Normal families of meromorphic functions sharing one function

Abstract

Suppose $p(z)$ is a holomorphic function, the multiplicity of its zeros is at most d, $P(z)$ is a nonconstant polynomial. Let ℱ be a family of meromorphic functions in a domain D, all of whose zeros and poles have multiplicity at least $max\{\frac{k}{2}+d+1,k+d\}$. If for each pair of functions f and g in ℱ, $P(f)f^{(k)}$ and $P(g)g^{(k)}$ share a holomorphic function $p(z)$, then ℱ is normal in D. It generalizes and extends the results of Jiang, Gao and Wu, Xu.

MSC:30D35, 30D45.

1 Introduction and results

Let D be a domain in ℂ, let ℱ be a family of meromorphic functions in D. ℱ is said to be normal in D, in the sense of Montel, if for any sequence $\{f_n\}\in \mathcal{F}$ contains a subsequence $\{f_{nj}\}$ such that $f_{nj}$ converges spherically locally uniformly in D to a meromorphic function or ∞ [1–3].

Let $a\in \mathbb{C}\cup \{\mathrm{\infty }\}$, let f and g be two nonconstant meromorphic functions in D. If $f(z)-a$ and $g(z)-a$ have the same zeros (ignoring multiplicity), we say f and g share the value a in D.

In 1959, Hayman [1] proved that if f is a transcendental meromorphic function, then $f^n{f}^{\prime }$ assumes every finite nonzero complex number infinitely often for any positive integer $n\ge 3$. He [4] conjectured that this remains valid for $n=1$ and $n=2$. Further, the case of $n=2$ was confirmed by Mues [5] in 1979. The case $n=1$ was considered and settled by Clunie [6].

In 1994, Yang and Yang [7] proposed a conjecture: If f is an entire function and $k\ge 2$, then ${(f{f}^{(k)})}^n-a(z)$ ($a(z)\not\equiv 0$) has infinitely many zeros.

Zhang and Song [8] proved the following theorem.

Theorem A Suppose that f is a transcendental meromorphic function, n, k are two positive integers, then when $n\ge 2$, ${(f{f}^{(k)})}^n-a(z)$ has infinitely many zeros, where $a(z)\not\equiv 0$ is a small function of f.

In 2005, Wang [9] proved the following theorem.

Theorem B Let f be a transcendental meromorphic function, let $c(z)\not\equiv 0$ be a small function of f, and let n, k be two positive integers. If $n\ge 3$, then $f^n{f}^{(k)}-c(z)$ has infinitely many zeros.

In the case of $f{f}^{(k)}$, Yang and Yang [7] proposed a conjecture: If f is transcendental, then $f{f}^{(k)}$ assumes every finite nonzero complex number infinitely often. In 2006, Wang [10] proved that this conjecture holds when f has only zeros of multiplicity at least $k+1$ ($k\ge 2$).

In 2011, Meng and Hu [11] obtained the following theorem.

Theorem C Take a positive integer k and a nonzero complex number a. Let ℱ be a family of meromorphic functions in a domain $D\in \mathbb{C}$ such that each $f\in \mathcal{F}$ has only zeros of multiplicity at least $k+1$. For each pair $(f,g)\in \mathcal{F}$, if $f{f}^{(k)}$ and $g{g}^{(k)}$ share a, then ℱ is normal in D.

In 2011, Jiang and Gao [12] obtained the following theorem.

Theorem D Suppose that d (≥0) is an integer, $p(z)$ is an analytic function in D, and the multiplicity of its all zeros is at most d. Let ℱ be a family of meromorphic functions in D, let n be a positive integer. If $n\ge 2d+2$ and for each pair of functions f and g in ℱ, $f^n{f}^{\prime }$ and $g^n{g}^{\prime }$ share $p(z)$ in D, then ℱ is normal in D.

In 2012, Wu and Xu [13] got the following theorem.

Theorem E Let k be a positive integer, let $b\ne 0$ be a finite complex number, let P be a polynomial with either $degP\ge 3$ or $degP=2$ and P having only one distinct zero, and let ℱ be a family of meromorphic functions in D, all of whose zeros have multiplicity at least k. If for each pair of functions f and g in ℱ, $P(f)f^{(k)}$ and $P(g)g^{(k)}$ share b in D, then ℱ is normal in D.

It is natural to ask whether Theorem E can be improved by the idea of sharing a holomorphic function. In this paper, we study the problem and obtain the following theorems.

Theorem 1.1 Suppose that $d\ge 0$ is an integer, $p(z)\not\equiv 0$ is a holomorphic function in D, and the multiplicity of its all zeros is at most d. Let ℱ be a family of meromorphic functions in D, the multiplicity of all zeros and poles of $f\in \mathcal{F}$ is at least $max\{\frac{k}{2}+d+1,k+d\}$. If for each pair of functions f and g in ℱ, $f{f}^{(k)}$ and $g{g}^{(k)}$ share $p(z)$ in D, then ℱ is normal in D.

Remark 1.1 Theorem 1.1 still holds when $p(z)$ is a nonzero finite constant.

Theorem 1.2 Suppose that $d\ge 0$ is an integer, $p(z)\not\equiv 0$ is a holomorphic function in D, and the multiplicity of its all zeros is at most d. Let P be a nonconstant polynomial, ℱ be a family of meromorphic functions in D, the multiplicity of all zeros and poles of $f\in \mathcal{F}$ is at least $max\{\frac{k}{2}+d+1,k+d\}$. If for each pair of functions f and g in ℱ, $P(f)f^{(k)}$ and $P(g)g^{(k)}$ share $p(z)$ in D, then ℱ is normal in D.

2 Some lemmas

Lemma 2.1 (see [14])

Let k be a positive integer, let ℱ be a family of meromorphic functions in D such that each function $f\in \mathcal{F}$ has only zeros with multiplicities 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 ℱ is not normal at $z_0\in D$, then for each α, $0\le \alpha \le k$, there exists a sequence of complex numbers $z_n\in D$, $z_n\to z_0$, a sequence of positive numbers ${\rho }_n\to 0$, and a sequence of functions $f_n\in \mathcal{F}$ such that

$$g_n(\xi )=\frac{f_n(z_n+{\rho }_n\xi )}{{\rho }_n^{\alpha }}\to g(\xi )$$

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

Lemma 2.2 (see [15])

Let $f(z)$ be a meromorphic function and k be a positive integer. If $f^{(k)}\not\equiv 0$, then

$$N(r,\frac{1}{f^{(k)}})\le N(r,\frac{1}{f})+k\overline{N}(r,f)+S(r,f).$$

Lemma 2.3 (see [1])

Let $f_1(z)$, $f_2(z)$ be two meromorphic functions defined in $D=\{z:|z|<R\}$, then

$$N(r,f_1{f}_2)-N(r,\frac{1}{f_1{f}_2})=N(r,f_1)+N(r,f_2)-N(r,\frac{1}{f_1})-N(r,\frac{1}{f_2}).$$

Lemma 2.4 (see [16])

Let f be a transcendental meromorphic function, let $P_f(z)$, $Q_f(z)$ be two differential polynomials of f. If $f^n{P}_f=Q_f$ holds and the degree of $Q_f$ is at most n, then $m(r,P_f)=S(r,f)$.

Lemma 2.5 Let d (≥0) be an integer, let k be a positive integer, and let $p(z)=a_d{z}^d+a_{d-1}{z}^{d-1}+\cdots +a_{1}z+a_0$ be a polynomial, where $a_d\ne 0$, $a_{d-1},\dots ,a_0$ are constants. Suppose that f is a transcendental meromorphic function, all of whose zeros and poles have multiplicity at least 2, $p(z)$ is a small function of $f(z)$, then $f{f}^{(k)}(z)-p(z)$ has infinitely many zeros.

Proof Let

$$\psi (z)=f{f}^{(k)}-p(z).$$

(2.1)

Suppose $f{f}^{(k)}-p(z)$ has only finitely many zeros, then $N(r,\frac{1}{\psi (z)})=S(r,f)$. By (2.1), then

$${\left(\frac{\psi }{p}\right)}^{\prime }=\frac{f^{\prime }{f}^{(k)}}{p}+\frac{f{f}^{(k+1)}}{p}+{\left(\frac{1}{p}\right)}^{\prime }f{f}^{(k)}.$$

(2.2)

Let

$${\psi }_1=\frac{\psi }{p}.$$

Since the multiplicity of zeros of $f(z)$ is at least 2, we can get from (2.2) that

$$N(r,\frac{1}{f})\le N(r,\frac{1}{{({\psi }_1)}^{\prime }})+S(r,f).$$

(2.3)

By Lemma 2.2, we know that

$$N(r,\frac{1}{{({\psi }_1)}^{\prime }})\le N(r,\frac{1}{{\psi }_1})+\overline{N}(r,f)+S(r,f).$$

(2.4)

We can get from (2.2) that

$$\frac{f{f}^{(k)}}{p}\frac{{({\psi }_1)}^{\prime }}{{\psi }_1}-\frac{f^{\prime }{f}^{(k)}}{p}-\frac{f{f}^{(k+1)}}{p}-{\left(\frac{1}{p}\right)}^{\prime }f{f}^{(k)}=\frac{{({\psi }_1)}^{\prime }}{{\psi }_1},$$

i.e.,

$$f(\frac{f^{(k)}}{p}\frac{{({\psi }_1)}^{\prime }}{{\psi }_1}-\frac{f^{\prime }{f}^{(k)}}{fp}-\frac{f^{(k+1)}}{p}-{\left(\frac{1}{p}\right)}^{\prime }{f}^{(k)})=\frac{{({\psi }_1)}^{\prime }}{{\psi }_1}.$$

(2.5)

Let

$$fH=\frac{{({\psi }_1)}^{\prime }}{{\psi }_1},$$

(2.6)

where $H=\frac{f^{(k)}}{p}\frac{{({\psi }_1)}^{\prime }}{{\psi }_1}-\frac{f^{\prime }{f}^{(k)}}{fp}-\frac{f^{(k+1)}}{p}-{(\frac{1}{p})}^{\prime }{f}^{(k)}$. By Lemma 2.4, we get $m(r,H)=S(r,f)$.

From (2.5) and Lemma 2.3, we obtain that

$$\begin{array}{rcl}m(r,\frac{1}{f})& \le & m(r,H)+m(r,\frac{{\psi }_1}{{({\psi }_1)}^{\prime }})\\ \le & N(r,\frac{{({\psi }_1)}^{\prime }}{{\psi }_1})-N(r,\frac{{\psi }_1}{{({\psi }_1)}^{\prime }})+m(r,\frac{{({\psi }_1)}^{\prime }}{{\psi }_1})+S(r,f)\\ \le & N(r,{({\psi }_1)}^{\prime })+N(r,\frac{1}{{\psi }_1})-N(r,\frac{1}{{({\psi }_1)}^{\prime }})-N(r,{\psi }_1)+S(r,f)\\ \le & \overline{N}(r,f)+N(r,\frac{1}{\psi })-N(r,\frac{1}{{({\psi }_1)}^{\prime }})+S(r,f).\end{array}$$

(2.7)

We can get from (2.6) that

$$f{f}^{(k)}(\frac{{({\psi }_1)}^{\prime }}{p{\psi }_1}-\frac{f^{\prime }}{fp}-\frac{f^{(k+1)}}{p{f}^{(k)}}-{\left(\frac{1}{p}\right)}^{\prime })=\frac{{({\psi }_1)}^{\prime }}{{\psi }_1}.$$

(2.8)

Let

$$f{f}^{(k)}G=\frac{{({\psi }_1)}^{\prime }}{{\psi }_1},$$

(2.9)

where $G=\frac{{({\psi }_1)}^{\prime }}{p{\psi }_1}-\frac{f^{\prime }}{fp}-\frac{f^{(k+1)}}{p{f}^{(k)}}-{(\frac{1}{p})}^{\prime }$. By Lemma 2.4, then $m(r,G)=S(r,f)$.

By (2.9), we have that

$$\begin{array}{rcl}m(r,f^{(k)})& \le & m(r,\frac{{({\psi }_1)}^{\prime }}{{\psi }_1})+m(r,\frac{1}{f})+m(r,\frac{1}{G})\\ \le & m(r,\frac{1}{f})+N(r,G)-N(r,\frac{1}{G})+S(r,f).\end{array}$$

(2.10)

Since $\frac{{({\psi }_1)}^{\prime }}{{\psi }_1}$ has only simple poles, and by (2.9) we know that the poles of f are impossible G’s. Hence the poles of G are only possible from the zeros and poles of $p(z)$ or the zeros of ${\psi }_1$, f and $f^{(k)}$.

Hence by (2.8) and (2.9), we obtain that

$$\begin{array}{rcl}N(r,G)& \le & \overline{N}(r,\frac{1}{{\psi }_1})+\overline{N}(r,\frac{1}{f})+\overline{N}(r,\frac{1}{f^{(k)}})+N(r,\frac{1}{p})\\ \le & \overline{N}(r,\frac{1}{\psi })+\overline{N}(r,\frac{1}{f})+\overline{N}(r,\frac{1}{f^{(k)}})+S(r,f).\end{array}$$

(2.11)

Since $\frac{{({\psi }_1)}^{\prime }}{{\psi }_1}$ has only simple poles, so by (2.9) we know that

$$N(r,\frac{1}{G})\ge N(r,f)+N(r,f^{(k)})-\overline{N}(r,f).$$

(2.12)

Combining (2.7) and (2.10)-(2.12), we have

$$\begin{array}{rcl}m(r,f^{(k)})& \le & \{\overline{N}(r,\frac{1}{\psi })+\overline{N}(r,\frac{1}{f})+\overline{N}(r,\frac{1}{f^{(k)}})\}-\{N(r,f)+N(r,f^{(k)})-\overline{N}(r,f)\}\\ +\{\overline{N}(r,f)+N(r,\frac{1}{\psi })-N(r,\frac{1}{{({\psi }_1)}^{\prime }})\}+S(r,f).\end{array}$$

Hence

$$\begin{array}{rcl}T(r,f^{(k)})& \le & \overline{N}(r,\frac{1}{\psi })+\overline{N}(r,\frac{1}{f})+\overline{N}(r,\frac{1}{f^{(k)}})-N(r,f)+\overline{N}(r,f)\\ +\overline{N}(r,f)+N(r,\frac{1}{\psi })-N(r,\frac{1}{{({\psi }_1)}^{\prime }})+S(r,f).\end{array}$$

Since the multiplicity of the zeros and poles of $f(z)$ is at least 2, by an elementary calculation and combing with Lemma 2.2, (2.3) and (2.4), the above inequality yields

$$\begin{array}{rcl}T(r,f^{(k)})& \le & N(r,\frac{1}{f^{(k)}})+2N(r,\frac{1}{\psi })+S(r,f)\\ \le & N(r,\frac{1}{f})+k\overline{N}(r,f)+2N(r,\frac{1}{\psi })+S(r,f)\\ \le & (k+1)\overline{N}(r,f)+3N(r,\frac{1}{\psi })+S(r,f).\end{array}$$

(2.13)

Since the multiplicity of the poles of $f(z)$ is at least 2, we can get from (2.13) that

$$\begin{array}{rcl}T(r,f^{(k)})& \le & (1-\frac{1}{k+2})N(r,f^{(k)})+3N(r,\frac{1}{\psi })+S(r,f)\\ \le & (1-\frac{1}{k+2})N(r,f^{(k)})+S(r,f).\end{array}$$

This implies $T(r,f^{(k)})=S(r,f)$, then $f^{(k)}$ is a rational function, thus f is a rational function which contradicts with f is transcendental. Hence $f{f}^{(k)}(z)-p(z)$ has infinitely many zeros. □

Remark 2.1 When $p(z)$ is a nonzero finite constant or a small function of $f(z)$, similarly we can get the same conclusion.

Lemma 2.6 Let d (≥0) be an integer, let k be a positive integer, and let $p(z)=a_d{z}^d+a_{d-1}{z}^{d-1}+\cdots +a_{1}z+a_0$ be a polynomial, where $a_d\ne 0$, $a_{d-1},\dots ,a_0$ are constants. If $f(z)$ is a nonconstant polynomial, all of whose zeros and poles have multiplicity at least $k+d$, then $f{f}^{(k)}(z)-p(z)$ has at least two distinct zeros, and $f{f}^{(k)}(z)-p(z)\not\equiv 0$.

Proof We discuss the following two cases.

Case 1. If $f{f}^{(k)}-p(z)\ne 0$, then $f{f}^{(k)}-p(z)\equiv C$, where C is a nonzero constant. So $f{f}^{(k)}\equiv p(z)+C$. Since the multiplicity of all the zeros of f is at least $k+d$, thus $deg(f{f}^{(k)})\ge k+2d$, which contradicts with $deg(p(z))=d$.

Case 2. If $f{f}^{(k)}-p(z)$ has only one zero $z_0$, we assume $f{f}^{(k)}-p(z)\equiv A{(z-z_0)}^l$, where A is a nonzero constant, l is a positive integer.

We discuss the following two cases.

• (i) If $l\le d+1$, then $f{f}^{(k)}\equiv p(z)+A{(z-z_0)}^l$. Since $deg(f{f}^{(k)})\ge k+2d$, the degree of the right of the equation is at most $d+1$, which is smaller than the degree of the left of the equation. We get a contradiction.

• (ii) If $l>d+1$, then $f{f}^{(k)}\equiv p(z)+A{(z-z_0)}^l$. So ${(f{f}^{(k)})}^{(d)}\equiv a_d+Al\cdots (l-d+1){(z-z_0)}^{l-d}$. Since $a_d\ne 0$, so ${(f{f}^{(k)})}^{(d)}$ has only simple zeros, which contradicts with the multiplicity of all the zeros of f is at least $k+d$.

By Case 1 and Case 2, $f{f}^{(k)}-p(z)$ has at least two distinct zeros.

If $f{f}^{(k)}-p(z)\equiv 0$, then similar to the proof of Case 1, we get a contradiction. Hence $f{f}^{(k)}-p(z)\not\equiv 0$. □

Lemma 2.7 Let d (≥0) be an integer, let k be a positive integer, and let $p(z)=a_d{z}^d+a_{d-1}{z}^{d-1}+\cdots +a_{1}z+a_0$ be a polynomial, where $a_d\ne 0$, $a_{d-1},\dots ,a_0$ are constants. If $f(z)$ is a nonconstant rational function and not a polynomial, and the multiplicity of whose zeros and poles is at least $\frac{k}{2}+d+1$, then $f{f}^{(k)}(z)-p(z)$ has at least two distinct zeros, and $f{f}^{(k)}(z)-p(z)\not\equiv 0$.

Proof Since $f(z)$ is a nonconstant rational function and not a polynomial, then obviously $f{f}^{(k)}(z)-p(z)\not\equiv 0$. Let

$$f(z)=B\frac{{(z-{\alpha }_1)}^{m_1}{(z-{\alpha }_2)}^{m_2}\cdots {(z-{\alpha }_s)}^{m_s}}{{(z-{\beta }_1)}^{n_1}{(z-{\beta }_2)}^{n_2}\cdots {(z-{\beta }_t)}^{n_t}},$$

(2.14)

where B is a nonzero constant. Since the multiplicity of the zeros and poles of f is at least $\frac{k}{2}+d+1$, we have $m_i\ge \frac{k}{2}+d+1$ ($i=1,2,\dots ,s$), $n_j\ge \frac{k}{2}+d+1$ ($j=1,2,\dots ,t$). For simplicity, we denote

$$m_1+m_2+\cdots +m_s=m\ge (\frac{k}{2}+d+1)s,n_1+n_2+\cdots +n_t=n\ge (\frac{k}{2}+d+1)t.$$

By (2.19), we get

$$f^{(k)}(z)=B\frac{{(z-{\alpha }_1)}^{m_1-k}{(z-{\alpha }_2)}^{m_2-k}\cdots {(z-{\alpha }_s)}^{m_s-k}g(z)}{{(z-{\beta }_1)}^{n_1+k}{(z-{\beta }_2)}^{n_2+k}\cdots {(z-{\beta }_t)}^{n_t+k}},$$

(2.15)

where $g(z)=(m-n)(m-n-1)\cdots (m-n-k+1)z^{k(s+t-1)}+\cdots +c_{1}z+c_0$ is a polynomial, $c_i$ ($i=0,1$) are constants and $deg(g(z))\le k(s+t-1)$. Thus (2.14) together with (2.15) implies

$$f{f}^{(k)}(z)=B\frac{{(z-{\alpha }_1)}^{2m_1-k}{(z-{\alpha }_2)}^{2m_2-k}\cdots {(z-{\alpha }_s)}^{2m_s-k}g(z)}{{(z-{\beta }_1)}^{2n_1+k}{(z-{\beta }_2)}^{2n_2+k}\cdots {(z-{\beta }_t)}^{2n_t+k}}.$$

(2.16)

By (2.16), we obtain

$${(f{f}^{(k)}(z))}^{(d+1)}=B\frac{{(z-{\alpha }_1)}^{2m_1-k-d-1}{(z-{\alpha }_2)}^{2m_2-k-d-1}\cdots {(z-{\alpha }_s)}^{2m_s-k-d-1}{g}_1(z)}{{(z-{\beta }_1)}^{2n_1+k+d+1}{(z-{\beta }_2)}^{2n_2+k+d+1}\cdots {(z-{\beta }_t)}^{2n_t+k+d+1}},$$

(2.17)

where $deg(g_1(z))\le (k+d+1)(s+t-1)$.

Next, we discuss the following two cases.

Case 1. If $f{f}^{(k)}-p(z)$ has only one zero $z_0$, then let

$$f{f}^{(k)}(z)-p(z)=C\frac{{(z-z_0)}^l}{{(z-{\beta }_1)}^{2n_1+k}{(z-{\beta }_2)}^{2n_2+k}\cdots {(z-{\beta }_t)}^{2n_t+k}}.$$

(2.18)

Subcase 1.1. When $d\ge l$.

Combining (2.16) and (2.18), we get $d+2n+kt=deg(g(z))+2m-ks\le k(s+t-1)+2m-ks$. That is, $2(m-n)\ge k+d$, and then $m>n$.

Differentiating both sides of (2.18), we have

$$\begin{array}{r}{(f{f}^{(k)}(z))}^{(d+1)}-p^{(d+1)}(z)\\ =C\frac{g_2(z)}{{(z-{\beta }_1)}^{2n_1+k+d+1}{(z-{\beta }_2)}^{2n_2+k+d+1}\cdots {(z-{\beta }_t)}^{2n_t+k+d+1}},\end{array}$$

(2.19)

where $deg(g_2(z))\le t(d+1)+l-d-1$.

By (2.17) and (2.19), we know $2m-(k+d+1)s\le deg(g_2(z))\le t(d+1)+l-d-1$. Thus $2m-(k+d+1)s-t(d+1)\le l-d-1$. Since $2m-(k+d+1)s-t(d+1)\ge$ $2m-(k+d+1)\frac{m}{k/2+d+1}-(d+1)\frac{n}{k/2+d+1}>0$, then $0<l-d-1$, which contradicts with $d\ge l$.

Subcase 1.2. When $d<l$.

Differentiating both sides of (2.18), we have

$$\begin{array}{r}{(f{f}^{(k)}(z))}^{(d+1)}-p^{(d+1)}(z)\\ =D\frac{{(z-z_0)}^{l-d-1}{g}_3(z)}{{(z-{\beta }_1)}^{2n_1+k+d+1}{(z-{\beta }_2)}^{2n_2+k+d+1}\cdots {(z-{\beta }_t)}^{2n_t+k+d+1}},\end{array}$$

(2.20)

where $g_3(z)=(l-2n-kt)\cdots (l-2n-kt-d)z^{t(d+1)}+\cdots +d_{1}z+d_0$, where $d_i$ ($i=0,1$) are constants and $deg(g_3(z))\le t(d+1)$.

Differentiating both sides of (2.18) step by step for d times, we have $z_0$ is a zero of ${(f{f}^{(k)}(z))}^{(d)}-p^{(d)}(z)$, as $p^{(d)}(z)=a_d\ne 0$ and the multiplicity of all the zeros of $f(z)$ is at least $\frac{k}{2}+d+1$, thus ${\alpha }_i\ne z_0$ ($i=1,2,\dots ,s$). When $p(z)$ is a constant, from (2.18) we can also get ${\alpha }_i\ne z_0$ ($i=1,2,\dots ,s$).

Here, we discuss three subcases as follows.

Subcase 1.2.1. When $l<2n+kt+d$.

Combining (2.16) and (2.18), we get $d+2n+kt=deg(g(z))+2m-ks\le k(s+t-1)+2m-ks$. That is, $2(m-n)\ge k+d$, and then $m>n$.

Since ${\alpha }_i\ne z_0$ ($i=1,2,\dots ,s$), by (2.17) and (2.20), we have $t(d+1)\ge deg(g_3(z))\ge 2m-(k+d+1)s$. Thus $2m\le (k+d+1)s+t(d+1)\le (k+d+1)\frac{m}{k/2+d+1}+(d+1)\frac{n}{k/2+d+1}<2m$, which is impossible.

Subcase 1.2.2. When $l=2n+kt+d$.

If $m>n$, by a similar discussion to Subcase 1.2.1, we can get a contradiction. Thus $m\le n$. Since ${\alpha }_i\ne z_0$ ($i=1,2,\dots ,s$), by (2.17) and (2.20), we have $l-d-1\le deg(g_1(z))\le (k+d+1)(s+t-1)$, since $l=2n+kt+d$, thus $2n+kt+d-d-1\le (k+d+1)(s+t-1)$. Then $2n\le (k+d+1)s+(d+1)t-(k+d)<(k+d+1)\frac{m}{k/2+d+1}+(d+1)\frac{n}{k/2+d+1}\le 2n$, which is impossible.

Subcase 1.2.3. When $l>2n+kt+d$.

By (2.16) and (2.18), we get $l=deg(g(z))+2m-ks\le k(s+t-1)+2m-ks=2m+kt-k$. If $m>n$, by a similar discussion to Subcase 1.2.1, we get a contradiction. Thus $m\le n$.

Case 2. If $f{f}^{(k)}-p(z)$ has no zero. Then $l=0$ in (2.18), by a similar discussion to Subcase 1.1, we get a contradiction.

By Case 1 and Case 2, we get $f{f}^{(k)}-p(z)$ has at least two distinct zeros. □

3 Proof of Theorem 1.1

For any point $z_0$ in D, either $p(z_0)=0$ or $p(z_0)\ne 0$.

Case 1. When $p(z_0)=0$. We may assume $z_0=0$. Then $p(z)=a_d{z}^d+a_{d+1}{z}^{d+1}+\cdots =z^{d}h(z)$, where $a_d(\ne 0),a_{d+1},\dots$ are constants, $d\ge 1$, $h(z_0)\ne 0$, without loss of generality, let $h(z_0)=a_d$, where $h(z)$ is a holomorphic function.

Let ${\mathcal{F}}_1=\{F_j|F_j=\frac{f_j}{z^{d/2}},f_j\in \mathcal{F}\}$. If ${\mathcal{F}}_1$ is not normal at 0, then by Lemma 2.1, there exists a sequence of complex numbers $z_j\to 0$, a sequence of positive numbers ${\rho }_j\to 0$ and a sequence of functions $F_j\in {\mathcal{F}}_1$ such that $G_j(\xi )={\rho }_j^{-\frac{k}{2}}{F}_j(z_j+{\rho }_j\xi )\to G(\xi )$ spherically locally uniformly in ℂ, where $G(\xi )$ is a nonconstant meromorphic function in ℂ, and the multiplicity of the zeros and poles of $G(\xi )$ is at least $max\{\frac{k}{2}+d+1,k+d\}$. Here, we discuss two cases as follows.

Case 1.1. There exists a subsequence of $\frac{z_j}{{\rho }_j}$, we may denote it as $\frac{z_j}{{\rho }_j}$ such that $\frac{z_j}{{\rho }_j}\to c$, c is a finite complex number. Then

$${\varphi }_j(\xi )=\frac{f_j({\rho }_j\xi )}{{\rho }_j^{\frac{d+k}{2}}}=\frac{{({\rho }_j\xi )}^{\frac{d}{2}}{F}_j(z_j+{\rho }_j(\xi -\frac{z_j}{{\rho }_j}))}{{\rho }_j^{\frac{d}{2}}{\rho }_j^{\frac{k}{2}}}\to {\xi }^{\frac{d}{2}}G(\xi -c)=H(\xi )$$

spherically locally uniformly in ℂ, so

$${\varphi }_j(\xi ){\varphi }_j^{(k)}(\xi )-\frac{p({\rho }_j\xi )}{{\rho }_j^d}=\frac{f_j({\rho }_j\xi )f_j^{(k)}({\rho }_j\xi )-p({\rho }_j\xi )}{{\rho }_j^d}\to H(\xi )H^{(k)}(\xi )-a_d{\xi }^d$$

spherically locally uniformly in ℂ.

Since $\mathrm{\forall }f\in \mathcal{F}$, the multiplicity of whose zeros and poles is at least $max\{\frac{k}{2}+d+1,k+d\}$, then the multiplicity of all zeros and poles of H is at least $max\{\frac{k}{2}+d+1,k+d\}$, by Lemmas 2.5-2.7, we get $H(\xi )H^{(k)}(\xi )-a_d{\xi }^d\not\equiv 0$, and $H(\xi )H^{(k)}(\xi )-a_d{\xi }^d$ has at least two distinct zeros.

Suppose ${\xi }_0$, ${\xi }_0^{\ast }$ are two distinct zeros of $H(\xi )H^{(k)}(\xi )-a_d{\xi }^d$. We may choose a proper $\sigma >0$ such that $D({\xi }_0,\sigma )\cap D({\xi }_0^{\ast },\sigma )=\mathrm{\varnothing }$, where $D({\xi }_0,\sigma )=\{\xi ||\xi -{\xi }_0|<\sigma \}$, $D({\xi }_0^{\ast },\sigma )=\{\xi ||\xi -{\xi }_0^{\ast }|<\sigma \}$.

By Hurwitz’s theorem, there exists a subsequence of $f_j({\rho }_j\xi )f_j^{(k)}({\rho }_j\xi )-p({\rho }_j\xi )$, we may still denote it as $f_j({\rho }_j\xi )f_j^{(k)}({\rho }_j\xi )-p({\rho }_j\xi )$, then there exist points ${\xi }_j\in D({\xi }_0,\sigma )$ and points ${\xi }_j^{\ast }\in D({\xi }_0^{\ast },\sigma )$ such that for sufficiently large j, $f_j({\rho }_j{\xi }_j)f_j^{(k)}({\rho }_j{\xi }_j)-p({\rho }_j{\xi }_j)=0$, $f_j({\rho }_j{\xi }_j^{\ast })f_j^{(k)}({\rho }_j{\xi }_j^{\ast })-p({\rho }_j{\xi }_j^{\ast })=0$.

Since $f_j{f}_j^{(k)}$ and $g_j{g}_j^{(k)}$ share $p(z)$ in D, it follows that for any positive integer m, $f_m({\rho }_j{\xi }_j)f_m^{(k)}({\rho }_j{\xi }_j)-p({\rho }_j{\xi }_j)=0$, $f_m({\rho }_j{\xi }_j^{\ast })f_m^{(k)}({\rho }_j{\xi }_j^{\ast })-p({\rho }_j{\xi }_j^{\ast })=0$.

Fix m, let $j\to \mathrm{\infty }$ and note ${\rho }_j{\xi }_j\to 0$, ${\rho }_j{\xi }_j^{\ast }\to 0$, we obtain $f_m(0)f_m^{(k)}(0)-p(0)=0$.

Since the zeros of $f_m(0)f_m^{(k)}(0)-p(0)$ have no accumulation points, in fact when j is large enough, we have ${\rho }_j{\xi }_j={\rho }_j{\xi }_j^{\ast }=0$. Thus, when j is large enough, ${\xi }_0={\xi }_0^{\ast }=0$, which contradicts with $D({\xi }_0,\sigma )\cap D({\xi }_0^{\ast },\sigma )=\mathrm{\varnothing }$. Thus, ${\mathcal{F}}_1$ is normal at 0.

Case 1.2. There exists a subsequence of $\frac{z_j}{{\rho }_j}$, we may denote it as $\frac{z_j}{{\rho }_j}$ such that $\frac{z_j}{{\rho }_j}\to \mathrm{\infty }$. Then

$$\begin{array}{rcl}{f}_j(z_j+{\rho }_j\xi )f_j^{(k)}(z_j+{\rho }_j\xi )& =& {(z_j+{\rho }_j\xi )}^{\frac{d}{2}}{F}_j(z_j+{\rho }_j\xi )[{(z_j+{\rho }_j\xi )}^{\frac{d}{2}}{(F_j(z_j+{\rho }_j\xi ))}^{(k)}\\ +\sum _{i=1}^k{c}_i{(z_j+{\rho }_j\xi )}^{\frac{d}{2}-i}{(F_j(z_j+{\rho }_j\xi ))}^{(k-i)}]\\ =& {(z_j+{\rho }_j\xi )}^d{G}_j(\xi )G_j^{(k)}(\xi )+\sum _{i=1}^k{c}_i{(z_j+{\rho }_j\xi )}^{d-i}{\rho }_j^i{G}_j(\xi )G_j^{(k-i)}(\xi ),\end{array}$$

where $c_i=\frac{d}{2}(\frac{d}{2}-1)\cdots (\frac{d}{2}-i+1)C_{d/2}^i$ when $\frac{d}{2}\ge i$, and $c_i=0$ when $\frac{d}{2}<i$.

Thus, we have

$$\begin{array}{r}\frac{a_d{f}_j(z_j+{\rho }_j\xi )f_j^{(k)}(z_j+{\rho }_j\xi )}{p(z_j+{\rho }_j\xi )}-a_d\\ =(G_j(\xi )G_j^{(k)}(\xi )+\sum _{i=1}^k{c}_i\frac{G_j(\xi )G_j^{(k-i)}(\xi )}{{(\frac{z_j}{{\rho }_j}+\xi )}^i})\frac{a_d}{h(z_j+{\rho }_j\xi )}-a_d\\ \to G(\xi )G^{(k)}(\xi )-a_d,\end{array}$$

spherically locally uniformly in $\mathbb{C}-\{\xi |G(\xi )=\mathrm{\infty }\}$.

Since the multiplicity of all zeros and poles of G is at least $max\{\frac{k}{2}+d+1,k+d\}$ and by Lemmas 2.5-2.7, we have $G(\xi )G^{(k)}(\xi )-a_d\not\equiv 0$, and $G(\xi )G^{(k)}(\xi )-a_d$ has at least two distinct zeros.

Suppose ${\xi }_1$, ${\xi }_1^{\ast }$ are two distinct zeros of $G(\xi )G^{(k)}(\xi )-a_d$. We may choose a proper $\delta >0$ such that $D({\xi }_1,\delta )\cap D({\xi }_1^{\ast },\delta )=\mathrm{\varnothing }$, where $D({\xi }_1,\delta )=\{\xi ||\xi -{\xi }_1|<\sigma \}$, $D({\xi }_1^{\ast },\delta )=\{\xi ||\xi -{\xi }_1^{\ast }|<\delta \}$.

By Hurwitz’s theorem, there exists a subsequence of $a_d{f}_j(z_j+{\rho }_j\xi )f_j^{(k)}(z_j+{\rho }_j\xi )-a_{d}p(z_j+{\rho }_j\xi )$, we may still denote it as $a_d{f}_j(z_j+{\rho }_j\xi )f_j^{(k)}(z_j+{\rho }_j\xi )-a_{d}p(z_j+{\rho }_j\xi )$, then there exist points ${\xi }_j\in D({\xi }_1,\delta )$ and points ${\xi }_j^{\ast }\in D({\xi }_1^{\ast },\delta )$ such that for sufficiently large j, $a_d{f}_j(z_j+{\rho }_j\xi )f_j^{(k)}(z_j+{\rho }_j\xi )-a_{d}p(z_j+{\rho }_j\xi )=0$, $a_d{f}_j(z_j+{\rho }_j\xi )f_j^{(k)}(z_j+{\rho }_j\xi )-a_{d}p(z_j+{\rho }_j\xi )=0$.

Similar to the proof of Case 1.1, we get a contradiction. Then ${\mathcal{F}}_1$ is normal at 0.

By Case 1.1 and Case 1.2, we know ${\mathcal{F}}_1$ is normal at 0. Hence there exists ${\mathrm{△}}_{\rho }=\{z:|z|<\rho \}$ and a subsequence of $F_{jk}$ of $F_j$ such that $F_{jk}$ converges spherically locally uniformly to a meromorphic function $F(z)$ or ∞ ($k\to \mathrm{\infty }$) in ${\mathrm{△}}_{\rho }$.

Here, we discuss the following two cases.

Case i. When k is large enough, $f_{jk}\ne 0$. Then $F(0)=\mathrm{\infty }$. Thus, for ∀ constant $R>0$, $\mathrm{\exists }\sigma \in (0,\rho )$, we have $|F(z)|>R$ when $z\in {\mathrm{△}}_{\rho }$. Thus, for sufficiently large k, $|F_{jk}(z)|>\frac{R}{2}$, $\frac{1}{f_{jk}}$ is a holomorphic function in ${\mathrm{△}}_{\rho }$. Hence when $|z|=\frac{\sigma }{2}$,

$$\left|\frac{1}{f_{jk}}\right|=\left|\frac{1}{F_{jk}{z}^{d/2}}\right|\le \frac{2^{d/2+1}}{R{\sigma }^{d/2}}.$$

By the maximum principle and Montel’s theorem, ℱ is normal at $z=0$.

Case ii. There exists a subsequence of $f_{jk}$, we may still denote it as $f_{jk}$, such that $f_{jk}(0)=0$. Since $\mathrm{\forall }f\in \mathcal{F}$, the multiplicity of whose zeros is at least $max\{\frac{k}{2}+d+1,k+d\}$, then $F(0)=0$. Thus, there exists $0<r<\rho$ such that $F(z)$ is holomorphic in ${\mathrm{△}}_r=\{z:|z|<r\}$ and has a unique zero $z=0$ in ${\mathrm{△}}_r$. Then $F_{jk}$ converges spherically locally uniformly to a holomorphic function $F(z)$ in ${\mathrm{△}}_r$, $f_{jk}$ converges spherically locally uniformly to a holomorphic function $z^{\frac{d}{2}}F(z)$ in ${\mathrm{△}}_r$. Hence ℱ is normal at $z=0$.

By Case i and Case ii, we obtain ℱ is normal at $z=0$.

Case 2. When $p(z_0)\ne 0$.

Suppose that ℱ is not normal at $z_0$. Then by Lemma 2.1, there exists a sequence of complex numbers $z_t\to z_0$, a sequence of positive numbers ${\rho }_t\to 0$ and a sequence of functions $f_t\in \mathcal{F}$ such that $g_t(\xi )={\rho }_t^{-\frac{k}{2}}{f}_t(z_t+{\rho }_t\xi )\to g(\xi )$ spherically locally uniformly in ℂ, where $g(\xi )$ is a nonconstant meromorphic function in ℂ, and the multiplicity of the zeros and poles of $g(\xi )$ is at least $max\{\frac{k}{2}+d+1,k+d\}$.

Hence by Lemmas 2.5-2.7, we have $g(\xi )g^{(k)}(\xi )-p(z_0)\not\equiv 0$, and $g(\xi )g^{(k)}(\xi )-p(z_0)$ has at least two distinct zeros. Similar to the proof of Case 1.1, we get a contradiction. Thus, ℱ is normal at $z_0$.

Hence, ℱ is normal in D as $z_0$ is arbitrary. The proof is complete.

4 Proof of Theorem 1.2

Because $P(z)$ has at least one zero, we may assume, with no loss of generality, that $P(z)=z^n+a_{n-1}{z}^{n-1}+\cdots +a_q{z}^q$, where $q\ge 1$ is a positive integer and $a_q\ne 0$. Suppose that ℱ is not normal in D. Then similar to the proof of Theorem 1.1, we can get a contradiction. Hence ℱ is normal in D. The proof is complete.