# Fine Spectra of Upper Triangular Triple-Band Matrices over the Sequence Spaceℓp (0<p<∞)

Authors: Ali Karaisa; Feyzi Başar
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

The fine spectra of lower triangular triple-band matrices have been examined by
several authors (e.g., Akhmedov (2006), Başar (2007), and Furken et al. (2010)). Here we determine the fine spectra of upper triangular triple-band matrices over the sequence spaceℓp. The operator A(r,s,t) on sequence space on ℓp is defined by A(r,s,t)x=(rxk+sxk+1+txk+2)k=0∞, where x=(xk)∈ℓp, with 0<p<∞. In this paper we have obtained the results on the spectrum and point spectrum for the operator A(r,s,t) on the sequence space ℓp. Further, the results on continuous spectrum, residual spectrum, and fine spectrum of the operator A(r,s,t) on the sequence space ℓp are also derived. Additionally, we give the approximate point spectrum, defect spectrum, and compression spectrum of the matrix operator A(r,s,t) over the space ℓp and we give some applications.


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## Body

## 1. Introduction

In functional analysis, the spectrum of an operator generalizes the notion of eigenvalues for matrices. The spectrum of an operator over a Banach space is partitioned into three parts, which are the point spectrum, the continuous spectrum, and the residual spectrum. The calculation of these three parts of the spectrum of an operator is called calculating the fine spectrum of the operator.Over the years and different names the spectrum and fine spectra of linear operators defined by some triangle matrices over certain sequence spaces were studied.Byω we denote the space of all complex-valued sequences. Any vector subspace of ω is called a sequence space. We write ℓ∞, c, c0, and bv for the spaces of all bounded, convergent, null, and bounded variation sequences, respectively, which are the Banach spaces with the sup-norm ∥x∥∞=supk∈ℕ|xk| and ∥x∥bv=∑k=0∞|xk-xk+1|, respectively, where ℕ={0,1,2,…}. Also by ℓ1 and ℓp we denote the spaces of all absolutely summable and p-absolutely summable sequences, which are the Banach spaces with the norm ∥x∥p=(∑k=0∞|xk|p)1/p, respectively, where 1⩽p<∞.Several authors studied the spectrum and fine spectrum of linear operators defined by some triangle matrices over some sequence spaces. We introduce knowledge in the existing literature concerning the spectrum and the fine spectrum. Cesàro operator of order one on the sequence spaceℓp was studied by Gonzàlez [1], where 1<p<∞. Also, weighted mean matrices of operators on ℓp have been investigated by Cartlidge [2]. The spectrum of the Cesàro operator of order one on the sequence spaces bv0 and bv were investigated by Okutoyi [3, 4]. The spectrum and fine spectrum of the Rally operators on the sequence space ℓp were examined by Yıldırım [5]. The fine spectrum of the difference operator Δ over the sequence spaces c0 and c was studied by Altay and Başar [6]. The same authors also worked out the fine spectrum of the generalized difference operator B(r,s) over c0 and c, in [7]. Recently, the fine spectra of the difference operator Δ over the sequence spaces c0 and c have been studied by Akhmedov and Başar [8, 9], where bvp is the space consisting of the sequences x=(xk) such that x=(xk-xk-1)∈ℓp and introduced by Başar and Altay [10] with 1⩽p⩽∞. In the recent paper, Furkan et al. [11] have studied the fine spectrum of B(r,s,t) over the sequence spaces ℓp and bvp with 1<p<∞, where B(r,s,t) is a lower triangular triple-band matrix. Later, Karakaya and Altun have determined the fine spectra of upper triangular double-band matrices over the sequence spaces c0 and c, in [12]. Quite recently, Karaisa [13] has determined the fine spectrum of the generalized difference operator A(r~,s~), defined as an upper triangular double-band matrix with the convergent sequences r~=(rk) and s~=(sk) having certain properties, over the sequence space ℓp, where 1<p<∞.In this paper, we study the fine spectrum of the generalized difference operatorA(r,s,t) defined by a triple sequential band matrix acting on the sequence space ℓp (0<p<∞), with respect to Goldberg's classification. Additionally, we give the approximate point spectrum and defect spectrum and give some applications.


## 2. Preliminaries, Background, and Notation

LetX and Y be two Banach spaces and T:X→Y be a bounded linear operator. By R(T) we denote range of T, that is,
(1)R(T)={y∈Y:y=Tx,x∈X}.
By B(X) we also denote the set of all bounded linear operators on X into itself. If T∈B(X) then the adjoint T* of T is a bounded linear operator on the dual X* of X defined by (T*f)(x)=f(Tx) for all f∈X* and x∈X.LetX≠{θ} be a complex normed space and T:D(T)→X be a linear operator with domain D(T)⊆X. With T we associate the operator Tα=T-αI, where α is a complex number and I is the identity operator on D(T). If Tα has an inverse which is linear, we denote it by Tα-1, that is,
(2)Tα-1=(T-αI)-1
and call it the resolvent operator of T.Many properties ofTα and Tα-1 depend on α, and spectral theory is concerned with those properties. For instance, we shall be interested in the set of all α in the complex plane such that Tα-1 exists. The boundedness of Tα-1 is another property that will be essential. We shall also ask for what α the domain of Tα-1 is dense in X, to name just a few aspects For our investigation of T, Tα, and Tα-1, we need some basic concepts in spectral theory which are given as follows (see [14, pp. 370-371]).LetX≠{θ} be a complex normed space and T:D(T)→X be a linear operator with domain D(T)⊆X. A regular value α of T is a complex number such that(R1)
Tα-1 exists,(R2)
Tα-1 is bounded,(R3)
Tα-1 is defined on a set which is dense in X.The resolvent setρ(T) of T is the set of all regular values α of T. Its complement ℂ∖ρ(T) in the complex plane ℂ is called the spectrum of T. Furthermore, the spectrum σ(T) is partitioned into three disjoint sets as follows. The point spectrumσp(T) is the set such that Tα-1 does not exist. α∈σp(T) is called an eigenvalue of T. The continuous spectrumσc(T) is the set such that Tα-1 exists and satisfies (R3) but not (R2). The residual spectrumσr(T) is the set such that Tα-1 exists but does not satisfy (R3).In this section, following Appell et al. [15], we define the three more subdivisions of the spectrum called the approximate point spectrum, defect spectrum,  andcompression spectrum.Given a bounded linear operatorT in a Banach space X, we call a sequence (xk) in X as a Weyl sequence for T if ∥xk∥=1 and ∥Txk∥→0, as k→∞.In what follows, we call the set(3)σap(T,X):={α∈ℂ:thereexistsaWeylsequenceforαI-T}
the  approximate point spectrum of T. Moreover, the subspectrum
(4)σδ(T,X):={α∈ℂ:αI-Tisnotsurjective}
is called defect spectrum of T.The two subspectra given by (3) and (4) form a (not necessarily disjoint) subdivisions
(5)σ(T,X)=σap(T,X)∪σδ(T,X)
of the spectrum. There is another subspectrum
(6)σco(T,X)={α∈ℂ:R(αI-T)¯≠X}
which is often called compression spectrum in the literature.By the definitions given above, we can illustrate the subdivisions of spectrum in Table1.Table 1
Subdivisions of spectrum of a linear operator.
1
2
3
T
α
-
1 exists and is bounded
T
α
-
1 exists and is unbounded
T
α
-
1 does not exist
A
R
(
α
I
-
T
)
=
X
α
∈
ρ
(
T
,
X
)
—
α
∈
σ
p
(
T
,
X
)
α
∈
σ
ap
(
T
,
X
)
B
R
(
α
I
-
T
)
-
=
X
α
∈
ρ
(
T
,
X
)
α
∈
σ
c
(
T
,
X
)
α
∈
σ
p
(
T
,
X
)
α
∈
σ
ap
(
T
,
X
)
α
∈
σ
ap
(
T
,
X
)
α
∈
σ
δ
(
T
,
X
)
α
∈
σ
δ
(
T
,
X
)
C
R
(
α
I
-
T
)
-
≠
X
α
∈
σ
r
(
T
,
X
)
α
∈
σ
r
(
T
,
X
)
α
∈
σ
p
(
T
,
X
)
α
∈
σ
δ
(
T
,
X
)
α
∈
σ
ap
(
T
,
X
)
α
∈
σ
ap
(
T
,
X
)
α
∈
σ
δ
(
T
,
X
)
α
∈
σ
δ
(
T
,
X
)
α
∈
σ
co
(
T
,
X
)
α
∈
σ
co
(
T
,
X
)
α
∈
σ
co
(
T
,
X
)From Goldberg [16] if T∈B(X), X a Banach space, then there are three possibilities for R(T):(I)
R(T)=X,(II)
R(T)≠R(T)¯=X,(III)
R(T)¯≠X,and three possibilities for T-1:(1)
T-1 exists and is continuous,(2)
T-1 exists but is discontinuous,(3)
T-1 does not exist.If these possibilities are combined in all possible ways, nine different states are created. These are labelled byI1, I2, I3, II1, II2, II3, III1, III2 and III3. If α is a complex number such that Tα∈I1 or Tα∈II1, then α is in the resolvent set ρ(X,T) of T. The further classification gives rise to the fine spectrum of T. If an operator is in state II2, for example, then R(T)≠R(T)¯=X and T-1 exists but is discontinuous and we write α∈II2σ(X,T).Letμ and γ be two sequence spaces and let A=(ank) be an infinite matrix of real or complex numbers ank, where n,k∈ℕ={0,1,2,…}. Then, we say that A defines a matrix mapping from μ into γ and we denote it by writing A:μ→γ if for every sequence x=(xk)∈μ the sequence Ax={(Ax)n}, the A-transform of x is in γ, where
(7)(Ax)n=∑kankxkforeachn∈ℕ.
By (μ:γ), we denote the class of all matrices A such that A:μ→γ. Thus, A∈(μ:γ) if and only if the series on the right side of (7) converges for each n∈ℕ and every x∈μ, and we have Ax={(Ax)n}n∈ℕ∈γ for all x∈μ.Proposition 1 (see [15, Proposition 1.3, p. 28]).
Spectra and subspectra of an operatorT∈B(X) and its adjoint T*∈B(X*) are related by the following relations:(a)
σ(T*,X*)=σ(T,X),(b)
σc(T*,X*)⊆σap(T,X),(c)
σap(T*,X*)=σδ(T,X),(d)
σδ(T*,X*)=σap(T,X),(e)
σp(T*,X*)=σco(T,X),(f)
σco(T*,X*)⊇σp(T,X),(g)
σ(T,X)=σap(T,X)∪σp(T*,X*)=σp(T,X)∪σap(T*,X*).
The relations (c)–(f) show that the approximate point spectrum is in a certain sense dual to defect spectrum and the point spectrum dual to the compression spectrum.
The equality (g) implies, in particular, thatσ(T,X)=σap(T,X) if X is a Hilbert space and T is normal. Roughly speaking, this shows that normal (in particular, self-adjoint) operators on Hilbert spaces are most similar to matrices in finite dimensional spaces (see [15]).Lemma 2 (see [16, p. 60]).
The adjoint operatorT*of   T is onto if and only if T has a bounded inverse.Lemma 3 (see [16, p. 59]).
T has a dense range if and only if T* is one to one.Our main focus in this paper is on the triple-band matrixA(r,s,t), where
(8)A(r,s,t)=[rst0…0rst…00rs…000r…⋮⋮⋮⋮⋱].
We assume here and after that s and t are complex parameters which do not simultaneously vanish. We introduce the introduce the operator A(r,s,t) from ℓp to itself by
(9)A(r,s,t)x=(rxk+sxk+1+txk+2)k=0∞,wherex=(xk)∈ℓp.


## 3. Fine Spectra of Upper Triangular Triple-Band Matrices over the Sequence Spaceℓp (0<p⩽1)

In this section, we prove that the operatorA(r,s,t):ℓp→ℓp is a bounded linear operator and compute its norm. We essentially emphasize the fine spectrum of the operator A(r,s,t):ℓp→ℓp in the case 0<p⩽1.Theorem 4.
The operatorA(r,s,t):ℓp→ℓp is a bounded linear operator and
(10)∥A(r,s,t)∥(ℓp:ℓp)=|r|p+|s|p+|t|p.Proof.
Since the linearity of the operatorA(r,s,t) is trivial, so it is omitted. Let us take e(2)∈ℓp. Then A(r,s,t)e(2)=(t,s,r,0,…) and observe that
(11)∥A(r,s,t)∥(ℓp:ℓp)⩾∥A(r,s,t)e(2)∥p∥e(2)∥p=|r|p+|s|p+|t|p
which gives the fact that
(12)∥A(r,s,t)∥(ℓp:ℓp)⩾|r|p+|s|p+|t|p.
Let x=(xk)∈ℓp, where 0<p⩽1. Then, since (txk+2),(rxk), and (sxk+1)∈ℓp, it is easy to see by triangle inequality that
(13)∥A(r,s,t)x∥p=∑k=0∞|rxk+sxk+1+txk+2|p⩽∑k=0∞|rxk|p+∑k=0∞|sxk+1|p+∑k=0∞|txk+2|p=|r|p∑k=0∞|xk|p+|s|p∑k=0∞|xk+1|p+|t|p∑k=0∞|xk+2|p=|s|p∥x∥p+|r|p∥x∥p+|t|p∥x∥p=(|r|p+|s|p+|t|p)∥x∥p
which leads us to the result that
(14)∥A(r,s,t)∥(ℓp:ℓp)⩽|r|p+|s|p+|t|p.
Therefore, by combining the inequalities (12) and (14) we see that (10) holds which completes the proof.IfT:ℓp→ℓp is a bounded matrix operator with the matrix A, then it is known that the adjoint operator T*:ℓp*→ℓp* is defined by the transpose of the matrix A. The dual space of ℓp is isomorphic to ℓ∞, where 0<p<1.Before giving the main theorem of this section, we should note the following remark. In this work, here and in what follows, ifz is a complex number, then by z we always mean the square root of z with a nonnegative real part. If Re(z)=0, then z represents the square root of z with Im(z)>0. The same results are obtained if z represents the other square root.Theorem 5.
Lets be a complex number such that s2=-s and define the set D1 by
(15)D1={α∈ℂ:2|r-α|⩽|-s+s2-4t(r-α)|}.
Then, σc(A(r,s,t),ℓp)⊆D1.Proof.
Lety=(yk)∈ℓ∞. Then, by solving the equation Aα(r,s,t)*x=y for x=(xk) in terms of y, we obtain
(16)x0=y0r-α,x1=y1r-α+-sy0(r-α)2,x2=y2r-α+-sy1(r-α)2+[s2-t(r-α)]y0(r-α)3,(17)⋮
and if we denote a1=1/(r-α), a2=-s/(r-α)2, and a3=(s2-t(r-α))/(r-α)3, we have
(18)x0=a1y0,x1=a1y1+a2y0,x2=a1y2+a2y1+a3y0,⋮xn=a1yn+a2yn-1+⋯+an+1y0=∑k=0nan+1-kyk.
Now we must find an. We have yn=txn-2+sxn-1+(r-α)xn and if we use relation (18), we have
(19)yn=t∑k=0n-2an-1-kyk+s∑k=0n-1an-kyk+(r-α)∑k=0nan+1-kyk=y0(tan-1+san+(r-α)an+1)+y1(tan-2+san-1+(r-α)an)+⋯+yna1(r-α).
This implies that
(20)tan-1+san+(r-α)an+1=0,tan-2+san-1+(r-α)an=0,…,a1(r-α)=1.
In fact this sequence is obtained recursively by letting
(21)a1=1r-α,a2=-s(r-α)2,tan-2+san-1+(r-α)an=0,∀n⩾3.
The characteristic polynomial of the recurrence relation is (r-α)λ2+sλ+t=0. There are two cases.
Case 1. If Δ=s2-4t(r-α)≠0 whose roots are
(22)λ1=-s+Δ2(r-α),λ2=-s-Δ2(r-α),
elementary calculation on recurrent sequence gives that
(23)an=λ1n-λ2ns2-4t(r-α),∀n⩾1.
In this case xk=(1/Δ)∑k=0n‍(λ1n+1-k-λ2n+1-k)yk. Assume that |λ1|<1. So we have
(24)|1+4t(r-α)s2|<|2(r-α)-s|.
Since |1-z|⩽|1+z| for any z∈ℂ, we must have
(25)|1-4t(r-α)s2|<|2(r-α)-s|.
It follows that |λ2|<1. Now, for |λ1|<1 we can see that
(26)|xn|⩽1|Δ|∑k=0n|λ1n+1-k||yk|+∑k=0n|λ2n+1-k||yk|
for all n∈ℕ. Taking limit on the inequality (26) as n→∞, we get
(27)∥x∥∞⩽1-(|λ2|+|λ2|)|(1-|λ2||1-λ2|)Δ|∥y∥∞.
Thus for |λ1|<1, Aα(r,s,t)* is onto and by Lemma 2, Aα(r,s,t) has a bounded inverse. This means that
(28)σc(A(r,s,t),ℓp)⊆{α∈ℂ:2|r-α|⩽|-s+s2-4t(r-α)|}=D1.
Case 2. If Δ=s2-4t(r-α)=0, a calculation on recurrent sequence gives that
(29)an=(2n-s)[-s2(r-α)]n,∀n⩾1.
Now, for |-s|<2|r-α| we can see that
(30)|xn|⩽∑k=0n|an-kyk|
for all n∈ℕ. Taking limit on the inequality (30) as n→∞, we obtain that
(31)∥x∥∞⩽∥y∥∞∑k=0∞|ak|.∑k=0∞‍|ak| is convergent, since |-s|<2|r-α|. Thus for |-s|<2|r-α|, Aα(r,s,t)* is onto and by Lemma 2, Aα(r,s,t) has a bounded inverse. This means that
(32)σc(A(r,s,t),ℓp)⊆{α∈ℂ:2|r-α|⩽|-s|}⊆D1.Theorem 6.
σ
p
(
A
(
r
,
s
,
t
)
*
,
ℓ
p
*
)
=
∅.Proof.
ConsiderA(r,s,t)*f=αf with f≠θ=(0,0,0,…) in ℓp*=ℓ∞. Then, by solving the system of linear equations
(33)rf0=αf0,sf0+rf1=αf1,tf0+sf1+rf2=αf2,tf1+sf2+rf3=αf3,⋮tfk-2+sfk-1+rfk=αfk,⋮
we find that f0=0 if α≠r and f1=f2=⋯=0 if f0=0 which contradicts f≠θ. If fn0 is the first nonzero entry of the sequence f=(fn) and α=r, then we get tfn0-2+sfn0-1+rfn0=αfn0 which implies fn0=0 which contradicts the assumption fn0≠0. Hence, the equation A(r,s,t)*f=αf has no solution f≠θ.Theorem 7.
σ
p
(
A
(
r
,
s
,
t
)
,
ℓ
p
)
=
D
2, where D2={α∈ℂ:2|r-α|<|-s+s2-4t(r-α)|}.Proof.
LetA(r,s,t)x=αx for θ≠x∈ℓp. Then, by solving the system of linear equations
(34)rx0+sx1+tx2=αx0,rx1+sx2+tx3=αx1,rx2+sx3+tx4=αx2,⋮rxk-2+sxk-1+txk=αxk,⋮
and we have
(35)x2=-stx1-r-αtx0,x3=s2-t(r-α)t2x1+s(r-α)t2x0,⋮xn=an(r-α)ntn-1x1-an-1(r-α)ntn-1x0,∀n⩾2.
Assume thatα∈D2. Then, we choose x0=1 and x1=2(r-α)/(-s+s2-4t(r-α)). We show that xn=x1n for all n⩾2. Since λ1,λ2 are roots of the characteristic equation (r-α)λ2+sλ+t=0, we must have
(36)λ1λ2=tr-α,λ1-λ2=Δr-α.
Combining the fact x1=1/λ1 with relation (35), we can see that
(37)xn=an(r-α)ntn-1x1-an-1(r-α)ntn-1x0=(r-αt)n-1(r-α)(-an-1x0+anx1)=1(λ1λ2)n-1r-αΔ(-λ1n-1+λ2n-1+λ1n-1-λ2nλ1-1)=1λ1n-1λ2n-1(1λ1-λ2)λ2n-1(λ1-λ2λ1)=1λ1n=x1n.
The same result may be obtained in case Δ=0. Now x=(xk)∈ℓp, since |x1|<1. This shows that D2⊆σp(A(r,s,t),ℓp).
Now we assume thatα∉D2, that is, |λ1|⩽1. We must show that α∉σp(A(r,s,t),ℓp). Therefore we obtain from the relation (35) that
(38)xn+1xn=(r-αt)an-1an-2(-x0+(an/an-1)x1-x0+(an-1/an-2)x1).
Now we examine three cases.
Case 1  (|λ2|<|λ1|⩽1). In this case we have s2≠4t(r-α) and
(39)anan-1=λ1n+1-λ2n+1λ1n-λ2n=λ1[1-(λ2/λ1)n+1][1-(λ2/λ1)n].
Then, we have
(40)limn→∞|anan-1|p=limn→∞|an-1an-2|p=limn→∞|λ1|p|1-(λ2/λ1)n+1|p|1-(λ2/λ1)n|p=|λ1|p.
Now, if -x0+λ1x1=0, then we have (xn)=(x0/λ1n) which is not in ℓp. Otherwise,
(41)limn→∞|xn+1xn|p=1|λ1|p|λ2|p|λ1|p=1|λ2|p>1.
Case 2   (|λ2|=|λ1|<1). In this case we have s2=4t(r-α) and using the formula
(42)an=(2n-s)[-s2(r-α)]n∀n⩾1,
we obtain that
(43)limn→∞|anan-1|p=|-s2(r-α)|p=|λ1|p
which leads to
(44)limn→∞|xn+1xn|p=1|λ1|p|λ2|p|λ1|p=1|λ2|p>1.
Case 3  (|λ2|=|λ1|=1). In this case we have s2=4t(r-α) and so we have |-s/(2t)|=1. Assume that α∈σp(A(r,s,t),ℓp). This implies that x∈ℓp and x≠θ. Thus we again derive (35)
(45)xn=(-s2t)n-1[-(n-1)-s2tx0+nx1].
Since limn→∞xn=0, we must have x0=x1=0. But this implies that x=θ, a contradiction which means that α∉σp(A(r,s,t),ℓp). Thus σp(A(r,s,t),ℓp)⊆D2. This completes the proof.Theorem 8.
σ
r
(
A
(
r
,
s
,
t
)
,
ℓ
p
)
=
∅.Proof.
By Proposition1, σr(A(r,s,t),ℓp)=σp(A(r,s,t)*,ℓp*)∖σp(A(r,s,t),ℓp). Since by Theorem 6,
(46)σp(A(r,s,t)*,ℓp*)=∅,σr(A(r,s,t),ℓp)=∅.
This completes the proof.Theorem 9.
Lets be a complex number such that s2=-s and define the set D1 by
(47)D1={α∈ℂ:2|r-α|⩽|-s+s2-4t(r-α)|}.
Then, σ(A(r,s,t),ℓp)=D1.Proof.
By Theorem7, we get
(48){α∈ℂ:2|r-α|<|-s+s2-4t(r-α)|}⊆σ(A(r,s,t),ℓp).
Since the spectrum of any bounded operator is closed, we have
(49){α∈ℂ:2|r-α|⩽|-s+s2-4t(r-α)|}⊆σ(A(r,s,t),ℓp)
and again from Theorems 5, 7, and 8,
(50)σ(A(r,s,t),ℓp)⊆{s2-4t(r-α)α∈ℂ:2|r-α|⩽|-s+s2-4t(r-α)|}.
Combining (49) and (50), we obtain that σ(A(r,s,t),ℓp)=D1, where D1 is defined by (47).Theorem 10.
σ
c
(
A
(
r
,
s
,
t
)
,
ℓ
p
)
=
D
3, where D3={α∈ℂ:2|r-α|=|-s+s2-4t(r-α)|}.Proof.
Because the partsσc(A(r,s,t),ℓp), σr(A(r,s,t),ℓp), and σp(A(r,s,t),ℓp) are pairwise disjoint sets and the union of these sets is σ(A(r,s,t),ℓp), the proof immediately follows from Theorems 7, 8, and 9.Theorem 11.
Ifα∈D2, α∈σ(A(r,s,t),ℓp)I3.Proof.
From Theorem7, α∈σp(A(r,s,t),ℓp). Thus, (A(r,s,t)-αI)-1 does not exist. By Theorem 6A(r,s,t)*-αI is one to one, so A(r,s,t)-αI has a dense range in ℓp by Lemma 3.Theorem 12.
The following statements hold:(i)
σap(A(r,s,t),ℓp)=D1,(ii)
σδ(A(r,s,t),ℓp)=D3,(iii)
σco(A(r,s,t),ℓp)=∅.Proof.
(i)  Since from Table1,
(51)σap(A(r,s,t),ℓp)=σ(A(r,s,t),ℓp)∖σ(A(r,s,t),ℓp)III1,
we have by Theorem 8(52)σ(A(r,s,t),ℓp)III1=σ(A(r,s,t),ℓp)III2=∅.
Hence
(53)σap(A(r,s,t),ℓp)=D1.
(ii) Since the following equality(54)σδ(A(r,s,t),ℓp)=σ(A(r,s,t),ℓp)∖σ(A(r,s,t),ℓp)I3
holds from Table 1, we derive by Theorems 8 and 11 that σδ(A(r,s,t),ℓp)=D2.
(iii) From Table1, we have
(55)σco(A(r,s,t),ℓp)=σ(A(r,s,t),ℓp)III1∪σ(A(r,s,t),ℓp)III2∪σ(A(r,s,t),c0)III3.
By Theorem 6 it is immediate that σco(A(r,s,t),ℓp)=∅.


## 4. Fine Spectra of Upper Triangular Triple-Band Matrices over the Sequence Spaceℓp(1<p<∞)

In the present section, we determine the fine spectrum of the operatorA(r,s,t):ℓp→ℓp in case 1⩽p<∞. We quote some lemmas which are needed in proving the theorems given in Section 4.Lemma 13 (see [17, p. 253, Theorem 34.16]).
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 14 (see [17, p. 245, Theorem 34.3]).
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 rows of A is bounded.Lemma 15 (see [17, p. 254, Theorem 34.18]).
Let1<p<∞ and A∈(ℓ∞:ℓ∞)∩(ℓ1:ℓ1). Then, A∈(ℓp:ℓp).Theorem 16.
The operatorA(r,s,t):ℓp→ℓp is a bounded linear operator and
(56)(|r|p+|s|p+|t|p)1/p⩽∥A(r,s,t)∥(ℓp:ℓp)⩽|r|+|s|+|t|.Proof.
Since the linearity of the operatorA(r,s,t) is not hard, we omit the details. Now we prove that (56) holds for the operator A(r,s,t) on the space ℓp. It is trivial that A(r,s,t)e(2)=(t,s,r,0,…) for e(2)∈ℓp. Therefore, we have
(57)∥A(r,s,t)∥(ℓp:ℓp)⩾∥A(r,s,t)e(2)∥p∥e(2)∥p=(|r|p+|s|p+|t|p)1/p
which implies that
(58)∥A(r,s,t)∥(ℓp:ℓp)⩾(|r|p+|s|p+|t|p)1/p.
Let x=(xk)∈ℓp, where 1<p<∞. Then, since (txk+2), (rxk), and (sxk+1)∈ℓp, it is easy to see by Minkowski's inequality that
(59)∥A(r,s,t)x∥p=(∑k=0∞‍|rxk+sxk+1+txk+2|p)1/p⩽(∑k=0∞|rxk|p)1/p+(∑k=0∞|sxk+1|p)1/p+(∑k=0∞|txk+2|p)1/p=|r|(∑k=0∞|xk|p)1/p+|s|(∑k=0∞|xk+1|p)1/p+|t|(∑k=0∞|xk+2|p)1/p=|s|∥x∥p+|r|∥x∥p+|t|∥x∥p=(|r|+|s|+|t|)∥x∥p
which leads us to the result that
(60)(|r|p+|s|p+|t|p)1/p⩽∥A(r,s,t)∥(ℓp:ℓp)⩽|r|+|s|+|t|.
Therefore, by combining the inequalities in (58) and (59) we have (56), as desired.IfT:ℓp→ℓp is a bounded matrix operator with the matrix A, then it is known that the adjoint operator T*:ℓp*→ℓp* is defined by the transpose of the matrix A. The dual space of ℓp is isomorphic to ℓq, where 1<p<∞.Theorem 17.
Lets be a complex number such that s2=-s and define the set D1 by
(61)D1={α∈ℂ:2|r-α|⩽|-s+s2-4t(r-α)|}.
Then, σc(A(r,s,t),ℓp)⊆D1.Proof.
We will show thatAα(r,s,t)* is onto, for 2|r-α|>|-s+s2-4t(r-α)|. Thus, for every y∈ℓq, we find x∈ℓq. Aα(r,s,t)* is a triangle so it has an inverse. Also equation Aα(r,s,t)*x=y gives [Aα(r,s,t)*]-1y=x. It is sufficient to show that [Aα(r,s,t)*]-1∈(ℓq:ℓq). We calculate that A=(ank)=[Aα(r,s,t)*]-1 as follows:
(62)A=(ank)=[a100…a2a10…a3a2a1…⋮⋮⋮⋱],
where
(63)a1=1r-α,a2=-s(r-α)2,a3=s2-t(r-α)(r-α)3,⋮
It is known that from Theorem 5(64)an=λ1n-λ2ns2-4t(r-α),∀n⩾1,whereλ1=-s+Δr-α,λ2=-s-Δr-α.
Now, we show that [Aα(r,s,t)*]-1∈(ℓ1:ℓ1), for |λ1|<1. By Theorem 5, we know that if |λ1|<1, |λ2|<1. We assume that s2-4t(r-α)≠0 and |λ1|<1,
(65)∥[Aα(r,s,t)*]-1∥(ℓ1:ℓ1)=supn∑k=n∞|ak|=∑k=1∞|ak|⩽1|Δ|(∑k=1∞|λ1|k+∑k=1∞|λ2|k)<∞,
since |λ1|<1 and |λ2|<1. This shows that (Aα(r,s,t)*]-1∈(ℓ1:ℓ1). Similarly we can show that [Aα(r,s,t)*]-1∈(ℓ∞:ℓ∞).
Now assume thats2-4t(r-α)=0. Then,
(66)an=(2n-s)[-s2(r-α)]n
and simple calculation gives that (an)∈ℓq if and only if |-s|<2|r-α|,
(67)[(A(r,s,t)-αI)*]-1∈(ℓq:ℓq)forα∈ℂwith2|r-α|>|-s+s2-4t(r-α)|.
Hence, Aα(r,s,t)* is onto. By Lemma 2, Aα(r,s,t) has a bounded inverse. This means that σc(A(r,s,t),ℓp)⊆D1, where D1 is defined by (61).Theorem 18.
σ
p
(
A
(
r
,
s
,
t
)
*
,
ℓ
p
*
)
=
∅.Proof.
LetA(r,s,t)*f=αf with f≠θ=(0,0,0,…) in ℓp*=ℓq. Then, by solving the system of linear equations
(68)rf0=αf0,sf0+rf1=αf1,tf0+sf1+rf2=αf2,tf1+sf2+rf3=αf3,⋮tfk-2+sfk-1+rfk=αfk,⋮
we find that f0=0 if α≠r and f1=f2=⋯=0 if f0=0 which contradicts f≠θ. If fn0 is the first nonzero entry of the sequence f=(fn) and α=r, then we get tfn0-2+sfn0-1+rfn0=αfn0 which implies fn0=0 which contradicts the assumption fn0≠0. Hence, the equation A(r,s,t)*f=αf has no solution f≠θ.In the case1<p<∞, since the proof of the theorems, in Section 4, determining the spectrum and fine spectrum of the matrix operator A(r,s,t) on the sequence space ℓp is similar to the case 0<p⩽1; to avoid the repetition of similar statements, we give the results by the following theorem without proof.Theorem 19.
The following statements hold:(i)
σ(A(r,s,t),ℓp)=D1,(ii)
σr(A(r,s,t),ℓp)=∅,(iii)
σp(A(r,s,t),ℓp)=D2,(iv)
σc(A(r,s,t),ℓp)=D3,(v)
σap(A(r,s,t),ℓp)=D1,(vi)
σco(A(r,s,t),ℓp)=∅,(vii)
σδ(A(r,s,t),ℓp)=D3.


## 5. Some Applications

In this section, we give two theorems related to Toeplitz matrix.Theorem 20.
LetP be a polynomial that corresponds to the n-tuple a and let z1,z2,z3,…,zn-1 also be the roots of P. Define T as a Toeplitz matrix associated with P, that is,
(69)T=[a0a1a2…an000…0a0a1a2…an00…00a0a1a2…an0…⋮⋮⋱⋱⋱⋱⋱⋱⋱].
The resolvent operator T over ℓp with 1<p<∞, where the domain of the resolvent operator is the whole space ℓp, exists if and if only all the roots of the polynomial are outside the unit disc {z∈ℂ:|z|⩽1}. That is T-1∈(ℓp:ℓp) if and if only |zi|>1, 1⩽i⩽n-1. In this case the resolvent operator is represented by
(70)T-1=1an-1A-1(-z1,1)A-1(-z2,1)⋯A-1(-zn-1,1),whereA-1(-zi,1)=-[1/zi1/zi21/zi31/zi41/zi5…01/zi1/zi21/zi31/zi4…001/zi1/zi21/zi3…0001/zi1/zi2…00001/zi…⋮⋮⋮⋮⋮⋱].Proof.
Suppose all the roots of the polynomialP(z)=a0+a1z+⋯+an-1zn-1=an(z-z1)(z-z2)⋯(z-zn-1) are outside the unit disc. The Toeplitz matrix associated with P can be written as the product
(71)T=anA(-z1,1)A(-z2,1)⋯A(-zn-1,1).
Since multiplication of upper triangular Toeplitz matrices is commutative, we can see that
(72)T-1=1an-1A-1(-z1,1)A-1(-z2,1)⋯A-1(-zn-1,1)
is left inverse of T. Since all roots are outside the unit disc, then
(73)∥T-1(-zi,1)∥(ℓ∞:ℓ∞)=supn∑k=n∞1|zi|k+1-n=∑k=1∞1|zi|k<∞.
Therefore each T-1(-zi,1)∈(ℓ∞:ℓ∞), for 1⩽i⩽n-1. Similarly we can say that T-1(-zi,1)∈(ℓ1:ℓ1). So we have T-1(-zi,1)∈(ℓp:ℓp).Theorem 21.
The resolvent operator ofA(r,s,t) over ℓp with 1<p<∞, where the domain of the resolvent operator is the space ℓp, exists if and only if 2|r|>|-s+s2-4tr|. In this case, the resolvent operator is represented by the infinite banded Toeplitz matrix
(74)E(r,s,t)=1t[u1u12u13u14…0u1u12u13…00u1u12…000u1…⋮⋮⋮⋮⋱][u2u22u23u24…0u2u22u23…00u2u22…000u2…⋮⋮⋮⋮⋱],whereu1=-s+s2-4tr2r,u2=-s-s2-4tr2r.Proof.
By Theorem20, we can see that E(r,s,t) is inverse of the matrix of A(r,s,t). But this is not enough to say it is resolvent operator. By Lemmas 13, 14, and 15, E(r,s,t)∈(ℓp:ℓp), when 2|r|>|-s+s2-4tr|. That is for 2|r|>|-s+s2-4tr|, E(r,s,t) is a resolvent operator.



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*Source: 342682-2013-03-20.xml*