Gen. Math. Notes, Vol. 25, No. 2, December 2014, pp.19-30
c
ISSN 2219-7184; Copyright ICSRS
Publication, 2014
www.i-csrs.org
Available free online at http://www.geman.in
Some Double Sequence Spaces Defined
by a Modulus Function
Gülsen Kılınç1 and İhsan Solak2
1
Adıyaman University, Faculty of Education
Department of Primary Mathematics, Adıyaman, Turkey
E-mail: gkilinc@adiyaman.edu.tr.
2
Nevşehir University, Faculty of Art and Sciences
Department of Mathematics, Nevşehir, Turkey
E-mail: ihsansolak@nevsehir.edu.tr.
(Received: 15-4-14 / Accepted: 23-9-14)
Abstract
In the present paper we construct some new double sequence spaces defined
by a modulus function. Further we give its some topological and algebraic
properties.
Keywords: Double sequence space, Modulus Function, vector valued double sequence space.
1
Introduction
Some works on double sequences were studied by Hardy[5], Moricz[6], Moricz
and Rhoades[7]. Hardy introduced the notion of regular convergence for double sequences. Hill [8] was the first who applied methods of functional analysis
to doubles sequences. He described the topological dual of the space of all
regularly convergent double sequences and perfectness of matrices with respect
to the regular convergence. Türkmenoğlu [10] showed under which conditions
that C0p (t) , Lu and C0bp (t) are paranormed double sequence spaces, determined their duals and gave some inclusion relations between those spaces.
Boos, Leiger and Zeller [11] defined the concept of ν-SM method by the application domain of a matrix sequence A = A(ν) of infinite matrices, gave
the consistency theory for such type methods and introduced the notion of
20
Gülsen Kılınç et al.
Ce -convergence for double sequences. By using gliding hump method, Zeltser
[12] characterized the class of four-dimensional matrix mappings from λ into
µ, where λ, µ ∈ {Ce , Cbe }. By Ce and Cbe , we denote the spaces of all Ce convergent and of all bounded Ce -convergent double sequences, respectively.
Also employing the same arguments, Zeltser [13] gave the theorems determining the necessary and sufficient conditions for Ce -SM and Cbe -SM methods to
be concervative and coercive.
Modulus function introduced by Nakano [1] and used to solve some structural problems of the scalar FK-spaces theory. Ruckle [2] constructed a class
of scalar FK-spaces L (f ), where f is a modulus function. L (f ) is a generalization of the spaces lp , (0 < p ≤ 1).
Yılmaz [3] introduced and investigated the sequence space λ (Xk , r, f, s)
defined by a modulus function and constructed its FK-structure under some
conditions.
In the present paper we introduce some new double sequence spaces by
using a modulus function f and investigate some properties of these sequence
spaces.
Let Ω (X) be the space of all X-valued double sequences and X be a Banach
space. The topology of Ω (X) is a locally convex topology produced by the
family of all seminorms defined by
pij (x) = kxij k .
Since the family of {pij : i, j ∈ N × N} are countable, this topology is metrizable, where k, k is the norm on X.
Also, total paranorm generating this metric, which is constructed by the
Frechet combination of seminorms {pij : i, j ∈ N × N}, is given by
X 1
kxij k
g (x) =
.
i+j
2 1 + kxij k
A locally convex double sequence space (E, τ ) is said to be a DK-space, if
all the seminorms defined by
rkl : E → R
x = (xij ) → |xkl | ,
are continuous. A DK-space with a Frechet topology is called an FDK-space.
A normed FDK-space is called BDK-space [4].
For λ (X) ⊂ Ω (X) , a Frechet sequence space λ (X) is called an FDK-space,
if the coordinate maps
fkl : λ (X) → X
x → fkl (x) = xkl
are continuous.
21
Some Double Sequence Spaces Defined...
2
The Double Sequence Spaces Lu (f )
Let f be a modulus and let us define the double squence space Lu (f ) by
X
Lu (f ) = {x ∈ Ω :
f (|xmn |) < ∞}.
m,n∈N
It is easy to show that Lu (f ) is a linear space, where Ω denotes the space of
all real or complex valued double sequences.
Theorem 2.1 Lu (f ) is a paranormed space with the function
X
g (x) =
f (|xmn |) .
m,n∈N
Proof: It is obvious that g is well-defined by the definition of Lu (f ) .Let
us verify that gPprovides the paranorm conditions:
i) g (θ) =
f (| θ |) = 0.
m,n∈N
ii) For each x ∈ Lu (f ), it is clear that
g (x) = g (−x) .
iii) For all x, y ∈ Lu (f ) , we have
X
X
g (x + y) =
f (|xmn + ymn |) ≤
f (|xmn | + |ymn |)
m,n∈N
≤
X
m,n∈N
f (|xmn |) +
m,n∈N
X
f (|ymn |) = g (x) + g (y) .
m,n∈N
iv) (a) Suppose that λ is a scalar and g (x) → 0. Then we get
X
X
g (λx) =
f (|λ| |xmn |) ≤ K.
f (|xmn |) ,
m,n∈N
m,n∈N
where K is a positive integer such that |λ| ≤ K. So g (λx) → 0.
(b) Suppose that λr → 0 and x ∈ Lu (f ) . Then there exist positive
numbers ε and k such that
∞
X
ε
f (|xmn |) < ,
2
m,n=k+1
by virtue of the the fact f (|xmn |) < ∞. Now let us write
h (t) =
k
X
m,n=1
f (t |xmn |) .
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Gülsen Kılınç et al.
Then h is continuous at 0. Therefore there exists a number δ such that
0 < δ < 1, for each 0 < t < δ
ε
|h (t)| < .
2
Then there is a number N such that for each r > N, we get
|λr | < δ.
Since λr → 0, for each r > N, then we have
r
g (λ x) =
k
X
∞
X
r
f (|λ xmn |) +
m,n=1
f (|λr xmn |)
m,n=k+1
∞
X
ε
f (|xmn |)
+
2 m,n=k+1
ε ε
≤
+ = ε,
2 2
≤
which completes the proof.
It can be verified easily that g is total, i.e., for each x ∈ Lu (f ), then
g (x) = 0 ⇒ x = θ.
Theorem 2.2 The double sequence space Lu (f ) is a DK-space.
Proof: For each k, l ∈ N, the functions
Pkl : Lu (f ) → C
x → Pkl (x) = xkl
are continuous. For all ε > 0, there exists a δ > 0 such that g (x) < δ.
Choosing δ = f (ε) , we have
∞
X
f (|xkl |) < f (ε) ⇒ f (|xkl |) < f (ε) .
k,l
Since f is increasing, we obtain
|xkl | = |Pkl (x)| < ε.
Theorem 2.3 Lu (f ) is complete.
23
Some Double Sequence Spaces Defined...
Proof: Let xl be a Cauchy
sequence in Lu (f ). For each ε > 0, ∃n0 ∈ N
l
r
3 for each l, r > n0 , g x − x < ε. Since, for each
i, j, coordinate functions
fij (x) = xij are continuous on Lu (f ), then xlij is a Cauchy sequence in C,
(l > n0 ) . Because of the completeness of C, the Cauchy sequence converges to
a point. Let xij be such a point. Construct the double sequence x = (xij ) with
these limit points. Then we have
l
g x −x
r
< ε⇒
⇒ lim
r
X
m X
n
X
l
r f xij − xij < ε ⇒
f xlij − xrij < ε
i,j
m
n
XX
f xlij − xrij =
f xlij − xij < ε
i=0 j=0
i=0 j=0
l
i=0 j=0
m
n
XX
⇒ g x − x < ε (∀ l > n0 ) ⇒ xl → x.
Also, we get
∞ X
∞
X
f (|xij |) =
i=0 j=0
≤
m X
n
X
i=0 j=0
∞ X
∞
X
f xij − xlij − xlij m X
n
X
l f xij − xij +
f xlij < ε,
i=0 j=0
i=0 j=0
which implies x ∈ Lu (f ) .
Corollary 2.4 Lu (f ) is an FDK-space.
Theorem 2.5 Lu (f ) ⊆ Lu , where Lu = {x ∈ Ω :
∞
P
| xij |< ∞}.
i,j=1
Proof: Suppose that x ∈ Lu (f ) but x ∈
/ Lu . Then we have
∞
X
f (|xkl |) < ∞
k,l=1
and
∞
X
(|xkl |) = ∞.
k,l=1
From the last equation above, one can see that there is a double sequence
of natural numbers (ni , mj ) such that,


ni+1 mj+1
ni+1 mj+1
ni+1 mj+1
X X
X X
X X


(|xkl |) > 1 ⇒ f (1) < f
(|xkl |) =
f (|xkl |) .
k=ni l=mj
k=ni l=mj
k=ni l=mj
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Gülsen Kılınç et al.
Then we obtain
∞
X
f (|xkl |) < ∞
k,l=1
and
ni+1 mj+1
X
X
f (|xkl |) → 0, (i, j → ∞),
k=ni l=mj
which implies a contradiction such that f (1) = 0. Hence, we get x ∈ Lu .
This completes the proof.
We note that
f (x) = x ⇒ Lu (f ) = Lu .
If f is unbounded, Lu (f ) ⊂ Lu .
Theorem 2.6 The double sequence of unit vectors is bounded in Lu (f ) .
Proof: Let δ be the double sequence of unit vectors, i.e.,


0
0
...
0
...


0 0 ... 0 ...
e11 e12 ...


,
.
.
.
0
...
δ =  e21 e22 ... ,
eij = 



.
. ...
0 0 ... 1 ...
. . ... 0 ...
where 1 is term of (i, j).
If a ∈ C, then g(ae11 ) = ... = g(aeij ) = f (|a|). Let us choose a λ ∈ C such
that f (|λ|) < . Then, for 0 ≤ a ≤ λ, we write
g(ae11 ) = f (|a|) ≤ f (|λ|) < .
This implies that ae11 ∈ {x : g(x) < ε}. Since we have
g(aeij ) = g(aekl ) = f (|a|),
for (i, j) 6= (k, l), then we get aδ ∈ {x : g(x) < ε}, which shows that each
sphere centered at origin contains the sequence δ. Thus δ is a bounded double
sequnce in Lu (f ) .
3
The Double Sequence Space Lu (X, f )
Let X be a Banach Space. We define
Lu (X, f ) = {x ∈ Ω (X) :
X
f (kxmn k) < ∞},
m,n
where k, k is the norm of X. It is obvious that Lu (X, f ) is a linear space.
25
Some Double Sequence Spaces Defined...
Theorem 3.1 Lu (X, f ) is paranormed space with
X
p (x) =
f (kxmn k) .
i,j
Proof: i) If x = θ, for each i, j, xij = 0 and p (θ) = 0.
ii) It is obvious that p (−x) = p (x) .
iii) For x, y ∈ Lu (X, f ), we write
X
X
p (x + y) =
f (kxij + yij k) ≤
f (kxij k) + f (kyij k)
i,j
=
i,j
X
f (kxij k) +
X
i,j
f kyij k).
i,j
l
iv) Let x = x ∈ Lu (X, f ) be a sequence and λ = λl be a sequence
of scalars. Assume
that λ → λ0 and p xl − x0 → 0, (l → ∞) . So scalar
sequence λ = λl is convergent, for each l ∈ N. Then there exists a positive
number K such that | λl |≤ K. Therefore we have
X p λl xl − λ0 x0 =
f λl xlij − λ0 x0ij i,j
=
X
≤
X
≤
X
f λl xlij − λl x0ij + λl x0ij − λ0 x0ij i,j
f (λl xlij − λl x0ij ) + f λl x0ij − λ0 x0ij i,j
X f (λl xlij − x0ij ) +
f (λl − λ0 x0ij )
i,j
i,j
l
= K.p x − x
0
+
X
f (λl − λ0 x0ij ).
(1)
i,j
0
Since K is a constant and p xl − x → 0 (l → ∞) , the first term in the
right hand side of inequality in (1) tends to zero. Also, for each l ∈ N, there
exists a T ≥ 0 such that
T x0 = T x0ij ∈ Lu (X, f ) .
For ε > 0, there exist numbers i0 , j0 such that for each l ∈ N, we get

PP
j>j0
f (λl − λ0 x0ij )
 i>i0
i0
 P
l
0 0  + P∞
xij )

j=j0 +1 f ( λ − λ
 i=0∞
l

P Pj0
λ − λ0 kxij k)
+
f
(
j=0
i=i0 +1


PP
j>j0
f (T x0ij )

 i>i0
i0

 P
0

 + P∞
 ≤ 
j=j0 +1 f (T xij )

 i=0∞
0


P Pj0
xij )
+
f
(T
j=0
i=i0 +1
ε
ε ε ε
+ + =
<
6 6 6
2







26
Gülsen Kılınç et al.
On the other hand, since
lim
l→∞
j0
i0 X
X
j0
i0 X
X
l
0 0
f ( lim λl − λ0 x0ij ) = 0.
f (λ − λ xij ) =
i=0 j=0
i=0 j=0
l→∞
For the same ε, there exists an l0 such that for each l > l0
j0
i0 X
X
i=0
ε
f (λl − λ0 x0ij ) < .
2
j=0
Then we get
PP





X 
l
0 0 f ( λ − λ xij ) ≤ 

i,j



0
xij )
f
(T
j>j0
i>i0
i0 P
0
P
∞
+
j=j0 +1 f (T xij )
i=0
∞ P
0
P
j0
+
j=0 f (T xij )
i=i0 +1
i0 P
l
P
j0
0 0 xij )
+
j=0 f ( λ − λ
i=0




 ε ε ε ε

 < + + + = ε,
 6 6 6 2



for each ε > 0 and l > l0 . Thus the second term in inequality (1) tends to
zero, i.e.,
p λl xl − λ0 x0 → 0
(l → ∞) .
So the proof is completed.
Theorem 3.2 Lu (X, f ) is an FDK-space.
Proof: i) We know that Lu (X, f ) is a linear space and paranormed space.
ii) Let us proof continuity of coordinate maps pkl which are defined by
pkl : Lu (X, f ) → X
pkl (x) = xkl .
For each ε > 0, let us choose δ = f (ε) such that
p (x) =
X
f (kxkl k) < δ = f (ε) .
k,l
We get
f (kxkl k) < f (ε)
and
kxkl k = kpkl (x)k < ε.
27
Some Double Sequence Spaces Defined...
iii) We will show that Lu (X, f ) is complete. Let xl be a Cauchy sequence
in Lu (X, f ) . For each ε > 0, there exists an n0 ∈ N such that, for l, r > n0
p xl − xr < ε.
Since functions
pij (x) = xij
are continuous for each (i, j) ∈ N × N, xlij is a Cauchy sequence in X,
(for each (i, j) ∈ N × N and l > n0 ). Since X is a Banach space, the Cauchy
sequence converges to a point xij . Let’s set the double sequence x = (xij ) by
this limit points. Now we have
xl → x ⇒ p x l − xr < ε
X ⇒
f xlij − xrij < ε
i,j
⇒
m X
n
X
f xlij − xrij < ε
i=0 j=0
lim
r
m X
n
X
f xlij − xrij =
i=0 j=0
X
f xlij − xrij < ε
i,j
⇒ p xl − x < ε, for each l > n0 .
On the other hand we obtain
X
XX
f k xlij k =
f k xij − xlij + xlij k
i,j
i
≤
j
∞ X
∞
X
∞ X
∞
X
f k xij − xlij k +
f k xlij k
i=0 j=0
i=0 j=0
< ε
which shows that x ∈ Lu (X, f ).
4
Double Sequence Spaces Lu (X) and Mu (X)
Let’s define double sequence spaces Lu (X) and Mu (X) as follow:
Mu (X) = {x = (xmn ) ∈ Ω (X) : sup kxmn k < ∞},
m,n∈N
Lu (X) = {x = (xmn ) ∈ Ω (X) :
X
m,n
kxmn k < ∞}.
28
Gülsen Kılınç et al.
Theorem 4.1 The space sLu (X) and Mu (X) are BDK-spaces.
Proposition 4.2 For each modulus function f , Lu (X, f ) ⊂ Lu (X) .
Proof: We assume that x ∈ Lu (X, f ) and x ∈
/ Lu (X) . Then increasing
sequences (kn ) and (ln ) can be found such that
kX
n −1
lX
n −1
kxij k ≥ 1
i=kn−1 j=ln−1

⇒ f (1) ≤ f 
kX
n −1
lX
n −1

kxij k
i=kn−1 j=ln−1
≤
kX
n −1
lX
n −1
f (kxij k)
i=kn−1 j=ln−1
⇒
kX
n −1
lX
n −1
f (kxij k) < ∞
i=kn−1 j=ln−1
lim
n
kX
n −1
lX
n −1
f (kxij k) = 0
i=kn−1 j=ln−1
It means f (1) = 0. This contradicts with f being a modulus function. Then
x ∈ Lu (X) .
Proposition 4.3 For each x ∈ Lu (X), x can be written as follows:
X
x=
Iij (xij ) ,
i,j∈N×N
where
Iij : X → Lu (X)
Iij (t) = y 3 yij = t and ykl = 0, if (k, l) 6= (i, j) .
Note that the meaning of the above expression is that the net {SF (x) : F ∈ F}
converges to a point x according to norm topology of Lu (X), where F is a
family of all finite subsets of N × N. This family is directed with relation ⊆ .
Theorem 4.4 For each x ∈ Lu (X, f ) , x can be represented as follows:
X
x=
Iij (xij ) ,
(i,j)∈F
where
Iij : X → Lu (X, f )
Iij (t) = y 3 yij = t and ykl = 0, if (k, l) 6= (i, j) .
29
Some Double Sequence Spaces Defined...
Proof: We will show that for all given ε > 0, there exists an F0 = F0 (ε) ∈
F such that when F0 ⊆ F, p(x − SF (x)) < ε. Due to the definition of Iij , we
can define a function as follow:
SF : Lu (X, f ) → Lu (X, f )
SF (xij ) = xij , ((i, j) ∈ F )
SF (xij ) = 0.
((i, j) ∈
/ F)
Then we write
X
p (x − SF (x)) =
f (kxij − {SF (x)}i,j k)
i,j∈N×N
X
=
f (kxij k) ,
i,j∈N×N\F
which implies x ∈ Lu (X, f ). So
P
f (kxij k) < ∞.
P
Thus for ε > 0, there exists an F0 (ε) 3
f (kxij k) < ε. Therefore for all
i,j
i,j
F ⊇ F0 , we have
X
f (kxij k) = p (SF (x) − x) < ε.
i,j∈N×N\F
It means that the net (SF (x) : F ∈ F) converges to x. This completes the
proof.
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