# The Periodicity of Entire Functions with Finite Order

Authors: Zhiguo Ren; Guoqiang Dang
Journal: Journal of Mathematics
 (2021)

---

## Abstract

This paper is concerned with the periodicity of entire functions with finite growth order, and some sufficient conditions are given. Letf is a transcendental entire function with finite growth order, zero is a Picard exceptional value of f, and a given differential monomial Qf of f is periodic, then f is also periodic. We are also interested in finding the following: let f is a transcendental entire function with finite growth order, d is a Picard exceptional value of f and fzΔcnfz is a periodic function, then f is also a periodic function. These results extend Yang’s conjecture.


---

## Body

## 1. Introduction

The aim of this paper is to bring together and generalize recent research work by Wang and Hu [1], Liu and Yu [2], Deng and Yang [3], Liu et al. [4], Lü and Zhang [5], and Liu et al. [6] which is related to Yang’s conjecture [1, 2].Conjecture 1.
(Yang’s conjecture [1, 2]). Let f be a transcendental entire function and k be a positive integer. If ffk is a periodic function, then f is also a periodic function.
We assume that the readers are familiar with Nevanlinna’s theory [7–9]. Denote by Sr,f any quantify satisfying Sr,f=oTr,f, r⟶∞, outside of a possible exceptional set of finite logarithmic measure. For a meromorphic function fz, define its shift by fcz=fz+c and its difference operators by(1)△cfz=fz+c−fz,△cnfz=△cn−1△cfz.
For a nonconstant meromorphic functionfz, the order σ of fz is defined by(2)σ=σf=limsupr⟶∞logTr,flogr.
Wang and Hu [1] proved the following theorem.Theorem 1.
(see [1, Theorem 1]). Let f be a transcendental entire function and let k be a positive integer. If f2k is a periodic function, then f is also a periodic function.
Iff is a transcendental entire function with a nonzero Picard exceptional value, Liu and Yu [2] proved the following theorem in 2019.Theorem 2.
(see [2, Theorem 1.1]). Let f be a transcendental entire function with a nonzero Picard exceptional value and let k be a positive integer. If ffk is a periodic function, then f is also a periodic function.
In fact, Theorem1 shows that Yang’s conjecture is true when k=1. The transcendental entire function f2 cannot replaced by f in Theorem 1 [2], due to a counterexample which had been presented by Liu and Yu which shows that fz=ez+z is not a periodic function, but fkz is a periodic function [2]. They also depicted that the function f2 can be replaced by fn in Theorem 1 provided that n≥3 [2]. In Theorem 2, Liu and Yu [2, Theorem 1.1] investigated the periodicity of a transcendental entire function with a nonzero Picard exceptional. Furthermore, in Theorem 1, if we restrict f be a transcendental entire function with finite order and zero is a Picard exceptional value, we can obtain the following more refined theorem which shows that f has an explicit expression.Theorem 3.
Letf be a transcendental entire function with finite order, zero be a Picard exceptional value of f, and k be a positive integer. If f2k is a periodic function with a period of c, then f is also a periodic function with a period of 2c; furthermore, fz=eaz+b, where a≠0,b are constants and e2ac=1.
If we replacef2k by ff2k in Theorem 3, we can obtain the following theorem.Theorem 4.
Letf be a transcendental entire function with finite order and zero be a Picard exceptional value of f, and let k be a positive integer. If ff2k is a periodic function with a period of c, then f is also a periodic function with a period of 3c; furthermore, fz=eaz+b, where a≠0,b are constants and e3ac=1.
In 2020, under the conditions of Theorem 2, if the condition “f with a nonzero Picard exceptional value” replaced by “zero be a Picard exceptional value of f,” Deng and Yang [3] proved the following theorem.Theorem 5.
(see [3, Theorem 3]). Let f be a transcendental entire function with finite order and zero be a Picard exceptional value of f and let k be a positive integer. If ffk is a periodic function with a periodic of c, then fz is also a periodic function; furthermore, fz=eaz+b, where a≠0,b are constants, e2ac=1, and fz is with a period of 2c.
We generalize Theorem5 and obtain the following result.Theorem 6.
Letf be a transcendental entire function with finite order and zero be a Picard exceptional value of f and let k be a positive integer. If ffk2 is a periodic function with a period of c, then f is also a periodic function with a period of 3c; furthermore, fz=eaz+b, where a≠0,b are constants and e3ac=1.
The following generalized Yang’s conjecture has been considered by Liu et al. in 2020 [4].Conjecture 2.
(generalized Yang’s conjecture [4]). Let f be a transcendental entire function and n,k be positive integers. If fnfk is a periodic function, then f is also a periodic function.
Iff is a transcendental entire function, Liu et al. [4] proved that the aforementioned generalized Yang’s conjecture is true under the conditions that if k=1 or if fz=ehz, where hz is a nonconstant polynomial, or if fz has a nonzero Picard exceptional value and fz is of finite order. In 2021, Liu et al. [6] proved that the generalized Yang’s conjecture is true for meromorphic functions in the case of k=1, while for k≥2, the conjecture is true under some certain conditions even if n is a negative integer value. Baker [10] proved that if fz is a nonconstant entire function and pz is a polynomial with degp≥3, then fp is not a periodic function. Contrarily, when pf is periodic, where p is a nonconstant polynomial, it is interesting to discuss the periodicity of f. Let f be a transcendental entire function, k be a positive integer, ai1≤i≤n be constants, and an≠0; Liu and Yu [2, Theorem 1.5] proved that if a2f2+a1fkn≥2 is a periodic function, then f is also a periodic function. Staring form Theorems 3–6, we have a question that whether f is also periodic if we change the conditions f2k, ff2k, and ffk2 into a differential polynomial of fz. Furthermore, let f is a transcendental entire function with σ2f<1 and n≥2,k are two positive integers, and Nr,1/f=Sr,f, Liu and Yu ([2], Theorem 1.7) also proved that if anfn+⋯+a1fkn≥2 is a periodic function, an≠0, then f is also a periodic function, but the exact formula of fz is still unobtained [2]. In order to construct the explicit expression of f, it is necessary to put some restrictions of f. In fact, we obtain the following theorem.Theorem 7.
Letf be a transcendental entire function with finite order and zero be a Picard exceptional value of f, and let k be a positive integer. If Qf=fn0f′n1f″n2…fknk is a periodic function with a period of c, then f is also periodic with a period of γc; furthermore, fz=eaz+b, where a≠0,b are constants, eγac=1, and γ=n0+n1+n2+⋯+nk.
Noting that if we setn0=1,n1=⋯=nk−1=0 and nk=2, Theorem 7 will reduce to Theorem 6.
In 2020, Deng and Yang [3] studied the difference analogue of Yang’s conjecture and obtained the following theorem.Theorem 8.
(see [3, Theorem 4]). Let f be a transcendental entire function with finite order and d be a Picard exceptional value of f, and fzΔcfz be a periodic function; then, f is also a periodic function.
In Theorem3, if we replace f2k into Δcf2, we can obtain the following theorem.Theorem 9.
Letf be a transcendental entire function with finite order and d be a Picard exceptional value of f, and Δcf2 be a periodic function; then, f is also a periodic function.
Theorems8 and 9 partly show the internal relation of periodicity between a transcendental entire function and differences; in a more general setting, we can put the following theorem.Theorem 10.
Letf be a transcendental entire function with finite order and d be a Picard exceptional value of f, and fzΔcnfz be a periodic function; then f is also a periodic function.


## 2. Lemma

Lemma 1.
(see [11]). Let f be a transcendental entire function with finite order and c be a nonzero constant; then, Tr,fz+c=Tr,f+Sr,f.Lemma 2.
(see [7]). Let fjzj=1,2,…,n be meromorphic functions and fjz1≤j≤n−1 be nonconstant; ∑k=1nfjz≡1, and for any k1≤k≤n,(3)∑i=1nNr,1fj+n−1∑i=1nN¯r,fj<λ+o1Tr,fk,r⟶∞,r∉E,where E⊂1,∞ is a set of finite measure; then, fnz≡1.


## 3. Proofs of Theorems

### 3.1. Proof of Theorem3

Proof.
By Theorem1, it follows easily that f is periodic. We only need to show that f=eaz+b, where a≠0 and e2ac=1.
Suppose thatfz=ePz, where Pz is a polynomial, degPz≥1, Tr,P=Sr,f. Differentiate f2k times, f2k=e2Pzk=HPze2Pz, where HPz is a differential polynomial of Pz, HPz≡0. Otherwise, if HPz≡0, then f2k≡0; then, fz is a polynomial, but it is a contradiction with fz which is a transcendental entire function.
Sincef2k is a periodic function with a period of c, f2zk=f2z+ck; consequently,(4)HPze2Pz=HPz+ce2Pz+c,and in this way,(5)HPz+cHPz=e2Pze2Pz+c=1e2Pz+c−2Pz.
In the event ofdegPz≥2, then degPz+c−Pz≥1. By Lemma 1, we have Tr,HPz=Sr,f, Tr,HPz+c=Sr,f. But(6)Tr,e2Pz+c−2Pz=Tr,HPzHPz+c=Sr,e2Pz+c−2Pz,which is impossible. Therefore, degPz=1.
It follows thatPz=az+b,a≠0; then, fz=eaz+b, f2k=2ake2az+2b, 2ake2az+2b=2ake2az+2ac+2b; consequently, e2ac=1. Hence, fz+2c=ePz+2c=eaz+2c+b=eaz+b=fz.
This finishes the proof.


### 3.2. Proof of Theorem4

Proof.
According to the proof of Theorem3, we have f2k=e2Pzk=HPze2Pz, where HPz is a differential polynomial of Pz, degPz≥1, and HPz≢0.
Becauseff2k is a periodic function with a period of c, then(7)fzf2zk=fz+cf2z+ck.
It follows that(8)HPze3Pz=HPz+ce3Pz+c,and accordingly,(9)HPz+cHPz=e3Pze3Pz+c=1e3Pz+c−3Pz.
In the event ofdegPz≥2, degPz+c−Pz≥1. By Lemma 1, we have Tr,HPz=Sr,f, Tr,HPz+c=Sr,f. Therefore,(10)Tr,e3Pz+c−3Pz=Tr,HPzHPz+c=Sr,e3Pz+c−3Pz,which is impossible. Hence, degPz=1.
Assuming thatPz=az+b,a≠0. Then, fz=eaz+b, f2k=2ake2az+2b, and 2ake3az+3b=2ake3az+3ac+3b; consequently, e3ac=1. Hence, fz+3c=ePz+3c=eaz+3c+b=eaz+b=fz; then, fz is a periodic function with a period of 3c.
This finishes the proof.


### 3.3. Proof of Theorem6

Proof.
Suppose thatfz=ePz, where Pz is a polynomial, and degPz≥1 and Tr,P=Sr,f. Then, fk=ePzk=HPzePz, where HPz is a differential polynomial of Pz and HPz≡0. Otherwise, if HPz≡0, then fk≡0; therefore, fz is a polynomial, which is contradicting with fz which is a transcendental entire function with finite order.
Sinceffk2 is a periodic function with a period of c, then(11)fzfk2=fz+cfkz+c2,and in this way,(12)H2Pze3Pz=H2Pz+ce3Pz+c.
Taking the virtue of Lemma1, we have Tr,HPz=Sr,f, Tr,HPz+c=Sr,f. Therefore,(13)H2Pz+cH2Pz=e3Pze3Pz+c=1e3Pz+c−3Pz.
In the event ofdegPz≥2, degPz+c−Pz≥1. But(14)Tr,e3Pz+c−3Pz=Tr,H2PzH2Pz+c=Sr,e3Pz+c−3Pz,which is impossible. Hence, degPz=1.
Assuming thatPz=az+b,a≠0. So, fz=eaz+b, fk2=a2ke2az+2b, and by (11), we have a2ke3az+3b=a2ke3az+3ac+3b; therefore, e3ac=1. Hence, fz+3c=ePz+3c=eaz+3c+b=eaz+b=fz; consequently, fz is a periodic function with a period of 3c.
This finishes the proof.


### 3.4. Proof of Theorem7

Proof.
Suppose thatfz=ePz, where Pz is a polynomial, and degPz≥1 and Tr,P=Sr,f. Differentiate fk times; then fk=ePzk=HPzePz, where HPz is a differential polynomial of Pz and HPz≡0. Otherwise, if HPz≡0, then f2k≡0; then, fz is a polynomial, but it is contradicting with fz which is a transcendental entire function with finite order. Therefore,(15)Qf=fn0f′n1f″n2…fknk=en0PzH1n1Pzen1PzH2n2Pzen2Pz…HknkPzenkPz=H0Pzen0+n1+⋯+nkPz=H0PzeγPz,where H0Pz is a differential polynomial of Pz. Since Qf is a periodic function with a period of c, then Qfz=Qfz+c, and(16)H0PzeγPz=H0Pz+ceγPz+c.
Therefore,(17)H0Pz+cH0Pz=eγPzeγPz+c=1eγPz+c−γPz.
In the event ofdegPz≥2, degPz+c−Pz≥1. By Lemma 1, we have Tr,H0Pz=Sr,Qf, Tr,H0Pz+c=Sr,Qf. But(18)Tr,eγPz+c−γPz=Tr,H0PzH0Pz+c=Sr,eγPz+c−λPz,which is impossible. Hence, degPz=1.
Assuming thatPz=az+b,a≠0. So, fz=eaz+b, fknk=aknkenkaz+nkb, and by (16), we have aΓeγaz+b=aΓeγaz+ac+b, where Γ=n1+2n2+3n3+⋯+knk, so eγac=1. Therefore, fz+γc=ePz+γc=eaz+γc+b=eaz+b=fz, and hence fz is a periodic function with a period of γc.
This finishes the proof.


### 3.5. Proof of Theorem9

Proof.
Letfz be a transcendental entire function with finite order and d be a Picard exceptional value of fz; then, fz=d+ePz, where Pz is a polynomial of degPz≥1, and Tr,P=Sr,f. By simple computation, we have(19)Δcf2=e2Pz+c−e2Pz+2dePz+c−ePz=ePz+c−ePzePz+c+ePz+2d.(i)
Case 1. If d≠0.Case 1.1. If Δcf2z≡0, then ePz+c−ePz=0; therefore, f is a periodic function with a period of c.Case 1.2. If Δcf2z≡0. Assuming that λ is a period of Δcf2z, then(20)Δcf2z+λ=Δcf2z.Case 1.2.1. λ+c≠0.Byfz=d+epz and (20), we have(21)e2Pz+λ+c−e2Pz+λ+2dePz+λ+c−ePz+λ=e2Pz+c−e2Pz+2dePz+c−ePz.Consequently,(22)e2Pz+λ+c−e2Pz+c−e2Pz+λ+e2Pz+2dePz+λ+c−ePz+λ−ePz+c+ePz=0.Furthermore,(23)e2Pze2Pz+λ+c−2Pz−e2Pz+c−2Pz−e2Pz+λ−2Pz+1+2dePzePz+λ+c−Pz−ePz+λ−Pz−ePz+c−Pz+1=0.Therefore, we must have(24)e2Pze2Pz+λ+c−2Pz−e2Pz+c−2Pz−e2Pz+λ−2Pz+1=0,2dePzePz+λ+c−Pz−ePz+λ−Pz−ePz+c−Pz+1=0.It follows that(25)e2Pz+λ+c−2Pz−e2Pz+c−2Pz−e2Pz+λ−2Pz+1=0,(26)2dePz+λ+c−Pz−ePz+λ−Pz−ePz+c−Pz+1=0.If not, by Lemmas1 and 2, combining (25) and (26), we have(27)Tr,ePz=Tr,2d−ePz+λ+c−Pz+ePz+λ−Pz−1+ePz+c−Pz−e2Pz+λ+c−2Pz+e2Pz+c−2Pz−1+e2Pz+λ−2Pz=Sr,ePz,which is impossible.Case 1.2.1.1. degP≥2.Assuming thatPz=anzn+an−1zn−1+⋯+a0,an≠0,n≥2. By computation, we have ΔcPz=nanczn−1+p0z. Hence, Pz+λ+c−Pz, Pz+λ−Pz, Pz+c−Pz are polynomials with degree no more than n−1; by (25) and Lemma 2, we have −e2Pz+λ+c−2Pz≡1, or e2Pz+c−2Pz≡1, or e2Pz+λ−2Pz≡1, but it is impossible.Case 1.2.1.2. degP=1.ByPz=a1z+a0,ΔcPz=a1c and (25), we have ea1z+a0e2a1λ+c−e2a1c−e2a1λ+1=0. By Lemma 2, we have e2a1λ≡1, or e2a1c≡1, or −e2a1λ+c≡1; therefore, fz is a function of a period of 2λ, or 2c, or 2λ+cπi. Since d≠0, by (26), ea1λ+c−ea1λ−ea1c+1=0; furthermore ea1λ≡1, or ea1c≡1, or −ea1λ+c≡1, and hence fz is a periodic function with a period of λ or c or λ+cπi.Case 1.2.2. λ+c=0.Byfz=d+epz and (20), we have(28)e2Pz−e2Pz−c+2dePz−ePz−c=e2Pz+c−e2Pz+2dePz+c−ePz.Consequently,(29)2e2Pz−e2Pz+c−e2Pz−c+2d2ePz−ePz−c−ePz+c=0.Therefore, we must have(30)2e2Pz−e2Pz+c−e2Pz−c=0,2d2ePz−ePz−c−ePz+c=0.It follows that(31)e2Pz2−e2Pz+c−2Pz−e2Pz−c−2Pz=0,(32)2dePz2−ePz−c−Pz−ePz+c−Pz=0.If not, by Lemmas1 and 2, combining (31) and (32), we have(33)Tr,ePz=Tr,1−1/2ePz−c−Pz−1/2ePz+c−Pz1−1/2e2Pz+c−2Pz−1/2e2Pz−c−2Pz=Sr,ePz,which is impossible.Case 1.2.2.1. degP≥2.Assuming thatPz=anzn+an−1zn−1+⋯+a0,an≠0,n≥2. By computation, we have ΔcPz=nanczn−1+p0z. Hence, Pz−c−Pz, Pz+c−Pz are polynomials with degree no more than n−1; by (31) and Lemma 2, we have e2Pz+c−2Pz≡2, or e2Pz−c−2Pz≡2, but it is impossible.Case 1.2.2.2. degP=1.ByPz=a1z+a0,ΔcPz=a1c and (31), we have e2a1z+2a02−e2a1c−e−2a1c=0. By Lemma 2, we have e−2a1c≡2, or e2a1c≡2; therefore, fz is a function of a period of 2ln1/2c. If d≠0, by (32), we have 2−ePz−c−Pz−ePz+c−Pz=0, and it follows that 1/2e−a1c≡1, or 1/2ea1c≡1, and hence fz is a periodic function with a period of ln1/2c.(ii)
Case 2. If d=0.Case 2.1. If Δcf2z≡0, then e2Pz+c−e2Pz=0; therefore, f is a periodic function with a period of c.Case 2.2. If Δcf2z≡0. Assuming that λ is a period of Δcf2z, then(34)Δcf2z+λ=Δcf2z.Case 2.2.1. λ+c≠0.Byfz=epz and (34), we have(35)e2Pz+λ+c−e2Pz+λ=e2Pz+c−e2Pz.Consequently,(36)e2Pz+λ+c−e2Pz+c−e2Pz+λ+e2Pz=0.Furthermore,(37)e2Pze2Pz+λ+c−2Pz−e2Pz+c−2Pz−e2Pz+λ−2Pz+1=0.Therefore, we must have(38)e2Pz+λ+c−2Pz−e2Pz+c−2Pz−e2Pz+λ−2Pz+1=0.Case 2.2.1.1. degP≥2.Assuming thatPz=anzn+an−1zn−1+⋯+a0,an≠0,n≥2. By computation, we have ΔcPz=nanczn−1+p0z. Hence, Pz+λ+c−Pz, Pz+λ−Pz, Pz+c−Pz are polynomials with degree no more than n−1; by (38) and Lemma 2, we have −e2Pz+λ+c−2Pz≡1, or e2Pz+c−2Pz≡1, or e2Pz+λ−2Pz≡1, but it is impossible.Case 2.2.1.2. degP=1.ByPz=a1z+a0,ΔcPz=a1c and (38), we have e2a1λ+c−e2a1c−e2a1λ+1=0. By Lemma 2, we have e2a1λ≡1, or e2a1c≡1, or −e2a1λ+c≡1; therefore, fz is a function of a period of 2λ, or 2c, or 2λ+cπi.Case 2.2.2. λ+c=0.Byfz=epz and (34), we have(39)e2Pz−e2Pz−c=e2Pz+c−e2Pz.Consequently,(40)2e2Pz−e2Pz+c−e2Pz−c=0.It follows that(41)e2Pz2−e2Pz+c−2Pz−e2Pz−c−2Pz=0.Case 2.2.2.1. degP≥2.Assuming thatPz=anzn+an−1zn−1+⋯+a0,an≠0,n≥2. By computation, we have ΔcPz=nanczn−1+p0z. Hence, Pz−c−Pz, Pz+c−Pz are polynomials with degree no more than n−1; by (41) and Lemma 2, we have e2Pz+c−2Pz≡2, or e2Pz−c−2Pz≡2, but it is impossible.Case 2.2.2.2. degP=1.ByPz=a1z+a0,ΔcPz=a1c and (41), we have e2a1z+2a02−e2a1c−e−2a1c=0. By Lemma 2, we have e−2a1c≡2, or e2a1c≡2; therefore, fz is a function of a period of 2ln1/2c.This finishes the proof.


### 3.6. Proof of Theorem10

Proof.
fz is a transcendental entire function with finite order and d is its Picard exceptional value; then, fz=d+ePz, where Pz is a polynomial of degPz≥1, and Tr,P=Sr,f. Then,(42)Δcnfz=ePz∑j=0n−1−1jCnjePz+n−jc−Pz+−1nePz.(i)
Case 1. If Δcnfz≡0.We assert thatfz is a periodic function with a period of c. If degP is odd, by (15), we have(43)∑j=0n−1−1jCnjePz+n−jc−Pz−1=0.By Lemma2, for each j0≤j≤n−1, we have −1jCnjePz+n−jc−Pz≡1. Therefore, Pz+n−jc−Pz=0; then, fz=d+ePz=d+ePz+n−jc=fz+n−jc, and hence fz is a periodic function with a period of c. If degP is even, it is easy to prove that f is also periodic.(ii)
Case 2. If Δcnfz≡0.Assuming thatλ is a periodic of fzΔcnfz, then(44)fzΔcnfz=fz+λΔcnfz+λ.Case 2.1. λ+c≠0.Byfz=d+ePz and (15) and (16),(45)fzePz∑j=0n−1−1jCnjePz+n−jc−Pz+−1nePz=fz+λePz+λ∑j=0n−1−1jCnjePz+λ+n−jc−Pz+λ+−1nePz+λ.Furthermore,(46)d+ePzePz∑j=0n−1−1jCnjePz+n−jc−Pz+−1nePz=d+ePz+λePz+λ∑j=0n−1−1jCnjePz+λ+n−jc−Pz+λ+−1nePz+λ.It follows that(47)dePz∑j=0n−1−1jCnjePz+n−jc−Pz+−1n−ePz+λ−Pz∑j=0n−1−1jCnjePz+λ+n−jc−Pz+λ+−1n+e2Pz∑j=0n−1−1jCnjePz+n−jc−Pz+−1n−e2Pz+λ−2Pz∑j=0n−1−1jCnjePz+λ+n−jc−Pz+λ+−1n=0.Consequently,(48)d∑j=0n−1−1jCnjeΔn−jcP+−1n−eΔλP∑j=0n−1−1jCnjeΔλ+n−jcP+−1nePz+∑j=0n−1−1jCnjeΔn−jcP+−1n−e2ΔλP∑j=0n−1−1jCnjeΔλ+n−jcP+−1ne2Pz=0.Let(49)A≔d∑j=0n−1−1jCnjeΔn−jcP+−1n−eΔλP∑j=0n−1−1jCnjeΔλ+n−jcP+−1n,B≔∑j=0n−1−1jCnjeΔn−jcP+−1n−e2ΔλP∑j=0n−1−1jCnjeΔλ+n−jcP+−1n.Hence,A=0,B=0. Otherwise, by Lemmas 1 and 2, degΔn−jcP<degP,degΔλ+n−jcP<degP, so Tr,eP=Tr,A/B=Sr,eP, but it is impossible.Case 2.1.1. degP≥2.By simple computation, we haveΔcPz=mamczm−1+P0z. Furthermore,(50)Δn−jcPz=mamn−jczm−1+P1z,Δλ+n−jcPz=mamλ+n−jczm−1+P2z,P0z,P1z,P2z are polynomials of degree no more than m−2; furthermore, ΔλPz, Δn−jcPz, Δλ+n−jcPz are polynomials with degree of m−1, where j=0,…,n−1; by B=0 and Lemma 2.2, we have −1jCnjeΔn−jcP+−1n≡0, or −1n−e2ΔλP−1n≡0, or −e2ΔλP−1jCnjeΔλ+n−jcP+−1n≡0, but it is impossible.Case 2.1.2. degP=1.ByPz=a1z+a0,ΔcPz=a1c,Δn−jcPz=a1n−jc, Δλ+n−jcPz=a1λ+n−jc, on account of B=0, we have(51)∑j=0n−1−1jCnjea1n−jc+−1n−e2a1λ∑j=0n−1−1jCnjea1λ+n−jc−−1ne2a1λ=0.By Lemma2 and B=0, when n is even, we have −1jCnjea1n−jc+1≡0, or −1jCnjea1λ+n−jc+2a1λ≡1, j=0,…,n−1, or e2a1λ≡1. Therefore, fz is a periodic function with a period of 2λ or n−jcln−1j+1Cnj or 3λ+n−jcln−1jCnj. If d≠0, by A=0, then −1jCnjea1n−jc+1≡0, or ea1λ≡1, or −1jCnjea1λ+n−jc+a1λ≡1, j=0,…,n−1, and hence fz is a periodic function with a period of λ or n−jcln−1j+1Cnj or 2λ+n−jcln−1jCnj. The same proof for n is odd. Therefore, fz is a periodic function.Case 2.2. λ+c=0.This proof follows by the same method as in Case 2.1; just replaceλ by −c, so we omit it here.This finishes the proof.


## 3.1. Proof of Theorem3

Proof.
By Theorem1, it follows easily that f is periodic. We only need to show that f=eaz+b, where a≠0 and e2ac=1.
Suppose thatfz=ePz, where Pz is a polynomial, degPz≥1, Tr,P=Sr,f. Differentiate f2k times, f2k=e2Pzk=HPze2Pz, where HPz is a differential polynomial of Pz, HPz≡0. Otherwise, if HPz≡0, then f2k≡0; then, fz is a polynomial, but it is a contradiction with fz which is a transcendental entire function.
Sincef2k is a periodic function with a period of c, f2zk=f2z+ck; consequently,(4)HPze2Pz=HPz+ce2Pz+c,and in this way,(5)HPz+cHPz=e2Pze2Pz+c=1e2Pz+c−2Pz.
In the event ofdegPz≥2, then degPz+c−Pz≥1. By Lemma 1, we have Tr,HPz=Sr,f, Tr,HPz+c=Sr,f. But(6)Tr,e2Pz+c−2Pz=Tr,HPzHPz+c=Sr,e2Pz+c−2Pz,which is impossible. Therefore, degPz=1.
It follows thatPz=az+b,a≠0; then, fz=eaz+b, f2k=2ake2az+2b, 2ake2az+2b=2ake2az+2ac+2b; consequently, e2ac=1. Hence, fz+2c=ePz+2c=eaz+2c+b=eaz+b=fz.
This finishes the proof.


## 3.2. Proof of Theorem4

Proof.
According to the proof of Theorem3, we have f2k=e2Pzk=HPze2Pz, where HPz is a differential polynomial of Pz, degPz≥1, and HPz≢0.
Becauseff2k is a periodic function with a period of c, then(7)fzf2zk=fz+cf2z+ck.
It follows that(8)HPze3Pz=HPz+ce3Pz+c,and accordingly,(9)HPz+cHPz=e3Pze3Pz+c=1e3Pz+c−3Pz.
In the event ofdegPz≥2, degPz+c−Pz≥1. By Lemma 1, we have Tr,HPz=Sr,f, Tr,HPz+c=Sr,f. Therefore,(10)Tr,e3Pz+c−3Pz=Tr,HPzHPz+c=Sr,e3Pz+c−3Pz,which is impossible. Hence, degPz=1.
Assuming thatPz=az+b,a≠0. Then, fz=eaz+b, f2k=2ake2az+2b, and 2ake3az+3b=2ake3az+3ac+3b; consequently, e3ac=1. Hence, fz+3c=ePz+3c=eaz+3c+b=eaz+b=fz; then, fz is a periodic function with a period of 3c.
This finishes the proof.


## 3.3. Proof of Theorem6

Proof.
Suppose thatfz=ePz, where Pz is a polynomial, and degPz≥1 and Tr,P=Sr,f. Then, fk=ePzk=HPzePz, where HPz is a differential polynomial of Pz and HPz≡0. Otherwise, if HPz≡0, then fk≡0; therefore, fz is a polynomial, which is contradicting with fz which is a transcendental entire function with finite order.
Sinceffk2 is a periodic function with a period of c, then(11)fzfk2=fz+cfkz+c2,and in this way,(12)H2Pze3Pz=H2Pz+ce3Pz+c.
Taking the virtue of Lemma1, we have Tr,HPz=Sr,f, Tr,HPz+c=Sr,f. Therefore,(13)H2Pz+cH2Pz=e3Pze3Pz+c=1e3Pz+c−3Pz.
In the event ofdegPz≥2, degPz+c−Pz≥1. But(14)Tr,e3Pz+c−3Pz=Tr,H2PzH2Pz+c=Sr,e3Pz+c−3Pz,which is impossible. Hence, degPz=1.
Assuming thatPz=az+b,a≠0. So, fz=eaz+b, fk2=a2ke2az+2b, and by (11), we have a2ke3az+3b=a2ke3az+3ac+3b; therefore, e3ac=1. Hence, fz+3c=ePz+3c=eaz+3c+b=eaz+b=fz; consequently, fz is a periodic function with a period of 3c.
This finishes the proof.


## 3.4. Proof of Theorem7

Proof.
Suppose thatfz=ePz, where Pz is a polynomial, and degPz≥1 and Tr,P=Sr,f. Differentiate fk times; then fk=ePzk=HPzePz, where HPz is a differential polynomial of Pz and HPz≡0. Otherwise, if HPz≡0, then f2k≡0; then, fz is a polynomial, but it is contradicting with fz which is a transcendental entire function with finite order. Therefore,(15)Qf=fn0f′n1f″n2…fknk=en0PzH1n1Pzen1PzH2n2Pzen2Pz…HknkPzenkPz=H0Pzen0+n1+⋯+nkPz=H0PzeγPz,where H0Pz is a differential polynomial of Pz. Since Qf is a periodic function with a period of c, then Qfz=Qfz+c, and(16)H0PzeγPz=H0Pz+ceγPz+c.
Therefore,(17)H0Pz+cH0Pz=eγPzeγPz+c=1eγPz+c−γPz.
In the event ofdegPz≥2, degPz+c−Pz≥1. By Lemma 1, we have Tr,H0Pz=Sr,Qf, Tr,H0Pz+c=Sr,Qf. But(18)Tr,eγPz+c−γPz=Tr,H0PzH0Pz+c=Sr,eγPz+c−λPz,which is impossible. Hence, degPz=1.
Assuming thatPz=az+b,a≠0. So, fz=eaz+b, fknk=aknkenkaz+nkb, and by (16), we have aΓeγaz+b=aΓeγaz+ac+b, where Γ=n1+2n2+3n3+⋯+knk, so eγac=1. Therefore, fz+γc=ePz+γc=eaz+γc+b=eaz+b=fz, and hence fz is a periodic function with a period of γc.
This finishes the proof.


## 3.5. Proof of Theorem9

Proof.
Letfz be a transcendental entire function with finite order and d be a Picard exceptional value of fz; then, fz=d+ePz, where Pz is a polynomial of degPz≥1, and Tr,P=Sr,f. By simple computation, we have(19)Δcf2=e2Pz+c−e2Pz+2dePz+c−ePz=ePz+c−ePzePz+c+ePz+2d.(i)
Case 1. If d≠0.Case 1.1. If Δcf2z≡0, then ePz+c−ePz=0; therefore, f is a periodic function with a period of c.Case 1.2. If Δcf2z≡0. Assuming that λ is a period of Δcf2z, then(20)Δcf2z+λ=Δcf2z.Case 1.2.1. λ+c≠0.Byfz=d+epz and (20), we have(21)e2Pz+λ+c−e2Pz+λ+2dePz+λ+c−ePz+λ=e2Pz+c−e2Pz+2dePz+c−ePz.Consequently,(22)e2Pz+λ+c−e2Pz+c−e2Pz+λ+e2Pz+2dePz+λ+c−ePz+λ−ePz+c+ePz=0.Furthermore,(23)e2Pze2Pz+λ+c−2Pz−e2Pz+c−2Pz−e2Pz+λ−2Pz+1+2dePzePz+λ+c−Pz−ePz+λ−Pz−ePz+c−Pz+1=0.Therefore, we must have(24)e2Pze2Pz+λ+c−2Pz−e2Pz+c−2Pz−e2Pz+λ−2Pz+1=0,2dePzePz+λ+c−Pz−ePz+λ−Pz−ePz+c−Pz+1=0.It follows that(25)e2Pz+λ+c−2Pz−e2Pz+c−2Pz−e2Pz+λ−2Pz+1=0,(26)2dePz+λ+c−Pz−ePz+λ−Pz−ePz+c−Pz+1=0.If not, by Lemmas1 and 2, combining (25) and (26), we have(27)Tr,ePz=Tr,2d−ePz+λ+c−Pz+ePz+λ−Pz−1+ePz+c−Pz−e2Pz+λ+c−2Pz+e2Pz+c−2Pz−1+e2Pz+λ−2Pz=Sr,ePz,which is impossible.Case 1.2.1.1. degP≥2.Assuming thatPz=anzn+an−1zn−1+⋯+a0,an≠0,n≥2. By computation, we have ΔcPz=nanczn−1+p0z. Hence, Pz+λ+c−Pz, Pz+λ−Pz, Pz+c−Pz are polynomials with degree no more than n−1; by (25) and Lemma 2, we have −e2Pz+λ+c−2Pz≡1, or e2Pz+c−2Pz≡1, or e2Pz+λ−2Pz≡1, but it is impossible.Case 1.2.1.2. degP=1.ByPz=a1z+a0,ΔcPz=a1c and (25), we have ea1z+a0e2a1λ+c−e2a1c−e2a1λ+1=0. By Lemma 2, we have e2a1λ≡1, or e2a1c≡1, or −e2a1λ+c≡1; therefore, fz is a function of a period of 2λ, or 2c, or 2λ+cπi. Since d≠0, by (26), ea1λ+c−ea1λ−ea1c+1=0; furthermore ea1λ≡1, or ea1c≡1, or −ea1λ+c≡1, and hence fz is a periodic function with a period of λ or c or λ+cπi.Case 1.2.2. λ+c=0.Byfz=d+epz and (20), we have(28)e2Pz−e2Pz−c+2dePz−ePz−c=e2Pz+c−e2Pz+2dePz+c−ePz.Consequently,(29)2e2Pz−e2Pz+c−e2Pz−c+2d2ePz−ePz−c−ePz+c=0.Therefore, we must have(30)2e2Pz−e2Pz+c−e2Pz−c=0,2d2ePz−ePz−c−ePz+c=0.It follows that(31)e2Pz2−e2Pz+c−2Pz−e2Pz−c−2Pz=0,(32)2dePz2−ePz−c−Pz−ePz+c−Pz=0.If not, by Lemmas1 and 2, combining (31) and (32), we have(33)Tr,ePz=Tr,1−1/2ePz−c−Pz−1/2ePz+c−Pz1−1/2e2Pz+c−2Pz−1/2e2Pz−c−2Pz=Sr,ePz,which is impossible.Case 1.2.2.1. degP≥2.Assuming thatPz=anzn+an−1zn−1+⋯+a0,an≠0,n≥2. By computation, we have ΔcPz=nanczn−1+p0z. Hence, Pz−c−Pz, Pz+c−Pz are polynomials with degree no more than n−1; by (31) and Lemma 2, we have e2Pz+c−2Pz≡2, or e2Pz−c−2Pz≡2, but it is impossible.Case 1.2.2.2. degP=1.ByPz=a1z+a0,ΔcPz=a1c and (31), we have e2a1z+2a02−e2a1c−e−2a1c=0. By Lemma 2, we have e−2a1c≡2, or e2a1c≡2; therefore, fz is a function of a period of 2ln1/2c. If d≠0, by (32), we have 2−ePz−c−Pz−ePz+c−Pz=0, and it follows that 1/2e−a1c≡1, or 1/2ea1c≡1, and hence fz is a periodic function with a period of ln1/2c.(ii)
Case 2. If d=0.Case 2.1. If Δcf2z≡0, then e2Pz+c−e2Pz=0; therefore, f is a periodic function with a period of c.Case 2.2. If Δcf2z≡0. Assuming that λ is a period of Δcf2z, then(34)Δcf2z+λ=Δcf2z.Case 2.2.1. λ+c≠0.Byfz=epz and (34), we have(35)e2Pz+λ+c−e2Pz+λ=e2Pz+c−e2Pz.Consequently,(36)e2Pz+λ+c−e2Pz+c−e2Pz+λ+e2Pz=0.Furthermore,(37)e2Pze2Pz+λ+c−2Pz−e2Pz+c−2Pz−e2Pz+λ−2Pz+1=0.Therefore, we must have(38)e2Pz+λ+c−2Pz−e2Pz+c−2Pz−e2Pz+λ−2Pz+1=0.Case 2.2.1.1. degP≥2.Assuming thatPz=anzn+an−1zn−1+⋯+a0,an≠0,n≥2. By computation, we have ΔcPz=nanczn−1+p0z. Hence, Pz+λ+c−Pz, Pz+λ−Pz, Pz+c−Pz are polynomials with degree no more than n−1; by (38) and Lemma 2, we have −e2Pz+λ+c−2Pz≡1, or e2Pz+c−2Pz≡1, or e2Pz+λ−2Pz≡1, but it is impossible.Case 2.2.1.2. degP=1.ByPz=a1z+a0,ΔcPz=a1c and (38), we have e2a1λ+c−e2a1c−e2a1λ+1=0. By Lemma 2, we have e2a1λ≡1, or e2a1c≡1, or −e2a1λ+c≡1; therefore, fz is a function of a period of 2λ, or 2c, or 2λ+cπi.Case 2.2.2. λ+c=0.Byfz=epz and (34), we have(39)e2Pz−e2Pz−c=e2Pz+c−e2Pz.Consequently,(40)2e2Pz−e2Pz+c−e2Pz−c=0.It follows that(41)e2Pz2−e2Pz+c−2Pz−e2Pz−c−2Pz=0.Case 2.2.2.1. degP≥2.Assuming thatPz=anzn+an−1zn−1+⋯+a0,an≠0,n≥2. By computation, we have ΔcPz=nanczn−1+p0z. Hence, Pz−c−Pz, Pz+c−Pz are polynomials with degree no more than n−1; by (41) and Lemma 2, we have e2Pz+c−2Pz≡2, or e2Pz−c−2Pz≡2, but it is impossible.Case 2.2.2.2. degP=1.ByPz=a1z+a0,ΔcPz=a1c and (41), we have e2a1z+2a02−e2a1c−e−2a1c=0. By Lemma 2, we have e−2a1c≡2, or e2a1c≡2; therefore, fz is a function of a period of 2ln1/2c.This finishes the proof.


## 3.6. Proof of Theorem10

Proof.
fz is a transcendental entire function with finite order and d is its Picard exceptional value; then, fz=d+ePz, where Pz is a polynomial of degPz≥1, and Tr,P=Sr,f. Then,(42)Δcnfz=ePz∑j=0n−1−1jCnjePz+n−jc−Pz+−1nePz.(i)
Case 1. If Δcnfz≡0.We assert thatfz is a periodic function with a period of c. If degP is odd, by (15), we have(43)∑j=0n−1−1jCnjePz+n−jc−Pz−1=0.By Lemma2, for each j0≤j≤n−1, we have −1jCnjePz+n−jc−Pz≡1. Therefore, Pz+n−jc−Pz=0; then, fz=d+ePz=d+ePz+n−jc=fz+n−jc, and hence fz is a periodic function with a period of c. If degP is even, it is easy to prove that f is also periodic.(ii)
Case 2. If Δcnfz≡0.Assuming thatλ is a periodic of fzΔcnfz, then(44)fzΔcnfz=fz+λΔcnfz+λ.Case 2.1. λ+c≠0.Byfz=d+ePz and (15) and (16),(45)fzePz∑j=0n−1−1jCnjePz+n−jc−Pz+−1nePz=fz+λePz+λ∑j=0n−1−1jCnjePz+λ+n−jc−Pz+λ+−1nePz+λ.Furthermore,(46)d+ePzePz∑j=0n−1−1jCnjePz+n−jc−Pz+−1nePz=d+ePz+λePz+λ∑j=0n−1−1jCnjePz+λ+n−jc−Pz+λ+−1nePz+λ.It follows that(47)dePz∑j=0n−1−1jCnjePz+n−jc−Pz+−1n−ePz+λ−Pz∑j=0n−1−1jCnjePz+λ+n−jc−Pz+λ+−1n+e2Pz∑j=0n−1−1jCnjePz+n−jc−Pz+−1n−e2Pz+λ−2Pz∑j=0n−1−1jCnjePz+λ+n−jc−Pz+λ+−1n=0.Consequently,(48)d∑j=0n−1−1jCnjeΔn−jcP+−1n−eΔλP∑j=0n−1−1jCnjeΔλ+n−jcP+−1nePz+∑j=0n−1−1jCnjeΔn−jcP+−1n−e2ΔλP∑j=0n−1−1jCnjeΔλ+n−jcP+−1ne2Pz=0.Let(49)A≔d∑j=0n−1−1jCnjeΔn−jcP+−1n−eΔλP∑j=0n−1−1jCnjeΔλ+n−jcP+−1n,B≔∑j=0n−1−1jCnjeΔn−jcP+−1n−e2ΔλP∑j=0n−1−1jCnjeΔλ+n−jcP+−1n.Hence,A=0,B=0. Otherwise, by Lemmas 1 and 2, degΔn−jcP<degP,degΔλ+n−jcP<degP, so Tr,eP=Tr,A/B=Sr,eP, but it is impossible.Case 2.1.1. degP≥2.By simple computation, we haveΔcPz=mamczm−1+P0z. Furthermore,(50)Δn−jcPz=mamn−jczm−1+P1z,Δλ+n−jcPz=mamλ+n−jczm−1+P2z,P0z,P1z,P2z are polynomials of degree no more than m−2; furthermore, ΔλPz, Δn−jcPz, Δλ+n−jcPz are polynomials with degree of m−1, where j=0,…,n−1; by B=0 and Lemma 2.2, we have −1jCnjeΔn−jcP+−1n≡0, or −1n−e2ΔλP−1n≡0, or −e2ΔλP−1jCnjeΔλ+n−jcP+−1n≡0, but it is impossible.Case 2.1.2. degP=1.ByPz=a1z+a0,ΔcPz=a1c,Δn−jcPz=a1n−jc, Δλ+n−jcPz=a1λ+n−jc, on account of B=0, we have(51)∑j=0n−1−1jCnjea1n−jc+−1n−e2a1λ∑j=0n−1−1jCnjea1λ+n−jc−−1ne2a1λ=0.By Lemma2 and B=0, when n is even, we have −1jCnjea1n−jc+1≡0, or −1jCnjea1λ+n−jc+2a1λ≡1, j=0,…,n−1, or e2a1λ≡1. Therefore, fz is a periodic function with a period of 2λ or n−jcln−1j+1Cnj or 3λ+n−jcln−1jCnj. If d≠0, by A=0, then −1jCnjea1n−jc+1≡0, or ea1λ≡1, or −1jCnjea1λ+n−jc+a1λ≡1, j=0,…,n−1, and hence fz is a periodic function with a period of λ or n−jcln−1j+1Cnj or 2λ+n−jcln−1jCnj. The same proof for n is odd. Therefore, fz is a periodic function.Case 2.2. λ+c=0.This proof follows by the same method as in Case 2.1; just replaceλ by −c, so we omit it here.This finishes the proof.



---

*Source: 9995695-2021-05-06.xml*