Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2013 Article ID 342682 10 pageshttpdxdoiorg1011552013342682

Research ArticleFine Spectra of Upper Triangular Triple-Band Matrices overthe Sequence Space ℓ119901 (0 lt 119901 lt infin)

Ali Karaisa1 and Feyzi BaGar2

1 Department of Mathematics Necmettin Erbakan University Karacigan Mahallesi Ankara Caddesi 74 42060 Konya Turkey2Department of Mathematics Fatih University Hadımkoy Campus Buyukcekmece 34500 Istanbul Turkey

Correspondence should be addressed to Ali Karaisa alikaraisahotmailcom

Received 27 September 2012 Revised 28 December 2012 Accepted 31 December 2012

Academic Editor Simeon Reich

Copyright copy 2013 A Karaisa and F Basar This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited

The fine spectra of lower triangular triple-band matrices have been examined by several authors (eg Akhmedov (2006) Basar(2007) and Furken et al (2010)) Here we determine the fine spectra of upper triangular triple-band matrices over the sequencespace ℓ119901 The operator 119860(119903 119904 119905) on sequence space on ℓ119901 is defined by 119860(119903 119904 119905)119909 = (119903119909119896 + 119904119909119896+1 + 119905119909119896+2)

infin

119896=0 where 119909 = (119909119896) isin ℓ119901

with 0 lt 119901 lt infin In this paper we have obtained the results on the spectrum and point spectrum for the operator 119860(119903 119904 119905) on thesequence space ℓ119901 Further the results on continuous spectrum residual spectrum and fine spectrum of the operator 119860(119903 119904 119905) onthe sequence space ℓ119901 are also derived Additionally we give the approximate point spectrum defect spectrum and compressionspectrum of the matrix operator 119860(119903 119904 119905) over the space ℓ119901 and we give some applications

1 Introduction

In functional analysis the spectrum of an operator gener-alizes the notion of eigenvalues for matrices The spectrumof an operator over a Banach space is partitioned into threeparts which are the point spectrum the continuous spec-trum and the residual spectrum The calculation of thesethree parts of the spectrumof an operator is called calculatingthe fine spectrum of the operator

Over the years and different names the spectrum and finespectra of linear operators defined by some triangle matricesover certain sequence spaces were studied

By 120596 we denote the space of all complex-valued sequen-ces Any vector subspace of 120596 is called a sequence space Wewrite ℓinfin 119888 1198880 and 119887119907 for the spaces of all bounded con-vergent null and bounded variation sequences respectivelywhich are the Banach spaces with the sup-norm 119909infin =

sup119896isinN|119909119896| and 119909119887119907 = sum

infin

119896=0|119909119896 minus 119909119896+1| respectively where

N = 0 1 2 Also by ℓ1 and ℓ119901 we denote the spaces of allabsolutely summable and 119901-absolutely summable sequenceswhich are the Banach spaces with the norm 119909119901 =

(suminfin

119896=0|119909119896|119901)1119901 respectively where 1 ⩽ 119901 lt infin

Several authors studied the spectrum and fine spectrumof linear operators defined by some triangle matrices oversome sequence spaces We introduce knowledge in the exist-ing literature concerning the spectrum and the fine spectrumCesaro operator of order one on the sequence space ℓ119901 wasstudied by Gonzalez [1] where 1 lt 119901 lt infin Also weightedmean matrices of operators on ℓ119901 have been investigated byCartlidge [2] The spectrum of the Cesaro operator of orderone on the sequence spaces 1198871199070 and 119887119907 were investigatedby Okutoyi [3 4] The spectrum and fine spectrum of theRally operators on the sequence space ℓ119901 were examined byYıldırım [5] The fine spectrum of the difference operatorΔ over the sequence spaces 1198880 and 119888 was studied by Altayand Basar [6] The same authors also worked out the finespectrum of the generalized difference operator 119861(119903 119904) over1198880 and 119888 in [7] Recently the fine spectra of the differenceoperator Δ over the sequence spaces 1198880 and 119888 have beenstudied by Akhmedov and Basar [8 9] where 119887119907119901 is the spaceconsisting of the sequences 119909 = (119909119896) such that 119909 = (119909119896 minus119909119896minus1) isin ℓ119901 and introduced by Basar and Altay [10] with 1 ⩽119901 ⩽ infin In the recent paper Furkan et al [11] have studied thefine spectrum of119861(119903 119904 119905) over the sequence spaces ℓ119901 and 119887119907119901

2 Abstract and Applied Analysis

with 1 lt 119901 lt infin where 119861(119903 119904 119905) is a lower triangular triple-band matrix Later Karakaya and Altun have determined thefine spectra of upper triangular double-band matrices overthe sequence spaces 1198880 and 119888 in [12] Quite recently Karaisa[13] has determined the fine spectrum of the generalizeddifference operator 119860(119903 119904) defined as an upper triangulardouble-band matrix with the convergent sequences 119903 = (119903119896)and 119904 = (119904119896) having certain properties over the sequencespace ℓ119901 where 1 lt 119901 lt infin

In this paper we study the fine spectrum of the general-ized difference operator 119860(119903 119904 119905) defined by a triple sequen-tial band matrix acting on the sequence space ℓ119901 (0 lt 119901 ltinfin) with respect to Goldbergrsquos classification Additionally wegive the approximate point spectrum and defect spectrumand give some applications

2 Preliminaries Background and Notation

Let 119883 and 119884 be two Banach spaces and 119879 119883 rarr 119884 be abounded linear operator By 119877(119879) we denote range of 119879 thatis

119877 (119879) = 119910 isin 119884 119910 = 119879119909 119909 isin 119883 (1)

By119861(119883)we also denote the set of all bounded linear operatorson 119883 into itself If 119879 isin 119861(119883) then the adjoint 119879lowast of 119879 isa bounded linear operator on the dual 119883lowast of 119883 defined by(119879lowast119891)(119909) = 119891(119879119909) for all 119891 isin 119883lowast and 119909 isin 119883Let 119883 = 120579 be a complex normed space and 119879 119863(119879) rarr

119883 be a linear operator with domain 119863(119879) sube 119883 With 119879 weassociate the operator 119879120572 = 119879 minus 120572119868 where 120572 is a complexnumber and 119868 is the identity operator on 119863(119879) If 119879120572 has aninverse which is linear we denote it by 119879minus1

120572 that is

119879minus1

120572= (119879 minus 120572119868)

minus1 (2)

and call it the resolvent operator of 119879Many properties of 119879120572 and 119879

minus1

120572depend on 120572 and spectral

theory is concerned with those properties For instance weshall be interested in the set of all 120572 in the complex plane suchthat 119879minus1

120572exists The boundedness of 119879minus1

120572is another property

that will be essential We shall also ask for what 120572 the domainof 119879minus1120572

is dense in 119883 to name just a few aspects For ourinvestigation of 119879 119879120572 and 119879

minus1

120572 we need some basic concepts

in spectral theory which are given as follows (see [14 pp 370-371])

Let 119883 = 120579 be a complex normed space and 119879 119863(119879) rarr119883 be a linear operator with domain 119863(119879) sube 119883 A regularvalue 120572 of 119879 is a complex number such that

(R1) 119879minus1120572

exists(R2) 119879minus1

120572is bounded

(R3) 119879minus1120572

is defined on a set which is dense in 119883

The resolvent set 120588(119879) of 119879 is the set of all regular values 120572 of119879 Its complement C120588(119879) in the complex plane C is calledthe spectrum of 119879 Furthermore the spectrum 120590(119879) is parti-tioned into three disjoint sets as follows The point spectrum120590119901(119879) is the set such that 119879minus1

120572does not exist 120572 isin 120590119901(119879) is

called an eigenvalue of 119879 The continuous spectrum 120590119888(119879) isthe set such that 119879minus1

120572exists and satisfies (R3) but not (R2)

The residual spectrum 120590119903(119879) is the set such that 119879minus1120572

exists butdoes not satisfy (R3)

In this section following Appell et al [15] we define thethree more subdivisions of the spectrum called the approxi-mate point spectrum defect spectrum and compression spec-trum

Given a bounded linear operator 119879 in a Banach space 119883we call a sequence (119909119896) in119883 as aWeyl sequence for119879 if 119909119896 =1 and 119879119909119896 rarr 0 as 119896 rarr infin

In what follows we call the set

120590ap (119879119883)

=120572 isin C there exists a Weyl sequence for 120572119868minus119879(3)

the approximate point spectrum of 119879 Moreover the subspec-trum

120590120575 (119879119883) = 120572 isin C 120572119868 minus 119879 is not surjective (4)

is called defect spectrum of 119879The two subspectra given by (3) and (4) form a (not

necessarily disjoint) subdivisions

120590 (119879119883) = 120590ap (119879119883) cup 120590120575 (119879119883) (5)

of the spectrum There is another subspectrum

120590co (119879119883) = 120572 isin C 119877 (120572119868 minus 119879) =119883 (6)

which is often called compression spectrum in the literatureBy the definitions given above we can illustrate the sub-

divisions of spectrum in Table 1From Goldberg [16] if 119879 isin 119861(119883) 119883 a Banach space then

there are three possibilities for 119877(119879)

(I) 119877(119879) = 119883(II) 119877(119879) = 119877(119879) = 119883(III) 119877(119879) =119883

and three possibilities for 119879minus1

(1) 119879minus1 exists and is continuous(2) 119879minus1 exists but is discontinuous(3) 119879minus1 does not exist

If these possibilities are combined in all possible waysnine different states are created These are labelled by 1198681 11986821198683 1198681198681 1198681198682 1198681198683 1198681198681198681 1198681198681198682 and 1198681198681198683 If 120572 is a complex numbersuch that 119879120572 isin 1198681 or 119879120572 isin 1198681198681 then 120572 is in the resolvent set120588(119883 119879) of 119879 The further classification gives rise to the finespectrum of 119879 If an operator is in state 1198681198682 for example then119877(119879) = 119877(119879) = 119883 and 119879minus1 exists but is discontinuous and wewrite 120572 isin 1198681198682120590(119883 119879)

Let 120583 and 120574 be two sequence spaces and let 119860 = (119886119899119896)

be an infinite matrix of real or complex numbers 119886119899119896 where119899 119896 isin N = 0 1 2 Then we say that 119860 defines a matrix

Abstract and Applied Analysis 3

mapping from 120583 into 120574 and we denote it by writing 119860 120583 rarr120574 if for every sequence 119909 = (119909119896) isin 120583 the sequence 119860119909 =(119860119909)119899 the 119860-transform of 119909 is in 120574 where

(119860119909)119899 = sum

119896

119886119899119896119909119896 for each 119899 isin N (7)

By (120583 120574) we denote the class of all matrices 119860 such that119860 120583 rarr 120574 Thus 119860 isin (120583 120574) if and only if the series on theright side of (7) converges for each 119899 isin N and every 119909 isin 120583and we have 119860119909 = (119860119909)119899119899isinN isin 120574 for all 119909 isin 120583

Proposition 1 (see [15 Proposition 13 p 28]) Spectra andsubspectra of an operator119879 isin 119861(119883) and its adjoint119879lowast isin 119861(119883lowast)are related by the following relations

(a) 120590(119879lowast 119883lowast) = 120590(119879119883)(b) 120590119888(119879

lowast 119883lowast) sube 120590ap(119879119883)

(c) 120590ap(119879lowast 119883lowast) = 120590120575(119879119883)

(d) 120590120575(119879lowast 119883lowast) = 120590ap(119879119883)

(e) 120590119901(119879lowast 119883lowast) = 120590co(119879119883)

(f) 120590co(119879lowast 119883lowast) supe 120590119901(119879119883)

(g) 120590(119879119883) = 120590ap(119879119883) cup 120590119901(119879lowast 119883lowast) = 120590119901(119879119883) cup

120590ap(119879lowast 119883lowast)

The relations (c)ndash(f) show that the approximate pointspectrum is in a certain sense dual to defect spectrum and thepoint spectrum dual to the compression spectrum

The equality (g) implies in particular that 120590(119879119883) =

120590ap(119879119883) if 119883 is a Hilbert space and 119879 is normal Roughlyspeaking this shows that normal (in particular self-adjoint)operators onHilbert spaces aremost similar tomatrices in finitedimensional spaces (see [15])

Lemma 2 (see [16 p 60]) The adjoint operator 119879lowast of 119879 isonto if and only if 119879 has a bounded inverse

Lemma 3 (see [16 p 59]) 119879 has a dense range if and only if119879lowast is one to one

Our main focus in this paper is on the triple-band matrix119860(119903 119904 119905) where

119860 (119903 119904 119905) =

[[[[[[

[

119903 119904 119905 0

0 119903 119904 119905

0 0 119903 119904

0 0 0 119903

]]]]]]

]

(8)

We assume here and after that 119904 and 119905 are complex parameterswhich do not simultaneously vanish We introduce the intro-duce the operator 119860(119903 119904 119905) from ℓ119901 to itself by

119860 (119903 119904 119905) 119909 = (119903119909119896 + 119904119909119896+1 + 119905119909119896+2)infin

119896=0

where 119909 = (119909119896) isin ℓ119901(9)

3 Fine Spectra of Upper TriangularTriple-Band Matrices over the SequenceSpace ℓ119901 (0 lt 119901 ⩽ 1)

In this section we prove that the operator 119860(119903 119904 119905) ℓ119901 rarrℓ119901 is a bounded linear operator and compute its norm Weessentially emphasize the fine spectrum of the operator119860(119903 119904 119905) ℓ119901 rarr ℓ119901 in the case 0 lt 119901 ⩽ 1

Theorem 4 The operator 119860(119903 119904 119905) ℓ119901 rarr ℓ119901 is a boundedlinear operator and

119860 (119903 119904 119905)(ℓ119901 ℓ119901)= |119903|119901+ |119904|119901+ |119905|119901 (10)

Proof Since the linearity of the operator 119860(119903 119904 119905) is trivialso it is omitted Let us take 119890(2) isin ℓ119901 Then 119860(119903 119904 119905)119890(2) =(119905 119904 119903 0 ) and observe that

119860 (119903 119904 119905)(ℓ119901 ℓ119901)⩾

10038171003817100381710038171003817119860 (119903 119904 119905) 119890

(2)100381710038171003817100381710038171199011003817100381710038171003817119890(2)1003817100381710038171003817119901

= |119903|119901+ |119904|119901+ |119905|119901

(11)

which gives the fact that

119860 (119903 119904 119905)(ℓ119901 ℓ119901)⩾ |119903|119901+ |119904|119901+ |119905|119901 (12)

Let 119909 = (119909119896) isin ℓ119901 where 0 lt 119901 ⩽ 1Then since (119905119909119896+2) (119903119909119896)and (119904119909119896+1) isin ℓ119901 it is easy to see by triangle inequality that

119860 (119903 119904 119905) 119909119901 =

infin

sum

119896=0

1003816100381610038161003816119903119909119896 + 119904119909119896+1 + 119905119909119896+21003816100381610038161003816119901

⩽

infin

sum

119896=0

10038161003816100381610038161199031199091198961003816100381610038161003816119901+

infin

sum

119896=0

1003816100381610038161003816119904119909119896+11003816100381610038161003816119901+

infin

sum

119896=0

1003816100381610038161003816119905119909119896+21003816100381610038161003816119901

= |119903|119901infin

sum

119896=0

10038161003816100381610038161199091198961003816100381610038161003816119901+ |119904|119901infin

sum

119896=0

1003816100381610038161003816119909119896+11003816100381610038161003816119901+ |119905|119901infin

sum

119896=0

1003816100381610038161003816119909119896+21003816100381610038161003816119901

= |119904|119901119909119901 + |119903|

119901119909119901 + |119905|

119901119909119901

= (|119903|119901+ |119904|119901+ |119905|119901) 119909119901

(13)

which leads us to the result that

119860 (119903 119904 119905)(ℓ119901 ℓ119901)⩽ |119903|119901+ |119904|119901+ |119905|119901 (14)

Therefore by combining the inequalities (12) and (14) we seethat (10) holds which completes the proof

If 119879 ℓ119901 rarr ℓ119901 is a bounded matrix operator with thematrix 119860 then it is known that the adjoint operator 119879lowast ℓlowast

119901rarr ℓlowast

119901is defined by the transpose of thematrix119860The dual

space of ℓ119901 is isomorphic to ℓinfin where 0 lt 119901 lt 1Before giving themain theorem of this section we should

note the following remark In this work here and in whatfollows if 119911 is a complex number then by radic119911 we alwaysmean the square root of 119911 with a nonnegative real part IfRe(radic119911) = 0 then radic119911 represents the square root of 119911 withIm(radic119911) gt 0 The same results are obtained if radic119911 representsthe other square root

4 Abstract and Applied Analysis

Table 1 Subdivisions of spectrum of a linear operator

1 2 3119879minus1

120572exists and is bounded 119879

minus1

120572exists and is unbounded 119879

minus1

120572does not exist

A 119877(120572119868 minus 119879) = 119883 120572 isin 120588(119879119883) mdash 120572 isin 120590119901(119879119883)

120572 isin 120590ap(119879119883)

B 119877(120572119868 minus 119879) = 119883 120572 isin 120588(119879119883)

120572 isin 120590119888(119879119883) 120572 isin 120590119901(119879119883)

120572 isin 120590ap(119879119883) 120572 isin 120590ap(119879119883)

120572 isin 120590120575(119879119883) 120572 isin 120590120575(119879119883)

C 119877(120572119868 minus 119879) =119883

120572 isin 120590119903(119879119883) 120572 isin 120590119903(119879119883) 120572 isin 120590119901(119879119883)

120572 isin 120590120575(119879119883) 120572 isin 120590ap(119879119883) 120572 isin 120590ap(119879119883)

120572 isin 120590120575(119879119883) 120572 isin 120590120575(119879119883)

120572 isin 120590co(119879119883) 120572 isin 120590co(119879119883) 120572 isin 120590co(119879119883)

Theorem 5 Let 119904 be a complex number such that radic1199042 = minus119904and define the set1198631 by

1198631 = 120572 isin C 2 |119903 minus 120572| ⩽1003816100381610038161003816100381610038161003816minus119904 + radic1199042 minus 4119905 (119903 minus 120572)

1003816100381610038161003816100381610038161003816 (15)

Then 120590119888(119860(119903 119904 119905) ℓ119901) sube 1198631

Proof Let119910 = (119910119896) isin ℓinfinThen by solving the equation119860120572(119903119904 119905)lowast119909 = 119910 for 119909 = (119909119896) in terms of 119910 we obtain

1199090 =1199100

119903 minus 120572

1199091 =1199101

119903 minus 120572+

minus1199041199100

(119903 minus 120572)2

1199092 =1199102

119903 minus 120572+

minus1199041199101

(119903 minus 120572)2+[1199042minus 119905 (119903 minus 120572)] 1199100

(119903 minus 120572)3

(16)

(17)

and if we denote 1198861 = 1(119903 minus 120572) 1198862 = minus119904(119903 minus 120572)2 and 1198863 =

(1199042minus 119905(119903 minus 120572))(119903 minus 120572)

3 we have1199090 = 11988611199100

1199091 = 11988611199101 + 11988621199100

1199092 = 11988611199102 + 11988621199101 + 11988631199100

119909119899 = 1198861119910119899 + 1198862119910119899minus1 + sdot sdot sdot + 119886119899+11199100 =

119899

sum

119896=0

119886119899+1minus119896119910119896

(18)

Now we must find 119886119899 We have 119910119899 = 119905119909119899minus2 + 119904119909119899minus1 + (119903 minus 120572)119909119899and if we use relation (18) we have

119910119899 = 119905

119899minus2

sum

119896=0

119886119899minus1minus119896119910119896 + 119904

119899minus1

sum

119896=0

119886119899minus119896119910119896 + (119903 minus 120572)

119899

sum

119896=0

119886119899+1minus119896119910119896

= 1199100 (119905119886119899minus1 + 119904119886119899 + (119903 minus 120572) 119886119899+1)

+ 1199101 (119905119886119899minus2 + 119904119886119899minus1 + (119903 minus 120572) 119886119899) + sdot sdot sdot + 1199101198991198861 (119903 minus 120572)

(19)

This implies that

119905119886119899minus1 + 119904119886119899 + (119903 minus 120572) 119886119899+1 = 0

119905119886119899minus2 + 119904119886119899minus1 + (119903 minus 120572) 119886119899 = 0 1198861 (119903 minus 120572) = 1(20)

In fact this sequence is obtained recursively by letting

1198861 =1

119903 minus 120572 1198862 =

minus119904

(119903 minus 120572)2

119905119886119899minus2 + 119904119886119899minus1 + (119903 minus 120572) 119886119899 = 0 forall119899 ⩾ 3

(21)

The characteristic polynomial of the recurrence relation is (119903minus120572)1205822+ 119904120582 + 119905 = 0 There are two cases

Case 1 If Δ = 1199042 minus 4119905(119903 minus 120572) = 0 whose roots are

1205821 =minus119904 + radicΔ

2 (119903 minus 120572) 1205822 =

minus119904 minus radicΔ

2 (119903 minus 120572) (22)

elementary calculation on recurrent sequence gives that

119886119899 =120582119899

1minus 120582119899

2

radic1199042 minus 4119905 (119903 minus 120572) forall119899 ⩾ 1 (23)

In this case 119909119896 = (1radicΔ)sum119899

119896=0(120582119899+1minus119896

1minus 120582119899+1minus119896

2)119910119896 Assume

that |1205821| lt 1 So we have10038161003816100381610038161003816100381610038161003816100381610038161003816

1 + radic4119905 (119903 minus 120572)

1199042

10038161003816100381610038161003816100381610038161003816100381610038161003816

lt

10038161003816100381610038161003816100381610038161003816

2 (119903 minus 120572)

minus119904

10038161003816100381610038161003816100381610038161003816 (24)

Since |1 minus radic119911| ⩽ |1 + radic119911| for any 119911 isin C we must have10038161003816100381610038161003816100381610038161003816100381610038161003816

1 minus radic4119905 (119903 minus 120572)

1199042

10038161003816100381610038161003816100381610038161003816100381610038161003816

lt

10038161003816100381610038161003816100381610038161003816

2 (119903 minus 120572)

minus119904

10038161003816100381610038161003816100381610038161003816 (25)

It follows that |1205822| lt 1 Now for |1205821| lt 1 we can see that

10038161003816100381610038161199091198991003816100381610038161003816 ⩽

110038161003816100381610038161003816radicΔ10038161003816100381610038161003816

119899

sum

119896=0

10038161003816100381610038161003816120582119899+1minus119896

1

1003816100381610038161003816100381610038161003816100381610038161199101198961003816100381610038161003816 +

119899

sum

119896=0

10038161003816100381610038161003816120582119899+1minus119896

2

1003816100381610038161003816100381610038161003816100381610038161199101198961003816100381610038161003816 (26)

for all 119899 isin N Taking limit on the inequality (26) as 119899 rarr infinwe get

119909infin ⩽1 minus (

100381610038161003816100381612058221003816100381610038161003816 +100381610038161003816100381612058221003816100381610038161003816)

10038161003816100381610038161003816(1 minus

10038161003816100381610038161205822100381610038161003816100381610038161003816100381610038161 minus 1205822

1003816100381610038161003816)radicΔ10038161003816100381610038161003816

10038171003817100381710038171199101003817100381710038171003817infin (27)

Abstract and Applied Analysis 5

Thus for |1205821| lt 1 119860120572(119903 119904 119905)lowast is onto and by Lemma 2 119860120572(119903

119904 119905) has a bounded inverse This means that

120590119888 (119860 (119903 119904 119905) ℓ119901)

sube 120572 isin C 2 |119903 minus 120572| ⩽1003816100381610038161003816100381610038161003816minus119904 + radic1199042 minus 4119905 (119903 minus 120572)

1003816100381610038161003816100381610038161003816

= 1198631

(28)

Case 2 If Δ = 1199042 minus 4119905(119903 minus 120572) = 0 a calculation on recurrentsequence gives that

119886119899 = (2119899

minus119904) [

minus119904

2 (119903 minus 120572)]

119899

forall119899 ⩾ 1 (29)

Now for | minus 119904| lt 2|119903 minus 120572| we can see that

10038161003816100381610038161199091198991003816100381610038161003816 ⩽

119899

sum

119896=0

1003816100381610038161003816119886119899minus1198961199101198961003816100381610038161003816 (30)

for all 119899 isin N Taking limit on the inequality (30) as 119899 rarr infinwe obtain that

119909infin ⩽10038171003817100381710038171199101003817100381710038171003817infin

infin

sum

119896=0

10038161003816100381610038161198861198961003816100381610038161003816 (31)

suminfin

119896=0|119886119896| is convergent since | minus 119904| lt 2|119903 minus 120572| Thus for | minus 119904| lt

2|119903 minus 120572| 119860120572(119903 119904 119905)lowast is onto and by Lemma 2 119860120572(119903 119904 119905) has a

bounded inverse This means that

120590119888 (119860 (119903 119904 119905) ℓ119901) sube 120572 isin C 2 |119903 minus 120572| ⩽ |minus119904| sube 1198631

(32)

Theorem 6 120590119901(119860(119903 119904 119905)lowast ℓlowast

119901) = 0

Proof Consider 119860(119903 119904 119905)lowast119891 = 120572119891 with 119891 = 120579 = (0 0 0 ) inℓlowast

119901= ℓinfin Then by solving the system of linear equations

1199031198910 = 1205721198910

1199041198910 + 1199031198911 = 1205721198911

1199051198910 + 1199041198911 + 1199031198912 = 1205721198912

1199051198911 + 1199041198912 + 1199031198913 = 1205721198913

119905119891119896minus2 + 119904119891119896minus1 + 119903119891119896 = 120572119891119896

(33)

we find that 1198910 = 0 if 120572 = 119903 and 1198911 = 1198912 = sdot sdot sdot = 0 if 1198910 = 0which contradicts119891 = 120579 If 1198911198990 is the first nonzero entry of thesequence119891 = (119891119899) and120572 = 119903 thenwe get 1199051198911198990minus2+1199041198911198990minus1+1199031198911198990 =1205721198911198990

which implies 1198911198990 = 0which contradicts the assumption1198911198990

= 0 Hence the equation119860(119903 119904 119905)lowast119891 = 120572119891 has no solution119891 = 120579

Theorem 7 120590119901(119860(119903 119904 119905) ℓ119901) = 1198632 where 1198632 = 120572 isin C

2|119903 minus 120572| lt | minus 119904 + radic1199042 minus 4119905(119903 minus 120572)|

Proof Let119860(119903 119904 119905)119909 = 120572119909 for 120579 = 119909 isin ℓ119901Then by solving thesystem of linear equations

1199031199090 + 1199041199091 + 1199051199092 = 1205721199090

1199031199091 + 1199041199092 + 1199051199093 = 1205721199091

1199031199092 + 1199041199093 + 1199051199094 = 1205721199092

119903119909119896minus2 + 119904119909119896minus1 + 119905119909119896 = 120572119909119896

(34)

and we have

1199092 =minus119904

1199051199091 minus

119903 minus 120572

1199051199090

1199093 =1199042minus 119905 (119903 minus 120572)

11990521199091 +

119904 (119903 minus 120572)

11990521199090

119909119899 =119886119899(119903 minus 120572)

119899

119905119899minus11199091 minus

119886119899minus1(119903 minus 120572)119899

119905119899minus11199090 forall119899 ⩾ 2

(35)

Assume that 120572 isin 1198632 Then we choose 1199090 = 1 and 1199091 = 2(119903 minus120572)(minus119904+radic1199042 minus 4119905(119903 minus 120572)) We show that 119909119899 = 119909

119899

1for all 119899 ⩾ 2

Since 1205821 1205822 are roots of the characteristic equation (119903minus120572)1205822+

119904120582 + 119905 = 0 we must have

12058211205822 =119905

119903 minus 120572 1205821 minus 1205822 =

radicΔ

119903 minus 120572 (36)

Combining the fact 1199091 = 11205821 with relation (35) we can seethat

119909119899 =119886119899(119903 minus 120572)

119899

119905119899minus11199091 minus

119886119899minus1(119903 minus 120572)119899

119905119899minus11199090

= (119903 minus 120572

119905)

119899minus1

(119903 minus 120572) (minus119886119899minus11199090 + 1198861198991199091)

=1

(12058211205822)119899minus1

119903 minus 120572

radicΔ(minus120582119899minus1

1+ 120582119899minus1

2+ 120582119899minus1

1minus 120582119899

2120582minus1

1)

=1

120582119899minus11120582119899minus12

(1

1205821 minus 1205822)120582119899minus1

2(1205821 minus 1205822

1205821)

=1

1205821198991

= 119909119899

1

(37)

The same result may be obtained in case Δ = 0 Now 119909 =(119909119896) isin ℓ119901 since |1199091| lt 1 This shows that 1198632 sube 120590119901(119860(119903

119904 119905) ℓ119901)

6 Abstract and Applied Analysis

Now we assume that 120572 notin 1198632 that is |1205821| ⩽ 1 We mustshow that 120572 notin 120590119901(119860(119903 119904 119905) ℓ119901) Therefore we obtain from therelation (35) that

119909119899+1

119909119899= (

119903 minus 120572

119905)119886119899minus1

119886119899minus2(minus1199090 + (119886119899119886119899minus1) 1199091

minus1199090 + (119886119899minus1119886119899minus2) 1199091) (38)

Now we examine three cases

Case 1 (|1205822| lt |1205821| ⩽ 1) In this case we have 1199042 = 4119905(119903minus120572) and

119886119899

119886119899minus1=120582119899+1

1minus 120582119899+1

2

1205821198991minus 1205821198992

=1205821 [1 minus (12058221205821)

119899+1]

[1 minus (12058221205821)119899] (39)

Then we have

lim119899rarrinfin

10038161003816100381610038161003816100381610038161003816

119886119899

119886119899minus1

10038161003816100381610038161003816100381610038161003816

119901

= lim119899rarrinfin

10038161003816100381610038161003816100381610038161003816

119886119899minus1

119886119899minus2

10038161003816100381610038161003816100381610038161003816

119901

= lim119899rarrinfin

100381610038161003816100381612058211003816100381610038161003816119901100381610038161003816100381610038161 minus (12058221205821)

119899+110038161003816100381610038161003816

119901

100381610038161003816100381610038161 minus (12058221205821)

11989910038161003816100381610038161003816

119901=100381610038161003816100381612058211003816100381610038161003816119901

(40)

Now if minus1199090 + 12058211199091 = 0 then we have (119909119899) = (1199090120582119899

1) which

is not in ℓ119901 Otherwise

lim119899rarrinfin

10038161003816100381610038161003816100381610038161003816

119909119899+1

119909119899

10038161003816100381610038161003816100381610038161003816

119901

=1

10038161003816100381610038161205821100381610038161003816100381611990110038161003816100381610038161205822

1003816100381610038161003816119901

100381610038161003816100381612058211003816100381610038161003816119901=

1

100381610038161003816100381612058221003816100381610038161003816119901gt 1 (41)

Case 2 (|1205822| = |1205821| lt 1) In this case we have 1199042 = 4119905(119903 minus 120572)and using the formula

119886119899 = (2119899

minus119904) [

minus119904

2 (119903 minus 120572)]

119899

forall119899 ⩾ 1 (42)

we obtain that

lim119899rarrinfin

10038161003816100381610038161003816100381610038161003816

119886119899

119886119899minus1

10038161003816100381610038161003816100381610038161003816

119901

=

10038161003816100381610038161003816100381610038161003816

minus119904

2 (119903 minus 120572)

10038161003816100381610038161003816100381610038161003816

119901

=100381610038161003816100381612058211003816100381610038161003816119901 (43)

which leads to

lim119899rarrinfin

10038161003816100381610038161003816100381610038161003816

119909119899+1

119909119899

10038161003816100381610038161003816100381610038161003816

119901

=1

10038161003816100381610038161205821100381610038161003816100381611990110038161003816100381610038161205822

1003816100381610038161003816119901

100381610038161003816100381612058211003816100381610038161003816119901=

1

100381610038161003816100381612058221003816100381610038161003816119901gt 1 (44)

Case 3 (|1205822| = |1205821| = 1) In this case we have 1199042 = 4119905(119903minus120572) andso we have | minus 119904(2119905)| = 1 Assume that 120572 isin 120590119901(119860(119903 119904 119905) ℓ119901)This implies that 119909 isin ℓ119901 and 119909 = 120579 Thus we again derive (35)

119909119899 = (minus119904

2119905)

119899minus1

[minus (119899 minus 1)minus119904

21199051199090 + 1198991199091] (45)

Since lim119899rarrinfin119909119899 = 0 we must have 1199090 = 1199091 = 0 But thisimplies that 119909 = 120579 a contradiction which means that 120572 notin120590119901(119860(119903 119904 119905) ℓ119901) Thus 120590119901(119860(119903 119904 119905) ℓ119901) sube 1198632 This completesthe proof

Theorem 8 120590119903(119860(119903 119904 119905) ℓ119901) = 0

Proof By Proposition 1 120590119903(119860(119903 119904 119905) ℓ119901) = 120590119901(119860(119903 119904 119905)lowast ℓlowast

119901)

120590119901(119860(119903 119904 119905) ℓ119901) Since byTheorem 6

120590119901 (119860(119903 119904 119905)lowast ℓlowast

119901) = 0 120590119903 (119860 (119903 119904 119905) ℓ119901) = 0

(46)

This completes the proof

Theorem 9 Let 119904 be a complex number such that radic1199042 = minus119904and define the set1198631 by

1198631 = 120572 isin C 2 |119903 minus 120572| ⩽1003816100381610038161003816100381610038161003816minus119904 + radic1199042 minus 4119905 (119903 minus 120572)

1003816100381610038161003816100381610038161003816

(47)

Then 120590(119860(119903 119904 119905) ℓ119901) = 1198631

Proof ByTheorem 7 we get

120572 isin C 2 |119903 minus 120572| lt1003816100381610038161003816100381610038161003816minus119904 + radic1199042 minus 4119905 (119903 minus 120572)

1003816100381610038161003816100381610038161003816

sube 120590 (119860 (119903 119904 119905) ℓ119901)

(48)

Since the spectrum of any bounded operator is closed wehave

120572 isin C 2 |119903 minus 120572| ⩽1003816100381610038161003816100381610038161003816minus119904 + radic1199042 minus 4119905 (119903 minus 120572)

1003816100381610038161003816100381610038161003816

sube 120590 (119860 (119903 119904 119905) ℓ119901)

(49)

and again fromTheorems 5 7 and 8

120590 (119860 (119903 119904 119905) ℓ119901) sube 120572 isin C 2 |119903 minus 120572|

⩽1003816100381610038161003816100381610038161003816minus119904 + radic1199042 minus 4119905 (119903 minus 120572)

1003816100381610038161003816100381610038161003816

(50)

Combining (49) and (50) we obtain that 120590(119860(119903 119904 119905) ℓ119901) =1198631 where1198631 is defined by (47)

Theorem 10 120590119888(119860(119903 119904 119905) ℓ119901) = 1198633 where 1198633 = 120572 isin C

2|119903 minus 120572| = | minus 119904 + radic1199042 minus 4119905(119903 minus 120572)|

Proof Because the parts 120590119888(119860(119903 119904 119905) ℓ119901) 120590119903(119860(119903 119904 119905) ℓ119901)and 120590119901(119860(119903 119904 119905) ℓ119901) are pairwise disjoint sets and the unionof these sets is 120590(119860(119903 119904 119905) ℓ119901) the proof immediately followsfromTheorems 7 8 and 9

Theorem 11 If 120572 isin 1198632 120572 isin 120590(119860(119903 119904 119905) ℓ119901)1198683

Proof From Theorem 7 120572 isin 120590119901(119860(119903 119904 119905) ℓ119901) Thus (119860(119903119904 119905) minus 120572119868)

minus1 does not exist By Theorem 6119860(119903 119904 119905)lowast minus 120572119868 isone to one so 119860(119903 119904 119905) minus 120572119868 has a dense range in ℓ119901 byLemma 3

Theorem 12 The following statements hold

(i) 120590ap(119860(119903 119904 119905) ℓ119901) = 1198631(ii) 120590120575(119860(119903 119904 119905) ℓ119901) = 1198633(iii) 120590co(119860(119903 119904 119905) ℓ119901) = 0

Abstract and Applied Analysis 7

Proof (i) Since from Table 1

120590ap (119860 (119903 119904 119905) ℓ119901)=120590 (119860 (119903 119904 119905) ℓ119901)120590 (119860 (119903 119904 119905) ℓ119901) 1198681198681198681

(51)

we have byTheorem 8

120590 (119860 (119903 119904 119905) ℓ119901) 1198681198681198681 = 120590 (119860 (119903 119904 119905) ℓ119901) 1198681198681198682 = 0 (52)

Hence

120590ap (119860 (119903 119904 119905) ℓ119901) = 1198631 (53)

(ii) Since the following equality

120590120575 (119860 (119903 119904 119905) ℓ119901) = 120590 (119860 (119903 119904 119905) ℓ119901) 120590 (119860 (119903 119904 119905) ℓ119901) 1198683

(54)

holds from Table 1 we derive by Theorems 8 and 11 that120590120575(119860(119903 119904 119905) ℓ119901) = 1198632

(iii) From Table 1 we have

120590co (119860 (119903 119904 119905) ℓ119901)

= 120590 (119860 (119903 119904 119905) ℓ119901) 1198681198681198681 cup 120590 (119860 (119903 119904 119905) ℓ119901) 1198681198681198682

cup 120590 (119860 (119903 119904 119905) 1198880) 1198681198681198683

(55)

ByTheorem 6 it is immediate that 120590co(119860(119903 119904 119905) ℓ119901) = 0

4 Fine Spectra of Upper TriangularTriple-Band Matrices over the SequenceSpace ℓ119901 (1 lt 119901 lt infin)

In the present section we determine the fine spectrum ofthe operator 119860(119903 119904 119905) ℓ119901 rarr ℓ119901 in case 1 ⩽ 119901 lt infinWe quote some lemmas which are needed in proving thetheorems given in Section 4

Lemma 13 (see [17 p 253 Theorem 3416]) The matrix 119860 =(119886119899119896) gives rise to a bounded linear operator 119879 isin 119861(ℓ1) from ℓ1to itself if and only if the supremum of ℓ1 norms of the columnsof 119860 is bounded

Lemma 14 (see [17 p 245 Theorem 343]) The matrix 119860 =(119886119899119896) gives rise to a bounded linear operator 119879 isin 119861(ℓinfin) fromℓinfin to itself if and only if the supremum of ℓ1 norms of the rowsof 119860 is bounded

Lemma 15 (see [17 p 254 Theorem 3418]) Let 1 lt 119901 lt infinand 119860 isin (ℓinfin ℓinfin) cap (ℓ1 ℓ1) Then 119860 isin (ℓ119901 ℓ119901)

Theorem 16 The operator 119860(119903 119904 119905) ℓ119901 rarr ℓ119901 is a boundedlinear operator and

(|119903|119901+ |119904|119901+ |119905|119901)1119901⩽ 119860 (119903 119904 119905)(ℓ119901 ℓ119901)

⩽ |119903| + |119904| + |119905|

(56)

Proof Since the linearity of the operator119860(119903 119904 119905) is not hardwe omit the details Now we prove that (56) holds for theoperator 119860(119903 119904 119905) on the space ℓ119901 It is trivial that 119860(119903 119904119905)119890(2)= (119905 119904 119903 0 ) for 119890(2) isin ℓ119901 Therefore we have

119860 (119903 119904 119905)(ℓ119901 ℓ119901)⩾

10038171003817100381710038171003817119860 (119903 119904 119905) 119890

(2)100381710038171003817100381710038171199011003817100381710038171003817119890(2)1003817100381710038171003817119901

= (|119903|119901+ |119904|119901+ |119905|119901)1119901

(57)

which implies that

119860 (119903 119904 119905)(ℓ119901 ℓ119901)⩾ (|119903|119901+ |s|119901 + |119905|119901)1119901 (58)

Let 119909 = (119909119896) isin ℓ119901 where 1 lt 119901 lt infin Then since (119905119909119896+2)(119903119909119896) and (119904119909119896+1) isin ℓ119901 it is easy to see by Minkowskirsquos ine-quality that

119860 (119903 119904 119905) 119909119901

= (

infin

sum

119896=0

1003816100381610038161003816119903119909119896 + 119904119909119896+1 + 119905119909119896+21003816100381610038161003816119901)

1119901

⩽ (

infin

sum

119896=0

10038161003816100381610038161199031199091198961003816100381610038161003816119901)

1119901

+ (

infin

sum

119896=0

1003816100381610038161003816119904119909119896+11003816100381610038161003816119901)

1119901

+ (

infin

sum

119896=0

1003816100381610038161003816119905119909119896+21003816100381610038161003816119901)

1119901

= |119903| (

infin

sum

119896=0

10038161003816100381610038161199091198961003816100381610038161003816119901)

1119901

+ |119904| (

infin

sum

119896=0

1003816100381610038161003816119909119896+11003816100381610038161003816119901)

1119901

+ |119905| (

infin

sum

119896=0

1003816100381610038161003816119909119896+21003816100381610038161003816119901)

1119901

= |119904| 119909119901 + |119903| 119909119901 + |119905| 119909119901

= (|119903| + |119904| + |119905|) 119909119901

(59)

which leads us to the result that

(|119903|119901+ |119904|119901+ |119905|119901)1119901⩽ 119860 (119903 119904 119905)(ℓ119901 ℓ119901)

⩽ |119903| + |119904| + |119905|

(60)

Therefore by combining the inequalities in (58) and (59) wehave (56) as desired

If 119879 ℓ119901 rarr ℓ119901 is a bounded matrix operator with thematrix 119860 then it is known that the adjoint operator 119879lowast ℓlowast

119901rarr ℓlowast

119901is defined by the transpose of thematrix119860The dual

space of ℓ119901 is isomorphic to ℓ119902 where 1 lt 119901 lt infin

Theorem 17 Let 119904 be a complex number such that radic1199042 = minus119904and define the set1198631 by

1198631 = 120572 isin C 2 |119903 minus 120572| ⩽1003816100381610038161003816100381610038161003816minus119904 + radic1199042 minus 4119905 (119903 minus 120572)

1003816100381610038161003816100381610038161003816

(61)

Then 120590119888(119860(119903 119904 119905) ℓ119901) sube 1198631

8 Abstract and Applied Analysis

Proof We will show that 119860120572(119903 119904 119905)lowast is onto for 2|119903 minus 120572| gt

| minus 119904 + radic1199042 minus 4119905(119903 minus 120572)| Thus for every 119910 isin ℓ119902 we find 119909 isinℓ119902 119860120572(119903 119904 119905)

lowast is a triangle so it has an inverse Also equation119860120572(119903 119904 119905)

lowast119909 = 119910 gives [119860120572(119903 119904 119905)

lowast]minus1119910 = 119909 It is sufficient

to show that [119860120572(119903 119904 119905)lowast]minus1isin (ℓ119902 ℓ119902) We calculate that

119860 = (119886119899119896) = [119860120572(119903 119904 119905)lowast]minus1 as follows

119860 = (119886119899119896) =

[[[[

[

1198861 0 0

1198862 1198861 0

1198863 1198862 1198861

]]]]

]

(62)

where

1198861 =1

119903 minus 120572

1198862 =minus119904

(119903 minus 120572)2

1198863 =1199042minus 119905 (119903 minus 120572)

(119903 minus 120572)3

(63)

It is known that fromTheorem 5

119886119899 =120582119899

1minus 120582119899

2

radic1199042 minus 4119905 (119903 minus 120572) forall119899 ⩾ 1

where 1205821 =minus119904 + radicΔ

119903 minus 120572 1205822 =

minus119904 minus radicΔ

119903 minus 120572

(64)

Now we show that [119860120572(119903 119904 119905)lowast]minus1isin (ℓ1 ℓ1) for |1205821| lt 1

By Theorem 5 we know that if |1205821| lt 1 |1205822| lt 1 We assumethat 1199042 minus 4119905(119903 minus 120572) = 0 and |1205821| lt 1

10038171003817100381710038171003817[119860120572(119903 119904 119905)

lowast]minus110038171003817100381710038171003817(ℓ1 ℓ1)

= sup119899

infin

sum

119896=119899

10038161003816100381610038161198861198961003816100381610038161003816 =

infin

sum

119896=1

10038161003816100381610038161198861198961003816100381610038161003816

⩽1

|Δ|(

infin

sum

119896=1

100381610038161003816100381612058211003816100381610038161003816119896+

infin

sum

119896=1

100381610038161003816100381612058221003816100381610038161003816119896) lt infin

(65)

since |1205821| lt 1 and |1205822| lt 1 This shows that (119860120572(119903 119904 119905)lowast]minus1isin

(ℓ1 ℓ1) Similarly we can show that [119860120572(119903 119904 119905)lowast]minus1isin (ℓinfin

ℓinfin)Now assume that 1199042 minus 4119905(119903 minus 120572) = 0 Then

119886119899 = (2119899

minus119904) [

minus119904

2 (119903 minus 120572)]

119899

(66)

and simple calculation gives that (119886119899) isin ℓ119902 if and only if |minus119904| lt2|119903 minus 120572|

[(119860 (119903 119904 119905) minus 120572119868)lowast]minus1isin (ℓ119902 ℓ119902)

for 120572 isin C with 2 |119903 minus 120572| gt1003816100381610038161003816100381610038161003816minus119904 + radic1199042 minus 4119905 (119903 minus 120572)

1003816100381610038161003816100381610038161003816

(67)

Hence 119860120572(119903 119904 119905)lowast is onto By Lemma 2 119860120572(119903 119904 119905) has a

bounded inverse This means that 120590119888(119860(119903 119904 119905) ℓ119901) sube 1198631where1198631 is defined by (61)

Theorem 18 120590119901(119860(119903 119904 119905)lowast ℓlowast

119901) = 0

Proof Let 119860(119903 119904 119905)lowast119891 = 120572119891 with 119891 = 120579 = (0 0 0 ) in ℓlowast119901=

ℓ119902 Then by solving the system of linear equations

1199031198910 = 1205721198910

1199041198910 + 1199031198911 = 1205721198911

1199051198910 + 1199041198911 + 1199031198912 = 1205721198912

1199051198911 + 1199041198912 + 1199031198913 = 1205721198913

119905119891119896minus2 + 119904119891119896minus1 + 119903119891119896 = 120572119891119896

(68)

we find that 1198910 = 0 if 120572 = 119903 and 1198911 = 1198912 = sdot sdot sdot = 0 if 1198910 = 0which contradicts119891 = 120579 If 1198911198990 is the first nonzero entry of thesequence 119891 = (119891119899) and 120572 = 119903 then we get 1199051198911198990minus2 + 1199041198911198990minus1 +1199031198911198990

= 1205721198911198990which implies 1198911198990 = 0 which contradicts the

assumption1198911198990 = 0 Hence the equation119860(119903 119904 119905)lowast119891 = 120572119891 has

no solution 119891 = 120579

In the case 1 lt 119901 lt infin since the proof of the theoremsin Section 4 determining the spectrum and fine spectrumof the matrix operator 119860(119903 119904 119905) on the sequence space ℓ119901is similar to the case 0 lt 119901 ⩽ 1 to avoid the repetitionof similar statements we give the results by the followingtheorem without proof

Theorem 19 The following statements hold(i) 120590(119860(119903 119904 119905) ℓ119901) = 1198631(ii) 120590119903(119860(119903 119904 119905) ℓ119901) = 0(iii) 120590119901(119860(119903 119904 119905) ℓ119901) = 1198632(iv) 120590119888(119860(119903 119904 119905) ℓ119901) = 1198633(v) 120590ap(119860(119903 119904 119905) ℓ119901) = 1198631(vi) 120590co(119860(119903 119904 119905) ℓ119901) = 0(vii) 120590120575(119860(119903 119904 119905) ℓ119901) = 1198633

5 Some Applications

In this section we give two theorems related to Toeplitz ma-trix

Theorem 20 Let 119875 be a polynomial that corresponds to the 119899-tuple 119886 and let 1199111 1199112 1199113 119911119899minus1 also be the roots of 119875 Define119879 as a Toeplitz matrix associated with 119875 that is

119879 =

[[[[

[

1198860 1198861 1198862 119886119899 0 0 0

0 1198860 1198861 1198862 119886119899 0 0

0 0 1198860 1198861 1198862 119886119899 0

]]]]

]

(69)

Abstract and Applied Analysis 9

The resolvent operator 119879 over ℓ119901 with 1 lt 119901 lt infin where thedomain of the resolvent operator is the whole space ℓ119901 exists ifand if only all the roots of the polynomial are outside the unitdisc 119911 isin C |119911| ⩽ 1 That is 119879minus1 isin (ℓ119901 ℓ119901) if and if only|119911119894| gt 1 1 ⩽ 119894 ⩽ 119899 minus 1 In this case the resolvent operator isrepresented by

119879minus1=

1

119886119899minus1119860minus1(minus1199111 1) 119860

minus1(minus1199112 1) sdot sdot sdot 119860

minus1(minus119911119899minus1 1)

where

119860minus1(minus119911119894 1) =minus

[[[[[[[[[[[[

[

1119911119894 11199112

11989411199113

11989411199114

11989411199115

119894

0 1119911119894 11199112

11989411199113

11989411199114

119894

0 0 1119911119894 11199112

11989411199113

119894

0 0 0 1119911119894 11199112

119894

0 0 0 0 1119911119894

]]]]]]]]]]]]

]

(70)

Proof Suppose all the roots of the polynomial 119875(119911) = 1198860 +1198861119911 + sdot sdot sdot + 119886119899minus1119911

119899minus1= 119886119899(119911 minus 1199111)(119911 minus 1199112) sdot sdot sdot (119911 minus 119911119899minus1) are

outside the unit disc The Toeplitz matrix associated with 119875can be written as the product

119879 = 119886119899119860 (minus1199111 1) 119860 (minus1199112 1) sdot sdot sdot 119860 (minus119911119899minus1 1) (71)

Since multiplication of upper triangular Toeplitz matrices iscommutative we can see that

119879minus1=

1

119886119899minus1119860minus1(minus1199111 1) 119860

minus1(minus1199112 1) sdot sdot sdot 119860

minus1(minus119911119899minus1 1)

(72)

is left inverse of 119879 Since all roots are outside the unit discthen

10038171003817100381710038171003817119879minus1(minus119911119894 1)

10038171003817100381710038171003817(ℓinfin ℓinfin)= sup119899

infin

sum

119896=119899

1

10038161003816100381610038161199111198941003816100381610038161003816119896+1minus119899

=

infin

sum

119896=1

1

10038161003816100381610038161199111198941003816100381610038161003816119896lt infin

(73)

Therefore each 119879minus1(minus119911119894 1) isin (ℓinfin ℓinfin) for 1 ⩽ 119894 ⩽ 119899 minus 1Similarly we can say that 119879minus1(minus119911119894 1) isin (ℓ1 ℓ1) So we have119879minus1(minus119911119894 1) isin (ℓ119901 ℓ119901)

Theorem 21 The resolvent operator of 119860(119903 119904 119905) over ℓ119901 with1 lt 119901 lt infin where the domain of the resolvent operator is thespace ℓ119901 exists if and only if 2|119903| gt | minus 119904 + radic1199042 minus 4119905119903| In this

case the resolvent operator is represented by the infinite bandedToeplitz matrix

119864 (119903 119904 119905)

=1

119905

[[[[[[[[[[

[

1199061 1199062

11199063

11199064

1

0 1199061 1199062

11199063

1

0 0 1199061 1199062

1

0 0 0 1199061

]]]]]]]]]]

]

[[[[[[[[[[

[

1199062 1199062

21199063

21199064

2

0 1199062 1199062

21199063

2

0 0 1199062 1199062

2

0 0 0 1199062

]]]]]]]]]]

]

where 1199061 =minus119904 + radic1199042 minus 4119905119903

2119903 1199062 =

minus119904 minus radic1199042 minus 4119905119903

2119903

(74)

Proof By Theorem 20 we can see that 119864(119903 119904 119905) is inverseof the matrix of 119860(119903 119904 119905) But this is not enough to say itis resolvent operator By Lemmas 13 14 and 15 119864(119903 119904 119905) isin(ℓ119901 ℓ119901) when 2|119903| gt | minus 119904 + radic1199042 minus 4119905119903| That is for 2|119903| gt| minus 119904 + radic1199042 minus 4119905119903| 119864(119903 119904 119905) is a resolvent operator

Acknowledgments

The authors would like to thank the referee for pointingout some mistakes and misprints in the earlier version ofthis paper They would like to express their pleasure toMs Medine Yesilkayagil Department of Mathematics UsakUniversity 1 Eylul Campus Usak Turkey for many helpfulsuggestions and interesting comments on the revised form ofthe paper

References

[1] M Gonzalez ldquoThe fine spectrum of the Cesaro operator in ℓ119901(1 lt 119901 lt infin)rdquo Archiv der Mathematik vol 44 no 4 pp 355ndash358 1985

[2] P J CartlidgeWeighted meanmatrices as operators on ℓ119901 [PhDdissertation] Indiana University 1978

[3] J I Okutoyi ldquoOn the spectrum of C1 as an operator on 1198871199070rdquoAustralian Mathematical Society A vol 48 no 1 pp 79ndash861990

[4] J I Okutoyi ldquoOn the spectrum of C1 as an operator on bvrdquoCommunications A vol 41 no 1-2 pp 197ndash207 1992

[5] M Yıldırım ldquoOn the spectrum of the Rhaly operators on ℓ119901rdquoIndian Journal of Pure and Applied Mathematics vol 32 no 2pp 191ndash198 2001

[6] B Altay and F Basar ldquoOn the fine spectrum of the differenceoperator Δ on 1198880 and 119888rdquo Information Sciences vol 168 no 1ndash4pp 217ndash224 2004

[7] B Altay and F Basar ldquoOn the fine spectrum of the generalizeddifference operator 119861(119903 119904) over the sequence spaces 1198880 and119888rdquo International Journal of Mathematics and MathematicalSciences vol 2005 no 18 pp 3005ndash3013 2005

[8] A M Akhmedov and F Basar ldquoOn the fine spectra of the dif-ference operator Δ over the sequence space ℓ119901 (1 le 119901 lt infin)rdquoDemonstratio Mathematica vol 39 no 3 pp 585ndash595 2006

10 Abstract and Applied Analysis

[9] AMAkhmedov andF Basar ldquoThefine spectra of the differenceoperator Δ over the sequence space 119887119907119901 (1 le 119901 lt infin)rdquo ActaMathematica Sinica vol 23 no 10 pp 1757ndash1768 2007

[10] F Basar and B Altay ldquoOn the space of sequences of p-boundedvariation and related matrix mappingsrdquo Ukrainian Mathemati-cal Journal vol 55 no 1 pp 136ndash147 2003

[11] H Furkan H Bilgic and F Basar ldquoOn the fine spectrum of theoperator 119861(119903 119904 119905) over the sequence spaces ℓ119901 and 119887119907119901(1 lt 119901 ltinfin)rdquo Computers ampMathematics with Applications vol 60 no 7pp 2141ndash2152 2010

[12] V Karakaya and M Altun ldquoFine spectra of upper triangulardouble-band matricesrdquo Journal of Computational and AppliedMathematics vol 234 no 5 pp 1387ndash1394 2010

[13] A Karaisa ldquoFine spectra of upper triangular double-bandmatrices over the sequence space ℓ119901 (1 lt 119901 lt infin)rdquo DiscreteDynamics in Nature and Society vol 2012 Article ID 381069 19pages 2012

[14] E Kreyszig Introductory Functional Analysis with ApplicationsJohn Wiley amp Sons New York NY USA 1978

[15] J Appell E Pascale and A Vignoli Nonlinear Spectral Theoryvol 10 of de Gruyter Series in Nonlinear Analysis and Applica-tions Walter de Gruyter Berlin Germany 2004

[16] S Goldberg Unbounded Linear Operators Dover PublicationsNew York NY USA 1985

[17] B Choudhary and S Nanda Functional Analysis with Applica-tions John Wiley amp Sons New York NY USA 1989

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

2 Abstract and Applied Analysis

with 1 lt 119901 lt infin where 119861(119903 119904 119905) is a lower triangular triple-band matrix Later Karakaya and Altun have determined thefine spectra of upper triangular double-band matrices overthe sequence spaces 1198880 and 119888 in [12] Quite recently Karaisa[13] has determined the fine spectrum of the generalizeddifference operator 119860(119903 119904) defined as an upper triangulardouble-band matrix with the convergent sequences 119903 = (119903119896)and 119904 = (119904119896) having certain properties over the sequencespace ℓ119901 where 1 lt 119901 lt infin

In this paper we study the fine spectrum of the general-ized difference operator 119860(119903 119904 119905) defined by a triple sequen-tial band matrix acting on the sequence space ℓ119901 (0 lt 119901 ltinfin) with respect to Goldbergrsquos classification Additionally wegive the approximate point spectrum and defect spectrumand give some applications

2 Preliminaries Background and Notation

Let 119883 and 119884 be two Banach spaces and 119879 119883 rarr 119884 be abounded linear operator By 119877(119879) we denote range of 119879 thatis

119877 (119879) = 119910 isin 119884 119910 = 119879119909 119909 isin 119883 (1)

By119861(119883)we also denote the set of all bounded linear operatorson 119883 into itself If 119879 isin 119861(119883) then the adjoint 119879lowast of 119879 isa bounded linear operator on the dual 119883lowast of 119883 defined by(119879lowast119891)(119909) = 119891(119879119909) for all 119891 isin 119883lowast and 119909 isin 119883Let 119883 = 120579 be a complex normed space and 119879 119863(119879) rarr

119883 be a linear operator with domain 119863(119879) sube 119883 With 119879 weassociate the operator 119879120572 = 119879 minus 120572119868 where 120572 is a complexnumber and 119868 is the identity operator on 119863(119879) If 119879120572 has aninverse which is linear we denote it by 119879minus1

120572 that is

119879minus1

120572= (119879 minus 120572119868)

minus1 (2)

and call it the resolvent operator of 119879Many properties of 119879120572 and 119879

minus1

120572depend on 120572 and spectral

theory is concerned with those properties For instance weshall be interested in the set of all 120572 in the complex plane suchthat 119879minus1

120572exists The boundedness of 119879minus1

120572is another property

that will be essential We shall also ask for what 120572 the domainof 119879minus1120572

is dense in 119883 to name just a few aspects For ourinvestigation of 119879 119879120572 and 119879

minus1

120572 we need some basic concepts

in spectral theory which are given as follows (see [14 pp 370-371])

Let 119883 = 120579 be a complex normed space and 119879 119863(119879) rarr119883 be a linear operator with domain 119863(119879) sube 119883 A regularvalue 120572 of 119879 is a complex number such that

(R1) 119879minus1120572

exists(R2) 119879minus1

120572is bounded

(R3) 119879minus1120572

is defined on a set which is dense in 119883

The resolvent set 120588(119879) of 119879 is the set of all regular values 120572 of119879 Its complement C120588(119879) in the complex plane C is calledthe spectrum of 119879 Furthermore the spectrum 120590(119879) is parti-tioned into three disjoint sets as follows The point spectrum120590119901(119879) is the set such that 119879minus1

120572does not exist 120572 isin 120590119901(119879) is

called an eigenvalue of 119879 The continuous spectrum 120590119888(119879) isthe set such that 119879minus1

120572exists and satisfies (R3) but not (R2)

The residual spectrum 120590119903(119879) is the set such that 119879minus1120572

exists butdoes not satisfy (R3)

In this section following Appell et al [15] we define thethree more subdivisions of the spectrum called the approxi-mate point spectrum defect spectrum and compression spec-trum

Given a bounded linear operator 119879 in a Banach space 119883we call a sequence (119909119896) in119883 as aWeyl sequence for119879 if 119909119896 =1 and 119879119909119896 rarr 0 as 119896 rarr infin

In what follows we call the set

120590ap (119879119883)

=120572 isin C there exists a Weyl sequence for 120572119868minus119879(3)

the approximate point spectrum of 119879 Moreover the subspec-trum

120590120575 (119879119883) = 120572 isin C 120572119868 minus 119879 is not surjective (4)

is called defect spectrum of 119879The two subspectra given by (3) and (4) form a (not

necessarily disjoint) subdivisions

120590 (119879119883) = 120590ap (119879119883) cup 120590120575 (119879119883) (5)

of the spectrum There is another subspectrum

120590co (119879119883) = 120572 isin C 119877 (120572119868 minus 119879) =119883 (6)

which is often called compression spectrum in the literatureBy the definitions given above we can illustrate the sub-

divisions of spectrum in Table 1From Goldberg [16] if 119879 isin 119861(119883) 119883 a Banach space then

there are three possibilities for 119877(119879)

(I) 119877(119879) = 119883(II) 119877(119879) = 119877(119879) = 119883(III) 119877(119879) =119883

and three possibilities for 119879minus1

(1) 119879minus1 exists and is continuous(2) 119879minus1 exists but is discontinuous(3) 119879minus1 does not exist

If these possibilities are combined in all possible waysnine different states are created These are labelled by 1198681 11986821198683 1198681198681 1198681198682 1198681198683 1198681198681198681 1198681198681198682 and 1198681198681198683 If 120572 is a complex numbersuch that 119879120572 isin 1198681 or 119879120572 isin 1198681198681 then 120572 is in the resolvent set120588(119883 119879) of 119879 The further classification gives rise to the finespectrum of 119879 If an operator is in state 1198681198682 for example then119877(119879) = 119877(119879) = 119883 and 119879minus1 exists but is discontinuous and wewrite 120572 isin 1198681198682120590(119883 119879)

Let 120583 and 120574 be two sequence spaces and let 119860 = (119886119899119896)

be an infinite matrix of real or complex numbers 119886119899119896 where119899 119896 isin N = 0 1 2 Then we say that 119860 defines a matrix

Abstract and Applied Analysis 3

mapping from 120583 into 120574 and we denote it by writing 119860 120583 rarr120574 if for every sequence 119909 = (119909119896) isin 120583 the sequence 119860119909 =(119860119909)119899 the 119860-transform of 119909 is in 120574 where

(119860119909)119899 = sum

119896

119886119899119896119909119896 for each 119899 isin N (7)

By (120583 120574) we denote the class of all matrices 119860 such that119860 120583 rarr 120574 Thus 119860 isin (120583 120574) if and only if the series on theright side of (7) converges for each 119899 isin N and every 119909 isin 120583and we have 119860119909 = (119860119909)119899119899isinN isin 120574 for all 119909 isin 120583

Proposition 1 (see [15 Proposition 13 p 28]) Spectra andsubspectra of an operator119879 isin 119861(119883) and its adjoint119879lowast isin 119861(119883lowast)are related by the following relations

(a) 120590(119879lowast 119883lowast) = 120590(119879119883)(b) 120590119888(119879

lowast 119883lowast) sube 120590ap(119879119883)

(c) 120590ap(119879lowast 119883lowast) = 120590120575(119879119883)

(d) 120590120575(119879lowast 119883lowast) = 120590ap(119879119883)

(e) 120590119901(119879lowast 119883lowast) = 120590co(119879119883)

(f) 120590co(119879lowast 119883lowast) supe 120590119901(119879119883)

(g) 120590(119879119883) = 120590ap(119879119883) cup 120590119901(119879lowast 119883lowast) = 120590119901(119879119883) cup

120590ap(119879lowast 119883lowast)

The relations (c)ndash(f) show that the approximate pointspectrum is in a certain sense dual to defect spectrum and thepoint spectrum dual to the compression spectrum

The equality (g) implies in particular that 120590(119879119883) =

120590ap(119879119883) if 119883 is a Hilbert space and 119879 is normal Roughlyspeaking this shows that normal (in particular self-adjoint)operators onHilbert spaces aremost similar tomatrices in finitedimensional spaces (see [15])

Lemma 2 (see [16 p 60]) The adjoint operator 119879lowast of 119879 isonto if and only if 119879 has a bounded inverse

Lemma 3 (see [16 p 59]) 119879 has a dense range if and only if119879lowast is one to one

Our main focus in this paper is on the triple-band matrix119860(119903 119904 119905) where

119860 (119903 119904 119905) =

[[[[[[

[

119903 119904 119905 0

0 119903 119904 119905

0 0 119903 119904

0 0 0 119903

]]]]]]

]

(8)

We assume here and after that 119904 and 119905 are complex parameterswhich do not simultaneously vanish We introduce the intro-duce the operator 119860(119903 119904 119905) from ℓ119901 to itself by

119860 (119903 119904 119905) 119909 = (119903119909119896 + 119904119909119896+1 + 119905119909119896+2)infin

119896=0

where 119909 = (119909119896) isin ℓ119901(9)

3 Fine Spectra of Upper TriangularTriple-Band Matrices over the SequenceSpace ℓ119901 (0 lt 119901 ⩽ 1)

In this section we prove that the operator 119860(119903 119904 119905) ℓ119901 rarrℓ119901 is a bounded linear operator and compute its norm Weessentially emphasize the fine spectrum of the operator119860(119903 119904 119905) ℓ119901 rarr ℓ119901 in the case 0 lt 119901 ⩽ 1

Theorem 4 The operator 119860(119903 119904 119905) ℓ119901 rarr ℓ119901 is a boundedlinear operator and

119860 (119903 119904 119905)(ℓ119901 ℓ119901)= |119903|119901+ |119904|119901+ |119905|119901 (10)

Proof Since the linearity of the operator 119860(119903 119904 119905) is trivialso it is omitted Let us take 119890(2) isin ℓ119901 Then 119860(119903 119904 119905)119890(2) =(119905 119904 119903 0 ) and observe that

119860 (119903 119904 119905)(ℓ119901 ℓ119901)⩾

10038171003817100381710038171003817119860 (119903 119904 119905) 119890

(2)100381710038171003817100381710038171199011003817100381710038171003817119890(2)1003817100381710038171003817119901

= |119903|119901+ |119904|119901+ |119905|119901

(11)

which gives the fact that

119860 (119903 119904 119905)(ℓ119901 ℓ119901)⩾ |119903|119901+ |119904|119901+ |119905|119901 (12)

Let 119909 = (119909119896) isin ℓ119901 where 0 lt 119901 ⩽ 1Then since (119905119909119896+2) (119903119909119896)and (119904119909119896+1) isin ℓ119901 it is easy to see by triangle inequality that

119860 (119903 119904 119905) 119909119901 =

infin

sum

119896=0

1003816100381610038161003816119903119909119896 + 119904119909119896+1 + 119905119909119896+21003816100381610038161003816119901

⩽

infin

sum

119896=0

10038161003816100381610038161199031199091198961003816100381610038161003816119901+

infin

sum

119896=0

1003816100381610038161003816119904119909119896+11003816100381610038161003816119901+

infin

sum

119896=0

1003816100381610038161003816119905119909119896+21003816100381610038161003816119901

= |119903|119901infin

sum

119896=0

10038161003816100381610038161199091198961003816100381610038161003816119901+ |119904|119901infin

sum

119896=0

1003816100381610038161003816119909119896+11003816100381610038161003816119901+ |119905|119901infin

sum

119896=0

1003816100381610038161003816119909119896+21003816100381610038161003816119901

= |119904|119901119909119901 + |119903|

119901119909119901 + |119905|

119901119909119901

= (|119903|119901+ |119904|119901+ |119905|119901) 119909119901

(13)

which leads us to the result that

119860 (119903 119904 119905)(ℓ119901 ℓ119901)⩽ |119903|119901+ |119904|119901+ |119905|119901 (14)

Therefore by combining the inequalities (12) and (14) we seethat (10) holds which completes the proof

If 119879 ℓ119901 rarr ℓ119901 is a bounded matrix operator with thematrix 119860 then it is known that the adjoint operator 119879lowast ℓlowast

119901rarr ℓlowast

119901is defined by the transpose of thematrix119860The dual

space of ℓ119901 is isomorphic to ℓinfin where 0 lt 119901 lt 1Before giving themain theorem of this section we should

note the following remark In this work here and in whatfollows if 119911 is a complex number then by radic119911 we alwaysmean the square root of 119911 with a nonnegative real part IfRe(radic119911) = 0 then radic119911 represents the square root of 119911 withIm(radic119911) gt 0 The same results are obtained if radic119911 representsthe other square root

4 Abstract and Applied Analysis

Table 1 Subdivisions of spectrum of a linear operator

1 2 3119879minus1

120572exists and is bounded 119879

minus1

120572exists and is unbounded 119879

minus1

120572does not exist

A 119877(120572119868 minus 119879) = 119883 120572 isin 120588(119879119883) mdash 120572 isin 120590119901(119879119883)

120572 isin 120590ap(119879119883)

B 119877(120572119868 minus 119879) = 119883 120572 isin 120588(119879119883)

120572 isin 120590119888(119879119883) 120572 isin 120590119901(119879119883)

120572 isin 120590ap(119879119883) 120572 isin 120590ap(119879119883)

120572 isin 120590120575(119879119883) 120572 isin 120590120575(119879119883)

C 119877(120572119868 minus 119879) =119883

120572 isin 120590119903(119879119883) 120572 isin 120590119903(119879119883) 120572 isin 120590119901(119879119883)

120572 isin 120590120575(119879119883) 120572 isin 120590ap(119879119883) 120572 isin 120590ap(119879119883)

120572 isin 120590120575(119879119883) 120572 isin 120590120575(119879119883)

120572 isin 120590co(119879119883) 120572 isin 120590co(119879119883) 120572 isin 120590co(119879119883)

Theorem 5 Let 119904 be a complex number such that radic1199042 = minus119904and define the set1198631 by

1198631 = 120572 isin C 2 |119903 minus 120572| ⩽1003816100381610038161003816100381610038161003816minus119904 + radic1199042 minus 4119905 (119903 minus 120572)

1003816100381610038161003816100381610038161003816 (15)

Then 120590119888(119860(119903 119904 119905) ℓ119901) sube 1198631

Proof Let119910 = (119910119896) isin ℓinfinThen by solving the equation119860120572(119903119904 119905)lowast119909 = 119910 for 119909 = (119909119896) in terms of 119910 we obtain

1199090 =1199100

119903 minus 120572

1199091 =1199101

119903 minus 120572+

minus1199041199100

(119903 minus 120572)2

1199092 =1199102

119903 minus 120572+

minus1199041199101

(119903 minus 120572)2+[1199042minus 119905 (119903 minus 120572)] 1199100

(119903 minus 120572)3

(16)

(17)

and if we denote 1198861 = 1(119903 minus 120572) 1198862 = minus119904(119903 minus 120572)2 and 1198863 =

(1199042minus 119905(119903 minus 120572))(119903 minus 120572)

3 we have1199090 = 11988611199100

1199091 = 11988611199101 + 11988621199100

1199092 = 11988611199102 + 11988621199101 + 11988631199100

119909119899 = 1198861119910119899 + 1198862119910119899minus1 + sdot sdot sdot + 119886119899+11199100 =

119899

sum

119896=0

119886119899+1minus119896119910119896

(18)

Now we must find 119886119899 We have 119910119899 = 119905119909119899minus2 + 119904119909119899minus1 + (119903 minus 120572)119909119899and if we use relation (18) we have

119910119899 = 119905

119899minus2

sum

119896=0

119886119899minus1minus119896119910119896 + 119904

119899minus1

sum

119896=0

119886119899minus119896119910119896 + (119903 minus 120572)

119899

sum

119896=0

119886119899+1minus119896119910119896

= 1199100 (119905119886119899minus1 + 119904119886119899 + (119903 minus 120572) 119886119899+1)

+ 1199101 (119905119886119899minus2 + 119904119886119899minus1 + (119903 minus 120572) 119886119899) + sdot sdot sdot + 1199101198991198861 (119903 minus 120572)

(19)

This implies that

119905119886119899minus1 + 119904119886119899 + (119903 minus 120572) 119886119899+1 = 0

119905119886119899minus2 + 119904119886119899minus1 + (119903 minus 120572) 119886119899 = 0 1198861 (119903 minus 120572) = 1(20)

In fact this sequence is obtained recursively by letting

1198861 =1

119903 minus 120572 1198862 =

minus119904

(119903 minus 120572)2

119905119886119899minus2 + 119904119886119899minus1 + (119903 minus 120572) 119886119899 = 0 forall119899 ⩾ 3

(21)

The characteristic polynomial of the recurrence relation is (119903minus120572)1205822+ 119904120582 + 119905 = 0 There are two cases

Case 1 If Δ = 1199042 minus 4119905(119903 minus 120572) = 0 whose roots are

1205821 =minus119904 + radicΔ

2 (119903 minus 120572) 1205822 =

minus119904 minus radicΔ

2 (119903 minus 120572) (22)

elementary calculation on recurrent sequence gives that

119886119899 =120582119899

1minus 120582119899

2

radic1199042 minus 4119905 (119903 minus 120572) forall119899 ⩾ 1 (23)

In this case 119909119896 = (1radicΔ)sum119899

119896=0(120582119899+1minus119896

1minus 120582119899+1minus119896

2)119910119896 Assume

that |1205821| lt 1 So we have10038161003816100381610038161003816100381610038161003816100381610038161003816

1 + radic4119905 (119903 minus 120572)

1199042

10038161003816100381610038161003816100381610038161003816100381610038161003816

lt

10038161003816100381610038161003816100381610038161003816

2 (119903 minus 120572)

minus119904

10038161003816100381610038161003816100381610038161003816 (24)

Since |1 minus radic119911| ⩽ |1 + radic119911| for any 119911 isin C we must have10038161003816100381610038161003816100381610038161003816100381610038161003816

1 minus radic4119905 (119903 minus 120572)

1199042

10038161003816100381610038161003816100381610038161003816100381610038161003816

lt

10038161003816100381610038161003816100381610038161003816

2 (119903 minus 120572)

minus119904

10038161003816100381610038161003816100381610038161003816 (25)

It follows that |1205822| lt 1 Now for |1205821| lt 1 we can see that

10038161003816100381610038161199091198991003816100381610038161003816 ⩽

110038161003816100381610038161003816radicΔ10038161003816100381610038161003816

119899

sum

119896=0

10038161003816100381610038161003816120582119899+1minus119896

1

1003816100381610038161003816100381610038161003816100381610038161199101198961003816100381610038161003816 +

119899

sum

119896=0

10038161003816100381610038161003816120582119899+1minus119896

2

1003816100381610038161003816100381610038161003816100381610038161199101198961003816100381610038161003816 (26)

for all 119899 isin N Taking limit on the inequality (26) as 119899 rarr infinwe get

119909infin ⩽1 minus (

100381610038161003816100381612058221003816100381610038161003816 +100381610038161003816100381612058221003816100381610038161003816)

10038161003816100381610038161003816(1 minus

10038161003816100381610038161205822100381610038161003816100381610038161003816100381610038161 minus 1205822

1003816100381610038161003816)radicΔ10038161003816100381610038161003816

10038171003817100381710038171199101003817100381710038171003817infin (27)

Abstract and Applied Analysis 5

Thus for |1205821| lt 1 119860120572(119903 119904 119905)lowast is onto and by Lemma 2 119860120572(119903

119904 119905) has a bounded inverse This means that

120590119888 (119860 (119903 119904 119905) ℓ119901)

sube 120572 isin C 2 |119903 minus 120572| ⩽1003816100381610038161003816100381610038161003816minus119904 + radic1199042 minus 4119905 (119903 minus 120572)

1003816100381610038161003816100381610038161003816

= 1198631

(28)

Case 2 If Δ = 1199042 minus 4119905(119903 minus 120572) = 0 a calculation on recurrentsequence gives that

119886119899 = (2119899

minus119904) [

minus119904

2 (119903 minus 120572)]

119899

forall119899 ⩾ 1 (29)

Now for | minus 119904| lt 2|119903 minus 120572| we can see that

10038161003816100381610038161199091198991003816100381610038161003816 ⩽

119899

sum

119896=0

1003816100381610038161003816119886119899minus1198961199101198961003816100381610038161003816 (30)

for all 119899 isin N Taking limit on the inequality (30) as 119899 rarr infinwe obtain that

119909infin ⩽10038171003817100381710038171199101003817100381710038171003817infin

infin

sum

119896=0

10038161003816100381610038161198861198961003816100381610038161003816 (31)

suminfin

119896=0|119886119896| is convergent since | minus 119904| lt 2|119903 minus 120572| Thus for | minus 119904| lt

2|119903 minus 120572| 119860120572(119903 119904 119905)lowast is onto and by Lemma 2 119860120572(119903 119904 119905) has a

bounded inverse This means that

120590119888 (119860 (119903 119904 119905) ℓ119901) sube 120572 isin C 2 |119903 minus 120572| ⩽ |minus119904| sube 1198631

(32)

Theorem 6 120590119901(119860(119903 119904 119905)lowast ℓlowast

119901) = 0

Proof Consider 119860(119903 119904 119905)lowast119891 = 120572119891 with 119891 = 120579 = (0 0 0 ) inℓlowast

119901= ℓinfin Then by solving the system of linear equations

1199031198910 = 1205721198910

1199041198910 + 1199031198911 = 1205721198911

1199051198910 + 1199041198911 + 1199031198912 = 1205721198912

1199051198911 + 1199041198912 + 1199031198913 = 1205721198913

119905119891119896minus2 + 119904119891119896minus1 + 119903119891119896 = 120572119891119896

(33)

we find that 1198910 = 0 if 120572 = 119903 and 1198911 = 1198912 = sdot sdot sdot = 0 if 1198910 = 0which contradicts119891 = 120579 If 1198911198990 is the first nonzero entry of thesequence119891 = (119891119899) and120572 = 119903 thenwe get 1199051198911198990minus2+1199041198911198990minus1+1199031198911198990 =1205721198911198990

which implies 1198911198990 = 0which contradicts the assumption1198911198990

= 0 Hence the equation119860(119903 119904 119905)lowast119891 = 120572119891 has no solution119891 = 120579

Theorem 7 120590119901(119860(119903 119904 119905) ℓ119901) = 1198632 where 1198632 = 120572 isin C

2|119903 minus 120572| lt | minus 119904 + radic1199042 minus 4119905(119903 minus 120572)|

Proof Let119860(119903 119904 119905)119909 = 120572119909 for 120579 = 119909 isin ℓ119901Then by solving thesystem of linear equations

1199031199090 + 1199041199091 + 1199051199092 = 1205721199090

1199031199091 + 1199041199092 + 1199051199093 = 1205721199091

1199031199092 + 1199041199093 + 1199051199094 = 1205721199092

119903119909119896minus2 + 119904119909119896minus1 + 119905119909119896 = 120572119909119896

(34)

and we have

1199092 =minus119904

1199051199091 minus

119903 minus 120572

1199051199090

1199093 =1199042minus 119905 (119903 minus 120572)

11990521199091 +

119904 (119903 minus 120572)

11990521199090

119909119899 =119886119899(119903 minus 120572)

119899

119905119899minus11199091 minus

119886119899minus1(119903 minus 120572)119899

119905119899minus11199090 forall119899 ⩾ 2

(35)

Assume that 120572 isin 1198632 Then we choose 1199090 = 1 and 1199091 = 2(119903 minus120572)(minus119904+radic1199042 minus 4119905(119903 minus 120572)) We show that 119909119899 = 119909

119899

1for all 119899 ⩾ 2

Since 1205821 1205822 are roots of the characteristic equation (119903minus120572)1205822+

119904120582 + 119905 = 0 we must have

12058211205822 =119905

119903 minus 120572 1205821 minus 1205822 =

radicΔ

119903 minus 120572 (36)

Combining the fact 1199091 = 11205821 with relation (35) we can seethat

119909119899 =119886119899(119903 minus 120572)

119899

119905119899minus11199091 minus

119886119899minus1(119903 minus 120572)119899

119905119899minus11199090

= (119903 minus 120572

119905)

119899minus1

(119903 minus 120572) (minus119886119899minus11199090 + 1198861198991199091)

=1

(12058211205822)119899minus1

119903 minus 120572

radicΔ(minus120582119899minus1

1+ 120582119899minus1

2+ 120582119899minus1

1minus 120582119899

2120582minus1

1)

=1

120582119899minus11120582119899minus12

(1

1205821 minus 1205822)120582119899minus1

2(1205821 minus 1205822

1205821)

=1

1205821198991

= 119909119899

1

(37)

The same result may be obtained in case Δ = 0 Now 119909 =(119909119896) isin ℓ119901 since |1199091| lt 1 This shows that 1198632 sube 120590119901(119860(119903

119904 119905) ℓ119901)

6 Abstract and Applied Analysis

Now we assume that 120572 notin 1198632 that is |1205821| ⩽ 1 We mustshow that 120572 notin 120590119901(119860(119903 119904 119905) ℓ119901) Therefore we obtain from therelation (35) that

119909119899+1

119909119899= (

119903 minus 120572

119905)119886119899minus1

119886119899minus2(minus1199090 + (119886119899119886119899minus1) 1199091

minus1199090 + (119886119899minus1119886119899minus2) 1199091) (38)

Now we examine three cases

Case 1 (|1205822| lt |1205821| ⩽ 1) In this case we have 1199042 = 4119905(119903minus120572) and

119886119899

119886119899minus1=120582119899+1

1minus 120582119899+1

2

1205821198991minus 1205821198992

=1205821 [1 minus (12058221205821)

119899+1]

[1 minus (12058221205821)119899] (39)

Then we have

lim119899rarrinfin

10038161003816100381610038161003816100381610038161003816

119886119899

119886119899minus1

10038161003816100381610038161003816100381610038161003816

119901

= lim119899rarrinfin

10038161003816100381610038161003816100381610038161003816

119886119899minus1

119886119899minus2

10038161003816100381610038161003816100381610038161003816

119901

= lim119899rarrinfin

100381610038161003816100381612058211003816100381610038161003816119901100381610038161003816100381610038161 minus (12058221205821)

119899+110038161003816100381610038161003816

119901

100381610038161003816100381610038161 minus (12058221205821)

11989910038161003816100381610038161003816

119901=100381610038161003816100381612058211003816100381610038161003816119901

(40)

Now if minus1199090 + 12058211199091 = 0 then we have (119909119899) = (1199090120582119899

1) which

is not in ℓ119901 Otherwise

lim119899rarrinfin

10038161003816100381610038161003816100381610038161003816

119909119899+1

119909119899

10038161003816100381610038161003816100381610038161003816

119901

=1

10038161003816100381610038161205821100381610038161003816100381611990110038161003816100381610038161205822

1003816100381610038161003816119901

100381610038161003816100381612058211003816100381610038161003816119901=

1

100381610038161003816100381612058221003816100381610038161003816119901gt 1 (41)

Case 2 (|1205822| = |1205821| lt 1) In this case we have 1199042 = 4119905(119903 minus 120572)and using the formula

119886119899 = (2119899

minus119904) [

minus119904

2 (119903 minus 120572)]

119899

forall119899 ⩾ 1 (42)

we obtain that

lim119899rarrinfin

10038161003816100381610038161003816100381610038161003816

119886119899

119886119899minus1

10038161003816100381610038161003816100381610038161003816

119901

=

10038161003816100381610038161003816100381610038161003816

minus119904

2 (119903 minus 120572)

10038161003816100381610038161003816100381610038161003816

119901

=100381610038161003816100381612058211003816100381610038161003816119901 (43)

which leads to

lim119899rarrinfin

10038161003816100381610038161003816100381610038161003816

119909119899+1

119909119899

10038161003816100381610038161003816100381610038161003816

119901

=1

10038161003816100381610038161205821100381610038161003816100381611990110038161003816100381610038161205822

1003816100381610038161003816119901

100381610038161003816100381612058211003816100381610038161003816119901=

1

100381610038161003816100381612058221003816100381610038161003816119901gt 1 (44)

Case 3 (|1205822| = |1205821| = 1) In this case we have 1199042 = 4119905(119903minus120572) andso we have | minus 119904(2119905)| = 1 Assume that 120572 isin 120590119901(119860(119903 119904 119905) ℓ119901)This implies that 119909 isin ℓ119901 and 119909 = 120579 Thus we again derive (35)

119909119899 = (minus119904

2119905)

119899minus1

[minus (119899 minus 1)minus119904

21199051199090 + 1198991199091] (45)

Since lim119899rarrinfin119909119899 = 0 we must have 1199090 = 1199091 = 0 But thisimplies that 119909 = 120579 a contradiction which means that 120572 notin120590119901(119860(119903 119904 119905) ℓ119901) Thus 120590119901(119860(119903 119904 119905) ℓ119901) sube 1198632 This completesthe proof

Theorem 8 120590119903(119860(119903 119904 119905) ℓ119901) = 0

Proof By Proposition 1 120590119903(119860(119903 119904 119905) ℓ119901) = 120590119901(119860(119903 119904 119905)lowast ℓlowast

119901)

120590119901(119860(119903 119904 119905) ℓ119901) Since byTheorem 6

120590119901 (119860(119903 119904 119905)lowast ℓlowast

119901) = 0 120590119903 (119860 (119903 119904 119905) ℓ119901) = 0

(46)

This completes the proof

Theorem 9 Let 119904 be a complex number such that radic1199042 = minus119904and define the set1198631 by

1198631 = 120572 isin C 2 |119903 minus 120572| ⩽1003816100381610038161003816100381610038161003816minus119904 + radic1199042 minus 4119905 (119903 minus 120572)

1003816100381610038161003816100381610038161003816

(47)

Then 120590(119860(119903 119904 119905) ℓ119901) = 1198631

Proof ByTheorem 7 we get

120572 isin C 2 |119903 minus 120572| lt1003816100381610038161003816100381610038161003816minus119904 + radic1199042 minus 4119905 (119903 minus 120572)

1003816100381610038161003816100381610038161003816

sube 120590 (119860 (119903 119904 119905) ℓ119901)

(48)

Since the spectrum of any bounded operator is closed wehave

120572 isin C 2 |119903 minus 120572| ⩽1003816100381610038161003816100381610038161003816minus119904 + radic1199042 minus 4119905 (119903 minus 120572)

1003816100381610038161003816100381610038161003816

sube 120590 (119860 (119903 119904 119905) ℓ119901)

(49)

and again fromTheorems 5 7 and 8

120590 (119860 (119903 119904 119905) ℓ119901) sube 120572 isin C 2 |119903 minus 120572|

⩽1003816100381610038161003816100381610038161003816minus119904 + radic1199042 minus 4119905 (119903 minus 120572)

1003816100381610038161003816100381610038161003816

(50)

Combining (49) and (50) we obtain that 120590(119860(119903 119904 119905) ℓ119901) =1198631 where1198631 is defined by (47)

Theorem 10 120590119888(119860(119903 119904 119905) ℓ119901) = 1198633 where 1198633 = 120572 isin C

2|119903 minus 120572| = | minus 119904 + radic1199042 minus 4119905(119903 minus 120572)|

Proof Because the parts 120590119888(119860(119903 119904 119905) ℓ119901) 120590119903(119860(119903 119904 119905) ℓ119901)and 120590119901(119860(119903 119904 119905) ℓ119901) are pairwise disjoint sets and the unionof these sets is 120590(119860(119903 119904 119905) ℓ119901) the proof immediately followsfromTheorems 7 8 and 9

Theorem 11 If 120572 isin 1198632 120572 isin 120590(119860(119903 119904 119905) ℓ119901)1198683

Proof From Theorem 7 120572 isin 120590119901(119860(119903 119904 119905) ℓ119901) Thus (119860(119903119904 119905) minus 120572119868)

minus1 does not exist By Theorem 6119860(119903 119904 119905)lowast minus 120572119868 isone to one so 119860(119903 119904 119905) minus 120572119868 has a dense range in ℓ119901 byLemma 3

Theorem 12 The following statements hold

(i) 120590ap(119860(119903 119904 119905) ℓ119901) = 1198631(ii) 120590120575(119860(119903 119904 119905) ℓ119901) = 1198633(iii) 120590co(119860(119903 119904 119905) ℓ119901) = 0

Abstract and Applied Analysis 7

Proof (i) Since from Table 1

120590ap (119860 (119903 119904 119905) ℓ119901)=120590 (119860 (119903 119904 119905) ℓ119901)120590 (119860 (119903 119904 119905) ℓ119901) 1198681198681198681

(51)

we have byTheorem 8

120590 (119860 (119903 119904 119905) ℓ119901) 1198681198681198681 = 120590 (119860 (119903 119904 119905) ℓ119901) 1198681198681198682 = 0 (52)

Hence

120590ap (119860 (119903 119904 119905) ℓ119901) = 1198631 (53)

(ii) Since the following equality

120590120575 (119860 (119903 119904 119905) ℓ119901) = 120590 (119860 (119903 119904 119905) ℓ119901) 120590 (119860 (119903 119904 119905) ℓ119901) 1198683

(54)

holds from Table 1 we derive by Theorems 8 and 11 that120590120575(119860(119903 119904 119905) ℓ119901) = 1198632

(iii) From Table 1 we have

120590co (119860 (119903 119904 119905) ℓ119901)

= 120590 (119860 (119903 119904 119905) ℓ119901) 1198681198681198681 cup 120590 (119860 (119903 119904 119905) ℓ119901) 1198681198681198682

cup 120590 (119860 (119903 119904 119905) 1198880) 1198681198681198683

(55)

ByTheorem 6 it is immediate that 120590co(119860(119903 119904 119905) ℓ119901) = 0

4 Fine Spectra of Upper TriangularTriple-Band Matrices over the SequenceSpace ℓ119901 (1 lt 119901 lt infin)

In the present section we determine the fine spectrum ofthe operator 119860(119903 119904 119905) ℓ119901 rarr ℓ119901 in case 1 ⩽ 119901 lt infinWe quote some lemmas which are needed in proving thetheorems given in Section 4

Lemma 13 (see [17 p 253 Theorem 3416]) The matrix 119860 =(119886119899119896) gives rise to a bounded linear operator 119879 isin 119861(ℓ1) from ℓ1to itself if and only if the supremum of ℓ1 norms of the columnsof 119860 is bounded

Lemma 14 (see [17 p 245 Theorem 343]) The matrix 119860 =(119886119899119896) gives rise to a bounded linear operator 119879 isin 119861(ℓinfin) fromℓinfin to itself if and only if the supremum of ℓ1 norms of the rowsof 119860 is bounded

Lemma 15 (see [17 p 254 Theorem 3418]) Let 1 lt 119901 lt infinand 119860 isin (ℓinfin ℓinfin) cap (ℓ1 ℓ1) Then 119860 isin (ℓ119901 ℓ119901)

Theorem 16 The operator 119860(119903 119904 119905) ℓ119901 rarr ℓ119901 is a boundedlinear operator and

(|119903|119901+ |119904|119901+ |119905|119901)1119901⩽ 119860 (119903 119904 119905)(ℓ119901 ℓ119901)

⩽ |119903| + |119904| + |119905|

(56)

Proof Since the linearity of the operator119860(119903 119904 119905) is not hardwe omit the details Now we prove that (56) holds for theoperator 119860(119903 119904 119905) on the space ℓ119901 It is trivial that 119860(119903 119904119905)119890(2)= (119905 119904 119903 0 ) for 119890(2) isin ℓ119901 Therefore we have

119860 (119903 119904 119905)(ℓ119901 ℓ119901)⩾

10038171003817100381710038171003817119860 (119903 119904 119905) 119890

(2)100381710038171003817100381710038171199011003817100381710038171003817119890(2)1003817100381710038171003817119901

= (|119903|119901+ |119904|119901+ |119905|119901)1119901

(57)

which implies that

119860 (119903 119904 119905)(ℓ119901 ℓ119901)⩾ (|119903|119901+ |s|119901 + |119905|119901)1119901 (58)

Let 119909 = (119909119896) isin ℓ119901 where 1 lt 119901 lt infin Then since (119905119909119896+2)(119903119909119896) and (119904119909119896+1) isin ℓ119901 it is easy to see by Minkowskirsquos ine-quality that

119860 (119903 119904 119905) 119909119901

= (

infin

sum

119896=0

1003816100381610038161003816119903119909119896 + 119904119909119896+1 + 119905119909119896+21003816100381610038161003816119901)

1119901

⩽ (

infin

sum

119896=0

10038161003816100381610038161199031199091198961003816100381610038161003816119901)

1119901

+ (

infin

sum

119896=0

1003816100381610038161003816119904119909119896+11003816100381610038161003816119901)

1119901

+ (

infin

sum

119896=0

1003816100381610038161003816119905119909119896+21003816100381610038161003816119901)

1119901

= |119903| (

infin

sum

119896=0

10038161003816100381610038161199091198961003816100381610038161003816119901)

1119901

+ |119904| (

infin

sum

119896=0

1003816100381610038161003816119909119896+11003816100381610038161003816119901)

1119901

+ |119905| (

infin

sum

119896=0

1003816100381610038161003816119909119896+21003816100381610038161003816119901)

1119901

= |119904| 119909119901 + |119903| 119909119901 + |119905| 119909119901

= (|119903| + |119904| + |119905|) 119909119901

(59)

which leads us to the result that

(|119903|119901+ |119904|119901+ |119905|119901)1119901⩽ 119860 (119903 119904 119905)(ℓ119901 ℓ119901)

⩽ |119903| + |119904| + |119905|

(60)

Therefore by combining the inequalities in (58) and (59) wehave (56) as desired

If 119879 ℓ119901 rarr ℓ119901 is a bounded matrix operator with thematrix 119860 then it is known that the adjoint operator 119879lowast ℓlowast

119901rarr ℓlowast

119901is defined by the transpose of thematrix119860The dual

space of ℓ119901 is isomorphic to ℓ119902 where 1 lt 119901 lt infin

Theorem 17 Let 119904 be a complex number such that radic1199042 = minus119904and define the set1198631 by

1198631 = 120572 isin C 2 |119903 minus 120572| ⩽1003816100381610038161003816100381610038161003816minus119904 + radic1199042 minus 4119905 (119903 minus 120572)

1003816100381610038161003816100381610038161003816

(61)

Then 120590119888(119860(119903 119904 119905) ℓ119901) sube 1198631

8 Abstract and Applied Analysis

Proof We will show that 119860120572(119903 119904 119905)lowast is onto for 2|119903 minus 120572| gt

| minus 119904 + radic1199042 minus 4119905(119903 minus 120572)| Thus for every 119910 isin ℓ119902 we find 119909 isinℓ119902 119860120572(119903 119904 119905)

lowast is a triangle so it has an inverse Also equation119860120572(119903 119904 119905)

lowast119909 = 119910 gives [119860120572(119903 119904 119905)

lowast]minus1119910 = 119909 It is sufficient

to show that [119860120572(119903 119904 119905)lowast]minus1isin (ℓ119902 ℓ119902) We calculate that

119860 = (119886119899119896) = [119860120572(119903 119904 119905)lowast]minus1 as follows

119860 = (119886119899119896) =

[[[[

[

1198861 0 0

1198862 1198861 0

1198863 1198862 1198861

]]]]

]

(62)

where

1198861 =1

119903 minus 120572

1198862 =minus119904

(119903 minus 120572)2

1198863 =1199042minus 119905 (119903 minus 120572)

(119903 minus 120572)3

(63)

It is known that fromTheorem 5

119886119899 =120582119899

1minus 120582119899

2

radic1199042 minus 4119905 (119903 minus 120572) forall119899 ⩾ 1

where 1205821 =minus119904 + radicΔ

119903 minus 120572 1205822 =

minus119904 minus radicΔ

119903 minus 120572

(64)

Now we show that [119860120572(119903 119904 119905)lowast]minus1isin (ℓ1 ℓ1) for |1205821| lt 1

By Theorem 5 we know that if |1205821| lt 1 |1205822| lt 1 We assumethat 1199042 minus 4119905(119903 minus 120572) = 0 and |1205821| lt 1

10038171003817100381710038171003817[119860120572(119903 119904 119905)

lowast]minus110038171003817100381710038171003817(ℓ1 ℓ1)

= sup119899

infin

sum

119896=119899

10038161003816100381610038161198861198961003816100381610038161003816 =

infin

sum

119896=1

10038161003816100381610038161198861198961003816100381610038161003816

⩽1

|Δ|(

infin

sum

119896=1

100381610038161003816100381612058211003816100381610038161003816119896+

infin

sum

119896=1

100381610038161003816100381612058221003816100381610038161003816119896) lt infin

(65)

since |1205821| lt 1 and |1205822| lt 1 This shows that (119860120572(119903 119904 119905)lowast]minus1isin

(ℓ1 ℓ1) Similarly we can show that [119860120572(119903 119904 119905)lowast]minus1isin (ℓinfin

ℓinfin)Now assume that 1199042 minus 4119905(119903 minus 120572) = 0 Then

119886119899 = (2119899

minus119904) [

minus119904

2 (119903 minus 120572)]

119899

(66)

and simple calculation gives that (119886119899) isin ℓ119902 if and only if |minus119904| lt2|119903 minus 120572|

[(119860 (119903 119904 119905) minus 120572119868)lowast]minus1isin (ℓ119902 ℓ119902)

for 120572 isin C with 2 |119903 minus 120572| gt1003816100381610038161003816100381610038161003816minus119904 + radic1199042 minus 4119905 (119903 minus 120572)

1003816100381610038161003816100381610038161003816

(67)

Hence 119860120572(119903 119904 119905)lowast is onto By Lemma 2 119860120572(119903 119904 119905) has a

bounded inverse This means that 120590119888(119860(119903 119904 119905) ℓ119901) sube 1198631where1198631 is defined by (61)

Theorem 18 120590119901(119860(119903 119904 119905)lowast ℓlowast

119901) = 0

Proof Let 119860(119903 119904 119905)lowast119891 = 120572119891 with 119891 = 120579 = (0 0 0 ) in ℓlowast119901=

ℓ119902 Then by solving the system of linear equations

1199031198910 = 1205721198910

1199041198910 + 1199031198911 = 1205721198911

1199051198910 + 1199041198911 + 1199031198912 = 1205721198912

1199051198911 + 1199041198912 + 1199031198913 = 1205721198913

119905119891119896minus2 + 119904119891119896minus1 + 119903119891119896 = 120572119891119896

(68)

we find that 1198910 = 0 if 120572 = 119903 and 1198911 = 1198912 = sdot sdot sdot = 0 if 1198910 = 0which contradicts119891 = 120579 If 1198911198990 is the first nonzero entry of thesequence 119891 = (119891119899) and 120572 = 119903 then we get 1199051198911198990minus2 + 1199041198911198990minus1 +1199031198911198990

= 1205721198911198990which implies 1198911198990 = 0 which contradicts the

assumption1198911198990 = 0 Hence the equation119860(119903 119904 119905)lowast119891 = 120572119891 has

no solution 119891 = 120579

In the case 1 lt 119901 lt infin since the proof of the theoremsin Section 4 determining the spectrum and fine spectrumof the matrix operator 119860(119903 119904 119905) on the sequence space ℓ119901is similar to the case 0 lt 119901 ⩽ 1 to avoid the repetitionof similar statements we give the results by the followingtheorem without proof

Theorem 19 The following statements hold(i) 120590(119860(119903 119904 119905) ℓ119901) = 1198631(ii) 120590119903(119860(119903 119904 119905) ℓ119901) = 0(iii) 120590119901(119860(119903 119904 119905) ℓ119901) = 1198632(iv) 120590119888(119860(119903 119904 119905) ℓ119901) = 1198633(v) 120590ap(119860(119903 119904 119905) ℓ119901) = 1198631(vi) 120590co(119860(119903 119904 119905) ℓ119901) = 0(vii) 120590120575(119860(119903 119904 119905) ℓ119901) = 1198633

5 Some Applications

In this section we give two theorems related to Toeplitz ma-trix

Theorem 20 Let 119875 be a polynomial that corresponds to the 119899-tuple 119886 and let 1199111 1199112 1199113 119911119899minus1 also be the roots of 119875 Define119879 as a Toeplitz matrix associated with 119875 that is

119879 =

[[[[

[

1198860 1198861 1198862 119886119899 0 0 0

0 1198860 1198861 1198862 119886119899 0 0

0 0 1198860 1198861 1198862 119886119899 0

]]]]

]

(69)

Abstract and Applied Analysis 9

The resolvent operator 119879 over ℓ119901 with 1 lt 119901 lt infin where thedomain of the resolvent operator is the whole space ℓ119901 exists ifand if only all the roots of the polynomial are outside the unitdisc 119911 isin C |119911| ⩽ 1 That is 119879minus1 isin (ℓ119901 ℓ119901) if and if only|119911119894| gt 1 1 ⩽ 119894 ⩽ 119899 minus 1 In this case the resolvent operator isrepresented by

119879minus1=

1

119886119899minus1119860minus1(minus1199111 1) 119860

minus1(minus1199112 1) sdot sdot sdot 119860

minus1(minus119911119899minus1 1)

where

119860minus1(minus119911119894 1) =minus

[[[[[[[[[[[[

[

1119911119894 11199112

11989411199113

11989411199114

11989411199115

119894

0 1119911119894 11199112

11989411199113

11989411199114

119894

0 0 1119911119894 11199112

11989411199113

119894

0 0 0 1119911119894 11199112

119894

0 0 0 0 1119911119894

]]]]]]]]]]]]

]

(70)

Proof Suppose all the roots of the polynomial 119875(119911) = 1198860 +1198861119911 + sdot sdot sdot + 119886119899minus1119911

119899minus1= 119886119899(119911 minus 1199111)(119911 minus 1199112) sdot sdot sdot (119911 minus 119911119899minus1) are

outside the unit disc The Toeplitz matrix associated with 119875can be written as the product

119879 = 119886119899119860 (minus1199111 1) 119860 (minus1199112 1) sdot sdot sdot 119860 (minus119911119899minus1 1) (71)

Since multiplication of upper triangular Toeplitz matrices iscommutative we can see that

119879minus1=

1

119886119899minus1119860minus1(minus1199111 1) 119860

minus1(minus1199112 1) sdot sdot sdot 119860

minus1(minus119911119899minus1 1)

(72)

is left inverse of 119879 Since all roots are outside the unit discthen

10038171003817100381710038171003817119879minus1(minus119911119894 1)

10038171003817100381710038171003817(ℓinfin ℓinfin)= sup119899

infin

sum

119896=119899

1

10038161003816100381610038161199111198941003816100381610038161003816119896+1minus119899

=

infin

sum

119896=1

1

10038161003816100381610038161199111198941003816100381610038161003816119896lt infin

(73)

Therefore each 119879minus1(minus119911119894 1) isin (ℓinfin ℓinfin) for 1 ⩽ 119894 ⩽ 119899 minus 1Similarly we can say that 119879minus1(minus119911119894 1) isin (ℓ1 ℓ1) So we have119879minus1(minus119911119894 1) isin (ℓ119901 ℓ119901)

Theorem 21 The resolvent operator of 119860(119903 119904 119905) over ℓ119901 with1 lt 119901 lt infin where the domain of the resolvent operator is thespace ℓ119901 exists if and only if 2|119903| gt | minus 119904 + radic1199042 minus 4119905119903| In this

case the resolvent operator is represented by the infinite bandedToeplitz matrix

119864 (119903 119904 119905)

=1

119905

[[[[[[[[[[

[

1199061 1199062

11199063

11199064

1

0 1199061 1199062

11199063

1

0 0 1199061 1199062

1

0 0 0 1199061

]]]]]]]]]]

]

[[[[[[[[[[

[

1199062 1199062

21199063

21199064

2

0 1199062 1199062

21199063

2

0 0 1199062 1199062

2

0 0 0 1199062

]]]]]]]]]]

]

where 1199061 =minus119904 + radic1199042 minus 4119905119903

2119903 1199062 =

minus119904 minus radic1199042 minus 4119905119903

2119903

(74)

Proof By Theorem 20 we can see that 119864(119903 119904 119905) is inverseof the matrix of 119860(119903 119904 119905) But this is not enough to say itis resolvent operator By Lemmas 13 14 and 15 119864(119903 119904 119905) isin(ℓ119901 ℓ119901) when 2|119903| gt | minus 119904 + radic1199042 minus 4119905119903| That is for 2|119903| gt| minus 119904 + radic1199042 minus 4119905119903| 119864(119903 119904 119905) is a resolvent operator

Acknowledgments

The authors would like to thank the referee for pointingout some mistakes and misprints in the earlier version ofthis paper They would like to express their pleasure toMs Medine Yesilkayagil Department of Mathematics UsakUniversity 1 Eylul Campus Usak Turkey for many helpfulsuggestions and interesting comments on the revised form ofthe paper

References

[1] M Gonzalez ldquoThe fine spectrum of the Cesaro operator in ℓ119901(1 lt 119901 lt infin)rdquo Archiv der Mathematik vol 44 no 4 pp 355ndash358 1985

[2] P J CartlidgeWeighted meanmatrices as operators on ℓ119901 [PhDdissertation] Indiana University 1978

[3] J I Okutoyi ldquoOn the spectrum of C1 as an operator on 1198871199070rdquoAustralian Mathematical Society A vol 48 no 1 pp 79ndash861990

[4] J I Okutoyi ldquoOn the spectrum of C1 as an operator on bvrdquoCommunications A vol 41 no 1-2 pp 197ndash207 1992

[5] M Yıldırım ldquoOn the spectrum of the Rhaly operators on ℓ119901rdquoIndian Journal of Pure and Applied Mathematics vol 32 no 2pp 191ndash198 2001

[6] B Altay and F Basar ldquoOn the fine spectrum of the differenceoperator Δ on 1198880 and 119888rdquo Information Sciences vol 168 no 1ndash4pp 217ndash224 2004

[7] B Altay and F Basar ldquoOn the fine spectrum of the generalizeddifference operator 119861(119903 119904) over the sequence spaces 1198880 and119888rdquo International Journal of Mathematics and MathematicalSciences vol 2005 no 18 pp 3005ndash3013 2005

[8] A M Akhmedov and F Basar ldquoOn the fine spectra of the dif-ference operator Δ over the sequence space ℓ119901 (1 le 119901 lt infin)rdquoDemonstratio Mathematica vol 39 no 3 pp 585ndash595 2006

10 Abstract and Applied Analysis

[9] AMAkhmedov andF Basar ldquoThefine spectra of the differenceoperator Δ over the sequence space 119887119907119901 (1 le 119901 lt infin)rdquo ActaMathematica Sinica vol 23 no 10 pp 1757ndash1768 2007

[10] F Basar and B Altay ldquoOn the space of sequences of p-boundedvariation and related matrix mappingsrdquo Ukrainian Mathemati-cal Journal vol 55 no 1 pp 136ndash147 2003

[11] H Furkan H Bilgic and F Basar ldquoOn the fine spectrum of theoperator 119861(119903 119904 119905) over the sequence spaces ℓ119901 and 119887119907119901(1 lt 119901 ltinfin)rdquo Computers ampMathematics with Applications vol 60 no 7pp 2141ndash2152 2010

[12] V Karakaya and M Altun ldquoFine spectra of upper triangulardouble-band matricesrdquo Journal of Computational and AppliedMathematics vol 234 no 5 pp 1387ndash1394 2010

[13] A Karaisa ldquoFine spectra of upper triangular double-bandmatrices over the sequence space ℓ119901 (1 lt 119901 lt infin)rdquo DiscreteDynamics in Nature and Society vol 2012 Article ID 381069 19pages 2012

[14] E Kreyszig Introductory Functional Analysis with ApplicationsJohn Wiley amp Sons New York NY USA 1978

[15] J Appell E Pascale and A Vignoli Nonlinear Spectral Theoryvol 10 of de Gruyter Series in Nonlinear Analysis and Applica-tions Walter de Gruyter Berlin Germany 2004

[16] S Goldberg Unbounded Linear Operators Dover PublicationsNew York NY USA 1985

[17] B Choudhary and S Nanda Functional Analysis with Applica-tions John Wiley amp Sons New York NY USA 1989

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 3

mapping from 120583 into 120574 and we denote it by writing 119860 120583 rarr120574 if for every sequence 119909 = (119909119896) isin 120583 the sequence 119860119909 =(119860119909)119899 the 119860-transform of 119909 is in 120574 where

(119860119909)119899 = sum

119896

119886119899119896119909119896 for each 119899 isin N (7)

By (120583 120574) we denote the class of all matrices 119860 such that119860 120583 rarr 120574 Thus 119860 isin (120583 120574) if and only if the series on theright side of (7) converges for each 119899 isin N and every 119909 isin 120583and we have 119860119909 = (119860119909)119899119899isinN isin 120574 for all 119909 isin 120583

Proposition 1 (see [15 Proposition 13 p 28]) Spectra andsubspectra of an operator119879 isin 119861(119883) and its adjoint119879lowast isin 119861(119883lowast)are related by the following relations

(a) 120590(119879lowast 119883lowast) = 120590(119879119883)(b) 120590119888(119879

lowast 119883lowast) sube 120590ap(119879119883)

(c) 120590ap(119879lowast 119883lowast) = 120590120575(119879119883)

(d) 120590120575(119879lowast 119883lowast) = 120590ap(119879119883)

(e) 120590119901(119879lowast 119883lowast) = 120590co(119879119883)

(f) 120590co(119879lowast 119883lowast) supe 120590119901(119879119883)

(g) 120590(119879119883) = 120590ap(119879119883) cup 120590119901(119879lowast 119883lowast) = 120590119901(119879119883) cup

120590ap(119879lowast 119883lowast)

The relations (c)ndash(f) show that the approximate pointspectrum is in a certain sense dual to defect spectrum and thepoint spectrum dual to the compression spectrum

The equality (g) implies in particular that 120590(119879119883) =

120590ap(119879119883) if 119883 is a Hilbert space and 119879 is normal Roughlyspeaking this shows that normal (in particular self-adjoint)operators onHilbert spaces aremost similar tomatrices in finitedimensional spaces (see [15])

Lemma 2 (see [16 p 60]) The adjoint operator 119879lowast of 119879 isonto if and only if 119879 has a bounded inverse

Lemma 3 (see [16 p 59]) 119879 has a dense range if and only if119879lowast is one to one

Our main focus in this paper is on the triple-band matrix119860(119903 119904 119905) where

119860 (119903 119904 119905) =

[[[[[[

[

119903 119904 119905 0

0 119903 119904 119905

0 0 119903 119904

0 0 0 119903

]]]]]]

]

(8)

We assume here and after that 119904 and 119905 are complex parameterswhich do not simultaneously vanish We introduce the intro-duce the operator 119860(119903 119904 119905) from ℓ119901 to itself by

119860 (119903 119904 119905) 119909 = (119903119909119896 + 119904119909119896+1 + 119905119909119896+2)infin

119896=0

where 119909 = (119909119896) isin ℓ119901(9)

3 Fine Spectra of Upper TriangularTriple-Band Matrices over the SequenceSpace ℓ119901 (0 lt 119901 ⩽ 1)

In this section we prove that the operator 119860(119903 119904 119905) ℓ119901 rarrℓ119901 is a bounded linear operator and compute its norm Weessentially emphasize the fine spectrum of the operator119860(119903 119904 119905) ℓ119901 rarr ℓ119901 in the case 0 lt 119901 ⩽ 1

Theorem 4 The operator 119860(119903 119904 119905) ℓ119901 rarr ℓ119901 is a boundedlinear operator and

119860 (119903 119904 119905)(ℓ119901 ℓ119901)= |119903|119901+ |119904|119901+ |119905|119901 (10)

Proof Since the linearity of the operator 119860(119903 119904 119905) is trivialso it is omitted Let us take 119890(2) isin ℓ119901 Then 119860(119903 119904 119905)119890(2) =(119905 119904 119903 0 ) and observe that

119860 (119903 119904 119905)(ℓ119901 ℓ119901)⩾

10038171003817100381710038171003817119860 (119903 119904 119905) 119890

(2)100381710038171003817100381710038171199011003817100381710038171003817119890(2)1003817100381710038171003817119901

= |119903|119901+ |119904|119901+ |119905|119901

(11)

which gives the fact that

119860 (119903 119904 119905)(ℓ119901 ℓ119901)⩾ |119903|119901+ |119904|119901+ |119905|119901 (12)

Let 119909 = (119909119896) isin ℓ119901 where 0 lt 119901 ⩽ 1Then since (119905119909119896+2) (119903119909119896)and (119904119909119896+1) isin ℓ119901 it is easy to see by triangle inequality that

119860 (119903 119904 119905) 119909119901 =

infin

sum

119896=0

1003816100381610038161003816119903119909119896 + 119904119909119896+1 + 119905119909119896+21003816100381610038161003816119901

⩽

infin

sum

119896=0

10038161003816100381610038161199031199091198961003816100381610038161003816119901+

infin

sum

119896=0

1003816100381610038161003816119904119909119896+11003816100381610038161003816119901+

infin

sum

119896=0

1003816100381610038161003816119905119909119896+21003816100381610038161003816119901

= |119903|119901infin

sum

119896=0

10038161003816100381610038161199091198961003816100381610038161003816119901+ |119904|119901infin

sum

119896=0

1003816100381610038161003816119909119896+11003816100381610038161003816119901+ |119905|119901infin

sum

119896=0

1003816100381610038161003816119909119896+21003816100381610038161003816119901

= |119904|119901119909119901 + |119903|

119901119909119901 + |119905|

119901119909119901

= (|119903|119901+ |119904|119901+ |119905|119901) 119909119901

(13)

which leads us to the result that

119860 (119903 119904 119905)(ℓ119901 ℓ119901)⩽ |119903|119901+ |119904|119901+ |119905|119901 (14)

Therefore by combining the inequalities (12) and (14) we seethat (10) holds which completes the proof

If 119879 ℓ119901 rarr ℓ119901 is a bounded matrix operator with thematrix 119860 then it is known that the adjoint operator 119879lowast ℓlowast

119901rarr ℓlowast

119901is defined by the transpose of thematrix119860The dual

space of ℓ119901 is isomorphic to ℓinfin where 0 lt 119901 lt 1Before giving themain theorem of this section we should

note the following remark In this work here and in whatfollows if 119911 is a complex number then by radic119911 we alwaysmean the square root of 119911 with a nonnegative real part IfRe(radic119911) = 0 then radic119911 represents the square root of 119911 withIm(radic119911) gt 0 The same results are obtained if radic119911 representsthe other square root

4 Abstract and Applied Analysis

Table 1 Subdivisions of spectrum of a linear operator

1 2 3119879minus1

120572exists and is bounded 119879

minus1

120572exists and is unbounded 119879

minus1

120572does not exist

A 119877(120572119868 minus 119879) = 119883 120572 isin 120588(119879119883) mdash 120572 isin 120590119901(119879119883)

120572 isin 120590ap(119879119883)

B 119877(120572119868 minus 119879) = 119883 120572 isin 120588(119879119883)

120572 isin 120590119888(119879119883) 120572 isin 120590119901(119879119883)

120572 isin 120590ap(119879119883) 120572 isin 120590ap(119879119883)

120572 isin 120590120575(119879119883) 120572 isin 120590120575(119879119883)

C 119877(120572119868 minus 119879) =119883

120572 isin 120590119903(119879119883) 120572 isin 120590119903(119879119883) 120572 isin 120590119901(119879119883)

120572 isin 120590120575(119879119883) 120572 isin 120590ap(119879119883) 120572 isin 120590ap(119879119883)

120572 isin 120590120575(119879119883) 120572 isin 120590120575(119879119883)

120572 isin 120590co(119879119883) 120572 isin 120590co(119879119883) 120572 isin 120590co(119879119883)

Theorem 5 Let 119904 be a complex number such that radic1199042 = minus119904and define the set1198631 by

1198631 = 120572 isin C 2 |119903 minus 120572| ⩽1003816100381610038161003816100381610038161003816minus119904 + radic1199042 minus 4119905 (119903 minus 120572)

1003816100381610038161003816100381610038161003816 (15)

Then 120590119888(119860(119903 119904 119905) ℓ119901) sube 1198631

Proof Let119910 = (119910119896) isin ℓinfinThen by solving the equation119860120572(119903119904 119905)lowast119909 = 119910 for 119909 = (119909119896) in terms of 119910 we obtain

1199090 =1199100

119903 minus 120572

1199091 =1199101

119903 minus 120572+

minus1199041199100

(119903 minus 120572)2

1199092 =1199102

119903 minus 120572+

minus1199041199101

(119903 minus 120572)2+[1199042minus 119905 (119903 minus 120572)] 1199100

(119903 minus 120572)3

(16)

(17)

and if we denote 1198861 = 1(119903 minus 120572) 1198862 = minus119904(119903 minus 120572)2 and 1198863 =

(1199042minus 119905(119903 minus 120572))(119903 minus 120572)

3 we have1199090 = 11988611199100

1199091 = 11988611199101 + 11988621199100

1199092 = 11988611199102 + 11988621199101 + 11988631199100

119909119899 = 1198861119910119899 + 1198862119910119899minus1 + sdot sdot sdot + 119886119899+11199100 =

119899

sum

119896=0

119886119899+1minus119896119910119896

(18)

Now we must find 119886119899 We have 119910119899 = 119905119909119899minus2 + 119904119909119899minus1 + (119903 minus 120572)119909119899and if we use relation (18) we have

119910119899 = 119905

119899minus2

sum

119896=0

119886119899minus1minus119896119910119896 + 119904

119899minus1

sum

119896=0

119886119899minus119896119910119896 + (119903 minus 120572)

119899

sum

119896=0

119886119899+1minus119896119910119896

= 1199100 (119905119886119899minus1 + 119904119886119899 + (119903 minus 120572) 119886119899+1)

+ 1199101 (119905119886119899minus2 + 119904119886119899minus1 + (119903 minus 120572) 119886119899) + sdot sdot sdot + 1199101198991198861 (119903 minus 120572)

(19)

This implies that

119905119886119899minus1 + 119904119886119899 + (119903 minus 120572) 119886119899+1 = 0

119905119886119899minus2 + 119904119886119899minus1 + (119903 minus 120572) 119886119899 = 0 1198861 (119903 minus 120572) = 1(20)

In fact this sequence is obtained recursively by letting

1198861 =1

119903 minus 120572 1198862 =

minus119904

(119903 minus 120572)2

119905119886119899minus2 + 119904119886119899minus1 + (119903 minus 120572) 119886119899 = 0 forall119899 ⩾ 3

(21)

The characteristic polynomial of the recurrence relation is (119903minus120572)1205822+ 119904120582 + 119905 = 0 There are two cases

Case 1 If Δ = 1199042 minus 4119905(119903 minus 120572) = 0 whose roots are

1205821 =minus119904 + radicΔ

2 (119903 minus 120572) 1205822 =

minus119904 minus radicΔ

2 (119903 minus 120572) (22)

elementary calculation on recurrent sequence gives that

119886119899 =120582119899

1minus 120582119899

2

radic1199042 minus 4119905 (119903 minus 120572) forall119899 ⩾ 1 (23)

In this case 119909119896 = (1radicΔ)sum119899

119896=0(120582119899+1minus119896

1minus 120582119899+1minus119896

2)119910119896 Assume

that |1205821| lt 1 So we have10038161003816100381610038161003816100381610038161003816100381610038161003816

1 + radic4119905 (119903 minus 120572)

1199042

10038161003816100381610038161003816100381610038161003816100381610038161003816

lt

10038161003816100381610038161003816100381610038161003816

2 (119903 minus 120572)

minus119904

10038161003816100381610038161003816100381610038161003816 (24)

Since |1 minus radic119911| ⩽ |1 + radic119911| for any 119911 isin C we must have10038161003816100381610038161003816100381610038161003816100381610038161003816

1 minus radic4119905 (119903 minus 120572)

1199042

10038161003816100381610038161003816100381610038161003816100381610038161003816

lt

10038161003816100381610038161003816100381610038161003816

2 (119903 minus 120572)

minus119904

10038161003816100381610038161003816100381610038161003816 (25)

It follows that |1205822| lt 1 Now for |1205821| lt 1 we can see that

10038161003816100381610038161199091198991003816100381610038161003816 ⩽

110038161003816100381610038161003816radicΔ10038161003816100381610038161003816

119899

sum

119896=0

10038161003816100381610038161003816120582119899+1minus119896

1

1003816100381610038161003816100381610038161003816100381610038161199101198961003816100381610038161003816 +

119899

sum

119896=0

10038161003816100381610038161003816120582119899+1minus119896

2

1003816100381610038161003816100381610038161003816100381610038161199101198961003816100381610038161003816 (26)

for all 119899 isin N Taking limit on the inequality (26) as 119899 rarr infinwe get

119909infin ⩽1 minus (

100381610038161003816100381612058221003816100381610038161003816 +100381610038161003816100381612058221003816100381610038161003816)

10038161003816100381610038161003816(1 minus

10038161003816100381610038161205822100381610038161003816100381610038161003816100381610038161 minus 1205822

1003816100381610038161003816)radicΔ10038161003816100381610038161003816

10038171003817100381710038171199101003817100381710038171003817infin (27)

Abstract and Applied Analysis 5

Thus for |1205821| lt 1 119860120572(119903 119904 119905)lowast is onto and by Lemma 2 119860120572(119903

119904 119905) has a bounded inverse This means that

120590119888 (119860 (119903 119904 119905) ℓ119901)

sube 120572 isin C 2 |119903 minus 120572| ⩽1003816100381610038161003816100381610038161003816minus119904 + radic1199042 minus 4119905 (119903 minus 120572)

1003816100381610038161003816100381610038161003816

= 1198631

(28)

Case 2 If Δ = 1199042 minus 4119905(119903 minus 120572) = 0 a calculation on recurrentsequence gives that

119886119899 = (2119899

minus119904) [

minus119904

2 (119903 minus 120572)]

119899

forall119899 ⩾ 1 (29)

Now for | minus 119904| lt 2|119903 minus 120572| we can see that

10038161003816100381610038161199091198991003816100381610038161003816 ⩽

119899

sum

119896=0

1003816100381610038161003816119886119899minus1198961199101198961003816100381610038161003816 (30)

for all 119899 isin N Taking limit on the inequality (30) as 119899 rarr infinwe obtain that

119909infin ⩽10038171003817100381710038171199101003817100381710038171003817infin

infin

sum

119896=0

10038161003816100381610038161198861198961003816100381610038161003816 (31)

suminfin

119896=0|119886119896| is convergent since | minus 119904| lt 2|119903 minus 120572| Thus for | minus 119904| lt

2|119903 minus 120572| 119860120572(119903 119904 119905)lowast is onto and by Lemma 2 119860120572(119903 119904 119905) has a

bounded inverse This means that

120590119888 (119860 (119903 119904 119905) ℓ119901) sube 120572 isin C 2 |119903 minus 120572| ⩽ |minus119904| sube 1198631

(32)

Theorem 6 120590119901(119860(119903 119904 119905)lowast ℓlowast

119901) = 0

Proof Consider 119860(119903 119904 119905)lowast119891 = 120572119891 with 119891 = 120579 = (0 0 0 ) inℓlowast

119901= ℓinfin Then by solving the system of linear equations

1199031198910 = 1205721198910

1199041198910 + 1199031198911 = 1205721198911

1199051198910 + 1199041198911 + 1199031198912 = 1205721198912

1199051198911 + 1199041198912 + 1199031198913 = 1205721198913

119905119891119896minus2 + 119904119891119896minus1 + 119903119891119896 = 120572119891119896

(33)

we find that 1198910 = 0 if 120572 = 119903 and 1198911 = 1198912 = sdot sdot sdot = 0 if 1198910 = 0which contradicts119891 = 120579 If 1198911198990 is the first nonzero entry of thesequence119891 = (119891119899) and120572 = 119903 thenwe get 1199051198911198990minus2+1199041198911198990minus1+1199031198911198990 =1205721198911198990

which implies 1198911198990 = 0which contradicts the assumption1198911198990

= 0 Hence the equation119860(119903 119904 119905)lowast119891 = 120572119891 has no solution119891 = 120579

Theorem 7 120590119901(119860(119903 119904 119905) ℓ119901) = 1198632 where 1198632 = 120572 isin C

2|119903 minus 120572| lt | minus 119904 + radic1199042 minus 4119905(119903 minus 120572)|

Proof Let119860(119903 119904 119905)119909 = 120572119909 for 120579 = 119909 isin ℓ119901Then by solving thesystem of linear equations

1199031199090 + 1199041199091 + 1199051199092 = 1205721199090

1199031199091 + 1199041199092 + 1199051199093 = 1205721199091

1199031199092 + 1199041199093 + 1199051199094 = 1205721199092

119903119909119896minus2 + 119904119909119896minus1 + 119905119909119896 = 120572119909119896

(34)

and we have

1199092 =minus119904

1199051199091 minus

119903 minus 120572

1199051199090

1199093 =1199042minus 119905 (119903 minus 120572)

11990521199091 +

119904 (119903 minus 120572)

11990521199090

119909119899 =119886119899(119903 minus 120572)

119899

119905119899minus11199091 minus

119886119899minus1(119903 minus 120572)119899

119905119899minus11199090 forall119899 ⩾ 2

(35)

Assume that 120572 isin 1198632 Then we choose 1199090 = 1 and 1199091 = 2(119903 minus120572)(minus119904+radic1199042 minus 4119905(119903 minus 120572)) We show that 119909119899 = 119909

119899

1for all 119899 ⩾ 2

Since 1205821 1205822 are roots of the characteristic equation (119903minus120572)1205822+

119904120582 + 119905 = 0 we must have

12058211205822 =119905

119903 minus 120572 1205821 minus 1205822 =

radicΔ

119903 minus 120572 (36)

Combining the fact 1199091 = 11205821 with relation (35) we can seethat

119909119899 =119886119899(119903 minus 120572)

119899

119905119899minus11199091 minus

119886119899minus1(119903 minus 120572)119899

119905119899minus11199090

= (119903 minus 120572

119905)

119899minus1

(119903 minus 120572) (minus119886119899minus11199090 + 1198861198991199091)

=1

(12058211205822)119899minus1

119903 minus 120572

radicΔ(minus120582119899minus1

1+ 120582119899minus1

2+ 120582119899minus1

1minus 120582119899

2120582minus1

1)

=1

120582119899minus11120582119899minus12

(1

1205821 minus 1205822)120582119899minus1

2(1205821 minus 1205822

1205821)

=1

1205821198991

= 119909119899

1

(37)

The same result may be obtained in case Δ = 0 Now 119909 =(119909119896) isin ℓ119901 since |1199091| lt 1 This shows that 1198632 sube 120590119901(119860(119903

119904 119905) ℓ119901)

6 Abstract and Applied Analysis

Now we assume that 120572 notin 1198632 that is |1205821| ⩽ 1 We mustshow that 120572 notin 120590119901(119860(119903 119904 119905) ℓ119901) Therefore we obtain from therelation (35) that

119909119899+1

119909119899= (

119903 minus 120572

119905)119886119899minus1

119886119899minus2(minus1199090 + (119886119899119886119899minus1) 1199091

minus1199090 + (119886119899minus1119886119899minus2) 1199091) (38)

Now we examine three cases

Case 1 (|1205822| lt |1205821| ⩽ 1) In this case we have 1199042 = 4119905(119903minus120572) and

119886119899

119886119899minus1=120582119899+1

1minus 120582119899+1

2

1205821198991minus 1205821198992

=1205821 [1 minus (12058221205821)

119899+1]

[1 minus (12058221205821)119899] (39)

Then we have

lim119899rarrinfin

10038161003816100381610038161003816100381610038161003816

119886119899

119886119899minus1

10038161003816100381610038161003816100381610038161003816

119901

= lim119899rarrinfin

10038161003816100381610038161003816100381610038161003816

119886119899minus1

119886119899minus2

10038161003816100381610038161003816100381610038161003816

119901

= lim119899rarrinfin

100381610038161003816100381612058211003816100381610038161003816119901100381610038161003816100381610038161 minus (12058221205821)

119899+110038161003816100381610038161003816

119901

100381610038161003816100381610038161 minus (12058221205821)

11989910038161003816100381610038161003816

119901=100381610038161003816100381612058211003816100381610038161003816119901

(40)

Now if minus1199090 + 12058211199091 = 0 then we have (119909119899) = (1199090120582119899

1) which

is not in ℓ119901 Otherwise

lim119899rarrinfin

10038161003816100381610038161003816100381610038161003816

119909119899+1

119909119899

10038161003816100381610038161003816100381610038161003816

119901

=1

10038161003816100381610038161205821100381610038161003816100381611990110038161003816100381610038161205822

1003816100381610038161003816119901

100381610038161003816100381612058211003816100381610038161003816119901=

1

100381610038161003816100381612058221003816100381610038161003816119901gt 1 (41)

Case 2 (|1205822| = |1205821| lt 1) In this case we have 1199042 = 4119905(119903 minus 120572)and using the formula

119886119899 = (2119899

minus119904) [

minus119904

2 (119903 minus 120572)]

119899

forall119899 ⩾ 1 (42)

we obtain that

lim119899rarrinfin

10038161003816100381610038161003816100381610038161003816

119886119899

119886119899minus1

10038161003816100381610038161003816100381610038161003816

119901

=

10038161003816100381610038161003816100381610038161003816

minus119904

2 (119903 minus 120572)

10038161003816100381610038161003816100381610038161003816

119901

=100381610038161003816100381612058211003816100381610038161003816119901 (43)

which leads to

lim119899rarrinfin

10038161003816100381610038161003816100381610038161003816

119909119899+1

119909119899

10038161003816100381610038161003816100381610038161003816

119901

=1

10038161003816100381610038161205821100381610038161003816100381611990110038161003816100381610038161205822

1003816100381610038161003816119901

100381610038161003816100381612058211003816100381610038161003816119901=

1

100381610038161003816100381612058221003816100381610038161003816119901gt 1 (44)

Case 3 (|1205822| = |1205821| = 1) In this case we have 1199042 = 4119905(119903minus120572) andso we have | minus 119904(2119905)| = 1 Assume that 120572 isin 120590119901(119860(119903 119904 119905) ℓ119901)This implies that 119909 isin ℓ119901 and 119909 = 120579 Thus we again derive (35)

119909119899 = (minus119904

2119905)

119899minus1

[minus (119899 minus 1)minus119904

21199051199090 + 1198991199091] (45)

Since lim119899rarrinfin119909119899 = 0 we must have 1199090 = 1199091 = 0 But thisimplies that 119909 = 120579 a contradiction which means that 120572 notin120590119901(119860(119903 119904 119905) ℓ119901) Thus 120590119901(119860(119903 119904 119905) ℓ119901) sube 1198632 This completesthe proof

Theorem 8 120590119903(119860(119903 119904 119905) ℓ119901) = 0

Proof By Proposition 1 120590119903(119860(119903 119904 119905) ℓ119901) = 120590119901(119860(119903 119904 119905)lowast ℓlowast

119901)

120590119901(119860(119903 119904 119905) ℓ119901) Since byTheorem 6

120590119901 (119860(119903 119904 119905)lowast ℓlowast

119901) = 0 120590119903 (119860 (119903 119904 119905) ℓ119901) = 0

(46)

This completes the proof

Theorem 9 Let 119904 be a complex number such that radic1199042 = minus119904and define the set1198631 by

1198631 = 120572 isin C 2 |119903 minus 120572| ⩽1003816100381610038161003816100381610038161003816minus119904 + radic1199042 minus 4119905 (119903 minus 120572)

1003816100381610038161003816100381610038161003816

(47)

Then 120590(119860(119903 119904 119905) ℓ119901) = 1198631

Proof ByTheorem 7 we get

120572 isin C 2 |119903 minus 120572| lt1003816100381610038161003816100381610038161003816minus119904 + radic1199042 minus 4119905 (119903 minus 120572)

1003816100381610038161003816100381610038161003816

sube 120590 (119860 (119903 119904 119905) ℓ119901)

(48)

Since the spectrum of any bounded operator is closed wehave

120572 isin C 2 |119903 minus 120572| ⩽1003816100381610038161003816100381610038161003816minus119904 + radic1199042 minus 4119905 (119903 minus 120572)

1003816100381610038161003816100381610038161003816

sube 120590 (119860 (119903 119904 119905) ℓ119901)

(49)

and again fromTheorems 5 7 and 8

120590 (119860 (119903 119904 119905) ℓ119901) sube 120572 isin C 2 |119903 minus 120572|

⩽1003816100381610038161003816100381610038161003816minus119904 + radic1199042 minus 4119905 (119903 minus 120572)

1003816100381610038161003816100381610038161003816

(50)

Combining (49) and (50) we obtain that 120590(119860(119903 119904 119905) ℓ119901) =1198631 where1198631 is defined by (47)

Theorem 10 120590119888(119860(119903 119904 119905) ℓ119901) = 1198633 where 1198633 = 120572 isin C

2|119903 minus 120572| = | minus 119904 + radic1199042 minus 4119905(119903 minus 120572)|

Proof Because the parts 120590119888(119860(119903 119904 119905) ℓ119901) 120590119903(119860(119903 119904 119905) ℓ119901)and 120590119901(119860(119903 119904 119905) ℓ119901) are pairwise disjoint sets and the unionof these sets is 120590(119860(119903 119904 119905) ℓ119901) the proof immediately followsfromTheorems 7 8 and 9

Theorem 11 If 120572 isin 1198632 120572 isin 120590(119860(119903 119904 119905) ℓ119901)1198683

Proof From Theorem 7 120572 isin 120590119901(119860(119903 119904 119905) ℓ119901) Thus (119860(119903119904 119905) minus 120572119868)

minus1 does not exist By Theorem 6119860(119903 119904 119905)lowast minus 120572119868 isone to one so 119860(119903 119904 119905) minus 120572119868 has a dense range in ℓ119901 byLemma 3

Theorem 12 The following statements hold

(i) 120590ap(119860(119903 119904 119905) ℓ119901) = 1198631(ii) 120590120575(119860(119903 119904 119905) ℓ119901) = 1198633(iii) 120590co(119860(119903 119904 119905) ℓ119901) = 0

Abstract and Applied Analysis 7

Proof (i) Since from Table 1

120590ap (119860 (119903 119904 119905) ℓ119901)=120590 (119860 (119903 119904 119905) ℓ119901)120590 (119860 (119903 119904 119905) ℓ119901) 1198681198681198681

(51)

we have byTheorem 8

120590 (119860 (119903 119904 119905) ℓ119901) 1198681198681198681 = 120590 (119860 (119903 119904 119905) ℓ119901) 1198681198681198682 = 0 (52)

Hence

120590ap (119860 (119903 119904 119905) ℓ119901) = 1198631 (53)

(ii) Since the following equality

120590120575 (119860 (119903 119904 119905) ℓ119901) = 120590 (119860 (119903 119904 119905) ℓ119901) 120590 (119860 (119903 119904 119905) ℓ119901) 1198683

(54)

holds from Table 1 we derive by Theorems 8 and 11 that120590120575(119860(119903 119904 119905) ℓ119901) = 1198632

(iii) From Table 1 we have

120590co (119860 (119903 119904 119905) ℓ119901)

= 120590 (119860 (119903 119904 119905) ℓ119901) 1198681198681198681 cup 120590 (119860 (119903 119904 119905) ℓ119901) 1198681198681198682

cup 120590 (119860 (119903 119904 119905) 1198880) 1198681198681198683

(55)

ByTheorem 6 it is immediate that 120590co(119860(119903 119904 119905) ℓ119901) = 0

4 Fine Spectra of Upper TriangularTriple-Band Matrices over the SequenceSpace ℓ119901 (1 lt 119901 lt infin)

In the present section we determine the fine spectrum ofthe operator 119860(119903 119904 119905) ℓ119901 rarr ℓ119901 in case 1 ⩽ 119901 lt infinWe quote some lemmas which are needed in proving thetheorems given in Section 4

Lemma 13 (see [17 p 253 Theorem 3416]) The matrix 119860 =(119886119899119896) gives rise to a bounded linear operator 119879 isin 119861(ℓ1) from ℓ1to itself if and only if the supremum of ℓ1 norms of the columnsof 119860 is bounded

Lemma 14 (see [17 p 245 Theorem 343]) The matrix 119860 =(119886119899119896) gives rise to a bounded linear operator 119879 isin 119861(ℓinfin) fromℓinfin to itself if and only if the supremum of ℓ1 norms of the rowsof 119860 is bounded

Lemma 15 (see [17 p 254 Theorem 3418]) Let 1 lt 119901 lt infinand 119860 isin (ℓinfin ℓinfin) cap (ℓ1 ℓ1) Then 119860 isin (ℓ119901 ℓ119901)

Theorem 16 The operator 119860(119903 119904 119905) ℓ119901 rarr ℓ119901 is a boundedlinear operator and

(|119903|119901+ |119904|119901+ |119905|119901)1119901⩽ 119860 (119903 119904 119905)(ℓ119901 ℓ119901)

⩽ |119903| + |119904| + |119905|

(56)

Proof Since the linearity of the operator119860(119903 119904 119905) is not hardwe omit the details Now we prove that (56) holds for theoperator 119860(119903 119904 119905) on the space ℓ119901 It is trivial that 119860(119903 119904119905)119890(2)= (119905 119904 119903 0 ) for 119890(2) isin ℓ119901 Therefore we have

119860 (119903 119904 119905)(ℓ119901 ℓ119901)⩾

10038171003817100381710038171003817119860 (119903 119904 119905) 119890

(2)100381710038171003817100381710038171199011003817100381710038171003817119890(2)1003817100381710038171003817119901

= (|119903|119901+ |119904|119901+ |119905|119901)1119901

(57)

which implies that

119860 (119903 119904 119905)(ℓ119901 ℓ119901)⩾ (|119903|119901+ |s|119901 + |119905|119901)1119901 (58)

Let 119909 = (119909119896) isin ℓ119901 where 1 lt 119901 lt infin Then since (119905119909119896+2)(119903119909119896) and (119904119909119896+1) isin ℓ119901 it is easy to see by Minkowskirsquos ine-quality that

119860 (119903 119904 119905) 119909119901

= (

infin

sum

119896=0

1003816100381610038161003816119903119909119896 + 119904119909119896+1 + 119905119909119896+21003816100381610038161003816119901)

1119901

⩽ (

infin

sum

119896=0

10038161003816100381610038161199031199091198961003816100381610038161003816119901)

1119901

+ (

infin

sum

119896=0

1003816100381610038161003816119904119909119896+11003816100381610038161003816119901)

1119901

+ (

infin

sum

119896=0

1003816100381610038161003816119905119909119896+21003816100381610038161003816119901)

1119901

= |119903| (

infin

sum

119896=0

10038161003816100381610038161199091198961003816100381610038161003816119901)

1119901

+ |119904| (

infin

sum

119896=0

1003816100381610038161003816119909119896+11003816100381610038161003816119901)

1119901

+ |119905| (

infin

sum

119896=0

1003816100381610038161003816119909119896+21003816100381610038161003816119901)

1119901

= |119904| 119909119901 + |119903| 119909119901 + |119905| 119909119901

= (|119903| + |119904| + |119905|) 119909119901

(59)

which leads us to the result that

(|119903|119901+ |119904|119901+ |119905|119901)1119901⩽ 119860 (119903 119904 119905)(ℓ119901 ℓ119901)

⩽ |119903| + |119904| + |119905|

(60)

Therefore by combining the inequalities in (58) and (59) wehave (56) as desired

If 119879 ℓ119901 rarr ℓ119901 is a bounded matrix operator with thematrix 119860 then it is known that the adjoint operator 119879lowast ℓlowast

119901rarr ℓlowast

119901is defined by the transpose of thematrix119860The dual

space of ℓ119901 is isomorphic to ℓ119902 where 1 lt 119901 lt infin

Theorem 17 Let 119904 be a complex number such that radic1199042 = minus119904and define the set1198631 by

1198631 = 120572 isin C 2 |119903 minus 120572| ⩽1003816100381610038161003816100381610038161003816minus119904 + radic1199042 minus 4119905 (119903 minus 120572)

1003816100381610038161003816100381610038161003816

(61)

Then 120590119888(119860(119903 119904 119905) ℓ119901) sube 1198631

8 Abstract and Applied Analysis

Proof We will show that 119860120572(119903 119904 119905)lowast is onto for 2|119903 minus 120572| gt

| minus 119904 + radic1199042 minus 4119905(119903 minus 120572)| Thus for every 119910 isin ℓ119902 we find 119909 isinℓ119902 119860120572(119903 119904 119905)

lowast is a triangle so it has an inverse Also equation119860120572(119903 119904 119905)

lowast119909 = 119910 gives [119860120572(119903 119904 119905)

lowast]minus1119910 = 119909 It is sufficient

to show that [119860120572(119903 119904 119905)lowast]minus1isin (ℓ119902 ℓ119902) We calculate that

119860 = (119886119899119896) = [119860120572(119903 119904 119905)lowast]minus1 as follows

119860 = (119886119899119896) =

[[[[

[

1198861 0 0

1198862 1198861 0

1198863 1198862 1198861

]]]]

]

(62)

where

1198861 =1

119903 minus 120572

1198862 =minus119904

(119903 minus 120572)2

1198863 =1199042minus 119905 (119903 minus 120572)

(119903 minus 120572)3

(63)

It is known that fromTheorem 5

119886119899 =120582119899

1minus 120582119899

2

radic1199042 minus 4119905 (119903 minus 120572) forall119899 ⩾ 1

where 1205821 =minus119904 + radicΔ

119903 minus 120572 1205822 =

minus119904 minus radicΔ

119903 minus 120572

(64)

Now we show that [119860120572(119903 119904 119905)lowast]minus1isin (ℓ1 ℓ1) for |1205821| lt 1

By Theorem 5 we know that if |1205821| lt 1 |1205822| lt 1 We assumethat 1199042 minus 4119905(119903 minus 120572) = 0 and |1205821| lt 1

10038171003817100381710038171003817[119860120572(119903 119904 119905)

lowast]minus110038171003817100381710038171003817(ℓ1 ℓ1)

= sup119899

infin

sum

119896=119899

10038161003816100381610038161198861198961003816100381610038161003816 =

infin

sum

119896=1

10038161003816100381610038161198861198961003816100381610038161003816

⩽1

|Δ|(

infin

sum

119896=1

100381610038161003816100381612058211003816100381610038161003816119896+

infin

sum

119896=1

100381610038161003816100381612058221003816100381610038161003816119896) lt infin

(65)

since |1205821| lt 1 and |1205822| lt 1 This shows that (119860120572(119903 119904 119905)lowast]minus1isin

(ℓ1 ℓ1) Similarly we can show that [119860120572(119903 119904 119905)lowast]minus1isin (ℓinfin

ℓinfin)Now assume that 1199042 minus 4119905(119903 minus 120572) = 0 Then

119886119899 = (2119899

minus119904) [

minus119904

2 (119903 minus 120572)]

119899

(66)

and simple calculation gives that (119886119899) isin ℓ119902 if and only if |minus119904| lt2|119903 minus 120572|

[(119860 (119903 119904 119905) minus 120572119868)lowast]minus1isin (ℓ119902 ℓ119902)

for 120572 isin C with 2 |119903 minus 120572| gt1003816100381610038161003816100381610038161003816minus119904 + radic1199042 minus 4119905 (119903 minus 120572)

1003816100381610038161003816100381610038161003816

(67)

Hence 119860120572(119903 119904 119905)lowast is onto By Lemma 2 119860120572(119903 119904 119905) has a

bounded inverse This means that 120590119888(119860(119903 119904 119905) ℓ119901) sube 1198631where1198631 is defined by (61)

Theorem 18 120590119901(119860(119903 119904 119905)lowast ℓlowast

119901) = 0

Proof Let 119860(119903 119904 119905)lowast119891 = 120572119891 with 119891 = 120579 = (0 0 0 ) in ℓlowast119901=

ℓ119902 Then by solving the system of linear equations

1199031198910 = 1205721198910

1199041198910 + 1199031198911 = 1205721198911

1199051198910 + 1199041198911 + 1199031198912 = 1205721198912

1199051198911 + 1199041198912 + 1199031198913 = 1205721198913

119905119891119896minus2 + 119904119891119896minus1 + 119903119891119896 = 120572119891119896

(68)

we find that 1198910 = 0 if 120572 = 119903 and 1198911 = 1198912 = sdot sdot sdot = 0 if 1198910 = 0which contradicts119891 = 120579 If 1198911198990 is the first nonzero entry of thesequence 119891 = (119891119899) and 120572 = 119903 then we get 1199051198911198990minus2 + 1199041198911198990minus1 +1199031198911198990

= 1205721198911198990which implies 1198911198990 = 0 which contradicts the

assumption1198911198990 = 0 Hence the equation119860(119903 119904 119905)lowast119891 = 120572119891 has

no solution 119891 = 120579

In the case 1 lt 119901 lt infin since the proof of the theoremsin Section 4 determining the spectrum and fine spectrumof the matrix operator 119860(119903 119904 119905) on the sequence space ℓ119901is similar to the case 0 lt 119901 ⩽ 1 to avoid the repetitionof similar statements we give the results by the followingtheorem without proof

Theorem 19 The following statements hold(i) 120590(119860(119903 119904 119905) ℓ119901) = 1198631(ii) 120590119903(119860(119903 119904 119905) ℓ119901) = 0(iii) 120590119901(119860(119903 119904 119905) ℓ119901) = 1198632(iv) 120590119888(119860(119903 119904 119905) ℓ119901) = 1198633(v) 120590ap(119860(119903 119904 119905) ℓ119901) = 1198631(vi) 120590co(119860(119903 119904 119905) ℓ119901) = 0(vii) 120590120575(119860(119903 119904 119905) ℓ119901) = 1198633

5 Some Applications

In this section we give two theorems related to Toeplitz ma-trix

Theorem 20 Let 119875 be a polynomial that corresponds to the 119899-tuple 119886 and let 1199111 1199112 1199113 119911119899minus1 also be the roots of 119875 Define119879 as a Toeplitz matrix associated with 119875 that is

119879 =

[[[[

[

1198860 1198861 1198862 119886119899 0 0 0

0 1198860 1198861 1198862 119886119899 0 0

0 0 1198860 1198861 1198862 119886119899 0

]]]]

]

(69)

Abstract and Applied Analysis 9

The resolvent operator 119879 over ℓ119901 with 1 lt 119901 lt infin where thedomain of the resolvent operator is the whole space ℓ119901 exists ifand if only all the roots of the polynomial are outside the unitdisc 119911 isin C |119911| ⩽ 1 That is 119879minus1 isin (ℓ119901 ℓ119901) if and if only|119911119894| gt 1 1 ⩽ 119894 ⩽ 119899 minus 1 In this case the resolvent operator isrepresented by

119879minus1=

1

119886119899minus1119860minus1(minus1199111 1) 119860

minus1(minus1199112 1) sdot sdot sdot 119860

minus1(minus119911119899minus1 1)

where

119860minus1(minus119911119894 1) =minus

[[[[[[[[[[[[

[

1119911119894 11199112

11989411199113

11989411199114

11989411199115

119894

0 1119911119894 11199112

11989411199113

11989411199114

119894

0 0 1119911119894 11199112

11989411199113

119894

0 0 0 1119911119894 11199112

119894

0 0 0 0 1119911119894

]]]]]]]]]]]]

]

(70)

Proof Suppose all the roots of the polynomial 119875(119911) = 1198860 +1198861119911 + sdot sdot sdot + 119886119899minus1119911

119899minus1= 119886119899(119911 minus 1199111)(119911 minus 1199112) sdot sdot sdot (119911 minus 119911119899minus1) are

outside the unit disc The Toeplitz matrix associated with 119875can be written as the product

119879 = 119886119899119860 (minus1199111 1) 119860 (minus1199112 1) sdot sdot sdot 119860 (minus119911119899minus1 1) (71)

Since multiplication of upper triangular Toeplitz matrices iscommutative we can see that

119879minus1=

1

119886119899minus1119860minus1(minus1199111 1) 119860

minus1(minus1199112 1) sdot sdot sdot 119860

minus1(minus119911119899minus1 1)

(72)

is left inverse of 119879 Since all roots are outside the unit discthen

10038171003817100381710038171003817119879minus1(minus119911119894 1)

10038171003817100381710038171003817(ℓinfin ℓinfin)= sup119899

infin

sum

119896=119899

1

10038161003816100381610038161199111198941003816100381610038161003816119896+1minus119899

=

infin

sum

119896=1

1

10038161003816100381610038161199111198941003816100381610038161003816119896lt infin

(73)

Therefore each 119879minus1(minus119911119894 1) isin (ℓinfin ℓinfin) for 1 ⩽ 119894 ⩽ 119899 minus 1Similarly we can say that 119879minus1(minus119911119894 1) isin (ℓ1 ℓ1) So we have119879minus1(minus119911119894 1) isin (ℓ119901 ℓ119901)

Theorem 21 The resolvent operator of 119860(119903 119904 119905) over ℓ119901 with1 lt 119901 lt infin where the domain of the resolvent operator is thespace ℓ119901 exists if and only if 2|119903| gt | minus 119904 + radic1199042 minus 4119905119903| In this

case the resolvent operator is represented by the infinite bandedToeplitz matrix

119864 (119903 119904 119905)

=1

119905

[[[[[[[[[[

[

1199061 1199062

11199063

11199064

1

0 1199061 1199062

11199063

1

0 0 1199061 1199062

1

0 0 0 1199061

]]]]]]]]]]

]

[[[[[[[[[[

[

1199062 1199062

21199063

21199064

2

0 1199062 1199062

21199063

2

0 0 1199062 1199062

2

0 0 0 1199062

]]]]]]]]]]

]

where 1199061 =minus119904 + radic1199042 minus 4119905119903

2119903 1199062 =

minus119904 minus radic1199042 minus 4119905119903

2119903

(74)

Proof By Theorem 20 we can see that 119864(119903 119904 119905) is inverseof the matrix of 119860(119903 119904 119905) But this is not enough to say itis resolvent operator By Lemmas 13 14 and 15 119864(119903 119904 119905) isin(ℓ119901 ℓ119901) when 2|119903| gt | minus 119904 + radic1199042 minus 4119905119903| That is for 2|119903| gt| minus 119904 + radic1199042 minus 4119905119903| 119864(119903 119904 119905) is a resolvent operator

Acknowledgments

The authors would like to thank the referee for pointingout some mistakes and misprints in the earlier version ofthis paper They would like to express their pleasure toMs Medine Yesilkayagil Department of Mathematics UsakUniversity 1 Eylul Campus Usak Turkey for many helpfulsuggestions and interesting comments on the revised form ofthe paper

References

[1] M Gonzalez ldquoThe fine spectrum of the Cesaro operator in ℓ119901(1 lt 119901 lt infin)rdquo Archiv der Mathematik vol 44 no 4 pp 355ndash358 1985

[2] P J CartlidgeWeighted meanmatrices as operators on ℓ119901 [PhDdissertation] Indiana University 1978

[3] J I Okutoyi ldquoOn the spectrum of C1 as an operator on 1198871199070rdquoAustralian Mathematical Society A vol 48 no 1 pp 79ndash861990

[4] J I Okutoyi ldquoOn the spectrum of C1 as an operator on bvrdquoCommunications A vol 41 no 1-2 pp 197ndash207 1992

[5] M Yıldırım ldquoOn the spectrum of the Rhaly operators on ℓ119901rdquoIndian Journal of Pure and Applied Mathematics vol 32 no 2pp 191ndash198 2001

[6] B Altay and F Basar ldquoOn the fine spectrum of the differenceoperator Δ on 1198880 and 119888rdquo Information Sciences vol 168 no 1ndash4pp 217ndash224 2004

[7] B Altay and F Basar ldquoOn the fine spectrum of the generalizeddifference operator 119861(119903 119904) over the sequence spaces 1198880 and119888rdquo International Journal of Mathematics and MathematicalSciences vol 2005 no 18 pp 3005ndash3013 2005

[8] A M Akhmedov and F Basar ldquoOn the fine spectra of the dif-ference operator Δ over the sequence space ℓ119901 (1 le 119901 lt infin)rdquoDemonstratio Mathematica vol 39 no 3 pp 585ndash595 2006

10 Abstract and Applied Analysis

[9] AMAkhmedov andF Basar ldquoThefine spectra of the differenceoperator Δ over the sequence space 119887119907119901 (1 le 119901 lt infin)rdquo ActaMathematica Sinica vol 23 no 10 pp 1757ndash1768 2007

[10] F Basar and B Altay ldquoOn the space of sequences of p-boundedvariation and related matrix mappingsrdquo Ukrainian Mathemati-cal Journal vol 55 no 1 pp 136ndash147 2003

[11] H Furkan H Bilgic and F Basar ldquoOn the fine spectrum of theoperator 119861(119903 119904 119905) over the sequence spaces ℓ119901 and 119887119907119901(1 lt 119901 ltinfin)rdquo Computers ampMathematics with Applications vol 60 no 7pp 2141ndash2152 2010

[12] V Karakaya and M Altun ldquoFine spectra of upper triangulardouble-band matricesrdquo Journal of Computational and AppliedMathematics vol 234 no 5 pp 1387ndash1394 2010

[13] A Karaisa ldquoFine spectra of upper triangular double-bandmatrices over the sequence space ℓ119901 (1 lt 119901 lt infin)rdquo DiscreteDynamics in Nature and Society vol 2012 Article ID 381069 19pages 2012

[14] E Kreyszig Introductory Functional Analysis with ApplicationsJohn Wiley amp Sons New York NY USA 1978

[15] J Appell E Pascale and A Vignoli Nonlinear Spectral Theoryvol 10 of de Gruyter Series in Nonlinear Analysis and Applica-tions Walter de Gruyter Berlin Germany 2004

[16] S Goldberg Unbounded Linear Operators Dover PublicationsNew York NY USA 1985

[17] B Choudhary and S Nanda Functional Analysis with Applica-tions John Wiley amp Sons New York NY USA 1989

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

4 Abstract and Applied Analysis

Table 1 Subdivisions of spectrum of a linear operator

1 2 3119879minus1

120572exists and is bounded 119879

minus1

120572exists and is unbounded 119879

minus1

120572does not exist

A 119877(120572119868 minus 119879) = 119883 120572 isin 120588(119879119883) mdash 120572 isin 120590119901(119879119883)

120572 isin 120590ap(119879119883)

B 119877(120572119868 minus 119879) = 119883 120572 isin 120588(119879119883)

120572 isin 120590119888(119879119883) 120572 isin 120590119901(119879119883)

120572 isin 120590ap(119879119883) 120572 isin 120590ap(119879119883)

120572 isin 120590120575(119879119883) 120572 isin 120590120575(119879119883)

C 119877(120572119868 minus 119879) =119883

120572 isin 120590119903(119879119883) 120572 isin 120590119903(119879119883) 120572 isin 120590119901(119879119883)

120572 isin 120590120575(119879119883) 120572 isin 120590ap(119879119883) 120572 isin 120590ap(119879119883)

120572 isin 120590120575(119879119883) 120572 isin 120590120575(119879119883)

120572 isin 120590co(119879119883) 120572 isin 120590co(119879119883) 120572 isin 120590co(119879119883)

Theorem 5 Let 119904 be a complex number such that radic1199042 = minus119904and define the set1198631 by

1198631 = 120572 isin C 2 |119903 minus 120572| ⩽1003816100381610038161003816100381610038161003816minus119904 + radic1199042 minus 4119905 (119903 minus 120572)

1003816100381610038161003816100381610038161003816 (15)

Then 120590119888(119860(119903 119904 119905) ℓ119901) sube 1198631

Proof Let119910 = (119910119896) isin ℓinfinThen by solving the equation119860120572(119903119904 119905)lowast119909 = 119910 for 119909 = (119909119896) in terms of 119910 we obtain

1199090 =1199100

119903 minus 120572

1199091 =1199101

119903 minus 120572+

minus1199041199100

(119903 minus 120572)2

1199092 =1199102

119903 minus 120572+

minus1199041199101

(119903 minus 120572)2+[1199042minus 119905 (119903 minus 120572)] 1199100

(119903 minus 120572)3

(16)

(17)

and if we denote 1198861 = 1(119903 minus 120572) 1198862 = minus119904(119903 minus 120572)2 and 1198863 =

(1199042minus 119905(119903 minus 120572))(119903 minus 120572)

3 we have1199090 = 11988611199100

1199091 = 11988611199101 + 11988621199100

1199092 = 11988611199102 + 11988621199101 + 11988631199100

119909119899 = 1198861119910119899 + 1198862119910119899minus1 + sdot sdot sdot + 119886119899+11199100 =

119899

sum

119896=0

119886119899+1minus119896119910119896

(18)

Now we must find 119886119899 We have 119910119899 = 119905119909119899minus2 + 119904119909119899minus1 + (119903 minus 120572)119909119899and if we use relation (18) we have

119910119899 = 119905

119899minus2

sum

119896=0

119886119899minus1minus119896119910119896 + 119904

119899minus1

sum

119896=0

119886119899minus119896119910119896 + (119903 minus 120572)

119899

sum

119896=0

119886119899+1minus119896119910119896

= 1199100 (119905119886119899minus1 + 119904119886119899 + (119903 minus 120572) 119886119899+1)

+ 1199101 (119905119886119899minus2 + 119904119886119899minus1 + (119903 minus 120572) 119886119899) + sdot sdot sdot + 1199101198991198861 (119903 minus 120572)

(19)

This implies that

119905119886119899minus1 + 119904119886119899 + (119903 minus 120572) 119886119899+1 = 0

119905119886119899minus2 + 119904119886119899minus1 + (119903 minus 120572) 119886119899 = 0 1198861 (119903 minus 120572) = 1(20)

In fact this sequence is obtained recursively by letting

1198861 =1

119903 minus 120572 1198862 =

minus119904

(119903 minus 120572)2

119905119886119899minus2 + 119904119886119899minus1 + (119903 minus 120572) 119886119899 = 0 forall119899 ⩾ 3

(21)

The characteristic polynomial of the recurrence relation is (119903minus120572)1205822+ 119904120582 + 119905 = 0 There are two cases

Case 1 If Δ = 1199042 minus 4119905(119903 minus 120572) = 0 whose roots are

1205821 =minus119904 + radicΔ

2 (119903 minus 120572) 1205822 =

minus119904 minus radicΔ

2 (119903 minus 120572) (22)

elementary calculation on recurrent sequence gives that

119886119899 =120582119899

1minus 120582119899

2

radic1199042 minus 4119905 (119903 minus 120572) forall119899 ⩾ 1 (23)

In this case 119909119896 = (1radicΔ)sum119899

119896=0(120582119899+1minus119896

1minus 120582119899+1minus119896

2)119910119896 Assume

that |1205821| lt 1 So we have10038161003816100381610038161003816100381610038161003816100381610038161003816

1 + radic4119905 (119903 minus 120572)

1199042

10038161003816100381610038161003816100381610038161003816100381610038161003816

lt

10038161003816100381610038161003816100381610038161003816

2 (119903 minus 120572)

minus119904

10038161003816100381610038161003816100381610038161003816 (24)

Since |1 minus radic119911| ⩽ |1 + radic119911| for any 119911 isin C we must have10038161003816100381610038161003816100381610038161003816100381610038161003816

1 minus radic4119905 (119903 minus 120572)

1199042

10038161003816100381610038161003816100381610038161003816100381610038161003816

lt

10038161003816100381610038161003816100381610038161003816

2 (119903 minus 120572)

minus119904

10038161003816100381610038161003816100381610038161003816 (25)

It follows that |1205822| lt 1 Now for |1205821| lt 1 we can see that

10038161003816100381610038161199091198991003816100381610038161003816 ⩽

110038161003816100381610038161003816radicΔ10038161003816100381610038161003816

119899

sum

119896=0

10038161003816100381610038161003816120582119899+1minus119896

1

1003816100381610038161003816100381610038161003816100381610038161199101198961003816100381610038161003816 +

119899

sum

119896=0

10038161003816100381610038161003816120582119899+1minus119896

2

1003816100381610038161003816100381610038161003816100381610038161199101198961003816100381610038161003816 (26)

for all 119899 isin N Taking limit on the inequality (26) as 119899 rarr infinwe get

119909infin ⩽1 minus (

100381610038161003816100381612058221003816100381610038161003816 +100381610038161003816100381612058221003816100381610038161003816)

10038161003816100381610038161003816(1 minus

10038161003816100381610038161205822100381610038161003816100381610038161003816100381610038161 minus 1205822

1003816100381610038161003816)radicΔ10038161003816100381610038161003816

10038171003817100381710038171199101003817100381710038171003817infin (27)

Abstract and Applied Analysis 5

Thus for |1205821| lt 1 119860120572(119903 119904 119905)lowast is onto and by Lemma 2 119860120572(119903

119904 119905) has a bounded inverse This means that

120590119888 (119860 (119903 119904 119905) ℓ119901)

sube 120572 isin C 2 |119903 minus 120572| ⩽1003816100381610038161003816100381610038161003816minus119904 + radic1199042 minus 4119905 (119903 minus 120572)

1003816100381610038161003816100381610038161003816

= 1198631

(28)

Case 2 If Δ = 1199042 minus 4119905(119903 minus 120572) = 0 a calculation on recurrentsequence gives that

119886119899 = (2119899

minus119904) [

minus119904

2 (119903 minus 120572)]

119899

forall119899 ⩾ 1 (29)

Now for | minus 119904| lt 2|119903 minus 120572| we can see that

10038161003816100381610038161199091198991003816100381610038161003816 ⩽

119899

sum

119896=0

1003816100381610038161003816119886119899minus1198961199101198961003816100381610038161003816 (30)

for all 119899 isin N Taking limit on the inequality (30) as 119899 rarr infinwe obtain that

119909infin ⩽10038171003817100381710038171199101003817100381710038171003817infin

infin

sum

119896=0

10038161003816100381610038161198861198961003816100381610038161003816 (31)

suminfin

119896=0|119886119896| is convergent since | minus 119904| lt 2|119903 minus 120572| Thus for | minus 119904| lt

2|119903 minus 120572| 119860120572(119903 119904 119905)lowast is onto and by Lemma 2 119860120572(119903 119904 119905) has a

bounded inverse This means that

120590119888 (119860 (119903 119904 119905) ℓ119901) sube 120572 isin C 2 |119903 minus 120572| ⩽ |minus119904| sube 1198631

(32)

Theorem 6 120590119901(119860(119903 119904 119905)lowast ℓlowast

119901) = 0

Proof Consider 119860(119903 119904 119905)lowast119891 = 120572119891 with 119891 = 120579 = (0 0 0 ) inℓlowast

119901= ℓinfin Then by solving the system of linear equations

1199031198910 = 1205721198910

1199041198910 + 1199031198911 = 1205721198911

1199051198910 + 1199041198911 + 1199031198912 = 1205721198912

1199051198911 + 1199041198912 + 1199031198913 = 1205721198913

119905119891119896minus2 + 119904119891119896minus1 + 119903119891119896 = 120572119891119896

(33)

we find that 1198910 = 0 if 120572 = 119903 and 1198911 = 1198912 = sdot sdot sdot = 0 if 1198910 = 0which contradicts119891 = 120579 If 1198911198990 is the first nonzero entry of thesequence119891 = (119891119899) and120572 = 119903 thenwe get 1199051198911198990minus2+1199041198911198990minus1+1199031198911198990 =1205721198911198990

which implies 1198911198990 = 0which contradicts the assumption1198911198990

= 0 Hence the equation119860(119903 119904 119905)lowast119891 = 120572119891 has no solution119891 = 120579

Theorem 7 120590119901(119860(119903 119904 119905) ℓ119901) = 1198632 where 1198632 = 120572 isin C

2|119903 minus 120572| lt | minus 119904 + radic1199042 minus 4119905(119903 minus 120572)|

Proof Let119860(119903 119904 119905)119909 = 120572119909 for 120579 = 119909 isin ℓ119901Then by solving thesystem of linear equations

1199031199090 + 1199041199091 + 1199051199092 = 1205721199090

1199031199091 + 1199041199092 + 1199051199093 = 1205721199091

1199031199092 + 1199041199093 + 1199051199094 = 1205721199092

119903119909119896minus2 + 119904119909119896minus1 + 119905119909119896 = 120572119909119896

(34)

and we have

1199092 =minus119904

1199051199091 minus

119903 minus 120572

1199051199090

1199093 =1199042minus 119905 (119903 minus 120572)

11990521199091 +

119904 (119903 minus 120572)

11990521199090

119909119899 =119886119899(119903 minus 120572)

119899

119905119899minus11199091 minus

119886119899minus1(119903 minus 120572)119899

119905119899minus11199090 forall119899 ⩾ 2

(35)

Assume that 120572 isin 1198632 Then we choose 1199090 = 1 and 1199091 = 2(119903 minus120572)(minus119904+radic1199042 minus 4119905(119903 minus 120572)) We show that 119909119899 = 119909

119899

1for all 119899 ⩾ 2

Since 1205821 1205822 are roots of the characteristic equation (119903minus120572)1205822+

119904120582 + 119905 = 0 we must have

12058211205822 =119905

119903 minus 120572 1205821 minus 1205822 =

radicΔ

119903 minus 120572 (36)

Combining the fact 1199091 = 11205821 with relation (35) we can seethat

119909119899 =119886119899(119903 minus 120572)

119899

119905119899minus11199091 minus

119886119899minus1(119903 minus 120572)119899

119905119899minus11199090

= (119903 minus 120572

119905)

119899minus1

(119903 minus 120572) (minus119886119899minus11199090 + 1198861198991199091)

=1

(12058211205822)119899minus1

119903 minus 120572

radicΔ(minus120582119899minus1

1+ 120582119899minus1

2+ 120582119899minus1

1minus 120582119899

2120582minus1

1)

=1

120582119899minus11120582119899minus12

(1

1205821 minus 1205822)120582119899minus1

2(1205821 minus 1205822

1205821)

=1

1205821198991

= 119909119899

1

(37)

The same result may be obtained in case Δ = 0 Now 119909 =(119909119896) isin ℓ119901 since |1199091| lt 1 This shows that 1198632 sube 120590119901(119860(119903

119904 119905) ℓ119901)

6 Abstract and Applied Analysis

Now we assume that 120572 notin 1198632 that is |1205821| ⩽ 1 We mustshow that 120572 notin 120590119901(119860(119903 119904 119905) ℓ119901) Therefore we obtain from therelation (35) that

119909119899+1

119909119899= (

119903 minus 120572

119905)119886119899minus1

119886119899minus2(minus1199090 + (119886119899119886119899minus1) 1199091

minus1199090 + (119886119899minus1119886119899minus2) 1199091) (38)

Now we examine three cases

Case 1 (|1205822| lt |1205821| ⩽ 1) In this case we have 1199042 = 4119905(119903minus120572) and

119886119899

119886119899minus1=120582119899+1

1minus 120582119899+1

2

1205821198991minus 1205821198992

=1205821 [1 minus (12058221205821)

119899+1]

[1 minus (12058221205821)119899] (39)

Then we have

lim119899rarrinfin

10038161003816100381610038161003816100381610038161003816

119886119899

119886119899minus1

10038161003816100381610038161003816100381610038161003816

119901

= lim119899rarrinfin

10038161003816100381610038161003816100381610038161003816

119886119899minus1

119886119899minus2

10038161003816100381610038161003816100381610038161003816

119901

= lim119899rarrinfin

100381610038161003816100381612058211003816100381610038161003816119901100381610038161003816100381610038161 minus (12058221205821)

119899+110038161003816100381610038161003816

119901

100381610038161003816100381610038161 minus (12058221205821)

11989910038161003816100381610038161003816

119901=100381610038161003816100381612058211003816100381610038161003816119901

(40)

Now if minus1199090 + 12058211199091 = 0 then we have (119909119899) = (1199090120582119899

1) which

is not in ℓ119901 Otherwise

lim119899rarrinfin

10038161003816100381610038161003816100381610038161003816

119909119899+1

119909119899

10038161003816100381610038161003816100381610038161003816

119901

=1

10038161003816100381610038161205821100381610038161003816100381611990110038161003816100381610038161205822

1003816100381610038161003816119901

100381610038161003816100381612058211003816100381610038161003816119901=

1

100381610038161003816100381612058221003816100381610038161003816119901gt 1 (41)

Case 2 (|1205822| = |1205821| lt 1) In this case we have 1199042 = 4119905(119903 minus 120572)and using the formula

119886119899 = (2119899

minus119904) [

minus119904

2 (119903 minus 120572)]

119899

forall119899 ⩾ 1 (42)

we obtain that

lim119899rarrinfin

10038161003816100381610038161003816100381610038161003816

119886119899

119886119899minus1

10038161003816100381610038161003816100381610038161003816

119901

=

10038161003816100381610038161003816100381610038161003816

minus119904

2 (119903 minus 120572)

10038161003816100381610038161003816100381610038161003816

119901

=100381610038161003816100381612058211003816100381610038161003816119901 (43)

which leads to

lim119899rarrinfin

10038161003816100381610038161003816100381610038161003816

119909119899+1

119909119899

10038161003816100381610038161003816100381610038161003816

119901

=1

10038161003816100381610038161205821100381610038161003816100381611990110038161003816100381610038161205822

1003816100381610038161003816119901

100381610038161003816100381612058211003816100381610038161003816119901=

1

100381610038161003816100381612058221003816100381610038161003816119901gt 1 (44)

Case 3 (|1205822| = |1205821| = 1) In this case we have 1199042 = 4119905(119903minus120572) andso we have | minus 119904(2119905)| = 1 Assume that 120572 isin 120590119901(119860(119903 119904 119905) ℓ119901)This implies that 119909 isin ℓ119901 and 119909 = 120579 Thus we again derive (35)

119909119899 = (minus119904

2119905)

119899minus1

[minus (119899 minus 1)minus119904

21199051199090 + 1198991199091] (45)

Since lim119899rarrinfin119909119899 = 0 we must have 1199090 = 1199091 = 0 But thisimplies that 119909 = 120579 a contradiction which means that 120572 notin120590119901(119860(119903 119904 119905) ℓ119901) Thus 120590119901(119860(119903 119904 119905) ℓ119901) sube 1198632 This completesthe proof

Theorem 8 120590119903(119860(119903 119904 119905) ℓ119901) = 0

Proof By Proposition 1 120590119903(119860(119903 119904 119905) ℓ119901) = 120590119901(119860(119903 119904 119905)lowast ℓlowast

119901)

120590119901(119860(119903 119904 119905) ℓ119901) Since byTheorem 6

120590119901 (119860(119903 119904 119905)lowast ℓlowast

119901) = 0 120590119903 (119860 (119903 119904 119905) ℓ119901) = 0

(46)

This completes the proof

Theorem 9 Let 119904 be a complex number such that radic1199042 = minus119904and define the set1198631 by

1198631 = 120572 isin C 2 |119903 minus 120572| ⩽1003816100381610038161003816100381610038161003816minus119904 + radic1199042 minus 4119905 (119903 minus 120572)

1003816100381610038161003816100381610038161003816

(47)

Then 120590(119860(119903 119904 119905) ℓ119901) = 1198631

Proof ByTheorem 7 we get

120572 isin C 2 |119903 minus 120572| lt1003816100381610038161003816100381610038161003816minus119904 + radic1199042 minus 4119905 (119903 minus 120572)

1003816100381610038161003816100381610038161003816

sube 120590 (119860 (119903 119904 119905) ℓ119901)

(48)

Since the spectrum of any bounded operator is closed wehave

120572 isin C 2 |119903 minus 120572| ⩽1003816100381610038161003816100381610038161003816minus119904 + radic1199042 minus 4119905 (119903 minus 120572)

1003816100381610038161003816100381610038161003816

sube 120590 (119860 (119903 119904 119905) ℓ119901)

(49)

and again fromTheorems 5 7 and 8

120590 (119860 (119903 119904 119905) ℓ119901) sube 120572 isin C 2 |119903 minus 120572|

⩽1003816100381610038161003816100381610038161003816minus119904 + radic1199042 minus 4119905 (119903 minus 120572)

1003816100381610038161003816100381610038161003816

(50)

Combining (49) and (50) we obtain that 120590(119860(119903 119904 119905) ℓ119901) =1198631 where1198631 is defined by (47)

Theorem 10 120590119888(119860(119903 119904 119905) ℓ119901) = 1198633 where 1198633 = 120572 isin C

2|119903 minus 120572| = | minus 119904 + radic1199042 minus 4119905(119903 minus 120572)|

Proof Because the parts 120590119888(119860(119903 119904 119905) ℓ119901) 120590119903(119860(119903 119904 119905) ℓ119901)and 120590119901(119860(119903 119904 119905) ℓ119901) are pairwise disjoint sets and the unionof these sets is 120590(119860(119903 119904 119905) ℓ119901) the proof immediately followsfromTheorems 7 8 and 9

Theorem 11 If 120572 isin 1198632 120572 isin 120590(119860(119903 119904 119905) ℓ119901)1198683

Proof From Theorem 7 120572 isin 120590119901(119860(119903 119904 119905) ℓ119901) Thus (119860(119903119904 119905) minus 120572119868)

minus1 does not exist By Theorem 6119860(119903 119904 119905)lowast minus 120572119868 isone to one so 119860(119903 119904 119905) minus 120572119868 has a dense range in ℓ119901 byLemma 3

Theorem 12 The following statements hold

(i) 120590ap(119860(119903 119904 119905) ℓ119901) = 1198631(ii) 120590120575(119860(119903 119904 119905) ℓ119901) = 1198633(iii) 120590co(119860(119903 119904 119905) ℓ119901) = 0

Abstract and Applied Analysis 7

Proof (i) Since from Table 1

120590ap (119860 (119903 119904 119905) ℓ119901)=120590 (119860 (119903 119904 119905) ℓ119901)120590 (119860 (119903 119904 119905) ℓ119901) 1198681198681198681

(51)

we have byTheorem 8

120590 (119860 (119903 119904 119905) ℓ119901) 1198681198681198681 = 120590 (119860 (119903 119904 119905) ℓ119901) 1198681198681198682 = 0 (52)

Hence

120590ap (119860 (119903 119904 119905) ℓ119901) = 1198631 (53)

(ii) Since the following equality

120590120575 (119860 (119903 119904 119905) ℓ119901) = 120590 (119860 (119903 119904 119905) ℓ119901) 120590 (119860 (119903 119904 119905) ℓ119901) 1198683

(54)

holds from Table 1 we derive by Theorems 8 and 11 that120590120575(119860(119903 119904 119905) ℓ119901) = 1198632

(iii) From Table 1 we have

120590co (119860 (119903 119904 119905) ℓ119901)

= 120590 (119860 (119903 119904 119905) ℓ119901) 1198681198681198681 cup 120590 (119860 (119903 119904 119905) ℓ119901) 1198681198681198682

cup 120590 (119860 (119903 119904 119905) 1198880) 1198681198681198683

(55)

ByTheorem 6 it is immediate that 120590co(119860(119903 119904 119905) ℓ119901) = 0

4 Fine Spectra of Upper TriangularTriple-Band Matrices over the SequenceSpace ℓ119901 (1 lt 119901 lt infin)

In the present section we determine the fine spectrum ofthe operator 119860(119903 119904 119905) ℓ119901 rarr ℓ119901 in case 1 ⩽ 119901 lt infinWe quote some lemmas which are needed in proving thetheorems given in Section 4

Lemma 13 (see [17 p 253 Theorem 3416]) The matrix 119860 =(119886119899119896) gives rise to a bounded linear operator 119879 isin 119861(ℓ1) from ℓ1to itself if and only if the supremum of ℓ1 norms of the columnsof 119860 is bounded

Lemma 14 (see [17 p 245 Theorem 343]) The matrix 119860 =(119886119899119896) gives rise to a bounded linear operator 119879 isin 119861(ℓinfin) fromℓinfin to itself if and only if the supremum of ℓ1 norms of the rowsof 119860 is bounded

Lemma 15 (see [17 p 254 Theorem 3418]) Let 1 lt 119901 lt infinand 119860 isin (ℓinfin ℓinfin) cap (ℓ1 ℓ1) Then 119860 isin (ℓ119901 ℓ119901)

Theorem 16 The operator 119860(119903 119904 119905) ℓ119901 rarr ℓ119901 is a boundedlinear operator and

(|119903|119901+ |119904|119901+ |119905|119901)1119901⩽ 119860 (119903 119904 119905)(ℓ119901 ℓ119901)

⩽ |119903| + |119904| + |119905|

(56)

Proof Since the linearity of the operator119860(119903 119904 119905) is not hardwe omit the details Now we prove that (56) holds for theoperator 119860(119903 119904 119905) on the space ℓ119901 It is trivial that 119860(119903 119904119905)119890(2)= (119905 119904 119903 0 ) for 119890(2) isin ℓ119901 Therefore we have

119860 (119903 119904 119905)(ℓ119901 ℓ119901)⩾

10038171003817100381710038171003817119860 (119903 119904 119905) 119890

(2)100381710038171003817100381710038171199011003817100381710038171003817119890(2)1003817100381710038171003817119901

= (|119903|119901+ |119904|119901+ |119905|119901)1119901

(57)

which implies that

119860 (119903 119904 119905)(ℓ119901 ℓ119901)⩾ (|119903|119901+ |s|119901 + |119905|119901)1119901 (58)

Let 119909 = (119909119896) isin ℓ119901 where 1 lt 119901 lt infin Then since (119905119909119896+2)(119903119909119896) and (119904119909119896+1) isin ℓ119901 it is easy to see by Minkowskirsquos ine-quality that

119860 (119903 119904 119905) 119909119901

= (

infin

sum

119896=0

1003816100381610038161003816119903119909119896 + 119904119909119896+1 + 119905119909119896+21003816100381610038161003816119901)

1119901

⩽ (

infin

sum

119896=0

10038161003816100381610038161199031199091198961003816100381610038161003816119901)

1119901

+ (

infin

sum

119896=0

1003816100381610038161003816119904119909119896+11003816100381610038161003816119901)

1119901

+ (

infin

sum

119896=0

1003816100381610038161003816119905119909119896+21003816100381610038161003816119901)

1119901

= |119903| (

infin

sum

119896=0

10038161003816100381610038161199091198961003816100381610038161003816119901)

1119901

+ |119904| (

infin

sum

119896=0

1003816100381610038161003816119909119896+11003816100381610038161003816119901)

1119901

+ |119905| (

infin

sum

119896=0

1003816100381610038161003816119909119896+21003816100381610038161003816119901)

1119901

= |119904| 119909119901 + |119903| 119909119901 + |119905| 119909119901

= (|119903| + |119904| + |119905|) 119909119901

(59)

which leads us to the result that

(|119903|119901+ |119904|119901+ |119905|119901)1119901⩽ 119860 (119903 119904 119905)(ℓ119901 ℓ119901)

⩽ |119903| + |119904| + |119905|

(60)

Therefore by combining the inequalities in (58) and (59) wehave (56) as desired

If 119879 ℓ119901 rarr ℓ119901 is a bounded matrix operator with thematrix 119860 then it is known that the adjoint operator 119879lowast ℓlowast

119901rarr ℓlowast

119901is defined by the transpose of thematrix119860The dual

space of ℓ119901 is isomorphic to ℓ119902 where 1 lt 119901 lt infin

Theorem 17 Let 119904 be a complex number such that radic1199042 = minus119904and define the set1198631 by

1198631 = 120572 isin C 2 |119903 minus 120572| ⩽1003816100381610038161003816100381610038161003816minus119904 + radic1199042 minus 4119905 (119903 minus 120572)

1003816100381610038161003816100381610038161003816

(61)

Then 120590119888(119860(119903 119904 119905) ℓ119901) sube 1198631

8 Abstract and Applied Analysis

Proof We will show that 119860120572(119903 119904 119905)lowast is onto for 2|119903 minus 120572| gt

| minus 119904 + radic1199042 minus 4119905(119903 minus 120572)| Thus for every 119910 isin ℓ119902 we find 119909 isinℓ119902 119860120572(119903 119904 119905)

lowast is a triangle so it has an inverse Also equation119860120572(119903 119904 119905)

lowast119909 = 119910 gives [119860120572(119903 119904 119905)

lowast]minus1119910 = 119909 It is sufficient

to show that [119860120572(119903 119904 119905)lowast]minus1isin (ℓ119902 ℓ119902) We calculate that

119860 = (119886119899119896) = [119860120572(119903 119904 119905)lowast]minus1 as follows

119860 = (119886119899119896) =

[[[[

[

1198861 0 0

1198862 1198861 0

1198863 1198862 1198861

]]]]

]

(62)

where

1198861 =1

119903 minus 120572

1198862 =minus119904

(119903 minus 120572)2

1198863 =1199042minus 119905 (119903 minus 120572)

(119903 minus 120572)3

(63)

It is known that fromTheorem 5

119886119899 =120582119899

1minus 120582119899

2

radic1199042 minus 4119905 (119903 minus 120572) forall119899 ⩾ 1

where 1205821 =minus119904 + radicΔ

119903 minus 120572 1205822 =

minus119904 minus radicΔ

119903 minus 120572

(64)

Now we show that [119860120572(119903 119904 119905)lowast]minus1isin (ℓ1 ℓ1) for |1205821| lt 1

By Theorem 5 we know that if |1205821| lt 1 |1205822| lt 1 We assumethat 1199042 minus 4119905(119903 minus 120572) = 0 and |1205821| lt 1

10038171003817100381710038171003817[119860120572(119903 119904 119905)

lowast]minus110038171003817100381710038171003817(ℓ1 ℓ1)

= sup119899

infin

sum

119896=119899

10038161003816100381610038161198861198961003816100381610038161003816 =

infin

sum

119896=1

10038161003816100381610038161198861198961003816100381610038161003816

⩽1

|Δ|(

infin

sum

119896=1

100381610038161003816100381612058211003816100381610038161003816119896+

infin

sum

119896=1

100381610038161003816100381612058221003816100381610038161003816119896) lt infin

(65)

since |1205821| lt 1 and |1205822| lt 1 This shows that (119860120572(119903 119904 119905)lowast]minus1isin

(ℓ1 ℓ1) Similarly we can show that [119860120572(119903 119904 119905)lowast]minus1isin (ℓinfin

ℓinfin)Now assume that 1199042 minus 4119905(119903 minus 120572) = 0 Then

119886119899 = (2119899

minus119904) [

minus119904

2 (119903 minus 120572)]

119899

(66)

and simple calculation gives that (119886119899) isin ℓ119902 if and only if |minus119904| lt2|119903 minus 120572|

[(119860 (119903 119904 119905) minus 120572119868)lowast]minus1isin (ℓ119902 ℓ119902)

for 120572 isin C with 2 |119903 minus 120572| gt1003816100381610038161003816100381610038161003816minus119904 + radic1199042 minus 4119905 (119903 minus 120572)

1003816100381610038161003816100381610038161003816

(67)

Hence 119860120572(119903 119904 119905)lowast is onto By Lemma 2 119860120572(119903 119904 119905) has a

bounded inverse This means that 120590119888(119860(119903 119904 119905) ℓ119901) sube 1198631where1198631 is defined by (61)

Theorem 18 120590119901(119860(119903 119904 119905)lowast ℓlowast

119901) = 0

Proof Let 119860(119903 119904 119905)lowast119891 = 120572119891 with 119891 = 120579 = (0 0 0 ) in ℓlowast119901=

ℓ119902 Then by solving the system of linear equations

1199031198910 = 1205721198910

1199041198910 + 1199031198911 = 1205721198911

1199051198910 + 1199041198911 + 1199031198912 = 1205721198912

1199051198911 + 1199041198912 + 1199031198913 = 1205721198913

119905119891119896minus2 + 119904119891119896minus1 + 119903119891119896 = 120572119891119896

(68)

we find that 1198910 = 0 if 120572 = 119903 and 1198911 = 1198912 = sdot sdot sdot = 0 if 1198910 = 0which contradicts119891 = 120579 If 1198911198990 is the first nonzero entry of thesequence 119891 = (119891119899) and 120572 = 119903 then we get 1199051198911198990minus2 + 1199041198911198990minus1 +1199031198911198990

= 1205721198911198990which implies 1198911198990 = 0 which contradicts the

assumption1198911198990 = 0 Hence the equation119860(119903 119904 119905)lowast119891 = 120572119891 has

no solution 119891 = 120579

In the case 1 lt 119901 lt infin since the proof of the theoremsin Section 4 determining the spectrum and fine spectrumof the matrix operator 119860(119903 119904 119905) on the sequence space ℓ119901is similar to the case 0 lt 119901 ⩽ 1 to avoid the repetitionof similar statements we give the results by the followingtheorem without proof

Theorem 19 The following statements hold(i) 120590(119860(119903 119904 119905) ℓ119901) = 1198631(ii) 120590119903(119860(119903 119904 119905) ℓ119901) = 0(iii) 120590119901(119860(119903 119904 119905) ℓ119901) = 1198632(iv) 120590119888(119860(119903 119904 119905) ℓ119901) = 1198633(v) 120590ap(119860(119903 119904 119905) ℓ119901) = 1198631(vi) 120590co(119860(119903 119904 119905) ℓ119901) = 0(vii) 120590120575(119860(119903 119904 119905) ℓ119901) = 1198633

5 Some Applications

In this section we give two theorems related to Toeplitz ma-trix

Theorem 20 Let 119875 be a polynomial that corresponds to the 119899-tuple 119886 and let 1199111 1199112 1199113 119911119899minus1 also be the roots of 119875 Define119879 as a Toeplitz matrix associated with 119875 that is

119879 =

[[[[

[

1198860 1198861 1198862 119886119899 0 0 0

0 1198860 1198861 1198862 119886119899 0 0

0 0 1198860 1198861 1198862 119886119899 0

]]]]

]

(69)

Abstract and Applied Analysis 9

The resolvent operator 119879 over ℓ119901 with 1 lt 119901 lt infin where thedomain of the resolvent operator is the whole space ℓ119901 exists ifand if only all the roots of the polynomial are outside the unitdisc 119911 isin C |119911| ⩽ 1 That is 119879minus1 isin (ℓ119901 ℓ119901) if and if only|119911119894| gt 1 1 ⩽ 119894 ⩽ 119899 minus 1 In this case the resolvent operator isrepresented by

119879minus1=

1

119886119899minus1119860minus1(minus1199111 1) 119860

minus1(minus1199112 1) sdot sdot sdot 119860

minus1(minus119911119899minus1 1)

where

119860minus1(minus119911119894 1) =minus

[[[[[[[[[[[[

[

1119911119894 11199112

11989411199113

11989411199114

11989411199115

119894

0 1119911119894 11199112

11989411199113

11989411199114

119894

0 0 1119911119894 11199112

11989411199113

119894

0 0 0 1119911119894 11199112

119894

0 0 0 0 1119911119894

]]]]]]]]]]]]

]

(70)

Proof Suppose all the roots of the polynomial 119875(119911) = 1198860 +1198861119911 + sdot sdot sdot + 119886119899minus1119911

119899minus1= 119886119899(119911 minus 1199111)(119911 minus 1199112) sdot sdot sdot (119911 minus 119911119899minus1) are

outside the unit disc The Toeplitz matrix associated with 119875can be written as the product

119879 = 119886119899119860 (minus1199111 1) 119860 (minus1199112 1) sdot sdot sdot 119860 (minus119911119899minus1 1) (71)

Since multiplication of upper triangular Toeplitz matrices iscommutative we can see that

119879minus1=

1

119886119899minus1119860minus1(minus1199111 1) 119860

minus1(minus1199112 1) sdot sdot sdot 119860

minus1(minus119911119899minus1 1)

(72)

is left inverse of 119879 Since all roots are outside the unit discthen

10038171003817100381710038171003817119879minus1(minus119911119894 1)

10038171003817100381710038171003817(ℓinfin ℓinfin)= sup119899

infin

sum

119896=119899

1

10038161003816100381610038161199111198941003816100381610038161003816119896+1minus119899

=

infin

sum

119896=1

1

10038161003816100381610038161199111198941003816100381610038161003816119896lt infin

(73)

Therefore each 119879minus1(minus119911119894 1) isin (ℓinfin ℓinfin) for 1 ⩽ 119894 ⩽ 119899 minus 1Similarly we can say that 119879minus1(minus119911119894 1) isin (ℓ1 ℓ1) So we have119879minus1(minus119911119894 1) isin (ℓ119901 ℓ119901)

Theorem 21 The resolvent operator of 119860(119903 119904 119905) over ℓ119901 with1 lt 119901 lt infin where the domain of the resolvent operator is thespace ℓ119901 exists if and only if 2|119903| gt | minus 119904 + radic1199042 minus 4119905119903| In this

case the resolvent operator is represented by the infinite bandedToeplitz matrix

119864 (119903 119904 119905)

=1

119905

[[[[[[[[[[

[

1199061 1199062

11199063

11199064

1

0 1199061 1199062

11199063

1

0 0 1199061 1199062

1

0 0 0 1199061

]]]]]]]]]]

]

[[[[[[[[[[

[

1199062 1199062

21199063

21199064

2

0 1199062 1199062

21199063

2

0 0 1199062 1199062

2

0 0 0 1199062

]]]]]]]]]]

]

where 1199061 =minus119904 + radic1199042 minus 4119905119903

2119903 1199062 =

minus119904 minus radic1199042 minus 4119905119903

2119903

(74)

Proof By Theorem 20 we can see that 119864(119903 119904 119905) is inverseof the matrix of 119860(119903 119904 119905) But this is not enough to say itis resolvent operator By Lemmas 13 14 and 15 119864(119903 119904 119905) isin(ℓ119901 ℓ119901) when 2|119903| gt | minus 119904 + radic1199042 minus 4119905119903| That is for 2|119903| gt| minus 119904 + radic1199042 minus 4119905119903| 119864(119903 119904 119905) is a resolvent operator

Acknowledgments

The authors would like to thank the referee for pointingout some mistakes and misprints in the earlier version ofthis paper They would like to express their pleasure toMs Medine Yesilkayagil Department of Mathematics UsakUniversity 1 Eylul Campus Usak Turkey for many helpfulsuggestions and interesting comments on the revised form ofthe paper

References

[1] M Gonzalez ldquoThe fine spectrum of the Cesaro operator in ℓ119901(1 lt 119901 lt infin)rdquo Archiv der Mathematik vol 44 no 4 pp 355ndash358 1985

[2] P J CartlidgeWeighted meanmatrices as operators on ℓ119901 [PhDdissertation] Indiana University 1978

[3] J I Okutoyi ldquoOn the spectrum of C1 as an operator on 1198871199070rdquoAustralian Mathematical Society A vol 48 no 1 pp 79ndash861990

[4] J I Okutoyi ldquoOn the spectrum of C1 as an operator on bvrdquoCommunications A vol 41 no 1-2 pp 197ndash207 1992

[5] M Yıldırım ldquoOn the spectrum of the Rhaly operators on ℓ119901rdquoIndian Journal of Pure and Applied Mathematics vol 32 no 2pp 191ndash198 2001

[6] B Altay and F Basar ldquoOn the fine spectrum of the differenceoperator Δ on 1198880 and 119888rdquo Information Sciences vol 168 no 1ndash4pp 217ndash224 2004

[7] B Altay and F Basar ldquoOn the fine spectrum of the generalizeddifference operator 119861(119903 119904) over the sequence spaces 1198880 and119888rdquo International Journal of Mathematics and MathematicalSciences vol 2005 no 18 pp 3005ndash3013 2005

[8] A M Akhmedov and F Basar ldquoOn the fine spectra of the dif-ference operator Δ over the sequence space ℓ119901 (1 le 119901 lt infin)rdquoDemonstratio Mathematica vol 39 no 3 pp 585ndash595 2006

10 Abstract and Applied Analysis

[9] AMAkhmedov andF Basar ldquoThefine spectra of the differenceoperator Δ over the sequence space 119887119907119901 (1 le 119901 lt infin)rdquo ActaMathematica Sinica vol 23 no 10 pp 1757ndash1768 2007

[10] F Basar and B Altay ldquoOn the space of sequences of p-boundedvariation and related matrix mappingsrdquo Ukrainian Mathemati-cal Journal vol 55 no 1 pp 136ndash147 2003

[11] H Furkan H Bilgic and F Basar ldquoOn the fine spectrum of theoperator 119861(119903 119904 119905) over the sequence spaces ℓ119901 and 119887119907119901(1 lt 119901 ltinfin)rdquo Computers ampMathematics with Applications vol 60 no 7pp 2141ndash2152 2010

[12] V Karakaya and M Altun ldquoFine spectra of upper triangulardouble-band matricesrdquo Journal of Computational and AppliedMathematics vol 234 no 5 pp 1387ndash1394 2010

[13] A Karaisa ldquoFine spectra of upper triangular double-bandmatrices over the sequence space ℓ119901 (1 lt 119901 lt infin)rdquo DiscreteDynamics in Nature and Society vol 2012 Article ID 381069 19pages 2012

[14] E Kreyszig Introductory Functional Analysis with ApplicationsJohn Wiley amp Sons New York NY USA 1978

[15] J Appell E Pascale and A Vignoli Nonlinear Spectral Theoryvol 10 of de Gruyter Series in Nonlinear Analysis and Applica-tions Walter de Gruyter Berlin Germany 2004

[16] S Goldberg Unbounded Linear Operators Dover PublicationsNew York NY USA 1985

[17] B Choudhary and S Nanda Functional Analysis with Applica-tions John Wiley amp Sons New York NY USA 1989

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Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 5

Thus for |1205821| lt 1 119860120572(119903 119904 119905)lowast is onto and by Lemma 2 119860120572(119903

119904 119905) has a bounded inverse This means that

120590119888 (119860 (119903 119904 119905) ℓ119901)

sube 120572 isin C 2 |119903 minus 120572| ⩽1003816100381610038161003816100381610038161003816minus119904 + radic1199042 minus 4119905 (119903 minus 120572)

1003816100381610038161003816100381610038161003816

= 1198631

(28)

Case 2 If Δ = 1199042 minus 4119905(119903 minus 120572) = 0 a calculation on recurrentsequence gives that

119886119899 = (2119899

minus119904) [

minus119904

2 (119903 minus 120572)]

119899

forall119899 ⩾ 1 (29)

Now for | minus 119904| lt 2|119903 minus 120572| we can see that

10038161003816100381610038161199091198991003816100381610038161003816 ⩽

119899

sum

119896=0

1003816100381610038161003816119886119899minus1198961199101198961003816100381610038161003816 (30)

for all 119899 isin N Taking limit on the inequality (30) as 119899 rarr infinwe obtain that

119909infin ⩽10038171003817100381710038171199101003817100381710038171003817infin

infin

sum

119896=0

10038161003816100381610038161198861198961003816100381610038161003816 (31)

suminfin

119896=0|119886119896| is convergent since | minus 119904| lt 2|119903 minus 120572| Thus for | minus 119904| lt

2|119903 minus 120572| 119860120572(119903 119904 119905)lowast is onto and by Lemma 2 119860120572(119903 119904 119905) has a

bounded inverse This means that

120590119888 (119860 (119903 119904 119905) ℓ119901) sube 120572 isin C 2 |119903 minus 120572| ⩽ |minus119904| sube 1198631

(32)

Theorem 6 120590119901(119860(119903 119904 119905)lowast ℓlowast

119901) = 0

Proof Consider 119860(119903 119904 119905)lowast119891 = 120572119891 with 119891 = 120579 = (0 0 0 ) inℓlowast

119901= ℓinfin Then by solving the system of linear equations

1199031198910 = 1205721198910

1199041198910 + 1199031198911 = 1205721198911

1199051198910 + 1199041198911 + 1199031198912 = 1205721198912

1199051198911 + 1199041198912 + 1199031198913 = 1205721198913

119905119891119896minus2 + 119904119891119896minus1 + 119903119891119896 = 120572119891119896

(33)

we find that 1198910 = 0 if 120572 = 119903 and 1198911 = 1198912 = sdot sdot sdot = 0 if 1198910 = 0which contradicts119891 = 120579 If 1198911198990 is the first nonzero entry of thesequence119891 = (119891119899) and120572 = 119903 thenwe get 1199051198911198990minus2+1199041198911198990minus1+1199031198911198990 =1205721198911198990

which implies 1198911198990 = 0which contradicts the assumption1198911198990

= 0 Hence the equation119860(119903 119904 119905)lowast119891 = 120572119891 has no solution119891 = 120579

Theorem 7 120590119901(119860(119903 119904 119905) ℓ119901) = 1198632 where 1198632 = 120572 isin C

2|119903 minus 120572| lt | minus 119904 + radic1199042 minus 4119905(119903 minus 120572)|

Proof Let119860(119903 119904 119905)119909 = 120572119909 for 120579 = 119909 isin ℓ119901Then by solving thesystem of linear equations

1199031199090 + 1199041199091 + 1199051199092 = 1205721199090

1199031199091 + 1199041199092 + 1199051199093 = 1205721199091

1199031199092 + 1199041199093 + 1199051199094 = 1205721199092

119903119909119896minus2 + 119904119909119896minus1 + 119905119909119896 = 120572119909119896

(34)

and we have

1199092 =minus119904

1199051199091 minus

119903 minus 120572

1199051199090

1199093 =1199042minus 119905 (119903 minus 120572)

11990521199091 +

119904 (119903 minus 120572)

11990521199090

119909119899 =119886119899(119903 minus 120572)

119899

119905119899minus11199091 minus

119886119899minus1(119903 minus 120572)119899

119905119899minus11199090 forall119899 ⩾ 2

(35)

Assume that 120572 isin 1198632 Then we choose 1199090 = 1 and 1199091 = 2(119903 minus120572)(minus119904+radic1199042 minus 4119905(119903 minus 120572)) We show that 119909119899 = 119909

119899

1for all 119899 ⩾ 2

Since 1205821 1205822 are roots of the characteristic equation (119903minus120572)1205822+

119904120582 + 119905 = 0 we must have

12058211205822 =119905

119903 minus 120572 1205821 minus 1205822 =

radicΔ

119903 minus 120572 (36)

Combining the fact 1199091 = 11205821 with relation (35) we can seethat

119909119899 =119886119899(119903 minus 120572)

119899

119905119899minus11199091 minus

119886119899minus1(119903 minus 120572)119899

119905119899minus11199090

= (119903 minus 120572

119905)

119899minus1

(119903 minus 120572) (minus119886119899minus11199090 + 1198861198991199091)

=1

(12058211205822)119899minus1

119903 minus 120572

radicΔ(minus120582119899minus1

1+ 120582119899minus1

2+ 120582119899minus1

1minus 120582119899

2120582minus1

1)

=1

120582119899minus11120582119899minus12

(1

1205821 minus 1205822)120582119899minus1

2(1205821 minus 1205822

1205821)

=1

1205821198991

= 119909119899

1

(37)

The same result may be obtained in case Δ = 0 Now 119909 =(119909119896) isin ℓ119901 since |1199091| lt 1 This shows that 1198632 sube 120590119901(119860(119903

119904 119905) ℓ119901)

6 Abstract and Applied Analysis

Now we assume that 120572 notin 1198632 that is |1205821| ⩽ 1 We mustshow that 120572 notin 120590119901(119860(119903 119904 119905) ℓ119901) Therefore we obtain from therelation (35) that

119909119899+1

119909119899= (

119903 minus 120572

119905)119886119899minus1

119886119899minus2(minus1199090 + (119886119899119886119899minus1) 1199091

minus1199090 + (119886119899minus1119886119899minus2) 1199091) (38)

Now we examine three cases

Case 1 (|1205822| lt |1205821| ⩽ 1) In this case we have 1199042 = 4119905(119903minus120572) and

119886119899

119886119899minus1=120582119899+1

1minus 120582119899+1

2

1205821198991minus 1205821198992

=1205821 [1 minus (12058221205821)

119899+1]

[1 minus (12058221205821)119899] (39)

Then we have

lim119899rarrinfin

10038161003816100381610038161003816100381610038161003816

119886119899

119886119899minus1

10038161003816100381610038161003816100381610038161003816

119901

= lim119899rarrinfin

10038161003816100381610038161003816100381610038161003816

119886119899minus1

119886119899minus2

10038161003816100381610038161003816100381610038161003816

119901

= lim119899rarrinfin

100381610038161003816100381612058211003816100381610038161003816119901100381610038161003816100381610038161 minus (12058221205821)

119899+110038161003816100381610038161003816

119901

100381610038161003816100381610038161 minus (12058221205821)

11989910038161003816100381610038161003816

119901=100381610038161003816100381612058211003816100381610038161003816119901

(40)

Now if minus1199090 + 12058211199091 = 0 then we have (119909119899) = (1199090120582119899

1) which

is not in ℓ119901 Otherwise

lim119899rarrinfin

10038161003816100381610038161003816100381610038161003816

119909119899+1

119909119899

10038161003816100381610038161003816100381610038161003816

119901

=1

10038161003816100381610038161205821100381610038161003816100381611990110038161003816100381610038161205822

1003816100381610038161003816119901

100381610038161003816100381612058211003816100381610038161003816119901=

1

100381610038161003816100381612058221003816100381610038161003816119901gt 1 (41)

Case 2 (|1205822| = |1205821| lt 1) In this case we have 1199042 = 4119905(119903 minus 120572)and using the formula

119886119899 = (2119899

minus119904) [

minus119904

2 (119903 minus 120572)]

119899

forall119899 ⩾ 1 (42)

we obtain that

lim119899rarrinfin

10038161003816100381610038161003816100381610038161003816

119886119899

119886119899minus1

10038161003816100381610038161003816100381610038161003816

119901

=

10038161003816100381610038161003816100381610038161003816

minus119904

2 (119903 minus 120572)

10038161003816100381610038161003816100381610038161003816

119901

=100381610038161003816100381612058211003816100381610038161003816119901 (43)

which leads to

lim119899rarrinfin

10038161003816100381610038161003816100381610038161003816

119909119899+1

119909119899

10038161003816100381610038161003816100381610038161003816

119901

=1

10038161003816100381610038161205821100381610038161003816100381611990110038161003816100381610038161205822

1003816100381610038161003816119901

100381610038161003816100381612058211003816100381610038161003816119901=

1

100381610038161003816100381612058221003816100381610038161003816119901gt 1 (44)

Case 3 (|1205822| = |1205821| = 1) In this case we have 1199042 = 4119905(119903minus120572) andso we have | minus 119904(2119905)| = 1 Assume that 120572 isin 120590119901(119860(119903 119904 119905) ℓ119901)This implies that 119909 isin ℓ119901 and 119909 = 120579 Thus we again derive (35)

119909119899 = (minus119904

2119905)

119899minus1

[minus (119899 minus 1)minus119904

21199051199090 + 1198991199091] (45)

Since lim119899rarrinfin119909119899 = 0 we must have 1199090 = 1199091 = 0 But thisimplies that 119909 = 120579 a contradiction which means that 120572 notin120590119901(119860(119903 119904 119905) ℓ119901) Thus 120590119901(119860(119903 119904 119905) ℓ119901) sube 1198632 This completesthe proof

Theorem 8 120590119903(119860(119903 119904 119905) ℓ119901) = 0

Proof By Proposition 1 120590119903(119860(119903 119904 119905) ℓ119901) = 120590119901(119860(119903 119904 119905)lowast ℓlowast

119901)

120590119901(119860(119903 119904 119905) ℓ119901) Since byTheorem 6

120590119901 (119860(119903 119904 119905)lowast ℓlowast

119901) = 0 120590119903 (119860 (119903 119904 119905) ℓ119901) = 0

(46)

This completes the proof

Theorem 9 Let 119904 be a complex number such that radic1199042 = minus119904and define the set1198631 by

1198631 = 120572 isin C 2 |119903 minus 120572| ⩽1003816100381610038161003816100381610038161003816minus119904 + radic1199042 minus 4119905 (119903 minus 120572)

1003816100381610038161003816100381610038161003816

(47)

Then 120590(119860(119903 119904 119905) ℓ119901) = 1198631

Proof ByTheorem 7 we get

120572 isin C 2 |119903 minus 120572| lt1003816100381610038161003816100381610038161003816minus119904 + radic1199042 minus 4119905 (119903 minus 120572)

1003816100381610038161003816100381610038161003816

sube 120590 (119860 (119903 119904 119905) ℓ119901)

(48)

Since the spectrum of any bounded operator is closed wehave

120572 isin C 2 |119903 minus 120572| ⩽1003816100381610038161003816100381610038161003816minus119904 + radic1199042 minus 4119905 (119903 minus 120572)

1003816100381610038161003816100381610038161003816

sube 120590 (119860 (119903 119904 119905) ℓ119901)

(49)

and again fromTheorems 5 7 and 8

120590 (119860 (119903 119904 119905) ℓ119901) sube 120572 isin C 2 |119903 minus 120572|

⩽1003816100381610038161003816100381610038161003816minus119904 + radic1199042 minus 4119905 (119903 minus 120572)

1003816100381610038161003816100381610038161003816

(50)

Combining (49) and (50) we obtain that 120590(119860(119903 119904 119905) ℓ119901) =1198631 where1198631 is defined by (47)

Theorem 10 120590119888(119860(119903 119904 119905) ℓ119901) = 1198633 where 1198633 = 120572 isin C

2|119903 minus 120572| = | minus 119904 + radic1199042 minus 4119905(119903 minus 120572)|

Proof Because the parts 120590119888(119860(119903 119904 119905) ℓ119901) 120590119903(119860(119903 119904 119905) ℓ119901)and 120590119901(119860(119903 119904 119905) ℓ119901) are pairwise disjoint sets and the unionof these sets is 120590(119860(119903 119904 119905) ℓ119901) the proof immediately followsfromTheorems 7 8 and 9

Theorem 11 If 120572 isin 1198632 120572 isin 120590(119860(119903 119904 119905) ℓ119901)1198683

Proof From Theorem 7 120572 isin 120590119901(119860(119903 119904 119905) ℓ119901) Thus (119860(119903119904 119905) minus 120572119868)

minus1 does not exist By Theorem 6119860(119903 119904 119905)lowast minus 120572119868 isone to one so 119860(119903 119904 119905) minus 120572119868 has a dense range in ℓ119901 byLemma 3

Theorem 12 The following statements hold

(i) 120590ap(119860(119903 119904 119905) ℓ119901) = 1198631(ii) 120590120575(119860(119903 119904 119905) ℓ119901) = 1198633(iii) 120590co(119860(119903 119904 119905) ℓ119901) = 0

Abstract and Applied Analysis 7

Proof (i) Since from Table 1

120590ap (119860 (119903 119904 119905) ℓ119901)=120590 (119860 (119903 119904 119905) ℓ119901)120590 (119860 (119903 119904 119905) ℓ119901) 1198681198681198681

(51)

we have byTheorem 8

120590 (119860 (119903 119904 119905) ℓ119901) 1198681198681198681 = 120590 (119860 (119903 119904 119905) ℓ119901) 1198681198681198682 = 0 (52)

Hence

120590ap (119860 (119903 119904 119905) ℓ119901) = 1198631 (53)

(ii) Since the following equality

120590120575 (119860 (119903 119904 119905) ℓ119901) = 120590 (119860 (119903 119904 119905) ℓ119901) 120590 (119860 (119903 119904 119905) ℓ119901) 1198683

(54)

holds from Table 1 we derive by Theorems 8 and 11 that120590120575(119860(119903 119904 119905) ℓ119901) = 1198632

(iii) From Table 1 we have

120590co (119860 (119903 119904 119905) ℓ119901)

= 120590 (119860 (119903 119904 119905) ℓ119901) 1198681198681198681 cup 120590 (119860 (119903 119904 119905) ℓ119901) 1198681198681198682

cup 120590 (119860 (119903 119904 119905) 1198880) 1198681198681198683

(55)

ByTheorem 6 it is immediate that 120590co(119860(119903 119904 119905) ℓ119901) = 0

4 Fine Spectra of Upper TriangularTriple-Band Matrices over the SequenceSpace ℓ119901 (1 lt 119901 lt infin)

In the present section we determine the fine spectrum ofthe operator 119860(119903 119904 119905) ℓ119901 rarr ℓ119901 in case 1 ⩽ 119901 lt infinWe quote some lemmas which are needed in proving thetheorems given in Section 4

Lemma 13 (see [17 p 253 Theorem 3416]) The matrix 119860 =(119886119899119896) gives rise to a bounded linear operator 119879 isin 119861(ℓ1) from ℓ1to itself if and only if the supremum of ℓ1 norms of the columnsof 119860 is bounded

Lemma 14 (see [17 p 245 Theorem 343]) The matrix 119860 =(119886119899119896) gives rise to a bounded linear operator 119879 isin 119861(ℓinfin) fromℓinfin to itself if and only if the supremum of ℓ1 norms of the rowsof 119860 is bounded

Lemma 15 (see [17 p 254 Theorem 3418]) Let 1 lt 119901 lt infinand 119860 isin (ℓinfin ℓinfin) cap (ℓ1 ℓ1) Then 119860 isin (ℓ119901 ℓ119901)

Theorem 16 The operator 119860(119903 119904 119905) ℓ119901 rarr ℓ119901 is a boundedlinear operator and

(|119903|119901+ |119904|119901+ |119905|119901)1119901⩽ 119860 (119903 119904 119905)(ℓ119901 ℓ119901)

⩽ |119903| + |119904| + |119905|

(56)

Proof Since the linearity of the operator119860(119903 119904 119905) is not hardwe omit the details Now we prove that (56) holds for theoperator 119860(119903 119904 119905) on the space ℓ119901 It is trivial that 119860(119903 119904119905)119890(2)= (119905 119904 119903 0 ) for 119890(2) isin ℓ119901 Therefore we have

119860 (119903 119904 119905)(ℓ119901 ℓ119901)⩾

10038171003817100381710038171003817119860 (119903 119904 119905) 119890

(2)100381710038171003817100381710038171199011003817100381710038171003817119890(2)1003817100381710038171003817119901

= (|119903|119901+ |119904|119901+ |119905|119901)1119901

(57)

which implies that

119860 (119903 119904 119905)(ℓ119901 ℓ119901)⩾ (|119903|119901+ |s|119901 + |119905|119901)1119901 (58)

Let 119909 = (119909119896) isin ℓ119901 where 1 lt 119901 lt infin Then since (119905119909119896+2)(119903119909119896) and (119904119909119896+1) isin ℓ119901 it is easy to see by Minkowskirsquos ine-quality that

119860 (119903 119904 119905) 119909119901

= (

infin

sum

119896=0

1003816100381610038161003816119903119909119896 + 119904119909119896+1 + 119905119909119896+21003816100381610038161003816119901)

1119901

⩽ (

infin

sum

119896=0

10038161003816100381610038161199031199091198961003816100381610038161003816119901)

1119901

+ (

infin

sum

119896=0

1003816100381610038161003816119904119909119896+11003816100381610038161003816119901)

1119901

+ (

infin

sum

119896=0

1003816100381610038161003816119905119909119896+21003816100381610038161003816119901)

1119901

= |119903| (

infin

sum

119896=0

10038161003816100381610038161199091198961003816100381610038161003816119901)

1119901

+ |119904| (

infin

sum

119896=0

1003816100381610038161003816119909119896+11003816100381610038161003816119901)

1119901

+ |119905| (

infin

sum

119896=0

1003816100381610038161003816119909119896+21003816100381610038161003816119901)

1119901

= |119904| 119909119901 + |119903| 119909119901 + |119905| 119909119901

= (|119903| + |119904| + |119905|) 119909119901

(59)

which leads us to the result that

(|119903|119901+ |119904|119901+ |119905|119901)1119901⩽ 119860 (119903 119904 119905)(ℓ119901 ℓ119901)

⩽ |119903| + |119904| + |119905|

(60)

Therefore by combining the inequalities in (58) and (59) wehave (56) as desired

If 119879 ℓ119901 rarr ℓ119901 is a bounded matrix operator with thematrix 119860 then it is known that the adjoint operator 119879lowast ℓlowast

119901rarr ℓlowast

119901is defined by the transpose of thematrix119860The dual

space of ℓ119901 is isomorphic to ℓ119902 where 1 lt 119901 lt infin

Theorem 17 Let 119904 be a complex number such that radic1199042 = minus119904and define the set1198631 by

1198631 = 120572 isin C 2 |119903 minus 120572| ⩽1003816100381610038161003816100381610038161003816minus119904 + radic1199042 minus 4119905 (119903 minus 120572)

1003816100381610038161003816100381610038161003816

(61)

Then 120590119888(119860(119903 119904 119905) ℓ119901) sube 1198631

8 Abstract and Applied Analysis

Proof We will show that 119860120572(119903 119904 119905)lowast is onto for 2|119903 minus 120572| gt

| minus 119904 + radic1199042 minus 4119905(119903 minus 120572)| Thus for every 119910 isin ℓ119902 we find 119909 isinℓ119902 119860120572(119903 119904 119905)

lowast is a triangle so it has an inverse Also equation119860120572(119903 119904 119905)

lowast119909 = 119910 gives [119860120572(119903 119904 119905)

lowast]minus1119910 = 119909 It is sufficient

to show that [119860120572(119903 119904 119905)lowast]minus1isin (ℓ119902 ℓ119902) We calculate that

119860 = (119886119899119896) = [119860120572(119903 119904 119905)lowast]minus1 as follows

119860 = (119886119899119896) =

[[[[

[

1198861 0 0

1198862 1198861 0

1198863 1198862 1198861

]]]]

]

(62)

where

1198861 =1

119903 minus 120572

1198862 =minus119904

(119903 minus 120572)2

1198863 =1199042minus 119905 (119903 minus 120572)

(119903 minus 120572)3

(63)

It is known that fromTheorem 5

119886119899 =120582119899

1minus 120582119899

2

radic1199042 minus 4119905 (119903 minus 120572) forall119899 ⩾ 1

where 1205821 =minus119904 + radicΔ

119903 minus 120572 1205822 =

minus119904 minus radicΔ

119903 minus 120572

(64)

Now we show that [119860120572(119903 119904 119905)lowast]minus1isin (ℓ1 ℓ1) for |1205821| lt 1

By Theorem 5 we know that if |1205821| lt 1 |1205822| lt 1 We assumethat 1199042 minus 4119905(119903 minus 120572) = 0 and |1205821| lt 1

10038171003817100381710038171003817[119860120572(119903 119904 119905)

lowast]minus110038171003817100381710038171003817(ℓ1 ℓ1)

= sup119899

infin

sum

119896=119899

10038161003816100381610038161198861198961003816100381610038161003816 =

infin

sum

119896=1

10038161003816100381610038161198861198961003816100381610038161003816

⩽1

|Δ|(

infin

sum

119896=1

100381610038161003816100381612058211003816100381610038161003816119896+

infin

sum

119896=1

100381610038161003816100381612058221003816100381610038161003816119896) lt infin

(65)

since |1205821| lt 1 and |1205822| lt 1 This shows that (119860120572(119903 119904 119905)lowast]minus1isin

(ℓ1 ℓ1) Similarly we can show that [119860120572(119903 119904 119905)lowast]minus1isin (ℓinfin

ℓinfin)Now assume that 1199042 minus 4119905(119903 minus 120572) = 0 Then

119886119899 = (2119899

minus119904) [

minus119904

2 (119903 minus 120572)]

119899

(66)

and simple calculation gives that (119886119899) isin ℓ119902 if and only if |minus119904| lt2|119903 minus 120572|

[(119860 (119903 119904 119905) minus 120572119868)lowast]minus1isin (ℓ119902 ℓ119902)

for 120572 isin C with 2 |119903 minus 120572| gt1003816100381610038161003816100381610038161003816minus119904 + radic1199042 minus 4119905 (119903 minus 120572)

1003816100381610038161003816100381610038161003816

(67)

Hence 119860120572(119903 119904 119905)lowast is onto By Lemma 2 119860120572(119903 119904 119905) has a

bounded inverse This means that 120590119888(119860(119903 119904 119905) ℓ119901) sube 1198631where1198631 is defined by (61)

Theorem 18 120590119901(119860(119903 119904 119905)lowast ℓlowast

119901) = 0

Proof Let 119860(119903 119904 119905)lowast119891 = 120572119891 with 119891 = 120579 = (0 0 0 ) in ℓlowast119901=

ℓ119902 Then by solving the system of linear equations

1199031198910 = 1205721198910

1199041198910 + 1199031198911 = 1205721198911

1199051198910 + 1199041198911 + 1199031198912 = 1205721198912

1199051198911 + 1199041198912 + 1199031198913 = 1205721198913

119905119891119896minus2 + 119904119891119896minus1 + 119903119891119896 = 120572119891119896

(68)

we find that 1198910 = 0 if 120572 = 119903 and 1198911 = 1198912 = sdot sdot sdot = 0 if 1198910 = 0which contradicts119891 = 120579 If 1198911198990 is the first nonzero entry of thesequence 119891 = (119891119899) and 120572 = 119903 then we get 1199051198911198990minus2 + 1199041198911198990minus1 +1199031198911198990

= 1205721198911198990which implies 1198911198990 = 0 which contradicts the

assumption1198911198990 = 0 Hence the equation119860(119903 119904 119905)lowast119891 = 120572119891 has

no solution 119891 = 120579

In the case 1 lt 119901 lt infin since the proof of the theoremsin Section 4 determining the spectrum and fine spectrumof the matrix operator 119860(119903 119904 119905) on the sequence space ℓ119901is similar to the case 0 lt 119901 ⩽ 1 to avoid the repetitionof similar statements we give the results by the followingtheorem without proof

Theorem 19 The following statements hold(i) 120590(119860(119903 119904 119905) ℓ119901) = 1198631(ii) 120590119903(119860(119903 119904 119905) ℓ119901) = 0(iii) 120590119901(119860(119903 119904 119905) ℓ119901) = 1198632(iv) 120590119888(119860(119903 119904 119905) ℓ119901) = 1198633(v) 120590ap(119860(119903 119904 119905) ℓ119901) = 1198631(vi) 120590co(119860(119903 119904 119905) ℓ119901) = 0(vii) 120590120575(119860(119903 119904 119905) ℓ119901) = 1198633

5 Some Applications

In this section we give two theorems related to Toeplitz ma-trix

Theorem 20 Let 119875 be a polynomial that corresponds to the 119899-tuple 119886 and let 1199111 1199112 1199113 119911119899minus1 also be the roots of 119875 Define119879 as a Toeplitz matrix associated with 119875 that is

119879 =

[[[[

[

1198860 1198861 1198862 119886119899 0 0 0

0 1198860 1198861 1198862 119886119899 0 0

0 0 1198860 1198861 1198862 119886119899 0

]]]]

]

(69)

Abstract and Applied Analysis 9

The resolvent operator 119879 over ℓ119901 with 1 lt 119901 lt infin where thedomain of the resolvent operator is the whole space ℓ119901 exists ifand if only all the roots of the polynomial are outside the unitdisc 119911 isin C |119911| ⩽ 1 That is 119879minus1 isin (ℓ119901 ℓ119901) if and if only|119911119894| gt 1 1 ⩽ 119894 ⩽ 119899 minus 1 In this case the resolvent operator isrepresented by

119879minus1=

1

119886119899minus1119860minus1(minus1199111 1) 119860

minus1(minus1199112 1) sdot sdot sdot 119860

minus1(minus119911119899minus1 1)

where

119860minus1(minus119911119894 1) =minus

[[[[[[[[[[[[

[

1119911119894 11199112

11989411199113

11989411199114

11989411199115

119894

0 1119911119894 11199112

11989411199113

11989411199114

119894

0 0 1119911119894 11199112

11989411199113

119894

0 0 0 1119911119894 11199112

119894

0 0 0 0 1119911119894

]]]]]]]]]]]]

]

(70)

Proof Suppose all the roots of the polynomial 119875(119911) = 1198860 +1198861119911 + sdot sdot sdot + 119886119899minus1119911

119899minus1= 119886119899(119911 minus 1199111)(119911 minus 1199112) sdot sdot sdot (119911 minus 119911119899minus1) are

outside the unit disc The Toeplitz matrix associated with 119875can be written as the product

119879 = 119886119899119860 (minus1199111 1) 119860 (minus1199112 1) sdot sdot sdot 119860 (minus119911119899minus1 1) (71)

Since multiplication of upper triangular Toeplitz matrices iscommutative we can see that

119879minus1=

1

119886119899minus1119860minus1(minus1199111 1) 119860

minus1(minus1199112 1) sdot sdot sdot 119860

minus1(minus119911119899minus1 1)

(72)

is left inverse of 119879 Since all roots are outside the unit discthen

10038171003817100381710038171003817119879minus1(minus119911119894 1)

10038171003817100381710038171003817(ℓinfin ℓinfin)= sup119899

infin

sum

119896=119899

1

10038161003816100381610038161199111198941003816100381610038161003816119896+1minus119899

=

infin

sum

119896=1

1

10038161003816100381610038161199111198941003816100381610038161003816119896lt infin

(73)

Therefore each 119879minus1(minus119911119894 1) isin (ℓinfin ℓinfin) for 1 ⩽ 119894 ⩽ 119899 minus 1Similarly we can say that 119879minus1(minus119911119894 1) isin (ℓ1 ℓ1) So we have119879minus1(minus119911119894 1) isin (ℓ119901 ℓ119901)

Theorem 21 The resolvent operator of 119860(119903 119904 119905) over ℓ119901 with1 lt 119901 lt infin where the domain of the resolvent operator is thespace ℓ119901 exists if and only if 2|119903| gt | minus 119904 + radic1199042 minus 4119905119903| In this

case the resolvent operator is represented by the infinite bandedToeplitz matrix

119864 (119903 119904 119905)

=1

119905

[[[[[[[[[[

[

1199061 1199062

11199063

11199064

1

0 1199061 1199062

11199063

1

0 0 1199061 1199062

1

0 0 0 1199061

]]]]]]]]]]

]

[[[[[[[[[[

[

1199062 1199062

21199063

21199064

2

0 1199062 1199062

21199063

2

0 0 1199062 1199062

2

0 0 0 1199062

]]]]]]]]]]

]

where 1199061 =minus119904 + radic1199042 minus 4119905119903

2119903 1199062 =

minus119904 minus radic1199042 minus 4119905119903

2119903

(74)

Proof By Theorem 20 we can see that 119864(119903 119904 119905) is inverseof the matrix of 119860(119903 119904 119905) But this is not enough to say itis resolvent operator By Lemmas 13 14 and 15 119864(119903 119904 119905) isin(ℓ119901 ℓ119901) when 2|119903| gt | minus 119904 + radic1199042 minus 4119905119903| That is for 2|119903| gt| minus 119904 + radic1199042 minus 4119905119903| 119864(119903 119904 119905) is a resolvent operator

Acknowledgments

The authors would like to thank the referee for pointingout some mistakes and misprints in the earlier version ofthis paper They would like to express their pleasure toMs Medine Yesilkayagil Department of Mathematics UsakUniversity 1 Eylul Campus Usak Turkey for many helpfulsuggestions and interesting comments on the revised form ofthe paper

References

[1] M Gonzalez ldquoThe fine spectrum of the Cesaro operator in ℓ119901(1 lt 119901 lt infin)rdquo Archiv der Mathematik vol 44 no 4 pp 355ndash358 1985

[2] P J CartlidgeWeighted meanmatrices as operators on ℓ119901 [PhDdissertation] Indiana University 1978

[3] J I Okutoyi ldquoOn the spectrum of C1 as an operator on 1198871199070rdquoAustralian Mathematical Society A vol 48 no 1 pp 79ndash861990

[4] J I Okutoyi ldquoOn the spectrum of C1 as an operator on bvrdquoCommunications A vol 41 no 1-2 pp 197ndash207 1992

[5] M Yıldırım ldquoOn the spectrum of the Rhaly operators on ℓ119901rdquoIndian Journal of Pure and Applied Mathematics vol 32 no 2pp 191ndash198 2001

[6] B Altay and F Basar ldquoOn the fine spectrum of the differenceoperator Δ on 1198880 and 119888rdquo Information Sciences vol 168 no 1ndash4pp 217ndash224 2004

[7] B Altay and F Basar ldquoOn the fine spectrum of the generalizeddifference operator 119861(119903 119904) over the sequence spaces 1198880 and119888rdquo International Journal of Mathematics and MathematicalSciences vol 2005 no 18 pp 3005ndash3013 2005

[8] A M Akhmedov and F Basar ldquoOn the fine spectra of the dif-ference operator Δ over the sequence space ℓ119901 (1 le 119901 lt infin)rdquoDemonstratio Mathematica vol 39 no 3 pp 585ndash595 2006

10 Abstract and Applied Analysis

[9] AMAkhmedov andF Basar ldquoThefine spectra of the differenceoperator Δ over the sequence space 119887119907119901 (1 le 119901 lt infin)rdquo ActaMathematica Sinica vol 23 no 10 pp 1757ndash1768 2007

[10] F Basar and B Altay ldquoOn the space of sequences of p-boundedvariation and related matrix mappingsrdquo Ukrainian Mathemati-cal Journal vol 55 no 1 pp 136ndash147 2003

[11] H Furkan H Bilgic and F Basar ldquoOn the fine spectrum of theoperator 119861(119903 119904 119905) over the sequence spaces ℓ119901 and 119887119907119901(1 lt 119901 ltinfin)rdquo Computers ampMathematics with Applications vol 60 no 7pp 2141ndash2152 2010

[12] V Karakaya and M Altun ldquoFine spectra of upper triangulardouble-band matricesrdquo Journal of Computational and AppliedMathematics vol 234 no 5 pp 1387ndash1394 2010

[13] A Karaisa ldquoFine spectra of upper triangular double-bandmatrices over the sequence space ℓ119901 (1 lt 119901 lt infin)rdquo DiscreteDynamics in Nature and Society vol 2012 Article ID 381069 19pages 2012

[14] E Kreyszig Introductory Functional Analysis with ApplicationsJohn Wiley amp Sons New York NY USA 1978

[15] J Appell E Pascale and A Vignoli Nonlinear Spectral Theoryvol 10 of de Gruyter Series in Nonlinear Analysis and Applica-tions Walter de Gruyter Berlin Germany 2004

[16] S Goldberg Unbounded Linear Operators Dover PublicationsNew York NY USA 1985

[17] B Choudhary and S Nanda Functional Analysis with Applica-tions John Wiley amp Sons New York NY USA 1989

Submit your manuscripts athttpwwwhindawicom

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Decision SciencesAdvances in

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

6 Abstract and Applied Analysis

Now we assume that 120572 notin 1198632 that is |1205821| ⩽ 1 We mustshow that 120572 notin 120590119901(119860(119903 119904 119905) ℓ119901) Therefore we obtain from therelation (35) that

119909119899+1

119909119899= (

119903 minus 120572

119905)119886119899minus1

119886119899minus2(minus1199090 + (119886119899119886119899minus1) 1199091

minus1199090 + (119886119899minus1119886119899minus2) 1199091) (38)

Now we examine three cases

Case 1 (|1205822| lt |1205821| ⩽ 1) In this case we have 1199042 = 4119905(119903minus120572) and

119886119899

119886119899minus1=120582119899+1

1minus 120582119899+1

2

1205821198991minus 1205821198992

=1205821 [1 minus (12058221205821)

119899+1]

[1 minus (12058221205821)119899] (39)

Then we have

lim119899rarrinfin

10038161003816100381610038161003816100381610038161003816

119886119899

119886119899minus1

10038161003816100381610038161003816100381610038161003816

119901

= lim119899rarrinfin

10038161003816100381610038161003816100381610038161003816

119886119899minus1

119886119899minus2

10038161003816100381610038161003816100381610038161003816

119901

= lim119899rarrinfin

100381610038161003816100381612058211003816100381610038161003816119901100381610038161003816100381610038161 minus (12058221205821)

119899+110038161003816100381610038161003816

119901

100381610038161003816100381610038161 minus (12058221205821)

11989910038161003816100381610038161003816

119901=100381610038161003816100381612058211003816100381610038161003816119901

(40)

Now if minus1199090 + 12058211199091 = 0 then we have (119909119899) = (1199090120582119899

1) which

is not in ℓ119901 Otherwise

lim119899rarrinfin

10038161003816100381610038161003816100381610038161003816

119909119899+1

119909119899

10038161003816100381610038161003816100381610038161003816

119901

=1

10038161003816100381610038161205821100381610038161003816100381611990110038161003816100381610038161205822

1003816100381610038161003816119901

100381610038161003816100381612058211003816100381610038161003816119901=

1

100381610038161003816100381612058221003816100381610038161003816119901gt 1 (41)

Case 2 (|1205822| = |1205821| lt 1) In this case we have 1199042 = 4119905(119903 minus 120572)and using the formula

119886119899 = (2119899

minus119904) [

minus119904

2 (119903 minus 120572)]

119899

forall119899 ⩾ 1 (42)

we obtain that

lim119899rarrinfin

10038161003816100381610038161003816100381610038161003816

119886119899

119886119899minus1

10038161003816100381610038161003816100381610038161003816

119901

=

10038161003816100381610038161003816100381610038161003816

minus119904

2 (119903 minus 120572)

10038161003816100381610038161003816100381610038161003816

119901

=100381610038161003816100381612058211003816100381610038161003816119901 (43)

which leads to

lim119899rarrinfin

10038161003816100381610038161003816100381610038161003816

119909119899+1

119909119899

10038161003816100381610038161003816100381610038161003816

119901

=1

10038161003816100381610038161205821100381610038161003816100381611990110038161003816100381610038161205822

1003816100381610038161003816119901

100381610038161003816100381612058211003816100381610038161003816119901=

1

100381610038161003816100381612058221003816100381610038161003816119901gt 1 (44)

Case 3 (|1205822| = |1205821| = 1) In this case we have 1199042 = 4119905(119903minus120572) andso we have | minus 119904(2119905)| = 1 Assume that 120572 isin 120590119901(119860(119903 119904 119905) ℓ119901)This implies that 119909 isin ℓ119901 and 119909 = 120579 Thus we again derive (35)

119909119899 = (minus119904

2119905)

119899minus1

[minus (119899 minus 1)minus119904

21199051199090 + 1198991199091] (45)

Since lim119899rarrinfin119909119899 = 0 we must have 1199090 = 1199091 = 0 But thisimplies that 119909 = 120579 a contradiction which means that 120572 notin120590119901(119860(119903 119904 119905) ℓ119901) Thus 120590119901(119860(119903 119904 119905) ℓ119901) sube 1198632 This completesthe proof

Theorem 8 120590119903(119860(119903 119904 119905) ℓ119901) = 0

Proof By Proposition 1 120590119903(119860(119903 119904 119905) ℓ119901) = 120590119901(119860(119903 119904 119905)lowast ℓlowast

119901)

120590119901(119860(119903 119904 119905) ℓ119901) Since byTheorem 6

120590119901 (119860(119903 119904 119905)lowast ℓlowast

119901) = 0 120590119903 (119860 (119903 119904 119905) ℓ119901) = 0

(46)

This completes the proof

Theorem 9 Let 119904 be a complex number such that radic1199042 = minus119904and define the set1198631 by

1198631 = 120572 isin C 2 |119903 minus 120572| ⩽1003816100381610038161003816100381610038161003816minus119904 + radic1199042 minus 4119905 (119903 minus 120572)

1003816100381610038161003816100381610038161003816

(47)

Then 120590(119860(119903 119904 119905) ℓ119901) = 1198631

Proof ByTheorem 7 we get

120572 isin C 2 |119903 minus 120572| lt1003816100381610038161003816100381610038161003816minus119904 + radic1199042 minus 4119905 (119903 minus 120572)

1003816100381610038161003816100381610038161003816

sube 120590 (119860 (119903 119904 119905) ℓ119901)

(48)

Since the spectrum of any bounded operator is closed wehave

120572 isin C 2 |119903 minus 120572| ⩽1003816100381610038161003816100381610038161003816minus119904 + radic1199042 minus 4119905 (119903 minus 120572)

1003816100381610038161003816100381610038161003816

sube 120590 (119860 (119903 119904 119905) ℓ119901)

(49)

and again fromTheorems 5 7 and 8

120590 (119860 (119903 119904 119905) ℓ119901) sube 120572 isin C 2 |119903 minus 120572|

⩽1003816100381610038161003816100381610038161003816minus119904 + radic1199042 minus 4119905 (119903 minus 120572)

1003816100381610038161003816100381610038161003816

(50)

Combining (49) and (50) we obtain that 120590(119860(119903 119904 119905) ℓ119901) =1198631 where1198631 is defined by (47)

Theorem 10 120590119888(119860(119903 119904 119905) ℓ119901) = 1198633 where 1198633 = 120572 isin C

2|119903 minus 120572| = | minus 119904 + radic1199042 minus 4119905(119903 minus 120572)|

Proof Because the parts 120590119888(119860(119903 119904 119905) ℓ119901) 120590119903(119860(119903 119904 119905) ℓ119901)and 120590119901(119860(119903 119904 119905) ℓ119901) are pairwise disjoint sets and the unionof these sets is 120590(119860(119903 119904 119905) ℓ119901) the proof immediately followsfromTheorems 7 8 and 9

Theorem 11 If 120572 isin 1198632 120572 isin 120590(119860(119903 119904 119905) ℓ119901)1198683

Proof From Theorem 7 120572 isin 120590119901(119860(119903 119904 119905) ℓ119901) Thus (119860(119903119904 119905) minus 120572119868)

minus1 does not exist By Theorem 6119860(119903 119904 119905)lowast minus 120572119868 isone to one so 119860(119903 119904 119905) minus 120572119868 has a dense range in ℓ119901 byLemma 3

Theorem 12 The following statements hold

(i) 120590ap(119860(119903 119904 119905) ℓ119901) = 1198631(ii) 120590120575(119860(119903 119904 119905) ℓ119901) = 1198633(iii) 120590co(119860(119903 119904 119905) ℓ119901) = 0

Abstract and Applied Analysis 7

Proof (i) Since from Table 1

120590ap (119860 (119903 119904 119905) ℓ119901)=120590 (119860 (119903 119904 119905) ℓ119901)120590 (119860 (119903 119904 119905) ℓ119901) 1198681198681198681

(51)

we have byTheorem 8

120590 (119860 (119903 119904 119905) ℓ119901) 1198681198681198681 = 120590 (119860 (119903 119904 119905) ℓ119901) 1198681198681198682 = 0 (52)

Hence

120590ap (119860 (119903 119904 119905) ℓ119901) = 1198631 (53)

(ii) Since the following equality

120590120575 (119860 (119903 119904 119905) ℓ119901) = 120590 (119860 (119903 119904 119905) ℓ119901) 120590 (119860 (119903 119904 119905) ℓ119901) 1198683

(54)

holds from Table 1 we derive by Theorems 8 and 11 that120590120575(119860(119903 119904 119905) ℓ119901) = 1198632

(iii) From Table 1 we have

120590co (119860 (119903 119904 119905) ℓ119901)

= 120590 (119860 (119903 119904 119905) ℓ119901) 1198681198681198681 cup 120590 (119860 (119903 119904 119905) ℓ119901) 1198681198681198682

cup 120590 (119860 (119903 119904 119905) 1198880) 1198681198681198683

(55)

ByTheorem 6 it is immediate that 120590co(119860(119903 119904 119905) ℓ119901) = 0

4 Fine Spectra of Upper TriangularTriple-Band Matrices over the SequenceSpace ℓ119901 (1 lt 119901 lt infin)

In the present section we determine the fine spectrum ofthe operator 119860(119903 119904 119905) ℓ119901 rarr ℓ119901 in case 1 ⩽ 119901 lt infinWe quote some lemmas which are needed in proving thetheorems given in Section 4

Lemma 13 (see [17 p 253 Theorem 3416]) The matrix 119860 =(119886119899119896) gives rise to a bounded linear operator 119879 isin 119861(ℓ1) from ℓ1to itself if and only if the supremum of ℓ1 norms of the columnsof 119860 is bounded

Lemma 14 (see [17 p 245 Theorem 343]) The matrix 119860 =(119886119899119896) gives rise to a bounded linear operator 119879 isin 119861(ℓinfin) fromℓinfin to itself if and only if the supremum of ℓ1 norms of the rowsof 119860 is bounded

Lemma 15 (see [17 p 254 Theorem 3418]) Let 1 lt 119901 lt infinand 119860 isin (ℓinfin ℓinfin) cap (ℓ1 ℓ1) Then 119860 isin (ℓ119901 ℓ119901)

Theorem 16 The operator 119860(119903 119904 119905) ℓ119901 rarr ℓ119901 is a boundedlinear operator and

(|119903|119901+ |119904|119901+ |119905|119901)1119901⩽ 119860 (119903 119904 119905)(ℓ119901 ℓ119901)

⩽ |119903| + |119904| + |119905|

(56)

Proof Since the linearity of the operator119860(119903 119904 119905) is not hardwe omit the details Now we prove that (56) holds for theoperator 119860(119903 119904 119905) on the space ℓ119901 It is trivial that 119860(119903 119904119905)119890(2)= (119905 119904 119903 0 ) for 119890(2) isin ℓ119901 Therefore we have

119860 (119903 119904 119905)(ℓ119901 ℓ119901)⩾

10038171003817100381710038171003817119860 (119903 119904 119905) 119890

(2)100381710038171003817100381710038171199011003817100381710038171003817119890(2)1003817100381710038171003817119901

= (|119903|119901+ |119904|119901+ |119905|119901)1119901

(57)

which implies that

119860 (119903 119904 119905)(ℓ119901 ℓ119901)⩾ (|119903|119901+ |s|119901 + |119905|119901)1119901 (58)

Let 119909 = (119909119896) isin ℓ119901 where 1 lt 119901 lt infin Then since (119905119909119896+2)(119903119909119896) and (119904119909119896+1) isin ℓ119901 it is easy to see by Minkowskirsquos ine-quality that

119860 (119903 119904 119905) 119909119901

= (

infin

sum

119896=0

1003816100381610038161003816119903119909119896 + 119904119909119896+1 + 119905119909119896+21003816100381610038161003816119901)

1119901

⩽ (

infin

sum

119896=0

10038161003816100381610038161199031199091198961003816100381610038161003816119901)

1119901

+ (

infin

sum

119896=0

1003816100381610038161003816119904119909119896+11003816100381610038161003816119901)

1119901

+ (

infin

sum

119896=0

1003816100381610038161003816119905119909119896+21003816100381610038161003816119901)

1119901

= |119903| (

infin

sum

119896=0

10038161003816100381610038161199091198961003816100381610038161003816119901)

1119901

+ |119904| (

infin

sum

119896=0

1003816100381610038161003816119909119896+11003816100381610038161003816119901)

1119901

+ |119905| (

infin

sum

119896=0

1003816100381610038161003816119909119896+21003816100381610038161003816119901)

1119901

= |119904| 119909119901 + |119903| 119909119901 + |119905| 119909119901

= (|119903| + |119904| + |119905|) 119909119901

(59)

which leads us to the result that

(|119903|119901+ |119904|119901+ |119905|119901)1119901⩽ 119860 (119903 119904 119905)(ℓ119901 ℓ119901)

⩽ |119903| + |119904| + |119905|

(60)

Therefore by combining the inequalities in (58) and (59) wehave (56) as desired

If 119879 ℓ119901 rarr ℓ119901 is a bounded matrix operator with thematrix 119860 then it is known that the adjoint operator 119879lowast ℓlowast

119901rarr ℓlowast

119901is defined by the transpose of thematrix119860The dual

space of ℓ119901 is isomorphic to ℓ119902 where 1 lt 119901 lt infin

Theorem 17 Let 119904 be a complex number such that radic1199042 = minus119904and define the set1198631 by

1198631 = 120572 isin C 2 |119903 minus 120572| ⩽1003816100381610038161003816100381610038161003816minus119904 + radic1199042 minus 4119905 (119903 minus 120572)

1003816100381610038161003816100381610038161003816

(61)

Then 120590119888(119860(119903 119904 119905) ℓ119901) sube 1198631

8 Abstract and Applied Analysis

Proof We will show that 119860120572(119903 119904 119905)lowast is onto for 2|119903 minus 120572| gt

| minus 119904 + radic1199042 minus 4119905(119903 minus 120572)| Thus for every 119910 isin ℓ119902 we find 119909 isinℓ119902 119860120572(119903 119904 119905)

lowast is a triangle so it has an inverse Also equation119860120572(119903 119904 119905)

lowast119909 = 119910 gives [119860120572(119903 119904 119905)

lowast]minus1119910 = 119909 It is sufficient

to show that [119860120572(119903 119904 119905)lowast]minus1isin (ℓ119902 ℓ119902) We calculate that

119860 = (119886119899119896) = [119860120572(119903 119904 119905)lowast]minus1 as follows

119860 = (119886119899119896) =

[[[[

[

1198861 0 0

1198862 1198861 0

1198863 1198862 1198861

]]]]

]

(62)

where

1198861 =1

119903 minus 120572

1198862 =minus119904

(119903 minus 120572)2

1198863 =1199042minus 119905 (119903 minus 120572)

(119903 minus 120572)3

(63)

It is known that fromTheorem 5

119886119899 =120582119899

1minus 120582119899

2

radic1199042 minus 4119905 (119903 minus 120572) forall119899 ⩾ 1

where 1205821 =minus119904 + radicΔ

119903 minus 120572 1205822 =

minus119904 minus radicΔ

119903 minus 120572

(64)

Now we show that [119860120572(119903 119904 119905)lowast]minus1isin (ℓ1 ℓ1) for |1205821| lt 1

By Theorem 5 we know that if |1205821| lt 1 |1205822| lt 1 We assumethat 1199042 minus 4119905(119903 minus 120572) = 0 and |1205821| lt 1

10038171003817100381710038171003817[119860120572(119903 119904 119905)

lowast]minus110038171003817100381710038171003817(ℓ1 ℓ1)

= sup119899

infin

sum

119896=119899

10038161003816100381610038161198861198961003816100381610038161003816 =

infin

sum

119896=1

10038161003816100381610038161198861198961003816100381610038161003816

⩽1

|Δ|(

infin

sum

119896=1

100381610038161003816100381612058211003816100381610038161003816119896+

infin

sum

119896=1

100381610038161003816100381612058221003816100381610038161003816119896) lt infin

(65)

since |1205821| lt 1 and |1205822| lt 1 This shows that (119860120572(119903 119904 119905)lowast]minus1isin

(ℓ1 ℓ1) Similarly we can show that [119860120572(119903 119904 119905)lowast]minus1isin (ℓinfin

ℓinfin)Now assume that 1199042 minus 4119905(119903 minus 120572) = 0 Then

119886119899 = (2119899

minus119904) [

minus119904

2 (119903 minus 120572)]

119899

(66)

and simple calculation gives that (119886119899) isin ℓ119902 if and only if |minus119904| lt2|119903 minus 120572|

[(119860 (119903 119904 119905) minus 120572119868)lowast]minus1isin (ℓ119902 ℓ119902)

for 120572 isin C with 2 |119903 minus 120572| gt1003816100381610038161003816100381610038161003816minus119904 + radic1199042 minus 4119905 (119903 minus 120572)

1003816100381610038161003816100381610038161003816

(67)

Hence 119860120572(119903 119904 119905)lowast is onto By Lemma 2 119860120572(119903 119904 119905) has a

bounded inverse This means that 120590119888(119860(119903 119904 119905) ℓ119901) sube 1198631where1198631 is defined by (61)

Theorem 18 120590119901(119860(119903 119904 119905)lowast ℓlowast

119901) = 0

Proof Let 119860(119903 119904 119905)lowast119891 = 120572119891 with 119891 = 120579 = (0 0 0 ) in ℓlowast119901=

ℓ119902 Then by solving the system of linear equations

1199031198910 = 1205721198910

1199041198910 + 1199031198911 = 1205721198911

1199051198910 + 1199041198911 + 1199031198912 = 1205721198912

1199051198911 + 1199041198912 + 1199031198913 = 1205721198913

119905119891119896minus2 + 119904119891119896minus1 + 119903119891119896 = 120572119891119896

(68)

we find that 1198910 = 0 if 120572 = 119903 and 1198911 = 1198912 = sdot sdot sdot = 0 if 1198910 = 0which contradicts119891 = 120579 If 1198911198990 is the first nonzero entry of thesequence 119891 = (119891119899) and 120572 = 119903 then we get 1199051198911198990minus2 + 1199041198911198990minus1 +1199031198911198990

= 1205721198911198990which implies 1198911198990 = 0 which contradicts the

assumption1198911198990 = 0 Hence the equation119860(119903 119904 119905)lowast119891 = 120572119891 has

no solution 119891 = 120579

In the case 1 lt 119901 lt infin since the proof of the theoremsin Section 4 determining the spectrum and fine spectrumof the matrix operator 119860(119903 119904 119905) on the sequence space ℓ119901is similar to the case 0 lt 119901 ⩽ 1 to avoid the repetitionof similar statements we give the results by the followingtheorem without proof

Theorem 19 The following statements hold(i) 120590(119860(119903 119904 119905) ℓ119901) = 1198631(ii) 120590119903(119860(119903 119904 119905) ℓ119901) = 0(iii) 120590119901(119860(119903 119904 119905) ℓ119901) = 1198632(iv) 120590119888(119860(119903 119904 119905) ℓ119901) = 1198633(v) 120590ap(119860(119903 119904 119905) ℓ119901) = 1198631(vi) 120590co(119860(119903 119904 119905) ℓ119901) = 0(vii) 120590120575(119860(119903 119904 119905) ℓ119901) = 1198633

5 Some Applications

In this section we give two theorems related to Toeplitz ma-trix

Theorem 20 Let 119875 be a polynomial that corresponds to the 119899-tuple 119886 and let 1199111 1199112 1199113 119911119899minus1 also be the roots of 119875 Define119879 as a Toeplitz matrix associated with 119875 that is

119879 =

[[[[

[

1198860 1198861 1198862 119886119899 0 0 0

0 1198860 1198861 1198862 119886119899 0 0

0 0 1198860 1198861 1198862 119886119899 0

]]]]

]

(69)

Abstract and Applied Analysis 9

The resolvent operator 119879 over ℓ119901 with 1 lt 119901 lt infin where thedomain of the resolvent operator is the whole space ℓ119901 exists ifand if only all the roots of the polynomial are outside the unitdisc 119911 isin C |119911| ⩽ 1 That is 119879minus1 isin (ℓ119901 ℓ119901) if and if only|119911119894| gt 1 1 ⩽ 119894 ⩽ 119899 minus 1 In this case the resolvent operator isrepresented by

119879minus1=

1

119886119899minus1119860minus1(minus1199111 1) 119860

minus1(minus1199112 1) sdot sdot sdot 119860

minus1(minus119911119899minus1 1)

where

119860minus1(minus119911119894 1) =minus

[[[[[[[[[[[[

[

1119911119894 11199112

11989411199113

11989411199114

11989411199115

119894

0 1119911119894 11199112

11989411199113

11989411199114

119894

0 0 1119911119894 11199112

11989411199113

119894

0 0 0 1119911119894 11199112

119894

0 0 0 0 1119911119894

]]]]]]]]]]]]

]

(70)

Proof Suppose all the roots of the polynomial 119875(119911) = 1198860 +1198861119911 + sdot sdot sdot + 119886119899minus1119911

119899minus1= 119886119899(119911 minus 1199111)(119911 minus 1199112) sdot sdot sdot (119911 minus 119911119899minus1) are

outside the unit disc The Toeplitz matrix associated with 119875can be written as the product

119879 = 119886119899119860 (minus1199111 1) 119860 (minus1199112 1) sdot sdot sdot 119860 (minus119911119899minus1 1) (71)

Since multiplication of upper triangular Toeplitz matrices iscommutative we can see that

119879minus1=

1

119886119899minus1119860minus1(minus1199111 1) 119860

minus1(minus1199112 1) sdot sdot sdot 119860

minus1(minus119911119899minus1 1)

(72)

is left inverse of 119879 Since all roots are outside the unit discthen

10038171003817100381710038171003817119879minus1(minus119911119894 1)

10038171003817100381710038171003817(ℓinfin ℓinfin)= sup119899

infin

sum

119896=119899

1

10038161003816100381610038161199111198941003816100381610038161003816119896+1minus119899

=

infin

sum

119896=1

1

10038161003816100381610038161199111198941003816100381610038161003816119896lt infin

(73)

Therefore each 119879minus1(minus119911119894 1) isin (ℓinfin ℓinfin) for 1 ⩽ 119894 ⩽ 119899 minus 1Similarly we can say that 119879minus1(minus119911119894 1) isin (ℓ1 ℓ1) So we have119879minus1(minus119911119894 1) isin (ℓ119901 ℓ119901)

Theorem 21 The resolvent operator of 119860(119903 119904 119905) over ℓ119901 with1 lt 119901 lt infin where the domain of the resolvent operator is thespace ℓ119901 exists if and only if 2|119903| gt | minus 119904 + radic1199042 minus 4119905119903| In this

case the resolvent operator is represented by the infinite bandedToeplitz matrix

119864 (119903 119904 119905)

=1

119905

[[[[[[[[[[

[

1199061 1199062

11199063

11199064

1

0 1199061 1199062

11199063

1

0 0 1199061 1199062

1

0 0 0 1199061

]]]]]]]]]]

]

[[[[[[[[[[

[

1199062 1199062

21199063

21199064

2

0 1199062 1199062

21199063

2

0 0 1199062 1199062

2

0 0 0 1199062

]]]]]]]]]]

]

where 1199061 =minus119904 + radic1199042 minus 4119905119903

2119903 1199062 =

minus119904 minus radic1199042 minus 4119905119903

2119903

(74)

Proof By Theorem 20 we can see that 119864(119903 119904 119905) is inverseof the matrix of 119860(119903 119904 119905) But this is not enough to say itis resolvent operator By Lemmas 13 14 and 15 119864(119903 119904 119905) isin(ℓ119901 ℓ119901) when 2|119903| gt | minus 119904 + radic1199042 minus 4119905119903| That is for 2|119903| gt| minus 119904 + radic1199042 minus 4119905119903| 119864(119903 119904 119905) is a resolvent operator

Acknowledgments

The authors would like to thank the referee for pointingout some mistakes and misprints in the earlier version ofthis paper They would like to express their pleasure toMs Medine Yesilkayagil Department of Mathematics UsakUniversity 1 Eylul Campus Usak Turkey for many helpfulsuggestions and interesting comments on the revised form ofthe paper

References

[1] M Gonzalez ldquoThe fine spectrum of the Cesaro operator in ℓ119901(1 lt 119901 lt infin)rdquo Archiv der Mathematik vol 44 no 4 pp 355ndash358 1985

[2] P J CartlidgeWeighted meanmatrices as operators on ℓ119901 [PhDdissertation] Indiana University 1978

[3] J I Okutoyi ldquoOn the spectrum of C1 as an operator on 1198871199070rdquoAustralian Mathematical Society A vol 48 no 1 pp 79ndash861990

[4] J I Okutoyi ldquoOn the spectrum of C1 as an operator on bvrdquoCommunications A vol 41 no 1-2 pp 197ndash207 1992

[5] M Yıldırım ldquoOn the spectrum of the Rhaly operators on ℓ119901rdquoIndian Journal of Pure and Applied Mathematics vol 32 no 2pp 191ndash198 2001

[6] B Altay and F Basar ldquoOn the fine spectrum of the differenceoperator Δ on 1198880 and 119888rdquo Information Sciences vol 168 no 1ndash4pp 217ndash224 2004

[7] B Altay and F Basar ldquoOn the fine spectrum of the generalizeddifference operator 119861(119903 119904) over the sequence spaces 1198880 and119888rdquo International Journal of Mathematics and MathematicalSciences vol 2005 no 18 pp 3005ndash3013 2005

[8] A M Akhmedov and F Basar ldquoOn the fine spectra of the dif-ference operator Δ over the sequence space ℓ119901 (1 le 119901 lt infin)rdquoDemonstratio Mathematica vol 39 no 3 pp 585ndash595 2006

10 Abstract and Applied Analysis

[9] AMAkhmedov andF Basar ldquoThefine spectra of the differenceoperator Δ over the sequence space 119887119907119901 (1 le 119901 lt infin)rdquo ActaMathematica Sinica vol 23 no 10 pp 1757ndash1768 2007

[10] F Basar and B Altay ldquoOn the space of sequences of p-boundedvariation and related matrix mappingsrdquo Ukrainian Mathemati-cal Journal vol 55 no 1 pp 136ndash147 2003

[11] H Furkan H Bilgic and F Basar ldquoOn the fine spectrum of theoperator 119861(119903 119904 119905) over the sequence spaces ℓ119901 and 119887119907119901(1 lt 119901 ltinfin)rdquo Computers ampMathematics with Applications vol 60 no 7pp 2141ndash2152 2010

[12] V Karakaya and M Altun ldquoFine spectra of upper triangulardouble-band matricesrdquo Journal of Computational and AppliedMathematics vol 234 no 5 pp 1387ndash1394 2010

[13] A Karaisa ldquoFine spectra of upper triangular double-bandmatrices over the sequence space ℓ119901 (1 lt 119901 lt infin)rdquo DiscreteDynamics in Nature and Society vol 2012 Article ID 381069 19pages 2012

[14] E Kreyszig Introductory Functional Analysis with ApplicationsJohn Wiley amp Sons New York NY USA 1978

[15] J Appell E Pascale and A Vignoli Nonlinear Spectral Theoryvol 10 of de Gruyter Series in Nonlinear Analysis and Applica-tions Walter de Gruyter Berlin Germany 2004

[16] S Goldberg Unbounded Linear Operators Dover PublicationsNew York NY USA 1985

[17] B Choudhary and S Nanda Functional Analysis with Applica-tions John Wiley amp Sons New York NY USA 1989

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

Volume 2014

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Mathematical PhysicsAdvances in

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 7

Proof (i) Since from Table 1

120590ap (119860 (119903 119904 119905) ℓ119901)=120590 (119860 (119903 119904 119905) ℓ119901)120590 (119860 (119903 119904 119905) ℓ119901) 1198681198681198681

(51)

we have byTheorem 8

120590 (119860 (119903 119904 119905) ℓ119901) 1198681198681198681 = 120590 (119860 (119903 119904 119905) ℓ119901) 1198681198681198682 = 0 (52)

Hence

120590ap (119860 (119903 119904 119905) ℓ119901) = 1198631 (53)

(ii) Since the following equality

120590120575 (119860 (119903 119904 119905) ℓ119901) = 120590 (119860 (119903 119904 119905) ℓ119901) 120590 (119860 (119903 119904 119905) ℓ119901) 1198683

(54)

holds from Table 1 we derive by Theorems 8 and 11 that120590120575(119860(119903 119904 119905) ℓ119901) = 1198632

(iii) From Table 1 we have

120590co (119860 (119903 119904 119905) ℓ119901)

= 120590 (119860 (119903 119904 119905) ℓ119901) 1198681198681198681 cup 120590 (119860 (119903 119904 119905) ℓ119901) 1198681198681198682

cup 120590 (119860 (119903 119904 119905) 1198880) 1198681198681198683

(55)

ByTheorem 6 it is immediate that 120590co(119860(119903 119904 119905) ℓ119901) = 0

4 Fine Spectra of Upper TriangularTriple-Band Matrices over the SequenceSpace ℓ119901 (1 lt 119901 lt infin)

In the present section we determine the fine spectrum ofthe operator 119860(119903 119904 119905) ℓ119901 rarr ℓ119901 in case 1 ⩽ 119901 lt infinWe quote some lemmas which are needed in proving thetheorems given in Section 4

Lemma 13 (see [17 p 253 Theorem 3416]) The matrix 119860 =(119886119899119896) gives rise to a bounded linear operator 119879 isin 119861(ℓ1) from ℓ1to itself if and only if the supremum of ℓ1 norms of the columnsof 119860 is bounded

Lemma 14 (see [17 p 245 Theorem 343]) The matrix 119860 =(119886119899119896) gives rise to a bounded linear operator 119879 isin 119861(ℓinfin) fromℓinfin to itself if and only if the supremum of ℓ1 norms of the rowsof 119860 is bounded

Lemma 15 (see [17 p 254 Theorem 3418]) Let 1 lt 119901 lt infinand 119860 isin (ℓinfin ℓinfin) cap (ℓ1 ℓ1) Then 119860 isin (ℓ119901 ℓ119901)

Theorem 16 The operator 119860(119903 119904 119905) ℓ119901 rarr ℓ119901 is a boundedlinear operator and

(|119903|119901+ |119904|119901+ |119905|119901)1119901⩽ 119860 (119903 119904 119905)(ℓ119901 ℓ119901)

⩽ |119903| + |119904| + |119905|

(56)

Proof Since the linearity of the operator119860(119903 119904 119905) is not hardwe omit the details Now we prove that (56) holds for theoperator 119860(119903 119904 119905) on the space ℓ119901 It is trivial that 119860(119903 119904119905)119890(2)= (119905 119904 119903 0 ) for 119890(2) isin ℓ119901 Therefore we have

119860 (119903 119904 119905)(ℓ119901 ℓ119901)⩾

10038171003817100381710038171003817119860 (119903 119904 119905) 119890

(2)100381710038171003817100381710038171199011003817100381710038171003817119890(2)1003817100381710038171003817119901

= (|119903|119901+ |119904|119901+ |119905|119901)1119901

(57)

which implies that

119860 (119903 119904 119905)(ℓ119901 ℓ119901)⩾ (|119903|119901+ |s|119901 + |119905|119901)1119901 (58)

Let 119909 = (119909119896) isin ℓ119901 where 1 lt 119901 lt infin Then since (119905119909119896+2)(119903119909119896) and (119904119909119896+1) isin ℓ119901 it is easy to see by Minkowskirsquos ine-quality that

119860 (119903 119904 119905) 119909119901

= (

infin

sum

119896=0

1003816100381610038161003816119903119909119896 + 119904119909119896+1 + 119905119909119896+21003816100381610038161003816119901)

1119901

⩽ (

infin

sum

119896=0

10038161003816100381610038161199031199091198961003816100381610038161003816119901)

1119901

+ (

infin

sum

119896=0

1003816100381610038161003816119904119909119896+11003816100381610038161003816119901)

1119901

+ (

infin

sum

119896=0

1003816100381610038161003816119905119909119896+21003816100381610038161003816119901)

1119901

= |119903| (

infin

sum

119896=0

10038161003816100381610038161199091198961003816100381610038161003816119901)

1119901

+ |119904| (

infin

sum

119896=0

1003816100381610038161003816119909119896+11003816100381610038161003816119901)

1119901

+ |119905| (

infin

sum

119896=0

1003816100381610038161003816119909119896+21003816100381610038161003816119901)

1119901

= |119904| 119909119901 + |119903| 119909119901 + |119905| 119909119901

= (|119903| + |119904| + |119905|) 119909119901

(59)

which leads us to the result that

(|119903|119901+ |119904|119901+ |119905|119901)1119901⩽ 119860 (119903 119904 119905)(ℓ119901 ℓ119901)

⩽ |119903| + |119904| + |119905|

(60)

Therefore by combining the inequalities in (58) and (59) wehave (56) as desired

If 119879 ℓ119901 rarr ℓ119901 is a bounded matrix operator with thematrix 119860 then it is known that the adjoint operator 119879lowast ℓlowast

119901rarr ℓlowast

119901is defined by the transpose of thematrix119860The dual

space of ℓ119901 is isomorphic to ℓ119902 where 1 lt 119901 lt infin

Theorem 17 Let 119904 be a complex number such that radic1199042 = minus119904and define the set1198631 by

1198631 = 120572 isin C 2 |119903 minus 120572| ⩽1003816100381610038161003816100381610038161003816minus119904 + radic1199042 minus 4119905 (119903 minus 120572)

1003816100381610038161003816100381610038161003816

(61)

Then 120590119888(119860(119903 119904 119905) ℓ119901) sube 1198631

8 Abstract and Applied Analysis

Proof We will show that 119860120572(119903 119904 119905)lowast is onto for 2|119903 minus 120572| gt

| minus 119904 + radic1199042 minus 4119905(119903 minus 120572)| Thus for every 119910 isin ℓ119902 we find 119909 isinℓ119902 119860120572(119903 119904 119905)

lowast is a triangle so it has an inverse Also equation119860120572(119903 119904 119905)

lowast119909 = 119910 gives [119860120572(119903 119904 119905)

lowast]minus1119910 = 119909 It is sufficient

to show that [119860120572(119903 119904 119905)lowast]minus1isin (ℓ119902 ℓ119902) We calculate that

119860 = (119886119899119896) = [119860120572(119903 119904 119905)lowast]minus1 as follows

119860 = (119886119899119896) =

[[[[

[

1198861 0 0

1198862 1198861 0

1198863 1198862 1198861

]]]]

]

(62)

where

1198861 =1

119903 minus 120572

1198862 =minus119904

(119903 minus 120572)2

1198863 =1199042minus 119905 (119903 minus 120572)

(119903 minus 120572)3

(63)

It is known that fromTheorem 5

119886119899 =120582119899

1minus 120582119899

2

radic1199042 minus 4119905 (119903 minus 120572) forall119899 ⩾ 1

where 1205821 =minus119904 + radicΔ

119903 minus 120572 1205822 =

minus119904 minus radicΔ

119903 minus 120572

(64)

Now we show that [119860120572(119903 119904 119905)lowast]minus1isin (ℓ1 ℓ1) for |1205821| lt 1

By Theorem 5 we know that if |1205821| lt 1 |1205822| lt 1 We assumethat 1199042 minus 4119905(119903 minus 120572) = 0 and |1205821| lt 1

10038171003817100381710038171003817[119860120572(119903 119904 119905)

lowast]minus110038171003817100381710038171003817(ℓ1 ℓ1)

= sup119899

infin

sum

119896=119899

10038161003816100381610038161198861198961003816100381610038161003816 =

infin

sum

119896=1

10038161003816100381610038161198861198961003816100381610038161003816

⩽1

|Δ|(

infin

sum

119896=1

100381610038161003816100381612058211003816100381610038161003816119896+

infin

sum

119896=1

100381610038161003816100381612058221003816100381610038161003816119896) lt infin

(65)

since |1205821| lt 1 and |1205822| lt 1 This shows that (119860120572(119903 119904 119905)lowast]minus1isin

(ℓ1 ℓ1) Similarly we can show that [119860120572(119903 119904 119905)lowast]minus1isin (ℓinfin

ℓinfin)Now assume that 1199042 minus 4119905(119903 minus 120572) = 0 Then

119886119899 = (2119899

minus119904) [

minus119904

2 (119903 minus 120572)]

119899

(66)

and simple calculation gives that (119886119899) isin ℓ119902 if and only if |minus119904| lt2|119903 minus 120572|

[(119860 (119903 119904 119905) minus 120572119868)lowast]minus1isin (ℓ119902 ℓ119902)

for 120572 isin C with 2 |119903 minus 120572| gt1003816100381610038161003816100381610038161003816minus119904 + radic1199042 minus 4119905 (119903 minus 120572)

1003816100381610038161003816100381610038161003816

(67)

Hence 119860120572(119903 119904 119905)lowast is onto By Lemma 2 119860120572(119903 119904 119905) has a

bounded inverse This means that 120590119888(119860(119903 119904 119905) ℓ119901) sube 1198631where1198631 is defined by (61)

Theorem 18 120590119901(119860(119903 119904 119905)lowast ℓlowast

119901) = 0

Proof Let 119860(119903 119904 119905)lowast119891 = 120572119891 with 119891 = 120579 = (0 0 0 ) in ℓlowast119901=

ℓ119902 Then by solving the system of linear equations

1199031198910 = 1205721198910

1199041198910 + 1199031198911 = 1205721198911

1199051198910 + 1199041198911 + 1199031198912 = 1205721198912

1199051198911 + 1199041198912 + 1199031198913 = 1205721198913

119905119891119896minus2 + 119904119891119896minus1 + 119903119891119896 = 120572119891119896

(68)

we find that 1198910 = 0 if 120572 = 119903 and 1198911 = 1198912 = sdot sdot sdot = 0 if 1198910 = 0which contradicts119891 = 120579 If 1198911198990 is the first nonzero entry of thesequence 119891 = (119891119899) and 120572 = 119903 then we get 1199051198911198990minus2 + 1199041198911198990minus1 +1199031198911198990

= 1205721198911198990which implies 1198911198990 = 0 which contradicts the

assumption1198911198990 = 0 Hence the equation119860(119903 119904 119905)lowast119891 = 120572119891 has

no solution 119891 = 120579

In the case 1 lt 119901 lt infin since the proof of the theoremsin Section 4 determining the spectrum and fine spectrumof the matrix operator 119860(119903 119904 119905) on the sequence space ℓ119901is similar to the case 0 lt 119901 ⩽ 1 to avoid the repetitionof similar statements we give the results by the followingtheorem without proof

Theorem 19 The following statements hold(i) 120590(119860(119903 119904 119905) ℓ119901) = 1198631(ii) 120590119903(119860(119903 119904 119905) ℓ119901) = 0(iii) 120590119901(119860(119903 119904 119905) ℓ119901) = 1198632(iv) 120590119888(119860(119903 119904 119905) ℓ119901) = 1198633(v) 120590ap(119860(119903 119904 119905) ℓ119901) = 1198631(vi) 120590co(119860(119903 119904 119905) ℓ119901) = 0(vii) 120590120575(119860(119903 119904 119905) ℓ119901) = 1198633

5 Some Applications

In this section we give two theorems related to Toeplitz ma-trix

Theorem 20 Let 119875 be a polynomial that corresponds to the 119899-tuple 119886 and let 1199111 1199112 1199113 119911119899minus1 also be the roots of 119875 Define119879 as a Toeplitz matrix associated with 119875 that is

119879 =

[[[[

[

1198860 1198861 1198862 119886119899 0 0 0

0 1198860 1198861 1198862 119886119899 0 0

0 0 1198860 1198861 1198862 119886119899 0

]]]]

]

(69)

Abstract and Applied Analysis 9

The resolvent operator 119879 over ℓ119901 with 1 lt 119901 lt infin where thedomain of the resolvent operator is the whole space ℓ119901 exists ifand if only all the roots of the polynomial are outside the unitdisc 119911 isin C |119911| ⩽ 1 That is 119879minus1 isin (ℓ119901 ℓ119901) if and if only|119911119894| gt 1 1 ⩽ 119894 ⩽ 119899 minus 1 In this case the resolvent operator isrepresented by

119879minus1=

1

119886119899minus1119860minus1(minus1199111 1) 119860

minus1(minus1199112 1) sdot sdot sdot 119860

minus1(minus119911119899minus1 1)

where

119860minus1(minus119911119894 1) =minus

[[[[[[[[[[[[

[

1119911119894 11199112

11989411199113

11989411199114

11989411199115

119894

0 1119911119894 11199112

11989411199113

11989411199114

119894

0 0 1119911119894 11199112

11989411199113

119894

0 0 0 1119911119894 11199112

119894

0 0 0 0 1119911119894

]]]]]]]]]]]]

]

(70)

Proof Suppose all the roots of the polynomial 119875(119911) = 1198860 +1198861119911 + sdot sdot sdot + 119886119899minus1119911

119899minus1= 119886119899(119911 minus 1199111)(119911 minus 1199112) sdot sdot sdot (119911 minus 119911119899minus1) are

outside the unit disc The Toeplitz matrix associated with 119875can be written as the product

119879 = 119886119899119860 (minus1199111 1) 119860 (minus1199112 1) sdot sdot sdot 119860 (minus119911119899minus1 1) (71)

Since multiplication of upper triangular Toeplitz matrices iscommutative we can see that

119879minus1=

1

119886119899minus1119860minus1(minus1199111 1) 119860

minus1(minus1199112 1) sdot sdot sdot 119860

minus1(minus119911119899minus1 1)

(72)

is left inverse of 119879 Since all roots are outside the unit discthen

10038171003817100381710038171003817119879minus1(minus119911119894 1)

10038171003817100381710038171003817(ℓinfin ℓinfin)= sup119899

infin

sum

119896=119899

1

10038161003816100381610038161199111198941003816100381610038161003816119896+1minus119899

=

infin

sum

119896=1

1

10038161003816100381610038161199111198941003816100381610038161003816119896lt infin

(73)

Therefore each 119879minus1(minus119911119894 1) isin (ℓinfin ℓinfin) for 1 ⩽ 119894 ⩽ 119899 minus 1Similarly we can say that 119879minus1(minus119911119894 1) isin (ℓ1 ℓ1) So we have119879minus1(minus119911119894 1) isin (ℓ119901 ℓ119901)

Theorem 21 The resolvent operator of 119860(119903 119904 119905) over ℓ119901 with1 lt 119901 lt infin where the domain of the resolvent operator is thespace ℓ119901 exists if and only if 2|119903| gt | minus 119904 + radic1199042 minus 4119905119903| In this

case the resolvent operator is represented by the infinite bandedToeplitz matrix

119864 (119903 119904 119905)

=1

119905

[[[[[[[[[[

[

1199061 1199062

11199063

11199064

1

0 1199061 1199062

11199063

1

0 0 1199061 1199062

1

0 0 0 1199061

]]]]]]]]]]

]

[[[[[[[[[[

[

1199062 1199062

21199063

21199064

2

0 1199062 1199062

21199063

2

0 0 1199062 1199062

2

0 0 0 1199062

]]]]]]]]]]

]

where 1199061 =minus119904 + radic1199042 minus 4119905119903

2119903 1199062 =

minus119904 minus radic1199042 minus 4119905119903

2119903

(74)

Proof By Theorem 20 we can see that 119864(119903 119904 119905) is inverseof the matrix of 119860(119903 119904 119905) But this is not enough to say itis resolvent operator By Lemmas 13 14 and 15 119864(119903 119904 119905) isin(ℓ119901 ℓ119901) when 2|119903| gt | minus 119904 + radic1199042 minus 4119905119903| That is for 2|119903| gt| minus 119904 + radic1199042 minus 4119905119903| 119864(119903 119904 119905) is a resolvent operator

Acknowledgments

The authors would like to thank the referee for pointingout some mistakes and misprints in the earlier version ofthis paper They would like to express their pleasure toMs Medine Yesilkayagil Department of Mathematics UsakUniversity 1 Eylul Campus Usak Turkey for many helpfulsuggestions and interesting comments on the revised form ofthe paper

References

[1] M Gonzalez ldquoThe fine spectrum of the Cesaro operator in ℓ119901(1 lt 119901 lt infin)rdquo Archiv der Mathematik vol 44 no 4 pp 355ndash358 1985

[2] P J CartlidgeWeighted meanmatrices as operators on ℓ119901 [PhDdissertation] Indiana University 1978

[3] J I Okutoyi ldquoOn the spectrum of C1 as an operator on 1198871199070rdquoAustralian Mathematical Society A vol 48 no 1 pp 79ndash861990

[4] J I Okutoyi ldquoOn the spectrum of C1 as an operator on bvrdquoCommunications A vol 41 no 1-2 pp 197ndash207 1992

[5] M Yıldırım ldquoOn the spectrum of the Rhaly operators on ℓ119901rdquoIndian Journal of Pure and Applied Mathematics vol 32 no 2pp 191ndash198 2001

[6] B Altay and F Basar ldquoOn the fine spectrum of the differenceoperator Δ on 1198880 and 119888rdquo Information Sciences vol 168 no 1ndash4pp 217ndash224 2004

[7] B Altay and F Basar ldquoOn the fine spectrum of the generalizeddifference operator 119861(119903 119904) over the sequence spaces 1198880 and119888rdquo International Journal of Mathematics and MathematicalSciences vol 2005 no 18 pp 3005ndash3013 2005

[8] A M Akhmedov and F Basar ldquoOn the fine spectra of the dif-ference operator Δ over the sequence space ℓ119901 (1 le 119901 lt infin)rdquoDemonstratio Mathematica vol 39 no 3 pp 585ndash595 2006

10 Abstract and Applied Analysis

[9] AMAkhmedov andF Basar ldquoThefine spectra of the differenceoperator Δ over the sequence space 119887119907119901 (1 le 119901 lt infin)rdquo ActaMathematica Sinica vol 23 no 10 pp 1757ndash1768 2007

[10] F Basar and B Altay ldquoOn the space of sequences of p-boundedvariation and related matrix mappingsrdquo Ukrainian Mathemati-cal Journal vol 55 no 1 pp 136ndash147 2003

[11] H Furkan H Bilgic and F Basar ldquoOn the fine spectrum of theoperator 119861(119903 119904 119905) over the sequence spaces ℓ119901 and 119887119907119901(1 lt 119901 ltinfin)rdquo Computers ampMathematics with Applications vol 60 no 7pp 2141ndash2152 2010

[12] V Karakaya and M Altun ldquoFine spectra of upper triangulardouble-band matricesrdquo Journal of Computational and AppliedMathematics vol 234 no 5 pp 1387ndash1394 2010

[13] A Karaisa ldquoFine spectra of upper triangular double-bandmatrices over the sequence space ℓ119901 (1 lt 119901 lt infin)rdquo DiscreteDynamics in Nature and Society vol 2012 Article ID 381069 19pages 2012

[14] E Kreyszig Introductory Functional Analysis with ApplicationsJohn Wiley amp Sons New York NY USA 1978

[15] J Appell E Pascale and A Vignoli Nonlinear Spectral Theoryvol 10 of de Gruyter Series in Nonlinear Analysis and Applica-tions Walter de Gruyter Berlin Germany 2004

[16] S Goldberg Unbounded Linear Operators Dover PublicationsNew York NY USA 1985

[17] B Choudhary and S Nanda Functional Analysis with Applica-tions John Wiley amp Sons New York NY USA 1989

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

8 Abstract and Applied Analysis

Proof We will show that 119860120572(119903 119904 119905)lowast is onto for 2|119903 minus 120572| gt

| minus 119904 + radic1199042 minus 4119905(119903 minus 120572)| Thus for every 119910 isin ℓ119902 we find 119909 isinℓ119902 119860120572(119903 119904 119905)

lowast is a triangle so it has an inverse Also equation119860120572(119903 119904 119905)

lowast119909 = 119910 gives [119860120572(119903 119904 119905)

lowast]minus1119910 = 119909 It is sufficient

to show that [119860120572(119903 119904 119905)lowast]minus1isin (ℓ119902 ℓ119902) We calculate that

119860 = (119886119899119896) = [119860120572(119903 119904 119905)lowast]minus1 as follows

119860 = (119886119899119896) =

[[[[

[

1198861 0 0

1198862 1198861 0

1198863 1198862 1198861

]]]]

]

(62)

where

1198861 =1

119903 minus 120572

1198862 =minus119904

(119903 minus 120572)2

1198863 =1199042minus 119905 (119903 minus 120572)

(119903 minus 120572)3

(63)

It is known that fromTheorem 5

119886119899 =120582119899

1minus 120582119899

2

radic1199042 minus 4119905 (119903 minus 120572) forall119899 ⩾ 1

where 1205821 =minus119904 + radicΔ

119903 minus 120572 1205822 =

minus119904 minus radicΔ

119903 minus 120572

(64)

Now we show that [119860120572(119903 119904 119905)lowast]minus1isin (ℓ1 ℓ1) for |1205821| lt 1

By Theorem 5 we know that if |1205821| lt 1 |1205822| lt 1 We assumethat 1199042 minus 4119905(119903 minus 120572) = 0 and |1205821| lt 1

10038171003817100381710038171003817[119860120572(119903 119904 119905)

lowast]minus110038171003817100381710038171003817(ℓ1 ℓ1)

= sup119899

infin

sum

119896=119899

10038161003816100381610038161198861198961003816100381610038161003816 =

infin

sum

119896=1

10038161003816100381610038161198861198961003816100381610038161003816

⩽1

|Δ|(

infin

sum

119896=1

100381610038161003816100381612058211003816100381610038161003816119896+

infin

sum

119896=1

100381610038161003816100381612058221003816100381610038161003816119896) lt infin

(65)

since |1205821| lt 1 and |1205822| lt 1 This shows that (119860120572(119903 119904 119905)lowast]minus1isin

(ℓ1 ℓ1) Similarly we can show that [119860120572(119903 119904 119905)lowast]minus1isin (ℓinfin

ℓinfin)Now assume that 1199042 minus 4119905(119903 minus 120572) = 0 Then

119886119899 = (2119899

minus119904) [

minus119904

2 (119903 minus 120572)]

119899

(66)

and simple calculation gives that (119886119899) isin ℓ119902 if and only if |minus119904| lt2|119903 minus 120572|

[(119860 (119903 119904 119905) minus 120572119868)lowast]minus1isin (ℓ119902 ℓ119902)

for 120572 isin C with 2 |119903 minus 120572| gt1003816100381610038161003816100381610038161003816minus119904 + radic1199042 minus 4119905 (119903 minus 120572)

1003816100381610038161003816100381610038161003816

(67)

Hence 119860120572(119903 119904 119905)lowast is onto By Lemma 2 119860120572(119903 119904 119905) has a

bounded inverse This means that 120590119888(119860(119903 119904 119905) ℓ119901) sube 1198631where1198631 is defined by (61)

Theorem 18 120590119901(119860(119903 119904 119905)lowast ℓlowast

119901) = 0

Proof Let 119860(119903 119904 119905)lowast119891 = 120572119891 with 119891 = 120579 = (0 0 0 ) in ℓlowast119901=

ℓ119902 Then by solving the system of linear equations

1199031198910 = 1205721198910

1199041198910 + 1199031198911 = 1205721198911

1199051198910 + 1199041198911 + 1199031198912 = 1205721198912

1199051198911 + 1199041198912 + 1199031198913 = 1205721198913

119905119891119896minus2 + 119904119891119896minus1 + 119903119891119896 = 120572119891119896

(68)

we find that 1198910 = 0 if 120572 = 119903 and 1198911 = 1198912 = sdot sdot sdot = 0 if 1198910 = 0which contradicts119891 = 120579 If 1198911198990 is the first nonzero entry of thesequence 119891 = (119891119899) and 120572 = 119903 then we get 1199051198911198990minus2 + 1199041198911198990minus1 +1199031198911198990

= 1205721198911198990which implies 1198911198990 = 0 which contradicts the

assumption1198911198990 = 0 Hence the equation119860(119903 119904 119905)lowast119891 = 120572119891 has

no solution 119891 = 120579

In the case 1 lt 119901 lt infin since the proof of the theoremsin Section 4 determining the spectrum and fine spectrumof the matrix operator 119860(119903 119904 119905) on the sequence space ℓ119901is similar to the case 0 lt 119901 ⩽ 1 to avoid the repetitionof similar statements we give the results by the followingtheorem without proof

Theorem 19 The following statements hold(i) 120590(119860(119903 119904 119905) ℓ119901) = 1198631(ii) 120590119903(119860(119903 119904 119905) ℓ119901) = 0(iii) 120590119901(119860(119903 119904 119905) ℓ119901) = 1198632(iv) 120590119888(119860(119903 119904 119905) ℓ119901) = 1198633(v) 120590ap(119860(119903 119904 119905) ℓ119901) = 1198631(vi) 120590co(119860(119903 119904 119905) ℓ119901) = 0(vii) 120590120575(119860(119903 119904 119905) ℓ119901) = 1198633

5 Some Applications

In this section we give two theorems related to Toeplitz ma-trix

Theorem 20 Let 119875 be a polynomial that corresponds to the 119899-tuple 119886 and let 1199111 1199112 1199113 119911119899minus1 also be the roots of 119875 Define119879 as a Toeplitz matrix associated with 119875 that is

119879 =

[[[[

[

1198860 1198861 1198862 119886119899 0 0 0

0 1198860 1198861 1198862 119886119899 0 0

0 0 1198860 1198861 1198862 119886119899 0

]]]]

]

(69)

Abstract and Applied Analysis 9

The resolvent operator 119879 over ℓ119901 with 1 lt 119901 lt infin where thedomain of the resolvent operator is the whole space ℓ119901 exists ifand if only all the roots of the polynomial are outside the unitdisc 119911 isin C |119911| ⩽ 1 That is 119879minus1 isin (ℓ119901 ℓ119901) if and if only|119911119894| gt 1 1 ⩽ 119894 ⩽ 119899 minus 1 In this case the resolvent operator isrepresented by

119879minus1=

1

119886119899minus1119860minus1(minus1199111 1) 119860

minus1(minus1199112 1) sdot sdot sdot 119860

minus1(minus119911119899minus1 1)

where

119860minus1(minus119911119894 1) =minus

[[[[[[[[[[[[

[

1119911119894 11199112

11989411199113

11989411199114

11989411199115

119894

0 1119911119894 11199112

11989411199113

11989411199114

119894

0 0 1119911119894 11199112

11989411199113

119894

0 0 0 1119911119894 11199112

119894

0 0 0 0 1119911119894

]]]]]]]]]]]]

]

(70)

Proof Suppose all the roots of the polynomial 119875(119911) = 1198860 +1198861119911 + sdot sdot sdot + 119886119899minus1119911

119899minus1= 119886119899(119911 minus 1199111)(119911 minus 1199112) sdot sdot sdot (119911 minus 119911119899minus1) are

outside the unit disc The Toeplitz matrix associated with 119875can be written as the product

119879 = 119886119899119860 (minus1199111 1) 119860 (minus1199112 1) sdot sdot sdot 119860 (minus119911119899minus1 1) (71)

Since multiplication of upper triangular Toeplitz matrices iscommutative we can see that

119879minus1=

1

119886119899minus1119860minus1(minus1199111 1) 119860

minus1(minus1199112 1) sdot sdot sdot 119860

minus1(minus119911119899minus1 1)

(72)

is left inverse of 119879 Since all roots are outside the unit discthen

10038171003817100381710038171003817119879minus1(minus119911119894 1)

10038171003817100381710038171003817(ℓinfin ℓinfin)= sup119899

infin

sum

119896=119899

1

10038161003816100381610038161199111198941003816100381610038161003816119896+1minus119899

=

infin

sum

119896=1

1

10038161003816100381610038161199111198941003816100381610038161003816119896lt infin

(73)

Therefore each 119879minus1(minus119911119894 1) isin (ℓinfin ℓinfin) for 1 ⩽ 119894 ⩽ 119899 minus 1Similarly we can say that 119879minus1(minus119911119894 1) isin (ℓ1 ℓ1) So we have119879minus1(minus119911119894 1) isin (ℓ119901 ℓ119901)

Theorem 21 The resolvent operator of 119860(119903 119904 119905) over ℓ119901 with1 lt 119901 lt infin where the domain of the resolvent operator is thespace ℓ119901 exists if and only if 2|119903| gt | minus 119904 + radic1199042 minus 4119905119903| In this

case the resolvent operator is represented by the infinite bandedToeplitz matrix

119864 (119903 119904 119905)

=1

119905

[[[[[[[[[[

[

1199061 1199062

11199063

11199064

1

0 1199061 1199062

11199063

1

0 0 1199061 1199062

1

0 0 0 1199061

]]]]]]]]]]

]

[[[[[[[[[[

[

1199062 1199062

21199063

21199064

2

0 1199062 1199062

21199063

2

0 0 1199062 1199062

2

0 0 0 1199062

]]]]]]]]]]

]

where 1199061 =minus119904 + radic1199042 minus 4119905119903

2119903 1199062 =

minus119904 minus radic1199042 minus 4119905119903

2119903

(74)

Proof By Theorem 20 we can see that 119864(119903 119904 119905) is inverseof the matrix of 119860(119903 119904 119905) But this is not enough to say itis resolvent operator By Lemmas 13 14 and 15 119864(119903 119904 119905) isin(ℓ119901 ℓ119901) when 2|119903| gt | minus 119904 + radic1199042 minus 4119905119903| That is for 2|119903| gt| minus 119904 + radic1199042 minus 4119905119903| 119864(119903 119904 119905) is a resolvent operator

Acknowledgments

The authors would like to thank the referee for pointingout some mistakes and misprints in the earlier version ofthis paper They would like to express their pleasure toMs Medine Yesilkayagil Department of Mathematics UsakUniversity 1 Eylul Campus Usak Turkey for many helpfulsuggestions and interesting comments on the revised form ofthe paper

References

[1] M Gonzalez ldquoThe fine spectrum of the Cesaro operator in ℓ119901(1 lt 119901 lt infin)rdquo Archiv der Mathematik vol 44 no 4 pp 355ndash358 1985

[2] P J CartlidgeWeighted meanmatrices as operators on ℓ119901 [PhDdissertation] Indiana University 1978

[3] J I Okutoyi ldquoOn the spectrum of C1 as an operator on 1198871199070rdquoAustralian Mathematical Society A vol 48 no 1 pp 79ndash861990

[4] J I Okutoyi ldquoOn the spectrum of C1 as an operator on bvrdquoCommunications A vol 41 no 1-2 pp 197ndash207 1992

[5] M Yıldırım ldquoOn the spectrum of the Rhaly operators on ℓ119901rdquoIndian Journal of Pure and Applied Mathematics vol 32 no 2pp 191ndash198 2001

[6] B Altay and F Basar ldquoOn the fine spectrum of the differenceoperator Δ on 1198880 and 119888rdquo Information Sciences vol 168 no 1ndash4pp 217ndash224 2004

[7] B Altay and F Basar ldquoOn the fine spectrum of the generalizeddifference operator 119861(119903 119904) over the sequence spaces 1198880 and119888rdquo International Journal of Mathematics and MathematicalSciences vol 2005 no 18 pp 3005ndash3013 2005

[8] A M Akhmedov and F Basar ldquoOn the fine spectra of the dif-ference operator Δ over the sequence space ℓ119901 (1 le 119901 lt infin)rdquoDemonstratio Mathematica vol 39 no 3 pp 585ndash595 2006

10 Abstract and Applied Analysis

[9] AMAkhmedov andF Basar ldquoThefine spectra of the differenceoperator Δ over the sequence space 119887119907119901 (1 le 119901 lt infin)rdquo ActaMathematica Sinica vol 23 no 10 pp 1757ndash1768 2007

[10] F Basar and B Altay ldquoOn the space of sequences of p-boundedvariation and related matrix mappingsrdquo Ukrainian Mathemati-cal Journal vol 55 no 1 pp 136ndash147 2003

[11] H Furkan H Bilgic and F Basar ldquoOn the fine spectrum of theoperator 119861(119903 119904 119905) over the sequence spaces ℓ119901 and 119887119907119901(1 lt 119901 ltinfin)rdquo Computers ampMathematics with Applications vol 60 no 7pp 2141ndash2152 2010

[12] V Karakaya and M Altun ldquoFine spectra of upper triangulardouble-band matricesrdquo Journal of Computational and AppliedMathematics vol 234 no 5 pp 1387ndash1394 2010

[13] A Karaisa ldquoFine spectra of upper triangular double-bandmatrices over the sequence space ℓ119901 (1 lt 119901 lt infin)rdquo DiscreteDynamics in Nature and Society vol 2012 Article ID 381069 19pages 2012

[14] E Kreyszig Introductory Functional Analysis with ApplicationsJohn Wiley amp Sons New York NY USA 1978

[15] J Appell E Pascale and A Vignoli Nonlinear Spectral Theoryvol 10 of de Gruyter Series in Nonlinear Analysis and Applica-tions Walter de Gruyter Berlin Germany 2004

[16] S Goldberg Unbounded Linear Operators Dover PublicationsNew York NY USA 1985

[17] B Choudhary and S Nanda Functional Analysis with Applica-tions John Wiley amp Sons New York NY USA 1989

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 9

The resolvent operator 119879 over ℓ119901 with 1 lt 119901 lt infin where thedomain of the resolvent operator is the whole space ℓ119901 exists ifand if only all the roots of the polynomial are outside the unitdisc 119911 isin C |119911| ⩽ 1 That is 119879minus1 isin (ℓ119901 ℓ119901) if and if only|119911119894| gt 1 1 ⩽ 119894 ⩽ 119899 minus 1 In this case the resolvent operator isrepresented by

119879minus1=

1

119886119899minus1119860minus1(minus1199111 1) 119860

minus1(minus1199112 1) sdot sdot sdot 119860

minus1(minus119911119899minus1 1)

where

119860minus1(minus119911119894 1) =minus

[[[[[[[[[[[[

[

1119911119894 11199112

11989411199113

11989411199114

11989411199115

119894

0 1119911119894 11199112

11989411199113

11989411199114

119894

0 0 1119911119894 11199112

11989411199113

119894

0 0 0 1119911119894 11199112

119894

0 0 0 0 1119911119894

]]]]]]]]]]]]

]

(70)

Proof Suppose all the roots of the polynomial 119875(119911) = 1198860 +1198861119911 + sdot sdot sdot + 119886119899minus1119911

119899minus1= 119886119899(119911 minus 1199111)(119911 minus 1199112) sdot sdot sdot (119911 minus 119911119899minus1) are

outside the unit disc The Toeplitz matrix associated with 119875can be written as the product

119879 = 119886119899119860 (minus1199111 1) 119860 (minus1199112 1) sdot sdot sdot 119860 (minus119911119899minus1 1) (71)

Since multiplication of upper triangular Toeplitz matrices iscommutative we can see that

119879minus1=

1

119886119899minus1119860minus1(minus1199111 1) 119860

minus1(minus1199112 1) sdot sdot sdot 119860

minus1(minus119911119899minus1 1)

(72)

is left inverse of 119879 Since all roots are outside the unit discthen

10038171003817100381710038171003817119879minus1(minus119911119894 1)

10038171003817100381710038171003817(ℓinfin ℓinfin)= sup119899

infin

sum

119896=119899

1

10038161003816100381610038161199111198941003816100381610038161003816119896+1minus119899

=

infin

sum

119896=1

1

10038161003816100381610038161199111198941003816100381610038161003816119896lt infin

(73)

Therefore each 119879minus1(minus119911119894 1) isin (ℓinfin ℓinfin) for 1 ⩽ 119894 ⩽ 119899 minus 1Similarly we can say that 119879minus1(minus119911119894 1) isin (ℓ1 ℓ1) So we have119879minus1(minus119911119894 1) isin (ℓ119901 ℓ119901)

Theorem 21 The resolvent operator of 119860(119903 119904 119905) over ℓ119901 with1 lt 119901 lt infin where the domain of the resolvent operator is thespace ℓ119901 exists if and only if 2|119903| gt | minus 119904 + radic1199042 minus 4119905119903| In this

case the resolvent operator is represented by the infinite bandedToeplitz matrix

119864 (119903 119904 119905)

=1

119905

[[[[[[[[[[

[

1199061 1199062

11199063

11199064

1

0 1199061 1199062

11199063

1

0 0 1199061 1199062

1

0 0 0 1199061

]]]]]]]]]]

]

[[[[[[[[[[

[

1199062 1199062

21199063

21199064

2

0 1199062 1199062

21199063

2

0 0 1199062 1199062

2

0 0 0 1199062

]]]]]]]]]]

]

where 1199061 =minus119904 + radic1199042 minus 4119905119903

2119903 1199062 =

minus119904 minus radic1199042 minus 4119905119903

2119903

(74)

Proof By Theorem 20 we can see that 119864(119903 119904 119905) is inverseof the matrix of 119860(119903 119904 119905) But this is not enough to say itis resolvent operator By Lemmas 13 14 and 15 119864(119903 119904 119905) isin(ℓ119901 ℓ119901) when 2|119903| gt | minus 119904 + radic1199042 minus 4119905119903| That is for 2|119903| gt| minus 119904 + radic1199042 minus 4119905119903| 119864(119903 119904 119905) is a resolvent operator

Acknowledgments

The authors would like to thank the referee for pointingout some mistakes and misprints in the earlier version ofthis paper They would like to express their pleasure toMs Medine Yesilkayagil Department of Mathematics UsakUniversity 1 Eylul Campus Usak Turkey for many helpfulsuggestions and interesting comments on the revised form ofthe paper

References

[1] M Gonzalez ldquoThe fine spectrum of the Cesaro operator in ℓ119901(1 lt 119901 lt infin)rdquo Archiv der Mathematik vol 44 no 4 pp 355ndash358 1985

[2] P J CartlidgeWeighted meanmatrices as operators on ℓ119901 [PhDdissertation] Indiana University 1978

[3] J I Okutoyi ldquoOn the spectrum of C1 as an operator on 1198871199070rdquoAustralian Mathematical Society A vol 48 no 1 pp 79ndash861990

[4] J I Okutoyi ldquoOn the spectrum of C1 as an operator on bvrdquoCommunications A vol 41 no 1-2 pp 197ndash207 1992

[5] M Yıldırım ldquoOn the spectrum of the Rhaly operators on ℓ119901rdquoIndian Journal of Pure and Applied Mathematics vol 32 no 2pp 191ndash198 2001

[6] B Altay and F Basar ldquoOn the fine spectrum of the differenceoperator Δ on 1198880 and 119888rdquo Information Sciences vol 168 no 1ndash4pp 217ndash224 2004

[7] B Altay and F Basar ldquoOn the fine spectrum of the generalizeddifference operator 119861(119903 119904) over the sequence spaces 1198880 and119888rdquo International Journal of Mathematics and MathematicalSciences vol 2005 no 18 pp 3005ndash3013 2005

[8] A M Akhmedov and F Basar ldquoOn the fine spectra of the dif-ference operator Δ over the sequence space ℓ119901 (1 le 119901 lt infin)rdquoDemonstratio Mathematica vol 39 no 3 pp 585ndash595 2006

10 Abstract and Applied Analysis

[9] AMAkhmedov andF Basar ldquoThefine spectra of the differenceoperator Δ over the sequence space 119887119907119901 (1 le 119901 lt infin)rdquo ActaMathematica Sinica vol 23 no 10 pp 1757ndash1768 2007

[10] F Basar and B Altay ldquoOn the space of sequences of p-boundedvariation and related matrix mappingsrdquo Ukrainian Mathemati-cal Journal vol 55 no 1 pp 136ndash147 2003

[11] H Furkan H Bilgic and F Basar ldquoOn the fine spectrum of theoperator 119861(119903 119904 119905) over the sequence spaces ℓ119901 and 119887119907119901(1 lt 119901 ltinfin)rdquo Computers ampMathematics with Applications vol 60 no 7pp 2141ndash2152 2010

[12] V Karakaya and M Altun ldquoFine spectra of upper triangulardouble-band matricesrdquo Journal of Computational and AppliedMathematics vol 234 no 5 pp 1387ndash1394 2010

[13] A Karaisa ldquoFine spectra of upper triangular double-bandmatrices over the sequence space ℓ119901 (1 lt 119901 lt infin)rdquo DiscreteDynamics in Nature and Society vol 2012 Article ID 381069 19pages 2012

[14] E Kreyszig Introductory Functional Analysis with ApplicationsJohn Wiley amp Sons New York NY USA 1978

[15] J Appell E Pascale and A Vignoli Nonlinear Spectral Theoryvol 10 of de Gruyter Series in Nonlinear Analysis and Applica-tions Walter de Gruyter Berlin Germany 2004

[16] S Goldberg Unbounded Linear Operators Dover PublicationsNew York NY USA 1985

[17] B Choudhary and S Nanda Functional Analysis with Applica-tions John Wiley amp Sons New York NY USA 1989

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

10 Abstract and Applied Analysis

[9] AMAkhmedov andF Basar ldquoThefine spectra of the differenceoperator Δ over the sequence space 119887119907119901 (1 le 119901 lt infin)rdquo ActaMathematica Sinica vol 23 no 10 pp 1757ndash1768 2007

[10] F Basar and B Altay ldquoOn the space of sequences of p-boundedvariation and related matrix mappingsrdquo Ukrainian Mathemati-cal Journal vol 55 no 1 pp 136ndash147 2003

[11] H Furkan H Bilgic and F Basar ldquoOn the fine spectrum of theoperator 119861(119903 119904 119905) over the sequence spaces ℓ119901 and 119887119907119901(1 lt 119901 ltinfin)rdquo Computers ampMathematics with Applications vol 60 no 7pp 2141ndash2152 2010

[12] V Karakaya and M Altun ldquoFine spectra of upper triangulardouble-band matricesrdquo Journal of Computational and AppliedMathematics vol 234 no 5 pp 1387ndash1394 2010

[13] A Karaisa ldquoFine spectra of upper triangular double-bandmatrices over the sequence space ℓ119901 (1 lt 119901 lt infin)rdquo DiscreteDynamics in Nature and Society vol 2012 Article ID 381069 19pages 2012

[14] E Kreyszig Introductory Functional Analysis with ApplicationsJohn Wiley amp Sons New York NY USA 1978

[15] J Appell E Pascale and A Vignoli Nonlinear Spectral Theoryvol 10 of de Gruyter Series in Nonlinear Analysis and Applica-tions Walter de Gruyter Berlin Germany 2004

[16] S Goldberg Unbounded Linear Operators Dover PublicationsNew York NY USA 1985

[17] B Choudhary and S Nanda Functional Analysis with Applica-tions John Wiley amp Sons New York NY USA 1989

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of