THE THIRD HANKEL DETERMINANT FOR INVERSE COEFFICIENTS OF CONVEX FUNCTIONS

Abstract

The sharp bound for the third Hankel determinant for the coefficients of the inverse function of convex functions is obtained, thus answering a recent conjecture concerning invariance of coefficient functionals for convex functions.

1 Introduction

Let $\mathcal {A}$ denote the class of normalised analytic functions f in the open unit disc $\mathbb {D}:=\{ z:\vert z\vert <1,\ z\in \mathbb {C} \} $ with Taylor expansion

(1.1) $$ \begin{align} f(z)=z+\overset{\infty }{\sum_{n=2}}a_{n}z^{n},\ \ \ \ z\in \mathbb{D} \end{align} $$

and let $\mathcal {S}$ denote the subclass of functions in $\mathcal {A}$ which are univalent in $\mathbb {D}$ . The classes $\mathcal {S}^{\ast }$ of starlike functions and $\mathcal {C}$ of convex functions are subclasses of $ \mathcal {S}$ and are analytically defined respectively as

(1.2) $$ \begin{align} \mathcal{S}^{\ast } &=\bigg\{ f\in \mathcal{A}:{\mathrm{Re}}\bigg( \frac{zf^{\prime }(z)}{f(z)}\bigg)>0, \ z\in \mathbb{D}\bigg\}, \notag \\ \mathcal{C} &=\bigg\{ f\in \mathcal{A}:{\mathrm{Re}}\bigg( 1+\frac{zf^{\prime \prime }(z)}{f^{\prime }(z)}\bigg) >0, \ z\in \mathbb{D}\bigg\}. \end{align} $$

For any univalent function f, there exists an inverse function $f^{-1}$ defined on some disc $\vert w\vert \leq r_{0}(f),$ with Taylor series expansion

(1.3) $$ \begin{align} f^{-1}( w) =w+A_{2}w^{2}+A_{3}w^{3}+\cdots. \end{align} $$

The qth Hankel determinant for analytic functions $f\in \mathcal {A}$ was introduced by Pommerenke [11] and is given by

$$ \begin{align*} H_{q}( n) (f):=\left\vert \begin{array}{llll} a_{n} & a_{n+1} & \ldots & a_{n+q-1} \\ a_{n+1} & a_{n+2} & \ldots & a_{n+q} \\ \vdots & \vdots & \ldots & \vdots \\ a_{n+q-1} & a_{n+q} & \ldots & a_{n+2q-2} \end{array} \right\vert , \end{align*} $$

where $n\geq 1$ and $q\geq 1.$ Many authors have studied the Hankel determinants $H_{2}( 2) (f)=a_{2}a_{4}-a_{3}^{2}$ and $H_{2}( 3) (f)=a_{3}a_{5}-a_{4}^{2}$ of order 2 (see, for example, [2, 3, 9, 12, 14, 15]), whereas $H_{2}( 1) (f)=a_{2}^{2}-a_{3}$ is classical. Sharp bounds for

(1.4) $$ \begin{align} \vert H_{3}( 1) (f)\vert =\vert 2a_{2}a_{3}a_{4}-a_{3}^{3}-a_{4}^{2}+a_{3}a_{5}-a_{2}^{2}a_{5}\vert \end{align} $$

are more difficult to find, but some notable recent results have been found, for example, [1, 4, 5, 7, 8, 13]. In particular, Kowalczyk et al. [5] obtained the sharp bound for $\vert H_{3}( 1) (f)\vert $ for $f\in \mathcal {C}$ by showing that

$$ \begin{align*} \vert H_{3}( 1) (f)\vert \leq \tfrac{4}{135}. \end{align*} $$

In this paper, we find the sharp bound for $|H_{3}( 1) (f^{-1})|$ for the inverse function when $f\in \mathcal {C}.$ Our result demonstrates a noninvariance property for $f\in \mathcal {C}$ between corresponding functionals discussed in [5, 16], thus settling a conjecture made in [5].

Let $\mathcal {P}$ denote the class of analytic functions p defined for $ z\in \mathbb {D}$ by

(1.5) $$ \begin{align} p(z)=1+\sum_{n=1}^{\infty }c_{n}z^{n} \end{align} $$

with positive real part in $\mathbb {D}$ . We will use the following lemma concerning the coefficients of functions in $ \mathcal {P}$ .

Lemma 1.1 [6, 10].

Let $p\in \mathcal {P}$ be given by (1.5) with $c_{1}>0$ . Then,

(1.6) $$ \begin{align} 2c_{2} &=c_{1}^{2}+\delta (4-c_{1}^{2}),\end{align} $$

(1.7) $$ \begin{align}4c_{3} &=c_{1}^{3}+2(4-c_{1}^{2})c_{1}\delta -(4-c_{1}^{2})c_{1}\delta ^{2}+2(4-c_{1}^{2})(1-|\delta |^{2})\eta ,\end{align} $$

(1.8) $$ \begin{align}8c_{4} &=c_{1}^{4}+(4-c_{1}^{2})\delta (c_{1}^{2}(\delta ^{2}-3\delta +3)+4\delta )-4(4-c_{1}^{2})(1-|\delta |^{2})(c_{1}(\delta -1)\eta \nonumber \\ &\quad +\bar{\delta}\eta ^{2}-(1-|\eta |^{2})\rho ), \end{align} $$

for some $\delta $ , $\eta $ and $\rho $ such that $|\delta |\leq 1$ , $|\eta |\leq 1$ and $|\rho |\leq 1$ .

2 Main result

Theorem 2.1. Let $f\in \mathcal {C}$ be given by (1.1). Then,

$$ \begin{align*} |H_{3}(1)(f^{-1})|\leq \tfrac{1}{36}. \end{align*} $$

The inequality is sharp for the function $f_{0}\in \mathcal {C}$ given by

(2.1) $$ \begin{align} f_{0}(z)=\int_{0}^{z}\frac{dx}{( 1-x^{3}) ^{2/3}}=\hspace{0.5pt} {}_{2}{F}_{1}\bigg( \frac{1}{3},\frac{2}{3};\frac{4}{3};z^{3}\bigg) =z+\frac{1}{ 6}z^{4}+\frac{5}{63}z^{7}+\cdots. \end{align} $$

Proof. Let $f\in \mathcal {C}.$ Then from (1.2), there exists a function $p\in \mathcal {P}$ such that

(2.2) $$ \begin{align} 1+\frac{zf^{\prime \prime }(z)}{f^{\prime }(z)}=p(z). \end{align} $$

Since

(2.3) $$ \begin{align} 1+\frac{zf^{\prime \prime }( z) }{f^{\prime }( z) } &=1+2a_{2}z+( 6a_{3}-4a_{2}^{2}) z^{2}+( 12a_{4}-18a_{2}a_{3}+8a_{2}^{3}) z^{3} \notag \\ &\quad +( 20a_{5}-32a_{2}a_{4}-18a_{3}^{2}+48a_{3}a_{2}^{2}-16a_{2}^{4}) z^{4}+\cdots , \end{align} $$

we obtain by comparing coefficients from (1.5), (2.2) and (2.3),

(2.4) $$ \begin{align} a_{2}& =\tfrac{1}{2}c_{1},\end{align} $$

(2.5) $$ \begin{align}a_{3}& =-\tfrac{1}{6}c_{2}+\tfrac{1}{6}c_{1}^{2},\end{align} $$

(2.6) $$ \begin{align}a_{4}& =\tfrac{1}{24}c_{1}^{3}+\tfrac{1}{8}c_{1}c_{2}+\tfrac{1}{12}c_{3},\end{align} $$

(2.7) $$ \begin{align}a_{5}& =\tfrac{1}{120}c_{1}^{4}+\tfrac{1}{20}c_{2}c_{1}^{2}+\tfrac{1}{15} c_{1}c_{3}+\tfrac{1}{40}c_{2}^{2}+\tfrac{1}{20}c_{4}. \end{align} $$

Next note that since $f( f^{-1}(w)) =w$ , using (1.3), it follows that

(2.8) $$ \begin{align} A_{2}& =-a_{2},\end{align} $$

(2.9) $$ \begin{align}A_{3}& =2a_{2}^{2}-a_{3},\end{align} $$

(2.10) $$ \begin{align}A_{4}& =5a_{2}a_{3}-5a_{2}^{3}-a_{4},\end{align} $$

(2.11) $$ \begin{align}A_{5}& =14a_{2}^{4}-21a_{3}a_{2}^{2}+6a_{2}a_{4}+3a_{3}^{2}-a_{5}.\end{align} $$

Substituting (2.4)–(2.7) into (2.8)–(2.11), respectively, we obtain

(2.12) $$ \begin{align} A_{2}& =-\tfrac{1}{2}c_{1},\end{align} $$

(2.13) $$ \begin{align}A_{3}& =\tfrac{1}{3}c_{1}^{2}-\tfrac{1}{6}c_{2},\end{align} $$

(2.14) $$ \begin{align}A_{4}& =-\tfrac{1}{4}c^{3}+\tfrac{7}{24}c_{1}c_{2}-\tfrac{1}{12}c_{3},\end{align} $$

(2.15) $$ \begin{align}A_{5}& =\tfrac{1}{5}c^{4}-\tfrac{23}{60}c_{1}^{2}c_{2}+\tfrac{11}{60}c_{1}c_{3}+ \tfrac{7}{120}c_{2}^{2}-\tfrac{1}{20}c_{4}. \end{align} $$

Using (2.12)–(2.15) in (1.4),

$$ \begin{align*} H_{3}(1)(f^{-1}) & = \tfrac{1}{34560} (16c_{1}^{6}-96c_{1}^{4}c_{2}+48c_{1}^{3}c_{3}+156c_{1}^{2}c_{2}^{2}-144c_{1}^{2}c_{4} \\ & \quad + 144c_{1}c_{2}c_{3}-176c_{2}^{3}+288c_{2}c_{4}-240c_{3}^{2}), \end{align*} $$

which after simplification, noting that since both the class $\mathcal {C}$ and the functional $H_{3}(1)(f^{-1})$ are rotationally invariant (so that $ c\in \lbrack 0,2]$ ), and using (1.6)–(1.8) gives

$$ \begin{align*} H_{3}(1)(f^{-1})=\tfrac{1}{34560}( v_{1}(c,\delta )+v_{2}(c,\delta )\eta +v_{3}(c,\delta )\eta ^{2}+\psi (c,\delta ,\eta )\rho ) , \end{align*} $$

where $\delta ,\ \eta $ , $\rho \in \overline {\mathbb {D}}$ and

$$ \begin{align*} v_{1}(c,\delta )& :=\delta ^{2}(4-c^{2})^{2}(-16\delta +10c^{2}\delta +3c^{2}+3c^{2}\delta ^{2}), \\ v_{2}(c,\delta )& :=-12\delta (4-c^{2})^{2}(1-|\delta |^{2})c(\delta +1), \\ v_{3}(c,\delta )& :=-12(4-c^{2})^{2}(1-|\delta |^{2})(5+\delta ^{2}), \\ \psi (c,\delta ,\eta )& :=72\delta (4-c^{2})^{2}(1-|\delta |^{2})(1-|\eta |^{2}). \end{align*} $$

Next, using $|\delta |=x,|\eta |=y$ and the fact $|\rho |\leq 1$ , we obtain

$$ \begin{align*} H_{3}(1)(f^{-1}) & \leq \tfrac{1}{34560}( |v_{1}(c,x)|+|v_{2}(c,x)|y+|v_{3}(c,x)|y^{2}+|\psi (c,x,y)|) \\ & \leq G(c,x,y), \end{align*} $$

where

$$ \begin{align*} G(c,x,y):=\tfrac{1}{34560}( g_{1}(c,x)+g_{2}(c,x)y+g_{3}(c,x)y^{2}+g_{4}(c,x)(1-y^{2})) , \end{align*} $$

with

$$ \begin{align*} g_{1}(c,x)& :=x^{2}(4-c^{2})^{2}(16x+10c^{2}x+3c^{2}+3c^{2}x^{2}), \\ g_{2}(c,x)& :=12x(4-c^{2})^{2}(1-x^{2})c(x+1), \\ g_{3}(c,x)& :=12(4-c^{2})^{2}(1-x^{2})(5+x^{2}), \\ g_{4}(c,x)& :=72x(4-c^{2})^{2}(1-x^{2}). \end{align*} $$

Thus, we need to maximise $G(c,x,y)$ over the closed cuboid $\Lambda :[0,2]\times \lbrack 0,1]\times \lbrack 0,1]$ . To do this, we find the maximum values in the interior of $\Lambda $ , in the interior of the six faces, at the vertices and on the 12 edges.

I. The interior of $\Lambda $ . Let $(c,x,y)\in (0,2)\times (0,1)\times (0,1)$ . We can write

$$ \begin{align*} 34560\, G(c,x,y)=g_{1}(c,x)+g_{4}(c,x)+g_{2}(c,x)y+( g_{3}(c,x)-g_{4}(c,x)) y^{2}. \end{align*} $$

Since $g_{2}(c,x)>0$ and

$$ \begin{align*} ( g_{3}(c,x)-g_{4}(c,x)) =12( 4-c^{2}) ( 1-x^{2})>0, \end{align*} $$

for $c\in (0,2), x\in (0,1)$ ,

$$ \begin{align*} G(c,x,y) < G(c,x,1):=\tfrac{1}{34560}( g_{1}(c,x)+g_{2}(c,x)+g_{3}(c,x)) \end{align*} $$

for $c\in (0,2), x\in (0,1)$ and $y<1$ . Thus, $G(c,x,y)$ does not have a maximum for $(c,x,y)\in (0,2)\times (0,1)\times [0,1).$

II. The interiors of the faces of $\Lambda $ . We deal with each face in turn, noting that we do not need to consider the face $y=0$ after Step I.

On the face $c=0$ , $G(c,x,y)$ reduces to

$$ \begin{align*} k_{1}(x,y):=G(0,x,y)=\frac{3(1-x^{2})(x-1)(x-5)y^{2}-2x(7x^{2}-9)}{540}, \quad x, y\in (0,1), \end{align*} $$

and we see that $k_{1}$ has no critical points in $(0,1)\times (0,1)$ since

$$ \begin{align*} \frac{\partial k_{1}}{\partial y}=\frac{(1-x^{2})(x-1)(x-5)y}{90}\neq 0, \quad x, y\in (0,1). \end{align*} $$

On the face $c=2$ , $G(c,x,y)$ reduces to

$$ \begin{align*} G(2,x,y)=0, \quad x, y\in (0,1). \end{align*} $$

On the face $x=0$ , $G(c,x,y)$ reduces to

$$ \begin{align*} k_{2}(c,y):=G(c,0,y)=\frac{(4-c^{2})^{2}y^{2}}{576}, \quad c\in (0,2), y\in (0,1), \end{align*} $$

and we see that $k_{2}$ has no critical point in $(0,2)\times (0,1)$ since

$$ \begin{align*} \frac{\partial k_{2}}{\partial y}=\frac{(4-c^{2})^{2}y}{288}\neq 0, \quad c\in (0,2), y\in (0,1). \end{align*} $$

On the face $x=1$ , $G(c,x,y)$ reduces to

$$ \begin{align*} k_{3}(c,y):=G(c,1,y)=\frac{c^{6}-7c^{4}+8c^{2}+16}{2160}, \quad c\in (0,2). \end{align*} $$

Solving ${\partial k_{3}}/{\partial c}=0$ , we obtain a maximum value $ {25}/{2916}<{1}/{36}$ at $c:=c_{0}={\sqrt {6}}/{3}$ .

Finally, on the face $y=1$ , $G(c,x,y)$ reduces to

$$ \begin{align*} k_{4}(c,x):=G(c,x,1)=\frac{(4-c^{2})^{2}}{34560}\left( \begin{array}{@{}l@{}} 16x^{3}+10x^{3}c^{2}+3x^{2}c^{2}+3x^{4}c^{2}+12x^{2}c \\ -\,12x^{4}c+12xc-12x^{3}c+60-48x^{2}-12x^{4} \end{array} \right). \end{align*} $$

Differentiating $k_{4}(c,x)$ with respect to c and $x$ , we obtain

$$ \begin{align*} \frac{\partial k_{4}}{\partial x}=\frac{(4-c^{2})^{2}}{34560}\left( \begin{array}{@{}c@{}} (3c^{2}-12c-12)x^{4}+(10c^{2}-12c+16)x^{3} \\ +\,(12c-48+3c^{2})x^{2}+12xc+60 \end{array} \right) \end{align*} $$

and

$$ \begin{align*} \frac{\partial k_{4}}{\partial c}=\frac{-4(4-c^{2})}{34560}\left( \begin{array}{@{}c@{}} x^{2}(6x^{2}+20x+6)c^{5}+12x(1-x^{2})(x+1)c^{4} \\ -15x^{2}(x+3)(3x+1)c^{3}-84x(1-x^{2})(x+1)c^{2} \\ +\,(336x^{3}+60+48x^{2}+84x^{4})c+192x(1-x^{2})(x+1) \end{array} \right). \end{align*} $$

Now equating both partial derivatives to zero, a numerical calculation shows that the only real roots are at $(c,x)=(7.03237\ldots , -5.85791\ldots ), (-0.95398, 1.04715\ldots )$ and $(c,x)=(0,0).$ We see that $c=x=0$ gives a maximum value ${1}/{36}$ and all other critical points are saddle points so there does not exist a maximum inside $(0,2)\times (0,1)$ .

III. The vertices of $\Lambda $ . We have

$$ \begin{align*} G(0,0,0)& =0,~~~G(0,0,1)=\tfrac{1}{36},~~~G(0,1,0)=\tfrac{1}{135},~~~G(0,1,1)= \tfrac{1}{135}, \\ &G(2,0,0) =G(2,0,1)=G(2,1,0)=G(2,1,1)=0. \end{align*} $$

IV. The interiors of the edges of $\Lambda $ . Finally, we find the points of the maxima of $G(c,x,y)$ on the 12 edges of $ \Lambda $ :

$$ \begin{align*} G(c,0,0)=0; \end{align*} $$

$$ \begin{align*} G(c,0,1) =\frac{(4-c^{2})^{2}}{576}\leq G(0,0,1) =\frac{1}{36}, \quad c\in (0,2); \end{align*} $$

$$ \begin{align*} G(c,1,0) =\frac{c^{6}-7c^{4}+8c^{2}+16}{2160}\leq G(\lambda _{1},1,0) =\frac{25}{2916}<\dfrac{1}{36}, \quad c\in (0,2), \end{align*} $$

where

$$ \begin{align*} c:=\lambda _{1}=\frac{\sqrt{6}}{3}; \end{align*} $$

$$ \begin{align*} G(0,x,0)=\frac{-x(7x^{2}-9)}{270}\leq G\left(0,\frac{\sqrt{21}}{7},0\right)=\frac{\sqrt{ 21}}{315}<\dfrac{1}{36}, \quad x\in (0,1); \end{align*} $$

$$ \begin{align*} G(0,x,1)=\frac{-3x^{4}+4x^{3}-12x^{2}+15}{540}\leq G(0,0,1)=\frac{1}{36} , \quad x\in (0,1); \end{align*} $$

$$ \begin{align*} G(2,x,0)& =0, \quad x\in (0,1); \end{align*} $$

$$ \begin{align*} G(2,x,1)& =0, \quad x\in (0,1); \end{align*} $$

$$ \begin{align*}G(0,0,y)& =\tfrac{1}{36}y^{2}\leq \tfrac{1}{36}, \quad y\in (0,1); \end{align*} $$

$$ \begin{align*} G(0,1,y)& =\tfrac{1}{135}, \quad y\in (0,1); \end{align*} $$

$$ \begin{align*}G(2,0,y)& =0, \quad y\in (0,1); \end{align*} $$

$$ \begin{align*} G(2,1,y)& =0, \quad y\in (0,1). \end{align*} $$

Thus, there is only one maximum point at $( 0,0,1)$ where $G(0,0,1)={1}/{36}$ . We therefore conclude that

$$ \begin{align*} |H_{3}(1)(f^{-1})|\leq \tfrac{1}{36}. \end{align*} $$

The result is sharp for $f_{0}$ given in (2.1), which is equivalent to choosing $a_{2}=a_{3}=a_{5}=0$ and $a_{4}=\tfrac 16.$ From (2.8)–(2.11), we see that $A_{2}=A_{3}=A_{5}=0$ and $ A_{4}=\tfrac 16$ , which from (1.4) gives $|H_{3}(1)(f^{-1})|={1}/{36}.$ This completes the proof.