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
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
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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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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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.