# The Bound of the Hankel Determinant of the Third Kind for Starlike Functions

Open Access
Article

## Abstract

In the present paper, the estimate of the third Hankel determinant
\begin{aligned} \begin{aligned} H_{3,1}(f)&= \begin{vmatrix} a_{1}&a_{2}&a_{3} \\ a_{2}&a_{3}&a_{4} \\ a_{3}&a_{4}&a_{5} \end{vmatrix} \end{aligned} \end{aligned}
for the class of starlike functions, i.e., for the class of analytic functions f standardly normalized such that $${{\mathrm{Re}}}(zf'(z)/f(z)) > 0,\ z\in {{\mathbb {D}}}:=\{z \in {\mathbb {C}} : |z|<1\},$$ is improved.

## Keywords

Univalent functions Starlike functions Carathéodory functions Hankel determinant Fourth coefficient

Primary 30C45

## 1 Introduction

Let $${{\mathcal {H}}}$$ be a class of analytic functions in $${\mathbb {D}}:= \left\{ z \in {\mathbb {C}} : |z|<1 \right\}$$ and let $${{\mathcal {A}}}$$ be its subclass normalized by $$f(0):=0$$, $$f'(0):=1,$$ i.e., of the form
\begin{aligned} f(z) = \sum _{n=1}^{\infty }a_{n}z^{n},\quad a_1:=1,\ z\in {\mathbb {D}}. \end{aligned}
(1.1)
Let $${{\mathcal {S}}}^{*}$$ denote the class of starlike functions, namely, the subclass of $${{\mathcal {A}}}$$ consisting of functions f such that
\begin{aligned} {{\mathrm{Re}}}\frac{zf'(z)}{f(z)} > 0,\quad z\in {\mathbb {D}}. \end{aligned}
(1.2)
Given $$q,n \in {\mathbb {N}},$$ the Hankel determinants $$H_{q,n}(f)$$ of Taylor’s coefficients of functions $$f \in {{\mathcal {A}}}$$ of the form (1.1) are defined as
\begin{aligned} H_{q,n}(f) := \begin{vmatrix} a_{n}&a_{n+1}&\cdots&a_{n+q-1} \\ a_{n+1}&a_{n+2}&\cdots&a_{n+q} \\ \vdots&\vdots&\vdots&\vdots \\ a_{n+q-1}&a_{n+q}&\cdots&a_{n+2(q-1)} \end{vmatrix}. \end{aligned}
Particularly, the third Hankel determinant $$H_{3,1}(f)$$ is given by
\begin{aligned} H_{3,1}(f) := \begin{vmatrix} a_{1}&a_{2}&a_{3} \\ a_{2}&a_{3}&a_{4} \\ a_{3}&a_{4}&a_{5} \end{vmatrix} = a_{3}( a_{2}a_{4} - a_{3}^{2} ) - a_{4}( a_{4} - a_{2}a_{3} ) + a_{5}( a_{3} - a_{2}^{2} ). \end{aligned}
(1.3)
To find the growth of the Hankel determinant $$H_{q,n}(f)$$ dependent on q and n for the whole class $${\mathcal {S}}\subset {\mathcal {A}}$$ of univalent functions as well as for its subclasses is an interesting problem to study. For the class $${\mathcal {S}}$$ some important result was shown by Pommerenke . For fixed q and n the growth problem can be reduced to an estimate of the Hankel determinant for the selected subclasses of $${\mathcal {A}}.$$ Recently many authors examined the Hankel determinant $$H_{2,2}(f)=a_2a_4-a_3^2$$ of order 2 (see, e.g., [3, 4, 6, 8, 12]). Note also that $$H_{2,1}(f)=a_3-a_2^2.$$ Thus the Hankel determinant $$H_{2,1}(f)$$ reduces to the well-known coefficient functional which for $${\mathcal {S}}$$ was estimated in 1916 by Bieberbach (see, e.g., [5, Vol. I, p. 35]).
The problem to find the upper bound of the Hankel determinant $$H_{3,1}(f)$$ of order 3 is more sophisticated if we expect to get sharp result. From (1.3) by using the triangle inequality we get at once the following inequality
\begin{aligned} |H_{3,1}(f)|\le |a_3||H_{2,2}(f)|+|a_{4}||a_{4} - a_{2}a_{3}| + |a_{5}||H_{2,1}(f)|. \end{aligned}
(1.4)
This simple observation allowed to estimate of $$|H_{3,1}(f)|$$ for compact subclasses $${\mathcal {F}}$$ of $${\mathcal {A}}$$ by various authors (see, e.g., [2, 15, 16, 17, 18]). However, these results are far from sharpness. If case when a given subclass $${\mathcal {F}}$$ of $${\mathcal {A}}$$ has a representation with using the Carathéodory class $${\mathcal {P}}$$, i.e., the class of functions $$p \in {{\mathcal {H}}}$$ of the form
\begin{aligned} p(z) = 1 + \sum _{n=1}^{\infty }c_{n}z^{n}, \quad z\in {\mathbb {D}}, \end{aligned}
(1.5)
having a positive real part in $${\mathbb {D}},$$ the coefficients of functions in $${\mathcal {F}}$$ have a suitable representation expressed by the coefficients of functions in $${\mathcal {P}}.$$ Therefore to get the upper bound of each term in (1.4) cited authors based their computing on the well-known formulas on coefficient $$c_2$$ (e.g., [14, p. 166]) and on the formula $$c_3$$ due to Libera and Zlotkiewicz .

In order to improve the bound of $$|H_{3,1}(f)|$$ we have to use directly formula (1.3), where we need to apply a formula for $$c_4,$$ similar to the formulas (2.1) and (2.2). In a recent paper  the authors found such a formula for $$c_4.$$ According to the authors’ knowledge, formulas for the coefficients $$c_n$$ for $$n \ge 5$$ analogous to the formulas (2.1) and (2.2) are not known.

Basing on the formulas for $$c_2,\ c_3$$ and $$c_4,$$ we improve the known estimate of the Hankel determinant $$H_{3,1}(f)$$ in the class $${{\mathcal {S}}}^{*}$$ of starlike functions. We show that $$|H_{3,1}(f)|\le 8/9.$$ Estimating each term of the right hand of (1.4) Babalola  showed that $$|H_{3,1}(f)|\le 16.$$ In  Zaprawa by a suitable grouping and using Lemma 1 due to Livingston  proved that $$|H_{3,1}(f)|\le 1.$$

## 2 Main Result

The basis for proof of the main result is the following lemma. It contains the well-known formula for $$c_2$$ (e.g., [14, p. 166]), the formula for $$c_3$$ due to Libera and Zlotkiewicz [9, 10] and the formula for $$c_4$$ found by the authors .

### Lemma 2.1

If $$p \in {{\mathcal {P}}}$$ is of the form (1.5) with $$c_1\ge 0,$$ then
\begin{aligned} 2c_2= & {} c_1^2 + (4-c_1^2)\zeta , \end{aligned}
(2.1)
\begin{aligned} 4c_3= & {} c_1^3 +(4-c_1^2)c_1\zeta (2-\zeta ) + 2(4-c_1^2)( 1 - |\zeta |^2) \eta \end{aligned}
(2.2)
and
\begin{aligned} \begin{aligned} 8c_4&= c_1^4+(4-c_1^2)\zeta \left[ c_1^2(\zeta ^2-3\zeta +3)+4\zeta \right] \\&-\,4(4-c_1^2)(1-|\zeta |^2)\left[ c_1(\zeta -1)\eta +\overline{\zeta }\eta ^2-\left( 1-|\eta |^2\right) \xi \right] \end{aligned} \end{aligned}
(2.3)
for some $$\zeta ,\eta ,\xi \in \overline{{\mathbb {D}}}:=\{z\in {\mathbb {C}}:|z|\le 1 \}.$$

Now, we will estimate the third-order Hankel determinant $$H_{3,1}(f)$$ for $$f \in {{\mathcal {S}}}^{*}$$. To this end, the following propositions are required.

### Proposition 2.2

Let $$\Theta : [0,3] \times [0,1] \rightarrow {\mathbb {R}}$$ be a function defined by
\begin{aligned} \Theta (t,x) := 96\theta _{1}(x) - 8\theta _{2}(x) t + 3\theta _{3}(x) t^{2}, \end{aligned}
(2.4)
where for $$x\in [0,1],$$
\begin{aligned} \theta _{1}(x):= & {} 2 +8x -x^{2} -6x^{3}, \\ \theta _{2}(x):= & {} 16 + 67x -34x^{2} -53x^{3} + 2x^{4} \end{aligned}
and
\begin{aligned} \theta _{3}(x) := 12 + 19x - 21x^{2} - 17x^{3} + 3x^{4}. \end{aligned}
Then $$\Theta (t,x) > 0$$ for $$0 \le t \le 3$$ and $$0 \le x \le 1$$.

### Proof

At first, note that the polynomial $$\theta _{3}$$ has a unique zero $$x=:x_{1}\approx 0.9314$$ in (0, 1). Since $$x_1\in (0.92,0.95)$$ and for $$x\in (0.92,0.95),$$
\begin{aligned}&\theta _{2}(x)>16 + 67\cdot (0.92)-34\cdot (0.95)^2 -53\cdot (0.95)^3 + 2\cdot (0.92)^4\\&\quad =2.94691092>0, \end{aligned}
it follows that
\begin{aligned} \frac{\partial }{\partial t} \Theta (t,x_1) = -8 \theta _{2}(x_1) \not =0. \end{aligned}
For $$x \not = x_1$$, $$(\partial /\partial t) \Theta (t,x)=0$$ occurs at
\begin{aligned} t= \frac{ 4\theta _{2}(x) }{ 3\theta _{3}(x) } =: t_{0}(x). \end{aligned}
We have
\begin{aligned} \frac{\partial }{\partial x}\Theta (t,x) \Big |_{t=t_{0}(x)} = \frac{16 \theta _{4}(x)}{9\theta _{3}^{2}(x)}, \end{aligned}
where
\begin{aligned} \begin{aligned} \theta _{4}(x)&:= 54\theta _{1}'(x)\theta _{3}^{2}(x) - 6\theta _{2}'(x) \theta _{2}(x) \theta _{3}(x) + 3\theta _{3}'(x) \theta _{2}^{2}(x) \\&= -3\left( 128 + 31896x - 18709x^{2} -133828x^{3} -3737x^{4} +198602x^{5} \right. \\&\quad \left. +74185x^{6}-91136x^{7} -54071x^{8} -2774x^{9} +668x^{10} \right. \\&\quad \left. +48x^{11} \right) ,\quad x\in (0,1). \end{aligned} \end{aligned}
The polynomial $$\theta _{4}$$ has exactly two zeros in (0, 1),  namely, $$x=:x_2\approx 0.533701$$ and $$x=:x_3\approx 0.811327.$$ We will now show that
\begin{aligned} t_{0}(x)>3,\quad x\in [0.5,0.9]. \end{aligned}
(2.5)
Since $$x_1>0.9,$$ so $$\theta _3(x)>0,$$ for $$x\in [0.5,0.9]$$ and the inequality (2.5) is equivalent to
\begin{aligned} 4\theta _2(x)-9\theta _3(x)>0,\quad x\in [0.5,0.9]. \end{aligned}
The above one can be equivalently written as
\begin{aligned} 19x^4+59x^3-53x^2-97x+44<0,\quad x\in [0.5,0.9]. \end{aligned}
As the polynomial on the left hand of the above inequality has a unique zero $$x\approx 0.40928$$ in [0, 1],  the above inequality is true, so is the inequality (2.5). Thus the function $$\Theta$$ has no critical point in $$(0,3) \times (0,1).$$ Hence it is sufficient to show that $$\Theta > 0$$ on the boundary of $$[0,3] \times [0,1]$$. We can easily check that the following inequalities hold:
\begin{aligned} \Theta (t,0)= & {} 4(48-32t+9t^{2}) \ge \frac{704}{9}, \quad t\in [0,3], \\ \Theta (t,1)= & {} 4(72+4t-3t^{2}) \ge 228, \quad t\in [0,3], \\ \Theta (0,x)= & {} 96(2+8x-x^{2}-6x^{3}) \ge 192, \quad x\in [0,1], \end{aligned}
and
\begin{aligned} \begin{aligned} \Theta (3,x)&= 3\left( 44 -109x +51x^{2} +79x^{3} +11x^{4} \right) \\&= 3\left( 44(1-x)^{3} +x(23 -81x +123x^{2}) +11x^{4} \right) \\&\ge 3\left( 44(1-x)^{3} + \frac{1585}{164}x +11x^{4} \right) > 0, \quad x \in [0,1]. \end{aligned} \end{aligned}
Thus the proof of the proposition is completed. $$\square$$

### Proposition 2.3

Let $$\Psi : [1,4] \times [0,1] \rightarrow {\mathbb {R}}$$ be a function defined by
\begin{aligned} \Psi (t,x) := 16 \psi _{1}(x) + 8 \psi _{2}(x) t + 3 \psi _{3}(x) t^{2}, \end{aligned}
(2.6)
where for $$x\in [0,1],$$
\begin{aligned} \psi _{1}(x):= & {} -2 + 27x + 21x^{2} - 37x^{3} +x^{4}, \\ \psi _{2}(x):= & {} 10 - 12x - 9x^{2} + 20x^{3} +x^{4} \end{aligned}
and
\begin{aligned} \psi _{3}(x) := x(3-5x-x^{2}-x^{3}). \end{aligned}
Then $$\Psi (t,x) > 0$$ for $$1 \le t \le 4$$ and $$0 \le x \le 1$$.

### Proof

At first, note that the function $$\psi _{3}$$ has a unique zero $$x=:x_{1}\approx 0.51839$$ in (0, 1). Since $$x_1\in (0.5,0.6)$$ and for $$x\in (0.5,0.6),$$
\begin{aligned} \psi _{2}(x)> 10 -12\cdot (0.6) -9\cdot (0.6)^2 + 20\cdot (0.5)^3 + (0.5)^4=2.1225>0, \end{aligned}
(2.7)
it follows that
\begin{aligned} \frac{\partial }{\partial t} \Psi (t,x_1) = 8\psi _{2}(x_1) \not =0. \end{aligned}
For $$x \not = x_1$$, $$(\partial /\partial t) \Psi (t,x)=0$$ occurs at
\begin{aligned} t = \frac{-4\psi _{2}(x)}{3\psi _{3}(x)} =: t_{0}(x). \end{aligned}
We have
\begin{aligned} \frac{\partial }{\partial x}\Psi (t,x) \Big |_{t=t_{0}(x)} = \frac{16 \psi _{4}(x)}{3 \psi _{3}^{2}(x)}, \end{aligned}
where
\begin{aligned} \begin{aligned} \psi _{4}(x)&:= 3\psi _{1}'(x)\psi _{3}^{2}(x) - 2\psi _{2}'(x) \psi _{2}(x) \psi _{3}(x) + \psi _{3}'(x) \psi _{2}^{2}(x) \\&= 300 - 1000x + 1737x^{2} - 4912x^{3} + 2009x^{4} + 13706x^{5} - 17777x^{6} \\&\quad + 6596x^{7}- 1541x^{8} + 546x^{9} -184x^{10} + 16x^{11},\quad x\in (0,1). \end{aligned} \end{aligned}
The polynomial $$\psi _{4}$$ has a unique zero $$x=:x_{2}\approx 0.388025$$ in (0, 1). Since $$x_1>0.5,$$ so $$\psi _3(x)>0,$$ for $$x\in (0,0.5).$$ Additionally, since $$\psi _2$$ has no zero in (0, 1),  the inequality (2.7) is true on [0, 1]. Thus $$t_0(x)<0$$ for $$x\in (0,0.5)$$ and in consequence, the function $$\Psi$$ has no critical point in $$(1,4) \times (0,1).$$ Hence it is sufficient to show that $$\Psi >0$$ on the boundary of $$[1,4] \times [0,1]$$. We can easily check that the following inequalities hold:
\begin{aligned} \Psi (t,0)= & {} -32 + 80t \ge 48, \quad t \in [1,4], \\ \Psi (t,1)= & {} 160 + 80t -12t^{2} \ge 228, \quad t \in [1,4], \\ \Psi (4,x)= & {} 96(3+2x-2x^{2}) \ge 288, \quad x \in [0,1], \end{aligned}
and
\begin{aligned} \begin{aligned} \Psi (1,x)&= 3\left( 16 +115x +83x^{2} -145x^{3} +7x^{4} \right) \\&= 3\left( 16 + 53x^2 + 7x^{4} + 115x(1-x^{2}) + 30x^{2}(1-x) \right) \\&\ge 48, \quad x \in [0,1]. \end{aligned} \end{aligned}
Thus the proof of the proposition is completed. $$\square$$

### Proposition 2.4

Let $$\Phi : [3,4] \times [0,1] \rightarrow {\mathbb {R}}$$ be a function defined by
\begin{aligned} \Phi (t,x) := 48\phi _{1}(x) +8\phi _{2}(x)t -3\phi _{3}(x)t^{2}, \end{aligned}
(2.8)
where for $$x\in [0,1],$$
\begin{aligned} \phi _{1}(x):= & {} 1 +7x +x^{2} -3x^{3}, \\ \phi _{2}(x):= & {} 5 -19x +10x^{2} +5x^{3} +x^{4} \end{aligned}
and
\begin{aligned} \phi _{3}(x) := x(-3 +5x +x^{2} +x^{3}). \end{aligned}
Then $$\Phi (t,x) > 0$$ for $$3 \le t \le 4$$ and $$0 \le x \le 1$$.

### Proof

Since $$\phi _{3}=-\psi _3,$$ by the part of proof of Proposition 2.3, we at once have
\begin{aligned} \frac{\partial }{\partial t} \Phi (t,x_{1}) = 8\phi _{2}(x_{1}) \not =0. \end{aligned}
For $$x \not = x_1$$, $$(\partial /\partial t)\Phi (t,x)=0$$ occurs at
\begin{aligned} t= \frac{ 4\phi _{2}(x) }{ 3\phi _{3}(x) } =: t_{0}(x). \end{aligned}
We have
\begin{aligned} \frac{\partial }{\partial x}\Phi (t,x) \Big |_{t=t_{0}(x)} = \frac{16 \phi _{4}(x)}{3 \phi _{3}^{2}(x)}, \end{aligned}
where
\begin{aligned} \begin{aligned} \phi _{4}(x)&:= 9\phi _{1}'(x)\phi _{3}^{2}(x) + 2\phi _{2}'(x) \phi _{2}(x) \phi _{3}(x) - \phi _{3}'(x) \phi _{2}^{2}(x) \\&= 75 -250x +59x^{2} +532x^{3} -893x^{4} +558x^{5} -269x^{6} +844x^{7} \\&\quad -366x^{8} +16x^{9} -46x^{10} +4x^{11},\quad x\in (0,1). \end{aligned} \end{aligned}
The polynomial $$\phi _4$$ has exactly two zeros in (0, 1),  namely $$x=:x_2\approx 0.414034$$ and $$x=:x_3\approx 0.663886.$$ We have $$t_{0}(x_2) \approx 3.59845$$ and $$t_{0}(x_3) = -2.95522.$$ Therefore the function $$\Phi$$ has a unique critical point $$(t_0(x_2),x_2)$$ in $$(3,4) \times (0,1)$$. For $$(t,x)\in [3.58,3.61]\times [0.39,0.43]$$ by simple computing, we show that $$\Phi (t,x)>0.$$ Thus, particularly $$\Phi (t_0(x_2),x_2)>0.$$ Therefore it is sufficient to show that $$\Phi >0$$ on the boundary of $$[3,4] \times [0,1].$$ We can easily check that the following inequalities hold:
\begin{aligned} \Phi (t,0)= & {} 8(6+5t) \ge 168 , \quad t \in [3,4], \\ \Phi (t,1)= & {} 4\left( 72 +4t -3t^{2} \right) \ge 160, \quad t \in [3,4], \\ \Phi (4,x)= & {} 16\left( 13 -8x +8x^{2} -2x^{3} -x^{4} \right) \ge 16(2 +8x^{2}) \ge 32, \quad x \in [0,1], \end{aligned}
and
\begin{aligned} \Phi (3,x) = 3\left( 56 -13x +51x^{2} -17x^{3} -x^{4} \right) \ge 3(25+ 51x^{2}) \ge 75, \quad x \in [0,1]. \end{aligned}
Thus the proof of the proposition is completed. $$\square$$

Finally, we estimate now the third-order Hankel determinant $$H_{3,1}(f)$$ for $$f \in {{\mathcal {S}}}^{*}$$.

### Theorem 2.5

If $$f \in {{\mathcal {S}}}^{*}$$ is the form (1.1), then
\begin{aligned} |H_{3,1}(f)| \le \frac{8}{9}. \end{aligned}
(2.9)

### Proof

Let $$f \in {{\mathcal {S}}}^{*}$$ be of the form (1.1). Then by (1.2) we have
\begin{aligned} zf'(z) = f(z)p(z), \quad z\in {\mathbb {D}}, \end{aligned}
(2.10)
for some function $$p \in {{\mathcal {P}}}$$ of the form (1.5). Since the class $${{\mathcal {P}}}$$ is invariant under the rotations, we may assume that $$c:=c_1 \in [0,2]$$ (e.g., [5, Vol. I, p. 80, Theorem 3]). Putting the series (1.1) and (1.5) into (2.10) and by equating the coefficients we get
\begin{aligned} a_{2}=c, \quad a_{3}=\frac{1}{2} \left( c^{2} +c_{2} \right) , \quad a_{4}=\frac{1}{6}\left( c^{3} + 3cc_{2} + 2c_{3} \right) \end{aligned}
and
\begin{aligned} a_{5} = \frac{1}{24} \left( c^{4} + 6c^{2}c_{2} +8cc_{3} + 3c_{2}^{2} +6c_{4} \right) . \end{aligned}
Hence
\begin{aligned} \begin{aligned} H_{3,1}(f)&= -a_{3}^{3} + 2a_{2}a_{3}a_{4} - a_{4}^{2} -a_{2}^{2}a_{5} + a_{3}a_{5} \\&= \frac{1}{144} \left( -c^{6} + 3c^{4}c_{2} - 9c_{2}^{3} +8c^{3}c_{3} + 24cc_{2}c_{3} -16c_{3}^{2} \right. \\&\quad \left. +18c_{2}c_{4} -9c^{2}c_{2}^{2} -18c^{2}c_{4} \right) . \end{aligned} \end{aligned}
Now using the equalities (2.1)–(2.3), by straightforward algebraic computation we have
\begin{aligned} H_{3,1}(f) = \frac{1}{1152} (c^{2}-4) \left[ \gamma _{1}(c,\zeta ) + \gamma _{2}(c,\zeta )\eta + \gamma _{3}(c,\zeta )\eta ^{2} + \Gamma (c,\zeta ,\eta ) \xi \right] , \end{aligned}
(2.11)
where for $$\zeta ,\,\eta ,\,\xi \in \overline{{\mathbb {D}}},$$
\begin{aligned} \gamma _{1}(c,\zeta ):= & {} c^{2}\zeta \left[ -3c^{2} + \left( 44-5c^{2} \right) \zeta + \left( 40-c^{2} \right) \zeta ^{2} \right] - c^{2}\left( 4-c^{2} \right) \zeta ^{4}, \\ \gamma _{2}(c,\zeta ):= & {} -4c\left( 1- |\zeta |^{2} \right) \left[ 3c^{2} + 4\left( 5+c^{2} \right) \zeta - \left( 4-c^{2} \right) \zeta ^{2} \right] , \\ \gamma _{3}(c,\zeta ):= & {} 32 \left( 4-c^{2} \right) -28|\zeta |^{2} \left( 4-c^{2} \right) -36c^{2} {\overline{\zeta }} \left( 1-|\zeta |^{2} \right) - 4\left( 4-c^{2} \right) |\zeta |^{4}, \end{aligned}
and
\begin{aligned} \Gamma (c,\zeta ,\eta ) := 36\left[ c^{2} + \left( c^{2}-4 \right) \zeta \right] \left( 1-|\zeta |^{2} \right) \left( 1-|\eta |^{2} \right) . \end{aligned}
Setting $$x:=|\zeta | \in [0,1],$$ $$y:=|\eta | \in [0,1]$$ and taking into account that $$|\xi |\le 1,$$ from (2.11) we get
\begin{aligned} \begin{aligned}&|H_{3,1}(f)| \\&\quad \le \frac{1}{1152} \left( 4-c^{2} \right) \left[ | \gamma _{1}(c,\zeta ) | + | \gamma _{2}(c,\zeta ) | |\eta | + | \gamma _{3}(c,\zeta ) | |\eta |^{2} + |\Gamma (c,\zeta ,\eta )| \right] \\&\quad \le \frac{1}{1152} (4-c^2)F(c,x,y), \end{aligned} \end{aligned}
(2.12)
where
\begin{aligned} F(c,x,y) := f_{1}(c,x) + f_{2}(c,x)y + f_{3}(c,x)y^{2} + f_{4}(c,x)\left( 1-y^{2} \right) , \end{aligned}
(2.13)
with
\begin{aligned} f_{1}(c,x):= & {} c^{2}x \left[ 3c^{2} + (44-5c^{2})x + (40-c^{2}) x^{2} \right] + c^{2}(4-c^{2})x^{4}, \\ f_{2}(c,x):= & {} 4c(1-x^{2}) \left[ 3c^{2} + 4(5+c^{2})x + (4-c^{2})x^{2} \right] , \\ f_{3}(c,x):= & {} 32(4-c^{2}) + 28x^{2}(4-c^{2}) + 36c^{2}x(1-x^{2}) + 4(4-c^{2})x^{4} \end{aligned}
and
\begin{aligned} f_{4}(c,x) := 36 \left[ c^{2}+(4-c^{2})x \right] (1-x^{2}). \end{aligned}
Now, we will show that
\begin{aligned} (4-c^2)F(c,x,y) \le 1024 \end{aligned}
(2.14)
for $$c\in [0,2],\ x\in [0,1]$$ and $$y\in [0,1].$$
I. Assume first that $$c\in [1,2].$$ Then by (2.13) we have
\begin{aligned} \begin{aligned}&F(c,x,y) \\&\quad \le f_{1}(c,x) + cf_{2}(c,x)y + f_{3}(c,x)y^{2} + f_{4}(c,x)\left( 1-y^{2} \right) \\&\quad = f_{1}(c,x) + f_{4}(c,x) + cf_{2}(c,x)y + (f_{3}(c,x)-f_4(c,x))y^{2} \\&\quad =: F_1(c,x,y) . \end{aligned} \end{aligned}
(2.15)
(a) Consider the case $$f_{3}(c,x)\ge f_4(c,x)$$ in $$[1,2]\times [0,1].$$ Let
\begin{aligned} \Omega _{1} := \left\{ (c,x) \in [1,2]\times [0,1]: f_{3}(c,x) \ge f_{4}(c,x) \right\} . \end{aligned}
By (2.15) we get
\begin{aligned} \begin{aligned} F_1(c,x,y)&\le F_1(c,x,1)\\&= f_{1}(c,x) + cf_{2}(c,x) + f_{3}(c,x),\quad (c,x)\in \Omega _1,\ y\in [0,1]. \end{aligned} \end{aligned}
Set $$t:=4-c^2.$$ Clearly, $$t\in [0,3].$$ Define
\begin{aligned} {\tilde{F}}_{1}(t,x):=tF_1(\sqrt{4-t},x,1), \quad (\sqrt{4-t},x)\in \Omega _1. \end{aligned}
A simple computing yields
\begin{aligned} \begin{aligned} {\tilde{F}}_{1}(t,x)&= t \left\{ (4-t)x \left[ 12-3t+(24+5t)x+(36+t)x^2 \right] \right. \\&\quad + t(4-t)x^4+32t+28tx^2+36(4-t)x(1-x^2) + 4tx^4 \\&\left. \quad + 4(4-t)(1-x^2) \left[ 12-3t+4(9-t)x+tx^2 \right] \right\} \\&= 96(2 + 8x - x^{2} - 6x^{3})t - 4(16 + 67x - 34x^{2} - 53x^{3} + 2x^{4})t^{2} \\&\quad + (12+19x-21x^{2}-17x^{3} + 3x^{4})t^{3},\quad (\sqrt{4-t},x)\in \Omega _1. \end{aligned} \end{aligned}
Hence and by Proposition 2.2 we have
\begin{aligned} \frac{\partial }{\partial t}{\tilde{F}}_{1}(t,x) = \Theta (t,x)>0,\quad (\sqrt{4-t},x)\in \Omega _1, \end{aligned}
where the function $$\Theta$$ is defined by (2.4). Thus the function $$[0,3]\ni t\mapsto {\tilde{F}}_{1}(t,\cdot )$$ is increasing, and therefore we have
\begin{aligned} {\tilde{F}}_{1}(t,x) \le {\tilde{F}}_{1}(3,x) = 9\left( 36+45x+41x^2 -31x^3 +x^4 \right) < 1024,\quad x\in [0,1]. \end{aligned}
(2.16)
Indeed, the last inequality is true since, as easy to verify the inequality
\begin{aligned} -700+405x+369x^2-279x^3+9x^4<0,\quad x\in [0,1], \end{aligned}
holds. Thus the inequality (2.16) confirms the inequality (2.14).
(b) Consider the case $$f_{3}(c,x)< f_4(c,x)$$ in $$[1,2]\times [0,1].$$ Let
\begin{aligned} \Omega _{2} := \left\{ (c,x) \in [1,2]\times [0,1]: f_{3}(c,x) < f_{4}(c,x) \right\} . \end{aligned}
Since $$f_{2}(c,x)\ge 0$$ in $$[1,2]\times [0,1],$$ so
\begin{aligned} \sigma :=\frac{-c f_{2}(c,x)}{2(f_{3}(c,x) - f_{4}(c,x))}\ge 0,\quad (c,x)\in \Omega _2. \end{aligned}
If $$\sigma \ge 1,$$ i.e., if $$cf_{2}(c,x)+2(f_{3}(c,x) - f_{4}(c,x))\ge 0,$$ then
\begin{aligned} \begin{aligned} F_1(c,x,y)&\le F_1(c,x,1)\\&= f_{1}(c,x) + cf_{2}(c,x) + f_{3}(c,x),\quad (c,x)\in \Omega _2,\ y\in [0,1]. \end{aligned} \end{aligned}
and repeating the argumentation of Part (a) we get the inequality (2.14).
If $$\sigma <1,$$ i.e., if $$cf_{2}(c,x)+2(f_{3}(c,x) - f_{4}(c,x)) < 0,$$ then
\begin{aligned} \begin{aligned} F_1(c,x,y)&\le F_1(c,x,\sigma )=\frac{-c^2f_2^2(c,x)}{4(f_{3}(c,x) - f_{4}(c,x))}+f_{1}(c,x)+f_{4}(c,x)\\&\le \frac{\left[ -2(f_3(c,x)-f_4(c,x)) \right] ^2}{4(f_{3}(c,x) - f_{4}(c,x))}+f_{1}(c,x)+f_{4}(c,x)\\&\le f_{1}(c,x)+f_3(c,x)+2f_{4}(c,x)=:F_2(c,x),\quad (c,x)\in \Omega _2. \end{aligned} \end{aligned}
Set $$t:=c^2.$$ Clearly, $$t\in [1,4].$$ Define
\begin{aligned} {\tilde{F}}_{2}(t,x) := (4-t)F_2(\sqrt{t},x),\quad (\sqrt{t},x)\in \Omega _2. \end{aligned}
A simple computing yields
\begin{aligned} \begin{aligned} {\tilde{F}}_{2}(t,x)&= (4-t) \left\{ tx \left[ 3t+(44-5t)x+(40-t)x^2 \right] +t(4-t)x^4 \right. \\&\quad + 32(4-t)+28x^2(4-t)+36tx(1-x^2)+4(4-t)x^4\\&\quad \left. + 72\left[ t+(4-t)x \right] (1-x^2) \right\} \\&= -\left\{ -64\left( 8+18x+7x^2-18x^3+x^4 \right) \right. \\&\quad + 16 \left( -2 + 27x + 21x^{2} - 37x^{3} +x^{4} \right) t \\&\quad + 4 \left( 10 -12x -9x^{2} + 20x^{3} +x^{4} \right) t^2 \\&\quad \left. + x\left( 3-5x-x^{2}-x^{3} \right) t^3 \right\} ,\quad (\sqrt{t},x)\in \Omega _2. \end{aligned} \end{aligned}
Hence and by Proposition 2.3 we have
\begin{aligned} \frac{\partial }{\partial t}{\tilde{F}}_{2}(t,x) = -\Psi (t,x)<0,\quad (\sqrt{t},x)\in \Omega _2, \end{aligned}
where the function $$\Psi$$ is defined by (2.6). Thus the function $$[1,4]\ni t\mapsto {\tilde{F}}_{2}(t,\cdot )$$ is decreasing, and therefore we have
\begin{aligned} {\tilde{F}}_{2}(t,x) \le {\tilde{F}}_{2}(1,x) = 9\left( 56+85x+17x^2-71x^3+5x^4 \right) < 1024,\quad x\in [0,1]. \end{aligned}
(2.17)
Indeed, the last inequality is true since, as easy to verify the inequality
\begin{aligned} -520+765x+153x^2-639x^3+45x^4<0,\quad x\in [0,1], \end{aligned}
holds. Thus the inequality (2.17) confirms the inequality (2.14).
II. Assume that $$c\in (0,1).$$ Then by (2.13) we have
\begin{aligned} \begin{aligned}&F(c,x,y) \\&\quad \le f_{1}(c,x) + \frac{1}{c}f_{2}(c,x)y + f_{3}(c,x)y^{2} + f_{4}(c,x)\left( 1-y^{2} \right) \\&\quad = f_{1}(c,x) + f_{4}(c,x) + \frac{1}{c}f_{2}(c,x)y + (f_{3}(c,x)-f_4(c,x))y^{2} \\&\quad =: F_3(c,x,y). \end{aligned} \end{aligned}
(2.18)
(a) Consider the case $$f_{3}(c,x)\ge f_4(c,x)$$ in $$(0,1)\times [0,1].$$ Let
\begin{aligned} \Omega _{3} := \left\{ (c,x) \in (0,1)\times [0,1]: f_{3}(c,x) \ge f_{4}(c,x) \right\} . \end{aligned}
By (2.18) we get
\begin{aligned} \begin{aligned} F_3(c,x,y)&\le F_3(c,x,1)\\&= f_{1}(c,x) + \frac{1}{c}f_{2}(c,x) + f_{3}(c,x),\quad (c,x)\in \Omega _3,\ y\in [0,1]. \end{aligned} \end{aligned}
Set $$t:=4-c^2.$$ Clearly, $$t\in (3,4).$$ Define
\begin{aligned} {\tilde{F}}_{3}(t,x):=tF_3(\sqrt{4-t},x,1), \quad (\sqrt{4-t},x)\in \Omega _3. \end{aligned}
A simple computing yields
\begin{aligned} \begin{aligned} {\tilde{F}}_{3}(t,x)&= t \left\{ 3(4-t)^2x+(4-t)(24+5t)x^2+(4-t)(36+t)x^3+(4-t)tx^4 \right. \\&\quad + 4(1-x^2)\left[ 12-3t+4(9-t)x+tx^2\right] \\&\quad \left. + 32t+28tx^2+36(4-t)x(1-x^2)+4tx^4 \right\} \\&= 48 \left( 1 +7x +x^{2} -3x^{3} \right) t + 4\left( 5 -19x +10x^{2} +5x^{3} +x^{4} \right) t^2 \\&\quad - x\left( -3 +5x +x^{2} +x^{3} \right) t^3 ,\quad (\sqrt{4-t},x)\in \Omega _3. \end{aligned} \end{aligned}
Hence and by Proposition 2.4 we have
\begin{aligned} \frac{\partial }{\partial t}{\tilde{F}}_{3}(t,x) = \Phi (t,x)>0,\quad (\sqrt{4-t},x)\in \Omega _3, \end{aligned}
where the function $$\Phi$$ is defined by (2.8). Thus the function $$(3,4)\ni t\mapsto {\tilde{F}}_{1}(t,\cdot )$$ is increasing, and therefore we have
\begin{aligned} {\tilde{F}}_{3}(t,x) \le 512+320x+512x^2-320x^3 \le 1024. \end{aligned}
(2.19)
Indeed, the last inequality is true since so is the following one
\begin{aligned} -512+320x+512x^2-320x^3=(1-x^2)(320x-512)\le 0,\quad x\in [0,1]. \end{aligned}
Thus the inequality (2.19) confirms the inequality (2.14).
(b) Consider the case $$f_{3}(c,x)< f_4(c,x)$$ in $$(0,1)\times [0,1]$$ which is equivalent to
\begin{aligned} \begin{aligned}&32-8c^2+28x^2-7c^2x^2+9c^2x-9c^2x^3+4x^4-c^2x^4\\&\quad < 9c^2-9c^2x^2+36x-9c^2x-36x^3+9c^2x^3 \end{aligned} \end{aligned}
(2.20)
for $$c\in (0,1)$$ and $$x\in [0,1].$$ Note that
\begin{aligned} 17-18x-2x^2+18x^3+x^4>0,\quad x\in [0,1]. \end{aligned}
Thus the inequality (2.20) can be written as
\begin{aligned} c^2>\frac{32-36x+28x^2+36x^3+4x^4}{17-18x-2x^2+18x^3+x^4},\quad c\in (0,1),\ x\in [0,1]. \end{aligned}
(2.21)
However,
\begin{aligned} \frac{32-36x+28x^2+36x^3+4x^4}{17-18x-2x^2+18x^3+x^4}\ge 1,\quad x\in [0,1]. \end{aligned}
(2.22)
Indeed, the above inequality is equivalent to
\begin{aligned} 32-36x+28x^2+36x^3+4x^4\ge 17-18x-2x^2+18x^3+x^4,\quad x\in [0,1], \end{aligned}
which by simplifying is equivalent to the true inequality
\begin{aligned} (x-1)^4+10x^3+4(x-1)^2+6x\ge 0,\quad x\in [0,1]. \end{aligned}
Thus by (2.21) and (2.22) it follows that $$c\ge 1$$ which contradicts the assumption.
III. At the end assume that $$c=0.$$ Then by (2.13) we have
\begin{aligned} \begin{aligned} F(0,x,y)&= 16\left( (8-9x+7x^2+9x^3+x^4)y^2+9x(1-x^2)\right) \\&\le 16(8+7x^2+x^4) \le 256,\quad x\in [0,1],\ y\in [0,1]. \end{aligned} \end{aligned}
Summarizing, from all considering cases it follows that the inequality (2.14) holds which together with (2.12) shows (2.9). $$\square$$

### Remark 2.6

Although the constant 8 / 9 improves essentially the estimates found in  and , it is not the best possible. To find the sharp estimate of the Hankel determinant $$H_{3,1}(f)$$ for starlike functions is still an open problem.

## References

1. 1.
Babalola, K.O.: On $$H_3(1)$$ Hankel determinants for some classes of univalent functions. In: Cho, Y.J. (ed.) Inequality Theory and Applications, vol. 6, pp. 1–7. Nova Science Publishers, New York (2010)Google Scholar
2. 2.
Bansal, D., Maharana, S., Prajapat, J.K.: Third order Hankel determinant for certain univalent functions. J. Korean Math. Soc. 52(6), 1139–1148 (2015)
3. 3.
Cho, N.E., Kowalczyk, B., Kwon, O.S., Lecko, A., Sim, Y.J.: The bounds of some determinants for starlike functions of order alpha. Bull. Malay. Math. Sci. Soc. 41, 523–535 (2018)
4. 4.
Cho, N.E., Kowalczyk, B., Kwon, O.S., Lecko, A., Sim, Y.J.: The bound of the Hankel detrminant for strongly starlike functions of order alpha. J. Math. Inequal. 11(2), 429–439 (2017)
5. 5.
Goodman, A.W.: Univalent Functions. Mariner, Tampa (1983)
6. 6.
Janteng, A., Halim, S.A., Darus, M.: Coefficient inequality for a function whose derivative has a positive real part. J. Inequal. Pure Appl. Math. 7(2), 1–5 (2006)
7. 7.
Kwon, O.S., Lecko, A., Sim, Y.J.: On the fourth coefficient of functions in the Carathéodory class. Comp. Meth. Funct. Theory 18, 307–314 (2018)
8. 8.
Lee, S.K., Ravichandran, V., Supramanian, S.: Bound for the second Hankel determinant of certain univalent functions. J. Inequal. Appl. 2013(281), 1–17 (2013)
9. 9.
Libera, R.J., Zlotkiewicz, E.J.: Early coefficients of the inverse of a regular convex function. Proc. Am. Math. Soc. 85(2), 225–230 (1982)
10. 10.
Libera, R.J., Zlotkiewicz, E.J.: Coefficient bounds for the inverse of a function with derivatives in $${{\cal{P}}}$$. Proc. Am. Math. Soc. 87(2), 251–257 (1983)
11. 11.
Livingston, A.E.: The coefficients of multivalent close-to-convex functions. Proc. Am. Math. Soc. 21(3), 545–552 (1969)
12. 12.
Mishra, A.K., Gochhayat, P.: Second Hankel determinat for a class of analytic functions defined by fractional derivative. Int. J. Math. Math. Sci. Article ID 153280 (2008)Google Scholar
13. 13.
Pommerenke, C.: On the coefficients and Hankel determinant of univalent functions. J. Lond. Math. Soc. 41, 111–122 (1966)
14. 14.
Pommerenke, C.: Univalent Functions. Vandenhoeck & Ruprecht, Göttingen (1975)
15. 15.
Prajapat, J.K., Bansal, D., Singh, A., Mishra, A.K.: Bounds on third Hankel determinant for close-to-convex functions. Acta Univ. Sapientae Mathematica 7(2), 210–219 (2015)
16. 16.
Raza, M., Malik, S.N.: Upper bound of the third Hankel determinant for a class of analytic functions related with lemniscate of Bernoulli. J. Inequal. Appl. 2013(412), 8 (2013)
17. 17.
Shanmugam, G., Stephen, B.A., Babalola, K.O.: Third Hankel determinant for $$\alpha$$-starlike functions. Gulf J. Math. 2(2), 107–113 (2014)
18. 18.
Sudharsan, T.V., Vijayalaksmi, S.P., Sthephen, B.A.: Third Hankel determinant for a subclass of analytic functions. Malaya J. Math. 2(4), 438–444 (2014)Google Scholar
19. 19.
Zaprawa, P.: Third Hankel determinants for subclasses of univalent functions. Mediter. J. Math. 14(1), 10 (2017)

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