Continuous Operators for Unbounded Convergence in Banach Lattices

Abstract

Recently, continuous functionals for unbounded order (norm, weak and weak*) in Banach lattices were studied. In this paper, we study the continuous operators with respect to unbounded convergences. We first investigate the approximation property of continuous operators for unbounded convergence. Then we show some characterizations of the continuity of the continuous operators for $uo$, $un$, $uaw$ and $ua{w}^{*}$-convergence. Based on these results, we discuss the order-weakly compact operators on Banach lattices.

1. Introduction

A net ${\left(x_{\alpha }\right)}_{\alpha \in A}$ in a Riesz space E is order convergent to $x\in E$ (write $x_{\alpha }\stackrel{o}{\to }x$) if there exists a net $\left(y_{\beta }\right)$, possibly over a different index set, such that $y_{\beta }\downarrow 0$ and for each $\beta \in B$ there exists ${\alpha }_0\in A$ satisfying $|x_{\alpha }-x|\le y_{\beta }$ for all $\alpha \ge {\alpha }_0$. A net $\left(x_{\alpha }\right)$ in a Banach lattice E is unbounded order (resp. norm, absolute weak) convergent to some x, denoted by $x_{\alpha }\stackrel{uo}{\to }x$ (resp. $x_{\alpha }\stackrel{un}{\to }x$, $x_{\alpha }\stackrel{uaw}{\to }x$), if the net $(|x_{\alpha }-x|\wedge u)$ converges to zero in order (resp. norm, weak) for all $u\in E_{+}$. A net $\left(x_{\alpha }^{\prime }\right)$ in a dual Banach lattice $E^{\prime }$ is unbounded absolute weak* convergent to some $x^{\prime }$, denoted by $x_{\alpha }^{\prime }\stackrel{ua{w}^{*}}{\to }{x}^{\prime }$, if $|x_{\alpha }^{\prime }-x^{\prime }|\wedge u^{\prime }\stackrel{w^{*}}{\to }0$ for all $u^{\prime }\in E_{+}^{\prime }$. For the basic theory of $uo$, $un$, $uaw$ and $ua{w}^{*}$-convergence, we refer to [1,2,3,4].

It can be easily verified that, in $l_p(1\le p<\infty )$, $uo$, $un$ and $uaw$ and $ua{w}^{*}$-convergence of nets are the same as coordinate-wise convergence. In $L_p\left(\mu \right)(1\le p<\infty )$ for finite measure $\mu$, $uo$-convergence for sequences is the same as almost everywhere convergence, $un$ and $uaw$-convergence for sequences are the same as convergence in measure. In $L_p\left(\mu \right)(1<p<\infty )$ for finite measure $\mu$, $ua{w}^{*}$-convergence for sequences is also the same as convergence in measure.

In [5], we studied the continuity of the linear functionals for different types of unbounded convergences $(uo;un;uaw;ua{w}^{*})$ in Banach lattices. We also dicussed the “unbounded-norm” continuous operators in [6] and had the following.

Theorem 1

([6], Theorem 2.7). For Banach lattices E and F, let $B_E$ and $B_{F^{\prime }}$ be the unit balls of E and F. Then for a continuous operator $T:E\to F$, the following statements hold.

1. 

The following conditions are equivalent.

(a) 

T is M-weakly compact ($T{x}_n\to 0$ for every disjoint sequence $\left(x_n\right)$ in $B_E$).

(b) 

$T{x}_n\to 0$ for every $uaw$-null sequence $\left(x_n\right)$ in $B_E$.

2. 

For its adjoint operator $T^{\prime }:F^{\prime }\to E^{\prime }$, the following conditions are equivalent.

(a) 

$T^{\prime }:F^{\prime }\to E^{\prime }$ is a M-weakly compact operator.

(b) 

$T^{\prime }{y}_n^{\prime }\to 0$ for every $ua{w}^{*}$-null sequence $\left(y_n^{\prime }\right)$ in $B_{F^{\prime }}$.

(c) 

$T^{\prime }{y}_n^{\prime }\to 0$ for every $uaw$-null sequence $\left(y_n^{\prime }\right)$ in $B_{F^{\prime }}$.

(d) 

$T^{\prime }{y}_n^{\prime }\to 0$ for every $uo$-null sequence $\left(y_n^{\prime }\right)$ in $B_{F^{\prime }}$.

Now we study the continuous operators which map unbounded convergence sequence to unbounded convergence sequence. In the first part of the paper, we investigate the approximation property of continuous operators for unbounded convergence. Then we show some characterizations of the continuity of the continuous operators for $uo$, $un$, $uaw$ and $ua{w}^{*}$-convergence. As an application of these results, we conclude the paper with characterizations of the order-weakly compact operators on Banach lattices. Some related results are obtained as well.

Recall that a Riesz space E is an ordered vector space in which $x\vee y=sup\{x,y\}$ and $x\wedge y=inf\{x,y\}$ exists for every $x,y\in E$. The positive cone of E is denoted by $E_{+}$, $i.e.,E_{+}=\{x\in E:x\ge 0\}$. For any vector x in E define $x^{+}:=x\vee 0,x^{-}:=(-x)\vee 0,|x|:=x\vee (-x)$. A Riesz space is said to be Dedekind σ-complete if every countable subset that is bounded above has a supremum. An operator $T:E\to F$ between two Riesz spaces is said to be positive if $Tx\ge 0$ for all $x\ge 0$. A net $\left(x_{\alpha }\right)$ in a Riesz space is called disjoint whenever $\alpha \ne \beta$ implies $|x_{\alpha }|\wedge |x_{\beta }|=0$ (denoted by $x_{\alpha }\perp x_{\beta }$). A set A in E is said to be order bounded if there exists some $u\in E_{+}$ such that $|x|\le u$ for all $x\in A$. The solid hull $Sol\left(A\right)$ of A is the smallest solid set including A and it equals the set $Sol\left(A\right):=\{x\in E:\exists y\in A,|x|\le |y|\}.$ An operator $T:E\to F$ is called order bounded if it maps order bounded subsets of E to order bounded subsets of F. A Banach lattice E is a Banach space $(E,\parallel ·\parallel )$ such that E is a Riesz space and its norm satisfies the following property: for each $x,y\in E$ with $|x|\le |y|$, we have $\parallel x\parallel \le \parallel y\parallel$.

For undefined terminology, notation and basic theory of Riesz space, Banach lattice and linear operator, we refer to [7,8].

2. Results

A Riesz pseudonorm is a real-valued function defined on a Riesz space E ($\rho :E\to \mathbb{R}$) satisfying the following properties.

$\rho \left(x\right)\ge 0$ for all $x\in E$.

$\rho (x+y)\le \rho \left(x\right)+\rho \left(y\right)$ for all $x,y\in E$.

$\rho \left({\lambda }_{n}x\right)\to 0$ as ${\lambda }_n\to 0$ for all $x\in E$.

$\rho \left(x\right)\le \rho \left(y\right)$ whenever $|x|\le |y|$ holds in E.

A (not necessarily Hausdorff) linear topology $\tau$ on a Riesz space E is said to be locally solid if $\tau$ has a base at zero consisting of solid sets. Clearly, a linear topology on a Riesz space is locally solid iff it is generated by a family of Riesz pseudonorms. It is natural to consider the “unbounded” topology generated by a family of Riesz pseudonorms.

Proposition 1.

Let $(E,\parallel ·\parallel )$ be a Banach lattice, the following statements hold.

1. 

The map ${\rho }_u:E\to {\mathbb{R}}_{+}$ defined by ${\rho }_u\left(x\right)=\parallel |x|\wedge u\parallel$ is a Riesz pseudonorm for each $u\in E_{+}$. Moreover, the $un$-topology and the topology generated by ${\left({\rho }_u\right)}_{u\in E_{+}}$ coincide.

2. 

The map ${\rho }_{u,x^{\prime }}:E\to {\mathbb{R}}_{+}$ as ${\rho }_{u,x^{\prime }}\left(x\right)=x^{\prime }(|x|\wedge u)$ is a Riesz pseudonorm for each $u\in E_{+},x^{\prime }\in E_{+}^{\prime }$. Moreover, the $uaw$-topology and the topology generated by the family ${\left({\rho }_{u,x^{\prime }}\right)}_{u\in E_{+},x^{\prime }\in E_{+}^{\prime }}$ coincide.

3. 

The map ${\rho }_{u^{\prime },x}:E^{\prime }\to {\mathbb{R}}_{+}$ as ${\rho }_{u^{\prime },x}\left(x^{\prime }\right)=(|x^{\prime }|\wedge u^{\prime })\left(x\right)$ is a Riesz pseudonorm for each $u^{\prime }\in E_{+}^{\prime },x\in E_{+}$. Moreover, the $ua{w}^{*}$-topology and the topology generated by the family ${\left({\rho }_{u^{\prime },x}\right)}_{u^{\prime }\in E_{+}^{\prime },x\in E_{+}}$ coincide.

Proof of Proposition 1. 

We just prove $\left(3\right)$, the rest of proof is similar.

Clearly, ${\rho }_{u^{\prime },x}\left(x^{\prime }\right)\ge 0$ for all $x^{\prime }\in E^{\prime }$ and ${\rho }_{u^{\prime },x}\left(x^{\prime }\right)\le {\rho }_{u^{\prime },x}\left(y^{\prime }\right)$ whenever $|x^{\prime }|\le |y^{\prime }|$ holds in $E^{\prime }$. Since $|x^{\prime }+y^{\prime }|\wedge u^{\prime }\le (|x^{\prime }|+|y^{\prime }|)\wedge u^{\prime }\le |x^{\prime }|\wedge u^{\prime }+|y^{\prime }|\wedge u^{\prime }$, we have ${\rho }_{u^{\prime },x}(x^{\prime }+y^{\prime })\le {\rho }_{u^{\prime },x}\left(x^{\prime }\right)+{\rho }_{u^{\prime },x}\left(y^{\prime }\right)$. According to the inequality ${\rho }_{u^{\prime },x}\left({\lambda }_n{x}^{\prime }\right)=(|{\lambda }_n{x}^{\prime }|\wedge u^{\prime })\left(x\right)\le (|{\lambda }_n{x}^{\prime }|)\left(x\right)=|{\lambda }_n||x^{\prime }|\left(x\right)$, we have ${\rho }_{u^{\prime },x}\left({\lambda }_n{x}^{\prime }\right)\to 0$ whenever ${\lambda }_n\to 0$. Therefore, ${\rho }_{u,x^{\prime }}$ is a Riesz pseudonorm for all $u\in E_{+},x^{\prime }\in E_{+}^{\prime }$. It is easy to see that the $ua{w}^{*}$-topology and the topology generated by the family ${\left({\rho }_{u^{\prime },x}\right)}_{u^{\prime }\in E_{+}^{\prime },x\in E_{+}}$ coincide. □

The following theorems describe an approximation property of continuous linear operators for unbounded convergence.

Theorem 2.

Let $T:E\to F$ be a continuous operator from a Banach lattice E to a Banach lattice F, let A be a bounded solid subset of E, and let ${\left({\rho }_{\gamma }\right)}_{\gamma \in \Gamma }$ be a family of norm continuous Riesz pseudonorms on F. If ${\rho }_{\gamma }\left(T{x}_n\right)\to 0$ holds for every disjoint sequence $\left(x_n\right)\subset A$ and all $\gamma \in \Gamma$, then for each $ϵ>0,\gamma \in \Gamma$ there exists some $u\in E_{+}$ lying in the ideal generated by A (which is the smallest ideal that includes A) such that ${\rho }_{\gamma }\left(T{(|x|-u)}^{+}\right)<ϵ$ holds for all $x\in A$.

Proof of Theorem 2. 

If the claim is not true, then there exists some $ϵ>0,\gamma \in \Gamma$ such that, for any $u\ge 0$ in the ideal generated by A, we have ${\rho }_{\gamma }\left(T{(|x|-u)}^{+}\right)\ge ϵ$ for at least one $x\in A$. In particular, there exists a sequence $\left(x_n\right)\subset A$ such that, for all $n\in \mathbb{N}$, we have ${\rho }_{\gamma }\left(T(|x_{n+1}|-4^n{\sum }_{i=1}^n|x_i{|)}^{+}\right)\ge ϵ$.

Put $y={\sum }_{n=1}^{\infty }{2}^{-n}|x_n|$, $w_n=(|x_{n+1}|-4^n{\sum }_{i=1}^n|x_i{|)}^{+}$ and $v_n=(|x_{n+1}|-4^n{\sum }_{i=1}^n|x_i{|-2^{-n}y)}^{+}$. According to [7], $\left(v_n\right)$ is disjoint. $\left(v_n\right)\subset A$ since A is solid and $0\le v_n\le |x_{n+1}|$. By our hypothesis, we have ${\rho }_{\gamma }\left(T{v}_n\right)\to 0$. It follows form $0\le w_n-v_n\le 2^{-n}y$ that $\parallel w_n-v_n\parallel \le 2^{-n}\parallel y\parallel$. Since ${\rho }_{\gamma }$ is a norm continuous Riesz pseudonorm, hence ${\rho }_{\gamma }\left(T(w_n-v_n)\right)\to 0$. From ${\rho }_{\gamma }\left(T{w}_n\right)\le {\rho }_{\gamma }\left(T(w_n-v_n)\right)+{\rho }_{\gamma }\left(T{v}_n\right)$, we see that ${\rho }_{\gamma }\left(T{w}_n\right)={\rho }_{\gamma }\left(T(|x_{n+1}|-4^n{\sum }_{i=1}^n|x_i{|)}^{+}\right)\to 0$. This leads to a contradiction, and the proof is completed. □

Corollary 1.

Let $T:E\to F$ be a continuous operator from a Banach lattice E to a Banach lattice F and A a bounded solid subset of E, the following statements hold.

1. 

If $T{x}_n\stackrel{un}{\to }0$ (i.e., ${\rho }_v\left(T{x}_n\right)=\parallel |T{x}_n|\wedge v\parallel \to 0$ for all $v\in F_{+}$) holds for every disjoint sequence $\left(x_n\right)\subset A$, then for each $ϵ>0,v\in F_{+}$ there exists some $u\in E_{+}$ lying the ideal generated by A such that ${\rho }_v\left(T{(|x|-u)}^{+}\right)<ϵ$ holds for all $x\in A$.

2. 

If $T{x}_n\stackrel{uaw}{\to }0$ (i.e., ${\rho }_{v,y^{\prime }}\left(T{x}_n\right)=y^{\prime }(|T{x}_n|\wedge v)\to 0$ for all $v\in F_{+},y^{\prime }\in F_{+}^{\prime }$) holds for every disjoint sequence $\left(x_n\right)\subset A$, then for each $ϵ>0,v\in F_{+},y^{\prime }\in F_{+}^{\prime }$ there exists some $u\in E_{+}$ lying the ideal generated by A such that ${\rho }_{v,y^{\prime }}\left(T{(|x|-u)}^{+}\right)<ϵ$ holds for all $x\in A$.

3. 

If F is a dual Banach lattice (denoted by $F^{\prime }$ in here) and $T{x}_n\stackrel{ua{w}^{*}}{\to }0$ (i.e., ${\rho }_{v^{\prime },y}\left(T{x}_n\right)=(|T{x}_n|\wedge v^{\prime })\left(y\right)\to 0$ for all $v^{\prime }\in F_{+}^{\prime },y\in F_{+}$) holds for every disjoint sequence $\left(x_n\right)\subset A$, then for each $ϵ>0,v^{\prime }\in F_{+}^{\prime },y\in F_{+}$ there exists some $u\in E_{+}$ lying the ideal generated by A such that ${\rho }_{v^{\prime },y}\left(T{(|x|-u)}^{+}\right)<ϵ$ holds for all $x\in A$.

The following results show some characterizations of the continuity of the continuous operators for $uo$, $un$, $uaw$ and $ua{w}^{*}$-convergence.

Recall that an operator $T:E\to F$ from a Riesz space E to a Banach space F is called order-weakly compact if $T[-x,x]$ is relatively weakly compact for all $x\in E_{+}$. A continuous operator $T:E\to F$ between two Banach lattices is said to be preserve a sublattice isomorphic to L if there exists closed sublattices $U\subset E,V\subset F$ isomorphic to L such that the restriction of T to U acts as an isomorphism onto V.

Lemma 1.

For a continuous operator $T:E\to F$ from a Dedekind σ-complete Banach lattice E to a Banach lattice F, if $T{x}_n\stackrel{un}{\to }0$ for every disjoint sequence $\left(x_n\right)\subset B_E$, then T is order-weakly compact.

Proof of Lemma 1. 

Assume that $T:E\to F$ is not order-weakly compact, according to ([8], Corollary 3.4.5), T preserve a sublattice isomorphic to ${\ell }_{\infty }$. Let $\left(e_n\right)$ be the unit vectors of ${\ell }_{\infty }$, clearly, $\left(T{e}_n\right)$ is not $un$-null in ${\ell }_{\infty }$. Indeed, if $\left(T{e}_n\right)$ is $un$-null in ${\ell }_{\infty }$, then $T{e}_n\to 0$ in ${\ell }_{\infty }$ by ([2], Theorem 2.3), moreover $e_n\to 0$ since the restriction of T to ${\ell }_{\infty }$ is isomorphic. Hence, $\left(T{e}_n\right)$ is not $un$-null in F. This leads to a contradiction. Therefore, T is order weakly compact. □

Theorem 3.

Let E and F be Banach lattices, for a continuous operator $T:E\to F$, the following statements hold.

1. 

If E is Dedekind σ-complete, then $T{x}_n\stackrel{un}{\to }0$ for every disjoint sequence $\left(x_n\right)\subset B_E$ iff $T{x}_n\stackrel{un}{\to }0$ for every $uaw$-null sequence $\left(x_n\right)\subset B_E$.

2. 

If T is positive, then the following conditions are equivalent.

(a) 

$T{x}_n\stackrel{uaw}{\to }0$ for every $uaw$-null sequence $\left(x_n\right)\subset B_E$.

(b) 

$T{x}_n\stackrel{uaw}{\to }0$ for every disjoint sequence $\left(x_n\right)\subset B_E$.

3. 

For the adjoint $T^{\prime }:F^{\prime }\to E^{\prime }$ of positive operator T, the following conditions are equivalent.

(a) 

$T^{\prime }{y}_n^{\prime }\stackrel{ua{w}^{*}}{\to }0$ for every $ua{w}^{*}$-null sequence $\left(y_n^{\prime }\right)\subset B_{F^{\prime }}$.

(b) 

$T^{\prime }{y}_n^{\prime }\stackrel{ua{w}^{*}}{\to }0$ for every $uaw$-null sequence $\left(y_n^{\prime }\right)\subset B_{F^{\prime }}$.

(c) 

$T^{\prime }{y}_n^{\prime }\stackrel{uaw*}{\to }0$ for every $uo$-null sequence $\left(y_n^{\prime }\right)\subset B_{F^{\prime }}$.

(d) 

$T^{\prime }{y}_n^{\prime }\stackrel{ua{w}^{*}}{\to }0$ for every disjoint sequence $\left(y_n^{\prime }\right)\subset B_{F^{\prime }}$.

Proof of Theorem 3. 

$\left(1\right)\Leftarrow$. According to ([3], Lemma 2), every disjoint sequence is $uaw$-null.

$\left(1\right)\Rightarrow$. T is order-weakly compact by Lemma 1. It follows from Corollary 1 that, for each $ϵ>0,v\in F_{+}$, there exists some $u\in E_{+}$ such that ${\rho }_v\left(T{(|x|-u)}^{+}\right)={\rho }_v\left(T(|x|)-T(|x|\wedge u)\right)<ϵ$ holds for all $x\in B_E$.

For a $uaw$-null sequence $\left(x_n\right)\subset B_E$, clearly, $|x_n|\wedge u\stackrel{w}{\to }0$. Since T is order-weakly compact, hence $T(|x_n|\wedge u)\to 0$ by ([8], Corollary 3.4.9), moreover ${\rho }_v\left(T(|x_n|\wedge u)\right)\to 0$. Thus, $T|x_n|\stackrel{un}{\to }0$. Since $x_n\stackrel{uaw}{\to }$ iff $x_n^{\pm }\stackrel{uaw}{\to }0$ iff $|x_n|\stackrel{uaw}{\to }0$, therefore $T{x}_n\stackrel{un}{\to }0$.

$\left(2\right)\left(a\right)\Rightarrow \left(2\right)\left(b\right)$. Obvious.

$\left(2\right)\left(b\right)\Rightarrow \left(2\right)\left(a\right)$. It is similar to the proof os $\left(1\right)\left(c\right)\Rightarrow \left(1\right)\left(b\right)$ that, for each $ϵ>0,v\in F_{+},y^{\prime }\in F_{+}^{\prime }$ there exists some $u\in E_{+}$ such that ${\rho }_{v,y^{\prime }}\left(T{(|x|-u)}^{+}\right)<ϵ$ holds for all $x\in B_E$. Using T is weak-weak continuous and positive, the rest of proof is similar.

$\left(3\right)\left(a\right)\Rightarrow \left(3\right)\left(b\right)$ and $\left(3\right)\left(a\right)\Rightarrow \left(3\right)\left(c\right)$. Every $uo$ (resp. $uaw$-null) sequence is $ua{w}^{*}$-null.

$\left(3\right)\left(b\right)\Rightarrow \left(3\right)\left(d\right)$ and $\left(3\right)\left(c\right)\Rightarrow \left(3\right)\left(d\right)$. Clearly.

$\left(3\right)\left(d\right)\Rightarrow \left(3\right)\left(a\right)$. According $T^{\prime }$ is positive and weak*-weak* continuous, the proof is similar. □

Corollary 2.

Let E and F be Banach lattices, for a continuous operator $T:E\to F$, the following statements hold.

1. 

The following conditions are equivalent.

(a) 

$T{x}_n\stackrel{w}{\to }0$ for every $uaw$-null sequence $\left(x_n\right)\subset B_E$.

(b) 

$T{x}_n\stackrel{w}{\to }0$ for every disjoint sequence $\left(x_n\right)\subset B_E$.

2. 

For the adjoint $T^{\prime }:F^{\prime }\to E^{\prime }$ of operator T, the following conditions are equivalent.

(a) 

$T^{\prime }{y}_n^{\prime }\stackrel{w^{*}}{\to }0$ for every $ua{w}^{*}$-null sequence $\left(y_n^{\prime }\right)\subset B_{F^{\prime }}$.

(b) 

$T^{\prime }{y}_n^{\prime }\stackrel{w^{*}}{\to }0$ for every $uaw$-null sequence $\left(y_n^{\prime }\right)\subset B_{F^{\prime }}$.

(c) 

$T^{\prime }{y}_n^{\prime }\stackrel{w*}{\to }0$ for every $uo$-null sequence $\left(y_n^{\prime }\right)\subset B_{F^{\prime }}$.

(d) 

$T^{\prime }{y}_n^{\prime }\stackrel{w^{*}}{\to }0$ for every disjoint sequence $\left(y_n^{\prime }\right)\subset B_{F^{\prime }}$.

Recall that a closed sublattice L in Banach lattice E which is generated by disjoint sequence $\left(x_n\right)\subset E_{+}$ is called sup-stable if $x={\sum }_{n=1}^{\infty }{a}_n{x}_n$ for every $x\in E$ with $limsup|a_n|<\infty$. For the unit vectors $\left(e_n\right)$ of ${\ell }_{\infty }$, clearly, $c_0$ is a closed sublattice in ${\ell }_{\infty }$ generated by $(e_n\wedge (1,\frac{1}{2},$…$,\frac{1}{n},$…$){)}_{n=1}^{\infty }$ and ${\ell }_{\infty }$ is a closed sup-stable sublattice generated by $(e_n\wedge (1,1,1,$…$){)}_{n=1}^{\infty }$. It is natural to ask that whether the result of the example holds in more general situations. The following results confirm the conjecture.

Proposition 2.

Suppose that E is a Banach lattice and $\left(x_n\right)\subset B_E$ is a disjoint sequence but not un-null, the following statements hold.

1. 

The closed sublattice L generated by $\{|x_n|\wedge u:n\in \mathbb{N}\}$ is isomorphic to $c_0$ for some $u\in E_{+}$.

2. 

If E is Dedeking σ-complete, then the closed sup-stable sublattice L generated by $\{|x_n|\wedge u:n\in \mathbb{N}\}$ is isomorphic to ${\ell }_{\infty }$ for some $u\in E_{+}$.

Proof of Proposition 2. 

Since $\left(x_n\right)$ is not $un$-null, hence there exists some $ϵ>0$ and a Riesz pseudonorm ${\rho }_u$ such that ${\rho }_u\left(x_n\right)=\parallel |x_n|\wedge u\parallel >ϵ$ for all n. Without loss of generality we may assume that ${\rho }_u\left(x_n\right)=\parallel |x_n|\wedge u\parallel =1$.

$\left(1\right)$ For $n\in \mathbb{N}$ and $z={\left(a_i\right)}_{i=1}^n\in {\mathbb{R}}^n$. Since ${\parallel z\parallel }_{\infty }\le \parallel {\sum }_{i=1}^n{a}_i·(|x_n|\wedge u)\parallel \le {\parallel u\parallel ·\parallel z\parallel }_{\infty }$, therefore the closed sublattice L generated by $\{|x_n|\wedge u\}$ is isomorphic to $c_0$.

$\left(2\right)$ For any $0\le z={\left(a_i\right)}_{i=1}^{\infty }\in {\ell }_{\infty }$, let $j\left(z\right)=sup\{a_n·(|x_n|\wedge u)\}\in L$, the supremum exsits since E is Dedekind $\sigma$-complete. Clearly, $j:{\ell }_{+}^{\infty }\to L_{+}$ is additive and positively homogeneous. Thus, j extends to all of ${\ell }_{\infty }$ as a lattice and norm isomorphism. □

An operator $T:E\to F$ is called unbounded norm continuous ($un$-continuous, for short) if $T{x}_n\stackrel{un}{\to }0$ for every $un$-null sequence $\left(x_n\right)\subset B_E$. According to $|T{x}_n|\wedge v=T|x_n|\wedge Tu=T(|x_n|\wedge u)$ for all $u\in E_{+},v\in F_{+}$, $x_n\stackrel{un}{\to }0$ implies $T{x}_n\stackrel{un}{\to }0$, therefore every onto lattice homomorphism is $un$-continuous. For every measurable space $(\mathsf{\Omega },\Sigma )$, let $St\left(\Sigma \right)$ denote the collection of all measurable stepfuntions:$f={\sum }_{i=1}^r{a}_i{\chi }_{A_i}:\mathsf{\Omega }\to \mathbb{R}$ where $r\in \mathbb{N}$, $a_i\in \mathbb{R}$ and $A_i\in \Sigma$ for $i\in \{1,2,$…$r\}$.

Theorem 4.

Let E and F be Banach lattices, for a onto lattice homomorphism $T:E\to F$. Assume that $\left(T{x}_n\right)$ is not $un$-null in F for a disjoint sequence $\left(x_n\right)\subset B_E$, then there exists a subsequence ${\left(k\left(n\right)\right)}_1^{\infty }$ such that the following statements hold.

1. 

The closed sublattice L generated by $\{|x_{k\left(n\right)}|\wedge u:n\in \mathbb{N}\}$ is isomorphic to $c_0$ and T acts on L as a isomorphism for some $u\in E_{+}$.

2. 

If E is Dedekind σ-complete, then the closed sup-stable sublattice L generated by $\{|x_{k\left(n\right)}|\wedge u:n\in \mathbb{N}\}$ is isomorphic to ${\ell }_{\infty }$ and T acts on L as a isomorphism for some $u\in E_{+}$.

Proof of Theorem 4. 

Since $E^{″}$ is Dedekind complete and E is a sublattice of $E^{″}$, hence $\left(2\right)$ implies $\left(1\right)$. Therefore, we just prove $\left(2\right)$.

Since $\left(T{x}_n\right)$ is not $un$-null, so there exists a Riesz pseudonorm ${\rho }_v$ such that ${\rho }_v\left(T{x}_n\right)=\parallel |T{x}_n|\wedge v\parallel >1$ for all n and some $v\in F_{+}$. Since T is onto and lattice homomorphism, therefore $\left(x_n\right)$ is not $un$-null, that is, $|x_n|\wedge u\nrightarrow 0$ for some $u\in E_{+}$ satisfying $Tu=v$. For every $n\in N$, let $y_n^{\prime }\in B_{F^{\prime }}\cap F_{+}^{\prime }$ such that $y_n^{\prime }(|T{x}_n|\wedge v)>1$. We set $x_n^{\prime }=T^{\prime }{y}_n^{\prime }$, since $x_n^{\prime }(|x_n|\wedge u)=y_n^{\prime }(|T{x}_n|\wedge v)$, so $x_n^{\prime }(|x_n|\wedge u)>1$.

Let $0<ϵ<\frac{1}{4}$. Cleary, $(|x_n{|\wedge u)}_1^{\infty }\subset [0,u]\subset E_{+}$, ${sup}_n||x_n^{\prime }|\left(u\right)|<\infty$ and ${\sum }_1^{\infty }|x_n|\wedge u\le u$. It follows from ([8], Theorem 2.3.7) that there exists a subsequence ${\left(k\left(n\right)\right)}_1^{\infty }$ satisfying $||x_{k\left(n\right)}^{\prime }|(sup\{|x_{k\left(j\right)}|\wedge u:n\ne j\in \mathbb{N}\})|<ϵ$. We may assume that $k\left(n\right)=n$ for all n. For every finitely valued $z\in {\ell }_{\infty }$, there exists disjoint subsets $A_1,A_2,$…$,A_r\subset \mathbb{N}$ and $a_1,a_2,$…$,a_r\in \mathbb{R}$ such that $z={\sum }_{i=1}^r{a}_i·{\chi }_{A_i}\in St\left(\mathbb{N}\right)$. For all $i=1,2,$…$,r$, we define $v_i=sup\{|x_n|\wedge u:n\in A_i\}$ and $T_{1}z={\sum }_{i=1}^r{a}_{i}T{v}_i$.

Clearly, $T_1$ is well defined and linear on $St\left(\mathbb{N}\right)$ such that $|y^{\prime }\left(T_{1}z\right)|\le {\sum }_{i=1}^r|a_i|·|y^{\prime }\left(T{v}_i\right){|\le \parallel z\parallel }_{\infty }\parallel T\parallel \parallel u\parallel$ for all $y^{\prime }\in B_{F^{\prime }}$. Therefore, $\parallel T_1\parallel \le \parallel T\parallel \parallel u\parallel$.

We may assume that ${\parallel z\parallel }_{\infty }=|a_1|$. For every $n\in A_1$,

$$\begin{array}{cc}\hfill \parallel T_{1}z\parallel & \ge |y_n^{\prime }\left(T_{1}z\right)|=|x_n^{\prime }(\sum _{i=1}^r{a}_i·v_i)|\hfill \\ \hfill & \ge {\parallel z\parallel }_{\infty }·|x_n^{\prime }(|x_n{|\wedge u)|-\parallel z\parallel }_{\infty }·||x_n^{\prime }|(sup\{|x_j|\wedge u:j\ne n\})|\hfill \\ \hfill & \ge {\parallel z\parallel }_{\infty }(1-ϵ)\ge {\displaystyle \frac{{\parallel z\parallel }_{\infty }}{2}}.\hfill \end{array}$$

Since $St\left(\mathbb{N}\right)$ is dense in ${\ell }_{\infty }$, hence $T_1$ extends as an lattice and norm isomorphism to all of ${\ell }_{\infty }$. According to Proposition 2, the sup-stable sublattice generated by $(|x_n|\wedge u)$ is isomorphic to ${\ell }_{\infty }$. The proof is completed. □

Finally, we can immediately get the characterization of order-weakly compact operators.

Theorem 5.

Let E be a Dedekind σ-complete Banach lattice and F a Banach lattice, for a onto lattice homomorphism $T:E\to F$, the following conditions are equivalent.

1. 

$T{x}_n\stackrel{un}{\to }0$ for every $uo$-null sequence $\left(x_n\right)\subset B_E$.

2. 

$T{x}_n\stackrel{un}{\to }0$ for every $uaw$-null sequence $\left(x_n\right)\subset B_E$.

3. 

$T{x}_n\stackrel{un}{\to }0$ for every disjoint sequence $\left(x_n\right)\subset B_E$.

4. 

T is order-weakly compact.

Proof of Theorem 5. 

 

$\left(1\right)\iff \left(2\right)\iff \left(3\right)$ by Theorem 3.

$\left(3\right)\Rightarrow \left(4\right)$ by Lemma 1.

$\left(4\right)\Rightarrow \left(3\right)$. Assume that $\left(3\right)$ does not hold, then there exists a bounded disjoint sequence $\left(x_n\right)$ such that $\left(T{x}_n\right)$ is not $un$-null. According to Theorem 4, T preserves a sublattice isomorphic to ${\ell }_{\infty }$. It follows from ([8], Corollary 3.4.5) that T is not order-weakly compact. This leads to a contradiction.

$\left(4\right)\Rightarrow \left(1\right)$ and $\left(4\right)\Rightarrow \left(2\right)$ by $|T{x}_n|\wedge v=T|x_n|\wedge Tu=T(|x_n|\wedge u)$ for all $u\in E_{+},v\in F_{+}$ and ([8], Theorem 3.4.4 and Corollary 3.4.9). □

3. Discussion

This paper is the third article in our series of work on this subject. Through the previous research on unbounded convergence in Banach lattices, we studied continuous functionals and operators on Banach lattices. In [5], we showed the equivalent relationship of $uo$, $uaw$ and $ua{w}^{*}$-continuous functionals. We also found that $uo$, $uaw$ and $ua{w}^{*}$ to norm continuous operators is exactly M-weakly compact in many cases (in [6]). In this paper, we found again that $uo$, $uaw$ and $ua{w}^{*}$ to $un$, $uaw$ and $ua{w}^{*}$-continuous operators are equivalent in sometimes. To date, we can find that $un$ to norm continuous is special. In this regard, we will introduce and study weak L- and M-weakly compact operators by the type of $un$ to norm continuous operators in the following article as a continuous and in-depth study of this subject.