# The Improvement on the Boundedness and Norm of a Class of Integral Operators onL

Authors: Lifang Zhou; Jin Lu
Journal: Journal of Function Spaces
 (2015)

---

## Abstract

We prove the condition “c is neither 0 nor a negative integer” can be dropped on the boundedness of a class of integral operators S
a
,
b
,
c on L
p space, which improves the result by Krues and Zhu. Besides, the exact norm of S
a
,
b
,
c on L
p space is also obtained under the assumption c
=
n
+
1
+
a
+
b.


---

## Body

## 1. Introduction

LetB
n be the open unit ball in the complex space C
n. The measure,(1)
d
ν
t
=
1
-
z
2
t
d
ν
z
,denotes the weighted Lebesgue measure on B
n, where t is real parameter and ν is the normalized Lebesgue measure on B
n such that v
(
B
n
)
=
1. It is easy to know d
ν
t is finite if and only if t
>
-
1. Suppose 1
≤
p
<
∞; to simplify the notation, we write L
t
p
:
=
L
p
(
B
n
,
v
t
) for the weighted L
p-space under the measure ν
t on B
n and L
p
:
=
L
0
p for the usual L
p-space under the measure ν.Supposea
,
b
,
c are real numbers, and a class of integral operators is defined by(2)
S
a
,
b
,
c
f
z
=
1
-
z
2
a
∫
B
n
‍
1
-
w
2
b
1
-
z
,
w
c
f
w
d
ν
w
.The class of integral operators is introduced by Kures and Zhu [1]. And it is closely related to “maximal Bergman projection” and Berezin transform. In fact, the boundedness of Bergman projection on L
α
p comes from the boundedness of the operator(3)
P
α
♯
f
z
=
Γ
n
+
α
+
1
n
!
Γ
α
+
1
∫
B
n
‍
f
w
1
-
z
,
w
n
+
1
+
α
d
v
α
w
,
iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
α
>
-
1
,on L
α
p; see [2]. Therefore, we can call P
α
♯ by “maximal Bergman projection,” which is the particular case of S
a
,
b
,
c. Berezin transforms, whatever the case of the unit disk [3, page 141] or the case of unit ball ([4, page 76], [5, page 383]), are all concluded in the form of S
a
,
b
,
c with special a
,
b
,
c.In [1], Krues and Zhu gave the sufficient and necessary conditions of the boundedness of operator S
a
,
b
,
c.Theorem A (see [1]).
Supposec is neither 0 nor a negative integer. (1)
The operatorS
a
,
b
,
c is bounded on L
t
p
(
1
<
p
<
∞
) if and only if -
p
a
<
t
+
1
<
p
(
b
+
1
)
,
c
≤
n
+
1
+
a
+
b.
(2)
The operatorS
a
,
b
,
c is bounded on L
t
1 if and only if -
a
<
t
+
1
<
b
+
1
,
c
=
n
+
1
+
a
+
b or -
a
<
t
+
1
≤
b
+
1
,
c
<
n
+
1
+
a
+
b.The main purposes of this note contain two parts. One part is to prove the condition “c is neither 0 nor a negative integer” in Theorem A can be removed; see Section 3. The other part is to give the accurate norm of the operator S
a
,
b
,
c on L
t
p under the assumption c
=
n
+
1
+
a
+
b, which can be seen from the following two theorems.Theorem 1.
Supposec
=
n
+
1
+
a
+
b. If 1
≤
p
<
∞ and -
p
a
<
t
+
1
<
p
(
b
+
1
), then(4)
S
a
,
b
,
c
L
t
p
→
L
t
p
=
n
!
Γ
a
+
t
+
1
/
p
Γ
b
+
1
-
t
+
1
/
p
Γ
2
n
+
1
+
a
+
b
/
2
.Else, we also give the sufficient and necessary conditions of the operatorS
a
,
b
,
c on L
∞ and the accurate norm under c
=
n
+
1
+
a
+
b of this case, where L
∞ denotes the set of all essentially bounded and measurable functions under the measure ν
t on B
n.Theorem 2.
The operatorS
a
,
b
,
c is bounded on L
∞ if and only if a
>
0, b
>
-
1, and c
=
n
+
1
+
a
+
b or a
≥
0, b
>
-
1, and c
<
n
+
1
+
a
+
b. Moreover, when c
=
n
+
1
+
a
+
b, we have(5)
S
a
,
b
,
c
L
∞
→
L
∞
=
n
!
Γ
a
Γ
1
+
b
Γ
2
n
+
1
+
a
+
b
/
2
.NoticeS
a
,
b
,
c is the generalization of “maximal Bergman projection” and Berezin transform which was first introduced by Berezin [6]. The boundedness of Berezin transform of f
∈
L
1
(
D
) is a well-known fact; see [7, Proposition 2.2]. But the norm of it was not calculated out until 2008 by Dostanić; see [8, Corollary 2]. Recently, the result by Dostanić has been extended to several complex variables in [9, Theorem 1.1]. Thus, Theorems 1 and 2 promote the main results in [8, 9]. And they also imply the following corollary.Corollary 3.
Suppose1
≤
p
<
∞, α
>
-
1, and the norm of P
α
♯ on L
α
p can be(6)
P
α
♯
L
α
p
→
L
α
p
=
Γ
α
+
1
/
p
Γ
α
+
1
-
α
+
1
/
p
Γ
n
+
α
+
1
Γ
2
n
+
1
+
α
/
2
Γ
(
α
+
1
)
,which implies ∥
P
α
♯
∥
L
α
p
→
L
α
p grows at most like (
α
+
1
)
-
1 as α
→
-
1.Next, we will see that the boundedness of an operator called Berezin-type transform onL
t
p can also be obtained from our main results. The Berezin-type transform is defined by(7)
B
k
,
α
,
β
f
z
=
C
k
,
α
,
β
×
∫
B
n
‍
1
-
z
2
n
+
α
+
β
+
k
+
1
1
-
w
2
k
1
-
z
,
w
n
+
α
+
k
+
1
1
-
w
,
z
n
+
β
+
k
+
1
f
w
d
ν
w
,where(8)
C
k
,
α
,
β
=
Γ
n
+
α
+
k
+
1
Γ
n
+
β
+
k
+
1
Γ
n
+
1
Γ
k
+
1
Γ
n
+
α
+
β
+
k
+
1
,and n
+
α
+
β
>
0, n
+
α
>
0, n
+
β
>
0, and k
>
-
1. The transform was introduced by Li and Liu [10] when they discuss whether the mean-value property implies (
α
,
β
)-harmonicity for integrable functions on the unit ball in C
n. Notice that(9)
B
k
,
α
,
β
f
z
≤
C
k
,
α
,
β
S
a
,
b
,
c
f
zwith a
=
n
+
α
+
β
+
k
+
1, b
=
k, and c
=
n
+
1
+
a
+
b. And B
k
,
α
,
α
f
(
z
)
=
C
k
,
α
,
α
S
a
,
b
,
c
f
(
z
) as α
=
β. Therefore, the boundedness of Berezin-type transform B
k
,
α
,
β on L
t
p comes from the boundedness of the operator S
a
,
b
,
c on L
t
p. Thus, we have the following result, which extends Propositions 3.3 and 3.4 in [10] combining the fact of Lemma 2.4 in [10] therein.Corollary 4.
If1
≤
p
<
∞ such that -
p
(
n
+
α
+
β
+
k
+
1
)
<
t
+
1
<
p
(
k
+
1
), then the Berezin-type B
k
,
α
,
β is bounded on L
t
p and(10)
B
k
,
α
,
β
L
t
p
→
L
t
p
≤
λ
k
,
α
,
β
,
p
Γ
n
+
α
+
k
+
1
Γ
n
+
β
+
k
+
1
Γ
2
n
+
1
+
α
+
β
/
2
+
k
,where(11)
λ
k
,
α
,
β
,
p
=
Γ
n
+
α
+
β
+
k
+
1
+
t
+
1
/
p
Γ
k
+
1
-
t
+
1
/
p
Γ
n
+
α
+
β
+
k
+
1
Γ
k
+
1
.Moreover, the Berezin-type transform is bounded on L
∞, and(12)
B
k
,
α
,
β
L
∞
→
L
∞
≤
Γ
n
+
α
+
k
+
1
Γ
n
+
β
+
k
+
1
Γ
2
n
+
1
+
α
+
β
/
2
+
k
.


## 2. Preliminaries

A number of hypergeometric functions will appear throughout. We use the classical notation2
F
1
α
,
β
;
γ
;
z to denote(13)
2
F
1
α
,
β
;
γ
;
z
=
∑
k
=
0
∞
‍
α
k
β
k
γ
k
z
k
k
!
,with γ
≠
0
,
-
1
,
-
2
,
…, where(14)
α
0
=
1
,
α
k
=
α
α
+
1
⋯
α
+
k
-
1
for
k
≥
1
.And the hypergeometric series in (13) converges absolutely for all the value of |
z
|
<
1. Moreover, as |
z
|
→
1
-, it is easy to know that(15)
2
F
1
α
,
β
;
γ
;
z
≈
1
,
if
γ
-
α
-
β
>
0
;
log
⁡
1
1
-
z
,
if
γ
-
α
-
β
=
0
;
1
-
z
γ
-
α
-
β
,
if
γ
-
α
-
β
<
0
,where a
(
z
)
≈
b
(
z
) represents the ratio and a
(
z
)
/
b
(
z
) has a positive finite limit as |
z
|
→
1
-. Now we list a few formulas for easy reference (see [11, Chapter II]):(16)
2
F
1
α
,
β
;
γ
;
1
=
Γ
γ
Γ
γ
-
α
-
β
Γ
γ
-
α
Γ
γ
-
β
,
hh
Re
γ
-
α
-
β
>
0
,
(17)
2
F
1
α
,
β
;
γ
;
z
=
1
-
z
γ
-
α
-
β
2
F
1
γ
-
α
,
γ
-
β
;
γ
;
z
,
(18)
2
F
1
α
,
β
;
γ
;
z
=
Γ
γ
Γ
λ
Γ
γ
-
λ
×
∫
0
1
‍
t
λ
-
1
1
-
t
γ
-
λ
-
1
2
F
1
α
,
β
;
λ
;
t
z
d
t
,
Re
γ
>
Re
λ
>
0
;
arg
1
-
z
<
π
;
z
≠
1
.Lemma 5.
SupposeRe
δ
>
0 and Re
(
λ
+
δ
-
α
-
β
)
>
0. Then(19)
∫
0
1
‍
t
λ
-
1
1
-
t
δ
-
1
2
F
1
α
,
β
;
λ
;
t
d
t
=
Γ
λ
Γ
δ
Γ
λ
+
δ
-
α
-
β
Γ
λ
+
δ
-
α
Γ
λ
+
δ
-
β
.Proof.
Note that, under the assumption of the lemma, both sides of (18) are continuous at z
=
1. The lemma then follows by letting z
→
1 in (18) and applying (16).The following integral formulae concerning the hypergeometric function are significant for our main results. And all these formulae are contained in [12]. Now we list them.Lemma 6 (see [12, Corollary 2.4]).
Forα
∈
R and γ
>
-
1, we have(20)
∫
B
n
‍
1
-
w
2
γ
1
-
z
,
w
2
α
d
ν
w
=
n
!
Γ
1
+
γ
Γ
n
+
1
+
γ
2
F
1
α
,
α
;
n
+
1
+
γ
;
z
2
.Lemma6 is also contained implicitly in the proof of Theorem 1.4.10 in [13] (see the formula in page 19, line 5 of [13]).Lemma 7 (see [12, Corollary 2.5]).
Suppose thatα
,
β
>
0, γ
∈
R, and n
+
α
+
β
-
2
γ
>
0. Then (21)
∫
B
n
‍
z
2
β
1
-
z
2
α
-
1
∫
B
n
‍
1
-
w
2
β
-
1
1
-
z
,
w
2
γ
d
ν
w
d
ν
z
=
n
n
!
Γ
α
Γ
β
Γ
n
+
α
+
β
-
2
γ
Γ
2
n
+
α
+
β
-
γ
.Proof.
Using Lemma6 in the inner integral, we have (22)
n
!
Γ
β
Γ
n
+
β
∫
B
n
‍
z
2
β
1
-
z
2
α
-
1
hh
×


2
F
1
γ
,
γ
;
n
+
β
;
z
2
d
ν
z
=
n
n
!
Γ
β
Γ
n
+
β
∫
0
1
‍
r
n
+
β
-
1
1
-
r
α
-
1
hhhhh
×


2
F
1
γ
,
γ
;
n
+
β
;
z
2
d
r
.Then (19) gives the result.The following result, usually called Schur’s test, is a very effective tool in proving theL
p-boundedness of integral operators. See, for example, [3].Lemma 8.
Suppose that(
X
,
μ
) is a σ-finite measure space, K
(
x
,
y
) is a nonnegative measurable function on X
×
X, and T is the associated integral operator:(23)
T
f
x
=
∫
X
‍
K
x
,
y
f
y
d
μ
y
.Let 1
<
p
<
∞ and 1
/
p
+
1
/
q
=
1. If there exist a positive constant C and a positive measurable function u on X such that(24)
∫
X
‍
K
x
,
y
u
y
q
d
μ
y
≤
C
u
x
q
,for almost every x in X, and(25)
∫
X
‍
K
x
,
y
u
x
p
d
μ
x
≤
C
u
y
p
,for almost every y in X, then T is bounded on L
p
(
X
,
μ
) with ∥
T
∥
≤
C.


## 3. The Improvement

The section mainly proposes the condition “c is neither 0 nor a negative integer” can be omitted in Theorem A. Notice the condition is only used to give c
≤
n
+
1
+
a
+
b while proving the necessity for the boundedness of the operator S
a
,
b
,
c on L
t
p
(
1
≤
p
<
∞
); see [1, lemma 12]. Now we will give a new proof of the necessity for the boundedness of S
a
,
b
,
c on L
t
p in Propositions 9 and 11 to introduce the condition can be put off.Proposition 9.
Suppose the operatorS
a
,
b
,
c is bounded on L
t
p
(
1
<
p
<
∞
), and then -
p
a
<
t
+
1
<
p
(
b
+
1
)
,
c
≤
n
+
1
+
a
+
b.Proof.
Letq be the number such that 1
/
p
+
1
/
q
=
1. For any fixed ϵ
>
0, define (26)
g
ϵ
w
=
C
1
ϵ
1
-
w
2
ϵ
-
t
+
1
/
p
,
h
ϵ
z
=
C
2
ϵ
1
-
z
2
ϵ
-
t
+
1
/
q
z
2
b
+
1
+
ϵ
-
t
-
1
/
p
,where(27)
C
1
(
ϵ
)
=
Γ
ϵ
Γ
n
+
1
Γ
n
+
ϵ
-
1
/
p
,
(28)
C
2
(
ϵ
)
=
n
Γ
ϵ
Γ
n
+
q
b
+
1
+
q
ϵ
-
t
+
1
/
p
Γ
n
+
q
b
+
1
+
q
ϵ
-
t
+
1
/
p
+
ϵ
-
1
/
q
.Easy calculation shows ∥
g
ϵ
∥
p
,
t
=
∥
h
ϵ
∥
q
,
t
=
1. Notice the fact (29)
S
a
,
b
,
c
L
t
p
→
L
t
p
=
sup
⁡
∥
f
∥
p
,
t
=
1
∥
g
∥
q
,
t
=
1
×
1
-
w
2
b
-
t
1
-
z
,
w
c

1
-
w
2
b
-
t
1
-
z
,
w
c

∫
B
n
‍
1
-
w
2
b
-
t
1
-
z
,
w
c

∫
B
n
‍
1
-
z
2
a
iiiii
×
1
-
w
2
b
-
t
1
-
z
,
w
c

f
w
d
ν
t
w
g
z
¯
d
v
t
z
.Then the boundedness of the operator S
a
,
b
,
c on L
t
p leads to the integral(30)
∫
B
n
‍
∫
B
n
‍
1
-
z
2
a
1
-
w
2
b
-
t
1
-
z
,
w
c
g
ϵ
w
d
ν
t
w
×
h
ϵ
z
¯
d
ν
t
z
∫
B
n
‍
1
-
z
2
a
1
-
w
2
b
-
t
1
-
z
,
w
c
g
ϵ
w
d
ν
t
w
∫
B
n
‍
≤
S
a
,
b
,
c
L
t
p
→
L
t
p
<
+
∞
.Hence, using Lemma 7 with α
=
a
+
ϵ
/
q
+
(
t
+
1
)
/
p, β
=
b
+
1
+
(
ϵ
-
(
t
+
1
)
)
/
p, and γ
=
c
/
2, we can conclude that(31)
a
+
ϵ
q
+
t
+
1
p
>
0
,
b
+
1
+
ϵ
-
t
+
1
p
>
0
,
n
+
1
+
a
+
b
+
ϵ
-
c
>
0
.Then the arbitrariness of ϵ gives(32)
-
p
a
≤
t
+
1
≤
p
b
+
1
,
c
≤
n
+
1
+
a
+
b
.
Now, we will give the proof by dividing into the following two cases.
Whenc
=
n
+
1
+
a
+
b, by Lemma 7, the integral in (30) equals (33)
n
!
Γ
a
+
ϵ
/
q
+
t
+
1
/
p
Γ
b
+
1
+
ϵ
-
t
+
1
/
p
Γ
2
n
+
1
+
a
+
b
/
2
+
ϵ
×
Γ
n
+
ϵ
Γ
n
1
/
p
×
Γ
n
+
q
b
+
1
+
q
ϵ
-
t
+
1
/
p
+
ϵ
Γ
n
+
q
b
+
1
+
q
ϵ
-
t
+
1
/
p
1
/
q
.Then letting ϵ
→
0
+, by (30), we can know the limits(34)
0
≤
lim
ϵ
→
0
+
n
!
Γ
a
+
ϵ
/
q
+
t
+
1
/
p
Γ
b
+
1
+
ϵ
-
t
+
1
/
p
Γ
2
n
+
1
+
a
+
b
/
2
+
ϵ
≤
S
a
,
b
,
c
L
t
p
→
L
t
p
.Then the boundedness of the operator S
a
,
b
,
c gives -
p
a
<
t
+
1
<
p
(
b
+
1
).
Whenc
<
n
+
1
+
a
+
b, take the function(35)
f
λ
z
=
1
-
z
2
λ
,with λ
>
a. The condition (32) implies the function f
λ
∈
L
t
p. And using Lemma 6, we have (36)
S
a
,
b
,
c
f
λ
z
=
1
-
z
2
a
∫
B
n
‍
1
-
w
2
b
1
-
z
,
w
c
f
λ
w
d
v
w
=
1
-
z
2
a
n
!
Γ
1
+
b
+
λ
Γ
n
+
1
+
b
+
λ
×


2
F
1
c
2
,
c
2
;
n
+
1
+
b
+
λ
;
z
2
.According to (15), we can obtain that S
a
,
b
,
c
f
λ
(
z
)
≈
(
1
-
|
z
|
2
)
a. Thus the boundedness of the operator S
a
,
b
,
c on L
t
p
(
B
n
) gives that p
a
+
t
>
-
1; that is, -
p
a
<
t
+
1. Now we consider the adjoint operator S
a
,
b
,
c
∗ of the operator S
a
,
b
,
c; that is,(37)
S
a
,
b
,
c
∗
f
z
=
1
-
z
2
b
-
t
∫
B
n
‍
1
-
w
2
a
+
t
1
-
z
,
w
c
f
w
d
v
w
.The boundedness of S
a
,
b
,
c on L
t
p implies the boundedness of S
a
,
b
,
c
∗ on L
t
q. With the similar discussion above, we can obtain that q
(
b
-
t
)
+
t
>
-
1; that is, t
+
1
<
p
(
b
+
1
).Whenc
=
n
+
1
+
a
+
b, (34) implies the following result.Corollary 10.
Supposec
=
n
+
1
+
a
+
b and 1
<
p
<
∞, -
p
a
<
t
+
1
<
p
(
b
+
1
), and then(38)
S
a
,
b
,
c
L
t
p
→
L
t
p
≥
n
!
Γ
a
+
t
+
1
/
p
)
Γ
(
b
+
1
-
t
+
1
/
p
Γ
2
n
+
1
+
a
+
b
/
2
.Proposition 11.
The operatorS
a
,
b
,
c is bounded on L
t
1 if and only if -
a
<
t
+
1
<
b
+
1
,
c
=
n
+
1
+
a
+
b or -
a
<
t
+
1
≤
b
+
1
,
c
<
n
+
1
+
a
+
b. And when c
=
n
+
1
+
a
+
b, we have(39)
S
a
,
b
,
c
L
t
1
→
L
t
1
=
n
!
Γ
1
+
a
+
t
Γ
b
-
t
Γ
2
n
+
1
+
a
+
b
/
2
.When c
<
n
+
1
+
a
+
b
,
-
a
<
t
+
1
=
b
+
1, we have(40)
S
a
,
b
,
c
L
t
1
→
L
t
1
=
n
!
Γ
1
+
a
+
b
Γ
σ
Γ
2
n
+
1
+
a
+
b
+
σ
/
2
,where σ
=
(
n
+
1
+
a
+
b
)
-
c.Proof.
By Lemma6, we have(41)
S
a
,
b
,
c
L
t
1
→
L
t
1
=
S
a
,
b
,
c
∗
L
∞
→
L
∞
=
sup
⁡
z
∈
B
n
1
-
z
2
b
-
t
∫
B
n
‍
1
-
w
2
a
+
t
1
-
z
,
w
c
d
ν
w
=
n
!
Γ
1
+
a
+
t
Γ
n
+
1
+
a
+
t
sup
⁡
z
∈
B
n
1
-
z
2
b
-
t
×


2
F
1
c
2
,
c
2
;
n
+
1
+
a
+
t
;
z
2
,where S
a
,
b
,
c
∗ denotes the adjoint operator of S
a
,
b
,
c. Then, using (15), we can obtain that the operator S
a
,
b
,
c is bounded on L
t
1 if and only if(42)
1
+
a
+
t
>
0
,
b
-
t
>
0
,
n
+
1
+
a
+
t
-
c
≥
t
-
b
,or(43)
1
+
a
+
t
>
0
,
b
-
t
=
0
,
n
+
1
+
a
+
t
-
c
>
0
,which gives the first part of the proposition.
Now we will give the second part. Whenc
<
n
+
1
+
a
+
b and -
a
<
t
+
1
=
b
+
1, the hypergeometric function in (41) is increasing since its Taylor coefficients are all positive. Applying (16), we have (40). When c
=
n
+
1
+
a
+
b, (17) gives (44)
2
F
1
n
+
1
+
a
+
b
2
,
n
+
1
+
a
+
b
2
;
n
+
1
+
a
+
t
;
z
2
=
1
-
z
2
t
-
b
×


2
F
1
n
+
1
+
a
-
b
2
+
t
,
n
+
1
+
a
-
b
2
+
t
;
hh
n
+
1
+
a
+
t
;
z
2
n
+
1
+
a
-
b
2
.Thus (41), the increase of the last hypergeometric function, and (16) lead to (45)
S
a
,
b
,
c
L
t
1
→
L
t
1
=
S
a
,
b
,
c
∗
L
∞
→
L
∞
=
n
!
Γ
1
+
a
+
t
Γ
n
+
1
+
a
+
t
×


2
F
1
n
+
1
+
a
-
b
2
+
t
,
n
+
1
+
a
-
b
2
+
t
;
n
+
1
+
a
+
t
;
1
n
+
1
+
a
-
b
2
=
n
!
Γ
1
+
a
+
t
Γ
b
-
t
Γ
2
n
+
1
+
a
+
b
/
2
.


## 4. The Proof of Theorems1 and 2

Proof of Theorems1 and 2.
Since(46)
S
a
,
b
,
c
L
∞
→
L
∞
=
sup
⁡
z
∈
B
n
1
-
z
2
a
∫
B
n
‍
1
-
w
2
b
1
-
z
,
w
c
d
ν
w
,therefore Theorem 2 comes out as the same discussion as Proposition 11.
Next, we will concentrate on the proof of Theorem1. Remember the hypothesis c
=
n
+
1
+
a
+
b throughout the following proof. Since (39) gives the case of p
=
1, for the case 1
<
p
<
∞, Corollary 10 gives the lower bound of ∥
S
a
,
b
,
c
∥
L
t
p
→
L
t
p. Thus we only show the fact (47)
S
a
,
b
,
c
L
t
p
→
L
t
p
≤
n
!
Γ
a
+
t
+
1
/
p
Γ
b
+
1
-
t
+
1
/
p
Γ
2
n
+
1
+
a
+
b
/
2
.To this end, we will use Schur’s test (Lemma 8) with(48)
K
z
,
w
=
1
-
z
2
a
1
-
w
2
b
-
t
1
-
z
,
w
n
+
1
+
a
+
b
.Set(49)
u
t
z
=
1
-
z
2
-
t
+
1
/
p
q
,where q is the conjugate exponent of p such that 1
/
p
+
1
/
q
=
1. It then suffices to show(50)
1
-
z
2
a
∫
B
n
‍
1
-
w
2
b
-
t
1
-
z
,
w
n
+
1
+
a
+
b
u
t
w
q
d
ν
t
w
≤
n
!
Γ
a
+
t
+
1
/
p
Γ
b
+
1
-
t
+
1
/
p
Γ
2
n
+
1
+
a
+
b
/
2
u
t
z
q
,for all z
∈
B
n, and(51)
1
-
w
2
b
-
t
∫
B
n
‍
1
-
z
2
a
1
-
z
,
w
n
+
1
+
a
+
b
u
t
z
p
d
ν
t
z
≤
n
!
Γ
a
+
t
+
1
/
p
Γ
b
+
1
-
t
+
1
/
p
Γ
2
n
+
1
+
a
+
b
/
2
u
t
w
pfor all w
∈
B
n. We only prove (50), since (51) comes from the same way as (50). Applying Lemma 6 and (17), we have (52)
1
-
z
2
a
∫
B
n
‍
1
-
w
2
b
-
t
1
-
z
,
w
n
+
1
+
a
+
b
u
t
w
q
d
ν
t
w
=
1
-
z
2
a
n
!
Γ
b
+
1
-
t
+
1
/
p
Γ
n
+
b
+
1
-
t
+
1
/
p
×


2
F
1
n
+
1
+
a
+
b
2
,
n
+
1
+
a
+
b
2
;
hh
n
+
b
+
1
-
t
+
1
p
;
z
2
=
n
!
Γ
(
b
+
1
-
(
t
+
1
)
/
p
)
Γ
n
+
b
+
1
-
t
+
1
/
p
1
-
z
2
-
t
+
1
/
p
×


2
F
1
n
+
1
+
b
-
a
2
-
t
+
1
p
,
n
+
1
+
b
-
a
2
hh
-
t
+
1
p
;
n
+
1
+
b
-
t
+
1
p
;
z
2
.By (16), the last hypergeometric function is bounded from the above by (53)
2
F
1
n
+
1
+
b
-
a
2
-
t
+
1
p
,
n
+
1
+
b
-
a
2
-
t
+
1
p
;
n
+
1
+
b
-
t
+
1
p
;
1
=
Γ
n
+
1
+
b
-
t
+
1
/
p
Γ
a
+
t
+
1
/
p
Γ
2
n
+
1
+
a
+
b
/
2
,since it is increasing on the interval [
0,1
). This proves (50), which in turn implies (47). The proof is completed.


## 5. Remark

The topic on the exact norm of an operator is an interesting but difficult problem. In this note, we only give the accurate norm of the generalized operatorS
a
,
b
,
c on L
t
p under c
=
n
+
1
+
a
+
b. But for other cases, except the particular case (40), we can give an upper bound of ∥
S
a
,
b
,
c
∥
L
t
p
→
L
t
p by Theorem 1 according to the fact (54)
1
-
z
2
a
1
-
w
2
b
-
t
1
-
z
,
w
c
≤
2
σ
1
-
z
2
a
1
-
w
2
b
-
t
1
-
z
,
w
n
+
1
+
a
+
band a lower bound for one fixed ϵ
>
0 by (30) and Lemma 7; thus the problem of the norm of other cases may be left as an open problem to consider.



---

*Source: 362681-2015-04-06.xml*