The second Hankel determinant of the logarithmic coefficients of strongly starlike and strongly convex functions

Abstract

Sharp bounds are given for the second Hankel determinant of the logarithmic coefficients of strongly starlike and strongly convex functions.

1 Introduction

Denote by $\mathcal H$ the class of analytic functions in $\mathbb D:= \left\{ z \in \mathbb C: |z|<1 \right\}$ with Taylor expansion

$$\begin{aligned} f(z) = \sum _{n=1}^{\infty }a_{n}z^{n},\quad z\in \mathbb D, \end{aligned}$$

(1)

and let $\mathcal A$ be the subclass of f normalized by $f'(0)=1.$ Let $\mathcal S$ denote the subclass of univalent functions in $\mathcal A$.

For $f\in \mathcal S$, logarithmic coefficients $\gamma _n:=\gamma _n(f)$ of f are defined by

$$\begin{aligned} F_f(z):=\log \frac{f(z)}{z}=2\sum _{n=1}^\infty \gamma _n(f) z^n,\quad z\in \mathbb D,\ \log 1:=0. \end{aligned}$$

(2)

and play a crucial role in the theory of univalent functions, and in articular to prove the Milin conjecture ([19], see also [7, p. 155]). We note that for the class $\mathcal S$ sharp estimates are known only for $\gamma _1$ and $\gamma _2,$ namely,

$$\begin{aligned} |\gamma _1|\le 1,\quad |\gamma _2|\le \frac{1}{2}+\frac{1}{\textrm{e}^2}=0.635\dots \end{aligned}$$

Estimating the modulus of logarithmic coefficients for $f\in \mathcal S$ and various subclasses has been considered recently by several authors (e.g., [1, 2, 5, 8, 12, 24]).

For $q,n \in \mathbb N,$ the Hankel determinant $H_{q,n}(f)$ of $f\in \mathcal A$ of the form (1) is defined as

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

and in particular many authors have examined the second and the third Hankel determinants $H_{2,2}(f)$ and $H_{3,1}(f)$ over selected subclasses of $\mathcal A,$ (see e.g., [4, 11] with further references). We note that $H_{2,1}(f)=a_3-a_2^2$ is the well known coefficient functional which for $\mathcal S$ was studied first in 1916 by Bieberbach (see e.g., [9, Vol. I, p. 35]).

Based on the these ideas, in this paper and in [10] we propose research study of the Hankel determinants $H_{q,n}(F_f/2)$ which entries are logarithmic coefficients of f. We are therefore concerned with

$$\begin{aligned} H_{q,n}(F_f/2) = \begin{vmatrix} \gamma _{n}&\gamma _{n+1}&\cdots&\gamma _{n+q-1} \\ \gamma _{n+1}&\gamma _{n+2}&\cdots&\gamma _{n+q} \\ \vdots&\vdots&\vdots&\vdots \\ \gamma _{n+q-1}&\gamma _{n+q}&\cdots&\gamma _{n+2(q-1)} \end{vmatrix}. \end{aligned}$$

Differentiating (2) and using (1) we obtain

$$\begin{aligned} \gamma _{1}=\frac{1}{2}a_{2},\quad \gamma _{2}=\frac{1}{2}\left( a_{3}-\frac{1}{2}a_{2}^{2}\right) ,\quad \gamma _{3}=\frac{1}{2}\left( a_{4}-a_{2}a_{3}+\frac{1}{3}a_{2}^{3}\right) , \end{aligned}$$

(3)

and so

$$\begin{aligned} H_{2,1}(F_f/2)=\gamma _1\gamma _3-\gamma _2^2=\frac{1}{4}\left( a_2a_4-a_3^2+\frac{1}{12}a_2^4\right) . \end{aligned}$$

(4)

Note that when $f\in \mathcal S,$ then for $f_\theta (z):=\textrm{e}^{-\textrm{i}\theta }f(\textrm{e}^{\textrm{i}\theta }z),\ \theta \in \mathbb R,$

$$\begin{aligned} H_{2,1}(F_{f_\theta }/2)=\frac{\textrm{e}^{4\textrm{i}\theta }}{4}\left( a_2a_4-a_3^2+\frac{1}{12}a_2^4\right) =\textrm{e}^{4\textrm{i}\theta }H_{2,1}(F_f/2), \end{aligned}$$

(5)

so $|H_{2,1}(F_{f_\theta }/2)|$ is rotationally invariant.

In this paper we find sharp upper bounds for $H_{2,1}(F_f/2)$ in the case when f is strongly starlike or strongly convex function of order $\alpha ,$ defined respectively as follows. Given $\alpha \in (0,1],$ a function $f\in \mathcal A$ is called strongly starlike of order $\alpha$ if

$$\begin{aligned} \left| \arg \frac{zf'(z)}{f(z)}\right| <\alpha \frac{\pi }{2},\quad z\in \mathbb D,\ \arg 1:=0. \end{aligned}$$

(6)

Also, a function $f\in \mathcal A$ is called strongly convex of order $\alpha$ if

$$\begin{aligned} \left| \arg \left\{ 1+\frac{zf''(z)}{f'(z)}\right\} \right| <\alpha \frac{\pi }{2},\quad z\in \mathbb D,\ \arg 1:=0. \end{aligned}$$

(7)

We denote these classes by $\mathcal S^*_\alpha$ and $\mathcal S^c_\alpha$ respectively, noting that $\mathcal S^*_1=:\mathcal S^*$ and $\mathcal S^c_1=:\mathcal S^c$ are the classes of starlike and convex functions, respectively.

The class of strongly starlike functions was introduced by Stankiewicz [21, 22], and independently by Brannan and Kirwan [3] (see also [9, Vol. I, pp. 137-142]). Stankiewicz [22] found an external geometrical characterization of strongly starlike functions and Brannan and Kirwan gave a geometrical condition called $\delta$-visibility, which is sufficient for functions to be strongly starlike. Subsequently Ma and Minda [16] proposed an internal characterization of functions in $\mathcal S_\alpha ^*$ based on the concept of k-starlike domains. Further results regarding the geometry of strongly starlike functions were given in [14, Chapter IV], [15] and [23].

In view of (6) and (7) both classes $\mathcal S^*_\alpha$ and $\mathcal S^c_\alpha$ can be represented using the Carathéodory class $\mathcal P$, i.e., the class of analytic functions p in $\mathbb D$ of the form

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

(8)

having a positive real part in $\mathbb D.$ Thus the coefficients of functions in $\mathcal S^*_\alpha$ and $\mathcal S^c_\alpha$ have a convenient representation in terms of the coefficients of functions in $\mathcal P.$ Therefore obtaining the upper bound of $H_{2,1}(F_f/2),$ we base our analysis on well-known expressions for $c_2$ (e.g., [20, p. 166]), and $c_3$ (Libera and Zlotkiewicz [17, 18]), and $c_4$ obtained recently in [13], all of which are contained in the following lemma [13]. Let $\overline{\mathbb D}:=\{z\in \mathbb C:|z|\le 1\}$ and $\mathbb T:=\{z\in \mathbb C:|z|= 1\}.$

Lemma 1

If $p \in {\mathcal P}$ and is given by (6) with $c_1\ge 0,$ then

$$\begin{aligned} c_1=2\zeta _1, \end{aligned}$$

(9)

$$\begin{aligned} c_2=2\zeta _1^2+2(1-\zeta _1^2)\zeta _2 \end{aligned}$$

(10)

and

$$\begin{aligned} c_3=2\zeta _1^3+4(1-\zeta _1^2)\zeta _1\zeta _2-2(1-\zeta _1^2)\zeta _1\zeta _2^2+2(1-\zeta _1^2)(1-|\zeta _2|^2)\zeta _3. \end{aligned}$$

(11)

for some $\zeta _1\in [0,1]$ and $\zeta _2,\zeta _3\in \overline{\mathbb D}.$

For $\zeta _1\in \mathbb T,$ there is a unique function $p\in \mathcal P$ with $c_1$ as in (9), namely,

$$\begin{aligned} p(z)=\frac{1+\zeta _1z}{1-\zeta _1z},\quad z\in \mathbb D. \end{aligned}$$

For $\zeta _1\in \mathbb D$ and $\zeta _2\in \mathbb T,$ there is a unique function $p\in \mathcal P$ with $c_1$ and $c_2$ as in (9)–(10), namely,

$$\begin{aligned} p(z)=\frac{1+\left( \overline{\zeta _1}\zeta _2+\zeta _1\right) z+\zeta _2z^2}{1+\left( \overline{\zeta _1}\zeta _2-\zeta _1\right) z-\zeta _2z^2},\quad z\in \mathbb D. \end{aligned}$$

(12)

For $\zeta _1,\zeta _2\in \mathbb D$ and $\zeta _3\in \mathbb T,$ there is a unique function $p\in \mathcal P$ with $c_1,$ $c_2$ and $c_3$ as in (9)–(11), namely,

$$\begin{aligned} p(z)=\frac{1+\left( \overline{\zeta _2}\zeta _3+\overline{\zeta _1}\zeta _2+\zeta _1\right) z+\left( \overline{\zeta _1}\zeta _3+\zeta _1\overline{\zeta _2}\zeta _3+\zeta _2\right) z^2+\zeta _3z^3}{1+\left( \overline{\zeta _2}\zeta _3+\overline{\zeta _1}\zeta _2-\zeta _1\right) z+\left( \overline{\zeta _1}\zeta _3-\zeta _1\overline{\zeta _2}\zeta _3-\zeta _2\right) z^2-\zeta _3z^3},\quad z\in \mathbb D. \end{aligned}$$

(13)

We will also use the following lemma.

Lemma 2

[6] Given real numbers A, B, C,  let

$$\begin{aligned} Y(A,B,C):= \max \left\{ |A+Bz+Cz^2|+1-|z|^2: z\in \overline{\mathbb {D}}\right\} . \end{aligned}$$

I. If $AC\ge 0,$ then

$$\begin{aligned} Y(A,B,C)={\left\{ \begin{array}{ll} |A|+|B|+|C|, &{} |B|\ge 2(1-|C|),\\ 1+|A|+\dfrac{B^2}{4(1-|C|)}, &{} |B|<2(1-|C|). \end{array}\right. } \end{aligned}$$

II. If $AC<0,$ then

$$\begin{aligned} \begin{aligned}&Y(A,B,C)\\&\quad ={\left\{ \begin{array}{ll} 1-|A|+\dfrac{B^2}{4(1-|C|)}, &{} -4AC(C^{-2}-1)\le B^2 \wedge |B|<2(1-|C|),\\ 1+|A|+\dfrac{B^2}{4(1+|C|)}, &{} B^2<\min \left\{ 4(1+|C|)^2,-4AC(C^{-2}-1)\right\} ,\\ R(A,B,C), &{} \textrm{otherwise} , \end{array}\right. } \end{aligned} \end{aligned}$$

where

$$\begin{aligned} R(A,B,C):={\left\{ \begin{array}{ll} |A|+|B|-|C|, &{} |C|(|B|+4|A|)\le |AB|, \\ -|A|+|B|+|C|, &{} |AB|\le |C|(|B|-4|A|), \\ (|C|+|A|)\sqrt{1-\dfrac{B^2}{4AC}}, &{} \textrm{otherwise}. \end{array}\right. } \end{aligned}$$

2 Strongly starlike functions

We prove the following sharp inequality for $|H_{2,1}(F_f/2)|$ for the class $\mathcal S^*_\alpha .$

Theorem 1

If $f\in \mathcal S^*_\alpha ,$ $\alpha \in (0,1],$ then

$$\begin{aligned} |H_{2,1}(F_f/2)|=|\gamma _1\gamma _3-\gamma _2^2|\le \frac{1}{4}\alpha ^2. \end{aligned}$$

(14)

The inequality is sharp.

Proof

Fix $\alpha \in (0,1]$ and let $f \in \mathcal S^*_\alpha$ be given by (1). Then by (6),

$$\begin{aligned} zf'(z)=(p(z))^\alpha f(z),\quad z\in \mathbb D, \end{aligned}$$

(15)

for some $p \in \mathcal P$ given by (8). Substituting (1) and (8) into (15) and equating coefficients gives

$$\begin{aligned} \begin{aligned}&a_2 =\alpha c_1, \quad a_3=\frac{\alpha }{4}\left[ 2c_2+(3\alpha -1)c_1^2\right] ,\\&a_4 = \frac{\alpha }{36}\left[ 12c_3+6(5\alpha -2)c_1c_2+(17\alpha ^2-15\alpha +4)c_1^3\right] . \end{aligned} \end{aligned}$$

(16)

Since the class $\mathcal S^*_\alpha$ is invariant under the rotations and (5) holds, we may assume that $a_2\ge 0$, so by (16) that $c_1 \ge 0,$ i.e., in view of (9) that $\zeta _1\in [0,1].$ Hence from (4) and (9)–(11) we obtain

$$\begin{aligned} \begin{aligned}&\gamma _1\gamma _3-\gamma _2^2=\frac{1}{4}\left( a_2a_4-a_3^2+\frac{1}{12}a_2^4\right) \\&\quad =\frac{\alpha ^2}{576}\left[ 48c_1c_3-12(1-\alpha )c_1^2c_2-36c_2^2+(7+\alpha )(1-\alpha )c_1^4\right] \\&\quad =\frac{\alpha ^2}{36}\left[ (4-\alpha ^2)\zeta _1^4+6\alpha (1-\zeta _1^2)\zeta _1^2\zeta _2-3(3+\zeta _1^2)(1-\zeta _1^2)\zeta _2^2\right. \\&\left. \qquad +12(1-\zeta _1^2)(1-|\zeta _2|^2)\zeta _1\zeta _3\right] . \end{aligned} \end{aligned}$$

(17)

A. Suppose that $\zeta _1=1.$ Then by (17), for $\alpha \in (0,1],$

$$\begin{aligned} |\gamma _1\gamma _3-\gamma _2^2|=\frac{\alpha ^2(4-\alpha ^2)}{36}\le \frac{\alpha ^2}{4}. \end{aligned}$$

B. Suppose that $\zeta _1=0.$ Then by (17), for $\alpha \in (0,1],$

$$\begin{aligned} |\gamma _1\gamma _3-\gamma _2^2|=\frac{\alpha ^2}{4}|\zeta _2|^2\le \frac{\alpha ^2}{4}. \end{aligned}$$

C. Suppose that $\zeta _1\in (0,1).$ Then since $|\zeta _3|\le 1$ from (17) we obtain

$$\begin{aligned} \begin{aligned}&|\gamma _1\gamma _3-\gamma _2^2|\\&\quad \le \frac{\alpha ^2}{36}\left[ \left| (4-\alpha ^2)\zeta _1^4+6\alpha (1-\zeta _1^2)\zeta _1^2\zeta _2-3(3+\zeta _1^2)(1-\zeta _1^2)\zeta _2^2\right| \right. \\&\left. \qquad +12(1-\zeta _1^2)(1-|\zeta _2|^2)\zeta _1\right] \\&\quad \le \frac{\alpha ^2}{3}\zeta _1(1-\zeta _1^2)\left[ |A+B\zeta _2+C\zeta _2^2|+1-|\zeta _2|^2\right] , \end{aligned} \end{aligned}$$

(18)

where

$$\begin{aligned} A:=\frac{(4-\alpha ^2)\zeta _1^3}{12(1-\zeta _1^2)},\quad B:=\frac{1}{2}\alpha \zeta _1,\quad C:=-\frac{3+\zeta _1^2}{4\zeta _1}. \end{aligned}$$

Since $AC<0,$ we now apply Lemma 2 only for the case II.

C1. Note that the inequality

$$\begin{aligned} -4AC\left( \frac{1}{C^2}-1\right) -B^2=\frac{(4-\alpha ^2)\zeta _1^2(3+\zeta _1^2)}{12(1-\zeta _1^2)}\left( \frac{16\zeta _1^2}{(3+\zeta _1^2)^2}-1\right) -\frac{\alpha ^2}{4}\zeta _1^2\le 0 \end{aligned}$$

is equivalent to

$$\begin{aligned} -\frac{(4-\alpha ^2)(9-\zeta _1^2)}{3(3+\zeta _1^2)}-\alpha ^2\le 0, \end{aligned}$$

which evidently holds for $\zeta _1\in (0,1).$

However, the inequality $|B|<2(1-|C|)$ is equivalent to $\alpha \zeta _1^2<-(1-\zeta _1^2)(3-\zeta _1^2),$ which is false for $\zeta _1\in (0,1).$

C2. Since

$$\begin{aligned} 4(1+|C|)^2=\frac{(\zeta _1^2+4\zeta _1+3)^2}{4\zeta _1^2}>0 \end{aligned}$$

and

$$\begin{aligned} -4AC\left( \frac{1}{C^2}-1\right) =-\frac{(4-\alpha ^2)\zeta _1^2(9-\zeta _1^2)}{12(3+\zeta _1^2)}<0, \end{aligned}$$

a simple calculation shows that the inequality

$$\begin{aligned} \frac{\alpha ^2\zeta _1^2}{4}=B^2<\min \left\{ 4(1+|C|)^2,-4AC\left( \frac{1}{C^2}-1\right) \right\} =-\frac{(4-\alpha ^2)\zeta _1^2(9-\zeta _1^2)}{12(3+\zeta _1^2)} \end{aligned}$$

is false for $\zeta _1\in (0,1).$

C3. Next note that the inequality

$$\begin{aligned} |C|(|B|+4|A|)-|AB|=\frac{3+\zeta _1^2}{4\zeta _1}\left( \frac{1}{2}\alpha \zeta _1+\frac{(4-\alpha ^2)\zeta _1^3}{3(1-\zeta _1^2)}\right) -\frac{\alpha (4-\alpha ^2)\zeta _1^4}{24(1-\zeta _1^2)}\le 0 \end{aligned}$$

is equivalent to $(\alpha -1)(\alpha ^2-\alpha -8)\zeta _1^4-6(\alpha ^2+\alpha -4)\zeta _1^2+9\alpha \le 0.$ However the last inequality is false for $\zeta _1\in (0,1)$ since $(\alpha -1)(\alpha ^2-\alpha -8)\ge 0$ and $\alpha ^2+\alpha -4<0$ for $\alpha \in (0,1].$

C4. Note that the inequality

$$\begin{aligned} \begin{aligned}&|AB|-|C|(|B|-4|A|)\\&\quad =\frac{\alpha (4-\alpha ^2)\zeta _1^4}{24(1-\zeta _1^2)}-\frac{3+\zeta _1^2}{4\zeta _1}\left( \frac{1}{2}\alpha \zeta _1-\frac{(4-\alpha ^2)\zeta _1^3}{3(1-\zeta _1^2)}\right) \le 0 \end{aligned} \end{aligned}$$

(19)

is equivalent to

$$\begin{aligned} \delta (\zeta _1^2)\ge 0, \end{aligned}$$

(20)

where

$$\begin{aligned} \delta (t):=9\alpha -3(8+2\alpha -2\alpha ^2)t-(8+7\alpha -2\alpha ^2-\alpha ^3)t^2,\quad t\in (0,1). \end{aligned}$$

We see that for $\alpha \in (0,1],$

$$\begin{aligned} 8+2\alpha -2\alpha ^2>0,\quad 8+7\alpha -2\alpha ^2-\alpha ^3>0, \end{aligned}$$

(21)

and the discriminant $\varDelta :=144(4+4\alpha -\alpha ^3)>0$ for $\alpha \in (0,1].$ Thus we consider

$$\begin{aligned} t_{1,2}:=\frac{3(8+2\alpha -2\alpha ^2)\mp 12\sqrt{4+4\alpha -\alpha ^3}}{-2(8+7\alpha -2\alpha ^2-\alpha ^3)}. \end{aligned}$$

From (21) it follows that $t_2<0$ and so it remains to check if $0<t_1<1.$ The inequality $t_1>0$ is equivalent to $8\alpha +7\alpha ^2-2\alpha ^3-\alpha ^4>0$ which is true for $\alpha \in (0,1].$ Further, the inequality $t_1<1$ can be written as

$$256+256\alpha -100\alpha ^2-104\alpha ^3+5\alpha ^4+10\alpha ^5+\alpha ^6>0$$

which is true since

$$\begin{aligned} \begin{aligned}&256+256\alpha -100\alpha ^2-104\alpha ^3+5\alpha ^4+10\alpha ^5+\alpha ^6\\&\quad>52+256\alpha +5\alpha ^4+10\alpha ^5+\alpha ^6>0,\quad \alpha \in (0,1]. \end{aligned} \end{aligned}$$

Therefore (20), and so (19) is valid for $0< \zeta _1\le \zeta ':=\sqrt{t_1}.$ Then by (19), Lemma 2 and the fact that $\varphi$ decreases, we obtain

$$\begin{aligned} \begin{aligned} |\gamma _1\gamma _3-\gamma _2^2|&\le \frac{\alpha ^2}{3}\zeta _1(1-\zeta _1^2)(-|A|+|B|+|C|)\\&=\frac{\alpha ^2}{36}\varphi (\zeta _1)\le \frac{\alpha ^2}{36}\varphi (0)=\frac{\alpha ^2}{4}, \end{aligned} \end{aligned}$$

(22)

where

$$\begin{aligned} \varphi (u):=9-6(1-\alpha )u^2-(1+\alpha )(7-\alpha )u^4,\quad 0\le u\le \zeta '. \end{aligned}$$

C5. It remains to consider the last case in Lemma 2, which in view of C4, holds for $\zeta '<\zeta _1<1.$ Then by (18),

$$\begin{aligned} \begin{aligned} |\gamma _1\gamma _3-\gamma _2^2|&\le \frac{\alpha ^2}{3}\zeta _1(1-\zeta _1^2)(|C|+|A|)\sqrt{1-\dfrac{B^2}{4AC}}\\&=\frac{\alpha ^2}{18}\psi (\zeta _1)\le \frac{\alpha ^2}{18}\psi (\zeta '), \end{aligned} \end{aligned}$$

(23)

where

$$\begin{aligned} \psi (t):=\left[ 9-6t^2+(1-\alpha ^2)t^4\right] \sqrt{\frac{3+(1-\alpha ^2)t^2}{(4-\alpha ^2)(3+t^2)}},\quad \zeta '\le t<1. \end{aligned}$$

To see that the last inequality in (23) is true, note that the function $\psi$ is decreasing, since

$$\begin{aligned} \begin{aligned} \psi '(t)=&-\frac{t}{(4-\alpha ^2)(3+t^2)^2}\sqrt{\frac{(4-\alpha ^2)(3+t^2)}{3+(1-\alpha ^2)t^2}}\\&\times \left[ 4(9-(1-\alpha ^2)^2t^4)(3+t^2)+3\alpha ^2(3-(1-\alpha )t^2)(3-(1+\alpha )t^2)\right] <0 \end{aligned} \end{aligned}$$

for $\zeta '< t<1.$

Simple but tedious computations show that

$$\begin{aligned} \varphi (\zeta ')=\psi (\zeta '). \end{aligned}$$

Hence from (22) and (23) we see that

$$\begin{aligned} \frac{\alpha ^2}{18}\psi (\zeta ')\le \frac{\alpha ^2}{4}. \end{aligned}$$

D. Summarizing from parts A-C we see that inequality (14) follows.

Equality holds for the function $f\in \mathcal A$ given by (15), where

$$\begin{aligned} p(z):=\frac{1+z^2}{1-z^2},\quad z\in \mathbb {D}. \end{aligned}$$

(24)

Then $c_1=c_3=0$ and $c_2=2,$ so by (16), $a_2=a_4=0$ and $a_3=\alpha$, and therefore by (3), $\gamma _1=\gamma _3=0$ and $\gamma _2=\alpha /2,$ which completes the proof of the theorem. $\square$

For $\alpha =1$ we obtain the following result for the class $\mathcal S^*$ of starlike functions [10].

Corollary 1

If $f\in \mathcal S^*,$ then

$$\begin{aligned} |\gamma _1\gamma _3-\gamma _2^2|\le \frac{1}{4}. \end{aligned}$$

The inequality is sharp.

3 Strongly convex functions

We prove the following sharp inequality for $|H_{2,1}(F_f/2)|$ in the class $\mathcal S^c_\alpha .$

Theorem 2

If $f\in \mathcal S^c_\alpha ,$ $\alpha \in (0,1],$ then

$$\begin{aligned} |\gamma _1\gamma _3-\gamma _2^2|\le {\left\{ \begin{array}{ll}\dfrac{\alpha ^2}{36}, &{} 0<\alpha \le \dfrac{1}{3},\\ \dfrac{\alpha ^2(17+18\alpha +13\alpha ^2)}{144(4+6\alpha +\alpha ^2)}, &{} \dfrac{1}{3}<\alpha \le 1. \end{array}\right. } \end{aligned}$$

(25)

Both inequalities are sharp.

Proof

Fix $\alpha \in (0,1]$ and let $f \in \mathcal S^c_\alpha$ be given by (1). Then by (7),

$$\begin{aligned} f'(z)+zf''(z)=f'(z)(p(z))^\alpha ,\quad z\in \mathbb D, \end{aligned}$$

(26)

for some $p \in \mathcal P$ given by (8). Substituting (1) and (8) into (26) and equating coefficients we obatin

$$\begin{aligned} \begin{aligned}&a_2= \frac{1}{2}\alpha c_1, \quad a_3=\frac{\alpha }{12}\left[ 2c_2+(3\alpha -1)c_1^2\right] ,\\&a_4 = \frac{\alpha }{144}\left[ 12c_3+6(5\alpha -2)c_1c_2+(17\alpha ^2-15\alpha +4)c_1^3\right] . \end{aligned} \end{aligned}$$

(27)

As in the proof of Theorem 1 we may assume that $c_1 \ge 0,$ i.e., in view of (9) that $\zeta _1\in [0,1].$ Hence from (4) and (9)–(11) we have

$$\begin{aligned} \begin{aligned} \gamma _1\gamma _3-\gamma _2^2&=\frac{\alpha ^2}{2304}\left[ 24c_1c_3+4(3\alpha -2)c_1^2c_2-16c_2^2+(\alpha ^2-6\alpha +4)c_1^4\right] \\&=\frac{\alpha ^2}{144}\left[ (2+\alpha ^2)\zeta _1^4+6\alpha (1-\zeta _1^2)\zeta _1^2\zeta _2-2(1-\zeta _1^2)(2+\zeta _1^2)\zeta _2^2\right. \\&\quad \left. +6(1-\zeta _1^2)(1-|\zeta _2|^2)\zeta _1\zeta _3\right] . \end{aligned} \end{aligned}$$

(28)

A. Suppose that $\zeta _1=1.$ Then by (28), for $\alpha \in (0,1],$

$$\begin{aligned} |\gamma _1\gamma _3-\gamma _2^2|=\frac{\alpha ^2(2+\alpha ^2)}{144}. \end{aligned}$$

(29)

B. Suppose that $\zeta _1=0.$ Then from (28), for $\alpha \in (0,1],$

$$\begin{aligned} |\gamma _1\gamma _3-\gamma _2^2|=\frac{\alpha ^2}{36}|\zeta _2|^2\le \frac{\alpha ^2}{36}. \end{aligned}$$

(30)

C. Suppose that $\zeta _1\in (0,1).$ Since $|\zeta _3|\le 1$ from (28) we obtain

$$\begin{aligned} \begin{aligned}&|\gamma _1\gamma _3-\gamma _2^2|\\&\quad \le \frac{\alpha ^2}{144}\left[ \left| (2+\alpha ^2)\zeta _1^4+6\alpha (1-\zeta _1^2)\zeta _1^2\zeta _2-2(1-\zeta _1^2)(2+\zeta _1^2)\zeta _2^2\right| \right. \\&\qquad +\left. 6(1-\zeta _1^2)(1-|\zeta _2|^2)\zeta _1\right] \\&\quad =\frac{\alpha ^2}{24}\zeta _1(1-\zeta _1^2)\left[ |A+B\zeta _2+C\zeta _2^2|+1-|\zeta _2|^2\right] , \end{aligned} \end{aligned}$$

(31)

where

$$\begin{aligned} A:=\frac{(2+\alpha ^2)\zeta _1^3}{6(1-\zeta _1^2)},\quad B:=\alpha \zeta _1,\quad C:=-\frac{2+\zeta _1^2}{3\zeta _1}. \end{aligned}$$

Since $AC<0,$ we apply Lemma 2 only in the case II.

C1. Note that the inequality

$$\begin{aligned} -4AC\left( \frac{1}{C^2}-1\right) -B^2=\frac{2(2+\alpha ^2)\zeta _1^2(2+\zeta _1^2)}{9(1-\zeta _1^2)}\left( \frac{9\zeta _1^2}{(2+\zeta _1^2)^2}-1\right) -\alpha ^2\zeta _1^2\le 0 \end{aligned}$$

is equivalent to $-2(2+\alpha ^2)(4-\zeta _1^2)\le 9\alpha ^2(2+\zeta _1^2),$ which evidently holds for $\zeta _1\in (0,1).$

Moreover, the inequality $|B|<2(1-|C|)$ is equivalent to $3\alpha \zeta _1^2< -2(1-\zeta _1)(2-\zeta _1),$ which is false for $\zeta _1\in (0,1).$

C2. Since

$$\begin{aligned} 4(1+|C|)^2=\frac{4(\zeta _1^2+3\zeta _1+2)^2}{9\zeta _1^2}>0 \end{aligned}$$

and

$$\begin{aligned} -4AC\left( \frac{1}{C^2}-1\right) =-\frac{2(2+\alpha ^2)\zeta _1^2(4-\zeta _1^2)}{9(2+\zeta _1^2)}<0, \end{aligned}$$

we see that the inequality

$$\begin{aligned} \alpha ^2\zeta _1^2=B^2<\min \left\{ 4(1+|C|)^2,-4AC\left( \frac{1}{C^2}-1\right) \right\} =-\frac{2(2+\alpha ^2)\zeta _1^2(4-\zeta _1^2)}{9(2+\zeta _1^2)} \end{aligned}$$

is false for $\zeta _1\in (0,1).$

C3. Next observe that the inequality

$$\begin{aligned} |C|(|B|+4|A|)-|AB|=\frac{2+\zeta _1^2}{3\zeta _1}\left( \alpha \zeta _1+\frac{2(2+\alpha ^2)\zeta _1^3}{3(1-\zeta _1^2)}\right) -\frac{(2+\alpha ^2)\alpha \zeta _1^4}{6(1-\zeta _1^2)}\le 0 \end{aligned}$$

is equivalent to

$$\begin{aligned} \phi (\zeta _1^2)\le 0, \end{aligned}$$

(32)

where

$$\begin{aligned} \phi (t):=(-3\alpha ^3+4\alpha ^2-12\alpha +8)t^2+(8\alpha ^2-6\alpha +16)t+12\alpha ,\quad t\in (0,1). \end{aligned}$$

Note that $8\alpha ^2-6\alpha +16>0$ for $\alpha \in (0,1]$ and $-3\alpha ^3+4\alpha ^2-12\alpha +8\ge 0$ for $\alpha \in (0,\alpha _0],$ where $\alpha _0\approx 0.74858\dots .$ Thus for $\alpha \in (0,\alpha _0]$ inequality (32) is evidently false. If $\alpha \in (\alpha _0,1]$, then $\varDelta :=4\left( 52\alpha ^4-72\alpha ^3+217\alpha ^2-144\alpha +64\right) >0,$ and so we consider

$$\begin{aligned} t_{1,2}:=\frac{-4\alpha ^2+3\alpha -8\mp \sqrt{52\alpha ^4-72\alpha ^3+217\alpha ^2-144\alpha +64}}{-3\alpha ^3+4\alpha ^2-12\alpha +8}. \end{aligned}$$

Observe now that $t_1>1.$ Indeed, the inequality $t_1>1$ is equivalent to the evidently true inequality

$$\begin{aligned} \sqrt{52\alpha ^4-72\alpha ^3+217\alpha ^2-144\alpha +64}>3\alpha ^3-8\alpha ^2+15\alpha -16, \end{aligned}$$

since the right hand side is negative for all $\alpha \in (\alpha _0,1].$ Further, $t_2<0.$ Indeed this inequality is equivalent to $-3\alpha ^3+4\alpha ^2-12\alpha +8<0$ which clearly holds for $\alpha \in (\alpha _0,1].$ Thus we deduce that the inequality (32) is false.

C4. Note next that the inequality

$$\begin{aligned} |AB|-|C|(|B|-4|A|)=\frac{(2+\alpha ^2)\alpha \zeta _1^4}{6(1-\zeta _1^2)}-\frac{2+\zeta _1^2}{3\zeta _1}\left( \alpha \zeta _1-\frac{2(2+\alpha ^2)\zeta _1^3}{3(1-\zeta _1^2)}\right) \le 0 \end{aligned}$$

(33)

is equivalent to

$$\begin{aligned} \delta (\zeta _1^2)\le 0, \end{aligned}$$

(34)

where

$$\begin{aligned} \delta (s):=(3\alpha ^3+4\alpha ^2+12\alpha +8)s^2+2(4\alpha ^2+3\alpha +8)s-12\alpha ,\quad s\in (0,1), \end{aligned}$$

so that $\varDelta :=4\left( 52\alpha ^4+72\alpha ^3+217\alpha ^2+144\alpha +64\right) >0$ for $\alpha \in (0,1].$ Therefore $s_1<0,$ where

$$\begin{aligned} s_{1,2}:=\frac{-(4\alpha ^2+3\alpha +8)\mp \sqrt{52\alpha ^4+72\alpha ^3+217\alpha ^2+144\alpha +64}}{3\alpha ^3+4\alpha ^2+12\alpha +8}. \end{aligned}$$

Moreover $0<s_2<1$ holds. Indeed, both inequalities $s_2>0$ and $s_2<1$ are equivalent to the evidently true inequalities

$$\begin{aligned} 36\alpha ^4+48\alpha ^3+144\alpha ^2+96\alpha >0, \end{aligned}$$

and

$$\begin{aligned} 9\alpha ^6+48\alpha ^5+102\alpha ^4+264\alpha ^3+264\alpha ^2+336\alpha +192>0, \end{aligned}$$

respectively. Thus (34), and so (33) is valid only when

$$\begin{aligned} 0<\zeta _1\le \sqrt{s_2}=:\zeta '. \end{aligned}$$

Then by (31) and Lemma 2,

$$\begin{aligned} |\gamma _1\gamma _3-\gamma _2^2|\le \frac{1}{24}\alpha ^2\zeta _1(1-\zeta _1^2)(-|A|+|B|+|C|)=\varphi (\zeta _1), \end{aligned}$$

where

$$\begin{aligned} \varphi (u):=\frac{\alpha ^2}{144}\left[ -(\alpha ^2+6\alpha +4)u^4+2(3\alpha -1)u^2+4\right] ,\quad 0\le u\le \zeta '. \end{aligned}$$

Since

$$\begin{aligned} \varphi '(u)=-\frac{\alpha ^2u}{36}\left[ (\alpha ^2+6\alpha +4)u^2+1-3\alpha \right] ,\quad 0<u<\zeta ', \end{aligned}$$

we see that for $0<\alpha \le 1/3,$ the function $\varphi$ decreases and so

$$\begin{aligned} \varphi (u)\le \varphi (0)=\frac{\alpha ^2}{36},\quad 0\le u\le \zeta '. \end{aligned}$$

(35)

In the case $1/3<\alpha \le 1,$

$$\begin{aligned} 0<u_0:=\sqrt{\frac{3\alpha -1}{\alpha ^2+6\alpha +4}}<\zeta _1 \end{aligned}$$

(36)

is a unique critical point of $\varphi$, which is a maximum.

It remains therefore to establish the second inequality, i.e., $u_0<\zeta _1,$ which is equivalent to

$$\begin{aligned} \begin{aligned} r(\alpha ):=&117\alpha ^8+240\alpha ^7-149\alpha ^6-1212\alpha ^5-4344\alpha ^4\\&-6288\alpha ^3-4464\alpha ^2-1920\alpha -448<0,\quad \alpha \in (0,1], \end{aligned} \end{aligned}$$

and since

$$\begin{aligned} r(\alpha )\le -149\alpha ^6-1212\alpha ^5-4344\alpha ^4-6288\alpha ^3-4464\alpha ^2-1920\alpha -91<0 \end{aligned}$$

for $\alpha \in (0,1],$ we deduce that $u_0<\zeta _1.$

Thus for $1/3<\alpha \le 1,$ we have

$$\begin{aligned} \varphi (u)\le \varphi (u_0)=\dfrac{\alpha ^2(17+18\alpha +13\alpha ^2)}{144(4+6\alpha +\alpha ^2)},\quad 0\le u\le \zeta '. \end{aligned}$$

(37)

C5. We now consider the last case in Lemma 2, which in view of C4 holds for $\zeta '<\zeta _1<1.$ Then by (31),

$$\begin{aligned} |\gamma _1\gamma _3-\gamma _2^2|\le \frac{\alpha ^2}{24}\zeta _1(1-\zeta _1^2)(|C|+|A|)\sqrt{1-\dfrac{B^2}{4AC}}=\psi (\zeta _1)\le \psi (\zeta '), \end{aligned}$$

(38)

where

$$\begin{aligned} \psi (u):=\frac{\alpha ^2}{144}(\alpha ^2u^4-2u^2+4)\sqrt{\frac{13\alpha ^2+8+(4-7\alpha ^2)u^2}{2(2+\alpha ^2)(2+u^2)}},\quad \zeta '\le u\le 1. \end{aligned}$$

To show that the last inequality in (38) holds, observe that $\psi$ is decreasing. Indeed, by a simple computation,

$$\begin{aligned} \begin{aligned} \psi '(u)&=-\frac{\alpha ^2x}{288(2+\alpha ^2)(2+x^2)^2}\sqrt{\frac{2(2+\alpha ^2)(2+u^2)}{13\alpha ^2+8+(4-7\alpha ^2)u^2}}\\&\quad \times \left[ 4(1-\alpha ^2u^2)(2+u^2)\left( 13\alpha ^2+8+(4-7\alpha ^2)u^2\right) \right. \\&\quad \left. +27\alpha ^2(\alpha ^2u^4-2u^2+4)\right] , \end{aligned} \end{aligned}$$

for $\zeta '<u<1.$ Note that

$$\begin{aligned} 13\alpha ^2+8+(4-7\alpha ^2)u^2>0,\quad \zeta '<u<1, \end{aligned}$$

(39)

which is clearly true for $0<\alpha \le 2/\sqrt{7}.$ If $2/\sqrt{7}<\alpha \le 1,$ then

$$\begin{aligned} 13\alpha ^2+8+(4-7\alpha ^2)u^2=13\alpha ^2+8-(7\alpha ^2-4)u^2\ge 6\alpha ^2+12>0 \end{aligned}$$

for $\zeta '<u<1.$ Further

$$\begin{aligned} \alpha ^2u^4-2u^2+4\ge \alpha ^2u^4+2>0,\quad \zeta '<u<1. \end{aligned}$$

(40)

Thus from (39) and (40) it follows that $\psi '(u)<0$ for $\zeta '<u<1,$ so $\psi$ decreases and hence

$$\begin{aligned} \psi (u)\le \psi (\zeta '),\quad \zeta '\le u\le 1. \end{aligned}$$

(41)

Simple but tedious computations show that

$$\begin{aligned} \varphi (\zeta ')=\psi (\zeta '), \end{aligned}$$

and so from (41), (35) and (37) we deduce that for $\alpha \in (0,1/3],$

$$\begin{aligned} \psi (u)\le \frac{\alpha ^2}{36},\quad \zeta '\le u\le 1, \end{aligned}$$

and for $\alpha \in (1/3,1],$

$$\begin{aligned} \psi (u)\le \varphi (u_0),\quad \zeta '\le u\le 1. \end{aligned}$$

D. It remains to compare the bounds in (29), (30), (35) and (37). The inequality

$$\begin{aligned} \frac{\alpha ^2(2+\alpha ^2)}{144}\le \frac{\alpha ^2}{36},\quad \alpha \in (0,1], \end{aligned}$$

is trivial, and the inequality

$$\begin{aligned} \frac{\alpha ^2(2+\alpha ^2)}{144}\le \dfrac{\alpha ^2(17+18\alpha +13\alpha ^2)}{144(4+6\alpha +\alpha ^2)},\quad \alpha \in (1/3,1], \end{aligned}$$

is equivalent to

$$\begin{aligned} -\alpha ^4-6\alpha ^3+7\alpha ^2+6\alpha +9\le 0,\quad \alpha \in (1/3,1], \end{aligned}$$

which is clearly true, and the inequality

$$\begin{aligned} \frac{\alpha ^2}{36}\le \dfrac{\alpha ^2(17+18\alpha +13\alpha ^2)}{144(4+6\alpha +\alpha ^2)},\quad \alpha \in (1/3,1], \end{aligned}$$

is equivalent to the evidently true inequality $(3\alpha -1)^2\ge 0.$

Thus summarizing the results in parts A-C we see that (25) is established.

We finally show that the inequalities in (25) are sharp. When $\alpha \in (0,1/3],$ equality holds for the function $f\in \mathcal A$ given by (26) with p given by (24). In this case $c_1=c_3=0$ and $c_2=2,$ so by (27), $a_2=a_4=0$ and $a_3=\alpha /3$ and therefore $\gamma _1=\gamma _3=0$ and $\gamma _2=\alpha /6.$

When $\alpha \in (1/3,1],$ equality holds for the function $f\in \mathcal A$ given by (26), where p is given by (12) with $\zeta _1=u_0=:\tau ,$ and $u_0$ given by (36), $\zeta _2=-1$ and $\zeta _3=1,$ i.e.,

$$\begin{aligned} p(z):=\frac{1-z^2}{1-2\tau z+z^2},\quad z\in \mathbb D, \end{aligned}$$

which completes the proof of the theorem. $\square$

For $\alpha =1$ we obtain the sharp inequality for the class $\mathcal S^c$ of convex functions [10].

Corollary 2

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

$$\begin{aligned} |\gamma _1\gamma _3-\gamma _2^2|\le \frac{1}{33}. \end{aligned}$$

The inequality is sharp.