Some results on value distribution concerning Hayman's alternative

Xinyu Zhuang; Ruilin Zheng; Zhiying He; Mingliang Fang

doi:10.1515/math-2024-0117

Article Open Access

Some results on value distribution concerning Hayman's alternative

Xinyu Zhuang , Ruilin Zheng , Zhiying He and Mingliang Fang

Published/Copyright: February 4, 2025

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$Open Mathematics$

From the journal Open Mathematics Volume 23 Issue 1

Abstract

In this article, we study the value distribution of meromorphic functions concerning Hayman’s alternative. We extend and improve some results due to Zhu [A far-reaching form of Hayman’s inequality; fixed points of meromorphic functions and their derivatives, Kexue Tongbao (Chinese) 31 (1986), no. 11, 801–804], Hua and Chuang [On a conjecture of Hayman, Acta Math. Sinica (N.S.) 7 (1991), no. 2, 119–126], and Charak and Singh [A value distribution result related to Hayman’s alternative, Commun. Korean Math. Soc. 34 (2019), no. 2, 495–506].

Keywords: meromorphic functions; value distribution; differential polynomials; Hayman’s alternative

MSC 2010: 30D30; 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 [1–4]. 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 measure.

Let f be a meromorphic function, and let k be a positive integer. We denote by N k ) ( r , f ) the counting function for poles of f with multiplicity ≤ k , counting multiplicity and by N ¯ k ) ( r , f ) the corresponding one for which multiplicity is not counted. Let N ( k ( r , f ) be the counting function for poles of f with multiplicity ≥ k , counting multiplicity and by N ¯ ( k ( r , f ) be the corresponding one for which multiplicity is not counted.

Let f be a nonconstant meromorphic function, let a 1 , a 2 , … , a k be small functions of f , and let n 0 , n 1 , n 2 , … , n k be nonnegative integers. We define the differential monomial of f as follows:

M ( f ) = f n 0 ( f ′ ) n 1 … ( f ( k ) ) n k ,

and its degree γ M = n 0 + n 1 + … + n k . Let M 1 ( f ) , M 2 ( f ) , … , M n ( f ) be differential monomials. We define the differential polynomial of f as follows:

(1.1) H ( f ) = a 1 M 1 ( f ) + a 2 M 2 ( f ) + … + a n M n ( f ) ,

and define γ H = max { γ M 1 , γ M 2 , … , γ M n } , γ ̲ H = min { γ M 1 , γ M 2 , … , γ M n } by the degree and the lower degree of H , respectively.

In 1986, Zhu [5] proved the following result.

Theorem 1.1

Let f be a transcendental meromorphic function, and let φ ( ≢ 0 ) be a small function of f. Then

T ( r , f ) ≤ 8 N r , 1 f + 8 N ¯ r , 1 f ′ − φ + S ( r , f ) .

In this article, we improve Theorem 1.1 as follows.

Theorem 1.2

Let f be a transcendental meromorphic function, and let φ ( ≢ 0 ) be a small function of f. Then

T ( r , f ) ≤ 5 N r , 1 f + 5 N ¯ r , 1 f ′ − φ + S ( r , f ) .

In 1991, Hua and Chuang [6] proved the following result.

Theorem 1.3

Let f be a nonconstant meromorphic function, and let n , m be two positive integers. Assume that Q ( f ) = f m H ( f ) , where H ( f ) ( ≢ 0 ) is a differential polynomial defined by (1.1). Then for any nonzero complex number b :

If n ≥ 3 , then T ( r , f ) ≤ 2 m N ¯ r , 1 Q n Q ′ − b + S ( r , f ) ;
If n = 2 , then T ( r , f ) ≤ 1 m N ¯ ( r , f ) + 2 m N ¯ r , 1 Q 2 Q ′ − b + S ( r , f ) .

In this article, we prove the following result.

Theorem 1.4

If n ≥ 3 , then T ( r , f ) ≤ 1 m N ¯ r , 1 Q n Q ′ − b + S ( r , f ) ;
If n = 2 , then T ( r , f ) ≤ 1 m N ¯ ( r , f ) + 1 m N ¯ r , 1 Q 2 Q ′ − b + S ( r , f ) .

In 2019, Charak and Singh [7] proved the following result.

Theorem 1.5

Let f be a transcendental meromorphic function, let ϕ be a small function of f such that f and ϕ have no common poles, and let k be a positive integer. If f ≠ 0 and f ( k ) ≠ ϕ , then f ( k + 1 ) = ϕ and f ( k + 1 ) = ϕ ′ have infinitely many solutions.

Remark 1.6

Theorem 1.5 is not valid by the following example.

Example 1.7

Let f = e z , and let ϕ ≡ 0 . Obviously, f ≠ 0 and f ( k ) ≠ ϕ , but f ( k + 1 ) = ϕ does not have infinitely many solutions.

Although Theorem 1.5 is not valid, we have the following result.

Theorem 1.8

Let f be a transcendental meromorphic function, let ϕ be a small function of f, and let k be a positive integer. If f ≠ 0 and f ( k ) ≠ ϕ , then ϕ ≡ 0 . In addition, if k ≥ 2 , then f = e a z + b , where a ( ≠ 0 ) , b are constants; if k = 1 , then f ″ has infinitely many zeros, except f = e a z + b , where a ( ≠ 0 ) , b are constants.

Remark 1.9

f ≠ 0 refers to the fact that for any z ∈ C , it holds that f ( z ) ≠ 0 . f ( k ) ≠ ϕ refers to the fact that for any z ∈ C , it holds that f ( k ) ( z ) ≠ ϕ ( z ) .

The following examples show that two cases occur in Theorem 1.8.

Example 1.10

Let f ( z ) = e z , and let k be a positive integer. Obviously, f ≠ 0 , f ( k ) ≠ 0 .

Example 1.11

Let f ( z ) = e e z . Obviously, f ≠ 0 , f ′ ≠ 0 . We have f ″ = e e z e z ( e z + 1 ) . Thus, f ″ has infinitely many zeros.

2 Some lemmas

For the proof of our results, we need the following lemmas.

Lemma 2.1

[3] Let f be a nonconstant meromorphic function, and let k be a positive integer. Then

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

Lemma 2.2

[1,3] Let f be a nonconstant meromorphic function, let n be a positive integer, and let a 1 , a 2 , … , a n be distinct small functions of f. Then

m r , 1 f − a 1 + … + m r , 1 f − a n ≤ m r , 1 f − a 1 + … + 1 f − a n + S ( r , f ) .

It follows from the theorem in [8, p. 247] the following result.

Lemma 2.3

Let f be a transcendental meromorphic function. Then

1 4 ≤ T ( r , f ) T ( r , f ′ ) ≤ 3 e + 1 4 ,

as r → ∞ on a set of positive lower logarithmic density.

Lemma 2.4

[1,3] Let f 1 and f 2 be two nonconstant 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 2.5

Let f be a nonconstant meromorphic function, let b be a nonzero complex number, and let n be a positive integer. Then we have

n ≥ 3 , T ( r , f ) ≤ N ¯ r , 1 f n f ′ − b + S ( r , f ) ;
n = 2 , T ( r , f ) ≤ N ¯ ( r , f ) + N ¯ r , 1 f 2 f ′ − b + S ( r , f ) .

Proof

By Lemmas 2.1 and 2.2 and Nevanlinna’s first fundamental theorem, we have

(2.1) m r , 1 f n + 1 + m r , 1 f n f ′ − b ≤ m r , f n f ′ f n + 1 + m r , 1 f n f ′ + m r , 1 f n f ′ − b ≤ m r , 1 f n f ′ + 1 f n f ′ − b + S ( r , f ) ≤ m r , ( f n f ′ ) ′ f n f ′ + ( f n f ′ − b ) ′ f n f ′ − b + m r , 1 ( f n f ′ ) ′ + S ( r , f ) ≤ T ( r , ( f n f ′ ) ′ ) − N r , 1 ( f n f ′ ) ′ + S ( r , f ) ≤ T ( r , f n f ′ ) + N ¯ ( r , f ) − N r , 1 ( f n f ′ ) ′ + S ( r , f ) .

Then, by adding N r , 1 f n + 1 + N r , 1 f n f ′ − b to both sides of (2.1), we have

m r , 1 f n + 1 + m r , 1 f n f ′ − b + N r , 1 f n + 1 + N r , 1 f n f ′ − b ≤ N r , 1 f n + 1 + N r , 1 f n f ′ − b + T ( r , f n f ′ ) + N ¯ ( r , f ) − N r , 1 ( f n f ′ ) ′ + S ( r , f ) .

It follows

T r , 1 f n + 1 + T r , 1 f n f ′ − b ≤ N r , 1 f n + 1 + N r , 1 f n f ′ − b + T ( r , f n f ′ ) + N ¯ ( r , f ) − N r , 1 ( f n f ′ ) ′ + S ( r , f ) .

By Nevanlinna’s first fundamental theorem, we have

T r , 1 f n f ′ − b = T ( r , f n f ′ ) + O ( 1 ) .

Thus, we obtain

T r , 1 f n + 1 ≤ N r , 1 f n + 1 + N r , 1 f n f ′ − b + N ¯ ( r , f ) − N r , 1 ( f n f ′ ) ′ + S ( r , f ) .

From T r , 1 f n + 1 = ( n + 1 ) T ( r , f ) + S ( r , f ) , we obtain

( n + 1 ) T ( r , f ) ≤ N r , 1 f n + 1 + N r , 1 f n f ′ − b + N ¯ ( r , f ) − N r , 1 ( f n f ′ ) ′ + S ( r , f ) .

If z 0 is a zero of f with multiplicity l 1 , then z 0 is a zero of f n + 1 with multiplicity ( n + 1 ) l 1 . Hence, z 0 must be a zero of ( f n f ′ ) ′ with multiplicity ( n + 1 ) l 1 − 2 . Similarly, if z 0 is a zero of f n f ′ − b with multiplicity l 2 , then z 0 must be a zero of ( f n f ′ − b ) ′ with multiplicity l 2 − 1 , which yields z 0 is also a zero of ( f n f ′ ) ′ with multiplicity l 2 − 1 . It follows

N r , 1 f n + 1 + N r , 1 f n f ′ − b − N r , 1 ( f n f ′ ) ′ ≤ 2 N ¯ r , 1 f + N ¯ r , 1 f n f ′ − b − N 0 * r , 1 ( f n f ′ ) ′ ,

where N 0 * r , 1 ( f n f ′ ) ′ is the counting function for the zeros of ( f n f ′ ) ′ , which are not zeros of f n + 1 ( f n f ′ − b ) .

Hence, we have

(2.2) ( n + 1 ) T ( r , f ) ≤ N r , 1 f n + 1 + N r , 1 f n f ′ − b + N ¯ ( r , f ) − N r , 1 ( f n f ′ ) ′ + S ( r , f ) ≤ N ¯ ( r , f ) + 2 N ¯ r , 1 f + N ¯ r , 1 f n f ′ − b − N 0 * r , 1 ( f n f ′ ) ′ + S ( r , f ) ≤ N ¯ ( r , f ) + 2 N ¯ r , 1 f + N ¯ r , 1 f n f ′ − b + S ( r , f ) .

Next we consider two cases.

Case 1. n ≥ 3 . From (2.2), we obtain

( n − 2 ) T ( r , f ) ≤ N ¯ r , 1 f n f ′ − b + S ( r , f ) .

It follows

(2.3) T ( r , f ) ≤ N ¯ r , 1 f n f ′ − b + S ( r , f ) .

Case 2. n = 2 . From (2.2), we have

(2.4)□ T ( r , f ) ≤ N ¯ ( r , f ) + N ¯ r , 1 f 2 f ′ − b + S ( r , f ) .

Lemma 2.6

[6] Let f be a nonconstant meromorphic function, and let k be a positive integer. Then

N r , 1 f ( k ) ≤ N r , 1 f + k N ¯ ( r , f ) + S ( r , f ) .

Lemma 2.7

[9] Let f be a transcendental meromorphic function, let ϕ ( ≢ 0 ) be a small function of f, and let k be a positive integer. If N r , 1 f = S ( r , f ) , then N r , 1 f ( k ) − ϕ ≠ S ( r , f ) .

Lemma 2.8

[10] Let f be a nonconstant meromorphic function, and let k ( ≥ 2 ) be an integer. If f ≠ 0 , f ( k ) ≠ 0 , then either f = e a z + b or f = ( a z + b ) − n , where a ( ≠ 0 ) , b are constants and n is a positive integer.

Remark 2.9

f ≠ 0 refer to the fact that for any z ∈ C , it holds that f ( z ) ≠ 0 . f ( k ) ≠ 0 refer to the fact that for any z ∈ C , it holds that f ( k ) ( z ) ≠ 0 .

It follows from Theorem 2.2 in [11, p. 423] the following result.

Lemma 2.10

Let f be a nonconstant meromorphic function, and let k ( ≥ 2 ) be an integer. If f and f ( k ) have finitely many zeros, then f = R e P , where R is a rational function and P is a polynomial.

3 Proof of Theorem 1.2

Set L = f ′ φ f ″ φ ′ . By Lemmas 2.1 and 2.2, we obtain

(3.1) m r , 1 f + m r , 1 f ′ − φ ≤ m r , f ′ f + m r , 1 f ′ + m r , 1 f ′ − φ ≤ m r , 1 f ′ + 1 f ′ − φ + S ( r , f ) ≤ m r , L f ′ + L f ′ − φ + m r , 1 L + S ( r , f ) ≤ m r , L f ′ + m r , L f ′ − φ + m r , 1 L + S ( r , f ) .

From the definition of L , φ ( ≢ 0 ) is a small function of f , and from Lemma 2.1, we have

(3.2) m r , L f ′ = m r , φ ′ − φ f ″ f ′ ≤ m ( r , φ ′ ) + m ( r , φ ) + m r , f ″ f ′ + S ( r , f ) ≤ S ( r , f )

and

(3.3) m r , L f ′ − φ = m r , φ ′ − φ ( f ′ − φ ) ′ f ′ − φ ≤ m ( r , φ ′ ) + m ( r , φ ) + m r , ( f ′ − φ ) ′ f ′ − φ + S ( r , f ) ≤ S ( r , f ) .

By combining (3.1)–(3.3) and Nevanlinna’s first fundamental theorem, we obtain

(3.4) m r , 1 f + m r , 1 f ′ − φ ≤ m r , 1 L + S ( r , f ) ≤ T r , 1 L − N r , 1 L + S ( r , f ) ≤ T ( r , L ) − N r , 1 L + S ( r , f ) .

From the definition of L and (3.3), we have

(3.5) m ( r , L ) = m r , ( f ′ − φ ) φ ′ − φ f ″ − φ ′ f ′ − φ ≤ m ( r , f ′ − φ ) + m r , φ ′ − φ f ″ − φ ′ f ′ − φ + S ( r , f ) ≤ m ( r , f ′ − φ ) + S ( r , f ) .

Similarly,

(3.6) N ( r , L ) = N ( r , ( f ′ − φ ) φ ′ − ( f ″ − φ ′ ) φ ) ≤ N ( r , ( f ′ − φ ) ′ ) + S ( r , f ) ≤ N ( r , f ′ − φ ) + N ¯ ( r , f ′ − φ ) + S ( r , f ) ≤ N ( r , f ′ − φ ) + N ¯ ( r , f ) + S ( r , f ) .

By combining (3.4)–(3.6), we obtain

(3.7) m r , 1 f + m r , 1 f ′ − φ ≤ T ( r , L ) − N r , 1 L + S ( r , f ) ≤ m ( r , L ) + N ( r , L ) − N r , 1 L + S ( r , f ) ≤ m ( r , f ′ − φ ) + N ( r , f ′ − φ ) + N ¯ ( r , f ) − N r , 1 L + S ( r , f ) ≤ T ( r , f ′ − φ ) + N ¯ ( r , f ) − N r , 1 L + S ( r , f ) .

By adding N r , 1 f + N r , 1 f ′ − φ to both sides of (3.7), we have

m r , 1 f + m r , 1 f ′ − φ + N r , 1 f + N r , 1 f ′ − φ ≤ N r , 1 f + N r , 1 f ′ − φ + T ( r , f ′ − φ ) + N ¯ ( r , f ) − N r , 1 L + S ( r , f ) .

Hence, we obtain

T r , 1 f + T r , 1 f ′ − φ ≤ N r , 1 f + N r , 1 f ′ − φ + T ( r , f ′ − φ ) + N ¯ ( r , f ) − N r , 1 L + S ( r , f ) .

It follows from Nevanlinna’s first fundamental theorem that

T r , 1 f = T ( r , f ) + O ( 1 ) , T r , 1 f ′ − φ = T ( r , f ′ − φ ) + O ( 1 ) .

Therefore,

(3.8) T ( r , f ) ≤ N r , 1 f + N r , 1 f ′ − φ + N ¯ ( r , f ) − N r , 1 L + S ( r , f ) ≤ N ¯ ( r , f ) + N r , 1 f + N ¯ r , 1 f ′ − φ + S ( r , f ) .

It follows from (3.8) and N ¯ ( r , f ) = N ¯ 1 ) ( r , f ) + N ¯ ( 2 ( r , f ) that

T ( r , f ) ≤ N ¯ 1 ) ( r , f ) + N ¯ ( 2 ( r , f ) + N r , 1 f + N ¯ r , 1 f ′ − φ + S ( r , f ) ≤ N 1 ) ( r , f ) + 1 2 N ( 2 ( r , f ) + N r , 1 f + N ¯ r , 1 f ′ − φ + S ( r , f ) ≤ 1 2 N 1 ) ( r , f ) + 1 2 T ( r , f ) + N r , 1 f + N ¯ r , 1 f ′ − φ + S ( r , f ) .

Hence, we obtain

(3.9) T ( r , f ) ≤ N 1 ) ( r , f ) + 2 N r , 1 f + 2 N ¯ r , 1 f ′ − φ + S ( r , f ) .

Obviously, f ″ φ − φ ′ f ′ ≢ 0 . Set

P = f ( f ′ − φ ) f ″ φ − φ ′ f ′ , G = ( P φ ) ″ + ( P φ ′ ) ′ + φ .

Thus, we have

(3.10) P φ f ″ − f ( f ′ − φ ) ≡ P φ ′ f ′ .

Let z 0 be a simple pole of f , and satisfy that φ ( z 0 ) ≠ 0 , ∞ , φ ′ ( z 0 ) ≠ 0 . Obviously, P φ and P φ ′ are holomorphic at z 0 , so we obtain

(3.11) f ( z ) = c − 1 z − z 0 + c 0 + c 1 ( z − z 0 ) + c 2 ( z − z 0 ) 2 + … ,

where c − 1 ( ≠ 0 ) , c 0 , c 1 , c 2 , … are constants.

(3.12) φ ( z ) = φ ( z 0 ) + φ ′ ( z 0 ) ( z − z 0 ) + 1 2 φ ″ ( z 0 ) ( z − z 0 ) 2 + … ,

(3.13) P φ ( z ) = P φ ( z 0 ) + ( P φ ) ′ ( z 0 ) ( z − z 0 ) + 1 2 ( P φ ) ″ ( z 0 ) ( z − z 0 ) 2 + … ,

(3.14) P φ ′ ( z ) = P φ ′ ( z 0 ) + ( P φ ′ ) ′ ( z 0 ) ( z − z 0 ) + 1 2 ( P φ ′ ) ″ ( z 0 ) ( z − z 0 ) 2 + … .

By substituting (3.11)–(3.14) into (3.10), and comparing the coefficients of 1 z − z 0 , we obtain

( P φ ) ″ ( z 0 ) + ( P φ ′ ) ′ ( z 0 ) + φ ( z 0 ) = 0 .

That is, G ( z 0 ) = 0 .

Next we consider the following two cases.

Case 1. G ≢ 0 . Since φ is a small function of f , then by Nevanlinna’s first fundamental theorem and Lemma 2.1, we have

(3.15) N 1 ) ( r , f ) ≤ N r , 1 G + N ¯ r , 1 φ + N ¯ r , 1 φ ′ + N ¯ ( r , φ ) ≤ N r , 1 G + S ( r , f ) ≤ T ( r , G ) + S ( r , f ) ≤ m ( r , P ″ φ + 3 P ′ φ ′ + 2 P φ ″ ) + m ( r , φ ) + N ( r , P ″ φ + 3 P ′ φ ′ + 2 P φ ″ ) + N ( r , φ ) + S ( r , f ) ≤ m ( r , P ) + m r , P ″ φ + 3 P ′ φ ′ + 2 P φ ″ P + N ( r , P ″ ) + S ( r , f ) ≤ m ( r , P ) + N ( r , P ) + 2 N ¯ ( r , P ) + S ( r , f ) ≤ 3 T ( r , P ) + S ( r , f ) .

From Lemma 2.1, we have

(3.16) m r , 1 P = m r , f ″ φ − φ ′ f ′ f ( f ′ − φ ) ≤ m r , f ″ φ − φ ′ f ′ f ′ ( f ′ − φ ) + m r , f ′ f + S ( r , f ) = m r , f ′ − φ f ′ ′ f ′ − φ f ′ + m r , f ′ f + S ( r , f ) ≤ S ( r , f ) .

From the definition of P , we have

N r , 1 P = N r , f ″ φ − φ ′ f ′ f ( f ′ − φ ) .

Since φ ( ≢ 0 ) is a small function of f , we have T ( r , φ ) = S ( r , f ) , it follows S ( r , φ ) = S ( r , f ) . Hence, we obtain T ( r , φ ′ ) = S ( r , f ) .

Next we consider the following three subcases.

Case 1.1. Let z 0 be a pole of f with multiplicity l 1 ( ≥ 1 ) , but φ ( z 0 ) ≠ ∞ . It follows that z 0 is a pole of f ′ − φ and f ″ φ − φ ′ f ′ with multiplicity l 1 + 1 and l 1 + 2 , respectively. In this case, the multiplicity of the poles in the numerator is not greater than that in the denominator. Hence, z 0 is not a pole of 1 P .

Case 1.2. Let z 0 be a pole of f with multiplicity l 1 ( ≥ 1 ) and φ ( z 0 ) = ∞ . From N ( r , φ ) = S ( r , f ) , we can deduce that the zeros of this type of P is S ( r , f ) .

Case 1.3. Let z 0 be a zero of f ′ − φ with multiplicity l 2 ( ≥ 1 ) , but not the zero of f . It follows that z 0 is zero of f ″ − φ ′ with multiplicity l 2 − 1 . From

f ″ φ − φ ′ f ′ = f ″ φ − φ φ ′ + φ φ ′ − φ ′ f ′ = φ ( f ″ − φ ′ ) − φ ′ ( f ′ − φ ) ,

we have

1 P = f ″ φ − φ ′ f ′ f ( f ′ − φ ) = φ ( f ″ − φ ′ ) − φ ′ ( f ′ − φ ) f ( f ′ − φ ) .

It follows that z 0 is a simple pole of 1 P .

Thus, we have

(3.17) N r , 1 P = N r , f ″ φ − φ ′ f ′ f ( f ′ − φ ) + S ( r , f ) ≤ N r , 1 f + N ¯ r , 1 f ′ − φ + S ( r , f ) .

By (3.15)–(3.17), and Nevanlinna’s first fundamental theorem, we obtain

(3.18) N 1 ) ( r , f ) ≤ 3 N r , 1 f + 3 N ¯ r , 1 f ′ − φ + S ( r , f ) .

Case 2. G ≡ 0 . That is,

(3.19) P ″ φ + 3 P ′ φ ′ + 2 P φ ″ + φ ≡ 0 .

Let z 0 be a pole of P with multiplicity l . Then z 0 must be either a zero or a pole of φ .

In fact, if φ ( z 0 ) = a , where a ( ≠ 0 , ∞ ) is a constant, then z 0 is a pole of P ″ φ with multiplicity l + 2 , a pole of P ′ φ ′ with multiplicity at most l + 1 , and a pole of P φ ″ with multiplicity at most l . Thus, by (3.19), we know that z 0 is a pole of P ″ φ + 3 P ′ φ ′ + 2 P φ ″ + φ , a contradiction.

Hence,

N ¯ ( r , P ) ≤ N ¯ ( r , φ ) + N ¯ r , 1 φ ≤ S ( r , f ) .

We define N 0 * r , 1 f ″ φ − φ ′ f ′ is the counting function for the zeros of f ″ φ − φ ′ f ′ , which are not zeros of f ( f ′ − φ ) , and N ¯ 0 * r , 1 f ″ φ − φ ′ f ′ as the corresponding function where the multiplicity is not counted.

Let z 0 be a zero of f ″ φ − φ ′ f ′ , but not the zero of f ( f ′ − φ ) . Then z 0 is a pole of P . It follows from N ¯ ( r , P ) ≤ S ( r , f ) that

(3.20) N ¯ 0 * r , 1 f ″ φ − φ ′ f ′ ≤ N ¯ ( r , P ) ≤ S ( r , f ) .

Let z 1 be a pole of f with multiplicity l 1 ( ≥ 2 ) , but φ ( z 1 ) ≠ 0 , ∞ . Obviously, z 1 is a pole of f ( f ′ − φ ) and f ″ φ − φ ′ f ′ with multiplicity 2 l 1 + 1 and l 1 + 2 , respectively. From the definition of P , we deduce that z 1 must be a pole of P with multiplicity l 1 − 1 . Hence, we have

(3.21) N ¯ ( 2 ( r , f ) ≤ N ¯ ( r , P ) + N ¯ ( r , φ ) + N ¯ r , 1 φ ≤ S ( r , f ) .

Set

g = φ 3 ( f ′ − φ ) 3 ( f ″ φ − f ′ φ ′ ) 2 = φ ( f ′ − φ ) f ″ − φ ′ f ′ − φ − φ ′ φ 2 .

Let z 0 be a simple pole of f with φ ( z 0 ) ≠ 0 , ∞ , φ ′ ( z 0 ) ≠ 0 . Then by (3.11) and (3.12), we obtain

g = − φ ( z 0 ) c − 1 4 + λ 2 ( z − z 0 ) 2 + λ 3 ( z − z 0 ) 3 + … ,

where − φ ( z 0 ) c − 1 4 ≠ 0 , λ 2 , λ 3 , … are constants. Hence, g ( z 0 ) ≠ 0 , ∞ , and g ′ ( z 0 ) = 0 .

Next we consider two subcases.

Case 2.1. g ′ ≢ 0 . It follows from that φ is a small function of f that

(3.22) N 1 ) ( r , f ) ≤ N 0 r , 1 g ′ + N ¯ r , 1 φ + N ¯ r , 1 φ ′ + N ¯ ( r , φ ) ≤ N 0 r , 1 g ′ + S ( r , f ) ,

where N 0 r , 1 g ′ is the counting function for the zeros of g ′ , which are not the zeros of g .

By Nevanlinna’s first fundamental theorem and Lemma 2.1, we have

(3.23) N r , g g ′ − N r , g ′ g = T r , g g ′ − m r , g g ′ − T r , g ′ g + m r , g ′ g = − m r , g g ′ + S ( r , g ) + S ( r , f ) = − m r , g g ′ + S ( r , f ) .

Referring to line -8 and line -6 in [1, p. 57], we obtain

(3.24) N ( r , g ′ ) − N ( r , g ) = N ¯ ( r , g )

and

(3.25) N r , 1 g − N r , 1 g ′ = N ¯ r , 1 g − N 0 r , 1 g ′ .

From (3.24), (3.25), and Lemma 2.4, we have

(3.26) N r , g g ′ − N r , g ′ g = N ( r , g ) + N r , 1 g ′ − N ( r , g ′ ) − N r , 1 g = N 0 r , 1 g ′ − N ¯ r , 1 g − N ¯ ( r , g ) .

By (3.20), (3.21), and the definition of g , we have

(3.27) N ¯ ( r , g ) + N ¯ r , 1 g ≤ N ¯ ( 2 ( r , f ) + N ¯ r , 1 f + N ¯ r , 1 f ′ − φ + N ¯ 0 * r , 1 f ″ φ − φ ′ f ′ + S ( r , f ) ≤ N ¯ r , 1 f + N ¯ r , 1 f ′ − φ + S ( r , f ) .

Combining (3.22), (3.23), (3.26) with (3.27), we obtain

(3.28) N 1 ) ( r , f ) ≤ N 0 r , 1 g ′ + S ( r , f ) ≤ N ¯ r , 1 g + N ¯ ( r , g ) − m r , g g ′ + S ( r , f ) ≤ N ¯ r , 1 g + N ¯ ( r , g ) + S ( r , f ) ≤ N ¯ r , 1 f + N ¯ r , 1 f ′ − φ + S ( r , f ) .

Case 2.2. g ′ ≡ 0 . Then g ≡ A , where A is a constant.

If A = 0 , then φ ( f ′ − φ ) f ″ − φ ′ f ′ − φ − φ ′ φ 2 ≡ 0 . Hence, we obtain that either φ ≡ 0 or f ′ − φ ≡ 0 , a contradiction. Thus, A ≠ 0 . Hence, we have

(3.29) φ ( f ′ − φ ) ≡ A f ″ − φ ′ f ′ − φ − φ ′ φ 2 .

Set F = f ′ − φ φ . Then by (3.29), we obtain

(3.30) F 3 = A φ 2 ( F ′ ) 2 .

Obviously, F ≢ 0 . Thus, we have 1 F = A φ 2 1 F ′ 2 .

Set H = 1 F . Then we have H = B H ′ φ 2 , where B 2 = A ≠ 0 . Hence, H = h 2 , H ′ = 2 h h ′ , where h is a meromorphic function. It follows ( h ′ ) 2 = φ 2 B 2 . Hence, T ( r , h ′ ) = S ( r , f ) .

Since h 2 = φ f ′ − φ , we have

(3.31) 2 T ( r , h ) = T r , φ f ′ − φ = T ( r , f ′ ) + S ( r , f ) .

From (3.31), Lemma 2.3, and T ( r , h ′ ) = S ( r , f ) , we deduce that T ( r , f ′ ) ≤ S ( r , f ) , a contradiction.

By (3.18), (3.28), and (3.9), we obtain

T ( r , f ) ≤ 5 N r , 1 f + 5 N ¯ r , 1 f ′ − φ + S ( r , f ) .

This completes the proof of Theorem 1.2.

4 Proof of Theorem 1.4

Obviously,

(4.1) m r , 1 f γ Q ≤ m r , Q f γ Q + m r , 1 Q + S ( r , f ) ,

where γ Q = γ Q ( f ) = m + γ H .

By (4.1) and Nevanlinna’s first fundamental theorem, we have

(4.2) T ( r , f γ Q ) ≤ T ( r , Q ) + N r , 1 f γ Q − N r , 1 Q + m r , Q f γ Q + S ( r , f ) .

Since ( m + γ H ) T ( r , f ) = T ( r , f γ Q ) + S ( r , f ) , then by (4.2) and Lemma 2.1, we obtain

( m + γ H ) T ( r , f ) ≤ T ( r , Q ) + N r , 1 f γ Q − N r , 1 Q + m r , Q f γ Q + S ( r , f ) ≤ T ( r , Q ) + ( m + γ H ) N r , 1 f − N r , 1 Q + ( γ H − γ ̲ H ) m r , 1 f + S ( r , f ) ≤ T ( r , Q ) + ( m + γ H ) N r , 1 f − N r , 1 Q + ( γ H − γ ̲ H ) T r , 1 f − N r , 1 f + S ( r , f ) ≤ T ( r , Q ) + ( m + γ ̲ H ) N r , 1 f − N r , 1 Q + ( γ H − γ ̲ H ) T ( r , f ) + S ( r , f ) .

Thus, we have

( m + γ ̲ H ) T ( r , f ) ≤ T ( r , Q ) + ( m + γ ̲ H ) N r , 1 f − N r , 1 Q + S ( r , f ) ≤ T ( r , Q ) + ( m + γ ̲ H ) N r , 1 f − m N r , 1 f + S ( r , f ) ≤ T ( r , Q ) + γ ̲ H N r , 1 f + S ( r , f ) .

It follows

(4.3) T ( r , f ) ≤ 1 m T ( r , Q ) + S ( r , f ) .

By (4.3) and Lemma 2.5, we prove Theorem 1.4.

5 Proof of Theorem 1.8

Suppose ϕ ≢ 0 . By Lemma 2.7, we have N r , 1 f ( k ) − ϕ ≠ S ( r , f ) , a contradiction. Hence, ϕ ≡ 0 .

Next we consider two cases.

Case 1. k ≥ 2 .

Since f is a transcendental meromorphic function, then by Lemma 2.8, we deduce that f = e a z + b , where a ( ≠ 0 ) , b are constants.

Case 2. k = 1 .

Suppose that f ″ has finitely many zeros. Then by Lemma 2.10 and f ≠ 0 , we deduce that f = 1 Q e P , where P , Q are polynomials. It follows from f ′ ≠ 0 that

(5.1) f ′ = e P ( P ′ Q − Q ′ ) Q 2 ≠ 0 .

Obviously, P ′ Q − Q ′ ≢ 0 and P ′ ≢ 0 .

Next we consider two subcases.

Case 2.1. There exists z such that P ′ ( z ) Q ( z ) − Q ′ ( z ) = 0 .

Let z 1 be a zero of P ′ ( z ) Q ( z ) − Q ′ ( z ) . By (5.1), we have Q ( z 1 ) = 0 . Hence, Q ′ ( z 1 ) = 0 .

Thus, we obtain

(5.2) P ′ ( z ) Q ( z ) − Q ′ ( z ) = ( z − z 1 ) l 1 φ 1 ( z ) ,

and

(5.3) Q ( z ) = ( z − z 1 ) l 2 φ 2 ( z ) ,

where l 1 , l 2 ( ≥ 2 ) are positive integers and φ 1 ( z ) , φ 2 ( z ) are two polynomials with φ 1 ( z 1 ) ≠ 0 , φ 2 ( z 1 ) ≠ 0 . It follows from (5.3) that

(5.4) Q ′ ( z ) = ( z − z 1 ) l 2 − 1 φ 3 ( z ) ,

where φ 3 ( z ) = l 2 φ 2 ( z ) + ( z − z 1 ) φ ′ ( z ) . Obviously, φ 3 ( z 1 ) = l 2 φ 2 ( z 1 ) ≠ 0 .

By (5.3) and (5.4), we have

(5.5) P ′ ( z ) Q ( z ) − Q ′ ( z ) = P ′ ( z ) ( z − z 1 ) l 2 φ 2 ( z ) − ( z − z 1 ) l 2 − 1 φ 3 ( z ) = ( z − z 1 ) l 2 − 1 [ P ′ ( z ) ( z − z 1 ) φ 2 ( z ) − φ 3 ( z ) ] = ( z − z 1 ) l 2 − 1 φ 4 ( z ) ,

where φ 4 ( z ) = P ′ ( z ) ( z − z 1 ) φ 2 ( z ) − φ 3 ( z ) . Obviously, φ 4 ( z 1 ) = − φ 3 ( z 1 ) ≠ 0 .

It follows from (5.2) and (5.5) that l 1 = l 2 − 1 . Then by (5.2) and (5.4), we know that z 1 is a zero of both P ′ ( z ) Q ( z ) − Q ′ ( z ) and Q ′ ( z ) with the same multiplicities.

Suppose that the distinct zeros of P ′ ( z ) Q ( z ) − Q ′ ( z ) are z 1 , z 2 , … , z s with multiplicities are m 1 , m 2 , … , m s , where s , m 1 , m 2 , … , m s are positive integers. It follows

P ′ ( z ) Q ( z ) − Q ′ ( z ) = A ( z − z 1 ) m 1 ( z − z 2 ) m 2 … ( z − z s ) m s ,

where A is a nonzero constant and m 1 + m 2 + … + m s = deg ( P ′ Q − Q ′ ) = deg P ′ + deg Q ≥ deg Q . Furthermore, we have

Q ′ ( z ) = ( z − z 1 ) m 1 ( z − z 2 ) m 2 … ( z − z s ) m s φ 5 ( z ) ,

where φ 5 ( z ) is a polynomial. Thus, we obtain deg Q ′ ≥ m 1 + m 2 + … + m s ≥ deg Q . Then we deduce that Q is a nonzero constant.

Since f = 1 Q e P and f ′ ≠ 0 , we obtain f ′ = 1 Q e P P ′ ≠ 0 . Thus, we obtain P ( z ) = a z + b 1 , where a ( ≠ 0 ) , b 1 are constants. It follows that f = e a z + b , where b is a constant.

Case 2.2. P ′ Q − Q ′ ≠ 0 .

From P ′ Q − Q ′ ≠ 0 , Q ≢ 0 and P ′ ≢ 0 , we know that P ′ and Q are nonzero constants. Thus, P = a z + b 2 and Q = b 3 , where a ( ≠ 0 ) , b 2 and b 3 ( ≠ 0 ) are constants. It follows that f = e a z + b , where b is a constant.

This completes the proof of Theorem 1.8.

Acknowledgments

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

Funding information: This article was supported by the National Natural Science Foundation of China (Grant Nos. 12171127, 12371074).
Author contributions: All authors have accepted responsibility for the entire content of this manuscript and consented to its submission to the journal, reviewed all the results, and approved the final version of the manuscript.
Conflict of interest: The authors state no conflict of interest.
Data availability statement: Data sharing not applicable to this article as no datasets were generated or analyzed during the current study.

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Received: 2023-12-21

Revised: 2024-10-21

Accepted: 2024-12-06

Published Online: 2025-02-04

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https://doi.org/10.1515/math-2024-0117

Keywords for this article

meromorphic functions; value distribution; differential polynomials; Hayman’s alternative

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