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# A note on entire functions and their differences

## Abstract

In this paper, we prove that for a transcendental entire function $f(z)$ of finite order such that $λ(f−a(z))<σ(f)$, where $a(z)$ is an entire function and satisfies $σ(a(z))<1$, n is a positive integer and if $Δ η n f(z)$ and $f(z)$ share the function $a(z)$ CM, where η ($∈C$) satisfies $Δ η n f(z)≢0$, then

$a(z)≡0andf(z)=c e c 1 z ,$

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 [13]). In addition, we use the notation $λ(f)$ for the exponent of convergence of the sequence of zeros of a meromorphic function f, and $σ(f)$ to denote the order growth of f. For a nonzero constant η, the forward differences $Δ η n f(z)$ are defined (see [4, 5]) by

$Δ η f ( z ) = Δ η 1 f ( z ) = f ( z + η ) − f ( z ) and Δ η n + 1 f ( z ) = Δ η n f ( z + η ) − Δ η n f ( z ) , n = 1 , 2 , … .$

Throughout this paper, we denote by $S(r,f)$ any function satisfying $S(r,f)=o(T(r,f))$ as $r→∞$, possibly outside a set of r of finite logarithmic measure. A meromorphic function $α(z)$ is said to be a small function of $f(z)$ if $T(r,α(z))=S(r,f)$, 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 $a∈C$. We say that f and g share the value a CM (IM) provided that $f−a$ and $g−a$ 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 $a(z)∈S(f)∩S(g)$. Nevanlinna’s four values theorem [6] says that if two nonconstant meromorphic functions f and g share four values CM, then $f≡g$ 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 well-known conjecture by Brück [10].

Conjecture Let f be a nonconstant entire function such that hyper order $σ 2 (f)<∞$ and $σ 2 (f)$ is not a positive integer. If f and $f ′$ share the finite value a CM, then

$f ′ −a=c(f−a),$

where c is a nonzero constant.

The notation $σ 2 (f)$ denotes hyper-order (see [11]) of $f(z)$ which is defined by

$σ 2 (f)= lim ¯ r → ∞ log log T ( r , f ) log r .$

The conjecture has been verified in the special cases when $a=0$ [10], or when f is of finite order [12], or when $σ 2 (f)< 1 2$ [13].

Recently, many authors [1417] 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 $σ(f)<2$, and let $c∈C$. If $f(z)$ and $f(z+c)$ share the values a ($∈C$) andCM, then

$f(z+c)−a=τ ( f ( z ) − a )$

for some constant τ.

In [15], Heittokangas et al. give the example $f(z)= e z 2 +1$ which shows that $σ(f)<2$ cannot be relaxed to $σ(f)≤2$.

Theorem B (See [16])

Let f be a meromorphic function of finite order, let $c∈C$. If $f(z)$ and $f(z+c)$ share three distinct periodic functions $a 1 , a 2 , a 3 ∈ S ˆ (f)$ with period c CM (where $S ˆ (f)=S(f)∪{∞}$), then $f(z)=f(z+c)$ for all $z∈C$.

Recently, many results of complex difference equations have been rapidly obtained (see [1825]). 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 $Δ η f(z)=f(z+η)−f(z)$ (where η ($∈C$) is a constant satisfying $f(z+η)−f(z)≢0$) is regarded as the difference counterpart of $f ′$. So, Chen and Yi [20] considered the problem that $Δ η f(z)$ 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 ($∈C$) be a constant such that $f(z+η)≢f(z)$. If $Δ η f(z)=f(z+η)−f(z)$ and $f(z)$ share the value a CM, then

$a=0and f ( z + η ) − f ( z ) f ( z ) =A,$

where A is a nonzero constant.

Question 1 What can be said if we consider the forward difference $Δ η n f(z)$ 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 $λ(f−a(z))<σ(f)$, where $a(z)$ is an entire function and satisfies $σ(a)<1$. Let n be a positive integer. If $Δ η n f(z)$ and $f(z)$ share $a(z)$ CM, where η ($∈C$) satisfies $Δ η n f(z)≢0$, then

$a(z)≡0andf(z)=c e c 1 z ,$

where c, $c 1$ are two nonzero constants.

In the special case, if we take $a(z)≡a$ 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 $Δ η n f(z)$ and $f(z)$ share value a CM, where η ($∈C$) satisfies $Δ η n f(z)≢0$, then

$a=0andf(z)=c e c 1 z ,$

where c, $c 1$ are two nonzero constants.

Remark 1.1 From Corollary 1.1, we know that $Δ η n f ( z ) f ( z ) = ( e c 1 η − 1 ) n$ and it shows that the quotient of $Δ η n f(z)$ 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 $b ∗$, then, for any constant η satisfying $Δ η n f(z)≢0$, the value $b ∗$ is not shared CM by $Δ η n f(z)$ and $f(z)$. See the following example.

Example 1.1 Suppose that $f(z)= e z + b ∗$, where $b ∗$ is a nonzero finite value. Then f has a nonzero finite Borel exceptional value $b ∗$. For any $η≠2kπi$, $k∈Z$, the value $b ∗$ is not shared CM by $Δ η n f(z)$ and $f(z)$. Observe that $Δ η n f(z)= ∑ j = 0 n ( − 1 ) j C n j f(z+(n−j)η)$, where $C n j$ are the binomial coefficients. Thus, for any $η≠2kπi$, $k∈Z$, we have $Δ η n f(z)= ( e η − 1 ) n ⋅ e z$. Thus, we can see that $f(z)− b ∗ = e z$ has no zero, but $Δ η n f(z)− b ∗ = ( e η − 1 ) n e z − b ∗$ has infinitely many zeros. Hence, the value $b ∗$ is not shared CM by $Δ η n f(z)$ 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 $λ(f−a(z))<σ(f)$, where $a(z)$ is an entire function and satisfies $σ(a)<1$. If $Δ η f(z)=f(z+η)−f(z)$ and $f(z)$ share $a(z)$ CM, where η ($∈C$) satisfies $f(z+η)≢f(z)$, then

$a(z)≡0andf(z)=c e c 1 z ,$

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 $Δ η f(z)=f(z+η)−f(z)$ and $f(z)$ share value a CM, where η ($∈C$) satisfies $f(z+η)≢f(z)$, then

$a=0andf(z)=c e c 1 z ,$

where c, $c 1$ are two nonzero constants.

Remark 1.2 The Corollary 1.2 shows that if a nonzero polynomial $a(z)$ satisfies $λ(f−a)<σ(f)$, then $a(z)$ is not shared CM by $Δf(z)$ and $f(z)$. For example, if we take $a(z)≡z$, and $λ(f−z)<σ(f)$ holds, then $Δf(z)$ 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 $λ(f(z)−z)=0<1=σ(f)$ and has no fixed point. But for any $η≠2kπi$, $k∈Z$, the function $Δ η f(z)=f(z+η)−f(z)=( e η −1) e z +η$ has infinitely many fixed points by Milloux’s theorem (see [1, 3]). Hence, the nonzero polynomial $a(z)≡z$ is not shared CM by $Δ η f(z)$ 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 $A= e c 1 η −1$ (obviously, we can obtain $A≠−1$). 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 $ε>0$ be given, then there exists a subset $E⊂(1,∞)$ with finite logarithmic measure such that for all z satisfying $|z|=r∉E∪[0,1]$, we have

$exp { − r σ − 1 + ε } ≤| f ( z + η ) f ( z ) |≤exp { r σ − 1 + ε } .$

Lemma 2.2 (See [11, 26])

Suppose that $n≥2$ and let $f 1 (z),…, f n (z)$ be meromorphic functions and $g 1 (z),…, g n (z)$ be entire functions such that

1. (i)

$∑ j = 1 n f j (z)exp{ g j (z)}=0$;

2. (ii)

when $1≤j, $g j (z)− g k (z)$ is not constant;

3. (iii)

when $1≤j≤n$, $1≤h,

$T(r, f j )=o { T ( r , exp { g h − g k } ) } (r→∞,r∉E),$

where $E⊂(1,∞)$ has finite linear measure or logarithmic measure.

Then $f j (z)≡0$, $j=1,…,n$.

ε-set Following Hayman [1], we define an ε-set 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 ε-set, then the set of $r≥1$, 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 $argz=θ$ 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 ε-set E such that

uniformly in c for $|c|≤h$.

Lemma 2.4 (See [25])

Let f be a transcendental meromorphic solution of finite order ρ of a difference equation of the form

$U(z,f)P(z,f)=Q(z,f),$

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 $degQ(z,f)≤n$. Moreover, we assume that $U(z,f)$ contains just one term of maximal total degree in $f(z)$ and its shifts. Then, for each $ε>0$,

$m ( r , P ( z , f ) ) =O ( r ρ − 1 + ε ) +S(r,f),$

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 $α λ (z)$, satisfy $m(r, α λ )=S(r,f)$, then the same conclusion still holds.

Lemma 2.5 (See [27])

Let $P n (z),…, P 0 (z)$ be polynomials such that $P n P 0 ≢0$ and satisfy

$P n (z)+⋯+ P 0 (z)≢0.$
(2.1)

Then every finite order transcendental meromorphic solution $f(z)$ (0) of the equation

$P n (z)f(z+n)+ P n − 1 (z)f(z+n−1)+⋯+ P 0 (z)f(z)=0$
(2.2)

satisfies $σ(f)≥1$, and $f(z)$ assumes every nonzero value $a∈C$ infinitely often and $λ(f−a)=σ(f)$.

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 $F(z), P n (z),…, P 0 (z)$ be polynomials such that $F P n P 0 ≢0$. Then every finite order transcendental meromorphic solution $f(z)$ (0) of the equation

$P n (z)f(z+n)+ P n − 1 (z)f(z+n−1)+⋯+ P 0 (z)f(z)=F$
(2.3)

satisfies $λ(f)=σ(f)≥1$.

Remark 2.3 From the proof of Lemma 2.5 in [27], we can see that if we replace $f(z+j)$ by $f(z+jη)$ ($j=1,…,n$) 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 $lim ̲ r → ∞ T ( r , f ) r =0$. Then $g(z)=f(z+1)−f(z)$ and $G(z)= f ( z + 1 ) − f ( z ) f ( z )$ 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: $Δ η f(z)=f(z+η)−f(z)$ and $G(z)= Δ η f ( z ) f ( z ) = f ( z + η ) − f ( z ) f ( z )$ 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 $σ(H)<1$ and $c 1$ is a nonzero constant. If $Δ η n f(z)≢0$ for some constant η, and

$Δ η n f ( z ) f ( z ) =A$
(2.4)

holds, where A is a constant, then $H(z)$ is a constant.

Proof From $Δ η n f(z)≢0$, we can see that $A≠0$. In order to prove that $H(z)$ is a constant, we only need to prove $H ′ (z)≡0$. Substituting $f(z)=H(z) e c 1 z$ into (2.4), we can obtain

$∑ j = 0 n − 1 ( − 1 ) j C n j e ( n − j ) c 1 η H ( z + ( n − j ) η ) + ( ( − 1 ) n − A ) H(z)=0.$
(2.5)

First, we assert that the sum of all coefficients of equation (2.5) is equal to zero, that is,

$e n c 1 η − C n 1 e ( n − 1 ) c 1 η +⋯+ ( − 1 ) n − 1 C n n − 1 e c 1 η + ( ( − 1 ) n − A ) =0.$
(2.6)

On the contrary, we suppose that

$e n c 1 η − C n 1 e ( n − 1 ) c 1 η +⋯+ ( − 1 ) n − 1 C n n − 1 e c 1 η + ( ( − 1 ) n − A ) ≠0.$

Thus, applying Lemma 2.5 and Remarks 2.2-2.3 to (2.5), we have $σ(H)≥1$, a contradiction. Hence, (2.6) holds. Thus, by (2.6) and (2.5), we have

$∑ j = 0 n − 1 ( − 1 ) j C n j e ( n − j ) c 1 η ( H ( z + ( n − j ) η ) − H ( z ) ) =0.$
(2.7)

By Lemma 2.3, we see that there exists an ε-set E such that for $j=1,2,…,n$,

(2.8)

Substituting (2.8) into (2.7), we can get

(2.9)

where K is a constant and satisfies

$K=n e n c 1 η − C n 1 (n−1) e ( n − 1 ) c 1 η +⋯+ ( − 1 ) n − 2 C n n − 2 2 e 2 c 1 η + ( − 1 ) n − 1 C n n − 1 e c 1 η .$

Secondly, we assert that $K≠0$. If $n=1$, then $K= e c 1 η ≠0$; if $n≥2$, on the contrary, we suppose that $K=0$. Then, for $j=0,1,…,n−1$, noting that

$C n j ⋅(n−j)= n ! ⋅ ( n − j ) ( n − j ) ! j ! = ( n − 1 ) ! ⋅ n ( n − 1 − j ) ! j ! =n C n − 1 j ,$

we have

$∑ j = 0 n − 1 ( − 1 ) j C n j (n−j) e ( n − j ) c 1 η =n e c 1 η ( e c 1 η − 1 ) n − 1 =0.$

Thus, we can obtain from the equality above that $e c 1 η =1$ since $n−1≥1$. By (2.6) we have $A= ( e c 1 η − 1 ) n =0$, which contradicts $A≠0$. Hence $K≠0$ and (2.9) implies $H ′ (z)≢0$. 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 $λ(f−a(z))<σ(f)$, where $a(z)$ is an entire function and satisfies $σ(a)<1$. Let n be a positive integer. If $Δ η n f(z)≢0$ for some constant η ($∈C$), and

$Δ η n f ( z ) − a ( z ) f ( z ) − a ( z ) =A$
(2.10)

holds, where A is a constant, then

$a(z)≡0,A≠0andf(z)=c e c 1 z ,$

where c, $c 1$ are two nonzero constants.

Proof Since $f(z)$ is a transcendental entire function of finite order and satisfies $λ(f−a(z))<σ(f)$, we can write $f(z)$ in the form

$f(z)=a(z)+H(z) e h ( z ) ,$
(2.11)

where H (0) is an entire function, h is a polynomial with $degh=k$ ($k≥1$), H and h satisfy

$λ(H)=σ(H)=λ ( f − a ( z ) ) <σ(f)=degh.$
(2.12)

First, we assert that $a(z)≡0$. Substituting (2.11) into (2.10), we can get that

$Δ η n f ( z ) − a ( z ) f ( z ) − a ( z ) = ∑ j = 0 n ( − 1 ) j C n j H ( z + ( n − j ) η ) e h ( z + ( n − j ) η ) + b ( z ) H ( z ) e h ( z ) =A,$
(2.13)

where $b(z)= Δ η n a(z)−a(z)$. Rewrite (2.13) in the form

$∑ j = 0 n − 1 ( − 1 ) j C n j H ( z + ( n − j ) η ) e h ( z + ( n − j ) η ) − h ( z ) + ( ( − 1 ) n − A ) H(z)=−b(z) e − h ( z ) .$
(2.14)

Suppose that $b(z)≢0$. Then, from $σ(H(z+(n−j)η))=σ(H(z)) ($j=0,1,…,n−1$), $deg(h(z+(n−j)η)−h(z))=k−1$ and $σ(b(z))≤σ(a(z))<1≤k$, we can see that the order of growth of the left-hand side of (2.14) is less than k, and the order of growth of the right-hand side of (2.14) is equal to k. This is a contradiction. Hence, $b(z)≡ Δ η n a(z)−a(z)≡0$. Namely,

$a(z+nη)− C n 1 a ( z + ( n − 1 ) η ) +⋯+ ( − 1 ) n − 1 C n n − 1 a(z+η)+ ( ( − 1 ) n − 1 ) a(z)=0.$
(2.15)

Suppose that $a(z)≢0$. Note that the sum of all coefficients of (2.15) does not vanish. Then we can apply Lemma 2.5 and Remarks 2.2-2.3 to (2.15) and obtain $σ(a(z))≥1$, which contradicts our hypothesis. Hence, $a(z)≡0$. Thus, (2.13) can be rewritten as

$Δ η n f ( z ) f ( z ) = ∑ j = 0 n ( − 1 ) j C n j H ( z + ( n − j ) η ) e h ( z + ( n − j ) η ) − h ( z ) H ( z ) =A.$
(2.16)

Secondly, we prove that $A≠0$. In fact, if $A=0$, we obtain from (2.16) that $Δ η n f(z)≡0$, which contradicts our hypothesis.

Thirdly, we prove that $σ(f)=k=1$. On the contrary, we suppose that $σ(f)=k≥2$. Thus, we will deduce a contradiction for cases $A= ( − 1 ) n$ and $A≠ ( − 1 ) n$, 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) $n≥3$.

Subcase 1.1. Suppose that $n=1$. Then, by $A=−1$, we can obtain from (2.16) that

$e h ( z + η ) − h ( z ) =(1+A)⋅ H ( z ) H ( z + η ) ≡0,$

Subcase 1.2. Suppose that $n=2$. Then, by $A= ( − 1 ) 2 =1$ and (2.16), we have

$e h ( z + 2 η ) − h ( z + η ) = 2 H ( z + η ) H ( z + 2 η ) .$
(2.17)

Set $Q 1 (z)= 2 H ( z + 2 η ) H ( z + η )$. Then, from (2.17), we can know that $Q 1 (z)$ is a nonconstant entire function. Set $σ(H)= σ 1$. Then $σ 1 <σ(f)=k$. By Lemma 2.1, we see that for any given $ε 1$ ($0<3 ε 1 ), there exists a set $E 1 ⊂(1,∞)$ of finite logarithmic measure such that for all z satisfying $|z|=r∉[0,1]∪ E 1$, we have

$exp { − r σ 1 − 1 + ε 1 } ≤| 2 H ( z + η ) H ( z + 2 η ) |≤exp { r σ 1 − 1 + ε 1 } .$
(2.18)

Since $Q 1 (z)$ is an entire function, by (2.18), we have

$T ( r , Q 1 ( z ) ) =m ( r , Q 1 ( z ) ) ≤m ( r , 2 H ( z + η ) H ( z + 2 η ) ) +O(1)≤ r σ 1 − 1 + ε 1 ,$

so that $σ( Q 1 (z))≤ σ 1 −1+ ε 1 . Thus, by $deg(h(z+η)−h(z))=k−1$ and $σ( Q 1 ), we can see that the order of growth of the left-hand side of (2.17) is equal to $k−1$, and the order of growth of the right-hand side of (2.17) is less than $k−1$. This is a contradiction.

Subcase 1.3. Suppose that $n≥3$. Then we can obtain from (2.16) that

$∑ j = 0 n − 2 ( − 1 ) j C n j H ( z + ( n − j ) η ) H ( z + η ) e h ( z + ( n − j ) η ) − h ( z + η ) + ( − 1 ) n − 1 C n n − 1 =0.$
(2.19)

Set $Q 2 (z)= e h ( z + 2 η ) − h ( z + η )$. Then $Q 2 (z)$ is a transcendental entire function since $σ( Q 2 (z))=k−1≥1$. For $j=3,4,…,n$, we have

$e h ( z + j η ) − h ( z + η ) = Q 2 ( z + ( j − 2 ) η ) Q 2 ( z + ( j − 3 ) η ) ⋯ Q 2 (z).$

Thus, (2.19) can be rewritten as

$U 2 ( z , Q 2 ( z ) ) ⋅ Q 2 (z)= ( − 1 ) n C n n − 1 ,$
(2.20)

where

$U 2 ( z , Q 2 ( z ) ) = H ( z + n η ) H ( z + η ) Q 2 ( z + ( n − 2 ) η ) Q 2 ( z + ( n − 3 ) η ) ⋯ Q 2 ( z + η ) − C n 1 H ( z + ( n − 1 ) η ) H ( z + η ) Q 2 ( z + ( n − 3 ) η ) Q 2 ( z + ( n − 4 ) η ) ⋯ Q 2 ( z + η ) + ⋯ + ( − 1 ) n − 2 C n n − 2 H ( z + 2 η ) H ( z + η ) .$

Noting that $( − 1 ) n C n n − 1 ≠0$, we can see that $U 2 (z, Q 2 (z))≢0$. Set $σ(H)= σ 2$. Then $σ 2 . Since $Q 2 (z)$ is of regular growth and $σ( Q 2 (z))=k−1$, for any given $ε 2$ ($0<3 ε 2 ) and all $r> r 0$ (>0), we have

$T ( r , Q 2 ( z ) ) > r k − 1 − ε 2 .$
(2.21)

By Lemma 2.1, we see that for $ε 2$, there exists a set $E 2 ⊂(1,∞)$ of finite logarithmic measure such that for all z satisfying $|z|=r∉[0,1]∪ E 2$, we have

$exp { − r σ 2 − 1 + ε 2 } ≤| H ( z + j η ) H ( z + η ) |≤exp { r σ 2 − 1 + ε 2 } (j=2,3,…,n).$
(2.22)

Thus, from (2.21) and (2.22), we can get that for $j=2,3,…,n$,

that is,

$m ( r , H ( z + j η ) H ( z + η ) ) =S(r, Q 2 )(j=2,3,…,n).$
(2.23)

Noting that $deg Q 2 U(z, Q 2 )=n−2≥1$ and by Lemma 2.4 and Remark 2.1, we have

$T(r, Q 2 )=m(r, Q 2 )=S(r, Q 2 ),$

Case 2. Suppose that $A≠ ( − 1 ) n$. Thus, for a positive integer n, there are two subcases: (1) $n=1$; (2) $n≥2$.

Subcase 2.1. Suppose that $n=1$. Thus, (2.16) can be rewritten as

$H ( z + η ) H ( z ) = ( A − ( − 1 ) n ) e h ( z ) − h ( z + η ) =(A+1) e h ( z ) − h ( z + η ) .$

Noting the $A+1≠0$, we can use the same method as in the proof of Subcase 1.2 and deduce a contradiction.

Subcase 2.2. Suppose that $n≥2$. Then we can obtain from (2.16) that

$∑ j = 0 n − 1 ( − 1 ) j C n j H ( z + ( n − j ) η ) H ( z ) e h ( z + ( n − j ) η ) − h ( z ) + ( − 1 ) n −A=0.$
(2.24)

Set $Q 3 (z)= e h ( z + η ) − h ( z )$. Then $Q 3 (z)$ is a transcendental entire function since $σ( Q 3 (z))=k−1≥1$. For $j=1,2,…,n$, we have

$e h ( z + j η ) − h ( z ) = Q 3 ( z + ( j − 1 ) η ) Q 3 ( z + ( j − 2 ) η ) ⋯ Q 3 (z).$

Thus, (2.24) can be rewritten as

$U 3 ( z , Q 3 ( z ) ) ⋅ Q 3 (z)=A− ( − 1 ) n ,$
(2.25)

where

$U 3 ( z , Q 3 ( z ) ) = H ( z + n η ) H ( z ) Q 3 ( z + ( n − 1 ) η ) Q 3 ( z + ( n − 2 ) η ) ⋯ Q 3 ( z + η ) − C n 1 H ( z + ( n − 1 ) η ) H ( z ) Q 3 ( z + ( n − 2 ) η ) Q 3 ( z + ( n − 3 ) η ) ⋯ Q 3 ( z + η ) + ⋯ + ( − 1 ) n − 1 C n n − 1 H ( z + η ) H ( z ) .$

We can see that $U 3 (z, Q 3 (z))≢0$ since $A− ( − 1 ) n ≠0$. Noting that $deg Q 3 U 3 (z, Q 3 (z))=n−1≥1$, we can use the same method as in the proof of Subcase 1.3 and deduce a contradiction.

Thus, we have proved that $σ(f)=k=1$. And $f(z)$ can be written as

$f(z)=H(z) e c 1 z + c 0 = H ∗ (z) e c 1 z ,$
(2.26)

where $c 0$, $c 1$ (≠0) are two constants and $H ∗ (z)= e c 0 H(z)$ (0) is an entire function and satisfies

$σ ( H ∗ ( z ) ) =λ ( H ∗ ( z ) ) =λ(f)<σ(f)=1.$
(2.27)

Thus, by (2.26), (2.27), (2.16) and Lemma 2.8, we can get that $H ∗ (z)$ is a nonzero constant, and so, $f(z)$ can be written as

$f(z)=c e c 1 z ,$

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 $A≠−1$ in Lemma 2.9 when $n=1$ and Theorem C. Unfortunately, we cannot obtain $A≠ ( − 1 ) n$ when $n≥2$ in Lemma 2.9. This is because we can get a contradiction from the equality $e c 1 η −1=−1$, but we cannot obtain a contradiction from the equality $( e c 1 η − 1 ) n = ( − 1 ) n$ when $n≥2$.

## 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 $Δ η n f(z)$ and $f(z)$ share an entire function $a(z)$ CM, then

$Δ η n f ( z ) − a ( z ) f ( z ) − a ( z ) = ∑ j = 0 n ( − 1 ) n − j C n j H ( z + j η ) e h ( z + j η ) + b ( z ) H ( z ) e h ( z ) = e P ( z ) ,$
(3.1)

where $P(z)$ is a polynomial and $b(z)= Δ η n a(z)−a(z)$. Obviously, $σ(b(z))≤σ(a(z))<1$.

First step. We prove

$Δ η n f ( z ) − a ( z ) f ( z ) − a ( z ) =A,$
(3.2)

where A (≠0) is a constant. If $P(z)≡0$, then, by (3.1), we see that (3.2) holds and $A=1$.

Now suppose that $P(z)≢0$ and $degP(z)=s$. Set

$h(z)= a k z k + a k − 1 z k − 1 +⋯+ a 0 ,P(z)= b s z s + b s − 1 z s − 1 +⋯+ b 0 ,$
(3.3)

where $k=σ(f)≥1$, $a k (≠0), a k − 1 ,…, a 0$, $b s (≠0), b s − 1 ,…, b 0$ are constants. By (3.1), we can see that $0≤degP=s≤degh=k$.

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≤s, respectively.

Case 1. Suppose that $1≤s=k$. Thus, there are two subcases: (1) $a(z)≢0$; (2) $a(z)≡0$.

Subcase 1.1. Suppose that $a(z)≢0$. First we suppose that $b k ≠− a k$. Then (3.1) is rewritten as

$g 11 (z) e P ( z ) + g 12 e − h ( z ) + g 13 e h 0 ( z ) =0,$
(3.4)

where $h 0 (z)≡0$ and

$g 11 (z)=−H(z); g 12 (z)=b(z); g 13 (z)= ∑ j = 0 n ( − 1 ) n − j C n j H(z+jη) e h ( z + j η ) − h ( z ) .$

Since $σ(H), $σ(b)<1≤k$ and $deg(h(z+jη)−h(z))=k−1$ ($j=1,2,…,n$), we can see that $σ( g 1 m (z)) ($m=1,2,3$). On the other hand, by $b k ≠− a k$, 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 $σ( g 1 m ) ($m=1,2,3$), we can see that for $m=1,2,3$,

$T(r, g 1 m )=o ( T ( r , e P − ( − h ) ) ) =o ( T ( r , e P − h 0 ) ) =o ( T ( r , e − h − h 0 ) ) .$
(3.5)

Thus, applying Lemma 2.2 to (3.4), by (3.5), we can obtain $g 1 m (z)≡0$ ($m=1,2,3$). Clearly, this is a contradiction.

Now we suppose that $b k =− a k$. Then (3.1) is rewritten as

$[ H ( z ) e P ( z ) + h ( z ) − b ( z ) ] e − h ( z ) = ∑ j = 0 n ( − 1 ) n − j C n j H(z+jη) e h ( z + j η ) − h ( z ) .$
(3.6)

We affirm that $H(z) e P ( z ) + h ( z ) −b(z)≢0$. In fact, if $H(z) e P ( z ) + h ( z ) −b(z)≡0$, then, by (3.6), we can obtain

$∑ j = 1 n ( − 1 ) n − j C n j H(z+jη) e h ( z + j η ) − h ( z ) + ( − 1 ) n H(z)≡0,$
(3.7)

this is the special case of (2.14) when $b(z)≡0$ 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 $σ(f)=k=1$. Hence, substituting $h(z)= c 1 z+ c 0$ into (3.7), we have

$∑ j = 0 n ( − 1 ) j C n j e ( n − j ) c 1 η H ( z + ( n − j ) η ) =0.$
(3.8)

On this occasion, we assert that $( e c 1 η − 1 ) n =0$. On the contrary, we suppose that $( e c 1 η − 1 ) n ≠0$. Then the sum of all coefficients of (3.8) is $( e η − 1 ) n$, which does not vanish. By Lemma 2.5 and Remarks 2.2-2.3, we have $σ(H)≥1$, a contradiction. Hence, $( e c 1 η − 1 ) n =0$. Thus, $e c 1 η =1$. Substituting it into (3.8), we have

$∑ j = 0 n ( − 1 ) j C n j H ( z + ( n − j ) η ) =0.$
(3.9)

First, we suppose that $H(z)$ is transcendental. Then, noting that $σ(H)<1$ implies $lim ̲ r → ∞ T ( r , H ) r =0$, by Lemma 2.7 and Remark 2.4, we know that $Δ η H(z)=H(z+η)−H(z)$ is transcendental. Moreover, $σ( Δ η H(z))≤σ(H(z))<1$ implies $lim ̲ r → ∞ T ( r , Δ η H ) r =0$. Repeating the process above $n−1$ times, we can see that $Δ η n H(z)$ is transcendental. That is, the left-hand 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 $Δ η n a(z)−a(z)=b(z)$ and by Remark 2.3, we have $σ(a)≥1$, a contradiction. Hence, $H(z) e P ( z ) + h ( z ) −b(z)≢0$. Thus, since $deg(P+h)≤k−1$, $deg(−h)=k$, $deg(h(z+jη)−h(z))=k−1$ ($j=1,2,…,n$) and $σ(H), we see that the order of growth of the left-hand side of (3.6) is equal to k, and the order of growth of the right-hand side of (3.6) is less than k. This is a contradiction.

Subcase 1.2. Suppose that $a(z)≡0$. Then (3.1) is rewritten as

$H(z) e P ( z ) = ∑ j = 0 n ( − 1 ) n − j C n j H(z+jη) e h ( z + j η ) − h ( z ) .$
(3.10)

Since $H(z)≢0$, $σ(H), $degP=s=k$ and $deg(h(z+jη)−h(z))=k−1$ ($j=1,2,…,n$), we can see that the order of growth of the left-hand side of (3.10) is equal to k, and the order of growth of the right-hand side of (3.10) is less than k. This is a contradiction.

Case 2. Suppose that $1≤s. Thus, there are two subcases: (1) $a(z)≢0$; (2) $a(z)≡0$.

Subcase 2.1. Suppose that $a(z)≢0$. Then, by (3.1), we can obtain

$∑ j = 0 n ( − 1 ) n − j C n j H(z+jη) e h ( z + j η ) − h ( z ) −H(z) e P ( z ) =b(z) e − h ( z ) .$
(3.11)

We assert that $b(z)≢0$. In fact, if $b(z)≡0$, then (2.15) obviously holds. Hence, using the same method as in the proof of Lemma 2.9, by Lemma 2.5 and Remarks 2.2-2.3, we can get that $σ(a)≥1$, a contradiction. Hence, $b(z)≢0$. Since $degh=k$, $deg(h(z+jη)−h(z))=k−1$ ($j=1,2,…,n$), $degP=s and $σ(H), we see that the order of growth of the left-hand side of (3.11) is less than k, and the order of growth of the right-hand side of (3.11) is equal to k. This is a contradiction.

Subcase 2.2. Suppose that $a(z)≡0$. Then, by (3.1), we obtain

$∑ j = 1 n ( − 1 ) n − j C n j H ( z + j η ) H ( z ) e h ( z + j η ) − h ( z ) + ( − 1 ) n = e P ( z ) .$
(3.12)

Thus, there are two subcases: (1) $n=1$; (2) $n≥2$.

Subcase 2.2.1. Suppose that $n=1$. Then (3.12) can be rewritten as

$H ( z + η ) H ( z ) e h ( z + η ) − h ( z ) −1= e P ( z ) .$
(3.13)

By (3.13), we see that $H ( z + η ) H ( z )$ is a nonzero entire function. Set $σ(H)= σ 4$. Then $σ 4 <σ(f)=k$. By Lemma 2.1, we see that for any given $ε 4$ ($0<3 ε 4 ), there exists a set $E 4 ⊂(1,∞)$ of finite logarithmic measure such that for all z satisfying $|z|=r∉[0,1]∪ E 4$, we have

$exp { − r σ 4 − 1 + ε 4 } ≤| H ( z + η ) H ( z ) |≤exp { r σ 4 − 1 + ε 4 } .$
(3.14)

Since $H ( z + η ) H ( z )$ is an entire function, by (3.13), we have

$T ( r , H ( z + η ) H ( z ) ) =m ( r , H ( z + η ) H ( z ) ) ≤ r σ 4 − 1 + ε 4 ,$

so that

$σ ( H ( z + η ) H ( z ) ) ≤ σ 4 −1+ ε 4
(3.15)

Since $s, we can see that $degP≤k−1$. If $degP, then, by (3.15) and $deg(h(z+η)−h(z))=k−1$, we can see that the order of growth of the left-hand side of (3.13) is equal to $k−1$, and the order of growth of the right-hand side of (3.13) is equal to degP which is less than $k−1$. This is a contradiction.

If $degP=k−1$, then since $H ( z + η ) H ( z )$ is an entire function and $deg(h(z+η)−h(z))=k−1≥1$, by (3.15), we can see that the entire function $H ( z + η ) H ( z ) e h ( z + η ) − h ( z )$ has a Borel exceptional value 0, thus the value 1 must be not its Borel exceptional value. Hence, the left-hand side of (3.13), $H ( z + η ) H ( z ) e h ( z + η ) − h ( z ) −1$, has infinitely many zeros, but the right-hand side of (3.13), $e P ( z )$, has no zero. This is a contradiction.

Subcase 2.2.2. Now we suppose that $n≥2$. Thus, for s ($=degP$), there are two subcases: (1) $s; (2) $s=k−1$.

Subcase 2.2.2.1. Now we suppose that $s. Set $Q 5 (z)= e h ( z + η ) − h ( z )$. Since $σ( Q 5 )=k−1≥1$, $Q 5 (z)$ is a transcendental entire function. Thus, (3.12) can be rewritten as

$U 5 ( z , Q 5 ( z ) ) ⋅ Q 5 (z)= e P ( z ) − ( − 1 ) n ,$
(3.16)

where

$U 5 ( z , Q 5 ( z ) ) = H ( z + n η ) H ( z ) Q 5 ( z + ( n − 1 ) η ) Q 5 ( z + ( n − 2 ) η ) ⋯ Q 5 ( z + η ) − C n 1 H ( z + ( n − 1 ) η ) H ( z ) Q 5 ( z + ( n − 2 ) η ) Q 5 ( z + ( n − 3 ) η ) ⋯ Q 5 ( z + η ) + ⋯ + ( − 1 ) n − 1 C n n − 1 H ( z + η ) H ( z ) .$
(3.17)

Thus, using the same method as in the proof of Subcase 1.3 in the proof of Lemma 2.9 and noting that $σ( e P ( z ) − ( − 1 ) n )=degP, we have

$m ( r , e P ( z ) − ( − 1 ) n ) =S(r, Q 5 ).$

Noting that $n≥2$ and so $deg U 5 (z, Q 5 )=n−1≥1$. Using the same method as in the proof of Subcase 1.3 in the proof of Lemma 2.9, we can obtain

$T(r, Q 5 )=m(r, Q 5 )=S(r, Q 5 ).$

Subcase 2.2.2.2. Now we suppose that $s=k−1$. Thus, (3.12) is written as

$∑ j = 1 n ( − 1 ) n − j C n j H ( z + j η ) H ( z ) e T j ( z ) + ( − 1 ) n − e P ( z ) =0,$
(3.18)

where $T j (z)=h(z+jη)−h(z)$ ($j=1,2,…,n$). Thus, by (3.3), we have

$T j (z)=jkη a k z k − 1 + P k − 2 , j (z),$
(3.19)

where $P k − 2 , j (z)$ is a polynomial with degree at most $k−2$. Thus, we have

$T j (z)− T t (z)=(j−t)kη a k z k − 1 + P j , t (z)(1≤j≠t≤n),$

where $P j , t (z)$ is a polynomial with degree at most $k−2$.

First, we suppose that there is some $j 0$ ($1≤ j 0 ≤n$) such that $j 0 kη a k = b k − 1$, that is, $deg( T j 0 (z)−P(z))≤k−2$. Thus, (3.18) can be written as

$∑ 1 ≤ j ≤ n , j ≠ j 0 ( − 1 ) n − j C n j H ( z + j η ) H ( z ) e h ( z + j η ) − h ( z ) + B j 0 (z) e h ( z + j 0 η ) − h ( z ) = ( − 1 ) n + 1 ,$
(3.20)

where

$B j 0 (z)= ( − 1 ) n − j 0 C n n − j 0 H ( z + j 0 η ) H ( z ) − e P ( z ) + h ( z ) − h ( z + j 0 η ) .$

Set $Q 6 (z)= e h ( z + η ) − h ( z )$ and $σ(H)= σ 6$. Then (3.20) can be rewritten as

$U 6 ( z , Q 6 ( z ) ) ⋅ Q 6 (z)= ( − 1 ) n + 1 ,$
(3.21)

where

$U 6 ( z , Q 6 ( z ) ) = ∑ 1 ≤ j ≤ n , j ≠ j 0 ( − 1 ) n − j C n n − j H ( z + j η ) H ( z ) Q 6 ( z + ( j − 1 ) η ) Q 6 ( z + ( j − 2 ) η ) ⋯ Q 6 ( z + η ) + B j 0 ( z ) Q 6 ( z + ( j 0 − 1 ) η ) Q 6 ( z + ( j 0 − 2 ) η ) ⋯ Q 6 ( z + η ) ( j 0 ≥ 2 ) ,$
(3.22)

or

$U 6 ( z , Q 6 ( z ) ) = ∑ 2 ≤ j ≤ n ( − 1 ) n − j C n n − j H ( z + j η ) H ( z ) Q 6 ( z + ( j − 1 ) η ) Q 6 ( z + ( j − 2 ) η ) ⋯ Q 6 ( z + η ) + B j 0 ( z ) ( j 0 = 1 ) .$
(3.23)

Noting that $( − 1 ) n + 1 ≠0$, we can see that $U 6 (z, Q 6 (z))≢0$. Since $σ(H) and $σ( e P ( z ) + h ( z ) − h ( z + j 0 η ) )≤k−2, using the same method as in the proof of Subcase 1.3 in the proof of Lemma 2.9, we have

$m ( r , B j 0 ( z ) ) =S(r, Q 6 ).$
(3.24)

Noting that $n≥2$ and so $deg U 6 (z, Q 6 )=n−1≥1$. 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

$T(r, Q 6 )=m(r, Q 6 )=S(r, Q 6 ).$

Secondly, we suppose that $jkη a k ≠ b k − 1$ for any $1≤j≤n$. Thus, equation (3.18) can be rewritten as

$e P ( z ) = e b k − 1 z k − 1 ⋅ e P k − 2 ( z ) = ∑ j = 0 n ( − 1 ) n − j C n j H ( z + j η ) H ( z ) e h ( z + j η ) − h ( z ) ,$
(3.25)

where $P k − 2 (z)=P(z)− b k − 1 z k − 1 = b k − 2 z k − 2 + b k − 3 z k − 3 +⋯+ b 0$. For dealing with equation (3.25), we just compare $| b k − 1 |$ with $nk|η a k |$ since $nk|η a k |>(n−1)k|η a k |>⋯>k|η a k |$. Without loss of generality, we suppose that $nk|η a k |≤| b k − 1 |$. Let $arg b k − 1 = θ 1$, $arg(η a k )= θ 2$ and $σ(H)= σ 7 . Take $θ 0$ such that $cos((k−1) θ 0 + θ 1 )=1$. By Lemma 2.1, we see that for any given $ε 7$ ($0<3 ε 7 ), there exists a set $E 7 ⊂(1,∞)$ of finite logarithmic measure such that for all $z=r e i θ 0$ satisfying $|z|=r∉[0,1]∪ E 7$, we have

$exp { − r σ 7 − 1 + ε 7 } ≤| H ( z + j η ) H ( z ) |≤exp { r σ 7 − 1 + ε 7 } (j=1,…,n).$
(3.26)

Thus, noting that $e P k − 2 ( z )$ is of regular growth, we can deduce from (3.25) and (3.26) that

$| e b k − 1 z k − 1 | = | e P ( z ) e P k − 2 ( z ) | ≤ | ∑ j = 0 n ( − 1 ) j C n j H ( z + ( n − j ) η ) H ( z ) e h ( z + ( n − j ) η ) − h ( z ) | | e b k − 2 z k − 2 + b k − 3 z k − 3 + ⋯ + b 0 | ≤ ( n + 1 ) n ! exp { r σ 7 − 1 + ε 7 } exp { n k | η a k | cos ( ( k − 1 ) θ 0 + θ 2 ) r k − 1 + O ( r k − 2 ) } exp { | b k − 2 | 2 r k − 2 } ,$

that is,

$exp { | b k − 1 | r k − 1 } ≤ exp { n k | η a k | cos ( ( k − 1 ) θ 0 + θ 2 ) r k − 1 + r σ 7 − 1 + ε 7 + O ( r k − 2 ) − | b k − 2 | 2 r k − 2 } ≤ exp { n k | η a k | cos ( ( k − 1 ) θ 0 + θ 2 ) r k − 1 + o ( r k − 1 ) } .$
(3.27)

We assert that

$nk|η a k |cos ( ( k − 1 ) θ 0 + θ 2 ) <| b k − 1 |.$

In fact, if $nk|η a k |=| b k − 1 |$, then, by $b k − 1 ≠nkη a k$, we know that $cos((k−1) θ 0 + θ 2 )≠1$, that is, $cos((k−1) θ 0 + θ 2 )<1$, and hence $nk|η a k |cos((k−1) θ 0 + θ 2 ). If $nk|η a k |<| b k − 1 |$, then we have $nk|η a k |cos((k−1) θ 0 + θ 2 )≤nk|η a k |<| b k − 1 |$.

Thus, taking a positive constant $ε 8$ ($0< ε 8 < | b k − 1 | − n k | η a k | cos ( ( k − 1 ) θ 0 + θ 2 ) 3$), we can deduce from (3.27) that

$exp { | b k − 1 | r k − 1 } ≤ exp { n k | η a k | cos ( ( k − 1 ) θ 0 + θ 2 ) r k − 1 + o ( r k − 1 ) } ≤ exp { ( | b k − 1 | − ε 8 ) r k − 1 } ,$

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).

## Author information

Correspondence to Zong-Xuan Chen.

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The authors declare that they have no competing interests.

### Authors’ contributions

All authors contributed equally to the manuscript and read and approved the final manuscript.

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