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Normal families of meromorphicfunctions concerning shared valuesJian-Jun Ding a , Li-Wei Ding a & Wen-Jun Yuan aa School of Mathematics and Information Sciences , GuangzhouUniversity , Guangzhou 510006 , ChinaPublished online: 01 Jun 2011.

To cite this article: Jian-Jun Ding , Li-Wei Ding & Wen-Jun Yuan (2013) Normal families ofmeromorphic functions concerning shared values, Complex Variables and Elliptic Equations: AnInternational Journal, 58:1, 113-121, DOI: 10.1080/17476933.2011.555644

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Complex Variables and Elliptic Equations, 2013

Vol. 58, No. 1, 113–121, http://dx.doi.org/10.1080/17476933.2011.555644

Normal families of meromorphic functions concerning shared values

Jian-Jun Ding, Li-Wei Ding and Wen-Jun Yuan*

School of Mathematics and Information Sciences, Guangzhou University,Guangzhou 510006, China

Communicated by J. Du

(Received 10 July 2010; final version received 12 January 2011)

In this article, we prove the following normality criterion: let n� 2, m, k bethree positive integers, and a be a non-zero complex number. Let F be afamily of meromorphic functions in a domain D such that each f2F hasonly zeros of multiplicity at least max{k, 2}. If for each pair of f and g inF,f m( f (k))n and gm(g(k))n share the value a IM, then F is normal in D.This extends Hu and Meng’s result.

Keywords: meromorphic function; normal family; shared values

AMS Subject Classification: 30D35

1. Introduction and main result

Let F be a family of meromorphic functions defined in a domain D. F is said to benormal, in the sense of Montel, if for any sequence { fn} of F there exists asubsequence f fnjg converges spherically, locally and uniformly in D to a meromor-phic function or1 (see [1]). It is also interesting to find normality criterion from thepoint of view of shared values. Let f and g be two non-constant meromorphicfunctions in D. We say that f and g share the value a IM (ignoring multiplicity) in Dif f �1(a)¼ g�1(a) (see [2]). Hayman [3] proposed the following problem.

THEOREM A Let F be a family of meromorphic functions in D, n be a positive integerand a be a finite non-zero complex number. If each function f2F satisfies f nf 0 6¼ a,then F is normal in D.

Theorem A is given by Yang and Chang [4,5] (for n� 5 and for n� 2 in case thatF is a family of holomorphic functions), Gu [6] (for n¼ 3, 4), Oshkin [7](for holomorphic functions, n¼ 1; cf. [8]), Pang [9] (for n� 2; cf. [6]) and Chenand Fang [10] (for n� 1). In 2008, Zhang [11] extended the result, he proved that Fis also normal when each pair ( f, g) of F satisfies that f nf 0 and gng0 share a finitenon-zero complex number a IM for n� 2 (or see [12]).

*Corresponding author. Email: gzywj@tom.com

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Schwick [13] proved a theorem of normal families, that is, when n� kþ 3,( f n)(k) 6¼ 1 for each meromorphic function f2F, then F is normal in D. Recently,Li and Gu [14] further extended Schwick’s result as follows.

THEOREM B Let F be a family of meromorphic functions in D; k, n be two positiveintegers and a be a finite non-zero complex number. If n� kþ 2 and for each pair offunctions f and g in F, ( f n)(k) and (gn)(k) share a non-zero complex number a IM,then F is normal in D.

Hu and Meng [15] proved the following normality criterion.

THEOREM C Take positive integers n and k with n, k� 2 and take a non-zero complexnumber a. Let F be a family of meromorphic functions in the plane domain D such thateach f2F has only zeros of multiplicity at least k. For each pair ( f, g)2F, if f( f (k))n

and g(g(k))n share a IM, then F is normal in D.

We extend Hu and Meng’s result, and prove the following result.

THEOREM 1 Let n� 2, m, k be three positive integers, and a be a non-zero complexnumber. Let F be a family of meromorphic functions in a domain D such that eachf2F has only zeros of multiplicity at least max{k, 2}. If for each pair of f and g in F,f m( f (k))n and gm(g(k))n share the value a IM, then F is normal in D.

Example 1 Let n� 2, k� 2, m be three positive integers, D¼ {z : jzj51}, a be a non-zero complex number and

F ¼ flzk�1 : l ¼ 1, 2, . . .g:

Then, we have for each f, g in F, f m( f (k))n and gm(g(k))n share the value a IM, ButFis not normal in D.

This example shows that, the condition that f2F has only zeros of multiplicityat least max{k, 2} is sharp in theorem.

2. Preliminary lemmas

In order to prove our theorems, we need the following preliminary results.

LEMMA 1 [16,17] Let F be a family of functions meromorphic in the unit disc, all ofwhose zeros have multiplicity at least k. If F is not normal, there exist, for each0��5k,

(a) a number 05r51;(b) points zn, jznj5r;(c) functions fn2F and(d) positive numbers �n! 0

such that ���n fnðzn þ �n�Þ ¼ gnð�Þ ! gð�Þ locally uniformly with respect to the sphericalmetric, where g is a nonconstant meromorphic function in C. And all zeros of g havemultiplicity at least k. Moreover, the order of g is not greater than 2.

LEMMA 2 [18–21] Let f be a transcendental meromorphic function and m, n, k be threepositive integers. If n� 2, then f m( f (k))n assumes every finite non-zero complex numberinfinitely often.

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LEMMA 3 Let n and k be two positive integers with n� 2 and a be a finite non-zero

complex number. If f is a rational but not a polynomial function and f has only zeros of

multiplicity at least 2, then f m( f (k))n� a has at least two distinct zeros.

Proof Assume, to the contrary, that f m( f (k))n� a has at most one zero. Set

ð f ðzÞÞmð f ðkÞðzÞÞn ¼ Aðz� �1Þ

m1 � � � ðz� �sÞms

ðz� �1Þn1 � � � ðz� �tÞ

nt, ð1Þ

where A is a non-zero constant. Since f has only zeros with multiplicity at least 2, we

deduce

mi � 2 ði ¼ 1, 2, . . . , sÞ; nj � nðkþ 1Þ þm ð j ¼ 1, 2, . . . , tÞ:

From the above formulas, we have

M :¼ m1 þm2 þ � � � þms � 2s, ð2Þ

N :¼ n1 þ n2 þ � � � þ nt � ½nðkþ 1Þ þm�t: ð3Þ

Differentiating (1), we get

½ð f ðzÞÞmð f ðkÞðzÞÞn� 0 ¼ðz� �1Þ

m1�1 � � � ðz� �sÞms�1

ðz� �1Þn1þ1 � � � ðz� �tÞ

ntþ1gðzÞ, ð4Þ

where g is a polynomial such that deg(g)� sþ t� 1. Next we may distinguish

two cases.

Case 1 The function f m( f (k))n� a has exactly one zero. Now we can write

ð f ðzÞÞmð f ðkÞðzÞÞn ¼ aþBðz� z0Þ

l

ðz� �1Þn1 � � � ðz� �tÞ

nt¼

PðzÞ

QðzÞ, ð5Þ

where l is a positive integer, B is a non-zero constant, P and Q are polynomials of

degree M and N, respectively. Also P and Q have no common factors. Obviously, we

have z0 6¼ �i(i¼ 1, . . . , s) since a 6¼ 0. Differentiating (5), we obtain

½ð f ðzÞÞmð f ðkÞðzÞÞn� 0 ¼ðz� z0Þ

l�1g1ðzÞ

ðz� �1Þn1þ1 � � � ðz� �tÞ

ntþ1, ð6Þ

where g1 is a polynomial of the form

g1ðzÞ ¼ Bðl�NÞzt þ Bt�1zt�1 þ � � � þ B0

in which B0, . . . ,Bt�1 are constants.

Case 1.1 l 6¼N. By (5), we obtain deg(P)� deg(Q), that is M�N. Since z0 6¼ �i,Equations (4) and (6) imply that

Xsi¼1

ðmi � 1Þ ¼M� s � degð g1Þ ¼ t,

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and so M� sþ t. By using (2) and (3), we obtain

M � sþ t �M

2þ

N

nðkþ 1Þ þm�

1

2þ

1

nðkþ 1Þ þm

� �M,

which is a contradiction since n� 2, k� 1.

Case 1.2 l¼N. By (5), we obtain deg(P)� deg(Q), that is M�N.By (4) and (6), we obtain

l� 1 � degð gÞ � sþ t� 1,

and hence

N ¼ l � degð gÞ þ 1 � sþ t �M

2þ

N

nðkþ 1Þ þm

�1

2þ

1

nðkþ 1Þ þm

� �N5N,

which is a contradiction.

Case 2 The function f m( f (k))n� a has no zero. Now we can write

ð f ðzÞÞmð f ðkÞðzÞÞn ¼ aþB

ðz� �1Þn1 � � � ðz� �tÞ

nt, ð7Þ

where B is a non-zero constant. By (1) and (7), we obtain M¼N. Differentiating (7),

we get

½ð f ðzÞÞmð f ðkÞðzÞÞn� 0 ¼g2ðzÞ

ðz� �1Þn1þ1 � � � ðz� �tÞ

ntþ1, ð8Þ

where g2(z) is a polynomial with deg(g2)¼ t� 1. By (4) and (8), we obtain

Xsi¼1

ðmi � 1Þ ¼M� s � degð g2Þ ¼ t� 1:

From (2), (3) and the above formula, we have

M � sþ t� 15 sþ t5M

2þ

N

nðkþ 1Þ þm¼

1

2þ

1

nðkþ 1Þ þm

� �M,

which is a contradiction. Now Lemma 3 is proved.

LEMMA 4 Let f be a non-constant meromorphic function, n� 2, m, k be three positiveintegers and a be a finite non-zero complex number. If f has only zeros of multiplicity atleast max{k, 2}, then f m( f (k))n� a has at least two distinct zeros.

Proof We may distinguish three cases.

Case 1 f is transcendental. This is a direct consequence of Lemma 2.

Case 2 f is a rational but not a polynomial function, it follows from Lemma 3.

Case 3 f is a polynomial. We obtain immediately that f m( f (k))n has multiple zerossince f has only zeros of multiplicity at least max{k, 2} which means particularlydeg( f )�max{k, 2}, and hence f m( f (k))n� a has at least one zero. If f m( f (k))n� a

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has only a unique zero z0, then there exist a non-zero constant A and an integer l� 2

such that

ð f ðzÞÞmð f ðkÞðzÞÞn ¼ aþ Aðz� z0Þl,

which has only simple zeros since a 6¼ 0. This contradicts with the hypothesis of

Lemma 4.

LEMMA 5 Let f be a non-constant meromorphic function, m, n be two positive

integers, and a be a finite non-zero complex number. If n� 2, then f m( f 0)n� a has at

least one zero.

Proof We may distinguish three cases.

Case 1 f is transcendental. This is a direct consequence of Lemma 2.

Case 2 f is a non-constant polynomial. Obviously, f m( f 0)n� a is also a non-

constant polynomial, and hence it has at least one zero.

Case 3 f is a rational but not a polynomial function. We may assume that f is

rational with at least one pole. Write

f ðzÞ ¼ Aðz� �1Þ

m1 � � � ðz� �sÞms

ðz� �1Þn1 � � � ðz� �tÞ

nt, ð9Þ

where A is a non-zero constant, and mi(i¼ 1, 2, . . .), nj ( j¼ 1, 2, . . . , t) are positive

integers. For simplicity, we denote

M :¼ m1 þm2 þ � � � þms, ð10Þ

N :¼ n1 þ n2 þ � � � þ nt: ð11Þ

Differentiating (9), we obtain

f 0ðzÞ ¼P1ðzÞ

Q1ðzÞ, ð12Þ

where

P1ðzÞ ¼ ðz� �1Þm1�1 � � � ðz� �sÞ

ms�1hðzÞ,

Q1ðzÞ ¼ ðz� �1Þn1þ1 � � � ðz� �tÞ

ntþ1,

hðzÞ ¼ AðM�N Þzsþt�1 þ � � � :

It follows from (9) and (12) that

f mð f 0Þn ¼P

Q, ð13Þ

where

PðzÞ ¼ Amðz� �1ÞðnþmÞm1�n � � � ðz� �sÞ

ðnþmÞms�nhnðzÞ,

QðzÞ ¼ ðz� �1ÞðnþmÞn1þn � � � ðz� �tÞ

ðnþmÞntþn:

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Suppose, to the contrary, that f m( f 0)n� a has no zero. Then

f mð f 0Þn ¼ aþB

Q¼

P

Q, ð14Þ

where B is a non-zero constant, which implies particularly that P¼ aQþB; hence,deg(P)¼ deg(Q). Now, we claim that M4N, otherwise, if M�N, then

degðP1Þ ¼M� sþ degðhÞ �M� sþ sþ t� 15Nþ t ¼ degðQ1Þ:

From (12), (13) and the above formula, we deduce

degðPÞ ¼ n degðP1Þ þmM5 n degðQ1Þ þmN ¼ degðQÞ:

This is a contradiction, and so the claim is proved. Hence, deg(hn)¼ n(sþ t� 1).This and (13) imply that

degðPÞ ¼Xsi¼1

½ðnþmÞmi � n� þ nðsþ t� 1Þ ¼ ðnþmÞMþ nt� n, ð15Þ

degðQÞ ¼Xtj¼1

½ðnþmÞnj þ n� ¼ ðnþmÞNþ nt: ð16Þ

Therefore, by (15), (16) and deg(P)¼deg(Q), we obtain

M�N ¼n

mþ n:

This is impossible since M�N is an integer. Therefore, f m( f 0)n� a has at leastone zero.

3. Proof of Theorem 1

Suppose, to the contrary, that F is not normal in D. Without loss of generality, weassume that F is not normal at z0¼ 0. Then, by Lemma 1, there exist points zj! 0,positive numbers �j! 0 and functions fj2F such that

gjð�Þ ¼ �� nk

nþm

j fjðzj þ �j�Þ

converges uniformly to a non-constant meromorphic function g(�) in C with respectto the spherical metric. Moreover, g(�) is of order at most 2, and the zeros of g(�)have at least multiplicity max{k, 2}.

On every compact subset of C which contains no poles of g, we have uniformly

f mj ðzj þ �j�Þð fðkÞj ðzj þ �j�ÞÞ

n� a ¼ gmj ð�Þð g

ðkÞj ð�ÞÞ

n� a! gmð�Þð gðkÞð�ÞÞn � a ð17Þ

with respect to the spherical metric. If gm(g(k))n� a, then g has no zeros. Of course, galso has no poles. Since g is a non-constant meromorphic function of order at most2, there exist constants Ci(i¼ 0, 1, 2) such that (C1,C2) 6¼ (0, 0), and

gð�Þ ¼ eC0þC1�þC2�2

:

Obviously, this contradicts with the relation gm(g(k))n� a. Hence gm(g(k))n � a.

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By Lemma 4, the function gm(g(k))n� a has at least two distinct zeros. Let �0 and��0 be two distinct zeros of gm(g(k))n� a. We choose a positive number � small enough

such that D1\D2¼Ø and such that gm(g(k))n� a has no other zeros in D1[D2

except for �0 and ��0, where

D1 ¼ f�2C : j� � �0j5 �g, D2 ¼ f�2C : j� � ��0j5 �g:

By (17) and Hurwitz’s theorem, for sufficiently large j there exist points �j 2D1,

��j 2D2 such that

f mj ðzj þ �j�j Þð fðkÞj ðzj þ �j�j ÞÞ

n� a ¼ 0,

f mj ðzj þ �j��j Þð f

ðkÞj ðzj þ �j�

�j ÞÞ

n� a ¼ 0:

Since, by the hypothesis in Theorem 1, f m1 ð fðkÞ1 Þ

n and f mj ð fðkÞj Þ

n share a for each j, and

it follows that

f m1 ðzj þ �j�j Þð fðkÞ1 ðzj þ �j�j ÞÞ

n� a ¼ 0,

f m1 ðzj þ �j��j Þð f

ðkÞ1 ðzj þ �j�

�j ÞÞ

n� a ¼ 0:

Letting j!1, and noting zjþ �j�j! 0, zj þ �j��j ! 0, we obtain

f m1 ð0Þð fðkÞ1 ð0ÞÞ

n� a ¼ 0:

Since the zeros of f m1 ð fðkÞ1 Þ

n� a have no accumulation points, we have

zj þ �j�j ¼ 0, zj þ �j��j ¼ 0:

This contradicts with the facts that �j 2D1, ��j 2D2 and D1\D2¼Ø. Theorem 1 is

proved completely.

4. Notes

According to the proof of Theorem 1, we obtain the following result which extends

Theorem 1 and Theorem A.

THEOREM 2 Let n� 2, m and k be three positive integers, and a be a non-zero complex

number. Let F be a family of meromorphic functions in the plain domain D such that

each f2F has only zeros of multiplicity at least k. For each f2F, if

f m(z)( f (k)(z))n¼ a implies j f (k)(z)j �A for a positive number A, then F is normal in D.

Proof Suppose, to the contrary, that F is not normal in D. Without loss of

generality, we assume that F is not normal at z0¼ 0. Then, by Lemma 1, there exist

points zj! 0, positive numbers �j! 0 and functions fj2F such that

gjð�Þ ¼ �� nk

nþm

j fjðzj þ �j�Þ

converges uniformly to a non-constant meromorphic function g(�) in C with respect

to the spherical metric, and the zeros of g(�) have at least multiplicity k. Hence, the

function gm(g(k))n� a has at least one zero �0 by Lemmas 4 and 5. By Hurwitz’s

theorem, there exist a sequence {�j}D with �j! �0 ( j!1) and a subsequence of gj

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(which, renumbering, we continue to denote by gj) such that

f mj ðzj þ �j�j Þð fðkÞj ðzj þ �j�j ÞÞ

n¼ gj ð�j Þ

mð gðkÞj ð�j ÞÞ

n¼ a:

Thus,

j gðkÞj ð�j Þj ¼ �

mknþm

j j fjðzi þ �j�j Þj � A�mknþm

j :

So,

gðkÞð�0Þ ¼ limj!1

gðkÞj ð�j Þ ¼ 0:

This contradicts with gm(�0)(g(k)(�0))

n¼ a 6¼ 0. Theorem 2 is proved.

It is easy to see that Theorem 1 is also true, if f m( f (k))n (resp. gm(g(k))n) is replaced

by f m( f (k)þP[ f ])n (resp. gm(g(k)þP[g])n), where P is a differential polynomial.

In order to state this result, we introduce some notations: k2N¼ {1, 2, . . .},

and rj 2N0 ¼ NSf0g for j¼ 0, 1, . . . , k� 1, and put r¼ (r0, r1, . . . , rk�1).

Define Mr[u] by

Mr½u�ðzÞ :¼ ½uðzÞ�r0 ½u 0ðzÞ�r1 � � � ½u ðk�1ÞðzÞ�rk�1 for z2C:

A differential polynomial P[u] is an expression of the form

P½u�ðzÞ :¼Xr2I

arðzÞMr½u�ðzÞ for z2C,

where the ar is entire in C and I is a finite index set.

THEOREM 3 Let n� 2, m, k be three positive integers, a be a non-zero complex

number and P be a differential polynomial withPk�1

i¼0 ½nk� ðmþ nÞi�ri þmk4 0 for

each r2 I. Let F be a family of meromorphic functions in a domain D such that each

f2F has only zeros of multiplicity at least max{k, 2}. If for each pair of f and g in F,

f m( f (k)þP[ f ])n and gm(g(k)þP[g])n share the value a IM, then F is normal in D.

Proof By using the notations in the proof of Theorem 1, and now noting that, by

Hurwitzs theorem, the zeros of g(�) have at least multiplicity max{k, 2}. Since

�r :¼Xk�1i¼0

½nk� ðmþ nÞi�ri þm

nþm4 0 for r2 I,

we have

f mj ðzj þ �j�Þ½ fðkÞj ðzj þ �j�Þ þ P½ fj �ðzj þ �j�Þ�

n� a

¼ gmj ð�Þ gðkÞj ð�Þ þ

Xr2I

��rj arðzj þ �j�ÞYk�1i¼0

ð gðiÞj ð�ÞÞ

ri

" #n

� a

converges uniformly to gm(�)(g(k)(�))n� a on every compact subset of C which

contains no poles of g. Similar to the proof of Theorem 1, we can easily prove thatF

is normal.

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Acknowledgements

The authors would like to express their hearty thanks to Professor Ming-Liang Fang,Chun-Lin Lei and Cui-Ping Zeng for supplying their manuscript and to Professor De-GuiYang and Jun-Fan Chen for their helpful discussions and suggestions. This research issupported by the NSF of China (10771220); Doctorial Point Fund of National EducationMinistry (200810780002).

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