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BY 4.0 license Open Access Published by De Gruyter Open Access June 24, 2022

Unicity of meromorphic functions concerning differences and small functions

Zhiying He, Jianbin Xiao and Mingliang Fang
From the journal Open Mathematics

Abstract

In this paper, we study the unicity of meromorphic functions concerning differences and small functions and mainly prove two results: 1. Let f be a transcendental entire function of finite order with a Borel exceptional entire small function a ( z ) , and let η be a constant such that Δ η 2 f 0 . If Δ η 2 f and Δ η f share Δ η a CM, then a ( z ) is a constant a and f ( z ) = a + B e A z , where A , B are two nonzero constants; 2. Let f be a transcendental meromorphic function with ρ 2 ( f ) < 1 , let a 1 , a 2 be two distinct small functions of f , let L ( z , f ) be a linear difference polynomial, and let a 1 L ( z , a 2 ) . If δ ( a 2 , f ) > 0 , and f and L ( z , f ) share a 1 and CM, then L ( z , f ) a 1 f a 1 = c , for some constant c 0 . The results improve some results following C. X. Chen and R. R. Zhang [Uniqueness theorems related difference operators of entire functions, Chinese Ann. Math. Ser. A 42 (2021), no. 1, 11–22] and R. R. Zhang, C. X. Chen, and Z. B. Huang [Uniqueness on linear difference polynomials of meromorphic functions, AIMS Math. 6 (2021), no. 4, 3874–3888].

MSC 2010: 30D35

1 Introduction and main results

In this paper, we assume that the reader is familiar with the basic notions of Nevanlinna’s value distribution theory, see [1,2, 3,4]. In the following, a meromorphic function always means meromorphic in the whole complex plane.

By S ( r , f ) , we denote any quantity satisfying S ( r , f ) = o ( T ( r , f ) ) as r possible outside of an exceptional set E with finite logarithmic measure E d r / r < . A meromorphic function a is said to be a small function of f if it satisfies T ( r , a ) = S ( r , f ) .

Let f be a nonconstant meromorphic function. The order and the hyper-order of f are defined by

ρ ( f ) = lim ¯ r log + T ( r , f ) log r

and

ρ 2 ( f ) = lim ¯ r log + log + T ( r , f ) log r .

Let f be a transcendental meromorphic function, and let a be a small function of f . We define

λ ( f a ) = lim ¯ r log + N r , 1 f a log r , δ ( a , f ) = lim ̲ r m r , 1 f a T ( r , f ) = 1 lim ¯ r N r , 1 f a T ( r , f ) .

It is clear that 0 δ ( a , f ) 1 . If δ ( a , f ) > 0 , then a is called a deficient function of f and δ ( a , f ) is its deficiency. If a is a constant, then a is called a deficient value of f . In this paper, deficiency possible outside of an exceptional set E with finite logarithmic measure.

If

lim ¯ r log + N r , 1 f a log r < ρ ( f ) ,

for ρ ( f ) > 0 ; and N r , 1 f a = O ( log r ) for ρ ( f ) = 0 , then a is called a Borel exceptional function of f . If a is a constant, then a is called a Borel exceptional value of f .

Let f and g be two meromorphic functions, and let a be a small function of both f and g . We say that f and g share a small function a CM(IM) if f a and g a have the same zeros counting multiplicities(ignoring multiplicities).

Let η be a nonzero finite complex number, and let n be a positive integer. We define the difference operators of f as Δ η f ( z ) = f ( z + η ) f ( z ) and Δ η n f ( z ) = Δ η ( Δ η n 1 f ( z ) ) , n 2 .

Let η 1 , η 2 , , η n be distinct complex numbers, and let b i ( 0 ) ( i = 1 , 2 , , n ) be small functions of f . We define the linear difference polynomial of f as follows:

(1.1) L ( z , f ) = b 1 ( z ) f ( z + η 1 ) + b 2 ( z ) f ( z + η 2 ) + + b n ( z ) f ( z + η n ) .

Nevanlinna [4] proved the following famous five-value theorem.

Theorem A

Let f and g be two nonconstant 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 .

Li and Qiao [5] improved Theorem A as follows:

Theorem B

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

In 1986, Jank et al. [6] proved.

Theorem C

Let f be a nonconstant entire function, and let a be a nonzero finite complex number. If f , f and f share a CM, then f f .

Recently, the uniqueness in difference analogs of meromorphic functions has become a subject of some interests, see [7,8,9, 10,11,12, 13,14,15, 16,17,18].

Chen et al. [10] and Farissi et al. [11] obtained the difference analog to Theorem C and proved

Theorem D

[11] Let f be a nonconstant entire function of finite order, let η be a nonzero constant, and let a ( 0 ) be an entire small function of f satisfying a ( z + η ) = a ( z ) , If f , Δ η f and Δ η 2 f share a CM, then f Δ η f .

In 2021, Chen and Zhang [8] proved.

Theorem E

Let f be a transcendental entire function of finite order with λ ( f a ) < ρ ( f ) , where a ( z ) is an entire small function of f ( z ) satisfying ρ ( a ) < 1 , and let η be a nonzero constant such that Δ η 2 f 0 . If Δ η 2 f and Δ η f share Δ η a CM, where Δ η a is a small function of Δ η 2 f , then f ( z ) = a ( z ) + B e A z , where A , B are two nonzero constants and a ( z ) is reduced to a constant.

In [8], the authors pointed out that ρ ( a ) < 1 is reasonable. According to the aforementioned theorems, we naturally pose the following problem.

Problem 1

Whether ρ ( a ) < 1 can be deleted in Theorem E?

In this paper, we give a positive answer to Problem 1 and prove the following result.

Theorem 1

Let f be a transcendental entire function of finite order with a Borel exceptional entire small function a ( z ) , and let η be a constant such that Δ η 2 f 0 . If Δ η 2 f and Δ η f share Δ η a CM, then a ( z ) is a constant a and f ( z ) = a + B e A z , where A , B are two nonzero constants.

Remark

If λ ( f a ) < ρ ( f ) , then a ( z ) is a Borel exceptional function of f ( z ) . Hence, Theorem 1 improves and extends Theorem E.

The following example shows that there exists a transcendental entire function f satisfying Theorem 1.

Example 1

[8] Suppose f = e z ln 2 + 1 , then it is easy to obtain 1 is a Borel exceptional value of f . Let η = 1 , we obtain Δ η 2 f Δ η f . Thus, we see Δ η 2 f and Δ η f share 0 CM.

In 1996, Brück [19] posed the following conjecture.

Conjecture

Let f be a nonconstant entire function such that ρ 2 ( f ) < , which is not a positive integer. If f and f share one finite value a CM, then

f a f a = c ,

for some constant c 0 .

In 2009, Heittokangas et al. [20] proved the following result.

Theorem F

Let f be a meromorphic function with ρ ( f ) < 2 , let η be a nonzero complex number, and let a be a finite complex number. If f and f ( z + η ) share a and CM, then

f ( z + η ) a f ( z ) a = c ,

for some constant c 0 .

In 2021, Zhang et al. [18] proved

Theorem G

Let f be a transcendental meromorphic function with ρ 2 ( f ) < 1 , let a 1 , a 2 be two distinct small functions of f satisfying ρ ( a j ) < 1 ( j = 1 , 2 ) , and let L ( z , f ) be a linear difference polynomial of the form (1.1) with ρ ( b i ) < 1 ( i = 1 , 2 , , n ) and a 1 L ( z , a 2 ) . If δ ( a 2 , f ) > 0 , and f and L ( z , f ) share a 1 and CM, then

L ( z , f ) a 1 f a 1 = c ,

for some constant c . In particular, if the deficient function a 2 0 , then L ( z , f ) f .

Naturally, we pose the following problem.

Problem 2

Whether ρ ( a j ) < 1 ( j = 1 , 2 ) , ρ ( b i ) < 1 ( i = 1 , 2 , , n ) can be deleted or not in Theorem G?

In this paper, we give a positive answer to Problem 2 and prove the following result.

Theorem 2

Let f be a transcendental meromorphic function with ρ 2 ( f ) < 1 , let a 1 , a 2 be two distinct small functions of f , let L ( z , f ) be a linear difference polynomial of the form (1.1), and let a 1 L ( z , a 2 ) . If δ ( a 2 , f ) > 0 , and f and L ( z , f ) share a 1 and CM, then

L ( z , f ) a 1 f a 1 = c ,

for some constant c 0 . In particular, if the deficient function a 2 0 , then L ( z , f ) f .

The following example shows that there exists a transcendental meromorphic function f with ρ 2 ( f ) < 1 satisfying Theorem 2.

Example 2

[18] Let f = e π i z + 6 , and let L ( z , f ) = Δ 1 f = 2 e π i z . Then, we have L ( z , f ) and f share 4, CM and δ ( 6 , f ) = 1 > 0 . Thus,

L ( z , f ) 4 f 4 = 2 .

2 Lemmas

In order to prove our results, we need the following lemmas.

Lemma 1

[13] Let f be a nonconstant entire function of finite order. If a is a Borel exceptional entire small function of f , then δ ( a , f ) = 1 .

Lemma 2

[21,22,23] Let f be a nonconstant mermorphic function with ρ 2 ( f ) < 1 , and let η be a nonzero finite complex number. Then

m r , f ( z + η ) f ( z ) = S ( r , f ) .

If f is of finite order, then for any ε > 0 , we have

m r , f ( z + η ) f ( z ) = O ( r ρ ( f ) 1 + ε ) .

Lemma 3

[7] Let a be a finite complex number, let f be a transcendental meromorphic function of finite order with two Borel exceptional values a and , and let η be a nonzero constant such that Δ η f 0 . If f and Δ η f share a , CM, then a = 0 , f ( z ) = e A z + B , where A ( 0 ) and B are two constants.

Lemma 4

[21] Let f be a nonconstant meromorphic function of finite order, and let η be a nonzero finite complex number. Then

N ( r , f ( z + η ) ) = N ( r , f ( z ) ) + S ( r , f ) .

Lemma 5

Let η be a nonzero finite complex number, let n be a positive integer, and let f be a transcendental meromorphic function of finite order satisfying δ ( a , f ) = 1 , δ ( , f ) = 1 , where a is a small function of f . If Δ η n f 0 , then

  1. T ( r , Δ η n f ) = T ( r , f ) + S ( r , f ) ;

  2. δ ( Δ η n a , Δ η n f ) = δ ( , Δ η n f ) = 1 .

Proof

By Lemma 2 and Nevanlinna’s first fundamental theorem, we have

(2.1) m r , 1 f a = m r , Δ η n ( f a ) f a + m r , 1 Δ η n ( f a ) m r , 1 Δ η n ( f a ) + S ( r , f ) T r , 1 Δ η n ( f a ) + S ( r , f ) T ( r , Δ η n ( f a ) ) + S ( r , f ) T ( r , Δ η n f ) + S ( r , f ) .

It follows from (2.1) and Lemmas 2 and 4 that

m r , 1 f a T ( r , f ) T ( r , Δ η n f ) T ( r , f ) + S ( r , f ) T ( r , f ) , 1 = δ ( a , f ) lim ̲ r T ( r , Δ η n f ) T ( r , f ) + lim ¯ r S ( r , f ) T ( r , f ) lim ̲ r T ( r , Δ η n f ) T ( r , f ) lim ¯ r T ( r , Δ η n f ) T ( r , f ) lim ¯ r m ( r , Δ η n f ) + N ( r , Δ η n f ) T ( r , f ) lim ¯ r m r , Δ η n f f + m ( r , f ) + ( n + 1 ) N ( r , f ) T ( r , f ) lim ¯ r T ( r , f ) + n N ( r , f ) + S ( r , f ) T ( r , f ) 1 + lim ¯ r n N ( r , f ) T ( r , f ) + lim ¯ r S ( r , f ) T ( r , f ) = 1 .

Then we have T ( r , Δ η n f ) = T ( r , f ) + S ( r , f ) .

By (2.1) and lim r T ( r , Δ η n f ) T ( r , f ) = 1 , we obtain

1 = δ ( a , f ) lim ̲ r m r , 1 Δ η n ( f a ) T ( r , Δ η n f ) lim ¯ r T ( r , Δ η n f ) T ( r , f ) + lim ¯ r S ( r , f ) T ( r , f ) lim ̲ r m r , 1 Δ η n ( f a ) T ( r , Δ η n f ) = δ ( Δ η n a , Δ η n f ) 1 .

It follows that δ ( Δ η n a , Δ η n f ) = 1 .

Combining δ ( , f ) = 1 , N ( r , Δ η n f ) ( n + 1 ) N ( r , f ) with lim r T ( r , Δ η n f ) T ( r , f ) = 1 , we obtain δ ( , Δ η n f ) = 1 . □

Lemma 6

Let f be a meromorphic function of finite order, and let η , c , d be three nonzero finite complex numbers. If f ( z + η ) = c f ( z ) , then either T ( r , f ) d r for sufficiently large r or f is a constant.

Proof

In the following, we consider three cases.

Case 1. There exists z 0 such that f ( z 0 ) = . Without loss of generality, we assume that z 0 = 0 , and then we deduce that for all positive integers j , f ( j η ) = . Thus, for sufficiently large r and 2 n η r < ( 2 n + 1 ) η , we have

T ( r , f ) N ( r , f ) = 0 r n ( t , f ) n ( 0 , f ) t d t + n ( 0 , f ) log r j = 1 2 n 1 j j η ( j + 1 ) η d t t = j = 1 2 n 1 j log 1 + 1 j j = 1 2 n 1 j log 1 + 1 2 n 1 = n log 1 + 1 2 n 1 2 n 1 n log 2 > log 2 4 η r .

It follows that T ( r , f ) d r , where d = log 2 4 η .

Case 2. There exists z 0 such that f ( z 0 ) = 0 and f 0 . Set g = 1 f . Then by f ( z + η ) = c f ( z ) , we obtain that g ( z + η ) = 1 c g ( z ) and g ( z 0 ) = . Thus, by the proof of Case 1, we deduce that T ( r , f ) d r .

Case 3. f 0 , . Since f is of finite order, then f = e p , where p is a polynomial. If deg p 1 , then T ( r , f ) d r ; if deg p = 0 , then f is a nonzero constant.□

3 Proof of Theorem 1

Firstly, we prove ρ ( f ) > 0 . Suppose on the contrary that ρ ( f ) = 0 .

Set F ( z ) = f ( z ) a ( z ) . Since a ( z ) is a Borel exceptional entire small function of f ( z ) , we obtain

N r , 1 F = N r , 1 f a = O ( log r ) .

Hence, F has finitely many zeros. Thus, we assume that z 1 , z 2 , , z n are zeros of F , where n is a positive integer.

Hence, by ρ ( f ) = 0 , we deduce that F ( z z 1 ) ( z z 2 ) ( z z n ) = e p , where p is a constant.

It follows that F ( z ) = c ( z z 1 ) ( z z 2 ) ( z z n ) , where c is a nonzero constant. Thus, we have

T ( r , F ) = n log r + O ( 1 ) .

Since T ( r , a ) = S ( r , F ) , we obtain that a ( z ) is a constant and f ( z ) is a nonconstant polynomial, which contradicts with Δ η 2 f and Δ η f share Δ η a CM. It follows ρ ( f ) > 0 .

Obviously, δ ( , f ) = 1 . Since a ( z ) is a Borel exceptional entire small function of f ( z ) , then by Lemma 1, we obtain δ ( a , f ) = 1 .

By Lemma 5, we obtain

(3.1) δ ( Δ η a , Δ η f ) = 1 , δ ( Δ η 2 a , Δ η 2 f ) = 1 ,

(3.2) δ ( , Δ η f ) = 1 , δ ( , Δ η 2 f ) = 1 .

We claim that Δ η a Δ η 2 a . Otherwise, since Δ η 2 f and Δ η f share Δ η a CM, then by Nevanlinna’s second fundamental theorem and Lemma 5 and (3.1), we have

T ( r , f ) = T ( r , Δ η 2 f ) + S ( r , f ) N ¯ ( r , Δ η 2 f ) + N ¯ r , 1 Δ η 2 f Δ η a + N ¯ r , 1 Δ η 2 f Δ η 2 a + S ( r , f ) N ¯ r , 1 Δ η f Δ η a + S ( r , f ) S ( r , f ) ,

a contradiction.

Obviously, δ ( 0 , F ) = δ ( a , f ) = 1 , δ ( , F ) = 1 . Since f is a transcendental entire function and Δ η 2 f and Δ η f share Δ η a CM , we have Δ η 2 F and Δ η F share 0, CM.

It follows from (3.1) and (3.2) that

(3.3) δ ( 0 , Δ η F ) = 1 , δ ( 0 , Δ η 2 F ) = 1 ,

(3.4) δ ( , Δ η F ) = 1 , δ ( , Δ η 2 F ) = 1 .

Set

G = Δ η F .

Since Δ η 2 F and Δ η F share 0, CM, we obtain Δ η G and G share 0, CM. By (3.3), (3.4), we obtain

(3.5) δ ( 0 , G ) = 1 , δ ( 0 , Δ η G ) = 1 ,

(3.6) δ ( , G ) = 1 , δ ( , Δ η G ) = 1 .

By Lemma 5, we have

(3.7) T ( r , G ) = T ( r , f ) + S ( r , f ) .

Since a ( z ) is a Borel exceptional entire small function of f ( z ) , we obtain λ ( f a ) < ρ ( f ) . It follows that

(3.8) lim ¯ r log + N r , 1 F log r = lim ¯ r log + N r , 1 f a log r < ρ ( f ) .

By Nevanlinna’s first fundamental theorem, we have

(3.9) m r , 1 F m r , 1 Δ η F + m r , Δ η F F , T ( r , F ) N r , 1 F T ( r , Δ η F ) N r , 1 Δ η F + S ( r , F ) , N r , 1 Δ η F N r , 1 F + T ( r , Δ η F ) T ( r , F ) + S ( r , F ) N r , 1 F + m r , Δ η F F + m ( r , F ) m ( r , F ) + S ( r , F ) N r , 1 F + S ( r , F ) .

By Lemma 2, set ε = 1 2 , we obtain

(3.10) S ( r , F ) M r ρ ( f ) 1 2 ,

where M is a positive constant.

It follows from (3.8) that

(3.11) N r , 1 F < r ρ ( f ) + λ ( F ) 2 .

By (3.10) and (3.11), we obtain

(3.12) N r , 1 F + S ( r , F ) < ( 1 + M ) r M 1 ,

where M 1 = max ρ ( f ) 1 2 , ρ ( f ) + λ ( F ) 2 .

By (3.9) and (3.12), we obtain

log + N r , 1 Δ η F log r log ( 1 + M ) r M 1 log r M 1 + log ( 1 + M ) log r .

Thus, we have

(3.13) lim ¯ r log + N r , 1 G log r = lim ¯ r log + N r , 1 Δ η F log r M 1 < ρ ( f ) .

It follows from (3.7) and (3.13), we deduce that 0 is a Borel exceptional value of G .

By Lemma 3, we obtain G ( z ) = e A 1 z + B 1 , where A 1 ( 0 ) , B 1 are two constants. That is,

(3.14) F ( z + η ) F ( z ) = e A 1 z + B 1 .

By Hadamard’s factorization theorem, we have

(3.15) F ( z ) = α ( z ) e p 1 ( z ) ,

where α is an entire function such that ρ ( α ) = λ ( α ) < ρ ( F ) , and p 1 is a nonconstant polynomial with deg p 1 = ρ ( F ) .

Hence, we obtain

(3.16) T ( r , α ) = S ( r , e p 1 ) .

It follows from (3.14) and (3.15) that

(3.17) α ( z + η ) e p 1 ( z + η ) α ( z ) e p 1 ( z ) = e A 1 z + B 1 .

Next, we consider two cases.

Case 1. deg p 1 2 . By (3.17), we have

(3.18) α ( z + η ) e A 1 z + B 1 e p 1 ( z + η ) α ( z ) e A 1 z + B 1 e p 1 ( z ) 1 .

Obviously, T ( r , e A 1 z + B 1 ) = S ( r , e p 1 ) . It follows from (3.16), (3.18), and Nevanlinna’s second fundamental theorem that

T ( r , e p 1 ) T r , α e A 1 z + B 1 e p 1 + S ( r , e p 1 ) N ¯ r , α e A 1 z + B 1 e p 1 + N ¯ r , 1 α e A 1 z + B 1 e p 1 + N ¯ r , 1 α e A 1 z + B 1 e p 1 + 1 + S r , α e A 1 z + B 1 e p 1 S ( r , e p 1 ) ,

a contradiction.

Case 2. deg p 1 = 1 . Let p 1 ( z ) = m z + n , where m ( 0 ) and n are two complex numbers.

Now, we consider two subcases.

Case 2.1. A 1 m . Thus, by (3.17), we obtain

(3.19) c 1 α ( z + η ) e ( m A 1 ) z + c 2 α ( z ) e ( m A 1 ) z 1 ,

where c 1 = e m η + n B 1 and c 2 = e n B 1 .

Obviously, T ( r , α ) = S ( r , e ( m A 1 ) z ) . It follows from (3.19) and Nevanlinna’s second fundamental theorem that

T ( r , e ( m A 1 ) z ) T ( r , c 2 α e ( m A 1 ) z ) + S ( r , e ( m A 1 ) z ) N ¯ ( r , c 2 α e ( m A 1 ) z ) + N ¯ r , 1 c 2 α e ( m A 1 ) z + N ¯ r , 1 c 2 α e ( m A 1 ) z 1 + S ( r , c 2 α e ( m A 1 ) z ) S ( r , e ( m A 1 ) z ) ,

a contradiction.

Case 2.2. A 1 = m . Thus, by (3.17), we obtain

(3.20) c 1 α ( z + η ) + c 2 α ( z ) 1 ,

where c 1 = e m η + n B 1 , c 2 = e n B 1 .

Next, we consider two subcases.

Case 2.2.1. c 1 + c 2 = 0 . Hence,

e m η + n B 1 e n B 1 = e n B 1 ( e m η 1 ) = 0 .

It follows e m η = 1 .

By Lemma 6 and ρ ( α ) < ρ ( F ) = 1 , we deduce

α = z η c 1 + c 3 ,

where c 3 is a constant.

Hence, f ( z ) = a ( z ) + z η c 1 + c 3 e m z + n . It follows

(3.21) T ( r , f ) = T r , a + z η c 1 + c 3 e m z + n T ( r , e m z + n ) + S ( r , f ) m π r + S ( r , f ) .

Since Δ η a Δ η 2 a , then b Δ η b , where b = Δ η a . It follows b ( z + η ) 2 b ( z ) . By Lemma 6, we deduce that either T ( r , b ) d r or b ( z ) is a constant. If T ( r , b ) d r , by (3.21), we know that b ( z ) is not a small function of f ( z ) , a contradiction. Then b ( z ) is a constant, obviously b ( z ) 0 . It follows a ( z + η ) a ( z ) . By Lemma 6 and (3.21), we deduce that a ( z ) is a constant.

Thus, we have

Δ η f = z + η η c 1 + c 3 e m ( z + η ) + n z η c 1 + c 3 e m z + n = z η c 1 + c 3 + 1 c 1 e m z + n e m η z η c 1 + c 3 e m z + n = 1 c 1 e m z + n ,

and

Δ η 2 f = Δ η ( Δ η f ) = 1 c 1 e m ( z + η ) + n 1 c 1 e m z + n = 1 c 1 e m z + n e m η 1 c 1 e m z + n = 0 .

This contradicts with Δ η 2 f 0 . Hence, this case cannot occur.

Case 2.2.2. c 1 + c 2 0 .

By Lemma 6 and ρ ( α ) < ρ ( F ) = 1 , we deduce

α = c ,

where c is a constant.

It follows that f ( z ) = a ( z ) + c e m z + n . Obviously, c 0 , we obtain

(3.22) T ( r , f ) = T ( r , a + c e m z + n ) T ( r , e m z + n ) + S ( r , f ) m π r + S ( r , f ) .

Since Δ η a Δ η 2 a , by (3.22) and using the same argument as used in case 2.2.1, we can prove that a ( z ) is a constant a . Therefore, we have f ( z ) = a + B e A z , where A , B are nonzero constants.

This completes the proof of Theorem 1.

4 Proof of Theorem 2

Since f and L ( z , f ) share a 1 and CM, we obtain

(4.1) L ( z , f ) a 1 ( z ) f ( z ) a 1 ( z ) = h ( z ) ,

where h is a meromorphic function satisfying N ( r , h ) = S ( r , f ) , N r , 1 h = S ( r , f ) .

It follows from (4.1) that

(4.2) 1 a 1 L ( z , a 2 ) ( a 1 a 2 ) h L ( z , f a 2 ) f a 2 h = 1 f a 2 .

By Lemma 2 and Nevanlinna’s first fundamental theorem, we have

T ( r , h ) = m ( r , h ) + S ( r , f ) = m r , L ( z , f ) a 1 f a 1 + S ( r , f ) m r , L ( z , f a 1 ) f a 1 + m r , L ( z , a 1 ) a 1 f a 1 + S ( r , f ) m r , 1 f a 1 + S ( r , f ) T ( r , f ) + S ( r , f ) .

It follows

(4.3) S ( r , h ) = S ( r , f ) .

Since δ ( a 2 , f ) > 0 , we deduce that m r , 1 f a 2 c 1 T ( r , f ) for sufficiently large r , where c 1 is some positive constant. Then, by (4.2), we have

T ( r , f ) 1 c 1 m r , 1 f a 2 = 1 c 1 m r , 1 a 1 L ( z , a 2 ) ( a 1 a 2 ) h L ( z , f a 2 ) f a 2 h 1 c 1 m r , 1 a 1 L ( z , a 2 ) ( a 1 a 2 ) h + 1 c 1 m r , L ( z , f a 2 ) f a 2 + 1 c 1 m ( r , h ) + S ( r , f ) 2 c 1 T ( r , h ) + S ( r , f ) .

It follows

(4.4) S ( r , f ) = S ( r , h ) .

Since a 1 ( z ) a 2 ( z ) , we have

N r , 1 a 1 L ( z , a 2 ) a 1 a 2 h = N r , a 1 a 2 a 1 L ( z , a 2 ) ( a 1 a 2 ) h N r , 1 a 1 L ( z , a 2 ) ( a 1 a 2 ) h + S ( r , f ) = N r , 1 a 1 a 2 1 a 1 L ( z , a 2 ) a 1 a 2 h + S ( r , f ) N r , 1 a 1 L ( z , a 2 ) a 1 a 2 h + S ( r , f ) .

Thus, we have

(4.5) N r , 1 a 1 L ( z , a 2 ) ( a 1 a 2 ) h = N r , 1 a 1 L ( z , a 2 ) a 1 a 2 h + S ( r , f ) .

It follows from (4.3), (4.4), a 1 L ( z , a 2 ) , and Nevanlinna’s second fundamental theorem that

(4.6) T ( r , h ) N ¯ ( r , h ) + N ¯ r , 1 h + N ¯ r , 1 h a 1 L ( z , a 2 ) a 1 a 2 + S ( r , f ) N ¯ r , 1 h a 1 L ( z , a 2 ) a 1 a 2 + S ( r , f ) T ( r , h ) + S ( r , f ) .

By (4.5) and (4.6), we obtain

N r , 1 a 1 L ( z , a 2 ) ( a 1 a 2 ) h = T ( r , h ) + S ( r , f ) .

It follows

(4.7) m r , 1 a 1 L ( z , a 2 ) ( a 1 a 2 ) h = S ( r , f ) .

By (4.7), we have

(4.8) m r , h a 1 L ( z , a 2 ) ( a 1 a 2 ) h = m r , 1 a 2 a 1 + L ( z , a 2 ) a 1 a 2 a 1 1 a 1 L ( z , a 2 ) ( a 1 a 2 ) h m r , 1 a 2 a 1 + m r , L ( z , a 2 ) a 1 a 2 a 1 + m r , 1 a 1 L ( z , a 2 ) ( a 1 a 2 ) h S ( r , f ) .

It follows from (4.2), (4.7), (4.8), and Lemma 2 that

m r , 1 f a 2 = m r , 1 a 1 L ( z , a 2 ) ( a 1 a 2 ) h L ( z , f a 2 ) f a 2 h m r , 1 a 1 L ( z , a 2 ) ( a 1 a 2 ) h L ( z , f a 2 ) f a 2 + m r , h a 1 L ( z , a 2 ) ( a 1 a 2 ) h

m r , 1 a 1 L ( z , a 2 ) ( a 1 a 2 ) h + m r , L ( z , f a 2 ) f a 2 + m r , h a 1 L ( z , a 2 ) ( a 1 a 2 ) h S ( r , f ) ,

which contradicts with δ ( a 2 , f ) > 0 . Hence, h is a constant c . That is,

L ( z , f ) a 1 f a 1 = c ,

obviously c 0 .

Next, we consider the case: a 2 0 . Then, by (4.2) and h = c , we have

1 a 1 ( 1 c ) L ( z , f ) f c = 1 f .

We claim that c = 1 . Suppose on the contrary that c 1 , then we obtain

m r , 1 f = m r , 1 a 1 ( 1 c ) L ( z , f ) f c S ( r , f ) ,

which contradicts with δ ( 0 , f ) > 0 . Hence, c = 1 . That is, L ( z , f ) f .

Thus, Theorem 2 is proved.

Acknowledgments

We are very grateful to the anonymous referees for their careful review and valuable suggestions.

  1. Funding information: This work was supported by the National Natural Science Foundation of China (Grant No 12171127) and the Natural Science Foundation of Zhejiang Province (Grant No LY21A010012).

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

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Received: 2021-09-27
Revised: 2022-01-31
Accepted: 2022-03-23
Published Online: 2022-06-24

© 2022 Zhiying He et al., published by De Gruyter

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

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