# The Generalized Difference OperatorΔi3 of Order Three and Its Domain in the Sequence Spaces ℓ1 and bv

Authors: Orhan Tuğ
Journal: Journal of Mathematics
 (2022)

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## Abstract

Most recently, the generalized difference operatorΔi3 of order three was defined and its domain in Hahn sequence space h was calculated. In this paper, the spaces ℓ1Δi3 and bvΔi3 are introduced as the domain of generalized difference operator Δi3 of order three in the sequence spaces ℓ1 and bv. Then, some topological properties of ℓ1Δi3 and bvΔi3 are given, and some inclusion relations are shown. Additionally, algebraic dual, α−, β−, and γ− dual spaces of ℓ1Δi3 and bvΔi3 are computed. In the last section, the classes μΔi3:λ and λ:μΔi3 of matrix transformations are characterized, where μ=ℓ1,bv and λ=c,c0,ℓ1,ℓ∞,bs,cs,bv,h.


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## Body

## 1. Preliminaries and Notations

The set of all complex valued sequences is denoted byω and, according to the classification of ω each subset of ω, is called a sequence space. In the literature of sequences, the set ℓ∞ which is called the set of all bounded sequences, the set c which is called the set of all convergent sequences, and the set c0 which is called the set of all null sequences are called classical sequence spaces. If a sequence space μ is a complete metric space with continuous coordinates, then it is called FK-space. A normed FK-spaces is called a BK-space. Therefore, the classical sequence spaces ℓ∞, c, and c0 are BK-spaces with respect to the norm defined by x∞=supk∈ℕxk. Moreover, the spaces of all absolutely p− summable, convergent series, null series, and bounded series are denoted by ℓp, cs, cs0, and bs, respectively, where 1<p<∞.The space of absolutely summable sequences which is denoted byℓ1 and the space of all sequences of bounded variation which is denoted by bv are defined, respectively, as follows:(1)ℓ1=x=xk∈ω:∑k=0∞xk<∞,and it is a BK-space with its norm xℓ1=∑k=0∞xk<∞:(2)bv=x=xk∈ω:∑k=0∞xk−xk−1<∞,and it is a BK-space with respect to the norm xbv=x0+∑k=1∞xk−xk−1<∞. On the one hand, the sequence space bv is defined as the backward difference operator Δ domain on the sequence space ℓ1, where Δxk=xk−xk−1, for all k∈ℕ. On the other hand, we can also represent bv as(3)bv=x=xk∈ω:∑k=1∞xk−xk+1<∞,which is the forward difference operator Δ domain on the sequence space ℓ1, where Δxk=xk−xk+1, for all k∈ℕ. The space bv0=bv∩c0 and the inclusions ℓ1⊂bv0⊂bv⊂c are strictly held.The alpha-dualλα, beta-dual λβ, and gamma-dual λγ of a sequence space λ are defined by(4)λα≔x=xk∈ω:xy=xkyk∈ℓ1for ally=yk∈λ,λβ≔x=xk∈ω:xy=xkyk∈csfor ally=yk∈λ,λγ≔x=xk∈ω:xy=xkyk∈bsfor ally=yk∈λ.Theα−,β−,γ− duals of the spaces ℓ1 and bv are defined by(5)ℓ1α=ℓ1β=ℓ1γ=ℓ∞,bvα=bv0α=ℓ1,bvβ=cs,bv0β=bs,bvγ=bv0γ=bs.LetA=ankk,n∈ℕ be an infinite matrix and λ,μ∈ω. We write(6)yk=Axn=∑kankxk,and then, we say that A defines a matrix transformation from λ into μ as A:λ⟶μ if Ax=Axn∈μ, for every x∈λ. We denote the set of all infinite matrices that map the sequence space λ into the sequence space μ by λ:μ. Thus, A∈λ:μ if and only if the right side of (6) converges for every n∈ℕ, that is, An∈λβ, for all n∈ℕ, and we have Ax∈μ, for all x∈λ.If a normed sequence spaceλ contains a sequence bn with the following property that, for every x∈λ, there is a unique sequence of scalars αn such that(7)limn⟶∞x−α0b0+α1b1+⋯+αnbn=0,then bn is called a Schauder basis for λ. The series ∑kαkbk which has the sum x is then called the expansion of x with respect to bn and written as x=∑kαkbk.Ifλ is an FK-space, ϕ⊂λ, and ek is a basis for λ, then λ is said to have AK property, where ek is a sequence whose only term in kth place is 1; the others are zero, for each k∈ℕ and ϕ=spanek. If ϕ is dense in λ, then λ is called AD-space; thus, AK implies AD.Letλ be a sequence space and A=ankn,k∈ℕ be an infinite matrix. Then, the matrix domain λA of an infinite matrix A in the sequence space λ is defined by(8)λA=x=xk∈ω:∑kankxkn∈ℕexistandisinλ.Wilansky (Theorem 4.4.2, p. 66 of [1]) defined that if λ is a sequence space, then the continuous dual λA∗ of the space λA is given by(9)λA∗=f:f=g°A,g∈λ∗.It is well known thatℓ1∗=bv∗=ℓ∞ (see [2, 3]).


## 2. The New Difference Sequence Spacesℓ1Δi3 and bvΔi3

Now, we define the new difference sequence spacesℓ1Δi3 and bvΔi3 as the domain of generalized difference matrix Δi3 of order three in the sequence space ℓ1 and bv. Then, we show that ℓ1Δi3 and bvΔi3 are BK− spaces and they are linearly isomorphic to the sequence spaces ℓ1 and bv, respectively, and we show that Δi3 is a linear and bounded operator over the sequence spaces ℓ1 and bv to prove the inclusion relations among ℓ1Δi3 and ℓ1 and bvΔi3 and bv, respectively.The difference matrixΔ of order one was defined by Kizmaz [4] as Δxk=xk−xk+1, and he studied its domain on classical sequence spaces. The generalized difference operator Δ2 of order two was defined by Dutta and Baliarsing [5] as Δ2xk=xk−2xk−1+xk−2, and they studied its spectrum on the sequence space c0. Moreover, Baliarsing and Dutta [5] defined the generalized difference operator Δi2 of order two as Δi2xk=xk−xk−1+1/3xk−2 and studied its spectral subdivisions over the sequence spaces c0 and ℓ1. Then, again, Dutta and Baliarsing [6] defined generalized difference operator Δ3 of order three as Δ3xk=xk−3xk−1+3xk−2−xk−3 and studied its spectrum over the sequence spaces c0 and ℓ1.The generalized difference matrixΔi3=δnk of order three was defined by Malkowsky et al. [7] as(10)δnk=100000⋯−3210000⋯1−321000⋯−141−32100⋯0−141−3210⋯00−141−321⋯⋮⋮⋮⋱⋱⋱⋱.The matrixΔi3 transforms a sequence x by(11)Δi3xk=∑i=03−1ii+13ixk−i=xk−32xk−1+xk−2−14xk−3.Then, most recently, the generalized difference matrixΔi3=δnk domain in Hahn sequence space h was calculated and studied by Tug et al. [8].Now, we define the following difference sequence spaces as the set of all sequences whoseΔi3-transforms are in the sequence spaces ℓ1 and bv as follows:(12)ℓ1Δi3≔x=xk∈ω:∑k=0∞Δi3xk<∞,bvΔi3≔x=xk∈ω:∑k=0∞Δi3xk−Δi3xk−1<∞.We define the sequencey=yk by the Δi3− transform of the sequence x=xk as(13)yk=Δi3xk=∑i=03−1ii+13ixk−i=xk−32xk−1+xk−2−14xk−3,for all k∈ℕ. The generalized difference matrix Δi3 of order three is a triangle; then, it is invertable and the inverse is unique. Therefore, we obtain by considering the relation between the terms of x=xk and y=yk (13), and x0,x−1,x−2,… are zero terms that(14)xk=Δi3−1yk=yk+32yk−1−yk−2+14yk−3,where Δi3−1=B=bnk is defined by(15)bnk=100000⋯3210000⋯54321000⋯585432100⋯11658543210⋯−7321165854321⋯⋮⋮⋮⋱⋱⋱⋱.Theorem 1.
The sequence spacesℓ1Δi3 and bvΔi3 are BK-spaces with respect to the norm:(16)xℓ1Δi3=∑k=0∞Δi3xk,xbvΔi3=x0+∑k=1∞Δi3xk−Δi3xk−1,orxbvΔi3=∑k=1∞Δi3xk−Δi3xk+1,respectively.Proof.
Sinceℓ1 and bv are BK-spaces and Δi3 is a triangle matrix, we obtain from the Theorem 4.3.2 of Wilansky (p. 61 of [1]) that ℓ1Δi3 and bvΔi3 are also BK-spaces.Relation (14) between the terms of the sequences x=xk and y=yk is given by the following calculation:(17)y1=x1⟹x1=y1y2=x2−32x1⟹x2=y2+32y1y3=x3−32x2+x1⟹x3=y3+32y2+54y1⋮Δi3xk=yk=xk−32xk−1+xk−2−14xk−3xk=Δi3−1yk=yk+32yk−1−yk−2+14yk−3.Thus, the following equation which was derived from (17) is given by(18)λ3−32λ2+λ−14=0⟹4λ3−6λ2+4λ−1=0,and we calculate roots of equation (18) as one real and two complex as follows:(19)λ1=12,λ2=12+i12andλ3=12−i12.Then, we have the following simple calculations by random three rootsλ1,λ2, and λ3 of equation (18):(20)λ1+λ2+λ3=32,(21)λ1λ2λ3=14,(22)λ1λ2+λ2λ3+λ1λ3=1,(23)λ13−32λ12+λ1−14=0,(24)λ12+λ22−32λ1λ2+λ1λ2+1=0,(25)λ12+λ22+λ32+λ1λ2+λ2λ3+λ1λ3−32λ1+λ2+λ3+1=0,(26)λ1+λ2+λ3−32=0.Theorem 2.
The sequence spaceℓ1Δi3 is linearly isomorphic to the sequence spaces ℓ1, i.e., ℓ1Δi3≅ℓ1.Proof.
Suppose that the transformationT defined from the space ℓ1Δi3 onto ℓ1 as T:ℓ1Δi3⟶ℓ1 by x⟶y=Tx=Δi3x. The linearity of T is clear. Moreover, since Tx=0 gives xk=0, for all k∈ℕ.
Let us takey∈ℓ1 and consider the sequence x=xk with respect to relation (14) as(27)xk=∑j=0k∑i=0k−j∑v=0k−j−iλ1k−j−i−vλ2vλ3iyj,for every k∈ℕ. By considering equations (23)–(26), we have the following Δi3 transform of the sequence x as(28)Δi3xk=xk−32xk−1+xk−2−14xk−3=∑j=0k∑i=0k−j∑v=0k−j−iλ1k−j−i−vλ2vλ3iyj−32∑j=0k−1∑i=0k−j−1∑v=0k−j−i−1λ1k−j−i−v−1λ2vλ3iyj+∑j=0k−2∑i=0k−j−2∑v=0k−j−i−2λ1k−j−i−v−2λ2vλ3iyj−14∑j=0k−3∑i=0k−j−3∑v=0k−j−i−3λ1k−j−i−v−3λ2vλ3iyj=∑j=0k−3∑i=0k−j−3∑v=0k−j−i−3λ1k−j−i−v−3λ2vλ3iλ13−32λ22+λ3−14+λ3iλ2k−j−iλ12+λ22−32λ1λ2+λ1λ2+1+λ3k−jλ12+λ22+λ32+λ1λ2+λ2λ3+λ1λ3−32λ1+λ2+λ3+1yj+yk−2λ12+λ22+λ32+λ1λ2+λ2λ3+λ1λ3−32λ1+λ2+λ3+1+yk−1λ1+λ2+λ3−32+yk=yk,for all k∈ℕ. Thus, Tx=y, for all x=xk∈ℓ1Δi3; then, T is surjective. Moreover, for every x=xk∈ℓ1Δi3, we have(29)xℓ1Δi3=∑k=0∞Δi3xk=∑k=0∞yk=yℓ1.
Hence,x is an element of ℓ1Δi3, and clearly, T is surjective and preserves the norm. Therefore, ℓ1Δi3≅ℓ1. It completes the proof.Theorem 3.
The sequence spacebvΔi3 is linearly isomorphic to the sequence space bv, i.e., bvΔi3≅bv.Proof.
Suppose that the transformationT is defined from the space bvΔi3 onto bv as T:bvΔi3⟶bv by x⟶y=Tx=Δi3x. The linearity of T is clear. Moreover, since Tx=0 gives xk=0, for all k∈ℕ, the rest of the proof can be followed by equation (28) in the proof of Theorem 2, and thus, Tx=y, for all x=xk∈bvΔi3; then, T is surjective. Moreover, for every x=xk∈bvΔi3, we have(30)xbvΔi3=x0+∑k=1∞Δi3xk−Δi3xk−1=y0+∑k=1∞yk−yk−1=ybv.
Hence,x is an element of bvΔi3, and clearly, T is surjective and preserves the norm. Therefore, bvΔi3≅bv. It completes the proof.Theorem 4.
The inclusionℓ1Δi3⊂bvΔi3 strictly holds.Proof.
Suppose thatx∈ℓ1Δi3, then y=Δi3x∈ℓ1. Since ℓ1⊂bv, then y=Δi3x∈bv and it says x∈bvΔi3 and it shows the inclusion ℓ1Δi3⊂bvΔi3) holds. To show that the inclusion ℓ1Δi3⊂bvΔi3 is strict, let us define xk=e=1,1,…. Then, clearly, x∈bvΔi3, but x∉ℓ1Δi3, since Δi3xk⟶1/4↛0 as k⟶∞. This completes the proof.Lemma 1.
The matrixA=ank is a bounded linear operator, A∈Bℓ1, from ℓ1 to itself, if and only if the supremum of ℓ1 norms of the columns of A is bounded, i.e.,(31)supk∈ℕ∑n=1∞ank<∞.Theorem 5.
Δi3:ℓ1⟶ℓ1 is a bounded linear operator.Proof.
The linearity is clear. We should show thatΔi3∈Bℓ1 which means that Δi3 satisfies the conditions of Lemma 1 with δnk instead of ank, that is,(32)Δi3ℓ1;ℓ1=154.
This completes the proof.Lemma 2.
The matrixA=ank is a bounded linear operator, A∈Bbv, from bv to itself, if and only if(33)supk∑n=0∞∑i=k∞ani−an−1,i<∞.Theorem 6.
Δi3:bv⟶bv is a bounded linear operator.Proof.
The linearity is clear. We should show thatΔi3∈Bbv which means that Δi3 satisfies the conditions of Lemma 2 with δnk instead of ank, that is,(34)Δi3bv;bv=supk∑n=0∞∑i=k∞δni−δn−1,i=154.
This completes the proof.Theorem 7.
ℓ1=ℓ1Δi3.Proof.
Since the operatorΔi3 is a bounded and linear operator on the sequence space ℓ1 by Theorem 5, Δi3∈ℓ1,ℓ1 if and only if condition (31) is satisfied. Moreover, Δi3x∈ℓ1, for every x∈ℓ1. This shows that the inclusion ℓ1⊂ℓ1Δi3 holds.
Moreover, the matrixB=bnk, the inverse matrix of Δi3, which can be reduced from the inverse matrix B=bnk in the Theorem 2 of [2] by only choosing λ=0. Then, we write the following to calculate Δi3−1∈ℓ1;ℓ1. Therefore, the operator Δi3−1 has the following equation and the following calculations of its roots:(35)Δi3−1yk=yk+32yk−1−yk−2+14yk−3,where Δi3−1=B=bnk is defined by(36)bnk=100000⋯3210000⋯54321000⋯585432100⋯11658543210⋯−7321165854321⋯⋮⋮⋮⋱⋱⋱⋱.
The notationΔi3−1yk=yk+3/2yk−1−yk−2+1/4yk−3 gives us the following equation as(37)w3+32w2−w+14=0⟹4w3+6w2−4w+1=0.
Then, one real and two complex roots of equation (37) are(38)w1=−12−1236−267332/3+7336−2673,w2=−12+141+i336−267332/3+71−i3336−2673,w3=−12+141−i336−267332/3+71+i3336−2673.
Then, we have the following simple calculations by random three rootsw1,w2, and w3 of equation (37):(39)w1+w2+w3=−32,w1w2w3=−14,w1w2+w2w3+w1w3=−1,w13+32w12−w1+14=0,w12+w22+32w1+w2+w1w2−1=0,w12+w22+w32+w1w2+w2w3+w1w3+32w1+w2+w3−1=0,w1+w2+w3+32=0,bnk=∑j=0k∑i=0k−j∑v=0k−j−iw1k−j−i−vw2vw3i,0≤k≤n,0,k>n,for all n,k∈ℕ. We can show that B∈ℓ1,ℓ1, and if x∈ℓ1Δi3, then y=Δi3x∈ℓ1 and x=Δi3−1y=By∈ℓ1. Thus, we can say that ℓ1Δi3⊂ℓ1. This completes the proof.Theorem 8.
bv=bvΔi3.Proof.
Since the proof can be done similarly as in the proof of Theorem7, we omit it.Theorem 9.
Letαk=Δi3xk, for all k∈ℕ and μ=ℓ1,bv. Define the sequence uk=unkn∈ℕ in the sequence space μΔi3 as follows:(40)unk=∑j=0k∑i=0k−j∑v=0k−j−iλ1k−j−i−vλ2vλ3i,0≤k≤n,0,k>n,for every fixed k∈ℕ. Then, uk=unkn∈ℕ is a basis for μΔi3, and there is a unique representation of x∈μΔi3 as(41)x=∑kαkuk.Proof.
Since the proof can be done similarly for both sequence space, we consider to prove for the sequence spaceℓ1. First, we need to show that uk∈ℓ1Δi3, and it is enough to show Δi3uk∈h, for all k∈ℕ. To show this(42)Δi3uk=uk−32uk−1+uk−2−14uk−3=∑j=0k∑i=0k−j∑v=0k−j−iλ1k−j−i−vλ2vλ3i−32∑j=0k−1∑i=0k−j−1∑v=0k−j−i−1λ1k−j−i−v−1λ2vλ3i+∑j=0k−2∑i=0k−j−2∑v=0k−j−i−2λ1k−j−i−v−2λ2vλ3i−14∑j=0k−3∑i=0k−j−3∑v=0k−j−i−3λ1k−j−i−v−3λ2vλ3i=∑j=0k−3∑i=0k−j−3∑v=0k−j−i−3λ1k−j−i−v−3λ2vλ3iλ13−32λ12+λ1−14+λ3iλ2k−j−iλ12+λ22−32λ1+λ2+λ1λ2+1+λ3k−jλ12+λ22+λ32+λ1λ2+λ2λ3+λ1λ3−32λ1+λ2+λ3+1+λ12+λ22+λ32+λ1λ2+λ2λ3+λ1λ3−32λ1+λ2+λ3+1+λ1+λ2+λ3−32+1=1,for every k∈ℕ. Then, clearly, one can see that Δi3uk=ek∈ℓ1 and then uk∈ℓ1Δi3.
Let us take a sequencex∈ℓ1Δi3. Then, we obtain the following representation for every nonnegative integer m as(43)xm=∑kmαkuk.
Then, the following holds by taking theΔi3 transform of (43) that(44)Δi3xm=∑kmαkΔi3uk=∑kmΔi3xkek,and from (44), we have(45)Δi3x−xmn=0,0≤n≤m,Δi3xn,n>m.
Therefore, for every givenε>0, there exists an integer m0 such that(46)∑n=m∞Δi3xn<ε2,for all m≥m0. Hence,(47)x−xmℓ1Δi3=∑n=m∞Δi3xn≤∑n=m0∞Δi3xn<ε2,for all m≥m0. This proves that x∈ℓ1Δi3.


## 3. Dual Spaces of the Sequence Spacesℓ1Δi3 and bvΔi3

We begin this section by calculating the algebraic dual space ofℓ1Δi3 and bvΔi3, respectively.Theorem 10.
The algebraic dualℓ1Δi3∗ of the space ℓ1Δi3 is the sequence space ℓ∞.Proof.
Let us defineT:ℓ1Δi3∗⟶ℓ∞ with Tf=fuk which is a surjective linear map, and T is injective since uk is a basis for ℓ1Δi3. Let f∈ℓ1Δi3∗, and since x∈ℓ1Δi3, we can write(48)fx=f∑k=0∞Δi3xkuk=∑k=0∞Δi3xkfuk.
Then, we have(49)fx=f∑k=0∞Δi3xkuk=∑k=0∞Δi3xkfuk≤∑k=0∞Δi3xkfuk≤supk≥1fuk∑k=0∞Δi3xk=Tf∞xℓ1Δi3.
Thus,f≤Tf∞.
Moreover, sincefuk≤fuk=f, for every k∈ℕ, then Tf∞=supk≥1fuk=f. Therefore, f=Tf∞.Theorem 11.
The algebraic dualbvΔi3∗ of the space bvΔi3 is the sequence space ℓ∞.Proof.
Since the proof can be followed similarly as in Theorem10, we omit the repetition.Now, we start with the following lemma which is needed in proving our theorems. Here and after, we denote the collection of all finite subsets ofℕ by K.Lemma 3 (see [9]).
LetA=ank be an infinite matrix over the complex field. Then, the following statements hold:(i)
A∈ℓ1:ℓ∞ if and only if(50)supn,k∈ℕank<∞.(i)
A∈ℓ1:ℓ1 if and only if(51)supk∈ℕ∑nank<∞.(iii)
A∈ℓ1:c if and only if (50) holds, and(52)∃ak∈ℂsuch thatlimn⟶∞ank=akforallk∈ℕ.(iv)
A∈c:ℓ1 if and only if(53)supK∈ℕfinite∑n∑k∈Kank<∞.Lemma 4 (see [9]).
LetA=ank be an infinite matrix over the complex field. Then, the following statements hold:(i)
A∈bv:ℓ1 if and only if(54)supl∈ℕ∑n∑k=l∞ank<∞.(ii)
A∈bv:bs if and only if(55)supm,l∈ℕ∑n=0m∑k=l∞ank<∞.(iii)
A∈bv:cs if and only if (55) holds, and(56)∑nankconverges for eachk∈ℕ,(57)∑n∑kankconverges.(iv)
A∈bv:bv if and only if(58)supk∑n=0∞∑i=k∞ani−an−1,i<∞.(v)
A∈bv:ℓ∞ if and only if(59)supn,k∈ℕ∑i=k∞ani<∞.(vi)
A∈bv:c if and only if (57) and (59) hold and(60)Ae∈c.Theorem 12.
Let the setd1 be defined as(61)d1=a=ak∈ω:supj∈ℕ∑k=0∞∑j=0k∑i=0k−j∑v=0k−j−iλ1k−j−i−vλ2jλ3iak<∞.
Then,ℓ1Δi3α=d1.Proof.
Let the sequencea=ak be in ω. By relation (27), we have the following equality:(62)akxk=∑j=0k∑i=0k−j∑v=0k−j−iλ1k−j−i−vλ2jλ3iakyj=Ayk,where the matrix A=akj is defined via the sequence a=ak as(63)akj=∑j=0k∑i=0k−j∑v=0k−j−iλ1k−j−i−vλ2jλ3iak,1≤j≤k,0,j>k.
Now, clearly, we can say thatax=akxk∈ℓ1 whenever x=xk∈ℓ1Δi3 if and only if Ay∈ℓ1 whenever y=yk∈ℓ1. Therefore, A∈ℓ1;ℓ1 and the condition of Lemma 3 (ii) holds, that is,(64)supj∈ℕ∑k=0∞∑j=0k∑i=0k−j∑v=0k−j−iλ1k−j−i−vλ2jλ3iak<∞.
It completes the proof.Theorem 13.
ℓ1Δi3β=d2∩d3, where(65)d2=a=ak∈ω:supn,k∈ℕ∑j=kn∑i=0j−k∑v=0j−k−iλ1j−k−i−vλ2jλ3iaj<∞,(66)d3=a=ak∈ω:limn⟶∞∑j=kn∑i=0j−k∑v=0j−k−iλ1j−k−i−vλ2jλ3iajexists for eachk=1,2,3,….Proof.
Suppose thata=ak∈ω. Then, let us consider the following equality:(67)∑k=0nakxk=∑k=0nak∑j=0k∑i=0k−j∑v=0k−j−iλ1k−j−i−vλ2jλ3iyj∑k=0n∑j=kn∑i=0j−k∑v=0j−k−iλ1j−k−i−vλ2jλ3iajyk=Byn,for every n∈ℕ, where we define the matrix B=bnk as(68)bnk=∑j=kn∑i=0j−k∑v=0j−k−iλ1j−k−i−vλ2jλ3iaj,1≤k≤n,0,k>n,for all k,n∈ℕ. By the definition β− dual of a sequence space, we can say that ax=akxk∈cs wherever x=xk∈ℓ1Δi3 if and only if By∈c wherever y=yk∈ℓ1. Thus, B∈ℓ1;c, and the conditions of Lemma 3 (iii) hold. That is,(69)supn,k∈ℕ∑j=kn∑i=0j−k∑v=0j−k−iλ1j−k−i−vλ2jλ3iaj<∞limn⟶∞∑j=kn∑i=0j−k∑v=0j−k−iλ1j−k−i−vλ2jλ3iajexists for eachk=1,2,3,….
It completes the proof.Theorem 14.
The gamma-dualℓ1Δi3γ of the sequence space ℓ1Δi3 is the set d2.Proof.
The proof can be easily done by considering the similar path in the proof of Theorem13. We only state here that ax=akxk∈bs wherever x=xk∈ℓ1Δi3 if and only if By∈ℓ∞ wherever y=yk∈ℓ1. Thus, B∈ℓ1;ℓ∞, and the condition of the Lemma 3 (i) holds. That is,(70)supn,k∈ℕ∑j=kn∑i=0j−k∑v=0j−k−iλ1j−k−i−vλ2jλ3iaj<∞.
It completes the proof.Theorem 15.
Let the setS1 be defined as(71)S1=a=ak∈ω:supl∈ℕ∑n=1∞∑k=l∞∑j=0k∑i=0k−j∑v=0k−j−iλ1k−j−i−vλ2jλ3iak<∞.Then,bvΔi3α=S1.Proof.
Let the sequencea=ak be in ω. By relation (27), we have the following equality:(72)akxk=∑j=0k∑i=0k−j∑v=0k−j−iλ1k−j−i−vλ2jλ3iakyj=Ayk,where the matrix A=akj is defined via the sequence a=ak as(73)akj=∑j=0k∑i=0k−j∑v=0k−j−iλ1k−j−i−vλ2jλ3iak,1≤j≤k,0,j>k.
Now, clearly, we can say thatax=akxk∈ℓ1 whenever x=xk∈bvΔi3 if and only if Ay∈ℓ1 whenever y=yk∈bv. Therefore, A∈bv;ℓ1 and the condition of Lemma 4 (i) holds, that is,(74)supl∈ℕ∑n=1∞∑k=l∞∑j=0k∑i=0k−j∑v=0k−j−iλ1k−j−i−vλ2jλ3iak<∞.
It completes the proof.Theorem 16.
bvΔi3β=S2∩S3∩S4, where(75)S2=a=ak∈ω:∑n∑j∑j=kn∑i=0j−k∑v=0j−k−iλ1j−k−i−vλ2jλ3iajconverges,S3=a=ak∈ω:supn,v∈ℕ∑k=v∞∑j=kn∑i=0j−k∑v=0j−k−iλ1j−k−i−vλ2jλ3iaj<∞,S4=a=ak∈ω:limn⟶∞∑j=kn∑i=0j−k∑v=0j−k−iλ1j−k−i−vλ2jλ3iajexists for eachk=1,2,3,….Proof.
Suppose thata=ak∈ω. Then, let us consider the following equality:(76)∑k=0nakxk=∑k=0nak∑j=0k∑i=0k−j∑v=0k−j−iλ1k−j−i−vλ2jλ3iyj,=∑k=0n∑j=kn∑i=0j−k∑v=0j−k−iλ1j−k−i−vλ2jλ3iajyk=Byn,for every n∈ℕ, where we define the matrix B=bnk as(77)bnk=∑j=kn∑i=0j−k∑v=0j−k−iλ1j−k−i−vλ2jλ3iaj,1≤k≤n,0,k>n,for all k,n∈ℕ. By the definition β− dual of a sequence space, we can say that ax=akxk∈cs wherever x=xk∈bvΔi3 if and only if By∈c wherever y=yk∈bv. Thus, B∈bv;c and the conditions of Lemma 4 (vi) hold. That is,(78)∑n∑j∑j=kn∑i=0j−k∑v=0j−k−iλ1j−k−i−vλ2jλ3iajconverges,supm,n∈ℕ∑n∑k=0m∑j=kn∑i=0j−k∑v=0j−k−iλ1j−k−i−vλ2jλ3iaj<∞,limn⟶∞∑j=kn∑i=0j−k∑v=0j−k−iλ1j−k−i−vλ2jλ3iajexists for eachk=1,2,3,….
It completes the proof.Theorem 17.
The gamma-dualbvΔi3γ of the sequence space bvΔi3 is the set S3.Proof.
The proof can be easily done by considering the similar path in the proof of Theorem16. We only state here that ax=akxk∈bs wherever x=xk∈bvΔi3 if and only if By∈ℓ∞ wherever y=yk∈bv. Thus, B∈bv;ℓ∞ and the condition of Lemma 4 (v) holds. That is,(79)supn,v∈ℕ∑k=v∞∑j=kn∑i=0j−k∑v=0j−k−iλ1j−k−i−vλ2jλ3iaj<∞.
It completes the proof.


## 4. Matrix Transformations onℓ1Δi3 and bvΔi3

In this section, we characterize the matrix classesλΔi3;μ and μ;λΔi3, where λ∈ℓ1,bv and μ represents any sequence space. Throughout this section, we define the following matrices E=enk and D=dnk as(80)enk=∑j=k∞∑i=0j−k∑v=0j−k−iλ1j−k−i−vλ2jλ3ianj,dnk=ank−32an−1,k+an−2,k−14an−3,k,for all k,n∈ℕ.Theorem 18.
The infinite matrixA∈λΔi3;μ if and only if(81)An∈λΔi3β,(82)E∈λ;μ.Proof.
Suppose thatA∈λΔi3;μ. Then, Ax exists and is in the sequence space μ for every arbitrary sequence x=xk∈λΔi3, that is, y=Δi3x∈λ. Then, clearly ankk∈ℕ∈λΔi3β holds. Since Ax exists for every arbitrary sequence x=xk∈λΔi3, we consider the following equality derived from the mth partial sum of the series ∑kankxk:(83)∑k=0mankxk=∑k=0mank∑j=0k∑i=0k−j∑v=0k−j−iλ1k−j−i−vλ2jλ3iyj=∑k=0m∑j=km∑i=0k−j∑v=0k−j−iλ1k−j−i−vλ2jλ3ianjyk.
Thus, by lettingm⟶∞ in equality (83), we can see that(84)∑kankxk=∑kenkyk,for every n∈ℕ. Therefore, Ax=Ey, for every y∈λ, that is, E∈λ;μ.
Conversely, assume that conditions (81) and (82) hold and suppose that x=xk∈λΔi3. We can clearly say by (81) that Ax exists, and since the sequence spaces λΔi3 and h are isomorphic, then we have Δi3x=y∈h. We should show that Ax∈μ for every x=xk∈λΔi3. It can be easily seen by applying equation (83) that Ax=Ey, and since (82) holds, we can state here that A∈λΔi3;μ which complete the proof.Corollary 1.
The following statements hold for the infinite matrixA=ank:(i)
A∈ℓ1Δi3;ℓ∞ if and only if (50) holds with enk instead of ank(ii)
A∈ℓ1Δi3;ℓ1 if and only if (51) holds with enk instead of ank(iii)
A∈ℓ1Δi3;c if and only if (50) and (52) hold with enk instead of ankCorollary 2.
The following statements hold for the infinite matrixA=ank:(i)
A∈bvΔi3;ℓ1 if and only if (54) holds with enk instead of ank(ii)
A∈bvΔi3;bs if and only if (55) holds with enk instead of ank(iii)
A∈bvΔi3;cs if and only if (55)–(57) hold with enk instead of ank(iv)
A∈bvΔi3;bv if and only if (58) holds with enk instead of ank(v)
A∈bvΔi3;ℓ∞ if and only if (59) holds with enk instead of ank(vi)
A∈bvΔi3;c if and only if (57), (59), and (60) hold with enk instead of ankLemma 5 (see Corollary5 of [3]).
The infinite matrixA=ank∈bv;h if and only if(85)limn⟶∞∑k=1mank=0,(86)supk∈ℕ∑n=1∞n∑k=1mank−an+1,k<∞,(87)Ae∈h.Lemma 6 (see Theorem 4.15 of [3]).
We haveA∈ℓ1,h if and only if holds and(88)limn⟶∞ank=0,k=1,2,…,(89)Aℓ1,h=supk∑n=1∞nank−an+1,k<∞.Corollary 3.
The following statements hold for the infinite matrixA=ank.(i)
A∈bvΔi3,h if and only if the conditions in (85)–(87) hold with enk instead of ank(ii)
A∈ℓ1Δi3,h if and only if the conditions in (88) and (89) hold with enk instead of ankTheorem 19.
The infinite matrixA∈μ;λΔi3 if and only if(90)D∈μ;λ.Proof.
Suppose thatA∈μ;λΔi3. Then, Ax exists, and it is in λΔi3, that is, Δi3Ax∈λ. Then, we observe the following equality that(91)Δi3Axn=Axn−32Axn−1+Axn−2−14Axn−3=∑kank−32an−1,k+an−2,k−14an−3,kxk=Dxn.
Thus, it showsDx∈λ, for every x∈μ, i.e., D∈μ;λ. This completes the proof.Corollary 4.
The following statements hold for the infinite matrixA=ank.(i)
A∈ℓ1,ℓ1Δi3 if and only if (51) holds with dnk instead of ank(ii)
A∈c,ℓ1Δi3 if and only if (53) holds with dnk instead of ank(iii)
A∈bv,bvΔi3 if and only if (58) holds with dnk instead of ankLemma 7 (see [10]).
The infinite matrixA∈h:ℓ1 if and only if(92)∑n=1∞ankconverges,k=1,2,…,(93)supm1m∑n=1∞∑k=1mank<∞.Lemma 8 (see Theorem 4.3 of [3]).
The infinite matrixA=ank∈h;bv if and only if(94)supm∈ℕ1m∑n∑k=1mank−an−1,k<∞.Corollary 5.
The following statements hold for the infinite matrixA=ank:(i)
A∈h,ℓ1Δi3 if and only if the conditions in (92) and (93) hold with dnk instead of ank(ii)
A∈h,bvΔi3 if and only if the condition in (94) holds with dnk instead of ank


## 5. Conclusion

The difference operator and generalized difference operator of several orders were studied by several distinguished mathematicians (see [11–26]) as a matrix domain on several sequence spaces. The generalized difference operator Δi3 of order three was defined and the spectrum of Δi3 on the Hahn sequence space h calculated by Malkowsky et al. [7]. Then, the matrix domain of Δi3 in Hahn sequence space h was calculated by Tug et al. [8].In this research paper, we calculated the generalized difference operatorΔi3 of order three domain in the sequence spaces ℓ1 and bv. Then, we stated some topological properties of ℓ1Δi3 and bvΔi3, and we showed some inclusion relations. Moreover, we calculated the algebraic dual, α−, β−, and γ− dual spaces of ℓ1Δi3 and bvΔi3. Finally, we characterized the matrix classes μΔi3:λ and λ:μΔi3, where μ=ℓ1,bv and λ=c,c0,ℓ1,ℓ∞,bs,cs,bv, and we conclude the paper with some important results.As a natural continuation of this paper, the fine spectrum and its subdivisions of the operatorΔi3 over the sequence spaces ℓ1 and bv can be calculated. Moreover, the matrix domain of Δi3 on the generalized Hahn sequence spaces hd which was defined and studied by Malkowsky et al. [27] can also be calculated.



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