7_ahmed.dvi EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 2, No. 2, 2009, (250-267) ISSN 1307-5543 – www.ejpam.com Some Characterizations of Weighted Holomorphic Bloch Space R. A. Rashwan1, A. El-Sayed Ahmed2∗ and A. Kamal3 1 Assiut University, Faculty of Science, Department of Mathematics, Assiut, Egypt 2 Sohag University, Faculty of Science, Department of Mathematics, Sohag 82524, Egypt Current Address: Taif University, Faculty of Science, Mathematics Department, El-Taif P.O.Box 888, El-Hawiyah, Kingdom of Saudi Arabia 3 Institute of Computer Science, Al-Kawser city at Sohag Egypt Abstract. In this paper we introduce a new space, the so called QK,ω space of analytic func- tions on the unit disk in terms of nondecreasing functions. The relation between integral norm of QK,ω space and integral norm of the weighted Bloch spaceBαω is also given. AMS subject classifications: 30D45, 46E15 Key words: QK,ω spaces, weighted Bloch functions. ∗Corresponding author. Email addresses: r_rashwan�yahoo. om (R. Rashwan), ahsayed80�hotmail. om (A. Ahmed),alaa_mohamed1�yahoo. om (A. Kamal) http://www.ejpam.com 250 c© 2009 EJPAM All rights reserved. R. Rashwan, A. Ahmed and A. Kamal / Eur. J. Pure Appl. Math, 2 (2009), (250-267) 251 1. Introduction Let ∆= {z : |z|< 1} be the open unit disk in the complex plane C. Recall that the well known Bloch space (cf. [2]) is defined as follows: B = { f : f analytic in ∆ and sup z∈∆ (1− |z|2)| f ′(z)|<∞}; the little Bloch spaceB0 (cf. [2]) is a subspace ofB consisting of all f ∈B such that lim |z|→1− (1− |z|2)| f ′(z)|= 0. The Dirichlet space is defined by D = { f : f analytic in ∆ and ∫ ∆ � � f ′(z) � � 2 dσz <∞}, where dσz is the Euclidean area element d xd y. Let 0 < q <∞. Then the Besov-type spaces Bq = � f : f analytic in ∆ and sup a∈∆ ∫ ∆ � � f ′(z) � � q� 1− |z|2 �q−2 (1− |ϕa(z)| 2)2dσz <∞ � are introduced and studied intensively by Stroethoff (cf. [11]). Here, ϕa(z) stands for the Möbius transformation of ∆ given by ϕa(z) = a− z 1− āz , where a ∈∆. In 1994, Aulaskari and Lappan [2] introduced a class of holomorphic functions, the so called Qp-spaces as follows: Qp = � f : f analytic in ∆ and sup a∈∆ ∫ ∆ � � f ′(z) � � 2 g p(z, a)dσz <∞ � , where 0< p <∞ and the weight function g(z, a) = log � � � � 1− āz a− z � � � � R. Rashwan, A. Ahmed and A. Kamal / Eur. J. Pure Appl. Math, 2 (2009), (250-267) 252 is defined as the composition of the Möbius transformation ϕa and the fundamental solution of the two-dimensional real Laplacian. The weight function g(z, a) is actually Green’s function in ∆ with pole at a ∈∆. For 0 < p < ∞,−2 < q < ∞, we say that a function f analytic in ∆ belongs to the space QK(p, q) (cf. [14]), if ‖ f ‖K,p,q = sup a∈∆ ∫ ∆ � � f ′(z) � � p� 1− |z|2 �q K(g(z, a))dσz <∞. Recall that the analytic function f (z) = ∞ ∑ k akznk (with nk ∈ N ; for all k ∈ N= {1, 2, 3, . . . } ) is said to belong to the Hadamard gap class (also known as lacunary series) if there exists a constant c > 1 such that nk+1 nk ≥ c for all k ∈ N (see e.g. [17]). Two quantities A f and B f , both depending on an analytic function f on ∆, are said to be equivalent, written as A f ≈ B f , if there exists a finite positive constant C not depending on f such that for every analytic function f on ∆ we have: 1 C B f ≤ A f ≤ CB f . If the quantities A f and B f , are equivalent, then in particular we have A f <∞ if and only if B f <∞. Now, given a reasonable function ω : (0, 1]→ [0,∞), the weighted Bloch space Bω (see [4]) is defined as the set of all analytic functions f on ∆ satisfying (1− |z|)| f ′(z)| ≤ Cω(1− |z|), z ∈∆, for some fixed C = C f > 0. In the special case where ω ≡ 1,Bω reduces to the classical Bloch space B . Here, the word "reasonable" is a non-mathematical term; it was just intended to mean that the "not too bad" and the function satisfy some natural conditions. Now, we introduce the following definitions: R. Rashwan, A. Ahmed and A. Kamal / Eur. J. Pure Appl. Math, 2 (2009), (250-267) 253 Definition 1.1. For a given reasonable functionω : (0, 1]→ [0,∞) and for 0< α<∞. An analytic function f on ∆ is said to belong to the α−weighted Bloch space Bα ω if ‖ f ‖Bαω = sup z∈∆ (1− |z|)α ω(1− |z|) | f ′(z)| <∞. Definition 1.2. For a given reasonable functionω : (0, 1]→ [0,∞) and for 0< α<∞. An analytic function f on ∆ is said to belong to the little weighted Bloch space Bα ω,0 if ‖ f ‖Bαω,0 = lim |z|→1− (1− |z|)α ω(1− |z|) | f ′(z)|= 0. Throughout this paper and for some techniques we consider the case of ω 6≡ 0. Now, we introduce the following new definition: Definition 1.3. For a nondecreasing function K : [0,∞)→ [0,∞), 0 < p <∞, and for a given reasonable function ω : (0, 1]→ (0,∞), an analytic function f in ∆ is said to belong to the space QK,ω if ‖ f ‖pK,ω = sup a∈∆ ∫ ∆ � � f ′(z) � � p (1− |z|)p K(g(z, a)) ωp(1− |z|) dσz <∞. Remark 1.1. It should be remarked that our QK,ω classes are more general than many classes of analytic functions. If ω ≡ 1, we obtain QK(p, p) type spaces (cf. [14] and [15]). If p = 2, and ω(t) = t , we obtain QK spaces as studied recently in [5, 6, 9, 12, 13, 16] and others. If p = 2, ω(t) = t and K(t) = t p, we obtain Qp spaces as studied in [2, 3, 17] and others. If ω ≡ 1 and K(t) = t s, then QK,ω = F(p, p, s) classes (cf. [1,18]). In this paper, we characterize the weighted Bloch space Bα ω by our QK,ω spaces. One of the main results is a general Besov-type characterization forBα ω functions that extends and generalizes the Stroethoff’s theorem [11]. Also, we extend and improve some results due to Essén et. al [6] using our new definitions. R. Rashwan, A. Ahmed and A. Kamal / Eur. J. Pure Appl. Math, 2 (2009), (250-267) 254 2. Holomorphic QK ,ω Classes In this paper we show some relations between QK,ω norms and Bα ω norms for a nondecreasing function K , , also we give a general way to construct different spaces QK,ω1 and QK2,ω by using some functions K1 and K2. Before proving theorems we recall few facts about the Möbius function ϕa. First, the function ϕa is easily seen to be it own inverse under composition: (ϕa ◦ϕa)(z) = z for all z ∈∆ The following identity can be obtained by straight forward computation: 1− |ϕa(z)| 2 = (1− |a|2)(1− |z|2) |1− az|2 , (a, z ∈∆). A slightly different form in which we will apply the above identity is: 1− |ϕa(z)| 2 1− |z|2 = |ϕ′ a (z)| , (a, z ∈∆). (2.1) For a ∈ ∆, the substitution z = ϕa(w) results in the Jacobian change in measure given by dσw = |ϕ ′ a (z)|2dσz . For a Lebesgue integrable or a non-negative Lebesgue measurable function h on ∆ we thus have the following change-of-variable formula: ∫ ∆(0,r) h(ϕa(w))dσw = ∫ ∆(a,r) h(z) � 1− |ϕa(z)| 2 1− |z|2 �2 dσz . (2.2) We assume throughout this paper that ∫ 1 0 K � log 1 r � r (1− r2)2 dr <∞ . (2.3) We need the following lemmas in the sequel. Lemma 2.1. [17] Let α ∈ (0,∞) and suppose that f (z) = ∞ ∑ j=1 a jz n j belongs to Hadamard gap class. Then f ∈Bα if and only if sup j∈N |a j|n 1−α j <∞ , where N = {1, 2, 3, . . . } . R. Rashwan, A. Ahmed and A. Kamal / Eur. J. Pure Appl. Math, 2 (2009), (250-267) 255 Lemma 2.2. Let ω : (0, 1]→ (0,∞) be a nondecreasing function. Then there are two functions f1 , f2 ∈Bω such that | f ′ 1 (z)|+ | f ′ 2 (z)| ≈ ω(1− |z|) (1− |z|) , z ∈∆. (2.4) Proof. For a large number q ∈ N, choose a gap series: f1(z) = ∞ ∑ j=0 zq j , z ∈∆. Then, apply lemma 2.1 to infer that (1−|z|)| f ′1(z)| ω((1−|z|)) ≤ λ holds for all z ∈ ∆, where λ is a constant. Furthermore, let us verify (1− |z|)| f ′ 1 (z)| ω((1− |z|)) ≥ λ , 1− q−k ≤ |z| ≤ 1− q−(k+ 1 2 ) , k ∈ N. (2.5) And q−(k+ 1 2 ) ≤ 1− |z| ≤ q−k⇒ω(q−(k+ 1 2 ))≤ω(1− |z|)≤ω(q−k). Observe that for any z ∈∆, | f ′ 1 (z)| ≥ qk|z|q k − k−1 ∑ j=0 q j|z|q j − ∞ ∑ k+1 q j|z|q j = T1− T2− T3. And then, fix a z with |z| ∈ [1− q−k, 1− q−(k+ 1 2 )], k ∈ N, and put x = |z|q k . Thus (1− q−k)q k ≤ x ≤ [(1− q−(k+ 1 2 ))q k+ 1 2 ]q −1 2 . If q is large enough, then for k ≥ 1 one has 1 3 ≤ x ≤ ( 1 2 )q −1 2 , (2.6) and hence T1 ≥ qk 3 . Since it is easy to establish T2 ≤ k−1 ∑ j=0 q j ≤ qk q− 1 , R. Rashwan, A. Ahmed and A. Kamal / Eur. J. Pure Appl. Math, 2 (2009), (250-267) 256 it remains to deal with the third term T3. Noting that |z|q n(q−1) ≤ |z|q k+1(q−1), n ≥ k+ 1, namely, in T3 the quotient of two successive terms is not greater than the ratio of the first two terms, one finds that the series of T3 is controlled by the geometric series having the same first two terms. Accordingly (2.6) is applied to produce T3 ≤ qk+1|z|q k+1 ∞ ∑ j=0 � q|z|q k+2−qk+1 � j = qk+1|z|q k+1 1− q|z|(qk+2−qk+1) = qk qxq 1− qxq2−q ≤ qk q(1 2 )q 1 2 1− q(1 2 )q 3 2 − q 1 2 . The preceding estimates for T1, T2 and T3 imply | f ′ 1 (z)| ≥ qk 4 ω(1− |z|) ω(1− |z|) = qk+ 1 2 4q 1 2 ω(1− |z|) ω(1− |z|) ≥ ω(1− |z|) 4q 1 2 (1− |z|)×ω(1− |z|) ≥ ω(1− |z|) 4q 1 2ω(q−k)× (1− |z|) ; ω(q−k) 6→ ∞. Reaching (2.5). In a completely similar manner one can prove that if q is a large natural number, for example q = m2 where m is a large natural number, and if f2(z) = ∞ ∑ j=0 zq j , z ∈∆, then (1− |z|2)| f ′ 2 (z)| ≤ λ for all z ∈∆ (owing to Lemma 2.1) and (1− |z|)| f ′ 1 (z)| ω((1− |z|)) ≤ λ, 1− q−(k+ 1 2 ) ≤ |z| ≤ 1− q−(k+1), k ∈ N. (2.7) R. Rashwan, A. Ahmed and A. Kamal / Eur. J. Pure Appl. Math, 2 (2009), (250-267) 257 Of course, (2.5) and (2.7) yield (2.4) unless it occurs that f ′ 1 and f ′ 2 have common zero in {z ∈ ∆ : |z| < 1 − q−1} in which case one can replace f2 with f2(ζz) for appropriate ζ ∈ ∂∆, where ∂∆ is the boundary of the unit disk (note that f ′(0) = 1). Our lemma is therefore proved . Using the same steps of Lemma 2.2, it is not hard to prove the following lemma. Lemma 2.3. Let ω : (0, 1]→ (0,∞) be a nondecreasing function and let 1 ≤ α <∞. Then there are two functions f1 , f2 ∈B α ω such that | f ′ 1 (z)|+ | f ′ 2 (z)| ≈ ω(1− |z|) (1− |z|)α , z ∈∆. (2.8) Proof. The proof is very similar to the proof of Lemma 2.2 and lemma 3.1 in [7], so it will be omitted. Theorem 2.1. For each non-decreasing function K : [0,∞) → [0,∞), 0 < p < ∞ and for a given reasonable non-decreasing function ω : (0, 1]→ (0,∞) with ω(α t) ≈ ω(t), α > 0, we have that (i) QK,ω ⊂B p+2 p ω and (ii) QK,ω =B p+2 p ω , iff ∫ 1 0 K � log 1 r � r (1− r2)2 dr <∞. Proof. For a fixed r ∈ (0, 1) and a ∈∆, let E(a, r) = � z ∈∆ , |z− a| < r(1− |a|) � . We know that E(a, r) ⊂∆(a, r) and for any z ∈ E(a, r), we have (1− r)(1− |a|) ≤ 1− |z| ≤ (1+ r)(1− |a|), which means that 1− |z|2 ≃ 1− |a|2 for any z ∈ E(a, r). Denote Fω,p( f )(z) = � � f ′(z) � � p (1− |z|)p ωp(1− |z|) R. Rashwan, A. Ahmed and A. Kamal / Eur. J. Pure Appl. Math, 2 (2009), (250-267) 258 Then, we obtain ∫ ∆ Fω,p( f )(z)K � g(z, a) � dσz ≥ ∫ ∆(a,r) Fω,p( f )(z)K � g(z, a) � dσz ≥ K � log 1 r � ∫ ∆(a,r) Fω,p( f )(z) dσz ≥ K � log 1 r � ∫ E(a,r) Fω,p( f )(z) dσz. For every z ∈ E(a, r), we have that (1− r)(1− |a|) ≤ 1− |z| ≤ (1+ r)(1− |a|), Then, (1− |z|)p ≥ (1− r)p(1− |a|)p , ∀ p > 0. Now, since we assume that ω is non-decreasing, then we obtain that ∫ E(a,r) Fω,p( f )(z) dσz ≥ (1− r)p(1− |a|)p ωp((1− r)(1− |a|)) ∫ E(a,r) � � f ′(z) � � p dσz. Since | f ′(z)|p is a subharmonic function, then ∫ E(a,r) � � f ′(z) � � p dσz ≥ |E(a, r)| . | f ′(a) � � p = r2(1− |a|)2| f ′(a) � � p . Then we obtain ∫ ∆ Fω,p( f )(z)K � g(z, a) � dσz ≥ K � log 1 r � (1− r)p(1− |a|)p+2 ωp((1− r)(1− |a|)) | f ′(a) � � p ≥ λK � log 1 r � (1− r)p(1− |a|)p+2 ωp(1− |a|) | f ′(a) � � p where λ is a constant. If f ∈ QK,ω, then by the above estimate we have that sup a∈∆ (1− |a|)p+2| f ′(z)|p ωp(1− |a|) <∞. R. Rashwan, A. Ahmed and A. Kamal / Eur. J. Pure Appl. Math, 2 (2009), (250-267) 259 The proof of (i) is therefore completed. Now, we show that B p+2 p ω ⊂ QK,ω provided that K satisfies condition (2.3). For f ∈ B p+2 p ω , we have that, ∫ ∆ Fω,p( f )(z)K � g(z, a) � dσz ≤ f p B p+2 p ω ∫ ∆ (1− |z|2)−2K � g(z, a) � dσz = 2π f p B p+2 p ω ∫ 1 0 K � log 1 r � r (1− r2)2 dr <∞, which shows that B p+2 p ω ⊂ QK,ω. Now we assume thatB p+2 p ω = QK,ω and we verify (2.3) holds. From Lemma 2.3, for f1 and f2 inB p+2 p ω , we have that | f ′ 1 (z)|+ | f ′ 2 (z)| ≥ ω(1− |z|) (1− |z|) p+2 p . (2.9) Then f1, f2 ∈ QK,ω and ∞ > sup a∈∆ ∫ ∆ � � � f ′ 1 (z) � � p + � � f ′ 2 (z) � � p � (1− |z|)p K � g(z, a) � ωp(1− |z|) dσz ≥ ∫ ∆ � � � f ′ 1 (z) � �+ � � f ′ 2 (z) � � �p (1− |z|)p K � g(z, 0) � ωp(1− |z|) dσz (2.10) From (2.9) and (2.10), we obtain ∫ ∆ � � � f ′ 1 (z) � � p + � � f ′ 2 (z) � � p� (1− |z|)p K � g(z, 0) � ωp(1− |z|) dσz ≈ 2π ∫ 1 0 K � log 1 r � r (1− r2)2 dr. Thus (2.3) holds, and this completes the proof. 3. The Classes QK ,ω,0 and Bα ω,0 We say that f ∈ QK,ω,0 if lim |a|→1− ∫ ∆ � � f ′(z) � � p (1− |z|)p K(g(z, a)) ωp(1− |z|) dσz = 0. (3.1) R. Rashwan, A. Ahmed and A. Kamal / Eur. J. Pure Appl. Math, 2 (2009), (250-267) 260 Also, as a subspace ofBα ω , we define the little weighted Bloch spaceBα ω,0 as the space which consists of analytic functions f on ∆ such that lim |z|→1− (1− |z|)α| f ′(z)| ω(1− |z|) = 0 where 0< α <∞. Thus we can obtain the following theorem: Theorem 3.1. For each nondecreasing function K : [0,∞)→ [0,∞), 0 < p <∞, for a given reasonable non-decreasing function ω : (0, 1]→ (0,∞) withω(α t)≈ω(t), α > 0. Then (i) QK,ω,0 ⊂B p+2 p ω,0 and (ii) QK,ω,0 =B p+2 p ω,0 , if and only if (2.3) holds. Proof. Without loss of generality, we assume that K(1) > 0. From the proof of Theorem 2.1, we have that π(1 e )2 K(1) (1−|a|)p+2 ωp(1−|a|) | f ′(a)|p ≤ K(1) ∫ E(a) Fω,p( f )(z) dσz ≤ K(1) ∫ ∆(a, 1 e ) Fω,p( f )(z) dσz ≤ ∫ ∆ Fω,p( f )(z)K � g(z, a) � dσz , where E(a) = � z ∈∆ , |z− a| < 1 e (1− |a|) � . If f ∈ QK,ω,0, we obtain that lim |a|→1− (1− |a|)p+2| f ′(a)|p ωp(1− |a|) = 0. (ii) We only need to prove thatB p+2 p ω,0 ⊂QK,w,0. Assume that A= ∫ 1 0 K � log 1 r � r (1− r2)2 dr <∞. R. Rashwan, A. Ahmed and A. Kamal / Eur. J. Pure Appl. Math, 2 (2009), (250-267) 261 For a given ε > 0 there exists an r1, 0 < r1 < 1, such that ∫ 1 r1 K � log 1 r � r (1− r2)2 dr < ε. (3.2) Then we have that, ∫ ∆\∆(a,r1) � � f ′(z) � � p (1− |z|)p K(g(z, a)) ωp(1− |z|) dσz ≤ f p B p+2 p ω,0 ∫ ∆\∆(a,r1) K(g(z, a)) (1− |z|2)2 dσz = f p B p+2 p ω,0 ∫ r1<|w|<1 K � log 1 |w| � 1 (1− |w|2)2 dσw = f p B p+2 p ω,0 ∫ 1 r1 K � log 1 r � r (1− r2)2 dr ≤ 2πε f p B p+2 p ω,0 . (3.3) Similarly, if f ∈B p+2 p ω,0 , we obtain that | f ′(ϕa(w))| p (1− |ϕa(w)| 2) p+2 p ωp(1− |ϕa(w)|) −→ 0 converges uniformly for |w| ≤ r if |a| → 1−, where r is fixed and 0 < r < 1. Then, we obtain that lim |a|→1− ∫ ∆ � � f ′(z) � � p (1− |z|)p K � g(z, a) � ωp(1− |z|) dσz = lim |a|→1− ∫ |w| 0 that, t2β exp{−4t} t= β 2 = �β 2 �2β exp{−2β}. Then, there exists an integer k for 3 4 ≤ r < 1 such that β 2 ≤ 2k(1− r) < β+1 2 and 2βk exp{−2k+2(1− r)} = (1− r)−2β � 2k(1− r) �2β exp{−2k+2(1− r)} > � 1+ β 2 �2β (1− r)−2β exp{−2(β + 1)}. (3.8) For 3 4 ≤ r < 1 we define f0(z) = ∞ ∑ k=0 ak 2 2k p z2k , where ak = g � 1− (p+1) p 2k � , k = 0, 1, 2, . . . . By (3.7) and (3.8), we deduce that M2 2 (r, f ′ 0 ) = ∫ 2π 0 | f ′0(r eiθ )|2 dθ = 2π ∞ ∑ k=0 a2 k 2 2k(p+2) p z2k−2 ≥ 2πg 2 p (r) 2 2k(p+2) p exp{−2k+2(1− r)} ≥ λ g 2 p (r)(1− r) −2(p+2) p , (3.9) where λ is a constant. Since f0 is defined by a gap series with Hadamard condition, we have M2(r, f ′ 0 )≈ Mp(r, f ′ 0 ), where Mp(r, f ′ 0 ) = � ∫ 2π 0 | f ′0(r eiθ )|p dθ � 1 p . R. Rashwan, A. Ahmed and A. Kamal / Eur. J. Pure Appl. Math, 2 (2009), (250-267) 263 Therefore, sup a∈∆ ∫ ∆ � � f ′ 0 (z) � � p (1− |z|)p K(g(z, a)) ωp(1− |z|) dσz ≥ ∫ 1 0 M p p (r, f ′ 0 )(1− r2)pK � log 1 r � r dr ≈ ∫ 1 0 M p 2 (r, f ′ 0 )(1− r2)pK � log 1 r � r dr ≥ ∫ 1 3 4 K � log 1 r � g(r) (1− r2)2 r dr =∞. This means that f0 ∈ B p+2 p ω,0 \QK,w,0, which is a contraction. Hence (2.3) holds. This completes the proof of our theorem. 4. More Results on QK ,ω-spaces The following result means that the kernel function K can be chosen as bounded. Theorem 4.1. Assume that K(1) > 0. Let K1(r) = inf{K(r), K(1)}, then QK,w = QK1,w. Proof. Since K1 ≤ K and K1 is nondecreasing, it is clear that QK,ω ⊂ QK1,w. It remains to prove that QK1,ω ⊂ QK,ω. We note that g(z, a) > 1, z ∈∆(a, 1 e ) and g(z, a) ≤ 1, z ∈∆ \∆(a, 1 e ). Thus K(g(z, a)) = K1(g(z, a)) in ∆\∆(a, 1 e ). It suffices to deal with integrals over ∆(a, 1 e ). If f ∈QK1,ω and f is a weighted Bloch function i.e, f ∈Bω then by Theorem 2.1, it follows that ∫ ∆(a, 1 e ) | f ′(z)|p (1− |z|)p K � g(z, a) � ωp(1− |z|) dσz ≤ f p B p+2 p ω ∫ ∆(a, 1 e ) K � g(z, a) � 1 (1− |z|2)2 dσz = f p B p+2 p ω ∫ ∆(0, 1 e ) K � log 1 |w| � 1 (1− |z|2)2 dσw ≤ C f p B p+2 p ω R. Rashwan, A. Ahmed and A. Kamal / Eur. J. Pure Appl. Math, 2 (2009), (250-267) 264 Thus, f ∈QK,ω and Theorem 4.1 is proved. Corollary 4.1. Let 0< p <∞, ω : (0, 1]→ (0,∞). Then f ∈ QK,w if and only if sup a∈∆ ∫ ∆ | f ′(z)|p (1− |z|)p K(1− |ϕa(z)| 2) ωp(1− |z|) dσz <∞. For the application of the above results, we state the following lemma which is needed later. Lemma 4.1. Let K : [0,∞)→ [0,∞), 0< p <∞, for a given reasonable function ω : (0, 1]→ (0,∞). Then (i) f ∈B p+2 p ω if and only if there exists R ∈ (0, 1) such that sup a∈∆ ∫ ∆(a,R) | f ′(z)|p (1− |z|)p (1− |z|)K(g(z, a)) ωp(1− |z|) dσz <∞, (4.1) (ii) f ∈B p+2 p ω,0 if and only if there exists R ∈ (0, 1) such that lim |a|→1− ∫ ∆(a,R) | f ′(z)|p (1− |z|)p K(g(z, a)) ωp(1− |z|) dσz = 0. (4.2) Proof. (i) Assume f ∈B p+2 p ω . For any R ∈ (0, 1) and a ∈∆, we have ∫ ∆(a,R) | f ′(z)|p (1− |z|)p K(g(z, a)) ωp(1− |z|) dσz = ∫ ∆(0,R) | f ′(ϕa(z))| p (1− |ϕa(z)| 2)p+2 (1+ |ϕa(z)|)p+2 K � 1 |z| � (1− |z|2)2ωp(1− |z|) dσz ≤ ‖ f ‖p B p+2 p ω ∫ ∆(0,R) K � log 1 |z| � 1 (1− |z|2)2 dσz ≤ λ1‖ f ‖ p B p+2 p ω , where 1 < (1+|ϕa(z)|) p+2 < 2p+2 and λ1 is a constant. Conversely, suppose that (4.1) holds for some R, 0 < R< 1, by the proof of Theorem 2.1 (i) with 1− |a| ≈ 1− |z| on R. Rashwan, A. Ahmed and A. Kamal / Eur. J. Pure Appl. Math, 2 (2009), (250-267) 265 E(a, R) ; a, z ∈∆, we obtain ∫ ∆(a,R) | f ′(z)|p (1− |z|)p K(g(z, a)) ωp(1− |z|) dσz ≥ K(log 1 R ) ∫ ∆(a,R) | f ′(z)|p (1− |z|)p ωp(1− |z|) dσz ≥ λ2 K � log 1 R � ω−p(1− |a|) ∫ E(a,R) | f ′(z)|p (1− |z|)p dσz ≥ πλ2R2K � log 1 R � (1− |a|)p ωp(1− |a|) | f ′(a)|p , where λ2 is a constant. The last inequality shows that f ∈ B p+2 p ω The proof of (ii) is similar to proof (i) by taking the limit when |a| −→ 1− in (i), hence it can be omitted. Theorem 4.2. Let 0< p <∞, ω : (0, 1]→ (0,∞). Assume K1(r)≤ K2(r) for r ∈ (0, 1) and K1(r) K2(r) → 0 as r → 0. If the integral in (2.3) is divergent for K2, then QK2,ω $ QK1,ω . Proof. It is clear that QK2,ω ⊂ QK1,ω. Suppose that QK2,ω =QK1,ω. By the open mapping theorem (see [8]), we know that the identity map from one of these spaces into the other one is continuous. Thus there exists a constant C such that ‖ f ‖K2,ω ≤ C‖ f ‖K1,ω . Since K1(r) K2(r) → 0 as r → 0, then there exists r0 ∈ (0, 1) such that K1(r) ≤ (2C)−1K2(r) for 0 < r ≤ r0. Choose t0 = e−r0 and we deduce that if f ∈ QK2,ω, then sup a∈∆ ∫ ∆ | f ′(z)|p (1− |z|)p K2 � g(z, a) � ωp(1− |z|) dσz ≤ C sup a∈∆ ∫ ∆(a,t0) | f ′(z)|p (1− |z|)p K1 � g(z, a) � ωp(1− |z|) dσz + 1 2 sup a∈∆ ∫ ∆ | f ′(z)|p (1− |z|)p K2 � g(z, a) � ωp(1− |z|) dσz . Therefore, sup a∈∆ ∫ ∆ | f ′(z)|p (1−|z|)p K2 � g(z, a) � ωp(1− |z|) dσz ≤ 2C sup a∈∆ ∫ ∆(a,t0) | f ′(z)|p (1−|z|)p K1 � g(z, a) � ωp(1− |z|) dσz . REFERENCES 266 By Lemma 4.1 and for f ∈ QK2,ω, there exists a constant C1 such that sup a∈∆ ∫ ∆ | f ′(z)|p (1− |z|)p K2 � g(z, a) � ωp(1− |z|) dσz ≤ C1‖ f ‖ p B p+2 p ω . (4.3) If g ∈ B p+2 p ω and gr(z) = g(rz) , 0 < r < 1, then gr B p+2 p ω ≤ g B p+2 p ω . Since gr ∈ QK2,ω , 0 < r < 1, we can choose f = gr in the inequality (4.3). Using Fatou’s lemma (see [10]), we deduce that sup a∈∆ ∫ ∆ |g ′(z)|p (1− |z|)p K2(g(z, a)) ωp(1− |z|) dσz < C1 g p B p+2 p ω . 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