BULLETIN of the Bull. Malays. Math. Sci. Soc. (2) 27 (2004), 149−160 MALAYSIAN MATHEMATICAL SCIENCES SOCIETY

Normal Functions and Normal Families

YAN XU Department of , Nanjing Normal University, Nanjing 210097 e-mail: [email protected]

Abstract. In this paper, we prove the following theorem: Let F be a family of holomorphic functions in the unit disc D and let a be a nonzero complex number. If, for any f ∈ F , f (z) = a ⇒ f ′(z) = a , f ′(z) = a ⇒ f ′′(z) = a , then F is uniformly normal in D, that is,

2 there exists a positive constant M such that (1 − z ) f # (z) ≤ M for each f ∈ F and z ∈ D , where M is independently of f. This result improves related results due to [2], [8], and [3].

2000 Mathematics Subject Classification: 30D45

1. Introduction

Let f and g be two meromorphic functions, and let a be a complex number. If g(z) = a whenever f (z) = a , we denote it by f = a ⇒ g = a , and f = a ⇔ g = a means f (z) = a if and if only if g(z) = a . Let D denote the unit disk in the C. A f meromorphic in D is called a normal function, in the sense of [6], if there exist a constant M ( f ) such that (1 − z 2 ) f # (z) ≤ M ( f ) for each z ∈ D , where

| f ′(z) | f # (z) = 1+ | f (z) | 2 denotes the spherical derivative. Let F be a family of meromorphic functions defined in D. F is said to be normal in

D (see [9]), in the sense of Montel, if for any fn ∈ F there exists a subsequence f , such that f converges spherically, locally and uniformly in D, to a n j n j or ∞ .

150 Y. Xu

Suppose that F is a family of functions meromorphic in D such that each function of F is a normal function, then, for each function f ∈ F , there exists a constant M ( f ) such that

(1 − z 2 ) f # (z) ≤ M ( f ) for each z ∈ D . In general, M ( f ) is a constant dependent on f, and we can not conclude that {M ( f ) , f ∈ F} is bounded. If {M ( f ) , f ∈ F} is bounded, we give the definition as follows

Definition. Let F be a family of meromorphic functions in the unit disc D. If there exists a positive constant M such that

sup {(1 − z 2 ) f # (z) : z ∈ D , f ∈ F } < M , we call the family F a uniformly normal family in D.

Remark 1. The idea of this definition is suggested by Pang (see [7]).

Remark 2. A well-known result due to Marty (see [4], [9] and [11]) says that a family F of functions meromorphic in D is a normal family if and only if for each # compact subset K of D there exists a constant M K such that f (z) ≤ M K for each f ∈ F and for each z ∈ K . Clearly, by Marty’s criterion if F is a uniformly normal family in D , then F must be normal in D . However, it is obvious that the converse is not always true.

It is natural to ask: When is a normal family F in D also uniformly normal in D? (The question is first introduced by Bergweiler and Pang (see [7]).)

Schwick [10] discovered a connection between normality criteria and sharing values. He proved

Theorem A. Let F be a family of meromorphic functions in the unit disc D and let a1 , a2 and a3 be distinct complex numbers. If, for any f ∈ F ,

f (z) = ai ⇔ f ′(z) = ai (i = 1, 2, 3) , then F is normal in D.

Pang [7] proved that the family F in Theorem A is also uniformly normal, as follows

Theorem B. Let F be a family of meromorphic functions in the unit disc D and let a1 , a2 and a3 be distinct complex numbers. If, for any f ∈ F , Normal Functions and Normal Families 151

f (z) = ai ⇔ f ′(z) = ai (i = 1, 2, 3) , then F is uniformly normal in D, that is, there exists a positive constant M such that

(1 − z 2 ) f # (z) ≤ M for each f ∈ F and z ∈ D , where M is independent of f.

Remark 3. In fact, from the proof in [7], we see that Theorem B still remains true if

f (z) = ai ⇒ f ′(z) = ai (i = 1, 2, 3) for any f ∈ F .

Chen and Hua [2], Pang and Zalcman [8] proved the following normality criterion.

Theorem C. Let F be a family of holomorphic functions in the unit disc D and let a be a nonzero complex number. If, for any f ∈ F , f (z) = a ⇔ f ′ = a , f ′ = a ⇒ f ′′(z) = a , then F is normal in D.

In [3], Fang improved Theorem C as follows

Theorem D. Let F be a family of holomorphic functions in the unit disc D and let a be a nonzero complex number. If, for any f ∈ F, f (z) = a ⇒ f ' (z) = a, f ′(z) = a ⇒ f ′′(z) = a , then F is normal in D.

In this paper, by using a method different from that used in [3], we obtain the following stronger result.

Theorem 1. Let F be a family of holomorphic functions in the unit disc D and let a be a nonzero complex number. If, for any f ∈ F , f (z) = a ⇒ f ′(z) = a , f ′(z) = a ⇒ f ′′(z) = a , then F is uniformly normal in D, that is, there exists a positive constant M such that

(1 − z 2 ) f # (z) ≤ M for each f ∈ F and z ∈ D , where M is independent of f.

Remark 4. The following example (see [2] and [3]) shows that a ≠ 0 cannot be omitted in Theorem 1.

nz Let F = { fn(z) = e : n = 1, 2, 3"}, D = { z : z < 1}. Then, for every fn ∈ F, it is easy to see that fn(z) = 0 ⇒ fn′(z) = 0 ⇒ fn′′(z) = 0. However, # fn (0) = n / 2 → ∞ as n → ∞ , thus F is not uniformly normal in D.

We shall use the standard notations in Nevanlinna theory (see [4], [11]). 152 Y. Xu 2. Lemmas

For convenience, we define

⎛ f ′ ⎞ ⎛ f ′′ ⎞ ⎛ f ′ ⎞ LD(r, f ) : = c1m ⎜r, ⎟ + c2m ⎜r, ⎟ + c3m ⎜r, ⎟ ⎝ f ⎠ ⎝ f ′ ⎠ ⎝ f − a ⎠

⎛ f ′′ ⎞ ⎛ f ′′ ⎞ + c4m ⎜r, ⎟ + c5m ⎜r, ⎟ , (a ∈ C) ⎝ f − a ⎠ ⎝ f ′ − a ⎠

where c1, c2 , c3 , c4 , c5 are constants, which may have different values at different occurrences.

Lemma 1. Let f be a non-constant holomorphic functions on the unit disc D, and a be a nonzero complex number. Let

f ′(z) + f ′′(z) 2 f ′′(z) ψ (z) := ψ ()f (z) = − . f (z) − a f ′(z) − a

If f = a ⇒ f ′ = a , f ′ = a ⇒ f ′′ = a on D, and f (0) ≠ a , f ′(0) ≠ a , f ′′(0) ≠ 0 , f ′(0) ≠ f ′′(0) and ψ (0) ≠ 0 , then

f (0) − a T( r, f ) ≤ LD (r, f ) + O (1) + 3log f ′′(0) − f ′(0)

( f (0) − a) ( f ′(0) − a) 1 + log + 2 log . f ′′(0) ψ (0)

Proof. Let f (z0 ) = a . By the assumptions we may assume that, near z0

a f (z) = a + a (z − z ) + ()()z − z 2 + b z − z 3 + O ()(z − z )4 , 0 2 0 0 0

(3) where b = f ( z0 ) / 6 is a constant. Then

2 3 f ′(z) = a + a ()z − z0 + 3b ( z − z0 ) + O (( z − z0 ) ),

2 f ′′(z) = a + 6b ()z − z0 + O ()( z − z0 ) ,

Normal Functions and Normal Families 153 and thus

f ′(z) + f ′′(z) 2 6b = + + O (z − z0 ) , f (z) − a z − z0 a

2 f ′′(z) 2 6b = + + O (z − z0 ) . f ′(z) − a z − z0 a

Hence ψ (z0 ) = 0 , and

⎛ 1 ⎞ ⎛ 1 ⎞ 1 N ⎜r, ⎟ ≤ N ⎜r, ⎟ ≤ T ()r,ψ + log ⎝ f − a ⎠ ⎝ ψ ⎠ ψ (0) 1 ≤ N ()r,ψ + LD (r, f )+ log (2.1) ψ (0)

⎛ 1 ⎞ 1 = N0 ⎜r, ⎟ + LD ()r, f + log , ⎝ f ′ − a ⎠ ψ (0)

where N0( r,1/ ( f ′ − a)) is the counting function for the zeros of f ′ − a which are not zeros of f − a . Since f = a ⇒ f ′ = a, form (2.1) we get

⎛ 1 ⎞ ⎛ 1 ⎞ 1 2N ⎜r, ⎟ ≤ N ⎜r, ⎟ + LD ()r, f + log . (2.2) ⎝ f − a ⎠ ⎝ f ′ − a ⎠ ψ (0)

On the other hand, by the assumptions we have

⎛ ⎞ ⎜ ⎟ ⎛ 1 ⎞ ⎜ 1 ⎟ ⎛ f ′′ ⎞ f ′(0) N ⎜ r, ⎟ ≤ N r, ≤ T ⎜r, ⎟ + log + O (1) ⎝ f ′ − a ⎠ ⎜ f ′′ ⎟ ⎝ f ′ ⎠ f ′′(0) − f ′(0) ⎜ − 1⎟ ⎝ f ′ ⎠

⎛ f ′′ ⎞ f ′(0) = N ⎜r, ⎟ + log + LD ()r, f + O (1) (2.3) ⎝ f ′ ⎠ f ′′()0 − f ′(0)

⎛ 1 ⎞ f ′(0) = N ⎜r, ⎟ + log + LD ()r, f + O (1). ⎝ f ′ ⎠ f ′′(0) − f ′(0)

154 Y. Xu

1 Next we need the estimate of N (r , f ′ ) . Since

⎛ 1 ⎞ ⎛ 1 ⎞ m ⎜ r, ⎟ ≤ m ⎜r, ⎟ + LD ()r, f ⎝ f − a ⎠ ⎝ f ′ ⎠ ⎛ 1 ⎞ 1 ≤ T ()r, f ′ − N ⎜r, ⎟ + LD ()r, f + log ⎝ f ′ ⎠ f ′(0) ⎛ 1 ⎞ 1 ≤ T ()r, f − N ⎜ r, ⎟ + LD ()r, f + log ⎝ f ′ ⎠ f ′(0) ⎛ 1 ⎞ ⎛ 1 ⎞ f (0) − a ≤ T ⎜ r, ⎟ − N ⎜ r, ⎟ + LD ()r, f + O (1) + log ⎝ f − a ⎠ ⎝ f ′ ⎠ f ′(0) ⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎜ = m ⎜ r, ⎟ + N ⎜ r, ⎟ − N ⎜r, ⎟ + LD ( r, f ) + O (1) ⎝ f − a ⎠ ⎝ f − a ⎠ ⎝ f ′ ⎠ f (0) − a + log , f ′(0) we obtain

⎛ 1 ⎞ ⎛ 1 ⎞ f (0) − a N ⎜ r, ⎟ ≤ N ⎜ r, ⎟ + LD (r, f ) + O (1) + log . (2.4) ⎝ f ′ ⎠ ⎝ f − a ⎠ f ′(0)

Thus, from (2.2), (2.3) and (2.4) , we get

⎛ 1 ⎞ f (0) − a 1 N ⎜ r, ⎟ ≤ LD (r, f ) + O (1) + log + log , (2.5) ⎝ f − a ⎠ f ′′(0) − f ′(0) ψ (0)

⎛ 1 ⎞ f (0) − a 1 N ⎜ r, ⎟ ≤ LD ( r, f ) + O (1) + 2 log + log . (2.6) ⎝ f ′ − a ⎠ f ′′(0) − f ′(0) ψ (0)

Using Milloux’s inequality, we have

⎛ 1 ⎞ ⎛ 1 ⎞ T( r, f ) ≤ N ⎜ r, ⎟ + N ⎜ r, ⎟ + LD ( r, f ) + O (1) ⎝ f − a ⎠ ⎝ f ′ − a ⎠ ( f (0) − a) ( f ′(0) − a) + log . f ′′(0)

Substituting (2.5) and (2.6) in the above inequality yields the conclusion. Normal Functions an d Normal Families 155

Lemma 2. (Bureau [1]) Let b1 , b2 , and b3 be positive numbers and U (r) a nonnegative, increasing and on an interval [ r0 , R ) , R < ∞ . If

1 U (r) ≤ b + b log+ + b log+ U (ρ) 1 2 ρ − r 3

for any r0 < r < ρ < R , then

1 U (r) ≤ B + B log+ 1 2 R − r

for r0 ≤ r < R , where B1 and B2 depend on bi (1 = 1, 2, 3) only.

Lemma 3. (see Hiong [4]) If f (z) is meromorphic in a disk z < R such that f (0) ≠ 0 , ∞ , then, for o < r < ρ < R ,

⎛ (k) ⎞ ⎧ f ⎪ + + 1 + 1 + 1 m ⎜ r, ⎟ ≤ Ck ⎨ 1 + log log + log + log ⎝ f ⎠ ⎩⎪ f (0) r ρ − r ⎫ + log+ ρ + log+ T(ρ, f ) ⎬, ⎭

where Ck is a constant depending only on k.

The following is the wellknown Zalcman’s lemma [12].

Lemma 4. Let F be a family of functions meromorphic in a domain D. If F is not normal at z0 ∈ D , then there exist a sequence of points zn ∈ D , zn → z0 , a sequence of positive numbers ρn → 0 , and a sequence of functions fn ∈ F such that

gn (ζ ) = fn ( zn + ρnζ ) → g(ζ ) locally uniformly with respect to the spherical , where g is a non-constant meromorphic function on C.

3. Proof of Theorem 1

Proof. Suppose that F is not uniformly normal in D. Then, we can find

fn ∈ F , zn ∈ D , such that

156 Y. Xu

2 gn(z) = fn (zn + (1 − zn ) z ) satisfies

# 2 # gn (0) = (1 − zn ) fn (zn) → ∞

as n → ∞ . It follows that { gn (z)} is not normal at z = 0 . We distinguish two cases:

z (1) gn = gn′ for every n ∈ N then gn (z) = Cne , which is normal at z = 0.

(2) Consider the case that gn and gn′ are not identical. By Lemma 4, there exist a

subsequence gn (without loss generality, we may assume gn ), a sequence

ηn ∈ D,ηn → 0 , and a positive sequence ρn → 0 such that

2 2 Gn(ζ ) = gn(ηn + ρnζ ) = fn (zn + (1 − zn ) ηn + ( 1 − zn ) ρnζ ) converges uniformly to a non-constant entire function G(ζ ) on each compact subset of C. Thus, for any positive integer k,

k (k) 2 k (k) 2 2 Gn (ζ ) = ()1 − zn ρn fn ()zn + (1 − zn ) ηn + (1 − zn ) ρnζ . (3.1)

We claim that G(ζ ) is not a polynomial of degree less than 3. Indeed, if G(ζ ) is a polynomial, then there exists a point ζ 0 such that G(ζ 0 ) = a . By Hurwitz’ theorem, there is a sequence ζ n → ζ 0 such that

2 2 Gn(ζ n) = gn(ηn + ρnζ n) = fn (zn + (1 − zn ) ηn + (1 − zn ) ρnζ n ) = a for n sufficiently large. It follows from the hypotheses on F that

2 2 fn′ ( zn + (1 − zn )ηn + ( 1 − zn ) ρn ζ n)

2 2 = fn′′ (zn + ( 1 − zn )ηn + ( 1 − zn ) ρn ζ n) = a for n sufficiently large. On the other hand, by (3.1), we have

2 2 2 Gn′(ζ n) = (1 − zn ) ρn fn′ (zn + (1 − zn ) ηn + ( 1 − zn ) ρnζ n) → G′(ζ 0) ,

2 2 2 2 2 Gn′′(ζ n) = ()1 − zn ρn fn′′()zn + (1 − zn ) ηn + (1 − zn ) ρnζ n → G′′(ζ 0). Normal Functions an d Normal Families 157

Thus G′(ζ 0 ) = G′′(ζ 0 ) = 0.

Choose ζ1 with

G(ζ1) ≠ 0, a ; G′(ζ1) ≠ 0; G′′(ζ1) ≠ 0. (3.2)

Then 1

2 2 2 ρn ()1 − zn

2 2 fn ( zn + (1 − zn )ηn + (1 − zn ) ρnζ1 ) − a × 2 2 2 2 fn′′ ()zn + (1 − zn )ηn + (1 − zn ) ρnζ1 − fn′ ()zn + (1 − zn )ηn + (1 − zn ) ρnζ1

G (ζ ) → 1 , G′′(ζ1)

1

2 ρ n ()1 − zn

2 2 2 2 ()fn ()zn + (1 − zn )ηn + (1 − zn )ρnζ1 − a ( fn′ ( zn + (1 − zn ) ηn + (1 − zn ) ρnζ1 ) − a ) × 2 2 fn′′ ()zn + (1 − zn )ηn + (1 − zn ) ρnζ1

G′(ζ )(G(ζ ) − a) → 1 1 . G′′(ζ 1)

On the other hand, we claim that there are only finitely many fn such that ψ ( fn ) ≡ 0. Indeed, suppose that there is a subsequence { f } ⊂ { f } such that ψ ( f ) ≡ 0. n j n n j Then

2 2 2 f ′ ⎛z + (1 − z )η + (1 − z ) 2 ρ ζ ⎞ + f ′′ ⎛z + (1 − z ) η + (1 − z ) ρ ζ ⎞ n j ⎜ n j n j n j nj n j ⎟ n j ⎜ n j n j n j n j n j ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ 2 2 ⎞ f ⎜z + (1 − z )η + (1 − z ) ρ ζ ⎟ − a n j ⎝ n j n j n j n j n j ⎠

2 2 2 f ′′ ⎛z + (1 − z )η + (1 − z ) ρ ζ ⎞ n j ⎜ n j n j n j n j n j ⎟ = ⎝ ⎠ , ⎛ 2 2 ⎞ f ′ ⎜z + (1 − z )η + (1 − z ) ρ ζ ⎟ − a n j ⎝ n j n j n j n j n j ⎠

158 Y. Xu and thus

⎛ 2 ⎞ ⎛ 2 ⎞ ρ 1 − z G′ (ζ ) + G′′ (ζ ) 2ρ 1 − z G′′ (ζ ) n j ⎜ n j ⎟ n j n j n j ⎜ n j ⎟ n j ⎝ ⎠ = ⎝ ⎠ . G (ζ ) − a 2 n j ⎛ ⎞ Gn′ (ζ ) − aρn ⎜1 − zn ⎟ j j ⎝ j ⎠

Letting j → ∞, we get G′′(ζ ) ≡ 0, a contradiction. Then we may assume that

ψ ( fn ) ≡/ 0, for all n. Thus

2 2 2 G′′(ζ ) ρnψ n()zn + (1 − zn )ηn + (1 − zn ) ρnζ1 → , (3.3) G(ζ1) − a

where ψ n = ψ ( f n ). So we have

2 2 fn()zn + (1 − zn )ηn + (1 − zn ) ρnζ1 − a log 2 2 2 2 fn′′ ()zn + (1 − zn )ηn + (1 − zn ) ρnζ1 − fn′ ()zn + (1 − zn )ηn + (1 − zn ) ρnζ1 → −∞, (3.4)

2 2 2 2 ()fn()zn + (1 − zn )ηn + (1 − zn ) ρnζ1 − a ( fn′ ( zn + (1 − zn )ηn + (1 − zn ) ρnζ1 ) − a ) log 2 2 fn′′ ()zn + (1 − zn )ηn + (1 − zn ) ρnζ1

→ −∞ , (3.5)

1 and log → − ∞, (3.6) 2 2 ψn ()zn + (1 − zn )ηn + (1 − zn ) ρnζ1

as n → ∞. For n = 1, 2,", set

2 2 Pn(z) = fn ( zn + (1 − zn )ηn + (1 − zn ) ρn ζ1 + z ).

Let n be sufficiently large. Then Pn is defined and holomorphic on the disk 1 0 < z < 2 , since

2 2 zn + (1 − zn )ηn + (1 − zn ) ρn ζ1 → 0. Normal Functions an d Normal Families 159

By (3.2) and (3.3), we have

Pn(0) = Gn(ζ1) → G(ζ1) ≠ 0, a, (3.7)

1 Pn′(0) = Gn′ ()ζ1 → ∞, (3.8) 2 (1 − zn ) ρn

1 Pn′′(0) = Gn′′()ζ1 → ∞, (3.9) 2 2 2 (1 − zn ) ρn

2 2 ψ ()Pn(0) = ψn ( zn + (1 − zn )ηn + (1 − zn ) ρnζ1 ) → ∞. (3.10)

Therefore, by (3.7)−(3.10) we may apply Lemma 1 to Pn(z) , and using (3.4), (3.5) and (3.6) we obtain

T ()r, Pn ≤ LD (r, Pn ), for sufficiently large n. Hence by Lemma 2 and Lemma 3, we get

⎛ 1 ⎞ T ⎜ , Pn ⎟ ≤ M, ⎝ 4 ⎠

where M is a constant independent of n. It follows that fn(z) are bounded for 1 sufficiently large n and z < 8 . But, from

2 2 2 2 2 ()1 − zn ρn fn′′ ()zn + (1 − zn )ηn + (1 − zn ) ρnζ1 = Gn′′(ζ1) → G′′ (ζ1) ≠ 0

1 we know that fn (z) cannot be bounded in z < 8 . We arrive at a contradiction. This completes the proof of Theorem 1.

Acknowledgement. Supported in part by NSF of China (Grant 10171047) and NSF of Education Department of Jiangsu Province (03KJB110058 ).

References

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Keywords: , normal family, normal function, uniformly normal family.