††footnotetext:  File: main.tex, printed: 2024-8-30, 21.45

# Revisiting Bohr Inequalities with Analytic and Harmonic Mappings on unit
disk

Molla Basir Ahamed  Molla Basir Ahamed, Department of Mathematics, Jadavpur
University, Kolkata-700032, West Bengal,India.
<EMAIL_ADDRESS>and  Partha Pratim Roy  Partha Pratim Roy,
Department of Mathematics, Jadavpur University, Kolkata-700032, West
Bengal,India<EMAIL_ADDRESS>
###### Abstract.

In this paper, we study some improved and refined versions of the classical
Bohr inequality applicable to the class $\mathcal{B}$, which consists of self-
analytic mappings defined on the unit disk $\mathbb{D}$. First, we improve the
Bohr inequality for the class $\mathcal{B}$ of analytic self-maps,
incorporating the area measurements of sub-disks $\mathbb{D}_{r}$ of
$\mathbb{D}$. Secondly, we establish a sharp inequality with suitable setting
as an improved version of the classic Bohr inequality. Then we obtain a sharp
refined Bohr inequality in which the coefficients $|a_{k}|$ $(k=0,1,2,3)$ in
the majorant series $M_{f}(r)$ of $f$ are replaced by $|f^{(k)}(z)|/k!$.
Finally, for a certain class $\mathcal{P}^{0}_{\mathcal{H}}(M)$ of harmonic
mappings of the form $f=h+\overline{g}$, we generalize the Bohr inequality
incorporating a sequence $\\{\varphi_{n}(r)\\}_{n=0}^{\infty}$ of continuous
functions of $r$ in $[0,1)$ and give some applications.

###### Key words and phrases:

Bounded analytic functions, Bohr inequality, Bohr-Rogosinski inequality,
Schwarz-Pick lemma, Harmonic mappings, Sequence of continuous functions

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

Primary 30A10, 30H05, 30C35, 30C50 Secondary 30C45

##  1\. Introduction

Let $\mathbb{D}:=\\{z\in\mathbb{C}:|z|<1\\}$ denote the open unit disk in
$\mathbb{C}$. Herald Bohr in $1914$ (see [20]) states that if $H_{\infty}$
denotes the class of all bounded analytic functions $f$ on $\mathbb{D}$, then
the following inequality holds

(1.1) $\displaystyle
B_{0}(f,r):=|a_{0}|+\sum_{n=1}^{\infty}|a_{n}|r^{n}\leq||f||_{\infty}:=\displaystyle\sup_{z\in\mathbb{D}}|f(z)|\;\;\mbox{for}\;\;0\leq
r\leq\frac{1}{3},$

where $a_{k}=f^{(k)}(0)/k!$ for $k\geq 0$. The constant $1/3$ is called the
Bohr radius and the inequality in (1.1) is called Bohr inequality for the
class $\mathcal{B}$ of analytic self-maps on $\mathbb{D}$. Henceforth, if
there exists a positive real number $r_{0}$ such that an inequality of the
form (1.1) holds for every elements of a class $\mathcal{M}$ for $0\leq r\leq
r_{0}$ and fails when $r>r_{0}$, then we shall say that $r_{0}$ is an sharp
bound for $r$ in the inequality w.r.t. to class $\mathcal{M}$.

Initially, Bohr showed the inequality (1.1) holds for $|z|\leq 1/6$, and later
M. Riesz, I. Shur and F. Wiener, independently proved its validity on a wider
range $0\leq r\leq 1/3$ and the number $1/3$ is optimal as seen by analyzing
suitable members of the conformal automorphism of the unit disk $\mathbb{D}$.

This result has created enormous interest on Bohr’s inequality in various
settings (see [17, 15, 19, 21, 25, 37, 40]. Initial exploration of this
problem was conducted by Harald Bohr as he delved into the absolute
convergence of Dirichlet series of the form $\sum a_{n}n^{-s}$. However, in
recent years, it has evolved into a thriving domain of research in modern
function theory. The theorem has drawn considerable interest because it plays
a crucial role in solving the characterization problem for Banach algebras
that satisfy the von Neumann inequality (see [26]). However, this compelling
inequality has been substantiated by several distinct proofs presented in
various articles (see [57, 58, 59]). Boas and Khavinson (see [15]) were the
first to introduce the concept of the n-dimensional Bohr radius
${\mathcal{K}}^{\infty}_{n}$ which expedites the research what is called the
multidimensional Bohr radii.

In [52, 54, 55, 21, 38], the authors have delved into the Bohr radius in the
context of a more general class of power series. In this investigation, they
have established generalized Bohr inequalities in view of changing the basis
of sequence $\\{r^{n}\\}_{n=0}^{\infty}$ with a sequence
$\\{\varphi_{n}(r)\\}_{n=0}^{\infty}$, where $\varphi_{n}(r)$ is a non-
negative continuous function defined on the interval $[0,1)$. It is noteworthy
that the requirement for this new sequence is that
$\sum_{n=0}^{\infty}\varphi_{n}(r)$ should exhibit local uniform convergence
with respect to the parameter $r$ in the interval $[0,1)$. Recent
investigations led by Ahamed _et al._ [7] and Evdoridis _et al._ [29] have
been dedicated to exploring the Bohr Phenomenon for the class of analytic
functions defined on shifted disk $\Omega_{\gamma}$ ($0\leq\gamma<1$)
containing the unit disk $\mathbb{D}$. Furthermore, the research in [10] have
directed towards analyzing the Bohr radius for certain classes of starlike and
convex univalent functions in the unit disk $\mathbb{D}$.

Bohr’s inequality for holomorphic and pluriharmonic mappings with values in
complex Hilbert spaces are obtained in [32]. However, it can be noted that not
every class of functions has Bohr phenomenon, for example, Bénéteau _et al._
[14] showed that there is no Bohr phenomenon in the Hardy space
$H^{p}(\mathbb{D},X),$ where $p\in[1,\infty).$ In [46], Liu and Liu have shown
that Bohr’s inequality fails to hold for the class
$\mathcal{H}(\mathbb{D}^{2},\mathbb{D}^{2})$ of holomorphic functions
$f:\mathbb{D}^{2}\rightarrow\mathbb{D}^{2}$ having lacunary series expansion.
Over the past few years, Bohr’s inequality has gained a significant attention,
resulting in extensive research and extensions in various directions with
different settings (see [12, 16, 18, 15, 49, 50, 53, 59, 43, 45] and
references therein).

In this paper, our aim is two fold. First, we establish some improved and
refined Bohr inequalities for the class $\mathcal{B}$, and second, we obtain
generalized Bohr inequality for certain class of harmonic mappings in terms of
a sequence of continuous functions. The organization of the paper is as
follows: In Section 2, we study Bohr phenomenon for the class $\mathcal{B}$ of
bounded analytic functions on unit disk $\mathbb{D}$. In Section 3,
generalization of the Bohr inequality for certain classes of harmonic mappings
and its applications are discussed. The proof of the main results are
discussed in each section individually.

##  2\. Bohr inequalities for the class $\mathcal{B}$ of bounded analytic
functions on unit disk $\mathbb{D}$

Before we continue the discussion, we fix some notations.

###  2.1. Basic Notations

Let

$\displaystyle\mathcal{B}=\\{f\in H_{\infty}\;:\;||f||_{\infty}\leq 1\\}.$

Also, for $f(z)=\sum_{n=0}^{\infty}|a_{n}|z^{n}\in\mathcal{B}$ and
$f_{0}(z):=f(z)-f(0)$, we let for convenience

$\displaystyle B_{k}(f,r):=\sum_{n=k}^{\infty}|a_{n}|r^{n}\;\mbox{for }k\geq
0,\;||f_{0}||^{2}_{r}:=\sum_{n=1}^{\infty}|a_{n}|^{2}r^{2n}\;\mbox{and}\;||f_{1}||^{2}_{r}:=\sum_{n=2}^{\infty}|a_{n}|^{2}r^{2n}$

so that

$\displaystyle\begin{cases}B_{0}(f,r)=|a_{0}|+B_{1}(f,r),\\\
B_{0}(f,r)=|a_{0}|+|a_{1}|r+B_{2}(f,r),\\\
B_{0}(f,r)=|a_{0}|+|a_{1}|r+|a_{2}|r^{2}+B_{3}(f,r),\;\mbox{so
on}.\end{cases}$

Further, we define the quantity $A(f_{0},r)$ and $A(f_{1},r)$ by

$\displaystyle\begin{cases}A(f_{0},r):=\left(\dfrac{1}{1+|a_{0}|}+\dfrac{r}{1-r}\right)||f_{0}||^{2}_{r},\vspace{2mm}\\\
A(f_{1},r):=\left(\dfrac{1}{1+|a_{0}|}+\dfrac{r}{1-r}\right)||f_{1}||^{2}_{r}.\end{cases}$

Kayumov _et al._ [36, 42] have derived the Bohr-Rogosinski inequality for
class $\mathcal{B}$ based on the investigation of Rogosinski’s inequality and
Rogosinski’s radius as presented in [56].

######  Theorem 2.1.

[42] Suppose that $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}\in\mathcal{B}$. Then

$\displaystyle|f(z)|+B_{N}(f,r)\leq 1\;\;\mbox{for}\;\;r\leq R_{N},$

where $R_{N}$ is the positive root of the equation $2(1+r)r^{N}-(1-r)^{2}=0$.
The radius $R_{N}$ is the best possible. Moreover,

$\displaystyle|f(z)|^{2}+B_{N}(f,r)\leq 1\;\;\mbox{for}\;\;r\leq
R^{\prime}_{N},$

where $R^{\prime}_{N}$ is the positive root of the equation
$(1+r)r^{N}-(1-r)^{2}=0$. The radius $R^{\prime}_{N}$ is the best possible.

This result generates a significant amount of research activity on Bohr-
Rogosinki inequalities for different classes of functions. However, Liu _et
al._ [48] have proved the following result for functions in the class
$\mathcal{B}$, where $|a_{0}|$ and $|a_{1}|$ are replaced by $|f(z)|$ and
$|f^{\prime}(z)|$, respectively.

######  Theorem 2.2.

[48] Suppose that $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}\in\mathcal{B}$. Then

$\displaystyle|f(z)|+|f^{\prime}(z)|r+B_{2}(f,r)\leq
1\;\;\mbox{for}\;\;r\leq(\sqrt{17}-3)/4.$

The constant $(\sqrt{17}-3)/4$ is best possible.

Further, several refined Bohr inequality with suitable setting are
investigated in [47] and obtained the following result.

######  Theorem 2.3.

[47] Suppose that $f(z)\in\mathcal{B}$ and
$f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$. Then

(2.1) $\displaystyle|f(z)|+B_{1}(f,r)+A(f_{0},r)\leq 1$

for $|z|=r_{0}\leq(2)/(3+|a_{0}|+\sqrt{5}(1+|a_{0}|))$ and the constant
$r_{0}$ is the best possible and $r_{0}>\sqrt{5}-2$. Moreover,

(2.2) $\displaystyle|f(z)|^{2}+B_{1}(f,r)+A(f_{0},r)\leq 1,$

for $|z|=r\leq r^{\prime}_{0},$ where $r^{\prime}_{0}$ is the unique root in
$(0,1)$ of the equation

$\displaystyle(1-|a_{0}|^{3})r^{3}-(1+2|a_{0}|)r^{2}-2r+1=0$

and $r^{\prime}_{0}$ is the best possible and
$1/3<r^{\prime}_{0}<1/(2+|a_{0}|)$.

Moreover, Liu _et. al_ (see [47]) refined the inequalities (2.1) and (2.2) by
replacing the second coefficient $|a_{1}|$ in the majorant series $B_{0}(f,r)$
by the quantity $|f^{\prime}(z)|$ and obtained the following result.

######  Theorem 2.4.

[47] Suppose that $f(z)\in\mathcal{B}$ and
$f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$. Then

(2.3) $\displaystyle|f(z)|+|f^{\prime}(z)|r+B_{2}(f,r)+A(f_{0},r)\leq 1$

for $|z|=r\leq(\sqrt{17}-3)/4$ and the constant $(\sqrt{17}-3)/4$ is the best
possible. Moreover,

(2.4) $\displaystyle|f(z)|^{2}+|f^{\prime}(z)|r+B_{2}(f,r)+A(f_{0},r)\leq 1$

for $|z|=r\leq r_{0}$, where $r_{0}\approx 0.385795$ is the unique positive
root of the equation $1-2r-r^{2}-r^{3}-r^{4}=0$ and the constant $r_{0}$ is
the best possible.

After that, the inequality (2.3) is refined (see [3]) by replacing the second
coefficient $|a_{2}|$ by $|f^{\prime\prime}(z)|/2!$ and the following result
is obtained.

######  Theorem 2.5 ([3]).

Suppose that $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}\in\mathcal{B}$. Then

(2.5)
$\displaystyle|f(z)|+|f^{\prime}(z)|r+\frac{|f^{\prime\prime}(z)|}{2!}r^{2}+B_{3}(f,r)+\frac{|a_{1}|^{2}r^{3}}{1-r}+A(f_{1},r)\leq
1$

for $|z|=r\leq r_{0}\approx 0.287459$, where $r_{0}$ is the unique root in
$(0,1)$ of the equation

$\displaystyle-1+3r+r^{2}+r^{3}+4r^{4}+2r^{5}=0$

and the constant $r_{0}$ is best possible.

###  2.2. Sharp refined Bohr inequality for the class $\mathcal{B}$

An extensive research work on Bohr phenomenon are carried out by several
authors for the class $\mathcal{B}$ of analytic self maps on the unit disk
$\mathbb{D}$. Recently, the article [3] has presented a sharp refined form of
the inequality (2.3) mentioned in the article (see Theorem 2.4), which was
originally proposed by Liu _et al._ (refer to [47]). However, [3] omits any
elaboration on the significant refinement of the inequality (2.4). In light of
this, to continue the research, a natural query might emerge, and
investigating its response could be pertinent in this topic of research.

######  Question 2.1.

Can we establish a sharp refinement of the inequality (2.4) by replacing the
third coefficient $|a_{2}|$ by $|f^{\prime\prime}|/2!?$

######  Question 2.2.

Can we the improve the classical Bohr inequality replacing the initial
coefficients $|a_{0}|,|a_{1}|,|a_{2}|$ and $|a_{3}|$ in the majorant series by
$|f(z)|,|f^{\prime}(z)|,|f^{\prime\prime}(z)|/2!$ and
$|f^{\prime\prime\prime}(z)|/3!$ respectively?

In this section, we were inspired by their result to adopt the proof technique
of Theorem 2.5, aiming to provide answers to the Question 2.1 and 2.2.
Consequently, we have obtained the following two results, to address the
Questions 2.1 and 2.2 successfully.

######  Theorem 2.6.

Suppose that $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}\in\mathcal{B}$. Then we have

$\displaystyle\mathcal{D}_{f}(z,r):=|f(z)|^{2}+|f^{\prime}(z)|r+\frac{|f^{\prime\prime}(z)|}{2!}r^{2}+B_{3}(f,r)+\frac{|a_{1}|^{2}r^{3}}{1-r}+A(f_{1},r)\leq
1$

for $|z|=r\leq r_{0}\approx 0.393727$, where $r_{0}$ is the unique root in
$(0,1)$ of the equation $-1+2r+r^{2}+2r^{4}+r^{5}=0$ and the constant $r_{0}$
is best possible.

To present our next result concisely, we need to define a functional.
Henceforth, for $f\in\mathcal{B}$ and $p=1,2$, we define

$\displaystyle\mathcal{J}_{f,p}(z,r):$
$\displaystyle=|f(z)|^{p}+|f^{\prime}(z)|r+\frac{|f^{\prime\prime}(z)|}{2!}r^{2}+\frac{|f^{\prime\prime\prime}(z)|}{3!}r^{3}+B_{4}(f,r)$
$\displaystyle\quad+|a_{1}|^{2}\frac{r^{4}}{1-r}+A(f_{1},r).$

######  Theorem 2.7.

Suppose that $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}\in\mathcal{B}$. Then we have

  1. (i)

$\mathcal{J}_{f,1}(z,r)\leq 1$ for $|z|=r\leq R_{1}\approx 0.285086$, where
$r_{0}$ is the unique root of the equation
$-1+4r-2r^{2}+3r^{4}+2r^{5}-2r^{6}-2r^{7}=0$ in $(0,1)$. The constant $R_{1}$
is best possible. The equality $\mathcal{J}_{f,1}(z,r)=1$ is achieved for the
function $f_{a}$.

  2. (ii)

$\mathcal{J}_{f,2}(z,r)\leq 1$ for $|z|=r\leq R_{2}\approx 0.386055$, where
$r_{0}$ is the unique root of the equation
$-1+3r-r^{2}-r^{3}+2r^{4}+r^{5}-r^{6}-r^{7}=0$ in $(0,1)$. The constant
$R_{2}$ is best possible. The equality $\mathcal{J}_{f,2}(z,r)=1$ is achieved
for the function $f_{a}$.

###  2.3. Improved Bohr inequality for the class $\mathcal{B}$

Let $f$ be holomorphic in $\mathbb{D}$, and for $0<r<1$, let
$\mathbb{D}_{r}=\\{z\in\mathbb{C}:|z|<r\\}$. Throughout the paper,
$S_{r}:=S_{r}(f)$ denotes the planar integral

$\displaystyle S_{r}:=\int_{\mathbb{D}_{r}}|f^{\prime}(z)|^{2}dA(z).$

Kayumov and Ponnusamy [36] proved the following sharp inequality for functions
in the class $\mathcal{B}$, for the function
$f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$,

(2.6)
$\displaystyle\frac{S_{r}}{\pi}:=\sum_{n=1}^{\infty}n|a_{n}|^{2}r^{2n}\leq
r^{2}\frac{(1-|a_{0}|^{2})^{2}}{(1-|a_{0}|^{2}r^{2})^{2}}\;\mbox{for}\;0<r\leq
1/\sqrt{2}.$

In [35], Kayumov and Ponnusamy have established the following improved version
of Bohr’s inequality in terms of $S_{r}$. In fact, it is worth mentioning that
this result later play a significant role in studying improved Bohr
inequalities for certain classes of harmonic mappings (see [6, 5, 2, 4]),
operator-valued functions (see [13]).

######  Theorem 2.8.

[35] Suppose that $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}\in\mathcal{B}.$ Then

(2.7) $\displaystyle B_{0}(f,r)+\frac{16}{9}\left(\frac{S_{r}}{\pi}\right)\leq
1\quad\mbox{for}\quad r\leq\frac{1}{3},$

and the numbers $1/3$, $16/9$ cannot be improved. Moreover,

(2.8)
$\displaystyle|a_{0}|^{2}+B_{1}(f,r)+\frac{9}{8}\left(\frac{S_{r}}{\pi}\right)\leq
1\quad\mbox{for}\quad r\leq\frac{1}{2},$

and the numbers $1/2$, $9/8$ cannot be improved.

Ismagilov _et al._ [33] proved the next result showing an interesting fact
that if $|a_{0}|$ is replaced by $|f(z)|$ in equation (2.7) of Theorem 2.8,
then the Bohr radius will decrease.

######  Theorem 2.9.

[33] Suppose that $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}\in\mathcal{B}$. Then

$\displaystyle|f(z)|+B_{1}(f,r)+p\left(\frac{S_{r}}{\pi}\right)\leq
1\;\;\mbox{for}\;|z|=r\leq r_{0}=\sqrt{5}-2,$

where the constants $r_{0}=\sqrt{5}-2\approx 0.236068$ and $p=2(\sqrt{5}-1)$
are sharp.

Continuing the research, Lui _et al._ (see [47]) have refined the inequalities
(2.7) and (2.8) and established the following result.

######  Theorem 2.10.

[47] Suppose that $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}\in\mathcal{B}$. Then

(2.9) $\displaystyle
B_{0}(f,r)+A(f_{0},r)+\frac{8}{9}\left(\frac{S_{r}}{\pi}\right)\leq
1\quad\mbox{for}\quad r\leq\frac{1}{3},$

and the numbers $1/3$ and $8/9$ cannot be improved. Moreover,

(2.10)
$\displaystyle|a_{0}|^{2}+B_{1}(f,r)+A(f_{0},r)+\frac{9}{8}\left(\frac{S_{r}}{\pi}\right)\leq
1$

for $r\leq{1}/{(3-|a_{0}|)},$ and the constant $9/8$ cannot be improved.

Recently, the inequality (2.10) in [9] have improved by replacing the
coefficient $|a_{0}|^{2}$ by $|f(z)|^{2}$ by obtaining the following result.

######  Theorem 2.11.

[9] Suppose that $f(z)\in\mathcal{B}$ and
$f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$. Then

$\displaystyle|f(z)|^{2}+B_{1}(f,r)+A(f_{0},r)+\lambda\left(\frac{S_{r}}{\pi}\right)\leq
1$

for $|z|=r\leq 1/3,$ and the constant $\lambda=8/9$ cannot be improved.

The investigation into improving the Bohr inequalities and their different
sharp versions constitutes a thriving realm of the research in Bohr
phenomenon. Actually, researchers have formulated several improved versions of
the Bohr inequalities by incorporating the quantities $S_{r}/\pi$ and
$S_{r}/(\pi-S_{r})$ in the Bohr inequalities. This line of inquiry was
initiated by Kayumov and Ponnusamy (see [35, Theorem 1]), who established
improved version of the Bohr inequalities, denoted as $B_{0}(f,r)$ and
$|a_{0}|^{2}+B_{1}(f,r)$, by the incorporation of $S_{r}/\pi$ and $S_{r}/(\pi-
S_{r})$, respectively. Subsequently, Ismagilov _et al._ (see [33, Theorem 3])
further improved the primary inequality of Kayumov and Ponnusamy (refer to
[35, Theorem 1]) by substituting the initial coefficient $|a_{0}|$ with
$|f(z)|$. Following this, Liu _et al._ (see [47, Theorem 4]) not only refined
both inequalities introduced by Kayumov and Ponnusamy (in [35, Theorem 1]) but
also included an augmenting term $A(f_{0},r)$ into the inequalities, thereby
ushering in a significant advancement in this domain of study.

As a result of a comprehensive scrutiny of the results and the corresponding
proofs furnished by Liu _et al._ (as mentioned in [47]), a compelling question
emerges. Addressing this question could effectively bridge a gap within the
existing research.

######  Question 2.3.

Can the inequality (2.3), established by Liu _et al._ (as elaborated in [47]),
be further improved by the inclusion of the term $S_{r}/\pi$? If so, what can
we say about its sharpness?

We obtain the following result which addresses the Question 2.3 completely.

######  Theorem 2.12.

Suppose that $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}\in\mathcal{B}.$ Then

$\displaystyle\mathcal{G}_{1,f}(r)$
$\displaystyle:=|f(z)|+|f^{\prime}(z)|r+B_{2}(f,r)+A(f_{0},r)+\lambda\left(\frac{S_{r}}{\pi}\right)\leq
1$

for $|z|=r\leq(\sqrt{17}-3)/4$, where the numbers
$\lambda=(221-43\sqrt{17})/64$ and $(\sqrt{17}-3)/4$ cannot improved. The
equality $\mathcal{G}_{1,f}(r)=1$ is achieved for the function $f_{a}$.

Moving forward, we will delve into the evaluation of the upper bound for the
quantity $S_{r}/(\pi-S_{r})$. It’s worth noting that Ismagilov _et al._ [34]
have introduced an improved version of the Bohr inequality by recognizing that

(2.11) $\displaystyle
1-\frac{S_{r}}{\pi}\geq\frac{(1-r^{2})(1-r^{2}|a_{0}|^{4})}{(1-|a_{0}|^{2}r^{2})^{2}}.$

With the help of (2.6) and (2.11), the following inequality

(2.12) $\displaystyle\frac{S_{r}}{\pi-
S_{r}}\leq\frac{r^{2}(1-|a_{0}|^{2})^{2}}{(1-r^{2})(1-r^{2}|a_{0}|^{4})}.$

can be obtained easily.

In view of this new setting, Ismagilov _et al._ [34] have investigated Theorem
2.8 further and obtained the following sharp result.

######  Theorem 2.13.

[34] Suppose that $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}\in\mathcal{B}$. Then

$\displaystyle B_{0}(f,r)+\frac{16}{9}\left(\frac{S_{r}}{\pi-S_{r}}\right)\leq
1\quad\mbox{for}\quad r\leq\frac{1}{3},$

and the number $16/9$ cannot be improved. Moreover,

$\displaystyle|a_{0}|^{2}+B_{1}(f,r)+\frac{9}{8}\left(\frac{S_{r}}{\pi-
S_{r}}\right)\leq 1\quad\mbox{for}\quad r\leq\frac{1}{2},$

and the number $9/8$ cannot be improved.

In recent developments, an improved version of the inequality (2.9) mentioned
in [9] has been put forth. This improvement involves the substitution of the
coefficient $|a_{0}|$ with $|f(z)|^{2}$, while employing the context of
$S_{r}/(\pi-S_{r})$ instead of $S_{r}/\pi$, as detailed below.

######  Theorem 2.14.

[9] Suppose that $f(z)\in\mathcal{B}$ and
$f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$. Then

(2.13)
$\displaystyle|f(z)|^{2}+B_{1}(f,r)+A(f_{0},r)+\lambda\left(\frac{S_{r}}{\pi-
S_{r}}\right)\leq 1,$

for $|z|=r\leq 1/3,$ and the constant $\lambda=8/9$ cannot be improved.

It’s worth observing that in all the aforementioned discussions, when scholars
employ the context of $S_{r}/(\pi-S_{r})$ in place of $S_{r}/\pi$, the
parameter $\lambda$ remains consistent. To facilitate further exploration of
the subject, the following question naturally arises.

######  Question 2.4.

Can we improve the inequality (2.3) which was established by Lui _et al._ (see
[47]) by adding $S_{r}/(\pi-S_{r})$ instead of $S_{r}/\pi$ and it will
interesting to see that the same $\lambda$ will work or not?

We prove the following sharp result corresponding to Theorem 2.10 and it
completely addresses the Question 2.4.

######  Theorem 2.15.

Suppose that $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}\in\mathcal{B}$ and $S_{r}$
denotes the Riemann surface of the function $f^{-1}$ defined on the image of
the sub-disk $|z|\leq r$ under the mapping $f$. Then

$\displaystyle\mathcal{G}_{2,f}(r)$
$\displaystyle:=|f(z)|+|f^{\prime}(z)|r+B_{2}(f,r)+A(f_{0},r)+\lambda\left(\frac{S_{r}}{\pi-
S_{r}}\right)\leq 1$

for $r\leq(\sqrt{17}-3)/4$, where the numbers $\lambda=(221-43\sqrt{17})/64$
and $(\sqrt{17}-3)/4\;$ cannot be improved. The equality
$\mathcal{G}_{2,f}(r)=1$ is achieved for the function $f_{a}$.

######  Remark 2.1.

The values of $\lambda=(221-43\sqrt{17})/64$ and $(\sqrt{17}-3)/4$ remain
invariant in Theorems 2.12 and 2.15, thereby exemplifying the alignment of our
study with the core tenets of Bohr inequalities for the class $\mathcal{B}$.

###  2.4. Key Lemmas, Proof of Theorems 2.6 and 2.7

In this section, we present some necessary lemmas which will play key roles to
prove Theorems 2.6 and 2.7 for the class $\mathcal{B}$.

######  Lemma 2.1.

[37] Let $|a_{0}|<1$ and $0<r\leq 1$. Suppose that
$f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}\in\mathcal{B}$. Then the following sharp
inequality holds:

$\displaystyle\sum_{n=1}^{\infty}|a_{n}|^{2}r^{pn}\leq
r^{p}\dfrac{(1-|a_{0}^{2}|)^{2}}{1-|a_{0}|^{2}r^{p}}.$

######  Lemma 2.2.

[36] Suppose that $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}\in\mathcal{B}.$ Then the
following sharp inequality holds:

$\displaystyle\frac{S_{r}}{\pi}=\sum_{n=1}^{\infty}n|a_{n}|^{2}r^{2n}\leq
r^{2}\frac{(1-|a_{0}|^{2})^{2}}{(1-|a_{0}|^{2}r^{2})^{2}}\;\,\,\mbox{for}\;\;0<r\leq
1/\sqrt{2}.$

######  Lemma 2.3.

[47] Suppose that $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}\in\mathcal{B}$. Then for
any N$\in\mathbb{N}$, the following inequality holds:

$\displaystyle\sum_{n=N}^{\infty}$
$\displaystyle|a_{n}|r^{n}+sgn(t)\sum_{n=1}^{t}|a_{n}|^{2}\dfrac{r^{N}}{1-r}+\left(\dfrac{1}{1+|a_{0}|}+\dfrac{r}{1-r}\right)\sum_{n=t+1}^{\infty}|a_{n}|^{2}r^{2n}$
$\displaystyle\leq(1-|a_{0}|^{2})\dfrac{r^{N}}{1-r},$

for $r\in[0,1),$ where $t=\lfloor{(N-1)/2}\rfloor$.

######  Lemma 2.4.

[31] Suppose that $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}\in\mathcal{B}$. Then
$|a_{n}|\leq 1-|a_{0}|^{2}$ for all $n=1,2,...$

######  Lemma 2.5.

[24] Suppose that $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}\in\mathcal{B}$. Then,
for all $n=1,2,3.....$, we have

$\displaystyle|f^{(k)}(z)|\leq\frac{k!(1-|f(z)|^{2})}{(1-|z|^{2})^{k}}(1+|z|)^{k-1},\;\;|z|<1.$

###### Proof of Theorem 2.6.

Let $f(z)=\sum_{k=0}^{\infty}a_{k}z^{k}$$\in\mathcal{B}$ and $a:=|a_{0}|$ .
Since $f(0)=a_{0}$, by the Schwarz-Pick Lemma, for $z\in\mathbb{D},$ we easily
obtain that

(2.14)
$\displaystyle\dfrac{|f(z)-a_{0}|}{|1-\overline{a_{0}}f(z)|}\leq|z|,\;\mbox{and}\;|f^{(k)}(z)|\leq\frac{k!(1-|f(z)|^{2})}{(1-|z|^{2})^{k}}(1+|z|)^{k-1},\;\;|z|<1.$

Thus it follows from the above inequalities and Lemma 2.4 that, for
$z=re^{i\theta}\in\mathbb{D}$,

(2.15) $\displaystyle|f(z)|\leq\dfrac{r+a}{1+ra}\;\mbox{and}\;|a_{k}|\leq
1-a^{2}\;\mbox{for}\;k=1,2,\cdots$

In view of the inequalities (2.14), (2.15)and Lemma 2.3 (with $N=3$), a simple
computation shows that

$\displaystyle\mathcal{D}_{f}(z,r)$
$\displaystyle\leq|f(z)|^{2}+\frac{r}{1-r^{2}}\left(1-|f(z)|^{2}\right)+\frac{r^{2}(1+r)}{(1-r^{2})^{2}}\left(1-|f(z)|^{2}\right)+B_{3}(f,r)+\frac{|a^{2}_{1}|r^{3}}{1-r}$
$\displaystyle\quad+\frac{(1-a^{2})r^{3}}{(1-r)}$
$\displaystyle\leq\left(\frac{r+a}{1+ra}\right)^{2}+\frac{r}{1-r^{2}}\left(1-\left(\frac{r+a}{1+ra}\right)^{2}\right)+\frac{r^{2}(1+r)}{(1-r^{2})^{2}}\left(1-\left(\frac{r+a}{1+ra}\right)^{2}\right)$
$\displaystyle\quad+\left(1-a^{2}\right)\frac{r^{3}}{1-r}$
$\displaystyle\leq\left(\frac{r+a}{1+ra}\right)^{2}+\frac{(1-a^{2})r}{(1+ar)^{2}}+\frac{(1-a^{2})r^{2}}{(1-r)(1+ar)^{2}}+\frac{(1-a^{2})r^{3}}{(1-r)}$
$\displaystyle=1+\dfrac{(1-a)G_{1}(a,r)}{(1-r)^{2}(1+ar)^{2}},$

where $G_{1}(a,r):=-1+2r+r^{2}+2ar^{4}+a^{2}r^{5}$. For each fixed
$r\in[0,1]$, it is clear that $G_{1}(a,r)$ is a monotonically increasing
function of $a\in[0,1)$ and thus, $G_{1}(a,r)\leq
G_{1}(1,r)=-1+2r+r^{2}+2r^{4}+r^{5}=G_{1}(r)$. Therefore,
$\mathcal{G}_{f}(z)\leq 1$ if $G_{1}(r)\leq 0$. In fact, we see that
$G_{1}(r)\leq 0$ for $r\leq\gamma_{1}$, where $r_{0}$ is the unique root in
$(0,1)$ of the equation $G_{1}(r)=-1+2r+r^{2}+2r^{4}+r^{5}=0$. Consequently,
the inequality of the result is obtained.

The second part of the proof is to show that $r_{0}$ is best possible. In
order to do that, we consider the function $f_{a}$ defined by

(2.16) $\displaystyle
f_{a}(z)=\dfrac{a-z}{1-az}=a-(1-a^{2})\sum_{k=1}^{\infty}a^{k-1}z^{k}.$

For this function $f_{a}$ with $z=-r$, an easy computation shows that

$\displaystyle\mathcal{D}_{f_{a}}(z,r)$
$\displaystyle=|f_{a}(-r)|^{2}+|f_{a}^{\prime}(-r)|r+\frac{|f_{a}^{\prime\prime}(-r)|}{2!}r^{2}+B_{3}(f_{a},r)+|a^{2}_{1}|\frac{r^{3}}{1-r}+A(f_{1},r)$
$\displaystyle=\left(\frac{r+a}{1+ra}\right)^{2}+\frac{r(1-a^{2})}{(1+ra)^{2}}+\frac{r^{2}a(1-a^{2})}{(1+ra)^{3}}+\sum_{n=3}^{\infty}(1-a^{2})a^{n-1}r^{n}$
$\displaystyle\quad+\frac{r^{3}(1-a^{2})^{2}}{1-r}+\frac{(1+ra)}{(1+a)(1-r)}\sum_{n=2}^{\infty}\left((1-a^{2})a^{n-1}\right)^{2}r^{2n}$
$\displaystyle=\left(\frac{r+a}{1+ra}\right)^{2}+\frac{r(1-a^{2})}{(1+ra)^{2}}+\frac{r^{2}a(1-a^{2})}{(1+ra)^{3}}+\frac{r^{3}a^{2}(1-a^{2})^{2}}{(1-ar)}+\frac{r^{3}(1-a^{2})^{2}}{(1-r)}$
$\displaystyle\quad+\frac{(1+ra)}{(1+a)(1-r)}\frac{(1-a^{2})^{2}a^{2}r^{4}}{(1-a^{2}r^{2})}$
$\displaystyle=1+\dfrac{(1-a^{2})E_{1}(a,r)}{(1-r)(1+ar)^{3}}>1$

if, and only if, $(1-a^{2})E_{1}(a,r)>0$, where $E_{1}(a,r)$ is defined by

$\displaystyle E_{1}(a,r)$
$\displaystyle:=-1+2r-ar+3ar^{2}-ar^{3}+2ar^{4}+3a^{2}r^{5}+a^{3}r^{6}$
$\displaystyle\leq-1+2r+3ar^{2}-ar^{3}+2ar^{4}+3a^{2}r^{5}+a^{3}r^{6}=E^{*}_{1}(a,r).$

Indeed, it is evident that

$\displaystyle\lim_{a\to
1}E^{*}_{1}(a,r)=-1+2r+3r^{2}-r^{3}+2r^{4}+3r^{5}+r^{6}=E_{1}(r).$

Since $E_{1}(r)$ is a real-valued differentiable function on $(0,1)$
satisfying $E_{1}(0)=-1<0$ and $E_{1}(0)=9>0$ and $E_{1}(r)$ is increasing
function in $(0,1)$, then with the help of Intermediate value theorem, we can
say that $E_{1}(r)$ has unique real root $r_{1}\approx 0.333004$ in $(0,1)$
and $r_{1}<r_{0}$. Thus, it is clear that $E_{1}(r)>0$ for $r>r_{1}$ and
hence, $E_{1}(a,r)>0$ for all $r>r_{0}$ and $a\in[0,1),$ and this shows that
$r_{0}$ is the best possible. This proves the sharpness. ∎

###### Proof of Theorem 2.7.

Let $f(z)=\sum_{k=0}^{\infty}a_{k}z^{k}$$\in\mathcal{B}$ and $a:=|a_{0}|$.
Since $f(0)=a_{0}$, In view of the inequalities (2.14), (2.15)and Lemma 2.3
(with $N=4$), a simple computation shows that

$\displaystyle\mathcal{J}_{f,1}(z,r)$
$\displaystyle\leq\frac{r+a}{1+ra}+\frac{r}{1-r^{2}}\left(1-\left(\frac{r+a}{1+ra}\right)^{2}\right)+\frac{r^{2}(1+r)}{(1-r^{2})^{2}}\left(1-\left(\frac{r+a}{1+ra}\right)^{2}\right)$
$\displaystyle\quad+\frac{r^{3}(1+r)^{2}}{(1-r^{2})^{3}}\left(1-\left(\frac{r+a}{1+ra}\right)^{2}\right)+\left(1-a^{2}\right)\frac{r^{4}}{1-r}$
$\displaystyle\leq\frac{r+a}{1+ra}+\frac{(1-a^{2})r}{(1+ar)^{2}}+\frac{(1-a^{2})r^{2}}{(1-r)(1+ar)^{2}}+\frac{(1-a^{2})r^{3}}{(1-r)^{2}(1+ar)^{3}}+\frac{(1-a^{2})r^{4}}{(1-r)}$
$\displaystyle=1+\dfrac{(1-a)G_{2}(a,r)}{(1-r)^{2}(1+ar)^{2}},$

where

$\displaystyle G_{2}(a,r):$
$\displaystyle=r^{6}(1-r)a^{3}+[r^{5}\\{(1-r)+(1-r^{2})\\}]a^{2}+[r^{5}+2r^{4}\\{(1-r)$
$\displaystyle\quad\quad+(1-r^{2})\\}]a-r^{5}+r^{4}+2r^{3}-4r^{2}+4r-1.$

For each fixed r$\in[0,1]$, it can be shown that $G_{2}(a,r)$ is a
monotonically increasing function of $a\in[0,1)$ and thus, we have

$\displaystyle G_{2}(a,r)\leq
G_{2}(1,r)=-1+4r-2r^{2}+3r^{4}+2r^{5}-2r^{6}-2r^{7}:=G_{2}(r).$

Therefore, we see that $\mathcal{J}_{f,1}(z,r)\leq 1$ if $G_{2}(r)\leq 0$.
However, a simple computation yields that $G_{2}(r)\leq 0$ for $r\leq R_{1}$,
where $R_{1}$ is the unique root in $(0,1)$ of the equation $G_{2}(r)=0$. Thus
the desired inequality of the theorem is established.

The second part of the proof is to show that the radius $R_{1}$ is optimal. In
order to do that, we consider the function $f_{a}$ defined by (2.16). For such
$f_{a}$ with $z=-r$, an easy computation shows that

$\displaystyle\mathcal{J}_{f_{a},1}(z,r)$
$\displaystyle=|f_{a}(-r)|+|f_{a}^{\prime}(-r)|r+\frac{|f_{a}^{\prime\prime}(-r)|}{2!}r^{2}+\frac{|f_{a}^{\prime\prime\prime}(-r)|}{3!}r^{3}+B_{4}(f_{a},r)+|a_{1}|^{2}\frac{r^{4}}{1-r}$
$\displaystyle\quad+A(f_{1},r)$
$\displaystyle=\frac{r+a}{1+ra}+\frac{r(1-a^{2})}{(1+ra)^{2}}+\frac{r^{2}a(1-a^{2})}{(1+ra)^{3}}+\frac{r^{3}a^{2}(1-a^{2})}{(1+ra)^{4}}+\sum_{n=4}^{\infty}(1-a^{2})a^{n-1}r^{n}$
$\displaystyle\quad+\frac{r^{4}(1-a^{2})^{2}}{1-r}+\frac{(1+ra)}{(1+a)(1-r)}\sum_{n=2}^{\infty}\left((1-a^{2})a^{n-1}\right)^{2}r^{2n}$
$\displaystyle=\frac{r+a}{1+ra}+\frac{r(1-a^{2})}{(1+ra)^{2}}+\frac{r^{2}a(1-a^{2})}{(1+ra)^{3}}+\frac{r^{3}a^{2}(1-a^{2})}{(1+ra)^{4}}+\frac{r^{4}(1-a^{2})^{2}}{(1-r)}$
$\displaystyle\quad+\frac{r^{4}a^{3}(1-a^{2})^{2}}{(1-ar)}+\frac{(1+ra)}{(1+a)(1-r)}\frac{(1-a^{2})^{2}a^{2}r^{4}}{(1-a^{2}r^{2})}$
$\displaystyle=1+\dfrac{(1-a)E_{2}(a,r)}{(1-r)(1+ar)^{4}}>1$

if, and only if, $(1-a)E_{2}(a,r)>0$, where $E_{2}(a,r)$ is defined by

$\displaystyle E_{2}(a,r)$
$\displaystyle:=-1+3r-2ar-2r^{2}+8ar^{2}-6ar^{3}+6a^{2}r^{3}+2a^{3}r^{3}+r^{4}+ar^{4}-6a^{2}r^{4}$
$\displaystyle\quad-a^{3}r^{4}+4ar^{5}+4a^{2}r^{5}-a^{3}r^{5}+6a^{2}r^{6}+6a^{3}r^{6}+4a^{3}r^{7}+4a^{4}r^{7}+a^{4}r^{8}+a^{5}r^{8}$
$\displaystyle\leq-1+3r-2ar+8ar^{2}-6ar^{3}+6a^{2}r^{3}+2a^{3}r^{3}+r^{4}+ar^{4}-a^{3}r^{4}+4ar^{5}$
$\displaystyle\quad+4a^{2}r^{5}-a^{3}r^{5}+6a^{2}r^{6}+6a^{3}r^{6}+4a^{3}r^{7}+4a^{4}r^{7}+a^{4}r^{8}+a^{5}r^{8}$
$\displaystyle:=E^{*}_{2}(a,r).$

It is easy to see that

$\displaystyle\lim_{a\to
1}E^{*}_{2}(a,r)=-1+r+8r^{2}+2r^{3}+r^{4}+7r^{5}+12r^{6}+8r^{7}+2r^{8}:=E_{2}(r).$

Since, $E_{2}(r)$ is a real-valued continuously differentiable function on
$(0,1)$ satisfying $E_{2}(0)=-1<0$ and $E_{2}(0)=40>0$ and $E_{2}(r)$ is
increasing function in $(0,1)$, with the help of Intermediate value theorem,
it is evident that $E_{2}(r)$ has unique real $r_{2}\approx 0.283682$ root in
$(0,1)$ and $r_{2}<R_{1}$. Thus, it is clear that $E_{2}(r)>0$ for $r>R_{1}$,
Hence, it is easy to see that $E_{2}(a,r)>0$ for all $r>R_{1}$ and
$a\in[0,1).$ This shows that $R_{1}$ is the best possible. This proves the
sharpness.

By the similar method as applied above, the other part can be proved easily.
Hence, we omit the details of its proof. ∎

###### Proof of Theorem 2.12.

Let $|a_{0}|=a\in(0,1)$. Combining the Lemma 2.3 (with $N=2$) with the
classical inequality for $|f(z)|$ and Schwarz-Pick Lemma, we obtain the
following estimate

$\displaystyle\mathcal{G}_{1,f}(r)$
$\displaystyle:=|f(z)|+|f^{\prime}(z)|r+B_{2}(f,r)+A(f_{0},r)+\frac{221-43\sqrt{17}}{64}\left(\frac{S_{r}}{\pi}\right)$
$\displaystyle\leq\frac{r+a}{1+ra}+\frac{r}{1-r^{2}}\left(1-\left(\frac{r+a}{1+ra}\right)^{2}\right)+\frac{(1-a^{2})r^{2}}{1-r}$
$\displaystyle\quad+\frac{221-43\sqrt{17}}{64}\frac{(1-a^{2})^{2}r^{2}}{(1-a^{2}r^{2})^{2}}:=A_{1}(r).$

It is easy to see that $A_{1}(r)$ is an increasing function of $r$. Thus it
follows that

$\displaystyle A_{1}(r)$ $\displaystyle\leq
A_{1}\left(\frac{\sqrt{17}-3}{4}\right)$
$\displaystyle=\frac{(-3+\sqrt{17})+4a}{4+\left(-3+\sqrt{17}\right)a}+\frac{(-3+\sqrt{17})^{2}(1-a^{2})}{4(7-\sqrt{17})}+\frac{4(-3+\sqrt{17})(1-a^{2})}{\left(4+(-3+\sqrt{17})|a_{0}|\right)^{2}}$
$\displaystyle\quad+\frac{(-3+\sqrt{17})^{2}(221-43\sqrt{17})(1-a^{2})^{2}}{16\left(8+(-13+3\sqrt{17})a^{2}\right)^{2}}$
$\displaystyle=1-\frac{(1-a)F_{1}(a)}{(7-\sqrt{17})\left(-16+(26-6\sqrt{17})a^{2}\right)^{2}},$

where

$\displaystyle F_{1}(a)$
$\displaystyle:=(-23126+5658\sqrt{17})+2\left(-28769+6983\sqrt{17}\right)a+24\left(-2041+495\sqrt{17}\right)a^{2}$
$\displaystyle\quad+8\left(-2041+495\sqrt{17}\right)a^{3}.$

Moreover, it is not hard to show that $F_{1}(a)\geq 0$ $\forall\;a\in(0,1)$.
Consequently, we have $A_{1}\left({\sqrt{17}-3}/{4}\right)\leq 1$, and with
this the desired inequality of the result is established.

To prove the constant $(221-43\sqrt{17})/64$ is best possible, let us consider
the function defined as

(2.17) $\displaystyle
f^{*}_{a}(z):=\frac{a+z}{1+az}=a+(1-a^{2})\sum_{k=1}^{\infty}(-a)^{k-1}z^{k}.$

For this function $f^{*}_{a}$ with $z=r$, a straightforward calculation shows
that

$\displaystyle\mathcal{G}_{1,f^{*}_{a}}(r)$
$\displaystyle:=|f^{*}_{a}(r)|+|\left(f^{*}_{a}\right)^{\prime}(r)|r+B_{2}(f,r)+A(({f}^{*}_{a})_{\theta},r)+\lambda\left(\frac{S_{r}}{\pi}\right)$
$\displaystyle=\frac{r+a}{1+ra}+\frac{(1-a^{2})r}{(1+ar)^{2}}+\frac{(1-a^{2})ar^{2}}{(1-ar)}+\frac{(1+ar)}{(1+a)(1-r)}\frac{(1-a^{2})^{2}r^{2}}{1-a^{2}r^{2}}$
$\displaystyle\quad+\lambda\frac{(1-a^{2})^{2}r^{2}}{(1-a^{2}r^{2})^{2}}.$

In the case $r=(\sqrt{17}-3)/4$, the last expression becomes

(2.18) $\displaystyle
1+\frac{(1-a)^{2}F_{2}(a,\lambda)}{(7-\sqrt{17})\left(8+(-13+3\sqrt{17})a^{2}\right)^{2}},$

where

$\displaystyle F_{2}(a,\lambda)$
$\displaystyle:=(-1248+288\sqrt{17})+(-5456+1328\sqrt{17})a+(-9168+2224\sqrt{17})a^{2}$
$\displaystyle\quad+(-4082+990\sqrt{17})a^{3}+(-4082+990\sqrt{17})a^{4}+16\lambda(71-17\sqrt{17})(1+a)^{2}.$

The expression (2.18) is bigger then $1$ when $F_{2}(a,\lambda)>0$ for $a\to
1^{-}$ if

$\displaystyle\lambda>\lim_{a\to
1^{-}}\frac{M_{1}(a)}{16(71-17\sqrt{17})(1+a)^{2}}=\frac{221-43\sqrt{17}}{64},$

where
$M_{1}(a)=1248-288\sqrt{17}+(5456-1328\sqrt{17})a+(9168-2224\sqrt{17})a^{2}+(4082+990\sqrt{17})a^{3}+(4082-990\sqrt{17})a^{4}$.
This shows that the constant $(221-43\sqrt{17})/64$ is best possible and with
this the the proof is completed. ∎

###### Proof of Theorem 2.15.

We suppose that $|a_{0}|=a\in(0,1)$. Combining, Lemma 2.3 (with $N=2$) with
the inequality for $|f(z)|$ and Schwarz-Pick lemma, in view of (2.12), we
obtain

$\displaystyle\mathcal{G}_{2,f}(r)$
$\displaystyle:=|f(z)|+|f^{\prime}(z)|r+B_{2}(f,r)+A(f_{0},r)+\frac{221-43\sqrt{17}}{64}\left(\frac{S_{r}}{\pi-
S_{r}}\right)$
$\displaystyle\leq\frac{r+a}{1+ra}+\frac{r}{1-r^{2}}\left(1-\left(\frac{r+a}{1+ra}\right)^{2}\right)+\frac{(1-a^{2})r^{2}}{1-r}+\frac{221-43\sqrt{17}}{64}\times$
$\displaystyle\quad\frac{(1-a^{2})^{2}r^{2}}{(1-r^{2})(1-a^{4}r^{2})^{2}}:=A_{2}(r).$

It is easy to see that $A_{2}(r)$ is an increasing function of $r$.
Consequently, we see that

$\displaystyle A_{2}(r)$ $\displaystyle\leq A_{2}\left((\sqrt{17}-3)/4\right)$
$\displaystyle=\frac{(-3+\sqrt{17})+4a}{4+\left(-3+\sqrt{17}\right)a}+\frac{(-3+\sqrt{17})^{2}(1-a^{2})}{4(7-\sqrt{17})}+\frac{4(-3+\sqrt{17})(1-a^{2})}{\left(4+(-3+\sqrt{17})a\right)^{2}}$
$\displaystyle\quad+\frac{(-3+\sqrt{17})^{2}(221-43\sqrt{17})(1-a^{2})^{2}}{16\left(-5+3\sqrt{17})\right)(8+(-13+3\sqrt{17})a^{4})}$
$\displaystyle=1-\frac{(1-a)^{3}F_{3}(a)}{2(4+(-3+\sqrt{17})a)^{2}(52\sqrt{17}-172+(611-149\sqrt{17})a^{4})},$

where

$\displaystyle F_{3}(a)$
$\displaystyle=(-77164+18804\sqrt{17})+8\left(-1133+283\sqrt{17}\right)a+\left(134385-32567\sqrt{17}\right)a^{2}$
$\displaystyle\quad+\left(-21913+5327\sqrt{17}\right)a^{3}+4\left(-33267+8069\sqrt{17}\right)a^{4}$
$\displaystyle\quad+4\left(-17725+4299\sqrt{17}\right)a^{5}.$

It can be easily shown that $F_{3}(a)\geq 0$ $\forall\;a\in(0,1)$. Moreover,
it is easy to see that $A_{2}\left((\sqrt{17}-3)/4\right)\leq 1$, and with
this the desired inequality of the result established.

To prove the constant $(221-43\sqrt{17})/64$ is best possible, let us consider
the function $f^{*}_{a}$ which is given by (2.17). For this function, a
straightforward calculation shows that

$\displaystyle\mathcal{G}_{2,f^{*}_{a}}(r)$
$\displaystyle:=|f^{*}_{a}(r)|+|(f^{*}_{a})^{\prime}(r)|r+B_{2}(f^{*}_{a},r)+A(({f}^{*}_{a})_{\theta},r)+\lambda\left(\frac{S_{r}}{\pi-
S_{r}}\right)$
$\displaystyle=\frac{r+a}{1+ra}+\frac{(1-a^{2})r}{(1+ar)^{2}}+\frac{(1-a^{2})ar^{2}}{(1-ar)}+\frac{(1+ar)}{(1+a)(1-r)}\frac{(1-a^{2})^{2}r^{2}}{1-a^{2}r^{2}}$
$\displaystyle\quad+\frac{\lambda(1-a^{2})^{2}r^{2}}{(1-r^{2})(1-a^{4}r^{2})^{2}}$

In the case of when $r=(\sqrt{17}-3)/4$, the last expression becomes

$\displaystyle
1+\frac{(1-a)^{2}F_{4}(a,\lambda)}{(4+(-3+\sqrt{17})a)^{2}(52\sqrt{17}-172+(611-149\sqrt{17})a^{4})},$

where

$\displaystyle F_{4}(a,\lambda)$
$\displaystyle=(10464-2592\sqrt{17})+16(611-149\sqrt{17})a+2(5588-1356\sqrt{17})a^{2}$
$\displaystyle\quad+24(-1397+339\sqrt{17})a^{4}+4(-7771+1885\sqrt{17})a^{5}$
$\displaystyle\quad+2(-17725+4299\sqrt{17})a^{6}+16\lambda(1+a)^{2}[(284-68\sqrt{17})$
$\displaystyle\quad+4(-251+61\sqrt{17})a+(895-217\sqrt{17})a^{2}].$

The expression (2.18 ) is bigger then $1$ when $F_{4}(a,\lambda)>0$ and as
$a\to 1^{-}$ shows that

$\displaystyle\lambda$ $\displaystyle>\lim_{a\to
1^{-}}\frac{M_{2}(a)}{16(1+a)^{2}[(284-68\sqrt{17})+4(-251+61\sqrt{17})a+(895-217\sqrt{17})a^{2}]}$
$\displaystyle=\frac{(221-43\sqrt{17})}{64},$

where
$M_{2}(a)=96(-109+27\sqrt{17})+16(-611+149\sqrt{17})a+8(-1397+339\sqrt{17})a^{2}+24(1397-339\sqrt{17})a^{4}+2(17725-4299\sqrt{17})a^{6}$.
This shows that $\lambda>(221-43\sqrt{17})/64$ as $a\to 1^{-}$. This completes
the proof. ∎

##  3\. Generalization of the Bohr inequality for certain classes of harmonic
mappings and its applications

Let $\mathcal{H}(\Omega)$ be the class of complex-valued functions harmonic in
$\Omega$. It is well-known that functions $f$ in the class
$\mathcal{H}(\Omega)$ has the following representation $f=h+\overline{g}$,
where $h$ and $g$ both are analytic functions in $\Omega$. The famous Lewy’s
theorem [44] in $1936$ states that a harmonic mapping $f=h+\overline{g}$ is
locally univalent on $\Omega$ if, and only if, the determinant $|J_{f}(z)|$ of
its Jacobian matrix $J_{f}(z)$ does not vanish on $\Omega$, where

$|J_{f}(z)|:=|f_{z}(z)|^{2}-|f_{\bar{z}}(z)|^{2}=|h^{\prime}(z)|^{2}-|g^{\prime}(z)|^{2}\neq
0.$

In view of this result, a locally univalent harmonic mapping is sense-
preserving if $|J_{f}(z)|>0$ and sense-reversing if $|J_{f}(z)|<0$ in
$\Omega$. For detailed information about the harmonic mappings, we refer the
reader to [22, 28]. In [41], Kayumov _et al._ first established the harmonic
extension of the classical Bohr theorem, since then investigating on the Bohr-
type inequalities for certain class of harmonic mappings becomes an
interesting topic of research in geometric function theory.

Methods of harmonic mappings have been applied to study and solve the fluid
flow problems (see [8, 23]). In particular, univalent harmonic mappings having
special geometric properties such as convexity, starlikeness and close-to-
convexity arise naturally in planar fluid dynamical problems. For example, in
$2012$, Aleman and Constantin [8] established a connection between harmonic
mappings and ideal fluid flows. In fact, Aleman and Constantin have developed
ingenious technique to solve the incompressible two dimensional Euler
equations in terms of univalent harmonic mappings (see [23] for details).

Bohr phenomenon can be studied in view of the Euclidean distance and in this
paper, we study the same for certain classes of harmonic mappings. Before we
go into details, we recall here the following concepts.

######  Definition 3.1.

Let $f$ and $g$ be two analytic functions in the unit disc $\mathbb{D}$. We
say that $g$ is subordinate to $f$ if there is a function $\varphi$, analytic
in $\mathbb{D}$, $\varphi(\mathbb{D})\subset\mathbb{D}$ and $\varphi(0)=0$ so
that $g=f\circ\varphi$. In particular, when the function $f$ is univalent, $g$
is subordinate to $f$ when $g(\mathbb{D}\subset f(\mathbb{D}))$ and
$g(0)=f(0)$ (see [27, p. 190]). Consequently, when $g$ is subordinate to $f$,
$|g^{\prime}(0)|\leq|f^{\prime}(0)|$. The class of all function $g$
subordinate to a fixed function $f$ is denoted by $\mathcal{S}(f)$ and
$f(\mathbb{D})=\Omega$.

######  Definition 3.2.

[1] We say that $\mathcal{S}(f)$ has Bohr phenomenon if for any
$g=\sum_{n=0}^{\infty}b_{n}z^{n}\in\mathcal{S}(f)$ and
$f=\sum_{n=0}^{\infty}a_{n}z^{n}$ there is a $\rho^{*}_{0}$,
$0<\rho^{*}_{0}\leq 1$ so that $\sum_{n=1}^{\infty}|b_{n}z^{n}|\leq
d(f(0),\partial\Omega)$, for $|z|<\rho^{*}_{0}$. Notice that
$d(f(0),\partial\Omega)$ denote the Euclidean distance between $f(0)$ and the
boundary of a domain $\Omega$, $\partial\Omega$. In particular, when
$\Omega=\mathbb{D}$, $d(f(0),\partial\Omega)=1-|f(0)|$ and in this case
$\sum_{n=1}^{\infty}|a_{n}z^{n}|\leq d(f(0),\partial\Omega)$ reduces to
$\sum_{n=0}^{\infty}|a_{n}z^{n}|\leq 1$.

The classical Bohr inequality can be written as

(3.1)
$d\left(\sum_{n=0}^{\infty}|a_{n}z^{n}|,|a_{0}|\right)=\sum_{n=1}^{\infty}|a_{n}z^{n}|\leq
1-|f(0)|=d(f(0),\partial(\mathbb{D})),$

where $d$ is the Euclidean distance. More generally, a class $\mathcal{F}$ of
analytic functions $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$ mapping $\mathbb{D}$
into a domain $\Omega$ is said to satisfy a Bohr phenomenon if an inequality
of type (3.1) holds uniformly in $|z|\leq\rho_{0}$, where $0<\rho_{0}\leq 1$
for functions in the class $\mathcal{F}$. Similar definition makes sense for
harmonic functions (see [41]).

In [1], Abu-Muhanna have obtained the following result for subordination class
$\mathcal{S}(f)$ when $f$ is univalent.

######  Theorem 3.1.

[1] If $g=\sum_{n=0}^{\infty}b_{n}z^{n}\in\mathcal{S}(f)$ and
$f=\sum_{n=0}^{\infty}a_{n}z^{n}$ is univalent, then

$\displaystyle\sum_{n=1}^{\infty}|b_{n}|r^{n}\leq d(f(0),\partial\Omega),$

for $|z|=\rho^{*}_{0}\leq 3-\sqrt{8}\approx 0.17157$, where $\rho^{*}_{0}$ is
sharp for the Koebe function $f(z)=z/(1-z)^{2}$.

Let $\mathcal{H}$ be the class of all complex-valued harmonic functions
$f=h+\bar{g}$ defined on the unit disk $\mathbb{D}$, where $h$ and $g$ are
analytic in $\mathbb{D}$ with the normalization $h(0)=h^{\prime}(0)-1=0$ and
$g(0)=0$. Let $\mathcal{H}_{0}$ be defined by
$\mathcal{H}_{0}=\\{f=h+\bar{g}\in\mathcal{H}:g^{\prime}(0)=0\\}.$ Therefore,
each $f=h+\overline{g}\in\mathcal{H}_{0}$ has the following representation

(3.2)
$f(z)=h(z)+\overline{g(z)}=\sum_{n=1}^{\infty}a_{n}z^{n}+\overline{\sum_{n=1}^{\infty}b_{n}z^{n}}=z+\sum_{n=2}^{\infty}a_{n}z^{n}+\overline{\sum_{n=2}^{\infty}b_{n}z^{n}},$

where $a_{1}=1$ and $b_{1}=0$, since $a_{1}$ and $b_{1}$ have been appeared in
later results and corresponding proofs.

Let us recall the Bohr radius for the class of harmonic mappings.

######  Definition 3.3.

[41] Let $f\in\mathcal{H}_{0}$ be given by (3.2). Then the Bohr phenomenon is
to find a constant $R^{*}\in(0,1]$ such that the inequality
$r+\sum_{n=2}^{\infty}\left(|a_{n}|+|b_{n}|\right)r^{n}\leq
d\left(f(0),\partial\Omega\right)$ holds for $|z|=r\leq R^{*}$, where
$d\left(f(0),\partial\Omega\right)$ is the Euclidean distance between $f(0)$
and the boundary of $\Omega:=f(\mathbb{D})$. The largest such radius $R^{*}$
is called the Bohr radius for the class $\mathcal{H}_{0}$.

Let $\\{\varphi_{n}(r)\\}^{\infty}_{n=0}$ be a sequence of non-negative
continuous function in $[0,1)$ such that the series
$\sum_{n=0}^{\infty}\varphi_{n}(r)$ converges locally uniformly on the
interval $[0,1)$.

######  Theorem 3.2.

[39] Let $f\in\mathcal{B}$ with $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$ and
$p\in(0,2]$. If

$\displaystyle\varphi_{0}(r)>\frac{2}{p}\sum_{n=1}^{\infty}\varphi_{n}(r)\;\mbox{for}\;r\in[0,R),$

where $R$ is the minimal positive root of the equation
$\varphi_{0}(x)=\frac{2}{p}\sum_{n=1}^{\infty}\varphi_{n}(x),$ then the
following sharp inequality holds:

$\displaystyle
B_{f}(\varphi,p,r):=|a_{0}|^{p}\varphi_{0}(r)+\sum_{n=1}^{\infty}|a_{n}|\varphi_{n}(r)\leq\varphi_{0}(r)\;\mbox{for
all}\;r\leq R.$

In the case when $\varphi_{0}(r)<\frac{2}{p}\sum_{n=1}^{\infty}\varphi_{n}(r)$
in some interval $(R,R+\epsilon)$, the number $R$ cannot be improved.If the
functions $\varphi_{n}(x)\;(n\geq 0)$ are smooth functions then the last
condition is equivalent to the inequality
$\varphi^{\prime}_{0}(R)<\frac{2}{p}\sum_{n=1}^{\infty}\varphi^{\prime}_{n}(R).$

Using Theorem 3.2, the authors found the value of the Bohr radius of the
identity operator and the convolution operator with hypergeometric Gasussian
function. Let

$\displaystyle
F(z)=F(a,b,c,z)=\sum_{n=0}^{\infty}\gamma_{n}z^{n},\;\gamma_{n}:=\frac{(a)_{n}(b)_{n}}{(c)_{n}(1)_{n}},$

where $a,b,c>-1$, such that $\gamma_{n}\geq 0$,
$(a)_{n}:=a(a+1)\cdots(a+n-1)$, $(a)_{0}=1$. Then for $p\in(0,2]$

$\displaystyle|a_{0}|^{p}+\sum_{n=1}^{\infty}|\gamma_{n}||a_{n}|r^{n}\leq\;\mbox{for
all}\;r\leq R,$

where $R$ is the minimal positive root of the equation $|F(a,b,c,x)-1|=p/2$.
Recently, Chen _et al._ [21] established several new versions of Bohr-type
inequalities for bounded analytic functions in the unit disk $\mathbb{D}$ by
allowing $\\{\varphi_{n}(r)\\}^{\infty}_{n=0}$ in the place of
$\\{r^{n}\\}^{\infty}_{n=0}$ in the power series representations of the
functions involved with the Bohr sum and thereby introducing a single
parameter, which generalized several related results.

Inspired by the idea of the proof technique of [39, 21], our aim is to
establish refined Bohr inequality, finding the corresponding sharp radius for
the class $\mathcal{P}^{0}_{\mathcal{H}}(M)$ which has been studied by Ghosh
and Vasudevarao (see [30])

$\displaystyle\mathcal{P}^{0}_{\mathcal{H}}(M)=\\{f=h+\overline{g}\in\mathcal{H}_{0}:(zh^{\prime\prime}(z))>-M+|zg^{\prime\prime}(z)|,\;z\in\mathbb{D}\;\mbox{and
}\;M>0\\}.$

To study Bohr inequality and Bohr radius for functions in
$\mathcal{P}^{0}_{\mathcal{H}}(M)$, we require the coefficient bounds and
growth estimate of functions in $\mathcal{P}^{0}_{\mathcal{H}}(M)$. We have
the following result on the coefficient bounds and growth estimate for
functions in $\mathcal{P}^{0}_{\mathcal{H}}(M)$.

######  Lemma 3.1.

[30] Let $f=h+\overline{g}\in\mathcal{P}^{0}_{\mathcal{H}}(M)$ be given by
(3.2) for some $M>0$. Then for $n\geq 2,$

  1. (i)

$\displaystyle|a_{n}|+|b_{n}|\leq\frac{2M}{n(n-1)};$

  2. (ii)

$\displaystyle||a_{n}|-|b_{n}||\leq\frac{2M}{n(n-1)};$

  3. (iii)

$\displaystyle|a_{n}|\leq\frac{2M}{n(n-1)}.$

The inequalities are sharp with extremal function $f_{M}$ given by
$f_{M}^{\prime}(z)=1-2M\,\ln\,(1-z).$

######  Lemma 3.2.

[30] Let $f\in\mathcal{P}^{0}_{\mathcal{H}}(M)$ be given by (3.2). Then

$|z|+2M\sum\limits_{n=2}^{\infty}\dfrac{(-1)^{n-1}|z|^{n}}{n(n-1)}\leq|f(z)|\leq|z|+2M\sum\limits_{n=2}^{\infty}\dfrac{|z|^{n}}{n(n-1)}.$

Both inequalities are sharp for the function $f_{M}$ given by
$f_{M}(z)=z+2M\sum\limits_{n=2}^{\infty}\dfrac{z^{n}}{n(n-1)}.$

For the class $\mathcal{P}^{0}_{\mathcal{H}}(M)$, Allu and Halder (see [11])
studied the Bohr inequality in terms of distance formulations and obtained the
following sharp Bohr inequality.

######  Theorem 3.3.

[11, Theorem 2.9] Let $f\in\mathcal{P}^{0}_{\mathcal{H}}(M)$ be given by (3.2)
with $0<M<1/(2(\ln 4-1))$ . Then

$\displaystyle|z|+\sum_{n=2}^{\infty}\left(|a_{n}|+|b_{n}|\right)|z|^{n}\leq
d\left(f(0),\partial f(\mathbb{D})\right)$

holds for $|z|=r\leq r_{f}$, where $r_{f}$ is the unique root of

$\displaystyle
r+2M\sum_{n=2}^{\infty}\frac{r^{n}}{n(n-1)}=1+2M\sum_{n=2}^{\infty}\frac{(-1)^{n-1}}{n(n-1)}$

in $(0,1)$. The radius $r_{f}$ is best possible.

Let $\mathcal{F}$ denote the set of all sequences
$\\{\varphi_{n}(r)\\}^{\infty}_{n}$ of non negative continuous functions in
$[0,1)$ such that the series $\sum_{n=0}^{\infty}\varphi_{n}(r)$ converges
locally uniformly on the interval $[0,1)$
$\;\forall\;n\in\mathbb{N}\cup\\{0\\}$. It is natural to raise the following
question for further exploration of Bohr inequality for the class
$\mathcal{P}^{0}_{\mathcal{H}}(M)$.

######  Question 3.1.

In view of the general setting proposed in [39] with a change of basis from
$\\{r^{n}\\}_{n=0}^{\infty}$ to $\\{\varphi_{n}(r)\\}_{n=0}^{\infty}$, is it
possible to establish a general result or improve Theorem 3.3 in the setting
of one parameter family of Bohr sum?

We begin by presenting our result concerning the Bohr inequality for the class
$\mathcal{P}^{0}_{\mathcal{H}}(M)$ in terms of distance formulation, which is
formulated in response to Question 3.1. It is worth noticing that, if, in
particular, $\varphi_{n}(r)=r^{n}$, $n\in\mathbb{N}\cup\\{0\\}$ in Theorem
3.4, then Theorem 3.3 becomes a special case of Theorem 3.4.

######  Theorem 3.4.

Let $f=h+\overline{g}\in\mathcal{P}^{0}_{\mathcal{H}}(M)$ be given by (3.2)
and suppose that $\\{\varphi_{n}(r)\\}^{\infty}_{n=0}\in\mathcal{F}$ with
condition $\varphi_{0}(1)=1$ and

(3.3)
$\displaystyle\sum_{n=2}^{\infty}\frac{\varphi_{n}(0)}{n(n-1)}<\frac{1}{2M}+1-\ln
4.$

Then

$\displaystyle
A_{f}(r)=r\varphi_{0}(r)+\sum_{n=2}^{\infty}\left(|a_{n}|+|b_{n}|\right)\varphi_{n}(r)\leq
d\left(f(0),\partial f(\mathbb{D})\right),$

for $|z|=r\leq R_{f}(M),$ where $R_{f}(M)$ is the unique positive root in
$(0,1)$ of

$\displaystyle
r\varphi_{0}(r)+2M\sum_{n=2}^{\infty}\frac{\varphi_{n}(r)}{n(n-1)}=1+2M\sum_{n=2}^{\infty}\frac{(-1)^{n-1}}{n(n-1)}.$

The radius $R_{f}(M)$ is best possible.

###  3.1. Application of Theorem 3.4

Let $f=h+\overline{g}\in\mathcal{P}^{0}_{\mathcal{H}}(M)$ be given by (3.2)
for $0<M<1/2(\ln 4-1)$. A simple computation shows that

$\displaystyle\begin{cases}\displaystyle\sum_{n=2}^{\infty}\frac{nr^{n}}{n(n-1)}=-r\ln(1-r),\vspace{2mm}\\\
\displaystyle\sum_{n=2}^{\infty}\frac{n^{2}r^{n}}{n(n-1)}=\frac{r[r-(1-r)\ln(1-r)]}{1-r},\vspace{2mm}\\\
\displaystyle\sum_{n=2}^{\infty}\frac{n^{3}r^{n}}{n(n-1)}=r\left(\frac{(3-2r)r}{(1-r)^{2}}-\ln(1-r)\right).\end{cases}$

  1. (i)

For $f=h+\overline{g}\in\mathcal{P}^{0}_{\mathcal{H}}(M)$ and for
$\varphi_{0}(r)=1$ and $\varphi_{n}(r)=n^{\alpha}r^{n}\;(n\geq 1)$, we define
the following functional

$\displaystyle\mathcal{C}^{f}_{\alpha}(r):=r\varphi_{0}(r)+\sum_{n=2}^{\infty}\left(|a_{n}|+|b_{n}|\right)\varphi_{n}(r).$

We see that $\varphi_{0}(1)=1$ and the condition

$\displaystyle\sum_{n=2}^{\infty}\frac{\varphi_{n}(0)}{n(n-1)}<\frac{1}{2M}+1-\ln
4\;\mbox{coincides with}\;0<M<\frac{1}{2(\ln 4-1)}.$

Then, we have

    1. (a)

$\mathcal{C}^{f}_{1}(r)\leq d\left(f(0),\partial f(\mathbb{D})\right)$ for
$|z|=r\leq R_{1}(M)$, where $R_{1}(M)$ is the unique positive root in $(0,1)$
of the equation $r-2Mr\ln(1-r)=1+2M(1-\ln 4)$. The number $R_{1}(M)$ is best
possible.

    2. (b)

$\mathcal{C}^{f}_{2}(r)\leq d\left(f(0),\partial f(\mathbb{D})\right)$ for
$|z|=r\leq R_{2}(M)$, where $R_{2}(M)$ is the unique positive root in $(0,1)$
of the equation

$\displaystyle r+\frac{2Mr[r-(1-r)\ln(1-r)]}{1-r}=1+2M(1-\ln 4).$

The number $R_{2}(M)$ is best possible.

    3. (c)

$\mathcal{C}^{f}_{3}(r)\leq d\left(f(0),\partial f(\mathbb{D})\right)$ for
$|z|=r\leq R_{3}(M)$, where $R_{3}(M)$ is the unique positive root in $(0,1)$
of the equation

$\displaystyle r+2Mr\left(\frac{(3-2r)r}{(1-r)^{2}}-\ln(1-r)\right)=1+2M(1-\ln
4).$

The number $R_{3}(M)$ is best possible.

$M$ | $0.431$ | $0.862$ | $1.210$ | $1.271$ | $1.289$ | $1.292$ | $1.2935$ | $1.29421$ | $1.29433$
---|---|---|---|---|---|---|---|---|---
$R_{1}(M)$ | $0.443$ | $0.230$ | $0.057$ | $0.017$ | $0.0040$ | $0.0018$ | $0.00065$ | $0.00010$ | $0.000015$
$R_{2}(M)$ | $0.358$ | $0.189$ | $0.029$ | $0.016$ | $0.0040$ | $0.0017$ | $0.00065$ | $0.00010$ | $0.000015$
$R_{3}(M)$ | $0.277$ | $0.149$ | $0.044$ | $0.015$ | $0.0039$ | $0.0017$ | $0.00065$ | $0.00010$ | $0.000015$
Table 1. This table exhibits the the approximate values of the roots
$R_{\alpha}(M)$ for different values of $M$ and $\alpha=1,2,3$, where
$0<M<1/(2(\ln 4-1))\approx 1.29435$. Moreover, we see that
$\lim\limits_{M\rightarrow 1.29435}R_{\alpha}(M)\approx 0.000015$.

  2. (ii)

A simple computation shows that

$\displaystyle\begin{cases}\displaystyle\sum_{n=2}^{\infty}\frac{(n+1)r^{n}}{n(n-1)}=r+(1-2r)\ln(1-r),\vspace{2mm}\\\
\displaystyle\sum_{n=2}^{\infty}\frac{(n+1)^{2}r^{n}}{n(n-1)}=\frac{r+(1-5r+4r^{2})\ln(1-r)}{1-r},\vspace{2mm}\\\
\displaystyle\sum_{n=2}^{\infty}\frac{(n+1)^{3}r^{n}}{n(n-1)}=\frac{r+4r^{2}-4r^{3}+(1-r)^{2}(1-8r)\ln(1-r)}{(1-r)^{2}}.\end{cases}$

For $f=h+\overline{g}\in\mathcal{P}^{0}_{\mathcal{H}}(M)$ and for
$\varphi_{n}(r)=(n+1)^{\beta}r^{n}\;(n\geq 1)$, we define the following
functional

$\displaystyle\mathcal{D}^{f}_{\beta}(r):=r\varphi_{0}(r)+\sum_{n=2}^{\infty}\left(|a_{n}|+|b_{n}|\right)\varphi_{n}(r).$

We see that $\varphi_{0}(1)=1$ and the condition

$\displaystyle\sum_{n=2}^{\infty}\frac{\varphi_{n}(0)}{n(n-1)}<\frac{1}{2M}+1-\ln
4\;\mbox{coincides with}\;0<M<\frac{1}{2(\ln 4-1)}.$

Then, we have

    1. (a)

$\mathcal{D}^{f}_{1}(r)\leq d\left(f(0),\partial f(\mathbb{D})\right)$ for
$|z|=r\leq R^{*}_{1}(M)$, where $R^{*}_{1}(M)$ is the unique positive root in
$(0,1)$ of the equation $r+2M[r+(1-2r)\ln(1-r)]=1+2M(1-\ln 4).$ The number
$R_{1}^{*}(M)$ is best possible.

    2. (b)

$\mathcal{D}^{f}_{2}(r)\leq d\left(f(0),\partial f(\mathbb{D})\right)$ for
$|z|=r\leq R^{*}_{2}(M)$, where $R_{2}(M)$ is the unique positive root in
$(0,1)$ of the equation

$\displaystyle r+\frac{2M[r+(1-5r+4r^{2})\ln(1-r)]}{1-r}=1+2M(1-\ln 4).$

The number $R^{*}_{2}(M)$ is best possible.

    3. (c)

$\mathcal{D}^{f}_{3}(r)\leq d\left(f(0),\partial f(\mathbb{D})\right)$ for
$|z|=r\leq R^{*}_{3}(M)$, where $R^{*}_{3}(M)$ is the unique positive root in
$(0,1)$ of the equation

$\displaystyle
r+\frac{2M[r+4r^{2}-4r^{3}+(1-r)^{2}(1-8r)\ln(1-r)]}{(1-r)^{2}}=1+2M(1-\ln
4).$

The number $R^{*}_{3}(M)$ is best possible.

$M$ | $0.431$ | $0.862$ | $1.210$ | $1.271$ | $1.289$ | $1.292$ | $1.2935$ | $1.29421$ | $1.29433$
---|---|---|---|---|---|---|---|---|---
$R^{*}_{1}(M)$ | $0.404$ | $0.208$ | $0.054$ | $0.016$ | $0.0040$ | $0.0018$ | $0.00065$ | $0.00010$ | $0.000015$
$R^{*}_{2}(M)$ | $0.284$ | $0.147$ | $0.043$ | $0.015$ | $0.0039$ | $0.0017$ | $0.00065$ | $0.00010$ | $0.000015$
$R^{*}_{3}(M)$ | $0.203$ | $0.147$ | $0.043$ | $0.015$ | $0.0039$ | $0.0017$ | $0.00065$ | $0.00010$ | $0.000015$
Table 2. This table exhibits the the approximate values of the roots
$R^{*}_{\beta}(M)$ for different values of $M$ and $\beta=1,2,3$, where
$0<M<1/(2(\ln 4-1))\approx 1.29435$. Moreover, we see that
$\lim\limits_{M\rightarrow 1.29435}R^{*}_{\beta}(M)\approx 0.000015$.

We now discuss the proof of Theorem 3.4 in details.

###  3.2. Proof of Theorem 3.4

Let $f=h+\overline{g}\in\mathcal{P}^{0}_{\mathcal{H}}(M)$ be given by (3.2)
for $0<M<1/2(\ln 4-1)$. By the Lemma 3.2, it is evident that the Euclidean
distance between $f(0)$ and the boundary of $f(\mathbb{D})$ is

(3.4) $\displaystyle d\left(f(0),\partial f(\mathbb{D})\right)\geq
1+2M\sum_{n=2}^{\infty}\frac{(-1)^{n-1}}{n(n-1)}.$

Let $H_{M}:[0,1]\to\mathbb{R}$ be defined by

(3.5) $\displaystyle
H_{M}(r)=r\varphi_{0}(r)+2M\sum_{n=2}^{\infty}\frac{\varphi_{n}(r)}{n(n-1)}-1-2M\sum_{n=2}^{\infty}\frac{(-1)^{n-1}}{n(n-1)}.$

Clearly, $H_{M}$ is continuous in $[0,1]$ and differentiable in $(0,1)$.
Moreover, in view of the condition in (3.3), we note that

$\displaystyle
H_{M}(0)=2M\sum_{n=2}^{\infty}\frac{\varphi_{n}(0)}{n(n-1)}-1-2M\sum_{n=2}^{\infty}\frac{(-1)^{n-1}}{n(n-1)}<0.$

Since $\varphi_{0}(1)=1$ and $1-\ln 4\approx-0.3862<0$, we see that

$\displaystyle H_{M}(1)=$
$\displaystyle\varphi_{0}(1)+2M\sum_{n=2}^{\infty}\frac{\varphi_{n}(1)}{n(n-1)}-1-2M\sum_{n=2}^{\infty}\frac{(-1)^{n-1}}{n(n-1)}$
$\displaystyle=2M\left(\sum_{n=2}^{\infty}\frac{\varphi_{n}(1)}{n(n-1)}-(1-\ln
4)\right)>0.$

By the Intermediate Value Theorem, the function $H_{M}$ has a root in $(0,1)$,
$R_{f}(M)$ say. To show that $R_{f}(M)$ is unique, it is sufficient to show
that $H_{M}$ is a monotonic function on $[0,1]$. An easy computation shows
that

$\displaystyle
H_{M}^{\prime}(r)=r\varphi^{\prime}_{0}(r)+\varphi_{0}(r)+2M\sum_{n=2}^{\infty}\frac{\varphi^{\prime}_{n}(r)}{n(n-1)}>0\;\mbox{for
all}\;r\in(0,1).$

Thus, we conclude that $H_{M}$ is a strictly monotone increasing function on
$(0,1)$. However, we see that

(3.6) $\displaystyle
R_{f}(M)\varphi_{0}(R_{f}(M))+2M\sum_{n=2}^{\infty}\frac{\varphi_{n}(R_{f}(M))}{n(n-1)}=1+2M\sum_{n=2}^{\infty}\frac{(-1)^{n-1}}{n(n-1)}$

The Lemma 3.1 in view of (3.4) yields that

$\displaystyle A_{f}(r)$
$\displaystyle=r\varphi_{0}(r)+\sum_{n=2}^{\infty}\left(|a_{n}|+|b_{n}|\right)\varphi_{n}(r)$
$\displaystyle\leq
r\varphi_{0}(r)+2M\sum_{n=2}^{\infty}\frac{\varphi_{n}(r)}{n(n-1)}$
$\displaystyle\leq 1+2M\sum_{n=2}^{\infty}\frac{(-1)^{n-1}}{n(n-1)}$
$\displaystyle\leq d\left(f(0),\partial f(\mathbb{D})\right)$

for $|z|=r\leq R_{f}(M)$. Thus the desired inequality is established.

The next part of the proof is to show that the number $R_{f}(M)$ is best
possible. Henceforth, we consider the function $f_{M}$ given by

$\displaystyle f_{M}(z)=z+2M\sum_{n=2}^{\infty}\frac{z^{n}}{n(n-1)}.$

Clearly, we see that $f_{M}\in\in\mathcal{P}^{0}_{\mathcal{H}}(M)$. For
$r>R_{f}(M)$ and $f=f_{M}$, an easy computation in view of part-(i) of Lemma
3.1 and (3.6) shows that

$\displaystyle A_{f_{M}}(r)$
$\displaystyle=r\varphi_{0}(r)+\sum_{n=2}^{\infty}\left(|a_{n}|+|b_{n}|\right)\varphi_{n}(r)$
$\displaystyle=r\varphi_{0}(r)+2M\sum_{n=2}^{\infty}\frac{\varphi_{n}(r)}{n(n-1)}$
$\displaystyle>R_{f}(M)\varphi_{0}(R_{f}(M))+2M\sum_{n=2}^{\infty}\frac{\varphi_{n}(R_{f}(M))}{n(n-1)}$
$\displaystyle=1+2M\sum_{n=2}^{\infty}\frac{(-1)^{n-1}}{n(n-1)}$
$\displaystyle=d\left(f_{M}(0),\partial f_{M}(\mathbb{D})\right).$

This turns out that the number $R_{f}(M)$ is best possible.

Compliance of Ethical Standards

Conflict of interest The authors declare that there is no conflict of interest
regarding the publication of this paper.

Data availability statement Data sharing not applicable to this article as no
datasets were generated or analyzed during the current study.

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