Abstract and Applied Analysis

Volume 2010, Article ID 803230, 14 pages

http://dx.doi.org/10.1155/2010/803230

## On the Generalized Hardy Spaces

Department of Mathematics, College of Sciences, Shiraz University, Shiraz 71454, Iran

Received 8 September 2009; Accepted 11 January 2010

Academic Editor: Stevo Stević

Copyright © 2010 M. Fatehi. 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.

#### Abstract

We introduce new spaces that are extensions of the Hardy spaces and we investigate the continuity of the point evaluations as well as the boundedness and the compactness of the composition operators on these spaces.

#### 1. Introduction

Let be the open unit disk in the complex plane , its boundary, and the space of all analytic functions on the unit disk.

For an analytic function on the unit disk and , we define the delay function by . It is easy to see that the functions are continuous for each , hence they are in , where is the normalized arc length on the unit circle.

For , the Hardy space is the set of all such that

Also we recall that is the space of all bounded analytic functions defined on , with supremum norm . We know that for , is a Banach space (see, e.g., [1, page 37]).

By the Littlewood Subordination Theorem (see [2, Corollary ]), we see that the supremum in the above definition of is actually a limit, that is,

Another important result is Fatou's Radial Limit Theorem (see [1, Theorems and ]), which says, if is in for some , then the radial limit

exists for almost all and the mapping is an isometry of to a closed subspace of . Therefore,

We will also write for . If and are the th coefficients of in its Maclaurin series, then we have another representation for the norm of on as follows:

The formula above defines a norm that turns into a Hilbert space whose inner product is given by

for each (see, e.g., [2]).

Let be the point evaluation at , that is, . It is well known that point evaluations at the points of are all continuous on (see, e.g., [1, page 36]).

Let and be a Hilbert space of analytic functions on . If is a bounded linear functional on , then the Riesz Representation Theorem implies that there is a function (which is usually called ) in that induces this linear functional, that is, .

Let be an analytic self-map of the unit disk. The linear composition operator is defined by for . It is well known (see, e.g., [1, page 29] or [3, Theorem ]) that the composition operators are bounded on each of the Hardy spaces (). One of the first papers in this research area is [3], while Schwartz in [4] begun the research on compact composition operators on . Shapiro and Taylor in [5] have studied the role of angular derivative for compactness of in . For some other classical results, see [2, 6].

The boundedness and compactness of composition operators, as well as weighted composition operators and other natural extensions of them, on various spaces of analytic functions have been investigated by many authors; see, books [2, 6], and, for example, the following recent papers [7–28] and the references therein.

Throughout this paper, denotes the set of all analytic polynomials and for a function , denotes the range of .

#### 2. Generalized Hardy Spaces

In this section, we define new spaces and investigate some basic properties of these spaces.

*Definition 2.1. *Let be a linear operator such that if and only if , that is, is 1-1. For , the generalized Hardy space is defined to be the collection of all analytic functions on for which
Denote the th root of this supremum by Since is a subharmonic function, so by [2, Corollary ], we have
Therefore, if and only if and
It is easy to see that is a normed space with the norm .

From now on, unless otherwise stated, we assume that satisfies the conditions of Definition 2.1.

In this section, we first set some conditions on such that becomes a Banach space. In the following theorem, we obtain a necessary and sufficient condition for to be a Banach space.

Theorem 2.2. *Let and . Then is a subspace of if and only if is a Banach space.*

*Proof. *Suppose that . Since is a normed space, it suffices to show that it is complete. Let be a Cauchy sequence in and set . Then is a Cauchy sequence in . Since is complete, there is a such that
Since , there is an such that . Now we show that this is the -limit of . We have
Hence for sufficiently large positive integer , which implies that . So in as

Conversely, suppose that is a Banach space. If , then there is a such that is not in . Since the polynomials are dense in , there is a sequence in such that as . Let . Then is a Cauchy sequence in and so there is a such that as . Hence as . On the other hand, as . This shows that which is a contradiction.

*Example 2.3. *Let , where . It is not hard to see that and is not in . So by Theorem 2.2, is not a Banach space.

Proposition 2.4. *If , then is a Hilbert space.*

*Proof. *We define a scalar product on by
It is easy to show that this scalar product defines an inner product on .

There is a Banach space such that it does not satisfy the condition of Theorem 2.2. For example, let , for each . Then . By the following proposition, we see that although , is a Banach space.

Proposition 2.5. *Suppose , and for every . Then is a Banach space.*

*Proof. *If , then by Theorem 2.2, the proposition holds. Otherwise, let be a Cauchy sequence in . Seting , so is a Cauchy sequence in . Therefore, there is a such that as . If , then the proof is similar to the proof of Theorem 2.2.

Now suppose that is not in . Then there are , and such that
where , and . Therefore, we have
Since as , we have
Hence as . Since the point evaluation at is a bounded linear functional on , we have
So , which is a contradiction.

The set of all analytic polynomials is dense in , but this is not the case for each space . Also it is possible that . For example, let , and . Then is not in , for if , then , which is a contradiction.

In the following proposition, we will find a dense subset in , whenever .

Proposition 2.6. *Suppose and . Then *

*Proof. *It is clear that Suppose that . Then there is a sequence in such that as . Setting , we have
so the result follows.

Corollary 2.7. *Suppose , and for each . Then *

#### 3. Point Evaluations

In this section, we investigate the continuity of the point evaluations on . The idea behind Theorem 3.1 is similar to the one found in [29, page 51].

Theorem 3.1. *Suppose that and . Then we have the following.*(a)*If and , then is continuous on . *(b)*Let and . If for each and , , then is continuous on .*

*Proof. * By Proposition 2.4, is a Hilbert space. Since
for each , where , is an orthonormal set in . Also if and for each , , then for each , . Since is a basis for , . Therefore, is a basis for . Since , there is an such that . For each , we have
Hence and is continuous on .

Let . If , then
so is continuous on .

Let . Then for each , and so
Also by [1, Theorem ], as . Therefore, by Holder's inequality and the fact that , we have
as . So we obtain
Hence
for each . Now let . If , then
so the result follows.

Suppose that and satisfy the hypotheses in the first line of Theorem 3.1. It is easy to see that if is continuous on , then according to the proof of the previous theorem, .

Now we give two examples for the preceding theorem.

*Example 3.2. * Let and , where . It is easy to see that and for each , . Therefore, by Theorem 3.1, is continuous on for each .

Let and such that is a permutation of for some and for . Then by Theorem 3.1, for each point in the open unit disk, is continuous on . Also if and is the above permutation, then
Therefore, .

There is a Banach space such that it does not satisfy the conditions of Theorem 3.1, but for each , is continuous on . For example, let , and for each . Then is not in and . By the following theorem, we see that although the hypotheses of Theorem 3.1 do not hold, is continuous on for each .

Theorem 3.3. *Let , , , and for each , . Then is continuous on .*

*Proof. *We break the proof into two parts.

Let . If and is the circle of radius with center at the origin, then the Cauchy formula shows that for any in ,
It follows that
where . Now if tends to , converges uniformly to the bounded function and . Hence there is an such that
and the result follows.

Let . Then where , and . Let for each , it is easy to see that . Then by the preceding part, there is a constant such that
for each . So is continuous on .

There is an example of such that is not a Hilbert space and is continuous for some .

*Example 3.4. *Let and , where . We can show that for each , is continuous on . Suppose , we have
as desired. Also it is easy to see that is continuous on for each and .

#### 4. Continuity of the Composition Operators on

In the most important classical spaces, all analytic maps of the unit disk into itself induce bounded composition operators, but there are analytic self-maps of the unit disk which do not give bounded composition operators on some generalized Hardy spaces.

*Example 4.1. * Suppose that and for each , , where . Let for each . Then . We see that
So and is not bounded on .

Let , where and is given in [2, Example ]. If , then in [2, Example ] shows that is not bounded on .

In this section, we investigate the continuity of the composition operator on some spaces in terms of a Carleson measure criterion. This criterion has been used to characterize the boundedness and the compactness of the composition operators in different papers (see, e.g., [16, 30]).

*Definition 4.2. *A positive measure on is called a Carleson measure (in ) if there is a constant such that for all and , where .

Proposition 4.3. *Let and . If is a finite, positive Borel measure on , then the following conditions are equivalent.*(a)* is a Carleson measure in .*(b)*There is a constant such that
for all in .*

*Proof. *By [2, Theorem ], (a)(b) is clear. Now we prove that (b) implies (a). For each , there is a such that and
If , then there is a sequence in such that as . So
By (4.3), we obtain
for each . The right-hand side of (4.5) is bounded, hence there is a subsequence such that converges. Since point evaluations at the points of are all continuous on , the Principle of Uniform Boundedness implies that converges to uniformly on compact subsets of . In particular, for each , the convergence is uniform on . Therefore, for each , as . Since is a bounded sequence, there is an such that for each . Hence and . Now let be an increasing sequence which converges to 1 and let . By [31, page 32], there is an integer such that , where . Let . Then by [31, page 73], we have . So there is a such that for each , . Again apply [31, page 32] for . Therefore, there are such that
for each . Let . Therefore, for each and , . So by [32, Theorem ], we have
So by the preceding relation and (4.4) and (4.5), we have
for every . Thus the result follows from [2, Theorem ].

*Definition 4.4. *For on the unit circle and , let

From now on, unless otherwise stated, we assume that is a Banach space.

Proposition 4.5. *Let and . If is a finite, positive Borel measure on , then the following conditions are equivalent.*(a)*There is a constant such that for all and . *(b)*There is a constant such that
for all .*

*Proof. *The implication (a)(b) follows by Proposition 4.3 and is exactly the same as the proof of [2, Theorem ()()]. For the other direction, by Theorem 2.2, . Hence for each , we have
Then the result follows from [2, Theorem ()()].

In the following theorem, we use the techniques used in [2, Theorem part ()].

Theorem 4.6. *Suppose , and are analytic self-maps of . Let and for each , . Then is a bounded operator on if and only if for all in and , where , is normalized Lebesgue measure on the unit circle, and is a subset of the closed disk .*

*Proof. *Let . By [2, Theorem ] and [33, page 163], we obtain
So the result follows from Proposition 4.5.

*Remark 4.7. *If , , and satisfy the hypotheses in Theorem 4.6, then by [2, Theorem part ()], is a bounded operator on if and only if is a bounded operator on . Since we know that the composition operators are always bounded on each of the Hardy spaces , is a bounded operator on .

#### 5. Compactness of the Composition Operators on

In this section, we investigate the compactness of the composition operators on .

The idea of the proof of the following theorem is similar to the proof of Proposition in [2]. A detailed proof of another similar result can be found, for example, in [18].

Theorem 5.1. *Suppose and for each , is continuous. Also if as uniformly on compact subsets of , then in as . Then the following conditions are equivalent. *(a)* is compact on . *(b)*If is a bounded sequence in and as uniformly on compact subsets of , then as . *

*Proof. *The implication ()() follows exactly as the proof of [2, Theorem ].

Now we show that ()(). Let be a bounded sequence in . Since is continuous for each , is a normal family. So there is a function and a subsequence such that as uniformly on compact subsets of . Hence in as . It is easy to see that . Therefore, is a bounded sequence in such that it converges uniformly to on compact subsets of . By the hypotheses, we conclude that in as . Thus is a compact operator.

Corollary 5.2. *Let , , and for each . Let be an analytic self-map of and . Then is a compact operator on *

In the rest of this section, we investigate the relation between compactness of on and .

Theorem 5.3. *Let , , , and satisfy the hypotheses in Theorem 4.6. Then is compact on if and only if as uniformly in in .*

*Proof. *Let as uniformly in in and be a bounded sequence in . Then by [2, Theorem part ()], there is a subsequence such that converges in . Since , converges in . Therefore, is a compact operator on .

Conversely, let be a bounded sequence in . By Theorem 2.2, , so there is a sequence in such that . Since is compact on , we may extract a subsequence such that converges in . So is a compact operator on and [2, Theorem part ()] implies the result.

*Remark 5.4. *If , , and satisfy the hypotheses in Theorem 5.3, then by [2, Theorem part ()], is a compact operator on if and only if is a compact operator on .

Now we present an example for Remarks 4.7 and 5.4.

*Example 5.5. *Let , and , where and for every odd positive integer and every , and . It is easy to see that by Remark 4.7, is a bounded operator on and by Remark 5.4 and [2, Corollary ] is not a compact operator on .

#### Acknowledgments

The author would like to thank Dr. Bahram Khani Robati for his suggestions and comments. Also she appreciates professor Stevo Stevic for the helpful advice and notification to this work. This work is a part of the author’s doctoral thesis, written at Shiraz University under the direction of B. Khani Robati.

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