# A Note on the Growth of Meromorphic Functions with a Radially Distributed Value

Authors: Jun-Fan Chen
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

This note is to investigate the growth of transcendental meromorphic functions with radially distributed values. We generalize a more recent result of Chen et al. (2011). The paper is closely related to some previous results due to Fang and Zalcman (2008), and Xu et al. (2009).


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## Body

## 1. Introduction and Results

Letf: ℂ→ℂ^ be a meromorphic function, where ℂ is the whole complex plane and ℂ^=ℂ∪{∞}. We shall use the basic results and notations of Nevanlinna's value distribution theory of meromorphic functions (see [1–3]), such as T(r,f),  N(r,f), and m(r,f). Meantime, the Nevanlinna's deficiency δ(a,f) of f(z) with respect to a∈ℂ is defined by
(1.1)δ(a,f)=liminfr→∞m(r,1/(f-a))T(r,f)=1-limsupr→∞N(r,1/(f-a))T(r,f)
and δ(∞,f) is obtained by the above formula with m(r,f) in place of m(r,1/(f-a)), and N(r,f) in place of N(r,1/(f-a)), respectively. The lower order μ and order λ are defined in turn as follows:
(1.2)μ:=μ(f)=liminfr→∞logT(r,f)logr,λ:=λ(f)=limsupr→∞logT(r,f)logr.For an unbounded subsetX of ℂ and a∈ℂ^, we denote by n(r,X,f=a) the number of the zeros of f-a counting multiplicities in X∩{z:|z|≤r}.In 1992, Yang [4], cf. [5] posed the following interesting conjecture.Yang's Conjecture
LetP be a property (or a set of properties) such that any entire (or meromorphic in ℂ) function satisfying P must be a constant. Suppose that f is an entire (or meromorphic in ℂ) function of finite lower order μ, and that Lj:argz=θj(j=1,2,…,q; 0≤θ1<θ2<⋯<θq<2π; θq+1=θ1+2π) are a finite number of rays issuing from the origin. If f satisfies P in ℂ∖(⋃j=1qLj), then the order λ of f has the following estimation:
(1.3)λ≤max1≤j≤q{πθj+1-θj,θq+1=θ1+2π}.
Yang's conjecture implies that if meromorphic functions satisfy certain properties in the vicinities of a finite number of rays, then the growth of meromorphic functions will be restricted.A well-known result of Clunie [6], cf. [7] is that an entire function f which satisfies f(z)f′(z)≠1 in ℂ must be constant. In [5], L. Yang and C.-C. Yang chose the property P as f(z)f′(z)≠1 and verified the above conjecture. On the other hand, Fang and Zalcman [8] considered the value distribution of f+a(f′)n in ℂ, where a is a nonzero finite complex number and n≥2 is a positive integer. Actually, they [8] gave an affirmative answer to a question suggested by Ye [9]. Later on, Xu et al. [10] further generalized f+a(f′)n to f+a(f(k))n in ℂ for a positive integer k with n≥k+1 and investigated its value distribution.More recently, Chen et al. [11] chose another property P as f+a(f′)n to continue to study Yang's Conjecture and proved the following results.Theorem A.
Letf(z) be a transcendental meromorphic function with δ(∞,f′)>0 in ℂ and let Lj:argz=θj(j=1,2,…,q) be a finite number of rays issued from the origin such that
(1.4)-π≤θ1<θ2<⋯<θq<π,θq+1=θ1+2π
with ω=max{π/(θj+1-θj):1≤j≤q}. Set Y=ℂ∖(⋃j=1qLj). If f(z) satisfies
(1.5)limsupr→∞logn(r,Y,f+a(f′)n=b)logr≤ρ
with a positive number ρ, finite complex numbers a≠0 and b, for any positive integer n≥2, then the order λ of f(z) has the estimation λ≤max{ω,ρ}.Theorem B.
Letf(z) be a transcendental meromorphic function of finite lower order μ with δ(∞,f')>0 in ℂ. For q pairs of real numbers {αj,βj} such that
(1.6)-π≤α1<β1≤α2<β2≤⋯≤αq<βq≤π
with ω=max{π/(βj-αj):1≤j≤q}, suppose that
(1.7)limsupr→∞logn(r,Y,f+a(f′)n=b)logr≤ρ
with a positive number ρ, finite complex numbers a≠0 and b, and Y=⋃j=1q{z:αj≤argz≤βj}, for any positive integer n≥2, and that
(1.8)∑j=1q‍(αj+1-βj)<4σarcsinδ(∞,f')2,αq+1=α1+2π
with σ=max{ω,ρ,μ}. Then the order λ of f(z) has the estimation λ≤max{ω,ρ}.Now there arises a natural question.Question 1.
What can be said iff′ in Theorems A and B is replaced by the kth derivative f(k)?In this paper, we will prove the following results which generalize Theorems A and B.Theorem 1.1.
Letf(z) be a transcendental meromorphic function with δ(∞,f(k))>0 for a positive integer k in ℂ and let Lj:argz=θj(j=1,2,…,q) be a finite number of rays issued from the origin such that
(1.9)-π≤θ1<θ2<⋯<θq<π,θq+1=θ1+2π
with ω=max{π/(θj+1-θj):1≤j≤q}. Set Y=ℂ∖(⋃j=1qLj). If f(z) satisfies
(1.10)limsupr→∞logn(r,Y,f+a(f(k))n=b)logr≤ρ
with a positive number ρ, finite complex numbers a≠0 and b, for any positive integer n≥k+1, then the order λ of f(z) has the estimation λ≤max{ω,ρ}.Remark 1.2.
Letk=1. Then by Theorem 1.1 we get Theorem A.Corollary 1.3.
Letf(z) be a transcendental entire function, let the notations θj (j=1,2,…,q+1), ω, and Y be defined as in Theorem 1.1, and suppose that the function f(z) fulfills the same condition (1.10) as in Theorem 1.1. Then the order λ of f(z) has the estimation λ≤max{ω,ρ}.Theorem 1.4.
Letf(z) be a transcendental meromorphic function of finite lower order μ with δ(∞,f(k))>0 for a positive integer k in ℂ. For q pairs of real numbers {αj,βj} such that
(1.11)-π≤α1<β1≤α2<β2≤⋯≤αq<βq≤π
with ω=max{π/(βj-αj):1≤j≤q}, suppose that
(1.12)limsupr→∞logn(r,Y,f+a(f(k))n=b)logr≤ρ
with a positive number ρ, finite complex numbers a≠0 and b, and Y=⋃j=1q{z:αj≤argz≤βj}, for any positive integer n≥k+1, and that
(1.13)∑j=1q‍(αj+1-βj)<4σarcsinδ(∞,f(k))2,αq+1=α1+2π
with σ=max{ω,ρ,μ}. Then the order λ of f(z) has the estimation λ≤max{ω,ρ}.Remark 1.5.
Letk=1. Then by Theorem 1.4 we get Theorem B.Corollary 1.6.
Letf(z) be a transcendental entire function of finite lower order μ, let the notations αj, βj (j=1,2,…,q), ω, σ, and Y be defined as in Theorem 1.4, and suppose that the function f(z) fulfills the same conditions (1.12) and (1.13) as in Theorem 1.4. Then the order λ of f(z) has the estimation λ≤max{ω,ρ}.In order to prove our results, we require the Nevanlinna theory of meromorphic functions in an angular domain. For the sake of convenience, we recall some notations and definitions. Letf be a meromorphic function on the angular domain Ω¯(α,β)={z:α≤argz≤β}, where 0<β-α≤2π. Nevanlinna et al. [12, 13] introduced the following notations:
(1.14)Aα,β(r,f)=ωπ∫1r‍(1tω-tωr2ω){log+|f(teiα)|+log+|f(teiβ)|}dtt,Bα,β(r,f)=2ωπrω∫αβ‍log+|f(reiθ)|sinω(θ-α)dθ,Cα,β(r,f)=2∑1<|bm|<r‍(1|bm|ω-|bm|ωr2ω)sinω(θm-α),
where ω=π/(β-α) and bm=|bm|eiθm are the poles of f in Ω¯(α,β) appearing according to their multiplicities. The function Cα,β(r,f) is called the angular counting function (counting multiplicities) of the poles of f in Ω¯(α,β), and C¯α,β(r,f) is called the angular reduced counting function (ignoring multiplicities) of the poles of f in Ω¯(α,β). Further, Nevanlinna's angular characteristic function Sα,β(r,f) is defined as follows:
(1.15)Sα,β(r,f)=Aα,β(r,f)+Bα,β(r,f)+Cα,β(r,f).Throughout the paper, we denote byR(r,*) a quantity satisfying
(1.16)R(r,*)=O{log(rT(r,*))},∀r∉E,
where E denotes a set of positive real numbers with finite linear measure. It is not necessarily the same for every occurrence in the context.


## 2. Some Lemmas

In this section we present some lemmas which will be needed in the sequel.Lemma 2.1 (see [5, 12–14]).
Letf be meromorphic in ℂ. Then in Ω¯(α,β) for an arbitrary finite complex number a, we have
(2.1)Sα,β(r,1f-a)=Sα,β(r,f)+O(1),
and for each positive integer k, we have
(2.2)Aα,β(r,f(k)f)+Bα,β(r,f(k)f)=R(r,f).Lemma 2.2 (see [10]).
Letn≥2 and k be positive integers, let a be a nonzero finite complex number, and let p(z) be a polynomial. Then the solution of the differential equation a[ω(k)(z)]n+ω(z)=p(z) must be polynomial.Lemma 2.3 (see [1, Theorem 3.1]; [1, page 33, (2.1)]).
Letf be meromorphic in ℂ, let a1,a2,…,aq, where q≥2, be distinct complex numbers, δ>0, and suppose that |aμ-aν|≥δ for 1≤μ<ν≤q. Then, for each positive integer k, we have
(2.3)T(r,f(k))≤T(r,f)+kN¯(r,f)+O(logrT(r,f)),m(r,∑ν=1q‍1f-aν)≥∑ν=1q‍m(r,1f-aν)-qlog+3qδ-log2,∀r∉E.
Next we slightly modify the proof of Lemma 2.4 in [10] to give the following key lemma, which is an important generalization of Lemma 3 in [11].Lemma 2.4.
Letf be transcendental meromorphic in ℂ, let a≠0 and b be finite complex numbers, and let n and k be positive integers with n≥k+1. Then in Ω¯(α,β),
(2.4)Bα,β(r,f(k))≤(k+1)Cα,β(r,1f+a(f(k))n-b)+R(r,f).Proof.
Put(2.5)g=f+a(f(k))n-b,(2.6)ϕ=g(k)g.
Then ϕ≢0, for otherwise g would be a polynomial of degree at most k-1. This and (2.5) together with Lemma 2.2 imply that f must be a polynomial, a contradiction. By the Nevanlinna's basic reasoning, Lemmas 2.1, and 2.3, (2.5), and (2.6), we have
(2.7)(Aα,β+Bα,β)(r,ϕ)=R(r,g)=R(r,f).
Now a simple calculation for (2.5) shows that
(2.8)g(k)=f(k)+a((f(k))n)(k)=f(k)(1+Q(f(k))),
where Q(f(k)) is a homogeneous differential polynomial in f(k) of degree n-1 and of the form
(2.9)Q(f(k))=a(f(k))n-k-1×(n!(f(k+1))k(n-k)!+k(k-1)n!2!(n-k+1)!f(k)(f(k+1))k-2f(k+2)+⋯+n(f(k))k-1f(2k)).
Then by (2.5), (2.6), and (2.8), we get
(2.10)f(k)(1+Q(f(k)))=ϕ(f+a(f(k))n-b).
From (2.5)–(2.10), and Lemma 2.1, it thus follows that
(2.11)Cα,β(r,1f(k))+Cα,β(r,1Q(f(k))+1)≤Cα,β(r,1ϕ)+Cα,β(r,1f+a(f(k))n-b)≤Cα,β(r,ϕ)+Cα,β(r,1f+a(f(k))n-b)+R(r,f)≤kC¯α,β(r,f)+(k+1)Cα,β(r,1f+a(f(k))n-b)+R(r,f)≤kk+1Cα,β(r,f(k))+(k+1)Cα,β(r,1f+a(f(k))n-b)+R(r,f).
On the other hand, by the Nevanlinna's basic reasoning, Lemmas2.1 and 2.3, we deduce that
(2.12)(Aα,β+Bα,β)(r,1(f(k))n-1)+(Aα,β+Bα,β)(r,1Q(f(k))+1)≤(Aα,β+Bα,β)(r,1Q(f(k)))+(Aα,β+Bα,β)(r,Q(f(k))(f(k))n-1)+(Aα,β+Bα,β)(r,1Q(f(k))+1)≤(Aα,β+Bα,β)(r,1Q(f(k))+1Q(f(k))+1)+R(r,f)≤(Aα,β+Bα,β)(r,(Q(f(k)))′Q(f(k))+(Q(f(k))+1)′Q(f(k))+1)+(Aα,β+Bα,β)(r,1(Q(f(k)))′)+R(r,f)≤Sα,β(r,(Q(f(k)))′)-Cα,β(r,1(Q(f(k)))′)+R(r,f)≤Sα,β(r,Q(f(k)))+C¯α,β(r,f)-Cα,β(r,1(Q(f(k)))′)+R(r,f)≤Sα,β(r,Q(f(k)))+1k+1Cα,β(r,f(k))-Cα,β(r,1(Q(f(k)))′)+R(r,f).
This, together with Lemma 2.1, yields
(2.13)(n-1)Sα,β(r,f(k))≤1k+1Cα,β(r,f(k))+(n-1)Cα,β(r,1f(k))+Cα,β(r,1Q(f(k))+1)-Cα,β(r,1(Q(f(k)))′)+R(r,f).
Next we divide into two cases.
Case  1 (n≥k+2). Suppose that z0 is a zero of f(k) of multiplicity m. Then we can know that z0 is a zero of (Q(f(k)))′ of multiplicity at least (n-k-1)m+k(m-1)-1=(n-1)m-k-1. Thus, we have
(2.14)(n-1)Cα,β(r,1f(k))+Cα,β(r,1Q(f(k))+1)-Cα,β(r,1(Q(f(k)))′)=Cα,β(r,1(f(k))n-1)+Cα,β(r,1Q(f(k))+1)-Cα,β(r,1(Q(f(k)))′)≤(k+1)C¯α,β(r,1f(k))+C¯α,β(r,1Q(f(k))+1).
Substituting this into (2.13) gives
(2.15)(n-1)Sα,β(r,f(k))≤1k+1Cα,β(r,f(k))+(k+1)C¯α,β(r,1f(k))+C¯α,β(r,1Q(f(k))+1)+R(r,f).
From this and (2.11) it follows that
(2.16)(n-1)Sα,β(r,f(k))≤k2+k+1k+1Cα,β(r,f(k))+(k+1)2Cα,β(r,1f+a(f(k))n-b)+R(r,f),
implying that
(2.17)Sα,β(r,f(k))≤k2+k+1(n-1)(k+1)Cα,β(r,f(k))+(k+1)2n-1Cα,β(r,1f+a(f(k))n-b)+R(r,f).
Thereby, noting n≥k+2, we get
(2.18)Sα,β(r,f(k))≤k2+k+1(k+1)2Cα,β(r,f(k))+(k+1)Cα,β(r,1f+a(f(k))n-b)+R(r,f)≤Cα,β(r,f(k))+(k+1)Cα,β(r,1f+a(f(k))n-b)+R(r,f),
so that
(2.19)Bα,β(r,f(k))≤Aα,β(r,f(k))+Bα,β(r,f(k))≤(k+1)Cα,β(r,1f+a(f(k))n-b)+R(r,f).
This is the desired result.
Case  2 (n=k+1). Then, by (2.11) and (2.13), we have
(2.20)kSα,β(r,f(k))≤k2+1k+1Cα,β(r,f(k))+k(k+1)Cα,β(r,1f+a(f(k))n-b)+R(r,f),
which leads to
(2.21)Sα,β(r,f(k))≤k2+1k(k+1)Cα,β(r,f(k))+(k+1)Cα,β(r,1f+a(f(k))n-b)+R(r,f)Sα,β(r,f(k))≤Cα,β(r,f(k))+(k+1)Cα,β(r,1f+a(f(k))n-b)+R(r,f).
This yields
(2.22)Bα,β(r,f(k))≤Aα,β(r,f(k))+Bα,β(r,f(k))≤(k+1)Cα,β(r,1f+a(f(k))n-b)+R(r,f),
obtaining the desired result.
This completes the proof of Lemma2.4.The following auxiliary results regarding Pólya peaks and the spread relation are necessary in the proofs of our theorems.Lemma 2.5 (see [14–16]).
Letf be a transcendental meromorphic function of finite lower order μ and order λ(0<λ≤∞) in ℂ. Then, for an arbitrary positive number σ satisfying μ≤σ≤λ and any set E of finite linear measure, there exist Pólya peaks {rn} satisfying the following:(i)
rn∉E, limn→∞(rn/n)=∞;(ii)
liminfn→∞(logT(rn,f))/logrn≥σ;(iii)
T(t,f)<(1+o(1))(t/rn)σT(rn,f),t∈[rn/n,nrn].
A sequence of{rn} satisfying (i), (ii), and (iii) in Lemma 2.5 is called a Pólya peak of order σ of f outside E. Given a positive function Λ=Λ(r) on (0,∞) with Λ→0 as r→∞, we define
(2.23)DΛ(r,a)={θ∈[-π,π)∣log+1|f(reiθ)-a|>Λ(r)T(r,f)},DΛ(r,∞)={θ∈[-π,π)∣log+|f(reiθ)|>Λ(r)T(r,f)}.Lemma 2.6 (see [17]).
Letf be a transcendental meromorphic function of finite lower order μ and order λ(0<λ≤∞) in ℂ. Suppose that δ=δ(a,f)>0 for some a∈ℂ^. Then for an arbitrary Pólya peak {rn} of order σ(μ≤σ≤λ) and an arbitrary positive function Λ=Λ(r) with Λ→0 as r→∞, we have
(2.24)liminfn→∞measDΛ(rn,a)≥min{2π,4σarcsinδ2}.
Now a more precise estimation ofT(r,f) in terms of T(r,f(k)) is introduced as follows.Lemma 2.7 (see [18]).
Letf be transcendental meromorphic in ℂ. Then, for a positive integer k and a real number τ>1, we have
(2.25)T(r,f)<Kτ,kT(τr,f(k))+log(τr)+O(1),
where Kτ,k is a positive number depending on only τ and k.At last, we state the following results due to Edrei, Hayman, and Miles, respectively.Lemma 2.8 (see [19]).
Letf be a transcendental meromorphic function with δ=δ(∞,f)>0 in ℂ. Then, given ε>0, we have
(2.26)measE(r,f)>1Tε(r,f)[logr]1+ε,∀r∉F,
where
(2.27)E(r,f)={θ∈[-π,π):log+|f(reiθ)|>δ4T(r,f)}
and F is a set of positive real numbers with finite logarithmic measure (i.e., ∫F‍(dt/t)<∞) depending on ε only.Lemma 2.9 (see [20]).
Letf be a transcendental meromorphic function in ℂ. Then for each K>1 there exists a set M(K) of the lower logarithmic density at least d(K)=1-(2eK-1-1)-1>0, that is,
(2.28)logdens_M(K):=liminfr→∞1logr∫M(K)∩[1,r]‍dtt≥d(K),
such that, for every positive integer p, we have
(2.29)limsupr→∞,r∈M(K)T(r,f)T(r,f(p))≤3eK.


## 3. Proofs of Theorems1.1 and 1.4

In this section, we state the detailed proofs of Theorems1.1 and 1.4 by using the method in [14]. To begin with, we give the proof of Theorem 1.4. Finally the proof of Theorem 1.1 can be derived from Theorem 1.4.


### 3.1. Proof of Theorem1.4

Assume on the contrary that Theorem1.4 does not hold. Then λ(f)>max{ω,ρ}. Now by (1.12), we have
(3.1)n(r,Ω¯(αj,βj),f+a(f(k))n=b)≤rρ+ε,j=1,2,…,q
for arbitrarily small ε>0 and sufficiently large r≥r0. Let ξm be the zeros of f+a(f(k))n-b on Ω¯(αj,βj) appearing according to their multiplicities, and set ωj=π/(βj-αj). By the definition of Cα,β(r,*), we deduce that
(3.2)Cαj,βj(r,1f+a(f(k))n-b)≤2∑1<|ξm|<r‍1|ξm|ωj=2∫1r‍dn(t,Ω¯(αj,βj),f+a(f(k))n=b)tωj≤2n(r,Ω¯(αj,βj),f+a(f(k))n=b)rωj+2ωj∫1r‍n(t,Ω¯(αj,βj),f+a(f(k))n=b)tωj+1dt≤2rρ+ε-ωj+O(1)+2ωj∫r0r‍tρ+εtωj+1dt≤Kj,εrρ+ε-ωj+O(logr),
where Kj,ε is a positive number depending on only j and ε, which is not necessarily the same for every occurrence in the context. From Lemma 2.4, we have
(3.3)Bαj,βj(r,f(k))≤(k+1)Cαj,βj(r,1f+a(f(k))n-b)+R(r,f).
Thus, it follows by (3.2) and (3.3) that
(3.4)Bαj,βj(r,f(k))≤Kj,εrρ+ε-ωj+O(logrT(r,f)),∀r∉E,
where the exceptional set E associated with R(r,f) is of at most finite linear measure.Now we discuss two cases separately.Case 1 (λ(f)>μ(f)).
Then by the assumptionσ=max{ω,ρ,μ} and λ(f)>max{ω,ρ}, we have λ(f(k))=λ(f)>σ≥μ(f)=μ(f(k)). Now from (1.13), we can find a real number ε>0 such that
(3.5)∑j=1q‍(αj+1-βj+2ε)+2ε<4σ+2εarcsinδ(∞,f(k))2,αq+1=α1+2π,(3.6)λ(f(k))>σ+2ε>μ(f(k)).
Applying Lemma 2.5 to f(k) gives the existence of the Pólya peak {rn} of order σ+2ε of f(k) outside the set E. Then, noting that σ+2ε>ωj≥1/2 and 0<δ(∞,f(k))≤1, by applying Lemma 2.6 to the Pólya peak {rn}, for sufficiently large n we have
(3.7)measDΛ(rn,∞)>4σ+2εarcsinδ(∞,f(k))2-ε.
Without loss of generality, we can assume that (3.7) holds for all the n. Set
(3.8)Kn:=meas(DΛ(rn,∞)∩⋃j=1q(αj+ε,βj-ε)).
It then follows from (3.5) and (3.7) that
(3.9)Kn≥meas(DΛ(rn,∞))-meas([-π,π)∖⋃j=1q‍(αj+ε,βj-ε))=meas(DΛ(rn,∞))-meas(⋃j=1q‍(βj-ε,αj+1+ε))=meas(DΛ(rn,∞))-∑j=1q‍(αj+1-βj+2ε)>ε>0.
By (3.9), it is easy to see that there exists a j0 such that, for infinitely many n, we have
(3.10)meas(DΛ(rn,∞)∩(αj0+ε,βj0-ε))≥Knq>εq.
Without loss of generality, we can assume that (3.10) holds for all the n. Set En=DΛ(rn,∞)∩(αj0+ε,βj0-ε) and Λ(r)=[logr]-1. From the definition of DΛ(rn,∞), we deduce that
(3.11)∫αj0+εβj0-ε‍log+|f(k)(rneiθ)|dθ≥∫En‍log+|f(k)(rneiθ)|dθ≥εqT(rn,f(k))logrn.
On the other hand, by the definition ofBα,β(r,*) and (3.4), it follows that
(3.12)∫αj0+εβj0-ε‍log+|f(k)(rneiθ)|dθ≤πrnωj02ωj0sin(εωj0)Bαj0,βj0(rn,f(k))≤Kj0,ε(rnρ+ε+rnωj0log(rnT(rn,f))),
where rn∉E, ωj0=π/(βj0-αj0), and Kj0,ε is a positive number depending on only j0 and ε. Combining (3.11) with (3.12) gives
(3.13)T(rn,f(k))≤qKj0,εlogrnε(rnρ+ε+rnωj0log(rnT(rn,f))),
implying together with (iii) in Lemma 2.5 and Lemma 2.7 that
(3.14)logT(rn,f(k))≤3loglogrn+max{ρ+ε,ωj0}logrn+loglogT(rn,f(k))+O(1).
Thus, from (ii) in Lemma 2.5 for σ+2ε, we have
(3.15)σ+2ε≤limsupn→∞logT(rn,f(k))logrn≤max{ρ+ε,ωj0}≤σ+ε,
which is impossible.Case 2 (λ(f)=μ(f)).
Then by the assumptionσ=max{ω,ρ,μ} and λ(f)>max{ω,ρ}, we have σ=μ(f)=λ(f)=λ(f(k))=μ(f(k)). By the same argument as in Case 1 with all the σ+2ε replaced by σ=μ(f), we can derive
(3.16)μ(f)=σ≤max{ρ+ε,ωj0}<λ(f),
which is also impossible.This completes the proof of Theorem1.4.


### 3.2. Proof of Theorem1.1

By Theorem1.4, it suffices to prove that the lower order μ(f) of f is finite. As in the proof of Theorem 1.4, we have, for each j∈{1,2,…,q},
(3.17)Bθj,θj+1(r,f(k))≤Kj,εrρ+ε-ωj+O(logrT(r,f)),ωj=πθj+1-θj,∀r∉E,
where the exceptional set E associated with R(r,f) is of at most finite linear measure.ForF in Lemma 2.8 and E in (3.17), logdens¯(F∪E)=0 and hence for M(2) in Lemma 2.9 when K=2, logdens_(M(2)∖(F∪E))≥d(2)>0. Applying Lemma 2.8 to f(k) gives the existence of a sequence {rn} of positive numbers such that rn→∞(n→∞), rn∈M(2)∖(F∪E), and
(3.18)measE(rn,f(k))>1Tε(rn,f(k))[logrn]1+ε.
Set
(3.19)εn=12q+11Tε(rn,f(k))[logrn]1+ε.
Then, from (3.18) and (3.19), it follows that
(3.20)meas(E(rn,f(k))∩⋃j=1q‍(θj+εn,θj+1-εn))≥measE(rn,f(k))-meas(⋃j=1q‍(θj-εn,θj+εn))≥(2q+1)εn-2qεn=εn>0.
Hence, there exists a j∈{1,2,…,q} such that, for infinitely many n, we have
(3.21)meas(E(rn,f(k))∩(θj+εn,θj+1-εn))≥εnq.
Without loss of generality, we can assume that this holds for all the n. Let En=E(rn,f(k))∩(θj+εn,θj+1-εn). Thus, by the definition of E(r,f) and (3.21), it follows that
(3.22)∫θj+εnθj+1-εn‍log+|f(k)(rneiθ)|dθ≥∫En‍log+|f(k)(rneiθ)|dθ≥meas(En)δ(∞,f(k))4T(rn,f(k))≥εnδ(∞,f(k))4qT(rn,f(k)).On the other hand, by the definition ofBα,β(r,*) and (3.17), we have
(3.23)∫θj+εnθj+1-εn‍log+|f(k)(rneiθ)|dθ≤πrnωj2ωjsin(εnωj)Bθj,θj+1(rn,f(k))≤π2Kj,ε4ωj2εn(rnρ+ε+rnωjlog(rnT(rn,f))),
where rn∉E∪F, ωj=π/(θj+1-θj), and Kj,ε is a positive number depending on only j and ε. Combining (3.22) with (3.23) now yields
(3.24)εn2T(rn,f(k))≤qπ2Kj,εωj2δ(∞,f(k))(rnρ+ε+rnωjlog(rnT(rn,f))),
so that, together with (3.19) and Lemma 2.9, we have
(3.25)T1-2ε(rn,f(k))≤qπ2(2q+1)2[logrn]2+2εKj,εωj2δ(∞,f(k))(rnρ+ε+rnωj(logrn+logT(rn,f(k))+log(6e))).
Thus μ(f)=μ(f(k))≤max{ρ+ε,ωj}/(1-2ε)<∞ and so Theorem 1.1 follows from Theorem 1.4.This completes the proof of Theorem1.1.


## 3.1. Proof of Theorem1.4

Assume on the contrary that Theorem1.4 does not hold. Then λ(f)>max{ω,ρ}. Now by (1.12), we have
(3.1)n(r,Ω¯(αj,βj),f+a(f(k))n=b)≤rρ+ε,j=1,2,…,q
for arbitrarily small ε>0 and sufficiently large r≥r0. Let ξm be the zeros of f+a(f(k))n-b on Ω¯(αj,βj) appearing according to their multiplicities, and set ωj=π/(βj-αj). By the definition of Cα,β(r,*), we deduce that
(3.2)Cαj,βj(r,1f+a(f(k))n-b)≤2∑1<|ξm|<r‍1|ξm|ωj=2∫1r‍dn(t,Ω¯(αj,βj),f+a(f(k))n=b)tωj≤2n(r,Ω¯(αj,βj),f+a(f(k))n=b)rωj+2ωj∫1r‍n(t,Ω¯(αj,βj),f+a(f(k))n=b)tωj+1dt≤2rρ+ε-ωj+O(1)+2ωj∫r0r‍tρ+εtωj+1dt≤Kj,εrρ+ε-ωj+O(logr),
where Kj,ε is a positive number depending on only j and ε, which is not necessarily the same for every occurrence in the context. From Lemma 2.4, we have
(3.3)Bαj,βj(r,f(k))≤(k+1)Cαj,βj(r,1f+a(f(k))n-b)+R(r,f).
Thus, it follows by (3.2) and (3.3) that
(3.4)Bαj,βj(r,f(k))≤Kj,εrρ+ε-ωj+O(logrT(r,f)),∀r∉E,
where the exceptional set E associated with R(r,f) is of at most finite linear measure.Now we discuss two cases separately.Case 1 (λ(f)>μ(f)).
Then by the assumptionσ=max{ω,ρ,μ} and λ(f)>max{ω,ρ}, we have λ(f(k))=λ(f)>σ≥μ(f)=μ(f(k)). Now from (1.13), we can find a real number ε>0 such that
(3.5)∑j=1q‍(αj+1-βj+2ε)+2ε<4σ+2εarcsinδ(∞,f(k))2,αq+1=α1+2π,(3.6)λ(f(k))>σ+2ε>μ(f(k)).
Applying Lemma 2.5 to f(k) gives the existence of the Pólya peak {rn} of order σ+2ε of f(k) outside the set E. Then, noting that σ+2ε>ωj≥1/2 and 0<δ(∞,f(k))≤1, by applying Lemma 2.6 to the Pólya peak {rn}, for sufficiently large n we have
(3.7)measDΛ(rn,∞)>4σ+2εarcsinδ(∞,f(k))2-ε.
Without loss of generality, we can assume that (3.7) holds for all the n. Set
(3.8)Kn:=meas(DΛ(rn,∞)∩⋃j=1q(αj+ε,βj-ε)).
It then follows from (3.5) and (3.7) that
(3.9)Kn≥meas(DΛ(rn,∞))-meas([-π,π)∖⋃j=1q‍(αj+ε,βj-ε))=meas(DΛ(rn,∞))-meas(⋃j=1q‍(βj-ε,αj+1+ε))=meas(DΛ(rn,∞))-∑j=1q‍(αj+1-βj+2ε)>ε>0.
By (3.9), it is easy to see that there exists a j0 such that, for infinitely many n, we have
(3.10)meas(DΛ(rn,∞)∩(αj0+ε,βj0-ε))≥Knq>εq.
Without loss of generality, we can assume that (3.10) holds for all the n. Set En=DΛ(rn,∞)∩(αj0+ε,βj0-ε) and Λ(r)=[logr]-1. From the definition of DΛ(rn,∞), we deduce that
(3.11)∫αj0+εβj0-ε‍log+|f(k)(rneiθ)|dθ≥∫En‍log+|f(k)(rneiθ)|dθ≥εqT(rn,f(k))logrn.
On the other hand, by the definition ofBα,β(r,*) and (3.4), it follows that
(3.12)∫αj0+εβj0-ε‍log+|f(k)(rneiθ)|dθ≤πrnωj02ωj0sin(εωj0)Bαj0,βj0(rn,f(k))≤Kj0,ε(rnρ+ε+rnωj0log(rnT(rn,f))),
where rn∉E, ωj0=π/(βj0-αj0), and Kj0,ε is a positive number depending on only j0 and ε. Combining (3.11) with (3.12) gives
(3.13)T(rn,f(k))≤qKj0,εlogrnε(rnρ+ε+rnωj0log(rnT(rn,f))),
implying together with (iii) in Lemma 2.5 and Lemma 2.7 that
(3.14)logT(rn,f(k))≤3loglogrn+max{ρ+ε,ωj0}logrn+loglogT(rn,f(k))+O(1).
Thus, from (ii) in Lemma 2.5 for σ+2ε, we have
(3.15)σ+2ε≤limsupn→∞logT(rn,f(k))logrn≤max{ρ+ε,ωj0}≤σ+ε,
which is impossible.Case 2 (λ(f)=μ(f)).
Then by the assumptionσ=max{ω,ρ,μ} and λ(f)>max{ω,ρ}, we have σ=μ(f)=λ(f)=λ(f(k))=μ(f(k)). By the same argument as in Case 1 with all the σ+2ε replaced by σ=μ(f), we can derive
(3.16)μ(f)=σ≤max{ρ+ε,ωj0}<λ(f),
which is also impossible.This completes the proof of Theorem1.4.


## 3.2. Proof of Theorem1.1

By Theorem1.4, it suffices to prove that the lower order μ(f) of f is finite. As in the proof of Theorem 1.4, we have, for each j∈{1,2,…,q},
(3.17)Bθj,θj+1(r,f(k))≤Kj,εrρ+ε-ωj+O(logrT(r,f)),ωj=πθj+1-θj,∀r∉E,
where the exceptional set E associated with R(r,f) is of at most finite linear measure.ForF in Lemma 2.8 and E in (3.17), logdens¯(F∪E)=0 and hence for M(2) in Lemma 2.9 when K=2, logdens_(M(2)∖(F∪E))≥d(2)>0. Applying Lemma 2.8 to f(k) gives the existence of a sequence {rn} of positive numbers such that rn→∞(n→∞), rn∈M(2)∖(F∪E), and
(3.18)measE(rn,f(k))>1Tε(rn,f(k))[logrn]1+ε.
Set
(3.19)εn=12q+11Tε(rn,f(k))[logrn]1+ε.
Then, from (3.18) and (3.19), it follows that
(3.20)meas(E(rn,f(k))∩⋃j=1q‍(θj+εn,θj+1-εn))≥measE(rn,f(k))-meas(⋃j=1q‍(θj-εn,θj+εn))≥(2q+1)εn-2qεn=εn>0.
Hence, there exists a j∈{1,2,…,q} such that, for infinitely many n, we have
(3.21)meas(E(rn,f(k))∩(θj+εn,θj+1-εn))≥εnq.
Without loss of generality, we can assume that this holds for all the n. Let En=E(rn,f(k))∩(θj+εn,θj+1-εn). Thus, by the definition of E(r,f) and (3.21), it follows that
(3.22)∫θj+εnθj+1-εn‍log+|f(k)(rneiθ)|dθ≥∫En‍log+|f(k)(rneiθ)|dθ≥meas(En)δ(∞,f(k))4T(rn,f(k))≥εnδ(∞,f(k))4qT(rn,f(k)).On the other hand, by the definition ofBα,β(r,*) and (3.17), we have
(3.23)∫θj+εnθj+1-εn‍log+|f(k)(rneiθ)|dθ≤πrnωj2ωjsin(εnωj)Bθj,θj+1(rn,f(k))≤π2Kj,ε4ωj2εn(rnρ+ε+rnωjlog(rnT(rn,f))),
where rn∉E∪F, ωj=π/(θj+1-θj), and Kj,ε is a positive number depending on only j and ε. Combining (3.22) with (3.23) now yields
(3.24)εn2T(rn,f(k))≤qπ2Kj,εωj2δ(∞,f(k))(rnρ+ε+rnωjlog(rnT(rn,f))),
so that, together with (3.19) and Lemma 2.9, we have
(3.25)T1-2ε(rn,f(k))≤qπ2(2q+1)2[logrn]2+2εKj,εωj2δ(∞,f(k))(rnρ+ε+rnωj(logrn+logT(rn,f(k))+log(6e))).
Thus μ(f)=μ(f(k))≤max{ρ+ε,ωj}/(1-2ε)<∞ and so Theorem 1.1 follows from Theorem 1.4.This completes the proof of Theorem1.1.



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*Source: 217280-2012-11-11.xml*