RIEMANN-STIELTJES OPERATORS BETWEEN
BERGMAN-TYPE SPACES AND α-BLOCH SPACES
SONGXIAO LI
Received 8 December 2005; Revised 11 May 2006; Accepted 28 May 2006
z
z
We study the following integral operators: Jg f (z) = 0 f (ξ)g (ξ)dξ; Ig f (z) = 0 f (ξ)g(ξ)dξ,
where g is an analytic function on the open unit disk in the complex plane. The boundedness and compactness of Jg , Ig between the Bergman-type spaces and the α-Bloch spaces
are investigated.
Copyright © 2006 Hindawi Publishing Corporation. All rights reserved.
1. Introduction
Let D be the open unit disk in the complex plane. Denote by H(D) the class of all analytic
functions on D. An analytic function f in D is said to belong to the α-Bloch space Ꮾα , or
Bloch-type space, if
α f α = sup 1 − |z|2 f (z) < ∞.
(1.1)
z ∈D
The expression f α defines a seminorm while the natural norm is given by f Ꮾα =
| f (0)| + f α . It makes Ꮾα into a Banach space.
A positive continuous function φ on [0,1) is normal, if there exists 0 < s < t such that
(see [7])
φ(r)
↑ ∞,
(1 − r)t
φ(r)
↓ 0,
(1 − r)s
as r −→ 1− .
(1.2)
For 0 < p < ∞ and a normal function φ, let H(p, p,φ) denote the space of all analytic
functions f on D such that
f H(p,p,φ) =
p
f (z) p φ |z| dA(z) < ∞.
1 − |z|
D
(1.3)
Here dA denotes the normalized Lebesgue area measure on the unit disk D such that
A(D) = 1. We call H(p, p,φ) the Bergman-type space. If 1 ≤ p < ∞, H(p, p,φ) is a Banach
Hindawi Publishing Corporation
International Journal of Mathematics and Mathematical Sciences
Volume 2006, Article ID 86259, Pages 1–17
DOI 10.1155/IJMMS/2006/86259
2
Riemann-Stieltjes operators between Bergman and Bloch
space equipped with the norm f H(p,p,φ) . When 0 < p < 1, H(p, p,φ) is a Fréchlet space.
In particular, if φ(r) = (1 − r)1/ p , then H(p, p,φ) is the Bergman space A p .
For an analytic function f (z) on the unit disk D with the Taylor expansion f (z) =
∞
n
n=0 an z , the Cesàro operator acting on f is
Ꮿ f (z) =
∞
n =0
1 ak z n .
n + 1 k =0
n
(1.4)
The integral form of Ꮿ is
Ꮿ( f )(z) =
1
z
z
0
f (ζ)
1
1−ζ
dζ =
1
z
z
0
f (ζ) ln
1
1−ζ
dζ,
(1.5)
taking simply as a path the segment joining 0 and z, we have that
Ꮿ( f )(z) =
1
0
f (tz) ln
1−ζ 1
ζ =tz
dt.
(1.6)
The following operator:
zᏯ( f )(z) =
z
0
f (ζ)
dζ
1−ζ
(1.7)
is closely related to the previous operator and on many spaces the boundedness of these
two operators is equivalent. It is well known that Cesàro operator acts as a bounded linear
operator on various analytic function spaces (see, e.g., [6, 9, 13, 15, 16, 18, 20], and the
references therein).
Suppose that g : D → C1 is an analytic map, f ∈ H(D). A class of integral operator
introduced by Pommerenke is defined by (see [11])
Jg f (z) =
z
0
f dg =
1
0
f (tz)zg (tz)dt =
z
0
f (ξ)g (ξ)dξ,
z ∈ D.
(1.8)
The operator Jg can be viewed as a generalization of the Cesàro operator which was called
the Riemann-Stieltjes operator (see [21]).
In [11], Pommerenke showed that Jg is a bounded operator on the Hardy space H 2
if and only if g ∈ BMOA. Aleman and Siskakis showed that Jg is bounded (compact) on
the Hardy space H p , 1 ≤ p < ∞, if and only if g ∈ BMOA (g ∈ V MOA), and that Jg is
bounded (compact) on the Bergman space A p if and only if g ∈ Ꮾ (g ∈ Ꮾ0 ), see [2, 3].
Recently, Jg acting on various function spaces, including the Bloch space, the weighted
Bergman space, the BMOA, and VMOA spaces have been studied (see [1–3, 17, 22], and
the related references therein).
Another integral operator has recently been defined as the following (see [22]):
Ig f (z) =
z
0
f (ξ)g(ξ)dξ.
(1.9)
In this paper, we study the boundedness and compactness of the operators Jg , Ig between the Bergman-type space and the α-Bloch space.
Songxiao Li
3
Constants are denoted by C in this paper, they are positive and may differ from one
occurrence to the other. a b means that there is a positive constant C such that a ≤ Cb.
Moreover, if both a b and b a hold, then one says that a b.
2. Jg , Ig : H(p, p,φ) → Ꮾα
In this section, we consider the boundedness and compactness of Jg , Ig : H(p, p,φ) → Ꮾα .
First, let us state some useful lemmas.
Lemma 2.1. Assume that 0 < p < ∞ and φ is normal on [0,1). If f ∈ H(p, p,φ), then
f H(p,p,φ)
f (z) ≤ C 1/ p .
φ |z| 1 − |z|2
(2.1)
Proof. Let β(z,w) denote the Bergman metric between two points z and w in D. It is given
by
1 + ϕz (w)
1
.
β(z,w) = log
2
1 − ϕz (w)
(2.2)
For a ∈ D and r > 0, the set D(a,r) = {z ∈ D : β(a,z) < r } is a Bergman metric disk with
center a and radius r. It is well known that (see [25])
2
1 − |a|2
1
1
1
,
2 2 4
|1 − āz|
D(a,r)
1 − |z|2
1 − |a|2
(2.3)
when z ∈ D(a,r). For 0 < r < 1 and z ∈ D, by the subharmonicity of | f (z)| p and the
normality of φ, we get
f (z) p ≤ C
2
1 − |z|2
D(z,r)
f (a) p dA(a)
C
1 − |z|2 φ p |z|
≤
C
D(z,r)
1 − |z|2 φ p |z|
≤
−1
1 − |a|
−1
D
1 − |a|
p
φ p |a| f (a) dA(a)
p
C f H(p,p,φ)
p
,
φ |a| f (a) dA(a) ≤ 1 − |z|2 φ p |z|
p
(2.4)
from which we get the desired result.
The following lemma can be found in [7, Theorem 2].
Lemma 2.2. Assume that 0 < p < ∞ and φ is normal on [0,1). Then for f ∈ H(D),
p
f H(p,p,φ)
p
f (0) +
p
p
f (z) 1 − |z|2 p φ |z| dA(z).
1 − |z|
D
From the proof of Lemmas 2.1 and 2.2, we get the following result.
(2.5)
4
Riemann-Stieltjes operators between Bergman and Bloch
Lemma 2.3. Assume that 0 < p < ∞ and φ is normal on [0,1). If f ∈ H(p, p,φ) and z ∈ D,
then
f (z) ≤ C f H(p,p,φ)
1/ p+1 ,
φ |z| 1 − |z|2
(z ∈ D).
(2.6)
The following lemma can be found in [14].
Lemma 2.4. For β > −1 and m > 1 + β,
1
0
(1 − ρr)−m (1 − r)β dr ≤ C(1 − ρ)1+β−m ,
0 < ρ < 1.
(2.7)
The next lemma can be proved in a standard way (see [5]).
Lemma 2.5. The operator Jg (or Ig ): H(p, p,φ) → Ꮾα is compact if and only if for any
bounded sequence ( fk )k∈N in H(p, p,φ) which converges to zero uniformly on compact subsets of D, Jg fk (or Ig fk ) → 0 in Ꮾα as k → ∞.
Theorem 2.6. Assume that 0 < p < ∞, α > 0, and φ is normal on [0,1). Then the operator
Jg : H(p, p,φ) → Ꮾα is bounded if and only if
α−1/ p
1 − |z|2
sup
φ |z|
z ∈D
g (z) < ∞.
(2.8)
Moreover, the following relationship:
Jg α−1/ p
1 − |z|2
sup
α
H(p,p,φ)→Ꮾ
φ |z|
z ∈D
g (z)
(2.9)
holds.
Proof. By (1.8), it is easy to see that (Jg f ) (z) = f (z)g (z), (Jg f )(0) = 0. Let f (z) ∈
H(p, p,φ). We have
α
α 1 − |z|2
1 − |z|2 Jg f (z) ≤ C f H(p,p,φ) 1/ p g (z).
φ |z| 1 − |z|2
(2.10)
Taking supremum over the unit disk in this inequality, we obtain
Jg f α−1/ p
1 − |z|2
Ꮾα ≤ C f H(p,p,φ) sup
φ |z|
z ∈D
g (z).
(2.11)
Therefore, (2.8) implies that Jg : H(p, p,φ) → Ꮾα is bounded.
Conversely, suppose Jg is a bounded operator from H(p, p,φ) to Ꮾα . For w ∈ D, let
t+1
1 − |w|2
.
fw (z) = φ |w| (1 − wz)1/ p+t+1
(2.12)
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5
It is easy to see that
f (w) =
1
fw (w) = 1/ p ,
φ |w| 1 − |w|2
w
|w |
1
+ t + 1 1/ p+1 .
p
φ |w| 1 − |w|2
(2.13)
By [12], we get
Mp
t+1
1 − |w|2
fw ,r ≤ C t+1 .
φ |w| 1 − r |w|
(2.14)
Since φ is normal, by Lemma 2.4,
p
fw p,p,φ =
1
0
|w|
0
p
r(1 − r)−1 φ p (r)M p fw ,r dr (1 − r)
−1
p
1 − |w |2
p
1 − |w |2
φ p (r) 1 − |w|2
0
p(t+1)
(1 − r)−1 φ p (r)
dr
|w|
(1 − r) pt−1 dr
p(t+1)
1 − r |w|
0
1
(1 − r) pt−1 dr
p(t+1)
1 − r |w|
|w |
(1 − r)
+ 1 − |w|2
+ 1 − |w|2
p(t+1)
1 − |w|2
p(t+1) dr
φ p |w| 1 − r |w|
1
p(t+1) +
|w | 1 − r |w |
φp
0
1
−1
φp
p(t+1)− ps
p(t+1)− ps
φ p (r) 1 − |w|2
|w |
0
dr
p(t+1)
|w | 1 − r |w |
1
1
p(t+1)
(1 − r) ps−1 dr
p(t+1)
1 − r |w|
(1 − r) ps−1 dr
p(t+1) ≤ C.
1 − r |w|
(2.15)
Therefore, fw ∈ H(p, p,φ) (or see [23]). Moreover, there is a positive constant C such that
fw H(p,p,φ) ≤ C. Hence
α 1 − |z|2 fw (z)g (z) ≤ Jg fw Ꮾα ≤ Jg H(p,p,φ)−→Ꮾα fw H(p,p,φ)
(2.16)
for every z,w ∈ D.
From this and (2.13), we have
α
2 α
1/ p g (w) ≤ 1 − |w | fw (w)g (w) ≤ C Jg H(p,p,φ)→Ꮾα ,
φ |w| 1 − |w|2
1 − |w|2
(2.17)
from which (2.8) follows. Combining (2.11) with (2.17), we get (2.9).
Theorem 2.7. Assume that 0 < p < ∞, α > 0, and φ is normal on [0,1). Then Ig : H(p, p,φ)
→ Ꮾα is bounded if and only if
α−1/ p−1
1 − |z|2
sup
φ |z|
z ∈D
g(z) < ∞.
(2.18)
6
Riemann-Stieltjes operators between Bergman and Bloch
Moreover, the following relationship:
Ig H(p,p,φ)→Ꮾα
sup
z ∈D
α−1/ p−1
1 − |z|2
φ |z|
g(z)
(2.19)
holds.
Proof. Similar to the case of Jg , we have (Ig f ) = f (z)g(z), (Ig f )(0) = 0. Assume (2.18)
holds. Let f (z) ∈ H(p, p,φ). Then
2 α
1 − |z|
Ig f
(z) ≤ C f H(p,p,φ) 1 − |z|2
α
φ |z|
1/ p+1
1 − |z|2
g(z).
(2.20)
It follows that Ig : H(p, p,φ) → Ꮾα is bounded.
Conversely, if Ig : H(p, p,φ) → Ꮾα is bounded. For w ∈ D, let fw (z) be defined by
(2.12). From (2.3) and (2.13),
1
+t+1
p
2
|w |2
φ2 |w| 1 − |w|2
2
2(1/ p+1) g(w)
2
= fw (w)g(w) ≤ C
2
1 − |w|2
≤
≤C
D(w,r)
D(w,r)
C
2
1 − |w|2
D(w,r)
2 f (z) g(z)2 dA(z)
w
2 f (z) g(z)2 1 − |z|2 2α w
1
2α dA(z)
1 − |z|2
(2.21)
2 2
dA(z)
2 2α fw (z) g(z)
2α+2 sup 1 − |z|
1 − |z|2
z∈D(w,r)
2
C
2α Ig fw Ꮾα ,
2
1 − |w|
≤
that is,
α
|w | 1 − |w |2
1/ p+1 g(w) ≤ C Ig fw Ꮾα ≤ C Ig H(p,p,φ)→Ꮾα .
φ |w| 1 − |w|2
(2.22)
Taking supremum in the last inequality over the set 1/2 ≤ |w| < 1 and noticing that by
the maximum modulus principle there is a positive constant C independent of g ∈ H(D)
such that
α
α
1 − |w|2
|w | 1 − |w |2
1/ p+1 g(w) ≤ C sup
1/ p+1 g(w),
|w |≤1/2 φ |w | 1 − |w |2
1/2≤|w|<1 φ |w | 1 − |w |2
sup
(2.23)
the result follows. From (2.20) and (2.22), we obtain (2.19).
Songxiao Li
7
Theorem 2.8. Assume that 0 < p < ∞, α > 0, and φ is normal on [0,1). Then the operator
Jg : H(p, p,φ) → Ꮾα is compact if and only if
α−1/ p
1 − |z|2
lim
|z|→1
φ |z|
g (z) = 0.
(2.24)
Proof. First, we assume that (2.24) holds. In order to prove that Jg is compact, by
Lemma 2.5, it suffices to show that if { fn } is a bounded sequence in H(p, p,φ) that converges to 0 uniformly on compact subsects of D, then Jg fn Ꮾα → 0. Let { fn } be a sequence in H(p, p,φ) with fn H(p,p,φ) ≤ 1 and fn → 0 uniformly on compact subsets of
D. By the assumption, for any > 0, there is a constant δ, 0 < δ < 1, such that δ < |z| < 1
implies
α−1/ p
1 − |z|2
φ |z|
g (z) < .
(2.25)
2
Let K = {z ∈ D : |z| ≤ δ }. Note that K is a compact subsect of D and φ is normal, we have
Jg fn Ꮾα
α = sup 1 − |z|2 Jg fn (z)
z ∈D
α ≤ sup 1 − |z|2 g (z) fn (z) + sup
z ∈K
z∈D\K
α
1/ p g (z) f H(p,p,φ)
φ |z| 1 − |z|2
1 − |z|2
s+1/ p fn (z) + C ,
≤ CN sup 1 − |z|2
2
z ∈K
(2.26)
where
N = sup
z ∈D
α−1/ p
1 − |z|2
φ |z|
g (z).
(2.27)
By the assumption and Theorem 2.6, we obtain Jg fn Ꮾα → 0 as n → ∞. Therefore, Jg :
H(p, p,φ) → Ꮾα is compact.
Conversely, suppose Jg : H(p, p,φ) → Ꮾα is compact. Let {zn } be a sequence in D such
that |zn | → 1 as n → ∞. Let
2 t+1
1 − zn fn (z) = 1/ p+t+1 .
φ zn 1 − zn z
(2.28)
Then fn ∈ H(p, p,φ) and fn converges to 0 uniformly on compact subsets of D (see [7]).
Since Jg is compact, by Lemma 2.5, Jg fn Ꮾα → 0 as n → ∞. In addition,
Jg fn 2 α zn g zn 1
−
α
= sup 1 − |z|2 Jg fn (z) ≥ 2 1/ p ,
z ∈D
φ zn 1 − zn Ꮾα
from which the result follows.
(2.29)
8
Riemann-Stieltjes operators between Bergman and Bloch
Theorem 2.9. Assume that 0 < p < ∞, α > 0, and φ is normal on [0,1). Then Ig : H(p, p,φ)
→ Ꮾα is compact if and only if
α−1/ p−1
1 − |z|2
φ |z|
lim
|z|→1
g(z) = 0.
(2.30)
Proof. Suppose that Ig : H(p, p,φ) → Ꮾα is compact. Let {zn } be a sequence in D such that
|zn | → 1 as n → ∞. Let fn (z) be defined by (2.28). Then from the proof of Theorem 2.8
and the compactness of Ig , Ig fn Ꮾα → 0 as n → ∞. In addition,
Ig fn α = sup 1 − |z|2 Ig fn (z) ≥
Ꮾα
z ∈D
2 α 1 − zn g zn zn 1
+ t + 1 1+1/ p ,
2
p
φ zn 1 − zn (2.31)
from which we get the desired result.
Assume (2.30) holds, in order to prove that Ig is compact, it suffices to show that if { fn }
is a bounded sequence in H(p, p,φ) that converges to 0 uniformly on compact subsects
of D, then Ig fn Ꮾα → 0. Let { fn } be a sequence in H(p, p,φ) with fn H(p,p,φ) ≤ 1 and
fn → 0 uniformly on compact subsets of D. By the assumption, for any > 0, there is a
constant δ, 0 < δ < 1, such that δ < |z| < 1 implies
α 1 − |z|2 g(z)
φ |z| 1 − |z|2
1+1/ p <
2
.
(2.32)
Similar to the proof of Theorem 2.8, we have
Ig fn Ꮾα
−→ 0
as n −→ ∞.
(2.33)
Therefore, Ig : H(p, p,φ) → Ꮾα is compact.
From the introduction, we can easily get the following corollary.
Corollary 2.10. Let 0 < p < ∞, α ≥ 1 + 2/ p. Then
(1) Jg : A p → Ꮾα is bounded if and only if
sup 1 − |z|2
z ∈D
α−2/ p g (z) < ∞,
(2.34)
(2) Ig : A p → Ꮾα is bounded if and only if
sup 1 − |z|2
z ∈D
α−2/ p−1 g(z) < ∞,
(2.35)
(3) Jg : A p → Ꮾα is compact if and only if
lim 1 − |z|2
|z|→1
α−2/ p g (z) = 0,
(2.36)
Songxiao Li
9
(4) Ig : A p → Ꮾα is compact if and only if
lim 1 − |z|2
|z|→1
α−2/ p−1 g(z) = 0.
(2.37)
3. Jg ,Ig : H(p, p,φ) → Ꮾα0
In this section, we characterize the boundedness and compactness of Jg ,Ig : H(p, p,φ) →
Ꮾα0 . For this purpose, we need Lemma 3.1. When α = 1, Lemma 3.1 was proved in [8]. For
the general case, the proof is similar to the proof of the case α = 1. We omit the details.
Lemma 3.1. A closed set K in Ꮾα0 is compact if and only if it is bounded and satisfies
α lim sup 1 − |z|2 f (z) = 0.
|z|→1 f ∈K
(3.1)
Theorem 3.2. Assume that 0 < p < ∞, α > 0, and φ is normal on [0,1). Then the following
statements hold.
(i) Jg : H(p, p,φ) → Ꮾα0 is bounded if and only if g ∈ Ꮾα0 and Jg : H(p, p,φ) → Ꮾα is
bounded.
(ii) Jg : H(p, p,φ) → Ꮾα0 is compact if and only if
α−1/ p
1 − |z|2
lim
|z|→1
φ |z|
g (z) = 0.
(3.2)
Proof. (i) It is clear to see that g ∈ Ꮾα0 and Jg : H(p, p,φ) → Ꮾα is bounded if Jg : H(p, p,φ)
→ Ꮾα0 is bounded.
Conversely, suppose that Jg : H(p, p,φ) → Ꮾα is bounded and g ∈ Ꮾα0 . For any polynomial p(z), since g ∈ Ꮾα0 and
1 − |z|2
α α
Jg p (z) = 1 − |z|2 p(z)g (z),
(3.3)
we know that Jg p ∈ Ꮾα0 . For any f ∈ H(p, p,φ), there exists a sequence of polynomials
{ pn } such that f − pn H(p,p,φ) → 0 as n → ∞. Since Ꮾα0 is closed, we get
Jg f = lim Jg pn ∈ Ꮾα0 .
n→∞
(3.4)
In addition, Jg : H(p, p,φ) → Ꮾα is bounded. Therefore, Jg : H(p, p,φ) → Ꮾα0 is bounded.
(ii) If Jg : H(p, p,φ) → Ꮾα0 is compact, then by Theorem 2.8, we get (3.2).
Conversely, assume that (3.2) holds. It follows from Lemma 3.1 that Jg : H(p, p,φ) →
Ꮾα0 is compact if and only if
lim
sup
|z|→1 f H(p,p,φ) ≤1
α 1 − |z|2 Jg f (z) = 0.
(3.5)
In fact,
α α 1 − |z|2 g (z) 2 1/ p 1 − |z|2 Jg f (z) = f (z).
1/ p φ |z| 1 − |z|
φ |z| 1 − |z|2
By Lemma 2.1, the result follows.
(3.6)
10
Riemann-Stieltjes operators between Bergman and Bloch
Similarly, we have the following results.
Theorem 3.3. Assume that 0 < p < ∞, α > 0, and φ is normal on [0,1). Then the following
statements hold.
(i) Ig : H(p, p,φ) → Ꮾα0 is bounded if and only if Ig : H(p, p,φ) → Ꮾα is bounded and
lim g(z) 1 − |z|2
α
= 0.
(3.7)
g(z) = 0.
(3.8)
|z|→1
(ii) Ig : H(p, p,φ) → Ꮾα0 is compact if and only if
α−1/ p−1
1 − |z|2
φ |z|
lim
|z|→1
Corollary 3.4. Let 0 < p < ∞, α > 0. Then the following statements hold.
(i) Jg : A p → Ꮾα0 is bounded if and only if Jg : A p → Ꮾα is bounded and g ∈ Ꮾα0 .
(ii) Jg : A p → Ꮾα0 is compact if and only if
lim 1 − |z|2
|z|→1
α−2/ p g (z) = 0.
(3.9)
Corollary 3.5. Let 0 < p < ∞, α > 0. Then the following statements hold.
(i) Ig : A p → Ꮾα0 is bounded if and only if Ig : A p → Ꮾα is bounded and
lim g(z) 1 − |z|2
|z|→1
α
= 0.
(3.10)
(ii) Ig : A p → Ꮾα0 is compact if and only if
lim 1 − |z|2
|z|→1
α−2/ p−1 g(z) = 0.
(3.11)
Remark 3.6. In Corollary 3.5, if α ≤ 2/ p + 1, then by the maximum modulus principle, it
is easy to see that g ≡ 0.
4. Jg ,Ig : Ꮾα → H(p, p,φ)
The following lemma is well known(e.g., see [19]).
Lemma 4.1. Let f ∈ Ꮾα (D), 0 < α < ∞. Then
⎧
1−α
⎪
1 − 1 − |z|
⎪
⎪
α
,
⎨ f (0) + f Ꮾ
1−α
f (z) ≤
⎪
⎪
⎪
f (0) + f Ꮾα ln 2 ,
⎩
1 − |z|
α = 1,
(4.1)
α = 1.
The following lemma can be found in [10].
Lemma 4.2. Let μ be a positive measure on D and 0 < p < ∞. Let either 0 < α < ∞ and
n ∈ N, or 1 < α < ∞ and n = 0. Then
D
dμ(z)
αp < ∞
1 − |z|2
(4.2)
Songxiao Li
11
if and only if there is a positive constant C such that
D
(n) p f (z) 1 − |z|2 p(n−1) dμ(z)
1/ p
≤ C f Ꮾ α
(4.3)
for all analytic functions f in D, in particular, for all f ∈ Ꮾα .
Let 0 < p < ∞, let μ be a positive Borel measure on D. Define
p
D p (μ) = f ∈ H(D), f D p (μ) =
D
p
f (z) dμ(z) < ∞ .
(4.4)
Lemma 4.3. Let μ be a positive measure on D and 0 < p, α < ∞. Then the following statements are equivalent.
(1) i : Ꮾα → D p (μ) is bounded.
(2) i : Ꮾα → D p (μ) is compact.
(3) i : Ꮾα0 → D p (μ) is bounded.
(4) i : Ꮾα0 → D p (μ) is compact.
(5)
D
dμ(z)
αp < ∞.
1 − |z|2
(4.5)
Remark 4.4. The above lemma was obtained by Zhao when 0 < α ≤ 1 (see [24]). In fact,
his proof implies that the result also holds for α > 1. Partial results can also be found in
[4] when α = 1.
Theorem 4.5. Assume that 0 < α < ∞, 0 < p < ∞, and φ is normal on [0,1). Then the
following statements are equivalent.
(1) Ig : Ꮾα → H(p, p,φ) is bounded.
(2) Ig : Ꮾα → H(p, p,φ) is compact.
(3) Ig : Ꮾα0 → H(p, p,φ) is bounded.
(4) Ig : Ꮾα0 → H(p, p,φ) is compact.
(5)
p
g(z) p 1 − |z|2 p− pα φ |z| dA(z).
1 − |z|
D
(4.6)
Proof. Since
p p
2 p φ |z |
Ig f (z) 1 − |z|
dA(z)
H(p,p,φ) 1 − |z|
D
p p
p
p
2 p φ |z |
=
g(z)
f (z) 1 − |z|
dA(z) = f (z) dμ(z),
1
−
|
z
|
D
D
Ig f p
(4.7)
where
p
dμ(z) = g(z) 1 − |z|2
p φ p |z|
dA(z).
1 − |z|
(4.8)
12
Riemann-Stieltjes operators between Bergman and Bloch
By Lemma 4.3, we know that Ig : Ꮾα (Ꮾα0 ) → H(p, p,φ) is bounded (or compact) if and
only if
dμ
∞>
αp =
D 1 − |z |2
p
g(z) p 1 − |z|2 p− pα φ |z| dA(z).
1 − |z|
D
(4.9)
Theorem 4.6. Assume that α > 1, 0 < p < ∞, and φ is normal on [0,1). Then the following
statements are equivalent.
(i) Jg : Ꮾα → H(p, p,φ) is bounded.
(ii) Jg : Ꮾα → H(p, p,φ) is compact.
(iii)
p
p
g (z) 1 − |z|2 (2−α)p φ |z| dA(z) < ∞.
1 − |z|
D
(4.10)
Proof. Since
p p
2 p φ |z |
Jg f (z) 1 − |z|
dA(z)
H(p,p,φ) 1 − |z|
D
p
p
p
2 p φ |z |
=
g (z)
f (z) 1 − |z|
dA(z)
1 − |z|
D
f (z) p 1 − |z|2 − p dμ(z),
=
p
Jg f (4.11)
D
where
p
p
2 2p φ |z |
dA(z).
dμ = g (z) 1 − |z|
1 − |z|
(4.12)
Similar to the proof of Theorem 4.5, we get (i)⇔(iii) by Lemma 4.2.
(ii)⇒(i) is clear. Next we prove that (iii)⇒(ii). Assume (iii) holds, we obtain that Jg is
bounded and so g ∈ H(p, p,φ). In addition to this, we also find that for any ε > 0, there is
an r ∈ (0,1) such that
p
p
g (z) 1 − |z|2 (2−α)p φ |z| dA(z) < ε.
1 − |z|
|z|>r
(4.13)
Let { fk } be any sequence in the unit ball of Ꮾα and converges to 0 uniformly on compact
subsets of D. For the above ε, there exists a k0 > 0 such that sup|z|≤r | fk (z)| < ε as k > k0 .
Hence we have
Jg fk p
p
Jg fk (z) p 1 − |z|2 p φ |z| dA(z)
1 − |z|
|z|≤r
|z|>r
p
p
p
p
g (z) 1 − |z|2 (2−α)p φ |z| dA(z)
≤ Cεg H(p,p,φ) + fk Ꮾα
1 − |z|
|z|>r
p
p
≤ Cεg H(p,p,φ) + ε fk Ꮾα .
H(p,p,φ) +
(4.14)
Songxiao Li
13
In other words, we obtain limk→∞ Jg f H(p,p,φ) = 0 and so J p : Ꮾα → H(p, p,φ) is com
pact.
Theorem 4.7. Assume that 0 < α < 1, 0 < p < ∞, and φ is normal on [0,1). Then the following statements are equivalent.
(i) Jg : Ꮾα → H(p, p,φ) is bounded.
(ii) Jg : Ꮾα → H(p, p,φ) is compact.
(iii)
p
p
g (z) 1 − |z|2 p φ |z| dA(z) < ∞.
1 − |z|
D
(4.15)
Proof. (ii)⇒(i) is clear.
(i)⇒(iii). Assume that Jg : Ꮾα → H(p, p,φ) is bounded. Hence
p
p
p
2 p φ |z |
g (z)
f (z) 1 − |z|
dA(z).
H(p,p,φ) 1 − |z|
D
p
Jg f (4.16)
Taking f = 1, we get (iii).
Conversely, we assume that (iii) holds. Let f ∈ Ꮾα , then | f (z)| ≤ C f Ꮾα . Therefore,
by (4.16), we see that Jg : Ꮾα → H(p, p,φ) is bounded. Similar to the proof of
Theorem 4.6, we obtain that (iii)⇒(ii).
Theorem 4.8. Assume that 0 < p < ∞ and φ is normal on [0,1). Then the following statements hold.
(i) If the operator Jg : Ꮾ → H(p, p,φ) is bounded, then
1+1/ p g (z) ln
2
< ∞.
1 − |z|2
(4.17)
1−1/ p g (z) ln
2
< ∞,
1 − |z|2
(4.18)
sup φ |z| 1 − |z|2
z ∈D
(ii) If
sup φ |z| 1 − |z|2
z ∈D
then Jg : Ꮾ → H(p, p,φ) is bounded.
Proof. Assume that (4.18) holds. For any f ∈ Ꮾ, by Lemmas 2.2, 4.1 and the fact
(Jg f ) (z) = f (z)g (z), (Jg f )(0) = 0, we have
p p
2 p φ |z |
Jg f (z) 1 − |z|
dA(z)
H(p,p,φ) 1 − |z|
D
p
p
p
2 p φ |z |
=
g (z)
f (z) 1 − |z|
dA(z)
1 − |z|
D
p
p
2 p
p
2 p φ |z |
≤ C f Ꮾ g (z) | ln
1
−
|
z
|
dA(z)
1 − |z|2
1 − |z|
D
p
Jg f 14
Riemann-Stieltjes operators between Bergman and Bloch
p
p
p
≤ C f Ꮾ sup g (z) | ln
z ∈D
p
p
p
≤ C f Ꮾ sup g (z) | ln
z ∈D
2
1 − |z|2
1 − |z|2
p φ p |z|
dA(z)
1 − |z| D
p −1 p 2 1 − |z|2
φ |z| .
1 − |z|2
(4.19)
Therefore, Jg : Ꮾ → H(p, p,φ) is bounded.
Assume that Jg : Ꮾ → H(p, p,φ) is bounded. For w ∈ D, put fw (z) = ln2/(1 − wz).
Since
1 − |z|2 fw (z) ≤ 1 − |z|2
1 − |z|2
|w |
≤
≤ 2,
|1 − wz|
|1 − wz|
(4.20)
we have fw Ꮾ ≤ ln2 + 2. By the subharmonicity, we have
p
p p
p
C Jg ≥ C Jg fw Ꮾ ≥ C Jg fw H(p,p,φ)
p p
2 p φ |z |
≥C
Jg fw (z) 1 − |z|
dA(z)
1 − |z|
D
p
p
g (z) fw (z) p 1 − |z|2 p φ |z| dA(z)
≥C
1 − |z|
D(w,r)
(4.21)
p
p
p+1 p ≥ C g (w) fw (w) 1 − |w |2
φ |w|
p
p+1 p ≥ C 1 − |w |
φ |w| g (w) ln
2
1 − |w|2
p
.
Therefore, we get the desired result.
Remark 4.9. We use another method to prove the necessary condition of the boundedness
of Jg : Ꮾ → H(p, p,φ).
Since Jg f ∈ H(p, p,φ), by Lemma 2.3, we have
Jg f (z) ≤ C Jg f Jg H(p,p,φ)
Ꮾ→H(p,p,φ) f Ꮾ
1/ p+1 ≤ C 1/ p+1 .
1 − |z|2
φ |z|
1 − |z|2
φ |z|
(4.22)
For any w ∈ D, let fw (z) = ln 2/(1 − zw). Then we get
g (z) fw (z)φ |z| 1 − |z|2 1/ p+1 ≤ C Jg fw .
Ꮾ
(4.23)
1/ p+1
2
φ |w| 1 − |w|2
≤ C Jg Ꮾ→H(p,p,φ) fw Ꮾ .
2
1 − |w|
(4.24)
Ꮾ→H(p,p,φ)
Let z = w, we have
g (w) ln
Therefore, we get the desired result.
Songxiao Li
15
Theorem 4.10. Assume that 0 < p < ∞ and φ is normal on [0,1). Then the following statements hold.
(i) If the operator Jg : Ꮾ → H(p, p,φ) is compact, then
1+1/ p g (z) ln
2
= 0.
1 − |z|2
(4.25)
1−1/ p g (z) ln
2
= 0,
1 − |z|2
(4.26)
lim φ |z| 1 − |z|2
|z|→1
(ii) If
lim φ |z| 1 − |z|2
|z|→1
then Jg : Ꮾ → H(p, p,φ) is compact.
Proof. Suppose the operator Jg : Ꮾ → H(p, p,φ) is compact. Let zn be a sequence in D
such that |zn | → 1 as n → ∞. Take
2
fn (z) = ln
2
1 − zn −1 2
ln
1 − zn z
2
.
(4.27)
Then
2
fn (z) = 2 ln
2
1 − zn −1 2
zn
ln
.
1 − zzn 1 − zzn
(4.28)
Thus for any z ∈ D,
ln2/ 1 − zzn C + ln2/ 1 − zn 1
2 ≤4
1 − |z| fn (z) ≤ 2 1 − |z| 2 2 ≤ C.
ln 2/ 1 − zn 1 − |z|
ln2/ 1 − zn 2
(4.29)
On the other hand,
fn (0) ≤
2
ln
1 − zn |2
−1
(ln2)2 ≤ ln2.
(4.30)
Thus fn Ꮾ ≤ M, where M is a constant independent of n. Since for |z| = r < 1, we have
2
2
− r) + C
fn (z) = ln 2/ 1 −zzn ≤ ln2/(1
2 −→ 0
2
ln 2/ 1 − zn ln 2/ 1 − zn (n −→ ∞),
(4.31)
that is, fn → 0 uniformly on compact subsets of D as n → ∞. By the proof of Theorem 4.8,
we obtain
2 1+1/ p g zn ln
φ zn 1 − zn as n → ∞. Therefore, we get (4.25).
2
2 ≤ Jg fn −→ 0
1 − zn
(4.32)
16
Riemann-Stieltjes operators between Bergman and Bloch
From (4.26), for any ε > 0, there exists an r, 0 < r < 1, such that
φ |z| 1 − |z|2
1−1/ p g (z) ln
2
1 − |z|2
< ε,
(4.33)
when |z| > r. Also, from (4.26), we see that there exist C > 0 such that
sup φ |z| 1 − |z|2
1−1/ p g (z) < C.
(4.34)
|z|≤r
Let { fk } be any sequence in the unit ball of Ꮾ and converges to 0 uniformly on compact
subsets of D. For the above ε, there exists a k0 > 0 such that sup|z|≤r | fk (z)| < ε as k > k0 .
Hence we have
Jg fk p
H(p,p,φ) |z|≤r
+
|z|>r
p
≤ Cε + fk Ꮾ
p
Jg fk (z) p 1 − |z|2 p φ |z| dA(z)
1 − |z|
|z|>r
p
g (z) 1 − |z|2 p ln
p
≤ Cε + ε fk Ꮾ ,
as k > k0 , from which we get the desired result.
2
1 − |z|2
p
φ p |z|
dA(z)
1 − |z|
(4.35)
Acknowledgments
The author thanks the referee for his (her) useful comments. The author also would like
to express his sincere thanks to Professor Stevo Stević and Professor Kehe Zhu for their
help during the preparation of this paper. The author is partially supported by NNSF (no.
10371051, no. 10371069).
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Songxiao Li: Department of Mathematics, Shantou University, 515063 Shantou, Guangdong, China;
Department of Mathematics, Jiaying University, 514015 Meizhou, Guangdong, China
E-mail addresses: lsx@mail.zjxu.edu.cn; jyulsx@163.com