Document 10813132 Gen. Math. Notes, Vol. 6, No. 2, October 2011, pp.80-94
c
Publication, 2011
www.i-csrs.org
Available free online at http://www.geman.in
On Some New Almost Double Lacunary
∆m-Sequence Spaces Defined by
Orlicz Functions
Vakeel A. Khan1 and Sabiha Tabassum2
Department of Mathematics
A.M.U. Aligarh-202002 (INDIA)
1
E-mail: vakhan@math.com
2
E-mail: sabihatabassum@math.com
Abstract
In this paper we introduce a new concept for almost double lacunary ∆m sequence spaces defined by Orlicz function and give inclusion relations.The results here in proved are analogous to those by Ayhan Esi [General Mathematics
(2009),2(17) 53-66].
Keywords: Lacunary Sequence, Differene Double sequence; Orlicz Function,Strongly almost convergence.
1
Introduction
Let l∞ , c and c0 be the spaces of bounded, convergent and null sequences
x = (xk ), with complex terms, respectively, normed by kxk∞ = sup |xk |, where
k
k ∈ N.
A sequence x = (xk ) ∈ l∞ is said to be almost convergent if all Banach
limits of x = (xk ) coincide. in , it was shown that
ĉ =
n
1X
xk+s exists, uniformly in s .
x = (xk ) : lim
n→∞ n
k=1
On Some New Almost Double Lacunary...
81
In [16,17], Maddox defined a sequence x = (xk ) to be strongly convergent to a
number L if
n
1X
lim
|xk − L| = 0, uniformly in s
n→∞ n
k=1
By a lacunary sequence θ = (kr ), r=0,1,2,... where ko = 0, we mean an increasing sequence of non negative integers hr = (kr − kr−1 ) → ∞(r → ∞).
kr
The intervals determined by θ are denoted by Ir = (kr−1 , kr ] and ratio kr−1
will be denoted by qr .
The space of lacunary strongly convergent sequence Nθ was defined by Freedman et al. as follows
Nθ =
1 X
|xk − L| = 0, for some L .
x = (xk ) : lim
r→∞ hr
k∈I
r
The double lacunary sequence was defined by E.Savas and R.F.Patterson
as follows:
The double sequence θr,s = {(kr , ls )} is called double lacunary if there exist
two increasing sequence of integers such that
k0 = 0, hr = kr − kr−1 → ∞ as r → ∞
and
l0 = 0, h−
s = ls − ls−1 → ∞ as s → ∞.
The following intervals are determined by θ.
Ir = {(kr ) : kr−1 &lt; k &lt; kr }, Is = {(l) : ls−1 &lt; l &lt; ls }
Ir,s = {(k, l) : kr−1 &lt; k &lt; kr and ls−1 &lt; l &lt; ls },
ls
kr
, qs− = ls−1
and qr,s = qr qs− . We will denote the set of all lacunary
qr = kr−1
sequences by Nθr,s .
Let x = (xkl ) be a double sequence that is a double infinite array of elements
xkl . The space of double lacunary strongly convergent sequence is defined as
follows:
1 X
Nθr,s = x = (xk l) : lim
|xkl − L| = 0 for some L (see).
r,s hr,s
(k,l)∈Ir,s
82
Vakeel A. Khan et al.
Double sequences have been studied by V.A.Khan[8,9,10,11], Moricz and
In , Kizmaz defined the sequence spaces
Z(∆) = {x = (xk ) : (∆xk ) ∈ Z} for Z = l∞ , c, c0 , where ∆x = (∆xk ) =
(xk − xk+1 ). After Et. and Colak  generalized the difference sequence spaces
to the sequence spaces Z(∆m ) = {x = (xk ) : (∆m xk ) ∈ Z} for Z = l∞ , c, c0 ,
where m ∈ N, ∆0 x = (xk ),
∆x = (xk − xk+1 ), ∆m x = (∆m xk ) = (∆m−1 xk − ∆m−1 xk+1 ), and so that
m
∆ xk =
m
X
v=0
m
(−1)
.
v
v
An Orlicz Function is a function M : [0, ∞) → [0, ∞) which is continuous,
nondecreasing and convex with M (0) = 0, M (x) &gt; 0 for x &gt; 0 and M (x) →
∞, as x → ∞.
An Orlicz function M satisfies the ∆2 − condition (M ∈ ∆2 for short ) if
there exist constant k ≥ 2 and u0 &gt; 0 such that
M (2u) ≤ KM (u)
whenever |u| ≤ u0 .
An Orlicz function M can always be represented (see)in the integral
form
Rx
M (x) = q(t)dt, where q known as the kernel of M, is right differentiable for
0
t ≥ 0, q(t) &gt; 0 for t &gt; 0, q is non-decreasing and q(t) → ∞ as t → ∞.
Note that an Orlicz function satisfies the inequality
M (λx) ≤ λM (x) for all λ with 0 &lt; λ &lt; 1,
since M is convex and M (0) = 0.
Lindesstrauss and Tzafriri  used the idea of Orlicz sequence space;
lM
∞
X
|xk |
&lt; ∞, for some ρ &gt; 0
:= x ∈ w :
M
ρ
k=1
which is Banach space with the norm the norm
kxkM
∞
X
|xk |
= inf ρ &gt; 0 :
M
ρ
k=1
≤1 .
83
On Some New Almost Double Lacunary...
The space lM is closely related to the space lp , which is an Orlicz sequence
space with M (x) = xp for 1 ≤ p &lt; ∞. Orlicz function has been studied by
V.A.Khan[4,5,6,7] and many others.
The purpose of this paper is to introduce and study a concept of lacunary
almost generalized ∆m −convergence function and to examine these new sequence spaces which also generalize the well known Orlicz sequence space lM
and strongly summable sequence [C, 1, p], [C, 1, P ]0 and [C, 1, p]∞ (see).
Let M be an Orlicz function and p = (pk ) be any bounded sequence of strictly
positive real numbers. Ayhan Esi defined the following sequence spaces:
m
pk
∞ |∆ xk+m − L|
1X
= 0,
M
[ĉ, M, p](∆ ) = x = (xk ) : lim
n→∞ n
ρ
k=1
m
uniformly in m for some ρ &gt; 0 and L &gt; 0 .
m
[ĉ, M, p]0 (∆ ) =
m
pk
n X
|∆ xk+m |
x = (xk ) : lim
M
= 0,
n→∞
ρ
k=1
uniformly in m, for some ρ &gt; 0
.
m
pk
1X
|∆ xk+m |
[ĉ, M, p]∞ (∆ ) = x = (xk ) : sup
n M
&lt; ∞, for some ρ &gt; 0 .
ρ
n,m n
k=1
m
If x = (xk ) ∈ [ĉ, M, p](∆m ), we say that x = (xk ) is lacunary almost ∆m convergent to L with respect an Orlicz function M.
The folowing inequality will be used throughout the paper
|xkl + ykl |pkl ≤ K(|xkl |pkl + |ykl |pkl )
[1.1]
where xkl and ykl are complex numbers, K = max(1, 2H−1 ) and H = sup pkl &lt;
∞.
k,l
84
2
Vakeel A. Khan et al.
Main Results
In the following paper we introduce and examine the following spaces defined
by Orlicz function.
Definition 2.1. Let M be an Orlicz function and p = (pkl ) be any bounded
sequence of strictly positive real numbers. We have
m
pkl
1 X
|∆ xk+m,l+n − L|
θ
m
[ĉ2 , M, p] (∆ ) = x = (xkl ) : lim
M
= 0,
r,s hr,s
ρ
(k,l)∈Ir,s
uniformly in m and n, for some ρ &gt; 0 and L &gt; 0 .
[ĉ2 , M, p]θ0 (∆m )
=
1
x = (xkl ) : lim
r,s hr,s
X (k,l)∈Ir,s
|∆m xk+m,l+n |
M
ρ
pkl
= 0,
uniformly in m and n, for some ρ &gt; 0 .
[ĉ2 , M, p]θ∞ (∆m )
= x = (xkl ) : sup
1
r,s,m,n hr,s
X (k,l)∈Ir,s
|∆m xk+m,l+n |
M
ρ
pkl
&lt; ∞, for some ρ &gt; 0 .
where
∆m x = (∆m xkl ) = (∆m−1 xkl − ∆m−1 xk,l+1 − ∆m−1 xk+1,l + ∆m−1 xk+1,l+1 ),
(∆1 xkl ) = (∆xkl ) = (xkl − xk,l+1 − xk+1,l + xk+1,l+1 ),
∆0 x = (xk,l ),
for all k, l ∈ N,
and also this generalized difference double notion has the following binomial
representation:
m X
m
X
m
i+j m
∆ xkl =
(−1)
xk+i,l+j
i
j
m
i=0 j=0
.
If x = (xkl ) ∈ [ĉ2 , M, p]θ (∆m ), we say that x = (xkl ) is double lacunary
almost ∆m -convergent to L with respect an Orlicz function M.
85
On Some New Almost Double Lacunary...
In this section we prove some results involving the double sequence spaces
[ĉ2 , M, p]θ (∆m ), [ĉ2 , M, p]θ0 (∆m ) and [ĉ2 , M, p]θ∞ (∆m ).
Theorem 2.1. Let M be an Orlicz function and p = (pkl ) be a bounded sequence of strictly real numbers. Then [ĉ2 , M, p]θ (∆m ), [ĉ2 , M, p]θ0 (∆m ) and [ĉ2 , M, p]θ∞ (∆m )
are linear spaces over the set of complex numbers C.
Proof. Let x = (xkl ), y = (ykl ) ∈ [ĉ2 , M, p]θ0 (∆m ) and α, β ∈ C. Then there
exists positive ρ1 and ρ2 such that
m
pkl
1 X
|∆ xk+m,l+n |
lim
= 0, uniformly in m and n
M
r,s hr,s
ρ1
(k,l)∈Ir,s
and
1
lim
r,s hr,s
X |∆m yk+m,l+n | pkl
= 0, uniformly in m and n
M
ρ2
(k,l)∈Ir,s
Let ρ3 = max(2|α|ρ1 , 2|β|ρ2 ). Since M is non decreasing convex function, by
using equation [1.1], we have
pkl
P
|∆m αxk+m,l+n +βyk+m,l+n |
1
M
hr,s
ρ3
(k,l)∈Ir,s
pkl
1 X
|∆m αxk+m,l+n | |β∆m (yk+m,l+n )|
=
M
+
hr,s
ρ3
ρ3
(k,l)∈Ir,s
m
m
pkl
|∆ xk+m,l+n |
1 X
|∆ (yk+m,l+n )|
1 X
1
1
M
+K
M
≤K
hrs
2pkl
ρ1
hrs
2pkl
ρ2
(k,l)∈Ir,s
(k,l)∈Ir,s
m
m
pkl
|∆ xk+m,l+n |
1 X
|∆ (yk+m,l+n )|
1 X
≤K
M
+K
M
hrs
ρ1
hrs
ρ2
(k,l)∈Ir,s
(k,l)∈Ir,s
→ 0 as r, s → ∞ uniformly in m and n.
So αx + βy ∈ [ĉ2 , M, p]θ0 (∆m ). Hence [ĉ2 , M, p]θ0 (∆m ) is a linear space.
The proof for the cases [ĉ2 , M, p]θ (∆m ) and [ĉ2 , M, p]θ∞ (∆m ) are similar to the
above proof.
Theorem 2.2. For any Orlicz function M on a bounded double sequence
p = (pkl ) of strictly positive real numbers, [ĉ2 , M, p]θ0 (∆m ) is a topological linear space paranormed by
pkl
1
h(xkl ) = sup |xk1 |+sup |x1l |+inf ρ H :
hr,s
k
l
X (k,l)∈Ir,s
|∆m xk+m,l+n |
M
ρ
pkl H1
≤1
86
Vakeel A. Khan et al.
where H = max(1, sup pkl ) ≤ ∞.
k,l
Proof. Clearly h(xkl ) ≥ 0, for all x = (xkl ) ∈ [ĉ, M, p]θ0 (∆m )
Since M (0) = 0, we get h(0) = 0
h(−(xkl )) = h(xkl ).
Let (xkl ), (ykl ) ∈ [ĉ2 , M, p]θ0 (∆m ). Then there exist ρ1 &gt; 0 and ρ2 &gt; 0 such that
1
hr,s
X (k,l)∈Ir,s
and
1
hr,s
|∆m xk+m,l+n |
M
ρ
pkl H1
≤1
1
X |∆m yk+m,l+n | pkl H
M
≤1
ρ
(k,l)∈Ir,s
for each r, s, m and n.
Let ρ = ρ1 + ρ2 . Then we have,
1
hr,s
P
M
(k,l)∈Ir,s
≤
≤
1
hr,s
1
hr,s
|∆m xk+m,l+n +yk+m,l+n |
ρ
X (k,l)∈Ir,s
X (k,l)∈Ir,s
pkl H1
|∆m xk+m,l+n | + |∆m yk+m,l+n |
M
ρ1 + ρ2
pkl H1
m
|∆ xk+m,l+n |
ρ1
M
ρ1 + ρ2
ρ1
m
pkl H1
ρ2
|∆ yk+m,l+n |
+
M
ρ1 + ρ2
ρ2
By Minkowski’s Inequality
≤
+
ρ1
ρ1 + ρ2
ρ1
ρ1 + ρ2
1
hr,s
1
hr,s
1
X |∆m xk+m,l+n | pkl H
M
ρ1
(k,l)∈Ir,s
1
X |∆m yk+m,l+n | pkl H
M
≤1
ρ2
(k,l)∈Ir,s
Since ρ1 and ρ2 are non negative, so we have
h(xkl +ykl ) = sup |xk1 +yk1 |+sup |x1l +yy1 l |+inf ρ
k
l
pkl
H
:
1
hr,s
X (k,l)∈Ir,s
|∆m xkl + ykl |
M
ρ
pkl H1
≤1
87
On Some New Almost Double Lacunary...
pkl
1
≤ sup |xk1 | + sup |x1l | + inf ρ1 H :
hr,s
k
l
pkl
1
H
+ sup |yk1 | + sup |y1l | + inf ρ2 :
hr,s
k
l
X (k,l)∈Ir,s
|∆m xkl |
M
ρ1
pkl H1
≤1
1
X |∆m ykl | pkl H
≤1
M
ρ2
(k,l)∈Ir,s
= h(xkl ) + y(kl )
This implies that
h(xkl + ykl ) ≤ h(xkl ) + y(kl ).
Finally, we prove that the scalar multiplication is continuous. Let λ be any
complex number. By definition
h(λ(xkl )) = sup |λ(xk1) |+sup |λ(x1l )|+inf ρ
k
pkl
H
:
l
1
hr,s
pkl
1
= |λ| sup |xk1 |+|λ| sup |x1l |+inf (|λ|t) H :
hr,s
k
l
X (k,l)∈Ir,s
|∆m λxk+m,l+n |
M
ρ
(k,l)∈Ir,s
Theorem 2.3. Let M be an Orlicz function. If sup[M (x)]pkl &lt; ∞ for all fixed
k,l
x &gt; 0 then
[ĉ2 , M, p]θ0 (∆m ) ⊂ [ĉ2 , M, p]θ∞ (∆m ).
Proof. Let x = (xkl ) ∈ [ĉ2 , M, p]θ0 (∆m ). Then there exists some positive ρ1
such that
m
pkl
|∆ xk+m,l+n |
1 X
M
= 0, uniformly in m and n
lim
k,l hr,s
ρ1
(k,l)∈Ir,s
Define ρ = 2ρ1 . Since M is non decreasing and convex, by using (1) we have
pkl
P
|∆m xk+m,l+n |
1
sup hr,s
M
ρ
(k,l)∈Ir,s
1
= sup
r,s,m,n hr,s
X (k,l)∈Ir,s
|∆m xk+m,l+n − L + L|
M
ρ
≤1
1
X |∆m xk+m,l+n | pkl H
M
≤1
ρ
ρ
where t = |λ|
This complets the proof of the theorem.
r,s,m,n
pkl H1
pkl
88
Vakeel A. Khan et al.
1
≤ K sup
r,s,m,n hr,s
m
pkl
X 1
|∆ xk+m,l+n − L|
M
2pkl
ρ1
(k,l)∈Ir,s
1
+K sup
r,s,m,n hr,s
1
≤ K sup
r,s,m,n hr,s
pkl
X 1
|L|
M
p
kl
2
ρ1
(k,l)∈Ir,s
X (k,l)∈Ir,s
1
+K sup
r,s,m,n hr,s
|∆m xk+m,l+n − L|
M
ρ1
X (k,l)∈Ir,s
|L|
M
ρ1
pkl
pkl
&lt;1
Hence x = (xkl ) ∈ [ĉ2 , M, p]θ∞ (∆m ).
This complets the proof.
Theorem 2.4. Let 0 &lt; inf pk,l = h ≤ pk,l = H ≤ ∞ and M, M1 be Orlicz functions satisfying ∆2 -condition, then we have [ĉ2 , M1 , p]θ0 (∆m ) ⊂ [ĉ2 , M0 M1 , p]θ0 (∆m ), [ĉ2 , M1 , p]θ (∆m ) ⊂
[ĉ2 , M0 M1 , p]θ (∆m ) and [ĉ2 , M1 , p]θ∞ (∆m ) ⊂ [ĉ2 , M0 M1 , p]θ∞ (∆m ).
Proof. Let x = (xkl ) ∈ [ĉ2 , M0 M1 , p]θ0 (∆m ). Then we have
1
lim
r,s→∞ hr,s
X M1
(k,l)∈Ir,s
|∆m xk+m,l+n |
ρ
pkl
= 0, uniformly in m and n.
Let &gt; 0 andchoose δ with
0 &lt; δ &lt; 1 such that M (t) &lt; for 0 ≤ t ≤ δ.
|∆m xk+m,l+n |
for k, l, m, n ∈ N
Let yk,l = M1
ρ
We can write
1
hr,s
X
[M (yk,l )]pkl =
(k,l)∈Ir,s
1
hr,s
1
hr,s
X
[M (yk,l )]pkl +
(k,l)∈Ir,s ,yk,l≤δ
X
1
hr,s
X
(k,l)∈Ir,s ,yk,l&gt;δ
[M (yk,l )]pkl &lt; (k,l)∈Ir,s ,yk,l ≤δ
since M is continuous and M (t) &lt; for t ≤ δ.
For yk,l &gt; δ we use the fact that
yk,l
yk,l
&lt;1+
δ
δ
Since M is non decreasing and convex, it follows that
2 2 −1
−1
M (yk,l ) &lt; M (1 + δ yk,l ) = M
+ δ yk,l
2 2
yk,l &lt;
[M (yk,l )]pkl
[2.1]
89
On Some New Almost Double Lacunary...
1
1
&lt; M (2) + M (2δ −1 yk,l )
2
2
Since M satisfies ∆2 -condition, there is a constant K &gt; 2 such that
1
M (2δ −1 yk,l ) ≤ Kδ −1 yk,l M (2)
2
Hence
1
hr,s
X
pkl
[M (yk,l )]
(k,l)∈Ir,s ,yk,l &gt;δ
KM (2)
1
≤ max 1,
δ
hr,s
X
[(yk,l )]pkl
(k,l)∈Ir,s ,yk,l &gt;δ
→ 0 as r, s → ∞
[2.2]
By [2.1] and [2.2] we have x = (xk,l ∈ [ĉ2 , M0 M1 , p]θ0 (∆m ).
Similarly we can prove that
[ĉ2 , M1 , p]θ (∆m ) ⊂ [ĉ2 , M0 M1 , p]θ (∆m ) and [ĉ2 , M1 , p]θ∞ (∆m ) ⊂ [ĉ2 , M0 M1 , p]θ∞ (∆m ).
This complets the proof.
Taking M1 (x) in above theorem we have the following result.
Corollary 2.5. Let 0 &lt; inf pk,l = h ≤ pk,l = H ≤ ∞ and M be Orlicz function
satisfying ∆2 -condition, then we have [ĉ2 , p]θ0 (∆m ) ⊂ [ĉ2 , M, p]θ0 (∆m ), and [ĉ2 , p]θ∞ (∆m ) ⊂
[ĉ2 , M, p]θ∞ (∆m ).
Theorem 2.6. Let M be an Orlicz function. Then the following statements
are euivalent:
(i) [ĉ2 , p]θ∞ (∆m ) ⊂ [ĉ2 , M, p]θ∞ (∆m ).
(ii) [ĉ2 , p]θ0 (∆m ) ⊂ [ĉ2 , M, p]θ∞ (∆m ).
pkl
P
1
(iii) sup hr,s
M ρt
&lt; ∞ (t, ρ &gt; 0).
r,s
(k,l)∈Ir,s
Proof. (i) ⇒ (ii): It is obvious, since [ĉ2 , p]θ0 (∆m ) ⊂ [ĉ2 , p]θ∞ (∆m ).
(ii) ⇒ (iii): Let [ĉ2 , p]θ0 (∆m ) ⊂ [ĉ2 , M, p]θ∞ (∆m ).
Suppose that (iii) doesnot hold. Then for some t, ρ &gt; 0
pkl
t
1 X
M
= ∞.
sup
ρ
r,s hr,s
(k,l)∈Ir,s
90
Vakeel A. Khan et al.
and therefore we can find a subinterval Ir(i),s(j) of the set of interval Ir,s such
that
pkl
X
1
(ij)−1
&gt; ij
[2.3]
M
hr(i),s(j)
ρ
(k,l)∈Ir(i),s(j)
i = 1, 2, 3, ......, j = 1, 2, 3......
Define the double sequence x = (xkl ) by
(
(ij)−1
∆m xk+m,l+n =
0
(k, l) ∈ Ir(i),s(j)
(k, l) ∈
/ Ir(i),s(j) .
for all m, n ∈ N
/ [ĉ2 , M, p]θ∞ (∆m ).
Then x = (xkl ) ∈ [ĉ2 , p]θ0 (∆m ) but by equation [2.3] x = (xkl ) ∈
Which contradicts (ii). Hence (iii) must hold.
(iii) ⇒ (i): Let (iii) hold and x = (xkl ) ∈ [ĉ2 , p]θ∞ (∆m ).
Suppose that x = (xkl ) ∈
/ [ĉ2 , M, p]θ∞ (∆m ). Then
pkl
m
1 X
∆ xk+m,l+n
=∞
sup
M
ρ
r,s,m,n hr,s
[2.4]
(k,l)∈Ir,s
Let t = |∆m xk+m,l+n | for each k, l and fixed m, n then by equation [2.4]
pkl
1 X
t
sup
M
=∞
ρ
r,s,m,n hr,s
(k,l)∈Ir,s
Which contradicts (iii). Hence (i) must hold.
Theorem 2.7. Let 1 ≤ pkl ≤ sup pkl &lt; ∞ and M be an Orlicz function. Then
the following statements are equivalent:
(i) [ĉ2 , M, p]θ0 (∆m ) ⊂ [ĉ2 , p]θ0 (∆m ).
(ii) [ĉ2 , M, p]θ0 (∆m ) ⊂ [ĉ2 , p]θ∞ (∆m ).
pkl
P
1
(iii) inf hr,s
M ρt
&gt; 0 (t, ρ &gt; 0)
r,s
(k,l)∈Ir,s
Proof. (i) ⇒ (ii): It is obvious.
(ii) ⇒ (iii): Let (ii) hold. Suppose (iii) doesnot hold. Then
91
On Some New Almost Double Lacunary...
1
inf
r,s hr,s
X (k,l)∈Ir,s
pkl
t
M
= 0 (t, ρ &gt; 0)
ρ
So we can find a subinterval Ir(i),s(j) of the set of interval Ir,s such that
pkl
X
1
ij
&lt; (ij)−1
M
hr(i),s(j)
ρ
[2.5]
(k,l)∈Ir(i),s(j)
i = 1, 2, 3, ......, j = 1, 2, 3, .......
Define the double sequence x = (xkl ) by
(
(ij)−1
m
∆ xk+m,l+n =
0
(k, l) ∈ Ir(i),s(j)
(k, l) ∈
/ Ir(i),s(j) .
for all m, n ∈ N
Thus by equation [2.5], x = (xkl ) ∈ [ĉ2 , M, p]θ0 (∆m ) but by equation [2.3]
x = (xkl ) ∈
/ [ĉ2 , p]θ∞ (∆m ). Which contradicts (ii). Hence (iii) must hold.
(iii) ⇒ (i): Let (iii) hold and suppose that x = (xkl ) ∈ [ĉ2 , M, p]θ0 (∆m ), that is
1
lim
r,s→∞ hr,s
X |∆m xk+m,l+n pkl
M
= 0 uniformly in m and, for some ρ &gt; 0.
ρ
(k,l)∈Ir,s
[2.6]
Again suppose that x = (xkl ) ∈ [ĉ, p]θ0 (∆m ). Then for some &gt; 0 and a subinterval Ir(i),s(j) of the set interval Ir,s , we have |∆m xk+m,l+n | ≥ for all k, l ∈ N
and some i ≥ i0 , j ≥ j0 . Then, from the properties of the Orlicz function, we
can write
m
p
pkl
|∆ xk+m,l+n kl
M
≥M
ρ
ρ
and cosequently by equation [2.6]
lim
r,s→∞
1
hr(i),s(j)
X
(k,l)∈Ir(i),s(j)
pkl
t
M
=0
ρ
Which contradicts (iii). Hence (i) must hold.
Theorem 2.8. Let 0 &lt; pk,l ≤ qk,l for all k, l ∈ N and
Then,
[ĉ2 , M, q]θ (∆m ) ⊂ [ĉ2 , p]θ (∆m ).
qk,l
pk,l
be bounded.
92
Vakeel A. Khan et al.
Proof. Let x ∈ [ĉ2 , M, q]θ (∆m )
Write
tk,l
∆m xk+m,l+n
= M
ρ
pkl
p
k,l
and λk,l = qk,l
.
Since 0 &lt; pk,l ≤ qk,l , therefore 0 &lt; λk,l ≤ 1.
Take 0 &lt; λ ≤ λk,l .
Define
(
tk,l
uk,l =
0
vk,l
(
0
=
tk,l
tk,l ≥ 1
tk,l &lt; 1.
tk,l ≥ 1
tk,l &lt; 1.
So tk,l = uk,l + vk,l and
λ
λ
λ
tk,lk,l = uk,lk,l + vk,lk,l
Now it follows that
λ
λ
λ
uk,lk,l ≤ uk,l ≤ tk,l and vk,lk,l ≤ vk,l
Therfore
1
hr,s
λ
tk,lk,l
X
(k,l)∈Ir,s
1
≤
hr,s
X
(k,l)∈Ir,s
1
tk,l +
hr,s
X
λ
vk,l
(k,l)∈Ir,s
Hence x ∈ [ĉ2 , M, p]θ (∆m ).
By using above theorem it is easy to prove the following result.
Corollary 2.9(i). If 0 &lt; inf pk,l ≤ pk,l ≤ 1 for all k, l ∈ N then,
[ĉ2 , M, p]θ (∆m ) ⊂ [ĉ2 , p]θ (∆m ).
. (ii). If 0 ≤ pk,l ≤ sup pk,l ≤ ∞ for all k, l ∈ N then,
[ĉ2 , M ]θ (∆m ) ⊂ [ĉ2 , M, p]θ (∆m ).
Acknowledgements
The authors would like to record their gratitude to the reviewer for her careful
reading and making some useful corrections which improved the presentation
of the paper.
On Some New Almost Double Lacunary...
93
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