Statistical limit superior and limit inferior in intuitionistic fuzzy normed spaces

Abstract

Recently, the concepts of statistical convergence, ideal convergence and lacunary statistical convergence have been studied in intuitionistic fuzzy normed spaces. In this article, we study the concepts of statistical limit superior and statistical limit inferior in intuitionistic fuzzy normed spaces. We also give an example to compute these points in intuitionistic fuzzy normed spaces.

AMS Subject Classification (2000): 40A05; 40D25; 11B05; 60H10; 60B99; 26A03.

1 Introduction and preliminaries

The concept of statistical convergence was first introduced by Fast [1] which was extended for double sequences in [2, 3]. In particular, active researches on this topic were started after the study of Fridy [4]. 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 [5] introduced the concept of statistical limit points and Fridy and Orhan [6] 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 [7–9]. Quite recently, the idea of statistical convergence in intuitionistic fuzzy normed spaces for single sequences has been studied in [10, 11]; and for double sequences by Mursaleen and Mohiuddine [12, 13].

Recently, Saadati and Park [14] introduced the notion of intuitionistic fuzzy normed space and quite recently, in [15, 16] 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.

In this article, we study the concept of statistical limit superior and statistical limit inferior in intuitionistic fuzzy normed spaces. An example is demonstrated to determine these points in intuitionistic fuzzy normed space. We observe that our results are analogous to the results of Fridy and Orhan [6] but proofs are somewhat different when we deal with these concepts in intuitionistic fuzzy normed spaces.

We recall some basic definitions and notations.

Definition 1.1 [14]. A binary operation *: [0, 1] × [0, 1] → [0, 1] is said to be a continuous t-norm if it satisfies the following conditions:

• (a) * is associative and commutative,

• (b) * is continuous,

• (c) a * 1 = a for all a ∈ [0, 1],

• (d) a * b ≤ c * d whenever a ≤ c and b ≤ d for each a, b, c, d ∈ [0, 1].

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 binary operation ◊: [0, 1] × [0, 1] → [0, 1] is said to be a continuous t-conorm if it satisfies the conditions (a), (b), (d) as above and a◊0 = a for all a ∈[0, 1].

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

Definition 1.2 [14]. The five-tuple (X, μ, ν, *, ◊) is said to be an intuitionistic fuzzy normed space (for short, IFNS) if X is a vector space,* is a continuous t-norm, ◊ is a continuous t-conorm, and μ, ν are fuzzy sets on X × (0, ∞) satisfying the following conditions. For every x, y ∈ X and s, t > 0,

• (a) μ(x, t) + ν(x, t) ≤ 1,

• (b) μ(x, t) > 0,

• (c) μ(x, t) = 1 if and only if x = 0,

• (d) $\mu \left(\alpha x,t\right)=\mu \left(x,\frac{t}{\left|\alpha \right|}\right)$ for each α ≠ 0,

• (e) μ(x, t) * μ(y, s) ≤ μ(x + y, t + s),

• (f) μ(x, ·): (0, ∞) → [0, 1] is continuous,

• (g) $\underset{t\to \infty }{\text{lim}}\mu \left(x,t\right)=1$ and $\underset{t\to 0}{\text{lim}}\mu \left(x,t\right)=0$,

• (h) ν (x, t) < 1,

• (i) ν (x, t) = 0 if and only if x = 0,

• (j) $\nu \left(\alpha x,t\right)=\nu \left(x,\frac{t}{\left|\alpha \right|}\right)$ for each α ≠ 0,

• (k) ν (x, t)◊ν(y, s) ≥ ν(x + y, t + s),

• (l) ν (x, ·): (0, ∞) → [0, 1] is continuous,

• (m) $\underset{t\to \infty }{\text{lim}}\nu \left(x,t\right)=0$ and $\underset{t\to 0}{\text{lim}}\nu \left(x,t\right)=1$.

In this case (μ, ν) is called an intuitionistic fuzzy norm.

Example. Suppose that (X, || ||) is a normed space and let a*b = ab and a◊b = min{a+b, 1} for all a, b ∈ [0, 1]. For all x ∈ X and every t > 0, consider

$$\mu \left(x;t\right):=\frac{t}{t+∥x∥}\text{and}v\left(x;t\right):=\frac{∥x∥}{t+∥x∥}.$$

Then (X, μ, ν, *, ◊) is an intuitionistic fuzzy normed space.

Definition 1.3 [14]. Let (X, μ, ν, *, ◊) be an intuitionistic fuzzy normed space. Then a sequence x = (x _n ) is said to be convergent to L ∈ X with respect to the intuitionistic fuzzy norm (μ, ν) if for every ϵ > 0 and t > 0, there exists a positive integer k_o such that μ(x _n - L; t) > 1 - ϵ and ν(x _n - L; t) < ϵ, whenever n ≥ k_o. In this case, we write (μ, ν)-lim x = L or $x_n\stackrel{\left(\mu ,\nu \right)}{\to }L$ as n → ∞.

Definition 1.4 [14]. Let (X, μ, ν, *, ◊) be an intuitionistic fuzzy normed space. Then a sequence x = (x _n ) is said to be a Cauchy sequence with respect to the intuitionistic fuzzy norm (μ, ν) if for every ϵ > 0 and t > 0, there exists a positive integer k_o such that μ(x _n - x _m ; t) > 1 - ϵ and ν(x _n - x _m ; t) < ϵ for all n, m ≥ k_o.

Definition 1.5 [17]. If K is a subset of ℕ, then the natural density of K denoted by δ(K), is defined as

$$\delta \left(K\right):=\underset{n}{\text{lim}}\frac{1}{n}\left|\left\{k\le n:k\in K\right\}\right|,$$

where the vertical bars denote the cardinality of the enclosed set.

Definition 1.6 [4, 18]. A sequence x = (x _n ) of numbers is said to be statistically convergent to L if

$$\delta \left(\left\{k\le n:|x_k-L|\ge \mathit{ϵ}\right\}\right)=0$$

for every ϵ > 0. In this case we write st-lim x = L.

Definition 1.7 [5, 6]. A sequence x = (x _n ) of numbers is said to be statistically bounded if there is a number B such that

$$\delta \left(\left\{k\le n:\left|x_k\right|\ge B\right\}\right)=0.$$

Definition 1.8 [5]. If {x_k(j)} is a subsequence of x = (x _k ) and K: = {k(j): j ∈ ℕ}, then we abbreviate {x_k(j)} by {x} _K . If δ(K) = 0 then {x} _K is called a subsequence of density zero or a thin subsequence. On the other hand, {x} _K is a nonthin subsequence of x if K does not have density zero.

Definition 1.9 [10, 12]. Let (X, μ, ν, *, ◊) be an intuitionistic fuzzy normed space. We say that a sequence x = (x _k ) is statistically convergent to L ∈ X with respect to the intuitionistic fuzzy normed (μ, ν) provided that for every t > 0 and b ∈ (0, 1)

$$\delta \left(\left\{k\le n:\mu \left(x_k-L;t\right)<1-b\text{or}\nu \left(x_k-L;t\right)>b\right\}\right)=0.$$

In this case we write st_{(μ, ν)}- lim x = L.

2 Statistical limit superior and inferior in IFNS

In this section, we define limit point, statistical limit point, statistical cluster point, statistical limit superior, and statistical limit inferior in intuitionistic fuzzy normed spaces and demonstrate through an example how to compute these points in a IFN-spaces.

Definition 2.1. A sequence x in an intuitionistic fuzzy normed space (X, μ, ν, *, ◊) is said to be statistically bounded if there exists some t_o > 0 and b ∈ (0, 1) such that δ({k: μ(x _k ; t_o) > 1 - b or ν(x _k ; t_o) < b}) = 0.

Definition 2.2. Let (X, μ, ν,*,◊) be an intuitionistic fuzzy normed space. Then l ∈ X is called a limit point of the sequence x = (x _k ) with respect to the intuitionistic fuzzy norm (μ, ν) provided that there is a subsequence of x that converges to l with respect to the intuitionistic fuzzy norm (μ, ν). Let L_{(μ, ν)}(x) denotes the set of all limit points of the sequence x with respect to the intuitionistic fuzzy norm (μ, ν).

Definition 2.3. Let (X, μ, ν,*,◊) be an intuitionistic fuzzy normed space. Then ξ ∈ X is called a statistical limit point of the sequence x = (x _k ) with respect to the intuitionistic fuzzy norm (μ, ν) provided that there is a nonthin subsequence of x that converges to ξ with respect to the intuitionistic fuzzy norm (μ, ν). In this case we say ξ is a st_{(μ, ν)}-limit point of sequence x = (x _k ). Let Λ_{(μ, ν)}(x) denotes the set of all st_{(μ, ν)}-limit points of the sequence x.

Definition 2.4. Let (X, μ, ν,*,◊) be an intuitionistic fuzzy normed space. Then η ∈ X is called a statistical cluster point of the sequence x = (x _k ) with respect to the intuitionistic fuzzy norm (μ, ν) provided that for every t_o > 0 and a ∈ (0, 1),

$$\bar{\delta }\left(\left\{k\le n:\mu \left(x_k-\eta ;t_o\right)>1-a\text{or}v\left(x_k-\eta ;t_o\right)<a\right\}\right)=0.$$

In this case we say η is a st_{(μ, ν)}-cluster point of the sequence x. Let Γ_{(μ, ν)}(x) denotes the set of all st_{(μ, ν)}-cluster points of the sequence x.

Definition 2.5. For a sequence x in an intuitionistic fuzzy normed space (X, μ, ν,*,◊), we define the sets $B_x^{\left(\mu ,\nu \right)}$ and $A_x^{\left(\mu ,\nu \right)}$ by

$$B_x^{\left(\mu ,\nu \right)}:=\left\{b\in \left(0,1\right):\delta \left(\left\{k:\mu \left(x_k;t\right)<1-b\text{or}\nu \left(x_k;t\right)>b\right\}\right)\ne 0\right\}$$

$$A_x^{\left(\mu ,\nu \right)}:=\left\{a\in \left(0,1\right):\delta \left(\left\{k:\mu \left(x_k;t\right)>1-a\text{or}\nu \left(x_k;t\right)<a\right\}\right)\ne 0\right\}$$

If x is a real number sequence then the statistical limit superior of x with respect to the intuitionistic fuzzy norm (μ, ν) is defined by

$$s{t}_{\left(\mu ,\nu \right)}\text{ - }\text{lim }\text{sup }x:=\left\{\begin{array}{ccc}\text{sup}\hfill & B_x^{\left(\mu ,\nu \right)}\text{if}\hfill & B_x^{\left(\mu ,\nu \right)}\ne \mathrm{0̸},\hfill \\ 0\hfill & \hfill \text{if}\hfill & B_x^{\left(\mu ,\nu \right)}\ne \mathrm{0̸}.\hfill \end{array}\right.$$

And the statistical limit inferior of x with respect to the intuitionistic fuzzy norm (μ, ν) is defined by

$$s{t}_{\left(\mu ,\nu \right)}-\text{lim }\text{inf }x:=\left\{\begin{array}{ccc}\text{inf}\hfill & A_x^{\left(\mu ,\nu \right)}\text{if}\hfill & A_x^{\left(\mu ,\nu \right)}\ne \mathrm{0̸},\hfill \\ 1\hfill & \hfill \text{if}\hfill & A_x^{\left(\mu ,\nu \right)}\ne \mathrm{0̸}.\hfill \end{array}\right.$$

Example. A simple example will help to illustrate the concepts just defined. Let the sequence x = (x _k ) be defined by

$$x_k:=\left\{\begin{array}{ccc}2k,\hfill & \text{if}\hfill & k\text{is}\text{an}\text{odd}\text{square},\hfill \\ -1,\hfill & \text{if}\hfill & k\text{is}\text{an}\text{even}\text{square},\hfill \\ 1/2,\hfill & \text{if}\hfill & k\text{is}\text{an}\text{odd}\text{nonsquare},\hfill \\ 0,\hfill & \text{if}\hfill & k\text{is}\text{an}\text{even}\text{nonsquare}\text{.}\hfill \end{array}\right.$$

Let $\mu \left(x_k;t\right)=\frac{t}{t+\left|x_k\right|}$ and $\nu \left(x_k;t\right)=\frac{\left|x_k\right|}{t+\left|x_k\right|}$.

The above sequence is clearly unbounded with respect to (μ, ν). On the other hand, it is statistically bounded with respect to (μ, ν). For this,

$$\begin{array}{c}\delta \left(\left\{k\le n:\mu \left(x_k;t_o\right)<1-b\text{or}\nu \left(x_k;t_o\right)>b\right\}\right)\\ =\delta \left(\left\{k\le n:\frac{t_o}{t_o+\left|x_k\right|}<1-b\text{or}\frac{\left|x_k\right|}{t_o+\left|x_k\right|}>b\right\}\right),\\ =\delta \left(\left\{k\le n:\left|x_k\right|>\frac{b{t}_o}{1-b}\right\}\right).\end{array}$$

Since $0<b<1,\frac{1}{b}-1>0$. Choose $t_{\text{o}}=\frac{1-b}{3b}$. Then t_o > 0 and

$$\begin{array}{c}\delta \left(\left\{k\le n:\mu \left(x_k;t_{\text{o}}\right)<1-b\text{or}\nu \left(x_k;t_{\text{o}}\right)>b\right\}\right)=\delta \left(\left\{k\le n:\left|x_k\right|>\frac{b}{1-b}\times \frac{1-b}{3b}=\frac{1}{3}\right\}\right)\\ =\delta \left(\left\{k\le n:\left|x_k\right|>\frac{1}{3}\right\}\right)=\underset{n\to \infty }{\text{lim}}\frac{1}{n}\times \sqrt{n}=0\end{array}$$

Hence it is statistically bounded with respect to (μ, ν).

To find $B_x^{\left(\mu ,\nu \right)}$, we have to find those b ∈ (0, 1) such that

$$\delta \left(\left\{k\le n:\mu \left(x_k;t\right)<1-b\text{or}\nu \left(x_k;t\right)>b\right\}\right)\ne 0.$$

Now,

$$\begin{array}{c}\delta \left(\left\{k\le n:\mu \left(x_k;t\right)<1-b\text{or}\nu \left(x_k;t\right)>b\right\}\right)\\ =\delta \left(\left\{k\le n:\frac{t}{t+\left|x_k\right|}<1-b\text{or}\frac{\left|x_k\right|}{t+\left|x_k\right|}>b\right\}\right),\\ =\delta \left(\left\{k\le n:\left|x_k\right|>\frac{bt}{1-b}\right\}\right).\end{array}$$

We can easily choose any t > 0 as $t<\frac{1}{3}\left(\frac{1}{b}-1\right)$ for 0 < b < 1, so that

$$0<\frac{bt}{1-b}<\frac{b}{1-b}\times \frac{1-b}{3b}=\frac{1}{3}.$$

Therefore

$$\delta \left(\left\{k\le n:\left|x_k\right|>\frac{bt}{1-b}\right\}\right)=\delta \left(\left\{k\le n:\left|x_k\right|>r=\frac{bt}{1-b}\right\}\right),$$

and by the above condition r ∈ (0, 1). Now the number of members of the sequence which satisfy the above condition is always greater than $n-\frac{n}{2}$ or $n-\frac{n-1}{2}$ for the case n is even or odd, respectively. Therefore

$$\delta \left(\left\{k\le n:\left|x_k\right|>r=\frac{bt}{1-b}\right\}\right)>\underset{n\to \infty }{\text{lim}}\frac{1}{n}\times \frac{n}{2}=\frac{1}{2}or\underset{n\to \infty }{\text{lim}}\frac{1}{n}\times \frac{n+1}{2}=\frac{1}{2}.$$

Thus

$$\delta \left(\left\{k\le n:\left|x_k\right|>r=\frac{bt}{1-b}\right\}\right)\ne 0\text{for}\text{all}b\in \left(0,1\right).$$

Hence

$$B_x^{\left(\mu ,\nu \right)}=\left(0,1\right),$$

and

$$s{t}_{\left(\mu ,\nu \right)}-\text{lim}\text{sup }x=1.$$

The above sequence has two subsequences

$$x=\left(x_{n_i}\right)\text{where}{x}_{n_i}=1\text{for each}{n}_i\in \left\{3,5,7,11,13,...\right\},$$

and

$$x=\left(x_{n_j}\right)\text{where}{x}_{n_j}=0\text{for each}{n}_j\in \left\{2,6,8,10,12,...\right\},$$

i, j ∈ ℕ; which are of positive density and clearly convergent to 1 and 0, respectively. Therefore, x is not statistically convergent. Similarly, we have

$$A_x^{\left(\mu ,\nu \right)}=\left(0,1\right),$$

and

$$s{t}_{\left(\mu ,\nu \right)}-\text{lim}\text{inf }x=0.$$

Hence the set of statistical cluster points of x is {0, 1}, where st_(μ,_ν)- lim inf x = least element and st_(μ,_ν)- lim sup x = greatest element of the above set.

This observation suggests the main idea of our first theorem of the following section.

3 Main results

The following results are analogs of the results due to Fridy and Orhan [6], while the proofs are different which show the technique to work with IFN-spaces. We observe that in contrast to the real case here from the definition limit sup cannot be infinite, as it can be at most 1.

Theorem 3.1. Let b = st_(μ,_ν)- lim sup x. Then for every positive numbers t and γ

$$\left.\begin{array}{c}\delta \left(\left\{k:\mu \left(x_k;t\right)<1-b+\gamma \text{or}v\left(x_k;t\right)>b-\gamma \right\}\right)\ne 0,\text{and}\hfill \\ \delta \left(\left\{k:\mu \left(x_k;t\right)<1-b+\gamma \text{or}v\left(x_k;t\right)>b+\gamma \right\}\right)\ne 0.\hfill \end{array}\right\}$$

(1)

Conversely, if (1) holds for every positive t and γ then b = st_(μ,_ν)- lim sup x.

Proof. Let b = st_(μ,_ν)-lim sup x, where b be finite. Then

$$\delta \left(\left\{k:\mu \left(x_k;t\right)<\text{1}-b\text{or}\nu \left(x_k;t\right)>b\right\}\right)\ne 0.$$

(2)

Since μ(x _k ; t) < 1 - b + γ or ν(x _k ; t) > b - γ for every k and for any t, γ > 0,

$$\delta \left(\left\{k\le n:\mu \left(x_k;t\right)<\text{1}-b+\gamma \text{or}\nu \left(x_k;t\right)>b-\gamma \right\}\right)\ne 0.$$

Now applying the definition of st_(μ,_ν)- lim sup x we have 1 - b as the least value and b as the greatest value satisfying (2).

Now if possible,

$$\mu \left(x_k;t\right)<\text{1}-b-\gamma \text{or}\nu \left(x_k;t\right)>b+\gamma \text{for}\text{some}\gamma >0.$$

Then 1 - b - γ and b + γ are another values with 1 - b - γ < 1 - b and b + γ > b which satisfies (2). This observation contradicts the fact that 1 - b and b are least and greatest values, respectively, which satisfies the above condition.

Hence,

$$\delta \left(\left\{k\le n:\mu \left(x_k;t\right)<\text{1}-b-\gamma \text{or}\nu \left(x_k;t\right)>b+\gamma \right\}\right)=0\text{for}\text{every}\gamma >0.$$

Conversely, if (1) holds for every positive t and γ, then

$$\delta \left(\left\{k\le n:\mu \left(x_k;t\right)<\text{1}-b+\gamma \text{or}\nu \left(x_k;t\right)>b-\gamma \right\}\right)\ne 0.$$

and

$$\delta \left(\left\{k\le n:\mu \left(x_k;t\right)<\text{1}-b-\gamma \text{ or }\nu \left(x_k;t\right)>b+\gamma \right\}\right)=0.$$

Therefore

$$\delta \left(\left\{k\le n:\mu \left(x_k;t\right)\le \text{1}-b\text{ or }\nu \left(x_k;t\right)\ge b\right\}\right)\ne 0$$

and

$$\delta \left(\left\{k\le n:\mu \left(x_k;t\right)=\text{1}-b\text{ or }\nu \left(x_k;t\right)=b\right\}\right)=0.$$

That is

$$\delta \left(\left\{k\le n:\mu \left(x_k;t\right)<\text{1}-b\text{ or }\nu \left(x_k;t\right)>b\right\}\right)\ne 0\text{for every}t>0.$$

Hence b = st_(μ,_ν)- lim sup x.

This completes the proof of the theorem.

The dual statement for st_(μ,_ν)- lim inf x can also be proved similarly.

Theorem 3.1 '. Let a = st_(μ,_ν)- lim inf x. Then for every positive number t and γ

$$\left.\begin{array}{c}\delta \left(\left\{k\le n:\mu \left(x_k;t\right)>\text{1}-a-\gamma \text{ or }\nu \left(x_k;t\right)<a+\gamma \right\}\right)\ne 0,\text{ and}\hfill \\ \delta \left(k\le n:\mu \left(x_k;t\right)>\text{1}-a+\gamma \text{ or }\nu \left(x_k;t\right)<a-\gamma \right)=0.\hfill \end{array}\right\}$$

(1A)

Conversely, if (1') holds for every positive t and γ then a = st_(μ,_ν)- lim inf x.

Remark. From the definition of statistical cluster points we see that Theorems 3.1 and 3.1' can be interpreted as saying that st_(μ,_ν)- lim sup x and st_(μ,_ν)- lim inf x are the greatest and the least statistical cluster points of x, respectively.

Theorem 3.2. For any sequence x, st_(μ,_ν)- lim inf x ≤ st_(μ,_ν)- lim sup x.

Proof. First consider the case in which st_(μ,_ν)- lim sup x = 0, which implies that

$$B_x^{\left(\mu ,\nu \right)}=\mathrm{\varnothing }.$$

Then for every b ∈ (0, 1),

$$B_x^{\left(\mu ,\nu \right)}=\delta \left(\left\{k:\mu \left(x_k;t\right)<\text{1}-b\text{ or }\nu \left(x_k;t\right)>b\right\}\right)=0,$$

that is

$$\delta \left(\left\{k:\mu \left(x_k;t\right)\ge \text{1}-b\text{ or }\nu \left(x_k;t\right)\le b\right\}\right)=\text{1}.$$

Also for every a ∈ (0, 1), we have

$$\delta \left(\left\{k:\mu \left(x_k;t\right)>\text{1}-a\text{ or }\nu \left(x_k;t\right)<a\right\}\right)\ne 0.$$

Hence, st_(μ,_ν)- lim inf x = 0.

The case in which st_(μ,_ν)- lim sup x = 1, is trivial.

Suppose that b = st_(μ,_ν)- lim sup x, and a = st_(μ,_ν)- lim inf x; where a and b are finite.

Now for given any γ, we show that $1-b-\gamma \in A_x^{\left(\mu ,\nu \right)}$.

By Theorem 3.1,

$$\delta \left(\left\{k:\mu \left(x_k;t\right)<\text{1}-b-\frac{\gamma }{2}\text{ or }\nu \left(x_k;t\right)>b+\frac{\gamma }{2}\right\}\right)=0,\text{where 1}-b=\text{least upper bound of }{B}_x^{\left(\mu ,\nu \right)}.$$

Therefore

$$\delta \left(\left\{k:\mu \left(x_k;t\right)\ge \text{1}-b-\frac{\gamma }{2}\text{ or }\nu \left(x_k;t\right)\le b+\frac{\gamma }{2}\right\}\right)=\text{1},$$

which in turn gives

$$\delta \left(\left\{k:\mu \left(x_k;t\right)>\text{1}-b-\gamma \text{ or }\nu \left(x_k;t\right)<b+\gamma \right\}\right)=\text{1}.$$

Hence, $1-b-\gamma \in A_x^{\left(\mu ,\nu \right)}$.

By definition

$$a=\text{inf }{A}_x^{\left(\mu ,\nu \right)},$$

so we conclude that

$$\text{1}-b-\gamma \le \text{1}-a,$$

and since γ is arbitrary,

$$\text{1}-b\le \text{1}-a,$$

that is

$$-b\le -a,$$

$$a\le b.$$

This completes the proof of the theorem.

Theorem 3.3. In an intuitionistic fuzzy normed space (X, μ, ν, *, ◊), the statistically bounded sequence x is statistically convergent if and only if

$$s{t}_{\left(\mu ,\nu \right)}-\text{lim inf }x=s{t}_{\left(\mu ,\nu \right)}-\text{lim sup }x.$$

Proof. Let α, β be st_(μ, _ν)- lim inf x and st_(μ, _ν)- lim sup x, respectively. Now we assume that st_(μ,_ν)- lim x = L. Then for every ϵ > 0 and b ∈ (0, 1),

$$\delta \left(\left\{k:\mu \left(x_k;t\right)\le \text{1}-b\text{ or }\nu \left(x_k;t\right)\ge b\right\}\right)=0,$$

so that

$$\delta \left(\left\{k:\mu \left(x_k;\frac{t}{2}\right)*\mu \left(L;\frac{t}{2}\right)\le 1-b\text{ or }\nu \left(x_k;\frac{t}{2}\right)♢\nu \left(L;\frac{t}{2}\right)\ge b\right\}\right)=0.$$

Let for every t > 0,

$$\underset{t}{\text{sup}}\mu \left(x_k;\frac{t}{2}\right)=1-b_1\text{ and }\underset{t}{\text{sup}}\mu \left(L;\frac{t}{2}\right)=1-b_2,$$

or

$$\underset{t}{\text{inf}}\nu \left(x_k;\frac{t}{2}\right)=b_1\text{ and }\underset{t}{\text{inf}}\nu \left(L;\frac{t}{2}\right)=b_2$$

such that

$$\left(\text{1}-b_{\text{1}}\right)*\left(\text{1}-b_{\text{2}}\right)\le \text{1}-b\text{ or }{b}_{\text{1}}♢b_{\text{2}}\ge b.$$

(3.1)

Then

$$\delta \left(\left\{k:\mu \left(x_k;\frac{t}{2}\right)\le 1-b_1\text{ or }\nu \left(x_k;\frac{t}{2}\right)\ge b_1\right\}\right)=0,$$

(3.2)

and therefore

$$\delta \left(\left\{k:\mu \left(x_k;\frac{t}{2}\right)<1-b_1-\gamma \text{ or }\nu \left(x_k;\frac{t}{2}\right)>b_1+\gamma \right\}\right)=0\text{ for every }\gamma >0.$$

(3.3)

Now applying Theorem 3.1 and the definition of st_(μ,_ν)- lim sup x, we get

$$\delta \left(\left\{k:\mu \left(x_k;\frac{t}{2}\right)<1-\beta -\gamma \text{ or }\nu \left(x_k;\frac{t}{2}\right)>\beta +\gamma \right\}\right)=0\text{ for every }\gamma >0.$$

(3.4)

From (3.3) and (3.4) and by the definition of st_(μ,_ν)- lim sup x, we get

$$\text{1}-b_{\text{1}}-\gamma \le \text{1}-\beta -\gamma \text{ or }{b}_{\text{1}}+\gamma \ge \beta +\gamma ,$$

that is,

$$\beta \le b_1.$$

(3.5)

Now we find those k such that

$$\mu \left(x_k;\frac{t}{2}\right)>1-b_1\text{ + }\gamma \text{ or }\nu \left(x_k;\frac{t}{2}\right)<b_1-\gamma .$$

We can easily observe that no such k exists which satisfy (3.1) and above condition together.

Therefore this implies that

$$\delta \left(\left\{k:\mu \left(x_k;\frac{t}{2}\right)>1-b_1\text{ + }\gamma \text{ or }\nu \left(x_k;\frac{t}{2}\right)<b_1-\gamma \right\}\right)=0.$$

Since α = st_(μ,_ν)- lim inf x, by Theorem 3.1', we get

$$\delta \left(\left\{k:\mu \left(x_k;\frac{t}{2}\right)>1-\alpha \text{ + }\gamma \text{ or }\nu \left(x_k;\frac{t}{2}\right)<\alpha -\gamma \right\}\right)=0.$$

By the definition of st_{(μ, ν)}- lim inf x, we have

$$\text{1}-\alpha +\gamma \le \text{1}-b_{\text{1}}+\gamma \text{ or }\alpha -\gamma \ge b_{\text{1}}-\gamma ,$$

that is,

$$b_1\le \alpha .$$

(3.6)

From (3.4) and(3.5), we get β ≤ α. Now combining Theorem 3.2 and the above inequality, we conclude α = β.

Conversely, suppose that α = β and let sup _t μ(L, t) = 1 - α or inf _t ν(L, t) = α. Then for any γ > 0, Theorems 3.1 and 3.1^'will together imply that

$$\delta \left(\left\{k:\mu \left(x_k;\frac{t}{2}\right)<1-\alpha \text{ + }\frac{\gamma }{2}\text{ or }\nu \left(x_k;\frac{t}{2}\right)>\alpha -\frac{\gamma }{2}\right\}\right)=0,$$

(3.7)

and

$$\delta \left(\left\{k:\mu \left(x_k;\frac{t}{2}\right)>1-\alpha \text{ + }\frac{\gamma }{2}\text{ or }\nu \left(x_k;\frac{t}{2}\right)<\alpha -\frac{\gamma }{2}\right\}\right)=0.$$

(3.8)

Now

$$\text{1}-\alpha \ge \mu \left(L;t\right)=\mu \left(x_k-\left(x_k-L\right);t\right)\ge \mu \left(x_k;\frac{t}{2}\right)*\mu \left(x_k-L;\frac{t}{2}\right),$$

and

$$\alpha \le \nu \left(L;t\right)=\nu \left(x_k-\left(x_k-L\right);t\right)\le \nu \left(x_k;\frac{t}{2}\right)♢\nu \left(x_k-L;\frac{t}{2}\right).$$

Therefore

$$\mu \left(x_k;\frac{t}{2}\right)*\mu \left(x_k-L;\frac{t}{2}\right)\le 1-\alpha \text{ or }\nu \left(x_k;\frac{t}{2}\right)♢\nu \left(x_k-L;\frac{t}{2}\right)\ge \alpha .$$

(3.9)

Let ${\text{sup}}_t\left\{\mu \left(x_k-L;\frac{t}{2}\right)\right\}=1-a_1$ or ${\text{inf}}_t\left\{\nu \left(x_k-L;\frac{t}{2}\right)\right\}=a_1$, where a₁ ∈ (0, 1) and (3.7) and (3.9) hold. Then

$$\delta \left(\left\{k:\mu \left(x_k-L;\frac{t}{2}\right)<1-{\alpha }_1-\frac{\gamma }{2}\text{ or }\nu \left(x_k-L;\frac{t}{2}\right)>{\alpha }_1+\frac{\gamma }{2}\right\}\right)=0,$$

which is true for all γ > 0. Hence

$$\delta \left(\left\{k:\mu \left(x_k-L;\frac{t}{2}\right)\le 1-{\alpha }_1\text{ or }\nu \left(x_k-L;\frac{t}{2}\right)\ge {\alpha }_1\right\}\right)=0,$$

which is true for all a ≤ a₁ ∈ (0, 1), because 1 - a₁ is the least upper bound or a₁ is the greatest lower bound.

Now repeat the process by taking (3.8) and (3.9) instead of (3.7) and (3.9). If (3.8) and (3.9) are satisfied, then ${\text{inf}}_t\mu \left(x_k-L;\frac{t}{2}\right)=1-a_1$ or ${\text{sup}}_t\nu \left(x_k-L;\frac{t}{2}\right)=a_1$,

On contrary suppose that $1-a_1\ne {\text{inf}}_t\mu \left(x_k-L;\frac{t}{2}\right)$ or $a_1\ne {\text{sup}}_t\nu \left(x_k-L;\frac{t}{2}\right)$ and conditions (3.8) and (3.9) be satisfied. This implies that there exists some r ∈ (0, 1) such that either $1-r=\mu \left(x_k-L;\frac{t}{2}\right)$ or $r=\nu \left(x_k-L;\frac{t}{2}\right)$ for some t > 0 where 1 - a₁ > 1 - r or a₁ < r.

As (3.8) and (3.9) are satisfied, and let us suppose that ${\text{inf}}_t\mu \left(x_k-L;\frac{t}{2}\right)=1-a_2$ or ${\text{sup}}_t\nu \left(x_k-L;\frac{t}{2}\right)=a_2$.

Then

$$\text{1}-a_{\text{1}}>\text{1}-a_{\text{2}}\text{ or }{a}_{\text{1}}<a_{\text{2}},$$

(3.10)

and from (3.9), we get

$$\mu \left(x_k-L;\frac{t}{2}\right)*\left(1-a_2\right)\le 1-\alpha \text{ or }\nu \left(x_k-L;\frac{t}{2}\right)♢{\alpha }_2\ge \alpha .$$

Using (3.8), we get

$$\left(1-\alpha +\frac{\gamma }{2}\right)*\left(1-a_2\right)\le 1-\alpha \text{ or }\nu \left(\alpha -\frac{\gamma }{2}\right)♢\left({\alpha }_2\right)\ge \alpha \text{ for all }\gamma >0.$$

Clearly,

$$\left(1-\alpha +\frac{\gamma }{2}\right)*\left(1-a_2\right)\le 1-\alpha \text{ or }\left(\alpha +\frac{\gamma }{2}\right)♢\left({\alpha }_2\right)\ge \alpha \text{ for all }\gamma >0.$$

(3.11)

Now

$$1-a_1=\underset{t}{\text{sup}}\mu \left(x_k-L;\frac{t}{2}\right)\text{ or }{\alpha }_1=\underset{t}{\text{inf}}\nu \left(x_k-L;\frac{t}{2}\right)$$

where a₁ ∈ (0, 1) and which satisfy (3.7) and (3.9).

From (3.11) we conclude that 1 - a₂ is another value satisfying (3.7) and (3.9).

Hence

$$\text{1}-a_{\text{1}}<\text{1}-a_{\text{2}}\text{ or }{a}_{\text{2}}<a_{\text{1}}.$$

This contradicts (3.10). Hence $1-a_1={\text{inf}}_t\mu \left(x_k-L;\frac{t}{2}\right)$ or $a_1={\text{sup}}_t\nu \left(x_k-L;\frac{t}{2}\right)$ satisfying conditions (3.8) and (3.9).

Therefore the inequality becomes true for all a ≥ a₁ ∈ (0, 1), because 1 - a₁ is the greatest lower bound, and hence

$$\delta \left(\left\{k:\mu \left(x_k-L;\frac{t}{2}\right)\le 1-\alpha \text{ or }\nu \left(x_k-L;\frac{t}{2}\right)\le \alpha \right\}\right)=0,$$

for each t > 0 and a ∈ (0, 1). Therefore

$$s{t}_{\left(\mu ,\nu \right)}\text{ - lim }x=L.$$

This completes the proof of the theorem.