Paper-7-Acta24.2010. 117KB Jun 04 2011 12:03:14 AM

Acta Universitatis Apulensis
ISSN: 1582-5329

No. 24/2010
pp. 63-72

ON ORLICZ FUNCTIONS OF GENERALIZED DIFFERENCE
SEQUENCE SPACES

Ahmad H. A. Bataineh and Alaa A. Al-Smadi
Abstract. In this paper, we define the sequence spaces : [V, M, p, ∆nu , s],
[V, M, p, ∆nu , s]0 and [V, M, p, ∆nu , s]∞ , where for any sequence x = (xn ), the dif∞
ference sequence ∆x is given by ∆x = (∆xn )∞
n=1 = (xn − xn−1 )n=1 . We also examine some inclusion relations between these spaces and discuss some properties and
results related to them.
2000 Mathematics Subject Classification:40D05, 40A05.
1.Introduction and definitions
Let X be a linear space. A function p : X → R is called paranorm if the following
are satisfied :
(i) p(0) ≥ 0
(ii) p(x) ≥ 0 for all x ∈ X

(iii) p(x) = p(−x) for all x ∈ X
(iv) p(x + y) ≤ p(x) + p(y) for all x ∈ X ( triangle inequality )
(v) if (λn ) is a sequence of scalars with λn → λ (n → ∞) and (xn ) is a sequence
of vectors with p(xn −x) → 0 (n → ∞), then p(λn xn −λx) → 0 (n → ∞) ( continuity
of multiplication by scalars ).
A paranorm p for which p(x) = 0 implies x = 0 is called total. It is well known
that the metric of any linear metric space is given by some total paranorm (cf.[11]).
Let Λ = (λn ) a nondecreasing sequence of positive reals tending to infinity and
λ1 = 1 and λn+1 ≤ λn + 1.
The generalized de la Vall˙ee-Poussin means is defined by :
tn (x) =

1 X
xk ,
λn
k∈In

where In = [n − λn + 1, n]. A sequence x = (xk ) is said to be (V, λ)−summable to a
number l ( see [2] ) if tn (x) → l, as n → ∞.
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A.H. A. Bataineh, A. A. Al-Smadi - On Orlics functions of generalized difference...

We write
1 X
| xk |= 0}
n λn
k∈In
1 X
[V, λ] = {x = (xk ) : lim
| xk − le |= 0, for some l ∈ C}
n λn

[V, λ]0 = {x = (xk ) : lim

k∈In

and
[V, λ]∞ = {x = (xk ) : sup
n

1 X
| xk |< ∞}.
λn
k∈In

For the set of sequences that are strongly summable to zero, strongly summable
and strongly bounded by the de la Vall˙ee-Poussin method. If λn = n for n =
1, 2, 3, · · · , then these sets reduce to ω 0 , ω and ω ∞ introduced and studied by Maddox
[5].
Following Lidenstrauss and Tzafriri [4], we recall that an Orlicz function M is
continuous, convex, nondecreasing function defined for x ≥ 0 such that M (0) = 0
and M (x) ≥ 0 for x > 0.
If convexity of M is replaced by M (x + y) ≤ M (x) + M (y), then it is called a
modulus function, defined and studied by Nakano [8], Ruckle [10], Maddox [6] and
others.
An Orlicz function M is said to satisfy the ∆2 −condition for all values of u, if
there exist a constant K > 0 such that
M (2u) ≤ KM (u) (u ≥ 0).
It is easy to see that always K > 2. The ∆2 −condition is equivalent to the

satisfaction of the inequality
M (lu) ≤ KlM (u),
for all values of u and for l > 1.
Lidenstrauss and Tzafriri [4] used the idea of Orlicz function to construct the
Orlicz sequence space :
lM := {x = (xk ) :

∞
X
k=1

M(

| xk |
) < ∞, for some ρ > 0},
ρ

which is a Banach space with the norm :

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A.H. A. Bataineh, A. A. Al-Smadi - On Orlics functions of generalized difference...

k x kM = inf{ρ > 0 :

∞
X

M(

k=1

| xk |
) ≤ 1}.
ρ

xp , 1

If M (x) =
≤ p < ∞, the space lM coincide with the classical sequence

space lp .
Parashar and Choudhary [9] have introduced and examined some properties
of four sequence spaces defined by using an Orlicz function M, which generalized
the well-known Orlicz sequence space lM and strongly summable sequence spaces
[C, 1, p], [C, 1, p]0 and [C, 1, p]∞ .
Let M be an Orlicz function, p = (pk ) be any sequence of strictly positive real
numbers and u = (uk ) be any sequence such that uk 6= 0(k = 1, 2, · · · ) . Then
Alsaedi and Bataineh [1] defined the following sequence spaces :
P
|uk ∆xk −le| pk
[V, M, p, u, ∆] = {x = (xk ) : limn λ1n ∞
)] = 0,
k∈In [M (
ρ
for some l and ρ > 0},
[V, M, p, u, ∆]0 = {x = (xk ) : limn

0},
and

[V, M, p, u, ∆]∞

1
λn

P∞

k∈In [M (

|uk ∆xk | pk
)]
ρ

= 0, for some ρ >

∞
1 X
| uk ∆xk | pk
= {x = (xk ) : sup
[M (

)] < ∞, for some ρ > 0}.
λ
ρ
n
n
k∈In

Now, if n is a nonnegative integer and s is any real number such that s ≥ 0, then
we define the following sequence spaces :
n

[V, M, p, ∆nu , s] = {x = (xk ) : limn
for some l, ρ > 0 and s ≥ 0},

1
λn

P∞

k −s [M ( |∆u xρk −le| )]pk = 0,

[V, M, p∆nu , s]0 = {x = (xk ) : limn
for some ρ > 0 and s ≥ 0},
and

1
λn

P∞

k −s [M ( |∆uρxk | )]pk = 0,

[V, M, p, ∆nu , s]∞ = {x = (xk ) : supn
for some ρ > 0 and s ≥ 0},

k∈In

n

k∈In

1
λn

P∞

k∈In

n

k −s [M ( |∆uρxk | )]pk < ∞,

where u = (uk ) is any sequence such that uk 6= 0 for each k,and
∆0u x = uk xk ,
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A.H. A. Bataineh, A. A. Al-Smadi - On Orlics functions of generalized difference...

∆1u x = uk xk − uk+1 xk+1 ,
∆2u x = ∆(∆1u x),

..
.
∆nu x = ∆(∆n−1
u x),
so that
∆nu x

=

∆nuk xk

=

n
X
r=0

 
n
(−1)
uk+r xk+r .
r
r

If n = 0 and s = 0, then these gives the spaces of Alsaedi and Bataineh [1].
2.Main results
We prove the following theorems :
Theorem 1. For any Orlicz function M and any sequence p = (pk ) of strictly
positive real numbers, [V, M, p, ∆nu , s], [V, M, p, ∆nu , s]0 and [V, M, p, ∆nu , s]∞ are linear spaces over the set of complex numbers.
Proof. We shall prove only for [V, M, p, ∆nu , s]0 . The others can be treated similarly. Let x, y ∈ [V, M, p, ∆nu , s]0 and α, β ∈ C. In order to prove the result, we need
to find some ρ > 0 such that :
lim
n

1 X −s
| α∆nu xk + β∆nu yk | pk
)] = 0.
k [M (
λn
ρ
k∈In

Since x, y ∈ [V, M, p, ∆nu , s]0 , there exists some positive ρ1 and ρ2 such that :
lim
n

1 X −s
1 X −s
| ∆n xk | pk
| ∆n yk | pk
)] = 0 and lim
)] = 0.
k [M ( u
k [M ( u
n λn
λn
ρ1
ρ2
k∈In

k∈In

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A.H. A. Bataineh, A. A. Al-Smadi - On Orlics functions of generalized difference...

Define ρ = max(2 | α | ρ1 , 2 | β | ρ2 ). Since M is nondecreasing and convex,
| α∆nu xk + β∆nu yk | pk
1 X −s
)]
k [M (
λn
ρ
k∈In

≤

1 X −s
| α∆nu xk | | β∆nu yk | pk
+
)]
k [M (
λn
ρ
ρ

≤

| ∆n yk | pk
| ∆n xk |
1 X −s 1
) + M( u
)]
[M ( u
k
p
λn
2k
ρ1
ρ2

≤

| ∆n yk | pk
| ∆n xk |
1 X −s
) + M( u
)]
k [M ( u
λn
ρ1
ρ2

k∈In

k∈In

≤ K.

k∈In

1 X −s
1 X −s
| ∆n xk | pk
| ∆n yk | pk
k [M ( u
k [M ( u
)] + K.
)] → 0,
λn
ρ1
λn
ρ2
k∈In

k∈In

as n → ∞, where K = max(1, 2H−1 ), H = sup pk , so that αx+βy ∈ [V, M, p, ∆nu , s]0 .
This completes the proof.
Theorem 2. For any Orlicz function M and a bounded sequence p = (pk ) of
strictly positive real numbers, [V, M, p, ∆nu , s]0 is a total paranormed space with :
g(x) = inf{ρpn /H : (

1 X −s
| ∆n xk | pk 1/H
k [M ( u
)] )
≤ 1, n = 1, 2, 3, · · · },
λn
ρ
k∈In

where H = max(1, sup pk ).
Proof. Clearly g(x) = g(−x). By using Theorem 2.1, for α = β = 1, we get
g(x+y) ≤ g(x)+g(y). Since M (0) = 0, we get inf{ρpn /H } = 0 for x = 0. Conversely,
suppose g(x) = 0, then :
inf{ρpn /H : (

1 X −s
| ∆n xk | pk 1/H
)] )
≤ 1} = 0.
k [M ( u
λn
ρ
k∈In

This implies that for a given ǫ > 0, there exists some ρǫ (0 < ρǫ < ǫ) such that :
(

| ∆n xk | pk 1/H
1 X −s
k [M ( u
)] )
≤ 1.
λn
ρǫ
k∈In

Thus,
(

1 X −s
1 X −s
| ∆n xk | pk 1/H
| ∆n xk | pk 1/H
)] )
≤(
)] )
≤ 1, for each n.
k [M ( u
k [M ( u
λn
ǫ
λn
ρǫ
k∈In

k∈In

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A.H. A. Bataineh, A. A. Al-Smadi - On Orlics functions of generalized difference...
n

Suppose that xnm 6= 0 for some m ∈ In , then ( ∆u xǫ nm ) → ∞. It follows that :
(

1 X −s
| ∆n xn |
k [M ( u m )]pk )1/H → ∞
λn
ǫ
k∈In

which is a contradiction. Therefor xnm = 0 for all m. Finally we prove that scalar
multiplication is continuous. Let µ be any complex number, then by definition,
g(µx) = inf{ρpn /H : (

1 X −s
| µ∆nu xk | pk 1/H
)] )
≤ 1, n = 1, 2, 3, · · · }.
k [M (
λn
ρ
k∈In

Then
n
P
g(µx) = inf{(| µ | t)pn /H : ( λ1n k∈In k −s [M ( |µ∆tu xk | )]pk )1/H ≤ 1 ≤ 1, n = 1, 2, 3, · · · },
where t = ρ/ | µ | . Since | µ |pn ≤ max(1, | µ |sup pn ), we have
g(µx) ≤ (max(1, | µ |sup pn ))1/H . inf{(t)pn /H
1 X −s
| ∆n xk | pk 1/H
:(
k [M ( u
)] )
≤ 1, n = 1, 2, 3, · · · }
λn
t
k∈In

which converges to zero as x converges to zero in [V, M, p, ∆nu , s]0 .
Now suppose µm → 0 and x is fixed in [V, M, p, ∆nu , s]0 . For arbitrary ǫ > 0, let
N be a positive integer such that
| ∆n xk | pk
1 X −s
)] < (ǫ/2)H for some ρ > 0 and all n > N.
k [M ( u
λn
ρ
k∈In

This implies that
| ∆n xk | pk
1 X −s
)] < ǫ/2 for some ρ > 0 and all n > N.
k [M ( u
λn
ρ
k∈In

Let 0 N, we get
1 X −s
1 X −s
| µ∆nu xk | pk
| ∆n xk | pk
)] <
)] < (ǫ/2)H .
k [M (
k [| µ | M ( u
λn
ρ
λn
ρ
k∈In

k∈In

Since M is continuous everywhere in [0, ∞), then for n ≤ N,
f (t) =

1 X −s
| t∆nu xk | pk
)]
k [M (
λn
ρ
k∈In

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A.H. A. Bataineh, A. A. Al-Smadi - On Orlics functions of generalized difference...
is continuous at zero. So there exists 1 > δ > 0 such that | f (t) |< (ǫ/2)H for
0 < t < δ.
Let K be such that | µm |< δ for m > K and n ≤ N, then
(

1 X −s
| µ ∆n xk | pk 1/H
)] )
< ǫ/2.
k [M ( m u
λn
ρ
k∈In

Thus
(

| µ ∆n xk | pk 1/H
1 X −s
)] )
< ǫ,
k [M ( m u
λn
ρ
k∈In

for m > K and all n, so that g(µx) → 0 (µ → 0).
Theorem 3. For any Orlicz function M which satisfies the ∆2 −condition, we
have [V, λ, ∆nu , s] ⊆ [V, M, ∆nu , s], where
P
[V , λ, ∆nu , s] = {x = (x k ) : limn λ1n k∈In k −s | ∆nu xk −le |= 0 , for some l ∈C}.
Proof. Let x ∈ [V, λ, ∆nu , s]. Then
Tn =

1 X −s
k | ∆nu xk − le |→ 0 as n → ∞, for some l.
λn
k∈In

Let ǫ > 0 and choose δ with 0 < δ < 1 such that M (t) < ǫ for 0 ≤ t ≤ δ. Write
yk =| ∆nu xk − le | and consider
X X
1 X −s
k M (| yk |) =
+
,
λn
1

k∈In

2

where the first summation over yk ≤ δ and the second over yk > δ. Since M is
continuous,
X
< λn ǫ
1

and for yk < δ, we use the fact that yk < yk /δ < 1 + yk /δ. Since M is nondecreasing
and convex, it follows that
1
1
M (yk ) < M (1 + δ −1 yk ) < M (2) + M (2δ −1 yk ).
2
2
Since M satisfies the ∆2 −condition, there is a constant K > 2 such that M (2δ −1 yk ) ≤
−1
1
2 Kδ yk M (2), therefor
1
1
Kδ −1 yk M (2) + Kδ −1 yk M (2)
2
2
= Kδ −1 yk M (2).

M (yk ) <

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A.H. A. Bataineh, A. A. Al-Smadi - On Orlics functions of generalized difference...

Hence
X
2

which together with
the proof.

P

1

M (yk ) ≤ Kδ −1 M (2)λn Tn

≤ ǫλn yields [V, λ, ∆nu , s] ⊆ [V, M, ∆nu , s]. This completes

The method of the proof of Theorem 3 shows that for any Orlicz function
M which satisfies the ∆2 −condition, we have [V, λ, ∆nu , s]0 ⊆ [V, M, ∆nu , s]0 and
[V, λ, ∆nu , s]∞ ⊆ [V, M, ∆nu , s]∞ , where
[V, λ, ∆nu , s]0 = {x = (xk ) : lim

1 X −s
k | ∆nu xk |= 0},
λn

[V, λ, ∆nu , s]∞ = {x = (xk ) : sup

1 X −s
k | ∆nu xk |< ∞}.
λn

n

k∈In

and

n

k∈In

Theorem 4. Let 0 ≤ pk ≤ qk and (qk /pk ) be bounded. Then [V, M, q, ∆nu , s] ⊂
[V, M, p, ∆nu , s].
Proof. The proof of Theorem 4 used the ideas similar to those used in
proving Theorem 7 of Parashar and Choudhary [9].Mursaleen [7] introduced the
concept of statistical convergence as follows :
A sequence x = (xk ) is said to be λ−statistically convergent or sλ −statistically
convergent to L if for every ǫ > 0,
lim
n

1
| {k ∈ In :| xk − L |≥ ǫ} |= 0,
λn

where the vertical bars indicates the number of elements in the enclosed set. In this
case we write sλ −lim x = L or xk → L (sλ ) and sλ = {x : ∃L ∈ R:sλ −lim x = L}.In
a similar way, we say that a sequence x = (xk ) is said to be (λ, ∆nu )−statistically
convergent or sλ (∆nu )−statistically convergent to L if for every ǫ > 0,
lim
n

1
| {k ∈ In :| ∆nu xk − Le |≥ ǫ} |= 0,
λn

where the vertical bars indicates the number of elements in the enclosed set. In this
case we write sλ (∆nu ) − lim x = Le or ∆nu xk → Le (sλ ) and sλ (∆nu ) = {x : ∃L ∈
R:sλ (∆nu ) − lim x = Le}.
Theorem 5. For any Orlicz function M, [V, M, ∆nu , s] ⊂ sλ (∆nu ).

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A.H. A. Bataineh, A. A. Al-Smadi - On Orlics functions of generalized difference...
Proof. Let x ∈ [V, M, ∆nu , s] and ǫ > 0. Then
1 X −s
| ∆n xk − le |
k M( u
) ≥
λn
ρ
k∈In

≥

1
λn

X

k∈In ,|∆n
u xk −le|≥ǫ

M(

| ∆nu xk − le |
)
ρ

1
M (ǫ/ρ). | {k ∈ In :| ∆nu xk − le |≥ ǫ} |
λn

from which it follows that x ∈ sλ (∆nu ).
To show that sλ (∆nu ) strictly contain
[V, M, ∆nu , s], we proceed as in [7]. We
√
define x = (xk ) by (xk ) = k if n − [ λn ] + 1 ≤ k ≤ n and (xk ) = 0 otherwise. Then
x∈
/ l∞ (∆nu , s) and for every ǫ (0 < ǫ ≤ 1),
√
1
[ λn ]
n
| {k ∈ In :| ∆u xk − 0 |≥ ǫ} |=
→ 0 as n → ∞
λn
λn
i.e. x → 0 (sλ (∆nu )), where [ ] denotes the greatest integer function. On the other
hand,
| ∆n xk − 0 |
1 X −s
) → ∞ as n → ∞
k M( u
λn
ρ
k∈In

i.e. xk 9 0

[V, M, ∆nu , s].

This completes the proof.

References
[1] Alsaedi, Ramzi S. and Bataineh, Ahmad H. A., Some generalized difference
sequence spaces defined by a sequence of Orlicz functions, Aust. J. Math. Analy.
& Appli. (AJMAA), Volume 5, Issue 2, Article 2, pp. 1-9, 2008.
¨
[2] Leindler, L., Uber
de la Pousinsche summierbarkeit allgemeiner Orthogonalreihen, Acta Math. Hung. 16(1965),375-378.
[3] Lindenstrauss, J., Some aspects of the theory of Banach spaces, Advances in
Math., 5 ( 1970 ), 159-180.
[4] Lindenstrauss, J., and Tzafriri, L., On Orlicz sequence spaces, Israel J. Math.,
10 ( 3 ) ( 1971 ), 379-390.
[5] Maddox, I. J., Spaces of strongly summable sequences, Quart. J. Math. Oxford
Ser. (2), 18(1967), 345-355.
[6] Maddox, I. J., Sequence spaces defined by a modulus, Math. Proc. Camb. Phil.
Soc., 100(1986), 161-166.
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A.H. A. Bataineh, A. A. Al-Smadi - On Orlics functions of generalized difference...

[7] Mursaleen, λ-statistical convergence, Math. Slovaca 50, No. 1(2000), 111-115.
[8] Nakano, H., Concave modulus, J. Math. Soc. Japan 5(1953), 29-49.
[9] Parashar, S. D., and Choudhary, b., Sequence spaces defined by Orlicz functions,
Indian J. pure appl. Math., 25 ( 4 ) ( 1994 ), 419-428.
[10] Ruckle, W. H., FK spaces in which the sequence of coordinate vectors is bounded,
Can. J. Math., 25 ( 5 ) ( 1973 ), 973-978.
[11] Wilansky, A., Summability through Functional Analysis, North-Holland Math.
Stud., 85(1984).
Ahmad H. A. Bataineh and Alaa A. Al-Smadi
Faculty of Science
Department of Mathematics
Al al-Bayt University
P.O. Box : 130095 Mafraq, Jordan
email : ahabf2003@yahoo.ca

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