eu CYCLICITY OF WEIGHTED COMPOSITION OPERATORS ON SOME BK SPACE

We will investigate the cyclicity for the adjoint of a weighted composition operator acting on (l̂p(α)) . AMS Subject Classification: 47B37, 46A25


Introduction
We write ω for the set of all complex sequences x = (x k ) ∞ k=0 .Let φ, l ∞ and c 0 denote the set of all finite, bounded and null sequences.We write . By e (n) (n ∈ N 0 ), we denote the sequence with e x k e (k) be its n-section.Given any subset F of ω, we write F for the set of all formal power series f with f (z) = ∞ k=0 f k z k where f = (f k ) ∞ k=0 ∈ F , regardless of whether or not the series converges for any value of z.Let Mz : F → ω be defined by A BK space is a Banach sequence space with the property that convergence implies coordinatewise convergence.A BK space F containing φ is said to have , that is f = lim n→∞ f [n] ; it is said to have AD, if φ is dense in F .Given any subset F of ω, the set Let F be a normed sequence space and F be the space of formal power series with coefficients in F endowed with the norm of F .Then F and F are norm isomorphic.
We say that a vector x in a Banach space X is a cyclic vector of a bounded operator A on X if X = span{A n x : n = 0, 1, 2, . ..}.
for any subset F of ω.From now on we suppose that α = {α k } ∞ k=0 ∈ ω satisfying α 0 = 1 and α k = 0 for all k ≥ 1.Note that the space lp (α) is a reflexive Banach space and the dual of lp (α) is lq (α −1 ).
A complex valued function ϕ on Ω for which ϕ f ∈ F for every f ∈ F is called a multiplier of F and the collection of all these multipliers is denoted by M( F ).For some sources on sequence spaces, see [1][2][3][4][5][6].

Main Results
In this section we will investigate the cyclicity of the adjoint of weighted composition operators acting on lp (α) * .By U we mean the open unit disc.Lemma 1.A complex number λ is a bounded point evaluation on lp (α) if and only if {λ n } ∞ n=0 ∈ l q (α −1 ).
Proof.Note that λ is a bounded point evaluation on lp (α) if and only if the functional e(λ) is bounded on lp (α).But the dual of lp (α) is lq (α −1 ) and we can see that e(λ)((e (k)) ) = (e (k)) (λ) = λ k for all integers k ≥ 0. This completes the proof.
Theorem 2. Let each point of U is a bounded point evaluation on lp (α).Then a polynomial p is cyclic for Mz if and only if p vanishes at no point in U.
If there exists z 0 ∈ U satisfying ŵ(ϕ k (z 0 )) = 0 for all k ≥ 0 and if the set {ϕ k (z 0 ) : k ≥ 0} has limit point in U , then e(z 0 ) is a cyclic vector for the operator (M w C ϕ ) * acting on lq (α −1 ).Then for each f in l p (α),

Proof. By the property
Thus { ψn f } n is convergent in lp (α).Now by the continuity of point evaluations ψn f converges pointwise to ψ f on U .So A f is analytic and agree with ψ f on U .Hence A f = ψ f and A = M ψ .Therefore L is continuouse and there is a By the assumptions, clearly we get ĝ • φk (z 0 ) = 0 for all k ≥ 0. Since { φk (z 0 ) : k ≥ 0} has limit point in U , it should be ĝ = 0. Thus, e(z 0 ) is a cyclic vector for the operator ( M ŵ C φ) * acting on lq (α −1 ).This completes the proof.
k = n.For any sequence x = (x k ) ∞ k=0 , let x [n] = n k=0 each point of U is a bounded point evaluation and the space lp (α) consists of functions analytic in the open unit disc U .Let the map L : M( lp (α)) → B( lp (α)) be given by L( ψ) = M ψ .We prove that L is continuous.For this we use the closed graph theorem.Suppose ψn converges to ψ in M( lp (α)) and L( ψn ) = M ψn converges to A in B( lp (α)).