#### Abstract

The main idea of the present paper is to compute the spectrum and the fine spectrum of the generalized difference operator over the sequence spaces . The operator denotes a triangular sequential band matrix defined by with for , where or , ; the set nonnegative integers and 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 over the sequence spaces and . These results are more general and comprehensive than the spectrum of the difference operators , , , , and and include some other special cases such as the spectrum of the operators , , and over the sequence spaces or .

#### 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 CesÃ ro operator on the sequence space for has been studied by Gonzalez [1]. Okutoyi [2] computed the spectrum of the CesÃ ro operator over the sequence space . The fine spectra of the CesÃ ro operator over the sequence space have been determined by Akhmedov and BaÅŸar [3]. Akhmedov and BaÅŸar [4, 5] have studied the fine spectrum of the difference operator over the sequence spaces and , where . Altay and BaÅŸar [6] have determined the fine spectrum of the difference operator over the sequence spaces , for . The fine spectrum of the difference operator over the sequence spaces and was investigated by Kayaduman and Furkan [7]. Srivastava and Kumar [8] have examined the fine spectrum of the generalized difference operator over the sequence space . Recently, the spectrum of the generalized difference operator over the sequence spaces and has been studied by Dutta and Baliarsingh [9, 10], respectively. The main focus of this paper is to define the difference operator and establish its spectral characterization with respect to the Goldbergâ€™s classifications.

Let be either constant or strictly decreasing sequence of positive real numbers satisfying By , we denote the space of all sequences of real or complex numbers. Any subspace of is called a sequence space, and we write , , , , , and for the spaces of all bounded, convergent, null, absolutely summable, -summable, and -bounded variation sequences, respectively. The sequence spaces and are defined by The main goal of this article is to define a generalized difference operator as follows.

For a positive integer , we define the generalized difference operator by , where with for , , and . It is clear that the operator can be represented by a lower triangular sequential band matrix for , and Equivalently,

In particular, the operator generalizes several operators considered by earlier authors concerning the same literature. For instance, (i)for , it generalizes the operator , considered by Srivastava and Kumar [8]; (ii)for and , , it generalizes the difference operator , considered by Altay and BaÅŸar [11];(iii)for and , , it generalizes the difference operator , considered by Altay and BaÅŸar [12]; (iv)for and , , it generalizes the difference operator , considered by Akhmedov and El-Shabrawy [13]; (v)for and , , it generalizes the difference operator , considered by Panigrahi and Srivastava [14]; (vi)for and , it generalizes the 2nd order difference operator , considered by Dutta and Baliarsingh [15];(vii)for and , , , it generalizes the difference operator , considered by Furkan et al. [16];

Let and be Banach spaces and a bounded linear operator. By , we denote the range of ; that is,

By , we denote the set of all bounded linear operators on into itself. If is any Banach space and , then the *adjoint* â€‰â€‰ of is a bounded linear operator on the dual of defined by for all and with .

Let be a normed linear space over the complex field and a linear operator, where denotes the domain of . With , for a complex number , we associate an operator , where is called identity operator on , and if has an inverse, we denote it by ; that is,
and is called the *resolvent* operator of . Many properties of and depend on , and the spectral theory is concerned with those properties. We are interested in the set of all â€™s in the complex plane such that exists, is bounded, and domain of is dense in . 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]). *Let and be defined as previous. A regular value of is a complex number such that

(R1) exists;

(R2) is bounded;

(R3) is defined on a set which is dense in .

The *resolvent* set of is the set of all *regular* values of . Its complement in the complex plane is called the *spectrum *of . Furthermore, the spectrum is partitioned into three disjoint sets as follows. (I)*Point Spectrum*â€‰â€‰. It is the set of all such that (R1) does not hold. The elements of are called eigen values of . (II)*Continuous Spectrumâ€‰â€‰ *. It is the set of all such that (R1) holds and satisfies (R3) but does not satisfy (R2). (III)*Residual Spectrum* â€‰â€‰ . It is the set of all 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 operator has a dense range if and only if the adjoint is one to one.*

Lemma 3 (see [18, page 60]). *The adjoint operator is onto if and only if T has a bounded inverse.*

Let , be two nonempty subsets of the space of all real or complex sequences and an infinite matrix of complex numbers , where . For every and every positive integer , we write The sequence , if it exists, is called the transformation of by the matrix . Infinite matrix if and only if whenever .

Lemma 4 (see [19, page 126]). *The matrix gives rise to a bounded linear operator from to itself if and only if the supremum of norms of the columns of A is bounded.*

Lemma 5 (see [20, page 254]). *The matrix gives rise to a bounded linear operator from to itself if and only if the supremum of norms of the columns of A is bounded.*

Lemma 6 (see [19, page 174]). *Let and , then .*

Lemma 7 (see [5]). *Define the spaces and consisting all sequences normed by**
Then, and are isometrically isomorphic to and .*

The basis of the space is also constructed and given by the following lemma.

Lemma 8 (see [5]). *Define the sequence of the elements of the space , for every fixed , the set of positive integers, by
**
Then, the sequence is a basis for the space , and has a unique representation of the form
*

#### 2. The Spectrum and Fine Spectrum of over

In this section, we compute the point spectrum, the spectrum, the continuous spectrum, the residual spectrum, and the fine spectrum of the operator over the sequence space .

Theorem 9. *The operator is a linear operator and satisfies the following inequalities: *(i)*if is a constant sequence
*(ii)*if is a strictly decreasing sequence
** where
*

*Proof. *(i) Suppose is a constant sequence and for all . Linearity of the operator is trivial, hence omitted. Let for all and such that . Now, by using Minkowskiâ€™s inequality we get
Thus,
Suppose we denote as a sequence whose 1st entry is 1 and otherwise 0. Clearly, and
Thus,
Combining inequations (16) and (18), we complete the proof.

(ii) Suppose is a strictly decreasing sequence. Proof of this bit follows from (i) and with the fact that .

Theorem 10. *The spectrum of on the sequence space is given by
*

*Proof. * The proof of this theorem consists of two parts. *Part 1*. In the first part, we have to show that

Equivalently, we need to show that if with . Let with . Now is a triangle and hence has an inverse , where

and , , and are as follows:

Similarly,
In fact, for , one can calculate
and so on.

Now, let
Now,
where and denote the coefficients of for . As per the assumption, ; hence, . Now, is a sequence of positive real numbers and is convergent; this implies the boundedness of and hence . In a similar way it can be shown that . By using Lemma 6, we obtain that . Hence, we have
*Part 2.* In this part, we need to show that . Consider , for all and . Clearly is a triangle and hence exists, but is unbounded:

Again and , for all with which implies . Thus, is unbounded:

Finally, we prove the result under the assumptions and . Then, we have the following cases.*Case **1.* If is a constant sequence and , then we havewhich is not invertible. Thus, the condition (R1) fails.*Case **2. *If is a strictly decreasing sequence and , then
As is strictly decreasing sequence, then for fixed , , , , for all . This shows that is not one to one for , thus, we have for . Again, for , then is unbounded. So, the condition (R2) fails. Thus,
Combining (27) and (32), we conclude the proof.

Theorem 11. * The point spectrum of the operator over is given by
*

*Proof. * Consider for in , which gives the system of linear equations
As per the definition of , we have the following cases.*Case **1. *Suppose is a constant sequence. That is, for all . On solving system of (34), we obtain that , which contradicts to our assumption. Thus, .*Case **2*. Suppose is a strictly decreasing sequence and consider for in . On solving system of (34), we obtain that for , , and so on. Proceeding this way, it can be proved that for , has a nonzero solution , but , and if , then for each , which is a contradiction. Therefore, . However, for a better explanation, we take a counter example.*Example **1**2*. Suppose for . Clearly, is a strictly decreasing sequence with
Also,

In fact, ; hence, satisfies the conditions (1). Our claim is to show that . Supposing for the contrary that we take , equivalently, let . From the system of (34), we obtain that for
Therefore, , and hence for . Again for , the system of (34) reduces to
From the system of (38), it is clear that
Indeed, for each the ratio and . Supposing that as , now by the inductive hypothesis we obtain
By using inductive principle, it is trivial to prove that for all , and therefore, . Proceeding this way, it can be shown that and similarly for all , . Hence, .

Theorem 13. *The point spectrum of the dual operator of over is given by
*

*Proof. * Suppose for , where , and Now consider the system of linear equations ; that is,
On solving the previous system of equations, for each , we observe that
since for both the constant and strictly decreasing sequences satisfying (1). From (44), it is clear that for each , is an eigen vector that corresponds to the eigen value satisfying . This follows from the fact that
Therefore,
which implies that
As a consequence .

Conversely, it is trivial to show that if , then . Suppose for the contrary, we take and . From (44), it can be shown that for . This implies that , which is a contradiction, and this step completes the proof.

Theorem 14. *The residual spectrum of the operator over is given by
*

*Proof. * For , the operator has an inverse. By Theorem 13, the operator is not one to one for with . By using Lemma 2, we have . Hence,

Theorem 15. *The continuous spectrum of the operator over is given by
*

*Proof. * Since , , and are disjoint subsets of , and their union contributes to . Therefore, from Theorems 9, 10, 11, and 14, we obtain that

#### 3. The Spectrum of over

In this section, we determine the spectrum of the operator over the sequence space . Now, we give some theorems starting with the result concerning the boundedness and linearity of the operator .

Theorem 16. *Consider the following:
*

*Proof. * Supposing and by using Minkowskiâ€™s inequality we have
Thus,
This step concludes the proof.

Theorem 17. *The spectrum of the operator over is given by
*

*Proof. * Firstly, we need to prove that exists and is in for , and after that the operator is not invertible for .

Let . Since is a triangle, hence exists. Let ; this implies that . Solving the system equation , we obtain that
By using notation , we can write for each
Now,
where and
By Theorem 10, it is clear that and . Therefore, equivalently, . Thus,
Again for and or , is a triangle, and hence exists. Suppose ; then by Theorem 10, the series is divergent. Finally, for or , this shows that