Unicity of meromorphic functions concerning differences and small functions

Zhiying He; Jianbin Xiao; Mingliang Fang

doi:10.1515/math-2022-0033

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

https://doi.org/10.1515/math-2022-0033

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].

Keywords: meromorphic functions; Borel exceptional functions; unicity; differences

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.

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

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

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).
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

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

Unicity of meromorphic functions concerning differences and small functions

Abstract

1 Introduction and main results

Theorem A

Theorem B

Theorem C

Theorem D

Theorem E

Problem 1

Theorem 1

Remark

Example 1

Conjecture

Theorem F

Theorem G

Problem 2

Theorem 2

Example 2

2 Lemmas

Lemma 1

Lemma 2

Lemma 3

Lemma 4

Lemma 5

Proof

Lemma 6

Proof

3 Proof of Theorem 1

4 Proof of Theorem 2

Acknowledgments

References

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