Open Access
Issue
Wuhan Univ. J. Nat. Sci.
Volume 29, Number 1, February 2024
Page(s) 7 - 12
DOI https://doi.org/10.1051/wujns/2024291007
Published online 15 March 2024

© Wuhan University 2023

Licence Creative CommonsThis is an Open Access article distributed under the terms of the Creative Commons Attribution License (https://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

0 Introduction and Main Results

At the outset, it will be assumed that the standard definitions and terminologies of value distribution theory are known to the readers (see Ref.[1]). In the following, a meromorphic function always means meromorphic in the whole complex plane. By , we denote any quantity satisfying as outside of an exceptional set with finite logarithmic measure. A meromorphic function is said to be a small function of if it satisfies . We say two nonconstant meromrophic functions and share the valueCM (IM) if and have the same zeros counting multiplicities (ignoring multiplicities). Let be a positive integer, we denote by the counting function of a points of with multiplicity and by the counting function of a points of with multiplicity , where each a point is counted according to its multiplicity.

The following three theorems are classical and well-known results in the study of the uniqueness problems of meromorphic functions.

Theorem 1[2] If two nonconstant meromorphic functions and share five distinct values IM, then .

Theorem 2[2] If two nonconstant meromorphic functions and share four distinct values CM, then or , where is a Mobiüs transformation.

Theorem 3 [3] If two nonconstant meromorphic functions and share two values IM, and share two other values CM, then and share all values CM.

The assumption "4CM" in Theorem 2 has been improved to "2IM+2CM" by Gunderson [3]. However, Gunderson [4] gave an example to show that Theorem 2 is not true if the condition "4CM" is replaced by "4IM", and the question whether "4CM" in Theorem 2 can be replaced by "1CM+3IM" still remains open. Many researchers are devoted to studying the open problem. For partial progress on this, we refer the readers to survey [5]. Recently, Wang and Fang[6] obtained some results from another direction.

Theorem 4 [6] Let and be nonconstant meromorphic functions, let be a finite nonzero value, and let be a positive integer satisfying . If and share CM, and share CM, and , then either or , where .

Naturally, a question arises:

Question 1 Can "2CM" be replaced by "1CM+1IM" in Theorem 4?

First, we give a partially answer to Question 1.

Theorem 5   Let and be nonconstant meromorphic functions, and let be a positive integer satisfying . Suppose that and share 1 CM, and share IM, and , then either or , where .

Remark 1   Actually, in the proof of Theorem 5, if we replace and with and , then the condition that and share 1 CM can be improved to and share CM, where is a nonzero finite value.

To reduce the number of shared values quickly, many authors began to consider the case that and have some special relationship. For example, when is the shift of , Heittokangas et al [7] obtained some uniqueness results in 2009. The background for these considerations lies in the development of the difference version to the usual Nevanlinna theory, especially the difference-type logarithmic derivative lemma, which starts in the papers [8-10].

Theorem 6[7] Let be a meromorphic function of finite order and , and let be three distinct periodic functions with period . If and share CM and IM, then .

Following that, several researchers began to study shared values problems related to a meromorphic function and its difference operators, see Refs.[11-19]. Here, we recall the following results from Qi [18] and Chen et al[20].

Theorem 7 [1] Let be a meromorphic function of finite order, let be two finite nonzero values, and let be a positive integer satisfying . If and share CM, and share CM, then , where .

Theorem 8 [20] Let be a meromorphic function of finite order, , and let be a positive integer satisfying . If and share CM, and share CM, then , where , .

Theorem 7   and 8 both require that have finite order. As a result, we pose two questions:

Question 2 What can happen if is of infinite order in Theorem 7-8?

Question 3 Is it possible to widen the range of in Theorem 7-8?

In the following, as an application of Theorem 5, we give two uniqueness results about a meromorphic function related to its shifts and difference operators, which are answers to the two questions.

Theorem 9   Let be a meromorphic function, and let be a positive integer satisfying . If and share 1 CM, and share IM, and , then either , where .

Theorem 10   Let be a meromorphic function, and let be a positive integer satisfying and . If and share 1 CM, and share IM, , then , where .

1 Preliminary Lemmas

To prove our results of this paper, the following lemmas are required.

Lemma 1 [21] Let be a nonconstant meromorphic function in the complex plane. Let and be distinct finite complex numbers. Then

where .

Lemma 2   Let and be nonconstant meromorphic functions with and . Suppose that and share 1 CM, and and share IM. If

(1)

Then .

Proof   Set

(2)

From (2) and the lemma of logarithmic derivative, we have . If , then we see that the possible poles of can occur at the poles of and , and zeros of and . Therefore, we get

(3)

Since , we get

Then from the assumption that and share IM and , we have

If is a 1 point of with multiplicities , then by a short caculation with laurent series and (2) we see that is a zero of . Since and share 1 CM, we know the 1 points of and are not poles of . From (2) and Nevanlinna's first fundamental theorem, it is easy to deduce that

It is a contradiction.

Thus . From (2) we have

(4)

where is a nonconstant. By the assumption that , there exists such that , so .

Lemma 3   Let and be nonconstant meromorphic functions with and . Suppose that and share 1 CM, and and share IM. If is not a Mobiüs transformation of , then

(5)

where denotes the zeros of which are not zeros of .

Proof   Set

(6)

If , then from (6), we have is a Mobiüs transformation of , a contradiction. Thus, , and . First, we know

(7)

From Lemma 2, we deduce that . If not, then is a transformation of , a contradiction. Moreover, by a short calculation with Laurent series, it is obvious that the simple 1 points of must be zeros of .Therefore, we can deduce from (7) that

Next, we need to estimate the poles of . By a calculation with Laurent series, and share 1 CM, and and share ∞IM, we get that the poles of can only occur at the poles of and , the zeros of and , the zeros of which are not zeros of , and the zeros of of which are not zeros of . Hence, we can get

From the assumption that and share IM, , and , so we have

This completes the proof of Lemma 3.

2 Proof of Theorem 5

Proof   Set

(8)

Now we consider the following cases.

Case 1 . By (8) we get

(9)

where is a constant.

If , from (9) we get , and thus , where .

If , by (9) we obtain

(10)

Clearly, this leads to

(11)

By Lemma 1, (10) and the assumption that share IM, we get

Similarly, we have

Combining the above two inequalities, we have

which contradicts the condition .

Case 2 . If has a pole with multiplicity , is a pole of with multiplicity , then is a pole of with multiplicity , and it is also a pole of with multiplicity . By a short calculation, we deduce that is a zero of . Since and share IM, we have

(12)

Clearly, from (8), we see that . On the other hand, since and share 1 CM, by a calculation with Laurent series, we know the 1 points of and are not the poles of . Thus, the poles of can only occur at the zeros of and . Therefore, from (12) and the hypothesis that , we deduce

(13)

Next, we still discuss two cases.

Case 2.1 If is not a Mobiüs transformation of , then from Lemma 3, we get

(14)

and

(15)

Thus, by Lemma 1, we obtain

(16)

Similarly,

(17)

By (13)-(17), we get

It is a contradiction.

Case 2.2 If is a Mobi's transformation of , that is

(18)

where are constants, and .

Next we discuss following two cases.

Case 2.2.1 . Thus . From (18), we have . If , , then . By Nevanlinna's second fundamental theorem and (13), we get

It is a contradiction.

Hence , so . If does not have 1 point, it is easy to get a contradiction by Nevanlinna's second fundamental theorem. So there exists such that . Thus we get , that is , , where .

Case 2.2.2 . If , then from (18) and share IM, we obtain , . By Nevanlinna's second fundamental theorem, we get a contradiction.

So , then . From (18), we have , . It is easy to get and . If , we get , which contradicts with Nevanlinna's second fundamental theorem. So , then . By similar reasoning as in Case 2.2.1, we can get . Thus we get , , where . This completes the proof of Theorem 5.

3 Proof of Theorem 9

By Theorem 5, we get or , where . If ,. From the assumption that , share IM, we obtain and . Otherwise, if or , then there exists such that or , then or . This is a contradiction. Similarly, we have . Thus , where be a nonconstant entire function, and hence . Since , we can deduce that , where is a nonconstant entire function. Hence, we have .

That is,

(19)

Therefore,

(20)

where A is a constant.

By calculation the derivative of (20) and , we get

(21)

We assert that . If , then , where is a constant. Hence, , where be a nonzero constant. By integrating, we can get , , Consequently, we have

(22)

Since is a constant, we can get . This is a contradiction.

Therefore, . From the lemma of logarithmic derivative,

(23)

Thus, is a small function of . From (21) we know , and it is easy to get is a constant by Nevanlinna's second fundamental theorem. Hence is a constant. We can derive a contradiction by similar discussion as above.

Therefore, we get , where .This completes the proof of Theorem 9.

4 Proof of Theorem 10

By Theorem 5, we get or , where . If , that is

(24)

From (24) and the fact , share IM, we get and . Thus , where be a nonconstant entire function. By (24) and , we obtain

(25)

That is

(26)

Since , we easily get , and obviously . Then we get , then , where are constants. This is a contradiction. Thus, we get , where . This completes the proof of Theorem 10.

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