1. Introduction
In 1921, Bergman introduced the concept of the Bergman kernel function when he studied the orthogonal expansion on the domain of the complex plane. It is well known that there exists a unique Bergman kernel function for any bounded domain in
. But for which domains can the Bergman kernel function be calculated explicitly? This is a natural question. The variety of domains for which an explicit expression for the Bergman kernel function can be calculated is not large. However, Bergman kernel functions can be explicitly calculated for some special domains. For example, Loo-keng Hua obtained Bergman kernel functions with explicit formulas for four types of irreducible symmetric classical domains in [
1]. In this paper, we will use the first irreducible symmetric classical domain usually called the first Cartan domain. This domain is defined by
where
is the conjugate of the matrix
Z,
is the transpose of
Z, and
m,
n are positive integers.
Let be the open unit ball of . When , is the open unit disk denoted by . Since , can be regarded as a generalization of . For the sake of convenience, is written by .
Let
be the set of all holomorphic functions on
. For
, the weighted-type space
on
consists of all
such that
The little weighted-type space
on
consists of all
such that
If
, then
and
are denoted by
and
, respectively. The weighted-type spaces on the unit disk and the unit ball are frequently discussed in the literature, see [
2,
3,
4,
5].
Let
be the space of all holomorphic functions on
. The weighted Zygmund space on
denoted by
consists of all
such that
where
is the radial derivative
and
. It is well known that
is a seminorm of
. For each
, we define
. Then,
is a norm on
, and
is a Banach space with this norm. We also usually use this space defined on the unit disk (see [
6]). For composition and product-type operators on or between the weighted Zygmund spaces, see, for example, refs. [
7,
8,
9] and the references therein.
For
, we define
and
We say that
is in the weighted Zygmund space
, if
If
,
is called the Zygmund space denoted by
.
is a Banach space with the norm
On
we also can define the following quantity:
The quantity
is a seminorm on
, and
is a norm of
. From the proof of Theorem 3.1 in [
10], we see that these two norms are equivalent. Therefore, we no longer need to distinguish them, uniformly denoted by
. The little weighted Zygmund space on
denoted by
consists of all
such that
It is not difficult to see that is a closed subspace of .
Let
X be a function space on
and
a function defined on
. The function
is called a multiplier on
X, if
for all
. The operator
is usually called a multiplication operator on
X. Generally speaking, there may exist some function
such that
does not belong to
X. Now, we will explain this phenomenon. To this end, we consider the Bloch space
, which consists of all
such that
where
. On
define the function
. If we choose the function
then
f belongs to
. But, it follows from a direct calculation that
does not belong to
. This shows that
is not a multiplier on
.
Multipliers and multiplication operators on function spaces have been studied for a long time. For example, Taylor started the study of the multipliers in [
11] in 1966. Stegenga studied the multipliers of the Dirichlet space in [
12] in 1980. Now, multipliers and multiplication operators on holomorphic function spaces of the unit disk
and the unit ball
have been studied (see, [
13,
14,
15,
16]). In addition, there is a great interest in some related operators for multiplication operators such as weighted composition operators, see, [
17,
18,
19,
20]. Recently, Su et al. in [
21] obtained the necessary condition and sufficient condition for the boundedness and compactness of the composition operators from
u-Bloch space to
v-Bloch space on the first Hua domain. Su et al. in [
22] gave the necessary condition and sufficient condition for the boundedness and compactness of the composition operators from
p-Bloch space to
q-Bloch space on the first Cartan-Hartogs domain. The author characterized the bounded and compact weighted composition operators on the weighted Bers-type spaces of the Hua domains in [
23]. It must be mentioned that these domains are defined by the first Cartan domain. On the other hand, we do not find any result about the multiplication operators that are defined on weighted Zygmund spaces of the first Cartan domain. Therefore, motivated by the above-mentioned studies and facts, the natural tendency is to extend the related studies to the first Cartan domain. For this purpose, we study just multiplication operators that are defined on weighted Zygmund spaces of the first Cartan domain in this paper. We obtain some necessary conditions and sufficient conditions for the boundedness and compactness of the multiplication operators.
We write for . Throughout the paper, real positive constants are denoted by C, and they may vary from place to place.
2. Some Elementary Lemmas
First, we obtain the following result from a direct calculation.
Lemma 1. Let . Then for each and , the following statement holds. To arrive at the point evaluation estimate for the functions in
, we need the following result (see [
22]).
Lemma 2. Let . Then there exist two unitary matrices U and V such thatwhere and are eigenvalues of . From the calculations, we obtain the following result.
Lemma 3. (a) If , then Lemma 4. (a) If , then there exists a positive constant C independent of and such that - (b)
If , then there exists a positive constant C independent of and such that - (c)
If , then there exists a positive constant C independent of and such that
Proof. We prove all three statements simultaneously. If
, then the lemma obviously holds. Now, assume that
. It follows from Lemma 2 that there exist two unitary matrices
U and
V such that
where
and
are eigenvalues of
. From (
1), we have
Let
. Since
, for each
we have
In particular, from (
2) we have
From the facts
and
, we obtain
which shows
Then, from Lemma 3 and (
2)–(
4), it follows that
Then, from (
5) and the fact
it follows that
From (
6), the desired result follows. The proof is complete. □
In order to prove Lemma 6, we need the following result.
Lemma 5. (a) If , then Proof. (a). We divide into three cases to prove the statement (a).
Case 1. Assume that .
Since the limitexists, we see thatis a definite integral. From this, it follows that is a positive constant. Then, we have Case 2. Assume that .
Let .
We have Since the limitexists, is a definite integral. On the other hand, sinceimplies thatfor sufficiently large x,
for sufficiently large x. From (7), we obtain that is convergent. So, is finite. Then, we have Case 3. Assume that . Let . Then, we see thatis convergent. Write . Therefore, we have Combining the above three cases, we complete the proof of (a).
(b). From the calculations, it follows that
(c). Since
for all
, we have
□
Lemma 6. (a)
If , then there exists a positive constant C independent of and such that- (b)
If , then there exists a positive constant C independent of and such that - (c)
If , then there exists a positive constant C independent of and such that
Proof. We prove all three statements simultaneously. Similar to the proof of Lemma 4, for
s,
we have
From (
3), (
4), (
8) and Lemma 5, for each
and
, we have
where
C is the constant in (a) of Lemma 5. The proof is complete. □
Remark 1. In Lemmas 4 and 6, we note the presence of the parameter m which is necessary and cannot be avoided. This maybe is the biggest difference from the corresponding results on ([24]). Unfortunately, we do not find an effective method to avoid it. However, it is shown that Lemmas 4 and 6 can be regarded as the generalizations of the corresponding results on . Replacing by in the definitions of the spaces and , respectively, we obtain the weighted Bloch space and the little weighted Bloch space on , denoted by and , respectively. is usually called the Bloch space, denoted by .
Let , and be the eigenvalues of . Then from the proof of Lemma 4, we see thatwhich shows that Proposition 1. If , then implies .
Proof. Let
. From (a) in Lemma 4, there exists a positive constant
C such that
for all
. By (
9) and (
10),
for all
, which shows that
. On other hand, from (a) in Lemma 6, there exists a positive constant
C such that
for all
. By (
9) and (
11),
for all
, which shows that
. Therefore, we prove that
. The proof is complete. □
In order to characterize the compactness, we need the following result. Since the proof is similar to that of Proposition 3.11 in [
25], we do not provide proof anymore.
Lemma 7. Let . Then the bounded operator on is compact if and only if for every bounded sequence in such that uniformly on every compact subset of as , it follows that In the case of several complex variables, Loo-keng Hua found an inequality (usually called the Hua’s inequality) in 1955. In [
22], the authors obtained a generalization of the Hua’s inequality on the first Cartan-Hartogs domain:
Setting
and
in Theorem 1 in [
22], we obtain the following inequality.
On
we define the function
For the sake of convenience, write
and
From the derivation rule of determinant functions, we obtain the following result.
Lemma 9. For each , we haveand Next, the following result holds.
Lemma 10. There exists a positive constant C independent of such that Proof. Since
Z,
, we have
and
. Then, for each
, it follows that
So, we have
and
for
and
. Let
denote the elements of the determinant in
. From Lemma 9, we see that
for
and
. With this and the definition of determinant, we obtain
where
denotes the inverse ordinal of the arrangement
. Let
. Then, from (
12) the desired result follows. The proof is complete. □
We can similarly prove the next two results. Therefore, the proofs are omitted.
Lemma 11. There exists a positive constant C independent of such that Lemma 12. There exists a positive constant C independent of such that Let
S be a fixed matrix in
. If
, we define
and if
, we define
Next, we prove that and .
Lemma 13. (a) The function belongs to . Moreover, there exists a positive constant C such that - (b)
The function belongs to . Moreover, there exists a positive constant C such that
Proof. (a). From a direct calculation, we have
From (
16), it is easy to see that
Then, from (
17) and
for each
i and
j, we have
From (
18) and Lemmas 10 and 11, we have
It is easy to see that
. From this and (
19), it follows that
and (
13) holds.
The statement (b) and (
14) can be similarly proven, and the details are omitted. The proof is complete. □
Remark 2. Since converges to zero as , we see that and uniformly converge to zero on any compact subset of as .
3. Boundedness and Compactness of on
We begin to study the boundedness and compactness of the multiplication operators on . We have the following result about the boundedness.
Theorem 1. Let and . Then the following statements hold.
- (a)
For , if , then the operator is bounded on .
- (b)
For , if and
then the operator is bounded on .
- (c)
For , if and
then the operator is bounded on .
- (d)
For , if ,
andthen the operator is bounded on . - (e)
For , if ,
andthen the operator is bounded on . Proof. We prove the statement (a). For
and
, it follows from Proposition 1 that
. Then, for
, it follows from (a) in Lemmas 4 and 6 that
From (
20) and the basic fact
, we have
It follows from (
21) that the operator
is bounded on
.
Now, we prove the statement (b). From (b) in Lemma 4 and (a) in Lemma 6, we have
From (
22) and the assumption, it follows that the operator
is bounded on
.
Next, we prove the statement (c). From (c) in Lemma 4 and (a) in Lemma 6, we have
From (
23) and the assumption, it follows that the operator
is bounded on
.
We prove the statement (d). From (c) in Lemma 4 and (b) in Lemma 6, it follows that
From (
24) and the assumption, it follows that the operator
is bounded on
.
Finally, we prove the statement (e). From (c) in Lemmas 4 and 6, we have
From (
25) and the assumption, it follows that the operator
is bounded on
. The proof is complete. □
Next, we consider the compactness of the operator on .
Theorem 2. Let , and the operator be bounded on . Then the following statements hold.
- (a)
For , if , then the operator is compact on .
- (b)
For , if and
then the operator is compact on .
- (c)
For , if and
then the operator is compact on .
- (d)
For , if ,
andthen the operator is compact on . - (e)
For , if ,
andthen the operator is compact on . Proof. We first prove the statement (a). Let
be a bounded sequence in
and
uniformly on any compact subset of
as
. To prove that the bounded operator
is compact on
, by Lemma 7 we only need to prove that
Since
is bounded, we assume that
, where
M is a positive number. From (a) in Lemmas 4 and 6, it follows that
for all
. Since
, for arbitrary
there exists an
such that on
it follows that
for
, 2. Then, from (
32) and (
33), it follows that
It is obvious to see that
converges to zero uniformly on any compact subset of
as
implies that
and
also perform the same convergence as
. Since
is a compact subset of
and the obvious fact
as
, it follows from (
34) that
which shows that the operator
is compact on
.
Now, we prove the statement (b). Assume that
is a sequence in
such that
and
uniformly on any compact subset of
as
. Then by Lemma 7 we only need to prove that
First, since the sequence
is bounded, by (a) in Lemma 6 there exists a positive constant
such that
for all
. From the conditions, we see that for arbitrary
there exists an
such that on
it follows that
and
For above
and
, by using (
35)–(
37), (b) in Lemma 4 and (a) in Lemma 6, we have
It is obvious to see that
converges to zero uniformly on any compact subset of
as
implies that
and
also perform the same convergence as
. From (
38), the compactness of
, and the obvious fact
as
, it follows that
which shows that the operator
is compact on
.
Next, we prove the statement (c). Assume that
is a sequence in
such that
and
uniformly on any compact subset of
as
. Then by Lemma 7 we only need to prove that
Since
is bounded in
, by (a) in Lemma 6 there exists a positive constant
C such that
for all
. From (c) in Lemma 4, it follows that there exists a positive constant
such that
Since
and the assumption (
27) holds, for arbitrary
there exists an
such that on
it follows that
and
For above
and
, by using (
39), (
40), (c) in Lemma 4 and (a) in Lemma 6, we have
Since
converges to zero uniformly on any compact subset of
as
implies that
and
also perform the same convergence as
. From (
41), the compactness of
, and the obvious fact
as
, it follows that
which shows that the operator
is compact on
.
We prove the statement (d). Assume that
is a sequence in
such that
and
uniformly on any compact subset of
as
. Then by Lemma 7 we only need to prove that
Since
is bounded in
, by (b) in Lemma 6 there exists a positive constant
C such that
for all
and
. From (c) in Lemma 4, it follows that there exists a positive constant
such that
for all
and
. Since
and the assumptions (
28) and (
29) hold, for arbitrary
there exists an
such that on
it follows that
and
For above
and
, by using (
42)–(
45), (c) in Lemma 4 and (b) in Lemma 6, we have
Since
converges to zero uniformly on any compact subset of
as
implies that
and
also perform the same convergence as
. From (
46), the compactness of
, and the obvious fact
as
, it follows that
which shows that the operator
is compact on
.
Finally, we prove the statement (e). Assume that
is a sequence in
such that
and
uniformly on any compact subset of
as
. Then by Lemma 7 we only need to prove that
Since
is bounded in
, by (c) in Lemma 6 there exists a positive constant
C such that
for all
and
. From (c) in Lemma 4, it follows that there exists a positive constant
such that
for all
and
. Since
and the assumptions (
30) and (
31) hold, for arbitrary
there exists an
such that on
it follows that
and
For above
and
, by using (
47)–(
50), (c) in Lemmas 4 and 6, we have
Since
converges to zero uniformly on any compact subset of
as
implies that
and
also perform the same convergence as
. From (
51), the compactness of
, and the obvious fact
as
, it follows that
which shows that the operator
is compact on
. The proof is complete. □
We have the following necessary conditions for the compactness of on , which can be easily obtained by using the functions and .
Theorem 3. Let and . Then the following statements hold.
- (a)
For , if the operator is compact on , then - (b)
For , if the operator is compact on , then