volume 7, issue 3, article 93, 2006.
Received 21 December, 2005;
accepted 28 July, 2006.
Communicated by:H.M. Srivastava
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Journal of Inequalities in Pure and Applied Mathematics
MEROMORPHIC FUNCTIONS THAT SHARE ONE VALUE WITH THEIR DERIVATIVES
KAI LIU AND LIAN-ZHONG YANG
School of Mathematics & System Sciences Shandong University
Jinan, Shandong, 250100, P. R. China EMail:liuk@mail.sdu.edu.cn EMail:lzyang@sdu.edu.cn
URL:http://202.194.3.2/html/professor/ylzh/YangIndex.html
c
2000Victoria University ISSN (electronic): 1443-5756 373-05
Meromorphic Functions that Share One Value with their
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Abstract
In this paper, we deal with the problems of uniqueness of meromorphic func- tions that share one finite value with their derivatives and obtain some results that improve the results given by Rainer Brück and Qingcai Zhang.
2000 Mathematics Subject Classification:30D35.
Key words: Meromorphic functions, Uniqueness, Sharing values.
This work was supported by the NNSF of China (No.10371065) and the NSF of Shandong Province, China (No.Z2002A01).
Contents
1 Introduction and Main Results. . . 3
2 Some Lemmas. . . 8
3 Proof of Theorem 1.2 . . . 9
4 Proof of Theorem 1.4 . . . 14
5 Proof of Theorem 1.6 and Theorem 1.7 . . . 18 References
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1. Introduction and Main Results
In this paper, a meromorphic function will mean meromorphic in the finite com- plex plane. We say that two meromorphic functionsf andgshare a finite value aIM (ignoring multiplicities) whenf−aandg−ahave the same zeros. Iff−a andg−ahave the same zeros with the same multiplicities, then we say thatf andgshare the valueaCM (counting multiplicities). We say thatf andg share
∞ CM provided that1/f and 1/g share 0 CM. It is assumed that the reader is familiar with the standard symbols and fundamental results of Nevanlinna Theory, as found in [3,6].
Let f(z) be a meromorphic function. It is known that the hyper-order of f(z),denoted byσ2(f),is defined by
σ2(f) = lim sup
r→∞
log logT(r, f) logr . In 1996, R. Brück posed the following conjecture (see [1]).
Conjecture 1.1. Let f be a non-constant entire function such that the hyper- orderσ2(f)off is not a positive integer andσ2(f)<+∞.Iff andf0 share a finite valueaCM, then
f0 −a f −a =c, wherecis nonzero constant.
In [1], Brück proved that the conjecture holds when a = 0. In 1998, Gun- dersen and Yang [2] proved that the conjecture is true whenf is of finite order.
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In 1999, Yang [4] confirmed that the conjecture is also true whenf0is replaced byf(k)(k ≥2)andf is of finite order.
In 1996, Brück obtained the following result.
Theorem A ([1]). Letf be a nonconstant entire function. Iff andf0 share the value 1 CM, and ifN
r,f10
=S(r, f), then
f0−1 f−1 ≡c
for a non-zero constantc.
In 1998, Q. Zhang proved the next two results in [7].
Theorem B. Letf be a nonconstant meromorphic function. Iff andf0 share the value 1 CM, and if
N(r, f) +N
r, 1 f0
<(λ+o(1))T(r, f0),
0< λ < 1 2
,
then f0−1
f−1 ≡c for some non-zero constantc.
Theorem C. Let f be a nonconstant meromorphic function, k be a positive integer. Iff andf(k)share the value 1 CM, and if
2N(r, f) +N
r, 1 f0
+N
r, 1
f(k)
<(λ+o(1))T(r, f(k)), (0< λ <1),
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then
f(k)−1 f−1 ≡c for some non-zero constantc.
The above results suggest the following question: What results can be ob- tained if the condition that f and f0 share the value 1 CM is replaced by the condition thatf andf0 share the value 1 IM?
In this paper, we obtained the following results.
Theorem 1.2. Letf be a nonconstant meromorphic function, iff andf0 share the value 1 IM, and if
N(r, f) +N
r, 1 f0
<(λ+o(1))T(r, f0)
0< λ < 1 4
,
then f0−1
f−1 ≡c for some non-zero constantc.
Corollary 1.3. Let f be a nonconstant entire function. If f and f0 share the value 1 IM, and if
N
r, 1 f0
<(λ+o(1))T(r, f0),
0< λ < 1 4
,
then f0−1
f−1 ≡c for some non-zero constantc.
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Theorem 1.4. Let f be a nonconstant meromorphic function, k be a positive integer. Iff andf(k)share the value 1 IM, and if
(3k+ 6)N(r, f) + 5N
r, 1 f
<(λ+o(1))T(r, fk), (0< λ <1), then
f(k)−1 f−1 ≡c for some non-zero constantc.
Corollary 1.5. Let f be a nonconstant entire function. Iff andf(k) share the value 1 IM, and if
N
r, 1 f
<(λ+o(1))T(r, f),
0< λ < 1 10
, then
f(k)−1 f−1 ≡c for some non-zero constantc.
Theorem 1.6. Let f be a nonconstant meromorphic function, k be a positive integer. Iff andf(k)share the valuea 6= 0CM, and satisfy one of the following conditions,
(i) δ(0, f) + Θ(∞, f)> 2k+14k , (ii) N(r, f) +N
r, 1f
<(λ+o(1))T(r, f), 0< λ < 2k+12 ,
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(iii) k+12
N(r, f) + 32N
r,f1
<(λ+o(1))T(r, f), (0< λ <1).
Thenf ≡f(k).
Theorem 1.7. Letf be a nonconstant meromorphic function. Iff andf0share the valuea6= 0IM, and if
N(r, f) +N
r, 1 f
<(λ+o(1))T(r, f),
0< λ < 2 3
, thenf ≡f0.
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2. Some Lemmas
Lemma 2.1 ([7]). Letf be a nonconstant meromorphic function,kbe a positive integer. Then
N
r, 1 f(k)
< N
r, 1 f
+kN(r, f) +S(r, f), (2.1)
N
r, f(k) f
< kN(r, f) +kN
r, 1 f
+S(r, f), (2.2)
N
r,f(k) f
< kN(r, f) +N
r, 1 f
+S(r, f).
(2.3)
Suppose that f andg share the value a IM, and let z0 be aa-point of f of orderp, aa-point off(k) of orderq. We denote byNL
r,f(k)1−a
the counting function of thosea-points off(k)whereq > p.
Lemma 2.2. Letfbe a nonconstant meromorphic function. Iff andf(k)share the value 1 IM, then
(2.4) NL
r, 1 f(k)−1
< N
r, 1 f(k)
+N(r, f) +S(r, f).
Lemma 2.3 ([7]). Letf be a nonconstant meromorphic function,kbe a positive integer. Iff andf(k)share the value 1 IM, then
T(r, f)<3T(r, f(k)) +S(r, f), specially iff is a nonconstant entire function, then
T(r, f)<2T(r, f(k)) +S(r, f).
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3. Proof of Theorem 1.2
Let N1)
r,f−a1
denote the counting function of the simple zeros of f − a, N(2
r,f−a1
denote the counting function of the multiplea-points of f. Each point in these counting functions is counted only once. We denote byN2
r,f−a1
the counting function of the zeros off−a,where a simple zero is counted once and a multiple zero is counted twice. It follows that
(3.1) N2
r, 1 f−a
=N1)
r, 1 f−a
+ 2N(2
r, 1 f−a
. Set
F = f000
f00 − 2f00 f0−1−
f00
f0 − 2f0 f −1
.
We suppose thatF 6≡0. By the lemma of logarithmic derivatives, we have
(3.2) m(r, F) =S(r, f)
and
(3.3) N(r, F)≤N(r, f) +N
r, 1 f0
+N(2
r, 1 f0−1
+N0
r, 1
f00
+S(r, f),
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where N(2
r,f01−1
denotes the counting function of multiple 1-points off0, and each 1-point is counted only once; N0
r,f100
denotes the counting func- tions off00which are not the zeros off0andf0−1.
Sincef andf0 share the value 1 IM, we know thatf −1has only simple zeros. Iff0−1also has only simple zeros, thenf andf0share the value 1 CM, and Theorem1.2follows by the conclusion of TheoremB.
Now we assume thatf0−1has multiple zeros. By calculation, we know that the common simple zeros of f −1and f0 −1are the zeros of F; we denote by NE1)
r,f−11
the counting function of common simple zeros off −1 and f0−1. It follows that
(3.4) NE1)
r, 1 f−1
≤N
r, 1 F
≤T(r, F) =N(r, F) +S(r, f).
From (3.3) and (3.4), we have (3.5) NE1)
r, 1
f −1
≤N(r, f) +N
r, 1 f0
+N(2
r, 1 f0−1
+N0
r, 1
f00
+S(r, f).
Notice that
(3.6) N
r, 1
f0−1
=NE1)
r, 1 f −1
+N(2
r, 1
f0−1
.
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By the second fundamental theorem, we have (3.7) T(r, f0)< N(r, f) +N
r, 1
f0
+N
r, 1 f0−1
−N0
r, 1 f00
+S(r, f).
From Lemma2.2, (3.8) N(2
r, 1
f0−1
=NL
r, 1 f0−1
< N(r, f) +N
r, 1 f0
+S(r, f).
Combining (3.5), (3.6), (3.7) and (3.8), we obtain T(r, f0)≤N(r, f) +N
r, 1
f0
+NE1)
r, 1 f−1
+NL
r, 1 f0−1
−N0
r, 1 f00
+S(r, f)
≤N(r, f) +N
r, 1 f0
+N(r, f) +N
r, 1 f0
+ 2NL
r, 1 f0−1
+S(r, f)
≤4N(r, f) + 4N
r, 1 f0
+S(r, f),
which contradicts the condition of Theorem1.2. Therefore, we haveF ≡0.By integrating twice, we have
1
f−1 = A
f0−1+B,
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whereA6= 0andB are constants.
We distinguish the following three cases.
Case 1. IfB 6= 0,−1, then
f = (B+ 1)f0+ (A−B−1) Bf0+ (A−B) ,
f0 = (B −A)f+ (A−B−1 Bf −(B+ 1) , and so
N r, 1
f0+ A−BB
!
=N(r, f)
By the second fundamental theorem T(r, f0)< N(r, f0) +N
r, 1
f0
+N r, 1 f0+ A−BB
!
+S(r, f)
<2N(r, f) +N
r, 1 f0
+S(r, f), which contradicts the assumption of Theorem1.2.
Case 2. IfB =−1, then
f = A
−f0+ (A−1)
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and so
N
r, 1
f0−(A+ 1)
=N(r, f).
We also get a contradiction by the second fundamental theorem.
Case 3. IfB = 0, it follows that
f0−1 f −1 =A, and the proof of Theorem1.2is thus complete.
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4. Proof of Theorem 1.4
Let
F = f(k+2)
f(k+1) −2 f(k+1)
f(k)−1 −f00
f0 + 2 f0 f−1
We suppose thatF 6≡0.Since the common zeros (with the same multiplicities) off −1andf(k)−1are not the poles ofF,and the common simple zeros of f −1andf(k)−1are the zeros ofF,we have
(4.1) NE1)
r, 1 f−1
≤N
r, 1 F
≤T(r, F) =N(r, F) +S(r, f).
and
(4.2) N(r, F)≤N(r, f) +NL
r, 1 f −1
+NL
r, 1 f(k)−1
+N(2(
r, 1 f
+N(2
r, 1
f(k)
+N0
r, 1 f0
+N0
r, 1
f(k+1)
+S(r, f),
whereN0
r,f(k+1)1
denotes the counting function of the zeros off(k+1)which are not the zeros off(k) andf(k)−1, N0
r,f10
denotes the counting function
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of the zeros off0which are not the zeros off. Since N
r, 1 f(k)−1
=N
r, 1 f −1
(4.3)
=NE1)
r, 1 f −1
+N(2E
r, 1
f −1
+NL
r, 1 f−1
+NL
r, 1 f(k)−1
,
we obtain from (4.1), (4.2) and (4.3) that N
r, 1 f(k)−1
≤N(r, f) + 2NL
r, 1 f −1
+N(2E
r, 1
f−1
+N(2
r, 1 f
+N
0r, 1 f0
+ 2NL
r, 1 f(k)−1
+N(2
r, 1 f(k)
+N0
r, 1
f(k+1)
+S(r, f)
≤N(r, f) + 2N
r, 1 f0
+ 2NL
r, 1 f(k)−1
+N(2
r, 1 f(k)
+N0
r, 1
f(k+1)
+S(r, f),
whereN(2E
r,f−11
is the counting function of common multiple zeros off−1 andf(k)−1,each point is counted once. By the second fundamental theorem
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and Lemma2.2, we have
T(r, f(k))≤N(r, f) +N
r, 1 f(k)
+N
r, 1 f(k)−1
−N0
r, 1 f(k+1)
+S(r, f)
≤N(r, f) +N
r, 1 f(k)
+N(r, f) + 2N
r, 1 f0
+ 2NL
r, 1 f(k)−1
+N(2
r, 1
f(k)
+S(r, f)
≤2N(r, f) +N
r, 1 f(k)
+ 2N
(r, 1
f0
+ 2N
r, 1 f(k)
+ 2N(r, f) +S(r, f)
≤(3k+ 6)N(r, f) + 5N
r, 1 f
+S(r, f),
which contradicts the assumption of Theorem1.2. HenceF ≡0.
By integrating twice, we get 1
f −1 = C
f(k)−1 +D,
whereC 6= 0andDare constants. By arguments similar to the proof of Theo- rem1.2, Theorem1.4follows.
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Remark 1. Let f be a non-constant entire function. Then we obtain from Lemma2.3that
1
2T(r, f)≤T(r, f(k)) +S(r, f).
By Theorem1.4, Corollary1.5holds.
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5. Proof of Theorem 1.6 and Theorem 1.7
Suppose thatf 6≡f(k). Let
F = f f(k). Then
(5.1) T(r, F) =m
r, 1 F
+N
r, 1
F
=N
r,f(k) f
+S(r, f).
Sincef andf(k)share the valuea 6= 0CM, we have N
r, 1
f−a
≤N
r, 1 f−f(k)
(5.2)
≤N
r, 1 F −1
≤T(r, F) +O(1).
By the lemma of logarithmic derivatives and the second fundamental theorem, we obtain
(5.3) m
r, 1
f
+m
r, 1 f −a
< m
r, 1 f(k)
+S(r, f),
and
(5.4) T r, f(k)
< N
r, 1 f(k)
+N(r, f(k)) +N
r, 1 f(k)−a
+S(r, f),
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from (5.4), we have
(5.5) m
r, 1
f(k)
< N(r, f) +N
r, 1 f(k)−a
+S(r, f).
Combining with (5.1), (5.2), (5.3), (5.4), (5.5) and (2.2) of Lemma 2.1, we obtain
2T(r, f)≤m
r, 1 f(k)
+N
r, 1
f
+N
r, 1 f −a
+S(r, f)
≤N(r, f) +N
r, 1 f(k)−a
+N
r, 1
f
+N
r, 1 f −a
+S(r, f)
≤N(r, f) +N
r, 1 f
+ 2N
r, 1
f−a
+S(r, f)
≤N(r, f) +N
r, 1 f
+ 2N
r,f(k)
f
+S(r, f)
≤N(r, f) + 2(kN(r, f) +kN
r, 1 f
+N
r, 1
f
+S(r, f)
≤(2k+ 1)N(r, f) + (2k+ 1)N
r, 1 f
+S(r, f),
which contradicts the assumptions (i) and (ii) of Theorem1.6. Hencef ≡f(k). Similarly, by the above inequality and (2.3) of Lemma 2.1, and suppose that (iii) is satisfied, then we get a contradiction if f 6≡ f(k), and we complete the proof of Theorem1.6.
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Remark 2. For a nonconstant meromorphic function f, if f and f0 share the valuea 6= 0IM andf 6≡f(k), since aa-point off is not a zero off0, we know thatf−ahas only simple zeros, and we have
N
r, 1 f−a
≤N
r, 1 F −1
≤T(r, F) +O(1),
whereF =f /f0.By the arguments similar to the proof of Theorem1.6, Theo- rem1.7follows.
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[3] C.C. YANGANDH.-X. YI, Uniqueness Theory of Meromorphic Functions, Kluwer Academic Publishers, 2003.
[4] L.-Z. YANG, Solution of a differential equation and its applications, Kodai Math. J., 22 (1999), 458–464.
[5] H.-X. YI, Meromorphic function that share one or two value II, Kodai Math.
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[6] H.-X. YIANDC.C. YANG, Uniqueness Theory of Meromorphic Functions, Science Press, Beijing, 1995. [In Chinese].
[7] Q.-C. ZHANG, The uniqueness of meromorohic function with their deriva- tives, Kodai Math. J., 21 (1998), 179–184.