On the Periodicity of Entire Functions

Abstract

The purpose of this paper is mainly to prove that if f is a transcendental entire function of hyper-order strictly less than 1 and $f(z)^{n}+a_{1}f'(z)+\cdots +a_{k}f^{(k)}(z)$ is a periodic function, then f(z) is also a periodic function, where n,  k are positive integers, and $a_{1},\cdots ,a_{k}$ are constants. Meanwhile, we offer a partial answer to Yang’s Conjecture, theses results extend some previous related theorems.

1 Introduction and Main Results

Herein let f denote a non-constant meromorphic function and we assume that the reader is familiar with the fundamental results of Nevanlinna theory and its standard notation such as $m(r,f),\, N(r, f ),\, T(r,f),$ etc (see e.g., [4] and [11]). In the sequel, S(r, f) will be used to denote a quantity that satisfies $S(r,f)=o\bigl (T(r,f)\bigr )$ as $r\rightarrow \infty$, outside possibly an exceptional set of r values of finite linear measure, and a meromorphic function a is said to be a small function of f if $T(r,a)=S(r,f).$ We use $\rho (f)$ and $\rho _{2}(f)$ to denote the order and hyper-order of f respectively.

The convergence exponent of zeros of f is defined as

$$\begin{aligned} \tau (f)=\limsup _{r\rightarrow \infty }\frac{\log N(r,\frac{1}{f})}{\log r}=\limsup _{r\rightarrow \infty }\frac{\log n(r,\frac{1}{f})}{\log r}. \end{aligned}$$

In addition, a complex number a is said to be a Borel exceptional value of f if

$$\begin{aligned} \limsup _{r\rightarrow \infty }\frac{\log ^{+}n\Big (r,\frac{1}{f-a}\Big )}{\log r}<\rho (f). \end{aligned}$$

In this note, we mainly consider the periodicity of entire functions, namely, if $f(z)^{n}+a_{1}f'(z)+\cdots +a_{k}f^{(k)}(z)$ is a periodic function, then f(z) is also a periodic function.

The motivation of this paper arises from the study of the real transcendental entire solutions of the differential equation

$$\begin{aligned} f(z)f^{(k)}(z)=p(z)\sin ^{2}z, \end{aligned}$$

where p(z) is a non-zero polynomial. It seems to us that Titchmarsh [9] firstly proved that the differential equation $f(z)f''(z)=-\sin ^{2}z$ has no real entire solutions of finite order other than $f(z)=\pm \sin z.$ The follow-up works were due to Li, Lü and Yang in [8], where they considered the similar problem when f(z) is real and of finite order. They obtained $f(z)f''(z)=-\sin ^{2}z$ has entire solutions $f(z)=\pm \sin z$ and no other solutions. Recently, Yang proposed the following interesting conjecture, see e.g., [8] and [10].

Yang’s Conjecture. Let f be a transcendental entire function and $k\,(\ge 1)$ be an integer. If $f(z)f^{(k)}(z)$ is a periodic function, then f(z) is also a periodic function.

From then on, a number of papers have focused on Yang’s Conjecture, see e.g., [6, 7] and references therein.

Recently, regarding Yang’s Conjecture, Liu et al. [5] obtained the following result.

Theorem A

Let f be a transcendental entire function and n, k be positive integers. If $f(z)^{n}f^{(k)}(z)$ is a periodic function and one of the following conditions is satisfied

• (i) $k=1;$

• (ii) $f(z)=\mathrm{e}^{h(z)}$, where h is a non-constant polynomial;

• (iii) f has a non-zero Picard exceptional value and f is of finite order,

then f(z) is also a periodic function.

A natural question would arise: what will happen if we drop the condition “ finite order ” in Theorem A. In this note, by considering a different proofs, we obtain the following result, which offers a partial answer to Yang’s Conjecture, and improves Theorem A and references therein.

Theorem 1.1

Let f be a transcendental entire function of hyper-order strictly less than 1,  and $n,\,k$ be positive integers. Suppose that f(z) has a finite Borel exceptional value b,  and $f(z)^{n}f^{(k)}(z)$ is a periodic function, then f(z) is also a periodic function.

Remark 1.1

If b is a Picard exceptional of f,  then b is a Borel exceptional of f.

In addition, Liu et al. [5] also obtained the following result.

Theorem B

Let f be a transcendental entire function and $n\ge 2,\, k\ge 1$ be integers. If $f(z)^n +f^{(k)}(z)$ is a periodic function with period c and one of the following conditions is satisfied

• (i) $k=1;$

• (ii) $f(z + c)-f(z)$ has no zeros;

• (iii) the zeros multiplicity of $f(z +c)-f(z)$ is great than or equal to k;  then f(z) is also a periodic function with period c or 2c.

In this paper, we will prove the following result.

Theorem 1.2

Let f be a transcendental entire function of hyper-order strictly less than 1, and $n\,(\ge 2),$ $k\,(\ge 1)$ be integers. If $f(z)^{n}+a_{1}f'(z)+\cdots +a_{k}f^{(k)}(z)$ is a periodic function, where $a_{1},\cdots ,a_{k}$ are constants, then f(z) is also a periodic function.

Remark 1.2

• (i) The condition “ $n\ge 2$ ” in Theorem 1.2 is necessary. For example, let $f(z)=z\mathrm{e}^{-z}.$ Then

  $$\begin{aligned} f(z)+f'(z)+f''(z)+f'''(z)=2\mathrm{e}^{-z} \end{aligned}$$

  is a periodic function, however $f(z)=z\mathrm{e}^{-z}$ is not a periodic function.

• (ii) Carefully checking the proof of Theorem 1.2, we may find when $n=2$ or $n\ge 4,$ the hypothesis “ $\rho _{2}(f)<1$ ” can be removed from Theorem 1.2.

2 Lemmas

In order to prove our results, we need the following lemmas.

Lemma 2.1

(see, e.g., [3]). Let f be a non-constant meromorphic function with $\rho _2(f)<1,$ $c\in \mathbb {C}.$ Then

$$\begin{aligned} m\Big (r,\frac{f(z+c)}{ f (z)}\Big ) = S(r, f ), \end{aligned}$$

outside of a possible exceptional set with finite logarithmic measure.

It is pointed out that if f is of finite order, we have

Lemma 2.1’

(see, e.g., [2]). Let f be a meromorphic function with $\rho =\rho (f)<+\infty ,$ $c \,(\ne 0)\in \mathbb {C}.$ Then for each $\varepsilon >0,$ we have

$$\begin{aligned} m\Big (r,\frac{f(z+c)}{ f (z)}\Big ) =O(r^{\rho -1+\varepsilon }) . \end{aligned}$$

By applying Lemma 2.1 and the Logarithmic Derivative Lemma, we have the following result.

Lemma 2.2

Let f be a non-constant meromorphic function with $\rho _{2}(f)<1.$ Then for $c\in \mathbb {C}$ and any positive integer k,  we have

$$\begin{aligned} m\Big (r,\frac{f^{(k)}(z+c)}{f(z)}\Big )=S(r,f), \end{aligned}$$

outside of a possible exceptional set with finite logarithmic measure.

Lemma 2.3

([11], Lemma 5.1). Let f denote a non-constant periodic function. Then $\rho (f)\ge 1.$

Lemma 2.4

([1] ). Let g be a function transcendental and meromorphic in the plane of order less than 1, and $h>0.$ Then there exists an $\varepsilon$-set E such that

$$\begin{aligned} \frac{g'(z+c)}{g(z+c)}\rightarrow 0, \, \frac{g(z+c)}{g(z)}\rightarrow 1 \, \text{ as } \ z\rightarrow \infty \text{ in } \ \mathbb {C} \backslash E, \end{aligned}$$

uniformly in c for $|c|\le h.$ Further, E may be chosen so that for large z not in E the function g has no zeros or poles in $|\zeta -z|\le h.$

Remark 2.1

According to the works of Hayman (see, e.g., [4]), an $\varepsilon$ set E is defined to be a countable union of open discs not containing the origin and subtending angles at the origin whose sum is finite. Suppose that E is an $\varepsilon$ set, then the set of $r\ge 1$ for which the circle S(0, r) meets E has finite logarithmic measure and for almost all real $\theta$ the intersection of E with the ray $\arg z=\theta$ is bounded.

Lemma 2.5

( [11], Theorem 1.62) Suppose that $f_{j}(j=1,2,\cdots ,n)$ $(n\ge 3)$ are meromorphic functions which are not constants except for $f_{n}.$ Furthermore, let

$$\begin{aligned} \sum _{j=1}^{n}f_{j}=1. \end{aligned}$$

If $f_{n}\not \equiv 0$ and

$$\begin{aligned} \sum _{j=1}^{n}N(r,\frac{1}{f_{j}})+(n-1)\sum _{j=1}^{n}\overline{N}(r,f_{j})<(\lambda +o(1))T(r,f_{k}), \end{aligned}$$

where $r\in I$, I is a set whose linear measure is infinite, $k\in \{1,2,\cdots ,n-1\}$ and $\lambda <1,$ then $f_{n}\equiv 1.$

3 Proof of Theorem 1.1

Note that b is a finite Borel exceptional value of f. Next, two cases will be considered.

Case 1. If $b=0,$ by the Hadamard factorization theorem, we get

$$\begin{aligned} f(z)=Q(z)\mathrm{e}^{p(z)}, \end{aligned}$$

where Q is the canonical product of f formed with its zeros, and p is a non-constant entire function satisfying $\rho (p)<1.$ Using the facts (see., e.g. [11], Theorem 2.2 and Theorem 2.3 ), it is easy to deduce that

$$\begin{aligned} \rho (Q)=\tau (Q)=\tau (f)<\rho (f). \end{aligned}$$

Thus, $\rho (f)=\rho (\mathrm{e}^{p}).$ Besides, since $f(z)^{n}f^{(k)}(z)$ is a periodic function with period c,  then

$$\begin{aligned} f(z)^{n}f^{(k)}(z)=f(z+c)^{n}f^{(k)}(z+c). \end{aligned}$$

(3.1)

Substituting $f(z)=Q(z)\mathrm{e}^{p(z)}$ into (3.1), it follows without difficulty that

$$\begin{aligned} {[}Q(z)\mathrm{e}^{p(z)}]^{n}\mathrm{e}^{p(z)}H_{1}(z)=[Q(z+c)\mathrm{e}^{p(z+c)}]^{n}\mathrm{e}^{p(z+c)}H_{1}(z+c), \end{aligned}$$

(3.2)

where $H_{1}$ is a differential polynomial of Q and p,  namely,

$$\begin{aligned} H_{1}(z)=Q^{(k)}(z)+A_{1}Q^{(k-1)}(z)p'(z)+A_{2}Q^{(k-2)}(z)[p''(z)]^{2}+\cdots +Q(z)p^{(k)}(z) \end{aligned}$$

with constants $A_{i}\, (i=1,2,\cdots ).$

Thereby, $\rho (H_{1})\le \max \{\rho (Q),\rho (p)\}<\rho (f).$

Now, we can rewrite (3.2) as

$$\begin{aligned} \mathrm{e}^{(n+1)[p(z)-p(z+c)]}=\frac{H_{1}(z+c)}{H_{1}(z)}\frac{Q(z+c)^{n}}{Q(z)^{n}}. \end{aligned}$$

(3.3)

In addition, (3.3) shows that $\rho (\mathrm{e}^{p(z)-p(z+c)})<+\infty$ since $\rho (H_{1})<+\infty , \rho (Q)<+\infty .$ This implies $p(z)-p(z+c)$ is a polynomial, say $p(z)-p(z+c)=q_{0}z^{m}+\cdots +q_{m},$ where m is a natural number and $q_0,\,\cdots , q_m$ are constants.

If $m>1,$ then $p^{(m+1)}(z)-p^{(m+1)}(z+c)\equiv 0,$ which implies $p^{(m+1)}$ is a periodic function. Therefore, Lemma 2.3 and $\rho (p^{(m+1)})=\rho (p)<1$ show that $p^{(m+1)}(z)$ is a constant, this leads to p is a polynomial, say $p(z)=a_{0}z^{m+1}+\cdots +a_{m+1}.$ In this case, it is easy to see $\rho (f)= m+1,$ and $\rho (p)=0.$

Set $\rho (Q)=\sigma .$ Then $\rho (H_1)\le \sigma ,$ and $\sigma <m+1.$

Again, applying Lemma 2.1’ to (3.3), we obtain

$$\begin{aligned} m(r,\mathrm{e}^{(n+1)[p(z)-p(z+c)]})=m\Big (r,\frac{H_{1}(z+c)}{H_{1}(z)}\frac{Q(z+c)^{n}}{Q(z)^{n}}\Big ), \end{aligned}$$

which implies $r^{m}\le O(r^{\sigma -1+\varepsilon }).$ This is impossible since we can choose $\varepsilon >0$ small enough such that $\sigma -1+\varepsilon <m.$

Thus, $p(z)=a_0 z+a_1,$ where $a_0,\,a_1$ are constants. Furthermore, if we set $\mathrm{e}^{(n+1)a_0 c}=A,$ then

$$\begin{aligned} A=\frac{H_{1}(z+c)}{H_{1}(z)}\frac{Q(z+c)^{n}}{Q(z)^{n}}. \end{aligned}$$

On the other hand, by Lemma 2.4, there exists a $\varepsilon -$set E such that

$$\begin{aligned} \frac{H_{1}(z+c)}{H_{1}(z)}\rightarrow 1, \ \frac{Q(z+c)}{Q(z)}\rightarrow 1, \, \text{ as } \ z\rightarrow \infty \text{ in } \ \mathbb {C} \backslash E. \end{aligned}$$

Trivially, $A=1,$ and

$$\begin{aligned} \frac{H_{1}(z+c)}{H_{1}(z)}\frac{Q(z+c)^{n}}{Q(z)^{n}}=1. \end{aligned}$$

It means that $H_{1}(z)Q(z)^{n}$ is a periodic function. Hence $\rho (H_{1}(z)Q(z)^{n})\ge 1$ if $H_{1}(z)Q(z)^{n}$ is not a constant. It follows by $\rho (H_{1}(z)Q(z)^{n})<1$ that $H_{1}(z)Q(z)^{n}$ is a constant. Therefore, Q must be a constant. Thus, we conclude that f(z) must be a periodic function with period $\frac{2\pi \mathrm{i}}{a_0}.$

Case 2. If $b\ne 0,$ then by the Hadamard factorization theorem, we get

$$\begin{aligned} f(z)=Q(z)\mathrm{e}^{p(z)}+b, \end{aligned}$$

where Q is the canonical product of $f-b$ formed with its zeros, and p is a non-constant entire function satisfying $\rho (p)<1.$ Using the same methods as the proof in Case 1, $\rho (Q)=\tau (Q)=\tau (f-b)<\rho (f-b)=\rho (f)$ follows. Thus, $\rho (f)=\rho (\mathrm{e}^{p}).$

Since $f(z)^{n}f^{(k)}(z)$ is a periodic function with period c,  then

$$\begin{aligned} f(z)^{n}f^{(k)}(z)=f(z+c)^{n}f^{(k)}(z+c). \end{aligned}$$

(3.4)

Substituting $f(z)=Q(z)\mathrm{e}^{p(z)}+b$ into (3.4), we have

$$\begin{aligned} {[}Q(z)\mathrm{e}^{p(z)}+b]^{n}\mathrm{e}^{p(z)}H_{1}(z)=[Q(z+c)\mathrm{e}^{p(z+c)}+b]^{n}\mathrm{e}^{p(z+c)}H_{1}(z+c), \end{aligned}$$

where $H_{1}$ is a differential polynomial of Q and p,  namely,

$$\begin{aligned} H_{1}(z)=Q^{(k)}(z)+A_{1}Q^{(k-1)}(z)p'(z)+A_{2}Q^{(k-2)}(z)[p''(z)]^{2}+\cdots +Q(z)p^{(k)}(z) \end{aligned}$$

with constants $A_{i} (i=1,2,\cdots ).$ In this case, we conclude

$$\begin{aligned} \rho (H_{1})\le \max \{\rho (Q),\rho (p)\}<\rho (f). \end{aligned}$$

Besides, we find

$$\begin{aligned}&Q(z)^{n}\mathrm{e}^{(n+1)p(z)}+C_{n}^{1}bQ(z)^{n-1}\mathrm{e}^{np(z)}+\cdots +C_{n}^{n-1}b^{n-1}Q(z)\mathrm{e}^{2p(z)}+b^{n}\mathrm{e}^{p(z)}= \\&\quad H(z)[Q(z+c)^{n}\mathrm{e}^{(n+1)p(z+c)}+C_{n}^{1}bQ(z+c)^{n-1}\mathrm{e}^{np(z+c)}\\&\quad +\cdots +C_{n}^{n-1}b^{n-1}Q(z+c)\mathrm{e}^{2p(z+c)}+b^{n}\mathrm{e}^{p(z+c)}], \end{aligned}$$

where $H(z)=\frac{H_{1}(z+c)}{H_{1}(z)},$ and $\rho (H)<\rho (f).$

Dividing both sides of the above equation by $b^{n}\mathrm{e}^{p(z)}$ gives

$$\begin{aligned}&\frac{H(z)}{b^{n}}\mathrm{e}^{(n+1)p(z+c)-p(z)}Q(z+c)^{n}+\cdots +H(z)\mathrm{e}^{p(z+c)-p(z)}\nonumber \\&\quad -\frac{Q(z)^{n}}{b^{n}}\mathrm{e}^{np(z)}-\cdots -C_n^{n-1}\frac{Q(z)}{b}\mathrm{e}^{p(z)}=1. \end{aligned}$$

(3.5)

Next, we will prove $mp(z+c)-p(z)\,(m=2,\cdots ,n+1)$ are not constants. In fact, if p is a non-constant polynomial, it is obvious. Now, we assume that p is a transcendental entire function. In this case, if $mp(z+c)-p(z)=q,$ here q is a constant, then $mp'(z+c)=p'(z).$ Noting $\rho (p)=\rho (p')<1,$ we apply Lemma 2.4 to $p'$ and obtain $m=1,$ a contradiction. Thereby, $mp(z+c)-p(z)(m=2,\cdots ,n+1)$ can not be constants. To complete the proof, we now employ Lemma 2.5 to (3.5) and have $H(z)\mathrm{e}^{p(z+c)-p(z)}\equiv 1.$ It means that $H_{1}(z+c)\mathrm{e}^{p(z+c)}=H_{1}(z)\mathrm{e}^{p(z)},$ and

$$\begin{aligned} {[}b+Q(z)\mathrm{e}^{p(z)}]^{n}=[b+Q(z+c)\mathrm{e}^{p(z+c)}]^{n} \end{aligned}$$

follows. Thus $f(z)^{n}=f(z+c)^{n}$ shows that f is a periodic function with period c or nc.

This completes the proof of Theorem 1.1.

4 Proof of Theorem 1.2

By assumption, $f(z)^{n}+a_{1}f'(z)+\cdots +a_{k}f^{(k)}(z)$ is a periodic function with period c,  then

$$\begin{aligned} f(z+c)^{n}+a_{1}f'(z+c)+\cdots +a_{k}f^{(k)}(z+c)=f(z)^{n}+a_{1}f'(z)+\cdots +a_{k}f^{(k)}(z), \end{aligned}$$

and thus

$$\begin{aligned} f(z+c)^{n}-f(z)^{n}=a_{1}[f'(z)-f'(z+c)]+\cdots +a_{k}[f^{(k)}(z)-f^{(k)}(z+c)].\nonumber \\ \end{aligned}$$

(4.1)

Next, we consider three cases.

Case 1. $n=2.$ In this case, we can rewrite (4.1) as

$$\begin{aligned} \begin{aligned}&[f(z+c)+f(z)][f(z+c)-f(z)]\\&\quad =a_{1}[f'(z)-f'(z+c)]+\cdots +a_{k}[f^{(k)}(z)-f^{(k)}(z+c)]. \end{aligned} \end{aligned}$$

(4.2)

If $f(z+c)-f(z)\equiv 0,$ then f(z) is a periodic function with period c.

Next, we may assume that $f(z+c)-f(z)\not \equiv 0.$ In this case, (4.2) can be rewritten as

$$\begin{aligned} \begin{aligned} f(z+c)+f(z)&= -a_{1}\frac{f'(z)-f'(z+c)}{f(z)-f(z+c)}-\cdots -a_{k}\frac{f^{(k)}(z)-f^{(k)}(z+c)}{f(z)-f(z+c)}\\&= -a_{1}\frac{g'(z)}{g(z)}-\cdots -a_{k}\frac{g^{(k)}(z)}{g(z)}, \end{aligned} \end{aligned}$$

(4.3)

where

$$\begin{aligned} g(z)=f(z)-f(z+c). \end{aligned}$$

(4.4)

Define $p_{i}(z)=\frac{g^{(i)}(z)}{g(z)}$ $(i=1,2,\cdots ,k),$ and

$$\begin{aligned} H(z)=-a_{1}p_{1}(z)-\cdots -a_{k}p_{k}(z). \end{aligned}$$

(4.5)

Then (4.3) becomes

$$\begin{aligned} f(z+c)+f(z)=H(z). \end{aligned}$$

(4.6)

Besides, applying the Logarithmic Derivative Lemma to (4.3), we have

$$\begin{aligned} T(r,H)=m(r,H)\le m(r,p_{1})+\cdots +m(r,p_{k})+O(1)\le S(r,g). \end{aligned}$$

(4.7)

Combining (4.4) and (4.6) yields that

$$\begin{aligned} f(z)=\frac{1}{2}[H(z)+g(z)], f(z+c)=\frac{1}{2}[H(z)-g(z)]. \end{aligned}$$

Thus, a routine computation leads to

$$\begin{aligned} g(z)+g(z+c)=H(z)-H(z+c). \end{aligned}$$

(4.8)

Moreover, (4.8) results in

$$\begin{aligned} g^{(i)}(z)+g^{(i)}(z+c)=H^{(i)}(z)-H^{(i)}(z+c). \end{aligned}$$

(4.9)

Thereby, it follows by $g^{(i)}(z)=p_{i}(z)g(z)$ and (4.9) that

$$\begin{aligned} \left\{ \begin{aligned}&p_{1}(z)g(z)+p_{1}(z+c)g(z+c)= H'(z)-H'(z+c),\\&p_{2}(z)g(z)+p_{2}(z+c)g(z+c)= H''(z)-H''(z+c),\\&\cdots \cdots \\&p_{k}(z)g(z)+p_{k}(z+c)g(z+c)= H^{(k)}(z)-H^{(k)}(z+c). \end{aligned} \right. \end{aligned}$$

(4.10)

Now, combining (4.5) and (4.10) yields

$$\begin{aligned} H(z)g(z)+H(z+c)g(z+c)= & {} -a_{1}[H'(z)-H'(z+c)]\nonumber \\&-\cdots -a_{k}[H^{(k)}(z)-H^{(k)}(z+c)].\nonumber \\ \end{aligned}$$

(4.11)

Furthermore, substituting $g(z+c)=H(z)-H(z+c)-g(z)$ in (4.11), we find

$$\begin{aligned} \begin{aligned}&H(z)g(z)+H(z+c)[H(z)-H(z+c)-g(z)]\\&\quad =-a_{1}[H'(z)-H'(z+c)]-\cdots -a_{k}[H^{(k)}(z)-H^{(k)}(z+c)]. \end{aligned} \end{aligned}$$

If $H(z+c)\not \equiv H(z),$ we obtain

$$\begin{aligned} g(z)=\frac{-a_{1}[H'(z)-H'(z+c)]-\cdots -a_{k}[H^{(k)}(z)-H^{(k)}(z+c)]+H(z+c)^{2}-H(z+c)H(z)}{H(z)-H(z+c)}.\nonumber \\ \end{aligned}$$

(4.12)

It follows from (4.7), (4.12) and Lemma 2.2 that

$$\begin{aligned} T(r,g)\le S(r,g), \end{aligned}$$

which is impossible. Hence, $H(z+c)=H(z),$ and (4.8) gives $g(z)+g(z+c)=0,$ this implies that f is a periodic function with period 2c.

Case 2. $n=3.$ Now, we can rewrite (4.1) as

$$\begin{aligned} \begin{aligned}&{[}f(z+c)-f(z)][f(z+c)-\eta f(z)][f(z+c)-\eta ^{2}f(z)]\\&\quad =a_{1}[f'(z)-f'(z+c)]+\cdots +a_{k}[f^{(k)}(z)-f^{(k)}(z+c)], \end{aligned} \end{aligned}$$

(4.13)

where $\eta (\ne 1)$ is a cube-root of the unity.

If $f(z+c)-f(z)\equiv 0,$ then f is a periodic function with period c.

If $f(z+c)-f(z)\not \equiv 0,$ (4.13) can be rewritten as

$$\begin{aligned} \begin{aligned}&[f(z+c)-\eta f(z)][f(z+c)-\eta ^{2}f(z)]\\&\quad = -a_{1}\frac{f'(z)-f'(z+c)}{f(z)-f(z+c)}-\cdots -a_{k}\frac{f^{(k)}(z)-f^{(k)}(z+c)}{f(z)-f(z+c)}\\&\quad = -a_{1}\frac{g'(z)}{g(z)}-\cdots -a_{k}\frac{g^{(k)}(z)}{g(z)}, \end{aligned} \end{aligned}$$

(4.14)

where

$$\begin{aligned} g(z)=f(z)-f(z+c). \end{aligned}$$

(4.15)

Define $p_{i}(z)=\frac{g^{(i)}(z)}{g(z)},\,i=1,2,\cdots ,k,$ and $H(z)=-a_{1}p_{1}(z)-\cdots -a_{k}p_{k}(z).$ Then (4.14) becomes

$$\begin{aligned} {[}f(z+c)-\eta f(z)][f(z+c)-\eta ^{2}f(z)]=H(z). \end{aligned}$$

(4.16)

Besides, the Logarithmic Derivative Lemma gives

$$\begin{aligned} T(r,H)= & {} m(r,H)\le m(r,p_{1})+\cdots +m(r,p_{k})+O(1)\\= & {} O\Big (\log (rT(r,g))\Big )\,(r\rightarrow \infty ,r\not \in E_{0}), \end{aligned}$$

where $E_{0}$ is a set whose linear measure is not greater than 2.

Note that $\rho _2(f)<1,$ $T(r,f(z+c))=T(r,f(z))+S(r,f)$ ( see, e.g., [2] and [3] ). By making use of (4.15), it is easy to see that $T(r,g)\le O(T(r,f)).$ It clearly follows by $\rho _{2}(f)<1$ that $\log T(r,f)\le r^{\lambda },$ where $\lambda \,(<1)$ is a positive number. Hence, $T(r,H)\le O(\log rT(r,g))\le O(r^{\lambda }),$ which implies that $\rho (H)<1.$ In addition, by the Hadamard factorization theorem, (4.16) can be changed as

$$\begin{aligned} f(z+c)-\eta f(z)=\Pi _{1}(z)\mathrm{e}^{\alpha (z)} \end{aligned}$$

(4.17)

and

$$\begin{aligned} f(z+c)-\eta ^{2}f(z)=\Pi _{2}(z)\mathrm{e}^{-\alpha (z)}, \end{aligned}$$

(4.18)

where $\alpha$ is a non-constant entire function satisfying $\rho (\alpha )<1,$ $\Pi _{1}(z)$ is the canonical product of $f(z+c)-\eta f(z)$ formed with its zeros, $\Pi _{2}(z)$ is the canonical product of $f(z+c)-\eta ^{2}f(z)$ formed with its zeros, and $\Pi _{1}(z)$, $\Pi _{2}(z)$ satisfy

$$\begin{aligned} \Pi _{1}(z)\Pi _{2}(z)=H(z). \end{aligned}$$

(4.19)

Using Theorem 2.2 and Theorem 2.3 in [11] , it is easy to deduce that

$$\begin{aligned} \rho (\Pi _{1})=\tau (\Pi _{1})\le \tau (H)\le \rho (H)<1. \end{aligned}$$

By applying the same analysis, we can easily conclude the following result

$$\begin{aligned} \rho (\Pi _{2})<1. \end{aligned}$$

Combining (4.17) and (4.18) yields

$$\begin{aligned} f(z)= & {} \frac{\Pi _{1}(z)\mathrm{e}^{\alpha (z)}-\Pi _{2}(z)\mathrm{e}^{-\alpha (z)}}{\eta (\eta -1)}, \end{aligned}$$

(4.20)

$$\begin{aligned} f(z+c)= & {} \frac{\eta \Pi _{1}(z)\mathrm{e}^{\alpha (z)}-\Pi _{2}(z)\mathrm{e}^{-\alpha (z)}}{\eta -1}. \end{aligned}$$

(4.21)

Thus, a routine computation leads to

$$\begin{aligned} \eta ^{2}\Pi _{1}(z)\mathrm{e}^{\alpha (z)}-\eta \Pi _{2}(z)\mathrm{e}^{-\alpha (z)}=\Pi _{1}(z+c)\mathrm{e}^{\alpha (z+c)}-\Pi _{2}(z+c)\mathrm{e}^{-\alpha (z+c)}.\nonumber \\ \end{aligned}$$

(4.22)

Now, dividing (4.22) by $\eta ^{2}\Pi _{1}(z)\mathrm{e}^{\alpha (z)},$ we obtain

$$\begin{aligned} \frac{1}{\eta }\frac{\Pi _{2}(z)}{\Pi _{1}(z)}\mathrm{e}^{-2\alpha (z)}+\frac{1}{\eta ^{2}}\frac{\Pi _{1}(z+c)}{\Pi _{1}(z)}\mathrm{e}^{\alpha (z+c)-\alpha (z)} -\frac{1}{\eta ^{2}}\frac{\Pi _{2}(z+c)}{\Pi _{1}(z)}\mathrm{e}^{-\alpha (z+c)-\alpha (z)}=1.\nonumber \\ \end{aligned}$$

(4.23)

Since $\alpha$ is a non-constant entire function with $\rho (\alpha )<1,$ then $-\alpha (z+c)-\alpha (z)$ is not a constant. Otherwise, if $-\alpha (z+c)-\alpha (z)$ is a constant, then $\alpha '(z)$ is a periodic function, and $\rho (\alpha ')=\rho (\alpha )\ge 1,$ a contradiction. Now, applying Lemma 2.5 to (4.23) yields

$$\begin{aligned} \frac{1}{\eta ^{2}}\frac{\Pi _{1}(z+c)}{\Pi _{1}(z)}\mathrm{e}^{\alpha (z+c)-\alpha (z)}\equiv 1. \end{aligned}$$

(4.24)

On the other hand, dividing (4.22) by $\eta \Pi _{2}(z)\mathrm{e}^{-\alpha (z)}$ implies

$$\begin{aligned} \eta \frac{\Pi _{1}(z)}{\Pi _{2}(z)}\mathrm{e}^{2\alpha (z)}-\frac{1}{\eta }\frac{\Pi _{1}(z+c)}{\Pi _{2}(z)}\mathrm{e}^{\alpha (z+c)+\alpha (z)} +\frac{1}{\eta }\frac{\Pi _{2}(z+c)}{\Pi _{2}(z)}\mathrm{e}^{-\alpha (z+c)+\alpha (z)}=1.\nonumber \\ \end{aligned}$$

(4.25)

Obviously, $2\alpha (z)$ and $\alpha (z+c)+\alpha (z)$ are not constants. Armed with Lemma 2.5 and (4.25), we deduce

$$\begin{aligned} \frac{1}{\eta }\frac{\Pi _{2}(z+c)}{\Pi _{2}(z)}\mathrm{e}^{-\alpha (z+c)+\alpha (z)}\equiv 1. \end{aligned}$$

(4.26)

Combining (4.24) and (4.26) yields

$$\begin{aligned} \Pi _{1}(z+c)\Pi _{2}(z+c)=\Pi _{1}(z)\Pi _{2}(z). \end{aligned}$$

This suggests that $H(z+c)=H(z).$ We conclude from $\rho (H)<1$ that H must be a constant. Thus, $f(z+c)-\eta f(z),f(z+c)-\eta ^{2} f(z)$ have no zeros, which shows that $\Pi _{1}(z)$, $\Pi _{2}(z)$ are constants. It follows from (4.24) and (4.26) that

$$\begin{aligned} \mathrm{e}^{\alpha (z+c)-\alpha (z)}=\eta ^{2} \ \text{ and } \ \mathrm{e}^{-\alpha (z+c)+\alpha (z)}=\eta , \end{aligned}$$

this results in

$$\begin{aligned} \alpha (z+c)-\alpha (z)\equiv C, \end{aligned}$$

(4.27)

where C is a constant. Differentiating (4.27) yields $\alpha '(z+c)-\alpha '(z)\equiv 0$, namely $\alpha '(z)$ is a periodic function. Noting $\rho (\alpha ')=\rho (\alpha )<1,$ we know $\alpha '(z)$ must be a constant, say, A. Thus, $\alpha (z)=Az+B$ with a constant B. By (4.20), we see that f is a periodic function with period $\frac{2\pi \mathrm{i}}{A}.$

Case 3. $n\ge 4.$ To complete the proof, we rewrite (4.1) as

$$\begin{aligned} \begin{aligned}&{[}f(z+c)^{n-1}+f(z+c)^{n-2}f(z)+\cdots +f(z)^{n-1}][f(z+c)-f(z)]\\&\quad = a_{1}[f'(z)-f'(z+c)]+\cdots +a_{k}[f^{(k)}(z)-f^{(k)}(z+c)].\\ \end{aligned} \end{aligned}$$

(4.28)

If $f(z+c)-f(z)\equiv 0,$ then f is a periodic function with period c.

If $f(z+c)-f(z)\not \equiv 0,$ we change (4.28) into

$$\begin{aligned} \begin{aligned}&f(z+c)^{n-1}+f(z+c)^{n-2}f(z)+\cdots +f(z)^{n-1}\\&\quad = -a_{1}\frac{f'(z)-f'(z+c)}{f(z)-f(z+c)}-\cdots -a_{k}\frac{f^{(k)}(z)-f^{(k)}(z+c)}{f(z)-f(z+c)}\\&\quad = -a_{1}\frac{g'(z)}{g(z)}-\cdots -a_{k}\frac{g^{(k)}(z)}{g(z)}, \end{aligned} \end{aligned}$$

(4.29)

where

$$\begin{aligned} g(z)=f(z)-f(z+c). \end{aligned}$$

(4.30)

Define $p_{i}(z)=\frac{g^{(i)}(z)}{g(z)},\,i=1,2,\cdots ,k,$ and $H(z)=-a_{1}p_{1}(z)-\cdots -a_{k}p_{k}(z).$ Then (4.29) becomes

$$\begin{aligned} f(z+c)^{n-1}+f(z+c)^{n-2}f(z)+\cdots +f(z)^{n-1}=H(z). \end{aligned}$$

(4.31)

Now, using the Logarithmic Derivative Lemma, we have

$$\begin{aligned} T(r,H)=m(r,H)\le m(r,p_{1})+\cdots +m(r,p_{k})+O(1)\le S(r,g). \end{aligned}$$

By (4.30), we conclude

$$\begin{aligned} f(z)\Big (1-\frac{f(z+c)}{f(z)}\Big )=g(z). \end{aligned}$$

(4.32)

Moreover, (4.31) gives

$$\begin{aligned} f(z)^{n-1}\left( \frac{f(z+c)^{n-1}}{f(z)^{n-1}}+\frac{f(z+c)^{n-2}}{f(z)^{n-2}}+\cdots +\frac{f(z+c)}{f(z)}+1\right) =H(z).\nonumber \\ \end{aligned}$$

(4.33)

Set $\omega (z)=\frac{f(z+c)}{f(z)}.$ Obviously, $\omega \not \equiv 1.$ Combining (4.32) and (4.33) yields

$$\begin{aligned} \frac{(1-\omega (z))^{n-1}}{\omega (z)^{n-1}+\omega (z)^{n-2}+\cdots +\omega (z)+1}=\frac{g(z)^{n-1}}{H(z)}, \end{aligned}$$

and so

$$\begin{aligned} (n-1)T(r,\omega )=(n-1)T(r,g)+S(r,g). \end{aligned}$$

Therefore, equation (4.33) implies

$$\begin{aligned} \omega (z)^{n-1}+\omega (z)^{n-2}+\cdots +\omega (z)+1=H(z)\frac{1}{f(z)^{n-1}}, \end{aligned}$$

and, consequently

$$\begin{aligned} N\Big (r,\frac{1}{\omega ^{n-1}+\omega ^{n-2}+\cdots +\omega +1}\Big )=N\Big (r,\frac{1}{H}\Big )\le T(r,H)=S(r,g)=S(r,\omega ). \end{aligned}$$

Now, applying the second main theorem gives

$$\begin{aligned} \begin{aligned} (n-2)T(r,\omega )&\le N\Big (r,\frac{1}{\omega -1}\Big )+N\Big (r,\frac{1}{\omega ^{n-1}+\omega ^{n-2}+\cdots +\omega +1}\Big )+S(r,\omega )\\&\le N\Big (r,\frac{1}{\omega -1}\Big )+S(r,\omega )\\&\le T\Big (r,\frac{1}{\omega -1}\Big )+S(r,\omega ). \end{aligned} \end{aligned}$$

Thus, $\omega \,(\not \equiv 1 )$ must be a constant. It follows by (4.32) that $T(r,f)=T(r,g)+S(r,f).$ A contradiction follows by (4.33) and $(n-1)T(r,f)=T(r,H)+S(r,f)=S(r,g)=S(r,f)$ since $n\ge 4.$

This finishes the proof of Theorem 1.2.