# Differential inequalities, normality and quasi-normality

Xiaojun Liu, Shahar Nevo and Xuecheng Pang  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>Xuecheng Pang, Department of Mathematics, East China
Normal University, Shanghai 200241, P.R.China<EMAIL_ADDRESS>
###### Abstract.

We prove that if $D$ is a domain in $\mathbb{C}$, $\alpha>1$ and $C>0$, then
the family $\mathcal{F}$ of functions $f$ meromorphic in $D$ such that

$\frac{|f^{\prime}(z)|}{1+|f(z)|^{\alpha}}>C\quad\text{for every }z\in D$

is normal in $D$. For $\alpha=1$, the same assumptions imply quasi-normality
but not necessarily normality.

###### Key words and phrases:

Normal family, quasi-normal family, differential inequality

###### 2010 Mathematics Subject Classification:

30A10, 30D35

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

Research of third author supported by the NNSF of China Approved No.11071074.

##  1\. Introduction

Throughout we use the following notation $D$ denotes a domain in $\mathbb{C}$.
For $z_{0}\in\mathbb{C}$ and $r>0$, $\Delta(z_{0},r)=\\{z:|z-z_{0}|<r\\}$,
$\Delta^{\prime}(z_{0},r)=\\{z:0<|z-z_{0}|<r\\}$,
$\overline{\Delta}(z_{0},r)=\\{z:|z-z_{0}|\leq r\\}$,
$\Gamma(z_{0},r)=\\{z:|z-z_{0}|=r\\}$ and
$R(z_{0},R_{1},R_{2})=\\{z:R_{1}<|z-z_{0}|<R_{2}\\}$. 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. The spherical derivative is denoted by $f^{\\#}(z)$. We
shall also use the notion of $Q_{m}-$ normality. For this recall that given a
set $E\subset D$, then the derived set of order $m$ of $E$ with respect to $D$
is defined by induction: $E^{(1)}_{D}$ is the set of accumulation points of
$E$ in $D$. $E^{(m)}_{D}=\left(E^{(m-1)}_{D}\right)^{(1)}_{D}$. A family
$\mathcal{F}$ of functions meromorphic in $D$ is said to be $Q_{m}-$normal in
$D$ if every subsequence $\\{f_{n}\\}^{\infty}_{n=1}$ of functions from
$\mathcal{F}$ has a subsequence that converges uniformly with respect to
$\chi$ on $D\backslash E$, where $E^{(m)}_{D}=\emptyset$ (Here if $m=0$, then
$\mathcal{F}$ is in fact normal family and if $m=1$, then $\mathcal{F}$ is
quasi-normal family). If, in addition there exists some $\nu\in\mathbb{N}$,
such that $E$ can always be taken to satisfy
$\left|E^{(m-1)}_{D}\right|\leq\nu$, then $\mathcal{F}$ is said to be
$Q_{m}-$normal family of order at most $\nu$.

For more about $Q_{m}-$normality see [1]. This paper deals with the meaning of
some differential inequalities. A natural point of departure is the following
famous criterion of normality due to F. Marty.

######  Marty’s Theorem.

[6, p. 75] A family $\mathcal{F}$ of functions meromorphic in a domain $D$ is
normal if and only if $\\{f^{\\#}(z):f\in\mathcal{F}\\}$ is locally uniformly
bounded in $D$.

Following Marty’s Theorem, L. Royden proved the following generalization.

######  Theorem R.

[5] Let $\mathcal{F}$ be a family of meromorphic functions in $D$, with the
property that for each compact set $K\subset D$, there is a positive
increasing function $h_{K}$, such that

$|f^{\prime}(z)|\leq h_{K}(|f(z)|)$

for all $f\in\mathcal{F}$ and $z\in K$, then $\mathcal{F}$ is normal in $D$.

This result was significantly extended further in various directions, see [3],
[7] and [9].

In [2], J. Grahl and the second author proved a counterpart to Marty’s
Theorem.

######  Theorem GN.

Let $\mathcal{F}$ be a family of functions meromorphic in $D$ and let
$\varepsilon>0$. If $f^{\\#}(z)\geq\varepsilon$ for every $f\in\mathcal{F}$
and $z\in D$, then $\mathcal{F}$ is normal in $D$.

It is equivalent to say that local uniform boundedness of the spherical
derivatives from zero implies normality.

The proof uses mainly Gu’s criterion to normality, Zalcman’s Lemma and Pang-
Zalcman Lemma. N. Steinmetz [8] gave shorter proof of Theorem GN, using the
Schwarzian derivative and some well-known facts on linear differential
equations. Here in this paper, we prove a generalization of Theorem GN (with
much simpler proof) and also, for the first time we present a differential
inequality that distinguish between normality to quasi-normality.

######  Theorem 1.

Let $0\leq\alpha<\infty$ and $C>0$. Let $\mathcal{F}_{\alpha,C}(D)$ be the
family of all meromorphic functions $f$ in $D$, such that

$\frac{|f^{\prime}(z)|}{1+|f(z)|^{\alpha}}>C\quad\text{for every }z\in D.$

The the following hold:

  1. (1)

If $\alpha>1$, then $\mathcal{F}_{\alpha,C}(D)$ is normal in $D$;

  2. (2)

If $\alpha=1$, then $\mathcal{F}_{\alpha,C}(D)$ is quasi-normal in $D$, but
not necessarily normal.

In section 2, we prove Theorem 1. In section 3, we show that
$\mathcal{F}_{1,C}(D)$ can be of infinite order and discuss the validity of
Theorem 1 for $\alpha<1$. In section 4, we discuss the reverse inequality
$\frac{\displaystyle|f^{\prime}(z)|}{\displaystyle 1+|f(z)|^{\alpha}}<C$.

##  2\. Proof of Theorem 1

We first state explicity the famous lemma of Pang and Zalcman (that was
already mentioned). Observe that this lemma is “if and only if”.

######  Lemma 1.

[4] Let $\mathcal{F}$ be a family of meromorphic functions in a domain $D$,
all of whose zeros have multiplicity at least $m$, and all of whose poles have
multiplicity at least $p$, and let $-p<\alpha<m$. Then $\mathcal{F}$ is not
normal at some $z_{0}\in D$ if and only if there exist sequences
$\\{f_{n}\\}^{\infty}_{n=1}\subset\mathcal{F}$,
$\\{z_{n}\\}^{\infty}_{n=1}\subset D$,
$\\{\rho_{n}\\}^{\infty}_{n=1}\subset(0,1)$, such that $\rho_{n}\to 0^{+}$,
$z_{n}\to z_{0}$ and

$g_{n}(\zeta):=\rho^{-\alpha}_{n}f_{n}(z_{n}+\rho_{n}\zeta)\overset{\chi}{\Rightarrow}g(\zeta)\quad\text{on}\quad\mathbb{C},$

where $g$ is a nonconstant meromorphic function in $D$.

Here the “if” direction holds if $g_{n}(\zeta)$ converges in some open set
$\Omega\subset\mathbb{C}$ to a nonconstant meromorphic function $g$ in
$\Omega$. For a full proof of this lemma see [2].

###  2.1. Proof of (1) of Theorem 1

Let $\\{f_{n}\\}^{\infty}_{n=1}$ be a sequence of functions in
$\mathcal{F}_{\alpha,C}(D)$. Let $z_{0}\in D$ and assume by negation that
$\\{f_{n}\\}$ is not normal at $z_{0}$. Suppose that there exist $r>0$, such
that each $f_{n}$ is holomorphic in $\Delta(z_{0},r)$. We take
$\beta>\frac{\displaystyle 1}{\displaystyle\alpha-1}>0$. By Lemma 1, there
exist a subsequence of $\\{f_{n}\\}$, that without loss of generality will
also be denoted by $\\{f_{n}\\}^{\infty}_{n=1}$ and sequences $\rho_{n}\to
0^{+}$, $z_{n}\to z_{0}$, such that

(1)
$g_{n}(\zeta):=\rho^{\beta}_{n}f_{n}(z_{n}+\rho_{n}\zeta)\overset{\chi}{\Rightarrow}g(\zeta)\quad\text{on}\quad\mathbb{C},$

where $g$ is a nonconstant entire function in $\mathbb{C}$.

Taking $\zeta_{0}\in\mathbb{C}$, such that

(2) $g(\zeta_{0})\neq 0,\ \infty.$

Then by (1), (2) and the value of $\beta$, we get that for large enough $n$,

$\displaystyle\rho^{\beta+1}_{n}\left|f^{\prime}_{n}(z_{n}+\rho_{n}\zeta_{0})\right|=\rho^{1+\beta-\beta\alpha}_{n}\left|\frac{f^{\prime}_{n}(z_{n}+\rho_{n}\zeta_{0})}{f^{\alpha}_{n}(z_{n}+\rho_{n}\zeta_{0})}\right||\rho^{\beta}_{n}f_{n}(z_{n}+\rho_{n}\zeta_{0})|^{\alpha}$
$\displaystyle>\rho^{1+\beta-\beta\alpha}_{n}\cdot
C\frac{|g(\zeta_{0})|^{\alpha}}{2}\underset{n\to\infty}{\longrightarrow}\infty.$

We thus got a contradiction and the holomorphic case is proven.

Suppose now that there is no $r>0$, such that for infinitely many indices $n$,
$f_{n}$ is holomorphic in $\Delta(z_{0},r)$. Hence we deduce the existence of
some subsequence of $\\{f_{n}\\}^{\infty}_{n=1}$, that without loss of
generality will also be denoted by $\\{f_{n}\\}^{\infty}_{n=1}$, and a
sequence $z_{n}\to z_{0}$, such that $f_{n}(z_{n})=\infty$ (otherwise we are
again in the holomorphic case and we are done). We can also assume (after
moving to subsequence of $\\{f_{n}\\}^{\infty}_{n=1}$…) that there exist a
sequence $\widetilde{z}_{n}\to z_{0}$, such that $f_{n}(\widetilde{z}_{n})=0$.
Indeed, otherwise, for some $\delta>0$ and large enough $n$, $f_{n}\neq 0$ in
$\Delta(z_{0},\delta)$ and $|f^{\prime}_{n}|>C$ there. Then by Gu’s criterion
we deduce that $\\{f_{n}\\}$ is normal.

######  Claim.

$\left\\{\frac{\displaystyle f_{n}}{\displaystyle
f^{\prime}_{n}}\right\\}^{\infty}_{n=1}$ is normal in $D$.

Proof of Claim. If $|f_{n}(z)|\leq 1$, then $\left|\frac{\displaystyle
f^{\prime}_{n}(z)}{\displaystyle f_{n}(z)}\right|\geq|f^{\prime}_{n}(z)|>C$.
If $|f_{n}(z)|>1$, then $\left|\frac{\displaystyle
f^{\prime}_{n}(z)}{\displaystyle
f_{n}(z)}\right|\geq\frac{\displaystyle|f^{\prime}_{n}(z)|}{\displaystyle
1+|f_{n}(z)|}>\frac{\displaystyle|f^{\prime}_{n}(z)|}{\displaystyle
1+|f_{n}(z)|^{\alpha}}>C$. Thus in any case, $\left|\frac{\displaystyle
f^{\prime}_{n}(z)}{\displaystyle f_{n}(z)}\right|>C$ for every $n$ and every
$z\in D$. Hence $\left\\{\frac{\displaystyle f^{\prime}_{n}}{\displaystyle
f_{n}}\right\\}^{\infty}_{n=1}$ is normal and so is
$\left\\{\frac{\displaystyle f_{n}}{\displaystyle
f^{\prime}_{n}}\right\\}^{\infty}_{n=1}$.

According to the claim, we can assume (after moving to subsequence of
$\\{f_{n}\\}^{\infty}_{n=1}$…) that

$\frac{f_{n}(z)}{f^{\prime}_{n}(z)}\overset{\chi}{\Rightarrow}H(z)\quad\text{in}\quad
D.$

Since $\frac{\displaystyle f_{n}}{\displaystyle f^{\prime}_{n}}$ vanish at the
zeros and at the poles of $f_{n}$, we deduce that $H$ is holomorphic in $D$.
We have

$\left(\frac{\displaystyle f_{n}}{\displaystyle
f^{\prime}_{n}}\right)^{\prime}=1-\frac{f_{n}f^{\prime\prime}_{n}}{f^{{}^{\prime}2}_{n}}.$

Thus we have

$\left(\frac{\displaystyle f_{n}}{\displaystyle
f^{\prime}_{n}}\right)^{\prime}\Bigg{|}_{z=\widetilde{z}_{n}}=1.$

At the poles $z_{n}$ of $f_{n}$, the situation is different. Each $z_{n}$ is a
pole of order $k=k_{n}$ of $f_{n}$. This means that in some neighborhood of
$z_{n}$, we have

$f_{n}(z)=\frac{a_{-k}}{(z-z_{n})^{k}}+\frac{a_{-k+1}}{(z-z_{n})^{k-1}}+\cdots\quad\quad(a_{-k}\neq
0).$

Thus

$f^{\prime}_{n}(z)=\frac{-ka_{-k}}{(z-z_{n})^{k+1}}+\cdots,\quad\text{and}\quad
f^{\prime\prime}_{n}(z)=\frac{k(k+1)a_{-k}}{(z-z_{n})^{k+2}}+\cdots.$

We then get that

$\frac{f_{n}f^{\prime\prime}_{n}}{f^{\prime
2}_{n}}\Bigg{|}_{z=z_{n}}=\frac{k(k+1)a^{2}_{-k}}{(ka_{-k})^{2}}=\frac{k+1}{k}=1+\frac{1}{k},$

and so

$\left(\frac{\displaystyle f_{n}}{\displaystyle
f^{\prime}_{n}}\right)^{\prime}\Bigg{|}_{z=z_{n}}=1-\left(1+\frac{1}{k}\right)=-\frac{1}{k}.$

Since $z_{n}\to z_{0}$ and also $\widetilde{z}_{n}\to z_{0}$, we get a
contradiction to any possible value of $H^{\prime}(0)$. This completes the
proof of (1).

###  2.2. Proof of (2) of Theorem 1

The family $\\{nz:\ n\in\mathbb{N}\\}$ which is not normal at $z=0$, shows
that local uniform boundedness of
$\left\\{\frac{\displaystyle|f^{\prime}|}{\displaystyle 1+|f|}:\
f\in\mathcal{F}\right\\}$ does not imply in general normality. In order to
prove quasi-normality, observe first that for every
$f\in\mathcal{F}_{1,C}(D)$, we have $\left|\frac{\displaystyle
f^{\prime}}{\displaystyle f}\right|>C$ and also $|f^{\prime}|>C$. Thus both
$\\{f^{\prime}:\ f\in\mathcal{F}_{1,C}(D)\\}$ and $\\{f^{\prime}/f:\
f\in\mathcal{F}_{1,C}(D)\\}$ are normal in $D$.

Let us take now a sequence $\\{f_{n}\\}^{\infty}_{n=1}$ of functions from
$\mathcal{F}_{1,C}(D)$. If, by negation $\\{f_{n}\\}_{n}$ is not normal at
some $\widehat{z}_{0}\in D$, then we can assume (after moving to subsequence…)
that there exist $z_{n}\to\widehat{z}_{0}$, and $\rho_{n}\to 0^{+}$ and a
nonconstant function $g$, meromorphic in $\mathbb{C}$ such that

$g_{n}(\zeta)=\rho^{-\frac{1}{2}}_{n}f_{n}(z_{n}+\rho_{n}\zeta)\overset{\chi}{\Rightarrow}g(\zeta)\quad\text{on}\quad\mathbb{C}.$

Let $P_{g}$ denotes the set of poles of $g$ in $\mathbb{C}$. If $g$ is not of
the form $g(\zeta)=a\zeta+b$, then we get by differentiation,

$g^{\prime}_{n}(\zeta)=\rho^{\frac{1}{2}}_{n}f^{\prime}_{n}(z_{n}+\rho_{n}\zeta)\Rightarrow
g^{\prime}(\zeta)\quad\text{on}\quad\mathbb{C}\backslash P_{g}.$

The derivative $g^{\prime}$ is nonconstant, and thus by Lemma 1,
$\\{f^{\prime}_{n}\\}_{n=1}^{\infty}$ is not normal at $\widehat{z}_{0}$, a
contradiction.

Thus, we must have $g(\zeta)=a\zeta+b$ $(a\neq 0)$ and by Rouche’s Theorem,
for any neighborhood $U$ of $\widehat{z}_{0}$, $f_{n}$ has for large enough
$n$ a zero in $U$. This means that we can assume (after moving to
subsequence…) that there exists a sequence $z^{\ast}_{n}\to\widehat{z}_{0}$,
such that $f_{n}(z^{\ast}_{n})=0$.

Now, suppose by negation that $\\{f_{n}\\}^{\infty}_{n=1}$ is not quasinormal
at some $z_{0}\in D$. After moving to subsequence, that will also be called
$\\{f_{n}\\}^{\infty}_{n=1}$, we can assume that there exist a sequence
$\\{z_{k}\\}^{\infty}_{k=1}$ of distinct points in $D$, such that
$z_{k}\underset{k\to\infty}{\longrightarrow}z_{0}$ and each subsequence of
$\\{f_{n}\\}^{\infty}_{n=1}$ is not normal at each $z_{k}$. According to the
previous discussion, for every $k=1,2,\cdots$, there exists $n_{k}$ and a
sequence $\\{z_{k,n}\\}^{\infty}_{n=n_{k}}$,
$z_{k,n}\underset{n\to\infty}{\longrightarrow}z_{k}$, such that
$f_{n}(z_{k,n})=0$ for every $n\geq n_{k}$.

Hence for every $\delta>0$, and for every $N\in\mathbb{N}$, $f_{n}$ has in
$\Delta(z_{0},\delta)$ at least $N$ zeros for large enough $n$. Now, since
$\left\\{\frac{\displaystyle f_{n}}{\displaystyle
f^{\prime}_{n}}:n\in\mathbb{N}\right\\}^{\infty}_{n=1}$ is normal, we can also
assume (after moving to subsequence…) that

$\frac{f_{n}(z)}{f^{\prime}_{n}(z)}\Rightarrow H(z)\quad\text{in}\quad D,$

where $H$ is holomorphic in $D$. Each zero of $f_{n}$ is also a zero of
$f_{n}/f^{\prime}_{n}$, so by the above discussion the number of zeros of
$f_{n}$ in any neighborhood of $z_{0}$ tends to $\infty$, as $n\to\infty$, and
thus we conclude that $H\equiv 0$. Hence we have

$\left(\frac{f_{n}}{f^{\prime}_{n}}\right)^{\prime}\Rightarrow
0\quad\text{in}\quad D.$

But on the other hand,

$\left(\frac{f_{n}}{f^{\prime}_{n}}\right)^{\prime}\Bigg{|}_{z=z_{k,n}}=1.$

This is a contradiction and (2) of Theorem 1 is proven.

##  3\. Some remarks

###  3.1. The order of quasi-normality of $\mathcal{F}_{1,C}(D)$

We shall show now that the order of quasi-normality of $\mathcal{F}_{1,C}(D)$
can general be large as we we like. Since we can make a linear change of the
variable, it is enough if we construct in some specific domain $D$, a sequence
$\\{f_{n}\\}^{\infty}_{n=1}$ of functions such that every subsequence of
$\\{f_{n}\\}^{\infty}_{n=1}$ has the same infinite set of points of non-
normality in $D$, and

$\inf\limits_{z\in D}\frac{|f^{\prime}_{n}(z)|}{1+|f_{n}(z)|}\geq
C\quad\text{for some}\quad C>0.$

So let $D=\left\\{z:|\operatorname{Im}z|<1,\ |z-\pi k|>\frac{\displaystyle
1}{\displaystyle 2},\ k\in\mathbb{Z}\right\\}$, and define for every $n\geq
1$, $f_{n}(z)=n\cos z$. It is obvious that every subsequence of
$\\{f_{n}\\}^{\infty}_{n=1}$ is not normal exactly at the points
$z_{k}=\frac{\displaystyle\pi}{\displaystyle 2}+\pi k$, $k\in\mathbb{Z}$. Thus
$\\{f_{n}\\}^{\infty}_{n=1}$ is quasi-normal of infinite order in $D$. Because
of the periodicity of $\cos z$, there exist some $C>0$, such that

$\frac{|f^{\prime}_{n}(z)|}{1+|f_{n}(z)|}\geq C\quad\text{for every}\ n\
\text{and for every}\ z\in D.$

Hence $\mathcal{F}_{1,C}(D)$ is quasi-normal of infinite order in $D$. We
deduce that for every domain $D$, and for every $\nu\in\mathbb{N}$ there
exists $C_{D,\nu}>0,$ such that $\mathcal{F}_{1,C_{D,\nu}}(D)$ is quasi-normal
in $D$, but not quasi-normal of order at most $\nu$.

###  3.2. The case $0\leq\alpha<1$

In this case for every bounded domain $D$ and every $C>0$,
$\mathcal{F}_{\alpha,C}(D)$ has no degree of normality. To be more precise we
have the following theorem.

######  Theorem 2.

Let $0\leq\alpha<1$, $m\geq 0$, $C>0$ and $D$ a bounded domain in $D$. Then
$\mathcal{F}_{\alpha,C}(D)$ is not $Q_{m}-$normal in $D$.

###### Proof.

For a given $0\leq\alpha<1$, let us first prove the theorem for some specific
domain. Let $1<\varepsilon<3^{\frac{1-\alpha}{1+\alpha}}$. Consider the
polynomial functions $P_{n}(z)=z^{n}-3^{n}$ defined on the ring
$D_{\varepsilon}:=R\left(0,\frac{\displaystyle
3}{\displaystyle\varepsilon},3\varepsilon\right)$. Clearly every subsequence
of $\\{P_{n}\\}^{\infty}_{n=1}$ is not normal exactly at any point of
$\Gamma(0,3)$ ($\Gamma(0,3)$ is of power $\aleph$ and of course
$\left(\Gamma(0,3)\right)^{(m)}_{D_{\varepsilon}}=\Gamma(0,3)$ for every
$m\geq 1$).

######  Claim.

$\inf\limits_{z\in
D_{\varepsilon}}\frac{\displaystyle|P^{\prime}_{n}(z)|}{\displaystyle
1+|P_{n}(z)|^{\alpha}}\underset{n\to\infty}{\longrightarrow}\infty$.

Proof of Claim. For every $z\in D_{\varepsilon}$, we have

$\frac{|P^{\prime}_{n}(z)|}{1+|P_{n}(z)|^{\alpha}}=\frac{n|z|^{n-1}}{1+|z^{n}-3^{n}|^{\alpha}}>\frac{n\cdot\left(\frac{3}{\varepsilon}\right)^{n}\cdot\frac{\varepsilon}{3}}{1+(2\cdot(3\varepsilon)^{n})^{\alpha}}>\frac{n\cdot\left(\frac{3}{\varepsilon}\right)^{n}\cdot\frac{\varepsilon}{3}}{2(2\cdot(3\varepsilon)^{n})^{\alpha}}=\frac{n\varepsilon}{6\cdot
2^{1+\alpha}}\left(\frac{3^{1-\alpha}}{3^{1+\alpha}}\right)^{n}.$

Since $\varepsilon^{1+\alpha}<3^{1-\alpha}$, the last expression tends to
$\infty$, as $n\to\infty$, and this proves the claim.

Now, give $C>0$, we have by the claim that there exists $N$, such that

$\inf\limits_{z\in
D_{\varepsilon}}\frac{\displaystyle|P^{\prime}_{n}(z)|}{\displaystyle
1+|P_{n}(z)|^{\alpha}}>C\quad\text{for}\quad n\geq N,$

and thus
$\\{P_{n}\\}^{\infty}_{n=N}\subset\mathcal{F}_{\alpha,C}(D_{\varepsilon})$.
Since $\\{P_{n}\\}^{\infty}_{n=N}$ is not $Q_{m}-$normal in $D_{\varepsilon}$,
we proved the theorem for $D=D_{\varepsilon}$.

Now, let $D$ be some bounded domain. There is a ring $R(z_{0},R_{1},R_{2})$,
together with a linear transformation $\varphi$, $\varphi(z)=az+b$, such that
$\varphi:$ $R(z_{0},R_{1},R_{2})\longrightarrow D_{\varepsilon}$ is one to one
and onto (that is, $R(z_{0},R_{1},R_{2})$ and $D_{\varepsilon}$ are
conformally equivalent) and such that $\varphi^{-1}(\Gamma(0,3))\cap D$
contain an arc of a circle. Every subsequence of
$\\{P_{n}\circ\varphi\\}^{\infty}_{n=1}$ is not $Q_{m}-$normal in $D$, for
every $m\geq 1$. Also for every $C>0$, there exists $N$, such that
$\\{P_{n}\circ\varphi\\}^{\infty}_{n=N}$ is contained in
$\mathcal{F}_{\alpha,C}(D)$ and thus also $\mathcal{F}_{\alpha,C}(D)$ is not
$Q_{m}-$normal in $D$ for every $m\geq 1$. This completes the proof of the
theorem. ∎

###  3.3. The case $\alpha<0$

Consider the family $\mathcal{F}_{\alpha,C}(D)$. If $\alpha<0$ and $f(z)=0$,
then $|f(z)|^{\alpha}$ is not well-defined, so if we require in addition that
$f\neq 0$, then since $|f^{\prime}|>C$, we get by Gu’s criterion that
$\mathcal{F}_{\alpha,C}(D)$ is normal.

If we permit that $f(z)=0$, then consider $z_{0}$, such that $f(z_{0})=0$, we
have

$\lim\limits_{z\to z_{0}}\frac{|f^{\prime}(z)|}{1+|f(z)|^{\alpha}}=0,$

and so the condition

$\frac{|f^{\prime}(z)|}{1+|f(z)|^{\alpha}}>C$

cannot be satisfied.

##  4\. The reverse inequality $\frac{\displaystyle|f^{\prime}|}{\displaystyle
1+|f|^{\alpha}}<C$

Let $\alpha>0$, and let $\mathcal{F}$ be a family of functions meromorphic in
a domain $D$. By Theorem R, if
$\mathcal{F}_{\alpha}:=\left\\{\frac{\displaystyle f^{\prime}}{\displaystyle
1+|f|^{\alpha}}:f\in\mathcal{F}\right\\}$ is locally uniformly bounded in $D$,
then $\mathcal{F}$ is normal.

For $0\leq\alpha\leq 1$, the converse is false. Consider the family
$\mathcal{F}=\\{z^{n}:n\in\mathbb{N}\\}$, in $D=\Delta(3,1)$. Obviously,
$f_{n}(z)\Rightarrow\infty$ in $D$, but

$\frac{|f^{\prime}_{n}(z)|}{1+|f_{n}(z)|^{\alpha}}=\frac{\displaystyle
n|z|^{n-1}}{\displaystyle 1+|z|^{n\alpha}}.$

Thus, since $\alpha<1$, we get that

$\inf\limits_{z\in
D}\frac{|f^{\prime}_{n}(z)|}{1+|f_{n}(z)|^{\alpha}}\underset{n\to\infty}{\longrightarrow}\infty,$

and thus $\mathcal{F}_{\alpha}$ is not locally uniformly bounded.

For $\alpha\geq 2$, the converse holds.

Indeed, assume that $\mathcal{F}$ is normal in $D$. We have for every
$f\in\mathcal{F}$ and $z\in D$

$\frac{|f^{\prime}(z)|}{1+|f(z)|^{\alpha}}=\frac{|f^{\prime}(z)|}{1+|f(z)|^{2}}\cdot\frac{1+|f(z)|^{2}}{1+|f(z)|^{\alpha}}.$

By Marty’s Theorem, $\mathcal{F}_{2}$ is locally uniformly bounded in $D$. In
addition, $h(x)=\frac{\displaystyle 1+x^{2}}{\displaystyle 1+x^{\alpha}}$ is
bounded in $[0,+\infty)$, and there is some $M>0$, such that

$\frac{\displaystyle 1+|f(z)|^{2}}{\displaystyle 1+|f(z)|^{\alpha}}\leq M$

for every $f\in\mathcal{F}$, $z\in D$. We then deduce that
$\mathcal{F}_{\alpha}$ is locally uniformly bounded in $D$.

We are left with the case $1<\alpha<2$. We show now that for meromorphic
functions, normality does not imply local uniform boundedness, for every
$1<\alpha<2$. Take $\mathcal{F}=\left\\{\frac{\displaystyle 1}{\displaystyle
z}\right\\}$ (only a single function) in $\Delta$. We have

$\frac{|f^{\prime}(z)|}{1+|f(z)|^{\alpha}}\underset{z\to
0}{\longrightarrow}\infty.$

For holomorphic functions, we can approve the converse:

######  Theorem 3.

Let $1<\alpha<2$. Suppose that $\mathcal{F}$ is a normal family of holomorphic
functions in $D$. Then

$\mathcal{F}_{\alpha}=\left\\{\frac{|f^{\prime}(z)|}{1+|f(z)|^{\alpha}}:f\in\mathcal{F}\right\\}$

is locally uniformly bounded in $D$.

###### Proof.

Suppose to the contrary that $\mathcal{F}_{\alpha}$ is not locally uniformly
bounded in $D$. Then there exist $z_{0}\in D$, $z_{n}\to z_{0}$ and
$f_{n}\in\mathcal{F}$, such that

(3)
$\frac{|f^{\prime}_{n}(z_{n})|}{1+|f_{n}(z_{n})|^{\alpha}}\underset{n\to\infty}{\longrightarrow}\infty.$

The sequence $\\{f_{n}\\}^{\infty}_{n=1}$ has a uniform convergent subsequence
in $D$, that without loss of generality we also call
$\\{f_{n}\\}^{\infty}_{n=1}$. So we assume that

$f_{n}\Rightarrow f\quad\text{in}\quad D.$

Let us separate into two cases, according to the behavior of $f$.

Case (1) $f$ is holomorphic in $D$.

Then $f^{\prime}_{n}\Rightarrow f^{\prime}$ in $D$, and we easily get a
contradiction to (3).

Case (2) $f\equiv\infty$.

In particular, we have

$f_{n}(z_{0})\underset{n\to\infty}{\longrightarrow}\infty.$

We take $R>0$, such that $\overline{\Delta}(z_{0},R)\subset D$ and

(4) $0<\rho<R\frac{\sqrt{1+\alpha}-\sqrt{2}}{\sqrt{1+\alpha}+\sqrt{2}}.$

By Harnack’s inequality, for large enough $n$, we have for every
$z\in\overline{\Delta}(z_{0},\rho)$

(5)
$|f_{n}(z_{0})|^{\frac{R-\rho}{R+\rho}}\leq|f_{n}(z)|\leq|f_{n}(z_{0})|^{\frac{R+\rho}{R-\rho}}.$

By (4) we get that

(6) $\frac{R+\rho}{R-\rho}<\sqrt{\frac{1+\alpha}{2}}\quad\bigg{(}\text{and
thus}\quad\frac{R-\rho}{R+\rho}>\sqrt{\frac{2}{1+\alpha}}\,\bigg{)}.$

Now, by (5),(6) and Cauchy’s integral formula, we get that for every
$z\in\overline{\Delta}(z_{0},\rho/2)$ and large enough $n$,

$|f^{\prime}_{n}(z)|=\frac{1}{2\pi}\left|\displaystyle\int\limits_{|\zeta-
z_{0}|=\rho}\frac{f_{n}(\zeta)}{(\zeta-z)^{2}}d\zeta\right|\leq\frac{\rho}{(\rho/2)^{2}}\max\limits_{|\zeta-
z_{0}|=\rho}|f_{n}(\zeta)|\leq\frac{4}{\rho}|f_{n}(z_{0})|^{\sqrt{\frac{1+\alpha}{2}}}.$

Thus, by the last inequality, (4) and (5), we have for large enough $n$,

$\displaystyle\frac{|f^{\prime}_{n}(z_{n})|}{1+|f_{n}(z_{n})|^{\alpha}}$
$\displaystyle\leq\frac{\displaystyle\frac{4}{\rho}|f_{n}(z_{0})|^{\sqrt{\frac{1+\alpha}{2}}}}{\displaystyle
1+|f_{n}(z_{0})|^{\frac{\alpha}{\sqrt{(1+\alpha)/2}}}}\leq\frac{4}{\rho}|f_{n}(z_{0})|^{\sqrt{\frac{1+\alpha}{2}}-\frac{\alpha}{\sqrt{(1+\alpha)/2}}}$
$\displaystyle=\frac{4}{\rho}|f_{n}(z_{0})|^{\frac{1-\alpha}{2}\big{/}\sqrt{(1+\alpha)/2}}\underset{n\to\infty}{\longrightarrow}0.$

This is a contradiction to (3) and thus the Theorem follows. ∎

## References

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