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

.

4. References

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[2] H. Fast, “Sur la Convergence Statistique,” Colloqium

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[3] I. J. Schoenberg, “The Integrability of Certain Functions

and Related Summability Methods,” American Mathema-

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doi:10.1090/S0002-9939-97-04000-8

[5] S. Gähler, “2-Merische Räme und Ihre Topological Stru-

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[6] S. Gähler, “Linear 2-Normietre Räme,” Mathematische

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[9] M. Gurdal and S. Pehlivan, “Statistical Convergence in

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[10] A. R. Freedman and I. J. Sember, “Densities and Summa-

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[12] I. J. Maddox, “Sequence Spaces Defined by a Modulus,”

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