Second Hankel Determinant of Logarithmic Coefficients of Convex and Starlike Functions of Order Alpha

Abstract

In the present paper, we found sharp bounds of the second Hankel determinant of logarithmic coefficients of starlike and convex functions of order $\alpha$.

1 Introduction

Let ${\mathcal {H}}$ be the class of analytic functions in $\mathbb {D}:= \left\{ z \in \mathbb {C}: |z|<1 \right\}$ of the form

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

(1.1)

and ${\mathcal {A}}$ be its subclass of all f normalized by $f'(0)=1.$ By $\mathcal {S}$ we denote the subclass of ${\mathcal {A}}$ of univalent functions.

For $f\in \mathcal {S}$ let

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

(1.2)

The numbers $\gamma _n:=\gamma _n(f)$ are called logarithmic coefficients of f. It is well known that the logarithmic coefficients play a crucial role in Milin conjecture ([16], see also [8, p. 155]). It is surprising that for the whole class $\mathcal {S}$ the sharp estimates of single logarithmic coefficients are known only for $\gamma _1$ and $\gamma _2,$ namely

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

and are unknown for $n\ge 3.$ Logarithmic coefficients is one of the topic recently being of the research interest by various authors (e.g., [1, 2, 6, 9, 13, 20]).

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

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

Recently, many authors examined the second and the third Hankel determinants $H_{2,2}(f)$ and $H_{3,1}(f)$ over selected subclasses of ${\mathcal {A}},$ particularly of $\mathcal {S}$ (see e.g., [4, 11] for further references).

Based on the ideas mentioned above, in [12] was begun the research study of the Hankel determinant $H_{q,n}(F_f/2)$ which entries are logarithmic coefficients of f,  i.e.,

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

Due to the great importance of logarithmic coefficients, the proposed topic seems reasonable and interesting. The results which can be obtained on Hankel determinants $H_{q,n}(F_f/2)$ broaden the knowledge of logarithmic coefficients. In this paper, we continue this research dealing with $H_{2,1}(F_f/2)=\gamma _1\gamma _3-\gamma _2^2$ for the classes of starlike and convex functions of order $\alpha .$ Recall that $H_{2,1}(F_f/2)$ corresponds to the well-known functional $H_{2,1}(f)=a_3-a_2^2$ over the class $\mathcal {S}$ or its subclasses. For the class ${\mathcal {S}}$ the functional $H_{2,1}(f)$ was estimated in 1916 by Bieberbach (see e.g., [10, Vol. I, p. 35]).

Differentiating (1.2) and using (1.1) we get

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

(1.3)

Therefore,

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

(1.4)

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

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

(1.5)

The main goal of this paper is to find sharp upper bounds for $H_{2,1}(F_f/2)$ in case when f is a starlike or convex function of order $\alpha .$ Given $\alpha \in [0,1),$ a function $f\in {\mathcal {A}}$ is called starlike of order $\alpha$ if

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

(1.6)

Further, a function $f\in {\mathcal {A}}$ is called convex of order $\alpha$ if

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

(1.7)

Both classes usually denoted as $\mathcal {S}^*(\alpha )$ and $\mathcal {S}^c(\alpha ),$ respectively, were introduced by Robertson [19] (e.g., [10, Vol. I, p. 138]). The classes $\mathcal {S}^*(0)=:\mathcal {S}^*$ and $\mathcal {S}^c(0)=:\mathcal {S}^c$ consist of starlike and convex functions, respectively. Let us mention that in the class $\mathcal {S}^*(\alpha ),$ $|\gamma _n|\le (1-\alpha )/n$ and in the class $\mathcal {S}^c(\alpha ),$ $|\gamma _n|\le (1-\eta (\alpha ))/n$ for $n\in \mathbb {N},$ with sharpness (cf. [21, p. 263]), where

$$\begin{aligned} \eta (\alpha ):={\left\{ \begin{array}{ll} \dfrac{1-2\alpha }{4^{1-\alpha }(1-2^{2\alpha -1})},&{} \alpha \not =1/2,\\ \dfrac{1}{2\log 2},&{} \alpha =1/2. \end{array}\right. } \end{aligned}$$

In particular, for $n\in {\mathbb {N}},$ $|\gamma _n|\le 1/n$ in the class $\mathcal {S}^*$ of starlike functions and $|\gamma _n|\le 1/(2n)$ in the class of convex functions.

In view of (1.6) and (1.7) both classes ${\mathcal {S}}^*(\alpha )$ and ${\mathcal {S}}^c(\alpha )$ have a representation with using the Carathéodory class ${\mathcal {P}}$, i.e., the class of analytic functions p in $\mathbb {D}$ of the form

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

(1.8)

having a positive real part in $\mathbb {D}.$ Therefore, the coefficients of functions in ${\mathcal {S}}^*(\alpha )$ and ${\mathcal {S}}^c(\alpha )$ have a suitable representation expressed by coefficients of functions in ${\mathcal {P}}.$ Thus, to get the upper bound of $H_{2,1}(F_f/2),$ we based our computing on the well-known formulas on coefficient $c_2$ (e.g., [18, p. 166]) and the formula $c_3$ due to Libera and Zlotkiewicz [14, 15]; cf. [17, Proposition 6]. Further remarks related to extremal functions see [5].

Let $\mathbb {T}:=\{z\in \mathbb {C}:|z|=1\}.$

Lemma 1.1

If $p \in {{\mathcal {P}}}$ is of form (1.6) with $c_1\ge 0,$ then

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

(1.9)

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

(1.10)

and

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

(1.11)

for some $\zeta _1\in [0,1]$ and $\zeta _2,\zeta _3\in \overline{\mathbb {D}}:=\{z\in \mathbb {C}:|z|<1\}.$

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 (1.9)–(1.10), namely

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

(1.12)

We will also apply the following lemma.

Lemma 1.2

(Choi et al. [7]) 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

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

• II. If $AC<0,$ then

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

  where

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

2 Starlike functions of order $\alpha$

We now discuss $H_{2,1}(F_f/2)$ for the class $\mathcal {S}^*(\alpha ).$

Theorem 2.1

Let $\alpha \in [0,1).$ If $f\in \mathcal {S}^*(\alpha ),$ then

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

(2.1)

The inequality is sharp.

Proof

Fix $\alpha \in [0,1)$ and let $f \in {\mathcal {S}}^*(\alpha )$ be of form (1.1). Then, by (1.6),

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

(2.2)

for some $p \in {\mathcal {P}}$ of form (1.8). Since the class ${\mathcal {P}}$ is invariant under the rotations and (1.5) holds, we may assume that $c_1 \in [0,2]$ ([3], see also [10, Vol. I, p. 80, Theorem 3]), i.e., in view of (1.9) that $\zeta _1\in [0,1].$ Substituting series (1.1) and (1.8) into (2.2) and equating coefficients we get

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

Hence by using (1.4) and (1.9)–(1.11) we obtain

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

(2.3)

• A. Suppose that $\zeta _1=1.$ Then, by (2.3),

  $$\begin{aligned} |\gamma _1\gamma _3-\gamma _2^2|=\frac{1}{12}(1-\alpha )^2. \end{aligned}$$

• B. Suppose that $\zeta _1=0.$ Then, by (2.3),

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

• C. Suppose that $\zeta _1\in (0,1).$ By the fact that $|\zeta _3|\le 1$ from (2.3) we obtain

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

  (2.4)

  where

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

  Since $AC<0,$ we apply Lemma 1.2 only for the case II.

  • C1. Note that the inequality

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

    is equivalent to $-9(1-\zeta _1^2)\le 3(1-\zeta _1^2),$ which evidently holds for $\zeta _1\in (0,1).$ Moreover, the inequality $|B|<2(1-|C|)$ is equivalent to $2\zeta _1^2-4\zeta _1+3<0,$ which is false for $\zeta _1\in (0,1).$

  • C2. Since

    $$\begin{aligned} 4(1+|C|)^2=4\frac{(\zeta _1^2+4\zeta _1+3)^2}{16\zeta _1^2}>0,\quad -4AC\left( \frac{1}{C^2}-1\right) =-\frac{\zeta _1^2(9-\zeta _1^2)}{4(3+\zeta _1^2)}<0, \end{aligned}$$

    we see that the inequality

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

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

  • C3. Observe that the inequality

    $$\begin{aligned} |C|(|B|+4|A|)-|AB|=\frac{3+\zeta _1^2}{4\zeta _1}\left( \frac{1}{2}\zeta _1+\frac{\zeta _1^3}{1-\zeta _1^2}\right) -\frac{\zeta _1^4}{8(1-\zeta _1^2)}\le 0 \end{aligned}$$

    is equivalent to $3+4\zeta _1^2\le 0,$ which is false for $\zeta _1\in (0,1).$

  • C4. Note that the inequality

    $$\begin{aligned} |AB|-|C|(|B|-4|A|)=\frac{\zeta _1^4}{8(1-\zeta _1^2)}-\frac{3+\zeta _1^2}{4\zeta _1}\left( \frac{1}{2}\zeta _1-\frac{\zeta _1^3}{1-\zeta _1^2}\right) \le 0 \end{aligned}$$

    is equivalent to $4\zeta _1^4+8\zeta _1^2-3\le 0,$ which is true for $0< \zeta _1\le \zeta ':=\sqrt{\sqrt{7}/2-1}\approx 0.5682.$ Then, by (2.3) and Lemma 1.2,

    $$\begin{aligned} \begin{aligned} |\gamma _1\gamma _3-\gamma _2^2|&\le \frac{1}{3}(1-\alpha )^2\zeta _1(1-\zeta _1^2)(-|A|+|B|+|C|)\\&=\frac{1}{12}(1-\alpha )^2(3-4\zeta _1^4)\le \frac{1}{4}(1-\alpha )^2, \end{aligned} \end{aligned}$$

    (2.5)

    for $0<\zeta _1\le \zeta '.$

  • C5. It remains to consider the last case in Lemma 1.2, which taking into account C4 holds for $\zeta '<\zeta _1<1.$ Then, by (2.3),

    $$\begin{aligned} |\gamma _1\gamma _3-\gamma _2^2|&\le \frac{1}{3}(1-\alpha )^2\zeta _1(1-\zeta _1^2)(|C|+|A|)\sqrt{1-\dfrac{B^2}{4AC}}\nonumber \\&=(1-\alpha )^2\psi (\zeta _1)\le (1-\alpha )^2\psi (\zeta ')=(1-\alpha )^2\frac{5-\sqrt{7}}{3\sqrt{8+2\sqrt{7}}}, \end{aligned}$$

    (2.6)

    where

    $$\begin{aligned} \psi (t):=\frac{3-2t^2}{6\sqrt{3+t^2}},\quad \zeta '\le t\le 1. \end{aligned}$$

    Indeed, to see that the last inequality in (2.6) is true, observe that since

    $$\begin{aligned} \psi '(t)=-\frac{15t+2t^3}{6(3+t^2)^{3/2}}<0,\quad \zeta '< t<1, \end{aligned}$$

    the function $\psi$ decrease which yields $\psi (t)\le \psi (\zeta ')\approx 0.21525$ for $\zeta '< t<1.$

• D. Summarizing from parts A–C it follows inequality (2.1). Equality holds for the function $f\in {\mathcal {A}}$ given by

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

  for which $a_2=a_4=0$ and $a_3=1-\alpha .$ $\square$

For $\alpha =0$ we get the estimate for the class $\mathcal {S}^*$ of starlike functions [12].

Corollary 2.2

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 Convex functions of order $\alpha$

Now we deal with $H_{2,1}(F_f/2)$ for the class ${\mathcal {S}}^c(\alpha ).$

Theorem 3.1

Let $\alpha \in [0,1).$ If $f\in \mathcal {S}^c(\alpha ),$ then

$$\begin{aligned} |\gamma _1\gamma _3-\gamma _2^2|\le \frac{(1-\alpha )^2(\alpha ^2+4\alpha -16)}{48(\alpha ^2+2\alpha -11)}. \end{aligned}$$

(3.1)

The inequality is sharp.

Proof

Fix $\alpha \in [0,1)$ and let $f \in {\mathcal {S}}^c(\alpha )$ be of form (1.1). Then, by (1.7),

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

(3.2)

for some $p \in {\mathcal {P}}$ of form (1.8). As in the proof of Theorem 2.1 we may assume that $c_1 \in [0,2],$ i.e., in view of (1.9) that $\zeta _1\in [0,1].$ Substituting series (1.1) and (1.8) into (3.2) and equating coefficients we get

$$\begin{aligned} \begin{aligned}&a_2= \frac{1}{2}(1-\alpha )c_1, \quad a_3=\frac{1}{6}(1-\alpha )\left( c_2+(1-\alpha )c_1^2\right) ,\\&a_4 = \frac{1}{24}(1-\alpha )\left[ 2c_3+3(1-\alpha )c_1c_2+(1-\alpha )^2c_1^3\right] . \end{aligned} \end{aligned}$$

Hence by using (1.4) and (1.9)–(1.11) we obtain

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

(3.3)

• A. Suppose that $\zeta _1=1.$ Then, by (3.3),

  $$\begin{aligned} |\gamma _1\gamma _3-\gamma _2^2|=\frac{1}{144}(1-\alpha )^2(3-\alpha ^2). \end{aligned}$$

• B. Suppose that $\zeta _1=0.$ Then, by (3.3),

  $$\begin{aligned} |\gamma _1\gamma _3-\gamma _2^2|=\frac{1}{36}(1-\alpha )^2|\zeta _2|^2\le \frac{1}{36}(1-\alpha )^2. \end{aligned}$$

• C. Suppose that $\zeta _1\in (0,1).$ By the fact that $|\zeta _3|\le 1$ from (3.3) we obtain

  $$\begin{aligned} \begin{aligned}&|\gamma _1\gamma _3-\gamma _2^2|\\&\quad \le \frac{(1-\alpha )^2}{144}\left[ \left| (3-\alpha ^2)\zeta _1^4+(6-2\alpha )(1-\zeta _1^2)\zeta _1^2\zeta _2-2(1-\zeta _1^2)(2+\zeta _1^2)\zeta _2^2\right| \right. \\&\qquad + \left. 6(1-\zeta _1^2)(1-|\zeta _2|^2)\zeta _1\right] \\&\quad =\frac{1}{24}(1-\alpha )^2\zeta _1(1-\zeta _1^2)\left[ |A+B\zeta _2+C\zeta _2^2|+1-|\zeta _2|^2\right] , \end{aligned} \end{aligned}$$

  (3.4)

  where

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

  Since $AC<0,$ we apply Lemma 1.2 only for the case II.

  • C1. Note that the inequality

    $$\begin{aligned} \begin{aligned}&-4AC\left( \frac{1}{C^2}-1\right) -B^2\\&\quad =\frac{2(3-\alpha ^2)(2+\zeta _1^2)\zeta _1^2}{9(1-\zeta _1^2)}\left( \frac{9\zeta _1^2}{(2+\zeta _1^2)^2}-1\right) -\frac{(3-\alpha )^2}{9}\zeta _1^2\le 0 \end{aligned} \end{aligned}$$

    is equivalent to $-2(3-\alpha ^2)(4-\zeta _1^2)\le (3-\alpha )^2(2+\zeta _1^2),$ which holds for $\zeta _1\in (0,1).$ Moreover, the inequality $|B|<2(1-|C|)$ is equivalent to $(3-\alpha )\zeta _1^2\le -2(1-\zeta _1)(2-\zeta _1),$ which is false for $\zeta _1\in (0,1).$

  • C2. Since

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

    and

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

    we see that the inequality

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

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

  • C3. Observe that the inequality

    $$\begin{aligned} \begin{aligned}&|C|(|B|+4|A|)-|AB|\\&\quad =\frac{2+\zeta _1^2}{3\zeta _1}\left( \frac{3-\alpha }{3}\zeta _1+\frac{2(3-\alpha ^2)\zeta _1^3}{3(1-\zeta _1^2)}\right) -\frac{(3-\alpha ^2)(3-\alpha )\zeta _1^4}{18(1-\zeta _1^2)}\le 0 \end{aligned} \end{aligned}$$

    is equivalent to

    $$\begin{aligned} (1-\alpha )^2(\alpha +3)t^2+2(4\alpha ^2-\alpha -9)t-4(3-\alpha )\ge 0, \end{aligned}$$

    (3.5)

    where $t:=\zeta _1^2\in (0,1).$ Note that $\Delta :=4\left( 12\alpha ^4-39\alpha ^2-54\alpha +117\right) >0$ for $\alpha \in [0,1).$ Consider now

    $$\begin{aligned} t_{1,2}:=\frac{-4\alpha ^2+\alpha +9\mp \sqrt{12\alpha ^4-39\alpha ^2-54\alpha +117}}{(1-\alpha )^2(3+\alpha )}. \end{aligned}$$

    Observe first that $t_1<0.$ Indeed, since $-4\alpha ^2+\alpha +9>0$ for $\alpha \in [0,1),$ the inequality $t_1<0$ is equivalent to

    $$\begin{aligned} \sqrt{12\alpha ^4-39\alpha ^2-54\alpha +117}>-4\alpha ^2+\alpha +9, \end{aligned}$$

    which is equivalent to the true inequality

    $$\begin{aligned} 4\alpha ^4-8\alpha ^3-32\alpha ^2+72\alpha -36<0,\quad \alpha \in [0,1). \end{aligned}$$

    Further, the inequality $t_2>1$ is equivalent to the true inequality

    $$\begin{aligned} 12\alpha ^4-\alpha ^3-44\alpha ^2-48\alpha +123>0,\quad \alpha \in [0,1). \end{aligned}$$

    Thus, we state that inequality (3.5) is false.

  • C4. Note that the inequality

    $$\begin{aligned} \begin{aligned}&|AB|-|C|(|B|-4|A|)\\&\quad =\frac{(3-\alpha ^2)(3-\alpha )\zeta _1^4}{18(1-\zeta _1^2)}-\frac{2+\zeta _1^2}{3\zeta _1}\left( \frac{3-\alpha }{3}\zeta _1-\frac{2(3-\alpha ^2)\zeta _1^3}{3(1-\zeta _1^2)}\right) \le 0 \end{aligned} \end{aligned}$$

    is equivalent to

    $$\begin{aligned}&(\alpha ^3-7\alpha ^2-5\alpha +27)t^2+(-8\alpha ^2-2\alpha +30)t\nonumber \\&\quad +4\alpha -12\le 0,\quad \alpha \in [0,1), \end{aligned}$$

    (3.6)

    where $t:{=}\zeta _1^2\in (0,1).$ We have $\Delta :{=}4\left( 12\alpha ^4{+}48\alpha ^3{-}183\alpha ^2{-}198\alpha {+}549\right) >0$ for $\alpha \in [0,1).$ Since $4\alpha ^2+\alpha -15<0$ and $\alpha ^3-7\alpha ^2-5\alpha +27>0$ for $\alpha \in [0,1),$ so $s_1<0,$ where

    $$\begin{aligned} s_{1,2}:=\frac{4\alpha ^2+\alpha -15\mp \sqrt{12\alpha ^4+48\alpha ^3-183\alpha ^2-198\alpha +549}}{\alpha ^3-7\alpha ^2-5\alpha +27}. \end{aligned}$$

    Further, the condition $s_2>0$ is equivalent to

    $$\begin{aligned} 12\alpha ^4+48\alpha ^3-183\alpha ^2-198\alpha +549>(-4\alpha ^2-\alpha +15)^2, \end{aligned}$$

    which is equivalent to the true inequality

    $$\begin{aligned} \alpha ^4-10\alpha ^3+16\alpha ^2+42\alpha -81<0,\quad \alpha \in [0,1). \end{aligned}$$

    Moreover, the inequality $s_2<1$ is equivalent to the inequality

    $$\begin{aligned} -\alpha ^6+22\alpha ^5-97\alpha ^4-168\alpha ^3+705\alpha ^2+306\alpha -1215<0 \end{aligned}$$

    which is valid for $\alpha \in [0,1).$ Thus, inequality (3.6) holds only when

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

    Then, by (3.4) and Lemma 1.2,

    $$\begin{aligned} \begin{aligned} |\gamma _1\gamma _3-\gamma _2^2|&\le \frac{(1-\alpha )^2}{24}\zeta _1(1-\zeta _1^2)(-|A|+|B|+|C|)=\varphi (\zeta _1)\le \varphi (u_0)\\&=\frac{(1-\alpha )^2(\alpha ^2+4\alpha -16)}{48(\alpha ^2+2\alpha -11)}, \end{aligned} \end{aligned}$$

    (3.7)

    where

    $$\begin{aligned} \varphi (u):=\frac{(1-\alpha )^2}{144}\left[ (\alpha ^2+2\alpha -11)u^4+(4-2\alpha )u^2+4\right] ,\quad 0\le u\le \zeta _1', \end{aligned}$$

    and

    $$\begin{aligned} 0<u_0:=\sqrt{\frac{\alpha -2}{\alpha ^2+2\alpha -11}}<\zeta ',\quad \alpha \in [0,1), \end{aligned}$$

    (3.8)

    is a unique critical point, namely the maximum of $\varphi .$ Observe here that $u_0<\zeta '$ leads to

    $$\begin{aligned} \begin{aligned}&3\alpha ^8{-}12\alpha ^7{-}135\alpha ^6{+}522\alpha ^5-471\alpha ^4{+}1440\alpha ^3{-}489\alpha ^2-13950\alpha +18468\\&\quad>3\alpha ^8+522\alpha ^5+1440\alpha ^3+3411>0,\quad \alpha \in [0,1), \end{aligned} \end{aligned}$$

    and the last inequality is true.

  • C5. It remains to consider the last case in Lemma 1.2, which taking into account C4 holds for $\zeta '<\zeta _1<1.$ Then, by (3.4),

    $$\begin{aligned} \begin{aligned} |\gamma _1\gamma _3-\gamma _2^2|&\le \frac{1}{24}(1-\alpha )^2\zeta _1(1-\zeta _1^2)(|C|+|A|)\sqrt{1-\dfrac{B^2}{4AC}}=\psi (\zeta _1)\\&\le \psi (\zeta ')=\frac{(1-\alpha )^2(a_1-a_2\sqrt{b})}{144d^2}\sqrt{\frac{a_3-3(\alpha -1)^2\sqrt{b}}{a_4+2(3-\alpha ^2)\sqrt{b}}}, \end{aligned} \end{aligned}$$

    (3.9)

    where

    $$\begin{aligned} \psi (u):=\frac{(1-\alpha )^2}{144}\left[ (1-\alpha ^2)u^4-2u^2+4\right] \sqrt{\frac{3(-\alpha ^2-2\alpha +7)-3(1-\alpha )^2u^2}{2(3-\alpha ^2)(2+u^2)}} \end{aligned}$$

    for $\zeta '\le u\le 1,$ and

    $$\begin{aligned} \begin{aligned} a_1&:=-24\alpha ^6-120\alpha ^5+540\alpha ^4+924\alpha ^3-2904\alpha ^2-1572\alpha +4500,\\ a_2&:=8\alpha ^4+4\alpha ^3-52\alpha ^2-12\alpha +84,\\ a_3&:=-3\alpha ^5+3\alpha ^4+99\alpha ^3-159\alpha ^2-360\alpha +612,\\ a_4&:=-4\alpha ^5+20\alpha ^4+30\alpha ^3-138\alpha ^2-54\alpha +234,\\ b&:=12\alpha ^4+48\alpha ^3-183\alpha ^2-198\alpha +549,\\ d&:=\alpha ^3-7\alpha ^2-5\alpha +27 \end{aligned} \end{aligned}$$

    for $\alpha \in [0,1).$ To see that the last inequality in (3.9) holds observe that $\psi$ is decreasing. Indeed, we have

    $$\begin{aligned} \begin{aligned} \psi '(u)&=\frac{-3(1-\alpha )^2u}{288(3-\alpha ^2)(2+u^2)^2}\sqrt{\frac{2(3-\alpha ^2)(2+u^2)}{3(-\alpha ^2-2\alpha +7)-3(1-\alpha )^2u^2}}\\&\quad \times \left[ 4(2+u^2)\left( 1-(1-\alpha ^2)u^2\right) \left( 7-2\alpha -\alpha ^2-(1-\alpha )^2u^2\right) \right. \\&\quad +\left. \left( (1-\alpha ^2)u^4-2u^2+4\right) (\alpha -3)^2)\right] ,\quad \zeta '< u<1. \end{aligned} \end{aligned}$$

    Since for $\zeta '< u<1,$

    $$\begin{aligned} 7-2\alpha -\alpha ^2-(1-\alpha )^2u^2\ge 7-2\alpha -\alpha ^2-(1-\alpha )^2=6-2\alpha ^2>0 \end{aligned}$$

    and

    $$\begin{aligned} \begin{aligned} (1-\alpha ^2)u^4-2u^2+4&=4-u^2\left( 2-(1-\alpha )^2u^2\right) \ge 4-\left( 2-(1-\alpha )^2u^2\right) \\&=2+(1-\alpha )^2u^2>0, \end{aligned} \end{aligned}$$

    we deduce that $\psi '\le 0$ for $\zeta '\le u<1,$ which confirm that $\psi$ decreases. Simple however tedious computations which we omit show that

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

    for each $\alpha \in [0,1).$ Hence taking into account (3.7) and (3.9) we see that

    $$\begin{aligned} \psi (\zeta ')\le \varphi (u_0)=\frac{(1-\alpha )^2(\alpha ^2+4\alpha -16)}{48(\alpha ^2+2\alpha -11)}. \end{aligned}$$

• D. Summarizing from parts A–C it follows inequality (3.1). Equality holds for the function $f\in {\mathcal {A}}$ given by (3.2), where the function $p\in {\mathcal {P}}$ is of form (1.12) with $\zeta _1=u_0=:\tau ,$ with $u_0$ given by (3.8) and $\zeta _2=-1$, i.e.,

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

$\square$

For $\alpha =0$ we get the estimate for the class $\mathcal {S}^c$ of convex functions [12].

Corollary 3.2

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.