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Boundedness and Compactness of Weighted Composition Operators from α-Bloch Spaces to Bers-Type Spaces on Generalized Hua Domains of the First Kind

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04 September 2023

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05 September 2023

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Abstract
We address weighted composition operators ψCφ from α-Bloch spaces to Bers-type spaces of bounded holomorphic functions on Y, where Y is a generalized Hua domain of the first kind, and obtain some necessary and sufficient conditions for the boundedness and compactness of those operators.
Keywords: 
Subject: Computer Science and Mathematics  -   Analysis

MSC:  32A27; 47B33

1. Introduction

Let Ω be a bounded domain of C n and H ( Ω ) the class of all holomorphic functions on Ω . Then consider a holomorphic self-map ϕ of Ω and a function ψ H ( Ω ) . The linear operator
( ψ C ϕ f ) ( z ) = ψ ( z ) f ( ϕ ( z ) ) ,
is referred to as a weighted composition operator for f H ( Ω ) . If ψ ( z ) 1 , it reduces to the composition operator, whereas for ϕ ( z ) = z it becomes the multiplication operator. For any given holomorphic function f, ( ψ C ϕ f ) ( z ) 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 α -Bloch spaces B α to Bers-type spaces built on generalised Hua domains of the first kind. On GHE I the α -Bloch space B α consists of all f H ( GHE I ) such that
f B α : = | f ( 0 , 0 ) | + sup ( Z , ξ ) GHE I [ det ( I Z Z ¯ ) k ξ p 2 ] α | f ( Z , ξ ) | < ,
where
f ( Z , ξ ) = f ( Z , ξ ) z 11 , f ( Z , ξ ) z 12 , · · · , f ( Z , ξ ) z m n , f ( Z , ξ ) ξ 1 , · · · , f ( Z , ξ ) ξ r .
It is clear that B α ( GHE I ) 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:
I ( m , n ) : = Z C m × n : I m Z Z ¯ > 0 , II ( p ) : = Z C p ( p + 1 ) 2 : I m Z Z ¯ > 0 , Z = Z , III ( q ) : = Z C q ( q 1 ) 2 : I m Z Z ¯ > 0 , Z = Z , IV ( n ) : = z C n : 1 + | z z | 2 2 z z ¯ > 0 , 1 | z z | 2 > 0 .
where Z denotes the transpose of Z, Z ¯ denotes the conjugate of Z, and m , n , p , q 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:
GHE I ( N 1 , N 2 , , N r ; m , n ; p 1 , p 2 , , p r ; k ) = ξ j C N j , Z I ( m , n ) : j = 1 r | ξ j | 2 p j < det ( I Z Z ¯ ) k , j = 1 , 2 , , r GHE II ( N 1 , N 2 , , N r ; p ; p 1 , p 2 , , p r ; k ) = ξ j C N j , Z II ( p ) : j = 1 r | ξ j | 2 p j < det ( I Z Z ¯ ) k , j = 1 , 2 , , r GHE III ( N 1 , N 2 , , N r ; q ; p 1 , p 2 , , p r ; k ) = ξ j C N j , Z III ( q ) : j = 1 r | ξ j | 2 p j < det ( I + Z Z ¯ ) k , j = 1 , 2 , , r GHE IV ( N 1 , N 2 , , N r ; n ; p 1 , p 2 , , p r ; k ) = ξ j C N j , z IV ( n ) : j = 1 r | ξ j | 2 p j < ( 1 + | z z | 2 2 z z ¯ ) k , j = 1 , 2 , , r
where ξ j = ( ξ j 1 , , ξ j N j ) , j = 1 , , r , I ( m , n ) , II ( p ) , III ( q ) , IV ( n ) denote respectively the Cartan domains of the first type, second type, third type and fourth type, Z denotes the transpose of Z, Z ¯ denotes the conjugate of Z , N 1 , , N r , m , n , p , q are positive integers , and p 1 , , p r are positive real numbers. For k = 1 , m = 1 , p 1 = = p r = 1 , the generalized Hua domain of the first kind reduces to the unit ball. Without loss of generality, we may assume that N j = 1 , then ξ j C , j = 1 , , r , ξ = ( ξ 1 , , ξ r ) and ξ p 2 = j = 1 r | ξ j | 2 p j . We define
ξ , t p = ξ 1 , t 1 p 1 + ξ 2 , t 2 p 2 + + ξ r , t r p r .
We also write
| ξ , t p | | ξ 1 , t 1 p 1 | + | ξ 2 , t 2 p 2 | + + | ξ r , t r p r | | ξ 1 | p 1 | t 1 | p 1 + + | ξ r | p r | t r | p r = | α , β | | α | | β | = ξ p t p ,
where | ξ i | p i = α i , | t i | p i = β i ( i = 1 , , r ) , α = ( α 1 , , α r ) , β = ( β 1 , , β r ) .
For the sake of convenience, the four types of generalized Hua domains will be referred to as GHE I , GHE II , GHE III , GHE IV .
On GHE I , a Bers-type space A β consists of all f H ( GHE I ) such that
f A β : = sup ( Z , ξ ) GHE I [ det ( I Z Z ¯ ) k ξ p 2 ] β | f ( Z , ξ ) | < .
It is easy to see that A β ( GHE I ) 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 α -Bloch space and β -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 D ϕ , u n : B H α . 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 H and α -Bloch space, whereas for the open unit ball, Li and Stević [9] studied the boundedness and the compactness of the weighted composition operators between H 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 α -Bloch to A β 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 α -Bloch to A β on generalised Hua domain of the first kind by using a generalisation of Hua’s inequalities.

2. Preliminaries

Lemma 2.1.Let β > 0 , then
| f ( Z , ξ ) | f A β [ det ( I Z Z ¯ ) k ξ p 2 ] β .
for all ( Z , ξ ) GHE I and f A β ( GHE I ) .
Proof. 
By the very definition of Bers-type space A β , we know that
f A β = sup ( Z , ξ ) GHE I [ det ( I Z Z ¯ ) k ξ p 2 ] β | f ( Z , ξ ) | < ,
and so,
| f ( Z , ξ ) | f A β [ det ( I Z Z ¯ ) k ξ p 2 ] β .
Lemma 2.2. Let 0 < a 1 , 0 < b 1 and b a , q is a positive integer, then
a b q ( a 1 q b 1 q )
Proof. 
a b = ( a 1 q ) q ( b 1 q ) q = ( a 1 q b 1 q ) ( a 1 q × ( q 1 ) + a 1 q × ( q 2 ) b 1 q + + b 1 q × ( q 1 ) ) q ( a 1 q b 1 q ) .
Lemma 2.3. (see[12]) Let x 1 , if 0 < α 1 , then
( 1 + x ) α 1 + α x ,
if α < 0 or α > 1 , then
( 1 + x ) α 1 + α x ,
and " = " holds if and only if x = 0 or α = 1 .
Lemma 2.4. (see[12]) Let a k 0 , k = 1 , 2 , , m , then
( a 1 · a 2 · · · a m ) 1 m a 1 + a 2 + · · · + a m m ,
where the equality holds if and only if a 1 = a 2 = = a m .
Lemma 2.5. (see[12]) Let a k C , if p 1 , then
k = 1 n | a k | p k = 1 n | a k | p n p 1 k = 1 n | a k | p .
If 0 < p < 1 , then
k = 1 n | a k | p k = 1 n | a k | p n p 1 k = 1 n | a k | p ,
where the equality holds if and only if p > 1 , then | a 1 | = = | a n | . If p = 1 , the equality always holds. If 0 < p < 1 , then at most one of the a 1 , , a n is not zero.
Lemma 2.6. (see[13]) Let
Z = z 11 z 12 z 1 n z 21 z 22 z 2 n z m 1 z m 2 z m n
be an m × n matrix ( m n ) . Then, there exist an m × m unitary matrix U and an n × n unitary matrix V such that
Z = U λ 1 0 0 0 0 0 λ 2 0 0 0 0 0 λ m 0 0 V ( λ 1 λ 2 λ m 0 ) ,
and
Z Z ¯ = U λ 1 2 0 0 0 λ 2 2 0 0 0 λ m 2 U ¯ ,
where λ 1 2 , , λ m 2 are the characteristic values of Z Z ¯ . I Z Z ¯ > 0 λ 1 < 1 .
Lemma 2.7. (see[13]) Let
Λ 1 = λ 1 0 0 0 λ 2 0 0 0 0 λ m ( λ 1 λ 2 λ m 0 ) ,
Λ 2 = μ 1 0 0 0 μ 2 0 0 0 0 μ m ( μ 1 μ 2 μ m 0 ) ,
satisfying
λ j μ k < 1 ( j , k = 1 , , m ) ,
Then, there exists a square matrix P such that
inf U U ¯ = I , V V ¯ = I | det ( I Λ 1 U Λ 2 U ¯ V ) | = | det ( I Λ 1 P Λ 2 P ) | ,
and the minimum value is obtained for U = Θ P and V = I , where
Θ = e i θ 1 0 0 0 e i θ 2 0 0 0 0 e i θ m
.
Lemma 2.8. (see[12] Minkowski inequality of integration formula ) Let a k , b k 0 , k = 1 , 2 , n , then
k = 1 n ( a k + b k ) 1 n k = 1 n a k 1 n + k = 1 n b k 1 n ,
where the equal sign holds if and only if a k = c b k , k = 1 , 2 , , n .
Lemma 2.9.Let p i ( i = 1 , 2 , , r ) be positive integers, 0 < k m 1 , and t [ 0 , 1 ] , then
1 det ( I t 2 Z Z ¯ ) k + t ξ p 2 t 2 1 det ( I Z Z ¯ ) k + ξ p 2 ,
for ( Z , ξ ) GHE I .
Proof. 
Decomposition in polar coordinates gives
det ( I t Z Z ¯ ) k = i = 1 m ( 1 t λ i 2 ) k .
Given λ i 2 = i , i = 1 , 2 , , m , we may consider the function
f ( t ) = i = 1 m ( 1 t i ) k , t [ 0 , 1 ]
ln f ( t ) = k i = 1 m ln ( 1 t i ) ,
Upon differentiating with respect to t, we obtain
f ( t ) = f ( t ) k i = 1 m i 1 t i 0 , f ( t ) = f ( t ) k i = 1 m i 1 t i f ( t ) k i = 1 m i 2 ( 1 t i ) 2 = f ( t ) k 2 i = 1 m i 1 t i 2 f ( t ) k i = 1 m i 2 ( 1 t i ) 2 = f ( t ) k k i = 1 m i 1 t i 2 i = 1 m i 2 ( 1 t i ) 2 .
An application of (2.6) then gives
f ( t ) = f ( t ) k k i = 1 m i 1 t i 2 i = 1 m i 2 ( 1 t i ) 2 f ( t ) k ( k m 1 ) i = 1 m i 2 ( 1 t i ) 2 0 .
This shows that f ( t ) is a concave function. It follows that
g ( t ) = 1 f ( t ) = 1 i = 1 m ( 1 t i ) k , t [ 0 , 1 ] ,
is a convex function and we have
1 i = 1 m ( 1 t i ) k t [ 1 i = 1 m ( 1 i ) k ] .
The very definition of ξ p 2 shows that
t ξ p 2 = | t ξ 1 | 2 p 1 + | t ξ 2 | 2 p 2 + + | t ξ r | 2 p r t 2 ( | ξ 1 | 2 p 1 + | ξ 2 | 2 p 2 + + | ξ r | 2 p r ) = t 2 ξ p 2 .
Hence, by inequalities (2.9) and (2.10), we obtain
1 det ( I t 2 Z Z ¯ ) k + t ξ p 2 t 2 1 det ( I Z Z ¯ ) k + ξ p 2 .
Lemma 2.10.Let us consider 0 < m k 1 , some positive intergers p j ( j = 1 , 2 , , r ) , t [ 0 , 1 ] , ( Z , ξ ) GHE I , q = max { p 1 , p 2 , , p r } . Then, the following inequality holds
| ( Z , ξ ) ¯ | M 1 det ( I Z Z ¯ ) k q + ξ p 2 q ,
where M = max { q k , r 1 1 q } .
Proof. 
If t [ 0 , 1 ] , ( Z , ξ ) GHE I , then ( t Z , t ξ ) GHE I , | Z | 2 = tr ( Z Z ¯ ) = λ 1 2 + λ 2 2 + · · · + λ m 2 . By Lemma 2.4 and (2.3), we get
det ( I Z Z ¯ ) k q = i = 1 m ( 1 λ i 2 ) k q = i = 1 m ( 1 λ i 2 ) 1 m m k q 1 m ( m i = 1 m λ i 2 ) m k q = ( 1 1 m i = 1 m λ i 2 ) m k q 1 m k q · 1 m | Z | 2 = 1 k q | Z | 2 ,
then
| Z | 2 q k [ 1 det ( I Z Z ¯ ) k q ] .
Using (2.7) one has
ξ p 2 q = ( | ξ 1 | 2 p 1 + | ξ 2 | 2 p 2 + + | ξ r | 2 p r ) 1 q r 1 q 1 ( | ξ 1 | 2 p 1 q + | ξ 2 | 2 p 2 q + + | ξ r | 2 p r q ) r 1 q 1 ( | ξ 1 | 2 + | ξ 2 | 2 + + | ξ r | 2 ) = r 1 q 1 | ξ | 2 ,
then
| ξ | 2 r 1 1 q ξ p 2 q .
Therefore, by combining (2.12) and (2.11) we have
| ( Z , ξ ) ¯ | = | Z | 2 + | ξ | 2 q k [ 1 det ( I Z Z ¯ ) k q ] + r 1 1 q ξ p 2 q M 1 det ( I Z Z ¯ ) k q + ξ p 2 q ,
where M = max { q k , r 1 1 q } . □
Lemma 2.11.Given 0 < k m 1 , p j some positive integers ( j = 1 , 2 , , r ) , ( Z , ξ ) GHE I , q = max { p 1 , p 2 , , p r } and f a holomorphic function on B α ( GHE I ) , then there exists a constant C such that
| f ( Z , ξ ) | C f B α 0 < α < 1 C f B α ln 2 q det ( I Z Z ¯ ) k ξ p 2 α = 1 C f B α 1 [ det ( I Z Z ¯ ) k ξ p 2 ] α 1 α > 1
where ( Z , ξ ) = ( z 11 , z 12 , , z m n , ξ 1 , ξ 2 , , ξ r ) .
Proof. 
According to Lemmas 2.2, 2.9 and (2.13),
| f ( Z , ξ ) | = | f ( 0 , 0 ) + 0 1 f ( t Z , t ξ ) , ( Z , ξ ) ¯ d t | | f ( 0 , 0 ) | + 0 1 | f ( t Z , t ξ ) | | ( Z , ξ ) ¯ | d t = | f ( 0 , 0 ) | + | ( Z , ξ ) ¯ | 0 1 [ det ( I t 2 Z Z ¯ ) k t ξ p 2 ] α | f ( t Z , t ξ ) | [ det ( I t 2 Z Z ¯ ) k t ξ p 2 ] α d t 1 + 0 1 | ( Z , ξ ) ¯ | [ det ( I t 2 Z Z ¯ ) k t ξ p 2 ] α d t f B α = 1 + 0 1 | ( Z , ξ ) ¯ | [ 1 ( 1 det ( I t 2 Z Z ¯ ) k + t ξ p 2 ) ] α d t f B α 1 + M 0 1 1 det ( I Z Z ¯ ) k q + ξ p 2 q [ 1 t 2 ( 1 det ( I Z Z ¯ ) k + ξ p 2 ) ] α d t f B α 1 + M 0 1 1 1 q ( det ( I Z Z ¯ ) k ξ p 2 ) [ 1 t 2 ( 1 1 q ( det ( I Z Z ¯ ) k ξ p 2 ) ) ] α d t f B α = 1 + M 0 1 [ 1 t 2 2 ] α d t f B α = 1 + M 0 1 [ ( 1 t ) ( 1 + t ) ] α d t f B α 1 + M 0 1 ( 1 t ) α d t f B α ,
where = 1 1 q ( det ( I Z Z ¯ ) k ξ p 2 ) .
Case 1: 0 < α < 1 ,
| f ( Z , ξ ) | 1 + M 1 α [ 1 ( 1 ) 1 α ] f B α ( 1 + M 1 α ) f B α C f B α ,
where C = 1 + M 1 α .
Case 2: α = 1 ,
| f ( Z , ξ ) | 1 + M 0 1 1 t d t f B α = 1 + M ln 1 1 f B α = 1 + M ln 1 + ( 1 ) ( 1 + ) f B α
1 + M ln 2 1 2 f B α 1 ln 2 ln 2 1 2 + M ln 2 1 2 f B α 1 ln 2 + M ln 2 1 2 f B α = C f B α ln 2 q det ( I Z Z ¯ ) k ξ p 2 ,
where C = 1 ln 2 + M .
Case 3: α > 1 ,
| f ( Z , ξ ) | 1 + M α 1 1 ( 1 ) α 1 1 f B α C + C 1 ( 1 ) α 1 1 f B α = C f B α 1 ( 1 ) α 1 = C f B α ( 1 + ) α 1 [ ( 1 ) ( 1 + ) ] α 1 2 α 1 C f B α 1 ( 1 2 ) α 1 = C f B α 1 [ det ( I Z Z ¯ ) k ξ p 2 ] α 1
where C = ( 2 q ) α 1 C , C = max { 1 , M α 1 } .
By combining (2.15)(2.16) and (2.17), the proof of the Lemma is complete. □
Lemma 2.12.Let ϕ = ( ϕ 11 , ϕ 12 ϕ m n + r ) be a holomorphic self-map of GHE I and ψ H ( GHE I ) . The weighted composition operator ψ C ϕ : B α ( GHE I ) A β ( GHE I ) is compact if and only if ψ C ϕ is bounded and for any bounded sequence { f n } n 1 in B α ( GHE I ) converging to 0 uniformly on compact subsets of GHE I , ψ C ϕ f n A β 0 as n .
Proof. 
Assume that ψ C ϕ : B α ( GHE I ) A β ( GHE I ) is compact. Let { f n } n 1 be a bounded sequence in B α ( GHE I ) and f n 0 uniformly on compact subsets of GHE I as n .
If ψ C ϕ f n A β 0 as n , then there exists a subsequence { f n j } j 1 of { f n } n 1 such that
inf j N ψ C ϕ f n j A β > 0 .
Since ψ C ϕ is compact, there exists a subsequence of the bounded sequence { f n j } j 1 (without loss of generality, we still write { f n j } j 1 ), such that
lim j ψ C ϕ f n j f A β = 0 , f A β ( GHE I ) .
Let K be a compact subspace of GHE I . From Lemma 2.1, it follows that
| ( ψ C ϕ f n j f ) ( Z , ξ ) | ψ C ϕ f n j f A β [ det ( I Z Z ¯ ) k ξ p 2 ] β , j .
for ( Z , ξ ) K GHE I . Thus, ψ C ϕ f n j f 0 uniformly on K. This means that for arbitrary ε > 0 , N 1 > 0 , such that for j > N 1 , we have
| ψ ( Z , ξ ) f n j ( ϕ ( Z , ξ ) ) f ( Z , ξ ) | < ε .
for all ( Z , ξ ) K . Since f n j 0 on compact subsets of GHE I as j , also there exists a positive integer N 2 , | f n j ( ϕ ( Z , ξ ) ) | < ε for ( Z , ξ ) K whenever j > N 2 . Let N = max { N 1 , N 2 } and M = max ( Z , ξ ) K | ψ ( Z , ξ ) | , whenever j > N , we have
| f ( Z , ξ ) | | f n j ( ϕ ( Z , ξ ) ) | max ( Z , ξ ) K | ψ ( Z , ξ ) | + ε ( M + 1 ) ε , ( Z , ξ ) K .
From the arbitrariness of ε , we obtain f ( Z , ξ ) 0 , ( Z , ξ ) K . By the uniqueness theorem of analytic functions, we have f ( Z , ξ ) 0 , ( Z , ξ ) GHE I . This shows that lim j ψ C ϕ f n j A β = 0 , which contradicts the assumption inf j N ψ C ϕ f n j A β > 0 .
Conversely, suppose that { f n } n 1 is a bounded sequence in B α ( GHE I ) , then f n B α D 1 , for all n. Clearly { f n } n 1 is uniformly bounded on compact subsets of GHE I . By Montel’s theorem, there exists a subsequence { f n j } j 1 of { f n } n 1 such that f n j f uniformly on every compact subset of GHE I and f B α ( GHE I ) . For all ( Z 0 , ξ 0 ) GHE I , there exists a compact set K ( Z 0 , ξ 0 ) such that ( Z 0 , ξ 0 ) K ( Z 0 , ξ 0 ) GHE I . By Weierstrass’s theorem and because f n j f as j , for ( Z , ξ ) K ( Z 0 , ξ 0 ) , we obtain f n j f as j . Then, there exists a J 0 > 0 , such that for j > J 0 , we have | f n j ( Z , ξ ) f ( Z , ξ ) | < 1 , for ( Z , ξ ) K ( Z 0 , ξ 0 ) . In addition, | f ( Z 0 , ξ 0 ) | | f ( Z 0 , ξ 0 ) f n j ( Z 0 , ξ 0 ) | + | f n j ( Z 0 , ξ 0 ) | , which suffices to obtain
[ det ( I Z 0 Z 0 ¯ ) k ξ 0 p 2 ] α | f ( Z 0 , ξ 0 ) | [ det ( I Z 0 Z 0 ¯ ) k ξ 0 p 2 ] α | f ( Z 0 , ξ 0 ) f n j ( Z 0 , ξ 0 ) | + [ det ( I Z 0 Z 0 ¯ ) k ξ 0 p 2 ] α | f n j ( Z 0 , ξ 0 ) | 1 + f n j B α 1 + D 1 .
For all ( Z , ξ ) GHE I , [ det ( I Z Z ¯ ) k ξ p 2 ] α | f ( Z , ξ ) | 1 + D 1 . We thus have f B α 1 + D 1 and f n j f B α f n j B α + f B α 2 D 1 + 1 and f n j f 0 on every compact subset of GHE I as j . Consequently, we have
lim j ψ C ϕ ( f n j f ) A β = lim j ψ C ϕ f n j ψ C ϕ f A β = 0 ,
which shows that ψ C ϕ : B α ( GHE I ) A β ( GHE I ) is compact. □
Lemma 2.13.Let ( Z , ξ ) , ( S , t ) GHE I ,if 0 < k m 1 ,then
det ( I m Z Z ¯ ) k + det ( I m S S ¯ ) k 2 | det ( I m Z S ¯ ) k | ,
and " = " holds if and only if ( Z , ξ ) = ( S , t ) . If k m > 1 ,then
det ( I m Z Z ¯ ) k + det ( I m S S ¯ ) k 2 m k | det ( I m Z S ¯ ) k | .
Proof. 
For m = n , since ( Z , ξ ) , ( S , t ) GHE I , there exist two m × m unitary matrices U 1 , U 2 and two n × n unitary matrices V 1 , V 2 (by Lemma 2.6) such that
Z = U 1 λ 1 0 0 0 λ 2 0 0 0 λ m V 1 = U 1 Λ 1 V 1 ( 1 > λ 1 λ 2 λ m 0 )
S = U 2 μ 1 0 0 0 μ 2 0 0 0 μ m V 2 = U 2 Λ 2 V 2 ( 1 > μ 1 μ 2 μ m 0 ) .
Then, one has
det ( I Z S ¯ ) = det ( I U 1 Λ 1 V 1 V 2 ¯ Λ 2 ¯ U 2 ¯ ) = det ( U 1 U 1 ¯ U 1 Λ 1 V 1 V 2 ¯ Λ 2 ¯ U 2 ¯ ) = det U 1 det ( U 1 ¯ Λ 1 V 1 V 2 ¯ Λ 2 ¯ U 2 ¯ ) = det ( I Λ 1 V 1 V 2 ¯ Λ 2 ¯ V 2 V 1 ¯ V 1 V 2 ¯ U 2 ¯ U 1 ) .
By Lemma 2.7, there exists a square matrix P such that
| det ( I Z S ¯ ) | | det ( I Λ 1 P Λ 2 P ) | = i = 1 m ( 1 λ i μ k i ) ,
where k 1 , k 2 , , k m is a permutation of 1 , 2 , , m .
If 0 < k m 1 ,and using(2.7) and Lemma 2.8, we get
2 | det ( I Z S ¯ ) k | = 2 1 m k · 2 m k | det ( I Z S ¯ ) k | = 2 1 m k [ 2 m | det ( I Z S ¯ ) | ] k 2 1 m k 2 m i = 1 m ( 1 λ i μ k i ) k 2 1 m k { i = 1 m [ ( 1 λ i 2 ) + ( 1 μ k i 2 ) ] } 1 m m k 2 1 m k i = 1 m ( 1 λ i 2 ) 1 m + i = 1 m ( 1 μ k i 2 ) 1 m m k 2 1 m k · 2 m k 1 i = 1 m ( 1 λ i 2 ) 1 m × m k + i = 1 m ( 1 μ k i 2 ) 1 m × m k = i = 1 m ( 1 λ i 2 ) k + i = 1 m ( 1 μ k i 2 ) k = det ( I Z Z ¯ ) k + det ( I S S ¯ ) k .
If k m > 1 , by using (2.6) and Lemma 2.8, we get
2 m k | det ( I Z S ¯ ) k | = [ 2 m | det ( I Z S ¯ ) | ] k 2 m i = 1 m ( 1 λ i μ k i ) k { i = 1 m [ ( 1 λ i 2 ) + ( 1 μ k i 2 ) ] } 1 m m k
i = 1 m ( 1 λ i 2 ) 1 m + i = 1 m ( 1 μ k i 2 ) 1 m m k i = 1 m ( 1 λ i 2 ) 1 m × m k + i = 1 m ( 1 μ k i 2 ) 1 m × m k = i = 1 m ( 1 λ i 2 ) k + i = 1 m ( 1 μ k i 2 ) k = det ( I Z Z ¯ ) k + det ( I S S ¯ ) k .
For m < n , there exists a unitary matrix U ( n ) such that
Z = ( Z 1 ( m ) , 0 ) U , S = ( S 1 ( m ) , S 2 ) U .
According to (2.20), we have
2 | det ( I Z S ¯ ) k | = 2 | det ( I Z 1 S 1 ¯ ) k | det ( I Z 1 Z 1 ¯ ) k + det ( I S 1 S 1 ¯ ) k det ( I Z 1 Z 1 ¯ ) k + det ( I S 1 S 1 ¯ S 2 S 2 ¯ ) k = det ( I Z Z ¯ ) k + det ( I S S ¯ ) k .
Thus, the inequality
2 | det ( I Z S ¯ ) k | det ( I Z Z ¯ ) k + det ( I S S ¯ ) k ,
holds when m n , whereas the equal sign holds if and only if Z = S .
According to (2.21), we see that
2 m k | det ( I Z S ¯ ) k | = 2 m k | det ( I Z 1 S 1 ¯ ) k | det ( I Z 1 Z 1 ¯ ) k + det ( I S 1 S 1 ¯ ) k det ( I Z 1 Z 1 ¯ ) k + det ( I S 1 S 1 ¯ S 2 S 2 ¯ ) k = det ( I Z Z ¯ ) k + det ( I S S ¯ ) k .
Thus, the inequality
2 m k | det ( I Z S ¯ ) k | det ( I Z Z ¯ ) k + det ( I S S ¯ ) k ,
holds when m n , with the equality holding if and only if Z = S and m k = 1 .
Lemma 2.14.Assume ( Z , ξ ) , ( S , t ) GHE I and 0 < k m 1 , then
[ det ( I m Z Z ¯ ) k ξ p 2 ] + [ det ( I m S S ¯ ) k t p 2 ] 2 | | det ( I m Z S ¯ ) k | ξ p t p | ,
with equality that holds if and only if ( Z , ξ ) = ( S , t ) .
Proof. 
Starting from the inequality a 2 + b 2 2 a b , we obtain
ξ p 2 + t p 2 2 ξ p t p .
Then, by (2.18), we get
2 | | det ( I Z S ¯ ) k | ξ p t p | = 2 | det ( I Z S ¯ ) k | 2 ξ p t p det ( I Z Z ¯ ) k + det ( I S S ¯ ) k ξ p 2 t p 2 = [ det ( I Z Z ¯ ) k ξ p 2 ] + [ det ( I S S ¯ ) k t p 2 ] .
This completes the proof. □
Lemma 2.15.Assume ( Z , ξ ) , ( S , t ) GHE I and 0 < k m 1 , then
det ( I m Z Z ¯ ) k ξ p 2 det ( I m S S ¯ ) k t p 2 | | det ( I m Z S ¯ ) k | ξ p t p | 2 ,
Proof. 
By the elementary inequality a + b 2 a b and Lemma 2.14, we have
det ( I m Z Z ¯ ) k ξ p 2 det ( I m S S ¯ ) k t p 2 [ det ( I m Z Z ¯ ) k ξ p 2 ] + [ det ( I m S S ¯ ) k t p 2 ] 2 2 | | det ( I m Z S ¯ ) k | ξ p t p | 2
Lemma 2.16. (see[18]) Assume Z , S I ( m , n ) , then there exists a constant C such that
| det ( I m Z S ¯ ) | 1 g m 1 l n | tr [ ( I m Z S ¯ ) 1 I g l S ¯ ] | 2 1 2 C .
where I g l is an m × n matrix where the elements of the gth row and lth column are one and the other elements are zero.

3. Boundedness of ψ C ϕ : B α A β

Theorem 3.1.Assume that α = 1 , β > 0 , 0 < k m 1 , and that p j ( j = 1 , 2 , , r ) are positive integers. Let ϕ = ( ϕ 11 , ϕ 12 ϕ m n + r ) be a holomorphic self-map of GHE I , with ψ H ( GHE I ) and ( Z ϕ , ξ ϕ ) = ϕ ( Z , ξ ) . If
K 1 : = sup ( Z , ξ ) GHE I | ψ ( Z , ξ ) | [ det ( I Z Z ¯ ) k ξ p 2 ] β ln 2 q det ( I Z ϕ Z ϕ ¯ ) k ξ ϕ p 2 < ,
then the weighted composition operator ψ C ϕ : B α ( GHE I ) A β ( GHE I ) is bounded.
Conversely, if the weighted composition operator ψ C ϕ : B α ( GHE I ) A β ( GHE I ) is bounded, then
K 2 : = sup ( Z , ξ ) GHE I | ψ ( Z , ξ ) | [ det ( I Z Z ¯ ) k ξ p 2 ] β det ( I Z ϕ Z ϕ ¯ ) 1 k × ln 2 det ( I Z ϕ Z ϕ ¯ ) k ξ ϕ p 2 < .
Proof. 
Assume that (3.1) holds. By Lemma 2.11 and for f B α ( GHE I ) , we know that
[ det ( I Z Z ¯ ) k ξ p 2 ] β | ( ψ C ϕ f ) ( Z , ξ ) | = [ det ( I Z Z ¯ ) k ξ p 2 ] β | ψ ( Z , ξ ) | | f ( ϕ ( Z , ξ ) ) | C | ψ ( Z , ξ ) | [ det ( I Z Z ¯ ) k ξ p 2 ] β × ln 2 q det ( I Z ϕ Z ϕ ¯ ) k ξ ϕ p 2 f B α C K 1 f B α .
For all ( Z , ξ ) GHE I , we have
ψ C ϕ f A β = sup ( Z , ξ ) GHE I [ det ( I Z Z ¯ ) k ξ p 2 ] β | ( ψ C ϕ f ) ( Z , ξ ) | C K 1 f B α ,
which implies that ψ C ϕ : B α ( GHE I ) A β ( GHE I ) is bounded.
Conversely, assume that ψ C ϕ : B α ( GHE I ) A β ( GHE I ) is bounded. For any ( S , t ) GHE I , let us introduce a test function f ( S , t ) H ( GHE I ) such that
f ( S , t ) ( Z , ξ ) : = det ( I S S ¯ ) 1 k ln 2 det ( I Z S ¯ ) k ξ , t p .
This means that
f ( S , t ) z g l = k · det ( I Z S ¯ ) k 1 · det ( I S S ¯ ) 1 k det ( I Z S ¯ ) k ξ , t p × det ( I Z S ¯ ) tr [ ( I Z S ¯ ) 1 I g l S ¯ ] , 1 g m , 1 l n , f ( S , t ) ξ j = det ( I S S ¯ ) 1 k · p j ξ j p j 1 t j ¯ p j det ( I Z S ¯ ) k ξ , t p , j = 1 , , r .
In view of (2.18), it follows that
2 | det ( I Z S ¯ ) | 1 m det ( I Z Z ¯ ) 1 m + det ( I S S ¯ ) 1 m ,
then
2 m ( 1 k ) [ | det ( I Z S ¯ ) | 1 m ] m ( 1 k ) [ det ( I Z Z ¯ ) 1 m + det ( I S S ¯ ) 1 m ] m ( 1 k ) [ det ( I S S ¯ ) 1 m ] m ( 1 k ) ,
which means that
2 m ( 1 k ) | det ( I Z S ¯ ) | 1 k det ( I S S ¯ ) 1 k .
According to (3.3) and Lemmas 2.14, 2.16 , there exists a constant C 1 > 0 such that
[ det ( I Z Z ¯ ) k ξ p 2 ] | f ( S , t ) ( Z , ξ ) | = det ( I Z Z ¯ ) k ξ p 2 | det ( I Z S ¯ ) k ξ , t p | × det ( I S S ¯ ) 1 k × { k 2 | det ( I Z S ¯ ) k 1 | 2 × 1 g m 1 l n | det ( I Z S ¯ ) tr [ ( I Z S ¯ ) 1 I g l S ¯ ] | 2 + j = 1 r | p j ξ j p j 1 t j ¯ p j | 2 } 1 2 [ det ( I Z Z ¯ ) k ξ p 2 ] × det ( I S S ¯ ) 1 k | | det ( I Z S ¯ ) k | | ξ , t p | | × { k | det ( I Z S ¯ ) | k 1 × 1 g m 1 l n | det ( I Z S ¯ ) | 2 | tr [ ( I Z S ¯ ) 1 I g l S ¯ ] | 2 1 2 + j = 1 r | p j ξ j p j 1 t j ¯ p j | 2 1 2 } [ det ( I Z Z ¯ ) k ξ p 2 ] | | det ( I Z S ¯ ) k | ξ p t p | × k C 1 | det ( I Z S ¯ ) | k 1 × det ( I S S ¯ ) 1 k + j = 1 r | p j ξ j p j 1 t j ¯ p j | 2 1 2 × det ( I S S ¯ ) 1 k 2 [ det ( I Z Z ¯ ) k ξ p 2 ] [ det ( I Z Z ¯ ) k ξ p 2 ] + [ det ( I S S ¯ ) k t p 2 ] × 2 m ( 1 k ) k C 1 + j = 1 r | p j | 2 1 2 2 [ det ( I Z Z ¯ ) k ξ p 2 ] det ( I Z Z ¯ ) k ξ p 2 × 2 m ( 1 k ) k C 1 + j = 1 r | p j | 2 1 2 2 × 2 m ( 1 k ) k C 1 + j = 1 r | p j | 2 1 2 C .
Since f ( S , t ) ( 0 , 0 ) ln 2 , one has
f ( S , t ) B α = | f ( S , t ) ( 0 , 0 ) | + sup ( Z , ξ ) GHE I [ det ( I Z Z ¯ ) k ξ p 2 ] α | f ( S , t ) ( Z , ξ ) | C + ln 2 .
Therefore, we have
> ( C + ln 2 ) ψ C ϕ B α A β ψ C ϕ f ( S , t ) A β = sup ( Z , ξ ) GHE I [ det ( I Z Z ¯ ) k ξ p 2 ] β | ψ ( Z , ξ ) f ( S , t ) ( ϕ ( Z , ξ ) ) | | ψ ( Z , ξ ) | [ det ( I Z Z ¯ ) k ξ p 2 ] β det ( I S S ¯ ) 1 k × | ln 2 det ( I Z ϕ S ¯ ) k ξ ϕ , t p | .
Let us now consider
( S , t ) = ( Z ϕ , ξ ϕ ) = ϕ ( Z , ξ ) ,
so that
sup ( Z , ξ ) GHE I | ψ ( Z , ξ ) | [ det ( I Z Z ¯ ) k ξ p 2 ] β det ( I Z ϕ Z ϕ ¯ ) 1 k ln 2 det ( I Z ϕ Z ϕ ¯ ) k ξ ϕ p 2 < .
The proof is thus completed. □
Theorem 3.2.Assume that α > 1 , β > 0 , 0 < k m 1 , and that p j are positive integers ( j = 1 , 2 , , r ) . Let ϕ = ( ϕ 11 , ϕ 12 ϕ m n + r ) be a holomorphic self-map of GHE I , with ψ H ( GHE I ) and ( Z ϕ , ξ ϕ ) = ϕ ( Z , ξ ) . If
K 3 : = sup ( Z , ξ ) GHE I | ψ ( Z , ξ ) | [ det ( I Z Z ¯ ) k ξ p 2 ] β [ det ( I Z ϕ Z ϕ ¯ ) k ξ ϕ p 2 ] α 1 < ,
then the weighted composition operator ψ C ϕ : B α ( GHE I ) A β ( GHE I ) is bounded.
Conversely, if the weighted composition operator ψ C ϕ : B α ( GHE I ) A β ( GHE I ) is bounded, then
K 4 : = sup ( Z , ξ ) GHE I | ψ ( Z , ξ ) | [ det ( I Z Z ¯ ) k ξ p 2 ] β det ( I Z ϕ Z ϕ ¯ ) 1 k [ det ( I Z ϕ Z ϕ ¯ ) k ξ ϕ p 2 ] α 1 < .
Proof. 
Assume that (3.4) holds. By Lemma 2.11 and for f B α ( GHE I ) , we have
[ det ( I Z Z ¯ ) k ξ p 2 ] β | ( ψ C ϕ f ) ( Z , ξ ) | = [ det ( I Z Z ¯ ) k ξ p 2 ] β | ψ ( Z , ξ ) · ( C ϕ f ) ( Z , ξ ) | = [ det ( I Z Z ¯ ) k ξ p 2 ] β | ψ ( Z , ξ ) | | f ( ϕ ( Z , ξ ) ) | C | ψ ( Z , ξ ) | [ det ( I Z Z ¯ ) k ξ p 2 ] β [ det ( I Z ϕ Z ϕ ¯ ) k ξ ϕ p 2 ] α 1 f B α C K 3 f B α .
For all ( Z , ξ ) GHE I , we obtain
ψ C ϕ f A β = sup ( Z , ξ ) GHE I [ det ( I Z Z ¯ ) k ξ p 2 ] β | ( ψ C ϕ f ) ( Z , ξ ) | C K 3 f B α .
which implies that ψ C ϕ : B α ( GHE I ) A β ( GHE I ) is bounded.
Conversely, assume that ψ C ϕ : B α ( GHE I ) A β ( GHE I ) is bounded. For ( S , t ) GHE I , define a test function f ( S , t ) H ( GHE I ) such that
f ( S , t ) ( Z , ξ ) : = det ( I S S ¯ ) 1 k [ det ( I Z S ¯ ) k ξ , t p ] α 1 .
For the test function f, we have
f ( S , t ) z g l = k ( α 1 ) · det ( I Z S ¯ ) k 1 · det ( I S S ¯ ) 1 k [ det ( I Z S ¯ ) k ξ , t p ] α × det ( I Z S ¯ ) tr [ ( I Z S ¯ ) 1 I g l S ¯ ] , 1 g m , 1 l n , f ( S , t ) ξ j = ( α 1 ) p j ξ j p j 1 t j ¯ p j · det ( I S S ¯ ) 1 k [ det ( I Z S ¯ ) k ξ , t p ] α , j = 1 , , r .
From (3.3) and Lemmas 2.14, 2.16, there exists a constant C 1 > 0 such that
[ det ( I Z Z ¯ ) k ξ p 2 ] α | f ( S , t ) ( Z , ξ ) | = [ det ( I Z Z ¯ ) k ξ p 2 ] α | det ( I Z S ¯ ) k ξ , t p | α × det ( I S S ¯ ) 1 k × ( α 1 ) × { k 2 | det ( I Z S ¯ ) k 1 | 2 × 1 g m 1 l n | det ( I Z S ¯ ) tr [ ( I Z S ¯ ) 1 I g l S ¯ ] | 2 + j = 1 r | p j ξ j p j 1 t j ¯ p j | 2 } 1 2 [ det ( I Z Z ¯ ) | k ξ p 2 ] α × det ( I S S ¯ ) 1 k | | det ( I Z S ¯ ) | k | ξ , t p | | α × ( α 1 ) × { k | det ( I Z S ¯ ) | k 1 × 1 g m 1 l n | det ( I Z S ¯ ) | 2 | tr [ ( I Z S ¯ ) 1 I g l S ¯ ] | 2 1 2 + j = 1 r | p j ξ j p j 1 t j ¯ p j | 2 1 2 } [ det ( I Z Z ¯ ) k ξ p 2 ] α | | det ( I Z S ¯ ) | k ξ p t p | α × ( α 1 ) × { k C 1 det ( I S S ¯ ) 1 k | det ( I Z S ¯ ) | k 1 + j = 1 r | p j ξ j p j 1 t j ¯ p j | 2 1 2 det ( I S S ¯ ) 1 k } 2 [ det ( I Z Z ¯ ) k ξ p 2 ] [ det ( I Z Z ¯ ) k ξ p 2 ] + [ det ( I S S ¯ ) k ξ p 2 ] α ( α 1 ) × ( k C 1 2 m ( 1 k ) + C 2 ) 2 [ det ( I Z Z ¯ ) k ξ p 2 ] det ( I Z Z ¯ ) p ξ p 2 α ( α 1 ) × ( k C 1 2 m ( 1 k ) + C 2 ) 2 α ( α 1 ) C 3 = C 4 .
Since f ( S , t ) ( 0 , 0 ) 1 , we obtain
f ( S , t ) B α = | f ( S , t ) ( 0 , 0 ) | + sup ( Z , ξ ) GHE I [ det ( I Z Z ¯ ) k ξ p 2 ] α | f ( S , t ) ( Z , ξ ) | C 4 + 1 .
It follows that
> ( C 4 + 1 ) ψ C ϕ B α A β ψ C ϕ f ( S , t ) A β = sup ( Z , ξ ) GHE I [ det ( I Z Z ¯ ) k ξ p 2 ] β | ψ ( Z , ξ ) f ( S , t ) ( ϕ ( Z , ξ ) ) | | ψ ( Z , ξ ) | [ det ( I Z Z ¯ ) k ξ p 2 ] β det ( I S S ¯ ) 1 k | det ( I Z ϕ S ¯ ) k ξ ϕ , t p | α 1 .
We write ( S , t ) = ( Z ϕ , ξ ϕ ) = ϕ ( Z , ξ ) , then
sup ( Z , ξ ) GHE I | ψ ( Z , ξ ) | [ det ( I Z Z ¯ ) k ξ p 2 ] β det ( I Z ϕ Z ϕ ¯ ) 1 k [ det ( I Z ϕ Z ϕ ¯ ) k ξ ϕ | p 2 ] α 1 < .
This completes the proof of the theorem.
Corollary 1. 
For α > 1 , k = m = 1 , p 1 = = p r = 1 , we have the case of the unit ball B = { z C n + r : | z | 2 < 1 } and ψ C ϕ : B α ( B ) A β ( B ) is bounded if and only if
sup z B | ψ ( z ) | ( 1 | z | 2 ) β ( 1 | ϕ ( z ) | 2 ) α 1 <
when β = 0 . This result is equivalent to that obtained by Li and Stević in [9].

4. Compactness of ψ C ϕ : B α A β

Theorem 4.1.Assume that α = 1 , β > 0 , 0 < k m 1 , and that p j ( j = 1 , 2 , , r ) are positive integers. Let ϕ = ( ϕ 11 , ϕ 12 ϕ m n + r ) be a holomorphic self-map of GHE I , with ψ H ( GHE I ) and ( Z ϕ , ξ ϕ ) = ϕ ( Z , ξ ) . If ψ A β and
lim ϕ ( Z , ξ ) GHE I | ψ ( Z , ξ ) | [ det ( I Z Z ¯ ) k ξ p 2 ] β ln 2 q det ( I Z ϕ Z ϕ ¯ ) k ξ ϕ p 2 = 0 ,
then the weighted composition operator ψ C ϕ : B α ( GHE I ) A β ( GHE I ) is compact.
Conversely, if the weighted composition operator ψ C ϕ : B α ( GHE I ) A β ( GHE I ) is compact, then ψ A β and
lim ϕ ( Z , ξ ) GHE I | ψ ( Z , ξ ) | [ det ( I Z Z ¯ ) k ξ p 2 ] β det ( I Z ϕ Z ϕ ¯ ) 1 k × ln 2 det ( I Z ϕ Z ϕ ¯ ) k ξ ϕ p 2 = 0 .
Proof. 
Assume that (4.1) holds. We have
sup ( Z , ξ ) GHE I | ψ ( Z , ξ ) | [ det ( I Z Z ¯ ) k ξ p 2 ] β ln 2 q det ( I Z ϕ Z ϕ ¯ ) k ξ ϕ p 2 < .
If ψ C ϕ is bounded, consider the bounded sequence { f k } k 1 in B α ( GHE I ) , which converges to 0 uniformly on compact subsets of GHE I . Hence, there exists M 1 > 0 such that f k B α M 1 , k = 1 , 2 , . By (4.1), this means that ε > 0 , δ ( 0 , 1 ) , such that for dist ( ϕ ( Z , ξ ) , GHE I ) < δ , we have
| ψ ( Z , ξ ) | [ det ( I Z Z ¯ ) k ξ p 2 ] β ln 2 q det ( I Z ϕ Z ϕ ¯ ) k ξ ϕ p 2 < ε .
According to Lemma 2.11, we obtain
[ det ( I Z Z ¯ ) k ξ p 2 ] β | ( ψ C ϕ f k ) ( Z , ξ ) | = [ det ( I Z Z ¯ ) k ξ p 2 ] β | ψ ( Z , ξ ) · ( C ϕ f k ) ( Z , ξ ) | = [ det ( I Z Z ¯ ) k ξ p 2 ] β | ψ ( Z , ξ ) | | f k ( ϕ ( Z , ξ ) ) | C | ψ ( Z , ξ ) | [ det ( I Z Z ¯ ) k ξ p 2 ] β f k B α × ln 2 q det ( I Z ϕ Z ϕ ¯ ) k ξ ϕ p 2 C M 1 ε .
On the other hand, let us introduce the set
E δ : = { ( Z , ξ ) GHE I : dist ( ϕ ( Z , ξ ) , GHE I ) δ } ,
which is a compact subset of GHE I . By the assumptions, f k converges to 0 uniformly on any compact subset of GHE I . From this, and since ψ A β , for such ε , we have
[ det ( I Z Z ¯ ) k ξ p 2 ] β | ( ψ C ϕ f k ) ( Z , ξ ) | = [ det ( I Z Z ¯ ) k ξ p 2 ] β | ψ ( Z , ξ ) · ( C ϕ f k ) ( Z , ξ ) | = [ det ( I Z Z ¯ ) k ξ p 2 ] β | ψ ( Z , ξ ) | | f k ( ϕ ( Z , ξ ) ) | ψ A β ε .
Combining (4.4) and (4.5), we have
ψ C ϕ f k A β = sup ( Z , ξ ) GHE I [ det ( I Z Z ¯ ) k ξ p 2 ] β | ( ψ C ϕ f k ) ( Z , ξ ) | 0 , k .
Consequently, making use of Lemma 2.12, we finally have that ψ C ϕ : B α ( GHE I ) A β ( GHE I ) is compact.
Conversely, suppose ψ C ϕ : B α ( GHE I ) A β ( GHE I ) is compact. Let f 1 , we have
[ det ( I Z Z ¯ ) k ξ p 2 ] β | ψ ( Z , ξ ) | = [ det ( I Z Z ¯ ) k ξ p 2 ] β | ( ψ C ϕ f ) ( Z , ξ ) | < .
This shows that ψ A β . Consider now a sequence ( S i , t i ) = ϕ ( Z i , ξ i ) in GHE I such that ϕ ( Z i , ξ i ) GHE I as i . If such a sequence does not exist, then condition (4.2) obviously holds. Moreover, let us introduce the following sequence of test functions { f i } i 1 :
f i ( Z , ξ ) = ln 2 det ( I S i S i ¯ ) k t i p 2 1 × ln 2 det ( I Z S i ¯ ) k ξ , t i p 2 × det ( I S i S i ¯ ) 1 k .
Differentiation gives
f i z g l = 2 k × det ( I Z S i ¯ ) k 1 det ( I S i S i ¯ ) 1 k det ( I Z S i ¯ ) k ξ , t i p × ln 2 det ( I Z S i ¯ ) k ξ , t i p ln 2 det ( I S i S i ¯ ) k t i p 2 × det ( I Z S i ¯ ) × tr [ ( I Z S i ¯ ) 1 I g l S i ¯ ] , 1 g m , 1 l n , f i ξ j = 2 p j ξ j p j 1 t j ¯ p j × det ( I S i S i ¯ ) 1 k det ( I Z S i ¯ ) k ξ , t i p × ln 2 det ( I Z S i ¯ ) k ξ , t i p ln 2 det ( I S i S i ¯ ) k t i p 2 , j = 1 , , r . i = 1 , 2 ,
From (3.3) and Lemmas 2.14, 2.16, there exists a constant C 5 > 0 such that
[ det ( I Z Z ¯ ) k ξ | p 2 ] | f i ( Z , ξ ) | = [ det ( I Z Z ¯ ) k ξ p 2 ] det ( I S i S i ¯ ) 1 k | det ( I Z S i ¯ ) k ξ , t i p | × | ln 2 det ( I Z S i ¯ ) k ξ , t i p ln 2 det ( I S i S i ¯ ) k t i p 2 | × { 4 k 2 | det ( I Z S i ¯ ) k 1 | 2 × 1 g m 1 l n | det ( I Z S i ¯ ) tr [ ( I Z S i ¯ ) 1 I g l S i ¯ ] | 2 + 4 j = 1 r | p j ξ j p j 1 t j ¯ p j | 2 } 1 2 [ det ( I Z Z ¯ ) k ξ p 2 ] [ det ( I S i S i ¯ ) 1 k | | det ( I Z S i ¯ ) k | | ξ , t i p | | × | ln 2 | det ( I Z S i ¯ ) k ξ , t i p | | + π ln 2 det ( I S i S i ¯ ) k t i p 2 × { 2 k | det ( I Z S i ¯ ) | k 1 × 1 g m 1 l n | det ( I Z S i ¯ ) tr [ ( I Z S i ¯ ) 1 I g l S i ¯ ] | 2 1 2 + 2 j = 1 r | p j ξ j p j 1 t j ¯ p j | 2 1 2 } det ( I Z Z ¯ ) k ξ p 2 | | det ( I Z S i ¯ ) k | ξ p t i p | × | ln 2 | det ( I Z S i ¯ ) k ξ , t i p | | + π ln 2 det ( I S i S i ¯ ) k t i p 2 × 2 k C 5 | det ( I Z S i ¯ ) | k 1 det ( I S i S i ¯ ) 1 k + 2 j = 1 r | p j ξ j p j 1 t j ¯ p j | 2 1 2 det ( I S i S i ¯ ) 1 k 2 [ det ( I Z Z ¯ ) k ξ p 2 ] [ det ( I Z Z ¯ ) k ξ p 2 ] + [ det ( I S i S i ¯ ) k t p 2 ] × | ln 2 | det ( I Z S i ¯ ) k ξ , t i p | | + π ln 2 det ( I S i S i ¯ ) k t i p 2 × 2 k C 5 | det ( I Z S i ¯ ) | k 1 det ( I S i S i ¯ ) 1 k + C 6 2 [ det ( I Z Z ¯ ) k ξ p 2 ] det ( I Z Z ¯ ) k ξ p 2 × | ln 2 | det ( I Z S i ¯ ) k ξ , t i p | | + π ln 2 det ( I S i S i ¯ ) k t i p 2 × ( 2 k C 5 · 2 m ( 1 k ) + C 6 ) 2 × ( 2 k C 5 · 2 m ( 1 k ) + C 6 ) × | ln 2 | det ( I Z S i ¯ ) k ξ , t i p | | + π ln 2 det ( I S i S i ¯ ) k t i p 2 C 7 × | ln 2 | det ( I Z S i ¯ ) k ξ , t i p | | + π ln 2 det ( I S i S i ¯ ) k t i p 2 .
We have now two cases.
Case A 1 . If | det ( I Z S i ¯ ) k ξ , t i p | 2 , then
| ln 2 | det ( I Z S i ¯ ) k ξ , t i p | | + π ln 2 det ( I S i S i ¯ ) k t i p 2 ln 2 | det ( I Z S i ¯ ) k | | ξ , t i p | + π ln 2 det ( I S i S i ¯ ) k t i p 2 ln 2 | det ( I Z S i ¯ ) k | ξ p t i p + π ln 2 det ( I S i S i ¯ ) k t i p 2 ln 4 [ det ( I Z Z ¯ ) k | ξ p 2 ] + [ det ( I S i S i ¯ ) k t i 2 ] p + π ln 2 det ( I S i S i ¯ ) k t i p 2 ln 4 det ( I S i S i ¯ ) k t i p 2 + π ln 2 det ( I S i S i ¯ ) k t i p 2 2 + π ln 2 det ( I S i S i ¯ ) k t i p 2 C 8 ,
where C 8 = 2 + π ln 2 .
Case A 2 . If | det ( I Z S i ¯ ) k ξ , t i p | > 2 , then
| ln 2 | det ( I Z S i ¯ ) k ξ , t i p | | + π ln 2 det ( I S i S i ¯ ) k t i p 2 = | ln 2 ln | det ( I Z S i ¯ ) k ξ , t i p | | + π ln 2 det ( I S i S i ¯ ) k t i p 2
ln | det ( I Z S i ¯ ) k ξ , t i p | + π ln 2 det ( I S i S i ¯ ) k t i p 2 ln ( | det ( I Z S i ¯ ) k | + | ξ , t i p | ) + π ln 2 .
Since Z S i ¯ = C = ( c i j ) m × m , c i j = g = 1 n z i g s j g ( i , j = 1 , , m ) , we may write I C = D = ( d i j ) m × m with
d i j = 1 ( g = 1 n z i g s j g ) i = j g = 1 n z i g s j g i j
Using (4.8), we have det ( I C ) = j 1 j 2 j m ( 1 ) τ ( j 1 j 2 j m ) d 1 j 1 d 2 j 2 d m j m and
| det ( I C ) | m ! ( n + 1 ) m = G
Hence,
| ln 2 | det ( I Z S i ¯ ) k ξ , t i p | | + π ln 2 det ( I S i S i ¯ ) k t i p 2 ln ( G k + ξ p t i p ) + π ln 2 ln ( G k + 1 ) + π ln 2 C 9 .
By using both cases A 1 and A 2 , we have [ det ( I Z Z ¯ ) ξ p 2 ] | f i ( Z , ξ ) | Q C 7 and then f i B α Q C 7 , which means that f i B α is bounded, where Q = max { C 8 , C 9 } . It follows that { f i } i 1 B α ( GHE I ) and
| f i ( Z , ξ ) | = ln 2 det ( I S i S i ¯ ) k t i p 2 1 × | ln 2 det ( I Z S i ¯ ) k ξ , t i p | 2 × det ( I S i S i ¯ ) 1 k ln 2 det ( I S i S i ¯ ) k t i p 2 1 × det ( I S i S i ¯ ) 1 k × | ln 2 | det ( I Z S i ¯ ) k ξ , t i p | | + π 2 .
If | det ( I Z S i ¯ ) k ξ , t i p | 2 , then
| f i ( Z , ξ ) | ln 2 det ( I S i S i ¯ ) k t i p 2 1 × det ( I S i S i ¯ ) 1 k × ln 2 | det ( I Z S i ¯ ) k | | ξ , t i p | + π 2 ln 2 det ( I S i S i ¯ ) k t i p 2 1 × det ( I S i S i ¯ ) 1 k × ln 2 | det ( I Z S i ¯ ) k | ξ p t i p + π 2 ln 2 det ( I S i S i ¯ ) k t i p 2 1 × det ( I S i S i ¯ ) 1 k × ln 4 [ det ( I Z Z ¯ ) k ξ p 2 ] + [ det ( I S i S i ¯ ) k t i p 2 ] + π 2 ln 2 det ( I S i S i ¯ ) k t i p 2 1 × det ( I S i S i ¯ ) 1 k × ln 4 det ( I Z Z ¯ ) k ξ p 2 + π 2
Since 0 < det ( I S i S i ¯ ) 1 k 1 , we take i and obtain ( S i , t i ) GHE I . This implies det ( I S i S i ¯ ) k t i p 2 0 , then ln 2 det ( I S i S i ¯ ) k t i p 2 1 0 . Let us now consider a compact subset E of GHE I . For ( Z , ξ ) E ,it is easy to see that det ( I Z Z ¯ ) k ξ p 2 has a positive lower bound. Thus, we have f i ( Z , ξ ) 0 , i on all compact subsets of GHE I . If | det ( I Z S i ¯ ) k ξ , t i p | > 2 , then
| f i ( Z , ξ ) | ln 2 det ( I S i S i ¯ ) k t i p 2 1 × det ( I S i S i ¯ ) 1 k × | ln 2 ln ( | det ( I Z S i ¯ ) k ξ , t i p | ) | + π 2 ln 2 det ( I S i S i ¯ ) k t i p 2 1 × det ( I S i S i ¯ ) 1 k × ln ( | det ( I Z S i ¯ ) k | + | ξ , t i p | ) + π 2 ln 2 det ( I S i S i ¯ ) k t i p 2 1 × det ( I S i S i ¯ ) 1 k × { ln ( G k + 1 ) + π } 2
From 0 < det ( I S i S i ¯ ) 1 k 1 and ln 2 det ( I S i S i ¯ ) k t i p 2 1 0 as i , one concludes that f i ( Z , ξ ) 0 , i .
The above proof shows that f i ( Z , ξ ) 0 , i on all compact subsets of GHE I . By Lemma 2.12, this implies that ψ C ϕ f i A β 0 . Therefore, we conclude that
0 ψ C ϕ f i A β = sup ϕ ( Z , ξ ) GHE I [ det ( I Z Z ¯ ) k ξ p 2 ] β | ψ ( Z , ξ ) | ln 2 det ( I S i S i ¯ ) k t i p 2 1 × | ln 2 det ( I Z ϕ S i ¯ ) k ξ ϕ , t i p | 2 × det ( I S i S i ¯ ) 1 k [ det ( I Z i Z i ¯ ) k ξ i p 2 ] β | ψ ( Z i , ξ i ) | ln 2 det ( I S i S i ¯ ) k t i p 2 1 × ln 2 det ( I S i S i ¯ ) k t i p 2 2 × det ( I S i S i ¯ ) 1 k = | ψ ( Z i , ξ i ) | [ det ( I Z i Z i ¯ ) k ξ i p 2 ] β det ( I S i S i ¯ ) 1 k × ln 2 det ( I S i S i ¯ ) k t i p 2 .
Theorem 4.2.Assume that α > 1 , β > 0 , 0 < k m 1 , and that p j are some positive integers ( j = 1 , 2 , , r ) . Let ϕ = ( ϕ 11 , ϕ 12 ϕ m n + r ) be a holomorphic self-map of GHE I , with ψ H ( GHE I ) and ( Z ϕ , ξ ϕ ) = ϕ ( Z , ξ ) . If ψ A β and
lim ϕ ( Z , ξ ) GHE I | ψ ( Z , ξ ) | [ det ( I Z Z ¯ ) k ξ p 2 ] β [ det ( I Z ϕ Z ϕ ¯ ) k ξ ϕ p 2 ] α 1 = 0 ,
then the weighted composition operator ψ C ϕ : B α ( GHE I ) A β ( GHE I ) is compact.
Conversely, if the weighted composition operator ψ C ϕ : B α ( GHE I ) A β ( GHE I ) is compact, then ψ A β and
lim ϕ ( Z , ξ ) GHE I | ψ ( Z , ξ ) | [ det ( I Z Z ¯ ) k ξ p 2 ] β [ det ( I Z ϕ Z ϕ ¯ ) k ξ ϕ p 2 ] α 1 k = 0 .
Proof. 
Assume that (4.9) holds. We have
sup ( Z , ξ ) GHE I | ψ ( Z , ξ ) | [ det ( I Z Z ¯ ) k ξ p 2 ] β [ det ( I Z ϕ Z ϕ ¯ ) k ξ ϕ p 2 ] α 1 < .
From Theorem 3.2, it follows that ψ C ϕ : B α ( GHE I ) A β ( GHE I ) is bounded. Let { f k } k 1 be a bounded sequence in B α ( GHE I ) with f k that converges to 0 uniformly on compact subsets of GHE I . There exists M 2 > 0 such that f k B α M 2 , k = 1 , 2 , . By (4.9), for any ε > 0 , there is a constant δ ( 0 , 1 ) such that
| ψ ( Z , ξ ) | [ det ( I Z Z ¯ ) k ξ p 2 ] β [ det ( I Z ϕ Z ϕ ¯ ) k ξ ϕ p 2 ] α 1 < ε ,
for dist ( ϕ ( Z , ξ ) , GHE I ) < δ . Using Lemma 2.11 we have
[ det ( I Z Z ¯ ) k ξ p 2 ] β | ( ψ C ϕ f k ) ( Z , ξ ) | = [ det ( I Z Z ¯ ) k ξ p 2 ] β | ψ ( Z , ξ ) · ( C ϕ f k ) ( Z , ξ ) | = [ det ( I Z Z ¯ ) k ξ p 2 ] β | ψ ( Z , ξ ) | | f k ( ϕ ( Z , ξ ) ) | C | ψ ( Z , ξ ) | f k B α [ det ( I Z Z ¯ ) k ξ p 2 ] β [ det ( I Z ϕ Z ϕ ¯ ) k ξ ϕ p 2 ] α 1 C M 2 ε .
On the other hand, if we set
E δ : = { ( Z , ξ ) GHE I : dist ( ϕ ( Z , ξ ) , GHE I ) δ } ,
we have that E δ is a compact subset of GHE I . For ε defined in (4.11), f k converges to 0 uniformly on any compact subset of GHE I . For ψ A β , we have
[ det ( I Z Z ¯ ) k ξ p 2 ] β | ( ψ C ϕ f k ) ( Z , ξ ) | = [ det ( I Z Z ¯ ) k ξ p 2 ] β | ψ ( Z , ξ ) · ( C ϕ f k ) ( Z , ξ ) | = [ det ( I Z Z ¯ ) k ξ p 2 ] β | ψ ( Z , ξ ) | | f k ( ϕ ( Z , ξ ) ) | ψ A β ε .
According to inequalities (4.12) and (4.13), we see that
ψ C ϕ f k A β = sup ( Z , ξ ) GHE I [ det ( I Z Z ¯ ) k ξ p 2 ] β | ( ψ C ϕ f k ) ( Z , ξ ) | 0 , k .
Consequently, making use of Lemma 2.12, we have that ψ C ϕ : B α ( GHE I ) A β ( GHE I ) is compact.
Conversely, suppose that ψ C ϕ : B α ( GHE I ) A β ( GHE I ) is compact. Then, ψ C ϕ : B α ( GHE I ) A β ( GHE I ) is bounded. Let f 1 , we get
[ det ( I Z Z ¯ ) k ξ p 2 ] β | ψ ( Z , ξ ) | = [ det ( I Z Z ¯ ) k ξ p 2 ] β | ( ψ C ϕ f ) ( Z , ξ ) | < .
This shows that ψ A β . Consider now a sequence ( S i , t i ) = ϕ ( Z i , ξ i ) in GHE I such that ϕ ( Z i , ξ i ) GHE I as i . If such a sequence does not exist, then condition (4.10) obviously holds.
Moreover, let us introduce a sequence of test functions { f i } i 1 :
f i ( Z , ξ ) : = [ det ( I S i S i ¯ ) k t i p 2 ] 1 k 1 + α [ det ( I Z S i ¯ ) k ξ , t i p ] 2 α 1 .
Differentiation gives
f i z g l = ( 2 α 1 ) k · det ( I Z S i ¯ ) k 1 [ det ( I S i S i ¯ ) k t i p 2 ] 1 k 1 + α [ det ( I Z S i ¯ ) k ξ , t i p ] 2 α × det ( I Z S i ¯ ) tr [ ( I Z S i ¯ ) 1 I g l S i ¯ ] , f i ξ j = ( 2 α 1 ) p j ξ j p j 1 t j ¯ p j [ det ( I S i S i ¯ ) k t i p 2 ] 1 k 1 + α [ det ( I Z S i ¯ ) k ξ , t i p ] 2 α .
From (3.3) and Lemma 2.15, it follows that there exists a constant C 10 > 0 such that
[ det ( I Z Z ¯ ) k ξ p 2 ] α | f i ( Z , ξ ) | = ( 2 α 1 ) [ det ( I Z Z ¯ ) k ξ p 2 ] α [ det ( I S i S i ¯ ) k t i p 2 ] 1 k 1 + α | det ( I Z S i ¯ ) k ξ , t i p | 2 α × { k 2 | det ( I Z S i ¯ ) k 1 | 2 × 1 g m 1 l n | det ( I Z S i ¯ ) tr [ ( I Z S i ¯ ) 1 I g l S i ¯ ] | 2 + j = 1 r | p j ξ j p j 1 t j ¯ p j | 2 } 1 2 ( 2 α 1 ) [ det ( I Z Z ¯ ) k ξ p 2 ] α [ det ( I S i S i ¯ ) k t i p 2 ] α | det ( I Z S i ¯ ) k ξ , t i p | 2 α × { k | det ( I Z S i ¯ ) | k 1 × [ det ( I S i S i ¯ ) k t i p 2 ] 1 k 1 × 1 g m 1 l n | det ( I Z S i ¯ ) tr [ ( I Z S i ¯ ) 1 I g l S i ¯ ] | 2 1 2 + j = 1 r | p j ξ j p j 1 t j ¯ p j | 2 1 2 × [ det ( I S i S i ¯ ) k t i p 2 ] 1 k 1 } ( 2 α 1 ) × k · C 1 | det ( I Z S i ¯ ) | k 1 [ det ( I S i S i ¯ ) k ] 1 k 1 + C 10 ( 2 α 1 ) × k · C 1 | det ( I Z S i ¯ ) | k 1 det ( I S i S i ¯ ) 1 k + C 10 ( 2 α 1 ) × C 1 · k · 2 m ( 1 k ) + C 10 C .
This shows that f i B α ( GHE I ) , i = 1 , 2 , and
| f i ( Z , ξ ) | = [ det ( I S i S i ¯ ) k t i p 2 ] 1 k 1 + α | det ( I Z S i ¯ ) k ξ , t i p | 2 α 1 [ det ( I S i S i ¯ ) k t i p 2 ] 1 k 1 + α | | det ( I Z S i ¯ ) k | | ξ , t i p | | 2 α 1 [ det ( I S i S i ¯ ) k t i p 2 ] 1 k 1 + α | det ( I Z S i ¯ ) k | ξ p t i p | 2 α 1 2 2 α 1 [ det ( I S i S i ¯ ) k t i p 2 ] 1 k 1 + α [ det ( I Z Z ¯ ) k ξ p 2 + det ( I S i S i ¯ ) k t i p 2 ] 2 α 1 2 2 α 1 [ det ( I S i S i ¯ ) k t i p 2 ] 1 k 1 + α [ det ( I Z Z ¯ ) k ξ p 2 ] 2 α 1 .
Taking i , we have ( S i , t i ) GHE I . This implies that det ( I S i S i ¯ ) k t i p 2 0 . If E is a compact subset of GHE I , for ( Z , ξ ) E , we have that det ( I Z Z ¯ ) k ξ p 2 has a positive lower bound. Thus, we have f i ( Z , ξ ) 0 , i on all compact subsets of GHE I . According to Lemma 2.12, we have that ψ C ϕ f i A β 0 . Hence,
0 ψ C ϕ f i A β = sup ϕ ( Z , ξ ) GHE I [ det ( I Z Z ¯ ) k ξ p 2 ] β | ψ ( Z , ξ ) | [ det ( I S i S i ¯ ) k t i p 2 ] 1 k 1 + α | det ( I Z ϕ S i ¯ ) k ξ ϕ , t i p | 2 α 1 [ det ( I Z i Z i ¯ ) k ξ i p 2 ] β | ψ ( Z i , ξ i ) | [ det ( I Z ϕ i Z ϕ i ¯ ) k ξ ϕ i p 2 ] α 1 k .
Corollary 2. 
For α > 1 , k = m = 1 , p 1 = = p r = 1 , we are back to the case of the unit ball B = { z C n + r : | z | 2 < 1 } , and ψ C ϕ : B α ( B ) A β ( B ) is compact if and only if ψ A β and
lim ϕ ( z ) B | ψ ( z ) | ( 1 | z | 2 ) β ( 1 | ϕ ( z ) | 2 ) α 1 = 0
when β = 0 . Also in this case, the result is analogue to that obtained by Li and Stević in [9].

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