Boundedness Index of Sequences: A Statistical Approach

Abstract

Statistical boundedness weakens ordinary boundedness by allowing it to fail on a set of density zero, and this notion can be graded by an order $\rho \in (0,1]$. For a real or complex sequence $x=\left(x_k\right)$, we introduce the statistical boundedness index $\eta \left(x\right)$, defined as the infimum of all orders $\rho \in (0,1]$ for which x is statistically bounded of order $\rho$; when no such order exists, x is called index-free. For each $r\in [0,1]$, we study the space $X_r$ of indexed sequences with $\eta \left(x\right)\le r$, and show that these spaces increase strictly with r, forming a continuous chain that interpolates between the bounded sequences and the statistically bounded sequences. The relationship between $X_r$ and $S{b}^r$ is also clarified, and we establish the main structural and duality properties of each $X_r$. These results yield a fine-grained classification of the asymptotic behaviour of real and complex sequences.

1. Introduction and Preliminaries

The notion of statistical convergence goes back to a remark of Zygmund [1] in the first edition of his monograph on trigonometric series (Warsaw, 1935). It was Fast [2] and Steinhaus [3] who, independently in 1951, turned this remark into a systematic theory. The density-based summability ideas that underlie the definition were later developed in a different direction by Schoenberg [4]. The idea is simply to weaken the requirement that $|x_k-\ell |$ be small along every tail of $\mathbb{N}$: one asks only that the exceptional indices have natural density zero. Ordinary convergence then corresponds to the case where the exceptional set is finite. Šalát [5] gave the notion a convenient set-theoretic form, proving that a sequence is statistically convergent to ℓ exactly when it has a subsequence converging to ℓ in the usual sense, indexed by a set of density one. This characterisation explains why statistical convergence belongs to the family of density-type convergence notions, rather than being an isolated weakening of ordinary convergence. Fridy [6] later obtained an intrinsic Cauchy criterion—statistical convergence is equivalent to the statistical Cauchy property—together with Tauberian conditions under which the notion can be strengthened back to ordinary convergence. Connor [7] compared statistical convergence with strong summability methods and showed that strong p-Cesàro summability, $0<p<\infty$, always implies statistical convergence, with the two notions coinciding for bounded sequences. This last equivalence was later extended by Saavedra et al. [8] to the setting of a general modulus function f.

The natural density of a subset K of natural numbers is defined by

$$\delta \left(K\right)=\underset{n\to \infty }{\lim }{\displaystyle \frac{1}{n}}\mathcal{C}\left\{k\le n:k\in K\right\}$$

when the limit exists, where $\mathcal{C}\{·\}$ denotes the cardinality of $\{·\}$.

A sequence $x=\left(x_k\right)$ is said to be statistically convergent to ℓ provided that

$$\underset{n\to \infty }{\lim }{\displaystyle \frac{1}{n}}\mathcal{C}\left\{k\le n:\left|x_k-\ell \right|\ge \epsilon \right\}=0$$

for each $\epsilon >0$. We write this by $st-\lim x_k=\ell$. If $\ell =0$ then we call x a statistically null sequence.

Once density zero is accepted as the right notion of negligibility, it is natural to ask how finely it can be graded. An exponent-based grading was proposed by Gadjiev and Orhan [9], motivated by Korovkin-type approximation theorems. In such problems the rate at which the density of the set $\{k\le n:|x_k-\ell |\ge \epsilon \}$ decays to zero is what matters, and statistical convergence without an order parameter is too coarse a tool to detect this rate. Replacing the normalizing factor n by $n^{\rho }$, $\rho \in (0,1]$, yields statistical convergence of order $\rho$. This is strictly stronger for smaller $\rho$, and it reduces to ordinary statistical convergence when $\rho =1$. Shortly afterwards, Bhunia, Das and Pal [10] and Çolak [11] independently rediscovered the same refinement.

A sequence $x=\left(x_k\right)$ is statistically convergent to ℓ of order $\rho$ provided that

$$\underset{n\to \infty }{\lim }{\displaystyle \frac{1}{n^{\rho }}}\mathcal{C}\left\{k\le n:\left|x_k-\ell \right|\ge \epsilon \right\}=0$$

for each $\epsilon >0$.

An analogous refinement was later carried out for boundedness rather than convergence. The concept of statistical boundedness first appeared in the paper of Fridy and Orhan [12], as a natural counterpart of finiteness of the statistical limit superior and limit inferior of a sequence:

A sequence $x=\left(x_k\right)$ is said to be statistically bounded if there exists a number $L\ge 0$ such that $\delta \left(\{k:|x_k|>L\}\right)=0$.

Statistical boundedness and its variants have been studied further in recent years [13,14].

Bhardwaj and Gupta [15] graded this notion by order in the same way that Gadjiev–Orhan [9], Bhunia–Das–Pal [10], and Çolak [11] had graded statistical convergence, replacing n by $n^{\rho }$ in the underlying density computation.

A sequence $x=\left(x_k\right)$ is said to be statistically bounded of order $\rho$ if there exists a number $L\ge 0$ so that

$$\underset{n\to \infty }{\lim }{\displaystyle \frac{1}{n^{\rho }}}\mathcal{C}\left\{k\le n:\left|x_k\right|>L\right\}=0$$

holds where $\rho \in \left(0,1\right]$. We denote by $S{b}^{\rho }$ the space of all sequences that are statistically bounded of order $\rho$.

The statistical boundedness index introduced in Section 2 is a scalar invariant that records, for a given sequence x, its threshold order: the infimum of the orders at which x is statistically bounded. The term index for this invariant was inspired by the numerical index of a Banach space [16]; no mathematical relationship between the two notions is implied. Before proceeding, we collect some standard notions from the theory of sequence spaces. We refer to [17] for a detailed treatment. Throughout, $\omega$ denotes the space of all real (or complex) valued sequences.

Definition 1.

Let X be a sequence space, and let χ denote the space of all sequences containing only 0 or 1. The sequence space X is said to be:

• Solid (or normal): if

  $$\tilde{X}=\left\{\left(u_k\right)\in \omega :\exists \left(v_k\right)\in X\mathit{such}\mathit{that}|u_k|\le |v_k|\forall k\in \mathbb{N}\right\}\subset X.$$
• Monotone: if $uv\in X$ for all $u\in \chi$ and $v\in X$, where $uv=\left(u_k{v}_k\right)$ is the coordinatewise multiplication.
• A sequence algebra: if it is closed under coordinatewise multiplication, meaning $uv=\left(u_i{v}_i\right)\in X$ whenever $u,v\in X$.
• Symmetric: if $\left(u_k\right)\in X$ implies $\left(u_{\pi \left(k\right)}\right)\in X$ for any permutation π on $\mathbb{N}$.

For a sequence space X, the $\alpha$-dual, $\beta$-dual, and $\gamma$-dual of X are defined, respectively, by

$$X^{\alpha }=\left\{u=\left(u_k\right):\sum _k\left|u_k{x}_k\right|<\infty \mathit{for}\mathit{all}x\in X\right\},$$

$$X^{\beta }=\left\{u=\left(u_k\right):\sum _k{u}_k{x}_k\mathit{converges}\mathit{for}\mathit{all}x\in X\right\},$$

$$X^{\gamma }=\left\{u=\left(u_k\right):\underset{n}{\sup }\left|{\displaystyle \sum _{k=1}^n}{u}_k{x}_k\right|<\infty \mathit{for}\mathit{all}x\in X\right\}.$$

2. Main Results

In this section we introduce the statistical boundedness index and study its main properties. Given a sequence x, Definition 2 collects the orders $\rho \in (0,1]$ at which x is statistically bounded into a set $I_x$, and takes $\eta \left(x\right)$ to be its infimum; if $I_x$ is empty, x is called index-free. Lemma 1 shows that $I_x$ is an interval with right endpoint 1, so the index is well defined whenever it exists. The index behaves well: it is subadditive (Corollary 1, from Proposition 1), invariant under nonzero scalar multiplication (Proposition 2), and vanishes on every bounded sequence, though not only on those (Proposition 3).

These properties allow us to attach to each $r\in [0,1]$ the space $X_r$ of indexed sequences with $\eta \left(x\right)\le r$. Theorem 1 shows that $X_r$ is a sequence space, Lemma 2 identifies $X_1$ with $Sb$, and Theorem 2 shows that the family is strictly increasing in r; together, Theorem 3 assembles these facts into the chain

$${\ell }_{\infty }\subset X_0\subset X_{r_1}\subset X_{r_2}\subset X_1=Sb,0\le r_1<r_2\le 1,$$

interpolating between boundedness and statistical boundedness. Remark 1 and Theorem 4 also settle the relation with the classical order-$\rho$ spaces $S{b}^{\rho }$: the two families agree only at $r=1$, and $S{b}^r$ is strictly contained in $X_r$ for every $r\in (0,1)$. The section ends with the structural theory of $X_r$. Theorem 5 establishes that each $X_r$ is solid (hence monotone) for every $r\in [0,1]$, and a sequence algebra but not symmetric for every $r\in [0,1)$, with the endpoint $r=1$ following separately from known properties of $Sb$. Theorem 6 computes the $\alpha$-, $\beta$-, and $\gamma$-duals of $X_r$, all of which turn out to be the space $\phi$ of finitely nonzero sequences.

2.1. The Statistical Boundedness Index $\eta$

Definition 2.

Let $x\in \omega$. The statistical boundedness index set of x is defined by

$$I_x=\left\{\rho \in (0,1]:\mathit{there}\mathit{exists}L\ge 0\mathit{such}\mathit{that}\underset{n\to \infty }{\lim }{\displaystyle \frac{1}{n^{\rho }}}\mathcal{C}\left\{k\le n:\left|x_k\right|>L\right\}=0\right\}.$$

A sequence $x=\left(x_k\right)$ is said to be indexed if $I_x\ne \varnothing$, and index-free otherwise. The number $\eta \left(x\right)=\inf I_x$ is called the statistical boundedness index of x.

We write ${\mathcal{I}}_{\mathrm{ind}}$ for the class of all indexed sequences and ${\mathcal{I}}_{\mathrm{free}}$ for the class of all index-free sequences. These two classes partition $\omega$.

Before turning to examples in ${\mathcal{I}}_{\mathrm{ind}}$ and ${\mathcal{I}}_{\mathrm{free}}$, we record a lemma that will simplify several arguments below.

Lemma 1.

Let $x=\left(x_k\right)\in \omega$. If $\rho \in I_x$ and $\rho \le {\rho }^{\prime }\le 1$, then ${\rho }^{\prime }\in I_x$. Hence, if $I_x\ne \varnothing$, the index set is an interval with right endpoint 1: it equals $\left[\eta \right(x),1]$ when the infimum is attained, and $\left(\eta \right(x),1]$ otherwise. In particular, $1\in I_x$ for every $x\in {\mathcal{I}}_{\mathrm{ind}}$.

Proof. 

Since $\rho \in I_x$, there is $L\ge 0$ such that

$$\underset{n\to \infty }{\lim }{\displaystyle \frac{1}{n^{\rho }}}\mathcal{C}\{k\le n:|x_k|>L\}=0.$$

Fix ${\rho }^{\prime }$ with $\rho \le {\rho }^{\prime }\le 1$. For every $n\in \mathbb{N}$,

$$0\le {\displaystyle \frac{1}{n^{{\rho }^{\prime }}}}\mathcal{C}\{k\le n:|x_k|>L\}\le {\displaystyle \frac{1}{n^{\rho }}}\mathcal{C}\{k\le n:|x_k|>L\}.$$

Taking the limit as $n\to \infty$, we obtain ${\rho }^{\prime }\in I_x$. This proves the first assertion.

For the second, note that $I_x\subset (0,1]$ by definition, so taking ${\rho }^{\prime }=1$ above shows $1\in I_x$ whenever $I_x\ne \varnothing$. The first assertion also implies that $I_x$ contains every $\rho$ between any of its elements and 1. Together with $1\in I_x$, this means that $I_x$ is precisely the set of all $\rho \in (0,1]$ with $\rho \ge \eta \left(x\right)$, that is, $\left[\eta \right(x),1]$ or $\left(\eta \right(x),1]$, according as the infimum $\eta \left(x\right)=\inf I_x$ is attained or not.

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Example 1.

Define $x=\left(x_k\right)$ by

$$x_k=\left\{\begin{array}{cc}\sqrt{k},& k\mathit{is}\mathit{a}\mathit{square},\\ 0,& k\mathit{is}\mathit{an}\mathit{odd}\mathit{non}\text{-}\mathit{square},\\ 1,& \mathit{otherwise}.\end{array}\right.$$

Indeed, the inequality

$${\displaystyle \frac{\mathcal{C}\left\{k\le n:\left|x_k\right|>1\right\}}{n^{\rho }}}\le {\displaystyle \frac{\sqrt{n}-1}{n^{\rho }}}$$

shows that x is statistically bounded of order ρ for every $\rho \in \left({\displaystyle \frac{1}{2}},1\right]$. On the other hand,

$$\underset{n\to \infty }{\lim }{\displaystyle \frac{\mathcal{C}\left\{k\le n:\left|x_k\right|>L\right\}}{n^{\rho }}}\ne 0$$

for all $L\ge 0$ and $\rho \in \left(0,{\displaystyle \frac{1}{2}}\right]$ (see, e.g., [18] (Thm. 2)). Hence $I_x=\left({\displaystyle \frac{1}{2}},1\right]$, $\eta \left(x\right)={\displaystyle \frac{1}{2}}$, and $x\in {\mathcal{I}}_{\mathrm{ind}}$.

Example 2.

Take $x_k=\ln k$. For any $L\ge 0$,

$$\mathcal{C}\{k\le n:|x_k|>L\}=n-\lfloor e^L\rfloor ,$$

and therefore

$$\underset{n\to \infty }{\lim }{\displaystyle \frac{\mathcal{C}\{k\le n:|x_k|>L\}}{n}}=\underset{n\to \infty }{\lim }{\displaystyle \frac{n-\lfloor e^L\rfloor }{n}}=1,$$

where $\lfloor ·\rfloor$ denotes the greatest integer function. Consequently, x is not statistically bounded, so $x\notin S{b}^{\rho }$ for any $\rho \in (0,1]$. Hence $I_x=\varnothing$ and $x\in {\mathcal{I}}_{\mathrm{free}}$.

Proposition 1.

If $x=\left(x_k\right)$, $y=\left(y_k\right)\in {\mathcal{I}}_{\mathrm{ind}}$ then $x+y\in {\mathcal{I}}_{\mathrm{ind}}$ and $\eta (x+y)\le \max \left\{\eta \left(x\right),\eta \left(y\right)\right\}$. For non-negative sequences $\eta (x+y)=\max \left\{\eta \left(x\right),\eta \left(y\right)\right\}$.

Proof. 

We start by proving the first assertion. Let $\epsilon >0$ and set $d=\max \left\{\eta \right(x),\eta (y\left)\right\}$. By definition of the infimum there exist ${\rho }_1\in I_x$ and ${\rho }_2\in I_y$, with corresponding numbers $L_1,L_2\ge 0$, such that ${\rho }_1<\eta \left(x\right)+\epsilon$ and ${\rho }_2<\eta \left(y\right)+\epsilon$. Put $\sigma =\max \{{\rho }_1,{\rho }_2\}$; since ${\rho }_1,{\rho }_2\in (0,1]$ we have $\sigma \in (0,1]$ as well, and $\sigma <d+\epsilon$. Because ${\rho }_1\le \sigma \le 1$ and ${\rho }_1\in I_x$, Lemma 1 yields $\sigma \in I_x$ and $\sigma \in I_y$. Hence

$$\underset{n\to \infty }{\lim }{\displaystyle \frac{\mathcal{C}\left\{k\le n:|x_k|>L_1\right\}}{n^{\sigma }}}=0,\underset{n\to \infty }{\lim }{\displaystyle \frac{\mathcal{C}\left\{k\le n:|y_k|>L_2\right\}}{n^{\sigma }}}=0.$$

Since

$${\displaystyle \frac{\mathcal{C}\left\{k\le n:|x_k+y_k|>L_1+L_2\right\}}{n^{\sigma }}}\le {\displaystyle \frac{\mathcal{C}\left\{k\le n:|x_k|>L_1\right\}}{n^{\sigma }}}+{\displaystyle \frac{\mathcal{C}\left\{k\le n:|y_k|>L_2\right\}}{n^{\sigma }}},$$

taking the limit as $n\to \infty$ gives

$$\underset{n\to \infty }{\lim }{\displaystyle \frac{\mathcal{C}\left\{k\le n:|x_k+y_k|>L_1+L_2\right\}}{n^{\sigma }}}=0,$$

so $\sigma \in I_{x+y}$ and $\eta (x+y)\le \sigma <d+\epsilon$. Since $\epsilon >0$ was arbitrary, $\eta (x+y)\le d=\max \left\{\eta \right(x),\eta (y\left)\right\}$, and in particular $x+y\in {\mathcal{I}}_{\mathrm{ind}}$.

Now, let x and y be non-negative. We show that $I_{x+y}\subseteq I_x$. Suppose $\rho \notin I_x$; then

$$\underset{n\to \infty }{\lim }{\displaystyle \frac{\mathcal{C}\left\{k\le n:\left|x_k\right|>L\right\}}{n^{\rho }}}\ne 0$$

for all $L\ge 0$. Since $\left\{k\le n:\left|x_k\right|>L\right\}\subseteq \left\{k\le n:\left|x_k+y_k\right|>L\right\}$, we get

$$\underset{n\to \infty }{\lim }{\displaystyle \frac{\mathcal{C}\left\{k\le n:\left|x_k+y_k\right|>L\right\}}{n^{\rho }}}\ne 0$$

for all $L\ge 0$, so $\rho \notin I_{x+y}$. Thus, $I_{x+y}\subseteq I_x$ and $\eta \left(x\right)\le \eta (x+y)$. We also have $\eta \left(y\right)\le \eta (x+y)$ by the same argument. Therefore $\max \left\{\eta \left(x\right),\eta \left(y\right)\right\}\le \eta (x+y)$. Combining this with the first part, we obtain $\eta (x+y)=\max \left\{\eta \left(x\right),\eta \left(y\right)\right\}$ for non-negative sequences.

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Corollary 1.

The index function η is subadditive.

Proposition 2.

The statistical boundedness index is invariant under nonzero scalar multiplication, that is, for any nonzero scalar c and $x=\left(x_k\right)\in {\mathcal{I}}_{\mathit{ind}}$, $\eta \left(cx\right)=\eta \left(x\right)$.

Proof. 

The index set of $cx$ is

$$I_{cx}=\left\{\rho \in (0,1]:\exists L\ge 0,\underset{n\to \infty }{\lim }{\displaystyle \frac{1}{n^{\rho }}}\mathcal{C}\left\{k\le n:|c{x}_k|>L\right\}=0\right\}.$$

Since $c\ne 0$ and $|c{x}_k|>L\iff |x_k|>{\displaystyle \frac{L}{\left|c\right|}}$, by setting $L^{\prime }={\displaystyle \frac{L}{\left|c\right|}}$, which is non-negative, we have

$$\mathcal{C}\left\{k\le n:|c{x}_k|>L\right\}=\mathcal{C}\left\{k\le n:|x_k|>L^{\prime }\right\}$$

for each $n\in \mathbb{N}$. Hence $I_{cx}=I_x$. In particular, $\eta \left(cx\right)=\eta \left(x\right)$.

□

We note that the hypothesis $c\ne 0$ cannot be dropped: taking x as in Example 1, where $\eta \left(x\right)=\frac{1}{2}$, we have $\eta (0·x)=\eta \left(0\right)=0\ne \frac{1}{2}=\eta \left(x\right)$ by Proposition 3.

Proposition 3.

Every bounded sequence is indexed with an index of 0, i.e., $\eta \left(x\right)=0$. The converse is not true, in general.

Proof. 

The first assertion is immediate: every bounded sequence is statistically bounded of order $\rho$ for every $\rho \in (0,1]$.

For the converse, consider the unbounded sequence $x=\left(x_k\right)$ defined by

$$x_k=\left\{\begin{array}{cc}k,\hfill & \mathrm{if}k=2^m\mathrm{for}\mathrm{some}m\in \mathbb{N},\hfill \\ 0,\hfill & \mathrm{otherwise}.\hfill \end{array}\right.$$

Taking $L=1$, we have

$$\mathcal{C}\left\{k\le n:|x_k|>1\right\}=\lfloor {\log }_{2}n\rfloor$$

for each $n\in \mathbb{N}$. Hence, for every $\rho \in (0,1]$,

$$\underset{n\to \infty }{\lim }{\displaystyle \frac{\mathcal{C}\left\{k\le n:|x_k|>1\right\}}{n^{\rho }}}\le \underset{n\to \infty }{\lim }{\displaystyle \frac{{\log }_{2}n}{n^{\rho }}}=0.$$

Therefore $x\in S{b}^{\rho }$ for all $\rho \in (0,1]$, so that $I_x=(0,1]$ and $\eta \left(x\right)=\inf I_x=0$.

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2.2. The Index Spaces $X_r$ and Their Structure

In this subsection, we introduce a family of sequence spaces $X_r$ built from the statistical boundedness index. This family yields a fine-grained chain interpolating between the space of bounded sequences ${\ell }_{\infty }$ and the space of statistically bounded sequences $Sb$.

For $r\in [0,1]$, we define

$$X_r=\{x\in {\mathcal{I}}_{\mathrm{ind}}:\eta \left(x\right)\le r\}.$$

The following theorem shows that $X_r$ is a sequence space.

Theorem 1.

$X_r$ is a sequence space.

Proof. 

Let $x=\left(x_k\right),y=\left(y_k\right)\in X_r$ and let c be a scalar. By Proposition 1, we have $\eta (x+y)\le \max \left\{\eta \right(x),\eta (y\left)\right\}\le r$, which yields $x+y\in X_r$. If $c=0$, then $cx$ is the zero sequence, which is bounded and hence satisfies $\eta \left(cx\right)=0\le r$ by Proposition 3, so that $cx\in X_r$. If $c\ne 0$, then $\eta \left(cx\right)=\eta \left(x\right)\le r$ by Proposition 2, so again $cx\in X_r$. Hence $X_r$ is closed under addition and scalar multiplication, so it is a sequence space. □

Lemma 2.

$Sb=X_1$.

Proof. 

If $x\in Sb(=S{b}^1)$, then $1\in I_x$ by definition, so x is indexed and $\eta \left(x\right)=\inf I_x\le 1$, i.e. $x\in X_1$. Conversely, if $x\in X_1$, then $x\in {\mathcal{I}}_{\mathrm{ind}}$, so $I_x\ne \varnothing$; picking any $\rho \in I_x$, Lemma 1 with ${\rho }^{\prime }=1$ gives $1\in I_x$, so $x\in S{b}^1=Sb$. Hence $Sb=X_1$. □

Theorem 2.

Let $r_1,r_2\in [0,1]$ with $r_1<r_2$. Then $X_{r_1}{X}_{r_2}$.

Proof. 

The inclusion $X_{r_1}\subset X_{r_2}$ follows immediately: if $x\in X_{r_1}$, then $\eta \left(x\right)\le r_1<r_2$, so $x\in X_{r_2}$.

To see that the inclusion is strict, fix $r_1<r_2$ in $[0,1)$ and consider the sequence $x=\left(x_k\right)$ given by

$$x_k=\left\{\begin{array}{cc}k,\hfill & \mathrm{if}k=\lfloor m^{1/r_2}\rfloor \mathrm{for}\mathrm{some}m\in \mathbb{N},\hfill \\ 0,\hfill & \mathrm{otherwise}.\hfill \end{array}\right.$$

We observe that for every $n\in \mathbb{N}$,

$$\mathcal{C}\left\{k\le n:|x_k|>\frac{1}{2}\right\}=\lceil {(n+1)}^{r_2}\rceil -1,$$

where $\lceil ·\rceil$ denotes the least integer function. It follows from the defining property of $\lceil ·\rceil$ that

$${(n+1)}^{r_2}-1\le \lceil {(n+1)}^{r_2}\rceil -1<{(n+1)}^{r_2},$$

and therefore

$${\displaystyle \frac{{(n+1)}^{r_2}-1}{n^{\rho }}}\le {\displaystyle \frac{\mathcal{C}\left\{k\le n:|x_k|>\frac{1}{2}\right\}}{n^{\rho }}}<{\displaystyle \frac{{(n+1)}^{r_2}}{n^{\rho }}}.$$

Taking the limit as $n\to \infty$,

$$\underset{n\to \infty }{\lim }{\displaystyle \frac{\mathcal{C}\left\{k\le n:|x_k|>\frac{1}{2}\right\}}{n^{\rho }}}=0$$

exactly when $r_2<\rho \le 1$, so $I_x=(r_2,1]$. Therefore $\eta \left(x\right)=r_2$, and $x\in X_{r_2}$. Since $\eta \left(x\right)=r_2>r_1$, however, $x\notin X_{r_1}$.

To see that $X_r\subset X_1$ is strict for every $r\in [0,1)$, we use the equality $Sb=X_1$ established in Lemma 2. Consider the sequence $y=\left(y_k\right)$ given by

$$y_k=\left\{\begin{array}{cc}k,\hfill & k\in A,\hfill \\ 0,\hfill & \mathrm{otherwise},\hfill \end{array}\right.$$

where $A=\{\lfloor m\ln m\rfloor :m\in \mathbb{N},m\ge 2\}$. Writing $A_n=\{k\le n:k\in A\}$, for every $n\in \mathbb{N}$ we have $\{k\le n:|y_k|>\frac{1}{3}\}=A_n$, and the cardinality of $A_n$ is

$$\mathcal{C}{A}_n=\mathcal{C}\left\{m\ge 2:m\ln m<n+1\right\}.$$

By a logarithmic computation, the inequality

$${\displaystyle \frac{n+1}{2\ln (n+1)}}-2<\mathcal{C}{A}_n<{\displaystyle \frac{2(n+1)}{\ln (n+1)}}$$

(1)

holds for all sufficiently large n. Dividing by n and letting $n\to \infty$,

$$\underset{n\to \infty }{\lim }{\displaystyle \frac{\mathcal{C}\left\{k\le n:|y_k|>\frac{1}{3}\right\}}{n}}=\underset{n\to \infty }{\lim }{\displaystyle \frac{\mathcal{C}{A}_n}{n}}\le \underset{n\to \infty }{\lim }{\displaystyle \frac{2(n+1)}{n\ln (n+1)}}=0,$$

so $y\in Sb=X_1$.

Conversely, for any given threshold $L>0$, the set $\{k\le n:|y_k|>L\}$ is cofinite in $A_n$. Thus, we can write

$$\mathcal{C}\left\{k\le n:|y_k|>L\right\}=\mathcal{C}{A}_n-c\left(L\right),$$

where $c\left(L\right)$ is a non-negative integer constant depending only on L. Now, for an arbitrary $\rho \in (0,1)$, the lower bound established in (1) gives

$$\underset{n\to \infty }{\lim }{\displaystyle \frac{\mathcal{C}\left\{k\le n:|y_k|>L\right\}}{n^{\rho }}}=\underset{n\to \infty }{\lim }{\displaystyle \frac{\mathcal{C}{A}_n}{n^{\rho }}}\ge \underset{n\to \infty }{\lim }{\displaystyle \frac{1}{n^{\rho }}}\left({\displaystyle \frac{n+1}{2\ln (n+1)}}-2\right)=\infty .$$

Therefore, $y\notin S{b}^{\rho }$ for every $\rho \in (0,1)$. Hence $I_y=\left\{1\right\}$ and $\eta \left(y\right)=1$, and in particular $y\notin X_r$ for all $r\in [0,1)$. This completes the proof.

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Theorem 3.

For any $r_1,r_2\in [0,1]$ with $r_1<r_2$,

$${\ell }_{\infty }\subset X_0\subset X_{r_1}\subset X_{r_2}\subset X_1=Sb.$$

Proof. 

The inclusion ${\ell }_{\infty }\subset X_0$ follows from Proposition 3: every bounded sequence has index 0, hence belongs to $X_0$. The inclusions $X_0\subset X_{r_1}$ and $X_{r_1}\subset X_{r_2}$ follow from Theorem 2, and the equality $X_1=Sb$ is given in Lemma 2. □

This chain shows that the spaces $X_r$ classify the asymptotic behavior of sequences lying strictly between ordinary boundedness and statistical boundedness.

Remark 1.

At first glance, one might expect the space $S{b}^r$ and the statistical boundedness index space $X_r$ to coincide, since both are built on the threshold of order r. This is true for $r=1$, where $Sb=X_1$, but the two spaces are distinct for every other r. The distinction lies at the endpoint of the index set: a sequence belongs to $S{b}^r$ if the density vanishes at order $\rho =r$, while membership in $X_r$ only requires the index to be at most r, even if r itself is excluded from the index set. Thus $X_r$ contains a wider class of sequences than $S{b}^r$. This strict inclusion is established in the following theorem.

Theorem 4.

Let $r\in (0,1)$ be fixed. Then the inclusion $S{b}^r\subset X_r$ strictly holds.

Proof. 

Fix $r\in (0,1)$. If $x\in S{b}^r$, then by the definition of the index set, $r\in I_x$, which implies $\eta \left(x\right)=\inf I_x\le r$ and hence $x\in X_r$. Thus, $S{b}^r\subset X_r$.

Replacing $r_2$ with r in the definition of the sequence $x=\left(x_k\right)$ given in Theorem 2 yields

$$x_k=\left\{\begin{array}{cc}k,\hfill & \mathrm{if}k=\lfloor m^{1/r}\rfloor \mathrm{for}\mathrm{some}m\in \mathbb{N},\hfill \\ 0,\hfill & \mathrm{otherwise}.\hfill \end{array}\right.$$

For this sequence, the same computation as in Theorem 2 gives $I_x=(r,1]$ and $\eta \left(x\right)=r$, so $x\in X_r$; since $r\notin I_x$, however, $x\notin S{b}^r$. Hence $x\in X_r\setminus S{b}^r$, which shows that the inclusion $S{b}^r\subset X_r$ is strict. □

We would like to point out that since $X_r$ strictly contains $S{b}^r$ for every $r\in [0,1)$ (Theorem 4), the structural results established below do not follow automatically from the corresponding facts about $S{b}^r$: $x\in X_r$ is determined by the index $\eta \left(x\right)\le r$ rather than by a single fixed order, so the sequences needed to witness these results are necessarily more delicate than those sufficient for $S{b}^{\rho }$. This also clarifies the role of the index $\eta$: rather than working with the parametrized family ${\left\{S{b}^{\rho }\right\}}_{\rho \in (0,1]}$, it organizes the underlying sequences into the single continuous chain of Theorem 3.

As an immediate consequence of Theorem 4 and Theorem 2, together with Lemma 1, the chain $\left\{X_r\right\}$ is not the only way to produce a space strictly between $X_{r_1}$ and $X_{r_2}$: for $r_1<r_2$, the classical spaces $S{b}^{\rho }$ already do this.

Corollary 2.

Let $r_1,r_2\in [0,1]$ with $r_1<r_2$, and let $\rho \in (r_1,r_2)$. Then

$$X_{r_1}⊊S{b}^{\rho }⊊X_{r_2}.$$

Proof. 

The right-hand inclusion follows from $S{b}^{\rho }⊊X_{\rho }$ (Theorem 4) together with $X_{\rho }⊊X_{r_2}$ (Theorem 2, since $\rho <r_2$). For the left-hand inclusion, if $x\in X_{r_1}$ then $\eta \left(x\right)\le r_1<\rho$, so $\rho \in I_x$ by Lemma 1, and hence $x\in S{b}^{\rho }$; this shows $X_{r_1}\subseteq S{b}^{\rho }$. To see that the inclusion is strict, let $\sigma =(r_1+\rho )/2\in (r_1,\rho )$ and let x be a sequence with $\eta \left(x\right)=\sigma$, constructed as in the proof of Theorem 2. Since $\sigma <\rho$, Lemma 1 gives $x\in S{b}^{\rho }$; but $\eta \left(x\right)=\sigma >r_1$, so $x\notin X_{r_1}$. Thus $X_{r_1}⊊S{b}^{\rho }⊊X_{r_2}$ for every $\rho \in (r_1,r_2)$. □

In the following theorem, we investigate the structural behavior of the space $X_r$ with respect to solidness, symmetry, and its character as a sequence algebra.

Theorem 5.

Let $r\in [0,1]$. Then:

(i) 

$X_r$ is solid and so monotone;

(ii) 

if $r\in [0,1)$, $X_r$ is a sequence algebra;

(iii) 

if $r\in [0,1)$, $X_r$ is not symmetric.

Proof. 

(i)

Let $y=\left(y_k\right)\in {\tilde{X}}_r$. By definition, there is $x=\left(x_k\right)\in X_r$ with $|y_k|\le |x_k|$ for every k. Fix $\rho \in I_x$. By definition of $I_x$, there is $L\ge 0$ such that

$$\underset{n\to \infty }{\lim }{\displaystyle \frac{\mathcal{C}\left\{k\le n:|x_k|>L\right\}}{n^{\rho }}}=0.$$

Since $|y_k|\le |x_k|$, we have $\{k\le n:|y_k|>L\}\subseteq \{k\le n:|x_k|>L\}$, and therefore

$$\underset{n\to \infty }{\lim }{\displaystyle \frac{\mathcal{C}\left\{k\le n:|y_k|>L\right\}}{n^{\rho }}}=0.$$

Hence $\rho \in I_y$. Since $\rho \in I_x$ was arbitrary, $I_x\subseteq I_y$, so $\eta \left(y\right)\le \eta \left(x\right)\le r$. Thus $y\in X_r$, and ${\tilde{X}}_r\subseteq X_r$. Therefore $X_r$ is solid. Since solid spaces are monotone, $X_r$ is monotone as well.

(ii)

Let $r\in [0,1)$, and let $x=\left(x_k\right),y=\left(y_k\right)\in X_r$, so $\eta \left(x\right)\le r$ and $\eta \left(y\right)\le r$. Let $\epsilon >0$ be arbitrary with $r+\epsilon \le 1$. By definition of the infimum there exist ${\rho }_1\in I_x$ and ${\rho }_2\in I_y$, with corresponding numbers $L_1,L_2\ge 0$, such that ${\rho }_1<\eta \left(x\right)+\epsilon \le r+\epsilon$ and ${\rho }_2<\eta \left(y\right)+\epsilon \le r+\epsilon$. Put $\sigma =\max \{{\rho }_1,{\rho }_2\}$; then $\sigma <r+\epsilon \le 1$, and since ${\rho }_1\le \sigma \le 1$ and ${\rho }_2\le \sigma \le 1$, Lemma 1 gives $\sigma \in I_x$ and $\sigma \in I_y$, with the same $L_1$ and $L_2$. Thus

$$\underset{n\to \infty }{\lim }{\displaystyle \frac{\mathcal{C}\left\{k\le n:|x_k|>L_1\right\}}{n^{\sigma }}}=0\mathrm{and}\underset{n\to \infty }{\lim }{\displaystyle \frac{\mathcal{C}\left\{k\le n:|y_k|>L_2\right\}}{n^{\sigma }}}=0.$$

Since

$$\{k\le n:|x_k{y}_k|>L_1{L}_2\}\subseteq \{k\le n:|x_k|>L_1\}\cup \{k\le n:|y_k|>L_2\},$$

we obtain, for every $n\in \mathbb{N}$,

$${\displaystyle \frac{\mathcal{C}\left\{k\le n:|x_k{y}_k|>L_1{L}_2\right\}}{n^{\sigma }}}\le {\displaystyle \frac{\mathcal{C}\left\{k\le n:|x_k|>L_1\right\}}{n^{\sigma }}}+{\displaystyle \frac{\mathcal{C}\left\{k\le n:|y_k|>L_2\right\}}{n^{\sigma }}}.$$

Taking the limit as $n\to \infty$ and using the two limits established above,

$$\underset{n\to \infty }{\lim }{\displaystyle \frac{\mathcal{C}\left\{k\le n:|x_k{y}_k|>L_1{L}_2\right\}}{n^{\sigma }}}=0.$$

Hence $\sigma \in I_{xy}$ and $\eta \left(xy\right)\le \sigma <r+\epsilon$. Since $\epsilon >0$ with $r+\epsilon \le 1$ was arbitrary, letting $\epsilon \to 0^{+}$ gives $\eta \left(xy\right)\le r$, so $xy\in X_r$. Hence $X_r$ is a sequence algebra.

(iii)

To see that $X_r$ is not symmetric for $r\in [0,1)$, we distinguish two cases.

Case $r=0$. Consider the sequence $x=\left(x_k\right)\in X_0$ given in the proof of Proposition 3 by

$$x_k=\left\{\begin{array}{cc}k,\hfill & \mathrm{if}k=2^m\mathrm{for}\mathrm{some}m\in \mathbb{N},\hfill \\ 0,\hfill & \mathrm{otherwise}.\hfill \end{array}\right.$$

Define a rearrangement $y=\left(y_k\right)$ of x by

$$y_k=\left\{\begin{array}{cc}{2}^m,\hfill & \mathrm{if}k=2m-1\mathrm{for}\mathrm{some}m\in \mathbb{N},\hfill \\ 0,\hfill & \mathrm{if}k\mathrm{is}\mathrm{even}.\hfill \end{array}\right.$$

Fix $L\ge 0$. Since $2^m\to \infty$ as $m\to \infty$, the set $\{m\in \mathbb{N}:2^m\le L\}$ is finite; let $M\left(L\right):=\mathcal{C}\left\{m\in \mathbb{N}:2^m\le L\right\}$. Then, for every $n\in \mathbb{N}$,

$$\mathcal{C}\left\{k\le n:|y_k|>L\right\}\ge ⌈{\displaystyle \frac{n}{2}}⌉-M\left(L\right)\ge {\displaystyle \frac{n}{2}}-M\left(L\right).$$

Consequently,

$$\underset{n\to \infty }{\lim }{\displaystyle \frac{\mathcal{C}\left\{k\le n:|y_k|>L\right\}}{n}}\ge \underset{n\to \infty }{\lim }\left({\displaystyle \frac{1}{2}}-{\displaystyle \frac{M\left(L\right)}{n}}\right)={\displaystyle \frac{1}{2}}$$

for every $L\ge 0$, so $y\notin Sb=X_1$ by Lemma 2. Since $X_0\subseteq X_1$, it follows that $y\notin X_0$.

Case $r\in (0,1)$. Consider the sequence $x=\left(x_k\right)\in X_r$ given in Theorem 4 by

$$x_k=\left\{\begin{array}{cc}k,\hfill & \mathrm{if}k=\lfloor m^{1/r}\rfloor \mathrm{for}\mathrm{some}m\in \mathbb{N},\hfill \\ 0,\hfill & \mathrm{otherwise}.\hfill \end{array}\right.$$

Define a rearrangement $y=\left(y_k\right)$ of x by

$$y_k=\left\{\begin{array}{cc}\lfloor m^{1/r}\rfloor ,\hfill & \mathrm{if}k=2m-1\mathrm{for}\mathrm{some}m\in \mathbb{N},\hfill \\ 0,\hfill & \mathrm{if}k\mathrm{is}\mathrm{even}.\hfill \end{array}\right.$$

Fix $L\ge 0$. Since $\lfloor m^{1/r}\rfloor \to \infty$ as $m\to \infty$, the set $\{m\in \mathbb{N}:\lfloor m^{1/r}\rfloor \le L\}$ is finite; let $M\left(L\right):=\mathcal{C}\left\{m\in \mathbb{N}:\lfloor m^{1/r}\rfloor \le L\right\}$. Then, for every $n\in \mathbb{N}$,

$$\mathcal{C}\left\{k\le n:|y_k|>L\right\}\ge ⌈{\displaystyle \frac{n}{2}}⌉-M\left(L\right)\ge {\displaystyle \frac{n}{2}}-M\left(L\right).$$

Consequently,

$$\underset{n\to \infty }{\lim }{\displaystyle \frac{\mathcal{C}\left\{k\le n:|y_k|>L\right\}}{n}}\ge \underset{n\to \infty }{\lim }\left({\displaystyle \frac{1}{2}}-{\displaystyle \frac{M\left(L\right)}{n}}\right)={\displaystyle \frac{1}{2}}$$

for every $L\ge 0$. Hence $y\notin Sb=X_1$ by Lemma 2. Since $X_r\subseteq X_1$, it follows that $y\notin X_r$.

In either case, $X_r$ is not symmetric.

□

At the endpoint $r=1$, both omitted properties continue to hold, but for a different reason: since $X_1=Sb$ by Lemma 2, parts (ii) and (iii) at $r=1$ follow instead from the known facts that $Sb$ is a sequence algebra and $Sb$ is not symmetric, due to Bhardwaj and Gupta [15] (Thm. 2.15(a),(c)).

Theorem 6.

For every $r\in [0,1]$, $X_r^{\alpha }=X_r^{\beta }=X_r^{\gamma }=\varphi$, where ϕ denotes the space of all finitely nonzero scalar sequences.

Proof. 

The inclusion $\varphi \subset X_r^{\alpha }$ is explicit. We will just prove $X_r^{\alpha }\subset \varphi$ leading to $X_r^{\alpha }=\varphi$.

• Case $r=1$. In view of the fact that $X_1=Sb$ (Lemma 2), we can omit proving this case since it was already given by Bhardwaj and Gupta [15] (Prop. 2.16).

Let $\left(u_k\right)\notin \varphi$, so that the set $B=\{k:u_k\ne 0\}$ is infinite.

• Case $r=0$. Since B is unbounded, we may recursively choose a strictly increasing sequence $\left(b_j\right)$ of elements of B with $b_j\ge 2^j$ for every j. Define $x=\left(x_k\right)$ by

$$x_k=\left\{\begin{array}{cc}1/u_{b_j},\hfill & \mathrm{if}k=b_j\mathrm{for}\mathrm{some}j\in \mathbb{N},\hfill \\ 0,\hfill & \mathrm{otherwise}.\hfill \end{array}\right.$$

Then $\mathcal{C}\left\{k\le n:|x_k|>0\right\}\le {\log }_{2}n$ for all $n\ge 2$, so

$$\underset{n\to \infty }{\lim }{\displaystyle \frac{1}{n^{\rho }}}\mathcal{C}\left\{k\le n:|x_k|>0\right\}=0$$

for every $\rho \in (0,1]$; hence $I_x=(0,1]$, $\eta \left(x\right)=0$, and $x\in X_0$.

• Case $r\in (0,1)$. Similarly, choose a strictly increasing sequence $\left(b_j\right)$ of elements of B with $b_j\ge j^{1/r}$ for every j, and define $y=\left(y_k\right)$ in the same manner. Then

$$\mathcal{C}\left\{k\le n:|y_k|>0\right\}=\mathcal{C}\left\{j:b_j\le n\right\}\le n^r\mathrm{for}\mathrm{every}n.$$

Consequently, for every $\rho \in (r,1]$, we have

$${\displaystyle \frac{1}{n^{\rho }}}\mathcal{C}\left\{k\le n:|y_k|>0\right\}\le {\displaystyle \frac{n^r}{n^{\rho }}}=n^{r-\rho }⟶0,$$

so $(r,1]\subset I_y$, whence $\eta \left(y\right)\le r$ and $y\in X_r$.

Now let z denote the sequence constructed in either case (so $z=x$ when $r=0$ and $z=y$ when $r\in (0,1)$). Then

$$\sum _k|u_k{z}_k|\ge \sum _j|u_{b_j}{z}_{b_j}|=\sum _{j}1=\infty ,$$

so $\left(u_k\right)\notin X_r^{\alpha }$ for every $r\in [0,1)$. Since $\left(u_k\right)\notin \varphi$ was arbitrary, we have $X_r^{\alpha }\subset \varphi$.

Solidness of $X_r$, established in Theorem 5(i), implies that its $\alpha$-, $\beta$-, and $\gamma$-duals coincide [17]; hence the proof is complete.

□

3. Conclusions

In this paper we introduced the statistical boundedness index $\eta$, a scalar invariant that records, for a given sequence x, its threshold order: the infimum of the orders $\rho \in (0,1]$ at which x is statistically bounded of order $\rho$. Propositions 1, 2 and 3, together with Corollary 1, established the basic calculus of $\eta$: subadditivity, invariance under nonzero scalar multiplication, and its vanishing exactly on all bounded sequences, though not only on them. Theorems 1–5 then used $\eta$ to build the parameterized family ${\left\{X_r\right\}}_{r\in [0,1]}$ and to place it correctly relative to the classical scale ${\left\{S{b}^{\rho }\right\}}_{\rho \in (0,1]}$, leading to the strict inclusion chain

$${\ell }_{\infty }\subset X_0\subset X_{r_1}\subset X_{r_2}\subset X_1=Sb(0\le r_1<r_2\le 1).$$

This chain refines the usual hierarchy between bounded and statistically bounded sequences into a continuum indexed by r, and Theorem 5 shows that the refinement is structurally meaningful: every $X_r$ inherits solidness and the sequence-algebra property from the underlying order-$\rho$ boundedness sets. Symmetry fails throughout the chain: Bhardwaj and Gupta [15] already showed that $Sb(=X_1)$ itself is not symmetric, and we show in Theorem 5 that the same holds for every $X_r$ with $r\in [0,1)$. We also computed the $\alpha$-, $\beta$-, and $\gamma$-duals of $X_r$, all of which coincide with the space $\varphi$ of finitely nonzero sequences.

One direction for further work seems particularly natural: the index $\eta$ was defined relative to ordinary natural density; replacing $\delta$ by a weighted density, a deferred density, a lacunary density, or a general modulus function would produce a family of related indices whose mutual comparison seems worth investigating. We leave this question for future work.