Some multiplier difference sequence spaces defined by a sequence ...

General Mathematics Vol. 17, No. 3 (2009), 173–186Some multiplier difference sequence spacesdefined by a sequence of modulus functions 1Hemen DuttaAbstractIn this paper we introduce and investigate the multiplier differencesequence spaces c 0 (F , ∆ n (m) , Λ, p), c(F , ∆n (m) , Λ, p) and l ∞(F ,∆ n (m) , Λ, p) defined by a sequence F = (f k) of modulus functions andp = (p k ) be any bounded sequence of positive real numbers. Westudy their different properties like completeness, solidity, monotonicity,symmetricity etc. We also obtain some relations betweenthese spaces as well as prove some inclusion results.2000 Mathematics Subject Classification: 40A05, 46A45, 46E30.Key words and phrases: Difference sequence, Modulus function,paranorm, completeness, solidity, symmetricity, convergence free,monotone space.1 Received 9 December, 2008Accepted for publication (in revised form) 17 March, 2009173

174 Hemen Dutta1 IntroductionThroughout the paper w, l ∞ , c and c 0 denote the spaces of all, bounded,convergent and null sequences x = (x k ) with complex terms respectively.The zero sequence is denoted by θ = (0, 0, . . . ).The notion of difference sequence space was introduced by Kizmaz [4],who studied the difference sequence spaces l ∞ (∆), c(∆) and c 0 (∆). Thenotion was further generalized by Et and Colak [1] by introducing the spacesl ∞ (∆ n ), c(∆ n ) and c 0 (∆ n ). Another type of generalization of the differencesequence spaces is due to Tripathy and Esi [10], who studied the spacesl ∞ (∆ m ), c(∆ m ) and c 0 (∆ m ). Tripathy, Esi and Tripathy [11] generalizedthe above notions and unified these as follows:haveLet m, n be non-negative integers, then for Z a given sequence space weZ(∆ n m) = {x = (x k ) ∈ w : (∆ n mx k ) ∈ Z},where ∆ n mx = (∆ n mx k ) = (∆ n−1m x k − ∆ n−1m x k+m ) and ∆ 0 mx k = x k for allk ∈ N, which is equivalent to the following binomial representation:∆ n mx k =n∑( ) n(−1) v x k+mv .vv=0Taking m = 1, we get the spaces l ∞ (∆ n ), c(∆ n ) and c 0 (∆ n ) studied by Etand Colak [1]. Taking n = 1, we get the spaces l ∞ (∆ m ), c(∆ m ) and c 0 (∆ m )studied by Tripathy and Esi [10]. Taking m = n = 1, we get the spacesl ∞ (∆), c(∆) and c 0 (∆) introduced and studied by Kizmaz [4].Let Λ = (λ k ) be a sequence of non-zero scalars. Then for E a sequencespace, the multiplier sequence space E(Λ) associated with the multiplier

Some multiplier difference sequence spaces 175sequence Λ is defined asE(Λ) = {(x k ) ∈ w : (λ k x k ) ∈ E}The scope for the studies on sequence spaces was extended by using thenotion of associated multiplier sequences. Goes and Goes [2] defined thedifferentiated sequence space dE and integrated sequence space ∫ E for agiven sequence space E, using the multiplier sequences (k −1 ) and (k) respectively.A multiplier sequence can be used to accelerate the convergence ofthe sequences in some spaces. In some sense, it can be viewed as a catalyst,which is used to accelerate the process of chemical reaction. Sometimes theassociated multiplier sequence delays the rate of convergence of a sequence.A function f : [0, ∞) −→ [0, ∞) is called a modulus if(a) f(x) = 0 if and only if x = 0,(b) f(x + y) ≤ f(x) + f(y), for x ≥ 0, y ≥ 0,(c) f is increasing,(d) f is continuous from the right at 0.Hence f is continuous everywhere in [0, ∞).The following inequality will be used throughout the article. Let p =(p k ) be a positive sequence of real numbers with 0 < p ≤ sup p k = G,D = max(1, 2 G−1 ). Then for all a k , b k ∈ C for all k ∈ N, we have|a k + b k | p k≤ D{|a k | p k+ |b k | p k}and for λ ∈ C,|λ| p k≤ max(1, |λ| G )

176 Hemen DuttaThe studies on paranormed sequence spaces were initiated by Nakano [8]and Simons [9] at the initial stage. Later on it was further studied by Maddox[7], Lascardies [5], Lascardies and Maddox [6], Ghosh and Srivastava[3] and many others.2 Definitions and PreliminariesA sequence space E is said to be solid (or normal) if (x k ) ∈ E implies(α k x k ) ∈ E for all sequences of scalars (α k ) with |α k | ≤ 1 for all k ∈ N.A sequence space E is said to be monotone if it contains the canonicalpreimages of all its step spaces.A sequence space E is said to be symmetric if (x π(k) ) ∈ E, where π is apermutation on N.A sequence space E is said to be convergence free if (y k ) ∈ E whenever(x k ) ∈ E and y k = 0 whenever x k = 0.A sequence space E is said to be sequence algebra if (x k .y k ) ∈ E whenever(x k ) ∈ E and (y k ) ∈ E.Let p = (p k ) be any bounded sequence of positive real numbers andΛ = (λ k ) be a sequence of non-zero scalars. Let m, n be non-negativeintegers, then for a sequence F = (f k ) of modulus functions we define thefollowing sequence spaces:c 0 (F , ∆ n (m), Λ, p) = {x = (x k ) : limk→∞(f k (|∆ n (m)λ k x k |)) p k= 0},c(F , ∆ n (m) , Λ, p) = {x = (x k) : limk→∞(f k (|∆ n (m) λ kx k − L|)) p k = 0,

Some multiplier difference sequence spaces 177for some L ∈ C},l ∞ (F , ∆ n (m), Λ, p) = {x = (x k ) : sup(f k (|∆ n (m)λ k x k |)) p k< ∞},k≥1where (∆ n (m) λ kx k ) = (∆ n−1(m) λ kx k − ∆ n−1(m) λ k−mx k−m ) and ∆ 0 (m) λ kx k = λ k x kfor all k ∈ N, which is equivalent to the following binomial representation:n∑( ) n∆ n (m)λ k x k = (−1) v λ k−mv x k−mv .vv=0In the above expansion it is important to note that we take x k−mv = 0and λ k−mv = 0, for non-positive values of k − mv. Also it is obvious thatc 0 (F , ∆ n (m) , Λ, p) ⊂ c(F , ∆n (m) , Λ, p) ⊂ l ∞(F , ∆ n (m), Λ, p).The inclusions are strict follows from the following examples.Example 1 Let m = 2, n = 2, f k (x) = x 6 , for all k odd and f k (x) = x 2for all k even, for all x ∈ [0, ∞) and p k = 1 for all k ≥ 1. Consider thesequences Λ = (k 8 ) and x = ( 1 ). Then x belongs to c(F, ∆ 2 k 6 (2), Λ, p), but xdoes not belong to c 0 (F, ∆ 2 (2), Λ, p).Example 2 Let m = 2, n = 2, f k (x) = x 2 , for all k ≥ 1 and x ∈ [0, ∞)and p k = 2 for all k odd and p k = 3 for all k even. Consider the sequencesΛ = {1, 1, 1, ...} and x = {1, 3, 2, 4, 5, 7, 6, 8, 9, 11, 10, 12, ...}. Then x belongsto l ∞ (F, ∆ 2 (2) , Λ, p), but x does not belong to c(F, ∆2 (2), Λ, p).Lemma 1 If a sequence space E is solid, then E is monotone.3 Main ResultsIn this section we prove the main results of this article. The proof of thefollowing result is easy, so omitted.

178 Hemen DuttaProposition 1 The classes of sequences c 0 (F, ∆ n (m) , Λ, p), c(F, ∆n (m), Λ, p)and l ∞ (F, ∆ n (m), Λ, p) are linear spaces.Theorem 1 For Z = l ∞ , c and c 0 , the spaces Z(F, ∆ n (m), Λ, p) are paranormedspaces, paranormed bywhere H = max(1, sup p k ).k≥1g(x) = sup((f k (|∆ n (m)λ k x k |)) p k) 1 H ,k≥1Proof. Clearly g(x) = g(−x); x = θ implies g(θ) = 0.Let (x k ) and (y k ) be any two sequences belong to anyone of the abovespaces. Then we have,g(x + y) = sup(f k (|∆ n (m) λ kx k + ∆ n (m) λ ky k |)) p kHk≥1≤ sup(f k (|∆ n (m) λ kx k |)) p kH + sup(f k (|∆ n (m) λ ky k |)) p kHk≥1k≥1This implies thatg(x + y) ≤ g(x) + g(y).The continuity of the scalar multiplication follows from the followinginequality:in |α|.g(αx) = sup((f k (|∆ n (m) αλ kx k |)) p 1k ) Hk≥1= sup((f k (|α||∆ n (m) λ kx k |)) p 1k ) Hk≥1≤ (1 + [|α|])g(x), where [|α|] denotes the largest integer containedHence the spaces Z(F , ∆ n (m) , Λ, p), for Z = l ∞, c, c 0spaces paranormed by g.are paranormedTheorem 2 For Z = l ∞ , c and c 0 , the spaces Z(F, ∆ n (m), Λ, p) are completeparanormed spaces, paranormed by g as defined above.

Some multiplier difference sequence spaces 179Proof. We prove the result for the case l ∞ (F , ∆ n (m), Λ, p) and for the otherspaces it will follow on applying similar arguments.Let (x i ) be any Cauchy sequence in l ∞ (F , ∆ n (m), Λ, p). Let ε > 0 begiven and let ε 1 = sup(f k (ε)) p kH . Then there exists a positive integer n 0ksuch thatg(x i − x j ) < ε 1 , for all i, j ≥ n 0 .Using the definition of paranorm, we getsup((f k (|∆ n (m)λ k x i k − ∆ n (m)λ k x j k |))p k) 1 H < ε1 , for all i, j ≥ n 0 .k≥1It follows that|∆ n (m)λ k x i k − ∆ n (m)λ k x j k | < ε, for each k ≥ 1 and for all i, j ≥ n 0.Hence (∆ n (m) λ kx i k ) is a Cauchy sequence in C for all k ∈ N. This impliesthat (∆ n (m) λ kx i k ) is convergent in C for all k ∈ N. Let limi→∞ ∆n (m) λ kx i k = y kfor each k ∈ N.Let k = 1, we have(1) limi→∞∆ n (m)λ 1 x i 1 = limi→∞Similarly we have,n∑( ) n(−1) v λ 1−mv x i 1−mv = y 1vv=0(2) limi→∞∆ n (m)λ k x i k = limi→∞λ k x i k = y k , for k = 1, ..., nmThus from (1) and (2) we have lim x i 1+nm exists. Let lim x i 1+nm = x 1+nm .i→∞ i→∞Proceeding in this way inductively, we have lim x i k = x k exists for eachi→∞

180 Hemen Duttak ∈ N.Now we have for all i, j ≥ n 0 ,Then we havesup((f k (|∆ n (m)λ k x i k − ∆ n (m)λ k x j k |))p k) 1 Hk≥1< εlim [sup ((f k (|∆ n (m)λ k x i k − ∆ n (m)λ k x jj→∞k |))p k) 1 H ] < ε, for all i ≥ n0k≥1This implies thatsup((f k (|∆ n (m)λ k x i k − ∆ n (m)λ k x k |)) p k) 1 Hk≥1< εIt follows that(x i − x) ∈ l ∞ (F , ∆ n (m), Λ, p).Since (x i ) ∈ l ∞ (F , ∆ n (m) , Λ, p) and l ∞(F , ∆ n (m), Λ, p) is a linear space, so wehave x = x i − (x i − x) ∈ l ∞ (F , ∆ n (m), Λ, p).This completes the proof of the Theorem.Theorem 3 If 0 < p k ≤ q k < ∞ for each k, then Z(F, ∆ n (m), Λ, p) ⊆Z(F, ∆ n (m) , Λ, q), for Z = c 0 and c.Proof. We prove the result for the case Z = c 0 and for the other case itwill follow on applying similar arguments.Let (x k ) ∈ c 0 (F , ∆ n (m), Λ, p). Then we haveThis implies thatlim (f k(|∆ n (m)λ k x k |)) p k= 0.k→∞f k (|∆ n (m)λ k x k |) < ε(0 < ε ≤ 1) for sufficiently large k.

Some multiplier difference sequence spaces 181Hence we getlim (f k(|∆ n (m)λ k x k |)) q k≤ lim (f k (|∆ nk→∞ k→∞(m)λ k x k |)) p k= 0This implies that (x k ) ∈ c 0 (F , ∆ n (m), Λ, q)This completes the proof.The following result is a consequence of Theorem 3.Corollary 1 (a) If 0 < inf p k ≤ p k ≤ 1, for each k, then Z(F, ∆ n (m), Λ, p)⊆ Z(F, ∆ n (m) , Λ), for Z = c 0 and c.(b) If 1 ≤ p k ≤ sup p k < ∞, for each k, then Z(F, ∆ n (m) , Λ) ⊆ Z(F, ∆n (m), Λ, p),for Z = c 0 and c.Theorem 4 Z(F, ∆ n−1(m) , Λ, p) ⊂ Z(F, ∆n (m) , Λ, p) (in general Z(F, ∆i (m), Λ, p)⊂ Z(F, ∆ n (m) , Λ, p), for i=1,2,. . . ,n-1), for Z = l ∞, c and c 0 .Proof. We prove the result for Z = c 0 and for the other cases it will followon applying similar arguments.Let x = (x k ) ∈ c 0 (F , ∆ n−1(m), Λ, p). Then we have(3) limk→∞(f k (|∆ n−1(m) λ kx k |)) p k= 0Now we haveHence we havef k (|∆ n (m)λ k x k |) ≤ f k (|∆ n−1(m) λ kx k |) + f k (|∆ n−1(m) λ k−mx k−m |)(f k (|∆ n (m)λ k x k |)) p k≤ D{(f k (|∆ n−1(m) λ kx k |)) p k+ (f k (|∆ n−1(m) λ k−mx k−m |)) p k}

182 Hemen DuttaThen using (3), we getlim (f k(|∆ n (m)λ k x k |)) p k= 0k→∞Thus c 0 (F , ∆ n−1(m) , Λ, p) ⊂ c 0(F , ∆ n (m), Λ, p).The inclusion is strict follows from the following example.Example 3 Let m = 3, n = 2, f k (x) = x 10 , for all k ≥ 1 and x ∈ [0, ∞)and p k = 5 for all k odd and p k = 3 for all k even. Consider the sequencesΛ = ( 1k 3 ) and x = (k 4 ). Then ∆ 2 (3) λ kx k = 0, for all k ∈ N. Hencex ∈ c 0 (F, ∆ 2 (3) , Λ, p). Again we have ∆1 (3) λ kx k = −3, for all k ∈ N. Hencex does not belong to c 0 (F, ∆ 1 (3), Λ, p). Thus the inclusion is strict.Theorem 5 The spaces c 0 (F, ∆ n (m) , Λ, p), c(F, ∆n (m) , Λ, p) and l ∞(F, ∆ n (m) ,Λ, p) are not monotone and as such are not solid in general.Proof. The proof follows from the following example.Example 4 Let n = 2, m = 3, p k = 1 for all k odd and p k = 2 for all keven and f k (x) = x 4 , for all x ∈ [0, ∞) and k ∈ N. Then ∆ 2 (3) λ kx k =λ k x k −2λ k−3 x k−3 + λ k−6 x k−6 , for all k ∈ N. Consider the J th step space of asequence space E defined as, for (x k ), (y k ) ∈ E J implies that y k = x k for kodd and y k = 0 for k even. Consider the sequences Λ = (k 3 ) and x = ( 1k 2 ).Then x ∈ Z(F, ∆ 2 (3) , Λ, p) for Z = l ∞, c and c 0 , but its J th canonical preimagedoes not belong to Z(F, ∆ 2 (3) , Λ, p) for Z = l ∞, c and c 0 . Hence thespaces Z(F, ∆ 2 (3) , Λ, p) for Z = l ∞, c and c 0 are not monotone and as suchare not solid in general.

Some multiplier difference sequence spaces 183Theorem 6 The spaces c 0 (F, ∆ n (m) , Λ, p), c(F, ∆n (m) , Λ, p) and l ∞(F, ∆ n (m) ,Λ, p) are not symmetric in general.Proof. The proof follows from the following example.Example 5 Let n = 2, m = 2, p k = 2 for all k odd and p k = 3 for allk even and f k (x) = x 2 , for all x ∈ [0, ∞) and for all k ≥ 1. Then∆ 2 (2) λ kx k =λ k x k − 2λ k−2 x k−2 + λ k−4 x k−4 , for all k ∈ N. Consider thesequences Λ = (1, 1, 1, . . . ) and x = (x k ) defined as x k = k for k oddand x k = 0 for k even. Then ∆ 2 (2) λ kx k = 0, for all k ∈ N. Hence(x k ) ∈ Z(F, ∆ 2 (2) , Λ, p) for Z = l ∞, c and c 0 . Consider the rearranged sequence,(y k ) of (x k ) defined as(y k ) = (x 1 , x 3 , x 2 , x 4 , x 5 , x 7 , x 6 , x 8 , x 9 , x 11 , x 10 , x 12 , ...)Then (y k ) does not belong to Z(F, ∆ 2 (2) , Λ, p) for Z = l ∞, c and c 0 .Hence the spaces Z(F, ∆ n (m) , Λ, p) for Z = l ∞, c and c 0 are not symmetricin general.Theorem 7 The spaces c 0 (F, ∆ n (m) , Λ, p), c(F, ∆n (m) , Λ, p) and l ∞(F, ∆ n (m) ,Λ, p) are not convergence free in general.Proof. The proof follows from the following example.Example 6 Let m = 3, n = 1, p k = 6 for all k and f k (x) = x 3 , for keven and f k (x) = x, for k odd, for all x ∈ [0, ∞). Then ∆ 1 (3) λ kx k =λ k x k −λ k−3 x k−3 , for all k ∈ N. Let Λ = ( 7) and consider the sequences (x k k) and(y k ) defined as x k = 4k for all k ∈ N and y 7 k = 1 7 k3 for all k ∈ N. Then

184 Hemen Dutta(x k ) belongs to Z(F, ∆ 1 (3) , Λ, p), but (y k) does not belong to Z(F, ∆ 1 (3), Λ, p)for Z = l ∞ , c and c 0 . Hence the spaces Z(F, ∆ n (m) , Λ, p) for Z = l ∞, c andc 0 are not convergence free in general.Theorem 8 The spaces c 0 (F, ∆ n (m) , Λ, p), c(F, ∆n (m) , Λ, p) and l ∞(F, ∆ n (m) ,Λ, p) are not sequence algebra in general.Proof. The proof follows from the following examples.Example 7 Let n = 2, m = 1, p k = 1 for all k and f k (x) = x 22 , for eachk ∈ N and x ∈ [0, ∞). Then ∆ 2 (1) λ kx k =λ k x k −2λ k−1 x k−1 +λ k−2 x k−2 , for allk ∈ N. Consider Λ = ( 1k 4 ) and let x = (k 5 ) and y = (k 6 ). Then x, y belongto Z(F, ∆ 2 (1) , Λ, p) for Z = l ∞, c, but x.y does not belong to Z(F, ∆ 2 (1), Λ, p)for Z = l ∞ , c. Hence the spaces c(F, ∆ n (m) , Λ, p), l ∞(F, ∆ n (m), Λ, p) are notsequence algebra in general.Example 8 Let n = 2, m = 1, p k = 3 for all k and f k (x) = x 5 , for eachk ∈ N and x ∈ [0, ∞). Then ∆ 2 (1) λ kx k =λ k x k −2λ k−1 x k−1 +λ k−2 x k−2 , for allk ∈ N. Consider Λ = ( 1k 7 ) and let x = (k 8 ) and y = (k 8 ). Then x, y belongto c 0 (F, ∆ 2 (1) , Λ, p), but x.y does not belong to c 0(F, ∆ 2 (1) , Λ, p) for Z = l ∞, c.Hence the space c 0 (F, ∆ n (m), Λ, p) is not sequence algebra in general.References[1] M. Et and R. Colak, On generalized difference sequence spaces, SoochowJ. Math. 21, 1985, 147-169.

Some multiplier difference sequence spaces 185[2] G. Goes and S. Goes, Sequences of bounded variation and sequences ofFourier coefficients, Math. Zeift. 118, 1970, 93-107.[3] D. Ghosh and P. D. Srivastava, On some vector valued sequence spacesdefined using a modulus function, Indian J. pure appl. Math. 30, 1999,819-826.[4] H. Kizmaz, On certain sequence spaces, Canad. Math. Bull. 24, 1981,169-176.[5] C. G. Lascarides, A study of certain sequence spaces of Maddox andgeneralization of a theorem of Iyer, Pacific Jour. Math. 38, 1971, 487-500.[6] C. G. Lascarides and I. J. Maddox, Matrix transformation betweensome classes of sequences, Proc. Camb. Phil. Soc. 68, 1970, 99-104.[7] I. J. Maddox, Paranormed sequence spaces generated by infinite matrices,Proc. Camb. Phil. Soc. 64, 1968, 335-340.[8] H. Nakano, Modular sequence spaces, Proc. Japan Acad. 27, 1951, 508-512.[9] S. Simons, The sequence spaces and, Proc. London Math. Soc. 15, 1965,422-436.[10] B. C. Tripathy and A. Esi, A new type of difference sequence spaces,International Journal of Sci. and Tech. 1, 2006, 11-14.

186 Hemen Dutta[11] B .C. Tripathy, A. Esi and B. K. Tripathy, On a new type of generalizeddifference Cesàro Sequence spaces, Soochow J. Math. 31, 2005, 333-340.Hemen DuttaA.D.P. CollegeDepartment of MathematicsNagaon-782002, Assam, Indiae-mail: hemen − dutta08@rediffmail.com