1 Introduction

Let \({{\mathcal {H}}}\) be the class of analytic functions in \({\mathbb {D}}:= \left\{ z \in {\mathbb {C}} : |z|<1 \right\} \) and let \({\mathcal {A}}\) be its subclass of functions f 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)

Given \(\alpha \in [0,1),\) let \({\mathcal {S}}^*(\alpha )\) denote the subclass of \({\mathcal {A}}\) of functions f such that

$$\begin{aligned} {{\mathrm{Re}}}\frac{zf'(z}{f(z)} > \alpha ,\quad z\in {\mathbb {D}}, \end{aligned}$$
(1.2)

called starlike of order \(\alpha .\) In particular, \({\mathcal {S}}^*(0)=:\mathcal S^*\) is the class of starlike functions, i.e., the family of all univalent functions in \({\mathcal {A}}\) which map \({\mathbb {D}}\) onto starlike domains (with respect to the origin). Starlike functions of order \(\alpha \) were introduced by Robertson [19] (see also [7, Vol. I, p. 138]). An important role is played by the class \({\mathcal {S}}^*(1/2).\) One of the significant results belongs to Marx [15] and Strohhäcker [23]. They proved that

$$\begin{aligned} {\mathcal {S}}^c\subset {\mathcal {S}}^*(1/2) \end{aligned}$$
(1.3)

(see also [16, Theorem 2.6a, p. 57]), where \({\mathcal {S}}^c\) means the class of convex functions, i.e., the family of all univalent functions in \({\mathcal {A}}\) which map \({\mathbb {D}}\) onto convex domains. By the well known result due to Study ([24], see also [6, p. 42]) a function f is in \({\mathcal {S}}^c\) if and only if

$$\begin{aligned} {{\mathrm{Re}}}\left\{ 1+\frac{zf''(z)}{f'(z)}\right\} > 0,\quad z\in {\mathbb {D}}. \end{aligned}$$

What is interesting, a function

$$\begin{aligned} f(z):=\frac{z}{1-z},\quad z\in {\mathbb {D}}, \end{aligned}$$
(1.4)

is extremal for many computational problems in both these two classes, i.e., in \({\mathcal {S}}^c\) and \({\mathcal {S}}^*(1/2).\)

For \(q,n \in {\mathbb {N}},\) the Hankel determinant \(H_{q,n}(f)\) of function \(f \in {{\mathcal {A}}}\) of the form (1.1) is defined as

$$\begin{aligned} H_{q,n}(f) := \begin{vmatrix} a_{n}&\quad a_{n+1}&\quad \cdots&\quad a_{n+q-1} \\ a_{n+1}&\quad a_{n+2}&\quad \cdots&\quad a_{n+q} \\ \vdots&\quad \vdots&\quad \vdots&\quad \vdots \\ a_{n+q-1}&\quad a_{n+q}&\quad \cdots&\quad a_{n+2(q-1)} \end{vmatrix}. \end{aligned}$$

Given a subfamily \({\mathcal {F}}\) of \({\mathcal {A}},\)q and n,  computing the upper bound of \(H_{q,n}\) is an interesting problem to study. Recently many authors examined the Hankel determinant \(H_{2,2}(f)=a_2a_4-a_3^2\) of order 2 (see e.g., [4, 5, 8, 9, 12, 17]). Note also that \(H_{2,1}(f)=a_3-a_2^2\) is the well known coefficient functional which for \({\mathcal {S}}\) was estimated in 1916 by Bieberbach (see e.g., [7, Vol. I, p. 35]). To find the upper bound of the Hankel determinant

$$\begin{aligned} H_{3,1}(f) = \begin{vmatrix} a_1&\quad a_1&\quad a_3 \\ a_2&\quad a_3&\quad a_4 \\ a_3&\quad a_4&\quad a_5 \end{vmatrix} =a_3(a_2a_4 - a_3^2) - a_4(a_4 - a_2a_3) + a_5(a_3 - a_2^2)\nonumber \\ \end{aligned}$$
(1.5)

of the third kind, is more difficult if we expect to get sharp estimate. Results in this direction however not sharp were obtained by various authors, e.g., [1, 2, 4, 5, 20,21,22, 25].

In this paper, we found the sharp bound of the Hankel determinant \(H_{3,1}\) over the class \({\mathcal {S}}^*(1/2),\) namely, we proved that \(|H_{3,1}(f)|\le 1/9\) for \(f\in {\mathcal {S}}^*(1/2)\) and that the inequality is sharp. Since the class \({\mathcal {S}}^*(1/2)\) has a representation with using the Carathéodory class \({\mathcal {P}}\), i.e., the class of functions \(p \in {{\mathcal {H}}}\) of the form

$$\begin{aligned} p(z) = 1 + \sum _{n=1}^{\infty }c_{n}z^{n}, \quad z\in {\mathbb {D}}, \end{aligned}$$
(1.6)

having a positive real part in \({\mathbb {D}},\) the coefficients of functions in \({\mathcal {S}}^*(1/2)\) have a suitable representation expressed by coefficients of functions in \({\mathcal {P}}.\) Therefore to get the upper bound of \(H_{3,1},\) we based our computing on the well known formulas on coefficient \(c_2\) (e.g., [18, p. 166]), the formula \(c_3\) due to Libera and Zlotkiewicz [13, 14] and the formula for \(c_4\) recently found in [11].

At the end let us mention that in [10] the authors proved that \(|H_{3,1}(f)|\le 4/135=0.0296\ldots \) for \(f\in \mathcal S^c\) and that the result is sharp. Looking at the inclusion (1.3) we can state that the the corresponding bounds of \(H_{3,1}\) carry some information about the richness of classes. Classical estimates of coefficients does not necessarily include such a distinction, e.g., both in the class \({\mathcal {S}}^c\) and in the class \({\mathcal {S}}^*(1/2)\) modules of all coefficients are bounded by 1 (see [7, Theorem 7, p. 117; Theorem 2, p. 140]) with the extremal function given by (1.4).

2 Main Result

The basis for proof of the main result is the following lemma which contains the well known formula for \(c_2\) (e.g., [18, p. 166]), the formula for \(c_3\) due to Libera and Zlotkiewicz [13, 14] and the formula for \(c_4\) found in [11].

Lemma 2.1

If \(p \in {{\mathcal {P}}}\) is of the form (1.6) 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

$$\begin{aligned} 8c_4= & {} c_1^4+(4-c_1^2)\zeta \left[ c_1^2(\zeta ^2-3\zeta +3)+4\zeta \right] \nonumber \\&-4(4-c_1^2)(1-|\zeta |^2)\left[ c_1(\zeta -1)\eta +\overline{\zeta }\eta ^2-\left( 1-|\eta |^2\right) \xi \right] \end{aligned}$$
(2.3)

for some \(\zeta ,\eta ,\xi \in \overline{{\mathbb {D}}}:=\{z\in {\mathbb {C}}:|z|\le 1 \}.\)

We will now estimate the third order Hankel determinant \(H_{3,1}(f)\) for \(f \in {\mathcal {S}}^*(1/2)\).

Theorem 2.2

$$\begin{aligned} \max \left\{ |H_{3,1}(f)|: f\in {\mathcal {S}}^*(1/2)\right\} = \frac{1}{9} \end{aligned}$$
(2.4)

with the extremal function

$$\begin{aligned} f(z):=\frac{z}{\root 3 \of {1-z^3}},\quad z\in {\mathbb {D}},\ \root 3 \of {1}:=1. \end{aligned}$$
(2.5)

Proof

Let \(f \in {\mathcal {S}}^*(1/2)\) be of the form (1.1). Then by (1.2) we have

$$\begin{aligned} zf'(z)=\frac{1}{2}(p(z)+1)f(z),\quad z\in {\mathbb {D}}, \end{aligned}$$
(2.6)

for some function \(p \in {\mathcal {P}}\) of the form (1.6). Since the classes \({\mathcal {P}}\) and \({\mathcal {S}}^*(1/2)\) are invariant under the rotations, by Carathéodory Theorem we may assume that \(c:=c_1 \in [0,2]\) ([3], see also [7, Vol. I, p. 80, Theorem 3]). Putting the series (1.1) and (1.6) into (2.6) and equating coefficients we get

$$\begin{aligned} \begin{aligned} a_2&= \frac{1}{2}c, \quad a_3=\frac{1}{8}\left( 2c_2+c^2\right) ,\quad a_4 = \frac{1}{48}\left( 8c_3+6cc_2+c^3\right) ,\\ a_5&= \frac{1}{384}\left( 48c_4+32cc_3+12c_2^2+12c^2c_2+c^4\right) . \end{aligned} \end{aligned}$$

Hence and by (1.5) we have

$$\begin{aligned} H_{3,1}(f)= & {} \frac{1}{9216}\left( -c^6+6c^4c_2-72c_2^3+32c^3c_3\right. \nonumber \\&\left. +\,192cc_2c_3-256c_3^2-36c^2c_2^2+288c_2c_4-144c^2c_4\right) . \end{aligned}$$
(2.7)

To simplify computation, let \(t:=4-c^2.\) Thus formulas (2.1)-(2.3) we can rewrite as

$$\begin{aligned} \begin{aligned} c_2&= \frac{1}{2}(c^2+t\zeta ),\quad c_3=\frac{1}{4}\left( c^3+2ct\zeta -ct\zeta ^2+2t(1-|\zeta |^2)\eta \right) ,\\ c_4&= \frac{1}{8}\left[ c^4+3c^2t\zeta +(4-3c^2)t \zeta ^2+c^2t\zeta ^3+4t(1-|\zeta |^2)\left( c\eta -c\zeta \eta - {\overline{\zeta }}\eta ^2\right) \right. \\&\quad \left. +\,4t(1-|\zeta |^2)(1-|\eta |^2)\xi \right] . \end{aligned} \end{aligned}$$

Hence by straightforward algebraic computation we have

$$\begin{aligned} \begin{aligned} 6c^4c_2&= 3(c^6+c^4t\zeta ),\\ 72c_2^3&= 9\left[ c^6+3c^4t\zeta +3c^2t^2\zeta ^2+t^3\zeta ^3\right] ,\\ 32c^3c_3&= 8\left[ c^6+2c^4t\zeta -c^4t\zeta ^2+2c^3t(1-|\zeta |^2)\eta \right] ,\\ 192cc_2c_3&= 24\left[ c^6+3c^4t\zeta +2c^2t^2\zeta ^2-c^4t\zeta ^2-c^2t^2\zeta ^3\right. \\&\quad \left. +\,2t(c^3+ct\zeta )(1-|\zeta |^2)\eta \right] ,\\ 256c_3^2&= 16\left[ c^6+4c^4t\zeta +4c^4t^2\zeta ^2-2c^4t\zeta ^2-4c^2t^2\zeta ^3+c^2t^2\zeta ^4\right. \\&\quad \left. +\,4t(c^3+2ct\zeta -ct\zeta ^2)(1-|\zeta |^2)\eta +4t^2(1-|\zeta |^2)^2\eta ^2\right] ,\\ 36c^2c_2^2&= 9\left[ c^6+2c^4t\zeta +c^2t^2\zeta ^2\right] ,\\ 144(2c_2c_4-c^2c_4)&= 18\left[ c^4t\zeta +3c^2t^2\zeta ^2+(4-3c^2)t^2\zeta ^3+c^2t^2\zeta ^4\right. \\&\quad +\,4t^2c\zeta (1-\zeta )(1-|\zeta |^2)\eta \\&\quad \left. -\,4t^2(1-|\zeta |^2)|\zeta |^2\eta ^2+4t^2(1-|\zeta |^2)(1-|\eta |^2)\zeta \xi \right] . \end{aligned} \end{aligned}$$

Setting the above expression to (2.7) we get

$$\begin{aligned}&H_{3,1}(f)\nonumber \\&\quad = \frac{1}{9216}(4-c^2)^2\left[ \gamma _1(c,\zeta ) + \gamma _2(c,\zeta )\eta + \gamma _3(c,\zeta )\eta ^{2} + \gamma _4(c,\zeta ,\eta ) \xi \right] , \nonumber \\&\end{aligned}$$
(2.8)

where for \(\zeta ,\,\eta ,\,\xi \in \overline{{\mathbb {D}}},\)

$$\begin{aligned} \begin{aligned} \gamma _{1}(c,\zeta )&:= \zeta ^2\left[ 2c^2+(36-5c^2)\zeta +2c^2\zeta ^2 \right] ,\\ \gamma _{2}(c,\zeta )&:= -8c\zeta (1+\zeta )(1-|\zeta |^2),\\ \gamma _{3}(c,\zeta )&:= -8(8+|\zeta |^2)(1-|\zeta |^2), \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \gamma _4(c,\zeta ,\eta ) := 72(1-|\zeta |^2)(1-|\eta |^2)\zeta . \end{aligned}$$

Let \(x:=|\zeta | \in [0,1]\) and \(y:=|\eta | \in [0,1].\) Since \(|\xi |\le 1,\) from (2.8) we obtain

$$\begin{aligned} |H_{3,1}(f)|\le & {} \frac{1}{9216}(4-c^2)^2\Big [|\gamma _1(c,\zeta )|\nonumber \\&\left. +\,|\gamma _2(c,\zeta )||\eta |+|\gamma _3(c,\zeta )||\eta |^2 +|\gamma _4(c,\zeta ,\eta )|\right] \nonumber \\\le & {} \frac{1}{9216}(4-c^2)^2F(c,x,y), \end{aligned}$$
(2.9)

where

$$\begin{aligned} F(c,x,y) :=f_1(c,x)+f_4(c,x)+f_2(c,x)y +(f_3(c,x)-f_4(c,x))y^2, \end{aligned}$$

with

$$\begin{aligned} \begin{aligned} f_1(c,x)&:= x^2\left[ 2c^2+(36-5c^2)x+2c^2x^2\right] ,\\ f_2(c,x)&:= 8cx(1+x)(1-|x|^2),\\ f_3(c,x)&:= 8(8+x^2)(1-x^2) \end{aligned} \end{aligned}$$

and

$$\begin{aligned} f_{4}(c,x):=72(1-x^2)x. \end{aligned}$$

Now, we will show that

$$\begin{aligned} F(c,x,y) \le 64,\quad c\in [0,2],\ x\in [0,1],\ y\in [0,1]. \end{aligned}$$
(2.10)

Since \(f_2(c,x)>0\) and

$$\begin{aligned} f_3(c,x)-f_4(c,x)=8(1-x)(8-x)(1-x^2)>0 \end{aligned}$$

for \(c\in (0,2)\) and \(x\in (0,1),\) so for \(c\in (0,2)\) and \(x\in (0,1),\)

$$\begin{aligned} F(c,x,y)\le & {} F(c,x,1)\nonumber \\= & {} f_1(c,x)+f_2(c,x) +f_3(c,x)\nonumber \\= & {} x^2(x-2)(2x-1)c^2+8x(x+1)(1-x^2)c\nonumber \\&-\,4(2x^4-9x^3+14x^2-16)=: G(c,x). \end{aligned}$$
(2.11)

For \(x=1/2\) the function \((0,2)\ni c\mapsto G(c,1/2)\) has no critical point, obviously. When \(x\not =1/2,\) then \(\partial G/\partial c=0\) iff

$$\begin{aligned} c=\frac{4x(x+1)(1-x^2)}{x^2(2-x)(2x-1)}=:c_0\in (0,2), \end{aligned}$$

which holds only for \(x\in ((2+3\sqrt{2})/7,1).\) Thus

$$\begin{aligned} \frac{\partial G}{\partial x}(c_0,x)=0 \end{aligned}$$

iff

$$\begin{aligned} \begin{aligned}&4(8x^2-15x+4)(x+1)^2(1-x^2)^2\\&\quad +8(4x^3+3x^2-2x-1)(x+1)(1-x^2)(x-2)(2x-1)\\&\quad -x^2(8x^2-27x+28)(x-2)^2(2x-1)^2=0 \end{aligned} \end{aligned}$$

which after simplifying reduces to

$$\begin{aligned} -64x^7+320x^6-788x^5+1503x^4-1624x^3+760x^2-80x-36=0 \end{aligned}$$

for \(x\in ((2+3\sqrt{2})/7,1).\) As we can check the above equation has no solution in \(((2+3\sqrt{2})/7,1)\) (real solutions are \(x_1\approx -0.1513,\ x_2\approx 1.0622,\ x_3\approx 2.4952\)). Thus the function G has no critical point in \((0,2)\times (0,1).\)

For \(c=0\) and \(c=2\) both functions

$$\begin{aligned} g_1(x):=F(0,x,1)=4(-2x^4+9x^3-14x^2+16),\quad x\in [0,1], \end{aligned}$$

and

$$\begin{aligned} g_2(x):=F(2,x,1)=16(-x^4-2x^2+4),\quad x\in [0,1], \end{aligned}$$

are decreasing, so

$$\begin{aligned} g_i(x)\le g_i(0)=64,\quad i=1,2,\ x\in [0,1]. \end{aligned}$$
(2.12)

For \(x=0\) and \(x=1\) we have respectively,

$$\begin{aligned} F(c,0,1)=64,\quad c\in [0,2], \end{aligned}$$

and

$$\begin{aligned} F(c,1,1)=-c^2+36\le 36,\quad c\in [0,2]. \end{aligned}$$

Hence, by (2.12) and (2.11) it follows that the (2.9) holds. This together with (2.9) shows that \(|H_{3,1}(f)|\le 1/9.\)

For the function (2.5) which is in \({\mathcal {S}}^*(1/2),\) we have \(a_2=a_3=a_5=0\) and \(a_4=1/3.\) Thus \(H_{3,1}(f)=-1/9,\) which makes equality in (2.4). \(\square \)