Research Article | Open Access

# On Convexity of Composition and Multiplication Operators on Weighted Hardy Spaces

**Academic Editor:**Stevo Stević

#### Abstract

A bounded linear operator on a Hilbert space , satisfying for every , is called a convex operator. In this paper, we give necessary and sufficient conditions under which a convex composition operator on a large class of weighted Hardy spaces is an isometry. Also, we discuss convexity of multiplication operators.

#### 1. Introduction and Preliminaries

We denote by the space of all bounded linear operators on a Hilbert space . An operator is said to be *convex*, if

for each Note that if is a convex operator then the sequence forms a convex sequence for every Taking it is easily seen that is a convex operator if and only if

A weighted Hardy space is a Hilbert space of analytic functions on the open unit disc for which the sequence forms a complete orthogonal set of nonzero vectors. It is usually assumed that . Writing , this space is denoted by and its norm is given by

Let be an analytic map of the open unit disc into itself, and define whenever is analytic on . The function is called the symbol of the composition operator. For a positive integer , the th iterate of , denoted by is the function obtained by composing with itself times; also is defined to be the identity function. Denote the reproducing kernel at , for the space , by . Then for every . It is known that for all in The generating function for is the function given by

This function is analytic on . Moreover, if then and (see [1]).

Recently, there has been a great interest in studying operator theoretic properties of composition and weighted composition operators, see, for example, monographs [1, 2], papers [3–16], as well as the reference therein.

Isometric operators on weighted Hardy spaces, especially those that are composition operators are discussed by many authors. Isometries of the Hardy space among composition operators are characterized in [17, page 444], [18] and [12, page 66]. Indeed, it is shown that the only composition operators on that are isometries are the ones induced by inner functions vanishing at the origin. Bayart [5] generalized this result and showed that every composition operator on which is similar to an isometry is induced by an inner function with a fixed point in the unit disc. The surjective isometries of , that are weighted composition operators have been described by Forelli [19]. Carswell and Hammond [6] proved that the isometric composition operators of the weighted Bergman space are the rotations. Cima and Wogen [20] have characterized all surjective isometries of the Bloch space. Furthermore, the identification of all isometric composition operators on the Bloch space is due to Colonna [8]. Some related results can be found also in [3, 4, 6, 21–25].

Herein, we are interested in studying the convexity of composition and multiplication operators acting on a weighted Hardy space . First, we give some preliminary facts on convex operators. Next, we will offer necessary and sufficient conditions under which a convex composition operator may be isometry on a large class of weighted Hardy spaces containing Hardy, Bergman, and Dirichlet spaces. We also discuss on convexity of the adjoint of a composition operator. Finally, we will obtain similar results for multiplication operators and their adjoints. For a good reference on isometric multiplication operators the reader can see [3].

Throughout this paper, is assumed to be a bounded linear operator on a Hilbert space . It is easy to see that for every convex operator , the sequence forms an increasing sequence. We use this fact to prove the following theorem.

Theorem 1.1. *If is a convex operator then so is every nonnegative integer power of .*

*Proof. *We argue by using mathematical induction. The convexity of implies that the result holds for . Suppose that , then
So the result holds for .

Proposition 1.2. *If is a convex operator, then for every nonnegative integer ,
*

*Proof. *We give the assertion by using mathematical induction on . The result is clearly true for . Suppose that . Thus,
So the result holds for .

Proposition 1.3. *Let be a convex operator and let be such that . If , then .*

*Proof. *By applying Proposition 1.2, we observe that for every nonnegative integer ,
Letting , the positivity of implies that ; hence, .

Proposition 1.4. *Let be an orthonormal basis for and let be a convex operator satisfying Suppose that there is a nonnegative integer and a scalar with so that , then is an invariant subspace for .*

*Proof. *Using Proposition 1.2, we see that
for every . Let . Since is a positive operator, we conclude that . Consequently, Now, if then ; hence, .

#### 2. Composition Operators

Our purpose in this section is to discuss on convex composition operators on a weighted Hardy space. Recall that an operator in is an isometry, if . At first, we give an example of a nonisometric composition operator on a weighted Hardy space such that . For simplicity of notation, is denoted by .

*Example 2.1. *Consider the weighted Hardy space with weight sequence given by Define by . It is easily seen that , and an application of the closed graph theorem shows that is bounded. Now, a simple calculation shows that
for all ; besides
which is positive for all , and zero whenever . It follows that , but is not an isometry.

Proposition 2.2. *Suppose that is a convex operator satisfying and , then
**
is a nontrivial invariant subspace of .*

*Proof. *Clearly is a nontrivial closed subspace of . To show that is invariant for , apply Proposition 1.4 for the Hilbert space the orthonormal basis given by and .

*Example 2.3. *Consider the Bergman space consisting of all analytic functions on the open unit disc , for which
where is the normalized Lebesgue area measure on . If is represented by , then
Also, forms an orthogonal basis for . Fix nonnegative integers and , and observe that
Thus, Proposition 1.3 implies that if and only if is an isometry. In this case, taking and in Proposition 2.2, we conclude that ; thus, the Schwarz lemma implies that for all . On the other hand, if then
and so almost everywhere with respect to the area measure. Hence, for some .

*Example 2.4. *Consider the Hardy space . If is an analytic self-map of the unit disc, then induces a bounded composition operator, and for all nonnegative integers and . Thus, by Proposition 1.3, if and only if is an isometry.

Recall that the Dirichlet space is the set of all functions analytic on whose derivatives lie in the Bergman space . The Dirichlet norm is defined by

If is a univalent self-map of , then is bounded on [2, page 18]. Also, the area formula [1, page 36], shows that

where is, as usual, the counting function defined as the cardinality of the set .

In the next theorem, we characterize all convex composition operators on satisfying . Note that we cannot use Proposition 1.3 for the Dirichlet space, thanks to the fact that in general the positive powers of are not uniformly bounded on the 's.

Theorem 2.5. *If is convex on the Dirichlet space , then if and only if is an isometry.*

*Proof. *One implication is clear. Suppose that is a positive operator, and take in Proposition 2.2. Since the identity function is in the subspace we conclude that . Thus, in light of (2.9), to show that is an isometry it is sufficient to prove that
Let be any function in the Dirichlet space . Then
Furthermore,
By summing up these two relations we get
But , and so
This, in turn, implies that almost everywhere. Substituting this in (2.11), and then considering (2.12) the assertion will be completed.

Observe that if , almost everywhere, and is bounded on then it is convex. Indeed,

In the next theorem, we turn to the adjoint of a composition operator and give necessary and sufficient conditions under which a convex operator is an isometry.

Theorem 2.6. *Let be an analytic self-map of with . If is a convex operator on , then it is an isometry if and only if .*

*Proof. *Suppose that , and assume that is not the identity or an elliptic automorphism. By the Denjoy-Wolff theorem converges uniformly to zero on compact subsets of [1], and so for every ,
Proposition 1.2 coupled with the fact that implies that for all and all nonnegative integers ,
Furthermore, the positivity of shows that . Thus, in light of (2.16) and (2.17) we conclude that for all , and so for every positive integer . Consequently, for all . It follows that
This contradiction shows that is the identity or an elliptic automorphism. Thus, there is a so that for all . Now, if then

It follows that . But it is easily seen that for every . Hence, is an isometry. The converse is obvious.

#### 3. Multiplication Operators

This section deals with convex multiplication operators on a weighted Hardy space. Recall that a multiplier of is an analytic function on such that . The set of all multipliers of is denoted by . It is known that . In fact, if and is the constant function then for every positive integer and for every we have

Now, letting , we conclude that is bounded. This coupled with the fact that implies that If is a multiplier, then the multiplication operator , defined by , is bounded on . Also note that for each , .

In what follows, the operator is assumed to be convex. First, we present an example of a nonisometric convex multiplication operator with .

*Example 3.1. *Consider the weighted Hardy space with weight sequence given by Define the mapping on by . Obviously, is bounded. Furthermore, it is easy to see that for every nonnegative integer ,
Consequently, is convex but not an isometry. Besides, is a positive operator.

Theorem 3.2. *Let consist of all multipliers of , and let be such that . If or then if and only if is an isometry.*

*Proof. *Suppose that is or and . Define the linear mapping by . An application of the closed graph theorem implies that is bounded. Therefore, there is such that for all ,
It follows that for every and every nonnegative integer ,
Thus, for every Since for all , by a similar method one can show that for all . Therefore, the result follows from Proposition 1.3.

*Example 3.3. *Let be the Bergman space or the Hardy space and let be or its adjoint on . It is well known that So if is a multiplier with , then by applying the preceding theorem, we observe that if and only if is an isometry.

We remark herein that if and on the Dirichlet space , then it is easily seen that but is not an isometry.

#### Acknowledgments

The authors would like to thank Dr. Faghih Ahmadi for her assistance and the referee for a number of helpful comments and suggestions. This research was in part supported by a grant no. (88-GR-SC-27) from Shiraz University Research Council.

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#### Copyright

Copyright © 2009 Karim Hedayatian and Lotfollah Karimi. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.