Value Distributions and Uniqueness of Difference Polynomials

Abstract

We investigate the zeros distributions of difference polynomials of meromorphic functions, which can be viewed as the Hayman conjecture as introduced by (Hayman 1967) for difference. And we also study the uniqueness of difference polynomials of meromorphic functions sharing a common value, and obtain uniqueness theorems for difference.

1. Introduction

A meromorphic function means meromorphic in the whole complex plane. Given a meromorphic function , recall that , is a small function with respect to , if , where is used to denote any quantity satisfying , as outside a possible exceptional set of finite logarithmic measure. We use notations , to denote the order of growth of and the exponent of convergence of the poles of , respectively. We say that meromorphic functions and share a finite value IM (ignoring multiplicities) when and have the same zeros. If and have the same zeros with the same multiplicities, then we say that and share the value CM (counting multiplicities). We assume that the reader is familiar with standard notations and fundamental results of Nevanlinna Theory [1–3].

As we all know that a finite value is called the Picard exception value of , if has no zeros. The Picard theorem shows that a transcendental entire function has at most one Picard exception value, a transcendental meromorphic function has at most two Picard exception values. The Hayman conjecture [4], is that if is a transcendental meromorphic function and , then takes every finite nonzero value infinitely often. This conjecture has been solved by Hayman [5] for , by Mues [6] for , by Bergweiler and Eremenko [7] for . From above, it is showed that the Picard exception value of may only be zero. Recently, for an analog of Hayman conjecture for difference, Laine and Yang [8, Theorem  2] proved the following.

Theorem A.

Let be a transcendental entire function with finite order and be a nonzero complex constant. Then for , assumes every nonzero value infinitely often.

Remark 1.1.

Theorem A implies that the Picard exception value of cannot be nonzero constant. However, Theorem A does not remain valid for meromorphic functions. For example, , , . Thus, we get that never takes the value −1, and never takes the value 1.

As the improvement of Theorem A to the case of meromorphic functions, we first obtain the following theorem. In the following, we assume that and are small functions with respect of , unless otherwise specified.

Theorem 1.2.

Let be a transcendental meromorphic function with finite order and be a nonzero complex constant. If , then the difference polynomial has infinitely many zeros.

Remark 1.3.

The restriction of finite order in Theorem 1.2 cannot be deleted. This can be seen by taking , , is a nonconstant polynomial, and is a nonzero rational function. Then is of infinite order and has finitely many poles, while

(1.1)

has finitely many zeros. We have given the example when in Remark 1.1 to show that may have finitely many zeros. But we have not succeed in reducing the condition to in Theorem 1.2.

In the following, we will consider the zeros of other difference polynomials. Using the similar method of the proof of Theorem 1.2 below, we also can obtain the following results.

Theorem 1.4.

Let be a transcendental meromorphic function with finite order and be a nonzero complex constant. If , then the difference polynomial has infinitely many zeros.

Theorem 1.5.

Let be a transcendental meromorphic function with finite order and be a nonzero complex constant. If , , then the difference polynomial has infinitely many zeros.

Remark 1.6.

The above two theorems also are not true when is of infinite order, which can be seen by function , , where in Theorem 1.4 and in Theorem 1.5.

Theorem 1.7.

Let be a transcendental meromorphic function with finite order and be a nonzero complex constant. If , , then the difference polynomial has infinitely many zeros.

Corollary 1.8.

There is no transcendental finite order meromorphic solution of the nonlinear difference equation

(1.2)

where and , are rational functions.

Remark 1.9.

Some results about the zeros distributions of difference polynomials of entire functions or meromorphic functions with the condition can be found in [9–12]. Theorem 1.7 is a partial improvement of [11, Theorem  1.1] for is an entire function and is also an improvement of [13, Theorem  1.1] for the case of .

The uniqueness problem of differential polynomials of meromorphic functions has been considered by many authors, such as Fang and Hua [14], Qiu and Fang [15], Xu and Yi [16], Yang and Hua [17], and Lahiri and Rupa [18]. The uniqueness results for difference polynomials of entire functions was considered in a recent paper [15], which can be stated as follows.

Theorem B (see [19, Theorem  1.1]).

Let and be transcendental entire functions with finite order, and be a nonzero complex constant. If , and share CM, then for a constant that satisfies .

Theorem C (see [19, Theorem  1.2]).

Let and be transcendental entire functions with finite order, and be a nonzero complex constant. If , and share 1 CM, then or for some constants and that satisfy and .

In this paper, we improve Theorems B and C to meromorphic functions and obtain the following results.

Theorem 1.10.

Let and be transcendental meromorphic functions with finite order. Suppose that is a nonzero constant and . If , and share 1 CM, then or , where .

Theorem 1.11.

Under the conditions of Theorem 1.10, if , and share 1 IM, then or , where .

Remark 1.12.

Let and , . Thus, and share the value 1 CM. From above, the case , where may occur in Theorems 1.10 and 1.11.

From the proof of Theorem 1.11 and (2.7) below, we obtain easily the next result.

Corollary 1.13.

Let and be transcendental entire functions with finite order, and be a nonzero complex constant. If , and share 1 IM, then or , where .

2. Some Lemmas

The difference logarithmic derivative lemma of functions with finite order, given by Chiang and Feng [20, Corollary  2.5], Halburd and Korhonen [21, Theorem  2.1], plays an important part in considering the difference Nevanlinna theory. Here, we state the following version.

Lemma 2.1 (see [22, Theorem  5.6]).

Let be a transcendental meromorphic function of finite order, and let . Then

(2.1)

for all outside of a set of finite logarithmic measure.

Lemma 2.2 (see [20, Theorem  2.1]).

Let be a transcendental meromorphic function of finite order. Then,

(2.2)

For the proof of Theorem 1.4, we need the following lemma.

Lemma 2.3.

Let be a transcendental meromorphic function of finite order. Then,

(2.3)

Proof.

Assume that , then

(2.4)

Using the first and second main theorems of Nevanlinna theory and Lemma 2.1, we get

(2.5)

thus, we get the (2.3).

In order to prove Theorem 1.5 and Corollary 1.13, we also need the next result.

Lemma 2.4.

Let be a transcendental meromorphic function with finite order, . Then

(2.6)

If is a transcendental entire function with finite order, and , , then

(2.7)

Proof.

We deduce from Lemma 2.1 and the standard Valiron-Mohon'ko [23] theorem,

(2.8)

Thus, (2.6) follows from (2.8). If is entire and , , then from above, we get

(2.9)

Moreover, follows by Lemma 2.2. Thus (2.7) is proved.

Lemma 2.5 (see [17, Lemma  3]).

Let and be two nonconstant meromorphic functions. If and share 1 CM, then one of the following three cases holds:

• (i) ,

• (ii) ,

• (iii) ,

where denotes the counting function of zeros of such that simple zeros are counted once and multiple zeros are counted twice.

For the proof of Theorem 1.11, we need the following lemma.

Lemma 2.6 (see [16, Lemma  2.3]).

Let and be two nonconstant meromorphic functions, and and share 1 IM. Let

(2.10)

If , then

(2.11)

3. Proof of the Theorems

Proof of Theorem 1.2.

Since is a transcendental meromorphic function, assume that , then we can get

(3.1)

Using the second main theorem, we have

(3.2)

So the condition implies that must have infinitely many zeros.

Proof of Theorem 1.7.

Let

(3.3)

We proceed to prove that has infinitely many zeros, which implies that has infinitely many zeros. We first prove that

(3.4)

Applying the first main theorem and Lemma 2.2, we observe that

(3.5)

From (3.5), we easily obtain the inequality (3.4). Concerning the zeros and poles of , we have

(3.6)

(3.7)

Using the second main theorem, Lemma 2.2, (3.6) and (3.7), we get

(3.8)

Since , then (3.8) implies that has infinitely many zeros, completing the proof.

Remark 3.1.

It is easy to know that if , then (3.7) can be replaced by

(3.9)

which implies that in Theorem 1.7.

Proof of Theorem 1.10.

Let and . Thus, and share the value 1 CM. Suppose first that and . From the beginning of the proof of Theorem 1.2, we obtain

(3.10)

Moreover, from Lemma 2.2, it is easy to get

(3.11)

Using the second main theorem, we have

(3.12)

Thus,

(3.13)

Similarly, we obtain

(3.14)

Therefore, from (3.13) and (3.14), follows. From the definition of , we get

(3.15)

Similarly, we can get

(3.16)

(3.17)

(3.18)

Thus,

(3.19)

Then, from (3.10), and (3.19), we have

(3.20)

which is in contradiction with .Therefore, applying Lemma 2.5, we must have either or . If , thus, . Let . Assume that is not a constant. Then we get

(3.21)

Thus, from Lemma 2.2, we get

(3.22)

which is a contradiction with . Hence must be a constant, which implies that , thus, and .

If , implies that

(3.23)

Let , similar as above, must be a constant. Thus , follows; we have completed the proof.

Proof of Theorem 1.11.

Let and , let be defined in Lemma 2.6. Using the similar proof as the proof of Theorem 1.10 up to (3.18), combining with Lemma 2.6 and

(3.24)

we can get

(3.25)

which is in contradiction with . Thus, we get . The following is standard. For the convenience of reader, we give a complete proof here. By integratiing (2.10) twice, we have

(3.26)

which implies . From (3.10)-(3.11), thus,

(3.27)

(3.28)

In the following, we will prove that or .

Case 1 ().

If , then by (3.26), we get

(3.29)

Combining the Nevanlinna second main theorem with Lemma 2.4 and (3.27), we have

(3.30)

This implies , which is in contradiction with . Thus, , hence

(3.31)

Using the same method as above,

(3.32)

which is also a contradiction.

Case 2 (, ).

From (3.26), we have

(3.33)

Similarly, we also can get a contradiction. Thus, follows, implies that . Thus, we get and .

Case 3 (, ).

From (3.26), we obtain

(3.34)

Similarly, we get a contradiction, follows. Thus, we get also implies , . Thus, we have completed the proof.