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A Subset of the Space of the Orlicz Space of
Sequences
B.Baskaran
1, S.Thalapathiraj
2Professor, Department of Mathematics, SRM University, Tamilnadu, India1 Assistant Professor, Department of Mathematics, SRM University, Tamilnadu, India2
ABSTRACT: Let
M denote the space of all Orlicz space gai rate sequences. Let
M denote the space of all Orlicz space of analytic rate sequences. Mainly in this paper in studied the general properties of Sectional space
M Sof (
M ).KEYWORDS: Orlicz space, rate sequence, FK space, AK space
I.INTRODUCTION
A complex sequence, whose kth term is xk is denoted by {xk} or simply x. Let be the set of all finite sequences. A
sequences x = (xk) is said to be analytic if supk |xk|
1/k
< . The vector space of all analytic sequences will be denoted by
.
was discussed in Kamthan [11]. Matrix transformations involving
were characterized by sridhar [12]and Sirajudeen[13] . A sequence x is called entire sequence if
lim
!
0
1
k k
k
k
x
. The vector space of all gai of phi sequences will bedenoted by
.Kizmaz [17] defined the following defined the following difference sequence spaces Z( ∆ ) = {x = (xk) : ∆ x
Z }for Z = l, c, c0 , where ∆ x = (∆x) 1
k = (xk - xk+1) )
1
k and showed that these are Banach space with norm
x
x
x
1 .Later on Et and Colak[10] generalized the notion as follows :Let m
N, Z(∆ m) = {x = (xk) : ∆ mx
Z} forZ = l, c, c0 where m
N ,∆0
x= (xk), (∆x) = (xk - xk+1) , ∆
mx = (∆m
xk) )
1
k =( ∆
m-1
xk - ∆ m-1
xk+1)
1
k
The generalized difference has the following binomial representation : ∆m
xk =
m m
0(
1
)
xk+
,They proved that these are Banach spaces with the norm
x
x
x
mm
i i
1
Orlicz [4] used the idea of Orlicz function to construct the space (LM ) . Lindenstrauss and Tzafriri [1] investigated Orlicz sequence space in more detail, and they proved that every Orlicz sequence space lMcontains a subspace isomorphic to
lp(
1
p
. Sufebsequently ,the different classes of sequence spaces were de fined by Parashar and Choudhry [2] ,Mursaleen etel[3] , Bektas and Altin [4] , Tripathy et al.[5], Rao and subramanian[6],and many others. The Orlicz sequence spaces are the special cases of Orlicz spaces studied in [7].
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M(x)> 0, x> 0 and M (x) as x ..
If the convexity of Orlicz function M is replaced by M (x + y) < M (x) + M (y) . then this function is called modulus function , introduced by Nakano [18] and further discussed by Ruckle [8] and Maddox[9], and many others.
An Orlicz function M is said to satisfy ∆2 condition for all values of u , if there exists a constant K > 0 , such that M(2u) <
KM (u) (u > 0). The condition ∆2 is equivalent to M(lu) < KlM (u), for all values u and for l > 1.Lindestrauss and Tzafriri
[1] used the idea of Orlicz function to construct Orlicz sequence space
)
1
...(
0
,
:
1
forsome
x
M
w
x
k kM
Where w={all complex sequence}.The space
M with the norm)
2
...(
1
:
0
inf
1
kx
kM
x
becomes a Banach space which is called an Orlicz sequence space. For M(t) = tp,(1
p <
) the space
M coincide with the classical sequence space
p. Given sequence x={xk } its nth section is the sequence xn= {x1, x2,……… xn, 0, 0,…..}n
= (0,0,…….1,0,0…..).,1 in the nth place and zero's else where ; and
n= (0,0,…….1,-1,0,0…..).,1 in the nth place ,-1 inthe (n + 1)th place and zero's else where .
If X is a sequence space, we define
i. X,=continuous dual of X.
ii.
1
,
:
k k k
k
a
x
foreachx
X
a
a
X
iii.
1
,
:
k k k
k
a
x
isconverge
nt
foreachx
X
a
a
X
iv.
1 sup
,
:
k k k
n
k
a
x
foreachx
X
a
a
X
v. Let X be an FK space
. ThenX
f
f
n:
f
X
,
X, X, X are called the ( or K the – T eplitz) dual of X, - ( or generalized K the – T eplitiz) dual of X, - dual of X. Note that XX X . If X Y then Y X, for = , , or
An FK space (Frechet coordinate space ) is a Frec space which is made up of numerical sequences and has the property that the coordinate functional spk(x) = xk(k= 1,2,3,....) are continuous .
An FK - space is a locally convex Frechet space which is made up of sequences and has the property that coordinate projections are continuous. An metric space (X,d) is said to have AK (or sectional convergence) if and only if d(xn , x)
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II.DEFINITION AND PRELIMINARIES
II.1DEFINITION
Let w,
M and
M denote the spaces of all, Orlicz space of gai sequences and Orlicz space of bounded sequence respectively. Let w denote the set of all complex sequence x = (xk) 1
k and M : [0,) [0,) be an orlicz function. Let t
denote the sequence with k
k
k
x
t
1 for allk
N
denote k ɛN.Define the sets.
0
,
0
!
:
1
1
forsome
ask
t
K
M
w
x
k
k k k M
.
0
,
sup
:
1
1
forsome
t
M
w
x
k
k k k k
M
The space
M is a metric space with the metric
1
.
!
sup
:
0
inf
)
,
(
1 1
k k
k k k k
y
x
K
M
y
x
d
The space
M is a metric space with the metric
1
.
sup
:
0
inf
)
,
(
11
k k
k k k k
y
x
M
y
x
d
Let (𝑀𝜋)s =
x
x
k:
k
M
where is a
k
1
!
x
1
2
!
x
2
...
k
!
x
k for k=1,2,3,… and
M S=
y
y
k:
k
M
where
k
y
1
y
2
...
y
k for k=1,2,3,…..Then (𝑀 𝜋ISSN: 2319-8753
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:
1
,
2
,
3
,...
1
.
!
sup
:
0
inf
)
,
(
11
M
K
k
y
x
d
k k
k k k k
And
M Sis a metric space with the metric
:
1
,
2
,
3
,...
1
.
sup
:
0
inf
)
,
(
11
M
k
y
x
d
k k
k k k k
Let
M denote the vector space of all sequence
k
that
x
x
kk
is an orlicz space of gai sequence . We recall
that
cs
0 denotes the vector space of all sequencesx
x
k such that
k is a null sequence.II.2.LEMMA
Let X be an FK-space
.Then (i)X
X
f(ii)If X has AK .
X
X
f(iii)If X has AD,
X
X
We note that
f k MM M M
M
for
all
x
x
kandy
y
in
II.3.REMARK
x { xk} (𝑀
𝜋
)
k
Mk
a s k
M
k k k
k k
0
1 1
M
ask
k k
k k
0
1 1
,because
k
k
1
ask
1
x =
x
k
M S
Hence
M S=(𝑀𝜋) the cesaro space of order 1.II.4PROPOSITION
M S
(𝑀𝜋) Proof: S
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ask
M
k k k k0
1 1
But
x
k
k
k1
k k k kk
M
1 1!
k k k kM
1 1 +
k kM
1 1
!
1 1
k k k kk
M
ask
k
M
k k k k0
!
1 1 1
Therefore
ask
x
k
M
k k k k0
!
1 1
x
MHence
M S
(𝑀𝜋)Note:The above inclusion is strict. Take the sequence
1
MWe have 11
1
k kM
1
0
1
12
k kM
1
0
0
1
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……….
1
...
0
0
0
1
1
k k
k
M
k terms and so on.Now
forallk
M
k k
k k
1
1 1
. Hence
k k
k k
M
11
does not tend to zero as
ask
.So 1
M S.
Thus the inclusion
M S
(𝑀𝜋) is strict .This completes the proof.II.5.PROPOSITION
M S
has AK – property Proof:Let
x
x
k
S
M and takex
x
1,
x
2,...
x
n,
0
,...
forn
1
,
2
,
3
,...
n
Hence
inf
0
:
sup
1
,
11
k k
k n K K k
n
M
x
x
d
k k
n n n
k
M
11 1
1
sup
:
0
inf
,
1...
1
1
k k
k n K K
M
Therefore
x
n
xasn
in
S
M .Thus S
M has AK .This completes the proof.II.6PROPOSITION
M S
is a linear space over the field complex numbers. Proof: It is easy .Therefore omit the proof.II.7PROPOSITION
M S
is solidProof:Let
x
k
y
k withy
y
k
M S.
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S
M
is solid. This completes this proof.II.8PROPOSITION
S M
M
Proof:step1: By proposition II.4.,we have
M S
M .Hence. S
M
M S
But
M
. Therfore is
M S
….(4)Step 2: Let
y
y
k
M S
consider
1
)
(
K
x
ky
kx
f
with
S M
x
Take
x
n
n1
0
,
0
,...
1
,
1
,
0
,
0
,...
,1 in the n th place and zero’s elsewhere . Then f(
n
n1)=y
n
y
n1.
Hence
kk n
n
y
y
1 1
M
=
k k
n n
f
1 1
M
1
f
d
n
n1
,
0
f
.
1
.So,
kk n
n
y
y
1 1
M
is bounded.Consequently
Mk k
n
n
y
y
M
1 1
That is
M k
k n
y
1M
. Buty
y
n is originally in
M S
Therfore
M S
M
...(
5
)
From (4) and (5) We conclude that
M S
M
This completes proof.II.9PROPOSITION
The
dual space of
M Sis
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1
,
2
,
3
,...
,
!
)
(
1 1
n
n
n
n
k
M
k k
n k
Define
x
x
k byf or
n
x
k
M
kk k
k
1
)
!
(
1
k=k(n) and
0
)
!
(
1
kk k
x
k
M
otherwise Then x is in
M ,but for infinitely many k,1
)
!
(
1
kk k k
x
y
k
M
Consider the sequence z=
z
k where z1=1!x1-s withs
k
!
x
k;
andz
k
k
!
x
k
k
2
,
3
,....
Then z is a point of
S
M .But by the equation
S k k
k k
x
z
k
M
1!
does not converge. Thus the sequence y would not
be in
M S
.This contradiction proves that
S
M ……(7)
Step2:By proposition 1.8 we have
M S
….(8)From (7) and (8) it follows that the
dualspaceo
f
M S
is
.This completes the proof.II.10.PROPOSITION
M S
for
,
,
,
f
Proof:
M s has AK by Proposition 4.5.Hence by Lemma1.2(ii)ISSN: 2319-8753
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Hence
M S
……(9)Step 2: Since AK implies AD.Hence by Lemma 1.2(iii) we get
M S =
M STherefore
S
M …….(10)
Step 3:
M s
is normal by Proposition 1.7.Hence , by[15 proposition 1.7],we get
M S =
M S ….(11)From (9) , (10) and (11) we have
SM =
M S =
M S
=
M S
f
II.11.PROPOSITION
The dual space of
M S
is
. In other words
M s
*
Proof:We recall that S k has 1 in the k th place , -1 in the (k+1)th place and zero’s else where with
k k k k k
M
s
11
=
kk k
k k
k
k k k
k
M
M
M
M
1 1 1
1 1
1 2 1
1 1 1
)
0
(
,
)
0
(
...,
,...
)
0
(
,
)
0
(
=
..
,...
0
,
)
0
(
,
,...
0
,
0
,
0
11
kk k
M
Exists. Hence
s
k
M S.We have f(x)=
1
k
ky
kwith
M S
andf
M s
* ,where
M s
* is the dual space of
M S.Take
S M
and
*s M
f
,where
M s
* , is the dual of S
M .Take
S M K
s
Then
y
k
f
d
s
k,
0
forallk
……(12)Thus (yk) is a bounded sequence and hence an analytic sequence .In other words ,
y
.Therefore
M S
*
.This completes the proof. Lemma 2.12f f
X
Y
X
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II.12.PROPOSITION
Let Y be any FK-space
ThenY
M S if and only if the sequence
k weakly analytic. Proof :The following implications establish the result
Y
M S
Y
f
M S
f , Since S
M has AD and by lemma
fY
.Since
M S
f
foreachf
y
'
,the topological dual ofY
f
k is analytic
k is weakly analytic . This completes the proof.II.13.PROPOSITION
In
M weak converges does not imply strong convergence..Proof:Assume that weak convergence implies strong convergence in
S
M . Then we would have
M s = S
M . But
M s
. By proposition 1.4 S
M is a proper subspace of
MThus
M s
S
M .Hence weak convergence does not imply strong convergence S
M . This completes the proof.II.14.DEFINITION
Fix k=0,1,2,…….. given a sequence
x
K.put
,...
3
,
2
,
1
!
...
!
3
!
2
!
1
1 3 2
1
.
f or
x
k
p
x
k
x
k
x
k
M
M
k k
k p k
k k
k
.
Let
k,p:
p
1
,
2
,
3
,...
M uniformly in k=0,1,2,3,…….Then we call (xk) an “almost orlicz space of gai sequence”. The set of all almost orlicz space gai sequence is denoted by
MII.15.PROPOSITION
M
M
M
where
M is the set of all almost orlicz space of gai sequences. Proof:Put k=0.Then
0.p
M
kk
p
x
p
x
x
x
1 3 2
1
2
!
3
!
...
!
!
1
M
M
x
x
x
p
x
asp
p p
0
!
...
!
3
!
2
!
1
1 3
2 1
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M
x
x
x
p
x
asp
p p
0
!
...
!
3
!
2
!
1
1 3
2 1
….(13)
x
k
cs
0
M
cs
0
put k=1Then
pM p
x
p
x
x
x
3
!
4
!
...
!
!
2
1 3 4,
1
M
M
x
x
x
p
x
asp
P p
0
!
...
!
4
!
3
!
2
1 4
3 1
M
x
x
x
p
x
asp
k k
P p
0
!
...
!
4
!
3
!
2
1
1 4
3 1
…..(14)
0
!
...
!
4
!
3
1
1 4
3
kk
P p
x
p
x
x
M
……(15)0
!
...
!
4
1
1 4
kk
P p
x
p
x
M
…..(16) and so on.From (13) and (14) it follows that
kK
x
M
1
!
11
0
!
...
!
4
!
3
!
2
!
1
1 4 3 1
1
kk
p
x
p
x
x
x
x
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0
!
...
!
4
!
3
!
2
1 4 3
1
kk
p
x
p
x
x
x
M
Similarly we obtain
2
!
2
0
x
M
,3
!
3
0
x
M
and so on.Hence
M
denote the sequence (0,0,……)Thus we have proved that
M
M
and
M
In otherwords
M
M
MThis completes this proof
II.16.PROPOSITION
cs
0
M S MProof:By proposition 1.4
M S
MAlso, since every orlicz space of gai sequence
KM
exists,it follows that
k k K
M
1 exists. In otherwords0 1
cs
M
k k
K
.Thus
M S
cs
0
. Consequently ,
M S
M
cs
0 ….(17)On the otherhand, if 0
1
!
cs
x
k
M
Mk k
K
.then
1
1
!
k
k k
z
x
k
M
z
f
is an Orlicz space of an gai of ratefunction. But
M
k
!
1x
cs
0k k
K
.So,
0
!
...
!
4
!
3
!
2
!
1
)
1
(
1 1 3 1 4
kk
p
x
p
x
x
x
x
M
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Hence
1 11
1
k
k k
K
k
z
M
z
z
f
is also orlicz space of an gai function .
Hence
M k
k K
M
1 . So
x
x
k
M
s
(
)
but (xk) is arbitrary in M
cs
0
.Therefore
M
cs
0
S
M ….(18) From (17) and (18) we get
M S
M
cs
0 . This completes the proof.II.17.DEFINITION
Let
0
be not an integer Write
x
A
S
n
n n1 1
, Where
A
() denote the binomial coefficient
!
1
...
1
Then
x
n
M means that
M nn
A
S
1
II.18.PROPOSITION
Let
0
be a number which is not an integer.Then
M
M
where
denotes the sequence (0,0,……..,0)Proof:Since
x
n
(
M)
We have
M nn
A
S
1 .
This is equivalent to
S
n
M .This, in turn, is equivalent to the assertion that
11
nn n
z
S
z
f
is orlicz space of an gai function.Now
