# On limited and almost limited operators between Banach lattices

###### Abstract

We study (almost) limited operators in Banach lattices and their relations to
L-weakly compact, semi-compact, and Dunford-Pettis operators. Several further
related topics are investigated.

Safak Alpay1, Svetlana Gorokhova 2
$1$ Middle East Technical University, Turkey
$2$ Uznyj matematiceskij institut VNC RAN, Russia

Keywords: Banach lattice, limited set, L-weakly compact set, semi-compact
operator, Dunford-Pettis operator.

MSC2020: 46B25, 46B42, 46B50, 47B60

Various types of compactness in Banach spaces were in streamline of functional
analysis in the second half of the 20th century. At some point, in order to
diversify the classes of compact operators, the Banach lattice structure had
come in play by Meyer-Nieberg, Dodds, Fremlin, Aliprantis, Wickstead, and
others. In the present paper, we continue this line of study with main
emphasis on limited and L-weakly compact operators (see, also the recent
papers [GE23], [OM22], [EG23], and [AEG23]).

##  1 Introduction and Preliminaries

Throughout the paper, all vector spaces are real and operators are linear;
$X$, $Y$, and $Z$ denote Banach spaces; $E$, $F$, and $G$ denote Banach
lattices. A subset $A$ of $X$ is called bounded if $A$ is norm bounded. We
denote by $B_{X}$ the closed unit ball of $X$, by $\text{\rm L}(X,Y)$
($\text{\rm W}(X,Y)$, $\text{\rm K}(X,Y)$) the space of all bounded (weakly
compact, compact) operators from $X$ to $Y$, by $\text{\rm Id}_{X}$ the
identity operator on $X$, by $E_{+}$ the positive cone of $E$, by
$\text{sol}(A)=\bigcup_{a\in A}[-|a|,|a|]$ 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 $E$.

### 1.1

We shall use the following definition.

######  Definition 1.1.1.

A bounded subset $A$

  1. a)

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

  2. b)

of $X$ is called Dunford-Pettis or a DP set if each weakly null sequence in
$X^{\ast}$ is uniformly null on $A$.

  3. c)

of $E$ is called almost limited if each disjoint w∗-null sequence in
$E^{\ast}$ is uniformly null on $A$.

  4. d)

of $E$ is called an Lwc set if each disjoint sequence in ${\text{\rm sol}}(A)$
is norm-null.

  5. e)

of an ordered vector space $V$ is called order bounded if $A$ is contained in
a set of the form $[a,b]$, where $a,b\in V$.

  6. f)

of an ordered Banach space $X$ is called almost order bounded if, for each
$\varepsilon>0$ there exist $a,b\in X$ satisfying $A\subseteq[a,b]+\varepsilon
B_{X}$.

It is well known that: $B_{X}$ is not limited unless $\dim(X)<\infty$
(however, $B_{c_{0}}$ is a limited subset of $\ell^{\infty}$ by Phillip’s
Lemma); each limited set is a DP-set; $B_{c_{0}}$ is a DP-subset of $c_{0}$;
and $A\subseteq X$ is limited iff $T(A)$ is relatively compact for every
bounded operator $T:X\to c_{0}$ (cf. [BD84, p. 56]). For every Lwc subset $A$
of $E$, we have $A\subseteq E^{a}$. Indeed, otherwise, for some $a\in A$ with
$|a|\in E\setminus E^{a}$ there exists a disjoint sequence $(x_{n})$ in
$[0,|a|]\subseteq\text{\rm sol}(A)$ with $\|x_{n}\|\not\to 0$.

### 1.2

The next well known lemmas will be used below.

######  Lemma 1.2.1.

If, for each $\varepsilon>0$, there exists

  1. (i)

a relatively compact subset $A_{\varepsilon}\subseteq X$ such that $A\subseteq
A_{\varepsilon}+\varepsilon B_{E}$, then $A$ is relatively compact.

  2. (ii)

a relatively weakly compact $A_{\varepsilon}\subseteq X$ such that $A\subseteq
A_{\varepsilon}+\varepsilon B_{E}$, then $A$ is relatively weakly compact.

  3. (iii)

a limited $A_{\varepsilon}\subseteq E$ such that $A\subseteq
A_{\varepsilon}+\varepsilon B_{E}$, then $A$ is limited.

  4. (iv)

an almost limited $A_{\varepsilon}\subseteq E$ such that $A\subseteq
A_{\varepsilon}+\varepsilon B_{E}$, then $A$ is almost limited.

The item (i) of Lemma 1.2.1 is a standard fact. The item (ii) is
Grothendieck’s characterization (cf. [AB06, Thm. 3.44]) of weakly compact
sets. The items (iii) and (iv) are contained in [EMM13, Lm. 2.2].

######  Lemma 1.2.2.

(cf. [AB06, Thm. 4.59] and [MN91, 2.4.14]) The following are equivalent.

  1. (i)

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

  2. (ii)

$E^{\ast}$ is a KB-space.

  3. (iii)

Every disjoint bounded sequence in $E$ is weakly null.

### 1.3

The next definition will be used often.

######  Definition 1.3.1.

A Banach space $X$ has:

  1. a)

the Schur property (or $X\in\text{\rm(SP)}$) if each weakly null sequence in
$X$ is norm null (cf. [AB06, p. 207]);

  2. b)

the Dunford–Pettis property (or $X\in\text{\rm(DPP)}$) if $f_{n}(x_{n})\to 0$
for each weakly null $(x_{n})$ in $X$ and each weakly null $(f_{n})$ in
$X^{\ast}$ (cf. [AB06, p. 341]);

  3. c)

the Dunford–Pettis∗ property (or $X\in\text{\rm(DP${}^{\ast}$P)}$) if weakly
compact subsets of $X$ are limited.

  4. d)

the Gelfand–Phillips property (or $X\in\text{\rm(GPP)}$) if limited subsets of
$X$ are relatively compact (cf. [BD84]).

Separable Banach spaces and reflexive Banach spaces are in (GPP) by [BD84]. If
$E\in\text{\rm(DP${}^{\ast}$P)}$, then weakly compact operators $T:X\to E$ are
limited. Order intervals of a Banach lattice are not necessarily limited
(e.g., $[-{\mathbb{1}},{\mathbb{1}}]$ in $c$ is not limited, where
${\mathbb{1}}$ denotes constantly one sequence). The closed unit ball of
$\ell^{\infty}$ is almost limited, but the closed unit ball of $c_{0}$ is not
almost limited.

### 1.4

Here we collect various Schur type properties.

######  Definition 1.4.1.

A Banach lattice $E$ has:

  1. a)

the dual disjoint w∗-property (or $E\in$DDw∗P) if every disjoint bounded
sequence in $E^{\ast}$ is w∗-null.

  2. b)

the positive Schur property (or $E\in\text{\rm(PSP)}$) if each positive weakly
null sequence in $E$ is norm null (cf. [Wnu13]);

  3. c)

the dual positive Schur property (or $E\in\text{\rm(DPSP)}$) if each positive
w∗-null sequence in $E^{\ast}$ is norm null;

  4. d)

the dual disjoint Schur property (or $E\in\text{\rm(DDSP)}$) if each disjoint
w∗-null sequence in $E^{\ast}$ is norm null (cf. [EMM13, Def. 3.2)]);

  5. e)

the property (d) (or $E\in\text{\rm(d)}$) if, for every disjoint $\text{\rm
w}^{\ast}$-null $(f_{n})$ in $E^{\ast}$, the sequence $(|f_{n}|)$ is also
$\text{\rm w}^{\ast}$-null (cf. [Elb17, Def. 1]);

  6. f)

the Grothendieck property (shortly, $E\in(\text{\rm GP})$) if w∗-null
sequences in $E^{\ast}$ are weakly null (cf. [Wnu13, p. 760]);

  7. g)

the positive Grothendieck property (or $E\in\text{\rm(PGP)}$) if each positive
w∗-null sequence in $E^{\ast}$ is weakly null (cf. [Wnu13, p. 760]);

  8. h)

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

The following result illustrates importance of the dual disjoint w∗-property.

######  Theorem 1.4.2.

For every Banach lattice $E$ the following holds.

$E\in(\text{\rm DGP})\cap(\text{\rm DDw}^{\ast}\text{\rm P})\Longrightarrow
E^{\ast\ast}\text{\it\ is a \text{\rm KB}-space}\Longrightarrow E\in(\text{\rm
DDw}^{\ast}\text{\rm P}).$

###### Proof.

Let $E\in(\text{\rm DGP})\cap(\text{\rm DDw${}^{\ast}$P})$ but $E^{\ast\ast}$
is not a KB-space. Then $(f_{n})$ is not weakly null for some disjoint bounded
$(f_{n})$ in $E^{\ast}$ by Lemma 1.2.2. However
$f_{n}\stackrel{{\scriptstyle\text{\rm w}^{\ast}}}{{\to}}0$ because
$E\in(\text{\rm DDw${}^{\ast}$P})$, and hence
$f_{n}\stackrel{{\scriptstyle\text{\rm w}}}{{\to}}0$ as $E\in(\text{\rm
DGP})$. The obtained contradiction proves the first implication.

For the second implication, let $E^{\ast\ast}$ be a KB-space. Then, by Lemma
1.2.2, every disjoint bounded $(f_{n})$ in $E^{\ast}$ is weakly null and hence
$\text{\rm w}^{\ast}$-null. It follows $E\in(\text{\rm DDw${}^{\ast}$P})$. ∎

The disjoint Grothendieck property was studied in [MFMA19] under the name
“weak Grothendieck property”.

The property (d) is weaker than the sequential $\text{\rm
w}^{\ast}$-continuity of the lattice operations, e.g. in $\ell^{\infty}$. It
was proved in [Wnu13, Prop. 1.4], that every Dedekind $\sigma$-complete $E$
has the property (d), and it was observed in [Wnu13, Rem. 1.5] that
$\ell^{\infty}/c_{0}\in\text{\rm(d)}$ but $\ell^{\infty}/c_{0}$ is not
Dedekind $\sigma$-complete. The lattice ${\text{\rm}L}^{p}[0,1]$, $1\leq
p\leq\infty$, has the property (d), whereas $C[0,1]\not\in\text{\rm(d)}$ (see
Example 1.4.4).

The following lemma (see [EMM13, Thm. 3.3]) describes the impact of the
property (d) on dual Schur properties.

######  Lemma 1.4.3.

The following holds for Banach lattices$:$

  1. (i)

$(\text{\rm DDSP})\subseteq(\text{\rm DPSP})$;

  2. (ii)

$((\text{\rm DDSP})\cap\text{\rm(d)})=((\text{\rm DPSP})\cap\text{\rm(d)})$.

Clearly
$((\text{PGP})\cap\text{\rm(d)})\subseteq((\text{DGP})\cap\text{\rm(d)})$. For
this inclusion, the property (d) is essential, as
$C[0,1]\in\text{\rm(PGP)}\setminus\text{\rm(DGP)}$ by Example 1.4.4 below. We
have no example of $E\in(\text{DGP})\setminus(\text{PGP})$. The next example
shows that the property (d) is essential in Lemma 1.4.3(ii).

######  Example 1.4.4.

Consider the Banach lattice $E=C[0,1]$. Let $(f_{n})$ be a positive w∗-null
sequence in $E^{\ast}$. 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 Rademacher functions on $[0,1]$ and define a
disjoint w∗-null sequence in $E^{\ast}$ 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 a disjoint w∗-null sequence in $E^{\ast}$, yet $\|g_{n}\|=1$ for all
$n\in{\mathbb{N}}$. Therefore $C[0,1]\not\in(\text{DDSP})$, and hence by Lemma
1.4.3, $C[0,1]\not\in(\text{d})$. Note that the set
$\\{g_{n}:n\in{\mathbb{N}}\\}$ is even not relatively weakly compact. Indeed,
for

$y:=\sum_{n=1}^{\infty}r_{2n}\cdot\mathbb{1}_{[\frac{1}{2^{n+1}},\frac{1}{2^{n}}]}\in
L^{\infty}[0,1]\subseteq E^{\ast},$

$y(g_{n})\equiv 1$ for all $n\in{\mathbb{N}}$. It also shows
$C[0,1]\not\in(\text{DGP})$.

### 1.5

We use also the following definitions concerning operators.

######  Definition 1.5.1.

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

  1. a)

Dunford–Pettis (shortly, $T\in\text{\rm DP}(X,Y)$) if $T$ takes weakly null
sequences to norm null ones;

  2. b)

limited ($T\in\text{\rm Lim}(X,Y)$) if $T(B_{X})$ is a limited subset of $Y$.

The canonical embedding of $c_{0}$ into $\ell^{\infty}$ is limited, but not
compact.

######  Definition 1.5.2.

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

  1. a)

almost limited ($T\in\text{\rm aLim}(X,F)$) if $T(B_{X})$ is almost limited.

  2. b)

semi-compact ($T\in\text{\rm semi-K}(X,F)$) if it carries bounded subsets of
$X$ onto almost order bounded subsets of $F$.

  3. c)

L-weakly compact ($T\in\text{\rm Lwc}(X,F)$) if $T$ carries bounded subsets of
$X$ onto Lwc subsets of $F$;

  4. d)

almost L-weakly compact ($T\in\text{\rm aLwc}(X,F)$) if $T$ carries relatively
weakly compact subsets of $X$ onto Lwc subsets of $F$ [BLM18, Def. 2.1].

  5. e)

limitedly L-weakly compact ($T\in l\text{\rm-Lwc}(X,F)$) if $T$ maps limited
subsets of $X$ onto Lwc subsets of $F$ (see [AEG23, Def. 2.8 (b)]).

######  Definition 1.5.3.

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

  1. a)

AM-compact ($T\in\text{\rm AMc}(E,Y)$) if $T$ carries order intervals of $E$
onto relatively compact subsets of $Y$.

  2. b)

order weakly compact ($T\in\text{\rm owc}(E,Y)$) if $T$ carries order
intervals of $E$ onto relatively weakly compact subsets of $Y$ [Dod75, Def.
2.1].

  3. c)

M-weakly compact ($T\in\text{\rm Mwc}(E,Y)$) if $\|Tx_{n}\|\to 0$ for every
disjoint bounded $(x_{n})$ in $E$ [MN74, Def. 1 iv)];

  4. d)

almost M-weakly compact ($T\in\text{\rm aMwc}(E,Y)$) if $f_{n}(Tx_{n})\to 0$
for every weakly convergent $(f_{n})$ in $Y^{\ast}$ and every disjoint bounded
$(x_{n})$ in $E$ [BLM18, Def. 2.2].

  5. e)

almost Dunford–Pettis ($T\in\text{\rm aDP}(E,Y)$) if $T$ takes disjoint weakly
null sequences to norm null ones [Wnu94].

It is well known that $\text{\rm semi-K}(X,F)=\text{\rm Lwc}(X,F)$ for every
$X$, whenever the norm in $F$ is o-continuous. Also $\text{\rm
W}(E,Y)\subseteq\text{\rm owc}(E,Y)$, and the inclusion is proper in general
(cf. [AAT03, Example 2.6]).

### 1.6

Note that [AB06, Thm. 5.63] yields a method of producing many classes of
operators. For example, if we consider sets of the form $A=T(C)$ with any
weakly compact subset $C$ of $X$ and $B=B_{F^{\ast}}$ then we obtain almost
Lwc and almost Mwc operators [BLM18], whereas the choice of sets of the form
$A=T[0,x]$ with any order interval $[0,x]$ in $E$ and $B=B_{F^{\ast}}$
produces order Lwc and order Mwc operators [LBM21]. $l$-Lwc operators were
defined in [AEG23, Def. 2.8 (b)]. As $T\in l\text{\rm-Lwc}(X,F)$ iff
$(T^{\ast}f_{n})$ is w∗-null for each disjoint bounded sequence $(f_{n})$ in
$F^{\ast}$ by [AEG23, Thm. 2.7], then $l$-Lwc operators coincide with weak Lwc
operators defined in [OM22, Def. 4.1].

### 1.7

In Section 2, we study how the order continuity of norm and w∗-continuity of
lattice operations effect operators. Sections 3 is devoted to various
modifications of limited operators, which have appeared in the last decade.
For further unexplained terminology and notation, we refer to [AB06, BD84,
AAT03, AEG22, LBM21, BLM18, EAS20, MN91].

##  2 The effect of w∗-continuity of lattice operations in the dual

We start by recalling some earlier results. We collect together results of
[CW99, CCJ14] on compactness of Lwc and Mwc operators that motivated much of
this work.

######  Remark 2.0.1.

  1. (i)

For all $\Gamma$ and $E$, $\text{\rm aLim}(E,c_{0}(\Gamma))\subseteq\text{\rm
K}(E,c_{0}(\Gamma))$.

  2. (ii)

For all $\Gamma$ and $E$, if $T\in\text{\rm aLim}(E,c_{0}(\Gamma))$ then $|T|$
exists and is almost limited.

  3. (iii)

If $F^{a}=F$ then $\text{\rm Mwc}(E,F)\subseteq\text{\rm aLim}(E,F)$ for all
$E$.

  4. (iv)

Let $E^{\ast}\in\text{\rm(PSP)}$ and let $F$ be a KB-space. Then each regular
operator $T:E\to F$ is almost limited.

  5. (v)

If $E^{\ast}$ is a KB-space and $F\in\text{\rm(PSP)}$ then every regular
operator $T:E\to F$ is almost limited.

###### Proof.

(i) By [CCJ14, Thm. 2.6], $\text{\rm aLim}(E,c_{0}(\Gamma))=\text{\rm
Lwc}(E,c_{0}(\Gamma))$. By [CW99, Cor. 2.8], $\text{\rm
Lwc}(E,c_{0}(\Gamma))\subseteq\text{\rm K}(E,c_{0}(\Gamma))$.

(ii) Let $T\in\text{\rm aLim}(E,c_{0}(\Gamma))$. Then $T\in\text{\rm
Lwc}(E,c_{0}(\Gamma))$ by [CCJ14, Thm. 2.6]. Then $|T|$ exists and
$|T|\in\text{\rm Lwc}(E,c_{0}(\Gamma))$ by [CW99, Thm. 2.9].

(iii) In this case $\text{\rm Mwc}(E,F)=\text{\rm Lwc}(E,F)$ by the Dodds –
Fremlin theorem (cf. [AB06, Thm. 5.68]). The result now follows from [CCJ14,
Thm. 2.6].

(iv) By [CW99, Thm. 3.8], $T\in\text{\rm Lwc}(E,F)$, and then $T\in\text{\rm
aLim}(E,F)$ by [CCJ14, Thm. 2.6].

(v) By [CW99, Thm. 3.9], $T\in\text{\rm Lwc}(E,F)$, and then $T\in\text{\rm
aLim}(E,F)$ by [CCJ14, Thm. 2.6]. ∎

Recall that $F\in\text{\rm(DP${}^{\ast}$P)}$ if every weakly compact set in
$F$ is limited. Thus if $F\in\text{\rm(DP${}^{\ast}$P)}$ then $\text{\rm
W}(X,F)\subseteq\text{\rm Lim}(X,F)$.

Let $T:E\to F$ and suppose $F^{a}=F$. If $T$ is limited then it is L-weakly
compact (cf. [CCJ14, Thm. 2.6]) and therefore is weakly compact and semi-
compact. Then $T^{\ast}\in\text{\rm aDP}(F^{\ast},E^{\ast})\cap\text{\rm
owc}(F^{\ast},E^{\ast})$ (cf. [Elb17, Thm. 1]). If $F^{\ast}$ has
w∗-continuous lattice operations, then $\text{\rm Lim}(E,F)$ is an order ideal
in $\text{\rm Lwc}(E,F)$.

### 2.1

We begin this subsection with a useful summary.

######  Proposition 2.1.1.

Let $T\in\text{\rm semi-K}(X,F)$.

  1. (i)

If $F$ has limited order intervals, then $T\in\text{\rm Lim}(X,F)$.

  2. (ii)

If $F$ has compact order intervals, then $T\in\text{\rm K}(X,F)$.

  3. (iii)

If $F$ has o-continuous norm, then $T\in\text{\rm W}(X,F)$.

###### Proof.

(i) We need to show that $T(B_{X})$ is limited. Since $T\in\text{\rm
semi-K}(X,F)$ then, for every $\varepsilon>0$, there exist $a,b\in F$ with
$T(B_{X})\subseteq[a,b]+\varepsilon B_{F}$. As each order interval in $F$ is
limited, $T(B_{X})$ is limited by Lemma 1.2.1(iii).

(ii) By the assumption, $T(B_{X})$ is totally bounded, and hence
$T\in\text{\rm K}(X,F)$ by Lemma 1.2.1(i).

(iii) Since $T\in\text{\rm semi-K}(X,F)$ then, for every $\varepsilon>0$,
there exist $a,b\in F$ with $T(B_{X})\subseteq[a,b]+\varepsilon B_{F}$. As
$F^{a}=F$, $[a,b]$ is weakly compact. By Lemma 1.2.1(ii), $T(B_{X})$ is
relatively weakly compact. Thus $T\in\text{\rm W}(X,F)$. ∎

######  Corollary 2.1.2.

Let $E^{\ast}$ be a KB-space and let $F$ have limited order intervals. The
following hold.

  1. (i)

$\text{\rm DP}_{+}(E,F)\subseteq\text{\rm Lim}(E,F)$.

  2. (ii)

If $E\in\text{\rm(DPP)}$, then $\text{\rm W}_{+}(E,F)\subseteq\text{\rm
Lim}(E,F)$.

###### Proof.

(i) Let $T\in\text{\rm DP}_{+}(E,F)$. By [Wnu90, Thm. 3.7.10], $T\in\text{\rm
Mwc}(E,F)$. Thus $T\in\text{\rm semi-K}(E,F)$, and hence $T\in\text{\rm
Lim}(E,F)$ by Proposition 2.1.1(i).

(ii) Let $T\in\text{\rm W}_{+}(E,F)$. Then $T\in\text{\rm DP}(E,F)$ by the
Brace – Grothendieck theorem (cf. [AB06, Thm. 5.82]). The rest of the proof is
as in (i). ∎

Since L-weakly compact, regular M-weakly compact, and positive operators
dominated by semi-compact operators are semi-compact (see [AB06, Thm. 5.71,
Thm. 5.72]), they are (almost) limited independently of their ranges due to
Proposition 2.1.1. A semi-compact operator need not be compact, weakly
compact, L- or M-weakly compact. The identity operator on $\ell^{\infty}$ is
semi-compact, but it has none of the compactness properties mentioned.

######  Corollary 2.1.3.

Let $F$ have o-continuous norm and limited order intervals. Then

  1. (i)

for every $E$, each semi-compact operator $T:E\to F$ is compact ;

  2. (ii)

for every $E$, each order bounded Mwc operator $T:E\to F$ is compact;

  3. (iii)

for every $E$, each Lwc operator $T:E\to F$ is compact;

  4. (iv)

if $E^{\ast}$ has o-continuous norm, then each order bounded aDP operator
$T:E\to F$ is compact.

###### Proof.

(i) Let $T\in\text{\rm semi-K}(E,F)$. By Proposition 2.1.1, $T\in\text{\rm
Lim}(E,F)$, and so $T(B_{E})$ is limited. As $F^{a}=F$, then $F$ is Dedekind
$\sigma$-complete. Then $F\in\text{\rm(GPP)}$ by [Buh73]. Thus $T(B_{E})$ is
relatively compact, so $T\in\text{\rm K}(E,F)$.

(ii) This follows from (i) in view of [AB06, Thm. 5.72].

(iii) This follows from (i) in view of [AB06, Thm. 5.71].

(iv) Since $T\in\text{\rm aDP}(E,F)$ and $T$ is order bounded, $T\in\text{\rm
Mwc}(E,F)$ and therefore $T\in\text{\rm semi-K}(E,F)$. By (i), $T\in\text{\rm
K}(E,F)$. ∎

By [AE11, Prop. 2.2], if $E^{\ast}$ is a KB-space, $T\in\text{\rm aDP}(E,F)$
and is order bounded, then $T\in\text{\rm semi-K}(E,F)$. Thus it follows
easily that if, additionally, $S\in\text{\rm AMc}(F,G)$ then $ST\in\text{\rm
K}(E,F)$.

### 2.2

Note that not every weakly compact operator is AM-compact, e.g., the identity
operator $I_{L^{2}[0,1]}$ is weakly compact but it is not AM-compact.

######  Proposition 2.2.1.

If $E$ has limited order intervals then, for all $Y$,

  1. (i)

$\text{\rm W}(E,Y)\subseteq\text{\rm AMc}(E,Y)$;

  2. (ii)

$\text{\rm owc}(E,Y)\subseteq\text{\rm AMc}(E,Y)$.

###### Proof.

(i) Let $T\in\text{\rm W}(E,Y)$, and let $[-x,x]\subseteq E$. By the
Davis–Figiel–Johnson–Pelczynski theorem, $T$ factors over a reflexive $Z$ as
$T=SR$ with $R\in\text{\rm L}(E,Z)$ and $S\in\text{\rm L}(Z,Y)$ (cf. [AB06,
Thm. 5.38]). The set $R[-x,x]$ is limited in $Z$ as $[-x,x]$ is limited in
$E$. Since $Z\in(\text{\rm GPP})$ by [BD84], therefore $R[-x,x]$ is relatively
compact in $Z$, and hence $SR[-x,x]$ is relatively compact in $Y$. Thus,
$T\in\text{\rm AMc}(E,Y)$.

(ii) Let $T\in\text{\rm owc}(E,Y)$ and let $[-x,x]\subseteq E$. Then $T$
admits a factorization over some $F=F^{a}$ as $T=SQ$, where $Q:E\to F$ is a
lattice homomorphism and $S\in\text{\rm L}(F,Y)$ by [AB06, Thm. 5.58]. The set
$Q[-x,x]$ is limited as $[-x,x]$ is limited. Since $F$ has o-continuous norm,
it is Dedekind $\sigma$-complete, hence $F\in(\text{\rm GPP})$ by [Buh73].
Therefore $Q[-x,x]$ is relatively compact in $F$, and hence $SQ[-x,x]$ is
relatively compact in $Y$. Thus $T\in\text{\rm AMc}(E,Y)$. ∎

######  Corollary 2.2.2.

Let $E$ have limited order intervals, and let $F$ have o-continuous norm. Then
the following holds.

  1. (i)

$\text{\rm semi-K}_{+}(E,F)\subseteq\text{\rm AMc}(E,F)$.

  2. (ii)

$\text{\rm Lwc}(E,F)\subseteq\text{\rm AMc}(E,F)$.

  3. (iii)

$\text{\rm Mwc}(E,F)\subseteq\text{\rm AMc}(E,F)$.

  4. (iv)

Let $T:E\to F$ have o-continuous operator norm. Then $T\in\text{\rm
AMc}(E,F)$.

###### Proof.

(i) As $T\in\text{\rm semi-K}_{+}(E,F)$, for each $\varepsilon>0$, we can find
$u_{\varepsilon}\in F_{+}$ such that
$T(B_{E})\subseteq[-u_{\varepsilon},u_{\varepsilon}]+\varepsilon B_{F}$. Since
$F$ has o-continuous norm, the interval $[-u_{\varepsilon},u_{\varepsilon}]$
is relatively weakly compact. Consequently $T\in\text{\rm W}(E,F)$ by Lemma
1.2.1(ii), and hence $T\in\text{\rm AMc}(E,F)$ due to Proposition 2.2.1.

(ii) and (iii) follows from (i) as L- and M-weakly compact operators are
weakly compact.

(iv) By the Dodds – Fremlin theorem, $T\in\text{\rm Lwc}(E,F)$ (cf. [AB06,
Thm. 5.68]). The rest follows from (ii). ∎

The following result gives some conditions on a weakly compact operator to be
compact.

######  Theorem 2.2.3.

Suppose $E^{\ast}$ is a KB-space, $E$ has limited order intervals, and
$F\in(\text{\rm PSP})$. Then $\text{\rm W}_{+}(E,F)\subseteq\text{\rm
K}(E,F)$.

###### Proof.

Let $T\in\text{\rm W}_{+}(E,F)$. Then $T\in\text{\rm AMc}(E,F)$ by Proposition
2.2.1. On the other hand, $T\in\text{\rm Mwc}(E,F)$ by [CW99, Thm. 3.3]. As
$T$ is positive, it is semi-compact by [AB06, Thm. 5.72]. Since
$F\in(\text{\rm PSP})$, it has o-continuous norm. As $E^{\ast}$ and $F$ both
have o-continuous norms, $T\in\text{\rm K}(E,F)$ by [Wnu90, Thm. 125.5]
because of $T\in\text{\rm semi-K}(E,F)\cap\text{\rm AMc}(E,F)$. ∎

### 2.3

Consider operators that take order bounded sets onto limited sets.

######  Definition 2.3.1.

An operator $T:E\to Y$ is called order limited (briefly $T\in\text{\rm
o-Lim}(E,Y)$) if $T[a,b]$ is limited for each order interval $[a,b]$.

######  Theorem 2.3.2.

Let $E$ has limited order intervals. Then the following holds.

  1. (i)

Let $T:E\to F$ be a positive operator. If $E^{\ast}$ and $F$ both have
o-continuous norms then $T\in\text{\rm o-Lim}(E,F)\cap\text{\rm AMc}(E,F)$.

  2. (ii)

Let $T:E\to Y$ be a bounded operator. If $E^{\ast}$ is a KB-space and $Y$
doesn’t contain $c_{0}$, then $T\in\text{\rm o-Lim}(E,Y)\cap\text{\rm
AMc}(E,Y)$.

###### Proof.

(i) By [AB06, Exer. 15, p. 314], $T$ has a factorization $T=SR$ over a Banach
lattice $G$ with o-continuous norm. Let $[a,b]\subseteq E$. Then $R[a,b]$ is
limited in $G$. Since $G\in(\text{\rm GPP})$, $R[a,b]$ is relatively compact
in $G$, and hence $SR[a,b]$ is relatively compact in $F$. Since the interval
$[a,b]$ is arbitrary, $T\in\text{\rm o-Lim}(E,F)\cap\text{\rm AMc}(E,F)$.

(ii) By [AB06, Exer. 19, p. 315], the operator $T$ admits a factorization over
a reflexive Banach lattice $G$ as $T=SR$, where $R:E\to G$ and $S:G\to Y$ are
bounded. The rest of the proof is similar to proof of (i) and is omitted. ∎

### 2.4

The next result gives some conditions on an operator to be limited.

######  Theorem 2.4.1.

  1. (i)

If $E^{\ast}$ is a KB-space and $F^{\ast}$ has w∗-continuous lattice
operations, then $\text{\rm aDP}_{+}(E,F)\subseteq\text{\rm Lim}(E,F)$.

  2. (ii)

If $E^{\ast}$ has w∗-sequentially continuous lattice operations, then
$\text{\rm Mwc}(E,F)\subseteq\text{\rm Lim}(E,F)$.

###### Proof.

(i) Let $T\in\text{\rm aDP}_{+}(E,F)$. By [AE11, Prop. 2.2], $T\in\text{\rm
semi-K}(E,F)$. Thus, for each $\varepsilon>0$, there exists some
$u_{\varepsilon}\in F_{+}$ such that
$T(B_{E})\subseteq[-u_{\varepsilon},u_{\varepsilon}]+\varepsilon B_{F}$. Since
$F$ has limited order intervals, $T(B_{E})$ is limited by Lemma 1.2.1(iii),
and hence $T\in\text{\rm Lim}(E,F)$.

(ii) Let $(f_{n})\stackrel{{\scriptstyle w^{\ast}}}{{\to}}0$ in $F^{\ast}$. We
need to show $\|T^{\ast}f_{n}\|\to 0$. For this it suffices to show
$|T^{\ast}f_{n}|\stackrel{{\scriptstyle w^{\ast}}}{{\to}}0$ and
$T^{\ast}(f_{n})(x_{n})\to 0$ for each norm bounded disjoint sequence
$(x_{n})$ in $E_{+}$. Since $T$ is bounded,
$T^{\ast}f_{n}\stackrel{{\scriptstyle w}}{{\to}}0$ in $F^{\ast}$. Continuity
of lattice operations in $E^{\ast}$ gives
$|T^{\ast}f_{n}|\stackrel{{\scriptstyle w^{\ast}}}{{\to}}0$. Consider,

$|T^{\ast}(f_{n})(x_{n})|=|f_{n}(Tx_{n})|\leq\|f_{n}\|\|Tx_{n}\|.$

Since $(f_{n})$ is norm bounded and $T\in\text{\rm Mwc}(E,F)$, the claim
follows. ∎

### 2.5

The next example shows a limited operator need not be Mwc.

######  Example 2.5.1.

Let $E^{\ast}$ be not a KB-space and let $F\neq\\{0\\}$. Then there exists a
positive compact (and hence limited) operator $T:E\to F$ that is not M-weakly
compact. Indeed, pick $y\in F_{+}\setminus\\{0\\}$. Since the norm in
$E^{\ast}$ is not o-continuous, $E^{\ast}$ contains a sublattice isomorphic to
$\ell^{\infty}$ (cf. [AB06, Thm. 4.51]) and then $E$ contains a sublattice
isomorphic to $\ell^{1}$ (cf. [MN91, Prop. 2.3.12]). Let $i:\ell^{1}\to E$ be
a lattice embedding. Then $i(\ell^{1})$ is the range of a positive projection
$P:E\to E$, that is $i(\ell^{1})=P(E)$ (cf. [MN91, Prop. 2.3.11]). Define
$S:P(E)\to F$ by

$S\left(i\left(\sum\limits_{k=1}^{\infty}\beta_{k}e_{k}\right)\right):=\left(\sum\limits_{k=1}^{\infty}\beta_{k}\right)y,$

and let $T=SP$. Then $T:E\to F$ is positive. Since $S$ has rank one, $T$ is
compact (and hence limited). But $T\not\in\text{\rm Mwc}(E,F)$. This follows
from $T(i(e_{n}))=SP(i(e_{n}))=S(i(e_{n}))=y$ and so $\|T(i(e_{n}))\|\not\to
0$.

### 2.6

We give relations between limited and aMwc operators.

######  Theorem 2.6.1.

$E^{\ast}$ is a KB-space if one of the following conditions yields.

  1. (i)

$\text{\rm Lim}_{+}(E,F)\subseteq\text{\rm Mwc}(E,F)$ for every $F$.

  2. (ii)

$\text{\rm Lim}(E,F)\subseteq\text{\rm aMwc}(E,F)$ for every $F$.

###### Proof.

(i) Assume, in contrary, that $E^{\ast}$ is not a KB-space. Then, in view of
Lemma 1.2.2 and the Fremlin–Meyer-Nieberg theorem (cf. [AB06, Thm. 4.14]),
there exists a disjoint sequence $(f_{n})\subseteq[0,f]\subseteq E^{\ast}$
such that $\|f_{n}\|\not\to 0$. Pick $y\in E_{+}$ such that $\|y\|=1$ and a
functional $g\in E^{\ast}$ with $\|g\|=1$ and $g(y)=1$. Define a positive
operator $T:E\to E$ by $T(x)=f(x)y$ for $x\in E$. As $T$ is compact,
$T\in\text{\rm Lim}(E,E)$. We claim $T\not\in\text{\rm Mwc}(E,E)$. By the
Meyer-Nieberg theorem (cf. [AB06, Thm. 5.64]), it is enough to show
$T^{\ast}\not\in\text{\rm Lwc}(E^{\ast},E^{\ast})$. Let us observe that
$T^{\ast}(g)=g(y)f=f$. Thus $(f_{n})$ is a disjoint sequence in the solid hull
of $T^{\ast}(B_{E})$ with $\|f_{n}\|\not\to 0$. Therefore
$T^{\ast}\not\in\text{\rm Lwc}(E^{\ast},E^{\ast})$ and hence
$T\not\in\text{\rm Mwc}(E,E)$ by the Meyer-Nieberg theorem, violating the
assumption $\text{\rm Lim}_{+}(E,E)\subseteq\text{\rm Mwc}(E,E)$. A
contradiction.

(ii) Under the assumption, $\text{\rm K}(E,F)\subseteq\text{\rm aMwc}(E,F)$,
hence $E^{\ast}$ is a KB-space by [EAS20, Thm. 2]. ∎

######  Theorem 2.6.2.

Let $E^{\ast}$ be a KB-space. Then the following holds.

  1. (i)

If $F\in\text{\rm(GPP)}$ then $\text{\rm Lim}(E,F)\subseteq\text{\rm
Mwc}(E,F)$.

  2. (ii)

$\text{\rm Lim}_{+}(E,F)\subseteq\text{\rm aMwc}(E,F)$ for every $F$.

###### Proof.

(i) Let $T\in\text{\rm Lim}(E,F)$ and let $(x_{n})$ be a disjoint bounded
sequence in $E$. Then, by Lemma 1.2.2, $x_{n}\stackrel{{\scriptstyle
w}}{{\to}}0$ and hence $Tx_{n}\stackrel{{\scriptstyle w}}{{\to}}0$. Since
$F\in\text{\rm(GPP)}$ and the set $\\{Tx_{n}\\}_{n=1}^{\infty}$ is limited, it
is relatively compact. Then $\|Tx_{n}\|\to 0$, and hence $T\in\text{\rm
Mwc}(E,F)$.

(ii) Suppose $T\in\text{\rm Lim}_{+}(E,F)$. Then $T\in\text{\rm
aLim}_{+}(E,F)$, and hence $T^{\ast}\in\text{\rm aDP}_{+}(F^{\ast},E^{\ast})$
(cf. [Elb17, Thm. 1]). Then $T^{\ast}\in\text{\rm
aLwc}_{+}(F^{\ast},E^{\ast})$ by [BLM18, Prop. 2.4], and thus $T\in\text{\rm
aMwc}(E,F)$ in view of [BLM18, Thm. 2.5]. ∎

### 2.7

In general, a weakly compact operator need not be limited.

######  Theorem 2.7.1.

If $\text{\rm W}(E,Y)\subseteq\text{\rm Lim}(E,Y)$, then one of the following
holds$:$

  1. (i)

$E^{\ast}$ is a KB-space;

  2. (ii)

$Y\in(\text{\rm DPP})$.

###### Proof.

Assume that $E^{\ast}$ is not a KB-space and $Y\not\in\text{\rm(DPP)}$. We
construct $T\in\text{\rm W}(E,Y)$ and $T\not\in\text{\rm Lim}(E,Y)$.

Since $E^{\ast}$ is not a KB-space, there exists a disjoint sequence $(u_{n})$
in $(B_{E})_{+}$ which is not weakly null by Lemma 1.2.2. By passing to a
subsequence, we may assume that, for some $\theta\in E^{\ast}_{+}$ and
$\varepsilon>0$, $\theta(u_{n})>\varepsilon$ for all $n$. Then the components
$\theta_{n}$ of $\theta$ in the carriers $C_{n}$ of $u_{n}$ form an order
bounded disjoint sequence in $E^{\ast}_{+}$ such that
$\theta_{n}(u_{n})=\theta(u_{n})$ for all $n$ and $\theta_{n}(u_{m})=0$ if
$n\neq m$ by [Zaa83, Thm. 116.3]. We define an operator $P:E\to\ell^{1}$ by

$P(x)=\left(\frac{\theta_{k}(x)}{\theta(u_{k})}\right)_{k=1}^{\infty}\quad(x\in
E).$

The operator $P$ is well defined as

$\|Px\|_{1}=\sum_{k=1}^{\infty}\left|\frac{\theta_{k}(x)}{\theta(u_{k})}\right|\leq\sum_{k=1}^{\infty}\frac{|\theta_{k}(x)|}{\varepsilon}\leq\frac{1}{\varepsilon}\sum_{k=1}^{\infty}\theta_{k}(|x|)\leq\frac{1}{\varepsilon}\theta(|x|)$

holds for each $x\in E$.

As $Y\not\in\text{\rm(DPP)}$, there is a weakly null sequence $(y_{n})$ in $Y$
and a weakly null (hence w∗-null) sequence $(f_{n})$ in $Y^{\ast}$ such that
$(f_{n}(y_{n}))\not\to 0$. Passing to a subsequence, we may assume that, for
some $\epsilon>0$,

$|f_{n}(y_{n})|>\epsilon\quad\quad(\forall n\in{\mathbb{N}}).$ $None$

We define an operator $S:\ell^{1}\to Y$ by
$S\alpha=\sum_{k=1}^{\infty}\alpha_{k}y_{k}$. Since
$y_{n}\stackrel{{\scriptstyle w}}{{\to}}0$, the sequence $(y_{n})$ is bounded,
and hence the operator $S$ is well defined. By [AB06, Thm. 5.26], $S$ is
weakly compact.

Then $T=SP\in\text{\rm W}(E,Y)$. Suppose that $T$ is limited. Then the
$\text{\rm w}^{\ast}$-null sequence $(f_{n})$ is uniformly null on $T(B_{E})$.
Since

$Tu_{n}=SPu_{n}=Se_{n}=y_{n}\quad\quad(\forall n\in{\mathbb{N}}),$

it follows from (1) that $|f_{n}(Tu_{n})|>\epsilon$ for all $n$, contradicting
$f_{n}\rightrightarrows 0(T(B_{E}))$. Therefore $T\notin\text{\rm Lim}(E,Y)$,
as desired. ∎

### 2.8

Recall $E\in\text{\rm(DDw${}^{\ast}$P)}$ if every disjoint bounded sequence in
$E^{\ast}$ is w∗-null (see Definition 1.4.1).

######  Proposition 2.8.1.

The following holds.

  1. (i)

Let $F$ have o-continuous norm, $T\in\text{\rm semi-K}(X,F)$, and
$S\in\text{\rm DP}(F,G)$. Then $ST\in\text{\rm K}(X,G)$.

  2. (ii)

Suppose $E\in(\text{\rm DDw}^{\ast}\text{\rm P})$, then each bounded $T\in
l\text{\rm-Lwc}(E,F)$ (see Definition 1.5.2(e)).

###### Proof.

(i) Since $T\in\text{\rm semi-K}(X,F)$, for each $\varepsilon>0$, there exists
$u_{\varepsilon}\in F_{+}$ such that
$T(B_{X})\subseteq[-u_{\varepsilon},u_{\varepsilon}]+\varepsilon B_{F}$. Since
$F$ has o-continuous norm, $[-u_{\varepsilon},u_{\varepsilon}]$ is relatively
weakly compact and hence $T(B_{X})$ is relatively weakly compact by Lemma
1.2.1(ii). In order to show that $ST(B_{E})$ is relatively compact, take a
sequence $(STx_{n})$ in $ST(B_{X})$. As $T(B_{X})$ is relatively weakly
compact then, by the Eberlein – S̆mulian theorem (cf. [AB06, Thm. 3.40]),
there exists a weakly convergent subsequence $(Tx_{n_{k}})$. Since
$S\in\text{\rm DP}(F,G)$, the sequence $(STx_{n_{k}})$ converges in norm. As
the sequence $(STx_{n})$ in $ST(B_{X})$ is arbitrary, $ST(B_{X})$ is compact
in $G$, and hence $ST\in\text{\rm K}(X,G)$.

(ii) If $E\in\text{\rm(DDw}^{\ast}\text{\rm P)}$, then each bounded disjoint
sequence $(f_{n})$ in $E^{\ast}$ is w∗-null, and therefore $T^{\ast}f_{n}\to
0$ in $\sigma(E^{\ast},E)$, and $T\in l\text{\rm-Lwc}(E,F)$. ∎

######  Corollary 2.8.2.

Suppose $E^{\ast}$ is a KB-space, $F=F^{a}$, and $T\in\text{\rm
aDP}_{r}(E,F)$. If $S\in\text{\rm DP}(F,G)$, then $ST\in\text{\rm K}(E,G)$.

###### Proof.

In view of [AE11, Prop. 2.2] $T\in\text{\rm semi-K}(E,F)$. The rest follows
from Proposition 2.8.1. ∎

### 2.9

The square of a semi-compact operator is not compact in general (e.g.,
$\text{\rm Id}_{c}$ is semi-compact but $\text{\rm Id}_{c}^{2}=\text{\rm
Id}_{c}$ is not compact). This problem was studied in [ANZ06, Thm. 2.4]. We
give another sufficient condition.

######  Theorem 2.9.1.

Let $E$ have limited order intervals.

  1. (i)

If $E$ has o-continuous norm and $T\in\text{\rm semi-K}_{+}(E)$ then
$T^{2}\in\text{\rm K}(E)$.

  2. (ii)

If $T\in\text{\rm semi-K}(E)\cap\text{\rm W}(E)$ then $T^{2}\in\text{\rm
K}(E)$.

###### Proof.

(i) As $T\in L(E)$, then for each $\varepsilon>0$ there exists $\delta>0$ with
$\delta T(B_{E})\subseteq\varepsilon B_{E}$. Since $T\in\text{\rm semi-K}(E)$,
for $\delta>0$ there exists $u_{\delta}\in E_{+}$ such that
$T(B_{E})\subseteq[-u_{\delta},u_{\delta}]+\delta B_{E}$, which yields
$T^{2}(B_{E})\subseteq[-Tu_{\delta},Tu_{\delta}]+\varepsilon B_{E}$. Since $E$
has limited order intervals, each $[-Tu_{\varepsilon},Tu_{\varepsilon}]$ is
limited. Hence $T^{2}(B_{E})$ is limited by Lemma 1.2.1(iii).

By [Buh73], a Dedekind $\sigma$-complete Banach lattice has the Gelfand –
Phillips property iff it has o-continuous norm. Therefore $E\in(\text{\rm
GPP})$. Then $T^{2}(B_{E})$ is relatively compact, and hence
$T^{2}\in\text{\rm K}(E)$.

(ii) As $T\in\text{\rm W}(E)$, it factors over a reflexive space $G$ as
$T=SR$, where $R:E\to G$ and $S:G\to E$ are bounded (cf. [AB06, Thm. 5.38]).

Let $\varepsilon>0$. Since $T\in\text{\rm semi-K}(E)$, there exists
$u_{\varepsilon}\in E_{+}$ such that
$T(B_{E})\subseteq[-u_{\varepsilon},u_{\varepsilon}]+\varepsilon B_{E}$. The
interval $[-u_{\varepsilon},u_{\varepsilon}]$ is limited in $E$. Applying $R$
to $[-u_{\varepsilon},u_{\varepsilon}]$, we see that
$R[-u_{\varepsilon},u_{\varepsilon}]$ is limited in $G$. Being reflexive,
$G\in(\text{\rm GPP})$, and hence $R[-u_{\varepsilon},u_{\varepsilon}]$ is
relatively compact in $G$. Therefore, $SR[-u_{\varepsilon},u_{\varepsilon}]$
is relatively compact in $E$ and

$T^{2}(B_{E})\subseteq SR[-u_{\varepsilon},u_{\varepsilon}]+\varepsilon
SR(B_{E})\subseteq SR[-u_{\varepsilon},u_{\varepsilon}]+\varepsilon B_{E}.$

Since $\varepsilon>0$ is arbitrary, Lemma 1.2.1(i) implies that $T^{2}(B_{E})$
is relatively compact in $E$, and hence $T^{2}\in\text{\rm K}(E)$. ∎

### 2.10

We now consider limitedly Lwc (Mwc) operators.

######  Proposition 2.10.1.

$\text{\rm aLwc}(X,F)\subseteq l\text{\rm-Lwc}(X,F)$.

###### Proof.

Let $T\in\text{\rm aLwc}(X,F)$. Then $T(X)\subseteq F^{a}$. Since $F^{a}$ has
o-continuous norm, $F^{a}\in\text{(\rm DDw${}^{\ast}$P)}$ so that each limited
subset in $F^{a}$ is Lwc by [CCJ14, Thm. 2.6], and hence $T\in l\text{\rm-
Lwc}(X,F)$. ∎

An operator $T:E\to F$ is called limitedly M-weakly compact if $(Tx_{n})$ is
weakly null for each bounded disjoint sequence $(x_{n})$ in $E$ ([AEG23, Def.
2.13], cf. [OM22, Def. 3.1], where these operators called weak M-weakly
compact). If $E^{\ast}$ is a KB-space, each bounded sequence is weakly null,
consequently $\text{\rm L}(E,F)\subseteq\ $l$\text{\rm-Mwc}(E,F)$.

######  Proposition 2.10.2.

Suppose $F$ has the Schur property. Then, for every $E$, $l\text{\rm-
Mwc}(E,F)\subseteq\text{\rm owc}(E,F)$.

###### Proof.

Let $T\in l\text{\rm-Mwc}(E,F)$. Due to the Dodds theorem (cf. [AB06, Thm.
5.57]), we need to show $\|Tx_{n}\|\to 0$ for each order bounded disjoint
sequence $(x_{n})$ in $E$. The sequence $(x_{n})$ is weakly null as it is
order bounded and disjoint. Thus $(x_{n})$ is bounded, as $T\in l\text{\rm-
Mwc}(E,F)$, we get $Tx_{n}\stackrel{{\scriptstyle{\rm w}}}{{\to}}0$ and so
$\|Tx_{n}\|\to 0$ since $F\in(\text{\rm SP})$. ∎

##  3 Almost limited operators

A bounded subset $A$ of a Banach space $X$ is called almost limited (we write
aLim-set) if each disjoint w∗-null sequence in $X^{\ast}$ converges uniformly
to zero on $A$. Almost limited sets were introduced in [CCJ14]. Let us also
note that every order interval in $E$ is aLim iff $E\in{\text{\rm}(d)}$. An
operator $T:X\to F$ is called almost limited (we write $T\in\text{\rm
aLim}(X,F)$) if $T(B_{X})$ is an aLim-set. Almost limited sets and operators
were studied in [EMM13, CCJ14, EMM14]. We say that
$E\in\text{\rm(aDP${}^{\ast}$P)}$ if every weakly compact subset of $E$ is
aLim.

Recall that $F\in(\text{\rm PDSP})$ if every disjoint w∗-null sequence in
$F_{+}^{\ast}$ is norm null. It follows that whenever $F\in(\text{\rm PDSP})$
then $\text{\rm L}(X,F)\subseteq\text{\rm aLim}(X,F)$ for every $X$. Also
notice that since each Lwc-set is aLim by [CCJ14, Thm. 2.6], $\text{\rm
Lwc}(X,F)\subseteq\text{\rm aLim}(X,F)$ for all $X$ and $F$. If $T\in\text{\rm
aLwc}(X,F)$ then, for each $S\in\text{\rm W}(E,X)$ we have $TS\in\text{\rm
Lwc}(E,F)$, and therefore $TS\in\text{\rm aLim}(E,F)$.

Let $E^{\ast}$ and $F$ both have o-continuous norms. Then:

  1. $a)$

$\text{\rm DP}_{+}(E,F)\subseteq\text{\rm Lwc}(E,F)$, and hence $\text{\rm
span(DP}_{+}(E,F))\subseteq\text{\rm aLim}(E,F)$.

  2. $b)$

Each order bounded aLwc operator $T:E\to F$ is Lwc, by [EAS20, Cor. 4], and
therefore $T\in\text{\rm aLim}(E,F)$.

### 3.1

It was shown in [EMM13, Prop. 4.5] that each order bounded Mwc operator
$T:E\to F$ is almost limited whenever lattice operations in $E^{\ast}$ are
w∗-sequentially continuous or $F\in\text{\rm(d)}$. By [Elb17, Cor. 4], if
lattice operations in $E^{\ast}$ are w∗-sequentially continuous then
$\text{\rm aLim}(X,E)\subseteq\text{\rm Lim}(X,E)$.

######  Theorem 3.1.1.

Let $E^{\ast}$ be a KB-space. The following holds.

  1. (i)

If $F\in\text{\rm(d)}$ or the lattice operations in $E^{\ast}$ are
w∗-sequentially continuous then each order bounded aLwc operator $T:E\to F$ is
aLim.

  2. (ii)

If $E$ is reflexive then $\text{\rm aLwc}(E,F)\subseteq\text{\rm aLim}(E,F)$
for every $F$.

###### Proof.

(i) Take an order bounded $T\in\text{\rm aLwc}(E,F)$. For $T\in\text{\rm
aLim}(E,F)$, it suffices to show $f_{n}(Tx_{n})\to 0$ for every disjoint
w∗-null $(f_{n})$ in $F^{\ast}$ and disjoint $(x_{n})$ in $(B_{E})_{+}$ (see,
[EMM13, Thm. 2.7]). Let $(f_{n})$ be disjoint w∗-null in $F^{\ast}$ and let
$(x_{n})$ be disjoint in $(B_{E})_{+}$. Since $f_{n}\stackrel{{\scriptstyle
w^{\ast}}}{{\to}}0$, it is bounded. As $E^{\ast}$ is a KB-space,
$x_{n}\stackrel{{\scriptstyle w}}{{\to}}0$. Since $T\in\text{\rm aLwc}(E,F)$
then $f_{n}(Tx_{n})\to 0$ by [BLM18, Thm. 2.2].

(ii) Let $T\in\text{\rm aLwc}(E,F)$, and let $(f_{n})$ be disjoint w∗-null in
$B_{F^{\ast}}$. We need to show that $f_{n}\rightrightarrows 0(T(B_{E}))$.
Assume in contrary that $\sup\limits_{x\in B_{E}}|f_{n}(Tx)|\not\to 0$ as
$n\to\infty$. Then, for some $\varepsilon>0$, there exists a sequence
$(x_{n_{k}})$ in $B_{E}$ such that

$|f_{n_{k}}(Tx_{n_{k}})|\geq\varepsilon\quad\quad(\forall k\in{\mathbb{N}}).$
$None$

$B_{E}$ is weakly compact because $E$ is reflexive. So,
$x_{n_{k_{j}}}\stackrel{{\scriptstyle w}}{{\to}}x\in E$ for some further
subsequence $(x_{n_{k_{j}}})$ by the Eberlein – Smulian theorem. Since
$T\in\text{\rm aLwc}(E,F)$, then by [BLM18, Thm. 2.2],
$f_{n_{k_{j}}}(Tx_{n_{k_{j}}})\to 0$ which contradicts (2). Therefore
$f_{n}\rightrightarrows 0(T(B_{E}))$, and hence $T\in\text{\rm aLim}(E,F)$. ∎

######  Theorem 3.1.2.

Let $F\in\text{\rm(d)}$. Then the following holds.

  1. (i)

If $E^{\ast}\in\text{\rm(PSP)}$ then $\text{\rm
span(W}_{+}(E,F))\subseteq\text{\rm aLim}(E,F)$.

  2. (ii)

If $F\in\text{\rm(PGP)}$, then $\text{\rm span}(\text{\rm
aMwc}_{+}(E,F))\subseteq{\rm aLim}(E,F)$.

###### Proof.

(i) By [CW99, Thm. 3.3], $\text{\rm W}(E,F)\subseteq\text{\rm Mwc}(E,F)$.
Then, by [EMM13, Prop. 4.5], $\text{\rm span(W}_{+}(E,F))\subseteq\text{\rm
aLim}(E,F)$.

(ii) Let $T\in\text{\rm span}(\text{\rm aMwc}_{+}(E,F))$. WLOG, suppose $T$ is
positive. For proving $T\in\text{\rm aLim}(E,F)$, it suffices to show
$f_{n}(Tx_{n})\to 0$ for every disjoint w∗-null $(f_{n})$ in $F^{\ast}_{+}$
and disjoint $(x_{n})$ in $(B_{E})_{+}$ (see, [EMM13, Thm. 2.7]). Let
$(f_{n})$ be disjoint w∗-null in $F^{\ast}_{+}$ and let $(x_{n})$ be disjoint
in $(B_{E})_{+}$. As $F\in\text{\rm(PGP)}$, $f_{n}\stackrel{{\scriptstyle
w}}{{\to}}0$. By Definition 1.5.3 d), $f_{n}(Tx_{n})\to 0$. Consequently
$T\in\text{\rm aLim}(E,F)$. ∎

######  Theorem 3.1.3.

If $F\in\text{\rm(d)}$ then $l\text{\rm-Lwc}(E,F)\cap\text{\rm
Mwc}(E,F)\subseteq\text{\rm aLim}(E,F)$ for every $E$.

###### Proof.

Let $T\in l\text{\rm-Lwc}(E,F)\cap\text{\rm Mwc}(E,F)$. We need to show that
$T(B_{E})$ is almost limited. For this it suffices to show that
$f_{n}(Tx_{n})\to 0$ for every disjoint $(x_{n})$ in $(B_{E})_{+}$ and every
disjoint w∗-null $(f_{n})$ in $B_{F^{\ast}}$ [EMM13, Thm. 2.7]. Let $(x_{n})$
be disjoint in $(B_{E})_{+}$ and $(f_{n})$ be disjoint w∗-null in
$B_{F^{\ast}}$. Since $T\in l\text{\rm-Lwc}(E,F)$, then
$(T^{\ast}f_{n})\stackrel{{\scriptstyle w^{\ast}}}{{\to}}0$ in $E^{\ast}$ by
[AEG23, Thm. 2.7]. As $f_{n}\stackrel{{\scriptstyle w^{\ast}}}{{\to}}0$,
$\|f_{n}\|\leq M$ for all $n$. Since $T\in\text{\rm Mwc}$ then $\|Tx_{n}\|\to
0$, and hence

$|f_{n}(Tx_{n})|\leq\|f_{n}\|\cdot\|Tx_{n}\|\leq M\cdot\|Tx_{n}\|$

implies $f_{n}(Tx_{n})\to 0$ as desired. ∎

### 3.2

Consider some conditions on an operator to be L-weakly compact.

######  Theorem 3.2.1.

The following are equivalent.

  1. (i)

$F$ has o-continuous norm.

  2. (ii)

For every $X$, $\text{\rm aLim}(X,F)\subseteq\text{\rm Lwc}(X,F)$.

  3. (iii)

For every $X$, $\text{\rm Lim}(X,F)\subseteq\text{\rm Lwc}(X,F)$.

  4. (iv)

For every $E$, each positive rank one operator $T:E\to F$ is Lwc.

###### Proof.

(i)$\Longrightarrow$(ii): Let $T\in\text{\rm aLim}(X,F)$. Then $T(B_{X})$ is
an aLim-set in $F$. By [CCJ14, Thm. 2.6], $T(B_{X})$ is an Lwc-set, and hence
$T\in\text{\rm Lwc}(X,F)$.

(ii)$\Longrightarrow$(iii)$\Longrightarrow$(iv): It is trivial.

(iv)$\Longrightarrow$(i): Since, for each $y\in F_{+}$, there exists a
positive rank one operator $T:E\to F$ such that $T(E)=\text{\rm span}(y)$,
then $y\in F^{a}$ by [AB06, Thm. 5.66], and hence $F$ has o-continuous norm. ∎

### 3.3

Not every almost limited operator is Mwc. The identity of $\ell^{\infty}$ is
almost limited but it is neither M- nor L-weakly compact.

######  Theorem 3.3.1.

Suppose $E$ has o-continuous norm, $\dim(E)=\infty$, and $F\in\text{\rm(d)}$.
If $\text{\rm aLim}(E,F)\subseteq\text{\rm Mwc}(E,F)$, then $E^{\ast}$ and $F$
both have o-continuous norms.

###### Proof.

If $\text{\rm aLim}(E,F)\subseteq\text{\rm Mwc}(E,F)$ then $\text{\rm
K}(E,F)\subseteq\text{\rm Mwc}(E,F)$. o-Continuity of the norm in $E^{\ast}$
follows from [EAS20, Thm. 2].

Suppose now the norm of $F$ is not o-continuous. Then there exist some $u\in
F_{+}$ and a disjoint sequence $(u_{n})$ in $[0,u]$ such that
$\|u_{n}\|\not\to 0$. We may assume $\|u_{n}\|=1$ for all $n$. Since
$\dim(E)=\infty$, there exists a disjoint normalized sequence $(x_{n})$ in
$E_{+}$. By [Zaa83, Thm. 116.3 (iii)], there exists a disjoint sequence
$(g_{n})$ in $(B_{E^{\ast}})_{+}$ such that $\frac{1}{2}\leq g_{n}(x_{n})\leq
1$ for all $n$ and $g_{n}(x_{m})=0$ for $n\neq m$. Since $E$ has o-continuous
norm, then $g_{n}\stackrel{{\scriptstyle w^{\ast}}}{{\to}}0$ by [Zaa83, Thm.
116.2].

Define a positive operator $R:E\to c_{0}$ by $Rx=(g_{k}(x))_{k=1}^{\infty}$.
Clearly $R(B_{E})\subseteq B_{c_{0}}$. Define a positive operator $S:c_{0}\to
F$ by $S\alpha=\sum_{n}\alpha_{n}u_{n}$. Then $S(B_{c_{0}})\subseteq[-u,u]$,
and hence $SR(B_{E})\subseteq[-u,u]$. As order intervals in $F$ are almost
limited, $SR(B_{E})$ is almost limited, and hence $SR\in\text{\rm aLim}(E,F)$.
Since $SR(x_{n})=S(g_{n}(x_{n})\cdot e_{n})=g_{n}(x_{n})\cdot u_{n}$, then
$\|SR(x_{n})\|\geq|g_{n}(x_{n})|\geq\frac{1}{2}$ for all $n$. Therefore
$SR\not\in\text{\rm Mwc}(E,F)$. A contradiction. ∎

### 3.4

Positively limited operators are introduced in [AC23].

######  Definition 3.4.1.

$T:E\to F$ is called positively limited if $\|T^{\ast}(f_{n})\|\to 0$ for each
positive w∗-null sequence $(f_{n})$ in $F^{\ast}$.

We need the following lemma (cf. [AC23, Thm. 2.2]).

######  Lemma 3.4.2.

$T\in\text{\rm Lim}_{+}(E,F)$ iff $T^{\ast}(y^{\ast}_{n})(x_{n})\to 0$ for
each $y_{n}^{\ast}\stackrel{{\scriptstyle w^{\ast}}}{{\to}}0$ in
$F^{\ast}_{+}$ and every disjoint bounded sequence $(x_{n})$ in $E_{+}$.

######  Theorem 3.4.3.

Suppose $E^{\ast}$ and $F$ both have o-continuous norms, $F$ has limited order
intervals, and each disjoint bounded sequence of $E$ is order bounded. Then
each regular operator $T:E\to F$ is almost limited.

###### Proof.

WLOG, suppose that $T:E\to F$ is positive. $T$ admits a factorization over a
Banach lattice $G$ with $G$ and $G^{\ast}$ both have o-continuous norms, say
$T=SR$, where $R$ and $S$ are both positive as in Proposition 2.2.1. We claim
$R$ is positively limited. Let $(z_{n}^{\ast})$ be a positive w∗-null sequence
in $G^{\ast}$ and $(x_{n})$ be a positive disjoint bounded sequence in $E$. By
the assumption, there exists $x$ with $0\leq x_{n}\leq x$ for each $n$. As $R$
is positive, $0\leq Rx_{n}\leq Rx$. Thus $0\leq
R^{\ast}(z_{n}^{\ast})(x_{n})\leq z_{n}^{\ast}(Rx_{n})\leq z_{n}^{\ast}(Rx)$
which shows $R^{\ast}(z_{n}^{\ast})(x_{n})\to 0$. By Lemma 3.4.2, we conclude
that $R$ is positively limited.

Now we show $T=SR\in\text{\rm aLim}(E,F)$. Let $(u_{n}^{\ast})$ be disjoint
w∗-null in $F^{\ast}$. Since $F^{\ast}$ has w∗-continuous lattice operations,
$|u_{n}^{\ast}|\stackrel{{\scriptstyle w^{\ast}}}{{\to}}0$ in $F^{\ast}$. As
$R$ is positively limited, $\|R^{\ast}S^{\ast}|u^{\ast}_{n}|\|\to 0$. Since
$R^{\ast}|S^{\ast}u^{\ast}_{n}|\leq R^{\ast}S^{\ast}|u^{\ast}_{n}|$, we have

$|R^{\ast}S^{\ast}u^{\ast}_{n}|\leq R^{\ast}|S^{\ast}u^{\ast}_{n}|\leq
R^{\ast}S^{\ast}|u^{\ast}_{n}|.$

Resulting that $\|R^{\ast}S^{\ast}u^{\ast}_{n}\|=\|T^{\ast}u_{n}\|\to 0$. It
follows $T\in\text{\rm aLim}(E,F)$. ∎

It is worth noting that the both conditions on $E$ in Theorem 3.4.3 can be
replaced by a single condition that $E$ is an AM-space in which every disjoint
bounded sequence is order bounded. Indeed, by the Abramovich result, if $E$ is
not isomorphic to an AM-space then $E$ contains a positive disjoint normalized
sequence $(x_{n})$ such that $\|\vee_{n=1}^{m}x_{n}\|\to\infty$, and hence
such a sequence $(x_{n})$ can not be order bounded [Abr78]. Thus, $E$ must be
(isomorphic to) an AM-space, and hence the norm in $E^{\ast}$ is o-continuous.
Also note that this condition on $E$ is properly weaker than the condition
that $E$ is an AM-space with unit, as the space $C_{\omega}(\Gamma)$ of real-
valued bounded countably supported functions on an uncountable set $\Gamma$
still satisfies the condition that every disjoint bounded sequence is order
bounded, yet $C_{\omega}(\Gamma)$ has no even a weak unit.

######  Theorem 3.4.4.

Let $T:E\to E$ be a positive operator in the center of $E$. Suppose
$E\in\text{\rm(DPSP)}$. Then $T$ is positively limited.

###### Proof.

For simplicity suppose $0\leq T\leq I$. Let $(f_{n})$ be a positive w∗-null
sequence in $E^{\ast}$. Then $0\leq T^{\ast}\leq I$. Hence $0\leq
T^{\ast}f_{n}\leq f_{n}$ for all $n$. Since $E\in\text{\rm(DPSP)}$,
$\|f_{n}\|\to 0$ hence $T^{\ast}f_{n}\to 0$. ∎

### 3.5

It was shown in [EAS20, Thm. 1] that $\text{\rm K}(X,F)\subseteq\text{\rm
aLwc}(X,F)$ for all $X$ iff $F$ has o-continuous norm. We extend this to
almost limited operators.

######  Theorem 3.5.1.

$\text{\rm aLim}(X,F)\subseteq\text{\rm aLwc}(X,F)$ iff $F$ has o-continuous
norm.

###### Proof.

The necessity follows from [EAS20, Thm. 1] as $\text{\rm
K}(X,F)\subseteq\text{\rm aLim}(X,F)$.

For the sufficiency, suppose in contrary that $\text{\rm
aLim}(X,F)\subseteq\text{\rm aLwc}(X,F)$ but the norm in $F$ is not
o-continuous. Then there exists $y_{0}\in F\setminus F^{a}$. We choose
$x_{0}\in X$ and $h\in X^{\ast}$ such that $h(x_{0})=\|x_{0}\|=1$ and define
$Tx=h(x)y_{0}$. The operator $T$ is almost limited as it is compact. However
$T\not\in\text{\rm aLwc}(X,F)$ as $\\{Tx_{0}\\}=\\{y_{0}\\}$ is not an Lwc-
set. A contradiction. ∎

### 3.6

By [EAS20, Thm. 2], $\text{\rm K}(E,X)\subseteq\text{\rm aMwc}(E,X)$ iff
$E^{\ast}$ is a KB-space. We extend this result as follows.

######  Proposition 3.6.1.

If $E^{\ast}$ is a KB-space then $\text{\rm aLim}_{+}(E,F)\subseteq\text{\rm
aMwc}(E,F)$.

###### Proof.

Assume that $E^{\ast}$ is a KB-space and $T\in\text{\rm aLim}_{+}(E,F)$. Then
$T^{\ast}\in\text{\rm aDP}_{+}(F^{\ast},E^{\ast})$ [Elb17, Thm. 1].
Consequently $T^{\ast}\in\text{\rm aLwc}(F^{\ast},E^{\ast})$ by [BLM18, Prop.
2.4]. Then $T\in\text{\rm aMwc}(E,F)$ by [BLM18, Thm. 2.5]. ∎

Our next result gives some conditions for positive weakly compact operators to
be almost limited.

######  Proposition 3.6.2.

Suppose $E=E^{a}$ and $\text{\rm W}(X,E)\subseteq\text{\rm aLim}(X,E)$. Then
$E$ is a KB-space or $X\in\text{\rm(DPP)}$.

###### Proof.

Suppose $E$ is not a KB-space. We need to show $X\in\text{\rm(DPP)}$. Since
$E$ is not a KB-space, it contains a lattice copy (say $F$) of $c_{0}$. Then
$F$ is a closed linear span of some disjoint sequence $(y_{n})$ in $E_{+}$
such that $d=\inf_{n}\|y_{n}\|>0$ and $\sup_{m}\|\sum_{i=1}^{m}y_{i}\|<\infty$
by [Wnu90, Prop. 0.5.1]. The operator

$S\beta=\sum_{k=1}^{\infty}\beta_{k}y_{k}\quad\quad(\beta=(\beta_{k})_{k}\in
c_{0})$

is a lattice homomorphism from $c_{0}$ onto $F$. There exists a positive
disjoint $(f_{n})$ in $B_{E^{\ast}}$ such that $f_{n}(y_{n})\geq d$ for all
$n$ and $f_{n}(y_{m})=0$ if $n\neq m$. Take $B_{X}\ni
x_{n}\stackrel{{\scriptstyle w}}{{\to}}0$ and $B_{X^{\ast}}\ni
x^{\ast}_{n}\stackrel{{\scriptstyle w}}{{\to}}0$. Define $R:X\to c_{0}$ by

$Rx=(x^{\ast}_{k}(x))_{k}.$

Then $R\in\text{\rm W}(X,c_{0})$ by [AB06, Thm. 5.26]. Therefore
$T=SR\in\text{\rm W}(X,E)$ and hence $T\in\text{\rm aLim}(X,E)$ by the
hypothesis. Since $E$ has o-continuous norm, the bounded disjoint sequence
$(f_{n})$ is w∗-null and therefore $\|T^{\ast}f_{n}\|\to 0$ as $T\in\text{\rm
aLim}(X,E)$. Since $(x_{n})$ is bounded and
$|f_{n}(Tx_{n})|\leq\|x_{n}\|\cdot\|T^{\ast}f_{n}\|$, these yield
$|f_{n}(Tx_{n})|\to 0$. Thus, it follows from

$|f_{n}(Tx_{n})|=|f_{n}(S(R(x_{n})))|=\left|f_{n}\left(\sum_{k=1}^{\infty}[R(x_{n})]_{k}y_{k}\right)\right|=$
$|f_{n}([R(x_{n})]_{n}y_{n})|=|f_{n}([(x^{\ast}_{k}(x_{n}))_{k}]_{n}y_{n})|=|f_{n}(x^{\ast}_{n}(x_{n})y_{n})|=$
$|x^{\ast}_{n}(x_{n})f_{n}(y_{n})|=|x^{\ast}_{n}(x_{n})|\cdot|f_{n}(y_{n})|\geq|x^{\ast}_{n}(x_{n})|\cdot
d$

that $x^{\ast}_{n}(x_{n})\to 0$, and hence $X\in\text{\rm(DPP)}$. ∎

######  Corollary 3.6.3.

$\text{\rm W}(X,c_{0})\subseteq\text{\rm aLim}(X,c_{0})$ $\Longrightarrow$
$X\in\text{\rm(DPP)}$.

######  Theorem 3.6.4.

Suppose $E^{\ast}$ and $F$ both have o-continuous norms then $\text{\rm
span}(\text{\rm aLim}_{+}(E,F))\subseteq\text{\rm Mwc}(E,F)$.

###### Proof.

Let $T\in\text{\rm span}(\text{\rm aLim}_{+}(E,F))$, WLOG $T\geq 0$. Let
$(x_{n})$ be a disjoint sequence in $B_{E}$, we need to show $\|Tx_{n}\|\to
0$. Since $E^{\ast}$ is a KB-space, $x_{n}\stackrel{{\scriptstyle w}}{{\to}}0$
in $E$. Since $(x_{n})$ is disjoint, $|x_{n}|\stackrel{{\scriptstyle
w}}{{\to}}0$ (cf. [Wnu13, Prop. 1.3]), and hence
$T|x_{n}|\stackrel{{\scriptstyle w}}{{\to}}0$ in $F$. As $T$ is positive,
$|Tx_{n}|\leq T|x_{n}|$ and $|Tx_{n}|\stackrel{{\scriptstyle w}}{{\to}}0$ in
$F$. It remains to show $f_{n}(Tx_{n})\to 0$ for each disjoint bounded
sequence $(f_{n})$ in $F^{\ast}$. This follows from
$|f_{n}(Tx_{n})|\leq\|T^{\ast}f_{n}\|\|x_{n}\|$. ∎

######  Theorem 3.6.5.

Let $E$ be an AL-space and let $F$ have o-continuous norm. Then each
$T\in\text{\rm aLim}(E,F)$ has an almost limited modulus.

###### Proof.

As $F^{a}=F$ and $T$ is almost limited, then $T$ is Lwc. Hence $T$ has the
modulus $|T|$ that is Lwc by [CW99, Thm. 2.4]. Therefore $|T|$ is almost
limited by [CCJ14, Thm. 2.6]. ∎

However [CW99, Thm. 2.2] shows there is a regular compact (hence limited)
operator from $E=L^{2}[0,1]$ to $F=c(L^{2}[0,1])$ which is both M- and
L-weakly compact whose modulus does not exist.

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