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volume 3, issue 5, article 76, 2002.

Received 30 June, 2002;

accepted 12 July, 2002.

Communicated by:H.M. Srivastava

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Journal of Inequalities in Pure and Applied Mathematics

EXTENSIONS OF HIONG’S INEQUALITY

MING-LIANG FANG AND DEGUI YANG

Department of Mathematics, Nanjing Normal University, Nanjing, 210097,

People’s Republic of China EMail:[email protected] College of Sciences,

South China Agricultural University, Guangzhou, 510642,

People’s Republic of China EMail:[email protected]

2000c Victoria University ISSN (electronic): 1443-5756 079-02

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Extensions of Hiong’s Inequality Ming-liang Fang and Degui Yang

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J. Ineq. Pure and Appl. Math. 3(5) Art. 76, 2002

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Abstract

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

2000 Mathematics Subject Classification:Primary 30D35, 30A10 Key words: Inequality, Value distribution, Meromorphic function.

Contents

1 Introduction. . . 3 2 Proof of Theorems . . . 6

References

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Extensions of Hiong’s Inequality Ming-liang Fang and Degui Yang

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1. Introduction

Letf 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.

Theorem 1.1. Letf be a non-constant meromorphic function; leta,bandcbe three finite complex numbers such thatb 6= 0, c6= 0 andb 6= c; and let kbe 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 Theorem1.1as follows.

Theorem 1.2. Let f be a non-constant meromorphic function; and let b and cbe two distinct nonzero finite complex numbers; and let n, k be two positive

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integers. If φ(6≡ 0) is a meromorphic function satisfying T(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 integers n even iff is meromorphic.

Theorem 1.3. Let f be a non-constant meromorphic function; and let b and cbe two distinct nonzero finite complex numbers; and let n, k be two positive integers. Ifφ(6≡0) is a meromorphic 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).

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Extensions of Hiong’s Inequality Ming-liang Fang and Degui Yang

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In [6], the author proved

Theorem 1.4. Let f be a transcendental meromorphic function; and letn be a positive integer. Then eitherfnf0 −aorfnf0+ahas 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−bhas infinitely many zeros, where a(6≡0)andb(6≡0)are two meromorphic functions satisfyingT(r, a) =S(r, f) andT(r, b) = S(r, f).

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Extensions of Hiong’s Inequality Ming-liang Fang and Degui Yang

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2. Proof of Theorems

For the proofs of Theorem1.3and1.5, we require the following lemmas.

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

δ(K) = min{(2eK−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 Lemma2.1.

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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).

Hence T(r, f) ≤ n−12 T(r, fn−1f(k)) +S(r, f), Therefore, T(r, f) = S(r, f), which is a contradiction. Which completes the proof of this lemma.

Lemma 2.3. [1]. If f is a meromorphic function, and a1, a2, a3 are distinct meromorphic functions satisfyingT(r, aj) =S(r, f)forj = 1,2,3.Then

T(r, f)≤

3

X

j=1

N

r, 1 f −aj

+S(r, f).

Proof of Theorem1.3. By Lemma2.2, we haveφfn−1f(k) 6≡constant ifn and k 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)

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≤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).

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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 Theorem1.3.

Proof of Theorem1.5. 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)

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≤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 Lemma2.3, 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.

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References

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

[2] W.K. HAYMANANDJ. 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 of φ(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 deriva- tive, Kodai Math. J., 24(3) (2001), 339–343.

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