Second Hankel determinant of logarithmic coefficients of inverse strongly starlike functions

Abstract

The sharp bound of the second Hankel determinant of logarithmic coefficients of inverse functions of strongly starlike functions is computed.

1. Introduction

For $r \gt 0,$ let $\mathbb{D}_r:=\{z\in\mathbb{C}: |z| \lt r \}$, $\mathbb{D}:=\mathbb{D}_1$, $\overline{\mathbb{D}}:=\{z\in\mathbb{C}:|z|\leqslant 1\}$ and let $\mathbb{R} T:=\{z\in\mathbb{C}:|z|=1\}.$ Let ${\mathcal H}(\mathbb{D}_r)$ denote the class of all analytic functions f in $\mathbb{D}_r$ and let $\mathcal{H}:=\mathcal{H}(\mathbb{D}).$ Then $f\in {\mathcal H}(\mathbb{D}_r)$ has the following representation

(1.1)\begin{equation} f(z) = \sum_{n=0}^{\infty}a_n(f) z^n,\quad z\in\mathbb{D}_r. \end{equation}

Let $\mathcal {A}(\mathbb{D}_r)$ be the subclass of ${\mathcal H}(\mathbb{D}_r)$ of all f normalized by $f(0)=0=f'(0)-1$ and let ${\mathcal A}:={\mathcal A}(\mathbb{D}).$ By ${\mathcal S}$ we denote the subclass of all univalent (i.e. analytic and injective in $\mathbb{D}$) functions in ${\mathcal A}$.

Given $\alpha\in (0,1],$ let ${\mathcal S}^*_{\alpha}$ denote class of all functions $f\in {\mathcal A}$ such that

(1.2)\begin{equation} \left|\operatorname{Arg} \frac{zf'(z)}{f(z)} \right| \lt \alpha\frac{\pi}{2}, \quad z\in \mathbb{D}, \end{equation}

and the so-called strongly starlike of order $\alpha.$ For $\alpha:=1$ the class $\mathcal{S}_1^*=:\mathcal{S}^*$ is the well-known class of starlike functions, i.e. functions f which map univalently $\mathbb{D}$ onto a set which is star-shaped with respect to the origin. Then, the condition (1.2) can be written as

\begin{equation*} \operatorname{Re} \frac{zf'(z)}{f(z)} \gt 0,\quad z\in\mathbb{D}. \end{equation*}

The class of strongly starlike functions was introduced by Stankiewicz [18] and [19] and independently by Brannan and Kirwan [1] (see also [6, Vol. I, pp. 137–142]). Stankiewicz [19] presented an external geometrical characterization of strongly starlike functions. Brannan and Kirwan found a geometrical condition called δ-visibility which is sufficient for functions to be strongly starlike. In turn, Ma and Minda [15] gave the internal characterization of functions in $\mathcal{S}_\alpha^*$ basing on the concept of k-starlike domains. Further results regarding the geometry of strongly starlike functions were presented in [13, Chapter IV], [14] and [20]. Since $\mathcal{S}^*\subset \mathcal{S}$ (cf. [5, pp. 40–41]) and ${\mathcal S}^*_{\alpha}\subset \mathcal{S}^*$ for every $\alpha\in(0,1]$, it follows that ${\mathcal S}^*_{\alpha}\subset \mathcal{S}$ for every $\alpha\in(0,1].$

If $f\in\mathcal{S},$ then the inverse function $F:=f^{-1}$ is well-defined and analytic in $\mathbb{D}_{r(f)},$ where $r(f):=\sup(\{r \gt 0:\mathbb{D}_r\subset f(\mathbb{D})\}).$ Thus

(1.3)\begin{equation} F(w)=w+\sum_{n=2}^\infty A_nw^n,\quad w\in \mathbb{D}_{r(f)}, \end{equation}

where $A_n:=a_n(F).$ By Koebe one-quarter theorem (e.g. [5, p. 31]), it follows that $r(f) \geqslant 1/4$ for every $f\in\mathcal{S}.$

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

\begin{equation*} F_f(z) := \frac12\log\frac{f(z)}{z}=\sum_{n=1}^\infty\gamma_n z^n,\quad z\in\mathbb{D}, \end{equation*}

a logarithmic function associated with $f.$ The numbers $\gamma_n:=a_n(F_f)$ are called the logarithmic coefficients of $f.$ It is well-known that the logarithmic coefficients play a crucial role in Milin’s conjecture (see [16], [5, p. 155]).

Referring to the above idea, for $f\in\mathcal{S}$, there exists the unique function $F_{f^{-1}}$ analytic in $\mathbb{D}_{r(f)}$ such that

(1.4)\begin{equation} F_{f^{-1}}(w):=\frac12\log\frac{f^{-1}(w)}{w}=\sum_{n=1}^\infty \varGamma_n w^n,\quad w\in\mathbb{D}_{r(f)}, \end{equation}

where $\varGamma_n:=a_n\left(F_{f^{-1}}\right)$ are logarithmic coefficients of the inverse function $f^{-1}.$

It follows from Equation (1.3) that (e.g. [6, Vol. I, p. 57])

(1.5)\begin{equation} \begin{aligned} &A_2=-a_2,\quad A_3=-a_3+2a_2^2\quad \mathrm{and}\quad A_4=-a_4+5a_2a_3-5a_2^3, \end{aligned} \end{equation}

where $a_n:=a_n(f).$ Thus from Equation (1.4) we derive that

\begin{equation*} \varGamma_1=\frac12A_2,\quad \varGamma_2=\frac12A_3-\frac14A_2^2,\quad \varGamma_3=\frac12A_4-\frac12A_2A_3+\frac16A_2^3, \end{equation*}

and next using Equation (1.5) we obtain

(1.6)\begin{equation} \begin{aligned} &\varGamma_1=-\frac12a_2,\quad \varGamma_2=-\frac12a_3+\frac34a_2^2\quad \mathrm{and}\quad \varGamma_3=-\frac12a_4+2a_2a_3-\frac53a_2^3. \end{aligned} \end{equation}

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

(1.7)\begin{equation} H_{q,n}(f) := \left[\begin{matrix} 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{matrix}\right]. \end{equation}

In recent years, there has been a great deal of attention devoted to finding bounds for the modulus of the second and third Hankel determinants $\det H_{2,2}(f)$ and $\det H_{3,1}(f)$, when f belongs to various subclasses of $\mathcal{A}$ (see [2, 10, 11] for further references).

Based on these ideas, in [8] and [9], the authors started the study the Hankel determinant $\det H_{q,n}(F_f)$ whose entries are logarithmic coefficients of $f\in\mathcal{S},$ that is, a_n in Equation (1.7) are replaced by $\gamma_n.$ In this paper, we continue analogous research considering the Hankel determinant $\det H_{q,n}(F_{f^{-1}})$ whose entries are logarithmic coefficients of inverse functions, i.e. a_n in Equation (1.7) are now replaced by $\varGamma_n.$ We demonstrate the sharp estimates of

\begin{equation*} \left|\det H_{2,1}\left(F_{f^{-1}}\right)\right|=\left|\varGamma_1\varGamma_3-\varGamma_2^2\right|=\frac{1}{48}\left|13a_2^4-12a_3^2+12a_2a_4-12a_2^2a_3\right| \end{equation*}

in the classes $\mathcal{S}_\alpha^*$.

2. Preliminary lemmas

Denote by ${\mathcal P}$ the class of analytic functions $p\in\mathcal{H}$ with positive real part given by

(2.1)\begin{equation} p(z)=1+\sum_{n=1}^{\infty}c_n z^n,\quad z\in \mathbb{D}, \end{equation}

where $c_n:=a_n(p).$

In the proof of the main result, we will use the following lemma which contains the well-known formula for c ₂ (see, e.g. [17, p. 166]) and the formula for c ₃ (see [3, Lemma 2.4] with further remarks related to extremal functions).

Lemma 1. If $p \in {\mathcal P}$ is of the form (2.1), then

(2.2)\begin{equation} c_1 = 2\zeta_1, \end{equation}

(2.3)\begin{equation} c_2 = 2\zeta_1^2 + 2(1-|\zeta_1|^2)\zeta_2 \end{equation}

and

(2.4)\begin{equation} c_3 = 2\zeta_1^3+2(1-|\zeta_1|^2)(2\zeta_1-\overline{\zeta_1}\zeta_2)\zeta_2 + 2(1-|\zeta_1|^2)(1-|\zeta_2|^2)\zeta_3 \end{equation}

for some $\zeta_1,\zeta_2, \zeta_3 \in \overline{\mathbb{D}}.$

For $\zeta_1 \in \mathbb{T}$, there is a unique function $p \in {\mathcal P}$ with c ₁ as in Equation (2.2), namely,

\begin{equation*} p(z) = \frac{1+\zeta_1 z}{1-\zeta_1 z}, \quad z\in\mathbb{D}. \end{equation*}

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 Equations (2.2) and (2.3), namely,

(2.5)\begin{equation} p(z) = \frac{1+( \overline{\zeta}_1 \zeta_2 +\zeta_1 )z + \zeta_2 z^2}{1+( \overline{\zeta}_1 \zeta_2 -\zeta_1 )z - \zeta_2 z^2}, \quad z\in\mathbb{D}. \end{equation}

Lemma 2. ([4])

For real numbers A, B, C, let

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

1. I. If $AC\geqslant 0,$ then

   \begin{equation*} Y(A,B,C)=\left\{ \begin{array}{ll} |A|+|B|+|C|, & |B|\geqslant 2(1-|C|),\\ 1+|A|+\dfrac{B^2}{4(1-|C|)}, & |B| \lt 2(1-|C|). \end{array} \right. \end{equation*}

2. II. If $AC \lt 0,$ then

   \begin{equation*} Y(A,B,C)=\left\{ \begin{array}{lll} 1-|A|+\dfrac{B^2}{4(1-|C|)}, & -4AC(C^{-2}-1)\leqslant B^2 \wedge |B| \lt 2(1-|C|), \\ 1+|A|+\dfrac{B^2}{4(1+|C|)}, & B^2 \lt \min\left\{4(1+|C|)^2,-4AC(C^{-2}-1)\right\}, \\ R(A,B,C), & {\rm otherwise} , \end{array} \right. \end{equation*}

   where

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

We recall now Laguerre’s rule of counting zeros of polynomials in an interval (see [7], [12], [21, pp. 19–20]). We will apply Laguerre’s algorithm in the proof of the main theorem. Given a real polynomial

\begin{equation*} Q(u) := d_0u^n + d_1u^{n-1} + \cdots+ d_{n-1}u + d_n,\quad u\in\mathbb{R} ,\ d_0,\dots,d_n\in\mathbb{R} , \end{equation*}

consider a finite sequence $(q_k), k = 0, 1,\dots, n,$ of polynomials of the form

\begin{equation*} q_k(u) =\sum_{j=0}^k d_ju^{k-j},\quad u\in\mathbb{R} . \end{equation*}

For each $u_0\in\mathbb{R},$ let $N(Q; u_0)$ denote the number of sign changes in the sequence $(q_k(u_0)), k = 0, 1,\dots, n.$ Given an interval $I \subset \mathbb{R},$ denote by $Z(Q; I)$ the number of zeros of Q in I counted with their orders. Then the following theorem due to Laguerre holds.

Theorem 1. If a < b and $Q(a)Q(b)\neq 0,$ then

\begin{equation*} Z(Q; (a, b)) = N(Q; a) - N(Q; b) \end{equation*}

or

\begin{equation*}N(Q; a) - N(Q; b) - Z(Q; (a, b))\end{equation*}

is an even positive integer.

Note that

\begin{equation*} q_k(0) = d_k,\quad q_k(1) =\sum_{j=0}^k d_j. \end{equation*}

Thus, when $[a, b] := [0, 1],$ Theorem 1 reduces to the following useful corollary.

Corollary 1. If $Q(0)Q(1)\neq 0,$ then

\begin{equation*} Z(Q; (0, 1)) = N(Q; 0) - N(Q; 1) \end{equation*}

or

\begin{equation*} N(Q; 0) - N(Q; 1) - Z(Q; (0, 1)) \end{equation*}

is an even positive integer, where $N(Q; 0)$ and $N(Q; 1)$ are the numbers of sign changes in the sequence of polynomial coefficients $(d_k)$ and in the sequence of sums $(\sum_{j=0}^k d_j ),$ where $k = 0, 1,\dots , n,$ respectively.

3. Main result

The main result of this paper is the following.

Theorem 2. Let $\alpha\in(0,1].$ If $f\in {\mathcal{S}}^*_{\alpha},$ then

(3.1)\begin{equation} \left|\det H_{2,1}\left(F_{f^{-1}}\right)\right|=\left|\varGamma_1\varGamma_3-\varGamma_2^2\right|\leqslant \begin{cases} \dfrac{1}{4}\alpha^2,&0 \lt \alpha \lt \dfrac{1}{5},\\ \dfrac{\alpha^2(15\alpha^2+5\alpha+2)}{(35\alpha^2+30\alpha+7)},& \dfrac{1}{5}\leqslant \alpha\leqslant\alpha_0,\\ \dfrac{1}{36}\alpha^2(35\alpha^2+4),& \alpha_0 \lt \alpha\leqslant 1,\\ \end{cases} \end{equation}

where $\alpha_0\approx 0.39059 $ is the unique root in $(0,1]$ of the equation

(3.2)\begin{equation} 1225\alpha^4+1050\alpha^3-155\alpha^2-60\alpha-44=0. \end{equation}

All inequalities are sharp.

Proof. Let $f\in{\mathcal{S}}^*_{\alpha}$ be of the form (1.1). Then by Equation (1.2), there exists $p\in\mathcal{P}$ of the form (2.1) such that

(3.3)\begin{equation} \frac{zf'(z)}{f(z)}=(p(z))^{\alpha},\quad z\in \mathbb{D}. \end{equation}

Putting the series (1.1) and (2.1) into (3.3), by equating the coefficients we get

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

Hence and from Equation (1.6) we obtain

\begin{equation*} \begin{aligned} &\varGamma_1 =-\frac{1}{2}\alpha c_1, \quad \varGamma_2 =-\frac{1}{8}\alpha(2c_2-(3\alpha+1) c_1^2),\\ &\varGamma_3 =-\frac{1}{72}\alpha\left(12c_3-(42\alpha+12)c_1c_2+ (29\alpha^2+21\alpha+4)c_1^3\right), \end{aligned} \end{equation*}

and therefore

(3.5)\begin{equation} \varGamma_1\varGamma_3-\varGamma_2^2 = \frac{1}{576}\alpha^2(c_1^4(35\alpha^2+30\alpha+7)-12(5\alpha+1)c_1^2c_2+48c_1c_3-36c_2^2). \end{equation}

Since both the class ${\mathcal{S}}^*_{\alpha}$ and $|\det H_{2,1}\left(F_{f^{-1}}\right)|$ are rotationally invariant, without loss of generality we may assume that $a_2\geqslant 0,$ which in view of Equation (3.4) yields $ c_1 \geqslant 0,$ i.e. by Equation (2.2) that $\zeta_1\in[0,1].$ Thus substituting Equations (2.2)–(2.4) into Equation (3.5), we obtain

(3.6)\begin{equation} \begin{aligned} \varGamma_1\varGamma_3-\varGamma_2^2 = \frac{\alpha^2}{36}&\left((35\alpha^2+4)\zeta_1^4-30\alpha(1-\zeta_1^2)\zeta_1^2\zeta_2-3(1-\zeta_1^2)(\zeta_1^2+3)\zeta_2^2\right.\\ &\left.+12\zeta_1(1-\zeta_1^2)(1-|\zeta_2|^2)\zeta_3 \right) \end{aligned} \end{equation}

for some $\zeta_1\in[0,1]$ and $\zeta_2,\zeta_3 \in \overline{\mathbb{D}}$.

A. Suppose that $\zeta_1=0.$ Then from Equation (3.6),

\begin{equation*} \left|\varGamma_1\varGamma_3-\varGamma_2^2\right|=\frac{\alpha^2}{4}|\zeta_2|^2\leqslant \frac{\alpha^2}{4}. \end{equation*}

B. Suppose that $\zeta_1=1.$ Then from Equation (3.6),

\begin{equation*} \left|\varGamma_1\varGamma_3-\varGamma_2^2\right|=\frac{1}{36}\alpha^2(35\alpha^2+4). \end{equation*}

C. Suppose that $\zeta_1\in(0,1).$ Since $\zeta_3 \in \overline{\mathbb{D}}$, from Equation (3.6) we get

\begin{equation*} \left|\varGamma_1\varGamma_3-\varGamma_2^2\right| \leqslant \frac{1}{3}\alpha^2\zeta_1(1-\zeta_1^2) \varPhi(A,B,C), \end{equation*}

where

\begin{equation*} \varPhi(A,B,C) := \left| A +B\zeta_2 +C\zeta_2^2 \right| + 1 -|\zeta_2|^2, \end{equation*}

with

\begin{equation*} A:=\frac{\zeta_1^3(35\alpha^2+4)}{12(1-\zeta_1^2)},\quad B:=\frac{-5\alpha\zeta_1}{2},\quad C:=\frac{-(\zeta_1^2+3)}{4\zeta_1}. \end{equation*}

Observe that AC < 0 and therefore we apply only the part II of Lemma 2.

C1. Let’s consider the condition $|B| \lt 2(1-|C|),$ i.e.

\begin{equation*} \frac{5\alpha\zeta_1}{2} \lt 2\left(1-\frac{\zeta_1^2+3}{4\zeta_1}\right). \end{equation*}

The above inequality is equivalent to

(3.7)\begin{equation} \frac{\zeta_1^2(5\alpha+1)-4\zeta_1+3}{2\zeta_1} \lt 0, \end{equation}

which is equivalent to $(5\alpha+1)\zeta_1^2-4\zeta_1+3 \lt 0.$ However

\begin{equation*}(5\alpha+1)\zeta_1^2-4\zeta_1+3=5\alpha\zeta_1^2+(1-\zeta_1)(3-\zeta_1) \gt 0\end{equation*}

for $\zeta_1\in(0,1)$, which shows that the inequality (3.7) is false.

C2. Since

\begin{equation*} -4AC\left(\frac{1}{C^2}-1\right)=-\frac{(9-\zeta_1^2)(35\alpha^2+4)\zeta_1^2}{12(\zeta_1^2+3)} \lt 0 \end{equation*}

for $\zeta_1\in(0,1),$ we deduce that the condition $B^2 \lt \min\{4(1+|C|)^2,-4AC(C^{-2}-1)\}$ is equivalent to

(3.8)\begin{equation} \frac{\zeta_1^2[(10\alpha^2-1)\zeta_1^2+135\alpha^2+9]}{3(\zeta_1^2+3)} \lt 0, \end{equation}

which is equivalent to $(10\alpha^2-1)\zeta_1^2+135\alpha^2+9 \lt 0$ for $\zeta_1\in(0,1).$ However, in the case when $10\alpha^2-1\geqslant 0$ we have

\begin{equation*} (10\alpha^2-1)\zeta_1^2+135\alpha^2+9\geqslant 135\alpha^2+9 \gt 0, \end{equation*}

and in the case when $10\alpha^2-1 \lt 0$ we have

\begin{equation*} (10\alpha^2-1)\zeta_1^2+135\alpha^2+9\geqslant 145\alpha^2+8 \gt 0, \end{equation*}

for all $\zeta_1\in(0,1).$ Thus the inequality (3.8) is false.

C3. The inequality $|C|(|B|+4|A|)\leqslant|AB|$ is equivalent to

\begin{equation*} \frac{(175\alpha^3-70\alpha^2+35\alpha-8)\zeta_1^4-6(35\alpha^2-5\alpha+4)\zeta_1^2-45\alpha}{24(1-\zeta_1^2)}\geqslant 0 \end{equation*}

which is equivalent to

(3.9)\begin{equation} \varphi_\alpha(\zeta_1^2)\geqslant 0, \end{equation}

where for $t\in\mathbb{R} ,$

\begin{equation*} \varphi_\alpha(t):=(175\alpha^3-70\alpha^2+35\alpha-8)t^2-6(35\alpha^2-5\alpha+4)t-45\alpha. \end{equation*}

Observe that the equation $175\alpha^3-70\alpha^2+35\alpha-8= 0$ has only one real root α ₁ in $(0,1],$ where

\begin{equation*} \alpha_1:=\frac{1}{105}(13769+882\sqrt{445})^{1/3}-\frac{77}{15(13769+882\sqrt{445})^{1/3}}+\frac{2}{15}\approx 0.2758, \end{equation*}

and that the inequality (3.9) is false for $\alpha:=\alpha_1.$ Let now $\alpha\in(0,1]\setminus\{\alpha_1\}.$ For φ_α, we have $\Delta:=144(525\alpha^4 - 175\alpha^3 + 120\alpha^2 - 20\alpha + 4) \gt 0,$ which is true for all $\alpha\in(0,1]\setminus\{\alpha_1\}.$ Hence the square trinomial φ_α has two roots

\begin{equation*} t_{1,2}:=\frac{3(35\alpha^2-5\alpha+4)\pm 6\sqrt{525\alpha^4 - 175\alpha^3 + 120\alpha^2 - 20\alpha + 4}}{175\alpha^3-70\alpha^2+35\alpha-8}. \end{equation*}

Note that for all $\alpha\in(0,1]\setminus\{\alpha_1\}$ we have $-6(35\alpha^2-5\alpha+4) \lt 0.$ Now for $\alpha\in(\alpha_1,1]$ we have $175\alpha^3-70\alpha^2+35\alpha-8 \gt 0.$ Hence $t_2 \lt 0$ because the inequality

\begin{equation*} 3(35\alpha^2-5\alpha+4)-6\sqrt{525\alpha^4 - 175\alpha^3 + 120\alpha^2 - 20\alpha + 4} \lt 0 \end{equation*}

is equivalent to

\begin{equation*} -45\alpha(175\alpha^3 - 70\alpha^2 + 35\alpha - 8) \lt 0, \end{equation*}

which is true for all $\alpha\in(\alpha_1,1].$ On the other hand, the inequality $t_1 \gt 1$ is equivalent to

(3.10)\begin{equation} 6\sqrt{525\alpha^4 - 175\alpha^3 + 120\alpha^2 - 20\alpha + 4} \gt 5 (35\alpha^3 - 35 \alpha^2 + 10\alpha - 4), \end{equation}

which is evidently true for $\alpha\in(\alpha_1,\alpha_2],$ where $\alpha_2\approx 0.82155,$ since then the right hand side of Equation (3.10) is non-positive. For $\alpha\in(\alpha_2,1]$ by squaring both sides of Equation (3.10), we equivalently get the inequality

\begin{equation*} (5\alpha - 8) (6125\alpha^5 - 2450\alpha^4 + 1925\alpha^3 - 560\alpha^2 + 140\alpha - 32) \lt 0 \end{equation*}

which is true for $\alpha\in(\alpha_2,1].$ Thus we conclude that for $\alpha\in(\alpha_1,1]$ the inequality (3.9) is false.

Let $\alpha\in(0,\alpha_1).$ Then $175\alpha^3-70\alpha^2+35\alpha-8 \lt 0$ and therefore $t_1 \lt 0$ evidently. Moreover, the inequality $t_2 \lt 0$ is equivalent to

\begin{equation*} 3(35\alpha^2-5\alpha+4)-6\sqrt{525\alpha^4 - 175\alpha^3 + 120\alpha^2 - 20\alpha + 4} \gt 0 \end{equation*}

which is equivalent to

\begin{equation*} -45\alpha(175\alpha^3 - 70\alpha^2 + 35\alpha - 8) \gt 0, \end{equation*}

which is true for all $\alpha\in(0,\alpha_1).$ Thus we conclude that for $\alpha\in(0,\alpha_1)$ the inequality (3.9) is false.

C4. The inequality $|C|(|B|-4|A|)\geqslant|AB|$ is equivalent to

\begin{equation*} \frac{(175\alpha^3+70\alpha^2+35\alpha+8)\zeta_1^4+6\zeta_1^2(35\alpha^2+5\alpha+4)-45\alpha}{24(1-\zeta_1^2)}\leqslant 0 \end{equation*}

which is equivalent to

(3.11)\begin{equation} \gamma_\alpha(\zeta_1^2)\leqslant 0, \end{equation}

where for $t\in\mathbb{R} ,$

\begin{equation*} \gamma_\alpha(t):=(175\alpha^3+70\alpha^2+35\alpha+8)t^2+6(35\alpha^2+5\alpha+4)t-45\alpha. \end{equation*}

For γ_α we have $\Delta:=144 (525\alpha^4 + 175\alpha^3 + 120\alpha^2 + 20 \alpha + 4) \gt 0$ for all $\alpha\in(0,1].$ Hence the square trinomial γ_α has two roots

\begin{equation*} t_{3,4}:=\frac{-3(35\alpha^2+5\alpha+4)\pm6\sqrt{525\alpha^4 + 175\alpha^3 + 120\alpha^2 + 20 \alpha + 4}}{175\alpha^3+70\alpha^2+35\alpha+8}. \end{equation*}

Note that $t_4 \lt 0$ evidently. Observe now that $t_3 \gt 0.$ Indeed, this inequality is equivalent to

\begin{equation*} -3(35\alpha^2+5\alpha+4)+6\sqrt{525\alpha^4 + 175\alpha^3 + 120\alpha^2 + 20 \alpha + 4} \gt 0 \end{equation*}

which is equivalent to the evidently true inequality

\begin{equation*} 45\alpha(175\alpha^3 + 70\alpha^2 + 35\alpha + 8) \gt 0,\quad \alpha\in(0,1]. \end{equation*}

Moreover, $t_3 \lt 1$ is equivalent to

\begin{equation*} 6\sqrt{525\alpha^4 + 175\alpha^3 + 120\alpha^2 + 20 \alpha + 4} \lt 5 (35\alpha^3 + 35\alpha^2 + 10\alpha + 4) \end{equation*}

that after squaring both sides is equivalent to

\begin{equation*} (5\alpha + 8) (6125\alpha^5 + 2450\alpha^4 + 1925\alpha^3 + 560 \alpha^2 + 140\alpha + 32) \gt 0, \end{equation*}

which is true for all $\alpha\in(0,1].$ Therefore the inequality (3.11) is true for $\zeta_1\in(0,\zeta_1^0],$ where $\zeta_1^0:=\sqrt{t_3}.$

Applying Lemma 2 for $0 \lt \zeta_1\leqslant \zeta_1^0$, we get

\begin{equation*} \left|\varGamma_1\varGamma_3-\varGamma_2^2\right| \leqslant\frac{1}{3}\alpha^2\zeta_1(1-\zeta_1^2)(-|A|+|B|+|C|)=\rho_\alpha(\zeta_1), \end{equation*}

where

\begin{equation*} \rho_\alpha(t):= -\frac{1}{36}\alpha^2((35\alpha^2+30\alpha+7)t^4-6(5\alpha-1)t^2-9),\quad t\in\mathbb{R} . \end{equation*}

We have

\begin{equation*} \rho_\alpha(0)=\frac{1}{4}\alpha^2 \end{equation*}

and

\begin{equation*} \begin{aligned} \rho_\alpha(\zeta_1^0)=&\frac{2\alpha^2}{(175\alpha^3+70\alpha^2+35\alpha+8)^2}\\ &\times \left[-18375\alpha^6-16625\alpha^5-10150\alpha^4-3775\alpha^3 -1025\alpha^2-150\alpha-12\right.\\ &\left.+(1050\alpha^4+700\alpha^3+320\alpha^2+80\alpha+10)\right.\\ &\left.\times\sqrt{525\alpha^4+175\alpha^3+120\alpha^2+20\alpha+4}\right]. \end{aligned} \end{equation*}

Note that for $\alpha\in(0,1/5]$ the equation

(3.12)\begin{equation} \rho_\alpha'(t)=-\frac{1}{9}\alpha^2t((35\alpha^2+30\alpha+7)t^2-3(5\alpha-1))=0 \end{equation}

has no root in $(0,\zeta_1^0)$ and then evidently

\begin{equation*} \rho_\alpha(t)\leqslant \rho_\alpha(0)=\frac{1}{4}\alpha^2,\quad 0\leqslant t\leqslant \zeta_1^0. \end{equation*}

For $\alpha\in(1/5,1]$, Equation (3.12) has a unique positive root, namely

(3.13)\begin{equation} t_5:=\sqrt{\frac{3(5\alpha-1)}{35\alpha^2+30\alpha+7}}. \end{equation}

It remains to check the condition $t_5 \lt \zeta_1^0$ equivalently written as

\begin{equation*} \frac{10(105\alpha^4+70\alpha^3+32\alpha^2+8\alpha+1)}{35\alpha^2+30\alpha+7} \lt \sqrt{525\alpha^4+175\alpha^3+120\alpha^2+20\alpha+4}, \end{equation*}

which is equivalent to

\begin{equation*} \frac{(175\alpha^3+70\alpha^2+35\alpha+8)(2625\alpha^5-175\alpha^4-925\alpha^3-425\alpha^2-80\alpha-12)}{(35\alpha^2+30\alpha+7)^2} \lt 0. \end{equation*}

The last inequality is true for $\alpha\in(1/5,\alpha_3),$ where $\alpha_3\approx0.812678$ is the unique root in $(0,1)$ of the equation

\begin{equation*} 2625\alpha^5-175\alpha^4-925\alpha^3-425\alpha^2-80\alpha-12=0. \end{equation*}

Then ρ_α attains its maximum value on $(0,\zeta_1^0]$ at t ₅ with

\begin{equation*} \rho_\alpha(t_5)=\frac{\alpha^2(15\alpha^2+5\alpha+2)}{35\alpha^2+30\alpha+7}. \end{equation*}

If $\alpha\in[\alpha_3,1],$ then evidently,

\begin{equation*} \rho_\alpha(t)\leqslant\max\left(\{\rho_\alpha(0),\rho_\alpha(\zeta_1^0)\}\right)=\rho_\alpha(\zeta_1^0),\quad 0\leqslant t\leqslant \zeta_1^0. \end{equation*}

C5. Applying Lemma 2 for $\zeta_1^0 \lt \zeta_1 \lt 1$, we get

\begin{equation*} \left|\varGamma_1\varGamma_3-\varGamma_2^2\right| \leqslant\frac{1}{3}\alpha^2\zeta_1(1-\zeta_1^2)\left(|A|+|C| \right)\sqrt{1-\frac{B^2}{4AC}}=\psi_\alpha(\zeta_1), \end{equation*}

where for $t\in[0,1],$

\begin{equation*} \begin{aligned} \psi_\alpha(t):= \frac{1}{18}\alpha^2((35\alpha^2+1)t^4-6t^2+9)\sqrt{\frac{-(10\alpha^2-1)t^2+45\alpha^2+3}{(35\alpha^2+4)(t^2+3)}}. \end{aligned} \end{equation*}

We have

\begin{equation*} \psi_\alpha (\zeta_1^0)=-\frac{2\alpha^2(35\alpha^2+4)}{(175\alpha^3+70\alpha^2+35\alpha+8)^2}H(\alpha)\sqrt{-\frac{G(\alpha)}{K(\alpha)}}, \end{equation*}

where

\begin{equation*} \begin{aligned} H(\alpha):= &-1050\alpha^4-525\alpha^3-270\alpha^2-65\alpha-10+(35\alpha^2+10\alpha+3)\times\\ &\times\sqrt{525\alpha^4+175\alpha^3+120\alpha^2+20\alpha+4}, \end{aligned} \end{equation*}

\begin{equation*} \begin{aligned} G(\alpha):=&-2625\alpha^5-1400\alpha^4-750\alpha^3-195\alpha^2-30\alpha-4+(20\alpha^2-2)\times\\ &\times\sqrt{525\alpha^4+175\alpha^3+120\alpha^2+20\alpha+4} \end{aligned} \end{equation*}

and

\begin{equation*} K(\alpha):=(175\alpha^3+35\alpha^2+30\alpha+4+2\sqrt{525\alpha^4+175\alpha^3+120\alpha^2+20\alpha+4})(35\alpha^2+4). \end{equation*}

Note that

\begin{equation*} \psi_\alpha(1)=\frac{1}{36}\alpha^2(35\alpha^2+4). \end{equation*}

Differentiating ψ_α leads to the equation

\begin{equation*} \begin{aligned} \psi_\alpha'(t)=-\frac{1}{18}t\alpha^2\frac{Q(t^2)}{(35\alpha^2+4)(t^2+3)^2\sqrt{\dfrac{-(10\alpha^2-1)t^2+45\alpha^2+3}{(35\alpha^2+4)(t^2+3)}}}=0, \end{aligned} \end{equation*}

where for $s\in[0,1],$

\begin{equation*} \begin{aligned} Q(s):=&4(35\alpha^2+1)(10\alpha^2-1)s^3+3(175\alpha^4-315\alpha^2-4)s^2\\ &-18(1050\alpha^4+115\alpha^2-2)s+2295\alpha^2+108. \end{aligned} \end{equation*}

Now we describe the number of zeros of Q in the interval $(0,1)$ by combining Descartes’ and Laguerre’s rules. To apply Descartes’ rule, we check the numbers of sign changes of coefficients of the polynomial $Q.$ We have:

• • $ d_0(\alpha):=q_0(0)=4(35\alpha^2+1)(10\alpha^2-1) \gt 0$ iff $\alpha\in\left(1/\sqrt{10},1\right),$

• • $ d_1(\alpha):=q_1(0)=3(175\alpha^4-315\alpha^2-4) \lt 0$ iff $\alpha\in\left(0,1\right),$

• • $ d_2(\alpha):=q_2(0)=-18(1050\alpha^4+115\alpha^2-2) \gt 0$ iff $\alpha\in\left(0,\alpha_4\right),$ where

  \begin{equation*}\alpha_4:=\frac{1}{2}\sqrt{\frac{1}{105}(\sqrt{865}-23))}\approx 0.12355,\end{equation*}

• • $d_3(\alpha):=q_3(0)= 2295\alpha^2+108 \gt 0$ iff $\alpha\in\left(0,1\right).$

Thus there is one change of signs in $\left(0,1/\sqrt{10}\right)$, i.e. $N(Q,0)=1,$ and two changes of signs in $\left[1/\sqrt{10},1\right),$ i.e. $N(Q,0)=2.$ According to Descartes’ rule of signs, the polynomial Q has one positive real root in $\left(0,1/\sqrt{10}\right)$ and zero or two positive real roots in $\left[1/\sqrt{10},1\right).$

To apply Laguerres’ rule, it remains to compute the number $N(Q,1)$ of sign changes in the sequence of sums $\sum_{j=0}^ku_j(\alpha)$, where $k=0,1,2,3.$ We have

• • $d_0(\alpha)=4(35\alpha^2+1)(10\alpha^2-1) \gt 0$ iff $\alpha\in\left(1/\sqrt{10},1\right),$

• • $d_0(\alpha)+d_1(\alpha)=1925 \alpha^4 - 1045\alpha^2 - 16 \gt 0$ iff $\alpha\in\left(\alpha_5,1\right),$ where

  \begin{equation*} \alpha_5:=\sqrt{(209 + 3\sqrt{5401})/770}\approx 0.74683, \end{equation*}

• • $d_0(\alpha)+d_1(\alpha)+d_2(\alpha)= -5 (3395 \alpha^4 + 623\alpha^2 - 4) \gt 0$ iff $\alpha\in\left(0,\alpha_6\right),$ where

  \begin{equation*} \alpha_6:=\sqrt{(3\sqrt{49161} - 623)/6790}\approx 0.078806, \end{equation*}

• • $d_0(\alpha)+d_1(\alpha)+d_2(\alpha)+d_3(\alpha)= -(35\alpha^2 + 4) (485\alpha^2 - 32) \gt 0$ iff $\alpha\in\left(0,\alpha_7\right),$ where $\alpha_7:=4\sqrt{2/485}\approx 0.25686.$

Thus there are no changes of signs in $\left(\alpha_7,1/\sqrt{10}\right)$, i.e. $N(Q,1)=0$, and one change of sign in $\left(0,\alpha_7\right]\cup\left[1/\sqrt{10},1\right)$ i.e. $N(Q,1)=1$. According to Laguerre’s rule, the polynomial Q has one root in $[0,1]$ for $\alpha\in\left(\alpha_7,1\right)$, and no roots in $[0,1]$ for $\alpha\in\left(0,\alpha_7\right].$ Therefore, for $\alpha\in\left(0,\alpha_7\right]$, the function ψ_α is increasing for $\zeta_1^0 \lt t \lt 1$ and hence

\begin{equation*} \psi_\alpha(t)\leqslant \psi_\alpha(1),\quad \zeta_1^0 \lt t \lt 1. \end{equation*}

In turn, for $\alpha\in\left(\alpha_7,1\right)$, the function ψ_α has a unique critical point in $[0,1],$ where by using jointly Descartes’ and Laguerre’s rules we state that ψ_α attains its minimum value. Thus

\begin{equation*} \psi_\alpha(t)\leqslant \max\left(\{\psi_\alpha(\zeta_1^0),\psi_\alpha(1)\}\right),\quad \zeta_1^0 \lt t \lt 1. \end{equation*}

Now we summarize results of sections C4 and C5.

1. (i) For $\alpha\in\left(0,1/5\right),$ we compare $\psi_\alpha(1)$ and $\varrho_\alpha(0).$ Note that then $\varrho_\alpha(0)\geqslant \psi_\alpha(1)$ since it is equivalent to

   \begin{equation*}\frac14\alpha^2-\dfrac{1}{36}\alpha^2(35\alpha^2+4)=\frac{1}{36}\alpha^2(5-35\alpha^2)\geqslant 0.\end{equation*}

2. (ii) For $\alpha\in\left[1/5,\alpha_3\right),$ we compare $\psi_\alpha(1)$ and $\varrho_\alpha(t_5).$ Note that the inequality

   \begin{equation*} \frac{\alpha^2(15\alpha^2+5\alpha+2)}{35\alpha^2+30\alpha+7}\geqslant \frac{1}{36}\alpha^2(35\alpha^2+4) \end{equation*}

   is equivalent to

   \begin{equation*} \frac{1}{36}\alpha^2(1225\alpha^4+1050\alpha^3-155\alpha^2-60\alpha-44)\leqslant 0 \end{equation*}

   which is true for $\alpha\in[1/5,\alpha_0],$ where $\alpha_0\approx0.390595$ is the unique root in $(0,1]$ of Equation (3.2). Thus $\varrho_\alpha(t_5)\geqslant \psi_\alpha(1)$ for $\alpha\in[1/5,\alpha_0],$ and $\varrho_\alpha(t_5) \lt \psi_\alpha(1)$ for $\alpha\in(\alpha_0,\alpha_3).$

3. (iii) For $\alpha\in\left[\alpha_3,1\right),$ we compare $\psi_\alpha(1)$ and $\rho_\alpha(\zeta_1^0).$ Note that the inequality $\rho_\alpha(\zeta_1^0)\leqslant \psi_\alpha(1)$ is equivalent to

   \begin{equation*} \begin{aligned} &-18375\alpha^6-16625\alpha^5-10150\alpha^4-3775\alpha^3 -1025\alpha^2-150\alpha-12\\ &+(1050\alpha^4+700\alpha^3+320\alpha^2+80\alpha+10)\sqrt{525\alpha^4+175\alpha^3+120\alpha^2+20\alpha+4}\\ \leqslant& \frac{1}{72}(35\alpha^2+4)(175\alpha^3+70\alpha^2+35\alpha+8)^2, \end{aligned} \end{equation*}

   equivalently written as

   \begin{equation*} \begin{aligned} &(1050\alpha^4+700\alpha^3+320\alpha^2+80\alpha+10)\times\sqrt{525\alpha^4+175\alpha^3+120\alpha^2+20\alpha+4}\\ \leqslant& \frac{1}{72}\left(1071875\alpha^8+857500\alpha^7+2045750\alpha^6 +1564500\alpha^5+881475\alpha^4+322200\alpha^3\right.\\ &\left.+85420\alpha^2+13040\alpha+1120\right), \end{aligned} \end{equation*}

   which is equivalent to

   \begin{equation*} \begin{aligned} \frac{25}{5184}&(35\alpha^2+4)(175\alpha^3+70\alpha^2+35\alpha+8)^2\times\\ &\times \left(42875\alpha^8+34300\alpha^7+134750\alpha^6 +110460\alpha^5+17835\alpha^4 -7344\alpha^3\right.\\ &\left.-5036\alpha^2-1120\alpha-128\right)\geqslant 0 \end{aligned} \end{equation*}

   which is true for $\alpha\in\left[\alpha_3,1\right).$

D. We now show sharpness of all inequalities by using the formula (3.5). In the first inequality in Equation (3.1), the equality is attained by the function $f\in{\mathcal{S}}^*_{\alpha}$ given by Equation (3.3) with

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

for which $c_1=c_3=0$ and $c_2=-2.$

In the second inequality in Equation (3.1), the equality is attained by the function $f\in{\mathcal{S}}^*_{\alpha}$ given by Equation (3.3), where $p\in\mathcal{P}$ is defined by Equation (2.5) with $\zeta_1=t_5=:\tau$ and $\zeta_2=1,$ i.e.

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

Here t ₅ is described by Equation (3.13).

In the third inequality in Equation (3.1), the equality is attained by the function $f\in{\mathcal{S}}^*_{\alpha}$ given by Equation (3.3), where $p\in\mathcal{P}$ is defined by

\begin{equation*} p(z):=\frac{1+z}{1-z},\quad z\in \mathbb{D}, \end{equation*}

for which $c_1=c_2=c_3=2.$

This ends the proof of the theorem.

For α = 1, we have the following result:

Corollary 2. If $f\in {\mathcal{S}}^*,$ then

\begin{equation*} \left|\det H_{2,1}\left(F_{f^{-1}}\right)\right|=\left|\varGamma_1\varGamma_3-\varGamma_2^2\right|\leqslant\frac{13}{12}. \end{equation*}

The inequality is sharp.