Open Access

On the eleventh question of Allen Shields

Fixed Point Theory and Applications20112011:16

https://doi.org/10.1186/1687-1812-2011-16

Received: 28 March 2011

Accepted: 20 July 2011

Published: 20 July 2011

Abstract

In this article, we will give sufficient conditions for the boundedness of the analytic projection on the set of multipliers of the formal Laurent series spaces. This answers a question that has been raised by A. L. Shields. Also, we will characterize the fixed points of some weighted composition operators acting on weighted Hardy spaces.

AMS Subject Classification: Primary 47B37; Secondary 47B38.

Keywords

Banach space of Laurent series associated with a sequence β bounded point evaluation; spectral set weak operator topology

1 Introduction

Let be a sequence of positive numbers satisfying β(0) = 1. If 1 < p < ∞, then the space L p (β) consists of all formal Laurent series such that the norm is finite. When n just runs over {0}, the space L p (β) only contains formal power series , and it is usually denoted by H p (β). These spaces are also called as weighted Hardy spaces. If p = 2, such spaces were introduced by Allen L. Shields to study weighted shift operators in his article [1] which is one of basic studies in this area, and is a pretty large study that contains a number of interesting results, and indeed it is mainly of auxiliary nature. Actually, Shields showed a close relation between injective weighted shifts and the multiplication operator M z acting on L2(β) or H2(β) (see [[1], Proposition 7]). These are reflexive Banach spaces with the norm ||·|| β . Let . Hence, f k (z) = z k and {f k }kis a basis for L p (β) such that ||f k || = β(k). Clearly M z , the operator of multiplication by z on L p (β), shifts the basis {f k } k . The operator M z is bounded, if and only if {β(k + 1)/β (k)} k is bounded, and in this case for all n {0}.

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

Let X be a Banach space. It is convenient and helpful to introduce the notation < x, x* > to stand for x*(x), for x X and x* X*. Also, the set of bounded linear operators on X is denoted by B(X). If A B(X), then by σ(A) we mean the spectrum of A and by r(A) we mean the spectral radius of A.

By the same method used in [2], we can see that where . Also, if and , then clearly . For some sources on these topics, we refer to [114].

If Ω is a domain in the complex plane , then by H(Ω) and H(Ω) we mean the set of analytic functions and the set of bounded analytic functions on Ω, respectively. By ||·||Ω, we denote the supremum norm on Ω.

We denote the set of multipliers {φ L p (β) : φL p (β) L p (β)} by and the linear transformation of multiplication by φ on L p (β) by M φ . The space is a commutative Banach algebra with the norm ||φ|| = ||M φ ||.

By an analytic projection we mean the map which send each two-sided sequence into the corresponding one-sided sequence [1].

It is known that the analytic projection for the unweighted shift is unbounded ([[15], Prob. 9, Chapter 14]). In this paper, we want to investigate conditions under which the analytic projection on to be bounded. Also, we investigate the fixed points of some weighted composition operators acting on weighted Hardy spaces.

2 Main results

First we note that the multiplication operator M z on L p (β) is unitarily equivalent to an injective bilateral weighted shift and conversely, every injective bilateral weighted shift is unitarily equivalent to M z acting on L p (β) for a suitable choice of β. Throughout this article, M z is a bounded operator on L p (β).

Let be in where

Define the analytic projection by J(φ) = φ1. Then, the projection J is clearly bounded since ||J(φ)|| p = ||φ1|| p ≤ ||φ|| p ≤ ||φ||. The problem of boundedness raised when we consider the projection J from into . It is certainly bounded if M z is not invertible on L p (β), since in this case we will see that for in , it should be for all n < 0. However, this projection is not necessarily a bounded operator on . For example, as stated in [1], it is not bounded when β(n) = 1 for all n in (see also [[16], Chapter VII, equations (2.2) and (2.3)]). We want to investigate those weighted Hardy spaces that admit the analytic projection as a bounded linear operator on . This answers the following question that has been considered by Shields in [[1], p. 91, Question 11]:

Question. For which bilateral shifts is the analytic projection a bounded operator on ?

We will use the following notations:

If r01< r11, then by the same method used for the formal power series in [2], we can see that each point of Ω1 is a bounded point evaluation on L p (β).

Note that for the algebra B(X) of all the bounded operators on a Banach space X, the weak operator topology is the one such that A α A in the weak operator topology if and only if A α xAx weakly, x X.

For the proof of the main theorems, we need the following lemmas.

Lemma 1. If there exists a constant c > 0 such that for all Laurent polynomials p, then .

Proof. Let be an arbitrary element in and define
Indeed, the functions P n (φ) are the averages of the partial sums of φ. Now, clearly J(P n (φ)) = P n (J(φ)) and we can also see that . But P n (φ) is a Laurent polynomial; thus by the assumption, we have
for all n {0}. Hence, is a norm-bounded sequence in B(L p (β)). Now, since the unit ball of B(L p (β)) is compact in the weak operator topology, we may assume, by passing to a subsequence if necessary, that for some operator A in the weak operator topology. Put Q n = P n (J(φ)). Then, clearly we have . Thus, we get
where f0 and f k are considered, respectively, as elements of L p (β) and . Hence, we obtain

for all j and k (here f j L p (β) and . However, is bounded, and in the weak operator topology, and so A = MJ(φ)and indeed . Now, by applying the closed graph theorem and the continuity of the coefficient functionals on , we conclude that . □

Lemma 2. Let J1 = I - J and for all Laurent polynomials p, then .

Proof. By the same method used in the proof of Lemma 1, we can see that and this implies that J is also a bounded operator from into . Hence, the proof is complete. □

Theorem 3. Let M z be invertible on L p (β), r22< r33 and let for some d > 0, for all Laurent polynomials p. Then, .

Proof. First, note that since M z is invertible, then , and hence, the inequality r22< r33 implies that ||M z || σ(M z ). Thus, by the Cauchy integral formula we have
for all polynomials p. Therefore,
However, t → ||(e it ||M z || - M z )-1|| is continuous on [0, 2π] and so it should be constant, say c1. Thus, for all polynomials p we get
where c = c1||M z ||. Now, let q be a Laurent polynomial. Then,
where

We have . On the other hand, since q H3), by using the Lemma in [[1], page 81], we get where . By the assumption , thus . Therefore, ||MJ(q)|| ≤ cdc0||M q || for all Laurent polynomials q and so by Lemma 1 the proof is complete. □

Theorem 4. Let M z be invertible on L p (β), r23< r12, and let for some d > 0, the relation holds for all Laurent polynomials q. Then, .

Proof. It is sufficient to prove that J1 = I - J is a bounded operator on . Note that since r23< r12, thus . Now by using the relation (*) with replacing z by and M z by we get for all polynomials q in z-1. Now, let p be a Laurent polynomial. Then,
where
Hence, . Since p H3), by using the Lemma in [[1], p. 81], we get where . By the assumption we obtain

for all Laurent polynomials p. Now by Lemma 2, and so the proof is complete. □

Proposition 5. If M z is not invertible on L p (β), then .

Proof. Let . Since M z is not invertible, . Now by the relation (**) in the proof of Theorem 6, we have

Hence, for k = -1 we get for all . Multiplying by z we have , etc. Thus, for all n ≤ -1 and so J is identity, i.e., J(φ)(z) = φ(z) for all and so the proof is complete. □

Definition 6. Let X be a Banach space. A compact subset K of the plane is called a spectral set of A B(X), if it contains the spectrum of A, σ(A), and ||f(A)|| ≤ max{|f(z)| : z K} for all rational functions f with poles off K.

From now on suppose that M z is invertible on L p (β) and Ω2 is nonempty.

Theorem 7. If σ(M z ) is a spectral set for M z , then the analytic projection is a bounded operator on .

Proof. Suppose that denotes the analytic projection. Note that . Since σ(M z ) is a spectral set, we have for all polynomials p. Also, since Ω2 is nonempty, we have r12< r22. Now, let q(z) = q1(z) + q2(z) be a Laurent polynomial such that and . Thus, we have where (see the Lemma in [[1], p. 81]). Therefore,
But and each point of Ω2 is a bounded point evaluation on . Indeed, if , then the relation implies that
By taking k = n - m, we get for all m. Therefore,

and so where by the root test the series converges on Ω2. This implies that and each points of Ω2 is a bounded point evaluation on . Also, Ω2 is the largest open annulus such that (see [2] for the case of formal power series). Since is a commutative Banach algebra and e λ is multiplicative, it should be ||e λ || = 1 for all λ Ω2, and this implies that for all ψ in . Therefore, ||MJ(q)|| ≤ cc1||M q || for all Laurent polynomials q. So the proof is complete. □

Corollary 8. If for all polynomials p, then the analytic projection is a bounded operator on .

Proof. In the proof of Theorem 7, we only used the inequality in the definition of the spectral set for polynomials instead of rational functions. Hence, Theorem 7 is also consistent if we substitute the statement " for all polynomials p in z", instead of the statement "σ(M z ) be a spectral set for M z ". This completes the proof. □

Corollary 9. If for all polynomials q in z-1, then the analytic projection is a bounded operator on .

Proof. Put J1 = I - J where denotes the analytic projection. So for a Laurent polynomial
where and , we have J1(p) = p2. Now by the Lemma in [[1], p. 81], we have where . Thus, by this hypothesis, we get

Now, the result follows from Lemma 2. □

Now we consider the special case when p = 2.

Theorem 10. Let M z be invertible on L2(β). If 0 < r12 = r23< r22 = r33, then .

Proof. Note that for the Hilbert space L2(β), the Von Neumann's inequality holds and since r23< r33, there exists a number c > 0 such that for all Laurent polynomials p (see Proposition 23 in [[1], p. 82]). Also, note that Ω2 = Ω3.

Let
where
Thus, . Since p H3), we have where . On the other hand, since r12< r22, as we saw earlier, and for all φ in . Therefore, we get

Now by Lemma 1, the proof is complete. □

Studying the fixed points of weighted composition operators entails a study of the iterate behavior of holomorphic self-maps. The holomorphic self maps of the open unit disc U are divided into classes of elliptic and non-elliptic. The elliptic type is an automorphism and has a fixed point in U. The maps that are not elliptic are called of non-elliptic type. The iterate of a non-elliptic map can be characterized by the Denjoy-Wolff Iteration Theorem [3]. By ψ n we denote the nth iterate of ψ and by ψ' (w) we denote the angular derivative of ψ at w ∂U. Note that if w U, then ψ' (w) has the natural meaning of derivative.

Denjoy-Wolff Iteration Theorem. Suppose ψ is a holomorphic self-map of U that is not an elliptic automorphism.

(i) If ψ has a fixed point w U, then ψ n w uniformly on compact subsets of U, and '(w)| < 1.

(ii) If ψ has no fixed point in U, then there is a point w ∂U such that ψ n w uniformly on compact subsets of U, and the angular derivative of ψ exists at w, with 0 < ψ' (w) ≤ 1.

We call the unique attracting point w, the Denjoy-Wolff point of ψ.

Suppose that ψ is a holomorphic self-map of U such that the composition operator C ψ acts boundedly on a Banach space of formal power series H p (β), and φ belongs to , the set of multipliers of H p (β). Then, the weighted composition operator C ψ,φ acting on H p (β) is defined by C φ,ψ = M φ C ψ . Note that if , then H p (β) H(U), and each point of U is a bounded point evaluation on H p (β).

Theorem 11. Suppose that ψ is a holomorphic self-map of U such that ||ψ|| U < 1, and the composition operator C ψ acts boundedly on a Banach space of formal power series H p (β) with . Also suppose that φ is a multiplier of H p (β), and w is the Denjoy-Wolff point of ψ. If and φ(w) ≠ 0, then the operator CΦ,ψhas a nonzero fixed point on H p (β) where .

Proof. Note that w U since ||ψ|| U < 1. Assume that φ(w) ≠ 0. Choose δ with '(w)| < δ < 1. Without loss of generality, suppose that w = 0. Hence, (z)| < δ|z| when z is sufficiency near to zero. If K is a compact subset of U, then by the Denjoy-Wolff Theorem, ψ n → 0 uniformly on K and |ψn+k(z)| < δ k |ψ n (z)| for sufficiently large n and every k , and z K. This implies that converges uniformly on compact subsets of U. Since φ is bounded, an application of Schwarz's lemma shows that there exists a constant M > 0 such that (0) - φ(z)| < M|z| for every z U. However, φ(0) ≠ 0, thus
By substituting ψ i (z) instead of z in the above inequality, we get
This implies that and consequently converges uniformly on compact subsets of U. Set . Then g is a nonzero holomorphic function on U. Also, note that φ · gψ = φ(0)g. Thus, generally there is a function g H(U) such that
Hence
so that

Because ||ψ n || U < 1, the function gψ n H(U ); moreover, each factor of belongs to H(U) since φ belongs to H(U). Thus, g H (U) H p (β), and thus (**) shows that the operator CΦ,ψhas a nonzero fixed point on H p (β) as desired. □

Declarations

Acknowledgements

The authors would like to thank Professor Manuel De la Sen for correspondences that led to Theorem 11. This paper is dedicated to Ali ibn-e Hamzeh.

Authors’ Affiliations

(1)
Department of Mathematics, Payame Noor University
(2)
Department of Mathematics, Yasuj University

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Copyright

© Yousefi and Kashkooly; licensee Springer. 2011

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.