Normal Families of Holomorphic Functions Concerning Zero Numbers
1 Introduction and Main Results
Let F be a meromorphic function in C,and D be a domain in C.F is said to be normal in D if any sequence{fn}⊂F contains a subsequence fnjsuch that fnjconverges spherically locally uniformly in D,to a meromorphic function or ∞ (see[1]–[3]).
In 1959,Hayman[4]proved the following result.
Theorem 1.1 Let f be a meromorphic function in C,n≥5 be a positive integer,and a(0),b be two finite constants.If f′− afn
b,then f is a constant.
The following normality criterion corresponding to Hayman’s result was proved by Drasin[5]and Ye[6].
Theorem 1.2 Let n≥2 be a positive integer,a(0),b be two finite constants,and F be a family of Holomorphic functions in a domain D.If for each f ∈ F,f′−afn
b,then F is normal in D.
Recently,by the idea of concerning zero numbers,Deng et al.[7]proved the following result.
Theorem 1.3 Let m,n,k be three positive integers satisfying n≥m+1,a(0),b be two finite constants,and F be a family of Holomorphic functions in a domain D,all of whose zeros have multiplicity at least k.If for each function f∈F,f(k)−afn−b has at most mk distinct zeros in D,then F is normal in D.
A natural problem arises:what can we say if f(k)in Theorem 1.3 is replaced by the(f(k))d?In this paper,we prove the following result.
按实验方法对水系沉积物成分标准样品GBW07449、GBW07453,钨钼矿石成分标准样品GBW07241,钼矿石成分标准样品GBW07239进行处理,平行测定12次,对测定结果进行统计分析,结果见表5。由表5可知,测定值与认定值基本一致,相对标准偏差(RSD)均不大于6.8%。
Theorem 1.4 Let m,n,k,d be four positive integers satisfying n≥(m+1)d,a(0),b be two finite constants,and F be a family of holomorphic functions in a domain D,all of whose zeros have multiplicity at least k.If for each function f∈F,(f(k))d−afn−b has at most mdk distinct zeros in D,then F is normal in D.
Example 1.1 Let n,k,d be three positive integers,a be a nonzero finite constant,and F={fj=jzk−1:j=1,2,3,···},D={z:|z|< 1}.Then,for each f ∈ F,(f(k))d−afn−0 has just one distinct zero in D,but F is not normal in D.This shows that the zeros of function f∈F have multiplicity at least k is necessary in Theorem 1.4.
Example 1.2 Let n,k,d be three positive integers,a be a nonzero finite constant,and F={fj=jzk:j=1,2,3,···},D={z:|z|< 1}.Then,for each f ∈ F,(f(k))d −af(m+1)d−1−0 has exactly[(m+1)d−1]k ≥ mdk distinct zero in D,and(f(k))d−af(m+1)d−0 has exactly(m+1)dk≥mdk+1,but F is not normal in D.This shows that both n≥(m+1)d and(f(k))d−afn−b have at most mdk distinct zeros in Theorem 1.4 are best possible.
On the other hand,by(2.2),we have
2Some Lemmas
In order to prove our theorems,we require the following results.
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Lemma 2.1[8] Let F be a family of meromorphic functions on the unit disc Δ satisfying all zeros of functions in F have 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 a point z0if and only if there exist
(i)points zn∈Δ,zn→z0;
(ii)positive numbers ρn,ρn → 0;
(iii)functions fn∈F
such thatfn(zn+ ρnζ)→ g(ζ)spherically uniformly on each compact subset of C,where g(ζ)is a nonconstant meromorphic function satisfying the zeros of g are of multiplicities≥p and the poles of g are of multiplicities≥q.Moreover,the order of g is at most 2.If g is holomorphic,then g is of exponential type and the order of g is at most 1.
Lemma 2.2[9] Let f be a nonconstant meromorphic(entire)function in the complex plane,a(0)be a finite constant,and n be a positive integer with n≥4(n≥2).Then f′−afnhas at least two distinct zeros.
Lemma 2.3 Take f be an entire function,let a(/0)be a finite constant,and n,m,d,k be four positive integers satisfying n≥(m+1)d and mdk≥2.If all zeros of f have multiplicity at least k,then
![pagenumber_ebook=5,pagenumber_book=99](https://qkk.lw71.com/uploads/image/2018/2/8d0f91c358f076e7372e631ca2de09f9.png)
Proof.Set
![pagenumber_ebook=5,pagenumber_book=99](https://qkk.lw71.com/uploads/image/2018/2/daed71ca50142cf56defd445ca6b55ef.png)
Since all zeros of f have multiplicity at least k,it follows f(k)(z)0.Otherwise,f is a polynomial with degf ≤ k − 1,thus f is a constant,a contradiction.Hence Ψ(z)
0.
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By(2.1),we have
![pagenumber_ebook=5,pagenumber_book=99](https://qkk.lw71.com/uploads/image/2018/2/6c2498e9c59a8d6a066490669410af06.png)
Thus,we get
![pagenumber_ebook=5,pagenumber_book=99](https://qkk.lw71.com/uploads/image/2018/2/c38fb4c88b8bbf1a3725bbb30bdbd4b3.png)
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![pagenumber_ebook=5,pagenumber_book=99](https://qkk.lw71.com/uploads/image/2018/2/1270eb83b197f6777b3449eb23b3d509.png)
在GPS测绘技术中,布网质量对测绘结果的精准度具有直接影响。因此在实际测绘期间,工作人员须采用点连式及或边连式测绘的方式对三个交互点图像图形进行测绘。在将GPS测绘技术用于工程枢纽施工时,可通过增加网格的精准度及强度方式保障测绘数据的准确性。不仅如此,由于GPS测绘技术可对地区特殊地质条件及环境进行模拟,更加适用于多领域工程前期勘探中,对提升工程测绘质量及效率具有重要意义。
![pagenumber_ebook=5,pagenumber_book=99](https://qkk.lw71.com/uploads/image/2018/2/fb9fb8d7db7935739c012f3ffe3deaed.png)
Obviously,(g′)d−agnhas at least n ≥ (m+1)d=mdk+d≥ mdk+1 distinct zeros,this is a contradiction.
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![pagenumber_ebook=5,pagenumber_book=99](https://qkk.lw71.com/uploads/image/2018/2/a118b02d293c4af842486ae77fa1e8bb.png)
By(2.2),we have
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![pagenumber_ebook=5,pagenumber_book=99](https://qkk.lw71.com/uploads/image/2018/2/7f3531ec1335bc86206d754d0338f33a.png)
and Ψ (z) 1.Otherwise,we get
![pagenumber_ebook=5,pagenumber_book=99](https://qkk.lw71.com/uploads/image/2018/2/904de0985fe22e4edb1ac6f6a268ac23.png)
Because f is an entire function,we have
![pagenumber_ebook=6,pagenumber_book=100](https://qkk.lw71.com/uploads/image/2018/2/b71ef9b9d0a4d0209758a42db5e3a7c3.png)
Then
![pagenumber_ebook=6,pagenumber_book=100](https://qkk.lw71.com/uploads/image/2018/2/757a8b87768d77d6e61b9b0a2af327c4.png)
Thus
![pagenumber_ebook=6,pagenumber_book=100](https://qkk.lw71.com/uploads/image/2018/2/d2227169b5d087caa17b59454bb13255.png)
a contradiction.Hence Ψ (z) 1.
it follows that
Thus by(2.3)–(2.5)and Nevanlinna’s first and second fundamental theorems,we have
![pagenumber_ebook=6,pagenumber_book=100](https://qkk.lw71.com/uploads/image/2018/2/f5bd9f332b7b4ea53196c28452accdbc.png)
Since
![pagenumber_ebook=6,pagenumber_book=100](https://qkk.lw71.com/uploads/image/2018/2/597a18f6148e5a2d270804da7bab9868.png)
it follows from(2.6)–(2.8)that
![pagenumber_ebook=6,pagenumber_book=100](https://qkk.lw71.com/uploads/image/2018/2/bbd601f5d1b2ca1220e1c4257d7b1b21.png)
Lemma 2.3 is proved.
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3 Proof of Theorem 1.4
Considering n≥(m+1)d,by(3.1),we get
![pagenumber_ebook=6,pagenumber_book=100](https://qkk.lw71.com/uploads/image/2018/2/e80f4231d4b90474fa8d383fdb23c049.png)
spherically uniformly on compact subsets C,where g is a nonconstant holomorphic function on C and whose zeros have multiplicity at least k.Moreover,the order of g is at most 1.Obviously,(g(k)(ξ))d − agn(ξ) 0.Suppose not,since g is an entire function,we have
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![pagenumber_ebook=7,pagenumber_book=101](https://qkk.lw71.com/uploads/image/2018/2/ab516516ff10db49831833f90c23c196.png)
By(3.1)and noticing that md=2,k=1,we have
![pagenumber_ebook=7,pagenumber_book=101](https://qkk.lw71.com/uploads/image/2018/2/512c0b789d4cb451249b8461e7917466.png)
So g is a constant,a contradiction.
We claim that(g(k)(ξ))d−agn(ξ)has at most mdk distinct zeros.Suppose that(g(k)(ξ))d−agn(ξ)has mdk+1 distinct zeros ξi(i=1,2,···,mdk+1).We have
![pagenumber_ebook=7,pagenumber_book=101](https://qkk.lw71.com/uploads/image/2018/2/c6b77c842cd46cc136332a3647baf91e.png)
uniformly on compact subsets of C.By Hurwitz’s theorem,for j sufficiently large,there exist points ξj,i(i=1,2,···,mdk+1)such that ξj,i → ξiand
![pagenumber_ebook=7,pagenumber_book=101](https://qkk.lw71.com/uploads/image/2018/2/88f38c1664e3da8ba205366f95441a27.png)
However,− b have at most mdk distinct zeros in D,and zj+ ρjξj,i → z0,which is a contradiction.So we can get(g(k)(ξ))d − agn(ξ)has at most mdk distinct zeros.If mdk=1,we get m=1,d=1,k=1,and this is a contradiction with Lemma 2.2.
Next,we consider the case mdk≥2.
Suppose that(g(k))d−agnhas l(≤ mdk)distinct zeros.Since g is an entire function all zeros of g have multiplicity at least k,by Lemma 2.3,we can obtain
黄体功能不好、无排卵子宫出血等,这样的出血可能没有那么规律。可以通过监测基础体温、黄体中期的孕酮水平或者借助诊刮协助诊断。
![pagenumber_ebook=7,pagenumber_book=101](https://qkk.lw71.com/uploads/image/2018/2/b7d82e8b4140736e55cc750efc2989cb.png)
Suppose that F is not normal at z0.Then by Lemma 2.1,there exist fj∈F,zj→z0,and ρj→ 0+such that
![pagenumber_ebook=7,pagenumber_book=101](https://qkk.lw71.com/uploads/image/2018/2/915d0d33cb51f6569eddd2c7a5b12530.png)
Thus,we deduce that g is a nonconstant rational function satisfying
Next we consider two cases.
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Case 1 md≥2 and k≥1.
We consider two subcases again.
Case 1.1 md≥2 and k=1.
By(3.2),we get
![pagenumber_ebook=7,pagenumber_book=101](https://qkk.lw71.com/uploads/image/2018/2/efe23eebc85258c78a7e6653b8011566.png)
Next we consider two subcases again.
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Case 1.1.1 md=2.
In this case we have
![pagenumber_ebook=8,pagenumber_book=102](https://qkk.lw71.com/uploads/image/2018/2/b6c05ac4d0129672c06227d21f4eb8aa.png)
We deduce that degg≤2.Next we consider two subcases again.
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Case 1.1.1.1 degg=1.
We can write g(ξ)=Aξ+B,where A0.So,
![pagenumber_ebook=8,pagenumber_book=102](https://qkk.lw71.com/uploads/image/2018/2/c236504e37e6260ca2ecced7a6951e10.png)
whereis the counting function of zeros of both Ψ and f(k).It follows that
Case 1.1.1.2 degg=2.
We can write g(ξ)=B1(ξ− α1)(ξ− α2),where B10.
When α1= α2,we have
![pagenumber_ebook=8,pagenumber_book=102](https://qkk.lw71.com/uploads/image/2018/2/bddf009cc43cc4eafdbc9d47646f5d39.png)
Then
![pagenumber_ebook=8,pagenumber_book=102](https://qkk.lw71.com/uploads/image/2018/2/ea55a18d048d2ece8188786f69f15d48.png)
So,we know
![pagenumber_ebook=8,pagenumber_book=102](https://qkk.lw71.com/uploads/image/2018/2/e3c3d36b2a8c9e54b93a76a571a8a660.png)
this is a contradiction.
If α1 α2,then we are supposed to notice that n ≥ (m+1)d=2+d.If n ≥ 3+d,then by(3.1)and noticing that md=2,k=1,we get
![pagenumber_ebook=8,pagenumber_book=102](https://qkk.lw71.com/uploads/image/2018/2/4f73e65e46285b40ee8230897a1fda90.png)
which is a contradiction.
If n=2+d,then g(ξ)=B(ξ− α)(ξ− α),where B0 and α
α.Set φ =
112112 noticing that md=2,k=1,next we consider two subcases again.
Case 1.1.1.2.1 m=2,d=1,k=1.
In this case,we get
![pagenumber_ebook=8,pagenumber_book=102](https://qkk.lw71.com/uploads/image/2018/2/12d3b6333bb81a128a18c1526a1fc4e6.png)
Obviously,g′− ag3and φ′φ +a have the same zeros.
SetThen we get ψ′= φ′φ,and we obtain
![pagenumber_ebook=8,pagenumber_book=102](https://qkk.lw71.com/uploads/image/2018/2/caa8a243f71d25ef590b65aad6cccadd.png)
where is a non-zero constant.So we have
![pagenumber_ebook=8,pagenumber_book=102](https://qkk.lw71.com/uploads/image/2018/2/f0fd21524b75bfb7eb65cb77aa793405.png)
where q1(ξ)is a polynomial with degq1=1.
By Lemma 2.2,we get that g′− ag3has at least two distinct zeros,then φ′φ +a has at least two distinct zeros.Suppose that φ′φ +a has only two distinct zeros b1,b2,and set
![pagenumber_ebook=9,pagenumber_book=103](https://qkk.lw71.com/uploads/image/2018/2/14273d09c0f463840fc91fbc1a8229af.png)
where C is a non-zero constant,and l1+l2=6.So
![pagenumber_ebook=9,pagenumber_book=103](https://qkk.lw71.com/uploads/image/2018/2/df688c691df941058a34ef8e07983774.png)
where q2(ξ)is a polynomial with degq2 ≤ 2.
From(3.3),we also have
![pagenumber_ebook=9,pagenumber_book=103](https://qkk.lw71.com/uploads/image/2018/2/2d544ca9d6740cc2f1e35f1b9d84719e.png)
where q3(ξ)is a polynomial with degq3 ≤ 2.Noticing that α1,α2and b1,b2are distinct,from(3.4)and(3.5),we get l1+l2−2≤degq3≤2.Then l1+l2≤4,this contradicts with l1+l2=6.Then ψ′+a has at least three distinct zeros.So g′−ag3has at least 3=mdk+1 distinct zeros,this is a contradiction.
Case 1.1.1.2.2 m=1,d=2,k=1.
In this case,we have
![pagenumber_ebook=9,pagenumber_book=103](https://qkk.lw71.com/uploads/image/2018/2/2c28d401cb2258c24c2796a4c7e84277.png)
Obviously,(g′)2 − ag4and(φ′)2 − a have the same zeros.By a simple calculation,we get
![pagenumber_ebook=9,pagenumber_book=103](https://qkk.lw71.com/uploads/image/2018/2/7a385be8cd0fdddf77bb52c20a65f8fd.png)
where q4(ξ)=[2ξ− (α1+ α2)]2.Obviously,(φ′)2 − a has at least two distinct zeros.Suppose that(φ′)2 − a has only two distinct zeros b3,b4,and set
![pagenumber_ebook=9,pagenumber_book=103](https://qkk.lw71.com/uploads/image/2018/2/9f03e6ed4d29973961770c5f886cdceb.png)
where C is a non-zero constant,and l3+l4=8.Then
![pagenumber_ebook=9,pagenumber_book=103](https://qkk.lw71.com/uploads/image/2018/2/4eb7213f2f986b40482d9fd664a61aa1.png)
where q5(ξ)is a polynomial with degq5 ≤ 2.From(3.6),we also have
![pagenumber_ebook=9,pagenumber_book=103](https://qkk.lw71.com/uploads/image/2018/2/375f5c927a9ce4eace3e900720ebbace.png)
where q6(ξ)is a polynomial with degq6 ≤ 3.Noticing that α1,α2and b3,b4are distinct,from(3.7)and(3.8),we get
![pagenumber_ebook=9,pagenumber_book=103](https://qkk.lw71.com/uploads/image/2018/2/cb746b0c98e332302b748e039bc5314d.png)
i.e.,l3+l4 ≤ 5,this contradicts with l3+l4=8.Then(φ′)2− a has at least three distinct zeros,so(g′)2 − ag4has at least 3=mdk+1 distinct zeros,this is a contradiction.
Case 1.1.2 md≥3.
By(3.2),we obtain
![pagenumber_ebook=9,pagenumber_book=103](https://qkk.lw71.com/uploads/image/2018/2/0dfbc814f42c280b919f0d458e5bcee2.png)
So we get degg=1.As discussed in case 1.1.1.1,we get a contradiction.
Case 1.2 md≥2,k≥2.
By(3.2),we get
![pagenumber_ebook=10,pagenumber_book=104](https://qkk.lw71.com/uploads/image/2018/2/1b5a2af3e1ae14a6ddf3a4f52edf31be.png)
Then we know that g has at most one zero since all zeros of g have multiplicity at least k.Assume that g(ξ)=A(ξ−c)l,where k ≤ l≤ k+1.Then,
![pagenumber_ebook=10,pagenumber_book=104](https://qkk.lw71.com/uploads/image/2018/2/974db6617a4c0923985492c59bbe7f4b.png)
Obviously,(g(k))d−agnhas nl−(l−k)d≥mdk+dk distinct zeros,this is a contradiction.
Case 2 md=1,k≥2.
By(3.2),we get
![pagenumber_ebook=10,pagenumber_book=104](https://qkk.lw71.com/uploads/image/2018/2/d7a045fc792536f474e28ac6a78bb51e.png)
Next,we consider two subcases.
Case 2.1 k=2.
By(3.9),we get
![pagenumber_ebook=10,pagenumber_book=104](https://qkk.lw71.com/uploads/image/2018/2/c38db168212349cda3086ecd4d5444f4.png)
So degg≤4.We divide into two subcases again.
Case 2.1.1 degg=4.
We deduce that g has two distinct zeros at most since all zeros of g have multiplicity at least k=2.If g has only one zero,as discussed in Case 1.2,we get a contradiction.Then g has two distinct zeros.In the condition of n≥(m+1)d≥2,we discuss it in two conditions.If n≥3,then by(3.1)and noticing that md=1,k=2,we obtain
![pagenumber_ebook=10,pagenumber_book=104](https://qkk.lw71.com/uploads/image/2018/2/b61503d4981cc7ddf6917ef6889d05d5.png)
which contradicts with degg=4.If n=2,we can write g(ξ)=A(ξ− α1)2(ξ−α2)2,where A0 is a constant and α1
α2,and set
![pagenumber_ebook=10,pagenumber_book=104](https://qkk.lw71.com/uploads/image/2018/2/4e9eee13fd6b44b9322a4ae5b4f47d53.png)
where β2=A,then
![pagenumber_ebook=10,pagenumber_book=104](https://qkk.lw71.com/uploads/image/2018/2/5ccf055672b79d201d50eb0a0ac6ec1e.png)
Obviously,g′′− ag2and −2ψ′′ψ +6(ψ′)2 − a have the same zeros.From(3.10),we have
![pagenumber_ebook=10,pagenumber_book=104](https://qkk.lw71.com/uploads/image/2018/2/0076a9d0069ae14e07172f28d7393bcb.png)
where q7(ξ)= −B[2ξ−(α1+α2)],B=and
![pagenumber_ebook=10,pagenumber_book=104](https://qkk.lw71.com/uploads/image/2018/2/456f420ecd3489f71744149b5e5372e6.png)
where q8(ξ)is a polynomial with degq8 ≤ 2.From(3.10)–(3.12),we get
![pagenumber_ebook=11,pagenumber_book=105](https://qkk.lw71.com/uploads/image/2018/2/440f87b11b9dde296d169e0341a9e738.png)
where h(ξ)is a polynomial with degh ≤ 2.Obviously,ϕ−a has at least two distinct zeros.Suppose that ϕ−a has only two distinct zeros b1,b2,and set
![pagenumber_ebook=11,pagenumber_book=105](https://qkk.lw71.com/uploads/image/2018/2/8800cf24cba282f1630537933b9fb62a.png)
where C is a non-zero constant,and l1+l2=8.So
![pagenumber_ebook=11,pagenumber_book=105](https://qkk.lw71.com/uploads/image/2018/2/6282ac8cc5fc470590a7b262dc20f84c.png)
where q9(ξ)is a polynomial with degq9 ≤ 2.From(3.13),we also have
![pagenumber_ebook=11,pagenumber_book=105](https://qkk.lw71.com/uploads/image/2018/2/86d677fb1a9a07d1c2af5ef89b9f3e9b.png)
where q10(ξ)is a polynomial with degq10 ≤ 3.Noticing that α1,α2and b1,b2are distinct,from(3.14)and(3.15),we have
![pagenumber_ebook=11,pagenumber_book=105](https://qkk.lw71.com/uploads/image/2018/2/8a5e1c0cc6b88bfe375b225d4c0f8b77.png)
i.e.,l1+l2 ≤ 5,this contradicts with l1+l2=8.Then ϕ−a has at least three distinct zeros,so g′′− ag2has at least 3=mdk+1 distinct zeros,this is a contradiction.
Case 2.1.2 degg≤3.
We get that g has only one zero.As discussed in Case 1.2,we get a contradiction.
Case 2.2 k>2.
By(3.9),we get degg≤k+2.So we get that g has only one zero.As discussed in Case 1.2,we get a contradiction.Thus,we deduce that(g(k)(ξ))d − agn(ξ)has at least mdk+1 distinct zeros,this is a contradiction.This shows that F is normal at z0.Thus F is normal in D.
References
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[2]SchiffJ.Normal Families.Berlin:Springer-Verlin,1993.
[3]Yang L.Value Distribution Theory.Berlin:Springer-Verlag,1993.
[4]Hayman W K.Picard values of meromorphic functions and their derivatives.Ann.of Math.,1959,70(2):9–42.
[5]Drasin D.Normal families and Nevanlinna theory.Acta.Math.,1969,122:231–263.
[6]Ye Y S.A new criterion and its application(in Chinese).Chinese Ann.Math.Ser.A(Suppl.),1991,12:44–49.
[7]Deng B M,Qiu H L,Liu D,Fang M L.Hayman’s question on normal families concerning zero numbers.Complex Var Elliptic Equ.,2014,59(5):616–630.
[8]Zalcman L.Normal families:new perspectives.Bull.Amer.Math.Soc.,1998,35:215–230.
[9]Zhang Q C.Normal families of meromorphic functions concerning shared values.J.Math.Anal.Appl.,2008,338:545–551.
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