Some results on uniqueness and higher order difference equations

Jia An; Haiying Zhang; Ge Wang; Mingliang Fang

doi:10.1515/math-2024-0100

Article Open Access

Some results on uniqueness and higher order difference equations

Jia An , Haiying Zhang , Ge Wang and Mingliang Fang

Published/Copyright: December 17, 2024

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$Open Mathematics$

From the journal Open Mathematics Volume 22 Issue 1

Abstract

In this article, we mainly study the solution of higher order difference equations. We solve some questions posed by Chen and Xu [Uniqueness on entire functions and their nth order exact differences with two shared values, Open Math. 18 (2020), no. 1, 211–215] and give a negative answer to some conjectures posed by Gao et al. [Uniqueness of meromorphic functions sharing values with their nth order exact differences, Anal. Math. 45 (2019), no. 2, 321–334] and Ahamed [Class of meromorphic functions partially shared values with their differences or shifts, Kyungpook Math. J. 61 (2021), no. 4, 745–763]. Meanwhile, we extend some results due to Adud and Chakraborty [Some results related to differential-difference counterpart of the Brück conjecture, Commun. Korean Math. Soc. 39 (2024), no. 1, 117–125].

Keywords: difference equations; meromorphic functions; uniqueness; small functions

MSC 2010: 30D35

1 Introduction and main results

In this article, we assume that the reader is familiar with the basic notions of Nevanlinna’s value distribution theory, see [1–3]. 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 ( z ) is said to be a small function of f ( z ) if it satisfies T ( r , a ) = S ( r , f ) .

Let f ( z ) be a nonconstant meromorphic function. The order ρ ( f ) and the hyper-order ρ 2 ( f ) are defined by

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

and the difference operators are defined as

Δ η f = f ( z + η ) − f ( z ) and Δ η n f = Δ η ( Δ η n − 1 f ) ,

where n ( ≥ 2 ) is a positive integer and η is a nonzero constant.

Θ ( ∞ , f ) is defined as

Θ ( ∞ , f ) = 1 − lim ¯ r → ∞ N ¯ ( r , f ) T ( r , f ) .

Let n be a positive integer, let a i ( i = 0 , 1 , … , n ) be nonzero complex numbers, and let η i ( i = 0 , 1 , … , n ) be distinct constants. The difference polynomial of f ( z ) is defined as

(1.1) L ( f ( z ) ) = ∑ i = 0 n a i f ( z + η i ) .

Let k j ( j = 1 , 2 , … , n ) be nonnegative integers. The differential-difference monomial of f ( z ) is defined as

M ( f ) = f ( k 1 ) ( z + η 1 ) f ( k 2 ) ( z + η 2 ) … f ( k n ) ( z + η n ) ,

where n is the degree of M ( f ) , and we denote by deg M ( f ) = n , where η i ( i = 1 , 2 , … , n ) are finite values.

Let m be a positive integer, let d 1 , d 2 , … , d m be nonzero complex numbers, and let M 1 ( f ) , M 2 ( f ) , … , M m ( f ) be differential-difference monomials of f . The differential-difference polynomial of f is defined as

(1.2) Q ( f ) = d 1 M 1 ( f ) + d 2 M 2 ( f ) + … + d m M m ( f ) .

If deg M 1 ( f ) = deg M 2 ( f ) = … = deg M m ( f ) = l , we say that Q ( f ) is homogeneous differential-difference polynomial of f with degree deg Q ( f ) = l .

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

In 2013, Chen and Yi [4] proved.

Theorem A

Let f ( z ) be a transcendental meromorphic function whose order is finite noninteger, let η be a nonzero complex number, and let a , b be two distinct constants. If Δ η f ( z ) ( ≢ 0 ) and f ( z ) share a , b , ∞ CM, then Δ η f ( z ) ≡ f ( z ) .

In 2019, Gao et al. [5] proved.

Theorem B

Let f ( z ) be a transcendental meromorphic function of ρ 2 ( f ) < 1 , let η be a nonzero complex number and n a positive integer, and let a , b be two distinct constants. If Δ η n f ( z ) ( ≢ 0 ) and f ( z ) share a , b , ∞ CM, then Δ η n f ( z ) ≡ f ( z ) .

In 2020, Chen and Xu [6] proved.

Theorem C

Let f ( z ) be a transcendental entire function of ρ 2 ( f ) < 1 , and let η be a nonzero complex number and n a positive integer. If Δ η n f ( z ) ( ≢ 0 ) and f ( z ) share 0 CM and 1 IM, then Δ η n f ( z ) ≡ f ( z ) .

In this article, we extend Theorem C as follows:

Theorem 1

Let f ( z ) be a transcendental meromorphic function of ρ 2 ( f ) < 1 with Θ ( ∞ , f ) > 0 , let n be a positive integer, and let L ( f ( z ) ) be a difference polynomial of f ( z ) as (1.1). Suppose that b ( z ) ( ≢ 0 ) is a small function of f ( z ) . If L ( f ( z ) ) and f ( z ) share 0 , ∞ CM and b ( z ) IM, then L ( f ( z ) ) ≡ f ( z ) .

Gao et al. [5] and Chen and Xu [6] proposed the following conjecture.

Conjecture 1

In Theorem B, the condition ρ 2 ( f ) < 1 can be removed.

Remark 1

The following example shows that the condition ρ 2 ( f ) < 1 cannot be removed in Theorem B.

Example 1

Let f ( z ) = 3 e e z + 1 , and let η be a finite complex number with e η = − 1 . Obviously, ρ 2 ( f ) = 1 and Δ η 1 ( f ( z ) ) = 3 ( e e z − 1 ) e e z + 1 . Hence, we have Δ η 1 ( f ( z ) ) and f ( z ) share 1 , 3 , ∞ CM, but Δ η 1 ( f ( z ) ) ≢ f ( z ) .

In addition, Chen and Xu [6] posed the following question.

Question 1

If f ( z ) is a transcendental entire solution of the difference equation Δ η n f ( z ) = f ( z ) , then f must be of the form f ( z ) = e a z g ( z ) , where a = log ( − 2 ) η and g ( z ) satisfies g ( z + η ) = − g ( z ) ?

In this article, we study this question and prove the following theorem.

Theorem 2

Suppose that f ( z ) is a meromorphic solution of the difference equation Δ η n f ( z ) = f ( z ) , n = 1 , 2 . Then, there exists g ( z ) satisfying g ( z + η ) = − g ( z ) such that f ( z ) = e a z g ( z ) , where a = log ( − 2 ) η .

Remark 2

The following example shows that Theorem 2 is not valid for n ≥ 3 .

Example 2

Let n ( ≥ 3 ) be a positive integer, let t ( ≠ − 1 ) be a solution of

∑ i = 0 n ( − 1 ) i ( − 2 ) n − i C n i t n − i = 1 ,

and let f ( z ) = e a z g ( z ) , where a = log ( − 2 ) η and g ( z ) satisfies g ( z + η ) = t g ( z ) .

Set

(1.3) h ( t ) = ∑ i = 0 n ( − 1 ) i ( − 2 ) n − i C n i t n − i − 1 = ( − 2 ) n t n + n 2 t n − 1 + … + 1 2 n − 1 ( − 2 ) n .

Obviously, h ( t ) has n zeros (counting multiplicities) and h ( − 1 ) = 0 .

We claim that there exists t 0 satisfying t 0 ≠ − 1 and h ( t 0 ) = 0 . Otherwise, t = − 1 is the only zero of h ( t ) , and it follows

h ( t ) = ( − 2 ) n ( t + 1 ) n = ( − 2 ) n ( t n + n t n − 1 + … + 1 ) .

This contradicts with (1.3). Hence, there must exist t 0 satisfying t 0 ≠ − 1 and h ( t 0 ) = 0 . That is, t 0 is a solution of ∑ i = 0 n ( − 1 ) i ( − 2 ) n − i C n i t n − i = 1 .

Hence, for n ≥ 3 , by f ( z ) = e a z g ( z ) , a = log ( − 2 ) η and g ( z ) satisfies g ( z + η ) = t g ( z ) , we have

Δ η n f ( z ) − f ( z ) = ∑ i = 0 n ( − 1 ) i C n i f ( z + ( n − i ) η ) − f ( z ) = ∑ i = 0 n ( − 1 ) i C n i e a ( z + ( n − i ) η ) g ( z + ( n − i ) η ) − e a z g ( z ) = e a z g ( z ) ∑ i = 0 n ( − 1 ) i ( − 2 ) n − i C n i t n − i − 1 ≡ 0 .

Therefore, for n ≥ 3 , there exists g ( z ) such that f ( z ) = e a z g ( z ) is a meromorphic solution of the difference equation Δ η n f ( z ) = f ( z ) , but g ( z + η ) ≠ − g ( z ) .

In this article, we further study Question 1 and prove

Theorem 3

Suppose that f ( z ) is a meromorphic solution of the difference equation Δ η n f ( z ) = f ( z ) , n = 1 , 2 . Then, there exists g ( z ) satisfying g ( z + η ) = g ( z ) such that f ( z ) = e a z g ( z ) , where a = log 2 η .

Remark 3

The following example shows that Theorem 3 is not valid for n ≥ 3 .

Example 3

Let n ( ≥ 3 ) be a positive integer, let t ( ≠ 1 ) be a solution of

∑ i = 0 n ( − 1 ) i 2 n − i C n i t n − i = 1 ,

and let f ( z ) = e a z g ( z ) , where a = log 2 η and g ( z ) satisfies g ( z + η ) = t g ( z ) .

Set

(1.4) h ( t ) = ∑ i = 0 n ( − 1 ) i 2 n − i C n i t n − i − 1 = 2 n t n − n 2 t n − 1 + … + − 1 2 n − 1 2 n .

Obviously, h ( t ) has n zeros (counting multiplicities) and h ( 1 ) = 0 .

Similar to Example 2, we claim that there exists t 0 satisfying t 0 ≠ 1 and h ( t 0 ) = 0 . Using the same argument as used in Example 2, we deduce that for n ≥ 3 there exists g ( z ) such that f ( z ) = e a z g ( z ) is a meromorphic solution of the difference equation Δ η n f ( z ) = f ( z ) , but g ( z + η ) ≠ g ( z ) .

In 2019 and 2023, Gao et al. [5] and Ahamed [7] proved.

Theorem D

Suppose that f ( z ) is a meromorphic solution of the difference equation Δ η n f ( z ) = f ( z ) . Then,

f ( z ) = λ 1 z η g 1 ( z ) + λ 2 z η g 2 ( z ) + … + λ n z η g n ( z ) ,

where λ k = 1 + e 2 k π i n ( k = 1 , 2 , … , n ) and g k ( z ) ( k = 1 , 2 , … , n ) are meromorphic functions of period η .

In 2022, Barki et al. [8] proved.

Theorem E

Suppose that f ( z ) is a nonconstant meromorphic solution of the difference equation Δ η n f ( z ) = f ( z ) . Then, f ( z + n η ) ≡ 2 n f ( z ) .

Theorems D and E are not valid for n ≥ 3 and valid for n = 1 , 2 .

Theorem 4

Suppose that f ( z ) is a meromorphic solution of the difference equation Δ η n f ( z ) = f ( z ) , n = 1 , 2 . Then,

f ( z ) = 2 z η g ( z ) ,

where g ( z ) is a meromorphic function with period η .

Remark 4

Example 3 shows that Theorems D and 4 are not valid for n ≥ 3 . In fact, there exist f ( z ) = e a z g ( z ) , which is a solution of the difference equation Δ η n f ( z ) = f ( z ) , where a = log 2 η , but g ( z + η ) ≠ g ( z ) .

Theorem 5

Suppose that f ( z ) is a nonconstant meromorphic solution of the difference equation Δ η n f ( z ) = f ( z ) , n = 1 , 2 . Then, f ( z + n η ) = 2 n f ( z ) .

Remark 5

Example 2 shows that Theorems E and 5 are not valid for n ≥ 3 . In fact, let t ( ≠ − 1 ) be a solution of ∑ i = 0 n ( − 1 ) i ( − 2 ) n − i C n i t n − i = 1 , and let g ( z ) satisfy g ( z + η ) = t g ( z ) . Set f ( z ) = e a z g ( z ) , where a = log ( − 2 ) η . Then, f ( z ) is a solution of the difference equation Δ η n f ( z ) = f ( z ) . Obviously, t n ≠ ( − 1 ) n . Therefore, we obtain

f ( z + n η ) = e a ( z + n η ) g ( z + n η ) = ( − 2 t ) n e a z g ( z ) = ( − 2 t ) n f ( z ) ≠ 2 n f ( z ) .

Hence, for n ≥ 3 , Theorems E and 5 are not valid.

In 2019, Ahamed [9] proved.

Theorem F

Let L η ( f ( z ) ) = a 1 f ( z + η ) + a 0 f ( z ) , where a 1 , a 0 are nonzero complex numbers. Suppose that f ( z ) is a transcendental meromorphic solution of equation L η ( f ( z ) ) = f ( z ) . Then,

(1.5) f ( z ) = 1 − a 0 a 1 z η g ( z ) , a 0 + a 1 ≠ 1 ; g ( z ) , a 0 + a 1 = 1 ,

where g ( z ) is a meromorphic function with period η .

In 2021, Ahamed [10] posed the following conjecture.

Conjecture 2

Let

(1.6) L η n ( f ( z ) ) = a n f ( z + n η ) + … + a 1 f ( z + η ) + a 0 f ( z ) ,

where a i ( i = 0 , 1 , 2 , … , n ) and η are nonzero complex numbers. Suppose that f ( z ) is a meromorphic solution of the difference equation L η n ( f ( z ) ) = f ( z ) . Then,

f ( z ) = λ 1 z η g 1 ( z ) + λ 2 z η g 2 ( z ) + … + λ n z η g n ( z ) ,

where g i ( z ) ( i = 1 , 2 , … , n ) are meromorphic functions of period η , and λ i ( i = 1 , 2 , … , n ) are roots of equation c n z n + … + c 1 z + c 0 = 1 .

In 2023, Ahamed [7] claimed that he give a positive answer to Conjecture 2.

Theorem G

Let

L η n ( f ( z ) ) = a n f ( z + n η ) + … + a 1 f ( z + η ) + a 0 f ( z ) .

where a i ( i = 0 , 1 , 2 , … , n ) and η are nonzero complex numbers. Suppose that f ( z ) is a meromorphic solution of the difference equation L η n ( f ( z ) ) = f ( z ) . Then,

f ( z ) = λ 1 z η g 1 ( z ) + λ 2 z η g 2 ( z ) + … + λ n z η g n ( z ) ,

where g i ( z ) ( i = 1 , 2 , … , n ) are meromorphic function of period η , and λ i ( i = 1 , 2 , … , n ) are roots of equation c n z n + … + c 1 z + c 0 = 1 .

Theorem G is not valid for n ≥ 3 and n = 2 with a 0 ≠ 1 .

Theorem 6

Let L η n ( f ( z ) ) be a difference-polynomial of f ( z ) as (1.6). Suppose that f ( z ) is a meromorphic solution of the difference equation L η n ( f ( z ) ) = f ( z ) . Then,

n = 1 , a 0 = 1 , f ( z ) ≡ 0 ;
n = 1 , a 0 ≠ 1 , f ( z ) = λ z η g ( z ) , where λ = 1 − a 0 a 1 and g ( z ) is a meromorphic function with period η for a 1 ≠ 0 and f ( z ) ≡ 0 for a 1 = 0 ;
n = 2 , a 0 = 1 , f ( z ) = λ z η g ( z ) , where λ = − a 1 a 2 and g ( z ) is a meromorphic function with period η for a 2 ≠ 0 and f ( z ) ≡ 0 for a 2 = 0 .

Remark 6

Example 3 shows that Theorems G and 6 are not valid for n ≥ 3 , and the following example shows that Theorems G and 6 are not valid for n = 2 with a 0 ≠ 1 .

Example 4

Set a 2 = 1 , a 1 = − 3 , a 0 = 3 . Then,

L η 2 ( f ( z ) ) = f ( z + 2 η ) − 3 f ( z + η ) + 3 f ( z ) .

Let h ( t ) = 4 t 2 − 6 t + 2 . Obviously, h ( 1 2 ) = 0 . Let f ( z ) = e a z g ( z ) , where a = log 2 η and g ( z ) satisfies g ( z + η ) = 1 2 g ( z ) . Thus, f ( z ) = e a z g ( z ) is a solution of L η 2 ( f ( z ) ) = f ( z ) , but g ( z + η ) ≠ g ( z ) .

Remark 7

From Theorem 6, we obtain Theorem F.

In 2009, Heittokangas et al. [11] proved,

Theorem H

Let f ( z ) be a nonconstant meromorphic function such that ρ ( f ) < 2 , and let η ( ≠ 0 ) , a be two complex numbers. If f ( z + η ) and f ( z ) share a , ∞ CM, then

f ( z + η ) − a f ( z ) − a ≡ C ,

where C is a nonzero constant.

In 2018–2024, Huang and Zhang [12] and Adud and Chakraborty [13] investigated this question further.

In this article, we extend Theorem H and the results of [12,13].

Theorem 7

Let a be a complex number, and let f ( z ) and g ( z ) be two transcendental meromorphic functions with ρ ( f ) < 2 and for any ε > 0 , m r , g − a f − a < O ( r ρ ( f ) − 1 + ε ) . If f ( z ) and g ( z ) share a , ∞ CM, then

g ( z ) − a f ( z ) − a ≡ C ,

where C is a nonzero constant.

Theorem 8

Let f ( z ) be a transcendental meromorphic function with ρ ( f ) < 2 , let η be a complex number and n a positive integer, and let Q ( f ( z ) ) be a homogeneous differential-difference polynomial with deg ( Q ( f ) ) = n . If Q ( f ( z ) ) and f n ( z + η ) share 0 , ∞ CM, then

Q ( f ( z ) ) f n ( z + η ) ≡ C ,

where C is a nonzero constant.

Remark 8

From Theorem 8, we obtain Theorem H and the results of [12, 13].

2 Some lemmas

Lemma 1

[14, 15] Let f be a nonconstant meromorphic 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 ) .

Particularly, if ρ ( f ) < + ∞ , then for any ε > 0 , we have

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

Lemma 2

[2] Let f ( z ) be a meromorphic function. If f ( z ) ≠ 0 , ∞ , there exists an entire function α ( z ) such that f ( z ) = e α ( z ) .

Lemma 3

[16] Let f be a meromorphic function of finite order, and let η be a nonzero finite complex number. Then, for each positive integer k , ρ ( Δ η k f ) ≤ ρ ( f ) .

3 Proof of Theorem 1

Proof

Since f ( z ) and L ( f ( z ) ) share 0 , ∞ CM, by Lemma 2, there exists an entire function β such that

(3.1) L ( f ( z ) ) f ( z ) = e β .

By (1.1), (3.1), and Lemma 1, we have

(3.2) T ( r , e β ) = m ( r , e β ) = S ( r , f ) .

Next, we consider two cases.

Case 1. e β ≡ 1 . It follows from (3.1) that L ( f ( z ) ) ≡ f ( z ) .

Case 2. e β ≢ 1 .

Let z 0 satisfy f ( z 0 ) = b ( z 0 ) and b ( z 0 ) ≠ 0 , ∞ . Then, by L ( f ( z ) ) and f ( z ) share b ( z ) , we deduce that L ( f ) ( z 0 ) = b ( z 0 ) . It follows from (3.1) that e β ( z 0 ) = 1 . Since b ( z ) is a small function of f ( z ) , by (3.2) and Nevanlinna’s first fundamental theorem, we obtain

(3.3) N ¯ r , 1 f − b ≤ N r , 1 e β − 1 + N r , 1 b + N ( r , b ) ≤ T ( r , e β ) + 2 T ( r , b ) ≤ S ( r , f ) .

By (3.1), we have

(3.4) L ( f ) − b = e β ( f − b e − β ) .

It follows from L ( f ) and f share b IM, (3.3) and (3.4) that

(3.5) N ¯ r , 1 f − b e − β = N ¯ r , 1 L ( f ) − b = N ¯ r , 1 f − b ≤ S ( r , f ) .

Since Θ ( ∞ , f ) > 0 , there exists λ < 1 such that

(3.6) N ¯ ( r , f ) ≤ λ T ( r , f ) + S ( r , f ) .

By (3.3), (3.5), (3.6), and Theorem 2.5 (in [1], page 47), we have

(3.7) T ( r , f ) ≤ N ¯ ( r , f ) + N ¯ r , 1 f − b + N ¯ r , 1 f − b e − β + S ( r , f ) ≤ λ T ( r , f ) + S ( r , f ) .

It follows from λ < 1 that T ( r , f ) ≤ S ( r , f ) , which is a contradiction.

This completes the proof of Theorem 1.□

4 Proof of Theorem 2

Proof

We consider two cases.

Case 1. n = 1 . It follows from Δ η f ( z ) = f ( z ) that f ( z + η ) = 2 f ( z ) .

Set

g ( z ) = f ( z ) e a z .

By a = log ( − 2 ) η , we have

(4.1) g ( z + η ) = f ( z + η ) e a ( z + η ) = 2 f ( z ) e a z ⋅ e a η = − f ( z ) e a z = − g ( z ) .

It follows f ( z ) = e a z g ( z ) , where a = log ( − 2 ) η and g ( z ) satisfies g ( z + η ) = − g ( z ) .

Case 2. n = 2 . It follows from Δ η 2 f ( z ) = f ( z ) that

(4.2) f ( z + 2 η ) = 2 f ( z + η ) .

That is f ( z + η ) = 2 f ( z ) . Using the same argument as used in Case 1, we obtain f ( z ) = e a z g ( z ) , where a = log ( − 2 ) η and g ( z ) satisfies g ( z + η ) = − g ( z ) .

This completes the proof of Theorem 2.□

5 Proof of Theorem 3

Proof

We consider two cases.

Case 1. n = 1 . It follows from Δ η f ( z ) = f ( z ) that f ( z + η ) = 2 f ( z ) .

Set

g ( z ) = f ( z ) e a z .

By a = log 2 η , we have

(5.1) g ( z + η ) = f ( z + η ) e a ( z + η ) = 2 f ( z ) e a z ⋅ e a η = f ( z ) e a z = g ( z ) .

It follows f ( z ) = e a z g ( z ) , where a = log 2 η and g ( z ) satisfies g ( z + η ) = g ( z ) .

Case 2. n = 2 . It follows from Δ η 2 f ( z ) = f ( z ) that

(5.2) f ( z + 2 η ) = 2 f ( z + η ) .

That is f ( z + η ) = 2 f ( z ) . Using the same argument as used in Case 1, we obtain f ( z ) = e a z g ( z ) , where a = log 2 η and g ( z ) satisfies g ( z + η ) = g ( z ) .

This completes the proof of Theorem 3.□

By imitating the proof of Theorem 3, it is easy to prove Theorem 4. We omit the proof of Theorem 4.

6 Proof of Theorem 5

Proof

It follows from the condition of Theorem 5 that Δ η n f ( z ) ≡ f ( z ) .

Next, we consider two cases.

Case 1. n = 1 . It follows from Δ η f ( z ) ≡ f ( z ) that f ( z + η ) ≡ 2 f ( z ) .

Case 2. n = 2 . It follows from Δ η 2 f ( z ) ≡ f ( z ) that f ( z + 2 η ) ≡ 2 f ( z + η ) . That is, f ( z + η ) = 2 f ( z ) . Hence, we obtain f ( z + 2 η ) = 2 2 f ( z ) .

This completes the proof of Theorem 5.□

7 Proof of Theorem 6

Proof

We consider three cases.

Case 1. n = 1 , a 0 = 1 . It follows from L η 1 ( f ( z ) ) = f ( z ) that f ( z + η ) = 0 . That is, f ( z ) ≡ 0 .

Case 2. n = 1 , a 0 ≠ 1 . It follows from L η 1 ( f ( z ) ) = f ( z ) that a 1 f ( z + η ) = ( 1 − a 0 ) f ( z ) .

Set

g ( z ) = f ( z ) λ z η ,

where λ = 1 − a 0 a 1 . Then,

g ( z + η ) = f ( z + η ) λ z + η η = 1 − a 0 a 1 f ( z ) λ z η λ = g ( z ) .

It follows f ( z ) = λ z η g ( z ) , where λ = 1 − a 0 a 1 and g ( z ) satisfies g ( z + η ) = g ( z ) .

Case 3. n = 2 , a 0 = 1 . It follows from L η 2 ( f ( z ) ) = f ( z ) that a 2 f ( z + 2 η ) = − a 1 f ( z + η ) . That is a 2 f ( z + η ) = − a 1 f ( z ) . Using the same argument as used in Case 2, we obtain f ( z ) = λ z η g ( z ) , where λ = − a 1 a 2 and g ( z ) satisfies g ( z + η ) = g ( z ) .

This completes the proof of Theorem 6.□

8 Proof of Theorem 7

Proof

Since f ( z ) and g ( z ) share a , ∞ CM, by Lemma 2, we know that there exists an entire function γ ( z ) such that

(8.1) g ( z ) − a f ( z ) − a = e γ ( z ) .

Set ε = 1 − ρ ( f ) 2 . By (8.1), we have

(8.2) T ( r , e γ ) = m ( r , e γ ) = m r , g − a f − a ≤ O ( r ρ ( f ) − 1 + ε ) ≤ O r ρ ( f ) 2 .

It follows from ρ ( f ) < 2 that e γ is a constant. Therefore, we deduce

g ( z ) − a f ( z ) − a ≡ C ,

where C is a nonzero constant.

This completes the proof of Theorem 7.□

9 Proof of Theorem 8

Proof

By ρ ( f ) < 2 , Lemma 3, and Theorem 1.21 (in [3], page 36), we know that

(9.1) ρ ( Q ( f ) ) < 2 , ρ ( f n ( z + η ) ) < 2 .

Since Q ( f ( z ) ) and f n ( z + η ) share 0 , ∞ CM, by Lemma 2 we know that there exists an entire function δ ( z ) such that

(9.2) Q ( f ( z ) ) f n ( z + η ) = e δ ( z ) .

Set ε = 1 − ρ ( f n ( z + η ) ) 2 . By (9.2), Lemma 1, and logarithmic derivative lemma (Theorem 3.1 in [1], page 55), we have

(9.3) T ( r , e δ ) = m ( r , e δ ) = m r , Q ( f ( z ) ) f n ( z + η ) ≤ O ( r ρ ( f n ( z + η ) ) − 1 + ε ) ≤ O r ρ ( f n ( z + η ) ) 2 .

It follows from (9.1) and (9.3) that e δ is a constant. Therefore, we deduce

Q ( f ( z ) ) f n ( z + η ) ≡ C ,

where C is a nonzero constant.

This completes the proof of Theorem 8.□

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 Nos. 12171127 and 12371074).
Author contributions: All authors have accepted responsibility for the entire content of this manuscript and approved its submission.
Conflict of interest: The authors state no conflict of interest.
Ethical approval: The conducted research is not related to either human or animal use.
Data availability statement: Data sharing is not applicable to this article as no datasets were generated or analyzed during the current study.

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Received: 2024-05-26

Revised: 2024-10-20

Accepted: 2024-11-10

Published Online: 2024-12-17

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

Articles in the same Issue

https://doi.org/10.1515/math-2024-0100

Keywords for this article

difference equations; meromorphic functions; uniqueness; small functions

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