Applications for Unbounded Convergences in Banach Lattices

Abstract

Several recent papers investigated unbounded convergences in Banach lattices. The focus of this paper is to apply the results of unbounded convergence to the classical Banach lattice theory from a new perspective. Combining all unbounded convergences, including unbounded order (norm, absolute weak, absolute weak*) convergence, we characterize L-weakly compact sets, L-weakly compact operators and M-weakly compact operators on Banach lattices. For applications, we introduce so-called statistical-unbounded convergence and use these convergences to describe KB-spaces and reflexive Banach lattices.

1. Introduction

A net ${(x_{\alpha })}_{\alpha \in A}$ in a Riesz space E is order-convergent to $x\in E$ (write $x_{\alpha }\stackrel{o}{\to }x$) if there exists a net $(y_{\beta })$, 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 $\alpha \ge {\alpha }_0$. The unbounded order convergence was considered firstly by Nakano in [1] and introduced by Wickstead in [2]. A net $(x_{\alpha })$ 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 $({x_{\alpha }}^{\prime })$ 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 theory of $uo$, $un$, $uaw$ and $ua{w}^{*}$-convergence, we refer to [3,4,5,6,7,8,9].

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

In this paper, we characterize the L-weakly compactness in Banach lattices by $uo$, $uaw$ and $ua{w}^{*}$-convergence. Additionally, as applications, we present some characterizations of L- and M-weakly compact operators on Banach lattices. Combining these convergences, we introduce so-called statistical-unbounded convergence and use these convergences to describe $KB$ spaces and reflexive Banach lattices.

Recall that a Riesz space E is an ordered vector space in which $x\vee y=sup\{x,y\}(x\wedge y=inf\{x,y\})$ exists for every $x,y\in E$. The positive cone of E is denoted by $E_{+}$, i.e., $E_{+}=\{x\in L:x\ge 0\}$. For any vector x in E define $x^{+}:=x\vee 0,x^{-}:=(-x)\vee 0,x:=x\vee (-x)$. 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 sequence $(x_n)$ in a Riesz space is called disjoint whenever $n\ne m$ implies $x_n\wedge x_m=0$ (denoted by $x_n\perp x_m$). A set A in E is said order bounded if there exists some $u\in E_{+}$ such that $x\le u$ for all $x\in A$. An operator $E:L\to F$ is called order bounded if it maps order bounded subsets of E to order bounded subsets of F. Throughout this paper, E will stand for a Banach lattice. 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 lattices and linear operators, we refer the reader to [10,11].

2. Applications for Classical Banach Lattice Theory

An element $e\in E_{+}$ is called an atom of the Riesz space E if the principal ideal $E_e$ is one-dimensional. E is called an atomic Banach lattice if it is the band generated by its atoms.

Proposition 1. 

Let E be a Banach lattice. Then, the following statements hold.

1. 

E is atomic and order-continuous iff ${x_{\alpha }}^{\prime }\stackrel{w^{*}}{\to }0\Rightarrow {x_{\alpha }}^{\prime }\stackrel{ua{w}^{*}}{\to }0$ for any net $({x_{\alpha }}^{\prime })$ in $E^{\prime }$.

2. 

E is atomic and reflexive iff ${x_{\alpha }}^{\prime }\stackrel{ua{w}^{*}}{\to }0\iff {x_{\alpha }}^{\prime }\stackrel{w}{\to }0$ for any bounded net $({x_{\alpha }}^{\prime })$ in $E^{\prime }$.

Proof. 

$(1)\Rightarrow$ For a $w^{*}$-null net $({x_{\alpha }}^{\prime })\subset E^{\prime }$, it follows from [4] (Theorem 3.4) that ${x_{\alpha }}^{\prime }\stackrel{uo}{\to }0$; therefore, ${x_{\alpha }}^{\prime }\stackrel{ua{w}^{*}}{\to }0$.

$(1)\Leftarrow$ If E is not order-continuous, then there exists a bounded disjoint sequence $({x_n}^{\prime })$ which is not $w^{*}$-null by [11] (Corollary 2.4.3). Therefore, we can find some subnet $({x_{\alpha }}^{\prime })$ of $({x_n}^{\prime })$ which $w^{*}$-converges to a non-zero functional $x^{\prime }\ne 0$ on E. Hence, $({x_{\alpha }}^{\prime }-x^{\prime })\stackrel{ua{w}^{*}}{\to }0$. However, ${x_{\alpha }}^{\prime }\stackrel{ua{w}^{*}}{\to }0$ since ${x_n}^{\prime }\stackrel{ua{w}^{*}}{\to }0$. This yields a contradiction.

Assume that E is not atomic. Then, $E^{\prime }$ is not atomic by [12] (Corollary 2.3). We construct a sequence which is $w^{*}$-null but not $ua{w}^{*}$-null. It follows from [13] (Theorem 3.1) that there exists a sequence $({x_n}^{\prime })\subset E^{\prime }$ such that ${x_n}^{\prime }\stackrel{w^{*}}{\to }0$ and $|{x_n}^{\prime }|=x^{\prime }>0$ for all $n\in N$ and some $x^{\prime }\in E^{\prime }$. However, $({x_n}^{\prime })$ is not $ua{w}^{*}$-null since $|{x_n}^{\prime }|\wedge x^{\prime }=x^{\prime }\nrightarrow 0$ by $\sigma (E^{\prime },E)$, which is absurd.

$(2)$ by [8] (p. 87) and $(1)$. □

Recall that a bounded subset A in a Banach lattice E is said to be L-weakly compact if every disjoint sequence in the solid hull of A is convergent to zero. $E^a$ is the maximal ideal in Banach lattice E on which the induced norm is order-continuous as:

$$E^a=\{x\in E:\mathrm{every}\mathrm{monotone}\mathrm{sequence}\mathrm{in}[0,|x|]\mathrm{is}\mathrm{convergent}\}.$$

(Ref. [11] p. 71) Let E be Riesz space. For all non-empty subsets $A\subset E$ and $B\subset E^{\sim }$, we define the generated absolutely monotone seminorms:

$${\rho }_A(x^{\prime })=sup\{|x^{\prime }|(|x|):x\in A\},{\rho }_B(x)=sup\{|x^{\prime }|(|x|):x^{\prime }\in B\},$$

for $x^{\prime }\in E^{\sim }$ and $x\in E$.

The following results characterize L-weakly compactness in Banach lattices by $uo$, $uaw$ and $ua{w}^{*}$-convergence.

Theorem 1. 

Let E be a Banach lattice; then, the following statements hold.

1. 

For a non-empty bounded subset $A\subset E$, the following conditions are equivalent.

(a) 

A is L-weakly compact.

(b) 

${sup}_{x\in A}|{x_n}^{\prime }(x)|\to 0$ for every norm-bounded $ua{w}^{*}$-null sequence $({x_n}^{\prime })\subset E^{\prime }$.

(c) 

${sup}_{x\in A}|{x_n}^{\prime }(x)|\to 0$ for every norm-bounded $uaw$-null sequence $({x_n}^{\prime })\subset E^{\prime }$.

(d) 

${sup}_{x\in A}|{x_n}^{\prime }(x)|\to 0$ for every norm-bounded $uo$-null sequence $({x_n}^{\prime })\subset E^{\prime }$.

2. 

For a non-empty bounded subset $B\subset E^{\prime }$, B is L-weakly compact if and only if ${sup}_{x^{\prime }\in B}|x^{\prime }(x_n)|\to 0$ for every norm-bounded $uaw$-null sequence $(x_n)\subset E$.

Proof. 

$(1)(b)\Rightarrow (1)(c)$ Every $uaw$-null net is $ua{w}^{*}$-null.

$(1)(b)\Rightarrow (1)(d)$ Every $uo$-null net is $ua{w}^{*}$-null.

$(1)(d)\Rightarrow (1)(a)$ Let $({x_n}^{\prime })$ be an arbitrary disjoint sequence in $B_{E^{\prime }}$. To prove that A is L-weakly compact, we only need to show that ${\rho }_A({x_n}^{\prime })\to 0$ by [11] (Proposition 3.6.2(2)), where ${\rho }_A(f)$ is defined by:

$${\rho }_A(f)=sup\{|f|(|x|):x\in A\}=sup\{|g(x)|:|g|\le |f|,x\in A\}$$

for every $f\in E^{\prime }$. Assume by way of contradiction that ${\rho }_A({x_n}^{\prime })\nrightarrow 0$. Then, by passing to a subsequence if necessary, we can suppose that there would exist some $ϵ>0$ satisfying ${\rho }_A({x_n}^{\prime })=sup\{|{x_n}^{\prime }|(x):x\in A\}>ϵ$ for all n. For each n choose some $x_n\in A$ and some $({y_n}^{\prime })$ in $E^{\prime }$ with $|{y_n}^{\prime }|\le |{x_n}^{\prime }|$ such that $|{y_n}^{\prime }(x_n)|>ϵ$. Clearly, $({y_n}^{\prime })$ is likewise a norm-bounded disjoint sequence. Hence, ${y_n}^{\prime }\stackrel{uo}{\to }0$. Therefore, $({y_n}^{\prime })$ converges uniformly to zero on A. This leads to a contradiction.

$(1)(c)\Rightarrow (1)(a)$ is similar to $(1)(d)\Rightarrow (1)(a)$.

$(1)(a)\Rightarrow (1)(b)$ According to [11] (Proposition 3.6.2), for any $ϵ>0$, there exists some $x\in E_{+}^a$ such that $A\subset [-x,x]+ϵ·B_E$.

First of all, we prove that ${y_n}^{\prime }(x)\to 0$ for every disjoint sequence $({y_n}^{\prime })$ in $B_{E^{\prime }}$. Every monotone sequence in $[0,x]$ is norm convergent, since $x\in E_{+}^a$. According to [11] (Corollary 2.3.6), every disjoint sequence in $[-x,x]$ is norm convergent. It follows from [11] (Theorem 2.3.3) that ${\rho }_{[-x,x]}({x_n}^{\prime })\to 0$ for every disjoint sequence $({y_n}^{\prime })\subset B_{E^{\prime }}$. Therefore, ${y_n}^{\prime }(x)\to 0$.

Then, we claim that every bounded $ua{w}^{*}$-null sequence $({x_n}^{\prime })$ converges uniformly on $[-x,x]$. We can find a sequence of functionals $({z_n}^{\prime })$ such that ${y_n}^{\prime }(x)=|{z_n}^{\prime }(x)|$ with $|{z_n}^{\prime }|\le |{y_n}^{\prime }|$ since $f(x)=max\{|g(x)|:|g|\le |f|\}$. Clearly, $({z_n}^{\prime })$ is also disjoint. Hence, ${y_n}^{\prime }(x)\to 0$. When applying [10] (Theorem 4.36) to the seminorm $·(x)$, the identity operator and $B_{E^{\prime }}$, we have that, for any $ϵ>0$, there exists some $u^{\prime }\in {E_{+}}^{\prime }$ such that:

$$\underset{x^{\prime }\in B_{E^{\prime }}}{sup}(x^{\prime }-x^{\prime }\wedge u^{\prime })(x)=\underset{x^{\prime }\in B_{E^{\prime }}}{sup}\left({(x^{\prime }-u^{\prime })}^{+}\right)(x)<ϵ.$$

Clearly, $({x_n}^{\prime }\wedge u^{\prime })(|x|)\to 0$. Therefore, ${x_n}^{\prime }(|x|)\to 0$; moreover, ${sup}_{y\in [-x,x]}{x_n}^{\prime }(y)\to 0$.

Finally, according to the arbitrariness of $ϵ$, we have that ${sup}_{x\in A}|{x_n}^{\prime }(x)|\le {sup}_{x\in [-x,x]+ϵ·B_E}|{x_n}^{\prime }(x)|\to 0$.

$(2)\Leftarrow$ is similar to $(1)(d)\Rightarrow (1)(a)$.

$(2)\Rightarrow$ Using [11] (Proposition 3.6.2) and [10] (Theorem 4.36), the proof is similar to $(1)(a)\Rightarrow (1)(b)$. □

Corollary 1. 

Let E be a Banach lattice, A a bounded subset in E and B a bounded subset in $E^{\prime }$. The following statements hold.

1. 

A is L-weakly compact set iff ${x_n}^{\prime }(x_n)\to 0$ for every bounded $ua{w}^{*}$-null sequence $({x_n}^{\prime })$ in $E^{\prime }$ and every sequence $(x_n)$ in A iff ${x_n}^{\prime }(x_n)\to 0$ for every bounded $uaw$-null sequence $({x_n}^{\prime })$ in $E^{\prime }$ and every sequence $(x_n)$ in A iff ${x_n}^{\prime }(x_n)\to 0$ for every bounded $uo$-null sequence $({x_n}^{\prime })$ in $E^{\prime }$ and every sequence $(x_n)$ in A.

2. 

B is L-weakly compact set iff ${x_n}^{\prime }(x_n)\to 0$ for every bounded $uaw$-null sequence $(x_n)$ in E and every sequence $({x_n}^{\prime })$ in B.

Theorem 2. 

Let E be a Banach lattice, for bounded solid subsets A of E and B of $E^{\prime }$. The following statements hold.

1. 

If E has order-continuous norm, then the following conditions are equivalent.

(a) 

A is L-weakly compact set.

(b) 

For every positive disjoint sequence $(x_n)$ in A and each bounded $ua{w}^{*}$-null sequence $({x_n}^{\prime })$ in $E^{\prime }$, ${x_n}^{\prime }(x_n)\to 0$.

(c) 

For every positive disjoint sequence $(x_n)$ in A and each bounded $uaw$-null sequence $({x_n}^{\prime })$ in $E^{\prime }$, ${x_n}^{\prime }(x_n)\to 0$.

(d) 

For every positive disjoint sequence $(x_n)$ in A and each bounded $uo$-null sequence $({x_n}^{\prime })$ in $E^{\prime }$, ${x_n}^{\prime }(x_n)\to 0$.

2. 

If $E^{\prime }$ has order-continuous norm, then B is L-weakly compact iff ${x_n}^{\prime }(x_n)\to 0$ for every positive disjoint sequence $({x_n}^{\prime })\subset B$ and each bounded $uaw$-null sequence $(x_n)$ in E.

Proof. 

$(1)(a)\Rightarrow (1)(b)$ by Corollary 1.

$(1)(b)\Rightarrow (1)(a)$ Let $({x_n}^{\prime })$ be an arbitrary bounded $ua{w}^{*}$-null sequence in $E^{\prime }$. To finish the proof, we have to show that ${sup}_{x\in A}|{x_n}^{\prime }(x)|\to 0$. Assume by way of contradiction that ${sup}_{x\in A}|{x_n}^{\prime }(x)|\nrightarrow 0$. Then, by passing to a subsequence if necessary, we can suppose that there must exist some $ϵ>0$ such that ${sup}_{x\in A}|{x_n}^{\prime }(x)|>ϵ$ for all n. Note that the equality ${sup}_{x\in A}|{x_n}^{\prime }(x)|={sup}_{0\le x\in A}|{x_n}^{\prime }|(x)$ holds, since A is solid. $|{x_n}^{\prime }|\stackrel{w^{*}}{\to }0$ since $E^{\prime }$ is order-continuous. Let $n_1=1$. As $|{x_n}^{\prime }|(4x_{n_1})\to 0$, there exists some $1<n_2\in \mathbb{N}$ such that $|{x_{n_2}}^{\prime }|(4x_{n_1})<\frac{1}{2}$. It is easy to see that we can find a strictly increasing subsequence ${(n_k)}_{k=1}^{\infty }\subset \mathbb{N}$ such that $|{x_{n_{m+1}}}^{\prime }|(4^m{\sum }_{k=1}^m{x}_{n_k})<\frac{1}{m}$ for all m. Let:

$$x=\sum _{k=1}^{\infty }{2}^{-k}{x}_{n_k},y_m={(x_{n_{m+1}}-4^m\sum _{k=1}^m{x}_{n_k}-2^{-m}x)}^{+}.$$

According to [10] (Lemma 4.35), $(y_m)$ is a disjoint sequence in $A\cap E_{+}$. Now, we have:

$$\begin{array}{cc}\hfill |{x_{n_{m+1}}}^{\prime }|(y_m)& =|{x_{n_{m+1}}}^{\prime }|{(x_{n_{m+1}}-4^m\sum _{k=1}^m{x}_{n_k}-2^{-m}x)}^{+}\hfill \\ \hfill & \ge |{x_{n_{m+1}}}^{\prime }|(x_{n_{m+1}}-4^m\sum _{k=1}^m{x}_{n_k}-2^{-m}x)\hfill \\ \hfill & =|{x_{n_{m+1}}}^{\prime }|(x_{n_{m+1}})-|{x_{n_{m+1}}}^{\prime }|(4^m\sum _{k=1}^m{x}_{n_k})-2^{-m}|{x_{n_{m+1}}}^{\prime }|x\hfill \\ \hfill & >ϵ-\frac{1}{m}-2^{-m}|{x_{n_{m+1}}}^{\prime }|x.\hfill \end{array}$$

Let $m\to \infty$, it is clear that $2^{-m}|{x_{n_{m+1}}}^{\prime }|x\to 0$. Hence, $|{x_{n_{m+1}}}^{\prime }|(y_m)\nrightarrow 0$. This leads to a contradiction.

The rest of the proof is similar. □

Recall that a continuous operator $T:E\to F$ from a Banach space to a Banach lattice is said to be L-weakly compact whenever $T(B_E)$ isa L-weakly compact set in F.

As applications of Theorem 1 and Corollary 1, the following results characterize L-weakly compact operators on Banach lattices.

Theorem 3. 

Let E be a Banach space and F a Banach lattice. The following statements hold.

1. 

For a continuous operator $T:E\to F$, the following conditions are equivalent.

(a) 

T is L-weakly compact.

(b) 

${y_n}^{\prime }(T{x}_n)\to 0$ for each sequence $(x_n)$ in $B_E$ and every bounded $ua{w}^{*}$-null sequence $({y_n}^{\prime })$ in $F^{\prime }$.

(c) 

${y_n}^{\prime }(T{x}_n)\to 0$ for each sequence $(x_n)$ in $B_E$ and every bounded $uaw$-null sequence $({y_n}^{\prime })$ in $F^{\prime }$.

(d) 

${y_n}^{\prime }(T{x}_n)\to 0$ for each sequence $(x_n)$ in $B_E$ and every bounded $uo$-null sequence $({y_n}^{\prime })$ in $F^{\prime }$.

(2)

For a continuous operator $S:E\to F^{\prime }$, S is L-weakly compact iff $S{x}_n(y_n)\to 0$ for each sequence $(x_n)$ in $B_E$ and every bounded $uaw$-null sequence $(x_n)$ in F.

Theorem 4. 

Let E and F be Banach lattices. Then, the following statements hold.

1. 

For a positive operator $T:E\to F$, if F is order-continuous, then the following conditions are equivalent.

(a) 

T is L-weakly compact.

(b) 

${y_n}^{\prime }(T{x}_n)\to 0$ for each positive disjoint sequence $(x_n)\subset B_E$ and every bounded $ua{w}^{*}$-null sequence $({y_n}^{\prime })$ in $F^{\prime }$.

(c) 

${y_n}^{\prime }(T{x}_n)\to 0$ for each positive disjoint sequence $(x_n)\subset B_E$ and every bounded $uaw$-null sequence $({y_n}^{\prime })$ in $F^{\prime }$.

(d) 

${y_n}^{\prime }(T{x}_n)\to 0$ for each positive disjoint sequence $(x_n)\subset B_E$ and every bounded $uo$-null sequence $({y_n}^{\prime })$ in $F^{\prime }$.

2. 

For a positive operator $S:E\to F^{\prime }$, if $F^{\prime }$ is order-continuous norm, then S is L-weakly compact iff $S{x}_n(y_n)\to 0$ for each positive disjoint sequence $(x_n)$ in $B_E$ and every bounded $uaw$-null sequence $(x_n)$ in F.

Proof. 

$(1)(a)\Rightarrow (1)(b)$ by Theorem 3.

$(1)(b)\Rightarrow (1)(a)$ Let $({y_n}^{\prime })$ be an arbitrary bounded $ua{w}^{*}$-null sequence in $F^{\prime }$. ${y_n}^{\prime }\stackrel{w^{*}}{\to }0$ since F is order-continuous. Hence, $|T^{\prime }({y_n}^{\prime })|(z)={sup}_{y\in T[-z,z]}|{y_n}^{\prime }(y)|\to 0$ for each $z\in E_{+}$. Without loss of generality, ${y_n}^{\prime }\ge 0$ for all n. To finish the proof, we have to show that ${sup}_{x\in B_E}|{y_n}^{\prime }(Tx)|\to 0$. Assume by way of contradiction that ${sup}_{x\in B_E}|{y_n}^{\prime }(Tx)|\nrightarrow 0$. Then, by passing to a subsequence if necessary, we can suppose that there would exist some $ϵ>0$ such that ${sup}_{x\in B_E}|{y_n}^{\prime }(Tx)|>ϵ$ for all n. Note that the equality ${sup}_{x\in B_E}|{y_n}^{\prime }(Tx)|={sup}_{0\le x\in B_E}\{|{y_n}^{\prime }\circ T|(x)\}$ since A is solid. For every n, there exists $z_n$ in $B_E\cap E_{+}$ such that $|T^{\prime }({y_n}^{\prime })|(z_n)>ϵ$. It is similar to the proof of Theorem 2 that there exists a subsequence $(y_n)$ of $(z_n)$ and a subsequence $(g_n)$ of $({y_n}^{\prime })$ such that:

$$|g_n\circ T|(y_n)>ϵ,|g_{n+1}\circ T|(4^n\sum _{i=1}^n{y}_i)<\frac{1}{n}.$$

Let $x={\sum }_{i=1}^{\infty }{2}^{-i}{y}_i$ and $x_n={(y_{n+1}-4^n({\sum }_{i=1}^n{y}_i)-2^{-n}x)}^{+}$. According to [10] (Lemma 4.35), $(x_n)$ is positive and disjoint. Hence,

$$|g_{n+1}\circ T|(x_n)\ge |g_{n+1}\circ T|(y_{n+1}-4^n(\sum _{i=1}^n{y}_i)-2^{-n}x)>ϵ-\frac{1}{n}-2^{-n}|g_{n+1}\circ T|x.$$

Therefore, $|g_{n+1}\circ T|(x_n)\nrightarrow 0$. Clearly, there exists a sequence $(u_n)$ in E satisfying $|u_n|\le x_n$ such that $|g_{n+1}(T{u}_n)|=|g_{n+1}\circ T|(x_n)$. As applications of:

$$|g_{n+1}(T{u}_n^{+})|+|g_{n+1}(T{u}_n^{-})|\ge |g_{n+1}(T{u}_n)|=|g_{n+1}\circ T|(x_n)\nrightarrow 0,$$

we have $g_{n+1}(T{u}_n^{+})\nrightarrow 0$. This leads to a contradiction.

The rest of the proof is similar. □

An operator $T:E\to F$ from a Banach lattice to a Banach space is said to be M-weakly compact if $T{x}_n\to 0$ for every disjoint sequence $(x_n)$ in $B_E$.

The following result shows some characterizations of M-weakly compact operators on Banach lattices.

Theorem 5. 

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

1. 

T is M-weakly compact iff $T{x}_n\to 0$ for every $uaw$-null sequence $(x_n)$ 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 $({y_n}^{\prime })$ in $B_{F^{\prime }}$.

(c) 

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

(d) 

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

Proof. 

$(1)\Leftarrow$ Every disjoint sequence is $uaw$-null.

$(1)\Rightarrow$ For a bounded $uaw$-null sequence $(x_n)$ in E. According to [11] (Proposition 3.6.11), we can find that $T^{\prime }:F^{\prime }\to E^{\prime }$ is L-weakly compact. Hence, ${sup}_{x^{\prime }\in T^{\prime }({B_F}^{\prime })}|x^{\prime }(x_n)|\to 0$ for all $uaw$-null sequences $(x_n)$ in $B_E$ by Theorem 1. Therefore, $\parallel T{x}_n\parallel ={sup}_{x^{\prime }\in T^{\prime }({B_F}^{\prime })}|x^{\prime }(x_n)|\to 0$ since $x^{\prime }(x_n)=(T^{\prime }{y}^{\prime })(x_n)=(y^{\prime }\circ T)(x_n)$.

In other words, according to [10] (Theorem 5.60), for any $ϵ>0$, there exists some $u\in E_{+}$ such that:

$$\parallel T(x-x\wedge u)\parallel =\parallel T\left({(x-u)}^{+}\right)\parallel <ϵ$$

holds for all $x\in B_E$. Clearly, T is order-weakly compact. It follows from [11] (Corollary 3.4.9) that $T(x_n\wedge u)\to 0$. Hence, $T{x}_n\to 0$. Therefore, $T{x}_n\to 0$ since $x_n\stackrel{uaw}{\to }0$ iff $x_n^{\pm }\stackrel{uaw}{\to }0$ iff $x_n\stackrel{uaw}{\to }0$.

$(2)$ It is an application of Theorem 1 and ${sup}_{y\in T(B_E)}|{y_n}^{\prime }(y)|=\parallel T^{\prime }{y_n}^{\prime }\parallel$. □

The above results are not true if we replace these convergences to $un$-convergence.

Example 1. 

Every bounded $un$-null sequence in $\left(ba(2^{\mathbb{N}}\right){)}^{\prime }={({({\ell }_{\infty })}^{\prime })}^{\prime }$ and ${\ell }_{\infty }$ converges uniformly to zero on the unit ball of $ba(2^{\mathbb{N}})=({\ell }_{\infty }){}^{\prime }$ since $({\ell }_{\infty })$ and $\left(ba(2^{\mathbb{N}}\right){)}^{\prime }$ have order unit. However, $B_{ba(2^{\mathbb{N}})}$ is not L-weakly compact.

Every bounded $un$-null sequence in $\left(ba(2^{\mathbb{N}}\right){)}^{\prime }={({({\ell }_{\infty })}^{\prime })}^{\prime }$ and ${\ell }_{\infty }$ converges uniformly to zero on $I(B_{ba(2^{\mathbb{N}})})$ (I is the identity operator on $B_{ba(2^{\mathbb{N}})}$). However, I is not L-weakly compact.

The identity operator $\widehat{I}$ on ${\ell }_{\infty }$ maps a bounded $un$-null sequence to a norm-null sequence. However, $\widehat{I}$ is not M-weakly compact.

3. Applications for Statistical Convergence

Statistical convergence has been studied in functional analysis literature, and also been researched in Riesz spaces and Banach lattices recently in [14,15]. Let $K\subset N$ be the natural density of K defined by:

$$\delta (K)=\underset{n\to \infty }{lim}\frac{1}{n}|\{k\in K:k\le n\}|$$

where $|·|$ denotes the cardinality of the set. A sequence $(x_n)$ in Riesz space E is statistically monotonically increasing (resp. decreasing) if $(x_k)$ is increasing (resp. decreasing) for $k\in K$ with $\delta (K)=1$. We write $x_n{\uparrow }^{st}$ (resp. $x_n{\downarrow }^{st}$). An increasing or decreasing sequence will be called a statistical monotonic sequence; moreover, if ${sup}_{k\in K}{x}_k=x$ (resp. $in{f}_{k\in K}{x}_k=x$) for some $x\in E$, then $(x_n)$ is statistical monotonousness convergent to x. Hence, we write $x_n{\uparrow }^{st}x$ (resp. $x_n{\downarrow }^{st}x$). A sequence $(x_n)$ in E is statistical-order-convergent to $x\in E$ provided that there exists a sequence $p_n{\downarrow }^{st}0$ such that $|x_n-x|\le p_n$. We call x the statistical-order limit of $(x_n)$ ($x_n\stackrel{st-o}{\to }x$ for short). It is easy to get that $\delta \{n\in N:|x_n-x|\nleqq p_n\}=0$. If $(x_n)$ is a sequence such that $(x_n)$ satisfies property P for all n except a set of natural density zeros, then we say that $(x_n)$ satisfies property P for almost all n and abbreviate this by $a.a.n$. It is clear that $x_n\stackrel{st-o}{\to }x$ iff $x_n\stackrel{o}{\to }x$ for $a.a.n$. A sequence $(x_n)$ in Banach space E is statistical norm convergent to $x\in E$ provided that $\delta \{n\in N:\parallel x_n-x\parallel \ge ϵ\}=0$ for all $ϵ>0$. We write $x_n\stackrel{st-n}{\to }x$. We call x the statistical norm limit of $(x_n)$. A sequence $(x_n)$ in E is statistical weak convergent to $x\in E$ provided that $\delta \{n\in N:|f(x_n-x)|\ge ϵ\}=0$ for all $ϵ>0$, $f\in E^{{}^{\prime }}$. We write $x_n\stackrel{st-w}{\to }x$. We call x the statistically weak limit of $(x_n)$. A sequence $({x_n}^{\prime })$ in $E^{\prime }$ is statistical weak* convergent to $x^{\prime }\in E^{\prime }$ provided that $\delta \{n\in N:|({x_n}^{\prime }-x^{\prime })(x)|\ge ϵ\}=0$ for all $ϵ>0$, $x\in E$. We write ${x_n}^{\prime }\stackrel{st-w^{*}}{\to }x$. We call $x^{\prime }$ the statistically weak limit of $({x_n}^{\prime })$.

Definition 1. 

A sequence $(x_n)$ in a Banach lattice E is statistical-unbounded order (resp. norm, weak) convergent to $x\in E$ if the sequence $(|x_n-x|\wedge u)$ is statistical-order (resp. norm, weak) convergent to zero for all $u\in E^{+}$ (denoted by $x_n\stackrel{st-uo}{\to }x$, $x_n\stackrel{st-un}{\to }x$, $x_n\stackrel{st-uaw}{\to }x$). A sequence $({x_n}^{\prime })$ in a dual Banach lattice $E^{\prime }$ is said to be statistical-unbounded absolute weak* convergent to $x^{\prime }\in E^{\prime }$ if the sequence $({x_n}^{\prime }-x^{\prime }\wedge u^{\prime })$ is statistically weak* convergent to zero and denoted by $x_n\stackrel{st-ua{w}^{*}}{\to }x$.

Clearly, $x_n\stackrel{st-uo}{\to }x$ (resp. $x_n\stackrel{st-un}{\to }x$, $x_n\stackrel{st-uaw}{\to }x$) is equivalent to $x_n\stackrel{uo}{\to }x$ (resp. $x_n\stackrel{un}{\to }x$, $x_n\stackrel{uaw}{\to }x$) for $a.a.n$ and ${x_n}^{\prime }\stackrel{st-ua{w}^{*}}{\to }x$ is equivalent to ${x_n}^{\prime }\stackrel{ua{w}^{*}}{\to }{x}^{\prime }$ for $a.a.n$.

The following results shows the basic properties of these convergences.

Proposition 2. 

For the sequences $(x_n)$, $(y_n)$ and $(z_n)$ in a Riesz space E, the following hold:

1. 

The $st$–$uo$-limit is unique;

2. 

Suppose $x_n\stackrel{st-uo}{\to }x$ and $y_n\stackrel{st-uo}{\to }y$, then $a{x}_n+b{y}_n\stackrel{st-uo}{\to }ax+by$ for any $a,b\in R$;

3. 

$x_n\stackrel{st-uo}{\to }x$ iff $(x_n-x)\stackrel{st-uo}{\to }0$;

4. 

The lattice operations ∨, ∧ and · are $st$–$uo$ continuous;

5. 

$x_n\stackrel{st-uo}{\to }x$ iff $x_n^{\pm }\stackrel{st-uo}{\to }{x}^{\pm }$. In this case, $|x_n|\stackrel{st-uo}{\to }|x|$;

6. 

If $x_n\stackrel{st-uo}{\to }x$ and $x_n\le y_n\stackrel{st-uo}{\to }y$, then $x\le y$;

7. 

If $x_n\stackrel{uo}{\to }x$, then $x_n\stackrel{st-uo}{\to }x$;

8. 

If $x_n\stackrel{st-o}{\to }x$, then $x_n\stackrel{st-uo}{\to }x$;

9. 

Suppose that $x_n\le y_n\le z_n$ for all $n\in K\subset N$ with $\delta (K)=1$, if $x_n\stackrel{st-uo}{\to }x$ and $z_n\stackrel{st-uo}{\to }x$, then $y_n\stackrel{st-uo}{\to }x$;

10. 

Let $x_n\stackrel{st-uo}{\to }x$ and $x_n\perp y$ for $a.a.n$, then $x\perp y$.

Proof. 

$(1)$ Suppose that $x_n\stackrel{st-uo}{\to }x$ and $x_n\stackrel{st-uo}{\to }y$ for some $x,y\in E$. Then we can find $(p_n)$, $(q_n)$ such that $p_n{\downarrow }^{st}0$ and $q_n{\downarrow }^{st}0$, and sets $K_1=\{n_1^1,n_2^1,\dots \},K_2=\{n_1^2,n_2^2,\dots \}$ with $\delta (K_1)=\delta (K_2)=1$ such that:

$$|x_{n_k^1}-x|\wedge u\le p_{n_k^1}\downarrow 0,|x_{n_k^2}-y|\wedge u\le q_{n_k^2}\downarrow 0.$$

Let $K=K_1\cap K_2=\{n_1,n_2,\dots \}$, since $K_1\cap K_2={(K_1^c\cup K_2^c)}^c$, so $\delta (K)=1$. Thus, we get:

$$0\le |x-y|\wedge u\le |x_{n_k}-x|\wedge u+|x_{n_k}-y|\wedge u\le p_{n_k}+q_{n_k}$$

for every $u\in E_{+}$, which shows that $x=y$;

$(2)$ The scalar-multiplication is obvious. We show that $st-uo$ convergence is additive. It is similar to the proof of $(1)$ in that we can find a set $K=\{n_1,n_2,\dots \}$ with $\delta (K)=1$ such that:

$$x_{n_k}-x\wedge u\le p_{n_k}\downarrow 0,y_{n_k}-y\wedge u\le q_{n_k}\downarrow 0.$$

It follows from:

$$0\le x_{n_k}+y_{n_k}-x-y\wedge u\le x_{n_k}-x\wedge u+y_{n_k}-y\wedge u\le p_{n_k}+q_{n_k}$$

that $x_n+y_n\stackrel{st-uo}{\to }x+y$.

$(3)$ is obvious.

$(4)$ We show the ∨ is $st$–$uo$ continuous, the others are similar. That is, $x_n\vee y_n\stackrel{st-uo}{\to }x\vee y$ whenever $x_n\stackrel{st-uo}{\to }x$ and $y_n\stackrel{st-uo}{\to }y$. Since:

$$x_n\vee y_n-x\vee y\wedge u=x_n\vee y_n-x_n\vee y+x_n\vee y-x\vee y\wedge u\le x_n-x\wedge u+y_n-y\wedge u,$$

according to the proofs of $(1)$ and $(2)$, we have the result.

$(5)$ by $(4)$.

$(6)$ Suppose that $x_n\le y_n$ for all $n\in N$. Then, $y_n-x_n\in E_{+}$ for all $n\in N$. According to $(2)$, we have $y_n-x_n\stackrel{st-uo}{\to }y-x$. It follows form $(5)$ that we have $y-x\in E_{+}$, and hence $x\le y$.

$(7)$ and $(8)$ are obvious.

$(9)$ By assumption, there exist sequences $(p_n)$ and $(q_n)$ and sets $K_1,K_2\in N$ with $\delta (K_1)=\delta (K_2)=1$ such that:

$$x_n-x\wedge u\le p_n\downarrow 0,n\in K_1,$$

$$z_n-x\wedge u\le q_n\downarrow 0,n\in K_2.$$

Thus, we have:

$$y_n-x\wedge u\le x_n-x\wedge u+z_n-x\wedge u\le (p_n+q_n)\downarrow 0$$

for $n\in K\cap K_1\cap K_2\subset N$. That is, $y_n\stackrel{st-uo}{\to }x$.

$(10)$ Since $x_n\stackrel{st-uo}{\to }x$, we have $|x_n|\stackrel{st-uo}{\to }|x|$. According to $|x_n|\wedge |y|=0$ for $a.a.n$, and $|x_n|\wedge |y|\stackrel{st-uo}{\to }|x|\wedge |y|$, it follows that $|x|\wedge |y|=0$, i.e., $x\perp y$. □

Proposition 3. 

For the sequences $(x_n)$, $(y_n)$ and $(z_n)$ in a (dual) Banach lattice E, the following hold:

1. 

The $st$–$un$-limit, $st$–$uaw$-limit and $st$–$ua{w}^{*}$-limit are unique;

2. 

The $st$–$un$-limit, $st$–$uaw$-limit and $st$–$ua{w}^{*}$-limit are linear;

3. 

$x_n\stackrel{st-un}{\to }x$ (resp. $x_n\stackrel{st-uaw}{\to }x$, $x_n\stackrel{st-ua{w}^{*}}{\to }x$) iff $(x_n-x)\stackrel{st-un}{\to }0$ (resp. $(x_n-x)\stackrel{st-uaw}{\to }0$, $(x_n-x)\stackrel{st-ua{w}^{*}}{\to }0$);

4. 

The lattice operations ∨, ∧ and · are $st$–$un$, $st$–$uaw$ and $st$–$ua{w}^{*}$ continuous;

5. 

$x_n\stackrel{st-un}{\to }x$ (resp. $x_n\stackrel{st-uaw}{\to }x$, $x_n\stackrel{st-ua{w}^{*}}{\to }x$) iff $x_n^{\pm }\stackrel{st-un}{\to }x$ (resp. $x_n^{\pm }\stackrel{st-uaw}{\to }x$, $x_n^{\pm }\stackrel{st-ua{w}^{*}}{\to }x$). In this case, $|x_n|\stackrel{st-un}{\to }|x|$ (resp. $|x_n|\stackrel{st-uaw}{\to }|x|$, $|x_n|\stackrel{st-ua{w}^{*}}{\to }|x|$);

6. 

If $x_n\stackrel{st-un}{\to }x$ (resp. $x_n\stackrel{st-uaw}{\to }x$, $x_n\stackrel{st-ua{w}^{*}}{\to }x$) and $x_n\le y_n\stackrel{st-un}{\to }y$ (resp. $x_n\le y_n\stackrel{st-uaw}{\to }y$, $x_n\le y_n\stackrel{st-ua{w}^{*}}{\to }y$), then $x\le y$;

7. 

If $x_n\stackrel{un}{\to }x$ (resp. $x_n\stackrel{uaw}{\to }x$, $x_n\stackrel{ua{w}^{*}}{\to }x$), then $x_n\stackrel{st-un}{\to }x$ (resp. $x_n\stackrel{st-uaw}{\to }x$, $x_n\stackrel{st-ua{w}^{*}}{\to }x$);

8. 

Suppose that $x_n\le y_n\le z_n$ for all $n\in K\subset N$ with $\delta (K)=1$, if $x_n\stackrel{st-un}{\to }x$ (resp. $x_n\stackrel{st-uaw}{\to }x$, $x_n\stackrel{st-ua{w}^{*}}{\to }x$) and $z_n\stackrel{st-un}{\to }x$ (resp. $z_n\stackrel{st-uaw}{\to }x$, $z_n\stackrel{st-ua{w}^{*}}{\to }x$), then $y_n\stackrel{st-un}{\to }x$ (resp. $y_n\stackrel{st-uaw}{\to }x$, $y_n\stackrel{st-ua{w}^{*}}{\to }x$);

9. 

Let $x_n\stackrel{st-un}{\to }x$ (resp. $x_n\stackrel{st-uaw}{\to }x$, $x_n\stackrel{st-ua{w}^{*}}{\to }x$) and $x_n\perp y$ for $a.a.n$, then $x\perp y$.

The next examples show that a subsequence of a $st$–$uo$ (resp. $st$–$un$, $st$–$uaw$, $st$–$ua{w}^{*}$) convergent sequences is not necessarily convergent, and the converse of Proposition 2 (7), Proposition 2 (8) and Proposition 3 (7) does not hold.

Example 2. 

In $l_p(1\le p<+\infty )$, let $(e_n)$ be the standard basis, and

$$x_n=\left\{\begin{array}{cc}{\displaystyle \sum _1^k{e}_{i^2},}\hfill & n=k^2(k\in N);\hfill \\ e_n,\hfill & ortherwise.\hfill \end{array}\right.$$

Since $(e_n)$ is a disjoint sequence, by [5] (Corollary 3.6), $(e_n)$ is unbounded order-convergent to 0; hence, $(x_n)$ is statistical-unbounded order-convergent to 0, but the subsequence $(x_{k^2})$ does not. We can also find that $(x_n)$ is not $st$-order-convergent and $uo$-convergent.

Example 3. 

In $L_p[0,1](1<p<+\infty )$, let:

$$x_n(t)=\left\{\begin{array}{c}0,t\notin [\frac{1}{2^n},\frac{1}{2^{n-1}}];\hfill \\ 1,t\in [\frac{1}{2^n},\frac{1}{2^{n-1}}].\hfill \end{array}\right.$$

and:

$$y_n=\left\{\begin{array}{cc}{\displaystyle \sum _1^{n}i{x}_i,}\hfill & n=k^2(k\in N);\hfill \\ 2^n{x}_n,\hfill & ortherwise.\hfill \end{array}\right.$$

Since $(x_n)$ is disjoint sequence, by [6] (Proposition 3.5), $(x_n)$ is $un$-null (moreover it is $uaw$-null and $ua{w}^{*}$-null), and hence $(y_n)$ is $st$–$un$-null, $st$–$uaw$-null and $st$–$ua{w}^{*}$-null, but the subsequence $(y_{k^2})$ is not. We can also find that $(y_n)$ is not $un$, $uaw$ and $ua{w}^{*}$-convergent.

Let $(x_n)$ be a sequence in Riesz space E. $(x_n)$ is called statistical-order bounded if there exists an order interval $[x,y]$ such that $\delta \{n\in N:x_n\notin [x,y]\}=0$. It is clear that every order bounded sequence is statistical-order bounded, but the converse is not true in general.

Example 4. 

In $l_{\infty }$, let $(e_n)$ be the unit vectors, and:

$$x_n=\left\{\begin{array}{cc}n{e}_n,\hfill & n=k^2(k\in N);\hfill \\ e_n,\hfill & ortherwise.\hfill \end{array}\right.$$

$(x_n)$ is statistical-order bounded sequence, but it is not order bounded.

Then we study when the converse of Proposition 2 (7), Proposition 2 (8) and Proposition 3 (7) hold.

Theorem 6. 

Let $(x_n)$ be a sequence in a Riesz space E. Then, the following hold.

1. 

$x_n\stackrel{st-uo}{\to }x$ in E iff there exsits a sequence $(y_n)$ with $x_n=y_n$ for $a.a.n$ and $y_n\stackrel{uo}{\to }x$.

2. 

If $x_n\uparrow$ and $x_n\stackrel{st-uo}{\to }x$ for some $x\in E$, then $x_n\uparrow x$.

3. 

If $(x_n)$ is statistical-order bounded, then $x_n\stackrel{st-uo}{\to }x\iff x_n\stackrel{st-o}{\to }x$.

4. 

If $(x_n)$ is monotonous, then $x_n\stackrel{o}{\to }x\iff x_n\stackrel{st-o}{\to }x\iff x_n\stackrel{uo}{\to }x\iff x_n\stackrel{st-uo}{\to }x$.

Proof. 

$(1)\Rightarrow$ Obvious.

$(1)\Leftarrow$ Assume that $x_n=y_n$ for $a.a.n$ and $y_n\stackrel{uo}{\to }x$. Then there exists a sequence $(p_n)\downarrow 0$, and moreover $p_n{\downarrow }^{st}0$, and $|y_n-x|\wedge u\le p_n$ for all $u\in E_{+}$. Thus, we have:

$$\{n\le m:|x_m-x|\wedge u\nleqq p_m\}\subset \{n\le m:x_m\ne y_m\}\cup \{n\le m:|y_m-x|\wedge u\nleqq p_m\}.$$

Since $y_n\stackrel{uo}{\to }x$, the second set on the right side is empty. Hence, we have $\delta \{n\in N:|x_n-x|\wedge u\nleqq p_n\}=0$, $i.e$: $x_n\stackrel{st-uo}{\to }x$.

$(2)$ Suppose that $x_n\uparrow x$ and $x_n\stackrel{st-uo}{\to }x$. Fix arbitrary $n\in N$. Then we have $x_m-x_n\in E_{+}$ for all $m\ge n$. It follows from Proposition 2, we have $x_m-x_n\stackrel{st-uo}{\to }x-x_n\in E_{+}$ as $m\to \infty$. Therefore, $x\ge x_n$, we claim that $x_n\uparrow x$. Assume that $y\ge x$. Then, $y\ge x_n$ for all $n\in N$. Thus, $y-x_n\stackrel{st-uo}{\to }y-x\in E_{+}$—that is, $y\ge x$.

$(3)$ Since $x_n\stackrel{st-uo}{\to }x$, $x_n\stackrel{uo}{\to }x$ for $a.a.n$. Additionally, since $(x_n)$ is statistical-order bounded, then there exists $u\in E_{+}$ such that $|x_n|\le u$ for $a.a.n$. Then, by fixing the u, we have $|x_n-x|\wedge 2u=|x_n-x|\stackrel{o}{\to }x$ for $a.a.n$; therefore, $x_n\stackrel{st-ord}{\to }x$.

$(4)$ by $(2)$ and $(3)$. □

Theorem 7. 

Let $(x_n)$ be a sequence in a (dual) Banach lattice E. Then, the following hold.

1. 

$x_n\stackrel{st-un}{\to }x$ (resp. $x_n\stackrel{st-uaw}{\to }x$, $x_n\stackrel{st-ua{w}^{*}}{\to }x$) in E iff there exsits a sequence $(y_n)$ with $x_n=y_n$ for $a.a.n$ and $y_n\stackrel{un}{\to }x$ (resp. $y_n\stackrel{uaw}{\to }x$, $y_n\stackrel{ua{w}^{*}}{\to }x$).

2. 

If $x_n\uparrow$ and $x_n\stackrel{st-un}{\to }x$ (resp. $x_n\stackrel{st-uaw}{\to }x$, $x_n\stackrel{st-ua{w}^{*}}{\to }x$) for some $x\in E$, then $x_n\uparrow x$.

3. 

If $(x_n)$ is statistical-order bounded, then $x_n\stackrel{st-un}{\to }x\iff x_n\stackrel{st-n}{\to }x$.

4. 

If $(x_n)$ is monotonous, then $x_n\stackrel{un}{\to }x\iff x_n\stackrel{st-un}{\to }x$ (resp. $x_n\stackrel{uaw}{\to }x\iff x_n\stackrel{st-uaw}{\to }x$, $x_n\stackrel{ua{w}^{*}}{\to }x\iff x_n\stackrel{st-ua{w}^{*}}{\to }x$).

Proof. 

We prove the $st$–$un$-convergence, and the others are similar.

$(1)\Rightarrow$ Obvious.

$(1)\Leftarrow$ Assume that $x_n=y_n$ for $a.a.n$ and $y_n\stackrel{un}{\to }x$. For any $ϵ>0$, there exits some $n_0\in N$ such that $\parallel |y_n-x|\wedge u\parallel <ϵ$ for all $u\in E_{+}$ and $n\ge n_0$. Thus, we have

$$\{n\le m:\parallel |x_m-x|\wedge u\parallel \ge ϵ\}\subset \{n\le m:x_m\ne y_m\}\cup \{n\le m:\parallel |y_m-x|\wedge u\parallel \ge ϵ\}.$$

Since $y_n\stackrel{un}{\to }x$, the second set on the right side is empty. Hence, we have $\delta \{n\in N:\parallel |x_n-x|\wedge u\parallel \ge ϵ\}=0$, i.e., $x_n\stackrel{st-un}{\to }x$.

$(2)$ Since $x_m-x_n\in E_{+}$ for all $m\ge n$, it follows from Proposition 3 that we have $x_m-x_n\stackrel{st-un}{\to }x-x_n\in E_{+}$ as $m\to \infty$. Therefore, $x\ge x_n$. We claim that $x_n\uparrow x$. Assume that $y\ge x$, then $y\ge x_n$ for all $n\in N$. Thus, $y-x_n\stackrel{st-un}{\to }y-x\in E_{+}$; that is, $y\ge x$.

$(3)$ Since $x_n\stackrel{st-un}{\to }x$, $x_n\stackrel{un}{\to }x$ for $a.a.n$. Additionally, since $(x_n)$ is statistical-order bounded, there exists $u\in E_{+}$ such that $|x_n|\le u$ for $a.a.n$. Then, fixing the u, we have $|x_n-x|\wedge 2u=|x_n-x|\to x$ for $a.a.n$; therefore, $x_n\stackrel{st-n}{\to }x$.

$(4)$ Assume that $x_n\uparrow$ for all $n\in N$ and $x_n\stackrel{st-un}{\to }x$. According to $(2)$, we have $x_n\uparrow x$. We claim that $x_n\stackrel{un}{\to }x$. There exists a set $K=\{n_1,n_2\dots \}\subset N$ with $\delta (K)=1$ such that $x_{n_k}\uparrow x$ and $x_{n_k}\stackrel{un}{\to }x$. For any $ϵ>0$ and $u\in E_{+}$, choose a $n_0\in K$ such that $\parallel |x_{n_0}-x|\wedge u\parallel <ϵ$. Then, for any $n\ge n_0$, we have:

$$0\le |x-x_n|\wedge u\le |x-x_{n_0}|\wedge u.$$

This shows:

$$\parallel |x-x_n|\wedge u\parallel \le \parallel |x-x_{n_0}|\wedge u\parallel <ϵ.$$

Hence, $x_n\stackrel{un}{\to }x$.

The rest of proof is similar. □

A norm on a Banach lattice E is called order-continuous if $\parallel x_{\alpha }\parallel \to 0$ for $x_{\alpha }\downarrow 0$, and E is called $KB$ space if every monotone bounded sequence is convergent. A sequence $(x_n)$ in E is said to be statistical-unbounded order Cauchy, if ${(x_n-x_m)}_{(n,m)}$ is $st$–$uo$-convergent to zero. Let E be a Banach space. $(x_n)$ is called statistical-norm-bounded if for any $C\in R^{+}$ there exists a $\lambda >0$ such that $\delta \{n\in N:\parallel \lambda x_n\parallel >C\}=0$. It is clear that every norm bounded sequence is statistical-norm-bounded, but the converse is not true in general. Refer to Example 2.

The following result gives a equivalent description to $KB$ spaces by $st$–$uo$, $st$–$un$ and $st$–$uaw$-convergence.

Theorem 8. 

Let E be an order-continuous Banach lattice. Then, the following statements are equivalent.

1. 

E is a $KB$ space.

2. 

every statistical-norm-bounded $st$–$uo$-Cauchy sequence in E is $st$–$uo$-convergent.

3. 

every statistical-norm-bounded $st$–$uo$-Cauchy sequence in E is $st$–$un$-convergent.

4. 

every statistical-norm-bounded $st$–$uo$-Cauchy sequence in E is $st$–$uaw$-convergent.

Proof. 

$(1)\Rightarrow (2)$ Assume that E is $KB$ space. It follows form [3] (Theorem 4.7) that every norm bounded $uo$-Cauchy sequence in E is $uo$-convergent. Let $(x_n)$ be a statistical-norm-bounded $st$–$uo$-Cauchy sequence in E. Then, there exit sets $K_1,K_2\subset N$ with $\delta (K_1)=\delta (K_2)=1$ such that $(x_n)$ is norm bounded along the set $K_1$ and $uo$-Cauchy along the set $K_2$. Let $K=K_1\cap K_2$. Then, we have that $\delta (K)=1$ and $(x_n)$ is statistical-norm-bounded and $st$–$uo$-Cauchy along the set K. By assumption, $(x_n)$ is $uo$-convergent; therefore, it is $st$–$uo$-convergent.

$(2)\Rightarrow (1)$ Assume that E is not a $KB$ space. Then, E contains a sublattice lattice isomorphic to $c_0$ by [10] (Theorem 4.61). Let $(x_n)$ be a sequence in $c_0$, and

$$x_n=\left\{\begin{array}{cc}n{e}_n,\hfill & n=k^2(k\in N);\hfill \\ \sum _1^n{e}_i,\hfill & ortherwise.\hfill \end{array}\right.$$

$(x_n)$ is statistical-norm-bounded and $st$–$uo$-Cauchy in E, but it is not $st$–$uo$-convergent.

The rest of the proof is similar. □

The following results characterize reflexive Banach lattices by statistical-unbounded convergence.

Theorem 9. 

For a Banach lattice E, the following conditions are equivalent:

1. 

E is reflexive;

2. 

Every statistical-norm-bounded $st$–$uaw$-Cauchy sequence in E is statistically weak convergent;

3. 

Every statistical-norm-bounded $st$–$ua{w}^{*}$-null sequence in $E^{\prime }$ is statistically weakly null.

4. 

Every statistical-norm-bounded $st$–$uo$-null sequence in $E^{\prime }$ is statistically weakly null.

Proof. 

$(1)\Rightarrow (2)$ Since E is reflexive, E and $E^{{}^{\prime }}$ are $KB$ space by [10] (Theorem 4.70), so we have that every statistical-norm-bounded $st$–$uaw$-Cauchy sequence in E is statistically weakly Cauchy. Since E is $KB$ space, so it is statistically weakly convergent.

$(2)\Rightarrow (1)$ Assume that E is not reflexive. Then, $c_0$ or $l_1$ is lattice embeddable in E by [10] (Theorem 4.71). We claim that E contains no lattice copies of $c_0$ or $l_1$. Suppose that E contains a sublattice isomorphic to $l_1$. Let $(x_n)$ be the sequence in $l_1$ of Example 3. It is $st$–$uaw$-null and statistical-norm-bounded but does not have statistically weakly convergence. Suppose that E contains a sublattice isomorphic to $c_0$. The $(x_n)$ of $c_0$ which is in the proof of Theorem 4 is a $st$–$uaw$-Cauchy sequence, but it is not statistically weak convergent.

$(1)\Rightarrow (3)$ is similar to $(1)\Rightarrow (2)$.

$(3)\Rightarrow (1)$ We claim that $E^{\prime }$ and $E^{″}$ are $KB$ spaces. For any ${x_n}^{\prime }\downarrow 0$, we have $|{x_n}^{\prime }|\stackrel{w^{*}}{\to }0$, hence ${x_n}^{\prime }\stackrel{st-ua{w}^{*}}{\to }0$, by assumption, ${x_n}^{\prime }\stackrel{st-w}{\to }0$. It follows form [15] (Theorem 29) that ${x_n}^{\prime }\stackrel{w}{\to }0$. By Dini type theorem ([10] Theorem 3.52), ${x_n}^{\prime }\to 0$, and hence $E^{\prime }$ is $\sigma$-order-continuous. Additionally, since $E^{\prime }$ is Dedekind complete, $E^{\prime }$ is order-continuous by Nakano theorem ([10] Theorem 4.9); therefore, $E^{\prime }$ is $KB$ space by [10] (Theorem 4.59). We claim that $E^{″}$ is also $KB$ space. Suppose that $E^{″}$ is not $KB$ space. Then, $E^{\prime }$ has a lattice copy of $l_1$ by [11] (Theorem 2.4.14). Let $(x_n)$ be the sequence in $l_1$ of Example 3, $x_n\stackrel{st-ua{w}^{*}}{\to }0$ in $E^{\prime }$, but it is not $st$-w-null in $E^{\prime }$. It is contraction, so $E^{″}$ is $KB$ space. We can complete the proof.

$(1)\Rightarrow (4)$ Every $uo$-null sequence is $ua{w}^{*}$-null.

$(4)\Rightarrow (1)$ is similar to $(3)\Rightarrow (1)$. □

4. Discussion

This paper is the second 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 [16], we showed the equivalent relationship of $uo$, $uaw$ and $ua{w}^{*}$-continuous functionals. We also found that $uo$, $uaw$ and $ua{w}^{*}$ norm-continuous operators are exactly M-weakly compact in many cases in this paper. In the next paper, we show 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 the fourth article, we will introduce and study weak L- and M-weakly compact operators by the type of $un$ to norm continuous operators as part of our continuous and in-depth study of this subject.