1. Introduction
Let
be a bounded domain of
and
the class of all holomorphic functions on
. Then consider a holomorphic self-map
of
and a function
. The linear operator
is referred to as a weighted composition operator for
. If
, it reduces to the composition operator, whereas for
it becomes the multiplication operator. For any given holomorphic function
f,
represents a generalised composition/multiplication operator. The reader is referred to book[1] for an extensive introduction to the topic.
In this paper, we study the boundedness and the compactness of weighted composition operators from
-Bloch spaces
to Bers-type spaces built on generalised Hua domains of the first kind. On
the
-Bloch space
consists of all
such that
where
It is clear that is a Banach space.
In 1930, Cartan [2] was the the first to characterise the six types of irreducible bounded symmetric domains, which consist of four types of bounded symmetric classical domains, also referred to as Cartan domains, and two exceptional domains whose complex dimension are 16 and 27, respectively. The Cartan domains are defined as follows:
where
denotes the transpose of
Z,
denotes the conjugate of
Z, and
are positive integers. In 1998, building on the notion of bounded symmetric domains, Yin and Roos constructed a new type of domain called the Cartan-Hartogs domain[3], and Yin introduced the so-called Hua domains[4], which include the Cartan-Hartogs domains, the Cartan-Egg domains, the Hua domains, the generalized Hua domains, and the Hua construction. The generalized Hua domains are defined as follows:
where
,
denote respectively the Cartan domains of the first type, second type, third type and fourth type,
denotes the transpose of
Z,
denotes the conjugate of
Z ,
are positive integers , and
are positive real numbers. For
, the generalized Hua domain of the first kind reduces to the unit ball. Without loss of generality, we may assume that
, then
and
. We define
We also write
where
For the sake of convenience, the four types of generalized Hua domains will be referred to as .
On
, a Bers-type space
consists of all
such that
It is easy to see that is a Banach space with norm .
The boundedness and the compactness of weighted composition operators on (or between) spaces of holomorphic functions on various domains received a large attention. Wang and Liu [5] studied the boundedness and the compactness of the weighted composition operators on the Bers-type space on the open unit disc, whereas Zhou and Xu [6] characterised the boundedness and the compactness of the weighted composition operators between -Bloch space and -Bloch space, and Li [11] investigated the boundedness and the compactness of the weighted composition operators from Hardy space to Bers-type space, Zhu [19] characterised the boundedness and compactness of . For the unit poly-disk, Li and Stević [7][8] presented some necessary and sufficient conditions for the boundedness and the compactness of the weighted composition operators between and -Bloch space, whereas for the open unit ball, Li and Stević [9] studied the boundedness and the compactness of the weighted composition operators between and Bloch space [see also [14]-[17]].
Jiang[10] has charaterised the boundedness and the compactness of the weighted composition operators on the Bers-type space on the Hua domains. On the other hand, the boundedness and the compactness of the weighted composition operators from -Bloch to have not been studied in details. In this paper, we obtain some necessary conditions and sufficient conditions for the boundedness and the compactness of the weighted composition operators from -Bloch to on generalised Hua domain of the first kind by using a generalisation of Hua’s inequalities.
2. Preliminaries
Lemma 2.1. Let , then
for all and .
Proof. By the very definition of Bers-type space
, we know that
and so,
□
Lemma 2.2. Let and , q is a positive integer, then
Lemma 2.3. (see[12])
Let , if , then
and holds if and only if or .
Lemma 2.4. (see[12])
Let , then
where the equality holds if and only if .
Lemma 2.5. (see[12])
Let , if , then
where the equality holds if and only if , then . If , the equality always holds. If , then at most one of the is not zero.
be an matrix . Then, there exist an unitary matrix U and an unitary matrix V such that
where are the characteristic values of . .
Then, there exists a square matrix P such that
and the minimum value is obtained for and , where
.
Lemma 2.8. (see[12] Minkowski inequality of integration formula )
Let , then
where the equal sign holds if and only if ,.
Lemma 2.9. Let be positive integers, , and , then
for .
Proof. Decomposition in polar coordinates gives
Given
, we may consider the function
Upon differentiating with respect to
t, we obtain
An application of (2.6) then gives
This shows that
is a concave function. It follows that
is a convex function and we have
The very definition of
shows that
Hence, by inequalities (2.9) and (2.10), we obtain
□
Lemma 2.10. Let us consider , some positive intergers , , . Then, the following inequality holds
where .
Proof. If
,
, then
,
. By Lemma 2.4 and (2.3), we get
then
Therefore, by combining (2.12) and (2.11) we have
where
. □
Lemma 2.11. Given , some positive integers , , and f a holomorphic function on , then there exists a constant C such that
where .
Proof. According to Lemmas 2.2, 2.9 and (2.13),
where
.
Case 1:,
where
.
Case 2:,
where
.
Case 3:,
where
,
By combining (2.15)(2.16) and (2.17), the proof of the Lemma is complete. □
Lemma 2.12. Let be a holomorphic self-map of and . The weighted composition operator is compact if and only if is bounded and for any bounded sequence in converging to 0 uniformly on compact subsets of , as .
Proof. Assume that is compact. Let be a bounded sequence in and uniformly on compact subsets of as .
If
as
, then there exists a subsequence
of
such that
Since
is compact, there exists a subsequence of the bounded sequence
(without loss of generality, we still write
), such that
Let
K be a compact subspace of
. From Lemma 2.1, it follows that
for
Thus,
uniformly on
K. This means that for arbitrary
, such that for
, we have
for all
. Since
on compact subsets of
as
, also there exists a positive integer
,
for
whenever
. Let
and
, whenever
, we have
From the arbitrariness of , we obtain , . By the uniqueness theorem of analytic functions, we have . This shows that , which contradicts the assumption .
Conversely, suppose that
is a bounded sequence in
, then
, for all
n. Clearly
is uniformly bounded on compact subsets of
. By Montel’s theorem, there exists a subsequence
of
such that
uniformly on every compact subset of
and
. For all
, there exists a compact set
such that
. By Weierstrass’s theorem and because
as
, for
, we obtain
as
. Then, there exists a
, such that for
, we have
, for
. In addition,
, which suffices to obtain
For all
,
. We thus have
and
and
on every compact subset of
as
. Consequently, we have
which shows that
is compact. □
Lemma 2.13. Let ,if ,then
and holds if and only if . If ,then
Proof. For
, since
, there exist two
unitary matrices
and two
unitary matrices
(by Lemma 2.6) such that
By Lemma 2.7, there exists a square matrix
P such that
where
is a permutation of
.
If
,and using(2.7) and Lemma 2.8, we get
If
, by using (2.6) and Lemma 2.8, we get
For
, there exists a unitary matrix
such that
According to (2.20), we have
Thus, the inequality
holds when
, whereas the equal sign holds if and only if
.
According to (2.21), we see that
Thus, the inequality
holds when
, with the equality holding if and only if
and
.
□
Lemma 2.14. Assume and , then
with equality that holds if and only if .
Proof. Starting from the inequality
, we obtain
This completes the proof. □
Lemma 2.15. Assume and , then
Proof. By the elementary inequality
and Lemma 2.14, we have
□
Lemma 2.16. (see[18])
Assume , then there exists a constant C such that
where is an matrix where the elements of the gth row and lth column are one and the other elements are zero.
3. Boundedness of
Theorem 3.1.
Assume that , , and that () are positive integers. Let be a holomorphic self-map of , with and . If
then the weighted composition operator is bounded.
Conversely, if the weighted composition operator is bounded, then
Proof. Assume that (3.1) holds. By Lemma 2.11 and for
, we know that
For all
, we have
which implies that
is bounded.
Conversely, assume that
is bounded. For any
, let us introduce a test function
such that
In view of (2.18), it follows that
then
which means that
According to (3.3) and Lemmas 2.14, 2.16 , there exists a constant
such that
Since
, one has
Let us now consider
so that
The proof is thus completed. □
Theorem 3.2.
Assume that , , and that are positive integers . Let be a holomorphic self-map of , with and . If
then the weighted composition operator is bounded.
Conversely, if the weighted composition operator is bounded, then
Proof. Assume that (3.4) holds. By Lemma 2.11 and for
, we have
For all
, we obtain
which implies that
is bounded.
Conversely, assume that
is bounded. For
, define a test function
such that
For the test function
f, we have
From (3.3) and Lemmas 2.14, 2.16, there exists a constant
such that
Since
, we obtain
We write
, then
□
This completes the proof of the theorem.
Corollary 1.
For , we have the case of the unit ball and is bounded if and only if
when . This result is equivalent to that obtained by Li and Stević in [9].
4. Compactness of
Theorem 4.1.
Assume that , , and that are positive integers. Let be a holomorphic self-map of , with and . If and
then the weighted composition operator is compact.
Conversely, if the weighted composition operator is compact, then and
Proof. Assume that (4.1) holds. We have
If
is bounded, consider the bounded sequence
in
, which converges to 0 uniformly on compact subsets of
. Hence, there exists
such that
. By (4.1), this means that
,
, such that for
, we have
According to Lemma 2.11, we obtain
On the other hand, let us introduce the set
which is a compact subset of
. By the assumptions,
converges to 0 uniformly on any compact subset of
. From this, and since
, for such
, we have
Combining (4.4) and (4.5), we have
Consequently, making use of Lemma 2.12, we finally have that is compact.
Conversely, suppose
is compact. Let
, we have
This shows that
Consider now a sequence
in
such that
as
. If such a sequence does not exist, then condition (4.2) obviously holds. Moreover, let us introduce the following sequence of test functions
:
From (3.3) and Lemmas 2.14, 2.16, there exists a constant
such that
We have now two cases.
Case
. If
, then
where
.
Case
. If
, then
Since
, we may write
with
Using (4.8), we have
and
By using both cases
and
, we have
and then
, which means that
is bounded, where
It follows that
and
If
, then
Since
, we take
and obtain
. This implies
, then
. Let us now consider a compact subset
E of
. For
,it is easy to see that
has a positive lower bound. Thus, we have
on all compact subsets of
. If
, then
From and as , one concludes that .
The above proof shows that
on all compact subsets of
. By Lemma 2.12, this implies that
. Therefore, we conclude that
□
Theorem 4.2.
Assume that , , and that are some positive integers . Let be a holomorphic self-map of , with and . If and
then the weighted composition operator is compact.
Conversely, if the weighted composition operator is compact, then and
Proof. Assume that (4.9) holds. We have
From Theorem 3.2, it follows that
is bounded. Let
be a bounded sequence in
with
that converges to 0 uniformly on compact subsets of
. There exists
such that
. By (4.9), for any
, there is a constant
such that
for
. Using Lemma 2.11 we have
On the other hand, if we set
we have that
is a compact subset of
. For
defined in (4.11),
converges to 0 uniformly on any compact subset of
. For
, we have
According to inequalities (4.12) and (4.13), we see that
Consequently, making use of Lemma 2.12, we have that is compact.
Conversely, suppose that
is compact. Then,
is bounded. Let
, we get
This shows that Consider now a sequence in such that as . If such a sequence does not exist, then condition (4.10) obviously holds.
Moreover, let us introduce a sequence of test functions
:
From (3.3) and Lemma 2.15, it follows that there exists a constant
such that
This shows that
,
and
Taking
, we have
. This implies that
. If
E is a compact subset of
, for
, we have that
has a positive lower bound. Thus, we have
on all compact subsets of
. According to Lemma 2.12, we have that
. Hence,
Corollary 2.
For , we are back to the case of the unit ball , and is compact if and only if and
when . Also in this case, the result is analogue to that obtained by Li and Stević in [9].
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