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A GENERALIZATION OF ORLICZ SEQUENCE SPACES
`
BY CESARO
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 ℓ′M (C, Λ) and hM (C, Λ).
1. Preliminaries, Background and Notation
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By ω, we denote the space of all complex valued sequences. Any vector subspace of ω which
contains φ, the set of all finitely non–zero sequences is called a sequence 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-norm kxk∞ = 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 an
F K-space provided X is a complete linear metric space. An F K-space whose topology is normable
is called a BK-space.
A function M : [0, ∞) → [0, ∞) which is convex with M (u) ≥ 0 for u ≥ 0, and M (u) → ∞ as
u → ∞, is called as an Orlicz function. An Orlicz function M can always be represented in the
Received March 29, 2010; revised December 9, 2010.
2010 Mathematics 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.
following integral form
M (u) =
Z
u
p(t)dt,
0
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 and p(t) → ∞ as t → ∞ whenever Mu(u) ↑ ∞ as u ↑ ∞.
Consider the kernel p associated with the Orlicz function M and let
q(s) = sup{t : p(t) ≤ s}.
Then q possesses the same properties as the function p. Suppose now
Z x
q(s)ds.
Φ(x) =
0
Then, Φ is an Orlicz function. The functions M and Φ are called mutually complementary Orlicz
functions.
Now, we give the following well-known results.
Let M and Φ are mutually complementary Orlicz functions. Then, we have:
(i) For all u, y ≥ 0,
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(1.1)
(ii) For all u ≥ 0,
(1.2)
up(u) = M (u) + Φ(p(u)).
(iii) For all u ≥ 0 and 0 < λ < 1,
(1.3)
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uy ≤ M (u) + Φ(y),
M (λu) < λM (u).
(Young’s inequality).
An Orlicz function M is said to satisfy the ∆2 -condition for small u or at 0 if for each k ∈ N, there
exist Rk > 0 and uk > 0 such that M (ku) ≤ Rk M (u) for all u ∈ (0, uk ]. Moreover, an Orlicz
function M 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 constants α, β and
b such that
(1.4)
M1 (αu) ≤ M2 (u) ≤ M1 (βu)
for all u ∈ [0, b].
Orlicz used the Orlicz function to introduce the sequence space ℓM (see Musielak [10]; Lindenstrauss and Tzafriri [9]), as follows
)
(
X |xk |
< ∞ for some ρ > 0 .
ℓM = x = (xk ) ∈ ω :
M
ρ
k
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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 space E, the multiplier sequence space E(Λ) associated with the
multiplier sequence Λ is defined by
E(Λ) = {x = (xk ) ∈ ω : Λx = (λk xk ) ∈ E} .
The scope for the studies on sequence spaces was extended by using the notion of associated
multiplier sequences. G. RGoes and S. Goes 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) 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.
Let C = (cnk ) be the Ces`
aro matrix of order one defined by
1
, 0 ≤ k ≤ n,
n
+
1
cnk :=
0,
k > n,
for all k, n ∈ N.
P
Definition 1.1. Let M be any Orlicz function and δ(M, x) := k M (|xk |), where x = (xk ) ∈ ω.
Then, we define the sets ℓeM (C, Λ) and ℓeM by
Pk
j=0 λj xj
X
2n+1 .
Pk λj xj
Without loss of generality, we can assume that j=0 k+1
, y n ≥ 0. Now, we can define a sequence
n
P yk
z = (zk ) by zk = n 2n+1 for all k ∈ N. By the convexity of Φ, we have
!
l
X
1 n
1
yk2
ykl
0
1
Φ
≤
Φ(y
)
+
Φ
y
+
y
+
·
·
·
+
k
k
2n+1 k
2
2
2l−1
n=0
··· ≤
l
X
n=0
1
2n+1
Φ(ykn )
for any positive integer l. Hence, using the continuity of Φ, we have
XX 1
X
X 1
n
Φ(zk ) ≤
δ(Φ, z) =
Φ(y
)
≤
= 1.
k
2n+1
2n+1
n
n
k
But for every l ∈ N, it holds
!
Pk
X
j=0 λj xj
k+1
k
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zk ≥
Hence
P
k
Pk
λj xj
k+1
j=0
X
k
=
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k
Pk
l X
X
n=0 k
j=0
λj xj
!
l
X
1 n
y
n+1 k
2
n=0
!
ykn
j=0 λj xj
≥ l.
k+1
2n+1
k+1
Pk
zk diverges and this implies that x ∈
/ ℓM (C, Λ), a contradiction. This leads
us to the required result.
The preceding result encourages us to introduce the following norm k · kC
M on ℓM (C, Λ).
Proposition 2.3. The following statements hold:
(i) ℓM (C, Λ) is a normed linear space under the norm k · kC
M defined by
)
!
(
P
k
X
j=0 λj xj
C
(2.2)
yk : δ(Φ, y) ≤ 1 .
kxkM = sup
k+1
k
(ii) ℓM (C, Λ) is a Banach space under the norm defined by (2.2).
(iii) ℓM (C, Λ) is a BK 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 that k · kC
M is a norm on the space
ℓM (C, Λ).
C
If x = 0, then obviously kxkC
M = 0. Conversely, assume kxkM = 0. Then using the definition of
the norm given by (2.2), we have
(
!
)
X Pk λj xj
j=0
yk : δ(Φ, y) ≤ 1 = 0.
sup
k+1
k
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P Pkj=0 λj xj
yk = 0 for all y such that δ(Φ, y) ≤ 1. Now considering y = ek
This implies that k
k+1
if Φ(1) ≤ 1 otherwise considering y = ek /Φ(1) so that λk xk = 0 for all k ∈ N, where ek is a
sequence whose only non-zero term is 1 in k th place for each k ∈ N. Hence we have xk = 0 for all
k ∈ N, since (λk ) is a sequence of non-zero scalars. Thus, x = 0.
C
C
C
C
It is easy to show that kαxkC
M = |α|kxkM and kx + ykM ≤ kxkM + kykM for all α ∈ C and
x, y ∈ ℓM (C, Λ).
(ii) Let (xs ) be any Cauchy sequence in the space ℓM (C, Λ). Then for any ε > 0, there exists a
positive integer n0 such that kxs − xt kC
M < ε for all s, t ≥ n0 . Using the definition of norm given
by (2.2), we get
(
#
"
)
X Pk λj xs − xt
j
j
j=0
sup
yk : δ(Φ, y) ≤ 1 < ε
k+1
k
for all s, t ≥ n0 . This implies that
#
"
X Pk λj xs − xt
j
j
j=0
yk < ε
k+1
k
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for all y with δ(Φ, y) ≤ 1 and for all s, t ≥ n0 . Now considering y = ek if Φ(1) ≤ 1, otherwise
considering y = ek /Φ(1) we have (λk xsk ) is a Cauchy sequence in C for all k ∈ N. Hence, it is a
convergent sequence in C for all k ∈ N.
Let lims→∞ λk xsk = xk for each k ∈ N. Using the continuity of the modulus, we can derive for
all s ≥ n0 as t → ∞, that
#
)
(
"
X Pk λj xs − xj
j
j=0
yk : δ(Φ, y) ≤ 1 < ε.
sup
k+1
k
It follows that (xs − x) ∈ ℓM (C, Λ). Since xs is in the space ℓM (C, Λ) and ℓM (C, Λ) is a linear
space, we have x = (xk ) ∈ ℓM (C, Λ).
s
(iii) From the above proof, one can easily conclude that kxs kC
M → 0 implies that xk → 0 for
each s ∈ N which 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
Pk
j=0 λj xj
X
≤1 .
ρ>0:
M
(2.3)
kxkC
(M ) = inf
ρ(k + 1)
k
C
Proof. Clearly kxkC
(M ) = 0 if x = 0. Now, suppose that kxk(M ) = 0. Then, we have
Pk
j=0 λj xj
X
≤ 1 = 0.
M
inf ρ > 0 :
ρ(k + 1)
k
This yields the fact for a given ε > 0 that there exists some ρε ∈ (0, ε) such that
Pk
j=0 λj xj
≤1
sup M
ρε (k + 1)
k∈N
which implies that
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for all k ∈ N. Thus,
Pk
j=0 λj xj
≤1
M
ρε (k + 1)
Pk
Pk
j=0 λj xj
j=0 λj xj
≤M
≤1
M
ε(k + 1)
ρε (k + 1)
i λ x
i λ x
| nj=0
| nj=0
j j|
j j|
for all k ∈ N. Suppose
=
6
0
for
some
n
∈
N.
Then,
→ ∞ as ε → 0. It
i
n
+1
ε(n
i +1)
i
Pn
i λ x
| j=0
j j|
→ ∞ as ε → 0 for some ni ∈ N, which is a contradiction. Therefore,
follows that M
ε(k+1)
P
|
P
λj xj |
k+1
Pk
j=0
= 0 for all k ∈ N. It follows that λk xk = 0 for all k ∈ N. Hence x = 0, since (λk ) is a
sequence of non-zero scalars.
Let x = (xk ) and y = (yk ) be any two elements of ℓM (C, Λ). Then, there exist ρ1 , ρ2 > 0 such
that
X
k
Pk
Pk
j=0 λj yj
j=0 λj xj
X
≤ 1.
M
M
≤ 1 and
ρ1 (k + 1)
ρ2 (k + 1)
k
Let ρ = ρ1 + ρ2 . Then by the convexity of M , we have
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X
k
Pk
Pk
λj xj
j=0 λj (xj + yj )
X
j=0
ρ
1
≤
M
M
ρ(k + 1)
ρ1 + ρ 2
ρ1 (k + 1)
k
Pk
λ
y
X
j
j
j=0
ρ2
≤ 1.
M
+
ρ 1 + ρ2
ρ2 (k + 1)
k
Hence, we have
kx + ykC
(M )
Pk
j=0 λj (xj + yj )
X
≤1
= inf ρ > 0 :
M
ρ
k
Pk
j=0 λj xj
X
≤1
M
≤ inf ρ1 > 0 :
ρ1
k
Pk
j=0 λj yj
X
≤1
M
+ inf ρ2 > 0 :
ρ2
k
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C
C
which gives that kx + ykC
(M ) ≤ kxk(M ) + kyk(M ) .
Finally, let α be any scalar and define r by r = ρ/|α|. Then,
Pk
j=0 αλj xj
X
≤1
kαxkC
ρ>0:
M
(M ) = inf
ρ(k + 1)
k
Pk
j=0 λj xj
X
≤ 1 = |α|kxkC
M
= inf r|α| > 0 :
(M ) .
r(k + 1)
k
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This completes the proof.
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Proposition 2.4 inspires us to define the following sequence space.
Definition 2.5. For any Orlicz function M , we define
Pk
j=0 λj xj
X
< ∞ for some ρ > 0 .
M
ℓ′M (C, Λ) := x = (xk ) ∈ ω :
ρ(k + 1)
k
Now, we can give the corresponding proposition on the space ℓ′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 norm k · kC
(M ) defined by (2.3).
′
(ii) ℓM (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).
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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 − xt kC
(M ) < ε/rδ for all s, t ≥ n0 . Using the definition of the norm given by (2.3), we get
Pk
j=0 λj xsj − xtj
X
≤1 < ε
inf ρ > 0 :
M
rδ
ρ(k + 1)
k
for all s, t ≥ n0 . This implies that
X
k
Pk
j=0 λj xsj − xtj
≤1
M s
kx − xt kC
(M ) (k + 1)
for all s, t ≥ n0 . It follows that
Pk
j=0 λj xsj − xtj
≤1
M s
kx − xt kC
(M ) (k + 1)
for all s, t ≥ n0 and for all k ∈ N. For r > 0 with M (rδ/2) ≥ 1, we have
Pk
j=0 λj xsj − xtj
≤ M rδ
M s
2
kx − xt kC
(M ) (k + 1)
for all s, t ≥ n0 and for all k ∈ N. Since M is non-decreasing, we have
P
k
j=0 λj xsj − xtj
rδ ε
ε
≤
·
=
k+1
2 rδ
2
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for all s, t ≥ n0 and for all k ∈ N. Hence, (λk xsk ) is a Cauchy sequence in C for all k ∈ N which
implies that it is a convergent sequence in C for all k ∈ N.
Let lims→∞ λk xsk = xk for each k ∈ N. Using the continuity of an Orlicz function and modulus,
we can have
Pk
j=0 λj (xsj − xj )
X
≤1 0 :
ρ(k + 1)
k
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for all s ≥ n0 , as j → ∞. It follows that (xs − x) ∈ ℓ′M (C, Λ). Since xs is in the space ℓ′M (C, Λ)
and ℓ′M (C, Λ) is a linear space, we have x = (xk ) ∈ ℓ′M (C, Λ).
(iii) From the above proof, one can easily conclude that kxs kC
M → 0 as s → ∞, which implies
that xsk → 0 as k → ∞ for each s ∈ N. This leads us to the desired result.
Proposition 2.7. The inequality
P
k
M
|
Pk
|
j=0 λj xj
kxkC
(k+1)
(M )
≤ 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 function M , we define
Pk
j=0 λj xj
X
< ∞ for each ρ > 0 .
M
hM (C, Λ) := x = (xk ) ∈ ω :
ρ(k + 1)
k
Clearly hM (C, Λ) is a subspace of ℓ′M (C, Λ).
Here and after we shall write k · k instead of k · kC
(M ) provided it does not lead to any confusion.
The topology of hM (C, Λ) is induced by k · k.
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Proposition 2.9. Let M be an Orlicz function. Then, (hM (C, Λ), k · k) is an AK-BK space.
Proof. First we show that hM (C, Λ) is an AK space. Let x = (xk ) ∈ hM (C, Λ). Then for each
ε ∈ (0, 1), we can find n0 such that
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X
i≥n0
Pk
j=0 λj xj
≤ 1.
M
ε(k + 1)
Pn
Define the nth section x(n) of a sequence x = (xk ) by x(n) = k=0 xk ek . Hence for n ≥ n0 , it
holds
Pk
λ
x
X
j
j
j=0
≤1
M
x − x(n)
= inf ρ > 0 :
ρ(k + 1)
k≥n0
Pk
j=0 λj xj
X
≤ 1 < ε.
≤ inf ρ > 0 :
M
ρ(k + 1)
k≥n
Thus, we can conclude that hM (C, Λ) is an AK space.
Next to show that hM (C, Λ) is a BK space, it is enough to show hM (C, Λ) is a closed subspace
of ℓ′M (C, Λ). For this, let (xn ) be a sequence in hM (C, Λ) such that kxn − xk → 0 as n → ∞ where
x = (xk ) ∈ ℓ′M (C, Λ). To complete the proof we need to show that x = (xk ) ∈ hM (C, Λ), i.e.,
Pk
j=0 λj xj
X
< ∞ for all ρ > 0.
M
ρ(k + 1)
k
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There is l corresponding to ρ > 0 such that kxl − xk ≤ ρ/2. Then, using the convexity of M ,
we have by Proposition 2.7 that
X
k
Pk
j=0 λj xj
M
ρ(k + 1)
P
P
Pk
k
k
2 j=0 λj xlj − 2 j=0 λj xlj − j=0 λj xj
X
=
M
2ρ(k + 1)
k
Pk
Pk
l
l
2
2
−
x
λ
x
λ
x
X
X
j
j
j
j
j
j=0
j=0
1
+ 1
M
M
≤
2
ρ(k + 1)
2
ρ(k + 1)
k
k
Pk
Pk
l
l
2
2
−
x
λ
x
λ
x
X
X
j
j
j=0 j j
j=0 j
1
+ 1
≤
M
M
2
ρ(k + 1)
2
kxl − xk(k + 1)
k
k
< ∞.
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Hence, x = (xk ) ∈ hM (C, Λ) and consequently hM (C, Λ) is a BK space.
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Proposition 2.10. Let M be an Orlicz function. If M satisfies the ∆2 -condition at 0, then
ℓ′M (C, Λ) is an AK space.
Proof. We shall show that ℓ′M (C, Λ) = hM (C, Λ) if M satisfies the ∆2 -condition at 0. To do
this it is enough to prove that ℓ′M (C, Λ) ⊂ hM (C, Λ). Let x = (xk ) ∈ ℓ′M (C, Λ). Then for some
ρ > 0,
X
k
This implies that
Pk
j=0 λj xj
< ∞.
M
ρ(k + 1)
Pk
j=0 λj xj
= 0.
lim M
k→∞
ρ(k + 1)
(2.4)
Choose an arbitrary l > 0. If ρ ≤ l, then
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P
kM
|
λj xj |
l(k+1)
Pk
j=0
< ∞. 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 all x ∈ (0, r]. By (2.4), there exists a positive integer n1 such that
Pk
j=0 λj xj
< rp r
for all k ≥ n1 .
M
ρ(k + 1)
2
2
We claim that
and thus
|
λj xj |
ρ(k+1)
Pk
j=0
≤ r for all k ≥ n1 . Otherwise, we can find d > n1 with
Pd
Z |Pdj=0 λj xj |/ρ(d+1)
j=0 λj xj
r r
≥
p(t)dt > p
M
,
ρ(d + 1)
2
2
r/2
|
j=0 λj xj |
ρ(d+1)
Pd
>r
a contradiction. Hence our claim is true. Then, we can find that
Pk
Pk
j=0 λj xj
j=0 λj xj
X
X
≤R
.
M
M
l(k + 1)
ρ(k + 1)
k≥n1
k≥n1
Hence,
Pk
λ
x
X
j=0 j j
< ∞ for all l > 0.
M
l(k + 1)
k
This completes the proof.
Proposition 2.11. Let M1 and M2 be two Orlicz functions. If M
1 and′ M2 are equivalent,
C
then ℓ′M1 (C, Λ) = ℓ′M2 (C, Λ) and the identity map I : ℓ′M1 (C, Λ), k · kC
M1 → ℓM2 (C, Λ), k · kM2 is
a topological isomorphism.
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Proof. Let α, β and b be constants from (1.3). Since M1 and M2 are equivalent, it is obvious
that (1.3) holds. Let us take any x = (xk ) ∈ ℓ′M2 (C, Λ). Then,
Pk
j=0 λj xj
X
< ∞ for some ρ > 0.
M2
ρ(k + 1)
k
|
j=0 λj xj |
lρ(k+1)
Pk
≤ b for all k ∈ N. Therefore,
Pk
Pk
α j=0 λj xj
j=0 λj xj
X
X
≤
1 with (b/2)µp2 (b/2) ≥ 1, where p2 is the kernel associated with M2 . Hence,
Pk
j=0 λj xj
≤ b µp2 b
M2
for all k ∈ N.
kxk2 (k + 1)
2
2
This implies that
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P
k
j=0 λj xj
µkxk2 (k + 1)
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Therefore,
X
k
≤ b for all k ∈ N.
Pk
α j=0 λj xj
< 1.
M1
µkxk2 (k + 1)
Hence, kxk1 ≤ (µ/α)kxk2 . Similarly, we can show that kxk2 ≤ βγkxk1 by choosing γ with γβ > 1
such that γβ(b/2)p1 (b/2) ≥ 1. Thus, αµ−1 kxk1 ≤ kxk2 ≤ βγkxk1 which establish that I 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 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
k
X
|λj xj |
0 such that
PIfk y ∈ ℓ∞
P
| j=0 λj yj |
j=0 λj yj |
< ∞. That is to say that y ∈ ℓ′M (C, Λ).
≤ b for k ∈ N. Hence, k M
ρ(k+1)
ρ(k+1)
Pk
On the other hand, let y ∈ ℓ′M (C, Λ). Then for some ρ > 0, we have
Pk
j=0 λj yj
X
< ∞.
M
ρ(k + 1)
k
| kj=0 λj yj |
Therefore,
≤ K < ∞ for a constant K > 0 and for all k ∈ N which yields that
k+1
′
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
Pkj=0 λj yj
M (u0 ) ≥ 1. Let y ∈ ℓ∞ (C, Λ) and α = kyk∞ = supk∈N k+1 > 0. For every ε ∈ (0, α),
P
dj=0 λj yj
we can determine d with d+1 > α − ε and so
P
X
k
Pk
j=0 λj yj b
≥ M α − εb .
M
α(k + 1)
α
Pk
P
| j=0 λj yj |b
Since M is continuous, k M
≥ 1, and so kyk∞ ≤ bkyk, otherwise
α(k+1)
Pk
P
| j=0 λj yj |
> 1 which contradicts Proposition 2.7. Again,
kM
kyk(k+1)
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X
k
Pk
j=0 λj yj b
=0
M
α(k + 1)
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.
P
| kj=0 λj yj |
≤ εb for all sufficiently
For the last part, let y ∈ hM (C, Λ). Then for any ε > 0,
k+1
large k, where b is a positive number such that p(b) > 0. Hence, y ∈ c0 (C, Λ). Conversely,
P
| kj=0 λj yj |
let y ∈ c0 (C, Λ). Then, for any ρ > 0,
< b/2 for all sufficiently large k. Thus,
ρ(k+1)
P
k
M
|
λj yj |
ρ(k+1)
Pk
j=0
< ∞ 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 function M , Λx = (λk xk ) ∈ ℓ∞ whenever x = (xk ) ∈ ℓ′M (C, Λ).
Proposition 2.14. Let M be an Orlicz function and p be the corresponding kernel of M . Define
the sets D1 and D2 by
)
(
X
λ−1 ak < ∞
D1 := a = (ak ) ∈ ω :
k
k
and
D2 := b = (bk ) ∈ ω : sup |λk bk | < ∞ .
k∈N
If p(x) = 0 for all x in [0, d], where d is some positive number, then the following statements hold:
(i) K¨
othe-Toeplitz dual of ℓ′M (C, Λ) is the set D1 .
(ii) K¨
othe-Toeplitz dual of D1 is the set D2 .
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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).
Let a = (ak ) ∈ D1 and x = (xk ) ∈ ℓ′M (C, Λ). Then, since
X
X
X
ak λ−1 |λk xk | ≤ sup |λk xk | ·
ak λ−1 < ∞,
|ak xk | =
k
k
k
Close
Quit
k
applying Lemma 2.13, we have a =
(2.7)
k∈N
α
(ak ) ∈ {ℓ′M (C, Λ)} .
α
D1 ⊂ {ℓ′M (C, Λ)}
k
Hence, the inclusion
holds.
α
Conversely, suppose that a = (ak ) ∈ {ℓ′M (C, Λ)} . Then, (ak xk ) ∈ ℓ1 , the space of all absolutely
convergent series, for every x = (xk ) ∈ ℓ′M (C, Λ). So, we can take xk = λ−1
for all
k ∈ N
k
P
′
ak λ−1 =
2.12
whenever
(x
)
∈
ℓ
(C,
Λ).
Therefore,
because
(x
)
∈
ℓ
(C,
Λ)
by
Proposition
k
∞
k
M
k
k
P
k |ak xk | < ∞ and we have a = (ak ) ∈ D1 . This leads us to the inclusion
α
{ℓ′M (C, Λ)} ⊂ D1 .
(2.8)
α
By combining the inclusion relations (2.7) and (2.8), we have {ℓ′M (C, Λ)} = D1 .
αα
Proposition 2.14 (ii) shows that {ℓ′M (C, Λ)} 6= ℓ′M (C, Λ) which leads us to the consequence
that ℓ′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 .
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Proof. Let a = (ak ) ∈ D1 and x = (xk ) ∈ hM (C, Λ). Then, since
X
X
X
ak λ−1 |λk xk | ≤ sup |λk xk | ·
ak λ−1 < ∞,
|ak xk | =
k
k
k
k
k∈N
k
α
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we have a = (ak ) ∈ {hM (C, Λ)} . Hence, the inclusion
(2.9)
holds.
D1 ⊂ {hM (C, Λ)}
α
α
Conversely, suppose that a = (ak ) ∈ {hM (C, Λ)} \ D1 . Then, there exists a strictly increasing
sequence (ni ) of positive integers ni such that
ni+1
X
−1
|ak | |λk |
> i.
k=ni +1
Define x = (xk ) by
xk :=
λ−1
k sgn 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
|ak xk | =
k
n1
X
ni+1
|ak xk | + · · · +
k=n0 +1
=
X
n1
X
ak λ−1 + · · · + 1
k
i
k=n0 +1
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|ak xk | + · · ·
k=ni +1
ni+1
X
ak λ−1 + · · ·
k
k=ni +1
which contradicts the hypothesis. Hence, a = (ak ) ∈ D1 . This leads us to the inclusion
(2.10)
α
{hM (C, Λ)} ⊂ D1 .
By combining the inclusion relations (2.9) and (2.10), we obtain the desired result
α
{hM (C, Λ)} = D1 .
This completes the proof.
3. Conclusion
The difference Orlicz spaces ℓM (∆, Λ) and ℓeM (∆, 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 ℓM . Indeed, if we define the infinite matrix C(λ) = {cnk (λ)}
via the multiplier sequence Λ = (λk ) by
λk
,
(0 ≤ k ≤ n),
cnk (λ) :=
n+1
0,
(k > n),
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for all k, n ∈ N, then the sequence spaces ℓ′M (C, Λ), c0 (C, Λ) and ℓ∞ (C, Λ) represent the domain
of the matrix C(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 spaces ℓ′M (C, Λ), c0 (C, Λ) and ℓ∞ (C, Λ) are
linearly isomorphic to the spaces ℓM , c0 and ℓ∞ , respectively.
Although some algebraic and topological properties of these new spaces are investigated, the
following further suggestions remain open:
C
(i) What is the relation between the norms k · kC
M and k · k(M ) ? Are they equivalent?
(ii) What is the relation between the spaces ℓM (C, Λ) and ℓ′M (C, Λ)? Do they coincide?
(iii) What are the β- and γ-duals of the spaces ℓ′M (C, Λ) and hM (C, Λ)?
(iv) Under which conditions an infinite matrix transforms the sets ℓ′M (C, Λ) or hM (C, Λ) to the
sequence spaces ℓ∞ and c?
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.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
Chen S., Geometry of Orlicz spaces. Dissertationes Math. 356 (1996), 1–204.
Dutta H., On Orlicz difference sequence spaces. SDU Journal of Science 5 (2010), 119–136.
Gribanov Y., On the theory of ℓM spaces. Uc. Zap. Kazansk un-ta, 117 (1957), 62–65, (in Russian).
Goes G.and Goes S., Sequences of bounded variation and sequences of Fourier coefficients. Math. Z. 118 (1970),
93–102.
Kamthan P. K., Bases in a certain class of Frechet spaces. Tamkang J. Math. 7 (1976), 41–49.
Kamthan P. K. and Gupta M., Sequence Spaces and Series. Marcel Dekker Inc., New York and Basel, 1981.
Krasnoselskii M. A. and Rutickii Y. B., Convex functions and Orlicz spaces. Groningen, Netherlands, 1961.
K¨
othe G. and Toeplitz O., Linear Raume mit unendlichvielen koordinaten and Ringe unenlicher Matrizen. J.
Reine Angew. Math. 171 (1934), 193–226.
Lindenstrauss J. and Tzafriri L., On Orlicz sequence spaces. Israel J. Math. 10 (1971), 379–390.
Musielak J., Orlicz Spaces and Modular Spaces. Lecture Notes in Mathematics 1083, Springer–Verlag, 1983.
Maddox I. J., Elements of Functional Analysis. The University Press, 2nd ed. Cambridge, 1988.
Tripathy B. C., A class of difference sequences related to the p-normed space ℓp . Demonstratio Math. 36(4)
(2003), 867–872.
H. Dutta, Department of Mathematics, Gauhati University, Kokrajhar Campus, Assam, India,
e-mail: hemen dutta08@rediffmail.com
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¨
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: fbasar@fatih.edu.tr, feyzibasar@gmail.com
`
BY CESARO
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 ℓ′M (C, Λ) and hM (C, Λ).
1. Preliminaries, Background and Notation
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By ω, we denote the space of all complex valued sequences. Any vector subspace of ω which
contains φ, the set of all finitely non–zero sequences is called a sequence 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-norm kxk∞ = 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 an
F K-space provided X is a complete linear metric space. An F K-space whose topology is normable
is called a BK-space.
A function M : [0, ∞) → [0, ∞) which is convex with M (u) ≥ 0 for u ≥ 0, and M (u) → ∞ as
u → ∞, is called as an Orlicz function. An Orlicz function M can always be represented in the
Received March 29, 2010; revised December 9, 2010.
2010 Mathematics 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.
following integral form
M (u) =
Z
u
p(t)dt,
0
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 and p(t) → ∞ as t → ∞ whenever Mu(u) ↑ ∞ as u ↑ ∞.
Consider the kernel p associated with the Orlicz function M and let
q(s) = sup{t : p(t) ≤ s}.
Then q possesses the same properties as the function p. Suppose now
Z x
q(s)ds.
Φ(x) =
0
Then, Φ is an Orlicz function. The functions M and Φ are called mutually complementary Orlicz
functions.
Now, we give the following well-known results.
Let M and Φ are mutually complementary Orlicz functions. Then, we have:
(i) For all u, y ≥ 0,
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(1.1)
(ii) For all u ≥ 0,
(1.2)
up(u) = M (u) + Φ(p(u)).
(iii) For all u ≥ 0 and 0 < λ < 1,
(1.3)
Quit
uy ≤ M (u) + Φ(y),
M (λu) < λM (u).
(Young’s inequality).
An Orlicz function M is said to satisfy the ∆2 -condition for small u or at 0 if for each k ∈ N, there
exist Rk > 0 and uk > 0 such that M (ku) ≤ Rk M (u) for all u ∈ (0, uk ]. Moreover, an Orlicz
function M 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 constants α, β and
b such that
(1.4)
M1 (αu) ≤ M2 (u) ≤ M1 (βu)
for all u ∈ [0, b].
Orlicz used the Orlicz function to introduce the sequence space ℓM (see Musielak [10]; Lindenstrauss and Tzafriri [9]), as follows
)
(
X |xk |
< ∞ for some ρ > 0 .
ℓM = x = (xk ) ∈ ω :
M
ρ
k
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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 space E, the multiplier sequence space E(Λ) associated with the
multiplier sequence Λ is defined by
E(Λ) = {x = (xk ) ∈ ω : Λx = (λk xk ) ∈ E} .
The scope for the studies on sequence spaces was extended by using the notion of associated
multiplier sequences. G. RGoes and S. Goes 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) 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.
Let C = (cnk ) be the Ces`
aro matrix of order one defined by
1
, 0 ≤ k ≤ n,
n
+
1
cnk :=
0,
k > n,
for all k, n ∈ N.
P
Definition 1.1. Let M be any Orlicz function and δ(M, x) := k M (|xk |), where x = (xk ) ∈ ω.
Then, we define the sets ℓeM (C, Λ) and ℓeM by
Pk
j=0 λj xj
X
2n+1 .
Pk λj xj
Without loss of generality, we can assume that j=0 k+1
, y n ≥ 0. Now, we can define a sequence
n
P yk
z = (zk ) by zk = n 2n+1 for all k ∈ N. By the convexity of Φ, we have
!
l
X
1 n
1
yk2
ykl
0
1
Φ
≤
Φ(y
)
+
Φ
y
+
y
+
·
·
·
+
k
k
2n+1 k
2
2
2l−1
n=0
··· ≤
l
X
n=0
1
2n+1
Φ(ykn )
for any positive integer l. Hence, using the continuity of Φ, we have
XX 1
X
X 1
n
Φ(zk ) ≤
δ(Φ, z) =
Φ(y
)
≤
= 1.
k
2n+1
2n+1
n
n
k
But for every l ∈ N, it holds
!
Pk
X
j=0 λj xj
k+1
k
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zk ≥
Hence
P
k
Pk
λj xj
k+1
j=0
X
k
=
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k
Pk
l X
X
n=0 k
j=0
λj xj
!
l
X
1 n
y
n+1 k
2
n=0
!
ykn
j=0 λj xj
≥ l.
k+1
2n+1
k+1
Pk
zk diverges and this implies that x ∈
/ ℓM (C, Λ), a contradiction. This leads
us to the required result.
The preceding result encourages us to introduce the following norm k · kC
M on ℓM (C, Λ).
Proposition 2.3. The following statements hold:
(i) ℓM (C, Λ) is a normed linear space under the norm k · kC
M defined by
)
!
(
P
k
X
j=0 λj xj
C
(2.2)
yk : δ(Φ, y) ≤ 1 .
kxkM = sup
k+1
k
(ii) ℓM (C, Λ) is a Banach space under the norm defined by (2.2).
(iii) ℓM (C, Λ) is a BK 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 that k · kC
M is a norm on the space
ℓM (C, Λ).
C
If x = 0, then obviously kxkC
M = 0. Conversely, assume kxkM = 0. Then using the definition of
the norm given by (2.2), we have
(
!
)
X Pk λj xj
j=0
yk : δ(Φ, y) ≤ 1 = 0.
sup
k+1
k
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P Pkj=0 λj xj
yk = 0 for all y such that δ(Φ, y) ≤ 1. Now considering y = ek
This implies that k
k+1
if Φ(1) ≤ 1 otherwise considering y = ek /Φ(1) so that λk xk = 0 for all k ∈ N, where ek is a
sequence whose only non-zero term is 1 in k th place for each k ∈ N. Hence we have xk = 0 for all
k ∈ N, since (λk ) is a sequence of non-zero scalars. Thus, x = 0.
C
C
C
C
It is easy to show that kαxkC
M = |α|kxkM and kx + ykM ≤ kxkM + kykM for all α ∈ C and
x, y ∈ ℓM (C, Λ).
(ii) Let (xs ) be any Cauchy sequence in the space ℓM (C, Λ). Then for any ε > 0, there exists a
positive integer n0 such that kxs − xt kC
M < ε for all s, t ≥ n0 . Using the definition of norm given
by (2.2), we get
(
#
"
)
X Pk λj xs − xt
j
j
j=0
sup
yk : δ(Φ, y) ≤ 1 < ε
k+1
k
for all s, t ≥ n0 . This implies that
#
"
X Pk λj xs − xt
j
j
j=0
yk < ε
k+1
k
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for all y with δ(Φ, y) ≤ 1 and for all s, t ≥ n0 . Now considering y = ek if Φ(1) ≤ 1, otherwise
considering y = ek /Φ(1) we have (λk xsk ) is a Cauchy sequence in C for all k ∈ N. Hence, it is a
convergent sequence in C for all k ∈ N.
Let lims→∞ λk xsk = xk for each k ∈ N. Using the continuity of the modulus, we can derive for
all s ≥ n0 as t → ∞, that
#
)
(
"
X Pk λj xs − xj
j
j=0
yk : δ(Φ, y) ≤ 1 < ε.
sup
k+1
k
It follows that (xs − x) ∈ ℓM (C, Λ). Since xs is in the space ℓM (C, Λ) and ℓM (C, Λ) is a linear
space, we have x = (xk ) ∈ ℓM (C, Λ).
s
(iii) From the above proof, one can easily conclude that kxs kC
M → 0 implies that xk → 0 for
each s ∈ N which 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
Pk
j=0 λj xj
X
≤1 .
ρ>0:
M
(2.3)
kxkC
(M ) = inf
ρ(k + 1)
k
C
Proof. Clearly kxkC
(M ) = 0 if x = 0. Now, suppose that kxk(M ) = 0. Then, we have
Pk
j=0 λj xj
X
≤ 1 = 0.
M
inf ρ > 0 :
ρ(k + 1)
k
This yields the fact for a given ε > 0 that there exists some ρε ∈ (0, ε) such that
Pk
j=0 λj xj
≤1
sup M
ρε (k + 1)
k∈N
which implies that
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for all k ∈ N. Thus,
Pk
j=0 λj xj
≤1
M
ρε (k + 1)
Pk
Pk
j=0 λj xj
j=0 λj xj
≤M
≤1
M
ε(k + 1)
ρε (k + 1)
i λ x
i λ x
| nj=0
| nj=0
j j|
j j|
for all k ∈ N. Suppose
=
6
0
for
some
n
∈
N.
Then,
→ ∞ as ε → 0. It
i
n
+1
ε(n
i +1)
i
Pn
i λ x
| j=0
j j|
→ ∞ as ε → 0 for some ni ∈ N, which is a contradiction. Therefore,
follows that M
ε(k+1)
P
|
P
λj xj |
k+1
Pk
j=0
= 0 for all k ∈ N. It follows that λk xk = 0 for all k ∈ N. Hence x = 0, since (λk ) is a
sequence of non-zero scalars.
Let x = (xk ) and y = (yk ) be any two elements of ℓM (C, Λ). Then, there exist ρ1 , ρ2 > 0 such
that
X
k
Pk
Pk
j=0 λj yj
j=0 λj xj
X
≤ 1.
M
M
≤ 1 and
ρ1 (k + 1)
ρ2 (k + 1)
k
Let ρ = ρ1 + ρ2 . Then by the convexity of M , we have
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X
k
Pk
Pk
λj xj
j=0 λj (xj + yj )
X
j=0
ρ
1
≤
M
M
ρ(k + 1)
ρ1 + ρ 2
ρ1 (k + 1)
k
Pk
λ
y
X
j
j
j=0
ρ2
≤ 1.
M
+
ρ 1 + ρ2
ρ2 (k + 1)
k
Hence, we have
kx + ykC
(M )
Pk
j=0 λj (xj + yj )
X
≤1
= inf ρ > 0 :
M
ρ
k
Pk
j=0 λj xj
X
≤1
M
≤ inf ρ1 > 0 :
ρ1
k
Pk
j=0 λj yj
X
≤1
M
+ inf ρ2 > 0 :
ρ2
k
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C
C
which gives that kx + ykC
(M ) ≤ kxk(M ) + kyk(M ) .
Finally, let α be any scalar and define r by r = ρ/|α|. Then,
Pk
j=0 αλj xj
X
≤1
kαxkC
ρ>0:
M
(M ) = inf
ρ(k + 1)
k
Pk
j=0 λj xj
X
≤ 1 = |α|kxkC
M
= inf r|α| > 0 :
(M ) .
r(k + 1)
k
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This completes the proof.
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Proposition 2.4 inspires us to define the following sequence space.
Definition 2.5. For any Orlicz function M , we define
Pk
j=0 λj xj
X
< ∞ for some ρ > 0 .
M
ℓ′M (C, Λ) := x = (xk ) ∈ ω :
ρ(k + 1)
k
Now, we can give the corresponding proposition on the space ℓ′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 norm k · kC
(M ) defined by (2.3).
′
(ii) ℓM (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).
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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 − xt kC
(M ) < ε/rδ for all s, t ≥ n0 . Using the definition of the norm given by (2.3), we get
Pk
j=0 λj xsj − xtj
X
≤1 < ε
inf ρ > 0 :
M
rδ
ρ(k + 1)
k
for all s, t ≥ n0 . This implies that
X
k
Pk
j=0 λj xsj − xtj
≤1
M s
kx − xt kC
(M ) (k + 1)
for all s, t ≥ n0 . It follows that
Pk
j=0 λj xsj − xtj
≤1
M s
kx − xt kC
(M ) (k + 1)
for all s, t ≥ n0 and for all k ∈ N. For r > 0 with M (rδ/2) ≥ 1, we have
Pk
j=0 λj xsj − xtj
≤ M rδ
M s
2
kx − xt kC
(M ) (k + 1)
for all s, t ≥ n0 and for all k ∈ N. Since M is non-decreasing, we have
P
k
j=0 λj xsj − xtj
rδ ε
ε
≤
·
=
k+1
2 rδ
2
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for all s, t ≥ n0 and for all k ∈ N. Hence, (λk xsk ) is a Cauchy sequence in C for all k ∈ N which
implies that it is a convergent sequence in C for all k ∈ N.
Let lims→∞ λk xsk = xk for each k ∈ N. Using the continuity of an Orlicz function and modulus,
we can have
Pk
j=0 λj (xsj − xj )
X
≤1 0 :
ρ(k + 1)
k
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for all s ≥ n0 , as j → ∞. It follows that (xs − x) ∈ ℓ′M (C, Λ). Since xs is in the space ℓ′M (C, Λ)
and ℓ′M (C, Λ) is a linear space, we have x = (xk ) ∈ ℓ′M (C, Λ).
(iii) From the above proof, one can easily conclude that kxs kC
M → 0 as s → ∞, which implies
that xsk → 0 as k → ∞ for each s ∈ N. This leads us to the desired result.
Proposition 2.7. The inequality
P
k
M
|
Pk
|
j=0 λj xj
kxkC
(k+1)
(M )
≤ 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 function M , we define
Pk
j=0 λj xj
X
< ∞ for each ρ > 0 .
M
hM (C, Λ) := x = (xk ) ∈ ω :
ρ(k + 1)
k
Clearly hM (C, Λ) is a subspace of ℓ′M (C, Λ).
Here and after we shall write k · k instead of k · kC
(M ) provided it does not lead to any confusion.
The topology of hM (C, Λ) is induced by k · k.
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Proposition 2.9. Let M be an Orlicz function. Then, (hM (C, Λ), k · k) is an AK-BK space.
Proof. First we show that hM (C, Λ) is an AK space. Let x = (xk ) ∈ hM (C, Λ). Then for each
ε ∈ (0, 1), we can find n0 such that
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X
i≥n0
Pk
j=0 λj xj
≤ 1.
M
ε(k + 1)
Pn
Define the nth section x(n) of a sequence x = (xk ) by x(n) = k=0 xk ek . Hence for n ≥ n0 , it
holds
Pk
λ
x
X
j
j
j=0
≤1
M
x − x(n)
= inf ρ > 0 :
ρ(k + 1)
k≥n0
Pk
j=0 λj xj
X
≤ 1 < ε.
≤ inf ρ > 0 :
M
ρ(k + 1)
k≥n
Thus, we can conclude that hM (C, Λ) is an AK space.
Next to show that hM (C, Λ) is a BK space, it is enough to show hM (C, Λ) is a closed subspace
of ℓ′M (C, Λ). For this, let (xn ) be a sequence in hM (C, Λ) such that kxn − xk → 0 as n → ∞ where
x = (xk ) ∈ ℓ′M (C, Λ). To complete the proof we need to show that x = (xk ) ∈ hM (C, Λ), i.e.,
Pk
j=0 λj xj
X
< ∞ for all ρ > 0.
M
ρ(k + 1)
k
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There is l corresponding to ρ > 0 such that kxl − xk ≤ ρ/2. Then, using the convexity of M ,
we have by Proposition 2.7 that
X
k
Pk
j=0 λj xj
M
ρ(k + 1)
P
P
Pk
k
k
2 j=0 λj xlj − 2 j=0 λj xlj − j=0 λj xj
X
=
M
2ρ(k + 1)
k
Pk
Pk
l
l
2
2
−
x
λ
x
λ
x
X
X
j
j
j
j
j
j=0
j=0
1
+ 1
M
M
≤
2
ρ(k + 1)
2
ρ(k + 1)
k
k
Pk
Pk
l
l
2
2
−
x
λ
x
λ
x
X
X
j
j
j=0 j j
j=0 j
1
+ 1
≤
M
M
2
ρ(k + 1)
2
kxl − xk(k + 1)
k
k
< ∞.
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Hence, x = (xk ) ∈ hM (C, Λ) and consequently hM (C, Λ) is a BK space.
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Proposition 2.10. Let M be an Orlicz function. If M satisfies the ∆2 -condition at 0, then
ℓ′M (C, Λ) is an AK space.
Proof. We shall show that ℓ′M (C, Λ) = hM (C, Λ) if M satisfies the ∆2 -condition at 0. To do
this it is enough to prove that ℓ′M (C, Λ) ⊂ hM (C, Λ). Let x = (xk ) ∈ ℓ′M (C, Λ). Then for some
ρ > 0,
X
k
This implies that
Pk
j=0 λj xj
< ∞.
M
ρ(k + 1)
Pk
j=0 λj xj
= 0.
lim M
k→∞
ρ(k + 1)
(2.4)
Choose an arbitrary l > 0. If ρ ≤ l, then
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P
kM
|
λj xj |
l(k+1)
Pk
j=0
< ∞. 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 all x ∈ (0, r]. By (2.4), there exists a positive integer n1 such that
Pk
j=0 λj xj
< rp r
for all k ≥ n1 .
M
ρ(k + 1)
2
2
We claim that
and thus
|
λj xj |
ρ(k+1)
Pk
j=0
≤ r for all k ≥ n1 . Otherwise, we can find d > n1 with
Pd
Z |Pdj=0 λj xj |/ρ(d+1)
j=0 λj xj
r r
≥
p(t)dt > p
M
,
ρ(d + 1)
2
2
r/2
|
j=0 λj xj |
ρ(d+1)
Pd
>r
a contradiction. Hence our claim is true. Then, we can find that
Pk
Pk
j=0 λj xj
j=0 λj xj
X
X
≤R
.
M
M
l(k + 1)
ρ(k + 1)
k≥n1
k≥n1
Hence,
Pk
λ
x
X
j=0 j j
< ∞ for all l > 0.
M
l(k + 1)
k
This completes the proof.
Proposition 2.11. Let M1 and M2 be two Orlicz functions. If M
1 and′ M2 are equivalent,
C
then ℓ′M1 (C, Λ) = ℓ′M2 (C, Λ) and the identity map I : ℓ′M1 (C, Λ), k · kC
M1 → ℓM2 (C, Λ), k · kM2 is
a topological isomorphism.
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Proof. Let α, β and b be constants from (1.3). Since M1 and M2 are equivalent, it is obvious
that (1.3) holds. Let us take any x = (xk ) ∈ ℓ′M2 (C, Λ). Then,
Pk
j=0 λj xj
X
< ∞ for some ρ > 0.
M2
ρ(k + 1)
k
|
j=0 λj xj |
lρ(k+1)
Pk
≤ b for all k ∈ N. Therefore,
Pk
Pk
α j=0 λj xj
j=0 λj xj
X
X
≤
1 with (b/2)µp2 (b/2) ≥ 1, where p2 is the kernel associated with M2 . Hence,
Pk
j=0 λj xj
≤ b µp2 b
M2
for all k ∈ N.
kxk2 (k + 1)
2
2
This implies that
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P
k
j=0 λj xj
µkxk2 (k + 1)
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Therefore,
X
k
≤ b for all k ∈ N.
Pk
α j=0 λj xj
< 1.
M1
µkxk2 (k + 1)
Hence, kxk1 ≤ (µ/α)kxk2 . Similarly, we can show that kxk2 ≤ βγkxk1 by choosing γ with γβ > 1
such that γβ(b/2)p1 (b/2) ≥ 1. Thus, αµ−1 kxk1 ≤ kxk2 ≤ βγkxk1 which establish that I 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 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
k
X
|λj xj |
0 such that
PIfk y ∈ ℓ∞
P
| j=0 λj yj |
j=0 λj yj |
< ∞. That is to say that y ∈ ℓ′M (C, Λ).
≤ b for k ∈ N. Hence, k M
ρ(k+1)
ρ(k+1)
Pk
On the other hand, let y ∈ ℓ′M (C, Λ). Then for some ρ > 0, we have
Pk
j=0 λj yj
X
< ∞.
M
ρ(k + 1)
k
| kj=0 λj yj |
Therefore,
≤ K < ∞ for a constant K > 0 and for all k ∈ N which yields that
k+1
′
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
Pkj=0 λj yj
M (u0 ) ≥ 1. Let y ∈ ℓ∞ (C, Λ) and α = kyk∞ = supk∈N k+1 > 0. For every ε ∈ (0, α),
P
dj=0 λj yj
we can determine d with d+1 > α − ε and so
P
X
k
Pk
j=0 λj yj b
≥ M α − εb .
M
α(k + 1)
α
Pk
P
| j=0 λj yj |b
Since M is continuous, k M
≥ 1, and so kyk∞ ≤ bkyk, otherwise
α(k+1)
Pk
P
| j=0 λj yj |
> 1 which contradicts Proposition 2.7. Again,
kM
kyk(k+1)
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X
k
Pk
j=0 λj yj b
=0
M
α(k + 1)
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.
P
| kj=0 λj yj |
≤ εb for all sufficiently
For the last part, let y ∈ hM (C, Λ). Then for any ε > 0,
k+1
large k, where b is a positive number such that p(b) > 0. Hence, y ∈ c0 (C, Λ). Conversely,
P
| kj=0 λj yj |
let y ∈ c0 (C, Λ). Then, for any ρ > 0,
< b/2 for all sufficiently large k. Thus,
ρ(k+1)
P
k
M
|
λj yj |
ρ(k+1)
Pk
j=0
< ∞ 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 function M , Λx = (λk xk ) ∈ ℓ∞ whenever x = (xk ) ∈ ℓ′M (C, Λ).
Proposition 2.14. Let M be an Orlicz function and p be the corresponding kernel of M . Define
the sets D1 and D2 by
)
(
X
λ−1 ak < ∞
D1 := a = (ak ) ∈ ω :
k
k
and
D2 := b = (bk ) ∈ ω : sup |λk bk | < ∞ .
k∈N
If p(x) = 0 for all x in [0, d], where d is some positive number, then the following statements hold:
(i) K¨
othe-Toeplitz dual of ℓ′M (C, Λ) is the set D1 .
(ii) K¨
othe-Toeplitz dual of D1 is the set D2 .
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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).
Let a = (ak ) ∈ D1 and x = (xk ) ∈ ℓ′M (C, Λ). Then, since
X
X
X
ak λ−1 |λk xk | ≤ sup |λk xk | ·
ak λ−1 < ∞,
|ak xk | =
k
k
k
Close
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k
applying Lemma 2.13, we have a =
(2.7)
k∈N
α
(ak ) ∈ {ℓ′M (C, Λ)} .
α
D1 ⊂ {ℓ′M (C, Λ)}
k
Hence, the inclusion
holds.
α
Conversely, suppose that a = (ak ) ∈ {ℓ′M (C, Λ)} . Then, (ak xk ) ∈ ℓ1 , the space of all absolutely
convergent series, for every x = (xk ) ∈ ℓ′M (C, Λ). So, we can take xk = λ−1
for all
k ∈ N
k
P
′
ak λ−1 =
2.12
whenever
(x
)
∈
ℓ
(C,
Λ).
Therefore,
because
(x
)
∈
ℓ
(C,
Λ)
by
Proposition
k
∞
k
M
k
k
P
k |ak xk | < ∞ and we have a = (ak ) ∈ D1 . This leads us to the inclusion
α
{ℓ′M (C, Λ)} ⊂ D1 .
(2.8)
α
By combining the inclusion relations (2.7) and (2.8), we have {ℓ′M (C, Λ)} = D1 .
αα
Proposition 2.14 (ii) shows that {ℓ′M (C, Λ)} 6= ℓ′M (C, Λ) which leads us to the consequence
that ℓ′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 .
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Proof. Let a = (ak ) ∈ D1 and x = (xk ) ∈ hM (C, Λ). Then, since
X
X
X
ak λ−1 |λk xk | ≤ sup |λk xk | ·
ak λ−1 < ∞,
|ak xk | =
k
k
k
k
k∈N
k
α
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we have a = (ak ) ∈ {hM (C, Λ)} . Hence, the inclusion
(2.9)
holds.
D1 ⊂ {hM (C, Λ)}
α
α
Conversely, suppose that a = (ak ) ∈ {hM (C, Λ)} \ D1 . Then, there exists a strictly increasing
sequence (ni ) of positive integers ni such that
ni+1
X
−1
|ak | |λk |
> i.
k=ni +1
Define x = (xk ) by
xk :=
λ−1
k sgn 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
|ak xk | =
k
n1
X
ni+1
|ak xk | + · · · +
k=n0 +1
=
X
n1
X
ak λ−1 + · · · + 1
k
i
k=n0 +1
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> 1 + · · · + 1 + · · · = ∞,
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|ak xk | + · · ·
k=ni +1
ni+1
X
ak λ−1 + · · ·
k
k=ni +1
which contradicts the hypothesis. Hence, a = (ak ) ∈ D1 . This leads us to the inclusion
(2.10)
α
{hM (C, Λ)} ⊂ D1 .
By combining the inclusion relations (2.9) and (2.10), we obtain the desired result
α
{hM (C, Λ)} = D1 .
This completes the proof.
3. Conclusion
The difference Orlicz spaces ℓM (∆, Λ) and ℓeM (∆, 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 ℓM . Indeed, if we define the infinite matrix C(λ) = {cnk (λ)}
via the multiplier sequence Λ = (λk ) by
λk
,
(0 ≤ k ≤ n),
cnk (λ) :=
n+1
0,
(k > n),
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for all k, n ∈ N, then the sequence spaces ℓ′M (C, Λ), c0 (C, Λ) and ℓ∞ (C, Λ) represent the domain
of the matrix C(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 spaces ℓ′M (C, Λ), c0 (C, Λ) and ℓ∞ (C, Λ) are
linearly isomorphic to the spaces ℓM , c0 and ℓ∞ , respectively.
Although some algebraic and topological properties of these new spaces are investigated, the
following further suggestions remain open:
C
(i) What is the relation between the norms k · kC
M and k · k(M ) ? Are they equivalent?
(ii) What is the relation between the spaces ℓM (C, Λ) and ℓ′M (C, Λ)? Do they coincide?
(iii) What are the β- and γ-duals of the spaces ℓ′M (C, Λ) and hM (C, Λ)?
(iv) Under which conditions an infinite matrix transforms the sets ℓ′M (C, Λ) or hM (C, Λ) to the
sequence spaces ℓ∞ and c?
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.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
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Dutta H., On Orlicz difference sequence spaces. SDU Journal of Science 5 (2010), 119–136.
Gribanov Y., On the theory of ℓM spaces. Uc. Zap. Kazansk un-ta, 117 (1957), 62–65, (in Russian).
Goes G.and Goes S., Sequences of bounded variation and sequences of Fourier coefficients. Math. Z. 118 (1970),
93–102.
Kamthan P. K., Bases in a certain class of Frechet spaces. Tamkang J. Math. 7 (1976), 41–49.
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H. Dutta, Department of Mathematics, Gauhati University, Kokrajhar Campus, Assam, India,
e-mail: hemen dutta08@rediffmail.com
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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: fbasar@fatih.edu.tr, feyzibasar@gmail.com