- Research
- Open Access

# Normal families of meromorphic functions sharing one function

- Ling Qiu
^{1}and - FeiFei Hu
^{1}Email author

**2013**:288

https://doi.org/10.1186/1029-242X-2013-288

© Qiu and Hu; licensee Springer 2013

**Received:**7 November 2012**Accepted:**9 May 2013**Published:**6 June 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.

## Keywords

- meromorphic function
- normal family
- shared holomorphic function

## 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* $p(z)={a}_{d}{z}^{d}+{a}_{d-1}{z}^{d-1}+\cdots +{a}_{1}z+{a}_{0}$ *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

*i.e.*,

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 $m(r,H)=S(r,f)$.

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 $m(r,G)=S(r,f)$.

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 ${f}^{(k)}$.

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.

- (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

*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

where $deg({g}_{1}(z))\le (k+d+1)(s+t-1)$.

Next, we discuss the following two cases.

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 $m>n$.

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 $d<l$.

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.

*c*is a finite complex number. Then

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.

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 $\frac{d}{2}<i$.

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 ${\mathrm{\u25b3}}_{\rho}=\{z:|z|<\rho \}$ 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.

*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 2.5-2.7, 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.

## Declarations

### Acknowledgements

The authors wish to thank the referees and editors for their very helpful comments and useful suggestions.

## Authors’ Affiliations

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