The Rate of Sectional entire sequence spaces Key Words: Rate space ...

(3s.) v. 30 2 (2012): 63–69. ISSN-00378712 in press doi:10.5269/bspm.v30i2.14304

Bol. Soc. Paran. Mat. c

SPM –ISSN-2175-1188 on line SPM: www.spm.uem.br/bspm

The Rate of Sectional entire sequence spaces N. Gurumoorthy and K. Chandrasekhara Rao

abstract: The space (Γs )π is introduced. This paper deveoted a study of the general properties of sectional rate space (Γs )π of Γ.

Key Words: Rate space, entire sequences, analytic sequences, β dual.. Contents 1 Introduction

63

2 Definitions and Preliminaries

64

3 Main Results

65 1. Introduction

A complex sequence, whose k th terms xk is denoted by {xk } or simply x. Let φ be the set of all finite sequence. A sequence x = {xk } is said to be analytic if 1/k supk |xk | < ∞. A vector space of all analytic sequence will be denoted by Λ. 1/k A sequence x = {xk } is called entire if limk→∞ |xk | = 0. The vector space of entire sequence denoted by Γ. Kizmaz [22] defined the following difference squence spaces Z (∆) = {x = (xk ) : ∆x ∈ Z} ∞

∞

for Z = ℓ∞ , c, c0 , where ∆x = (∆x)k=1 = (xk − xk+1 )k=1 and showed that these are Banach space with norm kxk = |x1 | + k∆xk∞ . Later on Et and Colak [23] generalized the notion as follows : Let m ∈ N, Z (∆m ) = {x = (xk ) : ∆m x ∈ Z} for Z = ℓ∞ , c, c0 where m ∈ N,
∞ ∞ ∆0 x = (xk ) , ∆x = (xk − xk+1 ) , ∆m x = (∆m xk )k=1 = ∆m−1 xk − ∆m−1 xk+1 k=1 . The generalized differences has the following binomial representation:   Pm γ m xk+γ , ∆m xk = γ=0 (−1) γ They proved that these are Banach spaces with the norm kxk∆ =

Pm

i=1

|xi | + k∆m xk∞

2000 Mathematics Subject Classification: 46A45

63

Typeset by BSM P style. c Soc. Paran. de Mat.

64

N. Gurumoorthy and K. Chandrasekhara Rao

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 sk = {0, 0, 0, · · · , 1, −1, 0, 0, · · · } , 1 in the nth place and -1 in the (n + 1)th place and zero’s else where. If X is a sequence space, we define ′ (i)X = the continuous dual of X. P∞ (ii)X α = {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 (iv)X γ = 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. 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, ...) are continuous. We recall the following definitions[see [14]]. An FK-space is a locally convex Frechet space which is made up of sequences and has the property that coodinate 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 [17]]. 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]. 2. Definitions and Preliminaries Throughout the paper w, Γ and Λ denote the spaces of all entire and bounded 1/k sequence respectively. Let t denote the sequence with t = |xk | for all k ∈ N. Define the sets: Γ = {x ∈ w : tk → 0 as k → ∞} and Λ = {x ∈ w : supk tk < ∞} . The spaces Γ and Λ are metric space with the metric n o 1/k d (x, y) = supk |xk − yk | : k = 1, 2, 3, · · · o n   Let (Γs )π = x = (xk ) ∈ w : ξ = πξkk ∈ Γ ; where ξk = x1 + x2 + · · · + xk . Let n   o (Λs )π = x = (xk ) ∈ w : η = πηkk ∈ Λ ; where ηk = y1 +y2 +· · ·+yk . The spaces (Γs )π and (Λs )π are metric space with the metric   ξk −ηk 1/k d (x, y) = supk πk : k = 1, 2, 3, · · · Let σ (Γ) denote the vector space of all sequences x = (xk ) such that



ξk k



is entire

sequence. We recall that cs0 denotes the vector space of all sequences x = (xk ) such that (ξk ) is a null sequence. Lemma 2.1 (see ( [17], Theorem 7.2.7)) 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 .

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The Rate of Sectional entire sequence spaces

1/k   → 0 as k → ∞ ⇔ Remark 2.2 x = (xk ) ∈ σ (Γs ) ⇔ πξkkk ∈ Γ ⇔ πξkkk 1/k ξk → 0 as k → ∞, because k 1/k → 1 as k → ∞ ⇔ x = (xk ) ∈ (Γs )π . Hence πk (Γs )π = σ (Γs ) , the ces` aro space of order 1.

3. Main Results Proposition 3.1 (Γs )π ⊂ Γπ Proof: Let x ∈ (Γs )π ⇒ ξ ∈ Γπ 1/k ξk → 0 as k → ∞ πk

(3.1)

. But πxk = πξk − πξk−1 k−1 1/kk k 1/k 1/k xk ξk ξk−1 1/k ≤ πξkk − πξk−1 ≤ + → 0 as k → ∞ by using (3.1) πk πk πk−1 k−1 1/k Therefore πxkk → 0 as k → ∞ ⇒ x ∈ Γπ . Hence (Γs )π ⊂ Γπ . (1) Note:   The above inclusions is strict. Take the sequence δ ∈ Γπ . We have ξ1 =1  π1  ξ2 =1+0=1  π2  ξ3 π3 = 1 + 0 + 0 = 1 .. .  ξk = 1 + 0 + 0 + ···+ = 1 πk → k − terms   → ξk 1/k 1/k and so on. Now πk = 1 for all k. Hence πξkk does not tend to zero as

k → ∞. So δ (1) ∈ / (Γs )π . Thus the inclusion (Γs )π ⊂ Γπ is strict. This completes the proof. 2 Proposition 3.2 (Γs )π is a linear space over the field C of complex numbers. Proof: It is easy. Therefore omit the proof.

2

Proposition 3.3 Λπ ⊂ (Γs )π ⊂ Λπ (∆) β

β

Proof: Step 1. By Proposition 3.1, we have (Γs )π ⊂ Γπ . Hence (Γπ ) ⊂ [(Γs )π ] . β But (Γπ ) = Λπ . Therefore β Λπ ⊂ [(Γs )π ] (3.2)

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N. Gurumoorthy and K. Chandrasekhara Rao

P∞ β Step2: Let y = (yk ) ∈ [(Γs )π ] . Consider f (x) = k=1 xk yk with x ∈ (Γs )π . Take x = δ n − δ n+1 = (0, 0, 0, · · · , π, −π, 0, 0, · · · ) nth (n + 1)th place th where, for each fixed n = 1, 2, 3, · · · ; δ (n)
= (0, 0, · · · π, 0, · · · ) , π in the n place n n+1 and zero’s elsewhere. Then f δ − δ = yn − yn+1 . Hence


yn yn+1 n n+1 n ≤ kf k d δ − δ n+1 , 0 ≤ kf k · 1. πn − πn+1 = f δ − δ   yn yn+1 yn ∈ − is bounded. Consequently So, πynn − πyn+1 πn πn+1 ∈ Λπ That is πn n+1 β

Λπ (∆) . But y = (yn ) is originally in [(Γs )π ] . Therefore β

(3.3)

[(Γs )π ] ⊂ Λπ (∆) β

From (3.2) and (3.3) we conclude that Λπ ⊂ [(Γs )π ] ⊂ Λπ (∆) . This completes the proof. 2 Proposition 3.4 The β− dual space of (Γs )π is Λπ β

Proof: Step1. Let y = (yk ) be an arbitray point in [(Γs )π ] . If y is not in Λπ , then for each natural number n, we can find an index k(n) such that   yk(n) 1/k(n) > n1 , (n = 1, 2, 3, · · · ) πk(n)

Define x = (xk ) by   xk πk

= 1/nk , f or k = k(n); and



xk πk



= 0 otherwise

Then x is in Γπ , but for infinitely many k,   xk yk >1 πk Consider the sequence z = {zk }, where z1 = xk πk

x1 − π1  xk πk

(3.4) s with s =

P xk

πk ;

and zk =

(k = 2, 3, · · · ) Then z is a point of Γπ . Also = 0 Hence z is in (Γs )π . But P  zk xk  does not converge. Thus the sequence y would not by the equation(3.4), πk β

to be in [(Γs )π ] . This contradiction proves that β

[(Γs )π ] ⊂ Λπ

(3.5)

Step2. By (3.2) of Proposition 3.3, we have Λπ ⊂ [(Γs )π ]

β

(3.6)

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The Rate of Sectional entire sequence spaces

From (3.5) and (3.6) it follows that the β− dual space of [(Γs )π ] completes the proof. Proposition 3.5 Suppose that x ∈ Γπ , then x ∈ [(Γs )π ] ⇔ Proof: Let

∞ X xk

k=1

πk

P∞

xk k=1 πk

=0

β

is Λπ This 2

=0

(3.7)

Note that x ∈ Γπ implies that the  sum  on the left hand side of (3.7) must exists. Denote the sum by s. Clearly πǫkk → s as k → ∞. Hence it follows that if s 6= 0 then x ∈ / Γπ . Assume that s = 0 then 

ǫk πk



=−

 ∞  X xν πν

(3.8)

γ=k+1

 1/ν < ǫ for all ν ≥ k0 Let ǫ > 0 be given with ǫ < 1/2. There is a k0 such that πxνν P k+1 ∞ equation (3.8) gives us the results for k ≥ k0 . Now πǫkk < ν=k+1 ǫν = ǫ1−ǫ < ǫk , 1/k since ǫ < 1/2. Thus πǫkk < ǫ. Consequently x ∈ [(Γs )π ] . This completes the proof. 2 Proposition 3.6 [(Γs )π ] is closed in Γπ in the metric toplogy of Γπ Proof: If x ∈ Γπ − [(Γ P s )π ] then s = 0. Let ǫ > 0 be given. If d (x, y) < ǫ, then ∞ xk P∞ yk P∞ xk yk ǫ k ≤ − − k=1 πk k=1 πk k=1 πk πk ≤ ǫ = 1−ǫ . This can be made arbitrary small by the choice of ǫ. Thus however d (x, y) is sufficiently small, we have P∞ y k k=1 πk 6= 0.

So that by Theorem 3.5 y ∈ / [(Γs )π ] . Accordingly y ∈ Γπ − [(Γs )π ] for each y in the open ball B (x, ǫ) with centre x and radius ǫ. Hence x ∈ B (x, ǫ) ⊂ Γπ − [(Γs )π ] . Therefore each point x of Γπ − [(Γs )π ] is an interior point Γπ − [(Γs )π ] . In other words Γπ −[(Γs )π ] is open in Γ. Consequently [(Γs )π ] is closed in Γπ . This completes the proof. 2 Proposition 3.7 [(Γs )π ] is a complete metric space Proof: The metric d for Γπ is given by d (x, y) . The metric d for Γπ may be restricted to [(Γs )π ] then [(Γs )π , d] is a metric space. Where

68

N. Gurumoorthy and K. Chandrasekhara Rao

  xk −yk 1/k : k = 1, 2, 3, · · · d (x, y) = supk πk

But by Theorem (3.5), [(Γs )π , d] is a closed subspace of Γπ . Also Γπ is a complete 2 metric space [see [1]]. Hence (Γs )π is a complete metric space. Proposition 3.8 [(Γs )π , d] is separable. Proof: Write δ 1 = {π, 0, 0 · · · , } ; δ 2 = {0, π, 0, · · · , } ; .. . δ k = {0, 0, 0 · · · , π} ; π in the k th place and 0 elsewhere. Where π is the constant sequence {π, π,· · · , } with π > 0. Then A = σ 1 , σ 2 , · · · is a Schauder basis for Γπ correspondingly we have, 1 ρ = {π, −π, 0, · · · , } ; ρ2 = {0, π, −π, 0, · · · , } ; .. . th ρk = {0, 0, 0 · · · , π, −π} ; π in the k th place and −π in the (k + 1) place and 1 2 3 zero elsewhere. Then B = ρ , ρ , ρ , · · · , is Schauder basis in [(Γs )π , d] . Hence [(Γs )π , d] is separable. This completes the proof. 2 References 1. K.Chandrasekhara Rao, Functional Analysis Alpha Science , Oxford, (2006). 2. P.K.Kamthan and M.Gupta, Sequences and Series Maneel Dekker, New York. 3. A.Wilansky, Functional Analysis Blaisdell, New York, (1964).
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N. Gurumoorthy Research Scholar, Department of Mathematics Srinivasa Ramanujan Centre, SASTRA University E-mail address: [email protected] and K. Chandrasekhara Rao Department of Mathematics Srinivasa Ramanujan Centre, SASTRA University E-mail address: [email protected]