sup> 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.

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