Normality of Meromorphic Functions and Uniformly Discrete Exceptional Sets

Abstract

Let $k\in \mathbb{N }$ and $h(\not \equiv 0)$ be a function holomorphic on $D$. Let $\mathcal{F }$ be a family of meromorphic functions in $D$, all of whose zeros have multiplicity at least $k+3$. Suppose that the sets $\{E_f\}_{f\in \mathcal{F }}$ are locally uniformly discrete in $D$, where $E_f=\{z\in D:f(z)=0\}\cup \{z\in D:f^{(k)}(z)=h(z)\}$. Suppose additionally that at the common zeros of $f\in \mathcal{F }$ and $h$, the multiplicities $m_f$ for $f$ and $m_h$ for $h$ satisfy $m_f\ge m_h+k+1$ for $k>1$ and $m_f\ge 2m_h+3$ for $k=1$. Then, $\mathcal{F }$ is normal in $D$. The number $k+3$ can be replaced by $k+2$ if the set $E_f$ is independent of $f$, or in other words, for each pair of functions $f$ and $g$ in $\mathcal{F }$, $f$ and $g$ share the value $0$, and $f^{(k)}$ and $g^{(k)}$ share the function $h$. Examples are also given to show that the conditions are necessary and sharp.

1 Introduction and Main Results

A family $\mathcal{F }$ of meromorphic functions defined in a plane domain $D\subset \mathbb{C }$ is said to be normal in $D$, if each sequence $\{f_n\}\subset \mathcal{F }$ contains a subsequence which converges spherically locally uniformly in $D$ to a meromorphic function or $\infty$. See [6, 11, 15].

The Gu’s normality criterion [5] which was conjectured by Hayman [6] says that a family $\mathcal{F }$ of functions meromorphic on $D$ is normal if $f\not =0$ and $f^{(k)}\not =1$ for each $f\in \mathcal{F }$. The following generalization of Gu’s theorem was proved by Yang [14].

Theorem 1

Let $\mathcal{F }$ be a family of meromorphic functions on $D$, $k\in \mathbb{N }$ and $h(\not \equiv 0)$ be a holomorphic function on $D$. If for every $f\in \mathcal{F }$, $f\not =0$ and $f^{(k)}\ne h$ on $D$, then $\mathcal{F }$ is normal on $D$.

In recent years, following Schwick [12], many normality criteria concerning shared values or functions have been proved. We say that two functions $f$ and $g$ share a value or a function $\phi$ if the two equations $f(z)=\phi (z)$ and $g(z)=\phi (z)$ have the same solutions (ignoring multiplicity). Here, we want to generalize the following result of Fang and Zalcman [4] by replacing the constant 1 by a function.

Theorem 2

Let $k$ be a positive integer and let $\mathcal{F }$ a family of meromorphic functions on $D$, all of whose zeros have multiplicity at least $k+2$, such that for each pair of functions $f$ and $g$ in $\mathcal{F }$, $f$ and $g$ share the value $0$, and $f^{(k)}$ and $g^{(k)}$ share the value $1$. Then, the family $\mathcal{F }$ is normal.

In general, the constant 1 cannot be replaced by a function. For example, the family $\{f_n\}$, where $f_n(z)=nz^{k+2}$, is not normal at 0. However, each pair of functions $f_n$ and $f_m$ share the value $0$, and $f_n^{(k)}$ and $f_m^{(k)}$ share the function $z^2$.

So we need some additional conditions. We prove the following generalization of Theorem 2.

Theorem 3

Let $k\in \mathbb{N }$ and $h(\not \equiv 0)$ be a function holomorphic on $D$. Let $\mathcal{F }$ be a family of meromorphic functions in $D$, all of whose zeros have multiplicity at least $k+2$, such that for each pair of functions $f$ and $g$ in $\mathcal{F }$, $f$ and $g$ share the value $0$, and $f^{(k)}$ and $g^{(k)}$ share the function $h$. Suppose additionally that at each common zero of $f$ and $h$ for every $f\in \mathcal{F }$, the multiplicities $m_f$ for $f$ and $m_h$ for $h$ satisfy $m_f\ge m_h+k+1$ for $k>1$ and $m_f\ge 2m_h+3$ for $k=1$. Then, $\mathcal{F }$ is normal in $D$.

The above example shows that the assumption $m_f\ge m_h+k+1$ is necessary. The condition $m_f\ge 2m_h+3$ for $k=1$ is also necessary and sharp as showed by the following example.

Example 1

Let $\alpha$ be a positive integer and $h(z)=z^\alpha$. Let for $n\in \mathbb{N }$,

$$\begin{aligned} f_n(z)=\frac{z^{2\alpha +2}}{(\alpha +1)[z^{\alpha +1}-1/n]}. \end{aligned}$$

Then,

$$\begin{aligned} f_n^{\prime }(z)-h(z)=\frac{z^{3\alpha +2}-\frac{2}{n}z^{2\alpha +1}}{\left(z^{\alpha +1}-\frac{1}{n}\right)^2} -z^\alpha =-\frac{\frac{1}{n^2}z^{\alpha }}{\left(z^{\alpha +1}-\frac{1}{n}\right)^2}. \end{aligned}$$

Thus, each pair of functions $f_n$ and $f_m$ share the value $0$, and $f_n^{\prime }$ and $f_m^{\prime }$ share the function $h(z)$. However, the family $\{f_n\}$ is not normal at $0$.

Let us look at some more aspects. By fixing a function $f_0\in \mathcal{F }$ and letting $E=\{z\in D:f_0(z)=0\}\cup \{z\in D:f_0^{(k)}(z)=h(z)\}$, the shared condition of Theorem 3 is equivalent to saying that there exists a fixed set $E\subset D$, which is independent of $f\in \mathcal{F }$, such that $f\not =0$ and $f^{(k)}\not =h$ on $D\setminus E$ for every $f\in \mathcal{F }$. There are two cases for the set $E$. One which is trivial is that $E=D$. Then, every $f\in \mathcal{F }$ satisfies $f\equiv 0$ or $f^{(k)}\equiv h$, and hence the normality of $\mathcal{F }$ can be easily dealt with. The non-trivial case is that the exceptional set $E$ is (locally) discrete in $D$. We focus on the non-trivial case and consider here that the exceptional set $E$ is dependent on $f$. To state our result, we require the following definition.

Definition 1

The sets $\{E_\lambda \}_{\lambda \in \Lambda }$ are said to be locally uniformly discrete in $D$, if for each point $z_0\in D$, there exists $\delta >0$ such that every $E_\lambda$ has at most one point lying in the disk $\Delta (z_0,\delta )=\{z:\ |z-z_0|<\delta \}$.

For example, the sets $\{E_n\}_{n\in \mathbb{N }}$, where $E_n=\{\frac{m-1}{m}+\frac{1}{n}: m\in \mathbb{N }\}$, are locally uniformly discrete in the unit disk $\Delta (0,1)$, but not in the domains that contain the point 1.

Theorem 4

Let $k\in \mathbb{N }$ and $h(\not \equiv 0)$ be a function holomorphic on $D$. Let $\mathcal{F }$ be a family of meromorphic functions in $D$, all of whose zeros have multiplicity at least $k+2$ or $k+3$ and $h$ has simple zeros. Suppose that the sets $\{E_f\}_{f\in \mathcal{F }}$, where $E_f=\{z\in D:f(z)=0\}\cup \{z\in D:f^{(k)}(z)=h(z)\}$, are locally uniformly discrete in $D$. Suppose additionally that at the common zeros of $f\in \mathcal{F }$ and $h$, the multiplicities $m_f$ for $f$ and $m_h$ for $h$ satisfy $m_f\ge m_h+k+1$ for $k>1$ and $m_f\ge 2m_h+3$ for $k=1$. Then, $\mathcal{F }$ is normal in $D$.

The following example shows that it is necessary (and sharp) to assume that the multiplicity of zeros of $f\in \mathcal{F }$ is at least $k+3$ when $h$ has simple zeros.

Example 2

Let $h(z)=z$ and

$$\begin{aligned} f_n(z)=\frac{(z-1/n)^{k+2}}{(k+1)![z-(k+2)/n]},\ \ n\in \mathbb{N }. \end{aligned}$$

Then, $E_{f_n}=\{1/n\}$, so that $\{E_{f_n}\}$ are locally uniformly discrete in $\mathbb{C }$. In fact, we have

$$\begin{aligned} f_n(z)&= \frac{\left[\left(z-\frac{k+2}{n}\right)+\frac{k+1}{n}\right]^{k+2}}{(k+1)!\left(z-\frac{k+2}{n}\right)} =\frac{1}{(k+1)!}\left[\left(z-\frac{k+2}{n}\right)^{k+1}\right.\\&\left.+\frac{(k+1)(k+2)}{n}\left(z-\frac{k+2}{n}\right)^{k}+P(z)+ \frac{\left(\frac{k+1}{n}\right)^{k+2}}{\left(z-\frac{k+2}{n}\right)}\right], \end{aligned}$$

where $P$ is a polynomial of degree $<k$, so that

$$\begin{aligned} f_n^{(k)}(z)=z+\frac{(-1)^k\left(\frac{k+1}{n}\right)^{k+2}}{(k+1)\left(z-\frac{k+2}{n}\right)^{k+1}}\not =h(z). \end{aligned}$$

However, $\{f_n\}$ is not normal at $0$, as $f_n(1/n)=0$ while $f_n((k+2)/n)=\infty$. Throughout in this paper, we denote by $\mathbb{C }$ the complex plane, by $\mathbb{C }^*$ the punctured complex plane $\mathbb{C }\setminus \{0\}$, by $\Delta (z_0,r)$ the open disk $\{z: |z-z_0|<r\}$, and by $\Delta ^\circ (z_0,r)$ the punctured disk $\Delta (z_0,r)\setminus \{z_0\}=\{z: 0<|z-z_0|<r\}$, where $z_0\in \mathbb{C }$ and $r>0$.

2 Auxiliary Results

To prove our results, we require some preliminary results.

Lemma 5

[3] Let $\mathcal{F }$ be a family of meromorphic functions in a domain $D$, all of whose zeros have multiplicity at least $k$. Then, if $\mathcal{F }$ is not normal at $z_0$, there exist, for each $-1< \alpha < k$, points $z_n\in D$ with $z_n\rightarrow z_0$, functions $f_n\in \mathcal{F }$ and positive numbers $\rho _n\rightarrow 0$ such that $g_n(\zeta ):=\rho _{n} ^{-\alpha }f_n(z_n+\rho _n\zeta )$ converges locally uniformly with respect to the spherical metric in $\mathbb{C }$ to a non-constant meromorphic function $g$ of finite order, all of whose zeros have multiplicity at least $k$.

The original form ($\alpha =0$) of this rescaling lemma is due to Zalcman [16], while the case $-1<\alpha <1$ was proved by Pang [8, 9]. The present form is due to Chen and Gu [3]. This lemma also holds for $\alpha =k$ [10] under an additional condition.

Lemma 6

[6, 7] Let $k$ be a positive integer, $f$ be a transcendental meromorphic function and $P(\not \equiv 0)$ be a polynomial. Then, either $f$ or $f^{(k)}-P$ has infinitely many zeros. If $f$ is a non-constant rational function, then either $f$ or $f^{(k)}-1$ has at least one zero.

Lemma 7

[13] Let $k$ be a positive integer and let $f$ be a non-constant rational function such that $f^{(k)}\not =1$ on $\mathbb{C }$. Then, either $f$ is a polynomial of degree at most $k$ or $f(z)=z^k/k!+P_{k}(z)+a(z-b)^{-n}$, where $a(\not =0), b\in \mathbb{C }$ and $n\in \mathbb{N }$ are constants, and $P_{k}$ is a polynomial of degree less than $k$. Furthermore, $f$ has a zero whose multiplicity is at most $k+1$.

Lemma 8

[2, Lemma 4] Let $k$ be a positive integer and $f$ be a non-constant rational function. If $f(z)\not =0$ for $z\in \mathbb{C }$, then $f^{(k)}-1$ has at least $k+1$ distinct zeros on $\mathbb{C }$.

Lemma 9

Let $k$ be a positive integer and $f$ be a non-constant rational function. If $f(z)[f^{(k)}(z)-1]\not =0$ for $z\in \mathbb{C }\setminus \{z_0\}$, where $z_0\in \mathbb{C }$, then $z_0$ is a zero of $f$ with multiplicity at most $k+1$.

Proof

We claim that $f(z_0)=0$. For otherwise, we would have $f(z)\not =0$ for $z\in \mathbb{C }$ by the condition, and hence, by Lemma 8, $f^{(k)}-1$ has at least $k+1\ge 2$ distinct zeros, which contradicts that $f^{(k)}(z)\not =1$ for $z\not =z_0$.

We now assume that the zero $z_0$ of $f$ has multiplicity at least $k+2$. Then, $f^{(k)}(z_0)=0$. Hence by the condition, we have $f(z)\not =0$ for $z\in \mathbb{C }\setminus \{z_0\}$, and $f^{(k)}(z)\not =1$ for $z\in \mathbb{C }$. Thus by Lemma 7, $f$ has a zero whose multiplicity is at most $k+1$. Since $f(z)\not =0$ for $z\in \mathbb{C }\setminus \{z_0\}$, this zero coincides with $z_0$, which contradicts the assumption that the zero $z_0$ of $f$ has multiplicity at least $k+2$.

Thus, $z_0$ is a zero of $f$ with multiplicity at most $k+1$. $\square$

Lemma 10

Let $k,m$ be positive integers and $f$ be a non-constant rational function. If $f^{(k)}(z)\not =z^m$ for $z\in \mathbb{C }$ and if $f(z)\not =0$ for $z\not =z_0$, where $z_0\in \mathbb{C }$, then $m=1$, $z_0\not =0$ and either $f(z)=(z-z_0)^{k+1}/(k+1)!$ or

$$\begin{aligned} f(z)=\frac{1}{(k+1)!}\cdot \frac{(z-z_0)^{k+2}}{z-(k+2)z_0}. \end{aligned}$$

Proof

Consider first the case that $f$ is a non-constant polynomial. Then by $f(z)\not =0$ for $z\not =z_0$, $f(z)=C_1(z-z_0)^l$ for some constant $C_1\not =0$ and $l\in \mathbb{N }$; and by $f^{(k)}(z)\not =z^m$, $f^{(k)}(z)=z^m+C_2$ for some constant $C_2\not =0$. Thus, $C_1[(z-z_0)^l]^{(k)}=z^m+C_2$. It can be easily seen that $z_0\not =0$, $m=1$, $l=k+1$ and $C_1=1/(k+1)!$. Hence $f(z)=(z-z_0)^{k+1}/(k+1)!$.

Now we assume that $f$ is a non-polynomial rational function. By $f^{(k)}(z)\not =z^m$ for $z\in \mathbb{C }$, we have

$$\begin{aligned} \left[f(z)-\frac{m!}{(m+k)!}z^{m+k}+\frac{1}{k!}z^k\right]^{(k)}=f^{(k)}(z)-z^m+1\not =1. \end{aligned}$$

Thus by Lemma 7,

$$\begin{aligned} f(z)=\frac{m!}{(m+k)!}z^{m+k}+P_k(z)+\frac{a}{(z-b)^n}, \end{aligned}$$

(1)

where $P_k,a,b,n$ are stated as in Lemma 7. Since $f(z)\not =0$ for $z\not =z_0$, we also get $f(z)=C(z-z_0)^{l}(z-b)^{-n},$ for some constant $C\not =0$ and integer $l\ge 0$. This, combined with (1), yields that

$$\begin{aligned} \left[\frac{m!}{(m+k)!}z^{m+k}+P_{k}(z)\right](z-b)^n=C(z-z_0)^{l}-a. \end{aligned}$$

(2)

Comparing the degree and the coefficient of the leading term of (2) yields that $l=m+k+n$ and $C=\frac{m!}{(m+k)!}$. Further, since each zero of the right hand side of (2) is simple, we see that $n=1$. Thus, $l=m+k+1$ and we can deduce from (2) that

$$\begin{aligned} \left[\frac{m!}{(m+k)!}z^{m+k}+P_{k}(z)\right](z-b)=\frac{m!}{(m+k)!}(z-z_0)^{m+k+1}-a. \end{aligned}$$

(3)

Now by comparing the coefficients of the term $z^{m+k}$, we get $b=(m+k+1)z_0$.

We claim that $z_0\not =0$. In fact, if $z_0=0$, then $b=0$, and hence $a=0$ by taking $z=0$ in (3). This is a contradiction.

Next by comparing the coefficients of the term $z^{m+k-1}$, we see that $m=1$. Thus, $f$ has the second desired form. $\square$

Lemma 11

Let $k,m$ be positive integers and $f$ be a rational function. If $f(z)\not =0$ for $z\in \mathbb{C }$, and $f^{(k)}(z)\not =z^m$ for $z\not =z_0$, where $z_0\in \mathbb{C }$, then $f$ is a constant.

Proof

If $f$ is a polynomial, then by $f(z)\not =0$, $f$ must be constant. Now suppose that $f$ is a non-polynomial rational function. Then by Lemma 10, $f^{(k)}(z)-z^m$ must have at least one zero. Hence by the condition, $f^{(k)}(z_0)=z_0^m$. Thus by $f(z)\not =0$ for $z\in \mathbb{C }$ and $f^{(k)}(z)\not =z^m$ for $z\not =z_0$, we can write

$$\begin{aligned} f(z)=C_1\prod _{i=1}^n(z-z_i)^{-p_i},\ f^{(k)}(z)=z^m+C_2(z-z_0)^{l}\prod _{i=1}^n(z-z_i)^{-p_i-k}, \end{aligned}$$

(4)

where $C_1, C_2$ are non-zero constants, $l,n, p_i$ are positive integers, and $z_i, 0\le i\le n$ are distinct complex numbers. By the expression of $f$ in (4), we have

$$\begin{aligned} f^{(k)}(z)=P(z)\prod _{i=1}^n(z-z_i)^{-p_i-k}, \end{aligned}$$

(5)

where $P$ is polynomial of degree $(n-1)k$. Thus by the two expressions of $f^{(k)}$ in (4) and (5),

$$\begin{aligned} z^m\prod _{i=1}^n(z-z_i)^{p_i+k}+C_2(z-z_0)^{l}=P(z). \end{aligned}$$

(6)

Since

$$\begin{aligned} \deg (P)=(n-1)k<m+\sum _{i=1}^n(p_i+k)=\deg \left(z^m\prod _{i=1}^n(z-z_i)^{p_i+k}\right), \end{aligned}$$

by (6), we have

$$\begin{aligned} l=m+\sum _{i=1}^n(p_i+k)=m+kn+\sum _{i=1}^np_i \end{aligned}$$

(7)

and $C_2=-1$. By letting $z=1/t$ in (6), we get

$$\begin{aligned} \prod _{i=1}^n(1-z_it)^{p_i+k}=(1-z_0t)^l+t^{l}P(1/t)= (1-z_0t)^l\left[1+O(t^{m+k+\sum _{i=1}^np_i})\right]\nonumber \\ \end{aligned}$$

(8)

as $t$ goes to $0$. Thus by taking the logarithmic derivatives,

$$\begin{aligned} \sum _{i=1}^n\frac{(p_i+k)z_i}{1-z_it}-\frac{lz_0}{1-z_0t}=O(t^{m+k-1+\sum _{i=1}^np_i}). \end{aligned}$$

(9)

It follows that

$$\begin{aligned} \sum _{i=1}^n(p_i+k)z_i^j-lz_0^j=0\ \ \text{ for}\ \ 1\le j\le m+k-1+\sum _{i=1}^np_i. \end{aligned}$$

(10)

Since $m+k-1+\sum _{i=1}^np_i\ge n+1$, it follows that the system of linear equations

$$\begin{aligned} \sum _{i=0}^nz_i^jx_i=0,\ \ \ 1\le j\le n+1 \end{aligned}$$

(11)

has a non-zero solution $(x_0,x_1,\ldots ,x_n)=(-l,p_1+k,\ldots ,p_n+k)$. This is impossible, as all $z_i$ are distinct. $\square$

Lemma 12

Let $k,m$ be positive integers, and let $f$ be a non-constant rational function. If $f(z)[f^{(k)}(z)-z^m]\not =0$ for $z\not =0$, and the multiplicity is at least $m+k+1$ when $0$ is a zero of $f$, then $k=1$ and $0$ is a zero of $f$ with exact multiplicity $2m+2$.

Proof

First, we show that $f$ cannot be a polynomial. Suppose not, then by $f(z)\not =0$ for $z\not =0$, $f(z)=Cz^s$ for some constant $C\not =0$ and integer $s\in \mathbb{N }$. Further, by the condition, $s\ge m+k+1$. Thus, $f^{(k)}(z)-z^m=Az^{s-k}-z^m=Az^m(z^{s-k-m}-1/A)$, where $A\not =0$ is a constant. This contradicts that $f^{(k)}(z)-z^m\not =0$ for $z\not =0$.

Thus, $f$ is a non-polynomial rational function. By Lemma 11, $f$ has at least one zero, and hence by $f(z)\not =0$ for $z\not =0$, we must have $f(0)=0$. Thus, we can write

$$\begin{aligned} f(z)=C_1z^l\prod _{i=1}^n(z-z_i)^{-p_i}, \end{aligned}$$

(12)

where $C_1\not =0$ is constant, $z_i\in \mathbb{C }, 1\le i\le n$ are distinct and non-zero, and $n,l, p_i\in \mathbb{N }$ with $l\ge m+k+1$ (by the condition). Thus, 0 is a zero of $f^{(k)}$ with multiplicity $l-k\ge m+1$, and hence 0 is a zero of $f^{(k)}(z)-z^m$ with exact multiplicity $m$. Hence, since $f^{(k)}(z)-z^m\not =0$ for $z\not =0$, we have

$$\begin{aligned} f^{(k)}(z)=z^m+C_2z^m\prod _{i=1}^n(z-z_i)^{-p_i-k}, \end{aligned}$$

(13)

for some constant $C_2\not =0$. However, by (12), one can obtain by induction that

$$\begin{aligned} f^{(k)}(z)=C_1z^{l-k}P(z)\prod _{i=1}^n(z-z_i)^{-p_i-k}, \end{aligned}$$

(14)

where

$$\begin{aligned} P(z)=\prod _{j=0}^{k-1}\left(l-j-\sum _{i=1}^np_i\right)z^{nk}+\cdots \ (\not \equiv 0) \end{aligned}$$

(15)

is a polynomial of degree at most $nk$. Thus by the two expressions of $f^{(k)}$ in (13) and (14),

$$\begin{aligned} z^m\prod _{i=1}^n(z-z_i)^{p_i+k}+C_2z^m=C_1z^{l-k}P(z), \end{aligned}$$

and hence

$$\begin{aligned} \prod _{i=1}^n(z-z_i)^{p_i+k}=-C_2+C_1z^{l-k-m}P(z). \end{aligned}$$

(16)

Then, comparing the degrees of the both sides of (16) shows that

$$\begin{aligned} \sum _{i=1}^n(p_i+k)=l-k-m+\deg (P). \end{aligned}$$

(17)

Since $\deg (P)\le nk$, it follows from (17) that $l\ge m+k+\sum _{i=1}^np_i$. Thus by (15), $\deg (P)=nk$ and then by (17),

$$\begin{aligned} l=m+k+\sum _{i=1}^np_i. \end{aligned}$$

(18)

By (16), we also have

$$\begin{aligned} C_2=-\prod _{i=1}^n(-z_i)^{p_i+k}. \end{aligned}$$

(19)

Now by taking the logarithmic derivatives of (16), we have

$$\begin{aligned} \sum _{i=1}^n\frac{p_i+k}{z-z_i}=O(z^{l-k-m-1})\ \ \text{ as}\ \ z\rightarrow 0. \end{aligned}$$

(20)

Thus,

$$\begin{aligned} \sum _{i=1}^n\frac{p_i+k}{(z_i)^j}=0\ \ \text{ for}\ \ 1\le j\le l-k-m-1. \end{aligned}$$

(21)

It follows that the system of linear equations

$$\begin{aligned} \sum _{i=1}^n\frac{x_i}{(z_i)^j}=0,\ \ \ 1\le j\le l-k-m-1 \end{aligned}$$

(22)

has a non-zero solution $(x_1,\ldots ,x_n)=(p_1+k,\ldots ,p_n+k)$. Since $z_i$ are non-zero and distinct, we get $l-k-m-1<n$, and hence by (18), $\sum _{i=1}^np_i<n+1$. It follows that all $p_i=1$. Thus, $l=m+k+n$ and by (21),

$$\begin{aligned} \sum _{i=1}^n\frac{1}{(z_i)^j}=0,\ \ j=1,\ldots ,n-1. \end{aligned}$$

(23)

It follows from the well-known Newton’s formula that $\prod _{i=1}^n(z-z_i)=z^n-r$, where $r\not =0$ is a constant. Hence by (19)

$$\begin{aligned} C_2=-\prod _{i=1}^n(-z_i)^{p_i+k}=-(-r)^{k+1}. \end{aligned}$$

(24)

Thus by (12) and (13),

$$\begin{aligned} f(z)&= C_1z^{k+m+n}(z^n-r)^{-1}=C_1z^{k+m}(1-rz^{-n})^{-1},\\ \nonumber f^{(k)}(z)&= z^m-(-r)^{k+1}z^m(z^n-r)^{-k-1} \end{aligned}$$

(25)

$$\begin{aligned}&= z^m\left[1-(-rz^{-n})^{k+1}(1-rz^{-n})^{-k-1}\right]. \end{aligned}$$

(26)

By (25), we have

$$\begin{aligned} f(z)=C_1z^{k+m}\sum _{s=0}^\infty (rz^{-n})^s =C_1\sum _{s=0}^\infty r^sz^{k+m-sn} \end{aligned}$$

for $z$ satisfying $|rz^{-n}|<1$, and hence

$$\begin{aligned} f^{(k)}(z)&= C_1\sum _{s=0}^\infty r^s\prod _{j=1}^k(j+m-sn)z^{m-sn}\nonumber \\&= C_1z^m\sum _{s=0}^\infty \prod _{j=1}^k(j+m-sn)(rz^{-n})^s. \end{aligned}$$

(27)

Thus by (26) and (27) with writing $w=rz^{-n}$,

$$\begin{aligned} 1-(-w)^{k+1}(1-w)^{-k-1}=C_1\sum _{s=0}^\infty \prod _{j=1}^k(j+m-sn)w^s\ \ \text{ for}\ \ |w|<1. \end{aligned}$$

(28)

Now comparing the coefficients of (28) yields that $\prod _{j=1}^k(j+m-sn)=0$ for $1\le s\le k$. Thus, $sn-m\in \{1,2,\ldots , k\}$ for each $1\le s\le k$. In particular, $1\le n-m\le k$ and $1\le kn-m\le k$. It follows that $k=1$ and $n=m+1$, and hence $0$ is a zero of $f$ with exact multiplicity $l=m+k+n=2m+2$. $\square$

3 Proof of Theorem 4

Let $z_0\in D$ be a point and $\{f_n\}\subset \mathcal{F }$ be a sequence. We have to prove that $\{f_n\}$ has a subsequence which is normal at $z_0$. We may assume that $z_0=0$. Since $\{E_{f_n}\}$ are locally uniformly discrete in $D$, there exists $\delta _0>0$ such that $E_{f_n}\cap \Delta (0,\delta _0)$ contains at most one point $z_{f_n}$. That is to say, we have $f_n\not =0$ and $f_n^{(k)}\not =h$ on $\Delta (0,\delta _0)\setminus \{z_{f_n}\}$ for all $f_n$. The following considerations for $\{f_n\}$ are understood to always hold with respect to the disk $\Delta (0,\delta _0)$.

Case 1

There exists $0<\delta <\delta _0$ such that $|z_{f_n}|\ge \delta$ for all $f_n$ (with $n$ sufficiently large). Then, we have $f_n\not =0$ and $f_n^{(k)}\not =h$ on $\Delta (0,\delta )$ for all $f_n$. Thus by Theorem 1, $\{f_n\}$ is normal on $\Delta (0,\delta )$, and hence at $z_0=0$.

Case 2

There exists a subsequence of $\{z_{f_n}\}$, which we continue to call $\{z_{f_n}\}$, such that $z_{f_n}\rightarrow z_0=0$. Then by Theorem 1, $\{f_n\}$ is normal on $\Delta ^\circ (0,\delta _0)$.

Suppose that $\{f_n\}$ has no subsequence which is normal at $0$. Next we consider two cases according to whether $h(0)$ is $0$ or not.

Case 3

For the case $h(0)\not =0$, we may say that $h(0)=1$. Since $\{f_n\}$ is not normal at $0$, by Lemma 5, there exist a subsequence of $\{f_n\}$ which we continue to call $\{f_n\}$, a sequence of points $z_n\rightarrow 0$ and a sequence of positive numbers $\rho _n\rightarrow 0$ such that

$$\begin{aligned} g_n(\zeta ):=\rho _n^{-k}f_n(z_n+\rho _n\zeta )\rightarrow g(\zeta ) \end{aligned}$$

(29)

spherically locally uniformly on $\mathbb{C }$, where $g$ is non-constant and meromorphic on $\mathbb{C }$.

Claim 1

$g^{(k)}(\zeta )\not \equiv 1$. In fact, if $g^{(k)}(\zeta )\equiv 1$, then $g$ is a polynomial with exact degree $k$, so that $g$ has a zero $\zeta _0\in \mathbb{C }$ with multiplicity at most $k$, and hence by applying Hurwitz’s theorem to (29), $g_n$ (for sufficiently large $n$) has a zero $\zeta _n\rightarrow \zeta _0$ with multiplicity at most $k$. It follows that $f_n$ has a zero $z_n+\rho _n\zeta _n\rightarrow 0$ with multiplicity at most $k$. This contradicts the assumption that all zeros of $f_n$ have multiplicity at least $k+2$.

Claim 2

$g$ has at most one zero, and if it has, then the multiplicity is at least $k+2$.

The latter assertion follows from an argument similar to that in Claim 1. We now prove the former. Suppose that $g$ has at least two distinct zeros $\zeta _1,\zeta _2\in \mathbb{C }$. Then by applying Hurwitz’s theorem to (29), $g_n$ (for sufficiently large $n$) has two distinct zeros $\zeta _n^{(1)}, \zeta _n^{(2)}$ tending to $\zeta _1,\zeta _2$ respectively, and hence $f_n$ has two distinct zeros $z_n+\rho _n\zeta _n^{(1)}$ and $z_n+\rho _n\zeta _n^{(2)}$, both tending to 0. This contradicts that $f_n(z)\not =0$ for $z\not =z_{f_n}$.

Claim 3

$g^{(k)}-1$ has at most one zero.

By Claim 1 and the fact that

$$\begin{aligned} f_n^{(k)}(z_n+\rho _n\zeta )-h(z_n+\rho _n\zeta )=g_n^{(k)}(\zeta )-h(z_n+\rho _n\zeta )\rightarrow g^{(k)}(\zeta )-1(\not \equiv 0)\nonumber \\ \end{aligned}$$

(30)

locally uniformly on $\mathbb{C }\setminus g^{-1}(\infty )$, an argument similar to that in Claim 2 yields this claim.

Claim 4

Either $g\not =0$ or $g^{(k)}\not =1$.

Suppose not, say $g(\zeta _0^{(1)})=0$ and $g^{(k)}(\zeta _0^{(2)})=1$. Since $\zeta _0^{(1)}$ is a zero of $g$ with multiplicity $\ge k+2$, we have $g^{(k)}(\zeta _0^{(1)})=0$, and hence $\zeta _0^{(1)}\not =\zeta _0^{(2)}$. By applying Hurwitz’s theorem to (29) and (30), there exist points $\zeta _n^{(i)}\rightarrow \zeta _0^{(i)}$ such that $g_n(\zeta _n^{(1)})=0$ and $g_n^{(k)}(\zeta _n^{(2)})-h(z_n+\rho _n\zeta _n^{(2)})=0$, and hence $f_n(z_n+\rho _n\zeta _n^{(1)})=0$ and $f_n^{(k)}(z_n+\rho _n\zeta _n^{(2)})-h(z_n+\rho _n\zeta _n^{(2)})=0$. Since $f_n(z)\not =0$ and $f^{(k)}_n(z)\not =h(z)$ for $z\not =z_{f_n}$, it follows that $z_n+\rho _n\zeta _n^{(1)}=z_n+\rho _n\zeta _n^{(2)}(=z_{f_n})$ so that $\zeta _n^{(1)}=\zeta _n^{(2)}$, and hence $\zeta _0^{(1)}=\zeta _0^{(2)}$. This is a contradiction.

Thus by Claims 2–4, $g(g^{(k)}-1)$ has at most one zero, and hence by Lemma 6, $g$ is a rational function. Further by Lemma 9, $g$ has a zero with multiplicity at most $k+1$. This contradicts Claim 2 which says that the zero of $g$ has multiplicity at least $k+2$.

Case 4

Next we consider the case that $h(0)=0$. Then $h(z)=z^m\phi (z)$, where $m\in \mathbb{N }$ and $\phi$ is holomorphic with $\phi (0)\not =0$. We can say $\phi \not =0$ in $\Delta (0,\delta _0)$ with the normalization $\phi (0)=1$. Set for each $n$

$$\begin{aligned} F_n(z)=z^{-m}f_n(z). \end{aligned}$$

(31)

Then $F_n(z)\not =0$ for $z\not =z_{f_n}$, since $f_n(z)\not =0$ for $z\not =z_{f_n}$.

We first show that $\{F_n\}$ has no subsequence which is normal at 0. Suppose not, say $\{F_n\}$ is normal at 0. Then $\{F_n\}$ has a subsequence, which we continue to call $\{F_n\}$, such that $\{F_n\}$ converges spherically locally uniformly to $\psi$ which may be $\infty$ identically in some neighborhood $\Delta (0,\eta _0)$.

If $\psi (0)\not =\infty$, then for sufficiently large $n$, $f_n(z)\not =\infty$ and $|F_n(z)|\le M$ in some closed domain $\overline{\Delta }(0,\eta )$ with $\eta <\eta _0$, where $M>0$ is a constant. It follows that the functions $f_n$ are holomorphic and satisfy $|f_n(z)|\le M|z|^{m}$ on $\overline{\Delta }(0,\eta )$. By the Montel’s theorem, $\{f_n\}$ is normal at 0, which contradicts our assumption that $\{f_n\}$ is not normal at 0.

If $\psi (0)=\infty$, then for sufficiently large $n$, $|F_n(z)|>1$ in some closed domain $\overline{\Delta }(0,\eta )$ with $\eta <\eta _0$. We claim that $f_n(z_{f_n})\not =0$ for all (sufficiently large) $n$. In fact, if $f_n(z_{f_n})=0$, then as $F_n(z_{f_n})\not =0$, we see that $z_{f_n}=0$, and so by the assumption on the multiplicities of common zeros of $f_n$ and $h$, $z_{f_n}=0$ is a zero of $f_n$ with multiplicity at least $m+k+1>m$, and hence $F_n(0)=0$, which contradicts that $|F_n(z)|>1$. Thus, $f_n(z_{f_n})\not =0$. This, combined with the fact that $f_n(z)\not =0$ for $z\not =z_{f_n}$, shows that the functions $1/f_n$ are holomorphic. By $|F_n(z)|>1$, we have $|1/f_n(z)|<|z|^{-m}$ on $\overline{\Delta }(0,\eta )$. Now the maximum modulus principle implies that $|1/f_n(z)|\le \eta ^{-m}$ on $\overline{\Delta }(0,\eta )$ and hence $\{1/f_n\}$ is normal at 0 by Montel’s theorem. This again contradicts our assumption that $\{f_n\}$ is not normal at 0.

Thus, $\{F_n\}$ has no subsequence which is normal at 0.

Claim 5

If $z_{f_n}$ is a zero of $F_n$, then the multiplicity is at least $k+1$, or $k+2$ if $z_{f_n}\not =0$. In fact, suppose $F_n(z_{f_n})=0$, then $f_n(z_{f_n})=0$. Thus for the case $z_{f_n}\not =0$, the claim is true by the assumption on the multiplicities of the zeros of $f_n$, while for the case $z_{f_n}=0$, $z_{f_n}=0$ is a common zero of $f_n$ and $h$, and hence by the assumption on the multiplicities of the common zeros of $f_n$ and $h$, 0 is a zero of $f_n$ with multiplicity at least $k+m+1$, and then the claim follows.

Thus by Claim 5 and the fact that $F_n(z)\not =0$ for $z\not =z_{f_n}$, Lemma 5 can be applied, and so there exist a subsequence of $\{f_n\}$, which we continue to call $\{f_n\}$, a sequence of points $w_n\rightarrow 0$ and a sequence of positive numbers $\eta _n\rightarrow 0$ such that

$$\begin{aligned} \widehat{g}_n(\zeta ):=\eta _n^{-k}F_n(w_n+\eta _n\zeta ) =\eta _n^{-k}(w_n+\eta _n\zeta )^{-m}f_n(w_n+\eta _n\zeta )\rightarrow \widehat{g}(\zeta ) \end{aligned}$$

(32)

spherically locally uniformly on $\mathbb{C }$, where $\widehat{g}$ is non-constant and meromorphic on $\mathbb{C }$.

Claim 6

$\widehat{g}$ has at most one zero.

In fact, if $\widehat{g}$ has two distinct zeros $\zeta _1,\zeta _2\in \mathbb{C }$, then applying Hurwitz’s theorem to (32), there exist points $\zeta _{i,n}\rightarrow \zeta _i$ such that $\widehat{g}_n(\zeta _{i,n})=0$ and hence $F_n(w_n+\eta _n\zeta _{i,n})=0$. Since $F_n(z)\not =0$ for $z\not =z_{f_n}$, we get $w_n+\eta _n\zeta _{1,n}=w_n+\eta _n\zeta _{2,n}(=z_{f_n})$. It follows that $\zeta _{1,n}=\zeta _{2,n}$, and hence $\zeta _1=\zeta _2$. This is a contradiction.

Next we consider two subcases according to whether the sequence $\{w_n/\eta _n\}$ is bounded or unbounded.

Case 5

First assume that the sequence $\{w_n/\eta _n\}$ is unbounded. Then, there exists a subsequence, which we continue to call $\{w_n/\eta _n\}$, such that $w_n/\eta _n\rightarrow \infty$.

Claim 7

If $\widehat{g}$ has a zero, then the multiplicity is at least $k+2$. And hence, $\widehat{g}^{(k)}\not \equiv 1$.

In fact, if $\widehat{g}(\zeta _{0})=0$, then as above, there exist points $\zeta _{n}\rightarrow \zeta _0$ such that $F_n(w_n+\eta _n\zeta _{n})=0$, and hence $z_{f_n}=w_n+\eta _n\zeta _{n}$. As $w_n/\eta _n\rightarrow \infty$, we see that $z_{f_n}\not =0$, and hence by Claim 5, $z_{f_n}$ is a zero of $F_n$ with multiplicity at least $k+2$. Thus, $\zeta _n$ is a zero of $\widehat{g}_n$ with multiplicity at least $k+2$. The claim then follows.

Claim 8

$\widehat{g}^{(k)}-1$ has at most one zero.

This follows from a similar argument as above with the following fact that

$$\begin{aligned} \frac{f_n^{(k)}(w_n+\eta _n\zeta )}{h(w_n+\eta _n\zeta )}&= \frac{[(\eta _n)^{-k}f_n(w_n+\eta _n\zeta )]^{(k)}}{h(w_n+\eta _n\zeta )}= \frac{[(w_n+\eta _n\zeta )^m\widehat{g}_n(\zeta )]^{(k)}}{h(w_n+\eta _n\zeta )}\nonumber \\&= \frac{\left[\left(\zeta \!+\!\frac{w_n}{\eta _n}\right)^m\widehat{g}_n(\zeta )\right]^{(k)}}{\left(\zeta \!+\!\frac{w_n}{\eta _n}\right)^m\phi (w_n+\eta _n\zeta )} =\frac{\sum _{i=0}^k\genfrac(){0.0pt}{}{k}{i}\left[\left(\zeta \!+\!\frac{w_n}{\eta _n}\right)^{m}\right]^{(k-i)} \widehat{g}_n^{(i)}(\zeta )}{\left(\zeta +\frac{w_n}{\eta _n}\right)^m\phi (w_n+\eta _n\zeta )}\nonumber \\&= \frac{1}{\phi (w_n+\eta _n\zeta )}\sum _{i=0}^{k} \frac{C_i\widehat{g}_n^{(i)}(\zeta )}{\left(\zeta +\frac{w_n}{\eta _n}\right)^{k-i}} \rightarrow \widehat{g}^{(k)}(\zeta ) \end{aligned}$$

(33)

locally uniformly on $\mathbb{C }\setminus \widehat{g}^{-1}(\infty )$, where $C_i=m(m-1)\cdots (m-(k-i)+1)\genfrac(){0.0pt}{}{k}{i}$ are constants, and in particular, $C_k=1$.

Claim 9

Either $\widehat{g}\not =0$ or $\widehat{g}^{(k)}\not =1$.

Combined with the fact that $F_n(z)\not =0$ and $f_n^{(k)}(z)\not =h(z)$ for $z\not =z_{f_n}$, this can be shown by an argument similar to the proof of Claim 4 by applying Hurwitz’s theorem to (32) and (33).

However, as in Case 2.1, the above Claims 6–9 lead to a contradiction.

Case 6

Now we consider the case that $\{w_n/\eta _n\}$ is bounded. Then, there is a subsequence, which we continue to call $\{w_n/\eta _n\}$, such that $w_n/\eta _n\rightarrow \alpha \in \mathbb{C }$. It follows from (32) that

$$\begin{aligned} \frac{f_n(\eta _n\zeta )}{\eta _n^{k+m}\zeta ^m}=\widehat{g}_n\left(\zeta -\frac{w_n}{\eta _n}\right) \rightarrow \widehat{g}(\zeta -\alpha ) \end{aligned}$$

(34)

spherically locally uniformly on $\mathbb{C }$, and hence

$$\begin{aligned} G_n(\zeta ):=\frac{f_n(\eta _n\zeta )}{\eta _n^{k+m}}=\zeta ^m\cdot \frac{f_n(\eta _n\zeta )}{\eta _n^{k+m}\zeta ^m}\rightarrow G(\zeta ):=\zeta ^m \widehat{g}(\zeta -\alpha ) \end{aligned}$$

(35)

spherically locally uniformly on $\mathbb{C }^*=\mathbb{C }\setminus \{0\}$, or on $\mathbb{C }$ if $\widehat{g}(-\alpha )\not =\infty$. Obviously, $G$ is meromorphic on $\mathbb{C }$ and $G\not \equiv 0$.

Claim 10

$G$ is non-constant. Suppose not. Then, $\widehat{g}(\zeta -\alpha )=A\zeta ^{-m}$ for some non-zero constant $A$. Thus by (34), we see that

$$\begin{aligned} G_n(\zeta ):=\frac{f_n(\eta _n\zeta )}{\eta _n^{k+m}}\rightarrow A \end{aligned}$$

(36)

locally uniformly on $\mathbb{C }$. Hence $G^{(k)}_n(\zeta )\rightarrow 0$, so that

$$\begin{aligned} \eta _n^{-m}[f_n^{(k)}(\eta _n\zeta )-h(\eta _n\zeta )]=G_n^{(k)}(\zeta )-\zeta ^m\phi (\eta _n\zeta )\rightarrow -\zeta ^m \end{aligned}$$

(37)

locally uniformly on $\mathbb{C }$. Thus by applying Hurwitz’s theorem to (37), there exist exactly $m$ points $\zeta ^{(j)}_n\rightarrow 0$, $j=1,2,\ldots ,m$, such that $f_n^{(k)}(\eta _n\zeta ^{(j)}_n)=h(\eta _n\zeta ^{(j)}_n)$. It follows from $f_n^{(k)}(z)\not =h(z)$ for $z\not =z_{f_n}$ that $z_{f_n}=\eta _n\zeta ^{(j)}_n$ for all $j$. This shows that the $m$ points $\zeta ^{(j)}_n$ are coincide with $\zeta _n:=z_{f_n}/\eta _n$, and $z_{f_n}$ is a zero of $f_n^{(k)}(z)-h(z)$ with multiplicity $m$, and by (36), $f_n(z_{f_n})\not =0$. Thus, $f_n(z)\not =0$ for $z\in \Delta (0,\delta _0)$.

Since $\{f_n\}$ is normal on $\Delta ^\circ (0,\delta _0)$, but not normal at 0, it follows from $f_n\not =0$ on $\Delta (0,\delta _0)$ that there exists a subsequence of $\{f_n\}$, which we continue to call $\{f_n\}$, such that $f_n\rightarrow 0$ and hence $f_n^{(k)}\rightarrow 0$ locally uniformly on $\Delta ^\circ (0,\delta _0)$.

Thus by the argument principle, we see that

$$\begin{aligned} n\left(\frac{\delta _0}{2},f_n^{(k)}-h\right)-n\left(\frac{\delta _0}{2},\frac{1}{f_n^{(k)}-h}\right) \rightarrow n\left(\frac{\delta _0}{2},h\right)-n\left(\frac{\delta _0}{2},\frac{1}{h}\right)=-m,\quad \end{aligned}$$

(38)

where $n(r,f)$ is the number of poles of $f$ in $\Delta (0,r)$, and $n(r,1/f)$ is the number of zeros of $f$ in $\Delta (0,r)$. Since both hand sides of (38) are integers, we see that for sufficiently large $n$,

$$\begin{aligned} n\left(\frac{\delta _0}{2},f_n^{(k)}-h\right)=n\left(\frac{\delta _0}{2},\frac{1}{f_n^{(k)}-h}\right)-m=0. \end{aligned}$$

(39)

It follows from (39) that $f_n$ are holomorphic on $\Delta (0,\delta _0/2)$. Hence by $f_n\rightarrow 0$ locally uniformly on $\Delta ^\circ (0,\delta _0)$, we get $f_n\rightarrow 0$ locally uniformly on $\Delta (0,\delta _0)$. This contradicts that $\{f_n\}$ is not normal at 0.

Claim 11

$G$ has at most one zero on $\mathbb{C }^*$, or on $\mathbb{C }$ if $\widehat{g}(-\alpha )\not =\infty$.

This can be proved by applying Hurwitz’s theorem to (35) with the fact that $f_n(z)\not =0$ for $z\not =z_{f_n}$.

Claim 12

If $G$ has a zero on $\mathbb{C }^*$, then $z_{f_n}\not =0$ and the multiplicity of the zero of $G$ on $\mathbb{C }^*$ is at least $k+2$, or $k+3$ if $m=1$.

Suppose that $\zeta _0\in \mathbb{C }^*$ is a zero of $G$. Then by applying Hurwitz’s theorem to (35), $G_n$ has a zero $\zeta _n\rightarrow \zeta _0$. Since $\zeta _0\not =0$, $\zeta _n\not =0$ (for $n$ large enough). It follows that $\eta _n\zeta _n\rightarrow 0$ is a non-zero zero of $f_n$. Since $f_n(z)\not =0$ for $z\not =z_{f_n}$, we see that $z_{f_n}=\eta _n\zeta _n\not =0$. Hence by the assumption, $z_{f_n}$ is a zero of $f_n$ with multiplicity at least $k+2$, or $k+3$ if $m=1$. It follows that the multiplicity of the zero $\zeta _0$ of $G$ is at least $k+2$, or $k+3$ if $m=1$.

Claim 13

If $G(0)=0$, then $f_n(0)=0$ and hence $z_{f_n}=0$ (for sufficiently large $n$).

For otherwise, say $f_n(0)\not =0$. Then by (34), 0 is a pole of $\widehat{g}_n(\zeta -w_n/\eta _n)$ with multiplicity at least $m$, and hence is a pole of $\widehat{g}(\zeta -\alpha )$ with multiplicity at least $m$. This shows that $0$ is not a zero of $G(\zeta )=\zeta ^m\widehat{g}(\zeta -\alpha )$, which is a contradiction.

By Claims 12 and 13, we see that if $G(0)=0$, then $G(\zeta )\not =0$ for $\zeta \in \mathbb{C }^*$. Hence by Claim 11, $G$ has at most one zero on $\mathbb{C }$.

Claim 14

If $0$ is a zero of $G$, then the multiplicity is at least $m+k+1$ for $k>1$ and $2m+3$ for $k=1$.

In fact, if $0$ is a zero of $G$, then Claim 13 shows that $0$ is a common zero of $f_n$ and $h$. Let $m_{f_n}$ be the multiplicity of $0$ as a zero of $f_n$. Then by the assumption, $m_{f_n}\ge m+k+1$ for $k>1$ and $m_{f_n}\ge 2m+3$ for $k=1$. It follows from (34) that $-w_n/\eta _n$ is a zero of $\widehat{g}_n$ with multiplicity $m_{f_n}-m$, so that $-\alpha$ is a zero of $\widehat{g}$ with multiplicity at least $k+1$ for $k>1$ and $m+3$ for $k=1$. The assertion then follows.

Now by (35), we have

$$\begin{aligned} \eta _n^{-m}[f_n^{(k)}(\eta _n\zeta )-h(\eta _n\zeta )]=G_n^{(k)}(\zeta )-\zeta ^m\phi (\eta _n\zeta )\rightarrow G^{(k)}(\zeta )-\zeta ^m \end{aligned}$$

(40)

locally uniformly on $\mathbb{C }^*\setminus G^{-1}(\infty )$, or on $\mathbb{C }\setminus G^{-1}(\infty )$ if $\widehat{g}(-\alpha )\not =\infty$.

Claim 15

$G^{(k)}(\zeta )\not \equiv \zeta ^m$.

For otherwise, $G(\zeta )=c\zeta ^{m+k}+P(\zeta )$ for some constant $c\not =0$ and polynomial $P$ of degree at most $k-1$. Thus, $G$ has at least one zero in $\mathbb{C }$. Let $\zeta _0$ be a zero of $G$, then by the fact that $\zeta _0$ has multiplicity $>k$, we get $G^{(k)}(\zeta _0)=0$ and hence $\zeta _0=0$ by $G^{(k)}(\zeta )\equiv \zeta ^m$. It follows that $G(\zeta )=c^{\prime }\zeta ^{l}$ for some $c^{\prime }\in \mathbb{C }$ and $l\in \mathbb{N }$. Thus, $c\zeta ^{m+k}+P(\zeta )=c^{\prime }\zeta ^l$. By comparing the degrees and coefficients, we see that $P\equiv 0$. Thus, $G(\zeta )=c\zeta ^{m+k}$. This contradicts Claim 14 which says that the multiplicity is at least $m+k+1$ if 0 is a zero of $G$.

Claim 16

$G^{(k)}(\zeta )-\zeta ^m$ has at most one zero on $\mathbb{C }^*$, or on $\mathbb{C }$ if $\widehat{g}(-\alpha )\not =\infty$.

By Claim 15, this claim can be seen by applying Hurwitz’s theorem to (40) with the condition $f_n^{(k)}(z)\not =h(z)$ for $z\not =z_{f_n}$.

By Claims 10, 11 and 16, it follows from Lemma 6 that $G$ is a non-constant rational function.

Claim 17

If $G^{(k)}(0)=0$, then $G\not =0$ on $\mathbb{C }^*$.

Suppose that $G(\zeta _0)=0$ for some $\zeta _0\not =0$. Then $G$ is holomorphic at 0. Since we have proved before Claim 14 that $G$ has at most one zero on $\mathbb{C }$, we get $G(0)\not =0$. Thus by $G(0)\not =0,\infty$, it follows from $G(\zeta )=\zeta ^m \widehat{g}(\zeta -\alpha )$ that $\widehat{g}(-\alpha )=\infty$ and that $0$ is a pole of $\widehat{g}(\zeta -\alpha )$ with exact multiplicity $m$. Thus by (34), there exists a positive number $\delta$ such that $f_n(\eta _n\zeta )\not =\infty$ for $|\zeta |\le \delta$ (and $n$ large enough), and hence $G_n(\zeta )$ is holomorphic on $|\zeta |\le \delta$. Thus, on $|\zeta |\le \delta$, the convergence of (40) is uniformly, and then Hurwitz’s theorem can be applied to (40). Since $G^{(k)}(0)=0^m$ and $G^{(k)}(\zeta )\not \equiv \zeta ^m$ by Claim 15, there exist points $\zeta _n\rightarrow 0$ such that $G_n^{(k)}(\zeta _n)=\zeta _n^m\phi (\eta _n\zeta _n)$ so that $f_n^{(k)}(\eta _n\zeta _n)=h(\eta _n\zeta _n)$. Hence, by $f_n^{(k)}(z)\not =h(z)$ for $z\not =z_{f_n}$, we get $z_{f_n}=\eta _n\zeta _n$. On the other hand, since $G(\zeta _0)=0$, applying Hurwitz’s theorem to (35) yields that there exist points $\zeta ^{\prime }_n\rightarrow \zeta _0$ such that $G_n(\zeta ^{\prime }_n)=0$ so that $f_n(\eta _n\zeta ^{\prime }_n)=0$. Thus by $f_n(z)\not =0$ for $z\not =z_{f_n}$, we get $z_{f_n}=\eta _n\zeta ^{\prime }_n$. Hence $\eta _n\zeta _n=\eta _n\zeta ^{\prime }_n(=z_{f_n})$ and then $\zeta ^{\prime }_n=\zeta _n$. It follows that $\zeta _0=0$, which is a contradiction.

Claim 18

If $G^{(k)}(0)=0$, then $G^{(k)}(\zeta )-\zeta ^m\not =0$ on $\mathbb{C }^*$.

Suppose that $G^{(k)}(\zeta _0)=\zeta _0^m$ for some $\zeta _0\not =0$. Then $G$ is holomorphic at 0. If $G(0)=0$, then by Claim 14, $\widehat{g}(-\alpha )=0$, and hence by Claim 16, $G^{(k)}(\zeta )-\zeta ^m$ has at most one zero on $\mathbb{C }$. This contradicts that $G^{(k)}(0)=0^m$ and $G^{(k)}(\zeta _0)=\zeta _0^m$. Thus, $G(0)\not =0$. Now by an argument similar to that in the proof of Claim 17, there exists $\delta >0$ such that on $|\zeta |\le \delta$, the convergence of (40) is uniformly, and then Hurwitz’s theorem can be applied to (40). It follows from $G^{(k)}(0)=0^m$ that there exist points $\zeta _n\rightarrow 0$ such that $G_n^{(k)}(\zeta _n)=\zeta _n^m\phi (\eta _n\zeta _n)$ so that $f_n^{(k)}(\eta _n\zeta _n)=h(\eta _n\zeta _n)$. Thus by $f_n^{(k)}(z)=h(z)$ for $z\not =z_{f_n}$, we get $z_{f_n}=\eta _n\zeta _n$. On the other hand, since $G^{(k)}(\zeta _0)=\zeta _0^m$ with $\zeta _0\not =0$, by applying Hurwitz’s theorem to (40), there exist points $\zeta ^{\prime }_n\rightarrow \zeta _0$ such that $G_n^{(k)}(\zeta ^{\prime }_n)=(\zeta ^{\prime }_n)^m\phi (\eta _n\zeta ^{\prime }_n)$ so that $f_n^{(k)}(\eta _n\zeta ^{\prime }_n)=h(\eta _n\zeta ^{\prime }_n)$. Again by $f_n^{(k)}(z)=h(z)$ for $z\not =z_{f_n}$, we get $z_{f_n}=\eta _n\zeta ^{\prime }_n$. Thus $\eta _n\zeta _n=\eta _n\zeta ^{\prime }_n(=z_{f_n})$ and then $\zeta ^{\prime }_n=\zeta _n$. It follows that $\zeta _0=0$, which is a contradiction.

Thus by Claims 16 and 18, $G^{(k)}(\zeta )-\zeta ^m$ has at most one zero on $\mathbb{C }$.

Claim 19

At least one of $G$ and $G^{(k)}(\zeta )-\zeta ^m$ has no zero on $\mathbb{C }^*$.

Suppose that $G(\zeta _0^{(1)})=0$ and $G^{(k)}(\zeta _0^{(2)})=(\zeta _0^{(2)})^m$ for some $\zeta _0^{(1)},\zeta _0^{(2)}\in \mathbb{C }^*$. By Claim 12, $\zeta _0^{(1)}$ is a zero of $G$ with multiplicity $\ge k+2$, and hence $G^{(k)}(\zeta _0^{(1)})=0$. It follows that $\zeta _0^{(1)}\not =\zeta _0^{(2)}$. Applying Hurwitz’s theorem to (35) shows that there exist points $\zeta _n^{(1)}\rightarrow \zeta _0^{(1)}$ such that $G_n(\zeta _n^{(1)})=0$. Thus $f_n(\eta _n\zeta _n^{(1)})=0$, and hence by $f_n(z)\not =0$ for $z\not =z_{f_n}$, we get $z_{f_n}=\eta _n\zeta _n^{(1)}$. Applying Hurwitz’s theorem to (40) shows that there exist points $\zeta _n^{(2)}\rightarrow \zeta _0^{(2)}$ such that $f_n^{(k)}(\eta _n\zeta _n^{(2)})=h(\eta _n\zeta _n^{(2)})$, and hence by $f_n^{(k)}(z)\not =h(z)$ for $z\not =z_{f_n}$, we get $z_{f_n}=\eta _n\zeta _n^{(2)}$. Thus $\zeta _n^{(1)}=\zeta _n^{(2)}$ and hence $\zeta _0^{(1)}=\zeta _0^{(2)}$. This contradicts that $\zeta _0^{(1)}\not =\zeta _0^{(2)}$.

Now, according to Claim 19, we are in one of the following three cases. Note that we have proved before Claim 17 that $G$ is a non-constant rational function.

Case 7

Neither of $G$ and $G^{(k)}(\zeta )-\zeta ^m$ has a zero on $\mathbb{C }^*$.

Then by Lemma 11, $G$ has at least one zero on $\mathbb{C }$, and hence $G(0)=0$. It now follows from Claim 14 and Lemma 12 that Case A cannot occur.

Case 8

$G(\zeta _0)=0$ for some $\zeta _0\in \mathbb{C }^*$ and $G^{(k)}(\zeta )-\zeta ^m$ has no zero on $\mathbb{C }^*$.

Then $G(\zeta )\not =0$ for $\zeta \in \mathbb{C }\setminus \{\zeta _0\}$, since we have proved before Claim 14 that $G$ has at most one zero on $\mathbb{C }$. Also, by Claim 17, we have $G^{(k)}(0)\not =0$, and hence $G^{(k)}(\zeta )\not =\zeta ^m$ for $\zeta \in \mathbb{C }$. It now follows from Claim 12 and Lemma 10 that Case B cannot occur.

Case 9

$G^{(k)}(\zeta _0)=\zeta _0^m$ for some $\zeta _0\in \mathbb{C }^*$ and $G(\zeta )$ has no zero on $\mathbb{C }^*$.

Then $G^{(k)}(\zeta )\not =\zeta ^m$ for $\zeta \in \mathbb{C }\setminus \{\zeta _0\}$, since we have proved before Claim 19 that $G^{(k)}(\zeta )-\zeta ^m$ has at most one zero on $\mathbb{C }$. Hence by Lemma 11, $G$ has at least one zero on $\mathbb{C }$, so that we must have $G(0)=0$. It now follows from Claim 14 that $G^{(k)}(0)=0$, which contradicts that $G^{(k)}(\zeta )\not =\zeta ^m$ for $\zeta \in \mathbb{C }\setminus \{\zeta _0\}$.

The proof of Theorem 4 is completed.

4 Proof of Theorem 3

From the proof of Theorem 4, we have the following result, which shows that the multiplicity $k+3$ in Theorem 4 can be replaced by $k+2$ when the set $E_f$ is independent of $f$. Note that the assumption that multiplicity $\ge k+3$ in Theorem 4 is only used to rule out the Case B within the proof of Theorem 4.

Theorem 13

Let $k\in \mathbb{N }$ and $h(\not \equiv 0)$ be a function holomorphic on $D$. Let $\mathcal{F }$ be a family of meromorphic functions in $D$, all of whose zeros have multiplicity at least $k+2$. Suppose that there exists a point $z_0\in D$ such that on $D\setminus \{z_0\}$, $f\not =0$ and $f^{(k)}\not =h$ for all functions $f\in \mathcal{F }$. Suppose additionally that if $z_0$ is a common zero of $f\in \mathcal{F }$ and $h$, the multiplicities $m_f$ for $f$ and $m_h$ for $h$ satisfy $m_f\ge m_h+k+1$ for $k>1$ and $m_f\ge 2m_h+3$ for $k=1$. Then $\mathcal{F }$ is normal in $D$.

Proof

The proof is similar to, and more simple than, the proof of Theorem 4, since in this case, we only have to consider the case that $z_{f_n}=0$ for all $n$ in the proof of Theorem 4. In fact, since $z_{f_n}=0$, we see from Claim 12 that $G$ has no zeros on $\mathbb{C }^*$, and a similar argument by applying Hurwitz’s theorem to (40) can be used to show that $G^{(k)}(\zeta )-\zeta ^m$ has no zeros on $\mathbb{C }^*$. So only the Case A appears and is required to be ruled out. $\square$

Proof of Theorem 3

Let $z_0\in D$ and a sequence $\{f_n\}\subset \mathcal{F }$. We have to prove that $\{f_n\}$ is normal at $z_0$.

Suppose first that there exists a neighborhood $U_0$ of $z_0$ such that on $U_0\setminus \{z_0\}$, $f_1\not =0$ and $f_1^{(k)}\not =h$. Then by the condition, on $U_0\setminus \{z_0\}$, $f_n\not =0$ and $f_n^{(k)}\not =h$ for all functions $f_n$. Hence, by Theorem 13, $\{f_n\}$ is normal in $U_0$ and hence normal at $z_0$.

The other case is that there exists a sequence $z_n\rightarrow z_0,\ z_n\not =z_0$ such that either $f_1(z_n)=0$ or $f_1^{(k)}(z_n)=h(z_n)$. It follows that either $f_1(z)\equiv 0$ or $f_1^{(k)}(z)\equiv h(z)$. Thus by the condition, either $f_n(z)\equiv 0$ or $f_n^{(k)}(z)\equiv h(z)$ for all $n$.

It is obviously that $\{f_n\}$ is normal if $f_n(z)\equiv 0$. Thus, we consider the case that $f_n^{(k)}(z)\equiv h(z)$. If $\{f_n\}$ is not normal at $z_0$, then by Lemma 5, there exists points $z_n\rightarrow z_0$, positive numbers $\rho _n\rightarrow 0$ and functions $f_n\in \{f_n\}$ such that $g_n(\zeta )=f_n(z_n+\rho _n\zeta )\rightarrow g(\zeta )$ locally uniformly on $\mathbb{C }$, where $g$ is a non-constant meromorphic function, all of whose zeros have multiplicity at least $k+2$. Since $g_n^{(k)}(\zeta )=\rho _n^kf^{(k)}_n(z_n+\rho _n\zeta )=\rho _n^kh(z_n+\rho _n\zeta )\rightarrow 0$, we get $g^{(k)}(\zeta )\equiv 0$. Since all zeros of $g$ have multiplicity at least $k+2$, it follows that $g$ is a constant, which is a contradiction.

Thus, $\{f_n\}$ is normal at $z_0$. Theorem 3 is proved.