数理解析研究所講究録
第 1824 巻 2013 年 107-110

107

Radii of Starlikeness and Convexity for analytic
functions with bounded derivative
Rikuo Yamakawa
Abstract
We consider two families of analytic functions $|f^{l}(z)-1|<r(|z|<1)$ , and $|f’(z)-1|<$$i(|z|<R)$ , then investigate radii of starhkeness and convexity for these families

1Introduction

Let $\mathbb{D}_{R}=\{|z|<R\}(0<R\leq 1)$ , and for brevity we write
Let be the class of functions of the form$\mathbb{D}_{1}=\mathbb{D}.$$\mathcal{A}$(1.1)$f(z)=z+ \sum_{n=2}^{\infty}a_{n}z^{n}$that are analytic in . And let
and denote the subclass ofwhich are starlike and convex, respectively. i.e.
$\mathbb{D}$$\mathcal{S}^{*}$$f(z)\in \mathcal{S}^{*}$$f(z)\in C$

A function$f(z)\in \mathcal{A}$inin$\mathcal{C}$consisting of functions$\mathcal{A}$$\mathbb{D}_{R}\Leftrightarrow Rae\frac{zf’(z)}{f(z)}>0(z\in \mathbb{D}_{R})$,$\mathbb{D}_{R}\Leftrightarrow{\rm Re}\{1+\frac{zf’(z)}{f’(z)}\}>0(z\in \mathbb{D}_{R})$is said to be in the class$\mathcal{B}_{r}$.

We investigate the following three problems.
$R_{1}$of

$R$

s.t.

$f\in \mathcal{B}_{1}\Rightarrow f(z)\in \mathcal{S}^{*}$

Problem 2. Find the maximum value$R_{2}$of$r$$R_{3}(r)$ofin$\mathbb{D}_{R}$s.t.

$f\in \mathcal{B}_{r}\Rightarrow f(z)\in \mathcal{S}^{*}$

Problem 3. Find the maximum value.

(1.3)

if it satisfies$|f’(z)-1|<r(z\in \mathbb{D})$Problem 1. Find the maximum value

(1.2)

$R$

$f\in \mathcal{B}_{r}\Rightarrow f(z)\in C$

in$\mathbb{D}$s.t.
in$\mathbb{D}_{R}$2004 Mathematics Subject Classij’cation: Primary $30C45.$
Key Words and Phrases: Radius of starlikeness, radius of convexity.

(1.4)

108

Known results2For$R_{1}$, first T.MacGregor showed in [1] that
$R_{1}\geq\frac{2}{\sqrt{5}}=0.894\cdots$

Lator M.Nunokawa showed in [3]$R_{1}>0.926\cdots$

And Nunokawa, Fukui, Owa, Saitoh and Sekine improved in [4]$R_{1}>0.933\cdots$

On the other hand P.Mocanu showed in [2] that
$R_{1}<1$

Using the Mocanu’s method, the present author[5] have showed that
$R_{1}<0.9982$

For$R_{2}$, P. Mocanu[2] also proved$R_{2}\geq 0.894\cdots$

R.Yamakawa[5] have showed$R_{2}<0.9962$

For$R_{3}$, T.MacGregor showed that
$R_{3}(1)=\frac{1}{2}$Now, for3$R_{1}$and$R_{2}$, we improve little. And for$R_{3}$we investigate$R_{3}(r)$.

Results

Theorem 1.
(1) Suppose$R_{1}<0.99815$ and $R_{2}<0.9961$$f(z)\in \mathcal{A}$. i.e.

satisfies $|f’(z)-1|<1$ , and suppose $R<0.99815$ then${\rm Re} \frac{zf’(z)}{f(z)}>0 (z\in \mathbb{D}_{R})$.

(2) Suppose $r<0.9961$ , and suppose $f(z)\in A$ satisfies $|f’(z)-1|<r$ , then${\rm Re} \{1+\frac{zf}{f’(z)}\}>0$

Proof (1) We only have to show that there exist
such that$Re\frac{z_{0}f’(z_{0})}{f(z_{0})}<0$$(z\in \mathbb{D})$.

$f(z)\in \mathcal{B}_{1}$, and $z_{0}\in\{|z|=0.99815\}$. Let$f’(z)=1+z \frac{1+az}{a+z}(a>1)$ ,(3.1)

109then from $f(0)=0$ , we have$f(z)=(2-a^{2})z+ \frac{a}{2}z^{2}+a(a^{2}-1)\log(1+\frac{z}{a})$

.

(3.2)

Putting$a=1.06559, z_{0}=re^{i\theta}$

where$r=0.99815, \theta=\pi+\cos^{-1}0.9479$

we obtain${\rm Re} \frac{z_{0}f’(z_{0})}{f(z_{0})}=-1.77497\cross 10^{-6}<0$$\blacksquare$(2) Similarly we set$f(z)=(1+(1-a^{2}) R)z+\frac{aR}{2}z^{2}+a(a^{2}-1)$ Rlog $(1+ \frac{z}{a})$

.

(3.3)

Then
$|f’(z)-1|= R|z||\frac{1+az}{a+z}|<R(z\in \mathbb{D})$.

Putting$a=1.065, R=0.9961, z_{0}=0.949e^{i\theta}$

where$\theta=\pi+\cos^{-1}0.949.$

We have${\rm Re} \frac{z_{0}f’(z_{0})}{f(z_{0})}=-0.000159\cdots<0$So$R_{2}<0.9961$$\blacksquare$

Theorem 2.
$R_{3}(r)\{$ $== \alpha(r)\frac{1}{2r}$where$( \frac{1+\sqrt{5}}{4}<r\leq 1)(0<r\leq\frac{1+\sqrt{5}}{4})$$\alpha(r)=\{\frac{\sqrt{(1-r)(1+3r)}-(1-r)}{2r}\}^{\frac{1}{2}}$,Proof From$|f’(z)-1|<r (0<r\leq 1) (z\in \mathbb{D})$ ,we can write$f’(z)=1+rzg(z)$ .

(3.4)

Where $g(z)$ is analytic and$|g(z)|\leq 1$,$|g’(z)| \leq\frac{1-|g(z)|^{2}}{1-|z|^{2}}$

(3.5)

(3.6)

110
in . And evidently$\mathbb{D}$$| \frac{zfl’(z)}{f(z)}|=\frac{|rz\{g(z)+zg’(z)\}|}{|1+rzg(z)|}.$

So,from (3.6), setting$|z|=s,$$|g(z)|=t(0\leq s, t\leq 1)$

we have$| \frac{zf}{f’(z)}|\leq rs\frac{(1-s^{2})t+s(1-t^{2})}{(1-rst)(1-s^{2})}.$

Let$\phi(t)=rs^{2}t^{2}-2rs(1-s^{2})t+1-s^{2}-rs^{2},$then$\phi(t)>0\Rightarrow|\frac{zf}{f(z)}|<1.$

Since$| \frac{zf}{f(z)}|<1\Rightarrow$

距 $\{1+\frac{zf}{f(z)}\}>0,$$\phi(t)>0\Rightarrow f(z)\in C$

.

(3.7)
$\blacksquare$

References[1] T.MacGregor,

$A$

class

$0$

univalent functions, Proc. Amer.Math. Soc., 15 (1964), 311$-317.$[2] P.T.Mocanu, Some starlikeness conditions for analytic functions,Rev. Roumaine Math. Pures. Appl.,33(1988), 117–1264.
[3] M.Nunokawa, On the starlike boundary of univalent functions,Suugaku, 31(1979), 255–256 (Japanese).

[4] M.Nunokawa, S.Fukui, S.Owa, S.Satoh, T.Sekine, On the starlikeness bound of univalent functions,Math. Japonica, 33, No.5(1988), 763–767.

[5] R.Yamakawa, Notes for starlikeness conditions of analytic functions, Koukyuroku ofRIMS, 821 (1993), 112–116.

$e$

R. Ya-makawa: Emeritus Prof ofShibaura Institute of Technology-mail : yamakawa@sic. shibaura-it. $ac$ .jp