Some Realisation of the Banach Space of All Continuous Linear Functionals on ℓ₁ Approximated by Weakly Symmetric Continuous Linear Functionals

Abstract

A general notion of a weakly symmetric continuous linear functional on a Banach space, in the case where the space is ${\ell }_1$ (i.e., the space of all absolutely summable sequences of complex numbers), reduces to a continuous linear functional whose Riesz representation is a periodic sequence. We consider the completion of the space of all such linear continuous functionals on ${\ell }_1$ with periods of Riesz representations equal to powers of 2. It is known that this completion is a Banach space with a Schauder basis. In this work, we construct a sequence Banach space with the standard Schauder basis $\{e_m=(\underset{m-1}{\underbrace{0,\dots ,0}},1,0,\dots ){\}}_{m=1}^{\infty }$ that is isometrically isomorphic to this completion. Results of the work can be used to describe spectra of topological algebras of analytic functions on ${\ell }_1$ that can be approximated by weakly symmetric functions.

1. Introduction

The principal object of this study is a Banach space Y defined below. As it is shown in [1], the space Y is the closure of the vector space of all periodic sequences with periods equal to powers of 2 in the Banach space ${\ell }_{\infty }$ of all bounded complex sequences. This vector space is the space of Riesz representations of the so-called weakly symmetric (defined below) linear continuous functionals on the Banach space ${\ell }_1$ of all absolutely summable complex sequences. Consequently, the space Y can be considered as the Banach space of all continuous linear functionals on ${\ell }_1$ approximated by weakly symmetric continuous linear functionals.

The notion of symmetry of a function on a Banach space in the general setting was defined in [2] (see also [3,4,5]). The concept of weak symmetry generalizes this notion (see also [6,7,8] for other generalizations). A function f on a Banach space X is called weakly symmetric with respect to some fixed sequence of groups $G_1,G_2,\dots$ of operators on X such that $G_1\supset G_2\supset \dots$ if there exists $n\in \mathbb{N}$ such that f is symmetric with respect to $G_n,$ i.e., $f\left(g\right(x\left)\right)=f\left(x\right)$ for every $x\in X$ and $g\in G_n.$ The choice of the sequence of groups depends naturally on the structure of $X.$ For example, in classical works [9,10] the symmetry group associated with a sequence Banach space with a symmetric Schauder basis, e.g. ${\ell }_1,$ (see ([11], Definition 3.a.1, p. 113) for the definition of a symmetric basis) is defined as the group of operators of the form $(x_1,x_2,\dots )\mapsto \left(x_{\sigma \left(1\right)},x_{\sigma \left(2\right)},\dots \right),$ where $\sigma :\mathbb{N}\to \mathbb{N}$ is a bijection, whereas the symmetry group associated with a rearrangement-invariant space of functions over $[0,1]$ is defined as the group of operators of the form $x\mapsto x\circ \tau ,$ where $\tau$ is a measurable automorphism of $[0,1].$ These groups typically serve as the group $G_1$ within the framework of the previously mentioned sequence of groups associated with the respective space. Each of the groups $G_2,G_3,\dots$ is formed as some natural weakening of the preceding group. These groups are the so-called block-symmetry groups [12,13,14]. For example, in the case $X={\ell }_1,$ the group $G_2$ consists of elements of $G_1$ generated by bijections $\sigma :\mathbb{N}\to \mathbb{N}$ that permute elements of $\mathbb{N}$ by blocks of the length two. Strictly speaking, for every $k\in \mathbb{N}$ there exists $l\in \mathbb{N}$ such that $\sigma (2k-1)=2l-1$ and $\sigma \left(2k\right)=2l.$ Block-symmetry groups on certain Banach spaces are connected to classical symmetry groups on Cartesian products of these spaces. For example, as it is shown in [15], block-symmetry groups on Banach spaces of p-absolutely summable sequences can be constructed via symmetry groups on their Cartesian powers. Similar results for the case of block-symmetry groups on spaces of Lebesgue measurable functions were established in [16]. The obtained results enabled to describe algebraic bases of the algebras of continuous block-symmetric polynomials defined on these spaces in [17].

On the one hand, in contrast to symmetric and block-symmetric functions on Banach spaces, as shown in [1], weakly symmetric functions can approximate functions lacking any form of symmetry. On the other hand, spaces of weakly symmetric functions retain many of the advantages shared by spaces of symmetric functions on Banach spaces and on their Cartesian powers. For example, algebras of symmetric continuous polynomials on Cartesian products of Banach spaces typically admit finite [18] or countable [19] algebraic bases, which makes the corresponding algebras of symmetric analytic functions finitely or countably generated. Similar results for algebras of weakly symmetric continuous polynomials on some Banach spaces of Lebesgue measurable functions were established in [20].

The above-mentioned benefits of weakly symmetric functions on Banach spaces will, in future work, allow the techniques developed for symmetric functions to be applied to classes of functions that are not symmetric in general. One of the problems where such an approach may be used is the problem of describing the spectra (sets of all nontrivial continuous linear multiplicative functionals) of algebras of analytic functions on Banach spaces. The problem of explicitly describing the spectrum of the Fréchet algebra of entire functions of bounded type (bounded on bounded sets) on a Banach space has remained unsolved in the general case for many years [21,22,23,24]. However, countably generated subalgebras of this algebra admit an explicit description of their spectrum [25,26,27,28]. Among such subalgebras are the algebras of entire symmetric functions. One example is the Fréchet algebra of all entire symmetric functions of bounded type on the space ${\ell }_1$, for which the description of the spectrum was initiated in [29], continued in [30,31,32,33,34], and completed in [35].

Among the problems discussed above is the problem of the description of the spectrum of the algebra of entire functions of bounded type on the space ${\ell }_1,$ generated by continuous linear functionals that can be approximated by continuous weakly symmetric functionals. The solution to this problem relies heavily on knowledge of the dual space to $Y.$ Although a Schauder basis for the space Y was constructed in [1], this basis does not have the standard form, which complicates the identification of the dual space to Y via the basis. In this work, we construct a sequence Banach space ${\mathrm{cs}}_2$ with the standard Schauder basis $\{e_m=(\underset{m-1}{\underbrace{0,\dots ,0}},1,0,\dots ){\}}_{m=1}^{\infty }$ that is isometrically isomorphic to the space $Y.$ In future work, results of this work will enable the construction of the dual space to $Y,$ thereby providing a framework for describing the spectrum of the above-mentioned algebra.

2. Definitions and Preliminaries

Let us denote by $\mathbb{N}$ and ${\mathbb{Z}}_{+}$ the sets of all positive and all non-negative integers resp.

Let Y be the space of all bounded sequences of complex numbers $y=(y_1,y_2,\dots )$ that have the following property:

$$\mathrm{for} \mathrm{every} \epsilon >0 \mathrm{there} \mathrm{exists} n\in \mathbb{N}\mathrm{such} \mathrm{that} |y_j-y_{j+k{2}^n}|<\epsilon \mathrm{for} \mathrm{every}j\in \{1,2,3,\dots ,2^n\} \mathrm{and} k\in {\mathbb{Z}}_{+}.$$

(1)

We endow Y with the norm

$${\parallel y\parallel }_{\infty }=\underset{j\in \mathbb{N}}{sup}\left|y_j\right|.$$

Illustrative elements of the space Y are presented in ([1], Example 1). By ([1], Theorem 3), Y is a Banach space. The space Y is the space of all linear continuous functionals on ${\ell }_1$ that can be approximated by weakly symmetric functionals.

For given $n\in {\mathbb{Z}}_{+}$ and $j\in \{1,2,3,\dots ,2^n\},$ let

$$A_{n,j}=\left\{j+k{2}^n:k\in {\mathbb{Z}}_{+}\right\}.$$

(2)

For $A\subset \mathbb{N},$ let $1_A$ be the sequence $(x_1,x_2,\dots )$ such that

$$x_j=\left\{\begin{array}{cc}1,\hfill & \mathrm{if}j\in A,\hfill \\ 0,\hfill & \mathrm{otherwise}\hfill \end{array}\right.$$

for $j\in \mathbb{N}.$ Let the mapping $\varkappa :\mathbb{N}\to {\mathbb{Z}}_{+}$ be defined by

$$\varkappa \left(j\right)=\lceil {log}_{2}j\rceil ,$$

(3)

where $j\in \mathbb{N}.$ Let

$$b_j=1_{A_{\varkappa \left(j\right),j}}$$

(4)

for $j\in \mathbb{N},$ where $\varkappa$ is defined by (3).

For $j\in \mathbb{N},$ let ${\alpha }_j:Y\to \mathbb{C}$ be defined by

$${\alpha }_j\left(y\right)=\left\{\begin{array}{cc}{y}_1,\hfill & \mathrm{if}j=1,\hfill \\ y_j-y_{j-2^{\varkappa \left(j\right)-1}},\hfill & \mathrm{if}j\ge 2,\hfill \end{array}\right.$$

(5)

where $y=(y_1,y_2,\dots )\in Y.$ Note that the mapping ${\alpha }_j$ is linear for every $j\in \mathbb{N}.$

By ([1], Theorem 4), for every $y\in Y,$ the series ${\sum }_{j=1}^{\infty }{\alpha }_j\left(y\right)b_j,$ where $b_j$ and ${\alpha }_j$ are defined by (4) and (5) respectively, converges to y with respect to the norm ${\parallel ·\parallel }_{\infty }$. Consequently, ${\left\{b_j\right\}}_{j=1}^{\infty }$ is a Schauder basis of the space $Y.$

3. Results

3.1. The Space ${\mathrm{cs}}_2$

Let $n\in \mathbb{N}.$ Consider the binary expansion of n without zero addends:

$$n=2^{j_1}+2^{j_2}+\dots +2^{j_s},$$

where $s\in \mathbb{N}$ and $j_1,j_2,\dots ,j_s\in {\mathbb{Z}}_{+}$ are such that $j_1<j_2<\dots <j_s.$ Set

$${\mathsf{\Omega }}_n=\left\{0\right\}\cup \left\{2^{j_1}+2^{j_2}+\dots +2^{j_l}:l\in \{1,\dots ,s\}\right\}.$$

(6)

Also we set

$${\mathsf{\Omega }}_0=\left\{0\right\}.$$

(7)

Note that

$$n\in {\mathsf{\Omega }}_n$$

(8)

and

$${\mathsf{\Omega }}_n\setminus \left\{n\right\}\subset \{0,1,\dots ,n-1\}.$$

(9)

Let $\mathcal{J}$ be the set of all strictly increasing sequences of elements of ${\mathbb{Z}}_{+}$. For $J=(j_1,j_2,\dots )\in \mathcal{J}$, let us define the mapping ${\nu }_J:{\mathbb{Z}}_{+}\to {\mathbb{Z}}_{+}$ by

$$\begin{array}{c}\hfill {\nu }_J\left(m\right)=\left\{\begin{array}{cc}0,\hfill & \mathrm{if}m=0,\hfill \\ 2^{j_1}+\dots +2^{j_m},\hfill & \mathrm{if}m\in \mathbb{N}.\hfill \end{array}\right.\end{array}$$

(10)

Let ${\mathrm{cs}}_2$ be the set of all complex sequences $x=(x_0,x_1,\dots )$ such that the following two conditions are satisfied:

$$\mathrm{for} \mathrm{every} J\in \mathcal{J}, \mathrm{the} \mathrm{series} \sum _{m=0}^{\infty }{x}_{{\nu }_J\left(m\right)} \mathrm{is} \mathrm{convergent},$$

(11)

and

$$\begin{array}{c}\mathrm{for} \mathrm{every} \epsilon >0 \mathrm{there} \mathrm{exists} K\in \mathbb{N} \mathrm{such} \mathrm{that} \left|\sum _{\begin{array}{c}m=0\\ {\nu }_J\left(m\right)\ge K_1\end{array}}^{\infty }{x}_{{\nu }_J\left(m\right)}\right|<\epsilon \hspace{1em}\hspace{1em}\hspace{1em}\hfill \\ \hfill \mathrm{for} \mathrm{every} J\in \mathcal{J} \mathrm{and} K_1>K,\end{array}$$

(12)

where ${\nu }_J$ is defined by (10). It can be verified that ${\mathrm{cs}}_2,$ equipped with coordinate-wise addition and scalar multiplication, is a vector space.

Let us endow ${\mathrm{cs}}_2$ with the norm defined as the supremum of the absolute values of partial sums of the series from the condition (11):

$${\parallel x\parallel }_{{\mathrm{cs}}_2}=\underset{n\in {\mathbb{Z}}_{+}}{sup}\left|\sum _{m\in {\mathsf{\Omega }}_n}{x}_m\right|,$$

(13)

where $x=(x_0,x_1,\dots )\in {\mathrm{cs}}_2$ and ${\mathsf{\Omega }}_n$ is defined by (6) and (7). Let us show that ${\mathrm{cs}}_2$ is a normed space.

Lemma 1.

Let $x=(x_0,x_1,\dots )\in {\mathrm{cs}}_2$ be such that ${\parallel x\parallel }_{{\mathrm{cs}}_2}=0.$ Then $x=0.$

Proof. 

Since ${\parallel x\parallel }_{{\mathrm{cs}}_2}=0,$ by (13),

$$\sum _{m\in {\mathsf{\Omega }}_n}{x}_m=0$$

(14)

for every $n\in {\mathbb{Z}}_{+}.$ Let us show that $x_n=0$ for every $n\in {\mathbb{Z}}_{+}.$ We proceed by induction on $n.$ Consider the case $n=0.$ By (7),

$$\sum _{m\in {\mathsf{\Omega }}_0}{x}_m=x_0.$$

Therefore, taking into account (14), $x_0=0.$ Suppose $n\in \mathbb{N}$ is such that $x_k=0$ for every $k\in \{0,\dots ,n-1\}.$ Let us show that $x_n=0.$ As a consequence of (8),

$${\mathsf{\Omega }}_n=\left\{n\right\}\cup \left({\mathsf{\Omega }}_n\setminus \left\{n\right\}\right).$$

Therefore

$$\sum _{m\in {\mathsf{\Omega }}_n}{x}_m=x_n+\sum _{m\in {\mathsf{\Omega }}_n\setminus \left\{n\right\}}{x}_m.$$

(15)

By (9), every $m\in {\mathsf{\Omega }}_n\setminus \left\{n\right\}$ belongs to $\{0,1,\dots ,n-1\}.$ Consequently, by the induction hypothesis, $x_m=0$ for every $m\in {\mathsf{\Omega }}_n\setminus \left\{n\right\}.$ Therefore

$$\sum _{m\in {\mathsf{\Omega }}_n\setminus \left\{n\right\}}{x}_m=0.$$

(16)

By (14)–(16), $x_n=0.$ So, $x=0.$ This completes the proof. □

Lemma 2.

For every $x=(x_0,x_1,\dots )\in {\mathrm{cs}}_2$ and $\lambda \in \mathbb{C},$

$${\parallel \lambda x\parallel }_{{\mathrm{cs}}_2}={\left|\lambda \right|\parallel x\parallel }_{{\mathrm{cs}}_2},$$

where $\lambda x=(\lambda x_0,\lambda x_1,\dots ).$

Proof. 

Note that

$${\parallel \lambda x\parallel }_{{\mathrm{cs}}_2}=\underset{n\in {\mathbb{Z}}_{+}}{sup}\left|\sum _{m\in {\mathsf{\Omega }}_n}\lambda x_m\right|=\left|\lambda \right|\underset{n\in {\mathbb{Z}}_{+}}{sup}\left|\sum _{m\in {\mathsf{\Omega }}_n}{x}_m\right|={\left|\lambda \right|\parallel x\parallel }_{{\mathrm{cs}}_2}.$$

This completes the proof. □

Lemma 3.

For every $x=(x_0,x_1,\dots ),y=(y_0,y_1,\dots )\in {\mathrm{cs}}_2,$

$${\parallel x+y\parallel }_{{\mathrm{cs}}_2}\le {\parallel x\parallel }_{{\mathrm{cs}}_2}+{\parallel y\parallel }_{{\mathrm{cs}}_2},$$

where $x+y=(x_0+y_0,x_1+y_1,\dots ).$

Proof. 

By the triangle inequality,

$$\begin{array}{cc}\hfill {\parallel x+y\parallel }_{{\mathrm{cs}}_2}& =\underset{n\in {\mathbb{Z}}_{+}}{sup}\left|\sum _{m\in {\mathsf{\Omega }}_n}(x_m+y_m)\right|\hfill \\ & =\underset{n\in {\mathbb{Z}}_{+}}{sup}\left|\sum _{m\in {\mathsf{\Omega }}_n}{x}_m+\sum _{m\in {\mathsf{\Omega }}_n}{y}_m\right|\hfill \\ & \le \underset{n\in {\mathbb{Z}}_{+}}{sup}\left(\left|\sum _{m\in {\mathsf{\Omega }}_n}{x}_m\right|+\left|\sum _{m\in {\mathsf{\Omega }}_n}{y}_m\right|\right)\hfill \\ & \le \underset{n\in {\mathbb{Z}}_{+}}{sup}\left|\sum _{m\in {\mathsf{\Omega }}_n}{x}_m\right|+\underset{n\in {\mathbb{Z}}_{+}}{sup}\left|\sum _{m\in {\mathsf{\Omega }}_n}{y}_m\right|\hfill \\ & ={\parallel x\parallel }_{{\mathrm{cs}}_2}+{\parallel y\parallel }_{{\mathrm{cs}}_2}.\hfill \end{array}$$

This completes the proof. □

Lemmas 1–3 imply the following result.

Proposition 1.

The functional ${\parallel ·\parallel }_{{\mathrm{cs}}_2}$ defined by (13) is a norm on ${\mathrm{cs}}_2.$ So, ${\mathrm{cs}}_2$ endowed with this norm is a normed space.

Consider some examples of sequences that belong to ${\mathrm{cs}}_2.$

Example 1.

Let us construct an element of ${\mathrm{cs}}_2$ which is not summable. Let $x=(x_0,x_1,\dots ),$ where

$$x_n=\left\{\begin{array}{cc}0,\hfill & if n=0,\hfill \\ {\displaystyle \frac{1}{2^{\lfloor {log}_{2}n\rfloor }}},\hfill & if n\in \mathbb{N},\hfill \end{array}\right.$$

(17)

that is,

$$x_0=0,x_1=1,x_2=x_3={\displaystyle \frac{1}{2}},x_4=x_5=\dots =x_7={\displaystyle \frac{1}{2^2}},x_8=x_9=\dots =x_{15}={\displaystyle \frac{1}{2^3}},\dots$$

Evidently, the sequence x is not summable. Let us show that $x\in {\mathrm{cs}}_2.$ First, let us prove that

$$x_{{\nu }_J\left(m\right)}=\left\{\begin{array}{cc}0,\hfill & if m=0,\hfill \\ {\displaystyle \frac{1}{2^{j_m}}},\hfill & if m\in \mathbb{N}\hfill \end{array}\right.$$

(18)

for every $J=(j_1,j_2,\dots )\in \mathcal{J}$. Let $J=(j_1,j_2,\dots )\in \mathcal{J}$ and $m\in \mathbb{N}$. By (10),

$$2^{j_m}\le {\nu }_J\left(m\right)<2^{j_m+1},$$

(19)

therefore,

$$j_m\le {log}_2{\nu }_J\left(m\right)<j_m+1$$

and, consequently,

$$\lfloor {log}_2{\nu }_J\left(m\right)\rfloor =j_m.$$

Therefore, by (17),

$$x_{{\nu }_J\left(m\right)}={\displaystyle \frac{1}{2^{\lfloor {log}_2{\nu }_J\left(m\right)\rfloor }}}={\displaystyle \frac{1}{2^{j_m}}}.$$

(20)

By (10), ${\nu }_J\left(0\right)=0$. Therefore, by (17),

$$x_{{\nu }_J\left(0\right)}=0.$$

(21)

The equalities (20) and (21) imply the equality (18).

Let us show that the condition (11) holds. Let $J=(j_1,j_2,\dots )\in \mathcal{J}$. By (18),

$$\sum _{m=0}^{\infty }{x}_{{\nu }_J\left(m\right)}=\sum _{m=1}^{\infty }{\displaystyle \frac{1}{2^{j_m}}}.$$

(22)

The convergence of the series ${\sum }_{m=1}^{\infty }{\displaystyle \frac{1}{2^m}}$ and the inequality $0<{\displaystyle \frac{1}{2^{j_m}}}\le {\displaystyle \frac{1}{2^m}}$ imply the convergence of the series in the right-hand side of (22). Therefore, the series on the left-hand side of (22) is convergent. Thus, (11) holds.

Let us show that the condition (12) holds. Let $\epsilon >0.$ Choose $N\in \mathbb{N}$ such that

$$\sum _{n=N+1}^{\infty }{\displaystyle \frac{1}{2^n}}<\epsilon .$$

(23)

Set $K=2^{N+1}.$ Let $J=(j_1,j_2,\dots )\in \mathcal{J}$ and $K_1>K$. Let us show that the inequality stated in the condition (12) holds. Let

$$m_0=min\left\{m\in {\mathbb{Z}}_{+}:{\nu }_J\left(m\right)\ge K_1\right\}.$$

Since the function ${\nu }_J$ is monotone, it follows that

$$\begin{array}{c}\hfill \sum _{\begin{array}{c}m=0\\ {\nu }_J\left(m\right)\ge K_1\end{array}}^{\infty }{x}_{{\nu }_J\left(m\right)}=\sum _{m=m_0}^{\infty }{x}_{{\nu }_J\left(m\right)}.\end{array}$$

(24)

Since ${\nu }_J\left(0\right)=0$ and $K_1>0$, it follows that $m_0$ cannot be equal to 0, so, $m_0\in \mathbb{N}$. Consequently, by (18),

$$\sum _{m=m_0}^{\infty }{x}_{{\nu }_J\left(m\right)}=\sum _{m=m_0}^{\infty }{\displaystyle \frac{1}{2^{j_m}}}.$$

(25)

By (19),

$$2^{j_{m_0}+1}>{\nu }_J\left(m_0\right),$$

consequently,

$$2^{j_{m_0}+1}>K_1.$$

Therefore, taking into account that $K_1>K$ and $K=2^{N+1}$,

$$2^{j_{m_0}+1}>2^{N+1}.$$

Consequently, $j_{m_0}>N$. Therefore, taking into account that, by the definition of $\mathcal{J}$,

$$j_{m_0}<j_{m_0+1}<j_{m_0+2}<\dots ,$$

we have $j_{m_0+k}>N+k$ for every $k\in {\mathbb{Z}}_{+}$. Consequently,

$$\sum _{m=m_0}^{\infty }{\displaystyle \frac{1}{2^{j_m}}}<\sum _{n=N+1}^{\infty }{\displaystyle \frac{1}{2^n}}.$$

(26)

By (23)–(26), the inequality stated in the condition (12) holds. So, $x\in {\mathrm{cs}}_2$.

Example 2.

Let us construct a sequence which is summable and does not belong to ${\mathrm{cs}}_2.$ Let $x=(x_0,x_1,\dots ),$ where

$$x_n=\left\{\begin{array}{cc}{\displaystyle \frac{1}{s}},\hfill & if n=2^s-1 for some s\in \mathbb{N},\hfill \\ -{\displaystyle \frac{1}{s}},\hfill & if n=2^s for some s\in \mathbb{N},\hfill \\ 0,\hfill & otherwise\hfill \end{array}\right.$$

(27)

for $n\in {\mathbb{Z}}_{+},$ i.e.

$$\begin{array}{c}{x}_0=0,x_1=1,x_2=-1,x_3={\displaystyle \frac{1}{2}},x_4=-{\displaystyle \frac{1}{2}},x_5=x_6=0,x_7={\displaystyle \frac{1}{3}},x_8=-{\displaystyle \frac{1}{3}},\hspace{1em}\hspace{1em}\hspace{1em}\hfill \\ \hfill x_9=\dots =x_{14}=0,x_{15}={\displaystyle \frac{1}{4}},x_{16}=-{\displaystyle \frac{1}{4}},\dots \end{array}$$

Evidently, the sequence x is summable. Let us show that $x\notin {\mathrm{cs}}_2.$ Let us show that the condition (11) does not hold. Let $J=(0,1,2,\dots ).$ Evidently, $J\in \mathcal{J}$. By (10),

$$\begin{array}{cc}\hfill {\nu }_J\left(m\right)& =\left\{\begin{array}{cc}0,\hfill & if m=0,\hfill \\ 2^0+2^1+\dots +2^{m-1},\hfill & \mathrm{if}m\in \mathbb{N}.\hfill \end{array}\right.\hfill \\ & =\left\{\begin{array}{cc}0,\hfill & if m=0,\hfill \\ 2^m-1,\hfill & if m\in \mathbb{N}.\hfill \end{array}\right.\hfill \end{array}$$

Therefore, by (27),

$$x_{{\nu }_J\left(m\right)}=\left\{\begin{array}{cc}0,\hfill & if m=0,\hfill \\ {\displaystyle \frac{1}{m}},\hfill & if m\in \mathbb{N}.\hfill \end{array}\right.$$

Consequently,

$$\sum _{m=0}^{\infty }{x}_{{\nu }_J\left(m\right)}=\sum _{m=1}^{\infty }{\displaystyle \frac{1}{m}}.$$

Since the series on the right-hand side of the latter equality diverges, it follows that the series on the left-hand side of this equality also diverges. Thus, (11) does not hold. So, $x\notin {\mathrm{cs}}_2.$

Remark 1.

Let us compare the space ${\mathrm{cs}}_2$ with some classical Banach spaces. We denote by ${\ell }_1\left({\mathbb{Z}}_{+}\right),$ $cs\left({\mathbb{Z}}_{+}\right),$ and $c_0\left({\mathbb{Z}}_{+}\right)$ Banach spaces of sequences of complex numbers indexed by elements of ${\mathbb{Z}}_{+}$ that are absolutely summable, summable, and convergent to $0,$ respectively. It can be shown that every absolutely summable sequence belongs to ${\mathrm{cs}}_2.$ Also, the condition (12) implies the convergence to 0 of every sequence that belongs to ${\mathrm{cs}}_2.$ Example 1 presents a construction of a sequence that belongs to ${\mathrm{cs}}_2$ and does not belong to ${\ell }_1\left({\mathbb{Z}}_{+}\right)$ and $cs\left({\mathbb{Z}}_{+}\right).$ A sequence constructed in Example 2 does not belong to ${\mathrm{cs}}_2$ and belongs to $cs\left({\mathbb{Z}}_{+}\right)$ and $c_0\left({\mathbb{Z}}_{+}\right).$ So, ${\ell }_1\left({\mathbb{Z}}_{+}\right)\subset {\mathrm{cs}}_2\subset c_0\left({\mathbb{Z}}_{+}\right)$ and ${\ell }_1\left({\mathbb{Z}}_{+}\right)\ne {\mathrm{cs}}_2\ne c_0\left({\mathbb{Z}}_{+}\right).$ Also, ${\mathrm{cs}}_2\not\subset cs\left({\mathbb{Z}}_{+}\right)$ and ${\mathrm{cs}}_2\not\supset cs\left({\mathbb{Z}}_{+}\right).$

Let us prove a criterion for sequences to belong to the space ${\mathrm{cs}}_2.$

Lemma 4.

A sequence of complex numbers $x=(x_0,x_1,\dots )$ belongs to ${\mathrm{cs}}_2$ if and only if x satisfies the following condition:

$$\begin{array}{c}for every \epsilon >0 there exists L\in \mathbb{N} such that \left|\sum _{\begin{array}{c}m=1\\ 2^{l_1}+\dots +2^{l_m}\ge L_1\end{array}}^M{x}_{2^{l_1}+\dots +2^{l_m}}\right|<\epsilon for \hfill \\ \hfill every L_1>L, and every M\in \mathbb{N}, and l_1,l_2,\dots ,l_M\in {\mathbb{Z}}_{+} such that l_1<l_2<\dots <l_M.\end{array}$$

(28)

Proof. 

Let $x=(x_0,x_1,\dots )\in {\mathrm{cs}}_2.$ Let us show that x satisfies (28). Let $\epsilon >0.$ Let us find $L\in \mathbb{N}$ such that (28) holds. Since $x\in {\mathrm{cs}}_2,$ it follows that (12) holds. Consequently, there exists $K\in \mathbb{N}$ such that

$$\left|\sum _{\begin{array}{c}m=0\\ {\nu }_J\left(m\right)\ge K_1\end{array}}^{\infty }{x}_{{\nu }_J\left(m\right)}\right|<\epsilon /2$$

(29)

for every $J\in \mathcal{J}$ and $K_1>K$. Set $L=K$. Let us show that (28) holds. Let $L_1>L, M\in \mathbb{N}$ and $l_1, l_2, \dots , l_M\in {\mathbb{Z}}_{+}\mathrm{be} \mathrm{such} \mathrm{that} l_1<l_2<\dots <l_M.$ Let us consider two cases. Case 1: $L_1>2^{l_1}+\dots +2^{l_M}$. In this case, the set $\{m\in \{l,\dots ,M\}:2^{l_1}+\dots +2^{l_m}\ge L_1\}$ is empty since $2^{l_1}+\dots +2^{l_m}\le 2^{l_1}+\dots +2^{l_M}<L_1$ for every $m\in \{l,\dots ,M\}$ and, consequently,

$$\sum _{\begin{array}{c}m=1\\ 2^{l_1}+\dots +2^{l_m}\ge L_1\end{array}}^M{x}_{2^{l_1}+\dots +2^{l_m}}=0.$$

Therefore (28) holds. Case 2: $L_1\le 2^{l_1}+\dots +2^{l_M}.$ Let $J=(j_1,j_2,\dots )$ be an arbitrary element of $\mathcal{J}$ such that

$$j_m=l_m$$

(30)

for every $m\in \{1,\dots ,M\}.$ By substituting $K_1=L_1$ and $K_1=2^{l_1}+\dots +2^{l_M}+1$ into (29), we obtain

$$\left|\sum _{\begin{array}{c}m=0\\ {\nu }_J\left(m\right)\ge L_1\end{array}}^{\infty }{x}_{{\nu }_J\left(m\right)}\right|<\epsilon /2\mathrm{and}\left|\sum _{\begin{array}{c}m=0\\ {\nu }_J\left(m\right)\ge 2^{l_1}+\dots +2^{l_M}+1\end{array}}^{\infty }{x}_{{\nu }_J\left(m\right)}\right|<\epsilon /2$$

(31)

respectively. Note that the set $\{m\in {\mathbb{Z}}_{+}:{\nu }_J\left(m\right)\ge L_1\}$ is the disjoint union of the sets

$$A=\left\{m\in {\mathbb{Z}}_{+}:L_1\le {\nu }_J\left(m\right)<2^{l_1}+\dots +2^{l_M}+1\right\}$$

(32)

and

$$\left\{m\in {\mathbb{Z}}_{+}:{\nu }_J\left(m\right)\ge 2^{l_1}+\dots +2^{l_M}+1\right\}.$$

Consequently,

$$\sum _{\begin{array}{c}m=0\\ {\nu }_J\left(m\right)\ge L_1\end{array}}^{\infty }{x}_{{\nu }_J\left(m\right)}=\sum _{m\in A}{x}_{{\nu }_J\left(m\right)}+\sum _{\begin{array}{c}m=0\\ {\nu }_J\left(m\right)\ge 2^{l_1}+\dots +2^{l_M}+1\end{array}}^{\infty }{x}_{{\nu }_J\left(m\right)}.$$

Therefore

$$\sum _{m\in A}{x}_{{\nu }_J\left(m\right)}=\sum _{\begin{array}{c}m=0\\ {\nu }_J\left(m\right)\ge L_1\end{array}}^{\infty }{x}_{{\nu }_J\left(m\right)}-\sum _{\begin{array}{c}m=0\\ {\nu }_J\left(m\right)\ge 2^{l_1}+\dots +2^{l_M}+1\end{array}}^{\infty }{x}_{{\nu }_J\left(m\right)}$$

and, consequently, taking into account (31),

$$\left|\sum _{m\in A}{x}_{{\nu }_J\left(m\right)}\right|\le \left|\sum _{\begin{array}{c}m=0\\ {\nu }_J\left(m\right)\ge L_1\end{array}}^{\infty }{x}_{{\nu }_J\left(m\right)}\right|+\left|\sum _{\begin{array}{c}m=0\\ {\nu }_J\left(m\right)\ge 2^{l_1}+\dots +2^{l_M}+1\end{array}}^{\infty }{x}_{{\nu }_J\left(m\right)}\right|<\epsilon /2+\epsilon /2=\epsilon .$$

(33)

Let us show that

$$A=\left\{m\in \{1,\dots ,M\}:L_1\le 2^{l_1}+\dots +2^{l_m}\right\}.$$

(34)

By (10) and (30),

$${\nu }_J\left(m\right)=\left\{\begin{array}{cc}0,\hfill & \mathrm{if}m=0,\hfill \\ 2^{l_1}+\dots +2^{l_m},\hfill & \mathrm{if}m\in \{1,\dots ,M\},\hfill \\ 2^{l_1}+\dots +2^{l_M}+2^{j_{M+1}}+\dots +2^{j_m},\hfill & \mathrm{if}m\in \{M+1,M+2,\dots \}.\hfill \end{array}\right.$$

(35)

Since $L\in \mathbb{N}$ and $L_1>L,$ it follows that $L_1>0.$ Consequently, taking into account (35), the inequality $L_1\le {\nu }_J\left(m\right)$ holds only for nonzero $m.$ Also, by (35), the inequality ${\nu }_J\left(m\right)<2^{l_1}+\dots +2^{l_M}+1$ holds if and only if $m\le M.$ So, taking into account (32), the equality (34) holds. Therefore

$$\sum _{m\in A}{x}_{{\nu }_J\left(m\right)}=\sum _{\begin{array}{c}m=1\\ 2^{l_1}+\dots +2^{l_m}\ge L_1\end{array}}^M{x}_{2^{l_1}+\dots +2^{l_m}}.$$

Consequently, taking into account (33),

$$\left|\sum _{\begin{array}{c}m=1\\ 2^{l_1}+\dots +2^{l_m}\ge L_1\end{array}}^M{x}_{2^{l_1}+\dots +2^{l_m}}\right|<\epsilon .$$

So, (28) holds.

Let $x=(x_0,x_1,\dots )$ be a sequence of complex numbers that satisfies (28). Let us prove that $x\in {\mathrm{cs}}_2,$ i.e., that x satisfies (11) and (12).

Let us show that (11) holds. Let $J=(j_1,j_2,\dots )$ be an arbitrary element of $\mathcal{J}$ Let us show that the series ${\sum }_{m=0}^{\infty }{x}_{{\nu }_J\left(m\right)}$ is convergent. It is enough to show that the sequence of partial sums of this series is a Cauchy sequence, i.e., for every $\epsilon >0$ there exists $n_0\in \mathbb{N}$ so that

$$\left|\sum _{m=n_0+1}^n{x}_{{\nu }_J\left(m\right)}\right|<\epsilon$$

(36)

for every $n>n_0.$ By (28), there exists $L\in \mathbb{N}$ such that the following condition holds:

$$\begin{array}{c}\left|\sum _{\begin{array}{c}m=1\\ 2^{l_1}+\dots +2^{l_m}\ge L_1\end{array}}^M{x}_{2^{l_1}+\dots +2^{l_m}}\right|<\epsilon \mathrm{for} \mathrm{every}{L}_1>L,M\in \mathbb{N}\mathrm{and}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hfill \\ \hfill l_1,l_2,\dots ,l_M\in {\mathbb{Z}}_{+}\mathrm{such} \mathrm{that}{l}_1<l_2<\dots <l_M.\end{array}$$

(37)

Let $n_0$ be an arbitrary fixed positive integer such that ${\nu }_J\left(n_0\right)\ge L.$ Let $n\in \mathbb{N}$ be such that $n>n_0.$ By substituting $L_1={\nu }_J\left(n_0\right)+1,$ $M=n$ and $l_m=j_m$ for $m\in \{1,\dots ,n\}$ into (37), we obtain

$$\left|\sum _{\begin{array}{c}m=1\\ 2^{j_1}+\dots +2^{j_m}\ge {\nu }_J\left(n_0\right)+1\end{array}}^n{x}_{2^{j_1}+\dots +2^{j_m}}\right|<\epsilon ,$$

i.e., taking into account (10), we obtain

$$\left|\sum _{\begin{array}{c}m=1\\ {\nu }_J\left(m\right)\ge {\nu }_J\left(n_0\right)+1\end{array}}^n{x}_{{\nu }_J\left(m\right)}\right|<\epsilon .$$

(38)

Since the mapping ${\nu }_J$ is monotone, it follows that ${\nu }_J\left(m\right)\ge {\nu }_J\left(n_0\right)+1$ if and only if $m>n_0,$ i. e. $m\ge n_0+1.$ Therefore, by (38), the inequality (36) holds. So, the series ${\sum }_{m=0}^{\infty }{x}_{{\nu }_J\left(m\right)}$ is convergent. Thus, (11) holds.

Let us show that x satisfies (12). Let $\epsilon >0.$ Let us find $K\in \mathbb{N}$ such that (12) holds. By (28), there exists $L\in \mathbb{N}$ such that the following condition holds:

$$\begin{array}{c}\left|\sum _{\begin{array}{c}m=1\\ 2^{l_1}+\dots +2^{l_m}\ge L_1\end{array}}^M{x}_{2^{l_1}+\dots +2^{l_m}}\right|<\epsilon /2\mathrm{for} \mathrm{every}{L}_1>L,M\in \mathbb{N}\mathrm{and}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hfill \\ \hfill l_1,l_2,\dots ,l_M\in {\mathbb{Z}}_{+}, \mathrm{such} \mathrm{that} l_1<l_2<\dots <l_M.\end{array}$$

(39)

Set $K=L+1.$ Let us show that (12) holds. Let $J=(j_1,j_2,\dots )\in \mathcal{J}$ and $K_1>K$. By substituting $L_1=K_1$ and $l_m=j_m$ for $m\in \mathbb{N}$ into (39), taking into account (10), we obtain

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \left|\sum _{\begin{array}{c}m=1\\ {\nu }_J\left(m\right)\ge K_1\end{array}}^M{x}_{{\nu }_J\left(m\right)}\right|<\epsilon /2\end{array}\end{array}$$

(40)

for every $M\in \mathbb{N}$. Since (40) holds for every $M\in \mathbb{N}$ and the series ${\displaystyle \sum _{\begin{array}{c}m=1\\ {\nu }_J\left(m\right)\ge K_1\end{array}}^{\infty }}{x}_{{\nu }_J\left(m\right)}$ is convergent, it follows that

$$\begin{array}{c}\hfill {\displaystyle \underset{M\to \infty }{lim}}\left|{\displaystyle \sum _{\begin{array}{c}m=1\\ {\nu }_J\left(m\right)\ge K_1\end{array}}^M}{x}_{{\nu }_J\left(m\right)}\right|\le \epsilon /2.\end{array}$$

Therefore, taking into account that $\epsilon /2<\epsilon$ and

$$\begin{array}{c}\hfill {\displaystyle \underset{M\to \infty }{lim}}\left|{\displaystyle \sum _{\begin{array}{c}m=1\\ {\nu }_J\left(m\right)\ge K_1\end{array}}^M}{x}_{{\nu }_J\left(m\right)}\right|=\left|{\displaystyle \underset{M\to \infty }{lim}}{\displaystyle \sum _{\begin{array}{c}m=1\\ {\nu }_J\left(m\right)\ge K_1\end{array}}^M}{x}_{{\nu }_J\left(m\right)}\right|=\left|{\displaystyle \sum _{\begin{array}{c}m=1\\ {\nu }_J\left(m\right)\ge K_1\end{array}}^{\infty }}{x}_{{\nu }_J\left(m\right)}\right|,\end{array}$$

we have

$$\left|\sum _{\begin{array}{c}m=1\\ {\nu }_J\left(m\right)\ge K_1\end{array}}^{\infty }{x}_{{\nu }_J\left(m\right)}\right|<\epsilon .$$

Thus, taking into account that ${\nu }_J\left(0\right)=0<K_1$ and, consequently,

$$\sum _{\begin{array}{c}m=1\\ {\nu }_J\left(m\right)\ge K_1\end{array}}^{\infty }{x}_{{\nu }_J\left(m\right)}=\sum _{\begin{array}{c}m=0\\ {\nu }_J\left(m\right)\ge K_1\end{array}}^{\infty }{x}_{{\nu }_J\left(m\right)},$$

the property (12) holds. This completes the proof. □

3.2. Some Properties of the Space Y

Lemma 5.

For every $n\in \mathbb{N},$

$$\lceil {log}_2(n+1)\rceil =\lfloor {log}_{2}n\rfloor +1.$$

Proof. 

Let $k=\lfloor {log}_{2}n\rfloor .$ Then

$$k\le {log}_{2}n$$

(41)

and

$${log}_{2}n<k+1.$$

(42)

By (41) and by the inequality ${log}_{2}n<{log}_2(n+1),$

$$k<{log}_2(n+1).$$

Consequently,

$$k<\lceil {log}_2(n+1)\rceil .$$

(43)

Since both k and $\lceil {log}_2(n+1)\rceil$ are integers, the inequality (43) implies the following inequality:

$$\lceil {log}_2(n+1)\rceil -k\ge 1,$$

that is,

$$k+1\le \lceil {log}_2(n+1)\rceil .$$

(44)

By (42), $n<2^{k+1}.$ Consequently, since both n and $2^{k+1}$ are integers, it follows that

$$2^{k+1}-n\ge 1,$$

i.e., $n+1\le 2^{k+1}.$ Therefore, ${log}_2(n+1)\le k+1$ and, consequently, since $k+1$ is an integer,

$$\lceil {log}_2(n+1)\rceil \le k+1.$$

Consequently, taking into account (44),

$$\lceil {log}_2(n+1)\rceil =k+1,$$

that is,

$$\lceil {log}_2(n+1)\rceil =\lfloor {log}_{2}n\rfloor +1.$$

This completes the proof. □

The following result follows directly from (6) and (7).

Lemma 6.

For every $n\in \mathbb{N},$

$${\mathsf{\Omega }}_n=\left\{n\right\}\bigsqcup {\mathsf{\Omega }}_{n-2^s},$$

where

$$s=max\{j\in {\mathbb{Z}}_{+}:2^j\le n\}$$

and “⊔” is the disjoint union operation.

Lemma 7.

Let $y=(y_1,y_2,\dots )\in Y.$ For every $n\in {\mathbb{Z}}_{+},$

$$\sum _{m\in {\mathsf{\Omega }}_n}{\alpha }_{m+1}\left(y\right)=y_{n+1},$$

(45)

where ${\alpha }_{m+1}$ is defined by (5).

Proof. 

We proceed by induction on $n.$

In the case $n=0,$ we have ${\mathsf{\Omega }}_n=\left\{0\right\}$ and, consequently, the left-hand side of (45) equals ${\alpha }_1\left(y\right),$ which, by (5), is equal to $y_1.$ So, in this case, the equality (45) holds.

Suppose that for some $n\in \mathbb{N},$

$$\sum _{m\in {\mathsf{\Omega }}_k}{\alpha }_{m+1}\left(y\right)=y_{k+1},$$

(46)

for all $k=\{0,1,\dots ,n-1\}.$ Let us show that

$$\sum _{m\in {\mathsf{\Omega }}_n}{\alpha }_{m+1}\left(y\right)=y_{n+1}.$$

By Lemma 6,

$${\mathsf{\Omega }}_n=\left\{n\right\}\bigsqcup {\mathsf{\Omega }}_{n-2^s},$$

(47)

where

$$s=max\{j\in {\mathbb{Z}}_{+}:2^j\le n\}.$$

(48)

By (48),

$$2^s\le n<2^{s+1}.$$

Consequently,

$$s\le {log}_{2}n<s+1.$$

(49)

Since s is an integer, by (49),

$$s=\lfloor {log}_{2}n\rfloor .$$

(50)

By (47),

$$\sum _{m\in {\mathsf{\Omega }}_n}{\alpha }_{m+1}\left(y\right)={\alpha }_{n+1}\left(y\right)+\sum _{m\in {\mathsf{\Omega }}_{n-2^s}}{\alpha }_{m+1}\left(y\right).$$

(51)

By (5),

$${\alpha }_{n+1}\left(y\right)=y_{n+1}-y_{n+1-2^{\varkappa (n+1)-1}},$$

(52)

where ϰ is defined by (3). By Lemma 5, taking into account (3) and (50),

$$\varkappa (n+1)-1=\lceil {log}_2(n+1)\rceil -1=\lfloor {log}_{2}n\rfloor +1-1=\lfloor {log}_{2}n\rfloor =s.$$

Therefore, by (52),

$${\alpha }_{n+1}\left(y\right)=y_{n+1}-y_{n+1-2^s}.$$

(53)

By (46),

$$\sum _{m\in {\mathsf{\Omega }}_{n-2^s}}{\alpha }_{m+1}\left(y\right)=y_{n-2^s+1}.$$

(54)

By (51), (53), and (54),

$$\sum _{m\in {\mathsf{\Omega }}_n}{\alpha }_{m+1}\left(y\right)=y_{n+1}-y_{n+1-2^s}+y_{n+1-2^s}=y_{n+1}.$$

This completes the proof. □

Let us prove a criterion for sequences to belong to the space $Y.$

Lemma 8.

A sequence of complex numbers $y=(y_1,y_2,\dots )$ belongs to Y if and only if y satisfies the following condition:

$$\begin{array}{c}\mathit{for} \mathit{every} \epsilon >0 \mathit{there} \mathit{exists} n\in \mathbb{N} \mathit{such} \mathit{that} |y_{s_1}-y_{s_2}|<\epsilon \mathit{for}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hfill \\ \hfill \mathit{every}{s}_1,s_2\in \mathbb{N} \mathit{such} \mathit{that} s_1-s_2 \mathit{is} \mathit{divisible} \mathit{by} 2^n.\end{array}$$

(55)

Proof. 

Let $y=(y_1,y_2,\dots )\in Y.$ Let us show that y satisfies (55). Let $\epsilon >0.$ Let us find $n\in \mathbb{N}$ such that (55) holds. Since $y\in Y,$ it follows that (1) holds. Consequently, there exists $n\in \mathbb{N}$ such that

$$|y_j-y_{j+k{2}^n}|<\epsilon /2 \mathrm{for} \mathrm{every} j\in \{1,2,3,\dots ,2^n\} \mathrm{and} k\in \mathbb{N}.$$

(56)

Let $s_1,s_2\in \mathbb{N}$ be such that $s_1-s_2$ is divisible by $2^n.$ Let $r\in \{0,\dots ,2^n-1\}$ be the remainder when $s_1$ is divided by $2^n.$ Since $s_1-s_2$ is divisible by $2^n,$ it follows that r is also the remainder of $s_2$ divided by $2^n.$ So, there exist $q_1,q_2\in {\mathbb{Z}}_{+}$ such that

$$s_i=r+q_i{2}^n,i\in \{1,2\}.$$

(57)

Let

$$j=\left\{\begin{array}{cc}r,\hfill & \mathrm{if}r>0,\hfill \\ 2^n,\hfill & \mathrm{if}r=0\hfill \end{array}\right.$$

(58)

and

$$k_i=\left\{\begin{array}{cc}{q}_i,\hfill & \mathrm{if}r>0,\hfill \\ q_i-1,\hfill & \mathrm{if}r=0\hfill \end{array}\right.$$

(59)

for $i\in \{1,2\}.$ Then

$$s_i=j+k_i{2}^n,i\in \{1,2\}.$$

(60)

Since $r\in \{0,\dots ,2^n-1\},$ by (58), $j\in \{1,\dots ,2^n\}.$ Let us show that $k_i\in {\mathbb{Z}}_{+}$ for $i\in \{1,2\}.$ Let $i\in \{1,2\}.$ If $r>0,$ then, by (59), $k_i=q_i$ and, consequently, $k_i\in {\mathbb{Z}}_{+}$ since $q_i\in {\mathbb{Z}}_{+}.$ Consider the case $r=0.$ In this case, by (57), $s_i=q_i{2}^n$ and, consequently, $q_i\in \mathbb{N}$ since $s_i\in \mathbb{N}.$ Therefore, $q_i-1$ belongs to ${\mathbb{Z}}_{+}.$ Consequently, taking into account (59), $k_i\in {\mathbb{Z}}_{+}.$ Thus, $j\in \{1,\dots ,2^n\}$ and $k_i\in {\mathbb{Z}}_{+}$ for $i\in \{1,2\}.$ Therefore, by (56),

$$|y_j-y_{j+k_i{2}^n}|<\epsilon /2,i\in \{1,2\},$$

that is, taking into account (60),

$$|y_j-y_{s_i}|<\epsilon /2,i\in \{1,2\}.$$

Consequently, by the triangle inequality,

$$|y_{s_1}-y_{s_2}|=|y_{s_1}-y_j+y_j-y_{s_2}|\le |y_{s_1}-y_j|+|y_j-y_{s_2}|<\epsilon /2+\epsilon /2=\epsilon .$$

So, (55) holds.

Let $y=(y_1,y_2,\dots )$ be a sequence of complex numbers that satisfies (55). Let us show that $y\in Y,$ i.e., (1) holds. Let $\epsilon >0.$ By (55), there exists $n\in \mathbb{N}$ such that

$$|y_{s_1}-y_{s_2}|<\epsilon$$

(61)

for every $s_1,s_2\in \mathbb{N}$ such that $s_1-s_2$ is divisible by $2^n.$ Let $j\in \{1,\dots ,2^n\}$ and $k\in {\mathbb{Z}}_{+}$ be arbitrary numbers. Set $s_1=j$ and $s_2=j+k{2}^n.$ Since $s_1-s_2$ is divisible by $2^n,$ by (61), $|y_{s_1}-y_{s_2}|<\epsilon ,$ i.e.,

$$|y_j-y_{j+k{2}^n}|<\epsilon .$$

Thus, (1) holds. Therefore $y\in Y.$ This completes the proof. □

3.3. Isometrical Isomorphism Between Y and ${\mathrm{cs}}_2$

Let us define $\iota :Y\to {\mathrm{cs}}_2$ in the following way. For $y=(y_1,y_2,\dots )\in Y,$ let

$$\iota \left(y\right)=(x_0,x_1,\dots ),$$

(62)

with

$$x_j={\alpha }_{j+1}\left(y\right)$$

for $j\in {\mathbb{Z}}_{+},$ where ${\alpha }_{j+1}$ is defined by (5).

Lemma 9.

Let $q\in \mathbb{N}$ and $p_1,p_2,\dots ,p_q\in {\mathbb{Z}}_{+}$ be such that $p_1<p_2<\dots <p_q.$ Let $q_0\in \{0,\dots ,q-1\}.$ Let

$$r=\left\{\begin{array}{cc}{2}^{p_1}+2^{p_2}+\dots +2^{p_{q_0}},\hfill & \mathit{if} q_0\ge 1,\hfill \\ 0,\hfill & \mathit{if} q_0=0.\hfill \end{array}\right.$$

Then

$${\mathsf{\Omega }}_r\subset {\mathsf{\Omega }}_{2^{p_1}+2^{p_2}+\dots +2^{p_q}}$$

(63)

and

$${\mathsf{\Omega }}_{2^{p_1}+2^{p_2}+\dots +2^{p_q}}\setminus {\mathsf{\Omega }}_r=\left\{2^{p_1}+2^{p_2}+\dots +2^{p_m}:m\in \{q_0+1,q_0+2,\dots ,q\}\right\},$$

(64)

where sets of the form ${\mathsf{\Omega }}_n$ are defined by (6) and (7).

Proof. 

By (6) and (7),

$${\mathsf{\Omega }}_{2^{p_1}+2^{p_2}+\dots +2^{p_q}}=\left\{0\right\}\cup \left\{2^{p_1}+2^{p_2}+\dots +2^{p_m}:m\in \{1,\dots ,q\}\right\}.$$

and

$${\mathsf{\Omega }}_r=\left\{\begin{array}{cc}\left\{0\right\}\cup \left\{2^{p_1}+2^{p_2}+\dots +2^{p_m}:m\in \{1,\dots ,q_0\}\right\},\hfill & \mathrm{if} q_0\ge 1,\hfill \\ \left\{0\right\},\hfill & \mathrm{if} q_0=0.\hfill \end{array}\right.$$

Therefore, (63) and (64) hold. □

Lemma 10.

For every $y\in Y,$ the sequence $\iota \left(y\right),$ defined by (62), belongs to ${\mathrm{cs}}_2.$

Proof. 

Let $y=(y_1,y_2,\dots )\in Y.$ Let $x=\iota \left(y\right),$ i.e., $x=(x_0,x_1,\dots ),$ where

$$x_j={\alpha }_{j+1}\left(y\right)$$

(65)

for $j\in {\mathbb{Z}}_{+},$ where ${\alpha }_{j+1}$ is defined by (5). Let us show that $x\in {\mathrm{cs}}_2.$ By Lemma 4, it is enough to show that x satisfies (28). Let us prove this fact. Let $\epsilon >0.$ Let us find K such that (28) holds. Since $y\in Y,$ by Lemma 8, y satisfies (55). Therefore, there exists $n\in \mathbb{N}$ such that

$$|y_{s_1}-y_{s_2}|<\epsilon$$

(66)

for every $s_1,s_2\in \mathbb{N}$ such that $s_1-s_2$ is divisible by $2^n.$ Set $L=2^n.$ Let us show that (28) holds, i.e.,

$$\begin{array}{c}\left|\sum _{\begin{array}{c}m=1\\ 2^{l_1}+\dots +2^{l_m}\ge L_1\end{array}}^M{x}_{2^{l_1}+\dots +2^{l_m}}\right|<\epsilon \mathrm{for}\mathrm{every}{L}_1>L,M\in \mathbb{N}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hfill \\ \hfill \mathrm{and}{l}_1,l_2,\dots ,l_M\in {\mathbb{Z}}_{+}\mathrm{such}\mathrm{that}{l}_1<l_2<\dots <l_M.\end{array}$$

Let $L_1>L,$ $M\in \mathbb{N}$ and $l_1,l_2,\dots ,l_M\in {\mathbb{Z}}_{+}$ be arbitrary numbers such that $l_1<l_2<\dots <l_M.$ Let us show that

$$\left|\sum _{\begin{array}{c}m=1\\ 2^{l_1}+\dots +2^{l_m}\ge L_1\end{array}}^M{x}_{2^{l_1}+\dots +2^{l_m}}\right|<\epsilon .$$

(67)

If $2^{l_1}+\dots +2^{l_M}<L_1,$ then the number of terms in the sum in the left-hand side of (67) is zero, and therefore (67) holds. Consider the case $2^{l_1}+\dots +2^{l_M}\ge L_1.$ Let

$$M_0=min\left\{m\in \{1,\dots ,M\}:2^{l_1}+\dots +2^{l_m}\ge L_1\right\}.$$

(68)

Then, for $m\in \{1,\dots ,M\},$ we have $2^{l_1}+\dots +2^{l_m}\ge L_1$ if and only if $m\ge M_0.$ Consequently,

$$\sum _{\begin{array}{c}m=1\\ 2^{l_1}+\dots +2^{l_m}\ge L_1\end{array}}^M{x}_{2^{l_1}+\dots +2^{l_m}}=\sum _{m=M_0}^M{x}_{2^{l_1}+\dots +2^{l_m}}.$$

(69)

According to Lemma 9, by substituting $q=M,$ $p_1=l_1,\dots ,p_M=l_M$ and $q_0=M_0-1,$ we obtain:

$${\mathsf{\Omega }}_r\subset {\mathsf{\Omega }}_{2^{l_1}+2^{l_2}+\dots +2^{l_M}},$$

(70)

and

$${\mathsf{\Omega }}_{2^{l_1}+2^{l_2}+\dots +2^{l_M}}\setminus {\mathsf{\Omega }}_r=\left\{2^{l_1}+2^{l_2}+\dots +2^{l_m}:m\in \{M_0,M_0+1,\dots ,M\}\right\},$$

(71)

where

$$r=\left\{\begin{array}{cc}{2}^{l_1}+2^{l_2}+\dots +2^{l_{M_0-1}},\hfill & \mathrm{if} M_0\ge 2,\hfill \\ 0,\hfill & \mathrm{if} M_0=1.\hfill \end{array}\right.$$

(72)

By (71),

$$\sum _{m=M_0}^M{x}_{2^{l_1}+\dots +2^{l_m}}=\sum _{k\in {\mathsf{\Omega }}_{2^{l_1}+2^{l_2}+\dots +2^{l_M}}\setminus {\mathsf{\Omega }}_r}{x}_k.$$

(73)

By (70),

$$\sum _{k\in {\mathsf{\Omega }}_{2^{l_1}+2^{l_2}+\dots +2^{l_M}}\setminus {\mathsf{\Omega }}_r}{x}_k=\sum _{k\in {\mathsf{\Omega }}_{2^{l_1}+2^{l_2}+\dots +2^{l_M}}}{x}_k-\sum _{k\in {\mathsf{\Omega }}_r}{x}_k.$$

(74)

By Lemma 7, taking into account (65), we obtain

$$\sum _{k\in {\mathsf{\Omega }}_{2^{l_1}+2^{l_2}+\dots +2^{l_M}}}{x}_k=y_{2^{l_1}+2^{l_2}+\dots +2^{l_M}+1}\mathrm{and}\sum _{k\in {\mathsf{\Omega }}_r}{x}_k=y_{r+1}.$$

Therefore

$$\sum _{k\in {\mathsf{\Omega }}_{2^{l_1}+2^{l_2}+\dots +2^{l_M}}}{x}_k-\sum _{k\in {\mathsf{\Omega }}_r}{x}_k=y_{2^{l_1}+2^{l_2}+\dots +2^{l_M}+1}-y_{r+1}.$$

(75)

Let $s_1=2^{l_1}+2^{l_2}+\dots +2^{l_M}+1$ and $s_2=r+1.$ Let us show that $s_1-s_2$ is divisible by $2^n.$ By (72),

$$\begin{array}{cc}\hfill s_1-s_2& =2^{l_1}+2^{l_2}+\dots +2^{l_M}-r\hfill \\ \hfill & =\left\{\begin{array}{cc}{2}^{l_1}+2^{l_2}+\dots +2^{l_M}-\left(2^{l_1}+2^{l_2}+\dots +2^{l_{M_0-1}}\right),\hfill & \mathrm{if} M_0\ge 2,\hfill \\ 2^{l_1}+2^{l_2}+\dots +2^{l_M},\hfill & \mathrm{if} M_0=1.\hfill \end{array}\right.\hfill \\ \hfill & =\left\{\begin{array}{cc}{2}^{l_{M_0}}+2^{l_{M_0+1}}+\dots +2^{l_M},\hfill & \mathrm{if} M_0\ge 2,\hfill \\ 2^{l_1}+2^{l_2}+\dots +2^{l_M},\hfill & \mathrm{if} M_0=1.\hfill \end{array}\right.\hfill \\ \hfill & =2^{l_{M_0}}+2^{l_{M_0+1}}+\dots +2^{l_M}.\hfill \end{array}$$

(76)

Let us show that

$$l_{M_0}\ge n.$$

(77)

Suppose $l_{M_0}<n.$ Then, taking into account that $l_1,l_2,\dots ,l_{M_0}\in {\mathbb{Z}}_{+}$ and $l_1<l_2<\dots <l_{M_0},$

$$2^{l_1}+2^{l_2}+\dots +2^{l_{M_0}}\le 2^0+2^1+2^2+\dots +2^{l_{M_0}}<2^n.$$

On the other hand, by (68), taking into account that $L_1>L$ and $L=2^n,$

$$2^{l_1}+2^{l_2}+\dots +2^{l_{M_0}}\ge L_1>2^n.$$

A contradiction. Thus, (77) holds. Consequently, $2^{l_{M_0}}+2^{l_{M_0+1}}+\dots +2^{l_M}$ is divisible by $2^n.$ So, by (76), $s_1-s_2$ is divisible by $2^n.$ Therefore, by (66), $|y_{s_1}-y_{s_2}|<\epsilon ,$ i.e.,

$$|y_{2^{l_1}+2^{l_2}+\dots +2^{l_M}+1}-y_{r+1}|<\epsilon .$$

(78)

By (69), (73), (74), (75) and (78), the inequality (67) holds. Thus, $x\in {\mathrm{cs}}_2.$ This completes the proof. □

Lemma 11.

The mapping $\iota ,$ defined by (62), is surjective.

Proof. 

Let $x=(x_0,x_1,\dots )\in {\mathrm{cs}}_2.$ Let us construct $y\in Y$ such that $\iota \left(y\right)=x.$ Let $y=(y_1,y_2,\dots ),$ with

$$y_i=\sum _{s\in {\mathsf{\Omega }}_{i-1}}{x}_s$$

(79)

for $i\in \mathbb{N},$ where ${\mathsf{\Omega }}_{i-1}$ is defined by (6) and (7). Let us show that $y\in Y,$ i.e., y satisfies (1). Let $\epsilon >0.$ Let us find $n\in \mathbb{N}$ such that

$$|y_j-y_{j+k{2}^n}|<\epsilon \mathrm{for} \mathrm{every} j\in \{1,2,3,\dots ,2^n\} \mathrm{and} k\in {\mathbb{Z}}_{+}.$$

(80)

Since $x\in {\mathrm{cs}}_2,$ by Lemma 4, there exists $L\in \mathbb{N}$ such that

$$\begin{array}{c}\left|\sum _{\begin{array}{c}m=1\\ 2^{l_1}+\dots +2^{l_m}\ge L_1\end{array}}^M{x}_{2^{l_1}+\dots +2^{l_m}}\right|<\epsilon \mathrm{for} \mathrm{every} L_1>L, \mathrm{and} M\in \mathbb{N},\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hfill \\ \hfill \mathrm{and} l_1,l_2,\dots ,l_M\in {\mathbb{Z}}_{+} \mathrm{such} \mathrm{that} l_1<l_2<\dots <l_M.\end{array}$$

(81)

Set

$$n=min\{i\in \mathbb{N}:2^i>L\}.$$

(82)

Let us show that (80) holds. Let $j\in \{1,2,3,\dots ,2^n\}$ and $k\in {\mathbb{Z}}_{+}$ be an arbitrary number. Let us show that

$$|y_j-y_{j+k{2}^n}|<\epsilon .$$

(83)

If $k=0,$ then $y_j-y_{j+k{2}^n}=0$ and, consequently, (83) holds. Consider the case $k\ne 0,$ i.e., $k\in \mathbb{N}.$ By (79),

$$y_{j+k{2}^n}=\sum _{s\in {\mathsf{\Omega }}_{j-1+k{2}^n}}{x}_s\mathrm{and}{y}_j=\sum _{s\in {\mathsf{\Omega }}_{j-1}}{x}_s.$$

(84)

Since $k\in \mathbb{N},$ there exist $w\in \mathbb{N}$ and $k_1,k_2,\dots ,k_w\in {\mathbb{Z}}_{+}$ such that

$$k_1<k_2<\dots <k_w$$

(85)

and

$$k=2^{k_1}+2^{k_2}+\dots +2^{k_w}.$$

(86)

If $j=1,$ we set

$$q^{*}=w\mathrm{and}{l}_1^{*}=k_1+n,l_2^{*}=k_2+n,\dots ,l_w^{*}=k_w+n.$$

(87)

Otherwise, i.e., in the case $j\in \{2,3,\dots ,2^n\},$ we set

$$q^{*}=v+w\mathrm{and}{l}_1^{*}=j_1,l_2^{*}=j_2,\dots ,l_v^{*}=j_v,l_{v+1}^{*}=k_1+n,l_{v+2}^{*}=k_2+n,\dots ,l_{v+w}^{*}=k_w+n,$$

(88)

where $v\in \mathbb{N}$ and $j_1,j_2,\dots ,j_v\in {\mathbb{Z}}_{+}$ are such that

$$j_1<j_2<\dots <j_v$$

(89)

and

$$j-1=2^{j_1}+2^{j_2}+\dots +2^{j_v}.$$

(90)

In both cases, $l_1^{*},l_2^{*},\dots ,l_{q^{*}}^{*}\in {\mathbb{Z}}_{+}.$ Let us show that

$$l_1^{*}<l_2^{*}<\dots <l_{q^{*}}^{*}.$$

(91)

In the case $j=1,$ (85) and (87) imply (91). Consider the case $j\in \{2,3,\dots ,2^n\}.$ Since $j\le 2^n,$ it follows that $j-1<2^n$ and, consequently, taking into account (90),

$$j_v<n.$$

(92)

Note that (85), (88), (89), and (92) imply (91). So, in both cases, (91) holds.

Since (91) holds, we can substitute $q=q^{*},$ $p_1=l_1^{*},p_2=l_2^{*},\dots ,p_{q^{*}}=l_{q^{*}}^{*}$ and

$$q_0=\left\{\begin{array}{cc}0,\hfill & \mathrm{if} j=1,\hfill \\ v,\hfill & \mathrm{if} j\in \{2,3,\dots ,2^n\}\hfill \end{array}\right.$$

(93)

into Lemma 9. We obtain

$${\mathsf{\Omega }}_r\subset {\mathsf{\Omega }}_{2^{l_1^{*}}+2^{l_2^{*}}+\dots +2^{l_{q^{*}}^{*}}}$$

(94)

and

$${\mathsf{\Omega }}_{2^{l_1^{*}}+2^{l_2^{*}}+\dots +2^{l_{q^{*}}^{*}}}\setminus {\mathsf{\Omega }}_r=\left\{2^{l_1^{*}}+2^{l_2^{*}}+\dots +2^{l_m^{*}}:m\in \{q_0+1,q_0+2,\dots ,q^{*}\}\right\},$$

(95)

where

$$r=\left\{\begin{array}{cc}0,\hfill & \mathrm{if} q_0=0,\hfill \\ 2^{l_1^{*}}+2^{l_2^{*}}+\dots +2^{l_{q_0}^{*}},\hfill & \mathrm{if} q_0\ge 1.\hfill \end{array}\right.$$

(96)

Let us show that

$$r=j-1,$$

(97)

$$2^{l_1^{*}}+2^{l_2^{*}}+\dots +2^{l_{q^{*}}^{*}}=j-1+k{2}^n$$

(98)

and

$$\begin{array}{c}\left\{2^{l_1^{*}}+2^{l_2^{*}}+\dots +2^{l_m^{*}}:m\in \{q_0+1,q_0+2,\dots ,q^{*}\}\right\}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hfill \\ \hfill =\left\{j-1+2^{k_1+n}+2^{k_2+n}+\dots +2^{k_m+n}:m\in \{1,2,\dots ,w\}\right\}.\end{array}$$

(99)

Consider the case $j=1.$ By (93), $q_0=0$ and, consequently, by (96), $r=0.$ Therefore, taking into account that $j-1=0,$ the equality (97) holds. By (86) and (87),

$$2^{l_1^{*}}+2^{l_2^{*}}+\dots +2^{l_{q^{*}}^{*}}=2^{k_1+n}+2^{k_2+n}+\dots +2^{k_w+n}=\left(2^{k_1}+2^{k_2}+\dots +2^{k_w}\right)2^n=k{2}^n.$$

Therefore, taking into account that $j-1=0,$ the equality (98) holds. Since $q_0=0$ and $j-1=0,$ by (87), the equality (99) holds.

Consider the case $j\in \{2,3,\dots ,2^n\}.$ By (93), $q_0=v$ and, consequently, by (96), $r=2^{l_1^{*}}+2^{l_2^{*}}+\dots +2^{l_{q_0}^{*}},$ i.e., taking into account (88) and (90),

$$r=2^{j_1}+2^{j_2}+\dots +2^{j_v}=j-1.$$

Thus, the equality (97) holds. By (86), (87), and (90),

$$\begin{array}{c}{2}^{l_1^{*}}+2^{l_2^{*}}+\dots +2^{l_{q^{*}}^{*}}=2^{j_1}+2^{j_2}+\dots +2^{j_v}+2^{k_1+n}+2^{k_2+n}+\dots +2^{k_w+n}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hfill \\ \hfill =2^{j_1}+2^{j_2}+\dots +2^{j_v}+\left(2^{k_1}+2^{k_2}+\dots +2^{k_w}\right)2^n=j-1+k{2}^n.\end{array}$$

So, the equality (98) holds. Since $q_0=v,$ by (88) and (90),

$$\begin{array}{cc}\hfill \{2^{l_1^{*}}+2^{l_2^{*}}& +\dots +2^{l_m^{*}}:m\in \{q_0+1,q_0+2,\dots ,q^{*}\}\}\hfill \\ & =\left\{2^{l_1^{*}}+2^{l_2^{*}}+\dots +2^{l_m^{*}}:m\in \{v+1,v+2,\dots ,v+w\}\right\}\hfill \\ & =\left\{2^{j_1}+2^{j_2}+\dots +2^{j_v}+2^{k_1+n}+2^{k_2+n}+\dots +2^{k_m+n}:m\in \{1,2,\dots ,w\}\right\}\hfill \\ & =\left\{j-1+2^{k_1+n}+2^{k_2+n}+\dots +2^{k_m+n}:m\in \{1,2,\dots ,w\}\right\}.\hfill \end{array}$$

Thus, the equality (99) holds.

By (94) and (95), taking into account (97)–(99),

$${\mathsf{\Omega }}_{j-1}\subset {\mathsf{\Omega }}_{j-1+k{2}^n}$$

and

$${\mathsf{\Omega }}_{j-1+k{2}^n}\setminus {\mathsf{\Omega }}_{j-1}=A,$$

where

$$A=\left\{j-1+2^{k_1+n}+2^{k_2+n}+\dots +2^{k_m+n}:m\in \{1,2,\dots ,w\}\right\}.$$

Therefore, by (84),

$$y_{j+k{2}^n}-y_j=\sum _{s\in {\mathsf{\Omega }}_{j-1+k{2}^n}}{x}_s-\sum _{s\in {\mathsf{\Omega }}_{j-1}}{x}_s=\sum _{s\in {\mathsf{\Omega }}_{j-1+k{2}^n}\setminus {\mathsf{\Omega }}_{j-1}}{x}_s=\sum _{s\in A}{x}_s.$$

(100)

By (82), $2^n>L.$ Therefore, taking into account (91), we can substitute $L_1=2^n,$ $M=q^{*}$ and $l_1=l_1^{*},l_2=l_2^{*},\dots ,l_M=l_M^{*}$ into (81). We obtain

$$\left|\sum _{\begin{array}{c}m=1\\ 2^{l_1^{*}}+\dots +2^{l_m^{*}}\ge 2^n\end{array}}^{q^{*}}{x}_{2^{l_1^{*}}+\dots +2^{l_m^{*}}}\right|<\epsilon ,$$

that is,

$$\left|\sum _{s\in B}{x}_s\right|<\epsilon ,$$

(101)

where

$$B=\left\{2^{l_1^{*}}+\dots +2^{l_m^{*}}:m\in \{1,\dots ,q^{*}\}\mathrm{such}\mathrm{that}{2}^{l_1^{*}}+\dots +2^{l_m^{*}}\ge 2^n\right\}.$$

(102)

Let us show that $B=A.$

Consider the case $j=1.$ By (87),

$$B=\left\{2^{k_1+n}+\dots +2^{k_m+n}:m\in \{1,\dots ,w\}\mathrm{such}\mathrm{that}{2}^{k_1+n}+\dots +2^{k_m+n}\ge 2^n\right\}.$$

Therefore, taking into account that $2^{k_1+n}+\dots +2^{k_m+n}\ge 2^n$ for every $m\in \{1,\dots ,w\},$

$$B=\left\{2^{k_1+n}+\dots +2^{k_m+n}:m\in \{1,\dots ,w\}\right\}.$$

Therefore, taking into account that $j-1=0,$ $B=A.$

Consider the case $j\in \{2,3,\dots ,2^n\}.$ By (88), $q^{*}=v+w$ and, taking into account (90), for $m\in \{1,\dots ,v+w\},$

$$\begin{array}{cc}\hfill 2^{l_1^{*}}+\dots +2^{l_m^{*}}& =\left\{\begin{array}{cc}{2}^{j_1}+\dots +2^{j_m},\hfill & \mathrm{if} m\in \{1,\dots ,v\},\hfill \\ 2^{j_1}+\dots +2^{j_v}+2^{k_1+n}+\dots +2^{k_{m-v}+n},\hfill & \mathrm{if} m\in \{v+1,\dots ,v+w\}\hfill \end{array}\right.\hfill \\ \hfill & =\left\{\begin{array}{cc}{2}^{j_1}+\dots +2^{j_m},\hfill & \mathrm{if} m\in \{1,\dots ,v\},\hfill \\ j-1+2^{k_1+n}+\dots +2^{k_{m-v}+n},\hfill & \mathrm{if} m\in \{v+1,\dots ,v+w\}.\hfill \end{array}\right.\hfill \end{array}$$

(103)

If $m\in \{1,\dots ,v\},$ then, by (103) and (90), taking into account the inequality $j\le 2^n,$

$$2^{l_1^{*}}+\dots +2^{l_m^{*}}=j-1+2^{j_1}+\dots +2^{j_m}\le 2^{j_1}+\dots +2^{j_v}=j-1<2^n.$$

If $m\in \{v+1,\dots ,v+w\},$ then, by (103),

$$2^{l_1^{*}}+\dots +2^{l_m^{*}}=j-1+2^{k_1+n}+\dots +2^{k_{m-v}+n}\ge 2^{k_1+n}\ge 2^n.$$

Thus, $2^{l_1^{*}}+\dots +2^{l_m^{*}}\ge 2^n$ if and only if $m\in \{v+1,\dots ,v+w\}.$ Therefore, by (102) and (103),

$$\begin{array}{cc}\hfill B& =\left\{2^{l_1^{*}}+\dots +2^{l_m^{*}}:m\in \{v+1,\dots ,v+w\}\right\}\hfill \\ & =\left\{j-1+2^{k_1+n}+\dots +2^{k_{m-v}+n}:m\in \{v+1,\dots ,v+w\}\right\}\hfill \\ & =\left\{j-1+2^{k_1+n}+2^{k_2+n}+\dots +2^{k_m+n}:m\in \{1,2,\dots ,w\}\right\}\hfill \\ & =A.\hfill \end{array}$$

So, $B=A.$ Therefore, by (84) and (101), the inequality (83) holds. Consequently, (80) holds. So, $y\in Y.$

By (62),

$$\iota \left(y\right)=\left(x_0^{*},x_1^{*},\dots \right)$$

where

$$x_i^{*}={\alpha }_{i+1}\left(y\right)$$

(104)

for $i\in {\mathbb{Z}}_{+}.$ Let us show that

$$x_i^{*}=x_i$$

for every $i\in {\mathbb{Z}}_{+}.$ In the case $i=0,$ by (104), (5), (79) and (7),

$$x_0^{*}={\alpha }_1\left(y\right)=y_1=\sum _{s\in {\mathsf{\Omega }}_0}{x}_s=\sum _{s\in \left\{0\right\}}{x}_s=x_0.$$

Consider the case $i\ge 1.$ By (104), (5), and (79),

$$x_i^{*}={\alpha }_{i+1}\left(y\right)=y_{i+1}-y_{i+1-2^{\varkappa (i+1)-1}}=\sum _{s\in {\mathsf{\Omega }}_i}{x}_s-\sum _{s\in {\mathsf{\Omega }}_{i-2^{\varkappa (i+1)-1}}}{x}_s.$$

(105)

Let $u\in \mathbb{N}$ and $i_1,i_2,\dots ,i_u\in {\mathbb{Z}}_{+}$ be such that $i_1<i_2<\dots <i_u$ and

$$i=2^{i_1}+2^{i_2}+\dots +2^{i_u}.$$

(106)

Then

$$i+1>i\ge 2^{i_u}$$

and, since $i<2^{i_u+1},$ it follows that

$$i+1\le 2^{i_u+1}.$$

Consequently, by (3),

$$\varkappa (i+1)=\lceil {log}_2(i+1)\rceil =i_u+1.$$

Therefore

$$2^{\varkappa (i+1)-1}=2^{i_u+1-1}=2^{i_u}.$$

(107)

By substituting $q=u,$ $p_1=i_1,p_2=i_2,\dots ,p_u=i_u$ and $q_0=u-1$ into Lemma 9, we obtain

$${\mathsf{\Omega }}_r\subset {\mathsf{\Omega }}_{2^{i_1}+2^{i_2}+\dots +2^{i_u}}$$

(108)

and

$${\mathsf{\Omega }}_{2^{i_1}+2^{i_2}+\dots +2^{i_u}}\setminus {\mathsf{\Omega }}_r=\left\{2^{i_1}+2^{i_2}+\dots +2^{i_u}\right\},$$

(109)

where

$$r=\left\{\begin{array}{cc}{2}^{i_1}+2^{i_2}+\dots +2^{i_{u-1}},\hfill & \mathrm{if} u\ge 2,\hfill \\ 0,\hfill & \mathrm{if} u=1.\hfill \end{array}\right.$$

(110)

By (106), (107), and (110), $r=i-2^{\varkappa (i+1)-1}.$ Therefore, by (106), (108), and (109),

$${\mathsf{\Omega }}_{i-2^{\varkappa (i+1)-1}}\subset {\mathsf{\Omega }}_i$$

and

$${\mathsf{\Omega }}_i\setminus {\mathsf{\Omega }}_{i-2^{\varkappa (i+1)-1}}=\left\{i\right\}.$$

Consequently,

$$\sum _{s\in {\mathsf{\Omega }}_i}{x}_s-\sum _{s\in {\mathsf{\Omega }}_{i-2^{\varkappa (i+1)-1}}}{x}_s=\sum _{s\in {\mathsf{\Omega }}_i\setminus {\mathsf{\Omega }}_{i-2^{\varkappa (i+1)-1}}}{x}_s=\sum _{s\in \left\{i\right\}}{x}_s=x_i.$$

Therefore, taking into account (105), $x_i^{*}=x_i.$ Consequently, $\iota \left(y\right)=x.$ This completes the proof. □

Lemma 12.

For every $y\in Y,$

$${\parallel \iota \left(y\right)\parallel }_{{\mathrm{cs}}_2}={\parallel y\parallel }_{\infty },$$

where ι is defined by (62).

Proof. 

Let $y=(y_1,y_2,\dots )\in Y.$ Let $x=\iota \left(y\right).$ By (62), $x=(x_0,x_1,\dots ),$ with

$$x_j={\alpha }_{j+1}\left(y\right)$$

for $j\in {\mathbb{Z}}_{+},$ where ${\alpha }_{j+1}$ is defined by (5). Therefore, by (13), taking into account Lemma 7,

$${\parallel x\parallel }_{{\mathrm{cs}}_2}=\underset{n\in {\mathbb{Z}}_{+}}{sup}\left|\sum _{m\in {\mathsf{\Omega }}_n}{x}_m\right|=\underset{n\in {\mathbb{Z}}_{+}}{sup}\left|\sum _{m\in {\mathsf{\Omega }}_n}{\alpha }_{m+1}\left(y\right)\right|=\underset{n\in {\mathbb{Z}}_{+}}{sup}|y_{n+1}|=\underset{n\in \mathbb{N}}{sup}|y_n{|=\parallel y\parallel }_{\infty }.$$

So, ${\parallel \iota \left(y\right)\parallel }_{{\mathrm{cs}}_2}={\parallel y\parallel }_{\infty }.$ □

Theorem 1.

The mapping $\iota ,$ defined by (62), is an isometrical isomorphism between normed spaces Y and ${\mathrm{cs}}_2.$

Proof. 

By Lemma 10, the mapping ι is well defined. By (62), taking into account the linearity of mappings ${\alpha }_j$ defined by (5), the mapping ι is linear. It can be verified that ι is linear. By Lemma 12, ι is isometrical. Since ι is linear and isometrical, it follows that ι is injective. By Lemma 11, ι is surjective. Thus, ι is a linear isometrical bijection, i.e., ι is an isometrical isomorphism. This completes the proof. □

Corollary 1.

The normed space ${\mathrm{cs}}_2$ is a Banach space.

Proof. 

Since Y is a Banach space and, by Theorem 1, ι is an isometrical isomorphism between Y and ${\mathrm{cs}}_2,$ it follows that ${\mathrm{cs}}_2$ is a Banach space. □

Lemma 13.

Let $n\in \mathbb{N},$ $s\in \{1,\dots ,2^n\}$ and $q\in \mathbb{N}$ be such that $q\ge n.$ Let the set $A_{n,s}$ be defined by (2). Then

1. 

$a+2^q\in A_{n,s}$ for every $a\in A_{n,s};$

2. 

If $b\in \mathbb{N}$ is such that $b\notin A_{n,s},$ then $b+2^q\notin A_{n,s}.$

Proof. 

By (2),

$$A_{n,s}=\left\{s+k{2}^n:k\in {\mathbb{Z}}_{+}\right\}.$$

(111)

Let us prove the first statement. Let $a\in A_{n,s}.$ Then, by (111), there exists $k\in {\mathbb{Z}}_{+}$ such that $a=s+k{2}^n.$ Therefore

$$a+2^q=s+k{2}^n+2^q=s+k{2}^n+2^{q-n}{2}^n=s+(k+2^{q-n})2^n=s+k_1{2}^n,$$

where $k_1=k+2^{q-n}.$ Therefore, since $k_1\in {\mathbb{Z}}_{+},$ by (111), $a+2^q\in A_{n,s}.$

Let us prove the second statement. Let $b\in \mathbb{N}$ be such that $b\notin A_{n,s}.$ Suppose $b+2^q\in A_{n,s}.$ Then, by (111), there exists $k\in {\mathbb{Z}}_{+}$ such that

$$b+2^q=s+k{2}^n.$$

(112)

Consider the case $b<s.$ In this case, taking into account that $s\in \{1,2,3,\dots ,2^n\}$ and $b\in \mathbb{N},$

$$0<s-b<2^n.$$

On the other hand, by (112), $s-b=2^q-k{2}^n$. Therefore

$$0<2^q-k{2}^n<2^n$$

and, consequently,

$$0<2^{q-n}-k<1$$

The latter inequality cannot hold since $2^{q-n}-k$ is integer. Therefore, the assumption $b<s$ leads to a contradiction.

Consider the case $b\ge s,$ i.e., $b-s\ge 0.$ By (112), $b-s=k{2}^n-2^q.$ Therefore

$$k{2}^n-2^q\ge 0,$$

i.e.,

$$k-2^{q-n}\ge 0.$$

(113)

By (112),

$$b=s+k{2}^n-2^q=s+(k-2^q)2^n=s+k_2{2}^n,$$

(114)

where $k_2=k-2^q.$ By (113), $k_2\ge 0,$ therefore $k_2\in {\mathbb{Z}}_{+}.$ Consequently, by (111) and (114), $b\in A_{n,s},$ which contradicts $b\notin A_{n,s}.$ So, $b+2^q\notin A_{n,s}.$ This completes the proof. □

Theorem 2.

The sequence ${\left\{e_m\right\}}_{m=1}^{\infty }$ is a Schauder basis of the space ${\mathrm{cs}}_2,$ where

$$e_m=(\underset{m-1}{\underbrace{0,\dots ,0}},1,0,\dots )$$

for $m\in \mathbb{N}.$

Proof. 

Since the sequence ${\left\{b_m\right\}}_{m=1}^{\infty }$ is a Schauder basis of the space Y and, by Theorem 1, ι is an isometrical isomorphism between Y and ${\mathrm{cs}}_2,$ it follows that the sequence ${\left\{\iota \left(b_m\right)\right\}}_{m=1}^{\infty }$ is a Schauder basis of the space ${\mathrm{cs}}_2.$ Let $m\in \mathbb{N}.$ Let us show that

$$\iota \left(b_m\right)=e_m.$$

(115)

By (62),

$$\iota \left(b_m\right)=({\alpha }_1\left(b_m\right),{\alpha }_2\left(b_m\right),\dots ).$$

So, to prove (115), it is enough to show that

$${\alpha }_j\left(b_m\right)=\left\{\begin{array}{cc}1,\hfill & \mathrm{if} j=m,\hfill \\ 0,\hfill & \mathrm{otherwise}\hfill \end{array}\right.$$

(116)

for every $j\in \mathbb{N}.$ Let us prove (116).

By (4), $b_m=1_{A_{\varkappa \left(m\right),m}},$ that is,

$$b_m=(y_1,y_2,\dots ),$$

(117)

where

$$y_i=\left\{\begin{array}{cc}1,\hfill & \mathrm{if} i\in A_{\varkappa \left(m\right),m},\hfill \\ 0,\hfill & \mathrm{otherwise}\hfill \end{array}\right.$$

(118)

for $i\in \mathbb{N}.$ By (2),

$$A_{\varkappa \left(m\right),m}=\left\{m+k{2}^{\varkappa \left(m\right)}:k\in {\mathbb{Z}}_{+}\right\}.$$

(119)

Consider the case $j=1.$ By (5), (117), and (118), and taking into account (119),

$${\alpha }_1\left(b_m\right)=y_1=\left\{\begin{array}{cc}1,\hfill & \mathrm{if}1\in A_{\varkappa \left(m\right),m},\hfill \\ 0,\hfill & \mathrm{otherwise}.\hfill \end{array}\right.=\left\{\begin{array}{cc}1,\hfill & \mathrm{if}m=1,\hfill \\ 0,\hfill & \mathrm{otherwise}.\hfill \end{array}\right.$$

So, in the case $j=1,$ the equality (116) holds.

Consider the case $j\ge 2.$ By (5) and (117),

$${\alpha }_j\left(b_m\right)=y_j-y_{j-2^{\varkappa \left(j\right)-1}},$$

(120)

where ϰ is defined by (3).

Consider the subcase $2\le j<m,$ which is possible only if $m\ge 3.$ Since $j-2^{\varkappa \left(j\right)-1}<j$ and $j<m,$ it follows that $j-2^{\varkappa \left(j\right)-1}<m.$ So, since $j<m$ and $j-2^{\varkappa \left(j\right)-1}<m,$ by (119), $j\notin A_{\varkappa \left(m\right),m}$ and $j-2^{\varkappa \left(j\right)-1}\notin A_{\varkappa \left(m\right),m}$ resp. Therefore, by (118), $y_j=0$ and $y_{j-2^{\varkappa \left(j\right)-1}}=0.$ Consequently, by (120), ${\alpha }_j\left(b_m\right)=0.$ So, in this subcase, (116) holds.

Consider the subcase $j=m$ and $m\ge 2.$ By (120),

$${\alpha }_m\left(b_m\right)=y_m-y_{m-2^{\varkappa \left(m\right)-1}}.$$

(121)

By (119), $m\in A_{\varkappa \left(m\right),m}.$ Therefore, by (118), $y_m=1.$ Since $m-2^{\varkappa \left(m\right)-1}<m,$ by (119), $m-2^{\varkappa \left(m\right)-1}\notin A_{\varkappa \left(m\right),m}.$ Therefore, by (118), $y_{m-2^{\varkappa \left(m\right)-1}}=0.$ So, by (121), ${\alpha }_m\left(b_m\right)=1.$ Thus, in this subcase, (116) holds.

Consider the subcase $m<j\le 2^{\varkappa \left(m\right)},$ which is possible only if $m<2^{\varkappa \left(m\right)},$ i.e., if m is not a power of $2.$ By (3), $\varkappa \left(j\right)=\varkappa \left(m\right).$ Therefore, by (120),

$${\alpha }_j\left(b_m\right)=y_j-y_{j-2^{\varkappa \left(m\right)-1}}.$$

(122)

By (119), taking into account the inequality $m<j\le 2^{\varkappa \left(m\right)},$ $j\notin A_{\varkappa \left(m\right),m}$ and, consequently, by (118), $y_j=0.$ Let us show that $j-2^{\varkappa \left(m\right)-1}\notin A_{\varkappa \left(m\right),m}.$ Suppose $j-2^{\varkappa \left(m\right)-1}\in A_{\varkappa \left(m\right),m}.$ Then, since $j-2^{\varkappa \left(m\right)-1}<j\le 2^{\varkappa \left(m\right)},$ by (119), $j-2^{\varkappa \left(m\right)-1}=m,$ i.e., $j=m+2^{\varkappa \left(m\right)-1}.$ Therefore, since $j\le 2^{\varkappa \left(m\right)},$

$$m+2^{\varkappa \left(m\right)-1}\le 2^{\varkappa \left(m\right)},$$

consequently,

$$m\le 2^{\varkappa \left(m\right)}-2^{\varkappa \left(m\right)-1}=2^{\varkappa \left(m\right)-1}.$$

Therefore, by (3),

$$\varkappa \left(m\right)\le \varkappa \left(2^{\varkappa \left(m\right)-1}\right)=\varkappa \left(m\right)-1.$$

A contradiction. Thus, $j-2^{\varkappa \left(m\right)-1}\notin A_{\varkappa \left(m\right),m}.$ Therefore, by (118), $y_{j-2^{\varkappa \left(m\right)-1}}=0.$ Thus, by (122), ${\alpha }_j\left(b_m\right)=0.$ So, in this subcase, (116) holds.

Consider the subcase $j>2^{\varkappa \left(m\right)}.$ By (3), $\varkappa \left(j\right)>\varkappa \left(m\right).$ Therefore, taking into account that both $\varkappa \left(j\right)$ and $\varkappa \left(m\right)$ are integers, $\varkappa \left(j\right)\ge \varkappa \left(m\right)+1,$ i.e.,

$$\varkappa \left(j\right)-1\ge \varkappa \left(m\right).$$

(123)

If $j-2^{\varkappa \left(j\right)-1}\in A_{\varkappa \left(m\right),m},$ by substituting $n=\varkappa \left(m\right),$ $s=m,$ $q=\varkappa \left(j\right)-1$ and $a=j-2^{\varkappa \left(j\right)-1}$ into Lemma 13 Item 1, taking into account (123), $j\in A_{\varkappa \left(m\right),m}.$ Consequently, by (118), $y_j=1$ and $y_{j-2^{\varkappa \left(j\right)-1}}=1.$ Therefore, by (116), ${\alpha }_j\left(b_m\right)=0.$ If $j-2^{\varkappa \left(j\right)-1}\notin A_{\varkappa \left(m\right),m},$ by substituting $n=\varkappa \left(m\right),$ $s=m,$ $q=\varkappa \left(j\right)-1$ and $b=j-2^{\varkappa \left(j\right)-1}$ into Lemma 13 Item 2, taking into account (123), $j\in A_{\varkappa \left(m\right),m}.$ Consequently, by (118), $y_j=0$ and $y_{j-2^{\varkappa \left(j\right)-1}}=0.$ Therefore, by (116), ${\alpha }_j\left(b_m\right)=0.$ So, in this subcase, (116) holds.

Thus, (116) holds in all cases. Therefore (115) holds. Thus, the sequence ${\left\{e_m\right\}}_{m=1}^{\infty }$ is a Schauder basis of the space ${\mathrm{cs}}_2.$ □

4. Discussion and Conclusions

In this work we construct some sequence normed space ${\mathrm{cs}}_2.$ We show that ${\mathrm{cs}}_2$ is isometrically isomorphic to the Banach space Y of all linear continuous functionals on ${\ell }_1$ that can be approximated by weakly symmetric linear continuous functionals. Using the isomorphism and the known Schauder basis of the space $Y,$ we show that ${\mathrm{cs}}_2$ possesses a standard Schauder basis. This result will, in future work, facilitate the construction of the dual space to Y since the problem of describing the dual space of ${\mathrm{cs}}_2$ is more straightforward than that of $Y.$ Such a construction will provide a framework for describing the spectrum of the algebra of entire functions of bounded type on the space ${\ell }_1,$ generated by continuous linear functionals approximable by continuous weakly symmetric functionals since every element of the dual space to Y generates some element of the spectrum of this algebra, and conversely, each element of the spectrum corresponds to an element of the dual space. Results of the work can be generalised to spaces of continuous polynomials on ${\ell }_1$ that can be approximated by continuous weakly symmetric polynomials.