∎

11institutetext: S. Ghanizadeh Zare22institutetext: Department of Mathematics
and Applications, Faculty of Sciences, University of Mohaghegh Ardabili,
Ardabil, Iran.
22email<EMAIL_ADDRESS>33institutetext: K. Haghnejad
Azar44institutetext: Department of Mathematics and Applications, Faculty of
Sciences, University of Mohaghegh Ardabili, Ardabil, Iran.
44email<EMAIL_ADDRESS>55institutetext: M. Matin66institutetext:
Department of Mathematics and Applications, Faculty of Sciences, University of
Mohaghegh Ardabili, Ardabil, Iran.
66email<EMAIL_ADDRESS>77institutetext: S. Hazrati88institutetext:
Department of Mathematics and Applications, Faculty of Sciences, University of
Mohaghegh Ardabili, Ardabil, Iran.
88email<EMAIL_ADDRESS>
#  $\tilde{o}$rder-norm continuous operators and $\tilde{o}$rder weakly
compact operators

Sajjad Ghanizadeh Zare  Kazem Haghnejad Azar  Mina Matin  Somayeh Hazrati

(Received: date / Accepted: date)

###### Abstract

Let $E$ be a sublattice of a vector lattice $F$. A continuous operator $T$
from the vector lattice $E$ into a normed vector space $X$ is said to be
$\tilde{o}$rder-norm continuous whenever $x_{\alpha}\xrightarrow{Fo}0$ implies
$Tx_{\alpha}\xrightarrow{\|.\|}0$ for each $(x_{\alpha})_{\alpha}\subseteq E$.
Our mean from the convergence $x_{\alpha}\stackrel{{\scriptstyle
Fo}}{{\longrightarrow}}x$ is that there exists another net
$\left(y_{\alpha}\right)$ in $F$ with the same index set satisfying
$y_{\alpha}\downarrow 0$ in $F$ and $|x_{\alpha}-x|\leq y_{\alpha}$ for all
indexes $\alpha$. In this paper, we will study some properties of this new
class of operators and its relationships with some known classifications of
operators. We also define the new class of operators that named
$\tilde{o}$rder weakly compact operators. A continuous operator
$T:E\rightarrow X$ is said to be $\tilde{o}$rder weakly compact, if $T(A)$ in
$X$ is a relatively weakly compact set for each $Fo$-bounded $A\subseteq E$.
In this manuscript, we study some properties of this class of operators and
its relationships with $\tilde{o}$rder-norm continuous operators.

###### Keywords:

Vector latticeproperty $(F)$ $\tilde{o}$-convergence order-to-norm continuous
operator $\tilde{o}$rder-norm continuous operator $\tilde{o}$rder weakly
compact.

###### MSC:

47B65 46B40 46B42

##  1 Introduction and Preliminaries

Our motivation in writing this article is to communicate and expand the
concepts which have been introduced in the articles 8  and p1 . In these
papers some concepts such as $\tilde{o}$-convergence, the property $(F)$ and
order-to-norm continuous operators have been introduced, studied and authors
have been investigated some of their properties and the relationships with
other lattice properties. We introduce the new classes of operators as the set
of $\tilde{o}$rder-norm continuous operators and we study some of their
properties and their relationships with others knowns operators.

To state our results, we need to fix some notations and recall some
definitions. A net $(x_{\alpha})_{\alpha\in A}$ in a vector lattice $E$ is
said to be order convergent to $x\in E$ if there is a net
$(y_{\beta})_{\beta\in B}$ in $E$ such that $y_{\beta}\downarrow 0$ and for
every $\beta\in B$, there exists $\alpha_{0}\in A$ such that
$|x_{\alpha}-x|\leq y_{\beta}$ whenever $\alpha\geq\alpha_{0}$. in short, we
will denote this convergence by $x_{\alpha}\xrightarrow{o}x$ and write that
$x_{\alpha}$ is $o$-convergent to $x$. A net $(x_{\alpha})_{\alpha}$ in vector
lattice $E$ is unbounded order convergent to $x\in E$ if $|x_{\alpha}-x|\wedge
u\xrightarrow{o}0$ for all $u\in E^{+}$. We denote this convergence by
$x_{\alpha}\xrightarrow{uo}x$ and write that $x_{\alpha}$ $uo$-convergent to
$x$. It is clear that for order bounded nets, $uo$-convergence is equivalent
to $o$-convergence. A net $(x_{\alpha})\subseteq E$ is said to be
$\tilde{o}$rder convergent (in short, $\tilde{o}$-convergent) to $x$ if there
is a net $(y_{\beta})\subseteq F$, possibly over a different index set, such
that $y_{\beta}\downarrow 0$ in $F$ and for every $\beta$, there exists
$\alpha_{0}$ such that $|x_{\alpha}-x|\leq y_{\beta}$ whenever
$\alpha\geq\alpha_{0}$. We denote this convergence by
$x_{\alpha}\xrightarrow{Fo}x$ and write that $(x_{\alpha})$ is
$\tilde{o}$-convergent to $x$. It is clear that if $E$ is regular in $F$ and
$x_{\alpha}\xrightarrow{o}x$ in $E$, then $x_{\alpha}\xrightarrow{Fo}x$. The
converse is not true in general. For example, $c_{0}$ is a sublattice of
$\ell^{\infty}$ and $(e_{n})\subseteq c_{0}$.
$e_{n}\xrightarrow{\ell^{\infty}o}0$ in $c_{0}$, but it is not order
convergent to $0$ in $c_{0}$. A subset $A$ of $E$ is said to be $F$-order
bounded (in short, $Fo$-bounded), if there exist $x,y\in F$ that
$A\subseteq[x,y]$. A vector lattice $E$ is said to have the property $(F)$, if
$A\subseteq E$ is order bounded whenever $A$ is $Fo$-bounded. (see 8 ). A
Banach lattice $E$ is said to be an $AM$-space if we have
$\|x+y\|=max\\{\|x\|,\|y\|\\}$ for each $x,y\in E$ such that $|x|\wedge|y|=0$.
A Banach lattice $E$ is said to be an $AL$-space if we have
$\|x+y\|=\|x\|+\|y\|$ for each $x,y\in E$ such that $|x|\wedge|y|=0$. A Banach
lattice $E$ is said to be $KB$-space whenever each increasing norm bounded
sequence of $E^{+}$ is norm convergent. Let $E$ and $G$ be vector spaces. We
will denote $L(E,G)$ by the collections of operators from $E$ into $G$.
$L_{b}(E,G)$ is the all of order bounded operators in this manuscript. An
operator $T$ from a Banach space $X$ into a Banach space $Y$ is weakly compact
if $\overline{{T(B_{X})}}$ is weakly compact where $B_{X}$ is the closed unit
ball of $X$. A continuous operator $T$ from Banach lattice $E$ into Banach
space $X$ is called $M$-weakly compact if $\lim\|Tx_{n}\|=0$ holds for every
norm bounded disjoint sequence $(x_{n})_{n}$ of $E$. An operator $T:E\to F$ is
said to preserve disjointness whenever $x\perp y$ in $E$ implies $Tx\perp Ty$.
A subset $A$ of a vector lattice $E$ is called $b$-order bounded in $E$ if it
is order bounded in $E^{\sim\sim}$. If each $b$-order bounded subset of $E$ is
order bounded in $E$, then $E$ is said to have the property ($b$). Jalili,
Haghnejad and Moghimi characterized $L_{o\tau}(E,G)$ and
$L^{\sigma}_{o\tau}(E,G)$ spaces in p1 . An operator $T$ from a vector lattice
$E$ into topological vector space $G$ is said to be order-to-topology
continuous whenever $x_{\alpha}\xrightarrow{o}0$ implies
$Tx_{\alpha}\xrightarrow{\tau}0$ for each $(x_{\alpha})_{\alpha}\subseteq E$.
For each sequence $(x_{n})\subseteq E$, if $x_{n}\xrightarrow{o}0$ implies
$Tx_{n}\xrightarrow{\tau}0$, then $T$ is called $\sigma$-order-to-topology
continuous operator. The collection of all order-to-topology continuous
operators will be denoted by $L_{o\tau}(E,G)$; the subscript $o\tau$ is
justified by the fact that the order-to-topology continuous operators; that
is,

$L_{o\tau}(E,G)=\\{T\in L(E,G):~{}T~{}\text{is order-to-topology continuous
}\\}.$

Similarly, $L^{\sigma}_{o\tau}(E,G)$ represents the collection of all
$\sigma$-order-to-topology continuous operators, that is,

$L^{\sigma}_{o\tau}(E,G)=\\{T\in L(E,G):~{}T~{}\text{is}~{}\sigma-\text{order-
to-topology continuous }\\}.$

For a normed space $G$, $L_{on}(E,G)$ is collection of order-to-norm topology
continuous operators.

Let $E$, $G$ be two normed vector lattices. Recall that from 12o , a
continuous operator $T:E\to G$ is said to be $\sigma$-$uon$-continuous, if
each norm bounded $uo$-null sequence $(x_{n})\subseteq E$ implies
$Tx_{n}\xrightarrow{\|.\|}0$. Remember that an operator $T$ from Banach
lattice $E$ into Banach space $X$ is a $wun$-Dunford-Pettis whenever
$x_{n}\xrightarrow{wun}0$ in $E$ implies $Tx_{n}\xrightarrow{\|.\|}0$ in $X$
for each sequence $(x_{n})\subseteq E$ (See 44  for more information).

Recall that a Banach lattice $E$ is said to have the property $(P)$ if there
exists a positive contractive projection $P:E^{**}\to E$ where $E$ is
identified with a sublattice of its topological bidual $E^{**}$.

In a Banach lattice $E$, a subset $A$ is said to be almost order bounded if
for any $\epsilon>0$ there exists $u\in E^{+}$ such that
$A\subseteq[-u,u]+\epsilon B_{E}$ ($B_{E}$ is the closed unit ball of $E$).
One should observe the following useful fact, which can be easily verified
using Riesz decomposition Theorem, that $A\subseteq[-u,u]+\epsilon B_{E}$ if
and only if $\sup_{x\in A}\|(|x|-u)^{+}\|=\sup_{x\in A}\||x|-|x|\wedge
u\|\leq\epsilon$. By Theorems 4.9 and 3.44 of 1 , each almost order bounded
subset in order continuous Banach lattice is relatively weakly compact.
$A\subseteq L_{1}(\mu)$ is relatively weakly compact if and only if it is
almost order bounded (see4b ).
A vector subspace $G$ of an ordered vector space $E$ is majorizing, whenever
there exists some $y\in G$ with $x\leq y$ for each $x\in E$. A sublattice $G$
of a vector lattice $E$ is said to be order dense in $E$ whenever there exists
some $y\in G$ with $0<y\leq x$ for each $0<x\in E$. Recall that a Banach
lattice $E$ is said to have the positive Schur property (the dual positive
Schur property) if every positive $w$-null sequence in $E$ (positive
$w^{*}$-null sequence in $E^{*}$) is norm null. A vector lattice is called
laterally complete whenever every subset of pairwise disjoint positive vectors
has a supremum.

Unless otherwise stated, throughout this paper, $F$ is a vector lattice, $E$
is a sublattice of $F$, and $X$ is a normed vector space.

##  2 The property $(F)$ in vector lattices

In this section, we investigate on the property $(F)$ which is defined in 8 .
We will try to get new content from this property.

Let $E$ be regular in $F$ and order continuous in its own right and closed,
and let $(x_{\alpha})\subseteq E$ be an order bounded net and $F$ be a
Dedekind complete Banach lattice. By Lemma $4.5$ of 4b , it is clear that
$x_{\alpha}\xrightarrow{Fo}x$ in $E$ if and only if
$x_{\alpha}\xrightarrow{o}x$ in $E$.

###### Remark 1

  1. i)

Let $E^{**}$ be lattice isomorphic by one sublattice of $F$. If $E$ has the
property $(b)$, then $E$ has the property $(F)$. And if $E^{**}$ is lattice
isomorphic with $F$, then the properties $(b)$ and $(F)$ are equivalent.

  2. ii)

If $E$ is a majorizing sublattice of $F$, then $E$ has the property $(F)$. Let
$A\subseteq E$ be a $F$-order bounded set. Therefore, there exists a $u\in
F^{+}$ that $A\subseteq[-u,u]$. Since $E$ is majorizing sublattice of $F$,
therefore, there exists $v\in E^{+}$ that $u\leq v$ and $-v\leq-u$. So
$A\subseteq[-v,v]$. Hence $A$ is order bounded in $E$.

  3. iii)

Let $E$ be a sublattice of $F$ and $F$ be a sublattice of $G$. If $F$ has the
property $(G)$, necessarily $E$ does not have the property $(G)$. For example,
$c_{0}$ is a sublattice of $c$ and $c$ is a sublattice of $\ell^{\infty}$. $c$
has the property $(\ell^{\infty})$ while $c_{0}$ has not property
$(\ell^{\infty})$.

  4. iv)

Let $E$ be a sublattice of $F$ and $F$ be a sublattice of $G$. It is clear
that if $E$ has the property $(G)$, then it has the property $(F)$. If
$A\subseteq E$ and it is order bounded in $F$, so it is order bounded in $G$.
By assumption $A$ is order bounded in $E$.

###### Theorem 2.1

Let $E$ and $F$ be two Banach lattices, and let $E$ has the property $(F)$.
Then, by one of the following assertions, if $A\subseteq E$ is an almost order
bounded in $F$, $A$ is an almost order bounded in $E$.

  1. i)

$F$ has an order unit.

  2. ii)

$E$ is an ideal of $F$.

###### Proof

Let $F$ has an order unit and $A\subseteq E$ be an almost order bounded set in
$F$. By Theorem 4.21 of 1 , $A$ is an order bounded set in $F$. By the
assumption, $A$ is an order bounded set in $E$ and therefore, it is an almost
order bounded set in $E$.

Suppose that $E$ is an ideal of $F$. Let $A\subseteq E$ be an almost order
bounded set in $F$. It means that for each $\varepsilon>0$, there exists a
$u\in F^{+}$ that $A\subseteq[-u,u]+\varepsilon B_{F}$. For each $x\in A$, we
have $x=x_{1}+x_{2}$ that $x_{1}\in[-u,u]$ and $x_{2}\in\varepsilon B_{F}$. We
assume that $x\neq 0$. It is obvious that $\mid
x_{2}\mid\leq\mid\varepsilon\frac{x}{\|x\|}\mid$. Since $E$ is an ideal of
$F$, therefore, $x_{2}\in E$. Because $\|x_{2}\|\leq\varepsilon$, therefore,
$x_{2}\in\varepsilon B_{E}$. On the other hand, we have $x_{1}=x-x_{2}$. So
$x_{1}\in E$. Since $E$ has the property $(F)$, there exists a $v\in E^{+}$
that $x_{1}\in[-v,v]$. Therefore, $A\subseteq[-v,v]+\varepsilon B_{E}$. Hence
$A$ is almost order bounded in $E$, as regard.

###### Corollary 1

  1. i)

Let $F$ has an order unit, $E$ has the property $(F)$ with order continuous
norm and let $(x_{n})\subseteq E$ be disjoint and almost order bounded
sequence in F. Then $x_{n}\xrightarrow{\|.\|}0$ in $E$.

  2. ii)

Let $(x_{n})\subseteq E$ be a disjoint and almost order bounded sequence in
$F$, and $E$ be an ideal of $F$ and has the property $(F)$. If $E$ has order
continuous norm, then $x_{n}\xrightarrow{\|.\|}0$ in $E$.

###### Proof

  1. i)

By proof of Theorem 1, $(x_{n})$ is order bounded in $E$ and so by Theorem
4.14 of 1 , $x_{n}\xrightarrow{\|.\|}0$ in $E$.

  2. ii)

Since $(x_{n})$ is a disjoint sequence, therefore, by Corollary 3.6 of 5g , we
have $x_{n}\xrightarrow{uo}0$ in $E$. By Theorem 2.1, $(x_{n})$ is almost
order bounded in $E$. By Proposition 3.7 of 4b , $x_{n}\xrightarrow{\|.\|}0$
in $E$.

##  3 $\tilde{o}$rder-norm continuous operators

A continuous operator $T:E\rightarrow X$ is said to be $\tilde{o}$rder-norm
continuous (or, $\tilde{o}n$-continuous for short), if $(x_{\alpha})\subseteq
E$ is $\tilde{o}$-null in $E$, then $(Tx_{\alpha})$ in $X$ is convergent to
$0$ in norm. A continuous operator $T:E\rightarrow X$ is said to be
$\sigma$-$\tilde{o}$rder-norm continuous (or, $\sigma$-$\tilde{o}n$-continuous
for short), if $(x_{n})\subseteq E$ is $\tilde{o}$-null in $E$, then
$(Tx_{n})$ in $X$ is convergent to $0$ in norm.

The collection of all $\tilde{o}n$-continuous operators from a vector lattice
$E$ into a Banach space $X$ (resp, $\sigma$-$\tilde{o}n$-continuous) will be
denoted by $L_{\tilde{o}n}(E,X)$ (resp, $L_{\tilde{o}n}^{\sigma}(E,X)$.

It is clear that, if $T:E\to X$ is a $\tilde{o}n$-continuous, then $T$ is an
order-to-norm topology continuous. By the following example, the converse is
not true in general.

###### Example 1

The identity operator $I:c_{0}\to c_{0}$ is order-to-norm topology continuous.
Let $(x_{\alpha})\subseteq c_{0}$ is order-null. Since $c_{0}$ has order
continuous norm, therefore, $x_{\alpha}\xrightarrow{\|.\|}0$. Consider
$(e_{n})\subseteq c_{0}$. $e_{n}\xrightarrow{\ell^{\infty}o}0$ in $c_{0}$. But
$(I(e_{n}))$ is not convergent to zero in norm in $c_{0}$. Hence $I:c_{0}\to
c_{0}$ is not $\ell^{\infty}on$-continuous.

Obviously $L_{\tilde{o}n}(E,X)$ is a subspace of $L_{on}(E,X)$. In the
following, we have some examples from $\tilde{o}n$-continuous operators.

###### Example 2

  1. i)

If $E$ has the property $(F)$, $E^{*}$ has order continuous norm and $X$ has
the Schur property, then each continuous operator from $E$ to $X$ is a
$\sigma$-$\tilde{o}n$-continuous operator. Let $(x_{n})\subseteq E$ be
$\tilde{o}$-null sequence. Therefore, $(x_{n})$ is order null in $F$ and so it
is order bounded in $F$. Since $E$ has the property $(F)$, hence $(x_{n})$ is
order bounded in $E$. On the other hand $(x_{n})$ is $uo$-null in $F$ and so
it is $uo$-null in $E$. Because $E^{*}$ has order continuous norm, therefore,
by Theorem 6.4 of 55 , $x_{n}\xrightarrow{w}0$ in $E$. By continuity of $T$,
we have $Tx_{n}\xrightarrow{w}0$ in $X$. Since $X$ has the Schur property,
hence $Tx_{n}\xrightarrow{\|.\|}0$ in $X$.
The Banach lattice $c$ has the property $(\ell^{\infty})$, $c^{*}$ has order
continuous norm and $\ell^{1}$ has the Schur property, therefore, each
continuous operator $T:c\to\ell^{1}$ is $\sigma$-$\ell^{\infty}on$-continuous.

  2. ii)

Let $F$ be a Dedekind complete Banach lattice. If $E$ has the property $(F)$
and order continuous norm, then each continuous operator from $E$ to $X$ is a
$\tilde{o}n$-continuous operator. Let $(x_{\alpha})\subseteq E$ is
$\tilde{o}$-null net. Therefore, $(x_{\alpha})$ is order null in $F$ and so it
is order bounded in $F$. Since $E$ has property $(F)$erty, hence
$(x_{\alpha})$ is order bounded in $E$. On the other hand $(x_{\alpha})$ is
$uo$-null in $F$ and so by Lemma $4.5$ of 4b , it is $uo$-null in $E$. Since
$(x_{\alpha})$ is order bounded, hence $(x_{\alpha})$ is order-null in $E$.
Because $E$ has order continuous norm, we have
$x_{\alpha}\xrightarrow{\|.\|}0$ in $E$. So $Tx_{\alpha}\xrightarrow{\|.\|}0$
in $X$.

  3. iii)

If $T:F\to X$ is a $uon$-continuous operators, then $T|_{E}:E\to X$ is a
$\tilde{o}n$-continuous operator. Let $(x_{\alpha})\subseteq E$ be a
$\tilde{o}$-null net. It is clear that $x_{\alpha}\xrightarrow{uo}0$ in $F$.
By assumption, $Tx_{\alpha}\xrightarrow{\|.\|}0$ in $X$.

The class of $\tilde{o}n$-continuous operators differs from the class of order
continuous operators. Since the identity operator $I:c_{0}\to c_{0}$ is order
continuous while it is not $\ell^{\infty}on$-continuous (see Example 1.).

###### Proposition 1

  1. i)

Let $T\in L_{\tilde{o}n}(E,X)$, $S:E\to X$ be a continuous operator, and
$0\leq S\leq T$, then $S$ is a $\tilde{o}n$-continuous operator.

  2. ii)

Moreover, if $X$ also is a sublattice of $F$, $T\in L_{\tilde{o}n}(E,X)$ and
$S\in L_{\tilde{o}n}(X,Y)$, then $S\circ T\in L_{\tilde{o}n}(E,Y)$.

###### Proof

  1. i)

Let $(x_{\alpha})\subseteq E$ be a $\tilde{o}$-null net. It is obvious that
$|x_{\alpha}|\xrightarrow{Fo}0$ in $E$. We have
$|Sx_{\alpha}|\leq|S||x_{\alpha}|=S|x_{\alpha}|\leq T|x_{\alpha}|$. By
assumption, $T|x_{\alpha}|\xrightarrow{\|.\|}0$. So
$|Sx_{\alpha}|\xrightarrow{\|.\|}0$. It means that $S$ is a
$\tilde{o}n$-continuous operator.

  2. ii)

Let $(x_{\alpha})\subseteq E$ and $x_{\alpha}\xrightarrow{Fo}0$. By
assumption, we have $Tx_{\alpha}\xrightarrow{\|.\|}0$. Because each
$\tilde{o}n$-continuous operator is continuous, therefore,
$STx_{\alpha}\xrightarrow{\|.\|}0$. Hence $S\circ T\in L_{\tilde{o}n}(E,Y)$.

###### Remark 2

Let $T:E\to G$ be an order continuous lattice homomorphism from a Dedekind
complete vector lattice $E$ to an Archimedean laterally complete normed vector
lattice $G$. If $E$ is an order dense in the Archimedean vector lattice $F$,
then by Theorem 2.32 of 1 , $T$ exteneded from $F$ to $G$ that is an order
continuous lattice homomorphism. Now if $G$ has order continuous norm, then
$T$ is a $\tilde{o}n$-continuous.

###### Theorem 3.1

For an order bounded operator $T:E\to G$ between two Riesz spaces with $G$
Dedekind complete and $G$ has order continuous norm Banach lattice, the
following statements are equivalent.

  1. (1)

$T$ is $\tilde{o}n$-continuous.

  2. (2)

If $(x_{\alpha})\subseteq E$ and $x_{\alpha}\downarrow 0$ holds in $F$, then
$Tx_{\alpha}$ is norm convergent to $0$ in $G$.

  3. (3)

If $(x_{\alpha})\subseteq E$ and $x_{\alpha}\downarrow 0$ holds in $F$, then
$\inf\left\\{\|Tx_{\alpha}\|\right\\}\,=\,0$.

  4. (4)

$T^{+}$, $T^{-}$ are both $\tilde{o}n$-continuous.

  5. (5)

$|T|$ is $\tilde{o}n$-continuous.

###### Proof

$(1)\Rightarrow(2)$ and $(2)\Rightarrow(3)$ are obvious. Clearly, if $T$ is a
positive operator then (1), (2), and (3) are equivalent.
$(3)\Rightarrow(4)$ It is enough to show that $T^{+}$ is
$\tilde{o}n$-continuous. To this end, let $(x_{\alpha})\subseteq E$ and
$x_{\alpha}\downarrow 0$ in $F$. Let $\|T^{+}x_{\alpha}\|\downarrow z\,\geq
0$. We have to show that $z\,=\,0$. Fix some index $\beta$ and put
$x\,=\,x_{\beta}$.
Now for each $0\leq y\leq x$ and each $\alpha\succeq\beta$ we have

$0\,\leq\,y-y\wedge x_{\alpha}=y\wedge x-y\wedge
x_{\alpha}\,\leq\,x-x_{\alpha},$ (1)

and consequently

$Ty-T(y\,\wedge\,x_{\alpha})\,=T(y-y\,\wedge\,x_{\alpha})\leq
T^{+}(x-x_{\alpha})=T^{+}x-T^{+}x_{\alpha},$ (2)

from which it follows that

$0\,\leq\,z\,\leq\,T^{+}x_{\alpha}\,\leq\,T^{+}x+|T(y\,\wedge\,x_{\alpha})|-Ty$
(3)

holds for all $\alpha\succeq\beta$ and all $0\leq y\leq x$. Now since for each
fixed vector
$0\leq y\leq x$ we have
$y\,\wedge\,x_{\alpha}\downarrow_{\alpha\,\succeq\beta}0$ in $F$, it then
follows from our hypothesis that
$\inf\left\\{\|T(y\,\wedge\,x_{\alpha})\|\right\\}\,=\,0$, and hence from (3)
we see that $0\,\leq\,\|z\|\,\leq\,\|T^{+}x-Ty\|$ holds for all $0\leq y\leq
x$. In view of $T^{+}x\,=\,\sup\left\\{Ty:\,\,0\,\leq\,y\,\leq\,x\right\\}$
and by order continuity of $G$, the latter inequality yields $z\,=\,0$, as
desired.
$(4)\Rightarrow(5)$ The implication follows from the identity
$|T|\,=\,T^{+}+T^{-}$.
$(5)\Rightarrow(1)$ The implication follows easily from the lattice inequality
$|Tx|\,\leq\,|T||x|$.

###### Theorem 3.2

If $E$ and $G$ two Riesz spaces with $G$ Dedekind complete and $G$ has order
continuous norm Banach lattice, the following assertions are true.

  1. i)

The set of all order bounded $\tilde{o}n$-continuous operators from $E$ into
$G$ is a band of $L_{b}(E,G)$.

  2. ii)

The set of all order bounded $\sigma$-$\tilde{o}n$-continuous operators from
$E$ into $G$ is a band of $L_{b}(E,G)$.

###### Proof

  1. i)

Let $T:E\to G$ be an order bounded $\tilde{o}n$-continuous operator. Note that
if $|S|\,\leq\,|T|$ holds in $L_{b}(E,G)$, then by Theorem 2, $S$ is order
bounded and $\tilde{o}n$-continuous operator. That is, the set of all order
bounded $\tilde{o}n$-continuous operators from $E$ into $G$ is an ideal of
$L_{b}(E,G)$. Let $0\,\leq\,T_{\lambda}\uparrow T$ in $L_{b}(E,G)$ that
$T_{\lambda}$ is an $\tilde{o}n$-continuous operator for all $\lambda$, and
let $(x_{\alpha})\subseteq E$ that $0\,\leq\,x_{\alpha}\uparrow x$ in $F$.
Then for each fixed index $\lambda$ we have

$0\,\leq\,T(x-x_{\alpha})\,\leq\,(T-T_{\lambda})(x)+T_{\lambda}(x-x_{\alpha}),$
(4)

and $x-x_{\alpha}\downarrow 0$, in conjunction with $T_{\lambda}$ is an
$\tilde{o}n$-continuous operator, implies

$0\,\leq\,\inf_{\alpha}\left\\{\|T(x-x_{\alpha})\|\right\\}\,\leq\,\|(T-T_{\lambda})(x)\|$
(5)

for all $\lambda$. From $T-T_{\lambda}\downarrow 0$ and by order continuity of
$G$, we see that $\inf_{\alpha}\left\\{\|T(x-x_{\alpha})\|\right\\}\,=\,0$.
Thus, $T$ is $\tilde{o}n$-continuous operator, and the proof follows.

  2. ii)

The proof is similar to (i).

###### Theorem 3.3

Let $T:E\to G$ be an order bounded operator. Then the following assertions are
true.

  1. i)

If $G$ is Archimedean vector lattice and $T$ is a preserves disjointness and
$\tilde{o}n$-continuous, then $|T|$ exists and $|T|\in L_{\tilde{o}n}(E,G)$.

  2. ii)

If $E$ is a band in $F$, $G$ is an atomic with order continuous norm Banach
lattice and $T:E\to G$ is $\sigma$-$\tilde{o}n$-continuous, then $|T|$ exists
and $|T|\in L_{\tilde{o}n}^{\sigma}(E,G)$.

###### Proof

  1. i)

Let $(x_{\alpha})\subseteq E$ be a $\tilde{o}$-null net. By assumption we have
$Tx_{\alpha}\xrightarrow{\|.\|}0$. By Theorem 2.40 of 1 , $|T|$ exists and
$|T||x|=|T|x||=|Tx|$ for all $x\in E$. Since,
$||T|x_{\alpha}|\leq|T||x_{\alpha}|\xrightarrow{\|.\|}0$, therefore,
$|T|x_{\alpha}\xrightarrow{\|.\|}0$. Hence $|T|\in L_{\tilde{o}n}(E,G)$.

  2. ii)

Let $(x_{n})\subseteq E$ and $x_{n}\xrightarrow{o}0$ in $E$. It is clear that
$x_{n}\xrightarrow{Fo}0$ in $E$. By assumption, $Tx_{n}\xrightarrow{\|.\|}0$
in $G$. It is obvious that, $(x_{n})$ is order bounded and therefore,
$(Tx_{n})$ is order bounded. Hence by Lemma 5.1 of 55 ,
$Tx_{n}\xrightarrow{o}0$ in $G$. Hence $T$ is an $\sigma$-order continuous
operator. Note that since $G$ has order continuous norm, therefore, it is a
Dedekind complete. So by Theorem 1.56 of 1 , $|T|$ exists and it is an
$\sigma$-order continuous. Let $(x_{n})\subseteq E$ be a $\tilde{o}$-null net.
We have $|x_{n}|=P_{E}|x_{n}|\leq P_{E}(y_{m})$. We have
$x_{n}\xrightarrow{o}0$ in $E$. By assumption, $|T|x_{n}\xrightarrow{o}0$ in
$G$. Because $G$ has order continuous norm, therefore,
$|T|x_{n}\xrightarrow{\|.\|}0$ in $G$. Hence $|T|\in
L_{\tilde{o}n}^{\sigma}(E,G)$

###### Corollary 2

By part 2 of Theorem 3.3, if $E$ is a band in $F$, $G$ is an atomic with order
continuous norm Banach lattice. It follows that $T:E\to G$ is a
$\sigma$-$\tilde{o}n$-continuous operator if and only if it is a
$\sigma$-order continuous operator. Therefore, by Theorem 1.57 of 1 ,
$L_{\tilde{o}n}^{\sigma}(E,G)$ is a band of $L_{b}(E,G)$.

##  4 $\tilde{o}$rder weakly compact operator

A continuous operator $T:E\rightarrow X$ is said to be $\tilde{o}$rder weakly
compact (or, $\tilde{o}$-weakly compact for short), if $A\subseteq E$ is
$Fo$-bounded in $E$, then $T(A)$ in $X$ is a relatively weakly compact set.
The collection of all $\tilde{o}$-weakly compact operators from vector lattice
$E$ into Banach space $X$ will be denoted by $W_{\tilde{o}}(E,X)$.

A subset $A$ in a Banach lattice $E$ is $F$-almost order bounded if for any
$\epsilon>0$ there exists $u\in F^{+}$ such that $A\subseteq[-u,u]+\epsilon
B_{E}$.

As following remark, very weakly compact operator $T:E\rightarrow X$ is a
$\tilde{o}$-weakly compact operator, the converse holds whenever $E$ has an
order unit.

###### Remark 3

  1. i)

Let $T:E\rightarrow X$ be an weakly compact operator. If $A$ is a $F$-order
bounded set in $E$, so it is norm bounded set. Since $T$ is a weakly compact
operator, $T(A)$ is relatively weakly compact set in $X$. It means that $T$ is
a $\tilde{o}$-weakly compact operator.

  2. ii)

Let $E$ has an order unit and $T:E\to X$ is a $\tilde{o}$-weakly compact
operator. If $A$ is a norm bounded set in $E$, then it is order bounded set in
$E$ and therefore, it is $F$-order bounded. By assumption, $T(A)$ is
relatively weakly compact set in $X$. It means that $T$ is a weakly compact
operator.

###### Proposition 2

If $E$ has order continuous norm with property $(F)$, then the identity
operator $I:E\to E$ is $\tilde{o}$-weakly compact

###### Proof

Let $A\subseteq E$ be a $Fo$-bounded set. Since $E$ has the property $(F)$,
therefore, $E$ is an order bounded set in $E$. It is obvious that $A$ is
almost order bounded in $E$. Since $E$ has order continuous norm, $A$ is a
relatively weakly compact set in $E$ by Theorem 4.9(5) and Theorem 3.44 of 1 .
Hence $I(A)$ is a relatively weakly compact set in $E$. It means that $I$ is a
$\tilde{o}$-weakly compact operator.

###### Theorem 4.1

An operator $T:E\to X$ is $\tilde{o}$-weakly compact if and only if for each
disjoint and $Fo$-bounded sequence $(x_{n})\subseteq E$ implies
$Tx_{n}\xrightarrow{\|.\|}0$.

###### Proof

Let $T:E\to X$ be a $\tilde{o}$-weakly compact operator. Therefore, for each
$u\in F^{+}$, $T([-u,u])$ is relatively weakly compact. Let $I_{u}$ be the
ideal generated by $u$ in $E$. The operator $T|_{I_{u}}:I_{u}\to X$ is weakly
compact operator. Since $I_{u}$ is an $AM$-space with order unit, therefore,
$T\arrowvert_{I_{u}}:I_{u}\to F$ is $M$-weakly. Hence for each disjoint norm
bounded sequence $(x_{n})\subseteq I_{u}$, we have
$Tx_{n}\xrightarrow{\|.\|}0$.

Conversely, let $A\subseteq E$ be a $Fo$-bounded set. Then there exists $u\in
F^{+}$ such that $A\subseteq[-u,u]$. Let $I_{u}$ be the ideal generated by $u$
in $E$ and $(x_{n})\subseteq A$ be a disjoint sequence. It is clear that
$(x_{n})$ is norm bounded. By assumption, we have $Tx_{n}\xrightarrow{\|.\|}0$
in $F$. Therefore, $T:I_{u}\to X$ is $M$-weakly compact and by Theorem 5.61 of
1 , $T:I_{u}\to X$ is a weakly compact operator. Let $A\subseteq E$ be a
$Fo$-bounded set. Then there exists $u\in F^{+}$ that $A$ is norm bounded in
$I_{u}$ and $T:I_{u}\to X$ is weakly compact. Therefore, $T(A)$ is a
relatively weakly compact in $X$. So $T:E\to X$ is a $\tilde{o}$-weakly
compact operator.

###### Corollary 3

  1. i)

Let $T,S$ be two operators that $0\leq T\leq S$ and $S$ is a
$\tilde{o}$-weakly compact operator. If $(x_{n})\subseteq E$ is disjoint and
$Fo$-bounded, then by Theorem 4.1, $Sx_{n}\xrightarrow{\|.\|}0$. It follows
that $Tx_{n}\xrightarrow{\|.\|}0$. So $T$ is a $\tilde{o}$-weakly compact
operator.

  2. ii)

Let $T,S$ be two $\tilde{o}$-weakly compact operators. By Theorem 4.1, it is
clear that $S\circ T$ is a $\tilde{o}$-weakly compact operator.

It is obvious that, if $T:E\to X$ is a $\tilde{o}$-weakly compact operator,
then is order weakly compact. By following example the converse is not true in
general.

###### Example 3

The operator $T:\ell^{1}\rightarrow\ell^{\infty}$ defined by

$T(x_{1},x_{2},\ldots)=(\sum_{i=1}^{\infty}x_{i},\sum_{i=1}^{\infty}x_{i},\ldots)$

is an order weakly compact operator. Let $(x_{n})\subseteq\ell^{1}$ be
disjoint order bounded sequence. We have $x_{n}\xrightarrow{uo}0$ and order
bounded. Therefore, $x_{n}\xrightarrow{o}0$. Since $\ell^{1}$ has order
continuous norm, therefore, $(x_{n})$ is norm-null. Because $T$ is a
continuous operator, hence $Tx_{n}\xrightarrow{\|.\|}0$ in $\ell^{\infty}$. So
by Theorem 5.57 of 1 , $T$ is an order weakly compact operator.
If we consider $(e_{n})\subseteq\ell^{1}$ we have
$e_{n}\xrightarrow{\ell^{\infty}o}0$ in $\ell^{1}$. On the other hand,
$Te_{n}=(1,1,1,\ldots)$, therefore, $(Te_{n})$ is not convergent to zero in
norm topology. Thus, $T$ is not $\tilde{o}$-weakly compact.

###### Theorem 4.2

Let $G$ be a normed vector lattice that is a sublattice of normed vector
lattice $H$ and $T:E\to G$ be a $\tilde{o}$-weakly compact operator. By one of
the following conditions, the modulus of $T$ exists and it is a
$\tilde{o}$-weakly compact operator.

  1. i)

$E$ is an $AL$-space and $X$ has the property $(P)$ and the $H$-property.

  2. ii)

$E$ and $G$ have an order unit.

  3. iii)

$G$ is Archimedean Dedekind complete and $T$ is an order bounded preserves
disjointness.

###### Proof

  1. i)

Let $(x_{n})\subseteq E$ be a disjoint order bounded sequence. It is obvious
that $(x_{n})$ is $Fo$-bounded. By assumption and Theorem 4.1,
$Tx_{n}\xrightarrow{\|.\|}0$. Hence by Theorem 5.57 of 1 , $T$ is an order
weakly compact operator. Since $E$ is an $AL$-space and $G$ has the property
$(b)$, by Theorem 2.2 of 11 , $|T|$ exists and is an order weakly compact
operator. Since $G$ has the $H$-property, $|T|$ is a $\tilde{o}$-weakly
compact operator.

  2. ii)

Let $A$ be a norm bounded set in $E$. Since $E$ has an order unit, therefore,
$A$ is order bounded and so is $Fo$-bounded. By assumption, $T(A)$ is a
relatively weakly compact set in $G$. Hence $T$ is a weakly compact operator.
Since $G$ has an order unit, therefore, by Theorem 2.3 of 15 , the modulus of
$T$ exists and it is a weakly compact operator. It is obvious that $|T|$ is a
$\tilde{o}$-weakly compact operator.

  3. iii)

By Theorem 2.40 of 1 , $|T|$ exists and we have $|T||x|=|T|x||=|Tx|$ for all
$x$. If $(x_{n})\subseteq E$ is a $Fo$-bounded disjoint sequence, then by
assumption $Tx_{n}\xrightarrow{\|.\|}0$. We have
$|T||x_{n}|=|T|x_{n}||=|Tx_{n}|\xrightarrow{\|.\|}0$ in $G$ for each $n$. Now
by inequality $||T|x_{n}|\leq|T||x_{n}|$, we have
$|T|x_{n}\xrightarrow{\|.\|}0$. Hence $|T|$ is a $\tilde{o}$-weakly compact
operator.

The following examples shows that $\tilde{o}$-weakly compact operators do not
have the duality property.

###### Example 4

  1. i)

Consider the operator $T:C[0,1]\rightarrow c_{0}$, given by

$T(f)=(\int_{0}^{1}f(x)\sin xdx,\int_{0}^{1}f(x)\sin 2xdx,...).$

By Example 3.15 of 44 , $T$ is a $wun$-Dunford-Pettis and by Theorem 3.11 of
44 , $T$ is a weakly compact operator. Therefore, $T$ is a $\tilde{o}$-weakly
compact operator. We have $T^{*}:\ell^{1}\to(C[0,1])^{*}$ that

$T^{*}x_{n}(f)=\sum_{n=1}^{\infty}x_{n}(\int_{0}^{1}f(t)sinntdt).$

Note that $(e_{n})\subseteq\ell^{1}$ is $\ell^{\infty}$-order bounded and
disjoint. Put $f_{n}(t)=sinnt$ for all $n$. We have

$\|T^{*}e_{n}\|\geq\|T^{*}e_{n}(f_{n})\|=\int_{0}^{1}(sinnt)^{2}dt\nrightarrow
0.$

Thus by Theorem 4.1, $T^{*}$ is not a $\tilde{o}$-weakly compact operator.

  2. ii)

Consider the functional $f:\ell^{1}\rightarrow\mathbb{R}$ defined by

$f(x_{1},x_{2},...)=\sum_{1}^{\infty}x_{i}.$

$(e_{n})\subseteq\ell^{1}$ is $\ell^{\infty}$-order bounded and disjoint while
$f(e_{n})\nrightarrow 0$. Therefore, by Theorem 4.1, $f$ is not a
$\tilde{o}$-weakly compact operator, but it is obvious that
$f^{*}:\mathbb{R}\to\ell^{\infty}$ is a $\tilde{o}$-weakly compact operator.

In the following, under some conditions, we show that if an operator $T$ is
$\tilde{o}$-weakly compact, then its adjoint $T^{*}$ is also
$\tilde{o}$-weakly compact and vice versa.

###### Proposition 3

Let $G$ be a vector lattice such that $G^{*}\subseteq F$. Then the following
assertions are true.

  1. i)

If $E$ has an order unit and $T:E\to G$ is $\tilde{o}$-weakly compact, then
$T^{*}$ is $\tilde{o}$-weakly compact.

  2. ii)

If $G^{*}$ has an order unit and $T^{*}:G^{*}\to E^{*}$ is $\tilde{o}$-weakly
compact, then $T$ is $\tilde{o}$-weakly compact.

###### Proof

  1. i)

Let $E$ has an order unit and $T:E\to G$ is a $\tilde{o}$-weakly compact
operator. It is obvious that $T$ is a weakly compact operator. By Theorem 5.23
of 1 , $T^{*}$ is a weakly compact operator and therefore, $T^{*}$ is a
$\tilde{o}$-weakly compact operator.

  2. ii)

The proof is similar to (i)).

###### Theorem 4.3

Let $T:F\to X$ be an operator. $T\mid_{E}:E\to X$ is $\tilde{o}$-weakly
compact if and only if $T(A)$ is relatively weakly compact for each $F$-almost
order bounded set $A\subseteq E$.

###### Proof

If $T(A)$ is relatively weakly compact for each $F$-almost order bounded
subset $A$ of $E$, it is obvious that $T\mid_{E}$ is $\tilde{o}$-weakly
compact.

Conversely, let $A\subseteq E$ be a $F$-almost order bounded. Therefore, there
exists a $u\in F^{+}$ that $A\subseteq[-u,u]+\varepsilon B_{E}$. It is obvious
that $T(A)\subseteq T[-u,u]+\varepsilon T(B_{E})$. Since $T$ is
$\tilde{o}$-weakly compact, hence $T[-u,u]$ is a relatively weakly compact. By
Theorem 3.44 of 1 , $T(A)$ is a relatively weakly compact in $X$.

## References

  * (1) C. D. Aliprantis and O. Burkinshaw, Positive operators. vol. 119. Springer Science, Business Media (2006). $\textbullet$
  * (2) Ş. Alpay and B. Altinand C. Tonyali, On property (b) of vector lattice.  Positivity 7 (1) (2003) 135-139.
  * (3) B. Aqzzouz and J. H’michane, Some results on order weakly compact operators. Mathematica Bohemica, 134 (4) (2009) 359–367.
  * (4) K. Haghnejad Azar, A generalization of order convergence in the vector lattices, Facta Universitis (NIˇS) Ser. Math. Inform. 37 (3 ) (2022) 521–528.
  * (5) Y. Deng and M. O’Brien and V. G. Troitsky, Unbounded norm convergence in Banach lattices. Positivity. 21 (2017) 963–974.
  * (6) N. Gao and V.g. Troitsky and F. Xanthos, uo-Convergence and its applications to Cesàro means in Banach lattices.  Isr. J. Math.  220 (2017) 649–689.
  * (7) N. Gao and F. Xanthos, Unbounded order convergence and application to martingales without probability. J. Math. Anal. Appl. 415 (2014) 931–947.
  * (8) S. A. Jalili, K. Haghnejad Azar and M. B. Farshbaf Moghimi, Order-to-topology continuous operators, Positivity 25 (2021) 1313–1322.
  * (9) K. Haghnejad Azar, M. Matin and R. Alavizadeh, Unbounded order-norm continuous and unbounded norm continuous operators, Filomat 35 (13) (2021) 4417–4426.
  * (10) K. Haghnejad Azar, M. Matin and R. Alavizadeh, Weakly Unbounded Norm Topology and $wun$-Dunford-Pettis Operators, To appear in Rendiconti del Circolo Matematico di Palermo Series 2, (2022).
  * (11) K. D. Schmidt, On the modulus of weakly compact operators and strongly additive vector measures, Proc. Amer. Math. Soc.  102 (4) (1988) 862-866.