††footnotetext:  File: Ahamed-Allu-Halder-P4.tex, printed: 2024-8-27, 17.02

# Bohr inequalities for unimodular bounded functions on simply connected
domains

Molla Basir Ahamed  Molla Basir Ahamed, School of Basic Science, Indian
Institute of Technology Bhubaneswar, Bhubaneswar-752050, Odisha, India.
<EMAIL_ADDRESS>,  Vasudevarao Allu  Vasudevarao Allu, School of Basic
Science, Indian Institute of Technology Bhubaneswar, Bhubaneswar-752050,
Odisha, India<EMAIL_ADDRESS>and  Himadri Halder  Himadri Halder, School
of Basic Science, Indian Institute of Technology Bhubaneswar,
Bhubaneswar-752050, Odisha, India<EMAIL_ADDRESS>
###### Abstract.

Let $\mathcal{H}(\mathbb{D})$ be the class of analytic functions in the unit
disk $\mathbb{D}:=\\{z\in\mathbb{C}:|z|<1\\}$. The classical Bohr’s inequality
states that if a power series $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$ converges
in $\mathbb{D}$ and $|f(z)|<1$ for $z\in\mathbb{D}$, then

$\sum_{n=0}^{\infty}|a_{n}|r^{n}\leq 1\;\;\mbox{for}\;\;r\leq\frac{1}{3}$

and the constant $1/3$ cannot be improved. The constant $1/3$ is known as Bohr
radius. In this paper, we study Bohr phenomenon for analytic as well as
harmonic mappings on simply connected domains. We prove several sharp results
on improved Bohr radius for analytic functions as well as for harmonic
mappings on simply connected domains.

###### Key words and phrases:

Unimodular bounded functions, analytic functions, harmonic mappings; Simply
connected domains, improved Bohr inequality, Bohr radius.

###### AMS Subject Classification: Mathematics Subject Classification:

Primary 30C45, 30C50, 30C80

##  1\. Introduction

Let $\Omega\subseteq\mathbb{C}$ be a simply connected domain containing the
unit disk $\mathbb{D}:=\\{z\in\mathbb{C}:|z|<1\\}$ and $\mathcal{B}(\Omega)$
be the class of analytic functions $f$ in $\Omega$ such that $|f(z)|<1$ for
all $z\in\Omega$. The Bohr radius $B_{\Omega}$ for the class
$\mathcal{B}(\Omega)$ is defined by

$B_{\Omega}:=\sup\bigg{\\{}r\in(0,1):\sum_{n=0}^{\infty}|a_{n}|r^{n}\leq
1\;\text{for
all}\;f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}\in\mathcal{B}(\Omega),\;z\in\mathbb{D}\bigg{\\}}.$

For $f\in B_{\Omega}$ given by $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$ in
$\mathbb{D}$, its associated majorant series defined by
$M_{f}(r):=\sum_{n=0}^{\infty}|a_{n}|r^{n}$. For $\Omega=\mathbb{D}$, it is
proved that $B_{\mathbb{D}}=1/3$, which is the classical Bohr radius. The
classical result of Bohr [20], which inspired a lot in the recent years is in
the following form after subsequent improvements due to M. Riesz, I. Schur,
and N. Wiener.

######  Theorem 1.1.

[20] If $f\in\mathcal{B}(\mathbb{D})$ and $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$
for $z\in\mathbb{D}$, then

(1.2) $\sum_{n=0}^{\infty}|a_{n}|r^{n}\leq
1\;\;\mbox{for}\;\;|z|=r\leq\frac{1}{3}.$

The radius $1/3$ cannot be improved.

It is important to note that if $|f(z)|\leq 1$ in $\mathbb{D}$ and
$|f(z^{*})|=1$ for some $z^{*}$ in $\mathbb{D}$, then $f(z)$ reduces to a
unimodular constant. Thus, we restrict our attention to $|f(z)|<1$. It is
worth to point out that if $|a_{0}|$ in the equality (1.2) is replaced by
$|a_{0}|^{2}$, then the radius $1/3$ could be replaced by $1/2$. Furthermore,
if $a_{0}=0$ in Theorem 1.1, then the sharp radius can be improved to
$1/\sqrt{2}$. In a more convenient way, we can demonstrate these facts that a
minor change in the coefficients gives different sharp radius, and thus, sharp
coefficient estimates which play a vital role to obtain the sharp radius.
Therefore, proofs of these results and the inequality (1.2) relied on the
sharp coefficient inequalities which may be obtained as an application of
Pick’s invariant form of Schwarz’s lemma for $f\in\mathcal{H}(\mathbb{D})$

$|f^{\prime}(z)|\leq\frac{1-|f(z)|^{2}}{1-|z|^{2}}\;\;\mbox{for}\;\;z\in\mathbb{D}.$

In particular, $|f^{\prime}(0)|=|a_{1}|\leq 1-|f(0)|^{2}=1-|a_{0}|^{2}$ and
hence from this, the sharp inequality $|a_{n}|\leq 1-|a_{0}|^{2}$ follows for
$n\geq 1$.

Studying the Bohr inequalities in recent years become an interesting topic of
research in functions of one as well as several complex variables. The notion
of Bohr inequality has been generalized to several complex variables, to
planar harmonic mappings, to polynomials, to solutions of elliptic partial
differential equations, and more abstract settings. In $1997$, Boas and
Khavinson [19] extended the Bohr inequality (1.2) to several complex variables
by finding multidimensional Bohr radius. For more interesting aspects of Bohr
inequalities in this particular direction, we refer the reader to [6, 7, 17,
19]. For various forms of Bohr inequalities we refer to (see [25, 26]). Many
interesting multidimensional analogue of improved Bohr inequality have been
obtained by Liu et al. (see [37]). Recently, Bohr inequalities for functions
defined in a simply connected domain have been developed by Evdoridis et al.
[23] (also see [4]). For various research works on Bohr inequality as stated
above, we suggest the reader glance through the articles [5, 12, 14] and the
references therein.

Let $\mathbb{D}(a,r):=\\{z\in\mathbb{C}:|z-a|<1\\}$. For $0\leq\gamma<1$, we
consider the disk $\Omega_{\gamma}$ defined by

$\Omega_{\gamma}:=\bigg{\\{}z\in\mathbb{C}:\bigg{|}z+\frac{\gamma}{1-\gamma}\bigg{|}<\frac{1}{1-\gamma}\bigg{\\}}.$

It is easy to see that $\Omega_{\gamma}$ always contains the unit disk
$\mathbb{D}$. The following coefficient estimates for class
$\mathcal{B}(\Omega_{\gamma})$ are required to prove some of our results.

######  Lemma 1.3.

[23] For $\gamma\in[0,1)$, let

$\Omega_{\gamma}:=\bigg{\\{}z\in\mathbb{C}:\bigg{|}z+\frac{\gamma}{1-\gamma}\bigg{|}<\frac{1}{1-\gamma}\bigg{\\}},$

and let $f$ be an analytic function in $\Omega_{\gamma}$, bounded by $1$, with
the series representation $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$ in the unit
disk $\mathbb{D}.$ Then

$|a_{n}|\leq\frac{1-|a_{0}|^{2}}{1+\gamma}\quad\mbox{for}\quad n\geq 1.$

In $2010$, the notion of classical Bohr inequality
$\sum_{n=0}^{\infty}|a_{n}|r^{n}\leq 1$ has been generalized by Fournier and
Ruscheweyh [24] to the class $\mathcal{B}(\Omega_{\gamma})$. More precisely,

######  Theorem 1.4.

[24] For $0\leq\gamma<1$, let $f\in\mathcal{B}(\Omega_{\gamma})$, with
$f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$ in $\mathbb{D}$. Then,

$\sum_{n=0}^{\infty}|a_{n}|r^{n}\leq
1\;\;\text{for}\;\;r\leq\rho:=\frac{1+\gamma}{3+\gamma}.$

Moreover, $\sum_{n=0}^{\infty}|a_{n}|\rho^{n}=1$ holds for a function
$f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$ in $\mathcal{B}(\Omega_{\gamma})$ if, and
only if, $f(z)=c$ with $|c|=1$.

In $2010$, Fournier and Ruscheweyh [24] proved the following result.

######  Theorem 1.5.

[24] Let $\Omega$ be a simply connected domain which contains the unit disk
$\mathbb{D}$ and let

(1.6) $\lambda:=\lambda(\Omega)=\sup_{f\in\mathcal{B}(\Omega),\;n\geq
1}\bigg{\\{}\frac{|a_{n}|}{1-|a_{0}|^{2}}:a_{0}\neq
f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}\;\;\mbox{for}\;\;z\in\mathbb{D}\bigg{\\}}.$

Then $1/(1+2\lambda)\leq\mathcal{B}_{\Omega}$ and the equality
$\sum_{n=0}^{\infty}|a_{n}|(1/(1+2\lambda))^{n}=1$ holds for a function
$f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$ in $\mathcal{B}(\Omega)$ if, and only if,
$f\equiv c$ with $|c|=1$.

A complex-valued function $f$ in a simply connected domain
$\Omega\subseteq\mathbb{C}$ is called a harmonic in $\Omega$ if it satisfies
the Laplace equation $\Delta f=4f_{z\overline{z}}=0$. It is known that each
harmonic mapping $f$ has a canonical representation of the form
$f=h+\overline{g}$ where $h$ and $g$ are analytic in a simply connected domain
$\Omega\subseteq\mathbb{C}$ and this representation is unique up to an
additive constant. A locally univalent function $f$ is sens-preserving if its
Jacobian $J_{f}(z)>0$ in $\Omega$, where $J_{f}$ is defined by
$J_{f}(z)=|h^{\prime}(z)|^{2}-|g^{\prime}(z)|^{2}$. Every harmonic function
$f$ in $\mathbb{D}$ has the following representation

(1.7)
$f(z)=h(z)+\overline{g(z)}=\sum_{n=0}^{\infty}a_{n}z^{n}+\overline{\sum_{n=0}^{\infty}b_{n}z^{n}}\;\;\mbox{for}\;\;z\in\mathbb{D}.$

In the recent years, harmonic extensions of the classical Bohr radius have
been extensively studied by several authors and we refer the reader to glance
through the articles [22, 28, 33, 35, 36].

##  2\. Main Results

Before we state an improved version of Theorem 1.4 for class
$\mathcal{B}(\Omega_{\gamma})$, we prove the following lemma. Let $Q(w)$ be a
polynomial defined by

(2.1)
$Q(w)=c_{1}w+c_{2}w^{2}+\ldots+c_{m}w^{m}\;\;\mbox{for}\;\;c_{j}\in\mathbb{R}^{+}\;\;j=1,2,\cdots,m.$

######  Lemma 2.2.

Let $g:\mathbb{D}\rightarrow\overline{\mathbb{D}}$ be an analytic function and
$\gamma\in\mathbb{D}$ be such that
$g(z)=\sum_{n=0}^{\infty}\alpha_{n}(z-\gamma)^{n}$ for
$|z-\gamma|<1-|\gamma|$. Then

$\displaystyle\sum_{n=0}^{\infty}|\alpha_{n}|\rho^{n}+Q\left(\frac{S_{\rho}}{\pi}\right)\leq
1\;\;\text{for}\;\;\rho\leq\rho_{0}=\frac{1-|\gamma|^{2}}{3+|\gamma|},$

where $S_{\rho}$ denotes the area of the image of the disk
$\mathbb{D}(\gamma;\rho(1-|\gamma|))$ under the mapping $g$ and the non-
negative real coefficients $c_{1},c_{2},\cdots,c_{m}$ of the polynomial $Q(w)$
given by (2.1) satisfy

$8c_{1}\left(\frac{3}{8}\right)^{2}+24c_{2}\left(\frac{3}{8}\right)^{4}+\cdots+8(2m-1)c_{m}\left(\frac{3}{8}\right)^{2m}=1.$

In view of Lemma 2.2, we obtain the following improved version of Theorem 1.4.
In deed, with an additional polynomial $Q\left(S_{r(\gamma-1)}/\pi\right)$
with the majorant series of $f$, we prove that the sum is still less than or
equals to $1$ for the radius $r_{0}=(1+\gamma)/(3+\gamma)$. More precisely,

######  Theorem 2.3.

For $0\leq\gamma<1,$ let $f\in\mathcal{B}(\Omega_{\gamma})$ with
$f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$ in $\mathbb{D}$, then we have

$\sum_{n=0}^{\infty}|a_{n}|r^{n}+Q\left(\frac{S_{r(\gamma-1)}}{\pi}\right)\leq
1\;\;\mbox{for}\;\;r\leq r_{0}=\frac{1+\gamma}{3+\gamma},$

where the coefficients of $Q(w)$ connected by the relations

$8c_{1}\left(\frac{3}{8}\right)^{2}+24c_{2}\left(\frac{3}{8}\right)^{4}+\cdots+8(2m-1)c_{m}\left(\frac{3}{8}\right)^{2m}=1.$

Furthermore, the quantities $c_{1},c_{2},\cdots,c_{m}$ and
$(1+\gamma)/(3+\gamma)$ can not be improved.

Let $P(w)$ be a polynomial of degree $m$ defined by

(2.4)
$P(w)=k_{1}w+k_{2}w^{2}+\cdots+k_{m}w^{m},\;\;\mbox{where}\;\;k_{j}=\left(\frac{1+\lambda}{1+2\lambda}\right)^{2j}\;$

for $j=1,2,\cdots,m.$ Our next aim is to prove a more general version of
Theorem 2.3, where an improved Bohr inequality is obtained for the analytic
functions defined on a simply connected domain $\Omega\subseteq\mathbb{C}$.

######  Theorem 2.5.

Let $\Omega$ be a simply connected domain containing the unit disk
$\mathbb{D}$ and $f\in\mathcal{B}(\Omega)$ with
$f(z)=\displaystyle\sum_{n=0}^{\infty}a_{n}z^{n}$ for $z\in\mathbb{D}$. Then
for the polynomial $P(w)$

$\sum_{n=0}^{\infty}|a_{n}|r^{n}+P\left(\frac{S_{r}}{\pi}\right)\leq
1\;\;\mbox{for}\;\;r\leq\frac{1}{1+2\lambda},$

where $S_{r}$ denotes the area of the image of the disk $\mathbb{D}(0,r)$
under the mapping $f$ and $P(w)$ is a polynomial given by (2.4). The equality
$\sum_{n=0}^{\infty}|a_{n}|r^{n}+P\left({S_{r}}/{\pi}\right)=1$ holds for the
function $f\in\mathcal{B}(\Omega)$ if, and only if, $f$ is an unimodular
function i.e.$,f\equiv c$ with $|c|=1.$

Next, we show that if we replace $|a_{n}|$ by $|a_{n}|+\beta|a_{n}|^{2}$ in
Theorem 2.5, then we obtain the same radius $1/(1+2\lambda)$ as we obtained in
Theorem 2.5.

######  Theorem 2.6.

Let $\Omega$ be a simply connected domain containing the unit disk
$\mathbb{D}$ and $f\in\mathcal{B}(\Omega)$ with
$f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$ for $z\in\mathbb{D}$. Then for
$0\leq\beta\leq 1/\lambda$,

$|a_{0}|+\sum_{n=1}^{\infty}\bigg{(}|a_{n}|+\beta|a_{n}|^{2}\bigg{)}r^{n}\leq
1\;\;\mbox{for}\;\;r\leq\frac{1}{1+2\lambda}.$

The equality
$|a_{0}|+\sum_{n=1}^{\infty}\bigg{(}|a_{n}|+\beta|a_{n}|^{2}\bigg{)}(1/(1+2\lambda))^{n}=1$
holds for a function $f\in\mathcal{B}(\Omega)$ if, and only if, $f\equiv c$
with $|c|=1$.

By refining the coefficients of functions in the class $\mathcal{B}(\Omega)$,
we obtain the following interesting result.

######  Theorem 2.7.

Let $\Omega$ be a simply connected domain containing the unit disk
$\mathbb{D}$ and $f\in\mathcal{B}(\Omega)$ with
$f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$ for $z\in\mathbb{D}$. Then

$\sum_{n=0}^{\infty}|a_{n}|r^{n}+\left(\frac{1+\lambda}{2\lambda(1+|a_{0}|)}+\frac{2}{3}\frac{(1+\lambda)r}{1-r}\right)\sum_{n=1}^{\infty}|a_{n}|^{2}r^{2n}\leq
1\;\;\mbox{for}\;\;r\leq\frac{1}{1+2\lambda}.$

The equality

$\sum_{n=0}^{\infty}|a_{n}|\left(\frac{1}{1+2\lambda}\right)^{n}+\left(\frac{1+\lambda}{2\lambda(1+|a_{0}|)}+\frac{2}{3}\frac{(1+\lambda)}{2\lambda}\right)\sum_{n=1}^{\infty}|a_{n}|^{2}\left(\frac{1}{1+2\lambda}\right)^{2n}=1$

holds for a function $f\in\mathcal{B}(\Omega)$ if, and only if, $f$ is an
unimodular function.

######  Theorem 2.8.

Let $f=h+\overline{g}$ given by (1.7) be harmonic in $\Omega_{\gamma}$, with
$|h(z)|+|g(z)|\leq 1$ in $\Omega_{\gamma}$. Then

$|a_{0}|+\sum_{n=1}^{\infty}\bigg{(}|a_{n}|+|b_{n}|\bigg{)}r^{n}\leq
1\quad\mbox{for}\quad|z|=r\leq r_{0}:=\frac{1+\gamma}{3+\gamma}.$

######  Theorem 2.9.

Let $f=h+\overline{g}$ given by (1.7) be harmonic in $\Omega_{\gamma}$, with
$|h(z)|+|g(z)|\leq 1$ in $\Omega_{\gamma}$. Then

$\sqrt{|a_{0}|^{2}+|b_{0}|^{2}}+\sum_{n=1}^{\infty}\sqrt{|a_{n}|^{2}+|b_{n}|^{2}}r^{n}\leq
1\quad\mbox{for}\quad|z|=r\leq r_{0}:=\frac{1+\gamma}{3+\gamma}.$

##  3\. Proof of the main results

Proof of Lemma 2.2. Without loss of generality, we assume that
$\gamma\in[0,1)$. Then it is easy to see that
$z\in\mathbb{D}_{\gamma}:=\mathbb{D}(\gamma;1-\gamma)$ if, and only if,
$w=(z-\gamma)/(1-\gamma)\in\mathbb{D}.$ Then we have

$\displaystyle
g(z)=\sum_{n=0}^{\infty}\alpha_{n}(1-\gamma)^{n}\phi^{n}(z)=\sum_{n=0}^{\infty}b_{n}\phi^{n}(z):=G(\phi(z))$

for $z\in\mathbb{D}_{\gamma},$ where $b_{n}=\alpha_{n}(1-\gamma)^{n}$. A
simple computation shows that

(3.1)
$\displaystyle\frac{S_{\rho}}{\pi}=\frac{1}{\pi}\text{Area}\bigg{(}G(\mathbb{D}(0,\rho))\bigg{)}\leq(1-|b_{0}|^{2})^{2}\frac{\rho^{2}}{(1-\rho^{2})^{2}}=(1-|\alpha_{0}|^{2})^{2}\frac{\rho^{2}}{(1-\rho^{2})^{2}}.$

Therefore,

(3.2)
$\displaystyle\sum_{n=1}^{\infty}|\alpha_{n}|\rho^{n}\leq\frac{1-|\alpha_{0}|^{2}}{1+\gamma}\sum_{n=1}^{\infty}\left(\frac{\rho}{1-\gamma}\right)^{n}=\frac{1-|\alpha_{0}|^{2}}{1+\gamma}\left(\frac{\rho}{1-\gamma-\rho}\right).$

In view of (3.1) and (3.2), we obtain

$\displaystyle|\alpha_{0}|+\sum_{n=1}^{\infty}|\alpha_{n}|\rho^{n}+Q\left(\frac{S_{\rho}}{\pi}\right)$
$\displaystyle=|\alpha_{0}|+\frac{(1-|\alpha_{0}|^{2})\rho}{(1+\gamma)(1-\gamma-\rho)}+\sum_{j=0}^{m}c_{j}\bigg{(}\frac{(1-|\alpha_{0}|^{2})\rho}{(1-\rho^{2})}\bigg{)}^{2j}$
$\displaystyle=1+\Psi_{1}^{\gamma}(\rho),$

where

$\displaystyle\Psi_{1}^{\gamma}(\rho)=\frac{(1-|\alpha_{0}|^{2})\rho}{(1+\gamma)(1-\gamma-\rho)}+\sum_{j=0}^{m}c_{j}\bigg{(}\frac{(1-|\alpha_{0}|^{2})\rho}{(1-\rho^{2})}\bigg{)}^{2j}-(1-|\alpha_{0}|).$

Let $G_{m}(\rho)$ be defined by

$\displaystyle G_{m}(\rho)$
$\displaystyle=\left(\frac{c_{m-1}}{c_{m}}\right)\frac{\rho^{2m-2}}{(1-\rho^{2})^{2m-2}(1-|\alpha_{0}|^{2})^{2m-2}}+\cdots+\left(\frac{c_{m-j}}{c_{m}}\right)\frac{\rho^{2m-2j}}{(1-\rho^{2})^{2m-2j}(1-|\alpha_{0}|^{2})^{2j}}$
$\displaystyle\quad\quad+\cdots+\left(\frac{c_{1}}{c_{m}}\right)\frac{\rho^{2}}{(1-\rho^{2})^{2}(1-|\alpha_{0}|^{2})^{2m-2}}+\frac{\rho^{2m}}{(1-\rho^{2})^{2m}}.$

Then we can write $\Psi_{1}^{\gamma}(\rho)$ as

$\displaystyle\Psi_{1}^{\gamma}(\rho)$
$\displaystyle=\frac{1-|\alpha_{0}|^{2}}{2}\bigg{(}1+2c_{m}(1-|\alpha_{0}|^{2})^{2m-2}G_{m}(\rho)-\frac{2}{1+|\alpha_{0}|}$
$\displaystyle\quad\quad+\frac{1}{2\lambda(1-|\alpha_{0}|^{2})^{3}}\left(\frac{2\rho}{(1+\gamma)(1-\gamma-\rho)}-1\right)\bigg{)}.$

We suppose that $\rho\leq\rho_{0}$. Then it is easy to see that
$\Psi_{1}^{\gamma}(\rho)$ is an increasing function and hence
$\Psi_{1}^{\gamma}(\rho)\leq\Psi_{1}^{\gamma}(\rho_{0})$, where

$\displaystyle\frac{2\rho_{0}}{(1+\gamma)(1-\gamma-\rho_{0})}=1$

which is equivalent to

$\rho_{0}=\frac{1-\gamma^{2}}{3+\gamma}.$

A simple computation shows that

$\displaystyle\Psi_{1}^{\gamma}(\rho_{0})=\frac{1-|\alpha_{0}|^{2}}{2}\left(1+2F_{m}(|\alpha_{0}|)-\frac{2}{1+|\alpha_{0}|}\right):=\frac{1-|\alpha_{0}|^{2}}{2}J(|\alpha_{0}|),$

where

$\displaystyle F_{m}(|\alpha_{0}|)$
$\displaystyle=c_{m}(1-|\alpha_{0}|^{2})^{2m-1}A^{2m}(\gamma)+c_{m-1}(1-|\alpha_{0}|^{2})^{2m-3}A^{2m-1}(\gamma)+\cdots$
$\displaystyle\quad\quad+c_{m-j}(1-|\alpha_{0}|^{2})^{2m-2j-1}A^{2j-1}(\gamma)+\cdots+c_{1}(1-|\alpha_{0}|^{2})A^{2}(\gamma),$
$\displaystyle J(x)$ $\displaystyle=1+2F_{m}(x)-\frac{2}{1+x}\;\;\mbox{for}\;\
x\in[0,1]\;\;\text{and}\;\;$ $\displaystyle A(\gamma)$
$\displaystyle=\frac{(3+\gamma)(1-\gamma^{2})}{(3+\gamma)^{2}-(1-\gamma^{2})^{2}}.$

We note that $A(\gamma)>0$ for $\gamma\in[0,1)$. In order to show that
$\Psi_{1}^{\gamma}(\rho_{0})\leq 0$, it is enough to show that $J(x)\leq 0$
for $x\in[0,1]$ and $\gamma\in[0,1)$ so that $\Psi_{1}^{\gamma}(\rho_{0})\leq
0$. Further, a simple computation shows that

$\displaystyle J(0)$
$\displaystyle=2c_{m}A^{2m}(\gamma)+2c_{m-1}A^{2m-1}(\gamma)+\cdots+c_{m-j}A^{2j-1}(\gamma)+\cdots+c_{1}A^{2}(\gamma)-1$
and $\displaystyle\lim_{x\rightarrow 1^{-}}J(x)=0.$

It is easy to see that $A(\gamma)=(f_{1}\circ f_{2})(\gamma)$, where
$f_{1}(\rho)=\rho/(1-\rho^{2})$ and $f_{2}(\gamma)=(1-\gamma^{2})/(3+\gamma).$
Since $A^{\prime}(\gamma)=f^{\prime}_{1}(f_{2}(\gamma))f^{\prime}_{2}(\gamma)$
and

(3.3)
$f^{\prime}_{2}(\gamma)=-\left(\frac{\gamma^{2}+6\gamma+1}{(3+\gamma)^{2}}\right)<0,$

we show that $f_{1}(\rho)$ is an increasing function of $\rho$ in $(0,1)$, and
$f_{2}$ is a decreasing function of $\gamma$ in $[0,1)$ so that $A(\gamma)$ is
a decreasing function of $\gamma$ in $[0,1)$, with $A(0)=3/8$ and $A(1)=0$. It
can be seen that each $A^{2j}(\gamma)$ is a decreasing function on $[0,1)$,
for $j=1,2,\cdots,m$. Therefore, we have

$\displaystyle A^{2j}(\gamma)\leq
A^{2j}(0)=\left(\frac{3}{8}\right)^{2j}\;\;\mbox{for}\;\;j=1,2,\cdots,m.$

Since $x\in[0,1]$, a simple computation shows that

$\displaystyle x(1+x)^{2}A^{2}(\gamma)$ $\displaystyle\leq
4\left(\frac{3}{8}\right)^{2}$ $\displaystyle
x(1+x)^{2}(1-x^{2})^{2}A^{4}(\gamma)\vspace{0.2in}$ $\displaystyle\leq
4\left(\frac{3}{8}\right)^{4}$ $\displaystyle\vdots$ $\displaystyle
x(1+x)^{2}(1-x^{2})^{2m-2}A^{2m}(\gamma)$ $\displaystyle\leq
4\left(\frac{3}{8}\right)^{2m}.$

It is easy to see that

$\displaystyle J^{\prime}(x)$
$\displaystyle=\frac{2}{(1+x)^{2}}\bigg{(}1-2c_{1}x(1+x)^{2}A^{2}(\gamma)-6c_{2}x(1+x)^{2}(1-x^{2})^{2}A^{4}(\gamma)$
$\displaystyle\quad\quad-\cdots-2(2m-1)c_{m}x(1+x)^{2}(1-x^{2})^{2m-2}A^{2m}(\gamma)\bigg{)}$
$\displaystyle\geq\frac{2}{(1+x)^{2}}\left(1-\left(8c_{1}\left(\frac{3}{8}\right)^{2}+24c_{2}\left(\frac{3}{8}\right)^{4}+\cdots+8(2m-1)c_{m}\left(\frac{3}{8}\right)^{2m}\right)\right)$
$\displaystyle\geq
0,\;\;\;\;\;\text{if}\;\;8c_{1}\left(\frac{3}{8}\right)^{2}+24c_{2}\left(\frac{3}{8}\right)^{4}+\cdots+8(2m-1)c_{m}\left(\frac{3}{8}\right)^{2m}\leq
1.$

Therefore, $J(x)$ is an increasing function in $[0,1]$ for

$8c_{1}\left(\frac{3}{8}\right)^{2}+24c_{2}\left(\frac{3}{8}\right)^{4}+\cdots+8(2m-1)c_{m}\left(\frac{3}{8}\right)^{2m}\leq
1.$

Hence, $J(x)\leq 0$ for all $x\in[0,1]$ and $\gamma\in[0,1)$. This completes
the proof. ∎

Proof of Theorem 2.5. From (1.6), we have

(3.4) $|a_{n}|\leq\lambda(1-|a_{0}|^{2})\;\;\mbox{for}\;\;n\geq 1.$

A simple calculation using (3.4) shows that

(3.5) $\displaystyle\frac{S_{r}}{\pi}$
$\displaystyle=\displaystyle\frac{1}{\pi}\iint\limits_{|z|<r}|f^{\prime}(z)|^{2}dxdy\leq\sum_{n=1}^{\infty}n|a_{n}|^{2}r^{2n}$
$\displaystyle\leq\lambda^{2}(1-|a_{0}|^{2})^{2}\sum_{n=1}^{\infty}nr^{2n}$
$\displaystyle=\lambda^{2}(1-|a_{0}|^{2})\left(\frac{r}{1-r^{2}}\right)^{2}.$

For $r\leq 1/(1+2\lambda)$, it is easy to see that

(3.6) $\frac{r}{1-r^{2}}\leq\frac{1+2\lambda}{4(1+\lambda)\lambda}.$

In view of (3.4), (3.5) and (3.6), we obtain

$\displaystyle\sum_{n=0}^{\infty}|a_{n}|r^{n}+P\left(\frac{S_{r}}{\pi}\right)$
$\displaystyle=|a_{0}|+\sum_{n=1}^{\infty}|a_{n}|r^{n}+\sum_{j=1}^{m}\left(\frac{1+\lambda}{1+2\lambda}\right)^{2j}\left(\frac{S_{r}}{\pi}\right)^{j}$
$\displaystyle=1-\frac{(1-|a_{0}|^{2})}{16^{m}}\left(\frac{16^{m}}{1+|a_{0}|}-\frac{16^{m}}{2}-\sum_{j=1}^{m}16^{m-j}(1-|a_{0}|^{2})^{2j-1}\right)$
$\displaystyle:\;=1-\frac{(1-|a_{0}|^{2})}{16^{m}}J_{1}(|a_{0}|),$

where

$J_{1}(x)=\frac{16^{m}}{1+x}-\frac{16^{m}}{2}-\sum_{j=1}^{m}16^{m-j}(1-x^{2})^{2j-1}.$

In particular, we have

$J_{1}(0)=\frac{16^{m}}{2}-\frac{16^{m}-1}{15}=\frac{13}{30}16^{m}-1>0\;\;\mbox{and}\;\;J_{1}(1)=0.$

A simple computation shows that

$\displaystyle J_{1}^{\prime}(x)$
$\displaystyle=-\frac{16^{m}}{(1+x)^{2}}+2x\sum_{j=1}^{m}16^{m-j}(1-x^{2})^{2j-2}$
$\displaystyle\leq-\frac{16^{m}}{4}+2\frac{16^{m}-1}{16-1}$
$\displaystyle=-\left(\frac{7}{60}16^{m}+\frac{2}{15}\right)<0.$

Therefore, $J_{1}(x)$ is a decreasing function in $[0,1]$ and hence
$J_{1}(x)\geq J_{1}(1)=0$ for $x\in[0,1].$ Thus, we have

$\sum_{n=0}^{\infty}|a_{n}|r^{n}+P\left(\frac{S_{r}}{\pi}\right)\leq
1\;\;\mbox{for}\;\;r\leq\frac{1}{1+2\lambda}.$

It is easy to see that the equality

$\sum_{n=0}^{\infty}|a_{n}|\left(\frac{1}{1+2\lambda}\right)^{n}+P\left(\frac{S_{\frac{1}{1+2\lambda}}}{\pi}\right)=1$

holds for the function $f$ of the form $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$
if, and only if, $f\equiv c$ with $|c|=1.$ This completes the proof. ∎

Proof of Theorem 2.6. From (1.6), we know that

$|a_{n}|\leq\lambda(1-|a_{0}|^{2})\;\;\mbox{for}\;\;n\geq 1.$

Then for $r\leq 1/(1+2\lambda)$, a simple computation shows that

(3.7)
$\displaystyle|a_{0}|+\sum_{n=1}^{\infty}\bigg{(}|a_{n}|+\beta|a_{n}|^{2}\bigg{)}r^{n}$
$\displaystyle\leq|a_{0}|+\bigg{(}\lambda(1-|a_{0}|^{2})+\beta\lambda^{2}(1-|a_{0}|^{2})^{2}\bigg{)}\frac{r}{1-r}$
$\displaystyle\leq|a_{0}|+\frac{1-|a_{0}|^{2}}{2}\left(1+\lambda\beta(1-|a_{0}|^{2})\right)$
$\displaystyle:=1-\frac{1-|a_{0}|^{2}}{2}F_{1}(x),$

where $F_{1}(x)$ is defined by

$F_{1}(x)=\frac{2}{1+x}-1-\lambda\beta(1-x^{2}).$

Since $0\leq\beta\leq 1/\lambda$, it follows that $F_{1}(0)=1-\lambda\beta\geq
0$ and $F_{1}(1)=0$. Our aim is to show that $F_{1}(x)\geq 0$ for $x\in[0,1]$.
It is easy to see that

$F^{\prime}_{1}(x)=-\frac{2}{(1+x)^{2}}+2\beta\lambda
x\leq-2+2\beta\lambda\leq 0\;\;\mbox{for}\;\;0\leq\beta\leq\frac{1}{\lambda}.$

Therefore, $F_{1}(x)$ is a decreasing function on $[0,1]$ and hence we have
$F_{1}(x)\geq F_{1}(1)=0$ for $x\in[0,1]$. In view of (3.7), we obtain

$|a_{0}|+\sum_{n=1}^{\infty}\bigg{(}|a_{n}|+\beta|a_{n}|^{2}\bigg{)}r^{n}\leq
1\;\;\mbox{for}\;\;r\leq\frac{1}{1+2\lambda}.$

It is easy to see that the equality

$|a_{0}|+\sum_{n=1}^{\infty}\bigg{(}|a_{n}|+\beta|a_{n}|^{2}\bigg{)}\left(\frac{1}{1+2\lambda}\right)^{n}=1$

holds for the function $f$ of the form $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$
if, and only if, $f\equiv c$ with $|c|=1.$ This completes the proof. ∎

Proof of Theorem 2.7. From (1.6), we have

$|a_{n}|\leq\lambda(1-|a_{0}|^{2})\;\;\mbox{for}\;\;n\geq 1.$

Then for $r\leq 1/(1+2\lambda)$, a computation shows that

$\displaystyle
B_{1}(r):=\sum_{n=0}^{\infty}|a_{n}|r^{n}+\left(\frac{1+\lambda}{2\lambda(1+|a_{0}|)}+\frac{2}{3}\frac{(1+\lambda)r}{1-r}\right)\sum_{n=1}^{\infty}|a_{n}|^{2}r^{2n}$
$\displaystyle\leq|a_{0}|+\lambda(1-|a_{0}|^{2})\frac{r}{1-r}+\left(\frac{1+\lambda}{2\lambda(1+|a_{0}|)}+\frac{2}{3}\frac{(1+\lambda)r}{1-r}\right)\lambda^{2}(1-|a_{0}|^{2})^{2}\frac{r^{2}}{1-r^{2}}$
$\displaystyle\leq|a_{0}|+\frac{1-|a_{0}|^{2}}{2}+\left(\frac{1+\lambda}{2\lambda(1+|a_{0}|)}+\frac{2}{3}\frac{(1+\lambda)r}{1-r}\right)\frac{\lambda}{4(1+\lambda)}(1-|a_{0}|)^{2}(1+|a_{0}|)^{2}$
$\displaystyle=1-\frac{1-|a_{0}|^{2}}{2}\left(1-\frac{1+|a_{0}|}{2}-\left(\frac{1}{4(1+|a_{0}|)}+\frac{1}{6}\right)(1-|a_{0}|)^{2}(1+|a_{0}|)\right)$
$\displaystyle=1-\frac{1-|a_{0}|^{2}}{2}F_{2}(|a_{0}|),$

where $F_{2}(x)$ is defined by

$F_{2}(x)=\frac{2}{1+x}-1-\frac{1}{2}\left(\frac{1}{2(1+x)}+\frac{1}{3}\right)(1-x)^{2}(1+x).$

Figure 1. Graph of $F_{2}(x)$ in $[0,1]$. Figure 2. The graph of the function
$F_{2}^{\prime}(x)$ in $[0,1]$.

In order to prove that $B_{1}(r)\leq 1$, it is enough to show that
$F_{2}(x)\geq 0$ for $x\in[0,1]$. Clearly, $F_{2}(1)=0$ and

$\displaystyle
F_{2}(0)=1-\frac{1}{2}\left(\frac{1}{2(1+x)}+\frac{1}{3}\right)\geq\frac{7}{12}>0.$

A simple calculation shows that for $x\in[0,1]$,

$F_{2}^{\prime}(x)=-\frac{2}{(1+x)^{2}}+\frac{(1-x)^{2}}{4(1+x)}-\frac{1}{2}\left(\frac{1}{2(1+x)}+\frac{1}{3}\right)(3x^{2}-2x-1)<0.$

Therefore, the function $F_{2}(x)$ is a decreasing function and hence we have
$F_{2}(x)\geq F_{2}(1)=0$. As $F_{2}(x)\geq 0$ for $x\in[0,1]$ and we obtain

$\sum_{n=0}^{\infty}|a_{n}|r^{n}+\left(\frac{1+\lambda}{2\lambda(1+|a_{0}|)}+\frac{2}{3}\frac{(1+\lambda)r}{1-r}\right)\sum_{n=1}^{\infty}|a_{n}|^{2}r^{2n}\leq
1\;\;\mbox{for}\;\;r\leq\frac{1}{1+2\lambda}.$

Now the equality

$\sum_{n=0}^{\infty}|a_{n}|\left(\frac{1}{1+2\lambda}\right)^{n}+\left(\frac{1+\lambda}{2\lambda(1+|a_{0}|)}+\frac{2}{3}\frac{(1+\lambda)}{2\lambda}\right)\sum_{n=1}^{\infty}|a_{n}|^{2}\left(\frac{1}{1+2\lambda}\right)^{2n}=1$

holds for a function $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$ in
$\mathcal{B}(\Omega)$ if, and only if, $f\equiv c$ with $|c|=1$. This
completes the proof. ∎

Proof of Theorem 2.3. Let $f\in\mathcal{B}(\Omega_{\gamma})$ and
$g(z)=f((z-\gamma)/(1-\gamma))$. Then it is easy to see that
$g\in\mathcal{B}(\mathbb{D})$ and

$\displaystyle
g(z)=\sum_{n=0}^{\infty}\frac{a_{n}}{(1-\gamma)^{n}}(z-\gamma)^{n}.$

Using Lemma 2.2, we obtain

$\displaystyle\sum_{n=0}^{\infty}\frac{|a_{n}|}{(1-\gamma)^{n}}\rho^{n}+Q\left(\frac{S_{\rho}}{\pi}\right)\leq
1\;\;\text{for}\;\;\rho\leq\frac{1-\gamma^{2}}{3+\gamma}$

which is equivalent to

(3.8)
$\displaystyle\sum_{n=0}^{\infty}{|a_{n}|}\left(\frac{\rho}{(1-\gamma)}\right)^{n}+Q\left(\frac{S_{\rho}}{\pi}\right)\leq
1\;\;\text{for}\;\;\rho\leq\frac{1-\gamma^{2}}{3+\gamma}.$

Set $\rho=r(1-\gamma)$, then in view of (3.8), we obtain

(3.9)
$\displaystyle\sum_{n=0}^{\infty}{|a_{n}|}r^{n}+Q\left(\frac{S_{r(1-\gamma)}}{\pi}\right)\leq
1\;\;\text{for}\;\;r\leq\frac{1+\gamma}{3+\gamma}.$

To show the sharpness of the result, we consider the following function

$\displaystyle
f_{a}(z)=\frac{a-\gamma-(1-\gamma)z}{1-a\gamma-a(1-\gamma)}\;\;\text{for}\;\;z\in\Omega_{\gamma}\;\;\text{and}\;\;a\in(0,1).$

Define $\phi_{1}:\mathbb{D}\rightarrow\mathbb{D}$ by
$\phi_{1}(z)=(a-z)/(1-az)$ and
$\phi_{2}(z):\Omega_{\gamma}\rightarrow\mathbb{D}$ by
$\phi_{2}(z)=(1-\gamma)z+\gamma$. Then, the function
$f_{a}=\phi_{1}\circ\phi_{2}$ maps $\Omega_{\gamma}$ univalently onto
$\mathbb{D}.$ A simple computation shows that

$f_{a}(z)=\frac{a-\gamma-(1-\gamma)z}{1-a\gamma-a(1-\gamma)}=C_{0}-\sum_{n=1}^{\infty}C_{n}z^{n}\;\;\mbox{for}\;\;z\in\mathbb{D},$

where $a\in(0,1)$ and

$C_{0}=\frac{a-\gamma}{1-a\gamma}\;\;\text{and}\;\;C_{n}=\frac{1-a^{2}}{a(1-a\gamma)}\left(\frac{a(1-\gamma)}{1-a\gamma}\right)^{n}.$

A simple computation using (3.9) shows that

$\displaystyle\sum_{n=0}^{\infty}{|a_{n}|}r^{n}+Q\left(\frac{S_{r(1-\gamma)}}{\pi}\right)$
$\displaystyle=\frac{a-\gamma}{1-a\gamma}+\left(\frac{1-a^{2}}{1-a\gamma}\right)\frac{(1-\gamma)r}{1-a\gamma-
ar(1-\gamma)}+\frac{c_{1}r^{2}(1-a^{2})^{2}(1-\gamma)^{4}}{((1-a\gamma)^{2}-a^{2}r^{2}(1-\gamma)^{4})^{2}}$
$\displaystyle\quad\quad+\frac{c_{2}r^{4}(1-a^{2})^{4}(1-\gamma)^{8}}{((1-a\gamma)^{2}-a^{2}r^{2}(1-\gamma)^{4})^{4}}+\cdots+\frac{c_{m}r^{2m}(1-a^{2})^{2m}(1-\gamma)^{4m}}{((1-a\gamma)^{2}-a^{2}r^{2}(1-\gamma)^{4})^{2m}}$
$\displaystyle:\;=1-(1-a)\Phi^{\gamma}_{1}(r),$

where

$\displaystyle\Phi^{\gamma}_{1}(r)$
$\displaystyle=-\frac{(1+a)(1-\gamma)r}{(1-a\gamma-
ar(1-\gamma))(1-a\gamma)}-\frac{c_{1}r^{2}(1-a)(1+a)^{2}(1-\gamma)^{4}}{((1-a\gamma)^{2}-a^{2}r^{2}(1-\gamma)^{4})^{2}}$
$\displaystyle\quad\quad-\frac{c_{2}r^{4}(1-a)^{3}(1+a)^{4}(1-\gamma)^{8}}{((1-a\gamma)^{2}-a^{2}r^{2}(1-\gamma)^{4})^{4}}-\frac{c_{m}r^{2m}(1-a)^{2m-1}(1+a)^{2m}(1-\gamma)^{4m}}{((1-a\gamma)^{2}-a^{2}r^{2}(1-\gamma)^{4})^{2m}}$
$\displaystyle\quad\quad-\frac{1}{1-a}\left(\frac{a-\gamma}{1+a\gamma}-1\right).$

It is not difficult to show that $\Phi^{\gamma}_{1}(r)$ is strictly decreasing
function of $r$ in $(0,1)$. Therefore, for $r>r_{0}=(1+\gamma)/(3+\gamma)$, we
have $\Phi^{\gamma}_{1}(r)<\Phi^{\gamma}_{1}(r_{0})$. An elementary
calculation shows that

$\displaystyle\lim_{a\rightarrow
1}\Phi^{\gamma}_{1}(r_{0})=-\frac{2r_{0}}{(1-\gamma)(1-r_{0})}+\frac{1+\gamma}{1-\gamma}=0.$

Therefore $\Phi^{\gamma}_{1}(r)<0$ for $r>r_{0}$. Hence
$1-(1-a)\Phi^{\gamma}_{1}(r)>1$ for $r>r_{0},$ which shows that $r_{0}$ is the
best possible. ∎

###### Proof of Theorem 2.8.

Let $f=h+\overline{g}$ given by (1.7) be harmonic in $\Omega_{\gamma}$ with
$|h(z)|+|g(z)|\leq 1$ in $\Omega_{\gamma}$. Let

(3.10) $\psi_{1}(z)=h(z)+\epsilon g(z),$

where $\epsilon\;(|\epsilon|=1)$ be arbitrary. Note that $\psi_{1}$ is
analytic in $\Omega_{\gamma}$ and $|\psi_{1}(z)|\leq 1$ in $\Omega_{\gamma}$.
From (3.10), we have

$\psi_{1}(z)=a_{0}+\sum_{n=1}^{\infty}(a_{n}+b_{n})z^{n}\quad\mbox{for}\quad
z\in\mathbb{D}.$

In view of Lemma 1.3, we obtain

(3.11) $|a_{n}+\epsilon
b_{n}|\leq\frac{1-|a_{0}|^{2}}{1+\gamma}\quad\mbox{for}\quad n\geq 1.$

Since $\epsilon\;(|\epsilon|=1)$ is arbitrary, we have

(3.12) $|a_{n}+b_{n}|\leq\frac{1-|a_{0}|^{2}}{1+\gamma}\quad\mbox{for}\quad
n\geq 1.$

A simple computation using (3.12) shows that

(3.13)
$|a_{0}|+\sum_{n=1}^{\infty}\bigg{(}|a_{n}|+|b_{n}|\bigg{)}r^{n}\leq|a_{0}|+\frac{1-|a_{0}|^{2}}{1+\gamma}\sum_{n=1}^{\infty}r^{n}=|a_{0}|+\frac{1-|a_{0}|^{2}}{1+\gamma}\left(\frac{r}{1-r}\right).$

Therefore, the inequality (3.13) is less than or equal to $1$, provided

$|a_{0}|+\frac{1-|a_{0}|^{2}}{1+\gamma}\left(\frac{r}{1-r}\right)\leq
1\quad\mbox{for}\quad r\leq r_{0}:=\frac{1+\gamma}{3+\gamma}.$

Hence, we have

$|a_{0}|+\sum_{n=1}^{\infty}\bigg{(}|a_{n}|+|b_{n}|\bigg{)}r^{n}\leq
1\quad\mbox{for}\quad r\leq r_{0}:=\frac{1+\gamma}{3+\gamma}.$

This completes the proof. ∎

###### Proof of Theorem 2.9.

Let $f=h+\overline{g}$ given by (1.7) be harmonic in $\Omega_{\gamma}$ with
$|h(z)|+|g(z)|\leq 1$ in $\Omega_{\gamma}$. Consider the function

(3.14) $\psi_{2}(z)=h(z)+\epsilon g(z),$

where $\epsilon\;(|\epsilon|=1)$ is arbitrary. Note that $\psi_{2}$ is
analytic in $\Omega_{\gamma}$ and $|\psi_{2}|\leq 1$ in $\Omega_{\gamma}$.
From the equation (3.10), we have

$\psi_{2}(z)=a_{0}+\sum_{n=1}^{\infty}(a_{n}+b_{n})z^{n}\quad\mbox{for}\quad
z\in\mathbb{D}.$

In view of Lemma 1.3, we obtain

(3.15) $|a_{n}+\epsilon b_{n}|\leq\frac{1-|a_{0}+\epsilon
b_{0}|^{2}}{1+\gamma}\quad\mbox{for}\quad n\geq 1.$

Since $\epsilon\;(|\epsilon|=1)$ is arbitrary, we obtain

(3.16) $|a_{n}-\epsilon b_{n}|\leq\frac{1-|a_{0}-\epsilon
b_{0}|^{2}}{1+\gamma}\quad\mbox{for}\quad n\geq 1.$

From (3.15) and (3.16), we obtain

(3.17) $|a_{n}+\epsilon b_{n}|+|a_{n}-\epsilon
b_{n}|\leq\frac{2-|a_{0}+\epsilon b_{0}|^{2}-|a_{0}-\epsilon
b_{0}|^{2}}{1+\gamma}\;\;\mbox{for}\;\;n\geq 1.$

It is easy to see that (3.17) is equivalent to

(3.18) $|a_{n}+\epsilon b_{n}|+|a_{n}-\epsilon
b_{n}|\leq\frac{2\left(1-|a_{0}|^{2}-|b_{0}|^{2}\right)}{1+\gamma}\;\;\mbox{for}\;\;n\geq
1.$

From (3.18), we obtain

$\bigg{(}|a_{n}+\epsilon b_{n}|+|a_{n}-\epsilon
b_{n}|\bigg{)}^{2}\leq\frac{4\bigg{(}1-|a_{0}|^{2}-|b_{0}|^{2}\bigg{)}^{2}}{(1+\gamma)^{2}}$

which is equivalent to

(3.19)
$|a_{n}|^{2}+|b_{n}|^{2}+|a^{2}-e^{i2\theta}b_{n}^{2}|\leq\frac{2\bigg{(}1-|a_{0}|^{2}-|b_{0}|^{2}\bigg{)}^{2}}{(1+\gamma)^{2}}.$

By choosing $\theta=\pi/2+\arg(a_{n})-\arg(b_{n})$, it follows from (3.19)
that

$|a_{n}|^{2}+|b_{n}|^{2}\leq\frac{\bigg{(}1-|a_{0}|^{2}-|b_{0}|^{2}\bigg{)}^{2}}{\left(1+\gamma\right)^{2}}$

and hence

(3.20)
$\sqrt{|a_{n}|^{2}+|b_{n}|^{2}}\leq\frac{1-|a_{0}|^{2}-|b_{0}|^{2}}{1+\gamma}\quad\mbox{for}\quad
n\geq 1.$

A simple computation using (3.20) shows that

(3.21) $\displaystyle\sqrt{|a_{0}|^{2}+|b_{0}|^{2}}$
$\displaystyle+\sum_{n=1}^{\infty}\bigg{(}\sqrt{|a_{n}|^{2}+|a_{n}|^{2}}\bigg{)}r^{n}$
$\displaystyle\leq|a_{0}|+\frac{1-|a_{0}|^{2}}{1+\gamma}\sum_{n=1}^{\infty}r^{n}$
$\displaystyle=|a_{0}|+\frac{1-|a_{0}|^{2}}{1+\gamma}\left(\frac{r}{1-r}\right).$

Therefore, from (3.20) and (3.21), we obtain

$\displaystyle\sum_{n=0}^{\infty}(\sqrt{|a_{n}|^{2}+|b_{n}|^{2}})r^{n}$
$\displaystyle\leq\sqrt{|a_{0}|^{2}+|b_{0}|^{2}}+\frac{1-|a_{0}|^{2}-|b_{0}|^{2}}{1+\gamma}\left(\frac{r}{1-r}\right)$
$\displaystyle\leq\sqrt{|a_{0}|^{2}+|b_{0}|^{2}}+\frac{1-\sqrt{|a_{0}|^{2}+|b_{0}|^{2}}}{1+\gamma}\left(\frac{2r}{1-r}\right)$
$\displaystyle\leq 1,$

if $2r/(1+\gamma)(1-r)\leq 1$, that is if $r\leq r_{0}=(1+\gamma)/(3+\gamma).$
Therefore, we have

$|a_{0}|+\sum_{n=1}^{\infty}\bigg{(}|a_{n}|+|b_{n}|\bigg{)}r^{n}\leq
1\quad\mbox{for}\quad r\leq r_{0}=\frac{1+\gamma}{3+\gamma}.$

Hence, we have

$\sqrt{|a_{0}|^{2}+|b_{0}|^{2}}+\sum_{n=1}^{\infty}\sqrt{|a_{n}|^{2}+|b_{n}|^{2}}r^{n}\leq
1\quad\mbox{for}\quad|z|=r\leq r_{0}=\frac{1+\gamma}{3+\gamma}.$

This completes the proof. ∎

Acknowledgment: The first author is supported by the Institute Post Doctoral
Fellowship of IIT Bhubaneswar, India, the second author is supported by SERB-
MATRICS, and third author is supported by CSIR, India.

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# Syntax-driven Data Augmentation for Named Entity Recognition

Arie Pratama Sutiono
University of Arizona
Linguistics
<EMAIL_ADDRESS>
&Gus Hahn-Powell
University of Arizona
Linguistics
<EMAIL_ADDRESS>

###### Abstract

In low resource settings, data augmentation strategies are commonly leveraged
to improve performance. Numerous approaches have attempted document-level
augmentation (e.g., text classification), but few studies have explored token-
level augmentation. Performed naively, data augmentation can produce
semantically incongruent and ungrammatical examples. In this work, we compare
simple masked language model replacement and an augmentation method using
constituency tree mutations to improve the performance of named entity
recognition in low-resource settings with the aim of preserving linguistic
cohesion of the augmented sentences.

##  1 Introduction

Deep neural networks have proven effective for a wide variety of tasks in
natural language processing; however, these networks often require large
annotated datasets before they begin to outperform simpler models. Such data
is not always available or sufficiently diverse, and its collection and
annotation can be an expensive and slow process. The trend of fine-tuning
large-scale language models originally trained using self-supervision has
helped to alleviate the need for large annotated datasets, but this approach
relies on the dataset for fine-tuning being diverse enough to train a model
that generalizes well. Careful data augmentation can help to improve dataset
diversity and ultimately the model’s generalizability.

Data augmentation, a technique to generate data given training set
characteristics, continues to play a critical role in low-resource settings;
however, the majority of work on data augmentation focuses on improving
document-level tasks such as text classification. Far less attention has been
paid to token-level tasks (Feng et al., 2021).

Prevailing approaches to sequence tagging tasks such as named entity
recognition (NER) require token-level ground truth. Naive replacement-based
methods for augmentation may introduce noise in the form of sentences that are
ungrammatical, semantically vacuous, or semantically incongruent. Whether in
the form of insertions, deletions, or substitutions, care must be taken with
token-level augmentation to preserve linguistic cohesion.

There is evidence that large language models may possess some syntactic
knowledge (Hewitt and Manning, 2019; Wei et al., 2021). Work by Bai et al.
(2021) suggests that incorporating syntactic tasks into pre-training improves
the performance of large language models. Inspired by these findings, we
investigate how constituency trees might be used to guide data augmentation.
We use tree-based transformations to mutate sentences while minimizing
undesired side effects of syntactic manipulation (i.e., preserving linguistic
cohesion). Related work by Zhang et al. (2022) explores a similar approach in
different settings. They focus on the effect of constituency based replacement
in single classification and pair sentence classification tasks, while this
work examines token-level classification.

We compare our syntax-driven method with no augmentation (our baseline),
augmented data generated through cloze-style (Taylor, 1953) masked language
modeling using a BERT-based classifier, successful approaches introduced by
Dai and Adel (2021), and two of the top-performing augmentation strategies
according to past work: synonym-based replacement and mention-based
replacement. Following prior work, we We use the i2b2-2010 English language
dataset Uzuner et al. (2011) for NER.

##  2 Related work

A variety of approaches have been explored for document-level data
augmentation. The usage of backtranslation to generate augmented samples was
introduced by Kobayashi (2018). Wei and Zou (2019) explored synonym
replacement, random insertion, random swap, and random deletion for text
classification. Quteineh et al. (2020) introduced an approach using monte
carlo tree search (MCTS) to guide generation of synthetic data.

Augmentation for token-level sequence tagging, however, is understudied.
Simple approaches for token-level classification (e.g., synonym replacement,
mention replacement, shuffling, etc.) was investigated by Devlin et al.
(2019). They used a sample of 50, 150, and 500 sentences to simulate a low-
resource setting. Token linearization (TL) was introduced by Ding et al.
(2020). The main idea of TL is to incorporate NER tags inside the training
sentences themselves. Ding et al. (2020) experimented with various sizes of
training data across six languages.

A Similar idea to that presented here has been used in parallel research by
Zhang et al. (2022). Their approach, TreeMix, works similarly to our approach
in that it replaces a phrase with another phrase from another another training
instance by swapping phrases with the same constituent labels. Zhang et al.
(2022) demonstrated that TreeMix outperforms a method that selects a random
span of text as the replacement for a target phrase, suggesting that
syntactically=aware replacement can improve data augmentation for at least
some tasks.

##  3 Approach

###  3.1 Synonym Replacement (SR)

Dai and Adel (2021) experimented with replacing randomly selected tokens from
the training corpus with a multiword synonyms originating from WordNet
(Miller, 1992). In this case, if the replaced token is the beginning of a
mention (B-ENTITY), then the first token of the synonym will be tagged as
B-ENTITY and the rest will be considered as I-ENTITY. In cases where the
replaced token is in the middle of a mention (I-ENTITY), then all of the
synonym’s tokens will be assigned to I-ENTITY.

###  3.2 Mention Replacement (MR)

Dai and Adel (2021) described mention replacement as using a Bernoulli
distribution to decide whether each mention should be replaced. If yes, then
another mention which has the same entity type as the target mention from the
original training set is selected to replace the target mention. For instance,
if the mention "myelopathy / B-PROBLEM is selected for replacement, then we
can select one of {"C5-6", "COPD", …} which all have the same entity type
(PROBLEM).

###  3.3 Language Model (LM)

We experimented with token replacement using a masked language model. We
restrict the system to replace only non-mention tokens (tokens with category
O). This is because if we replace tokens with a named entity, we cannot
guarantee that the output from the masked language model will have the same
category, such that if we replace a token categorized as B-TEST, we could not
guarantee that the masked language model will replace it with a similar token
to those in B-TEST category.

We randomly select, without replacement, $n$ tokens as candidates to be
replaced. The selected tokens are masked from the original sentence. Next, we
the language model generates replacements for the masked tokens. We may repeat
this token generation up to $k$ times to generate different augmented
sentences. We use Allen AI’s SciBERT model from the Hugging Face model
repository.

###  3.4 Constituency Replacement (CR)

As a preprocessing step, we perform constituency parsing over all of the
training data using Stanza (Qi et al., 2020). Given an XP non-terminal, we
select $p$ non-terminals as candidates for replacement. For each non-terminal,
we find other non-terminals with the same category from the training data, to
replace the candidate. Assuming that we chose VP as the non-terminal taget
node for replacement, the algorithm will choose another VP from the set of
parsed sentences in the training corpus and mutate the whole subtree (VP root
and the nodes below it). We can repeat this process to generate $k$ augmented
sentences. Additionally, we target nodes that have NER mentions as one of its
children.

##  4 Experiments and Results

###  4.1 Dataset

We used the i2b2-2010 dataset (Uzuner et al., 2011), an English language NER
dataset. Similar to Dai and Adel (2021) we use 3 different sizes of dataset to
simulate low-resource settings. We select the first 50, 150, and 500 sentences
from training set and denote them as S, M, L. We used the default train-test
split and limit the augmentations to training set.

###  4.2 Model

Following Dai and Adel (2021), we model NER task as sequence-labeling. We used
same components for modeling: a neural encoder and a conditional random field
layer. For our neural encoder, We used SciBERT. This model has been proven to
work effectively with scientific and medical data, like i2b2-2010.

###  4.3 Experiments

Each experiment was repeated with 5 different random seeds to calculate
standard deviation.

For our SR and MR approaches, our hyperparameters are the replace ratio (0.3)
and the number of generated samples (1).

LM and CR hyperparameters are similar. Both will have the number of generated
samples and number of replaced tokens (only for LM) or non-terminals (only for
ST). In this work, we limit CR replacements to non-terminals (phrases) in the
following set: {NP, VP, ADJP, ADVP, PP}. We leave out FRAG (fragment) because
it has too low of a non-terminal count.

One question to be explored is whether more augmented data can result in
continued gains in model performance. To answer this, we experimented with {5,
10, 20} generated samples. For each number of generated samples, we also set
the number of replaced tokens for the LM and the number of replaced non-
terminals for CR to be {1, 3, 5}. We have described the distribution of non
terminals in Table 1. All of these settings were tested against the 27,625
sentences from our validation set.

Phrase | S | M | L
---|---|---|---
NP | 332 | 637 | 2562
VP | 93 | 189 | 881
PP | 54 | 130 | 690
ADJP | 31 | 42 | 189
ADVP | 16 | 27 | 126
FRAG | 2 | 2 | 4
Table 1:  Distribution of the number of phrases in the training data.

###  4.4 Results

Table 2 described the highest F1 scores for each augmentation strategy. The
best F1 scores were taken for each strategy, across multiple hyperparameters.
We found that synonym replacement still outperforms other augmentation
strategies in small and medium dataset sizes.

Experiment | S | M | L
---|---|---|---
NoA | $46.3\pm 0.5$ | $61.4\pm 0.1$ | $70.7\pm 0.1$
SR | $\bm{53.0\pm 0.2}$ | $\bm{65.7\pm 0.1}$ | $71.0\pm 0.0$
MR | $51.9\pm 0.2$ | $61.7\pm 0.1$ | $70.2\pm 0.0$
LM | $52.9\pm 0.1$ | $63.3\pm 0.1$ | $\bm{73.3\pm 0.2}$
CR-ADJP | $47.8\pm 0.2$ | $61.0\pm 0.1$ | $71.2\pm 0.1$
CR-ADVP | $50.5\pm 0.3$ | $61.9\pm 0.1$ | $71.3\pm 0.1$
CR-NP | $52.1\pm 0.3$ | $60.6\pm 0.1$ | $70.2\pm 0.1$
CR-PP | $52.1\pm 0.1$ | $62.4\pm 0.1$ | $71.9\pm 0.1$
CR-VP | $52.9\pm 0.2$ | $62.8\pm 0.1$ | $72.8\pm 0.1$

Table 2: Results for data augmentation experiments across different data set
sizes. Top results for each data partition are marked in bold.

All augmentation methods tested seem to improve performance in terms of F1 for
the small training set ( 50 sentences). When we look at the medium dataset,
however, some methods such as CR-ADJP or CR-NP, start to have a negative
impact compared to no augmentation settings. Even more augmentation strategies
begin to show diminishing or negative effects on performance for the larger
dataset (e.g., MR and CR-NP). This suggests that some of the augmented data
might be detrimental for the model fine-tuning process.

To understand how the augmented data may start to hurt the original model’s
performance, we consider one original sentence processed using CR-NP strategy.
For example, “Dr. Foutchner will arrange for an outpatient Holter monitor”. In
the case of the CR strategy, the augmentation algorithm draws an NP from
another training sentence, resulting in “Dr. Foutchner will arrange for a T2
signal change” or “Dr. Foutchner will arrange for 10 beats”. These augmented
sentences are grammatical, but they lack cohesion. This phenomenon may impact
the model negatively. Future work should explore strategies to control for
this drift. For instance, by fine-tuning a large-scale language model to
perform masked language modeling on sentences where a portion of tokens are
provided in terms of phrasal category (XP) or functional category (part of
speech tag), we might hybridize syntax-driven transformations and instantiate
syntactic templates using large-scale language models.

We observed that among CR strategies, CR-NP performance seems to be worse
compared to CR-VP or CR-PP, despite NP has the most occurrences in the
training data. We suspect that the effectiveness of this strategy will heavily
depend on the scope of the constituency tag. NPs are usually located low in
the constituency tree, while VPs are usually located toward the top.

The augmentation strategies explored in this work can be further divided into
two groups: strategies that produce new vocabularies and strategies that do
not produce new vocabularies. SR and LM methods fall into augmentation that
produce new vocabularies. SR uses the Penn Treebank (Marcus et al., 1993) to
generate synonyms of replaced tokens. LM use its word embedding to guess the
masked target token and may generate new words that do not exist in the
training data. The other strategies (MR and CR) rely solely on the current
training dataset. This phenomenon suggests augmentation strategies that
produce new vocabularies seem to be more effective. This is plausible since
new words will make the fine-tuned model more robust to unseen data. Although
CR does not generate new words like the LM and SR methods, it still performs
competitively in comparison. The delta between F1 scores produced by CR-VP and
LM with our best hyperparameters for all dataset sizes are remarkably small at
around 0 0.5 points. The effect of simpler data augmentation strategies, SR
and MR, seems to be diminishing as the data size increases; however, it is not
the case with the LM and CR-VP strategies. They seem to perform well when more
training data is available.

| S | M | L
---|---|---|---
| 5 | 10 | 20 | 5 | 10 | 20 | 5 | 10 | 20
LM | 50.6 | 51.5 | 50.6 | 61.2 | 60.8 | 61.0 | 71.1 | 70.8 | 70.6
CR-VP | 50.5 | 52.0 | 52.3 | 61.9 | 62.3 | 62.5 | 71.9 | 72.3 | 72.3

Table 3: Comparison between CR-VP and LM augmentation. CR-VP holds more
consistent performance across the number of generated sentences, while LM
performance drops when the number of generated sentences is low.

Looking at Table 3, the CR-VP augmentation strategy seems to show more
consistent performance growth compared to the LM strategy. Whether it is 5, 10
or 20 sentences generated, CR-VP consistently trends upward as the number of
augmented sentences increases (cf the instability of the LM). The average
performance of the CR strategy shows an increased F1 as the number of
synthetic sentences grows. In contrast, the average performance of the LM
strategy is inconsistent and trending downward as the number of synthetic
sentences increases.

Lastly, the performance of the CR strategy will also be affected by the
performance of constituency parser component itself. For one of our augmented
examples, the original sentence “She $[_{\textsc{VP}}$ had a workup by her
neurologist$]$ and an MRI $[_{\textsc{VP}}$ call with any fevers , chills ,
increasing weakness… $]$” was mutated into “She $[_{\textsc{VP}}$ had a workup
by her neurologist$]$ and an MRI $[_{\textsc{VP}}$ flare$]$”. Here, the word
flare was falsely predicted as a verb and thus erroneously predicted as a VP
constituent, while the word flare here should be a part of COPD flare and
classified as noun.

##  5 Conclusion and Future Work

In this work, we examined data augmentation with a large-scale language model
(LM) and constituency tree mutation (ST). We compared these augmentation
methods with a baseline and previously proposed strategies for data
augmentation: synonym replacement (SR) and mention replacement (MR). We found
that SR performance is still most effective, by a small margin, but the
performance degrades quickly as the data size increased. We have also observed
that both LM and CR retained their performance throughout larger dataset
sizes. We also showed that CR performance seems to be consistent in its
improvement as the augmented dataset size increases, while the LM showed
degrading performance with more augmented data.

Future work should include improvements that hybridize the syntactic
transformations with a large-scale language model. One possibility to increase
the performance of the baseline language model is to train it to recognize
phrase-level constituents and functional categories to understand more about
constituency tags by first randomly swapping a few tokens with part of speech
tags. For example, the original sentence is “I take my medicine.”, then the
pre-training sentence is “I VB my medicine.” and "I take my NN.". We
hypothesize that this pre-training will improve the prediction performance of
the baseline language model that we used for CR augmentation by attending to
functional categories. Another possibility is to assign different weights to
datapoints that inform the model how much to "trust" augmented data compared
to gold data. This weight could be in the form of different learning rate.

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