Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 7, Issue 3, Article 101, 2006
THE DUAL SPACES OF THE SETS OF DIFFERENCE SEQUENCES OF ORDER m
Ç.A. BEKTAŞ AND M. ET
D EPARTMENT OF M ATHEMATICS
F IRAT U NIVERSITY
E LAZIG , 23119, TURKEY.
cbektas@firat.edu.tr
mikailet@yahoo.com
Received 16 December, 2005; accepted 07 January, 2006
Communicated by A.G. Babenko
A BSTRACT. The idea of difference sequence spaces was introduced by Kızmaz [5] and the
concept was generalized by Et and Çolak [3]. Let p = (pk ) be a bounded sequence of positive
real numbers and v = (vk ) be any fixed sequence of non-zero complex numbers. If x = (xk ) is
any sequence of complex numbers we write ∆m
v x for the sequence of the m -th order differences
m
of x and ∆m
(X)
=
{x
=
(x
)
:
∆
x
∈
X}
for any set X of sequences. In this paper we
k
v
v
determine the α -, β - and γ - duals of the sets ∆m
v (X) which are defined by Et et al. [2] for
X = `∞ (p), c(p) and c0 (p). This study generalizes results of Malkowsky [9] in special cases.
Key words and phrases: Difference sequences, α−, β− and γ−duals.
2000 Mathematics Subject Classification. 40C05, 46A45.
1. I NTRODUCTION , N OTATIONS AND K NOWN R ESULTS
Throughout this paper 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 space of bounded, convergent and null
sequences with complex terms, respectively, normed by
kxk∞ = sup |xk | ,
k
where k ∈ N = {1, 2, 3, . . . }, the set of positive integers. Furthermore, let p = (pk ) be bounded
sequences of positive real numbers and
pk
`∞ (p) = x ∈ ω : sup |xk | < ∞ ,
k
n
o
pk
c(p) = x ∈ ω : lim |xk − l| = 0 , for some l ∈ C ,
k→∞
ISSN (electronic): 1443-5756
c 2006 Victoria University. All rights reserved.
247-04
2
Ç.A. B EKTA Ş AND M. E T
n
o
c0 (p) = x ∈ ω : lim |xk |pk = 0
k→∞
(for details see [6], [7], [11]).
Let x and y be complex sequences, and E and F be subsets of ω. We write
\
M (E, F ) =
x−1 ∗ F = {a ∈ ω : ax ∈ F for all x ∈ E }
[12].
x∈E
In particular, the sets
E α = M (E, l1 ),
E β = M (E, cs) and
E γ = M (E, bs)
are called the α−, β− and γ− duals of E, where l1 , cs and bs are the sets of all convergent,
absolutely convergent and bounded series, respectively. If E ⊂ F , then F η ⊂ E η for η =
α, β, γ. It is clear that E α ⊂ (E α )α = E αα . If E = E αα , then E is an α -space. In particular,
an α -space is called a Köthe space or a perfect sequence space.
Throughout this paper X will be used to denote any one of the sequence spaces `∞ , c and c0 .
Kızmaz [5] introduced the notion of difference sequence spaces as follows:
X(∆) = {x = (xk ) : (∆xk ) ∈ X}.
Later on the notion was generalized by Et and Çolak in [3], namely,
X(∆m ) = {x = (xk ) : (∆m xk ) ∈ X}.
Subsequently difference sequence spaces have been studied by Malkowsky and Parashar [8],
Mursaleen [10], Çolak [1] and many others.
Let v = (vk ) be any fixed sequence of non-zero complex numbers. Et and Esi [4] generalized
the above sequence spaces to the following ones
m
∆m
v (X) = {x = (xk ) : (∆v xk ) ∈ X},
Pm
i m
m−1
where ∆0v x = (vk xk ), ∆m
xk −∆m−1
xk+1 ) such that ∆m
v x = (∆v
v
v xk =
i=0 (−1) i vk+i xk+i .
Recently Et et al. [2] generalized the sequence spaces ∆m
v (X) to the sequence spaces
m
∆m
v (X(p)) = {x = (xk ) :(∆v xk )∈ X(p)}
and showed that these spaces are complete paranormed spaces paranormed by
m
X
pk /M
g(x) =
|xi vi | + sup |∆m
,
v xk |
i=1
k
where H = supk pk and M = max (1, H).
m
Let us define the operator D : ∆m
v X(p) → ∆v X(p) by Dx = (0, 0, . . . , xm+1 , xm+2 , . . . ),
where x = (x1 , x2 , x3 , . . . ). It is trivial that D is a bounded linear operator on ∆m
v X(p). Furthermore the set
m
D[∆m
v X(p)] = D∆v X(p)
= {x = (xk ) : x ∈ ∆m
v X(p), x1 = x2 = · · · = xm = 0}
m
is a subspace of ∆m
v X(p). D∆v X(p) and X(p) are equivalent as topological spaces, since
(1.1)
m
∆m
v : D∆v X(p) → X(p)
0
0
m
m
defined by ∆m
v x = y = (∆v xk ) is a linear homeomorphism. Let [X(p)] and [D∆v X(p)]
denote the continuous duals of X(p) and D∆m
v X(p), respectively. It can be shown that
0
0
T : [D∆m
v X(p)] → [X(p)] ,
−1
f∆ → f∆ ◦ (∆m
= f,
v )
0
0
is a linear isometry. So [D∆m
v X(p)] is equivalent to [X(p)] .
J. Inequal. Pure and Appl. Math., 7(3) Art. 101, 2006
http://jipam.vu.edu.au/
T HE D UAL S PACES OF THE S ETS OF D IFFERENCE S EQUENCES OF O RDER m
3
Lemma 1.1 ([5]). Let (tn ) be a sequence of positive numbers increasing monotonically to infinity, then
P∞
Pn
i) If sup
k=n+1 ak < ∞,
P n | i=1 ti ai | < ∞, then supn tn P
∞
ii) If k tk ak is convergent, then limn→∞ tn k=n+1 ak = 0.
2. M AIN R ESULTS
In this section we determine the α−, β− and γ− duals of ∆m
v X(p).
Theorem 2.1. For every strictly positive sequence p = (pk ), we have
α
α
(i) [∆m
v l∞ (p)] = D1 (p),
αα
(ii) [∆m
= D1αα (p)
v l∞ (p)]
where
(
)
∞
k−m
∞
X
X k − j − 1
\
N 1/pj < ∞
a = (ak ) :
|ak | |vk |−1
D1α (p) =
m
−
1
j=1
N =2
k=1
and


"k−m #−1
∞ 

X k − j − 1
[
1/pj
αα
<∞ .
a = (ak ) : sup |ak | |vk |
N
D1 (p) =


m−1
k≥m+1
j=1
N =2
Proof.
m
pn
(i) Let a ∈ D1α (p) and x ∈ ∆m
v l∞ (p). We choose N > max(1, supn |∆v an | ).
Since
k−m
m X k − j − 1
X
k−j−1
1/pj
N
>
N 1/pj ≥ 1
m
−
1
m
−
j
j=1
j=1
xj | ≤ M (1 ≤ j ≤ m) for some
for arbitrary N > 1 (k = 2m, 2m + 1, . . . ) and |∆m−j
v
α
constant M , a ∈ D1 (p) implies
∞
m X
X
k − j − 1 m−j −1
∆v xj < ∞.
|ak | |vk |
m−j
j=1
k=1
Then
∞
X
|ak xk |
k=1
=
∞
X
|ak | |vk |−1
j=1
k=1
≤
∞
X
k=1
!
m
k−m
X
X
k
−
j
−
1
k
−
j
−
1
m−j
m
m−j
m
∆
x
∆
x
+
(−1)
(−1)
j
v
v j
m−j
m−1
|ak | |vk |−1
k−m
X
j=1
j=1
∞
m X
X
k−j−1
k − j − 1 m−j 1/pj
−1
N
+
|ak | |vk |
∆v xj
m−1
m
−
j
j=1
k=1
< ∞.
Conversely let a ∈
/ D1α (p). Then we have
∞
k−m
X
X k − j − 1
−1
N 1/pj = ∞
|ak ||vk |
m−1
j=1
k=1
J. Inequal. Pure and Appl. Math., 7(3) Art. 101, 2006
http://jipam.vu.edu.au/
4
Ç.A. B EKTA Ş AND M. E T
for some integer N > 1. We define the sequence x by
k−m
X k − j − 1
−1
xk = vk
N 1/pj
(k = m + 1, m + 2, . . . ).
m
−
1
j=1
P
α
Then it is easy to see that x ∈ ∆m
|ak xk | = ∞. Hence a ∈
/ [∆m
v l∞ (p) and
v l∞ (p)] .
k
This completes the proof of (i).
α
α
(ii) Let a ∈ D1αα (p) and x ∈ [∆m
v l∞ (p)] = D1 (p), by part (i). Then for some N > 1, we
have
∞
X
|ak xk |
k=m+1
=
∞
X
|ak | |vk |
"k−m X k − j − 1
m−1
j=1
k=m+1

≤ sup |ak | |vk |
k≥m+1
×
N 1/pj
m−1
−1
|xk | |vk |
k−m
X
j=1
k=m+1
|xk | |vk |−1
"k−m X k − j − 1
j=1
"k−m X k − j − 1
j=1
∞
X
#−1
m−1
#
N 1/pj
#−1 
N 1/pj

k−j−1
N 1/pj
m−1
< ∞.
Conversely let a ∈
/ D1αα (p). Then for all integers N > 1, we have
"k−m #−1
X k − j − 1
sup |ak | |vk |
N 1/pj
= ∞.
m
−
1
k≥m+1
j=1
We recall that
k−m
X
j=1
k−j−1
yj = 0
m−1
(k < m + 1)
for arbitrary yj .
Hence there is a strictly increasing sequence (k(s)) of integers k(s) ≥ m + 1 such
that

−1
k(s)−m X k(s) − j − 1
|ak(s) | |v k(s) | 
s1/pj  > sm+1
(s = m + 1, m + 2, . . . ).
m
−
1
j=1
We define the sequence x by

 |ak(s) |−1 , (k = k(s))
xk =
 0,
(k =
6 k(s)) (k = m + 1, m + 2, . . . )
Then for all integers N > m + 1, we have
∞
k−m
∞
X
X k − j − 1
X
−1
1/pj
|xk | |vk |
N
<
s−(m+1) < ∞.
m−1
s=m+1
j=1
k=1
J. Inequal. Pure and Appl. Math., 7(3) Art. 101, 2006
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T HE D UAL S PACES OF THE S ETS OF D IFFERENCE S EQUENCES OF O RDER m
α
Hence x ∈ [∆m
and
v l∞ (p)]
The proof is completed.
P∞
k=1
|ak xk | =
P∞
N =1
5
αα
1 = ∞. Hence a ∈
/ [∆m
v l∞ (p)] .
Theorem 2.2. For every strictly positive sequence p = (pk ), we have
α
α
(i) [∆m
v c0 (p)] = M0 (p),
αα
(ii) [∆m
= M0αα (p)
v c0 (p)]
where
)
(
∞
∞
k−m
[
X
X k − j − 1
N −1/pj < ∞
M0α (p) =
a∈ω:
|ak | |vk |−1
m
−
1
j=1
N =2
k=1
and


#−1
"k−m ∞ 

\
X k − j − 1
αα
−1/pj
M0 (p) =
a ∈ ω : sup |ak | |vk |
N
<∞ .


m−1
k≥m+1
j=1
N =2
Proof.
pk
m
≤
(i) Let a ∈ M0α (p) and x ∈ ∆m
v c0 (p). Then there is an integer k0 such that sup |∆v xk |
k>k0
N −1 , where N is the number in M0α (p). We put
pk
M = max |∆m
v xk | ,
1≤k≤k0
n = min pk ,
1≤k≤k0
L = (M + 1)N
and define the sequence y by yk = xk · L−1/n (k = 1, 2, . . . ). Then it is easy to see that
pk
supk |∆m
≤ N −1 .
v yk |
Since
k−m
m X k − j − 1
X
k−j−1
−1/pj
N
>
N −1/pj
m−1
m−j
j=1
j=1
for arbitrary N > 1 (k = 2m, 2m + 1, . . . ), a ∈ M0α (p) implies
∞
m X
X
k−j−1
|ak | |vk |
|∆m−j
yj | < ∞.
v
m
−
j
j=1
k=1
Then
∞
X
|ak xk | = L1/n
k=1
∞
X
|ak yk |
k=1
∞
X
k−m
X
k−j−1
≤L
|ak | |vk |
N −1/pj
m
−
1
j=1
k=1
∞
m X
X
k−j−1
1/n
−1
|∆m−j
yj |
+L
|ak | |vk |
v
m
−
j
j=1
k=1
1/n
−1
< ∞.
α
α
m
α
So we have a ∈ [∆m
v c0 (p)] . Therefore M0 (p) ⊂ [∆v c0 (p)] .
Conversely, let a ∈
/ M0α (p). Then we can determine a strictly increasing sequence
(k(s)) of integers such that k(1) = 1 and
k(s+1)−1
k−m
X k − j − 1
X
−1
(s + 1)−1/pj > 1 (s = 1, 2, . . . ).
M (s) =
|ak | |vk |
m
−
1
j=1
k=k(s)
J. Inequal. Pure and Appl. Math., 7(3) Art. 101, 2006
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6
Ç.A. B EKTA Ş AND M. E T
We define the sequence x by


s−1 k(l+1)−1
k−m
X
X k − j − 1
X k − j − 1
xk = vk −1 
(l + 1)−1/pj +
(s + 1)−1/pj 
m
−
1
m
−
1
l=1
j=k(l)
j=k(s)
(k(s) ≤ k ≤ k(s + 1) − 1; s = 1, 2, . . . ).
1
pk
(k(s) ≤ k ≤ k(s + 1) − 1; s = 1, 2, . . . )
Then it is easy to see that |∆m
= s+1
v xk |
P
P
∞
∞
m
α
hence x ∈ ∆v c0 (p), and k=1 |ak xk | ≥ s=1 = ∞, i.e. a ∈
/ [∆m
v c0 (p)] .
(ii) Omitted.
Theorem 2.3. For every strictly positive sequence p = (pk ), we have
(
)
∞
k−m
X
X k − j − 1
α
α
α
a∈ω:
|ak | |vk |−1
<∞ .
[∆m
v c(p)] = M (p) = M0 (p) ∩
m−1
j=1
k=1
m
Proof. Let a ∈ M α (p) and x ∈ ∆m
v c(p). Then there is a complex number l such that |∆v xk −
pk
l| → 0 (k → ∞). We define y = (yk ) by
k−m
X k − j − 1
−1
m+1
(k = 1, 2, . . . ).
yk = xk + vk l(−1)
m−1
j=1
Then y ∈ ∆m
v c0 (p) and
∞
X
∞
X
k−m
X
k−j−1
|ak xk | ≤
|ak | |vk |
|∆m
v yj |
m−1
j=1
k=1
k=1
∞
m X
X
k−j−1
−1
|ak | |vk |
+
|∆m−j
yj |
v
m−j
j=1
k=1
−1
+ |l|
∞
X
−1
|ak | |vk |
k=1
k−m
X
j=1
k−j−1
<∞
m−1
by Theorem 2.2(i) and since a ∈ M α (p).
α
m
α
α
Now let a ∈ [∆m
v c(p)] ⊂ [∆v c0 (p)] = M0 (p) by Theorem 2.2(i). Since the sequence x
defined by
k−m
X k − j − 1
m −1
xk = (−1) vk
(k = 1, 2, . . . )
m
−
1
j=1
is in ∆m
v c(p), we have
∞
X
|ak |
k=1
k−m
X
j=1
k−j−1
< ∞.
m−1
Theorem 2.4. For every strictly positive sequence p = (pk ), we have
β
β
(i) [D∆m
v `∞ (p)] = M∞ (p),
γ
γ
(ii) [D∆m
v `∞ (p)] = M∞ (p)
J. Inequal. Pure and Appl. Math., 7(3) Art. 101, 2006
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T HE D UAL S PACES OF THE S ETS OF D IFFERENCE S EQUENCES OF O RDER m
7
where
(
β
(p)
M∞
\
=
a∈ω:
N >1
∞
X
ak vk−1
k−m
X
j=1
k=1
k−j−1
N 1/pj
m−1
∞
X
|bk |
k−m+1
X k=1
(
γ
(p)
M∞
\
=
a ∈ ω : sup |
n
N >1
and bk =
P∞
j=k+1
n
X
ak vk−1
k−m
X
j=1
k=1
converges
j=1
and
)
k−j−1
N 1/pj < ∞ ,
m−2
k−j−1
N 1/pj | < ∞,
m−1
)
∞
k−m+1
X
X k − j − 1
N 1/pj < ∞
|bk |
m
−
2
j=1
k=1
vj−1 aj (k = 1, 2, . . . ).
Proof.
(i) If x ∈ D∆m
v `∞ (p) then there exists a unique y = (yk ) ∈ `∞ (p) such that
k−m
X
−1
m k−j −1
xk = vk
(−1)
yj
m−1
j=1
for sufficiently large k, for instance k > m by (1.1). Then there
is an integer N >
−1
β
pk
(p),
and
suppose
that
= 1. Then we may
}.
Let
a
∈
M
x
|
max{1, supk |∆m
∞
v k
−1
write
!
k−m
n
n
X
X
X
k
−
j
−
1
ak vk−1
(−1)m
ak x k =
yj
m
−
1
j=1
k=1
k=1
n−m
k n−m
X
X
X
k+m−j−2
m
m n−j −1
= (−1)
bk+m−1
yj − bn
(−1)
yj .
m−2
m−1
j=1
j=1
k=1
Since
∞
X
|bk+m−1 |
k X
k+m−j−2
m−2
j=1
k=1
N 1/pj < ∞,
the series
∞
X
bk+m−1
k X
k+m−j−2
j=1
k=1
m−2
yj
is absolutely convergent. Moreover by Lemma 1.1(ii), the convergence of
∞
k−m
X
X k − j − 1
−1
ak vk
N 1/pj
m
−
1
j=1
k=1
implies
lim bn
n→∞
Hence
P∞
k=1
n−m
X
j=1
n−j−1
N 1/pj = 0.
m−1
m
β
ak xk is convergent for all x ∈ D∆m
v `∞ (p), so a ∈ [D∆v `∞ (p)] .
J. Inequal. Pure and Appl. Math., 7(3) Art. 101, 2006
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8
Ç.A. B EKTA Ş AND M. E T
P∞
β
Conversely let a ∈ [D∆m
v `∞ (p)] . Then
k=1 ak xk is convergent for each x ∈
m
D∆v `∞ (p). If we take the sequence x = (xk ) defined by

k≤m
 0,
xk =
 −1 Pk−m k−j−1 1/pj
vk
N
, k>m
j=1
m−1
then we have
∞
X
ak vk−1
k−m
X
j=1
k=1
Thus the series
P∞
k=1
∞
X
k−j−1
1/pj
N
=
ak xk < ∞.
m−1
k=1
Pk−m
ak vk−1
k−j−1
m−1
N 1/pj is convergent. This implies that
n−m
X n − j − 1
lim bn
N 1/pj = 0
n→∞
m−1
j=1
j=1
by Lemma 1.1(ii).
Pk−m+1
P∞
β
β
Now let a ∈ [D∆m
v `∞ (p)] − M∞ (p). Then
j=1
k=1 |bk |
divergent, that is,
∞
k−m+1
X
X k − j − 1
|bk |
N 1/pj = ∞.
m
−
2
j=1
k=1
We define the sequence x = (xk ) by

 0,
xk =
Pi−m+1
 −1 Pk−1
vk
j=1
i=1 sgn bi
k−j−1
m−2
N 1/pj is
k≤m
i−j−1
m−2
N 1/pj , k > m
where ak > 0 for all k or ak < 0 for all k. It is trivial that x = (xk ) ∈ D∆m
v `∞ (p).
Then we may write for n > m
n
X
ak x k = −
k=1
m
X
bk−1 ∆v xk−1 −
k=1
n−m
X
bk+m−1 ∆v xk+m−1 − bn xn vn .
k=1
Since (bn xn vn ) ∈ c0 , now letting n → ∞we get
∞
X
ak x k = −
k=1
∞
X
bk+m−1 ∆v xk+m−1
k=1
=
∞
X
|bk+m−1 |
k=1
k X
k+m−j−2
j=1
m−2
N 1/pj = ∞.
β
β
This is a contradiction to a ∈ [D∆m
v `∞ (p)] . Hence a ∈ M∞ (p).
(ii) Can be proved by the same way as above, using Lemma 1.1(i).
η
m
η
Lemma 2.5. [D∆m
v `∞ (p)] = [D∆v c(p)] for η = β or γ.
The proof is obvious and is thus omitted.
Theorem 2.6. Let c+
0 denote the set of all positive null sequences.
J. Inequal. Pure and Appl. Math., 7(3) Art. 101, 2006
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T HE D UAL S PACES OF THE S ETS OF D IFFERENCE S EQUENCES OF O RDER m
(a) We put
(
M3β (p)
=
a∈ω:
∞
X
k=1
ak vk−1
9
k−m
X
j=1
k−j−1
N 1/pj uj converges and
m−1
)
∞
k−m+1
X
X k − j − 1
|bk |
.
N 1/pj uj < ∞, ∀u ∈ c+
0
m
−
2
j=1
k=1
β
β
Then [D∆m
v c0 (p)] =M3 (p).
(b) We put
(
n
k−m
X
X k − j − 1
γ
−1
N 1/pj uj | < ∞,
M4 (p) = a ∈ ω : sup |
ak v k
m−1
n
j=1
k=1
)
k−m+1
∞
X k − j − 1
X
N 1/pj uj < ∞, ∀u ∈ c+
.
|bk |
0
m
−
2
j=1
k=1
γ
γ
Then [D∆m
v c0 (p)] =M4 (p).
Proof. (a) and (b) can be proved in the same manner as Theorem 2.4, using Lemma 1.1(i) and
(ii).
Lemma 2.7.
η
m
η
i) [∆m
v `∞ (p)] = [D∆v `∞ (p)] ,
η
η
m
ii) [∆m
v c(p)] = [D∆v c(p)] ,
η
m
m
iii) [∆v c0 (p)] = [D∆v c0 (p)]η
for η = β or γ.
The proof is omitted.
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J. Inequal. Pure and Appl. Math., 7(3) Art. 101, 2006
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