Barrelled Weakly Köthe–Orlicz Summable Sequence Spaces

Abstract

Let ${\displaystyle E}$ be a Hausdorff locally convex space. We investigate the space ${\displaystyle {\mathsf{\Lambda }}_{\phi }\left[E\right]}$ of weakly Köthe–Orlicz summable sequences in ${\displaystyle E}$ with respect to an Orlicz function ${\displaystyle \phi }$ and a perfect sequence space ${\displaystyle \mathsf{\Lambda }}$. We endow ${\displaystyle {\mathsf{\Lambda }}_{\phi }\left[E\right]}$ with a Hausdorff locally convex topology and determine the continuous dual of the so-obtained space in terms of strongly Köthe–Orlicz summable sequences from the dual space ${\displaystyle E^{\prime }}$ of ${\displaystyle E}$. Next, we give necessary and sufficient conditions for ${\displaystyle {\mathsf{\Lambda }}_{\phi }\left[E\right]}$ to be barrelled or quasi-barrelled. This contributes to the understanding of different spaces of vector-valued sequences and their topological properties.

1. Introduction

Let ${\displaystyle E}$ be a locally convex space. The spaces ${\displaystyle {\ell }^p\left(E\right)}$ and ${\displaystyle {\ell }^p\left\{E\right\}}$ of weakly ${\displaystyle p}$-summable and absolutely ${\displaystyle p}$-summable sequences in ${\displaystyle E}$, respectively, were introduced by Pietsch in [1]. The same author investigated applications of these spaces in the study of absolutely ${\displaystyle p}$-summing operators. In addition, he investigated the spaces ${\displaystyle \mathsf{\Lambda }\left\{E\right\}}$ and ${\displaystyle \mathsf{\Lambda }\left(E\right)}$ of absolutely ${\displaystyle \mathsf{\Lambda }}$-summable and weakly ${\displaystyle \mathsf{\Lambda }}$-summable sequences in ${\displaystyle E}$, respectively, where ${\displaystyle \mathsf{\Lambda }}$ is a sequence space endowed with its Köthe normal topology. Building upon Pietsch’s work, Rosier [2] extended the study to the general case, wherein ${\displaystyle \mathsf{\Lambda }}$ is equipped with a general polar topology (instead of the Köthe normal topology). Rosier obtained notable results, which included a comprehensive description of the dual space of ${\displaystyle \mathsf{\Lambda }{\left\{E\right\}}_r}$.

Employing the ${\displaystyle AK}$ property, Florencio and Paúl [3] determined a representation of the elements of ${\displaystyle \mathsf{\Lambda }{\tilde{\otimes }}_{\epsilon }E}$ (the completion of the injective tensor product ${\displaystyle \mathsf{\Lambda }{\otimes }_{\epsilon }E}$) as weakly ${\displaystyle \mathsf{\Lambda }}$-summable sequences in ${\displaystyle E}$.

Later, Oubbi and Ould Sidaty extended in [4] the concept of strong summability, initially introduced by Cohen [5] for normed spaces, to the locally convex spaces. This extension allowed them to obtain a description of the continuous dual space of ${\displaystyle \mathsf{\Lambda }{\left(E\right)}_r}$. Further results and properties for ${\displaystyle \mathsf{\Lambda }\left(E\right)}$ were obtained in [6,7,8]. Recently, Ould Sidaty investigated in [9] the nuclearity (as a convex bornological space) of ${\displaystyle {\mathsf{\Lambda }}_b\left(E\right)}$, i.e., the space of all totally ${\displaystyle \mathsf{\Lambda }}$-summable sequences within the context defined by [10], where ${\displaystyle E}$ represents a convex bornological space. Furthermore, Ghosh and Srivastava explored in [11] the notion of absolute ${\displaystyle \mathsf{\Lambda }}$-summability (using an Orlicz function ${\displaystyle \phi }$). They introduced and investigated the space ${\displaystyle F(E,\phi )}$, consisting of all sequences ${\displaystyle {\left(x_n\right)}_n}$ in a Banach space ${\displaystyle E}$ that satisfy the condition

$${\left(\phi \left(\frac{\parallel x_n{\parallel }_E}{\rho }\right)\right)}_n\in F$$

for some ${\displaystyle \rho >0}$, where ${\displaystyle F}$ denotes a normal sequence space.

It is worth noting that several kinds of sequence spaces have already been investigated in the literature. Descriptions of some of them rely on infinite Köthe matrices ${\displaystyle {\left(a_{i,j}\right)}_{i,j\in \mathbb{N}}}$, some others rely on Cesàro operators, and others rely on different kinds of convergence or summability (see [1,2,3,4,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20]).

Of course, Orlicz functions yield natural sequence spaces in the scalar-valued case. They are also used to construct vector-valued sequence spaces (see for example [11,17,19] and the references therein). The characterization of continuous dual or Köthe–Toeplitz dual are examples of the main issues authors are interested in (see, e.g., [21]). But first of all, a linear topology must be defined on the sequence space in consideration.

In this paper, for an Orlicz function ${\displaystyle \phi }$ and a locally convex space ${\displaystyle E}$, we introduce the notion of a weakly ${\displaystyle (\phi ,\mathsf{\Lambda })}$-summable sequence ${\displaystyle {\left(x_n\right)}_n}$ in ${\displaystyle E}$ and examine some properties of the linear space ${\displaystyle {\mathsf{\Lambda }}_{\phi }\left[E\right]}$ consisting of all such sequences. Actually, weakly ${\displaystyle (\phi ,\mathsf{\Lambda })}$-summable sequences and the corresponding sequence spaces were investigated in [8] for a Banach space ${\displaystyle E}$. There, the author gave necessary and sufficient conditions for ${\displaystyle {\mathsf{\Lambda }}_{\phi }\left[E\right]}$ to be reflexive. The situation in a locally convex space is quite complicated, for the topology is no more given by a single norm but by a family of infinitely many semi-norms, which means that a bounded neighborhood of ${\displaystyle 0}$ may not exist there.

The outcomes of this paper extend and improve some results in the literature, especially those in [8]. We first equip ${\displaystyle {\mathsf{\Lambda }}_{\phi }\left[E\right]}$ with a Hausdorff locally convex topology, and then we investigate the completeness and the continuity of projections of the so-obtained locally convex space. We embed ${\displaystyle E}$ in ${\displaystyle {\mathsf{\Lambda }}_{\phi }\left[E\right]}$ as a complemented subspace. In order to investigate the topological dual of ${\displaystyle {\mathsf{\Lambda }}_{\phi }\left[E\right]}$, we define the notion of strongly ${\displaystyle (\phi ,\mathsf{\Lambda })}$-summable sequences and the space ${\displaystyle {\mathsf{\Lambda }}_{\phi }⟨E⟩}$ of all such sequences. Actually, we prove that whenever ${\displaystyle {\mathsf{\Lambda }}_{\phi }\left[E\right]}$ is ${\displaystyle AK}$, its topological dual can be given in terms of strongly summable sequences. Next, we characterize the property of barrelledness in ${\displaystyle {\mathsf{\Lambda }}_{\phi }\left[E\right]}$. To address this issue, we examine equicontinuous sets of the dual space of ${\displaystyle {\mathsf{\Lambda }}_{\phi }\left[E\right]}$. For ample information on barrelled locally convex spaces, we refer to the monograph [22].

2. Preliminaries

Throughout this paper, ${\displaystyle \mathbb{K}}$ denotes the field of real or complex numbers, ${\displaystyle \mathbb{N}}$ is the set of positive integers, and ${\displaystyle (E,\tau )}$ is a Hausdorff locally convex space over ${\displaystyle \mathbb{K}}$, for which the continuous dual is denoted by ${\displaystyle E^{\prime }}$. If ${\displaystyle M}$ runs over the collection ${\displaystyle \mathcal{M}}$ of all ${\displaystyle \sigma (E^{\prime },E)}$-closed and equicontinuous discs of ${\displaystyle E^{\prime }}$, the topology ${\displaystyle \tau }$ is generated by the semi-norms

$$P_M\left(x\right):=sup\left\{\left|a\left(x\right)\right|,a\in M\right\},x\in E,M\in \mathcal{M}.$$

For any nonempty set ${\displaystyle X}$, ${\displaystyle X^{\mathbb{N}}}$ denotes the set of all sequences from ${\displaystyle X}$, and ${\displaystyle X^{\left(\mathbb{N}\right)}}$ is the subset of ${\displaystyle X^{\mathbb{N}}}$ consisting of all sequences with finite support. If ${\displaystyle \mathsf{\Omega }\subset E^{\mathbb{N}}}$ is a linear space, its Köthe dual, as defined in [23], is the set

$${\mathsf{\Omega }}^{*}:=\left\{{\left(a_n\right)}_n\subset E^{\prime }:\sum _{n=1}^{+\infty }\left|a_n\left(x_n\right)\right|<+\infty ,{\left(x_n\right)}_n\in \mathsf{\Omega }\right\}.$$

If ${\displaystyle t\in E}$, we write ${\displaystyle t{e}_n}$ to mean the sequence for which the entrees are all zero, but the ${\displaystyle n\mathrm{th}}$ one equals ${\displaystyle t}$. The ${\displaystyle k\mathrm{th}}$ finite section of a sequence ${\displaystyle x:={\left(x_n\right)}_n\in \mathsf{\Omega }}$ is defined by

$$x^{\left(k\right)}=\sum _{n=1}^k{x}_n{e}_n=(x_1,x_2,\dots ,x_k,0,0,\dots ).$$

If a topology is given on ${\displaystyle \mathsf{\Omega }}$, we denote by ${\displaystyle {\mathsf{\Omega }}_r}$ the linear subspace of ${\displaystyle \mathsf{\Omega }}$ consisting of those sequences ${\displaystyle x}$ such that ${\displaystyle x^{\left(k\right)}\in \mathsf{\Omega }}$ for all ${\displaystyle k\in \mathbb{N}}$, and ${\displaystyle x=\underset{k\to \infty }{lim}{x}^{\left(k\right)}}$ in ${\displaystyle \mathsf{\Omega }}$.

If ${\displaystyle \mathsf{\Lambda }}$ is a normal linear subspace of ${\displaystyle {\mathbb{K}}^{\mathbb{N}}}$, then ${\displaystyle \mathsf{\Lambda }}$ contains the set ${\displaystyle {\mathsf{\Lambda }}_F}$ of all finite sections of its elements. Unless the contrary is clearly stated, it is equipped with a polar topology ${\displaystyle {\tau }_{\mathcal{S}}}$ defined by a topologizing family ${\displaystyle \mathcal{S}\subset {\mathsf{\Lambda }}^{*}}$ consisting of normal closed and bounded discs with respect to the weak topology ${\displaystyle \sigma ({\mathsf{\Lambda }}^{*},\mathsf{\Lambda })}$. Such a topology is given by the semi-norms

$$P_S\left({\left({\alpha }_n\right)}_n\right):=sup\left\{\sum _{n=1}^{+\infty }\left|{\alpha }_n{\beta }_n\right|,{\left({\beta }_n\right)}_n\in S\right\},{\left({\alpha }_n\right)}_n\in \mathsf{\Lambda },S\in \mathcal{S}.$$

For a bounded disc ${\displaystyle A}$ in a Hausdorff topological vector space ${\displaystyle F}$, ${\displaystyle F_A}$ is the linear span of ${\displaystyle A}$. When no topology is specified on ${\displaystyle F_A}$, it is endowed with the gauge ${\displaystyle {\parallel .\parallel }_A}$ of ${\displaystyle A}$ as a norm, where ${\displaystyle {\parallel t\parallel }_A:=inf\{r>0,t\in rA\}}$, ${\displaystyle t\in F_A}$. We then consider without any further mention the spaces ${\displaystyle E_B}$, ${\displaystyle E_M^{\prime }}$, ${\displaystyle {\mathsf{\Lambda }}_R}$ and ${\displaystyle {\mathsf{\Lambda }}_S^{*}}$, where ${\displaystyle B\subset E}$, ${\displaystyle M\in \mathcal{M}}$, ${\displaystyle S\in \mathcal{S}}$, and ${\displaystyle R\subset \mathsf{\Lambda }}$ are bounded discs, with ${\displaystyle R}$ normal.

We refer to [23] for details concerning Köthe theory of sequence spaces and to [24] for the terminology and notations concerning the general theory of locally convex spaces.

We consider an Orlicz function ${\displaystyle \phi }$: this is any mapping ${\displaystyle \phi :[0,+\infty )\to [0,+\infty ]}$ that is convex, vanishes at 0, and is non-constant (see [17]). The complement of ${\displaystyle \phi }$ is the function

$${\phi }^{*}\left(y\right):=sup\{xy-\phi \left(x\right),x\in [0,+\infty )\}.$$

Let us observe that ${\displaystyle {\phi }^{*}}$ is also an Orlicz function. Clearly, ${\displaystyle \phi }$ and ${\displaystyle {\phi }^{*}}$ satisfy the Young inequality; namely,

$$xy\le \phi \left(x\right)+{\phi }^{*}\left(y\right),x,y\ge 0.$$

The function ${\displaystyle \phi }$ is said to satisfy ${\displaystyle {\mathsf{\Delta }}_2}$ for small ${\displaystyle x}$ (or at 0) if for each ${\displaystyle k>1}$ there exist ${\displaystyle R_k>0}$ and ${\displaystyle x_k>0}$ such that ${\displaystyle \phi \left(kx\right)\le R_k\phi \left(x\right)}$ for all ${\displaystyle x\in (0,x_k]}$. The Orlicz sequence class associated with ${\displaystyle \phi }$ is

$${\tilde{\ell }}_{\phi }=\left\{x={\left(x_n\right)}_n\in {\mathbb{K}}^{\mathbb{N}}:\delta (x,\phi ):=\sum _{n=1}^{+\infty }\phi \left(\right|x_n\left|\right)<+\infty \right\}.$$

We denote by ${\displaystyle {\tilde{B}}_{\phi }}$ the set ${\displaystyle \{x={\left(x_n\right)}_n\in {\mathbb{K}}^{\mathbb{N}}:\delta (x,\phi )\le 1\}}$.

The Orlicz sequence space associated with ${\displaystyle \phi }$ is

$${\ell }_{\phi }=\left\{x:={\left(x_n\right)}_n\in {\mathbb{K}}^{\mathbb{N}}:\sum _{n\ge 1}{x}_n{y}_n\mathrm{converges}\mathrm{for}\mathrm{all}y\in {\tilde{\ell }}_{{\phi }^{*}}\right\}.$$

This is a Banach space with respect to the norm

$$\begin{array}{cc}\hfill {\parallel x\parallel }_{\phi }& =sup\left\{\left|\sum _{n\ge 1}{x}_n{y}_n\right|:\delta (y,{\phi }^{*})\le 1\right\}\hfill \\ \hfill & =sup\left\{\sum _{n\ge 1}\left|x_n{y}_n\right|:\delta (y,{\phi }^{*})\le 1\right\}.\hfill \end{array}$$

Like in [18], if ${\displaystyle x\in {\ell }_{\phi }}$ and ${\displaystyle {∥x∥}_{\phi }\le 1}$, then ${\displaystyle x\in {\tilde{\ell }}_{\phi }}$ and ${\displaystyle \delta (x,\phi )\le {∥x∥}_{\phi }}$.

3. Weakly Köthe–Orlicz Summable Sequences

In this section, we introduce the notion of weakly Köthe–Orlicz summable sequences in a locally convex space ${\displaystyle E}$ and investigate some first properties of the linear space of all such sequences.

Definition 1.

A sequence ${\displaystyle x={\left(x_n\right)}_n\subset E}$ is said to be weakly Köthe–Orlicz summable with respect to ${\displaystyle \phi }$ and ${\displaystyle \mathsf{\Lambda }}$ (for short, weakly (${\displaystyle \phi ,\mathsf{\Lambda }}$)-summable) if the sequence ${\displaystyle {\left({\alpha }_{n}f\left(x_n\right)\right)}_n}$ belongs to ${\displaystyle {\ell }_{\phi }}$ for every ${\displaystyle f\in E^{\prime }}$ and every ${\displaystyle \alpha \in {\mathsf{\Lambda }}^{*}}$. The set of all such sequences is denoted by ${\displaystyle {\mathsf{\Lambda }}_{\phi }\left[E\right]}$.

Since ${\displaystyle {\mathsf{\Lambda }}^{*}={\left({\mathsf{\Lambda }}^{**}\right)}^{*}}$, we assume with no loss of generality that ${\displaystyle \mathsf{\Lambda }}$ is perfect, i.e., ${\displaystyle \mathsf{\Lambda }={\mathsf{\Lambda }}^{**}}$.

Here are some examples of Orlicz functions and the corresponding ${\displaystyle {\mathsf{\Lambda }}_{\phi }\left[E\right]}$.

Example 1.

 1.

Let ${\displaystyle \phi }$ be the identity map ${\displaystyle x\mapsto x}$. Then ${\displaystyle {\mathsf{\Lambda }}_{\phi }\left[E\right]}$ coincides with the space ${\displaystyle \mathsf{\Lambda }\left[E\right]}$ of weakly summable sequence in ${\displaystyle E}$ (see, e.g., [4]).

 2.

Assume ${\displaystyle \mathsf{\Lambda }={\ell }^1}$ and ${\displaystyle E=\mathbb{K}}$. Then ${\displaystyle {\mathsf{\Lambda }}_{\phi }\left[E\right]}$ is nothing but the classical Orlicz sequence space ${\displaystyle {\ell }_{\phi }}$.

 3.

Let ${\displaystyle \phi }$ be the Orlicz function defined by ${\displaystyle \phi \left(x\right):=+\infty }$ if ${\displaystyle x>1}$ and ${\displaystyle \phi \left(x\right):=0}$ if ${\displaystyle 0\le x\le 1}$. Let ${\displaystyle \mathsf{\Lambda }=c_0}$, the space of all scalar null sequences, and let ${\displaystyle E}$ be a Hausdorff locally convex space. We claim that ${\displaystyle {\left(c_0\right)}_{\phi }\left[E\right]}$ is the set ${\displaystyle c_b\left(E\right)}$ of all bounded sequences in ${\displaystyle E}$. Indeed, since ${\displaystyle {\left(c_0\right)}^{**}={\ell }^{\infty }}$, ${\displaystyle {\left(c_0\right)}_{\phi }\left[E\right]={\ell }^{\infty }\left[E\right]}$. Let ${\displaystyle x\in {\ell }_{\phi }^{\infty }\left[E\right]}$. Then for every ${\displaystyle f\in E^{\prime }}$ and ${\displaystyle \alpha \in {\ell }^1:={\left({\ell }^{\infty }\right)}^{*}}$, we have ${\displaystyle {\left({\alpha }_{n}f\left(x_n\right)\right)}_n\in {\ell }_{\phi }:={\ell }^{\infty }}$. Since ${\displaystyle \alpha }$ is arbitrary in ${\displaystyle {\ell }^1}$, the sequence ${\displaystyle {\left(f\left(x_n\right)\right)}_n}$ belongs to ${\displaystyle {\ell }^{\infty }}$. This means that the sequence ${\displaystyle {\left(x_n\right)}_n}$ is weakly bounded in ${\displaystyle E}$, for ${\displaystyle f}$ is arbitrary in ${\displaystyle E^{\prime }}$. Hence, ${\displaystyle {\left(x_n\right)}_n}$ belongs to ${\displaystyle c_b\left(E\right)}$. The inverse inclusion ${\displaystyle c_b\left(E\right)\subset {\ell }_{\phi }^{\infty }\left[E\right]}$ is trivial.

Notice that if for every ${\displaystyle \alpha \in {\mathsf{\Lambda }}^{*}}$ and ${\displaystyle f\in E^{\prime }}$, ${\displaystyle {\psi }_{\alpha ,f}}$ is the endomorphism of ${\displaystyle E^{\mathbb{N}}}$ defined by ${\displaystyle {\psi }_{\alpha ,f}\left({\left(x_n\right)}_n\right)={\left({\alpha }_{n}f\left(x_n\right)\right)}_n}$, then

$${\mathsf{\Lambda }}_{\phi }\left[E\right]=\bigcap \{{\psi }_{\alpha ,f}^{-1}\left({\ell }_{\phi }\right),\alpha \in {\mathsf{\Lambda }}^{*},f\in E^{\prime }\}.$$

This shows that ${\displaystyle {\mathsf{\Lambda }}_{\phi }\left[E\right]}$ is a linear space.

Lemma 1.

For every ${\displaystyle x={\left(x_n\right)}_n\in {\mathsf{\Lambda }}_{\phi }\left[E\right]}$ and ${\displaystyle S\in \mathcal{S}}$, the set ${\displaystyle A_S^{\phi }}$ below is bounded in ${\displaystyle E}$.

$$A_S^{\phi }=\left\{\sum _{n=1}^p{\alpha }_n{y}_n{x}_n:\alpha \in S,y\in {\tilde{B}}_{{\phi }^{*}},p\in \mathbb{N}\right\}.$$

Therefore, for every ${\displaystyle S\in \mathcal{S}}$ and ${\displaystyle M\in \mathcal{M}}$, a semi-norm ${\displaystyle {\epsilon }_{S,M}^{\phi }}$ is defined on ${\displaystyle {\mathsf{\Lambda }}_{\phi }\left[E\right]}$, where

$${\epsilon }_{S,M}^{\phi }\left(x\right):=\underset{\alpha \in S,f\in M}{sup}{\parallel {\left({\alpha }_{n}f\left(x_n\right)\right)}_n\parallel }_{\phi },x={\left(x_n\right)}_n\in {\mathsf{\Lambda }}_{\phi }\left[E\right].$$

Proof. 

Let ${\displaystyle x={\left(x_n\right)}_n\in {\mathsf{\Lambda }}_{\phi }\left[E\right]}$, ${\displaystyle \alpha \in S}$, ${\displaystyle y\in {\tilde{B}}_{{\phi }^{*}}}$, ${\displaystyle p\in \mathbb{N}}$, and ${\displaystyle f\in E^{\prime }}$ be given. Then

$$\left|f\left(\sum _{n=1}^p{\alpha }_n{y}_n{x}_n\right)\right|=\left|\sum _{n=1}^p{\alpha }_n{y}_{n}f\left(x_n\right)\right|\le {\parallel {\left({\alpha }_{n}f\left(x_n\right)\right)}_n\parallel }_{\phi }.$$

Define a linear mapping ${\displaystyle g_f:{\mathsf{\Lambda }}_S^{*}⟶{\ell }_{\phi }}$ by ${\displaystyle g_f\left(\beta \right)=\left({\beta }_{n}f\left(x_n\right)\right)}$. Since ${\displaystyle {\mathsf{\Lambda }}_S^{*}}$ is a Banach space ([4], Lemma 3), ${\displaystyle g_f}$ is continuous by the closed graph theorem. Therefore, it is bounded on ${\displaystyle S}$ by the norm ${\displaystyle \parallel g_f\parallel }$ of ${\displaystyle g_f}$. This is

$$\left|f\left(\sum _{n=1}^p{\alpha }_n{y}_n{x}_n\right)\right|\le \parallel {\left({\alpha }_{n}f\left(x_n\right)\right)}_n{\parallel }_{\phi }\le \parallel g_f\parallel .$$

Since ${\displaystyle f}$ was arbitrary in ${\displaystyle E^{\prime }}$, ${\displaystyle A_S^{\phi }}$ is weakly bounded and is then also bounded in ${\displaystyle E}$. The remainder is trivial. □

We denote by ${\displaystyle {\epsilon }_{\mathcal{S},\mathcal{M}}^{\phi }}$ the locally convex topology defined on ${\displaystyle {\mathsf{\Lambda }}_{\phi }\left[E\right]}$ by the family ${\displaystyle {\left({\epsilon }_{S,M}^{\phi }\right)}_{\genfrac{}{}{0pt}{}{S\in \mathcal{S},}{M\in \mathcal{M}}}}$ of semi-norms.

Example 2.

 1.

If ${\displaystyle \phi }$ is the identity of ${\displaystyle {\mathbb{R}}_{+}}$, the topology ${\displaystyle {\epsilon }_{\mathcal{S},\mathcal{M}}^{\phi }}$ of ${\displaystyle {\mathsf{\Lambda }}_{\phi }\left[E\right]}$ is nothing but the topology ${\displaystyle {\epsilon }_{\mathcal{S},\mathcal{M}}}$ given in [4].

 2.

In case ${\displaystyle \mathsf{\Lambda }={\ell }^1}$ and ${\displaystyle E=\mathbb{K}}$, the topology ${\displaystyle {\epsilon }_{\mathcal{S},\mathcal{M}}^{\phi }}$ coincides with the norm topology of ${\displaystyle {\ell }_{\phi }}$.

 3.

When ${\displaystyle \phi }$ is the Orlicz function in (3) of Example 1, ${\displaystyle {\epsilon }_{\mathcal{S},\mathcal{M}}^{\phi }}$ is given by the semi-norms

$${\epsilon }_M\left(x\right):=\underset{f\in M}{sup}{\parallel {\left(f\left(x_n\right)\right)}_n\parallel }_{\infty },x\in c_b\left(E\right),M\in \mathcal{M}.$$

Lemma 2.

The topology ${\displaystyle {\epsilon }_{\mathcal{S},\mathcal{M}}^{\phi }}$ is Hausdorff. Moreover:

 1.

For every ${\displaystyle n\in \mathbb{N}}$, the projection ${\displaystyle {\mathcal{I}}_n:x:={\left(x_k\right)}_k\mapsto x_n}$ is a continuous mapping from ${\displaystyle {\mathsf{\Lambda }}_{\phi }\left[E\right]}$ into ${\displaystyle E}$;

 2.

${\displaystyle {\mathsf{\Lambda }}_{\phi }{\left[E\right]}_r}$ is a closed subspace of ${\displaystyle {\mathsf{\Lambda }}_{\phi }\left[E\right]}$.

Proof. 

It is easily seen that ${\displaystyle {\epsilon }_{\mathcal{S},\mathcal{M}}^{\phi }}$ is Hausdorff. To show this:

• Fix ${\displaystyle n\in \mathbb{N}}$, ${\displaystyle M\in \mathcal{M}}$ and choose ${\displaystyle S\in \mathcal{S}}$ such that ${\displaystyle e_n\in S}$. For all ${\displaystyle x={\left(x_n\right)}_n\in {\mathsf{\Lambda }}_{\phi }\left[E\right]}$, we have

  $$\begin{array}{cc}\hfill P_M\left({\mathcal{I}}_n\left(x\right)\right)=P_M\left(x_n\right)& =\frac{1}{{∥e_n∥}_{\phi }}{∥P_M\left(x_n\right)e_n∥}_{\phi }\hfill \\ \hfill & \le \frac{1}{{∥e_n∥}_{\phi }}{\epsilon }_{S,M}^{\phi }\left(x\right).\hfill \end{array}$$

  Then ${\displaystyle {\mathcal{I}}_n}$ is continuous.
• Let ${\displaystyle x\in \overline{{\mathsf{\Lambda }}_{\phi }{\left[E\right]}_r}}$. Then for all ${\displaystyle \epsilon >0}$, ${\displaystyle M\in \mathcal{M}}$, and ${\displaystyle S\in \mathcal{S}}$, there is ${\displaystyle y\in {\mathsf{\Lambda }}_{\phi }{\left[E\right]}_r}$ such that ${\displaystyle {\epsilon }_{S,M}^{\phi }(x-y)\le \frac{\epsilon }{3}}$. Since ${\displaystyle y\in {\mathsf{\Lambda }}_{\phi }{\left[E\right]}_r}$, there is ${\displaystyle n_0\in \mathbb{N}}$ such that for all ${\displaystyle i\ge n_0}$, ${\displaystyle {\epsilon }_{S,M}^{\phi }(y^{\left(i\right)}-y)\le \frac{\epsilon }{3}}$. So for all ${\displaystyle i\ge n_0}$:

  $$\begin{array}{cc}\hfill {\epsilon }_{S,M}^{\phi }(x^{\left(i\right)}-x)& \le {\epsilon }_{S,M}^{\phi }(x^{\left(i\right)}-y^{\left(i\right)})+{\epsilon }_{S,M}^{\phi }(y^{\left(i\right)}-y)+{\epsilon }_{S,M}^{\phi }(x-y)\hfill \\ \hfill & \le {\epsilon }_{S,M}^{\phi }\left({(x-y)}^{\left(i\right)}\right)+{\epsilon }_{S,M}^{\phi }(y^{\left(i\right)}-y)+{\epsilon }_{S,M}^{\phi }(x-y)\hfill \\ \hfill & \le 2{\epsilon }_{S,M}^{\phi }(x-y)+{\epsilon }_{S,M}^{\phi }(y^{\left(i\right)}-y)\le \epsilon .\hfill \end{array}$$
  Then ${\displaystyle {\mathsf{\Lambda }}_{\phi }{\left[E\right]}_r}$ is closed.

${\displaystyle \square }$

Remark 1.

According to the proof above, for every ${\displaystyle S\in \mathcal{S}}$, the set ${\displaystyle \{{\mathcal{I}}_n,e_n\in S\}}$ is even equicontinuous. In particular, if ${\displaystyle \mathsf{\Lambda }}$ is a normed space so that ${\displaystyle \parallel e_n{\parallel }_{{\mathsf{\Lambda }}^{*}}\le 1}$ for every ${\displaystyle n}$, then ${\displaystyle \{{\mathcal{I}}_n,n\in \mathbb{N}\}}$ is equicontinuous and is then also equibounded. An instance where this occurs is ${\displaystyle \mathsf{\Lambda }={\ell }^p}$.

The following lemma shows that not only is ${\displaystyle E}$ (identified with) a subspace of ${\displaystyle {\mathsf{\Lambda }}_{\phi }\left[E\right]}$, but it is also complemented in it.

Lemma 3.

The space ${\displaystyle E}$ is complemented in both spaces ${\displaystyle {\mathsf{\Lambda }}_{\phi }\left[E\right]}$ and ${\displaystyle {\mathsf{\Lambda }}_{\phi }{\left[E\right]}_r}$.

Proof. 

Set ${\displaystyle \left[E\right]:=\{t{e}_1:t\in E\}}$ and consider the mapping ${\displaystyle p:{\mathsf{\Lambda }}_{\phi }\left[E\right]\to \left[E\right]}$ defined for all ${\displaystyle {\left(x_n\right)}_n\in {\mathsf{\Lambda }}_{\phi }\left[E\right]}$ by ${\displaystyle p\left({\left(x_n\right)}_n\right)=x_1{e}_1}$. This is a projection, and since

$${\epsilon }_{S,M}^{\phi }\left(p\left({\left(x_n\right)}_n\right)\right)\le {\epsilon }_{S,M}^{\phi }\left({\left(x_n\right)}_n\right),{\left(x_n\right)}_n\in {\mathsf{\Lambda }}_{\phi }\left[E\right],(S,M)\in \mathcal{S}\times \mathcal{M},$$

${\displaystyle p}$ is a continuous. Therefore, ${\displaystyle \left[E\right]}$ is complemented in ${\displaystyle {\mathsf{\Lambda }}_{\phi }\left[E\right]}$. Now, the mapping ${\displaystyle \varphi :t\mapsto t{e}_1}$ is a bicontinuous linear isomorphism from ${\displaystyle E}$ into ${\displaystyle \left[E\right]}$ because for all ${\displaystyle t\in E}$ and all ${\displaystyle (S,M)\in \mathcal{S}\times \mathcal{M}}$,

$${\epsilon }_{S,M}^{\phi }\left(t{e}_1\right)={∥e_1∥}_{\phi }{P}_S\left(e_1\right)P_M\left(t\right).$$

Identifying ${\displaystyle E}$ and ${\displaystyle \left[E\right]}$, ${\displaystyle E}$ is complemented in ${\displaystyle {\mathsf{\Lambda }}_{\phi }\left[E\right]}$.

The same proof also works for ${\displaystyle {\mathsf{\Lambda }}_{\phi }{\left[E\right]}_r}$. ${\displaystyle \square }$

The following theorem shows when ${\displaystyle {\mathsf{\Lambda }}_{\phi }\left[E\right]}$ is complete or sequentially complete.

Theorem 1.

The space ${\displaystyle {\mathsf{\Lambda }}_{\phi }\left[E\right]}$ is (sequentially) complete if and only if ${\displaystyle E}$ is (sequentially) complete.

Proof. 

This necessity is derived from Lemma 3. As to the sufficiency, assume ${\displaystyle E}$ is complete, and let ${\displaystyle {\left(x^i\right)}_{i\in I}}$ be a Cauchy net in ${\displaystyle {\mathsf{\Lambda }}_{\phi }\left[E\right]}$, with ${\displaystyle (I,\le )}$ being an upwardly directed ordered set. The continuity of the projection ${\displaystyle {\mathcal{I}}_n}$ implies that ${\displaystyle {\left(x_n^i\right)}_i}$ is a Cauchy net in ${\displaystyle E}$ for all ${\displaystyle n}$. Hence, it converges to some ${\displaystyle x_n\in E}$.

We claim that ${\displaystyle x:={\left(x_n\right)}_n}$ belongs to ${\displaystyle {\mathsf{\Lambda }}_{\phi }\left[E\right]}$. For every ${\displaystyle S\in \mathcal{S}}$, ${\displaystyle M\in \mathcal{M}}$, and ${\displaystyle \epsilon >0}$, choose ${\displaystyle k\in I}$ such that for all ${\displaystyle i,j>k}$, ${\displaystyle {\epsilon }_{S,M}^{\phi }(x^i-x^j)<\epsilon }$. Then, by normality of ${\displaystyle {\ell }_{\phi }}$, for every ${\displaystyle \alpha \in S}$, ${\displaystyle f\in M}$, and ${\displaystyle i,j>k}$, one has

$${∥{\left({\alpha }_{n}f\left(x_n^i\right)-{\alpha }_{n}f\left(x_n^j\right)\right)}_n∥}_{\phi }\le {\epsilon }_{S,M}^{\phi }(x^i-x^j)<\epsilon .$$

Therefore, ${\displaystyle {\left({\alpha }_{n}f\left(x_n^i\right)\right)}_i}$ is a Cauchy sequence in the Banach space ${\displaystyle {\ell }_{\phi }}$ for all ${\displaystyle n\in \mathbb{N}}$. Let ${\displaystyle \gamma :={\left({\gamma }_n\right)}_n}$ be its limit in ${\displaystyle {\ell }_{\phi }}$. Then for every ${\displaystyle n\in \mathbb{N}}$, we have

$${\alpha }_{n}f\left(x_n\right)={\alpha }_{n}f\left(\underset{i}{lim}{x}_n^i\right)=\underset{i}{lim}{\alpha }_{n}f\left(x_n^i\right)={\gamma }_n.$$

But for ${\displaystyle i,j\ge k}$, ${\displaystyle \alpha \in \mathcal{S}}$, and ${\displaystyle N\in \mathbb{N}}$, we have

$$\underset{\delta (y,{\phi }^{*})⩽1}{sup}\sum _{n=1}^N\left|y_n{\alpha }_{n}f(x_n^i-x_n^j)\right|\le {∥{\left({\alpha }_{n}f(x_n^i-x_n^j)\right)}_n∥}_{\phi }\le {\epsilon }_{S,M}^{\phi }(x^i-x^j)<\epsilon .$$

Passing to the limit on ${\displaystyle j}$, we get for all ${\displaystyle N\ge n_0}$

$$\underset{\delta (y,{\phi }^{*})⩽1}{sup}\sum _{n=1}^N\left|y_n{\alpha }_{n}f(x_n^i-x_n)\right|\le \epsilon ,$$

and then ${\displaystyle {\epsilon }_{S,M}^{\phi }(x^i-x)\le \epsilon }$ for every ${\displaystyle i\ge k}$. This shows at once that ${\displaystyle x}$ belongs to ${\displaystyle {\mathsf{\Lambda }}_{\phi }\left[E\right]}$ and that ${\displaystyle {\left(x^i\right)}_{i\in I}}$ converges to ${\displaystyle x}$ in ${\displaystyle {\mathsf{\Lambda }}_{\phi }\left[E\right]}$.

With a similar proof, one shows that ${\displaystyle {\mathsf{\Lambda }}_{\phi }\left[E\right]}$ is sequentially complete if and only if ${\displaystyle E}$ is sequentially complete. ${\displaystyle \square }$

Lemma 3 and Theorem 1 show that the three spaces ${\displaystyle E}$, ${\displaystyle {\mathsf{\Lambda }}_{\phi }\left[E\right]}$, and ${\displaystyle {\mathsf{\Lambda }}_{\phi }{\left[E\right]}_r}$ are simultaneously complete or simultaneously not complete.

Proposition 1.

If ${\displaystyle E}$ is fast-barrelled, then

$${\mathsf{\Lambda }}_{\phi }\left[E_{\beta }^{\prime }\right]=\{a={\left(a_n\right)}_n\subset E^{\prime }:{\left({\alpha }_n{a}_n\left(x\right)\right)}_n\in {\ell }_{\phi },x\in E,\alpha \in {\mathsf{\Lambda }}^{*}\}.$$

Moreover, the topology of ${\displaystyle {\mathsf{\Lambda }}_{\phi }\left[E_{\beta }^{\prime }\right]}$ is given by the semi-norms

$${\epsilon }_{S,B}^{\phi }\left(a\right)=\underset{\alpha \in S,x\in B}{sup}{\parallel {\left({\alpha }_n{a}_n\left(x\right)\right)}_n\parallel }_{\phi },$$

where ${\displaystyle S}$ runs over ${\displaystyle \mathcal{S}}$, and ${\displaystyle B}$ runs over the collection ${\displaystyle \mathcal{B}}$ of all closed and bounded discs in ${\displaystyle E}$.

Proof. 

If

$$\mathsf{\Delta }:=\{a={\left(a_n\right)}_n\subset E^{\prime }:{\left({\alpha }_n{a}_n\left(x\right)\right)}_n\in {\ell }_{\phi },x\in E,\alpha \in {\mathsf{\Lambda }}^{*}\},$$

then clearly, ${\displaystyle {\mathsf{\Lambda }}_{\phi }\left[E_{\beta }^{\prime }\right]\subset \mathsf{\Delta }}$.

Conversely, consider ${\displaystyle a:={\left(a_n\right)}_n\in \mathsf{\Delta }}$, ${\displaystyle f\in {\left(E_{\beta }^{\prime }\right)}^{\prime }}$, ${\displaystyle y\in {\tilde{B}}_{{\phi }^{*}}}$, and ${\displaystyle \beta \in {\mathsf{\Lambda }}^{*}}$. Choose ${\displaystyle x\in E}$. Then

$$\left|\sum _{n=1}^p{y}_n{\beta }_n{a}_n\left(x\right)\right|\le \sum _{n=1}^{+\infty }\left|y_n{\beta }_n{a}_n\left(x\right)\right|<+\infty ,p\in \mathbb{N}.$$

Therefore,

$$A:=\left\{\sum _{n=1}^p{y}_n{\beta }_n{a}_n,p\in \mathbb{N}\right\}$$

is ${\displaystyle \sigma (E^{\prime },E)}$-bounded. Since ${\displaystyle E}$ is fast-barrelled, ${\displaystyle A}$ is bounded in ${\displaystyle E_{\beta }^{\prime }}$. Hence, there is some ${\displaystyle K>0}$ such that

$$\sum _{n=1}^{+\infty }\left|y_n{\beta }_{n}f\left(a_n\right)\right|\le K.$$

Consequently,

$$a\in {\mathsf{\Lambda }}_{\phi }\left[E_{\beta }^{\prime }\right].$$

Now, let ${\displaystyle M}$ be a closed equicontinuous disc in ${\displaystyle {\left(E_{\beta }^{\prime }\right)}^{\prime }}$. Then the polar ${\displaystyle M^{\circ }}$ of ${\displaystyle M}$ is a ${\displaystyle 0}$-neighborhood in ${\displaystyle E_{\beta }^{\prime }}$. If ${\displaystyle B}$ is the polar in ${\displaystyle E}$ of ${\displaystyle M^{\circ }}$, then ${\displaystyle B}$ is a closed bounded disc in ${\displaystyle E}$ such that

$$M=M^{\circ \circ }\subset B^{\circ \circ }={\overline{B}}^{\sigma (E^{\prime \prime },E^{\prime })}.$$

Then for every ${\displaystyle a\in E^{\prime }}$, we have

$$\underset{f\in M}{sup}\left|f\left(a\right)\right|\le \underset{x\in B^{\circ \circ }}{sup}\left|a\left(x\right)\right|\le \underset{x\in B}{sup}\left|a\left(x\right)\right|.$$

In particular, for ${\displaystyle a=\sum _{n=1}^p{\alpha }_n{y}_n{a}_n\in E^{\prime }}$ with ${\displaystyle y\in {\tilde{B}}_{{\phi }^{*}}}$, ${\displaystyle \alpha \in S}$ and ${\displaystyle a\in {\mathsf{\Lambda }}_{\phi }\left[E_{\beta }^{\prime }\right]}$, we have

$$\underset{f\in M}{sup}\left|\sum _{n=1}^p{\alpha }_n{y}_{n}f\left(a_n\right)\right|\le \underset{x\in B^{\circ \circ }}{sup}\left|\sum _{n=1}^p{\alpha }_n{y}_n{a}_n\left(x\right)\right|\le \underset{x\in B}{sup}\left|\sum _{n=1}^p{\alpha }_n{y}_n{a}_n\left(x\right)\right|.$$

Passing to the supremum on ${\displaystyle p}$, first on ${\displaystyle y\in {\tilde{B}}_{{\phi }^{*}}}$ and then on ${\displaystyle \alpha \in S}$, we get

$${\epsilon }_{S,M}^{\phi }\left(a\right)\le {\epsilon }_{S,B}^{\phi }\left(a\right),$$

which completes the proof. ${\displaystyle \square }$

4. Continuous Dual Space of ${\displaystyle {\mathsf{\Lambda }}_{\mathbf{\phi }}\left[\mathit{E}\right]}$

In the literature, several kinds of duals are considered when dealing with sequence spaces: mainly the Köthe-dual or the ${\displaystyle \alpha }$-dual, the ${\displaystyle \beta }$-dual, the Köthe–Toeplitz dual, the algebraic dual and, whenever the sequence space is equipped with a linear topology, the continuous dual (see [4,8,21]). In order to determine the continuous dual space of ${\displaystyle {\mathsf{\Lambda }}_{\phi }\left[E\right]}$, we introduce the notion of strongly Köthe–Orlicz summable sequences.

Definition 2.

A sequence ${\displaystyle x=\left(x_n\right)\subset E}$ is said to be strongly Köthe–Orlicz summable with respect to ${\displaystyle \phi }$ and ${\displaystyle \mathsf{\Lambda }}$ (for short, strongly (${\displaystyle \phi ,\mathsf{\Lambda }}$)-summable), if for every ${\displaystyle M\in \mathcal{M}}$ and every ${\displaystyle a={\left(a_n\right)}_n\in {\left({\mathsf{\Lambda }}^{*}\right)}_{{\phi }^{*}}\left[E_M^{\prime }\right]}$, the sequence ${\displaystyle {\left(a_n\left(x_n\right)\right)}_n}$ belongs to ${\displaystyle {\ell }^1}$.

The set of all strongly (${\displaystyle \phi ,\mathsf{\Lambda }}$)-summable sequences is denoted by ${\displaystyle {\mathsf{\Lambda }}_{\phi }⟨E⟩}$.

Proposition 2.

Let ${\displaystyle S\in \mathcal{S}}$ and ${\displaystyle M\in \mathcal{M}}$. Then:

 1.

The space ${\displaystyle {\left({\mathsf{\Lambda }}_S^{*}\right)}_{{\phi }^{*}}\left[E_M^{\prime }\right]}$ is a Banach space for the norm ${\displaystyle {\epsilon }_{S^{\circ },M^{\circ }}^{{\phi }^{*}}}$ defined by

$${\epsilon }_{S^{\circ },M^{\circ }}^{{\phi }^{*}}\left(a\right):=\underset{f\in M^{\circ },\alpha \in S^{\circ }}{sup}{\parallel {\left({\alpha }_{n}f\left(a_n\right)\right)}_n\parallel }_{{\phi }^{*}},a:={\left(a_n\right)}_n\in {\left({\mathsf{\Lambda }}_S^{*}\right)}_{{\phi }^{*}}\left[E_M^{\prime }\right],$$

with ${\displaystyle S^{\circ }}$ being the polar of ${\displaystyle S}$ in ${\displaystyle \mathsf{\Lambda }}$. Moreover, the projections ${\displaystyle {\left(a_n\right)}_n\mapsto a_n}$ are continuous.

 2.

The mapping ${\displaystyle {\sigma }_{S,M}^{\phi }}$ is a semi-norm on ${\displaystyle {\mathsf{\Lambda }}_{\phi }⟨E⟩}$, where for all ${\displaystyle x\in {\mathsf{\Lambda }}_{\phi }⟨E⟩}$,

$${\sigma }_{S,M}^{\phi }\left(x\right)=sup\left\{\sum _{n=1}^{+\infty }\left|a_n\left(x_n\right)\right|;a={\left(a_n\right)}_n\in {\left({\mathsf{\Lambda }}_S^{*}\right)}_{{\phi }^{*}}\left[E_M^{\prime }\right],{\epsilon }_{S^{\circ },M^{\circ }}^{{\phi }^{*}}\left(a\right)\le 1\right\}.$$

Proof. 

1. If ${\displaystyle {\mathcal{S}}^{\prime }:=\{r{S}^{\prime },r\ge 0\}}$, where ${\displaystyle S^{\prime }}$ denotes the ${\displaystyle \sigma ({\left({\mathsf{\Lambda }}_S^{*}\right)}^{*},{\mathsf{\Lambda }}_S^{*})}$-closure of ${\displaystyle S^{\circ }}$ in ${\displaystyle {\left({\mathsf{\Lambda }}_S^{*}\right)}^{*}}$, then the norm topology of ${\displaystyle {\mathsf{\Lambda }}_S^{*}}$ is nothing but the ${\displaystyle {\mathcal{S}}^{\prime }}$-topology. Therefore, by Theorem 1, ${\displaystyle {\left({\mathsf{\Lambda }}_S^{*}\right)}_{{\phi }^{*}}\left[E_M^{\prime }\right]}$ is the Banach space. Moreover, by Lemma 2, the projections are continuous.

2. It suffices to show that ${\displaystyle {\sigma }_{S,M}^{\phi }\left(x\right)}$ is finite for every ${\displaystyle x\in {\mathsf{\Lambda }}_{\phi }⟨E⟩}$. Fix then such an ${\displaystyle x}$ and define a linear mapping ${\displaystyle T_x}$ from ${\displaystyle {\left({\mathsf{\Lambda }}_S^{*}\right)}_{{\phi }^{*}}\left[E_M^{\prime }\right]}$ into ${\displaystyle {\ell }^1}$ by ${\displaystyle T_x\left({\left(a_n\right)}_n\right)={\left(a_n\left(x_n\right)\right)}_n}$. Suppose that ${\displaystyle {\left(a^i\right)}_i\in {\left({\mathsf{\Lambda }}_S^{*}\right)}_{{\phi }^{*}}\left[E_M^{\prime }\right]}$ converges to ${\displaystyle a:={\left(a_n\right)}_n}$ and ${\displaystyle {\left(T_x\left(a^i\right)\right)}_i}$ converges in ${\displaystyle {\ell }^1}$ to ${\displaystyle {\left({\gamma }_n\right)}_n}$. By continuity of the projections, ${\displaystyle {\left(a_n^i\right)}_i}$ converges in ${\displaystyle E_M^{\prime }}$ to some ${\displaystyle a_n}$ for every ${\displaystyle n\in \mathbb{N}}$. Then ${\displaystyle {\left(a_n^i\left(x_n\right)\right)}_i}$ converges to ${\displaystyle a_n\left(x_n\right)}$ as well. It follows that ${\displaystyle {\left(a_n\left(x_n\right)\right)}_n={\left({\gamma }_n\right)}_n}$: hence, the closedness of the graph of ${\displaystyle T_x}$. Therefore, ${\displaystyle T_x}$ is continuous and is then bounded on the unit ball of ${\displaystyle {\left({\mathsf{\Lambda }}_S^{*}\right)}_{{\phi }^{*}}\left[E_M^{\prime }\right]}$. This yields ${\displaystyle {\sigma }_{S,M}^{\phi }\left(x\right)<+\infty }$. ${\displaystyle \square }$

The following lemma can be shown using a standard argument. Its proof is thus omitted.

Lemma 4.

If ${\displaystyle \gamma :={\left({\gamma }_n\right)}_n\in c_0}$, then ${\displaystyle \gamma x={\left({\gamma }_n{x}_n\right)}_n\in {\mathsf{\Lambda }}_{\phi }{\left[E\right]}_r}$ for every ${\displaystyle x={\left(x_n\right)}_n\in {\mathsf{\Lambda }}_{\phi }\left[E\right]}$.

For a continuous linear functional ${\displaystyle F}$ on ${\displaystyle {\mathsf{\Lambda }}_{\phi }\left[E\right]}$ (or on ${\displaystyle {\mathsf{\Lambda }}_{\phi }{\left[E\right]}_r}$), let ${\displaystyle F_n\left(t\right):=F\left(t{e}_n\right)}$ for ${\displaystyle n\in \mathbb{N}}$ and ${\displaystyle t\in E}$. The following lemma shows that in some sense, the topological dual space of ${\displaystyle {\mathsf{\Lambda }}_{\phi }{\left[E\right]}_r}$ is contained in ${\displaystyle {\left({\mathsf{\Lambda }}_{\phi }\left[E\right]\right)}^{*}}$.

Lemma 5.

Let ${\displaystyle F}$ be a continuous linear functional on ${\displaystyle {\mathsf{\Lambda }}_{\phi }\left[E\right]}$. Then:

 1.

There exists ${\displaystyle M\in \mathcal{M}}$ such that ${\displaystyle {\left(F_n\right)}_n\in E_M^{\prime }}$.

 2.

The sequence ${\displaystyle {\left(F_n\right)}_n}$ belongs to ${\displaystyle {\left({\mathsf{\Lambda }}_{\phi }\left[E\right]\right)}^{*}}$.

If, in addition, the family ${\displaystyle \{e_n,n\in \mathbb{N}\}}$ is ${\displaystyle {\tau }_{\mathcal{S}}}$-bounded, then ${\displaystyle {\left(F_n\right)}_n}$ is equicontinuous.

Proof. 

By continuity of ${\displaystyle F}$, for every ${\displaystyle x\in {\mathsf{\Lambda }}_{\phi }{\left[E\right]}_r}$, we have

$$F\left(x\right)=F\left(\sum _{n\ge 1}{x}_n{e}_n\right)=\sum _{n\ge 1}F\left(x_n{e}_n\right)=\sum _{n\ge 1}{F}_n\left(x_n\right).$$

Moreover, there exist ${\displaystyle S\in \mathcal{S}}$ and ${\displaystyle M\in \mathcal{M}}$ such that ${\displaystyle \left|F\left(x\right)\right|\le {\epsilon }_{S,M}^{\phi }\left(x\right)}$ for all ${\displaystyle x\in {\mathsf{\Lambda }}_{\phi }\left\{E\right\}}$. Fix ${\displaystyle n\in \mathbb{N}}$ and ${\displaystyle t\in E}$. We have

$$\left|F_n\left(t\right)\right|=\left|F\left(t{e}_n\right)\right|\le {\epsilon }_{S,M}^{\phi }\left(t{e}_n\right)={\parallel e_n\parallel }_{\phi }{P}_S\left(e_n\right)P_M\left(t\right).$$

(1)

It follows that ${\displaystyle F_n}$ belongs to ${\displaystyle E_M^{\prime }}$ and thus Condition 1 is proved.

For Condition 2, let ${\displaystyle x\in {\mathsf{\Lambda }}_{\phi }\left[E\right]}$ be arbitrary. For all ${\displaystyle \gamma \in c_0}$, ${\displaystyle \gamma x\in {\mathsf{\Lambda }}_{\phi }{\left[E\right]}_r}$. Choose a scalar sequence ${\displaystyle \lambda ={\left({\lambda }_n\right)}_n}$ such that ${\displaystyle \left|{\lambda }_n\right|=1}$ and ${\displaystyle \left|{\gamma }_n{F}_n\left(x_n\right)\right|={\lambda }_n{\gamma }_n{F}_n\left(x_n\right)}$ for all ${\displaystyle n\in \mathbb{N}}$. Since ${\displaystyle \gamma \lambda x\in {\mathsf{\Lambda }}_{\phi }{\left[E\right]}_r}$, we have

$$\sum _{n\ge 1}\left|{\gamma }_n{F}_n\left(x_n\right)\right|=\sum _{n\ge 1}{\gamma }_n{\lambda }_n{F}_n\left(x_n\right)=\sum _{n\ge 1}{F}_n\left({\gamma }_n{\lambda }_n{x}_n\right)=F\left(\lambda \gamma x\right)<+\infty .$$

As ${\displaystyle \gamma \in c_0}$ was arbitrary, this shows that

$$\sum _{n\ge 1}\left|F_n\left(x_n\right)\right|<+\infty .$$

Hence, ${\displaystyle {\left(F_n\right)}_n\in {\left({\mathsf{\Lambda }}_{\phi }\left[E\right]\right)}^{*}}$.

Now, if in addition, the family ${\displaystyle \{e_n,n\in \mathbb{N}\}}$ is ${\displaystyle {\tau }_{\mathcal{S}}}$-bounded, choose ${\displaystyle s>0}$ such that for every ${\displaystyle n\in \mathbb{N}}$, ${\displaystyle P_S\left(e_n\right)\le s}$, ${\displaystyle \parallel e_n{\parallel }_{\phi }\le s}$. We then get

$$\left|F_n\left(t\right)\right|\le {∥e_n∥}_{\phi }{P}_M\left(t\right)P_S\left(e_n\right)\le s^2{P}_M\left(t\right).$$

Therefore, ${\displaystyle {\left(F_n\right)}_n}$ is equicontinuous. ${\displaystyle \square }$

Now, we give a better description of continuous functionals on ${\displaystyle {\mathsf{\Lambda }}_{\phi }\left[E\right]}$.

Theorem 2.

If ${\displaystyle F}$ is a continuous functional on ${\displaystyle {\mathsf{\Lambda }}_{\phi }\left[E\right]}$, then there exist ${\displaystyle M\in \mathcal{M}}$ and ${\displaystyle S\in \mathcal{S}}$ such that the sequence ${\displaystyle {\left(F_n\right)}_n}$ is strongly ${\displaystyle ({\phi }^{*},{\mathsf{\Lambda }}_S^{*})}$-summable in ${\displaystyle E_M^{\prime }}$, i.e., ${\displaystyle {\left(F_n\right)}_n\in {\left({\mathsf{\Lambda }}_S^{*}\right)}_{{\phi }^{*}}⟨E_M^{\prime }⟩}$.

Proof. 

Let ${\displaystyle S\in \mathcal{S}}$ and ${\displaystyle M\in \mathcal{M}}$ be such that

$$\left|F\left(x\right)\right|\le {\epsilon }_{S,M}^{\phi }\left(x\right),x={\left(x_n\right)}_n\in {\mathsf{\Lambda }}_{\phi }\left[E\right].$$

By Lemma 5, ${\displaystyle {\left(F_n\right)}_n\subset E_M^{\prime }}$. Now, fix ${\displaystyle {\left(f_n\right)}_n\in {\left({\mathsf{\Lambda }}_S^{*}\right)}_{\phi }^{*}\left[{\left(E_M^{\prime }\right)}^{\prime }\right]}$. We claim that ${\displaystyle {\left(f_n\left(F_n\right)\right)}_n}$ belongs to ${\displaystyle {\ell }^1}$. Indeed, take an arbitrary ${\displaystyle k\in \mathbb{N}}$ and ${\displaystyle \delta >0}$, and denote by ${\displaystyle X}$ the completion of the normed space ${\displaystyle (E/M^{\perp },\overline{P_M})}$ and by ${\displaystyle B_k}$ the linear span of ${\displaystyle \{F_1,F_2,\dots ,F_k\}}$. Here, ${\displaystyle M^{\perp }}$ is the annihilator of ${\displaystyle M}$ in ${\displaystyle E^{\prime }}$, and as usual,

$$\overline{P_M}(x+M^{\perp }):=P_M\left(x\right).$$

Since ${\displaystyle E_M^{\prime }}$ is isometrically isomorphic to ${\displaystyle {(E/M^{\perp })}^{\prime }=X^{\prime }}$, we have ${\displaystyle B_k\subset X^{\prime }}$. But

$${\left(f_n\right)}_n\in {\left({\mathsf{\Lambda }}_S^{*}\right)}_{\phi }^{*}\left[{\left(E_M^{\prime }\right)}^{\prime }\right],$$

hence

$${\left(f_n\right)}_n\subset {\left(E_M^{\prime }\right)}^{\prime }=X^{\prime \prime }.$$

Let ${\displaystyle A_k}$ be the linear span of ${\displaystyle \left\{f_1,f_2,\dots ,f_k\right\}}$. By the principle of local reflexivity, there exists a continuous operator ${\displaystyle T_k:A_k⟶X}$ such that:

• ${\displaystyle ∥T_k∥\le 1+\delta }$ with ${\displaystyle ∥T_k∥=\underset{f\in M^{\circ }}{sup}{∥T_k\left(f\right)∥}_X}$;
• ${\displaystyle F_n\left(T_k{f}_n\right)=f_n\left(F_n\right),n\in \left\{1,2,\dots ,k\right\}}$.

Since ${\displaystyle E/M^{\perp }}$ is dense in ${\displaystyle X}$, for any

$$0<{\delta }_n\le \frac{\delta }{k(1+\parallel e_n{\parallel }_{\phi }{P}_S\left(e_n\right))},$$

there is ${\displaystyle x_n\in E}$ such that:

$$\overline{P_M}(x_n+M^{\perp }-T_k{f}_n)\le {\delta }_n.$$

Next, (1) implies that ${\displaystyle \parallel F_n{\parallel }_M\le {∥e_n∥}_{\phi }{P}_S\left(e_n\right)}$. Therefore, as ${\displaystyle F_n}$ is continuous,

$$\begin{array}{cc}\hfill \left|F_n(x_n+M^{\perp }-T_k{f}_n)\right|& \le \parallel F_n{\parallel }_M{P}_M(x_n-T_k{f}_n)\hfill \\ \hfill & \le \parallel e_n{\parallel }_{\phi }{P}_S\left(e_n\right)\frac{\delta }{k(1+\parallel e_n{\parallel }_{\phi }{P}_S\left(e_n\right))}\hfill \\ \hfill & \le \frac{\delta }{k}.\hfill \end{array}$$

Choose ${\displaystyle {\lambda }_n}$ in the unit complex circle so that ${\displaystyle \left|F\left(x_n{e}_n\right)\right|={\lambda }_{n}F\left(x_n{e}_n\right)}$. Then

$$\begin{array}{cc}\hfill \sum _{n=1}^k\left|f_n\left(F_n\right)\right|& =\sum _{n=1}^k\left|F_n\left(T_k{f}_n\right)\right|\hfill \\ \hfill & \le \sum _{n=1}^k|F_n(x_n+M^{\perp }-T_k{f}_n)|+\left|F\left(\sum _{n=1}^k{\lambda }_n{x}_n{e}_n\right)\right|\hfill \\ \hfill & \le \delta +{\epsilon }_{S,M}^{\phi }\left((x_1,x_2,\dots ,x_k,0,\dots )\right)\hfill \\ \hfill & =\delta +sup\left\{\left|\sum _{n=1}^k{y}_n{\alpha }_{n}a\left(x_n\right)\right|:{\left({\alpha }_n\right)}_n\in S,a\in M,y\in {\tilde{B}}_{{\phi }^{*}}\right\}.\hfill \end{array}$$

But for every ${\displaystyle {\left({\alpha }_n\right)}_n\in S}$, ${\displaystyle y\in {\tilde{B}}_{{\phi }^{*}}}$, and ${\displaystyle a\in M}$,

$$\begin{array}{cc}\hfill \left|\sum _{n=1}^k{y}_n{\alpha }_{n}a\left(x_n\right)\right|& \le \left|\sum _{n=1}^k{y}_n{\alpha }_{n}a(x_n+M^{\perp }-T_k{f}_n)\right|+\left|\sum _{n=1}^k{y}_n{\alpha }_{n}a\left(T_k{f}_n\right)\right|\hfill \\ \hfill & \le \sum _{n=1}^k|y_n{\alpha }_n||a(x_n+M^{\perp }-T_k{f}_n)|+\left|a\left(T_k\left(\sum _{n=1}^k{y}_n{\alpha }_n{f}_n\right)\right)\right|\hfill \\ \hfill & \le \sum _{n=1}^k\left|y_n{\alpha }_n\right|{\parallel a\parallel }_M{\delta }_n+{\parallel a\parallel }_M∥T_k∥\underset{x^{\prime }\in M}{sup}\left\{\left|\sum _{n=1}^k{y}_n{\alpha }_n{f}_n\left(x^{\prime }\right)\right|\right\}\hfill \\ \hfill & \le \delta +(1+\delta ){\epsilon }_{S,M}^{\phi }\left({\left(f_n\right)}_n\right).\hfill \end{array}$$

Consequently,

$$\sum _{n=1}^k\left|f_n\left(F_n\right)\right|\le 2\delta +(1+\delta ){\epsilon }_{S,M}^{\phi }\left({\left(f_n\right)}_n\right),k\in \mathbb{N}.$$

Hence, ${\displaystyle {\left(f_n\left(F_n\right)\right)}_n}$ belongs to ${\displaystyle {\ell }^1}$. ${\displaystyle \square }$

Remark 2.

Since in the proof of Theorem 2, ${\displaystyle \delta }$ is arbitrary, it follows that

$$\sum _{n=1}^{+\infty }|f_n\left(F_n\right)|\le {\epsilon }_{S,M}^{\phi }\left({\left(f_n\right)}_n\right).$$

Using the Hahn–Banach theorem, we get:

Corollary 1.

If ${\displaystyle F}$ is a continuous functional on ${\displaystyle {\mathsf{\Lambda }}_{\phi }{\left[E\right]}_r}$, then there exist ${\displaystyle M\in \mathcal{M}}$ and ${\displaystyle S\in \mathcal{S}}$ such that ${\displaystyle {\left(F_n\right)}_n\in {\left({\mathsf{\Lambda }}_S^{*}\right)}_{{\phi }^{*}}⟨E_M^{\prime }⟩}$.

The following proposition is interesting on its own.

Proposition 3.

Let ${\displaystyle S\in \mathcal{S}}$ and ${\displaystyle M\in \mathcal{M}}$. If ${\displaystyle {\left(a_n\right)}_n\in {\left({\mathsf{\Lambda }}_S^{*}\right)}_{{\phi }^{*}}⟨E_M^{\prime }⟩}$, then ${\displaystyle (\parallel y_n{a}_n{{\parallel }_M)}_n\in {\mathsf{\Lambda }}_S^{*}}$ for every ${\displaystyle y\in {\tilde{B}}_{{\phi }^{*}}}$.

Proof. 

Fix ${\displaystyle {\left(a_n\right)}_n\in {\left({\mathsf{\Lambda }}_S^{*}\right)}_{{\phi }^{*}}⟨E_M^{\prime }⟩}$ and ${\displaystyle y\in {\tilde{B}}_{{\phi }^{*}}}$, and let ${\displaystyle {\left({\alpha }_n\right)}_n\in \mathsf{\Lambda }}$ and ${\displaystyle \epsilon >0}$ be given. We have

$$\parallel y_n{\alpha }_n{a}_n{\parallel }_M=\underset{t\in M^{\circ }}{sup}\left|y_n{\alpha }_n{a}_n\left(t\right)\right|,n\in \mathbb{N}.$$

Hence, for every ${\displaystyle n\in \mathbb{N}}$, there is ${\displaystyle t_n\in M^{\circ }\subset E}$ such that

$$\parallel y_n{\alpha }_n{a}_n{\parallel }_M\le \frac{\epsilon }{2^n}+\left|y_n{\alpha }_n{a}_n\left(t_n\right)\right|.$$

Fix ${\displaystyle n\in \mathbb{N}}$ and ${\displaystyle a\in E_M^{\prime }}$ and define ${\displaystyle f_n\left(a\right):={\alpha }_{n}a\left(t_n\right)}$. Then

$$\left|f_n\left(a\right)\right|=\left|{\alpha }_{n}a\left(t_n\right)\right|\le {\parallel a\parallel }_M{P}_M\left(t_n\right)\left|{\alpha }_n\right|\le {\parallel a\parallel }_M\left|{\alpha }_n\right|.$$

Since ${\displaystyle a\in E_M^{\prime }}$, there is ${\displaystyle \mu >0}$ such that ${\displaystyle a\in \mu M}$. Therefore, ${\displaystyle \left|y_n{f}_n\left(a\right)\right|\le \mu {\parallel y\parallel }_{\infty }\left|{\alpha }_n\right|}$, and as ${\displaystyle \mathsf{\Lambda }}$ is normal, ${\displaystyle {\left(y_n{f}_n\left(a\right)\right)}_n\in \mathsf{\Lambda }}$. Hence, ${\displaystyle {\left(y_n{f}_n\left(a\right)\right)}_n\in {\left({\mathsf{\Lambda }}_S^{*}\right)}^{*}}$ for ${\displaystyle \mathsf{\Lambda }\subset {\left({\mathsf{\Lambda }}_S^{*}\right)}^{*}}$. Using Proposition 1, we come to

$${\left(f_n\right)}_n\in {\left({\mathsf{\Lambda }}_S^{*}\right)}_{\phi }^{*}\left[{\left(E_M^{\prime }\right)}^{\prime }\right].$$

Further, since ${\displaystyle {\left(a_n\right)}_n\in {\left({\mathsf{\Lambda }}_S^{*}\right)}_{{\phi }^{*}}⟨E_M^{\prime }⟩}$, the series

$$\sum _{n=1}^{+\infty }{f}_n\left(a_n\right)=\sum _{n=1}^{+\infty }{\alpha }_n{a}_n\left(t_n\right)$$

is absolutely convergent. As

$$\sum _{n=1}^{+\infty }\parallel y_n{\alpha }_n{a}_n{\parallel }_M\le \epsilon +\sum _{n=1}^{+\infty }\left|y_n{\alpha }_n{a}_n\left(t_n\right)\right|\le \epsilon +{\parallel y\parallel }_{\infty }\sum _{n=1}^{+\infty }\left|f_n\left(a_n\right)\right|,$$

the series

$$\sum _{n=1}^{+\infty }\left|{\alpha }_n\right|{\parallel y_n{a}_n\parallel }_M$$

is convergent. Hence, ${\displaystyle (\parallel y_n{a}_n{{\parallel }_M)}_n\in {\mathsf{\Lambda }}^{*}}$ because ${\displaystyle \alpha }$ was arbitrary in ${\displaystyle \mathsf{\Lambda }}$.

Now, if ${\displaystyle {\left({\alpha }_n\right)}_n\in S^{\circ }\subset \mathsf{\Lambda }}$, by Remark 2, we have:

$$\sum _{n=1}^{+\infty }\left|y_n{\alpha }_n{a}_n\left(t_n\right)\right|\le {\parallel y\parallel }_{\infty }\sum _{n=1}^{+\infty }|f_n\left(a_n\right){|\le \parallel y\parallel }_{\infty }{\epsilon }_{S,M}^{\phi }\left({\left(f_n\right)}_n\right).$$

But

$$\begin{array}{cc}\hfill {\epsilon }_{S,M}^{\phi }\left({\left(f_n\right)}_n\right)& =sup\left\{\sum _{n=1}^{+\infty }\left|z_n{\beta }_n{f}_n\left(a\right)\right|:{\left({\beta }_n\right)}_n\in S,a\in M,z\in {\tilde{B}}_{{\phi }^{*}}\right\}\hfill \\ \hfill & \le sup\left\{{\parallel a\parallel }_M{t}_{{\phi }^{*}}\sum _{n=1}^{+\infty }\left|{\beta }_n{\alpha }_n\right|:{\left({\beta }_n\right)}_n\in S,a\in M\right\}\hfill \\ \hfill & \le t_{{\phi }^{*}},\hfill \end{array}$$

where ${\displaystyle t_{{\phi }^{*}}:=sup\left\{t\in [0;+\infty ),{\phi }^{*}\left(t\right)\le 1\right\}}$. Consequently,

$$\sum _{n=1}^{+\infty }\left|y_n{\alpha }_n{a}_n\left(t_n\right)\right|\le {\parallel y\parallel }_{\infty }{t}_{{\phi }^{*}},$$

whereby

$$\sum _{n=1}^{+\infty }\parallel y_n{\alpha }_n{a}_n{\parallel }_M\le {\parallel y\parallel }_{\infty }{t}_{{\phi }^{*}}+\epsilon .$$

This means that

$$(\parallel y_n{a}_n{\parallel }_M{)}_n\in {(\parallel y\parallel }_{\infty }{t}_{{\phi }^{*}}+\epsilon )S^{\circ \circ }={(\parallel y\parallel }_{\infty }{t}_{{\phi }^{*}}+\epsilon )S;$$

hence ${\displaystyle (\parallel y_n{a}_n{{\parallel }_M)}_n\in {\mathsf{\Lambda }}_S^{*}}$. ${\displaystyle \square }$

Proposition 4.

For every ${\displaystyle S\in \mathcal{S}}$, ${\displaystyle M\in \mathcal{M}}$ and ${\displaystyle a={\left(a_n\right)}_n\in {\left({\mathsf{\Lambda }}_S^{*}\right)}_{{\phi }^{*}}⟨E_M^{\prime }⟩}$, the mapping

$$f_a:x⟼\sum _{n=1}^{+\infty }{a}_n\left(x_n\right)$$

defines a continuous linear functional on ${\displaystyle {\mathsf{\Lambda }}_{\phi }\left[E\right]}$.

Proof. 

Fix an arbitrary ${\displaystyle S\in \mathcal{S}}$, ${\displaystyle M\in \mathcal{M}}$, and ${\displaystyle a={\left(a_n\right)}_n\in {\left({\mathsf{\Lambda }}_S^{*}\right)}_{{\phi }^{*}}⟨E_M^{\prime }⟩}$, and for every ${\displaystyle t\in E}$, denote by ${\displaystyle \widehat{t}}$ the continuous linear map on ${\displaystyle E_M^{\prime }}$ defined by ${\displaystyle \widehat{t}\left(f\right):=f\left(t\right)}$. Next, for ${\displaystyle x={\left(x_n\right)}_n\in {\mathsf{\Lambda }}_{\phi }\left[E\right]}$, ${\displaystyle u\in E_M^{\prime }\subset E^{\prime }}$, and ${\displaystyle y\in {\tilde{B}}_{\phi }}$, we have

$${\left(y_n{\widehat{x}}_n\left(u\right)\right)}_n={\left(y_{n}u\left(x_n\right)\right)}_n\in \mathsf{\Lambda }\subset {\left({\mathsf{\Lambda }}_S^{*}\right)}^{*}.$$

So using Proposition 1, we get

$${\left({\widehat{x}}_n\right)}_n\in {\left({\mathsf{\Lambda }}_S^{*}\right)}_{\phi }^{*}\left[{\left(E_M^{\prime }\right)}_{\beta }^{\prime }\right].$$

Consequently,

$$\sum _{n=1}^{+\infty }{a}_n\left(x_n\right)=\sum _{n=1}^{+\infty }{\widehat{x}}_n\left(a_n\right)$$

is convergent, and therefore, ${\displaystyle f_a}$ is well-defined.

Further, observe also that the mapping ${\displaystyle {\psi }_a:{\left({\mathsf{\Lambda }}_S^{*}\right)}_{\phi }^{*}\left[{\left(E_M^{\prime }\right)}^{\prime }\right]⟶{\ell }^1}$, given by

$${\left(f_n\right)}_n⟼{\psi }_a\left({\left(f_n\right)}_n\right)={\left(f_n\left(a_n\right)\right)}_n,$$

is well-defined.

In fact, let ${\displaystyle {\left(f_n\right)}_n\in {\left({\mathsf{\Lambda }}_S^{*}\right)}_{\phi }^{*}\left[{\left(E_M^{\prime }\right)}^{\prime }\right]}$ be given. Since ${\displaystyle a={\left(a_n\right)}_n\in {\left({\mathsf{\Lambda }}_S^{*}\right)}_{{\phi }^{*}}⟨E_M^{\prime }⟩}$, the series

$$\sum _{n=1}^{+\infty }{f}_n\left(a_n\right)$$

is absolutely convergent; hence, ${\displaystyle {\left(f_n\left(a_n\right)\right)}_n\in {\ell }^1}$.

Since ${\displaystyle {\left({\mathsf{\Lambda }}_S^{*}\right)}^{*}}$ is perfect and ${\displaystyle {\left(E_M^{\prime }\right)}^{\prime }}$ is a Banach space, ${\displaystyle ({\left({\mathsf{\Lambda }}_S^{*}\right)}_{\phi }^{*}\left[{\left(E_M^{\prime }\right)}^{\prime }\right],{\epsilon }_{S,M}^{\phi })}$ is also a Banach space. Further, assume that ${\displaystyle {\left({\left(f_n\right)}_n^i\right)}_i}$ is a null sequence in ${\displaystyle {\left({\mathsf{\Lambda }}_S^{*}\right)}_{\phi }^{*}\left[{\left(E_M^{\prime }\right)}^{\prime }\right]}$ such that ${\displaystyle {\left({\psi }_a\left({\left(f_n\right)}_n^i\right)\right)}_i}$ converges in ${\displaystyle {\ell }^1}$ to ${\displaystyle {\left({\alpha }_n\right)}_n}$. As the projections ${\displaystyle {\left(f_n\right)}_n\mapsto f_n}$ are continuous, ${\displaystyle {\left(f_n^i\right)}_i}$ converges in ${\displaystyle {\left(E_M^{\prime }\right)}^{\prime }}$ to 0 for all ${\displaystyle n\in \mathbb{N}}$. Hence, the sequence ${\displaystyle {\left({\psi }_a\left({\left(f_n^i\right)}_n\right)\right)}_i={\left({\left(f_n^i\left(a_n\right)\right)}_n\right)}_i}$ converges to 0, whereby ${\displaystyle {\alpha }_n=0}$ for every ${\displaystyle n}$. By the closed graph theorem, ${\displaystyle {\phi }_a}$ is continuous. Therefore, there is ${\displaystyle K>0}$ such that for every ${\displaystyle {\left(f_n\right)}_n\in {\left({\mathsf{\Lambda }}_S^{*}\right)}_{\phi }^{*}\left[{\left(E_M^{\prime }\right)}^{\prime }\right]}$, we have the inequality

$${∥{\psi }_a\left({\left(f_n\right)}_n\right)∥}_1\le K{\epsilon }_{S,M}^{\phi }\left({\left(f_n\right)}_n\right),$$

which means that

$$\sum _{n=1}^{+\infty }\left|f_n\left(a_n\right)\right|\le K{\epsilon }_{S,M}^{\phi }\left({\left(f_n\right)}_n\right).$$

But ${\displaystyle {\left({\widehat{x}}_n\right)}_n\in {\left({\mathsf{\Lambda }}_S^{*}\right)}_{\phi }^{*}\left[{\left(E_M^{\prime }\right)}^{\prime }\right]}$; hence,

$$|f_a\left(x\right)|=\left|\sum _{n=1}^{+\infty }{\widehat{x}}_n\left(a_n\right)\right|\le K{\epsilon }_{S,M}^{\phi }\left({\left({\widehat{x}}_n\right)}_n\right)\le K{\epsilon }_{S,M}^{\phi }\left(x\right).$$

Consequently, ${\displaystyle f_a}$ is continuous. ${\displaystyle \square }$

Theorem 3.

The following equality is valid:

$${\left({\mathsf{\Lambda }}_{\phi }\left[E\right]\right)}_r{)}^{\prime }=\bigcup _{S\in \mathcal{S},M\in \mathcal{M}}{\left({\mathsf{\Lambda }}_S^{*}\right)}_{{\phi }^{*}}⟨E_M^{\prime }⟩.$$

Proof. 

By Proposition 4, for every ${\displaystyle S\in \mathcal{S}}$, ${\displaystyle M\in \mathcal{M}}$, and ${\displaystyle a:={\left(a_n\right)}_n\in {\left({\mathsf{\Lambda }}_S^{*}\right)}_{{\phi }^{*}}⟨E_M^{\prime }⟩}$, we have ${\displaystyle f_a\in {\left({\mathsf{\Lambda }}_{\phi }{\left[E\right]}_r\right)}^{\prime }}$. Therefore, the function

$$\varphi :\bigcup _{S\in \mathcal{S},M\in \mathcal{M}}{\left({\mathsf{\Lambda }}_S^{*}\right)}_{{\phi }^{*}}⟨E_M^{\prime }⟩⟶{\left({\mathsf{\Lambda }}_{\phi }{\left[E\right]}_r\right)}^{\prime }$$

given by

$$a⟼f_a,$$

is well-defined and linear. Clearly, ${\displaystyle \varphi }$ is injective.

Moreover, observe that if ${\displaystyle F\in {\left({\mathsf{\Lambda }}_{\phi }{\left[E\right]}_r\right)}^{\prime }}$, then Corollary 1 implies that there exist ${\displaystyle S\in \mathcal{S}}$, ${\displaystyle M\in \mathcal{M}}$ such that the sequence ${\displaystyle a:={\left(F_n\right)}_n}$ belongs to ${\displaystyle {\left({\mathsf{\Lambda }}^{*}\right)}_{{\phi }^{*}}⟨E_M^{\prime }⟩}$. Next, for each ${\displaystyle x\in {\mathsf{\Lambda }}_{\phi }{\left[E\right]}_r}$, by the continuity of ${\displaystyle F}$, we have

$$\begin{array}{cc}\hfill F\left(x\right)& =\underset{k}{lim}F\left(x^{\left(k\right)}\right)=\underset{k}{lim}\sum _{n=1}^{k}F\left(x_n{e}_n\right)\hfill \\ \hfill & =\sum _{n=1}^{+\infty }{F}_n\left(x_n\right)=f_a\left(x\right).\hfill \end{array}$$

This means that ${\displaystyle \varphi }$ is also surjective. Consequently ${\displaystyle \varphi }$ is an isomorphism. ${\displaystyle \square }$

In the following, we describe a fundamental base of equicontinuous subsets of ${\displaystyle {\left({\mathsf{\Lambda }}_{\phi }{\left[E\right]}_r\right)}^{\prime }}$. In order to establish it, let us denote for ${\displaystyle S\in \mathcal{S}}$ and ${\displaystyle M\in \mathcal{M}}$:

$$K_{S,M}^{\phi }=\left\{{\left(f_n\right)}_n\in {\mathsf{\Lambda }}_{\phi }\left[{\left(E_M^{\prime }\right)}^{\prime }\right]:{\left(y_n{f}_n\left(a\right)\right)}_n\in S^{\circ },a\in M,y\in {\tilde{B}}_{{\phi }^{*}}\right\}.$$

Theorem 4.

The family of sets of the form

$$S_{\phi }⟨M⟩=\left\{{\left(a_n\right)}_n\in {\left({\mathsf{\Lambda }}_S^{*}\right)}_{{\phi }^{*}}⟨E_M^{\prime }⟩:\sum _{n=1}^{+\infty }\left|f_n\left(a_n\right)\right|\le 1,{\left(f_n\right)}_n\in K_{S,M}^{\phi }\right\},$$

with ${\displaystyle S}$ running over ${\displaystyle \mathcal{S}}$ and ${\displaystyle M}$ over ${\displaystyle \mathcal{M}}$ yields a fundamental system of equicontinuous subsets of ${\displaystyle {\left({\mathsf{\Lambda }}_{\phi }{\left[E\right]}_r\right)}^{\prime }}$.

Proof. 

Let us first show that ${\displaystyle S_{\phi }⟨M⟩}$ is equicontinuous. If ${\displaystyle x={\left(x_n\right)}_n\in {\mathsf{\Lambda }}_{\phi }\left[E\right]}$ is such that ${\displaystyle {\epsilon }_{S,M}^{\phi }\left(x\right)\le 1}$, then, as in the proof of Proposition 4, one has

$$\sum _{n=1}^{+\infty }|y_n{\alpha }_n{\widehat{x}}_n\left(u\right)|=\sum _{n=1}^{+\infty }\left|y_n{\alpha }_{n}u\left(x_n\right)\right|\le {\epsilon }_{S,M}^{\phi }\left(x\right)\le 1$$

for all ${\displaystyle y\in {\tilde{B}}_{{\phi }^{*}}}$, ${\displaystyle u\in M}$ and ${\displaystyle \alpha \in S}$. Hence,

$${\left(y_n{\widehat{x}}_n\left(u\right)\right)}_n\in S^{\circ }.$$

Therefore, ${\displaystyle {\left({\widehat{x}}_n\right)}_n\in K_{S,M}^{\phi }}$. Moreover, if ${\displaystyle a={\left(a_n\right)}_n\in S_{\phi }⟨M⟩}$, then

$$\left|\sum _{n=1}^{+\infty }{\widehat{x}}_n\left(a_n\right)\right|=\left|\sum _{n=1}^{+\infty }{a}_n\left(x_n\right)\right|\le 1.$$

Consequently, ${\displaystyle S_{\phi }⟨M⟩}$ is equicontinuous.

Now, if ${\displaystyle \mathbb{H}\subset {\left({\mathsf{\Lambda }}_{\phi }{\left[E\right]}_r\right)}^{\prime }}$ is equicontinuous, then there are ${\displaystyle S\in \mathcal{S}}$ and ${\displaystyle M\in \mathcal{M}}$ such that:

$$\left|\sum _{n=1}^{+\infty }{a}_n\left(x_n\right)\right|\le {\epsilon }_{S,M}^{\phi }\left(x\right)$$

for all ${\displaystyle x={\left(x_n\right)}_n\in {\mathsf{\Lambda }}_{\phi }{\left[E\right]}_r}$ and ${\displaystyle a={\left(a_n\right)}_n\in \mathbb{H}}$. Let ${\displaystyle f={\left(f_n\right)}_n\in K_{S,M}^{\phi }}$. Then ${\displaystyle {\epsilon }_{S,M}^{\phi }\left(f\right)\le 1}$, and by Remark 2, we have:

$$\sum _{n=1}^{+\infty }\left|f_n\left(a_n\right)\right|\le {\epsilon }_{S,M}^{\phi }\left(f\right)\le 1.$$

Consequently, ${\displaystyle \mathbb{H}\subset S_{\phi }⟨M⟩}$. ${\displaystyle \square }$

Let us consider the collections:

$$\begin{array}{cc}\hfill {\mathcal{B}}^{\prime }& :=\{B^{\prime }\subset E^{\prime }:B^{\prime }\mathrm{is}\mathrm{a}\mathrm{closed}\mathrm{weak}*\text{-}\mathrm{bounded}\mathrm{disc}\},\hfill \\ \hfill \mathcal{R}& :=\{R\subset \mathsf{\Lambda }:R\mathrm{is}\mathrm{a}\mathrm{closed}\mathrm{bounded}\mathrm{and}\mathrm{normal}\mathrm{disc}\},\hfill \\ \hfill {\mathcal{R}}^{\prime }& :=\{R^{\prime }\subset {\mathsf{\Lambda }}^{*}:R^{\prime }\mathrm{is}\mathrm{a}\mathrm{closed}\mathrm{weak}*\text{-}\mathrm{bounded}\mathrm{and}\mathrm{normal}\mathrm{disc}\},\hfill \end{array}$$

and for every ${\displaystyle R^{\prime }\in {\mathcal{R}}^{\prime }}$ and ${\displaystyle B^{\prime }\in {\mathcal{B}}^{\prime }}$, the sets:

$$\begin{array}{cc}\hfill K_{R^{\prime },B^{\prime }}& :=\left\{{\left(f_n\right)}_n\in {\mathsf{\Lambda }}_{\phi }\left[{\left(E_{B^{\prime }}^{\prime }\right)}^{\prime }\right]:{\left(y_n{f}_n\left(a\right)\right)}_n\in {\left(R^{\prime }\right)}^{\circ },a\in B^{\prime },y\in {\tilde{B}}_{{\phi }^{*}}\right\},\hfill \\ \hfill R_{\phi }^{\prime }⟨B^{\prime }⟩& :=\left\{{\left(a_n\right)}_n\in {\left({\mathsf{\Lambda }}_{\phi }{\left[E\right]}_r\right)}^{\prime }:\sum _{n=1}^{+\infty }\left|f_n\left(a_n\right)\right|\le 1,{\left(f_n\right)}_n\in K_{R^{\prime },B^{\prime }}\right\}.\hfill \end{array}$$

The following theorem gives a necessary and sufficient condition for the space ${\displaystyle {\mathsf{\Lambda }}_{\phi }{\left[E\right]}_r}$ to be barrelled or quasi-barrelled.

Theorem 5.

Assume that ${\displaystyle \mathsf{\Lambda }}$ is barrelled (quasi-barrelled). Then ${\displaystyle {\mathsf{\Lambda }}_{\phi }{\left[E\right]}_r}$ is barrelled (resp. quasi-barrelled) if and only if the following two conditions are satisfied:

 (i)

${\displaystyle E}$ is barrelled (resp. quasi-barrelled).

 (ii)

For each weak* bounded (resp. strongly bounded) subset ${\displaystyle \mathbb{B}}$ of ${\displaystyle {\left({\mathsf{\Lambda }}_{\phi }{\left[E\right]}_r\right)}^{\prime }}$, there exist ${\displaystyle B^{\prime }\in {\mathcal{B}}^{\prime }}$ and ${\displaystyle R^{\prime }\in {\mathcal{R}}^{\prime }}$ such that ${\displaystyle \mathbb{B}\subset R_{\phi }^{\prime }⟨B^{\prime }⟩}$.

Proof. 

Let ${\displaystyle T}$ be a barrel (resp. bornivorous barrel) in ${\displaystyle {\mathsf{\Lambda }}_{\phi }{\left[E\right]}_r}$. Then ${\displaystyle T^{\circ }}$ is a weakly bounded (resp. strongly bounded) subset of ${\displaystyle {\left({\mathsf{\Lambda }}_{\phi }{\left[E\right]}_r\right)}^{\prime }}$. By ${\displaystyle \left(ii\right)}$, there exists ${\displaystyle R^{\prime }\in {\mathcal{R}}^{\prime }}$ and ${\displaystyle B^{\prime }\in {\mathcal{B}}^{\prime }}$ such that ${\displaystyle T^{\circ }\subset R_{\phi }^{\prime }⟨B^{\prime }⟩}$. Since ${\displaystyle E}$ is barrelled (resp. quasi-barrelled), ${\displaystyle B^{\prime }}$ is equicontinuous. Hence, it is contained in some ${\displaystyle M\in \mathcal{M}}$.

Similarly, since ${\displaystyle \mathsf{\Lambda }}$ is barrelled (resp. quasi-barrelled), there exists ${\displaystyle S\in \mathcal{S}}$ such that ${\displaystyle R^{\prime }\subset S}$. Hence, ${\displaystyle T^{\circ }\subset R_{\phi }^{\prime }⟨B^{\prime }⟩\subset S_{\phi }⟨M⟩}$. Therefore, ${\displaystyle T^{\circ }}$ is equicontinuous and consequently ${\displaystyle T}$ is a neighborhood of ${\displaystyle 0}$ in ${\displaystyle {\mathsf{\Lambda }}_{\phi }{\left[E\right]}_r}$.

Now, assume that ${\displaystyle {\mathsf{\Lambda }}_{\phi }{\left[E\right]}_r}$ is barrelled. By Lemma 3, ${\displaystyle E}$ is complemented in ${\displaystyle {\mathsf{\Lambda }}_{\phi }{\left[E\right]}_r}$. Therefore, ${\displaystyle E}$ is a barrelled (resp. quasi-barrelled) space, whereby ${\displaystyle \left(i\right)}$ is satisfied. Moreover, Let ${\displaystyle \mathbb{B}}$ be a weakly bounded (resp. strongly bounded) subset of ${\displaystyle {\left({\mathsf{\Lambda }}_{\phi }{\left[E\right]}_r\right)}^{\prime }}$. Then ${\displaystyle \mathbb{B}}$ is an equicontinuous subset of ${\displaystyle {\left({\mathsf{\Lambda }}_{\phi }{\left[E\right]}_r\right)}^{\prime }}$. By Theorem 4, there exist ${\displaystyle S\in \mathcal{S}}$ and ${\displaystyle M\in \mathcal{M}}$ such that ${\displaystyle \mathbb{B}\subset S_{\phi }⟨M⟩}$. Hence, ${\displaystyle \left(ii\right)}$ is satisfied, too. ${\displaystyle \square }$

Example 3.

 1.

If ${\displaystyle \phi }$ is the identity of ${\displaystyle {\mathbb{R}}_{+}}$, the continuous dual of ${\displaystyle {\mathsf{\Lambda }}_{\phi }{\left[E\right]}_r}$ is as given in [4].

 2.

In case ${\displaystyle \mathsf{\Lambda }={\ell }^1}$ and ${\displaystyle E=\mathbb{K}}$, the continuous dual of ${\displaystyle {\mathsf{\Lambda }}_{\phi }{\left[E\right]}_r}$ is ${\displaystyle {\ell }_{{\phi }^{*}}}$.

 3.

When ${\displaystyle \phi }$ is the Orlicz function in (3) of Example 1, the continuous dual of ${\displaystyle {\left(c_0\right)}_{\phi }{\left[E\right]}_r:=c_b{\left(E\right)}_r}$ is ${\displaystyle \bigcup _{M\in \mathcal{M}}{\ell }^1⟨E_M^{\prime }⟩}$.

In order to give further examples as applications of our results, we determine the duals of some concrete sequence spaces and characterize the barrelledness therein. For this, let ${\displaystyle p\ge 1}$ be a real number and ${\displaystyle q}$ its conjugate (i.e., ${\displaystyle \frac{1}{p}+\frac{1}{q}=1}$ if ${\displaystyle p\ne 1}$, and ${\displaystyle q=+\infty }$ if ${\displaystyle p=1}$) and let ${\displaystyle (E,\parallel .{\parallel }_E)}$ be a normed space. Then the topology of ${\displaystyle {\ell }_{{\phi }^{*}}^q⟨E^{\prime }⟩}$ is defined by the single norm ${\displaystyle {\sigma }_{R^{\prime },B^{\prime }}^{{\phi }^{*}}}$; it is also denoted by ${\displaystyle {\sigma }_{q,E^{\prime }}^{{\phi }^{*}}}$. Here, ${\displaystyle R^{\prime }}$ and ${\displaystyle B^{\prime }}$ are the closed unit bulls of ${\displaystyle {\ell }^q}$ and ${\displaystyle E^{\prime }}$, respectively.

We have the following proposition:

Proposition 5.

The topological dual of ${\displaystyle {\ell }_{\phi }^p{\left[E\right]}_r}$ is ${\displaystyle {\ell }_{{\phi }^{*}}^q⟨E^{\prime }⟩}$. Moreover, ${\displaystyle {\ell }_{\phi }^p{\left[E\right]}_r}$ is barrelled if and only if ${\displaystyle E}$ is barrelled.

Proof. 

The first assertion results immediately from Theorem 3.

For the second one, notice that since ${\displaystyle {\ell }^p}$ is a Banach space, it is barrelled. As ${\displaystyle {\ell }_{{\phi }^{*}}^q⟨E^{\prime }⟩}$ is a Banach space, it is sufficient to show that if ${\displaystyle E}$ is barrelled, then the unit ball ${\displaystyle \mathbb{B}}$ of ${\displaystyle {\ell }_{{\phi }^{*}}^q⟨E^{\prime }⟩}$ is contained in ${\displaystyle R_{\phi }^{\prime }⟨B^{\prime }⟩}$, where ${\displaystyle R^{\prime }}$ and ${\displaystyle B^{\prime }}$ are the unit balls of ${\displaystyle {\ell }^q}$ and ${\displaystyle E^{\prime }}$, respectively.

So choose an arbitrary ${\displaystyle {\left(a_n\right)}_n\in \mathbb{B}}$ and ${\displaystyle {\left(f_n\right)}_n\in K_{R^{\prime },B^{\prime }}}$. Then ${\displaystyle \left(y_n{f}_n\left(b\right)\right)\in {\left(R^{\prime }\right)}^{\circ }}$ for every ${\displaystyle b\in B^{\prime }}$ and every ${\displaystyle y\in {\tilde{B}}_{{\phi }^{*}}}$, whereby

$$\underset{\alpha \in R^{\prime },b\in B^{\prime }}{sup}\underset{\delta (y,{\phi }^{*})⩽1}{sup}\sum _{n\ge 1}\left|y_n{\alpha }_n{f}_n\left(b\right)\right|\le 1.$$

This shows that

$${\epsilon }_{p,E^{\prime \prime }}^{\phi }\left(f\right):={\epsilon }_{R^{\prime },B^{\prime }}^{\phi }\left(f\right)\le 1.$$

Hence,

$$\sum _{n\ge 1}\left|f_n\left(a_n\right)\right|\le {\sigma }_{q,E^{\prime }}^{{\phi }^{*}}\left({\left(a_n\right)}_n\right)\le 1,$$

and consequently, ${\displaystyle {\left(a_n\right)}_n\in R_{\phi }^{\prime }⟨B^{\prime }⟩}$. ${\displaystyle \square }$

In the special case where ${\displaystyle \phi }$ is the identity ${\displaystyle x\mapsto x}$, the space ${\displaystyle {\ell }_{\phi }^p{\left[E\right]}_r}$ is nothing but the space ${\displaystyle {\ell }^p\left[E\right]}$ introduced by H. Apiola [13]. We then obtain a characterization of barrelledness in such spaces.

Corollary 2.

${\displaystyle {\ell }^p{\left[E\right]}_r}$ is barrelled if and only if ${\displaystyle E}$ is barrelled.

5. Conclusions and Future Work

We introduce the notions of weakly (resp. strongly) ${\displaystyle (\phi ,\mathsf{\Lambda })}$-summable sequences in a locally convex space ${\displaystyle E}$ and investigate topological properties of the linear space ${\displaystyle {\mathsf{\Lambda }}_{\phi }\left[E\right]}$ of all such sequences endowed with the topology induced by an appropriate family of semi-norms. We obtain that ${\displaystyle E}$ is embedded as a complemented subspace in ${\displaystyle {\mathsf{\Lambda }}_{\phi }\left[E\right]}$. Whenever ${\displaystyle {\mathsf{\Lambda }}_{\phi }\left[E\right]}$ has the property ${\displaystyle AK}$, we characterize its continuous dual in terms of strongly ${\displaystyle (\phi ,\mathsf{\Lambda })}$-summable sequences in ${\displaystyle E^{\prime }}$, which is the continuous dual of ${\displaystyle E}$. We further provide necessary and sufficient conditions under which ${\displaystyle {\mathsf{\Lambda }}_{\phi }\left[E\right]}$ is barrelled or quasi-barrelled. To illustrate the proposed results, we have included as applications concrete examples of such spaces (see Proposition 5 and Corollary 2). The outcomes of our paper extend and improve known results: in particular, of [8]. Our work paves the way for further investigations of these sequence spaces: namely, for studying reflexivity and distinguishedness.