generalized difference sequence spaces defined

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DEFINED BY A MODULUS FUNCTION. IN A LOCALLY CONVEX SPACE. BY. YAVUZ ALTIN AND MIKAIL ET. Abstract. The idea of difference sequence spaces ...
SOOCHOW JOURNAL OF MATHEMATICS

Volume 31, No. 2, pp. 233-243, April 2005

GENERALIZED DIFFERENCE SEQUENCE SPACES DEFINED BY A MODULUS FUNCTION IN A LOCALLY CONVEX SPACE BY YAVUZ ALTIN AND MIKAIL ET

Abstract. The idea of difference sequence spaces was introduced by Kızmaz [9] and this concept was generalized by Et and C ¸ olak [5]. In this paper, we introduce some new sequence spaces with respect to a modulus function and for all q ∈ Q, where Q is the set of continuous seminorms. We give some relation related to these sequence spaces.

1. Introduction w denotes the space of all sequences and let ` ∞ , c and c0 be linear spaces of bounded, convergent and null sequences x = (x k ) with complex terms, respectively normed by kxk ∞ = supk |xk |, where k ∈ N = {1, 2, 3, . . .} , the set of positive integers. Kızmaz [9] defined the sequence spaces X(∆) = {x = (xk ) : (∆xk ) ∈ X} for X = `∞ , c or c0 , where ∆x = (xk − xk+1 ).

Let m denote an arbitrary positive integer throughout the paper. Then Et and C ¸ olak [5], generalized the above sequence spaces to the sequence

spaces X(∆m ) = {x = (xk ) : (∆m xk ) ∈ X} Received November 11, 2003; revised May 24, 2004. AMS Subject Classification. 40A05, 40C05, 46A45. Key words. difference sequence, statistical convergence, modulus function. 233

234

YAVUZ ALTIN AND MIKAIL ET

for X = `∞ , c or c0 , where m ∈ N, ∆0 x = (xk ), ∆x = (xk − xk+1 ), ∆m x = (∆m−1 xk − ∆m−1 xk+1 ), and so that ∆m xk =

m X

v=0

(−1)v

!

m xk+v . v

Subsequently difference sequence spaces have been studied by various authors: (Malkowsky and Parashar [10], Mursaleen [14], Et and Ba¸sarır [4], Et [3]). The notion of a modulus function was introduced in 1953 by Nakano [15]. We recall that a modulus f is a function from [0, ∞) to [0, ∞) such that (i) f (x) = 0 if and only if x = 0, (ii) f (x + y) ≤ f (x) + f (y), for all x ≥ 0, y ≥ 0, (iii) f is increasing, (iv) f is continuous from the right at 0. It follows that f must be continuous on [0, ∞). A modulus may be bounded or unbounded. Maddox [12] and Ruckle [18] used a modulus f to construct some sequence spaces. After then some sequence spaces, defined by a modulus function, were intro¨ urk and Bilgin [16]. duced and studied by Bilgin [2], Ozt¨ A sequence space E is said to be solid (or normal) if (α k xk ) ∈ E, whenever (xk ) ∈ E and for all sequences (αk ) of scalars with |αk | ≤ 1 for all k ∈ N. A sequence space E is said to be symmetric if (x k ) ∈ E implies (xπ(k) ) ∈ E, where π(k) is a permutation of N. 2. Main Results Let f be a modulus function, X be a locally convex Hausdorff topological linear space whose topology is determined by a set Q of continuous seminorms q and p = (pk ) be a sequence of positive real numbers. The symbol w(X) denotes the space of all sequences defined over X. We define the following sequences spaces: w(∆m , f, p, q) n 1X = x ∈ w(X) : [f (q(∆m xk − L))]pk → 0 as n → ∞ for some L , n k=1

(

)

GENERALIZED DIFFERENCE SEQUENCE SPACES DEFINED

235

n 1X w0 (∆ , f, p, q) = x ∈ w(X) : [f (q(∆m xk ))]pk → 0 as n → ∞ , n k=1

(

m

)

n 1X [f (q(∆m xk ))]pk < ∞ w∞ (∆ , f, p, q) = x ∈ w(X) : sup n n k=1

(

m

)

.

We denote w(∆m , f, p, q), w0 (∆m , f, p, q), w∞ (∆m , f, p, q) as w (∆m , p, q), w0 (∆m , p, q), w∞ (∆m , p, q) when f (x) = x. The following inequality will be used throughout this paper. Let p = (p k ) be a positive sequence of real numbers with 0 < p k ≤ supk pk = H, C = max (1, 2H−1 ). Then for ak , bk ∈ C for all k ∈ N, we have

|ak + bk |pk ≤ C {|ak |pk + |bk |pk } ,

(1)

(see for instance Maddox [11]). Theorem 2.1. Let p = (pk ) be bounded, then w(∆m , f, p, q), w0 (∆m , f, p, q) and w∞ (∆m , f, p, q) are linear spaces. Proof. We shall only prove for w0 (∆m , f, p, q). The others can be treated similary. Let x, y ∈ w0 (∆m , f, p, q) . For λ, µ ∈ C, there exists Mλ and Nµ integers

such that |λ| ≤ Mλ and |µ| ≤ Nµ . Since f is subadditive, q is a seminorm and

∆m is linear

n 1X [f (q(∆m (λxk + µyk )))]pk n k=1



n 1X [f (|λ| q(∆m xk )) + f (|µ| q(∆m yk ))]pk n k=1

≤ C (Mλ )H

n n 1X 1X [f (q(∆m xk ))]pk + C (Nµ )H [f (q(∆m yk ))]pk → 0. n k=1 n k=1

This proves that w0 (∆m , f, p, q) is a linear space. Theorem 2.2.

w0 (∆m , f, p, q) is a paranormed space (not totally par-

anormed), paranormed by g∆ (x) = sup n

(

n 1X [f (q(∆m xk ))]pk n k=1

)

1 M

236

YAVUZ ALTIN AND MIKAIL ET

where H = sup pk < ∞ and M = max (1, H) . Proof. Clearly g∆ (x) = g∆ (−x), x = θ¯ implies ∆m xk = θ and such as  q (θ) = 0 and f (0) = 0. Therefore g∆ θ¯ = 0. Since pk /M ≤ 1 and M ≥ 1, using

the Minkowski’s inequality and definition of f, for each n, we have 1 M

(

n 1X [f (q (∆m xk + ∆m yk ))]pk n k=1



(

n 1X [f (q (∆m xk )) + f (q (∆m yk ))]pk n k=1



(

n 1X [f (q (∆m xk ))]pk n k=1

)

1 M

+

(

)

)

1 M

n 1X [f (q (∆m yk ))]pk n k=1

)

1 M

.

Now it follows that g∆ is subadditive. Finally, to check the continuity of multiplication, let us take any complex λ. By definition of f we have g∆ (λx) = sup n

(

n 1X [f (q(∆m λxk ))]pk n k=1

)

1 M

H

≤ KλM g∆ (x)

where Kλ is a integer such that |λ| < Kλ . Now, let λ → 0 for any fixed x with

g∆ (x) 6= 0. By definition of f for |λ| < 1, we have

n 1X [f (q(λ∆m xk ))]pk < ε f or n > N (ε) . n k=1

(2)

Also, for 1 ≤ n ≤ N, taking λ small enough, since f is continuous we have n 1X [f (q(λ∆m xk ))]pk < ε n k=1

(3)

(2) and (3) together imply that g∆ (λx) → 0 as λ → 0. Theorem 2.3. Let f, f1 , f2 are modulus functions and 0 < h = inf p k ≤

pk ≤ supk pk = H < ∞. Then

(i) w0 (∆m , f1 , p, q) ⊆ w0 (∆m , f ◦ f1 , p, q) ,

(ii) w0 (∆m , f1 , p, q) ∩ w0 (∆m , f2 , p, q) ⊆ w0 (∆m , f1 + f2 , p, q) .

GENERALIZED DIFFERENCE SEQUENCE SPACES DEFINED

237

Proof. (i) Let (xk ) ∈ w0 (∆m , f1 , p, q) . Let ε > 0 and choose δ with 0 < δ < 1 such that f (t) < ε for 0 ≤ t ≤ δ. Write yk = f1 (q(∆m xk )) and consider n X

[f (yk )]pk =

X

[f (yk )]pk +

1

k=1

X

[f (yk )]pk

2

where the first summation is over yk ≤ δ and the second over yk > δ. Since f is continuous, we have X [f (yk )]pk < n max(εh , εH ) (4) 1

and for yk > δ we use the fact that

yk yk δ, yk