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The higher order derivatives of QK type spaces
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, China b 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, where ϕa (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 2 p 2 f Qp = sup f (z) g(a, z) dA(z) < ∞, a∈Δ
Δ
✩
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: [email protected] (H. Wulan), [email protected] (J. Zhou). 0022-247X/$ – see front matter © 2006 Elsevier Inc. All rights reserved. doi:10.1016/j.jmaa.2006.10.082
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where dA is an area measure on Δ normalized so that A(Δ) = 1. It is easy to check that the space Qp is Möbius invariant in the sense that f ◦ ϕa Qp = f Qp for every f ∈ Qp and a ∈ Δ. We know 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]) consisting of analytic functions f on Δ whose derivatives f are subject to the growth restriction f B = sup 1 − |z|2 f (z) < ∞. z∈Δ
When p = 0, the space Qp degenerates to the Dirichlet space D. See [11,14] for a summary of recent 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 generalization, the so-called QK (p, q) spaces, and we characterize the QK (p, q) spaces in terms of higher 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], respectively. Our proof to ensure the higher order derivatives characterization of QK (p, q) spaces is 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 p q p f K,p,q = sup f (z) 1 − |z|2 K g(z, a) dA(z) < ∞. a∈Δ
Δ
Equipped with the norm |f (0)| + f K,p,q , the space QK (p, q) is Banach when p > 1. The spaces QK (p, q) were introduced in [12] and some basic properties about QK (p, q) can be found there. Throughout this paper, we suppose that the nondecreasing function K is differentiable and satisfies that K(2t) ≈ K(t); that is, there exist constants C1 and C2 such that C1 K(2t) K(t) C2 K(2t). Also, we assume that 1
q 1 r dr < ∞. 1 − r 2 K log r
(1.1)
0
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 be assumed 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 occurrence. In order to obtain our main results in this paper, we define an auxiliary function as follows: ϕK (s) = sup K(st)/K(t),
0 < s < ∞.
0t1
We further assume that 1 ϕK (s) 0
ds < ∞, s
(1.2)
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and that
I = sup a∈Δ
Δ
(1 − |z|2 )p−2 1 K log dA(z) < ∞. |1 − az| ¯ p |z|
(1.3)
We know that (1.2) implies (1.3) for 1 < p < ∞. In fact, by [4, Lemma 2.1] the integral I is dominated by 1 1 1 K(log 1r ) K(1 − r) ds dr ≈ dr K(1) ϕK (s) < ∞ 1−r 1−r s 0
0
0
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. Then f ∈ QK (p, q) if and only if sup HK,p,q (f, a, n) < ∞, a∈Δ
where
HK,p,q (f, a, n) =
(n) p f (z) 1 − |z|2 np−p+q K g(z, a) dA(z).
Δ
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 measure if 1 − |z| sup dμ(z) < ∞. K |I | I ⊂∂Δ S(I )
Here and henceforward supI ⊂∂Δ means the supremum taken over all subarcs I of ∂Δ. Clearly, if K(t) = t p , 0 < p < ∞, then μ is a K-Carleson measure if and only if (1 − |z|2 )p dμ(z) is a p-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 following property; see [4]. Theorem A. Let K satisfy (1.2). A positive measure μ on Δ is K-Carleson if and only if 2 K 1 − ϕa (z) dμ(z) < ∞. sup a∈Δ
Δ
A geometric characterization of functions in QK (p, q) in terms of K-Carleson measure can be 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 positive integer. Then f ∈ QK (p, q) if and only if (n) p f (z) 1 − |z|2 np−p+q dA(z) is a K-Carleson measure.
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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 constant C = C(p, q) such that p f (z) 1 − |z|2 p+q K log 1 dA(z) |z| Δ p 1 2 q f (z) 1 − |z| K log C dA(z). |z| Δ
Proof. Suppose 1 p < ∞ and write z = reiθ ∈ Δ and ρ > r. By the Cauchy formula, 1 f (z) = 2πi
f (ζ ) dζ ρ = 2 2π (ζ − z)
|ζ |=ρ
2π 0
f (ρei(t+θ) )ei(t−θ) dt. (ρeit − r)2
By Minkowski’s inequality we have 1 Mp (r, f ) 2π
2π 0
where Mp (r, f ) =
1 2π
Mp (ρ, f ) Mp (ρ, f ) dt = 2 , ρ 2 − 2ρr cos t + r 2 ρ − r2
(2.1)
1/p 2π iθ p f re dθ . 0
For the case 0 < p < 1 we suppose f has no zero in Δ. Thus the function F (z) = (f (z))p is analytic and by (2.1), M1 (r, F )
M1 (ρ, F ) (Mp (ρ, f ))p = . ρ2 − r 2 ρ2 − r 2
By Hölder’s inequality, p 1 Mp (r, f ) = 2π
2π iθ p f re dθ 0
2π 1 1 p iθ 1−p iθ p F re F re dθ = 2π p 0
1−p 2π
p 2π iθ iθ 1 1 p dθ dθ F re F re 2π p 0 0 p 1−p 1 p M1 (r, F ) M1 (r, F ) , = p
(2.2)
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which together with (2.2) gives 1 −1 Mp (ρ, f ) 1 . Mp (r, f ) M1 (r, F ) p M1 (r, F ) p p(ρ 2 − r 2 ) If f has zeros, for a fixed s, 0 < s < 1, let fs (z) = f (sz). By Lemma 2 in [6, p. 79] there exist two nonvanishing functions f1 and f2 in H p such that fs (z) = f1 (z) + f2 (z),
fi p 2fs p ,
i = 1, 2.
Since f1 and f2 are nonzero functions, for ρ > r sMp (sr, f ) Mp r, f1 + Mp r, f2 1 Mp (ρ, f1 ) Mp (ρ, f2 ) + p ρ2 − r 2 ρ2 − r 2 2Mp (ρ, fs ) . p(ρ 2 − r 2 ) Let s → 1 and then 2Mp (ρ, f ) . Mp (r, f ) p(ρ 2 − r 2 ) Thus for all p, 0 < p < ∞, Mp (ρ, f ) 2 +1 Mp (r, f ) , ρ > r. p ρ2 − r 2 By the assumption of K we have 1 4 2 K log K log CK log . r 1+r 1+r Choosing ρ = 1+r 2 we obtain p f (z) 1 − |z|2 q+p K log 1 dA(z) |z| Δ
1 =2
q+p p 1 r dr Mp (r, f ) 1 − r 2 K log r
0
p 1 2 q ρ dρ C Mp (ρ, f ) 1 − ρ K log ρ 0 p 1 2 q f (z) 1 − |z| K log dA(z). C |z| 1
Δ
The proof is complete.
2
Lemma 4. Let f be analytic in Δ and 0 < p < ∞, −2 < q < ∞. Then np−p+q+2 (n) p f (a) CHK,p,q (f, a, n) 1 − |a|2 holds for all positive integers n and a ∈ Δ.
(2.3)
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Proof. Set
1 E(a, 1/e) = z ∈ Δ: |z − a| < 1 − |a| , e
a ∈ Δ.
It is easy to see that 1 1 1 − |a| 1 − |z| 1 + 1 − |a| 1− e e
(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 any a ∈ Δ, by (2.4) we have (n) p f (z) 1 − |z|2 np−p+q K g(z, a) dA(z) Δ
(n) p f (z) 1 − |z|2 np−p+q dA(z)
K(1) Δ(a, 1e )
(n) p f (z) 1 − |z|2 np−p+q dA(z)
K(1) E(a, 1e )
np−p+q CK(1) 1 − |a|2
(n) p f (z) dA(z)
E(a, 1e )
np−p+q+2 (n) p f (a) . C 1 − |a|2 This means that (2.3) holds.
2
Lemma 5. [7, p. 758] Let 1 k < ∞, μ > 0, δ > 0, and let h : (0, 1) → [0, ∞) be measurable. Then k
r 1 1 (1 − r)kμ−1 0
(r − t)δ−1 h(t) dt
C
0
(1 − r)kμ+kδ−1 hk (r) dr. 0
Lemma 6. [9, p. 626] Let f be analytic in Δ and 0 < r < 1. If p 1, then
r −1 Mp (r, f ) C f (0) + r Mp (s, f ) ds .
(2.5)
0
If 0 < p < 1, then
Mp (r, f ) C f (0) + r −1
1
r (r − s)
p−1
0
p Mp (s, f ) ds
p
.
(2.6)
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Lemma 7. Let 0 < p < ∞, −1 < q < ∞. Then there exists a constant C = C(p, q) such that f (z)p 1 − |z|2 q K log 1 dA(z) |z| Δ p p p+q 1 dA(z) K log C f (0) + f (z) 1 − |z|2 |z| Δ
holds for all analytic functions f in Δ. Proof. We first show that for all p, 0 < p < ∞, we have the following estimate: 1
p 1 dr (1 − r) Mp (r, f ) K log r q
0
p C f (0) +
1 (1 − r)
p+q
1 p Mp (r, f ) K log dr , r
0
which gives the proof of Lemma 7 by setting r = ρ 2 . To use (2.5) and (2.6) in our estimates, it is easy to see that the integrals 1
p 1 dr (1 − r) Mp (r, f ) K log r q
0
and 1
p 1 dr (1 − r)q r p Mp (r, f ) K log r
0
are comparable. For the case 1 p < ∞, by (2.5) and Lemma 5, 1
p 1 dr (1 − r) Mp (r, f ) K log r q
0
p C f (0) +
1
1 (1 − r)q K log r
0
p C f (0) + p C f (0) +
(1 − r)q
Mp (s, f ) ds
dr
1/p p 1 Mp (s, f ) K log ds dr s
0
1 (1 − r)
p+q
0
p
0
r
1 0
r
1 p Mp (r, f ) K log dr . r
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On the other hand, for the case 0 < p < 1, (2.6) and Lemma 5 give 1
p 1 (1 − r)q Mp (r, f ) K log dr r
0
p C f (0) +
r
1 (1 − r)
(r − s)
q
p−1
0
p C f (0) + p C f (0) +
r (1 − r)
(r − s)
q
0
0
1
1
p−1
1 p dr Mp (s, f ) ds K log r
1 p ds dr Mp (s, f ) K log s
0
1 p dr . (1 − r)p+q Mp (r, f ) K log r
0
Thus, as we mentioned above the proof is complete.
2
3. The proof of Theorem 1 Our goal is to prove that sup HK,p,q (f, a, n) < ∞ a∈Δ
⇔
sup HK,p,q (f, a, n + 1) < ∞. a∈Δ
We write HK,p,q (f, a, n + 1) p np+q = f (n+1) (z) 1 − |z|2 K g(z, a) dA(z) Δ
2 np+q+2 (n+1) p dA(z) 1 f ϕa (z) 1 − ϕa (z) K log |z| (1 − |z|2 )2 Δ = + {·}
=
|z|< 12
1 2 |z|<1
= I1 + I2 . On one hand, 2 np+q+2 (n+1) p dA(z) 1 f ϕa (z) 1 − ϕa (z) I1 = K log |z| (1 − |z|2 )2 |z|< 12
2 np+q+2 p sup f (n+1) ϕa (z) 1 − ϕa (z) |z|< 12
p np+q+2 C sup f (n+1) (a) 1 − |a|2 . a∈Δ
|z|< 12
dA(z) 1 K log |z| (1 − |z|2 )2
(3.1)
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By Lemma 4 and the following estimate (cf. [18]): p p np−p+q+2 np+q+2 ≈ sup f (n+1) (a) 1 − |a|2 , sup f (n) (a) 1 − |a|2 a∈Δ
(3.2)
a∈Δ
we have
p np−p+q+2 I1 C sup f (n) (a) 1 − |a|2 sup HK,p,q (f, a, n). a∈Δ
a∈Δ
On the other hand, 2 np+q+2 (n+1) p dA(z) 1 f I2 = ϕa (z) 1 − ϕa (z) K log |z| (1 − |z|2 )2 1 2 |z|<1
|f (n+1) (ϕa (z))|p 1 2 np+q dA(z). 1 − |z| K log |z| |1 − az| ¯ 2(np+q+2)
np+q+2 = 1 − |a|2
1 2 |z|<1
Consider the function f (n) (ϕa (z)) g(z) = 2(np−p+q+2) p (1 − az) ¯ and then p (1 − |a|2 )p |f (n+1) (ϕa (z))|p |f (n) (ϕa (z))|p g (z) + . C |1 − az| ¯ 2(np+q+2) |1 − az| ¯ p+2(np−p+q+2) Hence, by Lemma 3, p |f (n) (ϕa (z))|p 2 np−p+q+2 I2 C 1 − |a| g (z) + |1 − az| ¯ p+2(np−p+q+2) 1 2 |z|<1
1 × 1 − |z| dA(z) K log |z| p 1 2 np−p+q+2 2 np−p+q g(z) 1 − |z| dA(z) C 1 − |a| K log |z|
2 np+q
Δ
|f (n) (ϕa (z))|p (1 − |z|2 )np+q 1 dA(z) K log |z| |1 − az| ¯ p+2(np−p+q+2) Δ 2 np−p+q+2 (n) p dA(z) 1 C f ϕa (z) 1 − ϕa (z) K log |z| (1 − |z|2 )2 np−p+q+2 + 1 − |a|2
Δ
= CHK,p,q (f, a, n). Combining above estimates for I1 and I2 we get sup HK,p,q (f, a, n + 1) C sup HK,p,q (f, a, n). a∈Δ
a∈Δ
Conversely, consider the function g(z) =
f (n) (ϕa (z))(1 − |a|2 ) (1 − az) ¯
np−p+q+2 p
2(np−p+q+2) p
.
(3.3)
H. Wulan, J. Zhou / J. Math. Anal. Appl. 332 (2007) 1216–1228
Thus by Lemma 7, HK,p,q (f, a, n) 2 np−p+q+2 (n) p dA(z) 1 f = ϕa (z) 1 − ϕa (z) K log |z| (1 − |z|2 )2 Δ p np−p+q 1 dA(z) = g(z) 1 − |z|2 K log |z| Δ p p np+q 1 dA(z) . C g(0) + g (z) 1 − |z|2 K log |z| Δ
We finish the proof by showing p p 1 2 np+q g (z) 1 − |z| sup g(0) + dA(z) K log |z| a∈Δ Δ
C sup HK,p,q (f, a, n + 1). a∈Δ
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 p p np−p+q+2 sup g(0) = sup f (n) (a) 1 − |a|2 a∈Δ
a∈Δ
p np+q+2 ≈ sup f (n+1) (a) 1 − |a|2 a∈Δ
C sup HK,p,q (f, a, n + 1). a∈Δ
Note that p g (z) 1 − |z|2 np+q K log 1 dA(z) J1 + J2 , |z| Δ
where
J1 = Δ
and
J2 = Δ
Obviously,
|(f (n) (ϕa (z))) |p 1 2 np−p+q+2 2 np+q dA(z) 1 − |a| 1 − |z| K log |z| |1 − az| ¯ 2(np−p+q+2) |f (n) (ϕa (z))|p (1 − |a|2 )np−p+q+2 1 2 np+q dA(z). 1 − |z| K log |z| |1 − az| ¯ 2(np−p+q+2)+p
J1 C Δ
2 np+q+2 (n+1) p dA(z) 1 f ϕa (z) 1 − ϕa (z) K log |z| (1 − |z|2 )2
= CHK,p,q (f, a, n + 1).
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By Lemma 3 and the assumption (1.3) we have |f (n) (ϕa (z))|p (1 − |a|2 )np−p+q+2 1 2 np+q dA(z) 1 − |z| K log |z| |1 − az| ¯ 2(np−p+q+2)+p
J2 = Δ
2 np−p+q+2 (1 − |z|2 )p−2 (n) p 1 f dA(z) ϕa (z) 1 − ϕa (z) K log |1 − az| ¯ p |z|
= Δ
np−p+q+2 (n) p f (a) sup 1 − |a|2 a∈Δ
Δ
(1 − |z|2 )p−2 1 dA(z) K log |1 − az| ¯ p |z|
C sup HK,p,q (f, a, n + 1). a∈Δ
Combining this and the estimate for J1 we obtain sup HK,p,q (f, a, n) C sup HK,p,q (f, a, n + 1), a∈Δ
(3.4)
a∈Δ
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 integer n p 2 np−p+q sup f (n) (z) 1 − |z|2 (4.1) K 1 − ϕa (z) dA(z) < ∞. a∈Δ
Δ
We need only to prove by Theorem 1 that (4.1) is equivalent to p np−p+q K g(a, z) dA(z) < ∞. sup f (n) (z) 1 − |z|2 a∈Δ
(4.2)
Δ
Indeed, since K is nondecreasing and the fact that 1 − t 2 2 log 1t for 0 < t 1, (4.2) implies (4.1). We now show the converse. Let (4.1) hold and we divide the integral in (4.2) into two parts on Δ(a, 14 ) = {z ∈ Δ: |ϕa (z)| 14 } and Δ \ Δ(a, 14 ), respectively. Since 2 g(a, z) 4 1 − ϕa (z) ,
ϕa (z) 1/4
and
(n) p f (z) 1 − |z|2 np−p+q K g(a, z) dA(z)
Δ\Δ(a, 14 )
C Δ\Δ(a, 14 )
(n) p f (z) 1 − |z|2 np−p+q K 1 − ϕa (z)2 dA(z).
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On the other hand, (n) p f (z) 1 − |z|2 np−p+q K g(a, z) dA(z) Δ(a, 14 )
np−p+q+2 (n) p f (z) sup 1 − |z|2 z∈Δ
1 − |z|2
−2 K g(a, z) dA(z)
Δ(a, 14 )
np−p+q+2 (n) p f (z) sup 1 − |z|2 z∈Δ
−2 1 − |z|2 K g(0, z) dA(z)
Δ(0, 14 )
np−p+q+2 (n) p f (z) . C sup 1 − |z|2 z∈Δ
Similar to the proof of Lemma 4 we have np−p+q+2 (n) p f (z) sup 1 − |z|2 z∈Δ
C sup a∈Δ
(n) p f (z) 1 − |z|2 np−p+q K 1 − ϕa (z)2 dA(z).
Δ
Hence, (4.1) implies (4.2). The proof is complete. Acknowledgments The authors thank the referees for their helpful comments.
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[15] J. Xiao, Carleson measures, atomic decomposition and free interpolation from the Bloch space, Ann. Acad. Sci. Fenn. Ser. AI Math. 19 (1994) 35–46. [16] R. Zhao, On a general family of function spaces, Fenn. Ann. Aced. Sci. Fenn. Math. Diss. 105, 1996. [17] K. Zhu, Operator Theory in Function Spaces, Marcel Dekker, New York, 1990. [18] K. Zhu, Bloch type spaces of functions, Rocky Mountain J. Math. 23 (1993) 1143–1177.
The higher order derivatives of QK type spaces ✩ Hasi Wulan a,∗ , Jizhen Zhou b a Department of Mathematics, Shantou University, Shantou 515063, China b 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, where ϕa (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 2 p 2 f Qp = sup f (z) g(a, z) dA(z) < ∞, a∈Δ
Δ
✩
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: [email protected] (H. Wulan), [email protected] (J. Zhou). 0022-247X/$ – see front matter © 2006 Elsevier Inc. All rights reserved. doi:10.1016/j.jmaa.2006.10.082
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where dA is an area measure on Δ normalized so that A(Δ) = 1. It is easy to check that the space Qp is Möbius invariant in the sense that f ◦ ϕa Qp = f Qp for every f ∈ Qp and a ∈ Δ. We know 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]) consisting of analytic functions f on Δ whose derivatives f are subject to the growth restriction f B = sup 1 − |z|2 f (z) < ∞. z∈Δ
When p = 0, the space Qp degenerates to the Dirichlet space D. See [11,14] for a summary of recent 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 generalization, the so-called QK (p, q) spaces, and we characterize the QK (p, q) spaces in terms of higher 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], respectively. Our proof to ensure the higher order derivatives characterization of QK (p, q) spaces is 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 p q p f K,p,q = sup f (z) 1 − |z|2 K g(z, a) dA(z) < ∞. a∈Δ
Δ
Equipped with the norm |f (0)| + f K,p,q , the space QK (p, q) is Banach when p > 1. The spaces QK (p, q) were introduced in [12] and some basic properties about QK (p, q) can be found there. Throughout this paper, we suppose that the nondecreasing function K is differentiable and satisfies that K(2t) ≈ K(t); that is, there exist constants C1 and C2 such that C1 K(2t) K(t) C2 K(2t). Also, we assume that 1
q 1 r dr < ∞. 1 − r 2 K log r
(1.1)
0
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 be assumed 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 occurrence. In order to obtain our main results in this paper, we define an auxiliary function as follows: ϕK (s) = sup K(st)/K(t),
0 < s < ∞.
0t1
We further assume that 1 ϕK (s) 0
ds < ∞, s
(1.2)
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and that
I = sup a∈Δ
Δ
(1 − |z|2 )p−2 1 K log dA(z) < ∞. |1 − az| ¯ p |z|
(1.3)
We know that (1.2) implies (1.3) for 1 < p < ∞. In fact, by [4, Lemma 2.1] the integral I is dominated by 1 1 1 K(log 1r ) K(1 − r) ds dr ≈ dr K(1) ϕK (s) < ∞ 1−r 1−r s 0
0
0
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. Then f ∈ QK (p, q) if and only if sup HK,p,q (f, a, n) < ∞, a∈Δ
where
HK,p,q (f, a, n) =
(n) p f (z) 1 − |z|2 np−p+q K g(z, a) dA(z).
Δ
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 measure if 1 − |z| sup dμ(z) < ∞. K |I | I ⊂∂Δ S(I )
Here and henceforward supI ⊂∂Δ means the supremum taken over all subarcs I of ∂Δ. Clearly, if K(t) = t p , 0 < p < ∞, then μ is a K-Carleson measure if and only if (1 − |z|2 )p dμ(z) is a p-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 following property; see [4]. Theorem A. Let K satisfy (1.2). A positive measure μ on Δ is K-Carleson if and only if 2 K 1 − ϕa (z) dμ(z) < ∞. sup a∈Δ
Δ
A geometric characterization of functions in QK (p, q) in terms of K-Carleson measure can be 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 positive integer. Then f ∈ QK (p, q) if and only if (n) p f (z) 1 − |z|2 np−p+q dA(z) is a K-Carleson measure.
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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 constant C = C(p, q) such that p f (z) 1 − |z|2 p+q K log 1 dA(z) |z| Δ p 1 2 q f (z) 1 − |z| K log C dA(z). |z| Δ
Proof. Suppose 1 p < ∞ and write z = reiθ ∈ Δ and ρ > r. By the Cauchy formula, 1 f (z) = 2πi
f (ζ ) dζ ρ = 2 2π (ζ − z)
|ζ |=ρ
2π 0
f (ρei(t+θ) )ei(t−θ) dt. (ρeit − r)2
By Minkowski’s inequality we have 1 Mp (r, f ) 2π
2π 0
where Mp (r, f ) =
1 2π
Mp (ρ, f ) Mp (ρ, f ) dt = 2 , ρ 2 − 2ρr cos t + r 2 ρ − r2
(2.1)
1/p 2π iθ p f re dθ . 0
For the case 0 < p < 1 we suppose f has no zero in Δ. Thus the function F (z) = (f (z))p is analytic and by (2.1), M1 (r, F )
M1 (ρ, F ) (Mp (ρ, f ))p = . ρ2 − r 2 ρ2 − r 2
By Hölder’s inequality, p 1 Mp (r, f ) = 2π
2π iθ p f re dθ 0
2π 1 1 p iθ 1−p iθ p F re F re dθ = 2π p 0
1−p 2π
p 2π iθ iθ 1 1 p dθ dθ F re F re 2π p 0 0 p 1−p 1 p M1 (r, F ) M1 (r, F ) , = p
(2.2)
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which together with (2.2) gives 1 −1 Mp (ρ, f ) 1 . Mp (r, f ) M1 (r, F ) p M1 (r, F ) p p(ρ 2 − r 2 ) If f has zeros, for a fixed s, 0 < s < 1, let fs (z) = f (sz). By Lemma 2 in [6, p. 79] there exist two nonvanishing functions f1 and f2 in H p such that fs (z) = f1 (z) + f2 (z),
fi p 2fs p ,
i = 1, 2.
Since f1 and f2 are nonzero functions, for ρ > r sMp (sr, f ) Mp r, f1 + Mp r, f2 1 Mp (ρ, f1 ) Mp (ρ, f2 ) + p ρ2 − r 2 ρ2 − r 2 2Mp (ρ, fs ) . p(ρ 2 − r 2 ) Let s → 1 and then 2Mp (ρ, f ) . Mp (r, f ) p(ρ 2 − r 2 ) Thus for all p, 0 < p < ∞, Mp (ρ, f ) 2 +1 Mp (r, f ) , ρ > r. p ρ2 − r 2 By the assumption of K we have 1 4 2 K log K log CK log . r 1+r 1+r Choosing ρ = 1+r 2 we obtain p f (z) 1 − |z|2 q+p K log 1 dA(z) |z| Δ
1 =2
q+p p 1 r dr Mp (r, f ) 1 − r 2 K log r
0
p 1 2 q ρ dρ C Mp (ρ, f ) 1 − ρ K log ρ 0 p 1 2 q f (z) 1 − |z| K log dA(z). C |z| 1
Δ
The proof is complete.
2
Lemma 4. Let f be analytic in Δ and 0 < p < ∞, −2 < q < ∞. Then np−p+q+2 (n) p f (a) CHK,p,q (f, a, n) 1 − |a|2 holds for all positive integers n and a ∈ Δ.
(2.3)
H. Wulan, J. Zhou / J. Math. Anal. Appl. 332 (2007) 1216–1228
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Proof. Set
1 E(a, 1/e) = z ∈ Δ: |z − a| < 1 − |a| , e
a ∈ Δ.
It is easy to see that 1 1 1 − |a| 1 − |z| 1 + 1 − |a| 1− e e
(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 any a ∈ Δ, by (2.4) we have (n) p f (z) 1 − |z|2 np−p+q K g(z, a) dA(z) Δ
(n) p f (z) 1 − |z|2 np−p+q dA(z)
K(1) Δ(a, 1e )
(n) p f (z) 1 − |z|2 np−p+q dA(z)
K(1) E(a, 1e )
np−p+q CK(1) 1 − |a|2
(n) p f (z) dA(z)
E(a, 1e )
np−p+q+2 (n) p f (a) . C 1 − |a|2 This means that (2.3) holds.
2
Lemma 5. [7, p. 758] Let 1 k < ∞, μ > 0, δ > 0, and let h : (0, 1) → [0, ∞) be measurable. Then k
r 1 1 (1 − r)kμ−1 0
(r − t)δ−1 h(t) dt
C
0
(1 − r)kμ+kδ−1 hk (r) dr. 0
Lemma 6. [9, p. 626] Let f be analytic in Δ and 0 < r < 1. If p 1, then
r −1 Mp (r, f ) C f (0) + r Mp (s, f ) ds .
(2.5)
0
If 0 < p < 1, then
Mp (r, f ) C f (0) + r −1
1
r (r − s)
p−1
0
p Mp (s, f ) ds
p
.
(2.6)
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Lemma 7. Let 0 < p < ∞, −1 < q < ∞. Then there exists a constant C = C(p, q) such that f (z)p 1 − |z|2 q K log 1 dA(z) |z| Δ p p p+q 1 dA(z) K log C f (0) + f (z) 1 − |z|2 |z| Δ
holds for all analytic functions f in Δ. Proof. We first show that for all p, 0 < p < ∞, we have the following estimate: 1
p 1 dr (1 − r) Mp (r, f ) K log r q
0
p C f (0) +
1 (1 − r)
p+q
1 p Mp (r, f ) K log dr , r
0
which gives the proof of Lemma 7 by setting r = ρ 2 . To use (2.5) and (2.6) in our estimates, it is easy to see that the integrals 1
p 1 dr (1 − r) Mp (r, f ) K log r q
0
and 1
p 1 dr (1 − r)q r p Mp (r, f ) K log r
0
are comparable. For the case 1 p < ∞, by (2.5) and Lemma 5, 1
p 1 dr (1 − r) Mp (r, f ) K log r q
0
p C f (0) +
1
1 (1 − r)q K log r
0
p C f (0) + p C f (0) +
(1 − r)q
Mp (s, f ) ds
dr
1/p p 1 Mp (s, f ) K log ds dr s
0
1 (1 − r)
p+q
0
p
0
r
1 0
r
1 p Mp (r, f ) K log dr . r
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On the other hand, for the case 0 < p < 1, (2.6) and Lemma 5 give 1
p 1 (1 − r)q Mp (r, f ) K log dr r
0
p C f (0) +
r
1 (1 − r)
(r − s)
q
p−1
0
p C f (0) + p C f (0) +
r (1 − r)
(r − s)
q
0
0
1
1
p−1
1 p dr Mp (s, f ) ds K log r
1 p ds dr Mp (s, f ) K log s
0
1 p dr . (1 − r)p+q Mp (r, f ) K log r
0
Thus, as we mentioned above the proof is complete.
2
3. The proof of Theorem 1 Our goal is to prove that sup HK,p,q (f, a, n) < ∞ a∈Δ
⇔
sup HK,p,q (f, a, n + 1) < ∞. a∈Δ
We write HK,p,q (f, a, n + 1) p np+q = f (n+1) (z) 1 − |z|2 K g(z, a) dA(z) Δ
2 np+q+2 (n+1) p dA(z) 1 f ϕa (z) 1 − ϕa (z) K log |z| (1 − |z|2 )2 Δ = + {·}
=
|z|< 12
1 2 |z|<1
= I1 + I2 . On one hand, 2 np+q+2 (n+1) p dA(z) 1 f ϕa (z) 1 − ϕa (z) I1 = K log |z| (1 − |z|2 )2 |z|< 12
2 np+q+2 p sup f (n+1) ϕa (z) 1 − ϕa (z) |z|< 12
p np+q+2 C sup f (n+1) (a) 1 − |a|2 . a∈Δ
|z|< 12
dA(z) 1 K log |z| (1 − |z|2 )2
(3.1)
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By Lemma 4 and the following estimate (cf. [18]): p p np−p+q+2 np+q+2 ≈ sup f (n+1) (a) 1 − |a|2 , sup f (n) (a) 1 − |a|2 a∈Δ
(3.2)
a∈Δ
we have
p np−p+q+2 I1 C sup f (n) (a) 1 − |a|2 sup HK,p,q (f, a, n). a∈Δ
a∈Δ
On the other hand, 2 np+q+2 (n+1) p dA(z) 1 f I2 = ϕa (z) 1 − ϕa (z) K log |z| (1 − |z|2 )2 1 2 |z|<1
|f (n+1) (ϕa (z))|p 1 2 np+q dA(z). 1 − |z| K log |z| |1 − az| ¯ 2(np+q+2)
np+q+2 = 1 − |a|2
1 2 |z|<1
Consider the function f (n) (ϕa (z)) g(z) = 2(np−p+q+2) p (1 − az) ¯ and then p (1 − |a|2 )p |f (n+1) (ϕa (z))|p |f (n) (ϕa (z))|p g (z) + . C |1 − az| ¯ 2(np+q+2) |1 − az| ¯ p+2(np−p+q+2) Hence, by Lemma 3, p |f (n) (ϕa (z))|p 2 np−p+q+2 I2 C 1 − |a| g (z) + |1 − az| ¯ p+2(np−p+q+2) 1 2 |z|<1
1 × 1 − |z| dA(z) K log |z| p 1 2 np−p+q+2 2 np−p+q g(z) 1 − |z| dA(z) C 1 − |a| K log |z|
2 np+q
Δ
|f (n) (ϕa (z))|p (1 − |z|2 )np+q 1 dA(z) K log |z| |1 − az| ¯ p+2(np−p+q+2) Δ 2 np−p+q+2 (n) p dA(z) 1 C f ϕa (z) 1 − ϕa (z) K log |z| (1 − |z|2 )2 np−p+q+2 + 1 − |a|2
Δ
= CHK,p,q (f, a, n). Combining above estimates for I1 and I2 we get sup HK,p,q (f, a, n + 1) C sup HK,p,q (f, a, n). a∈Δ
a∈Δ
Conversely, consider the function g(z) =
f (n) (ϕa (z))(1 − |a|2 ) (1 − az) ¯
np−p+q+2 p
2(np−p+q+2) p
.
(3.3)
H. Wulan, J. Zhou / J. Math. Anal. Appl. 332 (2007) 1216–1228
Thus by Lemma 7, HK,p,q (f, a, n) 2 np−p+q+2 (n) p dA(z) 1 f = ϕa (z) 1 − ϕa (z) K log |z| (1 − |z|2 )2 Δ p np−p+q 1 dA(z) = g(z) 1 − |z|2 K log |z| Δ p p np+q 1 dA(z) . C g(0) + g (z) 1 − |z|2 K log |z| Δ
We finish the proof by showing p p 1 2 np+q g (z) 1 − |z| sup g(0) + dA(z) K log |z| a∈Δ Δ
C sup HK,p,q (f, a, n + 1). a∈Δ
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 p p np−p+q+2 sup g(0) = sup f (n) (a) 1 − |a|2 a∈Δ
a∈Δ
p np+q+2 ≈ sup f (n+1) (a) 1 − |a|2 a∈Δ
C sup HK,p,q (f, a, n + 1). a∈Δ
Note that p g (z) 1 − |z|2 np+q K log 1 dA(z) J1 + J2 , |z| Δ
where
J1 = Δ
and
J2 = Δ
Obviously,
|(f (n) (ϕa (z))) |p 1 2 np−p+q+2 2 np+q dA(z) 1 − |a| 1 − |z| K log |z| |1 − az| ¯ 2(np−p+q+2) |f (n) (ϕa (z))|p (1 − |a|2 )np−p+q+2 1 2 np+q dA(z). 1 − |z| K log |z| |1 − az| ¯ 2(np−p+q+2)+p
J1 C Δ
2 np+q+2 (n+1) p dA(z) 1 f ϕa (z) 1 − ϕa (z) K log |z| (1 − |z|2 )2
= CHK,p,q (f, a, n + 1).
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By Lemma 3 and the assumption (1.3) we have |f (n) (ϕa (z))|p (1 − |a|2 )np−p+q+2 1 2 np+q dA(z) 1 − |z| K log |z| |1 − az| ¯ 2(np−p+q+2)+p
J2 = Δ
2 np−p+q+2 (1 − |z|2 )p−2 (n) p 1 f dA(z) ϕa (z) 1 − ϕa (z) K log |1 − az| ¯ p |z|
= Δ
np−p+q+2 (n) p f (a) sup 1 − |a|2 a∈Δ
Δ
(1 − |z|2 )p−2 1 dA(z) K log |1 − az| ¯ p |z|
C sup HK,p,q (f, a, n + 1). a∈Δ
Combining this and the estimate for J1 we obtain sup HK,p,q (f, a, n) C sup HK,p,q (f, a, n + 1), a∈Δ
(3.4)
a∈Δ
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 integer n p 2 np−p+q sup f (n) (z) 1 − |z|2 (4.1) K 1 − ϕa (z) dA(z) < ∞. a∈Δ
Δ
We need only to prove by Theorem 1 that (4.1) is equivalent to p np−p+q K g(a, z) dA(z) < ∞. sup f (n) (z) 1 − |z|2 a∈Δ
(4.2)
Δ
Indeed, since K is nondecreasing and the fact that 1 − t 2 2 log 1t for 0 < t 1, (4.2) implies (4.1). We now show the converse. Let (4.1) hold and we divide the integral in (4.2) into two parts on Δ(a, 14 ) = {z ∈ Δ: |ϕa (z)| 14 } and Δ \ Δ(a, 14 ), respectively. Since 2 g(a, z) 4 1 − ϕa (z) ,
ϕa (z) 1/4
and
(n) p f (z) 1 − |z|2 np−p+q K g(a, z) dA(z)
Δ\Δ(a, 14 )
C Δ\Δ(a, 14 )
(n) p f (z) 1 − |z|2 np−p+q K 1 − ϕa (z)2 dA(z).
H. Wulan, J. Zhou / J. Math. Anal. Appl. 332 (2007) 1216–1228
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On the other hand, (n) p f (z) 1 − |z|2 np−p+q K g(a, z) dA(z) Δ(a, 14 )
np−p+q+2 (n) p f (z) sup 1 − |z|2 z∈Δ
1 − |z|2
−2 K g(a, z) dA(z)
Δ(a, 14 )
np−p+q+2 (n) p f (z) sup 1 − |z|2 z∈Δ
−2 1 − |z|2 K g(0, z) dA(z)
Δ(0, 14 )
np−p+q+2 (n) p f (z) . C sup 1 − |z|2 z∈Δ
Similar to the proof of Lemma 4 we have np−p+q+2 (n) p f (z) sup 1 − |z|2 z∈Δ
C sup a∈Δ
(n) p f (z) 1 − |z|2 np−p+q K 1 − ϕa (z)2 dA(z).
Δ
Hence, (4.1) implies (4.2). The proof is complete. Acknowledgments The authors thank the referees for their helpful comments.
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