# Weighted Composition Operator from Mixed Norm Space to Bloch-Type Space on the Unit Ball

Authors: Yu-Xia Liang; Ren-Yu Chen
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

We discuss the boundedness and compactness of the weighted composition operator from mixed norm space to Bloch-type space on the unit ball ofC
n.


---

## Body

## 1. Introduction

LetH
(
B
n
) be the class of all holomorphic functions on B
n and S
(
B
n
) the collection of all the holomorphic self-mappings of B
n, where B
n is the unit ball in the n-dimensional complex space C
n. Let d
v denote the Lebesegue measure on B
n normalized so that v
(
B
n
)
=
1 and d
σ the normalized rotation invariant measure on the boundary S
=
∂
B
n of B
n. For f
∈
H
(
B
n
), let
(1)
R
f
(
z
)
=
∑
j
=
1
n
‍
z
j
∂
f
∂
z
j
(
z
)
be the radial derivative of f.A positive continuous functionμ on [
0,1
) is called normal (see, e.g., [1]) if there exist three constants 0
≤
δ
<
1, and 0
<
a
<
b
<
∞, such that for r
∈
[
δ
,
1
)
(2)
μ
(
r
)
(
1
-
r
)
a
↓
0
,
μ
(
r
)
(
1
-
r
)
b
↑
∞
,
r
⟶
1
.
In the rest of this paper we always assume that μ is normal on [
0,1
), and from now on if we say that a function μ
:
B
n
→
[
0
,
∞
) is normal we will also suppose that it is radial on B
n, that is, μ
(
z
)
=
μ
(
|
z
|
) for z
∈
B
n.Let0
<
p
≤
∞, 0
<
q
≤
∞, and μ be normal on [
0,1
). f is said to belong to the mixed norm space L
(
p
,
q
,
μ
) if f is a measurable function on B
n and ∥
f
∥
p
,
q
,
μ
<
∞, where
(3)
∥
f
∥
p
,
q
,
μ
=
{
∫
0
1
‍
r
2
n
-
1
(
1
-
r
)
-
1
μ
p
(
r
)
M
q
p
(
r
,
f
)
d
r
}
1
/
p
(
0
<
p
<
∞
,
0
<
q
≤
∞
)
,
∥
f
∥
∞
,
q
,
μ
=
sup
⁡
0
≤
r
<
1
μ
(
r
)
M
q
(
r
,
f
)
,
M
∞
(
r
,
f
)
=
sup
⁡
ζ
∈
S
|
f
(
r
ζ
)
|
,
M
q
(
r
,
f
)
=
{
∫
S
‍
|
f
(
r
ζ
)
|
q
d
σ
(
ζ
)
}
1
/
q
,
(
0
<
q
<
∞
)
.
If 0
<
p
=
q
<
∞, then L
(
p
,
q
,
μ
) is just the space L
p
(
μ
)
=
{
f is measurable function on B
n
:
∫
B
n
‍
|
f
(
z
)
|
p
(
μ
p
(
z
)
/
(
1
-
|
z
|
)
)
d
ν
(
z
)
<
∞
}.LetH
(
p
,
q
,
μ
)
=
L
(
p
,
q
,
μ
)
∩
H
(
B
n
). If 0
<
p
=
q
<
∞, then H
(
p
,
q
,
μ
) is just the weighted Bergman space L
a
p
(
μ
). In particular, H
(
p
,
q
,
μ
) is Bergman space L
a
p
(
μ
) if 0
<
p
=
q
<
∞ and μ
(
r
)
=
(
1
-
r
)
1
/
p. Otherwise, if p
=
q
=
2 and μ
(
r
)
=
(
1
-
r
)
β
/
2  (
β
<
0
), then H
(
p
,
q
,
μ
(
r
)
) is the Dirichlet-type space.For0
<
p
,
q
<
∞, -
1
<
γ
<
1, let μ
(
r
)
=
r
-
(
2
n
-
1
)
/
p
(
1
-
r
)
(
γ
+
1
)
/
p; it is easy to see that the mixed norm space H
(
p
,
q
,
μ
), written by H
p
,
q
,
γ, consists of all f
∈
H
(
B
n
) such that
(4)
∥
f
∥
H
p
,
q
,
γ
=
{
∫
0
1
‍
M
q
p
(
f
,
r
)
(
1
-
r
)
γ
d
r
}
1
/
p
<
∞
.Nowf
∈
H
(
B
n
) is said to belong to Bloch-type space B
μ if
(5)
∥
f
∥
μ
,
1
=
sup
⁡
z
∈
B
n
μ
(
z
)
|
∇
f
(
z
)
|
<
∞
,
where ∇
f
(
z
)
=
(
∂
f
(
z
)
/
∂
z
1
,
…
,
∂
f
(
z
)
/
∂
z
n
) is the complex gradient of f.It is clear thatB
μ is a Banach space with norm ∥
f
∥
B
μ
=
|
f
(
0
)
|
+
∥
f
∥
μ
,
1. For f
∈
H
(
B
n
), we denote
(6)
∥
f
∥
μ
,
2
=
sup
⁡
z
∈
B
n
μ
(
z
)
|
R
f
(
z
)
|
,
∥
f
∥
μ
,
3
=
sup
⁡
z
∈
B
n
Q
f
μ
(
z
)
,
where
(7)
Q
f
μ
(
z
)
=
sup
⁡
u
∈
C
n
∖
{
0
}
|
〈
∇
f
(
z
)
,
u
¯
〉
|
G
z
μ
(
u
,
u
)
,
G
z
μ
(
u
,
u
)
=
1
μ
2
(
z
)
{
μ
2
(
z
)
σ
μ
2
(
|
z
|
)
|
u
|
2
+
(
1
-
μ
2
(
z
)
σ
μ
2
(
|
z
|
)
)
|
〈
z
,
u
〉
|
2
|
z
|
2
}
(
z
≠
0
)
,
G
0
μ
(
u
,
u
)
=
|
u
|
2
μ
2
(
0
)
,
1
σ
μ
(
t
)
=
1
μ
(
0
)
+
∫
0
t
‍
d
τ
(
1
-
τ
)
1
/
2
μ
(
τ
)
(
0
≤
t
<
1
)
.It was proved that∥
f
∥
μ
,
1, ∥
f
∥
μ
,
2, and ∥
f
∥
μ
,
3 are equivalent for f
∈
B
μ
(
B
n
) in [2, 3].Letφ
∈
S
(
B
n
), ψ
∈
H
(
B
n
); the composition operator C
φ induced by φ is defined by
(8)
(
C
φ
f
)
(
z
)
=
f
(
φ
(
z
)
)
,
f
∈
H
(
B
n
)
,
z
∈
B
n
,
and the weighted composition operator T
ψ
,
φ is defined by
(9)
T
ψ
,
φ
(
f
)
=
ψ
f
∘
φ
for f
∈
H
(
B
n
). We can regard this operator as a generalization of a multiplication operator M
ψ and a composition operator C
φ. That is, when φ
(
z
)
≡
z, we obtain T
ψ
,
φ
f
(
z
)
=
M
ψ
f
(
z
)
=
ψ
(
z
)
f
(
z
) and when ψ
(
z
)
≡
1 we obtain T
ψ
,
φ
f
(
z
)
=
C
φ
f
(
z
)
=
f
(
φ
(
z
)
).It is interesting to provide a function theoretic characterization whenψ and φ induce a bounded or compact weighted composition operator between some spaces of holomorphic functions on B
n. Recently, this operator is well studied by many papers; see, for example, [3–17] and their references therein. In particular, Stević [18] gave some conditions of weighted composition operators between mixed-norm spaces and H
α
∞ spaces on the unit ball. Zhou and Chen [19] discussed weighted composition operators from F
(
p
,
q
,
s
) to Bloch-type spaces on the unit ball. More recently, the weighted composition operator from Bers-type space to Bloch-type space on the unit ball was studied in [6]. Now in this paper, we will continue this line of research and characterize the boundedness and compactness of the weighted composition operator T
ψ
,
φ acting from mixed-norm spaces H
p
,
q
,
γ to Bloch-type space B
μ on the unit ball of C
n. The paper is organized as follows. In Section 2, we give some lemmas. The main results are given in Section 3.Throughout the remainder of this paper,C will denote a positive constant; the exact value of which will vary from one appearance to the next. The notation A
≍
B means that there is a positive constant C such that B
/
C
≤
A
≤
C
B.


## 2. Some Lemmas

Lemma 1.
Assume that0
<
p
,
q
<
∞, -
1
<
γ
<
∞, and f
∈
H
p
,
q
,
γ. Then there is a positive constant C which is independent of f such that
(10)
|
f
(
z
)
|
≤
C
∥
f
∥
H
p
,
q
,
γ
(
1
-
|
z
|
2
)
n
/
q
+
(
γ
+
1
)
/
p
,
(11)
|
R
f
(
z
)
|
≤
C
∥
f
∥
H
p
,
q
,
γ
(
1
-
|
z
|
2
)
n
/
q
+
1
+
(
γ
+
1
)
/
p
.Proof.
We first prove (10). By the monotonicity of the integral means and [20, Theorem 1.12] we have that
(12)
∥
f
∥
H
p
,
q
,
γ
p
≥
∫
(
1
+
|
z
|
)
/
2
(
3
+
|
z
|
)
/
4
‍
M
q
p
(
f
,
r
)
(
1
-
r
)
γ
d
r
≥
C
M
q
p
(
f
,
1
+
|
z
|
2
)
∫
(
1
+
|
z
|
)
/
2
(
3
+
|
z
|
)
/
4
‍
(
1
-
r
)
γ
d
r
≥
C
M
q
p
(
f
,
1
+
|
z
|
2
)
(
1
-
|
z
|
2
)
γ
+
1
≥
C
(
1
-
|
z
|
2
)
γ
+
1
+
(
p
n
)
/
q
|
f
(
z
)
|
p
,
from which the desired result (10) follows.
Next we prove (11). By the monotonicity of the integral means, using the well-known asymptotic formula (e.g., [21, Theorem 2]), we obtain that
(13)
∫
0
1
‍
M
q
p
(
f
,
r
)
(
1
-
r
)
γ
d
r
≍
|
f
(
0
)
|
p
+
∫
0
1
‍
M
q
p
(
R
f
,
r
)
(
1
-
r
)
γ
+
p
d
r
.
By [20, Theorem 1.12], it follows that
(14)
∥
f
∥
H
p
,
q
,
γ
p
≥
∫
(
1
+
|
z
|
)
/
2
1
‍
M
q
p
(
f
,
r
)
(
1
-
r
)
γ
d
r
≥
C
∫
(
1
+
|
z
|
)
/
2
1
‍
M
q
p
(
R
f
,
r
)
(
1
-
r
)
γ
+
p
d
r
≥
C
M
q
p
(
R
f
,
1
+
|
z
|
2
)
∫
(
1
+
|
z
|
)
/
2
1
‍
(
1
-
r
)
γ
+
p
d
r
≥
C
M
q
p
(
R
f
,
1
+
|
z
|
2
)
(
1
-
|
z
|
2
)
γ
+
1
+
p
≥
C
(
1
-
|
z
|
2
)
γ
+
1
+
p
+
(
p
n
)
/
q
|
R
f
(
z
)
|
p
.
Then the desired result (11) follows. This completes the proof.From the above lemma, whenf
∈
H
p
,
q
,
γ, then
(15)
f
∈
B
n
/
q
+
1
+
(
γ
+
1
)
/
p
,
∥
f
∥
B
n
/
q
+
(
γ
+
1
)
/
p
+
1
≤
C
∥
f
∥
H
p
,
q
,
γ
.
For z
∈
B
n,  u
∈
C
n, denote the Bergman metric of B
n by
(16)
H
z
(
u
,
u
)
=
(
1
-
|
z
|
2
)
|
u
|
2
+
|
〈
z
,
u
〉
|
2
(
1
-
|
z
|
2
)
2
.Lemma 2.
Letv
(
r
)
=
(
1
-
r
2
)
n
/
q
+
(
γ
+
1
)
/
p
+
1 and φ
∈
S
(
B
n
). Then
(17)
G
φ
(
z
)
v
(
J
φ
(
z
)
z
,
J
φ
(
z
)
z
)
≤
C
H
φ
(
z
)
(
J
φ
(
z
)
z
,
J
φ
(
z
)
z
)
(
1
-
|
φ
(
z
)
|
2
)
2
(
n
/
q
+
(
γ
+
1
)
/
p
)
for all z
∈
B
n, where J
φ
(
z
) denotes the Jacobian matrix of φ
(
z
) and
(18)
J
φ
(
z
)
z
=
(
∑
k
=
1
n
‍
∂
φ
1
∂
z
k
z
k
,
…
,
∑
k
=
1
n
‍
∂
φ
n
∂
z
k
z
k
)
T
.Proof.
Letα
=
n
/
q
+
(
γ
+
1
)
/
p. If φ
(
z
)
=
0, the desired result is obvious. If φ
(
z
)
≠
0, from the definition of σ
v,
(19)
1
σ
v
(
r
)
=
1
+
∫
0
r
‍
d
t
(
1
-
t
)
1
/
2
(
1
-
t
2
)
α
+
1
≍
(
1
-
r
2
)
1
/
2
v
(
r
)
,
0
≤
r
<
1
.
Thus
(20)
G
φ
(
z
)
v
(
J
φ
(
z
)
,
J
φ
(
z
)
z
)
=
1
v
2
(
|
φ
(
z
)
|
)
×
[
(
1
-
v
2
(
|
φ
(
z
)
|
)
σ
v
2
(
|
φ
(
z
)
|
)
)
|
〈
φ
(
z
)
,
J
φ
(
z
)
z
〉
|
2
|
φ
(
z
)
|
2
v
2
(
|
φ
(
z
)
|
)
σ
v
2
(
|
φ
(
z
)
|
)
|
J
φ
(
z
)
z
|
2
+
(
1
-
v
2
(
|
φ
(
z
)
|
)
σ
v
2
(
|
φ
(
z
)
|
)
)
|
〈
φ
(
z
)
,
J
φ
(
z
)
z
〉
|
2
|
φ
(
z
)
|
2
]
=
1
v
2
(
|
φ
(
z
)
|
)
×
[
v
2
(
|
φ
(
z
)
|
)
σ
v
2
(
|
φ
(
z
)
|
)
(
|
J
φ
(
z
)
z
|
2
-
|
〈
φ
(
z
)
,
J
φ
(
z
)
z
〉
|
2
|
φ
(
z
)
|
2
)
+
|
〈
φ
(
z
)
,
J
φ
(
z
)
z
〉
|
2
|
φ
(
z
)
|
2
(
|
J
φ
(
z
)
z
|
2
-
|
〈
φ
(
z
)
,
J
φ
(
z
)
z
〉
|
2
|
φ
(
z
)
|
2
)
]
≤
C
v
2
(
|
φ
(
z
)
|
)
×
[
(
1
-
|
φ
(
z
)
|
2
)
(
|
J
φ
(
z
)
z
|
2
-
|
〈
φ
(
z
)
,
J
φ
(
z
)
z
〉
|
2
|
φ
(
z
)
|
2
)
+
|
〈
φ
(
z
)
,
J
φ
(
z
)
z
〉
|
2
|
φ
(
z
)
|
2
(
|
J
φ
(
z
)
z
|
2
-
|
〈
φ
(
z
)
,
J
φ
(
z
)
z
〉
|
2
|
φ
(
z
)
|
2
)
]
=
C
v
2
(
|
φ
(
z
)
|
)
×
[
(
1
-
|
φ
(
z
)
|
2
)
(
|
J
φ
(
z
)
z
|
2
+
|
〈
φ
(
z
)
,
J
φ
(
z
)
z
〉
|
2
)
]
=
C
(
1
-
|
φ
(
z
)
|
2
)
2
v
2
(
|
φ
(
z
)
|
)
H
φ
(
z
)
(
J
φ
(
z
)
z
,
J
φ
(
z
)
z
)
=
C
H
φ
(
z
)
(
J
φ
(
z
)
z
,
J
φ
(
z
)
z
)
(
1
-
|
φ
(
z
)
|
2
)
2
(
n
/
q
+
(
γ
+
1
)
/
p
)
.
The desired result follows from (20). The proof is completed.The proof of the next lemma is standard; see, for example, [4, Proposition 3.11]. Hence, it is omitted.Lemma 3.
Assume that0
<
p
,
q
<
∞, -
1
<
γ
<
∞, μ is a normal function, and φ
∈
S
(
B
n
), ψ
∈
H
(
B
n
). Then T
ψ
,
φ
:
H
p
,
q
,
γ
→
B
μ is compact if and only if for any bounded sequence {
f
k
}
k
∈
N in H
p
,
q
,
γ which converges to zero uniformly on compact subsets of B
n as k
→
∞; then ∥
T
ψ
,
φ
f
k
∥
B
μ
→
0, as k
→
∞.Lemma 4.
Forβ
>
-
1 and m
>
1
+
β, one has
(21)
∫
0
1
‍
(
1
-
r
)
β
(
1
-
ρ
r
)
m
d
r
≤
C
(
1
-
ρ
)
1
+
β
-
m
,
0
<
ρ
<
1
.Proof.
(22)
∫
0
1
‍
(
1
-
r
)
β
(
1
-
ρ
r
)
m
d
r
=
∫
0
1
‍
(
1
-
r
)
β
(
1
-
ρ
r
)
m
-
β
(
1
-
ρ
r
)
β
d
r
≤
∫
0
1
‍
(
1
-
r
)
β
(
1
-
ρ
r
)
m
-
β
(
1
-
r
)
β
d
r
=
∫
0
1
‍
1
(
1
-
ρ
r
)
m
-
β
d
r
=
1
p
(
1
+
β
-
m
)
(
1
-
ρ
)
1
+
β
-
m
=
C
(
1
-
ρ
)
1
+
β
-
m
.
This completes the proof.


## 3. The Boundedness and Compactness ofT
ψ
,
φ
:
H
p
,
q
,
γ
→
B
μ

Theorem 5.
Assume that0
<
p
,
q
<
∞, -
1
<
γ
<
∞, μ is a normal function, and φ
∈
S
(
B
n
), ψ
∈
H
(
B
n
). Then T
ψ
,
φ
:
H
p
,
q
,
γ
→
B
μ is bounded if and only if
(23)
M
1
∶
=
sup
⁡
z
∈
B
n
μ
(
z
)
|
R
ψ
(
z
)
|
(
1
-
|
φ
(
z
)
|
2
)
n
/
q
+
(
γ
+
1
)
/
p
<
∞
,
(24)
M
2
≔
sup
⁡
z
∈
B
n
μ
(
z
)
|
ψ
(
z
)
|
(
1
-
|
φ
(
z
)
|
2
)
n
/
q
+
(
γ
+
1
)
/
p
×
{
H
φ
(
z
)
(
J
φ
(
z
)
z
,
J
φ
(
z
)
z
)
}
1
/
2
<
∞
.Proof
Sufficiency. Assume that (23) and (24) hold. Then for any f
∈
H
p
,
q
,
γ, if J
φ
(
z
)
z
≠
0 for z
∈
B
n, by Lemma 1 and Lemma 2, it follows that
(25)
∥
T
ψ
,
φ
f
(
z
)
∥
B
μ
=
sup
⁡
z
∈
B
n
μ
(
z
)
|
R
(
T
ψ
,
φ
f
)
(
z
)
|
≤
sup
⁡
z
∈
B
n
μ
(
z
)
|
R
ψ
(
z
)
|
|
f
(
φ
(
z
)
)
|
+
sup
⁡
z
∈
B
n
μ
(
z
)
|
ψ
(
z
)
|
|
R
(
f
∘
φ
)
(
z
)
|
≤
sup
⁡
z
∈
B
n
μ
(
z
)
|
R
ψ
(
z
)
|
∥
f
∥
H
p
,
q
,
γ
(
1
-
|
φ
(
z
)
|
2
)
n
/
q
+
(
γ
+
1
)
/
p
+
sup
⁡
z
∈
B
n
μ
(
z
)
|
ψ
(
z
)
|
|
〈
∇
f
(
φ
(
z
)
)
,
J
φ
(
z
)
z
¯
〉
|
≤
M
1
∥
f
∥
H
p
,
q
,
γ
+
sup
⁡
z
∈
B
n
(
G
φ
(
z
)
v
(
J
φ
(
z
)
z
,
J
φ
(
z
)
z
)
)
-
1
(
C
μ
(
z
)
|
ψ
(
z
)
|
{
H
φ
(
z
)
(
J
φ
(
z
)
z
,
J
φ
(
z
)
z
)
}
1
/
2
×
|
〈
∇
f
(
φ
(
z
)
)
,
J
φ
(
z
)
z
¯
〉
|
C
μ
(
z
)
|
ψ
(
z
)
|
{
H
φ
(
z
)
(
J
φ
(
z
)
z
,
J
φ
(
z
)
z
)
}
1
/
2
)
×
(
G
φ
(
z
)
v
(
J
φ
(
z
)
z
,
J
φ
(
z
)
z
)
(
1
-
|
φ
(
z
)
|
2
)
q
/
n
+
(
γ
+
1
)
/
p
×
G
φ
(
z
)
v
(
J
φ
(
z
)
z
,
J
φ
(
z
)
z
)
G
φ
(
z
)
v
(
J
φ
(
z
)
z
,
J
φ
(
z
)
z
)
.
)
-
1
)
≤
M
1
∥
f
∥
H
p
,
q
,
γ
+
C
M
2
∥
f
∥
B
(
1
-
r
2
)
q
/
n
+
(
γ
+
1
)
/
p
+
1
≤
C
∥
f
∥
H
p
,
q
,
γ
.
When J
φ
(
z
)
z
=
0 for z
∈
B
n. From (23) we can easily obtain
(26)
μ
(
z
)
|
R
(
T
ψ
,
φ
(
f
)
)
(
z
)
|
≤
M
1
∥
f
∥
H
p
,
q
,
γ
.
Combining (25) and (26), the boundedness of T
ψ
,
φ
:
H
p
,
q
,
γ
→
B
μ follows.
Necessity. Suppose that T
ψ
,
φ
:
H
p
,
q
,
γ
→
B
μ is bounded. Firstly, we assume that w
∈
B
n and φ
(
w
)
=
r
w
e
1, where r
w
=
|
φ
(
w
)
| and e
1
=
(
1,0
,
0
,
…
,
0
).
If(
1
-
r
w
2
)
(
|
η
2
|
2
+
⋯
+
|
η
n
|
2
)
≤
|
η
1
|, where J
φ
(
w
)
w
=
(
η
1
,
…
,
η
n
)
T, choose the function
(27)
f
w
(
z
)
=
z
1
-
r
w
1
-
r
w
z
1
(
1
-
r
w
2
(
1
-
r
w
z
1
)
2
)
n
/
q
+
(
γ
+
1
)
/
p
.
By [20, Theorem 1.12] and Lemma 4 we have that
(28)
M
q
(
f
w
,
r
)
=
(
∫
S
‍
|
f
w
(
r
ζ
)
|
q
d
σ
(
ζ
)
)
1
/
q
≤
(
∫
S
‍
(
1
-
r
w
2
(
1
-
r
w
r
ζ
1
)
2
)
n
+
q
(
γ
+
1
)
/
p
d
σ
(
ζ
)
)
1
/
q
≤
C
(
1
-
r
w
2
)
n
/
q
+
(
γ
+
1
)
/
p
(
1
-
r
r
w
2
)
n
/
q
+
2
(
γ
+
1
)
/
p
,
∥
f
w
∥
H
p
,
q
,
γ
p
=
∫
0
1
‍
M
q
p
(
f
w
,
r
)
(
1
-
r
)
γ
d
r
≤
C
(
1
-
r
w
2
)
p
n
/
q
+
γ
+
1
∫
0
1
‍
(
1
-
r
)
γ
(
1
-
r
r
w
2
)
p
n
/
q
+
2
(
γ
+
1
)
d
r
≤
C
(
1
-
r
w
2
)
p
n
/
q
+
γ
+
1
1
(
1
-
r
w
2
)
p
n
/
q
+
γ
+
1
≤
C
.
Then f
w
∈
H
p
,
q
,
γ and ∥
f
w
∥
H
p
,
q
,
γ
≤
C. Moreover, f
w
(
φ
(
w
)
)
=
0 and
(29)
∇
f
w
(
φ
(
w
)
)
=
(
1
(
1
-
r
w
2
)
n
/
q
+
(
γ
+
1
)
/
p
+
1
,
0
,
…
,
0
)
.
Thus
(30)
∥
T
ψ
,
φ
f
w
∥
B
μ
≥
μ
(
w
)
|
R
(
ψ
f
∘
φ
)
(
w
)
|
≥
μ
(
w
)
|
ψ
(
w
)
|
|
R
(
f
∘
φ
)
(
w
)
|
-
μ
(
w
)
|
R
ψ
(
w
)
|
|
f
w
(
φ
(
w
)
)
|
=
μ
(
w
)
|
ψ
(
w
)
|
|
〈
∇
f
w
(
φ
(
w
)
)
,
J
φ
(
w
)
w
¯
〉
|
=
μ
(
w
)
|
ψ
(
w
)
|
|
η
1
|
(
1
-
r
w
2
)
n
/
q
+
(
γ
+
1
)
/
p
+
1
.
By the definition of H
φ
(
w
)
(
J
φ
(
w
)
w
,
J
φ
(
w
)
w
) and (30) it follows that
(31)
μ
(
w
)
|
ψ
(
w
)
|
{
H
φ
(
w
)
(
J
φ
(
w
)
w
,
J
φ
(
w
)
w
)
}
1
/
2
(
1
-
|
φ
(
w
)
|
2
)
n
/
q
+
(
γ
+
1
)
/
p
=
(
{
(
1
-
|
φ
(
w
)
|
2
)
|
J
φ
(
w
)
w
|
2
+
|
〈
φ
(
w
)
,
J
φ
(
w
)
w
〉
|
2
}
1
/
2
μ
(
w
)
|
ψ
(
w
)
|
×
{
(
1
-
|
φ
(
w
)
|
2
)
|
J
φ
(
w
)
w
|
2
+
|
〈
φ
(
w
)
,
J
φ
(
w
)
w
〉
|
2
}
1
/
2
)
×
(
(
1
-
|
φ
(
w
)
|
2
)
n
/
q
+
(
γ
+
1
)
/
p
+
1
)
-
1
=
μ
(
w
)
|
ψ
(
w
)
|
{
(
1
-
r
w
2
)
(
|
η
2
|
2
+
⋯
+
|
η
n
|
2
)
+
|
η
1
|
2
}
1
/
2
(
1
-
|
φ
(
w
)
|
2
)
n
/
q
+
(
γ
+
1
)
/
p
+
1
≤
2
μ
(
w
)
|
ψ
(
w
)
|
|
η
1
|
(
1
-
r
w
2
)
n
/
q
+
(
γ
+
1
)
/
p
+
1
≤
C
∥
T
ψ
,
φ
f
w
∥
B
μ
≤
C
.
This shows that when (
1
-
r
w
2
)
(
|
η
2
|
2
+
⋯
+
|
η
n
|
2
)
≤
|
η
1
|, (24) follows.
On the other hand, if(
1
-
r
w
2
)
(
|
η
2
|
2
+
⋯
+
|
η
n
|
2
)
>
|
η
1
|. For j
=
2
,
…
,
n, let θ
j
=
arg
η
j and a
j
=
e
-
i
θ
j, when η
j
≠
0; otherwise a
j
=
0 when η
j
=
0. Take
(32)
f
w
(
z
)
=
a
2
z
2
+
⋯
+
a
n
z
n
(
1
-
r
w
z
1
)
n
/
q
+
(
γ
+
1
)
/
p
+
1
.
By [20, Theorem 1.12] and Lemma 4 we obtain that
(33)
M
q
(
f
w
,
r
)
≤
{
∫
S
‍
(
|
ζ
2
|
+
⋯
+
|
ζ
n
|
)
q
|
1
-
r
w
r
ζ
1
|
n
+
q
(
γ
+
1
)
/
p
+
q
d
σ
(
ζ
)
}
1
/
q
≤
{
∫
S
‍
C
(
|
ζ
2
|
2
+
⋯
+
|
ζ
n
|
2
)
q
/
2
|
1
-
r
w
r
ζ
1
|
n
+
q
(
γ
+
1
)
/
p
+
q
d
σ
(
ζ
)
}
1
/
q
=
{
∫
S
‍
C
(
1
-
|
ζ
1
|
2
)
q
/
2
|
1
-
r
w
r
ζ
1
|
n
+
q
(
γ
+
1
)
/
p
+
q
d
σ
(
ζ
)
}
1
/
q
≤
C
{
∫
S
‍
1
|
1
-
r
w
r
ζ
1
|
n
+
q
(
γ
+
1
)
/
p
+
q
/
2
d
σ
(
ζ
)
}
1
/
q
≤
C
(
1
-
r
r
w
2
)
(
γ
+
1
)
/
p
+
1
/
2
.
∥
f
w
∥
H
p
,
q
,
γ
p
=
∫
0
1
‍
M
q
p
(
f
w
,
r
)
(
1
-
r
)
γ
d
r
≤
C
∫
0
1
‍
(
1
-
r
)
γ
(
1
-
r
r
w
2
)
γ
+
1
+
p
/
2
d
r
≤
C
(
1
-
r
w
2
)
p
/
2
≤
C
.
Hence f
w
∈
H
p
,
q
,
γ and ∥
f
w
∥
H
p
,
q
,
γ
≤
C. Moreover f
w
(
φ
(
w
)
)
=
0 and
(34)
∇
f
w
(
φ
(
w
)
)
=
(
0
,
a
2
(
1
-
r
w
2
)
n
/
q
+
(
γ
+
1
)
/
p
+
1
,
…
,
a
n
(
1
-
r
w
2
)
n
/
q
+
(
γ
+
1
)
/
p
+
1
)
.
Similar to the proof of (30), we obtain that
(35)
μ
(
w
)
|
ψ
(
w
)
|
(
|
η
2
|
+
⋯
+
|
η
n
|
)
(
1
-
r
w
2
)
n
/
q
+
(
γ
+
1
)
/
p
+
1
≤
C
∥
T
ψ
,
φ
f
w
∥
B
μ
.
It follows from (35) that
(36)
μ
(
w
)
|
ψ
(
w
)
|
(
1
-
|
φ
(
w
)
|
2
)
n
/
q
+
(
γ
+
1
)
/
p
{
H
φ
(
w
)
(
J
φ
(
w
)
w
,
J
φ
(
w
)
w
)
}
1
/
2
=
(
{
(
1
-
|
φ
(
w
)
|
2
)
|
J
φ
(
w
)
w
|
2
+
|
〈
φ
(
w
)
,
J
φ
(
w
)
w
〉
|
2
}
1
/
2
μ
(
w
)
|
ψ
(
w
)
|
×
{
(
1
-
|
φ
(
w
)
|
2
)
|
J
φ
(
w
)
w
|
2

+
|
〈
φ
(
w
)
,
J
φ
(
w
)
w
〉
|
2
}
1
/
2

)
×
(
(
1
-
|
φ
(
w
)
|
2
)
n
/
q
+
(
γ
+
1
)
/
p
+
1
)
-
1
=
μ
(
w
)
|
ψ
(
w
)
|
{
(
1
-
r
w
2
)
(
|
η
2
|
2
+
⋯
+
|
η
n
|
2
)
+
|
η
1
|
2
}
1
/
2
(
1
-
|
φ
(
w
)
|
2
)
n
/
q
+
(
γ
+
1
)
/
p
+
1
≤
μ
(
w
)
|
ψ
(
w
)
|
{
2
(
1
-
r
w
2
)
(
|
η
2
|
2
+
⋯
+
|
η
n
|
2
)
}
1
/
2
(
1
-
|
φ
(
w
)
|
2
)
n
/
q
+
(
γ
+
1
)
/
p
+
1
≤
C
μ
(
w
)
|
ψ
(
w
)
|
2
(
1
-
r
w
2
)
(
|
η
2
|
+
⋯
+
|
η
n
|
)
(
1
-
r
w
2
)
n
/
q
+
(
γ
+
1
)
/
p
+
1
≤
∥
T
ψ
,
φ
f
w
∥
B
μ
.
That is, when (
1
-
r
w
2
)
(
|
η
2
|
2
+
⋯
+
|
η
n
|
2
)
>
|
η
1
|, (24) follows. Combining the above two cases, the desired result (24) holds.
For the general situation, we can use some unitary transformU
w to make φ
(
w
)
=
r
w
e
1
U
w and we can prove (11) by taking the function g
w
=
f
w
∘
U
w
-
1. By the linearity of the unitary transform U
w, |
ζ
|
=
|
U
w
-
1
ζ
|, and d
σ the normalized rotation invariant measure on the boundary S, we get that
(37)
∥
g
w
∥
H
p
,
q
,
γ
p
=
∫
0
1
‍
(
∫
S
‍
|
g
w
(
r
ζ
)
|
q
d
σ
(
ζ
)
)
p
/
q
(
1
-
r
)
γ
d
r
=
∫
0
1
‍
(
∫
S
‍
|
f
w
(
U
w
-
1
(
r
ζ
)
)
|
q
d
σ
(
ζ
)
)
p
/
q
(
1
-
r
)
γ
d
r
=
∫
0
1
‍
(
∫
S
‍
|
f
w
(
r
U
w
-
1
(
ζ
)
)
|
q
d
σ
(
ζ
)
)
p
/
q
(
1
-
r
)
γ
d
r
=
∫
0
1
‍
(
∫
S
‍
|
f
w
(
r
η
)
|
q
d
σ
(
η
)
)
p
/
q
(
1
-
r
)
γ
d
r
=
∥
f
w
∥
H
p
,
q
,
γ
p
.
Next we prove (23). Set the function
(38)
h
w
(
z
)
=
(
1
-
|
w
|
2
)
b
-
(
γ
+
1
)
/
p
(
1
-
〈
z
,
w
〉
)
n
/
q
+
b
for fixed w
∈
B
n and b
>
(
γ
+
1
)
/
p. Then,
(39)
M
q
(
h
w
(
z
)
,
r
)
=
(
∫
∂
B
n
‍
|
h
w
(
r
ζ
)
|
q
d
σ
(
ζ
)
)
1
/
q
=
(
∫
∂
B
n
‍
(
1
-
|
w
|
2
)
(
b
-
(
γ
+
1
)
/
p
)
q
|
1
-
〈
r
ζ
,
w
〉
|
(
n
/
q
+
b
)
q
d
σ
(
ζ
)
)
1
/
q
.
By [20, Theorem 1.12], it follows that
(40)
M
p
(
h
w
(
z
)
,
r
)
≤
(
1
-
|
w
|
2
)
b
-
(
γ
+
1
)
/
p
(
1
-
r
|
w
|
2
)
b
.
Applying Lemma 4 we have that
(41)
∥
h
w
∥
H
p
,
q
,
γ
q
=
∫
0
1
‍
M
q
p
(
h
w
,
r
)
(
1
-
r
)
γ
d
r
≤
C
∫
0
1
‍
(
1
-
|
w
|
2
)
p
b
-
(
γ
+
1
)
(
1
-
r
|
w
|
2
)
p
b
(
1
-
r
)
γ
d
r
=
C
(
1
-
|
w
|
2
)
p
b
-
(
γ
+
1
)
∫
0
1
‍
(
1
-
r
)
γ
(
1
-
r
|
w
|
2
)
p
b
d
r
≤
C
(
1
-
|
w
|
2
)
p
b
-
(
γ
+
1
)
(
1
-
|
w
|
2
)
(
γ
+
1
)
-
p
b
=
C
.
Therefore h
w
∈
H
p
,
q
,
γ, and sup
⁡
w
∈
B
n
∥
h
w
∥
H
p
,
q
,
γ
≤
C. Besides,
(42)
h
φ
(
w
)
(
φ
(
w
)
)
=
(
1
1
-
|
φ
(
w
)
|
2
)
n
/
q
+
(
γ
+
1
)
/
p
,
(43)
∇
h
φ
(
w
)
(
φ
(
w
)
)
=
(
n
q
+
b
)
(
φ
1
(
w
)
¯
(
1
-
|
φ
(
w
)
|
2
)
n
/
q
+
(
γ
+
1
)
/
p
+
1
,
…
,
φ
n
(
w
)
¯
(
1
-
|
φ
(
w
)
|
2
)
n
/
q
+
(
γ
+
1
)
/
p
+
1
)
.
Therefore,
(44)
∞
>
∥
T
ψ
,
φ
(
h
φ
(
w
)
)
∥
B
μ
≥
μ
(
w
)
|
R
(
ψ
h
φ
(
w
)
∘
φ
)
(
w
)
|
=
μ
(
w
)
|
R
ψ
(
w
)
h
φ
(
w
)
(
φ
(
w
)
)
+
ψ
(
w
)
R
(
h
φ
(
w
)
∘
φ
)
(
w
)
|
≥
μ
(
w
)
|
R
ψ
(
w
)
|
(
1
-
|
φ
(
w
)
|
2
)
n
/
q
+
(
γ
+
1
)
/
p
-
μ
(
w
)
|
ψ
(
w
)
|
|
R
(
h
φ
(
w
)
∘
φ
)
(
w
)
|
.
It follows from (43) and (24) that
(45)
μ
(
w
)
|
ψ
(
w
)
|
|
R
(
h
φ
(
w
)
∘
φ
)
(
w
)
|
=
μ
(
w
)
|
ψ
(
w
)
|
|
〈
∇
h
φ
(
w
)
(
φ
(
w
)
)
,
J
φ
(
w
)
w
¯
〉
|
=
(
n
q
+
b
)
μ
(
w
)
|
ψ
(
w
)
|
|
〈
φ
(
w
)
,
J
φ
(
w
)
w
〉
|
(
1
-
|
φ
(
w
)
|
2
)
n
/
q
+
(
γ
+
1
)
/
p
+
1
≤
(
n
q
+
b
)
μ
(
w
)
|
ψ
(
w
)
|
(
1
-
|
φ
(
w
)
|
2
)
n
/
q
+
(
γ
+
1
)
/
p
×
{
H
φ
(
w
)
(
J
φ
(
w
)
w
,
J
φ
(
w
)
w
)
}
1
/
2
≤
C
M
2
<
∞
.
Combining (44) and (45), the desired result (23) holds. This completes the proof.Theorem 6.
Assume that0
<
p
,
q
<
∞, -
1
<
γ
<
∞, μ is a normal function, and φ
∈
S
(
B
n
), ψ
∈
H
(
B
n
). Then T
ψ
,
φ
:
H
p
,
q
,
γ
→
B
μ is compact if and only if the followings are all satisfied:(a)
ψ
∈
B
μ and ψ
φ
l
∈
B
μ for l
∈
{
1
,
…
,
n
};
(b)
(46)
lim
⁡
|
φ
(
z
)
|
→
1
μ
(
z
)
|
R
ψ
(
z
)
|
(
1
-
|
φ
(
z
)
|
2
)
n
/
q
+
(
γ
+
1
)
/
p
=
0
;
(c)
(47)
lim
⁡
|
φ
(
z
)
|
→
1
μ
(
z
)
|
ψ
(
z
)
|
(
1
-
|
φ
(
z
)
|
2
)
n
/
q
+
(
γ
+
1
)
/
p
×
{
H
φ
(
z
)
(
J
φ
(
z
)
z
,
J
φ
(
z
)
z
)
}
1
/
2
=
0
.Proof
Sufficiency. Suppose that (
a
), (
b
), and (
c
) hold. Then for any ɛ
>
0, there is δ
>
0, such that
(48)
μ
(
z
)
|
R
ψ
(
z
)
|
(
1
-
|
φ
(
z
)
|
2
)
n
/
q
+
(
γ
+
1
)
/
p
<
ɛ
,
μ
(
z
)
|
ψ
(
z
)
|
(
1
-
|
φ
(
z
)
|
2
)
n
/
q
+
(
γ
+
1
)
/
p
{
H
φ
(
z
)
(
J
φ
(
z
)
z
,
J
φ
(
z
)
z
)
}
1
/
2
<
ɛ
,
when |
φ
(
z
)
|
>
δ.
Let{
f
k
}
k
∈
N be any sequence which converges to 0 uniformly on compact subsets of B
n satisfying ∥
f
k
∥
H
p
,
q
,
γ
≤
1. Then f
k and R
f
k converge to 0 uniformly on K
=
{
w
∈
B
n
:
|
w
|
≤
δ
}. Hence
(49)
sup
⁡
z
∈
B
n
μ
(
z
)
|
R
(
T
ψ
,
φ
f
k
)
(
z
)
|
=
sup
⁡
φ
(
z
)
∈
K
μ
(
z
)
|
R
(
T
ψ
,
φ
f
k
)
(
z
)
|
+
sup
⁡
φ
(
z
)
∈
B
n
∖
K
μ
(
z
)
|
R
(
T
ψ
,
φ
f
k
)
(
z
)
|
.
Ifφ
(
z
)
∈
B
n
∖
K and J
φ
(
z
)
z
≠
0, by Lemma 1 and Lemma 2, we have
(50)
μ
(
z
)
|
R
(
T
ψ
,
φ
f
k
)
(
z
)
|
≤
μ
(
z
)
|
ψ
(
z
)
|
|
R
(
f
k
∘
φ
)
(
z
)
|
+
μ
(
z
)
|
R
ψ
(
z
)
|
|
f
k
(
φ
(
z
)
)
|
≤
(
C
μ
(
z
)
|
ψ
(
z
)
|
{
H
φ
(
z
)
(
J
φ
(
z
)
z
,
J
φ
(
z
)
z
)
}
1
/
2
×
|
〈
∇
f
k
(
φ
(
z
)
)
,
J
φ
(
z
)
z
¯
〉
|
C
μ
(
z
)
|
ψ
(
z
)
|
{
H
φ
(
z
)
(
J
φ
(
z
)
z
,
J
φ
(
z
)
z
)
}
1
/
2
)
×
(
(
1
-
|
φ
(
z
)
|
2
)
n
/
q
+
(
γ
+
1
)
/
p
G
φ
(
z
)
v
(
J
φ
(
z
)
z
,
J
φ
(
z
)
z
)
)
-
1
+
ɛ
∥
f
k
∥
H
p
,
q
,
γ
≤
C
ɛ
∥
f
k
∥
B
(
1
-
r
2
)
n
/
q
+
(
γ
+
1
)
/
p
+
1
+
ɛ
∥
f
k
∥
H
p
,
q
,
γ
≤
C
ɛ
.
When J
φ
(
z
)
z
=
0,
(51)
μ
(
z
)
|
R
(
T
ψ
,
φ
f
k
)
(
z
)
|
≤
ɛ
∥
f
k
∥
H
p
,
q
,
γ
≤
ɛ
.
Combining (50) and (51) we obtain that
(52)
sup
⁡
φ
(
z
)
∈
B
n
∖
K
μ
(
z
)
|
R
(
T
ψ
,
φ
f
k
)
(
z
)
|
≤
C
ɛ
.
If φ
(
z
)
∈
K, by (a), we have that
(53)
μ
(
z
)
|
R
(
T
ψ
,
φ
f
k
)
(
z
)
|
≤
μ
(
z
)
|
ψ
(
z
)
|
|
R
(
f
k
∘
φ
)
(
z
)
|
+
μ
(
z
)
|
R
ψ
(
z
)
|
|
f
k
(
φ
(
z
)
)
|
≤
μ
(
z
)
|
ψ
(
z
)
|
|
〈
∇
f
k
(
φ
(
z
)
)
,
J
φ
(
z
)
z
¯
〉
|
+
|
f
k
(
φ
(
z
)
)
|
∥
ψ
∥
B
μ
≤
|
∇
f
k
(
φ
(
z
)
)
|
∑
l
=
1
n
‍
(
μ
(
z
)
|
ψ
(
z
)
|
|
R
φ
l
(
z
)
|
)
+
|
f
k
(
φ
(
z
)
)
|
∥
ψ
∥
B
μ
≤
|
∇
f
k
(
φ
(
z
)
)
|
×
∑
l
=
1
n
‍
(
μ
(
z
)
|
ψ
(
z
)
|
|
R
φ
l
(
z
)
|
-
μ
(
z
)
|
φ
l
(
z
)
|
|
R
ψ
(
z
)
|
+
μ
(
z
)
|
R
ψ
(
z
)
|
)
+
|
f
k
(
φ
(
z
)
)
|
∥
ψ
∥
B
μ
≤
|
∇
f
k
(
φ
(
z
)
)
|
×
∑
l
=
1
n
‍
(
μ
(
z
)
|
ψ
(
z
)
R
φ
l
(
z
)
+
R
ψ
(
z
)
φ
l
(
z
)
|
+
μ
(
z
)
|
R
ψ
(
z
)
|
)
+
|
f
k
(
φ
(
z
)
)
|
∥
ψ
∥
B
μ
≤
|
∇
f
k
(
φ
(
z
)
)
|
∑
l
=
1
n
‍
(
∥
ψ
φ
l
∥
B
μ
+
∥
ψ
∥
B
μ
)
+
|
f
k
(
φ
(
z
)
)
|
∥
ψ
∥
B
μ
⟶
0
,
k
⟶
∞
.
Combining (49), (52), (53), and Lemma 4, it follows that the T
ψ
,
φ
:
H
p
,
q
,
γ
→
B
μ is compact.
Necessity. Assume that T
ψ
,
φ
:
H
p
,
q
,
γ
→
B
μ is compact. It is obvious that T
ψ
,
φ
:
H
p
,
q
,
γ
→
B
μ is bounded. Then taking f
(
z
)
=
1
∈
H
p
,
q
,
γ and by the boundedness of T
ψ
,
φ
:
H
p
,
q
,
γ
→
B
μ, it follows that
(54)
∥
T
ψ
,
φ
f
(
z
)
∥
B
μ
=
sup
⁡
z
∈
B
n
μ
(
z
)
|
R
(
T
ψ
,
φ
f
)
(
z
)
|
=
sup
⁡
z
∈
B
n
μ
(
z
)
|
R
ψ
(
z
)
f
(
φ
(
z
)
)
+
ψ
(
z
)
R
(
f
∘
φ
)
(
z
)
|
=
sup
⁡
z
∈
B
n
μ
(
z
)
|
R
ψ
(
z
)
|
<
∞
.
This shows that ψ
∈
B
μ.
On the other hand, forl
∈
{
1
,
…
,
n
}, take the function f
(
z
)
=
z
l
∈
H
p
,
q
,
γ. By the boundedness of T
ψ
,
φ
:
H
p
,
q
,
γ
→
B
μ, we get that
(55)
∥
T
ψ
,
φ
f
(
z
)
∥
B
μ
=
sup
⁡
z
∈
B
n
μ
(
z
)
|
R
ψ
(
z
)
f
(
φ
(
z
)
)
+
ψ
(
z
)
R
(
f
∘
φ
)
(
z
)
|
=
sup
⁡
z
∈
B
n
μ
(
z
)
|
R
ψ
(
z
)
φ
l
(
z
)
+
ψ
(
z
)
R
φ
l
(
z
)
|
=
sup
⁡
z
∈
B
n
μ
(
z
)
|
R
(
ψ
φ
l
)
(
z
)
|
<
∞
.
That is, ψ
φ
l
∈
B
μ for l
∈
{
1
,
…
,
n
}. Hence we obtain (a).
Next we prove (b) and (c). Let{
z
k
}
k
∈
N be a sequence in B
n such that |
φ
(
z
k
)
|
→
1 as k
→
∞. We can still suppose φ
(
z
k
)
=
r
k
e
1, where r
k
=
|
φ
(
z
k
)
| and e
1 is the vector (
1,0
,
0
,
…
,
0
). That is, |
r
k
|
→
1, k
→
∞.
If(
1
-
r
k
2
)
(
|
η
2
|
2
+
⋯
+
|
η
n
|
2
)
≤
|
η
1
|, where J
φ
(
z
k
)
z
k
=
(
η
1
,
…
,
η
n
)
T. Let
(56)
f
k
(
z
)
=
z
1
-
r
k
1
-
r
k
z
1
{
1
-
r
k
2
(
1
-
r
k
z
1
)
2
}
n
/
q
+
(
γ
+
1
)
/
p
.
From Theorem 5 we know that f
k
∈
H
p
,
q
,
γ, and we notice that f
k converges to 0 uniformly on compact subsets of B
n when k
→
∞. By Lemma 3 we have lim
⁡
k
→
∞
∥
T
ψ
,
φ
f
k
(
z
)
∥
B
μ
=
0. Then by a similar proof of (30) in Theorem 5 we have
(57)
μ
(
z
k
)
|
ψ
(
z
k
)
|
|
η
1
|
(
1
-
r
k
2
)
n
/
q
+
(
γ
+
1
)
/
p
+
1
≤
∥
T
ψ
,
φ
f
k
(
z
)
∥
B
μ
⟶
0
,
k
⟶
∞
.
And similar to the proofs of (31) and (57) we get that
(58)
μ
(
z
k
)
|
ψ
(
z
k
)
|
(
1
-
|
φ
(
z
k
)
|
2
)
n
/
q
+
(
γ
+
1
)
/
p
{
H
φ
(
z
k
)
(
J
φ
(
z
k
)
z
k
,
J
φ
(
z
k
)
z
k
)
}
1
/
2
≤
2
μ
(
z
k
)
|
ψ
(
z
k
)
|
|
η
1
|
(
1
-
r
k
2
)
n
/
q
+
(
γ
+
1
)
/
p
+
1
⟶
0
,
k
⟶
∞
.
On the other hand, we consider the case of(
1
-
r
k
2
)
(
|
η
2
|
2
+
⋯
+
|
η
n
|
2
)
>
|
η
1
|. For j
=
2
,
…
,
n, let θ
j
=
arg
η
j and a
j
=
e
-
i
θ
j, when η
j
≠
0; otherwise a
j
=
0 when η
j
=
0. Take
(59)
f
k
(
z
)
=
(
a
2
z
2
+
⋯
+
a
n
z
n
)
(
1
-
r
k
2
)
(
1
-
r
k
z
1
)
n
/
q
+
(
γ
+
1
)
/
p
+
2
.
Then f
k
∈
H
p
,
q
,
γ,  k
∈
N, and f
k converges to 0 uniformly on compact subsets of B
n when k
→
∞. By Lemma 3 we have lim
⁡
k
→
∞
∥
T
ψ
,
φ
f
k
(
z
)
∥
B
μ
=
0. Notice that f
k
(
φ
(
z
k
)
)
=
0 and
(60)
∇
f
w
(
φ
(
z
k
)
)
=
(
0
,
a
2
(
1
-
r
k
2
)
n
/
q
+
(
γ
+
1
)
/
p
+
1
,
…
,
a
n
(
1
-
r
k
2
)
n
/
q
+
(
γ
+
1
)
/
p
+
1
)
.
By a similar proof of (30), it follows that
(61)
μ
(
z
k
)
|
ψ
(
z
k
)
|
(
|
η
2
|
+
⋯
+
|
η
n
|
)
(
1
-
r
k
2
)
α
+
1
≤
∥
T
ψ
,
φ
f
k
∥
B
μ
⟶
0
,
k
⟶
∞
.
And similar to the proofs of (31) and (61), we obtain
(62)
μ
(
z
k
)
|
ψ
(
z
k
)
|
(
1
-
|
φ
(
z
k
)
|
2
)
α
{
H
φ
(
z
k
)
(
J
φ
(
z
k
)
z
k
,
J
φ
(
z
k
)
z
k
)
}
1
/
2
≤
C
μ
(
z
k
)
|
ψ
(
z
k
)
|
2
(
1
-
r
k
2
)
(
|
η
2
|
+
⋯
+
|
η
n
|
)
(
1
-
r
k
2
)
n
/
q
+
(
γ
+
1
)
/
p
+
1
⟶
0
1111111111111111111111111111111111
k
⟶
∞
.
Combining (58) and (62), (47) holds under the two cases.
For the general situation, if there existsφ
(
z
k
) such that φ
(
z
k
)
≠
|
φ
(
z
k
)
|
e
1, then there is a unitary transformation U
k such that φ
(
z
k
)
=
r
k
e
1
U
k, k
∈
{
1,2
,
…
,
n
}. And we can prove (47) by taking the function sequence g
k
=
f
k
∘
U
k
-
1 and the details are omitted.
Next we prove (46). Let {
z
k
}
k
∈
N be a sequence in B
n such that |
φ
(
z
k
)
|
→
1 as k
→
∞. Choose
(63)
h
k
(
z
)
=
(
1
-
|
φ
(
z
k
)
|
2
)
b
-
(
γ
+
1
)
/
p
(
1
-
〈
z
,
φ
(
z
k
)
〉
)
n
/
q
+
b
.
Then h
k
∈
H
p
,
q
,
γ,  k
∈
N, and sup
⁡
k
∈
N
∥
h
k
∥
H
p
,
q
,
γ
≤
C. It is obvious that h
k
→
0 uniformly on compact subsets of B
n as k
→
∞. By Lemma 3 we have that lim
⁡
k
→
∞
∥
T
ψ
,
φ
(
h
k
)
(
z
)
∥
B
μ
=
0. Then by the similar proof of (44) we obtain
(64)
∥
T
ψ
,
φ
(
h
k
)
(
z
)
∥
B
μ
≥
μ
(
z
k
)
|
R
ψ
(
z
k
)
|
(
1
-
|
φ
(
z
k
)
|
2
)
n
/
q
+
(
γ
+
1
)
/
p
-
μ
(
z
k
)
|
ψ
(
z
k
)
|
|
R
(
h
k
∘
φ
)
(
z
k
)
|
.
From the similar proof of (45) it follows that
(65)
μ
(
z
k
)
|
ψ
(
z
k
)
|
|
R
(
h
k
∘
φ
)
(
z
k
)
|
≤
(
n
q
+
b
)
μ
(
z
k
)
|
ψ
(
z
k
)
|
(
1
-
|
φ
(
z
k
)
|
2
)
n
/
q
+
(
γ
+
1
)
/
p
×
{
H
φ
(
z
k
)
(
J
φ
(
z
k
)
z
k
,
J
φ
(
z
k
)
z
k
)
}
1
/
2
⟶
0
,
k
⟶
∞
.
Combining (64) and (65) we obtain (46). This completes the proof.Corollary 7.
Assume that0
<
p
,
q
<
∞,  -
1
<
γ
<
∞, μ is a normal function, and φ
∈
S
(
B
n
). Then C
φ
:
H
p
,
q
,
γ
→
B
μ is bounded if and only if
(66)
sup
⁡
z
∈
B
n
μ
(
z
)
{
H
φ
(
z
)
(
J
φ
(
z
)
z
,
J
φ
(
z
)
z
)
}
1
/
2
(
1
-
|
φ
(
z
)
|
2
)
n
/
q
+
(
γ
+
1
)
/
p
<
∞
.Corollary 8.
Assume that0
<
p
,
q
<
∞, -
1
<
γ
<
∞, μ is a normal function, and φ
∈
S
(
B
n
). Then C
φ
:
H
p
,
q
,
γ
→
B
μ is compact if and only if
(67)
lim
⁡
|
φ
(
z
)
|
→
1
μ
(
z
)
{
H
φ
(
z
)
(
J
φ
(
z
)
z
,
J
φ
(
z
)
z
)
}
1
/
2
(
1
-
|
φ
(
z
)
|
2
)
n
/
q
+
(
γ
+
1
)
/
p
=
0
.
And φ
l
∈
B
μ  for l
∈
{
1
,
…
,
n
}.Corollary 9.
Assume that0
<
p
,
q
<
∞, -
1
<
γ
<
∞, μ is a normal function, and ψ
∈
H
(
B
n
). Then M
ψ
:
H
p
,
q
,
γ
→
B
μ is bounded if and only if
(68)
sup
⁡
z
∈
B
n
μ
(
z
)
|
R
ψ
(
z
)
|
(
1
-
|
z
|
2
)
n
/
q
+
(
γ
+
1
)
/
p
<
∞
,
sup
⁡
z
∈
B
n
μ
(
z
)
|
ψ
(
z
)
|
(
1
-
|
z
|
2
)
n
/
q
+
(
γ
+
1
)
/
p
+
1
<
∞
.Corollary 10.
Assume that0
<
p
,
q
<
∞, -
1
<
γ
<
∞, μ is a normal function, and ψ
∈
H
(
B
n
). Then M
ψ
:
H
p
,
q
,
γ
→
B
μ is compact if and only if the following are all satisfied:(a)
ψ
∈
B
μ and ψ
z
l
∈
B
μ for any l
∈
{
1
,
…
,
n
};
(b)
(69)
lim
⁡
|
z
|
→
1
μ
(
z
)
|
R
ψ
(
z
)
|
(
1
-
|
z
|
2
)
n
/
q
+
(
γ
+
1
)
/
p
=
0
;
(c)
(70)
lim
⁡
|
z
|
→
1
μ
(
z
)
|
ψ
(
z
)
|
(
1
-
|
z
|
2
)
n
/
q
+
(
γ
+
1
)
/
p
+
1
=
0
.



---

*Source: 107560-2014-08-07.xml*