# Sarason’s Conjecture of Toeplitz Operators on Fock-Sobolev Type Spaces

Authors: Xiaofeng Wang; Jianjun Chen; Jin Xia
Journal: Journal of Function Spaces
 (2018)

---

## Abstract

In this note, we will solve Sarason’s conjecture on the Fock-Sobolev type spaces and give a well solution that if Toeplitz productT
u
T
v
¯, with entire symbols u and v, is bounded if and only if u
=
e
q, v
=
C
e
-
q, where q is a linear complex polynomial and C is a nonzero constant.


---

## Body

## 1. Introduction

LetC
n denote the complex n-space and d
v be the ordinary volume measure on C
n that is normalized so that ∫
C
n
e
-
z
2
d
v
(
z
)
=
1. If given any two points z
=
(
z
1
,
z
2
,
…
,
z
n
) and w
=
(
w
1
,
w
2
,
…
,
w
n
) in C
n, we denote (1)
z
·
w
¯
=
∑
j
=
1
n
z
j
w
j
¯
,
z
=
z
·
z
¯
.For every 0
<
p
<
∞, α
∈
R, we denote by L
α
p
(
C
n
) the space of measurable functions f such that (2)
f
L
α
p
=
∫
C
n
f
z
e
-
1
/
2
z
2
p
d
v
z
1
+
z
α
1
/
p
<
∞
.Let H
(
C
n
) be the set of entire functions on C
n. Then for a given 0
<
p
<
∞, the Fock-Sobolev type space F
α
p with the norm ·
F
α
p
=
·
L
α
p is defined as (3)
F
α
p
=
f
∈
H
C
n
∣
f
L
α
p
<
∞
.Obviously, the Fock-Sobolev type spaceF
α
2 equipped with the natural inner product defined by (4)
f
,
g
L
α
2
=
∫
C
n
f
z
g
z
¯
e
-
z
2
d
v
z
1
+
z
αis a reproducing kernel Hilbert space for every real α. As stated in [1], with respect to the above inner product, it is difficult to compute the reproducing kernel of F
α
2 explicitly. So we use the equivalent norm with respect to a new measure z
-
α
d
v
(
z
). In more detail, for α
≤
0, we will let (5)
f
,
g
α
=
∫
C
n
f
z
g
z
¯
e
-
z
2
d
v
z
z
α
,and for α
>
0 we let (6)
f
,
g
α
=
∫
C
n
f
α
/
2
-
z
g
α
/
2
-
z
¯
e
-
z
2
d
v
z
+
∫
C
n
f
α
/
2
+
z
g
α
/
2
+
z
¯
e
-
z
2
d
v
z
z
α
,where f
α
/
2
- is the Taylor expansion of f up to order α
/
2 and f
α
/
2
+
=
f
-
f
α
/
2
-. Now we can bravely make sure that the inner product ·
,
·
α generates a new Hilbert space norm on F
α
2 that is equivalent to the F
α
2 norm ·
F
α
2. In particular, if we define the norm ·
F
α
2
~ on F
α
2 by, when α
≤
0, (7)
f
F
α
2
~
=
∫
C
n
f
z
2
e
-
z
2
d
v
z
z
α
1
/
2and when α
>
0, (8)
f
F
α
2
~
=
∫
C
n
f
α
/
2
-
z
2
e
-
z
2
d
v
z
1
/
2
+
∫
C
n
f
α
/
2
+
z
2
e
-
z
2
d
v
z
z
α
1
/
2
,and then we have that both ·
F
α
2
~ and ·
F
α
2 are equivalent norms.As is well known,F
α
2 is indeed a reproducing kernel Hilbert space (see Lemma 2.1 of [1] for more details). Therefore its reproducing kernel is (9)
K
z
α
w
=
∑
β
ϕ
β
w
ϕ
β
z
¯
,where {
ϕ
β
} is any orthonormal basis for F
α
2 with respect to ·
,
·
α. Note that polynomials form a dense subset of F
α
2 (see Proposition  2.3 in [2]). Also note that monomials are mutually orthogonal, which means that {
z
β
/
z
β
,
z
β
α
} is an orthonormal basis for F
α
2. The arguments that are identical to the ones in the proof of Theorem  4.5 in [2] then give us that (10)
K
z
α
w
=
I
-
α
/
2
K
z
w
,
if
α
≤
0
;
I
-
α
/
2
K
z
w
+
K
z
α
/
2
-
w
,
if
α
>
0
.Here I
s is the fractional integration operator defined as (11)
I
s
f
z
=
∑
k
=
0
∞
Γ
n
+
k
Γ
n
+
s
+
k
f
k
z
,
if
s
≥
0
;
∑
k
>
s
∞
Γ
n
+
k
Γ
n
+
s
+
k
f
k
z
,
if
s
<
0
,where each f
k is a polynomial of degree k. Moreover, for s
>
0, f
s
+ is the tail part of the Taylor expansion of f of degree higher than s given by (12)
f
s
+
z
=
∑
k
>
s
f
k
zand we let f
s
-
=
f
-
f
s
+ (see [2] for more information on fractional differentiation and integration).Now it is easy to see that ifα
≤
0, (
F
α
2
,
·
F
α
2
~
) is a closed subspace of L
α
2 with respect to ·
,
·
α. In this case, let P
α denote the orthogonal projection, so that (13)
P
α
f
z
=
f
,
K
z
α
αfor any f
∈
L
α
2. Unfortunately, the inner product ·
,
·
α does not make sense on L
α
2 when α
>
0. That means we can not define the Toeplitz operator on F
α
2 in the usual way in terms of this inner product. However according to the ideas of  [1], it makes sense to define the Toeplitz operator with the symbols in F
α
2 by the following formula: (14)
T
φ
α
f
z
=
∫
C
n
f
w
φ
w
K
z
α
w
¯
e
-
w
2
d
v
w
w
α
,if α
≤
0 and (15)
T
φ
α
f
z
=
∫
C
n
φ
w
f
α
/
2
-
w
K
z
α
α
/
2
-
w
¯
e
-
w
2
d
v
w
+
∫
C
n
φ
w
f
α
/
2
+
w
K
z
α
α
/
2
+
w
¯
e
-
w
2
d
v
w
w
αif α
>
0, for any φ
,
f
∈
F
α
2. In the sequel, we can reasonably define the Berezin transform of Toeplitz operator on F
α
2 by (16)
T
φ
α
~
z
=
T
φ
α
k
z
α
,
k
z
α
α
=
∫
C
n
φ
w
k
z
α
w
2
e
-
w
2
d
v
w
w
αif α
≤
0, and (17)
T
φ
α
~
z
=
T
φ
α
k
z
α
,
k
z
α
α
=
∫
C
n
φ
w
k
z
α
α
/
2
-
w
2
e
-
w
2
d
v
w
+
∫
C
n
φ
w
k
z
α
α
/
2
+
w
2
e
-
w
2
d
v
w
w
αif α
>
0 for any φ
∈
F
α
2, where k
z
α
(
w
) is the normalization of the kernel K
z
α
(
w
), that is, k
z
α
(
w
)
=
K
z
α
(
w
)
/
K
z
α
(
z
).The original product problem, owed to Sarason firstly in [3], describes the pairs of outer functions g and h in the Hardy space such that the operator T
g
T
h
¯ is bounded on the Hardy space. Sequentially, this problem was partially researched for the Hardy space in [4] and for the Bergman space in [5–8]. Unluckily it turns out that the Sarason’s conjecture is not true for both Hardy space and Bergman space of unit disk. See [9, 10] for counterexamples.We will, in this note, give the equivalent conditions about the Sarason’s conjecture of Toeplitz product on Fock-Sobolev type spacesF
α
2. Our main result will be the following.Main Theorem.Suppose that
u
and
v
are two nonzero functions in Fock-Sobolev type spaces
F
α
2
. Then the following conditions are equivalent:
(1)
The Toeplitz product
T
u
α
T
v
¯
α
is bounded on Fock-Sobolev type spaces
F
α
2
.
(2)
There exists a complex linear polynomial
q
(
z
)
on
C
n
such that
u
=
e
q
and
v
=
C
e
-
q
, where
C
is a nonzero complex constant.
(3)
The product
u
2
~
v
2
~
is a bounded function on the complex space
C
n
.In 2014, Cho et al. studied the products of Toeplitz operators on the classical Fock space (see [11]). In the case of the Fock-Sobolev space, Chen et al. (see [12]) had already proven the same topics and obtained the similar results. What is more, they claimed that if f and g are two nonzero functions in the Fock(-Sobolev) space, then the Toeplitz product T
f
T
g
¯ is bounded if and only if f
=
e
q and g
=
C
e
-
q, where C is a nonzero constant and q is a linear polynomial. More properties about Toeplitz operators on Fock-Sobolev spaces are referred to in [13]. Sequentially, Bommier-Hato et al. in [14] continued to research Cho’s results on the general Fock-type space with the weight functions exp
⁡
(
-
·
2
m
). They took full advantage of the exact form of the reproducing kernel of the general Fock-type space and concluded that if u and v are two nonzero functions, then the Toeplitz product T
u
T
v
¯ is bounded if and only if u
=
e
g and v
=
C
e
-
g, where C is a nonzero constant and g is a polynomial of degree at most m. The similar techniques are founded in [15, 16]. However, the translations appearing to the classical Fock spaces are not suitable to the generalized Fock space. To tackle the main theorem, we have to use the main ideas of [14], that is, making good use of the explicit properties of the reproducing kernel K
z
α in Fock-Sobolev type spaces F
α
2 instead of the Weyl operators defined by translations on the complex plane.At last, it is remarked that, as stated in [1], the Fock-Sobolev type spaces F
α
2 are in fact very natural generalization of the Fock-Sobolev spaces and the Fock-Sobolev spaces of fractional order. For example, when α
=
0, F
0
2 is the classical Fock space F
2. Thus in this paper, we always omit discussing the case of α
=
0 and the similar result of this case is obtained in [11, 14].Throughout this paper we writeX
≲
Y or Y
≳
X for nonnegative quantities X and Y whenever there is a constant C
>
0 independent of X and Y such that X
≤
C
Y. Similarly we write X
≃
Y if X
≲
Y and Y
≲
X.


## 2. Proof of the Main Result

We begin with some properties of the Fock-Sobolev spacesF
α
2. See [1] for more information.Lemma 1.
Suppose thatf belongs to the Fock-Sobolev type space F
α
p for any real α. Then for any z
,
w
∈
C
n, we have (18)
f
z
p
e
-
p
/
2
z
2
1
1
+
z
α
≲
f
F
α
p
p
,and when α
<
0,(19)
K
z
α
w
≲
1
+
z
w
α
/
2
e
x
p
1
2
z
2
+
1
2
w
2
-
1
8
z
-
w
2
,when α
>
0,(20)
K
z
α
w
≲
1
+
w
·
z
¯
α
/
2
e
x
p
1
2
z
2
+
1
2
w
2
-
1
8
z
-
w
2
.More specifically, (21)
K
z
α
z
≃
1
+
z
α
e
z
2for any z
∈
C
n and there is a r
>
0 such that (22)
K
z
α
w
≳
1
+
z
α
e
x
p
1
2
z
2
+
1
2
w
2
,for any z
∈
B
(
w
,
r
)
.A consequence of the first estimate in Lemma1 is that, for any function u
∈
F
α
2, the Toeplitz operators T
u and T
u
¯ are both densely defined on F
α
2.Lemma 2.
If the functionu belongs to the Fock-Sobolev type space F
α
2, we then have (
T
u
α
)
∗
=
T
u
¯
α.Proof.
In views of the Lemma  3.4 in [1], we can calculate that, for any polynomial f and g
∈
F
∞ (see [1] for the definition of F
∞), (23)
T
u
α
∗
f
,
g
α
=
∫
C
n
u
¯
z
f
z
g
z
¯
e
-
z
2
z
α
d
v
z
=
T
u
¯
α
f
,
g
α
, if α
≤
0, and if α
>
0, (24)
T
u
α
∗
f
,
g
α
=
∫
C
n
u
¯
z
f
α
/
2
-
z
g
α
/
2
-
z
¯
e
-
z
2
d
v
z
+
∫
C
n
u
¯
z
f
α
/
2
+
z
g
α
/
2
+
z
¯
e
-
z
2
z
α
d
v
. Lastly the fact that the set of all holomorphic polynomials is dense in F
α
2 completes the proof.Lemma 3.
For givenu
,
v
∈
F
α
2, if T
u
α
T
v
¯
α is bounded on F
α
2, then T
u
α
T
v
¯
α
K
z
α
(
w
)
=
u
(
w
)
v
¯
(
z
)
K
z
α
(
w
) for any z
,
w
∈
C
n.Proof.
Whenα
≤
0, in view of reproducing properties of K
z
α
(
w
), Lemma 3.4, the claim (
d
)
⇒
(
a
) of Lemma  3.10 in [1], and Lemma 2, we see that (25)
T
u
α
T
v
¯
α
K
z
α
w
=
T
u
α
T
v
¯
α
K
z
α
,
K
w
α
α
=
v
¯
z
K
z
α
,
u
¯
w
K
w
α
α
=
u
w
v
¯
z
K
z
α
w
.
On the other side, whenα
>
0, we have to use the Lemma 3.4, the claim (
d
)
⇒
(
a
) of Lemma  3.10 in [1] to achieve that, if u
,
v
∈
F
α
2, T
u
α
T
v
¯
α is bounded, (26)
T
u
α
T
v
¯
α
K
z
α
,
K
w
α
α
=
∫
C
n
u
λ
T
v
¯
α
K
z
α
α
/
2
-
λ
K
w
α
α
/
2
-
λ
¯
e
-
λ
2
d
v
λ
+
∫
C
n
u
λ
T
v
¯
α
K
z
α
α
/
2
+
λ
K
w
α
α
/
2
+
λ
¯
e
-
λ
2
λ
α
d
v
λ
.Together with (
3.5
) in [1], Fubini’s theorem, and the reproducing property, we can see that (27)
∫
C
n
u
λ
T
v
¯
α
K
z
α
α
/
2
+
λ
K
w
α
α
/
2
+
λ
¯
e
-
λ
2
λ
α
d
v
λ
=
∫
C
n
∫
C
n
u
λ
v
¯
ξ
K
z
α
α
/
2
+
ξ
K
ξ
α
α
/
2
+
λ
K
w
α
α
/
2
+
λ
¯
e
-
λ
2
λ
α
e
-
ξ
2
ξ
α
d
v
ξ
d
v
λ
=
∫
C
n
u
λ
K
ξ
α
α
/
2
+
λ
K
w
α
α
/
2
+
λ
¯
e
-
λ
2
λ
α
d
v
λ
∫
C
n
v
¯
ξ
K
z
α
α
/
2
+
ξ
e
-
ξ
2
ξ
α
d
v
ξ
=
u
w
∫
C
n
v
¯
ξ
K
ξ
α
α
/
2
+
w
K
z
α
α
/
2
+
ξ
e
-
ξ
2
ξ
α
d
v
ξ
=
u
w
v
¯
z
K
z
α
α
/
2
+
w
.Similarly, we can achieve that (28)
∫
C
n
u
λ
T
v
¯
α
K
z
α
α
/
2
-
λ
K
w
α
α
/
2
-
λ
¯
e
-
λ
2
d
v
λ
=
u
w
v
¯
z
K
z
α
α
/
2
-
w
.Therefore, (29)
T
u
α
T
v
¯
α
K
z
α
w
=
T
u
α
T
v
¯
α
K
z
α
,
K
w
α
α
=
u
w
v
¯
z
K
z
α
w
.Theorem 4 ((1) ⇒ (2)).
If we give thatu and v are two nonzero functions in the Fock-Sobolev type space F
α
2 such that Toeplitz product T
u
α
T
v
¯
α is bounded on F
α
2; then there is a complex linear polynomial q
(
z
) on C
n such that u
=
e
q and v
=
C
e
-
q, where C is a nonzero complex constant.Proof.
This proof is similar to Theorem2.4 in [12] and here we only give its brief illustration.
If the condition holds that the Toeplitz productT
u
α
T
v
¯
α is bounded on F
α
2, by Lemmas 2 and 3 and the Cauchy-Schwarz inequality, we can see that (30)
u
z
v
z
=
T
u
α
T
v
¯
α
~
≤
T
u
α
T
v
¯
α
<
∞
. Sequentially, the local property of reproducing kernel K
z shows us that the module of function (31)
T
z
,
w
=
T
u
α
T
v
¯
α
K
w
α
,
K
z
α
α
K
z
α
z
K
w
α
w is equivalent to e
q
(
z
)
-
q
(
w
)
¯ when z
-
w
<
ϵ
0. It implies that T
z
,
w is bounded in that situation.
On the other side, we give the representation of quadratic polynomial in the case of real inner product as follows:q
2
(
z
)
=
A
z
,
z, where q
=
q
1
+
q
2, q
1 is linear, q
2 is a homogeneous polynomial of degree 2, and A
=
A
n
×
n is a complex matrix symmetric in the real sense. After we choose w
=
r
ξ and z
=
r
ξ
+
(
ϵ
0
/
2
)
η, where r is any real positive number, we achieve that (32)
e
q
z
-
q
w
¯
=
M
e
x
p
r
ϵ
0
A
ξ
,
η is not bounded as r
→
∞. This contradiction finishes the proof.Theorem 5 ((2) ⇒ (1)).
Ifu
=
e
q and v
=
e
-
q where q is a complex linear polynomial on C
n, then T
u
α
T
v
¯
α is bounded on the Fock-Sobolev type space F
α
2.Proof.
To prove the boundedness ofT
u
α
T
v
¯
α, we will sufficiently obtain that T
u
α
T
v
¯
α
f
F
α
2
~ is bounded by means of the idea of [1]. In fact we only discuss the case of α
>
0 because the other case is the same as the proof of Theorem  2.5 in [12]. Using the similar ways, our goal is to obtain that (33)
T
u
α
T
v
¯
α
f
F
α
2
~
=
∫
C
n
T
u
α
T
v
¯
α
f
α
/
2
-
z
e
-
1
/
2
z
2
2
d
v
z
1
/
2
+
∫
C
n
T
u
α
T
v
¯
α
f
α
/
2
+
z
e
-
1
/
2
z
2
z
-
α
/
2
2
d
v
z
1
/
2 is bounded for any f
∈
F
α
2 in view of the definition of the norm ·
F
α
2
~. To the end, we focus our attention on the integrands in it. By formulae (
3.4
)  (
3.5
) in [1] and the definition of Toeplitz operator, the integrands in the norm are (34)
T
u
α
T
v
¯
α
f
α
/
2
-
z
=
∫
C
n
v
¯
η
f
α
/
2
-
η
∫
C
n
u
w
K
w
α
α
/
2
-
η
¯
K
z
α
α
/
2
-
w
¯
e
-
w
2
d
v
w
e
-
η
2
d
v
η
,
T
u
α
T
v
¯
α
f
α
/
2
+
z
=
∫
C
n
v
¯
η
f
α
/
2
+
η
∫
C
n
u
w
K
w
α
α
/
2
+
η
¯
K
z
α
α
/
2
+
w
¯
e
-
w
2
w
α
d
v
w
e
-
η
2
η
α
d
v
η
.
By the reproducing property, the estimations of their module are, respectively, coming from(35)
T
u
α
T
v
¯
α
f
α
/
2
-
z
≲
∫
C
n
u
z
v
¯
η
f
α
/
2
-
η
K
z
α
α
/
2
-
η
e
-
η
2
d
v
η
, and then, similarly, (36)
T
u
α
T
v
¯
α
f
α
/
2
+
z
≲
∫
C
n
u
z
v
¯
η
f
α
/
2
+
η
K
z
α
α
/
2
+
η
e
-
η
2
η
α
d
v
η
.Therefore, using the Cauchy-Schwarz inequality, we have the estimation of the first term of the norm as follows: (37)
∫
C
n
T
u
α
T
v
¯
α
f
α
/
2
-
z
e
-
1
/
2
z
2
2
d
v
z
≲
∫
C
n
∫
C
n
H
α
1
w
,
z
d
v
w
∫
C
n
H
α
1
w
,
z
f
α
/
2
-
w
2
e
-
w
2
d
v
w
d
v
z
,where H
α
1
(
w
,
z
)
=
e
-
1
/
2
z
2
K
z
α
α
/
2
-
w
e
-
(
1
/
2
)
w
2
e
x
p
(
R
e
(
q
(
z
)−  q
(
w
)
¯
)
). Similarly the estimation of the second norm has been achieved that (38)
∫
C
n
T
u
α
T
v
¯
α
f
α
/
2
+
z
e
-
1
/
2
z
2
z
-
α
/
2
2
d
v
z
≲
∫
C
n
∫
C
n
H
α
2
w
,
z
d
v
w
∫
C
n
H
α
2
w
,
z
f
α
/
2
+
w
2
e
-
w
2
w
α
d
v
w
d
v
z
,where H
α
2
w
,
z
=
e
-
1
/
2
z
2
z
-
α
/
2
K
z
α
α
/
2
+
w
e
-
1
/
2
w
2 ·
w
-
α
/
2
exp
⁡
(
R
e
(
q
(
z
)
-
q
(
w
)
¯
)
).
If we can affirm both(39)
s
u
p
z
∈
C
n
∫
C
n
H
α
1
z
,
w
d
v
w
<
∞
,
s
u
p
z
∈
C
n
∫
C
n
H
α
2
z
,
w
d
v
w
<
∞
, we would finish the proof because (40)
T
u
α
T
v
¯
α
f
F
α
2
~
2
≲
∫
C
n
f
α
/
2
-
w
2
e
-
w
2
d
v
w
+
∫
C
n
f
α
/
2
+
w
2
e
-
w
2
w
α
d
v
w
. To the finish, in terms of Lemma 1 and transformation, we can assert that (41)
s
u
p
z
∈
C
n
∫
C
n
H
α
1
z
,
w
d
v
w
≲
s
u
p
z
∈
C
n
∫
C
n
1
+
w
·
z
¯
α
/
2
e
-
1
/
8
z
-
w
2
e
q
z
-
w
d
v
w
<
∞
,
s
u
p
z
∈
C
n
∫
C
n
H
α
2
z
,
w
d
v
w
≲
s
u
p
z
∈
C
n
∫
C
n
z
-
α
/
2
w
-
α
/
2
e
-
1
/
8
z
-
w
2
1
+
z
·
w
-
α
/
2
e
q
z
-
w
d
v
w
<
∞
.This implies that T
u
T
v
¯
f
F
α
2
~ is bounded and completes the proof.Theorem 6 ((1) ⇒ (3)).
Ifu and v are two functions in the Fock-Sobolev type space F
α
2, not identically zero, such that the operator T
u
α
T
v
¯
α is bounded on F
α
2, then u
2
~
(
z
)
v
2
~
(
z
) is a bounded function on the complex space.Proof.
We omit the proof here for it is analogous to Theorem2.6 in [12].Theorem 7 ((3) ⇒ (2)).
Suppose thatu and v are two functions in the Fock-Sobolev type space F
α
2, not identically zero, such that u
2
~
(
z
)
v
2
~
(
z
) is bounded on C
n. Then there is a complex linear polynomial q
(
z
) on C
n satisfying u
=
e
q and v
=
C
e
-
q, where C is a nonzero complex constant.Proof.
Now we only consider the case ofα
>
0 while the other case would be referred to in Theorem  2.7 in [12].
It is easy to see that, for anyu
∈
F
α
2, u
~
(
z
)
=
T
u
α
k
z
α
,
k
z
α
α
=
u
(
z
)
. When α
>
0, we use the triangle inequality and Hölder’s inequality to calculate (42)
u
z
2
≲
∫
C
n
u
w
k
z
α
α
/
2
-
w
2
e
-
w
2
d
v
w
∫
C
n
k
z
α
α
/
2
-
w
2
e
-
w
2
d
v
w
+
∫
C
n
u
w
k
z
α
α
/
2
+
w
2
e
-
w
2
w
α
d
v
w
∫
C
n
k
z
α
α
/
2
+
w
2
e
-
w
2
w
α
d
v
w
.Because k
z
α is a unit element, that is, (43)
k
z
α
F
α
2
~
=
∫
C
n
k
z
α
α
/
2
-
w
2
e
-
w
2
d
v
w
1
/
2
+
∫
C
n
k
z
α
α
/
2
+
w
2
e
-
w
2
w
α
d
v
w
1
/
2
,we can see that (44)
∫
C
n
k
z
α
α
/
2
-
w
2
e
-
w
2
d
v
w
≲
1
,
∫
C
n
k
z
α
α
/
2
+
w
2
e
-
w
2
w
α
d
v
w
≲
1
. From the above inequations, the estimate of u
z
2 turns into (45)
u
z
2
≲
∫
C
n
u
w
2
k
z
α
α
/
2
-
w
2
e
-
w
2
d
v
w
+
∫
C
n
u
w
2
k
z
α
α
/
2
+
w
2
e
-
w
2
w
α
d
v
w
=
u
2
~
z
. If u
2
~
(
z
)
v
2
~
(
z
) is a bounded function on C
n, v
2
~
(
z
)
u
2
(
z
) and u
2
(
z
)
v
2
(
z
) are both bounded on C
n. By Liouville’s theorem, the boundedness of u
2
(
z
)
v
2
(
z
) implies that there exists a constant C such that u
v
=
C. Since neither u nor v is identically zero, we have C
≠
0. That is, both u and v are nonvanishing. By Lemma 1, there exists a complex polynomial q
(
z
) on C
n with deg
⁡
(
q
)
≤
2 such that u
=
e
q and v
=
C
e
-
q.
On the other side, by the definition of Berezin transformation in this case,(46)
u
2
~
z
v
2
z
=
∫
C
n
u
w
v
¯
z
k
z
α
α
/
2
-
w
2
e
-
w
2
d
v
w
+
∫
C
n
u
w
v
¯
z
k
z
α
α
/
2
+
w
2
e
-
w
2
w
α
d
v
w
. Now giving a sufficiently small δ
>
0, we further obtain that (47)
u
2
~
z
v
2
z
≳
∫
w
>
δ
u
w
v
¯
z
k
z
α
α
/
2
-
w
2
e
-
w
2
d
v
w
+
∫
w
>
δ
u
w
v
¯
z
k
z
α
α
/
2
+
w
2
e
-
w
2
w
α
d
v
w
≳
∫
w
>
δ
u
w
v
¯
z
k
z
α
w
2
e
-
w
2
w
α
d
v
w
≳
∫
w
>
δ
1
+
z
-
α
/
2
e
-
1
/
2
z
2
K
z
α
w
e
-
1
/
2
w
2
2
e
q
w
-
q
z
w
α
/
2
2
d
v
w
. When choosing a constant ε
>
0 satisfying Lemma 1, we can see that (48)
u
2
~
z
v
2
z
≳
∫
z
-
w
<
ε
w
>
δ
1
+
z
-
α
/
2
e
-
1
/
2
z
2
K
z
α
w
e
-
1
/
2
w
2
2
e
q
w
-
q
z
w
α
/
2
2
d
v
w
+
∫
z
-
w
≥
ε
w
>
δ
1
+
z
-
α
/
2
e
-
1
/
2
z
2
K
z
α
w
e
-
1
/
2
w
2
2
e
q
w
-
q
z
w
α
/
2
2
d
v
w
.Using the similar method like the case α
<
0, we can get the desired results and the proof is finished at this moment.


## 3. Conclusions

In this content, we deal with the Sarason’s problem on the Fock-Sobolev type spaces and have a complete solution thatu
=
e
q
,
v
=
C
e
-
q, where q is a linear complex polynomial and C is a nonzero constant. As stated in [1], we know that the Fock-Sobolev type space F
α
2 clearly does not fall under the class of weighted Fock spaces F
φ
2. Therefore the Sarason’s problem of weighted Fock spaces F
φ
2 is still open. We will focus on this open problem in the future study.



---

*Source: 1571365-2018-05-02.xml*