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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. (3) n→∞ nα

Now we define

yn = ξ, if k xn − ξ k≤ r ξ − xn = xn + r , otherwise k xn − ξ k A NOTE ON ROUGH STATISTICAL CONVERGENCE OF ORDER α 5

Then we can write

k yn − ξ k = 0, if k xn − ξ k≤ r

= k xn − ξ k−r, otherwise

and by definition of yn, we have k xn − yn k≤ r for all n ∈ N. Hence by (3) and the 1 definition of yn we get lim α |{k ≤ n :k yk − ξ k≥ ε}| = 0. Which implies that the n→∞ n sequence {yn}n∈N is statistically covergent of order α to ξ.

The condition is sufficient.

Since {yn}n∈N is statistically convergent of order α to ξ, we have 1 lim α |{k ≤ n :k yk − ξ k≥ ε}| = 0 for all ε> 0. Also since for a given r > 0 and for n→∞ n the sequence x = {xn}n∈N k xn − yn k< r, the inclusion {k ≤ n :k xk − ξ k≥ r + ε}⊆ 1 {k ≤ n :k yk − ξ k≥ ε} holds. Hence we get lim α |{k ≤ n :k xk − ξ k≥ r + ε}| = 0. n→∞ n This completes the proof. 

Theorem 3.6. For an arbitrary c ∈ Γx, where Γx is the set of all rough statistical cluster points of a sequence x ∈ X, we have for a positive real number r, k ξ − c k≤ r rα for all ξ ∈ st-LIMx . Proof. Let 0 < α ≤ 1 be given. On the contrary let assume that there exists a point rα kξ−ck−r c ∈ Γx and ξ ∈ st-LIMx such that k ξ − c k> r. Choose ε = 3 . Then

{k ≤ n :k xk − ξ k≥ r + ε}⊇{k ≤ n :k xk − c k<ε} (4) 1 holds. Since c ∈ Γx, we have lim α |{k ≤ n :k xk − c k< ε}| 6= 0. Hence by (4) n→∞ n 1 we have lim α |{k ≤ n :k xk − ξ k< r + ε}| 6= 0. This is a contradictin to the fact n→∞ n rα  ξ ∈ st-LIMx . Theorem 3.7. Let x be sequence in the strictly convex space. Let r and α be two rα positive real numbers. If for any y1,y2 ∈ st-LIMx with k y1 − y2 k= 2r, then x is y1+y2 statistically convergent of order α to 2 . rα Proof. Let z ∈ Γx. Then for any y1,y2 ∈ st − LIMx implies

k y1 − z k≤ r and k y2 − z k≤ r. (5)

On the other hand we have

2r =k y1 − y2 k≤k y1 − z k + k y2 − z k . (6) 6 M. MAITY

Hence by (5) and (6) we get k y1 − z k=k y2 − z k= r. Since 1 1 (y2 − y1)= [(z − y1) + (y2 − z)] (7) 2 2

1 and k y1 −y2 k=2r, we get k 2 (y2 −y1) k= r. By strict convexity of the space and from 1 1 the equality(7) we get 2 (y1 − y2)= z − y1 = y2 − z, which implies that z = 2 (y1 + y2). Hence z is unique statistical cluster point of the sequence x. On the other hand, from rα rα the assumption y1,y2 ∈ st-LIMx implies that st-LIMx 6= ∅. So by Theorem 3.1 the sequence x is statistically bounded of order α. Since z is the unique statistical cluster point ot the statistically bounded sequence x of order α we have the sequence x is 1  statistically convergent to z = 2 (y1 + y2). Theorem 3.8. Let 0 < α ≤ 1 and x and y be two sequences. Then α α (i) if r -st- lim x = x0 and c ∈ R, then r -st- lim cx = cx0 α α α (ii)if r -st- lim x = x0 and r -st- lim y = y0 then r -st- lim(x + y)= x0 + y0. Proof. (i) If c = 0 it is trivial. Suppose that c 6= 0. Then the proof of (i) follows from 1 1 r+ε 1 nα |{k ≤ n :k cxk − cx0 k≥ r + ε}| ≤ nα |{k ≤ n :k xk − x0 k≥ |c| }| = nα |{k ≤ n :k r ε xk − x0 k≥ |c| + |c| }|. Since x is rough statistical convergent of order α, hence cx is also rough statistical convergent of order α. Again 1 1 |{k ≤ n :k (x + y ) − (x0 + y0) k≥ r + ε}| ≤ |{k ≤ n :k x − x0 k≥ r + ε}| nα k k nα k 1 + |{k ≤ n :k y − y0 k≥ r + ε}| nα k 

It is easy to see that every convergent sequence is rough statistical convergent of order α, but the converse is not true always. Example 3.1. Let us consider the following sequence of real numbers defined by, 3 xk = 1, if k = n = 0, otherwise Then it is easy to see that the sequence is rough statistical convergent of order α with α 1 rS - lim xk =0 for α> 3 , but it is not a convergent sequence. Theorem 3.9. Let 0 < α ≤ β ≤ 1. Then rSα ⊆ rSβ, where rSα and rSβ denote the set of all rough statistical convergent sequence of order α and β respectively. Proof. If 0 < α ≤ β ≤ 1 then 1 1 |{k ≤ n :k x − l k≥ r + ε}| ≤ |{k ≤ n :k x − l k≥ r + ε}| nβ k nα k A NOTE ON ROUGH STATISTICAL CONVERGENCE OF ORDER α 7 for every ε> 0 and some r> 0 with limit l. Clearly this shows that rSα ⊆ rSβ.  We do not know whether for a sequence x in X, r > 0 and for 0 < α < 1, rα diam(st-LIMx ) ≤ 2r is true or not. So we leave this above fact as an open problem. Open Problem 3.1. Is it true for a sequence x in X, r > 0 and 0 <α < 1, rα diam(st-LIMx ) ≤ 2r. Acknowledgement: I express my gratitude to Prof. Salih Aytar, Suleyman Demirel University, Turkey, for his paper entitled “Rough Statistical Convergence” which in- spired me to prepare and develope this paper. I also express my gratitude to Prof. Pratulananda Das, Jadavpur University, India and Prof. Prasanta Malik, Burdwan University, India for their advice in preparation of this paper. References

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