Yamakawa, Rikuo

Kyoto University Research Information Repository

TitleRADIUS OF STARLIKENESS AND CONVEXITY 2 (Studyon Applications for Fractional Calculus Operators in UnivalentFunction Theory)Author(s)Yamakawa, RikuoCitation数理解析研究所講究録 (2004), 1363: 131-135Issue Date2004-04URL http://hdl.handle.net/2433/25313RightType Departmental Bulletin PaperTextversionpublisherKyoto University131RADIUS OF STARLIKENESS AND CONVEXITY 2RIKUO YAMAKAWA [ ] ( )SHIBAURA INSTITUTE OF TECIflVOLOGY. JAPANABSTRACT. In this note, we will investigate radius of starlikenessand covexity for certain analytic functions on the unit disk.1. INTRODUCTION AND NOTATIONSLet $U=\{|z|<1\}$ and let $B$ denote the family of functions $f(z)=$$z+ \sum_{\mathrm{n}=2}^{\infty}a_{n}z^{n}$ which are analytic in $U$ , and which satisfy the condition$|4|\leq n(n=2,3,4, \cdots)$ . For functions $f\in B,$ we consider the radiusof univalence , starlikeness and convexity i.e.,$R_{U}$ $= \max${$r:f\in B\Rightarrow f$ is univalent in $|z|<r$ $<1$}$R_{S}$ $= \max${$r:f\in B\Rightarrow f$ is starlike in $|z|<r<1$}$R_{G}$ $=$ $\max$ {$r:f\in B\Rightarrow f$ is convex in $|z|<r<1$}For $R_{U}$ , Yong Chan Kim and Mamoru Nunokawa obtained in [2] thefollowing beautiful result:Theorem A. $R_{U}=0.164878\cdot\cdot$ , , where $R_{U}$ is the root of the equa-tion $2r^{3}-6r^{2}+7r-1=0.$And for $R_{S}$ , Shigeyoshi Owa and Nunokawa [1] showedTheorem B. $R_{S}>0.08998\cdots \mathrm{t}$The present author has showed in [3]$R_{S}>$ 0.104 $\cdots$ , and $R_{C}>$ 0.056 $\ldots$ .’s\mathit{3}}>0.08998\cdots${C}>0.056\cdots$In this note, we shall improve the above two results.2. MAIN RESULTSTheorem 1. $R_{S}>0.137788\cdots$Theorem 2. $R_{G}>$ 0.075450 $\ldots$Proof of theorem 1. Noting $[perp] fz$) is analytic in $\mathrm{U}$ , we will need tworesults. The first is the $\mathrm{f}\mathrm{o}11\mathrm{o}\mathrm{w}\mathrm{i}\mathrm{n}\mathrm{g}^{z}$lemma due to Owa and Nunokawa:1991 Mathematics Subject Classification, Primaly $30\mathrm{C}45;\mathrm{S}\mathrm{e}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{a}\mathrm{r}\mathrm{y}26\mathrm{A}33$ ,$30\mathrm{C}80$ .1363 2004 131-135132Lemma 1.(1) $\{$ $\mathrm{r}\frac{14r+2\mathrm{r}^{2}}{1-\mathrm{r}),0}$,$\leq|\frac{f(z)}{z}|\leq$ $\mathrm{f}1-\neg r)1$ $(0\leq|z|=r<R_{1})$$\leq|\frac{f(z)}{z}|\leq_{(1-r)}\neg 1$ $(R_{1}\leq r<1)$where $R_{1}=1-\tau_{2}^{1}=$ 0.29289 . . 1The second is Poisson-Jensen’s Formula.(2) $\log\frac{f(z)}{z}=\frac{1}{2\pi}\int_{0}$”$\log|\frac{f(\zeta)}{\zeta}|$ . $\frac{\zeta+z}{\zeta-z}\mathrm{d}\phi$where $\zeta=\rho e^{i\phi}$ , $z=re^{\theta}.\cdot$ and $0\leq r<\rho<1.$ By logarithmicdifferentiation, we find from (2) that(3) $R_{\mathrm{e}} \frac{zf’(z)}{f(z)}=1+\frac{1}{2\pi}\int_{0}^{2\pi}\log|\frac{f(\zeta)}{\zeta}|R_{e}\frac{2\zeta z}{(\zeta-z)^{2}}\mathrm{d}\phi$where$R_{e} \frac{2\zeta z}{(\zeta-z)^{2}}=2\rho r\frac{(\rho^{2}+r^{2})\cos(\phi-\theta)-2\rho r}{\{\rho^{2}+r^{2}-2\rho r\cos(\phi-\theta)\}^{2}}$..For brevity, putting(4) $\lambda=\frac{2\rho r}{\rho^{2}+r^{2}}$ , $\alpha=\phi-\theta$and(5) $\beta=\cos^{-1}\lambda$ , $g( \alpha)=\frac{\lambda(\cos\alpha-\lambda)}{(1-\lambda\cos\alpha)^{2}}$we have from (3)(6)$R_{e} \frac{zf’(z)}{f(z)}=1+\frac{1}{2\pi}$ $\mathrm{C}$ $g(\alpha)\log$ $| \frac{f(\zeta)}{\zeta}|\mathrm{d}\phi+\frac{1}{2\pi}I_{\beta}^{2\pi-\beta}!/(’)$ $\log$$| \frac{f(\zeta)}{\zeta}|\mathrm{d}\phi$.Since $R_{1}>R_{U}$ , we may assume $r<R_{1}$ . And since$\{$$g(\alpha)\geq 0$ $(-\beta\leq\alpha\leq\beta)$$g(\alpha)\leq 0$ ($\beta\leq$ a $\leq 2\pi\beta$),we deduce from (1) and (5)133$R_{e} \frac{zf(z)}{f(z)},\geq 1+\frac{1}{2\pi}\int_{-\beta}^{\beta}g(\alpha)\log\frac{1-4\rho+2\rho^{2}}{(1-\rho)^{2}}\mathrm{d}\alpha$$+ \frac{1}{2\pi}\int_{\beta}^{2\pi-\beta}g(\alpha)\log\frac{1}{(1-\rho)^{2}}\mathrm{d}\alpha$Evidently7$g( \alpha)d\alpha=\int\frac{\lambda(\cos\alpha-\lambda)}{(1-\lambda\cos\alpha)^{2}}d\alpha=\lambda\frac{\sin\alpha}{1-\lambda\cos\alpha}$,So we have(7) $R_{e} \frac{zf(z)}{f(z)},\geq 1-1^{\log\frac{(1-\rho)^{2}}{1-4\rho+2\rho^{2}}}+\log$ $\frac{1}{(1-\rho)^{2}}$ }. $\frac{\sin\sqrt}{1-\lambda\cos\sqrt}$It deduce$R_{e} \frac{zf’(z)}{f(z)}\geq 1-\frac{\lambda}{\pi\sqrt{1-\lambda}}$ . $\log\frac{1}{1-4\rho+2\rho^{2}}$ ,$R_{e} \frac{zf’(z)}{fz)}\geq 1-\frac{2\rho r}{\pi(\rho^{22}-r^{\tau})}\cdot\log\frac{1}{1-4\rho+2\rho^{2}}$ .We set$\phi(r, \rho)=1-\frac{2\rho r}{\pi(\rho^{2}-r^{2})}$ . 1Og $\frac{1}{1-4\rho+2\rho^{2}}$ .According to$\phi(r,\rho_{0})\nearrow$ ( for fixed $\rho_{0}$ )We deduce$\phi(r_{0}, \rho 0)\geq 0\Rightarrow\phi(r, \rho 0)\geq 0$ ( for $r\leq r_{0}$ ).On the other hand$\phi(r_{8},2_{3})$ $=$ 9.47656 $\mathrm{x}10^{-6}>0$So we have$R_{S}\geq r_{3}=$ 0.137788phi(r_{8}, \rho_{3})=9.47656\mathrm{x}10^{-6}>0$_{s}\geq r3=0.137788$Proof of theorem $B$ . The proof is almost similar. We begin with thefollowing lemma;Lemma 2.(8) $\{$$1-7r+6r^{2}-2r^{8}(1\tilde{-r})$ $\leq|f’(z)|\leq$ $\varpi-\mathrm{r}1+\mathrm{r}\mathrm{f}\text{ }\mathrm{f}$0 $\leq|f’(z)|\leq(\neg 1-r1+r_{\mathrm{F}}$$z)|\leq\varpi 1-\mathrm{r}1+r_{\mathrm{f}\mathrm{f}}$ $(0\leq r<R_{3})$$z)|\leq 7^{\frac{1+r}{1_{-r}}\mathrm{F}}$ $(R_{3}\leq r<1)$134where, $R_{3}=R_{U}=0.164878\cdot\cdot \mathrm{c}$ It ’s very easy to prove the lemma.First,$|f’(z)$ $| \leq\sum_{n=1}^{\infty}742-1=\frac{1+r}{(1-r)^{3}}$ .Next,$|f’(z)|$ $\geq$ $1- \sum_{n=2}^{\infty}n^{2}r^{n-1}$$\geq$ $2- \sum_{1r\iota=}^{\infty}n^{2}r^{n-1}=\frac{1-7r+6r^{2}-2r^{3}}{(1\cdot-r)^{3}}$$1- \sum_{n=2}n^{2}r^{n-1}$um_{n=1}^{\infty}n^{2}r^{n-1}=\frac{1-7r+6r^{2}-2r^{3} {(1\c }$From Poisson-Jensen’s formula, we obtain(9) $\log f’(z)=$ $\mathrm{A}$ $\int_{0}^{\mathrm{z}\pi}.\log|f’(\zeta)|$ . $\frac{\zeta+z}{\zeta-z}\mathrm{d}\phi$ ,which mplies that(10) $R_{e} \frac{zf(z)}{f(z)},"+1=1$ $+$ $\frac{1}{2\pi}\int_{-\beta}^{\beta}g(\alpha)\log|$ $7”((;)$ $|\mathrm{d}\alpha$$+$ $\frac{1}{2\pi}\int_{\beta}^{2\pi-\beta}g(\alpha)\log|f’(\zeta)|\mathrm{d}\alpha$ .By virtue of (8) and (10), we obtain that$R_{e} \frac{zf(z)}{f(z)},"+1\geq 1$ $-$ $\frac{\lambda}{\pi}\{\log\frac{(1-\rho)^{3}}{1-7\rho+6\rho^{2}-2\rho^{3}}$$+ \log\frac{1+\rho}{(1-\rho)^{3}}\}$ $. \frac{\sin\sqrt}{1-\lambda\cos\sqrt}$ .(11) $R_{e} \frac{zf’(z)}{f(z)},’+1\geq 1-\frac{2\rho r}{\pi(\rho^{2}-r^{2})}\cdot\log\frac{1+\rho}{1-7\rho+6\rho^{2}-2\rho^{3}}$We write the right hand side equation of (11) by $\psi(r,\rho)$ . Finally, ifwe put $r_{4}=$ 0.075450 and $\rho_{4}=$ 0.127, then $\psi(r_{4}, \rho_{4})=$ $2.23722$ $\cross 10^{-6}$It deduce that$R_{e} \frac{zf(z)}{f(z)},"+1>0$ $(0\leq r_{4})$ .So we have $R_{G}>$ 0.075450.(r, \rr_{4}=0.07 50$ \rho_{4}=0.12 $ , psi(r_{4}, \rho_{4})=2.237 2\cross 10^{-6}$135REFERENCES[1] S.Owa and M.Nunokawa, On some inverse properties for univalent functions,Kyoto Univ.Koukyuroku, 1164(2000),73-76.[2] Y.C.Kim and M.Nunokawa, On some radius results for ce tain analytic func-tions, KyungpOOk.Math.J.37(1997) ,61-65.[3] R.Yamakawa, Radius of starlikeness and convexity, Proceedings of the 2001international symposium on complex analysis and its applications, 177-180.

RADIUS OF STARLIKENESS AND CONVEXITY 2 (Study on Applications for Fractional Calculus Operators in Univalent Function Theory)

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RADIUS OF STARLIKENESS AND CONVEXITY 2 (Study on Applications for Fractional Calculus Operators in Univalent Function Theory)

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