# On some new generalized sequence spaces

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• J. Math. Anal. Appl. 277 (2003) 681688www.elsevier.com/locate/jmaa

On some new generalized sequence spacesigdem A. Bektas

Department of Mathematics, Firat University, Elazig, 23119, TurkeyReceived 29 January 2001

Submitted by A.C. Peterson

Abstract

In this paper we define the sequence sets (u,2,p), c(u,2,p) and c0(u,2,p), and give- and -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; - and -duals

1. Introduction

Let be the set of all complex sequences x = (xk)k=1. We write , c, c0 for the spacesof bounded, convergent and null sequences, respectively; and cs, 1 for the spaces of con-vergent and absolutely convergent series. p = (pk)k=1 will denote an arbitrary sequenceof positive real numbers.

Ahmad and Mursaleen  defined the sequence spacesX(p)= {x : x X(p)}

where x = (xk)k=1 = (xk xk+1)k=1 and X= , c or c0.Let U be the set of all sequences u = (uk)k=1 such that uk = 0 and complex for all

k = 1,2, . . . . Throughout the paper we write wk = 1/|uk|.After Malkowsky  defined the sequence spaces

X(u,)= {x : (uk(xk xk+1))k=1 X}where u U and X = , c or c0.

• 682 .A. Bektas / J. Math. Anal. Appl. 277 (2003) 681688

Recently Asma and olak  defined the sequence spacesX(u,,p)= {x : (ukxk)k=1 X(p)}

where u U and X = , c or c0.Given an arbitrary sequence p = (pk)k=1 of positive real numbers pk and given u U ,

we define the sets

(u,2,p)={x : (uk2xk)k=1 (p)

},

c(u,2,p)= {x : (uk2xk)k=1 c(p)},c0(u,

2,p)= {x : (uk2xk)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 to p and u:

(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 ).

(ii) If pk = 1, then (u,2,p)= (u,2), c(u,2,p)= c(u,2) and c0(u,2,p)=c0(u,2) (see Mursaleen ).

Let us define the operator S : (u,2,p) (u,2,p) by x Sx = (0,0, x3, x4,. . .). It is clear that S 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)] = [(u,2,p)] .

If p = (pk) is a fixed bounded sequence of positive real numbers and u U , then(u,2,p), c(u,2,p) and c0(u,2,p) are linear spaces, under the usual operations

x + y = (xn + yn) and x = (xn)where is any complex number.

Theorem 1.2. Let p = (pk)k=1 be a bounded sequence of strictly positive real numberspk and u U . Then c0(u,2,p) is a paranormed space with paranorm g(x) =supk |uk2xk|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. KtheToeplitz duals

Let z be any sequence and let Y be any subset of . Then we shall write

z1 Y = {x : zx = (zkxk) Y}.

• .A. Bektas / J. Math. Anal. Appl. 277 (2003) 681688 683

For any subset X of , the sets

X =xX

(x1 1) and X =xX

(x1 cs)

are called - and -duals of X. We shall write X = (X) and X = (X) .

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

(i)[(u,2,p)

] =M1(p)= N=2

{a :

k=1

|ak|k1j=1

j1i=1

N1/piwi

• 684 .A. Bektas / J. Math. Anal. Appl. 277 (2003) 681688

k=1

|ak|k1j=1

j1i=1

N1/piwi +k=1

|ak|2

j=1

(k j 1

2 j)|2jxj |

1. We define the sequence x by

xk =k1j=1

j1i=1

N1/piwi (k = 2,3, . . .).

Then it is easy to see that x (u,2,p) and k=1 |akxk| = . Hence a /[(u,2,p)] .(ii) Let a M2(p) and x [(u,2,p)] =M1(p), by part (i). Then for someN > 1,

we havek=3

|akxk| =k=3

|ak|(k1j=1

j1i=1

N1/piwi

)1|xk|

k1j=1

j1i=1

N1/piwi

[

supk3

|ak|(k1j=1

j1i=1

N1/piwi

)1] k=3

|xk|k1j=1

j1i=1

N1/piwi 1, we have

supk3

|ak|(k1j=1

j1i=1

N1/piwi

)1=.

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

|ak(s)|(k(s)1j=1

j1i=1

s1/piwi

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

We define the sequence x by

xk ={ |ak(s)|1 (k = k(s)),

0 (k = k(s), s = 2,3, . . .).Then for all integers N > 3, we have

k=1

|xk|k1j=1

j1i=1

N1/piwi s=3

|ak(s)|1k(s)1j=1

j1i=1

N1/piwi

• .A. Bektas / J. Math. Anal. Appl. 277 (2003) 681688 685

N1s=3

|ak(s)|1k(s)1j=1

j1i=1

N1/piwi +s=N

|ak(s)|1k(s)1j=1

j1i=1

s1/piwi

N1s=3

|ak(s)|1k(s)1j=1

j1i=1

N1/piwi +s=N

s3 k0 |uk2xk|pk N1, where N is the number in D1(p). We putM := max

1kk0|uk2xk|pk , m := min

1kk0pk, L := (M + 1)N,

and define the sequence y by yk := xkL1/m (k = 1,2, . . .). Then it is easy to see thatsupk |uk2yk|pk N1 and, as in (1) with N replaced by N1, we have

k=1

|akxk| = L1/mk=1

|akyk| 1 (s = 1,2, . . .).

We define the sequence x by

xk =s1l=1

k(l+1)1j=k(l)

j1i=1

(l + 1)1/piwi +k1

j=k(s)

j1i=1

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

Then it is easy to see that |uk2xk|pk = 1/(s+ 1) (k(s) k k(s+ 1) 1, s = 1,2, . . .).Hence x c0(u,2,p) and k=1 |akxk|s=1 1=, i.e., a / [c0(u,2,p)] .

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

EpN =

{a :

k=1

|ak|k1j=1

j1i=1

N1/piwi

• 686 .A. Bektas / J. Math. Anal. Appl. 277 (2003) 681688

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

(i) [c(u,2,p)] =M3(p),(ii) [S(u,2,p)] =M(p),

where

M3(p)=D1(p) {a :

k=1

|ak|k1j=1

j1i=1

wi

• .A. Bektas / J. Math. Anal. Appl. 277 (2003) 681688 687

Sincek=1

R(2)k |2xk| k=1

R(2)k N1/pkwk

• 688 .A. Bektas / J. Math. Anal. Appl. 277 (2003) 681688

Also, x = (xk)= (k1j=1j1i=1 N1/piwi) S(u,2,p) (u,2,p), and hence (ii)holds. Now, using (2), we have

An(s, x)=s2k=1

R(2)nk

2xk R(2)n,s1s2k=1

2xk +Rnss1k=1

xk.

Taking limit as s, and using condition (ii) together with Corollary 2 of , we have

limsAn(s, x)=An(x)=

k=1

R(2)nk

2xk =k=1

(R(2)nk

uk

)(uk

2xk).

Define the sequence y = (yk) by yk = uk2xk . Therefore, x (u,2,p) impliesy (p), and An(x) = Rn(y), where R(2)u1 = (Rn(y)), Rn(y) =k(R(2)nk /uk)yk .Hence A ((u,2,p), ) implies that R(2)u1 ((p), ), i.e., condition (iii)holds.

(Sufficiency) Let x (u,2,p). Define x = (xk) by

xk =x1, k = 1,x2, k = 2,x k, k > 2,

where x = (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 and Ax = R(2)u1 = y .Hence a ((u,2,p), ).

This completes the proof of the theorem.

References

 Z.U. Ahmad, Mursaleen, KtheToeplitz duals of some new sequence spaces and their matrix maps, Publ.Inst. Math. (Beograd) 42 (1987) 5761.

 . Asma, R. olak, On The KtheToeplitz duals of some generalized sets of difference sequences,Demonstratio Math. 4 (2000) 797803.

 M. Et, On some difference sequence spaces, Doga-Tr. J. Math. 17 (1993) 1824. H. Kizmaz, On certain sequence spaces, Canad. Math. Bull. 24 (1981) 169176. C.G. Lascarides, A study of certain sequence spaces of Maddox and a generalization of a theorem of Iyer,

Pacific J. Math. 38 (1971) 487500. E. Malkowsky, A note on the KtheToeplitz duals of generalized sets of bounded and convergent difference

sequences, J. Anal. 4 (1996) 8191. Mursaleen, Generalized spaces of difference sequences, J. Math. Anal. Appl. 203 (1996) 738745.

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