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Generalized difference sequence spaces and their dual spaces
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J. Math. Anal. Appl. 292 (2004) 423–432 www.elsevier.com/locate/jmaa
Generalized difference sequence spaces and their dual spaces Ç.A. Bekta¸s,∗ M. Et, and R. Çolak Department of Mathematics, Firat University, Elazig 23119, Turkey Received 8 July 2003 Submitted by S. Kaijser
Abstract The definition of the pα-, pβ- and pγ -duals of a sequence space was defined by Et [Internat. J. Math. Math. Sci. 24 (2000) 785–791]. In this paper we compute pα- and N-duals of the sequence m spaces ∆m v (X) for X = ∞ , c and c0 , and compute β- and γ -duals of the sequence spaces ∆v (X) for X = ∞ , c and c0 . 2004 Elsevier Inc. All rights reserved. Keywords: Difference sequence spaces; N -dual; pα-, pβ- and pγ -duals
1. Introduction ω denotes the space of all scalar sequences and any subspace of ω is called a sequence space. Let ∞ , c and c0 be the linear spaces of bounded, convergent and null sequences x = (xk ) with complex terms, respectively, normed by x∞ = sup |xk |, k
where k ∈ N = {1, 2, . . .}, the set of positive integers. Throughout the paper X will denote one of the sequence spaces ∞ , c or c0 . The notion of difference sequence spaces was introduced by Kızmaz [2]. It was generalized by Et and Çolak [4] as follows. Let m be a non-negative integer. Then * Corresponding author.
E-mail address: [email protected] (Ç.A. Bekta¸s). 0022-247X/$ – see front matter 2004 Elsevier Inc. All rights reserved. doi:10.1016/j.jmaa.2003.12.006
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∆m (X) = x = (xk ): ∆m xk ∈ X , where ∆m xk = ∆m−1 xk − ∆m−1 xk+1 for all k ∈ N. Let v = (vk ) be any fixed sequence of non-zero complex numbers. Et and Esi [5] generalized the above sequence spaces to the following sequence spaces: m ∆m v (X) = x = (xk ): ∆v xk ∈ X , m m−1 x − ∆m−1 x m i m where ∆m k k+1 , ∆v xk = v x k = ∆v v i=0 (−1) i vk+i xk+i for all k ∈ N. The sequence spaces ∆m v (X) are Banach spaces normed by x∆ =
m
|vi xi | + ∆m v x ∞.
i=1
Now we define ∆(m) v xk
=
m i=0
m (−1) vk−i xk−i . i i
(m)
Remark. (∆m v xk ) ∈ X if and only if (∆v xk ) ∈ X. (m)
Now for x ∈ ∆v (X) define
x∆ = sup ∆(m) xk . k
v
It can be shown that (∆(m) v (X), · ∆ ) is a BK-space and the norms x∆ and x∆ are equivalent. Let us define the operator m D : ∆m v (X) → ∆v (X)
by Dx = (0, 0, . . . , xm+1 , xm+2 , . . .), where x = (x1 , x2 , . . . , xm , . . .). It is trivial that D is a bounded linear operator on ∆m v (X). Furthermore the set m m D ∆v (X) = D∆m v (X) = x = (xk ): x ∈ ∆v (X), x1 = x2 = · · · = xm = 0 m m m is a subspace of ∆m v (X) and normed by x∆ = ∆v x∞ in D∆v (X). D∆v (X) and X are equivalent as topological spaces, since m m m ∆m (1) v : D∆v (X) → X, defined by ∆v x = y = ∆v xk , is a linear homeomorphism. Let X and [D∆m v (X)] denote the continuous duals of X and m D∆v (X), respectively. It can be shown that
T : D∆m v (X) → X ,
−1 f∆ → f∆ ◦ ∆m = f, v
is a linear isometry. So [D∆m v (X)] is equivalent to X .
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425
2. Dual spaces In this section we give pα-, N -, β- and γ -duals of ∆m v (X). The following results can be found in [1,4]. Let m be a positive integer. Then there exist positive constants M1 and M2 such that m+k m M1 k (2) M2 k m , k = 0, 1, 2, . . ., k n n+m−k−1 n+m n+m = = , (3) m−1 m n k=0
x ∈ ∆m v (∞ )
implies
sup k −m |vk xk | < ∞.
(4)
k
Lemma 2.1 [2]. Let (pn ) be a sequence of positive numbers increasing monotonically to infinity. (i) If supn | nv=1 pv av | < ∞, then supn |pn ∞ a | < ∞. ∞k=n+1 k (ii) If k pk ak is convergent, then limn pn k=n+1 ak = 0. Definition 2.2 [3]. Let E be a sequence space, p > 0 and define
pα p E = a = (ak ): |ak xk | < ∞, ∀x ∈ E , k
E pβ
p = a = (ak ): (ak xk ) is convergent, ∀x ∈ E , k
n
E pγ = a = (ak ): sup (ak xk )p < ∞, ∀x ∈ E ,
n
k=0 E N = a = (ak ): lim ak xk = 0, ∀x ∈ E .
k
E pα ,
E pβ ,
E pγ
and E N are called pα-, pβ-, pγ - and N -duals of E, respectively. It Then pα can be shown that E ⊂ E pβ ⊂ E pγ and if E ⊂ F , then F η ⊂ E η for η = pα, pβ, pγ and N . If we take p = 1 in this definition, then we obtain the α-, β- and γ -duals of E. Theorem 2.3. Let 0 < p < ∞. Then pα = [∆m (c)]pα = [∆m (c )]pα = U , (i) [∆m 1 v (∞ )] v v 0 pα (ii) U1 = U2 ,
where
U1 = a = (ak ):
∞ k=1
p k pm vk−1 |ak |p
<∞ ,
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U2 = a = (ak ): sup k −pm |vk |p |ak |p < ∞ . k
Proof. (i) Let a ∈ U1 and x ∈ ∆m v (∞ ). Then ∞
|ak xk |p =
k=1
∞
p k pm vk−1 |ak |p k −pm |vk |p |xk |p
k=1
sup k −pm |vk xk |p k
∞
p k pm ak vk−1 < ∞.
k=1
pα ∈ [∆m v (∞ )] .
Hence a pα and a ∈ / U1 . Then there exists a strictly inConversely suppose that a ∈ [∆m v (c0 )] creasing sequence (ni ) of positive integers ni with n1 < n2 < · · · , such that
ni+1
|vk |−p k pm |ak |p > i p .
k=ni +1
Define x ∈ ∆m v (c0 ) by
0, 1 k n1 , xk = m ak k vk sgn i , ni < k ni+1 . Then we have ∞
|ak xk |p =
n2 k=n1 +1
k=1
=
n2
ni+1
|ak xk |p + · · · +
|ak xk |p
k=ni +1
k
pm
|vk |
−p
k=n1 +1
> 1 +1 + ··· =
ni+1 1 pm |ak | + · · · + p k |vk |−p |ak |p i p
k=ni +1
∞
1 = ∞.
i=1 pα This contradicts to a ∈ [∆m v (c0 )] . Hence a ∈ U1 . This completes the proof of (i). (ii) Let a ∈ U2 and x ∈ U1 . Then ∞
|ak xk | = p
k=1
∞
p k pm vk−1 |ak |p k −pm |xk |p |vk |p
k=1 ∞
p sup k −pm |ak |p |vk |p k pm vk−1 |xk |p < ∞. k
pα ∈ U1 .
k=1
Hence a pα Now suppose that a ∈ U1 and a ∈ / U2 . Then we have sup k −pm |vk |p |ak |p = ∞. k
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427
Hence there is a strictly increasing sequence (k(i)) of positive integers k(i) such that −pm |vk(i) |p |ak(i) |p > i m . k(i) We define the sequence x by
|ak(i)|−p , k = k(i), xk = 0, k = k(i). Then we have ∞
∞ ∞
p pm k(i) k pm vk−1 |xk |p = |vk(i) |−p |ak(i)|−p i −m < ∞,
k=1
Hence x ∈ U1 and a ∈ U2 . 2
i=1
∞
k=1 |ak xk |
p
=
m 2.
i=1
∞
k=1 1
pα
= ∞. This contradicts to a ∈ U1 . Hence
If we take vk = 1, for all k ∈ N in Theorem 2.3 then we obtain the following corollary. Corollary 2.4. Let 0 < p < ∞. Then we have (i) [∆m (∞ )]pα = [∆m (c)]pα = [∆m (c0 )]pα = G1 , pα (ii) G1 = G2 , where
G1 = a = (ak ):
∞
k
pm
|ak | < ∞ , p
k=1
G2 = a = (ak ): sup k −pm |ak |p < ∞ . k
Lemma 2.5. If x ∈ ∆m v (c0 ), then
m+k −1 k
vk xk → 0 (k → ∞).
Proof. Proof follows from (2)–(4). 2 Theorem 2.6. Let m be a positive integer. Then m N N N ∆v (∞ ) = ∆m = M1 (v) and ∆m = M2 (v), v (c) v (c0 ) where M1 (v) = a = (an ): vn−1 nm an → 0, n → ∞ and
n
n + m − k − 1
M2 (v) = a = (an ): sup
vn−1 an < ∞ .
m−1 n
k=0
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N m N Proof. The proof of the part [∆m v (∞ )] = [∆v (c)] = M1 (v) is easy. We only show m N m that [∆v (c0 )] = M2 (v). Let a ∈ M2 (v) and x ∈ ∆v (c0 ). Then by Lemma 2.5 and (3) we obtain n n −1 n + m − k − 1 n + m − k − 1 −1 vn xn lim an xn = lim vn an n n m−1 m−1 k=0
k=0
= 0. N Hence a ∈ [∆m v (c0 )] . m Now let a ∈ [∆v (c0 )]N . Then limn an xn = 0 for all x ∈ ∆m v (c0 ). On the other hand, for each x ∈ ∆m (c ) there exists one and only one y = (y ) ∈ c such that 0 k 0 v
xn = vn−1
n n n+m−k−1 n+m−k−1 yk = vn−1 yk = 0, m−1 m−1 k=1
k=0
y0 = 0, by (1) and Remark given above. Hence n n + m − k − 1 −1 lim an xn = lim vn an yk = 0, n n m−1
∀y ∈ c0 .
k=0
If we take
n+m−k−1 −1 vn an , 1 k n, m−1 ank = 0, k > n, then we get lim n
∞ k=0
n n + m − k − 1 −1 vn an yk = 0, ank yk = lim n m−1 k=0
Hence A ∈ (c0 , c0 ) and so supn completes the proof. 2
n
k=0 |ank |
= supn
n k=0
∀y ∈ c0 .
n+m−k−1 −1 |vn ||an | < ∞. This m−1
Throughout the following theorems, we shall write bk instead of Theorem 2.7. (a) We put ∞ k−m k − j − 1 E1 (v) = a ∈ ω: ak vk−1 is convergent, m−1 k=1 j =1 ∞ k−m+1 k − j − 1 <∞ . |bk | m−2 k=1
β Then [D∆m v (∞ )] = E1 (v).
j =1
∞
−1 j =k+1 vj aj .
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429
(b) We put
n
k−m
k − j − 1
−1 ak vk E2 (v) = a ∈ ω: sup
< ∞, m−1
n
j =1
k=1
∞ k−m+1 k − j − 1 |bk | <∞ . m−2 j =1
k=1
Then
γ [D∆m v (∞ )]
= E2 (v).
Proof. (a) If x ∈ D∆m v (∞ ) then there exists one and only one y = (yk ) ∈ ∞ such that xk = vk−1
k−m j =1
k−j −1 yj , m−1
y1−m = y2−m = · · · = y0 = 0
for (1). Let a ∈ E1 (v), and suppose that −1sufficiently large k, for instance k > 2mrby = 1 (in some literature it is assumed that −1 k = 0 for k < 0). Then we may write n
ak xk =
k=1
n
ak
vk−1
j =1
k=1
= (−1)
m
n−m k=1
− bn
n−m j =1
k−j −1 yj (−1) m−1
k−m
m
k k+m−j −2 bk+m−1 yj m−2 j =1
n−j −1 yj . (−1)m m−1
(5)
k k+m−j −2 < ∞, the series ∞ yj k=1 bk+m−1 j =1 m−2 n−m n−j −1 is absolutely convergent. Moreover we have bn j =1 m−1 → 0 as n → ∞ by m β is convergent for all x ∈ D∆m Lemma 2.1(ii), hence ∞ v (∞ ), so a ∈ [D∆v (∞ )] . k=1 ak xk ∞ m β m Let a ∈ [D∆v (∞ )] . Then k=1 ak xk is convergent for each x ∈ D∆v (∞ ). For the sequence x = (xk ) defined by
0, k m, xk = −1 k−m k−j −1 vk j =1 m−1 , k > m, Since
∞
k=1 |bk+m−1 |
k
j =1
k+m−j −2 m−2
we may write ∞ k=1
ak vk−1
k−m j =1
∞
∞ k−j −1 ak xk . = m−1 k=1
−1 k−m k−j −1
Thus the series k=1 ak vk j =1 m−1 n−j −1 → 0 as n → ∞ by Lemma 2.1(ii). m−1
is convergent. This implies that bn
n−m j =1
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∞ k−m+1 k−j −1 β Now let a ∈ [D∆m v (∞ )] − E1 (v). Then k=1 |bk | j =1 m−2 is divergent, that ∞ k−m+1 k−j −1 is, k=1 |bk | j =1 = ∞. We define the sequence x = (x k ) by m−2
0, k m, i−m+1 i−j −1 xk = −1 k−1 vk i=1 sgn bi j =1 m−2 , k > m, where ak > 0 for all k or ak < 0 for all k. Since |∆m v xk | = 1 for k > m, it is trivial that ( ). Then we may write for n > m, x = (xk ) ∈ D∆m v ∞ n
ak xk = −
k=1
m
bk−1 ∆v xk−1 −
k=1
n−m
bk+m−1 ∆v xk+m−1 − bn xn vn .
k=1
Since (bn xn vn ) ∈ c0 , from (6) now letting n → ∞ we get ∞
ak xk =
k=1
∞ k=1
=
∞ k=1
bk+m−1 ∆v xk+m−1 k k+m−j −2 = ∞. |bk+m−1 | m−2 j =1
β This contradicts to a ∈ [D∆m v (∞ )] . Hence a ∈ E1 (v). (b) can be proved by the same way as above, using Lemma 2.1(i). 2 η m η Lemma 2.8. [D∆m v (∞ )] = [D∆v (c)] for η = β or γ .
Proof is easy. Theorem 2.9. Let c0+ denote the set of all positive null sequences. (a) We put k−m ∞ k − j − 1 uj converges and ak vk−1 E3 (v) = a ∈ ω: m−1 j =1
k=1
∞
|bk |
j =1
k=1 β [D∆m v (c0 )]
Then (b) We put
k−j −1 + uj < ∞, ∀u ∈ c0 . m−2
k−m+1
= E3 (v).
n
k−m
k − j − 1
−1 ak vk uj < ∞, E4 (v) = a ∈ ω: sup
m−1 n
j =1
k=1
∞ k=1
Then
γ [D∆m v (c0 )]
= E4 (v).
|bk |
k−j −1 + uj < ∞, ∀u ∈ c0 . m−2
k−m+1 j =1
(6)
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431
Proof. (a) and (b) can be proved by the same way as Theorem 2.4, using Lemma 2.1(i) and (ii). 2 The proof of the following result is a routine work. Lemma 2.10. η m η (i) [∆m v (∞ )] = [D∆v (∞ )] , m η m η (ii) [∆v (c)] = [D∆v (c)] , η m η (iii) [∆m v (c0 )] = [D∆v (c0 )]
for η = β or γ . 3. Matrix transformations Given any infinite matrix A = (ank )∞ n,k=1 of complex numbers and any sequence x = ∞ write A (x) = a x (n = 1, 2, . . .) and Ax = (An (x))∞ (xk ), we n nk k k=1 n=1 , provided the ∞ series k=1 ank xk are convergent for each n ∈ N. The proof of the following result is a routine work in view of Theorem 2.7(a). Theorem 3.1. Let G = ∞ or c and H = ∞ or c. Then A = (ank ) ∈ (∆m v (G), H ) if and only if (i) (anj )n ∈ H (1 j m), ∞ k−m k−j −1 −r (ii) j ∈ H, k=1 ank vk−1 =1 n k−m+1 k−j j−1 −rm−1 (iii) bnk j =1 ∈ (G, H ), m−2 j where bnk =
∞
−1 i=k+1 ani vi .
Theorem 3.2. Let G = ∞ or c and H = ∞ , c or c0 . Then A = (ank ) ∈ (G, ∆m v (H )) if and only if ∞ (i) k=1 |ank | < ∞ for each n, (ii) C ∈ (G, H ), where C = (cnk ) = (∆m−1 vn ank − ∆m−1 vn+1 an+1,k ). The proof is omitted. References [1] E. Malkowsky, S.D. Parashar, Matrix transformations in spaces of bounded and convergent difference sequences of order m, Analysis 17 (1997) 87–97. [2] H. Kızmaz, On certain sequence spaces, Canad. Math. Bull. 24 (1981) 169–176.
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[3] M. Et, On some topological properties of generalized difference sequence spaces, Internat. J. Math. Math. Sci. 24 (2000) 785–791. [4] M. Et, R. Çolak, On some generalized difference sequence spaces, Soochow J. Math. 21 (1995) 377–386. [5] M. Et, A. Esi, On Köthe–Toeplitz duals of generalized difference sequence spaces, Bull. Malaysian Math. Sc. Soc. (2) 23 (2000) 25–32.
J. Math. Anal. Appl. 292 (2004) 423–432 www.elsevier.com/locate/jmaa
Generalized difference sequence spaces and their dual spaces Ç.A. Bekta¸s,∗ M. Et, and R. Çolak Department of Mathematics, Firat University, Elazig 23119, Turkey Received 8 July 2003 Submitted by S. Kaijser
Abstract The definition of the pα-, pβ- and pγ -duals of a sequence space was defined by Et [Internat. J. Math. Math. Sci. 24 (2000) 785–791]. In this paper we compute pα- and N-duals of the sequence m spaces ∆m v (X) for X = ∞ , c and c0 , and compute β- and γ -duals of the sequence spaces ∆v (X) for X = ∞ , c and c0 . 2004 Elsevier Inc. All rights reserved. Keywords: Difference sequence spaces; N -dual; pα-, pβ- and pγ -duals
1. Introduction ω denotes the space of all scalar sequences and any subspace of ω is called a sequence space. Let ∞ , c and c0 be the linear spaces of bounded, convergent and null sequences x = (xk ) with complex terms, respectively, normed by x∞ = sup |xk |, k
where k ∈ N = {1, 2, . . .}, the set of positive integers. Throughout the paper X will denote one of the sequence spaces ∞ , c or c0 . The notion of difference sequence spaces was introduced by Kızmaz [2]. It was generalized by Et and Çolak [4] as follows. Let m be a non-negative integer. Then * Corresponding author.
E-mail address: [email protected] (Ç.A. Bekta¸s). 0022-247X/$ – see front matter 2004 Elsevier Inc. All rights reserved. doi:10.1016/j.jmaa.2003.12.006
424
Ç.A. Bekta¸s et al. / J. Math. Anal. Appl. 292 (2004) 423–432
∆m (X) = x = (xk ): ∆m xk ∈ X , where ∆m xk = ∆m−1 xk − ∆m−1 xk+1 for all k ∈ N. Let v = (vk ) be any fixed sequence of non-zero complex numbers. Et and Esi [5] generalized the above sequence spaces to the following sequence spaces: m ∆m v (X) = x = (xk ): ∆v xk ∈ X , m m−1 x − ∆m−1 x m i m where ∆m k k+1 , ∆v xk = v x k = ∆v v i=0 (−1) i vk+i xk+i for all k ∈ N. The sequence spaces ∆m v (X) are Banach spaces normed by x∆ =
m
|vi xi | + ∆m v x ∞.
i=1
Now we define ∆(m) v xk
=
m i=0
m (−1) vk−i xk−i . i i
(m)
Remark. (∆m v xk ) ∈ X if and only if (∆v xk ) ∈ X. (m)
Now for x ∈ ∆v (X) define
x∆ = sup ∆(m) xk . k
v
It can be shown that (∆(m) v (X), · ∆ ) is a BK-space and the norms x∆ and x∆ are equivalent. Let us define the operator m D : ∆m v (X) → ∆v (X)
by Dx = (0, 0, . . . , xm+1 , xm+2 , . . .), where x = (x1 , x2 , . . . , xm , . . .). It is trivial that D is a bounded linear operator on ∆m v (X). Furthermore the set m m D ∆v (X) = D∆m v (X) = x = (xk ): x ∈ ∆v (X), x1 = x2 = · · · = xm = 0 m m m is a subspace of ∆m v (X) and normed by x∆ = ∆v x∞ in D∆v (X). D∆v (X) and X are equivalent as topological spaces, since m m m ∆m (1) v : D∆v (X) → X, defined by ∆v x = y = ∆v xk , is a linear homeomorphism. Let X and [D∆m v (X)] denote the continuous duals of X and m D∆v (X), respectively. It can be shown that
T : D∆m v (X) → X ,
−1 f∆ → f∆ ◦ ∆m = f, v
is a linear isometry. So [D∆m v (X)] is equivalent to X .
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425
2. Dual spaces In this section we give pα-, N -, β- and γ -duals of ∆m v (X). The following results can be found in [1,4]. Let m be a positive integer. Then there exist positive constants M1 and M2 such that m+k m M1 k (2) M2 k m , k = 0, 1, 2, . . ., k n n+m−k−1 n+m n+m = = , (3) m−1 m n k=0
x ∈ ∆m v (∞ )
implies
sup k −m |vk xk | < ∞.
(4)
k
Lemma 2.1 [2]. Let (pn ) be a sequence of positive numbers increasing monotonically to infinity. (i) If supn | nv=1 pv av | < ∞, then supn |pn ∞ a | < ∞. ∞k=n+1 k (ii) If k pk ak is convergent, then limn pn k=n+1 ak = 0. Definition 2.2 [3]. Let E be a sequence space, p > 0 and define
pα p E = a = (ak ): |ak xk | < ∞, ∀x ∈ E , k
E pβ
p = a = (ak ): (ak xk ) is convergent, ∀x ∈ E , k
n
E pγ = a = (ak ): sup (ak xk )p < ∞, ∀x ∈ E ,
n
k=0 E N = a = (ak ): lim ak xk = 0, ∀x ∈ E .
k
E pα ,
E pβ ,
E pγ
and E N are called pα-, pβ-, pγ - and N -duals of E, respectively. It Then pα can be shown that E ⊂ E pβ ⊂ E pγ and if E ⊂ F , then F η ⊂ E η for η = pα, pβ, pγ and N . If we take p = 1 in this definition, then we obtain the α-, β- and γ -duals of E. Theorem 2.3. Let 0 < p < ∞. Then pα = [∆m (c)]pα = [∆m (c )]pα = U , (i) [∆m 1 v (∞ )] v v 0 pα (ii) U1 = U2 ,
where
U1 = a = (ak ):
∞ k=1
p k pm vk−1 |ak |p
<∞ ,
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U2 = a = (ak ): sup k −pm |vk |p |ak |p < ∞ . k
Proof. (i) Let a ∈ U1 and x ∈ ∆m v (∞ ). Then ∞
|ak xk |p =
k=1
∞
p k pm vk−1 |ak |p k −pm |vk |p |xk |p
k=1
sup k −pm |vk xk |p k
∞
p k pm ak vk−1 < ∞.
k=1
pα ∈ [∆m v (∞ )] .
Hence a pα and a ∈ / U1 . Then there exists a strictly inConversely suppose that a ∈ [∆m v (c0 )] creasing sequence (ni ) of positive integers ni with n1 < n2 < · · · , such that
ni+1
|vk |−p k pm |ak |p > i p .
k=ni +1
Define x ∈ ∆m v (c0 ) by
0, 1 k n1 , xk = m ak k vk sgn i , ni < k ni+1 . Then we have ∞
|ak xk |p =
n2 k=n1 +1
k=1
=
n2
ni+1
|ak xk |p + · · · +
|ak xk |p
k=ni +1
k
pm
|vk |
−p
k=n1 +1
> 1 +1 + ··· =
ni+1 1 pm |ak | + · · · + p k |vk |−p |ak |p i p
k=ni +1
∞
1 = ∞.
i=1 pα This contradicts to a ∈ [∆m v (c0 )] . Hence a ∈ U1 . This completes the proof of (i). (ii) Let a ∈ U2 and x ∈ U1 . Then ∞
|ak xk | = p
k=1
∞
p k pm vk−1 |ak |p k −pm |xk |p |vk |p
k=1 ∞
p sup k −pm |ak |p |vk |p k pm vk−1 |xk |p < ∞. k
pα ∈ U1 .
k=1
Hence a pα Now suppose that a ∈ U1 and a ∈ / U2 . Then we have sup k −pm |vk |p |ak |p = ∞. k
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Hence there is a strictly increasing sequence (k(i)) of positive integers k(i) such that −pm |vk(i) |p |ak(i) |p > i m . k(i) We define the sequence x by
|ak(i)|−p , k = k(i), xk = 0, k = k(i). Then we have ∞
∞ ∞
p pm k(i) k pm vk−1 |xk |p = |vk(i) |−p |ak(i)|−p i −m < ∞,
k=1
Hence x ∈ U1 and a ∈ U2 . 2
i=1
∞
k=1 |ak xk |
p
=
m 2.
i=1
∞
k=1 1
pα
= ∞. This contradicts to a ∈ U1 . Hence
If we take vk = 1, for all k ∈ N in Theorem 2.3 then we obtain the following corollary. Corollary 2.4. Let 0 < p < ∞. Then we have (i) [∆m (∞ )]pα = [∆m (c)]pα = [∆m (c0 )]pα = G1 , pα (ii) G1 = G2 , where
G1 = a = (ak ):
∞
k
pm
|ak | < ∞ , p
k=1
G2 = a = (ak ): sup k −pm |ak |p < ∞ . k
Lemma 2.5. If x ∈ ∆m v (c0 ), then
m+k −1 k
vk xk → 0 (k → ∞).
Proof. Proof follows from (2)–(4). 2 Theorem 2.6. Let m be a positive integer. Then m N N N ∆v (∞ ) = ∆m = M1 (v) and ∆m = M2 (v), v (c) v (c0 ) where M1 (v) = a = (an ): vn−1 nm an → 0, n → ∞ and
n
n + m − k − 1
M2 (v) = a = (an ): sup
vn−1 an < ∞ .
m−1 n
k=0
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N m N Proof. The proof of the part [∆m v (∞ )] = [∆v (c)] = M1 (v) is easy. We only show m N m that [∆v (c0 )] = M2 (v). Let a ∈ M2 (v) and x ∈ ∆v (c0 ). Then by Lemma 2.5 and (3) we obtain n n −1 n + m − k − 1 n + m − k − 1 −1 vn xn lim an xn = lim vn an n n m−1 m−1 k=0
k=0
= 0. N Hence a ∈ [∆m v (c0 )] . m Now let a ∈ [∆v (c0 )]N . Then limn an xn = 0 for all x ∈ ∆m v (c0 ). On the other hand, for each x ∈ ∆m (c ) there exists one and only one y = (y ) ∈ c such that 0 k 0 v
xn = vn−1
n n n+m−k−1 n+m−k−1 yk = vn−1 yk = 0, m−1 m−1 k=1
k=0
y0 = 0, by (1) and Remark given above. Hence n n + m − k − 1 −1 lim an xn = lim vn an yk = 0, n n m−1
∀y ∈ c0 .
k=0
If we take
n+m−k−1 −1 vn an , 1 k n, m−1 ank = 0, k > n, then we get lim n
∞ k=0
n n + m − k − 1 −1 vn an yk = 0, ank yk = lim n m−1 k=0
Hence A ∈ (c0 , c0 ) and so supn completes the proof. 2
n
k=0 |ank |
= supn
n k=0
∀y ∈ c0 .
n+m−k−1 −1 |vn ||an | < ∞. This m−1
Throughout the following theorems, we shall write bk instead of Theorem 2.7. (a) We put ∞ k−m k − j − 1 E1 (v) = a ∈ ω: ak vk−1 is convergent, m−1 k=1 j =1 ∞ k−m+1 k − j − 1 <∞ . |bk | m−2 k=1
β Then [D∆m v (∞ )] = E1 (v).
j =1
∞
−1 j =k+1 vj aj .
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429
(b) We put
n
k−m
k − j − 1
−1 ak vk E2 (v) = a ∈ ω: sup
< ∞, m−1
n
j =1
k=1
∞ k−m+1 k − j − 1 |bk | <∞ . m−2 j =1
k=1
Then
γ [D∆m v (∞ )]
= E2 (v).
Proof. (a) If x ∈ D∆m v (∞ ) then there exists one and only one y = (yk ) ∈ ∞ such that xk = vk−1
k−m j =1
k−j −1 yj , m−1
y1−m = y2−m = · · · = y0 = 0
for (1). Let a ∈ E1 (v), and suppose that −1sufficiently large k, for instance k > 2mrby = 1 (in some literature it is assumed that −1 k = 0 for k < 0). Then we may write n
ak xk =
k=1
n
ak
vk−1
j =1
k=1
= (−1)
m
n−m k=1
− bn
n−m j =1
k−j −1 yj (−1) m−1
k−m
m
k k+m−j −2 bk+m−1 yj m−2 j =1
n−j −1 yj . (−1)m m−1
(5)
k k+m−j −2 < ∞, the series ∞ yj k=1 bk+m−1 j =1 m−2 n−m n−j −1 is absolutely convergent. Moreover we have bn j =1 m−1 → 0 as n → ∞ by m β is convergent for all x ∈ D∆m Lemma 2.1(ii), hence ∞ v (∞ ), so a ∈ [D∆v (∞ )] . k=1 ak xk ∞ m β m Let a ∈ [D∆v (∞ )] . Then k=1 ak xk is convergent for each x ∈ D∆v (∞ ). For the sequence x = (xk ) defined by
0, k m, xk = −1 k−m k−j −1 vk j =1 m−1 , k > m, Since
∞
k=1 |bk+m−1 |
k
j =1
k+m−j −2 m−2
we may write ∞ k=1
ak vk−1
k−m j =1
∞
∞ k−j −1 ak xk . = m−1 k=1
−1 k−m k−j −1
Thus the series k=1 ak vk j =1 m−1 n−j −1 → 0 as n → ∞ by Lemma 2.1(ii). m−1
is convergent. This implies that bn
n−m j =1
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∞ k−m+1 k−j −1 β Now let a ∈ [D∆m v (∞ )] − E1 (v). Then k=1 |bk | j =1 m−2 is divergent, that ∞ k−m+1 k−j −1 is, k=1 |bk | j =1 = ∞. We define the sequence x = (x k ) by m−2
0, k m, i−m+1 i−j −1 xk = −1 k−1 vk i=1 sgn bi j =1 m−2 , k > m, where ak > 0 for all k or ak < 0 for all k. Since |∆m v xk | = 1 for k > m, it is trivial that ( ). Then we may write for n > m, x = (xk ) ∈ D∆m v ∞ n
ak xk = −
k=1
m
bk−1 ∆v xk−1 −
k=1
n−m
bk+m−1 ∆v xk+m−1 − bn xn vn .
k=1
Since (bn xn vn ) ∈ c0 , from (6) now letting n → ∞ we get ∞
ak xk =
k=1
∞ k=1
=
∞ k=1
bk+m−1 ∆v xk+m−1 k k+m−j −2 = ∞. |bk+m−1 | m−2 j =1
β This contradicts to a ∈ [D∆m v (∞ )] . Hence a ∈ E1 (v). (b) can be proved by the same way as above, using Lemma 2.1(i). 2 η m η Lemma 2.8. [D∆m v (∞ )] = [D∆v (c)] for η = β or γ .
Proof is easy. Theorem 2.9. Let c0+ denote the set of all positive null sequences. (a) We put k−m ∞ k − j − 1 uj converges and ak vk−1 E3 (v) = a ∈ ω: m−1 j =1
k=1
∞
|bk |
j =1
k=1 β [D∆m v (c0 )]
Then (b) We put
k−j −1 + uj < ∞, ∀u ∈ c0 . m−2
k−m+1
= E3 (v).
n
k−m
k − j − 1
−1 ak vk uj < ∞, E4 (v) = a ∈ ω: sup
m−1 n
j =1
k=1
∞ k=1
Then
γ [D∆m v (c0 )]
= E4 (v).
|bk |
k−j −1 + uj < ∞, ∀u ∈ c0 . m−2
k−m+1 j =1
(6)
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Proof. (a) and (b) can be proved by the same way as Theorem 2.4, using Lemma 2.1(i) and (ii). 2 The proof of the following result is a routine work. Lemma 2.10. η m η (i) [∆m v (∞ )] = [D∆v (∞ )] , m η m η (ii) [∆v (c)] = [D∆v (c)] , η m η (iii) [∆m v (c0 )] = [D∆v (c0 )]
for η = β or γ . 3. Matrix transformations Given any infinite matrix A = (ank )∞ n,k=1 of complex numbers and any sequence x = ∞ write A (x) = a x (n = 1, 2, . . .) and Ax = (An (x))∞ (xk ), we n nk k k=1 n=1 , provided the ∞ series k=1 ank xk are convergent for each n ∈ N. The proof of the following result is a routine work in view of Theorem 2.7(a). Theorem 3.1. Let G = ∞ or c and H = ∞ or c. Then A = (ank ) ∈ (∆m v (G), H ) if and only if (i) (anj )n ∈ H (1 j m), ∞ k−m k−j −1 −r (ii) j ∈ H, k=1 ank vk−1 =1 n k−m+1 k−j j−1 −rm−1 (iii) bnk j =1 ∈ (G, H ), m−2 j where bnk =
∞
−1 i=k+1 ani vi .
Theorem 3.2. Let G = ∞ or c and H = ∞ , c or c0 . Then A = (ank ) ∈ (G, ∆m v (H )) if and only if ∞ (i) k=1 |ank | < ∞ for each n, (ii) C ∈ (G, H ), where C = (cnk ) = (∆m−1 vn ank − ∆m−1 vn+1 an+1,k ). The proof is omitted. References [1] E. Malkowsky, S.D. Parashar, Matrix transformations in spaces of bounded and convergent difference sequences of order m, Analysis 17 (1997) 87–97. [2] H. Kızmaz, On certain sequence spaces, Canad. Math. Bull. 24 (1981) 169–176.
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[3] M. Et, On some topological properties of generalized difference sequence spaces, Internat. J. Math. Math. Sci. 24 (2000) 785–791. [4] M. Et, R. Çolak, On some generalized difference sequence spaces, Soochow J. Math. 21 (1995) 377–386. [5] M. Et, A. Esi, On Köthe–Toeplitz duals of generalized difference sequence spaces, Bull. Malaysian Math. Sc. Soc. (2) 23 (2000) 25–32.