# Some Spectral Aspects of the OperatorΔνr over the Sequence Spaces ℓp and bvp(1<p<∞)

Authors: S. Dutta; P. Baliarsingh
Journal: Chinese Journal of Mathematics
 (2013)

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## Abstract

The main idea of the present paper is to compute the spectrum and the fine spectrum of the generalized difference operatorΔνr over the sequence spaces bvp(1<p<∞). The operator Δνr denotes a triangular sequential band matrix (ank) defined by Δνr(x)=(Δνrx)k=∑i=0r‍(-1)i(ri)νk-ixk-i, with xk=νk=0 for k<0, where x∈ℓp or bvp, r,k∈ℕ0={0,1,2,3,...}; the set nonnegative integers and ν=(νk) is either a constant or strictly decreasing sequence of positive real numbers satisfying certain conditions. Finally, we obtain the spectrum, the point spectrum, the residual spectrum, and the continuous spectrum of the operator Δνr over the sequence spaces ℓp and bvp. These results are more general and comprehensive than the spectrum of the difference operators Δ, Δν, Δ2,  Δuv2, and  Δr and include some other special cases such as the spectrum of the operators B(r,s), B(r,s,t), and Δab over the sequence spaces ℓp or bvp(1<p<∞).


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## Body

## 1. Introduction, Preliminaries, and Definitions

In analysis, operator theory is one of the important branch of mathematics which has vast applications in the field applied science and engineering. Operator theory deals with the study related to different properties of operators such as their inverse, spectrum, and fine spectrum. Since the spectrum of a bounded linear operator generalizes the notion of eigen values of the corresponding matrix, therefore, the study of spectrum of an operator takes a prominent position in solving many scientific and engineering problems. Hence, mathematicians and researchers have  devoted their works in achieving new ideas and concepts in the concerned field. For instance, the fine spectrum  of the Cesàro operator on the sequence spaceℓp for 1<p<∞ has been studied by Gonzalez [1]. Okutoyi [2] computed the spectrum of the Cesàro operator over the sequence space bv. The fine spectra of the Cesàro operator over the sequence space bvp have been determined by Akhmedov and Başar [3]. Akhmedov and Başar [4, 5] have studied the fine spectrum of the difference operator Δ over the sequence spaces ℓp and bvp, where 1<p<∞. Altay and Başar [6] have determined the fine spectrum of the difference operator Δ over the sequence spaces ℓp, for 0<p<1. The fine spectrum of the difference operator Δ over the sequence spaces ℓ1 and bv was investigated by Kayaduman and Furkan [7]. Srivastava and Kumar [8] have examined the fine spectrum of the generalized difference operator Δν over the sequence space c0. Recently, the spectrum of the generalized difference operator Δνr over the sequence spaces c0 and ℓ1 has been studied by Dutta and Baliarsingh [9, 10], respectively. The main focus of this paper is to define the difference operator Δνr and establish its spectral characterization with respect to the Goldberg’s classifications.Letν=(νk) be either constant or strictly decreasing sequence of positive real numbers satisfying
(1)limk→∞νk=L>0,supkνk≤2L,wherek∈ℕ0.
By ω, we denote the space of all sequences of real or complex numbers. Any subspace of ω is called a sequence space, and we write ℓ∞, c,  c0,  ℓ1,  ℓp,  and bvp for the spaces of all bounded, convergent, null, absolutely summable, p-summable, and p-bounded variation sequences, respectively. The sequence spaces lp and bvp are defined by
(2)ℓp={x∈ω:∑k=1∞‍|xk|p<∞},bvp={x∈ω:∑k=1∞‍|xk-xk-1|p<∞}.
The main goal of this article is to define a generalized difference operator Δνr as follows.For a positive integerr, we define the generalized difference operator Δνr:ℓp→ℓp by Δνr(x)=(Δνr(x))k, where
(3)Δνr(x)k=∑i=0r‍(-1)i(ri)xk-iνk-i,
with xk=νk=0 for k<0, x∈ℓp, and k∈ℕ0. It is clear that the operator Δνr can be represented by a lower triangular sequential band matrix Δνr=(ank) for n,k∈ℕ0, and
(4)ank={νk,n=k,(-1)i(ri)νk-i,n-1≤k≤n-i,1≤i≤r,0,otherwise.
Equivalently,(5)(ank)=(ν0000··⋯-rν0ν100··⋯r(r-1)2!ν0-rν1ν20··⋯-r(r-1)(r-2)3!ν0r(r-1)2!ν1-rν2ν3··⋯······⋯(-1)rν0(-1)r-1rν1··νr0⋯0(-1)rν1··-rνrνr+1⋯⋮⋮⋮⋮⋮⋮⋱).In particular, the operatorΔνr generalizes several operators considered by earlier authors concerning the same literature. For instance,  (i)
forr=1, it generalizes the operator Δν, considered by Srivastava and Kumar [8];(ii)
forr=1 and νk=r, νk-1=-s, it generalizes the difference operator B(r,s), considered by Altay and Başar [11];(iii)
forr=1 and νk=1, νk-1=1, it generalizes the difference operator Δ, considered by Altay and Başar [12];(iv)
forr=1 and νk=ak,  νk-1=-bk, it generalizes the difference operator Δa,b, considered by Akhmedov and El-Shabrawy [13];(v)
forr=2 and νk=uk, νk-1=(1/2)vk-1, it generalizes the difference operator Δuv2, considered by Panigrahi and Srivastava [14];(vi)
forr=2 and νk=1, it generalizes the 2nd order difference operator Δ2, considered by Dutta and Baliarsingh [15];(vii)
forr=2 and νk=r, νk-1=-(1/2)s, νk-2=t, it generalizes the difference operator B(r,s,t), considered by Furkan et al. [16];LetX and Y be Banach spaces and T:X→Y a bounded linear operator. By ℛ(T), we denote the range of T; that is,
(6)ℛ(T)={y∈Y:y=Tx;x∈X}.ByB(X), we denote the set of all bounded linear operators on X into itself. If X is any Banach space and T∈B(X), then the adjoint    T* of T is a bounded linear operator on the dual X* of X defined by (T*ϕ)(x)=ϕ(Tx) for all ϕ∈X* and x∈X with ∥T∥=∥T*∥.LetX≠{0} be a normed linear space over the complex field and T:D(T)→X a linear operator, where D(T) denotes the domain of  T. With T, for a complex number λ, we associate an operator Tλ=(T-λI), where I  is called identity operator on D(T), and if Tλ has an inverse, we denote it by Tλ-1; that is,
(7)Tλ-1=(T-λI)-1
and is called the resolvent operator of  T. Many properties of  Tλ and Tλ-1 depend on λ, and the spectral theory is concerned with those properties. We are interested in the set of all λ’s in the complex plane such that Tλ-1 exists, Tλ-1 is bounded, and domain of Tλ-1 is dense in X. For our investigation, we need some basic concepts in spectral theory which are given as some definitions and lemmas.Definition 1 (see [17, page 371]).
LetX and T be defined as previous. A regular value of T is a complex number λ such that
(R1)Tλ-1 exists;
(R2)Tλ-1 is bounded;
(R3)Tλ-1 is defined on a set which is dense in X.Theresolvent set ρ(T,X) of T is the set of all regular values of T. Its complement σ(T,X)=ℂ∖ρ(T,X) in the complex plane ℂ is called the spectrumof T. Furthermore, the spectrum ρ(T,X) is partitioned into three disjoint sets as follows. (I)
Point Spectrum  σp(T,X). It is the set of all λ∈ℂ such that (R1) does not hold. The elements of σp(T,X) are called eigen values of  T.(II)
Continuous Spectrumσc(T,X). It is the set of all λ∈ℂ such that (R1) holds and satisfies (R3) but does not satisfy (R2).(III)
Residual Spectrum    σr(T,X). It is the set of all λ∈ℂ such that (R1) holds but does not satisfy (R3). The condition (R2) may or may not hold.Lemma 2 (see [18, page 59]).
A linear operatorT has a dense range if and only if the adjoint T* is one to one.Lemma 3 (see [18, page 60]).
The adjoint operatorT* is onto if and only if T has a bounded inverse.LetP, Q be two nonempty subsets of the space w of all real or complex sequences and A=(ank) an infinite matrix of complex numbers ank, where n,k∈ℕ0. For every x=(xk)∈P and every positive integer n, we write
(8)An(x)=∑k‍ankxk.
The sequence Ax=(An(x)), if it exists, is called the transformation of x by the matrix A. Infinite matrix A∈(P,Q) if and only if Ax∈Q whenever x∈P.Lemma 4 (see [19, page 126]).
The matrixA=(ank) gives rise to a bounded linear operator T∈B(ℓ1) from ℓ1 to itself if and only if the supremum of ℓ1 norms of the columns of A is bounded.Lemma 5 (see [20, page 254]).
The matrixA=(ank) gives rise to a bounded linear operator T∈B(ℓ∞) from ℓ∞ to itself if and only if the supremum of ℓ1 norms of the columns of A is bounded.Lemma 6 (see [19, page 174]).
Let1<p<∞ and A∈(ℓ∞,ℓ∞)∩(ℓ1,ℓ1), then A∈(ℓp,ℓp).Lemma 7 (see [5]).
Define the spacesd1 and dq consisting all sequences x=(xk) normed by(9)∥x∥d1=supn,k|∑j=kn‍xj|,∥x∥dq=(∑k=1∞‍|∑j=k∞‍xj|q)1/q,(1<q<∞).
Then, bv1* and bvp* are isometrically isomorphic to d1 and dq.The basis of the spacebvp is also constructed and given by the following lemma.Lemma 8 (see [5]).
Define the sequenceb(k)={bn(k)}n∈ℕ of the elements of the space bvp, for every fixed k∈ℕ, the set of positive integers, by
(10)bn(k)={0,(n<k),1,(n≥k).
Then, the sequence b(k)={bn(k)}n∈ℕ is a basis for the space bvp, and x∈bvp has a unique representation of the form
(11)x=∑k=1∞‍λkb(k),whereλk=xk-xk-1,∀k∈ℕ.


## 2. The Spectrum and Fine Spectrum ofΔνr over ℓp(1<p<∞)

In this section, we compute the point spectrum, the spectrum, the continuous spectrum, the residual spectrum, and the fine spectrum of the operatorΔνr over the sequence space ℓp(1<p<∞).Theorem 9.
The operatorΔνr:ℓp→ℓp is a linear operator and satisfies the following inequalities:  (i)
if(νk) is a constant sequence
(12)(∑i=0r‍|(ri)|p)1/p≤1L∥Δνr∥(ℓp:ℓp)≤(rr1/2)(r+1),(ii)
if(νk) is a strictly decreasing sequence
(13)(∑i=0r‍|(ri)|p)1/p≤1L∥Δνr∥(ℓp:ℓp)≤2(rr1/2)(r+1), where
(14)r1/2={r2,ifriseven,r-12,ifrisodd.Proof.
(i) Suppose(νk) is a constant sequence and νk=L for all k∈ℕ0. Linearity of the operator Δνr is trivial, hence omitted. Let x=(xk)∈ℓp for all k∈ℕ0 and 1<p<∞ such that ∥x∥=1. Now, by using Minkowski’s inequality we get
(15)∥Δνr(x)∥(ℓp)=L(∑k=0∞‍|∑i=0r‍(-1)i(ri)xk-i|p)1/p≤L(∑k=0∞‍|(r0)xk|p)1/p+(∑k=0∞‍|(r1)xk-1|p)1/p+⋯+(∑k=0∞‍|(rr)xk-r|p)1/p≤supi≤r(ri)L[(∑k=0∞‍|xk|p)1/p+(∑k=0∞‍|xk-1|p)1/p+⋯+(∑k=0∞‍|xk-r|p)1/p]=L(rr1/2)(r+1)∥x∥ℓp.
Thus,
(16)1L∥Δνr∥(ℓp)≤(rr1/2)(r+1).
Suppose we denote e1=(1,0,0,…) as a sequence whose 1st entry is 1 and otherwise 0. Clearly, e1∈ℓp and
(17)∥Δνr(e1)∥(ℓp)=L(∑i=0∞‍|(ri)|p)1/p≤∥Δνr∥(ℓp)∥e1∥ℓp.
Thus,
(18)(∑i=0∞‍|(ri)|p)1/p≤1L∥Δr∥(ℓp).
Combining inequations (16) and (18), we complete the proof.
(ii) Suppose(νk) is a strictly decreasing sequence. Proof of this bit follows from (i) and with the fact that supkνk≤2L.Theorem 10.
The spectrum ofΔνr on the sequence space ℓp is given by
(19)σ(Δνr,ℓp)={α∈ℂ:|1-αL|≤2r-1}.Proof.
The proof of this theorem consists of two parts.
Part 1. In the first part, we have to show that
(20)σ(Δνr,ℓp)⊆{α∈ℂ:|1-αL|≤2r-1}.
Equivalently, we need to show that ifα∈ℂ with |1-(α/L)|>2r-1⇒α∉σ(Δνr,ℓp). Let α∈ℂ with |1-(α/L)|>2r-1. Now (Δνr-αI)=(ank) is a triangle and hence has an inverse (Δνr-αI)-1=(bnk), where (21)(bnk)=(1(ν0-α)000⋯ν0r(ν0-α)(ν1-α)1(ν1-α)00⋯b20ν1r(ν1-α)(ν2-α)1(ν2-α)0⋯b30b31ν2r(ν2-α)(ν3-α)1(ν3-α)⋯⋮⋮⋮⋮⋱),
andb20, b30, and b31  are as follows:
(22)b20=r2ν0ν1(ν0-α)(ν1-α)(ν2-α)-r(r-1)ν02!(ν0-α)(ν2-α).
Similarly,(23)b31=r2ν1ν2(ν1-α)(ν2-α)(ν3-α)-r(r-1)ν12!(ν1-α)(ν3-α),b30=r3ν0ν1ν2(ν0-α)(ν1-α)(ν2-α)(ν3-α)-r2(r-1)ν0ν12!(ν0-α)(ν1-α)(ν3-α)-r2(r-1)ν0ν22!(ν0-α)(ν2-α)(ν3-α)+r(r-1)(r-2)ν03!(ν0-α)(ν3-α).
In fact, for k∈ℕ0, one can calculate
(24)bkk=1νk-α,bk+1,k=rνk(νk-α)(νk+1-α),bk+2,k=r2νkνk+1(νk-α)(νk+1-α)(νk+2-α)-r(r-1)νk2!(νk-α)(νk+2-α),bk+3,k=r3νkνk+1νk+2(νk-α)(νk+1-α)(νk+2-α)(νk+3-α)-r2(r-1)νkνk+12!(νk-α)(νk+1-α)(νk+3-α)-r2(r-1)νkνk+12!(νk-α)(νk+2-α)(νk+3-α)+r(r-1)(r-2)νk3!(νk-α)(νk+3-α),⋮
and so on.
Now, let(25)Sk=∑n=0∞‍|bnk|=|bkk|+|bk+1,k|+|bk+2,k|+⋯=|1νk-α|+|rνk(νk-α)(νk+1-α)|+|r2νkνk+1(νk-α)(νk+1-α)(νk+2-α)-r(r-1)νk2!(νk-α)(νk+2-α)|+|r3νkνk+1νk+2(νk-α)(νk+1-α)(νk+2-α)(νk+3-α)-r2(r-1)νkνk+12!(νk-α)(νk+1-α)(νk+3-α)-r2(r-1)νkνk+12!(νk-α)(νk+2-α)(νk+3-α)+r(r-1)(r-2)νk3!(νk-α)(νk+3-α)|+⋯.
Now,
(26)limkSk=limk|1(νk-α)|+limk|rνk(νk-α)(νk+1-α)|+⋯≤1|Lr|{|LrL-α|+|LrL-α|2+|LrL-α|3+(r-1)2!|LrL-α|2+|Lr|4|L-α|4+2·(r-1)2!|LrL-α|3+(r-1)(r-2)3!|LrL-α|2+⋯}=1|Lr|{|βr|+|βr|2+|βr|3+(r-1)2!|βr|2+|βr|4+2·(r-1)2!|βr|3+(r-1)(r-2)3!|βr|2+⋯}=1|Lr|{|βr|+|βr|2(∑i≥1‍n2(i))+|βr|3(∑i≥1‍n3(i))+|βr|4(∑i≥1‍n4(i))+⋯}=1|Lr|{|βr|+(2r-1r)|βr|2{(2r-1r)3}+(2r-1r)2|βr|3+(2r-1r)3|βr|4+⋯}≤1|L-α|{1+|L(2r-1)L-α|+|L(2r-1)L-α|2+|L(2r-1)L-α|3+|L(2r-1)L-α|4+⋯}=1|L-α|-|L(2r-1)|<∞,
where βr=Lr/(L-α) and nk(i) denote the coefficients of |βr|k for k≥2. As per the assumption, |L(2r-1)/(L-α)|<1; hence, limkSk<∞. Now, (Sk) is a sequence of positive real numbers and is convergent; this implies the boundedness of (Sk) and hence (Δνr-αI)-1∈B(ℓ1). In a similar way it can be shown that (Δνr-αI)-1∈B(ℓ∞). By using Lemma 6, we obtain that (Δνr-αI)-1∈B(ℓp). Hence, we have
(27)σ(Δνr,ℓp)⊆{α∈ℂ:|1-αL|≤2r-1}.Part 2. In this part, we need to show that {α∈ℂ:|1-(α/L)|≤2r-1}⊆σ(Δνr,ℓp). Consider α≠νk,  L for all k∈ℕ0 and |1-(α/L)|<2r-1. Clearly (Δνr-αI) is a triangle and hence (Δνr-αI)-1 exists, but (Sk) is unbounded:
(28)⟹(Δνr-αI)-1∉B(ℓp)with|1-αL|<2r-1.
Againα∈ℂ and α≠νk,  L for all k∈ℕ0 with |1-(α/L)|=2r-1 which implies limkSk=∞. Thus, (Sk) is unbounded:
(29)⟹(Δνr-αI)-1∉B(ℓp)with|1-αL|=2r-1.
Finally, we prove the result under the assumptionsα=L and α=νk. Then, we have the following cases.
Case
1. If (νk) is a constant sequence and νk=L, then we have(30)(Δνr-αI)=(0000⋯-rν0000⋯r(r-1)2!ν0-rν100⋯-r(r-1)(r-2)3!ν0r(r-1)2!ν1-rν20⋯⋮⋮⋮⋮⋱)which is not invertible. Thus, the condition (R1) fails.
Case
2.If (νk) is a strictly decreasing sequence and α=νk, then
(31)(Δνr-αI)x=((ν0-νk)x0-rν0x0+(ν1-νk)x1r(r-1)2!ν0x0-rν1x1+(ν2-νk)x2⋮).
As (νk) is strictly decreasing sequence, then for fixed k, (Δνr-αI)x=0⇒x0=0, x1=0, x2=0⋯xk-1=0, xn+1=(rνk/(νn+1-νk))xn for all n≥k. This shows that (Δνr-αI) is not one to one for α=νk, thus, we have νk∈σ(Δνr,ℓp) for k∈ℕ0. Again, for α=L, then (Sk) is unbounded. So, the condition (R2) fails. Thus,
(32){α∈ℂ:|1-αL|≤2r-1}⊆σ(Δνr,ℓp).
Combining (27) and (32), we conclude the proof.Theorem 11.
The point spectrum of the operatorΔνr over ℓp is given by
(33)σp(Δνr,ℓp)=∅.Proof.
ConsiderΔνrx=αx for x≠θ in ℓp, which gives the system of linear equations
(34)ν0x0=αx0,-rν0x0+ν1x1=αx1,r(r-1)2!ν0x0-rν1x1+ν2x2=αx2,-r(r-1)(r-2)3!ν0x0+r(r-1)2!ν1x1,-rν2x2+ν3x3=αx3,⋮
As per the definition of (νk), we have the following cases.
Case
1.Suppose (νk) is a constant sequence. That is, νk=L  for all  k∈ℕ0. On solving system of (34), we obtain that x=θ, which contradicts to our assumption. Thus, σp(Δνr,ℓp)=∅.
Case
2. Suppose (νk) is a strictly decreasing sequence and consider Δνrx=αx  for  x≠0=(0,0,0,…) in ℓp. On solving system of (34), we obtain that α=ν0 for x0≠0, x1=(rν0/(ν1-ν0))x0,x2=[r2ν1ν0/(ν1-ν0)(ν2-ν0)-r(r-1)ν0/2!(ν2-ν0)]x0 and so on. Proceeding this way, it can be proved that for x0≠0, (Δνr-αI)x=0 has a nonzero solution x=(xk), but x∉ℓp, and if x0=0, then xk=0 for each k>0, which is a contradiction. Therefore, σp(Δνr,ℓp)=∅. However, for a better explanation, we take a counter example.
Example
1
2. Suppose νk=(k+2)-[(k+2)(k+1)]1/2 for k∈ℕ0. Clearly, (νk) is a strictly decreasing sequence with
(35)ν0=2-2≈0.585786437,ν1=3-6≈0.550510257,ν2=4-12≈0.535898384and  so  on.
Also,
(36)limk→∞νk=L=limkk+2(k+2)+(k+1)(k+2)=12.
In fact,supkνk=ν0=2-2≤2L=2·(1/2)=1; hence, (νk) satisfies the conditions (1). Our claim is to show that σp(Δνr,ℓp)=∅. Supposing for the contrary that we take σp(Δνr,ℓp)≠∅, equivalently, let ν0∈σp(Δνr,ℓp). From the system of (34), we obtain that for r=1(37)-νkxk+νk+1xk+1=ν0xk+1,⟹|xk+1xk|=|νkνk+1-ν0|⟹limk→∞|xk+1xk|=|1/212-(2+2)|=|122-3|≈5.8284>1.
Therefore, x∉ℓp, and hence ν0∉σp(Δνr,ℓp) for r=1. Again for r=2, the system of (34) reduces to
(38)ν0x0=ν0x0,-2ν0x0+ν1x1=ν0x1,ν0x0-2ν1x1+ν2x2=ν0x2,⋮νk-2xk-2-2νk-1xk-1+νkxk=ν0xk,⋮
From the system of (38), it is clear that
(39)|x1x0|=|2ν0ν1-ν0|≈33.21144358,|x2x1|=|3ν1+ν02(ν2-ν0)|≈22.42337679,|x3x2|=|2ν2-ν1x1/x2ν3-ν0|≈18.92786968and so on.
Indeed, for each 0<j<k the ratio xj/xj-1<0 and |xj/xj-1|>1. Supposing that |xk/xk-1|>1 as k→∞, now by the inductive hypothesis we obtain
(40)limk→∞|xk+1xk|=limk→∞|2νk-νk-1xk-1/xkνk+1-ν0|=|1-0.5xk-1/xk1/2-2+2|≥|1+0.51/2-2+2|≈17.48528137>1.
By using inductive principle, it is trivial to prove that |xk+1/xk|>1 for all k∈ℕ0, and therefore, x∉ℓp. Proceeding this way, it can be shown that ν1∉σp(Δνr,ℓp) and similarly for all k∈ℕ0, νk∉σp(Δνr,ℓp). Hence, σp(Δνr,ℓp)=∅.Theorem 13.
The point spectrum of the dual operator(Δνr)* of Δνr over ℓp*≊ℓq is given by
(41)σp((Δνr)*,ℓq)={α∈ℂ:|1-αL|≤r}.Proof.
Suppose(Δνr)*f=αf for 0≠f∈ℓq, where 1<p<∞, 1/p+1/q=1 and (42)(Δνr)*=(ν0-rν0r(r-1)2!ν0-r(r-1)(r-2)3!ν0⋯0ν1-rν1r(r-1)2!ν1⋯00ν2-rν2⋯000ν3⋯⋮⋮⋮⋮⋱),f=(f0f1f2⋮).Now consider the system of linear equations (Δνr)*f=αf; that is,
(43)ν0f0-rν0f1+r(r-1)2!ν0f2+⋯=αf0,ν1f1-rν1f2+r(r-1)2!ν1f3+⋯=αf1,ν2f2-rν2f3+r(r-1)2!ν2f4+⋯=αf2,⋮
On solving the previous system of equations, for each k∈ℕ0, we observe that
(44)|fk|=|rνkνk-α(fk+1-r-12!fk+2+(r-1)(r-2)3!fk+3+⋯+(-1)rfk+r(r-1)(r-2)3!)rνkνk-α|,
since |rνk/(νk-α)|≥|rL/(L-α)| for both the constant and strictly decreasing sequences (νk) satisfying (1). From (44), it is clear that for each k∈ℕ0, f=(0,0,…,fk,fk+1,0,0,…) is an eigen vector that corresponds to the eigen value α satisfying |rL/(L-α)|≥1. This follows from the fact that (45)|fk|≥|rL1-α||fk+1|≥|fk+1|.
Therefore,
(46)limk→∞|fk+1fk|<1
which implies that
(47)∑k|fk+1fk|q<∞.
As a consequence  x∈ℓq.
Conversely, it is trivial to show that ifx∈ℓq, then |1-(α/L)|≤r. Suppose for the contrary, we take x∈ℓq and |1-(α/L)|>r. From (44), it can be shown that limk→∞|fk+1/fk|>1 for |1-(α/L)|>r. This implies that x∉ℓq, which is a contradiction, and this step completes the proof.Theorem 14.
The residual spectrum of the operatorΔνr over ℓp is given by
(48)σr(Δνr,ℓp)={α∈ℂ:|1-αL|≤r}.Proof.
For|1-(α/L)|<r, the operator Δνr-αI has an inverse. By Theorem 13, the operator (Δνr)*-αI is not one to one for α∈ℂ with |1-(α/L)|≤r. By using Lemma 2, we have R(Δνr-αI)¯≠ℓp. Hence,
(49)σr(Δνr,ℓp)={α∈ℂ:|1-αL|≤r}.Theorem 15.
The continuous spectrum of the operatorΔνr over ℓp is given by
(50)σc(Δνr,ℓp)={α∈ℂ:r<|1-αL|≤2r-1}.Proof.
Sinceσp(Δνr,ℓp), σr(Δνr,ℓp), and σc(Δνr,ℓp) are disjoint subsets of σ(Δνr,ℓp), and their union contributes to σ(Δνr,ℓp). Therefore, from Theorems 9, 10, 11, and 14, we obtain that
(51)σc(Δνr,ℓp)={α∈ℂ:r<|1-αL|≤2r-1}.


## 3. The Spectrum ofΔνr over bvp(1<p<∞)

In this section, we determine the spectrum of the operatorΔνr over the sequence space bvp(1<p<∞). Now, we give some theorems starting with the result concerning the boundedness and linearity of the operator Δνr.Theorem 16.
Consider the following:(52)Δνr∈B(bvp).Proof.
Supposingx=(xk)∈bvp and by using Minkowski’s inequality we have
(53)∥Δνr(x)∥(bvp)=(∑k=0∞‍|∑i=0r‍(-1)i(ri)(νk-ixk-i-νk-i-1xk-i-1)|p)1/p≤(∑k=0∞‍|(r0)(νkxk-νk-1xk-1)|p)1/p+(∑k=0∞‍|(r1)(νk-1xk-1-νk-2xk-2)|p)1/p+⋯+(∑k=0∞‍|(rr)(νk-rxk-r-νk-r-1xk-r-1)|p)1/p≤(rr1/2)(r+1)ν0∥x∥(bvp).
Thus,
(54)∥Δνr∥(bvp)≤(rr1/2)(r+1)ν0.
This step concludes the proof.Theorem 17.
The spectrum of the operatorΔνr over bvp is given by
(55)σ(Δνr,bvp)={α∈ℂ:|1-αL|≤2r-1}.Proof.
Firstly, we need to prove that(Δνr-αI)-1 exists and is in B(bvp) for |1-(α/L)|>2r-1, and after that the operator Δνr-αI is not invertible for |1-(α/L)|≤2^r-1.
Let|1-(α/L)|>2r-1. Since Δνr-αI is a triangle, hence (Δνr-αI)-1 exists. Let y=(yk)∈bvp; this implies that yk-yk-1∈ℓp. Solving the system equation Δνr-αI=y, we obtain that
(56)x0=y0ν0-α,x1=y1ν1-α+rν0y0(ν1-α)(ν0-α),x2=y2ν2-α+rν1y1(ν2-α)(ν1-α)+{r2ν1ν0(ν2-α)(ν1-α)(ν0-α){(r2)}-(r2)ν0(ν2-α)(ν0-α)}y0,⋮
By using notation bnk, we can write for each k∈ℕ0(57)xk=∑j=0k‍bk-j,kyk-j.
Now,
(58)xk-xk-1=bkkyk+(bk,k-1-bk-1,k-1)yk-1+(bk,k-2-bk-1,k-2)yk-2+⋯+(bk,0-bk-1,0)y0=(B-B*)y,
where B=(bnk) and
(59)B*=(00(bnk)0).
By Theorem 10, it is clear that B,B*∈(ℓp,ℓp) and y∈ℓp. Therefore, (xk-xk-1)∈ℓp equivalently, x=(xk)∈bvp. Thus,
(60)σ(Δνr,bvp)⊂{α∈ℂ:|1-αL|≤2r-1}.
Again for α∈{α∈ℂ:|1-(α/L)|≤2r-1} and α≠L or νk, Δνr-αI is a triangle, and hence (Δνr-αI)-1 exists. Suppose y=(1,1,1,…)∈bvp; then by Theorem 10, the series ∑k‍|xk-xk-1| is divergent. Finally, for α=L or νk(Δνr-αI)-1∉B(ℓp), this shows that
(61){α∈ℂ:|1-αL|≤2r-1}⊂σ(Δνr,bvp).
Combining (60) and (61), we complete the proof.Since the spectrum and fine spectrum of the operatorΔνr over the sequence space bvp are similar to that over the space ℓp, we avoid the repetition of similar statements as discussed in the last section and give the results via following theorems without proof.Theorem 18.
Consider the following:(i)
σ(Δνr,bvp)={α∈ℂ:|1-(α/L)|≤2r-1}.(ii)
σp(Δνr,bvp)=∅.(iii)
σr(Δνr,bvp)={α∈ℂ:|1-(α/L)|≤r}.(iv)
σc(Δνr,bvp)={α∈ℂ:r<|1-(α/L)|≤2r-1}.


## 4.  Conclusion

In the present article, we give a unifying approach to most of the difference operators defined earlier in the literature and determine the spectrum and fine spectrum of the generalized rth difference operatorsΔνr over the sequence spaces ℓp and bvp(1<p<∞). Therefore, the results and theorems presented in this article are more general and comprehensive than the works done by earlier authors. Now, by choosing r and the sequence (νk) suitably, our work generalizes various other known results studied by several authors (see [4–6, 16]).



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*Source: 286748-2013-11-07.xml*