New Type of Generalized Difference Sequence Space of ... - PMF Niš

Filomat 28:7 (2014), 1381–1392 DOI 10.2298/FIL1407381M

Published by Faculty of Sciences and Mathematics, University of Niˇs, Serbia Available at: http://www.pmf.ni.ac.rs/filomat

New Type of Generalized Difference Sequence Space of Non-Absolute Type and Some Matrix Transformations M. Mursaleena , Ab. Hamid Ganieb , Neyaz Ahmad Sheikhb a Department b Department

of Mathematics, Aligarh Muslim University, Aligrah 202002, India of Mathematics, National Institute of Technology Srinagar, 190006, India

Abstract. In the present paper, we introduce a new difference sequence space rqB (u, p) by using the Riesz q

mean and the B- difference matrix. We show rB (u, p) is a complete linear metric space and is linearly isomorphic to the space l(p). We have also computed its α-, β- and γ-duals. Furthermore, we have q q constructed the basis of rB (u, p) and characterize a matrix class (rB (u, p), l∞ ).

1. Introduction, Background and Notation We denote the set of all sequences (real or complex) by ω. Any subspace of ω is called the sequence space. Let N, R and C denotes the set of non-negative integers, of real numbers and of complex numbers, respectively. Let l∞ , c and c0 denote the space of all bounded, convergent and null sequences, respectively. Also, by cs, l1 and l(p) we denote the spaces of all convergent, absolutely and p-absolutely convergent series, respectively. Let X be a real or complex linear space, h be a function from X to the set R of real numbers. Then, the pair (X, h) is called a paranormed space and h is a paranorm for X, if the following axioms are satisfied : (pn.1) h(θ) = 0, (pn.2) h(−x) = h(x), (pn.3) h(x + y) ≤ h(x) + h(y), and (pn.4) scalar multiplication is continuous, that is, |αn − α| → 0 and h(xn − x) → 0 imply h(αn xn − αx) → 0 for all α’s in R and x’s in X, where θ is a zero vector in the linear space X. Assume here and after that (pk ) be a bounded sequence of strictly positive real numbers with supk pk = H and M = max{1, H}. Then, the linear space l(p) was defined by Maddox [10] as follows X l(p) = {x = (xk ) : |xk |pk < ∞} k

which is complete space paranormed by

2010 Mathematics Subject Classification. Primary 40C05; Secondary 46A45, 46H05 Keywords. Sequence space of non-absolute type; β- duals; matrix transformations. Received: 21 June 2013; Accepted: 04 February 2014 Communicated by Dragan S. Djordjevi´c Email addresses: [email protected] (M. Mursaleen), [email protected] (Ab. Hamid Ganie), [email protected] (Neyaz Ahmad Sheikh)

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 1/M X  p  k h1 (x) =  |xk |  . k

0

We shall assume throughout the text that p−1 + pk k

−1

= 1 provided 1 < in f pk ≤ H < ∞.

Let X, Y be two sequence spaces and let A = (ank ) be an infinite matrix of real or complex numbers ank , where n, k ∈ N. Then, the matrix A defines the A-transformation from X into Y, if for every sequence P x = (xk ) ∈ X the sequence Ax = {(Ax)n }, the A-transform of x exists and is in Y; where (Ax)n = ank xk . k

For simplicity in notation, here and in what follows, the summation without limits runs from 0 to ∞. By A ∈ (X : Y), we mean the characterizations of matrices from X to Y i.e., A : X → Y. A sequence x is said to be A-summable to l if Ax converges to l which is called as the A-limit of x. For a sequence space X, the matrix domain XA of an infinite matrix A is defined as XA = {x = (xk ) ∈ ω : Ax ∈ X}.

(1)

Let (qk ) be a sequence of positive numbers and let us write, Qn =

n P

qk for n∈ N. Then the matrix

k=0

q

Rq = (rnk ) of the Riesz mean (R, qn ) is given by

q rnk

q k    Qn , =  0,

if 0 ≤ k ≤ n, if k > n.

The Riesz mean (R, qn ) is regular if and only if Qn → ∞ as n → ∞ [17]. Recently, Neyaz and Hamid [18] introduced the sequence space rq (u, p) as pk     k X   X   1   rq (u, p) =  u q x x = (x ) ∈ ω : < ∞ , (0 < pk ≤ H < ∞).  k j j k     Q k   j=0 k Kizmaz [8] defined the difference sequence spaces Z(4) as follows Z(4) = {x = (xk ) ∈ ω : (4xk ) ∈ Z}, where Z ∈ {l∞ , c, c0 } and 4xk = xk − xk−1 . Altay and Bas¸ar [2] defined the sequence space of p-bounded variation bvp which is defined as     X     p bvp =  x = (x ) ∈ ω : |x − x | < ∞ , 1 ≤ p < ∞.  k k k−1     k

With the notation of (1), the space bvp can be re-defined as bvp = (lp )4 , 1 ≤ p < ∞ where, 4 denotes the matrix 4 = (4nk ) and is defined as

4nk

   (−1)n−k , =  0,

if n − 1 ≤ k ≤ n, if k < n − 1 or k > n.

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p

Neyaz and Hamid [19] introduced the space rq (4u ) as : pk     k   X X   1   p q r (4u ) =  x = (x ) ∈ ω : u q 4x < ∞ ,  j j k k     Q   k j=0

k

where (0 < pk ≤ H < ∞). In [3] the generalized difference matrix B = (bnk ) is defined as :

bnk

  r,     = s,    0,

if k = n, if k = n − 1 if 0 ≤ k < n − 1 or k > n,

for all n, k ∈ N, r, s ∈ R − {0}. The matrix B can be reduced to difference matrix 4 incase r = 1, s = −1. The approach of constructing a new sequence space by means of matrix domain of a particular limitation method has been studied by several authors viz., [1, 4–7, 13, 16, 18, 19].

q

2. The Riesz Sequence Space r (u, p) of Non-Absolute Type B

q

q

In this section, we define the Riesz sequence space rB (u, p), and prove that the space rB (u, p) is a complete paranormed linear space and show it is linearly isomorphic to the space l(p). q

Define the sequence y = (yk ), which will be frequently used, by the Ru B-transform of a sequence x = (xk ), i.e.,

1 yk (q) = Qk

    k−1   X     u (q .r + q .s)x + u q .r.x , (k ∈ N).  j+1 j k j k k k      j=0 

(2)

¨ urk Following Bas¸ar and Altay [1], Mursaleen et al [14, 15], Neyaz and Hamid [18, 19], Bas¸arir and Ozt ¨ q q [4], we define the sequence space rB (u, p) as the set of all sequences such that Ru B transform of it is in the space l(p), that is,  q rB (u, p) = x = (xk ) ∈ ω : yk (q) ∈ l(p) .

q

p

Note that if we take r = 1 and s = −1, the sequence spaces rB (u, p) reduces to rq (∆u ), introduced by Neyaz q q and Hamid [19]. Also, if (uk ) = e = (1, 1, ...), the sequence spaces rB (u, p) reduces to rB (p) Bas¸arir [3]. With the notation of (1) that q

rB (u, p) = {l(p)}Rqu . Now, we prove the following theorem which is essential in the text.

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q

Theorem 2.1. rB (u, p) is a complete linear metric space paranormed by hB , defined as  X 1  hB (x) =   Qk k

  pk  M1 k−1 X      uk (q j .r + q j+1 .s)x j + qk uk .rxk   ,     j=0

where supk pk = H and M = max{1, H}. q

Proof. The linearity of rB (u, p) with respect to the co-ordinatewise addition and scalar multiplication q follows from the inequalities which are satisfied for z, x ∈ rB (u, p) [11] pk  M1  P 1 k−1   P uk (q j .r + q j+1 .s)(x j + z j ) + qk .r.uk (xk + zk )   Qk  j=0 k pk  M1  pk  M1  k−1 k−1   X 1 X X 1 X     uk (q j .r + q j+1 .s)x j + qk .r.uk xk  +  uk (q j .r + q j+1 .s)z j + qk .r.uk zk  ≤      Qk Qk k

j=0

k

(3)

j=0

and for any α ∈ R [12] |α|pk ≤ max(1, |α|M ).

(4) q

It is clear that, hB (θ) = 0 and hB (x) = hB (−x) for all x ∈ rB (u, p). Again the inequality (3) and (4), yield the subadditivity of hB and

hB (αx) ≤ max(1, |α|)hB (x). q

n

Let {x } be any sequence of points of the space rB (u, p) such that hB (xn − x) → 0 and (αn ) is a sequence of scalars such that αn → α. Then, since the inequality, hB (xn ) ≤ hB (x) + hB (xn − x)

holds by subadditivity of hB , {hB (xn )} is bounded and we thus have pk  M1  k−1  X 1 X   hB (αn xn − αx) =  uk (q j .r + q j+1 .s)(αn xnj − αx j ) + uk qk .r(αn xnk − αxk )   Qk  k

j=0

1

1

≤ |αn − α| M hB (xn ) + |α| M hB (xn − x) which tends to zero as n → ∞. That is to say, that the scalar multiplication is continuous. Hence, hB is q paranorm on the space rB (u, p). q

It remains to prove the completeness of the space rB (u, p). Let {x j } be any Cauchy sequence in the space q rB (u, p), where xi = {xi0 , xi1 , ...}. Then, for a given  > 0 there exists a positive integer n0 () such that

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hB (xi − x j ) < 

(5)

for all i, j ≥ n0 (). Using definition of hB and for each fixed k ∈ N that  1 X q pk  M q q q i j i j  (Ru Bx )k − (Ru Bx )k ≤  (Ru Bx )k − (Ru Bx )k  <    k

q

q

for i, j ≥ n0 (), which leads us to the fact that {(Ru Bx0 )k , (Ru Bx1 )k , . . . } is a Cauchy sequence of real numbers q q for every fixed k ∈ N. Since R is complete, it converges, say, (Ru Bxi )k → ((Ru Bx)k as i → ∞. Using these q q q q infinitely many limits (Ru Bx)0 , (Ru Bx)1 , . . . , we define the sequence {(Ru Bx)0 , (Ru Bx)1 , ...}. From (5) for each m ∈ N and i, j ≥ n0 (),

m X pk q q i (Ru Bx )k − (Ru Bx j )k ≤ hB (xi − x j )M < M .

(6)

k=0

Take any i, j ≥ n0 (). First, let j → ∞ in (6) and then m → ∞ , we obtain hB (xi − x) ≤ . Finally, taking  = 1 in (6) and letting i ≥ n0 (1). we have by Minkowski’s inequality for each m ∈ N that  m 1 pk  M X q  (Ru Bx)k  ≤ hB (xi − x) + hB (xi ) ≤ 1 + hB (xi )   k=0

q rB (u, p).

which implies that x ∈ Since hB (x − xi ) ≤  for all i ≥ n0 (), it follows that xi → x as i → ∞, hence we q have shown that rB (u, p) is complete, hence the proof . If we take r = 1, s = −1 in the theorem 2.1, then we have the following result which was proved by Neyaz and Hamid [19]. p

Corollary 2.2. rq (4u ) is a complete linear metric space paranormed by h4 , defined as  X 1  h4 (x) =   Qk k

  pk  M1 k−1 X      uk (q j − q j+1 )x j + qk uk xk   ,     j=0

where supk pk = Hand M = max{1, H}. q

Note that one can easily see the absolute property does not hold on the spaces rB (u, p) , that is q q hB (x) , hB (|x|) for atleast one sequence in the space rB (u, p) and this says that rB (u, p) is a sequence space of non-absolute type. q

Theorem 2.3. The sequence space rB (u, p) of non-absolute type is linearly isomorphic to the space l(p), where 0 < pk ≤ H < ∞.

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Proof. To prove the theorem, we should show the existence of a linear bijection between the spaces q q rB (u, p) and l(p), where 0 < pk ≤ H < ∞. With the notation of (2), define the transformation T from rB (u, p) to l(p) by x → y = Tx. The linearity of T is trivial. Further, it is obvious that x = θ whenever Tx = θ and hence T is injective. Let y ∈ l(p) and define the sequence x = (xk ) by xk =

k−1 X

(−1)

k−n

n=0

! Qk u−1 yk sk−n−1 sk−n k −1 + Q u y + n n k k−n k−n+1 r.qk r qn+1 r qn

Then, pk  M1  k−1  X 1 X   hB (x) =  uk (q j .r + q j+1 .s)x j + uk qk .r.xk    Qk j=0

k

pk  M1  k  X X   =  δk j y j    k

j=0

 1 X pk  M  yk  = h1 (y) < ∞, =  k

where,

   1, δk j =   0,

if k = j, if k , j. q

Thus, we have x ∈ rB (u, p). Consequently, T is surjective and is paranorm preserving. Hence, T is a q linear bijection and this says us that the spaces rB (u, p) and l(p) are linearly isomorphic, hence the proof.

q

3. Duals and Basis of r (u, p) B

q

In this section, we compute α-, β- and γ-duals of rB (u, p) and construct its basis. Theorem 3.1. (i) Let 1 < pk ≤ H < ∞ for every k ∈ N. Define the sets D1 (u, p) and D2 (u, p) as follows D1 (u, p) =   p0 #  k   [ X X "   an −1   −1 Q B < ∞ a = (a ) ∈ ω : sup O (n, k)a Q + u   n u n k k n     r.q n  K∈F k n∈K B>1  and

M. Mursaleen et al. / Filomat 28:7 (2014), 1381–1392

D2 (u, p) =

1387

   p0    k n   X [ X         −1    ak   + O (n, k) a a = (a ) ∈ ω : Q B < ∞ ,   u i   k k         r.u q k k  B>1  i=k+1 k

where Ou (n, k) = (−1)

k−n

! sk−n sk−n−1 + u−1 . rk−n qn+1 rk−n+1 qn k

Then, h

iα hq iβ hq iβ q rB (u, p) = D1 (u, p), rB (u, p) = D2 (u, p) ∩ cs, and rB (u, p) = D2 (u, p).

(ii) Let 0 < pk ≤ 1 for every k ∈ N. Define the sets D3 (u, p) and D4 (u, p) as follows D3 (u, p) =

pk   #   X"   a   n −1 −1 u Q B < ∞ a = (a ) ∈ ω : sup sup O (n, k)a Q +   n u n k k n     r.qn K∈F k n∈K

and      pk n   X  ak               + O (n, k) D4 (u, p) =  a a = (a ) ∈ ω : sup Q < ∞ .  u i    k k        r.u q k k k i=k+1

Then, hq iα hq iβ hq iγ rB (u, p) = D3 (u, p), rB (u, p) = D4 (u, p) ∩ cs, and rB (u, p) = D4 (u, p).

For the proof of the Theorem 3.1, we need following lemmas. Lemma 3.2. [7] (i) Let 1 < pk ≤ H < ∞. Then A∈ (l(p) : l1 ) if and only if there exists an integer B > 1 such that p0 X X k −1 sup a B nk < ∞. K∈F k∈N n∈K

(ii) Let 0 < pk ≤ 1. Then A∈ (l(p) : l1 ) if and only if pk X −1 sup sup ank B < ∞. K∈F k∈N n∈K Lemma 3.3. [9] (i) Let 1 < pk ≤ H < ∞. Then, A∈ (l(p) : l∞ ) if and only if there exists an integer B > 1 such that

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sup

X

n

1388

0

|ank B−1 |pk < ∞.

(7)

k

(ii) Let 0 < pk ≤ 1 for every k ∈ N . Then A ∈ (l(p) : l∞ ) if and only if sup |ank |pk < ∞.

(8)

n,k∈N

Lemma 3.4. [9] Let 0 < pk ≤ H < ∞ for every k ∈ N. Then A ∈ (l(p) : c) if and only if (7) and (8) hold along with lim ank = βk f or k ∈ N.

(9)

n

Proof of Theorem 3.1. We consider the case 1 < pk ≤ H < ∞ for every k ∈ N. Let us take any a = (an ) ∈ ω. From (2) we can easily see that

an xn =

n−1 X

n

Ou (n, k)an Qk yk +

k=0

an Qn yn −1 X uk = cnk yk = (Cy)n , r.qn

(10)

k=0

where n ∈ N and C = (cnk ) is defined by   Ou (n, k)an Qk , if 0 ≤ k ≤ n − 1,            an Qn −1 if k = n, u , cnk =    r.qn k         0, if k > n, q

where k, n ∈ N. Thus, we deduce from (10) with Lemma 3.2 that ax = (an xn ) ∈ l1 whenever x = (xn ) ∈ rB (u, p) hq iα if and only if Cy ∈ l1 whenever y ∈ l(p). This shows that rB (u, p) = D1 (u, p). Further, consider the equation   n n  n X X X   ak −1  an xn = ai  Qk yk = (Dy)n ,  r.qk uk + Ou (n, k) k=0

k=0

i=k+1

where n ∈ N and D = (dnk ) is defined by !  n  P ak −1   u + O (n, k) a  u i Qk ,    r.qk k i=k+1 dnk =       0,

if 0 ≤ k ≤ n, if k > n,

(11)

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for all k, n ∈ N. Thus we deduce from (11) with Lemma 3.3 that ax = (an xn ) ∈ cs whenever x = (xn ) ∈ rB (u, p) if and only if Dy ∈ c whenever y ∈ l(p). Therefore, we derive from (11) that

  p0   n X  a X k      k −1  u−1 Q  B−1 < ∞, u + O (n, k) a u i  k k   r.qk k k i=k+1

(12)

hq iβ and lim dnk exists and hence shows that rB (u, p) = D2 (u, p) ∩ cs. n

q

As proved above, from Lemma 3.4 together with (12) that ax = (ak xk ) ∈ bs whenever x = (xn ) ∈ rB (u, p) if and only if Dy ∈ l∞ whenever y = (yk ) ∈ l(p). Therefore, we again obtain the condition (12) which means hq iγ that rB (u, p) = D2 (u, p) and this completes the proof. q

(k)

Theorem 3.5. Define the sequence b(k) (q) = {bn (q)} of the elements of the space rB (u, p) for every fixed k ∈ N by

     (k) bn (q) =    

Qk −1 r.qk uk

+ Ou (n, k)Qk ,

if 0 ≤ n ≤ k, if n > k.

0,

q

q

Then, the sequence {b(k) (q)} is a basis for the space rB (u, p) and for any x ∈ rB (u, p) has a unique representation of the form

x=

X

λk (q)b(k) (q)

(13)

k

q

where, λk (q) = (Ru Bx)k for all k ∈ N and 0 < pk ≤ H < ∞. q

Proof. It is clear that {b(k) (q)} ⊂ rB (u, p), since q

Ru Bb(k) (q) = e(k) ∈ l(p) f or k ∈ N

(14)

and 0 < pk ≤ H < ∞, where e(k) is the sequence whose only non-zero term is 1 in kth place for each k ∈ N. q

Let x ∈ rB (u, p) be given. For every non-negative integer m, we put x[m] =

m X

λk (q)b(k) (q).

k=0 q

Then, by applying Ru B to (15) and using (14), we obtain

(15)

M. Mursaleen et al. / Filomat 28:7 (2014), 1381–1392

q

Ru Bx[m] =

m X

q

λk (q)Ru Bb(k) (q) =

k=0

m X

1390

q

(Ru Bx)k e(k)

k=0

and



q Ru B



x − x[m]

 i

  0,    =    (Rq Bx)i , u

if 0 ≤ i ≤ m if i > m

where i, m ∈ N. Given ε > 0, there exists an integer m0 such that ∞  M1 X q  pk   < ε ,  |(R Bx) | i  u  2 i=m

for all m ≥ m0 . Hence, 

[m]

hB x − x



∞  M1 X q  p  k =  |(Ru Bx)i |  i=m

 ∞  M1  X q  p  ≤  |(Ru Bx)i | k  i=m0

ε 1 such that

M. Mursaleen et al. / Filomat 28:7 (2014), 1381–1392

  p0   k n X  a X     −1  nk   C(B) = sup + O (n, k) a Q B u ni   r.uk qk < ∞  k  n k i=k+1

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(16)

and {ank }k∈N ∈ cs for each n ∈ N. q  (ii) Let 0 < pk ≤ 1 for every k ∈ N. Then A ∈ rB (u, p) : l∞ if and only if  pk   n X  ank   −1 + Ou (n, k) ani  Qk  B < ∞ sup  k r.uk qk i=k+1

(17)

and {ank }k∈N ∈ cs for each n ∈ N. q  Proof. We will prove (i) and (ii) can be proved in a similar fashion. So, let A ∈ rB (u, p) : l∞ and q q 1 < pk ≤ H < ∞ for every k ∈ N. Then Ax exists for x ∈ rB (u, p) and implies that {ank }k∈N ∈ {rB (u, p)}β for each n ∈ N. Hence necessity of (16) holds. q q Conversely, suppose that the necessities (16) hold and x ∈ rB (u, p), since {ank }k∈N ∈ {rB (u, p)}β for every fixed n ∈ N, so the A-transform of x exists. Consider the following equality obtained by using the relation (11) that m X k=0

   n m  X X    ank  Q  y .   + O (n, k) a ank xk = u ni  k  k  r.uk qk

(18)

i=k+1

k

Taking into account the assumptions we derive from (18) as m → ∞ that

X k

   n X  a X   nk    ank xk = ani  Qk  yk .  r.uk qk + Ou (n, k) k

(19)

i=k+1

Now, by combining (19) and the following inequality which holds for any B > 0 and any complex numbers a, b   p0 |ab| ≤ B aB−1 + |b|p with p−1 +p0 −1 = 1 [10, 16], one can easily see that    n X X  a X  nk  + Ou (n, k) ani  Qk |yk | sup ank xk ≤ sup n∈N k r.uk qk n∈N k i=k+1 h i ≤ B C(B) + hB1 (y) < ∞. q

This shows that Ax ∈ l∞ whenever x ∈ rB (u, p). This completes the proof.

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