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Normal families of meromorphic functions sharing one function
Journal of Inequalities and Applications volume 2013, Article number: 288 (2013)
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
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 ${g}^{\mathrm{\u266f}}(\xi )\le {g}^{\mathrm{\u266f}}(0)=kA+1$. 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
Lemma 2.3 (see [1])
Let ${f}_{1}(z)$, ${f}_{2}(z)$ be two meromorphic functions defined in $D=\{z:z<R\}$, then
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 $p(z)={a}_{d}{z}^{d}+{a}_{d1}{z}^{d1}+\cdots +{a}_{1}z+{a}_{0}$ be a polynomial, where ${a}_{d}\ne 0$, ${a}_{d1},\dots ,{a}_{0}$ 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
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
Let
Since the multiplicity of zeros of $f(z)$ is at least 2, we can get from (2.2) that
By Lemma 2.2, we know that
We can get from (2.2) that
i.e.,
Let
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
We can get from (2.6) that
Let
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
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
Since $\frac{{({\psi}_{1})}^{\prime}}{{\psi}_{1}}$ has only simple poles, so by (2.9) we know that
Combining (2.7) and (2.10)(2.12), we have
Hence
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
Since the multiplicity of the poles of $f(z)$ is at least 2, we can get from (2.13) that
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 $p(z)={a}_{d}{z}^{d}+{a}_{d1}{z}^{d1}+\cdots +{a}_{1}z+{a}_{0}$ be a polynomial, where ${a}_{d}\ne 0$, ${a}_{d1},\dots ,{a}_{0}$ 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 ${(f{f}^{(k)})}^{(d)}\equiv {a}_{d}+Al\cdots (ld+1){(z{z}_{0})}^{ld}$. 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 $p(z)={a}_{d}{z}^{d}+{a}_{d1}{z}^{d1}+\cdots +{a}_{1}z+{a}_{0}$ be a polynomial, where ${a}_{d}\ne 0$, ${a}_{d1},\dots ,{a}_{0}$ 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
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
By (2.19), we get
where $g(z)=(mn)(mn1)\cdots (mnk+1){z}^{k(s+t1)}+\cdots +{c}_{1}z+{c}_{0}$ is a polynomial, ${c}_{i}$ ($i=0,1$) are constants and $deg(g(z))\le k(s+t1)$. Thus (2.14) together with (2.15) implies
By (2.16), we obtain
where $deg({g}_{1}(z))\le (k+d+1)(s+t1)$.
Next, we discuss the following two cases.
Case 1. If $f{f}^{(k)}p(z)$ has only one zero ${z}_{0}$, then let
Subcase 1.1. When $d\ge l$.
Combining (2.16) and (2.18), we get $d+2n+kt=deg(g(z))+2mks\le k(s+t1)+2mks$. That is, $2(mn)\ge k+d$, and then $m>n$.
Differentiating both sides of (2.18), we have
where $deg({g}_{2}(z))\le t(d+1)+ld1$.
By (2.17) and (2.19), we know $2m(k+d+1)s\le deg({g}_{2}(z))\le t(d+1)+ld1$. Thus $2m(k+d+1)st(d+1)\le ld1$. Since $2m(k+d+1)st(d+1)\ge $ $2m(k+d+1)\frac{m}{k/2+d+1}(d+1)\frac{n}{k/2+d+1}>0$, then $0<ld1$, which contradicts with $d\ge l$.
Subcase 1.2. When $d<l$.
Differentiating both sides of (2.18), we have
where ${g}_{3}(z)=(l2nkt)\cdots (l2nktd){z}^{t(d+1)}+\cdots +{d}_{1}z+{d}_{0}$, 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 $d+2n+kt=deg(g(z))+2mks\le k(s+t1)+2mks$. That is, $2(mn)\ge k+d$, 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 $ld1\le deg({g}_{1}(z))\le (k+d+1)(s+t1)$, since $l=2n+kt+d$, thus $2n+kt+dd1\le (k+d+1)(s+t1)$. 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 $l=deg(g(z))+2mks\le k(s+t1)+2mks=2m+ktk$. 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
spherically locally uniformly in ℂ, so
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 2.52.7, 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
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
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 2.52.7, 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 ${\mathrm{\u25b3}}_{\rho}=\{z:z<\rho \}$ 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 ${\mathrm{\u25b3}}_{\rho}$.
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 $z\in {\mathrm{\u25b3}}_{\rho}$. Thus, for sufficiently large k, ${F}_{jk}(z)>\frac{R}{2}$, $\frac{1}{{f}_{jk}}$ is a holomorphic function in ${\mathrm{\u25b3}}_{\rho}$. Hence when $z=\frac{\sigma}{2}$,
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 ${\mathrm{\u25b3}}_{r}=\{z:z<r\}$ and has a unique zero $z=0$ in ${\mathrm{\u25b3}}_{r}$. Then ${F}_{jk}$ converges spherically locally uniformly to a holomorphic function $F(z)$ in ${\mathrm{\u25b3}}_{r}$, ${f}_{jk}$ converges spherically locally uniformly to a holomorphic function ${z}^{\frac{d}{2}}F(z)$ in ${\mathrm{\u25b3}}_{r}$. 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 2.52.7, 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 $P(z)={z}^{n}+{a}_{n1}{z}^{n1}+\cdots +{a}_{q}{z}^{q}$, 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.
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The authors wish to thank the referees and editors for their very helpful comments and useful suggestions.
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LQ and FH performed and drafted manuscript. All authors read and approved the final manuscript.
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Qiu, L., Hu, F. Normal families of meromorphic functions sharing one function. J Inequal Appl 2013, 288 (2013). https://doi.org/10.1186/1029242X2013288
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Keywords
 meromorphic function
 normal family
 shared holomorphic function