A generalization of order convergence

TL;DR

This paper introduces and studies a new form of order convergence called $F$-order convergence, explores its properties, and defines $b$-order continuous operators, analyzing their characteristics in vector lattices.

Contribution

It generalizes order convergence by defining $F$-order convergence and introduces $b$-order continuous operators, expanding the theoretical framework of vector lattice analysis.

Findings

Properties of $Fo$-convergence nets are established.

Extension of results to the general case of $Fo$-convergence.

Introduction and analysis of $b$-order continuous operators.

Abstract

Let $E$ be a sublattice of a vector lattice $F$. $\left( x_\alpha \right)\subseteq E$ is said to be $ F $-order convergent to a vector $ x $ (in symbols $ x_\alpha \xrightarrow{Fo} x $), whenever there exists another net $ \left(y_\alpha\right) $ in $F $ with the some index set satisfying $ y_\alpha\downarrow 0 $ in $F$ and $ | x_\alpha - x | \leq y_\alpha $ for all indexes $ \alpha $. If $F=E^{\sim\sim}$, this convergence is called $b$-order convergence and we write $ x_\alpha \xrightarrow{bo} x$. In this manuscript, first we study some properties of $Fo$-convergence nets and we extend some results to the general case. In the second part, we introduce $b$-order continuous operators and we invistegate some properties of this new concept. An operator $T$ between two vector lattices $E$ and $F$ is said to be $b$-order continuous, if $ x_\alpha \xrightarrow{bo} 0 $ in $E$ implies $ Tx_\alpha \xrightarrow{bo} 0$ in $F$.

Equations

$$x=(x_1,x_2,x_3)<y=(y_1,y_2,y_3)$$

$$L_{b^{\sim }}\left(E,F\right):=\{T\in L\left(E,F\right):T\text{ is b-order bounded operator}\}.$$

$$x_{\alpha}=\left\{\begin{array}[]{ll}\alpha&\alpha\in A;\\ 0&\alpha\notin A\end{array}\right.$$

$$L_{n^{\sim }}\left(E,F\right):=\{T\in L_{b^{\sim }}\left(E,F\right):T\text{ is b-order continuous}\},$$

Full text

*∗*Corresponding author

1. Introduction and preliminaries

To state our result, we need to fix some notation and recall some definitions. Let us say that a vector subspace $G$ of an ordered vector space $E$ is majorizing $E$ whenever for each $x\in E$ there exists some $y\in G$ with $x\leq y$. A vector sublattice $G$ of vector lattice $E$ is said to be order dense in $E$ whenever for each $0<x\in E$ there exists some $y\in G$ with $0<y\leq x$. A Dedekind complete vector lattice $E$ is said to be a Dedekind completion of the vector lattice $G$ whenever $E$ is lattice isomorphism to a majorizing order dense sublattice of $E$. A subset $A$ of a vector lattice $E$ is said to be order closed if it follows from $\{x_{\alpha}\}\subseteq A$ and $x_{\alpha}\xrightarrow{o}x$ in $E$ that $x\in A$. A vector sublattice $G$ of vector lattice $E$ is said to be regular, if the embedding of $E$ into $F$ preserves arbitrary suprema and infima. Let $E$, $F$ be vector lattices. An operator $T:E\rightarrow F$ is said to be order bounded if it maps each order bounded subset of $E$ into order bounded subset of $F$. The collection of all order bounded operators from a vector lattice $E$ into a vector lattice $F$ will be denoted by $\mathcal{L}_{b}(E,F)$. The vector space $E^{\sim}$ of all order bounded linear functionals on vector lattice $E$ is called the order dual of $E$, i.e., $E^{\sim}=L_{b}(E,R)$. Let $A$ be a subset of vector lattice $E$ and $Q_{E}$ be the natural mapping from $E$ into $E^{\sim\sim}$. If $Q_{E}(A)$ is order bounded in $E^{\sim\sim}$, then $A$ is said to $b-$order bounded $E$, see [3]. It is clear that every order bounded subset of $E$ is $b-$order bounded. However, the converse is not true in general. For example, $A=\{e_{n}\mid n\in N\}$ $b-$order bounded in $c_{0}$ but $A$ is not order bounded in $c_{0}$. A linear operator between two vector lattices is order continuous (resp. $\sigma$-order continuous) if it maps order null nets (resp. sequences) to order null nets (resp. sequences). The collection of all order continuous (resp. $\sigma$-order continuous) linear operators from vector lattice $E$ into vector lattice $F$ will be denoted by $\mathcal{L}_{n}(E,F)$ (resp. $\mathcal{L}_{c}(E,F)$). For unexplained terminology and facts on Banach lattices and positive operators, we refer the reader to [1, 2].

2. $F$-order convergent on vector lattices

In all parts of this section $E$ is a vector sublattice of vector lattice $F$. Let $A\subseteq E$. We say that $\inf A$ exists in $E$ with respect to $F$, if $\inf A$ exits in $F$ and $\inf A\in E$, in this case we write $\inf_{F}A$ exists,. For a net $(x_{\alpha})_{\alpha}\subseteq E$ and $x\in E$, the notation $x_{\alpha}\downarrow_{F}x$ means that $x_{\alpha}\downarrow$ and $\inf\left(x_{\alpha}\right)=x$ holds in $F$. The meanings of $x_{\alpha}\uparrow$ and $x_{\alpha}\uparrow_{F}x$ are analogous. Obviously if $x_{\alpha}\downarrow_{F}0$, then $x_{\alpha}\downarrow 0$, but as following example the converse in general not holds.

Example 2.1**.**

Assume that $F$ is a set of real valued functions on $[0,1]$ of form $f=g+h$ where $g$ is continuous and $h$ vanishes except at finitely many point. Let $E=C([0,1])$ and $f_{n}(t)=t^{n}$ where $t\in[0,1]$. It is clear that $f_{n}\downarrow 0$ in $E$ and $\inf_{F}f_{n}$ not exists in $E$, but we have $f_{n}\downarrow\chi_{\{1\}}$ in $F$.

It is obvious that if $E$ is regular in $F$, then for each net $(x_{\alpha})_{\alpha}\subseteq E$ and $x\in E$, $x_{\alpha}\downarrow_{F}x$ if and only if $x_{\alpha}\downarrow x$.

The notation $x_{\alpha}\downarrow_{b}x$ means that $x_{\alpha}\downarrow$ and $\inf\left(x_{\alpha}\right)=x$ holds in $E^{\sim\sim}$. The meanings of $x_{\alpha}\uparrow_{b}x$ is analogous.

Definition 2.2**.**

$\left(x_{\alpha}\right)\subseteq E$ is said to be $F$-order convergent (in short $Fo$-convergent) to a vector $x\in E$ (in symbols $x_{\alpha}\xrightarrow{Fo}x$ ), whenever there exists another net $\left(y_{\alpha}\right)$ in $F$ with the some index set satisfying $y_{\alpha}\downarrow 0$ and $|x_{\alpha}-x|\leq y_{\alpha}$ for all indexes $\alpha$.

In the same way, a net $\left(x_{\alpha}\right)$ of $E$ is said to be $b$-order convergent (in short $bo$-convergent) to a vector $x$ (in symbols $x_{\alpha}\xrightarrow{bo}x$ ), whenever there exists another net $\left(y_{\alpha}\right)$ in $E^{\sim\sim}$ with the some index set satisfying $y_{\alpha}\downarrow 0$ and $|x_{\alpha}-x|\leq y_{\alpha}$ for all indexes $\alpha$.

It is clear that every order convergence net in vector lattice $E$ is $F$-order convergent, but as following example the converse in general not holds.

Example 2.3**.**

Suppose that $E=c_{0}$ and $\left(e_{n}\right)$ is the standard basis of $c_{0}$. We know that $\left(e_{n}\right)$ is not order convergence to zero, but $\left(e_{n}\right)$ is $\ell^{\infty}$-order (or $b$-order) convergent to zero.

It can easily be seen that a net in vector lattice $E$ can have at most one $F$-order limit. The basic properties of $Fo$-convergent are summarized in the next theorem.

Theorem 2.4**.**

Assume that the nets $\left(x_{\alpha}\right)$ and $\left(y_{\beta}\right)$ of a vector lattice $E$ satisfy $x_{\alpha}\xrightarrow{Fo}x$ and $y_{\beta}\xrightarrow{Fo}y$. Then we have

1. (1)

$|x_{\alpha}|\xrightarrow{Fo}|x|$; $x^{+}_{\alpha}\xrightarrow{Fo}x^{+}$ and $x^{-}_{\alpha}\xrightarrow{Fo}x^{-}$. 2. (2)

$\lambda x_{\alpha}+\mu y_{\beta}\overset{Fo}{\underset{(\alpha,\beta)}{\longrightarrow}}\lambda x+\mu y$* for all $\lambda,\mu\in\mathbb{R}$.* 3. (3)

$x_{\alpha}\vee y_{\beta}\overset{Fo}{\underset{(\alpha,\beta)}{\longrightarrow}}x\vee y$* and $x_{\alpha}\wedge y_{\beta}\overset{Fo}{\underset{(\alpha,\beta)}{\longrightarrow}}x\wedge y$.* 4. (4)

For each $z\in F$, if $x_{\alpha}\leq z$ for all $\alpha\geq\alpha_{1}$, then $x\leq z$. 5. (5)

If $0\leq x_{\alpha}\leq y_{\alpha}$, then $0\leq x\leq y$. 6. (6)

If $P$ is order projection, then $Px_{\alpha}\xrightarrow{Fo}Px$.

Definition 2.5**.**

$A\subseteq E$ is said to be $F$-order closed whenever $\left(x_{\alpha}\right)\subseteq A$ and $x_{\alpha}\xrightarrow{Fo}x$ imply $x\in A$.

The set $A\subseteq E$ is $F$-order closed means that $A$ is order closed with respect to vector lattice $F$. If $A\subseteq E$ is order closed in $E$, then it is clear that $A$ is $F$-order closed, but but the converse in general not holds. For example $c_{0}$ is order closed, but is not $\ell^{\infty}$-order closed.

Lemma 2.6**.**

Let $A\subseteq E$ be a solid subset of $F$. Then $A$ is $F$-order closed if and only if $\left(x_{\alpha}\right)\subseteq A$ and $0\leq x_{\alpha}\uparrow_{F}x$ imply $x\in A$.

Proof.

Suppose that $A$ is a $F$-order closed, and $\left(x_{\alpha}\right)\subseteq A$ and $0\leq x_{\alpha}\uparrow_{F}x$. Therefore $0\leq|x_{\alpha}-x|\leq x-x_{\alpha}\downarrow_{F}0$. It follows that $x_{\alpha}\xrightarrow{Fo}x$, and since $A$ is order closed, implies that $x\in A$.

Conversely assume that $\left(x_{\alpha}\right)\subseteq A$ and $x_{\alpha}\xrightarrow{Fo}x$. Set a net $\left(y_{\alpha}\right)$ in $F$ with same index net satisfying $y_{\alpha}\downarrow_{F}0$ and $|x_{\alpha}-x|\leq y_{\alpha}$ for each $\alpha$. Since $\left(|x|-y_{\alpha}\right)^{+}\leq|x_{\alpha}|$ for each $\alpha$ and $A$ is solid, follows that $(\left(|x|-y_{\alpha}\right)^{+})_{\alpha}\subseteq A$. Obviously that $0\leq\left(|x|-y_{\alpha}\right)^{+}\uparrow_{b}|x|$, and so by hypothesis we have $x\in A$. Its follows that $A$ is $F$-order closed. ∎

Definition 2.7**.**

1. (1)

$E$ is said to be $F$-Dedekind (or $F$-order) complete, if every nonempty $A\subseteq E$ that is bounded from above in $F$ has supermum in $E$. In case $F=E^{\sim\sim}$, we say that $E$ is $b$-Dedekind complete. 2. (2)

A subset $A$ of $E$ is called $F$-order bounded, if $A$ is order bounded in $F$. In case $F=E^{\sim\sim}$, we say that $A$ is $b$-order bounded. 3. (3)

If each $F$-order bounded subset of $E$ is order bounded in $E$, then $E$ is said to have the $F$-property. In case $F=E^{\sim\sim}$, we say that $E$ has $b$-property.

Remark 2.8*.*

Every majorizing sublattice $E$ of $F$ has the $F$-property. Since $E^{\sim}$ has $b$-property, $E^{\sim}$ is $b$-Dedekind complete. If $E$ is $F$-Dedekind complete, then $E$ is Dedekind complete, but the converse in general not holds, of course $c_{0}$ is Dedekind complete, but is not $\ell^{\infty}$-Dedekind complete. Let $K$ be a compact Hausdorff space and let $C(K)$ and $B(K)$ be vector lattices of real valued continuous and bounded functions on $K$, respectively, under pointwise order and algebric operations. By easy calculation, it is obvious that $C(K)$ is both $B(K)$-Dedekind complete and $b$-Dedekind complete. It is clear that $E$ is $F$-Dedekind complete if and only if $E$ is Dedekind complete with $F$-property. It is easy to show that a vector lattice $E$ has $F$-property if and only if for each net $(x_{\alpha})$ in $E$ with $x_{\alpha}\uparrow\leq y$ for some $y\in F$, $(x_{\alpha})$ is order bounded in $E$.

Let $E$ be a vector sublattice of $F$ and $I$ be an ideal in $E$. In general, $I$ is not an ideal in $F$. For example, set $F=\mathbb{R}^{3}$ and define the order on $F$ in the following way:

[TABLE]

whenever one of the following relations hold

1. (1)

$x_{1}<y_{1}$ or; 2. (2)

$x_{1}=y_{1},~{}x_{2}<y_{2}$ or; 3. (3)

$x_{1}=y_{1},~{}x_{2}=y_{2},~{}x_{3}<y_{3}$.

It is clear $F$ with this order is a vector lattice. Now if we take $E=\{(x,y,0):~{}x,y\in\mathbb{R}\}$ and $I=\{(0,y,0):~{}y\in\mathbb{R}\}$, then obviously that $I$ is an ideal in $E$, but is not ideal in $F$.

The above example shows that the property of being ideal thus depend on the space in which $I$ is embedded. Now if $E$ is $F$-Dedekind complete and order dense in $F$, the following theorem shows that $I$ is an ideal in $E$ if and only if $I$ is an ideal in $F$.

Theorem 2.9**.**

Assume that $E$ is $F$-Dedekind complete. The following statements hold.

1. (1)

Each $F$-order convergence net in $E$ is order convergent in $E$. 2. (2)

If $E$ is order dense in $F$, then $B$ is a band in $E$ if and only if $B$ is a band in $F$. 3. (3)

By assumption (2), if $F=E^{\sim\sim}$, then $E^{\delta}=E^{\sim\sim}$. 4. (4)

If $F=E^{\sim\sim}$ and $E^{\sim}=E_{n}^{\sim}$, then $E$ is perfect.

Proof.

1. (1)

Assume that $(x_{\alpha})_{\alpha}\subseteq E$ is $F$-order convergent to $x$ in $E$. Then there exists a net $(y_{\alpha})_{\alpha}\subseteq F$ such that $y_{\alpha}\downarrow 0$ and $|x_{\alpha}-x|\leqslant y_{\alpha}$ for all indexes $\alpha$. Set $z_{\alpha}=|x_{\alpha}-x|$ and take $w_{\alpha}=\bigvee_{\beta\geqslant\alpha}z_{\beta}$. Since $E$ is $F$-Dedekind complete, $(w_{\alpha})_{\alpha}\subseteq E$. It follows that $|x_{\alpha}-x|\leqslant w_{\alpha}\leqslant y_{\alpha}$ and $w_{\alpha}\downarrow 0$ in $E$. Thus $(x_{\alpha})_{\alpha}$ is order convergence to $x$ in $E$. 2. (2)

If $B$ is a band in $F$, it is clear that $B$ is a band in $E$. Now assume that $B$ is a band in $E$. First we prove that $I$ is an ideal in $F$. Let $x\in F$ and $y\in I$ such that $0<|x|<|y|$. Since $E$ is order dense in $F$, there is a $z\in E$ such that $0<z\leqslant x^{+}\leqslant|x|$, which follows that $z\in B$. Put $\sup\{z\in B:~{}0<z\leqslant x^{+}\}=w$. By $F$-Dedekind completeness of $E$, we have $w\in E$, and so $w\in B$. If $w<x^{+}$, then $0<x^{+}-w$. Since $E$ is order dense in $F$, there is $v\in E$ such that $0<v<x^{+}-w$, and so $0<w+v<x^{+}$. It follows that $w+v\in B$, which is impossible. Thus $w=x^{+}$ belong to $B$. In the same way $x^{-}\in B$, and so $x\in B$. This shows that $B$ is an ideal in $F$. Now since $E$ is $F$-Dedekind complete, by using Lemma 2.6, $B$ is order closed in $F$ and proof follows. 3. (3)

By using Proposition 8 from [7], proof follows. 4. (4)

by using Corollary 10 from [7], proof follows.

∎

3. $b$-order continuous operators

Let $E$ and $F$ be vector lattices. An operator $T:E\rightarrow F$ is called $b$-order bounded operator if it maps $b$-order bounded subset of $E$ into $b$-order bounded subset of $F$. The collection of $b$-order bounded operators will be denoted by:

[TABLE]

An order bounded operator between two vector lattices is $b$-order bounded, but as Example 2.4, [3], the converse, in general, not holds.

Proposition 3.1**.**

Let $E$ and $F$ be vector lattices. $T\in L_{b^{\sim}}\left(E,F\right)$ if and only if $|T|\in L_{b^{\sim}}\left(E,F\right)$

Proof.

Assume that $T\in L_{b^{\sim}}\left(E,F\right)$, we shows that $|T|\in L_{b^{\sim}}\left(E,F\right)$. Let $(x_{\alpha})_{\alpha}$ be a net in $E^{+}$ with $x_{\alpha}\uparrow x^{\prime\prime}$ for some $x^{\prime\prime}\in E^{\sim\sim}$. Let $A$ be the solid hull of $(x_{\alpha})_{\alpha}$ in $E$. Since $(x_{\alpha})_{\alpha}$ is order bounded in $E^{\sim\sim}$, $A$ is $b$-order bounded in $E$. Then $T(A)$ is $b$-order bounded in $F$. Since $F^{\sim\sim}$ is Dedekind complete, $\sup T(A)$ exists in $F^{\sim\sim}$. Let $y\in E$ with $|y|\leq x_{\alpha}$ for fix $\alpha\in I$. Then $y\in A$ and $T(y)\leq\sup T(A)$. It follows that $|T|(x_{\alpha})\leq\sup T(A)$. By Dedekind completeness of $F^{\sim\sim}$, $\sup_{\alpha}|T|(x_{\alpha})$ exists in $F^{\sim\sim}$. This shows that $|T|\in L_{b^{\sim}}\left(E,F\right)$. The converse by easy calculation follows. ∎

The above proposition shows that $L_{b^{\sim}}\left(E,F\right)$ is a lattice, and so is a vector lattice. So it is easy to shows that $L_{b^{\sim}}\left(E,F\right)$ is an ideal in $L\left(E,F\right)$.

Definition 3.2**.**

An operator $T:E\rightarrow F$ between two vector lattices is said to be $b$-order continuous, if $x_{\alpha}\xrightarrow{bo}0$ in $E$ implies $Tx_{\alpha}\xrightarrow{bo}0$ in $F$.

Proposition 3.3**.**

If $T$ is $b$-order continuous operator between two vector lattices $E$ and $F$. Then $T$ is $b$-order bounded.

Proof.

Suppose that $T:E\rightarrow F$ is a $b$-order continuous operator. Let $A=\left[0,x^{\prime\prime}\right]\cap E$ for some $x^{\prime\prime}\in E^{\sim\sim}$. Let $\Lambda=\{\beta:0\leq\beta\leq x^{\prime\prime}\}$ and we write $\alpha\preceq\beta$ if and only if $\alpha\geq\beta$. We consider a net $\left(x_{\alpha}\right)_{\alpha\in\Lambda}$ as follows

[TABLE]

Therefore $x_{\alpha}\xrightarrow{bo}0$, since if we set $y_{\alpha}=\alpha$ then $y_{\alpha}\downarrow_{b}0$ and $|x_{\alpha}|\leq y_{\alpha}$. By the $b$-order continuity of $T$, there exists a net $\left(z_{\alpha}\right)$ of $F^{\sim\sim}$ with the same index $\Lambda$ such that $|Tx_{\alpha}|\leq z_{\alpha}\downarrow_{b}0$. Consequently, if $\alpha\in\Lambda$ then we have $|Tx_{\alpha}|\leq z_{\alpha}\leq z_{x^{\prime\prime}}$ that $z_{x^{\prime\prime}}\in F^{\sim\sim}$, and this show that $T$ is a $b$-order bounded operator. ∎

As above proposition, the class of $b$-order continuous operators is a subspace of $L_{b^{\sim}}\left(E,F\right)$ and will be denoted by $L_{n^{\sim}}\left(E,F\right)$, that is

[TABLE]

Proposition 3.4**.**

Let $E$ and $F$ be vector lattices, $T$ and $S$ are operators from $E$ into $F$ and $0\leq S\leq T$. If $T\in L_{n^{\sim}}\left(E,F\right)$, then $S\in L_{n^{\sim}}\left(E,F\right)$.

Proof.

Let $\left(x_{\alpha}\right)$ be net in $E$ that $x_{\alpha}\xrightarrow{bo}0$. Since $T$ is $b$-order continuous, there exists $y_{\alpha}\in F^{\sim\sim}$ such that $|T|x_{\alpha}||=T|x_{\alpha}|\leq y_{\alpha}\downarrow_{b}0$. On the other hand, $|S\left(x_{\alpha}\right)|\leq S|x_{\alpha}|\leq T|x_{\alpha}|\leq y_{\alpha}\downarrow_{b}0$ for every $x_{\alpha}$ in $E$, and this yields that $S$ is $b$-order continuous. ∎

Lemma 3.5**.**

Let $E$ and $F$ be two vector lattices with Dedekin complete. Then $0<T\in L_{n^{\sim}}\left(E,F\right)$ if and only if for each net $\left(x_{\alpha}\right)$ in $E$, $x_{\alpha}\downarrow_{b}0$ implies $Tx_{\alpha}\downarrow_{b}0$.

Proof.

Assume that $0<T\in L_{n^{\sim}}\left(E,F\right)$ and $x_{\alpha}\downarrow_{b}0$. It follows that $x_{\alpha}\xrightarrow{bo}0$, and so $Tx_{\alpha}\xrightarrow{bo}0$. Then there is a net $\left(z_{\alpha}\right)$ in $F^{\sim\sim}$ such that $Tx_{\alpha}=|Tx_{\alpha}|\leq z_{\alpha}\downarrow 0$, which follows that $Tx_{\alpha}\downarrow_{b}0$.

Conversely, Let $\left(x_{\alpha}\right)\subseteq E$ such that $x_{\alpha}\xrightarrow{bo}0$ in $E$. Then there is a net $\left(y_{\alpha}\right)$ in $E^{\sim\sim}$ such that $|x_{\alpha}|\leq y_{\alpha}\downarrow 0$ in $E^{\sim\sim}$. Set $w_{\alpha}=\bigvee_{\beta\geq\alpha}|x_{\beta}|<z_{\alpha}$. Then we have $w_{\alpha}\downarrow_{b}0$, and so $Tw_{\alpha}\downarrow_{b}0$. Since $|Tx_{\alpha}|\leq T|x_{\alpha}|\leq Tw_{\alpha}$, $Tx_{\alpha}\xrightarrow{bo}0$ in $F$, and proof follows. ∎

As an application of Lemma 3.5, we have the following corollary, in which the techniques of this corollary has been similar argument like as Theorem 1.56 [1] and we omit its proof.

Corollary 3.6**.**

Let $E$ and $F$ be two vector lattices with Dedekind complete. Then the following assertions are equivalent.

1. (1)

$T\in L_{n^{\sim}}\left(E,F\right)$. 2. (2)

$x_{\alpha}\downarrow_{b}0$* implies $Tx_{\alpha}\downarrow_{b}0$.* 3. (3)

$x_{\alpha}\downarrow_{b}0$* implies $\inf_{b}|Tx_{\alpha}|=0$.* 4. (4)

$T^{-},~{}T^{+}~{}\text{and}~{}|T|$* belong to $L_{n^{\sim}}\left(E,F\right)$.*

Proposition 3.7**.**

Let $E$ and $F$ be both vector lattices. Then we have the following assertions.

1. (1)

If $F$ is a $b$-Dedekind complete, then $L_{b}(E,F)=L_{b^{\sim}}(E,F)$. 2. (2)

If $E$ and $F$ are both $b$-Dedekind complete, then $L_{n^{\sim}}\left(E,F\right)=L_{n}\left(E,F\right)$. 3. (3)

$L_{n^{\sim}}\left(E,F\right)$* is a band of $L_{b^{\sim}}\left(E,F\right)$.*

Proof.

1. (1)

Obviously that $L_{b}(E,F)\subseteq L_{b^{\sim}}(E,F)$. Now we prove $L_{b^{\sim}}(E,F)\subseteq L_{b}(E,F)$. Let $A$ be an order bounded subset of $E$ and $T\in L_{b^{\sim}}(E,F)$. Then $T(A)$ is an $b$-order bounded subset of $F$. Since $F$ be $b$-Dedekind complete, follows that $\sup_{b}T(A)$ exists in $E$. It follows that $T(A)$ is order bounded in $E$, and so proof follows. 2. (2)

Assume that $T\in L_{n}\left(E,F\right)$. Let $\left(x_{\alpha}\right)$ be a net in $E$ that $x_{\alpha}\xrightarrow{bo}0$. Since $E$ is $b$-Dedekind complete, by using Theorem 2.9, $x_{\alpha}\xrightarrow{o}0$. By assumption, we have $Tx_{\alpha}\xrightarrow{o}0$, which follows that $Tx_{\alpha}\xrightarrow{bo}0$, and so $T\in L_{n^{\sim}}\left(E,F\right)$.

Now let $T\in L_{n}\left(E,F\right)$ and $\left(x_{\alpha}\right)\subseteq E$ such that $x_{\alpha}\xrightarrow{o}0$. It follows that $x_{\alpha}\xrightarrow{bo}0$, and so $Tx_{\alpha}\xrightarrow{bo}0$. By Dedekind completeness of $F$ and another using Theorem 2.9, we have $Tx_{\alpha}\xrightarrow{bo}0$, which follows that $L_{n}\left(E,F\right)\subseteq L_{n^{\sim}}\left(E,F\right)$, and proof down. 3. (3)

Proof has the similar argument from Theorem 1.57 [1].

∎