1 Introduction and main results

Let \(D\subseteq \mathbb {C}\) be a domain, and \(\mathcal F\) be a family of meromorphic functions defined on D. \(\mathcal F\) is said to be normal on D, in the sense of Montel, if for each sequence \(\{f_n\} \subset \mathcal {F}\) there exists a subsequence \(\{f_{n_k}\}\), such that \(\{f_{n_k}\}\) converges spherically locally uniformly on D, to a meromorphic function or \(\infty \) (see [4, 9, 13]).

The following well-known normality criterion was conjectured by Hayman [4], and proved by Gu [3].

Theorem A

Let \(\mathcal {F}\) be a family of meromorphic functions defined in a domain D, and k be a positive integer. If for every function \(f\in \mathcal {F}\), \(f\ne 0\) and \(f^{(k)}\ne 1\) in D, then \(\mathcal {F}\) is normal in D.

This result has undergone various extensions and improvements. In [6] (cf. [8, 11]), Pang–Yang–Zalcman obtained.

Theorem B

Let k be a positive integer. Let \(\mathcal {F}\) be a family of meromorphic functions defined in a domain D, all of whose zeros have multiplicity at least \(k+2\) and whose poles are multiple. Let \(h(z)(\not \equiv 0)\) be a holomorphic functions on D. If for each \(f\in \mathcal {F}\), \(f^{(k)}(z)\ne h(z)\), then \(\mathcal {F}\) is normal in D.

When \(k=1\), an example [8, Example 1] shows that the condition on the multiplicity of zeros of functions in \(\mathcal F\) cannot be weakened. When \(k\ge 2\), Zhang–Pang–Zalcman [14] proved that the multiplicity of zeros of functions in \(\mathcal F\) can be reduced from \(k+2\) to \(k+1\) in Theorem C.

Theorem C

Let \(k\ge 2\) be a positive integer. Let \(\mathcal {F}\) be a family of meromorphic functions defined in a domain D, all of whose zeros have multiplicity at least \(k+1\) and whose poles are multiple. Let \(h(z)(\not \equiv 0)\) be a holomorphic functions on D. If for each \(f\in \mathcal {F}\), \(f^{(k)}(z)\ne h(z)\), then \(\mathcal {F}\) is normal in D.

In [12], Xu reduced the multiplicity of the zeros of functions in \(\mathcal {F}\) to k for the case \(h(z)=z\), but restricting the values \(f^{(k)}\) can take at the zeros of f, as follows.

Theorem D

Let \(k\ge 4\) be a positive integer, \(A>1\) be a constant. Let \(\mathcal F\) be a family of meromorphic functions in a domain D. If, for every function \(f\in \mathcal F\), f has only zeros of multiplicity at least k and satisfies the following conditions:

  1. (a)

    \(f(z)=0\Rightarrow |f^{(k)}(z)|\le A|z|\).

  2. (b)

    \( f^{(k)}(z)\ne z\).

  3. (c)

    All poles of f are multiple .

Then \(\mathcal F\) is normal in D.

Theorem E

Let \(A>1\) be a constant. Let \(\mathcal F\) be a family of meromorphic functions in a domain D. If, for every function \(f\in \mathcal F\), f has only zeros of multiplicity at least 3 and satisfies the following conditions:

  1. (a)

    \(f(z)=0\Rightarrow |f^{\prime \prime \prime }(z)|\le A|z|\).

  2. (b)

    \( f^{\prime \prime \prime }(z)\ne z\).

  3. (c)

    All poles of f have multiplicity at least 3.

Then \(\mathcal {F}\) is normal in D.

Also in [12], Xu gave the following example to show that the condition (c) in Theorem E is necessary and the number 3 is best possible.

Example 1

(See [12]) Let \(\Delta =\{z: |z|<1\}\), and let

$$\begin{aligned} \mathcal{F}=\left\{ f_n(z)=\frac{(z-1/n)^3(z+1/n)^3}{24z^2}\right\} . \end{aligned}$$

Clearly,

$$\begin{aligned} f^{\prime \prime \prime }_n(z)=z+\frac{1}{n^{6}z^5}\ne z. \end{aligned}$$

For each n, \(f_n\) has two zeros \(z_1=1/n\) and \(z_2=-1/n\) of multiplicity 3. It’s easy to see that

$$\begin{aligned} f^{\prime \prime \prime }_n\left( \frac{1}{n}\right) =\frac{2}{n},\ f^{\prime \prime \prime }_n\left( -\frac{1}{n}\right) =-\frac{2}{n}, \end{aligned}$$

and \(|f^{\prime \prime \prime }_n(z_i)|\le 2|z_i|(i=1,2),\) then \(f_n(z)=0\Rightarrow |f^{\prime \prime \prime }_n(z)|\le 2|z|\). However \(\mathcal F\) is not normal at 0 since \(f_n(1/n)=0\) and \(f_n(0)=\infty \).

In this paper, inspired by the idea in [1, 2], we prove the following result, which shows that the counterexample above is unique in some sense.

Theorem 1

Let \(A>1\) be a constant and \(\mathcal {F}\) be a family of meromorphic functions defined in D, all of whose zeros have multiplicity at least 3 and whose poles all are multiple, such that for each \(f\in \mathcal {F},f(z)=0\Rightarrow |f^{\prime \prime \prime }(z)|\le A|z|\), and \(f^{\prime \prime \prime }(z)\ne z\). If \(\mathcal {F}\) is not normal at \(z_{0}\in D\), then \(z_{0}=0\) and there exist \(r>0\) and \(\{f_{n}\}\subset \mathcal {F}\) such that

$$\begin{aligned} f_{n}(z)=\frac{\left( z-\xi ^1_n\right) ^{3}\left( z-\xi ^2_n\right) ^{3}}{(z-\eta _n)^2}\hat{f}_{n}(z) \end{aligned}$$

on \(\Delta _r=\{z: |z|<r \}\), where \(\xi ^i_n/\rho _{n}\rightarrow c_i\)(\(i=1,2\)) and \(\eta _n/\rho _{n}\rightarrow (c{_1}+c_2)/2\) for some sequence of positive numbers \(\rho _n\rightarrow 0\) and two distinct constants \(c_1\) and \(c_2\). Moreover, \(\hat{f}_{n}(z)\) is holomorphic and non-vanishing on \(\Delta _r\) such that \(\hat{f_{n}}(z)\rightarrow \hat{f}(z)\equiv 1/24\) locally uniformly on \(\Delta _r\).

In this paper, we denote \(\Delta _r=\{z: |z|<r\}\) and \(\Delta ^{\prime }_r=\{z:0<|z|<r\}\), and the number r may be different in different place. When \(r=1\), we drop the subscript.

2 Lemmas

To prove our results, we need the following lemmas.

Lemma 1

([7, Lemma 2]) Let k be a positive integer and let \(\mathcal {F}\) be a family of meromorphic functions in a domain D, all of whose zeros have multiplicity 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 \(\mathcal {F}\) is not normal at \(z_{0} \in D\), then for each \(\alpha \), \(0\le \alpha \le k\), there exist a sequence of complex numbers \(z_{n} \in D\), \(z_{n}\rightarrow z_{0}\), a sequence of positive numbers \(\rho _{n} \rightarrow 0\), and a sequence of functions \(f_{n} \in \mathcal {F}\) such that

$$\begin{aligned} g_{n}(\zeta )=\frac{f_{n}(z_{n}+\rho _{n}\zeta )}{\rho _{n}^{\alpha }}\rightarrow g(\zeta ) \end{aligned}$$

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

Lemma 2

([12, Lemma 6]) Let f be a transcendental meromorphic function of finite order \(\rho \), and let \(k(\ge 2)\) be a positive integer. If f has only zeros of multiplicity at least k, and there exists \(A>1\) such that \(f(z)=0\Rightarrow |f^{(k)}(z)|\le A|z|\), then \(f^{(k)}\) has infinitely many fix-points.

Lemma 3

([11, Lemma 8]) Let f be a non-polynomial rational function and k be a positive integer. If \(f^{(k)}(z)\ne 1\), then

$$\begin{aligned} f(z)=\frac{1}{k!}z^k+a_{k-1}z^{k-1}+\cdots +a_0+\frac{a}{(z-b)^m}, \end{aligned}$$

where \(a_{k-1},\ldots , a_0, a(\ne 0) ,b\) are constants and m is a positive integer.

Lemma 4

([12, Lemma 10]) Let \(k\ge 3\) be a positive integer, \(A>1\) be a constant. Let \(\mathcal F\) be a family of meromorphic functions in a domain D. Suppose that, for every \(f\in \mathcal F\), f has only zeros of multiplicity at least k, and satisfies the following conditions:

  1. (a)

    \(f(z)=0\Rightarrow |f^{(k)}(z)|\le A|z|\).

  2. (b)

    \(f^{(k)}(z)\ne z\).

  3. (c)

    all poles of f are multiple.

Then \(\mathcal F\) is normal in \(D\backslash \{0\}\).

Lemma 5

Let \(\mathcal F\) be a family of functions meromorphic on \(\Delta _r\), \(b\in \overline{\mathbb {C}}\) to be a constant which satisfies \(f(z)\ne b\) on \(\Delta _r\) for each \(f\in \mathcal F\). If \(\mathcal F\) is normal on \(\Delta ^{\prime }_r\), but not normal on \(\Delta _r\), then there exists a subsequence \(\{f_n\} \subset \mathcal F\) such that \(f_n (z)\overset{\chi }{\Rightarrow }b\) on \(\Delta ^{\prime }_r\).

Proof

Without loss of generality, we assume that \(b=0\). Since \(\mathcal F\) is normal on \(\Delta ^{\prime }_r\), then there exists a subsequence \(\{f_n\} \subset \mathcal F\) such that \(f_n (z)\rightarrow f(z)\) spherically locally uniformly on \(\Delta ^{\prime }_r\). Set \(g_n(z)=1/f_n(z).\) Thus \(g_n (z)\rightarrow g(z)=1/f(z)\) on \(\Delta ^{\prime }_r\). Noting that \(f_n(z)\ne 0\), it follows that \(f(z)\ne 0\) or \( f(z) \equiv 0\) by Hurwitz’s theorem and \(g_n(z)\) is holomorphic on \(\Delta _r\). If \(f(z)\ne 0\), then the maximum modulus principle implies that \(g_n (z)\rightarrow g(z)\) on \(\Delta _r\). Hence \(f_n (z)\rightarrow \) on \(\Delta _r\), a contradiction. So, \(f(z)\equiv 0\). This finishes the proof of Lemma 5. \(\square \)

Lemma 6

Let f be a rational function, all of whose zeros are of multiplicity at least 3. If \(f^{\prime \prime \prime }(z)\ne z\), then one of the following three cases must occur:

  1. (i)
    $$\begin{aligned} f(z)=\frac{(z+c)^{4}}{24}; \end{aligned}$$
    (2.1)
  2. (ii)
    $$\begin{aligned} f(z)=\frac{(z-c_1)^5}{24(z-b)}; \end{aligned}$$
    (2.2)
  3. (iii)
    $$\begin{aligned} f(z)=\frac{(z-c_1)^3(z-c_2)^3}{24[z-(c_1+c_2)/2]^2}, \end{aligned}$$
    (2.3)

where c is nonzero constant, \(b(\ne c_1)\) is a constant and \(c_1, \ c_2\) are two distinct constants.

Proof

First, suppose that f is a polynomial. Since \(f^{\prime \prime \prime }(z)\ne z\), then \(f^{\prime \prime \prime }(z)=z+c\), where \(c(\ne 0)\) is a constant. Thus,

$$\begin{aligned} f(z)=\frac{1}{24}z^4+\frac{c}{6}z^3+a_1z^2+a_2z+a_3 \end{aligned}$$

where \(a_1, a_2\) and \(a_3\) are three constants. Noting that f has only zeros of multiplicity at least 3, it follows that f has only one zero of multiplicity 4. Thus, f has the form (2.1).

Then, suppose that f is a non-polynomial rational function. Set

$$\begin{aligned} g(z)=f(z)-\frac{1}{24}z^{4}+\frac{1}{6}z^3. \end{aligned}$$

Then \(g^{\prime \prime \prime }(z)\ne 1\), so by Lemma 3

$$\begin{aligned} g(z)=\frac{1}{6}z^3+a_{2}z^{2}+a_1z+a_0+\frac{a}{(z-b)^m}, \end{aligned}$$

where \(a_{2}, a_1, a_0, a(\ne 0) ,b\) are constants and m is a positive integer. Thus

$$\begin{aligned} f(z)=p_{4}(z)+\frac{a}{(z-b)^m}=\frac{p_{4}(z)(z-b)^m+a}{(z-b)^m}, \end{aligned}$$
(2.4)

where

$$\begin{aligned} p_{4}(z)=\frac{1}{24}z^4+a_{2}z^{2}+a_1z+a_0. \end{aligned}$$

Let \(c_1, c_2,\dots , c_q \) be q distinct zeros of \(p_{4}(z)(z-b)^m+a\), with multiplicity \(n_1, n_2, \dots , n_q\). Clearly, \(n_i\ge 3\), \(c_i\ne b\), and \(c_i\) is a zero of \([p_{4}(z)(z-b)^m+a]^{\prime }\) with multiplicity \(n_i-1\ge 2 (1\le i\le q)\). Since

$$\begin{aligned} \left[ p_{4}(z)(z-b)^m+a\right] ^{\prime }=(z-b)^{m-1}\left[ p_{4}^{\prime }(z)(z-b)+mp_{4}(z)\right] , \end{aligned}$$
(2.5)

then \(c_i\) must be a zero of \(p_{4}^{\prime }(z)(z-b)+mp_{4}(z)\) with multiplicity \(n_i-1(\ge 2)\). Comparing the degree on both sides of (2.5), it follows that \(\deg [p_{4}^{\prime }(z)(z-b)+mp_{4}(z)]=4\). Now we divide two cases:

  1. (a)

    \(p_{4}^{\prime }(z)(z-b)+mp_{4}(z)\) has only one zero \(c_1\) with multiplicity 4;

  2. (b)

    \(p_{4}^{\prime }(z)(z-b)+mp_{4}(z)\) has two distinct zeros \(c_1\) and \(c_2\) with multiplicity 2.

For case (a), it follows that \(m=1\) and

$$\begin{aligned} p_4(z)(z-b)+a=\frac{1}{24}(z-c_1)^5. \end{aligned}$$

Thus, by (2.4), f has the form (2.2).

For case (b), it’s easy to see that \(m=2\) and

$$\begin{aligned} p^{\prime }_4 (z)(z-b)+2p_{4}(z)= & {} \frac{1}{4}(z-c_1)^2(z-c_2)^2,\nonumber \\ p_4 (z)(z-b)^2+a= & {} \frac{1}{24}(z-c_1)^3(z-c_2)^3. \end{aligned}$$

These, together with (2.5) give

$$\begin{aligned} z-b=\frac{1}{2}(z-c_1+z-c_2). \end{aligned}$$

Thus, \(b=(c_1+c_2)/2\). Hence, by (2.4), f has the form (2.3).

This completes the proof of Lemma 6. \(\square \)

3 Proof of Theorem 1

Since \(\mathcal {F}\) is not normal at \(z_0\), by Lemma 4, \(z_{0}=0\). Without loss of generality, we assume that \(\mathcal {F}\) is normal on \(\Delta '\) but not normal at the origin.

Consider the family

$$\begin{aligned} \mathcal {G}=\left\{ g(z)=\frac{f(z)}{z}:f\in \mathcal {F}\right\} . \end{aligned}$$

It’s easy to know that \(f(0)\ne 0\) for every \(f\in \mathcal {F}\). Thus, for each \(g\in \mathcal {G}\), \(g(0)=\infty \). Furthermore, all zeros of g(z) have multiplicity at least 3. On the other hand, by simple calculation, we have

$$\begin{aligned} g^{\prime \prime \prime }(z)={\frac{f^{\prime \prime \prime }(z)}{z}}-{\frac{3g^{\prime \prime }(z)}{z}}. \end{aligned}$$
(3.1)

Since \(f(z)=0\Rightarrow |f^{\prime \prime \prime }(z)|\le A|z|\), it follows that \(g(z)=0\Rightarrow |g^{\prime \prime \prime }(z)|\le A\).

Clearly, \(\mathcal {G}\) is normal on \(\Delta ^{\prime }\). We claim that \(\mathcal {G}\) is not normal at \(z=0\). Indeed, if \(\mathcal {G}\) is normal at \(z=0\), then \(\mathcal {G}\) is normal on the whole disk \(\Delta \) and hence equicontinuous on \(\Delta \) with respect to the spherical distance. Noting that \(g(0)= \infty \) for each \(g \in \mathcal {G}\), so there exists \(r>0\) such that for every \(g \in \mathcal {G}\) and \(|g(z)|\ge 1\) for every \(z\in \Delta _{r}\). Then \(f(z) \ne 0\) on \(\Delta _{r}\) for all \(f \in \mathcal {F}.\) Since \(\mathcal {F}\) is normal on \(\Delta ^{\prime }\) but not normal on \( \Delta \), there exists a sequence \(\{f_{n}\} \subset \mathcal {F}\) such that \(f_{n} \rightarrow 0\) on \(\Delta ^{\prime }_{r}\) according to Lemma 5. So does \(\{g_{n}\}\subset \mathcal {G}\), where \(g_{n}(z)=f_{n}(z)/z.\) However \(|g_{n}(z)| \ge 1\) for \(z \in \Delta _{r}\), a contradiction.

Then, by Lemma 1, there exist functions \(g_{n} \in \mathcal {G}\), points \(z_{n} \rightarrow 0\) and positive numbers \(\rho _{n} \rightarrow 0\) such that

$$\begin{aligned} G_{n}(\zeta )=\frac{g_{n}(z_{n}+\rho _{n}\zeta )}{\rho _{n}^3}\rightarrow G(\zeta ), \end{aligned}$$
(3.2)

converges spherically uniformly on compact subsets of \(\mathbb {C}\), where G is a non-constant meromorphic function on \(\mathbb {C}\) and of finite order, all zeros of G have multiplicity at least 3, and \(G^{\#}(\zeta )\le G^{\#}(0)=3A+1\) for all \(\zeta \in \mathbb {C}\).

By [12, pp. 480–482], we can assume that \(z_{n}/\rho _{n}\rightarrow \alpha ,\) a finite complex number. Then

$$\begin{aligned} \frac{g_{n}(\rho _{n}\zeta )}{\rho _{n}^{3}}=G_{n}(\zeta -z_{n}/\rho _{n})\overset{\chi }{\rightarrow } G(\zeta -\alpha )=\widetilde{G}(\zeta ) \end{aligned}$$

on \(\mathbb {C}\). Clearly, all zeros of \(\widetilde{G}\) have multiplicity at least 3, and all poles of \(\widetilde{G}\) are multiple, except possibly the pole at 0.

Set

$$\begin{aligned} H_{n}(\zeta )=\frac{f_{n}(\rho _{n}\zeta )}{\rho _{n}^{4}}. \end{aligned}$$
(3.3)

Then

$$\begin{aligned} H_{n}(\zeta )=\frac{f_{n}(\rho _{n}\zeta )}{\rho _{n}^{4}}=\zeta \frac{g_{n}(\rho _{n}\zeta )}{\rho _{n}^{3}}\rightarrow \zeta \widetilde{G}(\zeta )=H(\zeta ) \end{aligned}$$
(3.4)

spherically uniformly on compact subsets of \(\mathbb {C}\), and

$$\begin{aligned} H^{\prime \prime \prime }_{n}(\zeta )=\frac{f^{\prime \prime \prime }_{n}(\rho _{n}\zeta )}{\rho _{n}} \rightarrow H^{\prime \prime \prime }(\zeta ) \end{aligned}$$
(3.5)

locally uniformly on \(\mathbb {C}{\setminus } H^{-1}(\infty )\). Obviously, all zeros of H have multiplicity at least 3, and all poles of H are multiple. Since \(\widetilde{G}(0)=\infty \), \(H(0)\ne 0\).

Claim (I) \(H(\zeta )=0\Rightarrow |H^{\prime \prime \prime }(\zeta )|\le A|\zeta |\); (II) \( H^{\prime \prime \prime }(\zeta )\ne \zeta \).

If \(H(\zeta _0)=0\), by Hurwitz’s theorem and (3.4), there exist \(\zeta _n \rightarrow \zeta _0 \) such that \(f_n(\rho _n \zeta _n)=0\) for for n sufficiently large. By the assumption, \(|f^{\prime \prime \prime }_n(\rho _n\zeta _n)|\le A|\rho _n\zeta _n|\). Then, it follows from (3.5) that \(|H^{\prime \prime \prime }(\zeta _0)|\le A|\zeta _0|\). Claim (I) is proved.

Suppose that there exists \(\zeta _0\) such that \(H^{\prime \prime \prime }(\zeta _0)=\zeta _0\). By (3.5),

$$\begin{aligned} 0\ne \frac{f^{\prime \prime \prime }_{n}(\rho _{n}\zeta )-\rho _{n}\zeta }{\rho _{n}}=H^{\prime \prime \prime }_{n}(\zeta )-\zeta \rightarrow H^{\prime \prime \prime }(\zeta )-\zeta , \end{aligned}$$

uniformly on compact subsets of \(\mathbb {C}{\setminus } H^{-1}(\infty )\). Hurwitz’s theorem implies that \(H^{\prime \prime \prime }(\zeta )\equiv \zeta \) on \(\mathbb {C}{\setminus } H^{-1}(\infty )\), and then on \(\mathbb {C}\). It follows that H is a polynomial of degree 4. Since all zeros of H have multiplicity at least 3, we know that H has a single zero \(\zeta _1\) with multiplicity 4, so that \(H^{\prime \prime \prime }(\zeta _1)=0\), and hence \(\zeta _1=0\) since \(H^{\prime \prime \prime }(\zeta )\equiv \zeta \). But \(H(0)\ne 0\), we arrive at a contradiction. This proves claim (II).

Then, by Lemma 2, H must be a rational function. Since all poles of H are multiple, it derives from Lemma 6 that \(H(\zeta )=(\zeta +b)/24\) or

$$\begin{aligned} H(\zeta )=\frac{(\zeta -c_1)^3(\zeta -c_2)^3}{24[\ zeta-(c_1+c_2)/2]^2}, \end{aligned}$$

where b is a constant, \(c_1\) and \( c_2\) are two distinct constants. But, \(H(\zeta )=(\zeta +b)/24\) is impossible(for details, see [12, pp. 483–485]). By (3.3) and (3.4), it follows that

$$\begin{aligned} \frac{f_{n}(\rho _{n}\zeta )}{\rho _n^4} \rightarrow \frac{(\zeta -c_1)^3(\zeta -c_2)^3}{24[\zeta -(c_1+c_2)/2]^2}. \end{aligned}$$
(3.6)

Noting that all zeros of \(f_{n}\) have multiplicity at least 3, there exist \(\zeta ^1_n\rightarrow c_1\), \(\zeta ^2_n\rightarrow c_2\) and \(\zeta ^3_n\rightarrow (c_1+c_2)/2\) such that \(\xi ^1_{n}=\rho _{n}\zeta ^1_{n}\) and \(\xi ^2_{n}=\rho _{n}\zeta ^2_{n}\) are zeros of \(f_{n}\) with exact multiplicity 3, and \(\eta _{n}=\rho _{n}\zeta ^3_{n}\) is the pole of \(f_{n}\) with exact multiplicity 2.

Now write

$$\begin{aligned} f_{n}(z)=\frac{\left( z-\xi ^1_{n}\right) ^{3}\left( z-\xi ^2_{n}\right) ^{3}}{\left( z-\eta _{n}\right) ^2}\hat{f}_{n}(z) \end{aligned}$$
(3.7)

Then by (3.6) and (3.7), it follows that

$$\begin{aligned} \hat{f}_{n}(\rho _{n}\zeta )\rightarrow \frac{1}{24} \end{aligned}$$
(3.8)

on \(\zeta \in \mathbb {C}\).

Next, we complete our proof in three steps.

Step 1. Claim that there exists a\(r>0\)such that\(\hat{f}_{n}(z) \ne 0\)on\(\Delta _{r}\).

Suppose not, taking a sequence and renumbering if necessary, \( \hat{f}_{n}\) has zeros tending to 0. Assume \(\hat{z}_{n} \rightarrow 0 \) is the zero of \( \hat{f}_{n}\) with the smallest modulus. Then by (3.8), it’s easy to know that \(\hat{z}_{n}/\rho _{n} \rightarrow \infty .\)

Set

$$\begin{aligned} \widehat{f}_{n}^{*}(z)=\hat{f}_{n}(\hat{z}_{n}z). \end{aligned}$$
(3.9)

Thus, \(\widehat{f}_{n}^{*}(z)\) is well-defined on \(\mathbb {C}\) and non-vanishing on \(\Delta \). Moreover, \(\widehat{f}_{n}^{*}(1)=0.\)

Now let

$$\begin{aligned} M_{n}(z)=\frac{\left( z-\xi ^1_{n}/\hat{z}_{n}\right) ^{3}\left( z-\xi ^2_{n}/\hat{z}_{n}\right) ^{3}}{\left( z-\eta _{n}/\hat{z}_{n}\right) ^2}\widehat{f}_{n}^{*}(z). \end{aligned}$$
(3.10)

According to (3.7), (3.9) and (3.10), it follows that

$$\begin{aligned} M_{n}(z)=\frac{\left( z\hat{z}_{n}-\xi ^1_{n}\right) ^{3}\left( z\hat{z}_{n}-\xi ^2_{n}\right) ^{3}}{\left( z\hat{z}_{n}-\eta _{n}\right) ^2} \frac{\hat{f}_{n}\left( \hat{z}_{n}z\right) }{\left( \hat{z}_{n}\right) ^{4}} =\frac{f_{n}\left( \hat{z}_{n}z\right) }{\left( \hat{z}_{n}\right) ^{4}}. \end{aligned}$$

Obviously, all zeros of \(M_{n}(z)\) have multiplicity at least 3 and all poles of \(M_{n}(z)\) have multiplicity at least 2. Since \(f_n (z)=0\Rightarrow |f^{\prime \prime \prime }_n(z)|\le A|z|\), it follows that \(M_n (z)=0\Rightarrow |M^{\prime \prime \prime }_n(z)|\le A|z|\). Now that \(f^{\prime \prime \prime }_{n}(z)\ne z\), it derives that

$$\begin{aligned} M^{\prime \prime \prime }_{n}(z)-z=\frac{\left( f^{\prime \prime \prime }_{n}\hat{z}_{n}z\right) -\hat{z}_{n}z}{\hat{z}_{n}}\ne 0. \end{aligned}$$
(3.11)

Hence, by Lemma 4, \(\{M_{n}(z)\}\) is normal on \(\mathbb {C}^{*}=\mathbb {C}\backslash \{0\}.\)

Noting that

$$\begin{aligned} \frac{\xi ^{1}_{n}}{\hat{z}_{n}}= & {} \frac{\xi ^1_{n}}{\rho _{n}}\frac{\rho _{n}}{\hat{z}_{n}}\rightarrow 0,\\ \frac{\xi ^2_{n}}{\hat{z}_{n}}= & {} \frac{\xi ^2_{n}}{\rho _{n}}\frac{\rho _{n}}{\hat{z}_{n}}\rightarrow 0,\\ \hbox {and}\quad \frac{\eta _{n}}{\hat{z}_{n}}= & {} \frac{\eta _{n}}{\rho _{n}}\frac{\rho _{n}}{\hat{z}_{n}}\rightarrow 0, \end{aligned}$$

we deduce from (3.10) that \(\{\widehat{f}_{n}^{*}\}\) is also normal on \(\mathbb {C}^{*}.\) Thus by taking a subsequence, we assume that \(\widehat{f}_{n}^{*}\rightarrow \widehat{f}^{*}\) spherically locally uniformly on \(\mathbb {C}^{*}\). Clearly, \(\widehat{f}^{*}(z)\) has a zero at 1 with multiplicity at least 3 since \(\widehat{f}_{n}^{*}(1)=0\).

Set

$$\begin{aligned} L_{n}(z)=M^{\prime \prime \prime }_{n}(z)-z. \end{aligned}$$
(3.12)

Then \(L_n\ne 0\) from (3.11).

Now we prove that \(\hat{f}^{*}(z)\not \equiv 0.\) Otherwise \(\hat{f}_{n}^{*}(z)\rightarrow 0\), thus \(L_{n}(z)\rightarrow -z\) and \( L_{n}'(z)\rightarrow -1\) locally uniformly on \(\mathbb {C}^{*}.\) By the argument principle, it derives that

$$\begin{aligned} \left| n(1,L_{n})-n\left( 1,\frac{1}{L_{n}}\right) \right| =\frac{1}{2\pi }\left| \int _{|z|=1}\frac{L_{n}^{\prime }}{L_{n}}dz\right| \rightarrow \frac{1}{2\pi }\left| \int _{|z|=1}\frac{1}{z}dz\right| =1,\nonumber \\ \end{aligned}$$
(3.13)

where n(r, f) denotes the number of poles of f in \(\Delta _{r}\), counting multiplicity. It follows that \(n(1,L_{n})=1\). On the other hand, the poles of \(L_{n}(z)=M^{\prime \prime \prime }_{n}(z)-z\) have multiplicity at least 4. A contradiction.

Then \(\hat{f}_{n}^{*}\rightarrow \hat{f}^{*}\not \equiv 0\) spherically locally uniformly on \(\mathbb {C}^{*}\). Since \(\hat{f}_{n}^{*}\) is non-vanishing on \(\Delta \), then \(\hat{f}_{n}^{*}\rightarrow \hat{f}^{*}\) on \(\Delta \) by Lemma 5. Hence, \(\hat{f}_{n}^{*}\rightarrow \hat{f}^{*}\) on \(\mathbb {C}\).

By (3.10) and (3.12), we see that

$$\begin{aligned} L_{n}(z)\rightarrow L(z)=\left( z^{4}\widehat{f}^{*}(z)\right) ^{\prime \prime \prime }-z \end{aligned}$$

on \(\mathbb {C^*}{\setminus } (\widehat{f}^{*})^{-1}(\infty )\). Obviously, \(\{L_n(z)\}\) is normal on \(\Delta _r\). If not, Lemma 5 derives that \(L(z)=(z^{4}\widehat{f}^{*}(z))^{^{\prime \prime \prime }}-z\equiv 0 \) since \(L_n\ne 0\) on \(\mathbb {C}\). Thus,

$$\begin{aligned} \hat{f}^{*} (z)= \frac{z^4+a_1z^2+a_2z+a_3}{24z^4}, \end{aligned}$$

where \(a_1\), \(a_2\) and \(a_3\) be three constants. Now that the zeros of \(\hat{f}^{*} (z)\) have multiplicity at least 3 and \(\hat{f}^{*} (1)=0\), then

$$\begin{aligned} \hat{f}^{*} (z)= \frac{(z-1)^4}{24z^4}, \end{aligned}$$

which is impossible since \(z^4+a_1z^2+a_2z+a_3\ne (z-1)^4\). So \(L_{n}(z)\rightarrow L(z)\) on \(\mathbb {C}\).

Since \(L_n(z)\ne 0\), Hurwitz’s theorem implies that either \(L(z)\equiv 0\) or \(L(z)\ne 0\). \(\hat{f}^{*} (1)=0\) follows that \(L(z)\ne 0\). On the other hand, \(\hat{f}_{n}^{*}(0)=\hat{f}_{n}(0)\rightarrow \hat{f}^{*}(0) =1/24\), it follows that \(L(0)=0\), a contradiction. The claim is completed.

Step 2. Show that there exists a\(r > 0\)such that\(\hat{f}_{n} (z)\)is holomorphic on\( \Delta _{r} \).

Since \(\{f_{n}\}\) and hence \(\{\hat{f}_{n}\}\) is normal on \(\Delta ^{\prime },\) taking a subsequence and renumbering, we have \(\hat{f}_{n} \rightarrow \hat{f}\) spherically locally uniformly on \( \Delta ^{\prime }\).

It’s easy to see that \(\hat{f}(z)\not \equiv 0 \) on \( \Delta ^{\prime } \). Otherwise, we have \(f^{\prime \prime \prime }_{n}(z)\rightarrow 0 \) and \(f_{n}^{(4)}(z)\rightarrow 0\) locally uniformly on \( \Delta ^{\prime } \). Then the argument principle yields that

$$\begin{aligned}&\left| n\left( \frac{1}{2},f^{\prime \prime \prime }_{n}-z\right) {-}n\left( \frac{1}{2},\frac{1}{f^{\prime \prime \prime }_{n}-z}\right) \right| {=}\frac{1}{2\pi }\left| \int _{|z|{=}\frac{1}{2}}\frac{f_{n}^{(4)}{-}1}{f^{\prime \prime \prime }_{n}-z}dz\right| \rightarrow \frac{1}{2\pi }\left| \int _{|z|=\frac{1}{2}}\frac{1}{z}dz\right| \\&\quad =1. \end{aligned}$$

Now that \(f^{\prime \prime \prime }_{n}(z)\ne z\), it follows that \(n(\frac{1}{2},f^{\prime \prime \prime }_{n})=n(\frac{1}{2},f^{\prime \prime \prime }_{n}-z)=1,\) which is impossible. Thus, \(\hat{f}_{n}\rightarrow \hat{f}\not \equiv 0\).

Recalling that \(\hat{f}_{n}(z)\ne 0\), and by Lemma 5, it gives that \(\hat{f}_{n}\rightarrow \hat{f}\) spherically locally uniformly on \(\Delta .\) Since \(\hat{f}_{n}(0)\rightarrow 1/24\), then \(\hat{f}(0)=1/24\). Thus, there exists a positive number r such that \(\hat{f}\) is holomorphic on \(\Delta _r\). Hence \(\hat{f}_{n}\) is holomorphic on \(\Delta _{r}.\)

Step 3. Prove that there exists a\(r>0\)such that\(\hat{f}_n (z) \rightarrow \hat{f}(z) \equiv 1/24 \)on\(\Delta _{r}\).

By (3.7), we get \(f_{n}(z)\rightarrow z^{4}\hat{f}(z)\) on \(\Delta ^{\prime }\). Thus

$$\begin{aligned} f^{\prime \prime \prime }_{n}(z)-z\rightarrow \left[ z^{4}\hat{f}(z)\right] ^{\prime \prime \prime }-z, \end{aligned}$$
(3.14)

on \(\Delta ^{\prime }{\setminus } \hat{f}^{-1}(\infty ).\)

Hence there exists \(r>0\) such that \(f^{\prime \prime \prime }_{n}(z)-z\rightarrow [z^{4}\hat{f}(z)]^{\prime \prime \prime }-z\) on \(\Delta ^{\prime }_r\).

If \(\{f_{n}^{^{\prime \prime \prime }}(z)-z\}\) is not normal on \(\Delta _r\), combining \(f_{n}^{^{\prime \prime \prime }}(z)\ne z\) with Lemma 5, it follows that \([z^{4}\hat{f}(z)]^{^{\prime \prime \prime }}-z\equiv 0\) on \(\Delta ^{\prime }_r\). Hence

$$\begin{aligned} z^4\hat{f}(z)=\frac{1}{24}z^4+a_1 z^2+a_2 z+a_3 \end{aligned}$$

on \(\Delta ^{\prime }_r\). Recalling that \(\hat{f}_{n}\rightarrow \hat{f}\) on \(\Delta \) and \(\hat{f}(0)=1/24\), so \(\hat{f}_n (z)\rightarrow \hat{f}(z)\equiv 1/24\) on \(\Delta _{r}\).

If \(\{f^{\prime \prime \prime }_{n}(z)-z\}\) is normal on \(\Delta _r\), then either \([z^{4}\hat{f}(z)]^{\prime \prime \prime }-z\equiv 0\) or \([z^{4}\hat{f}(z)]^{\prime \prime \prime }-z\ne 0\) according to \(f^{\prime \prime \prime }_{n}(z)\ne z\) . Noting the fact that \([(z^{4}\hat{f}(z))^{\prime \prime \prime }-z]|_{z=0}=0\), it derives that \([z^{4}\hat{f}(z)]^{\prime \prime \prime }-z\equiv 0\). Similarly, it follows that \(\hat{f_n}(z)\rightarrow \hat{f}(z)\equiv 1/24\) on \(\Delta _{r}\).

The proof of Theorem 1 is finished. \(\square \)