GENERALIZED SEQUENCE SPACES ON SEMINORMED SPACES

Transcription

1 Acta Universitatis Apulensis ISSN: No. 26/2011 pp GENERALIZED SEQUENCE SPACES ON SEMINORMED SPACES Çiğdem A. Bektaş Abstract. In this paper we define the sequence space l M (u, p, q, s) on a seminormed complex linear space by using Orlicz function and we give various properties and some inclusion relations on this space. This study generalized some results of Bektaş and Altın [1]. Keywords: Orlicz functions, sequence spaces, convex function Mathematics Subject Classification: 40C05, 46A Introduction Let ω be the set of all sequences x = (x k ) with complex terms. Lindenstrauss and Tzafriri [3] used the idea of Orlicz function to construct the sequence space l M = {x ω : ) < for some > 0}. The space M( x k l M with the norm x = inf{ > 0 : ) 1} becomes a Banach space which is called an Orlicz sequence space. The space l M is closely related to the space l p which is an Orlicz sequence space with M(x) = x p for 1 p An Orlicz function is a function M : [0, ) [0, ), which is continuous, non-decreasing and convex with M(0) = 0, M(x) > 0 for x > 0 and M(x) as x. Remark 1.1. If M is a convex function and M(0) = 0, then M(λx) λm(x) for all λ with 0 λ 1. M( x k 245

2 Let X be a complex linear space with zero element θ and (X, q) be a seminormed space with the seminorm q. By S(X) we denote the linear space of all sequences x = (x k ) with (x k ) X and the usual coordinatewise operations: αx = (αx k ) and x + y = (x k + y k ) for each α C where C denotes the set of all complex numbers. If λ = (λ k ) is a scalar sequence and x S(X) then we shall write λx = (λ k x k ). Let U be the set of all sequences u = (u k ) such that u k 0 and complex for all k = 1, 2,.... Let p = (p k ) be a sequence of positive real numbers and M be an Orlicz function. Given u U. Then we define the sequence space l M (u, p, q, s) = {x S(X) : ))]p k <, for some > 0}. The following inequality and p = (p k ) sequence will be used frequently throughout this paper. a k + b k p k D{ a k p k + b k p k }, where a k, b k C, 0 < p k sup k p k = G, D = max(1, 2 G 1 )[4]. A sequence space E is said to be solid (or normal) if (α k x k ) E whenever (x k ) E for all sequences (α k ) of scalars with α k Main Results Theorem 2.1. The sequence space l M (u, p, q, s) is a linear space over the field C complex numbers. Proof. Let x, y l M (u, p, q, s) and α, β C. Then there exist some positive numbers 1 and 2 such that < 1 and k s [M(q( u ky k 2 Define 3 = max(2 α 1, 2 β 2 ). Since M is non-decreasing and convex, and since q is a seminorm, we have k s [M(q( u k(αx k + βy k ) 3 k s [M(q( αu kx k 3 ) + q( βu ky k )] p k 3 246

3 1 2 p k s k 1 1 D 1 )) + M(q( u ky k 2 (1 ) )) + M(q( u ky k 2 + D k s [M(q( u ky k 2 This proves that l M (u, p, q, s) is a linear space. Theorem 2.2. The space l M (u, p, q, s) is paranormed ( not necessarily totaly paranormed ) with g u (x) = inf{ pn/h : ( ))]p k ) 1/H 1, n = 1, 2, 3,... } where H = max(1, sup k p k ). Proof. Clearly g u (x) = g u ( x). The subadditivity of g u follows from (1 ), on taking α = 1 and β = 1. Since q(θ) = 0 and M(0) = 0, we get inf{ pn/h } = 0 for x = θ. Finally, we prove that the scalar multiplication is continuous. Let λ be any number. By definition, g u (λx) = inf{ pn/h : ( k s [M(q( λu kx k ) 1/H 1, n = 1, 2, 3,... }. Then g u (λx) = inf{(λr) pn/h : ( ) 1/H 1, n = 1, 2, 3,... } r where r = /λ. Since λ p k max(1, λ H ), then λ pk/h (max(1, λ H )) 1/H. Hence g u (λx) (max(1, λ H )) 1/H inf{(r) pn/h : ( ) 1/H 1, r n = 1, 2, 3,... } and therefore g u (λx) converges to zero when g u (x) converges to zero in l M (u, p, q, s). Now suppose that λ n 0 and x is in l M (u, p, q, s). For arbitrary ε > 0, let N be a positive integer such that ))]p k < ( ε 2 )H 247

4 for some > 0. This implies that ( ))]p k ) 1/H ε 2. Let 0 < λ < 1, then using Remark 1.1 we get k s [M(q( λu kx k < Since M is continuous everywhere in [0, ), then f(t) = k s [ λ M(q( u kx k ))]p k < ( ε 2 )H. N [ k s M(q( tu ] pk kx k )) is continuous at 0. So there is 1 > δ > 0 such that f(t) < ε 2 be such that λ n < δ for n > K, then for n > K we have for 0 < t < δ. Let K Since 0 < ε < 1 we have ( N k s [M(q( λ nu k x k ) 1/H < ε 2. ( k s [M(q( λ nu k x k ) 1/H < 1, for n > K. If we take limit on inf{ pn/h } we get g u (λx) Some Particular Cases We get the following sequence spaces from l M (u, p, q, s) on giving particular values to p and s. Taking p k = 1 for all k N, we have l M (u, q, s) = {x S(X) : If we take s = 0, then we have l M (u, p, q) = {x S(X) : ))] <, s 0, > 0}. ))]p k <, > 0}. 248

5 If we take p k = 1 for all k N and s = 0, then we have l M (u, q) = {x S(X) : ))] <, > 0}. If we take s = 0, q(x) = x and X = C, then we have l M (u, p) = {x S(X) : [M( u kx k )] p k <, > 0}. In addition to the above sequence spaces, we write l M (u, p, q, s) = l M (p) due to Parashar and Choudhary [5], on taking u k = 1 for all k N, s = 0, q(x) = x and X = C. If we take u k = 1 for all k N, we have l M (u, p, q, s) = l M (p, q, s) [1]. Theorem 3.1. (i) Let 0 < p k t k < for each k N. Then l M (u, p, q) l M (u, t, q). (ii) l M (u, q) l M (u, q, s). (iii) l M (u, p, q) l M (u, p, q, s). Proof. (i) Let x l M (u, p, q). Then there exists some > 0 such that ))]p k This implies that M(q( u ix i )) 1 for sufficiently large values of i, say i k 0 for some fixed k 0 N. Since M is non-decreasing, we get since ))]t k <, ))]t k k k 0 k k 0 ))]p k Hence x l M (u, t, q). The proof of (ii) and (iii) is trivial. Theorem 3.2. Let 0 < p k t k < for each k. Then l M (u, p) l M (u, t). Proof. Proof can be proved by the same way as Theorem 3.1(i). Theorem 3.3. (i) If 0 < p k 1 for all k N, then l M (u, p, q) l M (u, q). (ii) If p k 1 for all k N, then l M (u, q) l M (u, p, q). Proof. 249

6 (i) If we take t k = 1 for all k N, in Theorem 3.1(i), then l M (u, p, q) l M (u, q). (ii) If we take p k = 1 for all k N, in Theorem 3.1(i), then l M (u, q) l M (u, p, q). Proposition 3.4 For any two sequences p = (p k ) and t = (t k ) of positive real numbers and any two seminorms q 1 and q 2 we have l M (u, p, q 1, r) l M (u, t, q 2, s) for r, s > 0. Proof. Since the zero element belongs to l M (u, p, q 1, r) and l M (u, t, q 2, s), thus the intersection is nonempty. Theorem 3.5. The sequence space l M (u, p, q, s) is solid. Proof. Let (x k ) l M (u, p, q, s), i.e, ))]p k Let (α k ) be sequence of scalars such that α k 1 for all k N. Then we have k s [M(q( α ku k x k ))]p k Hence (α k x k ) l M (u, p, q, s) for all sequences of scalars (α k ) with α k 1 for all k N, whenever (x k ) l M (u, p, q, s). Therefore the space l M (u, p, q, s) is a solid sequence space. Corollary 3.6. (i) Let u k 1 for all k N. Then l M (p, q, s) l M (u, p, q, s). (ii) Let u k 1 for all k N. Then l M (u, p, q, s) l M (p, q, s). Proof. Proof is trivial. References [1] Ç. Bektaş and Y. Altın, The sequence space l M (p, q, s) on seminormed space, Indian J. Pure Appl. Math., 34 (4), (2003), [2] T. Bilgin, The sequence space l(p, f, q, s) on seminormed spaces, Bull. Cal. Math. Soc., 86, (1994), [3] J. Lindenstrauss and L. Tzafriri, On Orlicz sequence spaces, Israel J. Math. 10, (1971), [4] I. J. Maddox, Elements of Functional Analysis, Cambridge Univ. Press, (1970). [5] S. D. Parashar and B. Choudhary, Sequence spaces defined by Orlicz functions, Indian J. Pure. Appl. Math., 25, (1994), Çiğdem A. Bektaş Department of Mathematics Firat University, Elazig, 23119, Turkey. cigdemas78@hotmail.com 250