◭◭ ◭ ◮ ◮◮ Go back Full Screen
Close Quit
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
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 ℓ∞, c
andc0 for the classical sequence spaces of all bounded, convergent and null sequences which are Banach spaces with the sup-normkxk∞= supk∈N|xk|, 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 bypi(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.
◭◭ ◭ ◮ ◮◮
Go back Full Screen
Close Quit
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 Mu(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,
◭◭ ◭ ◮ ◮◮
Go back Full Screen
Close Quit
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 spaceℓM (see Musielak [10];
Linden-strauss and Tzafriri [9]), as follows
ℓM = (
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}.
◭◭ ◭ ◮ ◮◮
Go back Full Screen
Close Quit
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) :=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 set ℓM(C,Λ) by
ℓM(C,Λ) := (
x= (xk)∈ω: X
k Pk
j=0λjxj
k+ 1
!
yk converges for ally= (yk)∈eℓΦ )
which is called as Orlicz sequence space associated with the multiplier sequence Λ = (λk) and
◭◭ ◭ ◮ ◮◮
Go back Full Screen
Close Quit
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, then Yα ⊂Xα. It is clear that X ⊂Xαα. If X =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),eℓM(C,Λ),ℓ′M(C,Λ)
andhM(C,Λ), and investigate their certain algebraic and topological properties. Furthermore, it is
proved that the spacesℓ′
M(C,Λ) andhM(C,Λ) are topologically isomorphic to the spacesℓ∞(C,Λ)
and c0(C,Λ) when M(u) = 0 on some interval, respectively. Finally, the α-dual of the spaces ℓ′
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 are noted.
2. Main Results
In this section, we emphasize the sequence spaces ℓM(C,Λ), eℓM(C,Λ), ℓ′M(C,Λ) and hM(C,Λ),
and give their some algebraic and topological properties.
◭◭ ◭ ◮ ◮◮
Go back Full Screen
Close Quit
Proof. Let x = (xk) ∈ ℓeM(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)∈ℓΦ. Thus,e 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 eachn∈N, there existsynwithδ(Φ, yn)≤1
such that
X
k Pk
j=0λjxj
k+ 1
!
yn k >2
◭◭ ◭ ◮ ◮◮
Go back Full Screen
Close Quit
Without loss of generality, we can assume thatPkj=0λjxjk+1, yn≥0. Now, we can define a sequence
z= (zk) byzk=Pn y n 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(y0k) + Φ
y1k+
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
1 2n+1 = 1.
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
zkdiverges and this implies thatx /∈ℓM(C,Λ), a contradiction. This leads
us to the required result.
The preceding result encourages us to introduce the following normk · kC
◭◭ ◭ ◮ ◮◮
Go back Full Screen
Close Quit
(i) ℓM(C,Λ)is a normed linear space under the norm k · 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 thatℓM(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 space
ℓM(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 ally such thatδ(Φ, y)≤1. Now consideringy= ek
if Φ(1) ≤ 1 otherwise considering y = 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 havex
k = 0 for all
k∈N, since (λk) is a sequence of non-zero scalars. Thus,x= 0.
It is easy to show that kαxkC
M =|α|kxkCM and kx+ykMC ≤ kxkCM +kykCM for allα∈ C and
x, y∈ℓM(C,Λ).
(ii) Let (xs) be any Cauchy sequence in the spaceℓ
M(C,Λ). Then for anyε >0, there exists a
positive integern0 such thatkxs−xtkC
◭◭ ◭ ◮ ◮◮
Go back Full Screen
Close Quit
by (2.2), we get
sup(
X
k " Pk
j=0λj xsj−xtj
k+ 1
#
yk
:δ(Φ, y)≤1 )
< ε
for alls, t≥n0. This implies that
X
k " Pk
j=0λj xsj−xtj
k+ 1
#
yk < ε
for ally with δ(Φ, y)≤ 1 and for all s, t ≥ n0. Now considering y = ek if Φ(1) ≤1, otherwise
consideringy = ek/Φ(1) we have (λ
kxsk) is a Cauchy sequence in Cfor all k∈N. Hence, it is a
convergent sequence inCfor allk∈N.
Let lims→∞λkxsk=xk for eachk∈N. Using the continuity of the modulus, we can derive for
alls≥n0 ast→ ∞, 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). Sincexs is in the space ℓM(C,Λ) and ℓM(C,Λ) is a linear
space, we havex= (xk)∈ℓM(C,Λ).
(iii) From the above proof, one can easily conclude thatkxskC
M →0 implies that xsk → 0 for
eachs∈Nwhich leads us to the desired result.
◭◭ ◭ ◮ ◮◮
Go back Full Screen
Close Quit
Proposition 2.4. ℓM(C,Λ)is a normed linear space under the normk · kC(M) defined by
kxkC(M)= inf
ρ >0 :
X
k
M
Pkj=0λjxj
ρ(k+ 1)
≤1
.
(2.3)
Proof. ClearlykxkC
(M)= 0 ifx= 0. Now, suppose thatkxk
C
(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
◭◭ ◭ ◮ ◮◮
Go back Full Screen
Close Quit
for allk ∈ N. Suppose | Pni
j=0λjxj|
ni+1 6= 0 for some ni ∈N. Then,
|Pni j=0λjxj|
ε(ni+1) → ∞ as ε→ 0. It
follows thatM
|Pni j=0λjxj| ε(k+1)
→ ∞asε→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) and y= (yk) be any two elements ofℓM(C,Λ). Then, there existρ1, 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)
◭◭ ◭ ◮ ◮◮
Go back Full Screen
Close Quit
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 define rbyr=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.
◭◭ ◭ ◮ ◮◮
Go back Full Screen
Close Quit
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 some ρ >0
.
Now, we can give the corresponding proposition on the spaceℓ′
M(C,Λ) to the Proposition2.3.
Proposition 2.6. The following statements hold:
(i) ℓ′
M(C,Λ)is a normed linear space under the norm k · 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 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 (xs) be any Cauchy sequence in the space ℓ′
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
kxs−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)
◭◭ ◭ ◮ ◮◮
Go back Full Screen
Close Quit
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≥n0 and for all k∈N. Hence, (λkxs
k) is a Cauchy sequence inC for allk ∈Nwhich
implies that it is a convergent sequence inCfor allk∈N.
Let lims→∞λkxsk=xk for eachk∈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 xs is in the space ℓ′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
◭◭ ◭ ◮ ◮◮
Go back Full Screen
Close Quit
Proposition 2.7. The inequalityPkM
|Pk j=0λjxj|
kxkC (M)(k+1)
≤1 holds for allx= (xk)∈ℓ′M(C,Λ).
Proof. This is immediate from the definition of the normk · 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 · k instead ofk · kC
(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, (hM(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
Pkj=0λjxj
ε(k+ 1)
◭◭ ◭ ◮ ◮◮
Go back Full Screen
Close Quit
Define thenth sectionx(n) of a sequence x = (x
k) by x(n) =Pnk=0xkek. Hence for n ≥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.
Thus, we can conclude thathM(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
ofℓ′
M(C,Λ). For this, let (xn) be a sequence inhM(C,Λ) such thatkxn−xk →0 asn→ ∞where
x= (xk)∈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)
◭◭ ◭ ◮ ◮◮
Go back Full Screen
Close Quit
There isl corresponding toρ >0 such that kxl−xk ≤ρ/2. Then, using the convexity of M,
we have by Proposition2.7that
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. Let M be an Orlicz function. If Msatisfies the ∆2-condition at 0, then
ℓ′
M(C,Λ) is anAK space.
Proof. We shall show that ℓ′
M(C,Λ) =hM(C,Λ) ifM satisfies the ∆2-condition at 0. To do
this it is enough to prove thatℓ′
◭◭ ◭ ◮ ◮◮
Go back Full Screen
Close Quit
ρ >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 arbitrary l >0. If ρ ≤l, then PkM
|Pk j=0λjxj| l(k+1)
< ∞. Now, let l < ρ and put
k = ρ/l. Since M satisfies ∆2-condition at 0, there exist R ≡ Rk > 0 and r ≡ 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 find d > n1 with
|Pd j=0λjxj| ρ(d+1) > r and thus
M
Pdj=0λjxj
ρ(d+ 1)
≥
Z |Pd
j=0λjxj|/ρ(d+1)
r/2
p(t)dt > r 2p
r
2
◭◭ ◭ ◮ ◮◮
Go back Full Screen
Close Quit
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. Let M1 and M2 be two Orlicz functions. If M1 and M2 are equivalent, thenℓ′
M1(C,Λ) =ℓ
′
M2(C,Λ) and the identity map I: ℓ
′
M1(C,Λ),k · k C M1
→ ℓ′
M2(C,Λ),k · k C M2
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= (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)
◭◭ ◭ ◮ ◮◮
Go back Full Screen
Close Quit
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)
(2.6)
also holds.
By combining the inclusions (2.5) and (2.6), we conclude thatℓ′
M1(C,Λ) =ℓ
′
M2(C,Λ).
For simplicity in notation, let us writek · k1andk · k2instead ofk · kCM1 andk · k C
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, wherep2 is the kernel associated withM2. 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)
◭◭ ◭ ◮ ◮◮
Go back Full Screen
Close Quit
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−1kxk
1 ≤ 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 ℓ′
M(C,Λ) and hM(C,Λ) are topologically isomorphic to the spaces ℓ∞(C,Λ) and c0(C,Λ), respectively; where ℓ∞(C,Λ) and
c0(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 spaces ℓ∞(C,Λ) and c0(C,Λ) are the Banach spaces under the norm
kxkC
∞= supk∈N
|Pk j=0λjxj|
k+1 .
Proof. Let p(x) = 0 for all x in [0, b]. If y ∈ ℓ∞(C,Λ), then we can find ρ > 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′
M(C,Λ).
On the other hand, lety∈ℓ′
M(C,Λ). Then for some ρ >0, we have
X
k
M
Pkj=0λjyj
ρ(k+ 1)
◭◭ ◭ ◮ ◮◮
Go back Full Screen
Close Quit
Therefore, |
Pk j=0λjyj|
k+1 ≤ K < ∞ for a constant K > 0 and for all k ∈ N which 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 everyε∈(0, α),
we can determinedwith
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 Proposition2.7. 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, lety ∈ 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,
let y ∈ c0(C,Λ). Then, for any ρ > 0, |
Pk j=0λjyj|
◭◭ ◭ ◮ ◮◮
Go back Full Screen
Close Quit
P kM
|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 the α-dual of the spaces ℓ′
M(C,Λ) and
hM(C,Λ), we present the following easy lemma without proof.
Lemma 2.13. For any Orlicz functionM,Λx= (λkxk)∈ℓ∞ wheneverx= (xk)∈ℓ′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−k1ak
<∞
)
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 ofℓ′
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= (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 Lemma2.13, we havea= (ak)∈ {ℓ′M(C,Λ)} α
. Hence, the inclusion
D1⊂ {ℓ′M(C,Λ)} α
◭◭ ◭ ◮ ◮◮
Go back Full Screen
Close Quit
holds.
Conversely, suppose thata= (ak)∈ {ℓ′M(C,Λ)} α
. Then, (akxk)∈ℓ1, the space of all absolutely
convergent series, for every x = (xk) ∈ ℓ′M(C,Λ). So, we can take xk = λ−k1 for all k ∈ N
because (xk)∈ℓ′M(C,Λ) by Proposition2.12whenever (xk)∈ℓ∞(C,Λ). Therefore,Pk akλ−k1
=
P
k|akxk|<∞and we havea= (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 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 ofhM(C,Λ) is the setD1.
Proof. Leta= (ak)∈D1 andx= (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)
◭◭ ◭ ◮ ◮◮
Go back Full Screen
Close Quit
Conversely, suppose thata= (ak)∈ {hM(C,Λ)}α\D1. Then, there exists a strictly increasing
sequence (ni) of positive integers ni such that
niX+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 all k ∈ N. Then, since x = (xk) ∈ c0(C,Λ), x = (xk) ∈ hM(C,Λ) by Proposition 2.12.
Therefore, we have
X
k
|akxk|= n1 X
k=n0+1
|akxk|+· · ·+ niX+1
k=ni+1
|akxk|+· · ·
=
n1 X
k=n0+1 akλ−k1
+· · ·+1 i
niX+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.
◭◭ ◭ ◮ ◮◮
Go back
Full Screen
Close
Quit
3. Conclusion
The difference Orlicz spacesℓM(∆,Λ) and eℓM(∆,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) andℓ∞(C,Λ) represent the domain
of the matrixC(λ) in the sequence spaces ℓM, c0 and ℓ∞, respectively. Since cnn(λ)6= 0 for all
n∈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,c0and ℓ∞, 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 · kC
M andk · 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) orhM(C,Λ) to the
sequence spacesℓ∞ andc?
◭◭ ◭ ◮ ◮◮ Go back Full Screen
Close Quit
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 ofℓM spaces. Uc. Zap. Kazansk un-ta,117(1957), 62–65, (in Russian).
4. Goes G.and Goes S.,Sequences of bounded variation and sequences of Fourier coefficients. Math. Z.118(1970), 93–102.
5. Kamthan P. K.,Bases in a certain class of Frechet spaces. Tamkang J. Math.7(1976), 41–49.
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. 8. K¨othe G. and Toeplitz O.,Linear Raume mit unendlichvielen koordinaten and Ringe unenlicher Matrizen. J.
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, 2nd
ed. Cambridge, 1988.
12. Tripathy B. C.,A class of difference sequences related to the p-normed spaceℓp. Demonstratio Math.36(4) (2003), 867–872.
H. Dutta, Department of Mathematics, Gauhati University, Kokrajhar Campus, Assam, India,
e-mail:hemen dutta08@rediffmail.com
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,
