### Articles

Kyungpook Mathematical Journal 2016; 56(3): 763-776

**Published online** October 1, 2016

Copyright © Kyungpook Mathematical Journal.

### Uniqueness of Entire Functions that Share an Entire Function of Smaller Order with One of Their Linear Differential Polynomials

Xiao-Min Li^{1}, Hong-Xun Yi^{2}

Department of Mathematics, Ocean University Of China, Qingdao, Shandong 266100, P. R. China^{1}Department of Mathematics, Shandong University, Jinan, Shandong 250100, P. R. China^{2}

**Received**: December 29, 2011; **Accepted**: March 21, 2014

### Abstract

We prove a uniqueness theorem of entire functions sharing an entire function of smaller order with their linear differential polynomials. The results in this paper improve the corresponding results given by Gundersen-Yang[

**Keywords**: Entire functions, Shared values, Order of growth, Differential polynomials, Uniqueness theorems.

### 1. Introduction and main results

In this paper, by meromorphic functions we will always mean meromorphic functions in the complex plane. We adopt the standard notations in Nevanlinna theory of meromorphic functions as explained, e.g., in [5], [7] and [11]. It will be convenient to let

Let

### Definition 1.1

For a nonconstant entire function _{2}(

and

where and what follows,

In 1977, Rubel-Yang [8] proved that if an entire function _{2}(_{2}(_{2}(

**Theorem A**

([4], Theorem 1)

Later on, Chang-Zhu [3] proved the following result to improve Theorem A:

**Theorem B**

([3], Theorem 1)

Consider the following linear differential polynomial related to

where and what follows, _{0}, _{1}, · · · _{k}_{−1} are complex numbers.

We will prove the following result to improve Theorems A and B:

### Theorem 1.2

L [f ]– a =c (f – a ), where c is some nonzero constant f is a solution of the equation L [f ]– a = (f – a )e ^{p}^{1}^{z}^{+}^{p}^{0}such that σ (f ) =μ (f ) = 1, where not all a _{0}, a _{1}, · · ·,a _{k}_{−1}are zeros, p _{1}≠ 0and p _{0}are complex numbers.

From Theorem 1.2 we get the following corollary:

### Corollary 1.3

^{(}^{k}^{)}^{(}^{k}^{)}

Proceeding as in the proof of Theorem 1.2 in Section 3 of this paper, we get the following theorem:

### Theorem 1.4

From Theorem 1.4 we get the following corollary:

### Corollary 1.5

^{(}^{k}^{)}^{(}^{k}^{)}

### Example 1.6

Let ^{z}^{−}^{z}

### Example 1.7

([3]) Let ^{2}^{z}^{z}^{2}^{z}^{z}^{z}

In 1995, Yi-Yang[12] posed the following question:

### Question 1.8

([12], p.398) Let ^{(}^{n}^{)} and ^{(}^{m}^{)} share the value ^{(}^{n}^{)}?

Regarding Question 1.8, Gundersen-Yang [4] proved the following result:

**Theorem E**

([4], Theorem 2) ^{(}^{n}^{)}^{(}^{n}^{+1)}^{(}^{n}^{)}^{(}^{n}^{+1)}

We will prove the following result to improve and complement Theorem E:

### Theorem 1.9

f (z ) =γ _{1}e ^{z}, where γ _{1}≠ 0is a constant. f (z ) =γ _{2}e ^{cz}– [a (1– c )]/c, where n ≥ 2, γ _{2}≠ 0is a constant. L [f ] =f ′and f ′– a =c (f – a ), where c is some nonzero constant. f (z ) =γ _{3}e ^{cz}, where γ _{3}and c are two nonzero constants.

### 2. Preliminaries

In this section, we introduce some important results that will be used to prove the main results in this paper. First of all we introduce Wiman-Valiron theory. For this purpose, we first introduce the following notions: Let _{n}^{n}_{m}^{m}

### Lemma 2.1

([7], Corollary 2.3.4) ^{(}^{k}^{)}^{(}^{k}^{)}

### Lemma 2.2

([6], Satz 4.5)

### Lemma 2.3

(see [7], Lemma 1.1.2) _{0} > 0 ^{α}_{0}

### Lemma 2.4

([6], Satz 4.4)

### Lemma 2.5

([9]) _{0}^{(}^{k}^{)} + _{1}^{(}^{k}^{−1)} + · · · + _{k}_{0}, _{1}, · · ·, _{k}_{0} ≠ 0,

### Lemma 2.6

([7], Remark of Corollary 2.3.5 or [12], Corollary of Theorem 1.21) ^{(}^{n}^{)})

### Lemma 2.7

([10], Theorem 1.1) ^{cz}

### 3. Proof of Theorems

### Proof of Theorem 1.2

From the condition that

where

as ^{Q}

**Case 1**

Suppose that

Then from (

From the condition that

as

where _{r}^{iθ}^{(}^{r}^{)}, and _{j}^{r}^{iθ}^{(}^{r}^{)} (_{r}_{j}_{r}

as _{rn} = _{n}^{iθ(rn)} satisfying _{n}_{rn})|, where ^{+} is a subset with logarithmic measure

and

Since

from (

as _{n}

as _{n}

where _{0}_{1}, · · ·_{m}_{−1}_{m}_{m}_{m}^{m}_{0}_{0} is a sufficiently large positive number. Combining this with (

as |

i.e.,

as _{n}

as _{n}

On the other hand, by Lemma 2.4 we know that

Since |_{j}_{1} for all nonnegative integers _{1} > 0, we get from (

where _{1} > 0 is a suitable constant. From Lemma 2.2 and the condition

as _{n}

as _{n}

From (

From (

**Case 2**

Suppose that

Then from (

We discuss the following two subcases.

### Subcase 2.1

Suppose that

By (_{rn} = _{n}^{iθ(rn)} satisfying _{n}_{rn})|, where ^{+} is a subset with logarithmic measure

### Subcase 2.2

Suppose that

we will prove

Suppose that

Then from (^{Q}^{Q}

where _{rn} = _{n}^{iθ(rn)} satisfying _{n}_{rn})|, where ^{+} is a subset with logarithmic measure

as _{n}

as _{n}_{0}, where _{0} = (1_{0} is a sufficiently large positive number. From (

From (

By (

If ^{(}^{k}^{)} + _{k}_{−1}^{(}^{k}^{−1)} + · · · + _{1}

By Lemma 2.6 we know that

From (_{rn} = _{n}^{iθ(rn)} satisfying _{n}_{rn})|, where ^{+} is a subset with logarithmic measure

and

From (

as _{n}

From (

which implies _{1} = 0. Similarly we have _{j}^{(}^{k}^{)} = (1^{(}^{k}^{)}) =

From (^{Q}

where _{1} ≠ 0 and _{0} are complex numbers. First of all, we will prove

In fact, from (

Suppose that

From (_{n}

From (

From (

Since ^{Q}^{Q}^{Q}_{0}_{1}_{k}_{−2} and _{k}_{−1} are zero. Assume the contrary, i.e., suppose that _{j}

From Definition 1.1, Lemma 2.3 Definition 1.1.1 and Theorem 1.1.3 from [13], and the assumption _{rn} = _{n}^{iθ(rn)} satisfying _{n}_{rn})|, where ^{+} is a subset with logarithmic measure

as _{n}

as _{n}_{1} ≠ 0 we get a contradiction. Thus from (

### Proof of Theorem 1.3

From Theorem B and the assumptions of Theorem 1.3 we know that there exists some nonzero constant _{1} such that

where _{1} is a nonzero constant. If _{1} = 1, from (_{1} ≠ 1. From Theorem 1.2 we know that

**Case 1**

Suppose that there exists some nonzero constant _{2} such that

From (

where _{0}_{1}_{k}_{−3}_{k}_{−2} are constants. By substituting (

If

From (

From (

Suppose that

From (

Suppose that

From (_{1}(

By substituting (

If _{1} +_{0} −_{2} ≠ 0, from (_{1} −_{2})_{1} +_{0} −_{2}), from which we get _{1}+_{0}−_{2} = 0, and so it follows from (_{1} = _{2} and _{0} = 0. Thus (

**Case 2**

Suppose that

such that _{1} (≠ 0) and _{0} are two finite complex numbers.

Suppose that

From (

If _{1} + _{0} = 0, then (

If _{1}_{1}+_{0}), then (^{p}^{1}^{z}^{+}^{p}^{0} = _{1}+_{0}, which is impossible. Next we suppose that

From (_{1}_{1}+_{0}) share 0 CM. From the (_{1}_{1} + _{0}) ≠ 1. Thus by (

and so we have

From (

### Acknowledgements

The authors wish to express their thanks to the referee for his/her valuable suggestions and comments.

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