Statistical limit superior and Statistical limit inferior in non-Archimedean L -fuzzy normed spaces

Abstract

The purpose of this article is to study the notion of statistical limit superior(SLS) and statistical limit inferior(SLI) in non-Archimedean(NA) $L$ -fuzzy normed spaces( $L$ -FNS). The concept of SLS and SLI is examined and extended to SLS and SLI in NA $L$ -FNS. Moreover, the analogue of some results between SLS and SLI over NA $L$ -FNS have been discussed. And also, it is proved that a bounded sequence is statistically convergent over NA $L$ -FNS. Throughout this article, $K$ denotes a complete, non-trivially valued, non-Archimedean fields(NAF).

Keywords

Statistical limit superior and statistical limit inferior 𝔏-fuzzy normed space non-Archimedean fields

1 Introduction

The concept of statistical convergence was first introduced by Fast [1]. In particular, many of the results of the theory of ordinary convergence have been extended to the theory of statistical convergence by using the notion of density. For instance, Fridy and Orhan [2] introduced the statistical analogs of limit superior and limit inferior of a sequence of real numbers. Recently, statistical convergence and some of its related concepts for fuzzy numbers have been studied in [3-10]. Quite recently, the idea of statistical convergence in intuitionistic fuzzy normed spaces for single sequences has been studied in [11, 12]. Recently, Alghamdi [13] introduced the notion of intuitionistic fuzzy normed space and quite recently, in [14, 15] the concepts of intuitionistic fuzzy 2-normed and intuitionistic fuzzy 2-metric spaces have been introduced and studied. Certainly, there are some situations where the ordinary norm does not work and the concept of intuitionistic fuzzy norm seems to be more suitable in such cases.

Among other fields, a progressive development is made in the field of fuzzy topology. One of the problems in L-fuzzy topology is to obtain an appropriate concept of L-fuzzy metric spaces and L-fuzzy normed spaces. Deschrijver et al. [16] introduced and studied a notion of L-fuzzy Euclidean normed spaces and then Shakeri et al. [17] and Saadati et al. [18] presented and examined in the notion of non-Archimedean L-fuzzy normed spaces. Quite recently, the concept of non-Archimedean L-fuzzy normed spaces has been generalized and studied [19, 20].

L-fuzzy normed spaces are natural generalizations of normed spaces, fuzzy normed spaces and intuitionistic fuzzy normed spaces based on some logical algebraic structures, which also enrich the notion of a L-fuzzy metric space. L-fuzzy normed spaces have been used to study various topological properties such as Hausdorffness, local convexity, and local boundedness, stability of functional equations, statistical convergence of sequences, fixed point theory.

In this article, we study the concept of SLS and SLI in NA $L$ -FNS. An example is demonstrated to determine these points in NA $L$ -FNS. We recall some basic definitions and notations.

Let x = {x_n} be a sequence and $B_{x} = {b \in K : {n : x_{n} > b}}$ and $A_{x} = {a \in K : {n : x_{n} < a}}$ . Then SLS of x is given by, $stat - lim sup x = {\begin{matrix} sup B_{x}; & if B_{x} \neq φ; \\ - \infty; & if B_{x} = φ . \end{matrix}$

and the SLI of x is given by, $stat - lim \inf x = {\begin{matrix} \inf A_{x}; & if A_{x} \neq φ; \\ + \infty; & if A_{x} = φ . \end{matrix}$

If a binary operation * : [0, 1] × [0, 1] → [0, 1] satisfies the following axioms, then it is said to be a t-norm:

* is associative,

* is commutative,

a₁ * 1 = a₁ for all a₁ ∈ [0, 1] and

a₁ * b₁ ≤ c₁ * d₁ whenever a₁ ≤ c₁ and b₁ ≤ d₁ for each a₁, b₁, c₁, d₁ ∈ [0, 1]. [11]

For example, a ∗ b = max {a + b - 1, 0} , a ∗ b = ab and a ∗ b = min {a, b} on [0, 1] are t-norms.

A complete lattice is a partially ordered set in which every subset has both a supremum and an infimum.

Let U be a non-empty set known as the universe and let $L = (L, \leq_{L})$ be a complete lattice. A mapping $A : U \to L$ is said to be $L$ -fuzzy set in U. If for every u in U, $A (u)$ denotes the degree (in L) to which u is an element of $A$ .

For example, let L = [0, 1] × [0, 1] and operation ≤_L defined by,

L = {(a₁, a₂) : (a₁, a₂) ∈ [0, 1] × [0, 1] and a₁ + a₂ ≤ _L1} ,

(a₁, a₂) ≤ _L (b₁, b₂) iff a₁ ≤ b₁, a₂ ≥ b₂, for all a = (a₁, a₂) , b = (b₁, b₂) ∈ L.

Then (L, ≤ _L) is a complete lattice. In a complete lattice, we denote its units by 0_L = (0, 1) and 1_L = (1, 0) .

A mapping ∗_L : L² → L is defined as a triangular norm(t-norm) on $L$ , if it meets the following axioms:

$(for every (x) \in L) (x *_{L} 1_{L} = x)$ (: boundary condition);

(for every (x, y) ∈ L²) (x ∗ _Ly = y ∗ _Lx) (: commutativity);

(for every (x, y, z) ∈ L³) (x ∗ _L (y ∗ _Lz)) = ((x ∗ _Ly) ∗ _Lz) (:associativity);

(for every (x, y, z, w) ∈ L⁴) (x ≤ Lx′ and y ≤ Ly′ ⇒ x ∗ _Ly ≤ Lx′ ∗ Ly′) (: monotonicity).

If a t-norm ∗_L on

L

is continuous, for any

x, y \in L

and any sequences x_n and y_n that converges to x and y respectively,

lim_{n \to \infty} (x_{n} *_{L} y_{n}) = x *_{L} y .

2 Preliminaries

In this section, we define limit point, statistical LP, SLS and SLI in NA $L$ -FNS and demonstrate through an example how to compute these points in a NA $L$ -FNS.

2.1 Definition

Let $K$ be a complete field with a NA valuation | . |. The three-tuple $(J, Q, *_{L})$ is said to be a NA $L$ -FNS, if $J$ is a vector space over $K$ , ∗_L is a continuous t-norm on $L$ and $Q$ are $L$ -fuzzy sets on $J \times (0, + \infty)$ satisfying the following axioms:

$Q (x, t) >_{L} 0_{L}$ ;

$Q (x, t) = 1_{L}$ if and only if x = 0;

$Q (ζ x, t) = Q (x, \frac{t}{| ζ |})$ for all ζ ≠ 0;

$Q (x, t) *_{L} Q (y, s) \leq_{L} Q (x + y, \max {t, s})$ ;

$Q (x, .) : (0, \infty) \to L$ is continuous;

$lim_{t \to 0} Q (x, t) = 0_{L}$ and $lim_{t \to \infty} Q (x, t) = 1_{L}$ . [18]

Example

Suppose that (X, ∥ . ∥) be a normed space. Let ∗_L (a, b) = (min {a₁, b₁} , max {a₂, b₂}) for all a = (a₁, a₂) , b = (b₁, b₂) ∈ [0, 1] × [0, 1] and ι be a mapping defined by

$Q (x, t) = (\frac{t}{t + ∥ x ∥}, \frac{∥ x ∥}{t + ∥ x ∥})$ , for all $t \in K .$

Then, $(J, Q, *_{L})$ is a $L$ -fuzzy normed space.

2.2 Definition

Any non-increasing mapping $N : L \to L$ satisfying $N (0_{L}) = 1_{L}$ and $N (1_{L}) = 0_{L}$ is said to be a negator on $L$ .

Let $N$ be an involutive negator, if $N (N (x)) = x$ if for every x ∈ L.

2.3 Definition

Let a sequence x = {x_n} in an $L$ -FNS $(J, Q, *_{L})$ is said to be a Cauchy sequence, if for every $ɛ \in L - {0_{L}, 1_{L}}$ and t > 0, ∃ $n_{0} \in N$ so that for every $n, m \geq n_{0}, Q (x_{n} - x_{m}, t) >_{L} N (ɛ)$ , if $N$ is a negator on $L$ .

2.4 Definition

Let a sequence x = {x_n} is said to be convergent to $x \in J$ in the $L$ -FNS $(J, Q, *_{L})$ , if $Q (x_{n} - x, t) \to 1_{L}$ , where n→ + ∞ for every t > 0.

Suppose that every Cauchy sequence in $J$ is convergent, then $L$ -FNS $(J, Q, *_{L})$ be complete.

2.5 Definition

Let a sequence x = {x_n} is said to be statistically bounded if there is a number ‘B’ such that $lim_{n \to \infty} \frac{1}{n} | {n \in ℕ : | x_{n} | \geq B} | = 0 .$

2.6 Definition

Let the three-tuple $(J, Q, *_{L})$ be a NA $L$ -FNS over $K$ . A sequence x = {x_n} is defined as statistically convergent to a limit ‘l’ $\in J$ in respect to the NA $L$ -FNS if for all $ɛ \in L - {0_{L}, 1_{L}}$ and t > 0, $lim_{n \to \infty} \frac{1}{n} | {n \in ℕ : Q (x_{n} - l, t) >_{L} N (ɛ)} | = 1,$ or $lim_{n \to \infty} \frac{1}{n} | {n \in ℕ : Q (x_{n} - l, t) > /_{L} N (ɛ)} | = 0 .$ We write, ${stat}_{Q} - lim_{n \to \infty} x_{n} = l .$

2.7 Definition

Let the three-tuple $(J, Q, *_{L})$ be a NA $L$ -FNS over $K$ . A sequence x = {x_n} is defined as statistically Cauchy if for each ɛ > 0 and t > 0 ∃ $ℕ$ so that for each $n, m \geq ℕ$ . $lim_{n \to \infty} \frac{1}{n} | {n, m \in ℕ : Q (x_{n} - x_{m}, t) >_{L} N (ɛ)} | = 1,$ or $lim_{n \to \infty} \frac{1}{n} | {n, m \in ℕ : Q (x_{n} - x_{m}, t) > /_{L} N (ɛ)} | = 0 .$

2.8 Definition

Let x = {x_n} be a NA $L$ -FNS $(J, Q, *_{L})$ is defined as statistically bounded to a limit ‘l’ if ∃ some t₀ > 0 and $N (b) \in (0, 1)$ such that $lim_{n \to \infty} \frac{1}{n} | {n \in ℕ : Q (x_{n} - l, t_{0}) > /_{L} N (b)} | = 1,$ or $lim_{n \to \infty} \frac{1}{n} | {n \in N : Q (x_{n} - l, t_{0}) >_{L} N (b)} | = 0 .$

2.9 Definition

If $(J, Q, *_{L})$ is a NA $L$ -FNS over $K$ . Then, $l \in J$ is said to be a LP of the sequence x = {x_n} concerning the $L$ -fuzzy norm(FN) $(Q)$ given that a subsequence of x which converges to a limit l concerning the $L$ -FN $(Q)$ . Let $l_{Q} (x)$ represent the LP of the sequence x concerning the $L$ -FN $(Q)$ .

2.10 Definition

If $(J, Q, *_{L})$ is a NA $L$ -FNS over $K$ . Then, φ ∈ X is said to be statistical-LP of the sequence x = {x_n} in respect to $L$ -FN $(Q)$ , given that a subsequence of x which converges to φ concerning the $L$ -FN $(Q)$ . Then it is said that φ is a ${stat}_{Q}$ -LP of sequence x = {x_n}. Let $Λ_{Q} (x)$ represent ${stat}_{Q}$ -LP of the sequence x.

2.11 Definition

If a sequence x = {x_n} be a NA $L$ -FNS $(J, Q, *_{L})$ , then the set ${B_{x}}^{Q}$ and ${A_{x}}^{Q}$ given by, $\begin{matrix} {B_{x}}^{Q} = {N (b) \in (0, 1) : lim_{n \to \infty} \frac{1}{n} | {n : \\ Q (x_{n}; t) > /_{L} N (b^{'}) or Q (x_{n}; t) >_{L} N (b)} | \neq 0}, \\ {A_{x}}^{Q} = {N (a) \in (0, 1) : lim_{n \to \infty} \frac{1}{n} | {n : \\ Q (x_{n}; t) >_{L} N (a^{'}) or Q (x_{n}; t) > /_{L} N (a)} | \neq 0} . \end{matrix}$

If x = {x_n} be a sequence then the SLS of x in respect to $L$ -FN $(Q)$ given by,

${stat}_{Q} - lim sup x = {\begin{matrix} sup {B_{x}}^{Q}; & if {B_{x}}^{Q} \neq φ; \\ 0; & if {B_{x}}^{Q} \neq φ . \end{matrix}$

The SLI of x concerning the $L$ -FN $(Q)$ is, ${stat}_{Q} - lim \inf x = {\begin{matrix} \inf {A_{x}}^{Q}; & if {A_{x}}^{Q} \neq φ; \\ 1; & if {A_{x}}^{Q} \neq φ . \end{matrix}$

Example

Let the sequence x = {x_n} be defined by, $x_{n} = {\begin{matrix} n, & if n is an odd square; \\ 2; & if n is an even square; \\ 1; & if n is an odd nonsquare; \\ 0; & if n is an even nonsquare; \end{matrix}$ Here x is statistically bounded. Thus $B_{x} = (- \infty, 1)$ and stat - lim sup x = 1.

Example

Let a sequence be

$a = {0, 1, 0, \frac{1}{2}, 1, 0, \frac{1}{3}, \frac{2}{3}, 1, 0, \frac{1}{4}, \frac{2}{4}, \frac{3}{4}, 1, \dots}$ ,

and define $x_{2 n - 1} : = 0$ and $x_{2 n - 1} : = a_{n} .$

Then stat - lim sup x = 1 because $δ {n : x_{n} > 1 - ɛ} = \frac{ɛ}{2}$ . Hence, the greatest statistical limit point of x is zero, but stat - lim sup x = 1.

Example

Let x be the sequence given by $x_{n} = {\begin{matrix} \sqrt{n}, & if n is a square; \\ 0; & if n is an odd nonsquare; \\ 1; & if n is an even nonsquare . \end{matrix}$ Since, $δ {n : x_{n} = 0} = \frac{1}{2} = δ {n : x_{n} = 1}$ , it is clear that stat - lim inf x = 0 and stat - lim sup x = 1. Therefore, x is not statistically convergent. Also note that x is statistically bounded since δ {n : |x_n|<1} =0. It remains A₁ (x) has lim 1 = stat - lim sup x. Let N² denote the set of squares, and let N⁰ and N¹ denote, respectively, the set od odd and even nonsquares. Where [u] : = max {n : n ≤ u}, thus we get $\begin{matrix} {(A_{1} n)}_{m} & = \frac{1}{m} \sum_{m \in N_{m}^{0}} x_{n} + \frac{1}{m} \sum_{m \in N_{m}^{1}} x_{n} + \frac{1}{m} \sum_{m \in N_{m}^{2}} x_{n} \\ = 0 + \frac{1}{m} \frac{[m - [\sqrt{m}]]}{2} + \frac{1}{m} \sum_{i \leq \sqrt{m}} i \\ = 1 + o (1) \end{matrix}$

Example

Let the sequence x = {x_n} be defined by, $x_{n} = {\begin{matrix} 2 n, & if n is an odd square; \\ - 1; & if n is an even square; \\ \frac{1}{2}; & if n is an odd nonsquare; \\ 0; & if n is an even nonsquare; \end{matrix}$

Proof. This proof is in the appendix A1. □

3 Main results

In this section, we prove theorems concerning SLS and SLI in $L$ -FNS over NAF $K$ .

3.1 Theorem

If $N (b) = {stat}_{Q} - lim sup x$ . Then for any positive numbers t and ξ $\begin{array}{l} \lim_{n \to \infty} \frac{1}{n} | {n : Q (x_{n}; t) > /_{L} N (b^{'}) + ξ \\ o r Q (x_{n}; t) >_{L} N (b) - ξ} | \neq 0, a n d \\ \lim_{n \to \infty} \frac{1}{n} | {n : Q (x_{n}; t) > /_{L} N (b^{'}) - ξ \\ o r Q (x_{n}; t) >_{L} N (b) + ξ} | = 0. \end{array}}$ (1) on the contrary, if (1) holds for any positive numbers t and ξ, then $N (b) = {stat}_{Q} - lim sup x$ .

Proof. This proof is in the appendix A2. □

3.2 Theorem

If $N (a) = {stat}_{Q} - lim inf x$ . Then for any non-negative numbers t and ξ $\begin{array}{l} \lim_{n \to \infty} \frac{1}{n} | {n : Q (x_{n}; t) >_{L} N (a^{'}) - ξ \\ o r Q (x_{n}; t) > /_{L} N (a) + ξ} | \neq 0, a n d \\ \lim_{n \to \infty} \frac{1}{n} | {n : Q (x_{n}; t) >_{L} N (a^{'}) + ξ \\ o r Q (x_{n}; t) > /_{L} N (a) - ξ} | = 0. \end{array}}$ (2) on the contrary, for any positive numbers t and ξ if ((3)) holds, then $N (a) = {stat}_{Q} - lim inf x$ .

Proof. This proof is in the appendix A3. □

3.3 Theorem

For every sequence x = {x_n}, ${stat}_{Q} - lim inf x \leq_{L} {stat}_{Q} - lim sup x .$

Proof. This proof is in the appendix A4. □

3.4 Theorem

Let the $L$ -FNS be $(J, Q, *_{L})$ , the statistically bounded sequence x is statistically convergent iff ${stat}_{Q} - lim inf x = {stat}_{Q} - lim sup x .$ Proof. This proof is in the appendix A5. □

4 Conclusion

In this article, we have discussed the theory of SLS and SLI in NA $L$ -FNS and proved some results related to SLS and SLI. Also, it has been given that statistically bounded sequence is statistically convergent. The significance of the results obtained in this article are generalized and extended from intuitionistic fuzzy normed space to NA $L$ -FNS. These findings may be combined with the lattice structure and the normed space structure, allowing a wider range of topological vector spaces to benefit from the convenience afforded by a variant of the notion of norm.

In future work, we may extend the concept of SLS and SLI over NA $L$ -FNS as I₂-SLS and I₂-SLI for double sequences in intutionistic fuzzy normed space over NAF.

Acknowledgments

The authors are thankful to the area editor and referees for giving valuable comments and suggestions.

Disclosure statement

The authors declare that they have no competing interests.

Funding

There are no fundings of this article.

Appendix

A. Appendices

A.1. First appendix

Let $Q (x_{n}, t) = (\frac{t}{t + ∥ x_{n} ∥}, \frac{∥ x_{n} ∥}{t + ∥ x_{n} ∥})$ ,

The sequence is unbounded with respect to $Q$ . Also it is statistically bounded with respect to $Q$ , for this, $\begin{matrix} lim_{n \to \infty} \frac{1}{n} | {n \in ℕ : & Q (x_{n}; t_{0}) > /_{L} N (b^{'}) \\ or Q (x_{n}; t_{0}) >_{L} N (b)} | \end{matrix}$ $\begin{matrix} = lim_{n \to \infty} \frac{1}{n} | {n \in & ℕ : (\frac{t_{0}}{t_{0} + ∥ x_{n} ∥} > /_{L} N (b^{'})) \\ or (\frac{∥ x_{n} ∥}{t_{0} + ∥ x_{n} ∥}) >_{L} N (b)} | \end{matrix}$ $\begin{matrix} = lim_{n \to \infty} \frac{1}{n} | {n \in & ℕ : ∥ x_{n} ∥ >_{L} \frac{N (b) t_{0}}{N (b^{'})}} | . \end{matrix}$ Since, $0 < N (b) < 1, \frac{1}{N (b^{'})} > 0$ . Choose $t_{0} = \frac{N (b^{'})}{3 N (b)}$ . Then t₀ > 0 and $\begin{matrix} lim_{n \to \infty} \frac{1}{n} | {n \in ℕ : & Q (x_{n}; t_{0}) > /_{L} N (b^{'}) \\ or Q (x_{n}; t_{0}) >_{L} N (b)} | \end{matrix}$ $\begin{matrix} = lim_{n \to \infty} \frac{1}{n} | {n \in ℕ : ∥ x_{n} ∥ >_{L} \frac{N (b)}{N (b^{'})} & \times \frac{N (b^{'})}{3 N (b)} \\ = \frac{1}{3}} | \end{matrix}$ $\begin{matrix} = lim_{n \to \infty} \frac{1}{n} | {n \in ℕ : ∥ x_{n} ∥ >_{L} \frac{1}{3}} | \\ = lim_{n \to \infty} \frac{1}{n} \times \sqrt{n} = 0 . \end{matrix}$

Hence, it is statistically bounded with respect to $Q$ .

Now to find ${B_{x}}^{Q}$ , we have to find $N (b) \in (0, 1)$ such that $\begin{matrix} lim_{n \to \infty} \frac{1}{n} | {n \in ℕ : & Q (x_{n}; t) > /_{L} N (b^{'}) \\ or Q (x_{n}; t) >_{L} N (b)} | \neq 0 . \end{matrix}$ Now, $\begin{matrix} lim_{n \to \infty} \frac{1}{n} | {n \in ℕ : & Q (x_{n}; t) > /_{L} N (b^{'}) \\ or Q (x_{n}; t) >_{L} N (b)} | \end{matrix}$

$\begin{matrix} = lim_{n \to \infty} \frac{1}{n} | {n \in ℕ : & \frac{t}{t + ∥ x_{n} ∥} > /_{L} N (b^{'}) \\ or \frac{∥ x_{n} ∥}{t + ∥ x_{n} ∥} >_{L} N (b)} |, \end{matrix}$ $\begin{matrix} = lim_{n \to \infty} \frac{1}{n} | {n \in ℕ : ∥ x_{n} ∥ >_{L} \frac{N (b) t}{N (b^{'})}} | . \end{matrix}$

We can choose any t > 0 as $t > \frac{1}{3} (\frac{1}{N (b^{'})})$ for $0 < N (b) < 1$ , so that $0 > /_{L} \frac{N (b) t}{N (b^{'})} > /_{L} \frac{N (b)}{N (b^{'})} \times \frac{N (b^{'})}{3 N (b)} = \frac{1}{3} .$ Therefore, $lim_{n \to \infty} \frac{1}{n} | {n \in ℕ : ∥ x_{n} ∥ >_{L} \frac{N (b) t}{N (b^{'})}} |$

$\begin{matrix} = lim_{n \to \infty} \frac{1}{n} | {n \in ℕ : ∥ x_{n} ∥ >_{L} N (ɛ) = \frac{N (b) t}{N (b^{'})}} | . \end{matrix}$ By using the above condition $N (ɛ) \in (0, 1)$ . Now the number of members of the sequence which satisfy the above condition is always greater than $m - \frac{m}{2}$ or $m - \frac{m - 1}{2}$ for the case m is even or respectively. Thus $\begin{matrix} = lim_{n \to \infty} \frac{1}{n} | {n \in ℕ : ∥ x_{n} ∥ >_{L} N (ɛ) = \frac{N (b) t}{N (b^{'})}} | \\ >_{L} lim_{n \to \infty} \frac{1}{m} \times \frac{m}{2} = \frac{1}{2} \\ or lim_{n \to \infty} \frac{1}{m} \times \frac{m + 1}{2} = \frac{1}{2} . \end{matrix}$

Thus $\begin{matrix} = lim_{n \to \infty} \frac{1}{n} | {n \in ℕ : ∥ x_{n} ∥ >_{L} & N (ɛ) \\ = \frac{N (b) t}{N (b^{'})}} | \neq 0 \end{matrix}$ for all $N (b) \in (0, 1) .$ Hence, ${B_{x}}^{Q} = (0, 1)$ and ${stat}_{Q} - lim sup x = 1 .$

The above sequence has two subsequences

x = (x_{m
_j}) where x_{m
_j} = 1

for each m_j ∈ {3, 5, 7, 11, 13, …} and x = {x_{m
_k}} where x_{m
_j} = 0 for each m_k ∈ {2, 6, 8, 10, 12, …}, $j, k \in ℕ :$ are positive integers and it is convergent to 1 and 0, respectively. Thus, x is not statistically convergent.

Simillarly, we have ${A_{x}}^{Q} = (0, 1)$ and ${stat}_{Q} - lim inf x = 0 .$

Hence, the set of statistical points is (0, 1) where ${stat}_{Q} - lim inf x =$ least elements and ${stat}_{Q} - lim inf x =$ greatest elements of the above set.

A.2. Second appendix

Assume that $N (b) = {stat}_{Q} - lim sup x$ , $N (b)$ is finite. Therefore,

$\begin{matrix} lim_{n \to \infty} \frac{1}{n} | {n : & Q (x_{n}; t) > /_{L} N (b^{'}) \\ or Q (x_{n}; t) >_{L} N (b)} | \neq 0 . \end{matrix}$ (3)

Since

$Q (x_{n}; t) > /_{L} N (b^{'}) + ξ or Q (x_{n}; t) >_{L} N (b) - ξ$ if for all n and for every t, ξ > 0, $\begin{matrix} lim_{n \to \infty} \frac{1}{n} | {n : & Q (x_{n}; t) > /_{L} N (b^{'}) + ξ \\ or Q (x_{n}; t) >_{L} N (b) - ξ} | \neq 0 . \end{matrix}$

Using the definition of ${stat}_{Q} - lim sup x$ it satisfies (2) thus we get $N (b^{'})$ and $N (b)$ as the least value(LV) and as the greatest value(GV) respectively.

If it is possible,

$Q (x_{n}; t) > /_{L} N (b^{'}) - ξ or Q (x_{n}; t) >_{L} N (b) + ξ$ for any ξ > 0 .

Then $N (b^{'}) - ξ$ and $N (b) + ξ$ are additional values of $N (b^{'}) - ξ > /_{L} N (b^{'})$ and $N (b) + ξ >_{L} N (b)$ that satisfies (2). Since the investigation contradict that $N (b^{'})$ as the LV and $N (b)$ as the GV, that meets the above condition.

Thus, $\begin{matrix} lim_{n \to \infty} \frac{1}{n} | { & n : Q (x_{n}; t) > /_{L} N (b^{'}) - ξ \\ or Q (x_{n}; t) >_{L} N (b) + ξ} | = 0 . \end{matrix}$ if for any ξ > 0.

on the contrary, for any positive numbers t and ξ if (1) holds, thus $\begin{matrix} lim_{n \to \infty} \frac{1}{n} | { & n : Q (x_{n}; t) > /_{L} N (b^{'}) + ξ \\ or Q (x_{n}; t) >_{L} N (b) - ξ} | \neq 0 \end{matrix}$ and $\begin{matrix} lim_{n \to \infty} \frac{1}{n} | { & n : Q (x_{n}; t) > /_{L} N (b^{'}) - ξ \\ or Q (x_{n}; t) >_{L} N (b) + ξ} | = 0 . \end{matrix}$

Therefore, $\begin{matrix} lim_{n \to \infty} \frac{1}{n} | {n : & Q (x_{n}; t) \leq_{L} N (b^{'}) \\ or Q (x_{n}; t) \geq_{L} N (b)} | \neq 0 \end{matrix}$ and $\begin{matrix} lim_{n \to \infty} \frac{1}{n} | {n : & Q (x_{n}; t) = N (b^{'}) \\ or Q (x_{n}; t) = N (b)} | = 0 . \end{matrix}$ That is,

$\begin{matrix} lim_{n \to \infty} \frac{1}{n} | {n : & Q (x_{n}; t) > /_{L} N (b^{'}) \\ or Q (x_{n}; t) >_{L} N (b)} | \neq 0 \end{matrix}$ for any t > 0 .

consequently, $N (b) = {stat}_{Q} - lim sup x .$

Similarly, we can prove ${stat}_{Q} - lim inf x .$

A.3. Third appendix

Let $N (a) = {stat}_{Q} - lim sup x$ , $N (a)$ is finite. Therefore, $\begin{matrix} lim_{n \to \infty} \frac{1}{n} | {n : & Q (x_{n}; t) >_{L} N (a^{'}) \\ or Q (x_{n}; t) > /_{L} N (a)} | \neq 0 . \end{matrix}$ (4)

Since

$Q (x_{n}; t) >_{L} N (a^{'}) - ξ or Q (x_{n}; t) > /_{L} N (a) + ξ$ for any n and for every t, ξ > 0, $\begin{matrix} lim_{n \to \infty} \frac{1}{n} | {n : & Q (x_{n}; t) >_{L} N (a^{'}) - ξ \\ or Q (x_{n}; t) > /_{L} N (a) + ξ} | \neq 0 . \end{matrix}$

Using the definition of ${stat}_{Q} - lim sup x$ it satisfying (4) thus we get $N (a^{'})$ and $N (a)$ as the LV and the GV.

If it is possible,

$Q (x_{n}; t) >_{L} N (a^{'}) - ξ or Q (x_{n}; t) > /_{L} N (a) + ξ$ for any ξ > 0 .

Then $N (a^{'}) - ξ$ and $N (a) + ξ$ are additional values of $N (a^{'}) - ξ >_{L} N (a^{'})$ and $N (a) + ξ > /_{L} N (a)$ that satisfies (4). Since the investigation contradict that $N (a^{'})$ and $N (a)$ are LV and GV respectively, that satisfies the above condition.

Thus,

$\begin{matrix} lim_{n \to \infty} \frac{1}{n} | {n : & Q (x_{n}; t) >_{L} N (a^{'}) - ξ \\ or Q (x_{n}; t) > /_{L} N (a) + ξ} | = 0 \end{matrix}$ for any ξ > 0.

on the contrary, for any positive numbers t and ξ if (3) holds, then $\begin{matrix} lim_{n \to \infty} \frac{1}{n} | {n : & Q (x_{n}; t) >_{L} N (a^{'}) - ξ \\ or Q (x_{n}; t) > /_{L} N (a) + ξ} | \neq 0 \end{matrix}$ and $\begin{matrix} lim_{n \to \infty} \frac{1}{n} | {n : & Q (x_{n}; t) >_{L} N (a^{'}) + ξ \\ or Q (x_{n}; t) > /_{L} N (a) - ξ} | = 0 . \end{matrix}$

Hence, $\begin{matrix} lim_{n \to \infty} \frac{1}{n} | {n : & Q (x_{n}; t) \geq_{L} N (a^{'}) \\ or Q (x_{n}; t) \leq_{L} N (a)} | \neq 0 \end{matrix}$ and $\begin{matrix} lim_{n \to \infty} \frac{1}{n} | {n : & Q (x_{n}; t) = N (a^{'}) \\ or Q (x_{n}; t) = N (a)} | = 0 . \end{matrix}$ That is, $\begin{matrix} lim_{n \to \infty} \frac{1}{n} | {n : & Q (x_{n}; t) >_{L} N (a^{'}) \\ or Q (x_{n}; t) > /_{L} N (a)} | \neq 0 . \end{matrix}$ for any t > 0.

Therefore, $N (a) = {stat}_{Q} - lim inf x .$

Similarly, we can prove ${stat}_{Q} - lim sup x .$

A.4. Fourth appendix

Let us consider the case that ${stat}_{Q} - lim sup x = 0$ , which shows that ${B_{x}}^{Q} = φ .$ Then for all $N (b) \in (0, 1)$ , $\begin{matrix} {B_{x}}^{Q} = lim_{n \to \infty} \frac{1}{n} & | {n : Q (x_{n}; t) > /_{L} N (b^{'}) \\ or Q (x_{n}; t) >_{L} N (b)} | = 0, \end{matrix}$ which is $\begin{matrix} lim_{n \to \infty} \frac{1}{n} | {n : & Q (x_{n}; t) \geq_{L} N (b^{'}) \\ or Q (x_{n}; t) \leq_{L} N (b)} | = 1 . \end{matrix}$ Also for each $N (a) \in (0, 1)$ , we get $\begin{matrix} lim_{n \to \infty} \frac{1}{n} | {n : & Q (x_{n}; t) >_{L} N (a^{'}) \\ or Q (x_{n}; t) > /_{L} N (a)} | \neq 0 . \end{matrix}$ Hence, ${stat}_{Q} - lim inf x = 0 .$ If ${stat}_{Q} - lim sup x = 1$ is trivial.

Assume that $N (b) = {stat}_{Q} - lim sup x$ , and $N (a) = {stat}_{Q} - lim inf x$ , if $N (a)$ and $N (b)$ be finite.

If for each ξ, we prove that $N (b^{'}) - ξ \in {A_{x}}^{Q}$ .

By 3.1, $\begin{matrix} lim_{n \to \infty} \frac{1}{n} | {n : & Q (x_{n}; t) > /_{L} N (b^{'}) - \frac{ξ}{2} \\ or Q (x_{n}; t) >_{L} N (b) + \frac{ξ}{2}} | = 0, \end{matrix}$ if $N (b^{'}) =$ least upper bounded(LUB) of ${B_{x}}^{Q} .$ Therefore, $\begin{matrix} lim_{n \to \infty} \frac{1}{n} | {n : & Q (x_{n}; t) \geq_{L} N (b^{'}) - \frac{ξ}{2} \\ or Q (x_{n}; t) \leq_{L} N (b) + \frac{ξ}{2}} | = 1, \end{matrix}$ that gives, $\begin{matrix} lim_{n \to \infty} \frac{1}{n} | {n : & Q (x_{n}; t) >_{L} N (b^{'}) - ξ \\ or Q (x_{n}; t) > /_{L} N (b) + ξ} | = 1 . \end{matrix}$

Thus, $N (b^{'}) - ξ \in {A_{x}}^{Q} .$

By definition $N (a) = inf {A_{x}}^{Q} .$ Therefore, we say that $N (b^{'}) - ξ \leq_{L} N (a^{'}),$ and considering that ξ is arbitary, $N (b^{'}) \leq_{L} N (a^{'}),$ which is $N (b) \geq_{L} N (a),$ $N (a) \leq_{L} N (b) .$

A.5. Fifth appendix

Let ζ, η be ${stat}_{Q} - lim inf x$ and ${stat}_{Q} - lim sup x$ , respectively.

Let us suppose that ${stat}_{Q} - lim x = l$ , for any ɛ > 0 and $N (b) \in (0, 1),$ $\begin{matrix} lim_{n \to \infty} \frac{1}{n} | {n : & Q (x_{n}; t) \leq_{L} N (b^{'}) \\ or Q (x_{n}; t) \geq_{L} N (b)} | = 0, \end{matrix}$ so that $\begin{matrix} lim_{n \to \infty} \frac{1}{n} & | {n : Q (x_{n}; \frac{t}{2}) *_{L} Q (l; \frac{t}{2}) \leq_{L} N (b^{'}) \\ or Q (x_{n}; \frac{t}{2}) *_{L} Q (l; \frac{t}{2}) \geq_{L} N (b)} | = 0 . \end{matrix}$ If for any t > 0, $\begin{matrix} sup_{t \to \infty} Q (x_{n}; \frac{t}{2}) = N ({b_{1}}^{'}) and \\ sup_{t \to \infty} Q (l; \frac{t}{2}) = N ({b_{2}}^{'}) \end{matrix}$ or $\begin{matrix} inf_{t \to \infty} Q (x_{n}; \frac{t}{2}) = N (b_{1}) and \\ inf_{t \to \infty} Q (l; \frac{t}{2}) = N (b_{2}) \end{matrix}$ such that $\begin{matrix} N ({b_{1}}^{'}) *_{L} N ({b_{2}}^{'}) \leq_{L} N (b^{'}) or \\ N (b_{1}) *_{L} N (b_{2}) \leq_{L} N (b) . \end{matrix}$ (5) Then $\begin{matrix} lim_{n \to \infty} \frac{1}{n} | {n : & Q (x_{n}; \frac{t}{2}) \leq_{L} N ({b_{1}}^{'}) \\ or Q (x_{n}; \frac{t}{2}) \geq_{L} N (b_{1})} | = 0 . \end{matrix}$ (6)

And therefore, $\begin{matrix} lim_{n \to \infty} \frac{1}{n} & | {n : Q (x_{n}; \frac{t}{2}) > /_{L} N ({b_{1}}^{'}) - ξ \\ or Q (x_{n}; \frac{t}{2}) >_{L} N (b_{1}) + ξ} | = 0 \end{matrix}$ (7) for any ξ > 0.

Now using 3.1 and ${stat}_{Q} - lim sup x$ , we get $\begin{matrix} lim_{n \to \infty} \frac{1}{n} | {n : & Q (x_{n}; \frac{t}{2}) > /_{L} η^{'} - ξ \\ or Q (x_{n}; \frac{t}{2}) >_{L} η + ξ} | = 0 \end{matrix}$ (8) for every ξ > 0.

Form (7) and (8) and ${stat}_{Q} - lim sup x$ , we get

$\begin{matrix} N ({b_{1}}^{'}) - ξ \leq_{L} η^{'} - ξ or N (b_{1}) + ξ \geq_{L} η + ξ, \end{matrix}$ which is, $N ({b_{1}}^{'}) \leq_{L} η^{'} or N (b_{1}) \geq_{L} η .$ (9) Now we find those n such that

$\begin{matrix} Q (x_{n}; \frac{t}{2}) >_{L} & N ({b_{1}}^{'}) + ξ \\ or Q (x_{n}; \frac{t}{2}) > /_{L} N (b_{1}) - ξ . \end{matrix}$ It can be easily seen that there is no such n exists which satisfying (5) and above condition together.

Thus, this suggests that $\begin{matrix} lim_{n \to \infty} \frac{1}{n} | {n : & Q (x_{n}; \frac{t}{2}) >_{L} N ({b_{1}}^{'}) + ξ \\ or Q (x_{n}; \frac{t}{2}) > /_{L} N (b_{1}) - ξ} | = 0 . \end{matrix}$

Since $ζ = {stat}_{Q} - lim inf x$ , by 3.2, we get $\begin{matrix} lim_{n \to \infty} \frac{1}{n} | {n : & Q (x_{n}; \frac{t}{2}) >_{L} ζ^{'} + ξ \\ or Q (x_{n}; \frac{t}{2}) > /_{L} ζ - ξ} | = 0 . \end{matrix}$ Using the definition of ${stat}_{Q} - lim inf x$ , thus we get $\begin{matrix} ζ^{'} + ξ \leq_{L} N ({b_{1}}^{'}) + ξ or ζ - ξ \geq_{L} N (b_{1}) - ξ . \end{matrix}$ thus $ζ^{'} \leq_{L} N ({b_{1}}^{'}) or ζ \geq_{L} N (b_{1}) .$ (10) From (8) and (9), we get ζ′ ≤ _Lη′ and η ≤ _Lζ. Then combining the inequality above with the 3.2 and we obtain ζ = η.

On the contrary, assume that ζ = η and

if $sup_{t \to \infty} Q (l; t) = ζ^{'}$ (or) $inf_{t \to \infty} Q (l; t) = ζ$ . Then for every ξ > 0, 3.1 and (5) together shows that $\begin{matrix} lim_{n \to \infty} \frac{1}{n} | {n : & Q (x_{n}; \frac{t}{2}) > /_{L} ζ^{'} + \frac{ξ}{2} \\ or Q (x_{n}; \frac{t}{2}) >_{L} ζ - \frac{ξ}{2}} | = 0 . \end{matrix}$ (11) and $\begin{matrix} lim_{n \to \infty} \frac{1}{n} | {n : & Q (x_{n}; \frac{t}{2}) >_{L} ζ^{'} + \frac{ξ}{2} \\ or Q (x_{n}; \frac{t}{2}) > /_{L} ζ - \frac{ξ}{2}} | = 0 . \end{matrix}$ (12) Now $\begin{matrix} ζ^{'} \geq_{L} Q (l; t) = Q (x_{n} - (x_{n} & - l); t) \geq_{L} Q (x_{n}; \frac{t}{2}) \\ *_{L} Q (x_{n} - l; \frac{t}{2}), \end{matrix}$ $\begin{matrix} ζ \leq_{L} Q (l; t) = Q (x_{n} - (x_{n} & - l); t) \leq_{L} Q (x_{n}; \frac{t}{2}) \\ *_{L} Q (x_{n} - l; \frac{t}{2}) . \end{matrix}$ Therefore, $\begin{matrix} Q (x_{n}; \frac{t}{2}) *_{L} Q (x_{n} - l; \frac{t}{2}) \leq_{L} ζ^{'} \\ or Q (x_{n}; \frac{t}{2}) *_{L} Q (x_{n} - l; \frac{t}{2}) \geq_{L} ζ . \end{matrix}$ (13) Let $\begin{matrix} sup_{t \to \infty} Q (x_{n} - l; \frac{t}{2}) = N ({a_{1}}^{'}) \\ or inf_{t \to \infty} Q (x_{n} - l; \frac{t}{2}) = N (a_{1}), \end{matrix}$ where $N (a_{1}) \in (0, 1)$ and (10) and (12) holds. Then, $\begin{matrix} lim_{n \to \infty} \frac{1}{n} & | {n : Q (x_{n} - l; \frac{t}{2}) > /_{L} ζ^{'} - \frac{ξ}{2} \\ or Q (x_{n} - l; \frac{t}{2}) >_{L} ζ_{1} + \frac{ξ}{2}} | = 0, \end{matrix}$ that is true for all ξ > 0. Hence $\begin{matrix} lim_{n \to \infty} \frac{1}{n} | {n : & Q (x_{n} - l; \frac{t}{2}) \leq_{L} ζ^{'} \\ or Q (x_{n} - l; \frac{t}{2}) \geq_{L} ζ_{1}} | = 0, \end{matrix}$ is true for any $N (a) \leq N (a_{1}) \in (0, 1)$ , whereas $N ({a_{1}}^{'})$ is the LUB or $N (a_{1})$ is the GLB.

Then iterate the procedure by taking (11) and (12) instead of (10) and (12).

If (11) and (12) are true, then $\begin{matrix} inf_{t \to \infty} Q (x_{n} - l; \frac{t}{2}) = N ({a_{1}}^{'}) \\ or sup_{t \to \infty} Q (x_{n} - l; \frac{t}{2}) = N (a_{1}) . \end{matrix}$ On contrary suppose that

$N ({a_{1}}^{'}) \neq inf_{t \to \infty} Q (x_{n} - l; \frac{t}{2})$ or

$N (a_{1}) \neq sup_{t \to \infty} Q (x_{n} - l; \frac{t}{2})$ and conditions (11) and (12) be satisfied.

It shows that ∃ some $N (ɛ) \in (0, 1)$ thus any $N (r^{'}) = Q (x_{n} - l; \frac{t}{2})$ for some t > 0

if $N ({a_{1}}^{'}) >_{L} N (r^{'})$ or $N (a_{1}) > /_{L} N (ɛ)$ .

Since (11) and (12) are satisfied. If $\begin{matrix} inf_{t \to \infty} Q (x_{n} - l; \frac{t}{2}) = N ({a_{2}}^{'}) \\ or sup_{t \to \infty} Q (x_{n} - l; \frac{t}{2}) = N (a_{2}) \end{matrix}$ Then $N ({a_{1}}^{'}) >_{L} N ({a_{2}}^{'}) or N (a_{1}) > /_{L} N (a_{2}) .$ (14) and from (12), we get $\begin{matrix} Q (x_{n} - l; \frac{t}{2}) *_{L} & N ({a_{2}}^{'}) \leq_{L} ζ^{'} \\ or Q (x_{n} - l; \frac{t}{2}) *_{L} ζ_{2} \geq_{L} ζ . \end{matrix}$

Using (11), we get $\begin{matrix} (ζ^{'} + \frac{ξ}{2}) *_{L} & (N ({a_{2}}^{'})) \leq_{L} ζ^{'} \\ or Q (ζ - \frac{ξ}{2}) *_{L} ζ_{2} \geq ζ \end{matrix}$ if for every ξ > 0. Therefore clearly, $\begin{matrix} (ζ^{'} + \frac{ξ}{2}) *_{L} & (N ({a_{2}}^{'})) \leq_{L} ζ^{'} \\ or (ζ + \frac{ξ}{2}) *_{L} ζ_{2} \geq_{L} ζ \end{matrix}$ (15) for all ξ > 0. Now, $\begin{matrix} N ({a_{1}}^{'}) & = sup_{t \to \infty} Q (x_{n} - l; \frac{t}{2}) \\ or ζ_{1} = inf_{t \to \infty} Q (x_{n} - l; \frac{t}{2}) \end{matrix}$ where $N (a_{1}) \in (0, 1)$ and (10) and (12) are true. From (14) we show that $N ({a_{2}}^{'})$ is another value satisfying (10) and (12). Therefore $\begin{matrix} N ({a_{1}}^{'}) > /_{L} N ({a_{2}}^{'}) or N (a_{2}) > /_{L} N (a_{1}) . \end{matrix}$ This contradicts (13). Thus $\begin{matrix} N ({a_{1}}^{'}) = inf_{t \to \infty} Q (x_{n} - l; \frac{t}{2}) \\ or N (a_{1}) = sup_{t \to \infty} Q (x_{n} - l; \frac{t}{2}) \end{matrix}$ satisfying conditions (11) and (12). Thus, the inequality becomes true for each

$N (a) \geq_{L} N (a_{1}) \in (0, 1)$ , since $N ({a_{1}}^{'})$ is the greatest lower bound(GLB) and therefore, $\begin{matrix} lim_{n \to \infty} \frac{1}{n} | {n : & Q (x_{n} - l; \frac{t}{2}) \leq_{L} ζ^{'} \\ or Q (x_{n} - l; \frac{t}{2}) \leq_{L} ζ} | = 0, \end{matrix}$ for each t > 0 and $N (a) \in (0, 1) .$

Therefore, ${stat}_{Q} - lim x = l .$

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