in the geometric function theory) Author(s) Shigeyoshi; Darus, Maslina; Cho, Na Citation 数理解析研究所講究録 (2010), 1717: 95-99

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Maslina; Cho Na Citation 数理解析研究所講究録 (2010) 1717: 95-99 Issue Date 2010-10 URL http://hdl.

1 On another proof of Ozaki's Titlefor univalence (Extensions of theorem the h in the geometric function theory) Author(s) Nunokawa Mamoru; Hayami Toshio; U Shigeyoshi; Darus Maslina; Cho Na Citation 数理解析研究所講究録 (2010) 1717: Issue Date URL Right Type Departmental Bulletin Paper Textversion publisher Kyoto University

2 On another proof of Ozaki s theorem and a sufficient condition for univalence Mamoru Nunokawa Toshio Hayami Neslihan Uyanik Shigeyoshi Owa Maslina Darus and Nak Eun Cho Abstract In 1935 S. Ozaki (Sci. Rep. Tokyo Bunrika Daigaku 2 (1935)) has given the sufficient condirion for analytic functions to be at most p-valent in the convex domain. The object of the present paper is to discuss new proof of Ozaki s teorem. A sufficient condition for univalent functions is also considered. 1 Main theorems Theorem 1 Let $f(z)$ be analytic in a convex domain $D$ and suppose that ${\rm Re}(f^{(p)}(z))>0$ $(z\in D)$. Then $f(z)$ is at most p-valent in $D$. Proof. Applying the mathematical method of redutive absurdity we prove it. If $f(z)$ is not at most p-valent in $D$ then there exist $p+1$ points $z_{11}$ $z_{12}$ $z_{13}$ $\cdots$ $z_{1p}$ $z_{1p+1}$ which are different each other for which $f(z_{11})=f(z_{12})=f(z_{13})=\cdots=f(z_{1p})=f(z_{1p+1})=0$. Let us number the points in order of multitude of real part of the points but if some of them have same real part then let us rotate the z-plane suitably. Renumbering of $p+1$ points then without generalization we can suppose that all the line $\overline{z_{11}z_{12}}$ segments $\overline{z_{12}z_{13}}$ $\overline{z_{13}z_{14}}$ $\cdots$ $\overline{z_{1p-1jp}z}$ $\overline{z_{1p}z_{1p+1}}$ are not perpendicular with the real axis and therefore we can put the following ${\rm Re}(z_{11})<{\rm Re}(z_{12})<{\rm Re}(z_{13})<\cdots<{\rm Re}(z_{1p})<{\rm Re}(z_{1p+1})$. Then we have the followings: 2000 Mathematics Subject Classification: Primary $30C45$. Keywords and Phrases: Analytic at most p-valent univalent.

3 96 ${\rm Re}( \frac{f(z_{12})-f(z_{11})}{z_{12}-z_{11}})$ $=$ ${\rm Re}(f (z_{21}))=0$ ${\rm Re}( \frac{f(z_{13})-f(z_{12})}{z_{13}-z_{12}})$ $=$ ${\rm Re}(f (z_{22}))=0$ (1) ${\rm Re}( \frac{f(z_{14})-f(z_{13})}{z_{14}-z_{13}})$ $={\rm Re}(f (z_{23}))=0$ :. ${\rm Re}( \frac{f(z_{1p})-f(z_{1p-1})}{z_{1p}-z_{1p-1}})$ $=$ ${\rm Re}(f^{f}(z_{2p-1}))=0$ ${\rm Re}( \frac{f(z_{1p+1})-f(z_{1p})}{z_{1r\vdash 1}-z_{1p}})$ $=$ ${\rm Re}(f (z_{2p}))=0$ $z_{2k}=z_{1k}+\theta_{1k}(z_{1k+1}-z_{1k})$ $(0<\theta_{1k}<1$ and $k=123$ $\cdotsp)$ and the sequence $\{{\rm Re}(z_{2k})\}$ is From step (1) we have a strictly increasing sequence. $\frac{{\rm Re}(f (z_{22})-f (z_{21}))}{{\rm Re}(z_{22}-z_{21})}$ $=$ ${\rm Re}( \frac{\partial f (z_{31})}{\partial x})=0$ (2) $\frac{{\rm Re}(f (z_{23})-f (z_{22}))}{{\rm Re}(z_{23}-z_{22})}$ $=$ ${\rm Re}( \frac{\partial f (z_{32})}{\partial x})=0$ : $\frac{{\rm Re}(f (z_{2p})-f (z_{2p-1}))}{{\rm Re}(z_{2p}-z_{2p-1})}$ $=$ ${\rm Re}( \frac{\partial f (z_{3p-1})}{\partial x})=0$ $z_{3k}=z_{2k}+\theta_{2k}(z_{2k+1}-z_{2k})$ $(0<\theta_{2k}<1$ and $k=12$ $\cdots$ $p-1)$. Then the sequence $\{{\rm Re}(z_{3k})\}$ is also a strictly increasing sequence.

4 97 Form step (2) we have $\frac{{\rm Re}(\frac{\partial f (z_{32})}{\partial x}-\frac{\partial f (z_{31})}{\partial x})}{{\rm Re}(z_{32}-z_{31})}$ $={\rm Re}( \frac{\partial^{2}f (z_{41})}{\partial x^{2}})=0$ $\frac{{\rm Re}(\frac{\partial f (z_{3_{1}3})}{\partial x}-\frac{\partial f (z_{32})}{\partial x})}{{\rm Re}(z_{33}-z_{32})}$ $={\rm Re}( \frac{\partial^{2}f (z_{42})}{\partial x^{2}})=0$ : $\frac{{\rm Re}(\frac{\partial f (z_{3_{)}p-1})}{\partial x}-\frac{\partial f (z_{3p-2})}{\partial x})}{{\rm Re}(z_{3p-1}-z_{3p-2})}$ $={\rm Re}( \frac{\partial^{2}f (z_{4p-2})}{\partial x^{2}})=0$ $z_{4k}=z_{3k}+\theta_{3k}(z_{3k+1}-z_{3k})$ $(0<\theta_{3k}<1$ and $k=12$ $\cdots$ and $\{{\rm Re}(z_{4k})\}$ is a strictly increasing sequence. $p-2)$ Let us continue the same steps as the above then we have finally the following equality ${\rm Re}( \frac{\partial^{p-1}\prime f (z_{p+11})}{\partial x^{p-1}})=0$ $z_{p+11}=z_{p1}+\theta_{p1}(z_{p2}-z_{p1})\in D$ $(0<\theta_{p1}<1)$. On the other hand since $f(z)$ is analytic in $D$ we have ${\rm Re}( \frac{\partial^{p-1}f (z_{p+11})}{\partial\tau^{p-1}})={\rm Re}(f^{(p)}(z_{p+11}))=0$. This contradicts the hypothesis of the theorem and it completes the proof of the theorem. $\square$ Remark In the proof of the above if $f(z)$ has zero at $z_{11}$ of order 2 or $z_{11}=z_{12}$ and all another zeros are of order 1 then in the step (1) we put ${\rm Re}(f (z_{21}))$ $=$ ${\rm Re}(f (z_{11}))={\rm Re}(f (z_{12}))=0$ ${\rm Re}(f (z_{22}))$ $=$ $0$ ${\rm Re}(f^{f}(z_{23}))$ $=$ $0$ : ${\rm Re}(f^{f}(z_{2p}))$ $=$ $0$

5 98 $z_{21}=z_{11}=z_{12}$ $\sim 2k=z_{1k}+\theta_{1k}(z_{1k+1}-z_{1k})$ $(0<\theta_{1k}<1$ and $k=23$ $\cdotsp)$ the sequence $\{{\rm Re}(z_{1k})\}$ is not a strictly increasing sequence but the sequence $\{{\rm Re}(z_{2k})\}$ is a strictly increasing sequence. Continuing the same steps as the proof of Theorem 1 we have the same conclusion. For the cases $f(z)$ has zeros at many points of multiple orders then applying the same idea as the above we obtain the same conclusion. Theorem 2 Let $f(z)$ be analytic in a convex domain $D$ and suppose that there exists a $\alpha$ complex constant which satisfies $ \arg(-\alpha) \geqq\frac{\pi}{2}(1+\delta)$ $0\leqq\delta$ and suppose that Then $f(z)$ is univalent in $D$. $ \arg(f (z)-\alpha) <\frac{\pi}{2}(1+\delta)$ $(z\in D)$. Proof. If $f(z)$ is not univalent in $D$ then there exist two points $z_{1}\in D$ and $z_{2}\in D$ $z_{1}\neq z_{2}$ for which Then it follows that $f(z_{1})=f(z_{2})$. $(f(z_{2})-\alpha z_{2})-(f(z_{1})-\alpha z_{1})$ $=$ $\int_{z_{1}}^{z_{2}}(f (z)-\alpha)dz$ $=$ $(z_{2}-z_{1}) \int_{0}^{1}\{f (z_{1}+t(z_{2}-z_{1}))-\alpha\}dt$ and therefore we have $\frac{f(z_{2})-f(z_{1})}{z_{2}-z_{1}}-\alpha=\int_{0}^{1}\{f (z_{1}+t(z_{2}-z_{1}))-\alpha\}dt$. Then we have $\frac{\pi}{2}(1+\delta)\leqq \arg(-\alpha) $ $=$ $ \arg(\frac{f(z_{2})-f(z_{1})}{\sim 2\sim-Z_{1}}-\alpha) $ $=$ $ \arg\int_{0}^{1}\{f (z_{1}+t(z_{2}-z_{1}))-\alpha\}dt $ $<$ $\frac{\pi}{2}(1+\delta)$. This is a contradiction and therefore it completes the proof. $\square$

6 99 References [1] S. Ozaki On the theory of multivalent functions Sci. Rep. Tokyo Bunrika Daigaku A 2 (1935) Mamoru Nunokawa Emeritus Professor Univ. of Gunma Hoshikuki-Cho Chou-ward Chiba city Japan mamoru-nuno@doctor.nifty.jp Toshio Hayami Department of Mathematics Kinki University Higashi-Osaka Osaka Japan ha-ya-toll2@hotmail.com Neslihan Uyanik Department of Mathematics Kazim Karabekir Faculty of Education Ataturk University Erzurum T $\mathcal{i}hrkey$ nesuyan@yahoo.com Shigeyoshi Owa Department of Mathematics Kinki University Higashi-Os $aka$ Osaka Japan owa@math.kindai.ac.jp Maslina Darus School of Mathematical Sciences Faculty of Sciences and Technology Universiti Kebangsaan Malaysia Bangi Selangor Malaysia maslina@pkrisc.cc.ukm.my Nak Eun Cho Department of Applied Mathematics Pukyong National University Pusan Korea necho@pknu.ac.kr

Title Geometric Univalent Function Theory. Citation 数理解析研究所講究録 (2008), 1579:

Title Geometric Univalent Function Theory. Citation 数理解析研究所講究録 (2008), 1579: Title Coefficient conditions for certain Geometric Univalent Function Theory Author(s) Hayami, Toshio; Owa, Shigeyoshi Citation 数理解析研究所講究録 (2008), 1579: 17-24 Issue Date 2008-02 URL http://hdlhlenet/2433/81395