Research  Open  Published:
A note on entire functions and their differences
Journal of Inequalities and Applicationsvolume 2013, Article number: 587 (2013)
Abstract
In this paper, we prove that for a transcendental entire function $f(z)$ of finite order such that $\lambda (fa(z))<\sigma (f)$, where $a(z)$ is an entire function and satisfies $\sigma (a(z))<1$, n is a positive integer and if ${\mathrm{\Delta}}_{\eta}^{n}f(z)$ and $f(z)$ share the function $a(z)$ CM, where η ($\in \mathbb{C}$) satisfies ${\mathrm{\Delta}}_{\eta}^{n}f(z)\not\equiv 0$, then
where c, ${c}_{1}$ are two nonzero constants.
MSC:39A10, 30D35.
1 Introduction and results
In this paper, we assume that the reader is familiar with the fundamental results and the standard notations of Nevanlinna’s value distribution theory of meromorphic functions (see [1–3]). In addition, we use the notation $\lambda (f)$ for the exponent of convergence of the sequence of zeros of a meromorphic function f, and $\sigma (f)$ to denote the order growth of f. For a nonzero constant η, the forward differences ${\mathrm{\Delta}}_{\eta}^{n}f(z)$ are defined (see [4, 5]) by
Throughout this paper, we denote by $S(r,f)$ any function satisfying $S(r,f)=o(T(r,f))$ as $r\to \mathrm{\infty}$, possibly outside a set of r of finite logarithmic measure. A meromorphic function $\alpha (z)$ is said to be a small function of $f(z)$ if $T(r,\alpha (z))=S(r,f)$, and we denote by $S(f)$ the set of functions which are small compared to $f(z)$.
Let f and g be two nonconstant meromorphic functions, and let $a\in \mathbb{C}$. We say that f and g share the value a CM (IM) provided that $fa$ and $ga$ have the same zeros counting multiplicities (ignoring multiplicities), that f and g share the value ∞ CM (IM) provided that f and g have the same poles counting multiplicities (ignoring multiplicities). Using the same method, we can define that f and g share the function $a(z)$ CM (IM), where $a(z)\in S(f)\cap S(g)$. Nevanlinna’s four values theorem [6] says that if two nonconstant meromorphic functions f and g share four values CM, then $f\equiv g$ or f is a Möbius transformation of g. The condition ‘f and g share four values CM’ has been weakened to ‘f and g share two values CM and two values IM’ by Gundersen [7, 8], as well as by Mues [9]. But whether the condition can be weakened to ‘f and g share three values IM and another value CM’ is still an open question.
In the special case, we recall a wellknown conjecture by Brück [10].
Conjecture Let f be a nonconstant entire function such that hyper order ${\sigma}_{2}(f)<\mathrm{\infty}$ and ${\sigma}_{2}(f)$ is not a positive integer. If f and ${f}^{\prime}$ share the finite value a CM, then
where c is a nonzero constant.
The notation ${\sigma}_{2}(f)$ denotes hyperorder (see [11]) of $f(z)$ which is defined by
The conjecture has been verified in the special cases when $a=0$ [10], or when f is of finite order [12], or when ${\sigma}_{2}(f)<\frac{1}{2}$ [13].
Recently, many authors [14–17] started to consider sharing values of meromorphic functions with their shifts. Heittokangas et al. proved the following theorems.
Theorem A (See [15])
Let f be a meromorphic function with $\sigma (f)<2$, and let $c\in \mathbb{C}$. If $f(z)$ and $f(z+c)$ share the values a ($\in \mathbb{C}$) and ∞ CM, then
for some constant τ.
In [15], Heittokangas et al. give the example $f(z)={e}^{{z}^{2}}+1$ which shows that $\sigma (f)<2$ cannot be relaxed to $\sigma (f)\le 2$.
Theorem B (See [16])
Let f be a meromorphic function of finite order, let $c\in \mathbb{C}$. If $f(z)$ and $f(z+c)$ share three distinct periodic functions ${a}_{1},{a}_{2},{a}_{3}\in \stackrel{\u02c6}{S}(f)$ with period c CM (where $\stackrel{\u02c6}{S}(f)=S(f)\cup \{\mathrm{\infty}\}$), then $f(z)=f(z+c)$ for all $z\in \mathbb{C}$.
Recently, many results of complex difference equations have been rapidly obtained (see [18–25]). In the present paper, we utilize a complex difference equation to consider uniqueness problems.
The main purpose of this paper is to utilize a complex difference equation to study problems concerning sharing values of meromorphic functions and their differences. It is well known that ${\mathrm{\Delta}}_{\eta}f(z)=f(z+\eta )f(z)$ (where η ($\in \mathbb{C}$) is a constant satisfying $f(z+\eta )f(z)\not\equiv 0$) is regarded as the difference counterpart of ${f}^{\prime}$. So, Chen and Yi [20] considered the problem that ${\mathrm{\Delta}}_{\eta}f(z)$ and $f(z)$ share one value a CM and proved the following theorem.
Theorem C (See [20])
Let f be a finite order transcendental entire function which has a finite Borel exceptional value a, and let η ($\in \mathbb{C}$) be a constant such that $f(z+\eta )\not\equiv f(z)$. If ${\mathrm{\Delta}}_{\eta}f(z)=f(z+\eta )f(z)$ and $f(z)$ share the value a CM, then
where A is a nonzero constant.
Question 1 What can be said if we consider the forward difference ${\mathrm{\Delta}}_{\eta}^{n}f(z)$ and $f(z)$ share one value or one small function?
In this paper, we answer Question 1 and prove the following theorem.
Theorem 1.1 Let $f(z)$ be a finite order transcendental entire function such that $\lambda (fa(z))<\sigma (f)$, where $a(z)$ is an entire function and satisfies $\sigma (a)<1$. Let n be a positive integer. If ${\mathrm{\Delta}}_{\eta}^{n}f(z)$ and $f(z)$ share $a(z)$ CM, where η ($\in \mathbb{C}$) satisfies ${\mathrm{\Delta}}_{\eta}^{n}f(z)\not\equiv 0$, then
where c, ${c}_{1}$ are two nonzero constants.
In the special case, if we take $a(z)\equiv a$ in Theorem 1.1, we can get the following corollary.
Corollary 1.1 Let $f(z)$ be a finite order transcendental entire function which has a finite Borel exceptional value a. Let n be a positive integer. If ${\mathrm{\Delta}}_{\eta}^{n}f(z)$ and $f(z)$ share value a CM, where η ($\in \mathbb{C}$) satisfies ${\mathrm{\Delta}}_{\eta}^{n}f(z)\not\equiv 0$, then
where c, ${c}_{1}$ are two nonzero constants.
Remark 1.1 From Corollary 1.1, we know that $\frac{{\mathrm{\Delta}}_{\eta}^{n}f(z)}{f(z)}={({e}^{{c}_{1}\eta}1)}^{n}$ and it shows that the quotient of ${\mathrm{\Delta}}_{\eta}^{n}f(z)$ and $f(z)$ is related to η, n and ${c}_{1}$, but not related to c. On the other hand, Corollary 1.1 shows that if f has a nonzero finite Borel exceptional value ${b}^{\ast}$, then, for any constant η satisfying ${\mathrm{\Delta}}_{\eta}^{n}f(z)\not\equiv 0$, the value ${b}^{\ast}$ is not shared CM by ${\mathrm{\Delta}}_{\eta}^{n}f(z)$ and $f(z)$. See the following example.
Example 1.1 Suppose that $f(z)={e}^{z}+{b}^{\ast}$, where ${b}^{\ast}$ is a nonzero finite value. Then f has a nonzero finite Borel exceptional value ${b}^{\ast}$. For any $\eta \ne 2k\pi i$, $k\in \mathbb{Z}$, the value ${b}^{\ast}$ is not shared CM by ${\mathrm{\Delta}}_{\eta}^{n}f(z)$ and $f(z)$. Observe that ${\mathrm{\Delta}}_{\eta}^{n}f(z)={\sum}_{j=0}^{n}{(1)}^{j}{C}_{n}^{j}f(z+(nj)\eta )$, where ${C}_{n}^{j}$ are the binomial coefficients. Thus, for any $\eta \ne 2k\pi i$, $k\in \mathbb{Z}$, we have ${\mathrm{\Delta}}_{\eta}^{n}f(z)={({e}^{\eta}1)}^{n}\cdot {e}^{z}$. Thus, we can see that $f(z){b}^{\ast}={e}^{z}$ has no zero, but ${\mathrm{\Delta}}_{\eta}^{n}f(z){b}^{\ast}={({e}^{\eta}1)}^{n}{e}^{z}{b}^{\ast}$ has infinitely many zeros. Hence, the value ${b}^{\ast}$ is not shared CM by ${\mathrm{\Delta}}_{\eta}^{n}f(z)$ and $f(z)$.
In the special case, if we take $n=1$ in Theorem 1.1 and $n=1$ in Corollary 1.1, we can obtain the following corollaries.
Corollary 1.2 Let $f(z)$ be a finite order transcendental entire function such that $\lambda (fa(z))<\sigma (f)$, where $a(z)$ is an entire function and satisfies $\sigma (a)<1$. If ${\mathrm{\Delta}}_{\eta}f(z)=f(z+\eta )f(z)$ and $f(z)$ share $a(z)$ CM, where η ($\in \mathbb{C}$) satisfies $f(z+\eta )\not\equiv f(z)$, then
where c, ${c}_{1}$ are two nonzero constants.
Corollary 1.3 Let $f(z)$ be a finite order transcendental entire function which has a finite Borel exceptional value a. If ${\mathrm{\Delta}}_{\eta}f(z)=f(z+\eta )f(z)$ and $f(z)$ share value a CM, where η ($\in \mathbb{C}$) satisfies $f(z+\eta )\not\equiv f(z)$, then
where c, ${c}_{1}$ are two nonzero constants.
Remark 1.2 The Corollary 1.2 shows that if a nonzero polynomial $a(z)$ satisfies $\lambda (fa)<\sigma (f)$, then $a(z)$ is not shared CM by $\mathrm{\Delta}f(z)$ and $f(z)$. For example, if we take $a(z)\equiv z$, and $\lambda (fz)<\sigma (f)$ holds, then $\mathrm{\Delta}f(z)$ and $f(z)$ do not have any common fixed point (counting multiplicities). See the following example.
Example 1.2 Suppose that $f(z)={e}^{z}+z$. Then $f(z)$ satisfies $\lambda (f(z)z)=0<1=\sigma (f)$ and has no fixed point. But for any $\eta \ne 2k\pi i$, $k\in \mathbb{Z}$, the function ${\mathrm{\Delta}}_{\eta}f(z)=f(z+\eta )f(z)=({e}^{\eta}1){e}^{z}+\eta $ has infinitely many fixed points by Milloux’s theorem (see [1, 3]). Hence, the nonzero polynomial $a(z)\equiv z$ is not shared CM by ${\mathrm{\Delta}}_{\eta}f(z)$ and $f(z)$.
Remark 1.3 From Corollary 1.3, we can see that under the hypothesis of Theorem C, we can get the expression of $f(z)$, that is, $f(z)=c{e}^{{c}_{1}z}$. Thus, we know that the constant A in Theorem C is related to η and ${c}_{1}$, but not related to c. Actually, from the proof of Lemma 2.9, we have $A={e}^{{c}_{1}\eta}1$ (obviously, we can obtain $A\ne 1$). Hence, Corollary 1.3 contains and improves Theorem C. Obviously, Theorem 1.1 generalizes Theorem C.
2 Lemmas for the proof of theorems
Lemma 2.1 (See [21])
Let f be a meromorphic function with a finite order σ, η be a nonzero constant. Let $\epsilon >0$ be given, then there exists a subset $E\subset (1,\mathrm{\infty})$ with finite logarithmic measure such that for all z satisfying $z=r\notin E\cup [0,1]$, we have
Suppose that $n\ge 2$ and let ${f}_{1}(z),\dots ,{f}_{n}(z)$ be meromorphic functions and ${g}_{1}(z),\dots ,{g}_{n}(z)$ be entire functions such that

(i)
${\sum}_{j=1}^{n}{f}_{j}(z)exp\{{g}_{j}(z)\}=0$;

(ii)
when $1\le j<k\le n$, ${g}_{j}(z){g}_{k}(z)$ is not constant;

(iii)
when $1\le j\le n$, $1\le h<k\le n$,
$$T(r,{f}_{j})=o\left\{T(r,exp\{{g}_{h}{g}_{k}\})\right\}\phantom{\rule{1em}{0ex}}(r\to \mathrm{\infty},r\notin E),$$
where $E\subset (1,\mathrm{\infty})$ has finite linear measure or logarithmic measure.
Then ${f}_{j}(z)\equiv 0$, $j=1,\dots ,n$.
εset Following Hayman [1], we define an εset to be a countable union of open discs not containing the origin and subtending angles at the origin whose sum is finite. If E is an εset, 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 θ, the intersection of E with the ray $argz=\theta $ is bounded.
Lemma 2.3 (See [4])
Let f be a function transcendental and meromorphic in the plane of order <1. Let $h>0$. Then there exists an εset E such that
uniformly in c for $c\le h$.
Lemma 2.4 (See [25])
Let f be a transcendental meromorphic solution of finite order ρ of a difference equation of the form
where $U(z,f)$, $P(z,f)$, $Q(z,f)$ are difference polynomials such that the total degree $degU(z,f)=n$ in $f(z)$ and its shifts, and $degQ(z,f)\le n$. Moreover, we assume that $U(z,f)$ contains just one term of maximal total degree in $f(z)$ and its shifts. Then, for each $\epsilon >0$,
possibly outside of an exceptional set of finite logarithmic measure.
Remark 2.1 From the proof of Lemma 2.4 in [25], we can see that if the coefficients of $U(z,f)$, $P(z,f)$, $Q(z,f)$, namely ${\alpha}_{\lambda}(z)$, satisfy $m(r,{\alpha}_{\lambda})=S(r,f)$, then the same conclusion still holds.
Lemma 2.5 (See [27])
Let ${P}_{n}(z),\dots ,{P}_{0}(z)$ be polynomials such that ${P}_{n}{P}_{0}\not\equiv 0$ and satisfy
Then every finite order transcendental meromorphic solution $f(z)$ (≢0) of the equation
satisfies $\sigma (f)\ge 1$, and $f(z)$ assumes every nonzero value $a\in \mathbb{C}$ infinitely often and $\lambda (fa)=\sigma (f)$.
Remark 2.2 If equation (2.2) satisfies condition (2.1) and all ${P}_{j}(z)$ are constants, we can easily see that equation (2.2) does not possess any nonzero polynomial solution.
Lemma 2.6 (See [27])
Let $F(z),{P}_{n}(z),\dots ,{P}_{0}(z)$ be polynomials such that $F{P}_{n}{P}_{0}\not\equiv 0$. Then every finite order transcendental meromorphic solution $f(z)$ (≢0) of the equation
satisfies $\lambda (f)=\sigma (f)\ge 1$.
Remark 2.3 From the proof of Lemma 2.5 in [27], we can see that if we replace $f(z+j)$ by $f(z+j\eta )$ ($j=1,\dots ,n$) in equation (2.2) or (2.3), then the corresponding conclusion still holds.
Lemma 2.7 (See [4])
Let f be a function transcendental and meromorphic in the plane which satisfies ${\underline{lim}}_{r\to \mathrm{\infty}}\frac{T(r,f)}{r}=0$. Then $g(z)=f(z+1)f(z)$ and $G(z)=\frac{f(z+1)f(z)}{f(z)}$ are both transcendental.
Remark 2.4 From the proof of Lemma 2.7 in [4], we can see that, under the same hypotheses of Lemma 2.7, we can obtain the following conclusion: ${\mathrm{\Delta}}_{\eta}f(z)=f(z+\eta )f(z)$ and $G(z)=\frac{{\mathrm{\Delta}}_{\eta}f(z)}{f(z)}=\frac{f(z+\eta )f(z)}{f(z)}$ are both transcendental.
Lemma 2.8 Let $f(z)=H(z){e}^{{c}_{1}z}$, where $H(z)$ (≢0) is an entire function such that $\sigma (H)<1$ and ${c}_{1}$ is a nonzero constant. If ${\mathrm{\Delta}}_{\eta}^{n}f(z)\not\equiv 0$ for some constant η, and
holds, where A is a constant, then $H(z)$ is a constant.
Proof From ${\mathrm{\Delta}}_{\eta}^{n}f(z)\not\equiv 0$, we can see that $A\ne 0$. In order to prove that $H(z)$ is a constant, we only need to prove ${H}^{\prime}(z)\equiv 0$. Substituting $f(z)=H(z){e}^{{c}_{1}z}$ into (2.4), we can obtain
First, we assert that the sum of all coefficients of equation (2.5) is equal to zero, that is,
On the contrary, we suppose that
Thus, applying Lemma 2.5 and Remarks 2.22.3 to (2.5), we have $\sigma (H)\ge 1$, a contradiction. Hence, (2.6) holds. Thus, by (2.6) and (2.5), we have
By Lemma 2.3, we see that there exists an εset E such that for $j=1,2,\dots ,n$,
Substituting (2.8) into (2.7), we can get
where K is a constant and satisfies
Secondly, we assert that $K\ne 0$. If $n=1$, then $K={e}^{{c}_{1}\eta}\ne 0$; if $n\ge 2$, on the contrary, we suppose that $K=0$. Then, for $j=0,1,\dots ,n1$, noting that
we have
Thus, we can obtain from the equality above that ${e}^{{c}_{1}\eta}=1$ since $n1\ge 1$. By (2.6) we have $A={({e}^{{c}_{1}\eta}1)}^{n}=0$, which contradicts $A\ne 0$. Hence $K\ne 0$ and (2.9) implies ${H}^{\prime}(z)\not\equiv 0$. Thus, we can know that $H(z)$ is a nonzero constant.
Thus, Lemma 2.8 is proved. □
Lemma 2.9 Suppose that $f(z)$ is a finite order transcendental entire function such that $\lambda (fa(z))<\sigma (f)$, where $a(z)$ is an entire function and satisfies $\sigma (a)<1$. Let n be a positive integer. If ${\mathrm{\Delta}}_{\eta}^{n}f(z)\not\equiv 0$ for some constant η ($\in \mathbb{C}$), and
holds, where A is a constant, then
where c, ${c}_{1}$ are two nonzero constants.
Proof Since $f(z)$ is a transcendental entire function of finite order and satisfies $\lambda (fa(z))<\sigma (f)$, we can write $f(z)$ in the form
where H (≢0) is an entire function, h is a polynomial with $degh=k$ ($k\ge 1$), H and h satisfy
First, we assert that $a(z)\equiv 0$. Substituting (2.11) into (2.10), we can get that
where $b(z)={\mathrm{\Delta}}_{\eta}^{n}a(z)a(z)$. Rewrite (2.13) in the form
Suppose that $b(z)\not\equiv 0$. Then, from $\sigma (H(z+(nj)\eta ))=\sigma (H(z))<degh(z)=k$ ($j=0,1,\dots ,n1$), $deg(h(z+(nj)\eta )h(z))=k1$ and $\sigma (b(z))\le \sigma (a(z))<1\le k$, we can see that the order of growth of the lefthand side of (2.14) is less than k, and the order of growth of the righthand side of (2.14) is equal to k. This is a contradiction. Hence, $b(z)\equiv {\mathrm{\Delta}}_{\eta}^{n}a(z)a(z)\equiv 0$. Namely,
Suppose that $a(z)\not\equiv 0$. Note that the sum of all coefficients of (2.15) does not vanish. Then we can apply Lemma 2.5 and Remarks 2.22.3 to (2.15) and obtain $\sigma (a(z))\ge 1$, which contradicts our hypothesis. Hence, $a(z)\equiv 0$. Thus, (2.13) can be rewritten as
Secondly, we prove that $A\ne 0$. In fact, if $A=0$, we obtain from (2.16) that ${\mathrm{\Delta}}_{\eta}^{n}f(z)\equiv 0$, which contradicts our hypothesis.
Thirdly, we prove that $\sigma (f)=k=1$. On the contrary, we suppose that $\sigma (f)=k\ge 2$. Thus, we will deduce a contradiction for cases $A={(1)}^{n}$ and $A\ne {(1)}^{n}$, respectively.
Case 1. Suppose that $A={(1)}^{n}$. Thus, for a positive integer n, there are three subcases: (1) $n=1$; (2) $n=2$; (3) $n\ge 3$.
Subcase 1.1. Suppose that $n=1$. Then, by $A=1$, we can obtain from (2.16) that
a contradiction.
Subcase 1.2. Suppose that $n=2$. Then, by $A={(1)}^{2}=1$ and (2.16), we have
Set ${Q}_{1}(z)=\frac{2H(z+2\eta )}{H(z+\eta )}$. Then, from (2.17), we can know that ${Q}_{1}(z)$ is a nonconstant entire function. Set $\sigma (H)={\sigma}_{1}$. Then ${\sigma}_{1}<\sigma (f)=k$. By Lemma 2.1, we see that for any given ${\epsilon}_{1}$ ($0<3{\epsilon}_{1}<k{\sigma}_{1}$), there exists a set ${E}_{1}\subset (1,\mathrm{\infty})$ of finite logarithmic measure such that for all z satisfying $z=r\notin [0,1]\cup {E}_{1}$, we have
Since ${Q}_{1}(z)$ is an entire function, by (2.18), we have
so that $\sigma ({Q}_{1}(z))\le {\sigma}_{1}1+{\epsilon}_{1}<k1$. Thus, by $deg(h(z+\eta )h(z))=k1$ and $\sigma ({Q}_{1})<k1$, we can see that the order of growth of the lefthand side of (2.17) is equal to $k1$, and the order of growth of the righthand side of (2.17) is less than $k1$. This is a contradiction.
Subcase 1.3. Suppose that $n\ge 3$. Then we can obtain from (2.16) that
Set ${Q}_{2}(z)={e}^{h(z+2\eta )h(z+\eta )}$. Then ${Q}_{2}(z)$ is a transcendental entire function since $\sigma ({Q}_{2}(z))=k1\ge 1$. For $j=3,4,\dots ,n$, we have
Thus, (2.19) can be rewritten as
where
Noting that ${(1)}^{n}{C}_{n}^{n1}\ne 0$, we can see that ${U}_{2}(z,{Q}_{2}(z))\not\equiv 0$. Set $\sigma (H)={\sigma}_{2}$. Then ${\sigma}_{2}<k$. Since ${Q}_{2}(z)$ is of regular growth and $\sigma ({Q}_{2}(z))=k1$, for any given ${\epsilon}_{2}$ ($0<3{\epsilon}_{2}<k{\sigma}_{2}$) and all $r>{r}_{0}$ (>0), we have
By Lemma 2.1, we see that for ${\epsilon}_{2}$, there exists a set ${E}_{2}\subset (1,\mathrm{\infty})$ of finite logarithmic measure such that for all z satisfying $z=r\notin [0,1]\cup {E}_{2}$, we have
Thus, from (2.21) and (2.22), we can get that for $j=2,3,\dots ,n$,
that is,
Noting that ${deg}_{{Q}_{2}}U(z,{Q}_{2})=n2\ge 1$ and by Lemma 2.4 and Remark 2.1, we have
a contradiction.
Case 2. Suppose that $A\ne {(1)}^{n}$. Thus, for a positive integer n, there are two subcases: (1) $n=1$; (2) $n\ge 2$.
Subcase 2.1. Suppose that $n=1$. Thus, (2.16) can be rewritten as
Noting the $A+1\ne 0$, we can use the same method as in the proof of Subcase 1.2 and deduce a contradiction.
Subcase 2.2. Suppose that $n\ge 2$. Then we can obtain from (2.16) that
Set ${Q}_{3}(z)={e}^{h(z+\eta )h(z)}$. Then ${Q}_{3}(z)$ is a transcendental entire function since $\sigma ({Q}_{3}(z))=k1\ge 1$. For $j=1,2,\dots ,n$, we have
Thus, (2.24) can be rewritten as
where
We can see that ${U}_{3}(z,{Q}_{3}(z))\not\equiv 0$ since $A{(1)}^{n}\ne 0$. Noting that ${deg}_{{Q}_{3}}{U}_{3}(z,{Q}_{3}(z))=n1\ge 1$, we can use the same method as in the proof of Subcase 1.3 and deduce a contradiction.
Thus, we have proved that $\sigma (f)=k=1$. And $f(z)$ can be written as
where ${c}_{0}$, ${c}_{1}$ (≠0) are two constants and ${H}^{\ast}(z)={e}^{{c}_{0}}H(z)$ (≢0) is an entire function and satisfies
Thus, by (2.26), (2.27), (2.16) and Lemma 2.8, we can get that ${H}^{\ast}(z)$ is a nonzero constant, and so, $f(z)$ can be written as
where c, ${c}_{1}$ are two nonzero constants.
Thus, Lemma 2.9 is proved. □
Remark 2.5 From the proof of Lemma 2.9 or Remark 1.3, we can see that $A\ne 1$ in Lemma 2.9 when $n=1$ and Theorem C. Unfortunately, we cannot obtain $A\ne {(1)}^{n}$ when $n\ge 2$ in Lemma 2.9. This is because we can get a contradiction from the equality ${e}^{{c}_{1}\eta}1=1$, but we cannot obtain a contradiction from the equality ${({e}^{{c}_{1}\eta}1)}^{n}={(1)}^{n}$ when $n\ge 2$.
3 Proof of Theorem 1.1
By the hypotheses of Theorem 1.1, we can write $f(z)$ in the form (2.11), and (2.12) holds. Since ${\mathrm{\Delta}}_{\eta}^{n}f(z)$ and $f(z)$ share an entire function $a(z)$ CM, then
where $P(z)$ is a polynomial and $b(z)={\mathrm{\Delta}}_{\eta}^{n}a(z)a(z)$. Obviously, $\sigma (b(z))\le \sigma (a(z))<1$.
First step. We prove
where A (≠0) is a constant. If $P(z)\equiv 0$, then, by (3.1), we see that (3.2) holds and $A=1$.
Now suppose that $P(z)\not\equiv 0$ and $degP(z)=s$. Set
where $k=\sigma (f)\ge 1$, ${a}_{k}\phantom{\rule{0.25em}{0ex}}(\ne 0),{a}_{k1},\dots ,{a}_{0}$, ${b}_{s}\phantom{\rule{0.25em}{0ex}}(\ne 0),{b}_{s1},\dots ,{b}_{0}$ are constants. By (3.1), we can see that $0\le degP=s\le degh=k$.
In this case, we prove that $P(z)$ is a constant, that is, $s=0$. To this end, we will deduce a contradiction for the cases $s=k$ and $1\le s<k$, respectively.
Case 1. Suppose that $1\le s=k$. Thus, there are two subcases: (1) $a(z)\not\equiv 0$; (2) $a(z)\equiv 0$.
Subcase 1.1. Suppose that $a(z)\not\equiv 0$. First we suppose that ${b}_{k}\ne {a}_{k}$. Then (3.1) is rewritten as
where ${h}_{0}(z)\equiv 0$ and
Since $\sigma (H)<k$, $\sigma (b)<1\le k$ and $deg(h(z+j\eta )h(z))=k1$ ($j=1,2,\dots ,n$), we can see that $\sigma ({g}_{1m}(z))<k$ ($m=1,2,3$). On the other hand, by ${b}_{k}\ne {a}_{k}$, we can see that $deg(P(h))=deg(P{h}_{0})=deg(h{h}_{0})=k$. Since ${e}^{P(h)}$, ${e}^{P{h}_{0}}$ and ${e}^{h{h}_{0}}$ are of regular growth, and $\sigma ({g}_{1m})<k$ ($m=1,2,3$), we can see that for $m=1,2,3$,
Thus, applying Lemma 2.2 to (3.4), by (3.5), we can obtain ${g}_{1m}(z)\equiv 0$ ($m=1,2,3$). Clearly, this is a contradiction.
Now we suppose that ${b}_{k}={a}_{k}$. Then (3.1) is rewritten as
We affirm that $H(z){e}^{P(z)+h(z)}b(z)\not\equiv 0$. In fact, if $H(z){e}^{P(z)+h(z)}b(z)\equiv 0$, then, by (3.6), we can obtain
this is the special case of (2.14) when $b(z)\equiv 0$ and $A=0$. Hence, using the same method as in the proof of Case 2 in the proof of Lemma 2.9, we can get that $\sigma (f)=k=1$. Hence, substituting $h(z)={c}_{1}z+{c}_{0}$ into (3.7), we have
On this occasion, we assert that ${({e}^{{c}_{1}\eta}1)}^{n}=0$. On the contrary, we suppose that ${({e}^{{c}_{1}\eta}1)}^{n}\ne 0$. Then the sum of all coefficients of (3.8) is ${({e}^{\eta}1)}^{n}$, which does not vanish. By Lemma 2.5 and Remarks 2.22.3, we have $\sigma (H)\ge 1$, a contradiction. Hence, ${({e}^{{c}_{1}\eta}1)}^{n}=0$. Thus, ${e}^{{c}_{1}\eta}=1$. Substituting it into (3.8), we have
First, we suppose that $H(z)$ is transcendental. Then, noting that $\sigma (H)<1$ implies ${\underline{lim}}_{r\to \mathrm{\infty}}\frac{T(r,H)}{r}=0$, by Lemma 2.7 and Remark 2.4, we know that ${\mathrm{\Delta}}_{\eta}H(z)=H(z+\eta )H(z)$ is transcendental. Moreover, $\sigma ({\mathrm{\Delta}}_{\eta}H(z))\le \sigma (H(z))<1$ implies ${\underline{lim}}_{r\to \mathrm{\infty}}\frac{T(r,{\mathrm{\Delta}}_{\eta}H)}{r}=0$. Repeating the process above $n1$ times, we can see that ${\mathrm{\Delta}}_{\eta}^{n}H(z)$ is transcendental. That is, the lefthand side of (3.9) is a transcendental function. Hence (3.9) is impossible.
Secondly, we suppose that $H(z)$ is a nonzero polynomial. Then, noting that ${b}_{k}={a}_{k}$, we can see that ${e}^{p(z)+h(z)}$ is a nonzero constant. Thus, from $b(z)=H(z){e}^{p(z)+h(z)}$, we can know that $b(z)$ is a nonzero polynomial. Thus, applying Lemma 2.6 to the equation ${\mathrm{\Delta}}_{\eta}^{n}a(z)a(z)=b(z)$ and by Remark 2.3, we have $\sigma (a)\ge 1$, a contradiction. Hence, $H(z){e}^{P(z)+h(z)}b(z)\not\equiv 0$. Thus, since $deg(P+h)\le k1$, $deg(h)=k$, $deg(h(z+j\eta )h(z))=k1$ ($j=1,2,\dots ,n$) and $\sigma (H)<k$, we see that the order of growth of the lefthand side of (3.6) is equal to k, and the order of growth of the righthand side of (3.6) is less than k. This is a contradiction.
Subcase 1.2. Suppose that $a(z)\equiv 0$. Then (3.1) is rewritten as
Since $H(z)\not\equiv 0$, $\sigma (H)<k$, $degP=s=k$ and $deg(h(z+j\eta )h(z))=k1$ ($j=1,2,\dots ,n$), we can see that the order of growth of the lefthand side of (3.10) is equal to k, and the order of growth of the righthand side of (3.10) is less than k. This is a contradiction.
Case 2. Suppose that $1\le s<k$. Thus, there are two subcases: (1) $a(z)\not\equiv 0$; (2) $a(z)\equiv 0$.
Subcase 2.1. Suppose that $a(z)\not\equiv 0$. Then, by (3.1), we can obtain
We assert that $b(z)\not\equiv 0$. In fact, if $b(z)\equiv 0$, then (2.15) obviously holds. Hence, using the same method as in the proof of Lemma 2.9, by Lemma 2.5 and Remarks 2.22.3, we can get that $\sigma (a)\ge 1$, a contradiction. Hence, $b(z)\not\equiv 0$. Since $degh=k$, $deg(h(z+j\eta )h(z))=k1$ ($j=1,2,\dots ,n$), $degP=s<k$ and $\sigma (H)<k$, we see that the order of growth of the lefthand side of (3.11) is less than k, and the order of growth of the righthand side of (3.11) is equal to k. This is a contradiction.
Subcase 2.2. Suppose that $a(z)\equiv 0$. Then, by (3.1), we obtain
Thus, there are two subcases: (1) $n=1$; (2) $n\ge 2$.
Subcase 2.2.1. Suppose that $n=1$. Then (3.12) can be rewritten as
By (3.13), we see that $\frac{H(z+\eta )}{H(z)}$ is a nonzero entire function. Set $\sigma (H)={\sigma}_{4}$. Then ${\sigma}_{4}<\sigma (f)=k$. By Lemma 2.1, we see that for any given ${\epsilon}_{4}$ ($0<3{\epsilon}_{4}<k{\sigma}_{4}$), there exists a set ${E}_{4}\subset (1,\mathrm{\infty})$ of finite logarithmic measure such that for all z satisfying $z=r\notin [0,1]\cup {E}_{4}$, we have
Since $\frac{H(z+\eta )}{H(z)}$ is an entire function, by (3.13), we have
so that
Since $s<k$, we can see that $degP\le k1$. If $degP<k1$, then, by (3.15) and $deg(h(z+\eta )h(z))=k1$, we can see that the order of growth of the lefthand side of (3.13) is equal to $k1$, and the order of growth of the righthand side of (3.13) is equal to degP which is less than $k1$. This is a contradiction.
If $degP=k1$, then since $\frac{H(z+\eta )}{H(z)}$ is an entire function and $deg(h(z+\eta )h(z))=k1\ge 1$, by (3.15), we can see that the entire function $\frac{H(z+\eta )}{H(z)}{e}^{h(z+\eta )h(z)}$ has a Borel exceptional value 0, thus the value 1 must be not its Borel exceptional value. Hence, the lefthand side of (3.13), $\frac{H(z+\eta )}{H(z)}{e}^{h(z+\eta )h(z)}1$, has infinitely many zeros, but the righthand side of (3.13), ${e}^{P(z)}$, has no zero. This is a contradiction.
Subcase 2.2.2. Now we suppose that $n\ge 2$. Thus, for s ($=degP$), there are two subcases: (1) $s<k1$; (2) $s=k1$.
Subcase 2.2.2.1. Now we suppose that $s<k1$. Set ${Q}_{5}(z)={e}^{h(z+\eta )h(z)}$. Since $\sigma ({Q}_{5})=k1\ge 1$, ${Q}_{5}(z)$ is a transcendental entire function. Thus, (3.12) can be rewritten as
where
Thus, using the same method as in the proof of Subcase 1.3 in the proof of Lemma 2.9 and noting that $\sigma ({e}^{P(z)}{(1)}^{n})=degP<k1$, we have
Noting that $n\ge 2$ and so $deg{U}_{5}(z,{Q}_{5})=n1\ge 1$. Using the same method as in the proof of Subcase 1.3 in the proof of Lemma 2.9, we can obtain
Clearly, this is a contradiction.
Subcase 2.2.2.2. Now we suppose that $s=k1$. Thus, (3.12) is written as
where ${T}_{j}(z)=h(z+j\eta )h(z)$ ($j=1,2,\dots ,n$). Thus, by (3.3), we have
where ${P}_{k2,j}(z)$ is a polynomial with degree at most $k2$. Thus, we have
where ${P}_{j,t}(z)$ is a polynomial with degree at most $k2$.
First, we suppose that there is some ${j}_{0}$ ($1\le {j}_{0}\le n$) such that ${j}_{0}k\eta {a}_{k}={b}_{k1}$, that is, $deg({T}_{{j}_{0}}(z)P(z))\le k2$. Thus, (3.18) can be written as
where
Set ${Q}_{6}(z)={e}^{h(z+\eta )h(z)}$ and $\sigma (H)={\sigma}_{6}$. Then (3.20) can be rewritten as
where
or
Noting that ${(1)}^{n+1}\ne 0$, we can see that ${U}_{6}(z,{Q}_{6}(z))\not\equiv 0$. Since $\sigma (H)<k$ and $\sigma ({e}^{P(z)+h(z)h(z+{j}_{0}\eta )})\le k2<k1$, using the same method as in the proof of Subcase 1.3 in the proof of Lemma 2.9, we have
Noting that $n\ge 2$ and so $deg{U}_{6}(z,{Q}_{6})=n1\ge 1$. Combining (3.21)(3.24), using the same method as in the proof of Subcase 1.3 in the proof of Lemma 2.9, we can obtain
Clearly, this is a contradiction.
Secondly, we suppose that $jk\eta {a}_{k}\ne {b}_{k1}$ for any $1\le j\le n$. Thus, equation (3.18) can be rewritten as
where ${P}_{k2}(z)=P(z){b}_{k1}{z}^{k1}={b}_{k2}{z}^{k2}+{b}_{k3}{z}^{k3}+\cdots +{b}_{0}$. For dealing with equation (3.25), we just compare ${b}_{k1}$ with $nk\eta {a}_{k}$ since $nk\eta {a}_{k}>(n1)k\eta {a}_{k}>\cdots >k\eta {a}_{k}$. Without loss of generality, we suppose that $nk\eta {a}_{k}\le {b}_{k1}$. Let $arg{b}_{k1}={\theta}_{1}$, $arg(\eta {a}_{k})={\theta}_{2}$ and $\sigma (H)={\sigma}_{7}<k$. Take ${\theta}_{0}$ such that $cos((k1){\theta}_{0}+{\theta}_{1})=1$. By Lemma 2.1, we see that for any given ${\epsilon}_{7}$ ($0<3{\epsilon}_{7}<k{\sigma}_{7}$), there exists a set ${E}_{7}\subset (1,\mathrm{\infty})$ of finite logarithmic measure such that for all $z=r{e}^{i{\theta}_{0}}$ satisfying $z=r\notin [0,1]\cup {E}_{7}$, we have
Thus, noting that ${e}^{{P}_{k2}(z)}$ is of regular growth, we can deduce from (3.25) and (3.26) that
that is,
We assert that
In fact, if $nk\eta {a}_{k}={b}_{k1}$, then, by ${b}_{k1}\ne nk\eta {a}_{k}$, we know that $cos((k1){\theta}_{0}+{\theta}_{2})\ne 1$, that is, $cos((k1){\theta}_{0}+{\theta}_{2})<1$, and hence $nk\eta {a}_{k}cos((k1){\theta}_{0}+{\theta}_{2})<nk\eta {a}_{k}={b}_{k1}$. If $nk\eta {a}_{k}<{b}_{k1}$, then we have $nk\eta {a}_{k}cos((k1){\theta}_{0}+{\theta}_{2})\le nk\eta {a}_{k}<{b}_{k1}$.
Thus, taking a positive constant ${\epsilon}_{8}$ ($0<{\epsilon}_{8}<\frac{{b}_{k1}nk\eta {a}_{k}cos((k1){\theta}_{0}+{\theta}_{2})}{3}$), we can deduce from (3.27) that
a contradiction. Thus, we have proved that P is only a constant and (3.2) holds.
Second step. Applying Lemma 2.9 to (3.2), we can obtain the conclusion.
Thus, Theorem 1.1 is proved.
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Acknowledgements
This research was supported by the National Natural Science Foundation of China (Nos. 11171119, 11226090) and supported by the Natural Science Foundation of Guangdong Province, China (No. S2013040014347).
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Keywords
 complex difference
 meromorphic function
 Borel exceptional value
 sharing value