3221 5687, (P) 3221 568X
Sathya. . . (IJ0SER) December - 2014
Generalized Difference Sequence Space On
Seminormed Space By Orlicz Function
A.Sathyakala Assistant Professor
PPG Institute of Technology,Coimbatore,India
.
Abstract: In this paper,we define the sequence space
l
M(
,
u
,
p
,
q
,
s
)
on seminormed complex linear space,by using orlicz functions. We give various properties and also some inclusion relations,results.This study generalized some results ofl
M(
u
,
p
,
q
,
s
)
.Keywords: Orlicz function, difference sequence, solid spaces, sequence algebra,convex function.
1. INTRODUCTION :
Let
l
,
C
and
C
0 be the linear spaces of bounded,convergent. Let
be the set of all null sequencex
(
x
k)
withX
x
k)
(
with complex numbers.Lindenstrauss and Tzafriri[4] have used an orlicz function to construct the sequence space,
10
some
for
:
k k Mx
M
x
l
The spaces
l
M with norm kK
x
x
sup
where
k
N
becomes a Banach space which is called an orlicz sequence space.Let zero sequence is denoted by
(
0
,
0
,...)
where
is a zero element of X. The space are seminormed space by)
(
sup
)
(
k N Kx
q
x
g
where (X,q) is a seminormed spaces, seminormed by q.DEFINITION :
An orlicz function is a function
M
:
[
0
,
)
[
0
,
)
which is continuous, non-decreasing and convex with0
for x
0
)
(
,
0
)
0
(
M
x
M
andM(x)
as
x
.Let X be a complex linear space with zero element
(
0
,
0
,...)
andX
(
X
,
q
)
be a seminormed space with the seminorm q.)
(x
denote the linear space of all sequencex
(
x
k)
withX
x
k)
(
with the usual coordinate wise operations.
x
(
x
k)
and
x
y
(
x
k
y
k)
for eachC
whereC
denotes the set of all complex numbers. If
(
k)
is a scalar sequence and)
(x
x
then
x
(
kx
k)
.Let U be the set of all sequence
u
u
k such thatu
k
0
and complex for all k=1,2….Letp
p
k be a sequence of positive numbers and M be an orlicz function. Givenu
U
. Then, we definethe sequence space is
0
some
for
0,
s
,
x
u
q
M
:
)
(
)
,
,
,
(
k p k k 1
k s MK
X
S
x
s
q
p
u
l
In this section the space
l
M(
u
,
p
,
q
,
s
)
is extended to)
,
,
,
,
(
u
p
q
s
l
M
by introducing the difference operator.)
x
-(x
)
(x
k
k k1
0
some
for
0,
s
,
x
u
q
M
:
)
(
)
,
,
,
,
(
k p k k 1
k s MK
x
x
s
q
p
u
l
3221 5687, (P) 3221 568X
Sathya. . . (IJ0SER) December - 2014
The following inequalities will be used throughout this paper. Let
p
p
kbe a sequenceof strictly positive real numbers with
0
p
k
sup
kp
k
G
and letD
max
1
,
2
G1
.Then for all K and
k,
k
C
,
we have
k k
k p k p k p k k
D
. DEFINITION :A Sequence space E is said to be solid (or normal)
1 2
3
max
2
,
2
whenever for all sequence of scalar with
k
1
2.THEOREMS OF
l
M(
,
u
,
p
,
q
,
s
)
: Theorem: 2.1Let sequence space
l
M(
,
u
,
p
,
q
,
s
)
is a linear space over the field C of complex numbers.Proof:
Let
x
,
y
l
M(
,
u
,
p
,
q
,
s
)
and
,
C
In order to prove we need to find some
3such that
pk 3 k y k x q M 1 k u k s K Sincex
,
y
l
M(
,
u
,
p
,
q
,
s
)
There exists positive numbers
1,
2 such that
k p 1 k k 1x
u
q
M
k sK
k
p
2
)
y
(
q
M
1
k
ku
k
s
K
Define
3
max
2
1,
2
2
k
p
3
)
y
x
(
q
M
1
k k
ku
k
s
K
k pq
q
M
k
s
K
3
)
y
(
u
3
)
x
(
u
1
k k k k
k
p
2
2
)
k
y
(
q
1
2
)
k
x
(
q
M
1
k ku
u
k
s
K
k k p 2 k 1 p 1 k 1y
q
M
2
1
x
q
M
2
1
k s p k s pK
D
K
D
k kHence
x
y
l
M(
,
u
,
p
,
q
,
s
)
if x and y belongs to)
,
,
,
,
(
u
p
q
s
l
M
and
,
C
. WhereD
max(
1
,
2
H1)
Hence
l
M(
,
u
,
p
,
q
,
s
)
is a linear space.Theorem : 2.2
The space
l
M(
,
u
,
p
,
q
,
s
)
is a paranormed ( need not total paranormed) with
1,2,3.. n , 1 / 1 k p x q M 1 : / inf ) ( k H p u k s K H pn x u g k Where
max
1
,
sup
p
k kH
Proof :To prove that
g
u(
x
)
g
u(
x
)
. The subadditivity of given follow from equation on taking
1
,
1
.
sinceq
(
)
0
andM(0)=0,
we
get
for x
0
inf
H Pn .Finally, we prove that the scalar multiplication is continuous. Let
be any number, by definition
1,2,3..
n
,
1
x
u
q
M
:
inf
)
(
/ 1 p k k 1 / k H k s H pn uK
x
g
Then,3221 5687, (P) 3221 568X
Sathya. . . (IJ0SER) December - 2014
1,2,3..
n
,
1
x
u
q
M
:
)
(
inf
)
(
/ 1 p k k 1 / k H k s H pn ur
K
r
x
g
wh ere.
r
since
Pk
H
,
1
max
, then
H H k P H 1,
1
max
. Hence,
1,2,3..
n
,
1
x
u
q
M
:
inf
,
1
max
)
(
/ 1 p k k 1 / k H k s H pn H uK
r
x
g
Andtherefore
g
u( x
)
converges to zero wheng
u(x
)
converges to zeroin
l
M(
,
u
,
p
,
q
,
s
)
. Now suppose that
n
0
and x is in)
,
,
,
,
(
u
p
q
s
l
M
. For arbitrary
0
, let N be a positive integersuch that H H N k s
K
2
x
u
q
M
/ 1 p k k 1 k
For some
0
. This implies that2
x
u
q
M
/ 1 p k k 1 k
H N k sK
Let
0
1
, then using definition of convex isH N k s H N k s
K
K
2
x
u
q
M
x
u
q
M
k k p k k 1 / 1 p k k 1
Since M is continuous every where in
[
0
,
)
, then k p k k 1x
u
q
M
)
(
k sK
t
f
is continuous at zero. So there is
1
0
,
such that 2) (t
f for
0
t
. Let K be such thatk,
n
for
n then n>k , we have1
x
u
q
M
/ 1 p k k 1 k
H n k sK
for n>k. If we take limit on
H Pn
inf
we getg
u(
x
)
0
.3.SOME PARTICULAR CASES
We get the following sequence spaces from
l
M(
,
u
,
p
,
q
,
s
)
on giving particular values to p,q,s. TakingP
k
1
for allN
k
, we have
0
some
for
0,
s
,
x
u
q
M
:
)
(
)
,
,
,
(
k k 1
k s MK
x
x
s
q
u
l
If S = 0,then we have
0
some
for
,
x
u
q
M
:
)
(
)
,
,
,
(
1 p k k k
k Mx
x
q
p
u
l
If1
kP
for allk
N
and S = 0, then we have
0
some
for
,
x
u
q
M
:
)
(
)
,
,
(
1 k k
k Mx
x
q
u
l
If we take
s
0
,
q
(
x
)
x
andx
c
,then we have
0
some
for
,
x
u
M
:
)
(
)
,
,
(
1 p k k k
k Mx
x
p
u
l
In addition to the above sequence spaces, we have
l
M(
,
u
,
p
,
q
,
s
)
l
M(
p
)
due to Parshar and Choudhry[5],on takingN,
k
all
for
1
ku
we have)
,
,
(
)
,
,
,
,
(
u
p
q
s
l
p
q
s
l
M
M THEOREM 3.1Let
0
P
k
t
k
for eachk
N
. Then)
,
,
,
(
)
,
,
,
(
)
(
i
l
M
p
u
q
l
M
t
u
q
)
,
,
,
(
)
,
,
(
)
(
ii
l
M
u
q
l
M
u
q
s
)
,
,
,
,
(
)
,
,
,
(
)
(
iii
l
M
p
u
q
l
M
p
u
q
s
Proof :3221 5687, (P) 3221 568X
Sathya. . . (IJ0SER) December - 2014
Let x
l
M(
,
p
,
u
,
q
)
(i) To Prove that
x
l
M(
,
t
,
u
,
q
)
)
,
,
,
(
l
Let x
M
p
u
q
then there exist some
0
such that
1 P k k)
x
(
q
M
k ku
This implies that
M
q
u
ix
i
1
for sufficiently large values of i.sayi
k
0, for some fixedk
0
N.
Since M is non-decreasing we get
1 t k k)
x
(
q
M
k ku
Since
0 k 0 k P k t k)
x
(
q
M
)
x
(
q
M
k k k k k ku
u
Hencex
l
M(
,
t
,
u
,
q
)
. There forel
M(
,
p
,
u
,
q
)
l
M(
,
t
,
u
,
q
)
(ii) To Prove that
l
M(
,
u
,
q
)
l
M(
,
u
,
q
,
s
)
Let
x
l
M(
,
u
,
q
)
To Prove that
x
l
M(
,
u
,
q
,
s
)
Let
x
l
M(
,
u
,
q
)
then there exist some
0
such that
1 k)
x
(
q
M
k ku
This implies that
M
q
u
ix
i
1
for sufficiently large values of i.Since M is non-decreasing we get
1 k k 1 k k)
x
u
(
q
M
)
x
u
(
q
M
k k sK
Hencex
l
M(
,
u
,
q
,
s
)
. There forel
M(
,
u
,
q
)
l
M(
,
u
,
q
,
s
)
(iii) To Provel
M(
,
p
,
u
,
q
)
l
M(
,
p
,
u
,
q
,
s
)
)
,
,
,
(
l
Let x
M
p
u
q
then there exist some
0
such that
1 P k k)
x
(
q
M
k ku
This implies that
M
q
u
ix
i
1
for sufficiently large values of i.sayi
k
0, forSince M is non-decreasing we get
k P k sK
1 k kx
)
u
(
q
M
Since
0 k 0 k P k P k)
x
(
q
M
)
x
(
q
M
k k k k k k su
u
K
Hencex
l
M(
,
p
,
u
,
q
,
s
)
. There forel
M(
,
p
,
u
,
q
)
l
M(
,
p
,
u
,
q
,
s
)
Hence the proof.
THEOREM 3.2
Let
0
P
k
t
k
for each k. Then).
,
,
(
)
,
,
(
u
p
l
u
t
l
M
M
Proof :Let x
l
M(
,
u
,
p
)
(i) To Prove that
x
l
M(
,
u
,
t
)
Let x
l
M(
,
u
,
p
)
then there exist some
0
such that
1 P k kx
M
k ku
3221 5687, (P) 3221 568X
Sathya. . . (IJ0SER) December - 2014 This implies that
M
u
ix
i
1
for sufficiently large valuesof i.
Since M is non-decreasing we get,
1 t k k)
x
(
M
k ku
Since
1 P k 1 t k k k)
x
(
M
)
x
(
M
k k k ku
u
Therefore,
1 t k k)
x
(
M
k ku
Hencex
l
M(
,
u
,
t
)
.)
,
,
(
)
,
,
(
l
M
u
p
l
M
u
t
Theorem : 3.3(i) If
0
p
k
1
for each k , then)
,
,
(
)
,
,
,
(
u
p
q
l
u
q
l
M
M
(ii) If
p
k
1
for each k , then)
,
,
,
(
)
,
,
(
u
q
l
u
p
q
l
M
M
Proof : Letx
l
M(
,
u
,
p
,
q
)
To prove thatx
l
M(
,
u
,
q
)
Then there exist some
0
such that
k p k 1 k kx
)
u
(
q
M
This implies that
M
q
u
ix
i
1
for sufficiently large values of i. Since M is non-decreasing we get
1)
(
q
M
k k kx
u
Since,
k p k k k k k k k kx
u
x
u
0 0)
(
q
M
)
(
q
M
Hencex
l
M(
,
u
,
q
)
There forel
M(
,
p
,
q
)
l
M(
,
q
)
(ii) To prove that)
,
,
,
(
)
,
,
(
u
q
l
u
p
q
l
M
M
Letx
l
M(
,
u
,
q
)
then there exist some
0
such that
1)
(
q
M
k k kx
u
This implies that
M
q
u
ix
i
1
for sufficiently large values of i. Since M is non-decreasing we get
k p k k kx
u
1)
(
q
M
Therefore
1 1)
u
(
q
M
)
u
(
q
M
k k k p k k kx
x
k
Henc ex
l
M(
,
u
,
p
,
q
)
Hence the proof.Proposition 3.4
For any two sequence
p
(
p
k)
and
t
(
t
k)
of positive real numbers and any two seminormsq
1, q
2we have
,
,
,
,
)
(
,
,
,
,
)
(
u
p
q
1r
l
u
t
q
2s
l
M M for all,
n
N
m
andr,
s
0.
Proof:Since the zero element belongs to
l
M(
,
u
,
p
,
q
1,
r
)
and)
,
,
,
,
(
u
t
q
2s
3221 5687, (P) 3221 568X
Sathya. . . (IJ0SER) December - 2014 THEOREM : 3.5
The sequence space
l
M(
,
u
,
p
,
q
,
s
)
is solid. Proof: Letx
k
l
M(
,
u
,
p
,
q
,
s
).
k P k sK
1 k kx
)
u
(
q
M
Let
(
k)
be sequence of scalars such that
k
1
for all.
k
N
Then we have,
k k P k s P k sK
K
1 k k 1 k k k)
x
u
(
q
M
)
x
u
(
q
M
Hence
(
kx
k)
l
M(
,
u
,
p
,
q
,
s
)
for all sequence of scalars(
k)
with
k
1
for allk
N
, whenever)
,
,
,
,
(
)
(
x
k
l
M
u
p
q
s
.Therefore the space
l
M(
,
u
,
p
,
q
,
s
)
is solid sequence space.REFERENCES :
[1]. Cigdem A.Bektas,the sequence space
l
M(
u
,
p
,
q
,
s
)
on seminormed space, Indian J.Pure Appl.Math,245-250.[2]. C.Bekatas and Y.Altin, the sequence space
)
,
,
(
p
q
s
l
M on seminormed space, Indian J.PureAppl.Math., 34(4), (2003), 529-534.
[3]. I.J.Maddox, “Elements of Functional Analysis”, Cambridge univ. press, 1970.
[4]. J.Lindenstrauss and L.Tzafriri,”On Orlicz sequence space”, Israel J.Math.,10(1971),379-390.
[5]. S.D.Parashar and B.Choudhsry,”Sequence spaces defined by Orlicz functions”, Indian J.Pure
Appl.Math.,25(1994),419-428.
[6]. K.Indirani and A.Sathya kala “The sequence space
)
,
,
,
(
p
q
s
l
M
on seminormed space”,International Journal for Technological Research inEngineering.,volume1,issue7,2347-4718.
[7]. I.J.Maddox, “ Spaces of strongly sumable sequences”, Quart. jour. Math.2(18) (1976),345-355.
[8]. Et.M and R.Colak, “On generalized difference sequence spaces” Soochow jour.math.21(4) (1995).
[9]. I.J.Maddox, “Elements of functional analysis”,Camb Univ.Press,1970.
[10]. H.KIZMAZ, “On certain sequence spaces”,Canad.Math.Bull. [11]. 24(1981),169-176.
