# The sharp bound of the third Hankel determinant for inverse of convex
functions

Biswajit Rath1, K. Sanjay Kumar ${}^{2},$ D. Vamshee Krishna3 1.2.3Department
of Mathematics, GITAM School of Science, GITAM (Deemed to be University),
Visakhapatnam- 530 045, A.P., India
brath@gitam.edu1∗,skarri9<EMAIL_ADDRESS>
###### Abstract.

The objective of this paper is to find the best possible upper bound of the
third Hankel determinant for inverse of convex functions.

###### Key words and phrases:

Holomorphic function, univalent function, Hankel determinant, Inverse of
Convex, Carathéodory function

###### 1991 Mathematics Subject Classification:

30C45, 30C50

##  1\. Introduction

Denote $\mathcal{H}$ the family all analytic functions in unit disk
$\mathbb{D}=\\{z\in\mathbb{C}:|z|<1\\}$ and $\mathcal{A}$ be the subfamily of
functions $f$ normilized by the conditions
$f(0)=f^{\prime}(0)-1=0,$ i.e, of the type

(1.1) $f(z)=\sum_{n=1}^{\infty}a_{n}z^{n},~{}a_{1}:=1,$

and $\mathcal{S}$ be the subfamily of $\mathcal{A},$ possessing univalent
(schlicht) mappings. For $f\in\mathcal{S},$ has an inverse $f^{-1}$ given by

(1.2)
$f^{-1}(w)=w+\sum_{n=2}^{\infty}t_{n}w^{n},~{}~{}|w|<r_{o}(f);\left(r_{o}(f)\geq\frac{1}{4}\right).$

A typical problem in geometric function theory is to study a functional made
up of combination of the coefficients of the original functions. For the
positive integers $r,~{}n,$ Pommerenke [16] characterized the $r^{th}$\-
Hankel determinant of $n^{th}$-order for $f$ given in (1.1), defined as
follows:

(1.3) $H_{r,n}(f)=\begin{array}[]{|cccc|}a_{n}&a_{n+1}&\cdots&a_{n+r-1}\\\
a_{n+1}&a_{n+2}&\cdots&a_{n+r}\\\ \vdots&\vdots&\ddots&\vdots\\\
a_{n+r-1}&a_{n+r}&\cdots&a_{n+2r-2}\end{array}.$

The problem of finding sharp estimates of the third Hankel determinant
obtained for $r=3$ and $n=1$ in (1.2), given by

(1.4) $H_{3,1}(f):=\begin{array}[]{|ccc|}a_{1}=1&a_{2}&a_{3}\\\
a_{2}&a_{3}&a_{4}\\\
a_{3}&a_{4}&a_{5}\end{array}=2a_{2}a_{3}a_{4}-a_{3}^{3}-a_{4}^{2}+a_{3}a_{5}-a_{2}^{2}a_{5},$

is technically much tough than $r=n=2$.

In recent years, many authors are working on obtaining upper bounds (see [2,
8, 17, 18, 20]) and a few papers were devoted to the estimation of sharp upper
bound to $H_{3,1}(f),$ for certain subclasses of analytic functions (see [3,
6, 7, 9, 10, 19]). Recently Lecko et al. [6] obtained the sharp bound for the
class of convex function denoted as $\mathcal{S}^{c}$, defined by

(1.5)
$\text{Re}\left\\{1+\frac{zf^{\prime\prime}(z)}{f^{\prime}(z)}\right\\}>0.$

Motivated by these results, in this paper we obtain sharp estimate for
$H_{3,1}(f^{-1})$ when $f\in\mathcal{S}^{c}$ as $1/36$.

The collection $\mathcal{P},$ of all functions $p,$ each one called as
Carathéodory function [5] of the form,

(1.6) $p(z)=1+\sum_{t=1}^{\infty}c_{t}z^{t},$

having a positive real part in $\mathbb{D}$. In view of (1.4) and (1.5), the
coefficients of the functions in $\mathcal{S}^{c}$ can be expressed in terms
of coefficients of functions in $\mathcal{P}$. We then obtain the upper bound
of $H_{3,1}(f^{-1})$, buliding our analysis on the familiar formulas of
coefficients $c_{2}$ (see, [15, p. 166]), $c_{3}$ (see [11, 12]) and $c_{4}$
can be found in [10].

The foundation for proofs of our main results is the following lemma and we
adopt the procedure framed through Libera and Zlotkiewicz [12].

######  Lemma 1.1.

If $p\in\mathcal{P},$ is of the form (1.5) with $c_{1}\geq 0,$ such that
$c_{1}\in[0,2]$ then

$\begin{split}&2c_{2}=c_{1}^{2}+\nu\mu,\\\ &4c_{3}=c_{1}^{3}+2c_{1}\nu\mu-
c_{1}\nu\mu^{2}+2\nu\left(1-|\mu|^{2}\right)\rho,\\\ \text{and}\\\
&\begin{split}8c_{4}=~{}&c_{1}^{4}+3c_{1}^{2}\nu\mu+\left(4-3c_{1}^{2}\right)\nu\mu^{2}+c_{1}^{2}\nu\mu^{3}+4\nu\left(1-|\mu|^{2}\right)\left(1-|\rho|^{2}\right)\psi\\\
&+4\nu\left(1-|\mu|^{2}\right)\left(c_{1}\rho-c\mu\rho-\bar{\mu}\rho^{2}\right),\end{split}\end{split}$

where $\nu:=4-c_{1}^{2},$ for some $\mu$, $\rho$ and $\psi$ such that
$|\mu|\leq 1$, $|\rho|\leq 1$ and $|\psi|\leq 1$.

##  2\. Main result

######  Theorem 2.1.

If $f\in$ $\mathcal{S}^{c}$, then

$\big{|}H_{3,1}(f^{-1})\big{|}\leq\frac{1}{36}$

and the inequality is sharp for $p_{0}(z)=(1+z^{3})/(1-z^{3})$.

###### Proof.

For $f\in$ $\mathcal{S}^{c}$, there exists a holomorphic function
$p\in\mathcal{P}$ such that

(2.1)
$\left\\{1+\frac{zf^{\prime\prime}(z)}{f^{\prime}(z)}\right\\}=p(z)\Leftrightarrow\left\\{f^{\prime}(z)+zf^{\prime\prime}(z)\right\\}=p(z)f^{\prime}(z)$

Using the series representation for $f$ and $p$ in (2.1), a simple calculation
gives

(2.2)
$\begin{split}&a_{2}=\frac{c_{1}}{2},~{}a_{3}=\frac{c_{1}^{2}+c_{2}}{6},~{}a_{4}=\frac{1}{12}\left[\frac{1}{2}c_{1}^{3}+\frac{3}{2}c_{1}c_{2}+c_{3}\right]\\\
\text{and}~{}~{}&a_{5}=\frac{1}{20}\left[\frac{1}{6}c_{1}^{4}+c_{1}^{2}c_{2}+\frac{1}{2}c_{2}^{2}+\frac{4}{3}c_{1}c_{3}+c_{4}\right]\end{split}$

Now from the defination (1.2), we have

(2.3) $w=f(f^{-1})=f^{-1}(w)+\sum_{n=2}^{\infty}a_{n}(f^{-1}(w))^{n}.$

Further, we have

(2.4)
$w=f(f^{-1})=w+\sum_{n=2}^{\infty}t_{n}w^{n}+\sum_{n=2}^{\infty}a_{n}(w+\sum_{n=2}^{\infty}t_{n}w^{n})^{n}.$

Upon simplification, we obtain

$\displaystyle(t_{2}+a_{2})w^{2}+(t_{3}+2a_{2}t_{2}+a_{3})w^{3}+(t_{4}+2a_{2}t_{3}+a_{2}t_{2}^{2}+3a_{3}t_{2}+a_{4})w^{4}$
(2.5)
$\displaystyle+(t_{5}+2a_{2}t_{4}+2a_{2}t_{2}t_{3}+3a_{3}t_{3}+3a_{3}t_{2}^{2}+4a_{4}t_{2}+a_{5})w^{5}+......=0.$

Equating the coefficients of like power in (2.5), upon simplification, we
obtain

(2.6)
$\begin{split}&t_{2}=-a_{2};t_{3}=\\{-a_{3}+2a_{2}^{2}\\};t_{4}=\\{-a_{4}+5a_{2}a_{3}-5a_{2}^{3}\\};\\\
&t_{5}=\\{-a_{5}+6a_{2}a_{4}-21a_{2}^{2}a_{3}+3a_{3}^{2}+14a_{2}^{4}\\}.\end{split}$

Using the values of $a_{n}(n=2,3,4,5)$ from (2.2) in (2.6), upon
simplification, we obtain

(2.7)
$\begin{split}&t_{2}=-\frac{c_{1}}{2},~{}t_{3}=\frac{1}{6}\left(2c_{1}^{2}-c_{2}\right),~{}t_{4}=\frac{1}{24}\left(-6c_{1}^{3}+7c_{1}c_{2}-2c_{3}\right)\\\
\text{and}~{}&t_{5}=\frac{1}{120}\left(-6c_{4}+22c_{1}c_{3}-46c_{1}^{2}c_{2}+7c_{2}^{2}+24c_{1}^{4}\right).\end{split}$

Now,

(2.11) $\displaystyle H_{3,1}(f^{-1})$
$\displaystyle=\begin{array}[]{|ccc|}t_{1}=1&t_{2}&t_{3}\\\
t_{2}&t_{3}&t_{4}\\\ t_{3}&t_{4}&t_{5}\end{array}~{},$

Using the values of $t_{j},~{}(j=2,3,4,5)$ from (2.7) in (2.8), it simplifies
to give

(2.12)
$\begin{split}H_{3,1}(f^{-1})=\frac{1}{8640}&\left[4c_{1}^{6}-24c_{1}^{4}c_{2}+12c_{1}^{3}c_{3}+39c_{1}^{2}c_{2}^{2}-44c_{2}^{3}+36c_{1}c_{2}c_{3}\right.\\\
&\left.-36c_{1}^{2}c_{4}-60c_{3}^{2}+72c_{2}c_{4}\right].\end{split}$

In view of (2.9), using the values of $c_{2},~{}c_{3}~{}\text{and}~{}c_{4}$
from lemma 1.1, gives

(2.13)
$\begin{split}24c_{1}^{4}c_{2}=&12\left[c_{1}^{6}+c_{1}^{4}\nu\mu\right];\\\
12c_{1}^{3}c_{3}=&3\left[c_{1}^{6}+2c_{1}^{4}\nu\mu-
c_{1}^{4}\nu\mu^{2}+2c_{1}^{3}\nu(1-|\mu|^{2})\rho\right]\\\
44c_{2}^{3}=&\frac{11}{2}\left[c_{1}^{6}+3c_{1}^{4}\nu\mu+3c_{1}^{2}\nu^{2}\mu^{2}+\nu^{3}\mu^{3}\right];\\\
39c_{1}^{2}c_{2}^{2}=&\frac{39}{4}\left[c_{1}^{6}+2c_{1}^{4}\nu\mu+c_{1}^{2}\nu^{2}\mu^{2}\right];\\\
36c_{1}c_{2}c_{3}=&\frac{9}{2}\left[c_{1}^{6}+3c_{1}^{4}\nu\mu+2c_{1}^{2}\nu^{2}\mu^{2}-c_{1}^{4}\nu\mu^{2}-c_{1}^{2}\nu^{2}\mu^{3}\right.\\\
&\left.+2\nu\left(c_{1}^{3}+c_{1}\nu\mu\right)\left(1-|\mu|^{2}\right)\rho\right];\\\
60c_{3}^{2}=&\frac{15}{4}\left[c^{6}+4c^{4}\nu\mu+4c^{4}\nu^{2}\mu^{2}-2c^{4}\nu\mu^{2}-4c^{2}\nu^{2}\mu^{3}+c^{2}\nu^{2}\mu^{4}\right.\\\
&\left.+4\nu(c^{3}+2c\nu\mu-c\nu\mu^{2})(1-|\mu|^{2})\rho+4\nu^{2}(1-|\mu|^{2})^{2}\rho^{2}\right];\\\
72c_{2}c_{4}-36c_{1}^{2}c_{4}=&\frac{9}{2}\left[c_{1}^{4}\nu\mu+3c_{1}^{2}\nu^{2}\mu^{2}+\left(4-3c_{1}^{2}\right)\nu^{2}\mu^{3}+c_{1}^{2}\nu^{2}\mu^{4}\right.\\\
&+4\nu^{2}c_{1}\mu\left(1-\mu\right)\left(1-|\mu|^{2}\right)\rho-4\nu^{2}\left(1-|\mu|^{2}\right)|\mu|^{2}\rho^{2}\\\
&\left.+4\nu^{2}\left(1-|\mu|^{2}\right)\left(1-|\rho|^{2}\right)\mu\psi\right].\end{split}$

Imputting the values from (2.10) in the expression (2.9), after simplifying,
we get

(2.14)
$\begin{split}H_{3,1}(f^{-1})=\frac{1}{8640}&\left[\frac{3}{4}c_{1}^{2}\nu^{2}\mu^{2}-3c_{1}^{2}\nu^{2}\mu^{3}+\frac{3}{4}c_{1}^{2}\nu^{2}\mu^{4}-\frac{11}{2}\nu^{3}\mu^{3}+18\nu^{2}\mu^{3}\right.\\\
&-\left(3c_{1}\nu^{2}\mu+3c_{1}\nu^{2}\mu^{2}\right)\left(1-|\mu|^{2}\right)\rho-3\nu^{2}\left(5+|\mu|^{2}\right)\left(1-|\mu|^{2}\right)\rho^{2}\\\
&\left.+18\nu^{2}\mu\left(1-|\mu|^{2}\right)\left.(1-|\rho|^{2}\right)\psi\right].\end{split}$

Putting $u:=c_{1}$ and taking $\nu=\left(4-u^{2}\right)$ in (2.11), we obtain

(2.15)
$\begin{split}H_{3,1}(f^{-1})=\frac{\left(4-u^{2}\right)^{2}}{8640}&\left[\frac{3}{4}u^{2}\mu^{2}+\frac{3}{2}u^{2}\mu^{3}+\frac{3}{4}u^{2}\mu^{4}-(4-u^{2})\mu^{3}\right.\\\
&-3u\mu\left(1+\mu\right)\left(1-|\mu|^{2}\right)\rho-3\left(5+|\mu|^{2}\right)\left(1-|\mu|^{2}\right)\rho^{2}\\\
&\left.+18\mu\left(1-|\mu|^{2}\right)\left.(1-|\rho|^{2}\right)\psi\right].\end{split}$

Taking modulus on both sides of (2.12), using $|\mu|=v\in[0,1]$,
$|\rho|=w\in[0,1]$, $c_{1}=u\in[0,2]$ and $|\psi|\leq 1$, we obtain

(2.16)
$\bigg{|}H_{3,1}(f^{-1})\bigg{|}\leq\frac{\vartheta\left(u,v,w\right)}{8640},$

where $\vartheta:\mathbb{R}^{3}\rightarrow\mathbb{R}$ is defined as

(2.17)
$\begin{split}\vartheta\left(u,v,w\right)=\left(4-u^{2}\right)^{2}&\left[\frac{3}{4}u^{2}v^{2}+\frac{3}{2}u^{2}v^{3}+\frac{3}{4}u^{2}v^{4}+\left(4-u^{2}\right)v^{3}\right.\\\
&+3uv\left(1+v\right)\left(1-v^{2}\right)w+3\left(5+v^{2}\right)\left(1-v^{2}\right)w^{2}\\\
&\left.+18v\left(1-v^{2}\right)\left.(1-w^{2}\right)\right]\end{split}$

Now, we are making an attempt to maximize the function
$\vartheta\left(u,v,w\right)$ on
$\Omega:=[0,2]\times[0,1]\times[0,1]$.
A. On the vertices of $\Omega$, from (2.14), we get

$\begin{split}&\vartheta\left(0,0,0\right)=\vartheta\left(2,0,0\right)=\vartheta\left(2,1,0\right)=\vartheta\left(2,0,1\right)=\vartheta\left(2,1,1\right)=0,\\\
&\vartheta\left(0,0,1\right)=240,~{}\vartheta\left(0,1,0\right)=\vartheta\left(0,1,1\right)=64.\end{split}$

B. On the edges of $\Omega$, from (2.14), we have

(i) For the edge $u=0,~{}v=0,~{}0<w<1,$ we obtain.

$\vartheta\left(0,0,w\right)=240w^{2}\leq 240.$

(ii) For the edge $u=0,~{}v=1,~{}0<w<1$, we obtain

$\vartheta\left(0,1,w\right)=64.$

(iii) For $u=0,~{}w=0,~{}0<v<1$,

$\vartheta\left(0,v,0\right)=32v(9-7v^{2})\leq
192\sqrt{\frac{3}{7}},,~{}\text{for}~{}v=\sqrt{2}.$

(iv) For $u=0,~{}w=1,~{}0<v<1$,

$\vartheta\left(0,v,1\right)=240-192v^{2}+64v^{3}-48v^{4}\leq 240.$

(v) For $v=0,~{}w=1,0<u<2$,

$\vartheta\left(u,0,1\right)=15(4-u^{2})^{2}\leq 240.$

(vi) For the edges: $v=1,~{}w=0,0<u<2$ or $v=1,~{}w=1,0<u<2$, we have

$\vartheta\left(u,1,w\right)=(4-u^{2})^{2}(4+2u^{2})\leq 64.$

(vii) For the edges: $u=2,~{}v=0,~{}0<w<1$ or $u=2,~{}v=1,~{}0<w<1$ or

$u=2,~{}w=0,~{}0<v<1$ or $c=2,~{}w=1,~{}0<v<1$ or $v=0,~{}w=0,~{}0<u<2$,

we obtain

$\vartheta\left(2,v,w\right)=0.$

C. Now, we consider the six faces of $\Omega$.

(i) On the face $u=2$, from (2.14), we obtain

$\vartheta\left(2,v,w\right)=0.$

(ii) On the face $u=0,~{}v\in(0,1)~{}\text{and}~{}w\in(0,1)$ from (2.14), we
get

$\begin{split}\vartheta\left(0,v,w\right)&=288v-224v^{3}+(240-288v-192v^{2}+288v^{3}-48v^{4})w^{2}\\\
&=288v-224v^{3}+48(5-v)(-1+v)^{2}(1+v)w^{2}\\\ &\leq
288v-224v^{3}+48(5-v)(-1+v)^{2}(1+v)\\\ &=240-192v^{2}+64v^{3}-48v^{4}\leq
240.\end{split}$

(iii) On the face $v=0~{}u\in(0,2),~{}w\in(0,1)$, from (2.14), we obtain

$\vartheta\left(u,0,w\right)=15(4-u^{2})^{2}w^{2}\leq 15(4-u^{2})^{2}\leq
240.$

(iv) On the face $v=1,~{}u\in(0,2),~{}w\in(0,1)$, from (2.14), we observe that
the

function $\vartheta\left(u,1,w\right)$ is independent of $w$, from B(vi), we
have $\vartheta\left(u,1,w\right)\leq 240.$

(v) On the face $w=0,~{}u\in(0,2),~{}v\in(0,1)$, from (2.14), we obtain

$\begin{split}\vartheta\left(u,v,0\right)=(4-u^{2})^{2}&\left(\frac{3u^{2}v^{2}}{4}+\frac{3u^{2}v^{3}}{2}+(4-u^{2})v^{3}+\frac{3u^{2}v^{4}}{4}+18v(1-v^{2})\right)\\\
=(4-u^{2})^{2}&\left(18v-14v^{3}+u^{2}\left(\frac{3v^{2}}{4}+\frac{v^{3}}{2}+\frac{3v^{4}}{4}\right)\right)\\\
\leq(4-u^{2})^{2}&\left(12\sqrt{\frac{3}{7}}+2u^{2}\right)\leq
192\sqrt{\dfrac{3}{7}},~{}u\in(0,2).\end{split}$

(vi) On the face $w=1,$ in (2.14), we obtain

$\displaystyle\vartheta\left(u,v,1\right)=(4-u^{2})^{2}$
$\displaystyle\bigg{(}\frac{3}{4}u^{2}v^{2}+\frac{3}{2}u^{2}v^{3}+\frac{3}{4}u^{2}v^{4}+(4-u^{2})v^{3}$
$\displaystyle+3uv(1+v)(1-v^{2})+3(5+v^{2})\left(1-v^{2}\right)\bigg{)}$
$\displaystyle:=g_{3}(u,v)$
$\displaystyle,~{}\text{with}~{}(u,v)\in\mathbb{R}^{2}.$

Note that all real solutions (u,v) of the system of equation

$\displaystyle\frac{\partial g_{3}}{\partial u}=\frac{3}{2}(-4+u^{2})$
$\displaystyle\left[8(-1+v)v(1+v)^{2}-10u^{2}(-1+v)v(1+v)^{2}\right.$
$\displaystyle\left.+u^{3}v^{2}(3+2v+3v^{2})-4u(-10+9v^{2}-2v^{3}+3v^{4})\right]=0$

and

$\displaystyle\frac{\partial g_{3}}{\partial v}$
$\displaystyle=\frac{3}{2}(-4+u^{2})^{2}(-8v(2-v+v^{2})+u^{2}v(1+v+2v^{2})$
$\displaystyle+u(2+4v-6v^{2}-8v^{3}))=0$

by a numerical computation are the following

$(0,0),(-2.63625,-1.53087),(-1.0493,1.14045)~{}\text{and}~{}(\pm
2,x),x\in\mathbb{R}.$

Therefore, $g_{3}$ has no critical point in $(0,2)\times(0,1).$

D. Now, consider the interior portion of $\Omega$ i.e.
$(0,2)\times(0,1)\times(0,1).$

Differentiating $\vartheta(u,v,w)$ partially with respect $w$, we obtain

$\begin{split}\frac{\partial\vartheta}{\partial
w}=\frac{1}{2}(4-u^{2})^{2}&\left[60w^{2}+3v^{2}(u^{2}+4uw-16w^{2})+12v(6+uw-6w^{2})\right.\\\
&\left.+3v^{4}(u^{2}-4uw-4w^{2})+2v^{3}(-28+u^{2}-6uw+36w^{2})\right]\end{split}$

upon solving $\frac{\partial\vartheta}{\partial w}=0$, we get

$w_{0}=-\frac{uv(1+v)}{2(5-v)(1-v)}\notin(0,1)~{}\text{for}~{}(u,v)\in(0,2)\times(0,1)$

Hence $\vartheta(u,v,w)$ has no critical point in the interior of $\Omega$.
In review of cases A, B, C and D, we obtained

(2.18)
$\max\bigg{\\{}\vartheta(u,v,w):u\in[0,2],v\in[0,1],~{}w\in[0,1]\bigg{\\}}=240.$

From expression (2.13) and (2.15), we obtain

(2.19) $\Big{|}H_{3,1}(f^{-1})\Big{|}\leq\frac{1}{36}.$

For $p_{0}\in\mathcal{S}^{c}$, we obtain $t_{2}=t_{3}=t_{5}=0,~{}t_{4}=1/6$,
which follows the result. ∎

Data Availability: My manuscript has no associate data

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