Normal family of meromorphic functions concerning limited the numbers of zeros

Abstract

Let $k, n \in {\mathbb {N}}, l \in {\mathbb {N}}\backslash \left\{ 1 \right\} , m\in {\mathbb {N}}\cup \left\{ 0 \right\}$, and let $a(z)(\not \equiv 0)$ be a holomorphic function, all zeros of a(z) have multiplicities at most m. Let ${\mathcal {F}}$ be a family of meromorphic functions in D. If for each $f \in {\mathcal {F}}$, the zeros of f have multiplicity at least $k+m$, and for $f\in {\mathcal {F}}$, $f^{l}(f^{(k)})^{n}-a(z)$ has at most one zero in D, then ${\mathcal {F}}$ is normal in D.

1 Introduction and main results

Let f be a meromorphic function in ${\mathbb {C}}$ and we shall use the usual notations and classical results of Nevanlinna’s theory, such as $m(r,f),N(r,f),{\overline{N}}(r,f),$ $T(r,f),\ldots$.

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

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

Theorem A

[1] Let $n\in {\mathbb {N}}$, and $a\in {\mathbb {C}}\backslash \left\{ 0 \right\}$. let ${\mathcal {F}}$ be a family of meromorphic function in D. If $f^{n}f' \ne a$, for each $f \in {\mathcal {F}}$, then ${\mathcal {F}}$ is normal in D.

This normal conjecture was showed by Yang and Zhang [2] (for $n \ge 5$), Gu [3] $(for n = 4, 3),$ Pang [4] (for $n \ge 2$) and Chen and Fang [5] (for $n = 1$).

For the related results, see Zhang [6], Meng and Hu [7], Deng et al.[8].

Ding et al. [9] studied the general case of $f^{l}(f^{(k)})^{n}$ and and proved the following theorem.

Theorem B

 Let $k, l\in {\mathbb {N}}, n\in {\mathbb {N}}\backslash \left\{ 1 \right\} , a\in {\mathbb {C}}\backslash \left\{ 0 \right\}$. Let ${\mathcal {F}}$ be a family of meromorphic functions in D. If for each $f \in {\mathcal {F}}$, the zeros of f have multiplicity at least $max\{k,2\}$, and for $f, g \in {\mathcal {F}}$, $f^{l}(f^{(k)})^{n}$ and $g^{l}(g^{(k)})^{n}$ share a, then ${\mathcal {F}}$ is normal in D.

Recently, Meng et al. [10] considered the case of sharing a holomorphic function and and proved the following result.

Theorem C

Let $k,l \in {\mathbb {N}}, n\in {\mathbb {N}}\backslash \left\{ 1 \right\} , m\in {\mathbb {N}}\cup \left\{ 0 \right\}$, and let $a(z)(\not \equiv 0)$ be a holomorphic function, all zeros of a(z) have multiplicities at most m, which is divisible by $n+l$. Let ${\mathcal {F}}$ be a family of meromorphic functions in D. If for each $f \in {\mathcal {F}}$, the zeros of f have multiplicities at least $k+m+1$ and all poles of f are of multiplicity at least $m+1$, and for $f, g \in {\mathcal {F}}$, $f^{l}(f^{(k)})^{n}$ and $g^{l}(g^{(k)})^{n}$ share a(z), then ${\mathcal {F}}$ is normal in D.

By Theorem C, the following question arises naturally:

Question 1.1

Is it possible to omit the conditions: (1)“ m is divisible by $n+l$” and (2)“all poles of f have multiplicity at least $m+1$” ?

In this paper, we study this problem and obtain the following result.

Theorem 1.1

Let $k, n \in {\mathbb {N}}, l \in {\mathbb {N}}\backslash \left\{ 1 \right\} , m\in {\mathbb {N}}\cup \left\{ 0 \right\}$, and let $a(z)(\not \equiv 0)$ be a holomorphic function, all zeros of a(z) have multiplicities at most m. Let ${\mathcal {F}}$ be a family of meromorphic functions in D. If for each $f \in {\mathcal {F}}$, the zeros of f have multiplicity at least $k+m$, and for $f\in {\mathcal {F}}$, $f^{l}(f^{(k)})^{n}-a(z)$ has at most one zero in D, then ${\mathcal {F}}$ is normal in D.

Now we give some examples to show that the conditions in our results are necessary.

Example 1.1

Let ${D}=\{z: |z| <1\}$ and $a(z)\equiv 0$. Let ${\mathcal {F}}=\{f_j(z)\}$, where

$$\begin{aligned} f_j(z)=e^{jz},z\in D,j=1,2\ldots . \end{aligned}$$

Then $f_{j}^{l}\left( z \right) \left( f_{j}^{\left( k \right) } \right) ^n\left( z \right) -a(z)\ne 0$ in D, however ${\mathcal {F}}$ is not normal at $z = 0$. This shows that $a(z)\not \equiv 0$ is necessary in Theorem 1.1.

Example 1.2

Let ${D}=\{z: |z| <1\}$ and $a(z)=\frac{1}{z^{l+kn+n}}$. Let ${\mathcal {F}}=\{f_j(z)\}$, where

$$\begin{aligned} f_j(z)=\frac{1}{jz},z\in D,j=1,2\cdots ,j^{l+n}\ne [(-1)^{k}k!]^{n}. \end{aligned}$$

Then $f_{j}^{l}\left( z \right) \left( f_{j}^{\left( k \right) } \right) ^n\left( z \right) -a(z)\ne 0$ in D, however ${\mathcal {F}}$ is not normal at $z = 0$. This shows that Theorem  1.1 is not valid if a(z) is a meromorphic function in D.

Example 1.3

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

$$\begin{aligned} f_j(z)=jz^{k-1},z\in D,j=1,2\cdots . \end{aligned}$$

Then $f_{j}^{l}\left( z \right) \left( f_{j}^{\left( k \right) } \right) ^n\left( z \right) -a$, which has no zero in D , however ${\mathcal {F}}$ is not normal at $z=0$. This shows that the condition “ all zeros of f have multiplicity at least $k+m$ ” in Theorem 1.1 is sharp.

Example 1.4

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

$$\begin{aligned} f_j(z)=jz^{k},z\in D,j=1,2\ldots . \end{aligned}$$

Then $f_{j}^{l}\left( z \right) \left( f_{j}^{\left( k \right) } \right) ^n\left( z \right) -a=j^{l+n}(k!)^{n}z^{lk}-a$, which has at least $l\ge 2$ distinct zeros in D, however ${\mathcal {F}}$ is not normal at $z=0$. This shows that the condition “$f^{l}(f^{(k)})^{n}-a(z)$ has at most one zero” in Theorem 1.1 is necessary.

2 Some lemmas

Lemma 2.1

[11] Let ${\mathcal {F}}$ be a family of functions meromorphic in the unit disc $\Delta$, all of whose zeros have multiplicity at least k. Then if ${\mathcal {F}}$ is not normal in any neighbourhood of $z_{0}\in \Delta$, there exist, for each $\alpha$, $0\le \alpha <k$,

• (i) points $z_n$, $z_n\rightarrow z_0$, $z_0\in \Delta$;

• (ii) functions $f_n \in {\mathcal {F}}$; and

• (iii) positive numbers $\rho _n\rightarrow 0^ +$, such that $g_n (\xi ) = \rho _n^{ - \alpha } f_n (z_n + \rho _n \xi ) \rightarrow g(\xi )$ spherically uniformly on compact subsets of ${\mathbb {C}}$, where g is a non-constant meromorphic function, all of whose zeros have multiplicity at least k.

Lemma 2.2

[12]  Let $k,n\in {\mathbb {N}}$, $l\in {\mathbb {N}}\backslash \left\{ 1 \right\}$, $a\in {\mathbb {C}}{\setminus} \left\{ 0 \right\}$, and let f(z) be a non-constant meromorphic with all zeros that have multiplicity at least k. Then $f^{l}(z)(f^{(k)})^{n}(z)-a$ has at least two distinct zeros.

Using the idea of Chang [13], we get the following lemma.

Lemma 2.3

Let $k,l,n,m \in {\mathbb {N}}$, let q(z) be a polynomial of degree m, and let f(z) be a non-constant rational function with $f(z)\ne 0$. Then $f^{l}(z)(f^{(k)})^{n}(z) -q(z)$ has at least $l+kn+n$ distinct zeros.

The proof of Lemma 2.3is almost exactly the same with Lemma 11 in Deng etc. [14], here, we omit the details.

Lemma 2.4

[15] Let $f_{j}(j=1,2)$ be two nonconstant meromorphic functions, then

$$\begin{aligned} N(r,f_{1}f_{2})-N \left(r,\frac{1}{f_{1}f_{2}} \right)=N(r,f_{1})+N(r,f_{2})-N \left(r,\frac{1}{f_{1}} \right)-N \left(r,\frac{1}{f_{2}}\right). \end{aligned}$$

Lemma 2.5

Let $k, m, n \in {\mathbb {N}}$, $l\in {\mathbb {N}}\backslash \left\{ 1 \right\}$, let q(z) be a polynomial of degree m, and let f(z) be a non-constant meromorphic function in ${\mathbb {C}}$, the zeros of f(z) have multiplicities at least $k+m$.Then $(f(z))^{l}(f^{(k)})^{n}(z) -q(z)$ has at least two distinct zeros.

Proof

Since

$$\begin{aligned} \frac{1}{f^{l+n}} &= \left( \frac{f^{\left( k \right) }}{f} \right) ^n\cdot \frac{1}{q}-\frac{f^l\left( f^{\left( k \right) } \right) ^n-q}{qf^{l+n}} \\ &= \frac{f^l\left( f^{\left( k \right) } \right) ^n}{qf^{l+n}}-\frac{\left[ f^l\left( f^{\left( k \right) } \right) ^n \right] 'q-q'\left[ f^l\left( f^{\left( k \right) } \right) ^n \right] }{qf^{l+n}}\cdot \frac{f^l\left( f^{\left( k \right) } \right) ^n-q}{\left[ f^l\left( f^{\left( k \right) } \right) ^n \right] 'q-q'\left[ f^l\left( f^{\left( k \right) } \right) ^n \right] } . \end{aligned}$$

Noticing that $m(r,\frac{f^{(k)}}{f})=S(r,f), m(r,\frac{1}{q})=O(1)$, and $m(r,q)=mlog r + O(1)$. By Nevanlinna’s Fundamental Theorem, we get

$$\begin{aligned}\left( l+n \right) m\left( r,\frac{1}{f} \right) & =m\left( r,\frac{1}{f^{l+n}} \right) \nonumber \\& =m\left( r,\frac{f^l\left( f^{\left( k \right) } \right) ^n}{qf^{l+n}}-\frac{\left[ f^l\left( f^{\left( k \right) } \right) ^n \right] 'q-q'\left[ f^l\left( f^{\left( k \right) } \right) ^n \right] }{qf^{l+n}}\cdot \frac{f^l\left( f^{\left( k \right) } \right) ^n-q}{\left[ f^l\left( f^{\left( k \right) } \right) ^n \right] 'q-q'\left[ f^l\left( f^{\left( k \right) } \right) ^n \right] }\right) \nonumber \\& \le m \left(r,\frac{1}{q} \right)+ n~m \left(r,\frac{f^{(k)}}{f} \right)+ m\left( r,\frac{\left[ f^l\left( f^{\left( k \right) } \right) ^n \right] 'q-q'\left[ f^l\left( f^{\left( k \right) } \right) ^n \right] }{qf^{l+n}} \right) \nonumber \\& \quad + m\left( r,\frac{f^l\left( f^{\left( k \right) } \right) ^n-q}{\left[ f^l\left( f^{\left( k \right) } \right) ^n \right] 'q-q'\left[ f^l\left( f^{\left( k \right) } \right) ^n \right] } \right) + O(1)\nonumber \\& \le m\left( r,\frac{\left[ f^l\left( f^{\left( k \right) } \right) ^n \right] 'q-q'\left[ f^l\left( f^{\left( k \right) } \right) ^n \right] }{qf^{l+n}} \right) \nonumber \\& \quad +m \left( r,\frac{f^l\left( f^{\left( k \right) } \right) ^n-q}{\left[ f^l\left( f^{\left( k \right) } \right) ^n \right] 'q-q'\left[ f^l\left( f^{\left( k \right) } \right) ^n \right] } \right) +S\left( r,f \right) \nonumber \\& \le T\left( r,\frac{f^l\left( f^{\left( k \right) } \right) ^n-q}{\left[ f^l\left( f^{\left( k \right) } \right) ^n \right] 'q-q'\left[ f^l\left( f^{\left( k \right) } \right) ^n \right] }\right) \nonumber \\& \quad - N\left( r,\frac{f^l\left( f^{\left( k \right) } \right) ^n-q}{\left[ f^l\left( f^{\left( k \right) } \right) ^n \right] 'q-q'\left[ f^l\left( f^{\left( k \right) } \right) ^n \right] } \right) +S\left( r,f \right) \nonumber \\& \le T\left( r,\frac{\left[ f^l\left( f^{\left( k \right) } \right) ^n \right] 'q-q'\left[ f^l\left( f^{\left( k \right) } \right) ^n \right] }{f^l\left( f^{\left( k \right) } \right) ^n-q} \right) \nonumber \\& \quad -N\left( r,\frac{f^l\left( f^{\left( k \right) } \right) ^n-q}{\left[ f^l\left( f^{\left( k \right) } \right) ^n \right] 'q-q'\left[ f^l\left( f^{\left( k \right) } \right) ^n \right] } \right) +S\left( r,f \right) \nonumber \\& = m\left( r,\frac{\left[ f^l\left( f^{\left( k \right) } \right) ^n \right] 'q-q'\left[ f^l\left( f^{\left( k \right) } \right) ^n \right] }{f^l\left( f^{\left( k \right) } \right) ^n-q} \right) +N\left( r,\frac{\left[ f^l\left( f^{\left( k \right) } \right) ^n \right] 'q-q'\left[ f^l\left( f^{\left( k \right) } \right) ^n \right] }{f^l\left( f^{\left( k \right) } \right) ^n-q} \right) \nonumber \\& \quad - N\left( r,\frac{f^l\left( f^{\left( k \right) } \right) ^n-q}{\left[ f^l\left( f^{\left( k \right) } \right) ^n \right] 'q-q'\left[ f^l\left( f^{\left( k \right) } \right) ^n \right] } \right) +S\left( r,f \right) \nonumber \\& = m\left( r,q\cdot \frac{\left[ \frac{f^l\left( f^{\left( k \right) } \right) ^n-q}{q} \right] '}{\frac{f^l\left( f^{\left( k \right) } \right) ^n-q}{q}} \right) +N\left( r,\frac{\left[ f^l\left( f^{\left( k \right) } \right) ^n \right] 'q-q'\left[ f^l\left( f^{\left( k \right) } \right) ^n \right] }{f^l\left( f^{\left( k \right) } \right) ^n-q} \right) \nonumber \\& \quad - N\left( r,\frac{f^l\left( f^{\left( k \right) } \right) ^n-q}{\left[ f^l\left( f^{\left( k \right) } \right) ^n \right] 'q-q'\left[ f^l\left( f^{\left( k \right) } \right) ^n \right] } \right) \nonumber \\& \quad +S\left( r,f \right) .\nonumber \\& \le m\left( r,\frac{\left[ \frac{f^l\left( f^{\left( k \right) } \right) ^n-q}{q} \right] '}{\frac{f^l\left( f^{\left( k \right) } \right) ^n-q}{q}} \right) +N\left( r,\frac{\left[ f^l\left( f^{\left( k \right) } \right) ^n \right] 'q-q'\left[ f^l\left( f^{\left( k \right) } \right) ^n \right] }{f^l\left( f^{\left( k \right) } \right) ^n-q} \right) \nonumber \\& \quad - N\left( r,\frac{f^l\left( f^{\left( k \right) } \right) ^n-q}{\left[ f^l\left( f^{\left( k \right) } \right) ^n \right] 'q-q'\left[ f^l\left( f^{\left( k \right) } \right) ^n \right] } \right) \nonumber \\& \quad +m\left( r,q \right) +S\left( r,f \right) . \end{aligned}$$

(2.1)

By Lemma 2.4 applied to (2.1), we can get

$$\begin{aligned}&\left( l+n \right) m\left( r,\frac{1}{f} \right) \le m\left( r,\frac{\left[ \frac{f^l\left( f^{\left( k \right) } \right) ^n-q}{q} \right] '}{\frac{f^l\left( f^{\left( k \right) } \right) ^n-q}{q}} \right) +N\left( r,\frac{1}{f^l\left( f^{\left( k \right) } \right) ^n-q} \right) \\& \quad -N\left( r,f^l\left( f^{\left( k \right) } \right) ^n-q \right) +N\left( r,\left[ f^l\left( f^{\left( k \right) } \right) ^n \right] 'q-q'\left[ f^l\left( f^{\left( k \right) } \right) ^n \right] \right) \\& \quad -N\left( r,\frac{1}{\left[ f^l\left( f^{\left( k \right) } \right) ^n \right] 'q-q'\left[ f^l\left( f^{\left( k \right) } \right) ^n \right] } \right) +m\log r +S\left( r,f \right) . \end{aligned}$$

This is

$$\begin{aligned}&\left( l+n \right) m\left( r,\frac{1}{f}\right) \le {\overline{N}}(r,f) +N\left( r,\frac{1}{f^l\left( f^{\left( k \right) } \right) ^n-q} \right) \nonumber \\&\quad -N\left( r,\frac{1}{\left[ f^l\left( f^{\left( k \right) } \right) ^n \right] 'q-q'\left[ f^l\left( f^{\left( k \right) } \right) ^n \right] } \right) +m\log r +S\left( r,f \right) . \end{aligned}$$

(2.2)

We add $(l+n)N \left(r,\frac{1}{f} \right)$ to both sides in (2.2), then

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

(2.3)

Let $\xi$ be a zero of f with multiplicity $t (\ge k+m)$, then $\xi$ is a zero of $[f^{l}(f^{(k)})^{n}]'q-[f^{l}(f^{(k)})^{n}]q'$ with multiplicity at least $(l + n)t - kn -1$. Noticing that

$$\begin{aligned} \left[ f^l\left( f^{\left( k \right) } \right) ^n \right] 'q-q'\left[ f^l\left( f^{\left( k \right) } \right) ^n \right] =\left[ f^l\left( f^{\left( k \right) } \right) ^n -q\right] 'q-q'\left[ f^l\left( f^{\left( k \right) } \right) ^n -q\right] , \end{aligned}$$

which implies

$$\begin{aligned}&N\left( r,\frac{1}{\left[ f^l\left( f^{\left( k \right) } \right) ^n \right] 'q-q'\left[ f^l\left( f^{\left( k \right) } \right) ^n \right] } \right) \ge N\left( r,\frac{1}{f^l\left( f^{\left( k \right) } \right) ^n-q} \right) \\&\quad -{\overline{N}}\left( r,\frac{1}{f^l\left( f^{\left( k \right) } \right) ^n-q} \right) . \end{aligned}$$

Therefore, from (2.3), we get

$$\begin{aligned} \left( l+n \right) T\left( r,f\right) & \le (kn+1){\overline{N}}\left(r,\frac{1}{f} \right)+{\overline{N}}(r,f) +{\overline{N}}\left( r,\frac{1}{f^l\left( f^{\left( k \right) } \right) ^n-q} \right) +m\log r +S\left( r,f \right) \\& \le \frac{kn+1}{k+m}N\left(r,\frac{1}{f} \right)+{\overline{N}}(r,f) +{\overline{N}}\left( r,\frac{1}{f^l\left( f^{\left( k \right) } \right) ^n-q} \right) +m\log r +S\left( r,f \right) . \end{aligned}$$

i.e.,

$$\begin{aligned} M~T\left( r,f\right) \le {\overline{N}}\left( r,\frac{1}{f^l\left( f^{\left( k \right) } \right) ^n-q} \right) +m\log r +S\left( r,f \right) , \end{aligned}$$

(2.4)

where

$$\begin{aligned} M=l+n -1-\frac{kn+1}{k+m}=l-1+\frac{mn-1}{k+m}. \end{aligned}$$

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

Next, we consider two cases.

Case 1: $n\ge 2$. By the assumptions,

$$\begin{aligned} M\ge 1+\frac{1}{k+m}. \end{aligned}$$

From (2.4), we get

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

It follows that f(z) is a rational function of degree $<m+1$. Since the zeros of f(z) have multiplicities at least $k+m\ge m+1$, then we get $f(z)\ne 0$. Thus, by Lemma 2.3, we obtain that $f^{l}(z)(f^{(k)})^{n}(z) -q(z)$ has at least $l + kn + n\ge 6$ distinct zeros, which is a contradiction.

Case 2: $n=1$. Then $M=l-\frac{k+1}{k+m}$.

$\mathbf {Subcase \,\,2.1}$: $m\ge 2$. By the assumptions, $M> 1$ and from (2.4), we get

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

It follows that f(z) is a rational function of degree $<m+1$. Since the zeros of f(z) have multiplicities at least $k+m\ge m+1$, then we get $f(z)\ne 0$. Thus, by Lemma 2.3, we obtain that $f^{l}(z)(f^{(k)})^{n}(z) -q(z)$ has at least $l + k + 1\ge 4$ distinct zeros, which is a contradiction.

$\mathbf {Subcase \,\,2.2}$: $m=1$. From (2.4), we get

$$\begin{aligned} \left( l-1\right) T\left( r,f\right) \le {\overline{N}}\left( r,\frac{1}{f^l f^{\left( k \right) }-q} \right) +\log r +S\left( r,f \right) , \end{aligned}$$

$\mathbf {Subcase \,\,2.2.1}$: $f^{l}(z)f^{(k)}(z)-q(z)\ne 0$. From (2.4), we get

$$\begin{aligned} T\left( r,f\right) \le (l-1) T\left( r,f\right) \le \log r +S\left( r,f \right) . \end{aligned}$$

It follows that f(z) is a rational function of degree $\le 1$. Since the zeros of f(z) have multiplicities at least $k+1\ge 2$, then we get $f(z)\ne 0$. Thus, by Lemma 2.3, we obtain that $f^{l}(z)(f^{(k)})^{n}(z) -q(z)$ has at least $l + k + 1\ge 4$ distinct zeros, which is a contradiction.

$\mathbf {Subcase \,\,2.2.2}$: $f^{l}(z)f^{(k)}(z)-q(z)= 0$. By the assumptions, we get $f^{l}(z)f^{(k)}(z)-q(z)$ has only one zero. Then, from (2.4), we obtain

$$\begin{aligned} (l-1)T\left( r,f\right) \le 2 \log r +S\left( r,f \right) . \end{aligned}$$

$\mathbf {Subcase\,\,2.2.2.1}$: $l\ge 3$, from (2.4), we obtain

$$\begin{aligned} T\left( r,f\right) \le \log r +S\left( r,f \right) . \end{aligned}$$

It follows that f(z) is a rational function of degree $\le 1$. Since the zeros of f(z) have multiplicities at least $k+1\ge 2$, then we get $f(z)\ne 0$. Thus, by Lemma 2.3, we obtain that $f^{l}(z)(f^{(k)})^{n}(z) -q(z)$ has at least $l + k + 1\ge 5$ distinct zeros, which is a contradiction.

$\mathbf {Subcase \,\,2.2.2.2}$: $l=2$, from (2.4), we obtain

$$\begin{aligned} T\left( r,f\right) \le 2\log r +S\left( r,f \right) . \end{aligned}$$

It follows that f(z) is a rational function of degree $\le 2$.

$\mathbf {Subcase \,\,2.2.2.2.1}$: $k\ge 2$. Since the zeros of f(z) have multiplicities at least $k+1\ge 3$, then we get $f(z)\ne 0$. Thus, by Lemma 2.3, we obtain that $f^{l}(z)(f^{(k)})^{n}(z) -q(z)$ has at least $l + k + 1\ge 5$ distinct zeros, which is a contradiction.

$\mathbf {Subcase\,\,2.2.2.2.2}$: $k=1$. Then we get $f(z)\ne 0$ or f(z) has only one zero with multiplicity 2.

The former case can be ruled out from Lemma 2.3. Hence f(z) has the following forms:

$$\begin{aligned} \mathrm{(i)}\,\, f(z)&= A(z-z_{0})^{2}; ~~{\rm (ii)}\,\, f(z)=\frac{A(z-z_{0})^{2}}{(z-z_{1})};\\ \mathrm{(iii)}\,\, f(z)&= \frac{A(z-z_{0})^{2}}{(z-z_{1})^{2}}; ~~{\rm (iv)} \,\, f(z)=\frac{A(z-z_{0})^{2}}{(z-z_{1})(z-z_{2})}, \end{aligned}$$

where $A, z_{0}$ are nonzero constants, and $z_{1}, z_{2}$ are distinct constants. Clearly, $z_{0}\ne z_{1}, z_{0}\ne z_{2}$, and $T(r,f)=2 \log r + O(1)$.

$\mathrm{(i)}\,\, f(z)=A(z-z_{0})^{2}$. Obviously, ${\overline{N}}\left(r,\frac{1}{f} \right)\le \frac{1}{2} T(r,f)+ O(1)$. From (2.4), we obtain

$$\begin{aligned} 3T(r,f)\le 2 {\overline{N}}\left(r,\frac{1}{f} \right)+{\overline{N}}(r,f)+2 \log r + S(r,f). \end{aligned}$$

Then

$$\begin{aligned} T(r,f)\le \log r + S(r,f), \end{aligned}$$

a contradiction.

$\mathrm{(ii)} \,\, f(z)=\frac{A(z-z_{0})^{2}}{(z-z_{1})}$. Then, ${\overline{N}}\left(r,\frac{1}{f} \right)\le \frac{1}{2} T(r,f)+ O(1), {\overline{N}}(r,f) = \log r$. From (2.4), we obtain

$$\begin{aligned} 3T(r,f)\le 2 {\overline{N}} \left(r,\frac{1}{f} \right)+{\overline{N}}(r,f)+2 \log r + S(r,f). \end{aligned}$$

Then

$$\begin{aligned} T(r,f)\le \frac{4}{3}\log r + S(r,f), \end{aligned}$$

which is a contradiction.

$\mathrm{(iii)} \,\, f(z)=\frac{A(z-z_{0})^{2}}{(z-z_{1})^{2}}$. Then, ${\overline{N}}\left(r,\frac{1}{f} \right)\le \frac{1}{2} T(r,f)+ O(1), {\overline{N}}(r,f)\le \frac{1}{2} T(r,f) + O(1)$. From (2.4), we obtain

$$\begin{aligned} 3T(r,f)\le 2 {\overline{N}}\left(r,\frac{1}{f} \right)+{\overline{N}}(r,f)+2 \log r + S(r,f). \end{aligned}$$

Then

$$\begin{aligned} T(r,f)\le \frac{7}{6}\log r + S(r,f), \end{aligned}$$

we also get a contradiction.

$\mathrm{(iv)} \,\, f(z)=\frac{A(z-z_{0})^{2}}{(z-z_{1})(z-z_{2})}$. Then

$$\begin{aligned} f^2\left( z \right) f'\left( z \right) =\frac{A^3\left( z-z_0 \right) ^5\left[ \left( 2z_0-\left( z_1+z_2 \right) \right) z+2z_1z_2-z_0\left( z_1+z_2 \right) \right] }{\left( z-z_1 \right) ^4\left( z-z_2 \right) ^4}. \end{aligned}$$

(2.5)

Since $q(z)=Bz + C$, where $B\ne 0, C$ are constants, and $f^{l}(z)f^{(k)}(z)-q(z)$ has only one zero. Then we have

$$\begin{aligned} f^{2}(z)f'(z)=Bz+ C +\frac{d(z-\zeta )^{t}}{\left( z-z_1 \right) ^4\left( z-z_2 \right) ^4}. \end{aligned}$$

(2.6)

Obviously, By calculation, we get $d=-B, t=9$, and $\zeta \ne z_{0}$.

Differentiating (2.5)–(2.6) two times separately, we obtain

$$\begin{aligned}{}[f^{2}(z)f'(z)]^{''}=\frac{(z-z_{0})^{3}g(z)}{\left( z-z_1 \right) ^6\left( z-z_2 \right) ^6}, \end{aligned}$$

where g(z) is a polynomial of degree $\le 5$, and

$$\begin{aligned}{}[f^{2}(z)f'(z)]^{''}=\frac{(z-\zeta )^{7}h(z)}{\left( z-z_1 \right) ^6\left( z-z_2 \right) ^6}, \end{aligned}$$

where h(z) is a polynomial of degree $\le 4$.

Since $z_{0} \ne \zeta$, then $(z-\zeta )^{7}$ is a factor of g(z). Thus g(z) is a polynomial of degree $\ge 7$, which is impossible. $\square$

Lemma 2.6

Let $k, n \in {\mathbb {N}}, l\in {\mathbb {N}}\backslash \left\{ 1 \right\}$, and let ${\mathcal {F}}=\{f_{m}\}$ be a sequence of meromorphic functions, $g_{m}(z)$ be a sequence of holomorphic functions in D such that $g_{m}(z)\longrightarrow g(z)$, where $g(z)(\ne 0)$ be a holomorphic function. If all zeros of function $f_{m}(z)$ have multiplicity at least k, and $f_{m}^{l}(z)(f_{n}^{(k)}(z))^{n}-g_{n}(z)$ has at most one zero, then ${\mathcal {F}}$ is normal in D.

Proof

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

$$\begin{aligned} h_{m}(\xi ) &=\frac{f_{m}(z_{m}+\rho _{m}\xi )}{\rho _{m}^{\frac{kn}{l+n}}}\longrightarrow h(\xi ) \end{aligned}$$

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

For each $\xi \in {\mathbb {C}}/\{h^{-1}(\infty )\}$, we have

$$\begin{aligned}&h_{m}^{l}(\xi )(h_{m}^{(k)}(\xi ))^{n}-g_{m}(z_{m}+\rho _{m}\xi )=f_{m}^{l}(z_{m}+\rho _{m}\xi )(f_{m}^{(k)})^{n}(z_{m}+\rho _{m}\xi )\\&\quad - g_{m}(z_{m}+\rho _{m}\xi )\longrightarrow h^{l}(\xi )(h^{(k)})^{n}(\xi )-g(z_{0}). \end{aligned}$$

Obviously, $h^{l}(\xi )(h^{(k)})^{n}(\xi )-g(z_{0})\not \equiv 0$.

Suppose that $h^{l}(\xi )(h^{(k)})^{n}(\xi )-g(z_{0})\equiv 0$, then $h(\xi )\ne 0$ since $g(z_{0})\ne 0$. It follows that

$$\begin{aligned} \frac{1}{h^{l+n}(\xi )}\equiv \frac{1}{g(z_{0})} \left[\frac{h^{(k)}(\xi )}{h(\xi )} \right]^{n}. \end{aligned}$$

Thus

$$\begin{aligned} (l+n)m \left(r,\frac{1}{h} \right)=m \left(r,\frac{1}{g(z_{0})} \left[\frac{h^{(k)}(\xi )}{h(\xi )} \right]^{n} \right)=S(r,h). \end{aligned}$$

Then $T(r,h)=S(r,h)$ since $h\ne 0$. we can deduce that $h(\xi )$ is a constant, a contradiction.

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

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

$$\begin{aligned} f_{m}^{l}(z_{m}+\rho _{m}\xi _{1,m})(f_{m}^{(k)})^{^{n}}(z_{m}+\rho _{m}\xi _{1,m})-g_{m}(z_{m}+\rho _{m}\xi _{1,m})=0, \end{aligned}$$

and

$$\begin{aligned} f_{m}^{l}(z_{m}+\rho _{m}\xi _{2,m})(f_{m}^{(k)})^{^{n}}(z_{m}+\rho _{m}\xi _{2,m})-g_{m}(z_{m}+\rho _{m}\xi _{2,m})=0 \end{aligned}$$

Since $f_{m}^{l}(z)(f_{m}^{(k)}(z))^{n}-g_{m}(z)$ has at most one zero in D, then

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

this is

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

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

From Lemma 2.2, we get $h^{l}(z)(h^{(k)})^{n}(z)-g(z_{0})$ has at least two distinct zeros, a contradiction. Therefore ${\mathcal {F}}$ is normal in D. $\square$

3 Proof of Theorem

Proof of Theorem 1.1

Suppose that ${\mathcal {F}}$ is not normal at $z_{0}$. From Lemma 2.6, we obtain $a(z_{0})= 0$. Without loss of generality, we assume that $z_0 =0$ and $a(z)=z^tb(z)$, where $1\le t\le m$, $b(0)=1$. Then by Lemma 2.1, there exists $z_{j} \longrightarrow 0$, $f_{j} \in {\mathcal {F}}$ and $\rho _{j} \longrightarrow 0^{+}$ such that

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

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

Next, we discuss two cases.

Case 1.  Let $\frac{z_n}{\rho _n}\rightarrow \alpha ,\alpha \in {\mathbb {C}}$.

For each $\xi \in {\mathbb {C}}/\{g^{-1}(\infty )\}$, It can be easily calculated that

$$\begin{aligned}&g_j^{l}\left( \xi \right) (g_{j}^{\left( k \right) }\left( \xi \right) )^{n} -\left( \xi +\frac{z_j}{\rho _j} \right) ^tb\left( z_j+\rho _j\xi \right) \\& \quad =\frac{f^{l}_j\left( z_j+\rho _j\xi \right) (f_{j}^{\left( k \right) }\left( z_j+\rho _j\xi \right) )^{n} -a\left( z_j+\rho _j\xi \right) }{\rho _{j}^{t}}\longrightarrow g^{l}\left( \xi \right) ( g^{\left( k \right) }\left( \xi \right) )^{n} -\left( \xi +\alpha \right) ^t. \end{aligned}$$

Since for sufficiently large j, $f^{l}_j\left( z_j+\rho _j\xi \right) (f_{j}^{\left( k \right) }\left( z_j+\rho _j\xi \right) )^{n} -a\left( z_j+\rho _j\xi \right)$ has one zero, from the proof Lemma  2.6, we can deduce that $g^{l}\left( \xi \right) (g^{\left( k \right) }\left( \xi \right) )^{n} -\left( \xi +\alpha \right) ^t$ has at most one distinct zero.

By Lemma 2.5, $g^{l}\left( \xi \right) (g^{\left( k \right) }\left( \xi \right) )^{n} -\left( \xi +\alpha \right) ^t$ have at least two distinct zeros. Thus $g(\xi )$ is a constant, we can get a contradiction.

Case 2. Let $\frac{z_n}{\rho _n}\rightarrow \infty$.

Set

$$\begin{aligned} F_{j}(\xi )=\frac{f_{j}(z_{j}+\rho _{j}\xi )}{\rho _{j}^{\frac{kn+t}{l+n}}}. \end{aligned}$$

It follows that

$$\begin{aligned} F^{l}_{j}(\xi )(F_{j}^{(k)}(\xi ))^{n}-(1+\xi )^{t}b(z_{j}+z_{j}\xi ) =\frac{{{f^{l}_j}({z_j} + {z _j}\xi )(f_j^{(k)}({z_j} + {z _j}\xi )})^{n}-{a({z_j} + {z _j}\xi )}}{z_{j}^{t}}. \end{aligned}$$

As the same argument as in Lemma 2.6, we can deduce that $F^{l}_{j}(\xi )(F_{j}^{(k)}(\xi ))^{n}-(1+\xi )^{t}b(z_{j}+z_{j}\xi )$ has at most one zero in $\Delta =\{\xi :|\xi |<1\}$.

Since all zeros of $F_{j}$ have multiplicity at least $k+m$, and $(1+\xi )^{t}b(z_{j}+z_{j}\xi )\rightarrow (1+\xi )^{t}\ne 0$ for $\xi \in \Delta$. Then by Lemma 2.6, $\{F_{n}\}$ is normal in $\Delta$.

Therefore, there exists a subsequence of $\{F_n(z)\}$(we still express it as $\{F_n(z)\}$) such that $\{F_n(z)\}$ converges spherically locally uniformly to a meromorphic function F(z) or $\infty$.

If $F(0)\ne \infty$, then, for each $\xi \in {\mathbb {C}}/\{g^{-1}(\infty )\}$, we have

$$\begin{aligned} g^{\left( k+m-1 \right) }\left( \xi \right)&= \lim _{j\rightarrow \infty }g_{j}^{\left( k+m-1 \right) }\left( \xi \right) =\lim _{j\rightarrow \infty }\frac{f_{j}^{\left( k+m-1 \right) }\left( z_j+\rho _j\xi \right) }{\rho _{j}^{\frac{kn+t}{l+n}-\left( k+m-1 \right) }} \\&= \underset{j\rightarrow \infty }{\lim }\left( \frac{\rho _j}{z_j} \right) ^{k+m-1-\frac{kn+t}{l+n}}F_{j}^{\left( k+m-1 \right) }\left( \frac{\rho _j}{z_j}\xi \right) =0. \end{aligned}$$

Hence $g^{(k+m-1)}\equiv 0$. It follows that g is a polynomial of degree $\le k+m-1$. Note that all zeros of g have multiplicity at least $k+m$, then we get that g is a constant, which is a contradiction.

If $F(0)=\infty$, then, for each $\xi \in {\mathbb {C}}/\{g^{-1}(0)\}$, we get

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

It follows that we have

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

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

Therefore ${\mathcal {F}}$ is normal at $z_0 = 0$. Hence ${\mathcal {F}}$ is normal in D.