Vol. LXXX, 2 (2011), pp. 185–200
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 spacesℓ′
M(C,Λ) andhM(C,Λ).
1. Preliminaries, Background and Notation
Byω, 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 asequence space. We write ℓ∞, c and c0 for the classical sequence spaces of all bounded,
convergent and null sequences which are Banach spaces with the sup-normkxk∞= supk∈N|xk|, where N = {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. A K-space X is
called anF K-spaceprovidedX is a complete linear metric space. AnF K-space whose topology is normable is called aBK-space.
A functionM: [0,∞)→[0,∞) which is convex withM(u)≥0 foru≥0, and
M(u)→ ∞ asu→ ∞, is called as anOrlicz function. An Orlicz function M can always be represented in the following integral form
M(u) =
Z u
0
p(t)dt,
where p, the kernel of M, is right differentiable fort ≥0,p(0) = 0, p(t)>0 for
t >0,pis non–decreasing andp(t)→ ∞ast→ ∞whenever Mu(u) ↑ ∞asu↑ ∞. Consider the kernelpassociated with the Orlicz functionM and let
q(s) = sup{t:p(t)≤s}.
Received March 29, 2010; revised December 9, 2010.
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 called mutually com-plementary 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)
An Orlicz functionM is said to satisfy the ∆2-condition for smalluor at 0 if for
eachk ∈N, there exist Rk >0 and uk >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 functions M1 and M2 are said to be equivalent if there are positive
constantsα, βandb such that
M1(αu)≤M2(u)≤M1(βu) for allu∈[0, b].
(1.4)
Orlicz used the Orlicz function to introduce the sequence spaceℓM (see Musielak
[10]; Lindenstrauss and Tzafriri [9]), as follows
ℓM =
(
x= (xk)∈ω:
X
k
M
|xk|
ρ
<∞ for someρ >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 spaceE(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 spacedE and integrated sequence spaceRE for a given sequence space
E, using the multiplier sequences (k−1) and (k) in [4], respectively. A multiplier
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. Let M be any Orlicz function and δ(M, x) := PkM(|xk|), wherex= (xk)∈ω. Then, we define the setsℓeM(C,Λ) andℓeM by
e
ℓM(C,Λ) :=
x= (xk)∈ω:bδC(M, x) = X
k
M
Pkj=0λjxj
k+ 1
<∞
and
e
ℓM :={x= (xk)∈ω:δ(M, x)<∞}.
Definition 1.2. LetM and Φ be mutually complementary functions. Then, we define the setℓM(C,Λ) by
ℓM(C,Λ) :=
(
x= (xk)∈ω:
X
k
Pk j=0λjxj
k+ 1
!
yk converges for ally= (yk)∈ℓeΦ )
which is called as Orlicz sequence space associated with the multiplier sequence Λ = (λk) and generated by Ces`aro matrix of order one.
Definition 1.3. Theα-dual or K¨othe-Toeplitz dualXαof a sequence spaceX
is defined by
Xα:=
(
a= (ak)∈ω:
X
k
|akxk|<∞for allx= (xk)∈X
) .
It is known that ifX ⊂Y, thenYα⊂Xα. It is clear thatX ⊂Xαα. IfX=Xαα,
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 spacesℓM(C,Λ),
e
ℓM(C,Λ), ℓ′M(C,Λ) and hM(C,Λ), and investigate their certain algebraic and
topological properties. Furthermore, it is proved that the spaces ℓ′
M(C,Λ) and
hM(C,Λ) are topologically isomorphic to the spacesℓ∞(C,Λ) andc0(C,Λ) when M(u) = 0 on some interval, respectively. Finally, theα-dual of the spacesℓ′
M(C,Λ)
and hM(C,Λ) are determined, and therefore the non-perfectness of the space
ℓ′
M(C,Λ) is showed whenM(u) = 0 on some interval, and some open problems
2. Main Results
In this section, we emphasize the sequence spacesℓM(C,Λ), ℓeM(C,Λ), ℓ′M(C,Λ)
andhM(C,Λ), and give their some algebraic and topological properties.
Proposition 2.1.For any Orlicz functionM, the inclusionℓeM(C,Λ)⊂ℓM(C,Λ)
holds.
Proof. Let x = (xk) ∈ eℓM(C,Λ). Then, since PkM
|Pk j=0λjxj|
k+1
< ∞ we
have from (1.1) that
X
k
Pk j=0λjxj
k+ 1
! yk
≤
X
k
Pk j=0λjxj
k+ 1
! yk
≤X
k
M
Pk j=0λjxj
k+ 1
!
+X
k
Φ(|yk|)<∞
for everyy= (yk)∈ℓeΦ. Thus,x= (xk)∈ℓM(C,Λ).
Proposition 2.2. For eachx= (xk)∈ℓM(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 each n∈N, there existsyn
withδ(Φ, yn)≤1 such that
X
k
Pk j=0λjxj
k+ 1
! yn
k
>2
n+1.
Without loss of generality, we can assume thatPkj=0λjxj
k+1, y
n≥0. Now, we can
define a sequencez= (zk) byzk =Pn yn
k
2n+1 for allk∈N. By the convexity of Φ,
we have
Φ
l
X
n=0
1 2n+1y
n k
!
≤ 1
2
Φ(y0
k) + Φ
y1
k+
y2
k
2 +· · ·+
yl k
2l−1
≤ · · · ≤ l
X
n=0
1 2n+1Φ(y
n 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
n k)≤
X
n
But for everyl∈N, it holds
X
k
Pk j=0λjxj
k+ 1
! zk ≥
X
k
Pk j=0λjxj
k+ 1
! l X
n=0
1 2n+1y
n k
=
l
X
n=0 X
k
Pk j=0λjxj
k+ 1
! yn
k
2n+1 ≥l.
HencePk
Pk j=0λjxj
k+1
zk diverges and this implies thatx /∈ℓM(C,Λ), a
contra-diction. This leads us to the required result.
The preceding result encourages us to introduce the following normk · kC M on
ℓM(C,Λ).
Proposition 2.3. The following statements hold:
(i) ℓM(C,Λ) is a normed linear space under the normk · kCM defined by
kxkCM = sup
(
X
k
Pk j=0λjxj
k+ 1
! yk
:δ(Φ, y)≤1 )
.
(2.2)
(ii) ℓM(C,Λ) is a Banach space under the norm defined by (2.2).
(iii) ℓM(C,Λ) is aBK space under the norm defined by (2.2).
Proof. (i) It is easy to verify that ℓM(C,Λ) is a linear space with respect to
the co-ordinatewise addition and scalar multiplication of sequences. Now we show thatk · kC
M is a norm on the spaceℓM(C,Λ).
Ifx= 0, then obviouslykxkC
M = 0. Conversely, assumekxkCM = 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
Pk
Pk j=0λjxj
k+1
yk
= 0 for all y such that δ(Φ, y) ≤1. Now
consideringy= ekif Φ(1)≤1 otherwise consideringy= ek/Φ(1) so thatλ kxk = 0
for all k∈N, where ek is a sequence whose only non-zero term is 1 inkth 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 thatkαxkC
M =|α|kxkCM and kx+ykCM ≤ kxkCM +kykCM for
allα∈Candx, y∈ℓM(C,Λ).
(ii) Let (xs) be any Cauchy sequence in the spaceℓM(C,Λ). Then for anyε >0,
there exists a positive integern0such thatkxs−xtkCM < εfor alls, t≥n0. Using
the definition of norm given by (2.2), we get
sup
(
X
k
" Pk
j=0λj xsj−xtj
k+ 1
# yk
:δ(Φ, y)≤1 )
for alls, t≥n0. This implies that
X
k
" Pk
j=0λj xsj−xtj
k+ 1
# yk
< ε
for allywithδ(Φ, y)≤1 and for alls, t≥n0. Now consideringy= ek if Φ(1)≤1,
otherwise consideringy= ek/Φ(1) we have (λ
kxsk) is a Cauchy sequence inCfor
allk∈N. Hence, it is a convergent sequence inCfor allk∈N.
Let lims→∞λkxsk =xk for each k ∈N. Using the continuity of the modulus,
we can derive for alls≥n0as t→ ∞, that
sup
(
X
k
" Pk
j=0λj xsj−xj
k+ 1
# yk
:δ(Φ, y)≤1 )
< ε.
It follows that (xs−x)∈ℓ
M(C,Λ). Sincexsis in the spaceℓM(C,Λ) andℓM(C,Λ)
is a linear space, we havex= (xk)∈ℓM(C,Λ).
(iii) From the above proof, one can easily conclude that kxskC
M → 0 implies
thatxs
k→0 for eachs∈Nwhich leads us to the desired result.
Therefore, the proof of the theorem is completed.
Proposition 2.4. ℓM(C,Λ) is a normed linear space under the norm k · kC(M) defined by
kxkC(M)= inf ρ >0 :
X
k
M
Pkj=0λjxj
ρ(k+ 1)
≤1
.
(2.3)
Proof. Clearly kxkC
(M)= 0 if x= 0. Now, suppose thatkxkC(M)= 0. Then, we
have
inf
ρ >0 :
X
k
M
Pkj=0λjxj 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
Pkj=0λjxj
ρε(k+ 1)
≤1
which implies that
M
Pkj=0λjxj
ρε(k+ 1)
≤1
for allk∈N. Thus,
M
Pkj=0λjxj
ε(k+ 1)
≤M
Pkj=0λjxj
ρε(k+ 1)
for allk∈N. Suppose | Pni
j=0λjxj|
ni+1 6= 0 for someni∈N. Then,
|Pni j=0λjxj|
ε(ni+1) → ∞as ε→0. It follows thatM
|Pni j=0λjxj|
ε(k+1)
→ ∞as ε→0 for someni ∈N, which is
a contradiction. Therefore, |
Pk j=0λjxj|
k+1 = 0 for allk∈N. It follows thatλkxk = 0
for allk∈N. Hencex= 0, since (λk) is a sequence of non-zero scalars.
Letx= (xk) andy= (yk) be any two elements ofℓM(C,Λ). Then, there exist
ρ1, ρ2>0 such that
X k M
Pkj=0λjxj ρ1(k+ 1)
≤1 and X
k M
Pkj=0λjyj ρ2(k+ 1)
≤1.
Letρ=ρ1+ρ2. Then by the convexity ofM, we have
X k M
Pkj=0λj(xj+yj)
ρ(k+ 1)
≤ ρ1
ρ1+ρ2 X k M
Pkj=0λjxj
ρ1(k+ 1)
+ ρ2
ρ1+ρ2 X k M
Pkj=0λjyj
ρ2(k+ 1)
≤1.
Hence, we have
kx+ykC
(M)= inf ρ >0 :
X k M
Pkj=0λj(xj+yj) ρ
≤1
≤ inf
ρ1>0 : X k M
Pkj=0λjxj ρ1
≤1
+ inf
ρ2>0 : X k M
Pkj=0λjyj
ρ2
≤1
which gives thatkx+ykC(M)≤ kxkC(M)+kykC(M).
Finally, letαbe any scalar and definerbyr=ρ/|α|. Then,
kαxkC(M)= inf ρ >0 :
X k M
Pkj=0αλjxj
ρ(k+ 1)
≤1
= inf
r|α|>0 : X k M
Pkj=0λjxj
r(k+ 1)
≤1
=|α|kxk
C
(M).
This completes the proof.
Definition 2.5. For any Orlicz functionM, we define
ℓ′M(C,Λ) :=
x= (xk)∈ω: X
k
M
Pkj=0λjxj
ρ(k+ 1)
<∞ for someρ >0
.
Now, we can give the corresponding proposition on the spaceℓ′
M(C,Λ) to the
Proposition 2.3.
Proposition 2.6. The following statements hold:
(i) ℓ′M(C,Λ)is a normed linear space under the normk · kC(M)defined by (2.3).
(ii) ℓ′
M(C,Λ) is a Banach space under the norm defined by (2.3).
(iii) ℓ′
M(C,Λ) is aBK 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 (xs) be any Cauchy sequence in the spaceℓ′
M(C,Λ). Letδ >0 be fixed
andr >0 be given such that 0< ε <1 andrδ≥1. Then, there exists a positive integern0 such thatkxs−xtkC(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
Pkj=0λj xsj−xtj
ρ(k+ 1)
≤1
<
ε rδ
for alls, t≥n0. This implies that
X
k
M
Pkj=0λj xsj−xtj kxs−xtkC
(M)(k+ 1) ≤1
for alls, t≥n0. It follows that
M
Pkj=0λj xsj−xtj kxs−xtkC
(M)(k+ 1) ≤1
for alls, t≥n0 and for allk∈N. Forr >0 withM(rδ/2)≥1, we have
M
Pkj=0λj xsj−xtj kxs−xtkC
(M)(k+ 1) ≤M
rδ
2
for alls, t≥n0 and for allk∈N. SinceM is non-decreasing, we have
Pkj=0λj xsj−xtj
k+ 1 ≤
rδ
2 ·
ε rδ =
ε
2
for alls, t≥n0and for allk∈N. Hence, (λkxsk) is a Cauchy sequence inCfor all
Let lims→∞λkxsk = xk for each k ∈ N. Using the continuity of an Orlicz
function and modulus, we can have
inf
ρ >0 :
X
k
M
Pkj=0λj(xsj−xj)
ρ(k+ 1)
≤1
< ε
for alls≥n0, asj → ∞. It follows that (xs−x)∈ℓ′M(C,Λ). Since xsis in the
spaceℓ′
M(C,Λ) andℓ′M(C,Λ) is a linear space, we havex= (xk)∈ℓ′M(C,Λ).
(iii) From the above proof, one can easily conclude thatkxskC
M →0 ass→ ∞,
which implies thatxs
k →0 ask→ ∞for eachs∈N. This leads us to the desired
result.
Proposition 2.7. The inequality PkM
|Pk j=0λjxj|
kxkC (M)(k+1)
≤1 holds for all x=
(xk)∈ℓ′M(C,Λ).
Proof. This is immediate from the definition of the norm k · kC
(M) defined by
(2.3).
Definition 2.8. For any Orlicz functionM, we define
hM(C,Λ) :=
x= (xk)∈ω: X
k
M
Pkj=0λjxj ρ(k+ 1)
<∞ for eachρ >0
.
ClearlyhM(C,Λ) is a subspace ofℓ′M(C,Λ).
Here and after we shall writek · kinstead ofk · kC
(M)provided it does not lead
to any confusion. The topology ofhM(C,Λ) is induced byk · k.
Proposition 2.9. Let M be an Orlicz function. Then,(hM(C,Λ),k · k) is an
AK-BK space.
Proof. First we show thathM(C,Λ) is anAK space. Letx= (xk)∈hM(C,Λ).
Then for eachε∈(0,1), we can findn0 such that
X
i≥n0 M
Pkj=0λjxj
ε(k+ 1)
≤1.
Define thenthsection x(n) of a sequence x= (x
k) by x(n) =Pnk=0xkek. Hence
forn≥n0, it holds
x−x(n)= inf
ρ >0 :
X
k≥n0 M
Pkj=0λjxj
ρ(k+ 1)
≤1
≤inf
ρ >0 :
X
k≥n
M
Pkj=0λjxj
ρ(k+ 1)
≤1
< ε.
Next to show thathM(C,Λ) is aBK space, it is enough to show hM(C,Λ) is
a closed subspace ofℓ′
M(C,Λ). For this, let (xn) be a sequence inhM(C,Λ) such
thatkxn−xk →0 asn→ ∞wherex= (x
k)∈ℓ′M(C,Λ). To complete the proof
we need to show thatx= (xk)∈hM(C,Λ), i.e.,
X
k
M
Pkj=0λjxj ρ(k+ 1)
<∞ for all ρ >0.
There is l corresponding to ρ > 0 such that kxl−xk ≤ ρ/2. Then, using the
convexity ofM, we have by Proposition 2.7 that
X
k
M
Pkj=0λjxj ρ(k+ 1)
=X
k
M 2
Pkj=0λjxlj
−2Pkj=0λjxlj
−Pkj=0λjxj
2ρ(k+ 1)
≤1
2
X
k
M 2
Pkj=0λjxlj
ρ(k+ 1)
+1
2
X
k
M 2
Pkj=0λj xlj−xj
ρ(k+ 1)
≤1
2
X
k
M 2
Pkj=0λjxlj
ρ(k+ 1)
+1
2
X
k
M 2
Pkj=0λj xlj−xj kxl−xk(k+ 1)
<∞.
Hence,x= (xk)∈hM(C,Λ) and consequentlyhM(C,Λ) is aBK space.
Proposition 2.10.LetM be an Orlicz function. IfMsatisfies the∆2-condition at 0, thenℓ′
M(C,Λ) is anAK space.
Proof. We shall show thatℓ′
M(C,Λ) =hM(C,Λ) ifM satisfies the ∆2-condition
at 0. To do this it is enough to prove thatℓ′
M(C,Λ)⊂hM(C,Λ). Let x= (xk)∈
ℓ′
M(C,Λ). Then for someρ >0,
X
k
M
Pkj=0λjxj
ρ(k+ 1)
<∞.
This implies that
lim
k→∞M
Pkj=0λjxj
ρ(k+ 1)
= 0.
(2.4)
Choose an arbitraryl > 0. If ρ ≤l, then PkM
|Pk j=0λjxj|
l(k+1)
<∞. Now, let
andr ≡rk >0 with M(kx)≤RM(u) for allx∈(0, r]. By (2.4), there exists a
positive integern1 such that
M
Pkj=0λjxj
ρ(k+ 1)
< r
2p
r
2
for all k≥n1.
We claim that |
Pk j=0λjxj|
ρ(k+1) ≤rfor allk≥n1. Otherwise, we can findd > n1 with
|Pd j=0λjxj|
ρ(d+1) > rand thus
M
Pdj=0λjxj ρ(d+ 1)
≥
Z |Pd
j=0λjxj|/ρ(d+1)
r/2
p(t)dt > r
2p
r
2
,
a contradiction. Hence our claim is true. Then, we can find that
X
k≥n1 M
Pkj=0λjxj
l(k+ 1)
≤R X
k≥n1 M
Pkj=0λjxj
ρ(k+ 1)
.
Hence,
X
k
M
Pkj=0λjxj
l(k+ 1)
<∞ for all l >0.
This completes the proof.
Proposition 2.11. LetM1 andM2 be two Orlicz functions. IfM1andM2are equivalent, thenℓ′
M1(C,Λ) =ℓ
′
M2(C,Λ)and the identity mapI: ℓ
′
M1(C,Λ),k · k
C M1
→ ℓ′M2(C,Λ),k · k
C M2
is a topological isomorphism.
Proof. Letα,β andbbe constants from (1.3). SinceM1andM2are equivalent,
it is obvious that (1.3) holds. Let us take anyx= (xk)∈ℓ′M2(C,Λ). Then,
X
k
M2
Pkj=0λjxj
ρ(k+ 1)
<∞ for some ρ >0.
Hence, for somel≥1, |
Pk j=0λjxj|
lρ(k+1) ≤bfor allk∈N. Therefore, X
k
M1 α
Pkj=0λjxj
lρ(k+ 1)
≤X
k
M2
Pkj=0λjxj
ρ(k+ 1)
<∞
which shows that the inclusion
ℓ′M2(C,Λ)⊂ℓ
′
M1(C,Λ)
(2.5)
holds. One can easily see in the same way that the inclusion
ℓ′
M1(C,Λ)⊂ℓ
′
M2(C,Λ)
By combining the inclusions (2.5) and (2.6), we conclude that ℓ′
M1(C,Λ) = ℓ′M2(C,Λ).
For simplicity in notation, let us writek · k1 andk · k2 instead of k · kCM1 and
k · kC
M2, respectively. Forx= (xk)∈ℓ
′
M2(C,Λ), we get
X
k
M2
Pkj=0λjxj
kxk2(k+ 1) ≤1.
One can findµ >1 with (b/2)µp2(b/2)≥1, wherep2is the kernel associated with M2. Hence,
M2
Pkj=0λjxj
kxk2(k+ 1) ≤ b
2µp2
b
2
for all k∈N.
This implies that
Pkj=0λjxj
µkxk2(k+ 1)
≤b for all k∈N.
Therefore,
X
k
M1 α
Pkj=0λjxj
µkxk2(k+ 1)
<1.
Hence,kxk1≤(µ/α)kxk2. Similarly, we can show thatkxk2≤βγkxk1by choosing γ withγβ >1 such thatγβ(b/2)p1(b/2)≥1. Thus,αµ−1kxk1≤ kxk2≤βγkxk1
which establish thatI is a topological isomorphism.
Proposition 2.12. Let M be an Orlicz function and p be the corresponding kernel. If p(x) = 0 for all xin [0, b], where b is some positive number, then the spacesℓ′
M(C,Λ)andhM(C,Λ)are topologically isomorphic to the spacesℓ∞(C,Λ)
andc0(C,Λ), respectively; where ℓ∞(C,Λ)andc0(C,Λ)are defined by
ℓ∞(C,Λ) =
x= (xk)∈ω: supk∈N
k
X
j=0
|λjxj|
k+ 1 <∞
and
c0(C,Λ) =
x= (xk)∈ω: limk→∞
k
X
j=0
|λjxj|
k+ 1 = 0
.
It is easy to see that the spacesℓ∞(C,Λ)andc0(C,Λ)are the Banach spaces under the normkxkC
∞= supk∈N
|Pk j=0λjxj|
k+1 .
Proof. Let p(x) = 0 for allxin [0, b]. Ify ∈ℓ∞(C,Λ), then we can findρ >0
such that |
Pk j=0λjyj|
ρ(k+1) ≤bfork∈N. Hence, P
kM
|Pk j=0λjyj|
ρ(k+1)
<∞. That is to
say thaty∈ℓ′
On the other hand, lety∈ℓ′
M(C,Λ). Then for some ρ >0, we have
X
k
M
Pkj=0λjyj
ρ(k+ 1)
<∞.
Therefore, |
Pk j=0λjyj|
k+1 ≤K < ∞ for a constantK > 0 and for all k ∈ Nwhich
yields that y ∈ ℓ∞(C,Λ). Hence, y ∈ ℓ∞(C,Λ) if and only if y ∈ ℓ′
M(C,Λ).
We can easily find b such that M(u0)≥ 1. Lety ∈ ℓ∞(C,Λ) and α=kyk∞ =
supk∈N
Pk j=0λjyj
k+1
> 0. For every ε ∈ (0, α), we can determine d with
Pd j=0λjyj
d+1
> α−εand so
X
k
M
Pkj=0λjyj b α(k+ 1)
≥M
α−ε
α b
.
SinceM is continuous,PkM
|Pk j=0λjyj|b
α(k+1)
≥1, and sokyk∞≤bkyk, otherwise
P
kM
|Pk j=0λjyj|
kyk(k+1)
>1 which contradicts Proposition 2.7. Again,
X
k
M
Pkj=0λjyj
b α(k+ 1)
= 0
which gives that kyk ≤ kyk∞/b. That is to say that the identity map
I: (ℓ′
M(C,Λ),k · k)→(ℓ∞(C,Λ),k · k) is a topological isomorphism.
For the last part, let y ∈hM(C,Λ). Then for any ε >0, |
Pk j=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, lety∈c0(C,Λ). Then, for anyρ >0, | Pk
j=0λjyj|
ρ(k+1) < b/2
for all sufficiently large k. Thus, PkM
|Pk j=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 theα-dual of the spaces
ℓ′
M(C,Λ) andhM(C,Λ), we present the following easy lemma without proof.
Lemma 2.13. For any Orlicz function M, Λx= (λkxk)∈ ℓ∞ whenever x= (xk)∈ℓ′M(C,Λ).
Proposition 2.14. Let M be an Orlicz function and p be the corresponding kernel ofM. Define the setsD1 andD2 by
D1:= (
a= (ak)∈ω:
X
k
λ−k1ak
<∞
and
D2:=
b= (bk)∈ω: sup k∈N
|λkbk|<∞
.
Ifp(x) = 0 for allxin [0, d], wheredis some positive number, then the following statements hold:
(i) K¨othe-Toeplitz dual ofℓ′
M(C,Λ)is the set D1.
(ii) K¨othe-Toeplitz dual ofD1 is the set D2.
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= (ak)∈D1andx= (xk)∈ℓ′M(C,Λ). Then, since
X
k
|akxk|=
X
k
akλ−k1
|λkxk| ≤sup k∈N
|λkxk| ·
X
k
akλ−k1
<∞,
applying Lemma 2.13, we havea= (ak)∈ {ℓ′M(C,Λ)} α
. Hence, the inclusion
D1⊂ {ℓ′M(C,Λ)} α
(2.7)
holds.
Conversely, suppose that a = (ak) ∈ {ℓ′M(C,Λ)} α
. Then, (akxk) ∈ ℓ1, the
space of all absolutely convergent series, for everyx= (xk) ∈ℓ′M(C,Λ). So, we
can take xk = λ−k1 for all k ∈ N because (xk) ∈ ℓ′M(C,Λ) by Proposition 2.12
whenever (xk)∈ℓ∞(C,Λ). Therefore,Pk
akλ−k1
=Pk|akxk|<∞and we have
a= (ak)∈D1. This leads us to the inclusion
{ℓ′M(C,Λ)} α
⊂D1.
(2.8)
By combining the inclusion relations (2.7) and (2.8), we have{ℓ′
M(C,Λ)} α
=D1.
Proposition 2.14 (ii) shows that {ℓ′
M(C,Λ)} αα
6
= ℓ′
M(C,Λ) which leads us to
the consequence thatℓ′
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 of M and the set D1 be defined as in the Proposition 2.14. If p(x) = 0 for all x in [0, b], where b is a positive number, then the K¨othe-Toeplitz dual of hM(C,Λ)is the set D1.
Proof. Leta= (ak)∈D1andx= (xk)∈hM(C,Λ). Then, since
X
k
|akxk|=
X
k
akλ−k1
|λkxk| ≤sup k∈N
|λkxk| ·
X
k
akλ−k1
<∞,
we havea= (ak)∈ {hM(C,Λ)}α. Hence, the inclusion
D1⊂ {hM(C,Λ)}α
(2.9)
Conversely, suppose thata = (ak)∈ {hM(C,Λ)}α\D1. Then, there exists a
strictly increasing sequence (ni) of positive integers ni such that nXi+1
k=ni+1
|ak| |λk|−1> i.
Definex= (xk) by
xk:=
λ−k1sgn ak/i , (ni< k≤ni+1),
0 , (0≤k < n0),
for allk∈N. Then, sincex= (xk)∈c0(C,Λ), x= (xk)∈hM(C,Λ) by
Proposi-tion 2.12. Therefore, we have
X
k
|akxk|= n1 X
k=n0+1
|akxk|+· · ·+ nXi+1
k=ni+1
|akxk|+· · ·
=
n1 X
k=n0+1 akλ−k1
+· · ·+1
i
nXi+1
k=ni+1 akλ−k1
+· · ·
>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.
3. Conclusion
The difference Orlicz spacesℓM(∆,Λ) andℓeM(∆,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 spaceℓM. Indeed, if we
define the infinite matrix C(λ) ={cnk(λ)} via the multiplier sequence Λ = (λk)
by
cnk(λ) :=
λk
n+ 1, (0≤k≤n), 0, (k > n),
for allk, n ∈ N, then the sequence spaces ℓ′
M(C,Λ), c0(C,Λ) and ℓ∞(C,Λ)
rep-resent the domain of the matrix C(λ) in the sequence spaces ℓM, c0 and ℓ∞,
respectively. Since cnn(λ)6= 0 for alln∈N, i.e., C(λ) is a triangle, it is obvious
that those spacesℓ′
M(C,Λ),c0(C,Λ) andℓ∞(C,Λ) are linearly isomorphic to the
spacesℓM,c0andℓ∞, respectively.
(i) What is the relation between the norms k · kC
M and k · kC(M)? Are they
equivalent?
(ii) What is the relation between the spacesℓM(C,Λ) andℓ′M(C,Λ)? Do they
coincide?
(iii) What are theβ- andγ-duals of the spacesℓ′
M(C,Λ) andhM(C,Λ)?
(iv) Under which conditions an infinite matrix transforms the setsℓ′
M(C,Λ) or
hM(C,Λ) to the sequence spacesℓ∞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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H. Dutta, Department of Mathematics, Gauhati University, Kokrajhar Campus, Assam, India, e-mail:hemen dutta08@rediffmail.com
