Cyclicity of Weighted Composition Operators on Some BK Space

International Journal of Pure and Applied Mathematics Volume 99 No. 2 2015, 201-204 AP ISSN: 1311-8080 (printed version); ISSN: 1314-3395 (on-line version) url: http://www.ijpam.eu doi: http://dx.doi.org/10.12732/ijpam.v99i2.7 ijpam.eu CYCLICITY OF WEIGHTED COMPOSITION OPERATORS ON SOME BK SPACE B. Yousefi1 § , S.M.A. Musavi2 1,2 Department of Mathematics Payame Noor University P.O. Box 19395-3697, Tehran, IRAN Abstract: We will investigate the cyclicity for the adjoint of a weighted composition operator acting on (lˆp (α))∗ . AMS Subject Classification: 47B37, 46A25 Key Words: AK space, BK space, weighted composition operator, multipliers, cyclic vector, bounded point evaluation 1. Introduction We write ω for the set of all complex sequences x = (xk )∞ k=0 . Let φ, l∞ and c0 denote the set of all finite, bounded and null sequences. We write lp = {x ∈ ω : ∞ X |xk |p < ∞} k=0 (n) for 1 ≤ p < ∞. By e(n) (n ∈ N0 ), we denote the sequence with en = 1 and Pn (n) (k) [n] = ek = 0 whenever k 6= n. For any sequence x = (xk )∞ k=0 xk e k=0 , let x be its n-section. Given any subset F of ω, we write F̂ for the set of all formal P k ∞ f power series fˆ with fˆ(z) = ∞ k=0 k z where f = (fk )k=0 ∈ F , regardless of whether or not the series converges for any value of z. Let M̂z : F̂ → ω̂ be P k+1 . defined by (M̂z fˆ) = ∞ f k=0 k z Received: November 13, 2014 § Correspondence author c 2015 Academic Publications, Ltd. url: www.acadpubl.eu 202 B. Yousefi, S.M.A. Musavi 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 if every sequence f = (fk )∞ k=0 ∈ F has a unique representation f = P∞ AK (k) [n] k=0 fk e , that is f = limn→∞ f ; it is said to have AD, if φ is dense in F . Given any subset F of ω, the set β F = {a ∈ ω : ∞ X ak f k converges for all f ∈ F} k=0 is called the β-dual of F. Let F be a normed sequence space and F̂ be the space of formal power series with coefficients in F endowed with the norm of F . Then F and F̂ 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{An x : n = 0, 1, 2, . . .}. ∞ Consider f = {fk }∞ k=0 and g = {gk }k=0 in ω and let E ⊂ ω. Define ∞ f g = {fk gk }k=0 and g−1 ⋆ E = {f ∈ ω : f g ∈ E}. If α = {αk }∞ k=0 ∈ ω is a given sequence with αk 6= 0 for all k, then by 1/α we −1 ⋆ F ) for any subset F of ω. From mean 1/α = {1/αk }∞ k=0 . Write F̂ (α) = (α ∞ now on we suppose that α = {αk }k=0 ∈ ω satisfying α0 = 1 and αk 6= 0 for all k ≥ 1. Note that the space ˆlp (α) is a reflexive Banach space and the dual of ˆlp (α) is ˆlq (α−1 ). If λ is a complex number, then e(λ) denotes the functional of evaluation at λ, defined on ˆlp (α) by e(λ)(fˆ) = fˆ(λ). A complex valued function ϕ on Ω for which ϕfˆ ∈ F̂ for every fˆ ∈ F̂ is called a multiplier of F̂ and the collection of all these multipliers is denoted by M(F̂ ). For some sources on sequence spaces, see [1–6]. 2. Main Results In this section we will investigate the cyclicity of the adjoint of weighted composition operators acting on ˆlp (α)∗ . By U we mean the open unit disc. Lemma 1. A complex number λ is a bounded point evaluation on ˆlp (α) if −1 and only if {λn }∞ n=0 ∈ lq (α ). Proof. Note that λ is a bounded point evaluation on ˆlp (α) if and only if the functional e(λ) is bounded on ˆlp (α). But the dual of ˆlp (α) is ˆlq (α−1 ) and we CYCLICITY OF WEIGHTED COMPOSITION... 203 can see that e(λ)((e(k)ˆ)) = (e(k)ˆ)(λ) = λk for all integers k ≥ 0. This completes the proof. Theorem 2. Let each point of U is a bounded point evaluation on ˆlp (α). Then a polynomial p̂ is cyclic for M̂z if and only if p̂ vanishes at no point in U. Proof. Let p̂(z) = (z − λ1 )...(z − λm ) be such that λi ∈ / U for i = 1, ..., m. Fix k ∈ {1, ..., m} and consider Mk ∈ ˆlq (α−1 ) satisfying Mk (M̂z )n (z − λk )) = 0 for all integers n ≥ 0. So there exists h ∈ lq (α−1 ) such that Mk fˆ =< fˆ, ĥ > for all f ∈ lp (α). Note that Mk (M̂z )n (z − λk )) = Mk (z n+1 − λk z n ) = hn+1 − λk hn for all integers n ≥ 0. Since Mk (M̂z )n (z − λk )) = 0, we get hn+1 = λk hn and (k) / lq (α−1 ) and h ∈ lq (α−1 ), hence h0 for all n ≥ 0. But {λnk }n ∈ so hn+1 = λn+1 k hn = 0 for all n and so Mk = 0. Thus z − λk is cyclic for k = 1, ..., m and so p̂(z) is a cyclic vector for M̂z . The converse case is clear. Theorem 3. Suppose that 1 < p < ∞, w ∈ M(lp (α)), M(ˆlp (α)) = H ∞ and ϕ is an analytic self-map of the open unit disc U satisfying ||ϕ||U < 1. ∞ P 1 1 Also, let α−q n < ∞ where p + q = 1. If there exists z0 ∈ U satisfying n=0 ŵ(ϕk (z0 )) 6= 0 for all k ≥ 0 and if the set {ϕk (z0 ) : k ≥ 0} has limit point in U , then e(z0 ) is a cyclic vector for the operator (Mw Cϕ )∗ acting on ˆlq (α−1 ). Proof. By the property ∞ P α−q n < ∞, each point of U is a bounded point n=0 evaluation and the space ˆlp (α) consists of functions analytic in the open unit disc U . Let the map L : M(ˆlp (α)) → B(ˆlp (α)) be given by L(ψ̂) = M̂ψ̂ . We prove that L is continuous. For this we use the closed graph theorem. Suppose ψ̂n converges to ψ̂ in M(ˆlp (α)) and L(ψ̂n ) = M̂ψ̂n converges to A in B(ˆlp (α)). Then for each f in lp (α), Afˆ = lim M̂ψˆn fˆ = lim ψˆn fˆ. n n Thus {ψˆn fˆ}n is convergent in ˆlp (α). Now by the continuity of point evaluations ψˆn fˆ converges pointwise to ψ̂ fˆ on U . So Afˆ is analytic and agree with ψ̂ fˆ on U . Hence Afˆ = ψ̂ fˆ and A = M̂ψ̂ . Therefore L is continuouse and there is a constant c such that ||M̂ψ̂ || ≤ c||ψ̂||U for all ψ̂ in M(F̂ ). But ||ψ̂|| ≤ ||M̂ψ̂ || for all ψ̂ in M(F̂ ). Thus ||ψ̂|| ≤ c||ψ̂||U for all ψ ∈ M(lp (α)). Since ϕ ∈ H ∞ and 204 B. Yousefi, S.M.A. Musavi M(ˆlp (α)) = H ∞ , we will use ϕ̂ instead of ϕ. Let f ∈ lp (α), then Cφ̂ fˆ = fˆoϕ̂ ∈ H ∞ since ||ϕ̂||U < 1. So ||fˆoϕ̂|| ≤ c||fˆoϕ̂||U ≤ c||fˆ||U , because ϕ̂(U ) ⊆ U . On the otherhand, note that for all f in lp (α), ||fˆ||U ≤ γ||fˆ|| ∞ P ˆ ˆ where γ = α−q n . Now we get ||Cϕ̂ f || ≤ cγ||f || which implies that Cϕ̂ and n=0 so M̂ŵ Cϕ̂ is bounded. Now, put A = M̂ŵ Cϕ̂ . To complete the proof we show that if for all k ≥ 0, < ĝ, (A∗ )k e(z0 ) >= 0, then ĝ should be the zero constant function. For this note that ! k−1 Y ∗ k < ĝ, (A ) e(z0 ) >= ŵ(ϕ̂i (z0 )) ĝ ◦ ϕ̂k (z0 ). i=0 By the assumptions, clearly we get ĝ ◦ ϕ̂k (z0 ) = 0 for all k ≥ 0. Since {ϕ̂k (z0 ) : k ≥ 0} has limit point in U , it should be ĝ = 0. Thus, e(z0 ) is a cyclic vector for the operator (M̂ŵ Cϕ̂ )∗ acting on ˆlq (α−1 ). This completes the proof. References [1] L. Bagheri and B. Yousefi, Reflexivity of the shift operator on some BK spaces, Rendiconti Del Circolo Matematico Di Palermo, Volume 2013, DOI 10.1007/s12215-013-0143-5 (2013). [2] M. Mursaleen and A. K. Noman, On some new difference sequence spaces of non-absolute type, Math. Comput. Modelling, 52 (2010), 603-617. [3] E. Malkowsky, Linear operators in certain BK spaces, Bolyai Society Mathematical Studies, 5 (1996), 259-273. [4] A. Wilansky, Summability through functional analysis, Mathematics Studies 85, North-Holland, 1984. [5] B. Yousefi, On the space ℓp (β), Rendiconti Del Circolo Matematico Di Palermo, Serie II, Tomo XLIX (2000) 115-120. [6] B. Yousefi, Unicellularity of the multiplication operators on Banach spaces of formal power series, Studia Math., 147 (2001) 201-209.