Extension of order bounded operators

TL;DR

This paper develops an extension norm for order dense majorizing sublattices of vector lattices and explores how lattice and topological properties extend to these larger spaces, especially for operators into Dedekind complete Banach lattices.

Contribution

It introduces an extension norm for order dense majorizing sublattices and studies the extension of lattice and topological properties of operators into Dedekind complete Banach lattices.

Findings

Extension norm $\

$T^t$ preserves operator norm and lattice homomorphism properties.

Positive operators maintain their norm under extension, and $T^t$ remains a lattice homomorphism.

Abstract

Assume that a normed lattice $E$ is order dense majorizing of a vector lattice $E^t$. There is an extension norm $\Vert.\Vert_t$ for $E^t$ and we extend some lattice and topological properties of normed lattice $(E,\Vert.\Vert)$ to new normed lattice $(E^t,\Vert.\Vert_t$). For a Dedekind complete Banach lattice $F$, $T^t$ is an extension of $T$ from $E^t$ into $F$ whenever $T$ is an order bounded operator from $E$ into $F$. For each positive operator $T$, we have $\Vert T\Vert=\Vert T^t\Vert$ and we show that $T^t$ is a lattice homomorphism from $E^t$ into $F$ and moreover $T^t\in \mathcal{L}_n(E^t,F)$ whenever $0\leq T\in \mathcal{L}_n(E,F)$ and $T(x\wedge y)=Tx \wedge Ty$ for each $0\leq x,y\in E$. We also extend some lattice and topological properties of $T\in \mathcal{L}_b(E,F)$ to the extension operator $T^t\in \mathcal{L}_b(E^t,F)$.

Equations

$$z=|z_1+z_2|\le |z_1|+|z_2|\le w_1+w_2\le |x|+|y|.$$

$$\Vert z\Vert \le \Vert w_1+w_2\Vert \le \Vert w_1\Vert +\Vert w_2\Vert \le \rho (x)+\rho (y).$$

$$\inf \{\Vert y\Vert :|x|\le y,y\in E^{+}\}=\sup \{\Vert z\Vert :z\le |x|,z\in E^{+}\}$$

$$\Vert x{\Vert }_t\ne \inf \{\Vert y\Vert :|x|\le y,y\in E^{+}\}.$$

$$\Vert x{\Vert }_t<\lambda <\inf \{\Vert y\Vert :|x|\le y,y\in E^{+}\}.$$

$$\inf \{\Vert y\Vert :|x|\le y,y\in E\}=\sup \{\Vert z\Vert :z\le |x|,z\in E\},$$

$$p(x)=\inf \{Ty:y\in E,x⩽y\},x\in E^t.$$

$$T^{t}y=(T^{+}{)}^t(y)-(T^{-}{)}^t(y)=T^{+}y-T^{-}y=Ty.$$

$${(T^t)}^{+}(x^t)$$

$$=\underset{0\le y^t\le x^t}{\sup }\underset{0\le z\le y^t}{\sup }{T}^{t}z$$

$$=\underset{0\le y\le x^t}{\sup }Ty$$

$$=\underset{0\le z\le x^t}{\sup }\underset{0\le y\le z}{\sup }Ty$$

$$=\underset{0\le z\le x^t}{\sup }{T}^{+}z$$

$$=(T^{+}{)}^t(x^t).$$

$$|T{|}^t=(T^{+}+T^{-}{)}^t=(T^{+}{)}^t+(T^{-}{)}^t=(T^t{)}^{+}+(T^t{)}^{-}=|T^t|.$$

$$S=S^{+}-S^{-}=(T^{+}{)}^t-(T^{-}{)}^t=(T^t{)}^{+}-(T^t{)}^{-}=T^t.$$

$$Sx=S{x}^{+}-S{x}^{-}=T^t{x}^{+}-T^t{x}^{-}=T^{t}x,$$

$$Ty=\inf \{T(x_{\alpha }\vee y)\}\le \inf T{x}_{\alpha }\le T^{t}x.$$

$$T^t(x\wedge y)$$

$$=\inf \{T(x_{\alpha })\}\wedge \inf \{T(y_{\beta })\}=T^{t}x\wedge T^{t}y.$$

$$A=\{y\in E^{+}:\exists \alpha \text{such that}{x}_{\alpha }\le y\}.$$

$$\Vert T^t{x}_n-s^{\ast }\Vert \le \Vert T^t{x}_n-T{x}_{n,m}\Vert \text{for each}n,m\in \mathbb{N}.$$

$$\Vert T^t{x}_n-s^{\ast }\Vert \le \Vert T^t{x}_n-T{y}_n\Vert +\Vert T{y}_n-s^{\ast }\Vert \to 0.$$

Full text

1 Introduction and Preliminaries

In the section 1.2 of 2 , authors have been studied extension operators on vector latticess. In this paper, we will study this problem in different way and we extend some results to general case. Assume that $E$ is a normed lattice and sublattice of $G$, and $E$ is order dense majorizing of a vector lattice $E^{t}\subseteq G$. The aim of this manuscript are in the following:

We extend the norm from $E$ to $E^{t}$. 2. 2.

Assume that $T$ is an order bounded operator from $E$ into Dedekind complete normed lattice $F$. $T^{t}$ is a linear extension of $T$, from $E^{t}$ into $F$, in the sense that if $S:E^{t}\rightarrow F$ is any operator that extends $T$ by same way, then $T^{t}=S$. 3. 3.

We also extend some lattice and topological properties from $E$ and $T$ for $E^{t}$ and $T^{t}$, respectively.

To state our result, we need to fix some notation and recall some definitions. A Banach lattice $E$ has order continuous norm if $\|x_{\alpha}\|\rightarrow 0$ for every decreasing net $(x_{\alpha})_{\alpha}$ with $\inf_{\alpha}x_{\alpha}=0$. Let $E$, $F$ be Riesz spaces. 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 Riesz space $E$ into a Riesz space $F$ will be denoted by $\mathcal{L}_{b}(E,F)$. A linear operator between two Riesz spaces 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. $\mathcal{L}_{c}(E,F)$). 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 lattice norm $\|.\|$ on a vector lattice $E$ is said to be a Fatou norm (or that $\|.\|$ satisfies the Fatou property) if $0\leq x_{\alpha}\uparrow x$ in $E$ implies $\|x_{\alpha}\|\uparrow\|\ x\|$. $\sigma$-Fatou norm has similar definition. An operator $T:E\rightarrow E$ on a vector lattice is said to be band preserving whenever $T$ leaves all bands of $E$ invariant, i.e., whenever $T(B)\subseteq B$ holds for each band $B$ of $E$. An operator $T:E\rightarrow F$ between two vector lattices is said to preserve disjointness whenever $x\perp y$ in $E$ implies $Tx\perp Ty$ in $F$. For a normed lattice $E$, $E^{\prime}$ is the its order dual and $\sigma(E,E^{\prime})$ is the weak topology for $E$. For unexplained terminology and facts on Banach lattices and positive operators, we refer the reader to 1 ; 2 .

2 Extension of norm to an order completion

Let $E$ be an Archimedean vector lattice. Then there exists a Dedekind complete vector lattice $E^{\delta}$ contains a majorizing order dense vector subspace that is Riesz isomorphism to $E$, which we identify with $E$. $E^{\delta}$ is called the Dedekind completion of $E$. Throughout this manuscript, we will assume that the vector lattices under consideration are Archimedean. $E$ and $G$ are normed lattice and vector lattice, respectively, in which $E$ is order dense majorizing of $G$. The universal completion of a vector lattice $E$ will be denoted by $E^{u}$. By Theorem 7.21 1 , every vector lattice has a unique universal completion.

Theorem 2.1

For each $x\in G$, $\rho(x)=\sup\{\|z\|:~{}z\leq|x|,~{}z\in E^{+}\}$ is a norm for $G$ and moreover $(G,\rho(x))$ is a normed lattice.

Proof

It is clear that $\rho(x)=0$ if and only if $x=0$, and $\rho(\lambda x)=|\lambda|\rho(x)$ for each real number $\lambda$ and $x\in G$. Now we prove that $\rho(x+y)\leq\rho(x)+\rho(y)$ whenever $x,y\in E^{t}$.

Let $x,y\in E^{t}$. Fix $z\in E^{+}$ such that $z\leq|x+y|$ . By Riesz Decomposition property, Theorem 1.10 1 , there are $z_{1},z_{2}\in G$ such that $|z_{1}|\leq|x|$, $|z_{2}|\leq|y|$ and $z=z_{1}+z_{2}$. Since $E$ is order dense in $G$, there are $w_{1},w_{2}\in E^{+}$ such that $|z_{1}|\leq w_{1}\leq|x|$ and $|z_{2}|\leq w_{2}\leq|y|$. It follows that

[TABLE]

Then we have

[TABLE]

Consequently, we have $\sup\{\|z\|:~{}{z\leq|x+y|}~{}\text{and}~{}z\in E^{+}\}\leq\rho(x)+\rho(y)$, which implies that $\rho(x+y)\leq\rho(x)+\rho(y)$.

For a normed lattice $(E,\|.\|)$, assume that $E^{\rho}$ is including members of $x\in G$ which satisfies in the following equality,

[TABLE]

then $\rho$ is defined a real function from $E^{\rho}$ into $[0,+\infty)$, in the sense $\rho(x)$ equal to $(1)$. Obviously that the function $\rho$ satisfies in the following conditions:

$\rho(x)=0$ iff $x=0$ 2. 2.

$\rho(\lambda x)=\lambda\rho(x)$ for each $\lambda\in\mathbb{R}$ and $x\in E^{\rho}$. 3. 3.

$\rho(x+y)\leq\rho(x)+\rho(y)$, whenever $x+y\in E^{\rho}$ for $x,y\in E^{\rho}$.

Example 1

Let $c$ be the collection of all real number sequences which convergence in $\mathbb{R}$ with $\ell^{\infty}$-norm. It is obvious that $c$ is order dense majorizing of $\ell^{\infty}$. By easy calculation, we can prove that $c^{\rho}=\ell^{\infty}$.

The above example make motivation to us for the following question.

Question 1

Is $E^{\rho}$ vector lattice and $(E^{\rho},\rho)$ normed lattice?

It is important to us to know that when $E^{\rho}$ is a vector lattice and moreover $(E^{\rho},\rho)$ is a normed lattice? Now in the following we define a vector lattice $E^{t}\subseteq G$ which is including normed lattice $E$ with extension norm of $E$.

Definition 1

Assume that $E\subseteq E^{t}$ is a vector sublattice of $G$ in which every element of $E^{t}$ satisfies in the equality $(1)$. Then $\rho(x)=\|x\|_{t}$ is defined a norm for $E^{t}$ in the sense equal to $(1)$ which is called $t$-norm for $E^{t}$.

It is easy to show that $\|.\|_{t}$ is a norm for $E^{t}$, and moreover $(E^{t},\|.\|_{t})$ is a normed lattice. Note that $E^{t}$ is not unique and we have $E\subseteq E^{t}\subseteq G$. The aim of this manuscript to find a vector lattice $E^{t}$ in the different from of $E$. So, when we say that $E^{t}$ exists, that is, $E$ is proper sublattice of $E^{t}$. Now in the Theorems 2.1 and 2.2 we show that $E^{t}=G$ whenever $E$ is a Dedekind complete or has order continuous norm, respectively.

Theorem 2.2

By one of the following conditions, the equality $(1)$ holds for each $x\in G$, that is, $E^{t}=G$, $(G,\|.\|_{t})$ is normed lattice and $\|y\|=\|y\|_{t}$ for each $y\in E$.

$E$* is a Dedekind complete.* 2. 2.

$E$* has order continuous norm.*

Proof

By Theorem 2.1, we know that $\|x\|_{t}=\sup\{\|z\|:~{}z\leq|x|,~{}z\in E^{+}\}$ is a norm for vector lattice $G$. By contradiction, assume that

[TABLE]

Consider $\lambda$ is a real number such that

[TABLE]

Let $A=\{y\in E^{+}:~{}|x|\leq y\}$. Since $E$ is order dense in $G$, $A$ is bounded below, and so $A$ has infimum in $E$, by Dedekind completeness of $E$. Take $\inf A=y_{0}$ where $y_{0}\in E$. It is clear that $|x|<y_{0}$ and $\lambda<\|y_{0}\|$. Let the natural number $n$ be enough large such that $\lambda<\|y_{0}\|-\frac{1}{n}\|y_{0}\|$. Put $z_{0}=(1-\frac{1}{n})y_{0}$. Consequently we have $z_{0}\in A$ and $\|z_{0}\|<\|y_{0}\|$, which is impossible. 2. 2.

First we show that

[TABLE]

holds whenever $x\in G$. Set $A=\{z\leq|x|:~{}z\in E^{+}\}$ and $B=\{y\geq|x|:~{}y\in E\}$. Since $E$ is order dense and majorizing of $G$, it follows that $A$ and $B$ are not empty and they are directed sets. We consider the set $A$ as a net $\{z_{\alpha}\}$, where $z_{\alpha}=\alpha$ for each $\alpha\in A$. In the same way we consider $B=\{y_{\beta}\}$, and by using Theorem 1.34 2 , we can write $z_{\alpha}\uparrow|x|$ and $y_{\beta}\downarrow|x|$. Since $z_{\alpha}\leq|x|\leq y_{\beta}$ for each $\alpha$ and $\beta$, it follows that $y_{\beta}-z_{\alpha}\downarrow 0$, and so $0\leq\|y_{\beta}\|-\|z_{\alpha}\|\leq\|y_{\beta}-z_{\alpha}\|\rightarrow 0$. It follows that $\|x\|_{t}=\inf\|y_{\beta}\|=\sup\|z_{\alpha}\|$. Obvious that $\|.\|_{t}$ is a norm for $G$ and $(G,\|.\|_{t})$ is a normed lattice.

In the Example 1, we see that $c^{t}=\ell^{\infty}$, but $c$ is not Dedekind complete and has not continuous norm. On the other hand, Theorem 2.2, make a motivation to us that we can extend the norm of $E$ for vector lattice $E^{t}$ in different cases. On the other hand, it is important to know that which time we have $({E^{t}})^{t}=E^{t}$. In the following example, we show that $E^{t}$ exists whenever $E$ satisfies in Fatou property. Note that by Example 4.3 and 4.4 from 1 , we know that every normed lattice with Fatou property, in general sense has not order continuous norm or Dedekind complete.

Example 2

By Theorem 4.12 1 , if $(E,\|.\|)$ satisfies the Fatou property, the Dedekind completion of $E$, $E^{\delta}$ is a normed space with $\delta-norm$. Let $E$ be the vector lattice of all real-valued functions defined on an infinite set $X$ whose range is finite, with the pointwise ordering and satisfies the Fatou property. It can be seen that $E$ is not Dedekind complete and $E^{\delta}=\ell^{\infty}(X)$.

Now in the following we introduce an important lemma that has many applications in every parts of this paper.

Lemma 1

Let $E$ has order continuous norm. For each $0\leq x\in E^{t}$, there are sequences $\{x_{n}\}\subseteq E^{+}$ and $\{y_{n}\}\subseteq E^{+}$ such that $x_{n}\uparrow x$, $x_{n}\xrightarrow{\|\|_{t}}x$, $y_{n}\downarrow x$ and $y_{n}\xrightarrow{\|\|_{t}}y$.

Proof

Choose $\{r_{n}\}\subseteq\mathbb{R}^{+}$ and $\{x_{n}\}\subseteq E^{+}$ satisfies in the following conditions:

$r_{n}\downarrow 0$, 2. 2.

$x_{n}\in\{z\in E:~{}z\leq x~{}\text{and}~{}\|x-z\|_{t}<r_{n}\}$, 3. 3.

$x_{n}\uparrow x$,

for each $n\in\mathbb{N}$. The reason of the above claim as follows:

By using Theorem 1.34 2 , take $A=\{z\leq x:~{}z\in E^{+}\}=\{z_{\alpha}\}$ and $B=\{y\geq x:~{}y\in E\}=\{y_{\beta}\}$ such that $z_{\alpha}\uparrow x$ and $y_{\beta}\downarrow x$. Then $z_{\alpha}\leq x\leq y_{\beta}$ holds for each $\alpha$ and $\beta$. Thus $\|x-y_{\beta}\|$, $\|x-z_{\alpha}\|\leq\|z_{\alpha}-y_{\beta}\|\rightarrow 0$. Let $0<r_{1}\in\mathbb{R}$. Then there exist $z_{1}\in\{z\in A:~{}\|x-z\|_{t}\leq r_{1}\}$ and $0<r_{2}<Min\{r_{1},~{}\|z_{1}-x\|_{t}\}$. We choose $z_{2},z_{3},...z_{n}$ and $z_{n+1}\in\{z\in A:~{}\|x-z_{n}\|_{t}\leq r_{n}\}$ where $0<r_{n}<Min\{r_{n-1},~{}\|z_{n-1}-x\|_{t}\}$. We define $x_{n}=\vee_{i=1}^{n}z_{i}$. Now, if $x_{n}\leqslant w\leqslant x$ for each $n\in\mathbb{N}$, then $0\leqslant x-w\leqslant x-x_{n}\leq x-z_{n}$. It follows that $\|x-w\|_{t}\leqslant\|x-x_{n}\|_{t}\leqslant\|x-z_{n}\|_{t}\leqslant r_{n}\downarrow 0$. Thus $x=w$, and so $\sup x_{n}=x$. Therefore $x_{n}\uparrow x$ and $\|x_{n}-x\|\rightarrow 0$. Existence of $\{y_{n}\}$ has the same argument.

Theorem 2.3

Let $E$ has order continuous norm. Then we have the following assertions.

If $E$ is a $KB$-pace, then so is $E^{t}$. 2. 2.

*If $E$ is an *$AL-$space, then so is $E^{t}$.

Proof

Assume that $\{x_{n}\}\subseteq(E^{t})^{+}$ is increasing sequence which $\sup\|x_{n}\|_{t}<+\infty$. By using Lemma 1, for each $n\in\mathbb{N}$, there are increasing and positive sequences $\{x_{n,m}\}_{m}$ such that $x_{n,m}\uparrow_{m}x_{n}$ and $\|x_{n}-x_{n,m}\|_{t}\xrightarrow{m}0$. Take $y_{n}=\bigvee_{i,j=1}^{n}x_{i,j}$. It follows that $0\leq y_{n}\uparrow$ and $\sup\|y_{n}\|\leq\sup_{i,j}\|x_{i,j}\|\leq\sup\|x_{n}\|<+\infty$. Since $E$ is a $KB$-pace, it follows that there exists $x\in E$ such that $\|y_{n}-x\|_{t}\rightarrow 0$. On the other hand, the inequalities $y_{n}\leq x_{n}\leq x$ implies that $\|x_{n}-x\|_{t}\leq\|y_{n}-x\|_{t}$ for each $n\in\mathbb{N}$. It follows that $\|x_{n}-x\|_{t}\rightarrow 0$ holds in $E^{t}$. 2. 2.

Since $E$ is an $AL-$space, then $E$ has order continuous norm. Now, let $0<x,y\in E^{t}$ with $x\wedge y=0$. By using Lemma 1, there are $\{x_{n}\}$ and $\{y_{n}\}$ in $E^{+}$ such that $x_{n}\uparrow x$, $y_{n}\uparrow y$, $\|x-x_{n}\|_{t}\rightarrow 0$ and $\|y-y_{n}\|_{t}\rightarrow 0$. It follows that $0\leqslant x_{n}\wedge y_{n}\uparrow x\wedge y=0$ implies that $x_{n}\wedge y_{n}=0$ for each $n\in\mathbb{N}$. Hence $\|x_{n}+y_{n}\|=\|x\|+\|y_{n}\|$ for each $n\in\mathbb{N}$. Then $\|x+y\|_{t}=\lim_{n}\|x_{n}+y_{n}\|=\lim_{n}\|x_{n}\|+\lim_{n}\|y_{n}\|=\|x\|_{t}+\|y\|_{t}$. Consequently, $E^{t}$ is an $AL-$space.

Theorem 2.4

For a normed lattice $E$ with order continuous norm, we have the following assertions

If $\hat{E}$ is a norm completion of $E$, then $E^{t}\subseteq\hat{E}=E^{u}$, and if $E$ is norm complete, then $E^{t}=E^{u}=E$. 2. 2.

For each $x\in E^{t}$ and $A\subseteq E$ with $\sup A=x$, we have $\|x\|_{t}=\sup_{z\in A}\|z\|$. 3. 3.

For each $x\in E^{t}$ and $A\subseteq E$ with $\inf A=x$, we have $\|x\|_{t}=\inf_{z\in A}\|z\|$. 4. 4.

$(E^{t},\|.\|_{t})$* has Fatou property and $B_{E^{t}}=\{x\in E^{t}:~{}\|x\|_{t}\leq 1\}$ is order closed.* 5. 5.

If $E$ is an ideal in $E^{t}$, then $\hat{E}=E^{t}$.

Proof

By Theorem 2.40 1 , $(\hat{E},\hat{\|.\|})$ is a normed lattice where $\hat{\|.\|}$ is unique extension of norm from $E$ into $\hat{E}$. Let $x\in E^{t}$. Then by Lemma 1, there exists $\{x_{n}\}$ in $E^{+}$ such that $x_{n}\uparrow x^{+}$ and $\|x^{+}-x_{n}\|_{t}\rightarrow 0$. Thus $\{x_{n}\}$ is a norm Cauchy sequence in $E$, and so convergence in $\hat{E}$. It follows that $x^{+}\in\hat{E}$. In the similar way $x^{-}\in\hat{E}$, which implies that $x\in\hat{E}$. Now by Theorem 7.51 of 1 , we conclude that $E^{t}\subseteq\hat{E}=E^{u}$ and $\|.\|_{t}=\hat{\|.\|}$. On the other hand if $E$ is norm complete, it is obvious that $E^{t}=E^{u}=E$ and $\|.\|=\|.\|_{t}=\hat{\|.\|}$. 2. 2.

By using Theorem 7.54 1 , $E^{u}$ has order continuous norm. since by part (1), we have $E^{t}\subseteq E^{u}$, it follows that $E^{t}$ has order continuous norm. Consider $A=(x_{\alpha})$ with $\sup A=x$. It follows that $x-x_{\alpha}\downarrow 0$ which implies that $\|x-x_{\alpha}\|_{t}\rightarrow 0$. Then by using inequalities $0\leq\|x\|_{t}-\|x_{\alpha}\|\leq\|x-x_{\alpha}\|_{t}$, we have $\sup_{\alpha}\|x_{\alpha}\|=\|x\|_{t}$. 3. 3.

Proof has the same argument such as (2). 4. 4.

By using Lemma 4.2 1 , $(E,\|.\|)$ has Fatou property. The proof of first statement has similar argument such as Theorem 2.3(1), we omit the proof. The second part, follows by Theorem 4.6 1 . 5. 5.

Proof follows by Theorem 3.8 1 .

3 Extension of order bounded operators

In every parts of this section, $T$ is order bounded operator from normed lattice $E$ into Dedekind complete normed lattice $F$.

Theorem 3.1

We have the following assertions.

$T$* has an order bounded extension $T^{t}$ from $E^{t}$ into $F$ such that $T^{t}(y)=Ty$ for each $y\in E$.* 2. 2.

For each positive continuous operator $T$, we have $\|T\|=\|T^{t}\|$, and if $T$ is norm continuous, then so is $T^{t}$. 3. 3.

$|T|^{t}=|T^{t}|$. 4. 4.

For each $T,S\in\mathcal{L}_{b}(E,F)$, we have $(T\vee S)^{t}=T^{t}\vee S^{t}$. 5. 5.

If $S:E^{t}\rightarrow F$ is an order bounded and norm continuous operator, then there exists order bounded and norm continuous operator $T:E\rightarrow F$ such that $T^{t}=S$. 6. 6.

Each order interval of $E^{t}$ is $\sigma(E^{t},(E^{t})^{\prime})$-compact.

Proof

Since $T$ is an order bounded operator and $F$ Dedekind complete, we can write $T=T^{+}-T^{-}$. So first we assume that $T$ is a positive operator from $E$ into $F$. By notes the proof of Theorem 1.32 2 , the mapping $p:E^{t}\rightarrow F$ defined via the formula

[TABLE]

is a monotone sublinear and $Ty\leq p(y)$ for each $y\in E$. So by using Theorem 1.5.7 4 , there is an extension of $T$, $T^{t}$, from $E^{t}$ into $F$ satisfying $T^{t}x\leq p(x^{+})$ for all $x\in E^{t}$, and $T^{t}y=Ty$ for all $y\in E$. Now we define $T^{t}=(T^{+})^{t}-(T^{-})^{t}$ in which $T^{t}$ is an extension of $T$ from $E^{t}$ into $F$ and for all $y\in E$, we have

[TABLE] 2. 2.

First assume that $T$ is a positive operators and $x\in E^{t}$. By part (1), we have $T^{t}x\leq p(x^{+})\leq Ty$ for all $y\geq x^{+}$, and so $\|T^{t}x\|\leq\|Ty\|$ for all $y\geq x^{+}$ which implies that $\|T^{t}x\|\leq\|T\|\inf_{y\geq x^{+}}\|y\|\leq\|T\|\|x^{+}\|_{t}\leq\|T\|\|x\|_{t}$. Then $\|T^{t}\|\leq\|T\|$. Since $B_{E}\subseteq B_{E^{t}}$, follows that $\|T\|\leqslant\|T^{t}\|$. Thus $\|T\|=\|T^{t}\|$. Thus $(T^{-})^{t}$ and $(T^{+})^{t}$ are norm continuous. This follows that $T^{t}$ is a norm continuous operator from $E^{t}$ into $F$. 3. 3.

In this part, we assume that $x,y,z$ are members of $E$ and $x^{t},y^{t},z^{t}$ are members of $E^{t}$ when there is not any confused. Now let $x^{t}\geq 0$. Since $E$ is order dense in $E^{t}$, we have the following equalities

[TABLE]

In the same way ${(T^{t})}^{-}(x^{t})=(T^{-})^{t}(x^{t})$ for all $x^{t}\geq 0$. It is obvious that for each $x^{t}\in E^{t}$, we have ${(T^{t})}^{+}x^{t}=(T^{+})^{t}(x^{t})$ and ${(T^{t})}^{-}x^{t}=(T^{-})^{t}(x^{t})$. Thus

[TABLE] 4. 4.

By using equality $T\vee S=\frac{1}{2}(T+S+|T-S|)$ and part (3), proof follows. 5. 5.

First let $0\leq x\in E^{t}$. By Lemma 1, there exists $\{x_{n}\}$ in $E^{+}$ such that $x_{n}\uparrow x^{+}$ and $\|x^{+}-x_{n}\|_{t}\rightarrow 0$. Since $S^{+}x_{n}\uparrow$ and $\|x^{+}-x_{n}\|\rightarrow 0$, follows that $S^{+}x_{n}\uparrow S^{+}x$. We define $T={S|}_{E}$ (restriction of $S$ on $E$), which follows that $T^{-}={S^{-}}|_{E}$ and $T^{+}=S^{+}|_{E}$ . Obviously $(T^{-})^{t}=S^{-}$ and $(T^{+})^{t}=S^{+}$, and so by part (3), we have the following equalities

[TABLE]

Thus $S=T^{t}$ on $E^{-}$ and $E^{+}$, which follows that

[TABLE]

for each $x\in E^{t}$. 6. 6.

Consider $a,b\in(E^{t})^{+}$ and $a<b$. By Lemma 1, take $\{x_{n}\}$ and $\{y_{n}\}$ in $E^{+}$ such that $x_{n}\uparrow a$, $y_{n}\downarrow b$, $\|a-x_{n}\|_{t}\rightarrow 0$ and $\|y_{n}-b\|_{t}\rightarrow 0$. Since $E$ has order continuous norm, $[x_{n},y_{n}]\cap E$ is $\sigma(E,E^{\prime})$-compact subset of $E$ for each $n\in\mathbb{N}$. It follows that $[a,b]\cap E$ is $\sigma(E,E^{\prime})$-compact subset of $E$. Now, if $V=\{s\in E:~{}x^{\prime}(s)<r~{}\text{and}~{}x^{\prime}\in E^{\prime}\}$, then by using part (5), the order density of $V$ is $V^{t}=\{s\in E^{t}:~{}(x^{\prime})^{t}(s)<r~{}\text{and}~{}(x^{\prime})^{t}\in(E^{t})^{\prime}\}$. Thus obvious that $V\subseteq V^{t}$, and so $\sigma(E,E^{\prime})\subseteq\sigma(E^{t},(E^{t})^{\prime})$. Since $[a,b]\cap E$ is order dense in $[a,b]$, follows that $[a,b]$ is $\sigma(E^{t},(E^{t})^{\prime})$-compact subset of $E^{t}$.

Now in the following, we investigate on some properties of operator $T^{t}$ from $E^{t}$ into $F$. We show that $T^{t}$ is keeping some lattice and topological properties whenever $T$ was done.

Theorem 3.2

Let $0\leq T\in\mathcal{L}_{n}(E,F)$. Then we have the following assertions

If $0\leq x\leq E^{t}$ and $\{x_{\alpha}\}\subseteq E^{+}$ with $x_{\alpha}\downarrow x$ , then $Tx_{\alpha}\downarrow T^{t}x$. 2. 2.

If $T(x\wedge y)=Tx\wedge Ty$ for each $0\leq x,y\in E$, then $T^{t}$ is a lattice homomorphism from $E^{t}$ into $F$ and moreover $T^{t}\in\mathcal{L}_{n}(E^{t},F)$. 3. 3.

If $0\leq T:E\rightarrow E$ is a band preserving, then $T^{t}:E^{t}\rightarrow E^{t}$ is so. 4. 4.

If $T:E\rightarrow F$ is preserving disjointness, then $T^{t}$ is so. 5. 5.

Let $E$ has order continuous norm. $\{Tx_{n}\}$ is norm convergent in $F$ for each positive increasing norm bounded sequence $\{x_{n}\}$ in $E$ iff $\{T^{t}x_{n}\}$ is norm convergent in $F$ for each positive increasing $t$-norm bounded sequence $\{x_{n}\}$ in $E^{t}$.

Proof

Let $\{x_{\alpha}\}\subseteq E^{+}$ such that $x_{\alpha}\downarrow x$. If $y\in E^{+}$ such that $x\leq y$, then $y\vee x_{\alpha}\downarrow y$ holds in $E$, and so by order continuouty of $T:E\rightarrow F$ and Theorem 2.4 (3), we see that

[TABLE]

This easily implies that $Tx_{\alpha}\downarrow T^{t}x$. 2. 2.

Assume that $0\leq x,y\in E^{t}$. We prove that $T^{t}(x\wedge y)=T^{t}x\wedge T^{t}y$. By Theorem 1.34 2 , there are $\{x_{\alpha}\}$ and $\{y_{\beta}\}$ of $E^{+}$ such that $x_{\alpha}\downarrow x$ and $y_{\beta}\downarrow y$. It follows that $x_{\alpha}\wedge y_{\beta}\downarrow x\wedge y$. Then by order continuouty of $T:E\rightarrow F$ and Theorem 2.4 (3), we have the following equalities,

[TABLE]

Combining Theorem 1.10, Theorem 2.14 from 2 and Theorem 3.1, we see that the mapping $T^{t}:(E^{t})^{+}\rightarrow(F^{t})^{+}$ extends to $T^{t}:E^{t}\rightarrow F^{t}$ which is lattice homomorphism.

Now, we show that $T^{t}\in\mathcal{L}_{n}(E^{t},F)$. Let $\{x_{\alpha}\}\subseteq(E^{t})^{+}$ such that $x_{\alpha}\downarrow 0$. Put

[TABLE]

Since $E$ mejorizes $E^{t}$, it follows that $A$ is not empty. By using Theorem 3.1 since $T$ is positive, $T^{t}$ is positive. Thus $\inf T(A)\geq\inf T^{t}x_{\alpha}\geq 0$ holds in $F$. Since $A\downarrow 0$ and $T\in\mathcal{L}_{n}(E,F)$, it follows that $\inf T(A)=0$, and so $T^{t}x_{\alpha}\downarrow 0$. 3. 3.

Let $x,y\in E^{t}$ satisfying $|x|\wedge|y|=0$. Assume that $(x_{\alpha}),(y_{\beta})\subseteq E^{+}$ such that $x_{\alpha}\uparrow|x|$ and $y_{\beta}\uparrow|y|$. It follows that $(x_{\alpha}\wedge y_{\beta})\uparrow|x|\wedge|y|=0$, and so $x_{\alpha}\wedge y_{\beta}=0$, which by using Theorem 2.36 2 implies that $|Tx_{\alpha}|\wedge y_{\beta}=0$ for each $\alpha$ and $\beta$. Since $|Tx_{\alpha}|\wedge y_{\beta}\uparrow|Tx|\wedge|y|$, we have $|Tx|\perp|y|$, and so by another using Theorem 2.36 2 , proof follows. 4. 4.

Let $x,y\in E^{t}$ satisfying $x\perp y$. Assume that $(x_{\alpha}),(y_{\beta})\subseteq E^{+}$ such that $x_{\alpha}\uparrow|x|$ and $y_{\beta}\uparrow|y|$. It follows that $(x_{\alpha}\wedge y_{\beta})\uparrow|x|\wedge|y|=0$. Now since $T$ preserve disjointness, follows that $Tx_{\alpha}\perp Tx_{\beta}$. From our hypothesis, we have $Tx_{\alpha}\wedge Tx_{\beta}\uparrow T|x|\wedge T|y|$ which follows that $T|x|\wedge T|y|=0$. Since $|Tx|\wedge|Ty|\leq T|x|\wedge T|y|$, we have $Tx\perp Ty$. 5. 5.

Since $T=T^{+}-T^{-}$, without loss generality, we assume that $T$ is a positive operator. Assume that $\{x_{n}\}\subseteq(E^{t})^{+}$ is increasing sequence with $\sup\|x_{n}\|_{t}<+\infty$. By using Lemma 1, for each $n\in\mathbb{N}$, there are positive increasing sequences $\{x_{n,m}\}_{m}$ with $x_{n,m}\uparrow_{m}x_{n}$ and $\|x_{n}-x_{n,m}\|_{t}\rightarrow 0$. Take $y_{n}=\bigvee_{i,j=1}^{n}x_{i,j}$. It follows that $0\leq y_{n}\uparrow$ and $\sup\|y_{n}\|\leq\sup_{i,j}\|x_{i,j}\|\leq\sup\|x_{n}\|<+\infty$. By assumption there is $s^{*}\in F$ such that $\|Ty_{n}-s^{*}\|\rightarrow 0$. Then by using Theorem 2.46 2 , $Ty_{n}\uparrow s^{*}$. By Theorem 3.1, we know that $T^{t}$ is norm continuous from $E^{t}$ into $F$. It follows that $\|T^{t}x_{n}-Tx_{n,m}\|\xrightarrow{m}0$ holds in $F$. The inequality $Tx_{n,m}\leq Ty_{n}\leq T^{t}x_{n}$ implies that

[TABLE]

Then

[TABLE]

Thus $T^{t}x_{n}\rightarrow s^{*}$, and the proof follows.

The converse is clear.

In the study of extension of normed lattice $E$ and linear operator $T$, we have some questions as follows. In the following questions, $\mathcal{K}(E,F)$ and $\mathcal{KW}(E,F)$ are the collections of compact operators and weakly compact operators from $E$ into $F$, respectively.

Question 2

Are $T^{t}\in\mathcal{K}(E^{t},F)$ and $T^{t}\in\mathcal{WK}(E^{t},F)$ whenever $T\in\mathcal{K}(E,F)$ and $T\in\mathcal{WK}(E,F)$, respectively? 2. 2.

Has $T^{t}$ Dunford-Pettis operator when $T$ has?

Question 3

When the equality $(T^{t})^{t}=T^{t}$ holds.