# A criterion of normality based on a single holomorphic function II

Xiaojun Liu1 and Shahar Nevo2 Xiaojun Liu, Department of Mathematics
University of Shanghai for Science and Technology, Shanghai 200093, P.R. China
<EMAIL_ADDRESS>Shahar Nevo, Department of Mathematics
Bar-Ilan University, 52900 Ramat-Gan, Israel<EMAIL_ADDRESS>
###### Abstract.

In this paper, we continue to discuss normality based on a singleholomorphic
function. We obtain the following result. Let $\mathcal{F}$ be a family of
functions holomorphic on a domain $D\subset\mathbb{C}$. Let $k\geq 2$ be an
integer and let $h(\not\equiv 0)$ be a holomorphic function on $D$, such that
$h(z)$ has no common zeros with any $f\in\mathcal{F}$. Assume also that the
following two conditions hold for every $f\in\mathcal{F}$:(a)
$f(z)=0\Longrightarrow f^{\prime}(z)=h(z)$ and (b)
$f^{\prime}(z)=h(z)\Longrightarrow|f^{(k)}(z)|\leq c$, where $c$ is a
constant. Then $\mathcal{F}$ is normal on $D$.

A geometrical approach is used to arrive at the result which significantly
improves the previous results of the authors, A criterion of normality based
on a single holomorphic function, Acta Math. Sinica, English Series (1) 27
(2011), 141–154 and of Chang, Fang, and Zalcman, Normal families of
holomorphic functions, Illinois Math. J. (1) 48 (2004), 319–337. We also deal
with two other similar criterions of normality. Our results are shown to be
sharp.

###### Key words and phrases:

Normal family, holomorphic functions, zero points

###### 2010 Mathematics Subject Classification:

30D35

1 Research supported by the NNSF of China Approved No.11071074 and also
supported by the Outstanding Youth Foundation of Shanghai No. slg10015.

2 Research supported by the Israel Science Foundation Grant No. 395/07

##  1\. Introduction

In [11], X.C. Pang and L. Zalcman proved the following theorem.

######  Theorem PZ.

Let $\mathcal{F}$ be a family of meromorphic functions on a domain
$D\subset\mathbb{C}$, all of whose zeros have multiplicity at least $k$, where
$k\geq 1$ is an integer. Suppose there exist constants $b\neq 0$ and $h>0$
such that, for every $f\in\mathcal{F}$, $f(z)=0\Longleftrightarrow
f^{(k)}(z)=b$ and $f(z)=0\Longrightarrow 0<|f^{(k+1)}(z)|\leq h$. Then
$\mathcal{F}$ is a normal family on $D$.

Then, in [1], J.M Chang, M.L. Fang, and L. Zalcman proved the following
result.

######  Theorem CFZ1.

[1, Theorem 4] Let $\mathcal{F}$ be a family of functions holomorphic on a
domain $D\subset\mathbb{C}$. Let $k\geq 2$ be an integer, and let $h(z)\neq 0$
be a function analytic in $D$. Assume also that the following two conditions
hold for every $f\in\mathcal{F}$:

  1. (a)

$f(z)=0\Longrightarrow f^{\prime}(z)=h(z)$; and

  2. (b)

$f^{\prime}(z)=h(z)\Longrightarrow|f^{(k)}(z)|\leq c$, where $c$ is a
constant.

Then $\mathcal{F}$ is normal on $D$.

And in [4], we replaced the condition $h(z)\neq 0$ with $h(z)\not\equiv 0$ and
obtained the following result.

######  Theorem LN.

Let $\mathcal{F}$ be a family of functions holomorphic on a domain
$D\subset\mathbb{C}$. Let $k\geq 2$ be an integer, and let $h(z)(\not\equiv
0)$ be a holomorphic function on $D$, all of whose zeros have multiplicity at
most $k-1$, that has no common zeros with any $f\in\mathcal{F}$. Assume also
that the following two conditions hold for every $f\in\mathcal{F}$:

  1. (a)

$f(z)=0\Longrightarrow f^{\prime}(z)=h(z)$ and

  2. (b)

$f^{\prime}(z)=h(z)\Longrightarrow|f^{(k)}(z)|\leq c$, where $c$ is a
constant.

Then $\mathcal{F}$ is normal on $D$.

We now pose the following question: can the restriction for the zeros of
$h(z)$ with multiplicity at most $k-1$ be dropped? In this paper, we continue
to study the above problem and obtain an affirmative answer.

######  Theorem 1.

Let $\mathcal{F}$ be a family of functions holomorphic on a domain
$D\subset\mathbb{C}$. Let $k\geq 2$ be an integer, and let $h(z)(\not\equiv
0)$ be a holomorphic function on $D$ that has no common zeros with any
$f\in\mathcal{F}$. Assume also that the following two conditions hold for
every $f\in\mathcal{F}$:

  1. (a)

$f(z)=0\Longrightarrow f^{\prime}(z)=h(z)$ and

  2. (b)

$f^{\prime}(z)=h(z)\Longrightarrow|f^{(k)}(z)|\leq c$, where $c$ is a
constant.

Then $\mathcal{F}$ is normal on $D$.

Also in [1], the case for the $k-$th derivative was considered and the
following result was proved .

######  Theorem CFZ2.

[1, Theorem 1] Let $\mathcal{F}$ be a family of functions holomorphic on a
domain $D\subset\mathbb{C}$, all of whose zeros have multiplicity at least
$k$, where $k\neq 2$ is a positive integer; and let $h(z)\neq 0$ be a function
analytic in $D$. Assume also that the following two conditions hold for every
$f\in\mathcal{F}$:

  1. (a)

$f(z)=0\Longrightarrow f^{(k)}(z)=h(z)$; and

  2. (b)

$f^{(k)}(z)=h(z)\Longrightarrow|f^{(k+1)}(z)|\leq c$, where $c$ is a constant.

Then $\mathcal{F}$ is normal on $D$.

For the case $k=2$, the following result was obtained.

######  Theorem CFZ3.

[1, Theorem 3] Let $\mathcal{F}$ be a family of functions holomorphic on a
domain $D\subset\mathbb{C}$, all of whose zeros are multiple, where $s\geq 4$
is an even integer; and let $h(z)\neq 0$ be a function analytic in $D$. Assume
also that the following two conditions hold for every $f\in\mathcal{F}$:

  1. (a)

$f(z)=0\Longrightarrow f^{\prime\prime}(z)=h(z)$; and

  2. (b)

$f^{\prime\prime}(z)=h(z)\Longrightarrow|f^{\prime\prime\prime}(z)|+|f^{(s)}(z)|\leq
c$, where $c$ is a constant.

Then $\mathcal{F}$ is normal on $D$.

In view of the improvement of Theorems CFZ1 and LN via Theorem 1, the question
that naturally arises concerning Theorem CFZ2 and CFZ3, is whether the
condition $h(z)\neq 0$, $z\in D$, can be relaxed to “$h\not\equiv 0$ ”. It
turns out that the answer is negative in both cases. It is negative even if
$h$ has no common zero with any $f\in\mathcal{F}$ (like in Theorem 1). To
construct the first example, concerning Theorem CFZ2, we first need to mention
the following famous result of F. Lucas.

######  Theorem Lu.

[5], [6, p. 22] Let $P(z)$ be a nonconstant polynomial. Then all the zeros of
$P^{\prime}(z)$ lie in the convex hull $H$ of the zeros of $P(z)$. Moreover,
there are no zeros of $P^{\prime}(z)$ on the boundary of $H$, unless this zero
is a multiple zero of $P(z)$ or the zeros of $P(z)$ are colinear.

######  Example 1.

Let $r\geq 1$ and $k\geq 3$ be integers, $D=\Delta$ be the unit disc and
$h(z)=z^{r}$. Define

$f_{n}(z)=a_{n}\left(z^{\ell}-\frac{\displaystyle 1}{\displaystyle
n^{\ell}}\right)^{k},$

where $\ell=k+r$ and $a_{n}=\frac{\displaystyle n^{(k-1)\ell}}{\displaystyle
k!\ell^{k}}$.

We have

$f_{n}(z)=a_{n}\prod\limits_{j=1}^{\ell}\left(z-\alpha^{(n)}_{j}\right)^{k},$

where $\alpha^{(n)}_{j}=\frac{\displaystyle\exp\left(i\frac{2\pi
j}{\ell}\right)}{\displaystyle n}$, for $1\leq j\leq\ell$.

By calculation,

$\displaystyle f_{n}^{(k)}\left(\alpha^{(n)}_{j}\right)$
$\displaystyle=k!a_{n}\prod\limits_{t=1,t\neq
j}^{\ell}\left(\alpha^{(n)}_{j}-\alpha^{(n)}_{t}\right)^{k}=k!a_{n}\left[\left(z^{\ell}-\frac{\displaystyle
1}{\displaystyle
n^{\ell}}\right)^{\prime}\Bigg{|}_{z=\alpha^{(n)}_{j}}\right]^{k}$
$\displaystyle=k!a_{n}\ell^{k}\left(\alpha^{(n)}_{j}\right)^{k(\ell-1)}.$

Thus,

(1)
$\arg\left[f_{n}^{(k)}\left(\alpha^{(n)}_{j}\right)\right]=(\ell-1)k\cdot\frac{\displaystyle
2\pi j}{\displaystyle\ell}=-\frac{\displaystyle 2\pi
kj}{\displaystyle\ell}=\frac{\displaystyle 2\pi
ri}{\displaystyle\ell}=\arg\left[z^{r}\Big{|}_{z=\alpha^{(n)}_{j}}\right].$

Here the equalities are modulo $2\pi$, and we used in the last equality that
$r+k=\ell$.

We have

(2) $\left|f_{n}^{(k)}\left(\alpha^{(n)}_{j}\right)\right|=\frac{\displaystyle
k!\ell^{k}n^{\ell(k-1)}}{\displaystyle k!\ell^{k}}\left(\frac{\displaystyle
1}{\displaystyle n}\right)^{k(\ell-1)}=\left(\frac{\displaystyle
1}{\displaystyle
n}\right)^{r}=\left|z^{r}\right|\Bigg{|}_{z=\alpha^{(n)}_{j}}.$

From (1) and (2) we have that $f_{n}(z)=0\Longrightarrow f^{(k)}_{n}(z)=h(z)$,
i.e., assumption (a) of Theorem CFZ2 holds.

In order to confirm (b) of Theorem CFZ2, set

$\widetilde{f}_{n}(z)=f_{n}(z)-\frac{\displaystyle
z^{\ell}}{\displaystyle\ell(\ell-1)\cdots(r+1)}.$

We have $f^{(k)}_{n}(z)=h(z)\Longleftrightarrow\widetilde{f}^{(k)}_{n}(z)=0$.

Now

(3) $\widetilde{f}_{n}(z)=0\Longleftrightarrow\frac{\displaystyle
n^{k(\ell-1)-r}}{\displaystyle k!\ell^{k}}\left(z^{\ell}-\frac{\displaystyle
1}{\displaystyle n^{\ell}}\right)^{k}=\frac{\displaystyle
z^{\ell}}{\displaystyle\ell(\ell-1)\cdots(r+1)}.$

Suppose by negation that there exist a sequence $\\{z_{n}\\}^{\infty}_{n=1}$
$(z_{n}\to 0)$ and a sequence of natural numbers $\\{k_{n}\\}^{\infty}_{n=1}$
$(k_{n}\underset{n\to\infty}{\longrightarrow}\infty)$, such that
$\widetilde{f}_{k_{n}}(z_{n})=0$. Then since
$\frac{\displaystyle(k_{n}z_{n})^{\ell}-1}{\displaystyle(k_{n}z_{n})^{\ell}}\underset{n\to\infty}{\longrightarrow}1$,
from (3) we get

(4) $\frac{\displaystyle k_{n}^{(k-1)\ell}(k_{n}z_{n})^{k\ell}}{\displaystyle
k_{n}^{k\ell}z^{\ell}_{n}}\underset{n\to\infty}{\longrightarrow}\frac{\displaystyle
k!\ell^{k}}{\displaystyle\ell(\ell-1)\cdots(r+1)}.$

But the left hand side of (4) tends to $\infty$, as $n\to\infty$, a
contradiction.

We deduce that there exists some $0<C_{1}<\infty$, such that every zero
$z_{n}$ of $\widetilde{f}_{n}$ satisfies $|z_{n}|\leq\frac{\displaystyle
C_{1}}{\displaystyle n}$. By Theorem Lu, we have also
$|\widehat{z}_{n}|\leq\frac{\displaystyle C_{1}}{\displaystyle n}$ for every
$\widehat{z}_{n}$, which is a zero of $\widetilde{f}^{(k)}_{n}$. But those
$\\{\widehat{z}_{n}\\}$ are exactly the points where $f^{(k)}_{n}(z)=h(z)$.

Hence $f^{(k)}_{n}(z)=h(z)$ implies that $|z|\leq\frac{\displaystyle
C_{1}}{\displaystyle n}$, and we have only to prove the following claim.

######  Claim 1.

There exists $0<C<\infty$, such that $|z|\leq\frac{\displaystyle
C_{1}}{\displaystyle n}$ implies $|f^{(k+1)}_{n}(z)|\leq C$.

###### Proof.

We have $f_{n}(z)=\frac{\displaystyle n^{(k-1)\ell}}{\displaystyle
k!\ell^{k}}\left(z^{\ell}-\frac{\displaystyle 1}{\displaystyle
n^{\ell}}\right)^{k}=\frac{\displaystyle n^{(k-1)\ell}}{\displaystyle
k!\ell^{k}}\sum\limits_{j=0}^{k}\binom{k}{j}z^{\ell
j}\left(\frac{\displaystyle 1}{\displaystyle n}\right)^{\ell(k-j)}(-1)^{k-j}$.
Thus, since $\ell j\geq k+1$ only for $j\geq 1$, we get that

$f^{(k+1)}_{n}(z)=\frac{\displaystyle n^{(k-1)\ell}}{\displaystyle
k!\ell^{k}}\sum\limits_{j=1}^{k}\binom{k}{j}\left(\frac{\displaystyle
1}{\displaystyle n}\right)^{\ell k-\ell j}(-1)^{k-j}\ell j(\ell
j-1)\cdots(\ell j-k-1)z^{\ell j-k-1}.$

Thus, if $|z|\leq\frac{\displaystyle C_{1}}{\displaystyle n}$, then

$\displaystyle|f^{(k+1)}_{n}(z)|$ $\displaystyle\leq\frac{\displaystyle
n^{(k-1)\ell}}{\displaystyle
k!\ell^{k}}\sum\limits_{j=1}^{k}\binom{k}{j}C^{\ell j-k-1}_{1}\ell j(\ell
j-1)\cdots(\ell j-k-1)n^{k+1-\ell j}\cdot n^{\ell j-\ell k}$
$\displaystyle=\frac{\displaystyle n^{k+1-\ell}}{\displaystyle
k!\ell^{k}}\sum\limits_{j=1}^{k}\binom{k}{j}C^{\ell j-k-1}_{1}\ell j(\ell
j-1)\cdots(\ell j-k-1)\leq C,$

where $C=\frac{\displaystyle 1}{\displaystyle
k!\ell^{k}}\sum\limits_{j=1}^{k}\binom{k}{j}C^{\ell j-k-1}_{1}\ell j(\ell
j-1)\cdots(\ell j-k-1)$. (Here we used that $k+1-\ell\leq 0$.) The Claim is
proved. ∎

Hence, $\\{f_{n}\\}$ with $h$ satisfy (a) and (b) of Theorem CFZ2, but
$\\{f_{n}\\}$ is not normal at $z=0$.

Observe that when $k=1$, then $a_{n}=\frac{\displaystyle
1}{\displaystyle\ell}\not\to\infty$, and we do not get a non-normal family, as
expected by Theorem 1.

The following example shows that the condition $h(z)\neq 0$ is essential also
forTheorem CFZ3.

######  Example 2.

(cf. [1, Ex. 4] Let $s\geq 4$ be an even integer and consider the family
$\mathcal{F}=\\{f_{n}(z)\\}^{\infty}_{n=1}$,

$f_{n}(z)=\frac{\displaystyle n^{s}}{\displaystyle
2s^{2}}\left(z^{s}-\frac{\displaystyle 1}{\displaystyle
n^{s}}\right)^{2}\quad\text{on}\quad\Delta.$

Let $h(z)=z^{s-2}$.

We have that

$f_{n}(z)=\frac{\displaystyle n^{s}}{\displaystyle
2s^{2}}\prod\limits_{j=1}^{s}\left(z-\alpha^{(n)}_{j}\right)^{2},$

where $\alpha^{(n)}_{j}=\frac{\displaystyle\exp(i2\pi j/s)}{\displaystyle n}$,
$1\leq j\leq s$.

By calculation we have

(5) $f^{\prime\prime}_{n}(z)=\frac{\displaystyle n^{s}}{\displaystyle
s}\left((2s-1)z^{s}-\frac{\displaystyle(s-1)}{\displaystyle
n^{s}}\right)z^{s-2},$ (6) $\displaystyle f^{\prime\prime\prime}_{n}(z)$
$\displaystyle=\frac{\displaystyle n^{s}}{\displaystyle
s}\left[(2s-1)(2s-2)z^{s}-\frac{\displaystyle(s-1)(s-2)}{\displaystyle
n^{s}}\right]z^{s-3}$
$\displaystyle=\frac{n^{s}}{s}(s-1)z^{s-3}\left[(4s-2)z^{s}-\frac{s-2}{n^{s}}\right],$

and

(7) $f^{(s)}_{n}(z)=\frac{\displaystyle n^{s}}{\displaystyle
s}\left[(2s-1)(2s-2)\cdots(s+1)z^{s}-\frac{\displaystyle(s-1)!}{\displaystyle
n^{s}}\right].$

Now, if $f_{n}(z)=0$, then $z=\alpha^{(n)}_{j}$ for some $1\leq j\leq s$, and
thus $z^{s}=\frac{\displaystyle 1}{\displaystyle n^{s}}$ and by (5),
$f^{\prime\prime}_{n}(z)=z^{s-2}=h(z)$.

If $f^{\prime\prime}_{n}(z)=z^{s-2}=h(z)$, then by (5), $z=0$ or
$z=\alpha^{(n)}_{j}$, $1\leq j\leq s$. By (6) and (7), we get

(8) $f^{(3)}_{n}(0)=0,\quad
f^{(s)}_{n}(0)=-\frac{\displaystyle(s-1)!}{\displaystyle n^{s}}$

and

(9) $f^{(3)}_{n}\left(\alpha^{(n)}_{j}\right)=3(s-1)\frac{\displaystyle
1}{\displaystyle n^{s-3}},\quad
f^{(s)}_{n}\left(\alpha^{(n)}_{j}\right)=\frac{\displaystyle 1}{\displaystyle
s}\left[\frac{\displaystyle(2s-1)!}{\displaystyle s!}-(s-1)!\right].$

From (8) and (9), we see that the family $\mathcal{F}$ with $h$ satisfy
assumption (a) and (b) of Theorem CFZ3, but $\mathcal{F}$ is not normal at
$z=0$. Indeed, the reason must be that $h(0)=0$.

In Example 1, we have that $f^{(k+1)}(z)\neq 0$ at the zero points of
$f^{(k)}(z)-h(z)$. If we strengthen condition (b) of Theorem CFZ2 to be
$f^{(k)}(z)=h(z)\Longrightarrow f^{(k+1)}(z)=0$, then we can obtain the
following normal criterion.

######  Theorem 2.

Let $\mathcal{F}$ be a family of functions holomorphic on a domain
$D\subset\mathbb{C}$, all of whose zeros have multiplicity at least $k$, where
$k\neq 2$ be a positive integer. Let $h(z)(\not\equiv 0)$ be a holomorphic
function on $D$, that has no common zeros with any $f\in\mathcal{F}$. Assume
also that the following two conditions hold for every $f\in\mathcal{F}$:

  1. (a)

$f(z)=0\Longrightarrow f^{(k)}(z)=h(z)$; and

  2. (b)

$f^{(k)}(z)=h(z)\Longrightarrow f^{(k+1)}(z)=0$.

Then $\mathcal{F}$ is normal on $D$.

Similarly, if we strengthen the condition (b) of Theorem CFZ3 to
$f^{\prime\prime}(z)=h(z)\Longrightarrow
f^{\prime\prime\prime}(z)=f^{(s)}(z)=0$, then we can also obtain the normality
criterion.

######  Theorem 3.

Let $\mathcal{F}$ be a family of functions holomorphic on a domain
$D\subset\mathbb{C}$, all of whose zeros are multiple, where $s\geq 2$ is an
even integer. Let $h(z)(\not\equiv 0)$ be a holomorphic function on $D$, that
has no common zeros with any $f\in\mathcal{F}$. Assume also that the following
two conditions hold for every $f\in\mathcal{F}$:

  1. (a)

$f(z)=0\Longrightarrow f^{\prime\prime}(z)=h(z)$; and

  2. (b)

$f^{\prime\prime}(z)=h(z)\Longrightarrow
f^{\prime\prime\prime}(z)=f^{(s)}(z)=0$.

Then $\mathcal{F}$ is normal on $D$.

Before we go to the proofs of the main results, let us set some notation.
Throughout, $D$ is a domain in $\mathbb{C}$. For $z_{0}\in\mathbb{C}$ and
$r>0$, $\Delta(z_{0},r)=\\{z:|z-z_{0}|<r\\}$ and
$\Delta^{\prime}(z_{0},r)=\\{z:0<|z-z_{0}|<r\\}$. The unit disc will be
denoted by $\Delta$ and $\mathbb{C}^{\ast}=\mathbb{C}\setminus\\{0\\}$. We
write $f_{n}(z)\overset{\chi}{\Rightarrow}f(z)$ on $D$ to indicate that the
sequence $\\{f_{n}\\}$ converges to $f$ in the spherical metric, uniformly on
compact subsets of $D$, and $f_{n}\Rightarrow f$ on $D$ if the convergence is
in the Euclidean metric. For a meromorphic function $f(z)$ in $D$ and
$a\in\widehat{\mathbb{C}}$, $\overline{E}_{f}(a):=\\{z\in D:f(z)=a\\}$. The
spherical derivative of the meromorphic function $f$ at the point $z$ is
denoted by $f^{\\#}(z).$

Frequently, given a sequence $\\{f_{n}\\}_{1}^{\infty}$ of functions, we need
to extract an appropriate subsequence; and this necessity may recur within a
single proof. To avoid the awkwardness of multiple indices, we again denote
the extracted subsequence by $\\{f_{n}\\}$ (rather than, say,
$\\{f_{n_{k}}\\})$ and designate this operation by writing “taking a
subsequence and renumbering,” or simply “renumbering”. The same convention
applies to sequences of constants.

The plan of the paper is as follows. In Section 2, we state a number of
preliminary results. Then in Section 3 we prove Theorem 1. Finally, in Section
4 we prove Theorem 2.

##  2\. Preliminary results

The following lemma is the local version of a well-known lemma of X. C. Pang
and L. Zalcman [11, Lemma 2]. For a proof see [4, Lemma 2], also cf.[9, Lemma
2], [14, pp. 216–217], [7, pp. 299–300], [8, p. 4].

######  Lemma 1.

Let $\mathcal{F}$ be a family of functions meromorphic in a domain $D$, all of
whose zeros have multiplicity at least $k$, and suppose that there exists
$A\geq 1$, such that $|f^{(k)}(z)|\leq A$ whenever $f(z)=0$. Then if
$\mathcal{F}$ is not normal at $z_{0}\in D$, there exist, for each
$0\leq\alpha\leq k$,

  1. (a)

points $z_{n}\to z_{0}$;

  2. (b)

functions $f_{n}\in\mathcal{F}$;and

  3. (c)

positive numbers $\rho_{n}\to 0^{+}$

such that
$g_{n}(\zeta):=\rho^{-\alpha}_{n}f_{n}(z_{n}+f_{n}\zeta)\overset{\chi}{\Rightarrow}g(\zeta)$
on $\mathbb{C}$, where $g$ is a nonconstant meromorphic function on
$\mathbb{C}$, such that for every $\zeta\in\mathbb{C}$, $g^{\\#}(\zeta)\leq
g^{\\#}(0)=kA+1$.

######  Lemma 2.

[1, Lemma 5] Let $f$ be a nonconstant entire function of order $\rho$,
$0\leq\rho\leq 1$, all of whose zeros have multiplicity at least $k$, where
$k\neq 2$ is a positive integer. And let $a\neq 0$ be a constant. If
$\overline{E}_{f}(0)\subset\overline{E}_{f^{(k)}}(a)\subset\overline{E}_{f^{(k+1)}}(0)$,
then

$f(z)=\frac{\displaystyle a(z-b)^{k}}{\displaystyle k!},$

where $b$ is a constant.

######  Lemma 3.

[1, Lemma 6] Let $f$ be a nonconstant entire function of order $\rho$,
$0\leq\rho\leq 1$, all of whose zeros are multiple. Let $s\geq 4$ be an even
integer and $a\neq 0$ be a constant. If
$\overline{E}_{f}(0)\subset\overline{E}_{f^{\prime\prime}}(a)\subset\overline{E}_{f^{\prime\prime\prime}}(0)\cap\overline{E}_{f^{(s)}}(0)$,
then

$f(z)=\frac{\displaystyle a(z-b)^{2}}{\displaystyle 2},$

where $b$ is a constant.

######  Lemma 4.

(see [2, pp. 118–119,122–123]) Let $f$ be a meromorphic function on
$\mathbb{C}$. If $f^{\\#}$ is uniformly bounded on $\mathbb{C}$, then the
order of $f$ is at most $2$. If $f$ is an entire function, then the order of
$f$ is at most $1$.

The following lemma is a slight generalization of Theorem CFZ2 for sequences.

######  Lemma 5.

(cf. [4, Lemma 5]) Let $\\{f_{n}\\}$ be a sequence of functions holomorphic on
a domain $D\subset\mathbb{C}$, all of whose zeros have multiplicity at least
$k$, and let $\\{h_{n}\\}$ be a sequence of functions analytic on $D$ such
that $h_{n}{(z)}\Rightarrow h{(z)}$ on $D$, where $h(z)\neq 0$ for $z\in D$
and $k\neq 2$ be a positive integer. Suppose that, for each $n$,
$f_{n}(z)=0\Longrightarrow f^{(k)}_{n}(z)=h_{n}(z)$ and
$f^{(k)}_{n}(z)=h_{n}(z)\Longrightarrow f^{(k+1)}_{n}(z)=0$. Then
$\\{f_{n}\\}$ is normal on $D$.

###### Proof.

Suppose to the contrary that there exists $z_{0}\in D$ such that $\\{f_{n}\\}$
is not normal at $z_{0}$. The convergence of $\\{h_{n}\\}$ to $h$ implies
that, in some neighborhood of $z_{0}$, we have
$f_{n}(z)=0\Rightarrow|f_{n}^{(k)}(z)|\leq|h(z_{0})|+1$ (for large enough
$n$). Thus we can apply Lemma 1 with $\alpha=k$ and $A$ such that
$kA+1>\max\Big{\\{}|h(z_{0})|+1,\frac{\displaystyle|h(z_{0})|}{\displaystyle(k-1)!},\frac{\displaystyle
k\cdot
k!}{\displaystyle|h(z_{0})|}\Big{\\}}=\max\Big{\\{}|h(z_{0})|+1,\frac{\displaystyle
k\cdot k!}{\displaystyle|h(z_{0})|}\Big{\\}}$. So we can take an appropriate
subsequence of $\\{f_{n}\\}$ (denoted also by $\\{f_{n}\\}$ after
renumbering), together with points $z_{n}\to z_{0}$ and positive numbers
$\rho_{n}\to 0^{+}$ such that

$g_{n}(\zeta)=\frac{f_{n}(z_{n}+\rho_{n}\zeta)}{\rho^{k}_{n}}\overset{\chi}{\Longrightarrow}g(\zeta)\quad\text{on}\quad\mathbb{C},$

where $g$ is a nonconstant entire function and $g^{\sharp}(\zeta)\leq
g^{\sharp}(0)=kA+1=k(|h(z_{0})|+1)+1$. We claim that

(10)
$\overline{E}_{g}(0)\subset\overline{E}_{g^{(k)}}(h(z_{0}))\subset\overline{E}_{g^{(k+1)}}(0).$

In fact, if there exists $\zeta_{0}\in\mathbb{C}$, such that $g(\zeta_{0})=0$,
then since $g(\zeta)\not\equiv 0$, there exist $\zeta_{n}$,
$\zeta_{n}\to\zeta_{0}$, such that if $n$ is sufficiently large,

$g_{n}(\zeta_{n})=\frac{f_{n}(z_{n}+\rho_{n}\zeta_{n})}{\rho^{k}_{n}}=0.$

Thus $f_{n}(z_{n}+\rho_{n}\zeta_{n})=0$, so that
$f^{(k)}_{n}(z_{n}+\rho_{n}\zeta_{n})=h_{n}(z_{n}+\rho_{n}\zeta_{n})$, i.e.,
that $g^{(k)}_{n}(\zeta_{n})=h_{n}(z_{n}+\rho_{n}\zeta_{n})$. Since
$g^{(k)}(\zeta_{0})=\lim\limits_{n\to\infty}g^{(k)}_{n}(\zeta_{n})=h(z_{0})$,
we have established the first part of the Claim that
$\overline{E}_{g}(0)\subset\overline{E}_{g^{(k)}}(h(z_{0}))$.

Now, suppose there exists $\zeta_{0}\in\mathbb{C}$, such that
$g^{(k)}(\zeta_{0})=h(z_{0})$. If $g^{(k)}(\zeta)\equiv h(z_{0})$, then
$g^{(k+1)}\equiv 0$ and we are done. Thus we can assume that $g^{(k)}$ is not
constant and since
$f^{(k)}_{n}(z_{n}+\rho_{n}\zeta)-h_{n}(z_{n}+\rho_{n}\zeta)\Rightarrow
g^{(k)}(\zeta)-h(z_{0})$, we get by Hurwitz’s Theorem that there exist
$\zeta_{n}$, $\zeta_{n}\to\zeta_{0}$, such that

$f^{(k)}_{n}(z_{n}+\rho_{n}\zeta_{n})-h_{n}(z_{n}+\rho_{n}\zeta_{n})=g^{(k)}_{n}(\zeta_{n})-h_{n}(z_{n}+\rho_{n}\zeta_{n})=0.$

Thus we have $f^{(k+1)}_{n}(z_{n}+\rho_{n}\zeta_{n})=0$ and
$g^{(k+1)}_{n}(\zeta_{n})=0$. Letting $n\to\infty$, we get that
$g^{(k+1)}(\zeta_{0})=0$. This completes the proof of the Claim. Now, by
Lemmas 4 and 2, we have $g(\zeta)=\frac{\displaystyle
h(z_{0})(\zeta-\zeta_{1})^{k}}{\displaystyle k!}$, where $\zeta_{1}$ is a
constant. Thus

$g^{\sharp}(0)=\frac{\displaystyle|h(z_{0})||\zeta_{1}|^{k-1}/(k-1)!}{\displaystyle
1+|h(z_{0})|^{2}|\zeta_{1}|^{2k}/k!^{2}}.$

Now, if $|\zeta_{1}|\leq 1$, then
$g^{\sharp}(0)\leq\frac{\displaystyle|h(z_{0})|}{\displaystyle(k-1)!}<kA+1$,
and if $|\zeta_{1}|>1$, then
$g^{\sharp}(0)\leq\frac{\displaystyle|h(z_{0})||\zeta_{1}|^{k-1}/(k-1)!}{\displaystyle|h(z_{0})|^{2}|\zeta_{1}|^{2k}/k!^{2}}\leq\frac{\displaystyle
k\cdot k!}{\displaystyle|h(z_{0})|}<kA+1$. In either case we get a
contradiction. ∎

Similarly, we can get a slight generalization of Theorem CFZ3 for sequences.

######  Lemma 6.

Let $\\{f_{n}\\}$ be a sequence of functions holomorphic on a domain
$D\subset\mathbb{C}$, all of whose zeros are multiple and $\\{h_{n}\\}$ be a
sequence of functions analytic on $D$ such that $h_{n}{(z)}\Rightarrow h{(z)}$
on $D$, where $h(z)\neq 0$ for $z\in D$ and $s\geq 2$ be an even integer.
Suppose that, for each $n$, $f_{n}(z)=0\Longrightarrow
f^{\prime\prime}_{n}(z)=h_{n}(z)$ and
$f^{\prime\prime}_{n}(z)=h_{n}(z)\Longrightarrow
f^{\prime\prime\prime}(z)=f^{(s)}_{n}(z)=0$, then $\\{f_{n}\\}$ is normal on
$D$.

The proof is very similar to the proof of Lemma 5. We start to argue the same
(with $2$ instead of $k$), and then instead of proving (10) we prove that

$\overline{E}_{g}(0)\subset\overline{E}_{g^{\prime\prime}}(h(z_{0}))\subset\overline{E}_{g^{(3)}}(0)\cap\overline{E}_{g^{(s)}}(0).$

The left inclusion is proved in the same manner. Concerning the right
inclusion, we now deduce from
$f^{\prime\prime}_{n}(z_{n}+\rho_{n}\zeta_{n})-h_{n}(z_{n}+\rho_{n}\zeta_{n})=0$
that
$f^{(3)}_{n}(z_{n}+\rho_{n}\zeta_{n})=f^{(s)}_{n}(z_{n}+\rho_{n}\zeta_{n})=0$.
Then, since $\rho_{n}f^{(3)}_{n}(z_{n}+\rho_{n}\zeta)\Rightarrow
g^{(3)}(\zeta)$ in $\mathbb{C}$ and
$\rho^{s-2}_{n}f^{(s)}_{n}(z_{n}+\rho_{n}\zeta)\Rightarrow g^{(s)}(\zeta)$ in
$\mathbb{C}$, we conclude that $g^{(3)}(\zeta_{0})=g^{(s)}(\zeta_{0})=0$. To
get the final contradiction, we apply now Lemmas 4 and 3 instead of Lemmas 4
and 2.

The following result will play an essential role in treating transcendental
functions which is used in the proofs of Theorems 2 and 3.

######  Theorem B.

$($[15] see also [2, p. 117]$)$ Let $f(z)$ be a function homomorphic
in$\\{z:R<|z|<\infty\\}$, with essential singularity at $z=\infty$. Then
$\varlimsup\limits_{|z|\to\infty}|z|f^{\\#}(z)=+\infty$.

For the proof of Theorem 2, we need also the following Lemma.

######  Lemma 7.

Let $h$ be a holomorphic function on $D,$ with a zero of order $\ell(\geq 1)$
at $z_{0}\in D.$ Let $\\{f_{n}\\}^{\infty}_{n=1}$ be a sequence of functions
with zeros of multiplicity at least $k$, such that $\\{f_{n}\\}$ and $h$
satisfy conditions (a) and (b) of Theorem 2. Let
$\\{\alpha_{n}\\}^{\infty}_{n=1}$ be a sequence of nonzero numbers such that
$\alpha_{n}\to 0$ as $n\to\infty$. Then
$\\{f_{n}(z_{0}+\alpha_{n}\zeta)/\alpha^{k+\ell}_{n}\\}^{\infty}_{n=1}$ is
normal in $\mathbb{C}^{\ast}$.

###### Proof.

Without loss of generality, we may assume that $z_{0}=0$. In a neighborhood of
the origin we have $h(z)=z^{\ell}b(z)$, where $b(z)$ is analytic, $b(0)\neq
0$. Define $r_{n}(\zeta)=\zeta^{\ell}b(\alpha_{n}\zeta)$. We will show that
the assumptions of Lemma 5 hold in $\mathbb{C}^{\ast}$ for the sequence
$\\{G_{n}(\zeta)\\}^{\infty}_{n=1}$,
$G_{n}(\zeta):=f_{n}(\alpha_{n}\zeta)/\alpha^{k+\ell}_{n}$ and
$\\{r_{n}(\zeta)\\}^{\infty}_{n=1}$. First, we have that
$r_{n}(\zeta)\Rightarrow b(0)\zeta^{\ell}$ on $\mathbb{C}$ and
$\zeta^{\ell}\neq 0$ in $\mathbb{C}^{\ast}$. Assume that $G_{n}(\zeta)=0$.
Then $f_{n}(\alpha_{n}\zeta)=0$ and
$f^{(k)}_{n}(\alpha_{n}\zeta)=(\alpha_{n}\zeta)^{\ell}b(\alpha_{n}\zeta)$, and
we get that $G^{(k)}_{n}(\zeta)=r_{n}(\zeta)$. Suppose now that
$G^{(k)}_{n}(\zeta)=r_{n}(\zeta)$. This means that
$f^{(k)}_{n}(\alpha_{n}\zeta)=h(\alpha_{n}\zeta)$ and thus
$f^{(k+1)}_{n}(\alpha_{n}\zeta)=0$. We have $G^{(k+1)}_{n}(\zeta)=0$, and thus
the assumptions of Lemma 5 hold. Hence we deduce that $\\{G_{n}(\zeta)\\}$ is
normal in $\mathbb{C}^{\ast}$, and the lemma is proved. ∎

The following lemma plays a similar role in the proof of Theorem 3, to the
role of Lemma 7 in the proof of Theorem 2.

######  Lemma 8.

Let $h$ be a holomorphic function on $D,$ with a zero of order $\ell(\geq 1)$
at $z_{0}\in D.$ Let $\\{f_{n}\\}^{\infty}_{n=1}$ be a sequence of functions
whose zeros are multiple, such that $\\{f_{n}\\}$ and $h$ satisfy conditions
(a) and (b) of Theorem 3. Let $\\{\alpha_{n}\\}^{\infty}_{n=1}$ be a sequence
of nonzero numbers such that $\alpha_{n}\to 0$ as $n\to\infty$. Then
$\\{f_{n}(z_{0}+\alpha_{n}\zeta)/\alpha^{2+\ell}_{n}\\}^{\infty}_{n=1}$ is
normal in $\mathbb{C}^{\ast}$.

The proof of this lemma is analogous to the proof of Lemma 7. Of course, we
use Lemma 6 instead of Lemma 5.

##  3\. Proof of Theorem 1

In this section, we do not use any of the preliminary results. The proof is
elementary.

By Theorem CFZ1, $\mathcal{F}$ is normal at every point $z_{0}\in D$ at which
$h(z_{0})\neq 0$(so immediately we get that $\mathcal{F}$ is quasinormal). So
let $z_{0}$ be a zero of $h$ of order $\ell(\geq 1)$. Without loss of
generality, we can assume that $z_{0}=0$, and then $h(z)=z^{\ell}b(z)$. Here
$b$ is an analytic function in $\Delta(0,\delta)$ and $b(z)\neq 0$ there. We
assume that $0<\delta<1$, and by taking a subsequence and renumbering, we can
assume that

(11) $f_{n}\Longrightarrow f\quad\text{in}\quad\Delta^{\prime}(0,\delta).$

Now, if $f$ is holomorphic in $\Delta^{\prime}(0,\delta)$, we deduce by the
maximum principle that $f_{n}\Rightarrow f$ on $\Delta(0,\delta)$, and we are
done. So let us assume that $f_{n}\Rightarrow\infty$ in
$\Delta^{\prime}(0,\delta)$. Fix $\eta$, $0<\eta<\delta$. By the minimum
principle (i.e., the maximum principle for $\\{1/f_{n}\\}$), there exists
$N=N(\eta)$, such that for every $n\geq N$, $f_{n}$ has $k_{n}(k_{n}\geq 1)$
simple zeros in $\overline{\Delta}(0,\eta)-\\{0\\}$, say $\alpha^{(n)}_{1}$,
$\alpha^{(n)}_{2}$, $\cdots$, $\alpha^{(n)}_{k_{n}}$ (otherwise we get that
$f_{n}\Rightarrow\infty$ in $\Delta(0,\eta)$ and we are done). Since
$f_{n}\Rightarrow\infty$ in $\Delta^{\prime}(0,\delta)$, we get that

(12) $\max\limits_{1\leq j\leq k_{n}}\\{|\alpha^{(n)}_{j}|\\}\to
0,\quad\text{as}\quad n\to\infty.$

We can write
$f_{n}(z)=t_{n}(z)\prod\limits_{i=1}^{k_{n}}\left(z-\alpha^{(n)}_{i}\right)$,
where $t_{n}(z)\neq 0$ for $z\in\overline{\Delta}(0,\eta)$ and $n\geq N$.
Since $\eta<1$, we get by (12) that $\frac{\displaystyle
t_{n}(z)}{\displaystyle b(z)}\Rightarrow\infty$ in
$\overline{\Delta}(0,\eta)$. By condition (a) of Theorem 1, we have, for
$n\geq N$,
$f^{\prime}_{n}(\alpha^{(n)}_{j})=\alpha^{(n)\ell}_{j}b(\alpha^{(n)}_{j})$,
$1\leq j\leq k_{n}$. By calculation,

$f^{\prime}_{n}(z)=t^{\prime}_{n}(z)\prod\limits_{i=1}^{k_{n}}\left(z-\alpha^{(n)}_{i}\right)+t_{n}(z)\left[\prod\limits_{i=1}^{k_{n}}\left(z-\alpha^{(n)}_{i}\right)\right]^{\prime},$

and so

(13)
$t_{n}\left(\alpha^{(n)}_{j}\right)\left[\prod\limits_{i=1}^{k_{n}}\left(z-\alpha^{(n)}_{i}\right)\right]^{\prime}\Bigg{|}_{z=\alpha^{(n)}_{j}}=\alpha^{(n)\ell}_{j}b\left(\alpha^{(n)}_{j}\right).$

Define, for $n\geq N$,

$M_{n}(z):=\frac{\displaystyle t_{n}(z)}{\displaystyle
b(z)}\left[\prod\limits_{i=1}^{k_{n}}\left(z-\alpha^{(n)}_{i}\right)\right]^{\prime}-z^{\ell}.$

By (13) we get that $M_{n}\left(\alpha^{(n)}_{j}\right)=0$ for $1\leq j\leq
k_{n}$, and so for $n\geq N,$ $M_{n}$ has at least $k_{n}$ zeros in
$\Delta^{\prime}(0,\eta)$, including multiplicities. Here we use the fact $h$
has no common zero with any $f_{n}.$ Since such a zero must be $z=0$ and would
be a zero of order $m$ (must be $m\geq 2$ by condition (a)) of $f_{n}$, and it
would be a zero of order $m-1$ of $M_{n}$ (if $\ell>m-1$) or even of order
$\ell<m-1$ (if $\ell<m-1$), then we would not know that the number of zeros
(including multiplicities) of $M_{n}$ is at least $k_{n}$. This fact, under
the assumption that there are no common zeros, will lead to the desired
contradiction.

######  Claim 2.

$\frac{\displaystyle t_{n}(z)}{\displaystyle
b(z)}\left[\prod\limits_{i=1}^{k_{n}}\left(z-\alpha^{(n)}_{i}\right)\right]^{\prime}\Rightarrow\infty$
in $\Delta^{\prime}(0,\eta)$.

###### Proof.

We write

(14) $\frac{\displaystyle t_{n}(z)}{\displaystyle
b(z)}\left[\prod\limits_{i=1}^{k_{n}}\left(z-\alpha^{(n)}_{i}\right)\right]^{\prime}=\sum\limits_{j=1}^{k_{n}}\frac{\displaystyle
t_{n}(z)}{\displaystyle b(z)}\prod\limits_{i=1,i\neq
j}^{k_{n}}\left(z-\alpha^{(n)}_{i}\right).$

For any $\varepsilon$, $0<\varepsilon<\eta$, we have that

(15) $\frac{\displaystyle t_{n}(z)}{\displaystyle
b(z)}\prod\limits_{i=2}^{k_{n}}\left(z-\alpha^{(n)}_{i}\right)\Longrightarrow\infty\quad\text{in}\quad\overline{R}_{\varepsilon,\eta}:=\\{z:\varepsilon\leq|z|\leq\eta\\}.$

Indeed, $\frac{\displaystyle t_{n}(z)}{\displaystyle
b(z)}\prod\limits_{i=2}^{k_{n}}\big{(}z-\alpha^{(n)}_{i}\big{)}=\frac{\displaystyle
f_{n}(z)}{\displaystyle b(z)\big{(}z-\alpha^{(n)}_{1}\big{)}}$, and since
$\eta<1$ and by (11) and (12), this term tends uniformly to $\infty$ in
$\overline{R}_{\varepsilon,\eta}$.

Now, for every $j$, $2\leq j\leq k_{n}$, we have that

$\frac{\displaystyle\frac{\displaystyle t_{n}(z)}{\displaystyle
b(z)}\prod\limits_{i=2}^{k_{n}}\left(z-\alpha^{(n)}_{i}\right)}{\displaystyle\frac{\displaystyle
t_{n}(z)}{\displaystyle b(z)}\prod\limits_{i=1,i\neq
j}^{k_{n}}\left(z-\alpha^{(n)}_{i}\right)}=\frac{\displaystyle
z-\alpha^{(n)}_{j}}{\displaystyle z-\alpha^{(n)}_{1}},$

and by (12) this term tends uniformly to $1$ as $n\to\infty$. This means, that
for every $1\leq j\leq k_{n}$, and $z\in\overline{R}_{\varepsilon,\eta}$,
$\frac{\displaystyle t_{n}(z)}{\displaystyle b(z)}\prod\limits_{i=1,i\neq
j}^{k_{n}}\big{(}z-\alpha^{(n)}_{i}\big{)}$ lies in the same quarter plane,
that is,

(16) $\Pi_{n,z}:=\left\\{z:\arg\left[\frac{\displaystyle
t_{n}(z)}{\displaystyle
b(z)}\prod\limits_{i=2}^{k_{n}}\left(z-\alpha^{(n)}_{i}\right)\right]-\frac{\displaystyle\pi}{\displaystyle
4}<\arg z<\arg\left[\frac{\displaystyle t_{n}(z)}{\displaystyle
b(z)}\prod\limits_{i=2}^{k_{n}}\left(z-\alpha^{(n)}_{i}\right)\right]+\frac{\displaystyle\pi}{\displaystyle
4}\right\\},$

for large enough $n$.

Now, if $a$ and $b$ are two complex numbers in the same quarter plane, then
$a+b$ also belongs to that quarter plane and $|a+b|\geq|a|$, $|b|$. We then
conclude by (16) that for each $z\in\overline{R}_{\varepsilon,\eta}$, we have
for large enough $n$,

$\left|\frac{\displaystyle t_{n}(z)}{\displaystyle
b(z)}\left[\prod\limits_{i=1}^{k_{n}}\left(z-\alpha^{(n)}_{i}\right)\right]^{\prime}\right|\geq\left|\frac{\displaystyle
t_{n}(z)}{\displaystyle
b(z)}\prod\limits_{i=2}^{k_{n}}\left(z-\alpha^{(n)}_{i}\right)\right|,$

and by (15) and (14), the Claim is proved.∎

Now, $\frac{\displaystyle t_{n}(z)}{\displaystyle
b(z)}\bigg{[}\prod\limits_{i=1}^{k_{n}}\big{(}z-\alpha^{(n)}_{i}\big{)}\bigg{]}^{\prime}$
has for large enough $n$ exactly $k_{n}-1$ zeros in $\Delta(0,\eta)$ (by
Theorem Lu). Then for large enough $n$ we have, for every $z$, $|z|=\eta$,

$\left|M_{n}(z)-\frac{\displaystyle t_{n}(z)}{\displaystyle
b(z)}\left[\prod\limits_{i=1}^{k_{n}}\left(z-\alpha^{(n)}_{i}\right)\right]^{\prime}\right|=|z^{\ell}|<\left|\frac{\displaystyle
t_{n}(z)}{\displaystyle
b(z)}\left[\prod\limits_{i=1}^{k_{n}}\left(z-\alpha^{(n)}_{i}\right)\right]^{\prime}\right|,$

and by Rouche’s Theorem, we get that $M_{n}$ has $k_{n}-1$ zeros in
$\Delta(0,\eta)$, a contradiction. Theorem 1 is proved. $\square$

##  4\. Proof of Theorem 2

This proof is similar to the proof of Theorem 1 in [4]. By our Theorem 1, we
need only to prove the case that $k\geq 3$. By Theorem CFZ2, $\mathcal{F}$ is
normal at every point $z_{0}\in D$ at which $h(z_{0})\neq 0$ (so that
$\mathcal{F}$ is quasinormal in $D$ ). Consider $z_{0}\in D$ such that
$h(z_{0})=0$. Without loss of generality, we can assume that $z_{0}=0$, and
then $h(z)=z^{\ell}b(z)$, where $\ell(\geq 1)$ is an integer, $b(z)\neq 0$ is
an analytic function in $\Delta(0,\delta)$. We take a subsequence
$\\{f_{n}\\}^{\infty}_{1}\subset\mathcal{F}$, and we want to prove that
$\\{f_{n}\\}$ is not normal at $z=0.$ Suppose by negation that $\\{f_{n}\\}$
is not normal at $z=0.$ Since $\\{f_{n}\\}$ is normal in
$\Delta^{\prime}(0,\delta),$ we can assume (after renumbering) that
$f_{n}\Rightarrow F$ on $\Delta^{\prime}(0,\delta)$. If
$F(z)\not\equiv\infty,$ then it is a holomorphic function; hence by the
maximum principle, $F$ extends to be analytic also at $z=0$, and so
$f_{n}\Rightarrow F$ on $\Delta(0,\delta)$, and we are done. Hence we assume
that

(17)
$f_{n}(z)\Longrightarrow\infty\quad\text{on}\quad\Delta^{\prime}(0,\delta).$

Define $\mathcal{F}_{1}=\left\\{F=\frac{\displaystyle f_{n}}{\displaystyle
h}:n\in\mathbb{N}\right\\}.$ It is enough to prove that $\mathcal{F}_{1}$ is
normal in $\Delta(0,\delta).$ Indeed, if (after renumbering)
$\frac{\displaystyle f_{n}(z)}{\displaystyle h}\Rightarrow H(z)$ on
$\Delta(0,\delta),$ then since $h\neq 0$ in $\Delta^{\prime}(0,\delta)$, it
follows from (17) that $H(z)\equiv\infty$ in $\Delta^{\prime}(0,\delta)$, and
thus $H(z)\equiv\infty$ also in $\Delta(0,\delta).$ In particular,
$\frac{\displaystyle f_{n}}{\displaystyle h}(z)\neq 0$ on each compact subset
of $\Delta(0,\delta)$ for large enough $n.$ Since $h\neq 0$ on
$\Delta^{\prime}(0,\delta)$ and since $f_{n}(0)\neq 0$ for every $n\geq 1$ by
assumptions of the theorem, we obtain $f_{n}(z)\neq 0$ on each compact subset
of $\Delta(0,\delta)$ for large enough $n.$ Then by the minimum principle, it
follows from (17) that $f_{n}(z)\Rightarrow\infty$ on $\Delta(0,\delta)$, and
this implies the normality of $\mathcal{F}.$ So suppose to the contrary that
$\mathcal{F}_{1}$ is not normal at $z=0$. By Lemma 1 and the assumptions of
Theorem 2, there exist (after renumbering) points $z_{n}\to 0$, $\rho_{n}\to
0^{+}$ and a nonconstant meromorphic function on $\mathbb{C}$, $g(\zeta)$ such
that

(18)
$g_{n}(\zeta)=\frac{F_{n}(z_{n}+\rho_{n}\zeta)}{\rho^{k}_{n}}=\frac{f_{n}(z_{n}+\rho_{n}\zeta)}{\rho^{k}_{n}h(z_{n}+\rho_{n}\zeta)}\overset{\chi}{\Longrightarrow}g(\zeta)\quad\text{on}\quad\mathbb{C},$

all of whose zeros have multiplicity at least $k$ and

(19) $\text{for every}\quad\zeta\in\mathbb{C},\quad g^{\sharp}(\zeta)\leq
g^{\sharp}(0)=kA+1,$

where $A>1$ is a constant. Here we have used Lemma 1 with $\alpha=k$. Observe
that $g_{n}(z)=0$ implies $g^{(k)}_{n}(\zeta)=1$ and so $A$ can be chosen to
be any number such that $A\geq 1.$ After renumbering we can assume that
$\\{z_{n}/\rho_{n}\\}^{\infty}_{n=1}$ converges. We separate now into two
cases.

Case (A)

(20) $\frac{z_{n}}{\rho_{n}}\to\infty.$

######  Claim 3.

$(1)$ $g(\zeta)=0\Longrightarrow g^{(k)}(\zeta)=1$; $(2)$
$g^{(k)}(\zeta)=1\Longrightarrow g^{(k+1)}(\zeta)=0$.

###### Proof.

Observe that from (18) and the fact that $h(z)\neq 0$ in
$\Delta^{\prime}(0,\delta),$ it follows that $g$ is an entire function.
Suppose that $g(\zeta_{0})=0$. Since $g(\zeta)\not\equiv 0$, there exist
$\zeta_{n}\to\zeta_{0}$, such that $g_{n}(\zeta_{n})=0$, and thus
$f_{n}(z_{n}+\rho_{n}\zeta_{n})=0$. Since $f_{n}$ and $h$ has no common zeros,
it follows by the assumption that $\zeta_{n}$ is a zero of multiplicity $k$ of
$g_{n}(\zeta)$. By Leibniz’s rule, and condition (a) of Theorem 2, it follows
that $g^{(k)}_{n}(\zeta_{n})=1$ and thus $g^{(k)}(\zeta_{0})=1$.

For the proof of the other part of the Claim, observe first that by (20) we
have

$\frac{\displaystyle
f_{n}(z_{n}+\rho_{n}\zeta)}{\displaystyle\rho_{n}^{k}z^{\ell}_{n}}\Rightarrow
g(\zeta)\quad\text{on}\quad\mathbb{C},$

and thus

$\frac{\displaystyle f^{(k)}_{n}(z_{n}+\rho_{n}\zeta)}{\displaystyle
z^{\ell}_{n}}\Rightarrow g^{(k)}(\zeta)\quad\text{on}\quad\mathbb{C},$

and then again by (19) we get that

$\frac{\displaystyle f^{(k)}_{n}(z_{n}+\rho_{n}\zeta)}{\displaystyle
h(z_{n}+\rho_{n}\zeta)}\Rightarrow
g^{(k)}(\zeta)\quad\text{on}\quad\mathbb{C}.$

Thus, if there exists $\zeta_{0}\in\mathbb{C}$, such that
$g^{(k)}(\zeta_{0})=1$, there exists a sequence $\zeta_{n}\to\zeta_{0}$, such
that $f^{(k)}_{n}(z_{n}+\rho_{n}\zeta_{n})=h(z_{n}+\rho_{n}\zeta)\neq 0$. By
assumption (b) of Theorem 2 we get that
$f^{(k+1)}_{n}(z_{n}+\rho_{n}\zeta_{n})=0$, and letting $n$ tend to $\infty$
we get that $g^{(k+1)}(\zeta_{0})=0$. The Claim is proved.∎

We conclude by Lemma 2 and by Lemma 4 that
$g(\zeta)=\frac{\displaystyle(\zeta-b)^{k}}{\displaystyle k!}$ for some
$b\in\mathbb{C}$ (observe that $g$ is holomorphic by (20)). By calculation we
get that

$g^{\sharp}(0)=\frac{\displaystyle|b|^{k-1}/(k-1)!}{\displaystyle
1+|b|^{2k}/k!^{2}}.$

Then if $|b|\leq 1$, we get that $g^{\sharp}(0)\leq\frac{\displaystyle
1}{\displaystyle(k-1)!}$, and if $|b|\geq 1$, then
$g^{\sharp}(0)\leq\frac{\displaystyle k}{\displaystyle 2}$. In either case, we
get a contradiction to (19).

Case (B)

(21) $\frac{z_{n}}{\rho_{n}}\to\alpha\in\mathbb{C}.$

As in Case (A), it follows that $g(\zeta_{0})=0\Longrightarrow
g^{(k)}(\zeta_{0})=1$. Now set

$G_{n}(\zeta)=\frac{f_{n}(\rho_{n}\zeta)}{\rho^{k+\ell}_{n}}.$

From (18) and (21) we have

(22) $G_{n}(\zeta)\Longrightarrow
G(\zeta)=g(\zeta-\alpha)\zeta^{\ell}b(0)\quad\text{on}\quad\mathbb{C}.$

Indeed,

$\frac{f_{n}(\rho_{n}\zeta)}{\rho_{n}^{k+\ell}}=\frac{f_{n}(\rho_{n}\zeta)}{\rho^{k}_{n}h(\rho_{n}\zeta)}\cdot\frac{h(\rho_{n}\zeta)}{\rho_{n}^{\ell}}=\frac{f_{n}\left(z_{n}+\rho_{n}\left(\zeta-\frac{\displaystyle
z_{n}}{\displaystyle\rho_{n}}\right)\right)}{\rho^{k}_{n}h\left(z_{n}+\rho_{n}\left(\zeta-\frac{\displaystyle
z_{n}}{\displaystyle\rho_{n}}\right)\right)}\frac{(\rho_{n}\zeta)^{\ell}b(\rho_{n}\zeta)}{\rho_{n}^{\ell}}$

(cf. [12, p. 7]). Since $g$ has a pole of order $\ell$ at $\zeta=-\alpha$
(here we use the fact that for every $n$, $h$ has no common zeros with
$f_{n}$) and since $\\{G_{n}\\}$ are analytic, we have

(23) $G(0)\neq 0,\ \infty.$

We now consider several subcases, depending on the nature of $G$.

Case (BI) $G$ is a polynomial.

Since $\\{f_{n}\\}$ is not normal at $z=0$, there exist (after renumbering) a
sequence $z^{\ast}_{n}\to 0$ such that

(24) $f_{n}(z^{\ast}_{n})=0.$

Otherwise, there is some $\delta^{\prime}$, $0<\delta^{\prime}<\delta$ such
that (before renumbering) $f_{n}(z)\neq 0$ in $\Delta(0,\delta^{\prime})$, and
since $f_{n}(z)\Rightarrow\infty$ on $\Delta^{\prime}(0,\delta)$ we would have
by the minimum principle that $f_{n}(z)\Rightarrow\infty$ on
$\Delta(0,\delta)$, a contradiction to the non-normality of $\\{f_{n}\\}$ at
$z=0$. We have that all the zeros of $g$ are of multiplicity exactly $k$. Then
by (22) and (23), it follows that all the zeros of $G$ are also of
multiplicity exactly $k$. We consider now two possibilities.

Case (BI1) $\operatorname{deg}(G)=0$.

We can assume that $z^{\ast}_{n}$ from (24) is the closest zero of $f_{n}$ to
the origin. Then we have

(25)
$\frac{f_{n}(\rho_{n}\zeta)}{\rho^{k+\ell}_{n}b(\rho_{n}\zeta)}\Longrightarrow\frac{G(0)}{b(0)}\quad\text{on}\quad\mathbb{C}.$

By (25) we have

(26) $\frac{z^{\ast}_{n}}{\rho_{n}}\to\infty.$

Define $t_{n}(\zeta)=f_{n}(z^{\ast}_{n}\zeta)/\left(z^{\ast
k+\ell}_{n}b(z^{\ast}_{n}\zeta)\right)$. We want to show that
$\\{t_{n}(\zeta)\\}$ is normal in $\mathbb{C}^{\ast}$. For this purpose set
$\tilde{t}_{n}(\zeta)=f_{n}(z^{\ast}_{n}\zeta)/z^{\ast k+\ell}_{n}$. Since
$b(0)\neq 0$, $\infty$ and $z^{\ast}_{n}\to 0$, the normality of $\\{t_{n}\\}$
is equivalent to the normality of $\\{\tilde{t}_{n}\\}$, and the latter
follows by Lemma 7. Now, if $\\{t_{n}\\}$ is not normal at $\zeta=0$, then we
can write (after renumbering) $t_{n}(\zeta)\Rightarrow\infty$ on
$\mathbb{C}^{\ast}$; but $t_{n}(1)=0$, so this is not possible. Hence
$\\{t_{n}(\zeta)\\}$ is normal at $\zeta=0$. By (25) and (26), $t_{n}(0)\to 0$
as $n\to\infty$; and thus since $t_{n}(\zeta)\neq 0$ in $\Delta(0,1/2),$ we
get by Hurwitz’s Theorem that $t_{n}(\zeta)\Rightarrow 0$ on $\mathbb{C}$. But
$t_{n}(1)=0$; so by assumption (b) of Theorem 2, we get that
$t^{(k)}_{n}(1)=1$, a contradiction.

Case (BI2) $G^{(k)}\equiv b(0)\zeta^{\ell}$.

Then we have $G^{(k-1)}(\zeta)=\frac{\displaystyle
b(0)\zeta^{\ell+1}}{\displaystyle\ell+1}+C$ and
$G^{(k-2)}(\zeta)=\frac{\displaystyle
b(0)\zeta^{\ell+2}}{\displaystyle(\ell+1)(\ell+2)}+C\zeta+D$, where $C$ and
$D$ are two constants. Since all zeros of $G$ have multiplicity exactly $k$,
then for any zero $\widehat{\zeta}$ of $G$, we have
$G^{(k-2)}(\widehat{\zeta})=G^{(k-1)}(\widehat{\zeta})=0$. So

(27)
$\frac{\displaystyle\widehat{\zeta}^{\ell+1}}{\displaystyle\ell+1}+C=0,\quad\text{and}\quad\frac{\displaystyle\widehat{\zeta}^{\ell+2}}{\displaystyle(\ell+1)(\ell+2)}+C\widehat{\zeta}+D=0.$

By calculation, we have
$\frac{\displaystyle(\ell+1)C}{\displaystyle\ell+2}\widehat{\zeta}=-D$. If
$CD=0$, then by (27), $\widehat{\zeta}=0$, a contradiction. If $CD\neq 0$,
then $\widehat{\zeta}=-\frac{\displaystyle(\ell+2)D}{\displaystyle(\ell+1)C}$,
which implies that $G$ has only one zero $\zeta_{0}$, and then

$G(\zeta)=\frac{\displaystyle
b(0)\zeta_{0}^{\ell}(\zeta-\zeta_{0})^{k}}{\displaystyle k!}.$

This contradicts $G^{(k)}\equiv b(0)\zeta^{\ell}$.

Case (BI3) $G$ is a nonconstant polynomial and $G^{(k)}\not\equiv
b(0)\zeta^{\ell}$.

Since all zeros of $G$ have multiplicity exactly $k$, we may assume that

$G=A\prod\limits_{j=1}^{t}(\zeta-\zeta_{j})^{k}.$

where $A\neq 0$ is a constant and $\zeta_{j}\neq 0$, $j=1,2,\cdots,t$.

######  Claim 4.

$G(\zeta)=0\Longrightarrow G^{(k)}(\zeta)=b(0)\zeta^{\ell}\Longrightarrow
G^{(k+1)}(\zeta)=0.$

###### Proof.

Suppose first that $G(\zeta_{0})=0$. Then there exists a sequence,
$\zeta_{n}\to\zeta_{0}$, such that $f_{n}(\rho_{n}\zeta_{n})=0$, and thus
$f_{n}^{(k)}(\rho_{n}\zeta_{n})=(\rho_{n}\zeta_{n})^{\ell}b(\rho_{n}\zeta_{n})$,
that is, $\frac{\displaystyle
f_{n}^{(k)}(\rho_{n}\zeta_{n})}{\displaystyle\rho^{\ell}_{n}}=\zeta^{\ell}_{n}b(\rho_{n}\zeta_{n})$.
In the last equation, the left hand side tends to $\zeta^{\ell}_{0}b(0)$ as
$n\to\infty$. This proves the first part of the Claim.

Suppose now that $G^{(k)}(\zeta_{0})=b(0)\zeta^{\ell}_{0}$. Since
$G^{(k)}(\zeta)\not\equiv b(0)\zeta^{\ell}$, there exists a sequence
$\zeta_{n}\to\zeta_{0}$, such that $\frac{\displaystyle
f_{n}^{(k)}(\rho_{n}\zeta_{n})}{\displaystyle\rho^{\ell}_{n}}=\zeta^{\ell}_{n}b(\rho_{n}\zeta_{n})$,
that is,
$f_{n}^{(k)}(\rho_{n}\zeta_{n})=(\rho_{n}\zeta_{n})^{\ell}b(\rho_{n}\zeta_{n})$,
and thus $f_{n}^{(k+1)}(\rho_{n}\zeta_{n})=0$. Since $\frac{\displaystyle
f^{(k+1)}_{n}(\rho_{n}\zeta)}{\displaystyle\rho^{\ell-1}_{n}}\Rightarrow
G^{(k+1)}(\zeta)$, we deduce that $G^{(k+1)}(\zeta_{0})=0$, and this completes
the proof of the Claim. ∎

It follows from Claim 4 that $G^{(k+1)}(\zeta_{j})=0$, for $1\leq j\leq t$.

If $t\geq 2$, we know that for every $1\leq j\leq t$,

$\displaystyle G^{(k+1)}(\zeta)$
$\displaystyle=A\left[\prod\limits_{j=1}^{t}(\zeta-\zeta_{j})^{k}\right]^{(k+1)}$
$\displaystyle=A\left\\{\sum\limits_{\mu=0}^{k+1}\binom{k+1}{\mu}\left[(\zeta-\zeta_{j})^{k}\right]^{(k+1-\mu)}\left[\prod\limits_{i=1,i\neq
j}^{t}(\zeta-\zeta_{i})^{k}\right]^{(\mu)}\right\\}$
$\displaystyle=A\left\\{(k+1)k!\left[\prod\limits_{i=1,i\neq
j}^{t}(\zeta-\zeta_{i})^{k}\right]^{\prime}+(\zeta-\zeta_{j})P_{j}(\zeta)\right\\},$

where $P_{j}$ is a polynomial. Thus, by Claim 4 we have

(28) $\left[\prod\limits_{i=1,i\neq
j}^{t}(\zeta-\zeta_{i})^{k}\right]^{\prime}\Bigg{|}_{\zeta_{j}}=0,\quad 1\leq
j\leq t.$

This means that for every $1\leq j\leq t$,

$\sum\limits_{i=1\atop i\neq
j}^{t}(\zeta-\zeta_{j})^{k-1}\prod\limits_{\ell=1\atop\ell\neq
i,j}^{t}(\zeta-\zeta_{\ell})^{k}\Bigg{|}_{\zeta_{j}}=0.$

Dividing in $\prod\limits_{\ell\neq j}(\zeta_{j}-\zeta_{\ell})^{k-1}$ gives

$\sum\limits_{i=1\atop i\neq j}^{t}\prod\limits_{\ell=1\atop\ell\neq
i,j}^{t}(\zeta_{j}-\zeta_{\ell})=0.$

Thus $T^{\prime\prime}(\zeta_{j})=0$ for $1\leq j\leq t$, where
$T(\zeta)=\prod\limits_{i=1}^{t}(\zeta-\zeta_{i})$.

Now, if $t\geq 3$, then $T^{\prime\prime}$ is of degree $t-2$, and vanishes at
$t$ different points, a contradiction. If $t=2$, we get from (28) that
$\left[(\zeta-\zeta_{2})^{k}\right]^{\prime}\Bigg{|}_{\zeta_{1}}=0$ and this
is also a contradiction. So $t=1$ and $G$ has only one zero $\zeta_{0}\
(\zeta_{0}\neq 0)$, which means that $G(\zeta)=\frac{\displaystyle
b(0)\zeta_{0}^{\ell}(\zeta-\zeta_{0})^{k}}{\displaystyle k!}.$

By Hurwitz’s Theorem, there exists a sequence $\zeta_{n,0}\to\zeta_{0}$, such
that $G_{n}(\zeta_{n,0})=0$. If there exists $\delta^{\prime}$,
$0<\delta^{\prime}<\delta$, such that for every $n$ (after renumbering),
$f_{n}(z)$ has only one zero $z_{n,0}=\rho_{n}\zeta_{n,0}$ in
$\Delta(0,\delta^{\prime})$.

Set

$H_{n}(z)=\frac{f_{n}(z)}{(z-z_{n,0})^{k}}.$

Since $H_{n}(z)$ is a nonvanishing holomorphic function in
$\Delta(0,\delta^{\prime})$ and $H_{n}(z)\Rightarrow\infty$ on
$\Delta^{\prime}(0,\delta)$, we can deduce as before by the minimum principle
that $H_{n}(z)\Rightarrow\infty$ on $\Delta(0,\delta^{\prime})$. But

$H_{n}(2z_{n,0})=\frac{f_{n}(2z_{n,0})}{z^{k}_{n,0}}=\frac{\displaystyle\rho^{\ell}_{n}G_{n}(2\zeta_{n,0})}{\displaystyle\zeta^{k}_{n,0}}\to
0,$

a contradiction. Thus, we can assume, after renumbering, that for every
$\delta^{\prime}>0$, $f_{n}$ has at least two zeros in
$\Delta(0,\delta^{\prime})$ for large enough $n$. Thus, there exists another
sequence of points $z_{n,1}=\rho_{n}\zeta_{n,1}$, tending to zero, where
$z_{n,1}$ is also a zero of $f_{n}(z)$ and $\zeta_{n,1}\to\infty$, as
$n\to\infty$. We can also assume that $z_{n,1}$ is the closest zero to the
origin of $f_{n}$, except $z_{n,0}$. Now set $c_{n}=z_{n,0}/z_{n,1}$ and
define $K_{n}(\zeta)=f_{n}(z_{n,1}\zeta)/z^{k+\ell}_{n,1}$. By Lemma 7,
$\\{K_{n}(\zeta)\\}$ is normal in $\mathbb{C}^{\ast}$. Now, if $\\{K_{n}\\}$
is normal at $\zeta=0$, then after renumbering we can assume that

$K_{n}(\zeta)\Longrightarrow K(\zeta)\quad\text{on}\quad\mathbb{C}.$

If $K(\zeta)\not\equiv$ const., then consider

$L_{n}(\zeta):=\frac{K_{n}(\zeta)}{(\zeta-c_{n})^{k}}.$

Since $c_{n}\underset{n\to\infty}{\longrightarrow}0$, then the sequence
$\\{L_{n}\\}^{\infty}_{1}$ is normal in $\mathbb{C}^{\ast}$. It is also normal
at $\zeta=0$. Indeed, $K_{n}(c_{n})=0$ (a zero of order $k$) and so $L_{n}$ is
a nonvanishing holomorphic function in $\Delta(0,1)$. Thus (after renumbering)

$L_{n}(\zeta)\Longrightarrow\frac{K(\zeta)}{\zeta^{k}}\quad\text{on}\quad\mathbb{C}.$

But

$L_{n}(0)=\frac{K_{n}(0)}{(-c_{n})^{k}}=\frac{G_{n}(0)}{\zeta^{\ell}_{n,1}(-\zeta_{n,0})^{k}}\underset{n\to\infty}{\longrightarrow}0,\quad(\text{since}\quad\zeta_{n,1}\underset{n\to\infty}{\longrightarrow}\infty),$

and $L_{n}(\zeta)\neq 0$ in $\Delta(0,1/2)$; thus $K(\zeta)/\zeta^{k}\equiv 0$
in $\mathbb{C}$, a contradiction. If, on the other hand, $K(\zeta)\equiv$
const., then $K(\zeta)\equiv 0$ and $K^{(k)}(1)=0$. But
$K^{(k)}(1)=\lim\limits_{n\to\infty}K^{(k)}_{n}(1)=\lim\limits_{n\to\infty}\frac{\displaystyle
f^{(k)}_{n}(z_{n,1})}{\displaystyle
z_{n,1}^{\ell}}=\lim\limits_{n\to\infty}\frac{\displaystyle
h(z_{n,1})}{\displaystyle
z_{n,1}^{\ell}}=\lim\limits_{n\to\infty}b(z_{n,1})=b(0)$, a contradiction.
Hence we can deduce that $\\{K_{n}\\}$ is not normal at $\zeta=0$, and since
$K_{n}(\zeta)$ is holomorphic in $\Delta$, then

$K_{n}(\zeta)\Longrightarrow\infty\quad\text{on}\quad\mathbb{C}^{\ast}.$

But $K_{n}(1)=0$, a contradiction.

Case (BII) $G(\zeta)$ is a transcendental entire function.

Consider the family

$\mathcal{F}(G)=\left\\{t_{n}(z):=\frac{G(2^{n}z)}{2^{n(k+\ell)}}:n\in\mathbb{N}\right\\}.$

By Claim 4, we deduce

  1. (i)

$t_{n}(z)=0\Longrightarrow t^{(k)}_{n}(z)=z^{\ell}$; and

  2. (ii)

$t^{(k)}_{n}(z)=z^{\ell}\Longrightarrow t^{(k+1)}_{n}(z)=0$.

We then get by Theorem CFZ2 that $\mathcal{F}(G)$ is normal in
$\mathbb{C}^{\ast}$. Thus there exists $M>0$ such that for every $z\in
R_{1,2}:=\\{z:1\leq|z|\leq 2\\}$

$t^{\\#}_{n}(z)=\frac{2^{n(k+\ell+1)}|G^{\prime}(2^{n}z)|}{2^{2n(k+\ell)}+|G(2^{n}z)|^{2}}\leq
M.$

Set $r(\zeta):=G(\zeta)/\zeta^{k+\ell}$. Then $r$ is a transcendental
meromorphic function, whose only pole is $\zeta=0$. For every $\zeta$,
$|\zeta|\geq 2$ there exists $n\geq 1$ and $z\in R_{1,2}$, such that

(29) $\zeta=2^{n}z.$

Calculation gives

$r^{\sharp}(\zeta)=\frac{|G^{\prime}(\zeta)\zeta^{k+\ell}-(k+\ell)\zeta^{k+\ell-1}G(\zeta)|}{|\zeta|^{2(k+\ell)}+|G(\zeta)|^{2}}.$

Thus, if $|\zeta|\geq 2$ satisfies (29) then

(30) $\displaystyle|\zeta r^{\sharp}(\zeta)|$
$\displaystyle=|2^{n}z|\frac{|G^{\prime}(2^{n}z)(2^{n}z)^{k+\ell}-(k+\ell)(2^{n}z)^{k+\ell-1}G(2^{n}z)|}{|2^{n}z|^{2(k+\ell)}+|G(2^{n}z)|^{2}}$
$\displaystyle\leq\frac{2^{k+\ell+1}\cdot
2^{n(k+\ell+1)}|G^{\prime}(2^{n}z)|}{2^{2n(k+\ell)}+|G(2^{n}z)|^{2}}+\frac{(k+\ell)2^{(n+1)(k+\ell)}|G(2^{n}z)|}{2^{2n(k+\ell)}+|G(2^{n}z)|^{2}}.$

By separating into two cases, depending on $|G(2^{n}z)|>2^{(n+1)(k+\ell)}$ or
$|G(2^{n}z)|\leq 2^{(n+1)(k+\ell)}$, we see that the last expression in (30)
is less or equal to

$2^{k+\ell+1}t^{\sharp}_{n}(z)+(k+\ell)2^{2(k+\ell)}.$

Thus, to every $|\zeta|\geq 2$,

$|\zeta r^{\sharp}(\zeta)|\leq M\cdot 2^{k+\ell+1}+(k+\ell)2^{2(k+\ell)}.$

But, according to Theorem B,
$\varlimsup\limits_{\zeta\to\infty}|\zeta|r^{\sharp}(\zeta)=\infty$, and we
thus have a contradiction (cf. [3, pp. 19-21]). Theorem 2 is proved. $\square$

##  5\. Proof of Theorem 3

By Theorem CFZ3, $\mathcal{F}$ is normal at every point $z_{0}\in D$ at which
$h(z_{0})\neq 0$ (so that $\mathcal{F}$ is quasinormal in $D$). Consider
$z_{0}\in D$ such that $h(z_{0})=0$. Without loss of generality, we can assume
that $z_{0}=0$, and then $h(z)=z^{\ell}b(z)$, where $\ell(\geq 1)$ is an
integer, $b(z)\neq 0$ is an analytic function in $\Delta(0,\delta)$. We take a
subsequence $\\{f_{n}\\}^{\infty}_{1}\subset\mathcal{F}$, and we only need to
prove that $\\{f_{n}\\}$ is not normal at $z=0.$

Define $\mathcal{F}_{2}=\left\\{F=\frac{\displaystyle f_{n}}{\displaystyle
h}:n\in\mathbb{N}\right\\}.$ It is enough to prove that $\mathcal{F}_{2}$ is
normal in $\Delta(0,\delta).$ Suppose to the contrary that $\mathcal{F}_{2}$
is not normal at $z=0$. By Lemma 1 and the assumptions of Theorem 3, there
exist (after renumbering) points $z_{n}\to 0$, $\rho_{n}\to 0^{+}$ and a
nonconstant meromorphic function on $\mathbb{C}$, $g(\zeta)$ such that

(31)
$g_{n}(\zeta)=\frac{F_{n}(z_{n}+\rho_{n}\zeta)}{\rho^{2}_{n}}=\frac{f_{n}(z_{n}+\rho_{n}\zeta)}{\rho^{2}_{n}h(z_{n}+\rho_{n}\zeta)}\overset{\chi}{\Longrightarrow}g(\zeta)\quad\text{on}\quad\mathbb{C},$

all of whose zeros are multiple and

(32) $\text{for every}\quad\zeta\in\mathbb{C},\quad g^{\sharp}(\zeta)\leq
g^{\sharp}(0)=2A+1,$

where $A>1$ is a constant. After renumbering we can assume that
$\\{z_{n}/\rho_{n}\\}^{\infty}_{n=1}$ converges. We separate now into two
cases.

Case (A) $\frac{\displaystyle z_{n}}{\displaystyle\rho_{n}}\to\infty$.

Similar to the proof of Theorem 2, we can prove that
$g(\zeta)=0\Longrightarrow g^{\prime\prime}(\zeta)=1$ and that
$g^{\prime\prime}(\zeta)=1\Longrightarrow
g^{\prime\prime\prime}(\zeta)=g^{(s)}(\zeta)=0$. Then by Lemmas 4 and 3, we
have

$g(\zeta)=\frac{\displaystyle(\zeta-b)^{2}}{\displaystyle 2},$

for some $b\in\mathbb{C}$. Thus
$g^{\sharp}(0)=\frac{\displaystyle|b|}{\displaystyle 1+|b|^{4}/4}$ and then
$g^{\sharp}(0)\leq 1$, which contradicts (32).

Case (B)

(33) $\frac{z_{n}}{\rho_{n}}\to\alpha\in\mathbb{C}.$

As in the proof of Theorem 2, we have $g(\zeta_{0})=0\Longrightarrow
g^{\prime\prime}(\zeta_{0})=1$. Now set $G_{n}(\zeta)=\frac{\displaystyle
f_{n}(\rho_{n}\zeta)}{\displaystyle\rho^{2+\ell}_{n}}.$ From (31) and (33) we
have

$G_{n}(\zeta)\Longrightarrow
G(\zeta)=b(0)g(\zeta-\alpha)\zeta^{\ell}\quad\text{on}\quad\mathbb{C}.$

Since $g$ has a pole of order $\ell$ at $\zeta=-\alpha$, $G(0)\neq 0,\
\infty.$

We now consider several subcases, depending on the nature of $G$.

Case (BI) $G$ is a polynomial.

By a similar method of proof used in the proof of Theorem 2 (and using Lemma 8
instead of Lemma 7 in the appropriate places), we can get

$G(\zeta)=\frac{\displaystyle
b(0)\zeta_{0}^{\ell}(\zeta-\zeta_{0})^{2}}{\displaystyle 2},$

and also we can arrive at a contradiction.

Case (BII) $G(\zeta)$ is a transcendental entire function.

Consider the family

$\mathcal{F}(G)=\left\\{t_{n}(z):=\frac{G(2^{n}z)}{2^{n(2+\ell)}}:n\in\mathbb{N}\right\\}.$

We have

  1. (i)

$t_{n}(z)=0\Longrightarrow t^{\prime\prime}_{n}(z)=z^{\ell}$; and

  2. (ii)

$t^{\prime\prime}_{n}(z)=z^{\ell}\Longrightarrow
t^{\prime\prime\prime}_{n}(z)=t^{(s)}_{n}(z)=0$.

We then get by Theorem CFZ3 that $\mathcal{F}(G)$ is normal in
$\mathbb{C}^{\ast}$. Set $r(\zeta):=G(\zeta)/\zeta^{2+\ell}$, and we have
that, for every $\zeta$, $|\zeta|\geq 2,$ there exists $n\geq 1$ and $z\in
R_{1,2}$, such that

$|\zeta r^{\sharp}(\zeta)|\leq M\cdot 2^{2+\ell+1}+(2+\ell)2^{2(2+\ell)}.$

But, according to Theorem B,
$\varlimsup\limits_{\zeta\to\infty}|\zeta|r^{\sharp}(\zeta)=\infty$, and we
thus have a contradiction (cf. [3, pp. 19-21]). Theorem 3 is proved. $\square$

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