Paper1-Acta26-2011. 144KB Jun 04 2011 12:03:15 AM

Acta Universitatis Apulensis
ISSN: 1582-5329

No. 26/2011
pp. 9-20

THE ORLICZ SPACE OF χπ

N. Subramanian, S.Krishnamoorthy, S. Balasubramanian
Abstract. In this paper we introduced the Orlicz space of χπ . We establish
some inclusion relations, topological results and we characterize the duals of the
Orlicz of χπ sequence spaces.
2000 Mathematics Subject Classification: 40A05,40C05,40D05.
1. Introduction
A complex sequence, whose k th terms is xk is denoted by {xk } or simply x. Let
w be the set of all sequences x = (xk ) and φ be the set of all finite sequences. Let
ℓ∞ , c, c0 be the sequence spaces of bounded, convergent and null sequences x = (xk )
respectively. In respect of ℓ∞ , c, c0 we have
sup

kxk = k |xk | , where x = (xk ) ∈ c0 ⊂ c ⊂ ℓ∞ . A sequence x = {xk } is said

to be analytic if supk |xk |1/k < ∞. The vector space of all analytic sequences will
be denoted by Λ. A sequence x is called entire sequence if limk→∞ |xk |1/k = 0.
The vector space of all entire sequences will be denoted by Γ.χ was discussed in
Kamthan [19]. Matrix transformation involving χ were characterized by Sridhar
[20] and Sirajiudeen [21]. Let χ be the set of all those sequences x = (xk ) such that
(k! |xk |)1/k → 0 as k → ∞. Then χ is a metric space with the metric
n
o
d (x, y) = supk (k! |xk − yk |)1/k : k = 1, 2, 3, · · ·

Orlicz [4] used the idea of Orlicz function to construct the space (LM ). Lindenstrauss and Tzafriri [5] investigated Orlicz sequence spaces in more detail, and
they proved that every Orlicz sequence space ℓM contains a subspace isomorphic
to ℓp (1 ≤ p < ∞). Subsequently different classes of sequence spaces defined by
Parashar and Choudhary[6], Mursaleen et al.[7], Bektas and Altin[8], Tripathy et
al.[9], Rao and subramanian[10] and many others. The Orlicz sequence spaces are
the special cases of Orlicz spaces studied in Ref[11].
9

N. Subramanian, S. Krishnamoorthy, S. Balasubramanian - The Orlicz space of χπ

Recall([4],[11]) an Orlicz function is a function M : [0, ∞) → [0, ∞) which is
continuous, non-decreasing and convex with M (0) = 0, M (x) > 0, for x > 0 and
M (x) → ∞ as x → ∞. If 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[12] and Maddox[13] 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 ∆2 − condition is
equivalent to M (ℓu) ≤ KℓM (u), for all values of u and for ℓ > 1. Lindenstrauss and
Tzafriri[5] used the idea of Orlicz function to construct Orlicz sequence space
(
)


∞
X
|xk |
M
< ∞, f orsome ρ > 0 .
(1)
ℓM = x ∈ w :

ρ
k=1

The space ℓM with the norm
(
kxk = inf

ρ>0:

∞
X
k=1

M



|xk |
ρ



≤1

)

(2)

becomes a Banach space which is called an Orlicz sequence space. For M (t) =
tp , 1 ≤ p < ∞, the space ℓM coincide with the classical sequence space ℓp · Given
a sequence x = {xk } its nth section is the sequence x(n) = {x1 , x2 , ..., xn , 0, 0, ...}
δ (n) = (0, 0, ..., 1, 0, 0, ...) , 1 in the nth place and zero’s else where; and s(k) =
(0, 0, ..., 1, −1, 0, ...) , 1 in the nth place,-1 in the (n + 1)th place and zero’s else where.
An FK-space (Frechet coordinate space) is a Frechet space which is made up of
numerical sequences and has the property that the coordinate functionals pk (x) =
xk (k = 1, 2, 3, . . .) are continuous. We recall the following definitions [see [15]].
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 x(n) , x → 0 as

n → ∞.[see[15]] The space is said to have AD (or) be an AD space if φ is dense in
X. We note that AK implies AD by [14].
If X is a sequence space, we define
′
(i)X = the continuous
dual of X.
P∞
α
(ii)X = {a = (ak ) : Pk=1 |ak xk | < ∞, f oreach x ∈ X} ;
(iii)X β = n
{a = (ak ) : ∞
k=1 ak xk is convergent, f oreach x ∈
o X} ;
sup Pn
γ
(iv)X = a = (ak ) : n | k=1 ak xk | < ∞, f oreach x ∈ X ;
n
o
′
(v)Let X be an FK-space⊃ φ. Then X f = f (δ (n) ) : f ∈ X .

X α , X β , X γ are called the α−(or K¨
o the-T o¨eplitz)dual of X, β− (or generalized
K¨
o the-T o¨eplitz)dual of X, γ−dual of X. Note that X α ⊂ X β ⊂ X γ . If X ⊂ Y
then Y µ ⊂ X µ , for µ = α, β, or γ.
10

N. Subramanian, S. Krishnamoorthy, S. Balasubramanian - The Orlicz space of χπ

Lemma 1.1. (See (15, T heorem7.27)) . Let X be an FK-space ⊃ φ. Then
(i) X γ ⊂ X f .
(ii) If X has AK, X β = X f .
(iii) If X has AD, X β = X γ .
2.Definitions and Prelimiaries
Let w denote the set of all complex double sequences x = (xk )∞
k=1 and M :
[0, ∞) → [0, ∞) be an Orlicz function, or a modulus function. Let
!!
!
(

(k! |xk |)1/k
π
χM = x ∈ w : limk→∞ M
= 0 f orsome ρ > 0 ,
1/k
πk ρ
(
!!
!
|xk |1/k
π
ΓM = x ∈ w : limk→∞ M
= 0 f orsome ρ > 0
1/k
πk ρ
 



|xk |1/k

π
and ΛM = x ∈ w : supk M
< ∞ f orsome ρ > 0
1/k
πk

ρ

The space χπM is a metric space with the metric
!!
)
(
(k! |xk − yk |)1/k
≤1
d (x, y) = inf ρ > 0 : supk M
1/k
πk ρ

(3)

The space ΓπM and ΛπM is a metric space with the metric
(
!!
)
|xk − yk |1/k
d (x, y) = inf ρ > 0 : supk M
≤1
1/k
πk ρ

(4)

3.Main Results
Proposition 3.1.

χπM

⊂

ΓπM ,

with the hypothesis that M



|xk |
1/k

πk

Proof. Let x ∈ χπM . Then we have the following implications
!
(k! |xk |)1/k
M
→ 0 as k → ∞
1/k
πk ρ

11

ρ



≤M



(k!|xk |)1/k
1/k

πk

ρ

(5)



N. Subramanian, S. Krishnamoorthy, S. Balasubramanian - The Orlicz space of χπ


|xk |





(k!|xk |)1/k



But M
≤M
, by our assumption, implies that
1/k
1/k
πk ρ
 πk ρ
|1/k
⇒ M |xk1/k
→ 0 as k → ∞, by (5).
πk

ρ

⇒ x ∈ ΓπM
⇒ χπM ⊂ ΓπM . This completes the proof.

Proposition 3.2. χπM has AK whereM is
function.
 a modulus
1/k
k |)
∈ χ, and hence
Proof. Let x = {xk } ∈ χπM , but then M (k!|x1/k
πk

(k! |xk |)1/k

supk≥n+1 M

d

x, x[n]




= supk≥n+1



(k!|xk |)1/k
1/k

πk

ρ

1/k

πk ρ


!

ρ

→ 0 as n → ∞.

(6)

→ 0 as n → ∞, by using (6)

⇒ x[n] → x as n → ∞, implying that χπM has AK. This completes the proof.
Proposition 3.3. χπM is solid.
π
|yk | and let
Proof. Let|xk | ≤ 
 y = (yk ) ∈ χM .
1/k
1/k
k |)
k |)
M (k!|x1/k
≤ M (k!|y1/k
, because M is non-decreasing.
πk ρ
πk ρ


1/k
k |)
But M (k!|y1/k
∈ χ, because y ∈ χπM .
πk ρ




(k!|yk |)1/k
(k!|xk |)1/k
That is M
→ 0 as k → ∞ and M
→ 0 as k → ∞. Therefore
1/k
1/k

x = {xk } ∈

πk ρ
π
χM . This

πk

ρ

completes the proof.

Proposition 3.4. Let M be an Orlicz function which satisfies ∆2 − condition.
Then χ ⊂ χπM .
Proof.Let
x∈χ

(7)

1/k
Then
large k and every ǫ > 0. But then by taking ρ ≥
 (k! |xk |) ≤ ǫ sufficiently

1/k
k |)
M (k!|x1/k
≤ M ρǫ ≤ M (2ǫ)(because M is non-decreasing)
πk

1
2

ρ

(k! |xk |)1/k

!

≤ KM (ǫ) by∆2 − condition, f or some K > 0 ≤ ǫ
(8)
1/k
πk ρ


(k!|xk |)1/k
⇒M
→ 0 as k → ∞ (by defining M (ǫ) < kǫ ). Hence x ∈ χπM . From (7)
1/k
M

πk

ρ

12

N. Subramanian, S. Krishnamoorthy, S. Balasubramanian - The Orlicz space of χπ

and since
x ∈ χπM

(9)

we get x ⊂ χπM . This completes the proof.
Proposition 3.5. If M is a modulus function, then χπM is linear set over the
set of complex numbers C
Proof. Let x, y ∈ χπM and α, β ∈ C. In order to prove the result we need to find
some ρ3 such that
!
(k! |αxk + βyk |)1/k
→ 0 as k → ∞.
(10)
M
1/k
πk ρ3
Since x, y ∈ χπM , there exists some positive ρ1 and ρ2 such that
!
!
(k! |xk |)1/k
(k! |yk |)1/k
M
→ 0 as k → ∞ and M
→ 0 as k → ∞.
1/k
1/k
πk ρ
πk ρ

(11)

Since

 modulus function, we have 

 M is a non decreasing
1/k
(k!|αxk |)1/k
|α|(k!|xk |)1/k
(k!|βyk |)1/k
|β|(k!|yk |)1/k
(k!|αxk +βyk |)
≤M
≤M
M
+
+
1/k
1/k
1/k
1/k
1/k
π k ρ3
π k ρ3
πk ρ3
n π k ρ3
o π k ρ3
1 1
1 1
Take ρ3 such that ρ13 = min |α|
ρ1 , |β| ρ2 . Then




1/k
1/k
+βyk |)1/k
k |)
k |)
≤ M (k!|x1/k
→ 0 (by(11)).
+ (k!|y1/k
M (k!|αxk1/k
π k ρ3
πk ρ2

 πk ρ 1
+βyk |)1/k
Hence M (k!|αxk1/k
→ 0 as k → ∞. So (αx + βy) ∈ χπM . Therefore χπM is
πk

ρ3

linear. This completes the proof.

Definition 3.6.
(pk ) 
be any sequence
 of positive real
 numbers. Then we
 Let p = 
1/k
(k!|x
|)
k
→ 0 as k → ∞ . Suppose that pk is
define χπM (p) = x = (xk ) : M
1/k
a constant for all k, then

χπM

(p) =

πk ρ
χπM .

Proposition 3.7. Let 0 ≤ pk ≤ qk and let
π
χM (p) .

n

qk
pk

o

be bounded. Then χπM (q) ⊂

Proof. Let

M

x ∈ χπM (q)
!!qk
(k! |xk |)1/k
→ 0 as k → ∞.
1/k
πk ρ
13

(12)
(13)

N. Subramanian, S. Krishnamoorthy, S. Balasubramanian - The Orlicz space of χπ

Let tk =



M



(k!|xk |)1/k
1/k

πk

ρ

0 < λ < λk . Define

qk

pk
qk .

and λk =

Since pk ≤ qk , we have 0 ≤ λk ≤ 1. Take

(
0 (tk ≥ 1)
and vk =
tk , (tk < 1)

(
tk , (tk ≥ 1)
uk =
0, (tk < 1)

(14)

tk = uk + vk ; tλk k = uλk k + vkλk . Now it follows that uλk k ≤ uk ≤ tk and vkλk ≤ vkλ . Since
tλk k = uλk k + vkλk , then tλk k ≤ tk + vkλ


M

⇒







(k!|xk |)1/k

M

1/k

πk





ρ

qk λk

(k!|xk |)1/k
1/k

πk

ρ
1/k

≤



M

qk pk /qk
pk



≤



(k!|xk |)1/k
1/k

πk



M




ρ

qk

(k!|xk |)1/k
1/k

πk
1/k

ρ

qk

qk

k |)
k |)
M (k!|x1/k
≤ M (k!|x1/k
πk ρ
πk ρ
qk
 
(k!|xk |)1/k
But M
→ 0 as k → ∞. (by (13))
1/k
πk ρ
 
pk
1/k
k |)
Therefore M (k!|x1/k
→ 0 as k → ∞. Hence

⇒

πk

ρ

x ∈ χπM (p)

(15)

From (12) and (15) we get χπM (q) ⊂ χπM (p) . This completes the proof.
Proposition 3.8. (a) Let 0 < inf pk ≤ pk ≤ 1. Then χπM (p) ⊂ χπM
(b) Let 1 ≤ pk ≤ suppk < ∞. Then χπM ⊂ χπM (p)
Proof. (a)Let x ∈ χπM (p)
M

(k! |xk |)1/k
1/k

πk ρ

!!pk

→ 0 as k → ∞.

(16)

Since 0 < inf pk ≤ pk ≤ 1
M

(k! |xk |)1/k
1/k

πk ρ

!!

≤

M

(k! |xk |)1/k
1/k

πk ρ

!!pk

(17)

From (16) and (17) it follows that x ∈ χπM .
Thus χπM (p) ⊂ χπM . We have thus proven (a). (b) Let pk ≥ 1 for each k and
14

N. Subramanian, S. Krishnamoorthy, S. Balasubramanian - The Orlicz space of χπ

suppk < ∞
Let x ∈ χπM

!!

(k! |xk |)1/k

M

1/k

πk ρ

→ 0 as k → ∞.

(18)

Since 1 ≤ pk ≤ suppk < ∞ we have
M

(k! |xk |)1/k
1/k

πk ρ


M



(k!|xk |)1/k
1/k

πk

ρ

pk

!!pk

≤

(k! |xk |)1/k

M

1/k

πk ρ

!!

(19)

→ 0 as k → ∞. by using (18).

Therefore x ∈ χπM (p) . This completes the proof.
Proposition 3.9. Let 0 < pk ≤ qk < ∞ for each k. Then χπM (p) ⊆ χπM (q).
Proof. Let x ∈ χπM (p)
(k! |xk |)1/k

!!pk

→ 0 as k → ∞.
1/k
πk ρ

 
(k!|xk |)1/k
≤ 1 for sufficiently large k.
This implies that M
1/k
M

πk

(20)

ρ

Since M is non-decreasing, we get
M

(k! |xk |)1/k
1/k

πk ρ
⇒



M



(k!|xk |)1/k
1/k

πk

ρ

qk

!!qk

≤

M

(k! |xk |)1/k
1/k

πk ρ

!!pk

(21)

→ 0 as k → ∞. (by using (20)).

x ∈ χπM (q)
Hence χπM (p) ⊆ χπM (q) . This completes the proof.
Proposition 3.10. χπM (p) is a r− convex for all r where 0 ≤ r ≤ inf pk .
Moreover if pk = p ≤ 1∀k, then they are p− convex.
Proof. We shall prove the Theorem for χπM (p) .
Let x ∈ χπM (p) and r ∈ (0, limn→∞ pn )
Then, there exists k0 such that r ≤ pk ∀k > k0 .
Now, define
(
!r
!pn )
(k! |xk − yk |)1/k
(k! |xk − yk |)1/k
∗
g (x) = inf ρ : M
+M
1/k
1/k
πk ρ
πk ρ
15

(22)

N. Subramanian, S. Krishnamoorthy, S. Balasubramanian - The Orlicz space of χπ

Since r ≤ pk ≤ 1∀k > k0
g ∗ is subadditive: Further, for 0 ≤ |λ| ≤ 1; |λ|pk ≤ |λ|r ∀k > k0 .
g ∗ (λx) ≤ |λ|r · g ∗ (x)

(23)

U = {x : g ∗ (x) ≤ δ} , which is an absolutely r − convex set, f or

(24)

|λ|r + |µ|r ≤ 1 x, y ∈ U

(25)

Now, for 0 < δ < 1,

Now
g ∗ (λx + µy) ≤ g ∗ (λx) + g ∗ (µy)
≤ |λ|r g ∗ (x) + |µ|r g ∗ (y)
≤ |λ|r δ + |µ|r δ using (23) and (24)
≤ (|λ|r + |µ|r ) δ
≤ 1 · δ, by using (25)
≤ δ. If pk = p ≤ 1∀k then for 0 < r < 1,
U = {x : g ∗ (x) ≤ δ} is an absolutely p− convex set.
This can be obtained by a similar analysis and there fore we omit the details. This
completes the proof.
Proposition 3.11. (χπM )β = Λ
Proof: Step 1: χπM ⊂ ΓπM by Proposition 3.1;
⇒ (ΓπM )β ⊂ (χπM )β . But (ΓπM )β = Λ
Λ ⊂ (χπM )β
Step 2: Let y ∈ (χπM )β we have f (x) =
π

1/k

P∞

k=1 xk yk

(26)
with x ∈ χπM .

th place and zero’s elsewhere, with
k
We recallthats(k) has 
k! in the k
n
 1/k 
o
1/k
k |)
x = s(k) , M (k!|x1/k
= 0, 0, · · · M (1)ρ
, 0, · · ·
πk ρ


which converges to zero. Hence s(k) ∈ χπM . Hence d s(k) , 0 = 1.


But |yk | ≤ kf k d s(k) , 0 < ∞∀k. Thus (yk ) is a bounded sequence and hence an
analytic sequence. In other words y ∈ Λ.

(χπM )β ⊂ Λ
Step 3 From (26) and (27) we obtain (χπM )β = Λ.
This completes the proof.
16

(27)

N. Subramanian, S. Krishnamoorthy, S. Balasubramanian - The Orlicz space of χπ
Proposition 3.12. (χπM )µ = Λ for µ = α, β, γ, f .
Proof. Step 1: χπM has AK by Proposition 3.2. Hence by Lemma 1.1 (i) we get
(χπM )β = (χπM )f . But (χπM )β = Λ. Hence
(χπM )f = Λ

(28)

Step 2: Since AK ⇒ AD. Hence by Lemma 1.1 (iii) we get (χπM )β = (χπM )γ .
Therefore
(29)
(χπM )γ = Λ
Step 3: χπM is normal by Proposition 3.3 Hence by Proposition 2.7 [16]. we get
(χπM )α = (χπM )γ = Λ.

(30)

From (28), (29) and 930) we have (χπM )α = (χπM )β = (χπM )γ = (χπM )f = Λ.
Proposition 3.13. The dual space of χπM is Λ. In other words (χπM )∗ = Λ.
π

1/k

Proof.Werecall that
s(k) has kk! has the k th place zero’s else where, with
n
 1/k 
o
1/k
k |)
= 0, 0, · · · M (1)ρ
x = s(k) , M (k!|x1/k
, 0, · · ·
πk ρ
P
(k)
π
π
π ∗
Hence s ∈ χM . We have f (x) = ∞
k=1 xk yk with x ∈ χM and f ∈ (χM ) , where
∗
π
π
(k)
π
(χM ) is the dual space of χM . Take x = s ∈ χM . Then


|yk | ≤ kf k d s(k) , 0 < ∞ f or all k.
(31)
Thus (yk ) is a bounded sequence and hence an analytic sequence. In other words,
y ∈ Λ. Therefore (χπM )∗ = Λ. This completes the proof.

Lemma 3.14. [15, Theorem 8.6.1] Y ⊃ X ⇔ Y f ⊂ X f where X is an AD-space
and Y an FK-space.
Proposition 3.15. Let Y be any FK-space⊃ φ. Then Y ⊃ χπM if and only if
the sequence s(k) is weakly analytic.
Proof. The following implications establish the result.
Y ⊃ X ⇔ Y f ⊂ (χπM )f , since χπM has AD and by Lemma 3.14.
⇔ Y f ⊂ Λ, since (χπM )f = Λ.


′
⇔ f or each f ∈ Y , the topological dual of Y . Therefore f s(k) ∈ Λ.


⇔ f s(k) is analytic
⇔ s(k) is weakly analytic.
This completes the proof.
17

N. Subramanian, S. Krishnamoorthy, S. Balasubramanian - The Orlicz space of χπ
π
Proposition3.16.
metric space
 χM is a complete

 under the metric
(k!|xk −yk |)1/k
d (x, y) = supk M
: k = 1, 2, 3, · · · where x = (xk ) ∈ χπM and
1/k
πk

y = (yk ) χπM .

ρ



Proof. Let x(n) be a Cauchy sequence in χπM .
Then given
any ǫ > 0 there exists a positive integer N depending on ǫ such that
d x(n) , x(m) < ǫ for all n ≥ N and for all m ≥ N. Hence
(
 1/k !)
 

supk

 (n) (m) 
k!xk −xk 

M

1/k

πk

Consequently

M

< ǫ for all n ≥ N and for all m ≥ N.

ρ
!!

 
 (n)  1/k
k!xk 
1/k

πk

ρ

is a Cauchy sequence in the metric space C of

complex numbers. But C is complete. So,
  
  
 (n)  1/k

  k! xk 
 → M
M 
1/k
πk ρ

(k! |xk |)1/k
1/k

πk ρ

!!

as n → ∞.

Hence there exists a positive integer n0 such that
(
 1/k !)
 
supk
(

M

M

1/k

πk



(k!|xk |)1/k
1/k

πk

∞. Thus

< ǫ for all n ≥ n0 . In particular, we have

1/k

πk ρ
 1/k !)
 

 (n)
k!xk −xk 

M




 (n)
k!xk −xk 

ρ

ρ



≤

< ǫ. Now

(

M
(

M

!)

 
(n )  1/k

k!xk −xk 0 
1/k

πk

+ M

ρ

(k! |xk |)1/k
1/k

πk ρ

(

!)

!)

 
 (n )  1/k
k!xk 0 
1/k

πk

ρ

< ǫ+0 as k →

< ǫ as k → ∞.

That is x ∈ χπM . Therefore, χπM is a complete metric space. This completes the
proof.

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N. Subramanian, S. Krishnamoorthy, S. Balasubramanian - The Orlicz space of χπ

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N. Subramanian, S. Krishnamoorthy, S. Balasubramanian - The Orlicz space of χπ

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N. Subramanian
Department of Mathematics, SASTRA University,
Thanjavur-613 401, India.
email: nsmaths@yahoo.com

S. Krishnamoorthy
Department of Mathematics, Govenment Arts College(Autonomus),
Kumbakonam-612 001, India.
email: drsk 01@yahoo.com
S. Balasubramanian
Department of Mathematics, Govenment Arts College(Autonomus),
Kumbakonam-612 001, India.
email: sbalasubramanian2009@yahoo.com

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