Continuous operators for unbounded convergence in Banach lattices

TL;DR

This paper explores the continuity of operators in Banach lattices concerning various types of unbounded convergence, providing new characterizations and insights into their approximation properties and compactness.

Contribution

It introduces new characterizations of operator continuity for unbounded convergences and examines order-weakly compact operators in Banach lattices.

Findings

Characterizations of operator continuity for uo, un, uaw, uaw* convergences

Approximation properties of continuous operators for unbounded convergence

Results on order-weakly compact operators in Banach lattices

Abstract

Recently, the functionals different types of unbounded convergences (uo, un, uaw, uaw*) 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 uaw*-convergence. Based on these results, we discuss the order-weakly compact operators on Banach lattices. Some related results are obtained as well.

Equations

$$\Vert T_{1}z\Vert$$

$$\displaystyle\geq\|z\|_{\infty}\cdot\lvert x^{\prime}_{n}(\lvert x_{n}\rvert\wedge u)\rvert-\|z\|_{\infty}\cdot\big{|}\lvert x^{\prime}_{n}\rvert(\sup\{\lvert x_{j}\rvert\wedge u:j\neq n\})\big{|}$$

$$\ge \Vert z{\Vert }_{\infty }(1-ϵ)\ge \frac{\Vert z{\Vert }_{\infty }}{2}.$$

Full text

1.. Introduction

A net $(x_{\alpha})$ in a Banach lattice $E$ is unbounded order (resp. norm, absolute weak) convergent to some $x$, denoted by $x_{\alpha}\xrightarrow{uo}x$ (resp. $x_{\alpha}\xrightarrow{un}x$, $x_{\alpha}\xrightarrow{uaw}x$), if the net $(|x_{\alpha}-x|\wedge u)$ converges to zero in order (resp. norm, weak) for all $u\in E_{+}$. A net $(x_{\alpha}^{\prime})$ in a dual Banach lattice $E^{\prime}$ is unbounded absolute weak* convergent to some $x^{\prime}$, denoted by $x_{\alpha}^{\prime}\xrightarrow{uaw^{*}}x^{\prime}$, if $|x^{\prime}_{\alpha}-x^{\prime}|\wedge u^{\prime}\xrightarrow{w^{*}}0$ for all $u^{\prime}\in E_{+}^{\prime}$. For the basic theory of $uo$, $un$, $uaw$ and $uaw^{*}$-convergence, we refer to [3, 4, 5, 6].

It can be easily verified that, in $l_{p}(1\leq p<\infty)$, $uo$, $un$ and $uaw$ and $uaw^{*}$-convergence of nets are the same as coordinate-wise convergence. In $L_{p}(\mu)(1\leq p<\infty)$ 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 $L_{p}(\mu)(1<p<\infty)$ for finite measure $\mu$, $uaw^{*}$-convergence for sequences is also the same as convergence in measure.

In [7], we studied the continuity of the linear functionals for different types of unbounded convergences $(uo;un;uaw;uaw^{*})$ in Banach lattices. The aims of the present paper are the continuity of linear operators for different types of unbounded convergences $(uo;un;uaw;uaw^{*})$ in Banach lattices. In [8, Theorem 2.7], we dicussed the ”unbounded-norm” continuous operators and had the following.

Theorem 1.1**.**

([8, Theorem 2.7]) Let $E$ and $F$ be Banach lattices, for a continuous operator $T:E\rightarrow F$, the following statements hold.

1. (1)

The following conditions are equivalent.

1. (a)

$T$* is a M-weakly compact.* 2. (b)

$Tx_{n}\rightarrow 0$* for every $uaw$-null sequence $(x_{n})$ in $B_{E}$.* 3. (c)

$Tx_{n}\rightarrow 0$* for every $uo$-null sequence $(x_{n})$ in $B_{E}$.* 2. (2)

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

1. (a)

$T^{\prime}:F^{\prime}\rightarrow E^{\prime}$* is a M-weakly compact operator.* 2. (b)

$T^{\prime}y^{\prime}_{n}\rightarrow 0$* for every $uaw^{*}$-null sequence $(y_{n}^{\prime})$ in $B_{F^{\prime}}$.* 3. (c)

$T^{\prime}y^{\prime}_{n}\rightarrow 0$* for every $uaw$-null sequence $(y_{n}^{\prime})$ in $B_{F^{\prime}}$.* 4. (d)

$T^{\prime}y^{\prime}_{n}\rightarrow 0$* for every $uo$-null sequence $(y_{n}^{\prime})$ in $B_{F^{\prime}}$.*

Now we study the ”unbounded-unbounded” continuous operators. 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 $uaw^{*}$-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 $E_{+}$, $i.e.,E_{+}=\{x\in E:x\geq 0\}$. For any vector $x$ in $E$ define $x^{+}:=x\vee 0,x^{-}:=(-x)\vee 0,|x|:=x\vee(-x)$. An operator $T:E\rightarrow F$ between two Riesz spaces is said to be positive if $Tx\geq 0$ for all $x\geq 0$. A net $(x_{\alpha})$ in a Riesz space is called disjoint whenever $\alpha\neq\beta$ 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 $|x|\leq u$ for all $x\in A$. The solid hull $Sol(A)$ of $A$ is the smallest solid set including $A$ and it equals the set $Sol(A):=\{x\in E:\exists y\in A,|x|\leq|y|\}.$ An operator $T:E\rightarrow F$ 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 $(E,\|\cdot\|)$ such that $E$ is a Riesz space and its norm satisfies the following property: for each $x,y\in E$ with $|x|\leq|y|$, we have $\|x\|\leq\|y\|$.

For undefined terminology, notation and basic theory of Riesz space, Banach lattice and linear operator, we refer to [1, 2].

2.. Results

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

1. (1)

$\rho(x)\geq 0$ for all $x\in E$. 2. (2)

$\rho(x+y)\leq\rho(x)+\rho(y)$ for all $x,y\in E$. 3. (3)

$\rho(\lambda_{n}x)\rightarrow 0$ as $\lambda_{n}\rightarrow 0$ for all $x\in E$. 4. (4)

$\rho(x)\leq\rho(y)$ whenever $\lvert x\rvert\leq\lvert y\rvert$ holds in $E$.

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 2.1**.**

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

1. (1)

The map $\rho_{u}:E\rightarrow\mathbb{R_{+}}$ defined by \rho_{u}(x)=\big{\|}\lvert x\rvert\wedge u\big{\|} is a Riesz pseudonorm for each $u\in E_{+}$. Moreover, the $un$-topology and the topology generated by $(\rho_{u})_{u\in E_{+}}$ coincide. 2. (2)

The map $\rho_{u,x^{\prime}}:E\rightarrow R_{+}$ as $\rho_{u,x^{\prime}}(x)=x^{\prime}(\lvert x\rvert\wedge u)$ 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 $(\rho_{u,f})_{u\in E_{+},x^{\prime}\in E^{\prime}_{+}}$ coincide. 3. (3)

The map $\rho_{u^{\prime},x}:E^{\prime}\rightarrow R_{+}$ as $\rho_{u^{\prime},x}(x^{\prime})=(\lvert x^{\prime}\rvert\wedge u^{\prime})(x)$ is a Riesz pseudonorm for each $u^{\prime}\in E^{\prime}_{+},x\in E_{+}$. Moreover, the $uaw^{*}$-topology and the topology generated by the family $(\rho_{u^{\prime},x})_{u^{\prime}\in E^{\prime}_{+},x\in E_{+}}$ coincide.

Proof.

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

Clearly, $\rho_{u^{\prime},x}(x^{\prime})\geq 0$ for all $x^{\prime}\in E^{\prime}$ and $\rho_{u^{\prime},x}(x^{\prime})\leq\rho_{u^{\prime},x}(y^{\prime})$ whenever $\lvert x^{\prime}\rvert\leq\lvert y^{\prime}\rvert$ holds in $E^{\prime}$. Since $\lvert x^{\prime}+y^{\prime}\rvert\wedge u^{\prime}\leq(\lvert x^{\prime}\rvert+\lvert y^{\prime}\rvert)\wedge u^{\prime}\leq\lvert x^{\prime}\rvert\wedge u^{\prime}+\lvert y^{\prime}\rvert\wedge u^{\prime}$, we have $\rho_{u^{\prime},x}(x^{\prime}+y^{\prime})\leq\rho_{u^{\prime},x}(x^{\prime})+\rho_{u^{\prime},x}(y^{\prime})$. According to the inequality $\rho_{u^{\prime},x}(\lambda_{n}x^{\prime})=(\lvert\lambda_{n}x^{\prime}\rvert\wedge u^{\prime})(x)\leq(\lvert\lambda_{n}x^{\prime}\rvert)(x)=\lvert\lambda_{n}\rvert|x^{\prime}|(x)$, we have $\rho_{u^{\prime},x}(\lambda_{n}x^{\prime})\rightarrow 0$ whenever $\lambda_{n}\rightarrow 0$. 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 $uaw^{*}$-topology and the topology generated by the family $(\rho_{u^{\prime},x})_{u^{\prime}\in E^{\prime}_{+},x\in E_{+}}$ coincide. ∎

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

Theorem 2.2**.**

Let $T:E\rightarrow F$ 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 $(\rho_{\gamma})_{\gamma\in\Gamma}$ be a family of norm continuous Riesz pseudonorms on $F$. If $\rho_{\gamma}(Tx_{n})\rightarrow 0$ holds for every disjoint sequence $(x_{n})\subset A$ and all $\gamma\in\Gamma$, then for each $\epsilon>0,\gamma\in\Gamma$ there exists some $u\in E_{+}$ lying the ideal generated by $A$ such that \rho_{\gamma}\big{(}T(\lvert x\rvert-u)^{+}\big{)}<\epsilon holds for all $x\in A$.

Proof.

If the claim is not true, then there exists some $\epsilon>0,\gamma\in\Gamma$ such that, for any $u\geq 0$ in the ideal generated by $A$, we have \rho_{\gamma}\big{(}T(\lvert x\rvert-u)^{+}\big{)}\geq\epsilon for at least one $x\in A$. In particular, there exists a sequence $(x_{n})\subset A$ such that, for all $n\in\mathbb{N}$, we have \rho_{\gamma}\big{(}T(\lvert x_{n+1}\rvert-4^{n}\sum_{i=1}^{n}\lvert x_{i}\rvert)^{+}\big{)}\geq\epsilon.

Put $y=\sum_{n=1}^{\infty}2^{-n}\lvert x_{n}\rvert$, $w_{n}=(\lvert x_{n+1}\rvert-4^{n}\sum_{i=1}^{n}\lvert x_{i}\rvert)^{+}$ and $v_{n}=(\lvert x_{n+1}\rvert-4^{n}\sum_{i=1}^{n}\lvert x_{i}\rvert-2^{-n}y)^{+}$. According to [1, Lemma 4.35], $(v_{n})$ is disjoint. $(v_{n})\subset A$ since $A$ is solid and $0\leq v_{n}\leq\lvert x_{n+1}\rvert$. By our hypothesis, we have $\rho_{\gamma}(Tv_{n})\rightarrow 0$. It follows form $0\leq w_{n}-v_{n}\leq 2^{-n}y$ that $\|w_{n}-v_{n}\|\leq 2^{-n}\|y\|$. Since $\rho_{\gamma}$ is a norm continuous Riesz pseudonorm, hence $\rho_{\gamma}(T(w_{n}-v_{n}))\rightarrow 0$. From $\rho_{\gamma}(Tw_{n})\leq\rho_{\gamma}(T(w_{n}-v_{n}))+\rho_{\gamma}(Tv_{n})$, we see that \rho_{\gamma}(Tw_{n})=\rho_{\gamma}\big{(}T(\lvert x_{n+1}\rvert-4^{n}\sum_{i=1}^{n}\lvert x_{i}\rvert)^{+}\big{)}\rightarrow 0. This leads to a contradiction, and the proof is completed. ∎

Corollary 2.3**.**

Let $T:E\rightarrow F$ 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. (1)

If $Tx_{n}\xrightarrow{un}0$ (i.e. $\rho_{v}(Tx_{n})=\|\lvert Tx_{n}\rvert\wedge v\|\rightarrow 0$ for all $v\in F_{+}$) holds for every disjoint sequence $(x_{n})\subset A$, then for each $\epsilon>0,v\in F_{+}$ there exists some $u\in E_{+}$ lying the ideal generated by $A$ such that \rho_{v}\big{(}T(\lvert x\rvert-u)^{+}\big{)}<\epsilon holds for all $x\in A$. 2. (2)

If $Tx_{n}\xrightarrow{uaw}0$ (i.e. $\rho_{v,y^{\prime}}(Tx_{n})=y^{\prime}(\lvert Tx_{n}\rvert\wedge v)\rightarrow 0$ for all $v\in F_{+},y^{\prime}\in F^{\prime}_{+}$) holds for every disjoint sequence $(x_{n})\subset A$, then for each $\epsilon>0,v\in F_{+},y^{\prime}\in F^{\prime}_{+}$ there exists some $u\in E_{+}$ lying the ideal generated by $A$ such that \rho_{v,y^{\prime}}\big{(}T(\lvert x\rvert-u)^{+}\big{)}<\epsilon holds for all $x\in A$. 3. (3)

If $F$ is a dual Banach lattice (denoted by $F^{\prime}$ in here) and $Tx_{n}\xrightarrow{uaw^{*}}0$ (i.e. $\rho_{v^{\prime},y}(Tx_{n})=(\lvert Tx_{n}\rvert\wedge v^{\prime})(y)\rightarrow 0$ for all $v^{\prime}\in F^{\prime}_{+},y\in F_{+}$) holds for every disjoint sequence $(x_{n})\subset A$, then for each $\epsilon>0,v^{\prime}\in F^{\prime}_{+},y\in F_{+}$ there exists some $u\in E_{+}$ lying the ideal generated by $A$ such that \rho_{v^{\prime},y}\big{(}T(\lvert x\rvert-u)^{+}\big{)}<\epsilon holds for all $x\in A$.

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

Recall that an operator $T:E\rightarrow F$ 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 $T:E\rightarrow F$ 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 2.4**.**

For a continuous operator $T:E\rightarrow F$ from a Dedekind $\sigma$-complete Banach lattice $E$ to a Banach lattice $F$, if $Tx_{n}\xrightarrow{un}0$ for every disjoint sequence $(x_{n})\subset B_{E}$, then $T$ is order-weakly compact.

Proof.

Assume that $T:E\rightarrow F$ is not order-weakly compact, according to [2, Corollary 3.4.5], $T$ preserve a sublattice isomorphic to $\ell_{\infty}$. Let $(e_{n})$ be the unit vectors of $\ell_{\infty}$, clearly, $(Te_{n})$ is not $un$-null in $\ell_{\infty}$. Indeed, if $(Te_{n})$ is $un$-null in $\ell_{\infty}$, then $Te_{n}\rightarrow 0$ in $\ell_{\infty}$ by [4, Theorem 2.3], moreover $e_{n}\rightarrow 0$ since the restriction of $T$ to $\ell_{\infty}$ is isomorphic. Hence, $(Te_{n})$ is not $un$-null in $F$. This leads to a contradiction. Therefore, $T$ is order weakly compact. ∎

Theorem 2.5**.**

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

1. (1)

If $E$ is Dedekind $\sigma$-complete, then the following conditions are equivalent.

1. (a)

$Tx_{n}\xrightarrow{un}0$* for every $uo$-null sequence $(x_{n})\subset B_{E}$.* 2. (b)

$Tx_{n}\xrightarrow{un}0$* for every $uaw$-null sequence $(x_{n})\subset B_{E}$.* 3. (c)

$Tx_{n}\xrightarrow{un}0$* for every disjoint sequence $(x_{n})\subset B_{E}$.* 2. (2)

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

1. (a)

$Tx_{n}\xrightarrow{uaw}0$* for every $uaw$-null sequence $(x_{n})\subset B_{E}$.* 2. (b)

$Tx_{n}\xrightarrow{uaw}0$* for every disjoint sequence $(x_{n})\subset B_{E}$.* 3. (3)

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

1. (a)

$T^{\prime}y_{n}^{\prime}\xrightarrow{uaw^{*}}0$* for every $uaw^{*}$-null sequence $(y_{n}^{\prime})\subset B_{F^{\prime}}$.* 2. (b)

$T^{\prime}y_{n}^{\prime}\xrightarrow{uaw^{*}}0$* for every $uaw$-null sequence $(y_{n}^{\prime})\subset B_{F^{\prime}}$.* 3. (c)

$T^{\prime}y_{n}^{\prime}\xrightarrow{uaw*}0$* for every $uo$-null sequence $(y_{n}^{\prime})\subset B_{F^{\prime}}$.* 4. (d)

$T^{\prime}y_{n}^{\prime}\xrightarrow{uaw^{*}}0$* for every disjoint sequence $(y_{n}^{\prime})\subset B_{F^{\prime}}$.*

Proof.

$(1)(a)\Rightarrow(1)(c)$ and $(1)(b)\Rightarrow(1)(c)$. According to [3, Corollary 3.6] and [5, Lemma 2], every disjoint sequence is $uo$-null and $uaw$-null.

$(1)(c)\Rightarrow(1)(a)$ and $(1)(c)\Rightarrow(1)(b)$. Assume that $(3)$ holds, $T$ is order-weakly compact by Lemma 2.4. It follows from Corollary 2.3 that, for each $\epsilon>0,v\in F_{+}$, there exists some $u\in E_{+}$ such that \rho_{v}\big{(}T(\lvert x\rvert-u)^{+}\big{)}=\rho_{v}\big{(}T(\lvert x\rvert)-T(\lvert x\rvert\wedge u)\big{)}<\epsilon holds for all $x\in B_{E}$.

For a $uo$-null (resp. $uaw$-null) sequence $(x_{n})\subset B_{E}$, clearly, $\lvert x_{n}\rvert\wedge u\xrightarrow{o(resp.w)}0$. Since $T$ is order-weakly compact, hence $T(\lvert x_{n}\rvert\wedge u)\rightarrow 0$ by [2, Theorem 3.4.4 and Corollary 3.4.9], moreover \rho_{v}\big{(}T(\lvert x_{n}\rvert\wedge u)\big{)}\rightarrow 0. Thus, $T\lvert x_{n}\rvert\xrightarrow{un}0$. Since $x_{n}\xrightarrow{uo(resp.uaw)}$ iff $x_{n}^{\pm}\xrightarrow{uo(resp.uaw)}0$ iff $\lvert x_{n}\rvert\xrightarrow{uo(resp.uaw)}0$, therefore $Tx_{n}\xrightarrow{un}0$.

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

$(2)(b)\Rightarrow(2)(a)$. It is similar to the proof os $(1)(c)\Rightarrow(1)(b)$ that, for each $\epsilon>0,v\in F_{+},y^{\prime}\in F^{\prime}_{+}$ there exists some $u\in E_{+}$ such that \rho_{v,y^{\prime}}\big{(}T(\lvert x\rvert-u)^{+}\big{)}<\epsilon holds for all $x\in B_{E}$. Using $T$ is weak-weak continuous and positive, the rest of proof is similar.

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

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

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

Recall that a closed sublattice $L$ in Banach lattice $E$ which is generated by disjoint sequence $(x_{n})\subset E_{+}$ is called sup-stable if $x=\sum_{n=1}^{\infty}a_{n}x_{n}$ for every $x\in E$ with $\lim\sup\lvert a_{n}\rvert<\infty$. For the unit vectors $(e_{n})$ of $\ell_{\infty}$, clearly, $c_{0}$ is a closed sublattice in $\ell_{\infty}$ generated by \big{(}e_{n}\wedge(1,\frac{1}{2},...,\frac{1}{n},...)\big{)}_{n=1}^{\infty} and $\ell_{\infty}$ is a closed sup-stable sublattice generated by \big{(}e_{n}\wedge(1,1,1,...)\big{)}_{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 hypothesis.

Proposition 2.6**.**

Suppose that $E$ is a Banach lattice and $(x_{n})\subset B_{E}$ is a disjoint sequence but not $un$-null, the following statements hold.

1. (1)

The closed sublattice $L$ generated by $\{\lvert x_{n}\rvert\wedge u:n\in\mathbb{N}\}$ is isomorphic to $c_{0}$ for some $u\in E_{+}$. 2. (2)

If $E$ is Dedeking $\sigma$-complete, then the closed sup-stable sublattice $L$ generated by $\{\lvert x_{n}\rvert\wedge u:n\in\mathbb{N}\}$ is isomorphic to $\ell_{\infty}$ for some $u\in E_{+}$.

Proof.

Since $(x_{n})$ is not $un$-null, hence there exists some $\epsilon>0$ and a Riesz pseudonorm $\rho_{u}$ such that \rho_{u}(x_{n})=\big{\|}\lvert x_{n}\rvert\wedge u\big{\|}>\epsilon for all $n$. Without loss of generality we may assume that \rho_{u}(x_{n})=\big{\|}\lvert x_{n}\rvert\wedge u\big{\|}=1.

$(1)$ For $n\in\mathbb{N}$ and $z=(a_{i})_{i=1}^{n}\in\mathbb{R}^{n}$. Since $\|z\|_{\infty}\leq\|\sum_{i=1}^{n}a_{i}\cdot(\lvert x_{n}\rvert\wedge u)\|\leq\|u\|\cdot\|z\|_{\infty}$, therefore the closed sublattice $L$ generated by $\{\lvert x_{n}\rvert\wedge u\}$ is isomorphic to $c_{0}$.

$(2)$ For any $0\leq z=(a_{i})_{i=1}^{\infty}\in\ell_{\infty}$, let $j(z)=\sup\{a_{n}\cdot(\lvert x_{n}\rvert\wedge u)\}\in L$, the supremum exsits since $E$ is Dedekind $\sigma$-complete. Clearly, $j:\ell^{\infty}_{+}\rightarrow L_{+}$ is additive and positively homogeneous. Thus, $j$ extends to all of $\ell_{\infty}$ as a lattice and norm isomorphism. ∎

An operator $T:E\rightarrow F$ is called unbounded norm continuous ($un$-continuous, for short) if $Tx_{n}\xrightarrow{un}0$ for every $un$-null sequence $(x_{n})\subset B_{E}$. According to $\lvert Tx_{n}\rvert\wedge v=T\lvert x_{n}\rvert\wedge Tu=T(\lvert x_{n}\rvert\wedge u)$ for all $u\in E_{+},v\in F_{+}$, $x_{n}\xrightarrow{un}0$ implies $Tx_{n}\xrightarrow{un}0$, therefore every onto lattice homomorphism is $un$-continuous. For every measurable space $(\Omega,\varSigma)$, let $St(\varSigma)$ denote the collection of all measurable stepfuntions: $f=\sum_{i=1}^{r}a_{i}\chi_{A_{i}}:\Omega\rightarrow\mathbb{R}$ where $r\in\mathbb{N}$, $a_{i}\in\mathbb{R}$ and $A_{i}\in\varSigma$ for $i\in\{1,2,...r\}$.

Theorem 2.7**.**

Let $E$ and $F$ be Banach lattices, for a onto lattice homomorphism $T:E\rightarrow F$. Assume that $(Tx_{n})$ is not $un$-null in $F$ for a disjoint sequence $(x_{n})\subset B_{E}$, then there exists a subsequence $(k(n))_{1}^{\infty}$ such that the following statements hold.

1. (1)

The closed sublattice $L$ generated by $\{\lvert x_{k(n)}\rvert\wedge u:n\in\mathbb{N}\}$ is isomorphic to $c_{0}$ and $T$ acts on $L$ as a isomorphism for some $u\in E_{+}$. 2. (2)

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

Proof.

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

Since $(Tx_{n})$ is not $un$-null, so there exists a Riesz pseudonorm $\rho_{v}$ such that \rho_{v}(Tx_{n})=\big{\|}\lvert Tx_{n}\rvert\wedge v\big{\|}>1 for all $n$ and some $v\in F_{+}$. Since $T$ is onto and lattice homomorphism, therefore $(x_{n})$ is not $un$-null, that is, $\lvert x_{n}\rvert\wedge u\nrightarrow 0$ 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 $y_{n}^{\prime}(\lvert Tx_{n}\rvert\wedge v)>1$. We set $x^{\prime}_{n}=T^{\prime}y_{n}^{\prime}$, since $x_{n}^{\prime}(\lvert x_{n}\rvert\wedge u)=y_{n}^{\prime}(\lvert Tx_{n}\rvert\wedge v)$, so $x_{n}^{\prime}(\lvert x_{n}\rvert\wedge u)>1$.

Let $0<\epsilon<\frac{1}{4}$. Cleary, $(\lvert x_{n}\rvert\wedge u)_{1}^{\infty}\subset[0,u]\subset E_{+}$, \sup_{n}\big{|}\lvert x_{n}^{\prime}\rvert(u)\big{|}<\infty and $\sum_{1}^{\infty}\lvert x_{n}\rvert\wedge u\leq u$. It follows from [2, Theorem 2.3.7] that there exists a subsequence $(k(n))_{1}^{\infty}$ satisfying \big{|}\lvert x^{\prime}_{k(n)}\rvert(\sup\{\lvert x_{k(j)}\rvert\wedge u:n\neq j\in\mathbb{N}\})\big{|}<\epsilon. We may assume that $k(n)=n$ for all $n$. For every finitely valued $z\in l_{\infty}$, there exists disjoint subsets $A_{1},A_{2},...,A_{r}\subset N$ and $a_{1},a_{2},...,a_{r}\in R$ such that $z=\sum_{i=1}^{r}a_{i}\cdot\chi_{A_{i}}\in St(\mathbb{N})$. For all $i=1,2,...,r$, we define $v_{i}=\sup\{\lvert x_{n}\rvert\wedge u:n\in A_{i}\}$ and $T_{1}z=\sum_{i=1}^{r}a_{i}Tv_{i}$.

Clearly, $T_{1}$ is well defined and linear on $St(\mathbb{N})$ such that $\lvert y^{\prime}(T_{1}z)\rvert\leq\sum_{i=1}^{r}\lvert a_{i}\rvert\cdot\lvert y^{\prime}(Tv_{i})\rvert\leq\|z\|_{\infty}\|T\|\|u\|$ for all $y^{\prime}\in B_{F^{\prime}}$. Therefore, $\|T_{1}\|\leq\|T\|\|u\|$.

We may assume that $\|z\|_{\infty}=\lvert a_{1}\rvert$. For every $n\in A_{1}$,

[TABLE]

Since $St(\mathbb{N})$ is dense in $\ell_{\infty}$, hence $T_{1}$ extends as an lattice and norm isomorphism to all of $l_{\infty}$. According to Theorem 2.6, the sup-stable sublattice generated by $(\lvert x_{n}\rvert\wedge u)$ is isomorphic to $l_{\infty}$. The proof is completed. ∎

Theorem 2.8**.**

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

1. (1)

$Tx_{n}\xrightarrow{un}0$* for every $uo$-null sequence $(x_{n})\subset B_{E}$.* 2. (2)

$Tx_{n}\xrightarrow{un}0$* for every $uaw$-null sequence $(x_{n})\subset B_{E}$.* 3. (3)

$Tx_{n}\xrightarrow{un}0$* for every disjoint sequence $(x_{n})\subset B_{E}$.* 4. (4)

$T$* is order-weakly compact.*

Proof.

$(1)\Leftrightarrow(2)\Leftrightarrow(3)$ by Theorem 2.5.

$(3)\Rightarrow(4)$ by Lemma 2.4.

$(4)\Rightarrow(3)$. Assume that $(3)$ does not hold, then there exists a bounded disjoint sequence $(x_{n})$ such that $(Tx_{n})$ is not $un$-null. According to Theorem 2.7, $T$ preserves a sublattice isomorphic to $l_{\infty}$. It follows from [2, Corollary 3.4.5] that $T$ is not order-weakly compact. This leads to a contradiction.

$(4)\Rightarrow(1)$ and $(4)\Rightarrow(2)$ by $\lvert Tx_{n}\rvert\wedge v=T\lvert x_{n}\rvert\wedge Tu=T(\lvert x_{n}\rvert\wedge u)$ for all $u\in E_{+},v\in F_{+}$ and [2, Theorem 3.4.4 and Corollary 3.4.9]. ∎

Acknowledgement. The research is supported by National Natural Science Foundation of China(NSFC:51875483).