# Normal Families of Meromorphic Functions Concerning Higher Derivative and Shared Values

Authors: Wei Chen; Wenjun Yuan; Honggen Tian
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We study the normal families related to a Hayman conjecture of higher derivative and concerning shared values and get two normal criteria. Our results improve the related theorems which were obtained independently, respectively by Fang and Yuan (2001), Yuan et al. ((2011) and (2012)), Wang et al. (2011), and Qiu et al. (2012). Meanwhile, some examples are given to show the sharpness of our results.


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## Body

## 1. Introduction and Main Results

Letf(z) and g(z) be two nonconstant meromorphic functions in a domain D⊆ℂ, and let a be a finite complex value. We say that f and g share a CM (or IM) in D provided that f-a and g-a have the same zeros counting (or ignoring) multiplicity in D. When a=∞, the zeros of f-a mean the poles of f (see [1]). It is assumed that the reader is familiar with the standard notations and the basic results of Nevanlinna’s value-distribution theory ([1] or [2–4]).It is very interesting to find normality criteria from the point of view of shared values. In this area, Schwick [5] first proved an interesting result that a family of meromorphic functions in a domain is normal in which every function shares three distinct finite complex numbers with their first derivative. And later, more results about normality criteria concerning shared values have emerged; for instance, see [6–8]. In recent years, this subject has attracted the attention of many researchers worldwide.We now first introduce a normality criterion related to a Hayman normal conjecture [9].Theorem 1.
Letℱ be a family of holomorphic (meromorphic) functions defined in a domain, n∈ℕ, a≠0, and b∈ℂ. If f′(z)+afn(z)-b does not vanish in D for each function f(z)∈ℱ and n≥2(n≥3), then ℱ is normal in D.The results for the holomorphic case are due to Drasin [10] for n≥3, Pang [11] for n=3, Chen and Fang [12] for n=2, Ye [13] for n=2, and Chen and Gu [14] for the generalized result with a and b replaced by meromorphic functions. The results for the meromorphic case are due to Li [15], Li [16], and Langley [17] for n≥5, Pang [11] for n=4, Chen and Fang [12] for n=3, and Zalcman [18] for n=3, obtained independently.Whenn=2 and ℱ is meromorphic, Theorem 1 is not valid in general. Fang and Yuan [19] gave an example to show this and got a special result below.Example 2.
The family of meromorphic functionsℱ={fj(z)=jz/(jz-1)2:j=1,2,…,} is not normal in D={z:|z|<1}. This is deduced by fj#(0)=j→∞, as j→∞ and Marty’s criterion [2], although for any fj(z)∈ℱ, fj′+fj2=j(jz-1)-4≠0.Heref#(ξ) denotes the spherical derivative
(1)f#(ξ)=|f′(ξ)|1+|f(ξ)|2.Theorem 3.
Letℱ be a family of meromorphic functions in a domain D, and a≠0, b∈ℂ. If f′(z)+af2(z)-b does not vanish in D and the poles of f(z) are of multiplicity ≥3 for each f(z)∈ℱ, then ℱ is normal in D.In 2008, by the ideas of shared values, Zhang [8] proved the following.Theorem 4.
Letℱ be a family of meromorphic (holomorphic) functions in D, let n be a positive integer and let a, b be two finite complex numbers such that a≠0. If n≥4(n≥2) and for every pair of functions f and g in ℱ, f′-afn and g′-agn share the value b IM, then ℱ is normal in D.Example 5 (see [8]).
The family of meromorphic functionsℱ={fj(z)=1/j(z-1/j):j=1,2,…,} is not normal in D={z:|z|<1}. Obviously fj′-fj3=-z/j(z-1/j)3. So for each pair of m, j, fj′-fj3 and fm′-fm3 share the value 0 in D, but ℱ is not normal at the point z=0, since fj#(0)=2(j)3/(1+j)→∞, as j→∞.Remark 6.
Example5 shows that Theorem 4 is not valid when n=3, and the condition n=4 is best possible for meromorphic case.In 2011, Yuan et al. [20] and Wang et al. [21] proved the following theorems, independently, respectively.Theorem 7 (see [20, 21]).
Letℱ be a family of meromorphic functions in D and a, b two finite complex numbers such that a≠0. Suppose that each f∈ℱ has no simple pole. If f′-af3 and g′-ag3 share the value b IM for every pair of functions f and g in ℱ, then ℱ is normal in D.Theorem 8 (see [20]).
Letℱ be a family of meromorphic functions in D and a and b two finite complex numbers such that a≠0. Suppose that each f∈ℱ admits zeros of multiple and the poles of multiplicity at least 3. If f′-af2 and g′-ag2 share the value b IM for every pair of functions f and g in ℱ, then ℱ is normal in D.Lately, Yuan et al. [22] and Qiu et al. [23] studied this result, independently, respectively, in which the derivative f′ was replaced by kth derivative f(k), and they got the following results.Theorem 9 (see [22]).
Letℱ be a family of meromorphic functions in D, and let k(≥2) and n(≥k+2) be two positive integers. Let a(≠0) and b be two finite complex numbers. If(i)
f(k)-afn and g(k)-agn share b IM in D for every pair of functions f and g in ℱ,(ii)
f has no simple pole and no zero of multiplicity less than k in D for every function f∈ℱ, then ℱ is normal in D.Theorem 10 (see [22, 23]).
Letℱ be a family of meromorphic functions in D, k(≥2) and n(≥k+3) be two positive integers. Let a(≠0) and b be two finite complex numbers. If(i)
f(k)-afn and g(k)-agn share b IM in D for every pair of functions f and g in ℱ,(ii)
f has no zero of multiplicity less than k in D for every function f∈ℱ, then ℱ is normal in D.It is natural to ask whether the conditionn≥k+2 or n≥k+3 in the previous theorems can be reduced. In this paper, we study this problem and get the following results.Theorem 11 (main theorem).
LetD be a domain in ℂ and let ℱ be a family of meromorphic functions in D. Let k,n,d∈ℕ+,n≥3,d≥(k+1)/(n-2) and let a, b be two finite complex numbers with a≠0. Suppose that every f∈ℱ has all its zeros of multiplicity at least k and all its poles of multiplicity at least d. If f(k)-afn and g(k)-agn share the value b IM for every pair of functions (f,g) of ℱ, then ℱ is a normal family in D.Remark 12.
Whenn≥k+2 or n≥k+3, we have d=1 or 2, respectively. It follows that Theorem 11 generalizes Theorems 4, 7, 9, and 10.Theorem 13 (main theorem).
LetD be a domain in ℂ and let ℱ be a family of meromorphic functions in D. Let k∈ℕ+ and a, b be two finite complex numbers with a≠0. Suppose that every f∈ℱ has all its zeros of multiplicity at least k+1 and all its poles of multiplicity at least k+2. If f(k)-af2 and g(k)-ag2 share the value b IM for every pair of functions (f,g) of ℱ, then ℱ is a normal family in D.Remark 14.
Whenk=1, it follows that Theorem 13 generalizes Theorem 8.Example 15 (see [23]).
Letn,k≥2 be two positive integers and a a nonzero complex constant. The family of meromorphic functions is ℱ={fj(z)=jzk-1:j=1,2,…,}, D={z:|z|<1}. Obviously, for each pair of m, j, fj(k)-afjn and fm(k)-afmn share the value 0 in D, but ℱ is not normal.Example 16.
Letk be a positive integer and a,b two nonzero complex constants such that bj=a/(-1)k((2k)!/k!)jk. The family of meromorphic functions is ℱ={fj(z)=1/(jz-bj)k+1:j=1,2,…}, D={z:|z|<1}. Obviously, for each pair of m,j,fj(k)-afj2 and fm(k)-afm2 share the value 0 in D, but ℱ is normal since f#(0)→∞ as j→∞.Remark 17.
Example15 shows that the condition that f(z) admits zeros of multiplicity at least k is best in Theorem 11. For the case k=1, n=3, Example 5 shows that the condition that f(z) admits poles of multiplicity at least d is sharp in Theorem 11. For the case n=2, Example 16 shows that the condition that f(z) admits poles of multiplicity at least k+2 is sharp in Theorem 13. For the case k=1, n=2, Example 2 shows that the condition that f(z) admits zeros of multiplicity at least k+1 in Theorem 13 is sharp.


## 2. Preliminary Lemmas

In order to prove our results, we need the following lemmas. The first is the extended version Zalcman’s [24] concerning normal families.Lemma 1 (see [25]).
Letℱ be a family of meromorphic functions on the unit disc satisfying all zeros of functions in ℱ having multiplicity ≥p and all poles of functions in ℱ having multiplicity ≥q. Let α be a real number satisfying -q<α<p. Then ℱ is not normal at 0 if and only if there exist (a)
a number0<r<1;(b)
pointsznwith|zn|<r;(c)
functionsfn∈ℱ;(d)
positive numbersρn→0such that gn(ζ):=ρn-αfn(zn+ρnζ) converges spherically uniformly on each compact subset of ℂ to a nonconstant meromorphic function g(ζ), whose all zeros have multiplicity ≥p and all poles have multiplicity ≥q and order is at most 2.Lemma 2.
Letf(z) be a meromorphic function such that f(k)(z)≢0 and c∈ℂ∖{0}, k,n,d∈ℕ+ with n≥3, d≥(k+1)/(n-2). If all zeros of f are of multiplicity at least k and all poles of f are of multiplicity at least d, then
(2)T(r,f)≤1kN(r,1f)+N¯(r,1f(k)-cfn)+S(r,f),
where S(r,f)=o(T(r,f)), as r→∞, possibly outside a set with finite linear measure.Proof.
Set(3)Φ(z):=f(k)(z)cfn(z).
Sincef(k)(z)≢0, we have Φ(z)≢0. Thus,
(4)fn(z)=f(k)(z)cΦ(z).
Hence,
(5)nm(r,f)=m(r,fn)≤m(r,f(k)Φ)+log+1|c|≤m(r,1Φ)+m(r,f(k))+log+1|c|≤m(r,1Φ)+m(r,f(k)f)+m(r,f)+log+1|c|.
So that
(6)(n-1)m(r,f)≤m(r,1Φ)+m(r,f(k)f)+log+1|c|.
On the other hand, (4) gives
(7)nN(r,f)≤N(r,fn)=N(r,f(k)Φ)≤N(r,f(k))+N(r,1Φ)-N¯(r,Φ=f(k)=0),
where N¯(r,Φ=f(k)=0) denotes the counting function of zeros of both Φ and f(k).
We obtain(8)nN(r,f)≤N(r,f)+kN¯(r,f)+N(r,1Φ)-N¯(r,Φ=f(k)=0),(n-1)N(r,f)≤kN¯(r,f)+N(r,1Φ)-N¯(r,Φ=f(k)=0).
By (4), we have
(9)N¯(r,φ)+N¯(r,1φ)≤N¯(r,1f)+N¯(r,f)+N¯(r,Φ=f(k)=0).
From (6)~(9), we obtain
(10)(n-1)T(r,f)≤kN¯(r,f)+T(r,1Φ)-N¯(r,Φ=f(k)=0)+S(r,f)≤kN¯(r,f)+T(r,Φ)-N¯(r,Φ=f(k)=0)+S(r,f)≤kN¯(r,f)+N¯(r,1Φ)+N¯(r,Φ)+N¯(r,1Φ-1)-N¯(r,Φ=f(k)=0)+S(r,f)≤(k+1)N¯(r,f)+N¯(r,1f)+N¯(r,1f(k)-cfn)+S(r,f).
Since all zeros and poles off are multiplicities at least k and d, we get
(11)N¯(r,f)≤1dN(r,f)≤1dT(r,f)≤n-2k+1T(r,f),N¯(r,1f)≤1kN(r,1f).
So that
(12)T(r,f)≤1kN(r,1f)+N¯(r,1f(k)-cfn)+S(r,f).Lemma 3.
Letf(z) be a transcendental meromorphic function such that f(k)(z)≢0. Let k∈ℕ+ and c∈ℂ∖{0}. If all zeros of f are of multiplicity at least k+1 and all poles of f are of multiplicity at least k+2, then f(k)-cf2 has infinitely many zeros.Proof.
Suppose thatf(k)-cf2 has only finitely many zeros; then N(r,1/(f(k)-cf2))=S(r,f). Clearly, an arbitrary zero of f is a zero of f(k)-cf2 since all zeros of f are of multiplicity at least k+1; then we can deduce that f has only finite zeros, so N(r,1/f)=O(logr)=S(r,f).
Set(13)Φ(z):=f(k)(z)cf2(z).
Similarly, with the proof of Lemma 2, we can get
(14)T(r,f)≤(k+1)N¯(r,f)+N¯(r,1f)+N¯(r,1f(k)-cf2)+S(r,f).
Since all poles off are multiplicities at least k+2, we obtain
(15)N¯(r,f)≤1k+2N(r,f)≤1k+2T(r,f),
so that
(16)T(r,f)≤(k+2)N¯(r,1f)+(k+2)N¯(r,1f(k)-cf2)+S(r,f)=S(r,f).
This is contradicting with the fact that f is transcendental.
Hence, Lemma3 is proved completely.Lemma 4.
Letf(z) be a nonconstant rational function such that f(k)(z)≢0. Let a∈ℂ∖{0}, and k,n,m∈ℕ+ with n≥2 and m≥(k+2)/(n-1). If f≠0 and all poles of f are of multiplicity at least m, then f(k)-afn has at least two zeros.Proof.
Suppose, to the contrary, thatf(k)-afn has at most one zero.
Sincef≠0, we get that f is a rational but not a polynomial.
Case  1. If f(k)-afn has only zero z0 with multiplicity l, set
(17)f(z)=A(z-z1)β1(z-z2)β2⋯(z-zt)βt,
where A is a nonzero constant and βi≥(k+2)/(n-1)(i=1,2,…,t).
For the sake of simplicity, we denote(18)β1+β2+⋯+βt=q.
From (17), we have
(19)f(k)=g(z)(z-z1)β1+k(z-z2)β2+k⋯(z-zt)βt+k,
where g(z) is a polynomial such that deg(g(z))≤k(t-1).
From (17) and (19), we get
(20)f(k)-afn=g(z)(z-z1)β1+k(z-z2)β2+k⋯(z-zt)βt+k-aAn(z-z1)nβ1(z-z2)nβ2⋯(z-zt)nβt=[g(z)(z-z1)(n-1)β1-k(z-z2)(n-1)β2-k⋯(z-zt)(n-1)βt-k-aAn]×((z-z1)nβ1(z-z2)nβ2⋯(z-zt)nβt)-1.
By the assumption thatf(k)-afn has exactly one zero z0 with multiply l, we have
(21)f(k)-afn=C(z-z0)l(z-z1)nβ1(z-z2)nβ2⋯(z-zt)nβt,
where C is a nonzero constant. Thus,
(22)C(z-z0)l≡g(z)(z-z1)(n-1)β1-k⋯(z-zt)(n-1)βt-k-aAn.
Differentiating (22), we obtain
(23)Cl(z-z0)l-1≡(z-z1)(n-1)β1-k-1⋯(z-zt)(n-1)βt-k-1×[∏j=1,j≠it‍g′(z)(z-z1)⋯(z-zt)+g(z)∑i=1t‍((n-1)βi-k)∏j=1,j≠it‍(z-zj)].
For the sake of simplicity, we denote(24)g1(z)=Cl(z-z0)l-1,g2(z)=(z-z1)(n-1)β1-k-1⋯(z-zt)(n-1)βt-k-1×[∏j=1,j≠it‍g′(z)(z-z1)⋯(z-zt)+g(z)∑i=1t‍((n-1)βi-k)∏j=1,j≠it‍(z-zj)].
Hence,
(25)g1(z)≡g2(z).
Since(n-1)βi-k-1≥1, we have g2(zi)=0. But g1(zi)≠0(i=1,2,…,t), a contradiction.
Case  2. If f(k)-afn has no zeros, then l=0 for (21). We have
(26)f(k)-afn=C(z-z1)nβ1(z-z2)nβ2⋯(z-zt)nβt,
where C is a nonzero constant. Thus,
(27)C≡g(z)(z-z1)(n-1)β1-k⋯(z-zt)(n-1)βt-k-aAn,g(z)(z-z1)(n-1)β1-k⋯(z-zt)(n-1)βt-k≡C+aAn.
Obviously,g(z)(z-z1)(n-1)β1-k⋯(z-zt)(n-1)βt-k is not a constant, a contradiction.
Lemma4 is proved.Lemma 5.
Letf(z) be a rational function and a∈ℂ∖{0}, and k,n,m∈ℕ+ with n≥2 and m≥(k+2)/(n-1). If all zeros of f are of multiplicity at least k+1 and all poles of f are of multiplicity at least m, then f(k)-afn has at least two distinct zeros.Proof.
Suppose, to the contrary, thatf(k)-afn has at most one zero.
Case  I. When f is a nonconstant polynomial, noting that all zeros of f have multiplicity at least k+1, we know that f(k)-afn must have zeros. We claim that f has exactly one zero. Otherwise, we can get that f(k)-afn has at least two zeros, which contradicts our assumption.
Set(28)f(z)=B(z-z0)s,
where s≥k+1, B is a nonzero constant. Then
(29)f(k)(z)-afn(z)=B(z-z0)s-k×[s(s-1)⋯(s-k+1)-a(z-z0)(n-1)s+k].
Sinces-k≥1, we obtain that s(s-1)⋯(s-k+1)-a(z-z0)s+k has at least one zero which is not z0 from (29). Therefore, f(k)-afn has at least two distinct zeros, a contradiction.
Case  II. When f is rational but not a polynomial, we consider two cases.
Case  1. Suppose that f(k)-afn has only zero z0 with multiplicity at least l. If f≠0, by Lemma 4, we get a contradiction. So f has zeros, and then we can deduce that z0 is the only zero of f. Otherwise, f(k)-afn has at least two distinct zeros, a contradiction.
We set(30)f(z)=A(z-z0)s(z-z1)β1(z-z2)β2⋯(z-zt)βt,
where A is a nonzero constant and s≥k+1, βi≥m≥((k+2)/(n-1))(i=1,2,…,t).
For the sake of simplicity, we denote(31)β1+β2+⋯+βt=q.
From (31), we have
(32)f(k)=A(z-z0)s-kg(z)(z-z1)β1+k(z-z2)β2+k⋯(z-zt)βt+k,
where g(z) is a polynomial with deg(g)≤k(s+t+1).
From (30) and (32), we get
(33)f(k)-afn=A(z-z0)s-kg(z)(z-z1)β1+k(z-z2)β2+k⋯(z-zt)βt+k-An(z-z0)ns(z-z1)nβ1(z-z2)nβ2⋯(z-zt)nβt=(A(z-z0)s-k×[g(z)(z-z1)(n-1)β1-k(z-z2)(n-1)β2-k⋯(z-zt)(n-1)βt-k-aAn-1×(z-z0)(n-1)s+k])×((z-z1)nβ1(z-z2)nβ2⋯(z-zt)nβt)-1.
By assumption thatf(k)-afn has exactly one zero z0 with multiplicity l, we have
(34)f(k)-afn=B(z-z0)l(z-z1)nβ1(z-z2)nβ2⋯(z-zt)nβt,
where B is a nonzero constant. Thus,
(35)B(z-z0)l≡A(z-z0)s-k×[g(z)(z-z1)(n-1)β1-k⋯(z-zt)(n-1)βt-k-aAn-1(z-z0)(n-1)s+k].Case  1.1. If l>s-k, from (35), we can deduce that z0 is a zero of (z-z1)(n-1)β1-k⋯(z-zt)(n-1)βt-k, a contradiction.
Case  1.2. If l=s-k, from (35), it follows that
(36)g(z)(z-z1)(n-1)β1-k⋯(z-zt)(n-1)βt-k-aAn-1(z-z0)(n-1)s+k≡BA.
Differentiating (36), we have
(37)(z-z1)(n-1)β1-k-1⋯(z-zt)(n-1)βt-k-1×[∏j=1,j≠it‍g′(z)(z-z1)⋯(z-zt)+g(z)∑i=1t‍((n-1)βi-k)∏j=1,j≠it‍(z-zj)]≡a((n-1)s+k)An-1(z-z0)(n-1)s+k-1.
For the sake of simplicity, we denote(38)g1(z)=(z-z1)(n-1)β1-k-1⋯(z-zt)(n-1)βt-k-1×[∏j=1,j≠it‍g′(z)(z-z1)⋯(z-zt)+g(z)∑i=1t‍((n-1)βi-k)∏j=1,j≠it‍(z-zj)],g2(z)=a((n-1)s+k)An-1(z-z0)(n-1)s+k-1.
Thus,
(39)g1(z)≡g2(z).
Since(n-1)βi-k-1≥1, we get g1(zi)=0, but g2(zi)≠0(i=1,2,…,t), a contradiction.
Case  2. If f(k)-afn has no zeros, then l=0 for (34). Similarly, as the proof of Case  1, we also have a contradiction.
The proof is completed.


## 3. Proofs of Theorems

Proof.
In Theorem11, suppose that ℱ is not normal in D. Then there exists at least one point z0 such that ℱ is not normal at the point z0. Without loss of generality, we assume that z0=0. By Lemma 1, there exist points zj→0, positive numbers ρj→0, and functions fj∈ℱ such that
(40)gj(ξ)=ρjk/(n-1)fj(zj+ρjξ)⟹g(ξ)
locally uniformly with respect to the spherical metric, where g is a nonconstant meromorphic function in ℂ and whose poles and zeros are of multiplicity at least d and k, respectively. Moreover, the order of g is at most 2.
From (40), we know that
(41)gj(k)(ξ)=ρjnk/(n-1)fj(k)(zj+ρjξ)⟹g(k)(ξ),(42)gj(k)(ξ)-agjn(ξ)-ρjnk/(n-1)b=ρjnk/(n-1)(fj(k)(zj+ρjξ)-afjn(zj+ρjξ)-b)⟹g(k)(ξ)-agn(ξ)
also locally uniformly with respect to the spherical metric.
Case  1. If k=1, by Theorems 4 and 7, the Theorem 11 assumes.
Case  2. If k=2, n≥4, by Theorems 9 and 10, the Theorem 11 assumes.
Case  3. If k=2, n=3 or k≥3, 3≤n≤k+1 (If n≥k+2, by Theorems 9 and 10, we can get Theorem 11).
Ifg(k)(ξ)-agn(ξ)≡0, since all poles of g have multiplicity at least d, we have
(43)nT(r,g)=T(r,gn)=T(r,g(k))+O(1)=m(r,g(k))+N(r,g(k))+O(1)≤m(r,g)+N(r,g)+kN¯(r,g)+S(r,g)≤T(r,g)+k(n-2)k+1T(r,g)+S(r,g)≤(n-1)T(r,g)+S(r,g).
Therefore, g(ξ) is a constant, a contradiction. So g(k)(ξ)-agn(ξ)≢0.
By Lemma2, we have
(44)T(r,g)≤1kN(r,1g)+N¯(r,1g(k)-agn)+S(r,g)≤1kT(r,1g)+N¯(r,1g(k)-agn)+S(r,g).
Then
(45)(1-1k)T(r,g)≤N¯(r,1g(k)-agn)+S(r,g),(46)T(r,g)≤(1+1k-1)N¯(r,1g(k)-agn)+S(r,g).
If g(k)(ξ)-agn(ξ)≠0, then (46) gives that g(ξ) is also a constant. Hence, g(k)(ξ)-agn(ξ) is a nonconstant meromorphic function and has at least one zero.
Next we prove thatg(k)(ξ)-agn(ξ) has just a unique zero. On the contrary, let ξ0 and ξ0* be two distinct zeros of g(k)(ξ)-agn(ξ), and choose δ(>0) small enough such that D(ξ0,δ)∩D(ξ0*,δ)=ϕ, where D(ξ0,δ)={ξ:|ξ-ξ0|<δ} and D(ξ0*,δ)={ξ:|ξ-ξ0*|<δ}. From (42), by Hurwitz’s theorem, there exist points ξj∈D(ξ0,δ), ξj*∈D(ξ0*,δ) such that for sufficiently large j(47)fj(k)(zj+ρjξj)-afjn(zj+ρjξj)-b=0,fj(k)(zj+ρjξj*)-afjn(zj+ρjξj*)-b=0.
By the hypothesis that for each pair of functionsf and g in ℱ, f(k)-afn and g(k)-agn share b in D, we know that for any positive integer m(48)fm(k)(zj+ρjξj)-afmn(zj+ρjξj)-b=0,fm(k)(zj+ρjξj*)-afmn(zj+ρjξj*)-b=0.
Fixm, take j→∞, and note zj+ρjξj→0, zj+ρjξj*→0; then fm(k)(0)-afmn(0)-b=0. Since the zeros of fm(k)-afmn-b have no accumulation point, so
(49)zj+ρjξj=0,zj+ρjξj*=0.
Hence, ξj=-zj/ρj, ξj*=-zj/ρj. This contradicts ξj∈D(ξ0,δ), ξj*∈D(ξ0*,δ) and D(ξ0,δ)∩D(ξ0*,δ)=ϕ. So g(k)(ξ)-agn(ξ) has just a unique zero, which can be denoted by ξ0.
Noting thatg has poles and zeros of multiplicities at least d and k, respectively, (46) deduces that g(ξ) is a rational function with degree at most 2.
Ifg(ξ) is a polynomial, noting that degg≤2 and the multiplicities of zeros are at least k, we have k=2 and n=3. Hence, there exist ξ1 and c≠0 such that g(ξ)=c(ξ-ξ1)2, and then g′′(ξ)-ag3(ξ)=c(2-ac2(ξ-ξ1)6) has 6 distinct zeros, a contradiction.
Suppose thatg(ξ) is not a polynomial, k=2 and n=3. Then the multiplicities of poles of g(ξ) are at least (k+1)/(n-2)=3, which implies that degg≥3, a contradiction.
Suppose thatg(ξ) is not a polynomial, k≥3 and 3≤n≤k+1; we distinguish two cases.
Case  i. If g(ξ) has zeros, since all zeros of g(ξ) have multiplicity at least k(≥3), it follows that degg≥3, a contradiction.
Case  ii. If g(ξ)≠0, then (46) should be as follows
(50)T(r,g)≤N¯(r,1g(k)-agn)+S(r,g).
From (50), we can see that g(ξ) is a rational function with degree at most 1. Since all poles of g(ξ) have multiplicity at least d(≥(k+1)/(n-2)≥(k+1)/(k-1)>1), which gives that degg≥2, a contradiction.
This completes the proof of Theorem11.Proof.
In Theorem13, suppose that ℱ is not normal in D. Then there exists at least one point z0 such that ℱ is not normal at the point z0. Without loss of generality, we assume that z0=0. By Lemma 1, there exist points zj→0, positive numbers ρj→0, and functions fj∈ℱ such that
(51)gj(ξ)=ρjkfj(zj+ρjξ)⟹g(ξ)
locally uniformly with respect to the spherical metric, where g is a nonconstant meromorphic function in ℂ and whose poles and zeros are of multiplicity at least k+2 and k+1, respectively. Moreover, the order of g is at most 2.
From (51), we know
(52)gj(k)(ξ)=ρj2kfj(k)(zj+ρjξ)⟹g(k)(ξ),(53)gj(k)(ξ)-agj2(ξ)-ρj2kb=ρj2k(fj(k)(zj+ρjξ)-afj2(zj+ρjξ)-b)⟹g(k)(ξ)-ag2(ξ)
also locally uniformly with respect to the spherical metric.
Ifg(k)(ξ)-ag2(ξ)≡0, since all poles of f have multiplicity at least k+2, we can deduce that g(ξ) is an entire function easily. Thus,
(54)2T(r,g)=T(r,g2)=T(r,g(k))+O(1)=m(r,g(k))+N(r,g(k))+O(1)≤m(r,g)+N(r,g)+kN¯(r,g)+S(r,g)≤T(r,g)+S(r,g).
Therefore, g(ξ) is a constant, a contradiction. So g(k)(ξ)-ag2(ξ)≢0. By Lemmas 3, 4, and 5, g(k)(ξ)-ag2(ξ) has at least two distinct zeros. Proceeding as in the later proof of Theorem 11, we will get a contradiction. The proof is completed.Similarly, as the proof of Theorem11, when f is a holomorphic function, we can get the following theorem which has been gotten in [23].Theorem 18 (see [23]).
LetD be a domain in ℂ and let ℱ be a family of holomorphic functions in D. Let k,n∈ℕ+, n≥2 and let a, b be two finite complex numbers with a≠0. Suppose that every f∈ℱ has all its zeros of multiplicity at least k. If f(k)-afn and g(k)-agn share the value b IM for every pair of functions (f,g) of ℱ, then ℱ is a normal family in D.



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*Source: 368321-2013-10-30.xml*