ISSN: 2229371X
Ab Hamid Ganie^{1*}, Mobin Ahmad^{2} and Neyaz Ahmad Sheikh^{3} 
Corresponding author: Ab Hamid Ganie, Email: 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 nonnegative 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 nonnegative linear functional such that L is invariant i.e., L(S_{x})=L(x) and L(e)=1, e=(1,1,1,…) [1]. 
Lorentz, called a sequence {x_{n}} almost convergent if all Banach limits of x, L(x), are same and this unique Banach limit is called Flimit of x [1]. In his paper, Lorentz proved the following criterian for almost convergent sequences. 
A sequence l∞ ∈ is almost convergent with Flimit L(x) if and only if 
where, uniformly in n≥0. 
We denote the set of almost convergent sequences by f. 
Several authors including Duran [2], Ganie et al. [37], 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 
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]. 
(triangle inequality) 
is a sequence of scalars with λ_{n}→λ (n→∞) and (x_{n}) 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=(p_{k}) 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]: 
The difference sequence spaces, 
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: 
Where 
and 
Recently, M. Et [19] defined the following: 
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: 
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: 
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,Báµ such that Since f is subadditive and is linear 
As n→∞, uniformly in m. This proves that is linear and the result follows. 
Theorem 2: Let g be any modulus function. Then 
Proof: We shall prove the result for and the second shall be proved on similar lines. Let 
Now, by definition of g, we have 
Thus, for any number L, there exists a positive integer KL such that we have 
Since, , we have x= ∈ ), and the proof of the result follows. 
Theorem 3: is a paranormed space with 
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 
which shows that is subadditive. Further, let α be any complex number. Therefore, we have by definition of g, we have 
where, S_{α} is an integer such that α<S_{α}. 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,…,r1 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 
Thus, Continuing in this way, we shall get for j=1,2,…,r1. The inclusion is strict. For this, we consider x=(k_{r}) and is in but does not belong to for f(x)=x and n=1. ( if x=(k_{r}), then and for all 
Theorem 5: 
Proof: The proof is obvious from Theorem 4 above. 
Theorem 6: Let g, g_{1} and g_{2} 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 
where the first summation is over and second summation is over y_{k+m}> δ. As g is continuous, we have (Equation 4) 
(4) 
and for y_{k+m}> δ, we use the fact that 
Now, by definition of g, we have for y_{k+m}> δ that 
Thus (Equation 5), 
(5) 
Consequently, we see from (4) and (5) that 
To prove (ii), we have from (1) that 
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, 
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 p_{k}=1 for all k and g(x)=x with r=1=n. Then, but when α_{k}=(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 p_{k}=1 for all k and g(x)=x with n=1. Then, Let the rearrangement of (x_{k}) be (y_{k}) where (y_{k}) is defined as follows, 
Then, and this proves the result. 
References 
