The Sharp Bound of the Hankel Determinant of the Third Kind for Starlike Functions with Real Coefficients

Abstract

Let ${\mathcal{SR}}^{*}$ be the class of starlike functions with real coefficients, i.e., the class of analytic functions f which satisfy the condition $f\left(0\right)=0=f^{\prime }\left(0\right)-1$, $\mathrm{Re}\{z{f}^{\prime }\left(z\right)/f\left(z\right)\}>0$, for $z\in \mathbb{D}:=\{z\in \mathbb{C}:|z|<1\}$ and $a_n:=f^{\left(n\right)}\left(0\right)/n!$ is real for all $n\in \mathbb{N}$. In the present paper, it is obtained that the sharp inequalities $-4/9\le H_{3,1}\left(f\right)\le \sqrt{3}/9$ hold for $f\in {\mathcal{SR}}^{*}$, where $H_{3,1}\left(f\right)$ is the third Hankel determinant of order 3 defined by $H_{3,1}\left(f\right)=a_3(a_2{a}_4-a_3^2)-a_4(a_4-a_2{a}_3)+a_5(a_3-a_2^2)$.

1. Introduction

Let $\mathcal{H}$ be the class of analytic functions in $\mathbb{D}:=\{z\in \mathbb{C}:|z|<1\}$ and let $\mathcal{A}$ be the class of functions $f\in \mathcal{H}$ normalized by $f\left(0\right)=0=f^{\prime }\left(0\right)-1$. That is, for $z\in \mathbb{D}$, $f\in \mathcal{A}$ has the following representation

$$f\left(z\right)=z+\sum _{n=2}^{\infty }{a}_n{z}^n.$$

(1)

For q, $n\in \mathbb{N}$, the Hankel determinant $H_{q,n}\left(f\right)$ of functions $f\in \mathcal{A}$ of the form (1) are defined by

$$H_{q,n}\left(f\right)=\left|\begin{array}{cccc}{a}_n& a_{n+1}& \cdots & a_{n+q-1}\\ a_{n+1}& a_{n+2}& \cdots & a_{n+q}\\ ⋮& ⋮& ⋮& ⋮\\ a_{n+q-1}& a_{n+q}& \cdots & a_{n+2q-2}\end{array}\right|.$$

(2)

Computing the upper bound of $H_{q,n}$ over subfamilies of $\mathcal{A}$ is an interesting problem to study. Note that $H_{2,1}\left(f\right)=a_3-a_2^2$ is the well-known functional which, for the class of univalent functions, was estimated by Bieberbach (see, e.g., [1] (Vol. I, p. 35)). Especially, the functional $H_{3,1}\left(f\right)$, Hankel determinant of order 3, is presented by

$$\begin{array}{cc}{H}_{3,1}\left(f\right)\hfill & =\left|\begin{array}{ccc}{a}_1& a_2& a_3\\ a_2& a_3& a_4\\ a_3& a_4& a_5\end{array}\right|\hfill \\ \hfill & =a_3(a_2{a}_4-a_3^2)-a_4(a_4-a_2{a}_3)+a_5(a_3-a_2^2).\hfill \end{array}$$

Let ${\mathcal{S}}^{*}$ be the class of starlike functions in $\mathcal{A}$. That is, the class ${\mathcal{S}}^{*}$ consists of all functions $f\in \mathcal{A}$ satisfying

$$\mathrm{Re}\left\{\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right\}>0,z\in \mathbb{D}.$$

(3)

The leading example of a function of class ${\mathcal{S}}^{*}$ is the Koebe function k, defined by

$$k\left(z\right)=z{(1-z)}^{-2}=z+2z^2+3z^3+\cdots ,z\in \mathbb{D}.$$

In [2], Janteng et al. obtained the sharp inequality $|H_{2,2}\left(f\right)|\le 1=|H_{2,2}\left(k\right)|$ for $f\in {\mathcal{S}}^{*}$. For the estimates on the Hankel determinant $H_{3,1}\left(f\right)$ over the class ${\mathcal{S}}^{*}$, Babalola [3] obtained the inequality $|H_{3,1}\left(f\right)|\le 16$. And Zaprawa [4] improved the result by proving $|H_{3,1}\left(f\right)|\le 1$. Next, Kwon et al. [5], recently found the inequality $|H_{3,1}\left(f\right)|\le 8/9$ and we conjectured that

$$|H_{3,1}\left(f\right)|\le 4/9,f\in {\mathcal{S}}^{*}.$$

(4)

The sharp bound of $|H_{3,1}\left(f\right)|$ over the class ${\mathcal{S}}^{*}$ is still open.

Let ${\mathcal{SR}}^{*}$ be the class of starlike functions in $\mathcal{A}$ with real coefficients. Hence, if $f\in \mathcal{A}$ belongs to the class ${\mathcal{SR}}^{*}$, then f has the form given by (1) with $a_n\in \mathbb{R}$, $n\in \mathbb{N}\setminus \left\{1\right\}$ and satisfies the condition (3).

In this paper, we will prove the following.

Theorem 1.

If$f\in {\mathcal{SR}}^{*}$is the form (1), then the following inequalities hold:

$$-\frac{4}{9}\le H_{3,1}\left(f\right)\le \frac{1}{9}\sqrt{3}.$$

(5)

The first inequality is sharp for the function$f=f_1\in {\mathcal{SR}}^{*}$, where

$$f_1\left(z\right):=z{(1-z^3)}^{-2/3}=z+\frac{2}{3}{z}^4+\frac{5}{9}{z}^7+\cdots ,z\in \mathbb{D}.$$

The second inequality is sharp for the function$f=f_2\in {\mathcal{SR}}^{*}$, where

$$\begin{array}{cc}{f}_2\left(z\right)\hfill & :=zexp\left(-{\int }_0^z\frac{(2/\sqrt{3})\zeta +2{\zeta }^3}{1+(2/\sqrt{3}){\zeta }^2+{\zeta }^4}\mathrm{d}\zeta \right)\hfill \\ \hfill & =z-\frac{z^3}{\sqrt{3}}+\frac{2z^7}{3\sqrt{3}}-\frac{7z^9}{18}+\cdots ,z\in \mathbb{D}.\hfill \end{array}$$

2. Preliminary Results

Let $\mathcal{P}$ be the class of functions $p\in \mathcal{H}$ of the form

$$p\left(z\right)=1+\sum _{n=1}^{\infty }{c}_n{z}^n,z\in \mathbb{D},$$

(6)

having a positive real part in $\mathbb{D}$, i.e., the Carathéodory class of functions. It is well known, e.g., [6] (p. 166), that for $p\in \mathcal{P}$ with the form given by (6),

$$2c_2=c_1^2+(4-c_1^2)\zeta ,$$

(7)

for some $\zeta \in \overline{\mathbb{D}}$. Moreover, the following lemma will be used for our investigation.

Lemma 1

([7]). The formula (7) with $c_1\in [0,2)$ and $\zeta \in \mathbb{T}$ holds only for the function $p\in \mathcal{P}$ defined by

$$p\left(z\right)=\frac{1+\tau (1+\zeta )z+\zeta z^2}{1-\tau (1-\zeta )z-\zeta z^2},z\in \mathbb{D},$$

where $\tau \in [0,1)$.

Let ${\mathcal{B}}_0$ be the subclass of $\mathcal{H}$ of all self-mappings $\omega$ of $\mathbb{D}$ of the form

$$\omega \left(z\right)=\sum _{n=1}^{\infty }{\beta }_n{z}^n,z\in \mathbb{D},$$

(8)

i.e., the class of Schwarz functions. It is well known that $\omega \in {\mathcal{B}}_0$ if and only if $p=(1+\omega )/(1-\omega )\in \mathcal{P}$. For coefficients of functions in ${\mathcal{B}}_0$, the following properties, which can be found in [1] (Vol. I, pp. 84–85 and Vol. II, p. 78) and [8] (p. 128), will be used for our proof.

Lemma 2.

If$\omega \in {\mathcal{B}}_0$is of the form given by (8), then

(1) 

$|{\beta }_1|\le 1$,

(2) 

$|{\beta }_2|\le 1-|{\beta }_1{|}^2$,

(3) 

$|{\beta }_3(1-|{\beta }_1{|}^2)+\overline{{\beta }_1}{\beta }_2^2|\le (1-|{\beta }_1{|}^2{)}^2-{\left|{\beta }_2\right|}^2$.

The following inequalities, which will be used, hold for the fourth coefficients for Schwarz functions with real coefficients.

Lemma 3 

([9]). If $\omega \in {\mathcal{B}}_0$ is the form (8), ${\beta }_n\in \mathbb{R},$ $n\in \mathbb{N},$ and ${\beta }_2^2\ne {(1-{\beta }_1^2)}^2$, then

$${\mathsf{\Psi }}_L\le {\beta }_4\le {\mathsf{\Psi }}_U,$$

(9)

where

$${\mathsf{\Psi }}_L:=\frac{1+{\beta }_1^4+{\beta }_2-{\beta }_2^2-{\beta }_2^3-2{\beta }_1^2-{\beta }_1^2{\beta }_2+2{\beta }_1{\beta }_2{\beta }_3-{\beta }_3^2}{-1+{\beta }_1^2-{\beta }_2}$$

(10)

and

$${\mathsf{\Psi }}_U:=\frac{1+{\beta }_1^4-{\beta }_2-{\beta }_2^2+{\beta }_2^3-2{\beta }_1^2+{\beta }_1^2{\beta }_2-2{\beta }_1{\beta }_2{\beta }_3-{\beta }_3^2}{1-{\beta }_1^2-{\beta }_2}.$$

(11)

For given a set A, let intA, clA and $\partial$A be the sets of interior, closure and boundary, respectively, points of A. And let $R=[0,1]\times [-1,1]$ be a rectangle in ${\mathbb{R}}^2$. From now, we obtain several inequalities for functions, defined in subsets of R, which will be used in the proof of Theorem 1.

Proposition 1.

Define a function$F_1$by

$$F_1(x,y)=\sum _{n=0}^4{b}_n\left(x\right)y^n,$$

(12)

where

$$\begin{array}{cc}\hfill b_4\left(x\right)& ={(1-x)}^2{(1+x)}^4,\hfill \\ \hfill b_3\left(x\right)& =-x{(1+x)}^3(10-11x+x^2),\hfill \\ \hfill b_2\left(x\right)& ={(1+x)}^2(7-16x+14x^3-5x^4),\hfill \\ \hfill b_1\left(x\right)& =x(10+9x-2x^2-6x^3-8x^4-3x^5),\hfill \\ \hfill b_0\left(x\right)& =-8+16x^2+6x^3-8x^4-6x^5.\hfill \end{array}$$

Then$F_1(x,y)<2\sqrt{3}$holds for all$(x,y)\in R$.

Proof. 

Let $(x,y)\in R$. Since $b_4\left(x\right)\ge 0$, we have $b_4\left(x\right)y^4\le b_4\left(x\right)y^2$ and

$$F_1(x,y)\le G(x,y),(x,y)\in R,$$

where

$$G(x,y)=b_3\left(x\right)y^3+(b_4\left(x\right)+b_2\left(x\right))y^2+b_1\left(x\right)y+b_0\left(x\right).$$

We will show that $G(x,y)<2\sqrt{3}$ holds for $(x,y)\in R$.

When $x=0$, we have $G(0,y)=-8(1-y^2)\le 0$, for $y\in [-1,1]$. And, when $x=1$, we have $G(1,y)\equiv 0$.

Now, let $x\in (0,1)$ be fixed and put $b_i=b_i\left(x\right)$ ($i\in \{0,1,2,3,4\}$). Then $b_3<0$. Define a function $g_x$ by $g_x\left(y\right)=G(x,y)$. Note that

$$g_x(-1)=0\mathrm{and}{g}_x\left(1\right)=4x^2(1-x^2)(5-2x^2)\le 0.$$

(13)

Also,

$$g_x^{\prime }\left(y\right)=3b_3{y}^2+2(b_4+b_2)y+b_1=0$$

(14)

occurs at $y={\zeta }_1$ or ${\zeta }_2$, where

$${\zeta }_i=\frac{-(b_4+b_2)+{(-1)}^{i+1}\sqrt{{(b_4+b_2)}^2-3b_1{b}_3}}{3b_3},i\in \{1,2\}.$$

It is trivial that ${\zeta }_1<0<{\zeta }_2$. Furthermore, since $b_3<0$, $g_x$ has the local minimum at $y={\zeta }_1$. Let $\alpha =0.322818\cdots$ be a zero of polynomial q, where

$$q\left(y\right)=8-10y-42y^2-14y^3+7y^4.$$

Note that ${\zeta }_2\ge 1$ holds for x satisfying

$$2(1-x^2)q\left(x\right)=b_1+2(b_4+b_2)+3b_3\ge 0.$$

Hence we obtain

$$\left\{\begin{array}{cc}{\zeta }_2\ge 1,\hfill & \mathrm{when}x\in (0,\alpha ],\hfill \\ {\zeta }_2\le 1,\hfill & \mathrm{when}x\in [\alpha ,1).\hfill \end{array}\right.$$

(a) When $x\in (0,\alpha ]$, since ${\zeta }_2\ge 1$, $g_x$ is convex in $[-1,1]$. So, it holds that

$$g_x\left(y\right)\le max\{g_x(-1),g_x\left(1\right)\},y\in [-1,1].$$

Hence, by (13), we get $g_x\left(y\right)\le 0<2\sqrt{3}$ for $y\in [-1,1]$.

(b) When $x\in [\alpha ,1)$, $g_x$ has its local maximum $g_x\left({\zeta }_2\right)$. Using the fact that ${\zeta }_2$ is a solution of the equation given by (14) leads us to

$$g_x\left({\zeta }_2\right)=\left(\frac{2}{3}{b}_1-\frac{2{(b_2+b_4)}^2}{9b_3}\right){\zeta }_2+\left(b_0-\frac{b_1(b_2+b_4)}{9b_3}\right).$$

We claim that $g_x\left({\zeta }_2\right)-3<0$ holds for all $x\in [\alpha ,1)$. A compuation gives

$$g_x\left({\zeta }_2\right)-3=\frac{1}{9b_3}(1-x){(1+x)}^3[-2(1-x)(1+x){\kappa }_1{\zeta }_2+x{\kappa }_2],$$

where

$${\kappa }_1=64-128x+204x^2+464x^3+249x^4-14x^5+7x^6$$

and

$${\kappa }_2=910-11x-1340x^2-414x^3+752x^4+398x^5-64x^6+12x^7.$$

Since $b_3<0$, $g_x\left({\zeta }_2\right)-3<0$ is equivalent to

$$2(1-x^2){\kappa }_1\sqrt{{(b_4+b_2)}^2-3b_1{b}_3}<-3x{\kappa }_2{b}_3-2(1-x^2){\kappa }_1(b_4+b_2).$$

(15)

We can see that the right-side of the above equation is positive for all $x\in [\alpha ,1)$. Thus, by squaring both sides of (15), we have $g_x\left({\zeta }_2\right)<0$ is equivalent to $\mathsf{\Psi }>0$, where

$$\mathsf{\Psi }={[3x{\kappa }_2{b}_3+2(1-x^2){\kappa }_1(b_4+b_2)]}^2-4{(1-x^2)}^2{\kappa }_1^2[{(b_4+b_2)}^2-3b_1{b}_3].$$

By a simple calculation we have

$$\mathsf{\Psi }=-27x^2{(10-x)}^2{(1-x)}^2{(1+x)}^6{\Lambda }_x,$$

(16)

where

$$\begin{array}{cc}{\Lambda }_x:=\hfill & 22528-90112x-143980x^2+177084x^3+333021x^4-21120x^5-258308x^6\hfill \\ \hfill & -143200x^7+452x^8+28728x^9+37512x^{10}+24288x^{11}+9748x^{12}+2720x^{13}\hfill \\ \hfill & +968x^{14}-48x^{15}+36x^{16}.\hfill \end{array}$$

Since ${\Lambda }_x<0$ holds for all $x\in [\alpha ,1)$, from (16), $\mathsf{\Psi }>0$, this implies

$$g_x\left({\zeta }_2\right)<3.$$

(17)

Finally, since

$$g_x\left(y\right)\le max\{g_x(-1),g_x\left(1\right),g_x\left({\zeta }_2\right)\},y\in [-1,1],$$

it follows from (13) and (17) that $g_x\left(y\right)<3<2\sqrt{3}$ holds for all $y\in [-1,1]$. Thus the proof of Proposition 1 is completed. □

Proposition 2.

Let

$$\Omega =\left\{(x,y)\in [0,1/2)\times [0,1):0\le x\le \frac{y}{1+y}\right\}\subset R.$$

Define a function$F_2:\Omega \to \mathbb{R}$by

$$F_2(x,y)=\frac{1-x}{8+y-x(17+y)}{H}_1(x,y),$$

(18)

where$H_1(x,y)={\sum }_{n=0}^3{d}_n\left(y\right)x^n$with

$$d_3\left(y\right)={(1+y)}^2(1-6y+y^2),d_2\left(y\right)=17+24y+10y^2-3y^4,$$

$$d_1\left(y\right)=-8-26y-y^2+12y^3+3y^4\mathrm{and}{d}_0\left(y\right)=y(8+y-8y^2-y^3).$$

Then$F_2(x,y)\le (2/9)\sqrt{3}$holds for all$(x,y)\in \Omega$.

Proof. 

First of all, we note that $F_2$ is well-defined, since $8+y-x(17+y)>0$ holds for all $(x,y)\in \Omega$.

Differentiating $F_2$ with respect to x twice gives

$$\frac{1}{2}{[8+y-x(17+y)]}^3\frac{{\partial }^2{F}_2}{\partial x^2}(x,y)=\sum _{n=0}^4{\tilde{d}}_n\left(y\right)x^n,$$

(19)

where

$$\begin{array}{cc}\hfill {\tilde{d}}_4\left(y\right)& =-3(1-6y+y^2){(17+18y+y^2)}^2,\hfill \\ \hfill {\tilde{d}}_3\left(y\right)& =-4(884+3197y+4605y^2+2062y^3-302y^4-75y^5-3y^6),\hfill \\ \hfill {\tilde{d}}_2\left(y\right)& =6(1024+2344y+2421y^2+956y^3-202y^4-60y^5-3y^6),\hfill \\ \hfill {\tilde{d}}_1\left(y\right)& =-12{(8+y)}^2(4+7y+5y^2+y^3-y^4),\hfill \\ \hfill {\tilde{d}}_0\left(y\right)& =512+1088y+960y^2-176y^3-83y^4-30y^5-3y^6.\hfill \end{array}$$

Fix now $y\in [0,1)$ and put $y_0=y/(1+y)\in [0,1/2)$. Let us define a function $g_y:[0,y_0]\to \mathbb{R}$ by $g_y\left(x\right)={\sum }_{n=0}^4{\tilde{d}}_n\left(y\right)x^n$. Then we have

$$g_y^{\prime }\left(x\right)=-12(1+y){[8+y-x(17+y)]}^2\phi \left(x\right),$$

(20)

where

$$\phi \left(x\right)=4+3y+2y^2-y^3+(1+y)(1-6y+y^2)x.$$

Since $-4\le 1-6y+y^2\le 1$, we have

$$\phi \left(x\right)\ge 4+3y+2y^2-y^3-4x(1+y)\ge 4-y+2y^2-y^3>0,x\in [0,y_0].$$

Thus, by (20), we get $g_y^{\prime }\left(x\right)<0$, when $x\in [0,y_0]$. So $g_y$ is decreasing on the interval $[0,y_0]$, which yields

$$g_y\left(x\right)\ge g_y\left(y_0\right)=\frac{64(1-y)(8-7y+2y^2+33y^3)}{{(1+y)}^2}\ge 0,x\in [0,y_0].$$

Since $8+y-x(17+y)>0$ holds for all $(x,y)\in \Omega$, by (19), $F_2(x,·)$ is convex on $[0,y_0]$. This gives us that

$$F_2(x,y)\le max\{F_2(0,y),F_2(y_0,y)\}=F_2(0,y)=y-y^3\le \frac{2}{9}\sqrt{3},(x,y)\in \Omega ,$$

as we asserted. □

Proposition 3.

Define a function$F_3$by

$$F_3(x,y)=\frac{9(1-x)(1+y)}{8-y+x(1+y)}{H}_2(x,y),$$

(21)

where$H_2(x,y)={\sum }_{n=0}^3{k}_n\left(y\right)x^n$with

$$\begin{array}{cc}\hfill k_3\left(y\right)& ={(1+y)}^3,k_2\left(y\right)=1+7y+3y^2-3y^3,\hfill \\ \hfill k_1\left(y\right)& =8-2y-15y^2+3y^{3}and{k}_0\left(y\right)=-y(8-9y+y^2).\hfill \end{array}$$

Then$F_3(x,y)\le 2\sqrt{3}$holds for all$(x,y)\in R$.

Proof. 

First of all, by simple calculations, the equation $(\partial F_3/\partial x)(x,y)=0$ gives us

$$(1-x)(8-y+x(1+y))\frac{\partial H_2}{\partial x}(x,y)=9H_2(x,y).$$

(22)

Also, the equation $(\partial F_3/\partial y)(x,y)=0$ holds when

$$-(1+y)(8-y+x(1+y))\frac{\partial H_2}{\partial y}(x,y)=9H_2(x,y).$$

(23)

Assume that the function $F_3$ has its critical point at $(x_0,y_0)\in \mathrm{int}R$. Since $8-y_0+x_0(1+y_0)\ne 0$, from (22) and (23), we have

$$(1-x_0)\frac{\partial H_2}{\partial x}(x_0,y_0)+(1+y_0)\frac{\partial H_2}{\partial y}(x_0,y_0)=0,$$

or, equivalently, $y_0=x_0/(1-x_0)$. However, it holds that

$$(1-x_0)(8-y_0+x_0(1+y_0))\frac{\partial H_2}{\partial x}(x_0,y_0)-9H_2(x_0,y_0)=64(1-x_0)\ne 0,$$

since $x_0\in (0,1)$. This contradicts to (22). Hence $F_3$ does not have any critical points in intR. Thus $F_3$ has its maximum on $\partial R$.

We now consider $F_3$ on $\partial R$.

(a) On the side $x=1$, we have $F_3(1,y)\equiv 0$.

(b) On the side $y=-1$, we have $F_3(x,-1)\equiv 0$.

(c) On the side $y=1$, we have

$$F_3(x,1)=\frac{-36x(3-7x+4x^3)}{7+2x}=:\phi \left(x\right),x\in [0,1].$$

(24)

Since the inequality $2(7+56x-126x^2+72x^4)>0$ holds for all $x\in [0,1]$, it follows that $\phi \left(x\right)<2$ ($x\in [0,1]$). This inequality with (24) implies $F_3(x,1)<2<2\sqrt{3}$ holds for $x\in [0,1]$.

(d) On the side $x=0$, we have

$$F_3(0,y)=-9y(1-y^2)=:\psi \left(y\right).$$

(25)

And the inequality $F_3(0,y)\le 2\sqrt{3}$ ($y\in [-1,1]$) comes directly from (25) and

$$\psi \left(y\right)\le \psi (-1/\sqrt{3})=2\sqrt{3},y\in [-1,1].$$

From (a)–(d), for all $(x,y)\in \partial R$, the inequality $F_3(x,y)\le 2\sqrt{3}$ holds. Thus the proof of Proposition 3 is completed. □

Proposition 4.

For$F_1$defined by (12), the inequality

$$F_1(x,y)\ge -8$$

holds for$(x,y)\in [0,1]\times [-1,0]$.

Proof. 

Define a function $G:[0,1]\times [0,1]\to \mathbb{R}$ by

$$G(x,y)=F(x,-y)-b_4\left(x\right)y^4+8=l_3\left(x\right)y^3+l_2\left(x\right)y^2+l_1\left(x\right)y+l_0\left(x\right),$$

where $l_3\left(x\right)=-b_3\left(x\right)$, $l_2\left(x\right)=b_2\left(x\right)$, $l_1\left(x\right)=-b_1\left(x\right)$ and $l_0\left(x\right)=b_0\left(x\right)+8$. Then we have

$$F(x,y)+8\ge G(x,-y),(x,y)\in [0,1]\times [-1,0].$$

We note that, when $x=0$, $G(0,y)=7y^2\ge 0$ holds for $y\in [-1,1]$. And, when $x=1$, $G(1,y)\equiv 8>0$.

Let $x\in (0,1)$ be fixed and put $l_i=l_i\left(x\right)$ ($i\in \{0,1,2,3\}$). Define a function $g_x:[0,1]\to \mathbb{R}$ by $g_x\left(y\right)=G(x,y)$. We will show that the inequality $g_x\left(y\right)\ge 0$ holds for all $y\in [0,1]$.

Note that $l_3>0$ and $l_1<0$. Let

$${\zeta }_i=\frac{-l_2+{(-1)}^i\sqrt{l_2^2-3l_1{l}_3}}{l_3},i=1,2$$

be the roots of the equation

$$g_x^{\prime }\left(y\right)=3l_3{y}^2+2l_{2}y+l_1=0.$$

Then it is easily seen that ${\zeta }_1<0<{\zeta }_2$. Moreover ${\zeta }_2<1$ holds. Indeed, ${\zeta }_2<1$ is equivalent to $l_1{l}_3+3l_3^2+2l_2{l}_3>0$. And a computation gives

$$l_1{l}_3+3l_3^2+2l_2{l}_3=-2x{(1-x)}^2{(1+x)}^4\phi \left(x\right),$$

(26)

where

$$\phi \left(x\right)=-70-73x-52x^2-34x^3-16x^4+2x^5.$$

Since $\phi \left(x\right)<0$, by (26), we get $l_1{l}_3+3l_3^2+2l_2{l}_3>0$ and ${\zeta }_2<1$. Therefore, we have

$$g_x\left(y\right)\ge g_x\left({\zeta }_2\right),y\in [0,1].$$

(27)

On the other hand, simple calculations give us that

$$\begin{array}{cc}{g}_x\left({\zeta }_2\right)\hfill & =\frac{1}{9l_3}[(6l_1{l}_3-2l_2^2){\zeta }_2+(9l_0{l}_3-l_1{l}_2)]\hfill \\ \hfill & =\frac{-1}{9l_3}(1-x){(1+x)}^3[2(1-x^2){\kappa }_1{\zeta }_2+x{\kappa }_2],\hfill \end{array}$$

where

$${\kappa }_1=49-126x+255x^2+472x^3+204x^4-24x^5+16x^6$$

and

$${\kappa }_2=-70+97x-1352x^2-429x^3+746x^4+401x^5-56x^6+15x^7.$$

Since $l_3>0$, $g_x\left({\zeta }_2\right)\ge 0$ holds, if

$$2(1-x^2){\kappa }_1{\zeta }_2+x{\kappa }_2\le 0.$$

(28)

Moreover (28) is equivalent to $\mathsf{\Psi }\ge 0$, where

$$\mathsf{\Psi }={[2(1-x^2){\kappa }_1{l}_2-3x{\kappa }_2{l}_3]}^2-4{(1-x^2)}^2{\kappa }_1^2(l_2^2-3l_1{l}_3).$$

We represent $\mathsf{\Psi }$ by

$$\mathsf{\Psi }=-27x^4{(10-x)}^2{(1-x)}^2{(1+x)}^6{\tilde{\Lambda }}_x,$$

(29)

where

$$\begin{array}{cc}{\tilde{\Lambda }}_x=\hfill & -17052+84812x-222415x^2-10212x^3+78990x^4-226456x^5\hfill \\ \hfill & -152793x^6+198120x^7+169280x^8-11796x^9-33413x^{10}+1068x^{11}\hfill \\ \hfill & +2790x^{12}-1008x^{13}+117x^{14}.\hfill \end{array}$$

(30)

Since ${\tilde{\Lambda }}_x<0$ holds for all $x\in (0,1)$, from (29), $\mathsf{\Psi }\ge 0$ is true. We thus have $g_x\left({\zeta }_2\right)\ge 0$. Finally, it follows from (27) that $g_x\left(y\right)\ge 0$ holds for all $y\in [0,1]$. The proof of Proposition 4 is completed. □

Proposition 5.

For a function$F_4$defined by

$$F_4(x,y)=F_1(-x,y),$$

(31)

where$F_1$is defined by (12), we have

$$F_4(x,y)\ge -8,(x,y)\in [0,1]\times [-1/3,1].$$

Proof. 

It is easily checked that $F_4(x,y)\ge -8$ holds for $y\in [-1/3,1]$ when $x=0$ or $x=1$. Let $x\in (0,1)$ be fixed and put $m_i=b_i(-x)$ ($i\in \{0,1,2,3,4\}$). Define a function $g_x:[-1/3,1]\to \mathbb{R}$ by $g_x\left(y\right)=F_4(x,y)$.

First, we will show that $g_x\left(y\right)\ge -8$ holds for $y\in [-1/3,0]$. Since $m_3>0$ and $m_4>0$, we have $m_4{y}^4\ge 0$ and $m_3{y}^3\ge -m_3{y}^2/3$ for $y\in [-1/3,0]$. Hence, we obtain

$$g_x\left(y\right)+8\ge {\phi }_x(-y),y\in [-1/3,0],$$

(32)

where ${\phi }_x:[0,1/3]\to \mathbb{R}$ is the function defined by

$${\phi }_x\left(y\right)=\left(-\frac{1}{3}{m}_3+m_2\right)y^2-m_{1}y+m_0+8,y\in [0,1/3].$$

Since $m_1<0$ and

$$-\frac{1}{3}{m}_3+m_2=\frac{1}{3}{(1-x^2)}^2(21-4x+14x^2)>0,x\in (0,1),$$

we get

$${\phi }_x^{\prime }\left(y\right)=2\left(-\frac{1}{3}{m}_3+m_2\right)y-m_1>0,y\in [0,1/3].$$

Therefore ${\phi }_x$ is increasing on $[0,1/3]$ and we get

$${\phi }_x\left(y\right)\ge {\phi }_x\left(0\right)=m_0+8=x^2(16-6x-8x^2+6x^3)\ge 0,y\in [0,1/3].$$

Thus, by (32), $g_x\left(y\right)\ge -8$ holds for $y\in [-1/3,0]$.

Next, we will show that $g_x\left(y\right)\ge -8$ holds for $y\in [0,1]$. For this, define a function ${\psi }_x:[0,1]\to \mathbb{R}$ by

$${\psi }_x\left(y\right)=g_x\left(y\right)-m_4{y}^4+8=m_3{y}^3+m_2{y}^2+m_{1}y+m_0+8.$$

It is sufficient to show that ${\psi }_x\left(y\right)\ge 0$ holds for $y\in [0,1]$, since

$$g_x\left(y\right)+8\ge {\psi }_x\left(y\right),y\in [0,1].$$

Let

$${\zeta }_i=\frac{-m_2+{(-1)}^i\sqrt{m_2^2-3m_1{m}_3}}{3m_3},i\in \{1,2\}$$

be the roots of the equation

$${\psi }_x^{\prime }\left(y\right)=3m_3{y}^2+2m_{2}y+m_1=0.$$

Clearly, ${\zeta }_1<0$. Thus we have

$${\psi }_x\left(y\right)\ge min\{{\psi }_x\left(1\right),{\psi }_x\left({\zeta }_2\right)\},y\in [0,1].$$

(33)

Since

$${\psi }_x\left(1\right)=7+2x-19x^2-4x^3+29x^4+2x^5-9x^6>0,x\in (0,1),$$

it is enough to show that ${\psi }_x\left({\zeta }_2\right)\ge 0$ holds. A similar argument with the proof of Proposition 4, for $x\in (0,1)$, ${\psi }_x\left({\zeta }_2\right)\ge 0$ holds if ${\tilde{\Lambda }}_{-x}<0$, where ${\tilde{\Lambda }}_x$ is the quantity defined by (30). It can be checked that ${\tilde{\Lambda }}_x<0$ holds for all $x\in (-1,0)$. Consequently, ${\psi }_x\left({\zeta }_2\right)\ge 0$, when $x\in (0,1)$, follows. Hence, by (33), ${\psi }_x\left(y\right)\ge 0$ holds for $y\in [0,1]$. It completes the proof of Proposition 5. □

3. The Proof of Thereom 1

By using all lemmas and propositions in Section 2, we can prove Theorem 1 as follows.

Proof of Theorem 1.

Let $f\in {\mathcal{SR}}^{*}$ be of the form (1). Then by (3) there exists a $\omega \in {\mathcal{B}}_0$ of the form (8) such that

$$\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}=\frac{1+\omega \left(z\right)}{1-\omega \left(z\right)}.$$

(34)

Substituting the series (1) and (8) into (34), by equating the coefficients we get

$$18H_{3,1}\left(f\right)=3{\beta }_1^4{\beta }_2+6{\beta }_1^3{\beta }_3+10{\beta }_1{\beta }_2{\beta }_3-8{\beta }_3^2-11{\beta }_1^2{\beta }_2^2+9({\beta }_2-{\beta }_1^2){\beta }_4.$$

(35)

Since $H_{3,1}\left(f\right)=H_{3,1}\left(\tilde{f}\right)$, where $\tilde{f}\left(z\right)=-f(-z)\in {\mathcal{SR}}^{*}$, we may assume that ${\beta }_1\in [0,1]$.

The inequality (5) will be proved case by case as in the following Table 1.

Table 1. An outline of the proof.

Cases	Conditions	Used Results for the Proof
I	${\beta }_1=1$	Schwarz’s lemma
II(a)	${\beta }_2=1-{\beta }_1^2$, ${\beta }_1\in [0,1)$	Lemma 1
II(b)	${\beta }_2={\beta }_1^2-1$, ${\beta }_1\in [0,1)$	Lemma 1
III(a)	$|{\beta }_2|\ne 1-{\beta }_1^2$, ${\beta }_1\in [0,1)$, ${\beta }_2\ge {\beta }_1^2$	Lemma 2 and 3, Proposition 1 and 2
III(b)	$|{\beta }_2|\ne 1-{\beta }_1^2$, ${\beta }_1\in [0,1)$, ${\beta }_2\le {\beta }_1^2$	Lemma 2 and 3, Proposition 1 and 3
IV(a)	$|{\beta }_2|\ne 1-{\beta }_1^2$, ${\beta }_1\in [0,1)$, ${\beta }_2\ge {\beta }_1^2$	Lemma 2 and 3, Proposition 5
IV(b)	$|{\beta }_2|\ne 1-{\beta }_1^2$, ${\beta }_1\in [0,1)$, ${\beta }_2\le {\beta }_1^2$	Lemma 2 and 3, Proposition 4 and 5

I. When ${\beta }_1=1$, then by Schwarz’s lemma, ${\beta }_n=0$ for all $n\ge 2$. Thus, by (35), $H_{3,1}\left(f\right)=0$.

II. When $\omega \in {\mathcal{B}}_0$ be such that $|{\beta }_2|=1-{\beta }_1^2$ and ${\beta }_1\in [0,1)$. Let $p=(1+\omega )/(1-\omega )\in \mathcal{P}$ be of the form (6). From the relations

$$c_1=2{\beta }_1\mathrm{and}{c}_2=2({\beta }_1^2+{\beta }_2),$$

it follows from that $c_1\in [0,2)$ and $2c_2=c_1^2+(4-c_1^2)\zeta$, where $\zeta =\pm 1\in \mathbb{T}$.

II(a) Assume that $\zeta =1$. Then, by Lemma 1, $p=p_1$, where

$$p_1\left(z\right)=\frac{1+2\tau z+z^2}{1-z^2}=1+2\tau z+2z^2+2\tau z^3+\cdots ,z\in \mathbb{D}$$

with $\tau \in [0,1)$. And, from $p=(1+\omega )/(1-\omega )$, we have

$${\beta }_1=\tau ,{\beta }_2=1-{\tau }^2,{\beta }_3=-\tau +{\tau }^3\mathrm{and}{\beta }_4={\tau }^2-{\tau }^4.$$

(36)

Substituting (36) into (35), we get

$$H_{3,1}\left(f\right)=-\frac{2}{9}{\tau }^2(5-7{\tau }^2+2{\tau }^4)=:g\left({\tau }^2\right),$$

(37)

where

$$g\left(x\right)=-\frac{2}{9}x(1-x)(5-2x).$$

It can be easily checked that $g\left(x\right)\le g\left(0\right)=0$, for $x\in [0,1)$. Moreover, since $g^{\prime }\left(x\right)=0$ occurs only when $x=x_1:=(7-\sqrt{19})/6=0.440184\cdots \in [0,1)$ and $g^{″}\left(x_1\right)=4\sqrt{19}/9>0$, it holds that

$$g\left(x\right)\ge g\left(x_1\right)=\frac{1}{243}(28-19\sqrt{19})\ge -\frac{4}{9},x\in [0,1).$$

So, from (37), the inequality (5) holds.

II(b) Now assume that $\zeta =-1$. Then, by Lemma 1 again, we get $p=p_2$, where

$$\begin{array}{cc}{p}_2\left(z\right)\hfill & =\frac{1-z^2}{1-2\tau z+z^2}\hfill \\ \hfill & =1+2\tau z+(-2+4{\tau }^2)z^2+(-6\tau +8{\tau }^3)z^3+(2-16{\tau }^2+16{\tau }^4)z^4+\cdots ,z\in \mathbb{D}\hfill \end{array}$$

with $\tau \in [0,1)$. Thus, we have

$${\beta }_1=\tau ,{\beta }_2={\tau }^2-1,{\beta }_3={\tau }^3-\tau \mathrm{and}{\beta }_4={\tau }^4-{\tau }^2.$$

(38)

Substituting (38) into (35), we get $H_{3,1}\left(f\right)=0$ and the inequality (5) holds.

III. Let now $|{\beta }_2|\ne 1-{\beta }_1^2$ and ${\beta }_1\ne 1$.

At first, we will show that the second inequality in (5) holds. Since ${\beta }_1$, ${\beta }_2$ and ${\beta }_3$ are real, by Lemma 2 for $s\in [0,1]$ and t, $u\in [-1,1]$ we have

$${\beta }_1=s,{\beta }_2=(1-s^2)t,{\beta }_3=(1-s^2)(u(1-t^2)-s{t}^2).$$

(39)

Substituting (39) into (10) and (11), we have

$${\mathsf{\Psi }}_U=(1-s^2)[1-u^2-u(u+2s)t-(1-u^2)t^2+{(u+s)}^2{t}^3]$$

(40)

and

$${\mathsf{\Psi }}_L=(1-s^2)[-1+u^2-u(u+2s)t+(1-u^2)t^2+{(u+s)}^2{t}^3].$$

(41)

We also have $(s,t)\notin C$, where C is a curve defined by

$$C=\left\{\right(s,t)\in R:s=1\mathrm{or}t=\pm 1\}\subset \partial R.$$

III(a) Consider the case ${\beta }_2\ge {\beta }_1^2$, i.e., $(s,t)\in {\Omega }_1$, where ${\Omega }_1$ is the set defined by

$${\Omega }_1=\left\{(s,t)\in [0,1/\sqrt{2})\times [0,1):\frac{s^2}{1-s^2}\le t<1\right\}$$

so that ${\Omega }_1\cap C=\varnothing$. In this case, by (40), we have

$$\begin{array}{cc}18H_{3,1}\left(f\right)\hfill & \le 3{\beta }_1^4{\beta }_2+6{\beta }_1^3{\beta }_3+10{\beta }_1{\beta }_2{\beta }_3-8{\beta }_3^2-11{\beta }_1^2{\beta }_2^2+9({\beta }_2-{\beta }_1^2){\mathsf{\Psi }}_U\hfill \\ \hfill & =-(1-s^2)(1+t)\Phi (s,t,u),(s,t,u)\in {\Omega }_1\times [-1,1],\hfill \end{array}$$

(42)

where

$$\Phi (s,t,u)={\Phi }_0+{\Phi }_{1}u+{\Phi }_2{u}^2$$

(43)

with

$$\begin{array}{cc}\hfill {\Phi }_0={\Phi }_0(s,t)& :=-9(1-t)t-s^{4}t(3+2t-t^2)+s^2(9+2t^2-t^3),\hfill \\ \hfill {\Phi }_1={\Phi }_1(s,t)& :=-2s(1-t)[(5-t)t+s^2(3+4t+t^2)],\hfill \\ \hfill {\Phi }_2={\Phi }_2(s,t)& :=(1-t^2)[8+t-s^2(17+t)].\hfill \end{array}$$

We note that ${\Phi }_2>0$, since

$$8+t-s^2(17+t)\ge \frac{8(1-t)}{1+t}>0,(s,t)\in {\Omega }_1.$$

Let $u_1=-{\Phi }_1/\left(2{\Phi }_2\right)$ be the root of the equation $(\partial \Phi /\partial u)(s,t,u)=0$. Then it can be seen that $u_1\ge -1$. Indeed, we note that $2{\Phi }_2-{\Phi }_1=2(1-t)\mathsf{{\rm Y}}(s,t)$, where $\mathsf{{\rm Y}}(s,t)={\lambda }_2\left(s\right)t^2+{\lambda }_1\left(s\right)t+{\lambda }_0\left(s\right)$, where

$${\lambda }_2\left(s\right)={(1-s)}^2(1+s),{\lambda }_1\left(s\right)=9+5s-18s^2+4s^3$$

and

$${\lambda }_0\left(s\right)=8-17s^2+3s^3.$$

Since ${\lambda }_i\left(s\right)\ge 0$ when $s\in [0,1/\sqrt{2})$ for $i\in \{1,2\}$, we have

$$\mathsf{{\rm Y}}(s,t)\ge \mathsf{{\rm Y}}\left(s,\frac{s^2}{1-s^2}\right)=\frac{8(1+s-s^2)}{1+s}\ge 0,(s,t)\in {\Omega }_1.$$

Hence, we get $2{\Phi }_2-{\Phi }_1\ge 0$ and it follows from ${\Phi }_2>0$ that $u_1\ge -1$.

(i) Assume that $u_1\ge 1$. Then we have

$$\Phi (s,t,u)\ge \Phi (s,t,1)={\Phi }_0+{\Phi }_1+{\Phi }_2,(s,t,u)\in {\Omega }_1\times [-1,1].$$

Therefore, by (42), it holds that

$$18H_{3,1}\left(f\right)\le -(1-s^2)(1+t)({\Phi }_0+{\Phi }_1+{\Phi }_2)=F_1(s,t),(s,t)\in {\Omega }_1,$$

(44)

where $F_1$ is the function defined by (12). From Proposition 1 and (44), we thus have $H_{3,1}\left(f\right)\le \sqrt{3}/9$.

(ii) Assume that $-1\le u_1\le 1$. Then we have

$$\Phi (s,t,u)\ge \Phi (s,t,u_1)={\Phi }_0-\frac{{\Phi }_1^2}{4{\Phi }_2},(s,t,u)\in {\Omega }_1\times [-1,1].$$

Therefore, by (42), it holds that

$$18H_{3,1}\left(f\right)\le -(1-s^2)(1+t)\left({\Phi }_0-\frac{{\Phi }_1^2}{4{\Phi }_2}\right)=9F_2(s^2,t),(s,t)\in {\Omega }_1,$$

where $F_2$ is the function defined by (18). Therefore, by Proposition 2, $H_{3,1}\left(f\right)\le \sqrt{3}/9$ holds.

III(b) Consider the case ${\beta }_2\le {\beta }_1^2$, i.e., $(s,t)\in {\Omega }_2$, where ${\Omega }_2$ is the set defined by ${\Omega }_2=\mathrm{cl}(R\setminus {\Omega }_1)\setminus C$. Then, from (41), we have

$$\begin{array}{cc}18H_{3,1}\left(f\right)\hfill & \le 3{\beta }_1^4{\beta }_2+6{\beta }_1^3{\beta }_3+10{\beta }_1{\beta }_2{\beta }_3-8{\beta }_3^2-11{\beta }_1^2{\beta }_2^2+9({\beta }_2-{\beta }_1^2){\mathsf{\Psi }}_L\hfill \\ \hfill & =-(1-s^2)(1+t)\widehat{\Phi }(s,t,u),(s,t,u)\in {\Omega }_2\times [-1,1],\hfill \end{array}$$

(45)

where

$$\widehat{\Phi }(s,t,u)={\widehat{\Phi }}_0+{\widehat{\Phi }}_{1}u+{\widehat{\Phi }}_2{u}^2$$

(46)

with

$$\begin{array}{cc}\hfill {\widehat{\Phi }}_0={\widehat{\Phi }}_0(s,t)& :=9(1-t)t-s^{4}t(3+2t-t^2)-s^2(9-20t^2+t^3),\hfill \\ \hfill {\widehat{\Phi }}_1={\widehat{\Phi }}_1(s,t)& :=-2s(1-t)[(5-t)t+s^2(3+4t+t^2)],\hfill \\ \hfill {\widehat{\Phi }}_2={\widehat{\Phi }}_2(s,t)& :={(1-t)}^2[8-t+s^2(1+t)].\hfill \end{array}$$

Using the inequality $s^2\ge t/(1+t)$, we have ${\widehat{\Phi }}_2\ge 8{(1-t)}^2>0$ for $(s,t)\in {\Omega }_2$. Let $u_2=-{\widehat{\Phi }}_1/\left(2{\widehat{\Phi }}_2\right)$ be the root of the equation $(\partial \widehat{\Phi }/\partial u)(s,t,u)=0$. Then, by a similar procedure with Part III(a), it can be seen that $u_2\ge -1$.

(i) Assume that $u_2\ge 1$. Then we have

$$\widehat{\Phi }(s,t,u)\ge \widehat{\Phi }(s,t,1)={\widehat{\Phi }}_0+{\widehat{\Phi }}_1+{\widehat{\Phi }}_2,(s,t,u)\in {\Omega }_2\times [-1,1].$$

Therefore, by (45), it holds that

$$18H_{3,1}\left(f\right)\le -(1-s^2)(1+t)({\widehat{\Phi }}_0+{\widehat{\Phi }}_1+{\widehat{\Phi }}_2)=F_1(s,t),(s,t)\in {\Omega }_2,$$

where $F_1$ is the function defined by (12). Thus, by Proposition 1, $H_{3,1}\left(f\right)\le \sqrt{3}/9$ holds.

(ii) Assume that $-1\le u_2\le 1$. Then we have

$$\widehat{\Phi }(s,t,u)\ge \widehat{\Phi }(s,t,u_2)={\widehat{\Phi }}_0-\frac{{\widehat{\Phi }}_1^2}{4{\widehat{\Phi }}_2},(s,t,u)\in {\Omega }_2\times [-1,1].$$

Therefore, by (45), it holds that

$$18H_{3,1}\left(f\right)\le -(1-s^2)(1+t)\left({\widehat{\Phi }}_0-\frac{{\widehat{\Phi }}_1^2}{4{\widehat{\Phi }}_2}\right)=F_3(s^2,t),(s,t)\in {\Omega }_2,$$

where $F_3$ is the function defined by (21). Therefore, by Proposition 3, we obtain $H_{3,1}\left(f\right)\le \sqrt{3}/9$.

Next, we will show that the first inequality in (5) holds.

IV(a) Consider the case ${\beta }_2\ge {\beta }_1^2$. Then we have

$$18H_{3,1}\left(f\right)\ge -(1-s^2)(1+t)\widehat{\Phi }(s,t,u),(s,t,u)\in {\Omega }_1\times [-1,1],$$

(47)

where $\widehat{\Phi }$ is the function defined by (46). Since ${\widehat{\Phi }}_1\le 0$ and ${\widehat{\Phi }}_2>0$, it holds that

$$\begin{array}{cc}\widehat{\Phi }(s,t,u)\hfill & \le max\{\widehat{\Phi }(s,t,-1),\widehat{\Phi }(s,t,1)\}\hfill \\ \hfill & =\widehat{\Phi }(s,t,-1)={\widehat{\Phi }}_2-{\widehat{\Phi }}_1+{\widehat{\Phi }}_0,(s,t,u)\in {\Omega }_1\times [-1,1].\hfill \end{array}$$

Hence, from (47), we obtain

$$H_{3,1}\left(f\right)\ge -(1-s^2)(1+t)({\widehat{\Phi }}_2-{\widehat{\Phi }}_1+{\widehat{\Phi }}_0)=F_4(s,t),(s,t)\in {\Omega }_1,$$

(48)

where $F_4$ is the function defined by (31). Thus, by Proposition 5 and (48), we get $H_{3,1}\left(f\right)\ge -4/9$.

IV(b) We consider the case ${\beta }_2\le {\beta }_1^2$. Then we have

$$18H_{3,1}\left(f\right)\ge -(1-s^2)(1+t)\Phi (s,t,u),(s,t,u)\in {\Omega }_2\times [-1,1],$$

where $\Phi$ is the function defined by (43).

For $t\in [-1/3,0]$, let

$$s_t=\frac{t^2-5t}{t^2+4t+3}$$

so that $0=s_0\le s_t\le s_{-1/3}=1$ holds for $t\in [-1/3,0]$. And let

$${\Omega }_3=\{(s,t)\in {\Omega }_2:s\le s_t\}\mathrm{and}{\Omega }_4=\{(s,t)\in {\Omega }_2:s\ge s_t\}.$$

We note that ${\Omega }_3\subset [0,1]\times [-1,0]$ and ${\Omega }_4\subset [0,1]\times [-1/3,1]$. Then ${\Phi }_1\ge 0$ when $(s,t)\in {\Omega }_3$, and ${\Phi }_1\le 0$ when $(s,t)\in {\Omega }_4$.

(i) For the case $(s,t)\in {\Omega }_3$, since ${\Phi }_1\ge 0$ and ${\Phi }_2\ge 0$, we have

$$\Phi (s,t,u)\le \Phi (s,t,1)={\Phi }_2+{\Phi }_1+{\Phi }_0,(s,t,u)\in {\Omega }_3\times [-1,1]$$

and, therefore, we get

$$18H_{3,1}\left(f\right)\ge -(1-s^2)(1+t)({\Phi }_2+{\Phi }_1+{\Phi }_0)=F_1(s,t),(s,t)\in {\Omega }_3,$$

where $F_1$ is the function defined by (12). Since ${\Omega }_3\subset [0,1]\times [-1,0]$, Proposition 4 gives us that $H_{3,1}\left(f\right)\ge -4/9$ holds.

(ii) For the case $(s,t)\in {\Omega }_4$, we have

$$18H_{3,1}\left(f\right)\ge -(1-s^2)(1+t)({\Phi }_2-{\Phi }_1+{\Phi }_0)=F_4(s,t),(s,t)\in {\Omega }_4,$$

where $F_4$ is the funciton defined by (31). Since ${\Omega }_4\subset [0,1]\times [-1/3,1]$, Proposition 5 gives us that $H_{3,1}\left(f\right)\ge -4/9$ holds. Thus the proof of Theorem 1 is now completed. □

4. Conclusions

In the present paper, we obtained that the sharp inequalities $-4/9\le H_{3,1}\left(f\right)\le \sqrt{3}/9$ hold for f in the class ${\mathcal{SR}}^{*}$, i.e., starlike functions with real coefficients. Therefore, it follows that $|H_{3,1}\left(f\right)|\le 4/9$ holds for $f\in {\mathcal{SR}}^{*}$ and this inequality is sharp with the extremal function $f_1\in {\mathcal{SR}}^{*}$, where $f_1\left(z\right)=z{(1-z^3)}^{-2/3}$. So it can be naturally expected that the sharp inequality $|H_{3,1}\left(f\right)|\le 4/9$ would hold for all $f\in {\mathcal{S}}^{*}$.