# How to Measure the Intelligence of Large Language Models?

Nils Körber, Silvan Wehrli, Christopher Irrgang
Centre for Artificial Intelligence in Public Health Research
Robert Koch Institute
Berlin
Germany
{KoerberN, WehrliS<EMAIL_ADDRESS>

With the release of ChatGPT and other large language models (LLMs) the
discussion about the intelligence, possibilities, and risks, of current and
future models have seen large attention [1, 2, 3]. This discussion included
much debated scenarios about the imminent rise of so-called “super-human” AI
[4], i.e., AI systems that are orders of magnitude smarter than humans. In the
spirit of Alan Turing, there is no doubt that current state-of-the-art
language models already pass his famous test [5]. Moreover, current models
outperform humans in several benchmark tests [6], so that publicly available
LLMs have already become versatile companions that connect everyday life,
industry and science. Despite their impressive capabilities, LLMs sometimes
fail completely at tasks that are thought to be trivial for humans. In other
cases, the trustworthiness of LLMs becomes much more elusive and difficult to
evaluate. Taking the example of academia, language models are capable of
writing convincing research articles on a given topic with only little input.
Yet, the lack of trustworthiness in terms of factual consistency or the
existence of persistent hallucinations in AI-generated text bodies has led to
a range of restrictions for AI-based content in many scientific journals. In
view of these observations, the question arises as to whether the same metrics
that apply to human intelligence can also be applied to computational methods
and has been discussed extensively [7]. In fact, the choice of metrics has
already been shown to dramatically influence assessments on potential
intelligence emergence [3]. Here, we argue that the intelligence of LLMs
should not only be assessed by task-specific statistical metrics, but
separately in terms of qualitative and quantitative measures.

## Quantitative and qualitative intelligence

To evaluate the intelligence performance of LLMs, the problem can be easier
accessed by separating the task into two separate metric classes: quantitative
intelligence and qualitative intelligence. With quantitative intelligence we
refer to the model’s inherent data storage and the model’s ability to
navigate, use and remix this information, analogous to the knowledge of a
human. In contrast, qualitative intelligence refers to the ability to analyze,
judge, and conclude from that data storage and novel information. At first
glance, this broad separation may seem obvious, especially in the context of
human intelligence. Nevertheless, there is still no standardized paradigm for
evaluating intelligence for LLMs and similar AI systems.

Figure 1: Quantitative (left) and qualitative (right) performance of LLMs.

Quantitative intelligence includes the combined fields of expertise that are
covered by an LLM. As the current LLMs are trained on massive amounts of data,
which mostly come from internet sources, the quantitative capacities are
enormous. A reasonable conversation with an LLM is possible in dozens of
languages and may cover topics from knitting tips via ancient literature to
quantum physics. At this stage, current models likely already can fall back on
more information than any human being (see Figure 1). To assess quantitative
intelligence, a systematic test should be carried out that covers a wide range
of topics and different levels of detail within each topic. Quantitative
measures are directly dependent on the training data and describe the areas of
specialization. Above that, the quantitative capacity of a model provides
better insight into whether the output of a model is a simple rearrangement of
training data or is actually concluded from context. In contrast to humans,
LLMs do not suffer from concentration loss and motivation and can be evaluated
quantitatively. Hence, instead of using a handful of questions as in a regular
examination test, a model might be analyzed on a larger scale with hundreds of
thousands of questions and tasks in a reasonably short time. Provided a
comprehensive question catalog covering the entire information spectrum from
elementary school, common knowledge to university curricula. The factual
accuracy of the individual responses and the corresponding metrics within each
topic, across topics and over varying levels of complexity draw a picture of
the model’s knowledge space. The massive multitask language understanding
(MMLU) test covering almost 16,000 questions separated over 57 tasks is a step
in the right direction. However, it does not explicitly differ between
knowledge retrievel and problem solving capabilities [8]. In order to examine
the quantitative capabilities in detail, the questions should be chosen to
focus exclusively information retrieval instead of reasoning, which should be
addressed in a separate qualitative test.

Current paradigms to evaluate human intelligence are based on the assumption
that a subject being able to complete a specific task is also able to complete
similar tasks with lower or comparable difficulty. If this generalizability
holds true for LLMs is of ongoing debate [9, 10] and could be represented as
what we claim as qualitative intelligence. With the term qualitative
intelligence, we refer to capabilities such as reasoning, strategic actions
and drawing meaningful conclusions from data that were never seen during
training. The last point distinguishes whether a model only rearranges seen
data or is truly able to solve unseen problems, which is claimed to be an
intelligence characteristic. Compared to quantitative intelligence,
qualitative measures require a whole different category of examination
techniques. However, given the partly closed source/weights of certain models
and the sheer amount of training data, that point is not easily addressable. A
noteworthy example is the current investigation on the persuasive capabilities
of LLMs in staged debates. By applying randomized control trials (RCTs), which
are often used in clinical settings or causal investigations, the machine
versus human persuasiveness can be assessed in both qualitative and
quantitative measures [11]. The assessment of the performance in a crowdsource
setting like the Chatbot Arena provides valuable tools for qualitative
evaluation [12]. Similar to standardised tools such as RCTs, a gold standard
for qualitative assessments of intelligence is urgently needed but still
missing. Nevertheless, comparing the results in reasoning tests of specialized
models, such as the first transformer models with their successor to current
LLMs, the improvements in terms of qualitative measures are only gradual [13].
This is particularly true when you consider the model size and the training
data, which have increased by multiple orders of magnitude over the same time
(Figure 1). For a fair comparison of general reasoning capabilites between
models, the quantitative aspect should be removed from the evaluation.

## Computational growth versus intelligence growth

The exponential growth of used training data with accompanying increase in
model size will likely put the models in the foreseeable future to a limit
regarding data availability. As a thought experiment, imagine training a model
on the complete information on the internet, all text ever written, and all
thoughts every human ever said. How smart could such a model potentially
become given the current self-supervised training paradigm predicting a word
in a sentence? Quantitatively, the model would encompass the entire human
knowledge. Qualitatively it potentially could outsmart every person ever
lived, being fed with countless human ideas. However, we argue that the
hypothetical qualitative improvement over humans would still be on a
comparable scale as the whole training basis inherently relies on human
thoughts and language from which the model can not escape. In the same way, a
perfect model for the language of an animal species, e.g., songbirds, would
have difficulty explaining Einstein’s theory of relativity. The basis of large
language models relies on their training data, which is based on the mental
capacity of the underlying species. While LLMs undoubtebly demonstrate
impressive linguistic precision, it remains unclear how that translates into
the broader cognitive capabilities often associated with language [14]. This
raises the question of whether language alone is a sufficient means for
acquiring such capabilities [15]. Furthermore, the debate persists as to how a
simple reorganisation of huge amounts of data might not be sufficient to
achieve results that resemble human behaviour[16].

Relating these thoughts to the notion of “super-human” AI, we see that current
and future state-of-the-art LLMs can exceed human intelligence quantitatively
with ease. This is especially the case, when AI tasks are combined with their
usually very short inference time. On the contrary, future LLMs that have a
qualitative improvement orders of magnitudes higher than humans seem still
very unlikely or even impossible with current learning paradigms. On top,
truly emergent intelligence properties of such models might even be invisible
to current measures. To ultimately prove or disprove emergent intelligence in
concrete settings or for specific LLMs, nuanced and combined attention to both
quantitative and qualitative intelligence measures will be crucial. The
development of frameworks that address these issues should receive dedicated
awareness by the research communities.

Undoubtedly, even without singular intelligence growth of LLMs, their societal
impact is already tremendous [6]. With growing quantitative intelligence,
models can be applied to a plethora of domains combining expertise from
various fields. Tasks that currently require a team with different backgrounds
could be handled by a single LLM in the future. Thus, a simple prompt can be
sufficient to launch a new campaign, create a movie, or plan a medical study.
These advancements are fundamental for accelerating existing processes. Access
to superior technology or new physical concepts, on the other hand, is not on
the immediate horizon. Accordingly, the creation of an uncontrollable
superintelligence may not be the most imminent threat, but large-scale job
losses, misinformation and election interference are.

In summary, current and future developments in LLMs will lead to models that
represent a significant part of the entire human knowledge, but which might
not quickly surpass the qualitative output of humans. To address this
bivalence and to reliably examine emergent intelligence properties or even
“super-human” behavior, the intelligence performance of models should be
accessed with separated quantitative and qualitative intelligence metrics.

## References

  * [1] Yoshua Bengio, Geoffrey Hinton, Andrew Yao, Dawn Song, Pieter Abbeel, Trevor Darrell, Yuval Noah Harari, Ya-Qin Zhang, Lan Xue, Shai Shalev-Shwartz, et al.  Managing extreme ai risks amid rapid progress.  Science, page eadn0117, 2024.
  * [2] Nanyi Fei, Zhiwu Lu, Yizhao Gao, Guoxing Yang, Yuqi Huo, Jingyuan Wen, Haoyu Lu, Ruihua Song, Xin Gao, Tao Xiang, et al.  Towards artificial general intelligence via a multimodal foundation model.  Nature Communications, 13(1):3094, 2022.
  * [3] Rylan Schaeffer, Brando Miranda, and Sanmi Koyejo.  Are emergent abilities of large language models a mirage?  Advances in Neural Information Processing Systems, 36, 2024.
  * [4] Melanie Mitchell.  How do we know how smart ai systems are?, 2023.
  * [5] Cameron R Jones and Benjamin K Bergen.  People cannot distinguish gpt-4 from a human in a turing test.  arXiv preprint arXiv:2405.08007, 2024.
  * [6] Ray Perrault and Jack Clark.  Artificial intelligence index report 2024.  2024\.
  * [7] Nicola Jones.  Ai now beats humans at basic tasks—new benchmarks are needed, says major report.  Nature, 628(8009):700–701, 2024.
  * [8] Dan Hendrycks, Collin Burns, Steven Basart, Andy Zou, Mantas Mazeika, Dawn Song, and Jacob Steinhardt.  Measuring massive multitask language understanding.  arXiv preprint arXiv:2009.03300, 2020.
  * [9] Yao Lu, Max Bartolo, Alastair Moore, Sebastian Riedel, and Pontus Stenetorp.  Fantastically ordered prompts and where to find them: Overcoming few-shot prompt order sensitivity.  arXiv preprint arXiv:2104.08786, 2021.
  * [10] Steve Yadlowsky, Lyric Doshi, and Nilesh Tripuraneni.  Pretraining data mixtures enable narrow model selection capabilities in transformer models.  arXiv preprint arXiv:2311.00871, 2023.
  * [11] Francesco Salvi, Manoel Horta Ribeiro, Riccardo Gallotti, and Robert West.  On the conversational persuasiveness of large language models: A randomized controlled trial.  arXiv preprint arXiv:2403.14380, 2024.
  * [12] Wei-Lin Chiang, Lianmin Zheng, Ying Sheng, Anastasios Nikolas Angelopoulos, Tianle Li, Dacheng Li, Hao Zhang, Banghua Zhu, Michael Jordan, Joseph E Gonzalez, et al.  Chatbot arena: An open platform for evaluating llms by human preference.  arXiv preprint arXiv:2403.04132, 2024.
  * [13] Marianna Nezhurina, Lucia Cipolina-Kun, Mehdi Cherti, and Jenia Jitsev.  Alice in wonderland: Simple tasks showing complete reasoning breakdown in state-of-the-art large language models.  arXiv preprint arXiv:2406.02061, 2024.
  * [14] Kyle Mahowald, Anna A Ivanova, Idan A Blank, Nancy Kanwisher, Joshua B Tenenbaum, and Evelina Fedorenko.  Dissociating language and thought in large language models.  Trends in Cognitive Sciences, 2024.
  * [15] David Silver, Satinder Singh, Doina Precup, and Richard S Sutton.  Reward is enough.  Artificial Intelligence, 299:103535, 2021.
  * [16] Melanie Mitchell and David C Krakauer.  The debate over understanding in ai’s large language models.  Proceedings of the National Academy of Sciences, 120(13):e2215907120, 2023.


# Generalizations of L- and M-weakly compact operators

###### Abstract

L- and M-weakly compact operators were introduced by Meyer-Nieberg in the
beginning of seventies in attempts of a diversification of the concept of
weakly compact operators via imposing Banach lattice structure on the range or
on the domain of operators. We investigate regularity and algebraic properties
of various generalizations of L- and M-weakly compact operators from a unified
point of view via using regularly P-operators.

Safak Alpay1, Eduard Emelyanov2, Svetlana Gorokhova 3
$1$ Middle East Technical University, Ankara, Turkey
$2$ Sobolev Institute of Mathematics, Novosibirsk, Russia
$3$ Uznyj matematiceskij institut VNC RAN, Vladikavkaz, Russia

Keywords: Banach lattice, regularly ${\cal P}$-operator, domination property,
limited set, bi-limited set, LW-set.

MSC2020: 46A40, 46B42, 46B50, 47B65

##  1 Introduction and Preliminaries

L- and M-weakly compact operators were introduced half a century ago in order
to diversify the concept of weakly compact operators by adding lattice
structure on the range/domain of operators. Since then, these operators
attract permanent attention and inspire further investigations. Apart of study
of already known classes of such operators, we introduce and investigate new
classes of limitedly LW\- and bi-limitedly LW-operators.

### 1.1

In the present paper: vector spaces are real; operators are linear; the
letters $X$, $Y$, $Z$ stand for Banach spaces, $E$, $F$, $G$ for Banach
lattices, and $V$, $W$ for ordered vector spaces. An operator $T:X\to Y$ (a
subset $A$ in $X$) is said to be bounded if $T$ is norm continuous (resp., if
$A$ is norm bounded). We denote by $B_{X}$ the closed unit ball of $X$; by
$\text{\rm L}(X,Y)$ ($\text{\rm K}(X,Y)$, $\text{\rm W}(X,Y)$) the space of
all bounded (compact, weakly compact) operators from $X$ to $Y$; by $E_{+}$
the positive cone of $E$; by $\text{\rm sol}(A)$ the solid hull of $A\subseteq
E$; and by $E^{a}=\\{x\in E:|x|\geq x_{n}\downarrow 0\Rightarrow\|x_{n}\|\to
0\\}$ the o-continuous part of $E$. An operator $T:V\to W$ is called regular
if $T=T_{1}-T_{2}$ for some positive operators $T_{1},T_{2}:V\to W$. We denote
by:

  1. ${\cal L}(V,W)$ the space of all linear operators from $V$ to $W$;

  2. ${\cal L}_{r}(V,W)$ the space of all regular operators from $V$ to $W$;

  3. ${\cal L}_{ob}(V,W)$ the space of all order bounded operators from $V$ to $W$.

Clearly, ${\cal L}_{r}(V,W)\subseteq{\cal L}_{ob}(V,W)$. The order dual
$E^{\sim}$ of $E$ coincides with the norm dual $E^{\prime}$. For any
$S\in{\cal L}_{ob}(E,F)$, the adjoint $S^{\prime}:F^{\prime}\to E^{\prime}$ is
order bounded and o-continuous (cf. [3, Thm.1.73]). Furthermore, let
$T\in{\cal L}_{ob}(E,F)$, then $T^{\prime\prime}\in{\cal
L}_{ob}(E^{\prime\prime},F^{\prime\prime})$. Since $F^{\prime\prime}$ is
Dedekind complete, $T^{\prime\prime}\in{\cal
L}_{r}(E^{\prime\prime},F^{\prime\prime})\subseteq\text{\rm
L}(E^{\prime\prime},F^{\prime\prime})$. Thus $T\in\text{\rm L}(E,F)$, and
hence ${\cal L}_{ob}(E,F)\subseteq\text{\rm L}(E,F)$. We shall also use

$\|f\|=\sup\\{|f(x)|:x\in B_{X}\\}=\sup\\{|y(f)|:y\in B_{X^{\prime\prime}}\\}\
\ \ \ (\forall f\in X^{\prime})$ (1)

and, as a consequence of (1),

$(f_{n})\ \text{\rm in}\ X^{\prime}\ \text{\rm converges uniformly on}\ B_{X}\
\text{\rm iff it converges uniformly on}\ B_{X^{\prime\prime}}$ (2)

under the identification of $f\in X^{\prime}$ with
$\hat{f}\in(X^{\prime})^{\prime\prime}$. By [4, Def.1.1]: a subset $A\subseteq
E$ is called b-order bounded whenever $\hat{A}$ is order bounded in
$E^{\prime\prime}$, where $E\hat{\to}E^{\prime\prime}$ is the natural
embedding of $E$ into $E^{\prime\prime}$. We call an operator $T:E\to F$
b-order bounded, if $T(A)$ is order bounded in $F$ for each b-order bounded
$A\subseteq E$. An operator $T:E\to Y$ is said to be b-w-compact (resp. o-w-
compact) if $T$ carries b-order bounded sets (resp. order interval) of $E$
onto a relatively w-compact subset of $Y$. We denote by $\text{\rm b-W}(E,Y)$
(resp. by $\text{\rm o-W}(E,Y)$) the space of all b-w-compact weakly (resp.
o-w-compact) operators from $E$ to $Y$.

### 1.2

The Meyer-Nieberg theorem (cf. [3, Thm.5.55]):

  1. ($\ast$)

if every disjoint sequence in the solid hull of $L\subseteq F$ is norm null,
then $\text{\rm sol}(L)$ $($and hence $L$ itself$)$ is relatively w-compact in
$F$

produces the following proper subclass of relatively w-compact subsets.

######  Definition 1.2.1.

([32, Def.1.iii)]) A bounded subset $L$ of $F$ is called an LW-set whenever
every disjoint sequence in $\text{\rm sol}(L)$ is norm null.

For every LW-subset $A$ of $E$, we have $A\subseteq E^{a}$. Indeed, otherwise
there is $a\in A$ with $|a|\in E\setminus E^{a}$, and then there exists a
disjoint sequence $(x_{n})$ in $[0,|a|]\subseteq\text{\rm sol}(A)$ with
$\|x_{n}\|\not\to 0$. It can be easily seen that $B_{E}$ is w-compact yet not
an LW-set in any infinite dimensional reflexive $E$. We need the following
characterization of LW-sets (see [33, Prop.3.6.2] and [13, Lm.2.4]):

######  Assertion 1.2.2.

For a nonempty bounded subset $A$ of $E$, TFAE.

  1. (i)

$A$ is an LW-set.

  2. (ii)

For every $\varepsilon>0$, there is $u_{\varepsilon}\in E_{+}^{a}$ such that
$A\subseteq[-u_{\varepsilon},u_{\varepsilon}]+\varepsilon B_{E}$.

  3. (iii)

$f_{n}(x_{n})\to 0$ for every $(x_{n})$ in $A$ and every disjoint bounded
$(f_{n})$ in $E^{\prime}$.

Recall that a subset $A$ of $E$ is almost order bounded whenever, for every
$\varepsilon>0$, there is $x\in E_{+}$ with $A\subseteq[-x,x]+\varepsilon
B_{E}$. By Assertion 1.2.2,

$\text{every almost order bounded set in an \text{\rm o}-continuous $E$ is an
\text{\rm LW}-set}.$ (3)

Clearly,

$\text{every relatively compact subset of $E$ is almost order bounded}.$ (4)

### 1.3

An LW-operator $T:X\to F$ is defined [32, Def.1.iii)] via the condition that
$T(B_{X})$ is an LW-set. Then $\text{\rm LW}(X,F)\subseteq\text{\rm W}(X,F)$,
and

$T\in\text{\rm LW}(X,F)\Longleftrightarrow\text{\rm$T$ takes bounded subsets
onto \text{\rm LW}-subsets.}$ (5)

In view of (3), and of (4):

$\text{if the norm on $F$ is \text{\rm o}-continuous then $\text{\rm
K}(X,F)\subseteq\text{\rm LW}(X,F)$}.$

The next important fact goes back to Meyer-Nieberg (see also [3, Thm.5.63] and
[15, Prop.2.2] for the more general setting). It is the key tool to various
generalizations of LW-operators.

######  Assertion 1.3.1.

Let $A\subseteq F$ and $B\subseteq F^{\prime}$ be nonempty bounded sets. TFAE.

  1. (i)

Each disjoint sequence in $\text{\rm sol}(A)$ is uniformly null on $B$.

  2. (ii)

Each disjoint sequence in $\text{\rm sol}(B)$ is uniformly null on $A$.

Assertion 1.3.1 yields a method of producing many clases of operators. For
example if we consider sets of the form $A=T(C)$ with any w-compact subset $C$
of $X$ and $B=B_{F^{\prime}}$ then we obtain almost LW\- and almost MW-
operators [13], whereas the choice of sets of the form $A=T[0,x]$ with any
order interval $[0,x]$ in $E$ and $B=B_{F^{\prime}}$ produces order LW\- and
order MW-operators [28]. We include a proof of the following certainly well
known fact.

######  Lemma 1.3.2.

Let $L$ be a nonempty bounded subset of $F^{\prime}$. TFAE.

  1. i)

$L$ is an LW-subset of $F^{\prime}$.

  2. ii)

Each disjoint bounded sequence in $F$ is uniformly null on $L$.

  3. iii)

Each disjoint bounded sequence in $F^{\prime\prime}$ is uniformly null on $L$.

###### Proof.

Keeping in mind (2) and the fact that the unit balls in Banach lattices are
solid, we apply Assertion 1.3.1 first to $A=B_{F}$ and $B=L$, and then to
$A=L$ and $B=B_{F^{\prime\prime}}$ in order to obtain that both ii) and iii)
are equivalent to the condition that every disjoint sequence in $\text{\rm
sol}(L)$ is norm null, which means that $L$ is an LW-subset of $F^{\prime}$. ∎

The next well known assertion is a direct consequence of Lemma 1.3.2.

######  Assertion 1.3.3.

Let $T:X\to F^{\prime}$ be a bounded operator. TFAE.

  1. (i)

$T\in\text{\rm LW}(X,F^{\prime})$.

  2. (ii)

$(T^{\prime}f_{n})$ is norm null for every disjoint bounded sequence $(f_{n})$
in $F$.

  3. (iii)

$(T^{\prime}f_{n})$ is norm null for every disjoint bounded sequence $(f_{n})$
in $F^{\prime\prime}$.

The condition (iii) had lead to the idea of the following definition.

######  Definition 1.3.4.

[32, Def.1.iv)] A bounded operator $S:E\to Y$ is called$:$

  1. an MW-operator, or $S\in\text{\rm MW}(E,Y)$, if $\|Sx_{n}\|\to 0$ for every disjoint bounded sequence $(x_{n})$ in $E$.

The following duality result [32, Satz.3] (cf. also [3, Them.5.64]) has
justified the success of introducing LW\- and MW-operators.

######  Assertion 1.3.5.

The following holds.

  1. (i)

$X\stackrel{{\scriptstyle T}}{{\rightarrow}}F$ is an LW-operator
$\Longleftrightarrow$ $F^{\prime}\stackrel{{\scriptstyle
T^{\prime}}}{{\rightarrow}}X^{\prime}$ is an MW-operator.

  2. (ii)

$E\stackrel{{\scriptstyle S}}{{\rightarrow}}Y$ is an MW-operator
$\Longleftrightarrow$ $Y^{\prime}\stackrel{{\scriptstyle
S^{\prime}}}{{\rightarrow}}E^{\prime}$ is an LW-operator.

### 1.4

Assertion 1.3.1 applied to $A=T(C)$ with any compact $C\subseteq X$ and
$B=B_{F^{\prime}}$ gives:

######  Definition 1.4.1.

  1. a)

An operator $T:X\to F$ is called a c-LW-operator if $T$ maps compact subsets
of $X$ into LW-subsets of $F$.

  2. b)

A bounded operator $S:E\to Y$ is called a c-MW-operator if
$S^{\prime}\in\text{\rm c-LW}(Y^{\prime},E^{\prime})$.

In view of Assertion 1.2.2, we may replace in Definition 1.4.1 a) compact
subsets of $X$ by singletons and get:

######  Assertion 1.4.2.

TFAE.

  1. (i)

$T\in\text{\rm c-LW}(X,F)$.

  2. (ii)

$\\{Tx\\}$ is an LW-subset of $F$ for each $x\in X$.

  3. (iii)

$T(X)\subseteq F^{a}$.

  4. (iv)

$f_{n}(Tx)\to 0$ for each $x\in X$ and for every disjoint bounded $(f_{n})$ in
$F^{\prime}$.

  5. (v)

$T^{\prime}$ takes disjoint bounded sequences of $F^{\prime}$ onto $\text{\rm
w}^{\ast}$-null sequences of $X^{\prime}$.

The next fact is a consequence of Lemma 1.3.2.

######  Assertion 1.4.3.

Let $T:X\to F^{\prime}$ be a bounded operator. TFAE.

  1. (i)

$T\in\text{\rm c-LW}(X,F^{\prime})$.

  2. (ii)

$(T^{\prime}f_{n})$ is $\text{\rm w}^{\ast}$-null in $X^{\prime}$ for every
disjoint bounded $(f_{n})$ in $F$.

  3. (iii)

$(T^{\prime}f_{n})$ is $\text{\rm w}^{\ast}$-null in $X^{\prime}$ for every
disjoint bounded $(f_{n})$ in $F^{\prime\prime}$.

By Assertion 1.4.2, the class of c-LW-operators is too large to be of any
reasonable interest.

b-LW-operators were introduced in [27]. They correspond to taking the pair
$(A,B)$ in Assertion 1.3.1: $A=T(U)$, with an arbitrary b-bounded $U\subseteq
E$, and $B=B_{F^{\prime}}$. We recall the following definition.

######  Definition 1.4.4.

A bounded subset $A$ of $X$ is called:

  1. a)

limited if each w∗-null sequence in $X^{\prime}$ is uniformly null on $A$ (cf.
[14]).

  2. b)

bi-limited if each w∗-null sequence in $X^{\prime\prime\prime}$ is uniformly
null on $A$.

It is well known that $B_{X}$ is not limited when $\dim(X)=\infty$, and

$A\ \text{\rm is compact}\ \Rightarrow A\ \text{\rm is limited}\ \Rightarrow
A\ \text{\rm is bi-limited}.$

Thus $B_{c_{0}}$ is not limited in $c_{0}$. However $B_{c_{0}}$ is bi-limited
in $c_{0}$ by Phillip’s lemma (cf. [3, Thm.4.67]).

b-LW-operators were introduced in [27]. They correspond to taking the pair
$(A,B)$ in Assertion 1.3.1: $A=T(U)$, with an arbitrary b-bounded $U\subseteq
E$, and $B=B_{F^{\prime}}$.

### 1.5

We recall the following definition.

######  Definition 1.5.1.

A bounded subset $A$ of $X$ is called:

  1. a)

limited if each w∗-null sequence in $X^{\prime}$ is uniformly null on $A$.

  2. b)

bi-limited if each w∗-null sequence in $X^{\prime\prime\prime}$ is uniformly
null on $A$.

It is well known that $B_{X}$ is not limited when $\dim(X)=\infty$, and

$A\ \text{\rm is relatively compact}\ \Rightarrow A\ \text{\rm is limited}\
\Rightarrow A\ \text{\rm is bi-limited}.$

Thus $B_{c_{0}}$ is not limited in $c_{0}$. However $B_{c_{0}}$ is bi-limited
in $c_{0}$ by Phillip’s lemma (cf. [3, Thm.4.67]). Every bi-limited subset of
a reflexive Banach space is relatively compact.

###  1.6 Regularly ${\cal P}$-operators and the domination property.

The following notion was introduced in [23, Def.2]. In the present paper, we
develope ideas of [23] for investigation algebraic and lattice properties of
the operators.

######  Definition 1.6.1.

Let ${\cal P}$ be a set of operators (the set of ${\cal P}$-operators) from
$V$ to $W$. An operator $T:V\to W$ is called a regularly ${\cal P}$-operator
(shortly, an r-${\cal P}$-operator), if $T$ is a difference of two positive
${\cal P}$-operators.

Given a set ${\cal P}$ of operators from $V$ to $W$. We denote by:

  1. ${\cal P}(V,W)$ the set of all ${\cal P}$-operators from $V$ to $W$;

  2. ${\cal P}_{+}(V,W)$ the set of all positive ${\cal P}$-operators from $V$ to $W$;

  3. ${\cal P}_{r}(V,W)$ the set of all regular operators in ${\cal P}(V,W)$;

  4. r-${\cal P}(V,W)$ the set of all r-${\cal P}$-operators in ${\cal P}(V,W)$.

${\cal P}$-operators are said to satisfy the domination property if it follows
from $0\leq S\leq T\in{\cal P}$ that $S$ is likewise a ${\cal P}$-operator. We
say that an operator $T\in{\cal L}(V,W)$ is ${\cal P}$-dominated if $\pm T\leq
U$ for some $U\in{\cal P}$ (cf. [1, p.3022]).

######  Proposition 1.6.2.

Let ${\cal P}\subseteq{\cal L}(V,W)$, ${\cal P}\pm{\cal P}\subseteq{\cal
P}\neq\emptyset$, and $T\in{\cal L}(V,W)$.

  1. i)

$T$ is an r-${\cal P}$-operator iff $T$ is a ${\cal P}$-dominated ${\cal
P}$-operator.

  2. ii)

Assume the modulus $|T|$ of $T$ exists in ${\cal L}(V,W)$ and suppose that
${\cal P}$-operators satisfy the domination property. Then $T$ is an r-${\cal
P}$-operator iff $|T|$ is a ${\cal P}$-operator.

###### Proof.

i) Let $T$ be an r-${\cal P}$-operator, say $T=T_{1}-T_{2}$ with two positive
${\cal P}$-operators $T_{1},T_{2}$. It follows from ${\cal P}\pm{\cal
P}\subseteq{\cal P}$ that $T$ and $U=T_{1}+T_{2}$ are both ${\cal
P}$-operators. Since $\pm T\leq U$ then $T$ is a ${\cal P}$-dominated ${\cal
P}$-operator.

Now, let $T$ be a ${\cal P}$-dominated ${\cal P}$-operator. Take a ${\cal
P}$-operator $U$ such that $\pm T\leq U$. Since $T=U-(U-T)$, and both $U$ and
$U-T$ are positive ${\cal P}$-operators, $T$ is an r-${\cal P}$-operator.

ii) First assume $|T|$ is a ${\cal P}$-operator. Since $T=T_{+}-T_{-}$, $0\leq
T_{\pm}\leq|T|$, and $|T|$ is a ${\cal P}$-operator, the domination property
implies that $T_{+}$ and $T_{-}$ are both positive ${\cal P}$-operators, and
hence $T=T_{+}-T_{-}$ is an r-${\cal P}$-operator.

Now, assume $T$ is an r-${\cal P}$-operator. Then there are two positive
${\cal P}$-operators $T_{1},T_{2}$ satisfying $T=T_{1}-T_{2}$. Since $0\leq
T_{+}\leq T_{1}$ and $0\leq T_{-}\leq T_{2}$, the domination property implies
that $T_{+}$ and $T_{-}$ are both ${\cal P}$-operators. Hence
$|T|=T_{+}+T_{-}$ is likewise a ${\cal P}$-operator. ∎

######  Proposition 1.6.3.

Let $F$ be Dedekind complete, and ${\cal P}$ a subspace in $\text{\rm
L}(E,F)$, satisfying the domination property. Then $\text{\rm r-}{\cal
P}(E,F)$ is an order ideal in the Dedekind complete vector lattice $\text{\rm
L}_{r}(E,F)$.

###### Proof.

Since $F$ is Dedekind complete, $\text{\rm L}_{r}(E,F)$ is a Dedekind complete
vector lattice. By Proposition 1.6.2 ii), $T\in\text{\rm r-}{\cal
P}(E,F)\Longrightarrow|T|\in\text{\rm r-}{\cal P}(E,F)$, and hence $\text{\rm
r-}{\cal P}(E,F)$ is a vector sublattice of $\text{\rm L}_{r}(E,F)$. Since
${\cal P}$ satisfies the domination property, $\text{\rm r-}{\cal P}(E,F)$ is
an order ideal in $\text{\rm L}_{r}(E,F)$. ∎

In general, quite a little is known about conditions on ${\cal P}$-operators
under that every regular ${\cal P}$-operator is an r-${\cal P}$-operator, even
for compact operators in a Banach lattice (see [16, 18] and references
therein). By [1, Thm.4(ii)], there exists a compactly dominated compact
operator on $L^{2}[0,1]\oplus c(L^{2}[0,1])$ without modulus. So, in terms of
Proposition 1.6.2 i), an r-compact operator need not to have modulus. By [17,
Thm.2.2], r-compact (resp. r-w-compact, r-LW-, r-MW-) operators from
$E=L^{2}[0,1]$ to $F=c(L^{2}[0,1])$ form a proper subspace of the space of
regular compact (resp. regular w-compact, regular LW-, regular MW-) operators
from $E$ to $F$. In view of Proposition 1.6.2 ii), the following fact is a
direct consequence of [26, Satz 7.1] and [17, Thm.2.9].

  1. If $F$ is an AM-space, then every compact (LW-) operator from $E$ to $F$ is r-compact (resp. r-LW-).

The domination property for compact, w-compact, LW-, MW-, and Dunford–Pettis
operators was well investigated in the last quarter of 20th century. Recently,
it was studied in [6, 23, 24] for KB and Levi operators between locally solid
lattices.

The following example goes back to [25] (see, also [3, Ex.5.6],[1],[9]).

######  Example 1.6.4.

Let $A_{n}$ be a $2^{n}\times 2^{n}$ matrix constructed inductively:

$A_{1}=\left[\begin{matrix}1&1\\\ 1&-1\end{matrix}\right]\quad\text{\rm
and}\quad A_{n+1}=\left[\begin{matrix}A_{n}&A_{n}\\\
A_{n}&-A_{n}\end{matrix}\right].$

For each $n$, let $T_{n}:\ell^{2}_{2^{n}}\to\ell^{2}_{2^{n}}$ be the operator,
whose matrix (with respect to the standard unit vectors of
$\ell^{2}_{2^{n}}={\mathbb{R}}^{2^{n}}$) is $2^{-\frac{n}{2}}A_{n}$. Since the
matrix $2^{-\frac{n}{2}}A_{n}$ is orthogonal, each $T_{n}$ is an isometry. On
the other hand, $|T_{n}|$ is the operator, whose matrix is a $2^{n}\times
2^{n}$ matrix with all of its entries equal to 1, and hence
$\||T_{n}|\|=2^{\frac{n}{2}}$ for all $n$. Consider the $c_{0}$-direct sum
$E:=(\oplus_{n=1}^{\infty}\ell^{2}_{2^{n}})_{0}$. Note that
$E^{\prime}=(\oplus_{n=1}^{\infty}\ell^{2}_{2^{n}})_{1}$, in particular, $E$
and $E^{\prime}$ both have o-continuous norms. By [20, Thm.4.5], compact
operators on $E$ satisfy the domination property. For any
$\alpha=(\alpha_{n})\in\ell^{\infty}$, define $T:E\to E$, by

$Tx=T(x_{1},x_{2},\dots):=(\alpha_{1}T_{1}x_{1},\alpha_{2}T_{2}x_{2},\dots).$
(6)

Then $\|T\|\leq\|\alpha\|_{\infty}$, and if $\alpha_{n}\in c_{0}$, then
$T\in\text{\rm K}(E)$. If $\alpha_{n}=2^{-\frac{n}{2}}$, the modulus $|T|$
exists and is given by
$|T|x=(\alpha_{1}|T_{1}|x_{1},\alpha_{2}|T_{2}|x_{2},\dots)$. As $|T|$ is not
compact, $T\in\text{\rm K}_{r}(E)\setminus\text{\rm r-K}(E)$ by Proposition
1.6.2 ii). For $\alpha_{n}=2^{-\frac{n}{3}}$, $T$ is compact but not order
bounded; so, $|T|$ does not exist.

###  1.7 Disjoint sequence properties.

In this subsection we discuss some facts and results on Banach lattices used
in the paper. The spaces of operators that we study here as well as the spaces
serving as domains and/or ranges to the operators have many interesting
disjointness properties. We begin with the next well known fact (cf. [3,
Thm.4.59]; [33, 2.4.14]) is used freely in the present paper.

######  Assertion 1.7.1.

TFAE.

  1. (i)

The norm on $E^{\prime}$ is o-continuous.

  2. (ii)

$E^{\prime}$ is a KB space.

  3. (iii)

Every disjoint bounded sequence in $E$ is w-null.

Recall that $E$ has sequentially w-continuous (w∗-continuous) lattice
operations if, for every w-null $(x_{n})$ in $E$ (w∗-null $(x_{n})$ in
$E^{\prime}$), the sequence $(|x_{n}|)$ is also w-null (w∗-null). It follows
from [3, Thm.4.34]) that, for any disjoint w-null $(x_{n})$ in $E$, the
sequence $(|x_{n}|)$ is w-null.

######  Definition 1.7.2.

([21, Def.1]). $E$ is said to have the property (d) (shortly,
$E\in\text{\rm(d)})$ if $(|f_{n}|)$ is also $\text{\rm w}^{\ast}$-null for
each disjoint $\text{\rm w}^{\ast}$-null $(f_{n})$ in $E^{\prime}$.

The property (d) is weaker than the sequential $\text{w}^{\ast}$-continuity of
the lattice operations, as $E=\ell^{\infty}$ shows. It was already studied in
[35], where no name was assigned to this property. In particular, it was
proved in [35, Prop.1.4] that every $\sigma$-Dedekind complete Banach lattice
$E$ has the property (d), and it was observed in [35, Rem.1.5] that
$\ell^{\infty}/c_{0}\in\text{\rm(d)}$, but $\ell^{\infty}/c_{0}$ is not
$\sigma$-Dedekind complete. Example 1.7.8 below shows that
$C[0,1]\not\in\text{\rm(d)}$.

######  Definition 1.7.3.

A Banach lattice $E$ is said to have:

  1. a)

the Schur property (shortly, $E\in(\text{\rm SP})$) if w-null sequences in $E$
are norm null (cf. [3, p.207]);

  2. b)

the positive Schur property (shortly, $E\in(\text{\rm PSP})$) if positive
w-null sequences in $E$ are norm null (cf. [35]);

  3. c)

the dual positive Schur property (shortly, $E\in(\text{\rm DPSP})$) if
positive w∗-null sequences in $E^{\prime}$ are norm null [8, Def.3.3];

  4. d)

the dual disjoint Schur property (shortly, $E\in(\text{\rm DDSP})$) if
disjoint w∗-null sequences in $E^{\prime}$ are norm null;

  5. e)

the Grothendieck property (shortly, $E\in(\text{\rm GP})$) if w∗-null
sequences in $E^{\prime}$ are w-null (cf. [35, p.760]);

  6. f)

the positive Grothendieck property (shortly, $E\in(\text{\rm PGP})$) if
positive w∗-null sequences in $E$ are w-null (cf. [35, p.760]);

  7. g)

the disjoint Grothendieck property (shortly, $E\in(\text{\rm DGP})$) if
disjoint w∗-null sequences in $E^{\prime}$ are w-null;

  8. h)

the dual disjoint w∗-property (shortly, $E\in(\text{\rm DDw${}^{\ast}$P})$) if
each disjoint bounded sequence in $E^{\prime}$ is w∗-null.

The disjoint Grothendieck property was introduced under the name the weak
Grothendieck property in [29, Def.4.8)].

######  Proposition 1.7.4.

For every Banach lattice $E$ the following holds.

$E\in(\text{\rm DGP})\cap(\text{\rm DDw}^{\ast}\text{\rm P})\Longrightarrow
E^{\prime\prime}\text{\it\ has \text{\rm o}-continuous norm}\Longrightarrow
E\in(\text{\rm DDw}^{\ast}\text{\rm P}).$

###### Proof.

Let $E\in(\text{\rm DGP})\cap(\text{\rm DDw${}^{\ast}$P})$ but the norm in
$E^{\prime\prime}$ is not o-continuous. Then $(f_{n})$ is not w-null for some
disjoint bounded $(f_{n})$ in $E^{\prime}$ by Assertion 1.7.1. However
$f_{n}\stackrel{{\scriptstyle\text{\rm w}^{\ast}}}{{\to}}0$ because
$E\in(\text{\rm DDw${}^{\ast}$P})$, and hence
$f_{n}\stackrel{{\scriptstyle\text{\rm w}}}{{\to}}0$ due to $E\in(\text{\rm
DGP})$. The obtained contradiction proves the first implication.

For the second implication, let the norm in $E^{\prime\prime}$ be
o-continuous. Then, by Assertion 1.7.1, every disjoint bounded $(f_{n})$ in
$E^{\prime}$ is w-null and hence $\text{\rm w}^{\ast}$-null. It follows
$E\in(\text{\rm DDw${}^{\ast}$P})$. ∎

The next important fact was proved in [35, Prop.2.3]. We do not know whether
or not in this assertion (DPSP) can be replaced by (PGP) and the norm null
condition by the w-null condition.

######  Assertion 1.7.5.

$E\in\text{\rm(DPSP)}$ iff every disjoint w∗-null sequence in $E^{\prime}_{+}$
is norm null.

Clearly, (DPSP)$\Longrightarrow$(PGP); and (DDSP)$\Longrightarrow$(DGP). In
AM-spaces, (PGP) agrees with (DPSP), by [35, Prop.4.1]. (PSP) and (SP)
coincide in the class of discrete Banach lattices [34, p.19]. By [33,
Thm.5.3.13], (PGP) and (GP) agree in the class of Banach lattices with the
interpolation property.

The following assertion [30, Thm.3.3] describes the impact of the property (d)
on dual Schur properties.

######  Assertion 1.7.6.

For every Banach lattice $E$ the following holds.

  1. i)

$E\in(\text{\rm DDSP})\ \Longrightarrow\ E\in(\text{\rm DPSP})$;

  2. ii)

if $E$ has the property (d) then
$E\in(\text{\rm DDSP})\ \Longleftrightarrow\ E\in(\text{\rm DPSP})$.

Clearly $(\text{PGP})\Longrightarrow(\text{DGP})$ for Banach lattices with the
property (d). For this implication, the property (d) is essential, as
$C[0,1]\in\text{\rm(PGP)}\setminus\text{\rm(DGP)}$ by Example 1.7.8 below. We
have no example of $E\in(\text{DGP})\setminus(\text{PGP})$. For the next
assertion, we refer to [33, Cor.3.6.8], [34, Thm.7].

######  Assertion 1.7.7.

For a Banach lattice $E$, TFAE.

  1. (i)

$E\in(\text{\rm PSP})$.

  2. (ii)

Each disjoint w-null sequence in $E$ is norm null.

  3. (iii)

Each disjoint w-null sequence in $E_{+}$ is norm null.

  4. (iv)

Every disjoint sequence in the solid hull of each relatively w-compact subset
of $E$ is norm-null.

  5. (v)

Almost order bounded subsets of $E$ coincide with relatively w-compact subsets
of $E$.

By Assertion 1.7.7, $\text{\rm if}\ F\in\text{\rm(PSP)}\ \text{\rm then}\
\text{\rm W}(X,F)\subseteq\text{\rm LW}(X,F).$ The next example shows that the
property (d) is essential in Assertion 1.7.6(ii).

######  Example 1.7.8.

Consider the Banach lattice $E=C[0,1]$. Let $(f_{n})$ be a positive w∗-null
sequence in $E^{\prime}$. It follows from
$\|f_{n}\|=f_{n}({\mathbb{1}}_{[0,1]})\to 0$ that $C[0,1]\in(\text{\rm
DPSP})$, and hence $C[0,1]\in(\text{\rm PGP})$.

Take the sequence $(r_{n})$ of the Rademacher functions on $[0,1]$ and define
a disjoint w∗-null sequence in $E^{\prime}$ as follows: $g_{n}=2^{n+1}\cdot
r_{2n}\cdot{\mathbb{1}}_{[\frac{1}{2^{n+1}},\frac{1}{2^{n}}]}$. Then $(g_{n})$
is disjoint w∗-null sequence in $E^{\prime}$, yet $\|g_{n}\|=1$ for all
$n\in{\mathbb{N}}$. Therefore $C[0,1]\not\in(\text{DDSP})$, and hence by
Assertion 1.7.6, $C[0,1]\not\in(\text{d})$. Note that $(g_{n})$ is even not
w-null. Indeed, for
$y=\sum_{n=1}^{\infty}r_{2n}\cdot{\mathbb{1}}_{[\frac{1}{2^{n+1}},\frac{1}{2^{n}}]}\in
L^{\infty}[0,1]\subseteq E^{\prime},$ $y(g_{n})\equiv 1$ for all
$n\in{\mathbb{N}}$. It also shows that $C[0,1]\not\in(\text{DGP})$.

For further unexplained terminology and notations, we refer to [3, 4, 5, 6,
12, 28, 13, 22, 33].

##  2 a-LW\- and l-LW-operators

Almost LW\- and almost MW-operators were introduced in [13]. Here, apart from
investigating them, we introduce and study limitedly LW\- and MW-operators.
They both respectively generalize LW\- and MW-operators. We also show below in
Proposition 2.3.4 that the property DDw∗P puts an end to further extensions as
when $F\in(\text{\rm DDw${}^{\ast}$P})$ then $\text{\rm l-LW}(X,F)=\text{\rm
L}(X,F)$ for any Banach space $X$.

### 2.1

We begin with the following definition.

######  Definition 2.1.1.

An operator $T:X\to F$ is called:

  1. i)

almost LW ($T$ is a-LW), if $T$ carries relatively w-compact subsets of $X$
onto LW-subsets of $F$ [13, Def.2.1];

  2. ii)

limitedly LW ($T$ is l-LW), if $T$ carries limited subsets of $X$ onto LW-
subsets of $F$.

  3. iii)

bi-limitedly LW ($T$ is bi-l-LW), if $T$ carries bi-limited subsets of $X$
onto LW-subsets of $F$.

Clearly, $\text{\rm LW}(X,F)\subseteq\text{\rm bi-l-LW}(X,F)\subseteq\text{\rm
l-LW}(X,F)\subseteq\text{\rm c-LW}(X,F)$. Furthermore, the above inclusion
turn to identities whenever $X$ is reflexive.

######  Example 2.1.2.

It was pointed out after Definition 2.1.3 that $B_{c_{0}}$ is not limited yet
is bi-limited in $c_{0}$. Since $B_{c_{0}}$ is not an LW-set in $c_{0}$,
$I_{c_{0}}\notin\text{\rm bi-l-LW}(c_{0})$. However, $I_{c_{0}}\in\text{\rm
l-LW}(c_{0})$ since each limited subset of $c_{0}$ is relatively compact and
relatively compact subsets of $c_{0}$ are LW-sets.

######  Definition 2.1.3.

A bounded subset $A$ of $X$ is called:

  1. a)

limited if each w∗-null sequence in $X^{\prime}$ is uniformly null on $A$.

  2. b)

bi-limited if each w∗-null sequence in $X^{\prime\prime\prime}$ is uniformly
null on $A$.

It is well known that $B_{X}$ is not limited when $\dim(X)=\infty$, and

$A\ \text{\rm is relatively compact}\ \Rightarrow A\ \text{\rm is limited}\
\Rightarrow A\ \text{\rm is bi-limited}.$

Thus $B_{c_{0}}$ is not limited in $c_{0}$. However $B_{c_{0}}$ is bi-limited
in $c_{0}$ by Phillip’s lemma (cf. [3, Thm.4.67]). Every bi-limited subset of
a reflexive Banach space is relatively compact.

### 2.2

We remind several facts on a-LW-operators. As w-compact sets are bounded, each
LW-operator is an a-LW-operator. By (5), if $S\in\text{\rm W}(Y,X)$ and
$T\in\text{\rm a-LW}(X,F)$ then $TS\in\text{\rm LW}(Y,F)$. Furthermore, each
a-LW-operator $T:X\to F$ is automatically bounded. Indeed, otherwise there
exists a norm null sequence $(x_{n})$ in $B_{X}$ satisfying $\|Tx_{n}\|\geq n$
for all $n\in\mathbb{N}$, violating the fact that the LW-set
$\\{Tx_{n}:n\in\mathbb{N}\\}$ is bounded. Since $\text{\rm
LW}(X,F)\subseteq\text{\rm W}(X,F)$, the identity operator
$I_{\ell^{1}}:\ell^{1}\to\ell^{1}$ is not LW. It was shown in [22, Prop.1]
that $T\in\text{\rm a-LW}(X,F)$ iff $T(X)\subseteq F^{a}$ and
$f_{n}(Tx_{n})\to 0$ for each disjoint bounded $(f_{n})$ in $F^{\prime}$ and
each w-null $(x_{n})$ in $X$.

Since ${\ell^{1}}$ has the Schur property, $I_{\ell^{1}}\in\text{\rm
a-LW}(\ell^{1})\setminus\text{\rm W}(\ell^{1})$. It was proved in [13,
Thm.2.4], that

$T\in\text{\rm a-LW}(X,F)\ \text{\rm iff}\ TS\in\text{\rm LW}(Y,F)\ \text{\rm
for every}\ Y\ \text{\rm and all}\ S\in\text{\rm W}(Y,X).$ (7)

######  Lemma 2.2.1.

For a pair of operators $T$ and $S$$:$

  1. (i)

if $S\in\text{\rm LW}(X,E)$ and $T\in\text{\rm a-LW}(E,F)$ then
$TS\in\text{\rm LW}(X,F)$;

  2. (ii)

if $S\in\text{\rm L}(X,Y)$ and $T\in\text{\rm a-LW}(Y,F)$ then $TS\in\text{\rm
a-LW}(Y,F)$.

###### Proof.

(i) follows from (7) because LW-operators are w-compact.

(ii) Because every bounded operator is w-continuous. ∎

######  Proposition 2.2.2.

For any Banach lattice $E$ the following hold.

  1. i)

$\text{\rm LW}(E)$ is a closed subalgebra of $\text{\rm W}(E)$ and is unital
iff the identity operator $I_{E}$ is LW.

  2. ii)

$\text{\rm a-LW}(E)$ is a closed right ideal in $\text{\rm L}(E)$ $($and hence
a subalgebra of $\text{\rm L}(E)$$)$, and is unital iff $I_{E}$ is a-LW.

###### Proof.

It is easy to see that both $\text{\rm LW}(E)$ and $\text{\rm a-LW}(E)$ are
closed subspaces of $\text{\rm L}(E)$ (see, for example, [13, Prop.2.1]), and
$\text{\rm LW}(E)\subseteq\text{\rm W}(E)$. The closeness of $\text{\rm
LW}(E)$ under composition follows from Lemma 2.2.1(i)(ii). The fact that
$\text{\rm a-LW}(E)$ is a right ideal in $\text{\rm L}(E)$ follows from Lemma
2.2.1(ii). The conditions on $I_{E}$, those make the algebras $\text{\rm
LW}(E)$ and $\text{\rm a-LW}(E)$ unital, are simply verified. ∎

As $\text{\rm a-LW}(E,F)$ satisfies the domination property by [2, Thm.1], the
next lemma follows from Proposition 1.6.2 ii).

######  Lemma 2.2.3.

Let an operator $T:E\to F$ possess the modulus. TFAE.

  1. i)

$T$ is a regularly a-LW-operator.

  2. ii)

$|T|$ is an a-LW-operator.

We also need the following fact [23, Lm.1].

######  Assertion 2.2.4.

Let ${\rm P}$ be a closed subspace of $\text{\rm L}(E)$ in operator norm. Then
the formula $\|T\|_{\text{\rm r-P}}=\inf\\{\|S\|:\pm T\leq S\in{\rm P}\\}$
defines a norm on the space $\text{\rm r-P}(E)$. 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.

######  Theorem 2.2.5.

For a Banach lattice $E$ the following hold.

  1. i)

The space $\text{\rm r-a-LW}(E)$ of regularly a-LW-operators on $E$ is a
subalgebra of $\text{\rm L}_{r}(E)$. Moreover, $\text{\rm r-a-LW}(E)=\text{\rm
L}_{r}(E)\ \Longleftrightarrow I_{E}\in\text{\rm a-LW}(E).$

  2. ii)

If, additionally $E$ is Dedekind complete then $(\text{\rm r-a-LW}(E),\
\|\cdot\|_{\text{\rm r-a-LW}})$ is a closed Riesz subalgebra and an order
ideal of the Banach lattice algebra $(\text{\rm L}_{r}(E),\ \|\cdot\|_{r})$.

###### Proof.

i) Since regular operators on $E$ are bounded, it follows from Lemma 2.2.1(ii)
that $\text{\rm r-a-LW}(E)$ is a right algebra ideal and hence is a subalgebra
of $\text{\rm L}_{r}(E)$. The condition on $I_{E}$ for $\text{\rm r-a-
LW}(E)=\text{\rm L}_{r}(E)$ is obvious.

ii) It follows from Lemma 2.2.3 that the subalgebra $\text{\rm r-a-LW}(E)$ is
closed under lattice operations in $\text{\rm L}_{r}(E)$, and hence is a Riesz
subalgebra of $\text{\rm L}_{r}(E)$. Proposition 1.6.3 implies that $\text{\rm
r-a-LW}(E)$ is an order ideal of $\text{\rm L}_{r}(E)$.

Let $S,T\in\text{\rm r-a-LW}(E)$ satisfy $|S|\leq|T|$. In order to show
$\|S\|_{\text{\rm r-a-LW}(E)}\leq\|T\|_{\text{\rm r-a-LW}(E)}$, observe that
$|S|,|T|\in\text{\rm a-LW}(E)$ by Lemma 2.2.3. Then

$\|S\|_{\text{\rm r-a-LW}(E)}=\|~{}|S|~{}\|\leq\|~{}|T|~{}\|=\|T\|_{\text{\rm
r-a-LW}(E)}.$

In particular $\|\cdot\|_{\text{\rm r-a-LW}(E)}$ coincide on $\text{\rm r-a-
LW}(E)$ with the regular norm $\|\cdot\|_{r}$, since $\|T\|_{r}=\|~{}|T|~{}\|$
for every $T\in\text{\rm L}_{r}(E)$ [3, p.255].

By Assertion 2.2.4, it follows from [13, Prop.2.1] that $(\text{\rm r-a-
LW}(E),\ \|\cdot\|_{\text{\rm r-a-LW}})$ is a Banach space. It remains to show
that the norm $\|\cdot\|_{\text{\rm r-a-LW}}$ is submultiplicative. Let
$S,T\in{\text{\rm r-a-LW}}(E)$. Since $\pm S\leq|S|\in\text{\rm a-LW}(E)$ and
$\pm T\leq|T|\in\text{\rm a-LW}(E)$ then $\pm ST\leq|ST|\leq|S|\cdot|T|$, and
hence

$\|ST\|_{\text{\rm r-a-
LW}(E)}\leq\|~{}|S||T|~{}\|\leq\|~{}|S|~{}\|\cdot\|~{}|T|~{}\|=\|S\|_{\text{\rm
r-a-LW}(E)}\cdot\|T\|_{\text{\rm r-a-LW}(E)},$

as desired. ∎

###  2.3 l-LW-operators.

Here we include basic properties of l-LW-operators.

######  Lemma 2.3.1.

TFAE.

  1. (i)

$T\in\text{\rm l-LW}(X,F)$.

  2. (ii)

$T^{\prime}f_{n}\stackrel{{\scriptstyle\text{\rm w}^{\ast}}}{{\to}}0$ in
$X^{\prime}$ for each disjoint bounded $(f_{n})$ in $F^{\prime}$.

###### Proof.

(i)$\Longrightarrow$(ii): Let $T\in\text{\rm l-LW}(X,F)$, $(f_{n})$ be
disjoint bounded in $F^{\prime}$, and $x\in X$. As $\\{Tx\\}$ is an LW-subset
of $F$ then $T^{\prime}f_{n}(x)=f_{n}(Tx)\to 0$ by Assertion 1.3.1. Since
$x\in X$ is arbitrary, $(T^{\prime}f_{n})$ is $\text{\rm w}^{\ast}$-null.

(ii)$\Longrightarrow$(i): Let $T^{\prime}f_{n}\stackrel{{\scriptstyle\text{\rm
w}^{\ast}}}{{\to}}0$ for each disjoint bounded $(f_{n})$ in $F^{\prime}$.
Suppose $T\not\in\text{\rm l-LW}(X,F)$. So, $T(L)$ is not an LW-subset of $F$
for some nonempty limited subset $L$ of $X$. Then there exists a disjoint
sequence $(g_{n})$ in $B_{F^{\prime}}$, and it is not uniformly null on $T(L)$
by Assertion 1.3.1. Therefore, $(T^{\prime}g_{n})$ is not uniformly null on
$L$ violating that $T^{\prime}g_{n}\stackrel{{\scriptstyle\text{\rm
w}^{\ast}}}{{\to}}0$ and that $L$ is limited subset of $X$. The obtained
contradiction proves the implication. ∎

######  Proposition 2.3.2.

Let $0\leq S\leq T\in\text{\rm l-LW}(E,F)$ then $S\in\text{\rm l-LW}(E,F)$.

###### Proof.

Let $(f_{n})$ be disjoint bounded in $F^{\prime}$. Then $(|f_{n}|)$ is also
disjoint bounded, and hence $T^{\prime}|f_{n}|\stackrel{{\scriptstyle\text{\rm
w}^{\ast}}}{{\to}}0$ by Lemma 2.3.1. It follows from

$|S^{\prime}f_{n}(x)|\leq S^{\prime}|f_{n}|(|x|)\leq T^{\prime}|f_{n}|(|x|)\to
0\ \ \ \ (\forall x\in E)$

that $S^{\prime}f_{n}\stackrel{{\scriptstyle\text{\rm w}^{\ast}}}{{\to}}0$.
Using Lemma 2.3.1 again, we conclude $S\in\text{\rm l-LW}(E,F)$. ∎

######  Lemma 2.3.3.

Let $\text{\rm l-LW}(X,F)\ni T_{n}\stackrel{{\scriptstyle\|\cdot\|}}{{\to}}T$.
Then $T\in\text{\rm l-LW}(X,F)$.

###### Proof.

Let $(f_{n})$ be disjoint bounded in $F^{\prime}$, and $x\in X$. By Lemma
2.3.1, we need to show $T^{\prime}f_{n}(x)\to 0$. Let $\varepsilon>0$. Pick
any $k\in\mathbb{N}$ with $\|T-T_{k}\|\leq\varepsilon$. Since
$T_{k}\in\text{\rm l-LW}(X,F)$ then $|T_{k}^{\prime}f_{n}(x)|\leq\varepsilon$
for $n\geq n_{0}$. As $\varepsilon>0$ is arbitrary, it follows from

$|T^{\prime}f_{n}(x)|\leq|T^{\prime}f_{n}(x)-T^{\prime}_{k}f_{n}(x)|+|T_{k}^{\prime}f_{n}(x)|\leq$
$\|T^{\prime}-T^{\prime}_{k}\|\|f_{n}\|\|x\|+|T_{k}^{\prime}f_{n}(x)|\leq(\|f_{n}\|\|x\|+1)\varepsilon\
\ \ \ (\forall n\geq n_{0})$

that $T^{\prime}f_{n}(x)\to 0$, as desired. ∎

######  Proposition 2.3.4.

TFAE.

  1. (i)

$F\in(\text{\rm DDw${}^{\ast}$P})$.

  2. (ii)

$I_{F}\in\text{\rm l-LW}(F)$.

  3. (iii)

Each limited subset of $F$ is an LW-set.

  4. (iv)

$\text{\rm l-LW}(F)=\text{\rm L}(F)$.

  5. (v)

$\text{\rm l-LW}(X,F)=\text{\rm L}(X,F)$ for each Banach space $X$.

###### Proof.

$\text{\rm(i)}\Longleftrightarrow\text{\rm(ii)}$: It follows from Lemma 2.3.1.

The implications
$\text{\rm(v)}\Longrightarrow\text{\rm(iv)}\Longrightarrow\text{\rm(ii)}\Longleftrightarrow\text{\rm(iii)}$
are trivial.

$\text{\rm(iii)}\Longrightarrow\text{\rm(v)}$: Let $T\in\text{\rm L}(X,F)$ and
let $L$ be limited subset of $X$. Then $T(L)$ is a limited subset of $F$, and
hence $T(L)$ is an LW-subset of $F$. Thus, $T\in\text{\rm l-LW}(X,F)$, as
desired. ∎

######  Proposition 2.3.5.

$\text{\rm l-LW}(E)$ is a closed right ideal in $\text{\rm L}(E)$ $($and hence
a subalgebra of $\text{\rm L}(E)$$)$, and it is unital iff $I_{E}$ is l-LW.

###### Proof.

$\text{\rm l-LW}(E)$ is a closed subspace of $\text{\rm L}(E)$ by Lemma 2.3.3.
Since each bounded operator maps limited sets onto limited sets, $\text{\rm
l-LW}(E)$ is a right ideal in $\text{\rm L}(E)$. The condition on $I_{E}$
making algebra $\text{\rm l-LW}(E)$ unital is trivial. ∎

In view of Proposition 2.3.2, the next lemma follows from Proposition 1.6.2
ii).

######  Lemma 2.3.6.

Let an operator $T:E\to F$ possess the modulus. TFAE.

  1. i)

$T$ is a regularly l-LW-operator.

  2. ii)

$|T|$ is an l-LW-operator.

######  Theorem 2.3.7.

The following statements hold.

  1. i)

$\text{\rm r-l-LW}(E)$ is a subalgebra of $\text{\rm L}_{r}(E)$. Moreover,

$\text{\rm r-l-LW}(E)=\text{\rm L}_{r}(E)\ \Longleftrightarrow
I_{E}\in\text{\rm l-LW}(E).$

  2. ii)

If $E$ is Dedekind complete then $(\text{\rm r-l-LW}(E),\ \|\cdot\|_{\text{\rm
r-l-LW}})$ is a closed Riesz subalgebra and an order ideal of $(\text{\rm
L}_{r}(E),\ \|\cdot\|_{r})$.

###### Proof.

i) It follows from Proposition 2.3.5 that $\text{\rm r-l-LW}(E)$ is a right
ideal and hence is a subalgebra of $\text{\rm L}_{r}(E)$. The condition on
$I_{E}$ under that $\text{\rm r-l-LW}(E)=\text{\rm L}_{r}(E)$ is trivial.

ii) It follows from Lemma 2.3.6 that $\text{\rm r-l-LW}(E)$ is a Riesz
subalgebra of $\text{\rm L}_{r}(E)$. The rest of the proof is similar to the
proof of Theorem 2.2.5. ∎

###  2.4 Notions of a-MW\- and l-MW-operators

######  Definition 2.4.1.

An operator $S:E\to Y$ is called:

  1. a)

an almost MW-operator ($S$ is a-MW), if $f_{n}(Sx_{n})\to 0$ for every
disjoint bounded $(x_{n})$ in $E$ and every w-convergent $(f_{n})$ in
$Y^{\prime}$ [13, Def.2.2];

  2. b)

a limitedly MW-operator ($S$ is l-MW), if
$Sx_{n}\stackrel{{\scriptstyle\text{\rm w}}}{{\to}}0$ for every disjoint
bounded $(x_{n})$ in $E$.

######  Remark 2.4.2.

It is easily seen that the identity operator $I_{E}$

  1. i)

in any infinite dimensional reflexive Banach lattice $E$ is not a-MW;

  2. ii)

is an l-MW-operator iff $E^{\prime}$ is a KB-space.

### 2.5

We remind several facts on a-MW-operators. The next important semi-duality
result was proved in [13, Thm.2.5].

######  Assertion 2.5.1.

The following holds.

  1. (i)

$T\in\text{\rm a-MW}(E,Y)\Longleftrightarrow T^{\prime}\in\text{\rm
a-LW}(Y^{\prime},E^{\prime})$.

  2. (ii)

$S^{\prime}\in\text{\rm a-MW}(F^{\prime},X^{\prime})\Longrightarrow
S\in\text{\rm a-LW}(X,F)$.

In general, $S\in\text{\rm a-LW}(X,F)\not\Longrightarrow
S^{\prime}\in\text{\rm a-MW}(F^{\prime},X^{\prime})$ (see [13, Rem.2.1]).
Since w-convergent sequences in $Y^{\prime}$ are bounded, $\text{\rm
MW}(E,Y)\subseteq\text{\rm a-MW}(E,Y)$. The last inclusion may be proper: e.g.
the identity operator $I_{c_{0}}:c_{0}\to c_{0}$ is a-MW yet not w-compact,
and therefore $I_{c_{0}}$ is not MW. In general, $\text{\rm
K}(E,F)\not\subseteq\text{\rm a-LW}(E,F)\cup\text{\rm a-MW}(E,F)$, as the next
example shows.

######  Example 2.5.2.

Define an operator $T\in\text{\rm K}(\ell^{1},c)$ by
$T\alpha=(\sum\limits_{k=1}^{\infty}\alpha_{k})\cdot{\mathbb{1}}_{\mathbb{N}}$.
As $\\{Te_{1}\\}=\\{{\mathbb{1}}_{\mathbb{N}}\\}$ is not an LW-subset of $c$,
$T\not\in\text{\rm a-LW}(\ell^{1},c)$. Furthermore, $T\not\in\text{\rm
a-MW}(\ell^{1},c)$. Indeed, for the disjoint bounded sequence $(e_{n})$ of
unit vectors in $B_{\ell^{1}}$ and for any $f\in c^{\prime}$ with
$f({\mathbb{1}}_{\mathbb{N}})=1$, the constant sequence $(f_{n})$,
$f_{n}\equiv f$, is w-convergent, yet $f_{n}(Te_{n})\equiv 1\not\to 0$.

Since $f_{n}(Tx_{n})\to 0$ for every disjoint bounded $(x_{n})$ in $\ell^{1}$
and every w-null $(f_{n})$ in $c^{\prime}$, a w-convergent sequence $(f_{n})$
in Definition 2.4.1 a) can not be replaced by w-null $(f_{n})$.

The implication below is straightforward.

$T\in\text{\rm a-MW}(E,X)\ \&\ S\in\text{\rm L}(X,Y)\ \Longrightarrow\
ST\in\text{\rm a-MW}(E,Y).$ (8)

###  2.6 Algebraic properties of a-MW-operators.

######  Lemma 2.6.1.

For a pair of operators $T$ and $S$$:$

  1. i)

if $T\in\text{\rm a-MW}(E,F)$ and $S\in\text{\rm MW}(F,Y)$ then
$ST\in\text{\rm MW}(E,Y)$;

  2. ii)

if $T\in\text{\rm a-MW}(E,X)$ and $S\in\text{\rm L}(X,Y)$ then $ST\in\text{\rm
a-MW}(E,Y)$.

###### Proof.

i) Let $T\in\text{\rm a-MW}(E,F)$ and $S\in\text{\rm MW}(F,Y)$. Then
$T^{\prime}\in\text{\rm a-LW}(F^{\prime},E^{\prime})$ by Assertion 2.5.1(i),
and $S^{\prime}\in\text{\rm LW}(Y^{\prime},F^{\prime})$ by Assertion 1.3.5(i).
It follows from Lemma 2.2.1(i), that
$(ST)^{\prime}=T^{\prime}S^{\prime}\in\text{\rm LW}(Y^{\prime},E^{\prime})$.
Hence $ST\in\text{\rm MW}(E,Y)$ by Assertion 1.3.5(i).
ii) Follows from (8). ∎

######  Proposition 2.6.2.

For any Banach lattice $E$ the following holds.

  1. $\text{\rm MW}(E)$ is a closed subalgebra of $\text{\rm W}(E)$ and is unital iff the identity operator $I_{E}$ is MW.

  2. $\text{\rm a-MW}(E)$ is a closed left ideal in $\text{\rm L}(E)$ $($and hence a subalgebra of $\text{\rm L}(E)$$)$, and is unital iff $I_{E}$ is a-MW.

###### Proof.

Both $\text{\rm MW}(E)$ and $\text{\rm a-MW}(E)$ are closed subspaces of
$\text{\rm W}(E)$ by Proposition 2.2.2 and Assertion 1.3.5. The closeness of
$\text{\rm MW}(E)$ under the composition follows from Lemma 2.6.1(i)(ii). The
fact that $\text{\rm a-MW}(E)$ is a right ideal in $\text{\rm L}(E)$ follows
from Lemma 2.6.1(ii). Conditions on $I_{E}$ making algebras $\text{\rm MW}(E)$
and $\text{\rm a-MW}(E)$ to be unital, are trivial. ∎

The following fact was established in [2, Thm.2].

######  Assertion 2.6.3.

For all Banach lattices $E$ and $F$ the a-MW-operators from $E$ to $F$ satisfy
the domination property.

In view of Assertion 2.6.3, the next lemma follows from Proposition 1.6.2 ii).

######  Lemma 2.6.4.

Let an operator $T:E\to F$ possess the modulus. TFAE.

  1. i)

$T$ is a regularly a-MW-operator.

  2. ii)

$|T|$ is an a-MW-operator.

######  Theorem 2.6.5.

The following statements hold.

  1. i)

$\text{\rm r-a-MW}(E)$ is a subalgebra of $\text{\rm L}_{r}(E)$. Moreover,

$\text{\rm r-a-MW}(E)=\text{\rm L}_{r}(E)\ \Longleftrightarrow
I_{E}\in\text{\rm a-MW}(E).$

  2. ii)

If $E$ is Dedekind complete then $(\text{\rm r-a-MW}(E),\ \|\cdot\|_{\text{\rm
r-a-MW}})$ is a closed Riesz subalgebra and an order ideal of $(\text{\rm
L}_{r}(E),\ \|\cdot\|_{r})$.

###### Proof.

i) It follows from Lemma 2.6.1(ii) that $\text{\rm r-a-MW}(E)$ is a left
algebra ideal and hence is a subalgebra of $\text{\rm L}_{r}(E)$. The
condition on $I_{E}$ under which $\text{\rm r-a-MW}(E)=\text{\rm L}_{r}(E)$ is
trivial.

ii) It follows from Lemma 2.6.4 that $\text{\rm r-a-MW}(E)$ is closed under
lattice operations, and hence is a Riesz subalgebra of $\text{\rm L}_{r}(E)$.

The rest of the proof is similar to the proof of Theorem 2.2.5. ∎

### 2.7

In certain cases a-LW\- or a-MW-operators may coincide with bounded operators.

######  Proposition 2.7.1.

For any Banach space $X$ the following hold.

  1. i)

If $F$ is an AL-space then $\text{\rm a-LW}(X,F)=\text{\rm L}(X,F)$.

  2. ii)

If $E$ is an AM-space then $\text{\rm a-MW}(E,Y)=\text{\rm L}(E,Y)$.

###### Proof.

i) Let $S:X\to F$ be a bounded operator and let $C$ be a relatively w-compact
subset of $X$. Then $S(C)$ is relatively w-compact subset of $F$. By [3,
Thm.5.56], $S(C)$ is an LW-subset of $F$, and hence $S\in\text{\rm
a-LW}(X,F)$.

ii) Let $T\in\text{\rm L}(E,Y)$. Since $E^{\prime}$ is an AL-space, it follows
from i) that $T^{\prime}\in\text{\rm a-LW}(Y^{\prime},E^{\prime})$. Then
$T\in\text{\rm a-MW}(E,Y)$ by Assertion 2.5.1(i). ∎

### 2.8

The next result describes some restrictions on $E$ and $F$ under those
$\text{\rm a-MW}_{+}(E,F)\subseteq\text{\rm MW}(E,F)$. Recall that a Banach
space $X$ has the Dunford–Pettis property whenever $f_{n}(x_{n})\to 0$ for
each w-null $(x_{n})$ in $X$ and each w-null $(f_{n})$ in $X^{\prime}$ (cf.
[3, p.341]). An operator $T:X\to Y$ is called a weak Dunford–Pettis operator
whenever $f_{n}(Tx_{n})\to f(x)$ for each
$f_{n}\stackrel{{\scriptstyle\text{\rm w}}}{{\to}}f$ in $Y^{\prime}$ and each
$x_{n}\stackrel{{\scriptstyle\text{\rm w}}}{{\to}}x$ in $X$ [3, p.349].

######  Proposition 2.8.1.

Let the norm in $E^{\prime}$ be o-continuous, $F$ be Dedekind complete, and
$\text{\rm a-MW}_{+}(E,F)\subseteq\text{\rm MW}(E,F)$. Then either
$E\in\text{\rm(PSP)}$ or $F$ has o-continuous norm.

###### Proof.

Suppose $E\not\in\text{\rm(PSP)}$ and $F$ has no o-continuous norm. It
suffices to construct a positive a-MW-operator from $E$ to $F$ that is not MW.

By Assertion 1.7.7, there exists a disjoint w-null sequence $(x_{n})$ in
$E_{+}$, that is not norm null. WLOG, $\|x_{n}\|=1$ for all $n$. There exists
a sequence $(f_{n})$ in $E^{\prime}_{+}$ with $\|f_{n}\|=1$, such that

$|f_{n}(x_{n})|\geq\frac{1}{2}\ \ \ \ (\forall n\in{\mathbb{N}}).$ (9)

Define a positive operator $T:E\to\ell^{\infty}$ by

$Tx:=[(f_{n}(x))_{n}]\ \ \ \ (\forall x\in E).$ (10)

Since $F$ is Dedekind complete and the norm in $F$ is not order continuous,
there exists a lattice embedding $S:\ell^{\infty}\to F$ (cf. [3, Thm.4.51]).
Then, for some $M>0$:

$\|Sa\|\geq M\|a\|_{\infty}\ \ \ (\forall a\in\ell^{\infty}).$ (11)

Since $\ell^{\infty}$ has the Dunford–Pettis property (cf. [3, Thm.5.85]), $S$
is a weak Dunford–Pettis operator [3, p.349], that is, for each
$g_{n}\stackrel{{\scriptstyle\text{\rm w}}}{{\to}}g$ in $F^{\prime}$ and each
w-null $(y_{n})$ in $\ell^{\infty}$:

$g_{n}(Sy_{n})\to 0.$ (12)

Let $(z_{n})$ be disjoint bounded sequence in $E$ and let
$g_{n}\stackrel{{\scriptstyle\text{\rm w}}}{{\to}}g$ in $F^{\prime}$. Since
$E^{\prime}$ has o-continuous norm, $(z_{n})$ is w-null in $E$ by Assertion
1.7.1. Then $(Tz_{n})$ is w-null in $\ell^{\infty}$. It follows from (11) that
$g_{n}(STz_{n})\to 0$. Therefore, $ST\in\text{\rm a-MW}(E,F)$.

By using of (9), (10), and (11), we obtain

$\|STx_{n}\|=\|S[(f_{k}(x_{n}))_{k}]\|\geq
M\|(f_{k}(x_{n}))_{k}\|_{\infty}\geq M|f_{n}(x_{n})|\geq\frac{M}{2}>0.$ (13)

It follows from (13) that $ST\not\in\text{\rm MW}(E,F)$, and we are done. ∎

The following result is an LW-version of Proposition 2.8.1.

######  Proposition 2.8.2.

If $\text{\rm a-LW}_{+}(E,F)\subseteq\text{\rm LW}(E,F)$, then either
$E^{\prime}$ has o-continuous norm or $F^{\prime}\in\text{\rm(PSP)}$.

###### Proof.

Suppose, in contrary, that the norm in $E^{\prime}$ is not o-continuous and
$F^{\prime}\not\in\text{\rm(PSP)}$. Then we can find a disjoint sequence
$(x^{\prime}_{n})$ in $E^{\prime}$ and $x^{\prime}\in E^{\prime}$ such that
$0\leq x^{\prime}_{n}\leq x^{\prime}$ and $\|x^{\prime}_{n}\|=1$ for all
$n\in{\mathbb{N}}$. By Assertion 1.7.7(iii), there is a disjoint w-null
sequence $(f_{n})$ in $F^{\prime}_{+}$ that is not norm null. By passing to a
subsequence and by scaling, we may assume that $\|f_{n}\|\geq 1$ for all
$n\in{\mathbb{N}}$. Choose a sequence $(y_{n})$ in $F_{+}$ with $\|y_{n}\|=1$
and $|f_{n}(y_{n})|\geq\frac{1}{2}$ for all $n\in{\mathbb{N}}$. We define
positive operators $T:E\to\ell^{1}$ and $S:\ell^{1}\to F$ by

$Tx:=(x^{\prime}_{k}(x))_{k}\ \ \&\ \ \ Sa:=\sum_{k=1}^{\infty}a_{k}y_{k}.$

Let $(z_{n})$ be a w-null sequence in $E$ and $(f_{n})$ be a disjoint sequence
in $B_{F^{\prime}}$. As $\ell^{1}$ has the Schur property, $\|Tz_{n}\|\to 0$,
and hence

$|f_{n}(STz_{n})|\leq\|f_{n}\|\cdot\|S\|\cdot\|Tz_{n}\|\to 0.$ (14)

By [22, Prop.1], (14) implies $ST\in\text{\rm a-LW}_{+}(E,F)$.

If $ST$ were to be an LW-operator, then $(ST)^{\prime}$ would be an MW-
operator. However,

$\|(ST)^{\prime}(f_{n})\|=\|T^{\prime}S^{\prime}(f_{n})\|=\|\sum_{k=1}^{\infty}f_{n}(y_{k})\|\geq|f_{n}(y_{n})|\geq\frac{1}{2}$

for each $n$, which shows $(ST)^{\prime}\notin\text{\rm
MW}(F^{\prime},E^{\prime})$ and hence $ST\notin\text{\rm LW}(E,F)$. This
contradiction completes the proof. ∎

######  Proposition 2.8.3.

Suppose the norm in $E^{\prime}$ is o-continuous, $F$ is Dedekind complete,
and $F^{\prime}$ has the Schur property. Then, for every order bounded $T:E\to
F$, $|T|$ is a-MW and $|T|^{\prime}$ is a-LW.

###### Proof.

Let $(x_{n})$ be a disjoint bounded sequence in $E$ and
$f_{n}\stackrel{{\scriptstyle\rm w}}{{\to}}f$ in $F^{\prime}$. By the Schur
property of $F^{\prime}$, we have $\|f_{n}-f\|\to 0$. Hence

$|(f_{n}-f)(|T|x_{n})|\leq\|f_{n}-f\|\cdot\||T|x_{n}\|\to 0.$ (15)

Since $E^{\prime}$ has o-continuous norm, $(x_{n})$ is w-null by Assertion
1.7.1 and hence $f(|T|x_{n})=|T|^{\prime}(f)(x_{n})\to 0$. It follows from
(15) that $f_{n}(|T|x_{n})\to 0$, and hence $|T|$ is a-MW. By Assertion 2.5.1
(i), $|T|^{\prime}\in\text{\rm a-LW}(F^{\prime},E^{\prime})$. ∎

###  2.9 Basic properties of l-MW-operators.

######  Proposition 2.9.1.

Let $0\leq S\leq T\in\text{\rm l-MW}(E,F)$ then $S\in\text{\rm l-MW}(E,F)$.

###### Proof.

Let $(x_{n})$ be disjoint bounded sequence in $E$. Then $(|x_{n}|)$ is also
disjoint bounded, and then $T|x_{n}|\stackrel{{\scriptstyle\text{\rm
w}}}{{\to}}0$. It follows from $|Sx_{n}|\leq S|x_{n}|$ that

$|f(Sx_{n})|\leq|f|(S|x_{n}|)\leq|f|(T|x_{n}|)\to 0\ \ \ \ (\forall f\in
F^{\prime}),$

and hence $Sx_{n}\stackrel{{\scriptstyle\text{\rm w}}}{{\to}}0$. Thus,
$S\in\text{\rm l-MW}(E,F)$. ∎

######  Lemma 2.9.2.

Let $\text{\rm l-MW}(E,Y)\ni T_{n}\stackrel{{\scriptstyle\|\cdot\|}}{{\to}}T$.
Then $T\in\text{\rm l-MW}(E,Y)$.

###### Proof.

Let $(x_{n})$ be a disjoint sequence in $B_{E}$, and $f\in Y^{\prime}$. Take
$\varepsilon>0$, and pick an $m\in\mathbb{N}$ with
$\|T-T_{m}\|\leq\varepsilon$. Since $T_{m}\in\text{\rm l-MW}(E,Y)$ then
$|f(T_{m}x_{n})|\leq\varepsilon$ for $n\geq n_{0}$. It follows from

$|f(Tx_{n})|\leq|f(Tx_{n})-f(T_{m}x_{n})|+|f(T_{m}x_{n})|\leq$
$\|T^{\prime}-T^{\prime}_{m}\|\|f\|\|x_{n}\|+|f(T_{m}x_{n})|\leq(\|f\|+1)\varepsilon\
\ \ \ (\forall n\geq n_{0}),$

that $f(Tx_{n})\to 0$, and hence $T\in\text{\rm l-MW}(E,Y)$. ∎

In view of Proposition 2.9.1, the next lemma follows from Proposition 1.6.2
ii).

######  Lemma 2.9.3.

Let an operator $T:E\to F$ possess the modulus. TFAE.

  1. i)

$T\in\text{\rm r-l-MW}(E,F)$.

  2. ii)

$|T|\in\text{\rm l-MW}(E,F)$.

######  Theorem 2.9.4.

The following statements hold.

  1. i)

$\text{\rm r-l-MW}(E)$ is a subalgebra of $\text{\rm L}_{r}(E)$. Moreover,

$\text{\rm r-l-MW}(E)=\text{\rm L}_{r}(E)\ \Longleftrightarrow
I_{E}\in\text{\rm l-MW}(E).$

  2. ii)

If $E$ is Dedekind complete then $(\text{\rm r-l-MW}(E),\ \|\cdot\|_{\text{\rm
r-l-MW}})$ is a closed Riesz subalgebra and an order ideal of $(\text{\rm
L}_{r}(E),\ \|\cdot\|_{r})$.

###### Proof.

i) Since each bounded operator is w-continuous, $\text{\rm r-l-MW}(E)$ is a
left ideal and hence is a subalgebra of $\text{\rm L}_{r}(E)$. The condition
on $I_{E}$ under that $\text{\rm r-l-MW}(E)=\text{\rm L}_{r}(E)$ is trivial.

ii) By Lemma 2.9.3, $\text{\rm r-l-MW}(E)$ is closed under lattice operations,
and hence is a Riesz subalgebra of $\text{\rm L}_{r}(E)$. It follows from
Assertion 2.2.4 and Lemma l-MW-closed that $(\text{\rm r-l-LW}(E),\
\|\cdot\|_{\text{\rm r-l-LW}})$ is a Banach space. The rest of the proof is
similar to the proof of Theorem 2.2.5. ∎

### 2.10

We include also the following semi-duality result. Recall that $X\in(\text{\rm
GP})$ if each w∗-null sequence in $X^{\prime}$ is w-null.

######  Proposition 2.10.1.

The following hold.

  1. i)

$T^{\prime}\in\text{\rm l-LW}(Y^{\prime},E^{\prime})\Longrightarrow
T\in\text{\rm l-MW}(E,Y)$, and if additionally $E^{\prime}\in(\text{\rm
DDw${}^{\ast}$P})$ then $T^{\prime}\in\text{\rm
l-LW}(Y^{\prime},E^{\prime})\Longleftrightarrow T\in\text{\rm l-MW}(E,Y)$.

  2. ii)

$S^{\prime}\in\text{\rm l-MW}(F^{\prime},X^{\prime})\Longrightarrow
S\in\text{\rm l-LW}(X,F)$, and if additionally $X\in(\text{\rm GP})$ then
$S^{\prime}\in\text{\rm l-MW}(F^{\prime},X^{\prime})\Longleftrightarrow
S\in\text{\rm l-LW}(X,F)$.

###### Proof.

i) Let $T^{\prime}\in\text{\rm l-LW}(Y^{\prime},E^{\prime})$. Take a disjoint
bounded sequence $(x_{n})$ in $E$. Then $(\hat{x}_{n})$ is disjoint bounded in
$E^{\prime\prime}$. By Lemma 2.3.1,
$T^{\prime\prime}\hat{x}_{n}\stackrel{{\scriptstyle\text{\rm
w}^{\ast}}}{{\to}}0$ in $Y^{\prime\prime}$, and hence

$y(Tx_{n})=(T^{\prime}y)x_{n}=\hat{x}_{n}(T^{\prime}y)=T^{\prime\prime}\hat{x}_{n}(y)\to
0\ \ \ \ (\forall y\in Y^{\prime}).$

Thus $Tx_{n}\stackrel{{\scriptstyle\text{\rm w}}}{{\to}}0$, and hence
$T\in\text{\rm l-MW}(E,Y)$.

Now, let $T\in\text{\rm l-MW}(E,Y)$ and $E^{\prime}\in(\text{\rm
DDw${}^{\ast}$P})$. Take a disjoint bounded sequence $(f_{n})$ in
$E^{\prime\prime}$. Since $E^{\prime}\in(\text{\rm DDw${}^{\ast}$P})$ then
$f_{n}\stackrel{{\scriptstyle\text{\rm w}^{\ast}}}{{\to}}0$ in
$E^{\prime\prime}$, and hence
$T^{\prime\prime}f_{n}\stackrel{{\scriptstyle\text{\rm w}^{\ast}}}{{\to}}0$ in
$Y^{\prime\prime}$. Lemma 2.3.1 implies $T^{\prime}\in\text{\rm
l-LW}(Y^{\prime},E^{\prime})$.

ii) Let $S^{\prime}\in\text{\rm l-MW}(F^{\prime},X^{\prime})$. Take a disjoint
bounded sequence $(f_{n})$ in $F^{\prime}$. Then
$S^{\prime}f_{n}\stackrel{{\scriptstyle\text{\rm w}}}{{\to}}0$ in
$X^{\prime}$, and hence $S^{\prime}f_{n}\stackrel{{\scriptstyle\text{\rm
w}^{\ast}}}{{\to}}0$ in $X^{\prime}$. It follows from Lemma 2.3.1 that
$S\in\text{\rm l-LW}(X,F)$.

Now, let $S\in\text{\rm l-LW}(X,F)$ and $X\in(\text{\rm GP})$. Take a disjoint
bounded sequence $(f_{n})$ in $F^{\prime}$. By Lemma 2.3.1,
$S^{\prime}f_{n}\stackrel{{\scriptstyle\text{\rm w}^{\ast}}}{{\to}}0$ in
$X^{\prime}$. Since $X\in(\text{\rm GP})$ then
$S^{\prime}f_{n}\stackrel{{\scriptstyle\text{\rm w}}}{{\to}}0$, and hence
$S^{\prime}\in\text{\rm l-MW}(F^{\prime},X^{\prime})$. ∎

##  3 Order LW\- (MW-) operators

Order LW\- (MW-) operators were introduced recently in [27, 28]. Here we
continue to investigate these classes of operators. The idea of definition of
an order LW-operator $T:E\to F$ lies in choosing in Assertion 1.3.1,

$\text{$A:=T[-x,x]$ for arbitrary $x\in E_{+}$; and $B:=B_{F^{\prime}}$}.$

###  3.1 Main definitions.

######  Definition 3.1.1.

A bounded operator $T:E\to F$ is called:

  1. a)

order LW (shortly, $T$ is o-LW) if $T[0,x]$ is an L-weakly compact subset of
$F$ for every $x\in E_{+}$ [28, Def.2.1];

  2. b)

order MW (shortly, $T$ is o-MW) if, for every disjoint bounded sequence
$(x_{n})$ in $E$ and every order bounded sequence $(f_{n})$ in $F^{\prime}$,
we have $f_{n}(Tx_{n})\to 0$ (see [27, Def.2.2] and [28, Def.2.2]).

Clearly,

$\text{\rm W}(E,Y)\subseteq\text{\rm o-W}(E,Y);$ $\text{\rm
LW}(E,F)\subseteq\text{\rm o-LW}(E,F);$ $\text{\rm MW}(E,F)\subseteq\text{\rm
o-MW}(E,F).$

Since the operators $\text{I}_{c_{0}}$ and $\text{I}_{c}$ are not w-compact,
one can directly derive from Definition 3.1.1 that

$\text{I}_{c_{0}}\in\text{\rm o-LW}(c_{0})\setminus\text{\rm LW}(c_{0})\ \
\text{\rm and}\ \ \text{I}_{c}\in\text{\rm o-MW}(c)\setminus\text{\rm MW}(c).$

The following two assertions (see [28, Thm.2.1 and Thm.2.2]) give important
characterizations of o-LW-operators.

######  Assertion 3.1.2.

For an operator $T:E\to F$, TFAE.

  1. (i)

$T\in\text{\rm o-LW}(E,F)$.

  2. (ii)

$(|T^{\prime}f_{n}|)$ is $\text{\rm w}^{\ast}$-null for every disjoint bounded
$(f_{n})$ in $F^{\prime}$.

  3. (iii)

$f_{n}(Tx_{n})\to 0$ for every order bounded $(x_{n})$ in $E$ and every
disjoint bounded $(f_{n})$ in $F^{\prime}$.

######  Assertion 3.1.3.

$S\in\text{\rm o-LW}(E^{\prime},F^{\prime})$ iff

$(Sf_{n})(y_{n})=y_{n}(Sf_{n})=(S^{\prime}y_{n})(f_{n})\to 0$

for every order bounded $(f_{n})$ in $E^{\prime}$ and every disjoint bounded
$(y_{n})$ in $F$.

The next semi-duality result [28, Thm.2.3] plays a key role in the
investigation of o-LW\- and o-MW-operators.

######  Assertion 3.1.4.

The following hold.

  1. (i)

$T\in\text{\rm o-MW}(E,F)\Longleftrightarrow T^{\prime}\in\text{\rm
o-LW}(F^{\prime},E^{\prime})$.

  2. (ii)

$S^{\prime}\in\text{\rm o-MW}(F^{\prime},E^{\prime})\Longrightarrow
S\in\text{\rm o-LW}(E,F)$.

It was also observed in [28, Rem.2.3] that, in general,

$S\in\text{\rm o-LW}(E,F)\not\Longrightarrow S^{\prime}\in\text{\rm
o-MW}(F^{\prime},E^{\prime}),$

as $I_{c_{0}}\in\text{\rm o-LW}(c_{0})$, but
$(I_{c_{0}})^{\prime}=I_{\ell^{1}}\not\in\text{\rm o-MW}(\ell^{1})$.

### 3.2

A rank one bounded operator need not to be o-LW. Consider the operator
$T:\ell^{2}\to\ell^{\infty}$ defined by
$Ta=(\sum\limits_{k=1}^{\infty}a_{k}){\mathbb{1}}_{\mathbb{N}}$. Then
$Te_{n}={\mathbb{1}}_{\mathbb{N}}$ and, as
${\mathbb{1}}_{\mathbb{N}}\not\in(\ell^{\infty})^{a}$, the operator $T$ is not
o-LW. The following result is an improvement of [28, Thm.2.5]:

######  Proposition 3.2.1.

For a Banach lattice $E$, TFAE.

  1. i)

The norm in $E^{\prime}$ is o-continuous.

  2. ii)

For every $F$, ${\cal L}_{ob}(E,F)\subseteq\text{\rm o-MW}(E,F)$.

  3. iii)

For every $F$, $\text{\rm L}_{r}(E,F)\subseteq\text{\rm o-MW}(E,F)$.

  4. iv)

For every $F$, $\text{\rm K}(E,F)\subseteq\text{\rm o-MW}(E,F)$.

  5. v)

Each rank one bounded operator $T:E\to E$ is o-MW.

###### Proof.

i) $\Longleftrightarrow$ ii) is in [28, Thm.2.5] and the implication ii)
$\Longrightarrow$ iii) is trivial.

iii) $\Longrightarrow$ i). If not, there is a disjoint bounded sequence
$(x_{n})$ in $E$ and a functional $f\in E^{\prime}$ such that $f(x_{n})\geq 1$
for all $n\in\mathbb{N}$ by Assertion 1.7.1. The identity operator $T=I:E\to
E$ is o-MW by $iii)$. However, for the order bounded sequence $(f_{n})$ in
$F^{\prime}$ with $f_{n}\equiv f$, it holds $f_{n}(Tx_{n})=f(x_{n})\geq 1$,
and hence $f_{n}(Tx_{n})\not\to 0$, violating Assertion 3.1.3. The obtained
contradiction proves the implication.

i) $\Longrightarrow$ iv). Let $T\in\text{\rm K}(E,F)$ and let $x\in
F^{\prime}_{+}$. Since $T^{\prime}\in\text{\rm K}(F^{\prime},E^{\prime})$,
then $T^{\prime}[0,x]$ is relatively compact and hence is almost order
bounded. As the norm in $E^{\prime}$ is o-continuous, $T^{\prime}[0,x]$ is an
LW-subset of $E^{\prime}$ by Assertion 1.2.2. This yields
$T^{\prime}\in\text{\rm o-LW}(F^{\prime},E^{\prime})$, and hence
$T\in\text{\rm o-MW}(E,F)$, by Assertion 3.1.4(i).

iv) $\Longrightarrow$ v). It is obvious.

v) $\Longrightarrow$ i). If the norm in $E^{\prime}$ is not o-continuous, take
$g\in E^{\prime}_{+}\setminus(E^{\prime})^{a}$. Let $T$ be a bounded rank one
operator defined by $T(x)=g(x)y_{0}$ with $y_{0}\in E_{+}\setminus\\{0\\}$.
Then $T\in\text{\rm o-MW}(E)$. Choose $f\in E^{\prime}$ with $f(y_{0})=1$.
Since $T^{\prime}f=f(y_{0})g=g\not\in(E^{\prime})^{a}$, $T^{\prime}[0,f]$ is
not an LW-subset of $E^{\prime}$ as each LW-subset of $E^{\prime}$ must lie in
$(E^{\prime})^{a}$. Therefore $T^{\prime}\not\in\text{\rm o-LW}(E^{\prime})$.
This yields $T\not\in\text{\rm o-MW}(E)$ by Assertion 3.1.4(i). The obtained
contradiction completes the proof. ∎

######  Corollary 3.2.2.

There is a Banach lattice $E$ with $\text{\rm L}_{r}(E)\subsetneqq\text{\rm
o-MW}(E)$.

###### Proof.

Let $E$ be as in Example 1.6.4. It follows from Proposition 3.2.1 that
$\text{\rm L}_{r}(E)\subseteq\text{\rm o-MW}(E)$. Consider the operator $T$
defined by (6) in Example 1.6.4 for $\alpha_{n}=2^{-\frac{n}{3}}$. As
$T\in\text{\rm K}(E)$, applying Proposition 3.2.1 again gives $T\in\text{\rm
o-MW}(E)$. However $|T|$ does not exist and hence $T\not\in\text{\rm
L}_{r}(E)$. ∎

######  Corollary 3.2.3.

Let $E$ and $F$ be Banach lattices with $\dim(E)\geq 1$ and $\dim(F)\geq 1$
then$:$

  1. i)

$E^{\prime}$ has o-continuous norm $\Longleftrightarrow$ every rank one
bounded operator $T:E\to F$ is o-MW;

  2. ii)

$F$ has o-continuous norm $\Longleftrightarrow$ every rank one bounded
operator $T:E\to F$ is o-LW.

###### Proof.

i) It is a part of Proposition 3.2.1.

ii) The implication ($\Longrightarrow$) is obvious.

ii) ($\Longleftarrow$) If not, there exist $y\in F_{+}$ and a disjoint
sequence $(y_{n})$ in $[0,y]$, that is not norm null. Choose $x\in E_{+}$,
$f\in E^{\prime}_{+}$ with $f(x)=1$ and define a rank one operator $T$ as
$Tz=f(z)y$. Then $Tx=y$ and $(y_{n})$ is a disjoint sequence in
$\text{sol}\,(T[0,x])$, and it is not norm null, violating that $T$ must be
o-LW. A contradiction. ∎

### 3.3

It follows directly from the definitions:

$\text{\rm LW}(E,F)\subseteq\text{\rm o-LW}(E,F)\cap\text{\rm a-LW}(E,F);$
$\text{\rm MW}(E,F)\subseteq\text{o-MW}(E,F)\cap\text{a-MW}(E,F).$

If $F$ has o-continuous norm, then ${\cal L}_{ob}(E,F)\subseteq\text{\rm
o-LW}(E,F)$, and if $E^{\prime}$ has o-continuous norm, then ${\cal
L}_{ob}(E,F)\subseteq\text{\rm o-MW}(E,F)$ by [28, Thm.2.5]. Clearly,

$\text{I}_{c_{0}}\in\text{\rm o-LW}(c_{0})\setminus\text{\rm a-LW}(c_{0})\
\text{\rm and}\ \text{I}_{\ell^{1}}\in\text{\rm
a-LW}(\ell^{1})\setminus\text{\rm a-MW}(\ell^{1}).$

It can be easily shown that $\text{I}_{\ell^{2}}\in\text{\rm
o-MW}(\ell^{2})\setminus\text{\rm a-MW}(\ell^{2})$ (cf. Remark 2.4.2). By
Proposition 2.7.1, for each $X$: $\text{\rm o-LW}(X,F)=\text{\rm
a-LW}(X,F)=\text{\rm L}(X,F)$ if $F$ is an AL-space; and $\text{\rm
o-MW}(E,X)=\text{\rm a-MW}(E,X)=\text{\rm L}(E,X)$ if $E$ is an AM-space. In
general, $\text{\rm a-MW}(E,Y)\not\subseteq\text{\rm o-MW}(E,Y)$. Indeed,

$\text{I}_{C[0,1]}\in\text{\rm a-MW}(C[0,1])\ \text{\rm(e.g., by
Proposition~{}\ref{prop 13}~{}ii))}.$ (16)

However,

$\text{I}_{C[0,1]}\not\in\text{\rm o-MW}(C[0,1])\ \text{\rm(e.g., by
Proposition~{}\ref{T is o-MW} below)}.$ (17)

It follows directly that $\text{\rm a-LW}(\ell^{2})\subseteq\text{\rm
o-LW}(\ell^{2})$. Then $\text{\rm a-MW}(\ell^{2})\subseteq\text{\rm
o-MW}(\ell^{2})$ by the duality. Since $\text{I}_{\ell^{2}}\in\text{\rm
o-MW}(\ell^{2})\setminus\text{\rm a-MW}(\ell^{2})$ then $\text{\rm
a-MW}(\ell^{2})\subsetneqq\text{\rm o-MW}(\ell^{2})$.

###  3.4 Algebraic properties of o-LW\- and o-MW\- operators.

The first of the following two elementary lemmas is a version of Lemma 2.2.1
for o-LW-operators.

######  Lemma 3.4.1.

If $S\in{\cal L}_{ob}(E,F)$ and $T\in\text{\rm o-LW}(F,G)$ then
$TS\in\text{\rm o-LW}(E,G)$.

Since o-LW-operators satisfy the domination property by [28, Cor.2.3], the
second lemma follows from Proposition 1.6.2 ii).

######  Lemma 3.4.2.

Let an operator $T\in{\cal L}(E,F)$ possess the modulus. TFAE.

  1. i)

$T$ is a regularly o-LW-operator.

  2. ii)

$|T|$ is an o-LW-operator.

We include a version of Theorem 2.2.5 for o-LW-operators.

######  Theorem 3.4.3.

The following statements hold.

  1. i)

$\text{\rm r-o-LW}(E)$ is a subalgebra of $\text{\rm L}_{r}(E)$. Moreover,

$\text{\rm r-o-LW}(E)=\text{\rm L}_{r}(E)\ \Longleftrightarrow
I_{E}\in\text{\rm o-LW}(E).$

  2. ii)

If $E$ is Dedekind complete then $(\text{\rm r-o-LW}(E),\ \|\cdot\|_{\text{\rm
r-o-LW}})$ is a closed Riesz subalgebra and an order ideal of $(\text{\rm
L}_{r}(E),\ \|\cdot\|_{r})$.

###### Proof.

i) Due to Lemma 3.4.1, $\text{\rm r-o-LW}(E)$ is a right algebra ideal and
hence is a subalgebra of $\text{\rm L}_{r}(E)$. The condition on $I_{E}$, that
makes $\text{\rm r-o-LW}(E)=\text{\rm L}_{r}(E)$, is clear.

ii) It follows from Lemma 3.4.2 that the subalgebra $\text{\rm r-o-LW}(E)$ is
closed under modulus, and hence it is Riesz subalgebra of $\text{\rm
L}_{r}(E)$.

As $\text{\rm o-LW}(E)$ is a closed subspace of $\text{\rm L}(E)$ by [28,
Prop.2.1], the rest of the proof is similar to the proof of Theorem 2.2.5. ∎

The following lemma is a version of Lemma 2.6.1 for o-MW-operators.

######  Lemma 3.4.4.

If $T\in\text{\rm o-MW}(E,F)$ and $S\in{\cal L}_{ob}(F,G)$, then
$ST\in\text{\rm o-MW}(E,G)$.

###### Proof.

Let $T\in\text{\rm o-MW}(E,F)$ and $S\in{\cal L}_{ob}(F,G)$. Then
$T^{\prime}\in\text{\rm o-LW}(F^{\prime},E^{\prime})$ by Assertion 3.1.4 (i).
It follows from Lemma 3.4.1 that
$(ST)^{\prime}=T^{\prime}S^{\prime}\in\text{\rm o-LW}(G^{\prime},E^{\prime})$.
Applying Assertion 3.1.4 (i) again, we get $ST\in\text{\rm o-MW}(E,G)$. ∎

It is straightforward that o-MW-operators satisfy the domination property.
Thus, the next lemma follows from Proposition 1.6.2 ii).

######  Lemma 3.4.5.

Let an operator $T\in{\cal L}(E,F)$ possess the modulus. TFAE.

  1. i)

$T$ is a regularly o-MW-operator.

  2. ii)

$|T|$ is an o-MW-operator.

The next result is a version of Proposition 2.6.2 for o-MW-operators. We omit
its proof as it is similar to the proof of Theorem 3.4.3, with replacement of
Lemma 3.4.2 by Lemma 3.4.5.

######  Theorem 3.4.6.

The following statements hold.

  1. i)

$\text{\rm r-o-MW}(E)$ is a subalgebra of $\text{\rm L}_{r}(E)$. Moreover,

$\text{\rm r-o-MW}(E)=\text{\rm L}_{r}(E)\ \Longleftrightarrow
I_{E}\in\text{\rm o-MW}(E).$

  2. ii)

If $E$ is Dedekind complete then $(\text{\rm r-o-MW}(E),\ \|\cdot\|_{\text{\rm
r-o-MW}})$ is a closed Riesz subalgebra and an order ideal of $(\text{\rm
L}_{r}(E),\ \|\cdot\|_{r})$.

So, in certain cases, $\text{\rm r-o-LW}(E)$ or $\text{\rm r-o-MW}(E)$ may
coincide with $\text{\rm L}_{r}(E)$. The following proposition develops this
observation further.

######  Proposition 3.4.7.

The following statements hold.

  1. i)

If $F$ is an AL-space then $\text{\rm r-o-LW}(E,F)=\text{\rm L}_{r}(E,F)$.

  2. ii)

If $E$ is an AM-space then $\text{\rm r-o-MW}(E,F)=\text{\rm L}_{r}(E,F)$.

###### Proof.

i) Let $S=S_{1}-S_{2}$ with both $S_{1},S_{2}:E\to F$ positive, and let $x\in
X_{+}$. Then $S_{1}([-x,x])$ and $S_{2}([-x,x])$ are both relatively w-compact
in $F$. By [3, Thm.5.56], the sets $S_{1}([-x,x])$ and $S_{2}([-x,x])$ are
both LW-subsets of $F$. Then $S_{1}$ and $S_{2}$ are positive o-LW-operators
and hence $S$ is a regularly o-LW-operator.

ii) Let $T\in\text{\rm L}_{r}(E,F)$. Since $E^{\prime}$ is an AL-space, it
follows from i) that $T^{\prime}\in\text{\rm r-o-LW}(F^{\prime},E^{\prime})$.
Then $T\in\text{\rm r-o-MW}(E,F)$ by Assertion 3.1.4 (i). ∎

###  3.5 Miscellanea.

An operator $T:E\to X$ is said to be AM-compact, if $T[0,x]$ is relatively
compact in $X$ for each $x\in E_{+}$. An operator $T:E\to X$ is said to be
b-AM-compact, if it carries each b-order bounded subset of $E$ into a
relatively compact subset of $X$ [7]. Denote by $\text{\rm AM}(E,X)$ (by
$\text{\rm b-AM}(E,X)$) the set of all AM-compact (resp. b-AM-compact)
operators from $E$ to $X$. Clearly, $\text{\rm b-AM}(E,X)\subseteq\text{\rm
AM}(E,X)$.

######  Theorem 3.5.1.

Let $F^{a}$ be discrete. Then $\text{\rm o-LW}(E,F)\subseteq\text{\rm
AM}(E,F)$.

###### Proof.

Let $T\in\text{\rm o-LW}(E,F)$. Take $x\in E_{+}$. Since $T[0,x]$ is an LW-
subset of $F$, for every $\varepsilon>0$, there is $u_{\varepsilon}\in
F_{+}^{a}$ with

$T[0,x]\subseteq[-u_{\varepsilon},u_{\varepsilon}]+\varepsilon B_{F},$ (18)

by Assertion 1.2.2. Let $\varepsilon>0$. Since $F^{a}$ is discrete,
$[-u_{\varepsilon},u_{\varepsilon}]$ is compact. As $\varepsilon>0$ is
arbitrary, (18) implies that $T[0,x]$ is totally bounded and hence is
relatively compact. Since $x\in E_{+}$ is arbitrary, $T\in\text{\rm AM}(E,F)$.
∎

While LW-operators are w-compact, o-LW-operators are generally not, e.g.
$\text{I}_{c_{0}}\in\text{\rm o-LW}(c_{0})\setminus\text{W}(c_{0})$. The
following corollary describes some conditions for w-compactness of o-LW-
operators.

######  Corollary 3.5.2.

Let $F^{a}$ be discrete and let either $E^{\prime}$ be $KB$ or $F$ be
reflexive. Then $\text{\rm o-LW}(E,F)\subseteq\text{\rm W}(E,F)$.

###### Proof.

Let $T\in\text{\rm o-LW}(E,F)$. By Theorem 3.5.1, $T\in\text{\rm AM}(E,F)$. In
view of [10, Thm.2.4], $T\in\text{\rm W}(E,F)$. ∎

######  Proposition 3.5.3.

Let $T\in\text{\rm L}_{r}(E,F)$ and $x\in E_{+}$. Then the restriction of
$T:(I_{x},\|.\|_{x})\to F$ is o-MW.

###### Proof.

We may suppose $T\geq 0$. The principal order ideal $I_{x}$ of $E$ is a Banach
lattice equipped with the norm $\|y\|_{x}=\inf\\{\lambda:|y|\leq\lambda x\\}$.
Let $(x_{n})$ be a disjoint order bounded sequence in $I_{x}$. Then $x_{n}\to
0$ in $\sigma(I_{x},I^{\prime}_{x})$. Since the lattice operations are
sequentially w-continuous (cf. [35, Prop.13]), we have $|x_{n}|\to 0$ in
$\sigma(I_{x},I^{\prime}_{x})$. Since $T$ is positive, $T:(I_{x},\|.\|_{x})\to
F$ is continuous, and hence $T$ is also continuous when the both spaces are
equipped with their w-topologies. Let a sequence $(f_{n})$ be order bounded in
$F^{\prime}$, say $|f_{n}|\leq f\in F^{\prime}$. Then

$|f_{n}(Tx_{n})|\leq|f_{n}|(|Tx_{n}|)\leq|f_{n}|(T|x_{n}|)\leq f(T|x_{n}|),$

and $|f_{n}(Tx_{n})|$ can be made arbitrary small, if $n$ is large enough.
Therefore $T\in\text{\rm o-MW}(I_{x},F)$. ∎

######  Definition 3.5.4.

An operator $T:X\to F$ is called semicompact if it takes bounded subsets of
$X$ onto almost order bounded subsets of $F$.

It is straightforward that $\text{\rm K}(X,F)\subseteq\text{\rm semi-K}(X,F)$,
where $\text{\rm semi-K}(X,F)$ denotes the space of all semicompact operators
from $X$ to $F$. It is also well known that $\text{\rm
semi-K}(X,F)\subseteq\text{\rm W}(X,F)$ for every $X$, whenever the norm in
$F$ is o-continuous. In view of Assertion 1.7.7(v), if $F\in(\text{\rm PSP})$
then $\text{\rm semi-K}(X,F)=\text{\rm W}(X,F)$ for every $X$.

######  Proposition 3.5.5.

Let $S\in\text{\rm semi-K}(E,F)$ and $T\in\text{\rm o-LW}(F,G)$. Then
$TS\in\text{\rm LW}(E,G)$.

###### Proof.

WLOG $\|T\|\leq 1$. Let $\varepsilon>0$ be arbitrary. Pick $u_{\varepsilon}\in
F_{+}$ such that
$S(B_{E})\subseteq[-u_{\varepsilon},u_{\varepsilon}]+\varepsilon B_{F}$. Then

$TS(B_{E})\subseteq T[-u_{\varepsilon},u_{\varepsilon}]+\varepsilon
T(B_{F})\subseteq T[-u_{\varepsilon},u_{\varepsilon}]+\varepsilon B_{G}.$

Since $T\in\text{\rm o-LW}(F,G)$, the set
$T[-u_{\varepsilon},u_{\varepsilon}]$ is an LW-subset of $G$. Since
$\varepsilon>0$ is arbitrary, $TS(B_{E})$ is an LW-subset of $G$. ∎

Finally, we include the following result. By [31], a bounded operator $T:E\to
Y$ is almost order weakly compact (shortly, $T\in\text{\rm ao-W}(E,Y)$),
whenever $T$ maps almost order bounded subsets of $E$ onto relatively weakly
compact subsets of $Y$.

######  Proposition 3.5.6.

$\text{\rm ao-W}(E,Y)=\text{\rm o-W}(E,Y)$.

###### Proof.

Clearly, $\text{\rm ao-W}(E,Y)\subseteq\text{\rm o-W}(E,Y)$.

Let $T\in\text{\rm o-W}(E,Y)$, and let $A$ be an almost order bounded subset
of $E$. Take $\varepsilon>0$. There exists $x\in E_{+}$ with
$A\subseteq[-x,x]+\varepsilon B_{E}$. Then

$T(A)\subseteq
T[-x,x]+\varepsilon\|T\|B_{Y}\subseteq\text{cl}_{\|\cdot\|}(T[-x,x])+\varepsilon\|T\|B_{Y},$

with a w-compact subset $\text{cl}_{\|\cdot\|}(T[-x,x])$ of $Y$. Since
$\varepsilon>0$ is arbitrary, [3, Thm.3.44] implies that $T(A)$ is relatively
w-compact, and hence $T\in\text{\rm ao-W}(E,Y)$. ∎

## References

  * [1] Y.A. Abramovich, A.W. Wickstead, Solutions of several problems in the theory of compact positive operators. Proc. Amer. Math. Soc. 123 (1995), 3021–3026.
  * [2] B. Akay, and Ö. Gök, On the Lattice Properties of Almost L-Weakly and Almost M-Weakly Compact Operators. J. Funct. Spaces (2021), Art. ID 1755373, 5 pp.
  * [3] C.D. Aliprantis, O. Burkinshaw, Positive Operators. Springer, Dordrecht, 2006.
  * [4] S. Alpay, B. Altın, C. Tonyali, On Property (b) of Vector Lattices. Positivity 7 (2003), 135–139.
  * [5] S. Alpay, E. Emelyanov, S. Gorokhova, Bibounded uo-convergence and b-property in vector lattices. Positivity 25 (2021), 1677–1684.
  * [6] Alpay, S., Emelyanov, E., Gorokhova, S. $o\tau$-Continuous, Lebesgue, KB, and Levi Operators Between Vector Lattices and Topological Vector Spaces. Results Math 77, 117 (2022), 25pp.
  * [7] B. Aqzzouz, J. H’Michane, The class of b-AM-compact operators. Quaest. Math. 36 (2013), 309–319.
  * [8] B. Aqzzouz, A. Elbour, A.W. Wickstead, Positive almost Dunford–Pettis operators and their duality. Positivity 15 (2011), 185–197.
  * [9] B. Aqzzouz, R. Nouira, The order $\sigma$-complete vector lattice or AM-compact operators. Czechoslovak Math. J. 59(134) (2009), no.3, 827–834.
  * [10] B. Aqzzouz, J. H’Michane, O. Aboutafail, Weak compactness of AM-compact operators. Bull. Belg. Math. Soc. Simon Stevin 19 (2012), no. 2, 329–338.
  * [11] B. Aqzzouz, R. Nouira, L. Zraoula, The duality problem for the class of AM-compact operators on Banach lattices. Canad. Math. Bull. 51, no.1 (2008), 15–20.
  * [12] H. Baklouti, M. Hajji, R. Moulahi, On the class of b-weakly compact operators. Positivity 26, 7 (2022). https://doi.org/10.1007/s11117-022-00892-3
  * [13] K. Bouras, D. Lhaimer, M. Moussa, On the class of almost L-weakly and almost M-weakly compact operators. Positivity 22 (2018), 1433–1443.
  * [14] J. Bourgain, J. Diestel, Limited operators and strict cosingularity. Math. Nachr. 119 (1984), 55–58.
  * [15] O. Burkinshaw, P. G. Dodds, Disjoint sequences, compactness and semireflexivity in locally convex Riesz spaces, Illinois J. Math. 21 (1977), 759–75.
  * [16] Z. L. Chen, A. W. Wickstead, Incompleteness of the linear span of the positive compact operators. Proc. Amer. Math. Soc. 125 (1997), 3381–3389.
  * [17] Z. L. Chen, A. W. Wickstead, L-weakly and M-weakly compact operators. Indag. Math. (N.S.) 10, no.3 (1999), 321–336.
  * [18] N. Cheng, Dedekind $\sigma$-complete vector lattice of b-AM-compact operators. Quaest. Math. 40 (2017), no. 3, 313–318.
  * [19] P. G. Dodds, o-weakly compact mappings of Riesz spaces. Trans. Amer. Math. Soc. 214 (1975), 389–402.
  * [20] P. G. Dodds, D. H. Fremlin, Compact operators in Banach lattices. Isr. J. Math. 34 (1979), 287–320.
  * [21] A. Elbour, Some characterizations of almost limited operators. Positivity 21 (2017), 865–874.
  * [22] A. Elbour, F. Afkir, M. Sabiri, Some properties of almost L-weakly and almost M-weakly compact operators. Positivity 24 (2020), 141–149.
  * [23] E. Emelyanov, Algebras of Lebesgue and KB regular operators on Banach lattices. https://arxiv.org/abs/2203.08326v3
  * [24] E. Emelyanov, S. Gorokhova, On Operators Dominated by Kantorovich-Banach Operators and Levy Operators in Locally Solid Lattices. Vladikavkaz Math. J. 24, Issue 3 (2022), 55–61.
  * [25] U. Krengel, Remark on the modulus of compact operators. Bull. Amer. Math. Soc. 72 (1966), 132–133.
  * [26] U. Krengel, Über den Absolutbetrag stetiger linearer Operatoren und seine Anwendung auf ergodische Zerlegungen. Math. Scand. 13 (1963), 151–187.
  * [27] D. Lhaimer, K. Bouras, M. Moussa, On the class of b-L-weakly and order M-weakly compact operators. Math. Bohem. 145, no. 3 (2020), 255–264.
  * [28] D. Lhaimer, K. Bouras, M. Moussa, On the class of order $L$\- and order $M$-weakly compact operators. Positivity 25 (2021), 1569–1578.
  * [29] N. Machrafi, K. El Fahri, M. Moussa, B. Altin, A note on weak almost limited operators. Hacet. J. Math. Stat. 48(3) (2019) 759–770.
  * [30] N. Machrafi, A. Elbour, M. Moussa, Some characterizations of almost limited sets and applications. https://arxiv.org/abs/1312.2770v2
  * [31] M. Matin, K. H. Azar, A. Ebadi, Almost order-weakly compact operators on Banach lattices. https://arxiv.org/abs/2201.02219v1
  * [32] P. Meyer-Nieberg, Über Klassen schwach kompakter Operatoren in Banachverbänden. Math. Z. 138 (1974), 145–159.
  * [33] P. Meyer-Nieberg, Banach Lattices. Universitext, Springer-Verlag, Berlin (1991).
  * [34] W. Wnuk, Banach lattices with properties of the Schur type: a survey. Conf. Sem. Mat. Univ. Bari Bari. 249 (1993), 1–25.
  * [35] W. Wnuk, On the dual positive Schur property in Banach lattices, Positivity 17 (2013), 759–773.