# A New Double Sequence Spacem

Authors: Birsen Sağır; Cenap Duyar; Oğuz Oğur
Journal: International Journal of Analysis
 (2014)

---

## Abstract

In this work we introduce new spacesm
2
(
F
,
ϕ
,
p
) of double sequences defined by a double sequence of modulus functions, and we study some properties of this space.


---

## Body

## 1. Introduction

In this work, byw and w
2, we denote the spaces of single complex sequences and double complex sequences, respectively. ℕ and ℂ denote the set of positive integers and complex numbers, respectively. If, for all ɛ
>
0, there is n
ɛ
∈
ℕ such that ∥
x
k
,
l
-
a
∥
X
<
ɛ where k
>
n
ɛ and l
>
n
ɛ, then a double sequence {
x
k
,
l
} is said to be converge (in terms of Pringsheim) to a
∈
ℂ. A real double sequence {
x
k
,
l
} is nondecreasing, if x
k
,
l
≤
x
p
,
q for (
k
,
l
)
<
(
p
,
q
). A double series is infinity sum ∑
k
,
l
=
1
∞
‍
x
k
,
l and its convergence implies the convergence by |
·
| of partial sums sequence {
S
n
,
m
}, where S
n
,
m
=
∑
k
=
1
n
‍
∑
l
=
1
m
‍
x
k
,
l (see [1–3]).For1
≤
p
<
∞, ℓ
p
(
2
) denote the space of sequences x
=
{
x
k
,
l
} such that
(1)
∑
k
,
l
=
1
∞
‍
|
x
k
,
l
|
p
<
∞
.
(see [4]).A double sequencex
=
{
x
k
,
l
} is said to be bounded if and only if sup
⁡
k
,
l
|
x
k
,
l
|
<
∞. The space of all bounded double sequences is denoted by ℓ
∞
(
2
). It is known that ℓ
∞
(
2
) is Banach space (see [5, 6]).A double sequence spaceE is said to be normal if (
y
k
l
)
∈
E whenever |
y
k
l
|
≤
|
x
k
l
| for all k
,
l
∈
ℕ and (
x
k
l
)
∈
E.The double sequence spaces in the various forms were introduced and studies by Khan and Tabassum in [7–14], by Khan in [15], and by Khan et al. in [16, 17].Now letφ
s be a family of subsets σ having most elements s in ℕ. Also φ
s
,
t denote the class of subsets σ
=
σ
1
×
σ
2 in ℕ
×
ℕ such that the elements of σ
1 and σ
2 are most s and t, respectively. Besides {
ϕ
k
,
l
} is taken as a nondecreasing double sequence of the positive real numbers such that
(2)
k
ϕ
k
+
1
,
l
≤
(
k
+
1
)
ϕ
k
,
l
,
l
ϕ
k
,
l
+
1
≤
(
l
+
1
)
ϕ
k
,
l
.
(see [18]).Letx
=
{
x
k
,
l
} be a double sequence. A set S
(
x
) is defined by
(3)
S
(
x
)
=
{
{
x
π
1
(
k
)
,
π
2
(
k
)
}
:
π
1
and
π
2
are
permutations
of
ℕ
}
.
A double sequence space E is said to be symmetric if u
=
(
u
k
l
)
∈
E and ∥
u
∥
=
∥
x
∥ whenever x
=
(
x
k
l
)
∈
E and u
∈
S
(
x
).A BK-space is a Banach sequence spaceE in which the coordinate maps are continuous.A functionf
:
[
0
,
∞
)
→
[
0
,
∞
) is said to be a modulus function if it satisfies the following:(1)
f
(
x
)
=
0 if and only if x
=
0;
(2)
f
(
x
+
y
)
≤
f
(
x
)
+
f
(
y
) for all x
,
y
∈
[
0
,
∞
);
(3)
f is increasing;
(4)
f is continuous from right at 0.It follows thatf is continuous on [
0
,
∞
). The modulus function may be bounded or unbounded. For example, if we take f
(
x
)
=
x
/
(
x
+
1
), then f
(
x
) is bounded. But, for 0
<
p
<
1, f
(
x
)
=
x
p is not bounded.The BK-spacesm
(
ϕ
), introduced by Sargent in [19], is in the form
(4)
m
(
ϕ
)
=
{
x
=
{
x
k
}
∈
w
:
∥
x
∥
m
(
ϕ
)
=
sup
⁡
s
≥
1
,
σ
∈
φ
s
1
ϕ
s
∑
k
∈
σ
‍
|
x
k
|
<
∞
}
.Sargent studied some properties of this space and examined relationship between this space andl
p-space.The spacem
(
ϕ
) was extended to m
(
ϕ
,
p
) by Tripathy and Sen [20] as follows:
(5)
m
(
ϕ
,
p
)
=
{
(
∑
k
∈
σ
‍
|
x
k
|
p
)
1
/
p
x
=
{
x
k
}
∈
w
:
∥
x
∥
m
(
ϕ
,
p
)
w
=
sup
⁡
s
≥
1
,
σ
∈
φ
s
1
ϕ
s
(
∑
k
∈
σ
‍
|
x
k
|
p
)
1
/
p
<
∞
}
.
Recently, Raj et al. [21] introduced and studied the following sequence space m
(
F
,
ϕ
,
p
).LetF
=
(
f
k
) be a sequence of modulus functions. Then the space m
(
F
,
ϕ
,
p
) is defined by
(6)
m
(
F
,
ϕ
,
p
)
=
{
x
=
{
x
k
}
∈
w
:
sup
⁡
s
≥
1
,
σ
∈
φ
s
1
ϕ
s
(
∑
k
∈
σ
‍
[
f
k
(
|
x
k
|
ρ
)
]
p
)
1
/
p
w
w
e
w
w
<
∞
,
for
some
ρ
>
0
(
∑
k
∈
σ
‍
[
f
k
(
|
x
k
|
ρ
)
]
p
)
1
/
p
}
.In this work we introduce sequence spacesm
2
(
F
,
ϕ
,
p
) defined by
(7)
m
2
(
F
,
ϕ
,
p
)
=
{
{
∑
i
∈
σ
1
‍
∑
j
∈
σ
2
‍
[
f
i
,
j
(
|
x
i
,
j
|
ρ
)
]
p
}
x
=
(
x
k
l
)
∈
w
2
:
(
ew
)
sup
⁡
(
s
,
t
)
≥
(
1,1
)
sup
⁡
σ
1
×
σ
2
∈
φ
s
t
1
ϕ
s
t
{
∑
i
∈
σ
1
‍
∑
j
∈
σ
2
‍
[
f
i
,
j
(
|
x
i
,
j
|
ρ
)
]
p
}
1
/
p
=
(
e
)
<
∞
for
ρ
>
0
{
∑
i
∈
σ
1
‍
∑
j
∈
σ
2
‍
[
f
i
,
j
(
|
x
i
,
j
|
ρ
)
]
p
}
}
,
where F
=
(
f
i
,
j
) is a double sequence of modulus functions.


## 2. Main Results

The result stated in the first theorem is not hard. So, we give it without proof.Theorem 1.
The sequence spacem
2
(
F
,
ϕ
,
p
) is a linear space.Theorem 2.
The sequence spacesm
2
(
F
,
ϕ
,
p
) are complete.Proof.
Let{
x
(
i
)
} be a double Cauchy sequence in m
2
(
F
,
ϕ
,
p
) such that x
(
i
)
=
{
x
k
,
l
(
i
)
}
k
,
l
=
1
∞ for all i
∈
ℕ. Then
(8)
sup
⁡
(
s
,
t
)
≥
(
1,1
)
sup
⁡
σ
1
×
σ
2
∈
φ
s
t
1
ϕ
s
t
{
∑
k
∈
σ
1
‍
∑
l
∈
σ
2
‍
[
f
k
,
l
(
|
x
k
,
l
(
i
)
|
ρ
)
]
p
}
1
/
p
<
∞
for some ρ
>
0 and for all i
∈
ℕ. For each ɛ
>
0, there exists a positive integer n
0 such that
(9)
∥
x
(
i
)
-
x
(
j
)
∥
m
2
(
F
,
ϕ
,
p
)
<
ɛ
for all i
,
j
≥
n
0. Hence
(10)
sup
⁡
(
s
,
t
)
≥
(
1,1
)
sup
⁡
σ
1
×
σ
2
∈
φ
s
t
1
ϕ
s
t
{
∑
k
∈
σ
1
‍
∑
l
∈
σ
2
‍
[
f
k
,
l
(
|
x
k
,
l
(
i
)
-
x
k
,
l
(
j
)
|
ρ
)
]
p
}
1
/
p
<
ɛ
for some ρ
>
0 and for all i
,
j
≥
n
0. This implies that
(11)
|
x
k
,
l
(
i
)
-
x
k
,
l
(
j
)
|
<
ε
ϕ
1,1
for all i
,
j
≥
n
0 and for each fixed (
k
,
l
)
∈
ℕ
×
ℕ. Hence {
x
(
i
)
} is a Cauchy sequence in ℂ.
Then, there existsx
k
,
l
∈
ℂ such that x
k
,
l
(
i
)
→
x
k
,
l as i
→
∞ and let us define x
=
(
x
k
,
l
). From (10), for each fixed (
s
,
t
),
(12)
{
∑
k
∈
σ
1
‍
∑
l
∈
σ
2
‍
[
f
k
,
l
(
|
x
k
,
l
(
i
)
-
x
k
,
l
(
j
)
|
ρ
)
]
p
}
<
ε
p
ϕ
s
t
p
for some ρ
>
0, for all i
,
j
≥
n
0 and σ
1
×
σ
2
∈
φ
s
t.
Lettingj
→
∞, we get
(13)
{
∑
k
∈
σ
1
‍
∑
l
∈
σ
2
‍
[
f
k
,
l
(
|
x
k
,
l
(
i
)
-
x
k
,
l
|
ρ
)
]
p
}
<
ε
p
ϕ
s
t
p
for some ρ
>
0, for all i
,
j
≥
n
0, and σ
1
×
σ
2
∈
φ
s
t. Thus we obtain
(14)
sup
⁡
(
s
,
t
)
≥
(
1,1
)
sup
⁡
σ
1
×
σ
2
∈
φ
s
t
1
ϕ
s
t
{
∑
k
∈
σ
1
‍
∑
l
∈
σ
2
‍
[
f
k
,
l
(
|
x
k
,
l
(
i
)
-
x
k
,
l
|
ρ
)
]
p
}
1
/
p
<
ɛ
for some ρ
>
0 and for all i
,
j
≥
n
0. This implies that x
(
i
)
-
x
∈
m
2
(
F
,
ϕ
,
p
) for all i
,
j
≥
n
0.
Hencex
=
x
(
n
0
)
+
x
-
x
(
n
0
)
∈
m
2
(
F
,
ϕ
,
p
). By (14),
(15)
∥
x
(
i
)
-
x
∥
m
2
(
F
,
ϕ
,
p
)
<
ɛ
for all i
≥
n
0. This means that x
(
i
)
→
x as i
→
∞. Hence m
2
(
F
,
ϕ
,
p
) is a Banach space.Theorem 3.
The spacem
2
(
F
,
ϕ
,
p
) is a BK-space.Proof.
Suppose that{
x
(
i
)
}
∈
m
2
(
F
,
ϕ
,
p
) with ∥
x
(
i
)
-
x
∥
m
2
(
F
,
ϕ
,
p
)
→
0 as i
→
∞. For each ɛ
>
0 there exists n
0
∈
N such that
(16)
∥
x
(
i
)
-
x
∥
m
2
(
F
,
ϕ
,
p
)
<
ɛ
for all i
≥
n
0. Thus
(17)
sup
⁡
(
s
,
t
)
≥
(
1,1
)
sup
⁡
σ
1
×
σ
2
∈
φ
s
t
1
ϕ
s
t
{
∑
k
∈
σ
1
‍
∑
l
∈
σ
2
‍
[
f
k
,
l
(
|
x
k
,
l
(
i
)
-
x
k
,
l
|
ρ
)
]
p
}
1
/
p
<
ɛ
for some ρ
>
0 and for all i
≥
n
0. Hence we obtain
(18)
|
x
k
,
l
(
i
)
-
x
k
,
l
|
<
ε
ϕ
1,1
for all i
≥
n
0 and for all (
k
,
l
)
∈
ℕ
×
ℕ. This implies |
x
k
,
l
(
i
)
-
x
k
,
l
|
→
0 as i
→
∞. This completes the proof.Corollary 4.
The spacem
2
(
F
,
ϕ
,
p
) is a symmetric space.Proof.
Letx
=
{
x
k
,
l
}
∈
m
2
(
F
,
ϕ
,
p
) and let y
=
{
y
k
,
l
}
∈
S
(
x
). Then we can write y
k
,
l
=
x
m
k
,
m
l. Thus we obtain
(19)
∥
x
∥
m
2
(
F
,
ϕ
,
p
)
=
∥
y
∥
m
2
(
F
,
ϕ
,
p
)
.Corollary 5.
The spacem
2
(
F
,
ϕ
,
p
) is a normal space.Proof.
It is obvious.Theorem 6.
Consider(20)
m
2
(
ϕ
)
⊆
m
2
(
F
,
ϕ
,
p
)
.Proof.
Suppose thatx
∈
m
2
(
ϕ
). Then we have
(21)
∥
x
∥
m
2
(
ϕ
)
=
sup
⁡
(
s
,
t
)
≥
(
1,1
)
sup
⁡
σ
1
×
σ
2
∈
φ
s
t
1
ϕ
s
t
{
∑
k
∈
σ
1
‍
∑
l
∈
σ
2
‍
|
x
k
,
l
|
}
∥
x
∥
m
2
(
ϕ
)
=
K
<
∞
.
Thus for each fixed (
s
,
t
) and for σ
1
×
σ
2
∈
φ
s
t,
(22)
∑
k
∈
σ
1
‍
∑
l
∈
σ
2
‍
|
x
k
,
l
|
≤
K
ϕ
s
t
for some ρ
>
0. Hence
(23)
sup
⁡
(
s
,
t
)
≥
(
1,1
)
sup
⁡
σ
1
×
σ
2
∈
φ
s
t
1
ϕ
s
t
{
∑
k
∈
σ
1
‍
∑
l
∈
σ
2
‍
[
f
k
,
l
(
|
x
k
,
l
|
ρ
)
]
p
}
1
/
p
≤
K
for some ρ
>
0. This implies that x
∈
m
2
(
F
,
ϕ
,
p
). Hence m
2
(
ϕ
)
⊆
m
2
(
F
,
ϕ
,
p
).Theorem 7.
m
2
(
F
,
ϕ
,
p
)
⊆
m
2
(
F
,
ψ
,
p
) if and only if sup
⁡
(
s
,
t
)
≥
(
1,1
)
(
ϕ
s
t
/
ψ
s
t
)
<
∞.Proof.
LetK
=
sup
⁡
(
s
,
t
)
≥
(
1,1
)
(
ϕ
s
t
/
ψ
s
t
)
<
∞. Then ϕ
s
t
≤
K
ψ
s
t for all (
s
,
t
)
≥
(
1,1
). If x
∈
m
2
(
F
,
ϕ
,
p
), then
(24)
sup
⁡
(
s
,
t
)
≥
(
1,1
)
sup
⁡
σ
1
×
σ
2
∈
φ
s
t
1
ϕ
s
t
{
∑
k
∈
σ
1
‍
∑
l
∈
σ
2
‍
[
f
k
,
l
(
|
x
k
,
l
|
ρ
)
]
p
}
1
/
p
<
∞
for some ρ
>
0. Thus
(25)
sup
⁡
(
s
,
t
)
≥
(
1,1
)
sup
⁡
σ
1
×
σ
2
∈
φ
s
t
1
K
ψ
s
t
{
∑
k
∈
σ
1
‍
∑
l
∈
σ
2
‍
[
f
k
,
l
(
|
x
k
,
l
|
ρ
)
]
p
}
1
/
p
<
∞
for some ρ
>
0. Hence x
∈
m
2
(
F
,
ψ
,
p
). This shows that m
2
(
F
,
ϕ
,
p
)
⊆
m
2
(
F
,
ψ
,
p
). Conversely, let m
2
(
F
,
ϕ
,
p
)
⊆
m
2
(
F
,
ψ
,
p
). We define γ
s
,
t
=
ϕ
s
t
/
ψ
s
t. Let sup
⁡
(
s
,
t
)
≥
(
1,1
)
γ
s
,
t
=
∞. Then there exists a subsequence {
γ
s
i
,
t
i
} of {
γ
s
,
t
} such that γ
s
i
,
t
i
→
∞ as i
→
∞. Then for x
∈
m
2
(
F
,
ϕ
,
p
) we have
(26)
sup
⁡
(
s
,
t
)
≥
(
1,1
)
sup
⁡
σ
1
×
σ
2
∈
φ
s
t
1
ψ
s
t
{
∑
k
∈
σ
1
‍
∑
l
∈
σ
2
‍
[
f
k
,
l
(
|
x
k
,
l
|
ρ
)
]
p
}
1
/
p
≥
sup
⁡
(
s
,
t
)
≥
(
1,1
)
sup
⁡
σ
1
×
σ
2
∈
φ
s
t
γ
s
,
t
ϕ
s
t
{
∑
k
∈
σ
1
‍
∑
l
∈
σ
2
‍
[
f
k
,
l
(
|
x
k
,
l
|
ρ
)
]
p
}
1
/
p
=
∞
for some ρ
>
0. This is a contradiction as x
∉
m
2
(
F
,
ψ
,
p
) and this completes the proof.Proposition 8.
Consider(27)
ℓ
p
(
2
)
⊆
m
2
(
F
,
ϕ
,
p
)
⊆
ℓ
∞
(
2
)
.Proof.
Clearly,ℓ
p
(
2
)
=
m
2
(
F
,
ψ
,
p
), where ψ
s
t
=
1 for s
,
t
=
1,2
,
… when f
k
,
l
(
x
)
=
x and sup
⁡
(
s
,
t
)
≥
(
1,1
)
(
ψ
s
t
/
ϕ
s
t
)
<
∞ by nondecreasing (
ϕ
s
t
). Then, by Theorem 7, first inclusion is obtained. Suppose x
∈
m
2
(
F
,
ϕ
,
p
). Then
(28)
sup
⁡
(
s
,
t
)
≥
(
1,1
)
sup
⁡
σ
1
×
σ
2
∈
φ
s
t
1
ϕ
s
t
{
∑
k
∈
σ
1
‍
∑
l
∈
σ
2
‍
[
f
k
,
l
(
|
x
k
,
l
|
ρ
)
]
p
}
1
/
p
=
K
<
∞
for some ρ
>
0. Hence we obtain
(29)
|
x
k
l
|
≤
K
ϕ
1,1
for all k
,
l
∈
ℕ. Thus x
∈
ℓ
∞
(
2
) and proof is completed.



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*Source: 684512-2014-02-11.xml*