Applied Mathematics, 2011, 2, 398-402
doi:10.4236/am.2011.24048 Published Online April 2011 (http://www.SciRP.org/journal/am)
Copyright © 2011 SciRes. AM
Statistically Convergent Double Sequence Spaces in
2-Normed Spaces Defined by Orlicz Function
Vakeel A. Khan, Sabiha Tabassum
Department of Mat hematics, Aligarh Muslim University, Aligarh, India
E-mail: vakhan@math.com, sabihatabassum@math.com
Received November 26, 2010; revised January 15, 2011; accepted Ja nu ary 18, 2011
Abstract
The concept of statistical convergence was introduced by Stinhauss [1] in 1951. In this paper, we study con-
vergence of double sequence spaces in 2-normed spaces and obtained a criteria for double sequences in 2-
normed spaces to be statistically Cauchy sequence in 2-normed spaces.1
Keywords: Double Sequence Spaces, Natural Density, Statistical Convergence, 2-Norm, Orlicz Function
1. Introduction
In order to extend the notion of convergence of sequen-
ces, statistical convergence was introduced by Fast [2]
and Schoenberg [3] independently. Later on it was fur-
ther investigated by Fridy and Orhan [4]. The idea de-
pends on the notion of density of subset of .
The concept of 2-normed spaces was initially intro-
duced by G
 ahler [5-7] in the mid of 1960’s. Since
then, many researchers have studied this concept and ob-
tained various results, see for instance [8].
Let
X
be a real vector space of dimension d,
where 2.d A 2-norm on
X
is a function
.,. :
X
XR which satisfies the following four
conditions:
1) 12
,=0xx if and only if 12
,
x
x are linearly de-
pendent;
2) 12 21
,=,
x
xxx:
3) 12 12
,= ,
x
xxx

, for any R
:
4) 22 2
,,,
x
xxxx xx


The pair

,.,.X is then called a 2-normed space
(see [9]).
Example 1.1. A standard example of a 2-normed
space is 2
R equipped wit h t he f oll owing 2-norm
,:=xy the area of the triangle having vertices 0,,.
x
y
Example 1.2. Let Y be a space of all bounded
real-valued functions on R. For ,
g in Y, define
,=0fg , if f, g are linearly dependent,
, =,
sup
tR
f
gftgt
if f, g are linearly independent.
Then .,. is a 2-norm on Y.
We recall some facts connecting with statistical con-
vergence. If K is subset of positive integers , then
n
K
denotes the set {:}.kKkn
The natural density
of K is given by

||
=,
lim n
n
K
Kn

where n
K
deno-
tes the number of elements in n
K
, provided this limit
exists. Finite subsets have natural density zero and
=1
c
K
K where =\
c
K
K, that is the comp-
lement of K. If 12
K
K and 1
K
and 2
K
have natu-
ral densities then

12
.KK Moreover, if
12
==1,KK then

12
=1KK (see [10]).
A real number sequence

=
j
x
x is statistically con-
vergent to L provided that for every >0
the set
:| |
j
nxL
 has natural density zero. The sequ-
ence
=
j
x
x is statustically Cauchy sequence if for
each >0
there is positive integer
=NN
such
that

:|= 0
jN
nxx
  (see [11]).
If
=
j
x
x is a sequence that satisfies some property
P for all n except a set of natural density zero, then we
say that
j
x
satisfies some property P for “almost all n”.
An Orlicz Function is a function
:0, 0,M 
which is continuous, nondecreasing and convex with
0=0M,
>0Mx for >0x and

Mx, as
x
.
If convexity of
M
is replaced by

M
xy Mx
M
y, then it is called a Modulus funtion (see Mad-
dox [12]). An Orlicz function may be bounded or un-
12000 Mathematics Subject Classification. 46E30, 46E40, 46B20.
V. A. KHAN ET AL.
Copyright © 2011 SciRes. AM
399
bounded. For example,
 
=0<1
p
Mx xp is un-
bounded and

=1
x
Mx x is bounded.
Lindesstrauss and Tzafriri [13] used the idea of Orlicz
sequence space;
=1
:=:<, for some >0
k
Mk
x
lxwM









which is Banach space with the norm
=1
||
=inf>0:1 .
k
Mk
x
xM






The space
M
l is closely related to the space
p
l,
which is an Orlicz sequence space with

=
p
M
xx for
1<.p
An Orlicz function
M
satisfies the 2condition
2
( f )
M
or short if there exist constant 2K and
0>0u such that
 
2
M
uKMu
whenever 0
||uu.
Note that an Orlicz function satisfies the inequality
 
for all with 0<<1.Mx Mx
 
Orlicz function has been studied by V. A. Khan [14-17]
and many others.
Throughout a double sequence

=kl
x
x is a double
infinite array of elements kl
x
for ,.kl
Double sequences have been studied by V. A. Khan
[18-20], Moricz and Rhoades [21] and many others.
A double sequence

=
j
k
x
x called statistically con-
vergent to L if

,
1,:, ,=0
lim jk
mn jkxLj mk n
mn
  
where the vertical bars indicate the number of elements
in the set. (see [19])
In this case we write 2lim =
jk
s
txL.
2. Definitions and Preliminaries
Let

j
x
be a sequence in 2-normed space
,.,.X.
The sequence

j
x
is said to be statistically convergent
to L, if for every >0
, the set

:,
j
jxLz

has natural density zero for each nonzero z in X, in other
words

j
x
statistically converges to L in 2-normed
space

,.,.X if

1:,=0
lim j
njx Lz
n

for each nonzero z in X. It means that for every zX
,
,< ...
j
x
Lz aan
In this case we write
,:=,.
lim j
n
s
tx LzLz


Example 2.1 Let 2
=
X
R be equiped with the 2-
norm by the formula

12211 21 2
,=,= ,,=,.
x
yxyxyxxxyyy
Define the
j
x
in 2-normed space

,.,.X by

2
1, if =,,
=
1
1, otherwise.
j
nnkkN
xn
n



and let
=1,1L and

12
=,zzz. If 1=0z then
=:,=
j
Kj xLz

for each z in 21
||
,:=,
z
Xj nkk


is a finite set,
so

1
2
2
1
:,
=:=,1 finite set.
j
jxLz
jjkk
z











Therefore,


1
2
2
1
1:,
1
:= ,101
j
jxLz
n
jjkk
zn




 






for each z in X. Hence,


:, =0
n
jxLz
 
for every >0
and zX
.
V. A. Khan and Sabiha Tabassum [20] defined a
double sequence
j
k
x
in 2-nor med space

,.,.X to
be Cauchy with respect to the 2-norm if
,,=0for every and ,.
lim jk pq
jp xxzzX kq


If every Cauchy sequence in
X
converges to some
,LX
then
X
is said to be complete with respect to
the 2-norm. Any complete 2-normed space is said to be
2-Banach space.
Example 2.2 Define the xi in 2-normed space
,.,.X
by

2
0, if =,,
=0,0 otherwise.
j
jjkkN
x
V. A. KHAN ET AL.
Copyright © 2011 SciRes. AM
400
and let

=0,0L and

12
=,zzz. If 1=0z then

2
:,1,4,9,16, , ;
j
jxLzj

We have that


:, =0
j
jxLz
  for every
>0
and zX. This implies that ,=
lim j
n
s
txz

,Lz. But the sequence
j
x
is not convergent to .L
A sequence which converges statistically need not be
bounded. This fact can be seen from Example [2.1] and
Example [2.2].
3. Main Results
In this paper we define a double sequence
j
k
x
in
2-normed space

,.,.X to be statistically Cauchy
with respect to the 2-norm if for ever y >0
and every
nonzero zX there exists a number

=,pp z
and

=,qqz
such that

,
1,:,,,=0
lim jk pq
mn jkNNxxzjmkn
mn
  
In this case we write 2lim, :=,
jk
s
tx LzLz .
Theorem 3.1. Let

j
k
x
be a double sequence in
2-normed space

,.,.X and ,LL X
. If
2lim, = ,
jk
s
txzLz and 2lim, =,
jk
s
txzLz
,
then =.LL
Proof. Assume =,LL
. Then =0LL
, so there
exists a zX, such that LL
and z are linearly in-
dependent. Therefore
,=2, with >0.LLz

Now

2= ,
,,.
jk jk
jk jk
LxxL z
x
LzxLz

 
So




,: ,<,: ,<
jk jk
jkxL zjkxL z

 
.
But



,:,<=0
jk
jkxL z
 . Contradicting the
fact that

.
jk
x
Lstat
Theorem 3.2. Let the double sequence

j
k
x
and

j
k
y in 2-normed space

,.,.X. If

j
k
y is a con-
vergent sequence such that =
j
kjk
x
y almost all n, then

j
k
x
is statistically convergent.
Proof. Suppose


(,): ==0
jk jk
jkNN xy

and ,,=,
lim jk
jk yz Lz
 . Then for every >0
and
zX.



,:,
,:=.
jk
jk jk
jkN NxLz
jkNNxy


Therefore







,:,
,:,
,:=.
jk
jk
jk jk
jkN NxLz
jkN NyLz
jkNNxy

 
 
(3.1)
Since ,=,
lim jk
nyzLz
 for every zX, the set

,:,
jk
jkN NyLz
 contains finite number
of integers. Hence,

,:,
jk
jkN NyLz

=0. Using inequality [3.1], we get

,:,=0
jk
jkN NxLz

for every >0
and .zX
Consequently,
2lim,=, .
jk
s
tx LzLz
Theorem 3.3. Let the double sequence
j
k
x
and
j
k
y in 2-normed space
,.,.X and ,LL X
and
a
.
If 2lim, = ,
jk
s
txzLz
and 2lim,=, ,
jk
s
tyzLz
for every nonzero zX
, then
1) 2lim, =,
jk jk
s
txyzLLz
 
, for each nonzero
zX
and
2) 2lim, =,
jk
s
taxzaLz, for each nonzero zX
.
Proof 1) Assume that 2lim,=, ,
jk
s
txzLz and
2lim, =,
jk
s
tyzLz
, for every nonzero zX
. Then
1=0K and
2=0K where
 
11
=:=,: ,
2
jk
KKjkNNx Lz


 
22
=:=,: ,
2
jk
KK jkNNyLz
 

for every >0
and zX
. Let
 
=:=,:(), .
jk jk
KKjk NNxyLLz

To prove that
=0K
, it is sufficient to prove that
12
K
KK. Suppose 00
,jk K. Then

000,
jk jk
o
xy LLz
  (3.2)
V. A. KHAN ET AL.
Copyright © 2011 SciRes. AM
401
Suppose to the contrary that 00 12
,jk K K. Then
00 1
,jk K and 002
,jkK. If 001
,jk K and
00 2
,jk K then
00 ,<
2
jk
xLz
and 00,<.
2
jk
xLz
Then, we get

000
000
,
,, <=
22
jk jk
o
jk jk
o
xy LLz
xLzyLz


 
which contradicts [3.2]. Hence 0012
,jk K K, that is,
12
K
KK.
2) Let 2lim,=, ,
jk
stxzL za and =0a.
Then

,:,=0.
jk
jkN NxLza










Then we have





,:,
=,: ,
=,: ,.
jk
jk
jk
jkN NaxaLz
jkNNa xLz
jkN NxLza
 
 


 



Hence, the right handside of above equality equals 0.
Hence, 2lim,=,,
jk
s
taxzaLz for every nonzero
.zX
From Theorem 1 of Fridy [11] we have
Theorem 3.4. Let

j
k
x
be statistically Cauchy se-
quence in a finite dimensional 2-normed space
,.,.X.
Then there exists a convergent double sequence
j
k
y
in

,.,.X such that =
j
kjk
x
y for almost all n.
Proof. See proof of Theorem 2.9 [9].
Theorem 3.5. Let

j
k
x
be a double sequence in 2-
normed space

,.,.X The double sequence ()
j
k
x
is
statistically convergent if and only if ()
j
k
x
is a statisti-
cally Cauchy sequence.
Proof. Assume that 2lim, = ,
jk
s
txzLz for every
nonzero zX and >0.
Then, for every zX,
,< almost all ,
2
jk
x
Lz n
and if
=,pp z
and
=,qq z
is chosen so that
,<,
2
pq
xLz
then, we have
,,,
< almost all .
22
= almost all .
jk pqjkpq
x
xzxLz Lxz
n
n


Hence,
j
k
x
is statistically Cauchy sequence.
Conversely, assume that
j
k
x
is a statistically Cauchy
sequence. By Theorem 3.4, we have 2lim, =
jk
s
txz
,Lz for each zX
.
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