Theory and Modern Applications

# Uniqueness and value distribution for difference operators of meromorphic function

## Abstract

We investigate the value distribution of difference operator for meromorphic functions. In addition, we study the sharing value problems related to a meromorphic function f (z) and its shift f (z + c).

## 1 Introduction and main results

A meromorphic function means meromorphic in the whole complex plane. We assume that the reader is familiar with standard symbols and fundamental results of Nevanlinna Theory . As usual, the abbreviation CM stands for "counting multiplicities", while IM means "ignoring multiplicities", and we denote the order of meromorphic function f by σ (f). For a non-constant meromorphic function f and a set S of complex numbers, we define the set E(S, f) = aS{z|f(z) - a = 0}, where a zero of f - a with multiplicity m counts m times in E(S, f).

We define difference operator as Δ c f = f (z + c) - f (z), where c is a non-zero constant. In particular, we denote by S (f) the family of all meromorphic functions a (z) that satisfy T(r, a) = S(r, f) = o(T(r, f)), where r → ∞ outside a possible exceptional set of finite logarithmic measure. For convenience, we set Ŝ(f) := S(f) {∞}.

The difference Nevanlinna theory and its applications to the uniqueness theory have become a subject of great interest , recently. With these fundamental results, Heittokangas et al. considered a meromorphic function f (z) sharing values with its shift f(z + c), we recall a key result from .

Theorem A [, Theorem 2]. Let f be a non-constant meromorphic function of finite order, let c , and let a, b, c Ŝ(f) be three distinct periodic functions with period c. If f (z) and f (z + c) share a, b CM and c IM, then f (z) = f (z + c) for all z .

Recently, Yang and Liu and one of the present authors  considered the case F = fn, where f is a meromorphic function, assuming value sharing with F and F (z + c):

Theorem B [, Theorem 1.4]. Let f be a non-constant meromorphic function of finite order, n ≥ 7 be an integer, let c , and let F = fn. If F (z) and F (z + c) share a S(f)\{0} andCM, then f (z) = ωf (z + c), for a constant ω that satisfies ωn= 1.

Next, we consider the problem that related to the Theorem B, and have the following result, where a is a periodic function with period c. However, our proof is different to the one in .

Theorem 1.1. Let f be a non-constant meromorphic function of finite order, let c , and let a S(f) \ {0} be a periodic function with period c. If f (z)nand f(z + c)nshare a andCM, and n ≥ 4 is an integer, then f (z) = ωf (z + c), for a constant ω that satisfies ωn= 1.

Remarks.

1. (1)

Theorem 1.1 is not true, if a = 0. This can be seen by considering $f\left(z\right)={e}^{{z}^{2}}$. Then f (z)nand f (z + c)nshare 0 and ∞ CM, however, f (z) ≠ ωf (z + c), where n is a positive integer.

2. (2)

Theorem 1.1 does not remain valid when n = 1. For example, f (z) = ez+ 1 and f (z + c) = ez+c+ 1, where c ≠ 2πi. Clearly, f(z) and f (z + c) share 1 and ∞ CM, however, f (z) ≠ ωf (z + c) for ωn= 1. Unfortunately, we have not succeeded in reducing the condition n ≥ 4 to n ≥ 2 in Theorem 1.1, and we also cannot give a counterexample when n = 2, 3 at present.

3. (3)

We give an example to show that the restriction of finite order in Theorem 1.1 cannot be deleted. This can be seen by taking $f\left(z\right)={e}^{{e}^{z}},n{e}^{c}=-1$. Then f (z)nand f (z + c)nshare 1 and ∞ CM, however, f(z) ≠ ωf (z + c), where n is a positive integer.

In 1976, Gross asked the following question [, Question 6]:

Question. Can one find (even one set) finite sets S j (j = 1, 2) such that any two entire functions f and g satisfying E(S j , f) = E(S j , g) (j = 1, 2) must be identical?

Since then, many results have been obtained for this and related topics (see ). We recall the following result given by Yi .

Theorem C [, Theorem 1]. Let S1 = {ω | ωn+ n- 1+ b = 0}, where n ≥ 7 is an integer, a and b are two non-zero constants such that the algebraic equation ωn+ n- 1+ b = 0 has no multiple roots. If f and g are two entire functions satisfying E(S1, f) = E(S1, g), then f = g.

Afterwards, Fang and Lahiri  got the result for meromorphic functions.

Theorem D [, Theorem 1]. Let S1 be defined as Theorem C and S2 = {∞}. Assume that f and g are two meromorphic functions satisfying E(S j , f) = E (S j , g) for j = 1,2. If f has no simple poles and n ≥ 7, then f = g.

Next, we give a difference analog of Theorem D that replacing g with f (z + c), and obtain the following result.

Theorem 1.2. Let S1 be defined as Theorem C and S2 = {∞}. Assume that f is a meromorphic function of finite order satisfying E(S j , f) = E(S j , f (z + c)) for j = 1,2. If n ≥ 6 and $\overline{N}\left(r,f\right)<\frac{n-3}{n-1}T\left(r,f\right)+S\left(r,f\right)$, then f (z) = f (z + c) for all z .

We investigate the value distribution of difference polynomials of meromorphic (entire) functions. Let f be a transcendental meromorphic function, and let n be a positive integer. Concerning to the value distribution of fnf", Hayman [, Corollary to Theorem 9] proved that fnf' takes every non-zero complex value infinitely often if n ≥ 3. Mues [, Satz 3] proved that f2f' - 1 has infinitely many zeros. Later on, Bergweiler and Eremenko [, Theorem 2] showed that ff' - 1 has infinitely many zeros also. As an analog result in difference, Laine and Yang  investigated the value distribution of difference products of entire functions, and obtained the following:

Theorem E [, Theorem 2]. Let f be a transcendental entire function of finite order, and let c be a non-zero complex constant. Then for n ≥ 2, f (z)nf (z + c) assumes every non-zero value a infinitely often.

In a recent article, one of the present authors considered the value distribution of f (z)n(f(z) - 1) f (z + c), the result may be stated as follows:

Theorem F [, Theorem 1]. Let f be a transcendental meromorphic function of finite order σ(f), let a ≠ 0 be a small function with respect to f, and let c be a non-zero complex constant. If the exponent of convergence of the poles of f satisfies $\lambda \left(\frac{1}{f}\right)<\sigma \left(f\right)$ and n ≥ 2, then f (z)n(f - 1) f (z + c) - a has infinitely many zeros.

In this article, we replace f (z + c) with Δ c f, and consider the value distribution of f (z)n(f(z) - 1)Δ c f. We get the following results:

Theorem 1.3. Let f be a transcendental meromorphic function of finite order σ(f) and Δ c f ≠ 0, let a ≠ 0 be a small function with respect to f, and let c be a non-zero complex constant. If the exponent of convergence of the poles off satisfies $\lambda \left(\frac{1}{f}\right)<\sigma \left(f\right)$ and n ≥ 3, then f (z)n(f - 1)Δ c f - a has infinitely many zeros.

Corollary 1.4. Let f be a transcendental entire function of finite order and Δ c f ≠ 0, let a ≠ 0 be a small function with respect to f, and let c be a non-zero complex constant. Then for n ≥ 3, f (z)n(f - 1)Δ c f - a has infinitely many zeros.

In particular, if a is a non-zero polynomial in Corollary 1.4, then Corollary 1.4 can be improved.

Theorem 1.5. Let f be a transcendental entire function of finite order and Δ c f ≠ 0, let a be a non-zero polynomial, and let c be a non-zero complex constant. Then for n ≥ 2, f (z)n(f - 1)Δ c f - a has infinitely many zeros.

## 2 Preliminary lemmas

Lemma 2.1. [, Theorem 2.1] Let f be a meromorphic function of finite order, and let c and δ (0, 1) . Then

$m\left(r,\frac{f\left(z+c\right)}{f\left(z\right)}\right)+m\left(r,\frac{f\left(z\right)}{f\left(z+c\right)}\right)=o\left(\frac{T\left(r,f\right)}{{r}^{\delta }}\right)=S\left(r,f\right).$

Chiang and Feng have obtained similar estimates for the logarithmic difference [, Corollary 2.5], and this study is independent from .

Lemma 2.2. [, Lemma 2.3] Let f be a meromorphic function of finite order and c . Then for any small function a S (f) with period c,

$m\left(r,\frac{{\Delta }_{c}f}{f-a}\right)=S\left(r,f\right).$

Lemma 2.3. [, Theorem 2.1] Let f be a meromorphic function of finite order σ (f), and let c be a non-zero constant. Then, for each ε > 0, we have

$T\left(r,f\left(z+c\right)\right)=T\left(r,f\left(z\right)\right)+O\left({r}^{\sigma \left(f\right)-1+\epsilon }\right)+O\left(\text{log}r\right).$

Lemma 2.4. [, Theorem 2.4.2] Let f be a transcendental meromorphic solution of

${f}^{n}A\left(z,f\right)=B\left(z,f\right),$

where A(z, f), B(z, f) are differential polynomials in f and its derivatives with small meromorphic coefficients a λ , in the sense of m(r, a λ ) = S(r, f) for all λ I. If the deg(B(z, f)) ≤ n, then m(r, A(z, f)) = S(r, f).

Lemma 2.5. Let f be a finite order entire function and Δ c f ≠ 0, and let c be a non-zero constant. Then

$m\left(r,f{f}^{\prime }{\Delta }_{c}f\right)\ge T\left(r,f\right)+S\left(r,f\right).$

Proof. Since f is an entire function with finite order, we deduce from Lemma 2.2 and the Lemma of logarithmic derivative that

$\begin{array}{c}3T\left(r,f\right)=T\left(r,{f}^{3}\right)=m\left(r,{f}^{3}\right)+S\left(r,f\right)\\ \le m\left(r,\frac{{f}^{3}}{f{f}^{\prime }{\Delta }_{c}f}\right)+m\left(r,f{f}^{\prime }{\Delta }_{c}f\right)+S\left(r,f\right)\\ =m\left(r,\frac{{f}^{2}}{{f}^{\prime }{\Delta }_{c}f}\right)+m\left(r,f{f}^{\prime }{\Delta }_{c}f\right)+S\left(r,f\right)\\ \le T\left(r,\frac{{f}^{\prime }}{f}\right)+T\left(r,\frac{{\Delta }_{c}f}{f}\right)+m\left(r,f{f}^{\prime }{\Delta }_{c}f\right)+S\left(r,f\right)\\ \le 2N\left(r,\frac{1}{f}\right)+m\left(r,f{f}^{\prime }{\Delta }_{c}f\right)+S\left(r,f\right)\\ \le 2T\left(r,f\right)+m\left(r,f{f}^{\prime }{\Delta }_{c}f\right)+S\left(r,f\right).\end{array}$

Hence, we get

$m\left(r,f{f}^{\prime }{\Delta }_{c}f\right)\ge T\left(r,f\right)+S\left(r,f\right).$
(1)

## 3 Proof of Theorem 1.1

Since f(z)nand f(z + c)nshare a and ∞ CM, we obtain that

$\frac{f{\left(z+c\right)}^{n}-a\left(z+c\right)}{f{\left(z\right)}^{n}-a\left(z\right)}={e}^{Q\left(z\right)},$
(2)

where Q(z) is a polynomial. From Lemma 2.1, we know that T (r, eQ(z)) = m (r, eQ(z)) = S (r, f). Rewrite (2) as

$f{\left(z+c\right)}^{n}={e}^{Q\left(z\right)}\left(f{\left(z\right)}^{n}-a\left(z\right)+a\left(z\right){e}^{-Q\left(z\right)}\right).$
(3)

Set

$G\left(z\right)=\frac{f{\left(z\right)}^{n}}{a\left(z\right)\left(1-{e}^{-Q\left(z\right)}\right)}.$

If eQ(z) 1, then we apply the Valiron-Mohon'ko theorem and the second main theorem to G (z), and get

$\begin{array}{ll}\hfill nT\left(r,f\right)& +S\left(r,f\right)=T\left(r,G\right)\le \overline{N}\left(r,\frac{1}{G}\right)+\overline{N}\left(r,G\right)+\overline{N}\left(r,\frac{1}{G-1}\right)+S\left(r,G\right)\phantom{\rule{2em}{0ex}}\\ \le \overline{N}\left(r,\frac{1}{f}\right)+\overline{N}\left(r,f\right)+\overline{N}\left(r,\frac{1}{f{\left(z\right)}^{n}-a\left(z\right)+a\left(z\right){e}^{-Q\left(z\right)}}\right)+S\left(r,f\right)\phantom{\rule{2em}{0ex}}\\ \le \overline{N}\left(r,\frac{1}{f}\right)+\overline{N}\left(r,f\right)+\overline{N}\left(r,\frac{1}{f\left(z+c\right)}\right)+S\left(r,f\right)\phantom{\rule{2em}{0ex}}\\ \le 2T\left(r,f\right)+T\left(r,f\left(z+c\right)\right)+S\left(r,f\right).\phantom{\rule{2em}{0ex}}\end{array}$
(4)

Combining (4) with Lemma 2.3, we get

$nT\left(r,f\right)\le 3T\left(r,f\right)+O\left({r}^{\sigma \left(f\right)-1+\epsilon }\right)+S\left(r,f\right),$

which contradicts that n ≥ 4. Therefore, eQ(z)≡ 1, that is, f (z)n= f (z + c)n, so we have f (z) = ω f (z + c), for a constant ω with ωn= 1.

## 4 Proof of Theorem 1.2

From the assumption of Theorem 1.2, we get that

$\frac{f{\left(z+c\right)}^{n}+af{\left(z+c\right)}^{n-1}+b}{f{\left(z\right)}^{n}+af{\left(z\right)}^{n-1}+b}={e}^{Q\left(z\right)},$
(5)

where Q(z) is a polynomial. Applying Lemma 2.1, we obtain that T (r, eQ(z)) = m (r, eQ(z)) = S (r, f). Rewrite (5) as

$f{\left(z+c\right)}^{n}+af{\left(z+c\right)}^{n-1}={e}^{Q\left(z\right)}\left(f{\left(z\right)}^{n}+af{\left(z\right)}^{n-1}+b-\frac{b}{{e}^{Q\left(z\right)}}\right).$
(6)

If eQ(z) 1, applying the second main theorem for three small functions, we get

$\begin{array}{ll}\hfill nT\left(r,f\right)+S\left(r,f\right)& =T\left(r,f{\left(z\right)}^{n}+af{\left(z\right)}^{n-1}\right)\phantom{\rule{2em}{0ex}}\\ \le \overline{N}\left(r,\frac{1}{f{\left(z\right)}^{n}+af{\left(z\right)}^{n-1}}\right)+\overline{N}\left(r,f{\left(z\right)}^{n}+af{\left(z\right)}^{n-1}\right)\phantom{\rule{2em}{0ex}}\\ +\overline{N}\left(r,\frac{1}{f{\left(z\right)}^{n}+af{\left(z\right)}^{n-1}+b-\frac{b}{{e}^{Q\left(z\right)}}}\right)+S\left(r,f\right)\phantom{\rule{2em}{0ex}}\\ \le \overline{N}\left(r,f\right)+\overline{N}\left(r,\frac{1}{f{\left(z+c\right)}^{n-1}\left(f\left(z+c\right)+a\right)}\right)\phantom{\rule{2em}{0ex}}\\ +\overline{N}\left(r,\frac{1}{f{\left(z\right)}^{n-1}\left(f\left(z\right)+a\right)}\right)+S\left(r,f\right)\phantom{\rule{2em}{0ex}}\\ \le 3T\left(r,f\right)+2T\left(r,f\left(z+c\right)\right)+S\left(r,f\right).\phantom{\rule{2em}{0ex}}\end{array}$
(7)

Combining (4.3) with Lemma 2.3, we get

$nT\left(r,f\right)\le 5T\left(r,f\right)+O\left({r}^{\sigma \left(f\right)-1+\epsilon }\right)+S\left(r,f\right),$

which contradicts n ≥ 6. Hence, eQ(z)≡ 1, we conclude by (5) that

$f{\left(z+c\right)}^{n}+af{\left(z+c\right)}^{n-1}=f{\left(z\right)}^{n}+af{\left(z\right)}^{n-1}.$
(8)

Set $G\left(z\right)=\frac{f\left(z\right)}{f\left(z+c\right)}$. If G (z) is non-constant, then we have from (8)

$f\left(z\right)=-\frac{aG\left({G}^{n-1}-1\right)}{{G}^{n}-1}=-a\frac{{G}^{n-1}+\cdots +G}{{G}^{n-1}+\cdots +1}.$
(9)

Making use of the standard Valiron-Mohon'ko lemma, we get from (9) that

$T\left(r,f\right)=\left(n-1\right)T\left(r,G\right)+S\left(r,f\right).$
(10)

Noting that n ≥ 6, we deduce that 1 is not a Picard value of Gn. Suppose that a j { \ 1} (j = 1, 2,..., n - 1) are the distinct roots of equation hn- 1 = 0. Applying the second main theorem to G, we conclude by (9) that

$\left(n-3\right)T\left(r,G\right)\le \sum _{j=1}^{n-1}\overline{N}\left(r,\frac{1}{G-{a}_{j}}\right)+S\left(r,G\right)=\overline{N}\left(r,f\right).$
(11)

From (10) and (11), we get $\overline{N}\left(r,f\right)\ge \frac{n-3}{n-1}T\left(r,f\right)+S\left(r,f\right)$, which contradicts the assumption.

So G(z) is a constant, and we get f (z) = tf (z + c), where t is a non-zero constant. From (8), we know t = 1, therefore, f = g.

## 5 Proof of Theorem 1.3

The main idea of this proof is from [, Theorem 1], while the details are somewhat different. For the convenience of the reader, we give a complete proof.

Set F(z) = fn(z) (f(z) - 1)Δ c f. Since f is a transcendental meromorphic function with finite order σ(f), we conclude by Lemma 2.3 that

$\begin{array}{ll}\hfill T\left(r,F\right)& \le T\left(r,{f}^{n}\left(z\right)\left(f\left(z\right)-1\right)\right)+T\left(r,{\Delta }_{c}f\right)+S\left(r,f\right)\phantom{\rule{2em}{0ex}}\\ \le \left(n+2\right)T\left(r,f\right)+T\left(r,f\left(z+c\right)\right)+S\left(r,f\right)\phantom{\rule{2em}{0ex}}\\ \le \left(n+3\right)T\left(r,f\right)+O\left({r}^{\sigma \left(f\right)-1+\epsilon }\right)+S\left(r,f\right).\phantom{\rule{2em}{0ex}}\end{array}$

Thus, we get S(r, F) = o(T(r, f)) = S(r, f). Moreover, we get

$\begin{array}{ll}\hfill T\left(r,{\Delta }_{c}f\right)& \le m\left(r,{\Delta }_{c}f\right)+O\left({r}^{\lambda \left(\frac{1}{f}\right)+\epsilon }\right)+S\left(r,f\right)\phantom{\rule{2em}{0ex}}\\ \le m\left(r,\frac{{\Delta }_{c}f}{f}\right)+m\left(r,f\right)+O\left({r}^{\lambda \left(\frac{1}{f}\right)+\epsilon }\right)+S\left(r,f\right)\phantom{\rule{2em}{0ex}}\\ \le T\left(r,f\right)+O\left({r}^{\lambda \left(\frac{1}{f}\right)+\epsilon }\right)+S\left(r,f\right).\phantom{\rule{2em}{0ex}}\end{array}$
(12)

On the other hand, we deduce by Lemma 2.2 that

$\begin{array}{ll}\hfill \left(n+2\right)T\left(r,f\right)& =T\left(r,{f}^{n+1}\left(f-1\right)\right)+S\left(r,f\right)\phantom{\rule{2em}{0ex}}\\ =m\left(r,{f}^{n+1}\left(f-1\right)\right)+O\left({r}^{\lambda \left(\frac{1}{f}\right)+\epsilon }\right)+S\left(r,f\right)\phantom{\rule{2em}{0ex}}\\ \le m\left(r,\frac{{f}^{n+1}\left(f-1\right)}{F}\right)+m\left(r,F\right)+O\left({r}^{\lambda \left(\frac{1}{f}\right)+\epsilon }\right)+S\left(r,f\right)\phantom{\rule{2em}{0ex}}\\ \le T\left(r,\frac{{\Delta }_{c}f}{f}\right)+m\left(r,F\right)+O\left({r}^{\lambda \left(\frac{1}{f}\right)+\epsilon }\right)+S\left(r,f\right)\phantom{\rule{2em}{0ex}}\\ \le m\left(r,\frac{{\Delta }_{c}f}{f}\right)+N\left(r,\frac{1}{f}\right)+m\left(r,F\right)+O\left({r}^{\lambda \left(\frac{1}{f}\right)+\epsilon }\right)+S\left(r,f\right)\phantom{\rule{2em}{0ex}}\\ \le T\left(r,f\right)+T\left(r,F\right)+O\left({r}^{\lambda \left(\frac{1}{f}\right)+\epsilon }\right)+S\left(r,f\right).\phantom{\rule{2em}{0ex}}\end{array}$

Hence

$\left(n+1\right)T\left(r,f\right)\le T\left(r,F\right)+O\left({r}^{\lambda \left(\frac{1}{f}\right)+\epsilon }\right)+S\left(r,f\right).$
(13)

The second main theorem yields

$\begin{array}{ll}\hfill T\left(r,F\right)& \le \overline{N}\left(r,F\right)+\overline{N}\left(r,\frac{1}{F}\right)+\overline{N}\left(r,\frac{1}{F-a}\right)+S\left(r,F\right)\phantom{\rule{2em}{0ex}}\\ \le \overline{N}\left(r,\frac{1}{F-a}\right)+\overline{N}\left(r,\frac{1}{f}\right)+\overline{N}\left(r,\frac{1}{f-1}\right)\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}+\overline{N}\left(r,\frac{1}{{\Delta }_{c}f}\right)+O\left({r}^{\lambda \left(\frac{1}{f}\right)+\epsilon }\right)+S\left(r,f\right)\phantom{\rule{2em}{0ex}}\\ \le \overline{N}\left(r,\frac{1}{F-a}\right)+2T\left(r,f\right)+T\left(r,{\Delta }_{c}f\right)+O\left({r}^{\lambda \left(\frac{1}{f}\right)+\epsilon }\right)+S\left(r,f\right).\phantom{\rule{2em}{0ex}}\end{array}$

From (12) and above inequality, we get that

$T\left(r,F\right)\le \overline{N}\left(r,\frac{1}{F-a}\right)+3T\left(r,f\right)+O\left({r}^{\lambda \left(\frac{1}{f}\right)+\epsilon }\right)+S\left(r,f\right).$
(14)

Combining (13) and (14), we have

$\left(n-2\right)T\left(r,f\right)\le \overline{N}\left(r,\frac{1}{F-a}\right)+O\left({r}^{\lambda \left(\frac{1}{f}\right)+\epsilon }\right)+S\left(r,f\right),$

which is a contradiction to the fact that f is of order σ (f) if F - a has finitely many zeros. The conclusion follows.

## 6 Proof of Theorem 1.5

Suppose that fn(f - 1)Δ c f - a admits finitely many zeros only. Then, there are two non-zero polynomials P(z), Q(z) such that

${f}^{n}\left(f-1\right){\Delta }_{c}f-a=P\left(z\right){e}^{Q\left(z\right)}.$
(15)

Differentiating (15) and eliminating eQ(z), we obtain

$\left({f}^{n}-{f}^{n-1}\right)F\left(z,f\right)={a}^{\prime }P\left(z\right)-a{P}^{*}\left(z\right)-P\left(z\right)f{\left(z\right)}^{n-1}{f}^{\prime }\left(z\right){\Delta }_{c}f,$
(16)

where

$F\left(z,f\right)=\left(n+1\right)P\left(z\right){f}^{\prime }\left(z\right){\Delta }_{c}f+P\left(z\right)f\left(z\right){\left({\Delta }_{c}f\right)}^{\prime }-{P}^{*}\left(z\right)f\left(z\right){\Delta }_{c}f$

and P*(z) = P'(z) + P(z)Q'(z).

First, we conclude that a'P (z) - aP*(z) 0. Otherwise, if a'P(z) - aP*(z) = 0, by integrating, then we have

$\frac{a}{P\left(z\right)}=A{e}^{Q\left(z\right)},$

where A is a non-zero constant. Hence, we get eQ(z)is a constant and

${f}^{n}\left(z\right)\left(f\left(z\right)-1\right){\Delta }_{c}f=BP\left(z\right)+a,$
(17)

where B is a non-zero constant. Then, from Lemma 2.3 and (17), we obtain that

$\left(n+1\right)T\left(r,f\right)\le 2T\left(r,f\right)+O\left({r}^{\sigma \left(f\right)-1+\epsilon }\right)+S\left(r,f\right),$

which is a contradiction when n ≥ 2.

If F(z, f) vanish identically, then

$a{P}^{*}\left(z\right)+P\left(z\right)f{\left(z\right)}^{n-1}{f}^{\prime }\left(z\right){\Delta }_{c}f-{a}^{\prime }P\left(z\right)\equiv 0.$
(18)

Rewrite (18), we get

${f}^{n-2}f{f}^{\prime }\left(z\right){\Delta }_{c}f=\frac{{a}^{\prime }P\left(z\right)-a{P}^{*}\left(z\right)}{P\left(z\right)},$

hence

${f}^{n-2}{f}^{2}{f}^{\prime }\left(z\right)\frac{{\Delta }_{c}f}{f}=\frac{{a}^{\prime }P\left(z\right)-a{P}^{*}\left(z\right)}{P\left(z\right)}.$
(19)

Then, combining Lemmas 2.2, 2.4 and Equation (19), we conclude that

$m\left(r,f{f}^{\prime }\left(z\right){\Delta }_{c}f\right)=S\left(r,f\right),$

It remains to consider the case that F (z, f) 0. We rewrite (16) in the form that

$\left(f{\left(z\right)}^{n+2}-f{\left(z\right)}^{n+1}\right)\frac{F\left(z,f\right)}{f{\left(z\right)}^{2}}={a}^{\prime }P\left(z\right)-a{P}^{*}\left(z\right)-P\left(z\right)f{\left(z\right)}^{n-1}{f}^{\prime }\left(z\right){\Delta }_{c}f$
(20)

and

$f{\left(z\right)}^{n+1}\left(\left(f\left(z\right)-1\right)\frac{F\left(z,f\right)}{f{\left(z\right)}^{2}}\right)={a}^{\prime }P\left(z\right)-a{P}^{*}\left(z\right)-P\left(z\right)f{\left(z\right)}^{n-1}\frac{{f}^{\prime }\left(z\right){\Delta }_{c}f}{f{\left(z\right)}^{2}}.$

By Lemmas 2.2 and 2.4, we know that

$m\left(r,\frac{F\left(z,f\right)}{f{\left(z\right)}^{2}}\right)=S\left(r,f\right)$

and

$m\left(r,\left(f\left(z\right)-1\right)\frac{F\left(z,f\right)}{f{\left(z\right)}^{2}}\right)=S\left(r,f\right).$

As f (z) is entire, we get that the poles of $\frac{F\left(z,f\right)}{f{\left(z\right)}^{2}}$ may be located only at the zeros of f (z). If $\frac{F\left(z,f\right)}{f{\left(z\right)}^{2}}$ has infinitely many poles, then from that a zero of f (z) with multiplicity t should be a pole of t + 1 of $\frac{F\left(z,f\right)}{f{\left(z\right)}^{2}}$. Since n ≥ 2, we know that the left side of (20) must have infinitely many zeros, which is a contradiction that a'P(z) - aP*(z) is a non-zero polynomial. We get

$N\left(r,\frac{F\left(z,f\right)}{f{\left(z\right)}^{2}}\right)=O\left(\text{log}r\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}N\left(r,\left(f\left(z\right)-1\right)\frac{F\left(z,f\right)}{f{\left(z\right)}^{2}}\right)=O\left(\text{log}r\right).$

Hence

$T\left(r,\frac{F\left(z,f\right)}{f{\left(z\right)}^{2}}\right)=S\left(r,f\right)$

and

$T\left(r,\left(f\left(z\right)-1\right)\frac{F\left(z,f\right)}{f{\left(z\right)}^{2}}\right)=S\left(r,f\right)$

as well. Combining these two estimates, we obtain

$T\left(r,f\right)=S\left(r,f\right)$

contradiction. This completes the proof of Theorem 1.5.

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## Acknowledgements

The authors thank the referee for his/her valuable suggestions to improve the present article. This study was supported by the NSF of Shandong Province, China (No. ZR2010AM030).

## Author information

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Correspondence to Xiaoguang Qi.

### Competing interests

The authors declare that they have no competing interests.

### Authors' contributions

XQ completed the main part of this article, JD and LY corrected the main theorems. All authors read and approved the final manuscript.

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Qi, X., Dou, J. & Yang, L. Uniqueness and value distribution for difference operators of meromorphic function. Adv Differ Equ 2012, 32 (2012). https://doi.org/10.1186/1687-1847-2012-32

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• DOI: https://doi.org/10.1186/1687-1847-2012-32

### Keywords

• Entire Function
• Meromorphic Function
• Finite Order
• Meromorphic Solution
• Small Function 