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J. Math. Anal. Appl. 277 (2003) 681–688

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On some new generalized sequence spaces

Çigdem A. Bekta¸s

Department of Mathematics, Firat University, Elazig, 23119, Turkey

Received 29 January 2001

Submitted by A.C. Peterson

Abstract

In this paper we define the sequence sets�∞(u,∆2,p), c(u,∆2,p) andc0(u,∆2,p), and giveα- andβ-duals of these sets. Further we investigate matrix transformations in the spaces and give acharacterization of the class(�∞(u,∆2,p), �∞). 2003 Elsevier Science (USA). All rights reserved.

Keywords: Difference sequences;α- andβ-duals

1. Introduction

Let ω be the set of all complex sequencesx = (xk)∞k=1. We write�∞, c, c0 for the spaces

of bounded, convergent and null sequences, respectively; andcs, �1 for the spaces of con-vergent and absolutely convergent series.p = (pk)

∞k=1 will denote an arbitrary sequence

of positive real numbers.Ahmad and Mursaleen [1] defined the sequence spaces

∆X(p) = {x ∈ ω: ∆x ∈ X(p)

}where∆x = (∆xk)

∞k=1 = (xk − xk+1)

∞k=1 andX = �∞, c or c0.

Let U be the set of all sequencesu = (uk)∞k=1 such thatuk �= 0 and complex for all

k = 1,2, . . . . Throughout the paper we writewk = 1/|uk|.After Malkowsky [6] defined the sequence spaces

X(u,∆) = {x ∈ ω:

(uk(xk − xk+1)

)∞k=1 ∈ X

}whereu ∈ U andX = �∞, c or c0.

E-mail address: [email protected].

0022-247X/03/$ – see front matter 2003 Elsevier Science (USA). All rights reserved.doi:10.1016/S0022-247X(02)00619-4

682 Ç.A. Bektas / J. Math. Anal. Appl. 277 (2003) 681–688

Recently Asma and Çolak [2] defined the sequence spaces

X(u,∆,p) = {x ∈ ω: (uk∆xk)

∞k=1 ∈ X(p)

}whereu ∈ U andX = �∞, c or c0.

Given an arbitrary sequencep = (pk)∞k=1 of positive real numberspk and givenu ∈ U ,

we define the sets

�∞(u,∆2,p) = {x ∈ ω: (uk∆

2xk)∞k=1 ∈ �∞(p)

},

c(u,∆2,p) = {x ∈ ω: (uk∆

2xk)∞k=1 ∈ c(p)

},

c0(u,∆2,p) = {x ∈ ω: (uk∆

2xk)∞k=1 ∈ c0(p)

}where∆2x = (∆2xk)

∞k=1 = (∆xk − ∆xk+1)

∞k=1.

We get the following sequence spaces from the above sequence spaces on givingparticular values top andu:

(i) If uk = 1 and pk = 1, then �∞(u,∆2,p) = �∞(∆2), c(u,∆2,p) = c(∆2) andc0(u,∆2,p) = c0(∆

2) (see Et [3]).(ii) If pk = 1, then�∞(u,∆2,p) = �∞(u,∆2), c(u,∆2,p) = c(u,∆2) andc0(u,∆2,p) =

c0(u,∆2) (see Mursaleen [7]).

Let us define the operatorS : �∞(u,∆2,p) → �∞(u,∆2,p) by x → Sx = (0,0, x3, x4,

. . .). It is clear thatS is a bounded linear operator on�∞(u,∆2,p) with ‖S‖ = 1. Further

S�∞(u,∆2,p) = {x = (xk): x ∈ �∞(u,∆2,p), x1 = x2 = 0

}.

Remark 1.1. [S�∞(u,∆2,p)]β = [�∞(u,∆2,p)]β .

If p = (pk) is a fixed bounded sequence of positive real numbers andu ∈ U , then�∞(u,∆2,p), c(u,∆2,p) andc0(u,∆2,p) are linear spaces, under the usual operations

x + y = (xn + yn) and αx = (αxn)

whereα is any complex number.

Theorem 1.2. Let p = (pk)∞k=1 be a bounded sequence of strictly positive real numbers

pk and u ∈ U . Then c0(u,∆2,p) is a paranormed space with paranorm g(x) =supk |uk∆

2xk|pk/M , where M = max{1,H = supk pk}. If infk pk > 0, then �∞(u,∆2,p)

and c(u,∆2,p) are paranormed spaces with the same norm as above.

The proof of this theorem is easy therefore we omit it.

2. Köthe–Toeplitz duals

Let z be any sequence and letY be any subset ofω. Then we shall write

z−1 ∗ Y = {x ∈ ω: zx = (zkxk) ∈ Y

}.

Ç.A. Bektas / J. Math. Anal. Appl. 277 (2003) 681–688 683

For any subsetX of ω, the sets

Xα =⋂x∈X

(x−1 ∗ �1) and Xβ =⋂x∈X

(x−1 ∗ cs)

are calledα- andβ-duals ofX. We shall writeXαα = (Xα)α andXββ = (Xβ)β .

Theorem 2.1. For every strictly positive sequence p = (pk), we have

(i)[�∞(u,∆2,p)

]α = M1(p) =∞⋂

N=2

{a ∈ ω:

∞∑k=1

|ak|k−1∑j=1

j−1∑i=1

N1/pi wi < ∞}

,

(ii)[�∞(u,∆2,p)

]αα = M2(p)

=∞⋃

N=2

{a ∈ ω: sup

k�3|ak|

(k−1∑j=1

j−1∑i=1

N1/pi wi

)−1

< ∞}

,

(iii )[c0(u,∆2,p)

]α = D1(p) =∞⋃

N=2

{a ∈ ω:

∞∑k=1

|ak|k−1∑j=1

j−1∑i=1

N−1/pi wi < ∞}

,

(iv)[c0(u,∆2,p)

]αα = D2(p)

=∞⋂

N=2

{a ∈ ω: sup

k�3|ak|

(k−1∑j=1

j−1∑i=1

N−1/pi wi

)−1

< ∞}

.

(We adopt the usual convention that∑m

j=1 yj = 0 (m < 1) for arbitrary yj .)

Proof. (i) Let a ∈ M1(p) andx ∈ �∞(u,∆2,p). We chooseN > max{1,supk |uk∆2xk|pk }.

Since

k−1∑j=1

j−1∑i=1

N1/pi wi >

2∑j=1

(k − j − 1

2− j

)N1/pi wi

for arbitraryN > 1 (k = 2,3, . . .) and|∆2−jxj | � Mwj (1 � j � 2) for some constantM,a ∈ M1(p) implies

∞∑k=1

|ak|2∑

j=1

(k − j − 1

2− j

)|∆2−j xj | < ∞.

Then

∞∑k=1

|akxk| =∞∑

k=1

|ak|∣∣∣∣∣

k−1∑j=1

j−1∑i=1

∆2xi +2∑

j=1

(−1)2−j

(k − j − 1

2− j

)∆2−j xj

∣∣∣∣∣


�∞∑

k=1

|ak|k−1∑j=1

j−1∑i=1

N1/pi wi +∞∑

k=1

|ak|2∑

j=1

(k − j − 1

2− j

)|∆2−jxj |

< ∞, (1)

and so we havea ∈ [�∞(u,∆2,p)]α . ThereforeM1(p) ⊂ [�∞(u,∆2,p)]α .Conversely, leta /∈ M1(p). Then we have

∞∑k=1

|ak|k−1∑j=1

j−1∑i=1

N1/pi wi = ∞

for some integerN > 1. We define the sequencex by

xk =k−1∑j=1

j−1∑i=1

N1/pi wi (k = 2,3, . . .).

Then it is easy to see thatx ∈ �∞(u,∆2,p) and∑∞

k=1 |akxk| = ∞. Hence a /∈[�∞(u,∆2,p)]α .

(ii) Let a ∈ M2(p) andx ∈ [�∞(u,∆2,p)]α = M1(p), by part (i). Then for someN > 1,we have

∞∑k=3

|akxk| =∞∑

k=3

|ak|(

k−1∑j=1

j−1∑i=1

N1/pi wi

)−1

|xk|k−1∑j=1

j−1∑i=1

N1/pi wi

�[

supk�3

|ak|(

k−1∑j=1

j−1∑i=1

N1/pi wi

)−1] ∞∑k=3

|xk|k−1∑j=1

j−1∑i=1

N1/pi wi < ∞.

Conversely, leta /∈ M2(p). Then for all integersN > 1, we have

supk�3

|ak|(

k−1∑j=1

j−1∑i=1

N1/pi wi

)−1

= ∞.

Hence there is a strictly increasing sequence(k(s)) of integersk(s) � 2 such that

|ak(s)|(

k(s)−1∑j=1

j−1∑i=1

s1/pi wi

)−1

> s3 (s = 2,3, . . .).

We define the sequencex by

xk ={ |ak(s)|−1 (k = k(s)),

0 (k �= k(s), s = 2,3, . . .).

Then for all integersN > 3, we have

∞∑k=1

|xk|k−1∑j=1

j−1∑i=1

N1/pi wi �∞∑

s=3

|ak(s)|−1k(s)−1∑j=1

j−1∑i=1

N1/pi wi


�N−1∑s=3

|ak(s)|−1k(s)−1∑j=1

j−1∑i=1

N1/pi wi +∞∑

s=N

|ak(s)|−1k(s)−1∑j=1

j−1∑i=1

s1/piwi

�N−1∑s=3

|ak(s)|−1k(s)−1∑j=1

j−1∑i=1

N1/pi wi +∞∑

s=N

s−3 < ∞.

Hencex ∈ [�∞(u,∆2,p)]α and∑∞

k=1 |akxk| =∑∞s=1 1= ∞, a /∈ [�∞(u,∆2,p)]αα .

(iii) Let a ∈ D1(p) and x ∈ c0(u,∆2,p). Then there is an integerk0 such thatsupk>k0

|uk∆2xk|pk � N−1, whereN is the number inD1(p). We put

M := max1�k�k0

|uk∆2xk|pk , m := min

1�k�k0pk, L := (M + 1)N,

and define the sequencey by yk := xkL−1/m (k = 1,2, . . .). Then it is easy to see that

supk |uk∆2yk|pk � N−1 and, as in (1) withN replaced byN−1, we have

∞∑k=1

|akxk| = L1/m∞∑

k=1

|akyk| < ∞.

Conversely, leta /∈ D1(p). Then we can determine a strictly increasing sequence(k(s))

of integers such thatk(1) = 1 and

k(s+1)−1∑k=k(s)

|ak|k−1∑j=1

j−1∑i=1

(s + 1)−1/piwi > 1 (s = 1,2, . . .).

We define the sequencex by

xk =s−1∑l=1

k(l+1)−1∑j=k(l)

j−1∑i=1

(l + 1)−1/pi wi +k−1∑

j=k(s)

j−1∑i=1

(s + 1)−1/piwi(k(s) � k � k(s + 1) − 1, s = 1,2, . . .

).

Then it is easy to see that|uk∆2xk|pk = 1/(s + 1) (k(s) � k � k(s + 1)− 1, s = 1,2, . . .).

Hencex ∈ c0(u,∆2,p) and∑∞

k=1 |akxk| �∑∞s=1 1= ∞, i.e.,a /∈ [c0(u,∆2,p)]α .

(iv) For N = 2,3, . . . , we put

EpN =

{a ∈ ω:

∞∑k=1

|ak|k−1∑j=1

j−1∑i=1

N−1/pi wi < ∞}

and

FpN =

{a ∈ ω: sup

k�2|ak|

(k−1∑j=1

j−1∑i=1

N−1/pi wi

)−1

< ∞}

.

By a well known result (cf. [5, Lemma 4(iv)]), we have to showF pN = (E

pN)α (N =

2,3, . . .). The proof of this is standard and therefore omitted.✷Now we shall give some new results:


Theorem 2.2. Let p = (pk) be a sequence of strictly positive real numbers. Then

(i) [c(u,∆2,p)]α = M3(p),(ii) [S�∞(u,∆2,p)]β = M∞(p),

where

M3(p) = D1(p) ∩{

a ∈ ω:∞∑

k=1

|ak|k−1∑j=1

j−1∑i=1

wi < ∞}

,

M∞(p) =⋂

N�2

{a ∈ ω:

∞∑k=1

ak

k−1∑j=1

N1/pi wi converges and

∞∑k=1

∣∣R(2)k

∣∣N1/pk wk < ∞}

,

and R(2)k =∑∞

v=k+1 Rv , Rk =∑∞v=k+1 av .

Proof. (i) Let a ∈ M3(p) andx ∈ c(u,∆2,p). Then there is a complex numberl such that|uk∆

2xk − l|pk → 0 (k → ∞). We define the sequencey by

yk = xk − l

k−1∑j=1

j−1∑i=1

u−1i (k = 1,2, . . .).

Theny ∈ c0(u,∆2,p) and

∞∑k=1

|akxk| �∞∑

k=1

|ak|k−1∑j=1

j−1∑i=1

|∆2yi | +∞∑

k=1

|ak|2∑

j=1

(k − j − 1

2− j

)|∆2−jyj |

+ |l|∞∑

k=1

|ak|k−1∑j=1

j−1∑i=1

wi < ∞

by Theorem 2.1(iii) and sincea ∈ M3(p).Now leta ∈ [c(u,∆2,p)]α . Since[c(u,∆2,p)]α ⊂ [c0(u,∆2,p)]α and[c0(u,∆2,p)]α

= D1(p) by Theorem 2.1(iii), thena ∈ D1(p). If we put xk = ∑k−1j=1

∑j−1i=1 wi (k =

1,2, . . .), thenx ∈ c(u,∆2,p) and therefore∑∞

k=1 |ak|∑k−1j=1

∑j−1i=1 wi < ∞. Thusa ∈

M3(p).(ii) Let a ∈ M∞(p) and x ∈ S�∞(u,∆2,p). Then (uk∆

2xk) ∈ �∞(p) and we canchoose an integerN > max{1,supk |uk∆

2xk|pk }. Now we have

n∑k=1

akxk =n−2∑k=1

R(2)k ∆2xk − R

(2)n−1

n−2∑k=1

∆2xk + Rn

n−1∑k=1

∆xk (n = 1,2, . . .). (2)


Since∞∑

k=1

∣∣R(2)k

∣∣|∆2xk| �∞∑

k=1

∣∣R(2)k

∣∣N1/pkwk < ∞,

it follows that the first term on the right of (2) is absolutely convergent. Now, by Corollary 2of [4], the convergence of

∑∞k=1 ak

∑k−1j=1

∑j−1i=1 N1/pi wi implies that the second and third

terms on the right of (2) tend to zero asn → ∞. Hence∑∞

k=1 akxk is convergent for eachx ∈ S�∞(u,∆2,p), and soa ∈ [S�∞(u,∆2,p)]β .

Conversely, leta ∈ [S�∞(u,∆2,p)]β . Then, forx ∈ S�∞(u,∆2,p),∑∞

k=1 akxk is con-

vergent. Now, sincex = (xk) = (∑k−1

j=1∑j−1

i=1 N1/pi wi)∞k=1 ∈ S�∞(u,∆2,p), it follows

that∑∞

k=1 ak

∑k−1j=1

∑j−1i=1 N1/pi wi is convergent. Further, by Corollary 2 of [4], we have

that the second and third terms on the right of (2) tend to zero. Therefore,∑∞

k=1 R(2)k ∆2xk

is convergent for allx ∈ S�∞(u,∆2,p). We write

∞∑k=1

R(2)k ∆2xk =

∞∑k=1

(R

(2)k

uk

)(uk∆

2xk),

so that∑∞

k=1 |R(2)k |N1/pkwk < ∞, since(uk∆

2xk)∞k=1 ∈ �∞(p). Hencea ∈ M∞(p).

By Remark 1.1,[�∞(u,∆2,p)]β = M∞(p). ✷

3. Matrix transformations

For any complex matrixA = (ank), we shall writeAn = (ank)k for the sequence in thenth row ofA. GivenA, we define the matrixB by

bnk = ank − an+1,k (n, k = 1,2, . . .).

Let X, Y be two subsets ofω. By (X,Y ) we denote the class of all matricesA such thatthe seriesAnx = ∑∞

k=1 ankxk converges for allx ∈ X (n = 1,2, . . .) and the sequenceAx = (Anx) is in Y for all x ∈ X.

Theorem 3.1. For every strictly positive sequence p, we have A ∈ (�∞(u,∆2,p), �∞) ifand only if the following three conditions hold:

(i) (an1) and (an2) ∈ �∞(u,∆2,p);(ii)

∑∞k=1 ank

∑k−1j=1

∑j−1i=1 N1/pi wi converges for all N > 1;

(iii) (R(2)u−1) ∈ (�∞(p), �∞) where R(2) = (R(2)nk ) = (

∑∞v=k+1 Rnv) and Rnv =∑∞

i=v+1 ani .

Proof. (Necessity) LetA ∈ (�∞(u,∆2,p), �∞). Then the seriesAn(x) =∑k ankxk con-

verges for eachn andAx = (An(x)) ∈ �∞ for eachx ∈ �∞(u,∆2,p). Conditions (i) and(ii) follow easily since the sequences(1,0,0, . . .) and(0,1,0, . . .) belong to�∞(u,∆2,p).


Also, x = (xk) = (∑k−1

j=1∑j−1

i=1 N1/pi wi) ∈ S�∞(u,∆2,p) ⊂ �∞(u,∆2,p), and hence (ii)holds. Now, using (2), we have

An(s, x) =s−2∑k=1

R(2)nk ∆2xk − R

(2)n,s−1

s−2∑k=1

∆2xk + Rns

s−1∑k=1

∆xk.

Taking limit ass → ∞, and using condition (ii) together with Corollary 2 of [4], we have

lims→∞An(s, x) = An(x) =

∞∑k=1

R(2)nk ∆2xk =

∞∑k=1

(R

(2)nk

uk

)(uk∆

2xk).

Define the sequencey = (yk) by yk = uk∆2xk. Therefore,x ∈ �∞(u,∆2,p) implies

y ∈ �∞(p), and An(x) = Rn(y), whereR(2)u−1 = (Rn(y)), Rn(y) = ∑k(R

(2)nk /uk)yk.

HenceA ∈ (�∞(u,∆2,p), �∞) implies thatR(2)u−1 ∈ (�∞(p), �∞), i.e., condition (iii)holds.

(Sufficiency) Letx ∈ �∞(u,∆2,p). Definex = (xk) by

xk =

x1, k = 1,

x2, k = 2,

x ′k, k > 2,

wherex ′ = (x ′k) ∈ S�∞(u,∆2,p). Again, by (2) we get

An(x) = an1x1 + an2x2 +∞∑

k=1

R(2)nk ∆2x ′

k

= an1x1 + an2x2 +∞∑

k=1

(R

(2)nk

uk

)(uk∆

2x ′k

).

Therefore, by conditions (i)–(iii),Ax = (An(x)) exists andAx = R(2)u−1 = y ∈ �∞.Hencea ∈ (�∞(u,∆2,p), �∞).

This completes the proof of the theorem.✷

References

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[2] Ç. Asma, R. Çolak, On The Köthe–Toeplitz duals of some generalized sets of difference sequences,Demonstratio Math. 4 (2000) 797–803.

[3] M. Et, On some difference sequence spaces, Doga-Tr. J. Math. 17 (1993) 18–24.[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 and a generalization of a theorem of Iyer,

Pacific J. Math. 38 (1971) 487–500.[6] E. Malkowsky, A note on the Köthe–Toeplitz duals of generalized sets of bounded and convergent difference

sequences, J. Anal. 4 (1996) 81–91.[7] Mursaleen, Generalized spaces of difference sequences, J. Math. Anal. Appl. 203 (1996) 738–745.