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J. Math. Anal. Appl. 332 (2007) 1216–1228
www.elsevier.com/locate/jmaa
The higher order derivatives of QK type spaces ✩
Hasi Wulan a,∗, Jizhen Zhou b
a Department of Mathematics, Shantou University, Shantou 515063, Chinab Department of Mathematics and Physics, Anhui University of Science and Technology,
Huainan, Anhui 232001, China
Received 2 July 2006
Available online 4 December 2006
Submitted by S. Ruscheweyh
Abstract
The higher order derivatives characterization of QK type spaces, the so-called QK(p,q) spaces, is given.Our result further means that the spaces QK(p,q) can be described by the K-Carleson measures.© 2006 Elsevier Inc. All rights reserved.
Keywords: Higher order derivatives; QK(p,q) spaces; K-Carleson measures
1. Introduction
In recent years a special class of Möbius invariant function spaces, the so-called Qp
spaces, has attracted a lot of attention. We recall some facts about Qp spaces. Let g(a, z) =log(1/|ϕa(z)|) be the Green function on the unit disk Δ in the complex plane C, whereϕa(z) = (a − z)/(1 − az) is the Möbius transformation of Δ. For 0 � p < ∞, the space Qp
consists of analytic functions f on Δ for which
‖f ‖2Qp
= supa∈Δ
∫Δ
∣∣f ′(z)∣∣2(
g(a, z))p
dA(z) < ∞,
✩ The authors are supported by NSF of China (Nos. 10371069, 10671115) and NSF of Guangdong Province of China(Nos. 04011000, 06105648).
* Corresponding author.E-mail addresses: wulan@stu.edu.cn (H. Wulan), hope189@tom.com (J. Zhou).
0022-247X/$ – see front matter © 2006 Elsevier Inc. All rights reserved.doi:10.1016/j.jmaa.2006.10.082
H. Wulan, J. Zhou / J. Math. Anal. Appl. 332 (2007) 1216–1228 1217
where dA is an area measure on Δ normalized so that A(Δ) = 1. It is easy to check that the spaceQp is Möbius invariant in the sense that ‖f ◦ ϕa‖Qp = ‖f ‖Qp for every f ∈ Qp and a ∈ Δ. Weknow that Q1 = BMOA, the space of all analytic functions of bounded mean oscillation [5,17],and for each p > 1, the space Qp is the Bloch space B [1] (for the case p = 2 see [15]) consistingof analytic functions f on Δ whose derivatives f ′ are subject to the growth restriction
‖f ‖B = supz∈Δ
(1 − |z|2)∣∣f ′(z)
∣∣ < ∞.
When p = 0, the space Qp degenerates to the Dirichlet space D. See [11,14] for a summary ofrecent research about Qp spaces.
There are a number of ways we can further generalize the Qp spaces; see, for example, [16]for F(p,q, s), and [3,10] for QK . In this paper we are concerned with a more general gener-alization, the so-called QK(p,q) spaces, and we characterize the QK(p,q) spaces in terms ofhigher order derivatives. The higher order derivatives characterization of Qp spaces was given in[2]. The corresponding problems for QK and F(p,q, s) spaces were studied in [8] and [13], re-spectively. Our proof to ensure the higher order derivatives characterization of QK(p,q) spacesis new even for above special cases.
Let K : [0,∞) → [0,∞) be a right-continuous and nondecreasing function. For 0 < p < ∞,−2 < q < ∞, the space QK(p,q) consists of analytic functions f defined in Δ satisfying
‖f ‖pK,p,q = sup
a∈Δ
∫Δ
∣∣f ′(z)∣∣p(
1 − |z|2)qK
(g(z, a)
)dA(z) < ∞.
Equipped with the norm |f (0)| + ‖f ‖K,p,q , the space QK(p,q) is Banach when p > 1. Thespaces QK(p,q) were introduced in [12] and some basic properties about QK(p,q) can befound there.
Throughout this paper, we suppose that the nondecreasing function K is differentiable andsatisfies that K(2t) ≈ K(t); that is, there exist constants C1 and C2 such that C1K(2t) � K(t) �C2K(2t). Also, we assume that
1∫0
(1 − r2)q
K
(log
1
r
)r dr < ∞. (1.1)
Otherwise, QK(p,q) contains constant functions only (cf. [12]). We know that QK(p,q) =QK1(p, q) for K1 = inf(K(r),K(1)) (see [12, Theorem 3.1]) and so the function K can beassumed to be bounded. To avoid QK(p,q) = F(p,q,0) we assume that K(0) = 0 in this paper;see [12, Corollary 3.5].
The letter C denotes a positive constant throughout the paper which may vary at each occur-rence.
In order to obtain our main results in this paper, we define an auxiliary function as follows:
ϕK(s) = sup0�t�1
K(st)/K(t), 0 < s < ∞.
We further assume that
1∫ϕK(s)
ds
s< ∞, (1.2)
0
1218 H. Wulan, J. Zhou / J. Math. Anal. Appl. 332 (2007) 1216–1228
and that
I = supa∈Δ
∫Δ
(1 − |z|2)p−2
|1 − az|p K
(log
1
|z|)
dA(z) < ∞. (1.3)
We know that (1.2) implies (1.3) for 1 < p < ∞. In fact, by [4, Lemma 2.1] the integral I isdominated by
1∫0
K(log 1r)
1 − rdr ≈
1∫0
K(1 − r)
1 − rdr � K(1)
1∫0
ϕK(s)ds
s< ∞
whenever (1.2) holds.One of our main results in the paper is the following.
Theorem 1. Let K satisfy (1.3) and 0 < p < ∞,−1 < q < ∞. Let n be a positive integer. Thenf ∈ QK(p,q) if and only if
supa∈Δ
HK,p,q(f, a,n) < ∞,
where
HK,p,q(f, a,n) =∫Δ
∣∣f (n)(z)∣∣p(
1 − |z|2)np−p+qK
(g(z, a)
)dA(z).
For a subarc I ⊂ ∂Δ, the boundary of Δ, let
S(I) = {rζ ∈ Δ: 1 − |I | < r < 1, ζ ∈ I
}.
If |I | � 1 then we set S(I) = Δ. A positive measure μ on Δ is said to be a K-Carleson measureif
supI⊂∂Δ
∫S(I)
K
(1 − |z|
|I |)
dμ(z) < ∞.
Here and henceforward supI⊂∂Δ means the supremum taken over all subarcs I of ∂Δ. Clearly,if K(t) = tp, 0 < p < ∞, then μ is a K-Carleson measure if and only if (1 − |z|2)p dμ(z) is ap-Carleson measure. Note that p = 1 gives the classical Carleson measure; see [5].
As a modification of p-Carleson measure our new K-Carleson measure has the followingproperty; see [4].
Theorem A. Let K satisfy (1.2). A positive measure μ on Δ is K-Carleson if and only if
supa∈Δ
∫Δ
K(1 − ∣∣ϕa(z)
∣∣2)dμ(z) < ∞.
A geometric characterization of functions in QK(p,q) in terms of K-Carleson measure canbe stated as follows, which is our second main result in this note.
Theorem 2. Let K satisfy (1.2), (1.3), 0 < p < ∞ and −1 < q < ∞. Suppose n is a positiveinteger. Then f ∈ QK(p,q) if and only if∣∣f (n)(z)
∣∣p(1 − |z|2)np−p+q
dA(z)
is a K-Carleson measure.
H. Wulan, J. Zhou / J. Math. Anal. Appl. 332 (2007) 1216–1228 1219
2. Preliminaries
The following lemmas will be used in the proof of our main theorems.
Lemma 3. Let 0 < p < ∞,−2 < q < ∞. If f is analytic in Δ, then there exists a constantC = C(p,q) such that∫
Δ
∣∣f ′(z)∣∣p(
1 − |z|2)p+qK
(log
1
|z|)
dA(z)
� C
∫Δ
∣∣f (z)∣∣p(
1 − |z|2)qK
(log
1
|z|)
dA(z).
Proof. Suppose 1 � p < ∞ and write z = reiθ ∈ Δ and ρ > r . By the Cauchy formula,
f ′(z) = 1
2πi
∫|ζ |=ρ
f (ζ ) dζ
(ζ − z)2= ρ
2π
2π∫0
f (ρei(t+θ))ei(t−θ)
(ρeit − r)2dt.
By Minkowski’s inequality we have
Mp(r,f ′) � 1
2π
2π∫0
Mp(ρ,f )dt
ρ2 − 2ρr cos t + r2= Mp(ρ,f )
ρ2 − r2, (2.1)
where
Mp(r,f ) ={
1
2π
2π∫0
∣∣f (reiθ
)∣∣p dθ
}1/p
.
For the case 0 < p < 1 we suppose f has no zero in Δ. Thus the function F(z) = (f (z))p isanalytic and by (2.1),
M1(r,F′) � M1(ρ,F )
ρ2 − r2= (Mp(ρ,f ))p
ρ2 − r2. (2.2)
By Hölder’s inequality,
(Mp(r,f ′)
)p = 1
2π
2π∫0
∣∣f ′(reiθ)∣∣p dθ
= 1
2π
(1
p
)p2π∫
0
∣∣F (reiθ
)∣∣1−p∣∣F ′(reiθ)∣∣p dθ
� 1
2π
(1
p
)p( 2π∫
0
∣∣F (reiθ
)∣∣dθ
)1−p( 2π∫0
∣∣F ′(reiθ)∣∣dθ
)p
=(
1)p(
M1(r,F ))1−p(
M1(r,F′))p
,
p1220 H. Wulan, J. Zhou / J. Math. Anal. Appl. 332 (2007) 1216–1228
which together with (2.2) gives
Mp(r,f ′) � 1
p
(M1(r,F )
) 1p
−1M1(r,F
′) � Mp(ρ,f )
p(ρ2 − r2).
If f has zeros, for a fixed s,0 < s < 1, let fs(z) = f (sz). By Lemma 2 in [6, p. 79] there existtwo nonvanishing functions f1 and f2 in Hp such that
fs(z) = f1(z) + f2(z), ‖fi‖p � 2‖fs‖p, i = 1,2.
Since f1 and f2 are nonzero functions, for ρ > r
sMp(sr, f ′) � Mp
(r, f ′
1
) + Mp
(r, f ′
2
)� 1
p
(Mp(ρ,f1)
ρ2 − r2+ Mp(ρ,f2)
ρ2 − r2
)
� 2Mp(ρ,fs)
p(ρ2 − r2).
Let s → 1 and then
Mp(r,f ′) � 2Mp(ρ,f )
p(ρ2 − r2).
Thus for all p,0 < p < ∞,
Mp(r, f ′) �(
2
p+ 1
)Mp(ρ,f )
ρ2 − r2, ρ > r.
By the assumption of K we have
K
(log
1
r
)� K
(log
4
1 + r
)� CK
(log
2
1 + r
).
Choosing ρ = 1+r2 we obtain∫
Δ
∣∣f ′(z)∣∣p(
1 − |z|2)q+pK
(log
1
|z|)
dA(z)
= 2
1∫0
(Mp(r,f ′)
)p(1 − r2)q+p
K
(log
1
r
)r dr
� C
1∫0
(Mp(ρ,f )
)p(1 − ρ2)q
K
(log
1
ρ
)ρ dρ
� C
∫Δ
∣∣f (z)∣∣p(
1 − |z|2)qK
(log
1
|z|)
dA(z).
The proof is complete. �Lemma 4. Let f be analytic in Δ and 0 < p < ∞,−2 < q < ∞. Then(
1 − |a|2)np−p+q+2∣∣f (n)(a)∣∣p � CHK,p,q(f, a,n) (2.3)
holds for all positive integers n and a ∈ Δ.
H. Wulan, J. Zhou / J. Math. Anal. Appl. 332 (2007) 1216–1228 1221
Proof. Set
E(a,1/e) ={z ∈ Δ: |z − a| < 1
e
(1 − |a|)}, a ∈ Δ.
It is easy to see that(1 − 1
e
)(1 − |a|) � 1 − |z| �
(1 + 1
e
)(1 − |a|) (2.4)
whenever z ∈ E(a,1/e). Since |f (n)(z)| is subharmonic in Δ and E(a,1/e) ⊂ Δ(a,1/e) for anya ∈ Δ, by (2.4) we have∫
Δ
∣∣f (n)(z)∣∣p(
1 − |z|2)np−p+qK
(g(z, a)
)dA(z)
� K(1)
∫Δ(a, 1
e)
∣∣f (n)(z)∣∣p(
1 − |z|2)np−p+qdA(z)
� K(1)
∫E(a, 1
e)
∣∣f (n)(z)∣∣p(
1 − |z|2)np−p+qdA(z)
� CK(1)(1 − |a|2)np−p+q
∫E(a, 1
e)
∣∣f (n)(z)∣∣p dA(z)
� C(1 − |a|2)np−p+q+2∣∣f (n)(a)
∣∣p.
This means that (2.3) holds. �Lemma 5. [7, p. 758] Let 1 � k < ∞, μ > 0, δ > 0, and let h : (0,1) → [0,∞) be measurable.Then
1∫0
(1 − r)kμ−1
{ r∫0
(r − t)δ−1h(t) dt
}k
� C
1∫0
(1 − r)kμ+kδ−1hk(r) dr.
Lemma 6. [9, p. 626] Let f be analytic in Δ and 0 < r < 1.If p � 1, then
Mp(r,f ) � C
(∣∣f (0)∣∣ + r−1
r∫0
Mp(s,f ′) ds
). (2.5)
If 0 < p < 1, then
Mp(r,f ) � C
{∣∣f (0)∣∣ + r−1
( r∫0
(r − s)p−1Mpp (s, f ′) ds
) 1p}
. (2.6)
1222 H. Wulan, J. Zhou / J. Math. Anal. Appl. 332 (2007) 1216–1228
Lemma 7. Let 0 < p < ∞,−1 < q < ∞. Then there exists a constant C = C(p,q) such that∫Δ
∣∣f (z)∣∣p(
1 − |z|2)qK
(log
1
|z|)
dA(z)
� C
(∣∣f (0)∣∣p +
∫Δ
∣∣f ′(z)∣∣p(
1 − |z|2)p+qK
(log
1
|z|)
dA(z)
)
holds for all analytic functions f in Δ.
Proof. We first show that for all p,0 < p < ∞, we have the following estimate:
1∫0
(1 − r)q(Mp(r,f )
)pK
(log
1
r
)dr
� C
(∣∣f (0)∣∣p +
1∫0
(1 − r)p+q(Mp(r,f ′)
)pK
(log
1
r
)dr
),
which gives the proof of Lemma 7 by setting r = ρ2.To use (2.5) and (2.6) in our estimates, it is easy to see that the integrals
1∫0
(1 − r)q(Mp(r,f )
)pK
(log
1
r
)dr
and
1∫0
(1 − r)qrp(Mp(r,f )
)pK
(log
1
r
)dr
are comparable. For the case 1 � p < ∞, by (2.5) and Lemma 5,
1∫0
(1 − r)q(Mp(r,f )
)pK
(log
1
r
)dr
� C
{∣∣f (0)∣∣p +
1∫0
(1 − r)qK
(log
1
r
)( r∫0
Mp(s,f ′) ds
)p
dr
}
� C
{∣∣f (0)∣∣p +
1∫0
(1 − r)q
( r∫0
Mp(s,f ′)(
K
(log
1
s
))1/p
ds
)p
dr
}
� C
{∣∣f (0)∣∣p +
1∫0
(1 − r)p+q(Mp(r,f ′)
)pK
(log
1
r
)dr
}.
H. Wulan, J. Zhou / J. Math. Anal. Appl. 332 (2007) 1216–1228 1223
On the other hand, for the case 0 < p < 1, (2.6) and Lemma 5 give
1∫0
(1 − r)q(Mp(r,f )
)pK
(log
1
r
)dr
� C
{∣∣f (0)∣∣p +
1∫0
(1 − r)q
( r∫0
(r − s)p−1(Mp(s,f ′))p
ds
)K
(log
1
r
)dr
}
� C
{∣∣f (0)∣∣p +
1∫0
(1 − r)q
r∫0
(r − s)p−1(Mp(s,f ′))p
K
(log
1
s
)ds dr
}
� C
{∣∣f (0)∣∣p +
1∫0
(1 − r)p+q(Mp(r,f ′)
)pK
(log
1
r
)dr
}.
Thus, as we mentioned above the proof is complete. �3. The proof of Theorem 1
Our goal is to prove that
supa∈Δ
HK,p,q(f, a,n) < ∞ ⇔ supa∈Δ
HK,p,q(f, a,n + 1) < ∞. (3.1)
We write
HK,p,q(f, a,n + 1)
=∫Δ
∣∣f (n+1)(z)∣∣p(
1 − |z|2)np+qK
(g(z, a)
)dA(z)
=∫Δ
∣∣f (n+1)(ϕa(z)
)∣∣p(1 − ∣∣ϕa(z)
∣∣2)np+q+2K
(log
1
|z|)
dA(z)
(1 − |z|2)2
=∫
|z|< 12
+∫
12 �|z|<1
{·}
= I1 + I2.
On one hand,
I1 =∫
|z|< 12
∣∣f (n+1)(ϕa(z)
)∣∣p(1 − ∣∣ϕa(z)
∣∣2)np+q+2K
(log
1
|z|)
dA(z)
(1 − |z|2)2
� sup|z|< 1
2
∣∣f (n+1)(ϕa(z)
)∣∣p(1 − ∣∣ϕa(z)
∣∣2)np+q+2∫
|z|< 12
K
(log
1
|z|)
dA(z)
(1 − |z|2)2
� C sup∣∣f (n+1)(a)
∣∣p(1 − |a|2)np+q+2
.
a∈Δ1224 H. Wulan, J. Zhou / J. Math. Anal. Appl. 332 (2007) 1216–1228
By Lemma 4 and the following estimate (cf. [18]):
supa∈Δ
∣∣f (n)(a)∣∣p(
1 − |a|2)np−p+q+2 ≈ supa∈Δ
∣∣f (n+1)(a)∣∣p(
1 − |a|2)np+q+2, (3.2)
we have
I1 � C supa∈Δ
∣∣f (n)(a)∣∣p(
1 − |a|2)np−p+q+2 � supa∈Δ
HK,p,q(f, a,n).
On the other hand,
I2 =∫
12 �|z|<1
∣∣f (n+1)(ϕa(z)
)∣∣p(1 − ∣∣ϕa(z)
∣∣2)np+q+2K
(log
1
|z|)
dA(z)
(1 − |z|2)2
= (1 − |a|2)np+q+2
∫12 �|z|<1
|f (n+1)(ϕa(z))|p|1 − az|2(np+q+2)
(1 − |z|2)np+q
K
(log
1
|z|)
dA(z).
Consider the function
g(z) = f (n)(ϕa(z))
(1 − az)2(np−p+q+2)
p
and then
(1 − |a|2)p|f (n+1)(ϕa(z))|p|1 − az|2(np+q+2)
� C
(∣∣g′(z)∣∣p + |f (n)(ϕa(z))|p
|1 − az|p+2(np−p+q+2)
).
Hence, by Lemma 3,
I2 � C(1 − |a|2)np−p+q+2
∫12 �|z|<1
(∣∣g′(z)∣∣p + |f (n)(ϕa(z))|p
|1 − az|p+2(np−p+q+2)
)
× (1 − |z|2)np+q
K
(log
1
|z|)
dA(z)
� C(1 − |a|2)np−p+q+2
∫Δ
∣∣g(z)∣∣p(
1 − |z|2)np−p+qK
(log
1
|z|)
dA(z)
+ (1 − |a|2)np−p+q+2
∫Δ
|f (n)(ϕa(z))|p(1 − |z|2)np+q
|1 − az|p+2(np−p+q+2)K
(log
1
|z|)
dA(z)
� C
∫Δ
∣∣f (n)(ϕa(z)
)∣∣p(1 − ∣∣ϕa(z)
∣∣2)np−p+q+2K
(log
1
|z|)
dA(z)
(1 − |z|2)2
= CHK,p,q(f, a,n).
Combining above estimates for I1 and I2 we get
supa∈Δ
HK,p,q(f, a,n + 1) � C supa∈Δ
HK,p,q(f, a,n). (3.3)
Conversely, consider the function
g(z) = f (n)(ϕa(z))(1 − |a|2) np−p+q+2p
2(np−p+q+2)p
.
(1 − az)
H. Wulan, J. Zhou / J. Math. Anal. Appl. 332 (2007) 1216–1228 1225
Thus by Lemma 7,
HK,p,q(f, a,n)
=∫Δ
∣∣f (n)(ϕa(z)
)∣∣p(1 − ∣∣ϕa(z)
∣∣2)np−p+q+2K
(log
1
|z|)
dA(z)
(1 − |z|2)2
=∫Δ
∣∣g(z)∣∣p(
1 − |z|2)np−p+qK
(log
1
|z|)
dA(z)
� C
(∣∣g(0)∣∣p +
∫Δ
∣∣g′(z)∣∣p(
1 − |z|2)np+qK
(log
1
|z|)
dA(z)
).
We finish the proof by showing
supa∈Δ
(∣∣g(0)∣∣p +
∫Δ
∣∣g′(z)∣∣p(
1 − |z|2)np+qK
(log
1
|z|)
dA(z)
)
� C supa∈Δ
HK,p,q(f, a,n + 1).
In fact, since∣∣g(0)∣∣p = ∣∣f (n)(a)
∣∣p(1 − |a|2)np−p+q+2
,
by Lemma 4 and the estimate (3.2) we have
supa∈Δ
∣∣g(0)∣∣p = sup
a∈Δ
∣∣f (n)(a)∣∣p(
1 − |a|2)np−p+q+2
≈ supa∈Δ
∣∣f (n+1)(a)∣∣p(
1 − |a|2)np+q+2
� C supa∈Δ
HK,p,q(f, a,n + 1).
Note that∫Δ
∣∣g′(z)∣∣p(
1 − |z|2)np+qK
(log
1
|z|)
dA(z) � J1 + J2,
where
J1 =∫Δ
|(f (n)(ϕa(z)))′|p
|1 − az|2(np−p+q+2)
(1 − |a|2)np−p+q+2(1 − |z|2)np+q
K
(log
1
|z|)
dA(z)
and
J2 =∫Δ
|f (n)(ϕa(z))|p(1 − |a|2)np−p+q+2
|1 − az|2(np−p+q+2)+p
(1 − |z|2)np+q
K
(log
1
|z|)
dA(z).
Obviously,
J1 � C
∫Δ
∣∣f (n+1)(ϕa(z)
)∣∣p(1 − ∣∣ϕa(z)
∣∣2)np+q+2K
(log
1
|z|)
dA(z)
(1 − |z|2)2
= CHK,p,q(f, a,n + 1).
1226 H. Wulan, J. Zhou / J. Math. Anal. Appl. 332 (2007) 1216–1228
By Lemma 3 and the assumption (1.3) we have
J2 =∫Δ
|f (n)(ϕa(z))|p(1 − |a|2)np−p+q+2
|1 − az|2(np−p+q+2)+p
(1 − |z|2)np+q
K
(log
1
|z|)
dA(z)
=∫Δ
∣∣f (n)(ϕa(z)
)∣∣p(1 − ∣∣ϕa(z)
∣∣2)np−p+q+2 (1 − |z|2)p−2
|1 − az|p K
(log
1
|z|)
dA(z)
� supa∈Δ
{(1 − |a|2)np−p+q+2∣∣f (n)(a)
∣∣p}∫Δ
(1 − |z|2)p−2
|1 − az|p K
(log
1
|z|)
dA(z)
� C supa∈Δ
HK,p,q(f, a,n + 1).
Combining this and the estimate for J1 we obtain
supa∈Δ
HK,p,q(f, a,n) � C supa∈Δ
HK,p,q(f, a,n + 1), (3.4)
which together with (3.3) shows that (3.1) holds. Theorem 1 is proved.
4. The proof of Theorem 2
By Theorem A it suffices to prove that f ∈ QK(p,q) if and only if for any positive inte-ger n
supa∈Δ
∫Δ
∣∣f (n)(z)∣∣p(
1 − |z|2)np−p+qK
(1 − ∣∣ϕa(z)
∣∣2)dA(z) < ∞. (4.1)
We need only to prove by Theorem 1 that (4.1) is equivalent to
supa∈Δ
∫Δ
∣∣f (n)(z)∣∣p(
1 − |z|2)np−p+qK
(g(a, z)
)dA(z) < ∞. (4.2)
Indeed, since K is nondecreasing and the fact that 1 − t2 � 2 log 1t
for 0 < t � 1, (4.2) im-plies (4.1). We now show the converse. Let (4.1) hold and we divide the integral in (4.2) into twoparts on Δ(a, 1
4 ) = {z ∈ Δ: |ϕa(z)| � 14 } and Δ \ Δ(a, 1
4 ), respectively. Since
g(a, z) � 4(1 − ∣∣ϕa(z)
∣∣2),
∣∣ϕa(z)∣∣ � 1/4
and ∫Δ\Δ(a, 1
4 )
∣∣f (n)(z)∣∣p(
1 − |z|2)np−p+qK
(g(a, z)
)dA(z)
� C
∫Δ\Δ(a, 1
4 )
∣∣f (n)(z)∣∣p(
1 − |z|2)np−p+qK
(1 − ∣∣ϕa(z)
∣∣2)dA(z).
H. Wulan, J. Zhou / J. Math. Anal. Appl. 332 (2007) 1216–1228 1227
On the other hand,∫Δ(a, 1
4 )
∣∣f (n)(z)∣∣p(
1 − |z|2)np−p+qK
(g(a, z)
)dA(z)
� supz∈Δ
(1 − |z|2)np−p+q+2∣∣f (n)(z)
∣∣p ∫Δ(a, 1
4 )
(1 − |z|2)−2
K(g(a, z)
)dA(z)
� supz∈Δ
(1 − |z|2)np−p+q+2∣∣f (n)(z)
∣∣p ∫Δ(0, 1
4 )
(1 − |z|2)−2
K(g(0, z)
)dA(z)
� C supz∈Δ
(1 − |z|2)np−p+q+2∣∣f (n)(z)
∣∣p.
Similar to the proof of Lemma 4 we have
supz∈Δ
(1 − |z|2)np−p+q+2∣∣f (n)(z)
∣∣p� C sup
a∈Δ
∫Δ
∣∣f (n)(z)∣∣p(
1 − |z|2)np−p+qK
(1 − ∣∣ϕa(z)
∣∣2)dA(z).
Hence, (4.1) implies (4.2). The proof is complete.
Acknowledgments
The authors thank the referees for their helpful comments.
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