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.
2000Mathematics Subject Classification: 40A05,40C05,40D05. 1. Introduction
A complex sequence, whosekth terms is xk is denoted by {xk} or simplyx. Let
w be the set of all sequences x = (xk) and φbe 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
kxk =supk |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 χbe the set of all those sequencesx= (xk) such that
(k!|xk|)1/k →0 as k→ ∞.Then χis a metric space with the metric
d(x, y) =supk n
(k!|xk−yk|)1/k:k= 1,2,3,· · · o
Orlicz [4] used the idea of Orlicz function to construct the space (LM). Linden-strauss and Tzafriri [5] investigated Orlicz sequence spaces in more detail, and they proved that every Orlicz sequence space ℓM contains a subspace isomorphic
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)→ ∞asx→ ∞.If convexity of Orlicz functionM is replaced byM(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 functionM is said to satisfy ∆2−condition for all values ofu,if there exists a constant K >0,such thatM(2u)≤KM(u)(u≥0).The ∆2−condition is equivalent toM(ℓu)≤KℓM(u),for all values ofu and forℓ >1.Lindenstrauss and Tzafriri[5] used the idea of Orlicz function to construct Orlicz sequence space
ℓM = (
x∈w:
∞ X
k=1 M
|xk|
ρ
<∞, f orsome ρ >0
)
. (1)
The spaceℓM 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 thenth place,-1 in the (n+ 1)thplace 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 ifd x(n), x
→0 as n→ ∞.[see[15]] The space is said to have AD (or) be an AD space ifφis dense in X. We note that AK implies AD by [14].
IfX is a sequence space, we define (i)X′ = the continuous dual ofX.
(ii)Xα ={a= (ak) :P∞k=1|akxk|<∞, f oreach x∈X};
(iii)Xβ ={a= (a
k) :P∞k=1akxkis convergent, f oreach x∈X};
(iv)Xγ =na= (ak) : sup
n |Pnk=1akxk|<∞, f oreach x∈X o
; (v)LetX be an FK-space⊃φ. ThenXf =nf(δ(n)) :f ∈X′o
.
Lemma 1.1. (See(15, T heorem7.27)).Let X be an FK-space ⊃φ. Then
(i)Xγ ⊂Xf.
(ii) IfX has AK,Xβ =Xf. (iii) IfX has AD,Xβ =Xγ.
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
χπM =
(
x∈w:limk→∞ M
(k!|xk|)1/k
πk1/kρ
!!
= 0f orsome ρ >0
!
,
ΓπM =
(
x∈w:limk→∞ M
|xk|1/k
π1k/kρ
!!
= 0f orsome ρ >0
!
and ΛπM =
x∈w:supk
M
|xk|1/k
π1k/kρ
<∞f orsome ρ >0
The spaceχπM is a metric space with the metric
d(x, y) =inf
(
ρ >0 :supk M
(k!|xk−yk|)1/k
π1k/kρ
!!
≤1
)
(3)
The space Γπ
M and ΛπM is a metric space with the metric
d(x, y) =inf
(
ρ >0 :supk M |xk−yk| 1/k
π1k/kρ
!!
≤1
)
(4)
3.Main Results
Proposition 3.1. χπM ⊂ΓπM,with the hypothesis that M
|xk|
πk1/kρ
≤M
(k!|xk|)1/k
π1k/kρ
Proof. Let x∈χπM.Then we have the following implications
M (k!|xk|) 1/k
π1k/kρ
!
But M
,by our assumption, implies that ⇒M
Proposition 3.2. χπM has AK where M is a modulus function. Proof. Let x={xk} ∈χπM,but then Proposition 3.3. χπM is solid.
Proof. Let |xk| ≤ |yk|and lety= (yk)∈χπM.
,because M is non-decreasing. But M
Proposition 3.4. Let M be an Orlicz function which satisfies ∆2− condition. Then χ⊂χπM.
≤M(2ǫ)(becauseM is non-decreasing)
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. Letx, y∈χπM and α, β∈C.In order to prove the result we need to find
Since M is a non decreasing modulus function, we have M linear. This completes the proof.
Definition 3.6. Let p= (pk) be any sequence of positive real numbers. Then we
suppk<∞
Let x∈χπM
M (k!|xk|) 1/k
πk1/kρ
!!
→0as k→ ∞. (18)
Since 1≤pk ≤suppk <∞ we have
M (k!|xk|) 1/k
π1k/kρ
!!pk
≤ M (k!|xk|) 1/k
π1k/kρ
!!
(19)
M
(k!|xk|)1/k
π1k/kρ
pk
→0as 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)
M (k!|xk|) 1/k
πk1/kρ
!!pk
→0as k→ ∞. (20)
This implies that
M
(k!|xk|)1/k
π1k/kρ
≤1 for sufficiently large k. Since M is non-decreasing, we get
M (k!|xk|) 1/k
πk1/kρ
!!qk
≤ M (k!|xk|) 1/k
πk1/kρ
!!pk
(21)
⇒
M
(k!|xk|)1/k
π1k/kρ
qk
→0as 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 forχπM(p). Let x∈χπM(p) and r∈(0, limn→∞pn)
Then, there exists k0 such thatr ≤pk∀k > k0. Now, define
g∗(x) =inf
(
ρ:M (k!|xk−yk|) 1/k
πk1/kρ
!r
+M (k!|xk−yk|) 1/k
πk1/kρ
!pn)
Since r ≤pk≤1∀k > k0
g∗ is subadditive: Further, for 0≤ |λ| ≤1;|λ|pk ≤ |λ|r∀k > k
0.
g∗(λx)≤ |λ|r·g∗(x) (23) Now, for 0< δ <1,
U ={x:g∗(x)≤δ}, which is an absolutely r−convex set, f or (24) |λ|r+|µ|r≤1x, y∈U (25) Now
g∗(λx+µy)≤g∗(λx) +g∗(µy) ≤ |λ|rg∗(x) +|µ|rg∗(y)
≤ |λ|rδ+|µ|rδ using (23) and (24) ≤(|λ|r+|µ|r)δ
≤1·δ,by using (25)
≤δ.If pk=p≤1∀kthen for 0< r <1,
U ={x:g∗(x)≤δ} is an absolutelyp−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; ⇒(Γπ
M)β ⊂(χπM)β.But (ΓπM)β = Λ
Λ⊂(χπM)β (26)
Step 2: Let y∈(χπM)β we havef(x) =P∞
k=1xkyk withx∈χπM.
We recall that s(k) has π 1/k k
k! in the kth place and zero’s elsewhere, with x=s(k),
M
(k!|xk|)1/k
πk1/kρ
=n0,0,· · ·M(1)ρ1/k,0,· · ·o which converges to zero. Hence s(k)∈χπM.Henced s(k),0
= 1. But |yk| ≤ kfkd s(k),0
< ∞∀k. Thus (yk) is a bounded sequence and hence an
analytic sequence. In other words y∈Λ.
(χπM)β ⊂Λ (27)
Step 3From (26) and (27) we obtain (χπ M)
β = Λ.
Proposition 3.12. (χπM)µ= Λfor µ=α, β, γ, f.
Proof. Step 1: χπM has AK by Proposition 3.2. Hence by Lemma 1.1 (i) we get (χπ
M)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 (χπ M)
β = (χπ M)
γ.
Therefore
(χπM)γ = Λ (29)
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 (χπ
M)
α = (χπ M)
β = (χπ M)
γ= (χπ M)
f = Λ.
Proposition 3.13. The dual space ofχπM is Λ.In other words (χπ M)
∗ = Λ.
Proof. We recall that s(k) has π 1/k k
k! has the kth place zero’s else where, with x=s(k),
M
(k!|xk|)1/k
πk1/kρ
=n0,0,· · ·M(1)ρ1/k,0,· · ·o
Hence s(k) ∈χπM.We havef(x) =P∞k=1xkyk with x∈χπM and f ∈(χπM)∗,where (χπM)∗ is the dual space of χπM.Take x=s(k)∈χπM.Then
|yk| ≤ kfkd
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 ⇔Yf ⊂Xf whereX 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 ⇔Yf ⊂(χπ
M)
f,sinceχπ
M has AD and by Lemma 3.14.
⇔Yf ⊂Λ,since (χπM)f = Λ.
⇔ f or each f ∈Y′,the topological dual ofY. Thereforef s(k) ∈Λ. ⇔f s(k) is analytic
Proposition 3.16. χπM is a complete metric space under the metric
Then given any ǫ > 0 there exists a positive integer N depending on ǫ such that d x(n), x(m)
is a Cauchy sequence in the metric space C of
complex numbers. ButC is complete. So,
Hence there exists a positive integern0 such that
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csT
d1, Communications of the Korean Mathematical Society, in press. N. Subramanian
Department of Mathematics, SASTRA University, Thanjavur-613 401, India.
email: [email protected]
S. Krishnamoorthy
Department of Mathematics, Govenment Arts College(Autonomus), Kumbakonam-612 001, India.
email: drsk [email protected]
S. Balasubramanian
Department of Mathematics, Govenment Arts College(Autonomus), Kumbakonam-612 001, India.
