Duals of Cesàro sequence vector lattices, Cesàro sums of Banach lattices, and their finite elements

Abstract

In this paper, we study the ideals of finite elements in special vector lattices of real sequences, first in the duals of Cesàro sequence spaces ${\text {ces}}_p$ for $p\in \{0\}\cup [1,\infty )$ and, second, after the Cesàro sum ${\text {ces}}_{p}{(\mathfrak {X})}$ of a sequence of Banach spaces is introduced, where $p=\infty$ is also allowed, we characterize their duals and the finite elements in these sums if the summed up spaces are Banach lattices. This is done by means of a remarkable extension of the corresponding result for direct sums.

1 Introduction

In the first part of the paper, guided by Bennett’s approach, we consider the dual spaces $d_q$ of the Cesàro sequence spaces ${\text {ces}}_p$ for $1< p < \infty$ where $1/q +1/p=1.$ Equipped with the appropriate norm and the coordinatewise order, they are Dedekind complete Banach lattices with order continuous norm. Then of interest are the finite elements in these spaces. In the last years, these classes of finite elements, which were introduced in [15], are studied thoroughly by many authors in different Banach lattices, in particular, in Cesàro sequence spaces by the authors in [11]. The spaces $d_q$ do not possess order units and all kinds of finite elements in them coincide with $c_{00}.$ In the second part, similarly to the classical ($c_0$-, $\ell _p$-, and $\ell _\infty$-) direct sums of Banach lattices in [18, §3.3.3] and [8, 9], we introduce the so-called Cesàro sum for a sequence of Banach spaces, study their dual space, and characterize the finite elements if the summed up spaces are Banach lattices. In this paper, the Theorem 4.3 essentially extends the corresponding result for direct sums in [9].

2 Preliminaries

The aim of this section is to provide some necessary definitions and facts. For unexplained terminology concerning the vector lattices theory, the reader can consult the books [2, 4, 5, 13].

An element $\varphi$ in an Archimedean vector lattice E is called finite whenever there exists a so-called majorant $z\in E$ such that for any $x\in E,$ there is a number $c_x>0$ with the property that $\left| x\right| \wedge n\left| \varphi \right| \le c_x \, z$ holds for any $n\in {\mathbb {N}}.$ If z is a finite element, then $\varphi$ is called totally finite and if $\left| \varphi \right|$ itself is a majorant, the element $\varphi$ is called selfmajorizing. The sets of all finite and totally finite elements of an Archimedean vector lattice E are denoted by $\Phi _1(E)$ and $\Phi _2(E),$ respectively. All positive selfmajorizing elements are denoted by $S_+(E)$ and $\Phi _3(E):=S_+(E)-S_+(E).$ The collections $\Phi _i(E)$ are order ideals in E for $i\in \{1,2,3\}$ (see [18, Chapt. 3]).

We need some more notions and facts the details of which can be found in [4, 5, 18].

Definition 2.1

• (a) An element $u\in E_+,$ $u\ne 0,$ of a vector lattice E is called an atom whenever $0\le x \le u,$ $0\le y \le u,$ and $x\wedge y = 0$ imply that either $x = 0$ or $y = 0.$

• (b) An element $u\in E_+,$ $u\ne 0,$ of a vector lattice E is called discrete, whenever $0 \le v \le u$ implies $v=\lambda u$ for some $\lambda \in {\mathbb {R}}_+.$

• (c) A vector lattice E is said to be atomic if for each $x>0,$ there exists an atom u such that $0<u\le x.$

In an Archimedean vector lattice E,  a positive element is an atom if and only if it is a discrete element. Also if u is an atom in E,  then $\{\lambda u:\lambda \in {\mathbb {R}}\}$ (the vector space generated by u) is a projection band. Each atom of a vector lattice is a totally finite element. Even more, for two elements a and x,  one has

$$\begin{aligned} \frac{1}{n}(\left| x\right| \wedge na)\le a, \end{aligned}$$

and if a is an atom, then $\left| x \right| \wedge na= \lambda _na\le \left| x\right|$ follows for some $\lambda _n\in {\mathbb {R}}_+.$ The Archimedean property implies $r_a(\left| x\right| )=\sup \{\lambda \in {\mathbb {R}}_+:\lambda a\le \left| x\right| \}<\infty$ for the atom a,  which finally yields $\left| x\right| \wedge na\le r_a(\left| x\right| )\,a,$ i.e., the element a is selfmajorizing.

For a normed vector lattice E,  denote by $\Gamma _E$ the set of all atoms of E with norm 1. Then $\Gamma _E$ is a subset of $\Phi _3(E).$ It consists of pairwise disjoint elements, and forms a linearly independent system. According to [18, Theorem 3.18] and the remark above, we have the following theorem.

Theorem 2.2

Let E be a Banach lattice with order continuous norm. Then

• (i) $\Phi _3(E)=\Phi _2(E)=\Phi _1(E)=\textrm{span}(\Gamma _E);$

• (ii) $\Phi _1(E)$ is closed in E if and only if $\Gamma _E$ is a finite set. In particular,  $\Phi _1(E)=E$ if and only if E is finite dimensional.

The following results will be used in the sequel.

Theorem 2.3

([18, Proposition 3.44]). If a vector lattice E has an order unit,  then $\Phi _i(E)=E,\; i\in \{1,2,3\}.$

Theorem 2.4

([18, Theorem 3.15]). Let E be a Banach lattice and $\varphi \in E.$ Then the following statements are equivalent : 

• (i) $\varphi$ is a finite element.

• (ii) The closed unit ball $B_{\{\varphi \}^{dd}}$ of $\{\varphi \}^{dd}$ is order bounded in E.

• (iii) $\{\varphi \}^{dd}$ has a generalized order unit,  i.e., there exists $0\le z\in E$ such that for any $x\in \{\varphi \}^{dd},$ there is a real number $\gamma _x>0$ with $\left| x\right| \le \gamma _x z.$

Theorem 2.5

([18, Theorem 3.28]). Let H be a projection band in a vector lattice E,  and $P_H$ the band projection from E onto H. Then $P_H(\Phi _1(E) ) = \Phi _1(E) \cap H = \Phi _1(H).$

3 The dual space of ${\text {ces}}_p$

The vector space ${\mathbb {R}}^{\mathbb {N}}$ which is partially ordered by the coordinatewise order is a Dedekind complete vector lattice, where the lattice operations are given coordinatewise: $x\vee y= (\max \{x_1,y_1\},\max \{x_2,y_2\},\ldots )$ and $x\wedge y= (\min \{x_1,y_1\},\min \{x_2,y_2\},\ldots ).$ As usual (cf. [4, Chapt. 13]) denote by $\ell _\infty ,$ $\ell _p,$ $c_0,$ and $c_{00}$ the spaces of bounded, p-summable, null, and finite sequences in ${\mathbb {R}}^{\mathbb {N}},$ respectively.

The Cesàro operator ${\mathcal {C}}:{\mathbb {R}}^{\mathbb {N}} \rightarrow {\mathbb {R}}^{\mathbb {N}}$ is given by

$$\begin{aligned} {\mathcal {C}}x=\left( \frac{1}{n}\sum _{k=1}^n x_k\right) _{n\in {\mathbb {N}}} \end{aligned}$$

for all $x=(x_n)_{n\in {\mathbb {N}}}$ in ${\mathbb {R}}^{\mathbb {N}}.$ By using Hardy’s inequality, for $x\in \ell _p\ (1<p<\infty ),$ we have

$$\begin{aligned} \left\| {\mathcal {C}}x\right\| _p\le \left\| {\mathcal {C}}|x|\right\| _p\le \frac{p}{p-1}\left\| x\right\| _p. \end{aligned}$$

So, the Cesàro operator ${{\mathcal {C}}}: \ell _p \rightarrow \ell _p$ is linear and continuous with operator norm $\frac{p}{p-1}.$ For $1\le p\le \infty ,$ the Cesàro sequence spaces are defined as

$$\begin{aligned} {\text {ces}}_p: =\{x\in {\mathbb {R}}^{\mathbb {N}}:{\mathcal {C}} |x|\in \ell _p\} \end{aligned}$$

with the norm $\left\| x\right\| _{{\text {ces}}_p}:=\left\| {\mathcal {C}}|x|\right\| _p.$

The space ${\text {ces}}_0$ is defined as

$$\begin{aligned} {\text {ces}}_0: =\{x\in {\mathbb {R}}^{\mathbb {N}}:{\mathcal {C}} |x|\in c_0\} \end{aligned}$$

with the norm $\left\| x\right\| _{{\text {ces}}_0}:=\left\| {\mathcal {C}}|x| \right\| _\infty .$

Actually, ${\text {ces}}_p$ and ${\text {ces}}_0$ spaces are Banach lattices. These spaces have been thoroughly studied by Leibowitz [14] and Shiue [17]. In particular, they showed that ${\text {ces}}_p$ is trivial if $p=1.$ Since they are order ideals in ${\mathbb {R}}^{\mathbb {N}},$ the Dedekind completeness of the latter implies that all these vector lattices are also Dedekind complete.

The characterization problem of the dual space ${\text {ces}}'_p$ of the Banach lattice $({\text {ces}}_p,\left\| \cdot \right\| _{{\text {ces}}_p})$ of Cesàro sequence spaces for $1<p<\infty ,$ which was offered by the Dutch Mathematical Society as a prijsvrage [1], was solved by Jagers [12]. A second solution, given by Bennet [6], provides a simple explicit isomorphic identification. To present Bennett’s approach, first consider for $0<q<\infty$ the set

$$\begin{aligned} d_{q}:= \left\{ a=(a_n)_{n\in {\mathbb {N}}}\in {\mathbb {R}}^{{\mathbb {N}}} :\left( \sup _{k\ge n}|a_k|\right) _{n\in {\mathbb {N}}}\in \ell _{q} \right\} . \end{aligned}$$

For $q=0$ and $q=\infty ,$ the above constructed sequences $\left( \sup _{k\ge n}|a_k|\right) _{n\in {\mathbb {N}}}$ are supposed to belong to $c_0$ and $\ell _\infty ,$ respectively. A simple calculation shows that $d_0=c_0$ and $d_\infty = \ell _\infty$ what is not of interest.

Each sequence $a=(a_n)_{n\in {\mathbb {N}}} \in {\mathbb {R}}^{{\mathbb {N}}}$ produces another sequence ${\hat{a}}=({\hat{a}}_n)_{n\in {\mathbb {N}}}$ with ${\hat{a}}_n:= \sup _{k\ge n}|a_k|$ called the least decreasing majorant of a. Clearly, $a \in d_q$ if and only if ${\hat{a}} \in \ell _q$ if and only if ${\hat{a}} \in d_q.$

Notice that $d_q$ is a proper subset of $\ell _q$ for $1\le q < \infty .$ For example, consider the sequence $a=(a_n)_{n\in {\mathbb {N}}},$ where

$$\begin{aligned} a_n = \left\{ \begin{array}{ll} \frac{1}{2^{k/q}} &{} \quad \text {if}\ n=2^k\ \text {for some}\ k\in {\mathbb {N}}, \\ 0 &{} \quad \text {otherwise}. \end{array} \right. \end{aligned}$$

For the case $q=1,$ there is $a=(0,\frac{1}{2^1},0,\frac{1}{2^2},0,0,0,\frac{1}{2^3},\ldots )$ and $a\in \ell _1$ is clear. On the other hand, ${\hat{a}}=(\frac{1}{2^1},\frac{1}{2^1},\frac{1}{2^2},\frac{1}{2^2},\frac{1}{2^3}, \frac{1}{2^3},\frac{1}{2^3}, \frac{1}{2^3},\frac{1}{2^{4}},\ldots ),$ i.e., ${\hat{a}}_n=\frac{1}{2^{m+1}}$ if $n>2$ and satisfies $2^m< n\le 2^{m+1},$ where $m\in {\mathbb {N}}.$ Then it is clear that

$$\begin{aligned} \sum \limits _{n=1}^\infty {\hat{a}}_n > \sum \limits _{m=1}^\infty \big (2^m\frac{1}{2^{m+1}}\big )=\sum \limits _{m=1}^\infty \frac{1}{2}=\infty , \hbox { i.e.}, {\hat{a}}\notin \ell _1 \hbox { and}, \hbox { hence } a\notin d_1. \end{aligned}$$

For $q\ge 1,$ under the norm

$$\begin{aligned} \left\Vert a\right\Vert _{d_{q}}:= \left\Vert {\hat{a}}\right\Vert _q =\left( \sum _{n=1}^\infty \sup _{k\ge n}|a_k|^q\right) ^{1/q}, \end{aligned}$$

$d_q$ is a Banach space. In particular, $\left\Vert a\right\Vert _{d_q}=\left\Vert {\hat{a}}\right\Vert _{q}=\left\Vert {\hat{a}}\right\Vert _{d_q}.$ Therefore, every eventually decreasing, non-negative sequence $a\in \ell _q$ is an element of $d_q$ because $a_n={\hat{a}}_n$ holds for all sufficiently large n. Hence, $c_{00}$ is a proper subset of $d_q,$ and so $c_{00}\subsetneqq d_q\subsetneqq \ell _q.$

The coordinatewise order makes the spaces $d_q$ into Dedekind complete Banach lattices.

Let $1<p,q < \infty$ and $\frac{1}{p}+\frac{1}{q}=1.$ Then there is a lattice isomorphism between ${\text {ces}}'_p$ and $d_q$ whose duality¹ with ${\text {ces}}_p$ is stated explicitly via

$$\begin{aligned} \left\langle a, x \right\rangle := \sum _{n=1}^\infty a_n x_n,\quad x \in {\text {ces}}_p, \ a \in d_q. \end{aligned}$$

Moreover, the identification is even isometric when ${\text {ces}}_p$ is endowed with another norm equivalent to $\left\Vert \cdot \right\Vert _{{\text {ces}}_p}.$ For details, see [6] and [7, Lemma 2.2 and Proposition 2.4]. Recently, Curbera and Ricker have proved the identification ${\text {ces}}_0^\prime = d_1$ with equality of norms [10, Lemma 6.2]. On the other hand, [3, Theorem 1] yields that ${\text {ces}}_0^{\prime \prime } = d_1^\prime ={\text {ces}}_\infty$ with equality of norms. Jagers [12] proved that ${\text {ces}}_p$ is reflexive for $1<p<\infty .$ Therefore, by Pettis’ theorem [16, Corollary 1.11.17], $d_q$ is reflexive as well, where $1<q<\infty .$ According to [4, Corollary 9.23], the norm of $d_q$ is order continuous for $1<q<\infty .$ However, the norm of ${\text {ces}}_\infty$ is not order continuous (see [11, Corollary 4.2]). Nevertheless, the following Lemma 3.2 shows that the norm of $d_1$ is order continuous.

Recently, the Banach lattice $d_q$ has been investigated in detail by Bonet and Ricker ([7], see also the references therein).

Lemma 3.1

Let $(x^{(n)})_{n\in {\mathbb {N}}}$ be a positive decreasing sequence and $1\le p < \infty .$ Then $x^{(n)} \downarrow \textbf{0}\,$ in $d_p$ if and only if ${\hat{x}}^{(n)} \downarrow \textbf{0}\,$ in $d_p.$

Proof

Let $x^{(n)} =(x^{(n)}_k)_{k\in {\mathbb {N}}}.$ The “if” part is clear from $|x_k^{(n)}| \le {\hat{x}}^{(n)}_k$ for all $n, k\in {\mathbb {N}}.$ Next, we prove the “only if” part. To this end, let $x^{(n)} \downarrow \textbf{0}\,$ in $d_p.$ Then $x^{(n)}_k \downarrow 0$ in ${\mathbb {R}}$ for each $k\in {\mathbb {N}}$ since $d_p$ is an ideal in ${\mathbb {R}}^{\mathbb {N}}.$ It suffices to show that ${\hat{x}}^{(n)}_k \downarrow 0$ in ${\mathbb {R}}$ for each $k\in {\mathbb {N}}.$ Since the sequence $({x}^{(n)})_{n\in {\mathbb {N}}}$ is decreasing (to $\textbf{0}\,$) in $d_p,$ we have $\Big (\sup _{i\ge k}|x^{(n)}_i|\Big )_{k\in {\mathbb {N}}}\in \ell _p,$ i.e., $({\hat{x}}^{(n)}_k)_{k\in {\mathbb {N}}}\in \ell _p$ and so $\sum _{k=1}^\infty |{\hat{x}}^{(n)}_k|^p<\infty .$ Given $\varepsilon >0,$ there exists $N\in {\mathbb {N}}$ such that

$$\begin{aligned} \sum _{k=N}^\infty \left| {\hat{x}}^{(n)}_k\right| ^p=\sum _{k=N}^\infty \sup _{i\ge k}\left| x^{(n)}_i\right| ^p \le \sum _{k=N}^\infty \sup _{i\ge k}\left| x^{(1)}_i\right| ^p < \varepsilon ^p \end{aligned}$$

for all $n\in {\mathbb {N}}.$ Moreover, for some $n_0 \in {\mathbb {N}},$ one has $0\le x^{(n)}_k \le \varepsilon$ for $n \ge n_0$ and $k\in \{1,\ldots ,N\}.$ Therefore, $0\le {\hat{x}}^{(n)}_k \le \varepsilon$ holds for all $n \ge n_0$ and $k\in {\mathbb {N}}.$ This completes the proof. $\square$

Lemma 3.2

The norm in the Banach lattice $d_p$ is order continuous for $1\le p < \infty .$

Proof

As mentioned above, the norm in the Banach lattice $d_p$ is order continuous for $1< p < \infty .$ By [5, Theorem 4.9], to complete the proof for the case $p=1,$ we must show that $x_n \downarrow \textbf{0}\,$ in $d_1$ implies $\left\Vert x_n\right\Vert _{d_1} \downarrow 0.$ Let $x_n \downarrow \textbf{0}\,$ in $d_1.$ Then it follows from Lemma 3.1 that ${\hat{x}}_n \downarrow \textbf{0}\,$ in $d_1.$ Also, ${\hat{x}}_n \downarrow \textbf{0}\,$ in $\ell _1$ since $d_1$ is an ideal in $\ell _1.$ By order continuity of the norm in $\ell _1,$ $\left\Vert {\hat{x}}_n\right\Vert _{1} \downarrow 0.$ The proof follows immediately from $\left\Vert x_n\right\Vert _{d_1}=\left\Vert {\hat{x}}_n\right\Vert _{1}.$ $\square$

Remark 3.3

Further on, let $e_n$ denote the sequence $(\underbrace{0,0,\ldots ,0}_{n-1},1,0,0,\ldots ).$ Note that ${\hat{e}}_n=(\underbrace{1,1,\ldots ,1}_n,0,\ldots )=\sum _{k=1}^n e_k$ and therefore², $e_n$ is an element of $d_p$ for each $n\in {\mathbb {N}}$ and $1\le p<\infty .$ It turns out that each $e_n$ is an atom in $d_p.$ Actually, there are no other atoms in $d_p$ except the non-zero multiples of the coordinate sequences $e_n.$ Therefore, $\Gamma _{d_p}=\{n^{-1/p}e_n:n\in {\mathbb {N}}\}.$ In particular, $d_p$ is an atomic vector lattice.

Theorem 3.4

Let $1\le p < \infty .$ Then

• (i) $\Phi _3(d_p)=\Phi _2(d_p)=\Phi _1(d_p)=c_{00},$

• (ii) the space $d_p$ has no order unit.

Proof

(i) Since $d_p$ is a Banach lattice with order continuous norm, it follows from Remark 3.3 and Theorem 2.2 that

$$\begin{aligned} \Phi _3(d_p)=\Phi _2(d_p)=\Phi _1(d_p)=\textrm{span}(\Gamma _{d_p})=\textrm{span}\{e_n:n\in {\mathbb {N}}\}=c_{00}. \end{aligned}$$

(ii) If $d_p$ had an order unit, then $\Phi _1(d_p)=d_p$ by Theorem 2.3. Due to part (i), there would be $d_p=c_{00},$ a contradiction since $c_{00}$ is a proper subspace of $d_p.$

4 Finite elements in Cesàro sums of Banach lattices

Denote by ${\mathfrak {X}}$ a sequence $(X_n, \left\| \cdot \right\| _n)_{n\in {\mathbb {N}}}$ of Banach spaces and let $p\in \{0\} \cup [1, \infty ].$ Then, similar to the construction of directed sums³ of Banach lattices in [18, §3.3.3], we define the p-Cesàro sums and $d_p$-sums of ${\mathfrak {X}}=(X_n)_{n\in {\mathbb {N}}}$ as follows

$$\begin{aligned} {\text {ces}}_p({\mathfrak {X}})&: =&\left\{ x=(x_n)_{n\in {\mathbb {N}}} :x_n \in X_n, \ \big (\left\Vert x_n\right\Vert _n\big )_{n\in {\mathbb {N}}}\in {\text {ces}}_p\right\} , \\ d_p({\mathfrak {X}})&: =&\left\{ x=(x_n)_{n\in {\mathbb {N}}} :x_n \in X_n, \ \big (\left\Vert x_n\right\Vert _n\big )_{n\in {\mathbb {N}}}\in d_p\right\} . \end{aligned}$$

Further on, in order to simplify the notation, we write $\left\| \cdot \right\|$ instead of $\left\| \cdot \right\| _n$ and $\textbf{0}\,$ for the zero vector in $X_n$ for each $n\in {\mathbb {N}}.$ Under the coordinatewise algebraic operations, these sets are vector spaces. With the norms defined by

$$\begin{aligned} {\left| \left| \left| x \right| \right| \right| }_{{\text {ces}}_p}{(\mathfrak {X})}=\left\Vert \big (\left\Vert x_n\right\Vert \big )_{n\in {\mathbb {N}}} \right\Vert _{{\text {ces}}_p} \hbox { and } \quad {\left| \left| \left| y \right| \right| \right| }_{d_p({\mathfrak {X}})}=\left\Vert \big (\left\Vert y_n\right\Vert \big )_{n\in {\mathbb {N}}} \right\Vert _{d_p} \end{aligned}$$

for $x\in {\text {ces}}_p({\mathfrak {X}})$ and $y \in d_p({\mathfrak {X}}),$ respectively, the spaces ${\text {ces}}_p({\mathfrak {X}})$ and $d_p({\mathfrak {X}})$ are Banach spaces as well.

Note that the equality⁴${\text {ces}}_1=\{\textbf{0}\,\}$ implies that ${\text {ces}}_1(\mathfrak {X})$ is trivial.

For $p=0$ and $1<p\le \infty ,$ define the map $J_j: X_j \rightarrow {\text {ces}}_p(\mathfrak {X})$ by

$$\begin{aligned} J_jx = (x_n)_{n\in {\mathbb {N}}}= (\textbf{0}\,,\ldots ,\textbf{0}\,,\underbrace{x}_{j\text {-th term}},\textbf{0}\,,\ldots ) = \left\{ \begin{array}{ll} \textbf{0}\,, &{}\quad n \ne j,\\ x, &{}\quad n=j, \end{array} \right. \end{aligned}$$

for $x\in X_j$ and $j\in {\mathbb {N}},$ which is an isomorphism into the space ${\text {ces}}_p(\mathfrak {X}).$ After some calculation, one obtains

$$\begin{aligned} {\left| \left| \left| J_jx \right| \right| \right| }_{{\text {ces}}_p({\mathfrak {X}})} =\left\{ \begin{array}{ll} \left\Vert x\right\Vert \left( \sum \limits _{i=j}^\infty \frac{1}{i^p}\right) ^{1/p}, &{}\quad 1<p <\infty ,\\ \frac{1}{j}\left\Vert x\right\Vert , &{}\quad p=0\ \text {or}\ \infty , \end{array} \right. \quad \text {for } x\in X_j. \end{aligned}$$

Let $x=(x_n)_{n\in {\mathbb {N}}}\in {\text {ces}}_p(\mathfrak {X})$ and $p\in \{0\} \cup (1,\infty ).$ Observe that

$$\begin{aligned} {\left| \left| \left| x-\sum _{n=1}^{N-1} J_n x_n \right| \right| \right| }_{{\text {ces}}_p}^p= \sum _{n=N}^\infty \left( \frac{1}{n} \sum _{k=N}^n \left\Vert x_k\right\Vert \right) ^p \le \sum _{n=N}^\infty \left( \frac{1}{n} \sum _{k=1}^n \left\Vert x_k\right\Vert \right) ^p \rightarrow 0 \end{aligned}$$

if $N\rightarrow \infty$ and

$$\begin{aligned} {\left| \left| \left| x-\sum _{n=1}^{N-1} J_n x_n \right| \right| \right| }_{{\text {ces}}_0}= \sup _{n\ge N} \left\{ \frac{1}{n} \sum _{k=N}^n \left\Vert x_k\right\Vert \right\} \le \sup _{n\ge N} \left\{ \frac{1}{n} \sum _{k=1}^n \left\Vert x_k\right\Vert \right\} \mathop {\longrightarrow }\limits _{N\rightarrow \infty } 0, \end{aligned}$$

which imply $x= \sum _{n=1}^\infty J_n x_n.$

Proposition 4.1

Let ${\mathfrak {X}}=(X_n)_{n\in {\mathbb {N}}}$ be a sequence of Banach spaces,  ${\mathfrak {X}}'=(X_n')_{n\in {\mathbb {N}}}$ the sequence of their dual spaces, and $1<p,q < \infty$ with $\frac{1}{p}+\frac{1}{q}=1.$ Then the mapping $x^\prime =(x_n^\prime )_{n\in {\mathbb {N}}} \mapsto f_{x^\prime }$ from $d_q({\mathfrak {X}}')$ to ${\text {ces}}'_p({\mathfrak {X}})$ defined by

$$\begin{aligned} f_{x'}(x):= \sum _{n=1}^\infty \langle x^\prime _n, x_n\rangle , \quad x=(x_n)_{n\in {\mathbb {N}}} \in {\text {ces}}_p(\mathfrak {X}), \end{aligned}$$

is a linear isomorphism from $d_q({\mathfrak {X}}')$ onto ${\text {ces}}'_p({\mathfrak {X}})$ and satisfies

$$\begin{aligned} \frac{1}{q} {\left| \left| \left| x^\prime \right| \right| \right| }_{d_q({\mathfrak {X}}')} \le \left\Vert f_{x'}\right\Vert \le (p-1)^{1/p} \ {\left| \left| \left| x^\prime \right| \right| \right| }_{d_q({\mathfrak {X}}')}\quad \hbox {for all } x' \in d_q({\mathfrak {X}}'). \end{aligned}$$

(4.1)

Similarly,  we have ${\text {ces}}^\prime _0(\mathfrak {X}) =d_1({\mathfrak {X}}')$ with equality of the norms,  i.e., $\left\| f_{x'}\right\| ={\left| \left| \left| x^\prime \right| \right| \right| }_{d_1({\mathfrak {X}}')}.$

Proof

We first demonstrate that the mapping $x^\prime \mapsto f_{x^\prime }$ is well defined. Fix $x^\prime =(x_n^\prime )_{n\in {\mathbb {N}}} \in d_q({\mathfrak {X}}').$ Then for each $x=(x_n)_{n\in {\mathbb {N}}} \in {\text {ces}}_p(\mathfrak {X}) ,$ we have $|\langle x'_n, x_n\rangle | \le \left\Vert x^\prime _n\right\Vert \left\Vert x_n\right\Vert$ for any $n\in {\mathbb {N}}$ and so by referring to the estimation established after the proof of [6, Theorem 12.3], we get

$$\begin{aligned} \sum _{n=1}^\infty |\langle x'_n,x_n\rangle | \le \sum _{n=1}^\infty \left\Vert x^\prime _n\right\Vert \left\Vert x_n\right\Vert \le (p-1)^{1/p} \ {\left| \left| \left| x^\prime \right| \right| \right| }_{d_q({\mathfrak {X}}')} {\left| \left| \left| x \right| \right| \right| }_{{\text {ces}}_p(\mathfrak {X})}. \end{aligned}$$

Thus, the formula $f_{x'}(x) = \sum _{n=1}^\infty \langle x'_n, x_n\rangle$ defines a continuous linear functional on ${\text {ces}}_p(\mathfrak {X})$ satisfying the relations

$$\begin{aligned} \left\Vert f_{x'}\right\Vert \le (p-1)^{1/p} \ {\left| \left| \left| x^\prime \right| \right| \right| }_{d_q({\mathfrak {X}}')} \quad \hbox {for all } x' \in d_q({\mathfrak {X}}'). \end{aligned}$$

(4.2)

Clearly, the mapping $x' \mapsto f_{x'}$ is a linear operator.

Now, let $f\in {\text {ces}}'_p(\mathfrak {X}).$ Define for any $n\in {\mathbb {N}},$ the linear functionals $x_n':X_n \rightarrow {\mathbb {R}}$ by $\langle x'_n, x_n\rangle =f( J_nx_n).$ Then $x'_n \in X_n'$ and, moreover, $f(x)=\sum _{n=1}^\infty \langle x_n',x_n\rangle$ holds for all $x=(x_n)_{n\in {\mathbb {N}}} \in {\text {ces}}_p(\mathfrak {X})$ since $x= \sum _{n=1}^\infty J_n x_n$ is true in ${\text {ces}}_p(\mathfrak {X})$ (as was mentioned just before the proposition). Fix $\alpha >1.$ For each n,  pick some $y_n \in X_n$ with $\left\| y_n\right\| =1$ and $\left\| x'_n\right\| \le \alpha \langle x'_n, y_n\rangle .$ Observe that for each k,  we have

$$\begin{aligned} \sum _{n=1}^k \left\| x'_n\right\| \left\| x_n\right\|\le & {} \sum _{n=1}^k \alpha \langle x'_n, y_n\rangle \left\| x_n \right\| = \alpha f\left( \sum _{n=1}^k J_n (\left\| x_n\right\| y_n) \right) \\ {}\le & {} \alpha \left\| f\right\| {\left| \left| \left| \sum _{n=1}^k J_n(\left\| x_n\right\| y_n) \right| \right| \right| }_{{\text {ces}}_p(\mathfrak {X})}\le \alpha \left\| f\right\| {\left| \left| \left| x \right| \right| \right| }_{{\text {ces}}_p(\mathfrak {X})}. \end{aligned}$$

Taking the limits as $k \rightarrow \infty$ and $\alpha \downarrow 1,$ we see that

$$\begin{aligned} \sum _{n=1}^\infty \left\Vert x^\prime _n\right\Vert \left\Vert x_n\right\Vert \le \left\| f\right\| {\left| \left| \left| x \right| \right| \right| }_{{\text {ces}}_p(\mathfrak {X})}. \end{aligned}$$

If $(\left\Vert x'_n\right\Vert )_{n\in {\mathbb {N}}}\in d_q,$ i.e., $x' \in d_q({\mathfrak {X}}'),$ then from [6, Corollary 12.17], it follows that $f= f_{x'}$ and ${\left| \left| \left| x^\prime \right| \right| \right| }_{d_q({\mathfrak {X}}')} \le q \left\Vert f_{x'}\right\Vert .$ Therefore, this together with (4.2) implies (4.1) and $x' \mapsto f_{x'}$ is a linear isomorphism from $d_q({\mathfrak {X}}')$ onto ${\text {ces}}^\prime _p(\mathfrak {X}).$ It remains to show that the relation $(\left\Vert x'_n\right\Vert )_{n\in {\mathbb {N}}}\in d_q$ holds⁵: Define the functional $\lambda :{\text {ces}}_p\rightarrow {\mathbb {R}}$ by $\lambda \big ((a_n)_{n\in {\mathbb {N}}}\big )=\sum _{n=1}^\infty \left\| x_n'\right\| a_n.$ By the above inequality, we have

$$\begin{aligned} \left| \lambda \big ((a_n)_{n\in {\mathbb {N}}}\big )\right|\le & {} \sum _{n=1}^\infty \left\| x_n'\right\| \left| a_n\right| = \sum _{n=1}^\infty \left\| x_n'\right\| \left\Vert \frac{a_n}{\left\| x_n\right\| } x_n\right\Vert \\ {}\le & {} \left\| f\right\| {\left| \left| \left| \Bigg ( \frac{a_n}{\left\| x_n\right\| }x_n\Bigg )_{n\in {\mathbb {N}}} \right| \right| \right| }_{{\text {ces}}_p({\mathfrak {X}})} = \left\| f\right\| \left\Vert (a_n)_{n\in {\mathbb {N}}}\right\Vert _{{\text {ces}}_p}. \end{aligned}$$

Therefore, $\lambda \in ({\text {ces}}_p)',$ i.e., $(\left\| x_n'\right\| )_{n\in {\mathbb {N}}}\in d_q.$

By the same method and [10, Lemma 6.2], one can show ${{\text {ces}}'_0(\mathfrak {X}) = d_1({\mathfrak {X}}')}$ even with equality of norms. $\square$

Remark 4.2

If ${\text {ces}}_p(\mathfrak {X})$ is endowed with a special norm equivalent to $\left\Vert \cdot \right\Vert _{{\text {ces}}_p},$ then by using [6, Theorem 4.5 and Corollary 12.17], it can be established that the mapping defined in Proposition 4.1 is an isometry.

Turn now to the case that each $X_n$ is a Banach lattice. Then with the coordinatewise defined lattice operations, the spaces ${\text {ces}}_p(\mathfrak {X})$ and $d_p({\mathfrak {X}})$ are even Banach lattices, $J_j$ is a lattice isomorphism from $X_j$ to ${\text {ces}}_p(\mathfrak {X}),$ and for any $j\in {\mathbb {N}},$ the set $J_jX_j$ is a projection band in ${\text {ces}}_p(\mathfrak {X}).$ The latter implies $\Phi _1(J_jX_j)=J_j\Phi _1(X_j).$ Denote by $P_j: {\text {ces}}_p({\mathfrak {X}})\rightarrow J_jX_j$ the band projection from ${\text {ces}}_p({\mathfrak {X}})$ onto $J_jX_j,$ where $P_j\big ((x_n)_{n\in {\mathbb {N}}}\big ) = J_jx_j.$ By Theorem 2.5 and in view of the just mentioned behaviour of $J_j\Phi _1(X_j),$ there hold the equalities

$$\begin{aligned} P_j\Big (\Phi _1\big ( {\text {ces}}_p(\mathfrak {X}) \big ) \Big )= \Phi _1\big ( {\text {ces}}_p(\mathfrak {X}) \big ) \cap J_jX_j= \Phi _1(J_jX_j) = J_j\Phi _1(X_j). \end{aligned}$$

(4.3)

The mapping $x^\prime \mapsto f_{x^\prime }$ defined in Proposition 4.1 is now an onto lattice isomorphism since both the map and its inverse are positive operators.

As the characterization of the finite elements in the Banach lattices ${\text {ces}}_p,$ where $1<p\le \infty$ or $p=0,$ we get a quite direct generalization of the results for the classical cases $X_n=c_0, \, \ell _p,$ and $\ell _\infty ,$ see [18, Theorem 3.33].

Theorem 4.3

The following statements hold : 

• (i) For $p=0$ and $1<p< \infty ,$ the element $\varphi =(\varphi _n)_{n\in {\mathbb {N}}}$ is finite in ${\text {ces}}_p(\mathfrak {X})$ if and only if $\varphi _n \in \Phi _1(X_n)$ for all $n\in {\mathbb {N}}$ and $\varphi _n=\textbf{0}\,$ for all but finitely many $n\in {\mathbb {N}}.$

• (ii) The element $\varphi =(\varphi _n)_{n\in {\mathbb {N}}}$ is finite in ${{\text {ces}}_\infty (\mathfrak {X})}$ if and only if there exist $w_n \in X_n^+$ such that $\left( n \left\Vert w_n\right\Vert \right) _{n\in {\mathbb {N}}} \in {\text {ces}}_\infty$ and

  $$\begin{aligned} B_{\{ \varphi _n\}^{dd}} \subset \left[ -w_n,\ w_n \right] . \end{aligned}$$

  In particular,  $\varphi _n \in \Phi _1(X_n)$ for all $n\in {\mathbb {N}}.$

Proof

(i) Let $\varphi =(\varphi _n)_{n\in {\mathbb {N}}}$ be such that $\varphi _n \in \Phi _1(X_n)$ for all $n\in {\mathbb {N}}$ and $\varphi _n=\textbf{0}\,$ for all but finitely many $n\in {\mathbb {N}}.$ Clearly, $\varphi \in {\text {ces}}_p(\mathfrak {X}).$ For all $n\in {\mathbb {N}},$ $P_n\varphi = J_n \varphi _n \in J_n\Phi _1( X_n)$ and (4.3) yield $P_n\varphi \in \Phi _1\big ({\text {ces}}_p(\mathfrak {X})\big ).$ From the linearity of the space $\Phi _1\big ({\text {ces}}_p(\mathfrak {X})\big ),$ it follows that $\varphi = \sum _{n=1}^{\infty } P_n\varphi \in \Phi _1\big ({\text {ces}}_p(\mathfrak {X})\big ).$ Observe that this sum has only finitely many non-zero terms.

For the converse, assume that $\varphi =(\varphi _n)_{n\in {\mathbb {N}}}$ is a finite element in ${\text {ces}}_p({\mathfrak {X}}).$ Then $J_n \varphi _n = P_n\varphi \in P_n\Big (\Phi _1\big ({\text {ces}}_p(\mathfrak {X})\big ) \Big )$ holds for all $n\in {\mathbb {N}}.$ Thus, using (4.3) again, one has $\varphi _n \in \Phi _1(X_n)$ for all $n\in {\mathbb {N}}.$ Next we claim that $\varphi _n=\textbf{0}\,$ for all but finitely many $n\in {\mathbb {N}}.$ To see this, assume by way of contradiction that $\varphi _{n_k}\ne \textbf{0}\,$ holds for some increasing sequence of natural numbers $(n_k)_{k\in {\mathbb {N}}}.$ Consider $\psi =(\psi _n)_{n\in {\mathbb {N}}},$ where

$$\begin{aligned} \psi _n = \left\{ \begin{array}{ll} \frac{1}{\left\| \varphi _n\right\| }|\varphi _{n}| &{} \quad \hbox {if } n=n_k \hbox { for some } k, \\ \textbf{0}\, &{} \quad \hbox {otherwise}. \end{array} \right. \end{aligned}$$

Note that $J_{n_k}\psi _{n_k}=\big (\textbf{0}\,,\ldots ,\textbf{0}\,,\frac{\left| \varphi _{n_k}\right| }{\left\| \varphi _{n_k}\right\| },\textbf{0}\,,\ldots \big )$ and consequently

$$\begin{aligned} {\left| \left| \left| J_{n_k}\psi _{n_k} \right| \right| \right| }^p_{{\text {ces}}_p(\mathfrak {X})}= \sum _{i=n_k}^\infty \frac{1}{i^p} \quad \hbox {and}\quad {\left| \left| \left| J_{n_k} \psi _{n_k} \right| \right| \right| }_{{\text {ces}}_0(\mathfrak {X})} = \frac{1}{n_k} \end{aligned}$$

if $1<p<\infty$ and $p=0,$ respectively. Therefore, if we define $t_{n_k}:=\big (\sum _{i=n_k}^\infty \frac{1}{i^p}\big )^{1/p}$ for $1<p< \infty$ and $t_{n_k}:=\frac{1}{n_k}$ for $p=0,$ then $\frac{1}{t_{n_k}} J_{n_k} \psi _{n_k} \in B_{\{ \varphi \}^{dd}} = B_{ {\text {ces}}_p(\mathfrak {X})} \cap \{ \varphi \}^{dd}$ for $p=0$ and $1<p< \infty .$ According to Theorem 2.4, there exists a positive $z=(z_n)_{n\in {\mathbb {N}}} \in {\text {ces}}_p(\mathfrak {X})$ such that $B_{\{ \varphi \}^{dd}} \subset \left[ -z,\ z \right] ,$ which implies that $0 \le \frac{1}{t_{n_k}} \psi _{n_k} \le z_{n_k},$ and so $\frac{1}{t_{n_k}} \le \left\| z_{n_k} \right\|$ for all $k\in {\mathbb {N}}.$ To complete the proof, it is enough to show that the sequence $a=(a_n)_{n\in {\mathbb {N}}}$ defined by

$$\begin{aligned} a_n = \left\{ \begin{array}{ll} \frac{1}{t_{n_k}} &{} \quad \hbox {if } n=n_k \hbox { for some } k, \\ 0 &{} \quad \hbox {otherwise}, \end{array} \right. \end{aligned}$$

does not belong to ${\text {ces}}_p(\mathfrak {X})$ for $p=0$ and $1<p< \infty ,$ which contradicts $z \in {\text {ces}}_p(\mathfrak {X})$ since ${\text {ces}}_p$ is an order ideal in ${\mathbb {R}}^{\mathbb {N}}.$ For $p=0,$ by the definition of a,  one has $a_{n_k} = n_k$ which yields $({\mathcal {C}}|a|)_{n_k}\ge 1$ for all $k\in {\mathbb {N}},$ i.e., a is not an element of ${\text {ces}}_0,$ whereas ${\mathcal {C}}(\left\| z_{n_k}\right\| )\in c_0.$

For $1<p< \infty ,$ there exists $c>0$ such that $t_{n_k} =\big (\sum _{i=n_k}^\infty \frac{1}{i^p}\big )^{1/p}\le c\, n_k^{-1/q}$ for each ${k\in {\mathbb {N}}}$ (with $\frac{1}{p}+\frac{1}{q}=1$). The latter estimation holds for each $n\in {\mathbb {N}}$ because of

$$\begin{aligned} \frac{1}{n^p}\le \frac{1}{n^{p-1}} \quad \hbox {and} \quad \sum _{i=n+1}^\infty \frac{1}{i^p}\le \int \limits _n^\infty \frac{dx}{x^p}=\frac{1}{(p-1)n^{p-1}}. \end{aligned}$$

Therefore, $\sum _{i=n}^\infty \frac{1}{i^p}\le \frac{1}{n^{p-1}}\cdot q,$ and, if n is replaced by $n_k,$ one has

$$\begin{aligned} n_k^{1/p} \,({\mathcal {C}}\left| a \right| )_{n_k}=n_k^{1/p} \left( \frac{1}{n_k}\sum _{i=1}^{n_k} a_i \right) \ge \frac{n_k^{1/p}}{n_kt_{n_k}} \ge \frac{n_k^{1/p}\ n_k^{1/q}}{cn_k}=\frac{1}{c}>0. \end{aligned}$$

Thus, $\big ((n_k)^{1/p} \,({\mathcal {C}}|a|)_{n_k}\big )_{k\in {\mathbb {N}}}$ does not converge to 0,  and so it follows from [10, Proposition 2.3 (iii)] that a is not an element of ${\text {ces}}_p(\mathfrak {X}).$

(ii) Let $\varphi =(\varphi _n)_{n\in {\mathbb {N}}}$ be a finite element in ${\text {ces}}_\infty (\mathfrak {X}).$ Again by Theorem 2.4, there is a positive $z=(z_n)_{n\in {\mathbb {N}}} \in {\text {ces}}_\infty (\mathfrak {X})$ such that $B_{\{ \varphi \}^{dd}} \subset \left[ -z,\ z \right] .$ For each $n\in {\mathbb {N}},$ put $w_n= \frac{z_n}{n},$ and note that $w_n \in X_n^+$ and $\left( n\,w_n\right) _{n\in {\mathbb {N}}} \in {\text {ces}}_\infty (\mathfrak {X}).$ Let $y \in B_{\{\varphi _n\}^{dd}}.$ Then, clearly, $J_n y \in \{ \varphi \}^{dd}$ and $\left\| J_{n} y \right\| = \frac{\left\| y \right\| }{n}\le \frac{1}{n},$ from where $J_{n} y \in \left[ -\frac{z}{n},\ \frac{z}{n} \right]$ follows for all ${n\in {\mathbb {N}}}.$ Thus, $y \in \left[ -w_n,\ w_n \right] ,$ i.e., $B_{\{ \varphi _n\}^{dd}} \subset \left[ -w_n,\ w_n \right] .$

For the converse, observe that

$$\begin{aligned} B_{\{\varphi \}^{dd}} \subset \{ (x_n)_{n\in {\mathbb {N}}} \in {\text {ces}}_\infty (\mathfrak {X}) :x_n \in n B_{\{ \varphi _n\}^{dd}} \text { for all } {n\in {\mathbb {N}}}\}. \end{aligned}$$

Since $nB_{\{ \varphi _n\}^{dd}}$ is order bounded and the order in ${\text {ces}}_\infty (\mathfrak {X})$ is coordinatewise, the proof of the theorem is completed by Theorem 2.4. $\square$