# Multiplication Operator with BMO Symbols and Berezin Transform

Authors: Xue Feng; Kan Zhang; Jianguo Dong; Xianmin Liu; Chi Guan
Journal: Journal of Function Spaces
 (2015)

---

## Abstract

We discuss multiplication operator with a special symbol on the weighted Bergman space of the unit ball. We give the necessary and sufficient conditions for the compactness of multiplication operator on the weighted Bergman space of the unit ball.


---

## Body

## 1. Introduction

LetB
n denote the unit ball in C
n, and let v be the normalized Lebesgue volume measure on B
n. For -
1
<
α
<
∞, we denote by v
α the measure on B
n defined by d
v
α
(
z
)
=
c
α
(
1
-
z
2
)
α
d
v
(
z
), where c
α
=
Γ
(
n
+
α
+
1
)
/
n
!
Γ
(
α
+
1
) is a normalizing constant such that v
α
(
B
n
)
=
1. For 1
≤
p
<
∞, we write ·
α
,
p for the norm on L
p
(
B
n
,
d
v
α
) and ·
,
·
α for the inner product on L
2
(
B
n
,
d
v
α
). The Bergman space A
α
2
(
B
n
) is the space of holomorphic functions which are square-integrable with respect to measure d
v
α on B
n. Reproducing kernels K
w
α and normalized reproducing kernels k
w
α in A
α
2
(
B
n
) are given by (1)
K
w
α
z
=
1
1
-
z
,
w
n
+
α
+
1
,
k
w
α
(
z
)
=
1
-
w
2
(
n
+
α
+
1
)
/
2
1
-
z
,
w
n
+
α
+
1
,respectively, for z
,
w
∈
B
n. For every h
∈
A
α
2
(
B
n
) we have h
,
K
w
α
α
=
h
(
w
), for all w
∈
B
n. The orthogonal projection P
α of L
2
(
B
n
,
d
v
α
) onto A
α
2
(
B
n
) is given by(2)
P
α
g
w
=
g
,
K
w
α
α
=
∫
B
n
‍
g
z
1
1
-
w
,
z
n
+
α
+
1
d
v
α
z
,for g
∈
L
2
(
B
n
,
d
v
α
) and w
∈
B
n.Givenf
∈
L
1
(
B
n
,
d
v
α
), the Toeplitz operator T
f
:
L
2
(
B
n
,
d
v
α
)
→
A
α
2
(
B
n
), the Hankel operator H
f
:
L
2
(
B
n
,
d
v
α
)
→
(
A
α
2
(
B
n
)
)
⊥, and the multiplication operator M
f
:
A
α
2
(
B
n
)
→
L
2
(
B
n
,
d
v
α
) are given by(3)
T
f
h
z
=
∫
B
n
‍
f
(
ω
)
h
(
ω
)
1
-
z
,
ω
n
+
α
+
1
d
v
α
(
ω
)
,
H
f
h
z
=
f
z
g
z
-
T
f
g
z
,
M
f
h
=
f
h
,respectively. For f
∈
L
1
(
B
n
,
d
v
α
), we define the Berezin transform of f to be the function f
~; that is,(4)
f
~
z
=
∫
B
n
‍
f
w
k
z
α
w
2
d
v
α
w
.If f is bounded, then f
~ is a bounded function on B
n. Since the kernels k
z
α converge weakly to zero as z tends ∂
B
n, we have that if f is compact, then f
~
→
0 as z
→
∂
B
n. The converse (in both cases) is not necessarily true. According to the definition of Berezin transform, the mean oscillation of f in the Bergman metric is the function MO
(
f
)
(
z
) defined on B
n by(5)
MO
(
f
)
(
z
)
=
f
2
~
(
z
)
-
f
~
z
2
.Forz
∈
B
n, let ψ
z be the automorphism of B
n such that ψ
z
(
0
)
=
z and ψ
z
=
(
ψ
z
)
-
1. Thus, we have the change-of-variable formula(6)
∫
B
n
‍
h
ψ
z
w
k
z
α
w
2
d
v
α
w
=
∫
B
n
‍
h
w
d
v
α
w
,for every h
∈
L
1
(
B
n
,
d
v
α
).Multiplication operators are one of the most widely studied classes of concrete operators. The study of their behavior on the Hardy and Bergman spaces has generated an extensive list of results in the operator theory and in the theory of function spaces [1–6]. One of the useful approaches is the use of the Berezin transform [7–11]. This method is motivated by its connections with quantum physics and noncommutative geometry.In general, Berezin transformf
~ plays important role in giving necessary and sufficient conditions for the boundedness and compactness of the Toeplitz operator [12, 13]. However Berezin transform f
2
~ or the mean oscillation MO
(
f
) is used to obtain the necessary and sufficient conditions for the boundedness and compactness of the Hankel operator or multiplication operator [14, 15]. This work is partially motivated by using Berezin transform f
~ to obtain necessary and sufficient conditions for the compactness of multiplication operator on the weighted Bergman space of the unit ball.Throughout the paper, we will use the letterc to denote a generic positive constant that can change its value at each occurrence.


## 2. Main Results

In this section, we give the necessary and sufficient conditions for the compactness of multiplication operator on the weighted Bergman space of the unit ball. We furthermore obtain the necessary and sufficient conditions for the compactness of Toeplitz operator and Hankel operator.Theorem 1.
Supposef
/
(
1
-
z
)
4
n
+
4
α
+
4 is bounded on B
n. Then M
f is compact operator on A
α
2
(
B
n
) if and only if f
~
(
z
)
→
0 as z
→
∂
B
n.Proof.
Supposef
~
(
z
)
→
0 as z
→
∂
B
n.
Since(7)
M
f
g
,
h
α
=
f
g
,
h
α
=
g
,
f
-
h
α
,it is clear that (
M
f
)
∗
=
M
f
-. It suffices to prove that the operator (
M
f
)
∗ is compact by showing that (
M
f
)
∗ can be approximated by compact operators in the operator norm.
Letg
∈
L
2
(
B
n
,
d
v
α
). Then (
M
f
)
∗
g
∈
A
α
2
(
B
n
), so we have (8)
M
f
∗
g
z
=
(
M
f
)
∗
g
,
K
z
α
α
=
∫
B
n
‍
g
w
f
w
K
z
α
w
¯
d
v
α
w
,for z
∈
B
n.
We define for0
<
r
<
1 an operator S
r by(9)
S
r
g
z
=
∫
B
n
‍
χ
r
B
n
z
g
w
f
w
K
z
α
w
¯
d
v
α
w
.Since f
/
(
1
-
z
)
4
n
+
4
α
+
4 is bounded on B
n, we prove that(10)
∫
B
n
∫
B
n
χ
r
B
n
z
f
w
K
z
α
w
¯
2
d
v
α
w
d
v
α
z
=
∫
r
B
n
‍
∫
B
n
‍
f
w
K
z
α
w
¯
2
d
v
α
w
d
v
α
z
<
+
∞
.Thus, the operator S
r is a Hilbert-Schmidt operator. Since(11)
M
f
∗
-
S
r
g
z
=
∫
B
n
‍
g
w
χ
B
n
∖
r
B
n
z
f
w
K
z
α
w
¯
d
v
α
w
,
(
M
f
)
∗
-
S
r is an integral operator with kernel K
r
f
(
z
,
w
)
=
χ
B
n
∖
r
B
n
(
z
)
f
(
w
)
K
z
α
(
w
)
¯.
By Schur’s test, whenever there exists a positive measurable functionh on B
n and constants c
1 and c
2 such that(12)
∫
B
n
‍
K
r
f
z
,
w
h
z
d
v
α
z
≤
c
1
h
w
,for all w in B
n, and(13)
∫
B
n
‍
K
r
f
z
,
w
h
w
d
v
α
w
≤
c
2
h
z
,for all z in B
n, we have(14)
M
f
∗
-
S
r
≤
c
1
c
2
.Let h
(
z
)
=
(
1
-
|
z
|
2
)
n
+
α
+
1
/
2. Since(15)
K
z
α
ψ
z
v
k
z
α
v
=
1
1
-
z
2
n
+
1
+
α
/
2
,
1
-
ψ
z
v
2
=
1
-
z
2
1
-
v
2
1
-
v
,
z
2
,and Hölder inequality, it is easy to prove that(16)
∫
B
n
‍
χ
B
n
∖
r
B
n
z
f
w
K
z
α
w
1
-
w
2
n
+
α
+
1
/
2
1
-
z
2
n
+
α
+
1
/
2
d
v
α
w
=
∫
B
n
‍
χ
B
n
∖
r
B
n
z
f
w
k
z
α
w
1
-
w
2
n
+
α
+
1
/
2
1
-
z
2
n
+
α
+
1
d
v
α
w
=
∫
B
n
‍
χ
B
n
∖
r
B
n
z
f
ψ
z
w
k
z
α
ψ
z
w
h
h
h
h
h
×
1
-
ψ
z
w
2
n
+
α
+
1
/
2
1
-
z
2
n
+
α
+
1
k
z
α
w
2
d
v
α
w
=
∫
B
n
‍
χ
B
n
∖
r
B
n
z
f
ψ
z
w
1
-
w
2
n
+
α
+
1
/
2
1
-
z
,
w
2
n
+
2
α
+
2
d
v
α
(
w
)
=
∫
B
n
‍
χ
B
n
∖
r
B
n
(
z
)
f
ψ
z
w
f
ψ
z
w
h
h
h
h
h
×
1
-
w
2
n
+
α
+
1
/
2
1
-
z
,
w
2
n
+
2
α
+
2
d
v
α
(
w
)
≤
χ
B
n
∖
r
B
n
z
∫
B
n
‍
f
ψ
z
w
d
v
α
w
1
/
2
×
∫
B
n
‍
f
ψ
z
w
1
-
w
2
n
+
α
+
1
1
-
z
,
w
4
n
+
4
α
+
4
d
v
α
w
1
/
2
≤
c
χ
B
n
∖
r
B
n
(
z
)
∫
B
n
‍
f
ψ
z
w
d
v
α
w
1
/
2
×
∫
B
n
‍
f
ψ
z
w
1
1
-
z
3
n
+
3
α
+
3
d
v
α
w
1
/
2
≤
c
f
~
z
1
/
2
=
c
1
,where c
1
=
c
[
f
~
(
z
)
]
1
/
2, r
<
z
<
1.
Since(17)
K
z
α
ψ
z
v
k
z
α
v
=
1
1
-
z
2
n
+
1
+
α
/
2
,
1
-
ψ
z
(
w
)
,
z
=
1
-
z
2
1
-
w
,
z
,then we obtain (18)
∫
B
n
‍
χ
B
n
∖
r
B
n
z
f
w
K
z
α
w
1
-
z
2
n
+
α
+
1
/
2
1
-
w
2
n
+
α
+
1
/
2
d
v
α
z
=
∫
B
n
‍
χ
B
n
∖
r
B
n
w
f
z
K
z
α
w
1
-
w
2
n
+
α
+
1
/
2
1
-
z
2
n
+
α
+
1
/
2
d
v
α
w
=
∫
B
n
‍
χ
B
n
∖
r
B
n
ψ
z
w
f
z
K
z
α
ψ
z
w
h
h
h
h
h
×
1
-
ψ
z
w
2
n
+
α
+
1
/
2
1
-
z
2
n
+
α
+
1
/
2
k
z
α
w
2
d
v
α
w
=
∫
B
n
‍
χ
B
n
∖
r
B
n
ψ
z
w
f
z
h
h
h
h
h
×
1
-
ψ
z
w
2
n
+
α
+
1
/
2
1
-
z
,
w
4
n
+
4
α
+
4
d
v
α
w
≤
c
2
,where c
2 is positive number.
By the above analysis, we get (12) and (13). By Schur’s test we get (
M
f
)
∗
-
S
r
≤
c
1
c
2, where c
1
→
0 as z
→
∂
B
n and c
2 does not depend on r. So f
~
(
z
)
→
0 as z
→
∂
B
n implies that M
f is compact on A
α
2
(
B
n
).
SupposeM
f is compact on A
α
2
(
B
n
).
Since the kernelsk
z
α converge weakly to zero as z tends ∂
B
n, then we have M
f
k
z
α
α
,
2 converges to zero as z tends ∂
B
n. So we obtain(19)
f
~
z
≤
f
2
~
z
1
/
2
=
M
f
k
z
α
α
,
2
⟶
0
,as z
→
∂
B
n.Letf
∈
L
1
(
B
n
,
d
v
α
) and let p
≥
1; we say that f
∈
BM
O
α
p
(
B
n
) whenever(20)
f
BM
O
α
p
=
sup
⁡
z
∈
B
n
f
∘
ψ
z
-
f
~
z
α
,
p
<
∞
.Note that ·
BM
O
α
p does not distinguish constants, while f
α
,
p
=
f
BM
O
α
p
+
f
~
0 is a norm in BM
O
α
p
(
B
n
). By Theorem  5 in [16], we know that BM
O
α
p
(
B
n
) is equivalent to BM
O
∂
p (see the definition in [16]).For anyp
≥
1, let VM
O
α
p denote the subspace of BM
O
α
p consisting of functions f such that(21)
lim
|
z
|
→
1
-
f
∘
ψ
z
-
f
~
z
α
,
p
=
0
.Theorem 2.
Supposef
∈
V
M
O
α
1 and f
/
(
1
-
z
)
4
n
+
4
α
+
4 is bounded on B
n. Then the following are equivalent:(a)
f
~
(
z
)
→
0 as z
→
∂
B
n;
(b)
M
f is compact operator on A
α
2
(
B
n
);
(c)
T
f is compact operator on A
α
2
(
B
n
).Proof.
It suffices to prove that(
a
)
⇔
(
b
) and (
a
)
⇔
(
c
).
(
a
)
⇔
(
b
). Since(22)
f
~
z
≤
f
~
z
,
f
~
z
-
f
~
z
=
∫
B
n
‍
f
w
-
f
~
z
1
-
z
2
n
+
α
+
1
1
-
z
,
w
2
n
+
2
α
+
2
d
v
α
w
≤
∫
B
n
‍
f
w
-
f
~
z
1
-
z
2
n
+
α
+
1
1
-
z
,
w
2
n
+
2
α
+
2
d
v
α
w
=
∫
B
n
‍
f
∘
ψ
z
w
-
f
~
z
d
v
α
w
=
f
∘
ψ
z
-
f
~
z
α
,
1
,then we obtain that f
~
(
z
)
→
0 as z
→
∂
B
n if and only if f
~
(
z
)
→
0 as z
→
∂
B
n. By Theorem 1, we obtain that M
f is compact operator on A
α
2
(
B
n
) if and only if f
~
(
z
)
→
0 as z
→
∂
B
n.
(
a
)
⇔
(
c
). It is clear that f
~
(
z
)
→
0 as z
→
∂
B
n if and only if  T
f with BM
O
α
1 symbol is compact operator on A
α
2
(
B
n
) in [12]. Since VM
O
α
1
⊂
BM
O
α
1, then it is clear that f
~
(
z
)
→
0 as z
→
∂
B
n if and only if T
f is compact operator on A
α
2
(
B
n
).Corollary 3.
Supposef
∈
V
M
O
α
1, f
/
(
1
-
z
)
4
n
+
4
α
+
4 is bounded on B
n, and H
f is compact operator on A
α
2
(
B
n
). Then M
f
-
f
~ is compact operator on A
α
2
(
B
n
).Proof.
SupposeH
f is compact operator on A
α
2
(
B
n
). So we obtain H
f
-
f
~ is compact operator on A
α
2
(
B
n
). Since f
-
f
~
~
(
z
)
→
0 as z
→
∂
B
n, then T
f
-
f
~ is compact operator on A
α
2
(
B
n
). Since(23)
M
f
-
f
~
g
=
T
f
-
f
~
g
+
H
f
-
f
~
g
,we obtain M
f
-
f
~ is compact operator on A
α
2
(
B
n
).By Lemma 17 and Theorem  19 in [16] and Theorem 2, we obtain the following theorem.Theorem 4.
Supposef
∈
V
M
O
α
1, f
~
(
0
)
=
0, and f
/
(
1
-
z
)
4
n
+
4
α
+
4 is bounded on B
n. Then the following are equivalent:(a)
M
f is compact operator on A
α
2
(
B
n
);
(b)
T
f is compact operator on A
α
2
(
B
n
);
(c)
H
f is compact operator on A
α
2
(
B
n
).



---

*Source: 754646-2015-04-06.xml*