GENERALIZED DIFFERENCE OPERATOR
B
(
r
,
s
)
OVER THE SEQUENCE SPACES
c
0AND
c
BIL ˆAL ALTAY AND FEYZ˙I BAS¸AR
Received 27 April 2005 and in revised form 15 July 2005
We determine the fine spectrum of the generalized difference operatorB(r,s) defined by a band matrix over the sequence spacesc0andc, and derive a Mercerian theorem. This generalizes our earlier work (2004) for the difference operator∆, and includes as other special cases the right shift and the Zweier matrices.
1. Preliminaries, background, and notation
LetX andY be the Banach spaces and letT:X→Y also be a bounded linear operator. ByR(T), we denote the range ofT, that is,
R(T)=y∈Y:y=Tx,x∈X. (1.1)
ByB(X), we also denote the set of all bounded linear operators onXinto itself. IfX is any Banach space andT∈B(X), then the adjointT∗ofTis a bounded linear operator on the dualX∗ofXdefined by (T∗f)(x)=f(Tx) for all f ∈X∗andx∈X.
LetX= {θ}be a nontrivial complex normed space andT:Ᏸ(T)→Xa linear operator defined on a subspaceᏰ(T)⊆X. We do not assume thatD(T) is dense inX, or thatThas closed graph{(x,Tx) :x∈D(T)} ⊆X×X. We mean by the expression “T isinvertible” that there exists a bounded linear operatorS:R(T)→Xfor whichST=IonD(T) and
R(T)=X; such thatS=T−1is necessarily uniquely determined, and linear; the bound-edness ofSmeans thatTmust bebounded below, in the sense that there isk >0 for which Tx ≥kxfor allx∈D(T). Associated with each complex number,αis the perturbed operator
Tα=T−αI, (1.2)
defined on the same domainD(T) asT. Thespectrumσ(T,X) consists of thoseα∈Cfor whichTαis not invertible, and theresolventis the mapping from the complementσ(T,X) of the spectrum into the algebra of bounded linear operators onXdefined byα→T−1
α . The name resolvent is appropriate sinceT−1
α helps to solve the equationTαx=y. Thus,
x=T−1
α yprovided thatTα−1exists. More important, the investigation of properties ofTα−1
Copyright©2005 Hindawi Publishing Corporation
will be basic for an understanding of the operatorTitself. Naturally, many properties of
TαandTα−1depend onα, and the spectral theory is concerned with those properties. For instance, we will be interested in the set of allα’s in the complex plane such thatT−1 α exists. Boundedness ofT−1
α is another property that will be essential. We will also ask for whatα’s the domain ofT−1
α is dense inX, to name just a few aspects. Aregular valueαof
Tis a complex number such thatT−1
α exists and is bounded and whose domain is dense inX. For our investigation ofT,Tα, andTα−1, we need some basic concepts in the spectral theory which are given as follows (see [8, pages 370–371]).
Theresolvent setρ(T,X) ofT is the set of all regular valuesαofT. Furthermore, the spectrumσ(T,X) is partitioned into the following three disjoint sets.
Thepoint (discrete) spectrumσp(T,X) is the set such thatTα−1does not exist. Aα∈
σp(T,X) is called aneigenvalueofT.
Thecontinuous spectrumσc(T,X) is the set such thatTα−1exists and is unbounded and the domain ofT−1
α is dense inX.
Theresidual spectrumσr(T,X) is the set such thatTα−1exists (and may be bounded or not) but the domain ofT−1
α is not dense inX.
To avoid trivial misunderstandings, let us say that some of the sets defined above may be empty. This is an existence problem which we will have to discuss. Indeed, it is well-known thatσc(T,X)=σr(T,X)= ∅and the spectrumσ(T,X) consists of only the set
σp(T,X) in the finite-dimensional case.
From Goldberg [6, pages 58–71], ifXis a Banach space andT∈B(X), then there are three possibilities forR(T) and forT−1:
(I)R(T)=X, (II)R(T)=R(T)=X, (III)R(T)=X, and
(1)T−1exists and is continuous, (2)T−1exists but is discontinuous, (3)T−1does not exist.
Applying Golberg’s classification toTα, we have three possibilities forTαand forTα−1: (I)Tαis surjective,
(II)R(Tα)=R(Tα)=X, (III)R(Tα)=X,
and
(1)Tαis injective andTα−1is continuous, (2)Tαis injective andT−1
α is discontinuous, (3)Tαis not injective.
If these possibilities are combined in all possible ways, nine different states are created. These are labeled byI1,I2,I3,II1,II2,II3,III1,III2, andIII3. Ifαis a complex number such thatTα∈I1 orTα∈II1, thenαis in the resolvent set ρ(T,X) of T. The further classification gives rise to thefine spectrumofT. If an operator is in stateII2for example, thenR(T)=R(T)=XandT−1exists but is discontinuous and we writeα∈II
We write∞,c,c0, andbvfor the spaces of all bounded, convergent, null, and bounded variation sequences, respectively. Also byp, we denote the space of allp-absolutely sum-mable sequences, where 1≤p <∞.
LetA=(ank) be an infinite matrix of complex numbersank, wheren,k∈N, and write
(Ax)n=
∞
k=0
ankxk, n∈N,x∈D00(A), (1.3)
whereD00(A) denotes the subspace ofwconsisting ofx∈wfor which the sum exists as a finite sum. More generally, ifµis a normed sequence space, we can writeDµ(A) forx∈w for which the sum in (1.3) converges in the norm ofµ. We will write
(λ:µ)=A:λ⊆Dµ(A) (1.4)
for the space of those matrices which send the whole of the sequence spaceλintoµin this sense. Our main focus in this note is on the band matrixA=B(r,s), where
B(r,s)=
r 0 0 ···
s r 0 ···
0 s r ···
..
. ... ... . ..
(s=0). (1.5)
We begin by determining when a matrixAinduces a bounded operator fromctoc.
Lemma 1.1 (cf. [14, Theorem 1.3.6, page 6]). The matrix A=(ank) gives rise to a bounded linear operatorT∈B(c)fromcto itself if and only if
(1)the rows ofAin1and their1norms are bounded, (2)the columns ofAare inc,
(3)the sequence of row sums ofAis inc.
The operator norm ofTis the supremum of the1norms of the rows.
Corollary1.2. B(r,s) :c→cis a bounded linear operator andB(r,s)(c:c)= |r|+|s|.
Lemma1.3 (cf. [14, Example 8.4.5A, page 129]). The matrixA=(ank)gives raise to a bounded linear operatorT∈B(c0)fromc0to itself if and only if
(1)the rows ofAin1and their1norms are bounded, (2)the columns ofAare inc0.
The operator norm ofTis the supremum of the1norms of the rows.
Corollary1.4. B(r,s) :c0→c0 is a bounded linear operator andB(r,s)(c0:c0)= B(r, s)(c:c).
bvhave also been investigated by Reade [11], Akhmedov and Bas¸ar [1], and Okutoyi [10], respectively. The fine spectrum of the Rhaly operators on the sequence spacesc0andchas been examined by Yıldırım [15]. Furthermore, Cos¸kun [4] has studied the spectrum and fine spectrum forp-Ces`aro operator acting on the spacec0. More recently, de Malafosse [5] and Altay and Bas¸ar [2] have, respectively, studied the spectrum and the fine spec-trum of the difference operator on the sequence spacessr andc0,c; wheresr denotes the Banach space of all sequencesx=(xk) normed by
xsr=sup
k∈N
xk
rk (r >0). (1.6)
In this work, our purpose is to determine the fine spectrum of the generalized diff er-ence operatorB(r,s) on the sequence spacesc0andc, and to give a Mercerian theorem. The main results of the present work are more general than those of Altay and Bas¸ar [2].
2. The spectrum of the operatorB(r,s)on the sequence spacesc0andc
In this section, we examine the spectrum, the point spectrum, the continuous spectrum, the residual spectrum, and the fine spectrum of the operatorB(r,s) on the sequence spacesc0andc. Finally, we also give a Mercerian theorem.
Theorem2.1. σ(B(r,s),c0)= {α∈C:|α−r| ≤ |s|}.
Proof. Firstly, we prove that (B(r,s)−αI)−1exists and is in (c
0:c0) for|α−r|>|s|, and secondly the operatorB(r,s)−αIis not invertible for|α−r| ≤ |s|.
Letα∈σ(B(r,s),c0). SinceB(r,s)−αI is triangle, (B(r,s)−αI)−1 exists and solving (B(r,s)−αI)x=yforxin terms ofygives the matrix (B(r,s)−αI)−1. Thenth row turns out to be
(−s)n−k
(r−α)n−k+1 (2.1)
in thekth place fork≤nand zero otherwise. Thus, we observe that
B(r,s)−αI−1
(c0:c0)=sup
n∈N
n
k=0
(r(−−αs))nn−−kk+1
=r 1
−α
sup
n∈N
n
k=0
r−sαk<∞, (2.2)
that is, (B(r,s)−αI)−1∈(c 0:c0).
Letα∈σ(B(r,s),c0) andα=r. SinceB(r,s)−αIis triangle, (B(r,s)−αI)−1exists but one can see by (2.2) that
B(r,s)−αI−1
(c0:c0)= ∞ (2.3)
Ifα=r, then the operatorB(r,s)−αI=B(0,s) is represented by the matrix
B(0,s)=
0 0 0 ···
s 0 0 ···
0 s 0 ···
..
. ... ... . ..
. (2.4)
SinceR(B(0,s))=c0,B(0,s) is not invertible. This completes the proof.
Theorem2.2. σp(B(r,s),c0)= ∅.
Proof. Suppose thatB(r,s)x=αxforx=θ=(0, 0, 0,...) inc0. Then, by solving the system of linear equations
rx0=αx0,
sx0+rx1=αx1,
sx1+rx2=αx2, .. .
sxk+rxk+1=αxk+1, .. .
(2.5)
we find that ifxn0is the first nonzero entry of the sequencex=(xn), thenα=rand
xn0+k=0 (2.6)
for allk∈N. This contradicts the fact thatxn0=0, which completes the proof.
IfT:c0→c0is a bounded linear operator with the matrixA, then it is known that the adjoint operatorT∗:c∗0 →c0∗is defined by the transposeAtof the matrixA. It should be noted that the dual spacec0∗ofc0is isometrically isomorphic to the Banach space1 of absolutely summable sequences normed byx =∞k=0|xk|.
Theorem2.3. σp(B(r,s)∗,c0∗)= {α∈C:|α−r|<|s|}.
Proof. Suppose thatB(r,s)∗x=αxforx=θ inc∗0 ∼=1. Then, by solving the system of linear equations
rx0+sx1=αx0,
rx1+sx2=αx1,
rx2+sx3=αx2, .. .
rxk+sxk+1=αxk, .. .
we observe that
xn=
α−r
s
n
x0. (2.8)
This shows thatx∈1if and only if|α−r|<|s|, as asserted.
Now, we may give the following lemma required in the proof of theorems given in the present section.
Lemma2.4 [6, page 59]. Thas a dense range if and only ifT∗is one to one.
Theorem2.5. σr(B(r,s),c0)= {α∈C:|α−r|<|s|}.
Proof. We show that the operatorB(r,s)−αIhas an inverse andR(B(r,s)−αI)=c0forα satisfying|α−r|<|s|. Forα=r, the operatorB(r,s)−αIis triangle, hence has an inverse. Forα=r, the operatorB(r,s)−αIis one to one, hence has an inverse. ButB(r,s)∗−αIis not one to one byTheorem 2.3. Now,Lemma 2.4yields the fact thatR(B(r,s)−αI)=c0
and this step concludes the proof.
Theorem2.6. Ifα=r, thenα∈III1σ(B(r,s),c0).
Proof. Since the operator B(r,s)−αI =B(0,s) for α=r, B(0,s)∈III1 or ∈III2 by Theorem 2.5. To verify the fact thatB(0,s) has a bounded inverse, it is enough to show thatB(0,s) is bounded below. Indeed, one can easily see for allx∈c0that
B(0,s)x≥|s|
2 x, (2.9)
which means thatB(0,s) is bounded below. This completes the proof.
Theorem2.7. Ifα=randα∈σr(B(r,s),c0), thenα∈III2σ(B(r,s),c0).
Proof. ByTheorem 2.5,B(r,s)−αI∈III1 or∈III2. Hence, by (2.2), the inverse of the operatorB(r,s)−αIis discontinuous. Therefore,B(r,s)−αIhas an unbounded inverse.
Theorem2.8. σc(B(r,s),c0)= {α∈C:|α−r| = |s|}.
Proof. For this, we prove that the operatorB(r,s)−αIhas an inverse and
RB(r,s)−αI=c0, (2.10)
ifα∈σc(B(r,s),c0). Since α=r,B(r,s)−αI is triangle and has an inverse. Therefore,
B(r,s)∗−αI is one to one by Theorem 2.3and (2.10) holds from Lemma 2.4. This is
what we wished to prove.
Theorem2.9. Ifα∈σc(B(r,s),c0), thenα∈II2σ(B(r,s),c0).
Proof. By (2.2), the inverse of the operatorB(r,s)−αI is discontinuous. Therefore,B(r,
s)−αIhas an unbounded inverse.
To verify that the operatorB(r,s)−αIis not surjective, it is sufficient to show that there is no sequencex=(xn) inc0such that (B(r,s)−αI)x=yfor somey∈c0. Let us consider the sequencey=(1, 0, 0,...)∈c0. For this sequence, we obtainxn= {s/(α−r)}n/(r−α). This yields thatx /∈c0, that is,B(r,s)−αIis not onto. This completes the proof.
Theorem2.10. σ(B(r,s),c)= {α∈C:|α−r| ≤ |s|}.
Proof. This is obtained in the similar way that is used in the proof ofTheorem 2.1.
Theorem2.11. σp(B(r,s),c)= ∅.
Proof. The proof may be obtained by proceeding as in provingTheorem 2.2. So, we omit
the details.
IfT:c→cis a bounded matrix operator with the matrixA, thenT∗:c∗→c∗acting onC⊕1has a matrix representation of the form
χ 0
b At
, (2.11)
whereχis the limit of the sequence of row sums ofAminus the sum of the limit of the columns ofA, andbis the column vector whosekth entry is the limit of thekth column ofAfor eachk∈N. ForB(r,s) :c→c, the matrixB(r,s)∗∈B(1) is of the form
B(r,s)∗=
r+s 0
0 B(r,s)t
. (2.12)
Theorem2.12. σp(B(r,s)∗,c∗)= {α∈C:|α−r|<|s|} ∪ {r+s}.
Proof. Suppose thatB(r,s)∗x=αxforx=θin1. Then by solving the system of linear equations
(r+s)x0=αx0,
rx1+sx2=αx1,
rx2+sx3=αx2, .. .
rxk+sxk+1=αxk, .. .
(2.13)
we obtain that
xn=
α−r
s
n−1
x1 (n≥2). (2.14)
Ifx0=0, thenα=r+s. So,α=r+sis an eigenvalue with the corresponding eigenvector
x=(x0, 0, 0,...). Ifα=r+s, thenx0=0 and one can see by (2.14) thatx∈1if and only
Theorem2.13. σr(B(r,s),c)=σp(B(r,s)∗,c∗).
Proof. The proof is obtained by the analogy with the proof ofTheorem 2.5.
Theorem2.14. σc(B(r,s),c)= {α∈C:|α−r| = |s|} \ {r+s}.
Proof. This is similar to the proof ofTheorem 2.8withα=r+ssuch that|α−r| = |s|.
Since the fine spectrum of the operatorB(r,s) onccan be derived by analogy to that spacec0, we omit the detail and give it without proof. Therefore, we have
B(r,s)−αI∈I1, α∈σ
B(r,s),c,
α∈III1σ
B(r,s),c, α=r,
α∈II2σ
B(r,s),c, α∈σcB(r,s),c,
α∈III2σB(r,s),c, α∈σrB(r,s),c\ {r}.
(2.15)
Theorem2.15. σ(B(r,s),∞)= {α∈C:|α−r| ≤ |s|}.
Proof. It is known by Cartlidge [3] that if a matrix operatorA is bounded onc, then
σ(A,c)=σ(A,∞). Now, the proof is immediate fromTheorem 2.10withA=B(r,s).
Subsequent to stating the concept of Mercerian theorem, we conclude this section by giving a Mercerian theorem. LetAbe an infinite matrix and the setcAdenotes the con-vergence field of that matrixA. A theorem which proves thatcA=cis called a Mercerian theorem, after Mercer, who proved a significant theorem of this type [9, page 186]. Now, we may give our final theorem.
Theorem2.16. Suppose thatαsatisfies the inequality|α(1−r) +r|>|s(1−α)|. Then the convergence field ofA=αI+ (1−α)B(r,s)isc.
Proof. Ifα=1, there is nothing to prove. Let us suppose thatα=1. Then, one can observe byTheorem 2.10and the choice ofα thatB(r,s)−[α/(α−1)]I has an inverse inB(c). That is to say, that
A−1= 1 1−α
B(r,s)−αα −1I
−1
∈B(c). (2.16)
SinceAis a triangle and is inB(c),A−1is also conservative which implies thatc
A=c; see
[14, page 12].
Acknowledgments
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Bilˆal Altay: Matematik E˘gitimi B¨ol¨um¨u, E˘gitim Fak¨ultesi, ˙In¨on¨u ¨Universitesi, 44280 Malatya, Turkey
E-mail address:[email protected]
Feyzi Bas¸ar: Matematik E˘gitimi B¨ol¨um¨u, E˘gitim Fak¨ultesi, ˙In¨on¨u ¨Universitesi, 44280 Malatya, Turkey
