# On the Maximum Term and Central Index of Entire Functions and Their Derivatives

Authors: Jianren Long; Zhigao Qin
Journal: Journal of Function Spaces
 (2018)

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## Abstract

We shall establish some criteria on entire series with finite logarithmic order in terms of maximum term and central index.


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## Body

## 1. Introduction

A functionf is called meromorphic, if it is nonconstant and analytic in the complex plane C except at possible isolated poles. If no poles occur, then f reduces to an entire function. In what follows, we assume that the reader is familiar with the standard notation and fundamental results in Nevanlinna theory of meromorphic functions; see [1, 2] or [3] for more details. We often use the order of growth and the lower order of growth to measure the growth of a meromorphic function. For a meromorphic function f in C, the order of growth and the lower order of growth of f are defined by (1)ρf=limsupr→∞log⁡Tr,flog⁡rand (2)λf=liminfr→∞log⁡Tr,flog⁡r, respectively. If f is entire function in C, the order and lower order of f are defined also by log⁡M(r,f); i.e., T(r,f) is replaced with log⁡M(r,f) in above equalities, where M(r,f)=maxz=rfz. By the following inequalities which can be found in [3, p. 10], then the order and lower order are same by definition of T(r,f) and log⁡M(r,f): (3)Tr,f≤log⁡Mr,f≤R+rR-rTR,f which hold for all r<R.The theory of meromorphic functions of finite positive order is fairly complete as compared to the theory of functions of order zero. Techniques that work well for functions of finite positive order often do not work for functions of order zero. In order to make some progress with functions of order zero, Chern introduced the concept of logarithmic order in [4]. For an entire function f of zero order, the logarithmic order of f is defined by(4)ρlogf=limsupr→∞log⁡log⁡Mr,flog⁡log⁡r.For an nonconstant entire function f, we must have ρlog(f)≥1, by the usual proof of Liouville’s theorem. It is easily seen that if f(z) has logarithmic order ρ then so has the function f(az+b) for a≠0. Furthermore, the function f(z)n is again of logarithmic order ρ, while f(zn) has logarithmic order ρ. It is clear that for a polynomial of degree k≥1 the logarithmic order is 1. There exists also transcendental entire series f(z) such that its logarithmic order is of one; for each positive number c>1, put rn=ecn, and set (5)fz=∑n=1∞1r1r2⋯rnzn, by a calculation [5, p. 6] we have ρlog(f)=1; the example can also be found in [6]. On the other hand, there exists transcendental entire series f(z) such that its logarithmic order is bigger than one. For each positive number ζ(>1), put k=ζ-1>0 and rn=en1/k, and set (6)fz=∑n=1∞1r1r2⋯rnzn,and by a direct calculation [5, p. 6] we have ρlog(f)=ζ>1; the example can also be found in [6]. Another case is of infinite logarithmic order; let q∈C and suppose that 0<q<1; then (7)fz=∑n=1∞qnlog⁡n2zn is of order zero, but its logarithmic order is infinite [7]. More results regarding logarithmic order can be found in [8–11].Wiman-Valiron theory is one of the important concepts in entire function theory; in the present paper, we study the properties of entire functions by Wiman-Valiron theory. To this end, we also need the following notations. Letf(z)=∑n=0∞anzn be a transcendental entire series in C. Then the maximum term μ(r,f) and central index ν(r,f) of f are denoted as μ(r)≡μ(r,f)=maxn≥0{anrn} and ν(r)≡ν(r,f)=max{n:μ(r,f)=anrn}.In [6], Chern and Kim consider some criteria conditions of logarithmic order with terms of maximum term and central index and proved the following consequence.Theorem A.
Letf be a transcendental entire series with finite logarithmic order; then the following statements are equivalent.(1)
T(r,f) has logarithmic order ρ.(2)
log⁡M(r,f) has logarithmic order ρ.(3)
log⁡μ(r,f) has logarithmic order ρ.(4)
ν(r,f) has logarithmic order ρ-1.Although, for any given entire seriesf of positive finite order, log⁡μ(r,f) and ν(r,f) both have the same order, the proof can be found in [12] or [2], but the situation is different for function of finite logarithmic order; from the Theorem A, we have(8)ρlogf-1=limsupr→∞log⁡log⁡μr,flog⁡log⁡r-1=limsupr→∞log⁡νr,flog⁡log⁡r.In [7], Berg and Pedersen described the logarithmic order by using Taylor coefficient of entire function and obtained the following result.Theorem B.
Letf(z)=∑n=0∞anzn be a transcendental entire series with finite logarithmic order; then its logarithmic order satisfies(9)ρlogf=1+limsupn→∞log⁡nlog⁡log⁡1/ann.Now the purpose of the paper is that the logarithmic order is described by using other forms in terms of maximum term and central index. To the end, we also need the following notations. Letμk(r)=μ(r,f(k)) and νk(r)=ν(r,f(k)), where f(k), k=1,2,…, denotes the kth-derivative of f. We reckon νk(r), k=0,1,2,⋯, from the first term of the series of f. For the uniformity in the notation we write μ0(r)=μ(r) and ν0(r)=ν(r). We denote the kth-derivative of μ(r) by μ(k)(r) at the point of its existence in (0,∞). It is easily seen that the functions μk(r) and νk(r), k=1,2,…, are positive, nondecreasing, and unbounded functions of r, having only ordinary discontinuities and νk(r)≥ν(r). For the entire function f, in the present paper, we find a precise measure of the rates of growth of μk(r)/μ(r), μ(k)(r)/μ(r), and νk(r)-ν(r) as r→∞ in terms of the parameters defined in (4). These results will be shown in Section 2.This paper is organized as follows. In Section2, we will state main results and prove them. In Section 3, we will discuss some further results.


## 2. Main Results and Proofs

In the proof of our theorems, the growth relationship between meromorphic functionf and its kth-derivative is needed. We prove the following result by using similar way in [[8], Theorem 3] and then omit the proof of details.Lemma 1.
Letf be a transcendental meromorphic function in C with zero order; then f and f(k), k=1,2,…, have the same logarithmic order.The following two consequences are due to G. Valiron [13], which can also be found in [12] or [2].Lemma 2.
Letf(z)=∑n=0∞anzn be an entire function in C. Then, for any 0<r0<r,(10)log⁡μr=log⁡μr0+∫r0rνttdt.Lemma 3.
Letf(z)=∑n=0∞anzn be an entire function in C. Then, for all r<R,(11)Mr,f<μrνR+RR-r.The first result is stated as follows.Theorem 4.
Letf(z)=∑n=0∞anzn be a transcendental entire series with finite logarithmic order ρ. Then(12)ρ-1=limsupr→∞logrμkr/μr1/kloglogr,k=1,2,….Proof.
By Lemma1, we know that the logarithmic order of a function and its derivative are the same. In view of (8), we have(13)ρ-1=limsupr→∞logνkr,floglogr,k=0,1,2,….Now, for k=0,1,2,…, let f(k)(z)=∑n=k∞Anzn-k, where An=n(n-1)⋯(n-k+1)an, and let νk(r)=N and νk+1(r)=N1; then (14)μk+1r=N1-kAN1rN1-k-1=N1-krAN1rN1-k≤νk+1r-krμkr. It follows that(15)rμk+1rμkr≤νk+1r-k,k=0,1,2,….On the other hand, for k=0,1,2,…, we have (16)μkr=ANrN-k=rN-kN-kANrN-k-1≤rνkr-kμk+1r. This implies(17)rμk+1rμkr≥νkr-k,k=0,1,2,….By using (15) and (17), we get(18)νkr-k≤rμk+1rμkr≤νk+1r-k,k=0,1,2,….By (18) and a simple calculation, we have (19)ν0r-k-1krk≤μkrμ0r≤νkrkrk,k=1,2,…. This implies(20)νr-k-1≤rμkrμr1/k≤νkr,k=1,2,….From (20), we get (21)logνr-k-1loglogr≤logrμkr/μr1/kloglogr≤logνkrloglogr,k=1,2,… Combining the inequality above and (13), we obtain(22)limsupr→∞log⁡rμkr/μr1/kloglogr=ρ-1,k=1,2,….The proof of Theorem 4 is completed.Theorem 5.
Letf(z)=∑n=0∞anzn be a transcendental entire series with finite logarithmic order ρ. Let ϕ(r,k)=νk(r)-ν(r), k=1,2,…. Then, for 0<r0<r, one has(23)ρ-1=limsupr→∞1kloglogr-1∫r0rϕt,ktdt,k=1,2,….Proof.
Letf(z)=∑n=0∞anzn, then, for k=1,2,…, we have(24)fkz=∑n=k∞nn-1⋯n-k+1anzn-kand(25)μkr=νkrνkr-1⋯νkr-k+1aνkrrνkr-k.The functions νk(r) and aνkr are constants in intervals and have at most an enumerable number of discontinuities and so their derivatives vanish almost everywhere except possibly at a set of measure zero. Taking logarithmic order of both the sides of (25), differentiating with respect to r, and denoting the derivative of μk(r) by μk′(r) at the point of its existence, we have for almost all values of r>r0>0(26)μk′rμkr=νkr-kr,k=1,2,….By (26), for sufficiently large r, we get(27)log⁡μkr=log⁡μkr0+∫r0rνkt-ktdt,k=1,2,….By using Lemma 2, for r>r0>0, we have(28)logμr=logμr0+∫r0rνttdt.Combining (27) and (28), for sufficiently large r, we have (29)log⁡μkrμr=O1+∫r0rνkt-k-νttdt,k=1,2,…; i.e.,(30)log⁡rμkrμr1/k=O1+1k∫r0rνkt-νttdt,k=1,2,….Thus, for k=1,2,…,(31)limsupr→∞log⁡rμkr/μr1/klog⁡log⁡r=limsupr→∞1/k∫r0rνkt-νt/tdtlog⁡log⁡r=limsupr→∞1klog⁡log⁡r-1∫r0rϕt,ktdt.In view of Theorem 4, (31) and (12) yield (23). This completes the proof of Theorem 5.Theorem 6.
Letf(z)=∑n=0∞anzn be a transcendental entire series with finite logarithmic order ρ. Then, for k=1,2,… and for almost all values of r satisfying 0<r0<r,(32)rμkrμk-1r=νr-k+1and(33)ρ-1=limsupr→∞,r∉Elogrμkr/μr1/klog⁡log⁡r,where E is set of r with zero measure which μ(k)(r) does not exist.Proof.
Sinceν(r) is step function with r, so μ(r)=aνrrν(r) is differential everywhere except at an enumerable set of points of discontinuities of aνr and ν(r). Hence, we have, at the points of existence of μ′(r), (34)μ′r=νraνrrνr-1=νrraνrrνr=νrrμr, for the derivatives of aνr and ν(r) vanish almost everywhere.
This implies(35)rμ′rμr=νr.By differentiating (35) at the points of existence of μ′(r) and μ′′(r), we get(36)μrμ′r+rμ′′r-rμ′r2=0.Combining (35) and (36), we have(37)rμ′′rμ′r=νr-1.On repeating the differentiation i times, we have(38)rμirμi-1r=νr-i+1.This proves (32).
Now, using (38), for i=1,2,…,k, and then multiplying the k-inequalities thus obtained give (39)rkμkrμr=νrνr-1⋯νr-k+1=νrk1-o1. Thus,(40)log⁡rμkr/μr1/klog⁡log⁡r+o1=logνrloglogr.Combining (40) and (13), we get (33), on proceeding to limits as r→∞ and r∉E. This completes the proof of Theorem 6.By the proof of Theorem6, we can get the following.Corollary 7.
Letf(z)=∑n=0∞anzn be a transcendental entire series. Then, for k=1,2,…, (41)μkrμk-1r~νrr and (42)μkrμr~νrrk as r→∞ and r∉E.


## 3. Further Discussion

In [4], Chern introduced the definition of logarithmic order of meromorphic function; however, there are not discussions of the lower logarithmic order. Hence, by using similar definition of lower order, we can define the lower logarithmic order and discuss some properties of entire functions in terms of lower logarithmic order.Definition 8.
Letf be an entire function with zero order. Then its lower logarithmic order is defined by(43)λlogf=liminfr→∞loglogMr,floglogr,where M(r,f)=maxz=rfz.In Theorem A, Chern obtained the growth relationship of logarithmic order by using maximum modulus, maximum term, and central index. In this section, we will try to find the growth relationship of lower logarithmic order by using maximum modulus, maximum term, and central index. To this end, letf(z)=∑n=0∞anzn be a transcendental entire series and set (44)l1=liminfr→∞loglogμr,floglogr, and (45)l2=liminfr→∞logνr,floglogr. Then we have the following consequences.Theorem 9.
Letf(z)=∑n=0∞anzn be a transcendental entire series with finite logarithmic order. Then λlog(f)=l1.Proof.
By using Cauchy inequality, we haveμ(r,f)≤M(r,f) for all r>0. Hence, l1≤λlog(f). Thus we just only prove l1≥λlog(f). By Lemma 2, there exists r0>0, such that, for all r>r0,(46)νrlog⁡2=νr∫r2r1tdt≤∫r2rνttdt≤log⁡μ2r.By Lemma 3 for R=2r, we get(47)Mr,f<μrν2r+2.Combining (46) and (47), we have(48)logMr,f<logμr+loglogμ4r+C,where C is positive constant. By (43) and (48), we have l1≥λlog(f).Theorem 10.
Letf(z)=∑n=0∞anzn be a transcendental entire series with finite logarithmic order. Then l1-1≤l2≤l1.Proof.
By using similar way of Theorem9, we get (46); hence l2≤l1 is obvious.
On the other hand, seta=maxn{an}; then, we get μ(r)≤arν(r). Hence,(49)log⁡log⁡μr≤log⁡νr+log⁡log⁡r+C,where C is positive constant number. This implies that l1-1≤l2. So, we have l1-1≤l2≤l1.Under Theorems9 and 10, we will consider similar results of Section 2 with respect to lower logarithmic order. To this end, let L∗ denote the class of functions f(z)=∑n=0∞anzn of transcendental entire function in C, satisfying (50)λlogf-1=liminfr→∞logνrloglogr.By using similar way to Section2, we can also prove the following consequences.Theorem 11.
Suppose thatf belongs to L∗. Then, for k=1,2,…, (51)λlogf-1=liminfr→∞log⁡rμkr/μr1/klog⁡log⁡r.Theorem 12.
Suppose thatf belong to L∗. Then, for 0<r0<r and k=1,2,…, (52)λlogf-1=liminfr→∞1klog⁡log⁡r-1∫r0rϕt,ktdt.Theorem 13.
Suppose thatf belongs to L∗. Then, for k=1,2,…, (53)λlogf-1=liminfr→∞,r∉Elog⁡rμkr/μr1/klog⁡log⁡r, where E is set of r with zero measure where μ(k)(r) does not exist.



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