Open Access
Issue
Wuhan Univ. J. Nat. Sci.
Volume 29, Number 4, August 2024
Page(s) 357 - 364
DOI https://doi.org/10.1051/wujns/2024294357
Published online 04 September 2024

© Wuhan University 2024

Licence Creative CommonsThis is an Open Access article distributed under the terms of the Creative Commons Attribution License (https://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

0 Introduction and Main Results

Let f(z): CC be a transcendental entire function in the complex plane C, and fm(z)=fο(fm-1)(z),mN denote the m-th iterate of f(z). The Fatou set F(f) and Julia set J(f) are defined by F(f)={zC|{fm(z)}m=1, which is normal at z and J(f)=C\F(f) respectively. Clearly, F(f) is open, and J(f) is closed and non-empty. For a basic understanding of complex dynamics, please refer to Ref.[1].

Suppose that f(z) is a transcendental entire function in C and argz=θ is a ray from the origin. The ray argz=θ, (θ[0,2π]) is said to be the limiting direction of J(f) if there exists an unbounded sequence {zn}J(f) such that limrnargzn=θ. Define Δ(f)={θ[0,2π)} the ray argz=θ is a limiting direction of J(f).

It is known that Δ(f) is closed and measurable, and we use measΔ(E) to stand for its linear measure.

The Nevanlinna theory is an important tool in this paper. We use some standard notations such as proximity function m(r,f), counting function of poles N(r,f), and Nevanlinna characteristic function T(r,f). The order ρ(f) and lower order μ(f) are defined by

ρ ( f ) = l i m s u p r l o g + l o g + M ( r , f ) l o g r ,

μ ( f ) = l i m i n f r l o g + l o g + M ( r , f ) l o g r ,

respectively, where M(r,f) denotes the maximum modulus of f on the circle |z|=r. And the deficiency of the values a defined by

δ ( a , f ) = l i m i n f r m ( r , 1 f - a ) T ( r , f ) .

We say that a is a Nevanlinna deficient value of f(z) if δ(a,f)>0. Here, when a=, we have

δ ( , f ) = l i m i n f r m ( r , f ) T ( r , f ) .

In addition, for a meromorphic function f(z), we use S(r,f) to denote any quantity satisfying S(r,f)=o(T(r,f)) for all r outside a possible exceptional set of finite logarithmic measure.

The Lebesgue linear measure of a set E[1,) is meas(E)=Edt , and the logarithmic measure of a set F[1,) is ml(F)=Fdtt. The upper and lower logarithmic densities of ml(F)=Fdtt are given by

l o g d e n s F ¯ : = l i m s u p r m 1 ( F [ 1 , r ) ) l o g   r

and

l o g d e n s F ̲ : = l i m i n f r m 1 ( F [ 1 , r ) ) l o g r ,

respectively.

Many observations on the radial distribution of Julia sets can be found in Refs.[2-6]. Baker[2] observed that, for a transcendental entire function f , J (f) cannot be contained in any finite set of straight lines. However, this is not true for transcendental meromorphic functions, for example J (tanz)=R. Qiao[3] showed that measΔ(f)=2π when μ(f)<1/2 and measΔ(f)π/μ(f) when μ(f)1/2, where f(z) is a transcendental entire function with finite lower order. Thus, a natural question arises: what can we say about the limit directions of entire functions with infinite lower order?

To answer this question, Huang and Wang[7,8] studied the radial distribution of Julia sets of solutions to complex linear differential equations and obtained the following results.

Theorem 1[7] Let {f1,f2,,fn} be a solution base of

f ( n ) + A ( z ) f = 0 (1)

where A(z) is a transcendental entire function with finite order, and denote E=f1f2fn.

Then

m e a s Δ ( E ) m i n { 2 π , π σ ( A ) } .

Remark 1   Actually, Huang and Wang[7] presented an example to illustrate that E(z) in Theorem 1 may occasionally have infinite lower order. In addition, Huang and Wang[8] directly studied the limiting direction of Julia sets of solutions of a class of higher order linear differential equations, and found that every non-trivial solution is of infinite lower order of these equations.

Theorem 2[8] Let Ai(z) (i=0,1,2,,n-1) be the entire functions of infinite order such that A0 is transcendental and m(r,Ai)=o(m(r,A0)) (i=1,2,,n-1) as r. Then every non-trivial solution f of the equation

f ( n ) + A n - 1 f ( n - 1 ) + + A 0 f = 0 (2)

satisfies measΔ(f)min{2π,πμ(A0)}.

Since then, the entire solutions of complex differential equations have attracted much attention; for references, please see Refs.[9-16]. For example, under the assumption of Theorem 2, Zhang et al[17] proved that meas(Δ(f)(Δ(f(k))))min{2π,π/μ(A0)}, where k is a positive integer.

Theorem 3[17] Let Ai(z) (i=0,1,2,,n-1) be the entire functions of finite lower order such that A0 is transcendental and m(r,Ai)=o(m(r,A0)) (i=1,2,,n-1) as r. Then every non-trivial solution f of Eq. (2) satisfies

m e a s ( Δ ( f ) ( Δ ( f ( k ) ) ) m i n { 2 π , π / μ ( A 0 ) } ,

where k is a positive integer.

To obtain a more precise relationship between T(r,f) and logM(r,f) of an entire function f, Petrenko introduced the so-called Petrenko's deviation as

β - ( , f ) = l i m i n f r l o g M ( r , f ) T ( r , f ) , β + ( , f ) = l i m s u p r l o g M ( r , f ) T ( r , f ) . (3)

If β-(,f)=β+(,f), then there exists a ν(0,1] such that

T ( r , f ) ~ ν l o g M ( r , f ) (4)

as r outside an exceptional set. An example f(z)=ez satisfies (4) with ν=1/π. Heittokangas[11] studied the oscillation of solutions of

f + A ( z ) f = 0 , (5)

where the coefficient A(z) is associated with Petrenko's deviation. In fact, he obtained the lower bound of the exponent of convergence of zeros of the product of two linearly independent solutions, which depends on Petrenko's deviation of the coefficient A(z). Similar to Ref.[11], let g(z) be entire and set

Ξ ( g ) : = { θ [ 0,2 π ) : l i m s u p r l o g + | g ( r e i θ | ) l o g r < } (6)

and

ξ ( g ) : = 1 2 π m e a s ( Ξ ( g ) ) .

Clearly, 0ξ(g)1.

Define the common limiting directions of the derivatives and primitives of an entire function f by L(f):=nΔ(f(n)), where f(n) denotes the n-th derivative or the n-th integral primitive of f for n0 or n<0, respectively. Combining the concept of Petrenko's deviation with the results of limiting directions of Julia set of solutions to complex differential equations, Zhang et al[17] proved the lower bound of the set of limiting directions of solutions to Eq. (1) has closed relations with the Petrenko's deviation of the coefficient A(z).

Theorem 4[17] Let ν(0,1] and A be a transcendental entire function that satisfies (4) as r outside a set G with logdens(G)<1. Then every nontrivial solution f of (2) satisfies

m e a s ( L ( f ) ) 2 π ν .

Moreover, let f1,f2,,fn be a solution base of Eq. (1), and denote E=f1f2fn. We have

m e a s ( L ( E ) ) 2 π ν .

Regarding Theorem 1-4 and the knowledge of limiting directions of complex differential equations, we aim to study the lower bound of the set of limiting directions of the following differential equation

F ( z ) f n ( z ) + P ( z , f ) = 0 , (7)

where F(z) is a transcendental entire function and it is associated with Petrenko's deviation,

P ( z , f ) = j = 1 s α j ( z ) f n 0 j ( f ' ) n 1 j ( f ( k ) ) n k j

is a differential polynomial in f(z) and its derivatives. The powers n0j,n1j,,nkj are non-negative integers and satisfy γp=min1js(i=0knij)n and the meromorphic fuctions αj(z) (j=1,2,,s) are small functions of F(z).

Theorem 5   Let ν(0,1] and F(z) be a transcendental entire function that satisfies (4) as r outside a set G with logdens(G)<1. Suppose that n,k are integers and that P(z,f) is a differential polynomial in f with γpn, where all coefficient αj(j=1,2,,s) are small functions of F(z). Then every non-trivial entire solution f(z) of Eq. (7) satisfies

m e a s ( L ( f ) ) 2 π ν . (8)

We recall the Jackson difference operator

D q f ( z ) = f ( q z ) - f ( z ) q z - z , z C \ { 0 } , q C \ { 0,1 } .

For kN{0}, the Jackson k-th difference operator is denoted by

D q 0 f ( z ) : = f ( z ) ,   D q k f ( z ) : = D q ( D q k - 1 f ( z ) ) .

Clearly, if f is differentiable,

l i m q 1 D q k f ( z ) = f ( k ) ( z ) .

Thus, a natural question arises: for Eq. (7), if we study the Jackson difference operators of f, does the conclusion meas(k{0}Δ(Dqkf(z)))2πv hold?

Set R(f)=k{0}Δ(Dqkf(z)), where q(0,+)\{1} and Dqkf(z) denotes the k-th Jackson difference operators of f(z). Our result can be stated as follows.

Theorem 6   Let ν(0,1] and F(z) be a transcendental entire function that satisfies (4) as r outside a set G with logdens(G)<1. Suppose that n,k are integers and that P(z,f) is a differential polynomial in f with γpn, where αj(j=1,2,,s) are small functions of F(z). Then we have

m e a s R ( f ) 2 π ν (9)

for every non-trivial entire solution f(z) of Eq. (7) .

In recent decades, due to the introduction of Nevanlinna theory in complex analysis, the properties of solutions of the Tumura-Clunie differential equation have been studied deeply. The original version of the Tumura-Clunie theory was stated by Tumura16], and the proof was completed by Clunie[18]. Next, we consider a general class of the Tumura-Clunie type non-linear differential equation

f n + A ( z ) P ( z , f ) = h ( z ) , ( n 2 ) , (10)

where A(z) and h(z) are entire functions, and P(z,f)=j=1sαj(z)fn0j(f')n1j(f(k))nkj is a differential polynomial in f(z) and its derivatives. The powers n0j,n1j,,nkj are non-negative integers and satisfy γp=min1js(i=0knij)n and the meromorphic functions αj(z) (j=1,2,,s) are small functions of h(z). Indeed, we obtain the following results.

Theorem 7   Let f be a nontrivial solution of Eq. (10), where A(z) is an entire function such that ξ(A)>0 and h(z) is an entire function with β-(,h)11-ξ(A). Then

m e a s ( L ( f ) ) m i n { 2 π , 2 π ( 1 β - ( , h ) + ξ ( A ) - 1 ) } .

For an entire function f(z)=n=0anzλn, if f(z) satisfies the gaps condition λnn as n, we call f(z) is an entire function with Fabry gaps. It satisfies

l o g L ( r , f ) ~ l o g M ( r , f ) , L ( r , f ) = m i n | z | = r | f ( z ) | (11)

as r outside a set of zero logarithmic density. We know that an entire function f with Fabry gaps satisfies β-(,f)=1, this yields the following immediate consequence of Theorem 3.

Theorem 8   Let f be a nontrivial solution of Eq. (10), where h(z) is a transcendental entire function with Fabry gaps. Then meas(L(f))2πξ(A).

2 Preliminary Lemmas

Before introducing lemmas and completing the proof of Theorems, we recall the Nevanlinna characteristic in an angle, see Refs.[10,14]. Assuming 0<α<β<2π, k=π/(β-α), we denote

Ω ( α , β ) = { z C | a r g z ( α , β ) } ,

Ω ( α , β , r ) = { z C | z Ω ( α , β ) , | z | < r } ,

Ω ( r , α , β ) = { z C | z Ω ( α , β ) , | z | > r } ,

and use Ω¯(α,β) to denote the closure of Ω(α,β).

Let f(z) be meromorphic on the angular Ω(α,β), we define

A α , β ( r , f ) = k π 1 r ( 1 t k - t k r 2 k ) { l o g + | f ( t e i α ) | + l o g + | f ( t e i β ) | } d t t , B α , β ( r , f ) = 2 k π r k α β l o g + | f ( r e i θ ) | s i n k ( θ - α ) d θ , C α , β ( r , f ) = 2 1 < | b v | < r ( 1 | b v | k - | b v | k r 2 k ) s i n k ( β v - α ) ,

where bv=|bv|eiβv(v=1,2,) are the poles of f(z) in Ω(α,β), counting multiplicities. The Nevanlinna angular characteristic function is defined by

S α , β ( r , f ) = A α , β ( r , f ) + B α , β ( r , f ) + C α , β ( r , f ) .

Especially, we use σα,β(f)=limsuprlogSα,β(r,f)logr to denote the order of Sα,β(r,f).

Lemma 1[19] If f is a transcendental entire function, then the Fatou set of f has no un-bounded multiply connected component.

Lemma 2[20] Suppose f(z) is analytic in Ω(r0,θ1,θ2), U is a hyperbolic domain and f :Ω(r0,θ1,θ2)U. If there exists a point aU\{} such that CU(a)>0, then there exists a constant d>0 such that for sufficiently small ε>0, we have

| f ( z ) | = O ( | z | d ) , z Ω ( r 0 , θ 1 + ε , θ 2 - ε ) , | z | .

Remark 2   The open set W is called a hyperbolic domain if C¯\W has at least two points. For an aC\W, we set

C W ( a ) = i n f { λ W ( z ) | z - a | : z W } ,

where λW(z) is the hyperbolic density on W. It is well known that if every component of W is simply connected, then CW(a)12. Before introducing the following lemma, we recall the definition of R-set. Suppose that the set B(zn,rn)={zC:|z-zn|<rn}, if n=1rn<,zn, then we call n=1B(zn,rn) a R-set. Obviously, {|z|:zn=1B(zn,rn)} is a set of the finite linear measure.

Lemma 3[8] Let z=rexp(iψ),r0+1<r and αψβ, where 0<β-α2π. Suppose that n(2) is an integer, and that f(z) is analytic in Ω(r0,α,β) with σα,β<. Choose α<α1<β1<β. Then, for every ε(0,βj-αj2) (j=1,2,,n-1) outside a set of linear measure zero with

α j = α + s = 1 j - 1 ε s a n d   β j = β + s = 1 j - 1 ε s ,   ( j = 2,3 , , n - 1 )

there exist K>0 and M>0 only depending on f, ε1,,εn-1 and Ω(αn-1,βn-1), and not depending on z such that

| f ' ( z ) f ( z ) | K r M ( s i n k ( ψ - α ) ) - 2

and

| f ( n ) ( z ) f ( z ) | K r M ( s i n k ( ψ - α ) j = 1 n - 1 s i n k j ( ψ - α j ) ) - 2

for all zΩ(αn-1,βn-1) outside an R-set H, where k=π/(β-α) and kεj=π/(βj-αj(j=1,2,,n-1)).

Remark 3   Ref.[20] proved that Lemma 3 holds when n=1, Wu[21] proved the case of n =2 and Huang and Wang[8] proved the case of n>2.

Lemma 4[16] Suppose that f(z) is a meromorphic function on Ω(α-ε,β+ε) for ε>0 and 0<α<β<2π. Then for r>1 possibly except a set with finite linear measure.

3 Proof of Theorem 5

For a sufficiently large positive constant M1, define D:={zC:|F(z)|>|z|M1} and H(r):={θ[0,2π): z=reiθD}. Then there exists some r1>0 such that if r>r1 , we have

2 π T ( r , F ) = H ( r ) l o g + | F ( r e i θ ) | d θ + l o g + | F ( r e i θ ) | d θ m e a s ( H ( r ) ) l o g M ( r , F ) + M 1 l o g r ( 2 π - m e a s ( H ( r ) ) ) (12)

Clearly, Eq. (12) leads to

2 π m e a s ( H ( r ) ) l o g M ( r , F ) T ( r , F ) + M 1 l o g r T ( r , F ) ( 2 π - m e a s ( H ( r ) ) ) . (13)

Since F(z) is transcendental and satisfies Eq. (4) outside G, we have

l i m i n f n m e a s ( H ( r ) ) 2 π ν . (14)

Therefore, there exists an infinite sequence {rn}(r1,+)\G such that

l i m i n f n m e a s ( H ( r n ) ) 2 π ν . (15)

We set j=1,2,,

B n : = j = n H ( r j ) .

It can be seen that Bn is monotone decreasing measurable set when n and meas(Bn)2π. Also, we set

H ˜ : = n = 1 B n ,

then H˜ is independent of r. Therefore, according to the monotone convergence Theorem and Eq. (15), we get

m e a s ( H ˜ ) = l i m n m e a s ( B n ) = l i m n m e a s ( j = n H ( r j ) ) 2 π ν . (16)

Suppose that meas(L(f))<2πν. Then meas(H˜\L(f))>0. Thus, we can choose a open interval I=(α,β) such that

I H ˜ , I L ( f ) = .

For every θI, argz=θ is not a limiting direction of the Julia set of some f(kθ)(z), where kθZ, only depending on θ. We can choose an angular domain Ω(θ-ζθ,θ+ζθ) such that

( θ - ζ θ , θ + ζ θ ) I   a n d   Ω ( r , θ - ζ θ , θ + ζ θ ) J ( f ( k θ ) ( z ) ) = (17)

where ζθ is a constant depending on θ. From Lemma 1, there exist a related r and an unbounded Fatou component U of (f(kθ)(z) such that Ω(r,θ-ζθ,θ+ζθ)U. Take an unbounded and connected closed section Γ on boundary U such that C\Γ is connected. From Remark 2, CC\Γ(a)1/2. Since f(kθ)(z):Ω(r,θ-ζθ,θ+ζθ)C\Γ is analytic, we have that for given sufficiently small ε>0, there is a constant d1>0 such that

| f ( k θ ) ( z ) | = O ( | z | d 1 ) a s | z | (18)

for zΩ(r,θ-ζθ+ε,θ+ζθ-ε).

Case 1 Let kθ0. Deriving from integral operation

| f ( k θ - 1 ) ( z ) | = 0 z | f ( k θ ) ( γ ) | | d γ | + c k θ , (19)

where ckθ is a constant, and the integration path is a straight line segment from 0 to z. From this and Eq. (18), we have |f(kθ-1)(z)|=O(|z|d1+1) for zΩ(r,θ-ζθ+ε,θ+ζθ-ε). By repeating the above discussion, it can be inferred that

| f ( z ) | = O ( | z | d 1 + k θ i j ) ,   z Ω ( r , θ - ζ θ + ε , θ + ζ θ - ε ) . (20)

Thus, from the definition of Nevanlinna angular characteristic, we have

S θ - ζ θ + ε , θ + ζ θ - ε ( r , f ) = O ( l o g r ) . (21)

Case 2 Let kθi<0. For any angular Ω(α,β), we get

S α , β ( f ( k θ + 1 ) ) S α , β ( r , f ( k θ + 1 ) f ( k θ ) ) + S α , β ( r , f ( k θ ) ) . (22)

By Lemma 4, we obtain

S α , β ( r , f ( k θ + 1 ) f ( k θ ) ) K 1 ( l o g + S α + ϵ , β - ϵ ( r , f ( k θ ) ) + l o g r + 1 ) , (23)

where ϵ=ε|kθ|, K1 is a positive constant. Combining Eq. (18), Eq. (22) and Eq. (23), we can get

S θ i j - ζ θ i j + 2 ε + ϵ , θ i j + ζ θ i j - 2 ε - ϵ ( r , f ( k θ + 1 ) ) = O ( l o g r ) . (24)

Similar to the above, repeating the discussion |kθ| times, we get

S θ - ζ θ + 3 ε , θ + ζ θ - 3 ε ( r , f ) = O ( l o g r ) . (25)

This means that whether kθ is positive or not, we always have

S θ i j - ζ θ i j + 3 ε , θ + ζ θ - 3 ε ( r , f ) = O ( l o g r ) . (26)

Thus, σθ-ζθ+3ε,θ+ζθ-3ε<. According to Lemma 3, there exist two constants K>0 and M2>0 such that

| f ( s ) ( z ) f ( z ) | K r M 2 , s = 1,2 , , k , (27)

for all zΩ(r,θ-ζθ+3ε,θ+ζθ-3ε) outside a R-set. From (7), we have

| z | M 1 < | F ( z ) | = | P ( z , f ) f n | (28)

and

| F ( z ) | = j = 1 s | α j ( z ) ( f ' f ) n 1 j ( f f ) n 2 j ( f ( k ) f ) n k j f n 0 j + n 1 j + + n k j - n | . (29)

Since n0j+n1j++nkj-n0, we get

f n 0 j + n 1 j + + n k j - n = O ( | z | d 1 )   a s   | z | . (30)

Combining Eq. (28), Eq. (29) and Eq. (30), it is found that

| z | M 1 < | F ( z ) | = | P ( z , f ) f n | K r M 3 . (31)

It is impossible since M1 can be taken sufficiently large and M3 is a finite positive constant. Therefore,

m e a s ( L ( f ) ) 2 π ν .

4 Proof of Theorem 6

Similar to the Theorem 5, we deduce that meas(H˜)2πν. Conversely, we assume that meas(R(f))<2πν. So meas(H˜\R(f))>0. We can therefore select finitely many open intervals I=(α,β) such that

I H ˜ ,   ( α , β ) R ( f ) = .

For every θI, argz=θ is not a limiting direction of the Julia set of Dqkf(z), where kN{0}. We can choose an angular domain Ω(θ-ϕθ,θ+ϕθ) such that

( θ - ϕ θ , θ + ϕ θ ) I ,     Ω ( r , θ - ϕ θ , θ + ϕ θ ) Δ ( D q k f ( z ) ) = (32)

where ϕθi is fixed based on θi. From Eq. (32) and Lemma 1, there is an unbounded Fatou component U of (Δ(Dqkf(z))) such that Ω(θ-ϕθ,θ+ϕθ)U. Take an unbounded and connected closed section Γ on boundary U such that C\Γ is connected. From Remark 2, CC\Γ(a)1/2. Since Dqkf(z):θ-ζϕ,θ+ϕθ)C\Γ is analytic, we have that for given sufficiently small enough ε > 0, there is a constant d2>0 such that

| D q k f ( z ) | = O ( | z | d 2 ) ,   z Ω ( α * , β * ) , (33)

where α*=θ-ϕθ+ε and β*=θ+ϕθ-ε .

According to the definition of Jackson k-th difference operator, we have

| D q k f ( z ) | = | D q k - 1 f ( q z ) - D q k - 1 f ( z ) | | q z - z | = O ( | z | d 2 ) ,   z Ω ( α * , β * ) . (34)

Thus,

| D q k - 1 f ( q z ) - D q k - 1 f ( z ) | = O ( | z | d 2 + 1 ) ,   z Ω ( α * , β * ) . (35)

Therefore, there exists a positive constant C such that

| D q k - 1 f ( q z ) - D q k - 1 f ( z ) | C ( | z | d 2 + 1 ) ,   z Ω ( α * , β * ) . (36)

There are two situations:

Case 1 Let q(0,1). If |z| is large enough, choose a positive integer r that satisfies (1q)r|z|(1q)r+1. In addition, 1|qrz|1q has been obtained. So there exists a constant M4 such that |Dqk-1f(zqt)|M4 where z{z|1|qrz|1q}. From Eq. (36), it can be concluded that

D q k - 1 f ( z ) | D q k - 1 f ( z ) - D q k - 1 f ( q z ) | + D q k - 1 f ( q z ) - D q k - 1 f ( q 2 z ) + + | D q k - 1 f ( q r - 1 z ) - D q k - 1 f ( q r z ) | + | D q k - 1 f ( q r z ) | C ( | z | d 2 + 1 ) + C ( | q z | d 2 + 1 ) + + C ( | q r - 1 z | d 2 + 1 ) + M 4 r C ( 1 + q d 2 + 1 + + q ( r - 1 ) ( d 2 + 1 ) ) | z | d 2 + 1 + M 4 = O ( | z | d 2 + 1 ) (37)

Thus,

| D q k - 1 f ( z ) | = O ( | z | d 2 + 1 ) ,   z Ω ( α * , β * ) . (38)

By repeating the discussion n times, it can be inferred that

| f ( z ) | = O ( | z | d 2 + k - 1 ) ,   z Ω ( α * , β * ) . (39)

Case 2 Let q(1,+). If |z| is large enough, choose a positive integer t that satisfies qt|z|qt+1. In addition, 1|zqt|q has been obtained. So there exists a normal number M5 such that |Dqk-1f(zqt)|M5 where z{z|1|zqt|q}. From (36), it can be concluded that

D q k - 1 f ( z ) | D q k - 1 f ( z ) - D q k - 1 f ( z q ) | + | D q k - 1 f ( z q ) - D q k - 1 f ( z q 2 ) | + + | D q k - 1 f ( z q t - 1 ) - D q k - 1 f ( z q t ) | + | D q k - 1 f ( z q t ) | C ( | z q | d 2 + 1 ) + C ( | z q 2 | d 2 + 1 ) + + C ( | z q t | d 2 + 1 ) + M 5 t C ( 1 q d 2 + 1 + 1 q 2 ( d 2 + 1 ) + + 1 q t ( d 2 + 1 ) ) | z | d 2 + 1 + M 5 (40)

Therefore,

| D q k - 1 f ( z ) | = O ( | z | d 2 + 1 ) ,   z Ω ( α * , β * ) . (41)

Similar to case 1, we have

| f ( z ) | = O ( | z | d 2 + k - 1 ) ,   z Ω ( α * , β * ) , (42)

which implies that

S α * , β * ( r , f ) = O ( l o g r ) (43)

Similar as Eq. (26) to Eq. (31), we can get a contradiction. Therefore,

m e a s R ( f ) 2 π ν .

5 Proof of Theorem 7

Since 1β-(,h)<11-ξ(A) and ξ(A)>0, there exist constants ε and d which satisfy

0 < ε < 1 β - ( , h ) - ( 1 - ξ ( A ) ) ,   2 2 + ε < d < 1 .

Therefore, we get

2 ( 1 - d ) d < ε < 1 β - ( , h ) - ( 1 - ξ ( A ) ) .

Define

I d ( r ) : = { θ [ 0,2 π ) : l o g | h ( r e i θ ) | ( 1 - d ) l o g M ( r , h ) } . (44)

Then,

2 π T ( r , h ) = I d ( r ) l o g + | h ( r e i θ ) | d θ + l o g + | h ( r e i θ ) | d θ m e a s ( I d ( r ) ) l o g M ( r , h ) + ( 2 π - m e a s ( I d ( r ) ) ) ( 1 - d ) l o g M ( r , h ) . (45)

Combining the definition of Eq. (3) yields

l i m s u p r m e a s ( I d ( r ) ) 2 π ( 1 d β - ( , h ) - 1 - d d ) . (46)

For the choice of ε and d, we deduce from Eq. (46) that there exists an infinite sequence {rn} such that

m e a s ( I d ( r n ) ) 2 π ( 1 d β - ( , h ) - 1 - d d ) - π ε 2 π d β - ( , h ) - 2 π ε 2 π d β - ( , h ) - 2 π ( 1 β - ( , h ) - ( 1 - ξ ( A ) ) ) 2 π ( 1 - ξ ( A ) ) . (47)

Set Dn:=n=jId(rj) and I˜d:=n=1Dn. Similar to Section 3, we have

m e a s ( I ˜ d ) = l i m r m e a s ( D n ) = l i m r m e a s ( n = j I d ( r j ) ) > 2 π ( 1 - ξ ( A ) ) .

Thus, we can conclude that

m e a s ( I ˜ d ) - 2 π ( 1 - ξ ( A ) ) > 2 π ( 1 d β - ( , h ) - 1 - d d ) - π ε - 2 π ( 1 - ξ ( A ) ) = 2 π [ ξ ( A ) - 1 d ( 1 - 1 β - ( , h ) ) - ε 2 ] > 2 π [ ξ ( A ) - 2 + ε 2 ( 1 - 1 β - ( , h ) ) - ε 2 ] = 2 π ( 1 β - ( , h ) - ( 1 - ξ ( A ) ) ) - π ε ( 2 - 1 β - ( , h ) ) . (48)

Since ε can be taken sufficiently small, we have

m e a s ( I ˜ d ) - 2 π ( 1 - ξ ( A ) ) 2 π ( 1 β - ( , h ) + ξ ( A ) - 1 ) > 0 .

Suppose that

m e a s ( L ( f ) ) < 2 π ( 1 β - ( , h ) + ξ ( A ) - 1 ) .

Then there exists an interval (α,β) such that

( α , β ) I ˜ d Ξ ( A ) ,   ( α , β ) L ( f ) = . (49)

By the similar arguments in Theorem 1, we deduce that for some integer nθ,

| f ( n θ ) ( z ) | = O ( | z | d 2 ) (50)

for zΩ(rθ,θ-ξθ+ε,θ+ξθ+ε)Ω(rθ,α,β) as |z|, where d2 is a positive constant, ε is a sufficiently small positive constant. Following the same discussion in Theorem 1, we have Sα*,β*(r,f)=O(logr), where

α * = θ - ξ + ε , β * = θ + ξ - ε

for nθ0, and

α * = θ - ξ + ε + ε ' , β * = θ + ξ - ε - ε '

for nθ<0.

This implies that σα*,β*(r,f)<. According to Lemma 3, there exist two constants K>0 and M6>0 such that

| f ( s ) ( z ) f ( z ) | K r M 6 , s = 1,2 , , k . (51)

From Eq. (18) to Eq. (20), Eq. (44), and Eq. (51), rewrite Eq. (10), for zΩ(rθ,α*+ε,β*+ε) outside an R-set, we have

| h ( z ) | | f n ( z ) | + | A ( z ) | j = 1 s | α j ( z ) ( f ' f ) n 1 j ( f f ) n 2 j ( f ( k ) f ) n k j f n 0 j + n 1 j + + n k j | .

This is impossible since h(z) is a transcendental entire function. Then the assertion follows.

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