The generalized multiplier space and its Köthe-Toeplitz and null duals

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1 MATHEMATICAL COMMUNICATIONS 273 Math. Commun ), The generalized multiplier space and its Köthe-Toeplitz and null duals Davoud Foroutannia and Hadi Roopaei Department of Mathematics, Vali-e-Asr University of Rafsanjan, P. O. Box Rafsanjan, Iran Received November 14, 2016; accepted February 18, 2017 Abstract. The purpose of the present study is to generalize the multiplier space for introducing the concepts of αb-, βb-, γb-duals and NB-duals, where B = b n,k ) is an infinite matrix with real entries. Moreover, these duals are computed for the sequence spaces X and X ), where X {l p,c,c 0} and 1 p. AMS subject classifications: 40A05, 40C05, 46A45 Key words: multiplier space, α-duals, β-duals, γ-duals, and N-duals, difference sequence spaces, matrix domains 1. Introduction Let ω denote the space of all real-valued sequences. Any vector subspace of ω is called a sequence space. For 1 p <, denote by l p the space of all real sequences x = x n ) ω such that x p = x n p ) 1/p <. For p =, x n p ) 1/p is interpreted as sup n 1 x n. We write c and c 0 for the spaces of all convergent and null sequences, respectively. Also, bs and cs are used for the spaces of all bounded and convergent series, respectively. Kizmaz [8, 9] defined the forward and backward difference sequence spaces. In this paper, we focus on the backward difference space X ) = {x = x k ) : x X}, for X {l,c,c 0 }, where x = x k x k 1 ), x 0 = 0. Observe that X ) is a Banach space with the norm x = sup x k x k 1. k 1 Corresponding author. addresses: foroutan@vru.ac.ir D.Foroutannia), h.roopaei@gmail.com H. Roopaei) c 2017 Department of Mathematics, University of Osijek

2 274 D. Foroutannia and H. Roopaei In summability theory, the β-dual of a sequence space is very important in connection with inclusion theorems. The idea of dual sequence space was introduced by Köthe and Toeplitz [10], and it is generalized to the vector-valued sequence spaces by Maddox [11]. For the sequence spaces X and Y, the set MX,Y) defined by MX,Y) = {z = z k ) ω : z k x k ) Y x = x k) X} is called the multiplier space of X and Y. With the above notation, the α-, β- γ and N-duals of a sequence space X, which are respectively denoted by X α, X β, X γ and X N, are defined by X α = MX,l 1 ), X β = MX,cs), X γ = MX,bs), X N = MX,c 0 ). For a sequence space X, the matrix domain XA) of an infinite matrix A is defined by XA) = {x = x n ) ω : Ax X}, 1) which is a sequence space. The new sequence space XA) generated by the limitation matrix A from a sequence space X can be the expansion or the contraction and the overlap of the original space X. In the past, several authors studied Köthe-Toeplitz duals of sequence spaces that arethe matrix domainsin classicalspacesl p, l, c and c 0. For instance, somematrix domains of the difference operator were studied in [4]. The domain of the backward difference matrix in the space l p was investigated for 1 p by Başar and Altay in [3] and was studied for 0 < p < 1 by Altay and Başar in [1]. Recently the Köthe- Toeplitz duals were computed for some new sequence spaces by Erfanmanesh and Foroutannia [5], [6] and Foroutannia [7]. For more details on the domain of triangle matrices in some sequence spaces, the reader may refer to Chapter 4 of [2]. In the present study, the concept of the multiplier space is generalized and the αb-, βb-, γb- and NB-duals are determined for the classical sequence spaces l p, c and c 0, where 1 p. Moreover, the B-dual are investigated for the difference sequence spaces X ), where X {l,c,c 0 } and {α,β,n}. 2. The αb-, βb-, γb- and NB-duals of sequence spaces In this section, we generalize the concept of multiplier space to introduce new generalizations of Köthe-Toeplitz duals and null duals of sequence spaces. Furthermore, we obtain these duals for the sequence spaces l p, c and c 0, where 1 p. Let A = a n,k ) and B = b n,k ) be two infinite matrices of real numbers and X and Y two sequence spaces. We write A n = a n,k ) for the sequence in the n-th row of A. We say that A defines a matrix mapping from X into Y, and denote it by A : X Y, if and only if A n X β for all n and Ax Y for all x X. If we conside the matrix AB t, where B t is the transpose of matrix B, then the matrix AB t defines a matrix mapping from X into Y, if and only if AB t ) n X β for all n and AB)x Y for all x X. Note that the condition AB t ) n X β implies that ) x k a n,i b i,k <. i=1

3 The generalized multiplier space 275 Based on this fact, we generalize the multiplier space MX,Y). Definition 1. Suppose that B = b n,k ) is an infinite matrix with real entries. For the sequence spaces X and Y, the set M B X,Y) defined by M B X,Y) = { z ω : b n,k z k <, n and x n is called the generalized multiplier space of X and Y. b n,k z k) Y, x X The αb-, βb-, γb- and NB-duals of a sequence space X, which are denoted by X αb, X βb, X γb and X NB, respectively, are defined by X αb =M B X,l 1 ), X βb =M B X,cs), X γb = M B X,bs), X NB = M B X,c 0 ). It should be noted that in the special case B = I, we have M B X,Y) = MX,Y). So X αb = X α, X βb = X β, X γb = X γ, X NB = X N. Theorem 1. If B = b n,k ) is an invertible matrix, then M B X,Y) MX,Y). Proof. With the map T : M B X,Y) MX,Y), which is defined by the proof is obvious. Tz = b n,k z k) We determine the generalized multiplier space for some sequence spaces. In order to do this, we state the following lemma which is essential in the study. Lemma 1. If X,Y,Z ω, then i) X Z implies M B Z,Y) M B X,Y), ii) Y Z implies M B X,Y) M B X,Z). Proof. The proof is elementary and so omitted. Remark 1. If B = I, we have Lemma 1.25 from [12]. Corollary 1. Suppose that X,Y ω and denotes either of the symbols α, β, γ or N. Then i) X αb X βb X γb ω; in particular, X B is a sequence space. ii) X Z implies Z B X B. Remark 2. If B = I, we have Corollary 1.26 from [12]., }

4 276 D. Foroutannia and H. Roopaei With the notation of 1), we can define the spaces XB) for X {l p,c,c 0 } and 1 p, as follows: { } XB) = x = x n ) ω : X. Theorem 2. We have the following statements. i) M B c 0,X) = l B), where X {l,c,c 0 }, ii) M B l,x) = c 0 B), where X {c,c 0 }, iii) M B c,x) = cb), where X {c,c 0 }. b n,k x k) Proof. i): Since c 0 c l, by applying Lemma 1ii), we have M B c 0,c 0 ) M B c 0,c) M B c 0,l ). So it is sufficient to verify l B) M B c 0,c 0 ) and M B c 0,l ) l B). Suppose that z l B) and x c 0. We have ) x n b n,k z k = 0. lim n This means that z M B c 0,c 0 ). Thus l B) M B c 0,c 0 ). Now we assume z l B). Then there is a subsequence the sequence b n,kz k ) such that b nj,kz k > j 2, for j = 1,2,. If the sequence x = x i ) is defined by { 1) j j x i = b i,k z k, if i = n j 0, otherwise, b n j,kz k ) of for i = 1,2,, we have x c 0 and x nj b n j,kz k = 1) j j, for all j. Hence x n l. b n,k z k) This shows M B c 0,l ) l B). ii): We have M B l,c 0 ) M B l,c), by applying Lemma 1ii). It is sufficient to prove c 0 B) M B l,c 0 ) and M B l,c) c 0 B). Suppose that z c 0 B). We have ) x n b n,k z k = 0, lim n

5 The generalized multiplier space 277 for all x l, that is, z M B l,c 0 ). Thus c 0 B) M B l,c 0 ). Now we assume z c 0 B). Then there are a real number as b > 0 and a subsequence b ) n j,kz k of the sequence b n,kz k ) such that b nj,kz k > b, for all j = 1,2,. If the sequence x = x i ) is defined by { 1) j x i = b i,kz k, if i = n j 0, otherwise, for all i N, then we have x l and x n b n,k z k) which implies z M B l,c). This shows that M B l,c) c 0 B). iii): Suppose that z cb). We deduce that lim n x n b n,kz k ) exists for all x c 0. So z M B c,c 0 ) and cb) M B c,c 0 ). Conversely, we assume z M B c,c). Let x = 1,1, ). It is obvious that x c and = x n c. b n,k z k) So z cb). This shows M B c,c) cb). c, b n,k z k) Remark 3. If B = I, we have Example 1.28 from [12]. Corollary 2. We have c NB 0 = l B), l NB = c 0B) and c NB = c 0 B). Below we recall the concept of normal and similarly to the Köthe-Toeplitz duals, we show that X αb = X βb = X γb when X is a normal set. Definition 2. A subset X of ω is said to be normal if y X and x n y n, for n = 1,2,, together imply x X. Example 1. The sequence spaces c 0 and l are normal, but c is not normal. Theorem 3. Let X be a normal subset of ω. We have X αb = X βb = X γb. Proof. Obviously, X αb X βb X γb, by Corollary 1i). To prove the statement, it is sufficient to verify X γb X αb. Let z X γb and x X be given. We define the sequence y such that ) y n = sgn b n,k z k x n,

6 278 D. Foroutannia and H. Roopaei for n = 1,2,. It is clear y n x n, for all n. Consequently, y X since X is normal. So ) n sup y n b n,k z k <. n Furthermore, by the definition of the sequence y, x n b n,k z k <. Since x X was arbitrary, z X αb. This finishes the proof of the theorem. Remark 4. If B = I and X is a normal subset of ω, we have hence Remark 1.27 from [12]. X α = X β = X γ, Now, we investigate the αb-, βb- and γb-duals for the sequence spaces l, c and c 0. Theorem 4. Suppose that denotes either of the symbols α, β or γ. We have c B 0 = c B = l B = l 1B). Proof. We only prove the statement for the case = β; the other cases are proved by Theorem 3. Obviously, l βb cβb c βb 0 by Corollary 1ii). So it is sufficient to show that l 1 B) l βb and c βb 0 l 1 B). Let z l 1 B) and x l be given. Hence x n b n,k z k sup x n b n,k z k <, 2) which shows x n b n,kz k ) cs. Thus z lβb and l 1B) l βb. Now let z l 1 B). We may choose an index subsequence n j ) in N with n 0 = 0 and n j 1 b n,k z k > j, j = 1,2,... n=n j 1 We define the sequence x c 0 such that { 1 x n = j sgn b n,kz k ), if n j 1 n < n j. 0, otherwise We get n j 1 n=n j 1 ) x n b n,k z k = 1 j n j 1 n=n j 1 for j = 1,2,. Therefore x n b n,kz k ) the proof of the theorem. b n,k z k > 1, cs, and z cβb 0. This completes

7 The generalized multiplier space 279 Remark 5. If B = I and denotes either of the symbols α, β or γ. We have hence Theorem 1.29 from [12]. c 0 = c = l = l 1, In the next theorem, we examine the αb-, βb- and γb-duals for the sequence space l p. Theorem 5. If 1 < p < and q = p/p 1), then l αb p = l βb p = l γb p = l q B). Moreover for p = 1, we have l αb 1 = l βb 1 = l γb 1 = l B). Proof. We only prove the statement for the case 1 < p < ; the case p = 1 is proved similarly. Let z l q B) be given. By Hölder s inequality, we have x k b k,j z j q b k,j z j 1/q x k p ) 1/p <, 3) for all x l p. This shows z lp βb and hence l q B) lp βb. Now, let z lp βb be given. We consider the linear functional f n : l p R defined by n n f n x) = x k b k,j z j, x l p, for n = 1,2,. Similarly to 3), we obtain q n n f n x) b k,j z j 1/q x k p ) 1/p for every x l p. So the linear functional f n is bounded and n n q f n b k,j z j for all n. We now prove the reverse of the above inequality. We define the sequence x = x k ) such that 1/q n n x k = sgn b k,j z j b k,j z j, q 1,,

8 280 D. Foroutannia and H. Roopaei for 1 k n, and put the remaining elements zero. Obviously, x l p, so n f n f nx) n b q q k,jz j n n = x p n n q) 1/p = b k,j z j b k,jz j for n = 1,2,. Since z lp βb, the map f z : l p R defined by f z x) = b k,j z j x k, x l p, is well-defined and linear, and also the sequence f n ) is pointwise convergent to f z. By using the Banach-Steinhaus theorem, it can be shown that f z sup n f n <, so q 1/q b k,j z j <, and z l q B). This establishes the proof of the theorem. Remark 6. If B = I and 1 < p < and q = p/p 1).Then we have l α p = lβ p = lγ p = l q. Moreover, for p = 1, we have l α 1 = lβ 1 = lγ 1 = l. 3. The αb-, βb- and NB-duals of sequence spaces X ) The purpose of this section is to compute the B-dual of the difference sequence spaces X ), where X {l,c,c 0 } and {α,β,n}. In order to do this, we first give a preliminary lemma. Lemma 2. i) If x l ), then sup k x k k <. ii) If x c ), then x k k ξk ), where x k ξk ). iii) If x c 0 ), then x k k 0k ). Proof. The proof is trivial and so omitted. For convenience of the notations, we use X B ) instead of X ) B, where denotes either of the symbols α, β or N. Theorem 6. Define the set as follows: { Then d 1 = z = z k ) : n b n,k z k) c NB ) = l NB ) = d 1. c 0 }. 1/q,

9 The generalized multiplier space 281 Proof. By using Corollary 1ii), we have l NB ) cnb ). So it is sufficient to show that d 1 l NB ) and c NB ) d 1. Let z d 1 and x l ). By Lemma 2 sup n x nn <, so lim n x n b n,k z k = lim n n b n,k z k x n n = 0. This implies that z l NB ). Now suppose that z c NB ), we have lim n x n b n,k z k = 0, for all x c ). If x = 1,2,3, ), we have x c ) and lim n b n,k z k = 0. n So z d 1 and the proof of the theorem is finished. Remark 7. If B = I, we have c N ) = l N ) = {z = z k ) : ka k ) c 0 }, [9]. Now, we recall the following theorem from [12] which is important to continue the discussion. Let A = a n,k ) be an infinite matrix of real numbers a n,k, where n,k N = {1,2, }. We consider the conditions sup n ) a n,k <, 4) lim n,k = 0, k = 1,2,..., n 5) lim n,k = l k, for some l k R,k = 1,2,... n 6) By X,Y), we denote the class of all infinite matrices A such that A : X Y. Theorem 7 see [13]). We have i) A c 0,c 0 ) if and only if conditions 4) and 5) hold; ii) A c 0,c) if and only if conditions 4) and 6) hold. Theorem 8. Define the set as follows: Then c NB 0 ) = d 2. d 2 = { z = z k ) : n b n,k z k) l }.

10 282 D. Foroutannia and H. Roopaei x Proof. Suppose that z d 2. For x c 0 ), by Lemma 2 we have lim nn n = 0. So lim x n b n,k z k = lim n x n b n,k z k n n n = 0, this implies that z c NB 0 ). Now let z c NB 0 ). We define the matrix D = d n,j ) by { d n,j = b n,kz k, if 1 j n 0, j > n, and prove that D c 0,c 0 ). To do this, we show that D n = d n,j ) cβ 0 for all n and moreover Dy c 0 for all y c 0. Since z c NB 0 ), we deduce that b n,kz k < for all n; hence for y c 0 n n ) d n,j y j = b n,k z k )y j = b n,k z k <, for all n, so D n c β 0 y j for all n. Moreover, z cnb 0 ) implies that lim x n b n,k z k = 0, n for all x c 0 ). There exists one and only one y = y k ) c 0 such that x n = n y j. So lim n d n,j y j = lim n n b n,k z k y j = lim n x n b n,k z k = 0, for all y c 0. Hence lim n D n y = 0 and Dy c 0 for all y c 0. By applying Theorem 7i) for D c 0,c 0 ), we obtain sup n n n b n,k z k = sup b n,k z k n = sup d n,j n <. This completes the proof of the theorem. Remark 8. If B = I, we have c N 0 ) = {z = z k) : kz k ) l }, Lemma 2 from [9]. Inwhatfollows,weconsidertheαB-dualforthesequencespacesc )andl ). Theorem 9. Define the set d 3 as follows: { d 3 = z = z k ) : n Then c αb ) = l αb ) = d 3. b n,k z k) l 1 }.

11 The generalized multiplier space 283 Proof. By applying Corollary 1ii), we have l αb ) cαb ). So it is sufficient to show that d 3 l αb ) and c αb ) d 3. Let z d 3 and x l ) be given. By Lemma 2 sup n x nn <, so x n b n,k z k sup x n n n b n,k z k <, which shows z c αb ) and d 3 l αb ). Now suppose that z c αb ). Since x = 1,2,3, ) c ), we conclude that n b n,k z k = x n b n,k z k <, So z d 3, and this completes the proof of the theorem. Remark 9. If B = I, we have c α ) = l α ) = {z = z k ) : kz k ) l 1 }. In order to investigate the βb-dual of the difference sequence space c 0 ), we need the following lemma. Lemma 3 see [9, Lemma 1]). If z l 1, x c 0 ) and lim k z k x k = L, then L = 0. For the next result we introduce the sequence R k ) given by Theorem 10. If then we have c βb 0 ) = d 4. R k = t=k b t,j z j. d 4 = {a = a k ) l 1 B) : R k ) l 1 c N 0 )}, Proof. Suppose that z d 4 and x c 0 ), by using Abel s summation formula, we have ) m m n m x n b n,k z k = b t,j z j x n x n+1 )+ b n,k z k )x m+1 This implies that = = t=1 m R 1 R n+1 )x n x n+1 )+R 1 R m+1 )x m+1 m+1 R n x n x n 1 ) R m+1 x m+1. 7) ) x n b n,k z k

12 284 D. Foroutannia and H. Roopaei is convergent, so z c βb 0 ). Let z c 0 ) βb, by applying Corollary1ii) and Theorem 4 we have c 0 ) βb c βb 0 = l 1 B); hence z l 1 B). If x c 0 ), then there exists y = y k ) c 0 such that x n = k y j. By applying Abel s summation formula m R n y n = = m n R n R n+1 ) m n y j y j + m R m+1 y n m b n,j z j + R m+1 y n. Thus m ) m m m b n,k z k x n = R n R m+1 )y n = b i,j z j y n. 8) i=n Now we define the matrix D = d n,k ) by { n d n,k = i=k b i,jz j, if 1 k n 0, k > n, and we prove that D c 0,c). To do this, we show that D n = d n,j ) cβ 0 n, and moreover Dy c for all y c 0. Since z c βb 0 ), we deduce that for all b n,k z k <, for all n; hence for y c 0 n n d n,k y k = b i,j z j y k <, i=k for all n. So D n c β 0 for all n. Moreover, z cβb 0 ) implies that ) x n b n,k z k is convergent for all x c 0 ). Hence by 8), we deduce that lim D ny = lim n n n d n,k y k = lim n x i n i=1 b i,j z j

13 The generalized multiplier space 285 exists. So Dy c 0 for all y c 0 and D c 0,c). This implies that n n sup d n,k = sup b i,j z j n n <, by Theorem 7ii). Thus we get i=k R k <. Furthermore, 7) implies that lim n R n+1 x n+1 exists for each x c 0 ); hence R n ) c N 0 ) by Lemma 3. This completes the proof of the theorem. Remark 10. If B = I, then we have c β 0 ) = {z = z k) l 1 : R k ) l 1 c N 0 )}, where the sequence R k ) given by R k = i=k z i, hence Lemma 3 from [9] is resulted. References [1] B.Altay, F. Başar, The fine spectrum and the matrix domain of the difference operator on the sequence space l p, 0 < p < 1), Commun. Math. Anal ), [2] F.Başar, Summability Theory and Its Applications, Bentham Science Publishers, Istanbul, [3] F.Başar, B.Altay, On the space of sequences of p-bounded variation and related matrix mappings, Ukrainian Math. J ), [4] V. K. Bhardwaj, S. Gupta, R. Karan, Köthe-Toeplitz duals and matrix transformations of Cesàro difference sequence spaces of second order, J. Math. Anal ), [5] S. Erfanmanesh, D. Foroutannia, Generalizations of Köthe-Toeplitz duals and null duals of new difference sequence spaces, J. Contemp. Math. Anal ), [6] S. Erfanmanesh, D. Foroutannia, Some new semi-normed sequence spaces of nonabsolute type and matrix transformations, Proceedings of IAM 42015), [7] D.Foroutannia, On the block sequence space l pe) and related matrix transformations, Turkish J. Math ), [8] H. Kizmaz, On certain sequence spaces I, Canad. Math. Bull ), [9] H. Kizmaz, On certain sequence spaces II, Int. J. Math. Math. Sci ), [10] G. Köthe, O. Toeplitz, Lineare Räume mit unendlich vielen Koordinaten und Ringe unendlicher Matrizen, J. Reine Angew. Math ), [11] I. J. Maddox, Generalized Köthe-Toeplitz Duals, Int. J. Math. Math. Sci ), [12] E. Malkowsky, V. Rakočević, An introductionm into the theory of sequence spaces and measures of noncompactness, Zb. Rad. Beogr ), [13] M.Stieglitz, H.Tietz, Matrix Transformationen von Folgenräumen Eine Ergebnisübersicht, Math. Z ), 1 16.