#### Abstract

We study the value distribution of the difference counterpart of and obtain an almost direct difference analogue of results of Hayman.

#### 1. Introduction and Results

Hayman proved the following Theorem A.

Theorem A (see [1]). If is a transcendental entire function, is an integer, and is a constant, then assumes all finite values infinitely often.

Recently, many papers (see [27]) have focused on complex differences, giving many difference analogues in value distribution theory of meromorphic functions.

It is well known that (where is a constant satisfying is regarded as the difference counterpart of , so that is regarded as the difference counterpart of , where is a constant.

Liu and Laine [7] obtain the following

Theorem B. Let be a transcendental entire function of finite order , not of period , where is a nonzero complex constant, and let be a nonzero function, small compared to . Then the difference polynomial has infinitely many zeros in the complex plane, provided that .

We use the basic notions of Nevanlinna's theory (see [8, 9]) and in addition use to denote the order of growth of the meromorphic and to denote the exponent of convergence of the zeros of .

In this paper, we consider the difference counterpart of Theorem A. When is an integer, we prove the following Theorem 1.1. Compared with Theorem B, Theorem 1.1 is an almost direct difference analogue of of Theorem A and gives an estimate of numbers of -points, namely, for every . Our method of the proof is also different from the method of the proof in Theorem B.

Theorem 1.1. Let be a transcendental entire function of finite order, and let be constants, with such that . Set , where and is an integer. Then assumes all finite values infinitely often, and for every one has .

Example 1.2. For , and , we have Here , which shows that Theorem 1.1 may fail for entire functions of infinite order.

Example 1.3. For , we have . Here , which shows that Theorem 1.1 may fail for and that the condition in Theorem 1.1 is sharp.

Example 1.4. For , we have , which assumes all finite values infinitely often.

What can we say about when ? We consider this question and obtain the following Theorems 1.5 and 1.6.

Theorem 1.5. Let be a transcendental entire function of finite order with a Borel exceptional value 0, and let be constants, with such that . Then assumes all finite values infinitely often, and for every one has .

Theorem 1.6. Let be a transcendental entire function of finite order with a finite nonzero Borel exceptional value , and let be constants, with such that . Then for every with , assumes the value infinitely often, and .

Remark 1.7. From Theorems 1.5 and 1.6, we see that if has the Borel exceptional value 0, then has not any finite Borel exceptional value, but if has a nonzero Borel exceptional value, then may have a finite Borel exceptional value. From Theorem 1.6, this possible Borel exceptional value is . Example 1.3 shows that this Borel exceptional value may arise, and thus the conclusion of Theorem 1.6 is sharp.

#### 2. Proof of Theorem 1.1

We need the following lemmas.

Lemma 2.1 (see [3, 4]). Let be a meromorphic function of finite order, and let . Then where .

Lemma 2.2 (see [3]). Let be a meromorphic function with order , and let be a nonzero constant. Then, for each , one has

Lemma 2.3. Suppose that , , , , satisfy the conditions of Theorem 1.1. If , then is transcendental.

Proof. Suppose that , where is a polynomial. Then By Lemma 2.2, for each , we have where . By an identity due to Valiron-Mohon'ko (see [10, 11]), we have This contradicts the fact that . Hence is transcendental.

Lemma 2.4. Suppose that , , , , satisfy the conditions of Theorem 1.1. Suppose also that , is a polynomial, and is an entire function with . If then

Proof. Suppose that Integrating (2.8) results in where is a constant. Therefore, by (2.6), (2.9), and the definition of , we obtain and so We must have . In fact, if , then by (2.11) and , we have that so is periodic. Then, write (2.8) as where Clearly, and . We obtain from (2.12) and (2.13) If , then by (2.15) and . Thus, by (2.12), we have , which contradicts our condition. If , then by (2.15), we have This is also a contradiction. Hence .
Differentiating (2.11), and then dividing by result in Therefore, by Lemma 2.1, we get that a contradiction for . Hence .

Halburd and Korhonen obtained the following difference analogue of the Clunie lemma [4, Corollary 3.3].

Lemma 2.5. Let be a nonconstant, finite order meromorphic solution of where , are difference polynomials in with small meromorphic coefficients, and let . If the degree of as a polynomial in and its shifts is at most , then for all outside an exceptional set of finite logarithmic measure.

Remark 2.6. If coefficients of , are satisfying , then using the same method as in the proof of Lemma 2.5 (see [4]), we have for all outside an exceptional set of finite logarithmic measure.

We are now able to prove Theorem 1.1. We only prove the case . For the case , we can use the same method in the proof. Suppose that and . Then, by Lemma 2.3, we see that is transcendental. Thus, can be written as where is a polynomial, is an entire function with .

Differentiating (2.22) and eliminating , we obtain where

By Lemma 2.4, we see that . Since and the total degree of as a polynomial in and its shifts, , by (2.23), Lemma 2.5, and Remark 2.6, we obtain that for for all outside of an exceptional set of finite logarithmic measure.

Thus, (2.25) give that for all outside of an exceptional set of finite logarithmic measure. This is a contradiction. Hence has infinitely many zeros and , which proves Theorem 1.1.

#### 3. Proof of Theorem 1.5

We need the following lemma.

Lemma 3.1 (see [12, page 69–70], [13, page 79–80], or [14]). Suppose that , and let , , be meromorphic functions and , , entire functions such that (i); (ii)when , is not constant; (iii)when, ,
where is of finite linear measure or finite logarithmic measure. Then , .

To prove Theorem 1.5, note first that has a Borel exceptional value 0, we can write as where is a constant, is an integer satisfying , and are entire functions such that , , .

First, we prove is transcendental. If , where is a polynomial, then Thus by Lemma 2.1 and an identity due to Valiron-Mohon'ko (see [10, 11]), we have By (3.4), we see that (3.3) is a contradiction.

Secondly, we prove . By the expression of , we have . Set . If , then by (3.2), we have Since and , we see that the left hand side of (3.5) is of order by applying the general form of the Valiron-Mohon'ko lemma in [10], a contradiction. So, .

Thirdly, we prove . If , then can be written as where is a constant, is an entire function satisfying . Thus by (3.2), (3.6), and , we have In (3.7), there are three cases for : (i) and ; (ii); (iii).

Applying Lemma 3.1 to (3.7), we have in case (i) in case (ii), we have in case (iii), we have Since and is transcendental, we see that any one of (3.8)–(3.10) is a contradiction. Hence .

#### 4. Proof of Theorem 1.6

Since has a nonzero Borel exceptional value , we can write as where is a constant, is an integer satisfying , and , are entire functions such that , , .

Using the same method as in the proof of Theorem 1.5, we can show that is transcendental and .

Now we show that . Set . If , then can be written as where is a constant and is an entire function satisfying . Thus by (4.1) and (4.3), we have In (4.4), there are three cases for : (i) and ; (ii); (iii).

Applying the same method as in the proof of Theorem 1.5 to these three cases, we obtain which contradicts our supposition that . Hence .

#### Acknowledgments

This research was supported by the National Natural Science Foundation of China (no. 10871076). The author is grateful to the referee for a number of helpful suggestions to improve the paper.