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Article

Product Type Operators Involving Radial Derivative Operator Acting between Some Analytic Function Spaces

1
School of Mathematics, Shri Mata Vaishno Devi University, Katra 182320, India
2
Department of Medical Research, China Medical University Hospital, China Medical University, Taichung 40402, Taiwan
3
Department of Mathematics, Aligarh Muslim University, Aligarh 202002, India
*
Author to whom correspondence should be addressed.
Mathematics 2021, 9(19), 2447; https://doi.org/10.3390/math9192447
Submission received: 21 August 2021 / Revised: 14 September 2021 / Accepted: 21 September 2021 / Published: 1 October 2021

Abstract

:
Let N denote the set of all positive integers and N 0 = N { 0 } . For m N , let B m = { z C m : | z | < 1 } be the open unit ball in the m dimensional Euclidean space C m . Let H ( B m ) be the space of all analytic functions on B m . For an analytic self map ξ = ( ξ 1 , ξ 2 , , ξ m ) on B m and ϕ 1 , ϕ 2 , ϕ 3 H ( B m ) , we have a product type operator T ϕ 1 , ϕ 2 , ϕ 3 , ξ which is basically a combination of three other operators namely composition operator C ξ , multiplication operator M ϕ and radial derivative operator R . We study the boundedness and compactness of this operator mapping from weighted Bergman–Orlicz space A σ Ψ into weighted type spaces H ω and H ω , 0 .

1. Introduction

Let z = ( z 1 , z 2 , , z m ) , w = ( w 1 , w 2 , , w m ) be points in C m and w ¯ = ( w 1 ¯ , w 2 ¯ , , w m ¯ ) , where w k ¯ is the complex conjugate of w k ( k = 1 , 2 , , m ) . We write
| z | = z , z = | z 1 | 2 + | z 2 | 2 + + | z m | 2
and
z , w = z 1 w 1 ¯ + z 2 w 2 ¯ + + z m w m ¯ .
Denote the open unit ball in C m by B m = { z C m : | z | < 1 } . Let H ( B m ) be the space of all analytic functions on B m , S m the boundary of B m called as the sphere in C m . Let d V be the Lebesgue measure on B m and d σ the normalized measure on S m . For σ > 1 , we write d V σ ( z ) = C σ ( 1 | z | 2 ) σ d V ( z ) , where C σ is such that V σ ( B m ) = 1 . For the result in settings of unit ball, refer to Ref. [1] and the references therein.
For 0 < p < and σ > 1 , the weighted Bergman space A σ p ( B m ) = A σ p , consists of all those functions f H ( B m ) for which we have the following norm
f A σ p p = B m | f ( z ) | p d V σ ( z ) < .
A non-zero function Ψ : [ 0 , ) [ 0 , ) is said to be a growth function if it is continuous and non-decreasing. Clearly, every growth function fixes origin, that is Ψ ( 0 ) = 0 . We say that the function Ψ is of positive upper type (respectively, negative lower type) for every s > 0 and t 1 , if there exist C > 0 and q > 0 (respectively, q < 0 ) such that Ψ ( s t ) C t q Ψ ( s ) . The class of all growth functions Ψ of positive upper type q, (for some q 1 ) for which the function t Ψ ( t ) / t is non-decreasing on ( 0 , ) is denoted by U q . Similarly, for every s > 0 and 0 < t 1 , a function Ψ is said to be of positive lower type (respectively, negative upper type) if there are C > 0 and p > 0 (respectively, p < 0 ) such that Ψ ( s t ) C t p Ψ ( s ) . The set of all growth functions Ψ of positive lower type r, (for some 0 < r 1 ) such that the function t Ψ ( t ) / t is non-increasing on ( 0 , ) is denoted by L r .
For a growth function Ψ , the weighted Bergman–Orlicz space A σ Ψ ( B m ) = A σ Ψ is the class of all functions f in H ( B m ) such that
f A σ Ψ = B m Ψ ( | f ( z ) | ) d V σ ( z ) < .
The quasi-norm on A σ Ψ is defined as follows:
f A σ Ψ l u x = inf λ > 0 : B m Ψ | f ( z ) | λ d V σ ( z ) 1 .
If Ψ U q or Ψ L r , then the quasi-norm on A σ Ψ is finite and called the Luxembourg norm. A quasi-norm on a linear space X is similar to a norm in that it satisfies the norm axioms, except that the triangle inequality is replaced by the quasi-triangle inequality, that is, x + y   C ( x + y ) , for some C > 0 and x , y X . The smallest C for which quasi-triangle inequality holds will be called the quasi-norm constant of ( X , · ) . The Luxembourg space equipped with the Luxembourg quasi-norm is really a quasi-normed function space with the same quasi-triangle constant as the one of the quasi-norm. If Ψ ( t ) = t p , for p > 0 , then we get the weighted Bergman space A σ p . If p 1 , then A σ Ψ p will be a Banach space which for 0 < p < 1 is a translation-invariant metric space such that d ( f , g ) = f g A σ p p . To know more about these spaces one may refer [2,3] and the references therein.
A positive continuous function ω on B m is called as a weight. The weight ω is called to be a standard weight, if for z B m , we have ω ( z ) 0 as | z | 1 . Further, for z B m , we call a weight ω to be radial, if ω ( z ) = ω ( | z | ) . For a weight ω the weighted-type space H ω ( B m ) = H ω is the space of all functions f H ( B m ) for which
f H ω = sup z B m ω ( z ) | f ( z ) | < .
The little weighted-type space H ω , 0 ( B m ) = H ω , 0 is the subspace of the space H ω and contains all those f H ( B m ) for which
lim | z | 1 ω ( z ) | f ( z ) | = 0 .
Clearly, H ω , 0 is closed in H ω . In particular, for ω ( z ) = ( 1 | z | 2 ) σ ; σ > 0 , the space H ω ( H ω , 0 ) will be the classical weighted-type spaces H σ (respectively, classical little weighted-type spaces H σ , 0 ). For ω 1 , the space H ω get reduced to the the space H of bounded analytic function on B m . The weighted-type spaces have been studied by various authors see e.g., [4,5,6] and the references therein.
Let ξ = ( ξ 1 , ξ 2 , , ξ m ) be a holomorphic self-map of B m and ϕ H ( B m ) . Then, the composition, multiplication, differential and weighted composition operator on H ( B m ) are respectively defined as
C ξ f ( z ) = ( f ξ ) ( z ) = f ( ξ ( z ) ) M ϕ f ( z ) = ϕ ( z ) f ( z ) D f ( z ) = f ( z ) W ϕ , ξ f ( z ) = ( M ϕ C ξ ) f ( z ) = ϕ ( z ) f ( ξ ( z ) ) , z B m ; f H ( B m ) .
W ϕ , ξ is a product-type operator as W ϕ , ξ = M ϕ C ξ . More results on weighted composition operators on class of holomorphic functions can be found in [7,8] and the references therein. The product-type operators W ϕ , ξ D and D W ϕ , ξ were respectively, considered in [2] and [3]. To characterize the product-type operators in a unified way, new product-type operator T ϕ 1 , ϕ 2 , ξ was introduced which can be found in [9,10] and the references therein.
For f H ( B m ) , the radial derivative is defined by
R f ( z ) = j = 1 m z j f z j ( z ) = f ( z ) , z ¯ ,
where f ( z ) denotes the gradient of f which is defined by
f ( z ) = f z 1 ( z ) , f z 2 ( z ) , , f z m ( z ) .
Let ξ = ( ξ 1 , ξ 2 , , ξ m ) be a holomorphic self map of B m , then
| ξ ( z ) | = j = 1 m | ξ j ( z ) | 2 R ξ ( z ) = R ξ 1 ( z ) , R ξ 2 ( z ) , , R ξ m ( z ) | R ξ ( z ) | = j = 1 m | R ξ j ( z ) | 2 .
For more information about the radial derivative operator we refer to [11,12] and the references therein. Using radial derivative operator and ϕ 1 , ϕ 2 , ϕ 3 H ( B m ) , Liu and Yu in [13] studied the extension of the operator T ϕ 1 , ϕ 2 , ξ which is denoted by T ϕ 1 , ϕ 2 , ϕ 3 , ξ and defined as
T ϕ 1 , ϕ 2 , ϕ 3 , ξ f ( z ) = ϕ 1 ( z ) f ( ξ ( z ) ) + ϕ 2 ( z ) R f ( ξ ( z ) ) + ϕ 3 ( z ) R ( f ξ ) ( z ) ,
where f H ( B m ) and z B m . Clearly, by fixing ϕ 1 , ϕ 2 , ϕ 3 in T ϕ 1 , ϕ 2 , ϕ 3 , ξ , all possible products of above defined operators can be obtained. In particular, by setting ϕ 2 ( z ) ϕ 3 ( z ) 0 , the operator T ϕ 1 , ϕ 2 , ϕ 3 , ξ get reduced to W ϕ 1 , ξ . Similarly, for ϕ 1 ( z ) ϕ 3 ( z ) 0 , the operator T ϕ 1 , ϕ 2 , ϕ 3 , ξ get reduced to W ϕ 2 , ξ R . The above defined operator can be found in [13] and the references therein. Various product-type operators have been studied for spaces of analytic functions on the unit disk. For this one can refer [14,15,16,17,18,19]. Product-type operators for the unit ball have also been considered by various experts which can be found in [5,13,20,21,22,23]. Operators involving radial derivative have been considered in many papers some of which are [6,24,25] and the references therein.
In this paper, we investigate the boundedness as well as the compactness of the operators T ϕ 1 , ϕ 2 , ϕ 3 , ξ . This paper is represented in a systematic manner. Introduction and literature part is kept in Section 1 and Section 2 consists of some auxiliary results which are used to derive the main results. In Section 3, we characterize the boundedness of operators T ϕ 1 , ϕ 2 , ϕ 3 , ξ : A σ Ψ H ω and T ϕ 1 , ϕ 2 , ϕ 3 , ξ : A σ Ψ H ω , 0 . Finally, in Section 4, the compactness of operators T ϕ 1 , ϕ 2 , ϕ 3 , ξ : A σ Ψ H ω and T ϕ 1 , ϕ 2 , ϕ 3 , ξ : A σ Ψ H ω , 0 is given. Throughout the paper, for any two positive quantities a and b, the notation a b means that a C b , for some constant C > 0 . The constant C may differ at each occurrence. Further, if both a b and b a hold, then we simply write a b .

2. Auxiliary Results

To obtain the desired results, we have used the following auxiliary results:
Lemma 1.
Let Ψ L p U q and σ > 1 . There is a constant C > 1 such that for any f A σ Ψ ,
| f ( z ) | C Ψ 1 1 ( 1 | z | 2 ) m + σ + 1 f σ , Ψ l u x .
Lemma 2.
Let 1 < σ < , w B m . Let K > 1 . Suppose that Ψ L p U q . Then, the following function is in A σ Ψ
f w ( z ) = Ψ 1 1 ( 1 | w | 2 ) m + σ + 1 1 | w | 2 1 z , w K ( m + σ + 1 ) .
Moreover, f w A σ Ψ l u x 1 .
For the proof of Lemma 1 and Lemma 2, we refer to [26,27].
Lemma 3.
[28] Let Ψ L p U q and σ > 1 . Then, there are two positive constants C 1 and C 2 such that for any f A σ Ψ ,
| f ( z ) | C 1 1 | z | 2 Ψ 1 C 2 ( 1 | z | 2 ) m + σ + 1 f σ , Ψ l u x , f o r   a n y z B m .
Lemma 4.
[20] Let ϕ 1 , ϕ 2 , ϕ 3 H ( B m ) , ξ = ( ξ 1 , ξ 2 , , ξ m ) denote an analytic self-map of B m . Then, T ϕ 1 , ϕ 2 , ϕ 3 , ξ : A σ Ψ H ω is compact if and only if T ϕ 1 , ϕ 2 , ϕ 3 , ξ : A σ Ψ H ω is bounded and for every sequence { f m } in A σ Ψ which is bounded and on compact subsets of B m uniformly converges to zero as m , we have T ϕ 1 , ϕ 2 , ϕ 3 , ξ f m H ω   0 as m .
The compactness of T ϕ 1 , ϕ 2 , ϕ 3 , ξ from a holomorphic space to H ω , 0 can be obtained by using the following lemma which is similar to Lemma 1 in [21]. So the proof is omitted.
Lemma 5.
A set A closed in H ω , 0 is compact if and only if it is bounded and satisfy the following condition
lim | z | 1 sup h A ω ( z ) | h ( z ) | = 0 .

3. Boundedness of Operator T ϕ 1 , ϕ 2 , ϕ 3 , ξ from Weighted Bergman–Orlicz Space to Weighted and Little Weighted Type Spaces

Theorem 1.
Let ξ = ( ξ 1 , ξ 2 , , ξ m ) denote an analytic self map of B m and ϕ 1 , ϕ 2 , ϕ 3 H ( B m ) . Then, the following statements hold true:
(a) 
If
sup z B m ω ( z ) | ϕ 1 ( z ) | Ψ 1 1 ( 1 | ξ ( z ) | 2 ) m + σ + 1 <
and
sup z B m ω ( z ) | ϕ 2 ( z ) ξ ( z ) + ϕ 3 ( z ) R ξ ( z ) | 1 | ξ ( z ) | 2 Ψ 1 C 2 ( 1 | ξ ( z ) | 2 ) m + σ + 1 < ,
then T ϕ 1 , ϕ 2 , ϕ 3 , ξ : A σ Ψ H ω is bounded.
(b) 
If T ϕ 1 , ϕ 2 , ϕ 3 , ξ is bounded then the condition (1) hold and
sup z B m ω ( z ) | ϕ 2 ( z ) ξ ( z ) + ϕ 3 ( z ) R ξ ( z ) , ξ ( z ) | 1 | ξ ( z ) | 2 Ψ 1 C 2 ( 1 | ξ ( z ) | 2 ) m + σ + 1 < .
Proof. 
( a ) First, suppose that conditions (1) and (2) are true. Then, for f A σ Ψ and any z B m , we have
R ( f ξ ) ( z ) = j = 1 m z j ( f ξ ) z j ( z ) = j = 1 m z j k = 1 m f z k ( ξ ( z ) ) ξ k z j ( z ) = k = 1 m f z k ( ξ ( z ) ) j = 1 m z j ξ k z j ( z ) = j = 1 m f z j ( ξ ( z ) ) R ξ j ( z ) = f ( ξ ( z ) ) , R ξ ( z ) ¯ .
Thus, we have
ω ( z ) | T ϕ 1 , ϕ 2 , ϕ 3 , ξ f ( z ) | = ω ( z ) | ϕ 1 ( z ) f ( ξ ( z ) ) + ϕ 2 ( z ) R f ( ξ ( z ) ) + ϕ 3 ( z ) R ( f ξ ) ( z ) | ω ( z ) | ϕ 1 ( z ) f ( ξ ( z ) ) | + ω ( z ) | ϕ 2 ( z ) R f ( ξ ( z ) ) + ϕ 3 ( z ) R ( f ξ ) ( z ) | = ω ( z ) | ϕ 1 ( z ) | | f ( ξ ( z ) ) | + ω ( z ) | ϕ 2 ( z ) f ( ξ ( z ) ) , ξ ( z ) ¯ + ϕ 3 ( z ) f ξ ( z ) ) , R ξ ( z ) ¯ | ω ( z ) | ϕ 1 ( z ) | | f ( ξ ( z ) ) | + ω ( z ) | f ( ξ ( z ) ) , ϕ 2 ( z ) ξ ( z ) + ϕ 3 ( z ) R ξ ( z ) ¯ | ω ( z ) | ϕ 1 ( z ) | | f ( ξ ( z ) ) | + ω ( z ) | ϕ 2 ( z ) ξ ( z ) + ϕ 3 ( z ) R ξ ( z ) | | f ( ξ ( z ) ) | C ω ( z ) | ϕ 1 ( z ) | Ψ 1 1 ( 1 | ξ ( z ) | 2 ) m + σ + 1 f σ , Ψ l u x + C | ϕ 2 ( z ) ξ ( z ) + ϕ 3 ( z ) R ξ ( z ) | 1 | ξ ( z ) | 2 Ψ 1 C 2 ( 1 | ξ ( z ) | 2 ) m + σ + 1 f σ , Ψ l u x = C [ ω ( z ) | ϕ 1 ( z ) | Ψ 1 1 ( 1 | ξ ( z ) | 2 ) m + σ + 1 + | ϕ 2 ( z ) ξ ( z ) + ϕ 3 ( z ) R ξ ( z ) | 1 | ξ ( z ) | 2 Ψ 1 C 2 ( 1 | ξ ( z ) | 2 ) m + σ + 1 ] f σ , Ψ l u x .
From (1), (2) and (4), it follows that T ϕ 1 , ϕ 2 , ϕ 3 , ξ : A σ Ψ H ω is bounded.
( b ) Conversely, suppose that T ϕ 1 , ϕ 2 , ϕ 3 , ξ : A σ Ψ H ω is bounded. Thus, there exist an independent constant C > 0 such that
T ϕ 1 , ϕ 2 , ϕ 3 , ξ f H ω C f A σ Ψ .
For w B m , let
f w ( z ) = a 0 1 | ξ ( w ) | 2 1 z , ξ ( w ) K ( m + σ + 1 ) + a 1 1 | ξ ( w ) | 2 1 z , ξ ( w ) K ( m + σ + 1 ) + 1 ,
where a 0 = K ( m + σ + 1 ) + 1 and a 1 = K ( m + σ + 1 ) . This implies
R f w ( z ) = a 0 ( 1 | ξ ( w ) | 2 ) K ( m + σ + 1 ) j = 1 m z j z j 1 1 z , ξ ( w ) K ( m + σ + 1 ) + a 0 ( 1 | ξ ( w ) | 2 ) K ( m + σ + 1 ) + 1 j = 1 m z j z j 1 1 z , ξ ( w ) K ( m + σ + 1 ) + 1
= a 0 K ( m + σ + 1 ) z , ξ ( w ) 1 z , ξ ( w ) ( 1 | ξ ( w ) | 2 ) 1 z , ξ ( w ) K ( m + σ + 1 ) + a 1 K ( m + σ + 1 ) + 1 z , ξ ( w ) 1 z , ξ ( w ) ( 1 | ξ ( w ) | 2 ) 1 z , ξ ( w ) K ( m + σ + 1 ) + 1 .
By Lemma 2, we have sup w B m f w A σ Ψ 1 . Using (5) and (6), we get
f w ( ξ ( w ) ) = 1 and R f w ( ξ ( w ) ) = 0 .
Since
f w ( ξ ( z ) ) = a 0 1 | ξ ( w ) | 2 1 ξ ( z ) , ξ ( w ) K ( m + σ + 1 ) + a 1 1 | ξ ( w ) | 2 1 ξ ( z ) , ξ ( w ) K ( m + σ + 1 ) + 1 .
Therefore,
R ( f w ξ ) ( z ) = a 0 K ( m + σ + 1 ) R ξ ( z ) , ξ ( w ) 1 ξ ( z ) , ξ ( w ) ( 1 | ξ ( w ) | 2 ) 1 ξ ( z ) , ξ ( w ) K ( m + σ + 1 ) + a 1 K ( m + σ + 1 ) + 1 R ξ ( z ) , ξ ( w ) 1 ξ ( z ) , ξ ( w ) ( 1 | ξ ( w ) | 2 ) 1 ξ ( z ) , ξ ( w ) K ( m + σ + 1 ) + 1 ,
which implies
R ( f w ξ ) ( w ) = 0 .
Using function f w , define a function as
f ( z ) = Ψ 1 1 ( 1 | ξ ( w ) | 2 ) m + σ + 1 f w ( z ) .
From conditions (7), (8) and (9), we obtain
R f ( ξ ( w ) ) = R ( f ξ ) ( w ) = 0 , f ( ξ ( w ) ) = Ψ 1 1 ( 1 | ξ ( w ) | 2 ) m + σ + 1 .
Using Lemma 2, we have f A σ Ψ and f A σ Ψ 1 . By Equation (10) and boundedness of the operator T ϕ 1 , ϕ 2 , ϕ 3 , ξ : A σ Ψ H ω , we have
ω ( w ) | ϕ 1 ( w ) | 1 ( 1 | ξ ( w ) | 2 ) m + σ + 1 C T ϕ 1 , ϕ 2 , ϕ 3 , ξ .
Thus,
sup w B m ω ( w ) | ϕ 1 ( w ) | 1 ( 1 | ξ ( w ) | 2 ) m + σ + 1 < .
This proves (1). For f ( z ) 1 A σ Ψ , we have
K 1 : = sup z B m ω ( z ) | ϕ 1 ( z ) | < .
Taking the function f j ( z ) = z j A σ Ψ , we obtain
sup z B m ω ( z ) | ϕ 1 ( z ) ξ j ( z ) + ϕ 2 ( z ) ξ j ( z ) + ϕ 3 ( z ) R ξ j ( z ) | < .
Using (11) and (12) with the fact that | ξ ( z ) | 1 , we get
sup z B m ω ( z ) | ϕ 2 ( z ) ξ ( z ) + ϕ 3 ( z ) R ξ ( z ) | = sup z B m j = 1 m | ϕ 2 ( z ) ξ j ( z ) + ϕ 3 ( z ) R ξ j ( z ) | 2 C + sup z B m j = 1 m | ϕ 1 ( z ) ξ j ( z ) | 2 C + sup z B m | ϕ 1 ( z ) | | ξ ( z ) | C + K 1 .
Thus,
K 2 : = sup z B m ω ( z ) | ϕ 2 ( z ) ξ ( z ) + ϕ 3 ( z ) R ξ ( z ) | < .
For a fixed w B m , define
g w ( z ) = 1 | ξ ( w ) | 2 1 z , ξ ( w ) K ( m + σ + 1 ) 1 | ξ ( w ) | 2 1 z , ξ ( w ) K ( m + σ + 1 ) + 1 .
This implies
R g w ( z ) = K ( m + σ + 1 ) z , ξ ( w ) 1 z , ξ ( w ) ( 1 | ξ ( w ) | 2 ) 1 z , ξ ( w ) K ( m + σ + 1 ) K ( m + σ + 1 ) + 1 z , ξ ( w ) 1 z , ξ ( w ) ( 1 | ξ ( w ) | 2 ) 1 z , ξ ( w ) K ( m + σ + 1 ) + 1 .
By Lemma 2, we have g w A σ Ψ ( B m ) and sup w B m g w A σ Ψ 1 . Conditions (14) and (15) implies that
g w ( ξ ( w ) ) = 0 , R w ( ξ ( w ) ) = | ξ ( w ) | 2 1 | ξ ( w ) | 2 .
Now, we have
R ( g w ξ ) ( z ) =   K ( m + σ + 1 ) R ξ ( z ) , ξ ( w ) 1 ξ ( z ) , ξ ( w ) ( 1 | ξ ( w ) | 2 ) 1 ξ ( z ) , ξ ( w ) K ( m + σ + 1 ) K ( m + σ + 1 ) + 1 R ξ ( z ) , ξ ( w ) 1 ξ ( z ) , ξ ( w ) ( 1 | ξ ( w ) | 2 ) 1 ξ ( z ) , ξ ( w ) K ( m + σ + 1 ) + 1 ,
which implies
R ( g w ξ ) ( w ) = R ξ ( w ) , ξ ( w ) 1 | ξ ( w ) | 2 .
Using function g w , define another function
g ( z ) = Ψ 1 C 2 ( 1 | ξ ( w ) | 2 ) m + σ + 1 g w ( z ) .
Therefore, by conditions (16), (17) and (18), we get
g ( ξ ( w ) ) = 0 , R g ( ξ ( w ) ) = | ξ ( w ) | 2 1 | ξ ( w ) | 2 Ψ 1 C 2 ( 1 | ξ ( w ) | 2 ) m + σ + 1 , R ( g ξ ) ( w ) = R ξ ( w ) , ξ ( w ) 1 | ξ ( w ) | 2 Ψ 1 C 2 ( 1 | ξ ( w ) | 2 ) m + σ + 1 .
For w B m , we have
C   T ϕ 1 , ϕ 2 , ϕ 3 , ξ g H ω ω ( w ) | ϕ 1 ( w ) g ( ξ ( w ) ) + ϕ 2 ( w ) R g ( ξ ( w ) ) + ϕ 3 ( z ) R ( g ξ ) ( w ) | = ω ( w ) | ϕ 2 ( w ) | ξ ( w ) | 2 + ϕ 3 ( w ) R ξ ( w ) , ξ ( w ) 1 | ξ ( w ) | 2 | Ψ 1 C 2 ( 1 | ξ ( w ) | 2 ) m + σ + 1 = ω ( w ) | ϕ 2 ( w ) ξ ( w ) + ϕ 3 ( w ) R ξ ( w ) , ξ ( w ) | 1 | ξ ( w ) | 2 Ψ 1 C 2 ( 1 | ξ ( w ) | 2 ) m + σ + 1 .
From (19), it follows that
sup w B m ω ( w ) | ϕ 2 ( w ) ξ ( w ) + ϕ 3 ( w ) R ξ ( w ) , ξ ( w ) | 1 | ξ ( w ) | 2 Ψ 1 C 2 ( 1 | ξ ( w ) | 2 ) m + σ + 1 < ,
which proves condition (3). This completes the theorem.  □
Corollary 1.
Let ξ = ( ξ 1 , ξ 2 , , ξ m ) be an analytic self map of B m , ϕ 1 , ϕ 2 , ϕ 3 H ( B m ) and suppose the functions ξ and ϕ 2 ξ + ϕ 3 R ξ are linearly dependent. Then, T ϕ 1 , ϕ 2 , ϕ 3 , ξ : A σ Ψ H ω is bounded if and only if it satisfies (1) and (2).
Proof. 
First suppose that T ϕ 1 , ϕ 2 , ϕ 3 , ξ : A σ Ψ H ω is bounded. Then, by condition ( b ) of Theorem 1, we see that (1) holds. So, we only need to prove (2). Using Cauchy-Schwartz inequality and (19), for w B m we get
C ω ( w ) | ϕ 2 ( w ) | ξ ( w ) | 2 + ϕ 3 ( w ) R ξ ( w ) , ξ ( w ) 1 | ξ ( w ) | 2 | Ψ 1 C 2 ( 1 | ξ ( w ) | 2 ) m + σ + 1 = ω ( w ) | ϕ 2 ( w ) ξ ( w ) + ϕ 3 ( w ) R ξ ( w ) , ξ ( w ) | 1 | ξ ( w ) | 2 Ψ 1 C 2 ( 1 | ξ ( w ) | 2 ) m + σ + 1 = ω ( w ) | ϕ 2 ( w ) ξ ( w ) + ϕ 3 ( w ) R ξ ( w ) | | ξ ( w ) | 1 | ξ ( w ) | 2 Ψ 1 C 2 ( 1 | ξ ( w ) | 2 ) m + σ + 1 .
Thus, for δ ( 0 , 1 ) , condition (20) implies that
sup δ < | ξ ( w ) | < 1 ω ( w ) | ϕ 2 ( w ) ξ ( w ) + ϕ 3 ( w ) R ξ ( w ) | 1 | ξ ( w ) | 2 Ψ 1 C 2 ( 1 | ξ ( w ) | 2 ) m + σ + 1 sup δ < | ξ ( w ) | < 1 ω ( w ) | ϕ 2 ( w ) ξ ( w ) + ϕ 3 ( w ) R ξ ( w ) | | ξ ( w ) | δ ( 1 | ξ ( w ) | 2 ) Ψ 1 C 2 ( 1 | ξ ( w ) | 2 ) m + σ + 1 C δ .
Therefore, by Equation (13), we have
sup | ξ ( w ) | δ ω ( w ) | ϕ 2 ( w ) ξ ( w ) + ϕ 3 ( w ) R ξ ( w ) | 1 | ξ ( w ) | 2 Ψ 1 C 2 ( 1 | ξ ( w ) | 2 ) m + σ + 1 sup | ξ ( w ) | δ ω ( w ) | ϕ 2 ( w ) ξ ( w ) + ϕ 3 ( w ) R ξ ( w ) | 1 δ 2 Ψ 1 C 2 ( 1 δ 2 ) m + σ + 1 K 2 1 δ 2 Ψ 1 C 2 ( 1 δ 2 ) m + σ + 1 < .
Hence, (2) follows from the combination of (21) and (22). This completes the proof.
Conversely suppose that (1) and (2) hold. Then, from condition ( a ) in Theorem 1 we get that T ϕ 1 , ϕ 2 , ϕ 3 , ξ : A σ Ψ H ω is bounded.  □
In Corollary 1, if we take ϕ 1 ( z ) = ϕ ( z ) and ϕ 2 ( z ) ϕ 3 ( z ) 0 , then the operator get reduced to the weighted composition operator W ϕ , ξ : A σ Ψ H ω . Thus we get the following corollary for the boundedness of W ϕ , ξ : A σ Ψ H ω as:
Corollary 2.
Let ξ = ( ξ 1 , ξ 2 , , ξ m ) be an analytic self map of B m and ϕ H ( B m ) . Then the operator W ϕ , ξ : A σ Ψ H ω is bounded if and only if
sup z B m ω ( z ) | ϕ ( z ) | Ψ 1 1 ( 1 | ξ ( z ) | 2 ) m + σ + 1 < .
Theorem 2.
Let ξ = ( ξ 1 , ξ 2 , , ξ m ) be an analytic self map of B m and ϕ 1 , ϕ 2 , ϕ 3 H ( B m ) . Then, the operator T ϕ 1 , ϕ 2 , ϕ 3 , ξ : A σ Ψ H ω , 0 is bounded if and only if T ϕ 1 , ϕ 2 , ϕ 3 , ξ : A σ Ψ H ω is bounded and satisfy the conditions
lim | z | 1 ω ( z ) | ϕ 1 ( z ) | = 0
and
lim | z | 1 ω ( z ) | ϕ 2 ( z ) ξ ( z ) + ϕ 3 ( z ) R ξ ( z ) | = 0 .
Proof. 
First suppose that T ϕ 1 , ϕ 2 , ϕ 3 , ξ : A σ Ψ H ω , 0 is bounded. This implies the boundedness of T ϕ 1 , ϕ 2 , ϕ 3 , ξ : A σ Ψ H ω . Thus for each f A σ Ψ , we have T ϕ 1 , ϕ 2 , ϕ 3 , ξ f H ω , 0 . On setting f ( z ) 1 A σ Ψ , we get that
lim | z | 1 ω ( z ) | ϕ 1 ( z ) | = 0 ,
which proves (23). Now, for k = 1 , 2 , , m take f k ( z ) = z k A σ Ψ . So, we get
lim | z | 1 ω ( z ) | ϕ 1 ( z ) ξ k ( z ) + ϕ 2 ( z ) ξ k ( z ) + ϕ 3 ( z ) R ξ k ( z ) | = 0 .
By using (23) and (25) and triangle inequality with condition | ξ k ( z ) | | ξ ( z ) | 1 , we get
lim | z | 1 ω ( z ) | ϕ 2 ( z ) ξ k ( z ) + ϕ 3 ( z ) R ξ k ( z ) | = 0 , k = 1 , 2 , , m .
Thus,
lim | z | 1 ω ( z ) | ϕ 2 ( z ) ξ ( z ) + ϕ 3 ( z ) R ξ ( z ) | = 0 ,
which proves (24).
Conversely, suppose that (23), (24) hold and T ϕ 1 , ϕ 2 , ϕ 3 , ξ : A σ Ψ H ω is bounded. Then, for each polynomial q and z B m , we obtain
| q ( z ) | q , and | q ( z ) | C < .
Therefore,
ω ( z ) | T ϕ 1 , ϕ 2 , ϕ 3 , ξ q ( z ) | = ω ( z ) | ϕ 1 ( z ) q ( ξ ( z ) ) + ϕ 2 ( z ) R q ( ξ ( z ) ) + ϕ 3 ( z ) R ( q ξ ) ( z ) | ω ( z ) | ϕ 1 ( z ) q ( ξ ( z ) ) | + ω ( z ) | ϕ 2 ( z ) R q ( ξ ( z ) ) + ϕ 3 ( z ) R ( q ξ ) ( z ) | = ω ( z ) | ϕ 1 ( z ) | | q ( ξ ( z ) ) | + ω ( z ) | ϕ 2 ( z ) q ( ξ ( z ) ) , ξ ( z ) ¯ + ϕ 3 ( z ) q ξ ( z ) ) , R ξ ( z ) ¯ | = ω ( z ) | ϕ 1 ( z ) | | q ( ξ ( z ) ) | + ω ( z ) | q ( ξ ( z ) ) , ϕ 2 ( z ) ξ ( z ) + ϕ 3 ( z ) R ξ ( z ) ¯ | ω ( z ) | ϕ 1 ( z ) | | q ( ξ ( z ) ) | + ω ( z ) | ϕ 2 ( z ) ξ ( z ) + ϕ 3 ( z ) R ξ ( z ) | | q ( ξ ( z ) ) | ω ( z ) | ϕ 1 ( z ) | q + C ω ( z ) | ϕ 2 ( z ) ξ ( z ) + ϕ 3 ( z ) R ξ ( z ) | 0 as | z | 1 .
From the above fact, we get that T ϕ 1 , ϕ 2 , ϕ 3 , ξ q H ω , 0 . As we know that the class of all the polynomial is dense in A σ Ψ , so for each f A σ Ψ , there will be a sequence { q k } k N such that
lim k q k f A σ Ψ = 0 .
By the boundedness of the operator T ϕ 1 , ϕ 2 , ϕ 3 , ξ : A σ Ψ H ω , we get that
T ϕ 1 , ϕ 2 , ϕ 3 , ξ q k T ϕ 1 , ϕ 2 , ϕ 3 , ξ f H ω T ϕ 1 , ϕ 2 , ϕ 3 , ξ q k f A σ Ψ 0 as k .
Thus, T ϕ 1 , ϕ 2 , ϕ 3 , ξ f H ω , 0 . As H ω , 0 H ω is closed, this implies T ϕ 1 , ϕ 2 , ϕ 3 , ξ ( A σ Ψ ) H ω , 0 . Hence, the boundedness of T ϕ 1 , ϕ 2 , ϕ 3 , ξ : A σ Ψ H ω , 0 follows from the boundedness of T ϕ 1 , ϕ 2 , ϕ 3 , ξ : A σ Ψ H ω . This completes the proof.  □
By taking ϕ 1 ( z ) = ϕ ( z ) and ϕ 2 ( z ) ϕ 3 ( z ) 0 in Theorem 2, we get W ϕ , ξ : A σ Ψ H ω , 0 . Thus, we get the following corollary for the boundedness of W ϕ , ξ : A σ Ψ H ω , 0 as:
Corollary 3.
Let ξ = ( ξ 1 , ξ 2 , , ξ m ) be an analytic self map of B m and ϕ H ( B m ) . Then, the operator W ϕ , ξ : A σ Ψ H ω , 0 is bounded if and only if W ϕ , ξ : A σ Ψ H ω is bounded and ϕ H ω , 0 .
Theorem 3.
Let ξ = ( ξ 1 , ξ 2 , , ξ m ) denotes an analytic self map of B m and ϕ 1 , ϕ 2 , ϕ 3 H ( B m ) . Then, the operator T ϕ 1 , ϕ 2 , ϕ 3 , ξ : A σ Ψ H ω , 0 is bounded if it satisfy the conditions
lim | z | 1 ω ( z ) | ϕ 1 ( z ) | Ψ 1 1 ( 1 | ξ ( z ) | 2 ) m + σ + 1 = 0
and
lim | z | 1 ω ( z ) | ϕ 2 ( z ) ξ ( z ) + ϕ 3 ( z ) R ξ ( z ) | 1 | ξ ( z ) | 2 Ψ 1 C 2 ( 1 | ξ ( z ) | 2 ) m + σ + 1 = 0 .
Proof. 
Suppose that (26) and (27) hold. It clearly implies that (1) and (2) hold. By using condition ( a ) in 1, we obtain that T ϕ 1 , ϕ 2 , ϕ 3 , ξ : A σ Ψ H ω is bounded. So, by taking f A σ Ψ and using (4), we get
ω ( z ) | T ϕ 1 , ϕ 2 , ϕ 3 , ξ f ( z ) | ω ( z ) | ϕ 1 ( z ) | | f ( ξ ( z ) ) | + ω ( z ) | ϕ 2 ( z ) R f ( ξ ( z ) ) + ϕ 3 ( z ) R ( f ξ ) ( z ) | C f A σ Ψ [ ω ( z ) | ϕ 1 ( z ) | Ψ 1 1 ( 1 | ξ ( z ) | 2 ) m + σ + 1 + ω ( z ) | ϕ 2 ( z ) ξ ( z ) + ϕ 3 ( z ) R ξ ( z ) | 1 | ξ ( z ) | 2 Ψ 1 C 2 ( 1 | ξ ( z ) | 2 ) m + σ + 1 ] 0 as | z | 1 ,
Thus, T ϕ 1 , ϕ 2 , ϕ 3 , ξ f H ω , 0 . Hence, the boundedness of the operator T ϕ 1 , ϕ 2 , ϕ 3 , ξ : A σ Ψ H ω , 0 follows from the boundedness of T ϕ 1 , ϕ 2 , ϕ 3 , ξ : A σ Ψ H ω . This completes the proof.  □

4. Compactness of the Operator T ϕ 1 , ϕ 2 , ϕ 3 , ξ from Weighted Bergman–Orlicz Space to Weighted and Little Weighted Type Spaces

Theorem 4.
Let ξ = ( ξ 1 , ξ 2 , , ξ m ) denote a holomorphic self-map of B m and ϕ 1 , ϕ 2 , ϕ 3 H ( B m ) . Then, the following statements hold true:
(a) 
If T ϕ 1 , ϕ 2 , ϕ 3 , ξ : A σ Ψ H ω is bounded and satisfy the conditions
lim | ξ ( z ) | 1 ω ( z ) | ϕ 1 ( z ) | Ψ 1 1 ( 1 | ξ ( z ) | 2 ) m + σ + 1 = 0
and
lim | ξ ( z ) | 1 ω ( z ) | ϕ 2 ( z ) ξ ( z ) + ϕ 3 ( z ) R ξ ( z ) | 1 | ξ ( z ) | 2 Ψ 1 C 2 ( 1 | ξ ( z ) | 2 ) m + σ + 1 = 0 ,
then, T ϕ 1 , ϕ 2 , ϕ 3 , ξ : A σ Ψ H ω is compact.
(b) 
If T ϕ 1 , ϕ 2 , ϕ 3 , ξ is compact then, it satisfy (28) along with the condition
lim | ξ ( z ) | 1 ω ( z ) | ϕ 2 ( z ) ξ ( z ) + ϕ 3 ( z ) R ξ ( z ) , ξ ( z ) | 1 | ξ ( z ) | 2 Ψ 1 C 2 ( 1 | ξ ( z ) | 2 ) m + σ + 1 = 0 .
Proof. 
( a ) First suppose that T ϕ 1 , ϕ 2 , ϕ 3 , ξ : A σ Ψ H ω is bounded and satisfy the conditions (28) and (29). Let { f k } A σ Ψ be a bounded sequence converging to zero uniformly on compact subset of B m as k . Let f k σ , Ψ l u x 1 .
Now, by conditions (28) and (29), we have that for any ε > 0 , there will be a δ ( 0 , 1 ) for which
ω ( z ) | ϕ 1 ( z ) | Ψ 1 1 ( 1 | ξ ( z ) | 2 ) m + σ + 1 < ε
and
ω ( z ) | ϕ 2 ( z ) ξ ( z ) + ϕ 3 ( z ) R ξ ( z ) | 1 | ξ ( z ) | 2 Ψ 1 C 2 ( 1 | ξ ( z ) | 2 ) m + σ + 1 < ε ,
when δ < | ξ ( z ) | < 1 . Since T ϕ 1 , ϕ 2 , ϕ 3 , ξ is bounded, so condition (1) and (3) hold. In addition, on compact subset of B m , the sequence { f k } A σ Ψ converges to zero. So, using the Cauchy’s estimate we get that on compact subsets of B m , the sequence f k uniformly converges to zero as k . This implies that lim k sup | W | δ | f k ( W ) | = 0 . Thus, there will be a K 0 N such that
sup | ξ ( z ) | δ ω ( z ) | T ϕ 1 , ϕ 2 , ϕ 3 , ξ f k ( z ) | sup | ξ ( z ) | δ ω ( z ) | ϕ 1 ( z ) | | f k ( ξ ( z ) ) | + sup | ξ ( z ) | δ ω ( z ) | ϕ 2 ( z ) R f k ( ξ ( z ) ) + ϕ 3 ( z ) R ( f k ξ ) ( z ) | K 1 sup | ξ ( z ) | δ | f k ( ξ ( z ) ) | + sup | ξ ( z ) | δ ω ( z ) | ϕ 2 ( z ) ξ ( z ) + ϕ 3 ( z ) R ξ ( z ) | | f k ( ξ ( z ) ) | K 1 sup | W | δ | f k ( W ) | + K 2 sup | W | δ | f k ( W ) | < C ε ,
where k > K 0 . On combining conditions (31), (32) and (33) with Lemma 1, we get
T ϕ 1 , ϕ 2 , ϕ 3 , ξ f k H ω = sup z B m ω ( z ) | T ϕ 1 , ϕ 2 , ϕ 3 , ξ f k ( z ) | sup | ξ ( z ) | δ ω ( z ) | T ϕ 1 , ϕ 2 , ϕ 3 , ξ f k ( z ) | + sup δ < | ξ ( z ) | < 1 ω ( z ) | T ϕ 1 , ϕ 2 , ϕ 3 , ξ f k ( z ) | < C ε + C sup δ < | ξ ( z ) | < 1 ω ( z ) | ϕ 1 ( z ) | Ψ 1 1 ( 1 | ξ ( z ) | 2 ) m + σ + 1 f k σ , Ψ l u x + C sup δ < | ξ ( z ) | < 1 ω ( z ) | ϕ 2 ( z ) ξ ( z ) + ϕ 3 ( z ) R ξ ( z ) | 1 | ξ ( z ) | 2 Ψ 1 C 2 ( 1 | ξ ( z ) | 2 ) m + σ + 1 f k σ , Ψ l u x < 3 C ε ,
when k > K 0 , which approaches to zero as k . Hence the operator T ϕ 1 , ϕ 2 , ϕ 3 , ξ : A σ Ψ H ω is compact.
( b ) Conversely, suppose that T ϕ 1 , ϕ 2 , ϕ 3 , ξ : A σ Ψ H ω is compact. This implies that T ϕ 1 , ϕ 2 , ϕ 3 , ξ is bounded. If ξ < 1 , then conditions (28) and (30) hold. Let ξ = 1 and { z k } B m be a sequence such that | ξ ( z k ) | 1 as k . Using sequence | ξ ( z k ) | define a function
f k ( z ) = Ψ 1 1 ( 1 | ξ ( z k ) | 2 ) m + σ + 1 f ξ ( z k ) ( z ) ,
where
f ξ ( z k ) ( z ) = K ( m + σ + 1 ) + 1 1 | ξ ( z k ) | 2 1 z , ξ ( z k ) k ( m + σ + 1 ) K ( m + σ + 1 ) 1 | ξ ( z k ) | 2 1 z , ξ ( z k ) k ( m + σ + 1 ) + 1 .
Clearly, f ξ ( z k ) A σ Ψ and sup k N f ξ ( z k ) A σ Ψ C . Therefore,
| f ξ ( z k ) ( z ) |   K ( m + σ + 1 ) + 1 | 1 | ξ ( z k ) | 2 1 z , ξ ( z k ) | k ( m + σ + 1 ) + K ( m + σ + 1 ) | 1 | ξ ( z k ) | 2 1 z , ξ ( z k ) | k ( m + σ + 1 ) + 1   K ( m + σ + 1 ) + 1 | 1 | ξ ( z k ) | 2 1 r | k ( m + σ + 1 ) + K ( m + σ + 1 ) | 1 | ξ ( z k ) | 2 1 r | k ( m + σ + 1 ) + 1 ,
when | z | r , we get that f ξ ( z k ) converges to zero uniformly on compact subsets of B m . Thus, f k converges to zero uniformly on compact subsets of B m . Therefore, by Lemma 4, it follows that the sequence { f k } uniformly converges to zero on any compact subsets of B m as k such that
lim k T ϕ 1 , ϕ 2 , ϕ 3 , ξ f k H ω = 0 .
In addition, we have
f k ( ξ ( z k ) ) = Ψ 1 1 1 ( | ξ ( z k ) | 2 ) m + σ + 1 a n d R f k ( ξ ( z k ) ) = R ( f k ξ ) ( z k ) = 0 .
Thus, we obtain
ω ( z k ) | ϕ 1 ( z k ) | Ψ 1 1 1 ( | ξ ( z k ) | 2 ) m + σ + 1 T ϕ 1 , ϕ 2 , ϕ 3 , ξ f k H ω 0 as k ,
which implies that (28) holds. In order to prove condition (30), we define another sequence of functions
g k ( z ) = Ψ 1 C 2 ( 1 | ξ ( z k ) | 2 ) m + σ + 1 g ξ ( z k ) ( z ) ,
where
g ξ ( z k ) ( z ) = 1 | ξ ( z k ) | 2 1 z , ξ ( z k ) k ( m + σ + 1 ) 1 | ξ ( z k ) | 2 1 z , ξ ( z k ) k ( m + σ + 1 ) + 1 .
Therefore,
( g ξ ( z ) k ξ ) ( z ) = 1 | ξ ( z k ) | 2 1 ξ ( z ) , ξ ( z k ) k ( m + σ + 1 ) 1 | ξ ( z k ) | 2 1 ξ ( z ) , ξ ( z k ) k ( m + σ + 1 ) + 1 ,
which implies
R g ξ ( z k ) ( z ) = K ( m + σ + 1 ) z , ξ ( z k ) 1 z , ξ ( z k ) 1 | ξ ( z k ) | 2 1 z , ξ ( z k ) k ( m + σ + 1 ) + K ( m + σ + 1 ) + 1 z , ξ ( z k ) 1 z , ξ ( z k ) 1 | ξ ( z k ) | 2 1 z , ξ ( z k ) k ( m + σ + 1 ) + 1
and
R ( g ξ ( z ) k ξ ) ( z ) = K ( m + σ + 1 ) R ξ ( z ) , ξ ( z k ) 1 ξ ( z ) , ξ ( z k ) 1 | ξ ( z k ) | 2 1 ξ ( z ) , ξ ( z k ) k ( m + σ + 1 ) + K ( m + σ + 1 ) + 1 R ξ ( z ) , ξ ( z k ) 1 ξ ( z ) , ξ ( z k ) 1 | ξ ( z k ) | 2 1 ξ ( z ) , ξ ( z k ) k ( m + σ + 1 ) + 1 .
Similar to the sequence f k , the sequence g ξ ( z k ) and hence the sequence g k converges to zero uniformly on compact subsets of B m , g k A σ Ψ and sup k N g k A σ Ψ C . By Lemma 4, we have
lim k T ϕ 1 , ϕ 2 , ϕ 3 , ξ g k H ω = 0 .
In addition, we have
g k ( ξ ( z k ) ) = 0 , R g k ( ξ ( z k ) ) = | ξ ( z k ) | 2 1 | ξ ( z k ) | 2 , R ( g k ξ ) ( z k ) = R ξ ( z k ) , ξ ( z k ) 1 | ξ ( z k ) | 2 .
Thus,
T ϕ 1 , ϕ 2 , ϕ 3 , ξ g k H ω = sup z B m ω ( z ) | T ϕ 1 , ϕ 2 , ϕ 3 , ξ g k ( z ) | ω ( z k ) | ϕ 1 ( z k ) g k ( ξ ( z k ) ) + ϕ 2 ( z k ) R g k ( ξ ( z k ) ) + ϕ 3 ( z k ) R ( g k ξ ) ( z k ) | = ω ( z k ) | ϕ 2 ( z k ) R g k ( ξ ( z k ) ) + ϕ 3 ( z k ) R ( g k ξ ) ( z k ) | = ω ( z k ) | ϕ 2 ( z k ) ξ ( z k ) + ϕ 3 ( z k ) R ξ ( z k ) , ξ ( z k ) | 1 | ξ ( z k ) | 2 .
By condition (36)
ω ( z k ) | ϕ 2 ( z k ) ξ ( z k ) + ϕ 3 ( z k ) R ξ ( z k ) , ξ ( z k ) | 1 | ξ ( z k ) | 2 0 as k ,
from which condition (30) follows.  □
Corollary 4.
Let ξ = ( ξ 1 , ξ 2 , , ξ m ) be an analytic self map of B m , ϕ 1 , ϕ 2 , ϕ 3 H ( B m ) and suppose the functions ξ and ϕ 2 ξ + ϕ 3 R ξ are linearly dependent. Then, we have the following equivalent statements:
(a) 
T ϕ 1 , ϕ 2 , ϕ 3 , ξ : A σ Ψ H ω is compact;
(b) 
T ϕ 1 , ϕ 2 , ϕ 3 , ξ : A σ Ψ H ω is bounded and, satisfy (28) and (29).
Proof. 
We omit the proof as it is easy to prove.  □
In Corollary 4 if we take ϕ 1 ( z ) = ϕ ( z ) and ϕ 2 ( z ) ϕ 3 ( z ) 0 , then we get the operator W ϕ , ξ : A σ Ψ H ω whose compactness can be given by the following corollary:
Corollary 5.
Let ξ = ( ξ 1 , ξ 2 , , ξ m ) be an analytic self map of B m and ϕ H ( B m ) . Then, the operator W ϕ , ξ : A σ Ψ H ω is compact if and only if W ϕ , ξ : A σ Ψ H ω is bounded and
lim | ξ ( z ) | 1 ω ( z ) | ϕ ( z ) | Ψ 1 1 ( 1 | ξ ( z ) | 2 ) m + σ + 1 = 0 .
Theorem 5.
Let ξ = ( ξ 1 , ξ 2 , , ξ m ) denote a holomorphic self-map of B m and ϕ 1 , ϕ 2 , ϕ 3 H ( B m ) . Then, the following statements hold true
(i) 
If (26) and (27) hold, then the operator T ϕ 1 , ϕ 2 , ϕ 3 , ξ : A σ Ψ H ω , 0 is compact;
(ii) 
If T ϕ 1 , ϕ 2 , ϕ 3 , ξ : A σ Ψ H ω , 0 is compact, then (26) hold and
lim | z | 1 ω ( z ) | ϕ 2 ( z ) ξ ( z ) + ϕ 3 ( z ) R ξ ( z ) , ξ ( z ) | 1 | ξ ( z ) | 2 Ψ 1 C 2 ( 1 | ξ ( z ) | 2 ) m + σ + 1 = 0 .
Proof. 
( i ) Initially, suppose that conditions (26) and (27) hold. Then, from Theorem 3 we get that T ϕ 1 , ϕ 2 , ϕ 3 , ξ : A σ Ψ H ω , 0 is bounded. Let | z | 1 in (4) and take supremum on the unit ball f A σ Ψ 1 , we get that
lim | z | 1 sup f A σ Ψ 1 ω ( z ) | T ϕ 1 , ϕ 2 , ϕ 2 , ξ f ( z ) | = 0 .
Thus, on applying lemma 5, we get that the operator T ϕ 1 , ϕ 2 , ϕ 3 , ξ : A σ Ψ H ω , 0 is compact.
( i i ) Assume that T ϕ 1 , ϕ 2 , ϕ 3 , ξ : A σ Ψ H ω , 0 is compact. This implies the compactness of T ϕ 1 , ϕ 2 , ϕ 3 , ξ : A σ Ψ H ω . Condition ( b ) in Theorem 4 implies that (28) and (30) hold. Thus, for ε > 0 , there exists a δ ( 0 , 1 ) for which (31) hold along with the condition
ω ( z ) | ϕ 2 ( z ) ξ ( z ) + ϕ 3 ( z ) R ξ ( z ) , ξ ( z ) | 1 | ξ ( z ) | 2 Ψ 1 C 2 ( 1 | ξ ( z ) | 2 ) m + σ + 1 < ε ,
where δ < | ξ ( z ) | < 1 . Further, the compactness of T ϕ 1 , ϕ 2 , ϕ 3 , ξ : A σ Ψ H ω , 0 implies that T ϕ 1 , ϕ 2 , ϕ 3 , ξ : A σ Ψ H ω , 0 is bounded. Using Theorem 2, we see that (23) and (24) hold. Thus, for any ε > 0 , there exists a η ( 0 , 1 ) such that
ω ( z ) | ϕ 1 ( z ) | < ε Ψ 1 1 ( 1 δ 2 ) m + σ + 1
and
ω ( z ) | ϕ 2 ( z ) ξ ( z ) + ϕ 3 ( z ) ( R ) ξ ( z ) | < ε 1 δ 2 Ψ 1 C 2 ( 1 δ 2 ) m + σ + 1 ,
when η < | z | < 1 . Therefore, with δ < | ξ ( z ) | < 1 and η < | z | < 1 condition (31) implies that
ω ( z ) | ϕ 1 ( z ) | Ψ 1 1 ( 1 | ξ ( z ) | 2 ) m + σ + 1 < ε .
Again, for | ξ ( z ) | δ and η < | z | < 1 condition (39) implies that
ω ( z ) | ϕ 1 ( z ) | Ψ 1 1 ( 1 | ξ ( z ) | 2 ) m + σ + 1 ω ( z ) | ϕ 1 ( z ) | Ψ 1 1 ( 1 δ 2 ) m + σ + 1 < ε .
Thus, (26) follows from (41) and (42). Similarly, by using Cauchy-Schwartz inequality, (38) and (40) we can obtain (37). This completes the proof.  □
Corollary 6.
Let ξ = ( ξ 1 , ξ 2 , , ξ m ) be an analytic self map of B m , ϕ 1 , ϕ 2 , ϕ 3 H ( B m ) and suppose the functions ξ and ϕ 2 ξ + ϕ 3 R ξ are linearly dependent. Then, we have the following equivalent statements:
(a) 
T ϕ 1 , ϕ 2 , ϕ 3 , ξ : A σ Ψ H ω , 0 is compact;
(b) 
(26) and (27) hold.
Proof. 
We omit the proof as it is easy to prove.  □
In Corollary 6 if we take ϕ 1 ( z ) = ϕ ( z ) and ϕ 2 ( z ) ϕ 3 ( z ) 0 , then we get the operator W ϕ , ξ : A σ Ψ H ω , 0 whose compactness can be given by the following corollary:
Corollary 7.
Let ξ = ( ξ 1 , ξ 2 , , ξ m ) be an analytic self map of B m and ϕ H ( B m ) . Then, the operator W ϕ , ξ : A σ Ψ H ω , 0 is compact if and only if
lim | z | 1 ω ( z ) | ϕ ( z ) | Ψ 1 1 ( 1 | ξ ( z ) | 2 ) m + σ + 1 = 0 .

5. Discussion and Conclusions

In this paper, we have considered the product type operators formed by the combination of composition, multiplication, differentiation and radial derivative operators acting between weighted Bergman–Orlicz spaces and weighted type spaces taken over the unit ball. We analysed these operators for basic properties including boundedness and compactness. The basic aim of this paper is to give the operator-theoretic characterization of these operators in terms of function–theoretic characterization of their including functions.

Author Contributions

M.D., K.R. and M.M. All authors have contributed equally to the conceptualization, design and implementation of this research work. All authors have read and agreed to the published version of the manuscript.

Funding

This research received no external funding.

Institutional Review Board Statement

Not applicable.

Informed Consent Statement

Not applicable.

Data Availability Statement

Not applicable.

Conflicts of Interest

The authors declare no conflict of interest.

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Devi, M.; Raj, K.; Mursaleen, M. Product Type Operators Involving Radial Derivative Operator Acting between Some Analytic Function Spaces. Mathematics 2021, 9, 2447. https://doi.org/10.3390/math9192447

AMA Style

Devi M, Raj K, Mursaleen M. Product Type Operators Involving Radial Derivative Operator Acting between Some Analytic Function Spaces. Mathematics. 2021; 9(19):2447. https://doi.org/10.3390/math9192447

Chicago/Turabian Style

Devi, Manisha, Kuldip Raj, and Mohammad Mursaleen. 2021. "Product Type Operators Involving Radial Derivative Operator Acting between Some Analytic Function Spaces" Mathematics 9, no. 19: 2447. https://doi.org/10.3390/math9192447

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