Yamakawa, Rikuo

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TitleRadii of Starlikeness and Convexity for analytic functions withbounded derivative (On Schwarzian Derivatives and ItsApplications)Author(s)Yamakawa, RikuoCitation数理解析研究所講究録 (2013), 1824: 107-110Issue Date2013-02URL http://hdl.handle.net/2433/194719RightType Departmental Bulletin PaperTextversionpublisherKyoto UniversityRadii of Starlikeness and Convexity for analyticfunctions with bounded derivativeRikuo YamakawaAbstractWe 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 families1 IntroductionLet $\mathbb{D}_{R}=\{|z|<R\}(0<R\leq 1)$ , and for brevity we write $\mathbb{D}_{1}=\mathbb{D}.$Let $\mathcal{A}$ be the class of functions of the form$f(z)=z+ \sum_{n=2}^{\infty}a_{n}z^{n}$ (1.1)that are analytic in $\mathbb{D}$ . And let $\mathcal{S}^{*}$ and $\mathcal{C}$ denote the subclass of $\mathcal{A}$ consisting of functionswhich are starlike and convex, respectively. i.e.$f(z)\in \mathcal{S}^{*}$ in $\mathbb{D}_{R}\Leftrightarrow Rae\frac{zf’(z)}{f(z)}>0(z\in \mathbb{D}_{R})$, (1.2)$f(z)\in C$ in $\mathbb{D}_{R}\Leftrightarrow{\rm Re}\{1+\frac{zf’(z)}{f’(z)}\}>0(z\in \mathbb{D}_{R})$ . (1.3)A function $f(z)\in \mathcal{A}$ is said to be in the class $\mathcal{B}_{r}$ if it satisfies$|f’(z)-1|<r(z\in \mathbb{D})$ . (1.4)We investigate the following three problems.Problem 1. Find the maximum value $R_{1}$ of $R$ s.t.$f\in \mathcal{B}_{1}\Rightarrow f(z)\in \mathcal{S}^{*}$ in $\mathbb{D}_{R}$Problem 2. Find the maximum value $R_{2}$ of $r$ s.t.$f\in \mathcal{B}_{r}\Rightarrow f(z)\in \mathcal{S}^{*}$ in $\mathbb{D}$Problem 3. Find the maximum value $R_{3}(r)$ of $R$ s.t.$f\in \mathcal{B}_{r}\Rightarrow f(z)\in C$ in $\mathbb{D}_{R}$2004 Mathematics Subject Classij’cation: Primary $30C45.$Key Words and Phrases: Radius of starlikeness, radius of convexity.1824 2013 107-110 1072 Known resultsFor $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, for $R_{1}$ and $R_{2}$ , we improve little. And for $R_{3}$ we investigate $R_{3}(r)$ .3 ResultsTheorem 1. $R_{1}<0.99815$ and $R_{2}<0.9961$ . i.e.(1) Suppose $f(z)\in \mathcal{A}$ 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$ $(z\in \mathbb{D})$ .Proof (1) We only have to show that there exist $f(z)\in \mathcal{B}_{1}$ , and $z_{0}\in\{|z|=0.99815\}$such that $Re\frac{z_{0}f’(z_{0})}{f(z_{0})}<0$ . Let$f’(z)=1+z \frac{1+az}{a+z}(a>1)$ , (3.1)108then 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$$\blacksquare$Theorem 2.$R_{3}(r)\{$ $== \alpha(r)\frac{1}{2r}$ $( \frac{1+\sqrt{5}}{4}<r\leq 1)(0<r\leq\frac{1+\sqrt{5}}{4})$,where $\alpha(r)=\{\frac{\sqrt{(1-r)(1+3r)}-(1-r)}{2r}\}^{\frac{1}{2}}$Proof From$|f’(z)-1|<r (0<r\leq 1) (z\in \mathbb{D})$ ,we can write$f’(z)=1+rzg(z)$ . (3.4)Where $g(z)$ is analytic and$|g(z)|\leq 1$ , (3.5)$|g’(z)| \leq\frac{1-|g(z)|^{2}}{1-|z|^{2}}$ (3.6)109in $\mathbb{D}$ . And evidentlySo,from (3.6), setting$| \frac{zfl’(z)}{f(z)}|=\frac{|rz\{g(z)+zg’(z)\}|}{|1+rzg(z)|}.$$|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$| \frac{zf}{f(z)}|<1\Rightarrow$ $\{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 univa-lent 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.R. Ya-makawa: Emeritus Prof ofShibaura Institute of Technology$e$-mail : yamakawa@sic. shibaura-it. $ac$.jp110

Radii of Starlikeness and Convexity for analytic functions with bounded derivative (On Schwarzian Derivatives and Its Applications)

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Radii of Starlikeness and Convexity for analytic functions with bounded derivative (On Schwarzian Derivatives and Its Applications)

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