Normal Families of Meromorphic Functions Concerning Higher Derivative and Shared Values

Abstract

We study the normal families related to a Hayman conjecture of higher derivative and concerning shared values and get two normal criteria. Our results improve the related theorems which were obtained independently, respectively by Fang and Yuan (2001), Yuan et al. ((2011) and (2012)), Wang et al. (2011), and Qiu et al. (2012). Meanwhile, some examples are given to show the sharpness of our results.

1. Introduction and Main Results

Let 𝑓(𝑧) and 𝑔(𝑧) be two nonconstant meromorphic functions in a domain 𝐷⊆ℂ, and let 𝑎 be a finite complex value. We say that 𝑓 and 𝑔 share 𝑎 CM (or IM) in 𝐷 provided that 𝑓−𝑎 and 𝑔−𝑎 have the same zeros counting (or ignoring) multiplicity in 𝐷. When 𝑎=∞, the zeros of 𝑓−𝑎 mean the poles of 𝑓 (see [1]). It is assumed that the reader is familiar with the standard notations and the basic results of Nevanlinna’s value-distribution theory ([1] or [2–4]).

It is very interesting to find normality criteria from the point of view of shared values. In this area, Schwick [5] first proved an interesting result that a family of meromorphic functions in a domain is normal in which every function shares three distinct finite complex numbers with their first derivative. And later, more results about normality criteria concerning shared values have emerged; for instance, see [6–8]. In recent years, this subject has attracted the attention of many researchers worldwide.

We now first introduce a normality criterion related to a Hayman normal conjecture [9].

Theorem 1. Let ℱ be a family of holomorphic (meromorphic) functions defined in a domain, 𝑛∈ℕ, 𝑎≠0, and 𝑏∈ℂ. If 𝑓^′(𝑧)+𝑎𝑓^𝑛(𝑧)−𝑏 does not vanish in 𝐷 for each function 𝑓(𝑧)∈ℱ and 𝑛≥2(𝑛≥3), then ℱ is normal in 𝐷.

The results for the holomorphic case are due to Drasin [10] for 𝑛≥3, Pang [11] for 𝑛=3, Chen and Fang [12] for 𝑛=2, Ye [13] for 𝑛=2, and Chen and Gu [14] for the generalized result with 𝑎 and 𝑏 replaced by meromorphic functions. The results for the meromorphic case are due to Li [15], Li [16], and Langley [17] for 𝑛≥5, Pang [11] for 𝑛=4, Chen and Fang [12] for 𝑛=3, and Zalcman [18] for 𝑛=3, obtained independently.

When 𝑛=2 and ℱ is meromorphic, Theorem 1 is not valid in general. Fang and Yuan [19] gave an example to show this and got a special result below.

Example 2. The family of meromorphic functions ℱ={𝑓_𝑗(𝑧)=𝑗𝑧/(𝑗𝑧−1)²∶𝑗=1,2,…,} is not normal in 𝐷={𝑧∶|𝑧|<1}. This is deduced by 𝑓^#_𝑗(0)=𝑗→∞, as 𝑗→∞ and Marty’s criterion [2], although for any 𝑓_𝑗(𝑧)∈ℱ, 𝑓^′_𝑗+𝑓²_𝑗=𝑗(𝑗𝑧−1)⁻⁴≠0.

Here 𝑓^#(𝜉) denotes the spherical derivative 𝑓^#(𝜉)=𝑓^′(𝜉)1+𝑓(𝜉)². (1)

Theorem 3. Let ℱ be a family of meromorphic functions in a domain 𝐷, and 𝑎≠0, 𝑏∈ℂ. If 𝑓^′(𝑧)+𝑎𝑓²(𝑧)−𝑏 does not vanish in 𝐷 and the poles of 𝑓(𝑧) are of multiplicity ≥3 for each 𝑓(𝑧)∈ℱ, then ℱ is normal in 𝐷.

In 2008, by the ideas of shared values, Zhang [8] proved the following.

Theorem 4. Let ℱ be a family of meromorphic (holomorphic) functions in 𝐷, let 𝑛 be a positive integer and let 𝑎, 𝑏 be two finite complex numbers such that 𝑎≠0. If 𝑛≥4(𝑛≥2) and for every pair of functions 𝑓 and 𝑔 in ℱ, 𝑓^′−𝑎𝑓^𝑛 and 𝑔^′−𝑎𝑔^𝑛 share the value 𝑏 IM, then ℱ is normal in 𝐷.

Example 5 (see [8]). The family of meromorphic functions ℱ={𝑓_𝑗(𝑧)=1/𝑗(𝑧−1/𝑗)∶𝑗=1,2,…,} is not normal in 𝐷={𝑧∶|𝑧|<1}. Obviously 𝑓^′_𝑗−𝑓³_𝑗=−𝑧/𝑗(𝑧−1/𝑗)³. So for each pair of 𝑚, 𝑗, 𝑓^′_𝑗−𝑓³_𝑗 and 𝑓^′_𝑚−𝑓³_𝑚 share the value 0 in 𝐷, but ℱ is not normal at the point 𝑧=0, since 𝑓^#_𝑗(0)=2(𝑗)³/(1+𝑗)→∞, as 𝑗→∞.

Remark 6. Example 5 shows that Theorem 4 is not valid when 𝑛=3, and the condition 𝑛=4 is best possible for meromorphic case.

In 2011, Yuan et al. [20] and Wang et al. [21] proved the following theorems, independently, respectively.

Theorem 7 (see [20, 21]). Let ℱ be a family of meromorphic functions in 𝐷 and 𝑎, 𝑏 two finite complex numbers such that 𝑎≠0. Suppose that each 𝑓∈ℱ has no simple pole. If 𝑓^′−𝑎𝑓³ and 𝑔^′−𝑎𝑔³ share the value 𝑏 IM for every pair of functions 𝑓 and 𝑔 in ℱ, then ℱ is normal in 𝐷.

Theorem 8 (see [20]). Let ℱ be a family of meromorphic functions in 𝐷 and 𝑎 and 𝑏 two finite complex numbers such that 𝑎≠0. Suppose that each 𝑓∈ℱ admits zeros of multiple and the poles of multiplicity at least 3. If 𝑓^′−𝑎𝑓² and 𝑔^′−𝑎𝑔² share the value 𝑏 IM for every pair of functions 𝑓 and 𝑔 in ℱ, then ℱ is normal in 𝐷.

Lately, Yuan et al. [22] and Qiu et al. [23] studied this result, independently, respectively, in which the derivative 𝑓^′ was replaced by 𝑘th derivative 𝑓^(𝑘), and they got the following results.

Theorem 9 (see [22]). Let ℱ be a family of meromorphic functions in 𝐷, and let 𝑘(≥2) and 𝑛(≥𝑘+2) be two positive integers. Let 𝑎(≠0) and 𝑏 be two finite complex numbers. If(i)𝑓^(𝑘)−𝑎𝑓^𝑛 and 𝑔^(𝑘)−𝑎𝑔^𝑛 share 𝑏 IM in 𝐷 for every pair of functions 𝑓 and 𝑔 in ℱ,(ii)𝑓 has no simple pole and no zero of multiplicity less than 𝑘 in 𝐷 for every function 𝑓∈ℱ, then ℱ is normal in 𝐷.

Theorem 10 (see [22, 23]). Let ℱ be a family of meromorphic functions in 𝐷, 𝑘(≥2) and 𝑛(≥𝑘+3) be two positive integers. Let 𝑎(≠0) and 𝑏 be two finite complex numbers. If(i)𝑓^(𝑘)−𝑎𝑓^𝑛 and 𝑔^(𝑘)−𝑎𝑔^𝑛 share 𝑏 IM in 𝐷 for every pair of functions 𝑓 and 𝑔 in ℱ,(ii)𝑓 has no zero of multiplicity less than 𝑘 in 𝐷 for every function 𝑓∈ℱ, then ℱ is normal in 𝐷.

It is natural to ask whether the condition 𝑛≥𝑘+2 or 𝑛≥𝑘+3 in the previous theorems can be reduced. In this paper, we study this problem and get the following results.

Theorem 11 (main theorem). Let 𝐷 be a domain in ℂ and let ℱ be a family of meromorphic functions in 𝐷. Let 𝑘,𝑛,𝑑∈ℕ⁺,𝑛≥3,𝑑≥(𝑘+1)/(𝑛−2) and let 𝑎, 𝑏 be two finite complex numbers with 𝑎≠0. Suppose that every 𝑓∈ℱ has all its zeros of multiplicity at least 𝑘 and all its poles of multiplicity at least 𝑑. If 𝑓^(𝑘)−𝑎𝑓^𝑛 and 𝑔^(𝑘)−𝑎𝑔^𝑛 share the value 𝑏 IM for every pair of functions (𝑓,𝑔) of ℱ, then ℱ is a normal family in 𝐷.

Remark 12. When 𝑛≥𝑘+2 or 𝑛≥𝑘+3, we have 𝑑=1 or 2, respectively. It follows that Theorem 11 generalizes Theorems 4, 7, 9, and 10.

Theorem 13 (main theorem). Let 𝐷 be a domain in ℂ and let ℱ be a family of meromorphic functions in 𝐷. Let 𝑘∈ℕ⁺ and 𝑎, 𝑏 be two finite complex numbers with 𝑎≠0. Suppose that every 𝑓∈ℱ has all its zeros of multiplicity at least 𝑘+1 and all its poles of multiplicity at least 𝑘+2. If 𝑓^(𝑘)−𝑎𝑓² and 𝑔^(𝑘)−𝑎𝑔² share the value 𝑏 IM for every pair of functions (𝑓,𝑔) of ℱ, then ℱ is a normal family in 𝐷.

Remark 14. When 𝑘=1, it follows that Theorem 13 generalizes Theorem 8.

Example 15 (see [23]). Let 𝑛,𝑘≥2 be two positive integers and 𝑎 a nonzero complex constant. The family of meromorphic functions is ℱ={𝑓_𝑗(𝑧)=𝑗𝑧^𝑘−1∶𝑗=1,2,…,}, 𝐷={𝑧∶|𝑧|<1}. Obviously, for each pair of 𝑚, 𝑗, 𝑓_𝑗^(𝑘)−𝑎𝑓^𝑛_𝑗 and 𝑓_𝑚^(𝑘)−𝑎𝑓^𝑛_𝑚 share the value 0 in 𝐷, but ℱ is not normal.

Example 16. Let 𝑘 be a positive integer and 𝑎,𝑏 two nonzero complex constants such that 𝑏_𝑗=𝑎/(−1)^𝑘((2𝑘)!/𝑘!)𝑗^𝑘. The family of meromorphic functions is ℱ={𝑓_𝑗(𝑧)=1/(𝑗𝑧−𝑏_𝑗)^𝑘+1∶𝑗=1,2,…}, 𝐷={𝑧∶|𝑧|<1}. Obviously, for each pair of 𝑚,𝑗,𝑓_𝑗^(𝑘)−𝑎𝑓²_𝑗 and 𝑓_𝑚^(𝑘)−𝑎𝑓²_𝑚 share the value 0 in 𝐷, but ℱ is normal since 𝑓^#(0)→∞ as 𝑗→∞.

Remark 17. Example 15 shows that the condition that 𝑓(𝑧) admits zeros of multiplicity at least 𝑘 is best in Theorem 11. For the case 𝑘=1, 𝑛=3, Example 5 shows that the condition that 𝑓(𝑧) admits poles of multiplicity at least 𝑑 is sharp in Theorem 11. For the case 𝑛=2, Example 16 shows that the condition that 𝑓(𝑧) admits poles of multiplicity at least 𝑘+2 is sharp in Theorem 13. For the case 𝑘=1, 𝑛=2, Example 2 shows that the condition that 𝑓(𝑧) admits zeros of multiplicity at least 𝑘+1 in Theorem 13 is sharp.

2. Preliminary Lemmas

In order to prove our results, we need the following lemmas. The first is the extended version Zalcman’s [24] concerning normal families.

Lemma 1 (see [25]). Let ℱ be a family of meromorphic functions on the unit disc satisfying all zeros of functions in ℱ having multiplicity ≥p and all poles of functions in ℱ having multiplicity ≥q. Let 𝛼 be a real number satisfying −𝑞<𝛼<𝑝. Then ℱ is not normal at 0 if and only if there exist (a)a number 0<𝑟<1;(b)points 𝑧_𝑛𝑤𝑖𝑡ℎ|𝑧_𝑛|<𝑟;(c)functions 𝑓_𝑛∈ℱ;(d)positive numbers 𝜌_𝑛→0such that 𝑔_𝑛(𝜁)∶=𝜌_𝑛^−𝛼𝑓_𝑛(𝑧_𝑛+𝜌_𝑛𝜁) converges spherically uniformly on each compact subset of ℂ to a nonconstant meromorphic function 𝑔(𝜁), whose all zeros have multiplicity ≥p and all poles have multiplicity ≥q and order is at most 2.

Lemma 2. Let 𝑓(𝑧) be a meromorphic function such that 𝑓^(𝑘)(𝑧)≢0 and 𝑐∈ℂ⧵{0}, 𝑘,𝑛,𝑑∈ℕ⁺ with 𝑛≥3, 𝑑≥(𝑘+1)/(𝑛−2). If all zeros of 𝑓 are of multiplicity at least 𝑘 and all poles of 𝑓 are of multiplicity at least 𝑑, then 𝑇≤1𝑟,𝑓𝑘𝑁1𝑟,𝑓+𝑁1𝑟,𝑓^(𝑘)−𝑐𝑓^𝑛+𝑆𝑟,𝑓, (2) where 𝑆(𝑟,𝑓)=𝑜(𝑇(𝑟,𝑓)), as 𝑟→∞, possibly outside a set with finite linear measure.

Proof. Set Φ(𝑧)𝑓∶=^(𝑘)(𝑧)𝑐𝑓^𝑛(𝑧). (3)
Since 𝑓^(𝑘)(𝑧)≢0, we have Φ(𝑧)≢0. Thus, 𝑓^𝑛(𝑧)=𝑓^(𝑘)(𝑧)(𝑧).𝑐Φ (4) Hence, 𝑛𝑚𝑟,𝑓=𝑚𝑟,𝑓^𝑛𝑓≤𝑚𝑟,^(𝑘)Φ+log⁺1|𝑐|1≤𝑚𝑟,Φ+𝑚𝑟,𝑓^(𝑘)+log⁺1|𝑐|1≤𝑚𝑟,Φ𝑓+𝑚𝑟,^(𝑘)𝑓+𝑚𝑟,𝑓+log⁺1|𝑐|. (5) So that ()𝑚1𝑛−1𝑟,𝑓≤𝑚𝑟,Φ𝑓+𝑚𝑟,^(𝑘)𝑓+log⁺1|𝑐|. (6)
On the other hand, (4) gives 𝑛𝑁𝑟,𝑓≤𝑁𝑟,𝑓^𝑛𝑓=𝑁𝑟,^(𝑘)Φ≤𝑁𝑟,𝑓^(𝑘)1+𝑁𝑟,Φ−𝑁𝑟,Φ=𝑓^(𝑘)=0, (7) where 𝑁(𝑟,Φ=𝑓^(𝑘)=0) denotes the counting function of zeros of both Φ and 𝑓^(𝑘).
We obtain 𝑛𝑁𝑟,𝑓≤𝑁𝑟,𝑓+𝑘𝑁1𝑟,𝑓+𝑁𝑟,Φ−𝑁𝑟,Φ=𝑓^(𝑘),()𝑁=0𝑛−1𝑟,𝑓≤𝑘𝑁1𝑟,𝑓+𝑁𝑟,Φ−𝑁𝑟,Φ=𝑓^(𝑘)=0. (8) By (4), we have 𝑁𝑟,𝜑+𝑁1𝑟,𝜑≤𝑁1𝑟,𝑓+𝑁𝑟,𝑓+𝑁𝑟,Φ=𝑓^(𝑘)=0. (9) From (6)~(9), we obtain (𝑛−1)𝑇𝑟,𝑓≤𝑘𝑁1𝑟,𝑓+𝑇𝑟,Φ−𝑁𝑟,Φ=𝑓^(𝑘)=0+𝑆𝑟,𝑓≤𝑘𝑁𝑟,𝑓+𝑇(𝑟,Φ)−𝑁𝑟,Φ=𝑓^(𝑘)=0+𝑆𝑟,𝑓≤𝑘𝑁𝑟,𝑓+𝑁1𝑟,Φ+𝑁(𝑟,Φ)+𝑁1𝑟,−Φ−1𝑁𝑟,Φ=𝑓^(𝑘)≤()=0+𝑆𝑟,𝑓𝑘+1𝑁𝑟,𝑓+𝑁1𝑟,𝑓+𝑁1𝑟,𝑓^(𝑘)−𝑐𝑓^𝑛+𝑆𝑟,𝑓. (10)
Since all zeros and poles of 𝑓 are multiplicities at least 𝑘 and 𝑑, we get 𝑁≤1𝑟,𝑓𝑑𝑁≤1𝑟,𝑓𝑑𝑇𝑟,𝑓≤𝑛−2𝑇,𝑘+1𝑟,𝑓𝑁1𝑟,𝑓≤1𝑘𝑁1𝑟,𝑓. (11) So that 𝑇≤1𝑟,𝑓𝑘𝑁1𝑟,𝑓+𝑁1𝑟,𝑓^(𝑘)−𝑐𝑓^𝑛+𝑆𝑟,𝑓. (12)

Lemma 3. Let 𝑓(𝑧) be a transcendental meromorphic function such that 𝑓^(𝑘)(𝑧)≢0. Let 𝑘∈ℕ⁺ and 𝑐∈ℂ⧵{0}. If all zeros of 𝑓 are of multiplicity at least 𝑘+1 and all poles of 𝑓 are of multiplicity at least 𝑘+2, then 𝑓^(𝑘)−𝑐𝑓² has infinitely many zeros.

Proof. Suppose that 𝑓^(𝑘)−𝑐𝑓² has only finitely many zeros; then 𝑁(𝑟,1/(𝑓^(𝑘)−𝑐𝑓²))=𝑆(𝑟,𝑓). Clearly, an arbitrary zero of 𝑓 is a zero of 𝑓^(𝑘)−𝑐𝑓² since all zeros of 𝑓 are of multiplicity at least 𝑘+1; then we can deduce that 𝑓 has only finite zeros, so 𝑁(𝑟,1/𝑓)=𝑂(log𝑟)=𝑆(𝑟,𝑓).
Set Φ(𝑧)𝑓∶=^(𝑘)(𝑧)𝑐𝑓²(𝑧). (13) Similarly, with the proof of Lemma 2, we can get 𝑇𝑟,𝑓≤(𝑘+1)𝑁𝑟,𝑓+𝑁1𝑟,𝑓+𝑁1𝑟,𝑓^(𝑘)−𝑐𝑓²+𝑆𝑟,𝑓. (14)
Since all poles of 𝑓 are multiplicities at least 𝑘+2, we obtain 𝑁≤1𝑟,𝑓𝑁≤1𝑘+2𝑟,𝑓𝑇,𝑘+2𝑟,𝑓 (15) so that 𝑇𝑟,𝑓≤(𝑘+2)𝑁1𝑟,𝑓+(𝑘+2)𝑁1𝑟,𝑓^(𝑘)−𝑐𝑓²+𝑆𝑟,𝑓=𝑆𝑟,𝑓. (16) This is contradicting with the fact that 𝑓 is transcendental.
Hence, Lemma 3 is proved completely.

Lemma 4. Let 𝑓(𝑧) be a nonconstant rational function such that 𝑓^(𝑘)(𝑧)≢0. Let 𝑎∈ℂ⧵{0}, and 𝑘,𝑛,𝑚∈ℕ⁺ with 𝑛≥2 and 𝑚≥(𝑘+2)/(𝑛−1). If 𝑓≠0 and all poles of 𝑓 are of multiplicity at least 𝑚, then 𝑓^(𝑘)−𝑎𝑓^𝑛 has at least two zeros.

Proof. Suppose, to the contrary, that 𝑓^(𝑘)−𝑎𝑓^𝑛 has at most one zero.
Since 𝑓≠0, we get that 𝑓 is a rational but not a polynomial.
Case  1. If 𝑓^(𝑘)−𝑎𝑓^𝑛 has only zero 𝑧₀ with multiplicity 𝑙, set 𝑓(𝑧)=𝐴𝑧−𝑧₁^𝛽1𝑧−𝑧₂^𝛽2⋯𝑧−𝑧_𝑡^𝛽𝑡, (17) where 𝐴 is a nonzero constant and 𝛽_𝑖≥(𝑘+2)/(𝑛−1)(𝑖=1,2,…,𝑡).
For the sake of simplicity, we denote 𝛽₁+𝛽₂+⋯+𝛽_𝑡=𝑞. (18)
From (17), we have 𝑓^(𝑘)=𝑔(𝑧)𝑧−𝑧₁^𝛽1+𝑘𝑧−𝑧₂^𝛽2+𝑘⋯𝑧−𝑧_𝑡^{𝛽𝑡+𝑘}, (19) where 𝑔(𝑧) is a polynomial such that deg(𝑔(𝑧))≤𝑘(𝑡−1).
From (17) and (19), we get 𝑓^(𝑘)−𝑎𝑓^𝑛=𝑔(𝑧)𝑧−𝑧₁^𝛽1+𝑘𝑧−𝑧₂^𝛽2+𝑘⋯𝑧−𝑧_𝑡^{𝛽𝑡+𝑘}−𝑎𝐴^𝑛𝑧−𝑧₁^𝑛𝛽1𝑧−𝑧₂^𝑛𝛽2⋯𝑧−𝑧_𝑡^𝑛𝛽𝑡=𝑔(𝑧)𝑧−𝑧₁^{(𝑛−1)𝛽1−𝑘}𝑧−𝑧₂^{(𝑛−1)𝛽2−𝑘}⋯𝑧−𝑧_𝑡^{(𝑛−1)𝛽𝑡−𝑘}−𝑎𝐴^𝑛×𝑧−𝑧₁^𝑛𝛽1𝑧−𝑧₂^𝑛𝛽2⋯𝑧−𝑧_𝑡^{𝑛𝛽𝑡−1}. (20)
By the assumption that 𝑓^(𝑘)−𝑎𝑓^𝑛 has exactly one zero 𝑧₀ with multiply 𝑙, we have 𝑓^(𝑘)−𝑎𝑓^𝑛=𝐶𝑧−𝑧₀^𝑙𝑧−𝑧₁^𝑛𝛽1𝑧−𝑧₂^𝑛𝛽2⋯𝑧−𝑧_𝑡^𝑛𝛽𝑡, (21) where 𝐶 is a nonzero constant. Thus, 𝐶𝑧−𝑧₀^𝑙≡𝑔(𝑧)𝑧−𝑧₁^{(𝑛−1)𝛽1−𝑘}⋯𝑧−𝑧_𝑡^{(𝑛−1)𝛽𝑡−𝑘}−𝑎𝐴^𝑛. (22)
Differentiating (22), we obtain 𝐶𝑙𝑧−𝑧₀^𝑙−1≡𝑧−𝑧₁^{(𝑛−1)𝛽1−𝑘−1}⋯𝑧−𝑧_𝑡^{(𝑛−1)𝛽𝑡−𝑘−1}×𝑔^′(𝑧)𝑧−𝑧₁⋯𝑧−𝑧_𝑡+𝑔(𝑧)^𝑡_𝑖=1(𝑛−1)𝛽_𝑖−𝑘^𝑡_{𝑗=1,𝑗≠𝑖}𝑧−𝑧_𝑗. (23)
For the sake of simplicity, we denote 𝑔₁(𝑧)=𝐶𝑙𝑧−𝑧₀^𝑙−1,𝑔₂(𝑧)=𝑧−𝑧₁^{(𝑛−1)𝛽1−𝑘−1}⋯𝑧−𝑧_𝑡^{(𝑛−1)𝛽𝑡−𝑘−1}×𝑔^′(𝑧)𝑧−𝑧₁⋯𝑧−𝑧_𝑡+𝑔(𝑧)^𝑡_𝑖=1(𝑛−1)𝛽_𝑖−𝑘^𝑡_{𝑗=1,𝑗≠𝑖}𝑧−𝑧_𝑗. (24) Hence, 𝑔₁(𝑧)≡𝑔₂(𝑧). (25)
Since (𝑛−1)𝛽_𝑖−𝑘−1≥1, we have 𝑔₂(𝑧_𝑖)=0. But 𝑔₁(𝑧_𝑖)≠0(𝑖=1,2,…,𝑡), a contradiction.
Case  2. If 𝑓^(𝑘)−𝑎𝑓^𝑛 has no zeros, then 𝑙=0 for (21). We have 𝑓^(𝑘)−𝑎𝑓^𝑛=𝐶𝑧−𝑧₁^𝑛𝛽1𝑧−𝑧₂^𝑛𝛽2⋯𝑧−𝑧_𝑡^𝑛𝛽𝑡, (26) where 𝐶 is a nonzero constant. Thus, 𝐶≡𝑔(𝑧)𝑧−𝑧₁^{(𝑛−1)𝛽1−𝑘}⋯𝑧−𝑧_𝑡^{(𝑛−1)𝛽𝑡−𝑘}−𝑎𝐴^𝑛,𝑔(𝑧)𝑧−𝑧₁^{(𝑛−1)𝛽1−𝑘}⋯𝑧−𝑧_𝑡^{(𝑛−1)𝛽𝑡−𝑘}≡𝐶+𝑎𝐴^𝑛. (27)
Obviously, 𝑔(𝑧)(𝑧−𝑧₁)^{(𝑛−1)𝛽1−𝑘}⋯(𝑧−𝑧_𝑡)^{(𝑛−1)𝛽𝑡−𝑘} is not a constant, a contradiction.
Lemma 4 is proved.

Lemma 5. Let 𝑓(𝑧) be a rational function and 𝑎∈ℂ⧵{0}, and 𝑘,𝑛,𝑚∈ℕ⁺ with 𝑛≥2 and 𝑚≥(𝑘+2)/(𝑛−1). If all zeros of 𝑓 are of multiplicity at least 𝑘+1 and all poles of 𝑓 are of multiplicity at least 𝑚, then 𝑓^(𝑘)−𝑎𝑓^𝑛 has at least two distinct zeros.

Proof. Suppose, to the contrary, that 𝑓^(𝑘)−𝑎𝑓^𝑛 has at most one zero.
Case  I. When 𝑓 is a nonconstant polynomial, noting that all zeros of 𝑓 have multiplicity at least 𝑘+1, we know that 𝑓^(𝑘)−𝑎𝑓^𝑛 must have zeros. We claim that 𝑓 has exactly one zero. Otherwise, we can get that 𝑓^(𝑘)−𝑎𝑓^𝑛 has at least two zeros, which contradicts our assumption.
Set 𝑓(𝑧)=𝐵𝑧−𝑧₀^𝑠, (28) where 𝑠≥𝑘+1, 𝐵 is a nonzero constant. Then 𝑓^(𝑘)(𝑧)−𝑎𝑓^𝑛(𝑧)=𝐵𝑧−𝑧₀^𝑠−𝑘×𝑠(𝑠−1)⋯(𝑠−𝑘+1)−𝑎𝑧−𝑧₀^{(𝑛−1)𝑠+𝑘}. (29)
Since 𝑠−𝑘≥1, we obtain that 𝑠(𝑠−1)⋯(𝑠−𝑘+1)−𝑎(𝑧−𝑧₀)^𝑠+𝑘 has at least one zero which is not 𝑧₀ from (29). Therefore, 𝑓^(𝑘)−𝑎𝑓^𝑛 has at least two distinct zeros, a contradiction.
Case  II. When 𝑓 is rational but not a polynomial, we consider two cases.
Case  1. Suppose that 𝑓^(𝑘)−𝑎𝑓^𝑛 has only zero 𝑧₀ with multiplicity at least 𝑙. If 𝑓≠0, by Lemma 4, we get a contradiction. So 𝑓 has zeros, and then we can deduce that 𝑧₀ is the only zero of 𝑓. Otherwise, 𝑓^(𝑘)−𝑎𝑓^𝑛 has at least two distinct zeros, a contradiction.
We set 𝑓(𝑧)=𝐴𝑧−𝑧₀^𝑠𝑧−𝑧₁^𝛽1𝑧−𝑧₂^𝛽2⋯𝑧−𝑧_𝑡^𝛽𝑡, (30) where 𝐴 is a nonzero constant and 𝑠≥𝑘+1, 𝛽_𝑖≥𝑚≥((𝑘+2)/(𝑛−1))(𝑖=1,2,…,𝑡).
For the sake of simplicity, we denote 𝛽₁+𝛽₂+⋯+𝛽_𝑡=𝑞. (31)
From (31), we have 𝑓^(𝑘)=𝐴𝑧−𝑧₀^𝑠−𝑘𝑔(𝑧)𝑧−𝑧₁^𝛽1+𝑘𝑧−𝑧₂^𝛽2+𝑘⋯𝑧−𝑧_𝑡^{𝛽𝑡+𝑘}, (32) where 𝑔(𝑧) is a polynomial with deg(𝑔)≤𝑘(𝑠+𝑡+1).
From (30) and (32), we get 𝑓^(𝑘)−𝑎𝑓^𝑛=𝐴𝑧−𝑧₀^𝑠−𝑘𝑔(𝑧)𝑧−𝑧₁^𝛽1+𝑘𝑧−𝑧₂^𝛽2+𝑘⋯𝑧−𝑧_𝑡^{𝛽𝑡+𝑘}−𝐴^𝑛𝑧−𝑧₀^𝑛𝑠𝑧−𝑧₁^𝑛𝛽1𝑧−𝑧₂^𝑛𝛽2⋯𝑧−𝑧_𝑡^𝑛𝛽𝑡=𝐴𝑧−𝑧₀^𝑠−𝑘×𝑔(𝑧)𝑧−𝑧₁^{(𝑛−1)𝛽1−𝑘}𝑧−𝑧₂^{(𝑛−1)𝛽2−𝑘}⋯𝑧−𝑧_𝑡^{(𝑛−1)𝛽𝑡−𝑘}−𝑎𝐴^𝑛−1×𝑧−𝑧₀^{(𝑛−1)𝑠+𝑘}×𝑧−𝑧₁^𝑛𝛽1𝑧−𝑧₂^𝑛𝛽2⋯𝑧−𝑧_𝑡^{𝑛𝛽𝑡−1}. (33)
By assumption that 𝑓^(𝑘)−𝑎𝑓^𝑛 has exactly one zero 𝑧₀ with multiplicity 𝑙, we have 𝑓^(𝑘)−𝑎𝑓^𝑛=𝐵𝑧−𝑧₀^𝑙𝑧−𝑧₁^𝑛𝛽1𝑧−𝑧₂^𝑛𝛽2⋯𝑧−𝑧_𝑡^𝑛𝛽𝑡, (34) where 𝐵 is a nonzero constant. Thus, 𝐵𝑧−𝑧₀^𝑙≡𝐴𝑧−𝑧₀^𝑠−𝑘×𝑔(𝑧)𝑧−𝑧₁^{(𝑛−1)𝛽1−𝑘}⋯𝑧−𝑧_𝑡^{(𝑛−1)𝛽𝑡−𝑘}−𝑎𝐴^𝑛−1𝑧−𝑧₀^{(𝑛−1)𝑠+𝑘}. (35)Case  1.1. If 𝑙>𝑠−𝑘, from (35), we can deduce that 𝑧₀ is a zero of (𝑧−𝑧₁)^{(𝑛−1)𝛽1−𝑘}⋯(𝑧−𝑧_𝑡)^{(𝑛−1)𝛽𝑡−𝑘}, a contradiction.
Case  1.2. If 𝑙=𝑠−𝑘, from (35), it follows that 𝑔(𝑧)𝑧−𝑧₁^{(𝑛−1)𝛽1−𝑘}⋯𝑧−𝑧_𝑡^{(𝑛−1)𝛽𝑡−𝑘}−𝑎𝐴^𝑛−1𝑧−𝑧₀^{(𝑛−1)𝑠+𝑘}≡𝐵𝐴. (36)
Differentiating (36), we have 𝑧−𝑧₁^{(𝑛−1)𝛽1−𝑘−1}⋯𝑧−𝑧_𝑡^{(𝑛−1)𝛽𝑡−𝑘−1}×𝑔^′(𝑧)𝑧−𝑧₁⋯𝑧−𝑧_𝑡+𝑔(𝑧)^𝑡_𝑖=1(𝑛−1)𝛽_𝑖−𝑘^𝑡_{𝑗=1,𝑗≠𝑖}𝑧−𝑧_𝑗≡𝑎((𝑛−1)𝑠+𝑘)𝐴^𝑛−1𝑧−𝑧₀^{(𝑛−1)𝑠+𝑘−1}. (37)
For the sake of simplicity, we denote 𝑔₁(𝑧)=𝑧−𝑧₁^{(𝑛−1)𝛽1−𝑘−1}⋯𝑧−𝑧_𝑡^{(𝑛−1)𝛽𝑡−𝑘−1}×𝑔^′(𝑧)𝑧−𝑧₁⋯𝑧−𝑧_𝑡+𝑔(𝑧)^𝑡_𝑖=1(𝑛−1)𝛽_𝑖−𝑘^𝑡_{𝑗=1,𝑗≠𝑖}𝑧−𝑧_𝑗,𝑔₂(𝑧)=𝑎((𝑛−1)𝑠+𝑘)𝐴^𝑛−1𝑧−𝑧₀^{(𝑛−1)𝑠+𝑘−1}. (38) Thus, 𝑔₁(𝑧)≡𝑔₂(𝑧). (39)
Since (𝑛−1)𝛽_𝑖−𝑘−1≥1, we get 𝑔₁(𝑧_𝑖)=0, but 𝑔₂(𝑧_𝑖)≠0(𝑖=1,2,…,𝑡), a contradiction.
Case  2. If 𝑓^(𝑘)−𝑎𝑓^𝑛 has no zeros, then 𝑙=0 for (34). Similarly, as the proof of Case  1, we also have a contradiction.
The proof is completed.

3. Proofs of Theorems

Proof. In Theorem 11, suppose 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. Without loss of generality, we assume that 𝑧₀=0. By Lemma 1, there exist points 𝑧_𝑗→0, positive numbers 𝜌_𝑗→0, and functions 𝑓_𝑗∈ℱ such that 𝑔_𝑗(𝜉)=𝜌_𝑗^{𝑘/(𝑛−1)}𝑓_𝑗𝑧_𝑗+𝜌_𝑗𝜉⟹𝑔(𝜉) (40) locally uniformly with respect to the spherical metric, where 𝑔 is a nonconstant meromorphic function in ℂ and whose poles and zeros are of multiplicity at least 𝑑 and 𝑘, respectively. Moreover, the order of 𝑔 is at most 2.
From (40), we know that 𝑔_𝑗^(𝑘)(𝜉)=𝜌_𝑗^{𝑛𝑘/(𝑛−1)}𝑓_𝑗^(𝑘)𝑧_𝑗+𝜌_𝑗𝜉⟹𝑔^(𝑘)(𝜉),𝑔_𝑗^(𝑘)(𝜉)−𝑎𝑔^𝑛_𝑗(𝜉)−𝜌_𝑗^{𝑛𝑘/(𝑛−1)}𝑏=𝜌_𝑗^{𝑛𝑘/(𝑛−1)}𝑓_𝑗^(𝑘)𝑧_𝑗+𝜌_𝑗𝜉−𝑎𝑓^𝑛_𝑗𝑧_𝑗+𝜌_𝑗𝜉−𝑏⟹𝑔^(𝑘)(𝜉)−𝑎𝑔^𝑛(𝜉) (41)(42) also locally uniformly with respect to the spherical metric.
Case  1. If 𝑘=1, by Theorems 4 and 7, the Theorem 11 assumes.
Case  2. If 𝑘=2, 𝑛≥4, by Theorems 9 and 10, the Theorem 11 assumes.
Case  3. If 𝑘=2, 𝑛=3 or 𝑘≥3, 3≤𝑛≤𝑘+1 (If 𝑛≥𝑘+2, by Theorems 9 and 10, we can get Theorem 11).
If 𝑔^(𝑘)(𝜉)−𝑎𝑔^𝑛(𝜉)≡0, since all poles of 𝑔 have multiplicity at least 𝑑, we have 𝑛𝑇𝑟,𝑔=𝑇𝑟,𝑔^𝑛=𝑇𝑟,𝑔^(𝑘)+𝑂(1)=𝑚𝑟,𝑔^(𝑘)+𝑁𝑟,𝑔^(𝑘)+𝑂(1)≤𝑚𝑟,𝑔+𝑁𝑟,𝑔+𝑘𝑁+𝑘()𝑟,𝑔+𝑆𝑟,𝑔≤𝑇𝑟,𝑔𝑛−2𝑇≤()𝑇.𝑘+1𝑟,𝑔+𝑆𝑟,𝑔𝑛−1𝑟,𝑔+𝑆𝑟,𝑔 (43) Therefore, 𝑔(𝜉) is a constant, a contradiction. So 𝑔^(𝑘)(𝜉)−𝑎𝑔^𝑛(𝜉)≢0.
By Lemma 2, we have 𝑇≤1𝑟,𝑔𝑘𝑁1𝑟,𝑔+𝑁1𝑟,𝑔^(𝑘)−𝑎𝑔^𝑛≤1+𝑆𝑟,𝑔𝑘𝑇1𝑟,𝑔+𝑁1𝑟,𝑔^(𝑘)−𝑎𝑔^𝑛+𝑆𝑟,𝑔. (44) Then 11−𝑘𝑇𝑟,𝑔≤𝑁1𝑟,𝑔^(𝑘)−𝑎𝑔^𝑛𝑇≤1+𝑆𝑟,𝑔,𝑟,𝑔1+𝑘−1𝑁1𝑟,𝑔^(𝑘)−𝑎𝑔^𝑛+𝑆𝑟,𝑔. (45)(46) If 𝑔^(𝑘)(𝜉)−𝑎𝑔^𝑛(𝜉)≠0, then (46) gives that 𝑔(𝜉) is also a constant. Hence, 𝑔^(𝑘)(𝜉)−𝑎𝑔^𝑛(𝜉) is a nonconstant meromorphic function and has at least one zero.
Next we prove that 𝑔^(𝑘)(𝜉)−𝑎𝑔^𝑛(𝜉) has just a unique zero. On the contrary, let 𝜉₀ and 𝜉^∗₀ be two distinct zeros of 𝑔^(𝑘)(𝜉)−𝑎𝑔^𝑛(𝜉), and choose 𝛿(>0) small enough such that 𝐷(𝜉₀,𝛿)∩𝐷(𝜉^∗₀,𝛿)=𝜙, where 𝐷(𝜉₀,𝛿)={𝜉∶|𝜉−𝜉₀|<𝛿} and 𝐷(𝜉^∗₀,𝛿)={𝜉∶|𝜉−𝜉^∗₀|<𝛿}. From (42), by Hurwitz’s theorem, there exist points 𝜉_𝑗∈𝐷(𝜉₀,𝛿), 𝜉^∗_𝑗∈𝐷(𝜉^∗₀,𝛿) such that for sufficiently large 𝑗𝑓_𝑗^(𝑘)𝑧_𝑗+𝜌_𝑗𝜉_𝑗−𝑎𝑓^𝑛_𝑗𝑧_𝑗+𝜌_𝑗𝜉_𝑗𝑓−𝑏=0,_𝑗(𝑘)𝑧_𝑗+𝜌_𝑗𝜉^∗_𝑗−𝑎𝑓^𝑛_𝑗𝑧_𝑗+𝜌_𝑗𝜉^∗_𝑗−𝑏=0. (47)
By the hypothesis that for each pair of functions 𝑓 and 𝑔 in ℱ, 𝑓^(𝑘)−𝑎𝑓^𝑛 and 𝑔^(𝑘)−𝑎𝑔^𝑛 share 𝑏 in 𝐷, we know that for any positive integer 𝑚𝑓_𝑚^(𝑘)𝑧_𝑗+𝜌_𝑗𝜉_𝑗−𝑎𝑓^𝑛_𝑚𝑧_𝑗+𝜌_𝑗𝜉_𝑗𝑓−𝑏=0,_𝑚(𝑘)𝑧_𝑗+𝜌_𝑗𝜉^∗_𝑗−𝑎𝑓^𝑛_𝑚𝑧_𝑗+𝜌_𝑗𝜉^∗_𝑗−𝑏=0. (48)
Fix 𝑚, take 𝑗→∞, and note 𝑧_𝑗+𝜌_𝑗𝜉_𝑗→0, 𝑧_𝑗+𝜌_𝑗𝜉^∗_𝑗→0; then 𝑓_𝑚^(𝑘)(0)−𝑎𝑓^𝑛_𝑚(0)−𝑏=0. Since the zeros of 𝑓_𝑚^(𝑘)−𝑎𝑓^𝑛_𝑚−𝑏 have no accumulation point, so 𝑧_𝑗+𝜌_𝑗𝜉_𝑗=0,𝑧_𝑗+𝜌_𝑗𝜉^∗_𝑗=0. (49) Hence, 𝜉_𝑗=−𝑧_𝑗/𝜌_𝑗, 𝜉^∗_𝑗=−𝑧_𝑗/𝜌_𝑗. This contradicts 𝜉_𝑗∈𝐷(𝜉₀,𝛿), 𝜉^∗_𝑗∈𝐷(𝜉^∗₀,𝛿) and 𝐷(𝜉₀,𝛿)∩𝐷(𝜉^∗₀,𝛿)=𝜙. So 𝑔^(𝑘)(𝜉)−𝑎𝑔^𝑛(𝜉) has just a unique zero, which can be denoted by 𝜉₀.
Noting that 𝑔 has poles and zeros of multiplicities at least 𝑑 and 𝑘, respectively, (46) deduces that 𝑔(𝜉) is a rational function with degree at most 2.
If 𝑔(𝜉) is a polynomial, noting that deg𝑔≤2 and the multiplicities of zeros are at least 𝑘, we have 𝑘=2 and 𝑛=3. Hence, there exist 𝜉₁ and 𝑐≠0 such that 𝑔(𝜉)=𝑐(𝜉−𝜉₁)², and then 𝑔^′′(𝜉)−𝑎𝑔³(𝜉)=𝑐(2−𝑎𝑐²(𝜉−𝜉₁)⁶) has 6 distinct zeros, a contradiction.
Suppose that 𝑔(𝜉) is not a polynomial, 𝑘=2 and 𝑛=3. Then the multiplicities of poles of 𝑔(𝜉) are at least (𝑘+1)/(𝑛−2)=3, which implies that deg𝑔≥3, a contradiction.
Suppose that 𝑔(𝜉) is not a polynomial, 𝑘≥3 and 3≤𝑛≤𝑘+1; we distinguish two cases.
Case  i. If 𝑔(𝜉) has zeros, since all zeros of 𝑔(𝜉) have multiplicity at least 𝑘(≥3), it follows that deg𝑔≥3, a contradiction.
Case  ii. If 𝑔(𝜉)≠0, then (46) should be as follows 𝑇𝑟,𝑔≤𝑁1𝑟,𝑔^(𝑘)−𝑎𝑔^𝑛+𝑆𝑟,𝑔. (50)
From (50), we can see that 𝑔(𝜉) is a rational function with degree at most 1. Since all poles of 𝑔(𝜉) have multiplicity at least 𝑑(≥(𝑘+1)/(𝑛−2)≥(𝑘+1)/(𝑘−1)>1), which gives that deg𝑔≥2, a contradiction.
This completes the proof of Theorem 11.

Proof. In Theorem 13, suppose 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. Without loss of generality, we assume that 𝑧₀=0. By Lemma 1, there exist points 𝑧_𝑗→0, positive numbers 𝜌_𝑗→0, and functions 𝑓_𝑗∈ℱ such that 𝑔_𝑗(𝜉)=𝜌^𝑘_𝑗𝑓_𝑗𝑧_𝑗+𝜌_𝑗𝜉⟹𝑔(𝜉) (51) locally uniformly with respect to the spherical metric, where 𝑔 is a nonconstant meromorphic function in ℂ and whose poles and zeros are of multiplicity at least 𝑘+2 and 𝑘+1, respectively. Moreover, the order of 𝑔 is at most 2.
From (51), we know 𝑔_𝑗^(𝑘)(𝜉)=𝜌_𝑗^2𝑘𝑓_𝑗^(𝑘)𝑧_𝑗+𝜌_𝑗𝜉⟹𝑔^(𝑘)(𝜉),𝑔_𝑗^(𝑘)(𝜉)−𝑎𝑔²_𝑗(𝜉)−𝜌_𝑗^2𝑘𝑏=𝜌_𝑗^2𝑘𝑓_𝑗^(𝑘)𝑧_𝑗+𝜌_𝑗𝜉−𝑎𝑓²_𝑗𝑧_𝑗+𝜌_𝑗𝜉−𝑏⟹𝑔^(𝑘)(𝜉)−𝑎𝑔²(𝜉) (52)(53) also locally uniformly with respect to the spherical metric.
If 𝑔^(𝑘)(𝜉)−𝑎𝑔²(𝜉)≡0, since all poles of 𝑓 have multiplicity at least 𝑘+2, we can deduce that 𝑔(𝜉) is an entire function easily. Thus, 2𝑇𝑟,𝑔=𝑇𝑟,𝑔²=𝑇𝑟,𝑔^(𝑘)+𝑂(1)=𝑚𝑟,𝑔^(𝑘)+𝑁𝑟,𝑔^(𝑘)+𝑂(1)≤𝑚𝑟,𝑔+𝑁𝑟,𝑔+𝑘𝑁.𝑟,𝑔+𝑆𝑟,𝑔≤𝑇𝑟,𝑔+𝑆𝑟,𝑔 (54) Therefore, 𝑔(𝜉) is a constant, a contradiction. So 𝑔^(𝑘)(𝜉)−𝑎𝑔²(𝜉)≢0. By Lemmas 3, 4, and 5, 𝑔^(𝑘)(𝜉)−𝑎𝑔²(𝜉) has at least two distinct zeros. Proceeding as in the later proof of Theorem 11, we will get a contradiction. The proof is completed.

Similarly, as the proof of Theorem 11, when 𝑓 is a holomorphic function, we can get the following theorem which has been gotten in [23].

Theorem 18 (see [23]). Let 𝐷 be a domain in ℂ and let ℱ be a family of holomorphic functions in 𝐷. Let 𝑘,𝑛∈ℕ⁺, 𝑛≥2 and let 𝑎, 𝑏 be two finite complex numbers with 𝑎≠0. Suppose that every 𝑓∈ℱ has all its zeros of multiplicity at least 𝑘. If 𝑓^(𝑘)−𝑎𝑓^𝑛 and 𝑔^(𝑘)−𝑎𝑔^𝑛 share the value 𝑏 IM for every pair of functions (𝑓,𝑔) of ℱ, then ℱ is a normal family in 𝐷.

Acknowledgments

This paper is supported by the Nature Science Foundation of China (11271090), Nature Science Foundation of Guangdong Province (S2012010010121), and Graduate Research and Innovation Projects of Xinjiang Province (XJGRI2013131). This work was supported by the Visiting Scholar Program of Chern Institute of Mathematics at Nankai University. The second author would like to express his hearty gratitude to Chern Institute of Mathematics which provided very comfortable research environments to him where he worked as a Visiting Scholar.