, as $F$ is discrete with order continuous norm, then it follows from [11,
Proposition 2.5.23] that the lattice operations in $F$ are sequentially weakly
continuous, and then the sequence
$\left(\left|T\left(x_{n}\right)\right|\wedge Tu\right)$ is order bounded
weakly null in $F^{+}$. It follows from [11, Corollary 3.4.9] that
$\left\|S\left(\left(\left|T\left(x_{n}\right)\right|\wedge
Tu\right)\right)\right\|\rightarrow 0$. This shows that
$\lim\sup\left|f_{n}\left(ST\left(x_{n}\right)\right)\right|\leq\varepsilon$ .
As $\varepsilon>0$ is arbitrary, we infer that
$f_{n}\left(ST\left(x_{n}\right)\right)\rightarrow 0$ as desired. ∎

Turning to the proof of the theorem, since the norm of $F$ is order continuous
then by [2, Theorem 4.9] each order interval of $F$ is weakly compact. Thus,
the identity operator $I:F\rightarrow F$ is order weakly compact. Now, it
follows from the preceding claim that $T=IT$ is a wDP operator as desired.

###### Corollary 3.1

Let $E$ be a Banach lattice such that the lattice operations in $E$ (resp. in
$E^{\prime}$) are sequentially weakly continuous. Then, $E$ has the wDP
property if and only if it has the DP property.

In case the range space is a discrete KB-space, the DP operators and their
three weak classes considered in this paper coincide on positive operators.
The details follow.

###### Corollary 3.2

Let $E$ and $F$ be two Banach lattices such that $F$ is a discrete KB-space.
Then, for a positive operator $T:E\rightarrow F$ the following statements are
equivalent:

  1. (1)

$T$ is an awDP operator.

  2. (2)

$T$ is a wDP operator.

  3. (3)

$T$ is an almost DP operator.

  4. (4)

$T$ is a DP operator.

###### Proof.

It suffices to show that a positive wDP operator $T:E\rightarrow F$ is DP. To
this end, let $\left(x_{n}\right)\subset E$ be a weakly null sequence. Since
$F$ is a discrete KB-space then it is a dual (see [11, Exercise 5.4.E2]), that
is, $F=G^{\prime}$ for some Banach lattice $G$. To see that
$\left\|Tx_{n}\right\|\rightarrow 0$ it suffices by [8, Corollary 2.7] to show
that $\left|Tx_{n}\right|\overset{w^{\ast}}{\rightarrow}0$ and
$\left(Tx_{n}\right)\left(y_{n}\right)\rightarrow 0$ for every disjoint
bounded sequence $\left(y_{n}\right)\subset G^{+}$. Note that, since the
lattice operations in $F$ are sequentially weakly continuous then we have
$\left|Tx_{n}\right|\overset{w}{\rightarrow}0$. Now, if
$\left(y_{n}\right)\subset G^{+}$ is a disjoint bounded sequence, then by
Theorem 2.4.14 of [11]$\ y_{n}\overset{w}{\rightarrow}0$ as the norm of
$G^{\prime}=F$ is order continuous. By the lattice embedding $G\hookrightarrow
G^{\prime\prime}$ we see that $y_{n}\overset{w}{\rightarrow}0$ in
$G^{\prime\prime}=F^{\prime}$. Since $T$ is a wDP operator, then
$\left(Tx_{n}\right)\left(y_{n}\right)=y_{n}\left(Tx_{n}\right)\rightarrow 0$
as desired. This completes the proof. ∎

In particular, we obtain a result noted by W. Wnuk in [15, Proposition 6].

###### Corollary 3.3

For a discrete KB-space $E$ the following statements are equivalent:

  1. (1)

$E$ has the wDP property.

  2. (2)

$E$ has the DP property.

  3. (3)

$E$ has the positive Schur property.

  4. (4)

$E$ has the Schur property.

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