# Normal Criteria of Function Families Related  to  a Result of Drasin

Authors: Zhu Bing; Lin Jianming; Yuan Wenjun
Journal: Discrete Dynamics in Nature and Society
 (2011)

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## Abstract

We study the normality of families of meromorphic functions related to a result of Drasin. We consider whether a
family meromorphic functionsℱ whose each function does not take
zero is normal in D, if for every pair of functions f and g in ℱ, f(z) and g(z) share ∞ or H(f)−1 and H(g)−1 share 0, where H(f):=f(k)(z)+ak−1f(k−1)(z)+⋯,a0f(z). Some examples show that the conditions in our results are best possible.


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## Body

## 1. Introduction and Main Result

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 (see [2–4] or [1]).Influenced from Bloch’s principle [5], every condition which reduces a meromorphic function in the plane ℂ to a constant makes a family of meromorphic functions in a domain D normal. Although the principle is false in general (see [6]), many authors proved normality criterion for families of meromorphic functions corresponding to Liouville-Picard type theorem (see [7] or [4]).It is also more interesting to find normality criteria from the point of view of shared values. In this area, Schwick [8] first proved an interesting result that a family of meromorphic functions in a domain is normal if in which every function shares three distinct finite complex numbers with its first derivative. And later, more results about normality criteria concerning shared values can be found, for instance, (see [9–13]) and so on. In recent years, this subject has attracted the attention of many researchers worldwide.The following result is due to  Schiff [14].Theorem 1.1.
Letℱ be a family holomorphic functions in D, and aj∈ℂ,j=0,1,…,k-1,k a positive integer. If f(z)≠0 and the zeros of H(f)-1 are of multiplicity ≥p>k+1 for each f(z)∈ℱ, where H(f):=f(k)(z)+ak-1f(k-1)(z)+⋯,a0f(z) and p is a positive integer, then ℱ is normal in D.In 2001, Fang and Yuan [15] extended Theorem 1.1 as follows.Theorem 1.2.
Letℱ be a family holomorphic functions in D, and aj∈ℂ,j=0,1,…,k-1,k a positive integer. If f(z)≠0 and the zeros of H(f)-1 are of multiplicity ≥2 for each f(z)∈ℱ, where H(f):=f(k)(z)+ak-1f(k-1)(z)+⋯,a0f(z), then ℱ is normal in D.It is natural to ask whether or not Theorems1.1 and 1.2 hold for meromorphic case or holomorphic case by the idea of shared values. In this paper, we answer above question and prove the following results.Theorem 1.3.
Letℱ be a family meromorphic functions in D, and aj∈ℂ,j=0,1,…,k-1,k a positive integer. If (i) f(z)≠0 and the zeros of H(f)-1 are of multiplicity ≥2 for each f(z)∈ℱ, (ii) f(z) and g(z) share ∞ for every pair of functions f(z) and g(z) in ℱ, where H(f):=f(k)(z)+ak-1f(k-1)(z)+⋯,a0f(z), then ℱ is normal in D.Theorem 1.4.
Letℱ be a family holomorphic functions in D, and aj∈ℂ,j=0,1,…,k-1,k a positive integer. If (i) f(z)≠0 for each f(z)∈ℱ, (ii) H(f)-1 and H(g)-1 share the value 0 for every pair of functions f(z) and g(z) in ℱ, where H(f):=f(k)(z)+ak-1f(k-1)(z)+⋯,a0f(z), then ℱ is normal in D.Example 1.5.
The family of holomorphic functionsℱ={fn(z)=nzez-nez:n=1,2,…} is not normal in D={z:|z|<1}. Obviously fn(k)-fn=kez≠0. On the other hand, fn(0)=-n,fn(1/n)=-1/2+(1/2n)+o(1/n)→-1/2, as n→∞. This implies that the family ℱ fails to be equicontinuous at 0, and thus ℱ is not normal at 0.Example 1.6.
The family of holomorphic functionsℱ={fn(z)=n(z+1)-1:n=1,2,…} is normal in D={z:|z|<1}. Obviously, fn(z)≠0 and fn′-fn=-nz+1. So for each pair m,n,fn′-fn and fm′-fm share the value 1 in D. Theorem 1.4 implies that the family ℱ is normal in D.Example 1.7.
The family of meromorphic functionsℱ={fn(z)=z/n-1:n=1,2,…} is normal in D={z:|z|<1}. The reason is the conditions of Theorems 1.3 and 1.4 hold, that is, fn(z)≠0 and fn′-fn=(1-z)/n+1≠1 for each fn(z)∈ℱ in D={z:|z|<1}.Example 1.8.
The family of meromorphic functionsℱ={fn(z)=n/z-1:n=1,2,…} is normal in D={z:|z|<1}. The reason is the conditions of Theorem 1.3 hold, that is, fn(z)≠0 and has only one pole 0 for each f(z)∈ℱ; fn(k)-fn-1=n((-1)k!-zk)/zk+1≠0 for each fn(z)∈ℱ in D={z:|z|<1}.Remark 1.9.
Example1.5 shows that H(g)-1 is not replaced by H(g) in Theorems 1.3 and 1.4. is not valid when n=1. All of Examples 1.6, 1.7 and 1.8 show that Theorems 1.3 and 1.4 occur.


## 2. Preliminary Lemmas

In order to prove our result, we need the following lemmas. The first one extends a famous result by Zalcman [16] concerning normal families.Lemma 2.1 (see [17]).
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 have multiplicity ≥q. Let α be a real number satisfying -p<α<q. Then, ℱ is not normal at 0 if and only if there exist(a)
a number0<r<1,(b)
pointszn with |zn|<r,(c)
functionsfn∈F,(d)
positive numbersρn→0such that gn(ζ):=ραfn(zn+ρnζ) converges spherically uniformly on each compact subset of ℂ to a nonconstant meromorphic function g(ζ), whose all zeros of functions in ℱ have multiplicity ≥p and all poles of functions in ℱ have multiplicity ≥q and order is at most 2.Remark 2.2.
Ifℱ is a family of holomorphic functions on the unit disc in Lemma 2.1, then g(ζ) is a nonconstant entire function whose order is at most 1.
The order ofg is defined by using the Nevanlinna’s characteristic function T(r,g)(2.1)ρ(g)=lim⁡r→∞sup⁡log⁡T(r,g)log⁡r.
Here, g#(ξ) denotes the spherical derivative
(2.2)g#(ξ)=|g′(ξ)|1+|g(ξ)|2.Lemma 2.3 (see [3]).
Letf(z) be a meromorphic function in ℂ, then
(2.3)T(r,f)≤(2+1k)N(r,1f)+(2+2k)N¯(r,1f(k)-1)+S(r,f),(2.4)T(r,f)≤N¯(r,f)+N(r,1f)+N¯(r,1f(k)-1)+S(r,f).Remark 2.4.
Both (2.3) and (2.4) are called Hayman inequality and Milloux inequality, respectively.


### 2.1. Proof of the Results

Proof of Theorem1.3.
Suppose thatℱ 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 2.1, there exist points zj→0, positive numbers ρj→0 and functions fj∈ℱ such that
(2.5)gj(ξ)=ρj-kfj(zj+ρjξ)⟹g(ξ),
locally uniformly with respect to the spherical metric, where g is a nonconstant meromorphic function in ℂ. Moreover, the order of g is less than 2.
Sincefj(z)∈ℱ, by Hurwitz’s theorem, we see g≠0.
From (2.5), we know
(2.6)gj(k)(ξ)=fj(k)(zj+ρjξ)⟹g(k)(ξ),(2.7)gj(k)(ξ)-1+∑i=1k-1ρjk-iaigj(i)(zj+ρjξ)=H(fj(zj+ρjξ))-1⟹g(k)(ξ)-1,
also locally uniformly with respect to the spherical metric.
Ifg(k)(ξ)-1≡0, then g is a polynomial with degree k. This contradicts g≠0.
Ifg(k)(ξ)-1≠0, then by Hayman inequality (2.3) of Lemma 2.3, we know that g is a constant, which is impossible.
Hence,g(k)(ξ)-1 is a nonconstant meromorphic function and has at least one zero. By (2.7) and Hurwitz’s theorem, we see that the zeros of g(k)(ξ)-1 are multiple.
Next, we prove thatg has at most one distinct pole. By contraries, let ξ0 and ξ0* be two distinct poles of g, 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 (2.5), by Hurwitz’s theorem, there exist points ξj∈D(ξ0,δ), ξj*∈D(ξ0*,δ) such that for sufficiently large j(2.8)fj-1(zj+ρjξj)=0,fj-1(zj+ρjξj*)=0.
By the hypothesis that for each pair of functionsf and g in ℱ, f-1 and g-1 share 0 in D, we know that for any positive integer m(2.9)fm-1(zj+ρjξj)=0,fm-1(zj+ρjξj*)=0.
Fixm, take j→∞, and note zj+ρjξj→0, zj+ρjξj*→0, then fm-1(0)=0. Since the zeros of fm-1 has no accumulation point, so.(2.10)zj+ρjξj=0,zj+ρjξj*=0.
Hence ξj=-zj/ρj,ξj*=-zj/ρj This contradicts with ξj∈D(ξ0,δ),ξj*∈D(ξ0*,δ) and D(ξ0,δ)∩D(ξ0*,δ)=ϕ. So, g has at most one distinct pole.
Forg applying Milloux inequality (2.4) of Lemma 2.3, we deduce
(2.11)T(r,g)≤N¯(r,g)+N(r,1g)+N¯(r,1g(k)-1)+S(r,g)≤log⁡r+12N(r,1g(k)-1)+S(r,g)≤12T(r,g)+log⁡r+S(r,g).
From (2.11), we know that g is a rational function of degree at most 2. Noting that g≠0, we can write g(ξ)=1/(aξ+b)m,1≤m≤2 where a≠0 and b are two finite complex numbers. Simple calculating shows that g(k)-1 has at least k+1 distinct simple zeros. This is impossible.
The proof of Theorem1.3 is complete.Proof of Theorem1.4.
Similarly with the proof of Theorem1.3, we have that (2.5) and (2.7) also hold. Moreover, g(ξ) is an entire function.
Ifg(k)(ξ)-1≡0, then g is a polynomial with degree k. This contradicts g≠0.
Ifg(k)(ξ)-1≠0, then by Hayman inequality (2.3) of Lemma 2.3 we know that g is a constant, which is impossible.
Hence,g(k)(ξ)-1 is a nonconstant entire function and has at least one zero.
Next, we prove thatg(k)(ξ)-1 has one distinct zero. By contraries, let ξ0 and ξ0* be two distinct zeros of g(k)(ξ)-1, 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 (2.7), by Hurwitz’s theorem, there exist points ξj∈D(ξ0,δ), ξj*∈D(ξ0*,δ) such that for sufficiently large j(2.12)H(fj(zj+ρjξj))-1=0,H(fj(zj+ρjξj*))-1=0.
By the hypothesis that for each pair of functionsf and g in ℱ, H(f)-1 and H(g)-1 share 0 in D, we know that for any positive integer m(2.13)H(fm(zj+ρjξj))-1=0,H(fm(zj+ρjξj*))-1=0.
Fixm, take j→∞, and note zj+ρjξj→0, zj+ρjξj*→0, then fm(k)(0)-1=0. Since the zeros of fm(k)-1 has no accumulation point, so(2.14)zj+ρjξj=0,zj+ρjξj*=0.
Hence ξj=-zj/ρj,ξj*=-zj/ρj This contradicts with ξj∈D(ξ0,δ),ξj*∈D(ξ0*,δ) and D(ξ0,δ)∩D(ξ0*,δ)=ϕ. So, g(k)(ξ)-1 has one distinct zero.
Forg applying Milloux inequality (2.4) of Lemma 2.3, we deduce(2.15)T(r,g)≤N¯(r,g)+N(r,1g)+N¯(r,1g(k)-1)+S(r,g)≤log⁡r+S(r,g).
From (2.15) and g≠0, we know that g is a polynomial of degree 1. We can write g(ξ)=(cξ+d), where c≠0 and d are two finite complex numbers. Hence, g(k)-1 has no only one distinct zero. This is impossible.
The proof of Theorem1.4 is complete.


## 2.1. Proof of the Results

Proof of Theorem1.3.
Suppose thatℱ 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 2.1, there exist points zj→0, positive numbers ρj→0 and functions fj∈ℱ such that
(2.5)gj(ξ)=ρj-kfj(zj+ρjξ)⟹g(ξ),
locally uniformly with respect to the spherical metric, where g is a nonconstant meromorphic function in ℂ. Moreover, the order of g is less than 2.
Sincefj(z)∈ℱ, by Hurwitz’s theorem, we see g≠0.
From (2.5), we know
(2.6)gj(k)(ξ)=fj(k)(zj+ρjξ)⟹g(k)(ξ),(2.7)gj(k)(ξ)-1+∑i=1k-1ρjk-iaigj(i)(zj+ρjξ)=H(fj(zj+ρjξ))-1⟹g(k)(ξ)-1,
also locally uniformly with respect to the spherical metric.
Ifg(k)(ξ)-1≡0, then g is a polynomial with degree k. This contradicts g≠0.
Ifg(k)(ξ)-1≠0, then by Hayman inequality (2.3) of Lemma 2.3, we know that g is a constant, which is impossible.
Hence,g(k)(ξ)-1 is a nonconstant meromorphic function and has at least one zero. By (2.7) and Hurwitz’s theorem, we see that the zeros of g(k)(ξ)-1 are multiple.
Next, we prove thatg has at most one distinct pole. By contraries, let ξ0 and ξ0* be two distinct poles of g, 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 (2.5), by Hurwitz’s theorem, there exist points ξj∈D(ξ0,δ), ξj*∈D(ξ0*,δ) such that for sufficiently large j(2.8)fj-1(zj+ρjξj)=0,fj-1(zj+ρjξj*)=0.
By the hypothesis that for each pair of functionsf and g in ℱ, f-1 and g-1 share 0 in D, we know that for any positive integer m(2.9)fm-1(zj+ρjξj)=0,fm-1(zj+ρjξj*)=0.
Fixm, take j→∞, and note zj+ρjξj→0, zj+ρjξj*→0, then fm-1(0)=0. Since the zeros of fm-1 has no accumulation point, so.(2.10)zj+ρjξj=0,zj+ρjξj*=0.
Hence ξj=-zj/ρj,ξj*=-zj/ρj This contradicts with ξj∈D(ξ0,δ),ξj*∈D(ξ0*,δ) and D(ξ0,δ)∩D(ξ0*,δ)=ϕ. So, g has at most one distinct pole.
Forg applying Milloux inequality (2.4) of Lemma 2.3, we deduce
(2.11)T(r,g)≤N¯(r,g)+N(r,1g)+N¯(r,1g(k)-1)+S(r,g)≤log⁡r+12N(r,1g(k)-1)+S(r,g)≤12T(r,g)+log⁡r+S(r,g).
From (2.11), we know that g is a rational function of degree at most 2. Noting that g≠0, we can write g(ξ)=1/(aξ+b)m,1≤m≤2 where a≠0 and b are two finite complex numbers. Simple calculating shows that g(k)-1 has at least k+1 distinct simple zeros. This is impossible.
The proof of Theorem1.3 is complete.Proof of Theorem1.4.
Similarly with the proof of Theorem1.3, we have that (2.5) and (2.7) also hold. Moreover, g(ξ) is an entire function.
Ifg(k)(ξ)-1≡0, then g is a polynomial with degree k. This contradicts g≠0.
Ifg(k)(ξ)-1≠0, then by Hayman inequality (2.3) of Lemma 2.3 we know that g is a constant, which is impossible.
Hence,g(k)(ξ)-1 is a nonconstant entire function and has at least one zero.
Next, we prove thatg(k)(ξ)-1 has one distinct zero. By contraries, let ξ0 and ξ0* be two distinct zeros of g(k)(ξ)-1, 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 (2.7), by Hurwitz’s theorem, there exist points ξj∈D(ξ0,δ), ξj*∈D(ξ0*,δ) such that for sufficiently large j(2.12)H(fj(zj+ρjξj))-1=0,H(fj(zj+ρjξj*))-1=0.
By the hypothesis that for each pair of functionsf and g in ℱ, H(f)-1 and H(g)-1 share 0 in D, we know that for any positive integer m(2.13)H(fm(zj+ρjξj))-1=0,H(fm(zj+ρjξj*))-1=0.
Fixm, take j→∞, and note zj+ρjξj→0, zj+ρjξj*→0, then fm(k)(0)-1=0. Since the zeros of fm(k)-1 has no accumulation point, so(2.14)zj+ρjξj=0,zj+ρjξj*=0.
Hence ξj=-zj/ρj,ξj*=-zj/ρj This contradicts with ξj∈D(ξ0,δ),ξj*∈D(ξ0*,δ) and D(ξ0,δ)∩D(ξ0*,δ)=ϕ. So, g(k)(ξ)-1 has one distinct zero.
Forg applying Milloux inequality (2.4) of Lemma 2.3, we deduce(2.15)T(r,g)≤N¯(r,g)+N(r,1g)+N¯(r,1g(k)-1)+S(r,g)≤log⁡r+S(r,g).
From (2.15) and g≠0, we know that g is a polynomial of degree 1. We can write g(ξ)=(cξ+d), where c≠0 and d are two finite complex numbers. Hence, g(k)-1 has no only one distinct zero. This is impossible.
The proof of Theorem1.4 is complete.



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