CYCLICITY ON SOME BK SPACES

In this paper, we give some sufficient conditions for cyclicity of adjoint of the multiplication operator acting on a space of formal power series with coefficients in some BK spaces with BK property. AMS Subject Classification: 47B37, 47A25


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 l p = {x ∈ ω : ∞ k=0 |x k | p < ∞} for 1 ≤ p < ∞.By e (n) (n ∈ N 0 ), we denote the sequence with e (n) n = 1 and e (n) k = 0 whenever k = n.For any sequence x = (x k ) ∞ k=0 , let x [n] = n k=0 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 k=0 ∈ F , regardless of whether or not the series converges for any value of z.Let Mz : F → ω be defined by f k e (k) , 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, . ..}.Here span{•} is the closed linear span of the set {•}.It is convenient and helpful to introduce the notation < x, x * > to stand for x * (x), for x ∈ X and x * ∈ X * .
In this paper, we study the cyclicity of adjoint of the multiplication operator acting on spaces of formal power series with coefficients in some BK spaces with AK property.For some sources on sequence spaces, see [1][2][3][4][5].

Main Results
In the main theorem of this paper we give some sufficient conditions for cyclicity of the adjoint of weighted composition operators on BK spaces of formal power series. Consider ∈ ω is a given sequence with α k = 0 for all k, then by 1/α we mean 1/α = {1/α k } ∞ k=0 .Write F (α) = (α −1 ⋆ F ) 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.
Lemma 2.1.Let 1 < p < ∞ and 1 p + 1 q = 1.Then (l p (α)) * = l q (α −1 ) where α Proof.Since l p is a BK space with AK property, thus l p (α) is also a Bk and AK space with respect to the norm f lp(α) = αf lp for all f ∈ l p (α).Furthermore, l * p and l β p are isomorphic.Thus we get (α Note that the space lp (α) is a reflexive Banach space and the dual of lp (α) is lq (α −1 ).Here for simplicity we use ||ĝ|| instead of ||ĝ|| lq(α −1 ) .Theorem 2.2.Let the sequence α = {α n } satisfy α n ≥ 1 for all n and f = {f n } ∞ n=0 be a vector in the Banach space l q (α −1 ) with infinitely many for all m ≥ 0, then f is a cyclic vector of M * z acting on lq (α −1 ).
Now by condition of the theorem lim is in M .Thus whenever f n+1 = 0, we have z + ∞ k=n+2 (f k /f n+1 )z k−n ∈ M .Now by induction we prove that z m ∈ M for all m: let 1, z, . . ., z m−1 ∈ M and note that For f m+n = 0, we have By letting n → ∞, we conclude that z m ∈ M .Thus indeed z m ∈ M for all m, hence M = l q (α −1 ) and so f is a cyclic vector of f is a cyclic vector of M * z .
space is a Banach sequence space with the property that convergence implies coordinatewise convergence.A BK space F containing φ is said to have AK if every sequence f = (f k ) ∞ url: www.acadpubl.eu§ Correspondence author ∞ k=0