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Fine spectrum of the generalized difference operator Δuv on sequence space l1
Applied Mathematics and Computation 218 (2012) 6407–6414
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Fine spectrum of the generalized difference operator Duv on sequence space l1 P.D. Srivastava ⇑, Sudhanshu Kumar Department of Mathematics, Indian Institute of Technology Kharagpur, Kharagpur 721302, India
a r t i c l e
i n f o
a b s t r a c t In this paper we obtained spectrum and point spectrum of the operator Duv on l1. The operator Duv on l1 is defined as Duv x ¼ ðun xn þ v n1 xn1 Þ1 n¼0 under certain conditions, where x1 = 0 and x = (xn) 2 l1. Further, the results on continuous spectrum, residual spectrum and fine spectrum of the operator Duv on the sequence space l1 are also derived. Crown Copyright Ó 2011 Published by Elsevier Inc. All rights reserved.
Keywords: Spectrum of an operator Generalized operator Duv Sequence space l1
1. Introduction Let u = (uk) and
v = (vk) be sequences such that
(i) u is either a constant sequence or sequence of distinct real numbers with U = limk?1uk, (ii) v is a sequence of nonzero real numbers with V = limk?1vk – 0, and (iii) jU ukj < jVj for each k 2 N0 ¼ f0; 1; 2; . . .g. We define the operator Duv on the sequence space l1 as follows:
Duv x ¼ ðun xn þ v n1 xn1 Þ1 n¼0 with x1 ¼ 0; where x ¼ ðxn Þ 2 l1 :
ð1:1Þ
It is easy to verify that the operator Duv can be represented by the matrix
0
u0
Bv B 0 Du v ¼ B B0 @ .. .
1
0
0
...
u1
0
v1
u2 ...
...C C C: ...C A .. .
.. .
ð1:2Þ
Spectrum of the Cesaro operator on sequence spaces l1 and lp (1 < p < 1) is investigated Reade [17] and Gonzalez [13], respectively. In 2006, Kayaduman and Furkan [14] obtained spectrum and fine spectrum of the difference operator D on sequence spaces l1 and bv, where Dx = (xn xn1). Fine spectrum of the difference operator D on the sequence space lp (0 < p < 1) is determined by Altay and Basar [6], while fine spectrum of the Zweier matrix Zs on sequence spaces l1 and bv is obtained by Altay and Karakus [7], where Zsx = (sxn + (1 s)xn1) and s is a real number with s – 0, 1. Fine spectrum of the operator B(r, s) on sequence spaces l1 and bv is studied by Furkan et al. [11], where B(r, s)x = (rxn + sxn1). In 2008, Bilgic and Furkan [9] obtained fine spectrum of the operator B(r, s) on sequence spaces lp and bvp (1 < p < 1). Recently, spectrum ⇑ Corresponding author. E-mail addresses: [email protected] (P.D. Srivastava), [email protected] (S. Kumar). 0096-3003/$ - see front matter Crown Copyright Ó 2011 Published by Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2011.12.010
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and fine spectrum of the operator B(r, s, t) on sequence spaces l1 and bv is studied by Bilgic and Furkan [8], where B(r, s, t)x = (rxn + sxn1 + txn2). Fine spectrum of the difference operator D on sequence spaces c0 and c is studied by Altay and Basar [4], and the generalization of these results are obtained by the same authors in [5] with the generalized difference operator B(r, s). In 2006 and 2007, Akhmedov and Basar [1 and 2] determined the fine spectrum of the difference operator D over the sequence spaces lp and bvp (1 6 p < 1). Recently, spectrum and fine spectrum of the operator B(r, s, t) on sequence spaces lp and bvp (1 < p < 1) have derived by Furkan et al. [10] which generalizes the main results of [1,2 and 9]. Quite recently, the fine spectrum of the operator Da,b over the sequence space c has been worked by Akhmedov and El-Shabrawy [3]. In this paper we determine spectrum, point spectrum, continuous spectrum and residual spectrum of the operator Duv on the sequence space l1. It is easy to verify that by choosing suitably u and v sequences, one can get easily the operators such as B(r, s), Zs, etc. Choosing u = (r), v = (s) and u = (s), v = (1 s), the operator Duv reduces to B(r, s) and Zs, respectively. Similarly, if u = (1), v = (1) and u = (0), v = (1), then the operator Duv reduces to D and right-shift operator, respectively. Thus, the results of this paper generalizes the corresponding results of many operator whose matrix representation has diagonal and postdiagonal elements studied by earlier authors. 2. Preliminaries and notation Let X and Y be Banach spaces and T : X ? Y be a bounded linear operator. The set of all bounded linear operators on X into itself is denoted by B(X). The adjoint T: Xw ? Xw of T is defined by
ðT /ÞðxÞ ¼ /ðTxÞ for all / 2 X H and x 2 X: Clearly, T is a bounded linear operator on the dual space Xw. Let X – {0} be a complex normed space and T : D(T) ? X be a linear operator with domain D(T) # X. With T, we associate the operator Ta = (T aI), where a is a complex number and I is the identity operator on D(T). The inverse of Ta (if exists) is denoted by T 1 a and known as the resolvent operator of T. Since the spectral theory is concerned with many properties of Ta 1 1 and T 1 a , which depend on a, so we are interested the set of those a in the complex plane for which T a exists or T a is bounded or domain of T 1 is dense in X. a Definition 2.1 [15, p. 371]. Let X – {0} be a complex normed space and T : D(T) ? X be a linear operator with domain D(T) # X. A regular value of T is a complex number a such that
(R1) T 1 a exists, (R2) T 1 a is bounded, (R3) T 1 a is defined on a set which is dense in X. Resolvent set q(T, X) of T is the set of all regular values a of T. Its complement rðT; XÞ ¼ C n qðT; XÞ in the complex plane C is called the spectrum of T. The spectrum r(T, X) is further partitioned into three disjoint sets namely point spectrum, continuous spectrum and residual spectrum as follows: Point spectrum rp(T, X) is the set of all a 2 C such that T 1 a does not exist, i.e., condition (R1) fails. The element of rp(T, X) is called eigenvalue of T. Continuous spectrum rc(T, X) is the set of all a 2 C such that conditions (R1) and (R3) hold but condition (R2) fails, i.e., T 1 a 1 exists, domain of T 1 a is dense in X but T a is unbounded. 1 Residual spectrum rr(T, X) is the set of all a 2 C such that T 1 a exists but do not satisfy condition (R3), i.e., domain of T a is not dense in X. The condition (R2) may or may not holds good. Definition 2.2 [16, pp. 220–221]. Let k, l be two nonempty subsets of the space W of all real or complex sequences and A = (ank) an infinite matrix of complex numbers ank, where n; k 2 N0 . For every x = (xk) 2 k and every integer n we write
An ðxÞ ¼
X
ank xk ;
k
where the sum without limits is always taken from k = 0 to k = 1. The sequence Ax = (An(x)), if it exists, is called the transformation of x by the matrix A. Infinite matrix A 2 (k, l) if and only if A x 2 l whenever x 2 k. Goldberg’s classification of operator Ta [12, p. 58]: Let X be a Banach space and Ta 2 B(X), where a is a complex number. Again, let R(Ta) and T 1 a denote the range and inverse of the operator Ta, respectively. Then the following possibilities may occur; (A) R(Ta) = X, (B) RðT a Þ – RðT a Þ ¼ X, (C) RðT a Þ – X,
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and (1) (2) (3)
Ta is injective and T 1 a is continuous, Ta is injective and T 1 a is discontinuous, Ta is not injective.
Remark 2.1. Combining (A), (B), (C) and (1), (2), (3); we get nine different cases. These are labeled by A1, A2, A3, B1, B2, B3, C1, C2 and C3 . The notation a 2 A2r(T, X) means the operator Ta 2 A2, i.e., R(Ta) = X and Ta is injective but T 1 a is discontinuous. Similarly others. Remark 2.2. If a is a complex number such that Ta 2 A1 or Ta 2 B1, then a belongs to the resolvent set q(T, X) of T on X. The other classification gives rise to the fine spectrum of T. Lemma 2.1 [18, p. 126]. The matrix A = (ank) gives rise to a bounded linear operator T 2 B(l1) from l1 to itself if and only if the supremum of l1 norms of the columns of A is bounded. Lemma 2.2 [12, p. 59]. T has a dense range if and only if T is one to one, where T denotes the adjoint operator of the operator T. Lemma 2.3 [12, p. 60]. The adjoint operator T of T is onto if and only if T has a bounded inverse.
3. Spectrum and point spectrum of the operator Duv on the sequence space l1 In this section we obtained spectrum and point spectrum of the operator Duv on l1. Theorem 3.1. The operator Duv : l1 ? l1 is a bounded linear operator and
kDuv kBðl1 Þ ¼ supðjuk j þ jv k jÞ: k
Proof. Proof is simple. So we omit. h Theorem 3.2. Spectrum of the operator Duv on the sequence space l1 is given by
rðDuv ; l1 Þ ¼ fa 2 C : jU aj 6 jVjg: Proof. Proof of this theorem is divided into two parts. In the first part, we show that which is equivalent to
rðDuv ; l1 Þ # fa 2 C : jU aj 6 jVjg,
a 2 C with jU aj > jVj implies a R rðDuv ; l1 Þ; i:e:; a 2 qðDuv ; l1 Þ: In the second part, we establish the reverse inclusion, i.e.,
fa 2 C : jU aj 6 jVjg # rðDuv ; l1 Þ: Part I: Let a 2 C with jU aj > jVj. Clearly, a – U and a – uk for each k 2 N0 as it does not satisfy this condition. Further, (Duv aI) = (ank) reduces to a triangle and hence has an inverse (Duv aI)1 = (bnk), where
0
1 ðu aÞ B 0 v 0 B B ðu0 aÞðu1 aÞ
0
0
1 ðu1 aÞ v 1 ðu1 aÞðu2 aÞ
0
ðbnk Þ ¼ B v0v1 B B ðu0 aÞðu1 aÞðu2 aÞ @ .. .. . .
1 ðu2 aÞ
.. .
...
1
C ...C C C: ...C C A .. .
ð3:1Þ
By Lemma 2.1, the operator (Duv aI)1 2 B(l1) if the supremum of l1 norms of the columns of (bnk) is bounded, i.e., P1 P P supk 1 1. In order to show supk 1 n¼0 jbnk j < P n¼0 jbnk j < 1, first we prove that the series n¼0 jbnk j is convergent for each 1 k 2 N0 . Let Sk ¼ n¼0 jbnk j. Then the series
S0 ¼
1 X n¼0
jbn0 j ¼
1 1 X v 0 v 1 v n1 þ u0 a ðu0 aÞðu1 aÞ ðun aÞ n¼1
is convergent because
bnþ1;0 ¼ lim v n ¼ V < 1 for a 2 C with jU aj > jVj: lim n!1 n!1 unþ1 a U a bn0
ð3:2Þ
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P Similarly, one can show the convergence of the series Sk ¼ 1 jb j for each k = 1, 2, 3, . . . n¼0 nk vk Since modulus function is continuous, so Now we claim that supkSk is finite. Let b ¼ limk!1 ukþ1 . a
V : b ¼ U a
ð3:3Þ
Clearly, 0 < b < 1 and
1 b : ¼ jVj U a
ð3:4Þ
We have
Sk ¼
1 vk v k v kþ1 þ þ : þ uk a ðuk aÞðukþ1 aÞ ðuk aÞðukþ1 aÞðukþ2 aÞ
ð3:5Þ
Taking limit both sides of equality (3.5) and using equalities (3.3) and (3.4), we get
lim Sk ¼
k!1
b 1 < 1: jVj 1 b
Since (Sk) is a sequence of positive real numbers and limk?1Sk < 1, so supkSk < 1. Thus,
ðDuv aIÞ1 2 Bðl1 Þ for a 2 C with jU aj > jVj:
ð3:6Þ
1
Next, we show that domain of the operator (Duv aI) is dense in l1 equivalent to say that range of the operator (Duv aI) is dense in l1, which follows immediately as (Duv aI)1 2 B(l1). Hence we have
rðDuv ; l1 Þ # fa 2 C : jU aj 6 jVjg:
ð3:7Þ
Part II: We now prove the reverse inclusion, i.e.,
fa 2 C : jU aj 6 jVjg # rðDuv ; l1 Þ:
ð3:8Þ
First we prove inclusion (3.8) under the assumption that a – U and a – uk for each k 2 N0 , i.e., we want to show that one of the conditions of Definition 2.1 fails. Let a 2 C with jU aj 6 jVj. Clearly, (Duv aI) is a triangle and hence (Duv aI)1 exists. So condition (R1) is satisfied but condition (R2) fails as can be seen below: Suppose a 2 C with jU aj < jVj. By equality (3.2), the series S0 is divergent because
bnþ1;0 ¼ lim v n ¼ V > 1: lim n!1 n!1 U a bn0 unþ1 a
So supkSk is unbounded. Hence
ðDuv aIÞ1 R Bðl1 Þ for a 2 C with jU aj < jVj:
ð3:9Þ
Next, we consider a 2 C with jU aj = jVj. Taking limit both sides of equality (3.5), we get
1 þ lim Sk ¼ k!1 U a
V2 1 1 1 þ þ þ ¼ 1: þ þ ¼ jVj jVj jVj ðU aÞ2 ðU aÞ3 V
This shows that supkSk is unbounded. Hence
ðDuv aIÞ1 R Bðl1 Þ for a 2 C with jU aj ¼ jVj:
ð3:10Þ
Finally, we prove the inclusion (3.8) under the assumption that a = U and a = uk for each k 2 N0 . Case (i): If (uk) is a constant sequence, say uk = U for all k 2 N0 , then
0
1 0 Bv x C B 0 0C C ðDuv UIÞx ¼ 0 ) B B v 1 x1 C ¼ 0 ) xi ¼ 0 for each i 2 N0 : @ A .. . This shows that the operator (Duv UI) is one to one, but R(Duv UI) is not dense in l1. So condition (R3) fails. Hence U 2 r(Duv, l1). Case (ii): If (uk) is a sequence of distinct real numbers, then the series S0 is divergent for each a = uk from equality (3.2) and consequently, supkSk is unbounded. Hence
ðDuv aIÞ1 R Bðl1 Þ for a ¼ uk :
ð3:11Þ
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So condition (R2) fails. Hence uk 2 r(Duv, l1) for all k 2 N0 . Again, taking limit both sides of equality (3.5), we get limk?1Sk = 1 for a = U. So supk Sk is unbounded. Hence
ðDuv aIÞ1 R Bðl1 Þ for a ¼ U:
ð3:12Þ
So U 2 r(Duv, l1). Thus, in this case also uk 2 r(Duv, l1) for all k 2 N0 and U 2 r(Duv, l1). Hence we have
fa 2 C : jU aj 6 jVjg # rðDuv ; l1 Þ:
ð3:13Þ
From inclusions (3.7) and (3.13), we get
rðDuv ; l1 Þ ¼ fa 2 C : jU aj 6 jVjg: This completes the proof. h Theorem 3.3. Point spectrum of the operator Duv on the sequence space l1 is
rp ðDuv ; l1 Þ ¼ ;: Proof. For the point spectrum of the operator Duv, we find those a in C such that the matrix equation Duv x = ax is satisfy for non-zero vector x = (xk) in l1. Consider Duvx = ax for x – 0 = (0, 0, . . .) in l1, which gives
u0 x0 ¼ ax0
v 0 x0 þ u1 x1 ¼ ax1 v 1 x1 þ u2 x2 ¼ ax2 .. .
9 > > > > > > > > > > =
> > > > > v k1 xk1 þ uk xk ¼ axk > > > > > .. ; .
ð3:14Þ
The proof of this Theorem is divided into two cases. Case (i): Suppose (uk) is a constant sequence, say uk = U for all k 2 N0 . Let xt be the first nonzero entry of the sequence x = (xn). So equation vt1xt1 + U xt = axt implies a = U, and from the equation vtxt + Uxt+1 = axt+1, we get xt = 0, which is a contradiction to our assumption. Hence rp(Duv, l1) = ;. Case (ii): Suppose (uk) is a sequence of distinct real numbers. Clearly,
xk ¼
v k1
a uk
xk1 for all k P 1:
If a = u0, then
xk V ¼ lim u U > 1 k!1 xk1 0 because jU u0j < jVj. So x R l1 for x0 – 0. Similarly, if a = uk for all k P 1, then xk1 = 0, xk2 = 0, . . . , x0 = 0 and
xnþ1 ¼
vn
uk unþ1
xn for all n P k:
This implies
xnþ1 V ¼ >1 lim n!1 xn uk U because jU ukj < jVj for all k P 1. So x R l1 for x0 – 0. If x0 = 0, then xk = 0 for all k P 1. Only possibility is x = 0 = (0, 0, . . .). Hence rp(Duv, l1) = ;. h 4. Residual and continuous spectrum of the operator Duv on the sequence space l1 Let T : X ? X be a bounded linear operator having matrix representation A and the dual space of X denoted by Xw. Again, let T be its adjoint operator on Xw. Then the matrix representation of T is the transpose of the matrix A.
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We need result for the point spectrum of D uv on the sequence space l1 for obtaining residual and continuous spectrum. So H we first determine point spectrum of the operator D uv on l1 . H
Theorem 4.1. Point spectrum of the adjoint operator D uv on l1 is
rp ðDuv ; lH1 Þ ¼ fa 2 C : jU aj 6 jVjg: H
Proof. Suppose D uv f ¼ af for 0 – f 2 l1 ffi l1 , where
0
u0
B0 B Duv ¼ B B0 @ .. .
1
0
1
v0
0
...
u1
v1
0 .. .
u2 .. .
Bf C ...C C B 1C C and f ¼ B C: B f2 C ...C A @ A .. ... .
f0
This gives
fk ¼
a uk1
v k1
fk1 ¼
ða uk1 Þða uk2 Þ ða u0 Þ f0 for all k P 1: v k1 v k2 v 0
ð4:1Þ
Hence
ða uk1 Þða uk2 Þ ða u0 Þ jf0 j for all k P 1: jfk j ¼ v k1 v k2 v 0
ð4:2Þ
Using equality (4.1), we get
fk ¼ lim a uk1 ¼ a U < 1 provided jU aj < jVj: lim k!1 fk1 k!1 V v k1 So f 2 l1 and consequently, f 2 l1. Again, for a 2 C with jU aj = jVj,
fk ¼ lim a uk1 ¼ a U ¼ 1: lim k!1 v V k!1 fk1 k1
Thus, the test fails. In this case, we take f = (fk) in such a way it is a decreasing sequence of positive real numbers and ratio limk!1 f fk ¼ 1. Clearly, f = (fk) is a convergent sequence and consequently, f 2 l1. Hence k1
jU aj 6 jVj ) sup jfk j < 1: k
Conversely, suppose supkjfkj < 1. From equality (4.2), it follows that
a uk1 6 1 for all k P m; where m is a positive integer: v k1
k1 So limk!1 au 6 1, i.e., jU aj 6 jVj. Hence v k1
sup jfk j < 1 ) jU aj 6 jVj: k
H This means that f 2 l1 if and only if f0 – 0, f = (fk) is a decreasing sequence of positive real numbers such that limk!1 f fk ¼ 1 k1 and jU aj 6 jVj. Thus, H rp ðDuv ; l1 Þ ¼ fa 2 C : jU aj 6 jVjg. h Theorem 4.2. Residual spectrum
rr ðDuv ; l1 Þ of the operator Duv on l1 is
rr ðDuv ; l1 Þ ¼ fa 2 C : jU aj 6 jVjg: Proof. The proof of this Theorem is divided into two cases. Case (i): Suppose (uk) is a constant sequence, say uk = U for all k 2 N0 . For a 2 C with jU aj 6 jVj, the operator (Duv aI) is a triangle except a = U and consequently, the operator (Duv aI) has an inverse. Further by Theorem 3.3, the operator (Duv aI) is one to one for a = U and hence has an inverse. But by Theorem 4.1, the operator (Duv aI) is not injective for a 2 C with jU aj 6jVj. Hence by Lemma 2.2, the range of the operator (Duv aI) is not dense in l1. Thus, rr ðDuv ; l1 Þ ¼ fa 2 C : jU aj 6 jVjg.
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Case (ii): Suppose (uk) is a sequence of distinct real numbers. For a 2 C with jU aj 6 jVj, the operator (Duv aI) is a triangle except a = uk for all k 2 N0 and consequently, the operator (Duv aI) has an inverse. Further by Theorem 3.3, the operator (Duv ukI) is one to one and hence (Duv ukI)1 exists for all k 2 N0 . On the basis of the argument given in Case (i), it is easy to verify that the range of the operator (Duv aI) is not dense in l1. Thus,
rr ðDuv ; l1 Þ ¼ fa 2 C : jU aj 6 jVjg: Theorem 4.3. Continuous spectrum
rc ðDuv ; l1 Þ of the operator Duv on l1 is
rc ðDuv ; l1 Þ ¼ ;: Proof. It is known that rp ðDuv ; l1 Þ, rr ðDuv ; l1 Þ and rc ðDuv ; l1 Þ are pairwise disjoint sets and union of these sets is rðDuv ; l1 Þ. But by Theorems 3.2 and 4.2, we get
rðDuv ; l1 Þ ¼ rr ðDuv ; l1 Þ; which shows that continuous spectrum is empty set.
h
5. Fine spectrum of the operator Duv on the sequence space l1 Theorem 5.1. If a satisfies jU aj > jVj, then (Duv a I) 2 A1. Proof. It is required to show that the operator (Duv aI) is bijective and has a continuous inverse for a 2 C with jU aj > jVj. Since a – U and a – uk for each k 2 N0 , therefore the operator (Duv aI) is a triangle. Hence it has an inverse. The operator (Duv aI)1 is continuous for a 2 C with jU aj > jVj by statement (3.6). Also the equation
ðDuv aIÞx ¼ y gives x ¼ ðDuv aIÞ1 y; i:e:; xn ¼ ððDuv aIÞ1 yÞn ;
n 2 N0 :
Thus, for every y 2 l1, we can find x 2 l1 such that
ðDuv aIÞx ¼ y; since ðDuv aIÞ1 2 Bðl1 Þ: This shows that the operator (Duv aI) is onto and hence (Duv aI) 2 A1. h Theorem 5.2. Let u be a constant sequence, say uk = U for all k 2 N0 . Then U 2 C 1 rðDuv ; l1 Þ. Proof. We have rr ðDuv ; l1 Þ ¼ fa 2 C : jU aj 6 jVjg. Clearly, U 2 rr(Duv, l1). It is sufficient to show that the operator (Duv H UI)1 is continuous. By Lemma 2.3, it is enough to show that (Duv UI) is onto, i.e., for given y ¼ ðyn Þ 2 l1 , we have to find H x ¼ ðxn Þ 2 l1 such that (Duv UI) x = y. Now (Duv UI) x = y, i.e.,
v 0 x1 ¼ y 0 v 1 x2 ¼ y 1 .. .
v i1 xi ¼ yi1 .. . Thus, vn1xn = yn1 for all n P 1 which implies supnjxnj < 1, since y 2 l1 and v = (vk) is a convergent sequence. This shows that the operator (Duv aI) is onto and hence U 2 C 1 rðDuv ; l1 Þ. h Theorem 5.3. Let u be a constant sequence, say uk = U for all k 2 N0 and a–U; a 2 rr ðDuv ; l1 Þ. Then a 2 C2 rðDuv ; l1 Þ. Proof. It is sufficient to show that the operator (Duv aI)1 is discontinuous for a – U and a 2 rr ðDuv ; l1 Þ. The operator (Duv aI)1 is discontinuous by statements (3.9) and (3.10) for U – a 2 C with jU aj 6 jVj. h Theorem 5.4. Let u be a sequence of distinct real numbers and a 2 rr ðDuv ; l1 Þ. Then a 2 C 2 rðDuv ; l1 Þ
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Proof. It is sufficient to show that the operator (Duv aI)1 is discontinuous for a 2 rr ðDuv ; l1 Þ. The operator (Duv aIÞ1 is discontinuous by statements (3.9)–(3.12) for a 2 C with jU aj 6 jVj. h References [1] A.M. Akhmedov, F. Basar, On the fine spectra of the difference operator D over the sequence space lp (1 6 p < 1), Demonst. Math. 39 (3) (2006) 585– 595. [2] A.M. Akhmedov, F. Basar, The fine spectra of the difference operator D over the sequence space bvp (1 6 p < 1), Acta Math. Sin., Engl. Ser. 23 (10) (2007) 1757–1768. [3] A.M. Akhmedov, S.R. El-Shabrawy, On the fine spectrum of the operator Da,b over the sequence space c, Comput. Math. Appl. 61 (10) (2011) 2994–3002. [4] B. Altay, F. Basar, On the fine spectrum of the difference operator D on c0 and c, Inform. Sci. 168 (1–4) (2004) 217–224. [5] B. Altay, F. Basar, On the fine spectrum of the generalized difference operator B(r, s) over the sequence spaces c0 and c, Int. J. Math. Math. Sci. 18 (2005) 3005–3013. [6] B. Altay, F. Basar, The fine spectrum and the matrix domain of the difference operator D on the sequence space lp (0 < p < 1), Commun. Math. Anal. 2 (2) (2007) 1–11. [7] B. Altay, M. Karakus, On the spectrum and the fine spectrum of the Zweier matrix as an operator on some sequence spaces, Thai J. Math. 3 (2) (2005) 153–162. [8] H. Bilgic, H. Furkan, On the fine spectrum of the operator B(r, s, t) over the sequence spaces l1 and bv, Math. Comput. Modell. 45 (7–8) (2007) 883–891. [9] H. Bilgic, H. Furkan, On the fine spectrum of the generalized difference operator B(r, s) over the sequence spaces lp and bvp (1 < p < 1), Nonlinear Anal. 68 (3) (2008) 499–506. [10] H. Furkan, H. Bilgic, F. Basar, On the fine spectrum of the operator B(r, s, t) over the sequence spaces lp and bvp (1 < p < 1), Comput. Math. Appl. 60 (7) (2010) 2141–2152. [11] H. Furkan, H. Bilgic, K. Kayaduman, On the fine spectrum of the generalized difference operator B(r, s) over the sequence spaces l1 and bv, Hokkaido Math. J. 35 (4) (2006) 893–904. [12] S. Goldberg, Unbounded Linear Operators, Dover Publications, Inc., New York, 1985. [13] M. Gonzalez, The fine spectrum of the Cesaro operator in lp (1 < p < 1), Arch. Math. 44 (1985) 355–358. [14] K. Kayaduman, H. Furkan, The fine spectra of the difference operator D over the sequence spaces l1 and bv, Int. Math. Forum 1 (21–24) (2006) 1153– 1160. [15] E. Kreyszig, Introductory Functional Analysis with Applications, John Wiley and Sons Inc., New York-Chichester-Brisbane-Toronato, 1978. [16] I.J. Maddox, Elements of Functional Analysis, Cambridge University Press, 1988. [17] J.B. Reade, On the spectrum of the Cesaro operator, Bull. London Math. Soc. 17 (3) (1985) 263–267. [18] A. Wilansky, Summability through functional analysis, North-Holland Mathematics Studies, North-Holland, Amsterdam, vol. 85, 1984.
Contents lists available at SciVerse ScienceDirect
Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc
Fine spectrum of the generalized difference operator Duv on sequence space l1 P.D. Srivastava ⇑, Sudhanshu Kumar Department of Mathematics, Indian Institute of Technology Kharagpur, Kharagpur 721302, India
a r t i c l e
i n f o
a b s t r a c t In this paper we obtained spectrum and point spectrum of the operator Duv on l1. The operator Duv on l1 is defined as Duv x ¼ ðun xn þ v n1 xn1 Þ1 n¼0 under certain conditions, where x1 = 0 and x = (xn) 2 l1. Further, the results on continuous spectrum, residual spectrum and fine spectrum of the operator Duv on the sequence space l1 are also derived. Crown Copyright Ó 2011 Published by Elsevier Inc. All rights reserved.
Keywords: Spectrum of an operator Generalized operator Duv Sequence space l1
1. Introduction Let u = (uk) and
v = (vk) be sequences such that
(i) u is either a constant sequence or sequence of distinct real numbers with U = limk?1uk, (ii) v is a sequence of nonzero real numbers with V = limk?1vk – 0, and (iii) jU ukj < jVj for each k 2 N0 ¼ f0; 1; 2; . . .g. We define the operator Duv on the sequence space l1 as follows:
Duv x ¼ ðun xn þ v n1 xn1 Þ1 n¼0 with x1 ¼ 0; where x ¼ ðxn Þ 2 l1 :
ð1:1Þ
It is easy to verify that the operator Duv can be represented by the matrix
0
u0
Bv B 0 Du v ¼ B B0 @ .. .
1
0
0
...
u1
0
v1
u2 ...
...C C C: ...C A .. .
.. .
ð1:2Þ
Spectrum of the Cesaro operator on sequence spaces l1 and lp (1 < p < 1) is investigated Reade [17] and Gonzalez [13], respectively. In 2006, Kayaduman and Furkan [14] obtained spectrum and fine spectrum of the difference operator D on sequence spaces l1 and bv, where Dx = (xn xn1). Fine spectrum of the difference operator D on the sequence space lp (0 < p < 1) is determined by Altay and Basar [6], while fine spectrum of the Zweier matrix Zs on sequence spaces l1 and bv is obtained by Altay and Karakus [7], where Zsx = (sxn + (1 s)xn1) and s is a real number with s – 0, 1. Fine spectrum of the operator B(r, s) on sequence spaces l1 and bv is studied by Furkan et al. [11], where B(r, s)x = (rxn + sxn1). In 2008, Bilgic and Furkan [9] obtained fine spectrum of the operator B(r, s) on sequence spaces lp and bvp (1 < p < 1). Recently, spectrum ⇑ Corresponding author. E-mail addresses: [email protected] (P.D. Srivastava), [email protected] (S. Kumar). 0096-3003/$ - see front matter Crown Copyright Ó 2011 Published by Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2011.12.010
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and fine spectrum of the operator B(r, s, t) on sequence spaces l1 and bv is studied by Bilgic and Furkan [8], where B(r, s, t)x = (rxn + sxn1 + txn2). Fine spectrum of the difference operator D on sequence spaces c0 and c is studied by Altay and Basar [4], and the generalization of these results are obtained by the same authors in [5] with the generalized difference operator B(r, s). In 2006 and 2007, Akhmedov and Basar [1 and 2] determined the fine spectrum of the difference operator D over the sequence spaces lp and bvp (1 6 p < 1). Recently, spectrum and fine spectrum of the operator B(r, s, t) on sequence spaces lp and bvp (1 < p < 1) have derived by Furkan et al. [10] which generalizes the main results of [1,2 and 9]. Quite recently, the fine spectrum of the operator Da,b over the sequence space c has been worked by Akhmedov and El-Shabrawy [3]. In this paper we determine spectrum, point spectrum, continuous spectrum and residual spectrum of the operator Duv on the sequence space l1. It is easy to verify that by choosing suitably u and v sequences, one can get easily the operators such as B(r, s), Zs, etc. Choosing u = (r), v = (s) and u = (s), v = (1 s), the operator Duv reduces to B(r, s) and Zs, respectively. Similarly, if u = (1), v = (1) and u = (0), v = (1), then the operator Duv reduces to D and right-shift operator, respectively. Thus, the results of this paper generalizes the corresponding results of many operator whose matrix representation has diagonal and postdiagonal elements studied by earlier authors. 2. Preliminaries and notation Let X and Y be Banach spaces and T : X ? Y be a bounded linear operator. The set of all bounded linear operators on X into itself is denoted by B(X). The adjoint T: Xw ? Xw of T is defined by
ðT /ÞðxÞ ¼ /ðTxÞ for all / 2 X H and x 2 X: Clearly, T is a bounded linear operator on the dual space Xw. Let X – {0} be a complex normed space and T : D(T) ? X be a linear operator with domain D(T) # X. With T, we associate the operator Ta = (T aI), where a is a complex number and I is the identity operator on D(T). The inverse of Ta (if exists) is denoted by T 1 a and known as the resolvent operator of T. Since the spectral theory is concerned with many properties of Ta 1 1 and T 1 a , which depend on a, so we are interested the set of those a in the complex plane for which T a exists or T a is bounded or domain of T 1 is dense in X. a Definition 2.1 [15, p. 371]. Let X – {0} be a complex normed space and T : D(T) ? X be a linear operator with domain D(T) # X. A regular value of T is a complex number a such that
(R1) T 1 a exists, (R2) T 1 a is bounded, (R3) T 1 a is defined on a set which is dense in X. Resolvent set q(T, X) of T is the set of all regular values a of T. Its complement rðT; XÞ ¼ C n qðT; XÞ in the complex plane C is called the spectrum of T. The spectrum r(T, X) is further partitioned into three disjoint sets namely point spectrum, continuous spectrum and residual spectrum as follows: Point spectrum rp(T, X) is the set of all a 2 C such that T 1 a does not exist, i.e., condition (R1) fails. The element of rp(T, X) is called eigenvalue of T. Continuous spectrum rc(T, X) is the set of all a 2 C such that conditions (R1) and (R3) hold but condition (R2) fails, i.e., T 1 a 1 exists, domain of T 1 a is dense in X but T a is unbounded. 1 Residual spectrum rr(T, X) is the set of all a 2 C such that T 1 a exists but do not satisfy condition (R3), i.e., domain of T a is not dense in X. The condition (R2) may or may not holds good. Definition 2.2 [16, pp. 220–221]. Let k, l be two nonempty subsets of the space W of all real or complex sequences and A = (ank) an infinite matrix of complex numbers ank, where n; k 2 N0 . For every x = (xk) 2 k and every integer n we write
An ðxÞ ¼
X
ank xk ;
k
where the sum without limits is always taken from k = 0 to k = 1. The sequence Ax = (An(x)), if it exists, is called the transformation of x by the matrix A. Infinite matrix A 2 (k, l) if and only if A x 2 l whenever x 2 k. Goldberg’s classification of operator Ta [12, p. 58]: Let X be a Banach space and Ta 2 B(X), where a is a complex number. Again, let R(Ta) and T 1 a denote the range and inverse of the operator Ta, respectively. Then the following possibilities may occur; (A) R(Ta) = X, (B) RðT a Þ – RðT a Þ ¼ X, (C) RðT a Þ – X,
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and (1) (2) (3)
Ta is injective and T 1 a is continuous, Ta is injective and T 1 a is discontinuous, Ta is not injective.
Remark 2.1. Combining (A), (B), (C) and (1), (2), (3); we get nine different cases. These are labeled by A1, A2, A3, B1, B2, B3, C1, C2 and C3 . The notation a 2 A2r(T, X) means the operator Ta 2 A2, i.e., R(Ta) = X and Ta is injective but T 1 a is discontinuous. Similarly others. Remark 2.2. If a is a complex number such that Ta 2 A1 or Ta 2 B1, then a belongs to the resolvent set q(T, X) of T on X. The other classification gives rise to the fine spectrum of T. Lemma 2.1 [18, p. 126]. The matrix A = (ank) gives rise to a bounded linear operator T 2 B(l1) from l1 to itself if and only if the supremum of l1 norms of the columns of A is bounded. Lemma 2.2 [12, p. 59]. T has a dense range if and only if T is one to one, where T denotes the adjoint operator of the operator T. Lemma 2.3 [12, p. 60]. The adjoint operator T of T is onto if and only if T has a bounded inverse.
3. Spectrum and point spectrum of the operator Duv on the sequence space l1 In this section we obtained spectrum and point spectrum of the operator Duv on l1. Theorem 3.1. The operator Duv : l1 ? l1 is a bounded linear operator and
kDuv kBðl1 Þ ¼ supðjuk j þ jv k jÞ: k
Proof. Proof is simple. So we omit. h Theorem 3.2. Spectrum of the operator Duv on the sequence space l1 is given by
rðDuv ; l1 Þ ¼ fa 2 C : jU aj 6 jVjg: Proof. Proof of this theorem is divided into two parts. In the first part, we show that which is equivalent to
rðDuv ; l1 Þ # fa 2 C : jU aj 6 jVjg,
a 2 C with jU aj > jVj implies a R rðDuv ; l1 Þ; i:e:; a 2 qðDuv ; l1 Þ: In the second part, we establish the reverse inclusion, i.e.,
fa 2 C : jU aj 6 jVjg # rðDuv ; l1 Þ: Part I: Let a 2 C with jU aj > jVj. Clearly, a – U and a – uk for each k 2 N0 as it does not satisfy this condition. Further, (Duv aI) = (ank) reduces to a triangle and hence has an inverse (Duv aI)1 = (bnk), where
0
1 ðu aÞ B 0 v 0 B B ðu0 aÞðu1 aÞ
0
0
1 ðu1 aÞ v 1 ðu1 aÞðu2 aÞ
0
ðbnk Þ ¼ B v0v1 B B ðu0 aÞðu1 aÞðu2 aÞ @ .. .. . .
1 ðu2 aÞ
.. .
...
1
C ...C C C: ...C C A .. .
ð3:1Þ
By Lemma 2.1, the operator (Duv aI)1 2 B(l1) if the supremum of l1 norms of the columns of (bnk) is bounded, i.e., P1 P P supk 1 1. In order to show supk 1 n¼0 jbnk j < P n¼0 jbnk j < 1, first we prove that the series n¼0 jbnk j is convergent for each 1 k 2 N0 . Let Sk ¼ n¼0 jbnk j. Then the series
S0 ¼
1 X n¼0
jbn0 j ¼
1 1 X v 0 v 1 v n1 þ u0 a ðu0 aÞðu1 aÞ ðun aÞ n¼1
is convergent because
bnþ1;0 ¼ lim v n ¼ V < 1 for a 2 C with jU aj > jVj: lim n!1 n!1 unþ1 a U a bn0
ð3:2Þ
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P Similarly, one can show the convergence of the series Sk ¼ 1 jb j for each k = 1, 2, 3, . . . n¼0 nk vk Since modulus function is continuous, so Now we claim that supkSk is finite. Let b ¼ limk!1 ukþ1 . a
V : b ¼ U a
ð3:3Þ
Clearly, 0 < b < 1 and
1 b : ¼ jVj U a
ð3:4Þ
We have
Sk ¼
1 vk v k v kþ1 þ þ : þ uk a ðuk aÞðukþ1 aÞ ðuk aÞðukþ1 aÞðukþ2 aÞ
ð3:5Þ
Taking limit both sides of equality (3.5) and using equalities (3.3) and (3.4), we get
lim Sk ¼
k!1
b 1 < 1: jVj 1 b
Since (Sk) is a sequence of positive real numbers and limk?1Sk < 1, so supkSk < 1. Thus,
ðDuv aIÞ1 2 Bðl1 Þ for a 2 C with jU aj > jVj:
ð3:6Þ
1
Next, we show that domain of the operator (Duv aI) is dense in l1 equivalent to say that range of the operator (Duv aI) is dense in l1, which follows immediately as (Duv aI)1 2 B(l1). Hence we have
rðDuv ; l1 Þ # fa 2 C : jU aj 6 jVjg:
ð3:7Þ
Part II: We now prove the reverse inclusion, i.e.,
fa 2 C : jU aj 6 jVjg # rðDuv ; l1 Þ:
ð3:8Þ
First we prove inclusion (3.8) under the assumption that a – U and a – uk for each k 2 N0 , i.e., we want to show that one of the conditions of Definition 2.1 fails. Let a 2 C with jU aj 6 jVj. Clearly, (Duv aI) is a triangle and hence (Duv aI)1 exists. So condition (R1) is satisfied but condition (R2) fails as can be seen below: Suppose a 2 C with jU aj < jVj. By equality (3.2), the series S0 is divergent because
bnþ1;0 ¼ lim v n ¼ V > 1: lim n!1 n!1 U a bn0 unþ1 a
So supkSk is unbounded. Hence
ðDuv aIÞ1 R Bðl1 Þ for a 2 C with jU aj < jVj:
ð3:9Þ
Next, we consider a 2 C with jU aj = jVj. Taking limit both sides of equality (3.5), we get
1 þ lim Sk ¼ k!1 U a
V2 1 1 1 þ þ þ ¼ 1: þ þ ¼ jVj jVj jVj ðU aÞ2 ðU aÞ3 V
This shows that supkSk is unbounded. Hence
ðDuv aIÞ1 R Bðl1 Þ for a 2 C with jU aj ¼ jVj:
ð3:10Þ
Finally, we prove the inclusion (3.8) under the assumption that a = U and a = uk for each k 2 N0 . Case (i): If (uk) is a constant sequence, say uk = U for all k 2 N0 , then
0
1 0 Bv x C B 0 0C C ðDuv UIÞx ¼ 0 ) B B v 1 x1 C ¼ 0 ) xi ¼ 0 for each i 2 N0 : @ A .. . This shows that the operator (Duv UI) is one to one, but R(Duv UI) is not dense in l1. So condition (R3) fails. Hence U 2 r(Duv, l1). Case (ii): If (uk) is a sequence of distinct real numbers, then the series S0 is divergent for each a = uk from equality (3.2) and consequently, supkSk is unbounded. Hence
ðDuv aIÞ1 R Bðl1 Þ for a ¼ uk :
ð3:11Þ
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So condition (R2) fails. Hence uk 2 r(Duv, l1) for all k 2 N0 . Again, taking limit both sides of equality (3.5), we get limk?1Sk = 1 for a = U. So supk Sk is unbounded. Hence
ðDuv aIÞ1 R Bðl1 Þ for a ¼ U:
ð3:12Þ
So U 2 r(Duv, l1). Thus, in this case also uk 2 r(Duv, l1) for all k 2 N0 and U 2 r(Duv, l1). Hence we have
fa 2 C : jU aj 6 jVjg # rðDuv ; l1 Þ:
ð3:13Þ
From inclusions (3.7) and (3.13), we get
rðDuv ; l1 Þ ¼ fa 2 C : jU aj 6 jVjg: This completes the proof. h Theorem 3.3. Point spectrum of the operator Duv on the sequence space l1 is
rp ðDuv ; l1 Þ ¼ ;: Proof. For the point spectrum of the operator Duv, we find those a in C such that the matrix equation Duv x = ax is satisfy for non-zero vector x = (xk) in l1. Consider Duvx = ax for x – 0 = (0, 0, . . .) in l1, which gives
u0 x0 ¼ ax0
v 0 x0 þ u1 x1 ¼ ax1 v 1 x1 þ u2 x2 ¼ ax2 .. .
9 > > > > > > > > > > =
> > > > > v k1 xk1 þ uk xk ¼ axk > > > > > .. ; .
ð3:14Þ
The proof of this Theorem is divided into two cases. Case (i): Suppose (uk) is a constant sequence, say uk = U for all k 2 N0 . Let xt be the first nonzero entry of the sequence x = (xn). So equation vt1xt1 + U xt = axt implies a = U, and from the equation vtxt + Uxt+1 = axt+1, we get xt = 0, which is a contradiction to our assumption. Hence rp(Duv, l1) = ;. Case (ii): Suppose (uk) is a sequence of distinct real numbers. Clearly,
xk ¼
v k1
a uk
xk1 for all k P 1:
If a = u0, then
xk V ¼ lim u U > 1 k!1 xk1 0 because jU u0j < jVj. So x R l1 for x0 – 0. Similarly, if a = uk for all k P 1, then xk1 = 0, xk2 = 0, . . . , x0 = 0 and
xnþ1 ¼
vn
uk unþ1
xn for all n P k:
This implies
xnþ1 V ¼ >1 lim n!1 xn uk U because jU ukj < jVj for all k P 1. So x R l1 for x0 – 0. If x0 = 0, then xk = 0 for all k P 1. Only possibility is x = 0 = (0, 0, . . .). Hence rp(Duv, l1) = ;. h 4. Residual and continuous spectrum of the operator Duv on the sequence space l1 Let T : X ? X be a bounded linear operator having matrix representation A and the dual space of X denoted by Xw. Again, let T be its adjoint operator on Xw. Then the matrix representation of T is the transpose of the matrix A.
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We need result for the point spectrum of D uv on the sequence space l1 for obtaining residual and continuous spectrum. So H we first determine point spectrum of the operator D uv on l1 . H
Theorem 4.1. Point spectrum of the adjoint operator D uv on l1 is
rp ðDuv ; lH1 Þ ¼ fa 2 C : jU aj 6 jVjg: H
Proof. Suppose D uv f ¼ af for 0 – f 2 l1 ffi l1 , where
0
u0
B0 B Duv ¼ B B0 @ .. .
1
0
1
v0
0
...
u1
v1
0 .. .
u2 .. .
Bf C ...C C B 1C C and f ¼ B C: B f2 C ...C A @ A .. ... .
f0
This gives
fk ¼
a uk1
v k1
fk1 ¼
ða uk1 Þða uk2 Þ ða u0 Þ f0 for all k P 1: v k1 v k2 v 0
ð4:1Þ
Hence
ða uk1 Þða uk2 Þ ða u0 Þ jf0 j for all k P 1: jfk j ¼ v k1 v k2 v 0
ð4:2Þ
Using equality (4.1), we get
fk ¼ lim a uk1 ¼ a U < 1 provided jU aj < jVj: lim k!1 fk1 k!1 V v k1 So f 2 l1 and consequently, f 2 l1. Again, for a 2 C with jU aj = jVj,
fk ¼ lim a uk1 ¼ a U ¼ 1: lim k!1 v V k!1 fk1 k1
Thus, the test fails. In this case, we take f = (fk) in such a way it is a decreasing sequence of positive real numbers and ratio limk!1 f fk ¼ 1. Clearly, f = (fk) is a convergent sequence and consequently, f 2 l1. Hence k1
jU aj 6 jVj ) sup jfk j < 1: k
Conversely, suppose supkjfkj < 1. From equality (4.2), it follows that
a uk1 6 1 for all k P m; where m is a positive integer: v k1
k1 So limk!1 au 6 1, i.e., jU aj 6 jVj. Hence v k1
sup jfk j < 1 ) jU aj 6 jVj: k
H This means that f 2 l1 if and only if f0 – 0, f = (fk) is a decreasing sequence of positive real numbers such that limk!1 f fk ¼ 1 k1 and jU aj 6 jVj. Thus, H rp ðDuv ; l1 Þ ¼ fa 2 C : jU aj 6 jVjg. h Theorem 4.2. Residual spectrum
rr ðDuv ; l1 Þ of the operator Duv on l1 is
rr ðDuv ; l1 Þ ¼ fa 2 C : jU aj 6 jVjg: Proof. The proof of this Theorem is divided into two cases. Case (i): Suppose (uk) is a constant sequence, say uk = U for all k 2 N0 . For a 2 C with jU aj 6 jVj, the operator (Duv aI) is a triangle except a = U and consequently, the operator (Duv aI) has an inverse. Further by Theorem 3.3, the operator (Duv aI) is one to one for a = U and hence has an inverse. But by Theorem 4.1, the operator (Duv aI) is not injective for a 2 C with jU aj 6jVj. Hence by Lemma 2.2, the range of the operator (Duv aI) is not dense in l1. Thus, rr ðDuv ; l1 Þ ¼ fa 2 C : jU aj 6 jVjg.
P.D. Srivastava, S. Kumar / Applied Mathematics and Computation 218 (2012) 6407–6414
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Case (ii): Suppose (uk) is a sequence of distinct real numbers. For a 2 C with jU aj 6 jVj, the operator (Duv aI) is a triangle except a = uk for all k 2 N0 and consequently, the operator (Duv aI) has an inverse. Further by Theorem 3.3, the operator (Duv ukI) is one to one and hence (Duv ukI)1 exists for all k 2 N0 . On the basis of the argument given in Case (i), it is easy to verify that the range of the operator (Duv aI) is not dense in l1. Thus,
rr ðDuv ; l1 Þ ¼ fa 2 C : jU aj 6 jVjg: Theorem 4.3. Continuous spectrum
rc ðDuv ; l1 Þ of the operator Duv on l1 is
rc ðDuv ; l1 Þ ¼ ;: Proof. It is known that rp ðDuv ; l1 Þ, rr ðDuv ; l1 Þ and rc ðDuv ; l1 Þ are pairwise disjoint sets and union of these sets is rðDuv ; l1 Þ. But by Theorems 3.2 and 4.2, we get
rðDuv ; l1 Þ ¼ rr ðDuv ; l1 Þ; which shows that continuous spectrum is empty set.
h
5. Fine spectrum of the operator Duv on the sequence space l1 Theorem 5.1. If a satisfies jU aj > jVj, then (Duv a I) 2 A1. Proof. It is required to show that the operator (Duv aI) is bijective and has a continuous inverse for a 2 C with jU aj > jVj. Since a – U and a – uk for each k 2 N0 , therefore the operator (Duv aI) is a triangle. Hence it has an inverse. The operator (Duv aI)1 is continuous for a 2 C with jU aj > jVj by statement (3.6). Also the equation
ðDuv aIÞx ¼ y gives x ¼ ðDuv aIÞ1 y; i:e:; xn ¼ ððDuv aIÞ1 yÞn ;
n 2 N0 :
Thus, for every y 2 l1, we can find x 2 l1 such that
ðDuv aIÞx ¼ y; since ðDuv aIÞ1 2 Bðl1 Þ: This shows that the operator (Duv aI) is onto and hence (Duv aI) 2 A1. h Theorem 5.2. Let u be a constant sequence, say uk = U for all k 2 N0 . Then U 2 C 1 rðDuv ; l1 Þ. Proof. We have rr ðDuv ; l1 Þ ¼ fa 2 C : jU aj 6 jVjg. Clearly, U 2 rr(Duv, l1). It is sufficient to show that the operator (Duv H UI)1 is continuous. By Lemma 2.3, it is enough to show that (Duv UI) is onto, i.e., for given y ¼ ðyn Þ 2 l1 , we have to find H x ¼ ðxn Þ 2 l1 such that (Duv UI) x = y. Now (Duv UI) x = y, i.e.,
v 0 x1 ¼ y 0 v 1 x2 ¼ y 1 .. .
v i1 xi ¼ yi1 .. . Thus, vn1xn = yn1 for all n P 1 which implies supnjxnj < 1, since y 2 l1 and v = (vk) is a convergent sequence. This shows that the operator (Duv aI) is onto and hence U 2 C 1 rðDuv ; l1 Þ. h Theorem 5.3. Let u be a constant sequence, say uk = U for all k 2 N0 and a–U; a 2 rr ðDuv ; l1 Þ. Then a 2 C2 rðDuv ; l1 Þ. Proof. It is sufficient to show that the operator (Duv aI)1 is discontinuous for a – U and a 2 rr ðDuv ; l1 Þ. The operator (Duv aI)1 is discontinuous by statements (3.9) and (3.10) for U – a 2 C with jU aj 6 jVj. h Theorem 5.4. Let u be a sequence of distinct real numbers and a 2 rr ðDuv ; l1 Þ. Then a 2 C 2 rðDuv ; l1 Þ
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Proof. It is sufficient to show that the operator (Duv aI)1 is discontinuous for a 2 rr ðDuv ; l1 Þ. The operator (Duv aIÞ1 is discontinuous by statements (3.9)–(3.12) for a 2 C with jU aj 6 jVj. h References [1] A.M. Akhmedov, F. Basar, On the fine spectra of the difference operator D over the sequence space lp (1 6 p < 1), Demonst. Math. 39 (3) (2006) 585– 595. [2] A.M. Akhmedov, F. Basar, The fine spectra of the difference operator D over the sequence space bvp (1 6 p < 1), Acta Math. Sin., Engl. Ser. 23 (10) (2007) 1757–1768. [3] A.M. Akhmedov, S.R. El-Shabrawy, On the fine spectrum of the operator Da,b over the sequence space c, Comput. Math. Appl. 61 (10) (2011) 2994–3002. [4] B. Altay, F. Basar, On the fine spectrum of the difference operator D on c0 and c, Inform. Sci. 168 (1–4) (2004) 217–224. [5] B. Altay, F. Basar, On the fine spectrum of the generalized difference operator B(r, s) over the sequence spaces c0 and c, Int. J. Math. Math. Sci. 18 (2005) 3005–3013. [6] B. Altay, F. Basar, The fine spectrum and the matrix domain of the difference operator D on the sequence space lp (0 < p < 1), Commun. Math. Anal. 2 (2) (2007) 1–11. [7] B. Altay, M. Karakus, On the spectrum and the fine spectrum of the Zweier matrix as an operator on some sequence spaces, Thai J. Math. 3 (2) (2005) 153–162. [8] H. Bilgic, H. Furkan, On the fine spectrum of the operator B(r, s, t) over the sequence spaces l1 and bv, Math. Comput. Modell. 45 (7–8) (2007) 883–891. [9] H. Bilgic, H. Furkan, On the fine spectrum of the generalized difference operator B(r, s) over the sequence spaces lp and bvp (1 < p < 1), Nonlinear Anal. 68 (3) (2008) 499–506. [10] H. Furkan, H. Bilgic, F. Basar, On the fine spectrum of the operator B(r, s, t) over the sequence spaces lp and bvp (1 < p < 1), Comput. Math. Appl. 60 (7) (2010) 2141–2152. [11] H. Furkan, H. Bilgic, K. Kayaduman, On the fine spectrum of the generalized difference operator B(r, s) over the sequence spaces l1 and bv, Hokkaido Math. J. 35 (4) (2006) 893–904. [12] S. Goldberg, Unbounded Linear Operators, Dover Publications, Inc., New York, 1985. [13] M. Gonzalez, The fine spectrum of the Cesaro operator in lp (1 < p < 1), Arch. Math. 44 (1985) 355–358. [14] K. Kayaduman, H. Furkan, The fine spectra of the difference operator D over the sequence spaces l1 and bv, Int. Math. Forum 1 (21–24) (2006) 1153– 1160. [15] E. Kreyszig, Introductory Functional Analysis with Applications, John Wiley and Sons Inc., New York-Chichester-Brisbane-Toronato, 1978. [16] I.J. Maddox, Elements of Functional Analysis, Cambridge University Press, 1988. [17] J.B. Reade, On the spectrum of the Cesaro operator, Bull. London Math. Soc. 17 (3) (1985) 263–267. [18] A. Wilansky, Summability through functional analysis, North-Holland Mathematics Studies, North-Holland, Amsterdam, vol. 85, 1984.