text:
\begin{document}
\bibliographystyle{amsplain}
\title[Refined Bohr inequality for bounded analytic functions]
{Refined Bohr inequality for bounded analytic functions}

\def\thefootnote{}
\footnotetext{ \texttt{\tiny File:~\jobname .tex,
          printed: \number\day-\number\month-\number\year,
          \thehours.\ifnum\theminutes<10{0}\fi\theminutes}
} \makeatletter\def\thefootnote{\@arabic\c@footnote}\makeatother

\author[G. Liu]{Gang Liu}
\address{G. Liu, College of Mathematics and Statistics
 (Hunan Provincial Key Laboratory of Intelligent Information Processing and Application),
Hengyang Normal University, Hengyang,  Hunan 421002, China}
\email{liugangmath@sina.cn}

\author[Z.H. Liu]{Zhihong Liu}
\address{Z.H. Liu, College of Science, Guilin University of Technology, Guilin, Guangxi 541004, China}
\email{liuzhihongmath@163.com}

\author[S. Ponnusamy]{Saminathan Ponnusamy

}

\address{S. Ponnusamy, Department of Mathematics,
Indian Institute of Technology Madras, Chennai-600 036, India}
\email{samy@iitm.ac.in}

\subjclass[2010]{Primary:   30A10, 30H05, 30C35; Secondary: 30C45}

\keywords{Bohr inequality, Bohr-Rogosinski inequality, Bohr radius, bounded analytic function, symmetric function.

}

\begin{abstract}
In this article, by combining appropriate refined Bohr's inequalities with some techniques concerning bounded analytic functions defined in the unit disk,
we generalize and improve several Bohr type inequalities for such functions.
\end{abstract}

\maketitle \pagestyle{myheadings}
\markboth{G. Liu, Z. H. Liu and S. Ponnusamy}{Refined Bohr inequality for bounded analytic functions}

\section{Introduction}  \label{sec1}
Throughout the paper, ${\mathcal B}$  denote the set of all analytic self-maps of the unit disk  $\mathbb{D}=\{z\in \IC:\, |z| < 1 \}$.
Let us start recalling the remarkable discovery of H. Bohr in 1914 (see \cite{boh}).

\begin{Thm}\label{thmA}
If $f\in {\mathcal B}$ and $f(z)=\sum_{n=0}^{\infty} a_n z^n$, then the following sharp inequality holds:
$ \sum_{n=0}^{\infty} |a_n|\, r^n \leq 1$ for  $r\leq 1/3$.
\end{Thm}

 We remark that if $|f(z)|\leq 1$ in $\ID$ and $|f(z_0)|=1$ for some point $z_0\in\ID$, then  $f(z)$ reduces a unimodular constant
and thus,  we restrict our attention to $f\in{\mathcal B}$.
The above inequality and the sharp constant $1/3$ are known as the classical Bohr inequality and the Bohr radius for the family ${\mathcal B}$, respectively.
Bohr \cite{boh}  showed it only for $r\leq1/6$ and this article includes Wiener's proof showing that $r=1/3$ is sharp.
Subsequently, some different proofs were given (see \cite{PPS,PS2004,PS2006,sid,tom} and also the recent survey chapters \cite{AAP} and \cite[Chapter 8]{GarMasRoss-2018}).
It is worth pointing out that if $|a_0|$ in Bohr inequality is replaced by $|a_0|^2$,
then the constant $1/3$ could be replaced by $1/2$.
Moreover, if $a_0=0$ in Theorem \Ref{thmA}, then the sharp Bohr radius can be improved to be $1/\sqrt{2}$
(see for example \cite{KP2017}, \cite[Corollary~2.9]{PPS} and the recent paper \cite{PW} for a general result).
For these results, one of the proofs relied on the sharp coefficient inequalities, i.e., $|a_n|\leq1-|a_0|^2$ $(n\geq1, f\in{\mathcal B})$.
By means of these inequalities, it is pointed out in \cite{KP2017} that the sharp result in Theorem \Ref{thmA} cannot be obtained in the
extremal case $|a_0|<1$.
However, the sharp version of Theorem \Ref{thmA} has been achieved for any individual function from ${\mathcal B}$ (see \cite{AKP})
and  for some  subclasses of univalent functions, we refer to \cite{abu,aiz}. In \cite{PVW} (see also \cite{PVW2}), the authors established the following
refined Bohr inequality by applying a refined version of the coefficient inequalities
carefully (see Lemma \ref{lem3}).

\begin{Thm}{ \rm(\cite[Theorem~1]{PVW})} \label{thmB}
Suppose that $f\in {\mathcal B}$ and $f(z)=\sum_{n=0}^{\infty} a_n z^n$.
Then
\begin{equation*}
 \sum_{n=0}^\infty |a_n|r^n+  \left(\frac{1}{1+|a_0|}+\frac{r}{1-r}\right)\sum_{n=1}^{\infty}|a_n|^2r^{2n} \leq 1 ~\mbox{for}~ r \leq \frac{1}{2+|a_0|}
\end{equation*}
and the numbers $\frac{1}{2+|a_0|}$ and $\frac{1}{1+|a_0|}$ cannot be improved.  Moreover,
\begin{equation*}
|a_0|^2+ \sum_{n=1}^\infty |a_n|r^n+ \left(\frac{1}{1+|a_0|}+\frac{r}{1-r}\right)\sum_{n=1}^{\infty}|a_n|^2r^{2n}\leq 1 ~\mbox{for}~ r \leq \frac{1}{2}
\end{equation*}
and the numbers $\frac{1}{2}$ and $\frac{1}{1+|a_0|}$ cannot be improved.
\end{Thm}

Besides these results, there are plenty of works about Bohr inequality for the family ${\mathcal B}$.

Based on the notion of Rogosinski's inequality and Rogosinski's radius investigated in \cite{LG,rog,SS}, Kayumov and Ponnusamy \cite{KP-pre} introduced and obtained
the following Bohr-Rogosinski inequality and Bohr-Rogosinski radius.

\begin{Thm}{ \rm(\cite[Theorem~1]{KP-pre})}\label{thmC}
Suppose that $f\in {\mathcal B}$ and $f(z)=\sum_{n=0}^{\infty} a_n z^n$. Then
$$
|f(z)|+\sum_{n=N}^{\infty}\left|a_{n}\right| r^{n} \leq 1 \quad \text {for} ~|z|=r\leq R_{N},
$$
where $R_{N}$ is the positive root of the equation  $2(1+r) r^{N}-(1-r)^{2}=0$.

The radius $R_{N}$ is best possible. Moreover,
$$
|f(z)|^{2}+\sum_{n=N}^{\infty}\left|a_{n}\right| r^{n} \leq 1 \quad \text {for} ~|z|=r\leq R_{N}^{\prime},
$$
where $R_{N}^{\prime}$ is the positive root of the equation $(1+r) r^{N}-(1-r)^{2}=0 .$ The radius $R_{N}^{\prime}$ is
best possible.
\end{Thm}

In order to determine the Bohr radius for the class of odd functions in the family ${\mathcal B}$, which was posed in \cite{ABS},
the authors in \cite{KP2017,KP2018-1} established a more general result consisting of functions of the form $f_{p,m}(z)=\sum_{n=0}^{\infty} a_{pn+m} z^{pn+m}$ in ${\mathcal B}$.
Note that $f_{p,1}$ is a $p-$symmetric function and $f_{2,1}$ is an odd function.

\begin{Thm}{ \rm(\cite[Theorem~1]{KP2018-1})}\label{thmD}
Let p $\in \mathbb{N}$ and 0 $\leq m \leq p$.
Suppose that $f\in {\mathcal B}$ and $f(z)=\sum_{n=0}^{\infty} a_{pn+m} z^{pn+m}$. Then
$$ \sum_{n=0}^{\infty}\left|a_{pn+m}\right| r^{pn+m} \leq 1 \quad \text {for} ~r \leq r_{p, m},
$$
where $r_{p, m}$ is the maximal positive root of the equation
$-6 r^{p-m}+r^{2(p-m)}+8 r^{2 p}+1=0.
$
The extremal function has the form $z^{m}\left(z^{p}-a\right) /\left(1-az^{p}\right),$ where
$$a=\left(1-\frac{\sqrt{1-r_{p, m}^{2 p}}}{\sqrt{2}}\right) \frac{1}{r_{p, m}^{p}}.
$$
\end{Thm}

 Besides these results,  few different formulations of improved Bohr inequalities were
obtained in \cite{KP2018-2} and also in the recent articles \cite{PVW,PW}.   Below we recall one of them which is associated
with area.

\begin{Thm}{ \rm(\cite[Theorem~1]{KP2018-2})} \label{thmE}
Suppose that $f\in {\mathcal B}$, $f(z)=\sum_{n=0}^{\infty} a_n z^n$ and $S_r$ denotes the area of the
Riemann surface of the function $f^{-1}$ defined on the image of the subdisk $|z|<r$ under the mapping $f$. Then
$$\sum_{n=0}^{\infty}\left|a_{n}\right| r^{n}+\frac{16}{9}\left(\frac{S_{r}}{\pi}\right) \leq 1~\text{for}~ r\leq\frac{1}{3}
$$
and the numbers $1/3$ and $16/9$ cannot be improved. Moreover,
$$
\left|a_{0}\right|^{2}+\sum_{n=1}^{\infty}\left|a_{n}\right| r^{n}+\frac{9}{8}\left(\frac{S_{r}}{\pi}\right) \leq 1 \quad \text {for} ~r\leq\frac{1}{2}
$$
and the constants $1/2$ and $9/8$ cannot be improved.
\end{Thm}

 Based on the initiation of Kayumov and Ponnusamy\cite{KP2018-2}, several forms of Bohr-type inequalities for the family ${\mathcal B}$
were considered in \cite{LSX} when the  Taylor coefficients of classical Bohr inequality are partly or completely replaced by higher order derivatives of $f$.
Here we only recall one of them.

\begin{Thm}{ \rm(\cite[Theorem~2.1]{LSX})} \label{thmF}
Suppose that $f\in {\mathcal B}$ and $f(z)=\sum_{n=0}^{\infty} a_n z^n$. Then the following sharp inequality holds:
$$
|f(z)|+\left|f^{\prime}(z)\right|r+\sum_{n=2}^{\infty}\left|a_{n}\right|r^{n} \leq 1 \quad \text {for}~|z|=r \leq \frac{\sqrt{17}-3}{4}.
$$

\end{Thm}

Besides these several authors have investigated some other extensions along with applications and connections with local Banach space theory and other topics.
See for instance \cite{Aizen-00-1,AAD,BDK5,BombBour-196,BombBour-2004,DF,DGM,DeGarM04,DMS,DP,GarMasRoss-2018} and the references therein.
In view of the recent developments on Bohr-type inequalities, it is natural to ask whether
we can further generalize or improve these results in the recent setting of \cite{PVW}. In this article, we give
an affirmative answer to this question.

The paper is organized as follows. In Section \ref{sec2}, we include some
preliminary results and a new lemma. In Section \ref{sec3}, we establish an improved version of the Bohr-Rogosinski inequality.
In Section \ref{sec4}, we obtain refined Bohr inequalities for symmetric functions.
Finally, in Section \ref{sec5}, we consider some refined Bohr type inequalities associated with
area, modulus of $f-a_0(f)$ and higher order derivatives of $f$ in part.

\section{Preliminary results}  \label{sec2}

In order to obtain our results, we need few lemmas.

\begin{lem} { \rm(\cite[Proof~ of~ Theorem ~1]{KP2017} and \cite{KP2018-1})} \label{lem1}
 Suppose that $f\in {\mathcal B}$ and $f(z)=\sum_{n=0}^{\infty} a_n z^n$. Then we have
$$ \sum_{n=1}^{\infty}\left|a_{n}\right| r^{n} \leq
\begin{cases}
\displaystyle r\frac{1-|a_0|^2}{1-r|a_0|}, &\, \text{for}\quad |a_0|\geq r,\\
 \displaystyle r\frac{\sqrt{1-|a_0|^2}}{\sqrt{1-r^2}}, &\, \text{for}\quad |a_0|<r.
\end{cases}
$$
\end{lem}

\begin{lem} { \rm(\cite[Lemma~1]{KP2018-2})} \label{lem2}
Suppose that $f\in {\mathcal B}$ and $f(z)=\sum_{n=0}^{\infty} a_n z^n$. Then the
following sharp inequality holds:
$$\sum_{n=1}^{\infty} n\left|a_{n}\right|^{2} r^{2n} \leq r^{2} \frac{\left(1-\left|a_{0}\right|^{2}\right)^{2}}{\left(1-\left|a_{0}\right|^{2} r^{2}\right)^{2}}
~\mbox{ for  $\ds 0<r\leq\frac{1}{\sqrt{2}}.$}
$$
\end{lem}

\begin{lem}  { \rm(\cite{Car} and \cite[Lemma~B]{PVW})} \label{lem3}

Suppose that $f\in {\mathcal B}$ and $f(z)=\sum_{n=0}^{\infty} a_n z^n$. Then the following inequalities hold.
\begin{enumerate}
	\item[{\rm (a)}] $|a_{2n+1}|\leq 1-|a_0|^2-\cdots - |a_n|^2,~\ n=0,1,\ldots$
	\item[{\rm (b)}] $|a_{2n}|\leq 1-|a_0|^2-\cdots -|a_{n-1}|^2 - \frac{|a_n|^2}{1+|a_0|} ,~\ n=1,2,\ldots$.
\end{enumerate}
Further, to have equality in {\rm (a)} it is necessary that $f$ is a rational function of the form
$$ f(z)=\frac{a_{0}+a_{1}z+ \cdots + a_{n}z^{n}+\epsilon z^{2n+1}}{1+(\overline{a_n}z^n+ \cdots +\overline{a_0}z^{2n+1})\epsilon},~|\epsilon|=1,
$$
and  to have equality in {\rm (b)} it is necessary that $f$ is a rational function of the form
$$ f(z)=\frac{a_{0}+a_{1}z+ \cdots + \frac{a_{n}}{1+|a_0|} z^{n}+ \epsilon z^{2n} }{1+\left(\frac{\overline{a_n}}{1+|a_0|}z^n+ \cdots +\overline{a_0}z^{2n}\right) \epsilon},
~|\epsilon|=1,
$$
where  $a_0 \overline{a_n}^2 \epsilon$ is non-positive real.
\end{lem}

In what follows, $\lfloor x\rfloor $ denotes the largest integer no more than $x$, where $x$ is a real number.

\begin{lem}\label{lem4}

Suppose that $f\in {\mathcal B}$ and $f(z)=\sum_{n=0}^{\infty} a_n z^n$.
Then for any $N\in\mathbb{N}$, the following inequality holds:
$$	\sum_{n=N}^{\infty}|a_n|r^n+sgn(t)\sum_{n=1}^{t}|a_n|^2\frac{r^N}{1-r}+ \left(\frac{1}{1+|a_0|}
+\frac{r}{1-r}\right)\sum_{n=t+1}^{\infty} |a_n|^2 r^{2n}\leq(1-|a_0|^2)\frac{r^N}{1-r},
$$
for $r\in[0,1)$, where $t=\lfloor(N-1)/2\rfloor$.
\end{lem}

\bpf
The proof is divided into two cases.
For the case of   even values of $N\in \IN$, we set $N=2m$ ($m\geq 1$).  It follows from Lemma \ref{lem3} that
\begin{align*}
\sum_{n=2m}^{\infty}|a_n|r^n=  \,&  \sum_{n=m}^{\infty}|a_{2n}|r^{2n}+\sum_{n=m}^{\infty}|a_{2n+1}|r^{2n+1}\\
\leq \, & \sum_{n=m}^{\infty}\left(1-\sum_{k=0}^{n-1}|a_k|^2 - \frac{|a_n|^2}{1+|a_0|}\right)r^{2n}
+\sum_{n=m}^{\infty}\left(1-\sum_{k=0}^n|a_k|^2\right)r^{2n+1}\\
=\, &\sum_{n=2m}^{\infty}r^n-\left(\sum_{n=0}^{m-1}|a_n|^2\right)\left(\sum_{n=2m}^{\infty}r^n\right)-\left(\frac{1}{1+|a_0|}
+\sum_{n=1}^{\infty}r^n\right)\sum_{n=m}^{\infty}|a_n|^2r^{2n}\\
=\, &\left(1-\sum_{n=0}^{m-1}|a_n|^2\right)\frac{r^{2m}}{1-r}-\left(\frac{1}{1+|a_0|}
+\frac{r}{1-r}\right)\sum_{n=m}^{\infty} |a_n|^2 r^{2n}
\end{align*}
so that
$$ \sum_{n=2m}^{\infty}|a_n|r^n  +\left (\sum_{n=1}^{m-1}|a_n|^2\right)\frac{r^{2m}}{1-r}+\left(\frac{1}{1+|a_0|}
+\frac{r}{1-r}\right)\sum_{n=m}^{\infty} |a_n|^2 r^{2n} \leq(1-|a_0|^2)\frac{r^N}{1-r},
$$
and the desired result follows easily.

For the case of  odd values of $N\in \IN$  associated with the  inequality in the statement,  we set $N=2m+1$ ($m\geq 0$).
It follows from Lemma \ref{lem3} that
\begin{align*}
\sum_{n=2m+1}^{\infty}|a_n|r^n=\,
& \sum_{n=2(m+1)}^{\infty}|a_{n}|r^{n}+|a_{2m+1}|r^{2m+1}\\
\leq \, & \left(1-\sum_{n=0}^{m}|a_n|^2\right)\frac{r^{2(m+1)}}{1-r}-\left(\frac{1}{1+|a_0|}
+\frac{r}{1-r}\right)\sum_{n=m+1}^{\infty} |a_n|^2 r^{2n}\\
& \hspace{1cm} +\left(1-\sum_{n=0}^m|a_n|^2\right)r^{2m+1}\\
=\, &\left(1-\sum_{n=0}^{m}|a_n|^2\right)\frac{r^{2m+1}}{1-r}-\left(\frac{1}{1+|a_0|}
+\frac{r}{1-r}\right)\sum_{n=m+1}^{\infty} |a_n|^2 r^{2n}.
\end{align*}
Simple translation gives the desired result. This completes the proof.
\epf

\section{Refined Bohr-Rogosinski inequalities}  \label{sec3}

Using Lemma \ref{lem4} and the similar proof of Theorem \Ref{thmC}, we can easily get the following result.
So we omit the details.

\begin{theorem} \label{thm1}
Suppose that $f\in {\mathcal B}$ and $f(z)=\sum_{n=0}^{\infty} a_n z^n$.  For $n\geq \IN$, let $t=\lfloor(N-1)/2\rfloor$.
Then $$|f(z)|+\sum_{n=N}^{\infty}|a_n|r^n+sgn(t)\sum_{n=1}^{t}|a_n|^2\frac{r^N}{1-r}+ \left(\frac{1}{1+|a_0|}
+\frac{r}{1-r}\right)\sum_{n=t+1}^{\infty} |a_n|^2 r^{2n} \leq 1
$$
for $|z|=r\leq R_{N}$, where $R_{N}$ is    as in Theorem {\rm \Ref{thmC}}.

The radius $R_{N}$ is best possible. Moreover,
$$ |f(z)|^{2}+\sum_{n=N}^{\infty}|a_n|r^n+sgn(t)\sum_{n=1}^{t}|a_n|^2\frac{r^N}{1-r}+ \left(\frac{1}{1+|a_0|}
+\frac{r}{1-r}\right)\sum_{n=t+1}^{\infty} |a_n|^2 r^{2n} \leq 1
$$
for $|z|=r\leq R_{N}^{\prime}$, where $R_{N}^{\prime}$ is  as in Theorem {\rm \Ref{thmC}}.

The radius $R_{N}^{\prime}$ is best possible.

\end{theorem}

For the case of $N=1$, it is easy to see that $R_1=\sqrt{5}-2$ and $R_1^{'}=1/3$.
However, the two constants can be improved  for any individual function in ${\mathcal B}$ (in the context of Theorem \Ref{thmC}
and Theorem \ref{thm1} with $N=1$).

\begin{theorem} \label{thm2}
Suppose that $f\in {\mathcal B}$ and $f(z)=\sum_{n=0}^{\infty} a_n z^n$. Then

$$A(z):=|f(z)|+\sum_{n=1}^{\infty}|a_n|r^n+ \left(\frac{1}{1+|a_0|}
+\frac{r}{1-r}\right)\sum_{n=1}^{\infty} |a_n|^2 r^{2n} \leq 1
$$
for $|z|=r\leq r_{a_0}= 2/\left(3+|a_0|+\sqrt{5}(1+|a_0|)\right).$
The radius $r_{a_0}$ is best possible and $r_{a_0}\geq \sqrt{5}-2$. Moreover,
$$B(z):= |f(z)|^2+\sum_{n=1}^{\infty}|a_n|r^n+ \left(\frac{1}{1+|a_0|}
+\frac{r}{1-r}\right)\sum_{n=1}^{\infty} |a_n|^2 r^{2n} \leq 1
$$
for $|z|=r\leq r_{a_0}'$,
where $r_{a_0}'$ is the unique positive root of the equation
$$(1-|a_0|^3)r^3-(1+2|a_0|)r^2-2r+1=0.
$$
The radius $r_{a_0}'$ is best possible.
Further, we have $1/3<r_{a_0}'<1/(2+|a_0|)$.
\end{theorem}

\bpf We consider the first part.
According to Lemma \ref{lem4} with $N=1$ and the classical inequality for $|f(z)|$ ($f\in {\mathcal B}$), we have

$$A(z)\leq\frac{r+|a_0|}{1+r|a_0|}+\frac{(1-|a_0|^2)r}{1-r}=1-\frac{(1-|a_0|) A_1(|a_0|,r)}{(1+|a_0|r)(1-r)},
$$
where
$$A_1(|a_0|,r)=(1-|a_0|-|a_0|^2)r^2-(3+|a_0|)r+1.
$$
To prove  the first inequality in Theorem \ref{thm2}, it suffices to prove that $A_1(|a_0|,r)\geq0$  for all $|a_0|\in[0,1)$ and $r\leq r_{a_0}$.
If $|a_0|=\frac{\sqrt{5}-1}{2}$, then we have $1-|a_0|-|a_0|^2=0$ and thus,  $A_1(|a_0|,r)\geq0$ is equivalent to
$$r\leq \frac{2}{3+|a_0|}=\frac{2}{5+\sqrt{5}}=\frac{2}{3+|a_0|+\sqrt{5}(1+|a_0|)}.
$$
For  $|a_0|\in [0,1)\backslash \{\frac{\sqrt{5}-1}{2}\}$, we have $1-|a_0|-|a_0|^2\neq0$ and  thus, we may write

$$A_1(|a_0|,r)=(1-|a_0|-|a_0|^2)\left(r-\frac{2}{3+|a_0|+\sqrt{5}(1+|a_0|)}\right)\left(r-\frac{2}{3+|a_0|-\sqrt{5}(1+|a_0|)}\right).
$$
Then the desired conclusion can be obtained by a simple analysis on two cases $0\leq |a_0|<\frac{\sqrt{5}-1}{2}$ and $\frac{\sqrt{5}-1}{2}<|a_0|<1$.

To prove the sharpness, we let $a\in[0,1)$ and consider the function
$$ \varphi_a(z)=\frac{a+z}{1+az}=a+(1-a^2)\sum_{n=1}^{\infty}(-a)^{n-1}z^n,\quad z\in\mathbb{D}.
$$
For this function, we find that
\begin{align*}

 A(r) =\, & \frac{r+a}{1+ra}+\frac{(1-a^2)r}{1-ar}+ \left(\frac{1}{1+a}+\frac{r}{1-r}\right)\frac{(1-a^2)^2r^2}{1-a^2r^2}\\
=\, &1-\frac{1-a}{(1+ar)(1-r)}A_1(a,r),
\end{align*}
where
$$A_1(a,r)=(1-a-a^2)r^2-(3+a)r+1.
$$
The above inequality is bigger than $1$ if and only if $A_1(a,r)<0$.
By the similar analysis in the previous proof, we observe that $A_1(a,r)<0$ if and only if $r>r_a=\frac{2}{3+a+\sqrt{5}(1+a)}$,
which implies the sharpness of the constant $r_{a_0}$ in the first part.

Next we prove the second part. Again, by Lemma \ref{lem4} and the classical inequality for $|f(z)|$, we have

$$ B(z) \leq\left(\frac{r+|a_0|}{1+r|a_0|}\right)^2+\frac{(1-|a_0|^2)r}{1-r}=1-\frac{ (1-|a_0|^2)A_2(|a_0|,r)}{(1+|a_0|r)^2(1-r)},
$$
where
$$A_2(|a_0|,r)=(1-|a_0|^2)r^3-(1+2|a_0|)r^2-2r+1.
$$
To prove  the second inequality in Theorem \ref{thm2}, it suffices to show that $A_2(|a_0|,r)\geq0$ for all $|a_0|\in[0,1)$, only in the case when $0\leq r\leq  r_{a_0}'$.
Elementary calculations show that

$$A_2\left(|a_0|,\frac{1}{3}\right)=\frac{1}{27}(1-|a_0|)(7+|a_0|)>0,
$$
$$A_2\left(|a_0|,\frac{1}{2+|a_0|}\right)=-\frac{(1-|a_0|)(1+|a_0|)^2}{(2+|a_0|)^3}<0
$$
and
$$\frac{\partial A_2}{\partial r}=-3|a_0|^2r^2-4|a_0|r-(1-r)(1+3r)-1<0.
$$
Therefore, the desired conclusion follows easily. The sharpness of the constant $r_{a_0}'$ can be established as in the previous case
and thus, we omit the details. The proof of the theorem is complete.
\epf

\section{Refined Bohr inequalities for symmetric functions}  \label{sec4}

\begin{theorem}  \label{thm3}
Suppose that $f\in {\mathcal B}$ and $f(z)=\sum_{n=0}^{\infty} a_{pn+m} z^{pn+m}$, where $p\in\mathbb{N}$ and $0\leq m\leq p$.
Then the following sharp inequality holds:
$$ C(r):=\sum_{n=0}^{\infty}|a_{pn+m}|r^{pn+m}+ \left(\frac{1}{1+|a_m|}
+\frac{r^p}{1-r^p}\right)\sum_{n=1}^{\infty} |a_{pn+m}|^2 r^{2pn+m}\leq1~\mbox{ for $r\leq r_{p,m,a_m}$},
$$
where $r_{p,m,a_m}$ is the unique positive root of  the equation
$$(1-|a_m|-|a_m|^2)r^{p+m}+r^p+|a_m|r^m-1=0.
$$
Further, we have $r_{p,m,a_m}\geq\sqrt[p]{1/(2+|a_m|)}$.
\end{theorem}

\bpf
Simple observation shows that the function $f$ can be represented as
$$f(z)=z^{m} g\left(z^{p}\right),
$$
where $g\in {\mathcal B}$ and $g(z)=\sum_{n=0}^{\infty} b_{n} z^{n}$ with $b_{n}=a_{pn+m} .$
It follows from Lemma \ref{lem4} with $N=1$ that
\begin{align*}
C(r)= \,& r^m\left[|b_0|+\sum_{n=1}^{\infty}|b_{n}|\left(r^{p}\right)^n+ \left(\frac{1}{1+|b_0|}
+\frac{r^p}{1-r^p}\right)\sum_{n=1}^{\infty} |b_{n}|^2 \left(r^{p}\right)^n\right]\\
\leq \, &r^m\left[|b_0|+(1-|b_0|^2)\frac{r^p}{1-r^p}\right]=r^m\left[|a_m|+(1-|a_m|^2)\frac{r^p}{1-r^p}\right]=:A_3(r)\\
\leq \, &A_3\left(r_{p,m,a_m}\right)=1 \quad \text{for}\quad r\leq r_{p,m,a_m}.
\end{align*}
Since $A_3(r)$ is monotonically increasing in $[0,1)$ and
$$A_3(\sqrt[p]{1/(2+|a_m|)})=(2+|a_m|)^{-m/p}\leq1,
$$
we find that $r_{p,m,a_m}\geq\sqrt[p]{1/(2+|a_m|)}$.

For the sharpness, we consider
$$f(z)=z^m\left (\frac{a-z^p}{1-az^p}\right ) =az^m-(1-a^2)\sum_{n=1}^{\infty}a^{n-1}z^{pn+m},\quad a\in[0,1).
$$
 For this function, we have
\begin{align*}

C(r)=\, & r^m\left[a+(1-a^2)\sum_{n=1}^{\infty}a^{n-1}r^{pn}+ \left(\frac{1}{1+a}
+\frac{r^p}{1-r^p}\right)(1-a^2)^2\sum_{n=1}^{\infty} a^{2(n-1)}r^{2pn}\right]\\
=\, &r^m\left[a+(1-a^2)\frac{r^p}{1-r^p}\right],
\end{align*}
which is bigger than $1$  if and only if $r>r_{p,m,a}$.
This completes the sharpness.
\epf

  If we set $m=0$ in Theorem \ref{thm3} and use its proof, then we have following results which improve \cite[Corollary~1]{KP2018-1}
and generalizes  Theorem \Ref{thmB}.

\begin{cor} \label{cor1}
If $f\in {\mathcal B}$ and $f(z)=\sum_{n=0}^{\infty} a_{pn} z^{pn}$ for some  $p\in\mathbb{N}$, then the following sharp inequalities hold:
\begin{enumerate}
	\item[{\rm (a)}] $\ds \sum_{n=0}^{\infty}|a_{pn}|r^{pn}+ \left(\frac{1}{1+|a_0|}
+\frac{r^p}{1-r^p}\right)\sum_{n=1}^{\infty} |a_{pn}|^2 r^{2pn}\leq1$ \quad \text{for}~~$\ds r\leq\frac{1}{\sqrt[p]{2+|a_0|}}.$
	\item[{\rm (b)}] $\ds |a_0|^2+\sum_{n=1}^{\infty}|a_{pn}|r^{pn}+ \left(\frac{1}{1+|a_0|}
+\frac{r^p}{1-r^p}\right)\sum_{n=1}^{\infty} |a_{pn}|^2 r^{2pn}\leq1$ \quad \text{for}~~$\ds r\leq\frac{1}{\sqrt[p]{2}}.$
\end{enumerate}
\end{cor}

\begin{cor} \label{cor2}
Suppose that $f\in {\mathcal B}$ and  $f(z)=\sum_{n=1}^{\infty} a_{pn} z^{pn}$ for some  $p\in\mathbb{N}$. Then the following sharp inequalities hold.
\begin{enumerate}
\item[{\rm (A)}] $\ds \sum_{n=1}^{\infty}|a_{pn}|r^{pn}+ \left(\frac{1}{1+|a_p|}
+\frac{r^p}{1-r^p}\right)\sum_{n=2}^{\infty} |a_{pn}|^2 r^{p(2n-1)}\leq1 \quad \text{for}~~r\leq\sqrt[p]{\frac{2}{\alpha(|a_p|)}},
$
where
$$\alpha(|a_p|)=1+|a_p|+\sqrt{(1-|a_p|)(5+3|a_p|)}.
$$
In particular, this inequality holds for $r\leq\sqrt[p]{3/5}$.

\item[{\rm (B)}] $\ds \sum_{n=1}^{\infty}|a_{pn}|r^{pn}+ \left(\frac{1}{1+|a_p|}
+\frac{r^p}{1-r^p}\right)\sum_{n=1}^{\infty} |a_{pn}|^2 r^{p(2n-1)}\leq1 \quad \text{for}~~r\leq\sqrt[p]{\frac{5-\sqrt{17}}{2}}.
$
\end{enumerate}
\end{cor}
\bpf The inequality (A) follows if we let $m=p$ in Theorem \ref{thm3}. Simple analysis shows that
$$\inf_{|a_p|\in[0,1)}\frac{2}{1+|a_p|+\sqrt{(1-|a_p|)(5+3|a_p|)}}=\frac{3}{5}, ~\mbox{ i.e. }~
 \sqrt[p]{\frac{2}{\alpha(|a_p|)}}\geq \sqrt[p]{\frac{3}{5}}.
$$

Next we prove (B). It follows from the proof of Theorem \ref{thm3} for the case $m=p$ that

\begin{align*}
&\sum_{n=1}^{\infty}|a_{pn}|r^{pn}+ \left(\frac{1}{1+|a_p|}
+\frac{r^p}{1-r^p}\right)\sum_{n=1}^{\infty} |a_{pn}|^2 r^{p(2n-1)}\\
=\,& \sum_{n=1}^{\infty}|a_{pn}|r^{pn}+ \left(\frac{1}{1+|a_p|}
+\frac{r^p}{1-r^p}\right)\sum_{n=2}^{\infty} |a_{pn}|^2 r^{p(2n-1)}+\left(\frac{1}{1+|a_p|}
+\frac{r^p}{1-r^p}\right)|a_p|^2r^{p}\\
\leq\, &r^p\left(|a_p|+(1-|a_p|^2)\frac{r^p}{1-r^p}+\frac{|a_p|^2}{1+|a_p|}
+\frac{|a_p|^2r^p}{1-r^p}\right)\\
=\,&|a_p|r^p+\frac{r^{2p}}{1-r^p}+\frac{|a_p|^2r^p}{1+|a_p|}.
\end{align*}
The remainder of the proof is similar to that of the proof of { \rm(\cite[Theorem~2 (b)]{PVW})} and it follows
just by replacing $|a_1|$ and $r$, respectively, by $|a_p|$ and $r^p$, and the extremal function turned out to be $z^p\left (\frac{a-z^p}{1-az^p}\right )$.
So we omit the details.
\epf

Corollary \ref{cor2}(A) not only generalizes  \cite[Theorem~2 (a)]{PVW}, but also improves upon it.
Furthermore, Corollary \ref{cor2}(B) generalizes  \cite[Theorem~2 (b)]{PVW}.

\section{Refined Bohr type inequalities}  \label{sec5}

The following result is a further refinement corresponding to Theorem \Ref{thmE}.

\begin{theorem} \label{thm4}
Suppose that $f\in {\mathcal B}$, $f(z)=\sum_{n=0}^{\infty} a_{n} z^{n}$ and $S_{r}$ denotes the Riemann surface of the
function $f^{-1}$ defined on the image of the subdisk $|z|<r$ under the mapping $f$.
Then
$$ D(r):=\sum_{n=0}^{\infty}\left|a_{n}\right| r^{n}+\left(\frac{1}{1+|a_0|}
+\frac{r}{1-r}\right)\sum_{n=1}^{\infty} |a_n|^2 r^{2n}+\frac{8}{9}\left(\frac{S_{r}}{\pi}\right) \leq 1 \text { for } r \leq \frac{1}{3}
$$
and the numbers $1/3$ and $8/9$ cannot be improved. Moreover,
$$ E(r):=|a_0|^2+\sum_{n=1}^{\infty}\left|a_{n}\right| r^{n}+\left(\frac{1}{1+|a_0|}
+\frac{r}{1-r}\right)\sum_{n=1}^{\infty} |a_n|^2 r^{2n}+\frac{9}{8}\left(\frac{S_{r}}{\pi}\right) \leq 1 \text { for } r \leq \frac{1}{3-a}
$$
and the constant $9/8$ cannot be improved.
\end{theorem}
\bpf
It follows from Lemma \ref{lem2} that
$$\frac{S_r}{\pi}=\frac{1}{\pi}\int\int_{|z|<r}|f'(z)|^2dxdy=\sum_{n=1}n|a_n|^2r^{2n}\leq\frac{(1-|a_0|^2)^2r^2}{(1-|a_0|^2r^2)^2}\quad \text{for}~~0\leq r\leq
\frac1{\sqrt{2}}.
$$
We now consider the first part. For the case $r\leq1/3$, by Lemma \ref{lem4}, we have
$$

D(r) \leq |a_0|+\frac{(1-|a_0|^2)r}{1-r}+\frac{8(1-|a_0|^2)^2r^2}{9(1-|a_0|^2r^2)^2}=: A_4(r)
$$
so that, because $A_4(r)$ is increasing,
$$D(r) \leq  A_4\left(\frac{1}{3}\right)=1-\frac{(1-|a_0|)^3}{2(9-|a_0|^2)^2}(5+|a_0|)(13+4|a_0|-|a_0|^2)\leq1.
$$
To prove that the constant $8/9$ is sharp, we consider the function
$$f(z)=\phi_a(z)=\frac{a-z}{1-az}.
$$
For this function, straightforward calculations show that
\begin{align*}
&\sum_{n=0}^{\infty}\left|a_{n}\right| r^{n}+\left(\frac{1}{1+|a_0|}
+\frac{r}{1-r}\right)\sum_{n=1}^{\infty} |a_n|^2 r^{2n}+\lambda\left(\frac{S_{r}}{\pi}\right)\\
=\, & a+\frac{(1-a^2)r}{1-ar}+ \frac{1+ar}{(1+a)(1-r)}\frac{(1-a^2)^2r^2}{1-a^2r^2}+\lambda\frac{(1-a^2)^2r^2}{(1-a^2r^2)^2}.
\end{align*}
For $r=1/3$, the right hand side of the last expression becomes
$$1+\frac{(1-a)^2}{2(9-a^2)^2}\{8(9\lambda-8)-8(9\lambda+4)(1-a)+6(3\lambda+2)(1-a)^2+4(1-a)^3-(1-a)^4\},
$$
which is easily seen to be bigger than $1$ in the case $\lambda>8/9$ and $a\rightarrow1$.

Again, if $r\leq1/(3-|a_0|)$, then as in the previous case it follows from Lemma \ref{lem4} that
\begin{align*}

E(r)~\leq& ~|a_0|^2+\frac{(1-|a_0|^2)r}{1-r}+\frac{9(1-|a_0|^2)^2r^2}{8(1-|a_0|^2r^2)^2}=: A_5(r)\\
~\leq& ~ A_5\left(\frac{1}{3-|a_0|}\right)=1-\frac{(1-|a_0|)^3(1+|a_0|)}{8(2-|a_0|)(3-2|a_0|)^2}[(54-39|a_0|)+|a_0|^2(6-|a_0|)]\\
~\leq& ~1.
\end{align*}
Finally, to prove that the constant $9/8$ is sharp in the second part of the statement of the theorem, we consider the function $f(z)=\phi_a(z)$
which is given above. Again, for this function, straightforward calculations show that
\begin{align*}
&|a_0|^2+\sum_{n=1}^{\infty}\left|a_{n}\right| r^{n}+\left(\frac{1}{1+|a_0|}
+\frac{r}{1-r}\right)\sum_{n=1}^{\infty} |a_n|^2 r^{2n}+\lambda\left(\frac{S_{r}}{\pi}\right)\\
=\, & a^2+\frac{(1-a^2)r}{1-ar}+ \frac{1+ar}{(1+a)(1-r)}\frac{(1-a^2)^2r^2}{1-a^2r^2}+\lambda\frac{(1-a^2)^2r^2}{(1-a^2r^2)^2}.
\end{align*}
For $r=1/(3-a)$, the last expression reduces to
$$1+\frac{(1-a)^2(1+a)}{9(3-2a)^2(2-a)} A_6(a,\lambda),
$$
which is again seen to be greater than $1$ when $\lambda>9/8$ and $a\rightarrow1$, where
$$A_6(a,\lambda)=(8\lambda-9)+12(\lambda-3)(1-a)+2(\lambda-18)(1-a)^2-3\lambda(1-a)^3-\lambda(1-a)^4.
$$
This completes the proof.
\epf

\begin{rems}
From the proof of Theorem \ref{thm4}, it is easy to see that the number $1/(3-a)$ in Theorem \ref{thm4} is not sharp for all $a\in[0,1)$.
However, it is sharp in the sense of approximation based on the following observation. On one hand, the number
$1/(3-a)$ approaches $1/2$ as $a\rightarrow1$.
On the other hand, suppose that $f\in {\mathcal B}$ and $f(z)=\sum_{n=0}^{\infty} a_{n} z^{n}$, we cannot find any $c>0$ such that
$$|a_0|^2+\sum_{n=1}^{\infty}\left|a_{n}\right| r^{n}+\left(\frac{1}{1+|a_0|}
+\frac{r}{1-r}\right)\sum_{n=1}^{\infty} |a_n|^2 r^{2n}+c\left(\frac{S_{r}}{\pi}\right) \leq 1 \text { for } r \leq \frac{1}{2};
$$
for instance, consider $f(z)=z$ and $r=1/2$.
\end{rems}

 Next, we consider another refined Bohr type inequality  with the square of the modulus of $f(z)-a_0(f)$, instead of area term.

\begin{theorem} \label{thm5}
 Suppose that $f\in {\mathcal B}$ and $f(z)=\sum_{n=0}^{\infty} a_{n} z^{n}$.  Then
$$ F(z):=\sum_{n=0}^{\infty}\left|a_{n}\right| r^{n}+\left(\frac{1}{1+|a_0|}
+\frac{r}{1-r}\right)\sum_{n=1}^{\infty} |a_n|^2 r^{2n}+|f(z)-a_0|^2\leq 1 \text { for } |z|=r \leq \frac{1}{3}
$$
if and only if $0\leq|a_0|\leq4\sqrt{2}-5\approx 0.656854$.
\end{theorem}
\bpf
Assume that $0\leq|a_0|\leq4\sqrt{2}-5$. Then $|a_0|^2+10|a_0|-7\leq0$.
If $4\sqrt{2}-5\geq|a_0|\geq r$ and $r\leq1/3$, then it follows from Lemma \ref{lem1} and Lemma \ref{lem4} with $N=1$ that
\begin{align*}

F(z)~\leq& ~|a_0|+\frac{(1-|a_0|^2)r}{1-r}+\left(\frac{(1-|a_0|^2)r}{1-|a_0|r}\right)^2=: A_7(r)\\
~\leq&~A_7\left(\frac{1}{3}\right)=1+\frac{(1-|a_0|)^2(|a_0|^2+10|a_0|-7)}{2(3-|a_0|)^2}\leq1.
\end{align*}
Again, if $0\leq|a_0|<r\leq1/3$, then it follows from Lemma \ref{lem1} and Lemma \ref{lem4} with $N=1$ that
\begin{align*}

F(z)~\leq& ~|a_0|+\frac{(1-|a_0|^2)r}{1-r}+\left(\frac{\sqrt{1-|a_0|^2}r}{\sqrt{1-r^2}}\right)^2=: A_8(r)\\
~\leq&~A_8\left(\frac{1}{3}\right)=|a_0|+\frac{1}{2}(1-|a_0|^2)+\frac{\sqrt{1-|a_0|^2}}{8}\\
~\leq&~\frac{1}{3}+\frac{1}{2}+\frac{1}{8}<1.
\end{align*}

To complete the proof, we choose $f(z)=\phi_a(z)=(a-z)/(1-az)$ and for this function, we have that
\begin{align*}

F(r)~=&~a+\frac{(1-a^2)r}{1-r}+\left(\frac{(1-a^2)r}{1-ar}\right)^2
\end{align*}
which for $r=1/3$ reduces to
$$1+\frac{(1-a)^2(a^2+10a-7)}{2(3-a)^2},
$$
and this is larger than $1$ if and only if $4\sqrt{2}-5<a<1$.
\epf

Clearly, Theorem \ref{thm5} improves \cite[Theorem~3]{KP2018-2} partly.
Surprisingly, if the square term $|f(z)-a_0|^2$ in Theorem \ref{thm5} is replaced by $|f(z)-a_0|$, then
we have the following result which works for all $|a_0|\in[0,1)$.

\begin{theorem} \label{thm6}
 Suppose that $f\in {\mathcal B}$ and $f(z)=\sum_{n=0}^{\infty} a_{n} z^{n}$.  Then
$$ G(z):=\sum_{n=0}^{\infty}\left|a_{n}\right| r^{n}+\left(\frac{1}{1+|a_0|}
+\frac{r}{1-r}\right)\sum_{n=1}^{\infty} |a_n|^2 r^{2n}+|f(z)-a_0|\leq 1 \text { for } |z|=r \leq \frac{1}{5}
$$
and the number $1/5$  cannot be improved. Moreover,
$$ H(z):=|a_0|^2+\sum_{n=1}^{\infty}\left|a_{n}\right| r^{n}+\left(\frac{1}{1+|a_0|}
+\frac{r}{1-r}\right)\sum_{n=1}^{\infty} |a_n|^2 r^{2n}+|f(z)-a_0|\leq 1
$$
$\text { for } |z|=r \leq \frac{1}{3}$,
and the constant $1/3$ cannot be improved.
\end{theorem}
\bpf
For the first part, we first consider $|a_0|\geq r$ and $r\leq1/5$. Then it follows from Lemma \ref{lem1} and  Lemma \ref{lem4} with $N=1$ that
\begin{align*}

G(z)~\leq&~|a_0|+\frac{(1-|a_0|^2)r}{1-r}+\frac{(1-|a_0|^2)r}{1-|a_0|r}=: A_9(r)\\
~\leq&~A_9\left(\frac{1}{5}\right)=1-\frac{(1-|a_0|)^2(11-|a_0|)}{4(5-|a_0|)}\leq1.
\end{align*}
If $|a_0|<r\leq1/5$, then it follows from Lemmas \ref{lem1} and   \ref{lem4} with $N=1$ that
\begin{align*}

G(z)~\leq&~|a_0|+\frac{(1-|a_0|^2)r}{1-r}+\frac{\sqrt{1-|a_0|^2}r}{\sqrt{1-r^2}}=: A_{10}(r)\\
~\leq&~A_{10}\left(\frac{1}{5}\right)=|a_0|+\frac{1}{4}(1-|a_0|^2)+\frac{\sqrt{1-|a_0|^2}}{2\sqrt{6}}\\
~\leq&~\frac{1}{5}+\frac{1}{4}+\frac{1}{4}<1.
\end{align*}
To check its sharpness, we choose $f(z)=\phi_a(z)=(a-z)/(1-az)$ and as before for this function, we find that
\begin{align*}

G(r)~=&~a+\frac{(1-a^2)r}{1-r}+\frac{(1-a^2)r}{1-ar}\\
~=&~1-\frac{(1-a)}{(1-r) (1-ar)}[r^2a^2+(3 r^2-3 r)a+(r^2-3 r+1)]
\end{align*}
which is larger than $1$ if and only if
$$ A_{11}(a,r):=r^2a^2+(3 r^2-3 r)a+(r^2-3 r+1)<0.
$$
If $1/5< r <(3-\sqrt{5})/2$, we let $a_r=\frac{3(1-r)-\sqrt{5r^2-6 r+5}}{2r}$.
Elementary calculations show that  $a_r\in(0,1)$ and $\frac{3r-3r^2}{2r^2}>1$, and thus
$A_{11}(a,r)<A_{11}(a_r,r)=0$ when $a_r<a<1$. This proves the sharpness.

For the proof of the second part of the theorem,  as in the previous case, we have

\begin{align*}

H(z)~\leq&~|a_0|^2+\frac{(1-|a_0|^2)r}{1-r}+\frac{(1-|a_0|^2)r}{1-|a_0|r}=: A_{12}(r)\\
~\leq&~A_{12}\left(\frac{1}{3}\right)=1-\frac{(1-|a_0|)^2(1+|a_0|)}{2(3-|a_0|)}\leq1
\end{align*}
when $|a_0|\geq r$ and $r\leq1/3$; and
\begin{align*}

H(z)~\leq&~|a_0|^2+\frac{(1-|a_0|^2)r}{1-r}+\frac{\sqrt{1-|a_0|^2}r}{\sqrt{1-r^2}}=: A_{13}(r)\\
\leq&A_{13}\left(\frac{1}{3}\right)=|a_0|^2+\frac{1}{2}(1-|a_0|^2)+\frac{\sqrt{2}}{4}\sqrt{1-|a_0|^2}\\
\leq&\frac{1}{9}+\frac{1}{2}+\frac{\sqrt{2}}{4}<1
\end{align*}
when $|a_0|<r\leq1/3$. To check its sharpness, we choose again $f(z)=\phi_a(z)=(a-z)/(1-az)$ and for this function we find that
\begin{align*}

H(r)~=&~a^2+\frac{(1-a^2)r}{1-r}+\frac{(1-a^2)r}{1-ar}\\
~=&~1-\frac{(1-a^2)}{(1-r) (1-ar)}[(1-3r+r^2)+(-r+2r^2)a]
\end{align*}
which is larger than $1$ if and only if
$ A_{14}(a,r):=(1-3r+r^2)+(-r+2r^2)a<0.
$
If $1/3< r <(3-\sqrt{5})/2$, then we let $a_r=\frac{r^2-3 r+1}{r-2 r^2}$ and we see that
$a_r\in(0,1)$ and $A_{14}(a,r)<A_{14}(a_r,r)=0$ when $a_r<a<1$.
This completes the sharpness.
\epf

Finally, we present an improved version of Theorem \Ref{thmF}.

\begin{theorem} \label{thm7}
Suppose that $f\in {\mathcal B}$ and $f(z)=\sum_{n=0}^{\infty} a_{n} z^{n}$.
Then
$$ I(z):=|f(z)|+\left|f'(z)\right|r+\sum_{n=2}^{\infty}|a_n|r^n+ \left(\frac{1}{1+|a_0|}
+\frac{r}{1-r}\right)\sum_{n=1}^{\infty} |a_n|^2 r^{2n} \leq 1
$$
for $|z|=r\leq\frac{\sqrt{17}-3}{4}$ and the constant $\frac{\sqrt{17}-3}{4}$ is best possible.
Moreover,
$$ J(z):=|f(z)|^2+\left|f'(z)\right|r+\sum_{n=2}^{\infty}|a_n|r^n+ \left(\frac{1}{1+|a_0|}
+\frac{r}{1-r}\right)\sum_{n=1}^{\infty} |a_n|^2 r^{2n} \leq 1
$$
for $|z|=r\leq r_0,$

where $r_0\approx0.385795$ is the unique positive root of the equation
$$1-2r-r^2-r^3-r^4=0
$$ and $r_0 $ is best possible.
\end{theorem}
\bpf
By simple calculation we know that $\frac{2r}{1-r^2}\leq1$ if $0\leq r\leq\sqrt{2}-1$.
Combining Lemma \ref{lem4}, the classical inequality for $|f(z)|$ and the Schwarz-Pick lemma, we have
\begin{align*}

I(z)~\leq&~|f(z)|+\frac{|z|}{1-|z|^2}(1-|f(z)|^2)+\frac{(1-|a_0|^2)r^2}{1-r}\\
~\leq&~ \frac{r+|a_0|}{1+r|a_0|}+\frac{r}{1-r^2}\left[1-\left(\frac{r+|a_0|}{1+r|a_0|}\right)^2\right]+\frac{(1-|a_0|^2)r^2}{1-r}\\
~=&~\frac{r+|a_0|}{1+r|a_0|}+\frac{r(1-|a_0|^2)}{(1+r|a_0|)^2} +\frac{(1-|a_0|^2)r^2}{1-r}\\
~=&~1+\frac{1-|a_0|}{(1+|a_0|r)^2(1-r)}[-1+3r-r^2+(2r^2+r^3)|a_0|+(2r^3+r^4)|a_0|^2+r^4|a_0|^3]\\
~\leq&~1+\frac{1-|a_0|}{(1+|a_0|r)^2(1-r)}[-1+3r-r^2+(2r^2+r^3)+(2r^3+r^4)+r^4]\\
~=&~1+\frac{2(1-|a_0|)(1+r^2)}{(1+|a_0|r)^2(1-r)}\left(r-\frac{\sqrt{17}-3}{4}\right)\left(r+\frac{\sqrt{17}+3}{4}\right)
\end{align*}
which is no more than $1$ for $0\leq r\leq \frac{\sqrt{17}-3}{4}<\sqrt{2}-1$. It is worth pointing out in the second inequality above, we have used the fact
that
$$\Phi(X)=X+\lambda (1-X^2)\leq \Phi(X_0) ~\mbox{ when $\ds X=|f(z)|\leq X_0=\frac{r+|a_0|}{1+r|a_0|}$ and $\ds \lambda =\frac{|z|}{1-|z|^2}$.}
$$
Clearly, $1-\frac{r}{1-r^2}\geq0$ if $0\leq r\leq\frac{1}{2}$.
Again, as in the previous case, we have
\begin{align*}

J(z)

~\leq&~ \left(\frac{r+|a_0|}{1+r|a_0|}\right)^2+\frac{r(1-|a_0|^2)}{(1+r|a_0|)^2}+\frac{(1-|a_0|^2)r^2}{1-r}\\
~=&~ 1-\frac{(1-r^2)(1-|a_0|^2)}{(1+r|a_0|)^2}+\frac{r(1-|a_0|^2)}{(1+r|a_0|)^2}+\frac{(1-|a_0|^2)r^2}{1-r}\\
~=&~1+\frac{1-|a_0|^2}{(1+|a_0|r)^2(1-r)}[-(1-r)(1-r-r^2)+r^2(1+r|a_0|)^2]\\

~\leq&~1+\frac{1-|a_0|^2}{(1+|a_0|r)^2(1-r)}[-1+2r+r^2+r^3+r^4]\leq1
\end{align*}
for $0\leq r\leq r_0<1/2$.

To check its sharpness, by choosing $f(z)=\varphi_a(z)=(a+z)/(1+az)$, we get
\begin{align*}

J(r)=\, & \left(\frac{r+a}{1+ra}\right)^2+\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)^2r^2}{1-a^2r^2}\\
=\, &1+\frac{1-a^2}{(1+ar)^2(1-r)}[-1+2r+r^2-r^3+2r^3a+r^4a^2].
\end{align*}
The last expression is larger than 1 if and only if
$$A_{15}(a,r):=-1+2r+r^2-r^3+2r^3a+r^4a^2>0
$$
for all $a\in[0,1)$ and all ~$r$~¡¡in some subset of $[0,1)$.
The equivalent condition implies $A_{15}(a,r)\geq0$ by $a\rightarrow1$.
It is easy to see that $A_{15}(1,r)$ is  greater than of equal to zero if and only if $r\geq r_0$.

This shows the sharpness.
\epf

\subsection*{Conflict of Interests}
The authors declare that there is no conflict of interests regarding the publication of this paper.

\subsection*{Acknowledgments}
The research of the first author was supported by Application-Oriented Characterized Disciplines,
Double First-Class University Project of Hunan Province (Xiangjiaotong [2018]469) and
the Science and Technology Plan Project of Hunan Province (No. 2016TP1020).

The second author was supported by the NSFs of China~(No. 11961013)
and the Natural Science Foundation of Guangxi (No. 2018GXNS-FAA050005).
The  work of the third author is supported by Mathematical Research Impact Centric Support (MATRICS) of
the Department of Science and Technology (DST), India  (MTR/2017/000367).

config: arxiv


TITLE: How to relate the spectrum of a self-adjoint unbounded operator to the spectrum of a compact solution operator
QUESTION [3 upvotes]: Apologies if this is a trivial question. I'm having some trouble getting this straight in my head and would appreciate either an explanation or a pointer to a good reference.
If this looks too long, here's a

tl;dr Say I have a densely defined, bounded below, symmetric operator $T$ defined in a Hilbert space $H$. After shifting by some $\sigma$ one can show there exists a compact solution operator $K$. Why does the spectrum of $T$ coincide with (a shift of) the spectrum of $K$?

Setup
Let $H$ be a Hilbert space with inner product $(\cdot,\cdot)$. Let $T$ be a densely defined symmetric operator bounded below by some constant $c$, with $D = \operatorname{dom}(T)$. Define the form $\mathfrak{t}$ on $D\times D$ by $\mathfrak{t}(u,v) = (Tu,v)$. By symmetry $\mathfrak{t}$ is sesquilinear. Because $T$ is bounded below, if we choose $\sigma > \max\{1,-c\}$, the form defined on $D\times D$ by
$$ \mathfrak{t}_\sigma(u,v) = \mathfrak{t}(u,v) + \sigma\cdot(u,v) $$
is a positive definite Hermitian inner product on $D$.
Define by $V$ the completion of $D$ with respect to $\mathfrak{t}_\sigma$. As $\mathfrak{t}_\sigma$ majorizes the norm on $H$ the injection $D\hookrightarrow H$ extends to a bounded embedding $\iota : V\hookrightarrow H$. Now assume that $V$ compactly embeds in $H$. (Here, if $T$ is not self-adjoint, take the Friedrichs extension.)
Define an eigenvalue of $\mathfrak{t}$ to be any $\mu\in\mathbb{C}$ such that there exists some $u\in V$ so for all $v\in V$ we have $\mathfrak{t}(u,v) = \mu(u,v)$. I do not assume eigenvectors of $\mathfrak{t}$ are elements of $D$.
By Riesz there exists a solution operator $S_\sigma: V^*\to V$. There is a bounded map $P:H\to V^*$ given by $u\mapsto (v\mapsto (u,v))$. (This is bounded by an application of Cauchy-Schwarz.) Composing these with the compact injection $\iota$ gives a compact map $K = S_\sigma P\iota:V\to V$. Note that $K$ has the property that for any $u\in V$,
$$ \mathfrak{t}_\sigma(Ku, v) = (u,v). $$
By the spectral theorem for compact operators, the spectrum of $K$ is discrete accumulating only at zero, and all nonzero elements of the spectrum are eigenvalues of finite multiplicity.
Suppose $u$ is an eigenvalue of $K$ with eigenvector $\mu$. Then for arbitrary $v\in V$,
\begin{align*}
\mathfrak{t}(u,v) &= \mathfrak{t}_\sigma(u,v) - \sigma(u,v) \\
  &= \frac{1}{\mu}\mathfrak{t}_\sigma(Ku, v) - \sigma(u,v) \\
  &= \bigg(\frac{1}{\mu}-\sigma\bigg)(u,v)
\end{align*}
and so we have that $\frac{1 - \mu\sigma}{\mu}$ is an eigenvalue of $\mathfrak{t}$ with eigenvector $u$.
Confusion

Does it follow that
  $$ \mbox{spectrum of } T = \bigg\{ \frac{1-\mu\sigma}{\mu}\ \bigg|\ \mu\in\mbox{spectrum of }K - \{0\} \bigg\}? $$

Thoughts
I've been worrying at this for a couple of days and trying to prove something like the following: If $\lambda$ is in the resolvent set of $T$, then $\frac{1}{\sigma+\lambda}$ is in the resolvent set of $K$. This would immediately imply what I want modulo showing that that eigenvalues of $\mathfrak{t}$ are contained in the spectrum of $T$. (This is not a concern, see the note below.) Some algebra and head-scratching later, nothing has panned out -- it looks promising, but I haven't found the zinger yet.
Maybe more to the point, I don't have enough intuition for where the algebra ought to go. I've started by assuming that $u$ is a $\mu$-eigenvector of $K$ in $H$ (after extending showing $K$ is symmetric and extending $K$ to a map $H\to H$ by density of $V$ and compactness), assuming that $\frac{1}{\lambda + \sigma}$ is in the resolvent set of $T$, and showing that we cannot have $\mu = \frac{1}{\lambda + \sigma}$, but this approach doesn't seem to rely on the assumption that the domain of $(T-\lambda)^{-1}$  is all of $H$, nor that it is bounded.
I'm aware that this closely follows standard reference materials (Gilbarg-Trudinger Ch 8, Evans ch 6.5). In fact in Evans 6.5, at the end of the proof of Theorem 1, Evans writes, "But observe as well that for $\eta\neq 0$, we have $Sw = \eta w$ if and only if $Lw = \lambda w$ for $\lambda = \frac{1}{\eta}$." You might interpret this question as, "why does the 'only if' follow?"
There are at least two other approaches I'm aware of to showing the spectrum of $T$ is discrete. One uses the Rayleigh quotient, shows there exists a lowest eigenvalue, and proceeds by induction to exhaust $H$ with orthonormal eigenspaces (e.g. Gilbarg-Trudinger, last section of ch 8). The other uses the theory of spectral measures (e.g. ch 13 of Green Rudin). I'd prefer to avoid those approaches if possible because I'd like to understand all three in as much detail as possible, in order to more thoroughly understand how they relate to each other.
(My intuition is that the exhaustion proof is related to how this proof would go if one unpacked the spectral theorem for compact operators, while the theory of spectral measures is "not homotopic," as it were, to this proof, but that's not yet clear to me.)
Anyway, I'm missing a piece of the puzzle here and I'm not sure where the missing piece fits to look more closely for it.
Notes
In the context of PDEs I'm happy to assume the facts bolded in the setup: $V$ is related to a Sobolev space $W^{1,1}$ so Rellich-Kondrachov comes into play, and elliptic regularity gives that the eigenvectors of $\mathfrak{t}$ are in fact eigenvectors of $T$.
If you've read this far, thanks for making it through my long-winded question :)

REPLY [0 votes]: Great question, Neal! I've been thinking a lot about this recently, so I may be able to help you.

The trick here is that the map $K$ should be the same map as used in the construction of the Friedrichs extension.

Let's briefly recap that construction, for details see these nice notes by Paul Garrett, which in turn follow the classic text of Riesz-Nagy.
To construct the Friedrichs extension of a positive operator $T$, we construct the form domain $V$. By your assumption $V$ combactly embeds into $H$. Call this embedding $\iota$. Call the isomorphism given by the Riesz representation theorem $H\leftrightarrow^{R_H}H^*$ Now let's deconstruct the map Garrett calls "$B$":
$$ B: H\xrightarrow{R_H}H^*\xrightarrow{\iota^*} V^* \xrightarrow{R_V} V. $$
This is almost the map you call $K$. One now proves that it is injective, hence is invertible on its range. Call the inverse "$A$". After a little more work, we see the map $A$ IS the Friedrichs extension of $T+\sigma$. So indeed the map $K$ is the inverse of the Friedrichs extension of $T+\sigma$.
Now, you restricted the domain of $K$ to $V$ in order to make it compact. However, by Schauder's theorem (see this MSE answer for proof) because we have assumed $\iota$ is compact, we have that $\iota^*$ is compact. Thus $B$ is compact without first restricting its domain to $V$!
Now we're almost done. Because we constructed the Friedrichs extension of $T$ shifted by a scalar, the map $B$ is in fact a compact resolvent of the original map $T$! So now the theory of compactly resolved unbounded operators comes into play, and we have that the spectrum of $T$ is discrete, accumulates at infinity, and in fact comprises eigenvectors. Ta-da!
As for this result's relationship with the theory of spectral measures and the functional calculus, that's not so clear to me right now. I look forward to helping you figure that out in the future!


TITLE: What happens to graviational force when separation between 2 objects is very small?
QUESTION [0 upvotes]: I recently learned about gravitational force and found out the equation for gravitational force on a object by an object according to Newton's law of universal gravitation.
$$ F = \frac{Gm_1m_2}{r^2}. $$
$r^2$ denotes the square of separation between two objects. So the question is what happens when this separation is very small. For example when someone is standing on ground, separation between him and Earth is very low (zero).
Does gravitational force $F$ becomes ∞ then?

REPLY [1 votes]: There are various issues with this thinking, I will ignore the issue of the radius not being 0 on the surface, as you covered this already.
1: While space is continuous over R, physics as a whole tends to break apart and collapse at a non-infinitesimal size, that being the Planck Length, which is around 1.6 x 10-35 metres. Beyond that we really don't know how things work.
2: At small enough distances, even before you reach the Planck Length, other forces take over. So gravity being considered as the primary force. Gravity becomes negligible. It is actually very weak in general, to the extent that there are many theories which attempt to understand why it is in such a completely different league than the other fundamental forces.
3: As also mentioned before, it is based on the distance to the centre of mass, which grows very, very small as you get to very close distances, as the object would have to become less massive in order to be smaller without collapsing into a black hole. This tends towards 0 as you get closer.
Sorry if I poorly explained anything.