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Volume 7, Issue 3, Article 101, 2006

THE DUAL SPACES OF THE SETS OF DIFFERENCE SEQUENCES OF ORDER m

Ç.A. BEKTA ¸S AND M. ET DEPARTMENT OFMATHEMATICS

FIRATUNIVERSITY

ELAZIG, 23119, TURKEY.

cbektas@firat.edu.tr mikailet@yahoo.com

Received 16 December, 2005; accepted 07 January, 2006 Communicated by A.G. Babenko

ABSTRACT. The idea of difference sequence spaces was introduced by Kızmaz [5] and the concept was generalized by Et and Çolak [3]. Letp= (pk)be a bounded sequence of positive real numbers andv= (vk)be any fixed sequence of non-zero complex numbers. Ifx= (xk)is any sequence of complex numbers we writemvxfor the sequence of them-th order differences ofxandmv (X) = {x= (xk) : ∆mvx X}for any setX of sequences. In this paper we determine theα-,β - andγ- duals of the setsmv(X)which are defined by Et et al. [2] for X =`(p),c(p)andc0(p).This study generalizes results of Malkowsky [9] in special cases.

Key words and phrases: Difference sequences,α−,β−andγ−duals.

2000 Mathematics Subject Classification. 40C05, 46A45.

1. INTRODUCTION, NOTATIONS ANDKNOWN RESULTS

Throughout this paperωdenotes the space of all scalar sequences and any subspace of ωis called a sequence space. Let `, candc0 be the linear space of bounded, convergent and null sequences with complex terms, respectively, normed by

kxk = sup

k

|xk|,

wherek∈N={1,2,3, . . .},the set of positive integers. Furthermore, letp= (pk)be bounded sequences of positive real numbers and

`(p) =

x∈ω : sup

k

|xk|pk <∞

, c(p) =n

x∈ω: lim

k→∞ |xk−l|pk = 0, for some l∈C o

,

ISSN (electronic): 1443-5756

247-04

(2)

c0(p) =n

x∈ω: lim

k→∞ |xk|pk = 0o (for details see [6], [7], [11]).

Letxandybe complex sequences, andE andF be subsets ofω. We write M(E, F) = \

x∈E

x−1 ∗F ={a∈ω :ax ∈F for allx∈E } [12].

In particular, the sets

Eα =M(E, l1), Eβ =M(E, cs) and Eγ =M(E, bs)

are called the α−, β− andγ− duals of E, where l1, cs andbs are the sets of all convergent, absolutely convergent and bounded series, respectively. If E ⊂ F, then Fη ⊂ Eη for η = α, β, γ. It is clear thatEα ⊂(Eα)α = Eαα. IfE =Eαα, thenE is anα-space. In particular, anα-space is called a Köthe space or a perfect sequence space.

Throughout this paperX will be used to denote any one of the sequence spaces`,candc0. Kızmaz [5] introduced the notion of difference sequence spaces as follows:

X(∆) ={x= (xk) : (∆xk)∈X}.

Later on the notion was generalized by Et and Çolak in [3], namely, X(∆m) = {x= (xk) : (∆mxk)∈X}.

Subsequently difference sequence spaces have been studied by Malkowsky and Parashar [8], Mursaleen [10], Çolak [1] and many others.

Letv = (vk)be any fixed sequence of non-zero complex numbers. Et and Esi [4] generalized the above sequence spaces to the following ones

mv (X) = {x= (xk) : (∆mv xk)∈X},

where∆0vx= (vkxk),∆mv x= (∆m−1v xk−∆m−1v xk+1)such that∆mv xk =Pm

i=0(−1)i mi

vk+ixk+i. Recently Et et al. [2] generalized the sequence spaces∆mv (X)to the sequence spaces

mv (X(p)) ={x= (xk) :(∆mv xk)∈X(p)}

and showed that these spaces are complete paranormed spaces paranormed by g(x) =

m

X

i=1

|xivi|+ sup

k

|∆mv xk|pk/M, whereH = supkpk andM = max (1, H).

Let us define the operatorD : ∆mv X(p) → ∆mv X(p) byDx = (0,0, . . . , xm+1, xm+2, . . .), wherex = (x1, x2, x3, . . .). It is trivial that Dis a bounded linear operator on ∆mv X(p). Fur- thermore the set

D[∆mv X(p)] = D∆mv X(p)

={x= (xk) :x∈∆mv X(p), x1 =x2 =· · ·=xm = 0}

is a subspace of∆mv X(p). D∆mv X(p)andX(p)are equivalent as topological spaces, since

(1.1) ∆mv :D∆mv X(p)→X(p)

defined by ∆mv x = y = (∆mv xk) is a linear homeomorphism. Let [X(p)]0 and [D∆mv X(p)]0 denote the continuous duals ofX(p)andD∆mv X(p), respectively. It can be shown that

T : [D∆mv X(p)]0 →[X(p)]0, f→f◦(∆mv )−1 =f, is a linear isometry. So[D∆mv X(p)]0 is equivalent to[X(p)]0.

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Lemma 1.1 ([5]). Let(tn)be a sequence of positive numbers increasing monotonically to in- finity, then

i) Ifsupn|Pn

i=1tiai|<∞, thensupn tnP

k=n+1ak <∞, ii) IfP

ktkakis convergent, thenlimn→∞tnP

k=n+1ak = 0.

2. MAINRESULTS

In this section we determine theα−,β−andγ−duals of∆mv X(p).

Theorem 2.1. For every strictly positive sequencep= (pk), we have (i) [∆mv l(p)]α =Dα1(p),

(ii) [∆mv l(p)]αα =D1αα(p) where

D1α(p) =

\

N=2

(

a= (ak) :

X

k=1

|ak| |vk|−1

k−m

X

j=1

k−j−1 m−1

N1/pj <∞ )

and

Dαα1 (p) =

[

N=2

a= (ak) : sup

k≥m+1

|ak| |vk|

"k−m X

j=1

k−j−1 m−1

N1/pj

#−1

<∞

 .

Proof. (i) Let a ∈ D1α(p) andx ∈ ∆mv l(p). We choose N > max(1,supn|∆mv an|pn).

Since

k−m

X

j=1

k−j −1 m−1

N1/pj >

m

X

j=1

k−j −1 m−j

N1/pj ≥1

for arbitraryN >1 (k = 2m,2m+ 1, . . .)and|∆m−jv xj| ≤M (1≤ j ≤m)for some constantM,a ∈D1α(p)implies

X

k=1

|ak| |vk|−1

m

X

j=1

k−j−1 m−j

m−jv xj <∞.

Then

X

k=1

|akxk|

=

X

k=1

|ak| |vk|−1

k−m

X

j=1

(−1)m

k−j −1 m−1

mv xj +

m

X

j=1

(−1)m−j

k−j−1 m−j

m−jv xj

!

X

k=1

|ak| |vk|−1

k−m

X

j=1

k−j−1 m−1

N1/pj +

X

k=1

|ak| |vk|−1

m

X

j=1

k−j−1 m−j

m−jv xj

<∞.

Conversely leta /∈Dα1(p). Then we have

X

k=1

|ak||vk|−1

k−m

X

j=1

k−j −1 m−1

N1/pj =∞

(4)

for some integerN >1. We define the sequencexby xk =vk−1

k−m

X

j=1

k−j −1 m−1

N1/pj (k =m+ 1, m+ 2, . . .).

Then it is easy to see thatx ∈ ∆mv l(p)andP

k

|akxk| =∞. Hencea /∈ [∆mv l(p)]α. This completes the proof of (i).

(ii) Leta∈ D1αα(p) andx∈ [∆mv l(p)]α =Dα1(p), by part (i). Then for someN >1, we have

X

k=m+1

|akxk|

=

X

k=m+1

|ak| |vk|

"k−m X

j=1

k−j −1 m−1

N1/pj

#−1

|xk| |vk|−1

"k−m X

j=1

k−j −1 m−1

N1/pj

#

≤ sup

k≥m+1

|ak| |vk|

"k−m X

j=1

k−j −1 m−1

N1/pj

#−1

×

X

k=m+1

|xk| |vk|−1

k−m

X

j=1

k−j −1 m−1

N1/pj

<∞.

Conversely leta /∈Dαα1 (p).Then for all integersN >1, we have sup

k≥m+1

|ak| |vk|

"k−m X

j=1

k−j−1 m−1

N1/pj

#−1

=∞.

We recall that

k−m

X

j=1

k−j −1 m−1

yj = 0 (k < m+ 1) for arbitraryyj.

Hence there is a strictly increasing sequence(k(s))of integers k(s) ≥ m+ 1such that

|ak(s)| |vk(s)|

k(s)−m

X

j=1

k(s)−j−1 m−1

s1/pj

−1

> sm+1 (s=m+ 1, m+ 2, . . .).

We define the sequencexby xk=

|ak(s)|−1, (k=k(s))

0, (k6=k(s)) (k =m+ 1, m+ 2, . . .) Then for all integersN > m+ 1, we have

X

k=1

|xk| |vk|−1

k−m

X

j=1

k−j−1 m−1

N1/pj <

X

s=m+1

s−(m+1) <∞.

(5)

Hencex ∈ [∆mv l(p)]α andP

k=1|akxk| =P

N=11 = ∞. Hencea /∈ [∆mv l(p)]αα. The proof is completed.

Theorem 2.2. For every strictly positive sequencep= (pk), we have

(i) [∆mv c0(p)]α =M0α(p), (ii) [∆mv c0(p)]αα =M0αα(p) where

M0α(p) =

[

N=2

(

a∈ω :

X

k=1

|ak| |vk|−1

k−m

X

j=1

k−j−1 m−1

N−1/pj <∞ )

and

M0αα(p) =

\

N=2

a ∈ω : sup

k≥m+1

|ak| |vk|

"k−m X

j=1

k−j−1 m−1

N−1/pj

#−1

<∞

 .

Proof.

(i) Leta∈M0α(p) andx∈∆mv c0(p). Then there is an integerk0 such thatsup

k>k0

|∆mv xk|pk ≤ N−1, whereN is the number inM0α(p). We put

M = max

1≤k≤k0|∆mv xk|pk, n= min

1≤k≤k0pk, L= (M+ 1)N

and define the sequencey byyk=xk·L−1/n (k= 1,2, . . .). Then it is easy to see that supk|∆mv yk|pk ≤N−1.

Since

k−m

X

j=1

k−j−1 m−1

N−1/pj >

m

X

j=1

k−j−1 m−j

N−1/pj for arbitraryN >1 (k= 2m,2m+ 1, . . .),a∈M0α(p) implies

X

k=1

|ak| |vk|

m

X

j=1

k−j−1 m−j

|∆m−jv yj|<∞.

Then

X

k=1

|akxk|=L1/n

X

k=1

|akyk|

≤L1/n

X

k=1

|ak| |vk|−1

k−m

X

j=1

k−j −1 m−1

N−1/pj

+L1/n

X

k=1

|ak| |vk|−1

m

X

j=1

k−j−1 m−j

|∆m−jv yj|

<∞.

So we havea∈[∆mv c0(p)]α. ThereforeM0α(p)⊂[∆mv c0(p)]α.

Conversely, let a /∈ M0α(p). Then we can determine a strictly increasing sequence (k(s))of integers such thatk(1) = 1and

M(s) =

k(s+1)−1

X

k=k(s)

|ak| |vk|−1

k−m

X

j=1

k−j−1 m−1

(s+ 1)−1/pj >1 (s = 1,2, . . .).

(6)

We define the sequencexby xk =vk−1

s−1

X

l=1

k(l+1)−1

X

j=k(l)

k−j−1 m−1

(l+ 1)−1/pj +

k−m

X

j=k(s)

k−j −1 m−1

(s+ 1)−1/pj

(k(s)≤k ≤k(s+ 1)−1;s = 1,2, . . .).

Then it is easy to see that |∆mv xk|pk = s+11 (k(s) ≤ k ≤ k(s+ 1)−1;s = 1,2, . . .) hencex∈∆mv c0(p), andP

k=1|akxk| ≥P

s=1 =∞, i.e. a /∈[∆mv c0(p)]α. (ii) Omitted.

Theorem 2.3. For every strictly positive sequencep= (pk), we have

[∆mv c(p)]α =Mα(p) =M0α(p)∩ (

a∈ω :

X

k=1

|ak| |vk|−1

k−m

X

j=1

k−j−1 m−1

<∞ )

.

Proof. Leta∈Mα(p) andx∈∆mv c(p). Then there is a complex numberl such that|∆mv xk− l|pk →0 (k → ∞). We definey= (yk) by

yk =xk+vk−1l(−1)m+1

k−m

X

j=1

k−j−1 m−1

(k= 1,2, . . .).

Theny ∈∆mv c0(p) and

X

k=1

|akxk| ≤

X

k=1

|ak| |vk|−1

k−m

X

j=1

k−j −1 m−1

|∆mv yj|

+

X

k=1

|ak| |vk|−1

m

X

j=1

k−j−1 m−j

|∆m−jv yj|

+|l|

X

k=1

|ak| |vk|−1

k−m

X

j=1

k−j−1 m−1

<∞ by Theorem 2.2(i) and sincea∈Mα(p).

Now leta ∈ [∆mv c(p)]α ⊂ [∆mv c0(p)]α = M0α(p)by Theorem 2.2(i). Since the sequencex defined by

xk= (−1)mv−1k

k−m

X

j=1

k−j −1 m−1

(k = 1,2, . . .) is in∆mvc(p), we have

X

k=1

|ak|

k−m

X

j=1

k−j−1 m−1

<∞.

Theorem 2.4. For every strictly positive sequencep= (pk), we have

(i) [D∆mv `(p)]β =Mβ(p), (ii) [D∆mv `(p)]γ =Mγ(p)

(7)

where

Mβ(p) = \

N >1

(

a ∈ω:

X

k=1

akvk−1

k−m

X

j=1

k−j−1 m−1

N1/pj converges and

X

k=1

|bk|

k−m+1

X

j=1

k−j−1 m−2

N1/pj <∞ )

,

Mγ(p) = \

N >1

(

a ∈ω: sup

n

|

n

X

k=1

akv−1k

k−m

X

j=1

k−j−1 m−1

N1/pj|<∞,

X

k=1

|bk|

k−m+1

X

j=1

k−j−1 m−2

N1/pj <∞ )

andbk =P

j=k+1v−1j aj (k = 1,2, . . .).

Proof.

(i) Ifx∈D∆mv `(p)then there exists a uniquey= (yk)∈`(p) such that xk=vk−1

k−m

X

j=1

(−1)m

k−j −1 m−1

yj

for sufficiently large k, for instance k > m by (1.1). Then there is an integer N >

max{1,supk|∆mv xk|pk}. Let a ∈ Mβ(p), and suppose that −1−1

= 1. Then we may write

n

X

k=1

akxk =

n

X

k=1

ak vk−1

k−m

X

j=1

(−1)m

k−j−1 m−1

yj

!

= (−1)m

n−m

X

k=1

bk+m−1 k

X

j=1

k+m−j−2 m−2

yj−bn

n−m

X

j=1

(−1)m

n−j−1 m−1

yj. Since

X

k=1

|bk+m−1|

k

X

j=1

k+m−j−2 m−2

N1/pj <∞, the series

X

k=1

bk+m−1

k

X

j=1

k+m−j−2 m−2

yj

is absolutely convergent. Moreover by Lemma 1.1(ii), the convergence of

X

k=1

akv−1k

k−m

X

j=1

k−j−1 m−1

N1/pj implies

n→∞lim bn

n−m

X

j=1

n−j−1 m−1

N1/pj = 0.

HenceP

k=1akxk is convergent for allx∈D∆mv `(p), soa∈[D∆mv `(p)]β.

(8)

Conversely let a ∈ [D∆mv `(p)]β. Then P

k=1akxk is convergent for each x ∈ D∆mv `(p). If we take the sequencex= (xk) defined by

xk =

0, k≤m

v−1k Pk−m j=1

k−j−1 m−1

N1/pj, k > m then we have

X

k=1

akvk−1

k−m

X

j=1

k−j−1 m−1

N1/pj =

X

k=1

akxk <∞.

Thus the seriesP

k=1akv−1k Pk−m j=1

k−j−1 m−1

N1/pj is convergent. This implies that

n→∞lim bn

n−m

X

j=1

n−j −1 m−1

N1/pj = 0 by Lemma 1.1(ii).

Now let a ∈ [D∆mv `(p)]β − Mβ(p). Then P

k=1|bk|Pk−m+1 j=1

k−j−1 m−2

N1/pj is divergent, that is,

X

k=1

|bk|

k−m+1

X

j=1

k−j−1 m−2

N1/pj =∞.

We define the sequencex= (xk) by xk =

0, k ≤m

vk−1Pk−1

i=1 sgnbiPi−m+1 j=1

i−j−1 m−2

N1/pj, k > m

where ak > 0for all k orak < 0 for all k. It is trivial thatx = (xk) ∈ D∆mv `(p).

Then we may write forn > m

n

X

k=1

akxk =−

m

X

k=1

bk−1vxk−1

n−m

X

k=1

bk+m−1vxk+m−1−bnxnvn. Since(bnxnvn)∈c0, now lettingn → ∞we get

X

k=1

akxk=−

X

k=1

bk+m−1vxk+m−1

=

X

k=1

|bk+m−1|

k

X

j=1

k+m−j−2 m−2

N1/pj =∞.

This is a contradiction toa∈[D∆mv `(p)]β. Hencea∈Mβ(p).

(ii) Can be proved by the same way as above, using Lemma 1.1(i).

Lemma 2.5. [D∆mv `(p)]η = [D∆mv c(p)]η forη=β orγ.

The proof is obvious and is thus omitted.

Theorem 2.6. Letc+0 denote the set of all positive null sequences.

(9)

(a) We put M3β(p) =

(

a∈ω :

X

k=1

akv−1k

k−m

X

j=1

k−j−1 m−1

N1/pj uj converges and

X

k=1

|bk|

k−m+1

X

j=1

k−j−1 m−2

N1/pj uj <∞,∀u∈c+0 )

. Then[D∆mv c0(p)]β =M3β(p).

(b) We put M4γ(p) =

(

a ∈ω : sup

n

|

n

X

k=1

akv−1k

k−m

X

j=1

k−j−1 m−1

N1/pj uj|<∞,

X

k=1

|bk|

k−m+1

X

j=1

k−j−1 m−2

N1/pj uj <∞,∀u∈c+0 )

. Then[D∆mv c0(p)]γ =M4γ(p).

Proof. (a) and (b) can be proved in the same manner as Theorem 2.4, using Lemma 1.1(i) and

(ii).

Lemma 2.7.

i) [∆mv `(p)]η = [D∆mv `(p)]η, ii) [∆mv c(p)]η = [D∆mv c(p)]η, iii) [∆mv c0(p)]η = [D∆mv c0(p)]η forη =β orγ.

The proof is omitted.

REFERENCES

[1] R. ÇOLAK, Lacunary strong convergence of difference sequences with respect to a modulus func- tion, Filomat, 17 (2003), 9–14.

[2] M. ET, H. ALTINOKANDY. ALTIN, On some generalized sequence spaces, Appl. Math. Comp., 154 (2004), 167–173.

[3] M. ETANDR. ÇOLAK, On some generalized difference sequence spaces, Soochow J. Math., 21(4) (1995), 377-386.

[4] M. ET AND A. ESI, On Köthe-Toeplitz duals of generalized difference sequence spaces, Bull.

Malays. Math. Sci. Soc., (2) 23 (2000), 25–32.

[5] H. KIZMAZ, On certain sequence spaces, Canad. Math. Bull., 24 (1981), 169–176.

[6] C.G. LASCARIDES AND I.J. MADDOX, Matrix transformations between some classes of se- quences, Proc. Cambridge Philos. Soc., 68 (1970), 99–104.

[7] I.J. MADDOX, Continuous and Köthe-Toeplitz duals of certain sequence spaces, Proc. Cambridge Philos. Soc., 65 (1967), 471–475.

[8] E. MALKOWSKYANDS.D. PARASHAR, Matrix transformations in spaces of bounded and con- vergent difference sequences of orderm, Analysis, 17 (1997), 87–97.

[9] E. MALKOWSKY, Absolute and ordinary Köthe-Toeplitz duals of some sets of sequences and matrix transformations, Publ. Inst. Math., (Beograd) (N.S.) 46(60) (1989), 97–103.

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[10] MURSALEEN, Generalized spaces of difference sequences, J. Math. Anal. Appl., 203 (1996), 738–745.

[11] S. SIMONS, The sequence spaces `(pv) andm(pv), Proc. London Math. Soc., (3) 15 (1965), 422–436.

[12] A. WILANSKY, Summability through Functional Analysis, North-Holland Mathematics Studies, 85 (1984).

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