R E S E A R C H
Open Access
A quasinormal criterion of meromorphic
functions and its application
Pai Yang
**Correspondence:
[email protected] Department of Mathematics, Nanjing University, Nanjing, 210093, P.R. China
College of Applied Mathematics, Chengdu University of Information Technology, Chengdu, 610225, P.R. China
Abstract
In this paper, we obtain a quasinormal criterion of meromorphic functions and give an example of an application in the value distribution theory. More than anything, we provide a general method to solve some problems in the value distribution theory. MSC: 30D35; 30D45
Keywords: elliptic function; quasinormal criterion; characteristic function
1 Introduction
We use the following notation. LetNdenote the set of positive integers. LetCbe complex plane andDbe a domain inC. Forz∈Candr> ,(z,r) ={z| |z–z|<r},(z,r) = {z| <|z–z|<r}and=(, ). Letn(D,f) denote the number of poles off(z) inD
(counting multiplicities) andn(D,f) denote the number of poles off(z) inD(not counting multiplicities). We writefn
χ
⇒finDto indicate that the sequence{fn}converges tof in the spherical metric uniformly on compact subsets ofDandfn⇒f inDif the convergence is in the Euclidean metric. Forf meromorphic inD, set
f#(z) := |f
(z)|
+|f(z)| and S(D,f) :=
π
D
f#(z)dxdy.
The Ahlfors-Shimizu characteristic is defined byT(r,f) =
r
S(t,f)
t dt. LetT(r,f) denote the usual Nevanlinna characteristic function. SinceT(r,f) –T(r,f) is bounded as a
func-tion ofr, we can replaceT(r,f) withT(r,f) in this paper.
Recall that an elliptic function [] is a meromorphic functionhdefined inCfor which there exist two nonzero complex numbersωandωwithω/ωnot real such thath(z+
ω) =h(z+ω) =h(z) for allzinC.
Recall that a familyF of functions meromorphic inDis said to be quasinormal inD [] if from each sequence{fn} ⊂F one can extract a subsequence{fnk}which converges
locally uniformly with respect to the spherical metric inD\E, where the setE(which may depend on{fnk}) has no accumulation points inD. IfE can always be chosen to satisfy |E| ≤ν,Fis said to quasinormal of orderνinD. Thus a family is quasinormal of order inDif and only if it is normal inD. The familyFis said to be (quasi)normal atz∈Dif it
is (quasi)normal in some neighborhood ofz. ThusFis quasinormal inDif and only if it
is quasinormal at each pointz∈D. On the other hand,Ffails to be quasinormal of order
νinDprecisely when there exist pointsz,z, . . . ,zν+inDand a sequence{fn} ⊂F such that no subsequence of{fn}is normal atzj(j= , , . . . ,ν+ ).
In , Nevoet al.proved the following quasinormal criterion.
Theorem A[, Theorem ] LetFbe a family of meromorphic functions in D,all of whose zeros have multiplicity at least k+ .If there exists a holomorphic function H univalent in D such that f(k)(z)=H(z)for all f ∈Fand all z∈D,thenFis quasinormal of orderin D.
Fork≥, we extend Theorem A in this paper. In fact, we obtain the following result.
Theorem . Let k≥be an integer,and letF be a family of meromorphic functions in D,all of whose zeros have multiplicity at least k+ .Let H be a nonconstant meromorphic function.Suppose that there existsν∈Nsuch that for each a∗∈C,n(D,H(z)–a∗)≤ν.If f(k)(z)=H(z)for all f ∈Fand all z∈D,thenFis quasinormal of orderνin D.
Remark . Fork≥, Theorem A is the special case of Theorem . withν= andH(z)= ,∞for allz∈D. Unfortunately, the restricted condition thatH(z)= ,∞largely restricts the applications of Theorem A, so it is important to remove the condition’s restriction.
For convenience, we give a generalized form of Theorem ..
Proposition . Let k≥be an integer,and let{fn}be a family of meromorphic functions in D,all of whose zeros have multiplicity at least k+ .Let H be a nonconstant meromorphic function,and there existsν∈Nsuch that for each a∗∈C,n(D,H(z)–a∗)≤ν.Let{hn}be a family of meromorphic functions in D such that hnand Hhave the same zeros and poles with the same multiplicity,and hn(z)
χ
⇒H(z)in D.If fn(k)(z)=hn(z)for all n∈Nand all z∈D,then{fn}is quasinormal of orderνin D.
Moreover,for each subsequence{fnk}of{fn},there exist a subsequence of{fnk}(still denoted
by{fnk})and a corresponding point set E which has no accumulation points in D such that: () fnk(z)
χ
⇒f(z)inD\E,wheref(z)is meromorphic or identically infinite there; () for eacha˜∈E,H(a)˜ = ∞and no subsequence of{fnk}is normal ata˜; () for eacha˜∈E,there existra˜ > andNa˜> such that for sufficiently largek,
n((a,˜ ra˜),f
nk) <Na˜,wherera˜andNa˜ only depend ona˜;and () for eacha˜∈E,f(z) =a˜zζ
˜
a · · ·
ζk– ˜
a H(ζk) dζkdζk–· · ·dζinD\E.
The value distribution theory of meromorphic functions occupies one of the central places in complex analysis which now has been applied to complex dynamics, complex differential and functional equations, Diophantine equations, and others.
In his excellent paper [], Hayman studied the value distribution of certain meromor-phic functions and their derivatives under various conditions. Among other important results, he proved that iff(z) is a transcendental meromorphic function in the plane, then eitherf(z) assumes every finite value infinitely often, or every derivative off(z) assumes every finite nonzero value infinitely often. This result is known as Hayman’s alternative. Thereafter, the value distribution of derivatives of transcendental functions continued to be studied.
Theorem B[, Theorem ] Let f be a meromorphic function of finite order inC.If f has infinitely many multiple zeros,then fassumes every finite nonzero value infinitely often.
In , Panget al.obtained the following result.
Theorem C[, Theorem ] Let f be a transcendental meromorphic function inC,all but finitely many of whose zeros are multiple,and let R(≡)be a rational function.Then f–R has infinitely many zeros.
Ris a small function compared withf in Theorem C. Specifically,T(r,R) =o{T(r,f)}as r→ ∞in Theorem C. A natural problem arises: What can we say if the rational function Rin Theorem C is replaced by a more general small functionα(z)? In this direction, we obtain the following result.
Theorem . Let k≥be an integer,let f(z)be a meromorphic function inC,and letα(z) = R(z)h(z) (≡),where h(z)is an elliptic function and R(z)is a rational function.Suppose that all but finitely many zeros of f have multiplicity at least k+ and T(r,α) =o{T(r,f)} as r→ ∞.Then the equation f(k)(z) =α(z)has infinitely many solutions(including the
possibility of infinitely many common poles of f(z)andα(z)).
2 Preliminary lemmas
Lemma .[, Corollary ] If h(z)is a nonconstant elliptic function with primitive periods ω,ω,whereω/ω is not real,then T(r,h) =Ar( +o())as r→ ∞,where A> is a
constant.
Lemma . LetF be a family of functions meromorphic in D, all of whose zeros have multiplicity at least k,and suppose that there exists A≥such that|f(k)(z)| ≤A whenever f(z) = .Then ifFis not normal at z,there exist,for each≤α≤k,
(a) pointszn,zn→z; (b) functionsfn∈F;and
(c) positive numbersρn→ such thatρ–α
n fn(zn+ρnζ) =gn(ζ)
χ
⇒g(ζ)inC,where g is a nonconstant meromorphic func-tion inCsuch that g#(ζ)≤
g#() =kA+ .In particular,g has order at most.
This is the local version of [, Lemma ] (cf.[, Lemma ]; [, pp.-]). The proof consists of a simple change of variable in the result cited from [];cf.[, pp.-].
Lemma .[, Lemma .] Let{fn}be a family of functions meromorphic in(z,r).
Suppose that fn
χ
⇒f in(z,r),where f is a nonconstant meromorphic function or f≡ ∞
in(z,r).If there exists M> such that for each n,n((z,r),fn) <M,then there exists
M> such that S((z,r/),fn) <M.
Lemma . Let{aj}be a sequence in D which has no accumulation points in D,and let
{ψn}be a family of holomorphic functions in D such thatψn⇒ψin D,whereψ= ,∞in D. Let{fn}be a family of meromorphic functions in D,all of whose zeros have multiplicity at least k+ ,such that fn(k)(z)=ψn(z)for all n∈Nand all z∈D.Suppose that:
(a) no subsequence of{fn}is normal ata; (b) fn(z)
χ
⇒f(z)inD\ {aj}∞j=.
Then
(c) there existsr> such thatfnhas a single(multiple)zero in(a,r)for sufficiently
largen;
(d) there existsr> such that for each <r<r,fnhas a single simple pole in(a,r)
for sufficiently largen;and
(e) f(z) =az
ζ
a· · ·
ζk–
a ψ(ζk) dζkdζk–· · ·dζforz∈D\ {aj} ∞
j=.Equivalently,
f extends to an analytic function inD\ {aj}∞j=such thatf(k)(z) =ψ(z)andf(j)(a) =
forj= , , , . . . ,k– .
Remark . Since Lemma . is not stated explicitly in [], let us indicate how it follows from the results of that paper. Suppose first that (c) has been shown to hold. By Lemma in [], (d) and (e) hold. Next, suppose that (c) do not hold. Similar to the treatment in Case (pp.-) of the proof of Theorem in [], we can finally derive a contradiction.
Lemma . Let k be a positive integer,and let{fn}be a family of meromorphic functions in D,all of whose zeros have multiplicity at least k+ .Let{hn}be a family of meromorphic functions in D such that hn(z)
χ
⇒H(z)in D,where H is a holomorphic univalent function in D.If fn(k)(z)=hn(z)for all n∈Nand all z∈D,then{fn}is quasinormal of orderin D.
Lemma . can be proved by an exactly analogous argument as in the proof of Theorem in []. In fact, there is no essential difference between Lemma . and Theorem in [], so we do not give the proof of Lemma ..
Lemma .[, Lemma ] Let R be a nonconstant rational function satisfying R(z)= inC.Then either R(z) =az+b or R(z) =(z+ac)n+b,where n∈Nand a(= ),b,c∈C.
Lemma .[, Theorem ] Let k be a positive integer,let f be a transcendental meromor-phic function inC,and let R(≡)be a rational function.If all but finitely many zeros of f have multiplicity at least k+ ,then f(k)–R has infinitely many zeros.
Remark . The proof of Lemma . is based, quite naturally, on a combination of ideas from [] and []. In fact, fully understanding the ideas and methods in [] and [], one can give the proof of Lemma . without difficulty.
Lemma .[, Lemma ] Let k and l be positive integers,and let R(z)be a rational func-tion,all of whose zeros have multiplicity at least k.If R(k)(z)= z–linC,then R(z)is a con-stant.
() fnandϕnare defined inRn,where the positive sequenceRnincreases to∞; () ϕn⇒inC;
() fn(k)(z)=ϕn(z)/zlforz∈Rn; () fn
χ
⇒f inC;and
() f() = .
Then f has a zero in.
Lemma . Let f be a nonconstant meromorphic function of finite order inC,all of whose zeros have multiplicity at least k+ .If f(k)(z)= inC,then
f(z) = k!
(z–a)k+
z–b
for some a and b(=a)inC.
This follows from results in [], specifically Lemma (whose proof depends in an es-sential fashion on Corollary of []) and Lemma . As an immediate consequence of Lemma ., we have the following result, which appears as Lemma of [].
Lemma .[, Lemma ] Let k be a positive integer,and let f be a meromorphic function of finite order inC,all of whose poles are multiple and whose zeros all have multiplicity at least k+ .If f(k)(z)=c for some constant c= and all z∈C,then f(z)is a constant.
3 Auxiliary lemmas
Lemma . Let{fn}and{ψn}be families of meromorphic functions in D,and let f(z)and ψ(z)be meromorphic functions in D.Suppose that:
(a) fn(z)
χ
⇒f(z)andψn(z)
χ
⇒ψ(z)inD,and
(b) fn(k)(z)=ψn(z)inD.
Then,either f(k)(z)≡ψ(z)or f(k)(z)=ψ(z)in D.
Proof Suppose thatf(k)(z)≡ψ(z) inD. SetA:=f–(∞)∪ψ–(∞)∪(f(k)–ψ)–(). By (a)
and (b), we have
∞ = fn(k)–ψn
⇒
f(k)–ψ inD\A.
Since fn(k)–ψn
is holomorphic inDandAhas no accumulation points inD, we have
fn(k)–ψn
⇒
f(k)–ψ inD. (.)
Thus
f(k)–ψ is a holomorphic function inDand thenf(k)–ψ= inD.
In order to show thatf(k)=ψinD, we need only show thatf andψ have no common
poles inD. Otherwise, we assume thatz∈Dis a pole of ordermoff and a pole of order
mofψ. Setm:=max{m+k,m}. Obviously,zis a zero of f(k)–ψ of order at mostm.
By (a) and Hurwitz’ theorem, there existsδ∗such that(z, δ∗)⊂Dand for eachδ∈
(,δ∗),fnandbnhave at leastm andm (counting multiplicities) poles respectively in
hencefn(k)–ψnhas at leastm+k+mpoles (counting multiplicities) in(z,δ). Sinceδ
can be made arbitrarily small,zis a zero of f(k)–ψ of order at leastm+k+mby (.).
Thusm+k+m>m. This is a contradiction.
Lemma . Let k be a positive integer,and let{fn}be a family of meromorphic functions in ,all of whose zeros have multiplicity at least k+ .Let{bn}be a sequence of meromorphic functions insuch that bn(z)
χ
⇒b(z)in,where b(≡)is a meromorphic function and b() = .Suppose that:
(a) bandbnhave the same zeros and poles with the same multiplicity;
(b) for alln∈Nand allz∈D,fn(k)(z)=bn(z);
(c) there exists pointszninsuch thatfn(zn) = andzn→;and
(d) fn(z)
χ
⇒f(z)in,wheref(z)is a meromorphic function in.
Then f(k)(z)≡b(z)in.
Proof SetFn(z) :=f
(k) n (z)
bn(z). By (a) and (b),f
(k)
n ()=bn() = , and henceFn() =∞. Since all zeros of{fn(z)}have multiplicity at leastk+ , we havefn()= . Hencezn= andFn(zn) = for sufficiently largen. SinceFn() =∞andFn(zn) = for sufficiently largen,{Fn(ζ)}is not equicontinuous at , and hence{fn(k)(z)
bn(z) – }is not normal at .
By Lemma ., we have eitherf(k)(z)≡b(z) orf(k)(z)=b(z) in. Suppose thatf(k)(z)=
b(z) in. By the assumptions, there existsδ> such thatf(z) has no poles on(,δ) and b(z) has no zeros on(,δ). Thus, we have
∞ = fn(k)(z)
bn(z) –
⇒
f(k)(z)
b(z) –
, z∈(,δ). (.)
By the maximum principle, (.) holds in(,δ), and then{f
(k) n (z)
bn(z) – }is normal at . This
is a contradiction. Thusf(k)(z)≡b(z) in.
Lemma . Let k be a positive integer,let{fn}be a family of meromorphic functions in D,and let{hn}be a family of meromorphic functions in D such that hn
χ
⇒h in D,where h≡,∞.If all n∈Nand all z∈D,fn(z)= and fn(k)(z)=hn(z),then{fn}is normal in D.
Proof By Lemma ., it suffices to prove that{fn}is normal at points whichhhas poles or zeros. Without loss of generality, we assume thatD=,h(z) =zlb(z), whereb(z)= ,∞ inandl(= ) is an integer. Then{fn}is normal in.
Suppose that{fn}is not normal at . Taking a subsequence and renumbering, we may assume that no subsequence of{fn}is normal at . Sincefn(z)= in, there existsr> such thatr⊂andfn⇒ inr. By the argument principle, we have, for sufficiently largen,
n
r, fn(k)–hn
–nr,fn(k)–hn = πi
|z|=r
fn(k+)–h n fn(k)–hn
dz= πi
|z|=r h
hdz=l.
Sincefn(k)(z)=hn(z), we have –n(r,fn(k)–hn) =l. Thusl< andfnhas poles (otherwise fn
χ
⇒ ∞in) which are different from the poles ofhn. Hencen(r,fn(k)–hn) > –l. This is a
Lemma . Let k be a positive integer,and let{fn}be a family of meromorphic functions in D,all of whose zeros have multiplicity at least k+ .Let{ψn}be a family of meromorphic functions in D such thatψn
χ
⇒ψ,whereψ(z)≡,∞in D.If fn(k)=ψnfor all n∈Nand all z∈D,then{fn}is quasinormal in D.
Proof It suffices to show that{fn}is quasinormal in a neighborhood of each point ofD. Let p∈D. There existst> such that(p,t)⊂Dandψis holomorphic and does not vanish in(p,t). By Lemma .,{fn}is quasinormal in(p,t).
Suppose now that {fn} is not quasinormal at p. Then there exist pointszj∈(p,δ) (j= , , . . .) and a subsequence of{fn}(still denoted by{fn}) such thatzj→pand no subse-quence of{fn}is normal at anyzj,j= , , . . . . SetE:={zj:j= , , . . .}. Taking a subsequence of{fn}(still denoted by{fn}), we may assume thatfn
χ
⇒Hin(p,δ)\E. By Lemma ., we haveH(k)≡ψ andH(z
j) = . It follows thatH is holomorphic in(p,t) andH(k)≡ψ there. Moreover, since ψ has no essential singularity atp, the same is true ofH. But thatH(zj) = forj= , , . . . implies thatH≡, and henceH(k)≡, which contradicts
H(k)≡ψ≡.
Lemma . Let k and l be positive integers,and let R be a rational function.If R(k)(z)=zl for all z∈C,then
R(z) =
n++l i= (z–αi)
(l+k)(l+k– )· · ·(l+ )(z–β)n+–k, (.)
where n(≥k– )is an integer andαi,β∈C(≤i≤l+n+ ).
Proof Obviously, (R(k–)(z) –zl+
l+)= . ThenR(k–)(z) –
zl+
l+ is a nonconstant rational
func-tion. By Lemma .,
R(k–)(z) = z l+
l+ +az+b or R
(k–)(z) = zl+
l+ + a (z+c)n+b,
wheren(≥k) is an integer anda(= ),b,c∈C. In fact, ifR(k–)(z) =zl+
l+ +
a
(z+c)n +b, then
z= –cis a pole ofR(z) of order at leastk, and hencen≥k. Now, we have
R(z) = z l+k
(l+k)· · ·(l+ )+Pk(z) or R(z) =
zl+k (l+k)· · ·(l+ )+
c
(z+c)n+–k +Pk–(z),
wherePk(z) is a polynomial of degreek,Pk–(z) is a polynomial of degree at mostk– ,
andcis a constant. ThusR(z) has the following form:
R(z) =
n++l i= (z–αi)
(l+k)(l+k– )· · ·(l+ )(z–β)n+–k,
wheren(≥k– ) is an integer andαi,β∈C(≤i≤l+n+ ).
d= .Iflimr→∞Tr(r,f)=∞,then there exist points an→ ∞and positive numbersδn→ such that
f(an) ad
n
→, f
(k)(a
n) ad
n
→ ∞ and S(an,δn),g → ∞.
Proof By standard results in Nevanlinna theory,T(r,f) =T(r,zdg)≤T(r,g) +T(r,zd) and T(r,zd) =O(logr) asr→ ∞. Thus,lim
r→∞T(rr,g) =∞, and thenlimr→∞Tr(r,g)=∞.
We claim that there existtn→ ∞andεn→ such that
S(tn,εn),g = π
|z–tn|<εn
g#(z)dxdy→ ∞. (.)
Otherwise there would existε> andM> such thatS((z,ε),g) <Mfor allz∈C.
From this follows
S(r,g) = π
|z|<r
g#(z)dxdy=Or .
Obviously,T(r,g) =
r
S(t)
t dt=O(r
). This contradicts the fact thatlim
r→∞Tr(r,g)=∞.
By (.), there exist pointsbnsuch that|bn–tn|<εnandg#(bn)→ ∞. Setgn(z) :=g(z+ bn). Clearly,gn#() =g#(bn)→ ∞, and hence{gn}is not normal at . Obviously, all zeros ofgn(z) have multiplicity at leastk+ infor sufficiently largen. Using Lemma . for α=k– /, there exist pointszn→, positive numbersρn→, and a subsequence of{gn} (still denoted by{gn}) such thatGn(ζ) =gn(ρznk–/+ρnζ)
n
χ
⇒G(ζ) inC, whereGis a nonconstant meromorphic function inC, all of whose zeros have multiplicity at leastk+ .
SinceG(k)(ζ) is not a constant (otherwise, eitherG(ζ) is a constant, or the zero ofG(ζ) have multiplicity at mostk), we may assumeζis not a zero or pole ofG(k)(ζ). Setan:= zn+ρnζ+bn. Now we have
g(i)(an) =gn(i)(zn+ρnζ) =ρ
k––i n G(ni)(ζ),
wherei= , , . . . ,k. Obviously,an→ ∞,g(k)(an)→ ∞, andg(i)(an)→ fori= , , . . . , k– .
Now, we havef(an)
adn =g(an)→ and
f(k)(a
n) ad
n =(z
dg(z))(k)
ad n
z=an
= m=k
m=
Tmamn–kg(m)(an)→ ∞,
where Tm (m= , , . . . ,k) are constants andTk = . Setδn:=εn+|an–tn|. Obviously, δn→ and(tn,εn)⊂(an,δn), and henceS((an,δn),g)→ ∞.
Lemma . Let k be a positive integer,and let{fn}be a family of meromorphic functions in D,all of whose poles are multiple and whose zeros all have multiplicity at least k+ .Let
{hn}be a family of meromorphic functions in D such that hn
χ
Proof Suppose that{fn}is not normal at a pointz∗∈D. Then by Lemma ., there exist pointszn→z∗, positive numbersρn→ and a subsequence of{fn}(still denoted by{fn}) such thatgn(ζ) =fn(znρ+ρknζ)
n
χ
⇒g(ζ) inC, wheregis a nonconstant meromorphic function inC, all of whose poles are multiple and whose zeros all have multiplicity at leastk+ . In particular,ghas order at most . Obviously,gn(k)(ζ) =fn(k)(zn+ρnζ)=hn(zn+ρnζ) and hn(zn+ρnζ)⇒h(z∗) (= ) inC. By Lemma ., we have eitherg(k)(z)≡h(z∗) orf(k)(z)= h(z∗) inC. Firstly, suppose thatg(k)(z)≡h(z∗) inC. It follows thatg(z) =a
kzk+ak–zk–+ · · ·+az+a. We arrive at a contradiction asg(z) is nonconstant and all zeros ofg(z) have
multiplicity at leastk+ . Secondly, suppose thatg(k)(z)=h(z∗) inC. By Lemma ., then
g(z) is a constant. A contradiction.
Lemma . Let k be a positive integer,and let{fn}be a family of meromorphic functions in D,all of whose poles are multiple and whose zeros all have multiplicity at least k+ .Let
{hn}be a family of meromorphic functions in D such that hn
χ
⇒h in D,where h(= )is a meromorphic function in D.Suppose that h and hnhave the same poles,all with the same multiplicity.If fn(k)(z)=hn(z)for all n∈Nand all z∈D,then{fn}is normal in D.
Lemma . can be proved by an exactly analogous argument as in the proof of Theorem in []. To facilitate the reading, Lemma . was proved in this paper.
Lemma . Let k≥be an integer,and let{fn}be a family of meromorphic functions in D,all of whose zeros have multiplicity at least k+ .Let{hn}be a family of meromorphic functions in D such that hn(z)
χ
⇒h(z)in D,where h(≡,∞)is a meromorphic function. Let E⊂D be a set which has no accumulation points in D.Suppose that:
(∗a) handhnhave the same zeros and poles with the same multiplicity;
(∗b) for alln∈Nand allz∈D,fn(k)(z)=hn(z);
(∗c) for eacha˜∈E,no subsequence of{fn}is normal ata˜;and
(∗d) fnk
χ
⇒f(z)inD\E,wheref(z)is meromorphic or identically infinite there.
Then
(∗e) for eacha˜∈E,h(a)˜ =∞;
(∗f ) for each a˜ ∈ E, there exist ra˜ > and Na˜ > such that for sufficiently large n, n((a,˜ ra˜),fn) <Na˜,wherera˜ andNa˜ only depend ona˜;and
(∗g) for eacha˜∈E,f(z) =a˜zζ ˜
a · · ·
ζk– ˜
a h(ζk) dζkdζk–· · ·dζinD\E.
4 Proof of Lemma 3.8
Proof Since normality is a local property, by Lemma ., we only need to prove that{fn}is normal at every pole ofh(z). Making standard normalizations, we may assumeD=and
h(z) = zl+
a–l+
zl– +· · ·=
φ(z)
zl (z∈),
wherelis a positive integer,φ() = , andφ(z)= ,∞for allz∈.
Sethn(z) :=φnzl(z). Sincehandhnhave the same poles, all with the same multiplicity and
hn
χ
⇒hinD, we haveφn(z)⇒φ(z) in.
Clearly, it is enough to show that{fn}is normal atz= . Suppose, on the contrary, that
renumbering, we may assume that no subsequence of{fn}is normal at . Our goal is to obtain a contradiction in the sequel.
We distinguish two cases. Case. ≤l≤k.
By Lemma ., there exist pointszn→, positive numbersρn→ and a subsequence of{fn}(still denoted by{fn}) such that
Fn(ζ) =
fn(zn+ρnζ) ρk–l
n
χ
⇒ F(ζ) inC,
whereF(ζ) is a nonconstant meromorphic function inC. By Hurwitz’s theorem, all poles ofF(ζ) are multiple and all zeros ofF(ζ) have multiplicity at leastk+ .
Taking a subsequence and renumbering, we may assumezn/ρn→α asn→ ∞, where α∈Corα=∞. Again we distinguish two subcases.
Subcase..zn/ρn→ ∞.
Setgn(ζ) :=znl–kfn(zn+znζ) =zln–kfn(zn( +ζ)). Obviously, all poles ofgn(ζ) are multiple and all zeros ofgn(ζ) have multiplicity at leastk+ ,
gn(k)(ζ) =zlnfn(k)zn( +ζ) =
φn(zn( +ζ)) ( +ζ)l and
φn(zn( +ζ)) ( +ζ)l
χ
⇒
( +ζ)l inC.
Then, by Lemma ., the family{gn}is normal in. Taking a subsequence and renumber-ing, we may assume thatgn(ζ)
χ
⇒g(ζ) in. Obviously, all zeros ofg(ζ) have multiplicity at leastk+ in.
We claim thatg(ζ) is a meromorphic function in. Otherwise, suppose thatg(ζ)≡ ∞ in. Then
Fn(ζ) =
fn(zn+zn(ρznnζ))
ρk–l n
=
zn ρn
k–lf
n(zn+zn(ρznnζ)) zk–l
n
=
zn ρn
k–l gn ρn zn ζ χ
⇒ ∞ inC.
Thus,F(ζ)≡ ∞inC. A contradiction.
We claim thatg()=∞. SinceF(ζ) is a nonconstant meromorphic function inC, there existζ∈Csuch thatF(ζ)=∞. Noting that ρznnζ→ asn→ ∞, we have
g() = lim n→∞gn
ρn zn ζ = lim n→∞
fn(zn+ρnζ)
ρk–l n
ρn zn
k–l =
⎧ ⎨ ⎩
F(ζ) ifl=k,
ifl<k.
For anyζ ∈C\F–(∞), we have
F(k–l)(ζ) = lim
n→∞F (k–l)
n (ζ) =nlim→∞fn(k–l)(zn+ρnζ) = lim n→∞f
(k–l)
n
zn+zn
ρn zn ζ = lim n→∞g
(k–l)
n ρn zn ζ
=g(k–l)().
This implies thatF(k–l)(ζ)≡g(k–l)() inC\F–(∞) and thusF(k)(ζ) is a constant inC. It
follows thatF(ζ) =akζk+· · ·+aζ+a. We arrive at a contradiction asF(ζ) is nonconstant
Subcase..zn/ρn→α(α∈C). Clearly, we have
Fn(k)(ζ) =ρnlfn(k)(zn+ρnζ)=ρnlhn(zn+ρnζ) and
ρnlhn(zn+ρnζ) =
ρnlφn(zn+ρnζ) (zn+ρnζ)l
χ
⇒
(α+ζ)l inC.
By Lemma ., eitherF(k)(ζ)= /(α+ζ)l orF(k)(ζ)≡/(α+ζ)l inC. The latter possibil-ity contradicts the fact that all poles ofF(k)(ζ) have multiplicity at leastk+ (>l). Thus
F(k)(ζ)= /(α+ζ)linC. It follows from Lemma . and Lemma . thatF(ζ) is a constant. A contradiction.
Case.l≥k+ . Set
G:=gn(z)|gn(z) =fn(z)/hn(z),z∈
.
Clearly, all poles ofgn(z) are multiple and all zeros ofgn(z) have multiplicity at leastk+ in. Sincefn(k)()=hn() =∞, we havefn()=∞. Thus, for eachn,gn() =fn()/hn() = . Obviously,gnhas a zero of order at leastlatz= for eachn.
We first prove thatGis normal in. Suppose thatGis not normal atz∈. Then by
Lemma ., there exist pointszn→z, positive numbersρn→, and a subsequence of
{gn}(still denoted by{gn}) such that
Gn(ζ) =
gn(zn+ρnζ) ρk
n
χ
⇒ G(ζ) inC,
whereG(ζ) is a nonconstant meromorphic function inC. In particular,G(ζ) has order at most . By Hurwitz’s theorem, all poles ofG(ζ) are multiple and all zeros ofG(ζ) have multiplicity at leastk+ .
Taking a subsequence and renumbering, we may assumezn/ρn→α asn→ ∞, where α∈Corα=∞. Again we distinguish two subcases.
Subcase..zn/ρn→ ∞. By simple calculation, we have
gn(k)(z) =f
(k)
n (z) hn(z)
–
n
gn(k–)(z)h
n(z) hn(z)
–
n
gn(k–)(z)h
n(z) hn(z)
–· · ·–gn(z) h(nk)(z)
hn(z) .
Then we have
Gn(k)(ζ) =gn(k)(zn+ρnζ)
=f
(k)
n (zn+ρnζ) hn(zn+ρnζ)
–
n
g(nk–)(zn+ρnζ)
hn(zn+ρnζ) hn(zn+ρnζ)
–· · ·
–gn(zn+ρnζ)
h(nk)(zn+ρnζ) hn(zn+ρnζ)
=f
(k)
n (zn+ρnζ) hn(zn+ρnζ)
–
n
g(nk–)(zn+ρnζ)
–l zn+ρnζ
+φ
n(zn+ρnζ) φn(zn+ρnζ)
–gn(zn+ρnζ)
(–)kl(l– )· · ·(l–k+ ) (zn+ρnζ)k
+
k
(–)k–l(l– )· · ·(l–k+ )
(zn+ρnζ)k–
φn(zn+ρnζ) φn(zn+ρnζ)
+· · ·+φ
(k)
n (zn+ρnζ) φn(zn+ρnζ)
=f
(k)
n (zn+ρnζ) hn(zn+ρnζ)
–
n
gn(k–)(zn+ρnζ) ρn
–l zn/ρn+ζ
+ρnφ
n(zn+ρnζ) φn(zn+ρnζ)
–· · ·
–gn(zn+ρnζ) ρk
n
(–)kl(l– )· · ·(l–k+ ) (zn/ρn+ζ)k
+
k
(–)k–l(l– )· · ·(l–k+ ) (zn/ρn+ζ)k–
ρnφn(zn+ρnζ) φn(zn+ρnζ)
+· · ·+ρ k nφ
(k)
n (zn+ρnζ) φn(zn+ρnζ)
.
On the other hand, we havelimn→∞zn/ρn+ζ = and then fori= , , . . . ,k,
ρnφn(i)(zn+ρnζ) φn(zn+ρnζ) ⇒
inC.
Sincegn(zn+ρnζ)/ρkn
χ
⇒G(ζ),gn(k–i)(ζˆ)/ρni is locally bounded inC\G–(∞). Thus,
fn(k)(zn+ρnζ) hn(zn+ρnζ)
χ
⇒ G(k)(ζ) inC\G–(∞).
Sincefn(k)(z)/hn(z)= , Hurwitz’s theorem yields that eitherG(k)(ζ)= orG(k)(ζ)≡ inC. If G(k)(ζ)= inC, then by Lemma .,G(ζ) is a constant. A contradiction. ThusG(k)(ζ)≡,
and thenG(ζ) =k!ζk+c
k–ζk–+· · ·+c. This contradicts the fact that all zeros ofG(ζ)
have multiplicity at leastk+ . Subcase..zn/ρn→α(α∈C).
gn(ρnζ) ρk
n
=gn(zn+ρn(ζ–zn/ρn)) ρk
n
=Gn(ζ–zn/ρn)
χ
⇒ G(ζ–α) =G(ζ) inC. (.)
Clearly, all poles ofG(ζ) are multiple and all zeros ofG(ζ) have multiplicity at leastk+ , andG(ζ) has a zero of order at leastlatζ= .
SetHn(ζ) :=ρnl–kfn(ρnζ). Then, we have
Hn(ζ) =ρnlhn(ρnζ)
fn(ρnζ) ρk
nhn(ρnζ)
=ρnlhn(ρnζ) gn(ρnζ)
ρk n
.
Noting thatρl nhn(ρnζ)
χ ⇒
ζl inC, we have
Hn(ζ)
χ
⇒
ζlG(ζ) =H(ζ) inC\ {}.
SinceG(ζ) has a zero atζ = , by (.), there existsδ∈(, ) such thatgn(ρnζ) is holo-morphic in(,δ) for sufficiently largen, and thusHn(ζ) is holomorphic in(,δ) for sufficiently largen. By the maximum principle, we have
Hn(ζ)
χ
⇒
Obviously, all poles ofH(ζ) are multiple and all zeros ofH(ζ) have multiplicity at least k+ . SinceG(ζ) has a zero of order at leastlatζ = , we haveH()=∞and thusH(ζ) is a meromorphic function inC. Noting that
Hn(k)(ζ) =ρnlfn(k)(ρnζ)= ρnlhn(ρnζ) and ρnlhn(ρnζ)
χ
⇒
ζl inC.
By Lemma ., eitherH(k)(ζ)= /ζlorH(k)(ζ)≡/ζlinC. The latter possibility contradicts the fact thatH()=∞. ThusH(k)(ζ)= /ζlinC. It follows from Lemma . and Lemma . thatH(ζ) is a constant. Next we will show that this is impossible. Indeed, suppose that H≡c. SinceGis not a constant,c= . Then we have
fn() = Hn()
ρl–k n
→ ∞ asn→ ∞. (.)
Suppose first that there exists δ ∈(, ) such that fn(z)= in (,δ) for sufficiently large n. By Lemma ., {fn} is normal in(,δ). But this contradicts our assumption that no subsequence of {fn} is normal at . Hence, taking a subsequence and renum-bering, we may assume thatz∗nis the zero offnof smallest modulus andz∗n→. Since Hn(ζ) =ρnl–kfn(ρnζ)⇒c(= ), we havez∗n/ρn→ ∞. Set
Hn∗(ζ) :=z∗n l–kfn
z∗nζ .
In view of the fact that
Hn∗(k)(ζ) =z∗n lfn(k)zn∗ζ =φn(z
∗
nζ) ζl
χ
⇒
ζl
andHn∗(ζ)= in, by Lemma .,{Hn∗(ζ)} is normal in. By Lemma ., {Hn∗(ζ)}is normal inC\ {}. Hence,{Hn∗(ζ)}is normal inC. Therefore, there exists a subsequence
{Hn∗(ζ)}(still denoted by{Hn∗(ζ)}) such thatHn∗⇒χ H∗inC. By the definition ofHn∗(ζ), we getHn∗() = and thusH∗() = . Since
Hn∗() =Hn()
z∗n ρn
l–k =c
z∗n ρn
l–k
→ ∞,
we get thatH∗() =∞. ThusH∗is nonconstant. However, sinceH∗() =∞andHn∗(ζ)= in, we haveH∗= inby Hurwitz’s theorem. However, Lemma . implies thatH∗(ζ) has a zero in. A contradiction.
ThusGis normal in. It remains to show that{fn}is normal at . SinceGis normal in, thenGis equicontinuous inwith respect to the spherical distance. On the other hand,gn() = for eachn, so there existsδ> such that|gn(z)| ≤ for allnin(,δ). It follows thatfn(z) is holomorphic in(,δ) for alln. Since{fn}is normal in, there exists a subsequence of{fn}(still denoted by{fn}) which converges locally uniformly in(,δ). The maximum modulus principle implies thatfnconverges locally uniformly in(,δ), and thus{fn}normal atz= , which contradicts our assumption that no subsequence of
5 Proof of Lemma 3.9
Proof It suffices to prove that each subsequence of{fn}has a subsequence which satisfies that (∗f ), and prove that (∗e) and (∗g) hold. So suppose that we have a subsequence of{fn}, which (to avoid complication in notation) we again call{fn}.
Without loss of generality, for eacha˜∈E, we may assume thata˜= ,⊂D,∩E=∅, and
h(z) =zl+al+zl++· · ·=zlh(z)ˆ
in, whereh() = andˆ h(z)ˆ = ,∞in. We consider the following three cases. Case.h() =∞.
We will derive a contradiction in the case, and hence (∗e) holds. For convenience, we set m:= –l, and then
h(z) = zm +
a–m+
zm– +· · ·= ˆ
h(z) zm
in, whereh(z)ˆ = ,∞in,h() = , andˆ mis a positive integer. Clearly, we haveh(z)= ,∞in,hn(z)= ,∞in, andh() =hn() =∞.
Subcase.. For sufficiently largen,fn()= .
We claim that for eachδ> , there exists at least one zero offnin(,δ) for sufficiently largen. Otherwise, there existδ∗ (> ) and a subsequence of{fn} (still denoted by{fn}) such thatfn(z)= in(,δ∗). By Lemma .,{fn}is normal at . This contradicts (∗c).
Taking a subsequence and renumbering, we may assume thatan(= ) is the zero of{fn} of smallest modulus andan→. SetFn(ζ) :=anm–kfn(anζ). We have:
(A) Fn(ζ)= in;
(A) all zeros ofFn(ζ)have multiplicity at leastk+ andFn() = ;
(A) Fn(k)(ζ)=amnhn(anζ)andamnhn(anζ)
χ ⇒
ζm inC.
By Lemma . and Lemma .,{Fn(ζ)}is normal inand quasinormal inC. Thus, there exist a subsequence of{Fn(ζ)}(still denoted by{Fn(ζ)}) andD⊂Csuch that:
(B) Dhas no accumulation points inC;
(B) for eachζ∈D, no subsequence of{Fn(ζ)}is normal atζ; (B) Fn(ζ)
χ
⇒F(ζ)inC\D.
Obviously,D∩=∅and all zeros ofF(ζ) have multiplicity at leastk+ inC\D.
Subcase... /∈D.
Obviously,F() =F(k)() = by (A). Letζ∈D. By Lemma .,F(k)(ζ) = ζm, which
contradictsF(k)() = . ThusD
is an empty set. SinceF() = ,F(ζ) is a meromorphic
function inC. By Lemma . and (A), eitherF(k)(ζ)≡
ζm orF(k)(ζ)=ζminC. IfF(k)(ζ)≡
ζm, thenF
(k)() = , which contradictsF(k)() = . IfF(k)(ζ)=
ζm, then by Lemma . and
Lemma .,Fis a constant function. SinceF() = ,F(ζ)≡ inC. Now,
Fn(ζ) =amn–kfn(anζ) ⇒ inC. (.)
thatfn(z) has no poles in(,δ∗). Sincefn(k)(z)=hn(z) andhn() =∞, we havef()=∞. Thusfn(z)=∞in(,δ∗). By Lemma .,{fn}is normal at . This contradicts (∗c).
Taking a subsequence and renumbering, we may assume thatyn(= ) is the pole offn(z) of smallest modulus andyn→. By Hurwitz’s theorem and (.), aynn →. SetGn(ζ) := ym–k
n fn(ynζ), we have:
(C) Gn(ζ)is holomorphic in;
(C) Gn() =∞;
(C) all zeros ofGn(ζ)have multiplicity at leastk+ ;
(C) G(nk)(ζ)=ymnhn(ynζ)andymnhn(ynζ)
χ ⇒
ζm inC.
By Lemma . and Lemma .,{Gn(ζ)}is normal inand quasinormal inC. Thus, there exist a subsequence of{Gn(ζ)}(still denoted by{Gn(ζ)}) andD⊂Csuch that:
(D) Dhas no accumulation points inC;
(D) for eachζ∈D, no subsequence of{Gn(ζ)}is normal atζ; (D) Gn(ζ)
χ
⇒G(ζ)inC\D.
Obviously,D∩=∅and all zeros ofG(ζ) have multiplicity at leastk+ inC\D.
Clearly,G() =limn→∞Gn(aynn) =ynm–kfn(an) = , soG(z) is meromorphic inC\D. By
Lemma . and (C), eitherG(k)(ζ)≡
ζm orG(k)(ζ)= ζm inC\D.
Subcase....Dis an empty set.
By (C), we haveG() =∞. IfG(k)(ζ)=
ζm inC, then by Lemma . and Lemma .,
G(ζ) is a constant function which contradicts thatG() =∞. IfG(k)(ζ) –
ζm ≡ inC, then
G() =∞, which contradictsG() = . Subcase....Dis not an empty set.
Letζ∈E. SinceD∩=∅, by Lemma ., we haveG(k)(ζ) =ζm inC\D–{}Clearly,
G(k)(ζ) and
ζm are meromorphic functions inC\D, so we haveG(k)(ζ) =ζm inC\D,
which contradictsG() = . Subcase... ∈D.
By Lemma ., we haveF(k)(ζ) =
ξm inC\D. Ifm≤k, thenF(ζ) is a multi-valued
function inC\D. A contradiction. Thus we havem>k.
We claim that D={}. Otherwise, there existsζsuch thatζ∈D andζ= , . By
Lemma .,
F(k–)(ζ) =
ζ
ξmdξ=
ζ–m+– –m+ ;
F(k–)(ζ) =
ζ
ξ–m+– –m+ dξ=
ζ–m+ (–m+ )(–m+ )–
ζ –m+ +
(–m+ );
(.)
F(k–)(ζ) =
ζ ζ
ξmdξ=
ζ–m+–ζ –m+
–m+ ;
F(k–)(ζ) =
ζ ζ
ξ–m+– –m+ dξ=
ζ–m+ (–m+ )(–m+ )–
ζ –m+ +
ζ
(m+ ).
(.)
By (.) and (.), we obtainζ= , which contradictsζ= . ThusD={}.
By Lemma .,
F(ζ) =
ζ
ζ
· · ·
ζk–
ζmdζkdζk–· · ·dζ
= +Pk–(ζ)ζ m–k
inC\D, wherePk–(ζ) is a polynomial of degreek– . By (.),
Fn(ζ)
χ
⇒ +Pk–(ζ)ζm–k
(–m+k)(–m+k– )· · ·(–m+ )ζm–k (.)
in . By Hurwitz’ theorem, there exist γn,i (i= , , . . . ,m–k) such thatγn,i→ and Fn(γn,i) =∞. Since Fn(k)(ζ)=amnhn(anζ), we have Fn()=∞, and henceγn,i= for i= , , . . . ,m–k.
SetUn(ξ) :=smn–kFn(snξ) andηn,i=γsnn,i, wheresnis one of{γn,,γn,, . . . ,γn,m–k}of largest modulus. Clearly,|ηn,i| ≤. Now, we have:
(E) for eachR> ,Un(ξ)= in(,R)for sufficiently largen;
(E) Un() =∞;
(E) Un(k)(ξ)=amnsmnhn(amnsmnξ)andamnsnmhn(amnsmnξ)
χ ⇒
ξm inC; (E) Un(ξ)has onlym–kpolesηn,ion, wherei= , , . . . ,m–k.
In fact, (E) holds by (.). By Lemma ., (E), and (E), we find thatUn(ξ) is normal inC. We assume thatUn(ξ)
χ
⇒U(ξ) inC. Obviously,U() =∞by (E). Subcase....U(ξ) is a meromorphic function inC.
By Lemma ., we have either U(k)(ζ)≡
ξm or U(k)(ξ)= ξm in C. Since U() =∞,
U(k)(ξ)=ξm inC. By Lemma . and Lemma .,U(ξ) is a constant function. This
con-tradictsU() =∞.
Subcase....U(ξ)≡ ∞inC. SetUn∗(ξ) :=Un(ξ)·
m–k
i= (ξ –ηn,i). By the maximum principle applied to U∗
n(ξ), we get
that
Un∗(ξ) ⇒ ∞χ inC. (.)
SetFn∗(ζ) :=Fn(ζ)·
m–k
i= (ζ–γn,i) =Fn(ζ)·
m–k
i= (ζ–snηn,i). By (.),Fn∗(ζ) has no poles in(,) for sufficiently largen. By the maximum principle,
Fn∗(ζ) ⇒χ +Pk–(ζ)ζ m–k
(–m+k)(–m+k– )· · ·(–m+ ) in
,
.
Hence,
Fn∗()→
(–m+k)(–m+k– )· · ·(–m+ ) asn→ ∞. (.)
On the other hand,
Fn∗(snξ) =Fn(snξ)· m–k
i=
(snξ–γn,i) =Fn(snξ)· m–k
i=
(snξ–snηn,i)
=smn–kFn(snξ)· m–k
i=
(ξ–ηn,i) =Un∗(ξ)
χ ⇒ ∞
inC, and henceFn∗()→ ∞, which contradicts (.).
Do as in Subcase .., we may assume thatyn(= ) is the pole offn(z) of smallest modulus andyn→. SetGn(ζ) :=ynm–kfn(ynζ). We have:
(F) Gn(ζ)is holomorphic function in;
(F) Gn() =∞;
(F) all zeros ofGn(ζ)have multiplicity at leastk+ ;
(F) G(nk)(ζ)=ymnhn(ynζ)andymnhn(ynζ)
χ ⇒
ζm inC.
By Lemma . and Lemma .,{Gn(ζ)}is normal inand quasinormal inC. Thus, there exist a subsequence of{Gn(ζ)}(still denoted by{Gn(ζ)}) andD⊂Csuch that:
(G) Dhas no accumulation points inC;
(G) for eachζ∈D, no subsequence of{Gn(ζ)}is normal atζ; (G) Gn(ζ)
χ
⇒G(ζ)inC\D.
Obviously,D∩=∅and all zeros ofG(ζ) have multiplicity at leastk+ inC\D.
Obviously,G() =limn→∞Gn() =ymn–kfn() = , soG(z) is a meromorphic function in
C\D. By Lemma . and (F), eitherG(k)(ζ)≡ζm orG(k)(ζ)=ζm inC\D.
Subcase...Dis an empty set.
By (F), G() =∞. If G(k)(ζ)=
ζm, then by Theorem . and Lemma ., G(ζ) is a
constant function, which contradicts that G() =∞. IfG(k)(ζ) –
ζm ≡, then we have
G() =∞, which contradictsG() = . Subcase...Dis not an empty set.
Letζ∈E. SinceD∩=∅, by Lemma ., we haveG(k)(ζ) =ζm, which contradicts
G() = .
Case.h() = .
In this case, we will show that (∗f ) and (∗g) hold. Clearly, we have h(z)= ,∞in, hn(z)= ,∞in, andh() =hn() = .
We claim that for eachδ> , there exists at least one zero offnin(,δ) for sufficiently largen. Otherwise, there existδ∗(> ) and a subsequence of{fn}(still denoted by{fn}) such thatfn(z)= in(,δ∗). Sincefn(k)(z)=hn(z) and all the zeros of{fn}have multiplicity at leastk+ , we havefn()= , and hencefn(z)= in(,δ∗). By Lemma .,{fn}is normal at , which contradicts (∗c).
Taking a subsequence and renumbering, we may assume thatan(= ) is the zero offnof smallest modulus andan→. SetFn(ζ) :=fn(anζ)
akn+l . We have: (a) Fn(ζ)= in;
(a) all zeros ofFn(ζ)have multiplicity at leastk+ andFn() = ;
(a) Fn(k)(ζ)=ζlh(aˆ nζ)andζlh(aˆ nζ)
χ
⇒ζlinC.
By Lemma . and Lemma .,{Fn(ζ)}is normal inand quasinormal inC. Thus, there exist a subsequence of{Fn(ζ)}(still denoted by{Fn(ζ)}) andD⊂Csuch that:
(b) Dhas no accumulation points inC;
(b) for eachζ∈D, no subsequence of{Fn(ζ)}is normal atζ; (b) Fn(ζ)
χ
⇒F(ζ)inC\D.
Obviously,D∩=∅and all zeros ofF(ζ) have multiplicity at leastk+ inC\D.
Subcase.. /∈D.
By (a),F() =F(k)() = , and henceF(ζ) is a meromorphic function inC\D .
We claim thatD=∅. Otherwise, letζ∈D. SinceD∩=∅, by Lemma .,F(k)(ζ) =
By Lemma . and (a), eitherF(k)(ζ)≡ζkorF(k)(ζ)=ζkinC. SinceF(k)() = , we have
F(k)(ζ)=ζkinC. By Lemma . and Lemma ., we have
F(ζ) =
t++l i= (ζ–αi)
(l+k)(l+k– )· · ·(l+ )(ζ–β)t+–k,
wheret≥kis an integer,β∈C, andαi= ,β(≤i≤t+ +l). Thus, we have
Fn(ζ)
χ ⇒
t++l i= (ζ–αi)
(l+k)(l+k– )· · ·(l+ )(ζ–β)t+–k inC. (.)
By Hurwitz’s theorem, there exist sequencesζn,i→αiandηn,j→β (counting multiplic-ities of zeros and poles, respectively) such that for sufficiently large n, Fn(ζn,i) = and Fn(ηn,j) =∞, wherei= , , . . . ,t+ +l andj= , , . . . ,t+ –k. setzn,i:=anζn,i. Thus, fn(zn,i) = andzn,i→, wherei= , , . . . ,t+ +l. Set
Bn:={zn,,zn,, . . . ,zn,t++l} (where the same elements are admissible).
Subcase... For eachδ> ,fnhas at leastt+ +lzeros (counting multiplicities) in (,δ) for sufficiently largen.
Taking a subsequence and renumbering, we may assume thatbn(= ) is the zero offnof smallest modulus in\Bnandbn→. Clearly,Fn(abnn) = andbann=ζn,ifori= , , . . . ,t+ +l. By Hurwitz’s theorem and (.), an
bn → asn→ ∞. SetGn(ζ) :=
fn(bnζ)
bkn+l . We have, for
sufficiently largen:
(c) Gn(ζ)has onlyt+ +lzerosan
ζn,i
bn in. Obviously,|
anζn,i
bn | →; (c) all zeros ofGn(ζ)have multiplicity at leastk+ andGn() = ;
(c) G(nk)(ζ)=ζlh(bˆ nζ)andζlh(bˆ nζ)
χ
⇒ζlinC.
By Lemma . and Lemma .,{Gn(ζ)}is normal inand quasinormal inC. Thus, there exist a subsequence of{Gn(ζ)}(still denoted by{Gn(ζ)}) andD⊂Csuch that:
(d) Dhas no accumulation points inC;
(d) for eachζ∈D, no subsequence of{Gn(ζ)}is normal atζ; (d) Gn(ζ)
χ
⇒G(ζ)inC\D.
Obviously,D∩=∅and all zeros ofG(ζ) have multiplicity at leastk+ inC\D.
Set
G∗n(ζ) :=Gn(ζ)
t+–k j= (ζ–
anηn,j
bn )
t++l i= (ζ–
anζn,i
bn )
and Fn∗(ζ) =Fn(ζ)
t+–k
j= (ζ–ηn,j)
t++l
i= (ζ–ζn,i) .
By (.),G∗n(anζ
bn) =F ∗
n(ζ)⇒(l+k)(l+k–)···(l+)inC. Hence
G∗n()→
(l+k)(l+k– )· · ·(l+ ). (.)
Subcase....G(ζ)≡ ∞inC\D.
Obviously,G∗n(ζ) has no zeros infor sufficiently largen. Applying the maximum prin-ciple to
G∗n(ζ), we see thatG ∗
n(ζ)
χ
