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 ,