Hindawi Publishing Corporation International Journal of Analysis Volume 2014, Article ID 786437, 7 pages http://dx.doi.org/10.1155/2014/786437 Research Article On the Spectral Properties of the Weighted Mean Difference Operator πΊ (π’, V; Δ) over the Sequence Space β1 P. Baliarsingh1 and S. Dutta2 1 Department of Mathematics, School of Applied Sciences, KIIT University, Bhubaneswar 751024, India 2 Department of Mathematics, Utkal University, Vanivihar, Bhubaneswar 751004, India Correspondence should be addressed to P. Baliarsingh; pb.math10@gmail.com Received 21 December 2013; Accepted 30 April 2014; Published 16 June 2014 Academic Editor: Baruch Cahlon Copyright © 2014 P. Baliarsingh and S. Dutta. 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. In the present work the generalized weighted mean difference operator πΊ (π’, V; Δ) has been introduced by combining the generalized weighted mean and difference operator under certain special cases of sequences π’ = (π’π ) and V = (Vπ ). For any two sequences π’ and V of either constant or strictly decreasing real numbers satisfying certain conditions the difference operator πΊ (π’, V; Δ) is defined by π (πΊ(π’, V; Δ)π₯)π = ∑π=0 π’π Vπ (π₯π − π₯π−1 ) with π₯π = 0 for all π < 0. Furthermore, we compute the spectrum and the fine spectrum of the operator πΊ (π’, V; Δ) over the sequence space β1 . In fact, we determine the spectrum, the point spectrum, the residual spectrum, and the continuous spectrum of this operator on the sequence space β1 . 1. Introduction, Preliminaries, and Definitions Let π’ = (π’π ) and V = (Vπ ) be two bounded sequences of either constant or strictly decreasing positive real numbers such that π’π =ΜΈ 0 and Vπ =ΜΈ 0 for all π, and lim π’π = π’, πσ³¨→∞ σ΅¨σ΅¨ V − V σ΅¨σ΅¨ σ΅¨ π+1 σ΅¨σ΅¨ lim σ΅¨σ΅¨σ΅¨ π σ΅¨=1 πσ³¨→∞ σ΅¨σ΅¨ Vπ−1 − Vπ σ΅¨σ΅¨σ΅¨ (1) (for strictly decreasing sequence) . (2) By β1 and πVπ , we denote the spaces of all absolutely summable and p-bounded variation series, respectively. Also, by β∞ , π, and π0 , we denote the spaces of all bounded, convergent, and null sequences, respectively. The main perception of this paper is to introduce the weighted mean difference operator πΊ(π’, V; Δ) as follows. Let π₯ = (π₯π ) be any sequence in β1 , and we define the weighted mean difference transform πΊ(π’, V; Δ)π₯ of π₯ by π (πΊ (π’, V; Δ) π₯)π = ∑π’π Vπ (π₯π − π₯π−1 ) π=0 (π ∈ N0 ) , (3) where N0 denotes the set of nonnegative integers and we assume throughout that any term with negative subscript is zero. Instead of writing (3), the operator πΊ(π’, V; Δ) can be expressed as a lower triangular matrix (πππ ), where πππ π’π Vπ , { { { = {π’π (Vπ − Vπ+1 ) , { { {0, (π = π) , (0 ≤ π ≤ π − 1) , (4) (π > 0) . Equivalently, in componentwise the triangle (πππ ) can be represented by 2 International Journal of Analysis π’0 V0 π’1 (V0 − V1 ) 0 π’1 V1 0 0 0 0 ( π’2 V2 0 (πππ ) = (π’2 (V0 − V1 ) π’2 (V1 − V2 ) π’3 (V0 − V1 ) π’3 (V1 − V2 ) π’3 (V2 − V3 ) π’3 V3 .. .. .. .. . . . . ( The main objective of this paper is to determine the spectrum of the operator πΊ(π’, V; Δ) over the basic sequence space β1 . The operator πΊ(π’, V; Δ) has been studied by Polat et al. [1] in detail by introducing the difference sequence spaces β∞ (π’, V; Δ), π0 (π’, V; Δ), and π(π’, V; Δ). In the existing literature several researchers have been actively engaged in finding the spectrum and fine spectrum of different bounded linear operators over various sequence spaces. The spectrum of weighted mean operator has been studied by Rhoades [2], whereas that of the difference operator Δ over the sequence spaces βπ for 0 < π < 1 and π, π0 has been studied by Altay and BasΜ§ar [3, 4]. Kayaduman and Furkan [5] have determined the fine spectrum of the difference operator Δ over the sequence spaces β1 and πVπ and on generalizing these results, Srivastava and Kumar [6, 7] have determined the fine spectrum of the operator Δ ] over the sequence spaces β1 and π0 , where (]π ) is a sequence of either constant or strictly deceasing sequence of reals satisfying certain conditions. Dutta and Baliarsingh [8– 10] have computed the spectrum of the operator Δπ] (π ∈ N0 ) and Δ2 over the sequence spaces β1 , π0 , and π0 , respectively. The fine spectrum of the generalized difference operators π΅(π, π ) and π΅(π, π , π‘) over the sequence spaces βπ , πVπ and β1 , πV has been studied by BilgicΜ§ and Furkan [11, 12], respectively. Recently, the spectrum of some particular limitation matrices in certain sequence spaces has been studied by [13–15] and many others. Let π and π be Banach spaces and π : π σ³¨→ π be a bounded linear operator. By R(π), we denote the range of π; that is, R (π) = {π¦ ∈ π : π¦ = ππ₯; π₯ ∈ π} . 0 ⋅⋅⋅ ) ). 0 ⋅⋅⋅ .. . d ) (5) concerned with those properties. We are interested in the set of all π’s in the complex plane such that ππ−1 exists/ππ−1 is bounded/domain of ππ−1 is dense in π. Now, we state the following results which are essential for our investigation. Definition 1 (see [16, page 371]). Let π =ΜΈ {0} be normed linear space over the complex field and π : π·(π) σ³¨→ π be a linear operator with domain π·(π) ⊆ π. A regular value of π is a complex number π such that (R1) ππ−1 exists; (R2) ππ−1 is bounded; (R3) ππ−1 is defined on a set which is dense in π. The resolvent set π(π, π) of π is the set of all regular values of π. Its complement π(π, π) = C \ π(π, π) in the complex plane C is called the spectrum of π. Furthermore, the spectrum π(π, π) is partitioned into three disjoint sets as follows. (I) Point Spectrum ππ (π, π). It is the set of all π ∈ C such that (R1) does not hold. The elements of ππ (π, π) are called eigen values of π. (II) Continuous Spectrum ππ (π, π). It is the set of all π ∈ C such that (R1) holds and satisfies (R3) but does not satisfy (R2). (III) Residual Spectrum ππ (π, π). It is the set of all π ∈ C such that (R1) holds but does not satisfy (R3). Condition (R2) may or may not hold. (6) By π΅(π), we denote the set of all bounded linear operators on π into itself. If π is any Banach space and π ∈ π΅(π) then the adjoint π∗ of π is a bounded linear operator on the dual π∗ of π defined by (π∗ π)(π₯) = π(ππ₯) for all π ∈ π∗ and π₯ ∈ π with βπβ = βπ∗ β. Let π =ΜΈ {0} be a normed linear space over the complex field and π : π·(π) σ³¨→ π be a linear operator, where π·(π) denotes the domain of π. With π, for a complex number π, we associate an operator ππ = (π − ππΌ), where πΌ is called identity operator on π·(π) and if ππ has an inverse, we denote it by ππ−1 , that is, ππ−1 = (π − ππΌ)−1 0 ⋅⋅⋅ 0 ⋅⋅⋅ (7) and is called the resolvent operator of π. Many properties of ππ and ππ−1 depend on π and the spectral theory is Goldberg’s Classification of the Operator ππ (see [17], page 58– 71). Let π be a Banach space and ππ = (π−ππΌ) ∈ π΅(π), where π is a complex number. Again, let π (ππ ) and ππ−1 denote the range and inverse of the operator ππ , respectively. Then the following possibilities may occur: (A) π (ππ ) = π; (B) π (ππ ) =ΜΈ π (ππ ) = π; (C) π (ππ ) =ΜΈ π and (1) ππ is injective and ππ−1 is continuous; (2) ππ is injective and ππ−1 is discontinuous; (3) ππ is not injective. International Journal of Analysis 3 Taking permutations (A), (B), (C) and (1), (2), (3), we get nine different states. These are labelled by π΄ 1 , π΄ 2 , π΄ 3 , π΅1 , π΅2 , π΅3 , πΆ1 , πΆ2 , and πΆ3 . If π is a complex number such that ππ ∈ π΄ 1 or ππ ∈ π΅1 , then π is in the resolvent set π(π, π) of π on π. The other classifications give rise to the fine spectrum of π. We use π ∈ π΅2 π(π, π) which means the operator ππ ∈ π΅2 , that is, π (ππ ) =ΜΈ π (ππ ) = π and ππ is injective but ππ−1 is discontinuous; similarly are the others. Lemma 2 (see [17, page 59]). A linear operator π has a dense range if and only if the adjoint π∗ is one to one. Lemma 3 (see [17, page 60]). The adjoint operator π∗ is onto if and and only if π has a bounded inverse. Let π, π be two nonempty subsets of the space π€ of all real or complex sequences and let π΄ = (πππ ) be an infinite matrix of complex numbers πππ , where π, π ∈ N0 . For every π₯ = (π₯π ) ∈ π, we write π΄ π (π₯) = ∑πππ π₯π (π ∈ N) . π Lemma 4 (see [18, page 126]). The matrix π΄ = (πππ ) gives rise to a bounded linear operator π ∈ π΅(β1 ) from β1 to itself if and only if the supremum β1 norms of the columns of π΄ is bounded. Lemma 5 (see [19], Theorem 2). If ππ (0) =ΜΈ 0 for all π ∈ N0 , then the inverse of the difference operator π΅(π[π]) is given by a lower triangular Toeplitz matrix πΆ = (πππ ) as follows: 1 , (π = π) , { { { π (0) π { { { { (−1)π−π { { { πππ = { ∏ππ=π ππ (0) { { { { (π) { { ×π·π−π (π [π]) , (0 ≤ π ≤ π − 1) , { { { (π > π) , {0, (π, π ∈ N0 ) , (8) The sequence π΄π₯ = (π΄ π (π₯)), if it exists, is called the transformation of π₯ by the matrix π΄. Infinite matrix π΄ ∈ (π, π) if and only if π΄π₯ ∈ π whenever π₯ ∈ π. where (π΅(π[π])π₯)π = ππ (0)π₯π + ππ−1 (1)π₯π−1 + ⋅ ⋅ ⋅ + ππ−π (π)π₯π−π , (π ∈ N0 ) and σ΅¨σ΅¨ π (1) ππ+1 (0) 0 ⋅⋅⋅ 0 0 ⋅⋅⋅ 0 σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨ π σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨ ππ (2) ππ+1 (1) ππ+2 (0) ⋅⋅⋅ 0 0 ⋅⋅⋅ 0 σ΅¨σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨σ΅¨ . .. .. .. .. .. .. .. σ΅¨σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨ .. . . . . . . . σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨ σ΅¨ ⋅⋅⋅ ππ−1 (1) ππ (0) ⋅ ⋅ ⋅ 0 σ΅¨σ΅¨σ΅¨σ΅¨ . π·π(π) (π [π]) = σ΅¨σ΅¨σ΅¨σ΅¨ππ (π) ππ+1 (π − 1) ππ+2 (π − 2) σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨ 0 ⋅⋅⋅ ππ−1 (2) ππ (1) ⋅ ⋅ ⋅ 0 σ΅¨σ΅¨σ΅¨ ππ+1 (π) ππ+2 (π − 1) σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨ .. .. .. .. .. .. .. σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨ . . . . . . d . σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨ 0 ⋅⋅⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ππ−1 (1)σ΅¨σ΅¨σ΅¨ 0 ⋅⋅⋅ ππ−π (π) 2. The Spectrum of the Operator πΊ(π’, V; Δ) over the Sequence β1 In this section, we compute the spectrum, the point spectrum, the continuous spectrum, and the residual spectrum of the difference matrix πΊ(π’, V; Δ) on the sequence space β1 . Theorem 6. The operator πΊ(π’, V; Δ) : β1 σ³¨→ β1 is a linear operator and (10) (i) If (π’π ) is a strictly decreasing sequence and (Vπ ) is a constant sequence of positive reals (say Vπ = V for all π), then βπΊ (π’, V; Δ)β(β1 :β1 ) = π’0 V. (12) (ii) If (π’π ) and (Vπ ) are strictly decreasing sequences of positive reals, then βπΊ (π’, V; Δ)β(β1 :β1 ) = π’0 V0 . (13) (iii) If (π’π ) and (Vπ ) are constant sequences of positive reals, then π−1 σ΅¨ σ΅¨ σ΅¨ σ΅¨ σ΅¨ σ΅¨ βπΊ(π’, V; Δ)β(β1 :β1 ) = sup σ΅¨σ΅¨σ΅¨π’π σ΅¨σ΅¨σ΅¨ ( ∑ σ΅¨σ΅¨σ΅¨Vπ − Vπ+1 σ΅¨σ΅¨σ΅¨ + σ΅¨σ΅¨σ΅¨Vπ σ΅¨σ΅¨σ΅¨) . π (9) βπΊ (π’, V; Δ)β(β1 :β1 ) = π’V. π=0 (11) Proof. Proof of this theorem follows from Lemma 4 and in particular cases. (14) (iv) If (π’π ) is a constant sequence and (Vπ ) is a strictly decreasing sequence of positive reals, then βπΊ (π’, V; Δ)β(β1 :β1 ) = π’V0 . (15) 4 International Journal of Analysis Theorem 7. The spectrum of πΊ(π’, V; Δ) over the sequence space β1 is given by Proof. The proof is divided into four parts as follows. Let (π’π ) and (Vπ ) be two bounded sequences of positive reals satisfying (1) and (2) and π ∈ C such that |(π − π’π Vπ+1 )/(π − π’π Vπ )| < 1 for all π ∈ N0 ; then (πΊ(π’, V; Δ) − ππΌ) is a triangle and hence has an inverse, that is, σ΅¨σ΅¨ π − π’ V σ΅¨σ΅¨ σ΅¨ π π+1 σ΅¨σ΅¨ π (πΊ (π’, V; Δ) , β1 ) = {π ∈ C : σ΅¨σ΅¨σ΅¨ σ΅¨ ≥ 1; (π ∈ N0 )} . σ΅¨σ΅¨ π − π’π Vπ σ΅¨σ΅¨σ΅¨ (16) (πΊ (π’, V; Δ) − ππΌ)−1 1 0 π’0 V0 − π 0 0 ( ( π’ (V − V1 ) 1 ( − 11 0 0 0 ( ( π’ V ∏π=0 (π’π Vπ − π) 1 1−π ( ( ( =( π’ (V − V2 ) π’2 (V0 − V1 ) (π − π’1 V2 ) 1 ( − 22 1 0 ( 2 π’ V ( (π’ V − π) (π’ V − π) ∏ ∏ 2 2−π π π π π π=0 π=1 ( ( ( ( π’3 (V0 − V1 ) (π − π’1 V2 ) (π − π’2 V3 ) π’3 (V1 − V2 ) (π − π’2 V3 ) π’ (V − V3 ) 1 (− − 33 2 3 3 ∏π=0 (π’π Vπ − π) ∏π=1 (π’π Vπ − π) ∏π=2 (π’π Vπ − π) π’3 V3 − π .. .. .. .. . . . . ( ⋅⋅⋅ ⋅⋅⋅ ) ) ) ) ) ) ) ) ). ) ) ) ) ) ) ) ) d ) ⋅⋅⋅ ⋅⋅⋅ (17) In general, by using Lemma 5, we observe that σ΅¨σ΅¨π’ (V − V ) (π’ V − π) 0 σ΅¨σ΅¨ π+1 π π+1 π+1 π+1 σ΅¨σ΅¨ σ΅¨σ΅¨π’ (V − V ) π’ (V − V ) (π’ V − π) σ΅¨σ΅¨ π+2 π π+1 π+2 π+1 π+2 π+2 π+2 σ΅¨σ΅¨ σ΅¨σ΅¨ (π) π·π (π’, V, π) = σ΅¨σ΅¨σ΅¨σ΅¨π’π+3 (Vπ − Vπ+1 ) π’π+3 (Vπ+1 − Vπ+2 ) π’π+3 (Vπ+2 − Vπ+3 ) σ΅¨σ΅¨ .. .. .. σ΅¨σ΅¨σ΅¨ . . . σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨ π’π (Vπ − Vπ+1 ) π’π (Vπ+1 − Vπ+2 ) π’π (Vπ+2 − Vπ+3 ) = (Vπ − Vπ+1 ) (Vπ+1 − Vπ+2 ) ⋅ ⋅ ⋅ π’π (Vπ−1 − Vπ ) × (π’π+1 − × (π’π+2 − ((πΊ (π’, V; Δ) − ππΌ)−1 )π,π π−1 (π − π’ V (Vπ − Vπ+1 ) π’π π π+1 ) . ∏ (π’π Vπ − π) (π’π Vπ − π) π=π+1 (π’π Vπ − π) By using Lemma 4, we obtain that 0 ⋅⋅⋅ 0 ⋅⋅⋅ 0 d .. . ⋅ ⋅ ⋅ π’π (Vπ−1 − Vπ ) σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨ (π’π+1 Vπ+1 − π) ) (Vπ+1 − Vπ+2 ) (π’π+2 Vπ+2 − π) (π’ V − π) ) × ⋅ ⋅ ⋅ × (π’π−1 − π−1 π−1 ). (Vπ+2 − Vπ+3 ) (Vπ−1 − Vπ ) Therefore, for all π = π, ((πΊ(π’, V; Δ) − ππΌ)−1 )π,π = 1/(π’π Vπ − π), for all π > π, ((πΊ(π’, V; Δ) − ππΌ)−1 )π,π = 0, and for all 0 ≤ π < π one can deduce that = ⋅⋅⋅ (19) σ΅© σ΅©σ΅© σ΅©σ΅©(πΊ (π’, V; Δ) − ππΌ)−1 σ΅©σ΅©σ΅© σ΅©(β1 :β1 ) σ΅© π σ΅¨ σ΅¨σ΅¨ σ΅¨ = sup ∑ σ΅¨σ΅¨σ΅¨((πΊ (π’, V; Δ) − ππΌ)−1 )π,π σ΅¨σ΅¨σ΅¨ σ΅¨ σ΅¨ π π=0 σ΅¨σ΅¨ σ΅¨σ΅¨ π−1 σ΅¨σ΅¨σ΅¨ (Vπ − Vπ+1 ) π’π 1 σ΅¨ σ΅¨σ΅¨ ≤ sup (σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨ + ∑ σ΅¨σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨ π’π Vπ − π σ΅¨σ΅¨ π=0 σ΅¨σ΅¨ (π’π Vπ − π) (π’π Vπ − π) π σ΅¨ (π − π’π Vπ+1 ) σ΅¨σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨) ×∏ σ΅¨σ΅¨ π=π+1 (π’π Vπ − π) σ΅¨σ΅¨ π−1 (18) International Journal of Analysis 5 σ΅¨ π−1 σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨ 1 σ΅¨σ΅¨ (Vπ − Vπ+1 ) π’π σ΅¨σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨ (1 + = sup σ΅¨σ΅¨σ΅¨ ∑ σ΅¨σ΅¨ σ΅¨σ΅¨ π σ΅¨σ΅¨ π’π Vπ − π σ΅¨σ΅¨ σ΅¨ (π’π Vπ − π) σ΅¨σ΅¨σ΅¨ π=0 σ΅¨ σ΅¨σ΅¨ π−1 σ΅¨σ΅¨σ΅¨ (π − π’ V σ΅¨ π π+1 ) σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨) × ∏ σ΅¨σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨ π=π+1 σ΅¨σ΅¨ (π’π Vπ − π) σ΅¨σ΅¨ σ΅¨ π−1 σ΅¨σ΅¨ σ΅¨σ΅¨ π’ σ΅¨σ΅¨ 1 σ΅¨ (Vπ − Vπ+1 ) σ΅¨σ΅¨σ΅¨ π−1 σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨ π σ΅¨σ΅¨ ∏ σ΅¨ππ σ΅¨) ≤ sup σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨ ( σ΅¨σ΅¨ σ΅¨σ΅¨ + ∑ σ΅¨σ΅¨σ΅¨σ΅¨ π σ΅¨σ΅¨ π’π Vπ − π σ΅¨σ΅¨ σ΅¨σ΅¨π’π σ΅¨σ΅¨ π=0 σ΅¨σ΅¨ (π’π Vπ − π) σ΅¨σ΅¨σ΅¨ π=π+1 σ΅¨ σ΅¨ σ΅¨σ΅¨ π π’ σ΅¨σ΅¨ 1 1 π−1σ΅¨σ΅¨ σ΅¨σ΅¨π−π−1 σ΅¨ σ΅¨ ≤ sup σ΅¨σ΅¨σ΅¨ π π σ΅¨σ΅¨σ΅¨ ( σ΅¨σ΅¨ + ) σ΅¨ σ΅¨ σ΅¨σ΅¨ππ σ΅¨σ΅¨σ΅¨ ∑ σ΅¨σ΅¨ππ σ΅¨σ΅¨ π σ΅¨σ΅¨ π’π Vπ − π σ΅¨σ΅¨ σ΅¨σ΅¨ππ π’π σ΅¨σ΅¨σ΅¨ σ΅¨ σ΅¨ π=0 σ΅¨ σ΅¨π−1 σ΅¨σ΅¨ π π’ σ΅¨σ΅¨ 1 − σ΅¨σ΅¨σ΅¨ππ σ΅¨σ΅¨σ΅¨ 1 1 σ΅¨ σ΅¨ + ( = sup σ΅¨σ΅¨σ΅¨ π π σ΅¨σ΅¨σ΅¨ [ σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨ σ΅¨ )] < ∞. π σ΅¨σ΅¨ π’π Vπ − π σ΅¨σ΅¨ σ΅¨σ΅¨ππ π’π σ΅¨σ΅¨σ΅¨ 1 − σ΅¨σ΅¨σ΅¨ππ σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨ππ σ΅¨σ΅¨ (20) For simplicity, we write ππ = supπ ((Vπ − Vπ+1 )/(π’π Vπ − π)) and ππ = supπ ((π − π’π Vπ+1 )/(π’π Vπ − π)). Since (π’π ) and (Vπ ) are two bounded sequences of positive reals, the quantity ππ is finite. As per the assumption, π ∈ C and |(π−π’π Vπ+1 )/(π−π’π Vπ )| < 1 for all π ∈ N0 . Therefore, |ππ | < 1. Now, we consider the four possible cases of the sequences (π’π ) and (Vπ ). Case 1. If (π’π ) and (Vπ ) are strictly decreasing sequences of positive reals, then from the above relation we have (πΊ(π’, V; Δ) − ππΌ)−1 ∈ (β1 : β1 ). Thus, σ΅¨σ΅¨σ΅¨ π − π’π Vπ+1 σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨ ≥ 1; (π ∈ N0 )} . π (πΊ (π’, V; Δ) , β1 ) ⊆ {π ∈ C : σ΅¨σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨ π − π’π Vπ σ΅¨σ΅¨ (21) Conversely, consider π =ΜΈ π’π Vπ and |(π−π’π Vπ+1 )/(π−π’π Vπ )| ≥ 1 for all π ∈ N0 , and clearly (πΊ(π’, V; Δ) − ππΌ) is a triangle and hence (πΊ(π’, V; Δ) − ππΌ)−1 exists, but β (πΊ(π’, V; Δ) − ππΌ)−1 β(β1 :β1 ) is not finite. Again, for π = π’π Vπ , the matrix (πΊ(π’, V; Δ) − π’π Vπ πΌ)(π ∈ N0 ) is not invertible. Thus, σ΅¨σ΅¨ π − π’ V σ΅¨σ΅¨ σ΅¨ π π+1 σ΅¨σ΅¨ {π ∈ C : σ΅¨σ΅¨σ΅¨ σ΅¨ ≥ 1; (π ∈ N0 )} ⊆ π (πΊ (π’, V; Δ) , β1 ) . σ΅¨σ΅¨ π − π’π Vπ σ΅¨σ΅¨σ΅¨ (22) Since π = 0, β(πΊ(π’, V; Δ) − ππΌ)−1 β(β1 :β1 ) = supπ |1/(π’π V − π)| < ∞ and the rest part of the proof is similar to that of Case 1. Case 3. If (π’π ) and (Vπ ) are constant sequences of positive reals, then the proof is similar to that of Case 2 and only the difference is β(πΊ(π’, V; Δ) − ππΌ)−1 β(β1 :β1 ) = |1/π’V − π| < ∞. Case 4. If (π’π ) is constant sequence and (Vπ ) is strictly decreasing sequence of positive reals, then the proof is similar to that of Case 1. Theorem 8. The point spectrum of the operator πΊ(π’, V; Δ) over β1 is given by π’π V, ππ (πΊ (π’, V; Δ) , β1 ) = { 0, (ππ (Vπ ) ππ π ππππ π‘πππ‘ π πππ’ππππ) (ππ‘βπππ€ππ π) . (24) Proof. Suppose π₯ ∈ β1 and consider the system of linear equations πΊ(π’, V; Δ)π₯ = ππ₯ for π₯ =ΜΈ 0 = {0, 0, 0, . . .} in β1 , π’0 V0 π₯0 = ππ₯0 π’1 (V0 − V1 ) π₯0 + π’1 V1 π₯1 = ππ₯1 π’2 (V0 − V1 ) π₯0 + π’2 (V1 − V2 ) π₯1 + π’2 V2 π₯2 = ππ₯2 .. . π’π (V0 − V1 ) π₯0 + π’π (V1 − V2 ) π₯1 + ⋅ ⋅ ⋅ + π’π Vπ π₯π = ππ₯π .. . (25) On solving the system of (25), we obtain that π = π’0 V0 whenever π₯0 =ΜΈ 0 and π₯1 = π’1 (V0 − V1 ) π₯ π’0 V0 − π’1 V1 0 Combining (21) and (22) we complete the proof for this case. π₯2 = π’2 (V0 − V1 ) (π’0 V0 − π’1 V2 ) π₯ and in general (π’0 V0 − π’1 V1 ) (π’0 V0 − π’2 V2 ) 0 Case 2. If (π’π ) is a strictly decreasing sequence and (Vπ ) is a constant sequence of positive reals, then π₯π = π’π (V0 − V1 ) π−1 π’0 V0 − π’π Vπ+1 π₯. ∏ π’0 V0 − π’π Vπ π=1 π’0 V0 − π’π Vπ 0 (πΊ (π’, V; Δ) − ππΌ)−1 1 π’0 V − π ( ( ( =( ( ( ( ( (26) Now, we have the following cases. 0 0 0 ⋅⋅⋅ 0 1 π’1 V − π 0 0 0 0 1 π’2 V − π 0 0 0 0 ⋅ ⋅ ⋅) ) ) ⋅ ⋅ ⋅) ). ) ) ⋅⋅⋅ .. . .. . .. . 1 π’3 V − π .. . d) (23) Case 1. If (π’π ) and (Vπ ) are strictly decreasing sequences of positive reals, then we need to show that |(π’0 V0 − π’π Vπ+1 )/(π’0 V0 − π’π Vπ )| ≥ 1. Suppose, for the contrary, if |(π’0 V0 − π’π Vπ+1 )/(π’0 V0 − π’π Vπ )| < 1, then Vπ+1 > Vπ which contradicts the fact that (Vπ ) is strictly decreasing sequence. Therefore, π₯ ∉ β1 and hence ππ (πΊ(π’, V; Δ), β1 ) = 0. Case 2. If (π’π ) is a strictly decreasing and (Vπ ) is a constant sequence of positive reals, then for all π ∈ N0 , π = π’π V is an eigen value corresponding to the eigen 6 International Journal of Analysis vector π₯ = {0, 0, . . . , 0, 1, 0, . . .} whose πth entry is 1. Thus, ππ (πΊ(π’, V; Δ), β1 ) = π’π V. Case 3. If (π’π ) and (Vπ ) are constant sequences of positive reals, then the proof is similar to that of Case 2 and ππ (πΊ(π’, V; Δ), β1 ) = π’V. Case 4. If (π’π ) is a constant and (Vπ ) is a strictly decreasing sequence of positive reals, then the proof is similar to that of Case 1. Theorem 9. The point spectrum of the dual operator (πΊ(π’, V; Δ))∗ over the sequence space β1∗ β β∞ is given by (ππ (Vπ ) ππ π ππππ π‘πππ‘ π πππ’ππππ) π’ V, { { π σ΅¨σ΅¨ π − π’ V σ΅¨σ΅¨ ππ ((πΊ (π’, V; Δ))∗ , β∞ ) = { π π+1 σ΅¨σ΅¨ {{π ∈ C : σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨ π − π’ V σ΅¨σ΅¨σ΅¨ > 1; (π ∈ N0 )} , (ππ‘βπππ€ππ π) . σ΅¨ { π π σ΅¨ Proof. Consider (πΊ(π’, V; Δ))∗ π = ππ and 0 =ΜΈ π ∈ β1 , and then the system of equations Vπ − Vπ+1 (π − π’π−1 Vπ−1 ) ( π − π’π ππ ) . π − π’π Vπ (Vπ−1 − Vπ ) π−1 (31) Therefore, by using ratio text, it is observed that π’1 V1 π1 + (V1 − V2 ) π’2 π2 + (V1 − V2 ) π’3 π3 + ⋅ ⋅ ⋅ = ππ1 σ΅¨σ΅¨ (V − V ) (π − π’ V ) σ΅¨σ΅¨ σ΅¨σ΅¨ π σ΅¨σ΅¨ σ΅¨ σ΅¨ σ΅¨ π+1 π−1 π−1 σ΅¨σ΅¨ σ΅¨σ΅¨ . lim σ΅¨σ΅¨σ΅¨ π σ΅¨σ΅¨σ΅¨ = lim σ΅¨σ΅¨σ΅¨σ΅¨ π πσ³¨→∞ σ΅¨σ΅¨ ππ−1 σ΅¨σ΅¨ πσ³¨→∞ σ΅¨σ΅¨ (Vπ−1 − Vπ ) (π − π’π Vπ+1 ) σ΅¨σ΅¨σ΅¨ Now, we have the following cases. π’2 V2 π2 + (V2 − V3 ) π’3 π3 + (V2 − V3 ) π’4 π4 + ⋅ ⋅ ⋅ = ππ2 .. . (32) Case 1. If (π’π ) and (Vπ ) are strictly decreasing sequences of positive reals, then by using (2), we conclude that limπσ³¨→∞ |ππ /ππ−1 | < 1 if and only if |(π − π’π−1 Vπ−1 )/(π − π’π Vπ+1 )| < 1 for all π ≥ 1. Therefore, ππ ((πΊ(π’, V; Δ))∗ , β∞ ) = {π ∈ C : |(π − π’π Vπ+1 )/(π − π’π Vπ )| > 1; (π ∈ N0 )}. π’π Vπ ππ + (Vπ − Vπ+1 ) π’π+1 ππ+1 + (Vπ − Vπ+1 ) π’π+1 ππ+1 + ⋅ ⋅ ⋅ = πππ .. . (28) From the system of linear equations (28), we obtain that ππ−1 = Combining (29) and (30), we have ππ = π’0 V0 π0 + (V0 − V1 ) π’1 π1 + (V0 − V1 ) π’2 π2 + ⋅ ⋅ ⋅ = ππ0 (27) Vπ−1 − Vπ (π’ π + π’π+1 ππ+1 + π’π+2 ππ+2 + ⋅ ⋅ ⋅ ) , π − π’π−1 Vπ−1 π π (29) V − Vπ+1 (π’ π + π’π+2 ππ+2 + π’π+3 ππ+3 + ⋅ ⋅ ⋅ ) ππ = π π − π’π Vπ π+1 π+1 (π ≥ 1) . (30) Case 2. If (π’π ) is a strictly decreasing and (Vπ ) is a constant sequence of positive reals, then from the system of (28), we observe that for all π ∈ N0 , π = π’π V is an eigen value corresponding to the eigen vector π = {0, 0, . . . , 0, 1, 0, . . .} whose πth entry is 1. Thus, ππ ((πΊ(π’, V; Δ))∗ , β∞ ) = π’π V. Case 3. If (π’π ) and (Vπ ) are constant sequences of positive reals, then the proof is similar to that of Case 2. Case 4. If (π’π ) is a constant and (Vπ ) is a strictly decreasing sequence of positive reals, then the proof is similar to that of Case 1. Theorem 10. The residual spectrum of the operator πΊ(π’, V; Δ) over the sequence space β1 is given by 0, { { σ΅¨σ΅¨ σ΅¨σ΅¨ ππ (πΊ (π’, V; Δ) , β1 ) = { {{π ∈ C : σ΅¨σ΅¨σ΅¨ π − π’π Vπ+1 σ΅¨σ΅¨σ΅¨ > 1; (π ∈ N0 )} , σ΅¨σ΅¨ π − π’ V σ΅¨σ΅¨ σ΅¨ { π π σ΅¨ Proof. Proof follows from Lemma 2 and Theorems 8 and 9. (ππ (Vπ ) ππ π ππππ π‘πππ‘ π πππ’ππππ) (ππ‘βπππ€ππ π) . (33) Theorem 11. The continuous spectrum of the operator πΊ(π’, V; Δ) over the sequence space β1 is given by International Journal of Analysis 7 0, (ππ (Vπ ) ππ π ππππ π‘πππ‘ π πππ’ππππ) { { σ΅¨σ΅¨ π − π’ V σ΅¨σ΅¨ ππ (πΊ (π’, V; Δ) , β1 ) = { π π+1 σ΅¨σ΅¨ {{π ∈ C : σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨ π − π’ V σ΅¨σ΅¨σ΅¨ = 1; (π ∈ N0 )} , (ππ‘βπππ€ππ π) . σ΅¨ { π π σ΅¨ Proof. The proof of this theorem follows from Theorems 7, 8, and 10 and along with the fact that [10] π (πΊ (π’, V; Δ) , β1 ) = ππ (πΊ (π’, V; Δ) , β1 ) ∪ ππ (πΊ (π’, V; Δ) , β1 ) (35) [11] ∪ ππ (πΊ (π’, V; Δ) , β1 ) . 3. 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