Xiaohuang Huang, Bingmao Deng and Mingliang Fang

# Entire functions that share two pairs of small functions

De Gruyter | Published online: May 3, 2021

# Abstract

In this paper, we study the unicity of entire functions and their derivatives and obtain the following result: let f be a non-constant entire function, let a 1 , a 2 , b 1 , and b 2 be four small functions of f such that a 1 b 1 , a 2 b 2 , and none of them is identically equal to . If f and f ( k ) share ( a 1 , a 2 ) CM and share ( b 1 , b 2 ) IM, then ( a 2 b 2 ) f ( a 1 b 1 ) f ( k ) a 2 b 1 a 1 b 2 . This extends the result due to Li and Yang [Value sharing of an entire function and its derivatives, J. Math. Soc. Japan. 51 (1999), no. 7, 781–799].

MSC 2010: 30D35; 39A32

## 1 Introduction and main results

In this paper, we use the general notations in the Nevanlinna value distribution theory, see ([1,2,3]).

Let f be a meromorphic function on the whole complex plane. For 0 < r < R , we define the following functions:

m ( r , f ) = 1 2 π 0 2 π log + f ( r e i θ ) d θ
is the average of the positive logarithmic f ( z ) on the circle z = r ;
N ( r , f ) = 0 r n ( t , f ) n ( 0 , f ) t d t + n ( 0 , f ) log r
is called the counting function of poles of f ( z ) , where n ( r , f ) denotes the number of poles of f ( z ) on the disc z t , multiple poles are counted according to their multiplicities, and n ( 0 , f ) denotes the multiplicity of pole of f ( z ) at the origin (if f ( 0 ) , then n ( 0 , f ) = 0 ). We denote N ¯ ( r , f ) the reduced counting function of f ( z ) whose the multiplicity of poles only counts once.
T ( r , f ) = m ( r , f ) + N ( r , f ) .
T ( r , f ) is said to be the characteristic function of f ( z ) which is obviously a non-negative function.

Let a be a complex number. Obviously, 1 f ( z ) a is meromorphic on the disc z R . Similar to the above definitions, we define the following functions.

m r , 1 f a = 1 2 π 0 2 π 1 log + f ( r e i θ ) a d θ
is the average of the positive logarithmic 1 f ( z ) a on the circle z = r .
N r , 1 f ( z ) a = 0 r n t , 1 f ( z ) a n 0 , 1 f ( z ) a t d t + n ( 0 , f ) log r ,
where n r , 1 f ( z ) a denotes the number of zeros of f ( z ) a on the disc z t , multiple poles are counted according to their multiplicities. n 0 , 1 f ( z ) a denotes the multiplicity of zeros of f ( z ) a at the origin. We denote N ¯ r , 1 f a the reduced counting function of f ( z ) a whose the multiplicity of zero only counts once.
T r , 1 f ( z ) a = m r , 1 f ( z ) a + N r , 1 f ( z ) a .
T r , 1 f ( z ) a is said to be the characteristic function of 1 f ( z ) a .

Obviously, T ( r , f ) is a non-decreasing function of r , and a convex function of log r .

We denote S ( r , f ) = o ( T ( r , f ) ) , as r outside of a possible exceptional set of finite linear measures. We say that a meromorphic function a is called a small function of f if it satisfies T ( r , a ) = S ( r , f ) .

Nevanlinna established two fundamental theorems, called the first fundamental theorem and the second fundamental theorem.

The first fundamental theorem. Let f be a meromorphic function on the whole complex plane. Then,

T ( r , f ) = T r , 1 f a + O ( 1 ) .

The second fundamental theorem. Let f be a meromorphic function on the whole complex plane and let a 1 , , a q ( q 3 ) be distinct complex values in the extended plane. Then,

( q 2 ) T ( r , f ) j = 1 q N ¯ r , 1 f a j + S ( r , f ) .

Let f and g be two meromorphic functions, and let a and b be two small functions of f and g . We say that f and g share a pair of small functions ( a , b ) CM (resp. IM) if f a and g b have the same zeros counting multiplicities (resp. ignoring multiplicities). When a = b , we say that f and g share a CM (resp. IM) if f a and g a have the same zeros counting multiplicities (resp. ignoring multiplicities).

In , Nevanlinna proved the following famous five-value theorem:

## Theorem A

Let f and g be two non-constant meromorphic functions, and let a j ( j = 1 , 2 , 3 , 4 , 5 ) be five distinct values in the extended complex plane. If f and g share a j ( j = 1 , 2 , 3 , 4 , 5 ) IM, then f g .

Brosch , Czubiak and Gundersen , Gundersen , Steinmetz , and Gundersen et al.  studied shared pairs of values. For five shared pairs of values, Gundersen et al.  proved the next result, which is the best possible:

## Theorem B

If f and g are two non-constant meromorphic functions that share two pairs of values CM and share three other pairs of values IM, then f is a Möbius transformation of g .

Li and Qiao  proved that Theorem A is still valid for five distinct small functions, and they proved the following:

## Theorem C

Let f and g be two non-constant meromorphic functions, and let a j ( j = 1 , 2 , 3 , 4 , 5 ) (one of them can be ) be five distinct small functions of f and g . If f and g share a j ( j = 1 , 2 , 3 , 4 , 5 ) IM, then f g .

Zhang and Yang  and Nguyen and Si  studied shared pairs of small functions.

Rubel and Yang  investigated the uniqueness of an entire function and its derivative and proved.

## Theorem D

Let f be a non-constant entire function, and let a , b be two finite distinct complex values. If f and f share a , b CM, then f f .

Zheng and Wang  improved Theorem D and proved the following: if f is a non-constant entire function, and f and f ( k ) share a , b CM, where a and b are two small functions of f , then f f ( k ) .

Li and Yang  improved the result of Zheng and Wang  and proved the following result:

## Theorem E

Let f be a non-constant entire function, and let a , b be two distinct small functions of f . If f and f ( k ) share a CM, and share b IM, then f f ( k ) .

In this paper, we study shared pair of small functions and extend Theorem E as follows.

## Theorem 1

Let f be a non-constant entire function, let a 1 , a 2 , b 1 , and b 2 be four small functions of f such that a 1 b 1 and a 2 b 2 , and none of them is identically equal to . If f and f ( k ) share ( a 1 , a 2 ) CM, and share ( b 1 , b 2 ) IM, then ( a 2 b 2 ) f ( a 1 b 1 ) f ( k ) a 2 b 1 a 1 b 2 .

## Remark 1

If a 1 a 2 , b 1 b 2 , then by Theorem 1, we obtain Theorem F.

The following example shows that the conclusion of Theorem 1 is not valid for meromorphic functions.

## Example 1

 Let f = h + h g q 1 , where

g = 1 3 e 2 z + 1 2 e z , h = 1 2 e z 1 3 e 2 z , q = e e z ,
and let a = h , b = g . By simple calculation, we obtain
T ( r , a ) = S ( r , f ) , T ( r , b ) = S ( r , f ) ,
f a = e 2 z ( f a ) ( f h ) , f b = e 2 z ( f b ) ( f g ) .
So f and f share ( a , a ) CM and ( b , b ) IM. However, f f .

The following example shows that there exist a transcendental entire function f and two pairs of small functions of f satisfying Theorem 1.

## Example 2

 Suppose f = e 3 z + 10 z , a 1 = 11 z , a 2 = 9 z , b 1 = 10 z , b 2 = 0 . Then it is an easy work to obtain

T ( r , a i ) = S ( r , f ) , T ( r , b i ) = S ( r , f )
for i = 1 , 2 , and
f a 1 = e 3 z z , f b 1 = 9 ( e 3 z z ) ,
f a 2 = e 3 z , f b 2 = 9 e 3 z .
Thus, we see that f and f share ( a 1 , b 1 ) CM and ( a 2 , b 2 ) IM. Furthermore, ( a 2 b 2 ) f ( a 1 b 1 ) f a 2 b 1 a 1 b 2 .

In 2020, Sahoo-Halder  proved.

## Theorem F

Let f be a non-constant entire function of finite order, let f S k ( a 1 , a 2 , b 1 , b 2 ) , and let a 1 , a 2 , b 1 and b 2 be four small functions of f such that a 1 b 1 and a 2 b 2 , none of them is identically equal to , and a i b i have at least one zero for i = 1 , 2 . If f and f ( k ) share ( a 1 , a 2 ) and ( b 1 , b 2 ) IM, then ( a 2 b 2 ) f ( a 1 b 1 ) f ( k ) a 2 b 1 a 1 b 2 .

The authors raised two questions as follows.

## Question 1

The condition of finite order of Theorem F and Corollary 1 in  can be removed in any way?

## Question 2

How far do the conclusions in Theorem F and Corollary 1 in  hold for a non-constant entire function?

In Theorem 1, we answer Question 1 in a setting with stronger conditions, and we answer that for any positive integers k Theorem F and Corollary 1 in  hold.

## Lemma 2.1

[1,2,3] Let f be a non-constant meromorphic function, and let k be a positive integer. Then

m r , f ( k ) f = S ( r , f ) .

## Lemma 2.2

 Let f be a meromorphic function on the complex plane, let n be a positive integer, and let ψ ( f ) = a 0 f + a 1 f + + a n f ( n ) , where a 0 , a 1 , , a n ( 0 ) are small functions of f . Then

m r , ψ ( f ) f = S ( r , f ) , T ( r , ψ ) T ( r , f ) + k N ¯ ( r , f ) + S ( r , f ) .

## Lemma 2.3

[1,2,3] Let f 1 , f 2 be two non-constant meromorphic functions, then

N ( r , f 1 f 2 ) N r , 1 f 1 f 2 = N ( r , f 1 ) + N ( r , f 2 ) N r , 1 f 1 N r , 1 f 2 .

## Lemma 2.4

Let f be a transcendental meromorphic function, let a 1 , a 2 , b 1 , and b 2 be four small functions of f such that a 1 b 1 and a 2 b 2 , and let

L ( f ) = f a 1 a 1 b 1 f a 1 a 1 b 1 , L ( f ( k ) ) = f ( k ) a 2 a 2 b 2 f ( k + 1 ) a 2 a 2 b 2 .
If f and f ( k ) share ( a 1 , a 2 ) CM, share ( b 1 , b 2 ) IM, then L ( f ) 0 , L ( f ( k ) ) 0 .

## Proof

Suppose that L ( f ) 0 . Then we get f a 1 f a 1 a 1 b 1 a 1 b 1 . It follows that f a 1 = c ( a 1 b 1 ) , where c is a nonzero constant. So T ( r , f ) = T ( r , c ( a 1 b 1 ) + a 1 ) = S ( r , f ) , a contradiction. Hence, L ( f ) 0 .

Since f is an entire function, f and f ( k ) share ( a 1 , a 2 ) CM, share ( b 1 , b 2 ) IM, we have

T ( r , f ) N ¯ r , 1 f a 1 + N ¯ r , 1 f b 1 + S ( r , f ) = N ¯ r , 1 f ( k ) a 2 + N ¯ r , 1 f ( k ) b 2 + S ( r , f ) 2 T ( r , f ( k ) ) + S ( r , f ) .
Hence, a 2 , b 2 are small functions of f ( k ) . Using the same argument as proving L ( f ) 0 , it is easy to obtain L ( f ( k ) ) 0 . Thus, the lemma is proved.□

## Lemma 2.5

Let k j ( j = 1 , 2 , , q ) be positive integers, let a 1 , b 1 be two distinct small functions of f and let d j = a 1 l j ( a 1 b 1 ) ( j = 1 , 2 , , q ). Then,

m r , L ( f ) f a 1 = S ( r , f ) , m r , L ( f ) f b 1 = S ( r , f )
and
(2.1) m r , L ( f ) f ( f d 1 ) ( f d 2 ) ( f d m ) = S ( r , f ) ,
where L ( f ) is defined as Lemma 2.4, and 2 m q .

## Proof

Since L ( f ) = ( a 1 b 1 ) ( f a 1 ) ( a 1 b 1 ) ( f a 1 ) , by Lemma 2.1, we have

m r , L ( f ) f a 1 m ( r , a 1 b 1 ) + m r , ( a 1 b 1 ) ( f a 1 ) f a 1 = S ( r , f ) .
Similarly, we have
m r , L ( f ) f b 1 = S ( r , f ) .
Obviously,
L ( f ) = f a 1 + l j ( a 1 b 1 ) a 1 b 1 f a 1 + l j ( a 1 b 1 ) a 1 b 1 = f d j a 1 b 1 f d j a 1 b 1 ,
L ( f ) f ( f d 1 ) ( f d 2 ) ( f d q ) = i = 1 q c i L ( f ) f d i ,
where c i ( i = 1 , 2 , , q ) are small functions of f .

Thus, by Lemma 2.2 and above two formulas, we obtain (2.1).□

## Lemma 2.6

Let f and g be two non-constant entire functions and let a 1 , a 2 , b 1 , and b 2 be four small functions of f and g such that a 1 b 1 , a 2 b 2 , and none of them is identically equal to . Suppose that

H = L ( f ) ( f a 1 ) ( f b 1 ) L ( g ) ( g a 2 ) ( g b 2 ) 0 ,
where
L ( f ) = ( a 1 b 1 ) ( f a 1 ) ( a 1 b 1 ) ( f a 1 ) ,
L ( g ) = ( a 2 b 2 ) ( g a 2 ) ( a 2 b 2 ) ( g 2 a 2 ) .

If f and g share ( a 1 , a 2 ) CM and share ( b 1 , b 2 ) IM, then either

(2.2) 2 T ( r , f ) N ¯ r , 1 f a 1 + N ¯ r , 1 f b 1 + S ( r , f )
or
(2.3) ( a 2 b 2 ) f ( a 1 b 1 ) g + a 1 b 2 a 2 b 1 0 .

## Proof

Since H 0 , it is easy to get

(2.4) g b 2 g a 2 = c f b 1 f a 1 ,
where c is a nonzero constant.

If c = 1 , then by (2.4), we get (2.3).

If c 1 , then by (2.4), we have

(2.5) a 2 b 2 g a 2 ( c 1 ) f ( c b 1 a 1 ) f a 1
and
T ( r , f ) = T ( r , g ) + S ( r , f ) + S ( r , g ) .

Obviously, c b 1 a 1 c 1 a 1 , c b 1 a 1 c 1 b 1 . It follows from (2.5) and the fact that f and g share ( a 1 , a 2 ) CM that N r , 1 f c b 1 a 1 c 1 = N r , 1 a 2 b 2 = S ( r , f ) . Thus by Nevanlinna’s second fundamental theorem, we have

2 T ( r , f ) N ¯ r , 1 f a 1 + N ¯ r , 1 f b 1 + N ¯ r , 1 f c b 1 a 1 c 1 + S ( r , f ) N ¯ r , 1 f a 1 + N ¯ r , 1 f b 1 + S ( r , f ) .

It follows that 2 T ( r , f ) N ¯ r , 1 f a 1 + N ¯ r , 1 f b 1 + S ( r , f ) .□

## 3 Proof of Theorem 1

We prove Theorem 1 by contradiction. Suppose that ( a 2 b 2 ) f ( a 1 b 1 ) f ( k ) + a 1 b 2 a 2 b 1 0 . Set

(3.1) f a 1 f ( k ) a 2 = H .
Since f is a transcendental entire function, and f and f ( k ) share ( a 1 , a 2 ) CM, then by ( 3.1), we have
(3.2) N ( r , H ) = S ( r , f ) , N r , 1 H = S ( r , f ) .

From the fact that f and f ( k ) share ( a 1 , a 2 ) CM and share ( b 1 , b 2 ) IM, and f is a transcendental entire function, by Nevanlinna’s second fundamental theorem and Lemma 2.1, we obtain

T ( r , f ) N ¯ r , 1 f a 1 + N ¯ r , 1 f b 1 + N ¯ ( r , f ) + S ( r , f ) = N ¯ r , 1 f ( k ) a 2 + N ¯ r , 1 f ( k ) b 2 + S ( r , f ) N r , 1 ( a 2 b 2 ) f ( a 1 b 1 ) f ( k ) + a 1 b 2 a 2 b 1 + S ( r , f ) T ( r , ( a 2 b 2 ) f ( a 1 b 1 ) f ( k ) + a 1 b 2 a 2 b 1 ) + S ( r , f ) m ( r , ( a 2 b 2 ) f ( a 1 b 1 ) f ( k ) + a 1 b 2 a 2 b 1 ) + S ( r , f ) m ( r , f ) + m r , ( a 2 b 2 ) f ( a 1 b 1 ) f ( k ) f + S ( r , f ) T ( r , f ) + S ( r , f ) .

Thus, we have

(3.3) T ( r , f ) = N ¯ r , 1 f a 1 + N ¯ r , 1 f b 1 + S ( r , f )
and
(3.4) T ( r , f ) = T ( r , ( a 2 b 2 ) f ( a 1 b 1 ) f ( k ) + a 1 b 2 a 2 b 1 ) + S ( r , f ) = N r , 1 ( a 2 b 2 ) f ( a 1 b 1 ) f ( k ) + a 1 b 2 a 2 b 1 + S ( r , f ) .

Obviously,

(3.5) T ( r , H ) = m r , 1 H + S ( r , f ) = m r , f ( k ) a 2 f a 1 + S ( r , f ) m r , 1 f a 1 + S ( r , f ) .
It follows from ( 3.1) and ( 3.4) that
(3.6) m r , 1 f a 1 = m r , ( a 1 b 1 ) H 1 + b 2 a 2 ( a 2 b 2 ) f ( a 1 b 1 ) f ( k ) + a 1 b 2 a 2 b 1 m r , 1 ( a 2 b 2 ) f ( a 1 b 1 ) f ( k ) + a 1 b 2 a 2 b 1 + m r , a 1 b 1 H + b 2 a 2 + S ( r , f ) T ( r , H ) + S ( r , f ) .
By ( 3.5) and ( 3.6), we get
(3.7) T ( r , H ) = m r , 1 f a 1 + S ( r , f ) .

On the other hand, (3.1) can be rewritten as

(3.8) ( a 2 b 2 ) f ( a 1 b 1 ) f ( k ) + a 1 b 2 a 2 b 1 f a 1 = [ ( a 1 b 1 ) H 1 + b 2 a 2 ] ,
which implies
(3.9) N ¯ r , 1 f b 1 N ¯ r , 1 ( a 1 b 1 ) H 1 + b 2 a 2 T ( r , H ) + S ( r , f ) .

Thus, by (3.3), (3.7), and (3.9)

m r , 1 f a 1 + N r , 1 f a 1 = N ¯ r , 1 f a 1 + N ¯ r , 1 f b 1 + S ( r , f ) N ¯ r , 1 f a 1 + N ¯ r , 1 ( a 1 b 1 ) H 1 + b 2 a 2 + S ( r , f ) N ¯ r , 1 f a 1 + m r , 1 f a 1 + S ( r , f ) .

Hence, by the above formula, (3.7), (3.9), we obtain

(3.10) N r , 1 f a 1 = N ¯ r , 1 f a 1 + S ( r , f ) ,
(3.11) N ¯ r , 1 f b 1 = T ( r , H ) + S ( r , f ) .
Set
(3.12) φ = L ( f ) ( ( a 2 b 2 ) f ( a 1 b 1 ) f ( k ) + a 1 b 2 a 2 b 1 ) ( f a 1 ) ( f b 1 )
and
(3.13) ψ = L ( f ( k ) ) ( ( a 2 b 2 ) f ( a 1 b 1 ) f ( k ) + a 1 b 2 a 2 b 1 ) ( f ( k ) a 2 ) ( f ( k ) b 2 ) .

Obviously, φ 0 . By (3.12), it is easy to know that N ( r , φ ) = S ( r , f ) . Thus, by Lemmas 2.1, 2.4, and 2.5, we have

T ( r , φ ) = m ( r , φ ) + N ( r , φ ) = m r , L ( f ) ( ( a 2 b 2 ) f ( a 1 b 1 ) f ( k ) + a 1 b 2 a 2 b 1 ) ( f a 1 ) ( f b 1 ) + S ( r , f ) m r , L ( f ) f ( f a 1 ) ( f b 1 ) + m r , ( a 2 b 2 ) f ( a 1 b 1 ) f ( k ) f + m r , L ( f ) ( a 1 b 2 a 2 b 1 ) ( f a 1 ) ( f b 1 ) + S ( r , f ) S ( r , f ) ,
that is
(3.14) T ( r , φ ) = S ( r , f ) .

Let d 1 = a 1 k ( a 1 b 1 ) ( k 0 , 1 ) and d 2 = a 2 k ( a 2 b 2 ) . It follows from Nevanlinna’s second fundamental theorem and (3.3) that

2 T ( r , f ) N ¯ r , 1 f a 1 + N ¯ r , 1 f b 1 + N ¯ r , 1 f d 1 + S ( r , f ) T ( r , f ) + N ¯ r , 1 f d 1 + S ( r , f ) 2 T ( r , f ) m r , 1 f d 1 + S ( r , f ) ,
which implies
(3.15) m r , 1 f d 1 = S ( r , f ) .

Rewrite (3.13) as

(3.16) ψ = a 2 d 2 a 2 b 2 L ( f ( k ) ) f ( k ) a 2 b 2 d 2 a 2 b 2 L ( f ( k ) ) f ( k ) b 2 ( a 2 b 2 ) ( f d 1 ) f ( k ) d 2 a 1 + b 1
and set
(3.17) ϕ = L ( f ) ( f a 1 ) ( f b 1 ) L ( f ( k ) ) ( f ( k ) a 2 ) ( f ( k ) b 2 ) .
Next, we consider two cases.

Case 1. ϕ 0 . Integrating both sides of (3.17) which implies

(3.18) f a 1 f b 1 = C f ( k ) a 2 f ( k ) b 2 ,
where C is a nonzero constant.

Then, by Lemma 2.6, we get

(3.19) 2 T ( r , f ) N ¯ r , 1 f a 1 + N ¯ r , 1 f b 1 + S ( r , f ) ,
which contradicts with ( 3.3).

Case 2. ϕ 0 . It is easy to obtain that m ( r , ϕ ) = S ( r , f ) , and all poles of ϕ must come from the zeros of f b 1 . Hence, we have

(3.20) T ( r , ϕ ) = m ( r , ϕ ) + N ( r , ϕ ) N ¯ r , 1 f b 1 + S ( r , f ) .

It follows from (3.4), (3.17), and (3.20) that

(3.21) m ( r , f ) = m ( r , ( a 2 b 2 ) f ( a 1 a 1 ) f ( k ) + a 1 b 2 a 2 b 1 ) + S ( r , f ) = m r , ϕ ( ( a 2 b 2 ) f ( a 1 a 1 ) f ( k ) + a 1 b 2 a 2 b 1 ) ϕ + S ( r , f ) = m r , φ ψ ϕ + S ( r , f ) T r , ϕ φ ψ + S ( r , f ) T ( r , φ ψ ) + T ( r , ϕ ) + S ( r , f ) T ( r , ψ ) + T ( r , ϕ ) + S ( r , f ) T ( r , ψ ) + N ¯ r , 1 f b 1 + S ( r , f ) .
On the other hand, by Lemma 2.4, ( 3.1), ( 3.8), and ( 3.11), we have
(3.22) T ( r , ψ ) = T r , L ( f ( k ) ) ( ( a 2 b 2 ) f ( a 1 a 1 ) f ( k ) + a 1 b 2 a 2 b 1 ) ( f ( k ) a 2 ) ( f ( k ) b 2 ) = m r , L ( f ( k ) ) ( ( a 2 b 2 ) f ( a 1 a 1 ) f ( k ) + a 1 b 2 a 2 b 1 ) ( f ( k ) a 2 ) ( f ( k ) b 2 ) + S ( r , f ) m r , L ( f ( k ) ) f ( k ) b 2 + m r , ( a 2 b 2 ) f ( a 1 a 1 ) f ( k ) + a 1 b 2 a 2 b 1 f ( k ) a 2 + S ( r , f ) m r , ( a 2 b 2 ) f ( a 1 a 1 ) f ( k ) + a 1 b 2 a 2 b 1 f a 1 f a 1 f ( k ) a 2 + S ( r , f ) m ( r , ( a 2 b 2 ) H + b 1 a 1 ) + S ( r , f ) m r , 1 f a 1 + S ( r , f ) = N ¯ r , 1 f b 1 + S ( r , f ) .
Thus, by ( 3.21) and ( 3.22), we obtain
(3.23) T ( r , f ) 2 N ¯ r , 1 f b 1 + S ( r , f ) .

If a 1 ( k ) a 2 , then by (3.1), (3.2), and Lemma 2.1, we get

(3.24) T ( r , H ) = m r , 1 H + S ( r , f ) = m r , f ( k ) a 1 ( k ) f a 1 + S ( r , f ) m r , f ( k ) a 1 ( k ) f a 1 + S ( r , f ) = S ( r , f ) .
It follows from ( 3.11), ( 3.23), and ( 3.24) that T ( r , f ) = S ( r , f ) , a contradiction.

If a 1 ( k ) b 2 , then by (3.11), (3.23), and

T ( r , f ) m r , 1 f a 1 + N ¯ r , 1 f b 1 + S ( r , f ) m r , 1 f ( k ) b 2 + N r , 1 f ( k ) b 2 + S ( r , f ) T ( r , f ( k ) ) + S ( r , f ) ,
which implies
(3.25) T ( r , f ) T ( r , f ( k ) ) + S ( r , f ) .
On the other hand, it follows from Lemma 2.2 that
(3.26) T ( r , f ( k ) ) T ( r , f ) + S ( r , f ) .
Thus,
(3.27) T ( r , f ) = T ( r , f ( k ) ) + S ( r , f ) .
Then, by Nevanlinna’s second fundamental theorem, Lemma 2.1, ( 3.3), and ( 3.27), we have
2 T ( r , f ) 2 T ( r , f ( k ) ) + S ( r , f ) N ¯ r , 1 f ( k ) a 2 + N ¯ r , 1 f ( k ) b 2 + N ¯ r , 1 f ( k ) d 2 + S ( r , f ) N ¯ r , 1 f a 1 + N ¯ r , 1 f b 1 + T r , 1 f ( k ) d 2 m r , 1 f ( k ) d 2 + S ( r , f ) T ( r , f ) + T ( r , f ( k ) ) m r , 1 f ( k ) d 2 + S ( r , f ) 2 T ( r , f ) m r , 1 f ( k ) d 2 + S ( r , f ) .
Thus,
(3.28) m r , 1 f ( k ) d 2 = S ( r , f ) .

It is easy to see that N ( r , ψ ) = S ( r , f ) . By Nevanlinna’s first fundamental theorem, Lemmas 2.1 and 2.3, (3.15), (3.27), (3.28), and the fact that f is a transcendental entire function, we obtain

m r , f d 1 f ( k ) d 2 = T r , f ( k ) d 2 f d 1 N r , f d 1 f ( k ) d 2 + O ( 1 ) m r , f ( k ) d 1 ( k ) f d 1 + m r , d 1 ( k ) d 2 f d 1 + N r , f ( k ) d 2 f d 1 N r , f d 1 f ( k ) d 2 + O ( 1 ) N r , 1 f d 1 N r , 1 f ( k ) d 2 + S ( r , f ) T ( r , f ) T ( r , f ( k ) ) + S ( r , f ) = S ( r , f ) .
It follows from Lemma 2.5 and ( 3.16) that
(3.29) T ( r , ψ ) = m ( r , ψ ) + N ( r , ψ ) = m r , a 2 d 2 a 2 b 2 L ( f ( k ) ) f ( k ) a 2 + m r , b 2 d 2 a 2 b 2 L ( f ( k ) ) f ( k ) b 2 + S ( r , f ) + m r , ( a 2 b 2 ) ( f d 1 ) f ( k ) d 2 a 1 + b 1 + S ( r , f ) = S ( r , f ) .
By ( 3.3), ( 3.10), ( 3.21), and ( 3.29), we get
(3.30) N r , 1 f a 1 = N ¯ r , 1 f a 1 + S ( r , f ) = S ( r , f ) .
Moreover, by ( 3.3), ( 3.27), and ( 3.30), we have
(3.31) m r , 1 ( f a 1 ) ( k ) = S ( r , f ) ,
which implies
(3.32) N ¯ r , 1 ( f a 1 ) ( k ) = m r , 1 f a 1 + S ( r , f ) m r , 1 ( f a 1 ) ( k ) + S ( r , f ) S ( r , f ) .
Then by ( 3.3), ( 3.30), and ( 3.32), we obtain T ( r , f ) = S ( r , f ) , a contradiction.

Hence, we have a 1 ( k ) a 2 and a 1 ( k ) b 2 . By (3.7), (3.11), and (3.23), we obtain

T ( r , f ) 2 m r , 1 f a 1 + S ( r , f ) 2 m r , 1 f ( k ) a 1 ( k ) + S ( r , f ) = 2 T ( r , f ( k ) ) 2 N r , 1 f ( k ) a 1 ( k ) + S ( r , f ) N ¯ r , 1 f ( k ) a 2 + N ¯ r , 1 f ( k ) b 2 + N ¯ r , 1 f ( k ) a 1 ( k ) 2 N r , 1 f ( k ) a 1 ( k ) + S ( r , f ) T ( r , f ) N r , 1 f ( k ) a 1 ( k ) + S ( r , f ) ,
which implies that
(3.33) N r , 1 f ( k ) a 1 ( k ) = S ( r , f ) .
In the following, we will prove T ( r , H ) = S ( r , f ) , then combing ( 3.11) with ( 3.23), we get T ( r , f ) = S ( r , f ) , which is impossible.

Rewrite (3.1) as

(3.34) f a 1 = H ( f ( k ) a 1 ( k ) ) + H ( a 1 ( k ) a 2 ) .
Set g = f ( k ) a 1 ( k ) . Differentiating ( 3.34) k times, we have
(3.35) g = H ( k ) g + k H ( k 1 ) g + + k H g ( k 1 ) + H g ( k ) + B ( k ) ,
where B = H ( a 1 ( k ) a 2 ) . It is easy to see that g 0 . Then, we rewrite ( 3.35) as
(3.36) 1 B ( k ) g = D H ,
where
(3.37) D = H ( k ) H + k H ( k 1 ) g H g + + k H g k 1 H g + g k g .
Note that N r , 1 f ( k ) a 1 ( k ) = N r , 1 g = S ( r , f ) . Then it follows from ( 3.2) that
(3.38) T ( r , D ) i = 1 k T r , H ( i ) H + T r , g ( i ) g + S ( r , f ) i = 1 k m r , H ( i ) H + N r , H ( i ) H + m r , g ( i ) g + N r , g ( i ) g + S ( r , f ) = S ( r , H ) + S ( r , f ) .
By ( 3.34) and Lemma 2.2, we get
(3.39) T ( r , H ) T ( r , f ) + T ( r , f ( k ) ) + S ( r , f ) 2 T ( r , f ) + S ( r , f ) .
Then it follows from ( 3.38) that T ( r , D ) = S ( r , f ) . Next, we discuss two subcases.

Subcase 2.1. H 1 D 0 . Rewrite (3.36) as

(3.40) g H ( H 1 D ) = B ( k ) .
We claim that D 0 . Otherwise, it follows from ( 3.40) that N r , 1 H 1 D = S ( r , f ) . Then applying Nevanlinna’s second fundamental theorem to H , we can obtain
(3.41) T ( r , H ) = T ( r , H 1 ) + O ( 1 ) N ¯ ( r , H 1 ) + N ¯ r , 1 H 1 + N ¯ r , 1 H 1 D + S ( r , H ) S ( r , H ) S ( r , f ) .
It follows from ( 3.11) and ( 3.23) that T ( r , f ) = S ( r , f ) , a contradiction. Thus, D 0 . Then by ( 3.40), we get
(3.42) g = B ( k ) .
Integrating ( 3.42), we get
(3.43) f = H ( a 1 ( k ) a 2 ) + P ( z ) + a 1 ,
where P ( z ) is a polynomial of degree at most k 1 . By ( 3.43), we have
(3.44) T ( r , f ) = T ( r , H ) + S ( r , f ) .
It follows from ( 3.3), ( 3.11), and ( 3.44) that
(3.45) N ¯ r , 1 f ( k ) a 2 = N ¯ r , 1 f a 1 = S ( r , f ) .
Then combining ( 3.33) with ( 3.45), and applying Nevanlinna’s second fundamental theorem to f ( k ) , we get
(3.46) T ( r , f ( k ) ) N ¯ r , 1 f ( k ) a 2 + N ¯ r , 1 f ( k ) a 1 ( k ) = S ( r , f ) .
Because f and f ( k ) share ( a 1 , a 2 ) CM and ( b 1 , b 2 ) IM, then ( 3.3) and ( 3.46) imply
(3.47) T ( r , f ) = N ¯ r , 1 f a 1 + N ¯ r , 1 f b 1 + S ( r , f ) = N ¯ r , 1 f ( k ) a 2 + N ¯ r , 1 f ( k ) b 2 + S ( r , f ) 2 T ( r , f ) ( k ) + S ( r , f ) = S ( r , f ) .
It is impossible.

Subcase 2.2 H 1 D 0 . Then by (3.38), we have T ( r , H ) = S ( r , f ) . It follows from (3.11) and (3.23) that T ( r , f ) = S ( r , f ) , but it is impossible.

This completes the proof of Theorem 1.

# Acknowledgments

The authors thank the referee for his careful reading of the paper and giving many valuable suggestions.

Funding information: This work was supported by the NNSF of China (Grant No. 11901119) and the Natural Science Foundation of Zhejiang Province (LY21A010012).

Conflict of interest: The authors state no conflict of interest.

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Received: 2020-09-12
Revised: 2021-01-11
Accepted: 2021-01-13
Published Online: 2021-05-03

© 2021 Xiaohuang Huang et al., published by DeGruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.