Abstract and Applied Analysis

Volume 2013, Article ID 240369, 9 pages

http://dx.doi.org/10.1155/2013/240369

## Uniqueness of Entire Functions concerning Difference Operator

College of Mathematics Science, Chongqing Normal University, Chongqing 401331, China

Received 1 July 2013; Accepted 27 September 2013

Academic Editor: Marco Sabatini

Copyright © 2013 Chun Wu. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

We deal with a uniqueness question of entire functions sharing a nonzero value with their difference operators and obtain some results, which improve the results of Qi et al. (2010) and Zhang (2011).

#### 1. Introduction and Main Results

In this paper, a meromorphic function will mean meromorphic in the whole complex plane. We will use the standard notations of Nevanlinna’s value distribution theory such as , , , and , as explained in Hayman [1], Yang [2], and Yang and Yi [3]. We denote by any quantity satisfying , as possibly outside a set of finite linear measures. For meromorphic in , denote by the family of all meromorphic functions that satisfy for outside a possible exceptional set of finite linear measure. In addition, we denote by and the order of and the hyperorder of [3, 4]. Moreover, we define difference operators by where is a nonzero constant. If , we use the usual difference notation .

Let and be two nonconstant meromorphic functions and be a finite complex number. We say that , share the value CM (counting multiplicities) if , have the same -points with the same multiplicities, and we say that , share the value IM (ignoring multiplicities) if we do not consider the multiplicities. We denote by the counting function for -points of both and about which has larger multiplicity than , with multiplicity not being counted. Similarly, we have the notation . Next, we denote by the counting function of those zeros of that are not the zeros of and denote by the counting function for common simple 1-point of both and . In addition, we need the following three definitions.

*Definition 1. *Let be a positive integer. Let and be two nonconstant meromorphic functions such that and share the value 1 IM. Let be a 1-point of with multiplicity and a 1-point of with multiplicity . We denote by the reduced counting function of those 1-points of and such that . is defined analogously.

*Definition 2 (see [5]). *Let be a nonnegative integer or infinity. For , we denote by the set of all -points of , where an -point of multiplicity is counted times if and times if . If , we say that share the value a with weight .

The definition implies that if , share a value a with weight , then is an -point of with multiplicity if and only if it is an -point of with multiplicity and is an -point of with multiplicity if and only if it is an -point of with multiplicity , where is not necessarily equal to .

We write that , share to mean that , share the value a with weight . Clearly if , share , then , share for any integer , . Also we note that , share a value IM or CM if and only if , share or , respectively.

*Definition 3. *Let be a nonconstant meromorphic function, and let be a positive integer and . Then, by , we denote the counting function of those -points of (counted with proper multiplicities) whose multiplicities are not greater than , and by , we denote the corresponding reduced counting function (ignoring multiplicities). By , we denote the counting function of those -points of (counted with proper multiplicities) whose multiplicities are not less than , and by , we denote the corresponding reduced counting function (ignoring multiplicities), where , , , and mean , , , and , respectively, if .

In 2010, Qi et al. [6] proved the following uniqueness theorem.

Theorem A. *Let and be transcendental entire functions of finite order, let be a nonzero complex constant, and let be an integer. If and share CM, then for a constant that satisfies .*

In 2011, Zhang et al. [7] complemented the above theorem and obtained the following result.

Theorem B. *Let and be nonconstant entire functions of finite order, and let be an integer. Suppose that is a nonzero complex constant such that and . If and share CM, and and share 0 CM then , where is a constant satisfying .*

In this paper, we complement Theorems A and B and obtain the following results which generalize the above theorems.

Theorem 4. *Let be a transcendental entire function of finite order and , let be a small function with respect to , and let be a nonzero complex constant. Then for , has infinitely many zeros.*

Theorem 5. *Let and be transcendental entire functions of , . Suppose that is a nonzero complex constant such that and . If and share 1 CM, then for a constant with .*

Theorem 6. *Let and be transcendental entire functions of , . is a nonzero complex constant such that and . If and share 1 IM, then for a constant with .*

#### 2. Some Lemmas

Lemma 7 (see [8]). *Let be a nonconstant meromorphic function of finite order , and let be a nonzero constant. Then, for each ,
*

Lemma 8 (see [9]). *Let be a meromorphic function of finite order, and let and . Then
*

Lemma 9 (see [10]). *Let , , and be nonconstant meromorphic functions such that . If , , and are linearly independent, then
**
where , , and denote a set of positive real numbers of finite linear measure.*

Lemma 10. *Let be transcendental entire functions of finite order, let be a nonzero complex constant, and set ; then
*

*Proof. *Since
then
That is,

Lemma 11 (see [11]). *Let and be two nonconstant meromorphic functions. If , where , , and are nonzero constants, then
*

Lemma 12 (see [12]). *Let be a nonconstant meromorphic function, and let be a positive integer. Suppose that ; then
*

Lemma 13 (see [13]). *Let , share . Then*(i)*,*(ii)*.*

Lemma 14. *Let and be two nonconstant entire functions. If and share 1 IM, then one of the following cases holds: *(i)*, the same inequality holding for ;*(ii)*, where A, B, C, and D are finite complex numbers satisfying .*

*Proof. *Let
Clearly . We consider the cases and .

If , then if is a common simple 1-point of and , substituting their Taylor series at into (10), we see that is a zero of . Thus, we have

Our assumptions are that has poles; all are simple only at zeros of and and poles of and , and 1-points of whose multiplicities are not equal to the multiplicities of the corresponding 1-points of . Thus, we deduce from (10) that
where is the counting function which only counts those points such that , but . By the second fundamental theorem, we have
since
Thus, we deduce from (11)–(14) that

From the definition of , we see that
The above inequality and Lemma 12 give
Substituting (17) in (15), we get
since
Similarly,

Combining the above inequalities, Lemma 13, and (18), we obtain
Thus, we obtain (i).

If , then by (10), we have
By integrating two sides of the above equality, we obtain
where , , , and are finite complex numbers satisfying . This proves the lemma.

Lemma 15 (see [14]). *Let be a nonconstant meromorphic function, , be two positive integers; then **
Clearly, .*

Lemma 16 (see [15]). *Let be polynomials such that ; let be constants and
**
where . If is a transcendental meromorphic solution of
**
then .*

#### 3. Proof of Theorems

##### 3.1. Proof of Theorem 4

Let . Since is a transcendental entire function of finite order, from Lemma 7, we have By the second main theorem, we deduce that According to (27) and (28), we have Noting that , we get that has infinitely many zeros.

This completes the proof of Theorem 4.

##### 3.2. Proof of Theorem 5

Since and share 1 CM, we have where is a polynomial. Set , ,

Next, we will prove that , , and are linearly dependent and either or is a constant.

Now, we suppose that neither nor is a constant and , , and are linearly independent; then by Lemma 9, we have Since are entire functions, by the above inequality, we get From (33) and the first main theorem, we have Assuming that is zero of (or ) with multiplicity , if is zero of (or ) with multiplicity , let , then is a zero of (or ) with multiplicity , and if is not zero of (or ), then is a zero of (or ) with multiplicity . Therefore, we get that since Therefore, from (34), (35), (36), (37), and Lemma 12, On the other hand, from (30), we have . Obviously, according to our assumptions, neither nor is a constant and , , and are linearly independent. Similarly, we have From (38) and (39), we obtain that which is a contradiction to .

Therefore, , , and are linearly dependent, and there exist constants , , which are not all equal to zero such that

Suppose that ; we have . If , we get ; that is, ; thus is a polynomial; it is impossible. Similarly, if , we also deduce a contradiction.

Suppose that , from (41); we know that . If , from (41), we have and , . That is, From Lemma 11, we have By the similar argument in (37), we have

From , if is zero of with multiplicity , then is a zero of with multiplicity , and we get According to (44), (45), and (46), we have which is a contradiction to .

Therefore, , , which gives . Similarly, we derive a contradiction by calculation.

Hence, we deduce that either or is a constant.

Suppose ; from , we have ; in the same manner as above, we get a contradiction. Therefore, ; that is, . Suppose ; similarly as above, we get ; that is, .

Therefore, we conclude that or .

If , since , we have . That is Since and and are transcendental entire functions with hyperorder less than one, we get that and have no zeros. Thus, where , are nonzero polynomials.

Substitute (49) into (48); we have Let , , , and . If , we have From (51), we know that ; If , then we have ; thus, must be a constant. By Lemma 16, we have ; thus, , which is a contradiction. If , then must be a constant; similarly, we also deduce a contradiction.

If , by calculation, we have If , then . If is transcendental entire, then we have which is a contradiction to being transcendental entire. If is a polynomial, from Lemma 16, which induces that , we get a contradiction. If , similar as above, we get a contradiction. For , using the similar Method as above, we also deduce a contradiction. Therefore, There are not transcendental entire functions and satisfying (48).

If , that is, , from (30), we get From (54), we have where is a polynomial of degree at most . Suppose ; then we get Therefore, from the second main theorem, we have Similarly, we have Therefore, which is a contradiction to . Thus, , which implies that

Let ; if is not a constant, then by (60), we have Thus, Combining , we obtain +, which is impossible.

Therefore, is a constant; then substituting into (60), we have . Hence , where is a constant and .

The proof of Theorem 5 is complete.

##### 3.3. Proof of Theorem 6

Let Then and share 1 IM, and , . By Lemma 10, we have Since is transcendental entire, by the definition of , we have Using the argument in (35), we have It follows from Lemma 12 and (66), (67), we have From Lemma 15, we have

Similarly, By Lemma 14, one of the following cases holds: (i), the same inequality holding for ;(ii). For case (i), we have Therefore, we get By (64) and Lemma 15, we have Similarly, By (70), (72), (73), and (74), we obtain which is a contradiction since .

For case (ii), we have where , , , and are finite complex numbers satisfying . Therefore, by the first fundamental theorem, .

Next, we consider three cases.

*Case 1. *; from (76), we get
By the second fundamental theorem and (69), we have
From (73), we obtain , contradicting to .

*Case 2. *, and . Then, .

If , by the second fundamental theorem and (69), (70), we have
Similarly,
From (73), (74), (79), and (80), we get
which is a contradiction to .

If , then . If , then ; that is, ; using the argument in (54) and noting that , we obtain , where is a constant and . If , by the condition that and share 1 IM, then and . we obtain then and . By the second fundamental theorem, we have
which is impossible.

*Case 3. *, and . Then, .

If , by the second fundamental theorem and (69), (70), we have
Similarly,
From (73), (74), (83), and (84), we get
which is a contradiction to .

If , then . If , then ; using the argument in (48) in Theorem 5 and noting that , we get a contradiction. If , by the condition that and share 1 IM, we obtain and . By the second fundamental theorem, we have
which is impossible.

The proof of Theorem 6 is complete.

#### Acknowledgments

This work is supported by Youth Foundation of Chongqing Normal University (12XWQ17), partially by Chongqing City Board of Education Science, and Technology Project (KJ130632), and partially by the fund of Chongqing Normal University (13XLB006).

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