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Abstract and Applied Analysis
VolumeΒ 2008Β (2008), Article IDΒ 404636, 9 pages
http://dx.doi.org/10.1155/2008/404636
Research Article

Some Sufficient Conditions for Analytic Functions to Belong to 𝒬𝐾,0(𝑝,π‘ž) Space

Department of Mathematics, Jia Ying University, Meizhou 514015, Guangdong, China

Received 31 May 2008; Accepted 7 June 2008

Academic Editor: StevoΒ Stevic

Copyright Β© 2008 Xiaoge Meng. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

This paper gives some sufficient conditions for an analytic function to belong to the space consisting of all analytic functions 𝑓 on the unit disk such lim|π‘Ž|β†’1βˆ«π”»|π‘“ξ…ž(𝑧)|𝑝(1βˆ’|𝑧|2)π‘žπΎ(𝑔(𝑧,π‘Ž))𝑑𝐴(𝑧)=0.

1. Introduction

Let 𝔻 be the open unit disk in the complex plane β„‚ and 𝐻(𝔻) the space of all analytic functions in 𝔻. For π‘Žβˆˆπ”», letπœ‘π‘Ž(𝑧)=π‘Žβˆ’π‘§1βˆ’π‘Žπ‘§(1.1)be the MΓΆbius transformation of 𝔻 and let𝑔(𝑧,π‘Ž)=log|1βˆ’π‘Žπ‘§|π‘Žβˆ’π‘§(1.2)be the Green's function on 𝔻. Let 𝔻(π‘Ž,π‘Ÿ) denote the pseudo-hyperbolic metric disk centered at π‘Žβˆˆπ”» with radius π‘Ÿβˆˆ(0,1), that is, 𝔻(π‘Ž,π‘Ÿ)={π‘§βˆˆπ”»βˆΆ|πœ‘π‘Ž(𝑧)|<π‘Ÿ}.

It is said that an analytic function βˆ‘π‘“(𝑧)=βˆžπ‘˜=1π‘Žπ‘˜π‘§π‘›π‘˜ is defined by a lacunary series ifπœ†=infπ‘˜βˆˆβ„•π‘›π‘˜+1π‘›π‘˜>1.(1.3)For some results in the topic, see, for example, [1–6] and the references therein.

Given a function 𝐾∢(0,∞)β†’[0,∞), we consider the space 𝒬𝐾(𝑝,π‘ž) of all functions π‘“βˆˆπ»(𝔻) such that‖𝑓‖𝑝𝒬𝐾(𝑝,π‘ž)=supπ‘Žβˆˆπ”»ξ€œπ”»|π‘“ξ…ž(𝑧)|𝑝(1βˆ’|𝑧|2)π‘žπΎ(𝑔(𝑧,π‘Ž))𝑑𝐴(𝑧)<∞.(1.4)By 𝒬𝐾,0(𝑝,π‘ž) we denote the space consisting of all π‘“βˆˆπ’¬πΎ(𝑝,π‘ž) such thatlim|π‘Ž|β†’1ξ€œπ”»|π‘“ξ…ž(𝑧)|𝑝(1βˆ’|𝑧|2)π‘žπΎ(𝑔(𝑧,π‘Ž))𝑑𝐴(𝑧)=0,(1.5)where 0<𝑝<∞, βˆ’2<π‘ž<∞, and 𝑑𝐴(𝑧)=(1/πœ‹)𝑑π‘₯𝑑𝑦=(1/πœ‹)π‘Ÿπ‘‘π‘Ÿπ‘‘πœƒ.

For 𝑝=2, π‘ž=0, the space 𝒬𝐾(𝑝,π‘ž) is reduced to 𝒬𝐾 (see, e.g., [7]). If 𝐾(𝑔(𝑧,π‘Ž))=(𝑔(𝑧,π‘Ž))𝑠, 0≀𝑠<∞, then 𝒬𝐾(𝑝,π‘ž)=𝐹(𝑝,π‘ž,𝑠) (see, e.g., [8, 9]).

Throughout the paper, we assume that the condition holds (see [7])ξ€œ10(1βˆ’π‘Ÿ2)π‘ž1𝐾(logπ‘Ÿ)<∞,(1.6)so that the space 𝒬𝐾(𝑝,π‘ž) we study is nontrivial. We also assume that 𝐾 as a nondecreasing function. An important tool in the study of 𝒬𝐾 spaces is the auxiliary function πœ‘πΎ defined by (see [10])πœ‘πΎ(𝑠)=sup0<𝑑≀1𝐾(𝑠𝑑)𝐾(𝑑),0<𝑠<∞.(1.7)The following conditionξ€œβˆž1πœ‘πΎ(𝑠)𝑑𝑠𝑠2<∞(1.8)is crucial in this paper. It has played an important role in the study of 𝒬𝐾 spaces during the last few years.

In this paper, we give some sufficient conditions for an analytic function 𝑓 to belong to the space 𝒬𝐾,0(𝑝,π‘ž).

The followings are our main results in this paper.

Theorem 1.1. Let π‘“βˆˆπ»(𝔻), 0<𝑝<∞, βˆ’2<π‘ž<∞, and let πœ‘ be a monotone increasing function in π‘Ÿ on (0,1) such that |𝑓′(𝑧)|β‰€πœ‘(π‘Ÿ), for |𝑧|=π‘Ÿ. If ξ€œ10πœ‘π‘(π‘Ÿ)(1βˆ’π‘Ÿ2)π‘žπΎ(1βˆ’π‘Ÿ2)π‘Ÿπ‘‘π‘Ÿ<∞,(1.9)then π‘“βˆˆπ’¬πΎ,0(𝑝,π‘ž).

Theorem 1.2. For 1≀𝑝<2, 0β‰€π‘ž<∞, and 1β‰€π‘βˆ’2π‘ž<3. If 𝐾 satisfies condition (1.7) and is a function with the property that 𝐾(𝑑)=𝐾(1) for 𝑑β‰₯1, then a lacunary series βˆ‘π‘“(𝑧)=βˆžπ‘˜=1π‘Žπ‘˜π‘§π‘›π‘˜ belongs to 𝒬𝐾,0(𝑝,π‘ž) if βˆžξ“π‘˜=1π‘›π‘˜π‘βˆ’π‘žβˆ’1|π‘Žπ‘˜|𝑝1𝐾(π‘›π‘˜)<∞.(1.10)

Throughout this paper, 𝐢 stands for a positive constant, whose value may differ from one occurrence to the other. The expression π‘Žβ‰ˆπ‘ means that there is a positive constant 𝐢 such that πΆβˆ’1π‘Žβ‰€π‘β‰€πΆπ‘Ž.

2. Main Results and Proofs

In this section, we give the proofs of Theorems 1.1 and 1.2. Before formulating the main results, we give some lemmas which are used in the proofs.

Lemma 2.1 (see [7]). Let 0<𝑝<∞, βˆ’2<π‘ž<∞,π‘“βˆˆπ»(𝔻). Then, π‘“βˆˆπ’¬πΎ,0(𝑝,π‘ž) if and only if lim|π‘Ž|β†’1ξ€œπ”»|π‘“ξ…ž(𝑧)|𝑝(1βˆ’|𝑧|2)π‘žπΎ(1βˆ’|πœ‘π‘Ž(𝑧)|2)𝑑𝐴(𝑧)=0.(2.1)

Lemma 2.2 (see [5]). Let 𝐾 be a function with the property that 𝐾(𝑑)=𝐾(1) for 𝑑β‰₯1. If 𝐾 satisfies condition (1.8), then there exists a constant 𝐢>0 such that 𝐾(2𝑑)β‰ˆπΎ(𝑑) for 𝑑>0.

Lemma 2.3 (see [5]). If 𝐾 satisfies condition (1.8), then we can find another nonnegative function πΎβˆ— such that 𝒬𝐾=π’¬πΎβˆ— and the new function πΎβˆ— has the following properties: (a)πΎβˆ— is nondecreasing on (0,∞);(b)πΎβˆ— satisfies condition (βˆ—);(c)πΎβˆ—(2𝑑)β‰ˆπΎβˆ—(𝑑) on (0,∞);(d)πΎβˆ— is differentiable (up to any given order) on (0,∞);(e)πΎβˆ— is concave on (0,∞);(f)πΎβˆ—(𝑑)=πΎβˆ—(1) for 𝑑β‰₯1;(g)πΎβˆ—(𝑑)β‰ˆπΎ(1) on (0,1].

Lemma 2.4 (see [5]). If 𝐾 satisfies condition (1.8), then for any 𝛼β‰₯1 and 0≀𝛽<1, one has ξ€œ10π‘Ÿπ›Όβˆ’11(logπ‘Ÿ)βˆ’π›½1𝐾(logπ‘Ÿ)π‘‘π‘Ÿβ‰ˆπΆ(𝛽)(1βˆ’π›½π›Ό)1βˆ’π›½πΎ(1βˆ’π›½π›Ό),(2.2)where 𝐢(𝛽) is a constant depending on 𝛽 alone.

Lemma 2.5 (see [11]). For 0<𝑝≀1, π‘Žβˆˆπ”», and 𝑧=π‘Ÿπ‘’π‘–πœƒβˆˆπ”», ξ€œ02πœ‹π‘‘πœƒ|1βˆ’π‘Žπ‘Ÿπ‘’π‘–πœƒ|2𝑝≀𝐢(1βˆ’|π‘Ž|π‘Ÿ)𝑝,(2.3)where 𝐢>0 is a constant.

Proof of Theorem 1.1. Let 𝑧=π‘Ÿπ‘’π‘–πœƒ. By Lemma 2.3, we may also assume that 𝐾 is concave, so that the following inequality true holds1ξ€œ2πœ‹02πœ‹πΎ(1βˆ’|πœ‘π‘Ž(π‘Ÿπ‘’π‘–πœƒ)|21)π‘‘πœƒβ‰€πΎ(ξ€œ2πœ‹02πœ‹(1βˆ’|πœ‘π‘Ž(π‘Ÿπ‘’π‘–πœƒ)|2)π‘‘πœƒ).(2.4)From the definition of πœ‘πΎ for 0<𝑠, 𝑑<1, we have that 𝐾(𝑠𝑑)β‰€πœ‘πΎ(𝑠)𝐾(𝑑). Using these facts and polar coordinates, it follows thatξ€œπΌ(π‘Ž)=𝔻|π‘“ξ…ž(𝑧)|𝑝(1βˆ’|𝑧|2)π‘žπΎ(1βˆ’|πœ‘π‘Ž(𝑧)|2=1)𝑑𝐴(𝑧)ξ€œ2πœ‹10ξ€œ02πœ‹|π‘“ξ…ž(π‘Ÿπ‘’π‘–πœƒ)|𝑝(1βˆ’π‘Ÿ2)π‘žπΎ(1βˆ’|πœ‘π‘Ž(π‘Ÿπ‘’π‘–πœƒ)|2ξ€œ)π‘Ÿπ‘‘π‘Ÿπ‘‘πœƒβ‰€210πœ‘π‘(π‘Ÿ)(1βˆ’π‘Ÿ2)π‘ž(1ξ€œ2πœ‹02πœ‹πΎ(1βˆ’|πœ‘π‘Ž(π‘Ÿπ‘’π‘–πœƒ)|2ξ€œ)π‘‘πœƒ)π‘Ÿπ‘‘π‘Ÿβ‰€210πœ‘π‘(π‘Ÿ)(1βˆ’π‘Ÿ2)π‘ž1𝐾(ξ€œ2πœ‹02πœ‹(1βˆ’|πœ‘π‘Ž(π‘Ÿπ‘’π‘–πœƒ)|2ξ€œ)π‘‘πœƒ)π‘Ÿπ‘‘π‘Ÿ=210πœ‘π‘(π‘Ÿ)(1βˆ’π‘Ÿ2)π‘ž1𝐾(ξ€œ2πœ‹02πœ‹1βˆ’|π‘Ž|2|1βˆ’π‘Žπ‘Ÿπ‘’π‘–πœƒ|2π‘‘πœƒ(1βˆ’π‘Ÿ2ξ€œ))π‘Ÿπ‘‘π‘Ÿ=210πœ‘π‘(π‘Ÿ)(1βˆ’π‘Ÿ2)π‘žπΎ(1βˆ’|π‘Ž|21βˆ’|π‘Ž|2π‘Ÿ2(1βˆ’π‘Ÿ2ξ€œ))π‘Ÿπ‘‘π‘Ÿβ‰€210πœ‘π‘(π‘Ÿ)(1βˆ’π‘Ÿ2)π‘žπΎ(1βˆ’π‘Ÿ2)πœ‘πΎ(1βˆ’|π‘Ž|21βˆ’|π‘Ž|2π‘Ÿ2)π‘Ÿπ‘‘π‘Ÿ.(2.5)The last integral exists for every π‘Žβˆˆπ”» in view of (1.9), and πœ‘πΎ((1βˆ’|π‘Ž|2)/(1βˆ’|π‘Ž|2π‘Ÿ2))≀1. Further, sincelim|π‘Ž|β†’1πœ‘πΎ(1βˆ’|π‘Ž|21βˆ’|π‘Ž|2π‘Ÿ2)=πœ‘πΎ(0)=0,(2.6)then the last integral tends to zero for every π‘Ÿβˆˆ(0,1) as |π‘Ž|β†’1. By Lebesgue's dominated convergence theorem, we obtain lim|π‘Ž|β†’1𝐼(π‘Ž)=0. By Lemma 2.1, we get π‘“βˆˆπ’¬πΎ,0(𝑝,π‘ž).

From Theorem 1.1, we have the following corollary. Here, we give a different and technical proof.

Corollary 2.6. Let π‘“βˆˆπ»(𝔻), 0<𝑝<∞, βˆ’2<π‘ž<∞, 0<𝑠≀1, and let πœ‘ be a monotone increasing function of π‘Ÿ in (0,1) such that |𝑓′(𝑧)|β‰€πœ‘(π‘Ÿ), for |𝑧|=π‘Ÿ. If ξ€œ10πœ‘π‘(π‘Ÿ)(1βˆ’π‘Ÿ2)π‘ž+π‘ π‘Ÿπ‘‘π‘Ÿ<∞,(2.7)then π‘“βˆˆπΉ0(𝑝,π‘ž,𝑠).

Proof. Let π‘Žβˆˆπ”». We haveξ€œπ½(π‘Ž)=𝔻|π‘“ξ…ž(𝑧)|𝑝(1βˆ’|𝑧|2)π‘ž(𝑔(𝑧,π‘Ž))π‘ ξ€œπ‘‘π΄(𝑧)=(𝔻⧡𝔻(π‘Ž,1/2)+ξ€œπ”»(π‘Ž,1/2))|π‘“ξ…ž(𝑧)|𝑝(1βˆ’|𝑧|2)π‘ž1(log|πœ‘π‘Ž)(𝑧)|𝑠𝑑𝐴(𝑧)=𝐼1(π‘Ž)+𝐼2(π‘Ž).(2.8)For π‘§βˆˆπ”»β§΅π”»(π‘Ž,1/2), 1/|πœ‘π‘Ž(𝑧)|≀2, hencelog|1βˆ’π‘Žπ‘§π‘Žβˆ’π‘§|<|1βˆ’π‘Žπ‘§π‘Žβˆ’π‘§|βˆ’1≀4(1βˆ’|π‘Žβˆ’π‘§1βˆ’|π‘Žπ‘§2)=4(1βˆ’|𝑧|2)(1βˆ’|π‘Ž|2)|1βˆ’π‘Žπ‘§|2.(2.9)For 0<𝑠≀1, by Lemma 2.5, we have𝐼1ξ€œ(π‘Ž)=𝔻⧡𝔻(π‘Ž,1/2)|π‘“ξ…ž(𝑧)|𝑝(1βˆ’|𝑧|2)π‘ž1(log|πœ‘π‘Ž)(𝑧)|𝑠𝑑𝐴(𝑧)≀4π‘ ξ€œπ”»β§΅π”»(π‘Ž,1/2)|π‘“ξ…ž(𝑧)|𝑝(1βˆ’|𝑧|2)π‘ž+𝑠(1βˆ’|π‘Ž|2)𝑠|1βˆ’π‘Žπ‘§|2𝑠≀4𝑑𝐴(𝑧)π‘ πœ‹ξ€œ10|π‘“ξ…ž(π‘Ÿπ‘’π‘–πœƒ)|𝑝(1βˆ’π‘Ÿ2)π‘ž+𝑠[ξ€œ02πœ‹(1βˆ’|π‘Ž|2)π‘ π‘‘πœƒ|1βˆ’π‘Žπ‘Ÿπ‘’π‘–πœƒ|2𝑠≀2]π‘Ÿπ‘‘π‘Ÿπ‘ 4π‘ πœ‹πΆξ€œ10πœ‘π‘(π‘Ÿ)(1βˆ’π‘Ÿ2)π‘ž+𝑠(1βˆ’|π‘Ž|)𝑠(1βˆ’|π‘Ž|π‘Ÿ)π‘ π‘Ÿπ‘‘π‘Ÿ.(2.10)The last integral exists since ∫10πœ‘π‘(π‘Ÿ)(1βˆ’π‘Ÿ2)π‘ž+π‘ π‘Ÿπ‘‘π‘Ÿ<∞ and (1βˆ’|π‘Ž|)/(1βˆ’|π‘Ž|π‘Ÿ)≀1. It is clear thatlim|π‘Ž|β†’1(1βˆ’|π‘Ž|)(1βˆ’|π‘Ž|π‘Ÿ)=0,(2.11)which implieslim|π‘Ž|β†’1ξ€œ10πœ‘π‘(π‘Ÿ)(1βˆ’π‘Ÿ2)π‘ž+𝑠(1βˆ’|π‘Ž|)𝑠(1βˆ’|π‘Ž|π‘Ÿ)π‘ π‘Ÿπ‘‘π‘Ÿ=0.(2.12)By Lebesgue's dominated convergence theorem, we getlim|π‘Ž|β†’1𝐼1(π‘Ž)=0.(2.13) Now, we consider the case π‘§βˆˆπ”»(π‘Ž,1/2). Note that|1βˆ’π‘Žπ‘§π‘Žβˆ’π‘§|>2(2.14)in the case. By a well-known inequality (see, e.g., [12, page 3]), we have that||𝑧|βˆ’|π‘Ž||≀11βˆ’|π‘Ž|𝑧|2(2.15)and consequently, for π‘§βˆˆπ”»(π‘Ž,1/2), we have|𝑧|≀1+2|π‘Ž|.2+|π‘Ž|(2.16)By the monotonicity of πœ‘, we have𝐼2ξ€œ(π‘Ž)=𝔻(π‘Ž,1/2)|π‘“ξ…ž(𝑧)|𝑝(1βˆ’|𝑧|2)π‘ž1(log|πœ‘π‘Ž)(𝑧)|𝑠𝑑𝐴(𝑧)β‰€πœ‘π‘(1+2|π‘Ž|)ξ€œ2+|π‘Ž|𝔻(π‘Ž,1/2)(1βˆ’|𝑧|2)π‘ž1(log|πœ‘π‘Ž)(𝑧)|𝑠𝑑𝐴(𝑧)=πœ‘π‘(1+2|π‘Ž|)ξ€œ2+|π‘Ž||𝑀|<1/2(1βˆ’|πœ‘π‘Ž(𝑀)|2)π‘ž1(log)|𝑀|𝑠(1βˆ’|π‘Ž|2)2|1βˆ’π‘Žπ‘€|4𝑑𝐴(𝑀)=πœ‘π‘(1+2|π‘Ž|)ξ€œ2+|π‘Ž||𝑀|<1/2(1βˆ’|π‘Ž|2)π‘ž+2(1βˆ’|𝑀|2)π‘ž1(log)|𝑀|𝑠1|1βˆ’π‘Žπ‘€|2π‘ž+4≀1𝑑𝐴(𝑀)πœ‹πœ‘π‘(1+2|π‘Ž|2+|π‘Ž|)(1βˆ’|π‘Ž|2)π‘ž+2ξ€œ01/2(1βˆ’π‘‘2)π‘ž1(log𝑑)𝑠[ξ€œ02πœ‹π‘‘πœƒ(1βˆ’|π‘Ž|𝑑)2π‘ž+4≀1]π‘‘π‘‘π‘‘πœ‹πœ‘π‘(1+2|π‘Ž|2+|π‘Ž|)2π‘ž+2(1βˆ’|π‘Ž|)π‘ž+2ξ€œ2πœ‹01/2(1βˆ’π‘‘2)π‘ž1(log𝑑)𝑠1(1βˆ’|π‘Ž|𝑑)2π‘ž+4π‘‘π‘‘π‘‘β‰€πœ‘π‘(1+2|π‘Ž|2+|π‘Ž|)2π‘ž+322π‘ž+4(1βˆ’|π‘Ž|)π‘ž+2ξ€œ01/2(1βˆ’π‘‘2)π‘ž1(log𝑑)𝑠𝑑𝑑𝑑.(1)This implies𝐼2(π‘Ž)β‰€πΆπœ‘π‘(1+2|π‘Ž|2+|π‘Ž|)(1βˆ’|π‘Ž|)π‘ž+2,(2.17)since the following integral existsξ€œ01/2(1βˆ’π‘‘2)π‘ž1(log𝑑)𝑠𝑑𝑑𝑑.(2.18)Choosing 𝑠=1, for every π‘Ÿβˆˆ(0,1), it follows thatξ€œ1π‘Ÿπœ‘π‘(𝑑)(1βˆ’π‘‘2)π‘ž+1𝑑𝑑𝑑β‰₯πœ‘π‘ξ€œ(π‘Ÿ)1π‘Ÿ(1βˆ’π‘‘2)π‘ž+1=1𝑑𝑑𝑑2πœ‘π‘1(π‘Ÿ)π‘ž+2(1βˆ’π‘Ÿ2)π‘ž+2.(2.19)This andξ€œ10πœ‘π‘(π‘Ÿ)(1βˆ’π‘Ÿ2)π‘ž+1π‘Ÿπ‘‘π‘Ÿ<∞(2.20)imply thatlimπ‘Ÿβ†’1πœ‘π‘(π‘Ÿ)(1βˆ’π‘Ÿ)π‘ž+2=0(2.21)orπœ‘π‘(1+2|π‘Ž|2+|π‘Ž|)(1βˆ’1+2|π‘Ž|)2+|π‘Ž|π‘ž+2=πœ‘π‘(1+2|π‘Ž|)2+|π‘Ž|(1βˆ’|π‘Ž|)π‘ž+2(2+|π‘Ž|)π‘ž+2β†’0,(2.22)as |π‘Ž|β†’1. Consequently,lim|π‘Ž|β†’1𝐼2(π‘Ž)≀𝐢lim|π‘Ž|β†’1πœ‘π‘(1+2|π‘Ž|2+|π‘Ž|)(1βˆ’|π‘Ž|)π‘ž+2=0.(2.23)Combining (2.8), (2.13) with (2.23) we see thatlim|π‘Ž|β†’1ξ€œπ”»|π‘“ξ…ž(𝑧)|𝑝(1βˆ’|𝑧|2)π‘ž(𝑔(𝑧,π‘Ž))𝑠𝑑𝐴(𝑧)=0,(2.24)which means π‘“βˆˆπΉ0(𝑝,π‘ž,𝑠). The proof is complete.

Proof of Theorem 1.2. Consider the monotone increasing functionπœ‘(π‘Ÿ)=βˆžξ“π‘˜=1π‘›π‘˜|π‘Žπ‘˜|π‘Ÿπ‘›π‘˜βˆ’1,0<π‘Ÿ<1.(2.25)For every πœƒβˆˆ[0,2πœ‹), we have|π‘“ξ…ž(π‘Ÿπ‘’π‘–πœƒ)|=|βˆžξ“π‘˜=1π‘›π‘˜π‘Žπ‘˜π‘Ÿπ‘›π‘˜βˆ’1|β‰€πœ‘(π‘Ÿ),(2.26)By Theorem 1.1, we only need to proveξ€œπΌ=10(βˆžξ“π‘˜=1π‘›π‘˜|π‘Žπ‘˜|π‘Ÿπ‘›π‘˜βˆ’1)𝑝(1βˆ’π‘Ÿ2)π‘žπΎ(1βˆ’π‘Ÿ2)π‘Ÿπ‘‘π‘Ÿ<∞.(2.27)By the inequality 1βˆ’π‘Ÿ2≀2log(1/π‘Ÿ), π‘Ÿβˆˆ(0,1), and Lemma 2.2, there exists a constant 𝐢 such that𝐾(1βˆ’π‘Ÿ21)≀𝐾(2logπ‘Ÿ1)β‰ˆπΆπΎ(logπ‘Ÿ).(2.28)Then for 0β‰€π‘ž<∞, we haveξ€œπΌβ‰€πΆ10(βˆžξ“π‘˜=1π‘›π‘˜|π‘Žπ‘˜|π‘Ÿπ‘›π‘˜)π‘π‘Ÿ1βˆ’π‘1(logπ‘Ÿ)π‘ž1𝐾(logπ‘Ÿ)π‘‘π‘Ÿ.(2.29)For 𝑝β‰₯1, assume 𝐼𝑛={π‘˜βˆΆ2π‘›β‰€π‘˜<2𝑛+1,π‘˜βˆˆβ„•}. Sinceβˆžξ“π‘›=02𝑛/2π‘Ÿ2𝑛≀2∞1/2𝑛=0ξ€œ2𝑛+12π‘›π‘‘βˆ’1/2π‘Ÿπ‘‘/2𝑑𝑑≀21/2ξ€œβˆž0π‘‘βˆ’1/2π‘Ÿπ‘‘/21𝑑𝑑=2Ξ“(21)(logπ‘Ÿ)βˆ’1/2,(2.30)which together with HΓΆlder's inequality gives[βˆžξ“π‘˜=1π‘›π‘˜|π‘Žπ‘˜|π‘Ÿπ‘›π‘˜]𝑝=[βˆžξ“π‘›=0ξ“π‘›π‘˜βˆˆπΌπ‘›π‘›π‘˜|π‘Žπ‘˜|π‘Ÿπ‘›π‘˜]𝑝≀[βˆžξ“π‘›=0ξ“π‘›π‘˜βˆˆπΌπ‘›π‘›π‘˜|π‘Žπ‘˜|π‘Ÿ2𝑛]𝑝=[βˆžξ“π‘›=0(2𝑛/2π‘Ÿ2𝑛)1βˆ’1/𝑝(π‘Ÿ2𝑛2(1βˆ’π‘)𝑛/2)1/π‘ξ“π‘›π‘˜βˆˆπΌπ‘›π‘›π‘˜|π‘Žπ‘˜|]𝑝≀[βˆžξ“π‘›=0π‘Ÿ2𝑛2((1βˆ’π‘)/2)𝑛(ξ“π‘›π‘˜βˆˆπΌπ‘›π‘›π‘˜|π‘Žπ‘˜|)𝑝][βˆžξ“π‘›=02𝑛/2π‘Ÿ2𝑛]π‘βˆ’11≀𝐢(logπ‘Ÿ)βˆžβˆ’(π‘βˆ’1)/2𝑛=0π‘Ÿ2𝑛2((1βˆ’π‘)/2)𝑛(ξ“π‘›π‘˜βˆˆπΌπ‘›π‘›π‘˜|π‘Žπ‘˜|)𝑝.(2.31)Hence,πΌβ‰€πΆβˆžξ“π‘›=0(ξ“π‘›π‘˜βˆˆπΌπ‘›π‘›π‘˜|π‘Žπ‘˜|)𝑝2((1βˆ’π‘)/2)π‘›ξ€œ10π‘Ÿ2𝑛+1βˆ’π‘1(logπ‘Ÿ)π‘žβˆ’((π‘βˆ’1)/2)1𝐾(logπ‘Ÿ)π‘‘π‘Ÿ.(2.32)For 1≀𝑝≀2, 1β‰€π‘βˆ’2π‘ž<3, by Lemma 2.4, choosing 𝛼=2𝑛+2βˆ’π‘, 𝛽=(π‘βˆ’2π‘žβˆ’1)/2, we obtainπΌβ‰€πΆβˆžξ“π‘›=0(ξ“π‘›π‘˜βˆˆπΌπ‘›π‘›π‘˜|π‘Žπ‘˜|)𝑝((1/2)(2π‘žβˆ’π‘+3)2𝑛)+2βˆ’π‘(2π‘žβˆ’π‘+3)/22((1βˆ’π‘)/2)𝑛𝐾((1/2)(2π‘žβˆ’π‘+3)2𝑛)+2βˆ’π‘β‰€πΆβˆžξ“π‘›=0(ξ“π‘›π‘˜βˆˆπΌπ‘›π‘›π‘˜|π‘Žπ‘˜|)𝑝(12𝑛)(2π‘žβˆ’π‘+3)/2(12𝑛)(π‘βˆ’1)/21𝐾(2𝑛)=πΆβˆžξ“π‘›=0(ξ“π‘›π‘˜βˆˆπΌπ‘›π‘›π‘˜|π‘Žπ‘˜|)𝑝(12𝑛)1+π‘ž1𝐾(2𝑛).(2.33)If π‘›π‘˜βˆˆπΌπ‘›, then π‘›π‘˜<2𝑛+1. The assumption that 𝐾 is nondecreasing and Lemma 2.2 give(12𝑛)1+π‘ž1𝐾(2𝑛1)≀(2𝑛)1+π‘ž121+π‘ž1𝐾(2𝑛+1)<π‘›π‘˜βˆ’(1+π‘ž)1𝐾(π‘›π‘˜).(2.34)Since 𝑓(𝑧) is a lacunary series, the Taylor series of 𝑓 has most [logπœ†2]+1 terms π‘Žπ‘˜π‘§π‘›π‘˜ such that π‘›π‘˜βˆˆπΌπ‘›. Combining this with the last inequality and HΓΆlder's inequality, we obtainπΌβ‰€πΆβˆžξ“π‘›=0(ξ“π‘›π‘˜βˆˆπΌπ‘›π‘›π‘˜1βˆ’((1+π‘ž)/𝑝)|π‘Žπ‘˜|𝐾1/𝑝(1π‘›π‘˜))π‘β‰€πΆβˆžξ“π‘›=0([logπœ†2]+1)π‘βˆ’1ξ“π‘›π‘˜βˆˆπΌπ‘›π‘›π‘˜π‘βˆ’π‘žβˆ’1|π‘Žπ‘˜|𝑝1𝐾(π‘›π‘˜)=𝐢([logπœ†2]+1)βˆžπ‘βˆ’1ξ“π‘˜=1π‘›π‘˜π‘βˆ’π‘žβˆ’1|π‘Žπ‘˜|𝑝1𝐾(π‘›π‘˜)<∞.(2.35)This shows that π‘“βˆˆπ’¬πΎ,0(𝑝,π‘ž).

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