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Spectra and fine spectra of certain lower triangular double-band matrices as operators on c 0

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Abstract

In this paper we determine the fine spectrum of the generalized difference operator Δ a , b defined by a lower triangular double-band matrix over the sequence space c 0 . The class of the operator Δ a , b contains as special cases many operators that have been studied recently in the literature. Illustrative examples showing the advantage of the present results are also given.

MSC:47A10, 47B37.

1 Introduction

Several authors have studied the spectrum and fine spectrum of linear operators defined by lower and upper triangular matrices over some sequence spaces [121].

Let X be a Banach space. By R(T), T , X , B(X), σ(T,X), σ p (T,X), σ r (T,X) and σ c (T,X), we denote the range of T, the adjoint operator of T, the space of all continuous linear functionals on X, the space of all bounded linear operators on X into itself, the spectrum of T on X, the point spectrum of T on X, the residual spectrum of T on X and the continuous spectrum of T on X, respectively. We shall write c and c 0 for the spaces of all convergent and null sequences, respectively. Also by l 1 we denote the space of all absolutely summable sequences.

We assume here some familiarity with basic concepts of spectral theory and we refer to Kreyszig [[22], pp.370-372] for basic definitions such as spectrum, point spectrum, residual spectrum, and continuous spectrum of linear operators in normed spaces. Also, we refer to Goldberg [[23], pp.58-71] for Goldberg’s classification of spectra.

Now, let ( a k ) and ( b k ) be two convergent sequences of nonzero real numbers with

lim k a k =aand lim k b k =b0.
(1)

We consider the operator Δ a , b : c 0 c 0 , which is defined as follows:

Δ a , b x= Δ a , b ( x k )= ( a k x k + b k 1 x k 1 ) k = 0 with  x 1 = b 1 =0.
(2)

It is easy to verify that the operator Δ a , b can be represented by a lower triangular double-band matrix of the form

Δ a , b = ( a 0 0 0 b 0 a 1 0 0 b 1 a 2 ) .
(3)

We begin by determining when a matrix A induces a bounded linear operator from c 0 to itself.

Lemma 1.1 (cf. [[24], p.129])

The matrix A=( a n k ) gives rise to a bounded linear operator TB( c 0 ) from c 0 to itself if and only if

  1. (1)

    the rows of A are in l 1 and their l 1 norms are bounded,

  2. (2)

    the columns of A are in c 0 .

The operator norm of T is the supremum of the l 1 norms of the rows.

As a consequence of the above lemma, we have the following corollary for the bounded linearity of the operator Δ a , b on the space c 0 .

Corollary 1.1 The operator Δ a , b : c 0 c 0 is a bounded linear operator with the norm Δ a , b c 0 = sup k (| a k |+| b k 1 |).

The rest of the paper is organized as follows. In Section 2, we analyze the spectrum of the operator Δ a , b on the sequence space c 0 . In Section 3 we give some illustrative examples. Finally, Section 4 concludes with remarks and some special cases.

2 Fine spectrum of the operator Δ a , b on c 0

In this section we examine the spectrum, the point spectrum, the residual spectrum and the continuous spectrum of the operator Δ a , b on the sequence space c 0 .

Theorem 2.1 Let D={λC:|λa||b|} and E={ a k :kN,| a k a|>|b|}. Then σ( Δ a , b , c 0 )=DE.

Proof First, we prove that ( Δ a , b λ I ) 1 exists and is in B( c 0 ) for λDE and then the operator Δ a , b λI is not invertible for λDE.

Let λDE. Then |λa|>|b| and λ a k , for all kN. So, ( Δ a , b λI) is triangle and hence ( Δ a , b λ I ) 1 exists. We can calculate that

( Δ a , b λ I ) 1 =( s k j )= ( 1 ( a 0 λ ) 0 0 b 0 ( a 0 λ ) ( a 1 λ ) 1 ( a 1 λ ) 0 b 0 b 1 ( a 0 λ ) ( a 1 λ ) ( a 2 λ ) b 1 ( a 1 λ ) ( a 2 λ ) 1 ( a 2 λ ) ) .

Now, for each kN, the series S k = j | s k j | is convergent since it is finite. Next, we prove that sup k S k is finite.

Since lim k | b k a k λ |=| b a λ |<1, then there exist k 0 N and q 0 <1 such that | b k a k λ |< q 0 , for all k k 0 . Then, for each k> k 0 ,

S k = 1 | a k λ | [ 1 + | b k 1 | | a k 1 λ | + | b k 1 | | b k 2 | | a k 1 λ | | a k 2 λ | + + | b k 1 | | b k 2 | | b k 0 | | a k 1 λ | | a k 2 λ | | a k 0 λ | + + | b k 1 | | b k 2 | | b k 0 + 1 | | b k 0 | | b 0 | | a k 1 λ | | a k 2 λ | | a k 0 + 1 λ | | a k 0 λ | | a 0 λ | ] 1 | a k λ | [ 1 + q 0 + q 0 2 + + q 0 k k 0 + q 0 k k 0 | b k 0 1 | | a k 0 1 λ | + q 0 k k 0 | b k 0 1 | | b k 0 2 | | a k 0 1 λ | | a k 0 2 λ | + + q 0 k k 0 | b k 0 1 | | b k 0 2 | | b 0 | | a k 0 1 λ | | a k 0 2 λ | | a 0 λ | ] .

Therefore

S k 1 | a k λ | [ 1 + q 0 + q 0 2 + + q 0 k k 0 m k 0 ] ,

where

m k 0 =1+ | b k 0 1 | | a k 0 1 λ | + | b k 0 1 | | b k 0 2 | | a k 0 1 λ | | a k 0 2 λ | ++ | b k 0 1 | | b k 0 2 | | b 0 | | a k 0 1 λ | | a k 0 2 λ | | a 0 λ | .

Then m k 0 >1 and so

S k m k 0 | a k λ | [ 1 + q 0 + q 0 2 + + q 0 k k 0 ] .

But there exist k 1 N and a real number q 1 < 1 | b | such that 1 | a k λ | < q 1 , for all k k 1 . Then

S k q 1 m k 0 1 q 0 ,

for all k>max{ k 0 , k 1 }. Thus sup k S k <.

Also, it is easy to see that lim k | s k j |=0, for all jN, since

lim k | s k + 1 , j s k , j |= lim k | b k a k + 1 λ |=| b a λ |<1.

So, the sequence ( s 0 j , s 1 j , s 2 j ,) converges to zero, for each jN. This shows that the columns of ( Δ a , b λ I ) 1 are in c 0 . Then, from Lemma 1.1, ( Δ a , b λ I ) 1 B( c 0 ) and so, λσ( Δ a , b , c 0 ). Thus σ( Δ a , b , c 0 )DE.

Conversely, suppose that λσ( Δ a , b , c 0 ). Then ( Δ a , b λ I ) 1 B( c 0 ). Since the ( Δ a , b λ I ) 1 transform of the unite sequence e 0 =(1,0,0,) is in c 0 , we have lim k | b k a k + 1 λ |=| b a λ |1 and λ a k , for all kN. Then {λC:|λa|<|b|}σ( Δ a , b , c 0 ) and { a k :kN}σ( Δ a , b , c 0 ). But σ( Δ a , b , c 0 ) is a compact set, and so it is closed. Then D={λC:|λa||b|}σ( Δ a , b , c 0 ) and E={ a k :kN,| a k a|>|b|}σ( Δ a , b , c 0 ). This completes the proof. □

Theorem 2.2 σ p ( Δ a , b , c 0 )=EK, where

K= { a j : j N , | a j a | = | b | , i = m b i 1 a j a i  diverges to zero for some  m N } .

Proof Suppose Δ a , b x=λx for any x c 0 . Then we obtain

( a 0 λ) x 0 =0and b k x k +( a k + 1 λ) x k + 1 =0,for all kN.

If the sequence ( a k ) is constant, then we can easily see that x=θ and so, σ p ( Δ a , b , c 0 )= and the result follows immediately. Now, if the sequence ( a k ) is not constant, then for all λ{ a k :kN}, we have x k =0 for all kN. So, λ σ p ( Δ a , b , c 0 ). Also, we can easily prove that a σ p ( Δ a , b , c 0 ). Thus σ p ( Δ a , b , c 0 ){ a k :kN}{a}. Now, we will prove that

λ σ p ( Δ a , b , c 0 )if and only ifλEK.

If λ σ p ( Δ a , b , c 0 ), then λ= a j a for some jN and there exists x c 0 , xθ such that Δ a , b x= a j x. Then

lim k | x k + 1 x k |=| b a a j |1.

Then λ= a j E or | a j a|=|b|. In the case when | a j a|=|b|, we have

x k = b k 1 b k 2 b m 1 ( a j a k ) ( a j a k 1 ) ( a j a m ) x m 1 = x m 1 i = m k b i 1 a j a i ,km.

Then i = m b i 1 a j a i diverges to 0, since x c 0 . Therefore λEK. Thus σ p ( Δ a , b , c 0 )EK.

Conversely, let λEK. If λE, then there exists iN such that λ= a i a and so we can take xθ such that Δ a , b x= a i x and

lim k | x k + 1 x k |=| b a a i |<1,

that is, x c 0 . Also, if λK, then there exists jN such that λ= a j a and | a j a|=|b|, i = m b i 1 a j a i diverges to 0, for some mN. Then we can take x c 0 , xθ such that Δ a , b x= a j x. Thus EK σ p ( Δ a , b , c 0 ). This completes the proof. □

Theorem 2.3 σ p ( Δ a , b , c 0 )={λC:|λa|<|b|}EH, where

H= { λ C : | λ a | = | b | , k = 0 | i = 0 k λ a i b i | < } .

Proof Suppose that Δ a , b f=λf for f=( f 0 , f 1 , f 2 ,)θ in c 0 l 1 . Then, by solving the system of linear equations

a 0 f 0 + b 0 f 1 = λ f 0 , a 1 f 1 + b 1 f 2 = λ f 1 , a 2 f 2 + b 2 f 3 = λ f 2 ,

we obtain

f k + 1 = λ a k b k f k ,kN.

Then we must take f 0 0 since otherwise we would have f=θ. It is clear that for all kN, the vector f=( f 0 , f 1 ,, f k ,0,0,) is an eigenvector of the operator Δ a , b corresponding to the eigenvalue λ= a k , where f 0 0 and f n = λ a n 1 b n 1 f n 1 , for all n=1,2,3,,k. Then { a k :kN} σ p ( Δ a , b , c 0 ). Also, if λ a k for all kN, then f k 0, for all k1 and k = 0 | f k |< if lim k | f k + 1 f k |=| λ a b |<1. Also, if |λa|=|b|, we can easily see that

| f k |=| ( λ a 0 ) ( λ a 1 ) ( λ a k 1 ) b 0 b 1 b k 1 || f 0 |=| f 0 | i = 0 k 1 | λ a i b i |,for all k1,

and so k = 0 | f k |< if k = 0 | i = 0 k λ a i b i |<. This implies that H σ p ( Δ a , b , c 0 ). Thus

{ λ C : | λ a | < | b | } EH σ p ( Δ a , b , c 0 ) .

The second inclusion can be proved analogously. □

The following lemma is required in the proof of the next theorem.

Lemma 2.1 [[23], p.59]

T has a dense range if and only if T is one to one.

Theorem 2.4 σ r ( Δ a , b , c 0 )= σ p ( Δ a , b , c 0 ) σ p ( Δ a , b , c 0 ).

Proof For λ σ p ( Δ a , b , c 0 ) σ p ( Δ a , b , c 0 ), the operator Δ a , b λI is one to one and hence has an inverse. But Δ a , b λI is not one to one. Now, Lemma 2.1 yields the fact that the range of the operator Δ a , b λI is not dense in c 0 . This implies that λ σ r ( Δ a , b , c 0 ). □

Theorem 2.5 σ r ( Δ a , b , c 0 )={λC:|λa|<|b|}(HK).

Proof The proof follows immediately from Theorems 2.2, 2.3, and 2.4. □

Theorem 2.6 σ c ( Δ a , b , c 0 )={λC:|λa|=|b|}H.

Proof Since σ( Δ a , b , c 0 ) is the disjoint union of the parts σ p ( Δ a , b , c 0 ), σ r ( Δ a , b , c 0 ) and σ c ( Δ a , b , c 0 ) we must have σ c ( Δ a , b , c 0 )={λC:|λa|=|b|}H. □

Also, we have the following result.

Theorem 2.7 σ c ( Δ a , b , c 0 )=σ( Δ a , b , c 0 ) σ p ( Δ a , b , c 0 ).

Proof The proof is obvious and so is omitted. □

With respect to Goldberg’s classification of the spectrum of an operator (see [[23], pp.58-71]), the spectrum is partitioned into nine states, which are I 1 , I 2 , I 3 , I I 1 , I I 2 , I I 3 , II I 1 , II I 2 , and II I 3 . For the operator Δ a , b : c 0 c 0 , we have

I 3 σ( Δ a , b , c 0 )=I I 3 σ( Δ a , b , c 0 )=,

since σ p ( Δ a , b , c 0 ) σ p ( Δ a , b , c 0 ). Also, I 2 σ( Δ a , b , c 0 )=, by the closed graph theorem. Thus we have to discuss the states I I 2 , II I 1 , II I 2 , and II I 3 .

Theorem 2.8 λ σ p ( Δ a , b , c 0 ) if and only if λII I 3 σ( Δ a , b , c 0 ).

Proof The proof is obvious and so is omitted. □

Theorem 2.9 λ σ c ( Δ a , b , c 0 ) if and only if λI I 2 σ( Δ a , b , c 0 ).

Proof Let λ σ c ( Δ a , b , c 0 ). By Theorem 2.7, Δ a , b λI is one to one. By Lemma 2.1, Δ a , b λI has dense range. Additionally, λ σ p ( Δ a , b , c 0 ) implies that the operator Δ a , b λI has inverse. Therefore, λI I 2 σ( Δ a , b , c 0 ) or λ I 2 σ( Δ a , b , c 0 ). But I 2 σ( Δ a , b , c 0 )=. Thus λI I 2 σ( Δ a , b , c 0 ). □

Theorem 2.10 λ σ r ( Δ a , b , c 0 ) if and only if λII I 1 σ( Δ a , b , c 0 )II I 2 σ( Δ a , b , c 0 ).

Proof Let λ σ r ( Δ a , b , c 0 ). By Theorem 2.4, Δ a , b λI is not one to one. By Lemma 2.1, Δ a , b λI has not a dense range. Additionally, λ σ p ( Δ a , b , c 0 ) implies that the operator Δ a , b λI has inverse. Therefore, λII I 1 σ( Δ a , b , c 0 )II I 2 σ( Δ a , b , c 0 ). □

3 Illustrative examples

In this section we provide some illustrative examples in support of our new results.

Example 3.1 Consider the sequences ( a k ) and ( b k ) defined by the following recurrence relations:

a 0 = 2 , a k + 1 = 2 + a k , b 0 = 2 , b k + 1 = 2 b k ,

for all kN. Then ( a k ) and ( b k ) are monotonically increasing sequences and lim k a k =a=2 and lim k b k =b=2. Also, a k b k for all kN. Thus, for all λC with |λ2|=2, one can prove that | λ a k b k |1 for all kN. This implies that H=. Also, we can prove that E=K=. Using Theorems 2.1, 2.2, 2.5, and 2.6, we have

σ ( Δ a , b , c 0 ) = { λ C : | λ 2 | 2 } , σ p ( Δ a , b , c 0 ) = , σ r ( Δ a , b , c 0 ) = { λ C : | λ 2 | < 2 } , σ c ( Δ a , b , c 0 ) = { λ C : | λ 2 | = 2 } .

Example 3.2 Let a k = k + 1 k + 2 and b k = k + 1 k + 3 for all kN. Then lim k a k =a=1 and lim k b k =b=1. Similarly, as in Example 3.1, we can prove that E=K=H= and so

σ ( Δ a , b , c 0 ) = { λ C : | λ 1 | 1 } , σ p ( Δ a , b , c 0 ) = , σ r ( Δ a , b , c 0 ) = { λ C : | λ 1 | < 1 } , σ c ( Δ a , b , c 0 ) = { λ C : | λ 1 | = 1 } .

Example 3.3 Consider the sequences ( a k ) and ( b k ) defined by the following recurrence relations:

a 0 = 4 , a 1 = 2 , a k = 1 for  k 2 , b 0 = 1 , b 1 = 1 , b k = ( k + 1 k ) 2 for  k 2 .

Therefore, lim k a k =a=1 and lim k b k =b=1. Then E={4}, K= and H={λC:|λ1|=1}, and so

σ ( Δ a , b , c 0 ) = { λ C : | λ 1 | 1 } { 4 } , σ p ( Δ a , b , c 0 ) = { 4 } , σ r ( Δ a , b , c 0 ) = { λ C : | λ 1 | 1 } , σ c ( Δ a , b , c 0 ) = .

Remark 3.1 From the above examples, we note that the spectrum of the operator Δ a , b on the space c 0 may include also a finite number of points outside a region enclosed by a circle. Also, we may have σ p ( Δ a , b , c 0 ).

Example 3.4 Let the sequences ( a k ) and ( b k ) be taken such that a k = b k = ( k + 2 ) 2 ( k + 2 ) 2 + ( k + 3 ) 2 , kN. Then we can prove that E=K=H= and so we have

σ ( Δ a , b , c 0 ) = { λ C : | λ 1 2 | 1 2 } , σ p ( Δ a , b , c 0 ) = , σ r ( Δ a , b , c 0 ) = { λ C : | λ 1 2 | < 1 2 } , σ c ( Δ a , b , c 0 ) = { λ C : | λ 1 2 | = 1 2 } .

4 Remarks and some special cases

In this section we are going to give some special cases of the operator Δ a , b which has been studied recently. More precisely, we show that special conditions on the sequences ( a k ) and ( b k ) characterize certain special cases of the operator Δ a , b .

The difference operator Δ: If a k =1 and b k =1 for all kN, then the operator Δ a , b reduces to the backward difference operator Δ (cf. [7]).

The generalized difference operator B(r,s): If a k =r and b k =s0 for all kN, then the operator Δ a , b reduces to the operator B(r,s) (cf. [8]).

The generalized difference operator Δ v : If a k = b k = v k for all kN, then the operator Δ a , b reduces to the operator Δ v (cf. [19]).

The generalized difference operator Δ u v : If ( a k ) is a sequence of positive real numbers such that a k 0 for all kN with lim k a k =U0 and ( b k ) is either constant or strictly decreasing sequence of positive real numbers with lim k b k =V0 and sup k a k <U+V, then the operator Δ a , b reduces to the operator Δ u v (cf. [12]).

Remark 4.1 If ( a k ) and ( b k ) are convergent sequences of nonzero real numbers such that

lim k a k =a>0,
(4)
lim k b k =b;|b|=a,
(5)

and

sup k a k a, b k 2 a k 2 for all kN,
(6)

then we can prove that H= and so we have:

σ ( Δ a , b , c 0 ) = { λ C : | λ a | a } { a k : k N , | a k a | > a } , σ p ( Δ a , b , c 0 ) = { a k : k N , | a k a | > a } , σ p ( Δ a , b , c 0 ) = { λ C : | λ a | < a } { a k : k N , | a k a | > a } , σ r ( Δ a , b , c 0 ) = { λ C : | λ a | < a } , σ c ( Δ a , b , c 0 ) = { λ C : | λ a | = a } .

It is immediate that our new results cover a wider class of linear operators which are represented by infinite lower triangular double-band matrices on the sequence space c 0 . For this reason, our study is more general and more comprehensive than the previous work. We note that our new results in this paper improve and generalize the results which have been stated in [3, 12].

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Acknowledgements

I wish to express my thanks to Prof. Ali M Akhmedov, Baku State University, Faculty of Mechanics & Mathematics, Baku, Azerbaijan, for his kind help, careful reading, and making useful comments on the earlier version of this paper. Also, I thank the editor and the anonymous referees for their careful reading and making some useful comments which improved the presentation of the paper.

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Correspondence to Saad R El-Shabrawy.

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Keywords

  • spectrum of an operator
  • infinite lower triangular matrices
  • sequence spaces