Reduced Order Models (ROMs) form essential tools across engineering domains by virtue of their function as surrogates for computationally intensive digital twinning simulators.
Although purely data-driven methods are available for ROM construction, schemes that allow to retain a portion of the physics tend to enhance the interpretability and generalization of ROMs.
However, physics-based techniques can adversely scale when dealing with nonlinear systems that feature parametric dependencies.
This study introduces a generative physics-based ROM that is suited for nonlinear systems with parametric dependencies and is additionally able to quantify the confidence associated with the respective estimates.
A main contribution of this work is the conditioning of these parametric ROMs to features that can be derived from monitoring measurements, feasibly in an online fashion.
This is contrary to most existing ROM schemes, which remain restricted to the prescription of the physics-based, and usually a priori unknown, system parameters.
Our work utilizes conditional Variational Autoencoders to continuously map the required reduction bases to a feature vector extracted from limited output measurements, while additionally allowing for a probabilistic assessment of the ROM-estimated Quantities of Interest.
An auxiliary task using a neural network-based parametrization of suitable probability distributions is introduced to re-establish the link with physical model parameters.
We verify the proposed scheme on a series of simulated case studies incorporating effects of geometric and material nonlinearity under parametric dependencies related to system properties and input load characteristics.

§ DATA AVAILABILITY

The data supporting this study's findings are available from the author KV upon reasonable request.

§ DECLARATION OF COMPETING INTEREST

The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper.

§ ACKNOWLEDGEMENTS

This research has received funding by the Sandia National Laboratories. Sandia National Laboratories is a multimission laboratory managed and operated by National Technology and Engineering Solutions of Sandia, LLC, a wholly owned subsidiary of Honeywell International Inc., for the U.S. Department of Energy’s National Nuclear Security Administration under contract DE-NA0003525.  This paper describes objective technical results and analysis. Any subjective views or opinions that might be expressed in the paper do not necessarily represent the views of the U.S. Department of Energy or the United States Government.

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§ CREDIT AUTHORSHIP CONTRIBUTION STATEMENT

Konstantinos Vlachas: Conceptualization, Methodology, Software, Validation, Writing.
Thomas Simpson: Software.
Anthony Garland: Methodology, Resources, Review & editing.
D. Dane Quinn: Methodology, Review & editing.
Charbel Farhat: Conceptualization, Methodology, Review & editing, Supervision.
Eleni Chatzi: Conceptualization,Methodology, Resources, Writing - review & editing, Supervision.


# Algebras of Lebesgue and KB regular operators in Banach lattices

###### Abstract

It is shown that the regularly Lebesgue and regularly (quasi) $KB$ operators
in a Banach lattice $E$ form operator algebras and, for Dedekind complete $E$,
even Banach lattice algebras. We also prove that a Banach lattice $E$ with
order continuous norm is a $KB$-space iff each positive compact operator in
$E$ is a $KB$ operator.

Eduard Emelyanov1
$1$ Sobolev Institute of Mathematics, Novosibirsk, Russia

Keywords: Banach lattice, regularly P-operator, Lebesgue operator, KB-operator

MSC2020: 46A40, 46B42, 47L05

##  1 Introduction

Several operator versions of Banach lattice properties like the property to be
a $KB$-space were introduced and studied in recent papers [2, 3, 6]. The main
idea of such operator versions lies in a redistribution of topological and
order properties between domain and range of operators. Since the order
convergence is not topological in infinite dimensional vector lattices [5],
the important operator versions emerge when order and norm convergences are
involved simultaneously. The case of regular operators acting in a Banach
lattice is of special interest.

### 1.1

In the present paper, $E$ is a real Banach lattice and $\text{\rm L}(E)$ is
the algebra of all continuous linear operators in $E$. The space $\text{\rm
L}_{r}(E)$ of regular operators in $E$ is an ordered subalgebra of $\text{\rm
L}(E)$. In general, $\text{\rm L}_{r}(E)$ is not a vector lattice. However,
$\text{\rm L}_{r}(E)$ is a Banach space under the regular norm
$\|T\|_{r}=\inf\\{\|S\|:\pm T\leq S\in\text{\rm L}_{r}(E)\\}$ [7, Proposition
1.3.6]. If $E$ is Dedekind complete, then $(\text{\rm
L}_{r}(E),\|\cdot\|_{r})$ is a Banach lattice algebra, and

$\|T\|_{r}=\|~{}|T|~{}\|\ \ \ \ \ (\forall T\in\text{\rm L}_{r}(E)).$ $None$

### 1.2

The following is an adopted variant of Definition 1.1 from [2] in the Banach
lattice setting.

######  Definition 1.

An operator $T\in\text{\rm L}(E)$ is said to be:

  1. $(i)$

Lebesgue ($\sigma$-Lebesgue) if $\|Tx_{\alpha}\|\to 0$ for every net
(sequence) $(x_{\alpha})_{\alpha}$ in $E$ such that $x_{\alpha}\downarrow 0$.
The set of all such operators is denoted by $\text{\rm L}_{Leb}(E)$ (resp.,
$\text{\rm L}^{\sigma}_{Leb}(E)$).

  2. $(ii)$

$on$-continuous ($on\sigma$-continuous) if $\|Tx_{\alpha}\|\to 0$ for every
net (sequence) $(x_{\alpha})_{\alpha}$ in $E$ such that
$x_{\alpha}\operatorname{\xrightarrow[]{o}}0$ (cf., [6, p.1315]). The set of
all such operators is denoted by $\text{\rm L}_{on}(E)$ (resp., $\text{\rm
L}^{\sigma}_{on}(E)$).

  3. $(iii)$

$KB$ ($\sigma$-$KB$) if, for every norm-bounded increasing net (sequence)
$(x_{\alpha})_{\alpha}$ in $E_{+}$, there exists (not necessarily unique)
$x\in E$ such that $\|Tx_{\alpha}-Tx\|\to 0$. The set of all such operators is
denoted by $\text{\rm L}_{KB}(E)$ (resp., $\text{\rm L}^{\sigma}_{KB}(E)$).

  4. $(iv)$

quasi $KB$ (quasi $\sigma$-$KB$) if $T$ takes norm-bounded increasing nets
(sequences) in $E_{+}$ to norm Cauchy nets. The set of all such operators is
denoted by $\text{\rm L}_{qKB}(E)$ (resp., $\text{\rm L}^{\sigma}_{qKB}(E)$)
(cf. [3, Definition 1.1]).

The identity operator in $c_{0}$ is Lebesgue yet not $KB$. The next example
shows that $KB$ operators also need not to be Lebesgue.

######  Example 1.

Let $E$ be a Banach lattice of all bounded real functions on $[0,1]$ that
differ from a constant on at most countable subset of $[0,1]$. Let $T:E\to E$
be an operator that assigns to each $f\in E$ the constant function $Tf$ on
$[0,1]$ such that the set $\\{x\in[0,1]:f(x)\neq(Tf)(x)\\}$ is at most
countable. Since $T$ is rank one and continuous, $T$ is $KB$. Consider the net
$(f_{\alpha})_{\alpha}$ in $E$, indexed by finite subsets of $[0,1]$, defined
by: $f_{\alpha}(x)=\left\\{\begin{array}[]{ccc}1&\text{ if }&x\not\in\alpha\\\
0&\text{ if }&x\in\alpha\end{array}\right..$ Then $f_{\alpha}\downarrow 0$ in
$E$, yet $\|f_{\alpha}\|_{\infty}=1$ for all $\alpha$. Thus $T$ is not a
Lebesgue operator. However, $T$ is $\sigma$-Lebesgue.

The following example shows that quasi $KB$ operators need not to be $KB$.

######  Example 2.

Let $T$ be an operator in $E=C[0,1]\oplus L_{1}[0,1]$ defined by
$T\bigl{(}(\phi,\psi)\bigl{)}=(0,\phi)$ for $\phi\in C[0,1]$ and $\psi\in
L_{1}[0,1]$. Clearly, $T$ is a quasi $KB$ operator. But $T$ is not $KB$.
Indeed, let a continuous function $\phi_{n}$ be equal to 1 on
$[0,\frac{1}{2}-\frac{1}{2^{n}}]$, to 0 on $[\frac{1}{2},1]$, and linear
otherwise. Let $f_{n}:=(\phi_{n},0)$. Then $f_{n}\uparrow$ in $E$ with
$\|f_{n}\|\leq 1$ for all $n\in\mathbb{N}$ and
$Tf_{n}\operatorname{\xrightarrow[]{\|\cdot\|}}(0,g)$, where $g\in L_{1}[0,1]$
is the indicator function of $[0,\frac{1}{2}]$. Since $g\not\in C[0,1]$, there
is no element $x$ of $E$ satisfying $Tx=(0,g)$, and hence $T$ is not $KB$.

### 1.3

The operators belonging to the linear span of positive compact operators
(r-compact operators) were investigated by Z.L. Chen and A.W. Wickstead in
[4]. The class of r-compact operators is closed under neither the operator nor
regular norm. However, there is a special norm on r-compact operators that
makes this class closed (see [4, Proposition 2.2]). This norm occurs in
application of Proposition 1.3.6 of [7] to compact rather than continuous
operators. The general case, that is a quite straightforward extension of [7,
Proposition 1.3.6], is described in Lemma 1 below. First, we need the
following notion.

######  Definition 2.

Given a set P of operators in $E$ $($briefly, the set P consists of
P-operators$)$. An operator $T:E\to E$ is called a regularly P-operator
$($shortly, an r-P-operator$)$, whenever there exist two positive P-operators
$T_{1},T_{2}$ such that $T=T_{1}-T_{2}$.

As examples of regularly P-operators one can consider: r-positive and
r-regular operators that are just regular; r-finite rank operators;
r-$\sigma$-Lebesgue; r-compact; r-$KB$ operators; etc.

The importance of regularly P-operators in the present paper is justified by
the fact that most of the key results below, like Propositions 1, 4, yields
for r-P-operators.

######  Lemma 1.

Let ${\rm P}$ be a closed in the operator norm subspace of $\text{\rm L}(E)$.
Then the formula

$\|T\|_{\text{\rm r-P}}=\inf\\{\|S\|:\pm T\leq S\in{\rm P}\\}$ $None$

defines a norm on the space $\text{\rm r-P}(E)$ of all r-P-operators in $E$.
Furthermore, $\|T\|_{\text{\rm r-P}}\geq\|T\|_{r}\geq\|T\|$ for all
$T\in\text{\rm r-P}(E)$, and $(\text{\rm r-P}(E),\|\cdot\|_{\text{\rm r-P}})$
is a Banach space.

###### Proof.

We skip routine checking that $\|\cdot\|_{\text{\rm r-P}}$ is a norm on
$\text{\rm r-P}(E)$ satisfying $\|\cdot\|_{\text{\rm
r-P}}\geq\|\cdot\|_{r}\geq\|\cdot\|$. Take any $\|\cdot\|_{\text{\rm
r-P}}$-Cauchy sequence $(T_{n})_{n}$ in $\text{\rm r-P}(E)$. Then
$T_{n}=G_{n}-R_{n}$ for some positive P-operators $G_{n},R_{n}$. We can assume
also that $\|T_{n+1}-T_{n}\|_{\text{\rm r-P}}<2^{-n}$ for all
$n\in\mathbb{N}$. As $\|\cdot\|_{\text{\rm r-P}}\geq\|\cdot\|$, there exists
$T\in\text{\rm L}(E)$ such that $\|T-T_{n}\|\to 0$. Since P is closed in the
operator norm, $T\in\text{\rm P}$. Pick $S_{n}\in\text{\rm P}$ with
$\|S_{n}\|<2^{-n}$ and $\pm(T_{n+1}-T_{n})\leq S_{n}$. Then

$T_{n+1}x^{+}-T_{n}x^{+}\leq Sx^{+}\ \ \text{and}\ \
-T_{n+1}x^{-}+T_{n}x^{-}\leq Sx^{-}$ $None$

for each $x\in E$. Summing up the inequalities in (3) gives
$T_{n+1}x-T_{n}x\leq S_{n}|x|$. Replacing $x$ by $-x$ gives
$T_{n}x-T_{n+1}x\leq S_{n}|x|$, and hence

$|(T_{n+1}-T_{n})x|\leq S_{n}|x|\ \ \ (\forall x\in E).$ $None$

As P is closed in the operator norm,
$Q_{n}:=\sum\limits_{k=n}^{\infty}S_{k}\in\text{\rm P}$ for all $n$. By (4),
we obtain

$|(T-T_{n})x|=\lim\limits_{k\to\infty}|(T_{k}-T_{n})x|\leq$
$\sum\limits_{k=n}^{\infty}|(T_{k+1}-T_{n})x|\leq Q_{n}|x|\ \ \ \ \ (x\in E),$

and hence $\pm(T-T_{n})\leq Q_{n}$. Then

$-Q_{n}\leq(T-T_{n})\leq Q_{n}\ \text{\rm and}\ 0\leq(T-T_{n})+Q_{n}$

for all $n\in\mathbb{N}$. Therefore

$T=[(T-T_{n})+Q_{n}]+[T_{n}-Q_{n}]=$
$[(T-T_{n})+Q_{n}+G_{n}]-[R_{n}+Q_{n}]\in\text{\rm r-P}(E),$

and hence $(T-T_{n})\in\text{\rm r-P}(E)$ for all $n\in\mathbb{N}$. We
conclude from $\|T-T_{n}\|_{\text{\rm r-P}}\leq\|Q_{n}\|<2^{1-n}$ that
$T_{n}\stackrel{{\scriptstyle\|\cdot\|_{\text{\rm
r-P}}}}{{\longrightarrow}}T$. ∎

##  2 Algebraic and lattice properties of regularly Lebesgue and KB operators

### 2.1

The following proposition contains a useful simple observation (cf. [6, Lemma
1] and [2, Lemma 2.1]).

######  Proposition 1.

Every r-Lebesgue $(resp.\ $r-$\sigma$-Lebesgue$)$ operator is order continuous
$($resp. $\sigma$-continuous$)$.

###### Proof.

Let $T$ be a regularly Lebesgue operator in $E$, and let $x_{\alpha}\downarrow
0$ in $E$. As $T=T_{1}-T_{2}$ with both $T_{1}$ and $T_{2}$ positive Lebesgue
operators, $T_{1}x_{\alpha}\downarrow$ and $\|T_{1}x_{\alpha}\|\to 0$. By [1,
Theorem 2.14], $T_{1}x_{\alpha}\downarrow 0$. Also, $T_{2}x_{\alpha}\downarrow
0$. Then $Tx_{\alpha}\operatorname{\xrightarrow[]{o}}0$. The
r-$\sigma$-Lebesgue case is similar. ∎

######  Lemma 2.

Let a net $(T_{\gamma})_{\gamma}$ in $\text{\rm L}_{Leb}(E)$ satisfy
$T_{\gamma}\stackrel{{\scriptstyle\|\cdot\|}}{{\longrightarrow}}T$ for some
$T\in\text{\rm L}(E)$. Then $T\in\text{\rm L}_{Leb}(E)$.

###### Proof.

Let $x_{\alpha}\downarrow 0$ in $E$. Passing to a tail of the net
$(x_{\alpha})_{\alpha}$, we may assume that $\|x_{\alpha}\|\leq M$ for some
$M>0$ and for all $\alpha$. Take an $\varepsilon>0$. Let
$\gamma_{\varepsilon}$ be such that
$\|T_{\gamma_{\varepsilon}}-T\|\leq\frac{\varepsilon}{M}$. Since
$T_{\gamma_{\varepsilon}}$ is Lebesgue,
$\lim_{\alpha}\|T_{\gamma_{\varepsilon}}x_{\alpha}\|=0$. There exists
$\alpha_{\varepsilon}$ with
$\|T_{\gamma_{\varepsilon}}x_{\alpha}\|\leq\varepsilon$ for
$\alpha\geq\alpha_{\varepsilon}$, and hence

$\|Tx_{\alpha}\|\leq\|T_{\gamma_{\varepsilon}}x_{\alpha}\|+\|T_{\gamma_{\varepsilon}}x_{\alpha}-Tx_{\alpha}\|\leq\varepsilon+\|T_{\gamma_{\varepsilon}}-T\|\|x_{\alpha}\|\leq
2\varepsilon$

for $\alpha\geq\alpha_{\varepsilon}$. Since $\varepsilon>0$ is arbitrary, $T$
is Lebesgue. ∎

Recall that P-operators satisfy the domination property if $0\leq S\leq
T\in\text{\rm P}$ implies $S\in\text{\rm P}$.

######  Theorem 1.

Let $E$ be a Banach lattice. Then $\text{\rm r-L}_{Leb}(E)$ and $\text{\rm
r-L}_{Leb}^{\sigma}(E)$ are both left algebraic ideals and hence subalgebras
of $\text{\rm L}_{r}(E)$. Furthermore$:$

  1. $i)$

$\text{\rm r-L}_{Leb}(E)$ $($resp., $\text{\rm r-L}_{Leb}^{\sigma}(E)$$)$
coincides with $\text{\rm L}_{r}(E)$ iff the norm of $E$ is order continuous
$($resp., $\sigma$-order continuous$)$.

  2. $ii)$

If $E$ is Dedekind complete then $\|T\|_{\text{\rm r-P}}=\|T\|_{r}$ for each
$T\in P:=\text{\rm L}_{Leb}(E)$, and

$(\text{\rm r-L}_{Leb}(E),\ \|\cdot\|_{\text{\rm r-P}})\ \ \ \ (\text{\rm
resp.,}\ (\text{\rm r-L}^{\sigma}_{Leb}(E),\ \|\cdot\|_{\text{\rm r-P}}))$

is a Banach lattice algebra.

###### Proof.

We consider only $\text{\rm r-L}_{Leb}(E)$, the case of $\text{\rm
r-L}^{\sigma}_{Leb}(E)$ is similar, and skip routine direct checking that
$\text{\rm r-L}_{Leb}(E)$ is a vector subspace of $\text{\rm L}_{r}(E)$. Let
$T\in\text{\rm r-L}_{Leb}(E)$ and $S\in\text{\rm L}_{r}(E)$. In order to show
that $\text{\rm r-L}_{Leb}(E)$ is a left algebraic ideal, we need to show that
$ST$ is r-Lebesgue. WLOG, we may assume $S,T\geq 0$. Let $x_{\alpha}\downarrow
0$ in $E$. Then $\|Tx_{\alpha}\|\to 0$, $\|S(Tx_{\alpha})\|\to 0$, and hence
$ST$ is r-Lebesgue. As each left ideal is a subalgebra, $\text{\rm
r-L}_{Leb}(E)$ is a subalgebra of $\text{\rm L}_{r}(E)$. Note that by the
similar argument, $\text{\rm r-L}_{Leb}(E)$ is a left ideal of $\text{\rm
L}(E)$.

$i)$ If $\text{\rm r-L}_{Leb}(E)=\text{\rm L}_{r}(E)$ then the identity
operator $I$ in $E$ is Lebesgue, and hence the norm in $E$ is order
continuous. Conversely, if the norm in $E$ is order continuous then
$I\in\text{\rm r-L}_{Leb}(E)$ and, since $\text{\rm r-L}_{Leb}(E)$ is a left
ideal of $\text{\rm L}_{r}(E)$ containing $I$ then $\text{\rm
r-L}_{Leb}(E)=\text{\rm L}_{r}(E)$.

$ii)$ By Lemmas 1 and 2, $(\text{\rm r-L}_{Leb}(E),\ \|\cdot\|_{\text{\rm
r-P}})$ is a Banach space. Let $T\in\text{\rm r-L}_{Leb}(E)$, say
$T=T_{1}-T_{2}$, with $0\leq T_{1},T_{2}\in\text{\rm L}_{Leb}(E)$. Then
$|T|\leq T_{1}+T_{2}$. Since the Lebesgue operators satisfy the domination
property, $|T|\in\text{\rm r-L}_{Leb}(E)$. Hence $\text{\rm r-L}_{Leb}(E)$ is
a vector lattice, and $\|T\|_{\text{\rm r-P}}=\|~{}|T|~{}\|$. By formula (1),
$\|T\|_{\text{\rm r-P}}=\|T\|_{r}$.

For proving that $(\text{\rm r-L}_{Leb}(E),\ \|\cdot\|_{\text{\rm r-P}})$ is a
Banach lattice, we need to show

$|S|\leq|T|\ \Rightarrow\ \|S\|_{\text{\rm r-P}}\leq\|T\|_{\text{\rm r-P}}\ \
\ (\forall S,T\in\text{\rm r-L}_{Leb}(E)).$ $None$

Let $S,T\in\text{\rm r-L}_{Leb}(E)$ satisfy $|S|\leq|T|$. Since
$|S|,|T|\in{\text{\rm r-}}{\text{\cal L}}_{Leb}(E)$ then $\|S\|_{\text{\rm
r-P}}=\|~{}|S|~{}\|\leq\|~{}|T|~{}\|=\|T\|_{\text{\rm r-P}}$, that proves (5).

It remains to show that the norm $\|\cdot\|_{\text{\rm r-P}}$ is
submultiplicative. Let $S,T\in\text{\rm r-L}_{Leb}(E)$. It follows from

$\pm STx\leq|STx|\leq|S||Tx|\leq|S||T|x\ \ \ (\forall x\in E_{+})$

that $\pm ST\leq|S||T|$. Thus,

$\|ST\|_{\text{\rm
r-P}}\leq\|~{}|S||T|~{}\|\leq\|~{}|S|~{}\|\cdot\|~{}|T|~{}\|=\|S\|_{\text{\rm
r-P}}\|T\|_{\text{\rm r-P}},$

as desired. ∎

### 2.2

The following fact is an adaptation of [6, Lemma 1] and [2, Lemma 2.1] for
regular operators. For convenience, we include its proof.

######  Proposition 2.

Let $E$ be a Banach lattice. Then

$\text{\rm r-L}_{Leb}(E)=\text{\rm r-L}_{on}(E)\ \ \text{and}\ \ \text{\rm
r-L}^{\sigma}_{Leb}(E)=\text{\rm r-L}^{\sigma}_{on}(E),$ $None$

where $\text{\rm r-L}_{on}(E)$ $($resp., $\text{\rm r-L}^{\sigma}_{on}(E)$$)$
is the space of all r-on-continuous $($resp., r-on$\sigma$-continuous$)$
operators in $E$.

###### Proof.

We restrict ourselves to the first equality. Since $\text{\rm
r-L}_{on}(E)\subseteq\text{\rm r-L}_{Leb}(E)$ is trivial, we only need to show
$\text{\rm r-L}_{Leb}(E)\subseteq\text{\rm r-L}_{on}(E)$. Let $T\in\text{\rm
r-L}_{Leb}(E)$, and let $x_{\alpha}\operatorname{\xrightarrow[]{o}}0$. WLOG,
suppose $T\geq 0$. There exists a net $y_{\beta}\downarrow 0$ in $E$ such
that, for every $\beta$, there exists $\alpha_{\beta}$ with $|x_{\alpha}|\leq
y_{\beta}$ (hence $T|x_{\alpha}|\leq Ty_{\beta}$) for
$\alpha\geq\alpha_{\beta}$. It follows $0\leq
T((x_{\alpha})^{+}),T((x_{\alpha})^{-})\leq Ty_{\beta}$ for
$\alpha\geq\alpha_{\beta}$. Then $\|Ty_{\beta}\|\to 0$ and hence
$\|T((x_{\alpha})^{\pm})\|\to 0$, that implies $\|Tx_{\alpha}\|\to 0$. ∎

In view of Proposition 1, one can replace in Theorem 1 regularly Lebesgue
$($$\sigma$-Lebesgue$)$ operators by regularly $on$-continuous
($on\sigma$-continuous) operators.

The author doesn’t know whether or not one can replace regularly Lebesgue
operators by Lebesgue regular operators in formula (6). More precisely, under
what conditions on $E$, the equalities

$(\text{\rm L}_{Leb})_{r}(E)=(\text{\rm L}_{on})_{r}(E)\ \ \text{\rm and}\ \
(\text{\rm L}^{\sigma}_{Leb})_{r}(E)=(\text{\rm L}^{\sigma}_{on})_{r}(E)$

hold true? By Proposition 1, the question has a positive answer for a Dedekind
complete Banach lattice $E$, where

$T\in(\text{\rm L}_{Leb})_{r}(E)\Longrightarrow|T|\in\text{\rm L}_{Leb}(E)$
$(\text{\rm resp.,}\ T\in(\text{\rm
L}^{\sigma}_{Leb})_{r}(E)\Longrightarrow|T|\in\text{\rm L}^{\sigma}_{Leb}(E)\
).$

### 2.3

Now we investigate algebraic structure of $KB$ and quasi $KB$ operators. We
begin with the following lemma, that is similar to Lemma 2 (cf. also [3,
Proposition 2.5]).

######  Lemma 3.

Let a net $(T_{\gamma})_{\gamma}$ in $\text{\rm L}_{qKB}(E)$ satisfies
$T_{\gamma}\stackrel{{\scriptstyle\|\cdot\|}}{{\longrightarrow}}T$ for some
$T\in\text{\rm L}(E)$. Then $T\in\text{\rm L}_{qKB}(E)$.

###### Proof.

Let $(x_{\alpha})_{\alpha}$ be an increasing norm bounded net in $E$. WLOG,
$\|x_{\alpha}\|\leq 1$ for all $\alpha$. Take $\varepsilon>0$ and choose
$\gamma_{0}$ satisfying $\|T_{\gamma_{0}}-T\|\leq\varepsilon$. Choose
$\alpha_{0}$ such that
$\|T_{\gamma_{0}}x_{\alpha^{{}^{\prime}}}-T_{\gamma_{0}}x_{\alpha^{{}^{\prime\prime}}}\|\leq\varepsilon$
for $\alpha^{{}^{\prime}},\alpha^{{}^{\prime\prime}}\geq\alpha_{0}$. Then

$\|Tx_{\alpha^{{}^{\prime}}}-Tx_{\alpha^{{}^{\prime\prime}}}\|\leq$
$\|Tx_{\alpha^{{}^{\prime}}}-T_{\gamma_{0}}x_{\alpha^{{}^{\prime}}}\|+\|T_{\gamma_{0}}x_{\alpha^{{}^{\prime}}}-T_{\gamma_{0}}x_{\alpha^{{}^{\prime\prime}}}\|+\|T_{\gamma_{0}}x_{\alpha^{{}^{\prime\prime}}}-Tx_{\alpha^{{}^{\prime\prime}}}\|\leq$
$\|T_{\gamma_{0}}-T\|\|x_{\alpha^{{}^{\prime}}}\|+\|T_{\gamma_{0}}x_{\alpha^{{}^{\prime}}}-T_{\gamma_{0}}x_{\alpha^{{}^{\prime\prime}}}\|+\|T_{\gamma_{0}}-T\|\|x_{\alpha^{{}^{\prime\prime}}}\|\leq
3\varepsilon$

for all $\alpha^{{}^{\prime}},\alpha^{{}^{\prime\prime}}\geq\alpha_{0}$, and
hence $T$ is quasi $KB$. ∎

By [2, Proposition 1.2], every quasi $\sigma$-$KB$ operator in $E$ is quasi
$KB$. The following theorem is a quasi KB version of Theorem 1 (cf. also [3,
Corollary 2.8]).

######  Theorem 2.

$\text{\rm r-L}_{KB}(E)$ $($resp., $\text{\rm r-L}_{qKB}(E)$$)$ is a left
$($resp., two sides$)$ algebraic ideal of $\text{\rm L}_{r}(E)$, and hence is
a subalgebra of $\text{\rm L}_{r}(E)$. Furthermore$:$

  1. $i)$

The algebra $\text{\rm r-L}_{KB}(E)$ is unital iff $\text{\rm r-L}_{qKB}(E)$
is unital iff $E$ is a KB-space.

  2. $ii)$

If $E$ is Dedekind complete then $(\text{\rm r-L}_{qKB}(E),\
\|\cdot\|_{\text{\rm r-P}})$ is a Banach lattice algebra for $\text{\rm
P}=\text{\rm L}_{qKB}(E)$.

###### Proof.

We omit the straightforward proof that $\text{\rm r-L}_{KB}(E)$ and $\text{\rm
r-L}_{qKB}(E)$ are both vector subspaces of $\text{\rm L}_{r}(E)$. We need to
show that $\text{\rm r-L}_{KB}(E)$ (resp., $\text{\rm r-L}_{qKB}(E)$) is
closed under left (two sides) multiplication by elements of $\text{\rm
L}_{r}(E)$.

A) Firstly consider $\text{\rm r-L}_{KB}(E)$. Let $T\in\text{\rm r-L}_{KB}(E)$
and $S\in\text{\rm L}_{r}(E)$. WLOG, suppose $T\geq 0$ and $S\geq 0$. If
$(x_{\alpha})_{\alpha}$ be an increasing norm bounded net in $E_{+}$, then
$\|Tx_{\alpha}-Tx\|\to 0$ for some $x\in E$ and, since $S$ is continuous, we
obtain $\|STx_{\alpha}-STx\|\to 0$. Then $ST\in\text{\rm r-L}_{KB}(E)$,
showing that $\text{\rm r-L}_{KB}(E)$ is a left ideal of $\text{\rm
L}_{r}(E)$.

B) Now, let $T\in\text{\rm r-L}_{qKB}(E)$ and $S\in\text{\rm L}_{r}(E)$, and
let $(x_{\alpha})_{\alpha}$ be a norm bounded increasing net in $E_{+}$. Then
the net $(Tx_{\alpha})_{\alpha}$ is $\|.\|$-Cauchy and, since $S$ is
continuous, the net $(STx_{\alpha})_{\alpha}$ is also $\|.\|$-Cauchy, showing
$ST\in\text{\rm r-L}_{qKB}(E)$.

Since the net $Sx_{\alpha}$ is increasing and norm bounded in $E_{+}$, we
obtain that the net $(TSx_{\alpha})_{\alpha}$ is $\|.\|$-Cauchy, showing that
$TS\in\text{\rm r-L}_{qKB}(E)$.

Thus, $\text{\rm r-L}_{qKB}(E)$ is a two sides algebraic ideal of $\text{\rm
L}_{r}(E)$.

$i)$ It is obvious since the three conditions are equivalent to the fact that
the identity operator in $E$ is KB.

$ii)$ Upon the fact that the KB operators satisfy the domination property [2,
Theorem 2.6], the proof is similar to the proof of $ii)$ of Theorem 1 via
replacement of Lemma 2 by Lemma 3. ∎

##  3 Relations between KB and compact operators

Here we discuss relations between KB and compact operators in $E$.

### 3.1

We begin with the following fact.

######  Proposition 3.

Every continuous finite rank operator on a Banach lattice $E$ is a regularly
$KB$ operator.

###### Proof.

Let $T\in\text{\rm L}(E)$ and $\dim(TE)<\infty$. Then

$T=\sum_{k=1}^{n}x_{k}\otimes f_{k}\ \text{for}\ x_{1},\dots,x_{n}\in E\
\text{and}\ f_{1},\dots,f_{n}\in E^{\ast}.$

WLOG, we can assume $T=x_{1}\otimes f_{1}$. Since the dual $E^{\ast}$ is
Dedekind complete, $f_{1}$ is regular, and hence $T$ is also regular. WLOG,
suppose $x_{1}\geq 0$ and $f_{1}\geq 0$. Then $T$ is positive, and hence
regular. Let $(x_{\alpha})_{\alpha}$ be an increasing norm bounded net in
$E_{+}$. Since $Tx_{\alpha}=(x_{1}\otimes
f_{1})(x_{\alpha})=f_{1}(x_{\alpha})x_{1}\uparrow$ and norm bounded, then
$Tx_{\alpha}\to y\in\text{\rm span}(x_{1})$ and
$Tx_{\alpha}\operatorname{\xrightarrow[]{\|\cdot\|}}Tx$ for some $x\in E$.
Hence $T$ is r-KB. ∎

It is worth to notice that, for a positive compact operator $T$ in the Banach
lattice $c$ of real convergent sequences, and, for a norm bounded increasing
net $(x_{\alpha})_{\alpha}$ in $c_{+}$, a vector $x\in c$, satisfying
$\|Tx_{\alpha}-Tx\|\to 0$, need not to be an upper bound of
$(x_{\alpha})_{\alpha}$. Indeed, let $f$ be a positive functional on $c$ that
assigns to any element of $c$ its limit. Define $T:c\to c$ by
$Tx:=f(x)\cdot\mathbb{1}$, where $\mathbb{1}$ is the real sequence identically
equal to $1$. Then $T$ is a $KB$ operator by Proposition 3. Take
$x_{n}:=(\underbrace{1,1,...,1}_{n},0,0,0,...)$ in $c_{+}$. Then
$x_{n}\uparrow$ and $\|x_{n}\|=1$ for all $n\in\mathbb{N}$. Yet, for every
$x\in c$ satisfying $x_{n}\uparrow\leq x$, we have $\|Tx\|\geq 1$ and hence
$\|Tx_{n}-Tx\|=\|Tx\|\geq 1$ for each $n$.

######  Proposition 4.

Every r-compact operator $T$ in a Banach lattice $E$ is regularly quasi $KB$.

###### Proof.

WLOG, we restrict ourselves to the case when $T$ is a positive compact
operator. Let a net $(x_{\alpha})_{\alpha}$ in $E$ be norm bounded and $0\leq
x_{\alpha}\uparrow$. Since $T$ is compact, the increasing net
$(Tx_{\alpha})_{\alpha}$ has a subnet $(Tx_{\alpha_{\beta}})_{\beta}$ with
$\left\|Tx_{\alpha_{\beta}}-y\right\|\to 0$ for some $y\in E$. In particular,
$(Tx_{\alpha})_{\alpha}$ has a $\|.\|$-Cauchy subnet, and hence
$(Tx_{\alpha})_{\alpha}$ is also $\|.\|$-Cauchy, showing that $T$ is quasi
$KB$, and since $T\geq 0$, $T$ is r-qKB. ∎

### 3.2

The next example shows that the inclusion

$\text{\rm r-L}_{KB}(E)\subseteq\text{\rm r-L}_{qKB}(E)$

can be proper.

######  Example 3.

Denote elements of $c_{0}$ by $x=\sum_{n=1}^{\infty}a_{n}\cdot e_{n}$, where
$e_{n}$ are orts placed at $n$ and $(a_{n})_{n}$ is a real sequence converging
to zero. Define an operator $T:c_{0}\to c_{0}$ by

$T\left(\sum_{n=1}^{\infty}a_{n}e_{n}\right)=\sum_{n=1}^{\infty}\frac{a_{n}}{n}e_{n}.$

Then $T$ is positive and compact, and hence a quasi $KB$ operator by
Proposition 4. Thus $T\in{\text{\rm r-}}{\text{\cal L}}_{qKB}(E)$. Take an
increasing norm bounded sequence $(x_{n})_{n}$ in $c_{0}$ by
$x_{n}=\sum\limits_{k=1}^{n}e_{k}$. The sequence
$(Tx_{n})_{n}=\bigl{(}\sum\limits_{k=1}^{n}\frac{1}{k}e_{k}\bigl{)}_{n}$ is
norm-Cauchy in $c_{0}$, however there is no element
$\sum\limits_{k=1}^{\infty}a_{k}e_{k}$ in $c_{0}$ satisfying

$T\left(\sum\limits_{k=1}^{\infty}a_{k}e_{k}\right)=\sum\limits_{k=1}^{\infty}\frac{1}{k}e_{k}=\|\cdot\|_{c_{0}}\text{-}\lim\limits_{n\to\infty}Tx_{n}.$

Therefore $T$ is not a $KB$-operator.

### 3.3

Our next result develops the idea of Example 3.

######  Theorem 3.

In a Banach lattice $E$ with order continuous norm the following are
equivalent$:$

  1. $1)$

$E$ is a $KB$-space;

  2. $2)$

every positive compact operator in $E$ is $KB$.

###### Proof.

The implication $1)\Rightarrow 2)$ is obvious, as any continuous operator on a
$KB$-space is a $KB$-operator.

$2)\Rightarrow 1)$ Assume, in contrary, that $E$ be not a $KB$-space. Then
there is a homeomorphic lattice embedding $c_{0}\stackrel{{\scriptstyle
i}}{{\longrightarrow}}E$, say
$i((a_{k})_{k=1}^{\infty})=\sum\limits_{k=1}^{\infty}a_{k}\cdot e_{k}$, where
$(a_{k})_{k=1}^{\infty}\in c_{0}$. WLOG, assume that the sequence
$(e_{k})_{k}$ in $E_{+}$ is pairwise disjoint and normalized. Let $f_{k}$ be
norm-one functionals on $E$ such that $f_{k}(e_{k})=1$ for all
$k\in\mathbb{N}$. Then $|f_{k}|$ are norm-one positive functionals on $E$ with
$|f_{k}|(e_{k})=1$ for all $k\in\mathbb{N}$. Indeed $\||f_{k}|\|=\|f_{k}\|=1$
and

$1=|f_{k}(e_{k})|\leq|f_{k}|(e_{k})=\sup\\{|f_{k}(x)|:|x|\leq e_{k}\\}\leq$
$\sup\\{|f_{k}(x)|:\|x\|\leq 1\\}=\|f_{k}\|=1$

for all $k\in\mathbb{N}$. Let $P_{k}$ be the band projection onto the
principle band $B_{e_{k}}$ generated by $e_{k}$. Define a positive operator
$T:E\to E$ by

$Tx=\sum_{k=1}^{\infty}\frac{|f_{k}|(P_{k}(x))}{k}e_{k}.$ $None$

Since $||f_{k}|(P_{k}(x))|\leq\||f_{k}|\|\cdot\|P_{k}(x)\|\leq\|x\|$ for all
$k\in\mathbb{N}$, the formula (7) implies

$T(B_{E})\subseteq i([-u,u]),$ $None$

where $u:=\bigl{(}\frac{1}{k}\bigl{)}_{k}\in c_{0}$. As order intervals of
$c_{0}$ are compact, the set $i([-u,u])$ is compact in $E$. It follows from
(8) that $T$ is a compact operator. By the assumption $2)$, $T$ is a
$KB$-operator. Consider an increasing norm bounded sequence
$y_{n}=\sum\limits_{k=1}^{n}e_{k}$ in $E$. Then there is some $y\in E$ such
that $\|Ty_{n}-Ty\|\to 0$. Since

$Ty_{n}=\sum_{k=1}^{n}\frac{|f_{k}|(P_{k}(y_{n}))}{k}e_{k}=\sum_{k=1}^{n}\frac{|f_{k}|(e_{k})}{k}e_{k}=\sum_{k=1}^{n}\frac{1}{k}e_{k}$

for all $n\in\mathbb{N}$, $Ty=\sum_{k=1}^{\infty}\frac{1}{k}e_{k}$. Let
$w_{n}=\sum_{k=1}^{n}P_{k}(y)$. Then $w_{n}\operatorname{\xrightarrow[]{o}}y$.
Since the norm in $E$ is order continuous, $\|w_{n}-y\|\to 0$. Then the
sequence $(w_{n})_{n}$ is norm-Cauchy. It follows from $\|Ty_{n}-Ty\|\to 0$,
$Ty_{n}=\sum_{k=1}^{n}\frac{1}{k}e_{k}$, and
$Ty=\sum_{k=1}^{\infty}\frac{1}{k}e_{k}$ that
$|f_{n}|(P_{n}(y_{n}))=|f_{n}|(P_{n}(y))$. Then

$1=|f_{n}|(e_{n})=|f_{n}|(P_{n}(y_{n}))=|f_{n}|(P_{n}(y))\leq\|P_{n}(y)\|,$

violating $\|P_{n}(y)\|=\|w_{n}-w_{n-1}\|\to 0$. The obtained contradiction
completes the proof. ∎

######  Corollary 1.

Let $E$ be a Banach lattice with order continuous norm. The following
conditions are equivalent.

  1. $1)$

$E$ is a $KB$-space.

  2. $2)$

$\text{\rm L}(E)=\text{\rm L}_{KB}(E)$.

  3. $3)$

$\text{\rm L}_{qKB}(E)=\text{\rm L}_{KB}(E)$.

  4. $4)$

$\text{\rm r-L}_{qKB}(E)=\text{\rm r-L}_{KB}(E)$.

  5. $5)$

$(\text{\rm L}_{qKB})_{+}(E)=(\text{\rm L}_{KB})_{+}(E)$.

  6. $6)$

Every positive compact operator in $E$ is $KB$.

###### Proof.

$1)\Rightarrow 2)\Rightarrow 3)$ is obvious.
$3)\Rightarrow 4)$ Let $T\in\text{\rm r-L}_{qKB}(E)$, say $T=T_{1}-T_{2}$ with
$T_{1},T_{2}\in(\text{\rm L}_{qKB})_{+}(E)$. Then $T_{1},T_{2}\in(\text{\rm
L}_{KB})_{+}(E)$, and hence $T\in\text{\rm r-L}_{KB}(E)$.
$4)\Rightarrow 5)$ is obvious.
$5)\Rightarrow 6)$ Let $T$ be a positive compact operator in $E$. Let
$(B_{E})_{+}\ni x_{\alpha}\uparrow$. By compactness of $T$, the net
$(Tx_{\alpha})_{\alpha}$ contains a norm convergent subnet, and, since
$Tx_{\alpha}\uparrow$ then the net $(Tx_{\alpha})_{\alpha}$ is norm
convergent. Thus $T\in(\text{\rm L}_{qKB})_{+}(E)$, and hence $T\in(\text{\rm
L}_{KB})(E)$.
$6)\Rightarrow 1)$ follows from Theorem 3. ∎

In view of Example 2 the assumption that the norm in $E$ is order continuous
is essential in Corollary 1. The author do not know whether or not the order
continuity in Corollary 1 can be replaced by $\sigma$-Dedekind completeness,
or by Dedekind completeness.

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