On the Ninth Coefficient of the Inverse of a Convex Function

Abstract

We consider the inverse function $z=g\left(w\right)=w+b_2{w}^2+\cdots$ of a normalized convex univalent function $w=f\left(z\right)=z+a_2{z}^2+\cdots$ on the unit disk in the complex plane. So far, it is known that $|b_n|\le 1$ for $n=2,3,\cdots ,8.$ On the other hand, the inequality $|b_n|\le 1$ is not valid for $n=10.$ It is conjectured that $|b_9|\le 1.$ The present paper offers the estimate $|b_9|<1.617.$

1. Introduction

An analytic function f on the unit disk $\mathbb{D}=\{z\in \mathbb{C}:|z|<1\}$ of the complex plane $\mathbb{C}$ is called convex if f maps $\mathbb{D}$ univalently onto a convex domain in $\mathbb{C}.$ Let $\mathcal{K}$ denote the class of convex functions f normalized so that $w=f\left(z\right)=z+a_2{z}^2+a_3{z}^3+\cdots .$ For a function $f\in \mathcal{K},$ we expand the inverse function $g=f^{-1}:f\left(\mathbb{D}\right)\to \mathbb{D}$ as a power series of the form

$$g\left(w\right)=w+b_2{w}^2+b_3{w}^3+\cdots .$$

It is known that $|a_n|\le 1$ for every $n\ge 2.$ This is sharp for every n and, indeed, the function $f_0\left(z\right)=z/(1-z)=z+z^2+z^3+\cdots$ satisfies the equality for all $n.$ Note that $g_0\left(w\right)=f_0^{-1}\left(w\right)=w/(1+w)=w-w^2+w^3-\cdots$ satisfies $|b_n|=1$ for all $n.$ Since $f\left(\mathbb{D}\right)$ contains the disk $\left|w\right|<1/2$ for every $f\in \mathcal{K},$ the radius of convergence of the above $g\left(w\right)$ is at least $1/2;$ namely, $lim\; sup|b_n{|}^{1/n}\le 2,$ and the number 2 is sharp. Thus we cannot expect small bounds for $b_n.$ Nevertheless, it has been proved so far that $|b_n|\le 1$ for $n=2,3,\cdots ,8$ (Libera-Złotkiewicz [1] for $n\le 7$ and Campschroer [2] for $n=8$). For clarity, we define the quantity

$$M_n=\underset{f\in \mathcal{K}}{sup}\left|b_n\right|=\underset{f\in \mathcal{K}}{sup}\frac{\left|{\left(f^{-1}\right)}^{\left(n\right)}\left(0\right)\right|}{n!}$$

for $n\ge 2.$ Then $M_n=1$ for $n=2,3,\cdots ,8.$ On the other hand, Kirwan and Schober [3] showed that $M_{10}>1.$ In the same paper, Kirwan and Schober also gave the estimate

$$M_n<\frac{2^n\mathsf{\Gamma }\left(\frac{n+1}{2}\right)}{\sqrt{\pi }n\mathsf{\Gamma }(\frac{n}{2}+1)}\sim \sqrt{\frac{2}{\pi }}\frac{2^n}{n^{3/2}}.$$

(1)

Moreover, for each $0<\epsilon <2,$ there is a number $n_{\epsilon }$ such that

$$M_n>\frac{2-\epsilon }{e^2}·\frac{2^n}{n^3},n\ge n_{\epsilon }.$$

Clunie [4] showed that $M_n=O(2^n{n}^{-3}logn)$ as $n\to \infty$ and conjectured that $M_n=O\left(2^n{n}^{-3}\right).$ The conjecture was confirmed by Campschroer [5].

It is believed that $M_9=1$ but this has neither been proved nor disproved so far. The estimate in Equation (1) gives in this case

$$M_9<\frac{2^9·4!}{9\sqrt{\pi }\mathsf{\Gamma }(11/2)}=\frac{131072}{2835\pi }\approx 14.7166.$$

The main purpose of this short note is to show the following.

Theorem 1.

$$M_9\le \frac{12223}{7560}\approx 1.6168.$$

The estimate is not optimal. We may find a better partition of the expression of $b_9$ for the proof in Section 3. However, it seems difficult to prove $|b_9|\le 1.$

2. Some Results on Carathéodory Functions

An analytic function P on the unit disk $\mathbb{D}$ is called Carathéodory if $ReP\left(z\right)>0$ for $z\in \mathbb{D}$ and $P\left(0\right)=1.$ We denote by $\mathcal{P}$ the class of Carathéodory functions. We expand $P\in \mathcal{P}$ in the forms

$$P\left(z\right)=1+\sum _{n=1}^{\infty }{d}_n{z}^n=1+2\sum _{n=1}^{\infty }{p}_n{z}^n,\left|z\right|<1.$$

The following general estimates are useful.

Lemma 1.

Let $P\in \mathcal{P}$ be expanded as above. Then

(i)

$|d_n|\le 2(n=1,2,\cdots ),$

(ii)

$|d_{n+k}-d_n{d}_k|\le 2(k,n=1,2,\cdots ),$

(iii)

$|d_{n+k}-d_n{d}_k/2|\le 2-|d_n{d}_k|/2(k,n=1,2,\cdots ).$

The inequalities in (i), (ii) and (iii) are due to Carathéodory [6], Livingston [7] and Campschroer [2], respectively. See also [8]. Note that (ii) follows from (iii); in other words, (iii) is a refinement of (ii). Let A and B be square matrices of order $n.$ We will say that A is majorized by B and write $A\ll B$ if the inequality $\left|Ax\right|\le \left|Bx\right|$ holds for each vector $x\in {\mathbb{C}}^n.$ Here, the norm of a vector $x={(x_1,\cdots ,x_n)}^T\in {\mathbb{C}}^n$ is defined by $\left|x\right|=\sqrt{|x_1{|}^2+\cdots +{\left|x_n\right|}^2}$ as usual. For $P\left(z\right)=1+d_{1}z+d_2{z}^2+\cdots ,$ we define two kinds of Toeplitz matrices of order n by

$$A_n=\left(\begin{array}{ccccc}{d}_1& d_2& d_3& \cdots & d_n\\ 0& d_1& d_2& \cdots & d_{n-1}\\ 0& 0& d_1& \cdots & d_{n-2}\\ & & & \ddots & \\ 0& 0& 0& \cdots & d_1\end{array}\right)\mathrm{and}{B}_n=\left(\begin{array}{ccccc}2& d_1& d_2& \cdots & d_{n-1}\\ 0& 2& d_1& \cdots & d_{n-2}\\ 0& 0& 2& \cdots & d_{n-3}\\ & & & \ddots & \\ 0& 0& 0& \cdots & 2\end{array}\right).$$

Then Campschroer [2] (Example 1.XIII) showed the following.

Lemma 2.

$A_n\ll B_n$ for a Carathédory function $P\left(z\right)=1+d_{1}z+d_2{z}^2+\cdots .$

For $n=6,$ we take $x={(0,0,d_1{d}_2-d_3,-d_2,0,1)}^T$ and apply the above lemma to get $|A_{n}x|\le |B_{n}x|$ for $P\in \mathcal{P};$ that is,

$$|d_1{d}_2{d}_3-d_3^2-d_2{d}_4+d_6|\le 2.$$

Similarly, for $n=8,x={(0,0,d_1{d}_4-d_5,-d_4,0,0,0,1)}^T,$ we obtain

$$|d_1{d}_3{d}_4-d_3{d}_5-d_4^2+d_8|\le 2.$$

The coefficient inequalities for $P\in \mathcal{P}$ are more convenient for the later use if we express them in terms of $p_n=d_n/2.$ Thus we summarize the above inequalities in the following form.

Lemma 3.

Let $P\left(z\right)=1+2p_{1}z+2p_2{z}^2+\cdots$ be a Carathédory function. Then the following inequalities hold.

(i)

$|p_n|\le 1(n=1,2,\cdots ),$

(ii)

$|p_{n+k}-2p_n{p}_k|\le 1(k,n=1,2,\cdots ),$

(iii)

$|p_{n+k}-p_n{p}_k|\le 1-|p_n{p}_k|(k,n=1,2,\cdots ),$

(iv)

$\left|R\right|\le 1,$ where $R=4p_1{p}_2{p}_3-2p_3^2-2p_2{p}_4+p_6,$

(v)

$\left|S\right|\le 1,$ where $S=4p_1{p}_3{p}_4-2p_3{p}_5-2p_4^2+p_8.$

Note that the above inequalities are all sharp because the function $P_0\left(z\right)=(1+z)/(1-z)=1+2z+2z^2+\cdots$ satisfies equalities.

3. Proof of the Theorem

We will show Theorem 1 in this section. Since computations are often involved, we need symbolic computations by computers. Suppose that a function $w=f\left(z\right)=z+a_2{z}^2+\cdots$ belongs to $\mathcal{K}$ and let $z=g\left(w\right)=f^{-1}\left(w\right)=w+b_2{w}^2+\cdots .$ Then, by a formal calculation, we have the formula

$$\begin{array}{cc}\hfill b_9& =1430a_2^8-5005a_2^6{a}_3+2002a_2^5{a}_4+5005a_2^4{a}_3^2-715a_2^4{a}_5-2860a_2^3{a}_3{a}_4+220a_2^3{a}_6\hfill \\ \hfill & -1430a_2^2{a}_3^3+330a_2^2{a}_4^2+660a_2^2{a}_3{a}_5-55a_2^2{a}_7+660a_2{a}_3^2{a}_4-110a_2{a}_4{a}_5-110a_2{a}_3{a}_6\hfill \\ \hfill & +10a_2{a}_8+55a_3^4-55a_3{a}_4^2+5a_5^2-55a_3^2{a}_5+10a_4{a}_6+10a_3{a}_7+10a_2{a}_8-a_9.\hfill \end{array}$$

It is well known (see [9] for example) that a normalized analytic function f on $\mathbb{D}$ is convex if and only if $1+z{f}^{\prime \prime }\left(z\right)/f^{\prime }\left(z\right)$ has a positive real part for each $z\in \mathbb{D}.$ Therefore, there is a function $P\in \mathcal{P}$ such that

$$1+\frac{z{f}^{\prime \prime }\left(z\right)}{f^{\prime }\left(z\right)}=\frac{1}{P\left(z\right)},z\in \mathbb{D}.$$

If we expand P in the form $P\left(z\right)=1+2p_{1}z+2p_2{z}^2+\cdots ,$ we have the following relations

$$\begin{array}{cc}\hfill a_2& =-p_1,3a_3=4p_1^2-p_2,6a_4=-12p_1^3+7p_1{p}_2-p_3,\hfill \\ \hfill 30a_5& =96p_1^4-92p_1^2{p}_2+9p_2^2+20p_1{p}_3-3p_4,\hfill \\ \hfill 90a_6& =-480p_1^5+652p_1^3{p}_2-172p_1^2{p}_3-157p_1{p}_2^2+39p_1{p}_4+34p_2{p}_3-6p_5,\hfill \\ \hfill 630a_7& =5760p_1^6-10224p_1^4{p}_2+3024p_1^3{p}_3+4184p_1^2{p}_2^2-828p_1^2{p}_4-1468p_1{p}_2{p}_3\hfill \\ \hfill & +192p_1{p}_5-225p_2^3+80p_3^2+165p_2{p}_4-30p_6,\hfill \\ \hfill 2520a_8& =-40320p_1^7+88848p_1^5{p}_2-28368p_1^4{p}_3-52760p_1^3{p}_2^2+8676p_1^3{p}_4+23368p_1^2{p}_2{p}_3\hfill \\ \hfill & -2424p_1^2{p}_5+7227p_1{p}_2^3-2060p_1{p}_3^2-4239p_1{p}_2{p}_4+570p_1{p}_6-1899p_2^2{p}_3\hfill \\ \hfill & +465p_3{p}_4+486p_2{p}_5-90p_7,\hfill \\ \hfill 22680a_9& =645120p_1^8-1703808p_1^6{p}_2+574848p_1^5{p}_3+1345136p_1^4{p}_2^2-189216p_1^4{p}_4\hfill \\ \hfill & -686368p_1^3{p}_2{p}_3+58944p_1^3{p}_5-320648p_1^2{p}_2^3+76304p_1^2{p}_3^2+156696p_1^2{p}_2{p}_4\hfill \\ \hfill & -16680p_1^2{p}_6+141880p_1{p}_2^2{p}_3-27768p_1{p}_3{p}_4-28944p_1{p}_2{p}_5+3960p_1{p}_7\hfill \\ \hfill & +11025p_2^4-12488p_2{p}_3^2+1575p_4^2-12810p_2^2{p}_4+3192p_3{p}_5+3360p_2{p}_6-630p_8.\hfill \end{array}$$

We substitute these relations into the above expression of $b_9$ to obtain

$$\begin{array}{cc}\hfill B& :=22680b_9=16p_1^8+224p_1^6{p}_2+1752p_1^5{p}_3-464p_1^4{p}_2^2+7020p_1^4{p}_4\hfill \\ \hfill & -3512p_1^3{p}_2{p}_3+12336p_1^3{p}_5+412p_1^2{p}_2^3-3764p_1^2{p}_3^2-2454p_1^2{p}_2{p}_4\hfill \\ \hfill & +10380p_1^2{p}_6+650p_1{p}_2^2{p}_3-4002p_1{p}_3{p}_4-36p_1{p}_2{p}_5+4140p_1{p}_7+p_2^4\hfill \\ \hfill & +158p_2{p}_3^2-441p_4^2+66p_2^2{p}_4-672p_3{p}_5+240p_2{p}_6+630p_8.\hfill \end{array}$$

We now write B as follows:

$$\begin{array}{cc}\hfill B& =336S+1882p_1^{2}R+232p_1^2{p}_2{(p_1^2-p_2)}^2\hfill \\ \hfill & +(8p_1^6+3386p_1^2{p}_4+36p_1{p}_5)(2p_1^2-p_2)+4696p_1{p}_4(p_1{p}_2-p_3)\hfill \\ \hfill & +650p_1{p}_3(p_2^2-p_4)+1752p_1^3{p}_3(p_1^2-p_2)+9288p_1^3(p_5-p_2{p}_3)\hfill \\ \hfill & +180p_1^2{p}_2^3+p_2^4+158p_2{p}_3^2+248p_1^4{p}_4+66p_2^2{p}_4+231p_4^2\hfill \\ \hfill & +2976p_1^3{p}_5+8498p_1^2{p}_6+240p_2{p}_6+4140p_1{p}_7+294p_8\hfill \end{array}$$

where R and S are given in Lemma 3. We now apply Lemma 3 to obtain

$$\begin{array}{cc}\hfill \left|B\right|& \le 336+1882t_1^2+232t_1^2{t}_2{(1-t_1^2)}^2\hfill \\ \hfill & +(8t_1^6+3386t_1^2{t}_4+36t_1{t}_5)+4696t_1{t}_4(1-t_1{t}_2)\hfill \\ \hfill & +650t_1{t}_3(1-t_2^2)+1752t_1^3{t}_3(1-t_1^2)+9288t_1^3(1-t_2{t}_3)\hfill \\ \hfill & +180t_1^2{t}_2^3+t_2^4+158t_2{t}_3^2+248t_1^4{t}_4+66t_2^2{t}_4+231t_4^2\hfill \\ \hfill & +2976t_1^3{t}_5+8498t_1^2{t}_6+240t_2{t}_6+4140t_1{t}_7+294t_8,\hfill \end{array}$$

where $t_j=\left|p_j\right|(j=1,2,\cdots ,8).$ Note that $0\le t_j\le 1$ by Lemma 3 (i). Hence,

$$\begin{array}{cc}\hfill \left|B\right|& \le 336+1882t_1^2+232t_1^2{t}_2{(1-t_1^2)}^2\hfill \\ \hfill & +(8t_1^6+3386t_1^2+36t_1)+4696t_1(1-t_1{t}_2)\hfill \\ \hfill & +650t_1(1-t_2^2)+1752t_1^3(1-t_1^2)+9288t_1^3\hfill \\ \hfill & +180t_1^2{t}_2^3+t_2^4+158t_2+248t_1^4+66t_2^2+231\hfill \\ \hfill & +2976t_1^3+8498t_1^2+240t_2+4140t_1+294\hfill \\ \hfill & =h(t_1,t_2),\hfill \end{array}$$

where

$$\begin{array}{cc}\hfill h(x,y)& =232x^{6}y+8x^6-1752x^5-464x^{4}y+248x^4+14016x^3+180x^2{y}^3\hfill \\ \hfill & -4464x^{2}y+13766x^2-650x{y}^2+9522x+y^4+66y^2+398y+861.\hfill \end{array}$$

Since

$$\begin{array}{cc}\hfill \frac{\partial h}{\partial x}& =x^5(1392y+48)-8760x^4+x^3(992-1856y)+42048x^2\hfill \\ \hfill & +x\left(360y^3-8928y+27532\right)-650y^2+9522\hfill \\ \hfill & \ge 48x^5-8760x^4-864x^3+42048x^2+18964x+8872>0\hfill \end{array}$$

for $0\le x,y\le 1,$ we observe that $h(x,y)$ is increasing in $0\le x\le 1$ for a fixed $y\in [0,1].$ Therefore, $h(x,y)\le h(1,y)=y^4+180y^3-584y^2-4298y+36669=:H\left(y\right).$ We compute $H^{\prime }\left(y\right)=2(2y^3+270y^2-584y-2149)\le 2(2+270-2149)<0$ and thus conclude that $H\left(y\right)\le H\left(0\right)=36669$ for $0\le y\le 1.$ In summary, we have obtained

$$|b_9|=\frac{\left|B\right|}{22680}\le \frac{36669}{22680}=\frac{12223}{7560}\approx 1.6167989.$$

The proof is now complete.