# Unicity of Meromorphic Function Sharing One Small Function with Its Derivative

Authors: Ang Chen; Xiuwang Wang; Guowei Zhang
Journal: International Journal of Mathematics and Mathematical Sciences
 (2010)

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## Abstract

We deal with the problem of uniqueness of a meromorphic function sharing one small function with its k's derivative and obtain some results.


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## Body

## 1. Introduction and Main Results

In this article, a meromorphic function means meromorphic in the open complex plane. We assume that the reader is familiar with the Nevanlinna theory of meromorphic functions and the standard notations such asT(r,f),m(r,f),N(r,f),N¯(r,f), and so on.Letf and g be two nonconstant meromorphic functions; a meromorphic function a(z)(≢∞) is called a small functions with respect to f provided that T(r,a)=S(r,f). Note that the set of all small function of f is a field. Let b(z) be a small function with respect to f and g. We say that f and g share b(z) CM(IM) provided that f-b and g-b have same zeros counting multiplicities (ignoring multiplicities).Moreover, we use the following notations.Letk be a positive integer. We denote by Nk)(r,1/(f-a)) the counting function for the zeros of f-a with multiplicity ≤k and by N¯k)(r,1/(f-a)) the corresponding one for which the multiplicity is not counted. Let N(k(r,1/(f-a)) be the counting function for the zeros of f-a with multiplicity ≥k, and let N¯(k(r,1/(f-a)) be the corresponding one for which the multiplicity is not counted. Set Nk(r,1/(f-a))=N¯(r·1/(f-a))+N¯(2(r,1/(f-a))+⋯+N¯(k(r,1/(f-a)). And we define(1.1)δp(a,f)=1-lim supr→+∞Np(r,1/(f-a))T(r,f).
Obviously, 1≥Θ(a,f)≥δp(a,f)≥δ(a,f)≥0. For more details, reader can see [1, 2].Brück (see [3]) considered the uniqueness problems of an entire function sharing one value with its derivative and proved the following result.Theorem 1 A.
Letfbe nonconstant entire function. If f and f′ share the value 1 CM and if N(r,1/f′)=S(r,f), then (f′-1)/(f-1)≡c  for some constant c∈ℂ∖{0}.Yang [4], Zhang [5], and Yu [6] extended Theorem A and obtained many excellent results.Theorem 1 B (see[5]).
Letf be a nonconstant meromorphic function and, let k be a positive integer. Suppose that  f andf(k) share 1 CM and
(1.2)2N¯(r,f)+N¯(r,1f′)+N(r,1f(k))<(λ+o(1))T(r,f(k)),
for r∈I, where I is a set of infinite linear measure and λ satisfies 0<λ<1, then (f(k)-1)/(f-1)≡c for some constant c∈ℂ∖{0}.Theorem 1 C (see[6]).
Letf be a nonconstant, nonentire meromorphic function and a(z)(≢0,∞) be a small function with respect tof. If(1)
f and a(z) have no common poles,(2)
f-a and f(k)-a share the value 0 CM,(3)
4δ(0,f)+2(k+8)Θ(∞,f)>2k+19, then f≡f(k), where k is a positive integer.In the same paper, Yu [6] posed four open questions. Lahiri and Sarkar [7] and Zhang [8] studied the problem of a meromorphic or an entire function sharing one small function with its derivative with weighted shared method and obtained the following result, which answered the open questions posed by Yu [6].Theorem 1 D (see[8]).
Letf be a non-constant meromorphic function and, let k be a positive integer. Also let  a(z)(≢0,∞) be a meromorphic function such that T(r,a)=S(r,f). Suppose that f-a and f(k)-a share 0 IM and
(1.3)4N¯(r,f)+3N2(r,1f(k))+2N¯(r,1(f/a)′)<(λ+o(1))T(r,f(k)),
for 0<λ<1,r∈I, and I is a set of infinite linear measure. Then (f(k)-a)∖(f-a)≡c for some constant c∈ℂ∖{0}.In this article, we will pay our attention to the value sharing off and [fn](k) that share a small function and obtain the following results, which are the improvements and complements of the above theorems.Theorem 1.1.
Letk(≥1),n(≥1) be integers and let  f be a non-constant meromorphic function. Also let a(z)(≢0,∞) be a small function with respect to f. If f and [fn](k) share a(z) IM and
(1.4)4N¯(r,f)+2N¯(r,1(f/a)′)+2N2(r,1(fn)(k))+N¯(r,1(fn)(k))≤(λ+o(1))T(r,(fn)(k)),
or f and [fn](k) share a(z) CM and
(1.5)2N¯(r,f)+N¯(r,1(f/a)′)+N2(r,1(fn)(k))≤(λ+o(1))T(r,(fn)(k)),
for 0<λ<1, r∈I, and I is a set of infinite linear measure, then (f-a)∖([fn](k)-a)≡c, for some constant c∈ℂ∖{0}.Theorem 1.2.
Letk(≥1),n(≥1) be integers and f be a non-constant meromorphic function. Also let a(z)(≢0,∞) be a small function with respect to f. If f and [fn](k) share a(z) IM and
(1.6)(2k+6)Θ(∞,f)+3Θ(0,f)+2δk+2(0,f)>2k+10,
or f and [fn](k) share a(z) CM and
(1.7)(k+3)Θ(∞,f)+δ2(0,f)+δk+2(0,f)>k+4,
then f≡(fn)(k).Clearly, Theorem1.1 improves and extends Theorems B and D, while 1.2 improves and extends Theorem C.


## 2. Some Lemmas

In this section, first of all, we give some definitions which will be used in the whole paper.Definition 2.1.
LetF and G be two meromorphic functions defined in ℂ; assume, that F and G share 1 IM; let z0 be a zero of F-1 with multiplicity p and a zero of G-1 with multiplicity q. We denote byNE1)(r,1/F-1)the counting function of the zeros of  F-1 where p=q=1 and byNE(2(r,1/F-1)the counting function of zeros of F-1 where p=q≥2. We denotes by NL(r,1/F-1) the counting function of the zeros of F-1 where p>q≥1; each point is counted according to its multiplicity, and N¯L(r,1/F-1) denote its reduced form. In the same way, we can defineNE1)(r,1/G-1),NE(2(r,1/G-1),N¯L(r,1/G-1), and so on.Definition 2.2.
In this paperN0(r,1/F′) denotes the counting function of the zeros ofF′ which are not the zeros of F andF-1, andN¯0(r,1/F′) denotes its reduced form. In the same way, we can define N0(r,1/G′) and N¯0(r,1/G′).Next we present some lemmas which will be needed in the sequel. LetF,G be two non-constant meromorphic functions defined in ℂ. We shall denote by H the following function:(2.1)H=(F′′F′-2F′F-1)-(G′′G′-2G′G-1).Lemma 2.3  (see[2]).
LetF,G be two nonconstant meromorphic functions defined in ℂ. If F and G are sharing 1 IM, then
(2.2)N(r,H)≤N¯(r,F)+N¯(2(r,1F)+N¯(2(r,1G)+N¯L(r,1F-1)+N¯L(r,1G-1)+N¯0(r,1F′)+N¯0(r,1G′)+S(r,f).
If F and G are sharing 1 CM, then
(2.3)N(r,H)≤N¯(r,F)+N¯(2(r,1F)+N¯(2(r,1G)+N¯0(r,1F′)+N¯0(r,1G′)+S(r,f).Lemma 2.4 (see[1]).
Letf be a meromorphic function and a is a finite complex number. Then(i)
T(r,1/(f-a))=T(r,f)+O(1),(ii)
m(r,f(k)/f(l))=S(r,f) for k>l≥0,(iii)
T(r,f)≤N¯(r,f)+N¯(r,1/(f-a1(z)))+N¯(r,1/(f-a2(z)))+S(r,f),where a1(z)a2(z) are two meromorphic functions such that T(r,ai)=S(r,f),(i=1,2).Lemma 2.5 (see[7]).
Letf be a non-constant meromorphic function, and k,p are two positive integers. Then
(2.4)Np(r,1f(k))≤Np+k(r,1f)+kN¯(r,f)+S(r,f).Lemma 2.6 (see[9]).
Letf be a non-constant meromorphic function and let n be a positive integer. P(f)=anfn+an-1fn-1+⋯+a1f where ai are meromorphic functions such that T(r,ai)=S(r,f)(i=1,2,…,n),  and an≢0. Then
(2.5)T(r,P(f))=nT(r,f)+S(r,f).


## 3. Proof of Theorem1.1

LetF=f(z)/a(z),G=(fn(z))(k)/a(z), then(3.1)F-1=f(z)-a(z)a(z),G-1=(fn(z))(k)-a(z)a(z).
From the definitions of F,G and recalling that F and G share value 1 IM(CM), we get(3.2)NE1)(r,1F-1)=NE1)(r,1G-1)+S(r,f),N¯E(2(r,1F-1)=N¯E(2(r,1G-1)+S(r,f),(3.3)N¯L(r,1F-1)≤N¯(r,1F)+N¯(r,F)+S(r,F),(3.4)N¯(r,1F-1)=N¯(r,1G-1)+S(r,F)=NE1)(r,1F-1)+N¯E(2(r,1F-1)+N¯L(r,1F-1)+N¯L(r,1G-1)+S(r,f).We will distinguish two cases below.Case 1 (H≢0).
From (2.1) it is easy to see that m(r,H)=S(r,f).Subcase 1.1.
Suppose thatf and (fn)(k) share a(z) IM. According to (3.1), F and G share 1 IM except the zeros and poles of a(z). By (3.1), we have
(3.5)N¯(r,F)=N¯(r,f)+S(r,f),N¯(r,G)=N¯(r,f)+S(r,f).
Letz0 be a simple zero of F-1 and G-1, but a(z0)≠0,∞. Through a simple calculation we know that z0 is a zero of H, so
(3.6)NE1)(r,1F-1)≤N(r,1H)+S(r,f)≤T(r,H)+S(r,f)≤N(r,H)+S(r,f).
From (3.4)–(3.6) and Lemma 2.3, we have
(3.7)N¯(r,1G-1)≤N¯(r,F)+2N¯L(r,1F-1)+2N¯L(r,1G-1)+N¯(2(r,1F)+N¯(2(r,1G)+N¯E(2(r,1F-1)+N¯0(r,1F′)+N¯0(r,1G′)+S(r,f)≤N¯(r,f)+2N¯(r,1F′)+2N¯L(r,1G-1)+N¯(2(r,1G)+N¯0(r,1G′)+S(r,f).
It follows by the second fundamental theorem, (3.5), and (3.7) that
(3.8)T(r,G)≤N¯(r,G)+N¯(r,1G)+N¯(r,1G-1)-N0(r,1G′)+S(r,G)≤2N¯(r,f)+2N¯(r,1F′)+2N¯(r,1G′)+N¯(r,1G)+S(r,f).
By Lemma 2.5, we have
(3.9)T(r,(fn)(k))≤4N¯(r,f)+2N¯(r,1(f/a)′)+2N2(r,1(fn)(k))+N¯(r,1(fn)(k))+S(r,f),
which contradicts (1.4).Subcase 1.2.
Suppose thatf and (fn)(k) share a(z) CM.
Letz0 be a simple zero of F-1 and G-1, but a(z0)≠0,∞. By a simple calculation, we can still get H(z0)=0. Therefore
(3.10)N1)(r,1F-1)≤N(r,1H)+S(r,f)≤N(r,H)+S(r,f).
Noting thatN1)(r,1/(F-1))=N1)(r,1/(G-1))+S(r,f), by (3.4) and Lemma 2.3, we can deduce
(3.11)N¯(r,1G-1)≤N¯(r,F)+N¯(2(r,1F)+N¯(2(r,1G)+N¯0(r,1F′)+N¯0(r,1G′)+N¯(2(r,1F-1)+S(r,f).
By the second fundamental theorem, (3.5), and (3.11), we have
(3.12)T(r,G)≤N¯(r,G)+N¯(r,1G)+N¯(r,1G-1)-N0(r,1G′)+S(r,G)≤2N¯(r,f)+N2(r,1G)+N¯(r,1F′)+S(r,f).
Taking into account (3.1), we have
(3.13)T(r,(fn)(k))≤2N¯(r,f)+N¯(r,1(f/a)′)+N2(r,1(fn)(k))+S(r,f).
This contradicts (1.5).Case 2 (H≡0).
Integration yields(3.14)1F-1≡AG-1+B,
where A,B are constants and A≠0. It is easy to see that F and G share 1 CM. Now we claim that B=0.
IfN¯(r,f)≠S(r,f), then by (3.14) we get B=0. So our claim holds. Hence we can assume that
(3.15)N¯(r,f)=S(r,f).
If B≠0, then we can rewrite (3.14) as
(3.16)1F-1≡B(G-1+A/B)G-1.
So
(3.17)N¯(r,1G-1+A/B)=N¯(r,F)=S(r,f).
If A≠B, then by Lemma 2.4 and (3.17) we have
(3.18)T(r,G)≤N¯(r,G)+N¯(r,1G)+N¯(r,1G-1+A/B)+S(r,f)≤N¯(r,1G)+S(r,f)≤T(r,G)+S(r,f).
Hence
(3.19)T(r,G)=N¯(r,1G)+S(r,f),
that is,
(3.20)T(r,(fn)(k))=N¯(r,1(fn)(k))+S(r,f).
This is a contradiction with (1.4) and (1.5). If A=B, then from (3.14) we get 1/(F-1)=AG/(G-1). We rewrite it as
(3.21)-a2fn(Af-a-aA)≡(fn)(k)fn.
So by Lemmas 2.4 and 2.6 and (3.15), we have
(3.22)(n+1)T(r,f)=T(r,(fn)(k)fn)+S(r,f)≤nN(r,1f)+kN¯(r,f)+S(r,f)≤nT(r,f)+S(r,f).
This implies that T(r,f)=S(r,f), since n≥1. This is impossible. Hence our claim is right. So (G-1)/(F-1)=A. Theorem 1.1 is, thus, completely proved.


## 4. Proof of Theorem1.2

The proof is similar to the proof of Theorem1.1. Let F and G be defined as in Theorem 1.1; hence, we have (3.1)–(3.5). We still distinguish two cases.Case 1.
H≢0Subcase 1.1.
Suppose thatf and (fn)(k) share a(z) IM, then we can still get (3.6) and (3.7). Then by the second fundamental theorem, Lemma 2.3, and (3.5) we have
(4.1)T(r,F)≤N¯(r,F)+N¯(r,1F)+N¯(r,1F-1)-N0(r,1F′)+S(r,F)≤2N¯(r,f)+2N¯(r,1G′)+2N¯(r,1F′)+N¯(r,1F)+S(r,f).
Applying Lemma 2.5 to the above inequality and noticing the definition of F,G, we get
(4.2)T(r,f)≤(2k+6)N¯(r,f)+3N¯(r,1f)+2Nk+2N(r,1f)+S(r,f)≤[(2k+6)(1-Θ(∞,f))+3-3Θ(0,f)+2-2δk+2(0,f)]T(r,f)+S(r,f).
This implies that
(4.3)(2k+6)Θ(∞,f)+3Θ(0,f)+2δk+2(0,f)≤2k+10.
This contradicts (1.6).Subcase 1.2.
Suppose thatf and (fn)(k) share a(z) CM. Similarly as above, we can easily obtain N1)(r,1/(F-1))=N1)(r,1/(G-1))+S(r,f); by Lemma 2.3, we can deduce
(4.4)N¯(r,1F-1)≤N¯(r,F)+N¯(2(r,1F)+N¯(2(r,1G)+N¯0(r,1F′)+N¯0(r,1G′)+N¯(2(r,1G-1)+S(r,f).
So by the second fundamental theorem, (4.4), and using Lemma 2.5 again, we have
(4.5)T(r,F)≤N¯(r,F)+N¯(r,1F)+N¯(r,1F-1)-N0(r,1F′)+S(r,f)≤2N¯(r,f)+N2(r,1f)+N¯(r,1G′)+S(r,f)≤[(k+5)-(k+3)Θ(∞,f)-δ2(0,f)-δk+2(0,f)]T(r,f)+S(r,f).
This implies that
(4.6)(k+3)Θ(∞,f)+δ2(0,f)+δk+2(0,f)≤k+4.
This contradicts (1.7).Case 2 (H≡0).
Similarly, we can also get (3.14). Next we claim that B=0. If N¯(r,f)≠S(r,f), then it follows that B=0 from (3.14). Hence, we may assume that (3.15) holds. If B≠0andB≠-1, then
(4.7)AG-1≡-BF-(B+1)F-1,
and so
(4.8)N¯(r,G)=N¯(r,1F-(B+1)/B).
Again by second fundamental theorem and (4.4) we have
(4.9)T(r,F)=N¯(r,1F)+S(r,f),
that is,
(4.10)T(r,f)≤N¯(r,1f)+S(r,f)≤T(r,f)+S(r,f).
Then we have T(r,f)=N¯(r,1/f), and it follows that Θ(0,f)=0 and from (3.15) we have Θ(∞,f)=1; then with (1.6) and (1.7) we may deduce δk+2(0,f)>1. It is impossible, and we can assume that B=-1; thus, we can get
(4.11)(fn)(k)a-(A+1)≡-A·a·1f.
It shows that T(r,f)=T(r,(fn)(k)).
IfA=-1, by (4.11), then we have f·(fn)(k)≡a2, which with the above equality may lead to T(r,f)=S(r,f), which is impossible. If A≠-1, then by second fundamental theorem, Lemma 2.5, (3.15), and (4.11) we have
(4.12)T(r,(fn)(k))≤N¯(r,1(fn)(k)-a(A+1))+N¯(r,1(fn)(k))+S(r,f),≤kN¯(r,f)+Nk+2(r,1f)+S(r,f)≤T(r,f)+S(r,f),
which with (3.15) may deduce Nk+2(r,1/f)=T(r,f)+S(r,f); so δk+2(o,f)=0, which with Θ(∞,f)=1 and (1.6) may deduce Θ(0,f)>1, which is impossible. Hence our claim holds.
Next we will prove thatA=1. From (3.17) we have G-1≡A(F-1). Then
(4.13)N¯(r,1G)=N¯(r,1F+1/A-1).
If A≠1, then we have
(4.14)T(r,F)≤N¯(r,F)+N¯(r,1F)+N¯(r,1G)+S(r,f).
By Lemma 2.5, we get
(4.15)T(r,f)≤(k+1)N¯(r,f)+N¯(r,1f)+Nk+2(r,1f)+S(r,f).
It implies that
(4.16)(k+1)Θ(∞,f)+Θ(0,f)+δk+2(0,f)≤k+2.
Combining (4.16) with (1.6) yields
(4.17)2(k+2)+Θ(0,f)≥2(k+3)Θ(∞,f)+3Θ(0,f)+2δ2+k(0,f)-4Θ(∞,f)>2k+6,
that is, Θ(0,f)>2. This is a contradiction.
Combining (4.16) with (1.7) yields
(4.18)k+2+2Θ(∞,f)≥(k+3)Θ(∞,f)+Θ(0,f)+δk+2(0,f)>k+4,
that is, Θ(∞,f)>1, which is also a contradiction. Hence A=1 and f≡(fn)(k). Now Theorem 1.2 has been completely proved.



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