Unbounded $\sigma$-order-to-norm continuous and $un$-continuous operators

TL;DR

This paper investigates classes of operators between vector and normed lattices that preserve unbounded order and unbounded norm convergence, exploring their properties and relationships.

Contribution

It introduces and analyzes unbounded $\sigma$-order-to-norm continuous and unbounded norm continuous operators, highlighting their properties and connections to existing operator classes.

Findings

Characterization of unbounded $\sigma$-order-to-norm continuous operators.

Relationship between unbounded norm continuous and other operator classes.

Properties and structural results of these operators.

Abstract

An operator $T $ from a vector lattice $E$ into a normed lattice $F$ is called unbounded $\sigma$-order-to-norm continuous whenever $x_{n}\xrightarrow{uo}0$ implies $\| Tx_{n}\|\rightarrow 0$, for each sequence $(x_{n})_n\subseteq E$. For a net $(x_{\alpha})_{\alpha}\subseteq E$, if $x_{\alpha}\xrightarrow{un}0$ implies $Tx_{\alpha}\xrightarrow{un}0$, then $T$ is called an unbounded norm continuous operator. In this manuscript, we study some properties of these classes of operators and their relationships with the other classes of operators.

Equations

$$T(x_1,x_2,\dots )=(\sum _{i=1}^{\infty }{x}_i,\sum _{i=1}^{\infty }{x}_i,\dots )$$

$$(1)\Rightarrow (2)\Rightarrow (3)\Rightarrow (4).$$

$$L_{un}(E,F)=\{T\in L(E,F):T\text{is unbounded norm continuous }\}.$$

$$L_{un}^{\sigma }(E,F)=\{T\in L(E,F):T\text{is σ-unbounded norm continuous }\}.$$

$$T(x_1,x_2,\dots )=(\sum _{i=1}^{\infty }{x}_i,\sum _{i=1}^{\infty }{x}_i,\dots ),$$

Full text

1 Introduction and Preliminaries

Let $E$ be a vector lattice and $F$ be a normed lattice. In the second section of this manuscript, we will study and investigate on operators $T:E\rightarrow F$ which carry every unbounded order convergent sequence into norm convergent sequence. The collection of all unbounded $\sigma$-order-to-norm continuous operators will be denoted by $L_{uon}^{\sigma}(E,F)$. We show that under some conditions the modulus of an unbounded $\sigma$-order-to-norm continuous operator exists and belongs to $L_{uon}^{\sigma}(E,F)$ and every $\sigma$-$uon$-continuous operator $T$ is a Dunford-Pettis operator. In the third section, we will introduce a new classification of operators named as $un$-continuous operators and we will investigate on some properties of them and their relationships with other classifications of operators.

To state our results, we need to fix some notations and recall some definitions. Throughout this paper, $E$ is a vector lattice, the subset $E^{+}=\{x\in E:x\geq 0\}$ is called the positive cone of $E$ and the elements of $E^{+}$ are called the positive elements of $E$. A subset $A\subseteq E$ is called order bounded if there exists $a,b\in E$ such that $A\subseteq[a,b]$ where $[a,b]=\{x\in E:~{}a\leq x\leq b\}$. An operator $T:E\rightarrow F$ between two vector lattices is said to be order bounded if it maps order bounded subsets of $E$ to order bounded subsets of $F$. If $E$ is a normed space, then $E^{\prime}$ is the topological dual space of $E$ and $T^{\prime}:F^{\prime}\to E^{\prime}$ is the adjoint of operator $T:E\to F$ between two normed space. A sequence $(x_{n})_{n}$ in a vector lattice $E$ is said to be disjoint whenever $n\neq m$ implies $x_{n}\perp x_{m}$. If $A$ is a nonempty subset of vector lattice $E$, then its disjoint complement $A^{d}$ is defined by $A^{d}=\{x\in E:\ x\perp y$ for all $y\in A\}$. An order closed ideal of $E$ is referred to as a band. A band $B$ in a vector lattice $E$ that satisfies $E=B\oplus B^{d}$ is referred to as a projection band. Let $B$ be a projection band in a vector lattice $E$. Thus every vector $x\in E$ has a unique decomposition $x=x_{1}+x_{2}$, where $x_{1}\in B$ and $x_{2}\in B^{d}$. Then it is easy to see that a projection $P_{B}:E\rightarrow E$ is defined via the formula $P_{B}(x)=x_{1}$. Clearly, $P_{B}$ is a positive projection. Any projection of the form $P_{B}$ is called a band projection.

Let $E$ be a vector lattice and $x\in E$. A net $(x_{\alpha})_{\alpha\in A}\subset E$ is said to be:

• •

order convergent to $x$ if there is a net $(z_{\beta})_{\beta\in B}$ in $E$ such that $z_{\beta}\downarrow 0$ and for every $\beta\in B$, there exists $\alpha_{0}\in A$ such that $|x_{\alpha}-x|\leq z_{\beta}$ whenever $\alpha\geq\alpha_{0}$. We denote this convergence by $x_{\alpha}\xrightarrow{o}x$ and write that $(x_{\alpha})_{\alpha}$ is $o$-convergent to $x$.

• •

unbounded order convergent to $x$ 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 $(x_{\alpha})_{\alpha}$ is $uo$-convergent to $x$. It was first introduced by Nakano in 14 and was later used by DeMarr in 4 .

• •

unbounded norm convergent to $x$ if $\||x_{\alpha}-x|\wedge u\|\rightarrow 0$ for all $u\in E^{+}$. We denote this convergence by $x_{\alpha}\xrightarrow{un}x$ and write that $(x_{\alpha})_{\alpha}$ is $un$-convergent to $x$. It was studied in 5 ; 12 .

It is clear that for order bounded nets, $uo$-convergence is equivalent to $o$-convergence. By Corollary 3.6 of 8 , every disjoint sequence in vector lattice $E$ is $uo$-null. In 15 , Wickstead characterized the spaces in which weak convergence of nets implies $uo$-convergence and vice versa and in 7 , Gao characterized the spaces $E$ such that in its dual space $E^{\prime}$, $uo$-convergence implies weak*∗*-convergence and vice versa.

Let $X$ and $Y$ be two Banach spaces and let $E$ and $F$ be two Banach lattices. An operator,

• •

$T:X\rightarrow Y$, is said to be Dunford-Pettis (or that $T$ has the Dunford-Pettis property) whenever $x_{n}\xrightarrow{w}0$ in $X$ implies $\|Tx_{n}\|\rightarrow 0$.

• •

$T:E\rightarrow X$, is said to be $M$-weakly compact if $T$ is continuous and $\lim\|Tx_{n}\|=0$ holds for every norm bounded disjoint sequence $(x_{n})_{n}$ of $E$.

• •

$T:E\rightarrow F$, is said to be disjointness preserving whenever $x\perp y$ in $E$ implies $Tx\perp Ty$ in $F$.

If for an operator $T:E\rightarrow F$ between two vector lattices, $T\vee(-T)$ exists we say its modulus $|T|$ exists. $e\in E^{+}$ is a strong unit when the ideal $I_{e}$ (generated by $e$) is equal to $E$. Equivalently, for every $x\geq 0$ there exists $n\in\mathbb{N}$ such that $x\leq ne$. $e\in E^{+}$ is also a quasi-interior point if the closure of $I_{e}$ equal with $E$; or equivalently, $x\wedge ne\xrightarrow{\|.\|}x$ for every $x\in E^{+}$. A non-zero element $a\in E^{+}$ is an atom, if $x\perp y$ and $x,y\in[0,a]$ imply either $x=0$ or $y=0$. $E$ is called an atomic Banach lattice if it is the band generated by its atoms. A Banach lattice $E$ is said to be $KB$-space whenever every increasing norm bounded sequence of $E^{+}$ is norm convergent. Recall that $E^{\sim}$ is the vector space of all order bounded linear functionals on $E$ and $E_{n}^{\sim}$ is the vector space of all order continuous linear functional on $E$. A sublattice $Y$ of a vector lattice $E$ is said to be regular if for every subset $A$ of $Y$, infimum of $A$ is the same in $E$ and in $Y$, whenever infimum of $A$ exists in $Y$. It is easy to see that $L_{p}[0,1]$ is an order dense regular sublattice of $L_{0}[0,1]$, whenever $0<p<\infty$. A vector lattice is called laterally complete whenever every subset of pairwise disjoint positive vectors has a supremum. For unexplained notation the reader is referred to 1 .

2 Unbounded $\sigma$-order-to-norm continuous operators

Let $E$ be a vector lattice and $F$ be a normed lattice. An operator $T:E\rightarrow F$ is said to be unbounded $\sigma$-order-to-norm continuous (or, $\sigma$-$uon$-continuous for short), if $x_{n}\xrightarrow{uo}0$ in $E$ implies $Tx_{n}\xrightarrow{\|.\|}0$ in $F$. The collection of all unbounded $\sigma$-order-to-norm continuous operators will be denoted by $L_{uon}^{\sigma}(E,F)$.

In the following example for $0\leq p\leq+\infty$, we show that every lattice homomorphism $T:L_{p}[0,1]\rightarrow\mathbb{R}$ is $\sigma$-$uon$-continuous operator.

Example 1

Let $0\leq p<\infty$ and let $T:L_{p}[0,1]\rightarrow\mathbb{R}$ be a lattice homomorphism, then $T$ is a $\sigma$-$uon$-continuous operator. First note that since $T$ is positive, $T$ is order bounded. Since $L_{p}[0,1]^{\sim}=L_{p}[0,1]^{\sim}_{n}$, $T$ is order continuous. If $p=0$, by Exercise 23 of page 214 from 1a , $L_{p}[0,1]$ is Dedekind complete and laterally complete. Therefore, by Theorem 3.2 of 12a , $T$ is $\sigma$-$uon$-continuous. Now let $0<p<\infty$, since $T$ is order continuous and lattice homomorphism, by Theorem 2.32 of 1 , there exists an extension $S$ of $T$ from $L_{0}[0,1]$ to $\mathbb{R}$, which is an order continuous lattice homomorphism. If $(x_{n})_{n}\subset L_{p}[0,1]$ and $x_{n}\xrightarrow{uo}0$ in $L_{p}[0,1]$, by Theorem 3.2 of 8 , $x_{n}\xrightarrow{uo}0$ in $L_{0}[0,1]$ and therefore by Theorem 3.2 of 12a , $x_{n}\xrightarrow{o}0$ in $L_{0}[0,1]$. Thus, $Tx_{n}=Sx_{n}\xrightarrow{o}0$ in $\mathbb{R}$. Hence $Tx_{n}\xrightarrow{\|.\|}0$ in $\mathbb{R}$.

Let $E$ be a vector lattice and $G$ and $F$ be two normed lattices. Then, it is obvious that for each $\sigma$-$uon$-continuous operator $T:E\rightarrow G$ and continuous operator $S:G\rightarrow F$, $ST:E\rightarrow F$ is a $\sigma$-$uon$-continuous operator. The following proposition shows that with some conditions, the combination of two $\sigma$-$uon$-continuous operators is $\sigma$-$uon$-continuous.

Proposition 1

Let $E$ be a vector lattice and $G$ and $F$ be two Banach lattices. If $E\xrightarrow{T}G\xrightarrow{S}F$ are $\sigma$-$uon$-continuous operators and the linear span of minimal ideals in $G$ is order dense in $G$, then $ST$ is likewise a $\sigma$-$uon$-continuous operator.

Proof

Let $(x_{n})_{n}\subseteq E$ and $x_{n}\xrightarrow{uo}0$ in $E$. Then by the assumption we have $Tx_{n}\xrightarrow{\|.\|}0$ in $G$, and so $Tx_{n}\xrightarrow{w}0$ in $G$. Now by Theorem 1 of 15 , we have $Tx_{n}\xrightarrow{uo}0$ in $G$. Therefore, $ST(x_{n})\xrightarrow{\|.\|}0$ in $F$.

Recall that an operator $T$ from vector lattice $E$ into normed vector lattice $F$ is said to be $\sigma$-order-to-norm continuous operator whenever $x_{n}\xrightarrow{o}0$ implies $Tx_{n}\xrightarrow{\|.\|}0$ for all sequence $(x_{n})_{n}\subseteq E$. This classification of operators has been introduce and studied by K. Haghnejad Azar, see 10 . Obviously, every $\sigma$-$uon$-continuous operator is a $\sigma$-order-to-norm continuous operator. However, the converse is not true in general, as shown in the following example.

Example 2

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

[TABLE]

is a $\sigma$-order-to-norm continuous operator ($\ell^{1}$ has order continuous norm and $T$ is a continuous operator). Now if $(e_{n})_{n}$ is the standard basis of $\ell^{1}$, then $e_{n}\xrightarrow{uo}0$ in $\ell^{1}$ and $T(e_{n})=(1,1,1,\ldots)$. Therefore, $\|T(e_{n})\|\nrightarrow 0$ in $\ell^{\infty}$. Thus $T$ is not $\sigma$-$uon$-continuous.

If a vector lattice $E$ is Dedekind $\sigma$-complete and laterally $\sigma$-complete, then every $\sigma$-order-to-norm continuous operator from $E$ into $F$ is $\sigma$-$uon$-continuous. Namely, if $(x_{n})_{n}$ is a sequence in $E$ such that $x_{n}\xrightarrow{uo}0$, then by Theorem 3.2 of 12a , $(x_{n})_{n}$ is order bounded and therefore $x_{n}\xrightarrow{o}0$ in $E$ and so $T(x_{n})\xrightarrow{\|.\|}0$ in $F$. Hence $T$ is a $\sigma$-$uon$-continuous operator.

Let $T:E\rightarrow F$ be a positive operator between two vector lattices. We say that an operator $S:E\rightarrow F$ is dominated by $T$ (or that $T$ dominates $S$) whenever $|Sx|\leq T|x|$ holds for each $x\in E$. Let $T$ be a $\sigma$-$uon$-continuous operator from vector lattice $E$ into normed lattice $F$. It is obvious that $S$ is $\sigma$-$uon$-continuous whenever $S$ is dominated by $T$.

Theorem 2.1

Let $E$ be a vector lattice and $F$ be a normed lattice. If $T:E\rightarrow F$ is an order bounded $\sigma$-$uon$-continuous operator, then by one of the following conditions, $|T|$ exists and belongs to $L_{uon}^{\sigma}(E,F)$.

1. $(1)$

$E$* is Dedekind $\sigma$-complete and laterally $\sigma$-complete and $F$ is atomic with order continuous norm.* 2. $(2)$

$T$* preserves disjointness.*

Proof

1. $(1)$

First we show that $T$ is a $\sigma$-order continuous operator. Let $(x_{n})_{n}\subseteq E$ and $x_{n}\xrightarrow{o}0$ in $E$. Then $x_{n}\xrightarrow{uo}0$ and by Theorem 3.2 of 12a , $(x_{n})_{n}$ is order bounded in $E$. Hence by the assumption, $(T(x_{n}))_{n}$ is order bounded and $T(x_{n})\xrightarrow{\|.\|}0$ in $F$. Now by Lemma 5.1 of 5 , $T(x_{n})\xrightarrow{o}0$ in $F$. Hence $T$ is a $\sigma$-order continuous operator. Note that by Theorem 4.10 of 1 , $F$ is Dedekind complete, therefore by using Theorem 1.56 of 1 , $|T|$ is a $\sigma$-order continuous operator. Now, assume that $(x_{n})_{n}\subseteq E$ and $x_{n}\xrightarrow{uo}0$ in $E$, since $E$ is Dedekind $\sigma$-complete and laterally $\sigma$-complete, $(x_{n})_{n}$ is order bounded. It follows that $x_{n}\xrightarrow{o}0$ in $E$ and $|T|(x_{n})\xrightarrow{o}0$ in $F$. Since $F$ has order continuous norm, we have $|T|(x_{n})\xrightarrow{\|.\|}0$. 2. $(2)$

It is obvious that $F$ is Archimedean. By Theorem 2.40 of 1 , $|T|$ exists and for all $x$, we have $|T|(|x|)=|T(|x|)|=|T(x)|$. If $(x_{n})_{n}\subseteq E$ and $x_{n}\xrightarrow{uo}0$, then for each $n$, $|T|(|x_{n}|)=|T(|x_{n}|)|=|T(x_{n})|\xrightarrow{\|.\|}0$ in $F$. Now by inequality $||T|(x_{n})|\leq|T||x_{n}|$, we have $|T|(x_{n})\xrightarrow{\|.\|}0$.

Recall that a vector lattice $E$ is said to be perfect whenever the natural embedding $x\mapsto\hat{x}$ from $E$ to $(E_{n}^{\sim})_{n}^{\sim}$ is one-to-one and onto. By Exercise 3 of page 74 of 1 , if $F$ is a perfect vector lattice, then $L_{b}(E,F)$ is likewise a perfect vector lattice for each vector lattice $E$.

Corollary 1

If the condition $(1)$ from above theorem holds, then

1. $(1)$

an order bounded operator $T:E\rightarrow F$ is a $\sigma$-order continuous operator if and only if it is $\sigma$-$uon$-continuous. It follows that $L_{uon}^{\sigma}(E,F)\cap L_{b}(E,F)$ is a band of $L_{b}(E,F)$. 2. $(2)$

if $F$ is a perfect vector lattice, then $L_{uon}^{\sigma}(E,F)\cap L_{b}(E,F)$ is likewise a perfect vector lattice for each vector lattice $E$.

Proof

1. $(1)$

Follows from Theorem 1.57 of 1 . 2. $(2)$

We will show that if $E$ is a perfect vector lattice and $B$ is a band of $E$, then $B$ is perfect vector lattice in its own right. Suppose $x,y\in B$ and $x\neq y$, by Theorem 1.71 of 1 , there exists $f\in E_{n}^{\sim}$ such that $f(x)\neq f(y)$. The restrection of $f$ to $B$ is order continuous and $f\mid_{B}(x)\neq f\mid_{B}(y)$. Therefore $B_{n}^{\sim}$ separates the points of $B$. Let $(x_{\alpha})_{\alpha}\subseteq B$, $0\leq x_{\alpha}\uparrow$ and $\sup\{f(x_{\alpha})\}<\infty$ for each $0\leq f\in B_{n}^{\sim}$. It is clear that $(x_{\alpha})_{\alpha}\subseteq E$ and $0\leq x_{\alpha}\uparrow$ in $E$. On the other hand, for each $f\in E_{n}^{\sim}$, $f\mid_{B}\in B_{n}^{\sim}$ and $f\mid_{B}(x_{\alpha})=f(x_{\alpha})$ for all $\alpha$, therefore $\sup\{|f(x_{\alpha})|\}<\infty$ for each $0\leq f\in E_{n}^{\sim}$. Thus by Theorem 1.71 of 1 , there exists some $x\in E$ satisfying $0\leq x_{\alpha}\uparrow x$. Since $B$ is a band of $E$, hence $x\in B$. Therefore by Theorem 1.71 of 1 , $B$ is a perfect vector lattice. Similarly $L_{uon}^{\sigma}(E,F)\cap L_{b}(E,F)$ is a perfect vector lattice.

Recall that a Banach space $E$ is said to be $AL$-space, if $\|x+y\|=\|x\|+\|y\|$ holds for all $x,y\in E^{+}$ with $x\wedge y=0$.

Theorem 2.2

Let $E$ be an $AL$-space, $F$ be a normed lattice and let $T:E\rightarrow F$ be a positive operator. Then for the following assertions:

1. $(1)$

$T$* is a $\sigma$-$uon$-continuous operator.* 2. $(2)$

T is a Dunford-Pettis operator. 3. $(3)$

for every relatively weakly compact net $(x_{\alpha})_{\alpha}\subseteq E$, $x_{\alpha}\xrightarrow{w}0$ implies $Tx_{\alpha}\xrightarrow{\|.\|}0$. 4. $(4)$

For every relatively weakly compact net $(x_{\alpha})_{\alpha}\subseteq E$, $x_{\alpha}\xrightarrow{uo}0$ implies $Tx_{\alpha}\xrightarrow{\|.\|}0$.

We have

[TABLE]

Proof

1. $(1)\Rightarrow(2)$ Let $(x_{n})_{n}\subseteq E$ be a norm bounded and disjoint sequence. Since $x_{n}\xrightarrow{uo}0$ in $E$, therefore $Tx_{n}\xrightarrow{\|.\|}0$ in $F$. Hence $T$ is a $M$-weakly compact operator. Now by Theorem 5.61 of 1 , $T$ is weakly compact and therefore by Theorems 5.85 and 5.82 of 1 , $T$ is a Dunford-Pettis operator.
2. $(2)\Rightarrow(3)$ Suppose $(x_{\alpha})_{\alpha}$ is relatively weakly compact and $x_{\alpha}\xrightarrow{w}0$. Suppose also $(Tx_{\alpha})_{\alpha}$ does not converge to 0 in norm. Then there exists $\epsilon>0$ such that for any $\alpha$, there exists $\beta(\alpha)\geq\alpha$ satisfying $\|Tx_{\beta(\alpha)}\|\geq\epsilon$. Thus by passing to the subnet $(Tx_{\beta(\alpha)})$, we may assume $\inf{\|Tx_{\alpha}\|}>0$. Since $0\in\overline{(x_{\alpha})}^{w}$, By Theorem 4.50 of 6 there exists a sequence $(y_{n})_{n}\subseteq\{x_{\alpha}:\alpha\}$ such that $y_{n}\xrightarrow{w}0$. The assumption $(2)$ now implies $Ty_{n}\xrightarrow{\|.\|}0$, which is a contradiction.
3. $(3)\Rightarrow(4)$ Suppose $(x_{\alpha})_{\alpha}$ is relatively weakly compact and $x_{\alpha}\xrightarrow{uo}0$. By Proposition 3.9 of 9 , $|x_{\alpha}|\xrightarrow{w}0$. Thus by assumption (3), $T|x_{\alpha}|\xrightarrow{\|.\|}0$. Now by inequality $|Tx_{\alpha}|\leq T|x_{\alpha}|$ we have $Tx_{\alpha}\xrightarrow{\|.\|}0$.

3 Unbounded norm continuous operators

An operator $T$ between two Banach lattices $E$ and $F$ is said to be unbounded norm continuous (or, $un$-continuous for short) whenever $x_{\alpha}\xrightarrow{un}0$ implies $Tx_{\alpha}\xrightarrow{un}0$, for $(x_{\alpha})_{\alpha}\subseteq E$. For sequence $(x_{n})_{n}\subseteq E$, if $x_{n}\xrightarrow{un}0$ implies $Tx_{n}\xrightarrow{un}0$, then $T$ is called a $\sigma$-unbounded norm continuous operator (or, $\sigma$-$un$-continuous for short). The collection of all unbounded norm continuous operators of $L(E,F)$ will be denoted by $L_{un}(E,F)$. That is,

[TABLE]

Similarly, $L_{un}^{\sigma}(E,F)$ will denote the collection of all operators from $E$ to $F$ that are $\sigma$-unbounded norm continuous. That is,

[TABLE]

Now in the following proposition, by using Theorem 5.3 from 5 and Theorem 2.3 from 12 , we show the relationship between the classifications of $\sigma$-unbounded norm continuous and unbounded $\sigma$-order-to-norm continuous operators.

Proposition 2

Let $E$ be a Banach lattice with order continuous norm and $F$ be a Banach lattice. Then we have the following assertions:

1. $(1)$

if $E$ is atomic, then $L_{uon}^{\sigma}(E,F)\subseteq L_{un}^{\sigma}(E,F)$. 2. $(2)$

if $F$ has a strong unit, then $L_{un}^{\sigma}(E,F)\subseteq L_{uon}^{\sigma}(E,F)$.

By using Theorem 4.3 of 12 , we also have the following proposition.

Proposition 3

Let $E$ and $F$ be two Banach lattices and Let $G$ be a sublattice of $E$ and $T\in L_{un}(E,F)$. Each of the following conditions implies that $T\in L_{un}(G,F)$.

1. $(1)$

$G$* is majorizing in $E$;* 2. $(2)$

$G$* is norm dense in $E$;* 3. $(3)$

$G$* is a projection band in $E$.*

The preceding proposition implies that if $T\in L_{un}(E^{\delta},F)$, then $T\in L_{un}(E,F)$ whenever $E$ and $F$ are Banach lattices and $E^{\delta}$ is a Dedekind completion of $E$. Deng, Brien and Troitsky in 5 , show that $un$-convergent is topological. For each $\epsilon>0$ and $x\in E^{+}$ the collections $V_{\epsilon,x}=\{y\in E:~{}\||y|\wedge x\|<\epsilon\}$ is a base of zero neighbourhoods for a topology, and convergence in this topology agrees with $un$-convergence. We will refer to this topology as $un$-topology. An operator $T:E\rightarrow F$ between two Banach lattices is $un$-continuous if and only if for each subset $A$ of $F$ that is $un$-open (resp, close), $T^{-1}(A)$ is $un$-open (resp, close) in $E$. Let $E,G$ and $F$ be Banach lattices. If $E\xrightarrow{T}G\xrightarrow{S}F$ are $un$-continuous operators, clearly $ST$ is likewise a $un$-continuous operator. Also, by using Theorem 2.40 of 1 , if $T:E\rightarrow F$ preserves disjointness and $un$-continuous operator, then $|T|$ exists and is a $un$-continuous operator. Now in the following, we give some examples of $un$-continuous operators.

Example 3

1. $(1)$

Let $B$ be a projection band of Banach lattice $E$ and $P_{B}$ the corresponding band projection. It follows easily from $0\leq P_{B}\leq I$(see Theorem 1.44 of 1 ) that if $x_{\alpha}\xrightarrow{un}0$ in $E$ then $P_{B}x_{\alpha}\xrightarrow{un}0$ in $B$. Therefore $P_{B}$ is a $un$-continuous operator. 2. $(2)$

Let $E$ and $F$ be two Banach lattices such that $E$ has a strong unit. Then, by Theorem 2.3 of 12 , each continuous operator $T:E\rightarrow F$ is $un$-continuous.

Recall that a topological space is said to be sequentially compact if every sequence has a convergent subsequence. An operator $T:E\rightarrow F$ between two Banach lattices is said to be (sequentially) $un$-compact if $TB_{E}$ ($B_{E}$ is closed unit ball of $E$) is relatively (sequentially) $un$-compact in $E$. Equivalently, for every bounded net $(x_{\alpha})_{\alpha}$ (respectively, every bounded sequence $(x_{n})_{n}$) its image has a subnet (respectively, subsequence), which is $un$-convergent. A net $(x_{\alpha})_{\alpha}$ is $un$-Cauchy if for every $un$-neighborhood $U$ of zero there exists $\alpha_{0}$ such that $x_{\alpha}-x_{\beta}\in U$ whenever $\alpha,\beta\geq\alpha_{0}$. The order continuous Banach lattice $E$ is $un$-complete if each $un$-Cauchy net $(x_{\alpha})_{\alpha}$ of $E$ is $un$-convergent to $x\in E$. These concepts have been introduce by Kandic, Marabeh and Troitsky, see 12 .

Clearly, every compact operator is both $un$-compact and sequentially $un$-compact. In general $un$-compact and sequentially $un$-compact operators are not $un$-continuous and vice versa as shown by the following example.

Example 4

1. $(1)$

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

[TABLE]

is clearly rank one, and so $T$ is a compact operator. It follows that $T$ is $un$-compact and sequentially $un$-compact. If $(e_{n})_{n}$ is the standard basis of $\ell^{1}$, by Proposition 3.5 of 12 , $e_{n}\xrightarrow{un}0$ in $\ell^{1}$. We have $T(e_{n})=(1,1,1,...)$, therefore $(T(e_{n}))_{n}$ is not $un$-convergent to 0. Hence $T$ is $un$-compact but is not a $un$-continuous operator. 2. $(2)$

Let $E=L_{1}[0,1]$. Clearly, the identity operator $I:E\rightarrow E$ is $un$-continuous. Since $E$ is a $KB$-space, by Theorem 6.4 of 12 , $B_{E}$ is $un$-complete. But since $E$ is not atomic, by Theorem 7.5 of 12 , $B_{E}$ is not $un$-compact. Hence $I$ is not $un$-compact.

A subset $A$ of Banach lattice $E$ is said to be $un$-bounded, if $A$ is bounded with respect to $un$-topology. An operator $T:E\rightarrow F$ between two Banach lattices is $un$-bounded if $T(A)$ is $un$-bounded in $F$ for each $un$-bounded subset $A$ of $E$.

The proof of the following lemma is similar to Theorem 1.15 (b) from 14a .

Lemma 1

Each $un$-compact subset of Banach lattice $E$ is $un$-bounded.

Also, we need the following lemma that is a specific case of Theorem 1.30 of 14a .

Lemma 2

Let $E$ be a Banach lattice and $A\subset E$. The following are equivalent:

1. $(1)$

$A$* is $un$-bounded.* 2. $(2)$

If $(x_{n})_{n}\subset A$ and $({\alpha}_{n})_{n}$ is a sequence of scalars such that $\alpha_{n}\rightarrow 0$ as $n\rightarrow\infty$, then ${\alpha}_{n}x_{n}\xrightarrow{un}0$ as $n\rightarrow\infty$.

Proposition 4

Let $T:E\rightarrow F$ be a $un$-continuous operator between two Banach lattices. Then we have the following assertions.

1. $(1)$

$T$* is a $un$-bounded operator.* 2. $(2)$

If $A\subseteq E$ is a $un$-compact set, then $T(A)$ is a $un$-compact set in $F$.

Proof

1. $(1)$

Let $A$ be a $un$-bounded subset of $E$. By Lemma 2 it is enough to show that for each $(Tx_{n})_{n}\subset T(A)$ and $({\alpha}_{n})_{n}\subset\mathbb{R}$ such that ${\alpha}_{n}\rightarrow 0$ as $n\rightarrow\infty$ we have ${\alpha}_{n}Tx_{n}\xrightarrow{un}0$ as $n\rightarrow\infty$. Let $(x_{n})_{n}\subset A$ and $({\alpha}_{n})_{n}\subset\mathbb{R}$ such that ${\alpha}_{n}\rightarrow 0$ as $n\rightarrow\infty$. It follows from $un$-boundedness of $A$ and Lemma 2 that ${\alpha}_{n}x_{n}\xrightarrow{un}0$ as $n\rightarrow\infty$. Since $T$ is $un$-continuous we have ${\alpha}_{n}T(x_{n})=T({\alpha}_{n}x_{n})\xrightarrow{un}0$. Therefore, the proof is complete. 2. $(2)$

Let $(y_{\alpha})_{\alpha}$ be a net in $T(A)$. Then there exists net $(x_{\alpha})_{\alpha}\subseteq A$ such that $T(x_{\alpha})=y_{\alpha}$ for all $\alpha$. It follows from $un$-compactness of $A$ that there exists a subnet $(x_{\alpha_{\beta}})$ of $(x_{\alpha})_{\alpha}$ with $x_{\alpha_{\beta}}\xrightarrow{un}x$ in $E$. Thus, $T(x_{\alpha_{\beta}})\xrightarrow{un}Tx$ in $F$. Hence $T(A)$ is a $un$-compact set.

Proposition 5

Assume that $E$ and $F$ are two Banach lattices and $E$ has quasi-interior point. $T:E\rightarrow F$ is a $un$-bounded operator if and only if $T$ is $\sigma$-$un$-continuous.

Proof

It follows from Proposition 4 that $T$ is a $un$-bounded operator whenever $T$ is a $\sigma$-$un$-continuous.

Conversely, let $(x_{n})_{n}\subseteq E$ and $x_{n}\xrightarrow{un}0$. By Theorem 3.2 from 12 , $E$ is metrizable, and so by Theorem 1.28(b) of 14a , there exists a sequence $(\alpha_{n})\subset\mathbb{R}^{+}$ such that $\alpha_{n}\rightarrow+\infty$ and $\alpha_{n}x_{n}\xrightarrow{un}0$. Obviously, $(\alpha_{n}x_{n})$ is $un$-bounded. Therefore $(T(\alpha_{n}x_{n}))$ is $un$-bounded. By Lemma 2 we have $Tx_{n}=\frac{1}{\alpha_{n}}T(\alpha_{n}x_{n})\xrightarrow{un}0$. Therefore, $T$ is $\sigma$-$un$-continuous.

There is a $un$-compact operator which is not $un$-bounded. Let $T:\ell^{1}\rightarrow\ell^{\infty}$ be the $un$-compact operator in Example 4. But, since $\ell^{1}$ has quasi-interior point and $T$ is not a $\sigma$-$un$-continuous operator, by Proposition 5, $T$ is not $un$-bounded.

Proposition 6

Assume that $E$ and $F$ are two Banach lattices.

1. $(1)$

Let $F$ has a strong unit and let $T:E\rightarrow F$ be a sequentially $un$-compact operator. Then $T^{\prime}$ is both sequentially $un$-compact and $un$-compact. 2. $(2)$

If $T^{\prime}:F^{\prime}\rightarrow E^{\prime}$ is a sequentially $un$-compact operator and $E^{\prime}$ has a strong unit, then $T$ is both sequentially $un$-compact and $un$-compact.

Proof

1. $(1)$

Let $(x_{n})_{n}$ be a bounded sequence in $E$. Then by the assumption $(T(x_{n}))_{n}$ has a subsequence, which is $un$-convergent. Now by Theorem 2.3 of 12 , $(T(x_{n}))_{n}$ is norm-convergent and therefore $T$ is a compact operator. Now by Theorem 5.2 of 1 , $T^{\prime}$ is compact and therefore it is both sequentially $un$-compact and $un$-compact. 2. $(2)$

Proof is similar to (1).

Recall that, for every ideal $I$ of a vector lattice $E$, the vector space $\dfrac{E}{I}$ is a vector lattice (see page 99 from 1 ).

Theorem 3.1

Let $T:E\rightarrow F$ be an operator between two Banach lattices and $T(E^{+})=F^{+}$. If $\ker(T)$ is an ideal of $E$, then $T$ is $un$-continuous.

Proof

At first it is clear that $T$ is surjective. On the other hand, as kernel of $T$ is an ideal of $E$, by Theorem 2.22 of 1 , the quotient vector space $\frac{E}{\ker(T)}$ is a vector lattice and the operator $S:E\rightarrow\frac{E}{\ker(T)}$ defined by $S(x)=x+\ker(T)$ is a Riesz homomorphism. Now we define the operator $K:\frac{E}{\ker(T)}\rightarrow F$ via $K(x+\ker(T))=Tx$. It is clear that $K$ is well defined, one-to-one operator and $T=KS$. Let $y\in F$ and $x\in E$ such that $y=T(x)=K(x+\ker(T))$. It follows that $K$ is a surjective operator. Now if $x+\ker(T)\geq 0$, then $|x+\ker(T)|=x+\ker(T)$. On the other hand, since $S$ is Riesz homomorphism, by Theorem 2.14 of 1 , we have $S(|x|)=|S(x)|$. Therefore $|x|+\ker(T)=|x+\ker(T)|=x+\ker(T)$ and hence $K(x+\ker(T))=K(|x|+\ker(T))=T(|x|)\geq 0$. Thus $K$ is a positive operator. Let $y\in F^{+}$. Since $T(E^{+})=F^{+}$, it follows that there exists $x\in E^{+}$ that $y=T(x)=K(x+\ker(T))$. It follows from $K^{-1}(y)=K^{-1}(K(x+\ker(T)))=x+\ker(T)$ and $|x+\ker(T)|=|x|+\ker(T)=x+\ker(T)$ that $K^{-1}$ is a positive operator. Now by Theorem 2.15 of 1 , $K$ is a Riesz homomorphism. It follows that $T$ is Riesz homomorphism. Now, the proof is complete by this fact that each surjective Riesz homomorphism is $un$-continuous.

Let $E$ be a vector lattice and $E^{\sim\sim}$ be the bidual of $E$. Recall that a subset $A$ of $E$ is $b$-order bounded in $E$ if $A$ is order bounded in $E^{\sim\sim}$.

Theorem 3.2

Let $E$ be a Banach lattice with order continuous norm and $F$ be a Dedekind $\sigma$-complete Banach lattice such that the norm of $F$ is not order continuous. If each operator $T:E\rightarrow F$ is $un$-continuous, then $E$ is $KB$-space.

Proof

By way of contradiction, suppose that $E$ is not a $KB$-space. Then by Lemmas 2.1 and 3.4 of 2 , there exists a $b$-order bounded disjoint sequence $(x_{n})_{n}$ of $E^{+}$ such that $\|x_{n}\|=1$ for all $n$, and there exists a positive disjoint $(g_{n})_{n}$ of $E^{\prime}$ with $\|g_{n}\|\leq 1$ such that $g_{n}(x_{n})=1$ for all $n$, and $g_{n}(x_{m})=0$ for $n\neq m$. Now we consider the operator $S:E\rightarrow\ell^{\infty}$ defined by $S(x)=(g_{k}(x))_{k=1}^{\infty}$ for all $x\in E$. Since $(x_{n})_{n}$ is disjoint, by Corollary 3.6 of 8 , it is $uo$-null. Hence $x_{n}\xrightarrow{uo}0$ in $E$ and therefore $x_{n}\xrightarrow{un}0$ in $E$. It follows that $\|S(x_{n})\|=\|(g_{k}(x_{n}))_{k=1}^{\infty}\|\geq g_{n}(x_{n})=1$. By Theorem 2.3 of 12 $S(x_{n})\nrightarrow 0$ with respect to $un$-topology in $\ell^{\infty}$. Therefore $S$ is not $un$-continuous. On the other hand, since the norm of $F$ is not order continuous and $F$ is Dedekind $\sigma$-complete, by Corollary 2.4.3 of 13 , $F$ contains a complemented copy of $\ell^{\infty}$. Now we consider the composed operator $T=I\circ S:E\rightarrow\ell^{\infty}\rightarrow F$, where $I$ is the canonical injection of $\ell^{\infty}$ into $F$. This operator is not $un$-continuous which is impossible, and so the proof follows.