Normality Criteria of Meromorphic Functions Sharing a Holomorphic Function

Abstract

Take three integers $m\ge 0,\,k\ge 1$, and $n\ge 2$. Let $a\ (\not \equiv 0)$ be a holomorphic function in a domain $D$ of $\mathbb {C}$ such that multiplicities of zeros of $a$ are at most $m$ and divisible by $n+1$. In this paper, we mainly obtain the following normality criterion: Let ${{{\fancyscript{F}}}}$ be the family of meromorphic functions on $D$ such that multiplicities of zeros of each $f\in {{\fancyscript{F}}}$ are at least $k+m$ and such that multiplicities of poles of $f$ are at least $m+1$. If each pair $(f,g)$ of ${{\fancyscript{F}}}$ satisfies that $f^{n}f^{(k)}$ and $g^{n}g^{(k)}$ share $a$ (ignoring multiplicity), then ${{\fancyscript{F}}}$ is normal.

1 Introduction

In this paper, we use the standard notations of the Nevanlinna theory as presented in [11, 17, 50, 52]. By definition, two meromorphic functions $F$ and $G$ are said to share $a$ IM if $F-a$ and $G-a$ assume the same zeros ignoring multiplicity. When $a=\infty$, the zeros of $F-a$ mean the poles of $F$.

Let $D$ be a domain in $\mathbb {C}$ and let ${{\fancyscript{F}}}$ be meromorphic functions defined in the domain $D$. Then ${{\fancyscript{F}}}$ is said to be normal in $D$, in the sense of Montel, if for any sequence $\{f_n\}\subset {{\fancyscript{F}}}$ there exists a subsequence $\{f_{n_j}\}$ such that $f_{n_j}$ converges spherically locally uniformly in $D$, to a meromorphic function or $\infty$ (cf. [15, 38]). For simplicity, we take $\rightarrow$ to stand for convergence and $\rightrightarrows$ for convergence spherically locally uniformly.

Let ${\fancyscript{M}}(D)$ (resp. ${\fancyscript{A}}(D)$) be the set of meromorphic (resp. holomorphic) functions on $D$. Let $n$ be an integer and take a positive integer $k$. We will study normality of the subset ${{\fancyscript{F}}}$ of ${\fancyscript{M}}(D)$ such that $f^nf^{(k)}$ satisfies some conditions for each $f\in {{\fancyscript{F}}}$.

First of all, we look at some background for the case $n=0$. Hayman [17] proved that if $F\in {\fancyscript{M}}(\mathbb {C})$ is transcendental, then either $F^{(k)}$ assumes every finite non-zero complex number infinitely often for any positive integer $k$ or $F$ assumes every finite complex number infinitely often. A normality criterion corresponding to Hayman’s theorem is obtained by Gu [14] which is stated as follows: If ${{\fancyscript{F}}}$ is the family in ${\fancyscript{M}}(D)$ such that each $f\in {{\fancyscript{F}}}$ satisfies $f^{(k)}\not =a$ and $f\not =b$, where $a,b$ are two complex numbers with $a\not =0$, then ${{\fancyscript{F}}}$ is normal in the sense of Montel. In particular, if ${{\fancyscript{F}}}\subset {\fancyscript{A}}(D)$, the normality criterion was conjectured by Montel (see [38], p. 125) for $k=1$, and proved by Miranda [30]. Further, Yang [51] and Schwick [40] confirmed that the normality criterion due to Gu is true if $a$ is replaced by a non-zero holomorphic function on $D$. In 2001, Jiang and Gao [22] proved that if ${{\fancyscript{F}}}$ is the family in ${\fancyscript{A}}(D)$ such that the multiplicities of zeros of each $f\in {{\fancyscript{F}}}$ are least $k+m+2$ for another non-negative integer $m$ and such that each pair $(f,g)$ of ${{\fancyscript{F}}}$ satisfies that $f^{(k)}$ and $g^{(k)}$ share $a$ IM (ignoring multiplicity), where $a\in {\fancyscript{A}}(D)$ and multiplicities of zeros of $a$ are at most $m$, then $F$ is normal in $D$, and obtained a similar result when ${{\fancyscript{F}}}\subset {\fancyscript{M}}(D)$. For other generations, see [3–5, 10, 23, 27, 28, 43, 44] and [46].

Next, we introduce some developments for the case $n\ge 1$ and $k=1$. In 1959, Hayman [16] proposed a conjecture: If $F\in {\fancyscript{M}}(\mathbb {C})$ is transcendental, then $F^nF'$ assumes every finite non-zero complex number infinitely often for any positive integer $n$. Hayman himself [16, 18] showed that it is true for $n\ge 3$, and for $n=2, F\in {\fancyscript{A}}(\mathbb {C})$. Mues [31] confirmed the conjecture for $n=2$ in 1979. Furthermore, the case of $n=1$ was considered by Clunie [9] when $F\in {\fancyscript{A}}(\mathbb {C})$, finally settled by Bergweiler and Eremenko [2], Chen and Fang [6]. Related to these results on value distribution, Hayman [18] conjectured that if ${{\fancyscript{F}}}$ is the family of ${\fancyscript{M}}(D)$ such that each $f\in {{\fancyscript{F}}}$ satisfies $f^nf'\not =a$ for a positive integer $n$ and a non-zero complex number $a$, then ${{\fancyscript{F}}}$ is normal. This conjecture has been confirmed by Yang and Zhang [54] (for $n\ge 5$, and for $n\ge 2$ with ${{\fancyscript{F}}}\subset {\fancyscript{A}}(D)$), Gu [13] (for $n=3, 4$), Pang [34] (for $n\ge 2$; cf. [12]) and Oshkin [32] (for $n=1$ with ${{\fancyscript{F}}}\subset {\fancyscript{A}}(D)$; cf. [24]). Finally, Pang [34] (or see [6, 55, 56]) indicated that the conjecture for $n=1$ is a consequence of his theorem and Chen-Fang’s theorem [6]. Recently, based on the ideas of sharing values, Zhang [58] proved that if ${{\fancyscript{F}}}$ is the family of ${\fancyscript{M}}(D)$ such that each pair $(f,g)$ of ${{\fancyscript{F}}}$ satisfies that $f^nf'$ and $g^ng'$ share a finite non-zero complex number $a$ IM for $n\ge 2$, then ${{\fancyscript{F}}}$ is normal. There are examples showing that this result is not true for the case $n=1$. Further, Jiang [22] concluded that if ${{\fancyscript{F}}}$ is the family of ${\fancyscript{M}}(D)$ such that each pair $(f,g)$ of ${{\fancyscript{F}}}$ satisfies that $f^nf'$ and $g^ng'$ share $a$ IM for $n\ge 2m+2$, where $a\in {\fancyscript{A}}(D)$ and multiplicities of zeros of $a$ are at most $m$, then ${{\fancyscript{F}}}$ is normal.

Similarly, we also have analogs related to some conditions of $f\left( f^{(k)}\right) ^l$ for a positive integer $l$. For example, Zhang and Song [60] announced that if $F\in {\fancyscript{M}}(\mathbb {C})$ is transcendental; $a(\not \equiv 0)$ a small function of $F$; $l\ge 2,$ then $F\left( F^{(k)}\right) ^{l}-a$ has infinitely many zeros. A simple proof was given by Alotaibi [1]. The normality criterion corresponding to this result was obtained by Jiang and Gao [21] which is stated as follows: Let $l, k\ge 2, m\ge 0$ be three integers such that $m$ is divisible by $l+1$ and suppose that $a (\not \equiv 0)$ is a holomorphic function in $D$ with zeros of multiplicity $m$. If ${{\fancyscript{F}}}$ is the family of ${\fancyscript{A}}(D)$ (resp. ${\fancyscript{M}}(D)$) such that multiplicities of zeros of each $f\in {{\fancyscript{F}}}$ are at least $k+m$ (resp. $\max \{k+m,2m+2\}$) and such that each pair $(f,g)$ of ${{\fancyscript{F}}}$ satisfies that $f\left( f^{(k)}\right) ^{l}$ and $g\left( g^{(k)}\right) ^{l}$ share $a$ IM, then ${{\fancyscript{F}}}$ is normal. For more results related to this topic, see Hennekemper [19], Hu and Meng [20], Li [25, 26], Schwick [39], Wang and Fang [42], Yang et al. [49].

Finally, we consider general cases of $n\ge 1$ and $k\ge 1$. In 1994, Zhang and Li [61] proved that if $F\in {\fancyscript{M}}(\mathbb {C})$ is transcendental, then $F^nL[F]-a$ has infinitely many zeros for $n\ge 2$ and $a\ne 0, \infty$, where

$$\begin{aligned} L[F]=a_{k}F^{(k)}+a_{k-1}F^{(k-1)}+\cdots +a_0F \end{aligned}$$

in which $a_{i}$ ($i=0,1,2,\cdots ,k$) are small functions of $F$. In 1999, Pang and Zalcman [36] obtained a corresponding normality criterion as follows: If ${{\fancyscript{F}}}$ is the family of ${\fancyscript{A}}(D)$ such that zeros of each $f\in {{\fancyscript{F}}}$ have multiplicities at least $k$ and such that each $f\in {{\fancyscript{F}}}$ satisfies $f^nf^{(k)} \not =a$ for a non-zero complex number $a$, then ${{\fancyscript{F}}}$ is normal. In 2005, Zhang [59] showed that when $n\ge 2$, this result is also true if $a$ is replaced by a non-vanishing holomorphic functions in $D$. For other related results, see Meng and Hu [29], Qi [37], Wang [41], Xu [45], Yang and Hu [48], Yang and Yang [53].

Take three integers $m\ge 0, k\ge 1$, and $n\ge 2$. Let $a\ (\not \equiv 0)$ be a holomorphic function in $D$ such that multiplicities of zeros of $a$ are at most $m$ and divisible by $n+1$. In this paper, we obtain the following normality criteria:

Theorem 1.1

Let ${{\fancyscript{F}}}$ be the family of ${\fancyscript{M}}(D)$ such that multiplicities of zeros of each $f\in {{\fancyscript{F}}}$ are at least $k+m$ and such that multiplicities of poles of $f$ are at least $m+1$ whenever $f$ have zeros and poles. If each pair $(f,g)$ of ${{\fancyscript{F}}}$ satisfies that $f^{n}f^{(k)}$ and $g^{n}g^{(k)}$ share $a$ IM, then ${{\fancyscript{F}}}$ is normal in $D$.

In special, if $a$ has no zeros, which means $m=0$, then Theorem 1.1 has the following form:

Corollary 1.1

Let ${{\fancyscript{F}}}$ be the family of ${\fancyscript{M}}(D)$ such that multiplicities of zeros of each $f\in {{\fancyscript{F}}}$ are at least $k$. If each pair $(f,g)$ of ${{\fancyscript{F}}}$ satisfies that $f^{n}f^{(k)}$ and $g^{n}g^{(k)}$ share $a$ IM, then ${{\fancyscript{F}}}$ is normal in $D$.

It is easy to see that this result extends above normality criteria due to Pang and Zalcman [36], and Zhang [59]. Furthermore, we can improve partially the normality criterion due to Jiang [22] as follows:

Theorem 1.2

If ${{\fancyscript{F}}}$ is the family of ${\fancyscript{M}}(D)$ such that each $f\in {{\fancyscript{F}}}$ satisfies that $f^{n}f'\ne a$, then ${{\fancyscript{F}}}$ is normal in $D$.

The condition $a(z)\not \equiv 0$ in Theorem 1.1 and 1.2 is necessary. This fact can be illustrated by the following example:

Example 1.1

Let $D=\{z\in \mathbb {C}\ |\ |z|<1\}$. Let $a(z)\equiv 0$ and

$$\begin{aligned} {{\fancyscript{F}}}=\left\{ \left. f_j(z)=e^{j(z-\frac{1}{2})}\ \right| \ j=1,2,\ldots \right\} \end{aligned}$$

Obviously, $f_i^{n}f_{i}^{(k)}$ and $f_j^{n} f_{j}^{(k)}$ share $a$ IM for distinct positive integers $i$ and $j$ $($resp. $f_j^{n} f_{j}'\ne a)$, but the family ${{\fancyscript{F}}}$ is not normal at $z=1/2$.

In Corollary 1.1, the condition that multiplicities of zeros of each $f\in {{\fancyscript{F}}}$ are at least $k$ is sharp. For example, we consider the following family:

Example 1.2

Denote $D$ as in Example 1.1. Let $a(z)=e^{z}$ and

$$\begin{aligned} {{\fancyscript{F}}}=\left\{ \left. f_j(z)=j\left( z-\frac{1}{2j}\right) ^{k-1}\ \right| \ j=1,2,\ldots \right\} . \end{aligned}$$

Any $f_{j} \in {{\fancyscript{F}}}$ has only a zero of multiplicity $k-1$ in $D$ and for distinct positive integers $i$ and $j, f_i^{n} f_{i}^{(k)}$ and $f_j^{n} f_{j}^{(k)}$ share $a$ IM. However, the family ${{\fancyscript{F}}}$ is not normal at $z=0$.

2 Preliminary Lemmas

In order to prove our results, we require the following Zalcman’s lemma (cf. [56]):

Lemma 2.1

Take a positive integer $k$. Let $\fancyscript{F}$ be a family of meromorphic functions in the unit disk $\triangle$ with the property that zeros of each $f\in \fancyscript{F}$ are of multiplicity at least $k$. If $\fancyscript{F}$ is not normal at a point $z_0\in \Delta$, then for $0\le \alpha < k$, there exist a sequence $\{z_n\}\subset \Delta$ of complex numbers with $z_n\rightarrow z_0$; a sequence $\{f_n\}$ of $\fancyscript{F}$; and a sequence $\{\rho _n\}$ of positive numbers with $\rho _n\rightarrow 0$ such that $g_n(\xi )=\rho _n^{-\alpha }f_n(z_n+\rho _n\xi )$ locally uniformly $($with respect to the spherical metric$)$ to a non-constant meromorphic function $g(\xi )$ on $\mathbb {C}$. Moreover, the zeros of $g(\xi )$ are of multiplicity at least $k$, and the function $g(\xi )$ may be taken to satisfy the normalization $g^{\sharp }(\xi )\le g^{\sharp }(0)=1$ for any $\xi \in \mathbb {C}$. In particular, $g(\xi )$ has at most order 2.

This result is Pang’s generalization (cf. [33, 35, 47]) to the Main Lemma in [55] (where $\alpha$ is taken to be $0$), with improvements due to Schwick [39], Chen and Gu [7]. In Lemma 2.1, the order of $g$ is defined using the Nevanlinna’s characteristic function $T(r,g)$:

$$\begin{aligned} \mathrm{ord}(g)=\limsup _{r\rightarrow \infty }\frac{\log T(r,g)}{\log r}. \end{aligned}$$

Here, as usual, $g^{\sharp }$ denotes the spherical derivative

$$\begin{aligned} g^{\sharp }(\xi )=\frac{|g'(\xi )|}{1+|g(\xi )|^2}. \end{aligned}$$

Lemma 2.2

Let $p\ge 0, k\ge 1$, and $n\ge 2$ be three integers, and let $a$ be a non-zero polynomial of degree $p$. If $f$ is a non-constant rational function which has only zeros of multiplicity at least $k+p$ and has only poles of multiplicity at least $p+1$, then $f^nf^{(k)}-a$ has at least one zero.

Proof

If $f$ is a polynomial, then $f^{(k)}\not \equiv 0$ since $f$ is non-constant and has only zeros of multiplicity at least $k+p$ which further means $\deg (f)\ge k+p$. Noting that $n\ge 2$, we immediately obtain that

$$\begin{aligned} \deg \left( f^nf^{(k)}\right) \ge n\deg (f)\ge n(k+p)>p=\deg (a). \end{aligned}$$

Therefore, it follows that $f^nf^{(k)}-a$ is also a non-constant polynomial, and hence $f^nf^{(k)}-a$ has at least one zero. Next, we assume that $f$ has poles. Set

$$\begin{aligned} f(z)=\frac{A(z-\alpha _1)^{m_1}(z-\alpha _2)^{m_2}\cdots (z-\alpha _{s})^{m_{s}}}{(z-\beta _1)^{n_1}(z-\beta _2)^{n_2}\cdots (z-\beta _{t})^{n_{t}}}, \end{aligned}$$

(2.1)

where $A$ is a non-zero constant, $\alpha _i$ distinct zeroes of $f$ with $s\ge 0$, and $\beta _j$ distinct poles of $f$ with $t\ge 1$. For simplicity, we put

$$\begin{aligned}&m_1+m_2+\cdots +m_{s}=M\ge (k+p)s,\end{aligned}$$

(2.2)

$$\begin{aligned}&n_1+n_2+\cdots +n_{t}=N\ge (p+1)t. \end{aligned}$$

(2.3)

From Eq. (2.1), we obtain

$$\begin{aligned} f^{(k)}(z)=\frac{(z-\alpha _1)^{m_1-k}(z-\alpha _2)^{m_2-k}\cdots (z-\alpha _{s})^{m_{s}-k}g(z)}{(z-\beta _1)^{n_1+k}(z-\beta _2)^{n_2+k}\cdots (z-\beta _{t})^{n_{t}+k}}, \end{aligned}$$

(2.4)

where $g$ is a polynomial of degree $\le k(s+t-1)$. From Eqs. (2.1) and (2.7), we get

$$\begin{aligned} f^n(z)f^{(k)}(z)=\frac{A^n(z-\alpha _1)^{M_1}(z-\alpha _2)^{M_2}\cdots (z-\alpha _s)^{M_s}g(z)}{(z-\beta _1)^{N_1}(z-\beta _2)^{N_2}\cdots (z-\beta _{t})^{N_t}}, \end{aligned}$$

(2.5)

in which

$$\begin{aligned} M_{i}&= (n+1)m_{i}-k, \quad \ i=1,2,\cdots ,s,\\ N_{j}&= (n+1)n_{j}+k, \quad \ j=1,2,\cdots ,t. \end{aligned}$$

Differentiating Eq. (2.5) yields

$$\begin{aligned} \left\{ f^nf^{(k)}\right\} ^{({p}+1)}(z)=\frac{(z-\alpha _1)^{M_1-p-1} (z-\alpha _2)^{M_2-p-1}\cdots (z-\alpha _s)^{M_s-p-1}g_{0}(z)}{(z-\beta _1)^{N_1+p+1}\cdots (z-\beta _t)^{N_{t+p}+1}}, \end{aligned}$$

(2.6)

where $g_{0}(z)$ is a polynomial of degree $\le (p+k+1)(s+t-1)$. We assume, to the contrary, that $f^{n}f^{(k)}-a$ has no zero, then

$$\begin{aligned} f^n(z)f^{(k)}(z)=a(z)+\frac{C}{(z-\beta _1)^{N_1}(z-\beta _2)^{N_2}\cdots (z-\beta _t)^{N_t}}, \end{aligned}$$

(2.7)

where $C$ is a non-zero constant. Subsequently, Eq. (2.12) yields

$$\begin{aligned} \left\{ f^nf^{(k)}\right\} ^{(p+1)}(z)=\frac{g_{1}(z)}{(z-\beta _1)^{N_1+p+1}\cdots (z-\beta _t)^{N_t+p+1}}, \end{aligned}$$

(2.8)

where $g_1(z)$ is a polynomial of degree $\le (p+1)(t-1)$.

Comparing Eq. (2.6) with Eq. (2.8), we get

$$\begin{aligned} (p+1)(t-1)\ge \deg (g_1)\ge (n+1)M-ks-(p+1)s, \end{aligned}$$

and hence

$$\begin{aligned} M < \frac{p+k+1}{n+1}s+\frac{p+1}{n+1}t. \end{aligned}$$

(2.9)

From Eqs. (2.5) and (2.7), we have

$$\begin{aligned} (n+1)N+kt+p=(n+1)M-ks+\deg (g). \end{aligned}$$

Since $\deg (g)\le k(s+t-1)$, we find

$$\begin{aligned} (n+1)N \le (n+1)M-ks+k(s+t-1)-kt-p, \end{aligned}$$

and thus

$$\begin{aligned} N < M. \end{aligned}$$

(2.10)

By Eqs. (2.9), (2.10) and noting that $M\ge (k+p)s, N\ge (p+1)t$, we deduce that

$$\begin{aligned} M < \frac{p+k+1}{n+1}s+\frac{p+1}{n+1}t \le \left\{ \frac{p+k+1}{(n+1)(k+p)}+\frac{1}{n+1}\right\} M. \end{aligned}$$

(2.11)

Note that $n \ge 2$ implies

$$\begin{aligned} \frac{p+k+1}{(n+1)(k+p)}+\frac{1}{n+1}=\frac{2(k+p)+1}{(n+1)(k+p)}\le 1. \end{aligned}$$

Hence it follows from Eq. (2.11) that $M<M$, which is a contradiction. Lemma 2.2 is proved.$\square$

Lemma 2.3

Let $p\ge 0, k\ge 1$, and $n\ge 2$ be three integers, and let $a$ be a non-zero polynomial of degree $p$. If $f$ is a non-constant rational function which has only zeros of multiplicity at least $k+p$ and has only poles of multiplicity at least $p+1$, then $f^nf^{(k)}-a$ has at least two distinct zeros.

Proof

Lemma 2.2 implies that $f^{n}f^{(k)}-a$ has at least one zero. Assume, to the contrary, that $f^{n}f^{(k)}-a$ has only one zero $z_{0}$. If $f$ is a polynomial, then we can write

$$\begin{aligned} f^{n}(z)f^{(k)}(z)-a(z)=A'(z-z_{0})^{d}, \end{aligned}$$

where $A'$ is a non-zero constant and $d$ is a positive integer. Since $f$ is a non-constant polynomial which has only zeros of multiplicity at least $k+p$, we find $f^{(k)}\not \equiv 0$, and hence

$$\begin{aligned} d= \deg (f^{n}f^{(k)}-a) \ge \deg (f^{n})\ge n(k+p) \ge 2p+2. \end{aligned}$$

By computing, we find

$$\begin{aligned} \left\{ f^nf^{(k)}\right\} ^{(p+1)}(z)=A'd(d-1)\ldots (d-p)(z-z_{0})^{{d}-p-1}, \end{aligned}$$

hence $\left\{ f^nf^{(k)}\right\} ^{(p+1)}$ has a unique zero $z_{0}$. Take a zero $\xi _{0}$ of $f$, then it is a zero of $f^{n}$ with multiplicity at least $2p+2$. It follows that $\xi _{0}$ is a common zero of $\left\{ f^nf^{(k)}\right\} ^{(p)}$ and $\left\{ f^nf^{(k)}\right\} ^{(p+1)}$, which further implies that $\xi _{0}=z_{0}$. Therefore, we obtain $\left\{ f^nf^{(k)}\right\} ^{(p)}(z_{0})=0.$

On the other hand, we get

$$\begin{aligned} \left\{ f^{n}f^{(k)}\right\} ^{(p)}(z)=a^{(p)}(z)+A'd(d-1)\ldots (d-p+1)(z-z_{0})^{{d-p}}, \end{aligned}$$

which means

$$\begin{aligned} \left\{ f^{n}f^{(k)}\right\} ^{(p)}(z_{0})=a^{(p)}(z_0)\ne 0 \end{aligned}$$

since $\deg (a)=p$. This is contradictory to $\left\{ f^{n}f^{(k)}\right\} ^{(p)}(z_{0})=0$.

If $f$ has poles, we can express $f$ by Eq. (2.1) again, and then find

$$\begin{aligned} f^n(z)f^{(k)}(z)=a(z)+\frac{C'(z-z_{0})^{l}}{(z-\beta _1)^{N_1}(z-\beta _2)^{N_2}\cdots (z-\beta _t)^{N_t}}, \end{aligned}$$

(2.12)

where $C'$ is a non-zero constant and $l$ is a positive integer. We distinguish two cases to deduce contradictions.

Case 1 $p \ge l$. Since $p\ge l$, the expression Eq. (2.5) together with Eq. (2.12) implies that

$$\begin{aligned} (n+1)N+kt+p=(n+1)M-ks+\deg (g). \end{aligned}$$

Therefore, we can also conclude Eq. (2.10), that is, $N<M$. Differentiating Eq. (2.12), we obtain

$$\begin{aligned} \left\{ f^nf^{(k)}\right\} ^{(p+1)}(z)= \frac{g_{2}(z)}{(z-\beta _1)^{N_1+p+1}\cdots (z-\beta _t)^{N_t+p+1}}, \end{aligned}$$

where $g_2(z)$ is a polynomial of degree at most $(p+1)t-(p-l+1)$, and hence

$$\begin{aligned} (p+1)t-(p-l+1)\ge \deg (g_2)\ge (n+1)M-ks-(p+1)s, \end{aligned}$$

where the last estimate follows from Eq. (2.6). Then we have

$$\begin{aligned} \frac{p-l}{n+1} < \frac{p+k+1}{n+1}s+\frac{p+1}{n+1}t -M\le \left\{ \frac{p+k+1}{(n+1)(k+p)}+\frac{1}{n+1}-1\right\} M\nonumber \\ \end{aligned}$$

(2.13)

since $M\ge (k+p)s,N\ge (p+1)t,M>N$. It follows that

$$\begin{aligned} \frac{p+k+1}{(n+1)(k+p)}+\frac{1}{n+1} \le 1 \end{aligned}$$

since $n \ge 2$. Therefore, from Eq. (2.13), we conclude that $p-l < 0$, a contradiction with the assumption $p\ge l$.

Case 2. $l > p$. The expression Eq. (2.12) yields

$$\begin{aligned} \left\{ f^nf^{(k)}\right\} ^{(p+1)}(z)=\frac{(z-z_{0})^{l-p-1}g_{3}(z)}{(z-\beta _1)^{N_1+p+1}\cdots (z-\beta _t)^{N_t+p+1}}, \end{aligned}$$

(2.14)

where $g_{3}(z)$ is a polynomial with $\deg (g_{3}) \le (p+1)t$. We claim that $z_0\not =\alpha _i$ for each $i$. Otherwise, if $z_0=\alpha _i$ for some $i$, then Eq. (2.12) yields

$$\begin{aligned} a^{(p)}(z_0)=\left\{ f^nf^{(k)}\right\} ^{(p)}(\alpha _i)=0 \end{aligned}$$

because each $\alpha _i$ is a zero of $f^nf^{(k)}$ of multiplicity $\ge n(k+p)\ge 2p+2$. This is impossible since $\deg (a)=p$. Hence $(z-z_{0})^{l-p-1}$ is a factor of the polynomial $g_0$ in Eq. (2.6). By Eqs. (2.6) and (2.14), we conclude that

$$\begin{aligned} (p+1)t \ge \deg (g_{3})\ge (n+1)M-ks-(p+1)s, \end{aligned}$$

which is equivalent to

$$\begin{aligned} M\le \frac{p+k+1}{n+1}s+\frac{p+1}{n+1}t. \end{aligned}$$

(2.15)

If  $l\ne (n+1)N+kt+p$, then Eq. (2.5) together with Eq. (2.12) implies

$$\begin{aligned} (n+1)N+kt+p \le (n+1)M-ks+\deg (g), \end{aligned}$$

so we get $N< M$ from $\deg (g)\le k(s+t-1)$. Therefore, using the facts $M\ge (k+p)s,N\ge (p+1)t$, Eq. (2.15) implies a contradiction

$$\begin{aligned} M<\left\{ \frac{p+k+1}{(n+1)(k+p)}+\frac{1}{n+1}\right\} M\le M. \end{aligned}$$

Hence $l= (n+1)N+kt+p$.

Now we must have $N\ge M$, otherwise, when $N<M$, we can deduce the contradiction $M<M$ from Eq. (2.15). Comparing Eq. (2.6) with Eq. (2.14), we find

$$\begin{aligned} (p+k+1)(s+t-1) \ge \deg (g_{0}) \ge l-p-1 \end{aligned}$$

since $(z-z_0)^{l-p-1}|g_0$, and hence

$$\begin{aligned} (n+1)N+kt+p = l \le (p+k+1)s+(p+k+1)t-k, \end{aligned}$$

which further yields

$$\begin{aligned} N<\frac{p+k+1}{n+1}s+\frac{p+1}{n+1}t. \end{aligned}$$

Since $M \ge (k+p)s$ and $N\ge (p+1)t$, it follows from Eq. (2.15) that

$$\begin{aligned} N< \frac{p+k+1}{(n+1)(k+p)}M+\frac{1}{n+1}N. \end{aligned}$$

Hence $N\ge M$ yields

$$\begin{aligned} N < \left\{ \frac{p+k+1}{(n+1)(k+p)}+\frac{1}{n+1}\right\} N. \end{aligned}$$

(2.16)

Since $n \ge 2$, we obtain consequently

$$\begin{aligned} \frac{p+k+1}{(n+1)(k+p)} + \frac{1}{n+1} \le 1. \end{aligned}$$

Hence Eq. (2.16) yields $N < N$. This is a contradiction. Proof of Lemma 2.3 is completed.$\square$

Lemma 2.4

Let $p\ge 0$ and $n\ge 2$ be two integers such that $p$ is divisible by $n+1$, and let $a$ be a non-zero polynomial of degree $p$. If $f$ is a non-constant rational function, then $f^nf'-a$ has at least one zero.

Proof

If $f$ is a non-constant polynomial, then $f'\not \equiv 0$. We consequently conclude that

$$\begin{aligned} \deg \left( f^nf'\right) =(n+1)\deg (f)-1\ne p \end{aligned}$$

since $p$ is divisible by $n+1$. It follows that $f^nf'-a$ is also a non-constant polynomial, so that $f^nf'-a$ has at least one zero.

If $f$ has poles, we can express $f$ by Eq. (2.1) again, and then by differentiating Eq. (2.1), we deduce that

$$\begin{aligned} f'(z)=\frac{(z-\alpha _1)^{m_1-1}(z-\alpha _2)^{{m}_2-1}\cdots (z-\alpha _s)^{{m_s}-1}h(z)}{(z-\beta _1)^{n_1+1}(z-\beta _2)^{n_2+1}\cdots (z-\beta _t)^{n_t+1}}, \end{aligned}$$

(2.17)

where $h(z)$ is a polynomial of form

$$\begin{aligned} h(z)=(M-N)z^{{s+t}-1}+\cdots . \end{aligned}$$

From Eqs. (2.1) and (2.17), we obtain

$$\begin{aligned} f^nf'=\frac{P}{Q}, \end{aligned}$$

in which

$$\begin{aligned}&\displaystyle P(z) = A^n(z-\alpha _1)^{(n+1){m}_1-1}(z-\alpha _2)^{(n+1){m}_2-1}\cdots (z-\alpha _s)^{(n+1){m}_s-1}h(z),\\&\displaystyle Q(z) = (z-\beta _1)^{(n+1)n_1+1}(z-\beta _2)^{(n+1)n_2+1}\cdots (z-\beta _t)^{(n+1)n_t+1}.\qquad \qquad \end{aligned}$$

We suppose, to the contrary, that $f^{n}f'-a$ has no zero. When $M\not =N$, we have

$$\begin{aligned} f^nf'=a+\frac{B}{Q}=\frac{P}{Q}, \end{aligned}$$

where $B$ is a non-zero constant. Therefore, we obtain

$$\begin{aligned} \deg (P)=\deg (Qa+B)=\deg (Q)+p. \end{aligned}$$

This implies that

$$\begin{aligned} (n+1)M-s+(s+t-1)= (n+1)N+t+p, \end{aligned}$$

or equivalently

$$\begin{aligned} M-N=\frac{p+1}{n+1}, \end{aligned}$$

in which $p$ is divisible by $n+1$. This is impossible since $M-N$ is an integer.

If $M=N$, we can rewrite Eq. (2.1) as follows

$$\begin{aligned} f(z)=A+ \frac{B'(z-\gamma _1)^{l_1}(z-\gamma _2)^{l_2}\cdots (z-\gamma _r)^{l_{r}}}{(z-\beta _1)^{n_1}(z-\beta _2)^{n_2}\cdots (z-\beta _t)^{n_t}}, \end{aligned}$$

where $B'$ is a non-zero constant, $\gamma _i$ are distinct with $l_i\ge 1, r\ge 0$, and

$$\begin{aligned} M'=l_1+\cdots +l_{r}<N. \end{aligned}$$

Thus we find

$$\begin{aligned} f'(z)= \frac{(z-\gamma _1)^{l_1-1}(z-\gamma _2)^{l_2-1}\cdots (z-\gamma _{r})^{l_r-1}\hbar (z)}{(z-\beta _1)^{n_1+1}(z-\beta _2)^{n_2+1}\cdots (z-\beta _t)^{n_t+1}} , \end{aligned}$$

where $\hbar (z)$ is a polynomial of form

$$\begin{aligned} \hbar (z)=(M'-N)z^{r+t-1}+\cdots . \end{aligned}$$

Similarly, since $\deg (P)=\deg (Q)+p$, we have

$$\begin{aligned} nM+M'-r+(r+t-1)=(n+1)N+t+p=(n+1)M+t+p, \end{aligned}$$

that is,

$$\begin{aligned} M'=M+p+1. \end{aligned}$$

This is impossible since $M'<N=M$. Therefore, $f^{n}f'-a$ has at least one zero.    $\square$

The following lemma is a direct consequence of a result from [61]:

Lemma 2.5

Let $n, k$ be two positive integers with $n\ge 2$, and let $a\ (\not \equiv 0)$ be a polynomial. If $f$ is a transcendental meromorphic function in $\mathbb {C}$, then $f^nf^{(k)}-a$ has infinitely zeros.

3 Proof of Theorem 1.1

Without loss of generality, we may assume that $D=\{z\in \mathbb {C}\ |\ |z|<1\}$. For any point $z_{0}$ in $D$, either $a(z_{0})=0$ or $a(z_{0})\ne 0$ holds. For simplicity, we assume $z_{0}=0$ and distinguish two cases.

Case 1 $a(0) \ne 0$. To the contrary, we suppose that ${{\fancyscript{F}}}$ is not normal at $z_{0}=0$. Then, by Lemma 2.1, there exist a sequence $\{z_j\}$ of complex numbers with $z_j\rightarrow 0$ $(j\rightarrow \infty )$; a sequence $\{f_j\}$ of ${{\fancyscript{F}}}$; and a sequence $\{\rho _j\}$ of positive numbers with $\rho _j\rightarrow 0$ $(j\rightarrow \infty )$ such that

$$\begin{aligned} g_j(\xi )=\rho _j^{-\frac{k}{n+1}}f_j(z_j+\rho _j\xi ) \end{aligned}$$

converges uniformly to a non-constant meromorphic function $g(\xi )$ in $\mathbb {C}$ with respect to the spherical metric. Moreover, $g(\xi )$ is of order at most 2. By Hurwitz’s theorem, the zeros of $g(\xi )$ have at least multiplicity $k+m$.

On every compact subset of $\mathbb {C}$ which contains no poles of $g$, we have uniformly

$$\begin{aligned}&f_j^{n}(z_j+\rho _j\xi ) f^{(k)}_j(z_j+\rho _j\xi ) -a(z_j+\rho _j\xi )\nonumber \\&\qquad = g_j^{n}(\xi )g^{(k)}_j(\xi )-a(z_j+\rho _j\xi )\rightrightarrows g^{n}(\xi ) g^{(k)}(\xi ) -a(0). \end{aligned}$$

(3.1)

If $g^{n}g^{(k)}\equiv a(0)$, then $g$ has no zeros and poles. Then there exist constants $c_i$ such that $(c_1,c_2)\not =(0,0)$, and

$$\begin{aligned} g(\xi )=e^{c_0+c_1\xi +c_2\xi ^2} \end{aligned}$$

since $g$ is a non-constant meromorphic function of order at most 2. Obviously, this is contrary to the case $g^{n}g^{(k)}\equiv a(0)$. Hence we have $g^{n}g^{(k)}\not \equiv a(0)$.

By Lemmas 2.3 and 2.5, the function $g^{n}g^{(k)}-a(0)$ has two distinct zeros $\xi _{0}$ and $\xi _{0}^{*}$. We choose a positive number $\delta$ small enough such that $D_1\cap D_2=\emptyset$ and such that $g^{n}g^{(k)}-a(0)$ has no other zeros in $D_1 \cup D_2$ except for $\xi _{0}$ and $\xi _{0}^{*}$, where

$$\begin{aligned} D_1=\{\xi \in \mathbb {C}\ |\ |\xi -\xi _0|<\delta \},\quad D_2=\{\xi \in \mathbb {C}\ |\ |\xi -\xi _0^*|<\delta \}. \end{aligned}$$

By Eq. (3.1) and Hurwitz’s theorem, there exist points $\xi _j\in D_1, \xi _j^*\in D_2$ such that

$$\begin{aligned} f^{n}_j(z_j+\rho _j\xi _j)f^{(k)}_j(z_j+\rho _j\xi _j)-a(z_j+\rho _j\xi _j)=0, \end{aligned}$$

and

$$\begin{aligned} f^{n}_j(z_j+\rho _j\xi _j^*)f^{(k)}_j(z_j+\rho _j\xi _j^*)-a(z_j+\rho _j\xi _j^{*})=0 \end{aligned}$$

for sufficiently large $j$.

By the assumption in Theorem 1.1, $f_1^{n} f_1^{(k)}$ and $f_j^{n} f_j^{(k)}$ share $a$ IM for each $j$. It follows

$$\begin{aligned} f_1^{n}(z_j+\rho _j\xi _j) f_1^{(k)}(z_j+\rho _j\xi _j) -a(z_j+\rho _j\xi _j)=0, \end{aligned}$$

and

$$\begin{aligned} f_1^{n}(z_j+\rho _j\xi _j^*) f_1^{(k)}(z_j+\rho _j\xi _j^*) -a(z_j+\rho _j\xi _j^{*})=0. \end{aligned}$$

By letting $j\rightarrow \infty$ and noting $z_j+\rho _j\xi _j\rightarrow 0, z_j+\rho _j\xi _j^*\rightarrow 0$, we obtain

$$\begin{aligned} f^{n}_1(0) f^{(k)}_1(0) -a(0)=0. \end{aligned}$$

Since the zeros of $f^{n}_1(\xi ) f^{(k)}_1(\xi ) -a(\xi )$ have no accumulation points, in fact we have

$$\begin{aligned} z_j+\rho _j\xi _j= 0, \quad \ z_j+\rho _j\xi _j^*= 0, \end{aligned}$$

or equivalently

$$\begin{aligned} \xi _j=-\frac{z_j}{\rho _j},\qquad \xi _j^*=-\frac{z_j}{\rho _j}. \end{aligned}$$

This contradicts with the facts that $\xi _j\in D_1, \xi _j^*\in D_2, D_1\cap D_2=\emptyset$. Thus ${{\fancyscript{F}}}$ is normal at $z_{0}=0$.

Case 2 $a(0)=0$. We assume that $z_{0}=0$ is a zero of $a$ of multiplicity $p$. Then we have $p\le m$ by the assumption. Write $a(z)=z^pb(z)$, in which $b(0)=b_p \ne 0$. Since multiplicities of all zeros of $a$ are divisible by $n+1$, then $d=p/(n+1)$ is just a positive integer. Thus we obtain a new family of ${\fancyscript{M}}(D)$ as follows

$$\begin{aligned} \fancyscript{H}= \left. \left\{ \frac{f(z)}{z^d}\ \right| \ f\in {{\fancyscript{F}}}\right\} . \end{aligned}$$

We claim that $\fancyscript{H}$ is normal at $0$.

Otherwise, if $\fancyscript{H}$ is not normal at $0$, then by lemma 2.1, there exist a sequence $\{z_j\}$ of complex numbers with $z_j\rightarrow 0$ $(j\rightarrow \infty )$; a sequence $\{h_j\}$ of $\fancyscript{H}$; and a sequence $\{\rho _j\}$ of positive numbers with $\rho _j\rightarrow 0$ $(j\rightarrow \infty )$ such that

$$\begin{aligned} g_j(\xi )=\rho _j^{-\frac{k}{n+1}}h_j(z_j+\rho _j\xi ) \end{aligned}$$

(3.2)

converges uniformly to a non-constant meromorphic function $g(\xi )$ in $\mathbb {C}$ with respect to the spherical metric, where $g^{\sharp }(\xi ) \le 1, \mathrm{ord}(g)\le 2$, and $h_j$ has the following form

$$\begin{aligned} h_j(z)=\frac{f_j(z)}{z^d}. \end{aligned}$$

We will deduce contradiction by distinguishing two cases.

Subcase 2.1 There exists a subsequence of $z_{j} / \rho _{j}$, for simplicity we still denote it as $z_{j} / \rho _{j}$, such that $z_{j} / \rho _{j}\rightarrow c$ as $j\rightarrow \infty$, where $c$ is a finite number. Thus we have

$$\begin{aligned} F_{j}(\xi )=\frac{f_{j}(\rho _{j}\xi )}{\rho _{j}^{\frac{k}{n+1}+d}}= \frac{(\rho _{j}\xi )^{d}h_{j}(z_{j}+\rho _{j}(\xi -\frac{z_{j}}{\rho _{j}}))}{(\rho _{j})^{d}(\rho _{j})^{\frac{k}{n+1}}} \rightrightarrows \xi ^{d} g(\xi -c)=h(\xi ), \end{aligned}$$

and

$$\begin{aligned} F_{j}^{n}(\xi )F_{j}^{(k)}(\xi )-\frac{a(\rho _{j}\xi )}{\rho _{j}^p}= \frac{f_{j}^{n}(\rho _{j}\xi )f_{j}^{(k)}(\rho _{j}\xi )-a(\rho _{j}\xi )}{\rho _{j}^p}\rightrightarrows h^{n}(\xi )h^{(k)}(\xi )-b_p\xi ^p. \end{aligned}$$

(3.3)

Noting $p\le m$, it follows from Lemmas 2.3 and 2.5 that $h^{n}(\xi )h^{(k)}(\xi )-b_{p}\xi ^{p}$ has two distinct zeros at least. Additionally, with similar discussion to the proof of Case 1, we can conclude that $h^{n}(\xi )h^{(k)}(\xi )-b_p\xi ^p \not \equiv 0$. Let $\xi _{1}$ and $\xi _{1}^{*}$ be two distinct zeros of $h^{n}(\xi )h^{(k)}(\xi )-b_p\xi ^p$. We choose a positive number $\gamma$ properly, such that $D_3\cap D_4=\emptyset$ and such that $h^{n}(\xi )h^{(k)}(\xi )-b_p\xi ^p$ has no other zeros in $D_3 \cup D_4$ except for $\xi _{1}$ and $\xi _{1}^{*}$, where

$$\begin{aligned} D_3=\{\xi \in \mathbb {C}\ |\ |\xi -\xi _1|<\gamma \},\quad D_4=\{\xi \in \mathbb {C}\ |\ |\xi -\xi _1^*|<\gamma \}. \end{aligned}$$

By Eq. (3.3) and Hurwitz’s theorem, there exist points $\zeta _j\in D_3, \zeta _j^*\in D_4$ such that

$$\begin{aligned} f^{n}_j(\rho _j\zeta _j)f^{(k)}_j(\rho _j\zeta _j)-a(\rho _j\zeta _j)=0, \end{aligned}$$

and

$$\begin{aligned} f^{n}_j(\rho _j\zeta _j^*)f^{(k)}_j(\rho _j\zeta _j^*)-a(\rho _j\zeta _j^{*})=0 \end{aligned}$$

for sufficiently large $j$. By the similar arguments in Case 1, we obtain a contradiction.

Subcase 2.2 There exists a subsequence of $z_{j} / \rho _{j}$, for simplicity we still denote it as $z_{j} / \rho _{j}$, such that $z_{j} / \rho _{j}\rightarrow \infty$ as $j\rightarrow \infty$. Then

$$\begin{aligned}&f_{j}^{(k)}(z_{j}+\rho _{j}\xi ) =\left\{ (z_{j}+\rho _{j}\xi )^{d}h_{j}(z_{j}+\rho _{j}\xi )\right\} ^{(k)}\\&\quad =(z_{j}+\rho _{j}\xi )^{d}h_{j}^{(k)}(z_{j}+\rho _{j}\xi ) +\sum _{i=1}^{k}a_{i}(z_{j}+\rho _{j}\xi )^{{d}-i}h_{j}^{(k-i)}(z_{j}+\rho _{j}\xi )\\&\quad =(z_{j}+\rho _{j}\xi )^{d}\rho _{j}^{-\frac{nk}{n+1}}g_{j}^{(k)}(\xi ) +\sum _{i=1}^{k}a_{i}(z_{j}+\rho _{j}\xi )^{{d}-i}\rho _{j}^{-\frac{nk}{n+1}+i} g_{j}^{(k-i)}(\xi ), \end{aligned}$$

in which $a_{i} (i=1,2,\cdots ,k)$ are all constants. Since $z_{j} / \rho _{j}\rightarrow \infty , b(z_{j}+\rho _{j}\xi )\rightarrow b_p$ as $j\rightarrow \infty$, it follows that

$$\begin{aligned}&b_p\frac{f_{j}^{n}(z_{j}+\rho _{j}\xi )f_{j}^{(k)}(z_{j} +\rho _{j}\xi )}{a(z_{j}+\rho _{j}\xi )}-b_p =b_p\frac{(z_{j}+\rho _{j}\xi )^{p}g_{j}^{n}(\xi )g_{j}^{(k)}(\xi )}{b(z_{j}+\rho _{j}\xi )(z_{j}+\rho _{j}\xi )^p}\nonumber \\&\quad +\sum _{i=1}^{k}b_p\frac{(z_{j}+\rho _{j}\xi )^pg_{j}^{n} (\xi )g_{j}^{(k-i)}(\xi )}{b(z_{j}+\rho _{j}\xi )(z_{j} +\rho _{j}\xi )^p}\left( \frac{\rho _{j}}{z_{j}+\rho _{j}\xi }\right) ^{i}\nonumber \\&\quad -b_p \rightrightarrows g^{n}(\xi )g^{(k)}(\xi )-b_p \end{aligned}$$

(3.4)

on every compact subset of $\mathbb {C}$ which contains no poles of $g$. Since all zeros of $f_{j}\in {{\fancyscript{F}}}$ have at least multiplicity $k+m$, then multiplicities of zeros of $g$ are at least $k$. Then from Lemmas 2.3 and 2.5, the function $g^{n}(\xi )g^{(k)}(\xi )-b_p$ has at least two distinct zeros. With similar discussion to the proof of Case 1, we can get a contradiction.

Hence the claim is proved, that is, $\fancyscript{H}$ is normal at $z_{0}=0$. Therefore, for any sequence $\{f_t\}\subset {{\fancyscript{F}}}$ there exist $\Delta _{r}=\{z: |z|< r\}$ and a subsequence $\{h_{t_{k}}\}$ of $\{h_t(z)=f_{t(z)}/z^{d}\}\subset \fancyscript{H}$ such that $h_{t_{k}} \rightrightarrows I$ or $\infty$ in $\Delta _{r}$, where $I$ is a meromorphic function. Next, we distinguish two cases.

Case A Assume $f_{t_{k}}(0)\ne 0$ when $k$ is sufficiently large. Then $I(0)=\infty$, and hence for arbitrary $R>0$, there exists a positive number $\delta$ with $0<\delta <r$ such that $|I(z)|> R$ when $z\in \Delta _{\delta }$. Hence when $k$ is sufficiently large, we have $|h_{t_{k}}(z)|> R/2$, which means that $1/ f_{t_{k}}$ is holomorphic in $\Delta _{\delta }$. In fact, when $|z|=\delta /2$,

$$\begin{aligned} \left| \frac{1}{f_{t_{k}}(z)}\right| =\left| \frac{1}{h_{t_{k}}(z)z^{d}}\right| \le M=\frac{2^{d+1}}{R\delta ^{d}}. \end{aligned}$$

By applying maximum principle, we have

$$\begin{aligned} \left| \frac{1}{f_{t_{k}}(z)}\right| \le M \end{aligned}$$

for $z\in \Delta _{\delta /2}$. It follows from Motel’s normal criterion that there exists a convergent subsequence of $\{f_{t_k}\}$, that is, ${{\fancyscript{F}}}$ is normal at $0$.

Case B There exists a subsequence of $f_{t_{k}}$, for simplicity we still denote it as $f_{t_{k}}$, such that $f_{t_{k}}(0)=0$. Then we get $I(0)=0$ since $h_{t_k}(z)=f_{t_{k}}(z)/ z^{d}\rightrightarrows I(z)$, and hence there exists a positive number $\rho$ with $0<\rho <r$ such that $I(z)$ is holomorphic in $\Delta _{\rho }$ and has a unique zero $z=0$ in $\Delta _{\rho }$. Therefore, we have $f_{t_{k}}(z)\rightrightarrows z^{d}I(z)$ in $\Delta _{\rho }$ since $h_{t_{k}}$ converges spherically locally uniformly to a holomorphic function $I$ in $\Delta _{\rho }$. Thus ${{\fancyscript{F}}}$ is normal at $0$.

Similarly, we can prove that ${{\fancyscript{F}}}$ is normal at arbitrary $z_{0}\in D$, and hence ${{\fancyscript{F}}}$ is normal in $D$.

4 Proof of Corollary 1.1

Using Lemmas 2.3 and 2.5, we find that if $f$ is a non-constant meromorphic function which has only zeros of multiplicity at least $k$, then $f^nf^{(k)}-a$ has at least two distinct zeros for a non-zero complex number $a$. Therefore, noting that $a$ has no zeroes, we can verify that ${{\fancyscript{F}}}$ is normal in $D$ by utilizing the same method in the proof of Theorem 1.1.

5 Proof of Theorem 1.2

Without loss of generality, we assume that $D=\{z\in \mathbb {C}\ |\ |z|<1\}$ and $z_{0}=0$. Now we distinguish two cases by either $a(0)=0$ or $a(0)\ne 0$.

Case 1 $a(0) \ne 0$. To the contrary, we suppose that ${{\fancyscript{F}}}$ is not normal at $0$. Using the notations in the proof of Theorem 1.1, we also obtain

$$\begin{aligned}&f_j^{n}(z_j+\rho _j\xi ) f_j'(z_j+\rho _j\xi ) -a(z_j+\rho _j\xi ) \nonumber \\&\quad = g_j^{n}(\xi )g_j'(\xi )-a(z_j+\rho _j\xi )\rightrightarrows g^{n}(\xi ) g'(\xi ) -a(0), \end{aligned}$$

(5.1)

where $g^{n}g^{(k)}\not \equiv a(0)$.

By Lemmas 2.4 and 2.5, the function $g^{n}g'-a(0)$ has a zero $\xi _{2}$. By Eq. (5.1) and Hurwitz’s theorem, there exist points $\eta _j \rightarrow \xi _{2}$ $(j\rightarrow \infty )$ such that for sufficiently large $j, z_j+\rho _j\eta _j \in D$ and

$$\begin{aligned} f^{n}_j(z_j+\rho _j\eta _j)f_j'(z_j+\rho _j\eta _j)-a(z_j+\rho _j\eta _j)=0, \end{aligned}$$

which contradicts the assumption that $f^{n}f'\ne a$.

Case 2 $a(0)=0$. Using the notations in the proof of Theorem 1.1, we also get the formulas Eqs. (3.1)–(3.4). Therefore, with the similar method in Case 1, we can prove that ${{\fancyscript{F}}}$ is normal at $z_{0}$, and hence ${{\fancyscript{F}}}$ is normal in $D$.