(1)JJ J I II Go back Full Screen Close Quit A GENERALIZATION OF ORLICZ SEQUENCE SPACES BY CES `ARO MEAN OF ORDER ONE H

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A GENERALIZATION OF ORLICZ SEQUENCE SPACES BY CES `ARO MEAN OF ORDER ONE

H. DUTTA and F. BAS¸AR

Abstract. In this paper, we introduce the Orlicz sequence spaces generated by Ces`aro mean of order one associated with a fixed multiplier sequence of non-zero scalars. Furthermore, we emphasize several algebraic and topological properties relevant to these spaces. Finally, we determine the K¨othe-Toeplitz dual of the spaces`⁰_M(C,Λ) andhM(C,Λ).

1. Preliminaries, Background and Notation

By ω, we denote the space of all complex valued sequences. Any vector subspace of ω which containsφ, the set of all finitely non–zero sequences is called a sequence space. We write `_∞, c andc₀ for the classical sequence spaces of all bounded, convergent and null sequences which are Banach spaces with the sup-normkxk_∞= sup_k∈N|x_k|, whereN={0,1,2, . . .}, the set of natural numbers. A sequence space X with a linear topology is called a K−space provided each of the maps pi: X → C defined by pi(x) = xi is continuous for all i ∈ N. AK-space X is called an F K-spaceprovidedXis a complete linear metric space. AnF K-space whose topology is normable is called aBK-space.

A functionM: [0,∞)→[0,∞) which is convex with M(u)≥0 foru≥0, and M(u)→ ∞as u→ ∞, is called as anOrlicz function. An Orlicz function M can always be represented in the

Received March 29, 2010; revised December 9, 2010.

2010Mathematics Subject Classification. Primary 46A45; Secondary 40C05, 40A05.

Key words and phrases. Orlicz function, Ces`aro mean of order one, sequence spaces, topological isomorphism, AK space, BK space.

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following integral form

M(u) = Z u

0

p(t)dt,

where p, the kernel of M, is right differentiable for t ≥ 0, p(0) = 0, p(t) > 0 for t > 0, p is non–decreasing andp(t)→ ∞as t→ ∞whenever ^M(u)_u ↑ ∞asu↑ ∞.

Consider the kernelpassociated with the Orlicz functionM and let q(s) = sup{t:p(t)≤s}.

Thenqpossesses the same properties as the functionp. Suppose now Φ(x) =

Z x 0

q(s)ds.

Then, Φ is an Orlicz function. The functionsM and Φ are calledmutually complementary Orlicz functions.

Now, we give the following well-known results.

LetM and Φ are mutually complementary Orlicz functions. Then, we have:

(i) For allu, y≥0,

uy≤M(u) + Φ(y), (Young’s inequality).

(1.1)

(ii) For allu≥0,

up(u) =M(u) + Φ(p(u)).

(1.2)

(iii) For allu≥0 and 0< λ <1,

M(λu)< λM(u).

(1.3)

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An Orlicz functionM is said to satisfy the ∆2-condition for smalluor at 0 if for eachk∈N, there exist Rk > 0 anduk > 0 such that M(ku) ≤ RkM(u) for all u∈ (0, uk]. Moreover, an Orlicz functionM is said to satisfy the ∆2-condition if and only if

lim sup

u→0⁺

M(2u) M(u) <∞.

Two Orlicz functionsM1 andM2 are said to be equivalent if there are positive constantsα, βand bsuch that

M₁(αu)≤M₂(u)≤M₁(βu) for allu∈[0, b].

(1.4)

Orlicz used the Orlicz function to introduce the sequence space`_M (see Musielak [10]; Linden- strauss and Tzafriri [9]), as follows

`M = (

x= (xk)∈ω:X

k

M |xk|

ρ

<∞ for some ρ >0 )

.

For simplicity in notation, here and in what follows, the summation without limits runs from 0 to

∞. For relevant terminology and additional knowledge on the Orlicz sequence spaces and related topics, the reader may refer to [1, 3,5,6,7,8,11,9,10] and [12].

Throughout the present article, we assume that Λ = (λk) is the sequence of non–zero complex numbers. Then for a sequence spaceE, the multiplier sequence space E(Λ) associated with the multiplier sequence Λ is defined by

E(Λ) ={x= (xk)∈ω: Λx= (λkxk)∈E}.

The scope for the studies on sequence spaces was extended by using the notion of associated multiplier sequences. G. Goes and S. Goes defined the differentiated sequence space dE and integrated sequence spaceR

E for a given sequence spaceE, using the multiplier sequences (k⁻¹)

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and (k) in [4], 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.

LetC= (cnk) be the Ces`aro matrix of order one defined by cnk:=





 1

n+ 1, 0≤k≤n, 0, k > n, for allk, n∈N.

Definition 1.1. LetMbe any Orlicz function andδ(M, x) :=P

kM(|xk|), wherex= (x_k)∈ω.

Then, we define the sets`e<sub>M</sub>(C,Λ) and`e_M by

e`_M(C,Λ) :=







x= (x_k)∈ω:bδ_C(M, x) =X

k

M





Pk j=0λjxj

k+ 1



<∞





 and

`eM :={x= (xk)∈ω:δ(M, x)<∞}.

Definition 1.2. LetM and Φ be mutually complementary functions. Then, we define the set

`M(C,Λ) by

`_M(C,Λ) :=

(

x= (x_k)∈ω:X

k

Pk j=0λjxj

k+ 1

!

y_k converges for ally= (y_k)∈e`_Φ )

which is called as Orlicz sequence space associated with the multiplier sequence Λ = (λk) and generated by Ces`aro matrix of order one.

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Definition 1.3. Theα-dual or K¨othe-Toeplitz dualX^αof a sequence spaceX is defined by

X^α:=

(

a= (a_k)∈ω:X

k

|akx_k|<∞for allx= (x_k)∈X )

.

It is known that ifX ⊂Y, then Y^α ⊂X^α. It is clear that X ⊂X^αα. If X =X^αα, thenX is called as anαspace. In particular, anαspace is called a K¨othe space or a perfect sequence space.

The main purpose of this paper is to introduce the sequence spaces`M(C,Λ),e`M(C,Λ),`<sup>0</sup><sub>M</sub>(C,Λ) andh<sub>M</sub>(C,Λ), and investigate their certain algebraic and topological properties. Furthermore, it is proved that the spaces`⁰_M(C,Λ) andh_M(C,Λ) are topologically isomorphic to the spaces`_∞(C,Λ) and c₀(C,Λ) when M(u) = 0 on some interval, respectively. Finally, the α-dual of the spaces

`<sup>0</sup><sub>M</sub>(C,Λ) and h<sub>M</sub>(C,Λ) are determined, and therefore the non-perfectness of the space `⁰_M(C,Λ) is showed whenM(u) = 0 on some interval, and some open problems are noted.

2. Main Results

In this section, we emphasize the sequence spaces `M(C,Λ), e`M(C,Λ), `⁰_M(C,Λ) and hM(C,Λ), and give their some algebraic and topological properties.

Proposition 2.1. For any Orlicz functionM, the inclusion `eM(C,Λ)⊂`M(C,Λ) holds.

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Proof. Let x = (xk) ∈ `eM(C,Λ). Then, since P

kM

|^Pkj=0λjxj|

k+1

< ∞ we have from (1.1) that

X

k

Pk j=0λjxj

k+ 1

! y_k

≤X

k

Pk j=0λjxj

k+ 1

! y_k

≤X

k

M

Pk j=0λjxj

k+ 1

!

+X

k

Φ(|yk|)<∞

for everyy= (y_k)∈`e<sub>Φ</sub>. Thus,x= (x<sub>k</sub>)∈`_M(C,Λ).

Proposition 2.2. For eachx= (xk)∈`M(C,Λ),

sup (

X

k

Pk j=0λjxj

k+ 1

! yk

:δ(Φ, y)≤1 )

<∞.

(2.1)

Proof. Suppose that (2.1) does not hold. Then for eachn∈N, there existsyⁿwithδ(Φ, yⁿ)≤1 such that

X

k

Pk j=0λjxj

k+ 1

! yⁿ_k

>2ⁿ⁺¹.

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Without loss of generality, we can assume thatPk j=0

λjxj

k+1, yⁿ≥0. Now, we can define a sequence z= (zk) byzk=P

n y_kⁿ

2ⁿ⁺¹ for allk∈N. By the convexity of Φ, we have Φ

l

X

n=0

1 2ⁿ⁺¹y_kⁿ

!

≤1 2

Φ(y⁰_k) + Φ

y¹_k+y²_k

2 +· · ·+ y_k^l 2^l−1

· · · ≤

l

X

n=0

1

2ⁿ⁺¹Φ(yⁿ_k)

for any positive integerl. Hence, using the continuity of Φ, we have δ(Φ, z) =X

k

Φ(zk)≤X

k

X

n

1

2ⁿ⁺¹Φ(y_kⁿ)≤X

n

1 2ⁿ⁺¹ = 1.

But for everyl∈N, it holds X

k

Pk j=0λjxj

k+ 1

!

z_k ≥X

k

Pk j=0λjxj

k+ 1

! _l X

n=0

1 2ⁿ⁺¹y_kⁿ

=

l

X

n=0

X

k

Pk j=0λjxj

k+ 1

! y_kⁿ 2ⁿ⁺¹ ≥l.

HenceP

k

_Pk

j=0λjxj

k+1

zkdiverges and this implies thatx /∈`M(C,Λ), a contradiction. This leads

us to the required result.

The preceding result encourages us to introduce the following normk · k^C_M on`_M(C,Λ).

Proposition 2.3. The following statements hold:

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(i) `M(C,Λ)is a normed linear space under the norm k · k^C_M defined by kxk^C_M = sup

(

X

k

Pk j=0λjxj

k+ 1

! yk

:δ(Φ, y)≤1 )

. (2.2)

(ii) `_M(C,Λ) is a Banach space under the norm defined by (2.2).

(iii) `M(C,Λ) is aBK space under the norm defined by (2.2).

Proof. (i) It is easy to verify that`M(C,Λ) is a linear space with respect to the co-ordinatewise addition and scalar multiplication of sequences. Now we show thatk · k^C_M is a norm on the space

`M(C,Λ).

Ifx= 0, then obviouslykxk^C_M = 0. Conversely, assumekxk^C_M = 0. Then using the definition of the norm given by (2.2), we have

sup (

X

k

Pk j=0λjxj

k+ 1

! yk

:δ(Φ, y)≤1 )

= 0.

This implies that

P

k

_Pk

j=0λ_jx_j k+1

yk

= 0 for ally such thatδ(Φ, y)≤1. Now consideringy= e^k if Φ(1) ≤ 1 otherwise considering y = e^k/Φ(1) so thatλkxk = 0 for all k ∈ N, where e^k is a sequence whose only non-zero term is 1 ink^th place for eachk∈N. Hence we havexk = 0 for all k∈N, since (λk) is a sequence of non-zero scalars. Thus,x= 0.

It is easy to show that kαxk^C_M =|α|kxk^C_M and kx+yk^C_M ≤ kxk^C_M +kyk^C_M for allα∈ C and x, y∈`M(C,Λ).

(ii) Let (x^s) be any Cauchy sequence in the space`M(C,Λ). Then for anyε >0, there exists a positive integern0 such thatkx^s−x^tk^C_M < ε for alls, t≥n0. Using the definition of norm given

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by (2.2), we get

sup (

X

k

" Pk

j=0λj x^s_j−x^t_j k+ 1

# y_k

:δ(Φ, y)≤1 )

< ε

for alls, t≥n₀. This implies that

X

k

" Pk

j=0λj x^s_j−x^t_j k+ 1

# yk

< ε

for ally with δ(Φ, y)≤ 1 and for all s, t ≥ n0. Now considering y = e^k if Φ(1) ≤1, otherwise consideringy = e^k/Φ(1) we have (λkx^s_k) is a Cauchy sequence in Cfor all k∈N. Hence, it is a convergent sequence inCfor allk∈N.

Let lims→∞λkx^s_k=xk for eachk∈N. Using the continuity of the modulus, we can derive for alls≥n₀ ast→ ∞, that

sup (

X

k

" Pk

j=0λj x^s_j−xj

k+ 1

# yk

:δ(Φ, y)≤1 )

< ε.

It follows that (x^s−x)∈ `<sub>M</sub>(C,Λ). Sincex<sup>s</sup> is in the space `_M(C,Λ) and `<sub>M</sub>(C,Λ) is a linear space, we havex= (xk)∈`M(C,Λ).

(iii) From the above proof, one can easily conclude thatkx^sk^C_M →0 implies that x^s_k → 0 for eachs∈Nwhich leads us to the desired result.

Therefore, the proof of the theorem is completed.

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Proposition 2.4. `M(C,Λ)is a normed linear space under the normk · k^C_(M) defined by

kxk^C_(M)= inf







ρ >0 :X

k

M





Pk

j=0λ_jx_j ρ(k+ 1)



≤1





 . (2.3)

Proof. Clearlykxk^C_(M)= 0 ifx= 0. Now, suppose thatkxk^C_(M)= 0. Then, we have

inf







ρ >0 :X

k

M





Pk

j=0λ_jx_j ρ(k+ 1)



≤1







= 0.

This yields the fact for a givenε >0 that there exists someρε∈(0, ε) such that

sup

k∈N

M





Pk

j=0λ_jx_j ρε(k+ 1)



≤1 which implies that

M





Pk

j=0λ_jx_j ρε(k+ 1)



≤1 for allk∈N. Thus,

M





Pk j=0λjxj

ε(k+ 1)



≤M





Pk j=0λjxj

ρε(k+ 1)



≤1

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for allk ∈ N. Suppose |^Pnij=0λjxj|

n_i+1 6= 0 for some ni ∈N. Then, |^Pnij=0λjxj|

ε(n_i+1) → ∞ as ε→ 0. It follows thatM

|^Pnij=0λjxj|

ε(k+1)

→ ∞asε→0 for someni∈N, which is a contradiction. Therefore,

|^Pkj=0λ_jx_j|

k+1 = 0 for allk∈N. It follows thatλ_kx_k = 0 for allk∈N. Hencex= 0, since (λ_k) is a sequence of non-zero scalars.

Letx= (xk) and y= (yk) be any two elements of`M(C,Λ). Then, there existρ1, ρ2>0 such that

X

k

M





Pk j=0λjxj

ρ₁(k+ 1)



≤1 and X

k

M





Pk j=0λjyj

ρ₂(k+ 1)



≤1.

Letρ=ρ1+ρ2. Then by the convexity ofM, we have

X

k

M





Pk

j=0λj(xj+yj) ρ(k+ 1)



≤ ρ1

ρ1+ρ2

X

k

M





Pk j=0λjxj

ρ1(k+ 1)





+ ρ₂ ρ1+ρ2

X

k

M





Pk j=0λjyj

ρ2(k+ 1)



≤1.

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Hence, we have

kx+yk^C_(M)= inf







ρ >0 :X

k

M





Pk

j=0λj(xj+yj) ρ



≤1







≤ inf







ρ₁>0 :X

k

M





Pk j=0λjxj

ρ1



≤1







+ inf







ρ₂>0 :X

k

M





Pk j=0λjyj

ρ2



≤1





 which gives thatkx+yk^C_(M)≤ kxk^C_(M)+kyk^C_(M).

Finally, letαbe any scalar and define rbyr=ρ/|α|. Then,

kαxk^C_(M)= inf







ρ >0 :X

k

M





Pk

j=0αλjxj

ρ(k+ 1)



≤1







= inf







r|α|>0 :X

k

M





Pk j=0λjxj

r(k+ 1)



≤1







=|α|kxk^C_(M).

This completes the proof.

Proposition2.4inspires us to define the following sequence space.

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Definition 2.5. For any Orlicz functionM, we define

`⁰_M(C,Λ) :=







x= (xk)∈ω:X

k

M





Pk j=0λjxj

ρ(k+ 1)



<∞ for some ρ >0





 .

Now, we can give the corresponding proposition on the space`⁰_M(C,Λ) to the Proposition2.3.

Proposition 2.6. The following statements hold:

(i) `⁰_M(C,Λ)is a normed linear space under the norm k · k^C_(M)defined by (2.3).

(ii) `⁰_M(C,Λ)is a Banach space under the norm defined by (2.3).

(iii) `⁰_M(C,Λ)is a BK space under the norm defined by (2.3).

Proof. (i) Since the proof is similar to the proof of Proposition 2.4, we omit the detail.

(ii) Let (x^s) be any Cauchy sequence in the space `⁰_M(C,Λ). Let δ > 0 be fixed and r > 0 be given such that 0 < ε < 1 and rδ ≥ 1. Then, there exists a positive integer n0 such that kx^s−x^tk^C_(M)< ε/rδ for alls, t≥n0. Using the definition of the norm given by (2.3), we get

inf







ρ >0 :X

k

M





Pk

j=0λ_j x^s_j−x^t_j ρ(k+ 1)



≤1







< ε rδ

for alls, t≥n0. This implies that

X

k

M





Pk

j=0λj x^s_j−x^t_j kx^s−x^tk^C_(M)(k+ 1)



≤1

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for alls, t≥n0. It follows that M





Pk

j=0λ_j x^s_j−x^t_j kx^s−x^tk^C_(M)(k+ 1)



≤1 for alls, t≥n₀ and for allk∈N. Forr >0 withM(rδ/2)≥1, we have

M





Pk

j=0λ_j x^s_j−x^t_j kx^s−x^tk^C_(M)(k+ 1)



≤M rδ

2

for alls, t≥n₀ and for allk∈N. SinceM is non-decreasing, we have

Pk

j=0λj x^s_j−x^t_j

k+ 1 ≤rδ

2 · ε rδ = ε

2

for alls, t≥n0 and for all k∈N. Hence, (λkx^s_k) is a Cauchy sequence inC for allk ∈Nwhich implies that it is a convergent sequence inCfor allk∈N.

Let lims→∞λkx^s_k=xk for eachk∈N. Using the continuity of an Orlicz function and modulus, we can have

inf







ρ >0 :X

k

M





Pk

j=0λj(x^s_j−xj) ρ(k+ 1)



≤1







< ε

for alls≥n₀, asj → ∞. It follows that (x^s−x)∈`<sup>0</sup><sub>M</sub>(C,Λ). Since x<sup>s</sup> is in the space `⁰_M(C,Λ) and`<sup>0</sup><sub>M</sub>(C,Λ) is a linear space, we havex= (x<sub>k</sub>)∈`⁰_M(C,Λ).

(iii) From the above proof, one can easily conclude thatkx^sk^C_M →0 ass→ ∞, which implies thatx^s_k→0 ask→ ∞for eachs∈N. This leads us to the desired result.

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Proposition 2.7. The inequalityP

kM

|^Pkj=0λjxj|

kxk^C_(M)(k+1)

≤1 holds for allx= (xk)∈`⁰_M(C,Λ).

Proof. This is immediate from the definition of the normk · k^C_(M)defined by (2.3).

Definition 2.8. For any Orlicz functionM, we define

hM(C,Λ) :=







x= (xk)∈ω:X

k

M





Pk j=0λjxj

ρ(k+ 1)



<∞ for eachρ >0





 .

Clearlyh_M(C,Λ) is a subspace of`⁰_M(C,Λ).

Here and after we shall writek · k instead ofk · k^C_(M)provided it does not lead to any confusion.

The topology ofhM(C,Λ) is induced byk · k.

Proposition 2.9. LetM be an Orlicz function. Then, (h_M(C,Λ),k · k)is an AK-BK space.

Proof. First we show that hM(C,Λ) is anAK space. Letx= (xk)∈hM(C,Λ). Then for each ε∈(0,1), we can findn0 such that

X

i≥n0

M





Pk j=0λjxj

ε(k+ 1)



≤1.

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Define then^th sectionx⁽ⁿ⁾ of a sequence x = (xk) by x⁽ⁿ⁾ =Pn

k=0xke^k. Hence for n ≥n0, it holds

x−x⁽ⁿ⁾ = inf







ρ >0 : X

k≥n0

M





Pk j=0λjxj

ρ(k+ 1)



≤1







≤inf







ρ >0 :X

k≥n

M





Pk j=0λjxj

ρ(k+ 1)



≤1







< ε.

Thus, we can conclude thath_M(C,Λ) is anAK space.

Next to show thathM(C,Λ) is aBK space, it is enough to showhM(C,Λ) is a closed subspace of`<sup>0</sup><sub>M</sub>(C,Λ). For this, let (x<sup>n</sup>) be a sequence inhM(C,Λ) such thatkx<sup>n</sup>−xk →0 asn→ ∞where x= (xk)∈`⁰_M(C,Λ). To complete the proof we need to show thatx= (xk)∈hM(C,Λ), i.e.,

X

k

M





Pk j=0λjxj

ρ(k+ 1)



<∞ for all ρ >0.

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There isl corresponding toρ >0 such that kx^l−xk ≤ρ/2. Then, using the convexity of M, we have by Proposition2.7that

X

k

M





Pk j=0λjxj

ρ(k+ 1)





=X

k

M



 2

Pk j=0λjx^l_j

−2

Pk j=0λjx^l_j

−

Pk j=0λjxj

2ρ(k+ 1)





≤1 2

X

k

M



 2

Pk j=0λjx^l_j

ρ(k+ 1)



+1 2

X

k

M



 2

Pk

j=0λj x^l_j−xj

ρ(k+ 1)





≤1 2

X

k

M



 2

Pk

j=0λ_jx^l_j ρ(k+ 1)



+1 2

X

k

M



 2

Pk

j=0λ_j x^l_j−x_j kx^l−xk(k+ 1)





<∞.

Hence,x= (x_k)∈h_M(C,Λ) and consequentlyh_M(C,Λ) is aBK space.

Proposition 2.10. Let M be an Orlicz function. If Msatisfies the ∆₂-condition at 0, then

`⁰_M(C,Λ) is anAK space.

Proof. We shall show that `<sup>0</sup><sub>M</sub>(C,Λ) =hM(C,Λ) ifM satisfies the ∆2-condition at 0. To do this it is enough to prove that`⁰_M(C,Λ) ⊂hM(C,Λ). Let x= (xk) ∈`⁰_M(C,Λ). Then for some

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ρ >0,

X

k

M





Pk

j=0λ_jx_j ρ(k+ 1)



<∞.

This implies that

lim

k→∞M





Pk j=0λjxj

ρ(k+ 1)



= 0.

(2.4)

Choose an arbitrary l >0. If ρ ≤l, then P

kM

|^Pk_j=0^λjx_j|

l(k+1)

< ∞. Now, let l < ρ and put k = ρ/l. Since M satisfies ∆₂-condition at 0, there exist R ≡ R_k > 0 and r ≡ r_k > 0 with M(kx)≤RM(u) for allx∈(0, r]. By (2.4), there exists a positive integern₁ such that

M





Pk j=0λjxj

ρ(k+ 1)



< r 2pr

2

for all k≥n1.

We claim that |^Pkj=0λjxj|

ρ(k+1) ≤rfor allk≥n₁. Otherwise, we can find d > n₁ with |^Pdj=0λjxj|

ρ(d+1) > r and thus

M





Pd j=0λjxj

ρ(d+ 1)



≥

Z |^Pdj=0λ_jx_j|^/ρ(d+1)

r/2

p(t)dt > r 2pr

2

,

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a contradiction. Hence our claim is true. Then, we can find that X

k≥n1

M





Pk j=0λjxj

l(k+ 1)



≤R X

k≥n1

M





Pk j=0λjxj

ρ(k+ 1)



.

Hence,

X

k

M





Pk j=0λjxj

l(k+ 1)



<∞ for all l >0.

This completes the proof.

Proposition 2.11. Let M₁ and M₂ be two Orlicz functions. If M₁ and M₂ are equivalent, then`⁰_M

1(C,Λ) =`⁰_M

2(C,Λ) and the identity map I: `⁰_M

1(C,Λ),k · k^C_M

1

→ `⁰_M

2(C,Λ),k · k^C_M

2

is a topological isomorphism.

Proof. Let α, β andb be constants from (1.3). Since M₁ and M₂ are equivalent, it is obvious that (1.3) holds. Let us take anyx= (x_k)∈`⁰_M

2(C,Λ). Then, X

k

M2





Pk j=0λjxj

ρ(k+ 1)



<∞ for some ρ >0.

Hence, for somel≥1, |^Pkj=0λjxj|

lρ(k+1) ≤bfor allk∈N. Therefore, X

k

M1



 α

Pk j=0λjxj

lρ(k+ 1)



≤X

k

M2





Pk j=0λjxj

ρ(k+ 1)



<∞

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which shows that the inclusion

`⁰_M

2(C,Λ)⊂`⁰_M

1(C,Λ) (2.5)

holds. One can easily see in the same way that the inclusion

`⁰_M

1(C,Λ)⊂`⁰_M

2(C,Λ) (2.6)

also holds.

By combining the inclusions (2.5) and (2.6), we conclude that`⁰_M

1(C,Λ) =`⁰_M

2(C,Λ).

For simplicity in notation, let us writek · k₁andk · k₂instead ofk · k^C_M

1 andk · k^C_M

2, respectively.

Forx= (x_k)∈`⁰_M

2(C,Λ), we get X

k

M2





Pk

j=0λ_jx_j kxk2(k+ 1)



≤1.

One can findµ >1 with (b/2)µp2(b/2)≥1, wherep2 is the kernel associated withM2. Hence, M₂





Pk j=0λjxj

kxk2(k+ 1)



≤ b 2µp₂

b 2

for all k∈N. This implies that

Pk j=0λjxj

µkxk2(k+ 1) ≤b for all k∈N. Therefore,

X

k

M1



 α

Pk j=0λjxj

µkxk2(k+ 1)



<1.

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Hence,kxk1≤(µ/α)kxk2. Similarly, we can show thatkxk2≤βγkxk1 by choosingγwithγβ >1 such that γβ(b/2)p1(b/2) ≥ 1. Thus, αµ⁻¹kxk1 ≤ kxk2 ≤ βγkxk1 which establish that I is a

topological isomorphism.

Proposition 2.12. LetM be an Orlicz function andpbe the corresponding kernel. Ifp(x) = 0 for all x in [0, b], where b is some positive number, then the spaces `<sup>0</sup><sub>M</sub>(C,Λ) and hM(C,Λ) are topologically isomorphic to the spaces `∞(C,Λ) and c0(C,Λ), respectively; where `∞(C,Λ) and c0(C,Λ) are defined by

`_∞(C,Λ) =







x= (xk)∈ω: sup

k∈N k

X

j=0

|λjxj| k+ 1 <∞





 and

c0(C,Λ) =







x= (xk)∈ω: lim

k→∞

k

X

j=0

|λjxj| k+ 1 = 0





 .

It is easy to see that the spaces `_∞(C,Λ) and c₀(C,Λ) are the Banach spaces under the norm kxk^C_∞= sup_k∈N|^Pkj=0λjxj|

k+1 .

Proof. Let p(x) = 0 for all x in [0, b]. If y ∈ `_∞(C,Λ), then we can find ρ > 0 such that

|^Pkj=0λ_jy_j|

ρ(k+1) ≤bfork∈N. Hence, P

kM

|^Pkj=0λ_jy_j|

ρ(k+1)

<∞. That is to say thaty∈`⁰_M(C,Λ).

On the other hand, lety∈`⁰_M(C,Λ). Then for some ρ >0, we have X

k

M





Pk j=0λjyj

ρ(k+ 1)



<∞.

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Therefore, |^Pkj=0λjyj|

k+1 ≤ K < ∞ for a constant K > 0 and for all k ∈ N which yields that y ∈`<sub>∞</sub>(C,Λ). Hence,y ∈`_∞(C,Λ) if and only if y ∈`<sup>0</sup><sub>M</sub>(C,Λ). We can easily find b such that M(u0)≥1. Lety ∈`∞(C,Λ) and α=kyk∞= sup_k∈N

Pk j=0λ_jy_j

k+1

>0. For everyε∈(0, α), we can determinedwith

Pd j=0λjyj

d+1

> α−εand so

X

k

M





Pk j=0λjyj

b α(k+ 1)



≥M

α−ε α b

.

SinceM is continuous,P

kM

|^Pk_j=0^λjy_j|^b

α(k+1)

≥1, and sokyk∞≤bkyk, otherwise P

kM

|^Pkj=0λjyj|

kyk(k+1)

>1 which contradicts Proposition2.7. Again,

X

k

M





Pk j=0λjyj

b α(k+ 1)



= 0

which gives that kyk ≤ kyk_∞/b. That is to say that the identity map I: (`<sup>0</sup><sub>M</sub>(C,Λ),k · k) → (`_∞(C,Λ),k · k) is a topological isomorphism.

For the last part, lety ∈ hM(C,Λ). Then for any ε >0, |^Pkj=0λjyj|

k+1 ≤ εb for all sufficiently large k, where b is a positive number such that p(b) > 0. Hence, y ∈ c0(C,Λ). Conversely, let y ∈ c0(C,Λ). Then, for any ρ > 0, |^Pkj=0λjyj|

ρ(k+1) < b/2 for all sufficiently large k. Thus,

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P

kM

|^Pkj=0λjyj|

ρ(k+1)

<∞for all ρ >0 and so y ∈ hM(C,Λ). Hence, hM(C,Λ) = c0(C,Λ) and

this step completes the proof.

Prior to giving our final two consequences concerning the α-dual of the spaces `⁰_M(C,Λ) and hM(C,Λ), we present the following easy lemma without proof.

Lemma 2.13. For any Orlicz function M,Λx= (λkxk)∈`<sub>∞</sub> wheneverx= (xk)∈`⁰_M(C,Λ).

Proposition 2.14. LetM be an Orlicz function andpbe the corresponding kernel ofM. Define the setsD1 andD2 by

D1:=

(

a= (ak)∈ω:X

k

λ⁻¹_k ak

<∞

)

and

D2:=

b= (bk)∈ω: sup

k∈N

|λkbk|<∞

.

Ifp(x) = 0for allxin[0, d], wheredis some positive number, then the following statements hold:

(i) K¨othe-Toeplitz dual of`⁰_M(C,Λ)is the setD1. (ii) K¨othe-Toeplitz dual ofD1 is the setD2.

Proof. Since the proof of Part (ii) is similar to that of the proof of Part (i), to avoid the repetition of the similar statements we prove only Part (i).

Leta= (a_k)∈D₁andx= (x_k)∈`⁰_M(C,Λ). Then, since X

k

|akxk|=X

k

akλ⁻¹_k

|λkxk| ≤sup

k∈N

|λkxk| ·X

k

akλ⁻¹_k <∞,

applying Lemma2.13, we havea= (ak)∈ {`<sup>0</sup><sub>M</sub>(C,Λ)}<sup>α</sup>. Hence, the inclusion D1⊂ {`⁰_M(C,Λ)}^α

(2.7)

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holds.

Conversely, suppose thata= (ak)∈ {`<sup>0</sup><sub>M</sub>(C,Λ)}<sup>α</sup>. Then, (akxk)∈`1, the space of all absolutely convergent series, for every x = (xk) ∈ `<sup>0</sup><sub>M</sub>(C,Λ). So, we can take xk = λ<sup>−1</sup><sub>k</sub> for all k ∈ N because (xk)∈`⁰_M(C,Λ) by Proposition2.12whenever (xk)∈`_∞(C,Λ). Therefore,P

k

akλ⁻¹_k = P

k|akxk|<∞and we havea= (ak)∈D1. This leads us to the inclusion {`⁰_M(C,Λ)}^α⊂D1.

(2.8)

By combining the inclusion relations (2.7) and (2.8), we have {`<sup>0</sup><sub>M</sub>(C,Λ)}<sup>α</sup>=D1. Proposition 2.14(ii) shows that{`⁰_M(C,Λ)}^αα 6=`<sup>0</sup><sub>M</sub>(C,Λ) which leads us to the consequence that`⁰_M(C,Λ) is not perfect under the given conditions.

Proposition 2.15. Let M be an Orlicz function and p be the corresponding kernel ofM and the setD₁ be defined as in the Proposition2.14. Ifp(x) = 0for allxin[0, b], where bis a positive number, then the K¨othe-Toeplitz dual ofh_M(C,Λ) is the setD₁.

Proof. Leta= (a_k)∈D₁ andx= (x_k)∈h_M(C,Λ). Then, since X

k

|akx_k|=X

k

a_kλ⁻¹_k

|λkx_k| ≤sup

k∈N

|λkx_k| ·X

k

a_kλ⁻¹_k <∞,

we havea= (ak)∈ {hM(C,Λ)}^α. Hence, the inclusion D1⊂ {hM(C,Λ)}^α (2.9)

holds.

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Conversely, suppose thata= (ak)∈ {hM(C,Λ)}^α\D1. Then, there exists a strictly increasing sequence (ni) of positive integers ni such that

n_i+1

X

k=ni+1

|ak| |λk|⁻¹> i.

Definex= (x_k) by

x_k:=

λ⁻¹_k sgn a_k/i , (n_i< k≤n_i+1), 0 , (0≤k < n₀),

for all k ∈ N. Then, since x = (x_k) ∈ c₀(C,Λ), x = (x_k) ∈ h_M(C,Λ) by Proposition 2.12.

Therefore, we have X

k

|akxk|=

n₁

X

k=n0+1

|akxk|+· · ·+

ni+1

X

k=ni+1

|akxk|+· · ·

=

n₁

X

k=n0+1

akλ⁻¹_k

+· · ·+1 i

ni+1

X

k=ni+1

akλ⁻¹_k +· · ·

>1 +· · ·+ 1 +· · ·=∞,

which contradicts the hypothesis. Hence,a= (ak)∈D1. This leads us to the inclusion {hM(C,Λ)}^α⊂D1.

(2.10)

By combining the inclusion relations (2.9) and (2.10), we obtain the desired result {hM(C,Λ)}^α=D1.

This completes the proof.

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3. Conclusion

The difference Orlicz spaces`<sub>M</sub>(∆,Λ) and e`_M(∆,Λ) were recently been studied by Dutta [2]. Of course, the sequence spaces introduced in this paper can be redefined as a domain of a suitable matrix in the Orlicz sequence space`_M. Indeed, if we define the infinite matrix C(λ) ={c_nk(λ)}

via the multiplier sequence Λ = (λ_k) by cnk(λ) :=





 λ_k

n+ 1, (0≤k≤n), 0, (k > n),

for allk, n∈N, then the sequence spaces`<sup>0</sup><sub>M</sub>(C,Λ), c0(C,Λ) and`∞(C,Λ) represent the domain of the matrixC(λ) in the sequence spaces `M, c0 and `∞, respectively. Since cnn(λ)6= 0 for all n∈N, i.e.,C(λ) is a triangle, it is obvious that those spaces`<sup>0</sup><sub>M</sub>(C,Λ),c<sub>0</sub>(C,Λ) and`_∞(C,Λ) are linearly isomorphic to the spaces`<sub>M</sub>,c<sub>0</sub>and `_∞, respectively.

Although some algebraic and topological properties of these new spaces are investigated, the following further suggestions remain open:

(i) What is the relation between the normsk · k^C_M andk · k^C_(M)? Are they equivalent?

(ii) What is the relation between the spaces`M(C,Λ) and`⁰_M(C,Λ)? Do they coincide?

(iii) What are theβ- andγ-duals of the spaces`⁰_M(C,Λ) andhM(C,Λ)?

(iv) Under which conditions an infinite matrix transforms the sets`<sup>0</sup><sub>M</sub>(C,Λ) orhM(C,Λ) to the sequence spaces`∞ andc?

Acknowledgment. The authors have benefited a lot from the referee’s report. So, they would like to express their pleasure to the anonymous reviewer for his/her careful reading and making some constructive comments on the main results of the earlier version of the manuscript which improved the presentation of the paper.

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1. Chen S.,Geometry of Orlicz spaces. Dissertationes Math.356(1996), 1–204.

2. Dutta H.,On Orlicz difference sequence spaces. SDU Journal of Science5(2010), 119–136.

3. Gribanov Y.,On the theory of`M spaces. Uc. Zap. Kazansk un-ta,117(1957), 62–65, (in Russian).

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6. Kamthan P. K. and Gupta M.,Sequence Spaces and Series. Marcel Dekker Inc., New York and Basel, 1981.

7. Krasnoselskii M. A. and Rutickii Y. B.,Convex functions and Orlicz spaces. Groningen, Netherlands, 1961.

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Reine Angew. Math.171(1934), 193–226.

9. Lindenstrauss J. and Tzafriri L.,On Orlicz sequence spaces. Israel J. Math.10(1971), 379–390.

10. Musielak J.,Orlicz Spaces and Modular Spaces. Lecture Notes in Mathematics 1083, Springer–Verlag, 1983.

11. Maddox I. J.,Elements of Functional Analysis. The University Press, 2^nd ed. Cambridge, 1988.

12. Tripathy B. C.,A class of difference sequences related to the p-normed space`^p. Demonstratio Math.36(4) (2003), 867–872.

H. Dutta, Department of Mathematics, Gauhati University, Kokrajhar Campus, Assam, India, e-mail:hemen [email protected]

F. Ba¸sar, Fatih ¨Universitesi, Fen- Edebiyat Fak¨ultesi, Matematik B¨ol¨um¨u, B¨uy¨uk¸cekmece Kamp¨us¨u 34500-Istanbul, T¨urkiye,

e-mail:[email protected], [email protected]