#  $S$-decomposable Banach lattices, optimal sequence spaces and interpolation

Sergey V. Astashkin  Astashkin: Department of Mathematics, Samara National
Research University, Moskovskoye shosse 34, 443086, Samara, Russian
Federation; Department of Mathematics, Bahcesehir University, 34353, Istanbul,
Turkey<EMAIL_ADDRESS>and  Per G. Nilsson  Nilsson: Stockholm, Sweden
<EMAIL_ADDRESS>
###### Abstract.

We investigate connections between upper/lower estimates for Banach lattices
and the notion of relative $s$-decomposability, which has roots in
interpolation theory. To get a characterization of relatively $s$-decomposable
Banach lattices in terms of the above estimates, we assign to each Banach
lattice $X$ two sequence spaces $X_{U}$ and $X_{L}$ that are largely
determined by the set of $p$, for which $l_{p}$ is finitely lattice
representable in $X$. As an application, we obtain an orbital factorization of
relative $K$-functional estimates for Banach couples $\vec{X}=(X_{0},X_{1})$
and $\vec{Y}=(Y_{0},Y_{1})$ through some suitable couples of weighted
$L_{{p}}$-spaces provided if $X_{i},Y_{i}$ are relatively $s$-decomposable for
$i=0,1$.

Also, we undertake a detailed study of the properties of optimal upper and
lower sequence spaces $X_{U}$ and $X_{L}$, and, in particular, prove that
these spaces are rearrangement invariant. In the Appendix, a description of
the optimal upper sequence space for a separable Orlicz space as a certain
intersection of some special Musielak-Orlicz sequence spaces is given.

###### Key words and phrases:

Banach lattice, $s$-relative decomposable couples, relative decomposable
couples, lower, upper estimates , interpolation, Calderón-Mityagin property

###### 2010 Mathematics Subject Classification:

Primary 46B70; Secondary 46B42, 15A15

The work of the first author was completed as a part of the implementation of
the development program of the Volga Region Scientific and Educational
Mathematical Center (agreement no. 075-02-2023-931.

##  1\. Introduction

This paper has roots in the classification problem of the interpolation theory
of operators, see for instance Peetre [25], i.e., the problem of
identification of equivalence classes of Banach couples with the ”same”
interpolation structure111For standard definitions and notation used in the
interpolation theory, see e.g. [3],[4], [5], [25].. Specifically, there are
close connections of the topic of this paper with the so-called Calderón-
Mityagin property of Banach couples, which often allows to describe
effectively the class of all interpolation spaces with respect to them. Let us
recall this notion.

Assuming that $\overrightarrow{X}=(X_{0},X_{1})$ and
$\overrightarrow{Y}=(Y_{0},Y_{1})$ are Banach couples, we consider the
following two properties of elements $x\in X_{0}+X_{1}$ and $y\in
Y_{0}+Y_{1}$:

(1.1) $y=Tx\text{ for some bounded linear operator }T:\,X_{i}\rightarrow
Y_{i},\;i=0,1,$

and

(1.2) $K(t,y;\overrightarrow{Y})\leq C\cdot K(t,x;\overrightarrow{X})\text{
for some constant }C\text{ and }t>0.$

Here, $K(t,x;\overrightarrow{X})$ is the Peetre $K$-functional defined for all
$x\in X_{0}+X_{1}$ and $t>0$ by

$K(t,x;\overrightarrow{X}):=\inf\\{\|x_{0}\|_{X_{0}}+t\|x_{1}\|_{X_{1}}:\,x=x_{0}+x_{1},\,x_{i}\in
X_{i}\\}.$

It is easy to check that condition (1.1) implies (1.2) with
$C=\max_{i=0,1}\|T\|_{{X_{i}}\rightarrow{Y_{i}}}$. If the converse implication
holds (for all $x\in X_{0}+X_{1}$ and $y\in Y_{0}+Y_{1}$), the couples
$\overrightarrow{X}$ and $\overrightarrow{Y}$ are said to have the relative
Calderón-Mityagin property (in brief, $\mathcal{C-M}$ property). In this case
all relative interpolation spaces with respect to these couples can be
described by using $K$-functionals in a precise quantitative manner, see for
instance [4, Theorem 4.1.11].

A result of this type related to the finite dimensional couples
$\left(l_{1}^{n},l_{\infty}^{n}\right)$ goes back to Hardy-Littlewood-Polya,
see [27, Theorem 5.5]. As the name of the term suggests, the first example of
a $\mathcal{C-M}$ couple in the non-discrete infinite dimensional setting was
obtained independently by Calderón [6] and Mityagin [24]. Their result that
the Banach couple $(L_{1},L_{\infty})$ over an arbitrary $\sigma$-finite
measure space has the $\mathcal{C-M}$ property marked the start of searching
for other such Banach couples. Early examples are pairs of weighted
$L_{\infty}$-spaces, Peetre [25], and $L_{1}$-spaces, Sedaev-Semenov, [28].
Over time, this property has been verified for a large number of Banach
couples (and also quasi-Banach couples, couples of normed abelian groups); cf.
[4, Sec. 4.7.2], [11, p. 2 onwards], [2].

In the paper [9], Cwikel have found a general condition, which guarantees that
two given couples of Banach lattices have the relative $\mathcal{C-M}$
property. Let $X$ and $Y$ be Banach lattices of measurable functions (possibly
having different underlying measure spaces). Then $X,Y$ are called relatively
decomposable if for any sequences $\\{x_{n}\\}_{n=1}^{\infty}\subset X$ and
$\\{y_{n}\\}_{n=1}^{\infty}\subset Y$ of elements with pair-wise disjoint
supports such that $\sum_{n=1}^{\infty}x_{n}\in X$ and
$\left\|y_{n}\right\|_{Y}\leq\left\|x_{n}\right\|_{X}$, $n\in N$, we have
$\sum_{n=1}^{\infty}y_{n}\in Y$ and

(1.3) $\left\|\sum_{n=1}^{\infty}y_{n}\right\|_{Y}\leq
D\left\|\sum_{n=1}^{\infty}x_{n}\right\|_{X},$

for some constant $D$ independent of $\\{x_{n}\\}_{n=1}^{\infty}$ and
$\\{y_{n}\\}_{n=1}^{\infty}$. As is proved in [9] (see also [4, Theorem
4.4.29]), couples of Banach lattices $\overrightarrow{X}=(X_{0},X_{1})$ and
$\overrightarrow{Y}=(Y_{0},Y_{1})$ have the relative Calderón-Mityagin
property whenever both pairs $X_{0}$, $Y_{0}$ and $X_{1}$, $Y_{1}$ are
relatively decomposable.

Let us mention that the latter notion is closely connected with
$L_{p}$-spaces. In particular, by an appropriate form of the Kakutani-
Bohnenblust-Tzafriri representation theorem, any decomposable lattice $X$ of
measurable functions on $\Omega$ (i.e., when $X=Y$), which is also
$\sigma$-order continuous and has the Fatou property, coincides, for some
$p\in[1,\infty)$, with an $L_{p}$-space of functions supported on some
measurable subset $\Omega^{\prime}$ of $\Omega$ for some suitable measure
defined on $\Omega^{\prime}$ (see Proposition 1.4 in [10, p. 58]). Observe
also that there is a simple sufficient condition for couples of Banach
lattices to be relatively decomposable. Namely, if $Y$ satisfies an upper
$p$-estimate and $X$ a lower $p$-estimate, then the Banach lattices $X$ and
$Y$ are relative decomposable.

The notion of relative decomposibility plays a central role in the paper [11],
where it was proved that all weighted couples modelled on two given couples
$\overrightarrow{X}=(X_{0},X_{1})$ and $\overrightarrow{Y}=(Y_{0},Y_{1})$ of
Banach lattices of measurable functions over $\sigma$ -finite measure spaces
possess the relative $\mathcal{C-M}$ property if and only if both pairs
$X_{0},Y_{0}$ and $X_{1},Y_{1}$ are relatively decomposable. Therefore, in the
case when $\overrightarrow{X}=\overrightarrow{Y}$ and $X_{0},X_{1}$ are
$\sigma$-order continuous and have the Fatou property, the latter property of
a couple $\overrightarrow{X}$ implies that both $X_{0}$ and $X_{1}$ are
$L_{p}$-spaces, which is a strong converse to the well-known result of Sparr
[30], asserting that each weighted $L_{p}$-couple has the relative
$\mathcal{C-M}$ property.

Of course, not all pairs of Banach couples possess the relative
$\mathcal{C-M}$ property and this motivated Cwikel to consider a weakened
version of that. Specifically, already in the papers [7] and [8], condition
(1.2) is replaced with the inequality

(1.4) $K(t,y;\overrightarrow{Y})\leq
w(t)K(t,x;\overrightarrow{X}),\;t>0,\;\;\mbox{with}\;\int\nolimits_{0}^{\infty}w(u)^{s}du/u<\infty,$

for some (fixed) function $w(u)\geq 0$ and $s\in[1,\infty]$ (if $s=\infty$,
after the usual modification of the condition imposed on $w$, we come
certainly to the definition of relative $\mathcal{C-M}$ property). In turn,
this led to the introduction of the following more general concept of
relatively $s$-decomposable pairs of Banach lattices.

Let $1\leq s\leq\infty$. A pair of Banach lattices $X$, $Y$ is said to be
relatively $s$-decomposable whenever for all sequences
$\\{x_{n}\\}_{n=1}^{\infty}\subset X$, $\\{y_{n}\\}_{n=1}^{\infty}\subset Y$
of pair-wise disjoint elements such that $\sum_{n=1}^{\infty}x_{n}\in X$ and
$\left\|y_{n}\right\|_{Y}\leq\left\|x_{n}\right\|_{X}$, $n\in N$, we have
$\sum_{n=1}^{\infty}y_{n}\in Y$ and

(1.5) $\left\|\sum_{n=1}^{\infty}\lambda_{n}y_{n}\right\|_{Y}\leq
D\left(\sum_{n=1}^{\infty}\left|\lambda_{n}\right|^{s}\right)^{1/s}\left\|\sum_{n=1}^{\infty}x_{n}\right\|_{X},$

for some constant $D$ and every sequence $\\{\lambda_{n}\\}_{n=1}^{\infty}\in
l_{s}$ (again with the usual modification in the case $s=\infty$ that gives
(1.3), i.e., the relative decomposibility). According to [9] (see also [4,
Remark 4.4.33]), if $\overrightarrow{X}=(X_{0},X_{1})$ and
$\overrightarrow{Y}=(Y_{0},Y_{1})$ are two couples of Banach lattices such
that the pairs $X_{0}$, $Y_{0}$ and $X_{1}$, $Y_{1}$ are relatively
$s$-decomposable, for every $x\in X_{0}+X_{1}$ and $y\in Y_{0}+Y_{1}$
satisfying condition (1.4) we have $y=Tx$ for some linear operator
$T:X_{i}\rightarrow Y_{i}$, $i=0,1$.

One of the main results of the paper [11] is a characterization of the
relative decomposability in the setting of Banach lattices of measurable
functions, implying that the above-mentioned trivial sufficient condition
expressed in terms of upper and lower estimates is also necessary. In the case
of relative $s$-decomposable pairs of Banach lattices $X$, $Y$ there is also a
simple sufficient condition, formulated in terms of upper estimates for $Y$
and lower estimates for $X.$ The main aim  of this paper is to prove that, in
a more general setting of abstract Banach lattices, this trivial sufficient
condition is also necessary (as in the case of relative decomposability, i.e.,
when $s=\infty$).

A pivotal role in the proof of our main result is played by the notions of
optimal upper and lower sequence spaces, introduced in this paper.
Specifically, we associate to every Banach lattice $X$ two sequence spaces
$X_{U}$ and $X_{L}$, which rather precisely reflect lattice properties of $X$,
in particular, encoding the optimal upper and lower estimate information,
respectively. Section 3 is devoted to a detailed study of the properties of
these spaces, which, as we believe, can be useful tools also when considering
other issues related to Banach lattices. As a result, in Section 4 we present
the proof of Theorem 2.8, which gives a solution of the above problem. On the
way, we obtain also other results related to optimal sequence spaces. We show
that if Banach lattices $X$ and $Y$ are relatively $s$-decomposable, then the
space of multiplicators from $X_{L}$ into $Y_{U}$ with respect to coordinate-
wise multiplication includes the space $l_{s}$ (see Proposition 4.1). Another
important ingredient in the proof of Theorem 2.8 is the relationship between
the construction of optimal sequence spaces $X_{U}$ and $X_{L}$ and the finite
lattice representability of $l_{r}$-spaces in $X$.

Section 5 contains some applications of Theorem 2.8 to the interpolation
theory. In particular, in Theorem 5.2 we prove that Banach lattices $X$ and
$Y$ being relatively $s$-decomposable admit an orbital factorization of
relative $K$-functional estimates through some suitable couples of weighted
$L_{{p}}$-spaces. In the next section we present the full proof of
rearrangement invariance of the optimal upper and lower sequence spaces (see
Theorem 3.3).

Finally, in the Appendix we identify the optimal upper sequence space for a
separable Orlicz space $L_{M}$ on $[0,1]$. Namely, in Theorem 6.14, we prove
that the space $(L_{M})_{U}$ can be described as a certain intersection of
some special Musielak-Orlicz sequence spaces.

It is worth noting that, in contrast to [9] and [4], in this paper we use a
weaker version of finite relative $s$-decomposibility, which involves
estimates (1.5) only for finite sequences. In a certain sense, this approach
seems to be more natural, since our main result reveals the relationship
between this property and upper/lower estimates of Banach lattices involved,
whose definition contains only finite sequences of elements of these lattices
as well. Observe that the above versions of the definition of relative
$s$-decomposibility coincide if $X$, $Y$ are Banach lattices of measurable
functions such that $Y$ has the Fatou property222A Banach lattice $X$ of
measurable functions on a $\sigma$-finite measure space
$\left(T,\Sigma,\mu\right)$ is said to have the Fatou property if the
conditions $x_{n}\in X$, $n=1,2,\dots$,
$\sup_{n=1,2,\dots}\|x_{n}\|_{X}<\infty$ and $x_{n}\to x$ a.e. on $T$ imply
that $x\in X$ and $\|x\|_{X}\leq\liminf_{n\to\infty}\|x_{n}\|_{X}$. (in this
case only, the concept of relative $s$-decomposibility is applied in [9] (see
also [4]) to the study of interpolation properties of Banach couples).

Some remarks around the history of this paper. Already in [11, Theorem 1.3, p.
98] the classification of decomposable pairs of Banach lattices on measure
spaces was addressed. In fact, the corresponding results for the general case
were also announced there (see [11, p. 100]). This problem was presented to
the second author by Michael Cwikel back in 2003. Some preliminary results
were presented by the second author during Jaak Peetre’s ”Summer Seminar” in
Lund, 2003. Time flies, and resulted the in the present paper 20 years later.
The authors thank Michael Cwikel for his insight and contributions to this
paper.

##  2\. Some preliminaries and statements of the main results.

We will assume that the reader is familiar with the definition of Banach
lattices, their basic properties and some basic terminology (see, for
instance, [20],[22],[27]). In particular, two elements $x,y$ from a Banach
lattice $X$ are said to be disjoint  if they satisfy
$\left|x\right|\wedge\left|y\right|=0$. For a given Banach space (lattice), we
set $B_{X}:=\left\\{x\in X:\,\left\|x\right\|_{X}\leq 1\right\\}$ and
$S_{X}:=\left\\{x\in X:\,\left\|x\right\|_{X}=1\right\\}$.

Recall that a lattice $X$ is said to be $\sigma$-order complete if every order
bounded sequence in $X$ has a least upper bound (see e.g. [20, Definition
1.a.3]). A Banach lattice has a $\sigma$-order continuous norm if every
positive, increasing, norm bounded sequence converges in norm (see e.g. [27,
Definition 5.12], [20, Definition 1.a.6]).

Let $l_{p}\left(I\right)$ denote the Banach space of all sequences, indexed by
the set $I,$ which are absolutely $p$-summable if $1\leq p<\infty$ (resp.
bounded if $p=\infty$). In the case $I=\mathbb{N}$ we simply write $l_{p}$. As
usual, by $c_{0}=c_{0}(\mathbb{N})$ will be denoted the space of all sequences
tending to zero as $n\rightarrow\infty$. For definiteness, all Banach spaces
and lattices considered in this paper are assumed to be real.

Let $F_{1}$ and $F_{2}$ be two positive functions (quasinorms). We write
$F_{1}\preceq F_{2}$ if we have $F_{1}\leq CF_{2}$ for some positive constant
$C$ that does not depend on the arguments of $F_{1}$ and $F_{2}$. In the case
when both $F_{1}\preceq F_{2}$ and $F_{2}\preceq F_{1}$ we write $F_{1}\asymp
F_{2}$. For a finite set $E\subset\mathbb{N}$ we denote by $|E|$ cardinality
of $E$. Finally, if $F:\,\Omega\rightarrow\mathbb{R}$ is a function (resp.
$a=\left\\{a_{i}\right\\}_{i=1}^{\infty}$ is a sequence of real numbers), then
$\mathrm{supp}\,F:=\\{\omega\in\Omega:\,F(\omega)\neq 0\\}$ (resp.
$\mathrm{supp}\,a:=\\{i\in\mathbb{N}:\,a_{i}\neq 0\\}$).

###  2.1. Relative decomposability of Banach lattices

The following definition generalizes the first of two definitions in [9, p.
44] (see also [4, Definition 4.4.26, p. 597]) and has roots in the
interpolation theory of operators (see e.g. [9, Theorem 2], [4, Theorem
4.4.29, p. 598] and Section 1). However, as was already mentioned in Section
1, in contrast to [9] (and also to [11], where the case $s=\infty$ is
covered), we will use a weaker version of this notion, which involves only
finite sums of elements of given lattices.

######  Definition 2.1.

Let $1\leq s\leq\infty.$ Banach lattices $X$ and $Y$ are said to be (finitely)
relative $s$-decomposable  if there exists a constant $D>0$ such that for each
$n\in\mathbb{N}$ and for all sequences of pair-wise disjoint non-zero elements
$\left\\{x_{i}\right\\}_{i=1}^{n}\subset X$ and
$\left\\{y_{i}\right\\}_{i=1}^{n}\subset Y$ it holds

$\left\|\sum_{i=1}^{n}y_{i}\right\|_{Y}\leq
D\left(\sum_{i=1}^{n}\left(\left\|y_{i}\right\|_{Y}/\left\|x_{i}\right\|_{X}\right)^{s}\right)^{1/s}\left\|\sum_{i=1}^{n}x_{i}\right\|_{X}$

(with the usual modification in the case $s=\infty$). Let
$D_{s}=D_{s}\left(X,Y\right)$ denote the infimum of constants $D$ satisfying
the above condition and we refer to $D_{s}$ as the relative
$s$-decomposibility constant of $X$ and $Y.$

In what follows, we will suppress the word ”finitely”, although this is a
change of terminology as compared to the previous use of this term, both in
[9] and [4, Definition 2.2.16] (see also a related discussion in Section 1).

Definition 2.1 can be also stated equivalently as follows. Let
$\left\\{x_{i}\right\\}_{i=1}^{n}\subseteq
S_{X},\left\\{y_{i}\right\\}_{i=1}^{n}\subseteq S_{Y}$ be two sequences of
pair-wise disjoint elements. Then for all sequences
$\left\\{a_{i}\right\\}_{i=1}^{n}$ and $\left\\{b_{i}\right\\}_{i=1}^{n}$ of
scalars it holds

$\left\|\sum_{i=1}^{n}a_{i}b_{i}y_{i}\right\|_{Y}\leq
D\left(\sum_{i=1}^{n}\left|a_{i}\right|^{s}\right)^{1/s}\left\|\sum_{i=1}^{n}b_{i}x_{i}\right\|_{X}.$

Note that in the case $s=\infty$ Definition 2.1 reduces to the following: if
$\left\\{x_{i}\right\\}_{i=1}^{n}\subseteq
X,\left\\{y_{i}\right\\}_{i=1}^{n}\subseteq Y$ are two sequences of pair-wise
disjoint elements with $\left\|y_{i}\right\|_{Y}\leq\left\|x_{i}\right\|_{X}$,
$i=1,\dots,n$, then

$\left\|\sum_{i=1}^{n}y_{i}\right\|_{Y}\leq
D\left\|\sum_{i=1}^{n}x_{i}\right\|_{X}$

Following [11], we will say in this case that Banach lattices $X$ and $Y$ are
relatively decomposable (see Section 1).

In particular, by Hölder inequality, we have

######  Example 2.2.

If $1\leq p,q,s\leq\infty$ then

$D_{s}\left(l_{q},l_{p}\right)<\infty\iff\frac{1}{p}\leq\frac{1}{q}+\frac{1}{s}$

and $D_{s}\left(l_{q},l_{p}\right)=1$ whenever this constant is finite.

###  2.2. Upper and lower estimates for disjoint elements in Banach lattices
and the Grobler-Dodds indices

Let us start with recalling the notions of lower and upper estimates in Banach
lattices; see [20, Definition 1.f.4]. A Banach lattice $X$ is said to satisfy
an upper (resp. a lower) $p$-estimate, where $p\in\left[1,\infty\right]$, if
for some constant $M$ and all finite sequences of pair-wise disjoint elements
$\left\\{x_{i}\right\\}_{i=1}^{n}\subseteq X$ it holds

$\left\|\sum_{i=1}^{n}x_{i}\right\|_{X}\leq
M\left(\sum_{i=1}^{n}\left\|x_{i}\right\|_{X}^{p}\right)^{1/p}$

(resp.

$\left(\sum_{i=1}^{n}\left\|x_{i}\right\|_{X}^{p}\right)^{1/p}\leq
M\left\|\sum_{i=1}^{n}x_{i}\right\|).$

The infimum of all $M$, satisfying the above inequality, is denoted by
$M^{\left[p\right]}\left(X\right)$ and $M_{\left[p\right]}\left(X\right)$,
respectively. Note that every Banach lattice admits (trivially) an upper
$1$-estimate and a lower $\infty$-estimate.

The fact that a Banach lattice satisfies an upper or a lower $p$-estimate can
be equivalently expressed in terms of relative decomposability. More
explicitly, since
$M_{\left[p\right]}\left(l_{p}\right)=M^{\left[p\right]}\left(l_{p}\right)=1$,
$1\leq p\leq\infty$, one can immediately check the following (see also [11]):

######  Proposition 2.3.

Let $X$ be a Banach lattice and $1\leq p\leq\infty.$ We have:

$\left(i\right)$ $X$ satisfies an upper $p$-estimate if and only if $l_{p}$
and $X$ are relatively decomposable;

$\left(ii\right)$ $X$ satisfies a lower $p$-estimate if and only if $X$ and
$l_{p}$ are relatively decomposable.

Moreover, we have

$M^{\left[p\right]}\left(X\right)=D_{\infty}\left(l_{p},X\right),M_{\left[p\right]}\left(X\right)=D_{\infty}\left(X,l_{p}\right).$

A direct application of the above definitions (see also Example 2.2) gives the
following useful result.

######  Proposition 2.4.

Assume that $X$ and $Y$ are Banach lattices such that $X$ satisfies a lower
$q$-estimate and $Y$ satisfies an upper $p$-estimate, where $1\leq
q,p\leq\infty$. If ${1}/{p}\leq{1}/{q}+{1}/{s}$, then $X,Y$ are relatively
$s$-decomposable and the following estimate holds:

$D_{s}\left(X,Y\right)\leq
M_{\left[q\right]}\left(X\right)M^{\left[p\right]}\left(Y\right).$

From the latter proposition and the trivial fact that every Banach lattice
satisfies an upper $1$-estimate and a lower $\infty$-estimate it follows that
any pair of Banach lattices is relatively $1$-decomposable. Hence, given a
pair of Banach lattices $X$ and $Y$, the set of all
$s\in\left[1,\infty\right]$ such that they are $s$-decomposable is always non-
empty and is of the form either $\left[1,s_{\max}\right]$ or
$\left[1,s_{\max}\right)$, where

$s_{\max}=s_{\max}(X,Y):=\sup\\{s\in\left[1,\infty\right]:\,X,Y\mbox{are
$s$-decomposable}\\}.$

Taking $X=l_{p}$, $Y=l_{1}$ and $s_{\max}$ so that $1/s_{\max}+1/p=1$, we
obtain an example of the first type because of $X$ and $Y$ are
$s$-decomposable if and only if $s\in\left[1,s_{\max}\right]$ (see Example
2.2). To get an example of the second type, let $Y$ be a Banach lattice that
satisfies, for a given $p_{\max}\in\left[1,\infty\right]$, an upper
$p$-estimate for all $p<p_{\max}$ but not for $p=p_{\max}.$ Then, taking
$l_{\infty}$ for $X$, we see that $X$ and $Y$ are $s$-decomposable if and only
if $s<p_{\max}$ (see Proposition 2.3(i)).

Recall that the Grobler-Dodds indices $\delta(X)$ and $\sigma(X)$ of a Banach
lattice $X$ are defined by

$\delta(X):=\sup\\{p\geq 1:\,X\;\text{satisfies an upper $p$-estimate}\\}$

and

$\sigma(X):=\inf\\{q\geq 1:\,X\;\text{satisfies a lower $q$-estimate}\\}.$

For every infinite-dimensional Banach lattice $X$ we have
$1\leq\delta(X)\leq\sigma(X)\leq\infty$. Moreover, the following duality
relations hold:

$\frac{1}{\delta(X)}+\frac{1}{\sigma(X^{\ast})}=1\;\;\mbox{and}\;\;\frac{1}{\sigma(X)}+\frac{1}{\delta(X^{\ast})}=1.$

######  Definition 2.5.

Let $1\leq p\leq\infty$. We say that $l_{p}$ is finitely lattice representable
in a Banach lattice $X$ whenever for every $n\in\mathbb{N}$ and each
$\varepsilon>0$ there exist pair-wise disjoint elements $x_{i}\in X$,
$i=1,2,\dots,n$, such that for any sequence $\left\\{a_{i}\right\\}_{i=1}^{n}$
of scalars we have

(2.1)
$\left(\sum_{i=1}^{n}\left|a_{i}\right|^{p}\right)^{1/p}\leq\left\|\sum_{i=1}^{n}a_{i}x_{i}\right\|_{X}\leq\left(1+\varepsilon\right)\left(\sum_{i=1}^{n}\left|a_{i}\right|^{p}\right)^{1/p}.$

Similarly, $l_{p}$ is said to be crudely finitely lattice representable in $X$
whenever instead of (2.1) it holds

$C^{-1}\left(\sum_{i=1}^{n}\left|a_{i}\right|^{p}\right)^{1/p}\leq\left\|\sum_{i=1}^{n}a_{i}x_{i}\right\|_{X}\leq
C\left(\sum_{i=1}^{n}\left|a_{i}\right|^{p}\right)^{1/p},$

where $C$ is a constant independent of $n\in\mathbb{N}$ and
$\left\\{a_{i}\right\\}_{i=1}^{n}$.

For the following result see, for instance, [20, Theorem 1.f.12.ii].

######  Proposition 2.6.

Let $X$ be a Banach lattice. Then $X$ admits a lower $p$-estimate for some
$p<\infty$ if and only if has $l_{\infty}$ fails to be finitely lattice
representable in $X$.

Moreover, in view of [20, Theorem 1.a.5, 1.a.7] and [13, p. 288], it follows

######  Proposition 2.7.

If a Banach lattice $X$ is not $\sigma$\- complete or has a not $\sigma$-order
continuous norm, then $l_{\infty}$ is finitely lattice representable in $X$.

###  2.3. The main result and its consequences

Now we are ready to state the main result of this paper, which gives a
characterization of relatively $s$-decomposable Banach lattices in terms of
their upper and lower estimates. This is an extension of results of the paper
[11], where the case $s=\infty$ was covered in a more restrictive setting of
Banach lattices of measurable functions. Note that the case when
$\delta\left(Y\right)\leq\sigma\left(X\right)$ is more interesting, because
then the lattices $X$ and $Y$ potentially may be not relatively decomposable.
As was mentioned in Section 1, a non-trivial part of the next theorem can be
also treated as the converse to Proposition 2.4.

######  Theorem 2.8.

Suppose $X$ and $Y$ are infinite dimensional Banach lattices. If
$\delta\left(Y\right)\leq\sigma\left(X\right)$ the following conditions are
equivalent:

$\left(i\right)$ $X$ and $Y$ are relatively $s$-decomposable;

$\left(ii\right)$ There exist $p,q$, with $1/p=1/q+1/s$, such that $X$
satisfies a lower $q$-estimate and $Y$ an upper $p$-estimate;

$\left(iii\right)$ There exist $p,q$ with $1/p=1/q+1/s$ such that $X,l_{q}$
and $l_{p},Y$ are relatively decomposable.

In addition, if

$F_{s}\left(X,Y\right):=\inf\left\\{M_{\left[q\right]}\left(X\right)M^{\left[p\right]}\left(Y\right):\frac{1}{s}=\frac{1}{p}-\frac{1}{q},1\leq
p\leq q\leq\infty\right\\},$

it holds

$D_{s}\left(X,Y\right)\leq F_{s}\left(X,Y\right)\leq
D_{s}\left(X,Y\right)^{2}.$

Moreover, $\sigma\left(X\right)\leq\delta\left(Y\right)$ if and only if
$s_{\max}(X,Y)=\infty$.

The proof of this theorem is presented in Section 4 below.

Now, by using standard arguments (see e.g. [20, Proposition 1.f.6]), one can
prove that Definition 2.1 is equivalent to the following assertion for general
sequences of elements in Banach lattices.

######  Corollary 2.9.

Infinite dimensional Banach lattices $X$ and $Y$ are relatively
$s$-decomposable if and only if there exists a constant $D_{s}>0$ such that
for each $n\in\mathbb{N}$ and all sequences
$\left\\{x_{i}\right\\}_{i=1}^{n}\subseteq S_{X}$,
$\left\\{y_{i}\right\\}_{i=1}^{n}\subseteq S_{Y}$ and
$\left\\{a_{i}\right\\}_{i=1}^{n}\subseteq\mathbb{R}$ we have

$\left\|\vee_{i=1}^{n}\left|a_{i}y_{i}\right|\right\|_{Y}\leq
D_{s}\left(\sum_{i=1}^{n}\left|a_{i}\right|^{s}\right)^{1/s}\left\|\vee_{i=1}^{n}\left|x_{i}\right|\right\|_{X}$

######  Corollary 2.10.

Suppose infinite dimensional Banach lattices $X$ and $Y$ are relatively
$s$-decomposable for some $s\in[1,\infty]$. Then, there exist equivalent norms
on $X$ and $Y$ such that $X$ and $Y$ are relatively $s$-decomposable Banach
lattices with constant one.

###### Proof.

By [22, Lemma $2.8.8$], every Banach lattice which satisfies a lower
$p$-estimate/an upper $q$-estimate admits an equivalent Banach lattice norm
such that the corresponding lower/upper estimate constant is equal to one.
Consequently, after a suitable renorming $X$ and $Y$, in the notation of
Theorem 2.8 we have $F_{s}\left(X,Y\right)=1$, which implies that
$D_{s}\left(X,Y\right)=1.$ ∎

###  2.4. Rearrangement invariant spaces

For a detailed theory of rearrangement invariant spaces we refer to the
monographs [20, 17].

Let $I=[0,1]$ or $(0,\infty)$ and let $m$ be the Lebesgue measure on $I$.
Given a measurable function $x(t)$ on $I$ we define its distribution function
by

$n_{x}(\tau):=m\\{t\in I:\,|x(t)|>\tau\\},\;\;\tau>0.$

Measurable functions $x(t)$ and $y(t)$ on $I$ are called equimeasurable if
$n_{x}(\tau)=n_{y}(\tau)$ for all $\tau>0$. In particular, each function
$x(t)$ on $I$ is equimeasurable with its non-increasing left-continuous
rearrangement $x^{\ast}(t)$ of $|x(t)|$, which defines by

$x^{\ast}(t):=\inf\\{\tau>0:\,n_{x}(\tau)<t\\},\;\;t\in I.$

######  Definition 2.11.

A Banach function space $X$ on $I$ is said to be rearrangement invariant (in
short, r.i.) (or symmetric) if the conditions $x\in X$ and $n_{y}(\tau)\leq
n_{x}(\tau)$ for all $\tau>0$ imply that $y\in X$ and
$\|y\|_{X}\leq\|x\|_{X}$.

Let $X$ be a r.i. space. If $I=[0,1]$ (resp. $I=(0,\infty)$) we have
$L_{\infty}\hookrightarrow X\hookrightarrow L_{1}$ (resp. $L_{\infty}\cap
L_{1}\hookrightarrow X\hookrightarrow L_{\infty}+L_{1}$).

The Köthe dual space $X^{\prime}$ consists of all measurable functions $y$
such that

$\|y\|_{X^{\prime}}:=\sup\\{\int_{I}|x(t)y(t)|\,dt:\,x\in X,\,\|x\|_{X}\leq
1\\}<\infty.$

Then, $X^{\prime}$ equipped with the norm $\|\cdot\|_{X^{\prime}}$ is a r.i.
space. Moreover, $X\subset X^{\prime\prime}$, and the isometric equality
$X=X^{\prime\prime}$ holds if and only if the norm in $X$ has the Fatou
property, that is, if the conditions $0\leq x_{n}\uparrow x$ a.e. on $I$ and
$\sup_{n\in{\mathbb{N}}}\|x_{n}\|<\infty$ imply $x\in X$ and
$\|x_{n}\|\uparrow\|x\|$.

The fundamental function $\phi_{X}$ of a r.i. space $X$ is defined by
$\phi_{X}(t):=\|\chi_{A}\|_{X}$, where $\chi_{A}$ is the characteristic
function of a measurable set $A\subset I$ with $m(A)=t$. The function
$\phi_{X}$ is quasi-concave (i.e., $\phi_{X}(0)=0$, $\phi_{X}$ increases and
$\phi_{X}(t)/t$ decreases).

Most important examples of r.i. spaces are the $L_{p}$-spaces, $1\leq
p\leq\infty$, and their natural generalization, the Orlicz spaces (for their
detailed theory we refer to the monographs [16, 26, 21]).

Let $M$ be an Orlicz function, that is, an increasing convex continuous
function on $[0,\infty)$ such that $M(0)=0$ and
$\lim_{t\rightarrow\infty}M(t)=\infty$. In what follows, we will assume also
that $M(1)=1$. Denote by $L_{M}:=L_{M}(I)$ the Orlicz space endowed with the
Luxemburg norm

$\|f\|_{L_{M}}:=\inf\left\\{\lambda>0\colon\int_{I}M\Big{(}\frac{|f(t)|}{\lambda}\Big{)}\,dt\leq
1\right\\}.$

In particular, if $M(u)=u^{p}$, $1\leq p<\infty$, we obtain $L_{p}$.

Note that the definition of an Orlicz function space $L_{M}[0,1]$ depends (up
to equivalence of norms) only on the behaviour of the function $M(t)$ for
large values of argument $t$. An easy calculation (see also formula (9.23) in
[16] on page 79 of the English version) shows that

(2.2) $\varphi_{L_{M}}(t)=\frac{1}{M^{-1}(1/t)},\;\;0<t\leq 1,$

where $M^{-1}$ is the inverse for $M$.

If $M$ is an Orlicz function, then the Young conjugate function $\tilde{M}$ is
defined by

$\tilde{M}(u):=\sup_{t>0}(ut-M(t)),\;\;u\geq 0.$

Moreover, $\tilde{M}$ is also an Orlicz function and the Young conjugate for
$\tilde{M}$ is $M$.

Every Orlicz space $L_{M}(I)$ has the Fatou property; $L_{M}[0,1]$ (resp.
$L_{M}(0,\infty)$) is separable if and only if the function $M$ satisfies the
$\Delta_{2}^{\infty}$-condition (resp. $\Delta_{2}$-condition), i.e.,
$\sup_{u\geq 1}{M(2u)}/{M(u)}<\infty$ (resp.
$\sup_{u>0}{M(2u)}/{M(u)}<\infty$). In this case we have
$L_{M}(I)^{\ast}=L_{M}(I)^{\prime}=L_{\tilde{M}}(I)$.

Let $1<q<\infty$, $1\leq r<\infty$. The Lorentz space $L_{q,r}=L_{q,r}(I)$
consists of all measurable functions $x$ such that

$\|x\|_{q,r}:=\Big{(}\frac{r}{q}\int_{I}(t^{1/q}x^{\ast}(t))^{r}\frac{dt}{t}\Big{)}^{1/r}<\infty.$

The functional $x\mapsto\|x\|_{q,r}$ is not subadditive, but it is equivalent
to the norm $x\mapsto\|x^{\ast\ast}\|_{q,r}$, where
$x^{\ast\ast}(t):=\frac{1}{t}\int_{0}^{t}x^{\ast}(s)\,ds$, $t>0$. Moreover,
$L_{q,r_{1}}\hookrightarrow L_{q,r_{2}}$, $1\leq r_{1}\leq r_{2}<\infty$ and
$L_{q,q}=L_{q}$ isometrically.

Rearrangement invariant (r.i.) sequence spaces are defined quite similarly. In
particular, the fundamental function of a r.i. sequence space $X$ is defined
by $\phi_{X}(n):=\|\sum_{k=1}^{n}e_{k}\|_{X}$, $n=1,2,\dots$. In what follows,
$e_{k}$ are the canonical unit vectors, i.e.,
$e_{k}=(e_{k}^{i})_{i=1}^{\infty}$, $e_{k}^{i}=0$ for $i\neq k$ and
$e_{k}^{k}=1$, $k,i=1,2,\dots$.

Recall that an Orlicz sequence space $\ell_{\psi}$, where $\psi$ is an Orlicz
function, consists of all sequences $(a_{k})_{k=1}^{\infty}$ such that

$\|(a_{k})\|_{\ell_{\psi}}:=\inf\left\\{u>0:\sum_{k=1}^{\infty}\psi\Big{(}\frac{|a_{k}|}{u}\Big{)}\leq
1\right\\}<\infty.$

Clearly, if $\psi(t)=t^{p}$, $p\geq 1$, then $\ell_{\psi}=\ell^{p}$
isometrically.

The fundamental function of an Orlicz sequence space $\ell_{\psi}$ can be
calculated by the formula: $\phi_{\ell_{\psi}}(n)=\frac{1}{\psi^{-1}(1/n)}$,
$n=1,2,\dots$ Furthermore, an Orlicz sequence space $\ell_{\psi}$ is separable
if and only if $\psi$ satisfies the $\Delta_{2}^{0}$-condition
($\psi\in\Delta_{2}^{0}$), that is,

$\sup_{0<u\leq 1}{\psi(2u)}/{\psi(u)}<\infty.$

In this case we have
$\ell_{\psi}^{*}=\ell_{\psi}^{\prime}=\ell_{\tilde{\psi}}$, with the Young
conjugate function $\tilde{\psi}$ for $\psi$.

Observe that the definition of an Orlicz sequence space $\ell_{\psi}$ depends
(up to equivalence of norms) only on the behaviour of $\psi$ near zero.

##  3\. Optimal Upper and Lower Sequence Lattices

In this section we introduce and study some specialized notions which will
play an important role in the proof of our main Theorem 2.8. They are a
special kind of sequence spaces which are generated via some appropriate
sequences of norms, defined on $\mathbb{R}^{n}$, $n\in\mathbb{N}$.

###  3.1. Definitions and general properties

######  Definition 3.1.

Let $X$ be a Banach lattice. For each integer $n,$ let
$\mathfrak{B}_{n}\left(X\right)$ denote the set of all sequences
$\left\\{x_{i}\right\\}_{i=1}^{n}\subseteq S_{X}$ of elements with pair-wise
disjoint support.

######  Lemma 3.2.

If $X$ is a Banach lattice of dimension at least $n,$ then the set
$\mathfrak{B}_{n}\left(X\right)$ is non-empty.

The proof of this lemma will be provided in Section 6.

Let now $X$ be an infinite dimensional Banach lattice. Based on $X$ we
associate two auxiliary constructions, which yield two sequence spaces $X_{U}$
and $X_{L}$ that satisfy the following norm one continuous embeddings:

(3.1)
$l_{1}\overset{1}{\hookrightarrow}X_{U}\overset{1}{\hookrightarrow}X_{L}\overset{1}{\hookrightarrow}l_{\infty}.$

We will call $X_{U}$ and $X_{L}$ the optimal upper and respectively optimal
lower sequence spaces generated by $X.$ Note that the construction, which
leads to the space $X_{L}$, is close to the one developed in the paper [14]
and related to the optimal cotype and summing properties of a Banach space.

To construct $X_{U}$ we define first, for each fixed integer $n$, the
following norm on $\mathbb{R}^{n}$ by

$\left\|\left\\{a_{i}\right\\}_{i=1}^{n}\right\|_{X_{U}\left(n\right)}:=\sup\left\\{\left\|\sum_{i=1}^{n}a_{i}x_{i}\right\|_{X}:\left\\{x_{i}\right\\}_{i=1}^{n}\in\mathfrak{B}_{n}\left(X\right)\right\\}.$

Let $X_{U}$ be the space of all real-valued sequences
$a=\left\\{a_{i}\right\\}_{i=1}^{\infty}$, for which

$\left\|a\right\|_{X_{U}}:=\sup_{n}\left\|\left\\{a_{i}\right\\}_{i=1}^{n}\right\|_{X_{U}\left(n\right)}<\infty.$

Since

$\left\|\left\\{a_{i}\right\\}_{i=1}^{n}\right\|_{X_{U}\left(n\right)}\leq\sum_{i=1}^{n}\left|a_{i}\right|,\;\;n\in\mathbb{N},$

it follows the left-hand side embedding in (3.1).

The first step in the definition of the space $X_{L}$ is the introduction of
the functionals $\Phi_{n}$, defined for
$a=\left\\{a_{i}\right\\}_{i=1}^{n}\in\mathbb{R}^{n}$, $n\in\mathbb{N}$, by

$\Phi_{n}\left(a\right):=\inf\left\\{\left\|\sum_{i=1}^{n}a_{i}x_{i}\right\|_{X}:\,\left\\{x_{i}\right\\}_{i=1}^{n}\in\mathfrak{B}_{n}\left(X\right)\right\\}.$

Next, we set

$\left\|a\right\|_{X_{L}\left(n\right)}:=\inf\left\\{\sum_{k\in
F}\Phi_{n}\left(a^{k}\right):\,F\subseteq\mathbb{N},\left|F\right|<\infty,a^{k}\in\mathbb{R}^{n},a=\sum_{k\in
F}a^{k}\right\\}.$

Note that $\sup_{1\leq i\leq n}\left|a_{i}\right|\leq\Phi_{n}\left(a\right)$
and hence
$\left\|a\right\|_{l_{\infty}^{n}}\leq\left\|a\right\|_{X_{L}\left(n\right)}$,
which implies that the mapping
$a\mapsto\left\|a\right\|_{X_{L}\left(n\right)}$ defines a norm on
$\mathbb{R}^{n}$. Finally, we define $X_{L}$ to be the space of all real-
valued sequences $a=\left\\{a_{i}\right\\}_{i=1}^{\infty}$, for which

$\left\|a\right\|_{X_{L}}:=\sup_{n}\left\|\left\\{a_{i}\right\\}_{i=1}^{n}\right\|_{X_{L}\left(n\right)}<\infty.$

Clearly, these definitions imply the second and third embeddings in (3.1).

The proof of the following important properties of the spaces $X_{U}$ and
$X_{L}$ we provide in Section 6 below.

######  Theorem 3.3.

Let $X$ be an infinite dimensional Banach lattice. Then $X_{L}$ is a r.i.
sequence space and $X_{U}$ is a Banach sequence lattice. Moreover, if
$l_{\infty}$ is not finitely representable in $X$, $X_{U}$ is a r.i. sequence
space as well.

Denote by $X_{L}^{0}$ (resp. $X_{U}^{0}$) the closed linear span of all
finitely supported sequences in $X_{L}$ (resp. $X_{U}$). Recall that $e_{m}$,
$m=1,2,\dots$, are the unit basis vectors in spaces of real-valued sequences.

######  Corollary 3.4.

If $X$ is an infinite dimensional Banach lattice, then $X_{L}^{0}$ is a Banach
space in which the vectors $e_{m}$, $m=1,2,\dots$, form a symmetric normed
basis. If $l_{\infty}$ is not finitely representable in $X$, the same
conclusion applies also to $X_{U}^{0}$.

In particular, if $\mathrm{supp}\,a\subset\\{1,2,\dots,n\\}$ for some
$n\in\mathbb{N}$, we have $\|a\|_{X_{L}}=\|a\|_{X_{L}(n)}$ and
$\|a\|_{X_{U}}=\|a\|_{X_{U}(n)}$.

######  Example 3.5.

We claim that $\left(c_{0}\right)_{U}=\left(c_{0}\right)_{L}=l_{\infty}$ and
hence $\left(c_{0}\right)_{U}^{0}=\left(c_{0}\right)_{L}^{0}=c_{0}.$

Indeed, first by (3.1), we have
$\left(c_{0}\right)_{U}\overset{1}{\hookrightarrow}\left(c_{0}\right)_{L}\overset{1}{\hookrightarrow}l_{\infty}.$
For the converse, fix an integer $n$ and let
$\left\\{x_{i}\right\\}_{i=1}^{n}\subseteq c_{0}$ be a positive unit norm
sequence with pair-wise disjoint support. Clearly, for all
$a_{i}\in\mathbb{R}$, $i=1,2,\dots,n$,

$\left|\sum_{i=1}^{n}a_{i}x_{i}\right|\leq\sup_{1\leq n\leq
n}\left|a_{i}\right|,$

and hence
$l_{\infty}^{n}\overset{1}{\hookrightarrow}\left(c_{0}\right)_{U}\left(n\right)$.
In consequence,
$l_{\infty}\overset{1}{\hookrightarrow}\left(c_{0}\right)_{U}.$ Thus, in view
of (3.1), everything is done.

The above example shows that the spaces $X_{U}$ and $X_{L}$ do not need to
have $\sigma$-order continuous norm even if $X$ has so. At the same time, the
construction of the optimal upper and lower sequence spaces ensures that they
always have the following somewhat weaker property.

Recall that a Banach function lattice $X$ on a measure space $(T,\mu)$ is
called order semi-continuous if the conditions $x_{n}\in X$, $n=1,2,\dots$,
$x\in X$ and $x_{n}\chi_{B}\rightarrow x\chi_{B}$ $\mu$-a.e. for each set
$B\subset T$ such that $\mu(B)<\infty$ imply that
$\|x\|_{X}\leq\liminf_{n\rightarrow\infty}\|x_{n}\|_{X}$.

In particular, Banach sequence lattice $E$ (in this case $T=\mathbb{N}$ with
the counting measure $\mu$) is order semi-continuous if
$\|a\|_{E}\leq\liminf_{n\rightarrow\infty}\|a^{n}\|_{E}$ whenever a sequence
$\\{a^{n}\\}_{n=1}^{\infty}\subset E$ converges coordinate-wise to $a\in E$.

######  Lemma 3.6.

$X_{U}$ and $X_{L}$ are order semi-continuous Banach sequence lattices for
each Banach lattice $X$.

###### Proof.

We prove this result only for $X_{U}$, because for $X_{L}$ this can be done in
the same way.

Assume that a sequence $\\{a^{n}\\}_{n=1}^{\infty}\subset X_{U}$ converges
coordinate-wise to an element $a\in X_{U}$. Let
$a^{n}=\\{a^{n}_{i}\\}_{i=1}^{\infty}$, $a=\\{a_{i}\\}_{i=1}^{\infty}$. For
arbitrary $\varepsilon>0$ select $m\in\mathbb{N}$ so that

$\left\|a\right\|_{X_{U}}\leq(1+\varepsilon)\left\|\left\\{a_{i}\right\\}_{i=1}^{m}\right\|_{X_{U}\left(m\right)}.$

Then, for all sufficiently large $n\in\mathbb{N}$ we have

$|a^{n}_{i}|\geq(1-\varepsilon)|a_{i}|,\;\;i=1,2,\dots,m,$

whence

$\left\|\left\\{a^{n}_{i}\right\\}_{i=1}^{m}\right\|_{X_{U}\left(m\right)}\geq(1-\varepsilon)\left\|\left\\{a_{i}\right\\}_{i=1}^{m}\right\|_{X_{U}\left(m\right)}.$

Combining the above inequalities, we get

$\left\|a\right\|_{X_{U}}\leq\frac{1+\varepsilon}{1-\varepsilon}\left\|\left\\{a^{n}_{i}\right\\}_{i=1}^{m}\right\|_{X_{U}\left(m\right)}$

if $n\in\mathbb{N}$ is sufficiently large. This implies that

$\left\|a\right\|_{X_{U}}\leq\frac{1+\varepsilon}{1-\varepsilon}\liminf_{n\rightarrow\infty}\|a^{n}\|_{X_{U}}.$

Since $\varepsilon>0$ is arbitrary, the desired result for $X_{U}$ is proved.
∎

For any sequence $a=\left\\{a_{i}\right\\}_{i=1}^{\infty}$ we have
$a_{i}=\left\langle a,e_{i}\right\rangle,$ where
$\left\langle\cdot,\cdot\right\rangle$ is the usual inner product. In what
follows, the properties of the optimal sequence spaces from the next
proposition, will play a crucial role.

######  Proposition 3.7.

Let $X$ be a Banach lattice.

$\left(i\right)$. For any $m\in\mathbb{N}$ and every pairwise disjoint
sequences $u_{k}\in X_{U}$, $k=1,2,\dots,m$, we have

$\left\|\sum_{k=1}^{m}u_{k}\right\|_{X_{U}}\leq\left\|\sum_{k=1}^{m}\left\|u_{k}\right\|_{X_{U}}e_{k}\right\|_{X_{U}};$

$\left(ii\right)$. For any $m\in\mathbb{N}$ and every pairwise disjoint
sequences $u_{k}\in X_{L}$, $k=1,2,\dots,m$,

$\left\|\sum_{k=1}^{m}\left\|u_{k}\right\|_{X_{L}}e_{k}\right\|_{X_{L}}\leq\left\|\sum_{k=1}^{m}u_{k}\right\|_{X_{L}}.$

###### Proof.

First, we prove $\left(i\right)$ assuming additionally that the elements
$u_{k}$, $k=1,2,\dots,m$, have finite support. Then, denoting
$u:=\sum_{k=1}^{m}u_{k}$ and $n:=|\mathrm{supp}\,u|$, we get

$u_{k}=\sum_{i\in\mathrm{supp}\,u_{k}}\left\langle u_{k},e_{i}\right\rangle
e_{i},\;\;k=1,2,\dots,m.$

Take $\left\\{x_{i}\right\\}_{i=1}^{n}\in\mathfrak{B}_{n}\left(X\right)$ and
put

$z_{k}:=\sum_{i\in\mathrm{supp}\,u_{k}}\left\langle u_{k},e_{i}\right\rangle
x_{i},\;\;k=1,2,\dots,m.$

Without any loss of generality, we may assume that $u_{k}\geq 0$ and
$u_{k}\neq 0$ for all $k$. Hence, $z_{k}\neq 0$ for all $k=1,2,\dots,m$ and

(3.2) $\left\|z_{k}\right\|_{X}\leq\left\|\sum_{i=1}^{n}\left\langle
u_{k},e_{i}\right\rangle
e_{i}\right\|_{X_{U}\left(n\right)}=\left\|u_{k}\right\|_{X_{U}}.$

Moreover, since $u_{k}$, $k=1,2,\dots,m$, are pairwise disjoint sequences and
$x_{i}$, $i=1,2,\dots,n$, are pairwise disjoint elements from $X$, we infer
that $z_{i}\wedge z_{j}=0$ if $i\neq j$. Thus,
$\left\\{z_{k}/\left\|z_{k}\right\|_{X}:\,1\leq k\leq
m\right\\}\in\mathfrak{B}_{m}\left(X\right)$. Consequently, from (3.2) it
follows

$\displaystyle\left\|\sum_{i=1}^{n}\left\langle u,e_{i}\right\rangle
x_{i}\right\|_{X}$ $\displaystyle=$
$\displaystyle\left\|\sum_{k=1}^{m}\sum_{i=1}^{n}\left\langle
u_{k},e_{i}\right\rangle
x_{i}\right\|_{X}=\left\|\sum_{k=1}^{m}\sum_{i\in\mathrm{supp}\,u_{k}}\left\langle
u_{k},e_{i}\right\rangle x_{i}\right\|_{X}$ $\displaystyle=$
$\displaystyle\left\|\sum_{k=1}^{m}z_{k}\right\|_{X}=\left\|\sum_{k=1}^{m}\left\|z_{k}\right\|_{X}\frac{z_{k}}{\left\|z_{k}\right\|_{X}}\right\|_{X}$
$\displaystyle\leq$
$\displaystyle\left\|\sum_{k=1}^{m}\left\|z_{k}\right\|_{X}e_{k}\right\|_{X_{U}\left(m\right)}\leq\left\|\sum_{k=1}^{m}\left\|u_{k}\right\|_{X_{U}}e_{k}\right\|_{X_{U}}.$

Hence, as a sequence
$\left\\{x_{i}\right\\}_{i=1}^{n}\in\mathfrak{B}_{n}\left(X\right)$ is
arbitrary, we conclude that

$\left\|u\right\|_{X_{U}}=\left\|\sum_{i=1}^{n}\left\langle
u,e_{i}\right\rangle
e_{i}\right\|_{X_{U}\left(n\right)}\leq\left\|\sum_{k=1}^{m}\left\|u_{k}\right\|_{X_{U}}e_{k}\right\|_{X_{U}},$

and for finitely supported sequences assertion $\left(i\right)$ is proved.

Let now $u_{k}\in X_{U}$, $k=1,2,\dots,m$, be arbitrary pairwise disjoint non-
negative elements. Denote by $u_{k}^{(n)}$ the truncations of $u_{k}$ to the
set $\left\\{1,\dots,n\right\\}$, that is,

(3.3) $u_{k}^{(n)}:=\sum_{1\leq j\leq
n,j\in\mathrm{supp}\,u_{k}}a_{j}e_{j},\;\;k=1,\dots,m,$

Since $u_{k}^{(n)}$, $k=1,2,\dots,m$, are pairwise disjoint sequences with
finite support, by the first part of the proof, we have

$\Big{\|}\sum_{k=1}^{m}u_{k}^{(n)}\Big{\|}_{X_{U}}\leq\Big{\|}\sum_{k=1}^{m}\|u_{k}^{(n)}\|_{X_{U}}e_{k}\Big{\|}_{X_{U}}\leq\Big{\|}\sum_{k=1}^{m}\|u_{k}\|_{X_{U}}e_{k}\Big{\|}_{X_{U}}.$

Therefore, taking into account that the sequence $\sum_{k=1}^{m}u_{k}^{(n)}$
tends coordinate-wise to $\sum_{k=1}^{m}u_{k}$ as $n\rightarrow\infty$, by
Lemma 3.6, we obtain

$\Big{\|}\sum_{k=1}^{m}u_{k}\Big{\|}_{X_{U}}\leq\liminf_{n\rightarrow\infty}\Big{\|}\sum_{k=1}^{m}u_{k}^{(n)}\Big{\|}_{X_{U}}\leq\Big{\|}\sum_{k=1}^{m}\|u_{k}\|_{X_{U}}e_{k}\Big{\|}_{X_{U}},$

which implies $\left(i\right)$ in the general case.

Proceeding with the proof of $\left(ii\right)$, we again consider first the
case when the elements $u_{k}$, $k=1,2,\dots,m$, have finite support. Let $u$,
$n\in\mathbb{N}$, $\left\\{x_{j}\right\\}_{i=1}^{n}$ and $z_{k}$,
$k=1,2,\dots,m$, be defined in the same way as in the beginning of the proof
of $\left(i\right)$. Assuming as above that $u_{k}\geq 0$ and $u_{k}\neq 0$,
$k=1,2,\dots,m$, we get $z_{k}\neq 0$, $k=1,2,\dots,m$, and $n\geq m$.
Consequently,
$\left\\{z_{k}/\left\|z_{k}\right\|_{X}\right\\}_{k=1}^{m}\in\mathfrak{B}_{m}\left(X\right).$
Moreover, by the definition of the norm in $X_{L}$, we have

$\left\|u_{k}\right\|_{X_{L}}\leq\Phi_{n}\left(u_{k}\right)\leq\left\|z_{k}\right\|_{X},\;\;k=1,2,\dots,m.$

Hence, by Theorem 3.3, it follows

$\displaystyle\left\|\sum_{k=1}^{m}\left\|u_{k}\right\|_{X_{L}}e_{k}\right\|_{X_{L}\left(m\right)}$
$\displaystyle\leq$
$\displaystyle\left\|\sum_{k=1}^{m}\left\|z_{k}\right\|_{X}e_{k}\right\|_{X_{L}\left(m\right)}\leq\left\|\sum_{k=1}^{m}\left\|z_{k}\right\|_{X}\frac{z_{k}}{\left\|z_{k}\right\|_{X}}\right\|_{X}$
$\displaystyle=$
$\displaystyle\left\|\sum_{k=1}^{m}z_{k}\right\|_{X}=\left\|\sum_{k=1}^{m}\sum_{j\in\sigma_{k}}\left\langle
u_{k},e_{j}\right\rangle x_{j}\right\|_{X}=\left\|\sum_{j=1}^{m}\left\langle
u,e_{j}\right\rangle x_{j}\right\|_{X}.$

Passing to the infimum over all
$\left\\{x_{j}\right\\}_{i=1}^{n}\in\mathfrak{B}_{n}\left(X\right)$, we obtain

(3.4)
$\left\|\sum_{k=1}^{m}\left\|u_{k}\right\|_{X_{L}}e_{k}\right\|_{X_{L}\left(m\right)}\leq\Phi_{n}\left(u\right).$

Next, let $u=\sum_{l\in F}v_{l}$ for some finite set $F$ of positive integers.
Clearly, we may assume that the supports of $v_{l}$ are contained in that of
$u$ and hence in the set $\cup_{k=1}^{m}{\mathrm{supp}\,u_{k}}$. Then, if

$v_{l,k}:=\sum_{i\in\mathrm{supp}\,u_{k}}\left\langle v_{l},e_{i}\right\rangle
e_{i},\;\;k=1,2,\dots,m,$

we have $v_{l}=\sum_{k=1}^{m}v_{l,k}$, $l\in F$, and $u_{k}=\sum_{l\in
F}v_{l,k}$, $k=1,2,\dots,m$. Furthermore, since $u_{k}^{l}$, $k=1,2,\dots,m$,
are pairwise disjoint and have finite support, applying (3.4) for $v_{l}$, we
infer

$\left\|\sum_{k=1}^{m}\left\|v_{l,k}\right\|_{X_{L}}e_{k}\right\|_{X_{L}\left(m\right)}\leq\Phi_{n}\left(v_{l}\right),\;\;l\in
F.$

Hence, by the triangle inequality,

$\displaystyle\left\|\sum_{k=1}^{m}\left\|u_{k}\right\|_{X_{L}}e_{k}\right\|_{X_{L}\left(m\right)}$
$\displaystyle\leq$ $\displaystyle\left\|\sum_{k=1}^{m}\sum_{l\in
F}\left\|v_{l,k}\right\|_{X_{L}}e_{k}\right\|_{X_{L}\left(m\right)}$
$\displaystyle\leq$ $\displaystyle\sum_{l\in
F}\left\|\sum_{k=1}^{m}\left\|v_{l,k}\right\|_{X_{L}}e_{k}\right\|_{X_{L}\left(m\right)}$
$\displaystyle\leq$ $\displaystyle\sum_{l\in F}\Phi_{n}\left(v_{l}\right).$

Since the above representation of $u$ is arbitrary, from Theorem 3.3 it
follows

$\left\|\sum_{k=1}^{m}\left\|u_{k}\right\|_{X_{L}}e_{k}\right\|_{X_{L}}=\left\|\sum_{k=1}^{m}\left\|u_{k}\right\|_{X_{L}}e_{k}\right\|_{X_{L}\left(m\right)}\leq\|u\|_{X_{L}\left(n\right)}=\|u\|_{X_{L}}.$

Thus, for sequences with finite support $\left(ii\right)$ is proved.

To extend the assertion $\left(ii\right)$ to the general case, assume that
$u_{k}\in X_{L}$, $k=1,\dots,m$, are pairwise disjoint and non-negative. Let
$n\in\mathbb{N}$ be arbitrary and $u_{k}^{(n)}$ be the truncations defined by
formula (3.3). Since $u_{k}^{(n)}$, $k=1,\dots,m$, are finitely supported, as
was already proved, it holds

$\Big{\|}\sum_{k=1}^{m}u_{k}\Big{\|}_{X_{L}}\geq\Big{\|}\sum_{k=1}^{m}u_{k}^{(n)}\Big{\|}_{X_{L}}\geq\Big{\|}\sum_{k=1}^{m}\|u_{k}^{(n)}\|_{X_{L}}e_{k}\Big{\|}_{X_{L}}.$

Observe that from Lemma 3.6 it follows
$\lim_{n\rightarrow\infty}\|u_{k}^{(n)}\|_{X_{L}}=\|u_{k}\|_{X_{L}}$ for each
$k=1,\dots,m$. In consequence, we have

$\lim_{n\rightarrow\infty}\Big{\|}\sum_{k=1}^{m}\|u_{k}^{(n)}\|_{X_{L}}e_{k}\Big{\|}_{X_{L}}=\Big{\|}\sum_{k=1}^{m}\|u_{k}\|_{X_{L}}e_{k}\Big{\|}_{X_{L}}.$

Combining this together with the preceding estimate, we infer that

$\Big{\|}\sum_{k=1}^{m}u_{k}\Big{\|}_{X_{L}}\geq\Big{\|}\sum_{k=1}^{m}\|u_{k}\|_{X_{L}}e_{k}\Big{\|}_{X_{L}},$

and so the proof is completed. ∎

By Theorem 3.3, both spaces $X_{U}$ and $X_{L}$ are Banach lattices and hence
$X_{U}$\- and $X_{L}$-constructions can be applied also to them. However, this
process terminates already on the second step, because of the following
result.

######  Proposition 3.8.

(a) For every Banach sequence lattice $E$ we have
$E\overset{1}{\hookrightarrow}E_{L}$. If additionally $E$ is order semi-
continuous, then $E_{U}\overset{1}{\hookrightarrow}E$.

(b) For every Banach lattice $X$ we have $(X_{L})_{L}=X_{L}$ and
$(X_{U})_{U}=X_{U}$ isometrically.

###### Proof.

(a) We show first that $E\overset{1}{\hookrightarrow}E_{L}$. Let
$n\in\mathbb{N}$ be arbitrary. Since
$\left\\{e_{i}\right\\}_{i=1}^{n}\in\mathfrak{B}_{n}\left(E\right)$, for every
$a=(a_{i})_{i=1}^{\infty}\in E$, we can write

$\|a\|_{E}\geq\Big{\|}\sum_{i=1}^{n}a_{i}e_{i}\Big{\|}_{E}\geq\Phi_{n}((a_{i})_{i=1}^{n})\geq\|(a_{i})_{i=1}^{n}\|_{E_{L}(n)}.$

Consequently, $\|a\|_{E}\geq\|a\|_{E_{L}}$, i.e.,
$E\overset{1}{\hookrightarrow}E_{L}$.

Assume now that $E$ is order semi-continuous. Then, for every
$a=(a_{i})_{i=1}^{\infty}\in E_{U}$ and $n\in\mathbb{N}$ we have

$\|(a_{i})_{i=1}^{n}\|_{E_{U}(n)}\geq\Big{\|}\sum_{i=1}^{n}a_{i}e_{i}\Big{\|}_{E},$

whence

$\|a\|_{E_{U}}\geq\Big{\|}\sum_{i=1}^{n}a_{i}e_{i}\Big{\|}_{E},\;\;n\in\mathbb{N}.$

Therefore, since $E$ is order semi-continuous, we get
$\|a\|_{E_{U}}\geq\|a\|_{E}$ for each $a\in E_{U}$. Thus, the proof of (a) is
completed.

(b) If $X$ is an arbitrary Banach lattice, then by Lemma 3.6, $X_{L}$ is an
order semi-continuous Banach sequence lattice. Hence, from the already proved
part (a) it follows that $X_{L}\overset{1}{\hookrightarrow}(X_{L})_{L}$. It
remains to check that $(X_{L})_{L}\overset{1}{\hookrightarrow}X_{L}$.

Suppose $n\in\mathbb{N}$ and $b=(b_{i})_{i=1}^{n}\in\mathbb{R}$ is arbitrary.
For every $\varepsilon>0$ there is a sequence
$\left\\{u_{i}\right\\}_{i=1}^{n}\in\mathfrak{B}_{n}\left(X_{L}\right)$ such
that

$\Phi_{n}(b)\geq(1-\varepsilon)\Big{\|}\sum_{i=1}^{n}b_{i}u_{i}\Big{\|}_{X_{L}}.$

Hence, by Proposition 3.7(ii), we obtain

$\Phi_{n}(b)\geq(1-\varepsilon)\Big{\|}\sum_{i=1}^{n}b_{i}\|u_{i}\|_{X_{L}}e_{i}\Big{\|}_{X_{L}}=(1-\varepsilon)\Big{\|}\sum_{i=1}^{n}b_{i}e_{i}\Big{\|}_{X_{L}}.$

Now, let $a=(a_{i})_{i=1}^{\infty}\in(X_{L})_{L}$ and $n\in\mathbb{N}$. Let
$\sum_{i=1}^{n}a_{i}e_{i}=\sum_{k\in F}b^{k}$, where $F\subset\mathbb{N}$ is a
finite set and $b^{k}=(b_{i}^{k})_{i=1}^{n}$, $k\in F$. Then, from the
preceding estimate and the triangle inequality it follows that

$\sum_{k\in F}\Phi_{n}((b_{i}^{k})_{i=1}^{n})\geq(1-\varepsilon)\sum_{k\in
F}\Big{\|}\sum_{i=1}^{n}b_{i}^{k}e_{i}\Big{\|}_{X_{L}}\geq(1-\varepsilon)\Big{\|}\sum_{i=1}^{n}a_{i}e_{i}\Big{\|}_{X_{L}},$

Hence, taking the infimum over all the above representations of
$(a_{i})_{i=1}^{n}$, implies

$\|a\|_{(X_{L})_{L}}\geq\|(a_{i})_{i=1}^{n}\|_{(X_{L})_{L}(n)}\geq(1-\varepsilon)\|(a_{i})_{i=1}^{n}\|_{X_{L}(n)}.$

Since $\varepsilon>0$ is arbitrary and $X_{L}$ is order semi-continuous, one
can easily get now that $\|a\|_{(X_{L})_{L}}\geq\|a\|_{X_{L}}$, and so the
proof of the equality $(X_{L})_{L}=X_{L}$ is completed.

The proof of the fact that $(X_{U})_{U}=X_{U}$ is very similar and simpler.
Again, in view of Lemma 3.6 and the part (a) of this proposition, it suffices
to show that $X_{U}\overset{1}{\hookrightarrow}(X_{U})_{U}$. Indeed, if
$a=(a_{i})_{i=1}^{\infty}\in X_{U}$ and $n\in\mathbb{N}$, then for each
sequence
$\left\\{u_{i}\right\\}_{i=1}^{n}\in\mathfrak{B}_{n}\left(X_{U}\right)$, by
Proposition 3.7(i), we have

$\Big{\|}\sum_{i=1}^{n}a_{i}u_{i}\Big{\|}_{X_{U}}\leq\Big{\|}\sum_{i=1}^{n}a_{i}\|u_{i}\|_{X_{U}}e_{i}\Big{\|}_{X_{U}}=\Big{\|}\sum_{i=1}^{n}a_{i}e_{i}\Big{\|}_{X_{U}}\leq\|a\|_{X_{U}}.$

Therefore,

$\|(a_{i})_{i=1}^{n}\|_{(X_{U})_{U}(n)}\leq\|a\|_{X_{U}}\;\;\mbox{for
all}\;n\in\mathbb{N},$

whence $\|a\|_{(X_{U})_{U}}\leq\|a\|_{X_{U}}$. Thus, the proof of the
proposition is completed. ∎

###  3.2. Optimal sequence spaces and upper/lower estimates of Banach lattices

As we will see in this section, properties of the spaces $X_{U}$ and $X_{L}$
are largely determined by the optimal upper and lower estimate information
related to the given Banach lattice $X$. The connections, revealed in the next
proposition, will play an important role in the proof of our main Theorem 2.8.
Recall that $\delta\left(X\right)$ and $\sigma\left(X\right)$ are the Grobler-
Dodds indices of a Banach lattice $X$ (see Section 2.2).

######  Proposition 3.9.

Let $X$ be a Banach lattice. Then,

(i) $X_{U}\overset{1}{\hookrightarrow}l_{\delta(X)}$ and $l_{p}\hookrightarrow
X_{U}$ for every $p<\delta(X)$;

(ii) $l_{\sigma(X)}\overset{1}{\hookrightarrow}X_{L}$ and
$X_{L}\hookrightarrow l_{q}$ for every $q>\sigma(X)$;

(iii) $X_{U}=l_{p}$ if and only if $p=\delta(X)$ and $X$ admits an upper
$\delta(X)$-estimate;

(iv) $X_{L}=l_{q}$ if and only if $q=\sigma(X)$ and $X$ admits a lower
$\sigma(X)$-estimate.

###### Proof.

(i) Let $p<\delta(X)$ and $a=(a_{k})_{k=1}^{\infty}\in l_{p}$. Since $X$
admits an upper $p$-estimate, then for every $n\in\mathbb{N}$ and
$\left\\{x_{k}\right\\}_{k=1}^{n}\in\mathfrak{B}_{n}\left(X\right)$ we have

$\left\|\sum_{k=1}^{n}a_{k}x_{k}\right\|_{X}\leq
C_{p}\left(\sum_{k=1}^{n}|a_{k}|^{p}\right)^{1/p},$

where $C_{p}$ depends only on $p$. Consequently, by the definition of $X_{U}$,
we get

$\|a\|_{X_{U}}=\sup_{n=1,2,\dots}\|(a_{k})_{k=1}^{n}\|_{X_{U}(n)}\leq
C_{p}\|a\|_{l_{p}}.$

Next, suppose $a=(a_{k})_{k=1}^{\infty}\in X_{U}$. By Schep’s result (see
[29]), $l_{\delta(X)}$ is finitely lattice representable in $X$, which implies
that for every $\varepsilon>0$ and $n\in\mathbb{N}$ there exists a sequence
$\left\\{y_{k}\right\\}_{k=1}^{n}\in\mathfrak{B}_{n}\left(X\right)$ such that
for any $a_{k}\in\mathbb{R}$, $k=1,2,\dots,n$, we have

$\left\|\sum_{k=1}^{n}a_{k}y_{k}\right\|_{X}\geq(1-\varepsilon)\|(a_{k})_{k=1}^{n}\|_{l_{\delta(X)}}.$

Hence,

$\|(a_{k})_{k=1}^{n}\|_{l_{\delta(X)}}\leq\frac{1}{1-\varepsilon}\left\|\sum_{k=1}^{n}a_{k}y_{k}\right\|_{X}\leq\frac{1}{1-\varepsilon}\|(a_{k})_{k=1}^{n}\|_{X_{U}(n)}\leq\frac{1}{1-\varepsilon}\|a\|_{X_{U}}.$

Since $n\in\mathbb{N}$ and $\varepsilon>0$ are arbitrary, we have
$\|a\|_{l_{\delta(X)}}\leq\|a\|_{X_{U}}$, which completes the proof of (i).

(ii) Let $q>\sigma(X)$, $n\in\mathbb{N}$ and
$a=(a_{k})_{k=1}^{n}\in\mathbb{R}^{n}$. Since $X$ admits a lower $q$-estimate,
then for every
$\left\\{x_{k}\right\\}_{k=1}^{n}\in\mathfrak{B}_{n}\left(X\right)$ we have

$\|a\|_{l_{q}^{n}}=\left(\sum_{k=1}^{n}|a_{k}|^{q}\right)^{1/q}\leq
C_{q}\left\|\sum_{k=1}^{n}a_{k}x_{k}\right\|_{X},$

where $C_{q}$ depends only on $q$. Passing to the infimum over all sequences
$\left\\{x_{k}\right\\}_{k=1}^{n}\in\mathfrak{B}_{n}\left(X\right)$, we come
to the inequality

$\|a\|_{l_{q}^{n}}\leq C_{q}\Phi_{n}(a).$

Next, if $a=\sum_{l\in F}b^{l}$, where $F\subset\mathbb{N}$ is finite and
$b^{l}\in\mathbb{R}^{n}$, we have

$\|a\|_{l_{q}^{n}}\leq\sum_{l\in F}\|b^{l}\|_{l_{q}^{n}}\leq C_{q}\sum_{l\in
F}\Phi_{n}(b^{l}).$

Passing to the infimum over all above representations of $a$, we obtain

$\|a\|_{l_{q}^{n}}\leq C_{q}\|a\|_{X_{L}(n)}.$

Since this holds for every $n\in\mathbb{N}$ and
$a=(a_{k})_{k=1}^{n}\in\mathbb{R}^{n}$, by using Lemma 3.6, we conclude that
$X_{L}\hookrightarrow l_{q}$.

Further, again appealing to [29], we have that $l_{\sigma(X)}$ is finitely
lattice representable in $X$. Therefore, for every $\varepsilon>0$ and
$n\in\mathbb{N}$ there exists a sequence
$\left\\{y_{k}\right\\}_{k=1}^{n}\in\mathfrak{B}_{n}\left(X\right)$ such that
for any $a=(a_{k})_{k=1}^{n}\in\mathbb{R}^{n}$ it holds

$\left\|\sum_{k=1}^{n}a_{k}y_{k}\right\|_{X}\leq(1+\varepsilon)\|a\|_{l_{\sigma(X)}}.$

In consequence, by the definition of the $X_{L}(n)$-norm,

$\|a\|_{X_{L}(n)}\leq\Phi_{n}(a)\leq(1+\varepsilon)\|a\|_{l_{\sigma(X)}},$

and hence again for each $a\in l_{\sigma(X)}$ and any $\varepsilon>0$ it
follows that

$\|a\|_{X_{L}}\leq(1+\varepsilon)\|a\|_{l_{\sigma(X)}}.$

Application of Lemma 3.6 again completes the proof.

(iii) If $X$ admits an upper $\delta(X)$-estimate, the same argument as in the
proof of (i) implies that $l_{\delta(X)}\hookrightarrow X_{U}$. Combining this
with the first embedding in (i), we get that $X_{U}=l_{\delta(X)}$.

Conversely, let $X_{U}=l_{p}$ for some $p\geq 1$. Then, from (i) it follows
immediately that $p$ should be equal to $\delta(X)$. It remains to show that
$X$ admits an upper $\delta(X)$-estimate.

Suppose that $x_{k}\in X$, $k=1,2,\dots,n$, are arbitrary pair-wise disjoint
elements. Then, by the definition of the $X_{U}(n)$-norm and the fact that
$X_{U}=l_{\delta(X)}$, we have

$\left\|\sum_{k=1}^{n}x_{k}\right\|_{X}\leq\left\|\sum_{k=1}^{n}\|x_{k}\|_{X}e_{k}\right\|_{X_{U}(n)}\leq
C\left(\sum_{k=1}^{n}\|x_{k}\|_{X}^{\delta(X)}\right)^{1/\delta(X)},$

where $C$ does not depend on $n$ and $x_{k}$. This means that $X$ admits an
upper $\delta(X)$-estimate.

(iv) If $X$ admits a lower $\sigma(X)$-estimate, then from (ii) it follows
immediately then $X_{L}=l_{\sigma(X)}$.

Conversely, if $X_{L}=l_{q}$, then, by (ii), $q=\sigma(X)$. Consequently, for
every pair-wise disjoint $x_{k}\in X$, $k=1,2,\dots,n$, by the definition of
the $X_{L}(n)$-norm, it follows

$\left(\sum_{k=1}^{n}\|x_{k}\|_{X}^{\sigma(X)}\right)^{1/\sigma(X)}\leq
C\left\|\sum_{k=1}^{n}\|x_{k}\|_{X}e_{k}\right\|_{X_{L}(n)}\leq
C\left\|\sum_{k=1}^{n}x_{k}\right\|_{X}.$

Therefore, $X$ admits a lower $\sigma(X)$-estimate, and the proof is complete.
∎

In some cases, an application of the last proposition allows to find
immediately the optimal sequence spaces.

######  Example 3.10.

Let $1<p<\infty$, $1\leq q<\infty$ and let $L_{p,q}=L_{p,q}(I)$ be the Lorentz
space, where $I=[0,1]$ or $(0,\infty)$ (see Section 2.4). It is well known
that $\delta(L_{p,q})=\min(p,q)$, $\sigma(L_{p,q})=\max(p,q)$, and moreover,
that $L_{p,q}$ admits an upper $\delta(L_{p,q})$-estimate and a lower
$\sigma(L_{p,q})$-estimate (see e.g. [12, Theorem 3]). Consequently, by
Proposition 3.9, $(L_{p,q})_{U}=l_{\min(p,q)}$ and
$(L_{p,q})_{L}=l_{\max(p,q)}$.

######  Corollary 3.11.

Let $X$ be a Banach lattice. Then, $\delta(X_{U})=\delta(X)$ and
$\sigma(X_{L})=\sigma(X)$.

###### Proof.

First, we claim that $X_{U}$ admits an upper $p$-estimate for every
$p<\delta(X)$. Indeed, if $p<\delta(X)$ and $u_{i}\in X_{U}$, $i=1,\dots,n$,
are disjoint, by Propositions 3.7 and 3.9(i), we have

$\Big{\|}\sum_{i=1}^{n}u_{i}\Big{\|}_{X_{U}}\leq\Big{\|}\sum_{i=1}^{n}\|u_{i}\|_{X_{U}}e_{i}\Big{\|}_{{X_{U}}}\leq
C_{p}\Big{(}\sum_{i=1}^{n}\|u_{i}\|_{{X_{U}}}^{p}\Big{)}^{1/p}.$

Since $p<\delta(X)$ is arbitrary, this inequality implies that
$\delta(X)\leq\delta(X_{U})$. It remains to prove the opposite inequality.

Suppose that $X_{U}$ admits an upper $p$-estimate with $p>\delta(X)$. Then,
there is a constant $C>0$ such that for every $n\in\mathbb{N}$ and any
$a_{k}\in\mathbb{R}$, $k=1,2,\dots,n$, we have

$\Big{\|}\sum_{k=1}^{n}a_{k}e_{k}\Big{\|}_{X_{U}}\leq
C\|(a_{k})_{k=1}^{n}\|_{l_{p}}.$

On the other hand, by Proposition 3.9(i),

$\Big{\|}\sum_{k=1}^{n}a_{k}e_{k}\Big{\|}_{X_{U}}\geq\|(a_{k})_{k=1}^{n}\|_{l_{\delta(X)}}$

for all $n\in\mathbb{N}$ and $a_{k}\in\mathbb{R}$, $k=1,2,\dots,n$. Since
$p>\delta(X)$, combining these inequalities, we come to a contradiction. Thus,
$\delta(X_{U})=\delta(X)$, as required.

Similarly, by using Propositions 3.7 and 3.9(ii), one can easily check that
the space $X_{L}$ has a lower $p$-estimate for every $p>\sigma(X)$. Therefore,
$\sigma(X_{L})\leq\sigma(X)$, and the equality $\sigma(X_{L})=\sigma(X)$ will
be proved, once we check that $X_{L}$ does not admit a lower $p$-estimate with
any $p<\sigma(X)$. To the contrary, assume that for some $p<\sigma(X)$ there
is a constant $C>0$ such that for every $n\in\mathbb{N}$ and any
$a_{k}\in\mathbb{R}$, $k=1,2,\dots,n$,

$\|(a_{k})_{k=1}^{n}\|_{l_{p}}\leq
C\left\|\sum_{k=1}^{n}a_{k}e_{k}\right\|_{X_{L}}.$

On the other hand, from Proposition 3.9(ii) it follows that

$\left\|\sum_{k=1}^{n}a_{k}e_{k}\right\|_{X_{L}}\leq\|(a_{k})_{k=1}^{n}\|_{l_{\sigma(X)}}$

for all $n\in\mathbb{N}$ and $a_{k}\in\mathbb{R}$, $k=1,2,\dots,n$. Since the
latter estimates imply a contradiction, everything is done. ∎

##  4\. Proof of Theorem 2.8

We start with some auxiliary assertions.

Our first result shows that relative $s$-decomposability of Banach lattices
$X$ and $Y$ implies that each sequence from the space $l_{s}$ can be treated
as a multiplicator, bounded from $X_{L}$ into $Y_{U}.$

######  Proposition 4.1.

Let $X$ and $Y$ be relatively $s$-decomposable Banach lattices. Then, we have

$X_{L}\cdot l_{s}\hookrightarrow Y_{U},$

i.e., the conditions $a=\left\\{a_{i}\right\\}_{i=1}^{\infty}\in X_{L}$,
$b=\left\\{b_{i}\right\\}_{i=1}^{\infty}\in l_{s}$ imply
$ab:=\left\\{a_{i}b_{i}\right\\}_{i=1}^{\infty}\in Y_{U}$ and

$\left\|ab\right\|_{Y_{U}}\leq
D_{s}\left(X,Y\right)\left\|b\right\|_{l_{s}}\left\|a\right\|_{X_{L}}.$

###### Proof.

Let $n$ be any positive integer and $D>D_{s}\left(X,Y\right).$ For arbitrary
$a=\left\\{a_{i}\right\\}_{i=1}^{\infty}\in X_{L}$,
$b=\left\\{b_{i}\right\\}_{i=1}^{\infty}\in l_{s}$ we put
$a^{(n)}:=\sum_{i=1}^{n}a_{i}e_{i},b^{(n)}:=\sum_{i=1}^{n}b_{i}e_{i}.$ Since
$X$ and $Y$ are relatively $s$-decomposable, for any
$\left\\{x_{i}\right\\}\in\mathfrak{B}_{n}\left(X\right)$ and
$\left\\{y_{i}\right\\}_{i=1}^{n}\in\mathfrak{B}_{n}\left(Y\right)$ it holds

$\left\|\sum_{i=1}^{n}a_{i}b_{i}y_{i}\right\|_{Y}\leq
D\left(\sum_{i=1}^{n}\left|b_{i}\right|^{s}\right)^{1/s}\left\|\sum_{i=1}^{n}a_{i}x_{i}\right\|_{X}.$

By taking the supremum over all sequences $\left\\{y_{i}\right\\}_{i=1}^{n}$
and the infimum over all sequences $\left\\{x_{i}\right\\}_{i=1}^{n}$ we infer

$\left\|a^{(n)}b^{(n)}\right\|_{Y_{U}\left(n\right)}\leq
D\left\|b\right\|_{l_{s}}\Phi_{n}\left(a^{(n)}\right)$

Next, write $a^{(n)}=\sum_{k\in F}a^{k}$ for some finite set
$F\subset\mathbb{N}$ and $a^{k}=\\{a^{k}_{i}\\}_{i=1}^{n}$, $k\in F$. Then
$a^{(n)}b^{(n)}=\sum_{k\in F}b^{(n)}a^{k}$ and the preceding estimate implies

$\left\|a^{(n)}b^{(n)}\right\|_{Y_{U}\left(n\right)}\leq\sum_{k\in
F}\left\|b^{(n)}a^{k}\right\|_{Y_{U}\left(n\right)}\leq
D\left\|b\right\|_{l_{s}}\left(\sum_{k\in
F}\Phi_{n}\left(a^{k}\right)\right).$

After taking the infimum over all such decompositions of $a^{(n)}$ we obtain

$\left\|a^{(n)}b^{(n)}\right\|_{Y_{U}\left(n\right)}\leq
D\left\|b\right\|_{l_{s}}\left\|a^{(n)}\right\|_{X_{L}\left(n\right)},\;\;n\in\mathbb{N},$

which implies the claimed result (see also Lemma 3.6). ∎

######  Proposition 4.2.

Suppose that $X$ and $Y$ are relatively $s$-decomposable Banach lattices for
some $1\leq s\leq\infty$, $l_{p}$ is finitely lattice representable in
$Y_{U}$, where $p\leq s$.

Then, $X$ satisfies a lower $q$-estimate for every $q$ such that
${1}/{p}\geq{1}/{q}+1/{s}$, and $M_{\left[q\right]}\left(X\right)\leq
D_{s}\left(X,Y\right)$.

###### Proof.

Let $n$ be any positive integer and $\varepsilon>0$ be arbitrary. By the
assumption, we can find pair-wise disjoint elements $u_{i}\in Y_{U}$,
$i=1,\dots,n$, satisfying

(4.1)
$\left\|b\right\|_{l_{p}^{n}}\leq\left\|\sum_{i=1}^{n}b_{i}u_{i}\right\|_{Y_{U}}\leq\left(1+\varepsilon\right)\left\|b\right\|_{l_{p}^{n}}$

for all sequences $b=\left\\{b_{i}\right\\}_{i=1}^{n}$ of scalars.

Let $\left\\{x_{i}\right\\}_{i=1}^{n}\in\mathfrak{B}_{n}\left(X\right)$ and
$D>D_{s}\left(X,Y\right).$ Then for any sequence
$\left\\{a_{i}\right\\}_{i=1}^{n}$ of scalars, by using (4.1), Proposition
3.7(i) and relative $s$-decomposability of $X$ and $Y$, we have

$\displaystyle\left(\sum_{i=1}^{n}\left|a_{i}b_{i}\right|^{p}\right)^{1/p}$
$\displaystyle\leq$
$\displaystyle\left\|\sum_{i=1}^{n}a_{i}b_{i}u_{i}\right\|_{Y_{U}}\leq\left\|\sum_{i=1}^{n}a_{i}b_{i}\left\|u_{i}\right\|_{Y_{U}}e_{i}\right\|_{Y_{U}}$
$\displaystyle=$
$\displaystyle\sup\left\\{\left\|\sum_{i=1}^{n}a_{i}b_{i}\left\|u_{i}\right\|_{Y_{U}}y_{i}\right\|_{Y}:\left\\{y_{i}\right\\}_{i=1}^{n}\in\mathfrak{B}_{n}\left(Y\right)\right\\}$
$\displaystyle\leq$ $\displaystyle
D\left(\sum_{i=1}^{n}\left|b_{i}\right|^{s}\right)^{1/s}\sup_{1\leq i\leq
n}\left\|u_{i}\right\|_{Y_{U}}\left\|\sum_{i=1}^{n}a_{i}x_{i}\right\|_{X}$

Consequently, since from (4.1) it follows $\left\|u_{i}\right\|_{Y_{U}}\leq
1+\varepsilon$, we get

$\left(\sum_{i=1}^{n}\left|a_{i}b_{i}\right|^{p}\right)^{1/p}\leq\left(1+\varepsilon\right)D\left(\sum_{i=1}^{n}\left|b_{i}\right|^{s}\right)^{1/s}\left\|\sum_{i=1}^{n}a_{i}x_{i}\right\|_{X}.$

Hence, by the reverse Hölder inequality,

$\left(\sum_{i=1}^{n}\left|a_{i}\right|^{q}\right)^{1/q}\leq\left(1+\varepsilon\right)D\left\|\sum_{i=1}^{n}a_{i}x_{i}\right\|_{X}$

whenever ${1}/{q}\leq{1}/{p}-1/{s}$. Thus, $X$ satisfies a lower $q$-estimate
and $M_{\left[q\right]}(X)\leq D_{s}\left(X,Y\right)$. ∎

######  Proposition 4.3.

Suppose that $X$ and $Y$ are relatively $s$-decomposable Banach lattices for
some $1\leq s\leq\infty$, $l_{q}$ is finitely lattice representable in $X_{L}$
and $1/q+1/s\leq 1$.

Then, $Y$ satisfies an upper $p$-estimate for every $p$ such that
${1}/{p}\geq{1}/{q}+1/{s}$, and $M^{\left[p\right]}\left(Y\right)\leq
D_{s}\left(X,Y\right)$.

###### Proof.

Let $n$ be a positive integer and $\varepsilon>0$ be arbitrary. By the
assumption, we can select pair-wise disjoint elements $u_{i}\in X_{L}$,
$i=1,\dots,n$, such that

(4.2)
$\left\|b\right\|_{l_{q}^{n}}\leq\left\|\sum_{i=1}^{n}b_{i}u_{i}\right\|_{X_{L}}\leq\left(1+\varepsilon\right)\left\|b\right\|_{l_{q}^{n}}$

for all sequences $b=\left\\{b_{i}\right\\}_{i=1}^{n}$ of scalars.

Suppose $\left\\{y_{i}\right\\}_{i=1}^{n}\in\mathfrak{B}_{n}\left(Y\right)$
and $D>D_{s}\left(X,Y\right).$ For each $b=\left\\{b_{i}\right\\}_{i=1}^{n}$
we write $b=\sum_{k\in F}b^{k}$, where $F\subset\mathbb{N}$ is a finite set
and $b^{k}=\\{b_{i}^{k}\\}_{i=1}^{n}$ are arbitrary. Then, for every sequences
$\left\\{x_{i}^{k}\right\\}_{i=1}^{n}\in\mathfrak{B}_{n}\left(X\right)$, $k\in
F$, and any sequence $a=\left\\{a_{i}\right\\}_{i=1}^{n}$ of scalars, by the
triangle inequality and relative $s$-decomposability of $X$ and $Y$, we have

$\left\|\sum_{i=1}^{n}a_{i}b_{i}y_{i}\right\|_{Y}\leq\sum_{k\in
F}\left\|\sum_{i=1}^{n}a_{i}b_{i}^{k}y_{i}\right\|_{Y}\leq
D\left(\sum_{i=1}^{n}\left|a_{i}\right|^{s}\right)^{1/s}\sum_{k\in
F}\left\|\sum_{i=1}^{n}b_{i}^{k}x_{i}^{k}\right\|_{X}.$

Therefore, taking the infimum over all sequences
$\left\\{x_{i}^{k}\right\\}_{i=1}^{n}\in\mathfrak{B}_{n}\left(X\right)$ for
each $k\in F$ implies that

$\left\|\sum_{i=1}^{n}a_{i}b_{i}y_{i}\right\|_{Y}\leq
D\left\|a\right\|_{l_{s}^{n}}\sum_{k\in F}\Phi_{n}\left(b^{k}\right),$

and hence

$\left\|\sum_{i=1}^{n}a_{i}b_{i}y_{i}\right\|_{Y}\leq
D\left\|a\right\|_{l_{s}^{n}}\left\|b\right\|_{X_{L}}.$

Thus, applying Proposition 3.7(ii) and inequalities (4.2), we obtain

$\displaystyle\left\|\sum_{i=1}^{n}a_{i}b_{i}y_{i}\right\|_{Y}$
$\displaystyle\leq$ $\displaystyle
D\left\|a\right\|_{l_{s}^{n}}\left\|\sum_{i=1}^{n}b_{i}e_{i}\right\|_{X_{L}}\leq
D\left\|a\right\|_{l_{s}^{n}}\left\|\sum_{i=1}^{n}b_{i}\frac{u_{i}}{\left\|u_{i}\right\|_{X_{L}}}\right\|_{X_{L}}$
$\displaystyle\leq$ $\displaystyle
D\left\|a\right\|_{l_{s}^{n}}\left\|\sum_{i=1}^{n}b_{i}{u_{i}}\right\|_{X_{L}}\leq
D\left(1+\varepsilon\right)\left\|a\right\|_{l_{s}^{n}}\left\|b\right\|_{l_{q}^{n}}.$

By the reverse Hölder inequality, this implies that

$\left\|\sum_{i=1}^{n}c_{i}y_{i}\right\|_{Y}\leq
D\left(1+\varepsilon\right)\left(\sum_{i=1}^{n}\left|c_{i}\right|^{p}\right)^{1/p},$

whenever $\frac{1}{p}\geq\frac{1}{q}+\frac{1}{s}$. As a result, $Y$ satisfies
an upper $p$-estimate and $M^{\left[p\right]}(Y)\leq D_{s}(X,Y)$. ∎

######  Proposition 4.4.

Suppose Banach lattices $X$ and $Y$ satisfy the following conditions:

$\left(a\right)$ $X$, $Y$ are relatively s-decomposable for some $1\leq
s\leq\infty$;

$\left(b\right)$ $l_{p}$ is finitely lattice representable in $Y_{U}$ ;

$\left(c\right)$ $l_{q}$ is finitely lattice representable in $X_{L}$.

Then, it holds

$\frac{1}{p}\leq\frac{1}{q}+\frac{1}{s}$

###### Proof.

Let $n$ be a positive integer and $\varepsilon>0$ be arbitrary. By assumption
$\left(b\right)$, there exist pair-wise disjoint elements $y_{i}\in Y_{U}$,
$i=1,\dots,n$, such that for all scalar sequences
$b=\left\\{b_{i}\right\\}_{i=1}^{n}$

$\left\|b\right\|_{l_{p}^{n}}\leq\left\|\sum_{i=1}^{n}b_{i}y_{i}\right\|_{Y_{U}(n)}\leq\left(1+\varepsilon\right)\left\|b\right\|_{l_{p}^{n}}.$

In the same manner, using $\left(c\right)$, we can select pair-wise disjoint
$x_{i}\in X_{L}$, $i=1,\dots,n$, such that for all scalar sequences
$a=\left\\{a_{i}\right\\}_{i=1}^{n}$

$\left\|a\right\|_{l_{q}^{n}}\leq\left\|\sum_{i=1}^{n}a_{i}x_{i}\right\|_{X_{L}(n)}\leq\left(1+\varepsilon\right)\left\|a\right\|_{l_{q}^{n}}.$

Applying these inequalities and Propositions 3.7(i), 4.1 and 3.7(ii), we
obtain

$\displaystyle\left\|ab\right\|_{l_{p}^{n}}$ $\displaystyle\leq$
$\displaystyle\left\|\sum_{i=1}^{n}a_{i}b_{i}y_{i}\right\|_{Y_{U}(n)}\leq\left\|\sum_{i=1}^{n}a_{i}b_{i}\left\|y_{i}\right\|_{Y_{U}}e_{i}\right\|_{Y_{U}(n)}$
$\displaystyle\leq$
$\displaystyle\left(1+\varepsilon\right)D_{s}\left(X,Y\right)\left\|b\right\|_{l_{s}^{n}}\left\|\sum_{i=1}^{n}a_{i}e_{i}\right\|_{X_{L}(n)}$
$\displaystyle\leq$
$\displaystyle\left(1+\varepsilon\right)D_{s}\left(X,Y\right)\left\|b\right\|_{l_{s}^{n}}\left\|\sum_{i=1}^{n}a_{i}x_{i}\right\|_{X_{L}(n)}$
$\displaystyle\leq$
$\displaystyle\left(1+\varepsilon\right)^{2}D_{s}\left(X,Y\right)\left\|b\right\|_{l_{s}^{n}}\left\|a\right\|_{l_{q}^{n}}.$

Since $n\in\mathbb{N}$, $b=\left\\{b_{i}\right\\}_{i=1}^{n}$ and
$a=\left\\{a_{i}\right\\}_{i=1}^{n}$ are arbitrary, the claim follows (see
also Example 2.2). ∎

Recall that

$s_{\max}=s_{\max}(X,Y):=\sup\\{s\in\left[1,\infty\right]:\,X,Y\mbox{are
$s$-decomposable}\\}.$

######  Proposition 4.5.

Let $X$ and $Y$ be Banach lattices such that $\delta(Y)\leq\sigma(X)$. Then,
we have

(4.3)
$\frac{1}{\delta\left(Y\right)}=\frac{1}{\sigma\left(X\right)}+\frac{1}{s_{\max}}.$

###### Proof.

Assume first that $s_{\max}>1$. Then, if $1\leq s<s_{\max}$, $X$ and $Y$ are
relatively $s$-decomposable. Moreover, by Schep’s result [29],
$l_{s\left(Y_{U}\right)}$ and $l_{\sigma\left(X_{L}\right)}$ are finitely
representable in $Y_{U}$ and $X_{L}$, respectively. Hence, all the conditions
of Proposition 4.4 are fulfilled and we conclude

$\frac{1}{\delta\left(Y_{U}\right)}\leq\frac{1}{\sigma\left(X_{L}\right)}+\frac{1}{s}.$

On the other hand, by Corollary 3.11, $s\left(Y_{U}\right)=s\left(Y\right)$
and $\sigma\left(X_{L}\right)=\sigma\left(X\right)$. Consequently, we have

$\frac{1}{\delta\left(Y\right)}\leq\frac{1}{\sigma\left(X\right)}+\frac{1}{s}.$

Since this holds for all $s<s_{\max}$, it follows

(4.4)
$\frac{1}{\delta\left(Y\right)}\leq\frac{1}{\sigma\left(X\right)}+\frac{1}{s_{\max}}.$

Observe that the same arguments work also in the case when $s_{\max}=1$,
because every Banach couples $X$ and $Y$ are relatively $1$-decomposable.
Therefore, we again get inequality (4.4).

For the opposite inequality, assume first that $s_{\max}=\infty$. Then, (4.4)
implies that $\sigma\left(X\right)\leq\delta\left(Y\right)$. Combining this
inequality with the assumption, we conclude that
$\sigma\left(X\right)=\delta\left(Y\right)$, and hence (4.4) becomes (4.3).

Let now $s_{\max}<\infty$. Assume that (4.3) fails, i.e.,

$\frac{1}{\delta\left(Y\right)}<\frac{1}{\sigma\left(X\right)}+\frac{1}{s_{\max}}.$

If $\delta\left(Y\right)>1$ and $\sigma\left(X\right)<\infty$, we can find
$1\leq p<\delta\left(Y\right)$, $q>\sigma\left(X\right)$ and $s>s_{\max}$ such
that $1/p=1/q+1/s$. Since $X$ satisfies a lower $q$-estimate and $Y$ an upper
$p$-estimate, from Proposition 2.4 it follows that $X$ and $Y$ are relatively
$s$-decomposable, which is impossible, since $s>s_{\max}$. Thus, in this case
(4.3) is proved.

If $\delta\left(Y\right)=1$ or $\sigma\left(X\right)=\infty$, the proof
follows by the same lines in view of the fact that each Banach lattice admits
an upper $1$-estimate and a lower $\infty$-estimate. ∎

###### Proof of Theorem 2.8.

We start with the case when $\delta(Y)\leq\sigma(X)$.

$\left(i\right)\Longrightarrow\left(ii\right)$. Assume first that
$s_{\max}>1$, $\delta(Y)>1$ and $\sigma(X)<\infty$.

Let $1\leq s<s_{\max}$. Then, by Proposition 4.5, we have

$\frac{1}{\delta\left(Y\right)}<\frac{1}{\sigma\left(X\right)}+\frac{1}{s}.$

Consequently, for some $1\leq p_{1}<\delta\left(Y\right)$ and
$q_{1}>\sigma\left(X\right)$ we obtain

$\frac{1}{p_{1}}=\frac{1}{\sigma\left(X\right)}+\frac{1}{s}\;\;\mbox{and}\;\;\frac{1}{\delta\left(Y\right)}=\frac{1}{q_{1}}+\frac{1}{s}.$

Since $\delta\left(Y_{U}\right)=\delta\left(Y\right)$ and
$\sigma\left(X_{L}\right)=\sigma\left(X\right)$ (see Corollary 3.11), by [29],
$l_{\delta\left(Y\right)}$ (resp. $l_{\sigma\left(X\right)}$) is finitely
lattice representable in $Y_{U}$ (resp. in $X_{L}$). Therefore, according to
Propositions 4.2 and 4.3, $X$ satisfies a lower $q_{1}$-estimate, $Y$
satisfies an upper $p_{1}$-estimate and
$M_{\left[q_{1}\right]}\left(X\right)\leq D_{s}\left(X,Y\right)$,
$M^{\left[p_{1}\right]}\left(Y\right)\leq D_{s}\left(X,Y\right)$. Next, if
$1/p=1/q+1/s$, where $p<\delta\left(Y\right)$ and $q>\sigma\left(X\right)$, we
have $p<p_{1}$ and $q>q_{1}$. Hence, $X$ satisfies a lower $q$-estimate, $Y$
satisfies an upper $p$-estimate and

$M_{\left[q\right]}\left(X\right)M^{\left[p\right]}\left(Y\right)\leq
M_{\left[q_{1}\right]}\left(X\right)M^{\left[p_{1}\right]}\left(Y\right)\leq
D_{s}\left(X,Y\right)^{2}.$

Suppose now that $X$ and $Y$ are relatively $s_{\max}$-decomposable. Since
$l_{\delta\left(Y\right)}$ is finitely lattice representable in $Y_{U}$, by
Propositions 4.2 and 4.5, $X$ satisfies a lower
$\sigma\left(X\right)$-estimate and
$M_{\left[\sigma\left(X\right)\right]}\left(X\right)\leq
D_{s}\left(X,Y\right)$. In the same manner, applying this time Proposition
4.3, we infer that $Y$ satisfies an upper $\delta\left(Y\right)$-estimate and
$M^{\left[\delta\left(Y\right)\right]}\left(Y\right)\leq
D_{s}\left(X,Y\right)$. Combining this together with equality (4.3), we come
to the desired result.

If $s_{\max}=1$, or $\delta\left(Y\right)=1$, or
$\sigma\left(X\right)=\infty$, we can use the same arguments, taking into
account that every Banach lattices $X$ and $Y$ are relatively $1$-decomposable
and each Banach couple satisfies an upper $1$-estimate and a lower
$\infty$-estimate.

$\left(ii\right)\Longrightarrow\left(i\right)$. This implication together with
the inequality

$D_{s}\left(X,Y\right)\leq
M_{\left[q\right]}\left(X\right)M^{\left[p\right]}\left(Y\right)$

is an immediate consequence of Proposition 2.4.

To complete the proof of the equivalence of $\left(i\right)$,
$\left(ii\right)$ and $\left(iii\right)$ it remains now to refer to
Proposition 2.3.

Finally, let us prove the equivalence of the conditions
$\sigma(X)\leq\delta(Y)$ and $s_{max}=\infty$.

If $\sigma\left(X\right)<\delta\left(Y\right)$, then $X$ satisfies a lower
$p$-estimate and $Y$ an upper $p$-estimate for
$p\in(\sigma\left(X\right),\delta\left(Y\right))$. Therefore, by Proposition
2.4, $X$, $Y$ are relatively decomposable. Hence, $s_{max}=\infty$. If
$\sigma\left(X\right)=\delta\left(Y\right)$, the same result follows from
Proposition 4.5.

On the contrary, assume that $s_{max}=\infty$. Then, $X$ and $Y$ are
relatively $s$-decomposable for each $s<\infty$. Therefore, since
$\delta\left(Y_{U}\right)=\delta\left(Y\right)$ and
$\sigma\left(X_{L}\right)=\sigma\left(X\right)$, by Proposition 4.4, we infer

$\frac{1}{\delta\left(Y\right)}\leq\frac{1}{\sigma\left(X\right)}+\frac{1}{s}.$

Tending $s\to\infty$, we get the required inequality, and so the proof is
completed. ∎

Recall that the main result of the paper [11] (see Theorem 1.3) reads that
Banach function lattices $X$, $Y$ are relatively decomposable (or
$\infty$-decomposable) if and only if there exists $p\geq 1$ such that $X$
satisfies a lower $p$-estimate and $Y$ an upper $p$-estimate. As an immediate
consequence of Theorem 2.8 and its proof we obtain the following extension of
this result to general Banach lattices.

######  Corollary 4.6.

Banach lattices $X$, $Y$ are relatively decomposable if and only if there
exists $p\geq 1$ such that $X$ satisfies a lower $p$-estimate and $Y$ an upper
$p$-estimate.

######  Remark 4.7.

In contrast to [11], our definition of relative decomposability (see
Definition 2.1) deals only with finite sums. Thanks to that, we need not to
impose on lattices $X$ and $Y$ any extra condition. In particular, if $X$ and
$Y$ are Banach lattices of measurable functions on a $\sigma$-finite measure
space we omit the assumption from [11, Theorem 1.3] that $Y$ has the Fatou
property.

From Proposition 3.9 and the proof of Theorem 2.8 we also deduce the following
result.

######  Corollary 4.8.

If Banach lattices $X$, $Y$ are relatively $s_{max}$-decomposable, then $X$
admits a lower $\sigma(X)$-estimate and $Y$ admits an upper
$\delta(Y)$-estimate (equivalently, $X_{L}=l_{\sigma(X)}$ and
$Y_{U}=l_{\delta(Y)}$).

##  5\. Applications to interpolation theory: Calderón-Mityagin couples of
type $s$.

In this section, we freely use notation and results from interpolation theory
as in [4], [3], [5].

Let $\left(X,\Sigma,\mu\right)$ be a $\sigma$-finite measure space. A
$\Sigma$-measurable function $\omega$ is called a weight if $\omega$ is non-
negative $\mu$-a.e. on $X$. Let $1\leq p\leq\infty$ and let
$L_{p}\left(\omega,\mu\right)$ be the Banach space of all (equivalence classes
of) $\Sigma$-measurable functions $f$ with $f\omega\in L_{p}\left(\mu\right).$
Given $1\leq p_{0},p_{1}\leq\infty$ put
$\overrightarrow{p}=\left(p_{0},p_{1}\right).$ A Banach couple
$\overrightarrow{U}=(U_{0},U_{1})$ of Banach lattices is called a
$L_{\overrightarrow{p}}$-couple if
$U_{i}=L_{p_{i}}\left(\omega_{i},\mu\right)$, $i=0,1$, for some measure space
$\left(X,\Sigma,\mu\right)$ and some weights $\omega_{0},\omega_{1}$ with
respect to this measure space.

Let $1\leq s_{0},s_{1}\leq\infty$ and $\overrightarrow{X},\overrightarrow{Y}$
be two Banach couples of Banach lattices such that $X_{i},Y_{i}$ are
relatively $s_{i}$-decomposable for $i=0,1$. Then, by Theorem 2.8, there exist
$1\leq p_{0},p_{1},q_{0},q_{1}\leq\infty$ with $1/p_{i}=1/q_{i}+1/s_{i}$,
$i=0,1$, such that for every $L_{\overrightarrow{q}}$-couple
$\overrightarrow{U}=(U_{0},U_{1})$ and $L_{\overrightarrow{p}}$-couple $\vskip
6.0pt plus 2.0pt minus 2.0pt\vskip 6.0pt plus 2.0pt minus
2.0pt\overrightarrow{V}=(V_{0},V_{1})$ both $X_{i},U_{i}$ and $V_{i},Y_{i}$
are relative decomposable for $i=0,1$.

Combining the last observation with the results of [11], we see that each of
the pairs of the couples $\overrightarrow{X},\overrightarrow{U}$ and
$\overrightarrow{V},\overrightarrow{Y}$ have the relative Calderón-Mityagin
property ($\mathcal{C-M}$ property). Hence, the $s$-decomposability relation
of couples of Banach lattices has some transitivity property, which is
manifested in factorization of this relation through the canonical
$s$-decomposability of suitable $L_{\overrightarrow{q}}$\- and
$L_{\overrightarrow{p}}$-couples. More precisely, we get the following result.

######  Theorem 5.1.

Let $\overrightarrow{X},\overrightarrow{Y}$ be two couples of Banach lattices
over a $\sigma$-finite measure space. If the spaces $X_{i},Y_{i}$ are relative
$s_{i}$-decomposable for $i=0,1$, where $1\leq s_{i}\leq\infty,$ then there
exist pairs $\overrightarrow{p}$, $\overrightarrow{q}$ of parameters such
that, for every $L_{\overrightarrow{q}}$-couple
$\overrightarrow{U}=(U_{0},U_{1})$ and every $L_{\overrightarrow{p}}$-couple
$\overrightarrow{V}=(V_{0},V_{1}),$ pairs of the couples
$\overrightarrow{X},\overrightarrow{U}$ and
$\overrightarrow{V},\overrightarrow{Y}$ have the relative $\mathcal{C-M}$
property and the spaces $U_{i},V_{i}$ are relatively $s_{i}$-decomposable,
$i=0,1$.

There are many pairs of Banach couples $\overrightarrow{X}$ and
$\overrightarrow{Y}$, which fail to have relative $\mathcal{C-M}$ property. In
[9], Cwikel introduced the following weaker condition that may be satisfied by
such a pair of Banach couples.

Let $\overrightarrow{X}=(X_{0},X_{1})$ and $\overrightarrow{Y}=(Y_{0},Y_{1})$
be two Banach couples. Given $1\leq s\leq\infty,$ define the relation $R_{s}$
for $(x,y)\in(X_{0}+X_{1})\times(Y_{0}+Y_{1})$ by

$xR_{s}y\iff\exists w\in
L_{s}\left((0,\infty),dt/t\right)\;\mbox{with}\;K(t,y;\overrightarrow{Y})\leq
w(t)\cdot K(t,x;\overrightarrow{X}),\;t>0.$

We say that the Banach couples $\overrightarrow{X},\overrightarrow{Y}$ are of
relative $\mathcal{C-M}$ type $s$ whenever the relation $xR_{s}y$ implies that
$y=Tx$ for some linear operator
$T:\,\overrightarrow{X}\rightarrow\overrightarrow{Y}$ (i.e.,
$T:\,X_{0}+X_{1}\rightarrow Y_{0}+Y_{1}$, and $T$ is bounded from $X_{i}$ into
$Y_{i}$, $i=0,1$).

Since each $K$-functional is a concave nondecreasing function in $t$, we can
assume that the function $w$ in this definition is continuous or constant on
each dyadic interval. From this observation it follows easily that if
$\overrightarrow{X},\overrightarrow{Y}$ are of relative $\mathcal{C-M}$ type
$s_{1}$ and $1\leq s_{2}\leq s_{1}$, then these couples are also of relative
$\mathcal{C-M}$ type $s_{2}$. Furthermore, it is known [7, Theorem 1] that
arbitrary couples $\overrightarrow{X},\overrightarrow{Y}$ are of relative
$\mathcal{C-M}$ type $1$. Hence, the set of real numbers $s$ in
$\left[1,\infty\right]$ such that $\overrightarrow{X},\overrightarrow{Y}$ are
of relative $\mathcal{C-M}$ type $s$ is an interval which includes $1.$ In [7]
and [9] one can find examples of Banach couples, for which this interval is
$[1,q]$, $1\leq q<\infty$, or $[1,\infty)$ (of course, it is
$\left[1,\infty\right]$ iff $\overrightarrow{X},\overrightarrow{Y}$ have the
relative $\mathcal{C-M}$ property).

Further, in [9], Cwikel proved that, if the couples
$\overrightarrow{X},\overrightarrow{Y}$ are mutually closed and $X_{i},Y_{i}$,
$i=0,1$, are relatively $s$-decomposable for some $1\leq s\leq\infty$, then
these couples are of relative $\mathcal{C-M}$ type $s$ (see also [4, p. 606]).
Let us show that, under some conditions, this implies the orbital
factorization of relative $K$-functional estimates for such couples through
suitable $L_{\overrightarrow{p}}$\- and $L_{\overrightarrow{q}}$-couples.

Given Banach couples $\overrightarrow{X},\overrightarrow{Y}$ the couple
$\overrightarrow{Y}$ is called $\overrightarrow{X}$-abundant, if for each
element $x\in X_{0}+X_{1}$ there exists $y\in Y_{0}+Y_{1}$ such that

$K(t,x;\overrightarrow{X})\asymp K(t,y;\overrightarrow{Y})$

with constants independent of $x\in X_{0}+X_{1}$ and $t>0$ (see e.g. [4,
Definition 4.4.8]). For instance, if a couple $\overrightarrow{X}$ is regular
(i.e., $X_{0}\cap X_{1}$ is dense in $X_{0}$ and $X_{1}),$ then the
$L_{\overrightarrow{p}}$-couples
$\left(l_{p_{0}}\left(\mathbb{Z},\left(1\right)_{n\in\mathbb{Z}}\right),l_{p_{1}}\left(\mathbb{Z},\left(2^{-n}\right)_{n\in\mathbb{Z}}\right)\right)$
and
$\left(L_{p_{0}}\left(\mathbb{R}_{+},dt/t\right),L_{p_{1}}\left(\mathbb{R}_{+},dt/t\right)\right)$
are $\overrightarrow{X}$-abundant for each pair
$\overrightarrow{p}=\left(p_{0},p_{1}\right)$ [4, Theorem 4.5.7]. With this
notation, we have the following version of Theorem 5.1.

######  Theorem 5.2.

Let $\overrightarrow{X}=(X_{0},X_{1})$ and $\overrightarrow{Y}=(Y_{0},Y_{1})$
be two Banach lattice couples over a $\sigma$-finite measure space such that
$X_{i},Y_{i}$, $i=0,1$, are relatively $s$-decomposable for some $1\leq
s\leq\infty$. Then, there are pairs $\overrightarrow{p}=(p_{0},p_{1})$ and
$\overrightarrow{q}=(q_{0},q_{1})$ of parameters such that for every
$L_{\overrightarrow{q}}$-couple $\overrightarrow{U}$, which is
$\overrightarrow{X}$-abundant, and every $L_{\overrightarrow{p}}$-couple
$\overrightarrow{V}$, which is $\overrightarrow{Y}$-abundant, we have the
following: If $x\in X_{0}+X_{1},y\in Y_{0}+Y_{1}$ satisfy the relation
$xR_{s}y$, then there exist linear operators
$T_{0}:\overrightarrow{X}\rightarrow\overrightarrow{U}$,
$T_{1}:\overrightarrow{U}\rightarrow\overrightarrow{V}$,
$T_{2}:\overrightarrow{V}\rightarrow\overrightarrow{Y}$ such that
$y=T_{2}T_{1}T_{0}x$.

###### Proof.

Applying first Theorem 5.1, we find parameters $1\leq p_{i},q_{i}\leq\infty$,
$1/p_{i}=1/q_{i}+1/s$, $i=0,1$, such that, if
$\overrightarrow{p}=(p_{0},p_{1})$, $\overrightarrow{q}=(q_{0},q_{1})$, then
for every $L_{\overrightarrow{q}}$-couple $\overrightarrow{U}=(U_{0},U_{1})$
and every $L_{\overrightarrow{p}}$-couple $\overrightarrow{V}=(V_{0},V_{1}),$
pairs of the couples $\overrightarrow{X},\overrightarrow{U}$ and
$\overrightarrow{V},\overrightarrow{Y}$ have relative $\mathcal{C-M}$ property
and the spaces $U_{i},V_{i}$, $i=0,1$, are relatively $s$-decomposable. Next,
assuming that $x\in X_{0}+X_{1},y\in Y_{0}+Y_{1}$ satisfy $xR_{s}y$, by using
the abundance assumption, we can select $u\in U_{0}+U_{1}$ and $v\in
V_{0}+V_{1}$ such that

$\displaystyle K(t,x;\overrightarrow{X})$ $\displaystyle\asymp$ $\displaystyle
K(t,u;\overrightarrow{U})$ $\displaystyle K(t,y;\overrightarrow{Y})$
$\displaystyle\asymp$ $\displaystyle K(t,v;\overrightarrow{V})$

with constants independent of $x\in X_{0}+X_{1}$, $y\in Y_{0}+Y_{1}$ and
$t>0$. Since the couples $\overrightarrow{X},\overrightarrow{U}$ and
$\overrightarrow{V},\overrightarrow{Y}$ have relative $\mathcal{C-M}$
property, we can find linear operators
$T_{0}:\overrightarrow{X}\rightarrow\overrightarrow{U}$ and
$T_{2}:\overrightarrow{V}\rightarrow\overrightarrow{Y}$ satisfying $u=T_{0}x$
and $y=T_{2}v.$ Moreover, as was above-mentioned (see [9]), the couples
$\overrightarrow{U}$ and $\overrightarrow{V}$ are of relative $\mathcal{C-M}$
type $s$. Hence, from the relation $xR_{s}y$ it follows the existence of a
linear operator $T_{1}:\overrightarrow{U}\rightarrow\overrightarrow{V}$ such
that $v=T_{1}u.$ ∎

Assume now that $\overrightarrow{X}=(X_{0},X_{1})$ and
$\overrightarrow{Y}=(Y_{0},Y_{1})$ are two Banach lattice couples such that
$X_{i},Y_{i}$ are relatively $\infty$-decomposable for $i=0,1$. Then, the
results of [11] imply that the couples $\overrightarrow{X}$ and
$\overrightarrow{Y}$ have relative $\mathcal{C-M}$ property. Arguing in the
same way as in the proof of Theorem 5.2, one can easily deduce the following
factorization result.

######  Theorem 5.3.

Let $\overrightarrow{X}=(X_{0},X_{1})$ and $\overrightarrow{Y}=(Y_{0},Y_{1})$
be two Banach lattice couples over a $\sigma$-finite measure space such that
$X_{0},Y_{0}$ and $X_{1},Y_{1}$ are relatively decomposable. Then, there is a
pair $\overrightarrow{p}=(p_{0},p_{1})$ of parameters such that for every
$L_{\overrightarrow{p}}$ couples $\overrightarrow{U}$ and $\overrightarrow{V}$
such that $\overrightarrow{U}$ is $\overrightarrow{X}$-abundant and
$\overrightarrow{V}$ is $\overrightarrow{Y}$-abundant we have the following:
If $x\in X_{0}+X_{1},y\in Y_{0}+Y_{1}$ satisfy

(5.1) $K(t,y;\overrightarrow{Y})\leq K(t,x;\overrightarrow{X}),\;\;t>0,$

then there exist linear operators
$T_{0}:\overrightarrow{X}\rightarrow\overrightarrow{U}$,
$T_{1}:\overrightarrow{U}\rightarrow\overrightarrow{V}$,
$T_{2}:\overrightarrow{V}\rightarrow\overrightarrow{Y}$ with
$y=T_{2}T_{1}T_{0}x.$

##  6\.  The proof of Theorem 3.3.

This proof will be broken down into a number of lemmas and propositions. The
main step is Proposition 6.1 showing under which conditions a scale of norms
on $\mathbb{R}^{n}$, $n\in\mathbb{N}$, generates a rearrangement invariant
Banach sequence lattice. The rest of this section is to secure that these
conditions are valid for both $X_{L}$\- and $X_{U}$-constructions.

Recall that a functional $\Psi$ (in particular, a norm $\left\|\cdot\right\|$)
defined on $\mathbb{R}^{n}$ is called lattice monotone or lattice norm if for
any elements
$a=\left\\{a_{i}\right\\}_{i=1}^{n},b=\left\\{b_{i}\right\\}_{i=1}^{n}\in\mathbb{R}^{n}$
such that $\left|a_{i}\right|\leq\left|b_{i}\right|,1\leq i\leq n$, it holds
$\Psi\left(a\right)\leq\Psi\left(b\right).$ This functional is said to be
symmetric if for any permutation $\sigma$ of the set
$\left\\{1,\dots,n\right\\}$ we have $\Psi\left(\sigma
a\right)=\Psi\left(a\right)$ where $\sigma
a=\left\\{a_{\sigma\left(i\right)}\right\\}_{i=1}^{n}.$ We introduce also the
operators

$\displaystyle I_{n}$ $\displaystyle:$
$\displaystyle\mathbb{R}^{n}\rightarrow\mathbb{R}^{n-1},\left\\{a_{i}\right\\}_{i=1}^{n}\mapsto\left\\{a_{i}\right\\}_{i=1}^{n-1}$
$\displaystyle Tr_{n}$ $\displaystyle:$
$\displaystyle\mathbb{R}^{n}\rightarrow\mathbb{R}^{n},\left\\{a_{i}\right\\}_{i=1}^{n}\mapsto\left\\{\begin{array}[]{c}a_{i}:i\neq
n\\\ 0:i=n\end{array}\right\\}$

As above, for any sequence $a=\left\\{a_{i}\right\\}_{i=1}^{\infty}$ of real
numbers and each integer $k$, by $a^{\left(k\right)}$ we will denote the
truncated sequence $a^{\left(k\right)}$ defined by
$a^{\left(k\right)}=\\{a_{i}^{(k)}\\}_{i=1}^{\infty}$, with
$a_{i}^{(k)}=a_{i}$ if $1\leq i\leq k$ and $a_{i}^{(k)}=0$ if $i>k$.

######  Proposition 6.1.

Let $\left\|\cdot\right\|_{n}$ be symmetric lattice norms on $\mathbb{R}^{n}$,
$n\in\mathbb{N}$. Assume that the restrictions $I_{n}$ are contractive with
respect to these norms. Denote by $Y$ the space of all sequences
$a=\left\\{a_{i}\right\\}_{i=1}^{\infty}$, for which the norm

$\left\|a\right\|_{Y}:=\sup_{n\geq
1}\left\|\left\\{a_{i}\right\\}_{i=1}^{n}\right\|_{n}$

is finite. If the space $Y$ is embedded into $c_{0},$ then $Y$ is a r.i.
Banach sequence lattice.

###### Proof.

First, one can easily check that the conditions
$a=\left\\{a_{i}\right\\}_{i=1}^{\infty}\in Y$ and
$\left|a_{i}\right|\leq\left|b_{i}\right|$, $i=1,2,\dots$, imply that
$b=\left\\{b_{i}\right\\}_{i=1}^{\infty}\in Y$ and $\|b\|_{Y}\leq\|a\|_{Y}$.
Consequently, $a\mapsto\left\|a\right\|_{Y}$ is a lattice norm on $Y$.

To prove the rearrangement invariance of $Y,$ assume that
$a=\left\\{a_{i}\right\\}_{i=1}^{\infty}\in Y$ and a sequence
$b=\left\\{b_{i}\right\\}_{i=1}^{\infty}$ is equi-measurable with $a.$ This
means that the sets $\left\\{i:\left|a_{i}\right|>t\right\\}$ and
$\left\\{i:\left|b_{i}\right|>t\right\\}$ have the same cardinality for every
$t>0$. Since $Y$ is embedded into $c_{0}$ these sets are finite and hence the
sets $A_{t}:=\left\\{i:\left|a_{i}\right|=t\right\\}$ and
$B_{t}:=\left\\{i:\left|b_{i}\right|=t\right\\}$ also have the same
cardinality for each $t>0.$ Put $t_{k}:=\left|b_{k}\right|$,
$B_{k}:=B_{t_{k}}$, $A_{k}:=A_{t_{k}}$, $k\in\mathbb{N}$.

Let $n\in\mathbb{N}$ be arbitrary. Take $u_{n}\in\mathbb{N}$ such that
$\cup_{k=1}^{n}A_{k}\subseteq\left\\{1,2,\dots,u_{n}\right\\}.$ Then, we have
$\left\\{\left|b_{k}\right|\right\\}_{k=1}^{n}\subseteq\left\\{\left|a_{k}\right|\right\\}_{k=1}^{u_{n}}$.
Indeed, if $1\leq k\leq n,$ then by construction $\left|b_{k}\right|=t_{k}\in
B_{k}$ and hence there exists $j\in A_{k}$ with
$\left|a_{j}\right|=\left|b_{k}\right|.$ Since
$A_{k}\subseteq\left\\{1,2,\dots,u_{n}\right\\}$, the conclusion follows.

Next, there is a permutation $\sigma$ of the set
$\left\\{1,2,\dots,u_{n}\right\\}$ with
$\left(\sigma\left|a\right|\right)_{k}=\left|b_{k}\right|$, $1\leq k\leq n.$
By the assumptions of the lemma, this implies the estimate

$\left\|\left\\{b_{k}\right\\}_{k=1}^{n}\right\|_{n}=\left\|\left\\{\left|b_{k}\right|\right\\}_{k=1}^{n}\right\|_{n}=\left\|\left\\{\sigma\left|a\right|\right\\}_{k=1}^{n}\right\|_{n}\leq\left\|\left\\{\sigma\left|a\right|\right\\}_{k=1}^{u_{n}}\right\|_{u_{n}}.$

Hence,

$\left\|\left\\{b_{k}\right\\}_{k=1}^{n}\right\|_{n}\leq\left\|a\right\|_{Y},\;\;n\in\mathbb{N},$

and so $b\in Y$ and $\left\|b\right\|_{Y}\leq\left\|a\right\|_{Y}.$ Similarly,
$\left\|a\right\|_{Y}\leq\left\|b\right\|_{Y},$ and thus
$\left\|a\right\|_{Y}=\left\|b\right\|_{Y}.$

By construction, $Y$ is a normed linear space of sequences. To prove
completeness of $Y$, take $\left\\{a^{n}\right\\}_{n=1}^{\infty}\subseteq Y$,
$a^{n}=\left\\{a_{i}^{n}\right\\}_{i=1}^{\infty}$, with
$\sum_{n=1}^{\infty}\left\|a^{n}\right\|_{Y}=C<\infty.$ Since $Y$ is embedded
in $c_{0}$, there exists $a\in c_{0}$ with $a=\sum_{n=1}^{\infty}a^{n}.$ Also,
for each integer $k$ we have

$\sum_{n=1}^{\infty}\|\\{a_{i}^{n}\\}_{i=1}^{k}\|_{k}\leq C.$

Hence, by completeness of the space $\mathbb{R}^{k}$ with respect to the norm
$\left\|\cdot\right\|_{k}$ and uniqueness of a representation of vectors by
using the canonical unit basis, we get
$a^{\left(k\right)}=\sum_{n=1}^{\infty}(a^{n})^{(k)}$ and
$\left\|a^{\left(k\right)}\right\|_{k}\leq C$ for all $k\in\mathbb{N}$.
Consequently, $a\in Y$ and $\left\|a\right\|_{Y}\leq C.$ The proposition is
proved. ∎

Next, we proceed with the postponed proof of Lemma 3.2 on the nonemptiness of
the sets $\mathfrak{B}_{n}\left(X\right)$, $n\in\mathbb{N}$. We will use the
notation $X_{+}$ for the positive cone $\left\\{x\in X:\,x\geq 0\right\\}$ of
a Banach lattice $X.$

###### Proof of Lemma 3.2.

If $l_{\infty}$ is finite lattice representable in $X$, then the desired
result follows immediately from Definition 2.5. Therefore, we can assume that
$l_{\infty}$ fails to be finite lattice representable in $X$, and hence, by
Proposition 2.7, $X$ is both $\sigma$-complete and $\sigma$-order continuous.
This implies that for each $x\in X_{+}$ we can define the contractive
projection $P_{x}:X\rightarrow X$ by $P_{x}\left(y\right)=\vee_{n\geq
1}\left(nx\wedge y\right),y\in X_{+},$ and then extend it by linearity to the
whole of $X$ (see e.g. [20]).

Suppose that $\left\\{x_{i}\right\\}_{i=1}^{m}$, where $m\in\mathbb{N}$, is a
maximal sequence of normalized positive pair-wise disjoint elements in a
Banach lattice $X$. We claim that $X$ has dimension not bigger than $m.$ We
will divide the proof of this fact into several parts.

$\left(i\right)$ Each element $x\in\left\\{x_{i}\right\\}_{i=1}^{m}$ is an
atom.

Assume that $x=y+z$ for some $y,z$ with
$\left|y\right|\wedge\left|z\right|=0.$ Since $x>0$, we have
$x=\left|y\right|+\left|z\right|$, and thus
$0\leq\left|y\right|,\left|z\right|\leq x.$ Hence, by maximality,
$\left|y\right|=x$ or $\left|z\right|=x$, i.e., $x$ is an atom.

$\left(ii\right)$ For every $x\in\left\\{x_{i}\right\\}_{i=1}^{m}$ the
projection $P_{x}$ has one dimensional range.

Recall that (see [20, p. 10])

(6.2) $\mathrm{Im}P_{x}=\left\\{z\in X:\,x\wedge y=0\;\mbox{for some}\;y\in
X_{+}\;\Longrightarrow\left|z\right|\wedge y=0\right\\}.$

Putting $z=P_{x}\left(y\right)$, where $y\in Y_{+}$, we have $z\geq 0$.
Without loss of generality, assume that $z>0.$ From (6.2) it follows that
$z\wedge x_{i}=0$ whenever $x_{i}\neq x.$ If $z\wedge x=0$ we get a
contradiction, because the set $\left\\{x_{i}\right\\}_{i=1}^{m}$ was selected
to be maximal. Hence, $0<z\wedge x\leq x$ and, since $x$ is an atom, we
conclude that $z\wedge x=\lambda x$ for some $\lambda>0.$ Observe that the set
$(\left\\{x_{i}\right\\}_{i=1}^{m}\smallsetminus\left\\{x\right\\})\cup\left\\{z/\|z\|_{X}\right\\}$
is also a maximal set of normalized positive pair-wise disjoint elements in
$X$. Consequently, from $\left(i\right)$ it follows that $z$ is an atom. Since
$\lambda x=x\wedge z\leq z$, this implies that $\lambda x=\mu z$ for some
scalar $\mu>0.$ Hence, $P_{x}$ has one dimensional range, generated by the
vector $x.$

$\left(iii\right)$ $X$ is the linear span of the sequence
$\left\\{x_{i}\right\\}_{i=1}^{m}.$

Put $x=\vee_{i=1}^{m}x_{i}$ and take $y\in X_{+}.$ Then, if
$z:=P_{x}\left(y\right)$, we have $x\wedge\left(y-z\right)=0.$ From the
inequalities $0\leq x_{i}\leq x$ and $0\leq z\leq y$ it follows that
$x_{i}\wedge\left(y-z\right)=0$ and hence, by maximality, we have
$y=z=P_{x}\left(y\right).$ Since $x\wedge y=\vee_{i=1}^{m}\left(x_{i}\wedge
y\right)$, we have for each integer $n$

$nx\wedge y=\vee_{i=1}^{m}\left(nx_{i}\wedge
y\right)=\sum_{i=1}^{m}nx_{i}\wedge
y\leq\sum_{i=1}^{m}P_{x_{i}}\left(y\right),$

which implies that

$y=P_{x}\left(y\right)\leq\sum_{i=1}^{m}P_{x_{i}}\left(y\right).$

By the decomposition property, we may write $y=\sum_{i=1}^{m}y_{i}$, where
$0\leq y_{i}\leq P_{x_{i}}\left(y\right)$, and hence
$y_{i}\in\mathrm{Im}P_{x_{i}}.$ Therefore, by $\left(ii\right)$,
$y_{i}=\lambda_{i}x_{i}$ for some scalars $\lambda_{i}$ and thus
$y=\sum_{i=1}^{m}\lambda_{i}x_{i}.$ As a result, the claim is proven and so
the lemma follows. ∎

######  Lemma 6.2.

Let $X$ be an infinite dimensional Banach lattice such that $l_{\infty}$ is
not finitely lattice representable in $X$. Then, for every sequence
$\left\\{x_{i}\right\\}_{i=1}^{n}\in\mathfrak{B}_{n}\left(X\right)$ and
$\varepsilon>0$ there exists a sequence
$\left\\{u_{i}\right\\}_{i=1}^{n+1}\in\mathfrak{B}_{n+1}\left(X\right)$ such
that either

$\left(i\right):$ $u_{i}=x_{i}$, $i=1,\dots,n$,

or

$\left(ii\right):$ there exists $k$ with $1\leq k\leq n$ and a bijection
$\psi:\left\\{1,..,n-1\right\\}\rightarrow\left\\{1,...,n\right\\}\smallsetminus\left\\{k\right\\}$
such that $u_{i}=x_{\psi\left(i\right)},1\leq i\leq n-1$ and $\alpha
u_{n}+\beta u_{n+1}=x_{k}$ for some positive scalars $\alpha$ and $\beta.$
Moreover, we have

$\left\|u_{n}-x_{k}\right\|_{X}\leq\varepsilon.$

###### Proof.

Given $\left\\{x_{i}\right\\}_{i=1}^{n}\in\mathfrak{B}_{n}\left(X\right)$ we
put $x=\vee_{i=1}^{n}x_{i}=\sum_{i=1}^{n}x_{i}.$ Take $y\in X_{+}$ and set
$z=P_{x}\left(y\right).$ Then $x\wedge\left(y-z\right)=0.$ If there exists $y$
such that $y\neq P_{x}\left(y\right)$, we define
$u_{n+1}:=\lambda\left(y-P_{x}\left(y\right)\right)$, where $\lambda$ is
selected so that $\left\|u_{n+1}\right\|=1.$ Then, setting $u_{i}=x_{i}$,
$i=1,\dots,n$, we see that the case $\left(i\right)$ holds.

Therefore, we may assume that $P_{x}\left(y\right)=y$ for each $y\in X_{+}.$
Hence, as in the proof of Lemma 3.2, it follows that $X$ is the direct sum of
the bands $P_{x_{i}}\left(X\right)$, $i=1,2,\dots,n$, and hence at least one
of them, say, $P_{x_{k}}\left(X\right)$, is infinite dimensional. Since
$x_{k}$ can not be an atom, we may write $x_{k}=u+v$, where $u,v\in X_{+}$ and
$u\wedge v=0.$ Further, at least one of the subspaces $P_{u}\left(X\right)$ or
$P_{v}\left(X\right)$ is again infinite dimensional. Arguing in the same way,
we conclude that, for any positive integer $m,$ $x_{k}$ is a sum of $m$ pair-
wise disjoint elements $w_{j}$, $j=1,2,\dots,m$. Without loss of generality,
assume that $\|w_{1}\|_{X}\geq\|w_{2}\|_{X}\geq\dots\geq\|w_{m}\|_{X}$. By the
assumption (see also Proposition 2.6), $X$ satisfies a lower $p$-estimate for
some $p<\infty.$ Consequently, we can estimate

$m^{1/p}\left\|w_{m}\right\|_{X}\leq\left(\sum_{i=1}^{m}\left\|w_{j}\right\|_{X}^{p}\right)^{1/p}\leq
M_{\left[p\right]}\left(x\right)\left\|x_{k}\right\|_{X}=M_{\left[p\right]}\left(X\right),$

whence $\lim_{m\rightarrow\infty}\left\|w_{m}\right\|_{X}=0.$

Put

$u_{n}=\left(x_{k}-w_{m}^{\left(m\right)}\right)/\left\|x-w_{m}^{\left(m\right)}\right\|_{X},\;\;u_{n+1}=w_{m}^{\left(m\right)}/\left\|w_{m}^{\left(m\right)}\right\|_{X}.$

Since $u_{n}\wedge u_{n+1}=0$, we have
$\left\\{u_{i}\right\\}_{i=1}^{n+1}\in\mathfrak{B}_{n+1}\left(X\right)$.
Moreover, by construction, $x_{k}=\alpha u_{n}+\beta u_{n+1}$, with
$\alpha=\|x_{k}-w_{m}^{\left(m\right)}\|_{X}$,
$\beta=\|w_{m}^{\left(m\right)}\|_{X}.$ Finally, since

$\displaystyle\left\|x_{k}-u_{n}\right\|_{X}$ $\displaystyle=$
$\displaystyle\left\|\frac{\left(\alpha-1\right)x_{k}+w_{m}^{\left(m\right)}}{\alpha}\right\|_{X}$
$\displaystyle\leq$
$\displaystyle\frac{|\|x_{k}-w_{m}^{\left(m\right)}\|_{X}-\|x_{k}\|_{X}|}{\alpha}\left\|x_{k}\right\|_{X}+\frac{\|w_{m}^{\left(m\right)}\|_{X}}{\alpha}$
$\displaystyle\leq$
$\displaystyle\frac{2}{\alpha}\left\|w_{m}^{\left(m\right)}\right\|_{X},$

we may select $m$ so that $\left\|x_{k}-u_{n}\right\|_{X}<\varepsilon.$ Thus,
all the conditions in $\left(ii\right)$ are fulfilled. ∎

######  Remark 6.3.

In the above proof we required that $l_{\infty}$ fails to be finitely lattice
representable in $X$. But, in fact, we need only a weaker property that if
$\left\\{x_{n}\right\\}_{n=1}^{\infty}$ is an infinite sequence of pair-wise
disjoint elements with decreasing norms in $X$, then
$\left\|x_{n}\right\|_{X}\downarrow 0.$ According the terminology from the
book [1], such a Banach lattice $X$ is said to have the Lebesgue property (see
[1, Theorem 3.22]).

######  Lemma 6.4.

Each of the functionals
$\left\|\cdot\right\|_{X_{U}\left(n\right)},\Phi_{n}\left(\cdot\right)$ and
$\left\|\cdot\right\|_{X_{L}\left(n\right)}$ defined on $\mathbb{R}^{n}$ is
lattice monotone and symmetric.

###### Proof.

Let $a=\left\\{a_{i}\right\\}_{i=1}^{n}$ and
$b=\left\\{b_{i}\right\\}_{i=1}^{n}$ be two sequences of scalars with
$\left|b_{i}\right|\leq\left|a_{i}\right|$, $1\leq i\leq n.$ Then, for every
sequence $\left\\{x_{i}\right\\}_{i=1}^{n}\in\mathfrak{B}_{n}\left(X\right)$
we have

(6.3)
$\left|\sum_{i=1}^{n}b_{i}x_{i}\right|=\sum_{i=1}^{n}\left|b_{i}\right|\left|x_{i}\right|\leq\sum_{i=1}^{n}\left|a_{i}\right|\left|x_{i}\right|=\left|\sum_{i=1}^{n}a_{i}x_{i}\right|.$

Also, let $\sigma$ be a permutation of the set $\left\\{1,\dots,n\right\\}$
and the sequence $c$ be defined by $c=\sigma
a:=\left\\{a_{\sigma\left(i\right)}\right\\}_{i=1}^{n}.$ Further, we prove the
desired claims for each functional separately.

$\left(i\right):\left\|\cdot\right\|_{X_{U}\left(n\right)}.$ From (6.3) it
follows that

$\left\|\sum_{i=1}^{n}b_{i}x_{i}\right\|_{X}\leq\left\|\sum_{i=1}^{n}a_{i}x_{i}\right\|_{X},$

which implies
$\left\|b\right\|_{X_{U}\left(n\right)}\leq\left\|a\right\|_{X_{U}\left(n\right)}$.
Consequently, $\left\|\cdot\right\|_{X_{U}\left(n\right)}$ is a lattice norm.

Next, since for every
$\left\\{x_{i}\right\\}_{i=1}^{n}\in\mathfrak{B}_{n}\left(X\right)$ and any
permutation $\pi$ of $\left\\{1,\dots,n\right\\}$ we have
$\left\\{x_{\pi(i)}\right\\}_{i=1}^{n}\in\mathfrak{B}_{n}\left(X\right)$,
denoting by $\sigma^{-1}$ the inverse permutation, we obtain

$\left\|\sum_{i=1}^{n}c_{i}x_{i}\right\|_{X}=\left\|\sum_{i=1}^{n}a_{\sigma\left(i\right)}x_{i}\right\|_{X}=\left\|\sum_{i=1}^{n}a_{i}x_{\sigma^{-1}\left(i\right)}\right\|_{X}\leq\left\|a\right\|_{X_{U}\left(n\right)}.$

Hence,
$\left\|c\right\|_{X_{U}\left(n\right)}\leq\left\|a\right\|_{X_{U}\left(n\right)}$,
and by symmetry we obtain that the norm
$\left\|\cdot\right\|_{X_{U}\left(n\right)}$ is symmetric.

$\left(ii\right):$ $\Phi_{n}\left(\cdot\right)$. In the same way, as above, we
have

$\Phi_{n}\left(b\right)\leq\left\|\sum_{i=1}^{n}b_{i}x_{i}\right\|_{X}\leq\left\|\sum_{i=1}^{n}a_{i}x_{i}\right\|_{X}.$

Thus, $\Phi_{n}\left(b\right)\leq\Phi_{n}\left(a\right)$, and so
$\Phi_{n}\left(\cdot\right)$ is a lattice functional. Also, arguing precisely
as in the case $\left(i\right)$, we obtain
$\Phi_{n}\left(c\right)\leq\Phi_{n}\left(a\right)$, and hence this functional
is symmetric.

$\left(iii\right):$ $\left\|\cdot\right\|_{X_{L}\left(n\right)}.$ Let
$a=\sum_{k\in F}a^{k}$, where $F\subseteq\mathbb{N}$ is finite and
$a^{k}=\left(a^{k}_{i}\right)_{i=1}^{n}$, $k\in F$. For each
$i\in\left\\{1,\dots,n\right\\}$ we have

$\left|b_{i}\right|\leq\left|a_{i}\right|\leq\sum_{k\in
F}\left|a^{k}_{i}\right|.$

One can readily select $b^{k}_{i}$ such that $b_{i}=\sum_{k\in F}b^{k}_{i}$,
$1\leq i\leq n$, and $\left|b^{k}_{i}\right|\leq\left|a^{k}_{i}\right|$ for
all $k$ and $i$. Then, setting $b^{k}=\left\\{b^{k}_{i}\right\\}_{i=1}^{n}$,
$k\in F$, we have $b=\sum_{k\in F}b^{k}$ and
$\Phi_{n}\left(b^{k}\right)\leq\Phi_{n}\left(a^{k}\right)$, which implies

$\left\|b\right\|_{X_{L}\left(n\right)}\leq\sum_{k\in
F}\Phi_{n}\left(b^{k}\right)\leq\sum_{k\in F}\Phi_{n}\left(a^{k}\right).$

In consequence,
$\left\|b\right\|_{X_{L}\left(n\right)}\leq\left\|a\right\|_{X_{L}\left(n\right)}$,
that is, the norm $\left\|\cdot\right\|_{X_{L}\left(n\right)}$ is lattice.

Next, note that

$\sigma a=\sum_{k\in F}\sigma a^{k},$

and hence, by $\left(ii\right)$,

$\left\|\sigma a\right\|_{X_{L}\left(n\right)}\leq\sum_{k\in
F}\Phi_{n}\left(\sigma a^{k}\right)=\sum_{k\in F}\Phi_{n}\left(a^{k}\right).$

Thus, $\left\|\sigma
a\right\|_{X_{L}\left(n\right)}\leq\left\|a\right\|_{X_{L}\left(n\right)}$ and
so $\left\|\cdot\right\|_{X_{L}\left(n\right)}$ is a symmetric norm. ∎

An immediate consequence of this lemma is the following

######  Corollary 6.5.

$\left\|\cdot\right\|_{X_{L}}$ and $\left\|\cdot\right\|_{X_{U}}$ are lattice
norms.

Our next two propositions state that the operators
$I_{n+1}:\mathbb{R}^{n+1}\rightarrow\mathbb{R}^{n}$ are contractions with
respect to each of the three functionals considered in the latter lemma.

######  Proposition 6.6.

Let $n\in\mathbb{N}$. For each infinite dimensional Banach lattice $X$ and all
$a\in\mathbb{R}^{n+1}$ we have

$\displaystyle\Phi_{n}\left(I_{n+1}a\right)$ $\displaystyle\leq$
$\displaystyle\Phi_{n+1}\left(a\right)$

and

$\displaystyle\left\|I_{n+1}a\right\|_{X_{L}\left(n\right)}$
$\displaystyle\leq$ $\displaystyle\left\|a\right\|_{X_{L}\left(n+1\right)}.$

###### Proof.

Put $a=\left\\{a_{i}\right\\}_{i=1}^{n+1}$ and
$b=I_{n+1}a=\left\\{a_{i}\right\\}_{i=1}^{n}$.

By Lemma 6.4, $\Phi_{n}$ is lattice monotone. Consequently, it suffices to
prove the result in the special case of $a_{n+1}=0.$ We set

$\mathfrak{B}_{n}^{\ast}\left(X\right):=\left\\{\left\\{x_{i}\right\\}_{i=1}^{n}:\,\mbox{there
is}\;x_{n+1}\in X\;\mbox{such
that}\;\left\\{x_{i}\right\\}_{i=1}^{n+1}\in\mathfrak{B}_{n+1}\left(X\right)\right\\}.$

Obviously,
$\mathfrak{B}_{n}^{\ast}\left(X\right)\subseteq\mathfrak{B}_{n}\left(X\right)$,
which implies, since $a_{n+1}=0,$ the following:

$\displaystyle\Phi_{n}\left(b\right)$ $\displaystyle=$
$\displaystyle\inf\left\\{\left\|\sum_{i=1}^{n}a_{i}x_{i}\right\|_{X}:\left\\{x_{i}\right\\}_{i=1}^{n}\in\mathfrak{B}_{n}\left(X\right)\right\\}$
$\displaystyle\leq$
$\displaystyle\inf\left\\{\left\|\sum_{i=1}^{n}a_{i}x_{i}\right\|_{X}:\left\\{x_{i}\right\\}_{i=1}^{n}\in\mathfrak{B}_{n}^{\ast}\left(X\right)\right\\}$
$\displaystyle=$
$\displaystyle\inf\left\\{\left\|\sum_{i=1}^{n}a_{i}x_{i}+a_{n+1}x_{n+1}\right\|_{X}:\left\\{x_{i}\right\\}_{i=1}^{n+1}\in\mathfrak{B}_{n+1}\left(X\right)\right\\}$
$\displaystyle=$ $\displaystyle\Phi_{n+1}\left(a\right),$

and the first inequality is proved.

To prove similar inequality for the norm
$\left\|\cdot\right\|_{X_{L}\left(n\right)}$, write $a=\sum_{k\in F}a^{k}$ for
some finite subset $F\subseteq\mathbb{N}$ and $a^{k}\in\mathbb{R}^{n+1}.$
Then, we have

$b=I_{n+1}Tr_{n+1}a=\sum_{k\in F}I_{n+1}Tr_{n+1}a^{k},$

and, by the first part,

$\left\|b\right\|_{X_{L}\left(n\right)}\leq\sum_{k\in
F}\Phi_{n}\left(I_{n+1}Tr_{n+1}a^{k}\right)\leq\sum_{k\in
F}\Phi_{n+1}\left(Tr_{n+1}a^{k}\right)\leq\sum_{k\in
F}\Phi_{n+1}\left(a^{k}\right).$

Hence,

$\left\|b\right\|_{X_{L}\left(n\right)}\leq\left\|a\right\|_{X_{L}\left(n+1\right)},$

and the proof is completed. ∎

######  Proposition 6.7.

Let $X$ be an infinite dimensional Banach lattice such that $l_{\infty}$ is
not finitely lattice representable in $X$. The following holds for every
positive integer $n$ and $a\in\mathbb{R}^{n+1}$

$\left\|I_{n+1}a\right\|_{X_{U}\left(n\right)}\leq\left\|a\right\|_{X_{U}\left(n+1\right)}.$

###### Proof.

We put again $a=\left\\{a_{i}\right\\}_{i=1}^{n+1}$ and
$b=I_{n+1}a=\left\\{a_{i}\right\\}_{i=1}^{n}$. As in the proof of Proposition
6.6, we may assume that $a_{n+1}=0$.

Let $\left\\{x_{i}\right\\}_{i=1}^{n}\in\mathfrak{B}_{n}\left(X\right)$ and
$\varepsilon>0.$ By Lemma 6.2, we can select a sequence
$\left\\{u_{n}\right\\}_{i=1}^{n+1}\in\mathfrak{B}_{n+1}\left(X\right)$ that
satisfies one of the conditions $\left(i\right)$ and $\left(ii\right)$ of that
lemma.

In the case when $\left(i\right)$ is fulfilled, we have

$\left\|\sum_{i=1}^{n}a_{i}x_{i}\right\|_{X}=\left\|\sum_{i=1}^{n+1}a_{i}u_{i}\right\|_{X}\leq\left\|a\right\|_{X_{U}\left(n+1\right)},$

and the desired result follows.

Assume now that we have $\left(ii\right)$ and let $k$, $\psi$ be as in the
statement of Lemma 6.2. Define the vector
$c=\left\\{c_{i}\right\\}_{i=1}^{n+1}\in\mathbb{R}^{n+1}$ by

$c_{i}=\left\\{\begin{array}[]{c}a_{\psi{\left(i\right)}}:i\neq k,1\leq i\leq
n-1\\\ c_{n}=a_{k}\\\ c_{n+1}=0\end{array}\right\\}$

Then, we have

$\left\|\sum_{i=1}^{n}a_{i}x_{i}\right\|_{X}\leq\left\|\sum_{i=1}^{n+1}c_{i}u_{i}\right\|_{X}+\left\|\sum_{i=1}^{n+1}c_{i}u_{i}-\sum_{i=1}^{n}a_{i}x_{i}\right\|_{X}.$

Since

$\displaystyle\sum_{i=1}^{n+1}c_{i}u_{i}-\sum_{i=1}^{n}a_{i}x_{i}$
$\displaystyle=$
$\displaystyle\sum_{i=1}^{n-1}a_{\psi\left(i\right)}x_{\psi\left(i\right)}+a_{k}u_{n}+0\cdot
u_{n+1}-\sum_{i=1}^{n}a_{i}x_{i}$ $\displaystyle=$
$\displaystyle\left(\sum_{i=1,i\neq
k}^{n}a_{i}x_{i}\right)+a_{k}u_{n}-a_{k}x_{k}-\sum_{i=1,i\neq
k}^{n}a_{i}x_{i}$ $\displaystyle=$ $\displaystyle
a_{k}\left(u_{n}-x_{k}\right),$

$\left|a_{k}\right|\leq\left\|a\right\|_{X_{U}\left(n+1\right)}$, $1\leq k\leq
n$, and $\left\|\cdot\right\|_{X_{U}\left(n+1\right)}$ is a symmetric norm, we
conclude

$\left\|\sum_{i=1}^{n}a_{i}x_{i}\right\|_{X}\leq\left\|\sum_{i=1}^{n+1}c_{i}u_{i}\right\|_{X}+\left\|u_{n}-x_{k}\right\|_{X}\leq\left\|c\right\|_{X_{U}\left(n+1\right)}+\varepsilon\left\|a\right\|_{X_{U}\left(n+1\right)}=(1+\varepsilon)\left\|a\right\|_{X_{U}\left(n+1\right)}.$

Thus, since $\varepsilon>0$ is arbitrary,
$\left\|b\right\|_{X_{U}\left(n\right)}\leq\left\|a\right\|_{X_{U}\left(n+1\right)}$,
what is required. ∎

######  Proposition 6.8.

Let $X$ be an infinite dimensional Banach lattice such that $X_{L}$ is not
contained in $c_{0}.$ Then, $l_{\infty}$ is finitely lattice representable in
$X$.

###### Proof.

By the assumption, there exists a sequence
$a=\left\\{a_{i}\right\\}_{i=1}^{\infty}\in X_{L}$ with
$\lim\sup_{i\rightarrow\infty}\left|a_{i}\right|>\delta$ for some $\delta>0$.
By scaling we may assume that $\delta=1.$ Define the sequence
$b=\left\\{b_{i}\right\\}_{i=1}^{\infty}$ by

$b_{i}=\left\\{\begin{array}[]{c}1:\left|a_{i}\right|>1\\\
0:\left|a_{i}\right|\leq 1\end{array}\right\\}$

Since $\left|b_{i}\right|\leq\left|a_{i}\right|$ for all $i=1,2,\dots$, by
Corollary 6.5, $b\in X_{L}$ and $\|b\|_{X_{L}}\leq\|a\|_{X_{L}}$.

Further, for each positive integer $m$ select $k_{m}$ such that the set

$U_{m}:=\left\\{i:\,1\leq i\leq k_{m},b_{i}=1\right\\}$

has cardinality $m.$ Note that for any sequence
$\left\\{y_{i}\right\\}_{i=1}^{k_{m}}\in\mathfrak{B}_{k_{m}}\left(X\right)$ it
holds

$\left\|\sum_{i\in
U_{m}}y_{i}\right\|_{X}=\left\|\sum_{i=1}^{k_{m}}b_{i}y_{i}\right\|_{X}\leq\Phi_{k_{m}}\left(b\right).$

Hence, if $b=\sum_{j\in F}b^{j}$ for some finite set $F$ and
$b^{j}\in\mathbb{R}^{k_{m}}$, by the triangle inequality, we have

$\left\|\sum_{i\in U_{m}}y_{i}\right\|=\left\|\sum_{i=1}^{k_{m}}\sum_{j\in
F}\left\langle b^{j},e_{i}\right\rangle y_{i}\right\|_{X}\leq\sum_{j\in
F}\left\|\sum_{i=1}^{k_{m}}\left\langle b^{j},e_{i}\right\rangle
y_{i}\right\|_{X}\leq\sum_{j\in F}\Phi_{k_{m}}\left(b^{j}\right).$

Consequently, for any sequence $\left\\{t_{i}\right\\}_{i\in U_{m}}$ of
scalars we obtain

$\sup_{i\in U_{m}}\left|t_{i}\right|\leq\left\|\sum_{i\in
U_{m}}t_{i}y_{i}\right\|_{X}\leq\sup_{i\in
U_{m}}\left|t_{i}\right|\left\|\sum_{i\in
U_{m}}y_{i}\right\|_{X}\leq\sup_{i\in
U_{m}}\left|t_{i}\right|\left\|b\right\|_{X_{L}(k_{m})}\leq\left\|a\right\|_{X_{L}}\sup_{i\in
U_{m}}\left|t_{i}\right|.$

Since the set $U_{m}$ has cardinality $m,$ which is arbitrary, and
$\left\|a\right\|_{X_{L}}$ is a constant that does not depend on $m$, the
latter inequality means that $l_{\infty}$ is crudely finitely lattice
representable in $X$ (see Section 2.2). Since the latter is equivalent to the
finite lattice representability of $l_{\infty}$ in $X$ [13, p. 288], the proof
is completed. ∎

######  Proposition 6.9.

Assume that $l_{\infty}$ is finitely lattice representable in a Banach lattice
$X$. Then $X_{L}$ coincides with $l_{\infty}$ isometrically.

###### Proof.

Fix a positive integer $n$ and $\varepsilon>0.$ By the assumption, there
exists a sequence $\left\\{x_{i}\right\\}_{i=1}^{n}$ of pair-wise disjoint
elements in $X$ such that

$\sup_{1\leq i\leq
n}\left|a_{i}\right|\leq\left\|\sum_{i=1}^{n}a_{i}x_{i}\right\|_{X}\leq\left(1+\varepsilon\right)\sup_{1\leq
i\leq n}\left|a_{i}\right|$

for any scalar sequence $a=\left\\{a_{i}\right\\}_{i=1}^{n}.$ Putting
$z_{i}:=x_{i}/\left\|x_{i}\right\|_{X}$, we get
$\left\\{z_{i}\right\\}_{i=1}^{n}\in\mathfrak{B}_{n}\left(X\right).$ Hence, in
view of embeddings (3.1), we have

$\displaystyle\left\|a\right\|_{l_{\infty}^{n}}$ $\displaystyle\leq$
$\displaystyle\left\|a\right\|_{X_{L}\left(n\right)}\leq\Phi_{n}\left(a\right)\leq\left\|\sum_{i=1}^{n}a_{i}z_{i}\right\|_{X}=\left\|\sum_{i=1}^{n}\frac{a_{i}}{\left\|x_{i}\right\|_{X}}x_{i}\right\|_{X}$
$\displaystyle\leq$ $\displaystyle\left(1+\varepsilon\right)\sup_{1\leq i\leq
n}\frac{\left|a_{i}\right|}{\left\|x_{i}\right\|_{X}}\leq\left(1+\varepsilon\right)\left\|a\right\|_{l_{\infty}^{n}}$

Since $\varepsilon>0$ is arbitrary, we conclude that
$\left\|a\right\|_{l_{\infty}^{n}}=\left\|a\right\|_{X_{L}\left(n\right)}$ for
all $n=1,2,\dots$, which implies that $X_{L}=l_{\infty}$ isometrically. ∎

Prove now the dual result.

######  Proposition 6.10.

If $l_{1}$ is finitely lattice representable in a Banach lattice $X$, then
$X_{U}$ coincides isometrically with $l_{1}.$

###### Proof.

For every positive integer $n$ and $\varepsilon>0$ we can select a sequence
$\left\\{x_{i}\right\\}_{i=1}^{n}$ of pair-wise disjoint elements such that

$\sum_{i=1}^{n}\left|a_{i}\right|\leq\left\|\sum_{i=1}^{n}a_{i}x_{i}\right\|_{X}\leq\left(1+\varepsilon\right)\sum_{i=1}^{n}\left|a_{i}\right|$

for any scalar sequence $a=\left\\{a_{i}\right\\}_{i=1}^{n}.$ As in the
preceding proof, we put $z_{i}=x_{i}/\left\|x_{i}\right\|_{X}$. Then
$\left\\{z_{i}\right\\}_{i=1}^{n}\in\mathfrak{B}_{n}\left(X\right)$, and since
$1\leq\left\|x_{i}\right\|_{X}\leq 1+\varepsilon$, $1\leq i\leq n$, we have

$\displaystyle\left\|a\right\|_{l_{1}^{n}}$ $\displaystyle\geq$
$\displaystyle\left\|a\right\|_{X_{U}\left(n\right)}\geq\left\|\sum_{i=1}^{n}a_{i}z_{i}\right\|_{X}=\left\|\sum_{i=1}^{n}\frac{a_{i}}{\left\|x_{i}\right\|_{X}}x_{i}\right\|_{X}$
$\displaystyle\geq$
$\displaystyle\sum_{i=1}^{n}\frac{\left|a_{i}\right|}{\left\|x_{i}\right\|_{X}}\geq\left(1+\varepsilon\right)^{-1}\sum_{i=1}^{n}\left|a_{i}\right|=\left(1+\varepsilon\right)^{-1}\left\|a\right\|_{l_{1}^{n}}$

This implies
$\left\|a\right\|_{l_{1}^{n}}=\left\|a\right\|_{X_{U}\left(n\right)}$ and thus
the proof is complete. ∎

As a result, all the pieces needed for the proof of Theorem 3.3 are in place.

###### Proof of Theorem 3.3.

Prove first the claim for $X_{L}.$ By Lemmas 6.4 and 6.6,
$\left\|\cdot\right\|_{X_{L}\left(n\right)}$ is a lattice, symmetric norm for
each positive integer $n$ and the operators $I_{n}$ are contractions with
respects to these norms. Hence, if $X_{L}$ is embedded in $c_{0}$, then
Proposition 6.1 may be applied and we conclude that $X_{L}$ is a r.i. Banach
sequence lattice. In the case when $X_{L}$ is not embedded in $c_{0},$ from
Propositions 6.8 and 6.9 it follows that $X_{L}$ coincides isometrically with
$l_{\infty}$, and hence it is a r.i. Banach sequence lattice as well.

Proceeding with the case of $X_{U}$, observe that, by the assumption,
$l_{\infty}$ fails to be finitely lattice representable in $X$, and so, using
Proposition 6.7, we have that the maps $I_{n}$ are contractive with respect to
these norms. Moreover, by Lemma 6.4,
$\left\|\cdot\right\|_{X_{U}\left(n\right)}$ is a lattice, symmetric norm for
each positive integer $n.$ Finally, from Proposition 2.6 it follows that $X$
satisfies a lower $p$-estimate for some $p<\infty.$ Hence, for every
$n\in\mathbb{N}$ and any sequences
$\left\\{x_{i}\right\\}_{i=1}^{n}\in\mathfrak{B}_{n}\left(X\right)$ and
$\left\\{a_{i}\right\\}_{i=1}^{n}$ of scalars it follows

$\left(\sum_{i=1}^{n}|a_{i}|^{p}\right)^{1/p}\leq
M_{\left[p\right]}\left(X\right)\left\|\sum_{i=1}^{n}a_{i}x_{i}\right\|_{X}\leq
M_{\left[p\right]}\left(X\right)\left\|\sum_{i=1}^{n}a_{i}e_{i}\right\|_{X_{U}},$

i.e., $X_{U}$ is embedded into $l_{p}$ and thus also in $c_{0}.$ Thus,
applying Proposition 6.1, we conclude that $X_{U}$ is a r.i. sequence Banach
lattice. ∎

## Appendix: A description of the optimal upper sequence lattices for Orlicz
spaces.

Recall that, according to Example 3.10, optimal upper and lower sequence
lattices for the $L_{p,q}$-spaces are just some $l_{r}$-spaces. As well known
(see e.g. [18, 19]), comparing with the Lorentz spaces, the structure of
disjoint sequences in Orlicz spaces is much more complicated. In particular,
in general, an Orlicz space $L_{M}$ need not to admit an upper
$\delta(L_{M})$-estimate or a lower $\sigma(L_{M})$-estimate (as above,
$\delta(X)$ and $\sigma(X)$ are the Grobler-Dodds indices of a Banach lattice
$X$). Therefore, we come to the problem of identification of optimal sequence
lattices for this class of r.i. spaces. In this section, we present a
description of optimal upper lattices for separable Orlicz spaces as
intersections of some special Musielak-Orlicz sequence spaces.

We start with an assertion that reduces the consideration of issues related to
pairwise disjoint functions to that of a simpler case of multiples of
characteristic functions of pairwise disjoint sets.

######  Proposition 6.11.

Let $M$ be an Orlicz function such that $M\in\Delta_{2}^{\infty}$ with the
constant $K$. For every $n\in\mathbb{N}$ and arbitrary pairwise disjoint
functions $y_{k}$, $k=1,\dots,n$, there exist two sequences
$\\{B_{k}\\}_{k=1}^{n}$ and $\\{B_{k}^{\prime}\\}_{k=1}^{2n}$ of pairwise
disjoint subsets of $[0,1]$, $r_{k}\in\mathbb{R}$, $k=1,\dots,n$, and
$r_{k}^{\prime}\in\mathbb{R}$, $k=1,\dots,2n$, such that for the functions
$h_{k}:=r_{k}\chi_{B_{k}}$, $k=1,\dots,n$, and
$f_{k}:=r_{k}^{\prime}\chi_{B_{k}^{\prime}}$, $k=1,\dots,2n$, we have

(6.4)
$\frac{1}{4}\|y_{k}\|_{L_{M}}\leq\|h_{k}\|_{L_{M}}\leq\|y_{k}\|_{L_{M}},\;\;\frac{1}{2}\|y_{k}\|_{L_{M}}\leq\|f_{k}\|_{L_{M}}\leq\frac{3}{2}\|y_{k}\|_{L_{M}},\;\;k=1,\dots,n,$

and

(6.5)
$\Big{\|}\sum_{k=1}^{n}h_{k}\Big{\|}_{L_{M}}\leq\Big{\|}\sum_{k=1}^{n}y_{k}\Big{\|}_{L_{M}}\leq(K+1)\Big{\|}\sum_{k=1}^{2n}f_{k}\Big{\|}_{L_{M}}.$

###### Proof.

Clearly, without loss of generality, we can assume that given functions
$y_{k}\in L_{M}$, $k=1,\dots,n$, are positive. Moreover, since
$M\in\Delta_{2}^{\infty}$, the space $L_{M}$ is separable, and, consequently,
it can be assumed also that $y_{k}$ are bounded functions.

For each $1\leq k\leq n$ we set

$c_{k}:=\frac{\|y_{k}\|_{L_{M}}}{2\varphi_{L_{M}}(m(\mathrm{supp}\,y_{k}))},u_{k}(t):=\begin{cases}y_{k}(t),&\mathrm{if}~{}y_{k}(t)\geq
c_{k},\\\
0,&\mathrm{if}~{}y_{k}(t)<c_{k}\end{cases}\;\;\mbox{and}\;\;g_{k}(t):=c_{k}\chi_{\mathrm{supp}y_{k}\setminus\mathrm{supp}u_{k}}(t).$

Then, it follows

(6.6)
$\sum_{k=1}^{n}u_{k}\leq\sum_{k=1}^{n}y_{k}\leq\sum_{k=1}^{n}u_{k}+\sum_{k=1}^{n}g_{k}.$

Observe also that

(6.7)
$\|g_{k}\|_{L_{M}}=c_{k}\varphi_{L_{M}}(m(\mathrm{supp}y_{k}\setminus\mathrm{supp}u_{k}))\leq\frac{1}{2}\|y_{k}\|_{L_{M}},\;\;k=1,\dots,n$

($\varphi_{L_{M}}$ is the fundamental function of $L_{M}$; see formula (2.2)),
and

(6.8)
$\frac{1}{2}\|y_{k}\|_{L_{M}}=\|y_{k}\|_{L_{M}}-\|c_{k}\chi_{\mathrm{supp}y_{k}}\|_{L_{M}}\leq\|u_{k}\|_{L_{M}}\leq\|y_{k}\|_{L_{M}},\;\;k=1,\dots,n.$

Next, we estimate the norm $\|\sum_{k=1}^{n}u_{k}\|_{L_{M}}$. To this end, we
show that there is $r_{k}\in[c_{k},\sup_{t}u_{k}(t)]$ such that

(6.9)
$M(r_{k})=M\left(\frac{r_{k}}{\|u_{k}\|_{L_{M}}}\right)\int_{0}^{1}M(u_{k}(t))\,dt.$

Indeed, let us consider the function

$H_{k}(t):=\frac{M(u_{k}(t))}{M\left(\frac{u_{k}(t)}{\|u_{k}\|_{L_{M}}}\right)},~{}t\in\mathrm{supp}\,u_{k}.$

From the equality $\int_{0}^{1}M(\frac{u_{k}(t)}{\|u_{k}\|_{L_{M}}})dt=1$ it
follows that

$\inf_{t\in\mathrm{supp}\,u_{k}}H_{k}(t)\leq\int_{0}^{1}M(u_{k}(t))\,dt\leq\sup\limits_{t\in\mathrm{supp}\,u_{k}}H_{k}(t).$

Thus, since $\inf\limits_{t\in\mathrm{supp}\,u_{k}}u_{k}(t)\geq c_{k}$, by
continuity of $M$, equality (6.9) holds for some $r_{k}$ such that
$r_{k}\in[c_{k},\sup_{t}u_{k}(t)]$.

Further, assuming as we can that the functions $M$ and $\varphi_{L_{M}}$ are
strictly increasing, define the real numbers $d_{k}\in[0,1]$,
$k=1,2,\ldots,n$, as follows:

$d_{k}:=\begin{cases}\varphi_{L_{M}}^{-1}\left(\frac{\|u_{k}\|_{L_{M}}}{r_{k}}\right),&\mathrm{if}~{}\|u_{k}\|_{L_{M}}\leq
r_{k}\varphi_{L_{M}}(m(\mathrm{supp}\,y_{k})),\\\
m(\mathrm{supp}\,y_{k}),&\mathrm{if}~{}\|u_{k}\|_{L_{M}}>r_{k}\varphi_{L_{M}}(m(\mathrm{supp}\,y_{k})).\end{cases}$

Clearly, the definition of $d_{k}$ implies that

(6.10) $r_{k}\varphi_{L_{M}}(d_{k})\leq\|u_{k}\|_{L_{M}}.$

Conversely, $r_{k}\varphi_{L_{M}}(d_{k})=\|u_{k}\|_{L_{M}}$ if
$\|u_{k}\|_{L_{M}}\leq r_{k}\varphi_{L_{M}}(m(\mathrm{supp}\,y_{k}))$.
Otherwise, since $r_{k}\geq c_{k},$ in view of (6.8) and the definition of
$c_{k}$, we obtain

$r_{k}\varphi_{L_{M}}(d_{k})\geq
c_{k}\varphi_{L_{M}}(m(\mathrm{supp}\,y_{k}))\geq\frac{1}{2}\|u_{k}\|_{L_{M}}.$

Thus, summing up, we conclude that

(6.11) $r_{k}\varphi_{L_{M}}(d_{k})\geq\frac{1}{2}\|u_{k}\|_{L_{M}}.$

Now, observe that from inequality (6.11) and formula (2.2) for the function
$\varphi_{L_{M}}$ it follows that

$d_{k}\geq\varphi_{L_{M}}^{-1}(\|u_{k}\|_{L_{M}}/(2r_{k})).$

and

$\varphi_{L_{M}}^{-1}(u)=\frac{1}{M(1/u)},\;\;0<t\leq 1,$

respectively. Hence, taking into account that $M\in\Delta_{2}^{\infty}$ with
constant $K$ and applying (6.9), we obtain

(6.12) $\displaystyle d_{k}M(r_{k})$ $\displaystyle\geq$
$\displaystyle\varphi_{L_{M}}^{-1}(\|u_{k}\|_{L_{M}}/(2r_{k}))M(r_{k})=\frac{M(r_{k})}{M\left(\frac{2r_{k}}{\|u_{k}\|_{L_{M}}}\right)}$
$\displaystyle\geq$
$\displaystyle\frac{1}{K}\frac{M(r_{k})}{M\left(\frac{r_{k}}{\|u_{k}\|_{L_{M}}}\right)}=\frac{1}{K}\int\limits_{0}^{1}M(u_{k}(t))\,dt.$

In the converse direction, from the equality
$1/d_{k}=M\left(1/\varphi_{L_{M}}(d_{k})\right)$ (see (2.2)), combined with (
6.10) and (6.9), it follows

(6.13)
$d_{k}M(r_{k})=\frac{M(r_{k})}{M(\frac{1}{\varphi_{L_{M}}(d_{k})})}\leq\frac{M(r_{k})}{M\left(\frac{r_{k}}{\|u_{k}\|_{L_{M}}}\right)}=\int\limits_{0}^{1}M(u_{k}(t))\,dt.$

Furthermore, by the definition of $d_{k}$, we have $d_{k}\leq
m(\mathrm{supp}\,y_{k})$. Therefore, we can define the following functions
$h_{k}(t):=r_{k}\chi_{B_{k}}(t),$ where $B_{k}\subset\mathrm{supp}\,y_{k}$ and
$m(B_{k})=d_{k}.$ Since $\|h_{k}\|_{L_{M}}=r_{k}\varphi_{L_{M}}(d_{k})$,
according to (6.10) and (6.11), it holds

$\frac{1}{2}\|u_{k}\|_{L_{M}}\leq\|h_{k}\|_{L_{M}}\leq\|u_{k}\|_{L_{M}},\;\;k=1,2,\ldots,n.$

Hence, from (6.8) it follows

(6.14)
$\frac{1}{4}\|y_{k}\|_{L_{M}}\leq\|h_{k}\|_{L_{M}}\leq\|y_{k}\|_{L_{M}},\;\;k=1,2,\ldots,n.$

Moreover, since the functions $h_{k}$ (respectively, $u_{k}$) are pairwise
disjoint, in view of estimate (6.13), we conclude that

$\displaystyle\int_{0}^{1}M\left(\sum_{k=1}^{n}h_{k}(t)\right)\,dt$
$\displaystyle=$
$\displaystyle\sum_{k=1}^{n}d_{k}M(r_{k})\leq\sum_{k=1}^{n}\int_{0}^{1}M(u_{k}(t))\,dt$
$\displaystyle\leq$
$\displaystyle\int_{0}^{1}M\left(\sum_{k=1}^{n}y_{k}(t)\right)\,dt.$

Conversely, by (6.12), we have

$\sum_{k=1}^{n}\int_{0}^{1}M(u_{k}(t))\,dt\leq
K\int_{0}^{1}M\left(\sum_{k=1}^{n}h_{k}(t)\right)\,dt.$

Therefore, since $M$ is convex and $K\geq 1$, it follows that

$\Big{\|}\sum_{k=1}^{n}h_{k}\Big{\|}_{L_{M}}\leq\Big{\|}\sum_{k=1}^{n}u_{k}\Big{\|}_{L_{M}}\leq
K\Big{\|}\sum_{k=1}^{n}h_{k}\Big{\|}_{L_{M}}.$

Noting that the collection $\\{g_{k},h_{k}\\}_{k=1}^{n}$ consists of $2n$
pairwise disjoint functions, we relabel them as $f_{k}$, $k=1,2,\dots,2n$.
Then, by (6.6) and the last inequality, we obtain

$\displaystyle\Big{\|}\sum_{k=1}^{n}h_{k}\Big{\|}_{L_{M}}\leq\Big{\|}\sum_{k=1}^{n}y_{k}\Big{\|}_{L_{M}}$
$\displaystyle\leq$
$\displaystyle\Big{\|}\sum_{k=1}^{n}u_{k}\Big{\|}_{L_{M}}+\Big{\|}\sum_{k=1}^{n}g_{k}\Big{\|}_{L_{M}}$
$\displaystyle\leq$ $\displaystyle
K\Big{\|}\sum_{k=1}^{n}h_{k}\Big{\|}_{L_{M}}+\Big{\|}\sum_{k=1}^{n}g_{k}\Big{\|}_{L_{M}}$
$\displaystyle\leq$
$\displaystyle(K+1)\Big{\|}\sum_{k=1}^{2n}f_{k}\Big{\|}_{L_{M}},$

and hence (6.5) is proved. Since inequalities (6.4) follow from (6.14) and
(6.7), the proof is completed. ∎

From Proposition 6.11 and its proof we obtain

######  Corollary 6.12.

Let $M$ be an Orlicz function such that $M\in\Delta_{2}^{\infty}$. Then,

$\|a\|_{(L_{M})_{U}}\asymp\sup\left\\{\Big{\|}\sum_{k=1}^{n}a_{k}\frac{\chi_{F_{k}}}{\varphi_{L_{M}}(m(F_{k}))}\Big{\|}_{L_{M}}:\,n\in\mathbb{N},F_{k}\subset[0,1]\;\mbox{pairwise
disjoint}\right\\},$

with constants independent of $a=(a_{k})_{k=1}^{\infty}$.

Recalling that the space $(L_{M})_{U}$ is rearrangement invariant (see Theorem
3.3), denote by $\phi_{U}$ its fundamental function, i.e.,
$\phi_{U}(n):=\|\sum_{k=1}^{n}e_{k}\|_{(L_{M})_{U}}$, $n\in\mathbb{N}$. Also,
let ${\Phi}_{g}$ be the dilation function of a function
$g:\,(0,\infty)\to(0,\infty)$ for large values of arguments defined by

${\Phi}_{g}(u):=\sup_{v\geq\max(1,1/u)}\frac{g(vu)}{g(v)},\;\;u>0.$

######  Corollary 6.13.

Let $M$ be an Orlicz function such that $M\in\Delta_{2}^{\infty}$. Then,

$\phi_{U}(n)\asymp{\Phi}_{M^{-1}}(n),\;\;n\in\mathbb{N},$

where $M^{-1}$ is the inverse function for $M$.

###### Proof.

By Corollary 6.12, we have

(6.15)
$\phi_{U}(n)\asymp\sup\left\\{\Big{\|}\sum_{k=1}^{n}\frac{\chi_{F_{k}}}{\varphi_{L_{M}}(m(F_{k}))}\Big{\|}_{L_{M}}:\,F_{k}\subset[0,1]\;\mbox{are
pairwise disjoint}\right\\}.$

Let $n\in\mathbb{N}$ and let $F_{k}\subset[0,1]$, $k=1,\dots,n$, be pairwise
disjoint. Then, by formula (2.2),

$\Big{\|}\sum_{k=1}^{n}\frac{\chi_{F_{k}}}{\varphi_{L_{M}}(m(F_{k}))}\Big{\|}_{L_{M}}=\inf\left\\{\lambda>0:\,\sum_{k=1}^{n}M\left(\frac{M^{-1}(1/m(F_{k}))}{\lambda}\right)m(F_{k})\leq
1\right\\}.$

Next, we write

$\displaystyle\sum_{k=1}^{n}M\left(\frac{M^{-1}(1/m(F_{k}))}{{\Phi}_{M^{-1}}(n)}\right)m(F_{k})$
$\displaystyle=$ $\displaystyle\sum_{k:\,m(F_{k})\geq
1/n}M\left(\frac{M^{-1}(1/m(F_{k}))}{{\Phi}_{M^{-1}}(n)}\right)m(F_{k})$
$\displaystyle+$
$\displaystyle\sum_{k:\,m(F_{k})<1/n}M\left(\frac{M^{-1}(1/m(F_{k}))}{{\Phi}_{M^{-1}}(n)}\right)m(F_{k})$
$\displaystyle=$ $\displaystyle(I)+(II).$

Observe that

$(I)\leq\sum_{k:\,m(F_{k})\geq
1/n}M\left(\frac{M^{-1}(n)}{{\Phi}_{M^{-1}}(n)}\right)m(F_{k})\leq M(1)=1$

and

$(II)\leq\sum_{k:\,m(F_{k})<1/n}M\left(\frac{M^{-1}(1/(m(F_{k})n))M^{-1}(1/m(F_{k}))}{M^{-1}(1/m(F_{k}))}\right)m(F_{k})=\sum_{k:\,m(F_{k})<1/n}\frac{1}{n}\leq
1.$

Summing up, we obtain

$\Big{\|}\sum_{k=1}^{n}\frac{\chi_{F_{k}}}{\varphi_{L_{M}}(m(F_{k}))}\Big{\|}_{L_{M}}\leq
2{\Phi}_{M^{-1}}(n),$

for every $n\in\mathbb{N}$ and all pairwise disjoint $F_{k}\subset[0,1]$,
$k=1,\dots,n$. Consequently, in view of (6.15), it follows

$\phi_{U}(n)\preceq{\Phi}_{M^{-1}}(n),\;\;n\in\mathbb{N}.$

Conversely, without loss of generality, assume that

${\Phi}_{M^{-1}}(n)=\frac{M^{-1}(nv_{n})}{M^{-1}(v_{n})}$

for some $v_{n}\geq 1$. Let $F_{k}\subset[0,1]$, $k=1,\dots,n$, be arbitrary
pairwise disjoint subsets of $[0,1]$ such that $m(F_{k})=(nv_{n})^{-1}$. Then,

$\displaystyle\phi_{U}(n)\succeq\Big{\|}\sum_{k=1}^{n}\frac{\chi_{F_{k}}}{\varphi_{L_{M}}(m(F_{k}))}\Big{\|}_{L_{M}}$
$\displaystyle=$
$\displaystyle\inf\left\\{\lambda>0:\,M\left(\frac{M^{-1}(nv_{n})}{\lambda}\right)\frac{1}{v_{n}}\leq
1\right\\}$ $\displaystyle=$
$\displaystyle\frac{M^{-1}(nv_{n})}{M^{-1}(v_{n})}={\Phi}_{M^{-1}}(n).$

∎

Recall that a family of Banach spaces