# A Note on Normal Families of Meromorphic Functions Concerning Shared Values

Authors: Yuan Wenjun; Wei Jinjin; Lin Jianming
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 Hayman conjecture. We consider whether a family of meromorphic functionsℱ is normal in D if, for every pair of functions f and g in ℱ, f′−afn and g′−agn share the value b for n=1,2, and 3, where a and b≠0 are two finite complex numbers. Some examples show that the conditions in our results are the best possible.


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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 ([2–4] or [1]).Bloch’s principle [5] states that every condition which reduces a meromorphic function in the plane ℂ to be a constant forces a family of meromorphic functions in a domain D normal. Although the principle is false in general (see [6]), many authors proved the 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 it in every function in it shares three distinct finite complex numbers with its first derivative. Later, more results about normality criteria concerning shared values can be found, for instance, see [9–11] and so on. 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 [12].Theorem 1.1.
Letℱ be a family of holomorphic (meromorphic) functions defined in a domain D, n∈ℕ,a≠0,b∈ℂ. If f′(z)+afn(z)-b≠0 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 [7] for n≥3, Pang [13] for n=3, and Chen and Fang [14] for n=2, Ye [15] for n=2, and Chen and Gu [16] for the generalized result with a and b replaced by meromorphic functions. The results for the meromorphic case are due to Li [17], Li [18], and Langley [19] for n≥5, Pang [13] for n=4, Chen and Fang [14] for n=3, and Zalcman [20] for n=3, obtained independently.Whenn=2 and ℱ is meromorphic, Theorem 1.1 is not valid in general. Fang and Yuan [21] gave an example to this, and moreover a result added other conditions below.Example 1.2.
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
(1.1)f#(ξ)=|f′(ξ)|1+|f(ξ)|2.Theorem 1.3.
Letℱ be a family of meromorphic functions in a domain D and a≠0,b∈ℂ. If f′(z)+a(f(z))2-b≠0 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 [11] proved the following.Theorem 1.4.
Letℱ be a family of meromorphic (holomorphic) functions in D, n a positive integer, and a, b 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, then ℱ is normal in D.Example 1.5 (see [11]).
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 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 1.6.
Example1.5 shows that Theorem 1.4 is not valid when n=3, and the condition n=4 is the best possible for the meromorphic case.It is natural to ask under what conditions Theorem1.4 holds for n≤3. In this paper, we answer the above question and prove the following results.Theorem 1.7.
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(z)∈ℱ has no simple pole. If f′-af3 and g′-ag3 share the value b for every pair of functions f and g in ℱ, then ℱ is normal in D.Remark 1.8.
Example1.5 shows that the condition added in Theorem 1.7 about the multiplicity of poles of f(z) is the best possible.Theorem 1.9.
Letℱ be a family of meromorphic functions in D and a and b two finite complex numbers such that a≠0. Suppose that f(z) admits the zeros of multiple and the poles of multiplicity ≥3 for each f(z)∈ℱ. If f′-af2 and g′-ag2 share the value b for every pair of functions f and g in ℱ, then ℱ is normal in D.Remark 1.10.
Example1.2 shows that the condition added in Theorem 1.9 about the multiplicity of poles and zeros of f(z) is the best possible.Theorem 1.11.
Letℱ be a family of meromorphic functions in D and a and b two nonzero finite complex numbers. Suppose that f(z)≠0 and that its poles are multiple for each f(z)∈ℱ. If f′-af and g′-ag share the value b for every pair of functions f and g in ℱ, then ℱ is normal in D.Corollary 1.12.
Letℱ be a family of holomorphic functions in D and a and b two finite complex numbers such that b≠0. Suppose that f(z)≠0 for each f(z)∈ℱ. If f′-af and g′-ag share the value b for every pair of functions f and g in ℱ, then ℱ is normal in D.Example 1.13.
The family of holomorphic functionsℱ={fj(z)=jzez-jez+j-b:j=1,2,…,} is not normal in D={z:|z|<1}. Obviously fj′-fj=j(ez-1)+b. So, for each pair m,j,fj′-fj and fm′-fm share the value b in D. On the other hand, fj(0)=-b,fj(1/j)=j(1+(1/j)+o(1))→∞, as j→∞. This implies that the family ℱ fails to be equicontinuous at 0, and thus ℱ is not normal at 0.Theorem 1.14.
Letℱ be a family of meromorphic functions in D and a and b two finite complex numbers such that b≠0. Suppose that f(z)≠0 and f′(z)-af(z)≠b for each f(z)∈ℱ. Then ℱ is normal in D.Example 1.15.
The family of holomorphic functionsℱ={fj(z)=j(z+1)-1:j=1,2,…,} is normal in D={z:|z|<1}. Obviously fj(z)≠0 and fj′-fj=-jz+1. So, for each pair m,j,fj′-fj and fm′-fm share the value 1 in D. Corollary 1.12 implies that the family ℱ is normal in D.Example 1.16.
The family of meromorphic functionsℱ={fj(z)=(z/j)-1:j=1,2,…,} is normal in D={z:|z|<1}. The reason is that the conditions of Theorem 1.14 hold that fj(z)≠0 and fj′-fj=((1-z)/j)+1≠1 in D={z:|z|<1}.Remark 1.17.
Example1.13 shows that Theorem 1.4 is not valid when n=1 and in the holomorphic case and the condition f(z)≠0 is necessary in Theorem 1.11, Corollary 1.12. Both Examples 1.15 and 1.16 tell us that Corollary 1.12 and Theorem 1.14 occur.


## 2. Preliminary Lemmas

In order to prove our result, we need the following lemmas. The first one extends a famous result by Zalcman [22] concerning normal families.Lemma 2.1 (see [23]).
Letℱ be a family of meromorphic functions on the unit disc satisfying all zeros of functions in ℱ  which have multiplicity ≥p and all poles of functions in ℱ which have 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):=ρ-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.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 Nevanlinna’s characteristic function T(r,g):
(2.1)ρ(g)=lim⁡r→∞sup⁡log⁡T(r,g)log⁡r.Lemma 2.3 (see [24] or [25]).
Letf(z) be a meromorphic function and c∈ℂ∖{0}. If f(z) has neither simple zero nor simple pole and f′(z)≠c, then f(z) is constant.Lemma 2.4 (see [26]).
Letf(z) be a transcendental meromorphic function of finite order in ℂ, with no simple zero; then f′(z) assumes every nonzero finite value infinitely often.Lemma 2.5 (see [3]).
Letf(z) be a meromorphic function in ℂ; then
(2.2)T(r,f)≤(2+1k)N(r,1f)+(2+2k)N¯(r,1f(k)-1)+S(r,f),(2.3)T(r,f)≤N¯(r,f)+N(r,1f)+N¯(r,1f(k)-1)+S(r,f).Remark 2.6.
Both (2.2) and (2.3) are called the Hayman inequality and Milloux inequality, respectively.


## 3. Proof of the Results

Proof of Theorem1.7.
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
(3.1)gj(ξ)=ρj1/2fj(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 no greater than 2 and the poles of g are of multiplicity ≥2.
From (3.1) we know that(3.2)gj′(ξ)=ρj3/2fj′(zj+ρjξ)⟹g′(ξ),(3.3)ρj3/2(fj′(zj+ρjξ)-afj3(zj+ρjξ)-b)=gj′(ξ)-agj3(ξ)-ρj3/2b⟹g′(ξ)-ag3(ξ)
in ℂ∖S locally uniformly with respect to the spherical metric, where S is the set of all poles of g(ξ).
Ifg′-ag3≡0, then -(1/2)(1/g2)≡aξ+c, where c is a constant. This contradicts with the idea that the poles of g are of multiplicity ≥2. So g′-ag3≢0.
Ifg′-ag3≠0, then g′/g3≠a. Set g=1/φ; then φφ′≠-a. By Lemma 2.3, φ is a constant, so g is also a constant which is a contradiction with g being a nonconstant. Hence, g′-ag3 is a nonconstant meromorphic function and has at least one zero.
Next we prove thatg′-ag3 has just a unique zero. By contraries, let ξ0 and ξ0* be two distinct zeros of g′-ag3, 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 (3.3), by Hurwitz's theorem, there exist points ξj∈D(ξ0,δ), ξj*∈D(ξ0*,δ) such that for sufficiently largej(3.4)fj′(zj+ρjξj)-afj3(zj+ρjξj)-b=0,fj′(zj+ρjξj*)-afj3(zj+ρjξj*)-b=0.
By the hypothesis that, for each pair of functionsf and g in ℱ, f′-af3 and g′-ag3 share b in D, we know that, for any positive integer m(3.5)fm′(zj+ρjξj)-afm3(zj+ρjξj)-b=0,fm′(zj+ρjξj*)-afm3(zj+ρjξj*)-b=0.
Fixm, take j→∞, and note that zj+ρjξj→0, zj+ρjξj*→0; then fm′(0)-afm3(0)-b=0. Since the zeros of fm′-afm3-b have no accumulation point,(3.6)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′-ag3 has just a unique zero, which can be denoted by ξ0.
Setg=1/φ again; then g′-ag3=-(φ′φ+a)/φ3. So(φ′φ+a)/φ3 has only a unique zero ξ0. Therefore, ξ0 is a multiple pole of φ, or else a zero of φ′φ+a. If ξ0 is a multiple pole of φ, since (φ′φ+a)/φ3 has only one zero ξ0, then φ′φ+a≠0. By Lemma 2.3 again, φ is a constant which contradicts with the idea that g is a nonconstant.
Soφ has no multiple pole and φ′φ+a has just a unique zero ξ0. By Lemma 2.3, φ is not any transcendental function.
Ifφ is a nonconstant polynomial, then φ′φ+a=A(ξ-ξ0)l, where A is a nonzero constant and l is a positive integer because the poles of g are of multiplicity ≥2. So the zeros of φ are of multiplicity ≥2, and thus, l≥3. Set ψ=(1/2)φ2; then ψ′=A(ξ-ξ0)l-a and ψ′′=Al(ξ-ξ0)l-1. Note that the zeros of ψ are of multiplicity ≥4. But ψ′′ has only one zero ξ0, and so ψ has only the same zero ξ0. Hence ψ′(ξ0)=0 which contradicts with ψ′(ξ0)=a≠0. Therefore φ and ψ are rational functions which are not polynomials and ψ′+a has just a unique zero ξ0.
Next we prove that there exists no rational function such asψ. Noting that ψ=(1/2)φ2, φ has no multiple pole, and the zeros of φ are of multiplicity ≥2, we can set(3.7)ψ(ξ)=A(ξ-ξ1)m1(ξ-ξ2)m2⋯(ξ-ξs)ms(ξ-η1)2(ξ-η2)2⋯(ξ-ηt)2,
where A is a nonzero constant, s≥1,t≥1,andmi≥4(i=1,2,…,s). For stating briefly, denote
(3.8)m=m1+m2+⋯+ms≥4s.
From (3.7),
(3.9)ψ′(ξ)=A(ξ-ξ1)m1-1(ξ-ξ2)m2-1⋯(ξ-ξs)ms-1h(ξ)(ξ-η1)3(ξ-η2)3⋯(ξ-ηt)3=p1(ξ)q1(ξ),
where
(3.10)h(ξ)=(m-2t)ξs+t-1+as+t-2ξs+t-2+⋯+a0,p1(ξ)=A(ξ-ξ1)m1-1(ξ-ξ2)m2-1⋯(ξ-ξs)ms-1h(ξ),q1(ξ)=(ξ-η1)3(ξ-η2)3⋯(ξ-ηt)3
are polynomials. Since ψ′(ξ)+a has only a unique zero ξ0, set
(3.11)ψ′(ξ)+a=B(ξ-ξ0)l(ξ-η1)3(ξ-η2)3⋯(ξ-ηt)3,
where B is a nonzero constant, and so
(3.12)ψ′′(ξ)=(ξ-ξ0)l-1p2(ξ)(ξ-η1)4(ξ-η2)4⋯(ξ-ηt)4,
where p2(ξ)=B(l-3t)ξt+bt-1ξt-1+⋯+b0 is a polynomial. From (3.9), we also have
(3.13)ψ′′(ξ)=(ξ-ξ1)m1-2(ξ-ξ2)m2-2⋯(ξ-ξs)ms-2p3(ξ)(ξ-η1)4(ξ-η2)4⋯(ξ-ηt)4,
where p3(ξ) is also a polynomial.
Letdeg(p) denote the degree of a polynomial p(ξ).
From (3.9) and (3.10),(3.14)deg⁡(h)≤s+t-1,deg⁡(p1)≤m+t-1,deg⁡(q1)=3t.
Similarly from (3.12) and (3.13) and noting (3.14),
(3.15)deg⁡(p2)≤t,(3.16)deg⁡(p3)≤deg⁡(p1)+t-1-(m-2s)≤2t+2s-2.
Note thatmi≥4(i=1,2,…,s). It follows from (3.9) and (3.11) that ψ′(ξi)=0(i=1,2,…,s) and ψ′(ξ0)=a≠0. Thus ξ0≠ξi(i=1,2,…,s), and then (ξ-ξ0)l-1 is a factor of p3(ξ). Hence we get that l-1≤deg(p3). Combining (3.12) and (3.13) we also have m-2s=deg(p2)+l-1-deg(p3)≤deg(p2). By (3.15) we obtain(3.17)m-2s≤deg⁡(p2)≤t.
Sincem≥4s, we know by (3.17) that(3.18)2s≤t.
Ifl≥3t, by (3.16), then 3t-1≤l-1≤deg(p3)≤2t+2s-2. Noting (3.18), we obtain 1≤0, a contradiction.
Ifl<3t, from (3.9) and (3.11), deg(p1)=deg(q1). Noting that deg(p1)=m+t-i,1≤i≤s+t,deg(q1)=3t, so m+t-i=3t,m=2t+i≠2t. From (3.10), then deg(h)=s+t-1, and then deg(p1)=m+t-1. Noting that deg(q1)=3t, m=2t+1. By (3.17), t≤2s-1. From (3.18), we obtain 1≤0, a contradiction.
The proof of Theorem1.7 is complete.Proof of Theorem1.9.
Similarly with the proof of Theorem1.7, we assume that ℱ is not normal at z0=0. Then, by Lemma 2.1, there are a sequence of complex numbers zj, zj→0(j→∞), a sequence of functions fj∈ℱ, and a sequence of positive numbers ρj→0+ such that gj(ξ)=ρjfj(zj+ρjξ) converges uniformly with respect to the spherical metric to a nonconstant mermorphic function g(ξ) and all zeros and poles of g(ξ) are multiple. Moreover, g(ξ) is of order 2 at most.
Thus(3.19)gj′(ξ)=ρj2fj′(zj+ρjξ)⟹g′(ξ),(3.20)ρj2(fj′(zj+ρjξ)-afj2(zj+ρjξ)-b)=gj′(ξ)-agj2(ξ)-ρj2b⟹g′(ξ)-ag2(ξ)
also locally uniformly with respect to the spherical metric.
Ifg′-ag2≡0, then -(1/g)≡aξ+c where c is a constant. This contradicts with the idea that the poles of g are multiple. So g′-ag2≢0.
Ifg′-ag2≠0, then (1/g)′≠-a. By Lemma 2.3, g is a constant which contradicts with our conclusion. Hence, g′-ag2 is a nonconstant meromorphic function and has at least one zero.
As the same argument in the proof of Theorem1.7, we obtain that g′-ag2 has only one distinct zero denoted by ξ0.
Setg=1/φ; then g′-ag2=-(φ′+a)/φ2. So (φ′+a)/φ2 has only a unique zero ξ0. Therefore ξ0 is a multiple pole of φ or a zero of φ′+a. If ξ0 is a multiple pole of φ, since (φ′+a)/φ2 has only one zero ξ0, then φ′+a≠0. By Lemma 2.3 again, φ is a constant, which is a contradiction.
Henceφ is an entire function with no simple zero and growth order at most 2 and φ′+a has just a unique zero ξ0. By Lemma 2.4, φ is not any transcendental function. Therefore φ is a nonconstant polynomial and has the form that φ′+a=C(ξ-ξ0)l, where C is a nonzero constant and l is a positive integer because the poles of g are of multiplicity ≥3. So the zeros of φ are of multiplicity ≥3; thus, l≥2,φ′′=Cl(ξ-ξ0)l-1. Note that φ′′ has only one zero ξ0, and so φ has only the same zero ξ0 too. Hence φ′(ξ0)=0 which contradicts with φ′(ξ0)=-a≠0.
The proof of Theorem1.9 is complete.Proof of Theorem1.11.
Similarly with the proof of Theorem1.7, we assume that ℱ is not normal at z0=0. Then, by Lemma 2.1, there are a sequence of complex numbers zj, zj→0(j→∞), a sequence of functions fj∈ℱ, and a sequence of positive numbers ρj→0+ such that gj(ξ)=ρj-1fj(zj+ρjξ) converges uniformly with respect to the spherical metric to a nonconstant meromorphic function g(ξ) whose poles are multiple and g(ξ)≠0. Moreover, g(ξ) is of order 2 at most.
Thus(3.21)gj′(ξ)=fj′(zj+ρjξ)⟹g′(ξ),(3.22)gj′(ξ)-aρjgj(ξ)-b=fj′(zj+ρjξ)-afj(zj+ρjξ)-b⟹g′(ξ)-b
also locally uniformly with respect to the spherical metric.
Ifg′-b≡0, then g=bξ+c, where c is a constant. This contradicts with g(ξ)≠0. So g′-b≢0.
Ifg′-b≠0, then by Milloux inequality (2.3) of Lemma 2.5 we have
(3.23)T(r,g)≤N¯(r,g)+N(r,1g)+N¯(r,1g′-b)+S(r,g)≤12N(r,g)+S(r,g)≤12T(r,g)+S(r,g).
From (3.23) we know that g is a constant which contradicts with our conclusion. Hence, g′-b is a nonconstant meromorphic function and has at least one zero.
As the same argument in the proof of Theorem1.7, we obtain that g′-b has only one distinct zero denoted by ξ0. Thus the Hayman inequality (2.2) of Lemma 2.5 implies that g is a rational function of degree 4 at most. Noting that g≠0 and has no simple pole, we obtain that g has at most two distinct poles. Using the Milloux inequality (2.3) of Lemma 2.5 again we get that g has at most one distinct pole. Hence we can write g(ξ)=1/(dξ+e)m,2≤m≤3, where d≠0 and e are two finite complex numbers. Simple calculation shows that g′-b has at least three distinct zeros. This is impossible.
The proof of Theorem1.11 is complete.Proof of Theorem1.14.
Similarly with the proof of Theorem1.11, we have that (3.22) also holds. Moreover, g(ξ)≠0 and g(ξ) is of order 2 at most.
Ifg′-b≡0, then g=bξ+c, where c is a constant. This contradicts with g(ξ)≠0.
Ifg′-b≠0, then by the Hayman inequality (2.2) of Lemma 2.5 we have
(3.24)T(r,g)≤3N(r,1g)+4N¯(r,1g′-b)+S(r,g)≤S(r,g).
From (3.24) we know that g is a constant which contradicts with our conclusion.
The proof of Theorem 1.8 is complete.



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