QK Type Spaces and Bloch Type Spaces on the Unit Ball

DOI: 10.4236/apm.2019.910042   PDF   HTML   XML   114 Downloads   264 Views  

Abstract

Different function spaces have certain inclusion or equivalence relations. In this paper, the author introduces a class of Möbius-invariant Banach spaces QK,0 (p,q) of analytic function on the unit ball of Cn, where K:(0,∞)→[0,∞) are non-decreasing functions and 0<P<, p/2-n-1<q<, studies the inclusion relations between QK,0 (p,q) and a class of B0α spaces which was known before, and concludes that QK,0 (p,q) is a subspace of B0(q+n+1)/p, and the sufficient and necessary condition on kernel function K(r) such that QK,0 (p,q)= B0(q+n+1)/p.

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Hu, R. (2019) QK Type Spaces and Bloch Type Spaces on the Unit Ball. Advances in Pure Mathematics, 9, 857-862. doi: 10.4236/apm.2019.910042.

1. Introduction

Q K spaces were first given by Hasi Wulan and Matts Essen around 2000. In recent years, Q K type spaces have caused extensive research (cf. [1] - [11] ). To study a new kind of function space, we usually need to establish the relationship between that and those known to all. The notion of the spaces Q K on the unit ball was defined by Xu Wen in his paper [4]. According to Hasi Wulan, Q K type spaces Q K ( p , q ) on unit disk were introduced and investigated, and the conditions on K such that Q K ( p , q ) become some known spaces were given (cf. [5] ). About multiple variables, the definition of Q K , 0 ( p , q ) on unit ball were given by Xu Wen (cf. [6] ), and the author has studied the inclusion relations between Q K ( p , q ) spaces and B q + n + 1 p spaces on the unit ball (cf. [7] ). In this paper, the author introduces the Q K , 0 ( p , q ) spaces and B 0 q + n + 1 p spaces on the unit ball of n , studies the inclusion relationship between them. Firstly, establish the relationship between the norm of the function which belongs to Q K , 0 ( p , q ) and the norm f B 0 α , proof that the Q K , 0 ( p , q ) is a subspace of B 0 q + n + 1 p ; and then obtain the necessary and sufficient condition of kernel functions K ( r ) when Q K , 0 ( p , q ) = B 0 q + n + 1 p .

2. Preliminaries

Let a B n and φ a be the involution of B n satisfied φ a ( 0 ) = a . d v ( z ) is the volume measure on B n , normalized so that v ( B n ) = 1 , and d λ = d v ( z ) ( 1 | z | 2 ) n + 1 is the Möbius invariant volume measure on B n (cf. [4] ), d σ is the normalized surface measure on S n , the measure v and σ are related by (cf. [12] )

B n f ( z ) d v ( z ) = 2 n 0 1 r 2 n 1 d r S n f ( r ζ ) d σ ( ζ ) . (1)

Let f ( z ) = ( f z 1 , f z 2 , , f z n ) denote the complex gradient of f, and ˜ f ( z ) = ( f φ z ) ( 0 ) is the invariant gradient of f (cf. [12] ). ˜ f ( z ) and f ( z ) are related by ( [12] )

( 1 | z | 2 ) | f ( z ) | | ˜ f ( z ) | ( 1 | z | 2 ) 1 2 | f ( z ) | . (2)

The Möbius invariant Green function is defined by G ( z , a ) = g ( φ a ( z ) ) , where

g ( z ) = n + 1 2 n | z | 1 ( 1 t 2 ) n 1 t 2 n + 1 d t . (3)

Definition 1 Let K : ( 0 , ) [ 0 , ) is a right-continuous, non-decreasing function, for 0 < p < , p 2 n 1 < q < , we say that a holomorphic function f belongs to the space Q K , 0 ( p , q ) if

lim | a | 1 B n | ˜ f ( z ) | p ( 1 | z | 2 ) q + n + 1 p K ( G ( z , a ) ) d λ ( z ) = 0 . (4)

Definition 2 B 0 α space is defined by

B 0 α = { f H ( B n ) : lim | a | 1 ( 1 | z | 2 ) α 1 | ˜ f ( z ) | = 0 } . (5)

The constant C can represent different values in different places in this paper.

3. Main Results

In this paper, the author demonstrates that Q K , 0 ( p , q ) is a subspace of B 0 q + n + 1 p as the first main result and it is of great help for the second one.

Theorem 1. Let 0 < p < , p 2 n 1 < q < , then Q K , 0 ( p , q ) B 0 q + n + 1 p .

Proof Let E ( a , r 0 ) = { z B n , | φ a ( z ) | < r 0 } , then

B n | ˜ f ( z ) | p ( 1 | z | 2 ) q + n + 1 p K ( G ( z , a ) ) d λ ( z ) E ( a , r 0 ) | ˜ f ( z ) | p ( 1 | z | 2 ) q + n + 1 p K ( g ( φ a ( z ) ) ) d λ ( z ) = | z | < r 0 | ˜ ( f φ a ) ( z ) | p ( 1 | φ a ( z ) | 2 ) q + n + 1 p K ( g ( z ) ) d λ ( z ) K ( g ( r 0 ) ) | z | < r 0 ( 1 | z | 2 ) p | ( f φ a ) ( z ) | p ( 1 | φ a ( z ) | 2 ) q + n + 1 p d v ( z ) ( 1 | z | 2 ) n + 1 C | z | < r 0 ( 1 | φ a ( z ) | 2 ) q + n + 1 p | ( f φ a ) ( z ) | p d v (z)

We have ( 1 | φ a ( z ) | 2 ) = ( 1 | z | 2 ) ( 1 | a | 2 ) | 1 z , a | 2 , when | z | r 0 , 1 r 0 2 ( 1 + r 0 ) 2 ( 1 | z | 2 ) | 1 z , a | 2 1 ( 1 r 0 ) 2 , and since | f ( z ) | p is subharmonic, that

B n | ˜ f ( z ) | p ( 1 | z | 2 ) q + n + 1 p K ( G ( z , a ) ) d λ ( z ) C ( 1 | a | 2 ) q + n + 1 p | z | < r 0 | ( f φ a ) ( z ) | p d v ( z ) = C ( 1 | a | 2 ) q + n + 1 p 0 r 0 r 2 n 1 d r S n | ( f φ a ) ( r ς ) | p d σ ( ς ) C ( 1 | a | 2 ) q + n + 1 p | ( f φ a ) ( z ) | p = C ( 1 | a | 2 ) q + n + 1 p | ˜ f ( a ) | p

Thus, we have lim | a | 1 ( 1 | a | 2 ) q + n + 1 p | ˜ f ( a ) | p = 0 when f Q K , 0 ( p , q ) , then f B 0 q + n + 1 p .

The following result is the further study on the equivalence between Q K , 0 ( p , q ) and B 0 q + n + 1 p .

Theorem 2. Let 0 < p < , p 2 n 1 < q < , Q K , 0 ( p , q ) = B 0 q + n + 1 p if and only if

0 1 ( 1 r 2 ) n 1 r 2 n 1 K ( g ( r ) ) d r < . (6)

Proof Sufficiency: By theorem 1, we only need to show that B 0 q + n + 1 p Q K , 0 ( p , q ) .

Since 0 1 ( 1 r 2 ) n 1 r 2 n 1 K ( g ( r ) ) d r < , for given ε > 0 , then there exists r 0 : 0 < r 0 < 1 , such that

r 0 1 ( 1 r 2 ) n 1 r 2 n 1 K ( g ( r ) ) d r < ε .

Let E ( a , r 0 ) = { z B n , | φ a ( z ) | < r 0 } , for any f B q + n + 1 p , z B n \ E ( a , r 0 ) , we have

B n \ E ( a , r 0 ) | ˜ f ( z ) | p ( 1 | z | 2 ) q + n + 1 p K ( G ( z , a ) ) d λ ( z ) f B q + n + 1 p p B n \ E ( a , r 0 ) K ( G ( z , a ) ) d λ ( z ) f B q + n + 1 p p r 0 < | z | < 1 ( 1 | z | 2 ) n 1 K ( g ( z ) ) d v ( z ) f B q + n + 1 p p r 0 1 ( 1 r 2 ) n 1 r 2 n 1 K ( g ( r ) ) d r S n d σ ( ς ) < ε f B q + n + 1 p p (7)

And when z E ( a , r 0 ) , we have

lim | a | 1 E ( a , r 0 ) | ˜ f ( z ) | p ( 1 | z | 2 ) q + n + 1 p K ( G ( z , a ) ) d λ ( z ) = lim | a | 1 | z | < r 0 | ˜ ( f φ a ) ( z ) | p ( 1 | φ a ( z ) | 2 ) q + n + 1 p K ( g ( z ) ) d λ ( z ) lim | a | 1 sup | z | < r 0 ( 1 | φ a ( z ) | 2 ) q + n + 1 p | ˜ ( f φ a ) ( z ) | p | z | < r 0 K ( g ( z ) ) ( 1 | z | 2 ) n 1 d V ( z ) = lim | a | 1 sup | z | < r 0 ( 1 | φ a ( z ) | 2 ) q + n + 1 p | ˜ ( f φ a ) ( z ) | p 2 n × 0 r 0 ( 1 r 2 ) n 1 r 2 n 1 K ( g ( r ) ) d r S n d σ ( ς ) C lim | a | 1 sup | z | < r 0 ( 1 | φ a ( z ) | 2 ) q + n + 1 p | ˜ ( f φ a ) ( z ) | p

( 1 | φ a ( z ) | 2 ) = ( 1 | z | 2 ) ( 1 | a | 2 ) | 1 z , a | 2 , and 1 r 0 2 ( 1 + r 0 ) 2 ( 1 | z | 2 ) | 1 z , a | 2 1 ( 1 r 0 ) 2 when | z | r 0 , so

lim | a | 1 ( 1 | φ a ( z ) | 2 ) q + n + 1 p p | ˜ ( f φ a ) ( z ) | = 0 ,

thus

lim | a | 1 E ( a , r 0 ) | ˜ f ( z ) | p ( 1 | z | 2 ) q + n + 1 p K ( G ( z , a ) ) d λ ( z ) = 0 ,

By formula(7), then we have lim | a | 1 B n | ˜ f ( z ) | p ( 1 | z | 2 ) q + n + 1 p K ( G ( z , a ) ) d λ ( z ) = 0 , i.e. f Q K , 0 ( p , q ) . It means B 0 q + n + 1 p Q K , 0 ( p , q ) .

Necessary: We only need to show that if 0 1 ( 1 r 2 ) n 1 r 2 n 1 K ( g ( r ) ) d r = , there exists a function f B 0 q + n + 1 p , but f Q K , 0 ( p , q ) .

Let α = ( α 1 , α 2 , , α n ) be an n-tuple of non-negative integers, and | α | = α 1 + α 2 + + α n satisfied | α | = 2 N where N is a integer. Let f = | α | q + n + 1 p p z α , it is easy to show that f B q + n + 1 p , and by the proof of theorem 3 in [7], we know that S n J ( r ς ) p 2 d σ ( ς ) C ( 1 r ) ( q + n + 1 ) + p 2 when r [ 3 4 , 1 ) , which

J ( r ς ) = r 2 | α | 2 | α | 2 ( q + n + 1 p ) p ( α 1 2 | ς 1 α 1 1 ς 2 α 2 ς n α n | 2 + + α n 2 | ς 1 α 1 ς 2 α 2 ς n α n 1 | 2 r 2 | α | 2 | ς α | 2 )

thus

B n | ˜ f ( z ) | p ( 1 | z | 2 ) q + n + 1 p K ( G ( z , a ) ) d λ ( z ) B n ( 1 | z | 2 ) p 2 ( J ( z ) ) p 2 ( 1 | z | 2 ) q + n + 1 p K ( g ( z ) ) d λ ( z ) = 2 n 0 1 ( 1 r 2 ) q p 2 r 2 n 1 K ( g ( r ) ) d r S n J ( r ς ) p 2 d σ ( ς ) C 3 4 1 ( 1 r 2 ) n 1 r 2 n 1 K ( g ( r ) ) d r

Since the conclusion of theorem 1 in [7], we have

0 3 4 ( 1 r 2 ) n 1 r 2 n 1 K ( g ( r ) ) d r C 0 1 ( 1 r 2 ) 2 n 1 r 2 n 1 K ( g ( r ) ) d r < ,

Then if 0 1 ( 1 r 2 ) n 1 r 2 n 1 K ( g ( r ) ) d r = , we can get

B n | ˜ f ( z ) | p ( 1 | z | 2 ) q + n + 1 p K ( G ( z , a ) ) d λ ( z ) = ,

which shows that f Q K , 0 ( p , q ) , the theorem is proved.

With the above conclusion, further study in this field of operator theory on Q K , 0 ( p , q ) can be conducted in the future.

Founding

Scientific Research Fund of Sichuan Provincial Education Department of China (18ZA0416).

Conflicts of Interest

The author declares no conflicts of interest regarding the publication of this paper.

References

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