# Sharp coefficients bounds for Starlike functions associated with Gregory
coefficients

Erhan Deniz${}^{1},$ Sercan Kazımoğlu1 and  H. M. Srivastava2,3,4,5,6
1Department of Mathematics, Faculty of Science and Letters, Kafkas University,
Kars-TURKEY
2Department of Mathematics and Statistics, University of Victoria, Victoria,
British Columbia V8W 3R4, Canada
3Department of Medical Research, China Medical University Hospital, China
Medical University, Taichung 40402, Taiwan
4Center for Converging Humanities, Kyung Hee University, 26 Kyungheedae-ro,
Dongdaemun-gu, Seoul 02447, Republic of Korea
5Department of Mathematics and Informatics, Azerbaijan University, 71 Jeyhun
Hajibeyli Street, AZ1007 Baku, Azerbaijan
6Section of Mathematics, International Telematic University Uninettuno,
I-00186 Rome, Italy<EMAIL_ADDRESS><EMAIL_ADDRESS>and
<EMAIL_ADDRESS>
###### Abstract.

In this paper we introduced the class $\mathcal{S}_{G}^{\ast}$ of analytic
functions which is related with starlike functions and generating function of
Gregory coefficients. By using bounds on some coefficient functionals for the
family of functions with positive real part, we obtain for functions in the
class $\mathcal{S}_{G}^{\ast}$ several sharp coefficient bounds on the first
six coeffcients and also further sharp bounds on the corresponding Hankel
determinants.

###### Key words and phrases:

Univalent functions, Starlike function, Gregory coefficients, Hankel
determinant.
2010 Mathematics Subject Classification. Priminary 30C45, Secondary 30C50,
30C80.

##  1\. Introduction

Let $\mathcal{A}$ be the class of functions $f$ which are analytic in the open
unit disc $\mathcal{U}=\left\\{z:\left|z\right|<1\right\\}$ and normalized by
the conditions $f\left(0\right)=f^{\prime}\left(0\right)-1=0.$ Let us denote
by $\mathcal{S}$ the subclass of $\mathcal{A}$ containing functions which are
univalent in $\mathcal{U}$. An analytic function $f$ is subordinate to an
analytic function $g$ (written as $f\prec g)$ if there exists an analytic
function $w$ with $w\left(0\right)=0$ and$\ \left|w\left(z\right)\right|<1$
for $z\in\mathcal{U}$ such that
$f\left(z\right)=g\left(w\left(z\right)\right).$ In particular, if $g$ is
univalent in $\mathcal{U},\ $then $f\left(0\right)=g\left(0\right)$ and
$f\left(\mathcal{U}\right)\subset g\left(\mathcal{U}\right).$

In 1992, Ma and Minda [30] gave a unified presentation of various subclasses
of starlike and convex functions by replacing the subordinate function
$(1+z)\diagup(1-z)$ by a more general analytic function $\varphi$ with
positive real part and normalized by the conditions $\varphi(0)=1,$
$\varphi^{\prime}(0)>0$ and $\varphi$ maps $\mathcal{U}$ onto univalently a
region starlike with respect to $1$ and symmetric with respect to the real
axis. They introduced the following general class that envelopes several well-
known classes as special cases:
$\mathcal{S}^{\ast}[\varphi]=\left\\{f\in\mathcal{A}:\text{
}zf^{\prime}\left(z\right)\diagup f\left(z\right)\prec\varphi(z)\right\\}.$ In
literature, the functions belonging to this class is called Ma-Minda starlike
function. For $-1\leq B<A\leq 1$,
$\mathcal{S}^{\ast}[\left(1+Az\right)\diagup\left(1+Bz\right)]:=\mathcal{S}^{\ast}[A,B]$
are called the Janowski starlike functions, introduced by Janowski [20]. The
class $\mathcal{S}^{\ast}(\beta)$ of starlike functions of order $\beta$
$\left(0\leq\beta<1\right)$ defined by taking
$\varphi(z)=(1+(1-2\beta)z)\diagup(1-z).$ Note that
$\mathcal{S}^{\ast}=\mathcal{S}^{\ast}(0)$ is the classical class of starlike
functions. By taking
$\varphi(z)=1+2\diagup\pi^{2}(\log\left((1+\sqrt{z})\diagup(1-\sqrt{z})\right)^{2},$
we obtain the class $\mathcal{S}_{p}$ of parabolic starlike functions,
introduced by Rønning [39].
$\mathcal{S}^{\ast}[\beta,-\beta]:=\widetilde{\mathcal{S}^{\ast}}(\beta)=\\{f\in\mathcal{A}:|zf^{\prime}\left(z\right)\diagup
f\left(z\right)-1|<\beta|zf^{\prime}\left(z\right)\diagup
f\left(z\right)+1|\\}$ has been studied in [1, 2].

Recently, the coefficient problem for $\mathcal{S}^{\ast}[\varphi]$ in the
case of generated functions of some well-known numbers of $\varphi$ studied by
[9, 14, 15, 25, 37]. For example, the case when $\varphi$ is defined by

$\varphi(z)=\frac{1+\tau^{2}z^{2}}{1-\tau z-\tau^{2}z^{2}}\text{ \ \
}(z\in\mathcal{\mathbb{C}},\;\tau=(1-\sqrt{5})/2),$

which generated function of Fibonacci numbers, coefficient problem studied by
Dziok and co-authors [14, 15]. Similarly, the function

$\varphi(z)=e^{e^{z}-1}\text{ \ \ }(z\in\mathcal{\mathbb{C}})$

is generated function of Bell numbers. The coefficient problem for this
function studied by Kumar and co-authors [25]. Recently, Mendiratta et al.
[31] and, Goel and Kumar [19] obtained the structural formula, inclusion
relations, coefficient estimates, growth and distortion results, subordination
theorems and various radii constants for the exponential function

$\varphi(z)=e^{z}\text{ \ \ }(z\in\mathcal{\mathbb{C}})$

and the Sigmoid function

$\varphi(z)=\frac{2}{1+e^{-z}}\text{ \ \ }(z\in\mathcal{\mathbb{C}})$

respectively. In 2021, Deniz [13] has studied sharp coefficient problem for
the function

$\varphi(z)=e^{z+\frac{\lambda}{2}z^{2}}\text{ \ \
}(z\in\mathcal{\mathbb{C}}\text{, }\lambda\geq 1),$

which is generated function of generalized telephone numbers.

Motivated by the above-cited works, we consider the function $\varphi$ for
which $\varphi(\mathcal{U})$ is starlike with respect to $1$ and whose
coefficients is the Gregory coefficients. Gregory coefficients are decreasing
rational numbers $1/2,1/12,$ $1/24,19/720,...,$ which are similar in their
role to the Bernoulli numbers and appear in a large number of problems,
especially in those related to the numerical analysis and to the number
theory. They first appeared in the works of the Scottish mathematician James
Gregory in 1671 and were subsequently rediscovered many times. Among the
famous mathematicians who rediscovered them, we find Laplace, Mascheroni,
Fontana, Bessel, Clausen, Hermite, Pearson and Fisher. Furthermore, Gregory’s
coefficients can rightly be considered one of the most frequently rediscovered
in mathematics (the last rediscovery dates back to our century), the reason
for which in literature we can find them under various names (e.g. reciprocal
logarithmic numbers, Bernoulli numbers of the second kind, Cauchy numbers,
etc.). For the same reason, their authorship is often attributed to various
mathematicians. In this paper, authors considered the generating function of
the Gregory coefficients $G_{n}$ (see [6, 35]) as follows:

$\frac{z}{\ln\left(1+z\right)}=\sum\limits_{n=0}^{\infty}G_{n}z^{n}\;\;\;(\left|z\right|<1).$

Clearly, $G_{n}$ for some values of $n$ as $G_{0}=1,$ $G_{1}=\frac{1}{2},$
$G_{2}=-\frac{1}{12},$ $G_{3}=\frac{1}{24},$ $G_{4}=-\frac{19}{720},$
$G_{5}=\frac{3}{160},$ and $G_{6}=-\frac{863}{60480}.$ We now consider the
function $\Psi(z):=\frac{z}{\ln\left(1+z\right)}$ with its domain of
definition as the open unit disk $\mathcal{U}$. In connection with this
function, we define the class

$\mathcal{S}_{G}^{\ast}:=\left\\{f:\;f\in\mathcal{S}\text{ and
}zf^{\prime}(z)/f(z)\prec\Psi(z)\right\\}.$

Graphic of $\Psi(\mathcal{U})$ as follows:

---

##  2\. Preliminary Results

The following lemmas is needed for the main results.

######  Lemma 2.1.

[28] If $p(z)=1+p_{1}z+p_{2}z^{2}+p_{3}z^{3}+\cdots\in\mathcal{P}$ $(p_{1}\geq
0),$ then

$\displaystyle 2p_{2}$ $\displaystyle=$ $\displaystyle
p_{1}^{2}+x(4-p_{1}^{2})$ (2.1) $\displaystyle 4p_{3}$ $\displaystyle=$
$\displaystyle
p_{1}^{3}+2(4-p_{1}^{2})p_{1}x-p_{1}(4-p_{1}^{2})x^{2}+2(4-p_{1}^{2})\left(1-\left|x\right|^{2}\right)y,$
(2.2)

for some $x,y\in\mathbb{C}$ with $\left|x\right|\leq 1\text{ and
}\left|y\right|\leq 1.$

######  Lemma 2.2.

If $p(z)=1+p_{1}z+p_{2}z^{2}+p_{3}z^{3}+\cdots\in\mathcal{P}$ $(p_{1}\geq 0),$
then

$\left|p_{n}\right|\leq 2~{}~{}\left(n\geq 1\right),$ (2.3)

and if $Q\in\left[0,1\right]$ and $Q\left(2Q-1\right)\leq R\leq Q,$ then

$\left|p_{3}-2Qp_{1}p_{2}+Rp_{1}^{3}\right|\leq 2.$ (2.4)

Also

$\displaystyle\left|p_{n+k}-\mu p_{n}p_{k}\right|$ $\displaystyle\leq$
$\displaystyle 2\max\left\\{1,\left|2\mu-1\right|\right\\}$ (2.7)
$\displaystyle=$ $\displaystyle 2\left\\{\begin{array}[]{ll}1,&\text{ for
}0\leq\mu\leq 1,\\\ \left|2\mu-1\right|,&\text{ otherwise
}\end{array}\right..$

The inequalities (2.3), (2.4) and (2.7) are taken from [8, 28] and [36],
respectively.

######  Lemma 2.3 (see [37]).

Let $\tau,~{}\psi,~{}\rho$ and $\varsigma$ satify the inequalities
$0<\tau<1,~{}0<\varsigma<1$ and

$\begin{split}8\varsigma\left(1-\varsigma\right)\left[\left(\tau\psi-2\rho\right)^{2}+\left(\tau\left(\varsigma+\tau\right)-\psi\right)^{2}\right]+\tau\left(1-\tau\right)\left(\psi-2\varsigma\tau\right)^{2}\\\
\leq
4\varsigma\tau^{2}\left(1-\tau\right)^{2}\left(1-\varsigma\right).\end{split}$
(2.8)

If $p(z)=1+p_{1}z+p_{2}z^{2}+p_{3}z^{3}+\cdots\in\mathcal{P}$ $(p_{1}\geq 0),$
then

$\left|\rho p_{1}^{4}+\varsigma p_{2}^{2}+2\tau p_{1}p_{3}-\frac{3}{2}\psi
p_{1}^{2}p_{2}-p_{4}\right|\leq 2.$ (2.9)

######  Lemma 2.4.

[34] Let $\overline{\mathcal{U}}=\left\\{z:\left|z\right|\leq 1\right\\}.$
Also, for any real numbers $a,$ $b$ and $c,$ let the quantity
$Y(a,b,c)=\max_{z\in\overline{\mathcal{U}}}\left\\{\left|a+bz+cz^{2}\right|+1-\left|z\right|^{2}\right\\}.$
If $ac\geq 0,$ then

$Y(a,b,c)=\left\\{\begin{array}[]{cc}\left|a\right|+\left|b\right|+\left|c\right|&\left|b\right|\geq
2(1-\left|c\right|)\\\
1+\left|a\right|+\frac{b^{2}}{4(1-\left|c\right|)}&\left|b\right|<2(1-\left|c\right|)\end{array}\right..$

Furthermore, if $ac<0,$ then

$Y(a,b,c)=\left\\{\begin{array}[]{cc}1-\left|a\right|+\frac{b^{2}}{4(1-\left|c\right|)}&\left(-4ac(c^{-2}-1)\leq
b^{2};\;\left|b\right|<2(1-\left|c\right|)\right)\\\
1+\left|a\right|+\frac{b^{2}}{4(1+\left|c\right|)}&b^{2}<\min\left\\{4(1+\left|c\right|)^{2},-4ac(c^{-2}-1)\right\\}\\\
R(a,b,c)&\left(\text{otherwise}\right)\end{array}\right.,$

where

$R(a,b,c)=\left\\{\begin{array}[]{cc}\left|a\right|+\left|b\right|-\left|c\right|&\left(\left|c\right|\left(\left|b\right|+4\left|a\right|\right)\leq\left|ab\right|)\right)\\\
-\left|a\right|+\left|b\right|+\left|c\right|&\left(\left|ab\right|\leq\left|c\right|\left(\left|b\right|-4\left|a\right|\right)\right)\\\
\left(\left|a\right|+\left|c\right|\right)\sqrt{1-\frac{b^{2}}{4ac}}&\left(\text{otherwise}\right)\end{array}\right..$

##  3\. Coefficient Estimates

Finding the upper bound for coefficients have been one of the central topic of
research in geometric function theory as it gives several properties of
functions. Therefore, we will be interested in the following problem in this
section.

Problem: Find $\sup\left|a_{n}\right|$ $(n=2,3,\cdots)$ for certain sublasses
of univalent functions. In particular, bound for the second coefficient gives
growth and distortion theorems for functions in the class. Another one is the
coefficient problem related with Hankel determinants. Similarly, using the
Hankel determinants (which also deals with the bound on coefficients), Cantor
[7] proved that “if ratio of two bounded analytic functions in $\mathcal{U}$,
then the function is rational”. The Hankel determinants
[32]$\;H_{q}(n)\;(n=1,2,...,\;q=1,2,...)$ of the function $f$ are defined by

$H_{q}(n)=\left|\begin{array}[]{cccc}a_{n}&a_{n+1}&...&a_{n+q-1}\\\
a_{n+1}&a_{n+2}&...&a_{n+q}\\\ \vdots&\vdots&\ddots&\vdots\\\
a_{n+q-1}&a_{n+q}&...&a_{n+2q-2}\end{array}\right|\text{ \ \ }(a_{1}=1).$

This determinant was discussed by several authors with $q=2$. For example, we
know that the functional $H_{2}(1):=a_{3}-a_{2}^{2}\;$is known as the Fekete-
Szegö functional and they consider the further generalized functional
$a_{3}-\mu a_{2}^{2},\;$where $\mu$ is some real number. Estimating for the
upper bound of $\left|a_{3}-\mu a_{2}^{2}\right|$ is known as the Fekete-Szegö
problem. In 1969, Keogh and Merkes [24] solved the Fekete-Szegö problem for
the class $\mathcal{S}^{\ast}$. The second Hankel determinant $H_{2}(2)$ is
given by $H_{2}(2):=\left|a_{2}a_{4}-a_{3}^{2}\right|.$ The bounds for the
second Hankel determinant $H_{2}(2)$ obtained for the class
$\mathcal{S}^{\ast}$ in [21]. Lee et al. [27] established the sharp bound to
$\left|H_{2}(2)\right|$ by generalizing their classes using subordination.
Moreover, the quantity given by
$H_{3}(1):=a_{3}(a_{2}a_{4}-a_{3}^{2})-a_{4}(a_{4}-a_{2}a_{3})+a_{5}(a_{3}-a_{2}^{2})$
is called third Hankel determinant. Zaprawa [47] proved that
$\left|H_{3}(1)\right|<1$ for $f\in\mathcal{S}^{\ast}.$ In 2019, Kwon, Lecko
and Sim [26] improved the result of Zaprawa as $\left|H_{3}(1)\right|<8\diagup
9.$ This result the best known upper bound of $\left|H_{3}(1)\right|$ for the
class $\mathcal{S}^{\ast}.$ Recently, Hankel determinants and Fekete-Szegö
problem have been considered in many papers of Srivastava and his co-authors
(see, for example, [40, 41, 42, 43, 44]) and Deniz and his co-authors (see,
[12, 23]).

In this paper, we seek sharp upper bounds for the coefficients $a_{j}$
$(j=2,3,4,5,6)$ and functionals $H_{2}(2)$ and $H_{3}(1)\;$for functions
$f\;$belonging to the class $\mathcal{S}_{G}^{\ast}.$

######  Theorem 3.1.

Let $f(z)=z+a_{2}z^{2}+a_{3}z^{3}+\cdots\in\mathcal{S}_{G}^{\ast}.$ Then,

$\displaystyle\left|a_{n}\right|$ $\displaystyle\leq$
$\displaystyle\frac{1}{2(n-1)}\text{ \ \ }(n=2,3,4,5)$
$\displaystyle\left|{a_{6}}\right|$ $\displaystyle\leq$
$\displaystyle\frac{13}{48}.$

All of the above estimates, except that on ${a_{6},}$ are sharp.

###### Proof.

Since $f\in\mathcal{S}_{G}^{\ast}$, there exists an analytic function $w$ with
$w(0)=0$ and $|w(z)|<1$ in $\mathcal{U}$ such that

$\frac{zf^{\prime}(z)}{f(z)}=\Psi(w(z))=\frac{w(z)}{\ln\left(1+w(z)\right)}\text{
\ \ }(z\in\mathcal{U}).$ (3.1)

Define the functions $p$ by

$p(z)=\frac{1+w(z)}{1-w(z)}=1+p_{1}z+p_{2}z^{2}+\cdots\text{ \ \
}(z\in\mathcal{U})$

or equivalently,

$\displaystyle w(z)$ $\displaystyle=$
$\displaystyle\frac{p(z)-1}{p(z)+1}=\frac{p_{1}}{2}z+\frac{1}{2}\left(p_{2}-\frac{p_{1}^{2}}{2}\right)z^{2}+\frac{1}{2}\left(p_{3}-p_{1}p_{2}+\frac{p_{1}^{3}}{4}\right)z^{3}$
$\displaystyle+\frac{1}{2}\left(p_{4}-p_{1}p_{3}+\frac{3p_{1}^{2}p_{2}}{4}-\frac{p_{2}^{2}}{2}-\frac{p_{1}^{4}}{8}\right)z^{4}$
$\displaystyle{+\frac{1}{2}\left(p_{5}-\frac{1}{2}p_{1}^{3}p_{2}+\frac{3}{4}p_{1}p_{2}^{2}+\frac{3}{4}p_{1}^{2}p_{3}-p_{2}p_{3}-p_{1}p_{4}+\frac{1}{16}p_{1}^{5}\right)z^{5}+}\cdots$

in $\mathcal{U}.$ Then $p$ is analytic in $\mathcal{U}$ with $p(0)=1$ and has
positive real part in $\mathcal{U}$. By using (3) together with
$\frac{w(z)}{\ln\left(1+w(z)\right)},$ it is evident that

$\displaystyle\Psi(w(z))$ $\displaystyle=$
$\displaystyle{1+\frac{p_{1}z}{4}+\frac{1}{48}\left(-7p_{1}^{2}+12p_{2}\right)z^{2}}{+\frac{1}{192}\left(17p_{1}^{3}-56p_{1}p_{2}+48p_{3}\right)z^{3}}$
(3.6)
$\displaystyle+\frac{1}{11520}\left(-649p_{1}^{4}+3060p_{1}^{2}p_{2}-3360p_{1}p_{3}-1680p_{2}^{2}+2880p_{4}\right)z^{4}$
$\displaystyle+\frac{1}{46080}\left(\begin{array}[]{l}1739p_{1}^{5}-10384p_{1}^{3}p_{2}+12240p_{1}^{2}p_{3}+12240p_{1}p_{2}^{2}\\\
-13440p_{2}p_{3}+13440p_{1}p_{4}+11520p_{5}\end{array}\right)z^{5}+\cdots.$

Since

$\displaystyle\frac{zf^{\prime}(z)}{f(z)}$ $\displaystyle{=}$
$\displaystyle{1+a_{2}z+\left(-a_{2}^{2}+2a_{3}\right)z^{2}+\left(a_{2}^{3}-3a_{2}a_{3}+3a_{4}\right)z^{3}}$
$\displaystyle{+\left(-a_{2}^{4}+4a_{2}^{2}a_{3}-2a_{3}^{2}-4a_{2}a_{4}+4a_{5}\right)z^{4}}$
$\displaystyle{+\left(a_{2}^{5}-5a_{2}^{3}a_{3}+5a_{2}a_{3}^{2}+5a_{2}^{2}a_{4}-5a_{3}a_{4}-5a_{2}a_{5}+5a_{6}\right)z^{5}+\cdots,}$

it follows by (3.1), (3.6) and (3) that

$\displaystyle a_{2}$ $\displaystyle=$ $\displaystyle{\frac{p_{1}}{4}},$ (3.8)
$\displaystyle a_{3}$ $\displaystyle=$
$\displaystyle{\frac{1}{24}\left(-p_{1}^{2}+3p_{2}\right)},$ (3.9)
$\displaystyle{a_{4}}$ $\displaystyle{=}$
$\displaystyle{\frac{1}{288}\left(4p_{1}^{3}-19p_{1}p_{2}+24p_{3}\right),}$
(3.10) $\displaystyle{a_{5}}$ $\displaystyle{=}$
$\displaystyle-\frac{1}{16}\left(\frac{71}{720}p_{1}^{4}+\frac{11}{24}p_{2}^{2}+\frac{5}{6}p_{1}p_{3}-\frac{85}{144}p_{1}^{2}p_{2}-p_{4}\right),$
(3.11) $\displaystyle{a_{6}}$ $\displaystyle{=}$
$\displaystyle{\frac{\left(\begin{array}[]{l}2267p_{1}^{5}-15677p_{1}^{3}p_{2}+21720p_{1}^{2}p_{3}+23370p_{2}^{2}p_{1}\\\
-29520p_{1}p_{4}-33120p_{2}p_{3}+34560p_{5}\end{array}\right)}{691200}.}$
(3.14)

Thus, we get

$\displaystyle\left|a_{2}\right|$ $\displaystyle\leq$
$\displaystyle\frac{1}{2}~{}~{}\left(\text{from (\ref{eq24})}\right),$
$\displaystyle\left|a_{3}\right|$ $\displaystyle=$
$\displaystyle\frac{1}{8}\left|p_{2}-\frac{1}{3}p_{1}^{2}\right|\leq\frac{1}{4}~{}~{}\left(\text{from
(\ref{eq26})}\right),$ $\displaystyle\left|a_{4}\right|$ $\displaystyle{=}$
$\displaystyle\frac{1}{12}\left|p_{3}-\frac{19}{24}p_{1}p_{2}+\frac{1}{6}p_{1}^{3}\right|\leq\frac{1}{6}~{}~{}\left(\text{from
(\ref{eq25})}\right),$ $\displaystyle\left|{a_{5}}\right|$ $\displaystyle{=}$
$\displaystyle\frac{1}{16}\left|\frac{71}{720}p_{1}^{4}+\frac{11}{24}p_{2}^{2}+\frac{5}{6}p_{1}p_{3}-\frac{85}{144}p_{1}^{2}p_{2}-p_{4}\right|\leq\frac{1}{8}~{}~{}\left(\text{from
(\ref{eq28})}\right).$

We now find the estimates on $\left|{a_{6}}\right|.$ Therefore, from the Lemma
2.3, $\left|p_{n}\right|\leq 2$ and (2.7), we find that

$\displaystyle\left|{a_{6}}\right|$ $\displaystyle{=}$
$\displaystyle\frac{\left|\begin{array}[]{l}2267p_{1}^{5}-15677p_{1}^{3}p_{2}+21720p_{1}^{2}p_{3}+23370p_{2}^{2}p_{1}\\\
-29520p_{1}p_{4}-33120p_{2}p_{3}+34560p_{5}\end{array}\right|}{691200}$
$\displaystyle\leq$
$\displaystyle\frac{41\left|p_{1}\right|}{960}\left|\frac{2267}{29520}p_{1}^{4}+\frac{779}{984}p_{2}^{2}+\frac{181}{246}p_{1}p_{3}-\frac{15677}{29520}p_{1}^{2}p_{2}-p_{4}\right|+\frac{1}{20}\left|p_{5}-\frac{23}{24}p_{2}p_{3}\right|$
$\displaystyle\leq$ $\displaystyle\frac{41}{240}+\frac{1}{10}=\frac{13}{48}.$

The first five results are sharp for the function
$f:\mathcal{U\rightarrow\mathbb{C}}$ given by

$\displaystyle f_{1}(z)$ $\displaystyle=$ $\displaystyle
z\exp\left(\int_{0}^{z}\frac{\Psi(t)-1}{t}dt\right)=z+\frac{1}{2}z^{2}+\cdots,$
(3.16) $\displaystyle f_{2}(z)$ $\displaystyle=$ $\displaystyle
z\exp\left(\int_{0}^{z}\frac{\Psi(t^{2})-1}{t}dt\right)=z+\frac{1}{4}z^{3}+\cdots,$
(3.17) $\displaystyle f_{3}(z)$ $\displaystyle=$ $\displaystyle
z\exp\left(\int_{0}^{z}\frac{\Psi(t^{3})-1}{t}dt\right)=z+\frac{1}{6}z^{4}+\cdots,$
(3.18) $\displaystyle f_{4}(z)$ $\displaystyle=$ $\displaystyle
z\exp\left(\int_{0}^{z}\frac{\Psi(t^{4})-1}{t}dt\right)=z+\frac{1}{8}z^{5}+\cdots,$
(3.19) $\displaystyle f_{5}(z)$ $\displaystyle=$ $\displaystyle
z\exp\left(\int_{0}^{z}\frac{\Psi(t^{5})-1}{t}dt\right)=z+\frac{1}{10}z^{6}+\cdots.$
(3.20)

This completes the proof of Theorem 3.1. ∎

######  Conjecture 3.2.

Let $f(z)=z+a_{2}z^{2}+a_{3}z^{3}+\cdots\in\mathcal{S}_{G}^{\ast}.$ As we see
from the extremal functions $f_{i}(z),~{}~{}i=1,2,3,4$, the first five
coeffitients are obtained as our sharp bound estimates for
$\left|{a_{n}}\right|,$ $n=2,3,4,5.$ By comparing with the extremal function
$f_{5}(z),$ the sixth coefficient is expected as the sharp bound estimate for
$\left|{a_{6}}\right|.$ So it is an open question whether the bound for
$\left|{a_{6}}\right|$ is sharp and if one can show the following estimate
$\left|{a_{6}}\right|\leq\frac{1}{10}$. Also, it is an open question the sharp
inequality $\left|a_{n}\right|\leq\frac{1}{2(n-1)}$ for all
$n\in\mathbb{N}\backslash\\{1\\}.$

In 2021, Deniz [13] defined a subclass of starlike functions which is related
with generalized telephone numbers. He obtained some sharp bounds of
coefficients for this class. These results are as follows:
$\left|a_{2}\right|\leq{1},$ $\left|a_{3}\right|\leq{1,}$
$\left|{a_{4}}\right|{\leq}{\frac{8}{9},}$
$\left|{a_{5}}\right|{\leq}\frac{107}{144}$ and
$\left|{a_{6}}\right|\leq\frac{2381}{3600}.$ When we compare these results
with Theorem 3.1, seen that our results are better.

######  Theorem 3.3.

Let $f(z)=z+a_{2}z^{2}+a_{3}z^{3}+\cdots\in\mathcal{S}_{G}^{\ast}.$ Then, the
following sharp estimates holds:

$\left|{a}_{3}-\mu{a_{2}^{2}}\right|\leq\frac{1}{4}\max\\{1,\left|\mu-\frac{1}{3}\right|\\},\text{
\ \ }\left(\mu\in\mathbb{C}\right)$
$\left|{a_{2}a}_{3}-a_{4}\right|\leq\frac{1}{6}$

and

$\left|{a_{2}a}_{4}-a_{3}^{2}\right|\leq\frac{1}{16}.$

###### Proof.

From (3.8) and (3.9), we have

${a}_{3}-\mu{a_{2}^{2}}={\frac{1}{24}\left(-p_{1}^{2}+3p_{2}\right)}-\mu{\frac{p_{1}^{2}}{16}}=\frac{1}{8}\left(p_{2}-\left(\frac{3\mu+2}{6}\right)p_{1}^{2}\right).$

If we take $\nu=\frac{3\mu+2}{6}$ in known result (see [24]):
$\left|{p}_{2}-\nu{p_{1}^{2}}\right|\leq 2\max\\{1,\left|2\nu-1\right|\\}$ for
$\mu,\nu\in\mathbb{C},$ we obtain that

$\left|{a}_{3}-\mu{a_{2}^{2}}\right|\leq\frac{1}{4}\max\\{1,\left|\mu-\frac{1}{3}\right|\\}.$

Similarly from (3.8), (3.9) and (3.10), we have

${a_{2}a}_{3}-a_{4}{=}-\frac{1}{288}\left(24p_{3}-28p_{1}p_{2}+7p_{1}^{3}\right)$

and so from (2.4)

$\left|{a_{2}a}_{3}-a_{4}\right|{\leq}\frac{1}{288}\left|24p_{3}-28p_{1}p_{2}+7p_{1}^{3}\right|=\frac{1}{12}\left|p_{3}-\frac{7}{6}p_{1}p_{2}+\frac{7}{24}p_{1}^{3}\right|\leq\frac{1}{6}.$

Now, we investigate last estimate in Theorem 3.3. From (3.8), (3.9) and (3.10)
again, we see that

${a_{2}a}_{4}-a_{3}^{2}=\frac{2p_{1}^{4}-7p_{1}^{2}p_{2}-18p_{2}^{2}+24p_{1}p_{3}}{1152}:=T,$

which, upon applying Lemma 2.1 and assuming that $s=p_{1}\in[0,2]$, we can
write

$T=\frac{4-s^{2}}{2304}\left(-s^{2}x-3\left(s^{2}+12\right)x^{2}+24s\left(1-\left|x\right|^{2}\right)y.\right)$

If $s=0,$ then $T=-\frac{1}{16}x^{2}.$ Thus, since $\left|x\right|\leq 1$, we
have

$\left|T\right|\leq\frac{1}{16}.$ (3.21)

If $s=2,$ then

$\left|T\right|=0.$ (3.22)

We now let $s\in(0,2).$ Then, we can write

$\displaystyle\left|T\right|$ $\displaystyle=$
$\displaystyle\left|\frac{4-s^{2}}{2304}\left(-s^{2}x-3\left(s^{2}+12\right)x^{2}+24s\left(1-\left|x\right|^{2}\right)y.\right)\right|$
$\displaystyle\leq$
$\displaystyle\frac{s\left(4-s^{2}\right)}{96}\left[\left|-\frac{s}{24}x-\frac{s^{2}+12}{8s}x^{2}\right|+1-\left|x\right|^{2}\right]$
$\displaystyle=$
$\displaystyle\frac{s\left(4-s^{2}\right)}{96}\left[\left|\widetilde{a}+\widetilde{b}x+\widetilde{c}x^{2}\right|+1-\left|x\right|^{2}\right]$

where

$\widetilde{a}=0,\;\widetilde{b}=-\frac{s}{24},\;\widetilde{c}=-\frac{s^{2}+12}{8s}.$

It follows that $\widetilde{a}\widetilde{c}=0.$ Also, we easily see that

$\left|\widetilde{b}\right|-2(1-\left|\widetilde{c}\right|)=\frac{7s^{2}-48s+72}{24s}>0.$

Therefore, we have

$\displaystyle\left|T\right|$ $\displaystyle\leq$
$\displaystyle\frac{s\left(4-s^{2}\right)}{96}\left(\left|\widetilde{a}\right|+\left|\widetilde{b}\right|+\left|\widetilde{c}\right|\right)=\frac{s\left(4-s^{2}\right)}{96}\left(\frac{s^{2}+9}{6s}\right)$
$\displaystyle=$ $\displaystyle\frac{-s^{4}-5s^{2}+36}{576}.$

Let $t=s^{2}$ $\left(t\in(0,4)\right).$ Then, we investigate maximum of the
function $H_{0}$ defined by

$H_{0}(t)=\frac{-t^{2}-5t+36}{576}.$

In that case, we have

$\left|T\right|\leq\left|H_{0}(t)\right|\leq\frac{-t^{2}-5t+36}{576}\leq\frac{1}{16}.$
(3.23)

∎

######  Remark 3.4.

By using Theorem 3.1 and Theorem 3.3, we have

$\left|H_{3}(1)\right|\leq\left|a_{3}\right|\left|a_{2}a_{4}-a_{3}^{2}\right|+\left|a_{4}\right|\left|a_{4}-a_{2}a_{3}\right|+\left|a_{5}\right|\left|a_{3}-a_{2}^{2}\right|\leq\frac{43}{576}.$

If we compare the bound $\frac{43}{576}$ with the results of Zaprawa [47] and
Deniz [13], see that this bound is better.

###  3.1. Logarithmic Coefficients

The following formula defines the logarithmic coefficients $\beta_{n}$ of
$f(z)=z+\sum_{n=2}^{\infty}a_{n}z^{n}$ that belongs to $\mathcal{S}$

$G_{f}\left(z\right):=\log\left(\frac{f\left(z\right)}{z}\right)=2\sum_{n=1}^{\infty}\beta_{n}z^{n}~{}~{}~{}\text{for}~{}z\in\mathcal{U}.$
(3.24)

In many estimations, these coefficients provide a significant contribution to
the concept of univalent functions. In 1985, De Branges [10] proved that

$\sum_{k=1}^{n}k\left(n-k+1\right)\left|\beta_{n}\right|^{2}\leq\sum_{k=1}^{n}\frac{n-k+1}{k}~{}~{}~{}\forall~{}n\geq
1$ (3.25)

and equality will be achieved if $f$ has the form
$z/\left(1-e^{i\theta}z\right)^{2}$ for some $\theta\in\mathbb{R}.$ In its
most comprehensive version, this inequality offers the famous Bieber-bach-
Robertson-Milin conjectures regarding Taylor coefficients of
$f\in\mathcal{S}.$ We refer to [5, 16, 17] for further details on the proof of
De Branges finding. By considering the logarithmic coefficients, Kayumov [22]
was able to prove Brennan’s conjecture for conformal mappings in 2005. For
your reference, we mention a few works that have made major contributions to
the research of the logarithmic coefficients. Andreev and Duren [4],
Alimohammadi et al. [3], Deng [11], Roth [38], Ye [46], Obradović et al. [33],
and finally the work of Girela [18] are the major contributions to the study
of logarithmic coefficients for different subclasses of holomorphic univalent
functions.

As stated in the definition, it is simple to determine that for
$f\in\mathcal{S},$ the logarithmic coefficients are computed by

$\beta_{1}=\frac{1}{2}a_{2},$ (3.26)
$\beta_{2}=\frac{1}{2}\left(a_{3}-\frac{1}{2}a_{2}^{2}\right),$ (3.27)
$\beta_{3}=\frac{1}{2}\left(a_{4}-a_{2}a_{3}+\frac{1}{3}a_{2}^{3}\right),$
(3.28)
$\beta_{4}=\frac{1}{2}\left(a_{5}-a_{2}a_{4}+a_{2}^{2}a_{3}-\frac{1}{2}a_{3}^{2}-\frac{1}{4}a_{2}^{4}\right).$
(3.29)

######  Theorem 3.5.

If $f\in\mathcal{S}^{*}_{G}$ and has the series representation
$f(z)=z+\sum_{n=2}^{\infty}a_{n}z^{n}$ then

$\left|\beta_{n}\right|\leq\frac{1}{4n}~{}~{}~{}\left(n=1,2,3,4\right).$
(3.30)

These bounds are sharp and can be obtained from the extremal functions
$f_{i}(z)~{}~{}i=1,2,3,4$ given by (3.16)-(3.19).

###### Proof.

Let $f\in\mathcal{S}^{*}_{G}.$ Then, putting (3.26), (3.27), (3.28) and (3.29)
in (3.8), (3.9), (3.10) and (3.11), we get

$\displaystyle\beta_{1}$ $\displaystyle=$ $\displaystyle{\frac{p_{1}}{8}},$
(3.31) $\displaystyle\beta_{2}$ $\displaystyle=$
$\displaystyle{\frac{1}{192}\left(-7p_{1}^{2}+12p_{2}\right)},$ (3.32)
$\displaystyle\beta_{3}$ $\displaystyle{=}$
$\displaystyle{\frac{1}{1152}\left(17p_{1}^{3}-56p_{1}p_{2}+48p_{3}\right),}$
(3.33) $\displaystyle\beta_{4}$ $\displaystyle{=}$
$\displaystyle-\frac{1}{92160}\left(649p_{1}^{4}+1680p_{2}^{2}+3360p_{1}p_{3}-3060p_{1}^{2}p_{2}-2280p_{4}\right).$
(3.34)

For $\beta_{1},$ using (2.3) in(3.31), we obtain

$\left|\beta_{1}\right|\leq\frac{1}{4}.$

For $\beta_{2},$ putting (2.7) in(3.32), we get

$\left|\beta_{2}\right|\leq\frac{1}{8}.$

For $\beta_{3},$ we can rewrite (3.33) as

$\left|\beta_{3}\right|=\frac{1}{24}\left|p_{3}-\frac{7}{6}p_{1}p_{2}+\frac{17}{48}p_{1}^{3}\right|,$

using (2.4) where $Q=\frac{7}{12}$ and $R=\frac{17}{48},$ we obtain

$\left|\beta_{3}\right|\leq\frac{1}{12}.$

For $\beta_{4},$ we can rewrite (3.34) as

$\beta_{4}=-\frac{1}{32}\left(\frac{649}{2880}p_{1}^{4}+\frac{7}{12}p_{2}^{2}+\frac{7}{6}p_{1}p_{3}-\frac{17}{16}p_{1}^{2}p_{2}-p_{4}\right).$
(3.35)

Comparing the right side of (3.35) with inequality (2.9), where
$\rho=\frac{649}{2880},$ $\varsigma=\frac{7}{12},$ $\tau=\frac{7}{12}$ and
$\psi=\frac{17}{24}.$ It follows that

$8\varsigma\left(1-\varsigma\right)\left[\left(\tau\psi-2\rho\right)^{2}+\left(\tau\left(\varsigma+\tau\right)-\psi\right)^{2}\right]+\tau\left(1-\tau\right)\left(\psi-2\varsigma\tau\right)^{2}=0.00442226$

and

$4\varsigma\tau^{2}\left(1-\tau\right)^{2}\left(1-\varsigma\right)=0.057435.$

Using (2.8) we deduce that

$\left|\beta_{4}\right|\leq\frac{1}{16}.$

∎

###  3.2. Inverse Coefficients

It is well-known that the function

$f(z)=z+\sum_{n=2}^{\infty}a_{n}z^{n}\in\mathcal{S}$

has an inverse $f^{-1},$ which is analytic in $\left|w\right|<1/4,$ as we know
from Koebe’s $1/4$-theorem. If $f\in\mathcal{S},$ then

$f^{-1}(w)=w+A_{2}w^{2}+A_{3}w^{3}+\cdots,~{}~{}~{}~{}\left|w\right|<1/4.$
(3.36)

Löwner [29] proved that, if $f\in\mathcal{S}$ and its inverse is given by
(3.36), then the sharp estimate

$\left|A_{n}\right|\leq\frac{\left(2n\right)!}{n!\left(n+1\right)!}$ (3.37)

holds. It has been shown that the inverse of the Koebe function
$k(z)=z/\left(1-z\right)^{2}$ provides the best bounds for all
$\left|A_{n}\right|~{}~{}\left(n=2,3,\ldots\right)$ in (3.37) over all members
of $\mathcal{S}.$ There has been a good deal of interest in determining the
behavior of the inverse coefficients of $f$ given in (3.36) when the
corresponding function $f$ is restricted to some proper geometric subclasses
of $\mathcal{S}.$ Alternate proofs of the inequality (3.37) have been given by
several authors but a simpler proof was given by Yang [45].
Since $f\left(f^{-1}(w)\right)=w,$ using (3.36) it’s very clear to see

$\begin{split}A_{2}=&-a_{2},\\\ A_{3}=&2a_{2}^{2}-a_{3},\\\
A_{4}=&-5a_{2}^{3}+5a_{2}a_{3}-a_{4}.\end{split}$ (3.38)

######  Theorem 3.6.

If $f\in\mathcal{S}^{*}_{G}$ and has the series representation
$f^{-1}(w)=w+A_{2}w^{2}+A_{3}w^{3}+\cdots$ then

$\begin{split}\left|A_{2}\right|&\leq\frac{1}{2},\\\
\left|A_{3}\right|&\leq\frac{5}{12},\\\
\left|A_{4}\right|&\leq\frac{31}{72}.\end{split}$ (3.39)

These bounds are sharp, except $A_{4}.$

###### Proof.

Let $f\in\mathcal{S}^{*}_{G}$ and $f^{-1}(w)=w+A_{2}w^{2}+A_{3}w^{3}+\cdots.$
If equations (3.8), (3.9) and (3.10) are written in (3.38), we obtain

$\displaystyle A_{2}$ $\displaystyle=-\frac{p_{1}}{4},$ (3.40) $\displaystyle
A_{3}$ $\displaystyle=\frac{p_{1}^{2}}{6}-\frac{p_{2}}{8},$ (3.41)
$\displaystyle A_{4}$
$\displaystyle=-\frac{1}{576}\left(48p_{3}-128p_{1}p_{2}+83p_{1}^{3}\right).$
(3.42)

Because of $\left|p_{1}\right|\leq 2,$ the result for
$\left|A_{2}\right|\leq\frac{1}{2}$ is trivial.
We now estimate $\left|A_{3}\right|$ by using (2.7) in equality (3.41). Thus,
we have

$\left|A_{3}\right|\leq\frac{1}{8}\left|p_{2}-\frac{4}{3}p_{1}^{2}\right|\leq\frac{5}{12}.$

From the relations Lemma 2.1 and (3.42), and by some simple calculations, we
have

$K=-\frac{4-\sigma^{2}}{24}\left(\frac{31\sigma^{3}}{24\left(4-\sigma^{2}\right)}-\frac{5\sigma}{3}x-\frac{\sigma}{2}x^{2}+\left(1-\left|x\right|^{2}\right)y\right),$

where $\sigma=p_{1}\in\left[0,2\right],$ $\left|x\right|\leq 1$ and
$\left|y\right|\leq 1.$ We now investigate upper bound of the $\left|K\right|$
according to $\sigma.$
A. If $\sigma=0,$ then $T=-\frac{1}{6}\left(1-\left|x\right|^{2}\right)y$ and
so we have $\left|T\right|\leq\frac{1}{6}.$
B. Let $\sigma=2.$ Then $T=-\frac{31}{72}$ and so we have
$\left|T\right|\leq\frac{31}{72}.$
C. We now assume that $\sigma\in\left(0,2\right).$ Then, we can write

$\begin{split}\left|K\right|=&\frac{4-\sigma^{2}}{24}\left|\frac{31\sigma^{3}}{24\left(4-\sigma^{2}\right)}-\frac{5\sigma}{3}x-\frac{\sigma}{2}x^{2}+\left(1-\left|x\right|^{2}\right)y\right|\\\
\leq&\frac{4-\sigma^{2}}{24}\left(\left|\widetilde{A}+\widetilde{B}x+\widetilde{C}x^{2}\right|+1-\left|x\right|^{2}\right),\end{split}$
(3.43)

where

$\widetilde{A}=\frac{31\sigma^{3}}{24\left(4-\sigma^{2}\right)},~{}~{}~{}\widetilde{B}=-\frac{5\sigma}{3}\text{
\ and \ }\widetilde{C}=-\frac{\sigma}{2}.$

For the rest of the proof, we use Lemma 2.4. Then

$\widetilde{A}\widetilde{C}=-\frac{31\sigma^{4}}{48\left(4-\sigma^{2}\right)}<0.$

C1. Note that the inequality

$-4\widetilde{A}\widetilde{C}\left(\widetilde{C}^{-2}-1\right)-\widetilde{B}^{2}=\frac{31\sigma^{4}}{12\left(4-\sigma^{2}\right)}\left(\frac{4}{\sigma^{2}}-1\right)-\frac{25\sigma^{2}}{9}\leq
0$

which evidently holds for $s\in\left(-\infty,\infty\right).$ Moreover, the
inequality
$\left|\widetilde{B}\right|<2\left(1-\left|\widetilde{C}\right|\right)$ is
equivalent to $\sigma<\frac{3}{4},$ which is true for
$\sigma\in\left(0,\frac{3}{4}\right).$ Then, by (3.43) and Lemma 2.4,

$\begin{split}\left|K\right|\leq&\frac{4-\sigma^{2}}{24}\left(1-\left|\widetilde{A}\right|+\frac{\widetilde{B}^{2}}{4\left(1-\left|\widetilde{C}\right|\right)}\right)\\\
=&\frac{7\sigma^{3}+128\sigma^{2}+288}{1728}\leq\frac{2581}{12288}\approx
0.210042\end{split}$ (3.44)

for $\sigma\in\left(0,\frac{3}{4}\right).$
C2. Since

$4\left(1+\left|\widetilde{C}\right|\right)^{2}=\sigma^{2}+4\sigma+4,\quad-4\widetilde{A}\widetilde{C}\left(\widetilde{C}^{-2}-1\right)=\frac{31\sigma^{2}}{12},$

for $\sigma\in\left[\frac{3}{4},2\right),$ we see that the inequality

$\frac{25\sigma^{2}}{9}=\widetilde{B}^{2}<\min\left\\{4\left(1+\left|\widetilde{C}\right|\right)^{2},~{}-4\widetilde{A}\widetilde{C}\left(\widetilde{C}^{-2}-1\right)\right\\}=\frac{31\sigma^{2}}{12}$

is equivalent to $\frac{7\sigma^{2}}{36}<0,$ which is false for
$\sigma\in\left[\frac{3}{4},2\right).$
C3. Observe that the inequality

$\left|\widetilde{C}\right|\left(\left|\widetilde{B}\right|+4\left|\widetilde{A}\right|\right)-\left|\widetilde{A}\widetilde{B}\right|=\frac{\sigma^{2}\left(29\sigma^{2}-240\right)}{72\left(\sigma^{2}-4\right)}\leq
0$

is false for $\sigma\in\left[\frac{3}{4},2\right).$
C4. Note that the inequality

$\left|\widetilde{A}\widetilde{B}\right|-\left|\widetilde{C}\right|\left(\left|\widetilde{B}\right|-4\left|\widetilde{A}\right|\right)=\frac{\sigma^{2}\left(401\sigma^{2}-240\right)}{72\left(4-\sigma^{2}\right)}\leq
0$

is true $\sigma\in\left[\frac{3}{4},4\sqrt{\frac{15}{401}}\right].$ Then, by
(3.43) and Lemma 2.4,

$\begin{split}\left|K\right|\leq&\frac{4-\sigma^{2}}{24}\left(-\left|\widetilde{A}\right|+\left|\widetilde{B}\right|+\left|\widetilde{C}\right|\right)\\\
=&\frac{1}{576}\left(208\sigma-83\sigma^{3}\right)\leq\frac{3968}{1203}\sqrt{\frac{5}{1203}}\approx
0.212646\end{split}$ (3.45)

for $\sigma\in\left[\frac{3}{4},4\sqrt{\frac{15}{401}}\right].$
C5. It remains to consider the last case in Lemma 2.4, for
$\sigma\in\left(4\sqrt{\frac{15}{401}},2\right).$ Then, by (3.43)

$\begin{split}\left|K\right|\leq&\frac{4-\sigma^{2}}{24}\left(\left|\widetilde{A}\right|+\left|\widetilde{C}\right|\right)\sqrt{1-\frac{\widetilde{B}^{2}}{4\widetilde{A}\widetilde{C}}}\\\
=&\frac{\sqrt{400-7\sigma^{2}}\left(19\sigma^{2}+48\right)}{576\sqrt{93}}\leq\frac{31}{72}\approx
0.430556\end{split}$ (3.46)

is true for $\sigma\in\left(4\sqrt{\frac{15}{401}},2\right).$ In addition, as
can be seen from the figure below, the function
$H_{1}(\sigma)=\frac{\sqrt{400-7\sigma^{2}}\left(19\sigma^{2}+48\right)}{576\sqrt{93}}$
is increasing in $\left(4\sqrt{\frac{15}{401}},2\right)$ and takes its maximum
value at the limit.

---
Figure 1. The graph of the function $H_{1}(\sigma)$ on $\left(0,2\right)$

From A-C, we have $\left|A_{3}\right|\leq\frac{31}{72}.$ ∎

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