Journal of Global Research in Computer Sciences
MenuISSN: 2229-371X
ISSN: 2229-371X
| Ab Hamid Ganie1*, Mobin Ahmad2 and Neyaz Ahmad Sheikh3 |
| Corresponding author: Ab Hamid Ganie, E-mail: ashamidg@rediffmail.com |
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The aim of the present paper is to introduce some new generalized difference sequence spaces with respect to modulus function involving strongly almost summable sequences. We give some topological properties and inclusion relations on these spacesa.
INTRODUCTION |
A sequence space is defined to be a linear space of real or complex sequences. Throughout the paper N, R and C denotes the set of non-negative integers, the set of real numbers and the set of complex numbers respectively. Letω denote the space of all sequences (real or complex). Let l∞ and c be Banach spaces of bounded and convergent sequences  supremum norm  . Let T denote the shift operator onω , that is,   and so on. A Banach limit L is defined on l∞ as a non-negative linear functional such that L is invariant i.e., L(Sx)=L(x) and L(e)=1, e=(1,1,1,…) [1]. |
| Lorentz, called a sequence {xn} almost convergent if all Banach limits of x, L(x), are same and this unique Banach limit is called F-limit of x [1]. In his paper, Lorentz proved the following criterian for almost convergent sequences. |
A sequence  l∞ ∈ is almost convergent with F-limit L(x) if and only if |
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where,  uniformly in n≥0. |
| We denote the set of almost convergent sequences by f. |
| Several authors including Duran [2], Ganie et al. [3-7], King [8], Lorentz [1] and many others have studied almost convergent sequences. Maddox [9,10] has defined x to be strongly almost convergent to a number α if |
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By [f] we denote the space of all strongly almost convergent sequences. It is easy to see that  |
| The concept of paranorm is related to linear matric spaces. It is a generalization of that of absolute value. Let X be a linear space. A function P:x→R is called a paranorm, if [11,12]. |
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 (triangle inequality) |
 is a sequence of scalars with λn→λ (n→∞) and (xn) is a sequence of vectors with  then    ),(continuity of multiplication of vectors). |
| A paranorm p for which p(x)=0 implies x=0 is called total. It is well known that the metric of any linear metric space is given by some total paranorm [10]. |
The following inequality will be used throughout this paper. Let p=(pk) be a sequence of positive real numbers with  and let  For  . We have that (Equation 1) [9,11]. |
 (1) |
| Nanda defined the following [13,14]: |
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| The difference sequence spaces, |
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| where X= ∞ l , C and C0, were studied by Kizmaz [15]. |
| It was further generalized by Ganie et al. [5], Et and Colak [16], Sengonul [17] and many others. |
| Further, it was Tripathy et al. [18] generalized the above notions and unified these as follows: |
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| Where |
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| and |
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| Recently, M. Et [19] defined the following: |
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| Following Maddox [20]and Ruckle [21], a modulus function g is a function from [0,∞) to [0,∞) such that |
| (i) g(x)=0 if and only if x=0, |
(ii) |
| (iii) g is increasing, |
| (iv) g if continuous from right at x=0. |
| Maddox [10] introduced and studied the following sets: |
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| of sequences that are strongly almost convergent to zero and strongly almost convergent. |
Let p=(pk) be a sequence of positive real numbers with  and H=max(1, M). |
MAIN RESULTS: |
In the present paper, we define the spaces  and  as follows: |
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| Where (pk) is any bounded sequence of positive real numbers. |
Theorem 1: Let (pk) be any bounded sequence and g be any modulus function. Then  and  are linear space over the set of complex numbers. |
Proof: We shall prove the result for and the others follows on similar lines. Let  Now for  , we can find positive numbers Aα,Báõàsuch that   Since f is subadditive and  is linear |
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As n→∞, uniformly in m. This proves that  is linear and the result follows. |
| Theorem 2: Let g be any modulus function. Then |
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Proof: We shall prove the result for  and the second shall be proved on similar lines. Let  |
| Now, by definition of g, we have |
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Thus, for any number L, there exists a positive integer KL such that  we have |
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Since,  , we have x= ∈ ), and the proof of the result follows. |
Theorem 3:  is a paranormed space with |
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Proof: From Theorem 2, for each   exists. Also, it is trivial that  and  for x=0. Since, h(0)=0, we have  for x=0. Since,  therefore, by Minkowski’s inequality and by definition of g for each n that |
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which shows that  is sub-additive. Further, let α be any complex number. Therefore, we have by definition of g, we have |
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where, Sα is an integer such that α<Sα. Now, let α→0 for any fixed x with  By definition of g for  we have for  that (Equation 2) |
 (2) |
| As g is continuous, we have, for 1≤n≤N and by choosing α so small that (Equation 3) |
 (3) |
Consequently, (2) and (3) gives that  as α→0. |
Theorem 4: Let X be any of the spaces [f,g], [f,g]0 and [f,g]∞. Then, is strict. In general,  all j=1,2,…,r-1 and the inclusion is strict. |
Proof: We give the proof for the space [f, g]∞ and others can be proved similarly. So, let  Then, we have  |
| Since, g is increasing function, we have |
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Thus,  Continuing in this way, we shall get  for j=1,2,…,r-1. The inclusion is strict. For this, we consider x=(kr) and is in  but does not belong to  for f(x)=x and n=1. ( if x=(kr), then  and  for all  |
Theorem 5:  |
| Proof: The proof is obvious from Theorem 4 above. |
| Theorem 6: Let g, g1 and g2 be any modulus functions. Then, |
(i)  |
(ii)  |
Proof: Let ε be given small positive number and choose δ with 0< δ<1 such that g(t)< ε for 0<t≤ δ. We put  and consider |
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where the first summation is over  and second summation is over yk+m> δ. As g is continuous, we have (Equation 4) |
 (4) |
| and for yk+m> δ, we use the fact that |
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| Now, by definition of g, we have for yk+m> δ that |
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| Thus (Equation 5), |
 (5) |
Consequently, we see from (4) and (5) that  |
| To prove (ii), we have from (1) that |
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Let  Consequently, by adding above inequality form k=1 to k=n, we have  and the result follows. |
| Theorem 7: Let g, g1 and g2 be any modulus functions. Then, |
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| Proof: The follows as a routine verification as of the Theorem 6. |
Theorem 8: The spaces  and  are not solid in general. |
Proof: To show that the spaces  and not solid in general, we consider the following example. |
Let pk=1 for all k and g(x)=x with r=1=n. Then, but  when αk=(-1)k for all  Hence is result follows. |
| From above Theorem, we have the following corollary. |
Corollary 9: The spaces  and  are not perfect. |
Theorem 10: The spaces  and  are not symmetric in general. |
Proof : To show that the spaces  and  are not perfect in general, to show this, let us consider pk=1 for all k and g(x)=x with n=1. Then,  Let the re-arrangement of (xk) be (yk) where (yk) is defined as follows, |
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Then,  and this proves the result. |
References |
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