STATISTICALLY CONVERGENT DOUBLE SEQUENCE SPACES IN n-normed SPACES

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1 STATISTICALLY CONVERGENT DOUBLE SEQUENCE SPACES IN n-normed SPACES Vakeel A. Khan*, Sabiha Tabassum** and Ayhan Esi*** Department of Mathematics A.M.U. Aligarh (INDIA) Department of Mathematics, University of Adiyaman, Altinsehir,02040, Adiyaman(Turkey) Abstract: The concept of statistical convergence was introduced by Stinhauss[2] in 95. M. Gurdal and S.Pehlivan[9] defined statistical convergence in 2-normed spaces in In this paper, we study statistical convergence of double sequence spaces in n-normed spaces. We show that some properties of statistical convergence of double sequences also hold in n-normed spaces. The results here in proved are analogus to those by Vakeel.A.Khan and Sabiha Tabassum [ Appllied Mathematics, Scientific Research Publishing, USA,2(4)(20): ]. Keywords and phrases : Double sequence spaces, Natural density, Statistical convergence, Statistically Cauchy sequence, n-norm Mathematics Subject Classification. 46E30, 46E40, 46B20. Introduction (X,.,.,...,. ) is then called an n-normed space. The notion of statistical convergence was introduced by Fast[] and Schoenberg[20] independently. Later on it was further investigated by Fridy and Orhan[4]. The idea depends on the notion of density of subset of N. Over the years and under different names statistical convergence has been discussed in the theory of fourier analysis, egrodic theory and number theory. Later on it was further investigated from the sequence space point of view and linked with summability theory by Fridy(985) and many others. In recent years, generalization of statistical convergence have appeared in the study of strong integral summability and the structure of ideals of bounded continuous functions on locally compact spaces. The concept of 2-normed spaces was initially introduced by Gahler[5,6,7] in the mid of 960 s. Since then, many researchers have studied this concept and obtained various results, see for instance[8]. Let n N and X be a real vector space of dimension d, where n d. A real valued function.,...,. on X n satisfying the following four conditions:. x, x 2,..., x n = 0 if and only if x, x 2,..., x n are linearly dependent; 2. x, x 2,..., x n is invariant under permutation: 3. αx, x 2,..., x n = α x, x 2,..., x n, for any α R: 4. x + x, x 2,..., x n x, x 2,..., x n + x, x 2,..., x n is called an n norm on X, and the pair Example As a standard example of a n-normed space we may take R n being equipped with the n-norm x, x 2,..., x n E = the volume of the n-dimensional parallelopiped spaned by the vectors x, x 2,..., x n which may be given explicitly by the formula x, x 2,..., x n E = det(x ij ), where x i = (x i, x i2,..., x in ) R n for each i =, 2,..., n. Example 2 Let (X,.,.,...,. ) be an n-normed space of dimension d n 2 and a, a 2,..., a n } be a linearly independent set in X. Then the following function.,.,...,. defined by x, x 2,..., x n = max x, x 2,..., x n, a i : i =, 2,..., n} defines an (n ) norm on X with respect to a, a 2,..., a n }. Example 3 Let n N and (X,.,. ) be a real inner product space of dimension d n. Then the following function.,.,...,. S on X..., X(n factor) defined by x, x 2,..., x n S = [det( x i, x j )] 2 A sequence (x j ) in an n-normed space (X,.,.,...,. ) is said to be converge to some L X in the n-norm if x j L, z,..., z n = 0, for every z,..., z n X. j The second author is supported by Maulana Azad National Fellowship under the University Grant Commision of India. 99

2 992 A sequence (x j ) in an n-normed space (X,.,.,...,. ) is said to be Cauchy with respect to the n-norm if x j x k, z,..., z n = 0, for every z,..., z n X. If every Cauchy sequence in X converges to some L X, then X is said to be complete with respect to the n-norm. Any complete n-normed space is said to be Banach space. We recall some facts connecting with statistical convergence. If K is subset of positive integers N, then K n denotes the set k K : k n}. The natural density of K is given by δ(k) = n K n n, where K n denotes the number of elements in K n, provided this it exists. Finite subsets have natural density zero and δ(k c ) = δ(k) where K c = N \ K, that is the complement of K. If K K 2 and K and K 2 have natural densities then δ(k ) δ(k 2 ). Moreover, if δ(k ) = δ(k 2 ) =, then δ(k K 2 ) = (see[2]). A real number sequence x = (x j ) is statistically convergent to L provided that for every ɛ > 0 the set n N : x j L ɛ} has natural density zero. The sequence x = (x j ) is statustically Cauchy sequence if for each ɛ > 0 there is positive integer N = N(ɛ) such that δ(n N : x j x N (ɛ)}) = 0.(see[3]) If x = (x j ) is a sequence that satisfies some property P for all n except a set of natural density zero, then we say that (x j ) satisfies some property P for almost all n. An Orlicz Function is a function M : [0, ) [0, ) which is continuous, nondecreasing and convex with M(0) = 0, M(x) > 0 for x > 0 and M(x), as x. If convexity of M is replaced by M(x + y) M(x) + M(y), then it is called a Modulus funtion (see Maddox [8]). An Orlicz function may be bounded or unbounded. For example, M(x) = x p (0 < p ) is unbounded and M(x) = x x+ is bounded. Lindesstrauss and Tzafriri [7] used the idea of Orlicz sequence space; ( ) xk l M := x w : M ρ k= which is Banach space with the norm ( ) xk x M = inf ρ > 0 : M ρ k= } <, for some ρ > 0 }. The space l M is closely related to the space l p, which is an Orlicz sequence space with M(x) = x p for p <. An Orlicz function M satisfies the 2 condition (M 2 for short ) if there exist constant K 2 and u 0 > 0 such that whenever u u 0. M(2u) KM(u) Note that an Orlicz function satisfies the inequality M(λx) λm(x) for all λ with 0 < λ <. Orlicz function has been studied by V.A.Khan[0,,2,3] and many others. Throughout a double sequence x = (x jk ) is a double infinite array of elements x jk for j, k N. Double sequences have been studied by V.A.Khan[4,5,6], Moricz and Rhoades[9] and many others. A double sequence x = (x jk ) called statistically convergent to L if mn (j, k) : x jk L ɛ, j m, k n = 0 where the vertical bars indicate the number of elements in the set.(see[5]) In this case we write st 2 x jk = L. Remark If x is statistically connvergent to the number l, then l is determined uniquely. 2 If x is bounded convergent double sequence the it is also statistically convergent to the same number. If x is unbounded convergent double sequence, then x is statistically convergent. 3 If x is statistically convergent, then x need not be convergent and it is not necessarily bounded. Example 4 Let x = (x jk ) be defined as k, j =, k, x jk = 0, otherwise. Then x is statistically convergent to 0 since m,n mn (j, k) : x jk 0 ɛ} m,n n mn = 0 Example 5 Let x = (x jk ) be defined as j, if j is square,, k, x jk = 2, otherwise. Then x is neither convergent nor bounded but statistically convergent to Main Results.

3 993 Recently V.A.Khan and Sabiha Tabassum[6] defined statistical convergence of double sequences in 2- normed spaces. In this section we introduce the notion of statistical covergence in n-normed spaces and give the main results of the paper. Definition Let (x jk ) be a double sequence in n-normed space (X,.,.,...,. ). The sequence (x jk ) is said to be statistically convergent to L, if for every ɛ > 0, (j, k) N N : mn x jk L, z j2, z j3,..., z jn ɛ, J m, k n = 0 for each nonzero z j2, z j3,..., z jn X. In this case we write st 2 x jk, z j2, z j3,..., z jn = L, z j2, z j3,..., z jn. Definition 2 Let (x jk ) be a double sequence in n-normed space (X,.,.,...,. ). The sequence (x jk ) is said to be statistically Cauchy sequence in X if for every ɛ > 0 and every nonzero z j2, z j3,..., z jn X, there exists a number p = p(ɛ, z j2, z j3,..., z jn ) and q = q(ɛ, z j2, z j3,..., z jn ) such that mn j, k N N : x jk x pq, z j2, z j3,..., z jn ɛ j m, k n = 0 V.A.Khan and Sabiha Tabassum[6] defined a double sequence (x jk ) in 2-normed space (X,., ) to be Cauchy with respect to the 2-norm if x jk x pq, z = 0 for every j,p z = (z jk ) X and k, q N. If every Cauchy sequence in X converges to some L X, then X is said to be complete with respect to the 2-norm. Any complete 2-normed space is said to be 2-Banach space. Theorem Let (x jk ) be a double sequence in n- normed space (X,.,.,...,. ) and L, L X. If st 2 x jk, z j2, z j3,..., z jn = L, z j2, z j3,..., z jn and st 2 x jk, z j2, z j3,..., z jn = L, z j2, z j3,..., z jn, then L = L. Proof Assume L L,. Then L L 0, so there exists a z j2, z j3,..., z jn X, such that L L and z j2, z j3,..., z jn are linearly independent. Therefore Now L L, z j2, z j3,..., z jn = 2ɛ, with ɛ > 0. 2ɛ = (L x jk ) + (x jk L ), z j2, z j3,..., z jn x jk L, z j2, z j3,..., z jn +x jk L, z j2, z j3,..., z jn. So (j, k) : x jk L, z j2, z j3,..., z jn < ɛ} (j, k) : x jk L, z j2, z j3,..., z jn < ɛ}. But δ((j, k) : x jk L, z j2, z j3,..., z jn < ɛ}) = 0. Contradicting the fact that x jk L (stat). Theorem 2 Let the double sequence (x jk ) and (y jk ) in n-normed space (X,.,.,...,. ). If (y jk ) is a convergent sequence such that x jk = y jk almost all n, then (x jk ) is statistically convergent. Proof Suppose δ((j, k) N N : x jk y jk }) = 0 and y jk, z j2, z j3,..., z jn = L, z. Then for every ɛ > 0 and z j2, z j3,..., z jn X. (j, k) N N : x jk L, z j2, z j3,..., z jn ɛ} (j, k) N N : x jk y jk }. Therefore δ((j, k) N N : x jk L, z j2, z j3,..., z jn ɛ}) δ((j, k) N N : y jk L, z j2, z j3,..., z jn ɛ}) +δ((j, k) N N : x jk y jk }). () Since y jk, z j2, z j3,..., z jn = L, z j2, z j3,..., z jn for n every z j2, z j3,..., z jn X, the set (j, k) N N : y jk L, z j2, z j3,..., z jn ɛ} contains finite number of integers. Hence, δ((j, k) N N : y jk L, z j2, z j3,..., z jn ɛ}) = 0. Using enequality (), we get δ((j, k) N N : x jk L, z j2, z j3,..., z jn ɛ}) = 0 for every ɛ > 0 and z j2, z j3,..., z jn X. Consequently, st 2 x jk L, z j2, z j3,..., z jn = L, z j2, z j3,..., z jn. Theorem 3 Let the double sequence (x jk ) and (y jk ) in 2-normed space (X,.,.,...,. ) and L, L X and a R. If st 2 x jk, z j2, z j3,..., z jn = L, z j2, z j3,..., z jn and st 2 y jk, z j2, z j3,..., z jn = L, z j2, z j3,..., z jn, for every nonzero z j2, z j3,..., z jn X, then (i) st 2 x jk + y jk, z j2, z j3,..., z jn = L + L, z j2, z j3,..., z jn, for each nonzero z j2, z j3,..., z jn X and (ii) st 2 ax jk, z j2, z j3,..., z jn = al, z j2, z j3,..., z jn, for each nonzero z j2, z j3,..., z jn X.

4 994 Proof(i) Assume that st 2 x jk, z j2, z j3,..., z jn = Theorem 5 Let (x jk ) be a double sequence in n-normed L, z j2, z j3,..., z jn, space (X,.,.,...,. ). The double sequence (x jk ) is statistically convergent if and only if (x jk ) is a statistically and st 2 y jk, z j2, z j3,..., z jn = L, z j2, z j3,..., z jn, for Cauchy sequence. every nonzero z j2, z j3,..., z jn X. Then δ(k ) = 0 and δ(k 2 ) = 0 where Proof K = K (ɛ) := (j, k) N N : x jk L, z j2, z j3,..., z jn ɛ } Assume that st 2 x jk, z j2, z j3,..., z jn = L, z j2, z j3,..., z jn for every nonzero z j2, z j3,..., z jn X 2and ɛ > 0. Then, K 2 = K 2 (ɛ) := (j, k) N N : y jk L, z j2, z j3,..., z jn ɛ } for every z j2, z j3,..., z jn X, 2 x jk L, z j2, z j3,..., z jn < ɛ almost all n, 2 for every ɛ > 0 and z j2, z j3,..., z jn X. Let and if p = p(ɛ, z j2, z j3,..., z jn ) and q = q(ɛ, z j2, z j3,..., z jn ) K = K(ɛ) := (j, k) N N : x jk +y jk (L+L ), z j2, z j3,..., isz chosen jn ɛ}. so that To prove that δ(k) = 0, it is sufficient to prove that K K K 2. Suppose j 0, k 0 K. Then x j0k o + y j0k 0 (L + L ), z j2, z j3,..., z jn ɛ} (2). Suppose to the contrary that j 0, k 0 / K K 2. Then j 0, k 0 / K and j 0, k 0 / K 2. If j 0, k 0 / K and j 0, k 0 / K 2 then x j0k 0 L, z j2, z j3,..., z jn < ɛ 2 and x j0k 0 L, z j2, z j3,..., z jn < ɛ 2. Then, we get x j0k o +y j0k 0 (L+L ), z j2, z j3,..., z jn x j0k o L, z j2, z j3,..., z jn + y j0k 0 L, z j2, z j3,..., z jn < ɛ 2 + ɛ 2 = ɛ which contradicts (2). Hence j 0, k 0 K K 2, that is, K K K 2. Acknowledgments. The authors would like to record their gratitude to the reviewer for his careful read- and making some useful corrections which improved (ii) Let st 2 x jk, z j2, z j3,..., z jn = L, z j2, z j3,..., z jn, aing R and a 0. Then the presentation of the paper. ( (j, k) N N : x jk L, z j2, z j3,..., z jn ɛ }) = 0. a REFERENCES Then we have. H.Fast: Sur la convergence statistique, Colloq.Math., (j, k) N N : ax jk al, z j2, z j3,..., z jn ɛ} = (j, k) N N 2, : a x jk L, (95). z j2, z j3,..., z jn ɛ} = (j, k) N N : x jk L, z j2, z j3,..., z jn ɛ }. 2. A.R.Freedman and I.J.Sember: Densities and a summability, Pasific J.Math., 95, (98). Hence, the right handside of above equality equals 0. Hence, st 2 ax jk, z j2, z j3,..., z jn = al, z j2, z j3,..., z jn, for every nonzero z j2, z j3,..., z jn X. From Theorem of Fridy[3] we have Theorem 4 Let (x jk ) be statistically Cauchy sequence in a finite dimensional n-normed space (X,.,.,...,. ). Then there exists a convergent double sequence (y jk ) for all in (X,.,.,...,. ) j, k N such that x jk = y jk for almost all n. Proof See proof of Theorem 2.9[9] then, we have x pq L, z j2, z j3,..., z jn < ɛ 2, x jk x pq, z j2, z j3,..., z jn x jk L, z j2, z j3,..., z jn + L x pq, z j2, z j3 < ɛ 2 + ɛ almost all n. 2 = ɛ almost all n. Hence, (x jk ) is statistically Cauchy sequence. Conversely, assume that x jk is a statistically Cauchy sequence. By Theorem 4, we have st 2 x jk, z j2, z j3,..., z jn = L, z j2, z j3,..., z jn for each z j2, z j3,..., z jn X. 3. J.A.Fridy: On statistical convergence, Analysis, 5, (985). 4. J.A.Fridy and C.Orhan: Statistical it superior and it inferior, Proc.Amer.Math.Soc., 25:2, (997). 5. S.Gähler: 2-merische Räme und ihre topological Struktur, Math. Nachr., 28, 5-48 (963). 6. S.Gähler: Linear 2-normietre Räme, Math. Nachr., 28, -43 (965). 7. S.Gähler: Uber der Uniformisierbarkeit 2-merische Räme, Math. Nachr., 28, (965).

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