数理解析研究所講究録
1062 巻 1998 年 62-68

62

Starlikeness of Libera transformation II
Mamoru Nunokawa and Yoshiaki Ota(布川護, 大田悦彰

群馬大学教育学部 )1. Introduction.
Let

$A$

denote the class of function$f(z)=z+ \sum_{n}\infty=2a_{n}Zn$that are analytic in $E=\{z:|z|<1\}$ .
A function $f(\approx)\in A$ is called to be starlike if and only if${\rm Re} \frac{zf’(Z)}{f(*)},>0$

in

$E$.

Similarly, $f(z)\in A$ is called to be convex if and only ifin

$1+{\rm Re} \frac{zf’’(Z)}{f(\wedge\sim)},>0$$E$.

The following interesting results are due to Libera [2].
Theorem A. Ifis starlike in , then so does the function $F(z)$ , defined by$E$$f(\approx)$$F(z)= \frac{2}{z}\int_{0}^{z}f(t)dt$

Theorem B. If $f(z)$ is convex in$E$.

, then so does the function $F(z)$ , defined by$F(z)= \frac{2}{z}\int_{0}^{z}f(t)dt$.

On the other hand, S.Singh and R.Singh [5, Theorem 1] and [4, Theorem 1] proved
the following Theorem
Theorem C. If$\mathrm{C}$and$f(\approx)\in \mathrm{A}$and Nunokawa [3] proved Theorem E.
and ${\rm Re} f^{;}(z)>0$ in , then the function$\mathrm{D}$$\mathrm{E}$,$F(z)= \frac{c+1}{z^{\mathrm{c}}}\int_{0}^{z}t^{\mathrm{C}-1}f(t)dt$

is starlike in$E$for all$\mathrm{c},- 1<\mathrm{c}\leq 0$.

63Theorem D. If $f(z)\in A$ and $|zf’’(z)/f’(z)|<3/2$ in$F(_{\sim}^{\sim}’)= \frac{2}{z}\int_{0}^{z}f(t)dt$Is$E_{i}$

then the function.

convex.
Theorem E. Let $f(z)\in A$ and suppose thatin${\rm Re} f^{f}(z)> \frac{1}{12}(\log\frac{4}{e})(\tan^{2*}\alpha\frac{\pi}{2}-3)$

$E$.

where
$-0.01759< \frac{1}{12}(\log\frac{4}{e})(\tan\alpha\frac{\pi}{2}2*-3)<-0.01751$

Then $F(z)$ is starlike in$E$.

, where$F(z)= \frac{2}{z}l^{z}f(t)dt$.

2. Prliminary.
Lemma. Let $f(z)\in A$ and $f(z)\neq 0$ in $0<|z|<1$ and suppose that$| \arg f’(Z)|<\frac{\pi}{2}\alpha$

in

$E$

where
$\alpha=\mathcal{B}+\frac{2}{\pi}\tan^{-1_{\frac{\beta}{2}}}$and$\beta>0$

Then we have$| \mathrm{a}x\mathrm{g}g’(z)|<\frac{\pi}{2}\alpha$

in

$E$

and
$| \arg\frac{g(z)}{z}|<\frac{\pi}{2}\gamma$

in

$E$

where
$0\prec\gamma$,$\beta=\gamma+\frac{2}{\pi}\tan^{-}’\gamma 1$

and$g(z)= \frac{2}{z}\int_{0}^{z}f(t)dt$

.

,.

64
Proof. From the hypothesis, it follows that$f(z)=g(z)+z_{\mathit{9}^{r}}(Z)$,$f’(z)=2g(\prime Z)+zg’(’)Z$and so$\arg f’(z)=\arg g’(Z)+\arg(2+\frac{zg’’(Z)}{g(_{\sim}^{\gamma})},)$.

Now, if there exists a point $z_{0}\in E$ such that$| \arg g’(Z)|<\frac{\pi}{2}\beta$

for$|z|<|z_{0}|$and$| \arg g(Z0)|=\frac{\pi}{2}\beta$

Then, from [3, Lemma], we have$, \frac{z_{0}g^{;}(\prime)z_{0}}{g(Z_{0})}=i\beta k$

where$k \geq\frac{1}{2}(a+\frac{1}{a}\rangle$

when$\arg g’(z_{0})=\frac{\pi}{2}\beta$$k \leq-\frac{1}{2}(a+\frac{1}{a})$when$\arg g’(z\mathrm{o})=-\frac{\pi}{2}\beta$

and

where
$p(Z0)^{\frac{1}{\beta}}=\pm ia$,$a>0$ .

If $\arg g’(z0)=\pi\beta/2_{!}$ then we have$\arg f’(z0)=\arg g’(z_{0})+\arg(2+,\frac{zg’’(_{\mathcal{Z}_{0}})}{g(z_{0})})$$= \frac{\pi}{2}\beta+\arg(2+i\beta k)$$\geq\frac{\pi}{2}(\beta+\frac{2}{\pi}\tan-1\frac{\prime\theta}{2})$.

65This contradicts the hypothesis. On the other hand, if $\arg g’(\sim 70)=-\pi\beta/2$ , then wehave$\arg f’(z_{0})$$=$$\leq$$=$$- \frac{\pi}{2}\beta+\arg(2+i\beta k)$$- \frac{\pi}{2}-\tan^{-1}\frac{\beta}{2}$$- \frac{\pi}{2}(\beta+\frac{2}{\pi}\tan-1\frac{\beta}{2})$.

This also contradicts the hypothesis. Therefore we must havein$| \arg g(Z)|<\frac{\pi}{2}\beta$$E$.

Putting$p(z)= \frac{g(z)}{z}$$p(0)=1$ ,,then we have$g’(z)=p(Z)+Zp’(_{Z)}$and$\arg g’(z)=\arg p(_{\mathcal{Z}})+\arg(1+\frac{zp’(z)}{p(z)})$.

If there exists a point $z_{0}\in E$ such that$| \arg p(Z)|<\frac{\pi}{2}\gamma$for$|z|<|z_{0}|$and$| \arg p(Z\mathrm{o})|=\frac{\pi}{2}\gamma$,

then we have$\frac{z_{0}p’(z0)}{p(z\mathrm{o})}=i\gamma k$

where$f_{v}^{\wedge} \geq\frac{1}{2}(a+\frac{1}{a})$when$\arg p(Z_{0}\rangle=\frac{\pi}{2}\gamma$

when$\arg p(Z0)=-\frac{\pi}{2}\gamma$

and$k \leq-\frac{1}{2}(a+\frac{1}{a})$

66
where
$p(z\mathrm{o})^{\frac{1}{\gamma}}.=\pm ia$$a>0$ .

,

Then, applying the same methed as the above: we have$| \arg g’(Z_{0})|<\frac{\pi}{2}(\gamma+\frac{2}{\pi}\tan-1\gamma)$.

This contradicts the hypothesis. Therefore we must havein$| \arg\frac{g(z)}{z}|<\frac{\pi}{2}\gamma$$E$.

3. Main result.
Theorem. Let $f(z)\in A,$ $f(z)\neq 0$ in $0<|z|<1$ ,(1)in$| \arg f’(Z)|<\frac{\pi}{2}(\frac{\check{\mathrm{o}}}{3}-\gamma_{0}^{\prime)}$where$\gamma_{0}$$E$,

is the smallest positive root of the equation$\frac{5}{6}=2\gamma+\frac{2}{\pi}\tan^{-1}\gamma+\frac{2}{\pi}\tan^{-1}\frac{1}{2}(\wedge(+\frac{2}{\pi}\tan^{-1}\gamma)$

and $0.266<\gamma_{0}<0.267$ .
Let us put(2)$g(z)= \frac{2}{z}\int_{0}^{z}f(t)dt$

Then $g(z)$ is starlike in$E$.

.

Proof. From (1) and Lemma, we easily havein$| \arg\frac{g(z)}{z}|<\frac{\pi}{2}\gamma_{0}$$E$.

From(2), we have

2 $f(z)=g(z)+zg’(Z)$and

2 $f’(z)=2g(\prime z)+zg’’(Z)$Let us put$\frac{zg’(z)}{g(z)}=\frac{1+w(z)}{1-w(z)}$then $w(z)$ is analytic in $E$ and $w(z)\neq 1$ in$E$.

,$w(0)=0$

67
Then it follows that

2 $f^{t}(z)=2g(_{Z)}’+.\vee‘,g’’(_{Z})$$= \frac{zg’(z)}{g(z)}[(\frac{1+w(z)}{1-w(z)})2+\frac{2zw’(Z)}{(1-w(_{\sim}\gamma))2}+\frac{1+w(z)}{1-w(z)}]$.

If there exists a point $z_{0}\in E$ such that$|^{\max_{z|\leq|}}z\mathrm{o}||w(z)|=|w(z_{0)1=}1$,then from Jack’s lemma [l,Lemma], we have$\frac{z_{0}w’(_{\sim 0}\backslash 7)}{w(Z_{0})}=k\geq 1$Putting$w(z\mathrm{o})=e^{i\theta}\backslash \prime 0\leq\theta<2\pi$

2 $f’(Z_{0})$, we have$=$$2g’(z_{0})+Z0g’’(z_{0})$$=$$\frac{g(z\mathrm{o})}{\sim 0}[(\frac{1+e^{i\theta}}{1-e^{i\theta}})^{2}+\frac{2ke^{i\theta}}{(_{\backslash }1-e^{i\theta})^{2}}7+^{\frac{1+e^{i\theta}}{1-e^{i\theta}}]}$$=$$\frac{g(Z_{0})}{z_{0}}[-\frac{\sin^{2}\theta}{(1-\cos\theta)^{2}}-\frac{k}{1-\cos\theta}+\frac{i\sin\theta}{1-\cos\theta}\mathrm{J}\rceil$.

Then we have$\arg f’(\mathcal{Z}_{0})$$\geq$$\arg(-\frac{\sin^{2}\theta+k(1-\cos\theta)}{(1-\cos\theta)^{2}}-\frac{i\sin(1-\cos\theta)}{(\mathrm{l}-\cos\theta)2})-|arg(\frac{g(Z_{0})}{\sim r_{0}})|$$\geq$$\arg(-\frac{sin^{2}\theta+1-\cos\theta}{(1-\cos\theta)^{\mathrm{z}}}-\frac{i\sin\theta(1-\cos\theta)}{(1-\cos\theta)^{2}})-\frac{\pi}{2}\gamma 0$$=$$\pi-\tan^{-1}\frac{|\sin\theta|(1-\cos\theta)}{\sin^{2}\theta+1-\cos\theta}-\frac{\pi}{2}\gamma 0$$=$$\pi-\tan^{-1}\frac{|\sin\theta|}{2+\cos\theta}-\frac{\pi}{2}\gamma 0$$\geq$$\pi-\tan^{-1}\frac{1}{\sqrt{3}}-\frac{\pi}{2}\gamma 0$$=$$\frac{5}{6}\pi-\frac{\pi}{2}\gamma 0$$=$$\frac{\pi}{2}(\frac{5}{3}-\gamma_{0})$.

This contradicts (1). Therefore we must have$|w(z)|<1$in$E$.

This show that
$Re \frac{\sim^{g’(Z)}7}{g(z)}>0$

or $g(z)$ is starlike in$E$

.

in

$E$.

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References[1] I. S. Jack, Functions starlike and convx of order474(1971).

$\alpha$. J. London Math. Soc., 3, 469-

[2] R. J. Libera, Some classes of regular univalent functions, Proc. Amer. Math. Soc.,
16, 755-758(1965).
[3] M. Nunokawa, On Starlikeness of Libera Transformation, Complex
79-83(1991).

$\mathrm{v}_{\mathrm{a}\mathrm{r}}\mathrm{i}\mathrm{a}\mathrm{b}\mathrm{l}\mathrm{e}\mathrm{s}_{j}17$,[4] R. Singh and S. Singh, Starlikeness and convexity of certain integrals, Ann Univ.
Mariac Curie-Sklodowska Sect. A 35, 45-47(1981).
[5] R. Singh and S. Singh, Starlikeness ofMath., 13, 190-194(1982).

$\mathrm{c}1_{\mathrm{o}\mathrm{s}\mathrm{e}-}\mathrm{t}\mathrm{o}-\mathrm{c}0\mathrm{n}\mathrm{V}\mathrm{e}\mathrm{x}$functions. Indian J. Pure Appl.

Mamoru Nunokawa
Yoshiaki Ota
Department of Mathematics
University of Gunma
Aramaki, Maebashi, Gunma 371, Japan