Int. Journal of Math. Analysis, Vol. 4, 2010, no. 30, 1465 - 1468

Conditions for Reflexivity on

Some Sequence Spaces

F. Ershad, B. Yousefi and M. Habibi

Department of Mathematics, Payame-Noor UniversityShahrake Golestan, P.O. Box: 71955-1368, Shiraz, Iran

bahmann@spnu.ac.irershad@spnu.ac.ir

mezbanhabibi@yahoo.com

Abstract

In this paper we will give sufficient conditions for the multiplictionoperator Mz to be reflexive on the weighted Hardy spaces.

Mathematics Subject Classification: 47B37; 47L10

Keywords: Banach space of Laurent series associated with a sequence β,bounded point evaluation, reflexive operator

1. Introduction

Let {β(n)}∞n=−∞ be a sequence of positive numbers satisfying β(0) = 1.If 1 < p < ∞, the space Lp(β) consists of all Laurent power series f(z) =

∞∑n=−∞

f(n)zn such that the norm

‖f‖p = ‖f‖pβ =∞∑

n=−∞|f(n)|pβ(n)p

is finite. These are reflexive Banach spaces with the norm ‖ · ‖β. Let fk(n) =δk(n). So fk(z) = zk and then {fk}k is a basis for Lp(β) such that ‖fk‖ = β(k).We denote the set of multipliers {ϕ ∈ Lp(β) : ϕLp(β) ⊆ Lp(β)} by Lp∞(β)and the linear operator of multiplication by ϕ on Lp(β) by Mϕ.

We say that a complex number λ is a bounded point evaluation on Lp(β)if the functional e(λ) : Lp(β) −→ C defined by e(λ)(f) = f(λ) is bounded.

1466 F. Ershad, B. Yousefi and M. Habibi

It is well known that Lp(β)∗ = Lq(βpq ), where 1

p+ 1

q= 1. Also if f(z) =

∑n

f(n)zn ∈ Lp(β) and g(z) =∑n

g(n)zn ∈ Lq(βpq ), then clearly

< f, g >=∑

n

f(n)g(n)β(n)p

where the notation < f, g > stands for g(f). For a source in formal series, werefer the reader to [1– 12].

Recall that if E is a separable Banach space and A ∈ B(E), then Lat(A)is by definition the lattice of all invariant subspaces of A, and AlgLat(A) isthe algebra of all operators B in B(E) such that Lat(A) ⊂ Lat(B). For thealgebra B(E), the weak operator topology is the one induced by the family ofseminorms px∗,x(A) = | < Ax, x∗ > | where x ∈ E, x∗ ∈ E∗ and A ∈ B(E).Hence Aα −→ A in the weak operator topology if and only if Aαx −→ Axweakly. Also similarly Aα −→ A in the strong operator topology if and onlyif Aαx −→ Ax in the norm topology. An operator A in B(E) is said to bereflexive if AlgLat(A) = W (A), where W (A) is the smallest subalgebra ofB(E) that contains A and the identity I and is closed in the weak operatortopology.

2. Main Result

In this section we will investigate the reflexivity of the operator Mz actingon Lp(β).

From now on we suppose that Ω1 �= ∅ and Mz is invertible on Lp(β). Also,we will use the following notations:

r01 = limβ(−n)−1n , Ω01 = {z ∈ C : |z| > r01}

r11 = limβ(n)1n , Ω11 = {z ∈ C : |z| < r11}

Ω1 = Ω01 ∩ Ω11

Theorem. If Lp(β) = Lp∞(β), then Mz is reflexive on Lp(β).Proof. First note that since Lp(β) = Lp∞(β), |f(λ)| ≤ ‖f‖ for all λ ∈ Ω1,

so ‖f‖Ω1 ≤ ‖f‖ where the second norm is that of Lp(β).Put M = H∞(Ω11) ∩ Lp∞(β). Then M �= ∅, since 1 ∈ M. To prove

that M is closed, choose a sequence {fn} in M such that fn converges tof in Lp(β). Then ‖fn‖Ω1 ≤ c for a constant c. By the continuity of pointevaluations, fn(λ) → f(λ) for every λ ∈ Ω1. Because ‖fn‖Ω11 = ‖fn‖Ω1 ≤ c([4]), the sequence {fn} is a normal family and by passing to a subsequence, ifnecessary, we may assume that fn → g uniformly on compact subsets of Ω11.Therefore g is bounded and analytic on Ω11. From the pointwise convergence

Conditions for Reflexivity on Some Sequence Spaces 1467

on Ω1 of fn → f we conclude that f = g ∈ H∞(Ω11). Thus indeed f ∈ M,since Lp(β) = Lp∞(β), so M is closed.

Now we show that L : (M, ‖.‖Ω11) −→ B(Lp(β)) be given by L(ϕ) = Mϕ

is continuous. Suppose that the sequence {ϕn}n converges to ϕ in M andL(ϕn) = Mϕn converges to A in B(Lp(β)). Then for each f in Lp(β),

Af = limnMϕnf = lim

nϕnf

and so {ϕnf}n is convergent in Lp(β). Note that by the continuity of pointevaluations ϕnf converges pointwise to ϕf . Thus Af is analytic and agreewith ϕf on Ω1. Hence A = Mϕ and so L is continuous. This implies thatthere is a constant d > 0 such that ‖Mϕ‖ ≤ d‖ϕ‖Ω1 for all ϕ in M.

Let A ∈ AlgLat(Mz). Then A = Mψ for some ψ ∈ Lp∞(β). Now clearlyM ∈ Lat(Mz), thus AM ⊂ M. Since 1 ∈ M we get A1 = ϕ ∈ M =H∞(Ω11). But Ω11 is a Caratheodory domain and so there is a sequence {pn}nof polynomials converging to ψ such that for all n, ‖pn‖Ω11 ≤ e for some e > 0.So we obtain

‖Mpn‖ ≤ d‖pn‖Ω11 ≤ de

for all n. Since Lp(β) is reflexive, the unit ball of Lp(β) is weakly compact.Therefore ball B(Lp(β)) is compact in the weak operator topology and so bypassing to a subsequence if necessary, we may assume that for some X ∈B(Lp(β)), Mpn −→ X in the weak operator topology. Using the fact thatM∗

pn−→ X∗ in the weak operator topology and by acting these operators on

e(λ) we obtain that

pn(λ)e(λ) = M∗pne(λ) −→ X∗e(λ)

weakly. Since pn(λ) −→ ψ(λ) we see that X∗e(λ) = ψ(λ)e(λ). Because theclosed linear span of {e(λ) : λ ∈ Ω1} is dense in Lq(β), where 1

p+ 1

q= 1, we

conclude that X = Mψ = A. This implies that A ∈ W (Mz) and so Mz isreflexive. This completes the proof. �

Acknowledgment. This research was partially supported by a grantfrom Research Council of Shiraz Payame Noor University and the authorsgratefully acknowledge this support.

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1468 F. Ershad, B. Yousefi and M. Habibi

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Received: January, 2010