数理解析研究所講究録
第 1878 巻 2014 年 85-9385

Univalence and starlikeness of a function
defined by convolution of analytic function
and hypergeometric function${}_{3}F_{2}$Yutaka Shimoda, Yayoi Nakamura, and Shigeyoshi Owa
Abstract

We consider functions defined by a eondition of functions in the subclassofanalytic functions with generalized Gauss hypergeometric functions. In this paper, wegive a condition of the parameter for which the function to be univalent and starlike.
$\mathcal{U}(\lambda)$$\lambda$

1

IntroductionLet$\mathcal{A}$denote the class of functions $f(z)$ of the form

(1.1)

$f(z)=z+ \sum_{n=2}^{\infty}a_{n}z^{n}$that are analytic in the open unit disk $\mathbb{U}=\{z\in \mathbb{C} : |z|<1\}$ , and let$A$ consisting of $f(z)$that are univalent in$S$

be the subclass of$\mathbb{U}.$Obradovi\v{c} and Ponnusamy define in [4] the class(1.2)$\mathcal{U}(\lambda)$$|( \frac{z}{f(z)})^{2}f’(z)-1|\leq\lambda.

for some real
We set$f(z)\in \mathcal{A}$satisfing the condition(z\in \mathbb{U})$, where $f’$ denotes the derivative of with respect to the variable. It is easy to see that the the condition (1.2) is equivalent to$\lambda>0$$\mathcal{U}(1)=u$of$f$$|z^{2}( \frac{1}{f(z)}-\frac{1}{z})’|\leq\lambda (z\in \mathbb{U})$

$z.$

.

If $f(z)\in S$ maps onto a starlike domain (with respect to the origin), i.e. if, then we say that is a starlike function. The class of allwhenever $t\in[O, 1]$ andssary and sufficient condition forstarlike functions is denoted byto bestarlike is that the inequality$\mathbb{U}$$tw\in f(\mathbb{U})$$w\in f(\mathbb{U})$$\mathcal{S}^{*}.$$f$$Anec\cdot,e_{\iota}$$f(z)\in \mathcal{A}$2010 Mathematics Subject Classification: Primary $30C45.$
Key Words and Phrases : Analytic, starlike, Convolution(Hadamard Product)

86$R\epsilon(\frac{zf(z)}{f(z)})>0$$(z\in U)$holds.
For these facts, the following lemmas hold.

IfLemma 1 ([3])$f(z)\in \mathcal{U}(\lambda),$$a;= \frac{|f(0)|}{2}\leq 1$

and$0 \leq\lambda\leq\frac{\sqrt{2-a^{2}}-a}{2}$, then$f(z)\in \mathcal{S}^{*}.$

Lemma 2 ([7])

If $f(z)=z+a_{n+1}z^{n+i}+\cdots(n\geq 2)$ belongs to$\mathcal{U}(\lambda)$

and$0 \leq\lambda\leq\frac{n-1}{\sqrt{(n-1)^{2}+1}},$then $f(z)\in S^{*}.$For analytic functions $f(z)$ and $g(z)$ on

power seriesFor$\sum_{n=0}^{\infty}a_{m}b_{n}z^{n}$

$U$

with

$f(z)= \sum_{n=0}^{\infty}a_{n}z^{n}$and$g(z)= \sum_{n=0}^{\infty}b_{n}z^{n}$, theis said the convolution of $f(z)$ and $g(z)$ , denoted by $f*g$ (cf ([5])]).

$f(z)=z+ \sum_{n=2}^{\infty}a_{n}z^{n}$in$\mathcal{A}$, we have a natural convolution operator defined by$zF(a, b;c;z)*f(z):= \sum_{n=1}^{\infty}\frac{(a)_{n-1}(b)_{n-1}}{(c)_{n-1}(1)_{n-1}}a_{n}z^{n},$$c\in\{-1, -2, -3, \cdots\},$$z\in \mathbb{U},$denotes the Pochhammer symbol $(a)_{0}=1,$ $(a)_{n}=a(a+1)\cdots(a+n-1)$ for. Here $F(a, b;c;z)$ denotes the Gauss hypergeometric function which is analytic in
As a special case of the Euler integral representation for the hypergeometric function, one
haswhere$(a)_{n}$$\mathbb{U}.$$n\in \mathbb{N}$$F(1, b;c;z)= \frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)}\int_{0}^{1}\frac{1}{1-tz}t^{\triangleright 1}(1-t)^{c-\triangleright i}dt,$$z\in U,$${\rm Re} c>Reb>0.$

Using this representation, we have, for $f(z)\in A,$$zF( i, c;c+1;z)*f(z)=z(F(1, c;c+1;z)*\frac{f(z)}{z})=zc.\int_{0}^{1}\frac{f(tz)}{tz}t^{c-1}dt,$ $z\in U,$

Obradovi\v{c} and Ponnusamy have shown the following result.

Theorem $A$ ([5])
andLet$f\in \mathcal{U}(\lambda)$$c\in \mathbb{C}$with ${\rm Re} c>0$ such that$( \frac{z}{f(z)})*F(1,c;c+1;z)\neq 0$

in$z\in U,$${\rm Re} c>0.$87
and $G(z)=G_{f}^{c}(z)$ be the transformed function defined by$G(z)= \frac{z}{(\frac{z}{f(z)})*F(1,c;c+1;z)} (z\in U)$.

Then we have the following;(1)$G \in \mathcal{U}(\frac{\lambda|c|}{|c+2|})$

whenever$G\in \mathcal{U}$(2)$G\in \mathcal{S}$2. The result is sharp especially when$| \frac{f"(0)}{2}|\leq 1-\lambda$. In particular,$0< \lambda\leq|\frac{c+2}{c}|.$whenever $0< \lambda\leq\frac{|c+2|}{2|c|}(\sqrt{2-A^{2}}-A)$ with $A=| \frac{c\Gamma(0)}{c+12}|\leq 1.$

Main Result

For the generalized hypergeometric function
Theoreml
Let $f(z)\in u(\lambda)$ . Let${\rm Re}\alpha\geq 0,$$\alpha,$$\beta\in \mathbb{C}$${\rm Re}\beta\geq 0,$${}_{3}F_{2}(1, \alpha, \beta;\alpha+1, \beta+1;z)$, we obtainsatisfying$\frac{1}{|\alpha+\beta|}(\frac{|\alpha||\beta|}{|\beta+2|}+\frac{|\beta||\alpha|}{|\alpha+2|})<1$and$|\alpha+\beta|>|\alpha\beta|$

and$\frac{z}{f(z)}*{}_{3}F_{2}(1, \alpha, \beta;\alpha+1, \beta+1;z)\neq 0, z\in \mathbb{U}.$Denote by$G(z)=G_{f}^{\alpha,\beta}(z)$(2.1)

the function defined by$G(z)= \frac{z}{\frac{z}{f(z)}*{}_{3}F_{2}(1,\alpha,\beta;\alpha+1,\beta+1;z)}, z\in \mathbb{U},$where
the following:${}_{3}F_{2}(1, \alpha, \beta;\alpha+1, \beta+1;z)$(1)$G(z) \in \mathcal{U}(\frac{\lambda|\alpha+\beta|}{|\alpha+\beta+4|})$is the generalized hypergeometric function. Then we have. The result is sharp especially when

In particular, $G(z)\in u$ whenever$| \frac{f"(0)}{2}|\leq 1-\lambda.$$0< \lambda\leq\frac{|\alpha+\beta+4|}{|\alpha+\beta|}$(2) $G(z)\in S^{*}$whenever$0< \lambda\leq\frac{|\alpha+\beta+4|}{2|\alpha+\beta|}(\sqrt{2-A^{2}}-A)$

with$A=| \frac{\alpha\beta}{(\alpha+1)(\beta+1)}\frac{f"(0)}{2}|\leq 1.$

Proof.
Since(2.2)${}_{3}F_{2}(1, \alpha, \beta;\alpha+1, \beta+1;z)=\sum_{n=0}^{\infty}\frac{\alpha\beta}{(\alpha+n)(\beta+n)}z^{n}=1+\sum_{n=1}^{\infty}\frac{\alpha\beta}{(\alpha+n)(\beta+n)}z^{n},$88
we have$\frac{z}{f(z)}*{}_{3}F_{2}(1, \alpha, \beta;\alpha+1, \beta+1;z)=1-\frac{\alpha\beta a_{2}}{(\alpha+1)(\beta+1)}z+\frac{\alpha\beta(a_{2}^{2}-a_{3})}{(\alpha+2)(\beta+2)}z^{2}+\cdots$$= \{1-\frac{\alpha a_{2}}{\alpha+1}z+\frac{\alpha(a_{2}^{2}-a_{3})}{\alpha+2}z^{2}+\cdots\}*\{1-\frac{\prime\theta a_{2}}{\beta+1}z+\frac{\beta(a_{2}^{2}-a_{3})}{\beta+2}z^{2}+\cdots\}$$= \{\frac{z}{f(z)}*F(1, \alpha;\alpha+1;z)\}*F(1,\beta;\beta+1;z)$.

Thus $G(z)$ can be written as$G(z)= \frac{z}{\{\frac{z}{f(z)}*F(1,\alpha;\alpha+1;z)\}*F(1,\beta;\beta+1;z)}.$In the same manner, $G(z)$ can be also written as$G(z)= \frac{z}{\{\frac{z}{f(z)}*F(1,\beta;\beta+1;z)\}*F(1,\alpha;\alpha+1;z)}.$Put$h_{1}(z)= \frac{z}{\frac{z}{f(z)}*F(1,\alpha;\alpha+1;z)}, h_{2}(z)=\frac{z}{\frac{z}{f(z)}*F(1,\beta;\beta+1;z)}.$then$\frac{z}{f(z)}*F(1, \alpha;\alpha+1;z)=\frac{z}{h_{1}(z)}, \frac{z}{f(z)}*F(1,\beta;\beta+1;z)=\frac{z}{h_{2}(z)}$By the Theorem A in the introduction, we have. .

$|( \frac{z}{h_{1}(z)})^{2}h_{1}’(z)-1|<\frac{\lambda|\alpha|}{|\alpha+2|}$. .

$|( \frac{z}{h_{q}(z)})^{2}h_{2}’(z)-1|<\frac{\lambda|_{j}\theta|}{|\beta+2|}.$$h_{1}(z) \in \mathcal{U}(\frac{\lambda|\alpha|}{|\alpha+2|})$$i$$h_{2}(z) \in u(\frac{\lambda|_{j}9|}{|\beta+2|})$

$i$

$e$

and
$e$

Since
$\frac{z}{G(z)}=\frac{z}{h_{1}(z)}*F(1, \beta;\beta+1;z) (z\in \mathbb{U})$,we have(2.3)$( \beta+1)\frac{z}{G(z)}-(\frac{z}{G(z)})^{2}G’(z)=\beta\frac{z}{G(z)}+z(\frac{z}{G(z)})’$On the other hand,$\frac{z}{G(z)}$can be also written as$\frac{z}{G(z)}=\frac{z}{h_{2}(z)}*F(1, \alpha;\alpha+1;z) (z\in \mathbb{U})$,89
we have(2.4)$( \beta+1)\frac{z}{G(z)}-(\frac{z}{G(z)})^{2}G’(z)=\beta\frac{z}{G(z)}+z(\frac{z}{G(z)})’$

Then we have
(2.5)$( \alpha+1)\frac{z}{G(z)}-(\frac{z}{G(z)})^{2}G’(z)=\alpha\frac{z}{h_{2}(z)} (z\in \mathbb{U})$and(2.6)$( \beta+1)\frac{z}{G(z)}-(\frac{z}{G(z)})^{2}G’(z)=\beta\frac{z}{h_{1}(z)} (z\in \mathbb{U})$.

Set$p(z)=( \frac{z}{G(z)})^{2}G’(z)$Then $p(z)$ is analytic on(2.7)$\mathbb{U}$.

with $p(O)=1$ and $p’(O)=0$ , and$p(z)=( \alpha+1)\frac{z}{G(z)}-\alpha\frac{z}{h_{2}(z)}$and(2.8)$p(z)=( \beta+1)\frac{z}{G(z)}-\beta\frac{z}{h_{1}(z)}.$From (2.3), (2.4), (2.5), (2.6), (2.7) and (2.8) one then obtain that$\alpha p(z)+zp^{l}(z)$$=$$( \alpha+1)\alpha\frac{z}{G(z)}+(\alpha+1)z(\frac{z}{G(z)})’-\alpha^{2}\frac{z}{h_{2}(z)}-\alpha z(\frac{z}{h_{2}(z)})’$$= \alpha[(\alpha+1)\frac{z}{h_{2}(z)}-\alpha\frac{z}{h_{2}(z)}-z(\frac{z}{h_{2}(z)})’]$$= \alpha[\frac{z}{h_{2}(z)}-z(\frac{z}{h_{2}(z)})’]$$= \alpha(\frac{z}{h_{2}(z)})^{2}h_{2}’(z)$and$\beta p(z)+zp’(z)$$=$$( \beta+1)\beta\frac{z}{G(z)}+(\beta+1)z(\frac{z}{G(z)})’-\beta^{2}\frac{z}{h_{1}(z)}-\beta z(\frac{z}{h_{1}(z)})’$$= \beta[(\beta+1)\frac{z}{h_{1}(z)}-\beta\frac{z}{h_{1}(z)}-z(\frac{z}{h_{1}(z)})’]$$= \beta[\frac{z}{h_{1}(z)}-z(\frac{z}{h_{1}(z)})’]$$= \beta(\frac{z}{h_{1}(z)})^{2}h_{1}’(z)$

.

90
Since
$( \alpha+\beta)p(z)+2zp’(z)=\alpha(\frac{z}{hq(Z)})^{2}h_{2}’(z)+\beta(\frac{z}{h_{1}(z)})^{2}h_{1}(z)$,we have.

$p(z)+ \frac{2}{\alpha+\beta}zp’(z)=\frac{\alpha}{\alpha+\beta}(\frac{z}{h_{2}(z)})^{2}h_{2}’(z)+\frac{\beta}{\alpha+\beta}(\frac{z}{h_{1}(z)})^{2}h_{1}’(z)$Now, as$h_{1}(z) \in u(\frac{\lambda|\alpha|}{|\alpha+2|})$$|p(z)+ \frac{2}{\alpha+\beta}zp’(z)-1|$

$=$

and

$h_{2}(z) \in \mathcal{U}(\frac{\lambda|\beta|}{|\beta+2|})$, it follows that$| \frac{\alpha}{\alpha+\beta}\{(\frac{z}{h_{q}(z)})^{2}h_{q}’(z)-1\}+\frac{\beta}{\alpha+\beta}\{(\frac{z}{h_{1}(z)})^{2}h_{1}’(z)-1\}|$$\leq |\frac{\alpha}{\alpha+\beta}||(\frac{z}{h_{2}(z)})^{2}h_{2}’(z)-1|+|\frac{\beta}{\alpha+\beta}||(\frac{z}{h_{1}z)})^{2}h_{1}’(z)-1|$$< \underline{|\alpha|}\underline{\lambda|\beta|}\underline{|\beta|}\underline{\lambda|\alpha|}+$$|\alpha+\beta||\beta+2| |\alpha+\beta||\alpha+2|$$= \lambda\{\frac{1}{|\alpha+\beta|}(\frac{|\alpha||\beta|}{|\beta+2|}+\frac{|\beta||\alpha|}{|\alpha+2|})\}.$

By the assumption, we have
(2.9)$|p(z)+ \frac{2}{\alpha+\beta}zp’(z)-1|<\lambda.$

From the work of Hallenbeck and Rusheweyh ([2],[6]), we deduce that
(2.10)$|p(z)-1| \leq\frac{\lambda|\alpha+\beta|}{|\alpha+_{f}/f+4|} (z\in U)$

Thus we have$G(z) \in u(\frac{\lambda|\alpha+\beta|}{|\alpha+\beta+4|})$.

.

To prove the sharpness, we consider functions $f(z)$ in$\mathcal{U}(\lambda)$of the form$f(z)= \frac{z}{1-a_{2}z+\lambda z^{2}},$whereand. Since, so that $1-a_{2}z+\lambda z^{2}\neq 0$ for all
and, it follows that $|\alpha+2|>|\alpha+1|>|\alpha|$ and $|\beta+2|>|\beta+1|>|\beta|$ and, therefore$a_{2}= \frac{f"(0)}{2}$${\rm Re}\beta\geq 0$$|a_{2}|\leq 1-\lambda$$z\in \mathbb{U}$${\rm Re}\alpha\geq 0$$|1-a_{2} \frac{\alpha\beta}{(\alpha+1)(\beta+1)}z+\lambda\frac{\alpha\sqrt{}}{(\alpha+2)(\beta+2)}z^{2}|\neq 0$

for all $z\in U$ , provided
we have$|a_{2}|\leq 1-\lambda$. By the series expantion (2.2) of${}_{3}F_{2}(1, \alpha, \beta;\alpha+1, \beta+1;z)$$G(z)= \frac{z}{1-\frac{a_{2}\alpha\beta}{(\alpha+1)(\beta+1)}z+\frac{\lambda(\alpha\beta)}{(\alpha+2)(\beta+2)}z^{2}}.$,

91
Obviously, $G(z)$ is analytic on$\mathbb{U}$and$\frac{z}{G(z)}\neq 0$

on U. Since$( \frac{z}{G(z)})^{2}G’(z)-1=-\frac{\lambda\alpha\beta}{(\alpha+2)(\beta+2)}z^{2},$we have that
(2.11)$|( \frac{z}{G(z)})^{2}G’(z)-1|\leq\frac{\lambda|\alpha\beta|}{|(\alpha+2)(\beta+2)|}.$

Now, let us compare the right hand sides of (2.10) and (2.11). Firstly, since $|\alpha+\beta+4|<$$|(\alpha+2)(\beta+2)|,$then$\frac{1}{|(\alpha+2)(\beta+2)|}<\frac{1}{|\alpha+\beta+4|}$. From the assumption, we see$\frac{|\alpha\beta|}{|(\alpha+2)(\beta+2)|}<\frac{|\alpha+\beta|}{|(\alpha+2)(\beta+2)|}<\frac{|\alpha+\beta|}{|\alpha+\beta+4|}.$

Then, we have that$|( \frac{z}{G(z)})^{2}G’(z)-1|\leq\frac{\lambda|\alpha\beta|}{|(\alpha+2)(\beta+2)|}<\frac{|\alpha+\beta|}{|\alpha+\beta+4|}.$Thus, we have that the bound$\frac{|\alpha+\beta|}{|\alpha+\beta+4|}$

is sharp. We conclude that the first assertion ofTheorem 1.

The second assertion is a direct consequence of Lemma 1. In fact, obviously$A= \frac{G"(0)}{2}=\frac{\alpha\beta}{(\alpha+1)(\beta+1)}\frac{f"(0)}{2}$is smaller than or equal to 1.
Theorem 2
For a fixed $n\geq 2$ , let $f(z)=z+a_{n+1}z^{n+1}+\cdots$ belong to$\mathcal{U}(\lambda)$. Let$\alpha,$$\beta\geq 0$

and${\rm Re} \alpha\geq 0, {\rm Re}\beta\geq 0, \frac{1}{|\alpha+\beta|}(\frac{|\alpha||\beta|}{|\beta+n|}+\frac{|\alpha||\beta|}{|\alpha+n|})<1,$

and$\frac{z}{f(z)}*{}_{3}F_{2}(1,\alpha, \beta;\alpha+1,\beta+1;z)\neq 0, z\in \mathbb{U}.$

and
(1)

$G(z)=G_{f}^{\alpha_{!}\beta}(z)$be the transform function defined by (2.1). Then we have the following:$G(z) \in \mathcal{U}(\frac{\lambda|\alpha,+\beta|}{|\alpha+,9+2n|})$(2) $G(z)\in S^{*}$ whenever. In paticular,$G(z)\in \mathcal{U}$whenever$0< \lambda\leq\frac{|\alpha+\beta+2n|}{|\alpha+\beta|}$$0< \lambda\leq\frac{(n-1)|\alpha+\beta+2n|}{|\alpha+\beta|\sqrt{(n-1)^{2}+1}}.$

Proof. Using the Gaussian hypergeometric function, $G(z)$ can be written as

92$G(z)= \frac{z}{\{\frac{z}{f(z)}*F(1,a;\alpha+1;z)\}*F(1,\beta;\beta+1;z)}$and$G(z)= \frac{z}{\{\frac{z}{f(z)}*F(1,\beta;\beta+1;z)\}*F(1,\alpha;\alpha+1;z)}.$Put$h_{3}(z)= \frac{z}{\frac{z}{f(z)}*F(1,\alpha;\alpha+1;z)}, h_{4}(z)=\frac{z}{\frac{z}{f(z)}*F(1,\beta;\beta+1;z)}.$Then$\frac{z}{f(z)}*F(1,\alpha;\alpha+1;z)=\frac{z}{h_{3}(z)}, \frac{z}{f(z)}*F(1,\beta;\beta+1;z)=\frac{z}{h_{4}(z)},$We see. .

$|( \frac{z}{h_{3}(z)})^{2}h_{3}’(z)-1|<\frac{\lambda|\alpha|}{|\alpha+n|}$. .

$|( \frac{z}{h_{4}(z)})^{2}h_{4}(z)-1|<\frac{\lambda|\beta|}{|\beta+n|}.$$h_{3}(z) \in \mathcal{U}(\frac{\lambda|\alpha|}{|\alpha+n|})$$i$$h_{4}(z) \in u(\frac{\lambda|\beta|}{|\beta+n|})$

$i$

$e$

and
$e$

Since
$\frac{z}{f(z)}=\frac{1}{1+a_{n+i}z^{n}+}=1-a_{n+1}z^{n}+\cdots,$so that$\frac{z}{f(z)}*{}_{3}F_{2}(1,\alpha, \beta;\alpha+1, \beta+1;z)=1-a_{n+1}\{\frac{\alpha\beta}{(\alpha+n)(\beta+n)}\}z^{n}+\cdots$Thus, $G(z)$ can be written in the form
$G(z)=z+a_{n+1} \{\frac{\alpha\beta}{(\alpha+n)(\beta+n)}\}z^{n+1}+\cdots$

Therefore, as in the proof of Theoreml, the function $p(z)$ defined by$p(z)=( \frac{z}{G(z)})^{2}G’(z)=1+(n-1)a_{n+1}\{\frac{\alpha\beta}{(\alpha+n)(\beta+n)}\}z^{n}+\cdots$

is analytic in$U$

and $p(O)=1,$ $p^{\int}(O)=\cdots=p^{(n-1)}(0)=0.$ $p(z)$ can be written as$p(z)=( \alpha+1)\frac{z}{G(z)}-\alpha\frac{z}{h_{3}(z)}$and$p(z)=( \beta+1)\frac{z}{G(z)}-\beta\frac{z}{h_{4}(z)}.$

By the same argument of proof of Therorem 1 using$p(z)$ satisfies (2.9). Consequentry, we obtain that$h_{3}(z)$and$h_{4}(z)$instead of$h_{i}(z)$and$h_{2}(z),$$|p(z)-1| \leq\frac{\lambda|\alpha+\beta||z|^{n}}{|\alpha+_{r’}f+2n|} (z\in \mathbb{U})$,and the proof of part(l) is complete. The second part is a direct consequence of Lemma 2.

93References[1] P. L. Duren, Univalent Functions, Springer-Verlag, New York, Berlin, Heiderberg, Tokyo,

1983[2] D.J.Hallenbeck, St Ruschweyh, Subordination by convex functions, Proc. Amer. Math.
Soc. $52(1975)191-195$[3] M. Obradvi\v{c}, S. Ponnusamy, V. Singh, P. Vasundhra, Univalency starlikenss and convexity applied to certain classes of rational functions, Analysis (Munich) 22 (3) (2002)

225-242[4] M. Obradovi\v{c} and S. Ponnusamy, Radius properties for subclasses of univalent functions,Analysis 25(2005), 183–188[5] M. Obradovi\v{c}, S. Ponnusamy, Univalence and starlikeness of certain transforms definedby convolution of analytic functions, J. Math. Anal. Appl. $336(2007)758-767$[6] S. Ponnusamy, Differntial subordination and Bazilevic functions, Proc. Ind. Acad. Sci.
Math. Sci $105(1995)169-186$[7] S. Ponnusamy, P. Sahoo,
properties of certain linear transforms, Bul. Belg.
Math. Soc. Simon Stevin $12(2005)95-108$$Geometr\dot{b}\mathcal{C}$

Department of Mathematics
Kinki University
Higashi-Osaka, Osaka 577-8502
Japan-Mail: yutakashimoda119@gmail.com
yayoi@math.kindai.ac.jp$E$shige21@ican.zaq.ne.jp