Gen. Math. Notes, Vol. 6, No. 2, October 2011, pp.33-44
c
ISSN 2219-7184; Copyright ICSRS
Publication, 2011
www.i-csrs.org
Available free online at http://www.geman.in
On Lacunary Strongly Convergent
Difference Sequence Spaces
Defined by a Sequence of ϕ-Functions
Metin Başarır1 and Selma Altundağ2
1,2
Department of Mathematics, Sakarya University
54187, Sakarya/TURKEY
1
E-mail: basarir@sakarya.edu.tr
2
E-mail: scaylan@sakarya.edu.tr
(Received: 10-12-10/ Accepted: 9-10-11)
Abstract
In this paper, we introduce the new sequence spaces with lacunary strong
convergence using by a sequence of modulus functions and a sequence of ϕfunctions. We also study some connections between lacunary (A, ϕk , ∆m
u) m
statistically convergence and lacunary strong (A, ϕk , ∆u ) - convergence .
Keywords: Difference sequence, modulus function, ϕ- function, lacunary
sequence, statistical convergence.
1
Introduction
Let w be the set of all sequences of real or complex numbers and l∞ , c and c0
be, respectively, the Banach spaces of bounded, convergent and null sequences
x = (xk ) with the usual norm kxk = sup |xk | .
k
By a lacunary θ = (kr ); r = 0, 1, 2, . . .where k0 = 0, we shall mean an
increasing sequence of non-negative integers with kr − kr−1 → ∞ as r → ∞
. The intervals determined by θ will be denoted by Ir = (kr−1 , kr ] and hr =
kr
will be denoted by qr . In [1], the space of lacunary
kr − kr−1 .The ratio kr−1
34
Metin Başarır et al.
strongly convergent sequences Nθ was defined as follows:


Nθ = x = (xi ) : r→∞
lim h−1
r
X
i∈Ir


|xi − s| = 0 for some s .
A modulus function f is a function from acting [0, ∞) to [0, ∞) such that
(i) f (x) = 0 if and only if x = 0,
(ii) f (x + y) ≤ f (x) + f (y) for all x, y ≥ 0,
(iii) f increasing,
(iv) f is continuous from at the right zero.
Since |f (x) − f (y)| ≤ f (|x − y|), it follows from condition (iv) that f is
continuous on [0, ∞). Furthermore, we have f (nx) ≤ nf (x) for all n ∈ N ,
from condition (ii) and so
x
1
f (x) = f (nx ) ≤ nf ( ).
n
n
Hence, for all n ∈ N
1
x
f (x) ≤ f ( ).
n
n
A modulus may be bounded or unbounded. For example, f (x) = xp , for
x
is bounded. Ruckle [9] and Maddox
0 < p ≤ 1 is unbounded, but f (x) = 1+x
[10], used a modulus f to construct some sequence spaces.
Furthermore, modulus function has been discussed in [5], [11], [12], [13]
and [14] and many others.
The difference sequence space X(∆) was first introduced by Kızmaz [2] as
follows:
X(∆) = {x = (xk ) ∈ w : (∆xk ) ∈ X}
for X = l∞ , c and c; where ∆xk = xk − xk+1 for all k ∈ N.
The notion of difference sequence spaces was further generalized by Et and
Colak [3] as follows:
X(∆m ) = {x = (xk ) ∈ w : (∆m xk ) ∈ X}
for X = l∞ , c and c; where ∆m xk = ∆m−1 xk − ∆m−1 xk+1 and ∆0 xk = xk for
all k ∈ N . Taking X = l∞ (p), c(p) and c0 (p), these sequence spaces has been
generalized by Et and Başarır [4].
The generalized difference has the following binomial representation:
m
∆ xk =
m
X
v=0
for all k ∈ N .
(−1)
v
m
v
!
xk+v
On Lacunary Strongly Convergent...
35
Subsequently, difference sequence spaces have been discussed by several
authors [18], [14], [6] and [7].
By a ϕ-function we understood a continuous non-decreasing function ϕ(v)
defined for v ≥ 0 and such that ϕ(0) = 0, ϕ(v) > 0 for v > 0 and ϕ(v) → ∞
as v → ∞.
In [15], [16], [17] and [19]; some sequence spaces was studied using by ϕfunction.
Let ϕ = (ϕk ) and ψ = (ψk ) be sequences of ϕ-functions. A sequence of
ϕ-functions ϕ is called non weaker than a sequence of ϕ-function ψ and we
write ψ ≺ ϕ (or ψk ≺ ϕk for all k) if there are constants c, b, n, l > 0 such that
cψk (lv) ≺ bϕk (nv) (for all, large or small v, respectively).
Two sequences of ϕ-functions ϕ and ψ are called equivalent and we write
ϕ ∼ ψ (or ψk ≺ ϕk for all k) if there are positive constants b1 , b2 , c, k1 , k2 , l such
that b1 ϕk (k1 v) ≤ cψk (lv) ≤ b2 ϕk (k2 v) (for all, large or small v, respectively).
A sequence of ϕ-functions ϕ is said to satisfy the ∆2 -condition (for all,
large or small v, respectively) if for some constant l > 1 there is satisfied
the inequality ϕk (2v) ≤ lϕk (v) for all k. For a ϕ-function satisfying the
∆2 -condition, there is L > 0 such that
ϕk (cv) ≤ Lϕk (v)
(1)
for v large enough. Indeed, for every c > 0 there is an integer s such that
c ≤ 2s and
ϕk (cv) ≤ ϕk (2s v) ≤ ls ϕk (v)
(2)
for v large enough.
Let A = (ank ) be an infinite matrix such that;
a) A is non-negative, i.e. ank ≥ 0 for n, k = 1, 2, ... ,
b) for an arbitrary positive integer n (or k) there exists a positive integer
k0 (or n0 ) such that ank 6= 0 (or an0 k 6= 0), respectively,
c) there exists lim
ank = 0 for k = 1, 2, . . . , ,
n
d) sup
n
∞
P
ank < ∞,
k=1
e) sup ank → 0 as k → ∞.
n
In the present paper, we introduce and study some properties of the following difference sequence space that is defined by using a sequence of ϕ-functions
and a sequence of modulus functions.
2
Main Results
Let θ = (kr ) be a lacunary sequence, ϕ = (ϕk ) and f = (fn ) be given a
sequence of ϕ-functions and a sequence of modulus functions, respectively, m
36
Metin Başarır et al.
be a positive integer and u = (uk ) be any sequence such that uk 6= 0 for all k.
Moreover, let a matrix A = (ank ) be given in the above. Then we define,



∞

X
1 X
m
Vθ0 ((A, ϕk , ∆m
)
,
f
)
=
x
=
(x
)
∈
w
:
lim
f
a
ϕ
(|∆
x
|)
=
0
n
k
n
nk k
u
u k

r h

r n∈Ir
k=1
!
where ∆m
uk ∆m xk = (uk ∆m−1 xk − uk+1 ∆m−1 xk+1 ) such that ∆m
u xk = !
u xk =
m
P
m
(−1)n
uk+n xk+n , ∆0u xk = (uk xk ) and ∆u xk = (uk xk − uk+1 xk+1 ).
n
n=0
Throughout this paper, the sequence of modulus functions f = (fn ) satisfy
the condition lim+ sup fn (v) = 0.
v→0
n
If x ∈ Vθ0 ((A, ϕk , ∆m
u ) , fn ) then the sequence x is said to be lacunary
)convergent
to zero with respect to a sequence of modulus
strongly (A, ϕk , ∆m
u
f.
If we take θ = (2r ) then we have
k
∞
X
1X
fn
ank ϕk (|∆m
= x ∈ w : lim
u xk |) =0 .
k k
n=1
k=1
!
(
V
0
((A, ϕk , ∆m
u ) , fn )
)
When ϕk (x) = x for all x and k, uk = 1 for all k, we obtain
Vθ0 ((A, ∆m ) , fn ) =


x ∈ w : lim
r

1
hr
X
n∈Ir
fn
∞
X
!


ank (|∆m xk |) = 0 .

k=1
If fn (x) = x for all x and n, uk = 1 for all k, we write



∞

1 X X
Vθ0 (A, ϕk , ∆m ) = x ∈ w : lim
ank ϕk (|∆m xk |) = 0 .

r h

r n∈Ir k=1
!
When A = I and uk = 1 for all k, we get the following sequence space,




1 X
m
Vθ0 ((I, ϕk , ∆m ) , fn ) = x ∈ w : lim
f
(ϕ
(|∆
x
|))
=
0
.
n
n
n
r h

r n∈Ir
If we take A = I, ϕk (x) = x for all x and k and uk = 1 for all k then we
have




1 X
Vθ0 ((I, ∆m ) , fn ) = x ∈ w : lim
fn (|∆m xn |) = 0 .

r h

r n∈Ir
37
On Lacunary Strongly Convergent...
If we take A = I, ϕk (x) = x for all x and k and uk = 1 for all k, fn (x) =
f (x) for all x and n
Vθ0 ((I, ∆m ) , f ) =


x ∈ w : lim

r
1
hr
X


f (|∆m xn |) = 0 .

n∈Ir
If we take A = I, ϕk (x) = x for all x and k, fn (x) = x for all x and n and
uk = 1 for all k then we have




1 X
m
(|∆
x
|)
=
0
.
Vθ0 (I, ∆m ) = x ∈ w : lim
n

r h
r n∈Ir
If we define the matrix A = (ank ) as follows:
1
for n ≥ k and ank = 0 for n < k
n
then we have the sequence space,
ank =



n

1X
1 X
0
m
m
Vθ ((C, ϕk , ∆u ) , fn ) = x ∈ w : lim
f
ϕ
n
k (|∆u xk |) = 0 .
r h

n k=1
r n∈Ir
!
Now we have,
Theorem 2.1 Let us suppose that ϕ = (ϕk ) and ψ = (ψk ) be two sequences
of ϕ-functions and ψ = (ψk (v)) satisfies the ∆2 -condition for large v.
0
m
(i) If ψ ≺ ϕ then Vθ0 ((A, ϕk , ∆m
u ) , fn ) ⊂ Vθ ((A, ψk , ∆u ) , fn ).
(ii) If two sequences of ϕ-functions (ϕk (v)) and (ψk (v))
are
equivalent for large v and they satisfy the ∆2 -condition for large v then
0
m
Vθ0 ((A, ϕk , ∆m
u ) , fn ) = Vθ ((A, ψk , ∆u ) , fn ).
Proof. (i) Let x = (xk ) ∈ Vθ0 ((A, ϕk , ∆m
u ) , fn ). Then
∞
X
1 X
lim
fn
ank ϕk (|∆m
u xk |) = 0.
r h
r n∈Ir
k=1
!
By assumption, ψ ≺ ϕ, we have
ψk (|xk |) ≤ bϕk (c |xk |)
(3)
0
00
for b,c >0, all k, and |xk | > v0 . Let us denotes x = x +x , where for all m,
0
0
m 0
m
m
m 0
x = ∆m
u xk and ∆u xk = ∆u xk for |∆u xk | < v0 and ∆u xk = 0 remaining
38
Metin Başarır et al.
0
values of k. It is easy to see that x ∈ Vθ0 ((A, ψk , ∆m
u ) , fn ). Furthermore, by
the assumptions and the inequality (3) we get
∞
∞
X
X
1 X
1 X
00
00
|)
≤
x
fn
ank ψk (|∆m
f
b
ank ϕk (c |∆m
n
u k
u xk |)
hr n∈Ir
h
r
n∈Ir
k=1
k=1
!
∞
X
1 X
00
≤
fn bL
ank ψk (|∆m
u xk |)
hr n∈Ir
k=1
!
!
∞
X
K X
00
fn
ank ϕk (|∆m
≤
u xk |)
hr n∈Ir
k=1
!
where the constants K and L are connected with properties of f and ϕ
functions. We recall that a ϕ−function satisfying the ∆2 −condition implies
(1) and (2).
00 00
Finally, we obtain x = xk ∈ Vθ0 ((A, ψk , ∆m
u ) , fn ) and in consequence
m
0
x ∈ Vθ ((A, ψk , ∆u ) , fn ).
m
0
(ii) The identity Vθ0 ((A, ϕk , ∆m
u ) , fn ) = Vθ ((A, ψk , ∆u ) , fn ) is proved by
using the same argument.
Theorem 2.2 Let the sequence ϕ = (ϕk (v)) of ϕ-functions satisfies the
∆2 -condition for all k and for large v then Vθ0 ((A, ϕk , ∆m
u ) , fn ) is linear space.
Proof. Firstly we prove that if x = (xk ) ∈ Vθ0 ((A, ϕk , ∆m
u ) , fn ) and α is
0
m
an arbitrary number then αx ∈ Vθ ((A, ϕk , ∆u ) , fn ). Let us remark that for
0 < α < 1 we get
∞
∞
X
X
1 X
1 X
fn
ank ϕk (|∆m
αx
|)
≤
f
ank ϕk (|∆m
k
n
u
u xk |) .
hr n∈Ir
h
r
n∈Ir
k=1
k=1
!
!
Moreover, if α > 1 then we may find a positive number s such that α < 2s
and we obtain
∞
∞
X
X
1 X
1 X
s
fn
ank ϕk (|∆m
αx
|)
≤
f
d
ank ϕk (|∆m
k
n
u
u xk |)
hr n∈Ir
h
r
n∈Ir
k=1
k=1
!
!
∞
X
K X
≤
fn
ank ϕk (|∆m
u xk |)
hr n∈Ir
k=1
!
where d and K are constants connected with the properties of ϕ and f functions. We recall that a ϕ−function satisfying the ∆2 −condition implies (1)
and (2). Hence we obtain αx ∈ Vθ0 ((A, ϕk , ∆m
u ) , fn ).
39
On Lacunary Strongly Convergent...
Secondly, let x, y ∈ Vθ0 ((A, ϕk , ∆m
u ) , fn ) and α, β arbitrary numbers. We
0
will show that αx + βy ∈ Vθ ((A, ϕk , ∆m
u ) , fn ).
∞
∞
X
X
1 X
K1 X
fn
ank ϕk (|∆m
(αx
+
βy
)|)
≤
f
ank ϕk (|∆m
k
k
n
u
u xk |)
hr n∈Ir
h
r n∈Ir
k=1
k=1
!
!
∞
X
K2 X
+
fn
ank ϕk (|∆m
u yk |)
hr n∈Ir
k=1
!
where the constants K1 and K2 are defined as above. In consequence,
αx + βy ∈ Vθ0 ((A, ϕk , ∆m
u ) , fn ) .
Now, we give the following Proposition that is necessary for proof
of the Theorem 2.4.
Proposition 2.3 ([5]) Let f be a modulus and let 0 < δ < 1. Then for each
v ≥ δ we have f (v) ≤ 2f (1)δ −1 v.
Theorem 2.4 Let ϕ = (ϕk ) and f = (fn ) be given a sequence of ϕfunctions and a sequence of modulus functions, respectively and sup fn (1) < ∞.
n
0
m
Then Vθ0 (A, ϕk , ∆m
u ) ⊂ Vθ ((A, ϕk , ∆u ) , fn ).
Proof. Let x ∈ Vθ0 (A, ϕk , ∆m
u ) and put sup fn (1) = M . For a given ε > 0
n
we choose 0 < δ < 1 such that fn (x) < ε for every x ∈ [0, δ] and for all n. We
can write
∞
X
1 X
fn
ank ϕk (|∆m
u xk |) = S1 + S2
hr n∈Ir
k=1
!
where
∞
X
1 X
S1 =
fn
ank ϕk (|∆m
u xk |)
hr n∈Ir
k=1
!
and this sum is taken over
∞
X
ank ϕk (|∆m
u xk |) ≤ δ
k=1
and
∞
X
1 X
fn
S2 =
ank ϕk (|∆m
u xk |)
hr n∈Ir
k=1
!
and this sum is taken over
40
Metin Başarır et al.
∞
X
ank ϕk (|∆m
u xk |) > δ.
k=1
By the definition of the modulus f we have
S1 ≤
1 X
1
fn (δ) <
(hr ε) = ε
hr n∈Ir
hr
and moreover
∞
1 1 XX
S2 ≤ 2M
ank ϕk (|∆m
u xk |)
δ hr n∈Ir k=1
by Proposition 2.3. Finally we have x ∈ Vθ0 ((A, ϕk , ∆m
u ) , fn ). This completes the proof.
Theorem 2.5 Let ϕ = (ϕk ) and f = (fn ) be given a sequence of ϕfunctions and a sequence of modulus functions, respectively. If lim inf fnv(v) >
v→∞ n
0 then
0
m
Vθ0 ((A, ϕk , ∆m
u ) , fn ) = Vθ (A, ϕk , ∆u ) .
Proof. If lim inf fnv(v) > 0 then there exists a number c > 0 such that
v→∞ n
fn (v) > cv for v > 0 and n ∈ N . Let x ∈ Vθ0 ((A, ϕk , ∆m
u ) , fn ) . Clearly
∞
∞
X
X
1 X
1 X
ank ϕk (|∆m
ank ϕk (|∆m
fn
x
|)
≥
c
u k
u xk |)
hr n∈Ir
h
r n∈Ir
k=1
k=1
!
!
∞
c X X
=
ank ϕk (|∆m
u xk |) .
hr n∈Ir k=1
!
Therefore x ∈ Vθ0 (A, ϕk , ∆m
u ) . By using Theorem 2.4, the proof is completed.
Theorem 2.6 Let θ = (kr ) be a lacunary sequence and f = (fn ) be a
sequence of modulus functions.
0
m
(i) If lim inf qr > 1 then V 0 ((A, ϕk , ∆m
u ) , fn ) ⊂ Vθ ((A, ϕk , ∆u ) , fn ) .
0
m
(ii) If lim sup qr < ∞ then Vθ0 ((A, ϕk , ∆m
u ) , fn ) ⊂ V ((A, ϕk , ∆u ) , fn ) .
0
m
(iii) If 1 < lim inf qr ≤ lim sup qr < ∞ then Vθ0 ((A, ϕk , ∆m
u ) , fn ) = V ((A, ϕk , ∆u ) , fn ) .
Proof. This can be proved by using the same techniques in [11] and hence
we omit the proof.
The next result follows from Theorem 2.5 and Theorem 2.6.
Corollary 2.7 If lim inf fnv(v) > 0 and 1 < lim inf qr ≤ lim sup qr < ∞
v→∞ n
then Vθ0 (A, ϕk , ∆m
)
=
V 0 ((A, ϕk , ∆m
u
u ) , fn ).
41
On Lacunary Strongly Convergent...
3
Sθ0 (A, ϕk , ∆m
u ) -Statistical Convergence
Let the matrix A = (ank ) be given as previously, θ = (kr ) be a lacunary
sequence, the sequence of ϕ-functions ϕ = (ϕk ) and a positive number ε > 0
be given. We write,
(
Kθr
((A, ϕk , ∆m
u ) , ε)
= n ∈ Ir :
∞
X
)
ank ϕk (|∆m
u xk |)
≥ε .
k=1
The sequence x is said to be lacunary (A, ϕk , ∆m
u )- statistically convergent
to a number zero if for every ε > 0
1
µ (Kθr ((A, ϕk , ∆m
u ) , ε)) = 0
r h
r
where µ (Kθr (((A, ϕk , ∆m
u ) , ε))) denotes the number of element belonging to
r
m
Kθ (((A, ϕk , ∆u ) , ε)). We denote by Sθ0 (A, ϕk , ∆m
u ), the set of sequences x =
m
(xk ) which are lacunary (A, ϕk , ∆u )-statistically convergent to a number zero.
We write
lim
Sθ0 (A, ϕk , ∆m
u ) = x = (xk ) : lim
r
1
µ (Kθr ((A, ϕk , ∆m
u ) , ε)) = 0 .
hr
0
m
When we take θ = (2r ), Sθ0 (A, ϕk , ∆m
u ) reduces to S (A, ϕk , ∆u ).
0
If we take A = I and ϕk (x) = x for all k and x, then Sθ (A, ϕk , ∆m
u ) reduces
m
0
to Sθ (∆u ) defined by
Sθ0 (∆m
u ) = x = (xk ) : lim
r
1
µ ({k ∈ Ir : (|∆m
u xk |) ≥ ε}) = 0 .
hr
Now we have,
Theorem 3.1 Let θ = (kr ) be a lacunary sequence, ϕ = (ϕk (v)) and ψ =
(ψk (v)) are two sequences of ϕ-functions.
(i) If ψ ≺ ϕ and ϕk satisfies the ∆2 -condition for large v and for all k then
0
m
(A, ψk , ∆m
u ) ⊂ Sθ (A, ϕk , ∆u ).
(ii) If ϕ ∼ ψ and ϕk and ψk satisfy the ∆2 -condition for large v and for
0
m
all k then Sθ0 (A, ψk , ∆m
u ) = Sθ (A, ϕk , ∆u ).
Sθ0
Proof. (i) Let x ∈ Sθ0 (A, ψk , ∆m
By
assumption
u ).
m
m
ψk (|∆u xk |) ≤ bϕk (c |∆u xk |) and we have for all n and m,
∞
X
k=1
ank ψk (|∆m
u xk |) ≤ b
∞
X
k=1
ank ϕk (c |∆m
u xk |) ≤ K
∞
X
k=1
we
ank ϕk (|∆m
u xk |)
have
42
Metin Başarır et al.
for b, c > 0, where the constant K is connected with properties of ϕ func∞
P
tions. Thus the condition
the condition
ank ψk (|∆m
u xk |) ≥ ε implies
∞
P
k=1
k=1
ank ϕk (|∆m
u xk |)
≥ ε and in consequence we get
r
m
Kθr ((A, ϕk , ∆m
u ) , ε) ⊂ Kθ ((A, ψk , ∆u ) , ε)
and
lim
r
1
1
µ (Kθr ((A, ϕk , ∆m
µ (Kθr ((A, ψk , ∆m
u ) , ε)) ≤ lim
u ) , ε)) .
r
hr
hr
This completes the proof.
m
0
(ii) The identity Sθ0 (A, ψk , ∆m
u ) = Sθ (A, ϕk , ∆u ) is proved by using the
same argument.
Theorem 3.2 Let f = (fn ) be given a sequence of modulus functions. If
inf
fn (v) > 0 then
n
0
m
Vθ0 ((A, ϕk , ∆m
u ) , fn ) ⊂ Sθ (A, ϕk , ∆u ) .
Proof. If inf fn (v) > 0 then there exists a number α > 0 such that
n
fn (v) ≥ α for v > 0 and n ∈ N . Let x ∈ Vθ0 ((A, ϕk , ∆m
u ) , fn ) .
∞
X
1 X
1
ank ϕk (|∆m
fn
u xk |) ≥
hr n∈Ir
hr
k=1
!
X
∞
P
fn
n∈Ir
∞
X
!
ank ϕk (|∆m
u xk |)
k=1
ank ϕk (|∆m
u xk |)≥ε
k=1
α
≥
hr
(
n
∈ Ir :
∞
X
ank ϕk (|∆m
u xk |)
k=1
≥
)
ε and it follows that x ∈ Sθ0 (A, ϕk , ∆m
u ).
Theorem 3.3 Let f = (fn ) be given a sequence of modulus functions.
0
m
If sup sup fn (v) < ∞ then Sθ0 (A, ϕk , ∆m
u ) ⊂ Vθ ((A, ϕk , ∆u ) , fn ) .
v
n
Proof.
We suppose T (v) = sup fn (v) and
T = sup T (v) .Let x ∈
n
v
Sθ0 (A, ϕk , ∆m
u ). Since fn (v) ≤ T for n ∈ N and v > 0, we have
∞
X
1
1 X
fn
ank ϕk (|∆m
u xk |) ≥
hr n∈Ir
hr
k=1
!
X
∞
P
k=1
n∈Ir
ank ϕk (|∆m
u xk |)≥ε
fn
∞
X
k=1
!
ank ϕk (|∆m
u xk |)
43
On Lacunary Strongly Convergent...
1
+
hr
X
∞
P
fn
n∈Ir
∞
X
!
ank ϕk (|∆m
u xk |)
k=1
ank ϕk (|∆m
u xk |)<ε
k=1
T
≤
hr
(
n
∈ Ir :
∞
X
k=1
ank ϕk (|∆m
u xk |)
≥
)
ε + T (ε).
Taking the limit as ε → 0, it follows that x ∈ Vθ0 ((A, ϕk , ∆m
u ) , fn ).
Corollary 3.4 Let f = (fn ) be given a sequence of modulus functions.
If inf fn (v) > 0 (v > 0) and sup sup fn (v) < ∞ then Sθ0 (A, ϕk , ∆m
u) =
n
Vθ0
v
n
((A, ϕk , ∆m
u ) , fn ).
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