© Copyright 2014 | Centre for Info Bio Technology (CIBTech) 51
C RATE SEQUENCE SPACES OF DIFFERENCE SEQUENCE DEFINED BY A MODULUS FUNCTION
Cumali ÇATAL, İlhan DAĞADUR* and Burcu ÖZALTIN
Mersin Unıversıty, Faculty of Scıence and Lıterature, Department of Mathematıcs, 33343 Mersin- Turkey
*Author for Correspondence
ABSTARCT
Recall that a C method is obtained by deleting a set of rows from the Cesáro matrix C1. The purpose of this article is to introduce the sequence spaces C0(p,f,s,), Cc(p,f,s,) and C(p,f,s,) using a modulus function f. Several properties of this spaces, and some inclusion relations have been examined.
Keywords: Modulus Function, Rate Sequence Spaces, Difference Sequence, Paranorm, C-Summability Method
AMS Clasification: 40C05, 40D25, 40G05, 42A05, 42A10
INTRODUCTION
The notion of modulus function was introduced by Nakano
12 and further investigated by Ruckle
14, Maddox
9, Tripathy and Chandra
15 and many others. A function f :
0,
0, is called a modulus if i f(x)0 if and only if x0, ii f(xy)f(x)f(y), f
iii)
( is increasing,
iv f is continuous from the right at 0.
It is immediate from ii and iv that f is continuous everywhere on
0,. The idea of difference sequence was first introduced by Kizmaz
6 write xkxkxk1 for k1,2,3,.... Let denote the space of all comlex-valued sequences, : be the difference defined by x xk k1. Let denote the space of all real or complex-valued sequence. It can be topologized with the seminorms pn(x) xn,,...), 2 , 1
(n any vector subspace X of is a sequence space. A sequence space X with a vector space topology , is a K-space provided that the inclusion map i : X , , i x x, is continuous. If, in addition, is complete, metrizable and locally convex then X, is an FK-space. So an FK-space is a complete, metrizable locally convex topological vector space of sequences for which the coordinate functionals Pn(x)xn, (n1,2,...) , are continuous. The basic properties of FK-spaces may be found in
3 ,
16,
17 and
19 .Ruckle
14 used the idea of a modulus function f to construct a class of FK spaces. ) ( : ) ( ) (
1
k
k
k f x
x x f L
Let (n) be a sequence of positive numbers i.e, n0,nN and X an FKspace. We shall consider the sets of sequences x(xn)
}.
:
{ x X
w x X
n n
The set X may be considered as FK-space. We shall call them as rate spaces ( see,
4 and
5).Let F be an infinite subset of N and F as the range of a strictly increasing sequence of positive integers, say© Copyright 2014 | Centre for Info Bio Technology (CIBTech) 52
n n1
F . The Cesáro submethod C is defined as
, 1,2,...,
1
1
n n x
x
C k
n n k
where xk is a sequence of a real or complex numbers. Therefore, the C-method yields a subsequence of the Cesáro method C1, and hence it is regular for any . C is obtained by deleting a set of rows from Cesáro matrix. The basic properties of Cmethod can be found in
1 and
13. We need the following inequality throughout the paper. Let p(pk) be a sequence of positive real numbers with Gsupkpk and).
2 , 1 max( 1
G
D Then, it is well known that for all ak, bkC, the field of complex numbers, for all kN,
k k
. 1k p
k p k p k
k b D a b
a
Also for any complex ,
2 )
, 1 (
max G
pk
see in
11. Motivating by Maddox
9, Jürimäe
4 and Başarır
2 we make the following definitions: The following sequence spaces) , ( sup 1 : )
(
) (
1
i k
i k
x w k
x
C
, ) ( ) 0
( : 1 )
(
) (
1 0
x k
w k x C
i k
i
, 0 some for ), ( ) 0
( : 1 )
(
) (
1
x L k L
w k x C
i k
i c
) , ( sup 1 : )
, (
) (
1
pk
i k
i k
x w k
x p
C
, ) ( ) 0
( : 1 )
, (
) (
1 0
x k
w k x p C
pk
i k
i
, 0 ),
( ) 0
( : 1 )
, (
) (
1
x L k for someL
w k x p C
pk
i k
i c
, 0 ) ,
( sup 1 : )
, , (
) (
1
x s
k k w x s p C
pk
i k
i s
k
0( ), 0
) ( : 1 )
, , (
) (
1
0 x k s
k k w x s p C
pk
i k
i s
and
. 0 ,
0 ), ( ) 0
( : 1 )
, , (
) (
1
x L k s for someL
k k w x s p C
pk
i k
i s c
RESULTS
In this section, Crate sequence spaces of difference sequence is defined by a modulus function, and several theorems on this subject are given.
Definition 1. Let f be a modulus function. Then we define the following sets of sequences
© Copyright 2014 | Centre for Info Bio Technology (CIBTech) 53
, 0 ) ,
( sup 1 : )
, , , (
) (
1
x s
k f k w x s f p C
pk
i k
i s
k
0( ), 0
) ( : 1 )
, , , (
) (
1
0 x k s
k f k w x s f p C
pk
i k
i s
and
, 0 ,
0 ), ( ) 0
( : 1 )
, , , (
) (
1
x L k s for someL
k f k w x s f p C
pk
i k
i s c
where
x k xkk xkk11.Theorem 1.
i For any modulus function f, C0(p,f,s,),Cc(p,f,s,) and C(p,f,s,) are linear spaces over the complex field C.
ii Let f be any modulus. Then C0(p,f,s,)Cc(p,f,s,)C(p,f,s,).
Proof. i Let x,yC0(p,f,s,). For any ,C, there exist integers M and N such that M and
.
N By definition of modulus function and inequality (1) we have
) . ( 1 )
( 1
) ( 1 )
( 1
) ( 1
) (
1 )
( 1
) (
1 )
( 1
) (
1
k k
k k
p
i k
i s
G p
i k
i s
G
p
i k
i i k
i s
p
i k
i s
f y k k
N x D
k f k M D
f y k f x
k k
y f x
k k
This implies that xyC0(p,f,s,), and completes the proof. The others cases are routine works in view of the above theorem.
ii The proof of the inclusion C0(p,f,s,)Cc(p,f,s,) is routine verification. So, we leave it to the reader. Let xCc(p,f,s,). Then there is some L such that
. 0 ), ( ) 0
( 1 ( )
1
x L k s
k f k
pk
i k
i s
From inequality (1), we get
.) ( 1 ) ( 1 )
( 1
) (
1 ) (
1 )
( 1
k k
k k
s p p
i k
i s
p
i k
i s p
i k
i s
L f k D x L
k f k D
L x L
k f x k
k f k
There exissts an integer M such that LM. Therefore we have
© Copyright 2014 | Centre for Info Bio Technology (CIBTech) 54
1
.) ( 1 ) ( 1
) (
1 ) (
1
s G p
i k
i s
p
i k
i s
Mf k D x L
k f k D
f x k k
k k
This shows that xC(p,f,s,), and completes the proof.
Theorem 2. C0(p,f,s,),Cc(p,f,s,) and C(p,f,s,) are linear topological spaces paranormed by
) , ( sup 1 ) (
/ )
( 1
M p
i k
i s
k
k
f x k k
x
g
where M max(1, Gsupk pk).
The proof follows by using standart techniques and the fact that every paranormed space is a topological linear space
18 p, .37
. So we omit the details.Theorem 3. C0(p,f,s,),Cc(p,f,s,) and C(p,f,s,) are complete in their paranorm topologies.
Proof. Let
x j be a Cauchy sequence in C0(p,f,s,), where
x j
x1j,x2j,...
,jN.Hence for a given 0 there exists N that g
x j x t
for all j,tN, that is
3 .
, all for ) ,
( sup 1
/ )
( 1
N t x j
f x k k
M p
i t
i k j
i s k
k
This implies that for each fixed i,
x j i
x t i . So
x j i
is a Cauchy sequence in C, but C is complete so for iN we have
x j i
x i(j). From (3) we get
. , all for ) ,
( sup 1
) (
1
N t x j
f x
k k M
p
i t
i k j
i s k
k
So for each fixed k
. , all for ) ,
( 1 ( )
1
N t x j
f x
k k M
p
i t
i k j
i s
k
By taking t in the above inequality we obtain
. all for ) ,
( 1 ()
1
N x j
f x
k k M
p
i i k j
i s
k
Since k is arbitrary we have
, all for ) ,
( sup 1
) (
1
N x j
f x
k k M
p
i i k j
i s k
k
this implies g
x(j)x
, for jN.Now we have to show that xC0(p,f,s,). Since pk/M1 and M1, using Minkowski's inequality and definition of modulus function, we can get