# The Normality Criteria of Meromorphic Functions Concerning Shared Fixed-Points

Authors: Wei Chen; Qi Yang; Wen-jun Yuan; Hong-gen Tian
Journal: Discrete Dynamics in Nature and Society
 (2014)

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## Abstract

We study the normality criteria of meromorphic functions concerning shared fixed-points; we obtain the following: Letℱ be a family of meromorphic functions defined in a domain D and k≥2 a positive integer. For every f∈ℱ, all zeros of f are of  multiplicity at least k+2 and all poles of f are multiple. If ff(k) and gg(k) share z in D for each pair of functions f and g, then ℱ is normal.


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## Body

## 1. Introduction and Main Results

Letℱ be a family of meromorphic functions defined in the domain D⊆ℂ. If any sequence {fn}⊂ℱ contains a subsequence fnj that converges spherically locally uniformly in D, to a meromorphic function or ∞, we say that ℱ is normal in D (see [1, 2]).Letf(z) be a meromorphic function in a domain D, and z∈ℂ. We say z is a fixed-point of f(z) when f(z)=z. Let f(z) and g(z) be two meromorphic functions in D; if f(z)-z and g(z)-z have the same zeros (ignoring multiplicity), then we say f(z) and g(z) share fixed-points.In2009, Lu and Gu [3] proved the following results.Theorem 1.
Letk be a positive integer and f a transcendental. If all zeros of f have multiplicity at least k+2, then ff(k) assumes every finite nonzero value infinitely often.Theorem 2.
Letf be a family of meromorphic functions defined in a domain D. Suppose that k is a positive integer and a≠0 is a finite complex number. If, for each f∈ℱ, all zeros of f are of multiplicity at least k+2, and ff(k)≠a, then ℱ is normal in D.In 2011, Hu and Meng [4] extended Theorem 2 as follows.Theorem 3.
Letℱ be a family of meromorphic functions defined in a domain D. Let k be a positive integer and a≠0 be a finite complex number. If, for every f∈ℱ, ff(k) and gg(k) share a in D, and, for every pair of functions f, g∈ℱ, all zeros of f are of multiplicity at least k+1, then ℱ is normal in D.A natural question is the following: what can be said if the finite complex numbera in Theorem 3 is replaced by the fixed-point z? In this paper, we answer this question by proving the following theorems.Theorem 4 (main theorem).
Letℱ be a family of meromorphic functions defined in a domain D. Let k be a positive integer. If, for every f∈ℱ such that f≠0, ff(k) and gg(k) share z in D for every pair of function f, g∈ℱ, then ℱ is normal in D.Theorem 5 (main theorem).
Letℱ be a family of meromorphic functions defined in a domain D. Let k≥2 be a positive integer. For every f∈ℱ, all zeros of f have multiplicity at least k+2 and all poles of f are multiple. If ff(k) and gg(k) share z in D for every pair of function f, g∈ℱ, then ℱ is normal in D.Theorem 6 (main theorem).
Letℱ be a family of meromorphic functions defined in a domain D. Let k be a positive integer. For every f∈ℱ, f has only zeros with multiplicity at least k+1 and f has poles at most [(k-1)/2]. If ff(k) and gg(k) share z in D for every pair of function f, g∈ℱ, then ℱ is normal in D.


## 2. Some Lemmas

In order to prove our theorems, we require the following results.Lemma 7 (see [5, 6]).
Letℱ be a family of meromorphic functions in a domain D, and let k be 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
(1)gn(ζ)=fn(zn+ρnς)ρnα⟶g(ζ)
locally uniformly with respect to the spherical metric in ℂ, where g is a nonconstant meromorphic function, whose zeros all have multiplicity at least k, such that g#(ζ)≤g#(0)=kA+1. Moreover, g has order at most 2.
Here as usual,g#(ζ)=|g′(ζ)|/(1+|g(ζ)|2) is the spherical derivative.Lemma 8.
Letk be a integer and f a nonconstant rational meromorphic function such that f≠0; then ff(k)-z has at least two distinct zeros.Proof.
Sincef≠0, set
(2)f(z)=A(z-β1)n1(z-β2)n2⋯(z-βt)nt,
where A is a nonzero constant and nj(j=1,2,…,t) are positive integers.
For simplicity, we denoteN:=n1+n2+⋯+nt. Obviously, N≥t.
From (2), we have
(3)f(k)=P(z)(z-β1)n1+k(z-β2)n2+k⋯(z-βt)nt+k,ff(k)=P(z)(z-β1)2n1+k(z-β2)2n2+k⋯(z-βt)2nt+k,
where p(z)≢0 is a polynomial and deg⁡⁡(P)=k(t-1).
Differentiating (3), we get
(4)(ff(k))′′=P1(z)(z-β1)2n1+k+2(z-β2)2n2+k+2⋯(z-βt)2nt+k+2,
where P1(z)≢0 is a polynomial and deg⁡⁡(P1)=(k+2)(t-1).
Next, we assume, to the contrary, thatff(k)-z has at most one zero. We distinguish two cases.
Case
1. If ff(k)-z has exactly one zero z0. From (3), we obtain
(5)ff(k)-z=B(z-z0)2N+kt+1(z-β1)2n1+k(z-β2)2n2+k⋯(z-βt)2nt+k,
where B is a nonconstant. Differentiating (5), we have
(6)(ff(k))′′=P2(z)(z-z0)2N+kt-1(z-β1)2n1+k+2(z-β2)2n2+k+2⋯(z-βt)2nt+k+2,
where P2(z)≢0 is a polynomial and deg⁡(P2)≤2t.
From (4) and (6), we obtain
(7)(k+2)(t-1)≥2N+kt-1≥2t+kt-1;
then -k≥1; this is a contradiction.
Case
2. If ff(k)-z has no zero. Then we have 2N+kt+1=0 from (5), which is a contradiction.
The proof is complete.Lemma 9.
Letk≥2 be an integer and f a nonconstant rational meromorphic function. If f has only zeros with multiplicity at least k+2 and all poles of f are multiple, then ff(k)-z has at least two distinct zeros.Proof.
Assume, to the contrary, thatff(k)-z has at most one zero.
Case
1. If f is a polynomial, since all zeros of f are of multiplicity at least k+2, then we can know that ff(k) has at least one zero with multiplicity k+4. So ff(k)-z has at least one zero and (ff(k))′ has zeros with multiplicity at least k+3. According to the assumption, we obtain that ff(k)-z has only a zero z0; then there exists a nonzero constant A and an integer l≥2 such that
(8)ff(k)=z+A(z-z0)l.
So we have
(9)(ff(k))′=1+Al(z-z0)l-1,
which, however, has only simple zeros. This is a contraction.
Case
2. If f is a rational but not a polynomial. We see
(10)f(z)=A(z-α1)m1(z-α2)m2⋯(z-αt)ms(z-β1)n1(z-β2)n2⋯(z-βt)nt,
where A is a nonzero constant and mj≥(k+2)  (j=1,2,…,s).
For simplicity, we denote(11)M=m1+m2+⋯+ms≥(k+2)s,N=n1+n2+⋯+nt≥2t.
From (10), we have
(12)f(k)=(z-α1)m1-k(z-α2)m2-k⋯(z-αt)ms-kg(z)(z-β1)n1+k(z-β2)n2+k⋯(z-βt)nt+k,(13)ff(k)=(z-α1)2m1-k(z-α2)2m2-k⋯(z-αt)2ms-kg(z)(z-β1)2n1+k(z-β2)2n2+k⋯(z-βt)2nt+k=P(z)Q(z),
where g(z)≢0 is a polynomial and deg⁡⁡(g)≤k(s+t-1).
Differentiating (13), we have
(14)(ff(k))′=(z-α1)2m1-k-1(z-α2)2m2-k-1⋯(z-αt)2ms-k-1g1(z)(z-β1)2n1+k+1(z-β2)2n2+k+1⋯(z-βt)2nt+k+1,(15)(ff(k))′′=(z-α1)2m1-k-2(z-α2)2m2-k-2⋯(z-αt)2ms-k-2g2(z)(z-β1)2n1+k+2(z-β2)2n2+k+2⋯(z-βt)2nt+k+2,
where g1(z)≢0,  g2(z)≢0 are polynomials and deg⁡⁡(g1)≤(k+1)(s+t-1), deg⁡⁡(g2)≤(k+2)(s+t-1).
Now we distinguish two subcases.
Case
2.1. Supposing that ff(k)-z has exactly one zero z0, from (13), we obtain
(16)ff(k)=z+B(z-z0)l(z-β1)2n1+k(z-β2)2n2+k⋯(z-βt)2nt+k=P(z)Q(z).
Differentiating (16), we have
(17)(ff(k))′=1+(z-z0)l-1g1*(z)(z-β1)2n1+k+1(z-β2)2n2+k+1⋯(z-βt)2nt+k+1,(18)(ff(k))′′=(z-z0)l-2g2*(z)(z-β1)2n1+k+2(z-β2)2n2+k+2⋯(z-βt)2nt+k+2,
where g1*(z)≢0,  g2*(z)≢0 are polynomials, deg⁡⁡(g1*)≤t, and deg⁡⁡(g2*)≤2t. From (14) and (17), we have z0≠αj(j=1,2,…,s).
Further, we distinguish two subcases.
Case
2.1.1.
l
≠
2
N
+
k
t
-
1. From (16), it is easily obtained that deg⁡⁡(P)≥deg⁡⁡(Q). Thus, (13) implies
(19)deg⁡⁡(Q)=2N+kt≤deg⁡⁡(P)=2M-ks+deg⁡⁡(g)≤2M-ks+k(s+t-1)=2M+kt-k.
So M≥N+k/2; that is, M>N. From (13) and (16), noting that z0≠αj(j=1,2,…,s), we have
(20)2M-(k+2)s≤deg⁡⁡(g2*)≤2t.
It follows that 2M≤(k+2)s+2t; combining this inequality with (11), we obtain
(21)2M≤(k+2)s+2t≤(k+2)Mk+2+N<2M,
which is impossible.
Case
2.1.2. l=2N+kt-1. Next we distinguish two subcases: M>N and M≤N.
WhenM>N, similar to the Case 2.1.1, it follows that 2M<2N; from (13) and (16), we get a contradiction.
IfM≤N, by combining (15) and (18), we may give the following inequality:
(22)l-2≤deg⁡⁡(g2*)≤(k+2)(s+t-1),
and hence
(23)2N=l-kt+1≤(k+2)(s+t-1)-kt+1≤(k+2)s+2t-(k1)<(k+2)Mk+2+N≤2N.
This is a contradiction.
Case
2.2. Suppose that ff(k)-z has no zero; then l=0 for (16). Similarly with the proof of Case 2.1, we also obtain a contradiction.Lemma 10.
Letk be an integer and f a nonconstant rational meromorphic function. If f has only zeros with multiplicity at least k+1 and f has poles at most [(k-1)/2], then ff(k)-z has at least two distinct zeros.Proof.
Assume, to the contrary, thatff(k)-z has at most one zero.
Case
1. If f is a polynomial, since all zeros of f are of multiplicity at least k+1, then we can know that ff(k) has at least one zero with multiplicity k+2. So we see that ff(k)-z has at least one zero and (ff(k))′ has zeros with multiplicity at least k+1. According to the assumption, we obtain that ff(k)-z has only a zero z0; then there exists a nonzero constant A and an integer l≥2 such that
(24)ff(k)=z+A(z-z0)l.
So we have
(25)(ff(k))′=1+Al(z-z0)l-1,
which, however, has only simple zeros. This is a contraction.
Case
2. If f is a rational but not a polynomial, we set
(26)f(z)=A(z-α1)m1(z-α2)m2⋯(z-αt)ms(z-β1)n1(z-β2)n2⋯(z-βt)nt,
where A is a nonzero constant and mj≥(k+2)(j=1,2,…,s).
For simplicity, we denote(27)M=m1+m2+⋯+ms≥(k+1)s,N=n1+n2+⋯+nt≥t(t≤[k-12]).
From (26), we have
(28)f(k)=(z-α1)m1-k(z-α2)m2-k⋯(z-αt)ms-kg(z)(z-β1)n1+k(z-β2)n2+k⋯(z-βt)nt+k,(29)ff(k)=(z-α1)2m1-k(z-α2)2m2-k⋯(z-αt)2ms-kg(z)(z-β1)2n1+k(z-β2)2n2+k⋯(z-βt)2nt+k,
where g(z)≢0 is a polynomial and deg⁡⁡(g)≤k(s+t-1).
Differentiating (29), we have
(30)(ff(k))′=(z-α1)2m1-k-1(z-α2)2m2-k-1⋯(z-αt)2ms-k-1g1(z)(z-β1)2n1+k+1(z-β2)2n2+k+1⋯(z-βt)2nt+k+1=P1(z)Q1(z),(31)(ff(k))′′=(z-α1)2m1-k-2(z-α2)2m2-k-2⋯(z-αt)2ms-k-2g2(z)(z-β1)2n1+k+2(z-β2)2n2+k+2⋯(z-βt)2nt+k+2,
where g1(z)≢0, g2(z)≢0 are polynomials and deg⁡⁡(g1)≤(k+1)(s+t-1) and deg⁡⁡(g2)≤(k+2)(s+t-1).
Now we distinguish two subcases.
Case
2.1. Supposing that ff(k)-z has exactly one zero z0, from (13), we obtain
(32)ff(k)=z+B(z-z0)l(z-β1)2n1+k(z-β2)2n2+k⋯(z-βt)2nt+k.
Differentiating (32), we have
(33)(ff(k))′=1+(z-z0)l-1g1*(z)(z-β1)2n1+k+1(z-β2)2n2+k+1⋯(z-βt)2nt+k+1=P1(z)Q1(z),(34)(ff(k))′′=(z-z0)l-2g2*(z)(z-β1)2n1+k+2(z-β2)2n2+k+2⋯(z-βt)2nt+k+2,
where g1*(z)≢0, g2*(z)≢0 are polynomials, deg⁡(g1*)≤t, and deg⁡(g2*)≤2t. From (30) and (33), we have z0≠αj(j=1,2,…,s). From (31) and (34), we see
(35)2M-(k+2)s≤deg⁡⁡(g2*)≤2t.
So M≤((k+2)s+2t)/2. Since M≥(k+1)s, we have (k+1)s≤M≤((k+2)s+2t)/2, so s≤2t/k≤[k-1]/k<1.
Therefore,f has no zeros. According to Lemma 8, we have a contradiction.
Case
2.2. Suppose that ff(k)-z has no zero; then l=0 for (32). Similarly with the proof of Case 2.1, we also obtain a contradiction.Lemma 11 (see [7]).
Letf be a transcendental meromorphic function, and let φ(z) be a small function such that T(r,φ)=S(r,f); then
(36)4T(r,f)<4N¯(r,1f)+4N¯(r,1φff(k)-1)+3N(k(r,1f)+3kN¯(k+1(r,1f)+S(r,f).Lemma 12.
Letf be a transcendental meromorphic function; let k be a positive integer; let P(z)(≢0) be a polynomial. If all zeros of f have multiplicity at least k+1, then f(z)f(k)(z)-P(z) has infinitely many zeros.Proof.
We denote(37)ψ(z)=f(z)f(k)(z)-P(z);
then
(38)ψ(z)P(z)=f(z)f(k)(z)P(z)-1=1P(z)f(z)f(k)(z)-1.
Since p(z) is a polynomial, we see T(r,1/P(z))=s(r,f). Now we distinguish two cases as follows.(i)
Ifk≤4, we have
(39)N¯(r,1f)=N¯(k+1(r,1f)+S(r,f).By Lemma 11, we have
(40)4T(r,f)<4N¯(r,1f)+4N¯(r,1(1/P)ff(k)-1)+3kN¯(k+1(r,1f)+S(r,f)=(3k+4)N¯(k+1(r,1f)+4N¯(r,1(1/P)ff(k)-1)+S(r,f)≤3k+4k+1N(r,1f)+4N¯(r,1(1/P)ff(k)-1)+S(r,f)≤72T(r,1f)+4N¯(r,1(1/P)ff(k)-1)+S(r,f).
Thus,
(41)T(r,f)<8N¯(r,1(1/P)ff(k)-1)+s(r,f).(ii)
Ifk≥5, then
(42)N¯(r,1f)≤N¯(6(r,1f)+s(r,f)≤16N(r,1f)+S(r,f).By Lemma 11, we have
(43)4T(r,f)<23N(r,1f)+4N¯(r,1(1/P)ff(k)-1)+3N¯(k+1(r,1f)+S(r,f)=113N(k+1(r,1f)+4N¯(r,1(1/P)ff(k)-1)+S(r,f)≤113T(r,1f)+4N¯(r,1(1/P)ff(k)-1)+S(r,f).
So(44)T(r,f)<12N¯(r,1(1/P)ff(k)-1)+S(r,f).
From (41) and (44), we can deduce that (1/P)ff(k)-1 has infinitely many zeros; thus, ff(k)-P has infinitely many zeros.Lemma 13 (see [4]).
Take a positive integerk and a nonzero complex number a. If f is a nonconstant meromorphic function such that f has only zeros of multiplicity at least k+1, then ff(k)-a has at least two distinct zeros.


## 3. Proof of Theorems

Proof of Theorem5.
Consider the following.
Case
1. For z0=0, let ℱ1={Fj:Fj(z)=fj(z)/z1/2∣fj∈ℱ}. If ℱ1 is not normal at 0, by Lemma 7, there exist a sequence {zj} of complex numbers with zj→0 and a sequence {ρj} of positive numbers with ρj→0 such that
(45)gj(ξ)=ρj-k/2Fj(zj+ρjξ)⟶g(ξ)
locally uniformly on compact subsets of ℂ, where g(ξ) is a nonconstant meromorphic function in ℂ.
Here we distinguish two cases.
Case
1.1. Supposing that zj/ρj→c, c is a finite complex number. Then
(46)ϕj(ξ)=fj(ρjξ)ρj(1+k)/2=Fj(zj+ρj(ξ-(zj/ρj)))ρjk/2(ρjξ)1/2ρj1/2⟶ξ1/2g(ξ-c)=H(ξ)
locally uniformly on compact subsets of ℂ disjoint from the poles of g, where H(ξ) is a nonconstant meromorphic function in ℂ, all of whose zeros have multiplicity at least k+2 and all poles of which have multiplicity at least 2. Hence
(47)ϕj(ξ)ϕj(k)(ξ)-ρjξρj=fj(ρjξ)fj(k)(ρjξ)-(ρjξ)ρj⟶H(ξ)H(k)(ξ)-ξ
spherically locally uniformly in C disjoint from the poles of g.
IfH(ξ)H(k)(ξ)≡ξ, since H(ξ) has zeros with multiplicity at least k+2, obviously there is a contradiction. Hence, H(ξ)H(k)(ξ)≢ξ. Since the multiplicity of all zeros of H(ξ) is at least k+2, so by Lemmas 9 and 12, H(ξ)H(k)(ξ)-ξ has at least two distinct zeros.
Suppose thatξ0, ξ0* are two distinct zeros of H(ξ)H(k)(ξ)-ξ. We choose a positive number δ small enough such that D1∩D2=∅ and such that H(ξ)H(k)(ξ)-ξ has no other zeros in D1∪D2 except for ξ0 and ξ0*, where
(48)D1={ξ∈C∣|ξ-ξ0|<δ},D2={ξ∈C∣|ξ-ξ0*|<δ}.
By Hurwitz’s Theorem, for sufficiently largej, there exist points ξj∈D1, ξj*∈D2 such that
(49)fj(ρjξj*)fj(k)(ρjξj*)-(ρjξj*)=0,fj(ρjξj)fj(k)(ρjξj)-(ρjξj)=0.
By the assumption in Theorem5 that fmfm(k) and fjfj(k) share z, it follows that
(50)fm(ρjξj*)fm(k)(ρjξj*)-(ρjξj*)=0,fm(ρjξj)fm(k)(ρjξj)-(ρjξj)=0.
Fix m, take j→∞, and note ρjξj→0, ρjξj*→0; we obtain
(51)fm(z0)fm(k)(0)=0.
Since the zeros of fj(ξ)fj(k)(ξ)-ξ has no accumulation point, for sufficiently large j, we have
(52)ρjξj=0,ρjξj*=0.
Therefore, when j is large enough, ξ0=ξ0*. This contradicts with the facts ξj∈D1, ξj*∈D2, D1∩D2=∅. Thus ℱ1 is normal at 0.
Case
1.2. We may suppose that zj/ρj→∞. We have
(53)fj(k)(z)=z1/2Fj(k)(z)+∑l=1k‍ckl(z1/2)(l)Fj(k-l)(z)=z1/2Fj(k)(z)+∑l=1k‍clz(1/2)-lFj(k-l)(z),
where cl=(1/2)((1/2)-1)⋯((1/2)-l)Ckl. Thus we have
(54)fj(z)fj(k)(z)=fj(z)z1/2Fj(k)(z)+fj(z)∑l=1k‍clz(1/2)-lFj(k-l)(z)=zFj(z)Fj(k)(z)+z1/2Fj(z)∑l=1k‍clz(1/2)-lFj(k-l)(z),fj(z)fj(k)(z)z=(Fj(k)(z)Fj(z)+∑l=1k‍clFj(k-l)(z)Fj(z)zl).
Since gj(k-l)(ξ)=ρj(k/2)-lFj(k-l)(zj+ρjξ), l=0,1,…,k, we have
(55)fj(zj+ρjξ)fj(k)(zj+ρjξ)zj+ρjξ=(gj(k)(ξ)gj(ξ)+∑l=1k‍clgj(k-l)(ζ)gj(ξ)((zj/ρj)+ξ)l).
On the other hand, for l=1,2,…,k, we have
(56)lim⁡j→∞cl((zj/ρj)+ξ)l=0.
Thus, we have(57)fj(zj+ρjξ)fj(k)(zj+ρjξ)zj+ρjξ⟶g(ζ)g(k)(ξ)
spherically locally uniformly in ℂ disjoint from the poles of g.
Ifg(ξ)g(k)(ξ)≡1, then g has no zeros. Of course, g also has no poles. Since g is a nonconstant meromorphic function of order at most 2, then there exists constant ci(i=1,2), (c1,c2)≠(0,0), and g(ξ)=ec0+c1ξ+c2ξ2; obviously, this is contrary to the case g(ξ)g(k)(ξ)≡1. Hence, g(ξ)g(k)(ξ)≢1.
Since the multiplicity of all zeros ofg is at least k+1 and all poles of which have multiplicity at least 2, thus, by Lemma 13, g(ξ)g(k)(ξ)-1 has at least two distinct zeros.
Suppose thatξ1, ξ1* are two distinct zeros of g(ξ)g(k)(ξ)-1. We choose a positive number δ small enough such that D1∩D2=∅ and such that g(ξ)g(k)(ξ)-1 has no other zeros in D1∪D2 except for ξ1 and ξ1*, where
(58)D1={ξ∈C∣|ξ-ξ1|<δ},D2={ξ∈C∣|ξ-ξ1*|<δ}.
By Hurwitz’s Theorem, for sufficiently largej there exist points ξj^∈D1, ξj~∈D2 such that
(59)fj(zj+ρjξj^)fj(k)(zj+ρjξj^)-(zj+ρjξj^)=0,fj(zj+ρjξj~)fj(k)(zj+ρjξj~)-(zj+ρjξj~)=0.
Similar to the proof of Case 1.1, we get a contradiction. Then,ℱ1 is normal at 0.
From Cases 1.1 and 1.2, we know thatℱ1 is normal at 0; there exists Δ={z:|z|<ρ} and a subsequence of Fj, and we may still denote it as Fj, such that Fj converges spherically locally uniformly to a meromorphic function F(z) or ∞ in Δ.
Here we distinguish two cases.
Case
i. When j is large enough, fj(0)≠0. Then F(0)=∞. Thus, for each Fj(z)∈ℱ1, there exists δ>0 such that if F(z)∈ℱ1, then |F(z)|>1 for all z∈Δδ={z:|z|<δ}. Thus, for sufficiently large j, |Fj(z)|≥1, 1/fj is holomorphic in Δδ. Therefore, for all fj∈ℱ, when |z|=δ/2, we have
(60)|1fj|=|1Fj(z)z1/2|≤(2δ)1/2.
By maximum Principle and Montel’s Theorem, ℱ is normal at z=0.
Case
ii. There exists a subsequence of fj, and we may still denote it as fj such that fj(0)=0. Since f∈ℱ, the multiplicity of all zeros of f is at least k+2; then Fj(0)=0. Thus, there exists 0<r<ρ such that Fj(z) is holomorphic in Δr={z:|z|<r} and has a unique zero z=0 in Δr. Then Fj converges spherically locally uniformly to a holomorphic function F(z) in Δr; fj converges spherically locally uniformly to a holomorphic function F(z)z1/2 in Δr. Hence ℱ is normal at z=0.
By Casesi and ii, ℱ is normal at z=0.
Case
2. For z0≠0, suppose, to the contrary, that F is not normal in D. Then there exists at least one point z0 such that F is not normal at the point z0. Then by Lemma 7, there exist a sequence {znj} of complex numbers with znj→z0 and a sequence {ρn} of positive numbers with ρn→0 such that
(61)gn(ξ)=ρnk/2fn(zn+ρnξ)⟶g(ξ)
locally uniformly on compact subsets of C, where g(ξ) is a nonconstant meromorphic function in C, whose zeros all have multiplicity at least k+2. Moreover, g(ξ) has order at most 2.
From (61) we get
(62)gn(ξ)gn(k)(ξ)-(zn+ρnξ)=fn(zn+ρnξ)fn(k)(zn+ρnξ)-(zn+ρnξ)⟶g(ξ)g(k)(ξ)-z0
also locally uniformly with respect to the spherical metric besides the poles of g(ξ).
Ifgg(k)≡z0, then g has no zeros. Of course, g also has no poles. Since g is a nonconstant meromorphic function of order at most 2, then there exists constant ci(i=1,2)≠0, and g(ξ)=ec0+c1ξ+c2ξ2. Obviously, this is contrary to the case gg(k)≡z0. Hence gg(k)≢z0.
By Lemmas12 and 13, we deduce that gg(k)-z0 has at least two distinct zeros. Next we show that it is impossible. Let ξ0 and ξ0* be two distinct zeros of gg(k)-z0. We choose a positive number δ small enough such that D1∩D2=∅ and such that gg(k)-z0 has no other zeros in D1∪D2 expect for ξ0 and ξ0*, where
(63)D1={ξ∈C∣|ξ-ξ0|<δ},D2={ξ∈C∣|ξ-ξ0*|<δ}.
Similar to the proof of Case 1, we get a contradiction. This finally completes the proof of Theorem5.Similar to the proof of Theorem5, combining with Lemmas 8 and 10 and Lemma 12, we can prove Theorems 4 and 6 easily; here we omit the proof.



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*Source: 654294-2014-04-06.xml*