Normal Families and Shared Functions Concerning Hayman’s Question

Abstract

In this paper, we studied a normality criterion concerning Hayman’s question and proved: let $n({\ge }2),k({\ge }1), m({\ge }0)$ be three integers, let $h(z)({\not \equiv } 0)$ be a holomorphic function in a domain D with all zeros that have multiplicity at most m, and let $\mathcal {F}$ be a family of functions meromorphic in a domain D, all of whose zeros have multiplicity at least $k+m$. If, for any two functions $f, g\in \mathcal {F}$, $f^nf^{(k)}$ and $g^ng^{(k)}$ share h(z) in D, then $\mathcal {F}$ is normal in D. The result gets rid of two conditions “all zeros of h(z) have multiplicity divisible by $n+1$” and “all poles of f(z) have multiplicity at least $m+1$” in the result due to Meng and Hu (Bull Malays Math Sci Soc 38:1331–1347, 2015).

1 Introduction

In this paper, we assume the reader is familiar with Nevanlinna theory of meromorphic functions. Let D be a domain in $\mathbb {C}$ and let $\mathcal {F}$ be a family of meromorphic functions in D. We say that $\mathcal {F}$ is normal in D (in the sense of Montel) if each sequence $\{f_n\}$ in $\mathcal {F}$ has a subsequence $\{f_{n_j}\}$ that converges locally uniformly in D, with respect to the spherical metric, to a meromorphic function or $\infty$ (see [7, 15, 17]).

For simplicity, we take $\rightarrow$ to stand for convergence and $\Rightarrow$ for convergence spherically locally uniformly.

Let f(z) and g(z) be two meromorphic functions in a domain D, and let h(z) be a holomorphic function in D. If $f(z)-h(z)$ and $g(z)-h(z)$ have the same zeros ignoring multiplicity (counting multiplicity), then we say that f(z) and g(z) share h(z) IM (CM) in D.

The following normality criterion was conjectured by Hayman [8] and proved by several authors (see [1, 4, 6, 10, 16]).

Theorem 1

Let n be a positive integer, and let $\mathcal {F}$ be a family of meromorphic functions in D. If, for each $f\in \mathcal {F}$, $f^nf' \ne 1$, then $\mathcal {F}$ is normal in D.

For other related results, see Bergweiler and Langley [2], Pang and Zalcman [11], Wu and Xu [14] and Tan et al. [13].

In 2008, Zhang [18] considered the case of shared value and obtained.

Theorem 2

Let $\mathcal {F}$ be a family of meromorphic functions in D, and let $n({\ge }2)$ be a positive integer. If, for any two functions $f, g\in \mathcal {F}$, $f^nf'$ and $g^ng'$ share a nonzero value a IM in D, then $\mathcal {F}$ is normal in D.

In 2015, Meng and Hu [9] studied the case of $f^nf^{(k)}(n\ge 2)$ sharing a holomorphic function and obtained

Theorem 3

Let $k({\ge }1),n({\ge }2), m({\ge }0)$ be three integers, let $h(z)({\not \equiv }0)$ be a holomorphic function in a domain D with all zeros that have multiplicity at most m and divisible by $n + 1$, and let $\mathcal {F}$ be a family of meromorphic functions in domain D such that each $f \in \mathcal {F}$ has zeros of multiplicity at least $k + m$ and poles of multiplicity at least $m + 1$. If, for any two functions $f, g\in \mathcal {F}$, $f^n(z)f^{(k)}(z)$ and $g^n(z)g^{(k)}(z)$ share h(z) IM in D, then $\mathcal {F}$ is normal in D.

By Theorems 2 and 3, it is nature to ask that: can we get rid of the condition “all zeros of h(z) have multiplicity divisible by $n+1$” and “all poles of f have multiplicity at least $m+1$ in Theorem 3”?

In this paper, we studied the question and gave an affirmative answer to the question.

Theorem 4

Let $k({\ge }1),n({\ge }2), m({\ge }0)$ be three integers, let $h(z)({\not \equiv }0)$ be a holomorphic function in a domain D with all zeros that have multiplicity at most m, and let $\mathcal {F}$ be a family of meromorphic functions in domain D such that each $f \in \mathcal {F}$ has zeros of multiplicity at least $k + m$. If, for any two functions $f, g\in \mathcal {F}$, $f^n(z)f^{(k)}(z)$ and $g^n(z)g^{(k)}(z)$ share h(z) IM in D, then $\mathcal {F}$ is normal in D.

In fact, we proved the following more general result:

Theorem 5

Let $k({\ge }1),n({\ge }2), m({\ge }0)$ be three integers, let $h(z)({\not \equiv }0)$ be a holomorphic function in a domain D with all zeros that have multiplicity at most m, and let $\mathcal {F}$ be a family of meromorphic functions in a domain D such that each $f \in \mathcal {F}$ has zeros of multiplicity at least $k + m$. If, for any two functions $f, g\in \mathcal {F}$, $f^n(z)f^{(k)}(z)-h(z)$ has at most one distinct zero in D, then $\mathcal {F}$ is normal in D.

The following examples show that the conditions in Theorem 5 are necessary.

Example 1

[9] Let $D =\{z\in \mathbb {C}|\ |z|< 1\}$, let $h(z)\equiv 0$ and let

$$\begin{aligned} \mathcal {F}=\left\{ f_j(z)=e^{jz} \mid j=1,2,\ldots \right\} . \end{aligned}$$

Obviously, $f^n_j(z)f_j^{(k)}(z)-h(z)$ does not have zero in D for each positive integer j. But the family $\mathcal {F}$ is not normal at $z =0$. This shows that $h(z)\not \equiv 0$ is necessary Theorem 5.

Example 2

Let $D =\{z\in \mathbb {C}|\ |z|< 1\}$, let $h(z)=\frac{1}{z^{n+k+1}}$ and let

$$\begin{aligned} \mathcal {F}=\left\{ f_j(z)=\frac{1}{jz} \mid j=1,2,\ldots , \text {and} \ j^{n+1}\ne (-1)^k k!\right\} . \end{aligned}$$

Obviously, $f^n_j(z)f_j^{(k)}(z)-h(z)$ does not have zero in D for each positive integer j. But the family $\mathcal {F}$ is not normal at $z =0$. This shows that Theorem 5 is not valid if h(z) is a meromorphic function in D.

Example 3

Let $D =\{z\in \mathbb {C}|\ |z|< 1\}$, let $h(z)=1$ and let

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

Then $f^n_j(z)f_j^{(k)}(z)-h(z)$ does not have zero in D for each positive integer j. But the family $\mathcal {F}$ is not normal at $z =0$. This shows that the condition “all zeros of f have multiplicity at least $k + m$ ” in Theorem 5 is best.

Example 4

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

$$\begin{aligned} \mathcal {F}=\left\{ f_j(z)=jz \mid j=1,2,\ldots \right\} . \end{aligned}$$

Obviously, $f^2_j(z)f'_j(z)-h(z)=j^3z^2-1$ have exactly two distinct zeros in D for each positive integer j. But the family $\mathcal {F}$ is not normal at $z =0$. This shows that the condition “$f^n(z)f^{(k)}(z)-h(z)$ has at most one distinct zero” in Theorem 5 is necessary.

2 Some Lemmas

For the proofs of our theorems, we require the following results.

Lemma 1

[12, 17] Let $\mathcal {F}$ be a family of meromorphic functions in the unit disk $\varDelta$ such that all zeros of functions in $\mathcal {F}$ have multiplicity $\ge l$. Let $\alpha$ be a real number satisfying $-l<\alpha <1$. Then $\mathcal {F}$ is not normal in any neighborhood of $z_0\in \varDelta$ if and only if there exist

• (a) points $z_j\in \varDelta ,$$z_j \rightarrow z_0;$

• (b) functions $f_j\in {\mathcal {F}};$ and

• (c) positive numbers $\rho _j\rightarrow 0$

such that $g_j(\xi )=\rho _{j} ^{\alpha }f_j(z_j+\rho _j\xi )\Rightarrow g(\xi )$ spherically uniformly on compact subsets of $\mathbb {C}$, where $g(\xi )$ is a non-constant meromorphic function in $\mathbb {C}$ satisfying that all zeros of g have multiplicity at least l.

Lemma 2

[15] Let $f_1$ and $f_2$ be two non-constant meromorphic functions in $\mathbb {C}$, then

$$\begin{aligned} N\left( r,f_1f_2\right) -N\left( r,\frac{1}{f_1f_2}\right) =N\left( r,f_1\right) +N\left( r,f_2\right) -N\left( r,\frac{1}{f_1}\right) -N\left( r,\frac{1}{f_2}\right) . \end{aligned}$$

The following Lemma was proved by Zhang and Li [19] when f is a transcendental meromorphic function, and by Meng and Hu [9] when f is a rational function.

Lemma 3

Let $n({\ge }2), k({\ge }1)$ be three integers, let $a\ne 0$ be a finite complex number, and let f(z) be a non-constant meromorphic in $\mathbb {C}$ with all zeros that have multiplicity at least k. Then $f^n(z)f^{(k)}(z)-a$ have at least two distinct zeros.

Lemma 4

Let $n({\ge }1), k({\ge }1), M({\ge }1)$ be three integers, let p(z) be a polynomial with $\deg p=M$, and let f(z) be a non-constant rational function in $\mathbb {C}$ with $f(z)\ne 0$. Then $f^n(z)f^{(k)}(z)-p(z)$ has at least $n+k+1$ distinct zeros.

The proof of Lemma 4 is almost the same with Chang [3] and Lemma 11 in Deng etc. [5], we omit the detail.

Lemma 5

Let $n({\ge }2), k({\ge }1), m({\ge }1)$ be three integers, let p(z) be a polynomial with $\deg p=m$, and let f(z) be a non-constant meromorphic in $\mathbb {C}$ with all zeros that have multiplicity at least $k+m$. Then $f^n(z)f^{(k)}(z)-p(z)$ has at least two distinct zeros.

Proof

Set

$$\begin{aligned} \frac{1}{f^{n+1}}=\frac{f^nf^{(k)}}{pf^{n+1}}-\frac{p[f^nf^{(k)}]'-p'f^nf^{(k)}}{pf^{n+1}}\frac{f^nf^{(k)}-p}{p[f^nf^{(k)}]'-p'f^nf^{(k)}}. \end{aligned}$$

Then by $m(r,\frac{f^{(i)}}{f})=S(r,f)(i\ge 1)$, $m(r,p)=m\log r+O(1)$, $m\big (r,\frac{1}{p}\big )=O(1)$, Lemma 2 and Nevanlinna’s elementary theory, we get

$$\begin{aligned} \begin{aligned} \left( n+1\right) m\left( r,\frac{1}{f}\right)&\le m\left( r,\frac{f^nf^{\left( k\right) }}{pf^{n+1}}\right) +m\left( r,\frac{p\left( f^nf^{\left( k\right) }\right) '-p'f^nf^{\left( k\right) }}{pf^{n+1}}\right) \\&\quad +\,m\left( r,\frac{f^nf^{\left( k\right) }-p}{p\left[ f^nf^{\left( k\right) }\right] '-p'f^nf^{\left( k\right) }}\right) +S\left( r,f\right) \\&\le T\left( r,\frac{f^nf^{\left( k\right) }-p}{p\left[ f^nf^{\left( k\right) }\right] '-p'f^nf^{\left( k\right) }}\right) \\&\quad -\,N\left( r,\frac{f^nf^{\left( k\right) }-p}{p\left[ f^nf^{\left( k\right) }\right] '-p'f^nf^{\left( k\right) }}\right) +S\left( r,f\right) \\&=m\left( r,\frac{p\left[ f^nf^{\left( k\right) }\right] '-p'f^nf^{\left( k\right) }}{f^nf^{\left( k\right) }-p}\right) \\&\quad +\,N\left( r,\frac{p\left[ f^nf^{\left( k\right) }\right] '-p'f^nf^{\left( k\right) }}{f^nf^{\left( k\right) }-p}\right) \\&\quad -\,N\left( r,\frac{f^nf^{\left( k\right) }-p}{p\left[ f^nf^{\left( k\right) }\right] '-p'f^nf^{\left( k\right) }}\right) +S\left( r,f\right) \\&=m\left( r,\frac{p\left[ \frac{f^nf^{\left( k\right) }}{p}-1\right] '}{\frac{f^nf^{\left( k\right) }}{p}-1}\right) \\&\quad +\,N\left( r,p\left[ f^nf^{\left( k\right) }\right] '-p'f^nf^{\left( k\right) }\right) +N\left( r,\frac{1}{f^nf^{\left( k\right) }-p}\right) \\&\quad -\,N\left( r,\frac{1}{p\left[ f^nf^{\left( k\right) }\right] '-p'f^nf^{\left( k\right) }}\right) \\&\quad -\,N\left( r,f^nf^{\left( k\right) }-p\right) +S\left( r,f\right) \\&\le \overline{N}\left( r,f\right) +N\left( r,\frac{1}{f^nf^{\left( k\right) }-p}\right) \\&\quad -\,N\left( r,\frac{1}{p\left[ f^nf^{\left( k\right) }\right] '-p'f^nf^{\left( k\right) }}\right) +m\log r+S\left( r,f\right) . \end{aligned} \end{aligned}$$

Let $z_1$ is a zero of f with multiplicity $l_1\ge k+m$. Then $z_1$ is a zero of $p[f^nf^{(k)}]'-p'f^nf^{(k)}$ with multiplicity at least $(n+1)l_1-k-1$.

Let $z_2$ is a zero of $f^nf^{(k)}-p$ with multiplicity $l_2$. Obviously, we have

$$\begin{aligned} p[f^nf^{(k)}]'-p'f^nf^{(k)}=p[f^nf^{(k)}-p]'-p'[f^nf^{(k)}-p]. \end{aligned}$$

Then $z_2$ is a zero of $p[f^nf^{(k)}]'-p'f^nf^{(k)}$ with multiplicity at least $l_2-1$.

Hence, we have

$$\begin{aligned} \left( n+1\right) T\left( r,f\right)\le & {} \overline{N}\left( r,f\right) + \left( n+1\right) N\left( r,\frac{1}{f}\right) +N\left( r,\frac{1}{f^nf^{\left( k\right) }-p}\right) \nonumber \\&\quad +\,m\log r-N\left( r,\frac{1}{p\left[ f^nf^{\left( k\right) }\right] '-p'f^nf^{\left( k\right) }}\right) +S\left( r,f\right) \nonumber \\\le & {} \overline{N}\left( r,f\right) + \left( k+1\right) \overline{N}\left( r,\frac{1}{f}\right) +\overline{N}\left( r,\frac{1}{f^nf^{\left( k\right) }-p}\right) \nonumber \\&\quad +\,m\log r+S\left( r,f\right) \nonumber \\\le & {} \overline{N}\left( r,f\right) + \frac{k+1}{k+m}N\left( r,\frac{1}{f}\right) +\overline{N}\left( r,\frac{1}{f^nf^{\left( k\right) }-p}\right) \nonumber \\&\quad +\,m\log r+S\left( r,f\right) . \end{aligned}$$

(2.1)

Suppose that $f^n(z)f^{(k)}(z)-p(z)$ has at most one distinct zero.

Next we consider two cases.

Case 1$m\ge 2$. Then by (2.1), we have

$$\begin{aligned} T(r,f)<\left( n-\frac{k+1}{k+m}\right) T(r,f)\le (m+1)\log r+S(r,f). \end{aligned}$$

Thus, f is a rational function with $\deg f<m+1$. Since all zeros of f have multiplicity at least $k+m\ge 1+m$, we deduce that $f(z)\ne 0$. Then by Lemma 4, we obtain that $f^n(z)f^{(k)}(z)-p(z)$ has at least $n+k+1>2$ distinct zeros, a contradiction.

Case 2$m=1$.

If $f^n(z)f^{(k)}(z)-p(z)\ne 0$, then by (2.1), we get $T(r,f)\le \log r+S(r,f)$. It follows that f is a rational function with $\deg f\le 1$. We deduce that $f(z)\ne 0$, since all zeros of f have multiplicity at least $k+m\ge 2$, Then by Lemma 4, we get $f^n(z)f^{(k)}(z)-p(z)$ has at least $n+k+1>2$ distinct zeros, a contradiction.

Thus $f^n(z)f^{(k)}(z)-p(z)$ has exactly one distinct zero. By (2.1), we have

$$\begin{aligned} nT(r,f)\le 2\log r+\overline{N}(r,f)+S(r,f). \end{aligned}$$

(2.2)

If $n\ge 3$, by (2.2), we obtain $T(r,f)\le \log r+S(r,f)$. It follows that f is a rational function with $\deg f \le 1$. Since all zeros of f have multiplicity at least $k+m\ge 2$, we obtain $f(z)\ne 0$, then by Lemma 4, we get $f^n(z)f^{(k)}(z)-p(z)$ has at least $n+k+1>2$ distinct zeros, a contradiction.

Thus $n=2$. By (2.2) again, we get $T(r,f)\le 2\log r+S(r,f)$. It follows that f is a rational function with $\deg f \le 2$. If $k\ge 2$, since all zeros of f have multiplicity at least $k+m\ge 3$, we get $f(z)\ne 0$, then by Lemma 4, we get a contradiction. Hence $k=1$, then f has one zero with multiplicity 2 at most. If f has no zero, then by Lemma 4, a contradiction. Thus, f(z) has exactly one distinct zero with multiplicity 2, and of the following forms:

$$\begin{aligned} A_1{:}\,f(z)= & {} a(z-\alpha )^2;\quad A_2{:}\,f(z)=a\frac{(z-\alpha )^2}{z-\beta };\\ A_3{:}\,f(z)= & {} \frac{a(z-\alpha )^2}{(z-\beta _1)(z-\beta _2)};\quad A_4{:}\, f(z)=a\frac{(z-\alpha )^2}{(z-\beta )^2}. \end{aligned}$$

If f(z) has the form $A_1$ or $A_2$ or $A_4$, we have $\overline{N}(r,f) \le \log r =1/2T(r,f)+O(1)$. Then by (2.2), we get $T(r,f)\le 4/3\log r+S(r,f)$, this contradicts with $T(r,f)=2\log r +O(1)$.

Then

$$\begin{aligned} f(z)=\frac{a(z-\alpha )^2}{(z-\beta _1)(z-\beta _2)}. \end{aligned}$$

(2.3)

It follows from (2.3) that

$$\begin{aligned} f'\left( z\right) =\frac{a\left( z-\alpha \right) \left[ \left( 2\alpha -\beta _1-\beta _2\right) z+2\beta _1\beta _2-\alpha \left( \beta _1+\beta _2\right) \right] }{\left( z-\beta _1\right) ^2\left( z-\beta _2\right) ^2}. \end{aligned}$$

(2.4)

By (2.3) and (2.4), we get

$$\begin{aligned} f^2(z)f'(z)=\frac{a^3(z-\alpha )^5\left[ \left( 2\alpha -\beta _1-\beta _2\right) z+2\beta _1\beta _2-\alpha (\beta _1+\beta _2)\right] }{(z-\beta _1)^4(z-\beta _2)^4}. \end{aligned}$$

(2.5)

Since $\deg p=m=1$, we may set $p(z)=b(z-z_0)$, where $b\ne 0$ is a constant. Since $f^n(z)f^{(k)}(z)-p(z)$ has exactly one distinct zero, by (2.5), we may set

$$\begin{aligned} f^2(z)f'(z)=b(z-z_0)-\frac{b(z-w)^9}{(z-\beta _1)^4(z-\beta _2)^4}, \end{aligned}$$

(2.6)

where $w\ne \alpha$. Otherwise, if $w= \alpha$, then by (2.5), we get $\alpha$ is a zero of $(f^2(z)f'(z))''$ with multiplicity 3. But from (2.6), we get $\alpha$ is a zero of $(f^2(z)f'(z))''$ with multiplicity 7, a contradiction.

Differentiating (2.5) two times, we obtain,

$$\begin{aligned}{}[f^2(z)f'(z)]''=\frac{(z-\alpha )^3g_1(z)}{(z-\beta _1)^6(z-\beta _2)^6}, \end{aligned}$$

(2.7)

where $g_1(z)$ is a polynomial with $\deg g_1\le 5$.

On the other hand, differentiating (2.6) two times, we obtain,

$$\begin{aligned}{}[f^2(z)f'(z)]''=\frac{(z-w)^7g_2(z)}{(z-\beta _1)^6(z-\beta _2)^6}, \end{aligned}$$

(2.8)

where $g_2(z)$ is a polynomial with $\deg g_2\le 4$.

From (2.7)–(2.8), and $w\ne \alpha$, we get $7\le \deg g_1\le 5$, a contradiction.

This completes the proof of Lemma 5.

Lemma 6

Let $n({\ge }2), k({\ge }1)$ be three integers, and let $\{f_j\}$ be a sequence of meromorphic functions in domain D, $\{h_j(z)\}$ be a sequence of holomorphic functions in D such that $h_j(z)\Rightarrow h(z)$, where $h(z)\not = 0$ be a holomorphic function. If, for each $j\in N^+$, all zeros of function $f_j(z)$ have multiplicity at least k, and $f^n_j(z)f_j^{(k)}(z)-h_j(z)$ has at most one distinct zero in D, then $\{f_j\}$ is normal in D.

Proof

Suppose that $\{f_j\}$ is not normal at $z_0\in D$. By Lemma 1, there exists a sequence $z_j$ of complex numbers $z_j\rightarrow z_0$, a sequence $\rho _j$ of positive numbers $\rho _j\rightarrow 0$ , and a subsequence of $\{f_j\}$ (we may still denote by $\{f_j\}$) such that

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

locally uniformly on compact subsets of $\mathbb {C}$, where $g(\xi )$ is a non-constant meromorphic function in $\mathbb {C}$. By Hurwitz’s theorem, all zeros of $g(\xi )$have multiplicity at least k. Then, we have

$$\begin{aligned} g^n_j(\xi )g_j^{(k)}(\xi )-h_j(z_j+\rho _j\xi )= & {} f^n_j(z_j+\rho _j\xi )f_j^{(k)}(z_j+\rho _j\xi )-h_j(z_j+\rho _j\xi )\\\Rightarrow & {} g^n(\xi )g^{(k)}(\xi )-h(z_0). \end{aligned}$$

for all $\xi \in \mathbb {C}/\{g^{-1}(\infty )\}$.

Obviously, $g^n(\xi )g^{(k)}(\xi )-h(z_0)\not \equiv 0$. Otherwise, suppose that

$$\begin{aligned} g^n(\xi )g^{(k)}(\xi )-h(z_0) \equiv 0, \end{aligned}$$

(2.9)

then we have $g(\xi )\ne 0$ since $h(z_0)\ne 0$.

It follows from (2.9) that

$$\begin{aligned} \frac{1}{g^{n+1}(\xi )}\equiv \frac{g^{(k)}(\xi )}{h(z_0)g(\xi )}. \end{aligned}$$

Then, we get

$$\begin{aligned} (n+1)m\left( r,\frac{1}{g}\right) =m\left( r, \frac{g^{(k)}}{h(z_0)g}\right) =S(r,g). \end{aligned}$$

It follows that $T(r,g)=S(r,g)$ since $g\ne 0$. Hence g is a constant, a contradiction.

We claim that $g^n(\xi )g^{(k)}(\xi )-h(z_0)$ has at most one distinct zero. Otherwise, suppose that $\xi _1, \xi _2$ are two distinct zeros of $g^n(\xi )g^{(k)}(\xi )-h(z_0)$. We choose a positive number $\sigma$ small enough such that $D_1\cap D_2=\emptyset$ and $g^n(\xi )g^{(k)}(\xi )-h(z_0)$ has no other zeros in $D_1\bigcup D_2$ except for $\xi _1$ and $\xi _2$, where $D_1=\{\xi {:}\,\mid \xi -\xi _1\mid < \sigma \}$ and $D_2=\{\xi {:}\mid \xi -\xi _2\mid < \sigma \}$.

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

$$\begin{aligned}&f^n_j(z_j+\rho _j\xi _{1,j})f_j^{(k)}(z_j+\rho _j\xi _{1,j})-h_j(z_j+\rho _j\xi _{1,j})=0;\\&f^n_j(z_j+\rho _j\xi _{2,j})f_j^{(k)}(z_j+\rho _j\xi _{2,j})-h_j(z_j+\rho _j\xi _{2,j})=0. \end{aligned}$$

By the assumption in Lemma 6, $f^n_jf_j^{(k)}(z)-h_j(z)$ has at most one zero in D, it follows that $z_j+\rho _j\xi _{1,j}=z_j+\rho _j\xi _{2,j}$, that is $\xi _{1,j}=\xi _{2,j}=(z_0-z_j)/\rho _j$, which contradicts with the facts $D_1\cap D_2=\emptyset$.

The claim is proved. On the other hand, it follows from Lemma 3 that $g^n(\xi )g^{(k)}(\xi )-h(z_0)$ has at least two distinct zeros, a contradiction. Thus $\{f_j\}$ is normal in D.

3 Proof of Theorems

Proof of Theorem 5

By Lemma 6, it is enough to prove that $\mathcal {F}$ is normal at the point $z_0$, where $h(z_0)=0$. By making standard normalization, we may assume that $z_0=0$, and $h(z)=z^tb(z)$ where $1\le t\le m$ is a positive integer, and $b(0)= 1$.

Suppose that $\mathcal {F}$ is not normal at $z_0=0$. By Lemma 1, there exists a sequence $z_j$ of complex numbers $z_j\rightarrow 0$, a sequence $\rho _j$ of positive numbers $\rho _j\rightarrow 0$, and a sequence of functions $\{f_j\}\subseteq \mathcal {F}$ such that

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

(3.1)

locally uniformly on compact subsets of $\mathbb {C}$, where $g(\xi )$ is a non-constant meromorphic function in $\mathbb {C}$. By Hurwitz’s theorem, all zeros of $g(\xi )$ have multiplicity at least $k+m$. Next we consider two cases.

Case 1$z_j/ \rho _j \rightarrow \infty$. Set

$$\begin{aligned} F_j(\xi )=\frac{f_j(z_j+z_j\xi )}{z_j^{\frac{k+t}{n+1}}}. \end{aligned}$$

Then, we have

$$\begin{aligned}&F^n_j(\xi )F_j^{(k)}(\xi )-(1+\xi )^tb(z_j+z_j\xi )\\&\quad =\frac{f^n_j(z_j+z_j\xi )f_j^{(k)}(z_j+z_j\xi )-h(z_j+z_j\xi )}{z_j^t}. \end{aligned}$$

As the same argument as in Lemma 6, we deduce that $F_j^n(\xi )F_j^{(k)}(\xi )-(1+\xi )^tb(z_j+z_j\xi )$ has at most one distinct zero in $\varDelta =\{\xi : \mid \xi \mid <1\}$.

Since all zeros of $F_j$ have multiplicity at least $k+m$, and $(1+\xi )^tb(z_j+z_j\xi )\rightarrow (1+\xi )^t \ne 0$ when $\xi \in \varDelta$. Then by Lemma 6, $\{F_j\}$ is normal in $\varDelta$.

So, there exists a subsequence of functions [we still denote as ${F_j (\xi )}$] and a function $F(\xi )$ (a meromorphic function or $\infty$), such that $F_j(\xi ){\Rightarrow }F(\xi ).$

If $F(0)\ne \infty$, then it follows from $k+m-1-\frac{k+t}{n+1}>0$ that

$$\begin{aligned} g^{\left( k+m-1\right) }\left( \xi \right)= & {} \underset{j\rightarrow \infty }{\lim }g_j^{\left( k+m-1\right) }\left( \xi \right) =\underset{j\rightarrow \infty }{\lim }\frac{f_j^{\left( k+m-1\right) }\left( z_j+\rho _j\xi \right) }{\rho _j^{\frac{k+t}{n+1}-\left( k+m-1\right) }}\\= & {} \underset{j\rightarrow \infty }{\lim }\left( \frac{\rho _j}{z_j}\right) ^{k+m-1-\frac{k+t}{n+1}}F_j^{\left( k+m-1\right) }\left( \frac{\rho _j}{z_j}\xi \right) =0, \end{aligned}$$

for all $\xi \in \mathbb {C}/\{g^{-1}(\infty )\}$.

Thus we deduce that $g^{(k+m-1)}\equiv 0$. Hence g is a polynomial of degree at most $k+m-1$. Since all zeros of g have multiplicity at least $k+m$, it follows that $g(\xi )$ is a constant, a contradiction.

If $F(0)= \infty$, then by

$$\begin{aligned} \frac{1}{F_j\left( \frac{\rho _j}{z_j}\xi \right) }=\frac{z_j^{\frac{k+t}{n+1}}}{f_j\left( z_j+\rho _j\xi \right) }\rightarrow \frac{1}{F(0)}= 0, \end{aligned}$$

when $\xi \in \mathbb {C}/\{g^{-1}(0)\}$, we obtain that,

$$\begin{aligned} \frac{1}{g\left( \xi \right) }=\underset{j\rightarrow \infty }{\lim }\frac{\rho _j^{\frac{k+t}{n+1}}}{f_j\left( z_j+\rho _j\xi \right) }=\underset{j\rightarrow \infty }{\lim }\left( \frac{\rho _j}{z_j}\right) ^{\frac{k+t}{n+1}}\frac{z_j^{\frac{k+t}{n+1}}}{f_j\left( z_j+\rho _j\xi \right) }=0. \end{aligned}$$

Thus $g(\xi ) \equiv \infty$, which contradicts that $g(\xi )$ is a non-constant meromorphic function.

Case 2$z_j/ \rho _j \rightarrow \alpha$, where $\alpha$ is a finite complex number. Then by (3.1), we have

$$\begin{aligned}&g^n_j\left( \xi \right) g_j^{\left( k\right) }\left( \xi \right) -\left( \xi +\frac{z_j}{\rho _j}\right) ^tb\left( z_j+\rho _j\xi \right) \\&\quad =\frac{f^n_j\left( z_j+\rho _j\xi \right) f_j^{\left( k\right) }\left( z_j+\rho _j\xi \right) -h\left( z_j+\rho _j\xi \right) }{\rho _j^t} \\&\quad \Rightarrow g^n\left( \xi \right) g^{\left( k\right) }\left( \xi \right) -\left( \xi +\alpha \right) ^t \end{aligned}$$

for all $\xi \in \mathbb {C}/ \{g^{-1}(\infty )\}$.

Since for sufficiently large j, $f^n_j(z_j+\rho _j\xi )f_j^{(k)}(z_j+\rho _j\xi )-h(z_j+\rho _j\xi )$ has one distinct zero, it follows from the proof of Lemma 6 that $g^n(\xi )g^{(k)}(\xi )-(\xi +\alpha )^t$ has at most one distinct zero.

But from Lemma 5, $g^n(\xi )g^{(k)}(\xi )-(\xi +\alpha )^t$ have at least two distinct zeros. Hence $g(\xi )$ is a constant, a contradiction.

This completes the proof of Theorem 5.

Proof of Theorem 4

Let $z_0\in D$. We show that $\mathcal {F}$ is normal at $z_0$. Let $f\in \mathcal {F}$.

We consider two cases.

Case 1$f^n(z_0)f^{(k)}(z_0)\ne h(z_0)$. Then there exists a disk $D_\delta (z_0)=\{z{:}\,\mid z- z_0\mid < \delta \}$ such that $f^n(z)f^{(k)}(z)\ne h(z)$ in $D_\delta (z_0)$. Since for each pair of functions $(f,g)\in \mathcal {F}$, $f^n(z)f^{(k)}(z)$ and $g^n(z)g^{(k)}(z)$ share h(z) in D. Thus, for every $g\in \mathcal {F}$, $g^n(z)g^{(k)}\ne h(z)$ in $D_\delta (z_0)$. By Theorem 5, $\mathcal {F}$ is normal in $D_\delta (z_0)$. Hence $\mathcal {F}$ is normal at $z_0$.

Case 2$f^n(z_0)f^{(k)}(z_0)= h(z_0)$. Then there exists a disk $D_\delta (z_0)=\{z{:}\,\mid z- z_0\mid < \delta \}$ such that $f^n(z)f^{(k)}(z)\ne h(z)$ in $D^{0}_\delta (z_0)$. Since for each pair of functions $(f,g)\in \mathcal {F}$, $f^n(z)f^{(k)}(z)$ and $g^n(z)g^{(k)}$ share h(z) in D. Thus, for every $g\in \mathcal {F}$, $g^n(z)g^{(k)}\ne h(z)$ in $D^{0}_\delta (z_0)$ and $g^n(z_0)g^{(k)}(z_0)= h(z_0)$. So, $g^n(z)g^{(k)}- h(z)$ have only distinct zero in $D_\delta (z_0)$. By Theorem 5, $\mathcal {F}$ is normal in $D_\delta (z_0)$. Hence $\mathcal {F}$ is normal at $z_0$.

This completes the proof of Theorem 4.