#### Abstract

We study the boundedness and compactness of the weighted composition operators as well as integral-type operators between weighted Hardy spaces on the unit ball.

#### 1. Introduction

Let denote the open unit ball of the -dimensional complex vector space , its boundary, and let denote the space of all holomorphic functions on . For and we define the *weighted Hardy space* as follows:
where is the normalized Lebesgue measure on (see, also [1], as well as [2], for an equivalent definition of the space). Note that for the weighted Hardy space becomes the Hardy space . We define the norm on this space as follows:
With this norm is a Banach space when . For a related space on the unit polydisk; see [3]. In this paper, we investigate two types of operators acting between weighted Hardy spaces.

Let be a holomorphic self-map of and . Then and induce a *weighted composition operator* on which is defined by . This type of operators has been studied on various spaces of holomorphic functions in , by many authors; see, for example, [4], recent papers [5–17], and the references therein.

Let and be a holomorphic self-map of the open unit disk in the complex plane. Products of integral and composition operators on were introduced by S. Li and S. Stević in a private communication (see [18–21], as well as papers [22] and [23] for closely related operators) as follows:

In [24] the first author of this paper has extended the operator in (1.4) in the unit ball settings as follows (see also [25, 26]). Assume , and is a holomorphic self-map of , then we define an operator on the unit ball as follows: If , then and , so that for some . By the change of variable , it follows that Thus the operator (1.5) is a natural extension of operator in (1.4). For related operators see [27–33] as well as the references therein.

In this paper we study the boundedness and compactness of the weighted composition operators as well as the integral-type operator , between different weighted Hardy spaces on the unit ball.

Throughout this paper, constants are denoted by , they are positive and may differ from one occurrence to the other. The notation means that there is a positive constant such that . Moreover, if both and hold, then one says that .

#### 2. Weighted Composition Operators

This section is devoted to studying weighted composition operators between weighted Hardy spaces. Weighted composition operators between different Hardy spaces on the unit ball were previously studied in [15, 34], while the composition operators on the unit ball were studied in [35, 36]. For the case of the unit disk see also [37].

Before we formulate the main results in this section we quote several auxiliary results which will be used in the proofs of these ones.

Lemma 2.1. *Let and . Suppose that and is a holomorphic self-map of . Then for each **
where .*

*Proof. *Fix . Fatou's lemma shows that
Hence we have the desired inequality.

Recall that an has the homogeneous expansion where is a multi-index, and . For the homogeneous expansion of and any integer , let and where is the identity operator. Note that is compact operator on for each

Lemma 2.2. *If , then converges to pointwise in the Hardy space as .*

*Proof. *See [34, Corollary 3.4] .

Lemma 2.2 and the uniform boundedness principle show that is an uniformly bounded sequence in .

The following lemma is proved similar to [4, Lemma 3.16]. We omit its proof.

Lemma 2.3. *If is bounded from into , then
**
where and denote the essential norm and the operator norm, respectively.*

Lemma 2.4. *Let . Suppose that is a positive Borel measure on which satisfies
**
for some positive constant . Then there exists a positive constant which depends only on and the dimension such that
**
for any . Here .*

*Proof. *See [38, page 13, Theorem] or [34, Lemma 2.1] .

Let . For each , a holomorphic self-map of and , we define a positive Borel measure on by for all Borel sets of . By the change of variables formula from measure theory, we can verify for each nonnegative measurable function in .

Theorem 2.5. *Let and . Suppose that and is a holomorphic self-map of . Then is bounded if and only if
*

*Proof. *For we put
Then we see that and moreover . By a straightforward calculation, we have
for all . Hence if is bounded, then and satisfy the condition

Next we assume

Fix and , respectively. For and , we put and . Since the function , which is defined by (2.11) for this , satisfies

for all , we have
By the same argument, the function gives the following estimate:
Now we need to prove that there exists a positive constant such that
for all and . By the estimate (2.16), we see that the inequality (2.18) is true for all . Thus we assume . By the same argument as in [36, pages 241-242, proof of Theorem 1.1] , we see that the inequality (2.17) shows that there exists a positive constant which depends only on the dimension such that
Hence for , we have the inequality in (2.18).

For the dilate function belongs to the ball algebra, and so is in the Hardy space . Hence Lemma 2.4 gives

for some positive constant and all . This implies that
and so we have
for all . By Lemma 2.1 we have
This completes the proof.

The following proposition is proved in a standard way; see, for example, the proofs of the corresponding results in [4, 32, 33, 39]. Hence we omit its proof.

Proposition 2.6. *Let and . Suppose that and is a holomorphic self-map of which induce the bounded operator . Then is compact if and only if for every bounded sequence in which converges to uniformly on compact subsets of , converges to in . *

In the proof of Theorem 2.8, we need the following lemma.

Lemma 2.7. *Let , , and be the family of test functions defined in (2.11). Then weakly in as .*

*Proof. *The family is bounded in and uniformly on compact subsets of as . By the definitions of the space and the norm , we see that is a subspace of the weighted Bergman space and
for some positive constant which depends on and . This inequality implies that the family is also bounded in . Note also that the family converges to uniformly on compact subsets of as . Hence weakly in as .

In order to prove that weakly in as , we take an arbitrary bounded linear functional on . By the Hahn-Banach theorem, can be extended to a bounded linear functional on so that for all . Since weakly in as , we have as , and so weakly in as .

Theorem 2.8. *Let and . Suppose that and is a holomorphic self-map of such that is bounded. Then the th power of the essential norm is comparable to
**
Hence is compact if and only if
*

*Proof. *To prove a lower estimate
we consider the test functions defined in (2.11). The family is bounded in , say by , and uniformly on compact subsets of as . Thus by Lemma 2.7 we have that weakly in as , so that as for every compact operator . Hence
This inequality and (2.12) give the lower estimate for .

Next we prove an upper estimate. Take with . Fix and put

Then we can choose such that
for with . Fix and . By the same argument as in the proof of inequality (2.20) in Theorem 2.5, we obtain that
where the positive constant is independent of , and a positive integer . Since is in the ball algebra, Lemma 2.2 gives
Combining this with (2.31), we have
and so we have
Letting , by Lemma 2.1, we obtain
Since is arbitrary, this estimate and Lemma 2.3 imply
which completes the proof.

*Remark 2.9. *In the above proof, we used Lemma 2.2. This lemma required the assumption . Hence we cannot have an upper estimate for in the case . However, Proposition 2.6 shows that the compactness of is equivalent to

#### 3. Integral-Type Operators

Here we study the boundedness and compactness of the integral-type operators between weighted Hardy spaces on the unit ball.

For with the Taylor expansion , let be the radial derivative of .

The following lemma was proved in [24] (see also [25]).

Lemma 3.1. *Assume that is a holomorphic self-map of , and Then for every one holds
**A positive continuous function on the interval is called normal [40] if there is a and and , such that**If one that a function is normal, one will also assume that it is radial.*

Lemma 3.2. *Assume that , is a positive integer and is normal. Then for every **
where
*

*Proof. *The proof of the lemma in the case can be found in [27, Theorem 2]. However, due to an overlook, the proof for the case has a gap. Hence we will give a correct proof here in the case.

We may assume that , otherwise we can consider the functions Also we may assume that , to avoid some minor technical difficulties.

By [27, Lemma 1], for each fixed , there is a positive constant depending only on and the dimension such that

for every and such that .

From (3.5) and the fact that is normal, we have

By [40, page 291, Lemma 6] there exists a positive constant such that
for every . Combining this with (3.6), we have

The reverse inequality is proved by the following inequality:

and the fact that for normal (see [27]). Hence, we obtain the result for the case

For it should be only noticed that is still normal, that for every integer , and use the method of induction.

Theorem 3.3. *Let and . Suppose that with and is a holomorphic self-map of . Then is bounded if and only if
*

*Proof. *Take with . Since the function for and is normal, Lemma 3.2 gives
The assumption implies and Lemma 3.1 shows . Hence we obtain
and so we obtain . This implies that the boundedness of is equivalent to the boundedness of . So Theorem 2.5 shows that the condition
is a necessary and sufficient condition for the boundedness of . This completes the proof.

The next proposition is proved similar to Proposition 2.6.

Proposition 3.4. *Let , and . Suppose that with and is a holomorphic self-map of which induce the bounded operator . Then is compact if and only if for every bounded sequence in which converges to uniformly on compact subsets of , converges to in .*

Theorem 3.5. *Let and . Suppose that with and is a holomorphic self-map of which induce the bounded operator . Then is compact if and only if
*

*Proof. *First we assume that condition (3.14) holds. Take a bounded sequence which converges to uniformly on compact subsets of . Theorem 2.8 and the remark in Section 2 show that is compact. Thus Proposition 2.6 implies that
From (3.15) and since , we have that as . By Proposition 3.4, we see that is compact.

To prove the necessity of the condition in (3.14), we consider the family of test functions which is defined by (2.11). Hence we have

for all . Since is a bounded sequence in and uniformly on compact subsets of as , the compactness of and Proposition 3.4 show that as . This fact along with (3.16) implies the condition in (3.14), finishing the proof of the theorem.

Theorem 3.6. *Let and . Suppose that with and is a holomorphic self-map of which induce the bounded operator . Then the th power of the essential norm of is comparable to
*

*Proof. *To prove a lower estimate, we take an arbitrary compact operator . Since Lemma 2.7 implies that the family of functions defined by (2.11) converges to weakly in as , we obtain
Combining this with (3.16), we have
which is a lower estimate.

By some modification of Lemma 2.3 and the application of Lemmas 3.1 and 3.2, we get

As in the proof of Theorem 2.8, we obtain
and so we have an upper estimate for .

#### 4. The Case

When and , we define the weighted-type space as follows: It is easy to see that if and only if , so we define the norm on by this supremum.

Furthermore we consider the subspace defined by

Theorem 4.1. *Let . Suppose that with and is a holomorphic self-map of . Then is bounded if and only if
**
In this case, the operator norm is comparable to the above supremum.*

*Proof. *By the definition of the space , satisfies the growth condition
so it follows from Lemma 3.1 and Lemma 3.2 that
for every .

Hence we obtain

Now we prove the reverse inequality. For , we put

Note that for each and moreover .

When , we have

for all . Letting in (4.8), we have

For the constant function we obtain

Inequality (4.10) shows that the estimate in (4.9) also holds when .

Hence, from (4.9) we obtain

which along with the obvious inequality
completes the proof of the theorem.

For the compactness of , we can also prove the following proposition which is similar to Proposition 2.6.

Proposition 4.2. *Let . Suppose that with and is a holomorphic self-map of which induce the bounded operator . Then is compact if and only if for every bounded sequence in which converges to uniformly on compact subsets of , converges to in .*

Theorem 4.3. *Let . Suppose that with and is a holomorphic self-map of such that is bounded. Then the essential norm is comparable to
**
In particular, is compact if and only if
*

*Proof. *First we consider the family where
We can easily check that , for all and uniformly on compact subsets of as . Hence [40, page 296, Theorem 2] implies that weakly in as .

If , then as in the proof of [26, Theorem 3] it can be seen that the operator is compact, so that

On the other hand, the limit in (4.13) is vacuously equal to zero, from which the result follows in this case. If then take a sequence in with as and put for each . Then is a bounded sequence in and converges to weakly in as . Hence for every compact operator we have as . So we have
for all compact operators . Taking the infimum over the set of all compact operators , we obtain
Combining this with the estimate , we have

Next we prove an upper estimate. Assume that is a sequence which increasingly converges to . For this , we define the operators defined by

As in the proof of [26, Theorem 3], Proposition 4.2 shows that is compact for each .

Put

and fix . Then we can choose such that
if . Take with and an integer . By Lemma 3.1 and Lemma 3.2, we have