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Two results on the value distribution of meromorphic functions

  • Degui Yang , Zhiying He and Dan Liu EMAIL logo
Published/Copyright: April 5, 2024
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Open Mathematics
From the journal Open Mathematics Volume 22 Issue 1

Abstract

In this article, we prove two results on the value distribution of meromorphic functions. Using the theorem of Yamanoi, the first result gives a precise estimation of the relationship between the characteristic function of a meromorphic function and its k th derivative in a concise form. This result extends and improves some results of Shan, Singh, Gopalakrishna, Edrei, Weitsman, Yang, Wu and Wu, etc. The second result answers a conjecture posed by C. C. Yang. This conjecture turned to be false by a counter-example, but it will be true with an additional condition.

Keywords: meromorphic function; maximal deficiency sum; characteristic function; derivatives
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 [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 measure.

Let f be a nonconstant meromorphic function. The order of f is defined by

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

Let a be a complex number (finite or infinite). According to R. Nevanlinna

δ ( a , f ) = lim ̲ r → ∞ m r , 1 f − a T ( r , f ) = 1 − lim ¯ r → ∞ N r , 1 f − a T ( r , f ) ,

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

It is clear that 0 ≤ δ ( a , f ) ≤ 1 (or 0 ≤ Θ ( a , f ) ≤ 1 ). If δ ( a , f ) > 0 (or Θ ( a , f ) > 0 ), then a is called a deficient value with respect to f and δ ( a , f ) (or Θ ( a , f ) ) is its deficiency. The most fundamental result of Nevanlinna theory can be stated as follows.

Any transcendental meromorphic function f in the finite plane has countable deficient values at most and the total deficiency does not exceed 2, i.e.,

∑ a ∈ C ˆ δ ( a , f ) ≤ ∑ a ∈ C ˆ Θ ( a , f ) ≤ 2 , where C ˆ = C ⋃ { ∞ } .

It is the famous Defect Relation (or Deficient Relation). In the general case, the upper bound 2 is sharp. If ∑ a ∈ C ˆ δ ( a , f ) = 2 or ∑ a ∈ C ˆ Θ ( a , f ) = 2 , then we say that f has maximum deficiency sum.

Let f be a meromorphic function in C . The characteristic function of derivative of f with maximum deficiency sum was studied by Shan, Singh, Gopalakrishna, Edrei, and Weitsman (see [4–10]). Edrei [4] and Weitsman [8] proved the following.

Theorem A

Let f be a transcendental meromorphic function of finite order and satisfying ∑ a ∈ C δ ( a , f ) + δ ( ∞ , f ) = 2 . Then

lim r → ∞ T ( r , f ′ ) T ( r , f ) = 2 − δ ( ∞ , f ) .

Yang [10] considered the case of higher order derivatives and proved the following.

Theorem B

Let k be a positive integer, and let f be a transcendental meromorphic function of finite order and satisfying ∑ a ∈ C δ ( a , f ) + δ ( ∞ , f ) = 2 . Then

lim r → ∞ T ( r , f ( k ) ) T ( r , f ) = 1 , f o r δ ( ∞ , f ) = 1 ; lim r → ∞ T ( r , f ( k ) ) T ( r , f ) = k + 1 , f o r δ ( ∞ , f ) = 0 .

If ∑ a ∈ C δ ( a , f ) + δ ( ∞ , f ) = 2 is replaced by ∑ a ∈ C Θ ( a , f ) + Θ ( ∞ , f ) = 2 , Singh-Gopalakrishna [6] and Singh-Kulkarni [7] proved the following.

Theorem C

Let f be a transcendental meromorphic function of finite order and satisfying ∑ a ∈ C Θ ( a , f ) + Θ ( ∞ , f ) = 2 . Then

lim r → ∞ T ( r , f ′ ) T ( r , f ) = 2 − Θ ( ∞ , f ) .

Wu and Wu [9] considered the case of higher order derivatives and proved the following.

Theorem D

Let k be a positive integer, and let f be a transcendental meromorphic function of finite order and satisfying ∑ a ∈ C Θ ( a , f ) + Θ ( ∞ , f ) = 2 . Then

lim r → ∞ T ( r , f ( k ) ) T ( r , f ) = 1 , f o r Θ ( ∞ , f ) = 1 ; lim r → ∞ T ( r , f ( k ) ) T ( r , f ) = k + 1 , f o r Θ ( ∞ , f ) = 0 .

Remark 1

There exists a gap in the proof of Theorem D (at page 1120 from line 3 to line 4).

In this article, we extend and improve the aforementioned results and obtain the following result.

Theorem 1

Let k be a positive integer, and let f be a transcendental meromorphic function of finite order and satisfying ∑ a ∈ C Θ ( a , f ) + Θ ( ∞ , f ) = 2 . Then

lim r → ∞ , r ∉ E T ( r , f ( k ) ) T ( r , f ) = k + 1 − k Θ ( ∞ , f ) ,

where E is a set of logarithmic density 0.

From Theorem 1, we obtain

Corollary 2

Let k be a positive integer, and let f be a transcendental meromorphic function of finite order and satisfying ∑ a ∈ C δ ( a , f ) + δ ( ∞ , f ) = 2 . Then

lim r → ∞ , r ∉ E T ( r , f ( k ) ) T ( r , f ) = k + 1 − k δ ( ∞ , f ) ,

where E is a set of logarithmic density 0.

In 1959, Hayman proved the following result about value distribution of meromorphic function.

Theorem E

[11] Let f be a nonconstant meromorphic function in C , a , b ( ≠ 0 ) be finite values, and k be a positive integer. Then either f = a or f ( k ) = b has at least one root. Moreover, if f is transcendental, then either f = a or f ( k ) = b has infinitely many zeros.

In 2004, Yang proved the following theorem.

Theorem F

[12] Let f be a transcendental meromorphic function satisfying

N ( r , 1 ∕ f ) = S ( r , f ) ,

where S ( r , f ) denotes any quantity that satisfy S ( r , f ) = o ( 1 ) T ( r , f ) as r → + ∞ , except a set of finite linear length. Then, for any n ≥ 1 and any small function b ( z ) ( ≢ 0 ) of f ,

N r , 1 f ( n ) − b ≠ S ( r , f ) .

In the same article, Yang posed a conjecture as follows.

Conjecture

[12] If α is a nonconstant entire function, and z 0 is an arbitrary point in C , then F ( z ) = ∫ z 0 z e α ( z ) d z must have infinitely many zeros.

In this article, we will give a counter-example to show that the conjecture is not true.

Example

Let α ( z ) = z + ∫ 0 z e z − 1 z d z , F ( z ) = ∫ 0 z e α ( z ) d z , and let G ( z ) = z ⋅ e ∫ 0 z e z − 1 z d z . It is easy to prove that α ( z ) is an entire function, and

F ′ ( z ) = e α ( z ) = e z + ∫ 0 z e z − 1 z d z ,

G ′ ( z ) = e ∫ 0 z e z − 1 z d z + z ⋅ e ∫ 0 z e z − 1 z d z ⋅ e z − 1 z = e z + ∫ 0 z e z − 1 z d z .

Therefore, F ′ ( z ) = G ′ ( z ) , which means F ( z ) = G ( z ) + C , and C is a constant. Since F ( 0 ) = G ( 0 ) = 0 , we have C = 0 . It follows that F ( z ) = G ( z ) .

For z = 0 is the only zero of G ( z ) , we can deduce that F ( z ) has only one zero.

This example shows that it could be possible for functions of the form ∫ 0 z e α ( z ) d z have only finitely many zeros. Thus, the conjecture of C. C. Yang is not true. But this conjecture will be correct if α ′ ( z ) has finitely number of zeros. In fact, we have proved the following result.

Theorem 3

Let α ( z ) be an entire function, and α ′ ( z ) has finitely many zeros, then F ( z ) = ∫ 0 z e α ( z ) d z has infinitely many zeros.

By Theorem 3, we naturally have the following corollary.

Corollary 4

Let α ( z ) be a nonconstant polynomial, then F ( z ) = ∫ 0 z e α ( z ) d z has infinitely many zeros.

2 Some lemmas

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

Lemma 1

[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

[13] Let a 1 , a 2 , … , a q be distinct finite complex numbers, where q ≥ 2 ; let f be a transcendental meromorphic function in the complex plane; and let k ≥ 2 be an integer and ε > 0 . Then

(2.1) ( k − 1 ) N ¯ ( r , f ) + ∑ i = 1 q N 1 r , 1 f − a i ≤ N r , 1 f ( k ) + ε T ( r , f ) ,

for all r > e outside a set E ⊂ ( e , ∞ ) of logarithmic density 0. Here E depends on f , k , ε , and

N 1 r , 1 f − a i = N r , 1 f − a i − N ¯ r , 1 f − a i , i = 1 , 2 , … , q .

Lemma 3

[1, p. 33, (2.1)] Suppose that f ( z ) is a nonconstant meromorphic function in ∣ z ∣ ≤ r . Let a 1 , a 2 , … , a q be distinct finite complex numbers, where q ≥ 2 . Then

∑ i = 1 q m r , 1 f − a i ≤ m r , ∑ i = 1 q 1 f − a i + O ( 1 ) .

Lemma 4

Let f be a transcendental meromorphic function in the complex plane, and let k be an integer. Then

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

Proof

By Nevanlinna’s first fundamental theorem and Lemma 1, we have

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

Lemma 5

[14] Let the function f be meromorphic in C . If f ( m ) and f ( n ) have finitely many zeros, where 0 ≤ m ≤ n − 2 , then f has finitely many poles and finite order, i.e.,

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

and f ( m ) = Re P with R a rational function and P a polynomial.

By Lemma 5, we naturally have the following result.

Lemma 6

Let f be a meromorphic function in C . If f and f ″ both have finitely many zeros, then

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

which means f is a meromorphic function of finite order.

Lemma 7

Let α ( z ) be a nonconstant polynomial, then F ( z ) = ∫ 0 z e α ( z ) d z must have infinitely many zeros.

Proof

We will prove this lemma by contradiction. Suppose that F ( z ) has finite number of zeros, by Hadamard Fractional Theorem,

F ( z ) = p ( z ) e β ( z ) ,

where p ( z ) is a polynomial, and β ( z ) is a nonconstant polynomial.

Therefore,

F ′ ( z ) = e α ( z ) = p ′ ( z ) e β ( z ) + p ( z ) β ′ ( z ) e β ( z ) = ( p ′ ( z ) + p ( z ) β ′ ( z ) ) e β ( z ) .

Hence,

p ′ ( z ) + p ( z ) β ′ ( z ) = e α ( z ) − β ( z ) .

Comparing the order of both sides of the above equation, we have α ( z ) − β ( z ) = c and p ′ ( z ) + p ( z ) β ′ ( z ) = e c . Thus, p ( z ) is a constant, and α ( z ) , β ( z ) are polynomials of degree 1.

Assume that α ( z ) = a z + b , where a ( ≠ 0 ) , b are constants. Then

F ( z ) = ∫ 0 z e a z + b d z = 1 a ( e a z + b − e b ) .

It is obvious that F ( z ) has infinite number of zeros since e b is a nonzero constant, which is a contradiction.

Thus, Lemma 7 is proved.□

3 Proof of Theorem 1

Firstly, we prove that

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

Without loss of generality, we assume that f has infinitely many finite deficient values a 1 , a 2 , … , a n , … . It follows from Lemma 1, Lemma 3, Lemma 4, and Nevanlinna’s first fundamental theorem that for every n ∈ N ,

(3.2) ∑ i = 1 n m r , 1 f − a i ≤ m r , ∑ i = 1 n 1 f − a i + O ( 1 ) ≤ m r , ∑ i = 1 n f ′ f − a i + m r , 1 f ′ + S ( r , f ) = T ( r , f ′ ) − N r , 1 f ′ + S ( r , f ) ≤ T ( r , f ) + N ¯ ( r , f ) − N r , 1 f ′ + S ( r , f ) .

Hence by Nevanlinna’s first fundamental theorem, we obtain

(3.3) n T ( r , f ) ≤ T ( r , f ) + N ¯ ( r , f ) + ∑ i = 1 n N r , 1 f − a i − N r , 1 f ′ + S ( r , f ) ≤ T ( r , f ) + N ¯ ( r , f ) + ∑ i = 1 n N ¯ r , 1 f − a i + S ( r , f ) .

Thus, we have

n − 1 ≤ N ¯ ( r , f ) T ( r , f ) + ∑ i = 1 n N ¯ r , 1 f − a i T ( r , f ) + S ( r , f ) T ( r , f ) .

It follows that

n − 1 ≤ lim ̲ r → ∞ N ¯ ( r , f ) T ( r , f ) + ∑ i = 1 n lim ¯ r → ∞ N ¯ r , 1 f − a i T ( r , f ) + lim ¯ r → ∞ S ( r , f ) T ( r , f ) .

Thus, we have

(3.4) ∑ i = 1 n Θ ( a i , f ) ≤ 1 + lim ̲ r → ∞ N ¯ ( r , f ) T ( r , f ) ≤ 1 + lim r → ∞ ¯ N ¯ ( r , f ) T ( r , f ) = 2 − Θ ( ∞ , f ) .

By the condition of the theorem, we have

(3.5) ∑ i = 1 ∞ Θ ( a i , f ) = 2 − Θ ( ∞ , f ) .

Let n → ∞ in (3.4), and by (3.5) we obtain (3.1).

By Lemma 1, Lemma 3, and Nevanlinna’s first fundamental theorem, we obtain

(3.6) ∑ i = 1 n m r , 1 f − a i ≤ m r , 1 f ( k ) + S ( r , f ) ≤ T ( r , f ( k ) ) − N r , 1 f ( k ) + S ( r , f ) .

It follows from (3.6), Lemma 2, and Nevanlinna’s first fundamental theorem that

(3.7) n T ( r , f ) ≤ ∑ i = 1 n N r , 1 f − a i + T ( r , f ( k ) ) − N r , 1 f ( k ) + S ( r , f ) = ∑ i = 1 n N ¯ r , 1 f − a i + T ( r , f ( k ) ) + ∑ i = 1 n N r , 1 f − a i − N ¯ r , 1 f − a i − N r , 1 f ( k ) + S ( r , f ) ≤ ∑ i = 1 n N ¯ r , 1 f − a i + T ( r , f ( k ) ) − ( k − 1 ) N ¯ ( r , f ) + ε T ( r , f ) + S ( r , f ) ,

for all r > e outside of a set E ⊂ ( e , + ∞ ) of logarithmic density 0.

Thus, we obtain

(3.8) n ≤ ∑ i = 1 n N ¯ r , 1 f − a i T ( r , f ) + T ( r , f ( k ) ) T ( r , f ) − ( k − 1 ) N ¯ ( r , f ) T ( r , f ) + ε + S ( r , f ) T ( r , f ) .

It follows from (3.1) and (3.8) that

(3.9) ∑ i = 1 n Θ ( a i , f ) ≤ lim ̲ r → ∞ , r ∉ E T ( r , f ( k ) ) T ( r , f ) + ( 1 − k ) ( 1 − Θ ( ∞ , f ) ) + ε ,

where E is a set of logarithmic density 0.

Let n → ∞ , ε → 0 in (3.9), we obtain

(3.10) lim ̲ r → ∞ , r ∉ E T ( r , f ( k ) ) T ( r , f ) ≥ 1 + k − k Θ ( ∞ , f ) ,

where E is a set of logarithmic density 0.

On the other hand, by Lemma 4, we obtain

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

It follows that

(3.11) lim ¯ r → ∞ T ( r , f ( k ) ) T ( r , f ) ≤ 1 + k − k Θ ( ∞ , f ) .

Thus by (3.10) and (3.11), we obtain

lim r → ∞ , r ∉ E T ( r , f ( k ) ) T ( r , f ) = 1 + k − k Θ ( ∞ , f ) ,

where E is a set of logarithmic density 0.

Thus, Theorem 1 is proved.

4 Proof of Theorem 3

Suppose that F ( z ) has finitely many zeros. Obviously,

F ″ ( z ) = ( F ′ ( z ) ) ′ = ( e α ( z ) ) ′ = α ′ ( z ) e α ( z ) .

It is easy to see that F ″ ( z ) has finitely many zeros since α ′ ( z ) has finitely zeros. By Lemma 6, F ( z ) is a function of finite order. Hence, α ( z ) is a polynomial. By Lemma 7, F ( z ) has infinitely many zeros, which leads to a contradiction to our assumption.

Therefore, F ( z ) has infinitely many zeros. Thus, Theorem 3 is proved.

Acknowledgements

We are very grateful to the anonymous reviewers 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. Author contributions: All authors have accepted responsibility for the entire content of this manuscript and approved its submission.

  3. Conflict of interest: The authors state no conflicts of interest.

  4. 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: 2023-09-21
Revised: 2023-12-12
Accepted: 2024-03-07
Published Online: 2024-04-05

© 2024 the author(s), published by De Gruyter

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

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  114. On the existence of tripartite graphs and n-partite graphs
  115. Classifying pentavalent symmetric graphs of order 12pq
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  129. Erratum
  130. Erratum to “Infinitesimals via Cauchy sequences: Refining the classical equivalence”
  131. Corrigendum
  132. Corrigendum to “Matrix stretching”
  133. Corrigendum to “A comprehensive review of the recent numerical methods for solving FPDEs”
https://doi.org/10.1515/math-2024-0004
Keywords for this article
meromorphic function; maximal deficiency sum; characteristic function; derivatives
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