# Normal Family of Meromorphic Functions concerning Shared Values

Authors: Wei Chen; Honggen Tian; Yingying Zhang; Wenjun Yuan
Journal: Journal of Complex Analysis
 (2013)

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## Abstract

We obtain a normal criterion of meromorphic functions concerning, shared values. Letℱ be a family of meromorphic functions in a domain D and let k,n≥k+2 be positive integers. Let a≠0,b be two finite complex constants. If, for each f∈ℱ, all zeros of f have multiplicity at least k+1 and f+a(f(k))n and g+a(g(k))n share b in D for every pair of functions f,g∈ℱ, then ℱ is normal in D. This result generalizes the related theorem according to Xu et al. and Qi et al., respectively. There is a gap in the proofs of Lemma 3 by Wang (2012) and Theorem 1 by Zhang (2008), respectively. They did not consider the case of f(z) being zerofree. We will fill the gap in this paper.


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## Body

## 1. Introduction and Main Results

We useC to denote the open complex plane, C^(=C∪{∞}) to denote the extended complex plane, and D to denote a domain in C. A family ℱ of meromorphic functions defined in D⊂C is said to be normal, if for any sequence {fn}⊂ℱ contains a subsequence which converges spherically, and locally, uniformly in D to a meromorphic function or ∞. Clearly, ℱ is said to be normal in D if and only if it is normal at every point of D (see [1, 2]).Leth be a meromorphic function in a domain D⊆C. We say that h is a normal function if there exists a positive number M such that h♯(z)≥M for all z∈D, where h♯(z)=|h′(z)|/(1+h(z)|2) denotes the spherical derivative of h.LetD be a domain in C and let f and g be two nonconstant meromorphic functions in D. Let a and b be two complex numbers. If g(z)=b whenever f(z)=a, we write that
(1)f(z)=a⇒g(z)=b.
If f(z)=a⇒g(z)=b and (z)=b⇒f(z)=a, we write that
(2)f(z)=a⟺g(z)=b.Iff(z)=a⇔g(z)=a, we say that f and g share a (ignoring multiplicities) on D. When a=∞,f=a means 1/f=0 (see [2]).Influenced from Bloch's principle [3], 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 [4]), many authors proved normality criterion for families of meromorphic functions corresponding to Liouville-Picard type theorem (see [2, 5, 6]).It is also more interesting to find normality criteria from the point of view of shared values. In this area, Schwick [7] first proved an interesting result that a family of meromorphic functions in a domina is normal if in which every function shares three distinct finite complex numbers with its first derivative. And later, Sun [8] proved that a family of meromorphic functions in a domina is normal if in which each pair of functions share three fixed distinct values, which is an improvement of the famous Montel's Normal Criterion [9] by the ideas of shared values. More results about normality criteria concerning shared values can be found, for instance, see [10–13] and so on.In 1959, Hayman [14] proved that let f be a meromorphic function in C, if f′-afn≠b, where n is a positive integer and a, b are two finite complex numbers such that n≥5 and a≠0, then f is a constant. On the other hand, Mues [15] showed that for n=3,4, the conclusion is not valid.The following theorem confirmed a Hayman's well-known conjecture about normal families in [16].Theorem A.
Letℱ be a family of meromorphic functions in D, n a positive integer, and a, b two finite complex numbers such that a≠0. If n≥3 and for each function f∈ℱ, f′-afn≠b, then ℱ is normal in D.In 2008, by the ideas of shared values, Zhang [13] proved the following result.Theorem B.
Letℱ be a family of meromorphic functions in D, n a positive integer, and a, b two finite complex numbers such that a≠0. If n≥4 and for every pair of functions f and g in ℱ, if f′-afn and g′-agn share the value b, then ℱ is normal in D.In 1994, Ye [17] considered a similar problem and obtained that if f is a transcendental meromorphic function and a is a nonzero finite complex number, then f+a(f′)n assumes every finite complex value infinitely often for n≥3. Ye [17] also asked whether the conclusion remains valid for n=2.In 2008, Fang and Zalcman [18] solved this problem and obtained the following result.Theorem C.
Letf be a transcendental meromorphic function and let a be a nonzero complex number. Then, f+a(f′)n assumes every complex value infinitely often for each positive integer n≥2.Remark 1.
By a special example, Fang and Zalcman [18] showed Theorem C is not valid for n=1.On the basis of the previous results, Fang and Zalcman [18] obtained the normality criterion corresponding to Theorem C.Theorem D.
Letℱ be a family of meromorphic functions in D. Let n(n≥2) be an integer and let a(a≠0),b be two finite complex numbers. If, for each f∈ℱ, all zeros of f are multiple and f+a(f′)n≠b on D, then ℱ is normal in D.With the ideas of shared values, Qi and Zhu [19] and Wang [20] extended Theorem D and obtained the following result.Theorem E.
Letℱ be a family of meromorphic functions in D, all of whose zeros are multiple. Let n(n≥2) be an integer and a, b two nonzero finite complex numbers. If f+a(f′)n and g+a(g′)n share b on D for every pair of functions f,g∈ℱ, then ℱ is normal in D.It is natural to ask whether the conditionf′ can be replaced by f(k) in the previous theorems.In 2009, Xu et al. [21] considered the case that f′ is replaced by the kth derivative f(k) and proved the following results.Theorem F.
Letf be a transcendental meromorphic function on C, a be a nonzero finite complex number, and n and k two positive integers. If n≥k+1, then f+a(f(k))n assumes each value b∈C infinitely often.Theorem G.
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 f has only zeros of multiplicity at least k+1 and f+a(f(k))n≠b in D for every function f∈ℱ, then ℱ is normal.In this paper, we study the previous problem and get the following results.Theorem 2.
Letℱ be a family of meromorphic functions in the plane domain D and k,n≥k+2 positive integers. Let a≠0,b be two finite complex constants. If all zeros of f have multiplicity at least k+1 for each f∈ℱ and f+a(f(k))n and g+a(g(k))n share b in D for every pair of functions f,g∈ℱ, then ℱ is normal in D.Theorem 3.
Letℱ be a family of meromorphic functions in the plane domain D and k,n≥k+2 positive integers. Let a≠0,b,c be finite complex constants. If all zeros of f have multiplicity at least k and f(z)=0⇒(f(z))(k)=c for each f∈ℱf+a(f(k))n and g+a(g(k))n share b in D for every pair of functions f,g∈ℱ, then ℱ is normal in D.Example 4.
LetD={z:|z|<1} and ℱ={fn(z)}, where
(3)fn(z)=n(k+1)!zk+1,z∈D,n=1,2,…,
then
(4)fn(z)+a(fn(k)(z))k+1=(n(k+1)!+ank+1)zk+1.
Clearly, for each pair of functions, f,g, f+a(f(k))k+1 and g+a(g(k))k+1 share the value 0 in D; however, ℱ is not normal in D.Example 5.
Letk=1,a=1, D={z:|z|<1}, and ℱ={fn(z)}, where
(5)fn(z)=nz-n3,z∈D,n=1,2,…,
then
(6)fn(z)+a(fn′(z))3=nz.
Clearly, for each pair of functions, f,g, f+a(f′)n and g+a(g′)n share the value 0 in D; however, ℱ is not normal in D.Remark 6.
Example4 shows that the condition n≥k+2 in Theorem 2 is sharp. And Example 5 shows that the condition in which all zeros of f have multiplicity at least k+1 in Theorem 2 is sharp.Remark 7.
There is a gap in the proofs of Theorems 1.2 and 1.5 by Zhang [13] and Wang [20], respectively. They did not consider the case of f(z) being zero free in the proof of both Lemma 3 in [20] and Theorem 1 in [13], respectively. In Section 2, both Lemmas 9 and 11 fill the gap in their proofs of both Lemma 3 in [20] and Theorem 1 in [13], respectively.


## 2. Preliminary Lemmas

In order to prove our theorems, we need the following lemmas.First, we need the following well-known Pang-Zalcman lemma, which is the local version of [11, 22].Lemma 8.
Letℱ be a family of meromorphic functions in a domain D and k a positive integer, such that each function f∈ℱ has only zeros of multiplicity at least k, and suppose that there exists A≥1 such that |f(k)(z)|≤A whenever f(z)=0,f∈ℱ. If ℱ is not normal at z0∈D, then for each 0≤α≤k, there exist a sequence of points zn∈D,zn→z0, a sequence of positive numbers ρn→0+, and a subsequence of functions fn∈ℱ such that
(7)gn(ζ)=fn(zn+ρnς)ρnα⟶g(ζ),
locally uniformly with respect to the spherical metric in C. Here, g is a nonconstant meromorphic function such that g♯(ζ)≤g♯(0)=kA+1 and all of whose zeros have multiplicity at least k. Moreover, g has an order at most 2. If g is a holomorphic function, then g is of exponential type and has an order at most 1.Lemma 9 (see [23]).
Letf be a nonconstant zero-free rational function, n≥2, k two positive integers, and a≠0,b two complex constants. Then, the function f+a(f(k))n-b has at least nk+1 distinct zeros in C.Lemma 10.
Letk,n≥k+2 be two positive integers and let a≠0 be a constant. Let f be a nonconstant rational meromorphic function, all zeros of f have multiplicity at least k+1, then f+a(f(k))n has at least two distinct zeros.Proof.
Whenf is a nonconstant polynomial, noting that n≥k+2 and all zeros of f have multiplicity at least k+1, we know that f+a(f(k))n must have zero. Hence, f+a(f(k))n has exactly one zero z0. Set that
(8)f+a(f(k))n=A(z-z0)l,
where A is a nonzero constant and l is a positive integer such that l≥k+1.
It follows from (8) that
(9)f(k)+a((f(k))n)(k)=f(k)(1+aF(z))=Al(l-1)⋯(l-k+1)(z-z0)l-k,
where F(z)=n(n-1)⋯(n-k+1)(f(k))n-k-1(f(k+1))k+⋯+n(f(k))n-1f(2k). Since all zeros of f have multiplicity at least k+1 and f(k) has only one zero z0, we can deduce from (9) that f has only zero z0. Writing that f(z)=B(z-z0)m, where m is an integer such that m≥k+1, then
(10)f+a(f(k))n=B(z-z0)m+a[(B(z-z0)m)(k)]n=B(z-z0)m[1+aBn-1Cn(z-z0)n(m-k)-m],
where C=m(m-1)⋯(m-k+1). Equation (9) combing inequality that n(m-k)-m≥(n-1)(k+1)-nk=n-k-1≥1 gives that f+a(f(k))n has at least two distinct zeros, a contradiction.
Whenf is rational but not a polynomial, we consider two cases.
Case  1. If f+a(f(k))n has exactly one zero, then suppose that f+a(f(k))n has only zero z0 with multiplicity at least l. If f≠0, by Lemma 9, we get a contradiction. So f has zeros, then we can deduce that z0 is the only zero of f. Otherwise, f+a(f(k))n has at least two zeros. Set that
(11)f(z)=A(z-z0)m(z-z1)β1(z-z2)β2⋯(z-zt)βt,
where A is a nonzero constant and m≥k+1. For simplicity, we denote that
(12)β1+β2+⋯+βt=q.
From (11), we have
(13)f(k)=A(z-z0)m-kg(z)(z-z1)β1+k(z-z2)β2+k⋯(z-zt)βt+k,
where g(z)=(m-q)(m-q-1)⋯(m-q-k+1)zkt+⋯+C is polynomial and C is a constant.
By (11) and (13), we have
(14)f+a(f(k))n=A(z-z0)m(z-z1)β1(z-z2)β2⋯(z-zt)βt+aAn(z-z0)n(m-k)gn(z)(z-z1)n(β1+k)(z-z2)n(β2+k)⋯(z-zt)n(βt+k)=A(z-z0)m(z-z1)(n-1)β1+nk⋯(z-zt)(n-1)βt+nk(z-z1)n(β1+k)(z-z2)n(β2+k)⋯(z-zt)n(βt+k)+aAn(z-z0)n(m-k)gn(z)(z-z1)n(β1+k)(z-z2)n(β2+k)⋯(z-zt)n(βt+k).
Sincen≥k+2,m≥k+1, we deduce that n(m-k)-m≥(n-1)(k+1)-nk=n-k-1≥1, and then
(15)f+a(f(k))n=A(z-z0)m[(z-z1)(n-1)β1+nk⋯(z-zt)(n-1)βt+nk+aAn-1(z-z0)n(m-k)-mgn(z)]×((z-z1)n(β1+k)(z-z2)n(β2+k)⋯(z-zt)n(βt+k))-1.
By the assumption thatf+a(f(k))n has exactly one zero z0 with multiply l and (15), we get
(16)f+a(f(k))n=C(z-z0)l(z-z1)n(β1+k)(z-z2)n(β2+k)⋯(z-zt)n(βt+k),
where C is a nonzero constant. Both (15) and (16) imply that
(17)C(z-z0)l≡A(z-z0)m[(z-z1)(n-1)β1+nk⋯(z-zt)(n-1)βt+nk+aAn-1(z-z0)n(m-k)-mgn(z)].
Case  1.1. If l>m, from (17), we can deduce that z0 is a zero of (z-z1)(n-1)β1+nk⋯(z-zt)(n-1)βt+nk, a contradiction.
Case  1.2. If l=m, from (17), it follows that
(18)(z-z1)(n-1)β1+nk⋯(z-zt)(n-1)βt+nk+aAn-1(z-z0)n(m-k)-mgn(z)=CA.
For simplicity, we denote that(19)g1(z)=(z-z1)(n-1)β1+nk⋯(z-zt)(n-1)βt+nk,g2(z)=aAn-1(z-z0)n(m-k)-mgn(z).
Case  1.2.1. If q≥m-k, then
(20)deg(g1)=(n-1)(β1+β2+⋯+βt)+nkt=(n-1)q+nkt≥(n-1)(m-k)+nkt,(21)deg(g2)=n(m-k)-m+ndeg(g)≤n(m-k)-m+nkt.
Thus,deg(g1)-deg(g2)≥(n-1)(m-k)-n(m-k)+m=k. Therefore, g1+g2 is a non-constant polynomial, a contradiction.
Case  1.2.2. If q<m-k, then
(22)deg(g1)=(n-1)(β1+β2+⋯+βt)+nkt=(n-1)q+nkt≤(n-1)(m-k-1)+nkt,deg(g2)=n(m-k)-m+ndeg(g)=n(m-k)-m+nkt.
Hence,
(23)deg(g2)-deg(g1)≥n(m-k)-m-(n-1)(m-k-1)=n-k-1≥1.
Therefore,g1+g2 is a non-constant polynomial, a contradiction.
Case  2. If f+a(f(k))n has no zeros, then l=0 for (16). It is the same argument as the proof of Case  1 that we have a contradiction.
The proof is complete.Lemma 11 (see [24]).
Letf be a nonconstant zero-free rational function and k a positive integer. Then, f(k)-1 has at least k+1 distinct zeros.


## 3. The Proofs of Theorems

Proof of Theorem2.
Without loss of generality, we may assume thatD=Δ={z:|z|<1}. Suppose, on the contrary, that ℱ is not normal at z0.
Case  1. When b=0, by Lemma 8, there exist a sequence {zj} of complex numbers with zj→z0 and a sequence {ρj} of positive numbers with ρj→0 such that
(24)gj(ξ)=ρj(-nk)/(n-1)fj(zj+ρjξ)→g(ξ),
locally uniformly on compact subsets of C, where g(ξ) is a non-constant meromorphic function in C, all of whose zeros have multiplicity at least k+1. Moreover, g(ξ) has an order at most 2.
Ifg+a(g(k))n≡0, then g has no poles, so g is an entire function. Since all zeros of g have multiplicity at least k+1, if z0 is a zero of g with multiply q(≥k+1), then z0 is a zero of (g(k))n with multiplicity n(q-k), by g+a(g(k))n≡0. It follows that g=-a(g(k))n, so q=(q-k)n which implies that nk≥(n-1)(k+1), then n≤k+1, which contradicts with n≥k+2. So g has no zeros, then we can deduce that g=ecξ+d, where c≠0,d are constants. Then,
(25)g+a(g(k))n=ecξ+d+a(ckecξ+d)n=ecξ+d(1+acnke(n-1)(cξ+d))≡0.
Hence,
(26)e(n-1)(cξ+d)≡-1acnk,
which is impossible because n≥k+2≥2.
Theorems F, G, and Lemma10 imply that g+a(g(k))n has at least two distinct zeros; we may assume that there exist two zeros ξ0, ξ0*. We choose a positive number δ small enough such that D1∩D2=∅ and f+a(f(k))n has no other zeros in D1∪D2 except for ξ0 and ξ0*, where
(27)D1={ξ∈C∣|ξ-ξ0|<δ},D2={ξ∈C∣|ξ-ξ0*|<δ}.
Since
(28)gj+a(gj(k))n=ρj(-nk)/(n-1)×[(fj(k)(zj+ρjξj))nfj(zj+ρjξj)555+a(fj(k)(zj+ρjξj))n]→g+a(g(k))n,
by Hurwitz's theorem, we know that for sufficiently large j there exist points ξj∈D1,ξj*∈D2 such that
(29)fj(zj+ρjξj*)+a[fj(k)(zj+ρjξj*)]n=0,fj(zj+ρjξj)+a[fj(k)(zj+ρjξj)]n=0.
Note that fmfm(k) and fjfj(k) share b, it follows that
(30)fm(zj+ρjξj*)+a[fm(k)(zj+ρjξj*)]n=0,fm(zj+ρjξj)+a[fm(k)(zj+ρjξj)]n=0.
Fix m, let j→∞, and note that zj+ρjξj→z0,zj+ρjξj*→z0, then we obtain
(31)fj(z0)+a[fj(k)(z0)]n=0.
Since the zeros of fj(ξ)+a[fj(k)(ξ)]n have no accumulation point, for sufficiently large j, we have
(32)zj+ρjξj=z0,zj+ρjξj*=z0.
Hence,
(33)ξj=z0-zjρj,ξj*=z0-zjρj.
This contradicts with the factsξj∈D1,ξj*∈D2,D1∩D2=∅.
Case  2. When b≠0, by Lemma 8, there exist a sequence {zj} of complex numbers with zj→z0 and a sequence {ρj} of positive numbers with ρj→0 such that
(34)gj(ξ)=ρj-kfj(zj+ρjξ)→g(ξ),
locally uniformly on compact subsets of C, where g(ξ) is a non-constant meromorphic function in C, all of whose zeros have multiplicity at least k+1. Moreover, g(ξ) has an order at most 2.
Case  2.1. If a(g(k))n≡b, then g is a polynomial with a degree at most k, this contradicts with the fact that g has only zeros with multiplicity at least k+1.
Case  2.2. If a(g(k))n≠b, let c1,c2,…,cn be the (distinct) solutions of wn=b/a, then g(k)≠ci(i=1,2,…,n), and then g(k) is a constant by Picard theorem. It follows that g is a polynomial with a degree at most k, this contradicts with the fact that g has only zeros with multiplicity at least k+1. Therefore, a(g(k))n=b has solutions.
We claim thata(g(k))n-b has just one zero. Otherwise, we may assume that there exist two zeros ξ0, ξ0*. We choose a positive number δ small enough such that D1∩D2=∅ and such that f+a(f(k))n has no other zeros in D1∪D2 except for ξ0 and ξ0*, where
(35)D1={ξ∈C∣|ξ-ξ0|<δ},D2={ξ∈C∣|ξ-ξ0*|<δ}.
Combing
(36)ρjkgj+a(gj(k))n-b=fj(zj+ρjξj)+a(fj(k)(zj+ρjξj))n-b→a(g(k))n-b
and Hurwitz's theorem, we have that for sufficiently large j there exist points ξj∈D1,ξj*∈D2 such that
(37)fj(zj+ρjξj*)+a[fj(k)(zj+ρjξj*)]n-b=0,fj(zj+ρjξj)+a[fj(k)(zj+ρjξj)]n-b=0.
From the hypothesis thatfmfm(k) and fjfj(k) share b, it follows that
(38)fm(zj+ρjξj*)+a[fm(k)(zj+ρjξj*)]n-b=0,fm(zj+ρjξj)+a[fm(k)(zj+ρjξj)]n-b=0.
By letting m→∞ and noting that zj+ρjξj→z0,zj+ρjξj*→z0, we obtain
(39)fj(z0)+a[fj(k)(z0)]n-b=0.
Since the zeros of fj(ξ)+a[fj(k)(ξ)]n have no accumulation point, for sufficiently large j, we have
(40)zj+ρjξj=z0,zj+ρjξj*=z0.
Hence,
(41)ξj=z0-zjρj,ξj*=z0-zjρj,
which contradicts with the facts ξj∈D1, ξj*∈D2, D1∩D2=∅.
Therefore,a(g(k))n-b has just one zero. Thus, g(k)-ci has at most one zero for the only one of i∈{1,2,…,n}. We may assume that g(k)-c1 has at most one zero, then g(k)≠ci(i=2,3,…,n). From the second fundamental theorem of Nevanlinna,
(42)T(r,g(k))≤N¯(r,g(k))+N¯(r,1g(k)-c2)+N¯(r,1g(k)-c3)+⋯+N¯(r,1g(k)-cn)+S(r,g(k))≤N¯(r,g(k))+S(r,g(k))≤12N(r,g(k))+S(r,g(k))≤12T(r,g(k))+S(r,g(k)).
It follows that T(r,g(k))=S(r,g(k)), which is impossible.
The proof of Theorem2 is proved.Proof of Theorem3.
Suppose, to the contrary, thatℱ is not normal in Δ. 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. Then, by Lemma 8, we can find that fj∈ℱ,zj→0, and ρj→0, such that
(43)gj(ξ)=ρj(-nk)/(n-1)fj(zj+ρjξ)→g(ξ)
converges locally uniformly with respect to the spherical metric to a nonconstant meromorphic function g(ζ), and all zeros of g(ζ) are of multiplicity at least k, and the spherical derivative of g(ζ) is bounded.
Then, we have(44)gj(k)(ξ)=ρj(-k)/(n-1)(fj(zj+ρjξ))(k)→g(k)(ξ),
also locally uniformly with respect to the spherical metric.
We claim that all zeros ofg(ξ) are of multiplicity at least k+1. Suppose that ξ0 is a zero of g(ξ), then obviously, g(ξ)≢0. From Hurwitz's theorem, there exist ξj→ξ0 such that
(45)fj(zj+ρjξj)=0,
then we have (fj(zj+ρjξj))(k)=c by the condition that f(z)=0⇒(f(z))(k)=c.
Ifc=0, then all zeros of each f∈ℱ are of multiplicity at least k+1 near 0 by the hypothesis. Hence, Theorem 2 tells us that ℱ is normal at 0, which contradicts our supposition.
Ifc≠0, then
(46)(g(ξ0))(k)=limj→∞gj(k)(ξj)=limj→∞ρj(-k)/(n-1)(fj(zj+ρjξj))(k)=∞,
which is also a contradiction.
The proof of Theorem3 is completed.



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