Applied Mathematics, 2011, 2, 398-402 doi:10.4236/am.2011.24048 Published Online April 2011 (http://www.SciRP.org/journal/am)
Statistically Convergent Double Sequence Spaces in 2-Normed Spaces Defined by Orlicz Function Vakeel A. Khan, Sabiha Tabassum Department of Mathematics, Aligarh Muslim University, Aligarh, India E-mail:
[email protected],
[email protected] Received November 26, 2010; revised January 15, 2011; accepted January 18, 2011
Abstract The concept of statistical convergence was introduced by Stinhauss [1] in 1951. In this paper, we study convergence of double sequence spaces in 2-normed spaces and obtained a criteria for double sequences in 2normed 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
f , g = sup f t g t , if f, g are linearly independent. t R
In order to extend the notion of convergence of sequences, statistical convergence was introduced by Fast [2] and Schoenberg [3] independently. Later on it was further investigated by Fridy and Orhan [4]. The idea depends on the notion of density of subset of . The concept of 2-normed spaces was initially intro ahler [5-7] in the mid of 1960’s. Since duced by G then, many researchers have studied this concept and obtained 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 X R which satisfies the following four conditions: 1) x1 , x2 = 0 if and only if x1 , x2 are linearly dependent; 2) x1 , x2 = x2 , x1 : 3) x1 , x2 = x1 , x2 , for any R : 4)
x x , x2 x, x2 x , x2
The pair X , .,. is then called a 2-normed space (see [9]). Example 1.1. A standard example of a 2-normed space is R 2 equipped with the following 2-norm x, y := 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 f , g in Y , define f , g = 0 , if f, g are linearly dependent, 1
2000 Mathematics Subject Classification. 46E30, 46E40, 46B20.
Copyright © 2011 SciRes.
Then .,. is a 2-norm on Y . We recall some facts connecting with statistical convergence. If K is subset of positive integers , then K n denotes the set {k K : k n}. The natural density |K | of K is given by K = lim n , where K n denon n tes the number of elements in K n , provided this limit exists. Finite subsets have natural density zero and K c = 1 K where K c = \ K , that is the complement of K. If K1 K 2 and K1 and K 2 have natural densities then K1 K 2 . Moreover, if K1 = K 2 = 1, then K1 K 2 = 1 (see [10]).
A real number sequence x = x j is statistically convergent to L provided that for every > 0 the set n :| x j L | has natural density zero. The sequ-
ence x = x j is statustically Cauchy sequence if for each > 0 there is positive integer N = N such
that n :| x j xN = 0 (see [11]).
If x = x j is a sequence that satisfies some property P for all n except a set of natural density zero, then we say that x j satisfies some property P for “almost all n”. An Orlicz Function is a function M : 0, 0, which is continuous, nondecreasing and convex with M 0 = 0 , M x > 0 for x > 0 and M x , as x. If convexity of M is replaced by M x y M x M y , then it is called a Modulus funtion (see Maddox [12]). An Orlicz function may be bounded or unAM
V. A. KHAN ET AL.
bounded. For example, M x = x p 0 < p 1 is unx bounded and M x = is bounded. x 1 Lindesstrauss and Tzafriri [13] used the idea of Orlicz sequence space; x lM := x w : M k < , for some > 0 k =1
which is Banach space with the norm x
M
| x | = inf > 0 : M k 1 . k =1
The space lM is closely related to the space l p , which is an Orlicz sequence space with M x = x p for 1 p < . An Orlicz function M satisfies the 2 condition ( M 2 for short ) if there exist constant K 2 and u0 > 0 such that M 2u KM u
whenever | u | u0 . Note that an Orlicz function satisfies the inequality M x M x for all with 0 < < 1.
Orlicz function has been studied by V. A. Khan [14-17] and many others. Throughout a double sequence x = xkl is a double infinite array of elements xkl for k , l . Double sequences have been studied by V. A. Khan [18-20], Moricz and Rhoades [21] and many others. A double sequence x = x jk called statistically convergent to L if 1 j, k : x jk L , j m, k n = 0 lim m , n mn where the vertical bars indicate the number of elements in the set. (see [19]) In this case we write st2 lim x jk = L .
2. Definitions and Preliminaries Let x j be a sequence in 2-normed space X , .,. . The sequence x j is said to be statistically convergent to L, if for every > 0 , the set
j: x
j
L, z
has natural density zero for each nonzero z in X, in other words x j statistically converges to L in 2-normed space X , .,. if
1 j : x j L, z n n lim
=0
for each nonzero z in X. It means that for every z X , Copyright © 2011 SciRes.
399
x j L, z < a.a.n.
In this case we write st lim x j L, z := L, z . n
Example 2.1 Let X = R 2 be equiped with the 2norm by the formula x, y = x1 y2 x2 y1 , x = x1 , x2 , y = y1 , y2 .
Define the
x j
X , .,.
in 2-normed space
by
1, n if n = k , k N , x j = n 1 1, otherwise. n 2
and let L = 1,1 and z = z1 , z2 . If z1 = 0 then
K = j : x j L, z =
| z | for each z in X , j : n = k 2 , k 1 is a finite set, so
j: x
j
L, z
1 2 2 = j : j = k ,k 1 finite set . z 1
Therefore,
1 j : x j L, z n
1 2 1 2 j : j = k , k 1 0 1 z1 n
for each z in X. Hence, j : xn L, z = 0 for every > 0 and z X . V. A. Khan and Sabiha Tabassum [20] defined a double sequence x jk in 2-normed space X , .,. to be Cauchy with respect to the 2-norm if lim x jk x pq , z = 0 for every z X and k , q .
j , p
If every Cauchy sequence in X converges to some L X , 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 0, j xj = 0, 0
if j = k 2 , k N , otherwise. AM
V. A. KHAN ET AL.
400
j, k N N :
and let L = 0, 0 and z = z1 , z2 . If z1 = 0 then
j : x j L, z 1, 4,9,16, , j ;
We have that
j : x
j
2
L, z
= 0 for every
> 0 and z X . This implies that st lim x j , z = n
L, z . But the sequence x j 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].
In this paper we define a double sequence x jk in 2-normed space X , .,. to be statistically Cauchy with respect to the 2-norm if for every > 0 and every nonzero z X there exists a number p = p , z and q = q , z such that
lim
1 j, k N N : x jk x pq , z , j m, k n = 0 mn
In this case we write st2 lim x jk L, z := L, z .
Theorem 3.1. Let x jk be a double sequence in 2-normed space X , .,. and L, L X . If st2 lim x jk , z = L, z and st2 lim x jk , z = L, z ,
then L = L. Proof. Assume L = L , . Then L L = 0 , so there exists a z X , such that L L and z are linearly independent. Therefore L L, z = 2 , with > 0.
Therefore
j, k N N : x L, z j , k N N : y L, z
jk
(3.1)
jk
j, k N N : x
Since lim y jk , z = L, z
j, k N N :
for every z X , the set
y jk L, z
of integers. Hence,
= y jk .
jk
contains finite number
j, k N N : y
jk
L, z
= 0. Using inequality [3.1], we get
j, k N N : x
jk
L, z
= 0
for every > 0 and z X . Consequently, st2 lim x jk L, z = L, z .
Theorem 3.3. Let the double sequence x jk and y jk in 2-normed space X , .,. and L, L X and a . If st2 lim x jk , z = L, z
and st2 lim y jk , z = L, z , for every nonzero z X , then 1) st2 lim x jk y jk , z = L L, z , for each nonzero
2 = L x jk x jk L , z x jk L, z x jk L, z .
st2 lim y jk , z = L, z , for every nonzero z X . Then
< = 0 .
j, k : x jk L, z <
But
z X and 2) st2 lim ax jk , z = aL, z , for each nonzero z X . Proof 1) Assume that st2 lim x jk , z = L, z , and
Now
So
j , k N N : x jk = y jk .
n
3. Main Results
m, n
x jk L, z
j, k : x
jk
L, z
j, k : x jk L, z < . Contradicting the
fact that x jk L stat .
Theorem 3.2. Let the double sequence x jk and in 2-normed space X , .,. . If y jk is a conjk vergent sequence such that x jk = y jk almost all n, then x jk is statistically convergent.
y
Proof. Suppose ( j , k ) N N : x jk = y jk = 0
and lim y jk , z = L, z . Then for every > 0 and j , k
z X . Copyright © 2011 SciRes.
K1 = 0 and K 2 = 0 where
K1 = K1 := j , k N N : x jk L, z 2 K 2 = K 2 := j , k N N : y jk L, z 2 for every > 0 and z X . Let K = K :=
j, k N N : x
jk
y jk ( L L), z .
To prove that K = 0 , it is sufficient to prove that K K1 K 2 . Suppose j0 , k0 K . Then
x
j0 ko
y j0 k0 L L , z
(3.2)
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V. A. KHAN ET AL.
Suppose to the contrary that j0 , k0 K1 K 2 . Then j0 , k0 K1 and j0 , k0 K 2 . If j0 , k0 K1 and j0 , k0 K 2 then x j0 k0 L, z
