## On Orlicz Difference Sequence Spaces. Hemen Dutta

SDU Journal of Science (E-Journal), 2010, 5 (1): 119-136 ___________________________________________________________________ On Orlicz Difference Seq...
Author: Alberta Watts
SDU Journal of Science (E-Journal), 2010, 5 (1): 119-136 ___________________________________________________________________

On Orlicz Difference Sequence Spaces Hemen Dutta Gauhati University, Department of Mathematics, Kokrajhar Campus, Assam, INDIA e-mail: [email protected] Received:13 November 2008, Accepted: 9 February 2010 Abstract: The main aim of this article is to generalize the famous Orlicz sequence space by using difference operators and a sequence of non-zero scalars and investigate some topological structure relevant to this generalized space. Key words: Difference sequence space, multiplier sequence space, Orlicz function, AK-BK space, topological isomorphism and Köthe-Toeplitz dual.

Orlicz Fark Dizi Uzayları Üzerine Özet: Bu makalenin amacı, sıfırdan farklı skalerlerden oluşan bir diziyi ve fark operatörlerini kullanarak Orlicz dizi uzaylarını genelleştirmek ve bu yeni tanımladığımız uzayın topolojik yapısını incelemektir. Anahtar kelimeler: Fark dizi uzayı, çok indisli dizi uzayı, Orlicz fonksiyonu, AK-BK uzayı, toplojik izomorfizm, Köthe-Toeplitz duali. 2000 Mathematics Subject Classification: 40A05, 40C05, 46A45.

1. Introduction Throughout this paper w, l ∞ , ℓ1, c and c° denote the spaces of all, bounded, absolutely summable, convergent and null sequences x = ( xk ) with complex terms respectively. The notion of difference sequence space was introduced by Kizmaz [1], who studied the difference sequence spaces l ∞ (∆ ) , c(∆ ) and c0 ( ∆ ) , where

Z ( ∆ ) = { x = ( xk ) ∈ w : ( ∆xk ) ∈ Z } ,

where ∆x = ( ∆xk ) = ( xk − xk +1 ) and ∆ 0 xk = xk for all k, for Z= l ∞ , c and c0 . An Orlicz function M :[0, ∞) → [0, ∞) is a function, which is continuous, non-decreasing and convex with M (0) = 0 , M ( x) > 0 , for x > 0 and M ( x) → ∞ , as x → ∞ . An Orlicz function M can always be represented in the following integral form: x

M(x) = ∫ p (t ) dt , 0

where p, known as kernel of M, is right differentiable for t ≥0, p(0) = 0, p(t) > 0 for t > 0, p is non-decreasing, and p (t ) → ∞ as t → ∞ .

119

H. Dutta

Consider the kernel p(t) associated with the Orlicz function M(t), and let q(s) = sup{t: p(t) ≤ s } Then q possesses the same properties as the function p. Suppose now x

Φ ( x) = ∫ q ( s ) ds 0

Then Φ is an Orlicz function. The functions M and Φ are called mutually complementary Orlicz functions. Now we state the following well known results which can be found in [2]. Let M and F are mutually complementary Orlicz functions. Then we have (Young’s inequality) (1) (i) For x, y ≥ 0, xy ≤ M(x) + Φ ( y ) We also have (2) (ii) For x≥ 0, xp(x) = M(x) + Φ ( p ( x) ) (iii) M(λx) < λM(x) (3)

for all x ≥ 0 and λ with 0< λ0 there exist Rk>0 and xk>0 such that M(kx) ≤ RkM(x)

for all x ∈ (0, xk].

Moreover an Orlicz function M is said to satisfy the ∆2-condition if and only if M (2 x) 0  . k =1    ρ  For more details about Orlicz functions and sequence spaces associated with Orlicz functions one may refer to [2-5].

Let Λ = (λk) be a sequence of non-zero scalars. Then for a sequence space E, the multiplier sequence space E(Λ), associated with the multiplier sequence Λ is defined as E(Λ) =

{(xk ) ∈ w : (λk xk ) ∈ E}.

120

SDU Journal of Science (E-Journal), 2010, 5 (1): 119-136 ___________________________________________________________________

The scope for the studies on sequence spaces was extended by using the notion of associated multiplier sequences. Goes and Goes [6] defined the differentiated sequence space dE and integrated sequence space ∫ E for a given sequence space E, using the

multiplier sequences (k-1) and (k) respectively. A multiplier sequence can be used to accelerate the convergence of the sequences in some spaces. In some sense, it can be viewed as a catalyst, which is used to accelerate the process of chemical reaction. Sometimes the associated multiplier sequence delays the rate of convergence of a sequence. Thus it also covers a larger class of sequences for study. In the present article we shall consider a general multiplier sequence Λ = (λk) of non-zero scalars.

The notion of duals of sequence spaces was introduced by Köthe and Toeplitz [7]. Later on it was studied by Kizmaz [1], Kamthan [8] and many others. Let E and F be two sequence spaces. Then the F dual of E is defined as

EF = {(xk)∈ w : (xkyk)∈ F for all(yk)∈ E }. For F = ℓ1, the dual is termed as Köthe-Toeplitz or α-dual of E and denoted by Eα. More precisely, we have the following definition of Köthe Toeplitz dual of E:   E α = a = (ak ) : ∑ ak xk < ∞, for all x ∈ E  . k   It is known that if X Ì Y , then Yα ⊂ Xα. If EFF=E, where EFF= (EF)F, then E is said to be F-reflexive or F-perfect. In particular, if Eαα = E, then E is also said to be a Köthe space. Let Λ = (λk) be a sequence of non-zero scalars. Then we define the following spaces.

Definition 1.1. Let M be any Orlicz function. Then we define ∞   l% M ( ∆, Λ ) =  x ∈ w : δ ∆Λ ( M , x ) = ∑ M ( ∆λk xk ) < ∞  , k =1   where ∆λk xk = λk xk − λk +1 xk +1 for all k ≥ 1. ~ We can write l% M ∆ 0 , Λ = l M (Λ ) and if λk= 1 for all k ≥ 1, then we write ~ l% M ∆ 0 , Λ = l M .

(

)

(

)

Similarly we can define l% M ( ∇, Λ ) , where ∇λk xk = λk xk − λk −1 xk −1 for all k ≥ 1. Definition 1.2. Let M and Φ be mutually complementary functions. Then we define ∞   l M ( ∆, Λ ) =  x ∈ w : ∑ (∆λk xk ) yk converges for all y ∈ l% Φ  . k =1   We call this sequence space as Orlicz difference sequence space associated with the multiplier sequence Λ = (λk).

We can write l M ( ∆ 0 , Λ ) = l M (Λ ) and if λk= 1 for all k ≥ 1, then we write

121

H. Dutta

(

)

l M ∆0 , Λ = l M . Similarly we can define l M ( ∇, Λ ) where ∇λk xk = λk xk − λk −1 xk −1 for all k ≥ 1. One can easily observe in the special case M(x) = xp with 0 0, there exists a positive integer n0 such that

124

SDU Journal of Science (E-Journal), 2010, 5 (1): 119-136 ___________________________________________________________________ ∆

xi − x j

M

inf 0 : ρ M = λ x +   ≤ 1 , ∑ 1 1 (M ) k =1   ρ   ∆

(11)

(ii) l M ( ∇, Λ ) is a normed linear space under the norm . ( M ) defined by ∇

x

Proof. (i) Clearly x

∆ (M )

∞  ∇λk xk    > = inf 0 : ρ M   ≤ 1 . ∑ (M ) k =1   ρ  

=0 if x=θ. Next suppose x

∆ (M )

(12)

=0. Then from (11) we have

λ1 x1 =0 and so λ1 x1 = 0 .

(13)

∞  ∆λk xk    Again inf  ρ > 0 : ∑ M   ≤ 1 =0. This implies that for a given ε > 0 , there k =1   ρ   exists some ρ ε (0 < ρ ε < ε ) such that

 ∆λk xk  sup M   ≤ 1. ρ k ε    ∆λk xk  This implies that M   ≤ 1 for all k≥ 1. Thus  ρε   ∆λk xk   ∆λk xk  M ≤M  ≤1 ε ρ ε     for all k≥ 1. ∆λni xni → ∞ . It follows that Suppose ∆λni xni ≠ 0 , for some i. Let ε → 0 , then

ε

 ∆λn xn i i M  ε 

  → ∞ as ε → 0 for some ni ∈ N . This is a contradiction. Therefore   ∆λk xk =0

(14)

126

SDU Journal of Science (E-Journal), 2010, 5 (1): 119-136 ___________________________________________________________________

for all k≥ 1. Thus, by (13) and (14), it follows that λ k x k =0 for all k≥ 1. Hence x = θ , since (λk) is a sequence of non-zero scalars. Let x = (xk) and y = (yk) be any two elements of l M ( ∆, Λ ) . Then there exist ρ1 , ρ 2 >0 such that  ∆λk xk   ∆λk yk  sup M   ≤ 1 and sup M   ≤ 1. k k  ρ1   ρ2  Let ρ = ρ1 + ρ 2 . Then by convexity of M, we have

 ∆λk ( xk + yk ) sup M   ρ k 

  ∆λk xk   ∆λk yk  ρ1 ρ2 + ≤ sup M  sup M    ≤1.  ρ1 + ρ 2 k  ρ1  ρ1 + ρ 2 k  ρ2  

Hence we have

x+ y

  ∆λk ( xk + yk ) + + = ( ) inf λ x y  ρ > 0 : sup M  1 1 1 (M ) ρ k   ∆

   ≤ 1   

 ∆λk xk    ≤ λ1 x1 + inf  ρ1 > 0 : sup M   ≤ 1 + λ1 y1 k   ρ1     ∆λk yk   + inf  ρ 2 > 0 : sup M   ≤ 1 . k  ρ 2    This implies x + y

∆ (M )

≤ x

∆ (M )

+ x

∆ (M )

.

Finally, let ν be any scalar. Then

νx

∆ (M )

 ∆νλk xk    = νλ1 x1 + inf  ρ > 0 : sup M   ≤ 1 ρ   k    ∆λk xk    = ν λ1 x1 + inf r ν > 0 : sup M   ≤ 1 k  r    =ν x

where r =

∆ (M )

ρ . This completes the proof. ν

(ii) Proof is easy than part (i). ∆

Remark. It is obvious that the norms . ( M ) and . ( M ) are equivalent. Proposition 2.7. For x∈ l M ( ∇, Λ ) , we have

127

H. Dutta

 ∇λk xk  M ∑  || x ||(∆M−1 ) k =1  ∞

  ≤1.  

Proof. Proof is immediate from (12). ∇

Now we show that the norms . ( M ) and . M are equivalent. To prove this some other results are required. First we prove those results.

Proposition 2.8. Let x ∈ l M ( ∇, Λ ) with

(

δ Φ, { p ( ∇λn xn

x

)}) ≤1.

M

{ p ( ∇λ x )} ∈ l%

≤1. Then

n n

Φ

and

~ Proof. For any z ∈ lΦ , we may write if δ (Φ, z ) ≤ 1

∇ || x ||M (∇λi xi ) zi ≤  ∑ ∇ i =1 δ (Φ, z ) || x ||M ∞

Let now x∈ l M ( ∇, Λ ) with x

M

if δ (Φ, z ) > 1

.

(15)

≤1. Also x(n) = (x1,… xn, 0,0, …..) ∈ l M ( ∇, Λ ) for n ≥1.

We observe that

x

≥ M

∑ (∇λ x ) y

(n) i

i i

i =1

=

∇ M

( n) i i

i =1

~ for every y ∈ lΦ with δ (Φ, y ) ≤1 and thus x( n)

∑ (∇λ x

≤ x

∇ M

Since n

(

∑ Φ p ( ∇λi xi i =1

We find that

{ p ( ∇λ x )} ∈ ~l (n) i i

))

) yi ,

≤ 1.

( (

= ∑ Φ p ∇λi xi( n ) i =1

n ≥1

)) .

for each n ≥1. Let l ≥1 be an integer such that

Φ l

∑ Φ ( p ( ∇λ x ) ) >1. ∑ Φ ( p ( ∇λ x ∞

Then

i =1

i

i i

i =1

)) >1. Using (2), we have Φ ( p ( ∇ λ x ) ) < M ( ∇λ x ) + Φ ( p ( ∇ λ x ) ) = ∇ λ x p ( ∇λ x ) (l )

i

(l )

i

(l )

i

i

l i i

for all i, l ≥1. So by (15), we get

∑ Φ ( p ( ∇λ x ∞

i =1

i

(l ) i

)) < || x

(l )

(l )

i

i

i

l i i

( {(

||∇M δ Φ, p ∇λi xil

)}) .

128

SDU Journal of Science (E-Journal), 2010, 5 (1): 119-136 ___________________________________________________________________

This implies that || x (l ) ||∇M >1, a contradiction. This contradiction implies that l

∑ Φ ( p ( ∇λ x ) ) ≤1

{

for all l ≥1. Hence p ( ∇λi xi

)} ∈ l%

i i

i =1

Φ

( {

)}) ≤1.

and δ Φ, p ( ∇λi xi

x∈ l% M ( ∇, Λ )

and

~ Proof. Let y= p ( ∇λi xi ) / sgn(∇λi xi ) . Then from Proposition 2.8, y ∈ lΦ

and

Proposition

2.9.

δ ∇Λ ( M , x) ≤ x

∇ M

x∈ l M ( ∇, Λ )

Let

with

x

∇ M

≤1.

Then

.

{

}

δ (Φ, y ) ≤1. By (2), we get ∞

∑ M ( ∇λ x ) ≤ ∑ M ( ∇ λ x ) + ∑ Φ ( p ( ∇ λ x ) ) i i

i =1

i i

i =1 ∞

= ∑ ∇λi xi p ( ∇λi xi i =1 ∞

=

∑ (∇λ x ) y i i

i =1

This implies that δ ∇Λ ( M , x) ≤ x

∇ M

i

i i

i =1

≤ x

) ∇ M

.

. ∞

 ∇λk xk  ≤1. ∇  M  

∑ M  || x ||

Proposition 2.10. For x∈ l M ( ∇, Λ ) , we have

k =1

Proof. Proof is immediate from Proposition 2.9. Theorem 2.11. For x ∈ l M ( ∇, Λ ) , || x ||∇( M ) ≤ || x ||∇M ≤2 || x ||∇( M ) . Proof. We have ∞  ∇λk xk    inf 0 : > M ρ =   ≤ 1 . ∑ (M ) k =1   ρ   Then using Proposition 2.10, we get || x ||∇( M ) ≤ || x ||∇M .

x

Let us suppose that x∈ l M ( ∇, Λ ) with x

∇ (M )

≤1. Then x∈ l% M ( ∇, Λ ) and δ ∇Λ ( M , x) ≤1.

Indeed, 1 || x ||∇( M )

∑ M ( ∇λi xi ) ≤ i =1

 ∇λi xi   ≤ 1 , ∇ M ( )  

∑ M  || x || i =1

by Proposition 2.7.

129

H. Dutta

 x % ( ∇, Λ ) with δ  M , x ∈ l M  || x ||∇( M ) || x ||∇( M )  arbitrary z∈ l% M ( ∇, Λ ) , Thus

  ≤1. We further observe that for an 

 ∞  || z ||∇M = sup  ∑ (∇λi zi ) yi : δ ( Φ, y ) ≤ 1 ≤ 1+ δ ∇Λ ( M , z )  i =1  x using (1). Hence taking z = , we have || x ||∇( M )

x || x ||∇( M )

∞  x  ≤ 1+ ∑ M  ≤2  || x ||∇( M )  i =1   M

by Proposition 2.7. Thus || x ||∇M ≤ 2 || x ||∇( M ) . This completes the proof. Proposition 2.12. For any Orlicz function M, l M ( ∇, Λ ) = l /M ( ∇, Λ ) , where ∞   ∇λk xk  l /M ( ∇, Λ ) =  x ∈ w : ∑ M   0  . 

Proof. Proof follows from Proposition 2.10. In view of above Proposition we give the following definition. Definition 2.13. For any Orlicz function M, ∞    ∇λk xk  hM ( ∇, Λ ) =  x ∈ w : ∑ M   < ∞ , for each ρ > 0  . k =1    ρ 

Clearly hM ( ∇, Λ ) is a subspace of l M ( ∇, Λ ) . Henceforth we shall write ||.|| instead of . ( M ) provided it does not lead to any confusion. The topology of hM ( ∇, Λ ) is the one it ∇

inherits from ||.||. Proposition 2.14. Let M be an Orlicz function. Then ( hM ( ∇, Λ ) ,||.||) is an AK-BK space. Proof. First we show that hM ( ∇, Λ ) is an AK space. Let x∈ hM ( ∇, Λ ) . Then for each ε, 0< ε < 1, we can find an n0 such that  ∇λi xi   ≤1.  ε 

∑M 

i ≥ n0

Hence for n ≥ n0,    ∇λi xi    ∇λi xi   ||x-x(n)|| = inf  ρ > 0 : ∑ M   ≤ 1 ≤ inf  ρ > 0 : ∑ M   ≤ 1 < ε . i ≥ n +1 i≥n  ρ    ρ    

130

SDU Journal of Science (E-Journal), 2010, 5 (1): 119-136 ___________________________________________________________________

Thus we can conclude that hM ( ∇, Λ ) is an AK space. Next to show hM ( ∇, Λ ) is an BK space it is enough to show hM ( ∇, Λ ) is a closed subspace of hM ( ∇, Λ ) . For this let {xn} be a sequence in hM ( ∇, Λ ) such that ||xn-x|| →0, where x∈ hM ( ∇, Λ ) . To complete the proof we need to show that x∈ hM ( ∇, Λ ) , i.e.,  ∇λi xi  0. To ρ>0 there corresponds an l such that ||xl-x|| ≤ convexity of M,

(

 2 ∇λi xil − 2 ∇λi xil − ∇λi xi  ∇λi xi  M  =∑M  ∑ 2ρ ρ i ≥1   i ≥1   2 ∇λi xil 1 ≤ ∑M  2 i ≥1  ρ   2 ∇λi xil 1 ≤ ∑M  2 i ≥1  ρ 

2

. Then using

)   

 1  2 ∇λi ( xil − xi )  + ∑M   2 i ≥1  ρ    1  2 ∇λi ( xil − xi )  + ∑M   2 i ≥1  || x l − x ||  

     < ∞  

by proposition 2.7. Thus x ∈ hM ( ∇, Λ ) and consequently hM ( ∇, Λ ) is a BK space.

Proposition 2.15. Let M be an Orlicz function. If M satisfies the ∆2-condition at 0, then l M ( ∇, Λ ) is an AK space. Proof. In fact we shall show that if M satisfies the ∆2-condition at 0, then l M ( ∇, Λ ) = hM ( ∇, Λ ) and the result follows. Therefore it is enough to show that l M ( ∇, Λ ) ⊂ hM ( ∇, Λ ) . Let x ∈ l M ( ∇, Λ ) , then ρ > 0,

 ∇λi xi   0. If ρ ≤ l, then ∑ M   < ∞ . Let now l < ρ and put k= . l i ≥1  l  Since M satisfies ∆2-condition at 0, there exist R ≡ Rk>0 and r ≡ rk > 0 with M(kx)≤RM(x) for all x ∈ (0, r]. By (16) there exists a positive integer n1 such that  ∇λi xi  1  r  M  < rp   ρ  2 2

131

H. Dutta

for all i ≥ n1. We claim that ∇λ j x j

∇λi xi

ρ

≤ r for all i ≥ n1. Otherwise, we can find j > n1 with

> r, and thus

ρ

∇λ j x j  ∇λ j x j  1 r  ≥ ∫ ρ p(t ) dt > rp  M  ρ  r /2 2 2   Is a contradiction. Hence our claim is true. Then we can find that  ∇λi xi   ∇λi xi  M  ≤∑M  , ∑ i ≥ n1  l  i ≥ n1  ρ  and hence  ∇λi xi  M  0. This completes our proof.

Proposition 2.16. Let M1 and M2 be two Orlicz functions. If M1 and M2 are equivalent then l M1 ( ∇, Λ ) = l M 2 ( ∇, Λ ) and the identity map

(

I: l M1 ( ∇, Λ ) , . M ∇

1

) → (l

( ∇, Λ ) , . M ∇

M2

2

)

is a topological isomorphism. Proof. Let M1 and M2 are equivalent and so satisfy (4). Suppose x ∈ l M 2 ( ∇, Λ ) , then  ∇λi xi   < ∞ i =1  ρ  ∇λi xi for some ρ > 0. Hence for some l ≥1, ≤ x0 for all i ≥1. Therefore, lρ ∞

∑M

2

 α ∇λi xi  ∞  ∇λi xi  M  ≤ ∑ M2   < ∞. ∑ 1 i =1  l ρ  i =1  ρ  ∞

Thus l M 2 ( ∇, Λ ) ⊂ l M1 ( ∇, Λ ) . Similarly l M1 ( ∇, Λ ) ⊂ l M 2 ( ∇, Λ ) . Let us abbreviate here . M and . M by . 1 and . 2 , respectively. For x ∈ l M 2 ( ∇, Λ ) , ∇

1

2

 ∇λi xi  M ≤ 1. ∑ 2  x  i =1 2   x  x  One can find µ >1 with  0  µp2  0  ≥1, where p2 is the kernel associated with M2. 2 2 Hence  ∇λi xi   x0  x  M2  ≤   µp2  0    x  2 2 2   ∞

132

SDU Journal of Science (E-Journal), 2010, 5 (1): 119-136 ___________________________________________________________________

for all i ≥1. This implies that

∇λi xi

µ x

≤ x0 for all i ≥1. Therefore

2 ∞

∑M i =1

1

 α ∇λi xi   µ x2

  1 α  x  x  such that γβ  0  p1  0  ≥1. Thus αµ −1 x 1 ≤ x 2 ≤ βγ x 1 which establishes that I is a 2 2 topological isomorphism. Proposition 2.17. (i) l M ( Λ ) ⊂ l M ( ∇, Λ ) ,

(ii) l M ( Λ ) ⊂ l M ( ∆, Λ ) .

Proof. (i) Proof follows from the following inequality:  ∇λi xi  1 ∞  λi xi  1 ∞  λi −1 xi −1 ≤ M M    + ∑M  ∑ ∑ i =1  2 ρ  2 i =1  ρ  2 i =1  ρ ∞

 , 

(ii) Proof is similar to that of part (i). Proposition 2.18. Let M be an Orlicz function and p the corresponding kernel. If p(x) = 0 for all x in [0, x0] where x0 is some positive number, then l M ( ∇, Λ ) is

topologically isomorphic to l ∞ ( ∇, Λ ) and hM ( ∇, Λ ) is topologically isomorphic to c0 ( ∇, Λ ) .

Proof. Let p(x) = 0 for all x in [0, x0]. If y ∈ l ∞ ( ∇, Λ ) , then we can find a ρ > 0 such  ∇λi yi   < ∞, giving thus y ∈ l M ( ∇, Λ ) . On ρ i =1  ρ  ∞  ∇λi yi  the other hand let y ∈ l M ( ∇, Λ ) , then ∑ M   < ∞, for some ρ > 0 and so i =1  ρ  ∇λi yi 0. (It is easy to show that ||y||∞= sup ( ∇λi yi i

i

)

is a norm on

l ∞ ( ∇, Λ ) ). For every ε, 0< ε < α, we can determine yj with ∇λ j y j > α-ε and so ∞

 ∇λi yi x1   (α − ε ) x1  . ≥M α α    

∑M  i =1

133

H. Dutta

 ∇λi yi x1   ≥1, and so ||y||∞≤ x1||y||, for otherwise α i =1   ∞ ∞  ∇λi yi   ∇λi yi x0  M  >1 is a contradiction by Proposition 2.7. Again, ∑ M   =0 ∑ α i =1 i =1  || y ||    1 and it follows that || y ||≤ || y ||∞ . Thus the identity map x0

Since M is continuous, we find

∑M 

I: ( l M ( ∇, Λ ) , . ) → ( l ∞ ( ∇, Λ ) , . )

is a topological isomorphism. For the last part, let y ∈ hM ( ∇, Λ ) , then for any ε > 0, ∇λi yi ≤ εx1, for all sufficiently large i, where x1 is some positive number with p(x1) > 0. Hence y ∈ c0 ( ∇, Λ ) . Next let y ∈ c0 ( ∇, Λ ) . Then for any ρ>0,

∇λi yi

ρ

1 < x0 for all sufficiently large i. Thus 2

 ∇λi yi  M  0 and so y ∈ hM ( ∇, Λ ) . Hence hM ( ∇, Λ ) = c0 ( ∇, Λ ) and we  ρ  are done.

Corollary 2.19. Let M be an Orlicz function and p the corresponding kernel. If p(x) = 0 for all x in [0, x0] where x0 is some positive number, then l M ( ∇, Λ ) is topologically

isomorphic to l ∞ and hM ( ∇, Λ ) is topologically isomorphic to c0 . Proof. Let us define the mapping for Z = l ∞ , c0

T: Z ( ∇, Λ ) → Z

by Tx = ( ∇λk xk ) , for every x∈ Z ( ∇, Λ ) . Then clearly T is a linear homeomorphism. Hence the proof follows from Proposition 2.18. Lemma 2.20. Let M be an Orlicz function. Then x ∈ l M ( ∆, Λ ) implies ( k −1λk xk ) ∈ l ∞ . Proof. Let x ∈ l M ( ∆, Λ ) . Then, one can easily prove that ( ∆λk xk ) ∈ l ∞ which gives the

result ( k −1λk xk ) ∈ l ∞ .

Proposition 2.21. Let M be an Orlicz function and p be the corresponding kernel of M. If p(x) = 0 for all x in [0, x0], where x0 is some positive number, then (i) Köthe-Toeplitz dual of l M ( ∆, Λ ) is D1, where ∞   D1= (ak ) : ∑ k λk−1ak < ∞  , k =1  

134

SDU Journal of Science (E-Journal), 2010, 5 (1): 119-136 ___________________________________________________________________

(ii) Köthe-Toeplitz dual of D1 is D2, where

{

}

D2= (bk ) : sup k −1 λk bk < ∞ . k

Proof. (i) Let a ∈ D1 and x ∈ l M ( ∆, Λ ) . Then ∞

∑ a k xk = k =1

∑ k λk−1ak k −1 λk xk ≤ sup k −1 λk xk

∑k λ

k

k =1

−1 k k

k =1

α

a < ∞.

α

Hence a ∈  l M ( ∆, Λ )  . Thus, the inclusion D1 ⊂  l M ( ∆, Λ )  holds. α

Conversely suppose that a ∈  l M ( ∆, Λ )  . Then

∑a x k =1

k

k

< ∞ for every x ∈ l M ( ∆, Λ ) .

So we can take xk = λk−1k for all k ≥1, because then (xk) ∈ l ∞ ( ∆, Λ ) and hence

(xk) ∈ l M ( ∆, Λ ) as shown in Proposition 2.18. Now

k =1

k =1

α

∑ k λk−1ak = ∑ ak xk < ∞ and thus a∈ D1. Hence, the inclusion l M ( ∆, Λ ) ⊂ D1

holds. (ii) Proof follows by similar arguments used in the prove of case (i). Proposition 2.22. Let M be an Orlicz function and p be the corresponding kernel of M. If p(x) = 0 for all x in [0, x0], where x0 is some positive number, then Köthe-Toeplitz dual of hM ( ∆, Λ ) is D1, where D1 is defined as in Proposition 2.21. Proof. Let a ∈ D1 and x ∈ hM ( ∆, Λ ) . Then ∞

∑ a k xk = k =1

∑ k λk−1ak k −1 λk xk ≤ sup k −1 λk xk k

k =1

∑k λ

−1 k k

k =1

α

a < ∞.

α

Hence a ∈  hM ( ∆, Λ )  , that is the inclusion D1 ⊂  hM ( ∆, Λ )  holds. α

Conversely suppose that a ∈  hM ( ∆, Λ )  and a ∉ D1 . Then there exists a strictly increasing sequence (ni) of positive integers such that n1 < n2 i.

Define (xk) by , 0 xk =  −1 k λk sgn ak / i ,

1 ≤ k ≤ n1 ni < k ≤ ni +1

Then (xk) ∈ c0 ( ∆, Λ ) and so by Proposition 2.18, (xk) ∈ hM ( ∆, Λ ) . Then we have

135

H. Dutta

∑ a k xk = k =1

n2

∑ ak xk +…+

k = n1 +1 n2

=

k = n1 +1

α

ni +1

∑a

k = ni +1

k λk−1ak +…+

k

x k +…

1 ni+1 k λk−1ak +…> 1+1+…= ∞. ∑ i k = ni +1 α

This contradicts to a Î  hM ( ∆, Λ )  . Hence a ∈ D1, i.e. the inclusion  hM ( ∆, Λ )  ⊂ D1 also holds. This completes the proof. References [1] Kizmaz H., 1981. On certain sequence spaces, Canadian Mathematical Bulletin, 24 (2): 169-176. [2] Kamthan P.K., Gupta M., 1981. Sequence Spaces and Series, Marcel Dekker Inc., New York, USA, p. 368. [3] Lindenstrauss J., Tzafriri L., 1971. On Orlicz sequence spaces, Israel Journal of Mathematics, 10: 379-390. [4] Gribanov Y., 1957. On the theory of ℓM-spaces(Russian), Ucenyja Zapiski Kazansk un-ta, 117: 62-65. [5] Krasnoselskii M.A., Rutitsky Y.B., 1961. Convex functions and Orlicz spaces, Groningen, Netherlands, p. 249. [6] Goes G., Goes S., 1970. Sequences of bounded variation and sequences of Fourier coefficients, Mathematische Zeitschrift, 118 (2): 93-102. [7] Köthe G., Toeplitz O., 1934. Linear Raume mit unendlichvielen koordinaten and Ringe unendlicher Matrizen, Journal Für Die Reine und Angewandte Mathematik, 1934 (171): 193-226. [8] Kamthan P.K., 1976. Bases in a certain class of Frechet spaces, Tamkang Journal of Mathematics, 7 (1): 41-49. [9] Başar F., Altay B., 2003. On the space of sequences of p-bounded variation and related matrix mappings, Ukrainian Mathematical Journal, 55 (1): 136-147. [10] Altay B., Başar F., 2007. The fine spectrum and the matrix domain of the difference operator ∆ on the sequence space l p , ( 0 < p < 1) , Communications in Mathematical Analysis, 2 (2): 1-11.

136