Abstract

We give sufficient conditions under which a weighted composition operator on a Hilbert space of analytic functions is not weakly supercyclic. Also, we give some necessary and sufficient conditions for hypercyclicity and supercyclicity of weighted composition operators on the space of analytic functions on the open unit disc.

1. Introduction

Let denote the open unit disc in the complex plane. Let be a Hilbert space of analytic functions defined on such that , and for each in , the linear functional of point evaluation at given by is bounded. By a Hilbert space of analytic functions we mean the one satisfying the above conditions.

For , let denote the linear functional of point evaluation at on ; that is, for every in . Since is a bounded linear functional, the Riesz representation theorem states that for some

A well-known example of a Hilbert space of analytic functions is the weighted Hardy space. Let be a sequence of positive numbers with The weighted Hardy space is defined as the space of functions analytic on such that

Since the function is in , we see that These spaces are Hilbert spaces with the inner product for every and in

Let for every nonnegative integer . Then is an orthogonal basis. As a particular consequence of this fact, the polynomials are dense in It is clear that . We call the set the standard basis for The sequence of weights allows us to consider the generating function which is analytic on Take . It is easy to see that for any function in , where ; moreover,

The classical Hardy space, the Bergman space, and the Dirichlet space are weighted Hardy spaces with weights, respectively, given by , and . Weighted Bergman and Dirichlet spaces are also weighted Hardy spaces.

Let be an automorphism of the disc. Recall that is elliptic if it has one fixed point in the disc and the other in the complement of the closed disc, hyperbolic if both of its fixed points are on the unit circle, and parabolic if it has one fixed point on the unit circle (of multiplicity two).

Recall that a multiplier of is an analytic function on such that . The set of all multipliers of is denoted by . If is a multiplier, then the multiplication operator , defined by , is bounded on . Also, note that for each , . It is known that . In fact, since the constant functions are in , every function in is analytic on . Moreover, if , then which implies that for every , and so

In what follows, suppose that and that is an analytic self-map of such that is in for every . An application of the closed graph theorem shows that the weighted composition operator defined by is bounded. The mapping is called the composition map and is called the weight. For a positive integer the th iterate of is denoted by also is the identity function. We note that for all and . Moreover, for every and . The weighted composition operators come up naturally. In 1964, Forelli [1] showed that every isometry on for and is a weighted composition operator. Recently, there has been a great interest in studying composition and weighted composition operators on the unit disc, polydisc, or the unit ball; see, for example, monographs [2, 3], papers [418].

Suppose that and is an operator on . The set , denoted by , is called the orbit of under . If there is a vector whose orbit is (weakly) dense in , then is called a (weakly) hypercyclic operator and is called a (weakly) hypercyclic vector for . The operator is called (weakly) supercyclic if the set of scalar multiples of the elements of is (weakly) dense and is called cyclic if the linear span of is dense. In each case, the vector is called, respectively, (weakly) supercyclic vector or cyclic vector for .

Hypercyclicity of operators is studied a lot in literature. The classical hypercyclic operator is on the space , where is the backward shift [19]. It is proved that many famous operators are hypercyclic. For instance, certain operators in the classes of composition operators [20], the adjoint of subnormal, hyponormal, and multiplication operators [21, 22] and weighted shift operators [23] are hypercyclic. As a good source on hypercyclicity, supercyclicity, weak hypercyclicity, and weak supercyclicity of operators, one can see [24].

Recently the hypercyclicity of the adjoint of a weighted composition operator on a Hilbert space of analytic functions is discussed in [25]. Chan and Sanders [26] showed that a weakly hypercyclic operator may fail to be norm hypercyclic. Furthermore, Sanders in [27] proved that there exists a weakly supercyclic operator that fails to be norm supercyclic. In this paper, we discuss weak supercyclicity and weak hypercyclicity of a weighted composition operator.

2. Weighted Composition Operators on

The following proposition limits the kinds of maps that can produce weakly supercyclic weighted composition operators on .

Proposition 2.1. Suppose that is weakly supercyclic. Then(i) for all (ii) is univalent.

Proof. Suppose that is a weakly supercyclic vector for .
(i) Let , and let be a positive number. Since the set is a weak neighborhood of , there exist and a scalar such that ; that is, Now, if , then . Since is arbitrary, we get a contradiction.
(ii) Suppose that and put For any , the set is a weak neighborhood of , and so there exist and a scalar such that ; that is, Thus, , and so ; this implies that . Consequently, , which, in turn, implies that .

Hereafter, we may assume that is univalent and for all . Take , and consider an automorphism of defined by . A Hilbert space of analytic functions is called automorphism invariant if for every whenever and whenever Spaces such as the Hardy and the Bergman spaces are automorphism invariant.

Theorem 2.2. Suppose that is automorphism invariant. If is weakly supercyclic and is a fixed point of , then is an unbounded sequence for every .

Proof. First, suppose that Let be a weakly supercyclic vector for Since the constant function 1 is in , the point evaluation is surjective. Therefore, the continuity of implies that the set is dense in hence, should be nonzero.
Let and . Put Since is a weak neighborhood of , there exist and a scalar such that . Therefore, On the other hand, since , the Schwarz lemma yields that for every positive integer ; so there is a constant such that for all . Put and assume, on the contrary, that there is a constant such that for all . Thus, (2.8) imply that for every , which contradicts the fact that .
Now, suppose that and put The operators and are similar; indeed, Thus, is weakly supercyclic. If then for every positive integer . Moreover, and ; thus, by the previous step the sequence is unbounded, and so is the sequence .

The following corollaries are direct consequences of Theorem 2.2.

Corollary 2.3. Let be automorphism invariant and have a fixed point in If there is a point and a positive integer such that for all then is not weakly supercyclic on .

Example 2.4. Suppose that , where and Then for all Therefore, is not weakly supercyclic on

Corollary 2.5. Suppose that   has a fixed point . If and is a bounded sequence, for some ; then is not weakly supercyclic on .

Example 2.6. For in , define and . Since and , is bounded on the Hardy space and the Bergman space An easy computation shows that is an elliptic automorphism with a fixed point and . Also, for the even values of j, and so ; moreover, for the odd values of j, and . Thus, for all and . Therefore, is a bounded sequence and so is not weakly supercyclic.

By taking in Corollary 2.3, the following result will be obtained.

Corollary 2.7. If is automorphism invariant and has a fixed point in , then is not weakly supercyclic.

Proposition 2.8. Suppose that polynomials are dense in . If is weakly supercyclic, then w is a cyclic vector for the operator .

Proof. Since is weakly supercyclic, its range is weakly dense. The convexity of the range of implies that it is norm dense, and so is the range of . Suppose that and . Then there is a vector in such that . Moreover, there is a polynomial such that Thus, This shows that is a cyclic vector for .

Corollary 2.9. If is weakly supercyclic on the Hardy space then w is an outer function.

Proof. By Beurling's theorem, the only cyclic vectors for on are the outer functions.

A necessary condition for the supercyclicity of an operator, called the angle criterion, was introduced by A. Montes-Rodríguez and H. N. Salas. They used this criterion to construct the non-supercyclic vectors for certain weighted shifts. We need the Hilbert space setting version of this criterion which is presented here, for the sake of completeness.

Theorem 2.10 (the angle criterion [24]). Let be a Hilbert space, and suppose that T is bounded operator on . If a vector is supercyclic for T, then for every nonzero vector .

We say that a sequence of complex numbers is not a Blaschke sequence if there exists such that for all , and diverges.

Theorem 2.11. Suppose that is not an elliptic automorphism and has a fixed point If there exist and a positive integer so that for all , then is not supercyclic.

Proof. If is the identity map, then . The point spectrum of contains more than one point; in fact, for all . Thus, is not supercyclic [24, page 13]. So, we may assume that is not the identity map.
Assume that is a supercyclic operator on and . Since , the set is open in . We know that the set of supercyclic vectors for is dense in [24, page 9]. So, there exists a supercyclic vector for in . If is not a Blaschke sequence, then . On the other hand, we have Therefore, However since and , the left side of the above inequality tends to zero, but by the angle criterion, which is a contradiction.
Now, suppose that is a Blaschke sequence. If , then there exists a sequence such that as . So as Thus, which is a contradiction.

The following theorem is a kind of the angle criterion for the weak supercyclicity which is called the weak angle criterion.

Theorem 2.12 (the weak angle criterion [24]). Let be a Hilbert space, and let be a bounded linear operator on . Also, let . Assume that one can find some nonzero such that Then, x is not a weakly supercyclic vector for .

An operator is bounded below if there exists such that for all . It is well known that is bounded below if and only if it has a closed range and . It can be easily seen that ; so a weighted composition operator is bounded below if and only if its range is closed.

Proposition 2.13. Suppose that has a fixed point If then is not weakly supercyclic.

Proof. If , then and Consequently, , and theorem follows by the weak angle criterion.

Corollary 2.14. Let be an inner function of the disc with a fixed point . Suppose that is nonconstant and there exists such that almost everywhere on . If , then is not weakly supercyclic on the Hardy space .

Proof. Since is an inner function, for all [2, page 123]. So and the result follows by applying Proposition 2.13.

In the last part of this section, we give some necessary conditions for the weak hypercyclicity of a weighted composition operator.

Proposition 2.15. If one of the following conditions holds, then is not weakly hypercyclic.(i), and .(ii)If a is the Denjoy-Wolff point of , then is continuous at a with (iii); moreover, if is the Denjoy-Wolff point of , then is continuous at and .(iv)The function has a fixed point in .

Proof. Assume, on the contrary, that has a weakly hypercyclic vector . If and , then Suppose that holds. Put in (2.23). Since we have However, the mapping is weakly continuous [28, page 167], and so the sequence is dense in , which is impossible.
If holds and , then by the Denjoy-Wolff theorem as Since , there is such that for all . Hence, by (2.23), we have Thus, the sequence is bounded; now, similar to part , we get a contradiction.
Suppose that holds. If , then and so as Since , ; equivalently, Now, by (2.23), the sequence is bounded and similar to part ; again we get a contradiction.
At last, suppose that holds. Let be a fixed point of . Since , the point spectrum of is nonempty, which is a contradiction [24, page 11].

We remark that if =, then ; moreover, if , then Indeed, for every and , we have

3. Weighted Composition Operators on

In this section, we give some necessary and sufficient conditions for hypercyclicity and supercyclicity of weighted composition operators on the space of analytic functions . If is a compact subset of , for , define Then is a family of seminorms that makes into a locally convex space. In fact, this topology is the topology of uniform convergence on compact subsets of the open unit disc, the usual topology for which is an F-space.

Definition 3.1. A sequence of self-maps of is said to be a run-away sequence if for every compact set , there exists a positive integer such that .

For example, if is an automorphism, then is a run-away sequence if and only if has no fixed point in ; see [10].

To prove the next theorem, we apply Birkhoff's transitivity theorem.

Theorem 3.2 (Birkhoff's transitivity Theorem [24]). Let be a separable F-space, and let be a bounded linear operator in . The following statements are equivalent.(i) is hypercyclic;(ii) is topologically transitive; that is, for each pair of nonempty open subsets of , there exists a positive integer such that

Theorem 3.3. If is a run-away sequence and is in such that for all , then is a hypercyclic operator on .

Proof. Let and be two nonempty open subsets of There exist two compact subsets, and , of and two functions such that Put Therefore, Since is compact, we can choose two closed discs, , in such that Since is a run-away sequence, there exists a positive integer such that Now, consider the map , defined by , if and by if The sets are two disjoint closed subsets of ; moreover, since by convention, and so are injective, the complement of is connected. Then by Runge's theorem, there exists a polynomial such that for all So which implies that . Also, which implies that Therefore, Then which, in turn, implies that is nonempty. Thus, Birkhoff's transitivity theorem implies that is hypercyclic.

Proposition 3.4. Suppose that is hypercyclic on and is not a run-away sequence. Then .

Proof. Assume that for all or for all . Let be a hypercyclic vector for , and choose a subsequence so that as . Since is not a run-away sequence there exists a compact subset of such that for every positive integer ; so one can find a sequence satisfying . Since is a compact set, there exists such that (passing, if necessary, to a subsequence of ), and so . Now, if for all , then which implies that or, equivalently, as . In the case that for all , using a similar argument, we obtain as ; so in both cases we have contradictions, due to the fact that

It is worthy of attention to note that by a similar argument as in the proof of Theorem 2.2, one can show that it can also be applied on the space with a further assumption that . Since the space is automorphism invariant, we have the following result.

Corollary 3.5. Suppose that is a univalent self-map of the disc. Then the following statements are equivalent:(a) has no fixed point in .(b) is weakly hypercyclic on for every complex number with .(c) is weakly supercyclic on .

Proof. If (a) holds, then by Proposition of [25], is hypercyclic on for every complex number with . Clearly (b) implies (c) and by the preceding discussion (c) implies (a).

Proposition 3.6. Suppose that is a self-map of the disc with the Denjoy-Wollf point in satisfying . Then is weakly supercyclic if and only if is weakly supercyclic on .

Proof. By Proposition . of [25], is an eigenvalue for with an eigenvector which has no zero in ; so the operator is invertible. On the other hand, thus, is similar to . Now, the result follows from the comparison principle [24, Page 10].

Corollary 3.7. Suppose that is a univalent self-map of the disc with the Denjoy-Wollf point in such that . Then is weakly hypercyclic if and only if is weakly hypercyclic on .

Proof. Corollary 3.5 shows that is weakly hypercyclic if and only if is weakly hypercyclic. Now, the proof is the same as the preceding proposition.

Acknowledgment

This research was in part supported by a grant from Shiraz University Research Council.