A new type of difference sequence spaces
Vakeel A. Khan
Abstract. In this article we introduce a new sequence space denoted by
m(4uv , φ, p). We give some topological properties and inclusion relations
on this space. The results herein proved are analogous to those from [1].
M.S.C. 2000: 40C05, 46A45.
Key words: Difference sequence; solid spaces; symmetric space.
1. Introduction
Let `0 be the set of all complex sequences and l∞ , c and c0 be the sets of all
bounded, convergent and null sequences x = (xk ) with complex terms, respectively,
normed by
||x||∞ = sup |xk |, where k ∈ N = {1, 2, · · · }.
k
A sequence space X with linear topology is called a K-space if each of the maps
Pk : X → C defined by Pk (x) = xk is continuous for k = 1, 2, · · · . A K- space X is
called a BK-space provided X is a Banach space.
The idea of difference sequence space was introduced by Kizmaz [12]. In 1981,
Kizmaz [12] defined the sequence spaces:
l∞ (4) = {x = {xk } ∈ `0 : (4xk ) ∈ l∞ },
c(4) = {x = {xk } ∈ `0 : (4xk ) ∈ c},
and
c0 (4) = {x = {xk } ∈ `0 : (4xk ) ∈ c0 },
where 4x = (xk − xk+1 ). These are Banach spaces with the norm
||x||4 = |x1 | + ||4x||∞ .
Et and Colak [5] generalized the above sequence spaces to the sequence spaces
X(4r ) = {x = {xk } ∈ `0 : 4r xk ∈ X},
for X = l∞ , c and c0 , where r ∈ N,
40 x = (xk ), 4x = (xk − xk+1 ), 4r x = (4r xk − 4r xk+1 )
Applied Sciences, Vol.12, 2010, pp.
102-108.
c Balkan Society of Geometers, Geometry Balkan Press 2010.
°
A new type of difference sequence spaces
and so that
r
4 xk =
r
X
i=0
103
· ¸
r
(−1)
xk+i .
i
i
Difference sequence spaces have been studied by Colak and Et [2], Et [4], Et and Esi
[6], Vakeel A. Khan [8,9,10,11] and many others.
Let X, Y ⊂ `0 . Then we shall write
\
x−1 ∗ Y = {a ∈ `0 : ax ∈ Y for all x ∈ X}.
M (X, Y ) =
x∈X
The set
X α = M (X, l1 )
is called Köthe - Toeplitz dual space or α - dual of X(see [16]).
Let X be a sequence space. Then X is called
(i) solid (or normal), if (αk xk ) ∈ X, whenever (xk ) ∈ X for all sequences of scalars
(αk ) with |αk | ≤ 1 for all k ∈ N.
(ii) symmetric, if (xk ) ∈ X implies (xπ(k) ) ∈ X, where π(k) is a permutation of N.
(iii)perfect, if X = X αα .
(iv) sequence algebra, if x.y ∈ X, whenever x, y ∈ X.
It is well known that if X is perfect then X is normal (see [7]).
Let C denote the space whose elements are finite sets of distinct positive integers.
Given any element σ of C, we denote by c(σ) the sequence {cn (σ)} which is such that
cn (σ) = 1 if n ∈ σ, cn (σ) = 0 otherwise. Further,
(
)
∞
X
cn (σ) ≤ s (cf [13]),
Cs = σ ∈ C :
n=1
the set of those σ whose support has cardinality at most s, and
½
µ ¶
¾
φk
0
Φ = φ = {φk } ∈ ` : φ1 > 0, ∆φk ≥ 0 and ∆
≤ 0 (k = 1, 2, . . .) ,
k
where ∆φk = φk − φk−1 ; and `0 is the set of all real sequences.
For φ ∈ Φ, we define the following sequence space, introduce in [14],
!
)
(
Ã
1 X
0
| xk | < ∞ .
m(φ) = x = {xk } ∈ ` : sup sup
φs
s≥1 σ∈Cs
k∈σ
Recently the space m(φ) was extended to m(φ, p) by Tripathy and Sen [15] as follows:
(
)
1 X
0
p
m(φ, p) := x = {xk } ∈ ` : sup sup
| xk | < ∞ .
s≥1 σ∈Cs φs
k∈σ
It is easy to see that:
Ã
k x km(φ,p) = sup sup
s≥1 σ∈Cs
1 X
| xk |p
φs
k∈σ
! p1
.
104
Vakeel A. Khan
Remark 1. The space m(φ, p) is a Banach space with the norm
Ã
k x km(φ,p) = sup sup
s≥1 σ∈Cs
1 X
| xk |p
φs
! p1
.
k∈σ
Remark 2. As in [14], we have
(i) If φn = 1 for all n ∈ N then m(φ, p) = lp . Moreover
lp ⊆ m(φ, p) ⊆ l∞ .
(ii) If p = 1, then m(φ, p) = m(φ). Also
m(φ) ⊆ m(φ, p).
³ ´
φs
(iii) m(φ, p) ⊆ m(ψ, p) if and only if sups≥1 ψ
< ∞.
s
2. Main Results
Let u be a fixed positive integer and v = (vk ) be any fixed sequence of non zero
complex numbers (see [3]). Now we define the sequence space m(4uv , φ, p) as follows:
(
)
1 X
0
u
u
p
m(4v , φ, p) := x = {xk } ∈ ` : sup sup
(|4v xk | ) < ∞,
0≤p<∞ .
s≥1 σ∈Cs φs
k∈σ
where
40v xk = (vk xk ),
4v xk = (vk xk − vk+1 xk+1 ),
4uv xk = (4u−1
xk − 4u−1
xk+1 )
v
v
such that
4uv xk
· ¸
u
X
i u
=
(−1)
vk+i xk+i .
i
i=0
It is clear that if u = 0, v = (1, 1, 1, · · · ) and p = 1, we have m(φ), which was defined
by Sargent [13].
Theorem 2.1. The sequence space m(4uv , φ, p) is a Banach space for φ ∈ Φ
normed by
(2.1.1)
||x||41 =
u
X
|xi | + sup sup
i=1
s≥1 σ∈Cs
1 X
1/p
((|4uv xk |p )) ,
φs
1 ≤ p < ∞,
k∈σ
and a complete p-normed space by p-norm
(2.1.2)
||x||42 =
u
X
i=1
|xi |p + sup sup
s≥1 σ∈Cs
1 X
(|4uv xk |p ) ,
φs
0 < p < 1.
k∈σ
Proof. It is clear that m(4uv , φ, p) is a normed linear space normed by (2.1.1) for
1 ≤ p < ∞ and a p - normed space by p - norm (2.1.2) for 0 < p < 1. We need to
A new type of difference sequence spaces
105
show that m(4uv , φ, p) is complete. Let {x(l) } be a Cauchy sequence in m(4uv , φ, p)
where xl = (xlk )k = (xl1 , xl2 , · · · ) ∈ m(4uv , φ, p) for each l ∈ N. Then for given ² > 0
there exists n0 ∈ N such that
l
t
||x −x ||41 =
u
X
|xli −xti |+sup sup
s≥1 σ∈Cs
i=1
¢¢1/p
1 X ¡¡ u l
|4v (xk − xtk )|p
< ², for all l, t > n0 .
φs
k∈σ
Now we obtain
|xlk − xtk | → 0, as
l, t → ∞, for each k ∈ N.
Therefore (xlk )l = (x1k , x2k , · · · ) is a Cauchy sequence in C for each k. Since C is
complete, it is convergent
lim xlk = xk
(say) for each k ∈ N.
l
Taking limit as t → ∞ in (2.1.3), we get
u
X
|xli − xi | + sup sup
s≥1 σ∈Cs
i=1
¢1/p
1 X¡ u l
|4v (xk − xk )|p
< ²,
φs
for all n > n0 .
k∈σ
Hence (xlk − xi ) ∈ m(4uv , φ, p). We know that m(4uv , φ, p) is a linear space and (xlk ),
(xlk − xk ) are in m(4uv , φ, p), it follows that
(xk ) = (xlk ) − (xlk − xk ) ∈ m(4uv , φ, p).
Hence m(4uv , φ, p) is complete. Similarly, we can show that m(4uv , φ, p) is complete
space p-normed by (2.1.1) for 0 < p < 1.
¤
Theorem 2.2. For any φ ∈ Φ the space m(4uv , φ, p) is a K - space.
The proof is straightforward.
Theorem 2.3. m(4uv , φ) ⊂ m(4uv , φ, p), for any φ ∈ Φ.
Proof. Let x ∈ m(4uv , φ). Then there exist a positive number K such that
X
|4uv xk |p ≤ Kφs ,
σ ∈ φs
for each fixed s.
k∈σ
Hence
X
|4uv xk |p < Kφs ,
σ ∈ φs
for each p ¿ 0 and σ ∈ φs .
k∈σ
Thus x ∈ m(4uv , φ, p).
¤
Theorem 2.4. For any two sequences (φs ) and (ψs ) of real numbers, we have
m(4uv , φ, p) ⊂ m(4uv , ψ, p) if and only if
µ ¶
φs
sup
< ∞.
ψs
s≥1
106
Vakeel A. Khan
P
Let x ∈ m(4uv , φ, p). Then sups≥1 supσ∈Cs φ1s k∈σ |4uv xk |p < ∞.
³ ´
φs
Suppose that sups≥1 ψ
< ∞. Then φs ≤ Kψs and so that ψ1s ≤ φKs for some
s
positive number K and for all s. Therefore we have
Proof.
1 X u p
K X u p
|4v xk | ≤
|4v xk |
ψs
φs
k∈σ
for each s.
k∈σ
Now
1 X u p
1 X u p
|4v xk | ≤ K sup sup
|4v xk | .
s≥1 σ∈Cs ψs
s≥1 σ∈Cs φs
k∈σ
k∈σ
P
Hence sups≥1 supσ∈Cs ψ1s k∈σ |4uv xk |p < ∞. Therefore x ∈ m(4uv , ψ, p).
³ ´
φs
Conversely, let m(4uv , φ, p) ⊆ m(4uv , ψ, p) and suppose that sup ψ
= ∞. Then
s
s≥1
³ ´
φ
there exists aN increasing sequence (si ) of natural numbers such that lim ψssi = ∞.
sup sup
i
Now for every B ∈ R+ (the set of positive real numbers), there exists i0 ∈ N such
φ
that ψssi > B for all si ≥ i0 . Hence ψ1s > ψBs and
i
i
i
B X u p
1 X u p
|4v xk | >
|4v xk |
ψsi
ψsi
k∈σ
k∈σ
for all si ≥ i0 . Now taking supremum over si ≥ i0 and σ ∈ Cs we get
(2.2.1)
sup sup
si ≥i0 σ∈Cs
1 X u p
1 X u p
|4v xk | > B sup sup
|4v xk | .
ψsi
si ≥i0 σ∈Cs φsi
k∈σ
k∈σ
Since (2.2.1) holds for all B ∈ R+ (we may take B sufficiently large ) we have
sup sup
si ≥i0 σ∈Cs
1 X u p
|4v xk | = ∞
ψsi
k∈σ
when x ∈ m(4uv , φ, p) with
0 < sup sup
si ≥i0 σ∈Cs
1 X u p
|4v xk | < ∞.
φsi
k∈σ
u
u
u
Therefore
³ ´ x 6∈ m(4v , ψ, p). This contradicts to m(4v , φ, p) ⊆ m(4v , ψ, p), whence
φs
¤
sup ψs < ∞.
s≥1
From Theorem 2.4, we get the following result.
Corollary 2.1. m(4uv , φ, p) = m(4uv , ψ, p) if and only if
µ ¶
µ ¶
φs
φs
0 < inf
≤ sup
< ∞.
s≥1 ψs
ψs
s≥1
Theorem 2.5. m(4u−1
, φ, p) ⊂ m(4uv , φ, p) and the inclusion is strict.
v
A new type of difference sequence spaces
107
Proof. The proof follows from the following inequality and Minkowski’s inequality
|4uv x| = |4u−1
xk − 4u−1
xk+1 | ≤ |4u−1
xk | + |4u−1
xk+1 |.
v
v
v
v
To show that the inclusion is strict consider the following example.
Example 2.1. Let φn = 1 for all n ∈ N, x = (k u−1 ) and v = (1, 1, 1, · · · ), then
).
x ∈ lp (4uv ) \ lp (4u−1
v
Theorem 2.6. The sequence space m(4uv , φ, p) is not sequence algebra, is not
solid and is not symmetric, for u ≥ 1.
Proof. For the proof of this theorem, consider the following examples:
Example 2.2. Let x = (k u−1 ), y = (k u−1 ) and v = (1, 1, 1, · · · ). Then x, y ∈
m(4uv , φ, p), but x.y ∈
/ m(4uv , φ, p). Hence m(4uv , φ, p) is not sequence algebra.
Example 2.3. Let x = (k u−1 ), v = (1, 1, 1, · · · ) and αk = (−1)k . Then x =
u−1
(k
) ∈ m(4uv , φ, p), but
(αk xk ) ∈
/ m(4uv , φ, p) for α = (αk ) = (−1)k .
Hence m(4uv , φ, p) is not solid.
Example 2.4. Let x = (k u−1 ) and v = (1, 1, 1, · · · ). Let (yk ) be a arrangement
of (xk ) which is defined as follows :
yk = {x1 , x2 , x4 , x3 , x9 , x5 , x16 , x6 , x25 , x7 , x36 , x8 , x49 , x10 , · · · }.
Then y ∈
/ m(4uv , φ, p). Hence m(4uv , φ, p) is not symmetric.
The following result is a consequence of Theorem 2.6.
Corollary 2.2. The sequence space m(4uv , φ, p) is not perfect.
Theorem 2.7. lp (4uv ) ⊆ m(4uv , φ, p) ⊆ l∞ (4uv ).
Proof. Since m(4uv , φ, p) = lp (4uv ) for φn = 1, for all nN, then
lp (4uv ) ⊆ m(4uv , φ, p).
Now suppose that x ∈ m(4uv , φ, p). Then we have
sup sup
s≥1 σ∈Cs
1 X u p
|4v xk | < ∞,
φs
k∈σ
and hence
|4uv xk | < Kφ1 , for all n ∈ N and for some positive integer K.
Thus x ∈ l∞ (4uv ). This completes the proof of Theorem.
Corollary 2.3. If 0 < p < q, then m(4uv , φ, p) ⊆ m(4uv , φ, q).
Proof. For the proof of this theorem follows from the following inequality
à n
! q1
à n
! p1
X
X
q
p
|xk |
≤
|xk |
, (0 < p < q).
k=1
k=1
¤
108
Vakeel A. Khan
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Author’s address:
Vakeel A. Khan
Department of Mathematics, A.M.U. Aligarh, India.
E-mail: vakhan@math.com