A NOTE ON ROUGH STATISTICAL CONVERGENCE OF ORDER α MANOJIT MAITY 25 Teachers Housing Estate, P.O.- Panchasayar, Kolkata- 700094, West Bengal, In- dia. Email: [email protected] Abstract. In this paper, in the line of Aytar[1] and C¸olak [2], we introduce the notion of rough statistical convergence of order α in normed linear spaces and study some properties of the set of all rough statistical limit points of order α. Key words and phrases : Rough statistical convergence of order α, rough statistical limit points of order α, AMS subject classification (2010) : 40A05, 40G99 . 1. Introduction: The concept of statistical convergence was introduced by Steinhaus [9] and Fast [3] and later it was reintroduced by Schoenberg [8] independently. Over the years a lot of works have been done in this area. The concept of rough statistical convergence of single sequences was first introduced by S. Aytar [1]. Later the concept of statistical convergence of order α was introduced by R. C¸olak [2]. If x = {xn}n∈N is a sequence in some normed linear space (X, k . k) and r is a nonnegative real number then x is said to rough statistical convergent to ξ ∈ X if for 1 any ε> 0, lim |{k ≤ n :k xk − ξ k≥ r + ε}| = 0, [1]. n→∞ n For r = 0, rough statistical convergence coincides with statistical convergence. In this paper following the line of Aytar [1] and C¸olak [2] we introduce the notion of rough statistical convergence of order α in normed linear spaces and prove some properties of the set of all rough statistical limit points of order α. arXiv:1603.00183v1 [math.FA] 1 Mar 2016 2. Basic Definitions and Notations Definition 2.1. Let K be a subset of the set of positive integers N. Let Kn = {k ∈ |Kn| K : k ≤ n}. Then the natural density of K is given by lim , where |Kn| denotes n→∞ n the number of elements in Kn. 1 2 M. MAITY Definition 2.2. Let K be a subset of the set of positive integers N and α be any real number with 0 < α ≤ 1. Let Kn = {k ∈ K : k ≤ n}.Then the natural density of order |Kn| α of K is given by lim α , where |Kn| denotes the number of elements in Kn. n→∞ n Note 2.1. Let x = {xn}n∈N be a sequence. Then x satisfies some property P for all k except a set whose natural density is zero. Then we say that the sequence x satisfies P for almost all k and we abbreviated this by a.a.k. Note 2.2. Let x = {xn}n∈N be a sequence. Then x satisfies some property P for all k except a set whose natural density of order α is zero. Then we say that the sequence x satisfies P for almost all k and we abbreviated this by a.a.k(α). Definition 2.3. Let x = {xn}n∈N be a sequence in a normed linear space (X, k . k) and r be a nonnegative real number. Let 0 < α ≤ 1 be given. Then x is said to be st rα rough statistical convergent of order α to ξ ∈ X, denoted by x −→− ξ if for any ε> 0, 1 lim α |{k ≤ n :k xk − ξ k≥ r + ε}| = 0, that is a.a.k.(α) k xk − ξ k< r + ε for every n→∞ n ε> 0 and some r> 0. In this case ξ is called a rα-st-limit of x. The set of all rough statistical convergent sequences of order α will be denoted by rSα for fixed r with 0 < r ≤ 1. Throughout this paper X will denote a normed linear space and r will denote a nonnegative real number and x will denote the sequence x = {xn}n∈N in X. In general, the rα-st-limit point of a sequence may not be unique. So we consider α α rα st−r r -st-limit set of a sequence x, which is defined by st-LIMx = {ξ ∈ X : xn −→ ξ}. α rα The sequence x is said to be r -statistical convergent provided that st-LIMx 6= ∅. r For unbounded sequence rough limit set LIMx = ∅. rα But in case of rough statistical convergence of order α st-LIMx 6= ∅ even though the sequence is unbounded. For this we consider the following example. Example 2.1. Let X = R. We define a sequence in the following way, n 2 xn = (−1) : i 6= n , α =1 = n, otherwise Then rα st-LIMx = ∅, if r< 1 = [1 − r, r − 1], otherwise. rα and LIMx = ∅ for all r ≥ 0. Definition 2.4. A sequence {xn}n∈N is said to be statistically bounded if there exists 1 a positive real number M such that lim |{k ≤ n :k xk k≥ M}| =0 n→∞ n A NOTE ON ROUGH STATISTICAL CONVERGENCE OF ORDER α 3 Definition 2.5. Let 0 < α ≤ 1 be given. A sequence {xn}n∈N is said to be statistically 1 bounded of order α if there exists a positive real number M such that lim α |{k ≤ n→∞ n n :k xk k≥ M}| =0 3. Main Results Theorem 3.1. Let x be a sequence in X. Then x is statistically bounded of order α if rα and only if there exists a nonnegative real number r such that st-LIMx 6= ∅. Proof. The condition is necessary. Since the sequence x is statistically bounded, there exists a positive real number M 1 such that lim α |{k ≤ n :k xk k≥ M}| = 0. Let K = {k ∈ N :k xk k≥ M}. Define n→∞ n ′ ′ c r α r = sup{k xk k: k ∈ K }. Then the set st-LIM contains the origin of X. Hence ′ x r α st-LIMx 6= ∅. The condition is sufficient. rα rα Let st-LIMx 6= ∅ for some r ≥ 0. Then there exists l ∈ X such that l ∈ st-LIMx . 1 Then lim α |{k ≤ n :k xk − l k≥ r + ε}| = 0 for each ε> 0. Then we say that almost n→∞ n all xk’s are contained in some ball with any radius greater than r. So the sequence x is statistically bounded. ′ rα Theorem 3.2. If x = {xnk }k∈N is a subsequence of x = {xn}n∈N then st-LIMx ⊆ rα st LIM ′ - x . Proof. The proof is straight forward. So we omit it. rα Theorem 3.3. st-LIMx , the rough statistical limit set of order α of a sequence x is closed. rα rα Proof. If st-LIMx = ∅ then there is nothing to prove. So we assume that st-LIMx 6= rα ∅. We can choose a sequence {yk}k∈N ⊆ st-LIMx such that yk −→ y∗ for k → ∞. It rα suffices to prove that y∗ ∈ st-LIMx . N ε Let ε > 0. Since yk −→ y∗ there exists kε ∈ such that k yk − y∗ k< 2 for k > kε. N ε Now choose k0 ∈ such that k0 > kε. Then we can write k yk0 − y∗ k< . Again since α α 2 r r {yk}k∈N ⊆ st-LIMx we have yk0 ∈ st-LIMx . This implies 1 ε lim |{k ≤ n :k xk − yk0 k≥ r + }| =0. (1) n→∞ nα 2 Now we show the inclusion ε {k ≤ n :k x − y k< r + }⊆{k ≤ n :k x − y k< r + ε} (2) k k0 2 k ∗ 4 M. MAITY holds. ε ε Choose j ∈{k ≤ n :k xk − yk0 k< r + 2 }. Then we have k xj − yk0 k< r + 2 and hence k xj − y∗ k≤k xj − yk0 k + k yk0 − y∗ k< r + ε which implies j ∈{k ≤ n :k xk − y∗ k< r + ε}, which proves the inclusion (2). From (1) we can say that the set on the right hand side of(2) has natural density 1. Then the set on the left hand side of (2) must have natural density 1. Hence we get 1 lim α |{k ≤ n :k xk − y∗ k≥ r + ε}| = 0. n→∞ n This completes the proof. Theorem 3.4. Let x be a sequence in X. Then the rough statistical limit set of order rα α st-LIMx is convex. rα Proof. Choose y1,y2 ∈ st-LIMx and let ε > 0. Define K1 = {k ≤ n :k xk − y1 k≥ rα r + ε} and K2 = {k ≤ n :k xk − y2 k≥ r + ε}. Since y1,y2 ∈ st-LIMx , we have 1 1 lim α |K1| = lim α |K2| = 0. Let λ be any positive real number with 0 ≤ λ ≤ 1. n→∞ n n→∞ n Then k xk − [(1 − λ)y1 + λy2] k=k (1 − λ)(xk − y1)+ λ(xk − y2) k< r + ε c c 1 c c for each k ∈ K1 ∩ K2 . Since lim α |K1 ∩ K2 | = 1, we get n→∞ n 1 lim |{k ≤ n :k xk − [(1 − λ)y1 + λy2] k≥ r + ε}| =0 n→∞ nα that is rα [(1 − λ)y1 + (λ)y2] ∈ st-LIMx rα which proves the convexity of the set st-LIMx . Theorem 3.5. Let x be a sequence in X and r > 0. Then the sequence x is rough statistical convergent of order α to ξ ∈ X if and only if there exists a sequence y = {yn}n∈N in X such that y is statistically convergent of order α to ξ and k xn − yn k≤ r for all n ∈ N. Proof. The condition is necessary. st-rα Let xn −→ ξ. Choose any ε> 0. Then we have 1 lim |{k ≤ n :k xk − ξ k≥ r + ε}| = 0 for some r> 0.