A normal criterion concerning zero numbers

Abstract

Let $n \ge 4$ be a positive integer, $\mathcal {F}$ be a family of meromorphic functions in D and let $a(z)(\not \equiv 0), b(z)$ be two holomorphic functions in D. If, for any function $f \in \mathcal { F}$, (1)$f(z) \ne \infty$ when $a(z)=0$, (2) $f'(z)-a(z)f^{n}(z)-b(z)$ has at most one zero in D, then $\mathcal {F}$ is normal in D.

1 Introduction and main results

Let D be a domain in ${\mathbb {C}}$ and ${\mathcal {F}}$ be a family of meromorphic functions in D. A family ${\mathcal {F}}$ is said to be normal in D, in the sense of Montel, if each sequence $\left\{ f_n \right\}$ has a subsequence $\left\{ f_{n_k}\right\}$ that converges spherically locally uniformly in D to a meromorphic function or to the constant $\infty$.

The following well-known normal conjecture was proposed by Hayman in 1967.

Theorem A

[1] Let $n \ge 5$ be a positive integer, ${\mathcal {F}}$ be a family of meromorphic functions in D and let $a(\ne 0), b$ be two finite complex numbers. If, for any function $f \in {\mathcal {F}}, f'(z)-af^{n}(z)\ne b$ in D, then ${\mathcal {F}}$ is normal in D.

The conjecture was proved by Li [2], Li [3], Langley [4] (for $n \ge 5$), Pang [5] (for $n = 4$), Chen and Fang [6], Bergweiler and Eremenko [7] (for $n = 3$).

Up to now, this result has undergone various extensions and improvements (see [8,9,10,11,12]).

The following generalization of Theorem A was proved by Yang et al. [13].

Theorem B

Let $n \ge 4$ be a positive integer, ${\mathcal {F}}$ be a family of meromorphic functions in D and let $a(z)(\not \equiv 0), b(z)$ be two holomorphic functions in D. If, for any function $f \in {\mathcal {F}}$, (1)$f(z) \ne \infty$ when $a(z)=0$, (2)$f'(z)-a(z)f^{n}(z)\ne b(z)$ in D, then ${\mathcal {F}}$ is normal in D.

It is natural to ask what can be said if $f'(z)-a(z)f^{n}(z)= b(z)$ has solutions in Theorem B.

In this article, we study this problem and prove the following result.

Theorem 1.1

Let $n \ge 4$ be a positive integer, ${\mathcal {F}}$ be a family of meromorphic functions in D and let $a(z)(\not \equiv 0), b(z)$ be two holomorphic functions in D. If, for any function $f \in {\mathcal {F}}$, (1)$f(z) \ne \infty$ when $a(z)=0$, (2) $f'(z)-a(z)f^{n}(z)-b(z)$ has at most one zero in D, then ${\mathcal {F}}$ is normal in D.

The following examples show that both $n \ge 4$ and the conditions (1), (2) in Theorem 1.1 are necessary and sharp.

Example 1.1

Let ${\Delta }=\{z:| z| <1\}$ and $a(z)=-2z^{3}, b(z)=0$. Let ${\mathcal {F}}=\{f_j(z)\}$, where

$$\begin{aligned} f_j(z)=\frac{1}{z^{2}-\frac{1}{j}},z\in {\Delta },j=1,2\ldots . \end{aligned}$$

Then $f'_{j}(z)-a(z)f^{3}_{j}(z)-b(z)= \frac{\frac{2}{j}z}{(z^{2}-\frac{1}{j})^{3}}$, which has a zero in $\Delta$. Clearly,$f \ne \infty$ when $a(z)=0$, however ${\mathcal {F}}$ is not normal at 0.

Example 1.2

Let ${\Delta }=\{z:| z| <1\}, a(z)=-z, b(z)=0$. Let ${\mathcal {F}}=\{f_j(z)\}$, where

$$\begin{aligned} f_j(z)=\frac{1}{j+z},z\in \Delta ,j=1,2\ldots . \end{aligned}$$

Then $f'_{j}(z)-a(z)f^{4}_{j}(z)-b(z)= \frac{-z^{2}+(1-2j)z-j^{2}}{(j+z)^{4}}$, which has exactly two distinct zeros in $\Delta$. Clearly,$f \ne \infty$ when $a(z)=0$, however ${\mathcal {F}}$ is not normal at 0.

Example 1.3

Let ${\Delta }=\{z:| z| <1\}, a(z)=-z, b(z)=0$. Let ${\mathcal {F}}=\{f_j(z)\}$, where

$$\begin{aligned} f_j(z)=\frac{1}{jz},z\in \Delta ,j=2,3\ldots . \end{aligned}$$

Then $f'_{j}(z)-a(z)f^{4}_{j}(z)-b(z)= \frac{1-j^{3}z}{j^{4}z^{3}}$, which has a zero in $\Delta$. But ${\mathcal {F}}$ is not normal at 0.

The next example shows that Theorem 1.1 is not valid if a(z) is a meromorphic function in D.

Example 1.4

Let ${\Delta }=\{z:| z| <1\}, a(z)=z^{-4}, b(z)=0$. Let ${\mathcal {F}}=\{f_j(z)\}$, where

$$\begin{aligned} f_j(z)=jz,z\in \Delta ,j=2,3\ldots . \end{aligned}$$

Then $f'_{j}(z)-a(z)f^{4}_{j}(z)-b(z)= j-j^4$, which has no zero in $\Delta$. But ${\mathcal {F}}$ is not normal at 0.

Remark 1.1

Clearly Theorem 1.1 improves and extends Theorem B.

2 Some lemmas

Lemma 2.1

[14] Let k be a positive integer and let ${\mathcal {F}}$ be a family of functions meromorphic in the unit disc $\Delta$, all of whose zeros have multiplicity at least k. If ${\mathcal {F}}$ is not normal in any neighbourhood of $z_{0}\in \Delta$, then for each $\alpha , 0\le \alpha <k$ there exist a sequence of complex numbers $z_n, z_n\rightarrow z_0, z_0\in \Delta$, a sequence of positive numbers $\rho _n\rightarrow 0$, and a sequence of functions $f_{n}\in {\mathcal {F}}$ such that $g_n (\xi ) = \rho _n^{ - \alpha } f_n (z_n + \rho _n \xi ) \rightarrow g(\xi )$ spherically uniformly on compact subsets of ${\mathbb {C}}$, where g is a non-constant meromorphic function, all of whose zeros have multiplicity at least k. Moreover, $g(\xi )$ has order at most 2.

We remark that one can take $-k < \alpha \le 0$ in the above lemma if all poles of each $f_{n}\in {\mathcal {F}}$ have multiplicity at least k (see [15]).

Lemma 2.2

[16, Theorem 4, p. 381] Let f(z) be a transcendental meromorphic function with finite order, all of whose zeros are of multiplicity at least 2, and let $P(z)(\not \equiv 0)$ be a polynomial, then $f'(z)-P(z)$ has infinitely many zeros.

Lemma 2.3

[17, Lemma 6] Let k be a positive integer, f(z) be a transcendental meromorphic function with finite order, all of whose zeros have multiplicity at least $k+1$, then $f^{(k)}-1$ has infinitely many zeros.

Lemma 2.4

[18, Lemma 5] Let k be a positive integer, f(z) be a nonconstant rational function, all of whose zeros have multiplicity at least $k+2$, and all of whose poles have multiplicity at least 3, then $f^{(k)}(z)-1$ has at least two distinct zeros.

Lemma 2.5

[12, Lemma 3] Let $n\ge 4$ be a positive integer, $a\ne 0$ be a finite complex number and let f(z) be a nonconstant meromorphic function, then $f'(z)-af^{n}(z)$ has at least two distinct zeros.

Lemma 2.6

[19, Lemma 7(iv), p. 261] Let k, l be two positive integers, Q(z) be a rational function, all of whose zeros have multiplicity at least $k+2$ and all of whose poles have multiplicity at least 2 with the possible exception of $z=0$, then $Q^{(k)}(z)= z^l$ has a solution in ${\mathbb {C}}$.

Lemma 2.7

[20, Lemma 11] Let k, m be two positive integers, f(z) be a rational function. If $f(z)\ne 0$ for $z\in {\mathbb {C}}$, and $f^{(k)}(z)\ne z^{m}$ for $z\ne z_{0}$, where $z_{0}\in {\mathbb {C}}$, then f(z) is a constant.

Lemma 2.8

Let $n\ge 4$ be a positive integer, ${\mathcal {F}}=\{f_j\}$ be a family of meromorphic functions in a domain D, and let $a_j(z),b_j(z)$ be two sequences of analytic functions in D such that $a_j(z)\longrightarrow a(z)\ne 0, b_j(z)\longrightarrow b(z)$. If $f'_j(z)-a_j(z)f_{j}^{n}(z)-b_{j}(z)$ has at most one zero, then ${\mathcal {F}}$ is normal in D.

Proof

Suppose that ${\mathcal {F}}$ is not normal at $z_{0}\in D$. By Lemma 2.1, there exists $z_{j}\rightarrow z_{0}, \rho _{j}\rightarrow 0^{+}$, and $f_{j}\in {\mathcal {F}}$ such that

$$\begin{aligned} g_{j}(\xi )={\rho _{j}^{\frac{1}{n-1}}}{f_{j}(z_{j}+\rho _{j}\xi )}\longrightarrow g(\xi ) \end{aligned}$$

locally uniformly on compact subsets of ${\mathbb {C}}$, where $g(\xi )$ is a non-constant meromorphic function in ${\mathbb {C}}$.

For each $\xi \in {\mathbb {C}}/\{g^{-1}(\infty )\}$, we have $g_{j}'(\xi )-a_{j}(z_{j}+\rho _{j}\xi )g_{j}^{n}(\xi )-\rho _{j}^{\frac{n}{n-1}}b_{j}(z_{j}+\rho _{j}\xi ) \longrightarrow g'(\xi )-a(z_{0})g^{n}(\xi ).$

Obviously, $g'(\xi )-a(z_{0})g^{n}(\xi )\not \equiv 0$.

Suppose that $g'(\xi )-a(z_{0})g^{n}(\xi )\equiv 0$, then $g(\xi )$ must be an entire function. Hence, since $n\ge 3$ and $a(z_{0})\ne 0$, we have

$$\begin{aligned} nT(r,g)=T(r,g^{n})=T(r,\frac{g'}{a(z_{0})})=T(r,g')+log^{+}\frac{1}{|a(z_{0})|} \le T(r,g)+S(r,g). \end{aligned}$$

It follows that we have $T(r,g)=S(r,g)$. But this is impossible since $g(\xi )$ is a non-constant meromorphic function.

Claim: $g'(\xi )-a(z_{0})g^{n}(\xi )$ has at most one zero.

Suppose this is not the case, and $g'(\xi )-a(z_{0})g^{n}(\xi )$ has two distinct zeros $\xi _{1}$, and $\xi _{2}$. We choose a positive number $\delta$ small enough such that $D_{1}\cap D_{2}=\emptyset$ and $g'(\xi )-a(z_{0})g^{n}(\xi )$ has no other zeros in $D_{1}\cup D_{2}$ except for $\xi _{1}$ and $\xi _{2}$, where $D_{1}=\{\xi :|\xi -\xi _{1}|<\delta \}$ and $D_{2}=\{\xi :|\xi -\xi _{2}|<\delta \}$.

By Hurwitz’s theorem, for sufficiently large j, there exist points $\xi _{1,j}\rightarrow \xi _{1}$ and $\xi _{2,j}\rightarrow \xi _{2}$ such that

$$\begin{aligned} f_{j}'(\xi _{1,j})-a_{j}(z_{j}+\rho _{j}\xi _{1,j})f_{j}^{n}(z_{j}+\rho _{j}\xi _{1,j})-b_{j}(z_{j}+\rho _{j}\xi _{1,j})=0 \end{aligned}$$

and

$$\begin{aligned} f_{j}'(\xi _{2,j})-a_{j}(z_{j}+\rho _{j}\xi _{2,j})f_{j}^{n}(z_{j}+\rho _{j}\xi _{2,j})-b_{j}(z_{j}+\rho _{j}\xi _{2,j})=0 \end{aligned}$$

Since $f'_j(z)-a_j(z)f_{j}^{n}(z)-b_{j}(z)$ has at most one zero in D, then

$$\begin{aligned} z_{j}+\rho _{j}\xi _{1,j}=z_{j}+\rho _{j}\xi _{2,j}, \end{aligned}$$

this is

$$\begin{aligned} \xi _{1,j}=\xi _{2,j}=\frac{z_{0}-z_{j}}{\rho _{j}}, \end{aligned}$$

which contradicts the fact $D_{1}\cap D_{2}=\emptyset$. The claim is proved.

From Lemma 2.5, we have $g'(\xi )-a(z_{0})g^{n}(\xi )$ has at least two distinct zeros, this contradicts Claim which says that $g'(\xi )-a(z_{0})g^{n}(\xi )$ has at most one zero. Therefore ${\mathcal {F}}$ is normal in D. $\square$

3 Proof of theorems

Proof of Theorem 1.1

Suppose that ${\mathcal {F}}$ is not normal at $z_{0}$. From Lemma 2.8, we obtain $a(z_{0})= 0$. Without loss of generality, we assume that $z_0 =0$ and $a(z)=z^m\phi (z)$, where $m\in {\mathbb {N}}$, and $\phi$ is holomorphic with $\phi (0)\ne 0$. We can say $\phi \ne 0$ in $\Delta (0, \delta _{0})=\{z:|z|<\delta _{0}\}$ with the normalization $\phi (0) = 1$.

Since $f(z)\ne \infty$ when $a(z)=0$ and $f'(z)-a(z)f^{n}(z)-b(z)$ has at most one zero in $\Delta$, then $\frac{f'(z)}{a(z)f^{n}(z)}-\frac{b(z)}{a(z)f^{n}(z)}-1$ has at most one zero in $\Delta$. Consider the family as follows

$$\begin{aligned} {\mathcal {G}}=\{g(z)=\frac{1}{a(z)f^{n-1}(z)},f\in {\mathcal {F}}\}, \end{aligned}$$

(3.1)

where all zeros and poles of g(z) have multiplicity at least $n-1(\ge 3)$, except possibly the pole at 0, which has order at least m. Hence ${\mathcal {G}}$ is not normal at $z_0=0$ in $\Delta$. Then, by Lemma 2.1, there exists $z_{j} \longrightarrow 0,g_{j} \in {\mathcal {G}}$ and $\rho _{j} \longrightarrow 0^{+}$ such that

$$\begin{aligned} G_{j}(\xi )=\frac{g_{j}(z_{j}+\rho _{j}\xi )}{\rho _{j}}\longrightarrow G(\xi ) \end{aligned}$$

(3.2)

locally uniformly on compact subsets of ${\mathbb {C}}$, where $G(\xi )$ is a non-constant meromorphic functions in ${\mathbb {C}}$, and all zeros and poles of $G(\xi )$ have multiplicity at least $n-1(\ge 3)$, except possibly a pole, which has order at least m.

Next we consider two cases according to whether the sequence ${\frac{z_{n}}{\rho _{n}}}$ is bounded or unbounded.

Case  1. First assume that the sequence ${\frac{z_{n}}{\rho _{n}}}$ is unbounded. Then, there exists a subsequence, which we continue to call ${\frac{z_{n}}{\rho _{n}}}$, such that ${\frac{z_{n}}{\rho _{n}}}\rightarrow \infty$.

By simple calculation, from (3.1) and (3.2), we have

$$\begin{aligned} G'_{j}(\xi )= & {} \left[ \frac{g_{j}(z_{j}+\rho _{j}\xi )}{\rho _{j}}\right] '=\left[ \frac{1}{\rho _{j}a(z_{j}+\rho _{j}\xi )f^{n-1}_{j}(z_{j}+\rho _{j}\xi )}\right] '\\= & {} -\frac{a'(z_{j}+\rho _{j}\xi )}{a^{2}(z_{j}+\rho _{j}\xi )f^{n-1}_{j}(z_{j}+\rho _{j}\xi )}-\frac{(n-1)f_{j}'(z_{j}+\rho _{j}\xi )}{a(z_{j}+\rho _{j}\xi )f^{n}_{j}(z_{j}+\rho _{j}\xi )}\\= & {} -\frac{a'(z_{j}+\rho _{j}\xi )}{a^{2}(z_{j}+\rho _{j}\xi )f^{n-1}_{j}(z_{j}+\rho _{j}\xi )}-\frac{(n-1)b(z_{j}+\rho _{j}\xi )}{a(z_{j}+\rho _{j}\xi )f^{n}_{j}(z_{j}+\rho _{j}\xi )}\\&-(n-1)\left[ \frac{f_{j}'(z_{j}+\rho _{j}\xi )}{a(z_{j}+\rho _{j}\xi )f^{n}_{j}(z_{j}+\rho _{j}\xi )}-\frac{b(z_{j}+\rho _{j}\xi )}{a(z_{j}+\rho _{j}\xi )f^{n}_{j}(z_{j}+\rho _{j}\xi )}\right] . \end{aligned}$$

Noting that

$$\begin{aligned}&\frac{a'(z_{j}+\rho _{j}\xi )}{a^{2}(z_{j}+\rho _{j}\xi )f^{n-1}_{j}(z_{j}+\rho _{j}\xi )}\\&\quad =\frac{m(z_{j}+\rho _{j}\xi )^{m-1}\rho _{j}\phi (z_{j}+\rho _{j}\xi )+(z_{j}+\rho _{j}\xi )^{m}\rho _{j}\phi '(z_{j}+\rho _{j}\xi )}{(z_{j}+\rho _{j}\xi )^{2m}\phi ^{2}(z_{j}+\rho _{j}\xi )}\\&\qquad g_{j}(z_{j}+\rho _{j}\xi )(z_{j}+\rho _{j}\xi )^{m}\phi (z_{j}+\rho _{j}\xi )\\&\quad =\frac{m\rho _{j}}{z_{j}+\rho _{j}\xi }g_{j}(z_{j}+\rho _{j}\xi )+\frac{\rho _{j}\phi '(z_{j}+\rho _{j}\xi )}{\phi (z_{j}+\rho _{j}\xi )}g_{j}(z_{j}+\rho _{j}\xi )\rightarrow 0,\hbox { and }\\&\left[ \frac{b(z_{j}+\rho _{j}\xi )}{a(z_{j}+\rho _{j}\xi )f^{n}_{j}(z_{j}+\rho _{j}\xi )}\right] ^{n-1}\\&\quad =\rho ^{n}_{j}a(z_{j}+\rho _{j}\xi )b^{n-1}(z_{j}+\rho _{j}\xi )\left[ \frac{1}{\rho _{j}a(z_{j}+\rho _{j}\xi )f^{n-1}_{j}(z_{j}+\rho _{j}\xi )}\right] ^{n}\\&\quad =\rho ^{n}_{j}a(z_{j}+\rho _{j}\xi )b^{n-1}(z_{j}+\rho _{j}\xi )G^{n}_{j}(\xi )\rightarrow 0 \end{aligned}$$

uniformly on compact subsets of ${\mathbb {C}}$ disjoint from the poles of G. we deduce that $-(n-1)\left[ \frac{f_{j}'(z_{j}+\rho _{j}\xi )}{a(z_{j}+\rho _{j}\xi )f^{n}_{j}(z_{j}+\rho _{j}\xi )}-\frac{b(z_{j}+\rho _{j}\xi )}{a(z_{j}+\rho _{j}\xi )f^{n}_{j}(z_{j}+\rho _{j}\xi )}\right] \rightarrow G'(\xi )$

uniformly on compact subsets of ${\mathbb {C}}$ disjoint from the poles of G.

If $G'(\xi )\equiv -(n-1)=1-n$, then $G(\xi )=(1-n)\xi +A$, where A is a constant, which contradicts the fact that all zeros of $G(\xi )$ have multiplicity at least $n-1$. Hence, we have $G'(\xi )\not \equiv -(n-1)$.

Using an argument similar to Claim in Lemma 2.8, we can obtain $G'(\xi )+(n-1)$ has at most one zero.

On the other hand, by Lemmas 2.3 and 2.4, we have $G'(\xi )+(n-1)$ has at least two distinct zeros. Hence $G(\xi )$ is a constant, a contradiction.

Case 2. Now we consider the case that ${\frac{z_{n}}{\rho _{n}}}$ is bounded. Then, there is a subsequence, which we continue to call ${\frac{z_{n}}{\rho _{n}}}$, such that ${\frac{z_{n}}{\rho _{n}}}\rightarrow \alpha \in {\mathbb {C}}$.

It follows from (3.2) that

$$\begin{aligned} \frac{g_{j}(\rho _{j}\xi )}{\rho _{j}}=\frac{g_{j}(z_{j}+\rho _{j}(\xi -\frac{z_{j}}{\rho _{j}}))}{\rho _{j}}= G_{j}(\xi -\frac{z_{j}}{\rho _{j}})\rightarrow G(\xi -\alpha )={\widehat{G}}(\xi ) \end{aligned}$$

(3.3)

spherically uniformly on compact subsets of ${\mathbb {C}}$. Clearly, all zeros and all poles of ${\widehat{G}}(\xi )$ have multiplicity at least $n-1$, and $\xi = 0$ is a pole of ${\widehat{G}}(\xi )$ with multiplicity at least m.

Set

$$\begin{aligned} H_{j}(\xi )=\frac{1}{\rho ^{m+1}_{j}f^{n-1}_{j}(\rho _{j}\xi )}. \end{aligned}$$

It follows from (3.2) and (3.3) that

$$\begin{aligned} H_{j}(\xi )=a(\rho _{j}\xi )\frac{g_{j}(z_{j}+\rho _{j}(\xi -\frac{z_{j}}{\rho _{j}}))}{\rho _{j}}\rightarrow H(\xi )=\xi ^{m}G(\xi -\alpha )=\xi ^{m}{\widehat{G}}(\xi ) \end{aligned}$$

(3.4)

spherically locally uniformly on ${\mathbb {C}}\backslash \{0\}$, or on ${\mathbb {C}}$ if ${G}(-\alpha )\ne \infty$. Obviously, all zeros and all poles of $H(\xi )$ have multiplicity at least $n-1$, and $H(0)\ne 0$ since $f(\xi )\ne \infty$ when $a(\xi )=0$.

Noting that

$$\begin{aligned}&H'_{j}(\xi )+(n-1)\frac{b(\rho _{j}\xi )}{\rho ^{m}_{j}f^{n}_{j}(\rho _{j}\xi )}=\left[ \frac{1}{\rho ^{m+1}_{j}f^{n-1}_{j}(\rho _{j}\xi )}\right] '+(n-1)\frac{b(\rho _{j}\xi )}{\rho ^{m}_{j}f^{n}_{j}(\rho _{j}\xi )} \\&\quad =-(n-1)\xi ^{m}\phi (\rho _{j}\xi )\left[ \frac{f'_{j}(\rho _{j}\xi )}{a(\rho _{j}\xi )f^{n}_{j}(\rho _{j}\xi )}-\frac{b(\rho _{j}\xi )}{a(\rho _{j}\xi )f^{n}_{j}(\rho _{j}\xi )}\right] . \end{aligned}$$

On the other hand,

$$\begin{aligned}&\left[ \frac{b(\rho _{j}\xi )}{\rho ^{m}_{j}f^{n}_{j}(\rho _{j}\xi )}\right] ^{n-1}=\frac{\rho ^{n}_{j}b^{n-1}(\rho _{j}\xi )a^{n}(\rho _{j}\xi )}{\rho ^{m(n-1)}_{j}}\frac{1}{\rho ^{n}_{j}a^{n}(\rho _{j}\xi )f^{n(n-1)}_{j}(\rho _{j}\xi )} \\&\quad =\frac{\rho ^{n}_{j}b^{n-1}(\rho _{j}\xi )a^{n}(\rho _{j}\xi )}{\rho ^{m(n-1)}_{j}}G^{n}_{j}(\rho _{j}\xi )=\frac{\rho ^{n}_{j}b^{n-1}(\rho _{j}\xi )a^{n}(\rho _{j}\xi )}{\rho ^{m(n-1)}_{j}}G^{n}_{j}(z_{j}+\rho _{j}(\xi -\frac{z_{j}}{\rho _{j}})) \\&\quad =\frac{\rho ^{n(m+1)}_{j}b^{n-1}(\rho _{j}\xi )\xi ^{nm}\phi ^{n}(\rho _{j}\xi )}{\rho ^{m(n-1)}_{j}}G^{n}_{j}(z_{j}+\rho _{j}(\xi -\frac{z_{j}}{\rho _{j}})) \\&\quad =\rho ^{m+n}_{j}b^{n-1}(\rho _{j}\xi )\xi ^{nm}\phi ^{n}(\rho _{j}\xi )G^{n}_{j}(z_{j}+\rho _{j}(\xi -\frac{z_{j}}{\rho _{j}}))\rightarrow 0 \end{aligned}$$

uniformly on compact subsets of ${\mathbb {C}}$ disjoint from the poles of G. Hence

$$\begin{aligned} -(n-1)\xi ^{m}\phi (\rho _{j}\xi )\left[ \frac{f'_{j}(\rho _{j}\xi )}{a(\rho _{j}\xi )f^{n}_{j}(\rho _{j}\xi )}-\frac{b(\rho _{j}\xi )}{a(\rho _{j}\xi )f^{n}_{j}(\rho _{j}\xi )}\right] \rightarrow H'(\xi ) \end{aligned}$$

uniformly on compact subsets of ${\mathbb {C}}$ disjoint from the poles of G.

If $H'(\xi )\equiv -(n-1)\xi ^{m}$, then $H(\xi )=\frac{(1-n)\xi ^{m+1}}{m+1}+B$, where B is a constant. Thus, $H(\xi )$ has at least one zero in ${\mathbb {C}}$. Let $\xi _{0}$ be a zero of $H(\xi )$, then by the fact that $\xi _{0}$ has multiplicity at least $n-1(\ge 3)$, we get $H'(\xi _{0})=0$ and hence $\xi _{0} = 0$ by $H'(\xi ) \equiv -(n-1)\xi ^{m}$. It follows that $H(\xi )=c~\xi ^{l}, c\in {\mathbb {C}}$ and $l\in {\mathbb {N}}$. Thus, $\frac{(1-n)\xi ^{m+1}}{m+1}+B=c~\xi ^{l}$. By comparing the degrees and coefficients, we see that $B=0$. Thus, $H(\xi )=\frac{(1-n)\xi ^{m+1}}{m+1}$. It follows that $H(0)=0$, which contradicts that $H(0)\ne 0$.

Using an argument similar to Claim in Lemma 2.8, we can obtain $H'(\xi )+(n-1)\xi ^{m}$ has at most one zero.

Next we consider two subcases according to whether $H'(\xi )+(n-1)\xi ^{m}$ has zero or not.

Case 2.1 $H'(\xi )\ne (n-1)\xi ^{m}$. It now follows from Lemmas 2.2 and 2.6 that Case 2.1 cannot occur.

Case 2.2 $H'(\xi )= (n-1)\xi ^{m}$. Then $H'(\xi )\ne (n-1)\xi ^{m}$ for $\xi \ne \xi _{0}$, where $\xi _{0}\in {\mathbb {C}}$.

Case 2.2.1 $\xi _{0}=0$. Thus $H'(0)=0$. It follows that $H(0)=0$, which contradicts that $H(0)\ne 0$.

Case 2.2.2 $\xi _{0}\ne 0$.

If $H(\xi _{0})=0$, then $H'(\xi _{0})=0$ since all zeros of $H(\xi )$ have multiplicity at least $n-1$. Thus $\xi _{0}=0$, which contradicts that $\xi _{0}\ne 0$.

If $H(\xi )\ne 0, \xi \in {\mathbb {C}}$. Combined with the fact that $H'(\xi )\ne (n-1)\xi ^{m}$ for $\xi \ne \xi _{0}$, where $\xi _{0}\in {\mathbb {C}}$. Hence by Lemma 2.7$H(\xi )$ is a constant, a contradiction.

If $H(\xi )=0$, and $H(\xi _{0})\ne 0, \xi \in {\mathbb {C}}\backslash \{\xi _{0}\}$. Combined with the fact that $H'(\xi )\ne (n-1)\xi ^{m}$ for $\xi \in {\mathbb {C}}\backslash \{\xi _{0}\}$. It now follows from Lemmas 2.2 and 2.6 that this case cannot occur. Hence we show that ${\mathcal {G}}$ is normal at $z_{0}=0$.

Next, we show that ${\mathcal {F}}$ is normal at $z_0=0$. Since ${\mathcal {G}}$ is normal at $z_0=0$, let $g_j\longrightarrow g$ in a neighborhood of 0, then there exist ${\Delta _\delta } = \left\{ {z:\left| z \right| < \delta } \right\}$ and a subsequence of $\{g_j\}$ such that $\{g_j\}$ converges uniformly to a meromorphic function or $\infty$. Noting $g(0)=\infty$, we can find a $\varepsilon$ with $0<\varepsilon < \delta$ and $M>0$ such that $|g(z)|>M, z\in \Delta _\varepsilon$. So, for sufficiently large j, we get $|g_j(z)|\ge \frac{M}{2}$, hence $|a(z)f^{n-1}_{j}(z)|\le \frac{2}{M}$. Therefore $f_j(z)\ne \infty$ for sufficiently large j and $z\in \Delta _\varepsilon$.

Hence ${f_j(z)}$ is analytic in $\Delta _\varepsilon$. Choosing $\varepsilon$ small enough that $|a(z)|\ge \frac{|z|^{m}}{M}$, it follows that, for sufficiently large j, we have

$$\begin{aligned} |f_j(z)|=\left| {\frac{1}{a(z)g_{j}(z)}}\right| ^{\frac{1}{n-1}} \le {\left( {\frac{2^{m+1}}{\varepsilon ^{m} }} \right) ^\frac{1}{n-1}},\quad \left| z \right| = \frac{\varepsilon }{2}. \end{aligned}$$

By the Maximum Principle and Montel’s theorem, ${\mathcal {F}}$ is normal at $z_0=0$. The complete proof of Theorem 1.1 is given. $\square$