Normal families and fixed-points of meromorphic functions

Abstract

In this paper, we obtain some normality criteria of families of meromorphic functions, which improve and generalize the related results of Gu, Pang-Yang-Zalcman, and Zhang-Pang-Zalcman, respectively. Some examples are given to show the sharpness of our results.

1 Introduction and main results

Let D be a domain in the complex plane $\mathbb {C}$, and $\mathcal F$ be a family of meromorphic functions defined on D. $\mathcal F$ is said to be normal on D, in the sense of Montel, if for any sequence $\{f_n\} \subset \mathcal F$ there exists a subsequence $\{f_{n_k}\}$, such that $\{f_{n_k}\}$ converges spherically locally uniformly on D, to a meromorphic function or $\infty$ (see [3],[8],[13]).

The following well-known normality criterion was conjectured by Hayman[3], and proved by Gu [2].

Theorem A

Let k be a positive integer. Let $\mathcal {F}$ be a family of meromorphic functions defined in a domain D. If for each $f\in \mathcal {F}$, $f\ne 0$ and $f^{(k)}\ne 1$, then $\mathcal {F}$ is normal in D.

This result has undergone various extensions and improvements. In [5] (cf. [6, 11]), Pang-Yang-Zalcman obtained.

Theorem B

Let k be a positive integer. Let $\mathcal {F}$ be a family of meromorphic functions defined in a domain D, all of whose zeros have multiplicity at least $k+2$ and whose poles are multiple. Let $h(z)(\not \equiv 0)$ be a holomorphic functions on D. If for each $f\in \mathcal {F}$, $f^{(k)}(z)\ne h(z)$, then $\mathcal {F}$ is normal in D.

When $k=1$, an example [19, Example 1] (cf. [6]) shows that the condition on the multiplicity of zeros of functions in $\mathcal F$ cannot be weakened. Zhang-Pang-Zalcman[14] proved that when $k\ge 2$ the multiplicity of zeros of functions in $\mathcal F$ can be reduced from $k+2$ to $k+1$ in Theorem B.

Theorem E

Let $k\ge 2$ be a positive integer. Let $\mathcal {F}$ be a family of meromorphic functions defined in a domain D, all of whose zeros have multiplicity at least $k+1$ and whose poles are multiple. Let $h(z)(\not \equiv 0)$ be a holomorphic functions on D. If for each $f\in \mathcal {F}$, $f^{(k)}(z)\ne h(z)$, then $\mathcal {F}$ is normal in D.

Also in [14], they indicated that one cannot further reduce the assumption on the multiplicity of the zeros from $k+1$ to k, by considering the following example.

Example 1

(see [14])   Let $\Delta =\{z: |z|<1\}$, $h(z)=z$, and let

$$\begin{aligned} \mathcal{F}=\left\{ f_n(z)=nz^k\right\} . \end{aligned}$$

Clearly, all zeros of $f_n$ are of multiplicity k, and $f^{(k)}_n(z)=nk!\ne z$ on $\Delta$. However, $\mathcal F$ fails to be equicontinuous at 0, and then $\mathcal F$ is not normal in $\Delta$.

In this paper, we consider the case $h(z)=z$, then $f^{(k)}(z)\ne h(z)$ means that $f^{(k)}$ has no fixed-points. We reduce the multiplicity of zeros of functions in $\mathcal F$ to k, but restricting the values $f^{(k)}$ can take at the zeros of f, as follows.

Theorem 1

Let $k\ge 4$ be a positive integer, $A>1$ be a constant. Let $\mathcal F$ be a family of meromorphic functions in a domain D. If, for every function $f\in \mathcal F$, f has only zeros of multiplicity at least k and satisfies the following conditions:

• (a) $f(z)=0\Rightarrow |f^{(k)}(z)|\le A|z|$.

• (b) $f^{(k)}(z)\ne z$.

• (c) All poles of f are multiple.

Then $\mathcal F$ is normal in D.

For the case $k=2$ or 3, the multiplicity of poles of $f\in \mathcal F$ need be at least three.

Theorem 2

Let $k=2$ or 3, $A>1$ be a constant. Let $\mathcal F$ be a family of meromorphic functions in a domain D. If, for every function $f\in \mathcal F$, f has only zeros of multiplicity at least k and satisfies the following conditions:

• (a) $f(z)=0\Rightarrow |f^{(k)}(z)|\le A|z|$.

• (b) $f^{(k)}(z)\ne z$.

• (c) All poles of f have multiplicity at least 3.

Then $\mathcal F$ is normal in D.

Example 1 shows that condition (a) in Theorems 1 and 2 cannot be removed. For the case $k=1$, the above theorems are no longer true even if the multiplicities of poles of $f\in \mathcal F$ are large enough, as is shown by the next example.

Example 2

Let j be a positive integer, $\Delta =\{z: |z|<1\}$, and let

$$\begin{aligned} \mathcal{F}=\left\{ f_n(z)=\frac{z^{j+2}-1/n^{j+2}}{2z^{j}}\right\} . \end{aligned}$$

Clearly,

$$\begin{aligned} f'_n(z)=z+\frac{j}{2n^{j+2}z^{j+1}}\ne z. \end{aligned}$$

For each n, $f_n$ has one pole $z=0$ with multiplicity j, and $j+2$ simple zeros $z_m=\frac{1}{n}e^{i\frac{2m\pi }{j+2}}(m=0,1,\ldots ,j+1)$ in $\Delta$. We have

$$\begin{aligned} f'_n(z_m)=z_m+\frac{j}{2n^{j+2}z_m^{j+1}}=\frac{j+2}{2n}e^{i\frac{2m\pi }{j+2}}, \end{aligned}$$

and then

$$\begin{aligned} |f'_n(z_m)|\le \frac{j+2}{2}|z_m|, \end{aligned}$$

that is, $f_n(z)=0\Rightarrow |f'_n(z)|\le \frac{j+2}{2}|z|$. But, since $f_n(1/n)=0$ and $f_n(0)=\infty$, $\mathcal F$ fails to be equicontinuous at $z=0$, and then $\mathcal F$ is not normal in $\Delta$.

The following example shows that condition (c) in Theorem 2 is necessary, and the number 3 is best possible.

Example 3

Let $\Delta =\{z: |z|<1\}$, and let

$$\begin{aligned} \mathcal{F}=\left\{ f_n(z)=\frac{(z-1/n)^3(z+1/n)^3}{24z^2}\right\} . \end{aligned}$$

Clearly,

$$\begin{aligned} f^{(3)}_n(z)=z+\frac{1}{n^{6}z^5}\ne z. \end{aligned}$$

For each n, $f_n$ has two zeros $z_1=1/n$ and $z_2=-1/n$ of multiplicity 3. We have

$$\begin{aligned} f^{(3)}_n(\frac{1}{n})=\frac{2}{n},\quad f^{(3)}_n(-\frac{1}{n})=-\frac{2}{n}, \end{aligned}$$

and $|f^{(3)}_n(z_i)|\le 2|z_i|(i=1,2),$ then $f_n(z)=0\Rightarrow |f^{(3)}_n(z)|\le 2|z|$. However $\mathcal F$ is not normal at 0 since $f_n(1/n)=0$ and $f_n(0)=\infty$.

The next example shows that condition (c) cannot be omitted in Theorem 1.

Example 4

Let k be a positive integer, $\Delta =\{z: |z|<1\}$ and

$$\begin{aligned} \mathcal {F}=\left\{ f_{n}(z)=\frac{1}{(k+1)!}\frac{(z-1/n)^{k+2}}{z-(k+2)/n}\right\} . \end{aligned}$$

Clearly, the zero of $f_{n}$ is of multiplicity $k+2$, so that $f(z)=0\Rightarrow |f^{(k)}(z)|\le |z|$; the pole of $f_{n}$ is simple. On the other hand, since

$$\begin{aligned} f_{n}(z)=\frac{1}{(k+1)!}\left( z^{k+1}+P_{k-1}(z)+\frac{a}{z-(k+2)/n}\right) , \end{aligned}$$

where $P_{k-1}(z)$ is a polynomial of degree $k-1$ and a is a nonzero constant, we have $f_{n}^{(k)}(z)\ne z$. But $\mathcal {F}$ is not normal at 0 since $f_n(1/n)=0$ and $f_n((k+2)/n)=\infty$.

In this paper, we write $\Delta =\{z: |z|<1\}$ and $\Delta '=\{z: 0<|z|<1\}$. For $z_0\in \mathbb {C}$ and $r>0$, we write $\Delta (z_0,r)=\{z: |z-z_0|<r\}$, and $\Delta '(z_0,r)=\{z: 0<|z-z_0|<r\}$.

2 Preliminary results

To prove our results, we need the following lemmas.

Lemma 1

[4, Lemma 2] Let k be a positive integer and let $\mathcal {F}$ be a family of meromorphic functions in a domain D, all of whose zeros have multiplicity at least k, and suppose that there exists $A\ge 1$ such that $|f^{(k)}(z) |\le A$ whenever $f(z)=0, f\in \mathcal {F}$. If $\mathcal {F}$ is not normal at $z_{0} \in D$, then for each $\alpha$, $0\le \alpha \le k$, there exist a sequence of complex numbers $z_{n} \in D$, $z_{n}\rightarrow z_{0}$, a sequence of positive numbers $\rho _{n} \rightarrow 0$, and a sequence of functions $f_{n} \in \mathcal {F}$ such that

$$\begin{aligned}g_{n}(\zeta )=\frac{f_{n}(z_{n}+\rho _{n}\zeta )}{\rho _{n}^{\alpha }}\rightarrow g(\zeta )\end{aligned}$$

locally uniformly with respect to the spherical metric, where g is a nonconstant meromorphic function on $\mathbb {C}$, all of whose zeros have multiplicity at least k , such that $g^{\#}(\zeta ) \le g^{\#}(0)= kA+1$. Moreover, $g(\zeta )$ has order at most 2.

Here, as usual, $g^{\#}(\zeta )=|g'(\zeta )|/(1+|g(\zeta )|^{2})$ is the spherical derivative.

Lemma 2

[11, Lemma 5] Let f be a transcendental meromorphic function, $k(\ge 2), \ell$ be positive integers. If f has only zeros of order at least 3, then $f^{(k)}-z^{\ell }$ has infinitely many zeros.

The next is a generalization of Hayman inequality, which is due to Yang [12].

Lemma 3

Let f be a transcendental meromorphic function, $\varphi$ be a small meromorphic function of f, and $k\in \mathbb {N}$. Then

$$\begin{aligned} T(r,f) \le 3N\left( r,\frac{1}{f}\right) +4N\left( r,\frac{1}{f^{(k)}-\varphi }\right) +S(r,f). \end{aligned}$$

Lemma 4

[1, Corollary 2] Let f be meromorphic in $\mathbb {C}$ and of finite order $\rho$ and E be the set of its critical values. If f has at most $2\rho +card E'$ asymptotic values, where $E'$ is the derived set of E.

Lemma 5

[7, Lemma 2.2] Let f be meromorphic in $\mathbb {C}$ and suppose that the set of all finite critical and asymptotic values of f is bounded. Then there exists $R>0$ such that if $|z|>R$ and $|f(z)|>R$, then

$$\begin{aligned} |f'(z)|\ge \frac{|f(z)|\log |f(z)|}{16 \pi |z|}. \end{aligned}$$

Lemma 6

Let f be a transcendental meromorphic function of finite order $\rho$, and let $k(\ge 2)$ be a positive integer. If f has only zeros of multiplicity at least k, and there exists $A>1$ such that $f(z)=0\Rightarrow |f^{(k)}(z)|\le A|z|$, then $f^{(k)}$ has infinitely many fix-points.

Proof

Suppose that $f^{(k)}$ has finitely many fix-points. Lemma 3 implies that f has infinitely many zeros, say $z_n$($n=1,2,\ldots$). Clearly, $z_n\rightarrow \infty$. Now set

$$\begin{aligned} g(z)=\frac{z^2}{2}-f^{(k-1)}(z). \end{aligned}$$

Then g is also of finite order $\rho$, and $g'(z)=z-f^{(k)}(z)$ has only finitely many zeros. By Lemma 4 or Denjoy-Carleman-Ahlfors’ theorem, g has at most $2\rho$ asymptotic values, and then satisfies the hypotheses of Lemma 5 for some $R>0$. It follows that

$$\begin{aligned} \frac{|z_ng'(z_n)|}{|g(z_n)|}\ge \frac{\log |g(z_n)|}{16\pi } \end{aligned}$$

for large n. Since $g(z_n)=z_n^{2}/2$ and $|g'(z_n)|=|z_n-f^{(k)}(z_n)|\le (A+1)|z_n|$, we have

$$\begin{aligned} 2(A+1)\ge \frac{1}{16\pi }[2\log |z_n|-\log 2]\rightarrow \infty \end{aligned}$$

as $n\rightarrow \infty ,$ a contradiction. Lemma 6 is proved. $\square$

Lemma 7

[10, Lemma 5] Let f be meromorphic in $\mathbb {C}$ and of finite order, and let $k\ge 2$ be a positive integer and K be a positive number. Suppose that f has only zeros of multiplicity at least k, $|f^{(k)}(z)|<K$ whenever $f(z)=0$, and $f^{(k)}(z)\ne 1$. Then one of the following two cases must occur:

• (1)
  $$\begin{aligned} f(z)=\alpha (z-\beta )^k, \end{aligned}$$

  (1)

  where $\alpha , \beta \in \mathbb {C}$, and $\alpha \cdot k!\ne 1.$

• (2) If $k=2$, then

  $$\begin{aligned} f(z)=\frac{(z-c_1)^2(z-c_2)^2}{2(z-c)^2}, \end{aligned}$$

  (2)

  or

  $$\begin{aligned} f(z)=\frac{(z-c_1)^3}{2(z-c)}. \end{aligned}$$

  (3)

  If $k\ge 3$, then

  $$\begin{aligned} f(z)=\frac{1}{k!}\frac{(z-c_1)^{k+1}}{(z-c)}. \end{aligned}$$

  (4)

  Here $c_1,c_2,c$ are distinct complex numbers.

Lemma 8

[9, Lemma 8] Let f be a non-polynomial rational function and k be a positive integer. If $f^{(k)}(z)\ne 1$, then

$$\begin{aligned} f(z)=\frac{1}{k!}z^k+a_{k-1}z^{k-1}+\cdots +a_0+\frac{a}{(z-b)^m}, \end{aligned}$$

where $a_{k-1},\ldots , a_0, a(\ne 0) ,b$ are constants and m is a positive integer.

Lemma 9

Let $k(\ge 2)$ be a positive integer, and f be a rational function, all of whose zeros are of multiplicity at least k. If $f^{(k)}(z)\ne z$, then one of the following three cases must occur:

• (1)
  $$\begin{aligned} f(z)=\frac{(z+c)^{k+1}}{(k+1)!}; \end{aligned}$$

  (5)
• (2)
  $$\begin{aligned} f(z)=\frac{(z-c_1)^{k+2}}{(k+1)!(z-b)}; \end{aligned}$$

  (6)
• (3)
  $$\begin{aligned}&f(z)=\frac{(z-c_1)^2(z-c_2)^3}{6(z-b)^2} \ (for\ k=2),\end{aligned}$$

  (7)
  $$\begin{aligned}&f(z)=\frac{(z-c_1)^3(z-c_2)^3}{24(z-b)^2} \ (for\ k=3), \end{aligned}$$

  (8)

where c is nonzero constant, and $c_1, c_2$ and b are distinct constants.

Proof

Suppose first that f is a polynomial. Then $f^{(k)}(z)=z+c$, where $c(\ne 0)$ is a constant, so that

$$\begin{aligned} f^{(k-1)}(z)=\frac{z^2}{2}+cz+d \end{aligned}$$

where d is a constant. If f vanishes at $z_0$, then $f^{(k-1)}(z_0)=z_0^2/2+cz_0+d=0$ since f has only zeros of multiplicity at least k. It follows that f has at most two zeros. So f has either only one zero of multiplicity $k+1$ or two distinct zeros of multiplicity exactly k. If f has two distinct zeros of multiplicity exactly k, then $\deg f=2k$ and $\deg f^{(k)}=k$, which contradicts the fact that $f^{(k)}(z)=z+c$ and $k\ge 2$. Thus, f has only one zero of multiplicity $k+1$, and hence f has the form (5).

Suppose then that f is a nonpolynomial rational function. Set

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

Then $g^{(k)}(z)\ne 1$, so by Lemma 8

$$\begin{aligned} g(z)=\frac{1}{k!}z^k+a_{k-1}z^{k-1}+\cdots +a_0+\frac{a}{(z-b)^m}, \end{aligned}$$

where $a_{k-1},\ldots , a_0, a(\ne 0) ,b$ are constants and m is a positive integer. Thus

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

(9)

where

$$\begin{aligned} p(z)=\frac{1}{(k+1)!}z^{k+1}+a_{k-1}z^{k-1}+\cdots +a_0. \end{aligned}$$

Let $c_1, c_2,\cdots , c_q$ be q distinct zeros of $p(z)(z-b)^m+a$, with multiplicity $n_1, n_2, \cdots , n_q$. Clearly, $n_i\ge k$, $c_i\ne b$, and $c_i$ is a zero of $(p(z)(z-b)^m+a)'$ with multiplicity $n_i-1\ge k-1(1\le i\le q)$. Since

$$\begin{aligned} \left( p(z)(z-b)^m+a\right) '=(z-b)^{m-1}\left( p'(z)(z-b)+mp(z)\right) , \end{aligned}$$

(10)

then $c_i$ must be a zero of $p'(z)(z-b)+mp(z)$ with multiplicity $n_i-1(\ge k-1)$. Note that $\deg [p'(z)(z-b)+mp(z)]=k+1$. Now we divide into three cases.

Case 1. $k=2.$

Then $\deg [p'(z)(z-b)+mp(z)]=3$, and hence

• (a) $p'(z)(z-b)+mp(z)$ has three simple zeros $c_1, c_2$, and $c_3$; or

• (b) $p'(z)(z-b)+mp(z)$ has one simple zero $c_1$ and one zero $c_2$ with multiplicity 2; or

• (c) $p'(z)(z-b)+mp(z)$ has only one zero $c_1$ with multiplicity 3.

For case (a), we deduce that $m=3$, and

$$\begin{aligned} p'(z)(z-b)+3p(z)=(z-c_1)(z-c_2)(z-c_3), \end{aligned}$$

$$\begin{aligned} p(z)(z-b)^3+a=\frac{1}{6}(z-c_1)^2(z-c_2)^2(z-c_3)^2. \end{aligned}$$

These, together with (10) give

$$\begin{aligned} (z-b)^2= \frac{1}{3}[(z-c_1)(z-c_2)+(z-c_1)(z-c_3)+(z-c_2)(z-c_3)]. \end{aligned}$$

Equating coefficients, we have $b=(c_1+c_2+c_3)/3$ and $b^2=(c_1c_2+c_1c_3+c_2c_3)/3$, so that

$$\begin{aligned} c_1^2+c_2^2+c_2^2=c_1c_2+c_1c_3+c_2c_3, \end{aligned}$$

that is,

$$\begin{aligned} (c_1-c_2)^2+(c_1-c_3)^2+(c_2-c_3)^2=0, \end{aligned}$$

and hence $c_1=c_2=c_3$, a contradiction. Thus case (1) is ruled out.

For case (b), we deduce that $m=2$ and

$$\begin{aligned} p(z)(z-b)^2+a=\frac{1}{6}(z-c_1)^2(z-c_2)^3. \end{aligned}$$

Then, by (9), f has the form (7).

For case (c), we can deduce that $m=1$ and

$$\begin{aligned} p(z)(z-b)+a=\frac{1}{6}(z-c_1)^4, \end{aligned}$$

This, together with (9), gives that f has the form (6).

Case 2. $k=3$.

Since $\deg [p'(z)(z-b)+mp(z)]=4$, $p'(z)(z-b)+mp(z)$ has two zeros $c_1, c_2$ with multiplicity 2 or one zero $c_1$ with multiplicity 4. It follows that $m=2$ and

$$\begin{aligned} p(z)(z-b)^2+a=\frac{1}{24}(z-c_1)^3(z-c_2)^3 \end{aligned}$$

or $m=1$ and

$$\begin{aligned} p(z)(z-b)+a=\frac{1}{24}(z-c_1)^5. \end{aligned}$$

Then, by (9), f has the form (6) or (8).

Case 3. $k\ge 4$.

Noting that $\deg [p'(z)(z-b)+mp(z)]=k+1$, we conclude that $p'(z)(z-b)+mp(z)$ has only one zero $c_1$ with multiplicity $k+1$. In fact, if $p'(z)(z-b)+mp(z)$ has at least two zeros $c_1, c_2$ with multiplicity $n_1, n_2,\ge k-1$, then $2(k-1)\le k+1$, and thus $k\le 3$, a contradiction. Thus $m=1$ and $p(z)(z-c)+b=\frac{1}{k!}(z-c_1)^{k+2}$, and hence f has the form (6). This completes the proof of Lemma 9. $\square$

Lemma 10

Let $k\ge 3$ be a positive integer, $A>1$ be a constant. Let $\mathcal F$ be a family of meromorphic functions in a domain D. Suppose that, for every $f\in \mathcal F$, f has only zeros of multiplicity at least k, and satisfies the following conditions:

• (a) $f(z)=0\Rightarrow |f^{(k)}(z)|\le A|z|$.

• (b) $f^{(k)}(z)\ne z$.

• (c) all poles of f are multiple.

Then $\mathcal F$ is normal in $D\backslash \{0\}$.

Proof

Suppose that $\mathcal F$ is not normal at a point $z_0\in D\backslash \{0\}$. Giving a small $r>0$ such that $\Delta (z_0,r)\subset D\backslash \{0\}$ and $f(z)=0\Rightarrow |f^{(k)}(z)|\le A|z_0|+1$ for $f\in \mathcal F$ and $z\in \Delta (z_0,r)$. Then by Lemma 1, for $\alpha =k$, there exist a sequence of functions $f_n\in \mathcal F$, a sequence of complex numbers $z_n\rightarrow z_0$ and a sequence of positive numbers $\rho _n \rightarrow 0$, such that

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

converges sphericaly uniformly on compact subsets of ${\mathbb {C}}$, where g is a non-constant meromorphic functionon on $\mathbb {C}$, all zeros of g have multiplicity at least k, and

$$\begin{aligned}g^{\#}(\zeta )\le g^{\#}(0)=k(A|z_0|+1)+1\end{aligned}$$

for all $\zeta \in \mathbb {C}$. Moreover, , g is of finite order. By Hurwitz’s theorem, all poles of g are multiple.

We claim: (1) $g=0\Rightarrow |g^{(k)}|\le A|z_0|$; (2) $g^{(k)}(\zeta )\ne z_0$.

Let $\zeta _0$ be a zero of $g(\zeta )$. Then there exist $\zeta _n, \zeta _n \rightarrow \zeta _0$, such that $g_n(\zeta _n)=\rho _n^{-k}f_n(z_n+\rho _n \zeta _n)=0$ for n sufficiently large. Thus $f_n(z_n+\rho _n \zeta _n)=0$, so that $|f^{(k)}_n(z_n+\rho _n \zeta _n)|\le A|z_n+\rho _n \zeta _n|$ for sufficiently large n. Since

$$\begin{aligned} g_n^{(k)}(\zeta _n)=f_n^{(k)}(z_n+\rho _n \zeta _n)\rightarrow g^{(k)}(\zeta _0), \end{aligned}$$

we have $|g^{(k)}(\zeta _0)|\le A|z_0|$. We have proved (i).

Suppose that there exists $\zeta _0$ such that $g^{(k)}(\zeta _0)=z_0$. Since

$$\begin{aligned} 0\ne f^{(k)}_n(z_n+\rho _n \zeta )-(z_n+\rho _n \zeta )=g^{(k)}_n(\zeta )-(z_n+\rho _n \zeta )\rightarrow g^{(k)}(\zeta )-z_0, \end{aligned}$$

Hurwitz’s theorem implies that $g^{(k)}(\zeta )\equiv z_0$. Note that g has only zeros of multiplicity at least k, we have

$$\begin{aligned} g(\zeta )=\frac{z_0}{k!}(z-\alpha )^k,\quad \alpha \in \mathbb {C}. \end{aligned}$$

A simple calculation shows that

$$\begin{aligned} g^{\#}(0)\le \left\{ \begin{array}{ll} k/2 \ &{} \quad \mathrm {if}\ |\alpha |\ge 1; \\ |z_0|\ &{} \quad \mathrm {if}\ |\alpha |<1. \end{array} \right. \end{aligned}$$

But this contradicts $g^{\#}(0)=k(A|z_0|+1)+1$, and thus (2) is proved.

By Lemma 7, g has the form (1) or (4) in Lemma 7. Similarly as above, we exclude the case that g has the form (1), so that g has the form (4). But g has only multiple poles, a contradiction. This completes the proof of Lemma 10. $\square$

Lemma 11

Let $\mathcal F$ be a family of meromorphic functions in a domain D, $A>1$ be a constant. Suppose that, for every $f\in \mathcal F$, f has only zeros of multiplicity at least k, and satisfies the following conditions:

• (a) $f(z)=0\Rightarrow |f''(z)|\le A|z|$.

• (b) $f''(z)\ne z$.

• (c) all poles of f are of multiplicity at least 3.

Then $\mathcal F$ is normal in $D\backslash \{0\}$.

This lemma can be proved almost the same as Lemma 10. We omit the details here.

3 Proof of Theorems 1 and 2

Proof of Theorem 1

Since normality is a local property, by Lemma 10, we only need to prove that $\mathcal {F}$ is normal at $z=0$. Without loss of generality, we may assume $D=\Delta$. Suppose, on the contrary, $\mathcal {F}$ is not normal at the origin. Our goal is to obtain a contradiction in the sequel.

Consider the family

$$\begin{aligned} \mathcal {G}=\left\{ g(z)=\frac{f(z)}{z}:f\in \mathcal {F}\right\} . \end{aligned}$$

We claim that $f(0)\ne 0$ for every $f\in \mathcal {F}$. Otherwise, if $f(0)=0$, by the assumption of Theorem 1, $|f^{(k)}(0)|\le 0$, and then $f^{(k)}(0)=0$. But $f^{(k)}(z)\ne z$, a contradiction. Thus, for each $g\in \mathcal {G}$, $g(0)=\infty$. Furthermore, all zeros of g(z) have multiplicity at least k. On the other hand, by simple calculation, we have

$$\begin{aligned} g^{(k)}(z)=\frac{f^{(k)}(z)}{z}-\frac{kg^{(k-1)}(z)}{z}. \end{aligned}$$

(11)

Since $f(z)=0\Rightarrow |f^{(k)}(z)|\le A|z|$, we deduce that $g(z)=0\Rightarrow |g^{(k)}(z)|\le A$.

We first prove that $\mathcal {G}$ is normal at 0. Suppose not; by Lemma 1, there exist functions $g_{n}\in \mathcal {G}$, points $z_{n}\rightarrow 0$ and positive numbers $\rho _{n}\rightarrow 0$ such that

$$\begin{aligned} G_{n}(\zeta )=\frac{g_{n}(z_{n}+\rho _{n}\zeta )}{\rho _{n}^k}\rightarrow G(\zeta ), \end{aligned}$$

(12)

converges spherically uniformly on compact subsets of ${\mathbb {C}}$, where G is a non-constant meromorphic functionon on $\mathbb {C}$ and of finite order, all zeros of G have multiplicity at least k, and $G^{\#}(\zeta )\le G^{\#}(0)=kA+1$ for all $\zeta \in \mathbb {C}$.

We distinguish two cases.

Case 1.   $z_{n}{/}\rho _{n}\rightarrow \infty$. Since $G_{n}(-z_{n}{/}\rho _{n})=g_{n}(0){/}\rho _{n}^k$, the pole of $G_n$ corresponding to that of $g_n$ at 0 drifts to infity. Then, by Hurwitz’s theorem, G has only mutiple poles. By (11) and (12), we have

$$\begin{aligned} G_n^{(k)}(\zeta )= & {} g_n^{(k)}(z_n+\rho _n\zeta )\\= & {} \frac{f_n^{(k)}(z_n+\rho _n\zeta )}{z_n+\rho _n\zeta }- k\frac{g_n^{(k-1)}(z_n+\rho _n\zeta )}{\rho _n}\frac{\rho _n}{z_n+\rho _n\zeta }.\\ \end{aligned}$$

Noting that

$$\begin{aligned} \frac{\rho _n}{z_n+\rho _n\zeta }\rightarrow 0 \end{aligned}$$

uniformly on compact subsets of $\mathbb {C}$, and $g_n^{(k-1)}(z_n+\rho _n\zeta ){/}\rho _n$ is locally bounded on $\mathbb {C}\backslash G^{-1}(\infty )$ since $g_{n}(z_{n}+\rho _{n}\zeta ){/}\rho _{n}^k\rightarrow G(\zeta )$. Thus

$$\begin{aligned} \frac{f_n^{(k)}(z_n+\rho _n\zeta )}{z_n+\rho _n\zeta }\rightarrow G^{(k)}(\zeta ), \end{aligned}$$

(13)

uniformly on compact subsets of $\mathbb {C}{\backslash } G^{-1}(\infty )$.

Claim: (I) $G(\zeta )=0\Rightarrow |G^{(k)}(\zeta )|\le A$; (II) $G^{(k)}(\zeta )\ne 1$.

Indeed, if $G(\zeta _0)=0$, Hurwitz’s theorem and (12) imply that there exist $\zeta _n, \zeta _n \rightarrow \zeta _0$, such that $g_n(z_n+\rho _n \zeta _n)=0$, and then $f_n(z_n+\rho _n \zeta _n)=0$ for n sufficiently large. By assumption, $|f_n^{(k)}(z_n+\rho _n\zeta _n)|\le A|z_n+\rho _n\zeta _n|$. It follows from (13) that $|G^{(k)}(\zeta _0)|\le A$. Claim (I) is proved.

Since $f_n^{(k)}(z)\ne z$, Hurwitz’s theorem and (13) yield that either $G^{(k)}(\zeta )\ne 1$ or $G^{(k)}(\zeta )\equiv 1$ for any $\zeta \in \mathbb {C}\backslash G^{-1}(\infty )$. Clearly, these also hold for all $\zeta \in \mathbb {C}$. If $G^{(k)}(\zeta )\equiv 1$, noting that all zeros of G have multiplicity at least k, we have $G(\zeta )=(\zeta -\alpha )^k{/}k!$($\alpha \in \mathbb {C}$). As in the proof of Lemma 10,

$$\begin{aligned} G^{\#}(0)\le \left\{ \begin{array}{ll} k/2 \ &{} \quad \mathrm {if}\ |\alpha |\ge 1; \\ 1\ &{} \quad \mathrm {if}\ |\alpha |<1. \end{array} \right. \end{aligned}$$

which contradicts $G^{\#}(0)=kA+1$. Then Claim (II) is proved. Then by Lemma 7, G has the form (1) or (4) in Lemma 7. The form (1) can be ruled out similarly as above. Thus

$$\begin{aligned} G(\zeta )=\frac{1}{k!}\frac{(\zeta -c_1)^{k+1}}{(\zeta -c)}, \end{aligned}$$

where $c_1, c$ are distinct complex numbers. But, this contradicts that G has only mutiple poles.

Case 2.   

. Taking subsequence, we can assume that $z_{n}/\rho _{n}\rightarrow \alpha ,$ a finite complex number. Then

$$\begin{aligned} \frac{g_{n}(\rho _{n}\zeta )}{\rho _{n}^{k}}=G_{n}(\zeta -z_{n}/\rho _{n})\overset{\chi }{\rightarrow } G(\zeta -\alpha )=\widetilde{G}(\zeta ) \end{aligned}$$

on $\mathbb {C}$. Clearly, all zeros of $\widetilde{G}$ have multiplicity at least k, and all poles of $\widetilde{G}$ are multiple, except possibly the pole at 0.

Set

$$\begin{aligned} H_{n}(\zeta )=\frac{f_{n}(\rho _{n}\zeta )}{\rho _{n}^{k+1}}. \end{aligned}$$

(14)

Then

$$\begin{aligned} H_{n}(\zeta )=\frac{f_{n}(\rho _{n}\zeta )}{\rho _{n}^{k+1}}=\zeta \frac{g_{n}(\rho _{n}\zeta )}{\rho _{n}^{k}}\rightarrow \zeta \widetilde{G}(\zeta )=H(\zeta ) \end{aligned}$$

(15)

spherically uniformly on compact subsets of ${\mathbb {C}}$, and

$$\begin{aligned} H_{n}^{(k)}(\zeta )=\frac{f_{n}^{(k)}(\rho _{n}\zeta )}{\rho _{n}} \rightarrow H^{(k)}(\zeta ) \end{aligned}$$

(16)

locally uniformly on $\mathbb {C}\setminus H^{-1}(\infty )$. Obviously, all zeros of H have multiplicity at least k, and all poles of H are multiple. Since $\widetilde{G}(0)=\infty$, $H(0)\ne 0$.

Claim: (III) $H(\zeta )=0\Rightarrow |H^{(k)}(\zeta )|\le A|\zeta |$; (IV) $H^{(k)}(\zeta )\ne \zeta$.

If $H(\zeta _0)=0$, by Hurwitz’s theorem and (15), there exist $\zeta _n \rightarrow \zeta _0$ such that $f_n(\rho _n \zeta _n)=0$ for for n sufficiently large. By the assumption, $|f_n^{(k)}(\rho _n\zeta _n)|\le A|\rho _n\zeta _n|$. Then, it follows from (16) that $|H^{(k)}(\zeta _0)|\le A|\zeta _0|$. Claim (III) is proved.

Suppose that there exists $\zeta _0$ such that $H^{(k)}(\zeta _0)=\zeta _0$. By (16),

$$\begin{aligned} 0\ne \frac{f_{n}^{(k)}(\rho _{n}\zeta )-\rho _{n}\zeta }{\rho _{n}}=H_{n}^{(k)}(\zeta )-\zeta \rightarrow H^{(k)}(\zeta )-\zeta , \end{aligned}$$

uniformly on compact subsets of $\mathbb {C}{\setminus } H^{-1}(\infty )$. Hurwitz’s theorem implies that $H^{(k)}(\zeta )\equiv \zeta$ on $\mathbb {C}{\setminus } H^{-1}(\infty )$, and then on $\mathbb {C}$. It follows that H is a polynomial of degree $k+1$. Since all zeros of H have multiplicity at least k, and noting that $k\ge 4$, we know that H has a single zero $\zeta _1$ with multiplicity $k+1$, so that $H^{(k)}(\zeta _1)=0$, and hence $\zeta _1=0$ since $H^{(k)}(\zeta )\equiv \zeta$. But $H(0)\ne 0$, we arrive at a contradiction. This proves claim (IV).

Then, by Lemma 6, H must be a rational function, and thus Lemma 9 implies that H has the form (5) or (6) in Lemma 9. The form (6) can be excluded since all poles of H are multiple. Thus we have

$$\begin{aligned} H(\zeta )=\frac{(\zeta +c)^{k+1}}{(k+1)!} \end{aligned}$$

(17)

where $c(\ne 0)$ is a constant.

Next we will show that (17) is impossible. Indeed, combining (15) and (17) gives

$$\begin{aligned} \frac{f_{n}(\rho _{n}\zeta )}{\rho _{n}^{k+1}}\rightarrow \frac{(\zeta +c)^{k+1}}{(k+1)!}. \end{aligned}$$

(18)

Note that all zeros of $f_n$ have multiplicity at least k and $k\ge 4$, there exist points $\zeta _{n,0}\rightarrow -c$ such that $z_{n,0}=\rho _{n}\zeta _{n,0}$ is a zero of $f_{n}$ with multiplicity $k+1$.

We now consider two subcases.

Case 2.1 There exists $0<\delta \le 1$ such that the functions $f_{n}(z)$ (for large n) are all holomorphic on $\Delta (0,\delta )$.

Since $\{f_{n}\}$ is normal on $\Delta '(0,\delta )$, but not normal at 0, it follows from the maximum modulus principle that $f_{n}\rightarrow \infty$ locally uniformly on $\Delta '(0,\delta ).$

Suppose that there exists $0<\sigma <\delta$ such that each $f_{n}$ has only one zero $z_{n,0}$ in $\Delta (0,\sigma ).$ Set

$$\begin{aligned} K_{n}(z)=\frac{f_{n}(z)}{(z-z_{n,0})^{k+1}}. \end{aligned}$$

(19)

Then $\{K_{n}\}$ is a sequence of nonvanishing holomorphic functions on $\Delta (0,\sigma )$, and $K_{n}(z)\rightarrow \infty$ locally uniformly on $\Delta '(0,\sigma )$. It follows that $\{1{/}K_{n}\}$ is holomorphic on $\Delta (0,\sigma )$, and $1{/}K_{n}(z)\rightarrow 0$ locally uniformly on $\Delta '(0,\sigma )$, and hence on $\Delta (0,\sigma )$ by the maximum modulus principle. So $K_{n}(z)\rightarrow \infty$ locally uniformly on $\Delta (0,\sigma )$. In particular, $K_{n}(2z_{n,0})\rightarrow \infty .$ But, by (18) and (19),

$$\begin{aligned} K_{n}(2z_{n,0})=\frac{f_{n}(2z_{n,0})}{z_{n,0}^{k+1}} =\frac{f_{n}(2\rho _{n}\zeta _{n,0})}{\rho _{n}^{k+1}\zeta _{n,0}^{k+1}}\rightarrow \frac{1}{(k+1)!}, \end{aligned}$$

a contradiction.

Hence, taking a subsequence if necessary, for any $0<\sigma <\delta$, $f_{n}$ has at least two distinct zeros in $\Delta (0,\sigma )$ for sufficiently large n. We assume that $z_{n,1}$ is a zero of $f_{n}$ on $\Delta (0,\sigma )\backslash \{z_{n,0}\}$. Clearly, $z_{n,1}\rightarrow 0$. Let $\zeta _{n,1}=z_{n,1}/\rho _{n}$, it follows froms (18) that $\zeta _{n,1}\rightarrow \infty$. Hence $z_{n,0}/z_{n,1}=\zeta _{n,0}/\zeta _{n,1}\rightarrow 0$. Set

$$\begin{aligned} L_{n}(z)=\frac{f_{n}(z_{n,1}z)}{z_{n,1}^{k+1}}. \end{aligned}$$

Then, for sufficiently large n, $\{L_{n}\}$ is well-defined and holomorphic on each bounded set of $\mathbb {C}$, and all of whose zeros have multiplicity at least k. By the assumption, we have $L_{n}(z)=0\Rightarrow |L_{n}^{(k)}(z)|\le A|z|$, and $L_{n}^{(k)}(z)\ne z$. By Lemma 10, $\{L_{n}\}$ is normal on the punctured complex plane $\mathbb {C}^{*}=\mathbb {C}\backslash \{0\}.$ We claim that $\{L_{n}\}$ is also normal at 0. Otherwise, the maximum modulus principle implies that $L_{n}\rightarrow \infty$ locally uniformly on $\mathbb {C}^{*}$. But, this is impossible since $L_{n}(1)=0$. Hence $\{L_{n}\}$ is normal on the whole plane $\mathbb {C}.$

Taking a subsequence and renumbering, we assume that

$$\begin{aligned} L_{n}(z)\rightarrow L(z), \end{aligned}$$

and then

$$\begin{aligned} L_{n}^{(k)}(z)\rightarrow L^{(k)}(z) \end{aligned}$$

(20)

locally uniformly on $\mathbb {C}$, where L is entire, all zeros of L have multiplicity at least k. Clearly, $L(1)=0$. On the other hand, $L_{n}(z_{n,0}/z_{n,1})=0$ and $z_{n,0}/z_{n,1}\rightarrow 0$, we get that $L(0)=0$. Since $L_{n}(z)=0\Rightarrow |L_{n}^{(k)}(z)|\le A|z|$, an argument similar to that in Claim III yields that $L(z)=0\Rightarrow |L^{(k)}(z)|\le |z|$. So it follows from $L(0)=0$ that $L^{(k)}(0)=0$. Since $L_{n}^{(k)}(z)\ne z$, Hurwitz’s theorem and (20) imply that $L^{(k)}(z)\equiv z$. Note that all zeros of L have multiplicity at least k and $L(0)=0$, we deduce that $L(z)=z^{k+1}/(k+1)!$. But, this in impossible since $L(1)=0$.

Case 2.2 By taking a subsequence, if necessary, for any $\delta >0$, $f_{n}$ has at least one pole on $\Delta (0,\delta )$ for all n.

Then there exist points $z_{n,\infty }\rightarrow 0$ such that $f_n(z_{n,\infty })=\infty$. We may assume that $z_{n,\infty }$ is the pole of $f_{n}$ of smallest modulus. Let $\zeta _{n,\infty }=z_{n,\infty }{/}\rho _{n}$. It follows from (18) that $\zeta _{n,\infty }\rightarrow \infty$, and then $z_{n,0}/z_{n,\infty }=\zeta _{n,0}/\zeta _{n,\infty }\rightarrow 0$. Now set

$$\begin{aligned} M_{n}(z)=\frac{f_{n}(z_{n,\infty }z)}{z_{n,\infty }^{k+1}}. \end{aligned}$$

Then, for sufficiently large n, $\{M_{n}\}$ is well-defined for each $z\in \mathbb {C}$, all of whose zeros have multiplicity at least k and whose poles are are multiple. Moreover, $\{M_{n}\}$ is holomorphic on $\Delta$ for sufficiently large n. By the assumption, we have $M_{n}(z)=0\Rightarrow |M_{n}^{(k)}(z)|\le A|z|$, and $M_{n}^{(k)}(z)\ne z$. Lemma 10 implies that $\{M_{n}\}$ is normal on $\mathbb {C}^{*}.$ We claim that $\{M_{n}\}$ is also normal at 0. Otherwise, $\{M_{n}\}$ is normal on $\Delta '$, but not normal at 0. Since $\{M_{n}\}$ is holomorphic on $\Delta$, he maximum modulus principle implies that $M_n\rightarrow \infty$. But $M_{n}(z_{n,0}/z_{n,\infty })=0$ and $z_{n,0}/z_{n,\infty }\rightarrow 0$. This contradiction proves our claim. Hence, $\{M_{n}\}$ is normal on $\mathbb {C}.$

Then, taking a subsequence and renumbering,

$$\begin{aligned} M_{n}(z)\rightarrow M(z) \end{aligned}$$

spherically uniformly on compact subsets of ${\mathbb {C}}$, where M is meromorphic, all of whose zeros have multiplicity at least k. Clearly, $M(1)=\infty$. On the other hand, $M_{n}(z_{n,0}/z_{n,\infty })=0$ and $z_{n,0}/z_{n,\infty }\rightarrow 0$, we obtain $M(0)=0$. Arguing as in Case 2.1 (for L(z)), we have $M(z)=z^{k+1}/(k+1)!$. But, $M(1)=\infty$, a contradiction. Then we have shown that (17) is impossible.

We thus have proved that $\mathcal G$ is normal at 0.

We now turn to show that is normal at $z=0$. Since $\mathcal G$ is normal at 0, then the family $\mathcal G$ is equicontinuous at 0 with respect to the spherical distance. On the other hand, $g(0)=\infty$ for each $g\in \mathcal G$, so there exists $\delta >0$ such that $|g(z)|\ge 1$ for all $g\in \mathcal G$ and each $z\in \Delta (0,\delta )$. It follows that $f(z)\ne 0$ for all $f\in \mathcal F$ and $z\in \Delta (0,\delta )$. Since $\mathcal F$ is normal on $\Delta '$ but not normal at $z=0$, the family $1/\mathcal{F}={\{}1/f: f\in \mathcal{F}{\}}$ is holomorphic in $D_{\delta }$ and normal on $\Delta '(0,\delta )$, but not normal at $z=0$. Thus there exists a sequence $\{ 1/f_n \}\subset 1/\mathcal F$ which converges locally uniformly in $\Delta '(0,\delta )$, but not on $\Delta (0,\delta )$. The maximum modulus principle implies that $1{/}f_n\rightarrow \infty$ in $\Delta '(0,\delta )$. Thus $f_n\rightarrow 0$ converges locally uniformly in $\Delta '(0,\delta )$, and hence so does $\{ g_n \}\subset \mathcal G$, where $g_n(z)=f_n(z)/z$. But $|g_n(z)|\ge 1$ for $z\in \Delta (0,\delta )$, a contradiction. This completes the proof of Theorem 1. $\square$

Proof of Theorem 2

Using the same argument as in the proof of Theorem 1, we can prove Theorem 2. We here omit the details. $\square$