INTRODUCTION Let f be a nonconstant meromorphic function in the whole complex plane

http://jipam.vu.edu.au/

Volume 3, Issue 5, Article 76, 2002

EXTENSIONS OF HIONG’S INEQUALITY

MING-LIANG FANG AND DEGUI YANG DEPARTMENT OFMATHEMATICS,

NANJINGNORMALUNIVERSITY, NANJING, 210097, PEOPLE’SREPUBLIC OFCHINA

mlfang@pine.njnu.edu.cn COLLEGE OFSCIENCES,

SOUTHCHINAAGRICULTURALUNIVERSITY, GUANGZHOU, 510642,

PEOPLE’SREPUBLIC OFCHINA

yangde@macs.biu.ac.il

Received 15 July, 2002; accepted 25 July, 2002 Communicated by H.M. Srivastava

ABSTRACT. In this paper, we treat the value distribution ofφfⁿ⁻¹f^(k), wheref is a transcen- dental meromorphic function,φis a meromorphic function satisfyingT(r, φ) =S(r, f), nand kare positive integers. We generalize some results of Hiong and Yu.

Key words and phrases: Inequality, Value distribution, Meromorphic function.

2000 Mathematics Subject Classification. Primary 30D35, 30A10.

1. INTRODUCTION

Let f be a nonconstant meromorphic function in the whole complex plane. We use the following standard notation of value distribution theory,

T(r, f), m(r, f), N(r, f), N(r, f), . . .

(see Hayman [1], Yang [4]). We denote byS(r, f)any function satisfying S(r, f) = o{T(r, f)},

asr →+∞, possibly outside of a set with finite measure.

In 1956, Hiong [3] proved the following inequality.

ISSN (electronic): 1443-5756 c

2002 Victoria University. All rights reserved.

Supported by the National Nature Science Foundation of China (Grant No. 10071038) and “Qinglan Project”of the Educational Department of Jiangsu Province.

079-02

Theorem 1.1. Let f be a non-constant meromorphic function; let a, b and c be three finite complex numbers such thatb 6= 0,c6= 0andb 6=c; and letkbe a positive integer. Then

T(r, f)≤N

r, 1 f −a

+N

r, 1

f^(k)−b

+N

r, 1 f^(k)−c

−N

r, 1 f^(k+1)

+S(r, f).

Recently, Yu [5] extended Theorem 1.1 as follows.

Theorem 1.2. Let f be a non-constant meromorphic function; and letb andcbe two distinct nonzero finite complex numbers; and letn, k be two positive integers. Ifφ(6≡ 0) is a meromor- phic function satisfyingT(r, φ) = S(r, f),n= 1orn≥k+ 3, then

(1.1) T(r, f)≤N

r, 1 f

+ 1

n

N

r, 1

φfⁿ⁻¹f^(k)−b

+N

r, 1

φfⁿ⁻¹f^(k)−c

− 1 n

N(r, f) +N

r, 1

(φfⁿ⁻¹f^(k+1))⁰

+S(r, f).

Iff is entire, then (1.1) is valid for all positive integersn(6= 2).

In [5], the author expected that (1.1) is also valid forn = 2iff is entire.

In this note, we prove that (1.1) is valid for all positive integersneven iff is meromorphic.

Theorem 1.3. Let f be a non-constant meromorphic function; and letb andcbe two distinct nonzero finite complex numbers; and letn, k be two positive integers. Ifφ(6≡ 0) is a meromor- phic function satisfyingT(r, φ) = S(r, f), then

(1.2) T(r, f)≤N

r, 1 f

+ 1

n

N

r, 1

φfⁿ⁻¹f^(k)−b

+N

r, 1

φfⁿ⁻¹f^(k)−c

−N(r, f)− 1 n

(k−1)N(r, f) +N

r, 1

(φfⁿ⁻¹f^(k+1))⁰

+S(r, f).

In [6], the author proved

Theorem 1.4. Letf be a transcendental meromorphic function; and letnbe a positive integer.

Then either fⁿf⁰ −a or fⁿf⁰ +a has infinitely many zeros, wherea(6≡ 0)is a meromorphic function satisfyingT(r, a) =S(r, f).

In this note, we will prove

Theorem 1.5. Letf be a transcendental meromorphic function; and letnbe a positive integer.

Then eitherfⁿf⁰ −aorfⁿf⁰ −b has infinitely many zeros, wherea(6≡0)andb(6≡ 0)are two meromorphic functions satisfyingT(r, a) =S(r, f)andT(r, b) =S(r, f).

2. PROOF OFTHEOREMS

For the proofs of Theorem 1.3 and 1.5, we require the following lemmas.

Lemma 2.1. [2]. Iff is a transcendental meromorphic function andK >1, then there exists a setM(K)of upper logarithmic density at most

δ(K) = min{(2e^K−1−1)⁻¹,(1 +e(K−1)) exp(e(1−K))}

such that for every positive integerk,

(2.1) lim sup

r→∞,r /∈M(K)

T(r, f)

T(r, f^(k)) ≤3eK.

Lemma 2.2. If f is a transcendental meromorphic function and φ(6≡ 0) is a meromorphic function satisfyingT(r, φ) =S(r, f). Thenφfⁿ⁻¹f^(k) 6≡constant for every positive integern.

Proof. Suppose that φfⁿ⁻¹f^(k) ≡ constant. If n = 1, then φf^(k) ≡ constant. Therefore, T(r, f^(k)) =S(r, f), which implies that

lim sup

r→∞,r /∈M(K)

T(r, f)

T(r, f^(k)) =∞.

This is contradiction to Lemma 2.1.

Ifn ≥2,thenT(r, fⁿ⁻¹f^(k)) = S(r, f).On the other hand, nT(r, f)≤T(r, fⁿ⁻¹f^(k)) +T

r, f

f^(k)

+S(r, f)

≤T(r, fⁿ⁻¹f^(k)) +T

r,f^(k) f

+S(r, f)

≤T(r, fⁿ⁻¹f^(k)) +N

r,f^(k) f

+S(r, f)

≤T(r, fⁿ⁻¹f^(k)) +N(r, 1

f) +N(r, fⁿ⁻¹f^(k)) +S(r, f)

≤2T(r, fⁿ⁻¹f^(k)) +T(r, f) +S(r, f).

HenceT(r, f) ≤ _n−1² T(r, fⁿ⁻¹f^(k)) +S(r, f),Therefore, T(r, f) = S(r, f),which is a con-

tradiction. Which completes the proof of this lemma.

Lemma 2.3. [1]. Iff is a meromorphic function, anda1, a2, a3 are distinct meromorphic func- tions satisfyingT(r, a_j) = S(r, f)forj = 1,2,3.Then

T(r, f)≤

3

X

j=1

N

r, 1 f−a_j

+S(r, f).

Proof of Theorem 1.3. By Lemma 2.2, we have φfⁿ⁻¹f^(k) 6≡constant if n andk are positive integers. By (4.17) of [1], we have

m

r, 1 fⁿ

+m

r, 1

φfⁿ⁻¹f^(k)−b

+m

r, 1

φfⁿ⁻¹f^(k)−c (2.2)

≤m

r, 1

φfⁿ⁻¹f^(k)

+m

r,f^(k) f

+m

r, 1

φfⁿ⁻¹f^(k)−b

+m

r, 1

φfⁿ⁻¹f^(k)−c

+S(r, f)

≤m

r, 1

φfⁿ⁻¹f^(k)

+m

r, 1

φfⁿ⁻¹f^(k)−b

+m

r, 1

φfⁿ⁻¹f^(k)−c

+S(r, f)

≤m

r, 1

(φfⁿ⁻¹f^(k))⁰

+S(r, f)

≤T(r,(φfⁿ⁻¹f^(k))⁰)−N

r, 1

(φfⁿ⁻¹f^(k))⁰

+S(r, f)

≤T(r, φfⁿ⁻¹f^(k)) +N(r, f)−N

r, 1

(φfⁿ⁻¹f^(k))⁰

+S(r, f) By (2.2), we have

T(r, fⁿ) +T(r, φfⁿ⁻¹f^(k))

≤N

r, 1 fⁿ

+N

r, 1

φfⁿ⁻¹f^(k)−b

+N

r, 1

φfⁿ⁻¹f^(k)−c

+N(r, f)−N

r, 1

(φfⁿ⁻¹f^(k))⁰

+S(r, f).

Therefore,

nT(r, f)≤nN

r, 1 f

+N

r, 1

φfⁿ⁻¹f^(k)−b

+N

r, 1

φfⁿ⁻¹f^(k)−c

+N(r, f)−N

r, 1

(φfⁿ⁻¹f^(k))⁰

−N(r, fⁿ⁻¹f^(k)) +S(r, f)

≤nN

r, 1 f

+N

r, 1

φfⁿ⁻¹f^(k)−b

+N

r, 1

φfⁿ⁻¹f^(k)−c

−nN(r, f)−(k−1)N(r, f)−N

r, 1

(φfⁿ⁻¹f^(k))⁰

+S(r, f),

thus we get (1.2). This completes the proof of Theorem 1.3.

Proof of Theorem 1.5. By Nevanlinna’s first fundamental theorem, we have 2T(r, f) = T

r, f f⁰· f f⁰

≤T(r, f f⁰) +T

r, f f⁰

+S(r, f)

≤T(r, f f⁰) +T

r,f⁰ f

+S(r, f)

≤T(r, f f⁰) +N

r,f⁰ f

+S(r, f)

=T(r, f f⁰) +N

r, 1 f

+N(r, f) +S(r, f)

≤T(r, f f⁰) +T(r, f) + 1

3N(r, f f⁰) +S(r, f).

Thus we get

T(r, f)≤ 4

3T(r, f f⁰) +S(r, f).

Hence we getT(r, a) = S(r, f f⁰)andT(r, b) =S(r, f f⁰).

By Lemma 2.3, we have

T(r, f f⁰)≤N(r, f) +N

r, 1 f f⁰−a

+N

r, 1

f f⁰−b

+S(r, f f⁰)

≤ 1

3N(r, f f⁰) +N

r, 1 f f⁰−a

+N

r, 1

f f⁰ −b

+S(r, f f⁰).

Hence we get

T(r, f)≤ 3 2

N

r, 1

f f⁰ −a

+N

r, 1 f f⁰−b

+S(r, f f⁰).

Thus we know that eitherf f⁰ −aorf f⁰−bhas infinitely many zeros.

REFERENCES

[1] W.K. HAYMAN, Meromorphic Functions, Clarendon Press, Oxford, 1964.

[2] W.K. HAYMAN AND J. MILES, On the growth of a meromorphic function and its derivatives, Complex Variables, 12(1989), 245–260.

[3] K.L. HIONG, Sur la limitation deT(r, f)sans intervention des pôles, Bull. Sci. Math., 80 (1956), 175–190.

[4] L. YANG, Value Distribution Theory, Springer-Verlag, Berlin, 1993.

[5] K.W. YU, On the distribition ofφ(z)fⁿ⁻¹(z)f^(k)(z), J. of Ineq. Pure and Appl. Math., 3(1) (2002), Article 8. [ONLINE:http://jipam.vu.edu.au/v3n1/037_01.html]

[6] K.W. YU, A note on the product of a meromorphic function and its derivative, Kodai Math. J., 24(3) (2001), 339–343.