Global Journal of Pure and Applied Mathematics.
ISSN 0973-1768 Volume 13, Number 8 (2017), pp. 4265–4273
© Research India Publications
http://www.ripublication.com/gjpam.htm
Some Lacunary Difference Sequence Spaces
Defined by a Modulus Function
Mushir A. Khan
Department of Mathematics,
Aligarh Muslim University,
Aligarh-202002 (India).
Abstract
In this paper, we introduce the sequence spaces [Nθ , f, p]1 (r , E), [Nθ , f, p]0 (r , E),
[Nθ , f, p]∞ (r , E) and Sθ (r , E), where E is any Banach space.We discuss some
toplogical properties and establish some inclusion relations between these spaces.
AMS subject classification: 40A05, 40C05, 46A45.
Keywords: Laucunary sequence, Difference sequence, Modulus function.
1.
Introduction
By a Lacunary sequence θ = (kr ); r = 0, 1, 2, . . ., where k0 = 0, we shall mean
an increasing sequence of non-negative integers with kr − kr−1 → ∞. The intervals
determined by θ will be denoted by Ir = (kr−1 , kr ], and we let hr = kr − kr−1 . The ratio
kr /kr−1 will be denoted by qr . The space of lacunary strongly convergent sequences Nθ
was defined by Freedman et al. [6] as follows:
| xk − L |= 0 for some L}
Nθ = {x = (xk ) : lim h−1
r
r→∞
k∈Ir
The space Nθ is a BK-space with the norm
⎛
xθ = sup ⎝h−1
r
r
k∈Ir
⎞
| xk |⎠ .
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Mushir A. Khan
Nθ0 denotes the subset of those sequences in Nθ for which L = 0. (Nθ0 , · θ ) is also
a BK-space. There is a strong connection [6] between Nθ and the space w of strongly
Cesaro summable sequences, which is defined by
−1
w = {x = (xk ) : lim n
n→∞
n
| xk − L |= 0 for some L}.
k=1
In the special case when θ = (2r ), we have Nθ = w.
The idea of difference sequence spaces was introduced by kizmaz [8]. In 1981,
Kizmaz [8] defined the sequence spaces
X() = {x = (xk ) : x ∈ X}
for X = l∞ , c and c0 , where x = (xk − xk+1 ). Then Et and Colak [ ] generalized the
above sequence spaces to the sequence spaces
X(r ) = {x = (xk ) : r x ∈ X}
I , 0 x = (xk ), x = (xk − xk+1 ), r x =
for X = l∞ , c and c0 , where r ∈ N
r
r
r
r
ν r
( xk − xk+1 ), and so xk =
(−1)
xk+ν .
ν
ν=0
Later on difference sequence spaces were studied by Malkowsky and Parashar [12],
Et and Basarir [3], Mursaleen, Mushir A. Khan and Qamaruddin [15] and many others.
We recall that a modulus f is a funtion from [0, ∞) to [0, ∞) such that
(i) f (x) = 0 if and only if x = 0,
(ii) f (x + y) ≤ f (x) + f (y) for x, y ≥ 0,
(iii) f is increasing
(iv) f is continuous from the right at 0.
It follows that f must be continuous everywhere on [0, ∞). A modulus may be
bounded or unbounded. Ruckle [16] and Maddox [11], used a modulus f to construct
some sequence spaces.
Subsequently modulus function has been discussed in [1], [13], [18] and others have
also used a modulus function to construct some sequence spaces.
The following inequality will be used throughout this paper.
| ak + bk |pk ≤ C{| ak |pk + | bk |pk },
where ak , bk ∈ C, 0 < pk ≤ sup pk = H , C = max(1, 2H −1 ) [10].
k
(1)
Lacunary Difference Sequence Spaces Defined by a Modulus Function
2.
4267
Main Results
In this section we introduce the sequence space [Nθ , f, p]1 (r , E), [Nθ , f, p]0 (r , E),
[Nθ , f, p]∞ (r , E) and Sθ (r , E). We give some topological properties and establish
some inclusion relations between these spaces.
Definition 2.1. Let E be a Banach space. We define w(E) to be the vector space of all
E-valued sequences that is w(E) = {x = (xk ) : xk ∈ E}. Let f be a modulus function
and p = (pk ) be any sequence of strictly positive real numbers. We define the following
sequence sets
[Nθ , f, p]1 (r , E) = {x ∈ w(E) : lim h−1
r
r
[f (r xk −L)]pk = 0, for some L},
k∈Ir
[Nθ , f, p]0 (r , E) = {x ∈ w(E) : lim h−1
r
r
[Nθ , f, p]∞ (r , E) = {x ∈ w(E) : sup h−1
r
r
[f (r xk )]pk = 0},
k∈Ir
[f (r xk )]pk < ∞}.
k∈Ir
If x ∈ [Nθ , f, p]1 (r , E) then we will write xk → L[Nθ , f, p]1 (r , E) and L will
be called Lacunary difference limit of x with respect to the modulus f .
Throughout this paper Z will denote any one of the notation 0, 1 or ∞.
I , we shall write [Nθ ]Z (r , E) and
In the case f (x) = x, pk = 1 for all k ∈ N
[Nθ , f ]Z (r , E) instead of [Nθ , f, p]Z (r , E) respectively.
Theorem 2.2. Let the sequence (pk ) be bounded. Then the sequence spaces
[Nθ , f, p]Z (r , E) are linear spaces.
Proof. We shall prove only for [Nθ , f, p]0 (r , E). The others can be treated similarly.
Let x, y ∈ [Nθ , f, p]0 (r , E) and α, β ∈ C. Then there exist positive numbers Mα and
Nβ such that | α |≤ Mα and | β |≤ Nβ . Since f is subadditive and r is linear
h−1
r
[f (r (αxk + βyk ))]pk
k∈Ir
≤ h−1
r
≤
[f (| α | r xk ) + f (| β | r yk )]pk
k∈Ir
C(Mα )H h−1
r
k∈Ir
[f (r xk )]pk + C(Nβ )H h−1
r
[f (r yk )]pk → 0
k∈Ir
as r → ∞. This proves that [Nθ , f, p]0 (r , E) is a linear space.
Theorem 2.3. Let f be a modulus function, then
[Nθ , f, p]0 (r , E) ⊂ [Nθ , f, p]1 (r , E) ⊂ [Nθ , f, p]∞ (r , E).
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Mushir A. Khan
Proof. The first inclusion is obvious. We establish the second inclusion. Let x ∈
[Nθ , f, p]1 (r , E). By definition of f we have
[f (r xk )]pk = h−1
[f (r xk − L + L)]pk
h−1
r
r
k∈Ir
k∈Ir
≤ Ch−1
r
[f (r xk − L)]pk + Ch−1
r
[f (L)]pk .
k∈Ir
k∈Ir
There exists a positive integer KL such that L ≤ KL . Hence we have
H
[f (r xk )]pk = Ch−1
[f (r xk − L)]pk + Ch−1
h−1
r
r [KL f (1)] hr .
r
k∈Ir
k∈Ir
Since x ∈ [Nθ , f, p]1 (r , E) we have x ∈ [Nθ , f, p]∞ (r , E) and this completes the
proof.
Theorem 2.4. [Nθ , f, p]0 (r , E) is a paranormed (need not total paranorm) space with
⎛
g (x) =
sup ⎝h−1
r
r
⎞1
M
pk ⎠
[f ( xk )]
r
k∈Ir
where M = max(1, sup pk ).
Proof. Clearly g (x) = g (−x). It is trivial that r xk = 0 for x = 0. Since f (o) = 0,
we get g (x) = 0 for x = 0. Since pk /M ≤ 1 and M ≥ 1, using the Minkowski’s
inequality and definition of f , for each r, we have g (x + y) ≤ g (x) + g (y). Finally,
to check the continuity of multiplication, let us take any complex number β. By definition
of f we have
⎛
⎞1
M
−1
r
p
[f ( (βxk ))] k ⎠
g (βx) = sup ⎝hr
r
k∈Ir
H
≤ KβM g (x)
where Kβ is a positive integer such that | β |< Kβ . Now let β → 0 for any fixed x with
g (x) = 0. By definition of f for | β |< 1, we have
[f (βr xk )]pk < for r > r0 ().
(2)
h−1
r
k∈Ir
Also, for 1 ≤ r ≤ r0 , taking β small enough, since f is continuous we have
[f (βr xk )]pk < .
h−1
r
k∈Ir
(3)
Lacunary Difference Sequence Spaces Defined by a Modulus Function
(2) and (3) together imply that g (βx) → 0 as β → 0.
4269
Theorem 2.5. If r ≥ 1, then the inclusion [Nθ , f ]Z (r−1 , E) ⊂ [Nθ , f ]Z (r , E) is
strict. In general [Nθ , f ]Z (i , E) ⊂ [Nθ , f ]Z (r , E) for all i = 1, 2, . . . , r − 1 and
the inclusion is strict.
Proof. We give the proof for Z = ∞ only. Other can be proved in a similar way for
Z = 0 and Z = 1. Let x ∈ [Nθ , f ]∞ (r−1 , E). Then we have
sup h−1
[f (r−1 xk )] < ∞
r
r
k∈Ir
By definition of f , we have
h−1
[f (r xk )] ≤ h−1
[f (r−1 xk )] + h−1
[f (r−1 xk+1 )] < ∞
r
r
r
k∈Ir
k∈Ir
k∈Ir
Thus [Nθ , f ]∞ (r−1 , E) ⊂ [Nθ , f ]∞ (r , E). Proceeding in this way one will
have [Nθ , f ]∞ (r , E) ⊂ [Nθ , f ]∞ (r , E) for i = 1, 2, · · · , r − 1. Let E = C,
I . Then the sequence x = (k r ), for example, belongs
and hr = r for each r ∈ N
r
to [Nθ , f ]∞ ( , E), but does not belong to [Nθ , f ]∞ (r−1 , E) forf (x) = x. (If
r −1
x = (k r ), then r xk = (−1)r r! and r−1 xk = (−1)r+1 r! k +
for all k ∈ N
I ).
2
The proof of the following result is a routine work.
Proposition 2.6. [Nθ , f, p]1 (r−1 , E) ⊂ [Nθ , f, p]0 (r , E).
Theorem 2.7. Let f, f1 , f2 be modulus functions. Then we have
(i) [Nθ , f1 , p]Z (r , E) ⊂ [Nθ , f of1 , p]Z (r , E),
(ii) [Nθ , f1 , p]Z (r , E) ∩ [Nθ , f2 , p]Z (r , E) ⊂ [Nθ , f1 + f2 , p]Z (r , E).
Proof. (i). We shall only prove (i). Let > 0 and choose δ with 0 < δ < 1 such that
f (t) < for 0 ≤ t ≤ δ. Write yk = f1 (r xk ) and consider
[f (yk )]pk =
[f (yk )]pk +
[f (yk )]pk
1
k∈Ir
2
where the first summation is over yk ≤ δ and second summation is over yk > δ. Since
f is continuous, we have
[f (yk )]pk < hr H
(4)
1
and for yk > δ, we use the fact that
yk <
yk
yk
≤ 1+ .
δ
δ
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Mushir A. Khan
By the definition of f we have for yk > δ,
f (yk ) < 2f (1)
Hence
h−1
r
yk
.
δ
[f (yk )]pk ≤ max(1, (2f (1)δ −1 )H )h−1
yk .
r
2
(5)
k∈Ir
from (4) and (5), we obtain [Nθ , f, p]0 (r ) ⊂ [Nθ , f of1 , p]0 (r ).
The proof of (ii) follows from the following inequality
[(f1 + f2 )(r xk )]pk ≤ C[f1 (r xk )]pk + C[f2 (r xk )]pk .
The following result is a consequence of Theorem 2.7 (i).
Proposition 2.8. Let f be a modulus function. Then
[Nθ , p]Z (r , E) ⊂ [Nθ , f, p]Z (r , E).
3.
Statistical Convergence
The idea of statistical convergence was introduced by Fast [5] and studied by various
authors ([2],[7],[9],[14] and [17]).
In this section we give some inclusion relations between Sθ (r , E) and
[Nθ , f, p](r , E).
Definition 3.1. A sequence x = (xk ) is said to be Sθ (r , E)-statistically convergent to
the number L if for every > 0,
{k ∈ Ir : r xk − L ≥ } = 0.
lim h−1
r
r
In this case we write Sθ (r , E) − lim x = L or xk → LSθ (r , E). In the case
hr = r and L = 0 we shall write S(r , E) and Sθ0 (r , E) instead of Sθ (r , E).
Theorem 3.2.
(i) If xk → L[Nθ ]1 (r , E) then xk → LSθ (r , E),
(ii) If x ∈ ∞ (r , E) and xk → LSθ (r , E), then xk → L[Nθ ]1 (r , E),
(iii) Sθ (r , E) ∩ ∞ (r , E) = [Nθ ]1 (r , E) ∩ ∞ (r , E).
where
∞ (r , E) = {x ∈ w(E) : sup r xk < ∞}.
k
Lacunary Difference Sequence Spaces Defined by a Modulus Function
4271
Proof.
(i) Let > 0 and xk → L[Nθ ]1 (r , E). Then we have
r xk − L ≥ {k ∈ Ir : r xk − L ≥ }
k∈Ir
Hence xk → LSθ (r , E).
(ii) Suppose that xk → LSθ (r , E) and x ∈ ∞ (r , E) say r xk −L ≤ M. Given
> 0, we have
h−1
r
r xk −L = h−1
r
k∈Ir
r xk −L+h−1
r
k∈Ir
r xk −L
k∈Ir
r xk −L≥
r xk −L<
r
≤ Mh−1
r {k ∈ Ir : xk − L ≥ } + Hence x is Sθ (r , E)-statistically convergent to the number L.
(iii) This follows from (i) and (ii).
Theorem 3.3. If lim inf
hr
> 0, then S(r , E) ⊆ Sθ (r , E).
r
Proof. For given > 0, we get
{k ≤ r : r xk − L ≥ } ⊃ {k ∈ Ir : r xk − L ≥ }.
Hence
1 1
{k ≤ r : r xk − L ≥ } ≥ {k ≤ r : r xk − L ≥ }
r
r
≥
hr −1
hr |{k ∈ Ir : r xk − L ≥ }|
r
Hence x ∈ Sθ (r , E).
Theorem 3.4. Let f be a modulus function and sup pk = H . Then [Nθ , f, p]1 (r , E) ⊂
Sθ (r , E).
k
Proof. Let x ∈ [Nθ , f, p]1 (r , E) and > 0 be given. Let
1
denote the sum over
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Mushir A. Khan
k ≤ r such that r xk − L < . Then
r
pk
−1
r
pk
−1
h−1
[f
(
x
−
L)]
=
h
[f
(
x
−
L)]
+
h
[f (r xk − L)]pk
k
k
r
r
r
k∈Ir
1
2
[f (r xk − L)]pk
≥ h−1
r
1
[f ()]pk
≥
h−1
r
≥
h−1
r
≥
1
−1 hr {k
1
min([f ()]inf pk , [f ()]H )
∈ Ir : r xk − L
≥ }| min([f ()]inf pk , [f ()]H ).
Hence x ∈ Sθ (r , E).
Theorem 3.5. Let f be bounded and 0 < m = inf pk ≤ pk ≤ sup pk = H < ∞. Then
k
k
Sθ ( , E) ⊂ [Nθ , f, p]1 ( , E).
r
r
Proof. Suppose that f is bounded. Let > 0 be given and
and
1
be in previous
2
theorem. Since f is bounded there exists an integer K such that f (x) < K, for all
x ≥ 0. Then
pk
r
pk
−1
r
−1
+
h
[f (r xk − L)]pk
[f
(|
x
−
L
|)]
[f
(
x
−
L)]
=
h
h−1
k
k
r
r
r
1
k∈Ir
≤
h−1
r
2
m
max(K , K
H
) + h−1
r
1
[f ()]pk
2
≤ max(K , K
m
H
)h−1
r
m
| {k ∈ Ir : r xk − L
≥ } | + max(f () , f ()H )
Hence x ∈ [Nθ , f, p]1 (r , E).
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