# Normal Families of Zero-Free Meromorphic Functions

Authors: Yuntong Li
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

Leta(≠0),b∈ℂ, and n and k be two positive integers such that n≥2. Let
ℱ be a family of zero-free meromorphic functions defined in a domain 𝒟 such
that for each f∈ℱ, f+a(f(k))n−b has at most nk zeros, ignoring multiplicity.
Then ℱ is normal in 𝒟.


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## Body

## 1. Introduction and Main Results

Let𝒟 be a domain in ℂ, and let ℱ be a family of meromorphic functions defined in the domain 𝒟. ℱ is said to be normal in 𝒟, in the sense of Montel, if for every sequence {fn}⊆ℱ contains a subsequence {fnj} such that fnj converges spherically uniformly on compact subsets of 𝒟 (see [1, Definition  3.1.1]).ℱ is said to be normal at a point z0∈𝒟 if there exists a neighborhood of z0 in which ℱ is normal. It is well known that ℱ is normal in a domain 𝒟 if and only if it is normal at each of its points (see [1, Theorem  3.3.2]).Letf be a meromorphic function in the complex plane. We use the standard notations and results of value distribution theory as presented in [2–4]. In particular, T(r,f) is Nevanlinna’s characteristic function and S(r,f) denotes a function with the property S(r,f)=o(T(r,f)) as r→∞ (outside an exceptional set of finite linear measure).In 1959, Hayman [5] proved the following well-known result.Theorem A.
Letf be a transcendental meromorphic function on the complex plane C, let a be a nonzero finite complex number, and let n be a positive integer. If n≥5, then f′+afn assumes each value b∈C infinitely often.There are some examples constructed by Mues [6] which show that Theorem A is not true when n=3,4. Corresponding to Theorem A, Ye [7, Theorem  2.1] proved the following interesting result.Theorem B.
Letf be a transcendental meromorphic function. If a≠0 is a finite complex number and n≥3 is an positive integer, then f+af′n assumes all finite complex number infinitely often.In [7, Theorem  2.2], Ye also obtained the following result, which may be considered as a normal family analogue of Theorem B.Theorem C.
Letℱ be a family of meromorphic functions defined in a domain 𝒟, f≠b and f+af′n≠b for every f∈ℱ, where n≥2 is an integer and a≠0,b are two finite complex numbers. Then, ℱ is normal.Ye [7] asked whether Theorem B remains valid for n=2. Recently, Fang and Zalcman showed that Theorem B holds for n=2. In [8], the condition in Theorem C that f≠b can be relaxed to that all zeros of each function in ℱ are of multiplicity at least 2. Actually. they obtained the following results.Theorem D.
Letf be a transcendental meromorphic function. If a≠0 is a finite complex number and n≥2 is an positive integer, then f+af′n assumes all finite complex number infinitely often.Theorem E.
Letℱ be a family of meromorphic functions on the plane domain 𝒟, let n≥2 be a positive integer, and let a≠0,b be complex numbers. If, for each f∈ℱ, all zeros of f are multiple and f+af′n≠b on D, then ℱ is normal on D.A natural problem arises: what can we say iff′ in Theorems E is replaced by the kth derivative f(k)? In [9], Xu et al. proved the following result.Theorem F.
Leta(≠0),b∈ℂ and n and k be two positive integers such that n≥k+1. Let ℱ be a family of meromorphic functions defined on a domain 𝒟. If, for every function f∈ℱ, f has only zeros of multiplicity at least k+1, and f+a(f(k))n≠b in D, then ℱ is normal.Xu et al. [9] asked whether Theorem F remains valid for n=2. We partially answer this question. If f≠0, we generalize Theorem F by allowing f+a(f(k))n-b to have zeros but restricting their numbers.Theorem 1.1.
Leta(≠0),b∈ℂ, and n and k be two positive integers such that n≥2. Let ℱ be a family of zero-free meromorphic functions defined in a domain 𝒟 such that for each f∈ℱ, f+a(f(k))n-b has at most nk zeros, ignoring multiplicity. Then, ℱ is normal in 𝒟.Remark 1.2.
Here,f≠0 can be replaced by f≠c, where c is any finite complex numbers.Example 1.3.
Let𝒟={z:|z|<1}. Let ℱ={fm}, where fm:=emz. Then, fm+afm′=(1+am)emz≠0 in 𝒟 for every function f∈ℱ. However, it is easily obtained that ℱ is not normal at the point z=0.Example 1.4.
Let𝒟={z:|z|<1}. Let ℱ={fm}, where fm:=1/mz. Then, fm+a(fm′)2=(mz3+1)/m2z4 has 3 zeros in 𝒟 for every function f∈ℱ. However, it is easily obtained that ℱ is not normal at the point z=0.Example 1.5.
Let𝒟={z:|z|<1}. Let ℱ={fm}, where fm:=mz. It follows that fm+a(fm′)2=mz+m2 has no zero in 𝒟 for every function f∈ℱ. However, it is easily obtained that ℱ is not normal at the point z=0.Examples1.3 and 1.4 show that the conditions that n≥2 and f+a(f(k))n-b have at most nk distinct zeros in Theorem 1.1 are shape. Example 1.5 shows the condition that f≠0 cannot be omitted.


## 2.  Some Lemmas

To prove our results, we need some preliminary results.Lemma 2.1 ([9], Lemma 2.2).
Letn≥2,k be positive integers, let a be a nonzero constant and let P(z) be a polynomial. Then, the solution of the differential equation a(W(k)(z))n+W(z)=P(z) must be polynomial.Lemma 2.2.
Letf be a nonzero transcendental meromorphic function. If a be a nonzero finite complex number and let n≥2 and k be two positive integers. Then, f+a(f(k))n assumes each value b∈ℂ infinitely often.Proof.
Set(2.1)F=f+a(f(k))n-b,(2.2)ϕ=F′F=f′+an(f(k))n-1f(k+1)f+a(f(k))n-b,(2.3)ψ=nf(k+1)f(k)-F′F=nf(k+1)f-bnf(k)-f′f(k)f(k)(f+a(f(k))n-b).
We claim thatϕψ≢0. If ϕ≡0, then F≡0. We can deduce that F≡c, where c is a finite complex number. We conclude from (2.1) and Lemma 2.1 that, f must be a polynomial, which is a contradiction.
Ifψ≡0, from (2.3), we can obtain
(2.4)c(f(k))n=f+a(f(k))n-b,
where c is a finite complex number, that is,
(2.5)(a-c)(f(k))n+f=b.Ifa-c=0, we can get that f≡b, which is a contradiction.Ifa-c≠0, we conclude from (2.5) and Lemma 2.1 that f must be a polynomial, which is a contradiction.
By elementary Nevanlinna theory and (2.1), we have T(r,F)=O(T(r,f)). Thus, from (2.2) and (2.3), we have
(2.6)m(r,ϕ)=S(r,f),m(r,ψ)=S(r,f).
It follows from (2.2), (2.3) and Nevanlinna’s First Fundamental Theorem that
(2.7)N(r,1ϕ)⩽m(r,ϕ)+N(r,ϕ)-m(r,1ϕ)+O(1)⩽N(r,ϕ)+S(r,f)⩽N-(r,f)+N-(r,1F)+S(r,f),(2.8)N(r,1ψ)⩽m(r,ψ)+N(r,ψ)-m(r,1ψ)+O(1)⩽N(r,ψ)+S(r,f)⩽N-(r,1f(k))+N(r,1F)+S(r,f).
By (2.2) and (2.3), we get
(2.9)ϕ(f-b)-f′=a(f(k))nψ.
We have by (2.6)-(2.7)
(2.10)T(r,ϕ(f-b)-f′)=T(r,(f-b)(ϕ-f′f-b))⩽T(r,f-b)+T(r,ϕ-f′f-b)+S(r,f)⩽m(r,f-b)+N(r,f-b)+m(r,ϕ-f′f-b)+N(r,ϕ-f′f-b)+S(r,f)⩽m(r,f)+N(r,f)+m(r,ϕ)+m(r,f′f-b)+N(r,ϕ-f′f-b)+S(r,f)⩽T(r,f)+N-(r,f)+N-(r,1F)+S(r,f).
It follows from (2.6)–(2.10) that
(2.11)nT(r,f(k))⩽T(r,ψ)+T(r,ϕ(f-b)-f′)+S(r,f)⩽m(r,ψ)+N(r,ψ)+T(r,f)+N-(r,f)+N(r,1F)+S(r,f)⩽N-(r,1f(k))+N(r,1F)+m(r,1f)+N(r,1f)+N-(r,f)+N-(r,1F)+S(r,f)⩽N-(r,1f(k))+2N(r,1F)+m(r,f(k)f)+m(r,1f(k))+N(r,1f)+N-(r,f)+S(r,f)⩽T(r,1f(k))+2N(r,1F)+N(r,1f)+N-(r,f)+S(r,f)⩽T(r,f(k))+2N(r,1F)+N(r,1f)+N-(r,f)+S(r,f).
So, we have(2.12)(n-1)T(r,f(k))⩽2N(r,1F)+N(r,1f)+N-(r,f)+S(r,f).
We have(2.13)(n-1)T(r,f(k))≥(n-1)N(r,f(k))≥(n-1)N(r,f)+(n-1)N-(r,f).
Sincef≠0, if f+a(f(k))n assumes the value b only finitely often, we by (2.12) can get
(2.14)N(r,f)=S(r,f).
Hence,
(2.15)(n-1)T(r,f(k))⩽2N(r,1F)+S(r,f).
So f+a(f(k))n assumes each value b∈ℂ infinitely often.
We complete the proof of Lemma2.2.Using the method of Chang [10, Lemma  4], we obtain the following lemma.Lemma 2.3.
Letf be a nonconstant zero-free rational function, n≥2,  let k be two positive integers, and a≠0,b be two complex constants. Then, the function f+a(f(k))n-b has at least nk+1 distinct zeros in ℂ.Proof.
Sincef(z) is a nonconstant zero-free rational function, f(z) is not a polynomial, and hence it has at least one finite pole. Thus, we can write
(2.16)f(z)=C1∏i=1m‍(z+zi)pi,
where C1 is a nonzero constant, m and pi are positive integers, the zi (when 1≤i≤m) are distinct complex numbers, and denote p=∑i=1m‍pi.
By induction, we deduce from (2.16) that
(2.17)f(k)(z)=P(m-1)k∏i=1m‍(z+zi)pi+k,
where P(m-1)k is polynomial of degree (m-1)k.
So the degree of numerator of the functionf+a(f(k))n is equal to ∑i=1m‍(n-1)pi+nk. By calculation, f+a(f(k))n-b has at least one zero in ℂ. Thus, we can write
(2.18)f+a(f(k))n-b=C2∏i=1s‍(z+αi)li∏i=1m‍(z+zi)n(pi+k),
where C2 is a nonzero constant, li are positive integers, αi (when 1≤i≤s), and zi (when 1≤i≤m) are distinct complex numbers. Thus, by (2.16), (2.17), and (2.18), we get
(2.19)C1∏i=1m‍(z+zi)(n-1)pi+nk+a(P(m-1)k)n=b∏i=1m‍(z+zi)n(pi+k)+C2∏i=1s‍(z+αi)li.Case  1. If b=0, it follows that ∑i=1m‍[(n-1)pi+nk]=∑i=1s‍li and C1=C2. Thus, it follows from (2.19) that
(2.20)∏i=1m‍(1+zit)(n-1)pi+nk-∏i=1s‍(1+αit)li=t(n-1)p+nkQ(t),
where Q(t)=(-a/C1)t(m-1)nk(P(m-1)k(1/t))n is a polynomial. Then, Q(t) is a polynomial of degree less than (m-1)nk, and it follows that
(2.21)∏i=1m‍(1+zit)(n-1)pi+nk∏i=1s‍(1+αit)li=1+t(n-1)p+nkQ(t)∏i=1s‍(1+αit)li=1+O(t(n-1)p+nk)
as t→0.
Logarithmic differentiation of both sides of (2.21) shows that
(2.22)∑i=1m‍((n-1)pi+nk)zi1+zit-∑i=1s‍liαi1+αit=O(t(n-1)p+nk-1)
as t→0.
Comparing the coefficient of (2.22) for tj, j=0,1,…,(n-1)p+nk-2, we have
(2.23)∑i=1m‍((n-1)pi+nk)zij-∑i=1s‍liαij=0
for j=1,…,(n-1)p+nk-1.
Setzm+i=-αi when 1≤i≤s. Noting that ∑i=1m‍[(n-1)pi+nk]=∑i=1s‍li, then it follows from (2.23) that the system of linear equations,
(2.24)∑i=1m+s‍zijxi=0,
where 0≤j≤(n-1)p+nk-1, has a nonzero solution
(2.25)(x1,…,xm,xm+1,…,xm+s)=((n-1)p1+nk,…,(n-1)pm+nk,l1,…,ls).
If(n-1)p+nk≥m+s, then the determinant det(zij)(m+s)×(m+s) of the coefficients of the system of (2.24), where 0≤j≤(n-1)p+nk-1, is equal to zero, by Cramer’s rule (see, e.g., [11]). However, the zi are distinct complex numbers when 1≤i≤m+s, and the determinant is a Vandermonde determinant, so it cannot be 0 (see [11]), which is a contradiction.
Hence, we conclude that(n-1)p+nk<m+s. Noting that n≥2, it follows from this and p=∑i=1m‍pi≥m that s≥nk+1.Case  2. If b≠0, set
(2.26)b∏i=1m‍(z+zi)n(pi+k)-C1∏i=1m‍(z+zi)(n-1)pi+nk=b∏i=1m‍(z+zi)(n-1)pi+nk∏i=1q‍(z+βi)ti,
where ti are positive integers. It follows that βi (when 1≤i≤q) and zi (when 1≤i≤m) are distinct complex numbers, and ∑i=1q‍ti=p.By (2.19), we have
(2.27)b∏i=1m‍(z+zi)(n-1)pi+nk∏i=1q‍(z+βi)ti+C2∏i=1s‍(z+αi)li=a(P(m-1)k)n.It follows that(2.28)∑i=1m‍[(n-1)pi+nk]+∑i=1q‍ti=np+nmk=∑i=1s‍li,
and C2=-b. Thus, by (2.27),
(2.29)∏i=1m‍(1+zit)(n-1)pi+nk∏i=1q‍(1+βit)ti-∏i=1s‍(1+αit)li=tn(p+k)Q(t),
where Q(t)=(a/b)t(m-1)nk(P(m-1)k(1/t))n is a polynomial. Then, Q(t) is a polynomial of degree less than (m-1)nk, and it follows that
(2.30)∏i=1m‍(1+zit)(n-1)pi+nk∏i=1q‍(1+βit)ti∏i=1s‍(1+αit)li=1+tn(p+k)Q(t)∏i=1s‍(1+αit)li=O(tn(p+k))
as t→0.Thus, by taking logarithmic derivatives of both sides of (2.12), we get
(2.31)∑i=1m‍((n-1)pi+nk)zi1+zit+∑i=1q‍tiβi1+βit-∑i=1s‍liαi1+αit=O(tn(p+k)-1).We consider two cases.Subcase 2.1 ({α1,…,αs}∩{β1,…,βq}=∅). Applying the reasoning of Case 1 and noting that p≥q, we deduce that s≥nk.Subcase 2.2 ({α1,…,αs}∩{β1,…,βq}≠∅). Without loss of generality, we may assume that αq-i=βi,for(1≤i≤M). Denote
(2.32)zi={zifor1≤i≤m,βi-mform+1≤i≤m+q,αM+i-m-qform+q+1≤i≤m+q+s-M,Ni={(n-1)pi+nkfor1≤i≤m,ti-mform+1≤i≤m+s-M,ti-m-li-m-s+Mform+s-M+1≤i≤m+q,li-m-q+Mform+q+1≤i≤m+q+s-M.The formula (2.31) can be rewritten:
(2.33)∑i=1m+q+s-M‍Nizi1+zit=O(tn(p+k)-1).Applying the reasoning of Case 1, and noting thatp≥q, we deduce that s≥nk+1.This completes the proof of Lemma2.3.Lemma 2.4 ([10], Lemma 4).
Letf be a nonconstant zero-free rational function, let a≠0 be a complex constant, and let k be a positive integer. Then f(k)-a has at least k+1 distinct zeros in ℂ.Lemma 2.5 (see [12], Lemma  2, Zalcman’s lemma).
Letℱ be a family of functions meromorphic on a domain 𝒟, all of whose zeros have multiplicity at least k. Suppose that there exists A⩾1 such that |f(k)(z)|⩽A whenever f(z)=0. Then, if ℱ is not normal at z0∈𝒟, there exist, for each 0⩽α⩽k,(a)
pointszn,zn→z0;(b)
functionsfn∈ℱ;(c)
positive numbersρn→0+; such that ρn-αfn(zn+ρnξ)=gn(ξ)→g(ξ) locally uniformly with respect to the spherical metric, where g(ξ) is a nonconstant meromorphic function on ℂ, all of whose zeros of g(ξ) are of multiplicity at least k, such that g#(ξ)≤g#(0)=kA+1.Here, as usual,g#(ξ)=|g′(ξ)|/(1+|g(ξ)|2) is the spherical derivative.


## 3. Proof of Theorem

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∈𝒟. Without loss of generality, we assume that z0=0. We consider two cases.Case 1 (b=0).
By Zalcman’s lemma, there exist:(a)
pointszn,zn→z0;(b)
functionsfn∈ℱ;(c)
positive numbersρn→0+; such that
(3.1)gj(ξ)=ρj-nk/(n-1)fj(zj+ρjξ)⟶g(ξ),
spherically uniformly on compact subsets of ℂ, where g(ξ) is a nonconstant meromorphic function in ℂ. Since fj≠0, by Hurwitz’s theorem, it implies that g(ξ)≠0.
On every compact subset ofℂ which contains no poles of g, from (3.1), we get
(3.2)gj(ξ)+a(gjk(ξ))n=ρj-nk/(n-1)(fj(zj+ρjξ)+a(fjk(zj+ρjξ))n)⟶g(ξ)+a(gk(ξ))n,
also locally uniformly with respect to the spherical metric.
We claim thatg(ξ)+a(gk(ξ))n has at most nk distinct zeros.
Suppose thatg(ξ)+a(gk(ξ))n has nk+1 distinct zeros ξi, 1≤i≤nk+1, and choose δ(>0) small enough such that ⋂i=1nk+1‍D(ξi,δ)=∅, where D(ξ0,δ)={ξ∣|ξ-ξi|<δ}.
From (3.2), by Hurwitz’s theorem, there exist points ξij∈D(ξi,δ) (1≤i≤nk+1) such that for sufficiently large j,
(3.3)fj(zj+ρjξij)+a(fjk(zj+ρjξij))n=0,
for 1≤i≤nk+1.
Sincezj→0 and ρj→0+, we have zj+ρjξij∈D(0,σ)  (σ is a positive constant) for sufficiently large j, so fj(z)+a(fjk(z))n has nk+1 distinct zeros, which contradicts the fact that fj(z)+a(fjk(z))n has at most nk zero.
However, by Lemmas2.2 and 2.3, there do not exist nonconstant meromorphic functions that have the above properties. This contradiction shows that ℱ is normal in 𝒟.Case 2 (b≠0).
By Zalcman’s lemma, there exist:(a)
pointszn,zn→z0;(b)
functionsfn∈ℱ;(c)
positive numbersρn→0+; such that
(3.4)gj(ξ)=ρj-kfj(zj+ρjξ)⟶g(ξ)
spherically uniformly on compact subsets of ℂ, where g(ξ) is a nonconstant meromorphic function in ℂ. Since fj≠0, by Hurwitz’s theorem, it implies that g(ξ)≠0.
On every compact subset ofℂ which contains no poles of g, from (3.4), we get
(3.5)ρjkgj(ξ)+a(gjk(ξ))n-b⟶a(gk(ξ))n-b
also locally uniformly with respect to the spherical metric.
Noting that(3.6)ρjkgj(ξ)+a(gjk(ξ))n-b=fj(zj+ρjξ)+a(fjk(zj+ρjξ))n-b,
we claim that a(gk(ξ))n-b has at most nk distinct zeros.
Suppose thatg(ξ)+a(gk(ξ))n-b has nk+1 distinct zeros ξi, 1≤i≤nk+1, and choose δ(>0) small enough such that ⋂i=1nk+1‍D(ξi,δ)=∅, where D(ξ0,δ)={ξ∣|ξ-ξi|<δ}.
From (3.2), by Hurwitz’s theorem, there exist points ξij∈D(ξi,δ) (1≤i≤nk+1) such that for sufficiently large j(3.7)fj(zj+ρjξij)+a(fjk(zj+ρjξij))n-b=0,
for 1≤i≤nk+1.
Sincezj→0 and ρj→0+, we have zj+ρjξij∈D(0,σ)  (σ is a positive constant) for sufficiently large j, so fj(z)+a(fjk(z))n-b has nk+1 distinct zeros, which contradicts the fact that fj(z)+a(fjk(z))n-b has at most nk zero.
Denotec1,c2,…,cn by the different roots of ωn=b/a, then
(3.8)a(gk(ξ))n-b=a∏i=1n‍(gk(ξ)-ci).Subcase 2.1 (If g(ξ) is a rational function). By Lemma 2.4 and (3.8), we can deduce that a(gk(ξ))n-b has at least nk+n distinct zeros. This contradicts the claim that a(gk(ξ))n-b has at most nk distinct zeros.Subcase 2.2 (If g(ξ) is a transcendental meromorphic function). By Nevanlinnas second main theorem, we have
(3.9)T(r,g(k))≤N-(r,g(k))+∑i=1n‍N-(r,1g(k)-ci)+S(r,g(k))=N-(r,g(k))+N-(r,1a(g(k))n-b)+S(r,g(k))≤1k+1N(r,g(k))+S(r,g(k))≤1k+1T(r,g(k))+S(r,g(k)).It follows thatT(r,g(k))≤S(r,g(k)), which is a contradiction. This contradiction shows that ℱ is normal in 𝒟.Hence, Theorem1.1 is proved.



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