Multiplication operators on weighted Bloch spaces of the first Cartan domains

Let ℜI(m,n)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$\Re_{I}(m,n)$\end{document} be the first Cartan domain. Motivated by some results of the multiplication operators on the holomorphic function spaces on the unit ball of Cn\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}${\mathbb {C}}^{n}$\end{document}, we study multiplication operators on weighted Bloch spaces of the first Cartan domain and obtain some necessary or sufficient conditions for the boundedness and compactness in this paper.

Let B = {z ∈ C n : |z| < 1} be the open unit ball of C n . It is obvious that B = I (1, n), which shows that I (m, n) is a generalization of B. The weighted Bloch space on B, usually It is well known that the quantity b(f ) is a seminorm of B α (B) and under the norm is a Banach space. The weighted Bloch spaces also frequently appear in the literature; see, for example, [2,[13][14][15]27]. Similarly, the weighted Bloch space B α ( I ) on I consists of all f ∈ H( I ) such that Under the norm Let X and Y be two function spaces on I . If it follows that ψf ∈ Y for all f ∈ X, then ψ is called a multiplier from X to Y , and usually M ψ : f → ψf is called a multiplication operator from X to Y . In general, ψf cannot necessarily belong to Y for some f ∈ X. In order to explain this fact, we need to introduce the Bloch space B(U 2 ) (see [24]), where U 2 = {z = (z 1 , z 2 ) : |z 1 | < 1, |z 2 | < 1} is the unit polydisk in C 2 . We say a holomorphic function f belongs to B(U 2 ) if f satisfies the condition If we consider , it is easy to see that ψf does not belong to B(U 2 ) for ψ(z) = z 1 . Hence, a natural problem is to find some conditions when ψ satisfies M ψ f ∈ Y for all f ∈ X. It is well known that the theory of the multipliers or multiplication operators on function spaces has been studied for a long time. In 1966, Talyor started an investigation of the multipliers on D α in [21]. Later on, for example, Stegenga considered the multipliers of the Dirichlet space in [10]. Now multipliers or multiplication operators between or on various function spaces of the classical domains have been studied by many authors (see, for example, [5,7,23,26]). But beyond that, there is a great interest in some generalizations of the multiplication operators on classical domains (for example, see [1,4,8,15,18] for the weighted composition operators). However, we do not find any result of multipliers or multiplication operators on the holomorphic function spaces of the first Cartan domain. In this paper, we study the multiplication operators on weighted Bloch spaces of the first Cartan domain and obtain some necessary or sufficient conditions for the boundedness and compactness.
For Z = (z ij ) m×n ∈ C m×n , let |Z| 2 = 1≤i≤m 1≤j≤n |z ij | 2 . Constants are denoted by C, they are positive and may differ from one occurrence to the next. The notation a b means that there exists a positive constant C independent of the essential variables in the quantities a and b such that a ≤ Cb.

Some lemmas
First, we have the following obvious result.

Lemma 1
Let Z = (z ij ) m×n ∈ I , then |z ij | < 1 for all i and j.
By a direct calculation, we obtain the following formula.
Lemma 2 Let f ∈ H( I ). Then, for all Z ∈ I , it follows that We need the following result (see [20]) to obtain the point evaluation estimate for the Bloch functions.

Lemma 3 Let Z ∈ I . Then there exist two unitary matrices U and V such that
Lemma 4 Let α > 0. Then there exists a positive constant C independent of f ∈ B α ( I ) and Z ∈ I such that , mα = 1, Proof If Z = 0, then the result holds obviously. Now, assume that Z = (z ij ) m×n = 0. It follows from Lemma 3 that there exist two unitary matrices U and V such that where 1 > λ 1 ≥ λ 2 ≥ · · · ≥ λ m ≥ 0 and λ 2 1 , . . . , λ 2 m are eigenvalues of ZZ T . By (2), we have It follows from (3) that By this, we have From the facts Then, from (4) and (5) we obtain where (7) is obtained by (6) by using the elementary fact This completes the proof.
Remark 1 In Lemma 4, there exists a parameter m which maybe is the biggest difference from the weighted Bloch spaces on the unit ball. Unfortunately, we do not find an effective method to avoid it. However, this also shows that this result is a generalization of the corresponding result on B α (B).
In order to study the compactness of the operator M ψ on B α ( I ), we need the following result which is similar to Proposition 3.11 in [3].
Since M ψ is compact on B α ( I ), there exist a function g ∈ B α ( I ) and a subsequence of {f n j } (without loss of generality, still written by {f n j }) such that Let K be a compact subset of I . From Lemma 4, it follows that M ψ f n jg → 0 uniformly on K as j → ∞. From this, for ε > 0, there exists a positive integer N 1 such that for all Z ∈ K , whenever j > N 1 . Since f n j → 0 uniformly on K as j → ∞, also there exists a positive integer N 2 such that |f n j (Z)| < ε for all Z ∈ K , whenever j > N 2 . Let N = max{N 1 , N 2 } and M = max Z∈K |ψ(Z)|. From (9), we have for all Z ∈ K , whenever j > N . From (10) and the arbitrariness of ε, we obtain g(Z) = 0 for all Z ∈ K , which leads to g ≡ 0 on I . This shows that lim j→∞ M ψ f n j B α ( I ) = 0 which contradicts (8). Now suppose that {f n } is a bounded sequence in B α ( I ). Then it is locally uniformly bounded on I , which shows that there exists a subsequence {f n j } of {f n } such that f n j → f uniformly on every compact subset of I as j → ∞. From this, we have f n jf → 0 uniformly on every compact subset of I as j → ∞. Consequently, we obtain In the studies of the several complex variables, the mathematician Loo-Keng Hua found the following matrix inequality in 1955.

Lemma 6 Let I -AA T and I -BB T be two Hermitian and positive definite matrices. Then
By the way, as an easy application of Lemma 6, we see that, for each Z, S ∈ I , the matrix I -ZS T is reversible. We also have the following result (see [20]).

Lemma 7 There exists a positive constant C independent of all Z, S ∈ I such that
where I ij is an m × n matrix whose element of the ith row and the jth column is 1, and the other elements are 0. To prove that f S belongs to B 1 2 ( I ) (or g S belongs to B a ( I ) for α = 1 2 ), let us recall the definition of the function matrix derivative. Let and each y ij (x) be differentiable on the interval I. Then the well-known derivative of Y (x) is defined as If we regard det(I -ZS T ) as a function of Z, we have the following result.

Lemma 8 Let S be a fixed point in I . Then on I it follows that
Proof Let A(x) be a differentiable function matrix and det A(x) = 0 for each x ∈ I. Then by the formula (see [9]) By the definition of the function matrix derivative, it is easy to see that From (13) and (14), the desired result follows.
Then we obtain By Lemmas 6-8 and (17), we obtain From (18), it follows that f S ∈ B 1 2 ( I ) and (15) holds. Next, we prove (ii). Obviously, we have Then Also by , we have From (20), it follows that g S ∈ B α ( I ) and (16) holds.
Remark 2 It is easy to see that f S and g S uniformly converge to zero on any compact subset of I as S → ∂ I .

Boundedness and compactness of M ψ on B α ( I )
First, we have the following result of the operator M ψ on B α ( I ).
then the operator M ψ is bounded on B α ( I ). (iii) For mα > 1, if ψ ∈ H ∞ ( I ) and then the operator M ψ is bounded on B α ( I ).
Proof We first prove (i). For f ∈ B α ( I ), by Lemma 4, we have Then from (23) we obtain From the assumption and (24), it follows that the operator M ψ is bounded on B α ( I ).
We prove statement (ii). For each f ∈ B α ( I ), by Lemmas 2 and 4, we have By (25), we obtain which shows that the operator M ψ is bounded on B α ( I ).
Statement (iii) can be obtained similarly. Here we omit. Now, we begin to prove (iv). Choose f (Z) ≡ 1 on I . Then by the boundedness of the operator M ψ on B α ( I ), we have which shows that is, ψ ∈ B α ( I ). Again applying the boundedness of the operator M ψ on B α ( I ) to the function g(Z) = z 11 , by Lemma 1 and (27) we obtain By (28), we have that is, ψ ∈ H ∞ α ( I ). Combining (27) and (29), we obtain ψ ∈ H ∞ α ( I ) ∩ B α ( I ).
Next, we discuss the compactness of the operator M ψ on B α ( I ).
Theorem 2 Let α > 0 and ψ be the holomorphic function on I . Then the following statements hold.
then the operator M ψ is compact on B α ( I ).
Proof Here we only prove statement (ii). Statements (i) and (iii) can be similarly proved. By Lemma 5 we only need to prove that, if {f i } is a sequence in B α ( I ) such that sup i∈N f i B α ( I ) ≤ M and f i → 0 uniformly on any compact subset of I as i → ∞, then lim i→∞ M ψ f i B α ( I ) = 0. We observe that ψ ∈ H ∞ 0 ( I ) and condition (30) imply that, for every ε > 0, there exists σ > 0 such that on K = {Z ∈ I : dist(Z, ∂ I ) < σ } it follows that ψ(Z) < ε and det I -ZZ T α ∇ψ(Z) log 2 For such ε and σ , by using (32) and Lemma 4, we have It is obvious to see that if f i → 0 uniformly on a compact subset of I as i → ∞, then We can similarly prove statement (ii) and the details are omitted.