Zweier I-convergent difference sequence spaces defined by a sequence of moduli

J. Math. Comput. Sci. 6 (2016), No. 4, 588-596 ISSN: 1927-5307

ZWEIER I-CONVERGENT DIFFERENCE SEQUENCE SPACES DEFINED BY A SEQUENCE OF MODULI

KHALID EBADULLAH

College Of Science and Theoretical Studies, Saudi Electronic University, Kingdom of Saudi Arabia

Copyright c2016 Khalid Ebadullah. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract. In this paper, we introduce the sequence spacesZ₀I(F,4)andZ_∞I(F,4)for the sequence of modulii

F= (fk)and study some inclusion relations that arise on the said spaces.

Keywords:difference sequence spaces, sequence of moduli, I-Convergence, Zweier sequences.

2010 AMS Subject Classification:40A05, 40A35, 40C05.

1. Introduction

LetN,RandCbe the sets of all natural, real and complex numbers respectively. We write

ω ={x= (xk):xk∈R or C},

the space of all real or complex sequences.

Let`_∞, c andc0 be the linear spaces of bounded, convergent and null sequences respectively,

normed by

||x||_∞=sup

k

|x_k|, wherek∈N.

Received March 03, 2016

The approach of constructing new sequence spaces by means of the matrix domain of a partic-ular limitation method have been recently employed by Altay, Bas¸ar and Mursaleen[1], Bas¸ar and Altay[2], Malkowsky[11], Ng and Lee[13], and Wang[15].

S¸eng¨on¨ul[14] defined the sequencey= (yi)which is frequently used as theZptransform of the

sequencex= (xi)i.e,

y_i=px_i+ (1−p)x_i−1

wherex₋₁=0,p6=1, 1<p<∞andZpdenotes the matrixZp= (zik)defined by

z_ik=

      

     

p, (i=k),

1−p, (i−1=k),(i,k∈N) 0otherwise

Following Bas¸ar and Altay[2], S¸eng¨on¨ul[14] introduced the Zweier sequence spacesZ and

Z0as follows

Z ={x= (xk)∈ω :Zpx∈c}

Z0={x= (x_k)∈ω :Zpx∈c0}.

The idea of difference sequence spaces was introduced by Kizmaz [10] as

`∞(4) ={x= (xk)∈ω :(4xk)∈`∞}, c(4) ={x= (x_k)∈ω :(4xk)∈c},

and

c₀(4) ={x= (x_k)∈ω :(4xk)∈c0},

where4x= (x_k−x_k+1)and40_{x= (x}

k).

These are Banach spaces with the norm

||x||4=|x1|+||4x||∞.

The idea of modulus was structured by Nakano[12]. A function f : [0,∞)−→[0,∞) is called a modulus if

(2) f(t+u)≤ f(t)+ f(u) for all t, u≥0, (3) f is increasing, and

(4) f is continuous from the right at zero.

If X be a non- empty set, then a family of setI⊂P(X)(P(X)denoting the power set of X) is called an ideal in X if and only if

(a)φ ∈I;

(b) For eachA,B∈I, we haveA∪B∈I; (c) For eachA∈IandB⊂Awe haveB∈I.

If X be a non- empty set. A non- empty family of setsF ⊂P(X)(P(X)denoting the power set of X) is called a filter on X if and only if

(a)φ ∈/F;

(b) For eachA,B∈F, we haveA∩B∈F; (c) For eachA∈F andA⊂Bwe haveB∈F.

Recently Khan, Ebadullah and Yasmeen[9] introduced the following classes of sequences.

ZI

0 ={x= (xk)∈ω :I−limZpx=0},

ZI _{={x= (x}

k)∈ω:I−limZpx=Lfor some L∈C},

ZI

∞={x= (xk)∈ω : sup

k

|Zpx|<∞}.

In [6] for a modulus function f

ZI

0(f) ={(xk)∈ω : for a givenε>0,{k∈N: f(|x/k|)≥ε} ∈I},

ZI_{(f) ={(x}

k)∈ω :∃L∈Csuch that for a givenε>0,{k∈N: f(|xk/−L|)≥ε} ∈I},

ZI

∞(f) ={(xk)∈ω :{k∈N: f(|x

/

where(x/_k) = (Zpx)

In [8] for a sequence of moduliiF= (f_k)

ZI

0(F) ={(xk)∈ω :{k∈N: fk(|x /

k|)≥ε} ∈I},

ZI_{(F) ={(x}

k)∈ω:{k∈N: fk(|x /

k−L|)≥ε, for some L∈C} ∈I},

ZI

∞(F) ={(xk)∈ω :{k∈N: fk(|x

/

k|)≥M, for each fixed M>0} ∈I}.

We need the following results in order to establish some of the results of this article.

Lemma 1.1.[3, Lemma 1.2.] The condition sup

k

f_k(t)<∞, t>0 holds if and only if there is

a pointt₀>0 such that sup

k

f_k(t0)<∞.

Lemma 1.2.[3, Lemma 1.3.] The condition inf

k fk(t)>0 holds if and only if there exists a

pointt0>0 such that inf

k fk(t0)>0.

Theorem 1.3.[14, Theorem 2.2.] The sequence spacesZ andZ0are linearly isomorphic to the spacescandc₀respectively, i.eZ ∼=candZ0∼=c₀

Theorem 1.4.[14, Theorem 2.3.] The inclusionsZ0⊂Z strictly hold for p6=1.

c.f. ([3],[4],[5],[7],[9]).

2. MAIN RESULTS.

In this article we introduce the following classes of sequence spaces.

ZI

0(F,4) ={x= (xk)∈ω : I-limfk(|4x /

k|) =0};

ZI

∞(F,4) ={x= (xk)∈ω: sup

k

f_k(|4x/_k|)<∞}.

Theorem 2.1.For a sequenceF= (f_k)of moduli, the following statements are equivalent: (a)Z_∞I(4)⊆ZI

∞(F,4)

(b)Z₀I(4)⊂ZI

∞(F,4)

(c) sup

k

f_k(t)<∞, (t>0)

Proof.(a) implies (b) is obvious.

(b) implies (c). LetZ₀I(4)⊂ZI

∞(F,4).

Suppose that (c) is not true. Then by Lemma 1.1 sup

k

f_k(t) =∞for allt>0, and, therefore there is a sequence (ki) of positive

integers such that

f_ki(1

i)>ifor i=1,2,3... [2.1]

Definex= (xk)as follows

x_k=

 



1

i, ifk=ki,i=1,2,3...;

0, otherwise.

Thenx∈ZI

0(4)but by [2.1],x∈/Z∞I(F,4)which contradicts (b). Hence (c) must hold.

(c) implies (a). Let (c) be satisfied andx∈ZI

∞(4).

If we suppose thatx∈/ZI

∞(F,4)then

sup

k

If we take t =|4x|then sup

k

f_k(t) =∞which contradicts (c).

HenceZ_∞I(4)⊆ZI

∞(F,4).

Theorem 2.2.IfF = (f_k)is a sequence of moduli, then the following statements are equiva-lent:

(a)Z₀I(F,4)⊆Z₀I(4),

(b)Z₀I(F,4)⊂Z_∞I(4),

(c) inf

k fk(t)>0, (t>0).

Proof.(a) implies (b) is obvious.

(b) implies (c). LetZ₀I(F,4)⊂ZI

∞(4).

Suppose that (c) does not hold. Then, by lemma 1.2 ,

inf

k

f_k(t) =0,(t>0), [2.2] and therefore there is a sequence(ki)of positive integers such that

f_ki(i2)< 1

i fori=1,2, ...

Define the sequencex= (x_k)by

x_k=

 



i2,ifk=ki, i=1,2,3...;

0, otherwise.

By [2.2]x∈ZI

Hence (c) must hold.

(c) implies (a). Let (c) holds andx∈ZI

0(F,4)that is

lim

k fk(|4xk|) =0

.

Suppose thatx∈/Z₀I(4).

Then for someε0>0 and positive integerk0we have|4xk| ≤ε0fork≥ko.

Therefore f_k(ε₀)≥ f_k(|4x_k|)fork≥k₀and hence lim

k fk(ε0)>0 which contradictsx∈/Z I

0(4).

ThusZ₀I(F,4)⊆ZI

0(4).

Theorem 2.3.The inclusionZ_∞I(F,4)⊆ZI

0(4)holds if and only if

lim

k fk(t) =∞fort>0. [2.3] Proof.LetZ_∞I(F,4)⊆ZI

0(4)such that lim

k fk(t) =∞for t>0 does not hold.

Then there is a numbert0>0 and a sequence(ki)of positive integers such that

f_ki(t0)≤M<∞. [2.4]

Define the sequencex= (xk)by

x_k=

 



t0, ifk=ki, i=1,2,3...;

0, otherwise.

Thusx∈Z_∞I(F,4), by [2.4]. Butx∈/ZI

0(4), so that [2.3] must hold IfZ∞I(F,4)⊆Z

I

0(4).

Conversely, let [2.3] hold. Ifx∈ZI

∞(F,4), then fk(|4xk|)≤M<∞

for k = 1,2,3...Suppose thatx∈/ZI

0(4).

Then for some ε0 >0 and positive integer k0 we have |4xk| <ε0 for k ≥ k0. Therefore

Hencex∈ZI

0(4).

Theorem 2.4.The inclusionZ_∞I(4)⊆ZI

0(F,4)holds, if and only if

lim

k fk(t) =0 fort>0. [2.5] Proof.Suppose thatZ_∞I(4)⊆ZI

0(F,4)but [2.5] does not hold.

Then

lim

k fk(t0) =l6=0. [2.6]

for somet₀>0.

Define the sequencex= (xk) by

x_k=t₀

k−1

∑

v=0

(−1)





k−v

k−v





for k = 1,2,3... Thenx∈/ZI

0(F,4), by [2.6].

Hence [2.5] must hold. Conversly , letx∈ZI

∞(4)and suppose that [2.5] holds. Then|4x_k| ≤M<∞for k = 1,2,3...

Therefore f_k(|4x_k|)≤ f_k(M)for k = 1,2,3... and lim

k fk(|4xk|)≤limk fk(M) =0, by [2.5].

Hencex∈ZI

0(F,4).

Conflict of Interests

The author declare that there is no conflict of interests.

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