数理解析研究所講究録 1341 巻 2003 年 94-101

94

Some Criteria for univalence of certain integral operators
Virgil Pescar and Shigeyoshi Owa
Abstract
In this work, we derive some criteria for univalence of certain integral operators for analytic
functions in the open unit disk.

1Introduction
Let $A$ be the class of the functions $f(z)$ which are analytic in the open unit disk$=\{z \in \mathbb{C}:|z|<1\}$ and $f(0)=f’(0)-1=0$ .
We denote by the subclass of $A$ consisting of functions $f(z)$ $\in A$ which are univalent inU. Miller and Mocanu [1] have considered many integral operators for functions $f(z)$ belonging
to the class $A$ . In this paper, we consider the following integral operators$\mathrm{u}$$S$($F_{\alpha}(z)= \{\frac{1}{\alpha}\int_{0}^{z}f(u)^{\mathrm{A}}\alpha u^{-1}du\}^{\alpha}$

for $f(z)\in A$ and for somewheredenots the subclass of$\alpha\in \mathrm{C}$$S^{*}$. It is well-k own that

$S$$z$: U)(1.1)

for $f(z)\in S^{*}$ and $\alpha>0$ ,consisting of all starlike functions $f(z)$ in U.
$F_{\alpha}(z)\in S$

2PRELIMINARY RESULTS
To discuss about our integral operators, we need the following theorems.

Theorem 2.1 ([3])

satisfiesLet$\alpha$be a complex number with ${\rm Re}(\alpha)>0$ , and $f(z)\in A$ .

If $f(z)$

,

(2.1)

$\frac{1-|z|^{2{\rm Re}(\alpha)}}{{\rm Re}(\alpha)}|\frac{zf^{lJ}(z)}{f(z)},|\leqq 1$

for all$z\in \mathrm{u}$, then the following integral operator$G_{\alpha}(z)= \{\alpha\int_{\mathit{0}}^{z}u^{\alpha-1}f’(u)du\}^{\frac{1}{a}}$is in the class$S$.

2000 Mathematics Subject Classification. Primary$30\mathrm{C}45$.

Key Words and Phrases. Analytic function, univalent function,
integral operator, Schwarz lemma

(2.1)

95
Theorem 2.2 ([4])
s atisfies

Let$\alpha$be a complex number with $Re(\alpha)>0$ and $f(z)\in A$ . If $f(z)$(2.3)$\frac{1-|z|^{2R\mathrm{e}(\alpha)}}{Re(\alpha)}|\frac{zf’(z)}{f’(z)}|\leqq 1$

for all$z\in \mathrm{u}$, then, for any complex number$\beta$with${\rm Re}(\beta)\geqq{\rm Re}(\alpha)$, the integral operator(14)$G_{\beta}(z)= \{\beta\int_{0}^{z}u^{\beta-1}f’(u)du\}^{F}1$

is in the class $S$.

Example 2.3 Defining the function $f(z)$ by$f(z)$

with${\rm Re}(\alpha)\geqq 1$$= \int_{0}^{z}(\frac{1+u^{\mathrm{R}*(\alpha)}}{1-u^{{\rm Re}(\alpha)}})^{2}1$

du, we have that$\frac{1-z^{2{\rm Re}(\alpha)}}{{\rm Re}(\alpha)}(\frac{zf’(z)}{f(z)},)=z^{{\rm Re}(\alpha)-1}$

Thus the function $f(z)$ satisfies the condition of Theorem 2.2. Therefore, for${\rm Re}(\beta)\geqq \mathrm{R}ae(\alpha)$,$G_{\beta}(z)= \{\beta\int_{0}^{z}u^{\beta-1}(\frac{1+u^{{\rm Re}(\alpha)}}{1-u^{{\rm Re}(\alpha)}})^{\frac{1}{2}}du\}^{7}1$

is in the class S.
Thorem 2.4 [2]satisfies

If the function $g(z)$ is regular in , then, for all ($\mathrm{u}$$\in \mathrm{u}$and$z\in \mathrm{u}$, $g(z)$(2.5)$| \frac{g(\xi)-g(z)}{1-\overline{g(z)}g(\xi)}|\leqq|\frac{\xi-z}{1-\overline{z}\xi}|$

and

(2.6)

$| \oint(z)|\leqq$

The equalities hold only in the case$g(z)= \epsilon\frac{z+u}{1+\overline{u}z}$, where$|\epsilon|=1$

and $|u|<1$ .

96Remark 2.5 ([2])

For

$z$

$=0$

, from inequality (2.5)(2.7)$| \frac{g(\xi)-g(0)}{1-\overline{g(0)}g(\xi)}|\leqq|\xi|$and, hence$|g( \xi)|\leqq\frac{|\xi|+|g(0)|}{1+|g(0)||\xi|}$

.

(2.8)

Considering $g(0)=a$ and $\xi=z$ , we see that

(2.9)$|g(z)| \leqq\frac{|z|+|a|}{1+|a||z|}$

for all$z\in \mathrm{u}$.

Schwarz Lemma ([2]), thenz

If the function $g(z)$ is regular in$\mathrm{U},\mathrm{p}(0)=0$

and $|g(z)|\leqq 1$ for all$\in \mathrm{u}$$|g(z)|\leqq|z|$

for all
where$z\in \mathrm{u}$,(2.10), $and|g’(0)|\leqq 1$ . The equality in (2.10) for $z\neq 0$ holds only in the case$|\epsilon|=1$.

$\mathrm{g}\{\mathrm{z})=\epsilon z$,

3Main results
Theorem 3.1

Let$\alpha$be a complex number with${\rm Re}( \frac{1}{\alpha})=a>0$

and the function $g(z)\in A$$sat\dot{1}S\ovalbox{\tt\small REJECT}$$| \frac{zg’(z)}{g(z)}-1|\leqq 1$

TAen. for$(z\in \mathrm{U})$

.

(3.1)

97
$| \alpha|\geqq\frac{2}{(2a+1)^{\frac{2a+1}{2u}}}$

(3.2)

,

the integral operator$F_{\alpha}(z)= \{\frac{1}{\alpha}\int_{0}^{z}g(u)^{1}au^{-1}du\}^{\alpha}$

is in the class $S$

Prvof

Let

(3.3)

.

$\frac{1}{\alpha}=\beta$. Then we have$F \#(z)=\{\beta\int_{0}^{z}u^{\beta-1}(\frac{g(u)}{u})^{\beta}du\}^{\#}$(3.4)

Let us consider the function$f(z)= \int_{0}^{z}(\frac{g(u)}{u})^{\beta}$

(3.5)

du.

Then the function,(3.6)$h(z)=( \frac{1}{|\beta|})\frac{zf’(z)}{f(z)}$is regular in$\mathrm{u}$and the constant$|\beta|$satisfies the inequality(3.7)$| \beta|\leqq\frac{(2a+1)}{2}\underline{2}\mathrm{g},\llcorner 1a$

Prom (3.5) and (3.6), we have that$h(z)$ $= \frac{\beta}{|\beta|}(\frac{zg’(z)}{g(z)}-1)$

.

(3.8)

Using (3.8) and (3.1), we obtain$|h(z)|\leqq 1$$(z\in \mathrm{U})$

.

(3.9)

Noting that $h(0)=0$ and applying Schwarz-Lemma for $h(z)$ , we get$\frac{1}{|\beta|}|\frac{zf’(z)}{f’(z)}|\leqq|z|$$(z\in \mathrm{U})$,(3.10)

98
and hence, we obtain$\frac{1-|z|^{2a}}{a}|\frac{zf’(z)}{f’(z)}|\leqq|\beta|(\frac{1-|z|^{2a}}{a})|z|$$(z\in \mathrm{U})$.

(3.11)

Because,$\max_{|z|\leq 1}(\frac{1-|z|^{2a}}{a}|z|)=\frac{2}{(2a+1)^{\frac{u\neq 1}{2a}}}$from (3.11) and (3.7), we have

(3.12)$\frac{1-|z|^{2a}}{a}|\frac{zf’(z)}{f’(z)}|\leqq 1$

for$z\in \mathrm{u}$. From (3.12) and Theorem 2.1, it follows that
(3.13)$G_{\beta}(z)= \{\beta\int_{0}^{z}u^{\betarightarrow 1}f’(u)du\}^{\#}$

belongs to the class .
By means of (3.13) and (3.5), we have the integtal operator $F\#(z)$ is in the class $S$ , andhence, we conclude that the integral operator
is in the class .
$S$

$S$

$F_{\alpha}(z)$Example 3.2

If we take the function $g(z)=ze^{z}$ and a$= \frac{1}{a}>0$, then$g(z)=z+a_{2}z^{2}+a_{S}z^{3}+\cdots$is analytic in$\mathrm{u}$and$| \frac{zg’(z)}{g(z)}-1|=|z|$$<1$$(z\in \mathrm{U})$.

Since the function $g(z)$ satisfies the condition of Theorem 3.1, we have$T_{\alpha}(z)= \{\frac{1}{\alpha}\int_{0}^{z}e^{[perp]}\alpha uu^{l_{-1}}\alpha du\}^{\alpha}\in S$

Theorem 3.3$g(z)\in A$

satishLet$\alpha,\beta$

be complex numbers with$| \frac{zg’(z)-g(z)}{zg(z)}|\leqq 1$.

${\rm Re}(\beta)\geqq{\rm Re}(\alpha)>0$$(z\in \mathrm{U})$.

and the function(3.14)

se 99Then, for$| \alpha|\geqq\max_{|z|\leqq 1}\{(\frac{1-|z|^{2{\rm Re}(\alpha)}}{{\rm Re}(\alpha)})|z|(\frac{|z|+|a_{2}|}{1+|a_{2}||z|})\}$,(3.15)the integral operator(3.16)$F_{\alpha,\beta}(z)= \{\beta\int_{0}^{z}g(u)^{1}\alpha u^{\beta_{\alpha}-1}-[perp] du\}^{F}1$

is in the class $S$

Proof.

We have$F_{\alpha_{1}\beta}(z)= \{\beta\int_{0}^{z}u^{\beta-1}(\frac{g(u)}{u})^{a}4_{-}du\}^{\#}$.

(3.17)Let us consider the function$f(z)= \int_{0}^{z}(\frac{g(u)}{u})^{\frac{1}{\alpha}}$(3.18)

du.

which is regular in U. The function$p(z)=| \alpha|\frac{f’(z)}{f’(z)}$where the constant
we obtain$|\alpha|$,(3.19)

satisfies the inequality (3.15), is regular in U. From (3.19) and (3.18),(3.20)$p(z)= \frac{|\alpha|}{\alpha}\{\frac{zg’(z)-g(z)}{zg(z)}\}$

and using (3.14) we have$|p(z)|<1$(3.21)$(z \in \mathrm{U})$and $|p(0)|=|a_{2}|$ . Applying Remark 2.5, we obtain$| \alpha\frac{f’(z)}{f’(z)}|\leq\frac{|z|+|a_{2}|}{1+|a_{2}||z|}$$(z \in \mathrm{U})$.

(3.22)

100
It follows that
(3.23)$\frac{1-|z|^{2{\rm Re}(\alpha)}}{{\rm Re}(\alpha)}|\frac{zf’(z)}{f(z)},|\leqq(\frac{1}{|\alpha|})(\frac{1-|z|^{2{\rm Re}(\alpha)}}{{\rm Re}(\alpha)})|z|(\frac{|z|+|a_{2}|}{1+|a_{2}||z|})$for all$z\in \mathrm{U}$. Let us consider the function$(x=|z|;x\in[0,1])$$Q(x)=( \frac{1-x^{2\mathrm{R}\epsilon(\alpha)}}{{\rm Re}(\alpha)})x(\frac{x+|a_{2}|}{1+|a_{2}|x})$.

Because $Q( \frac{1}{2})>0$ , $Q(x)$ satisfies(3.24)$\max_{[]\in[0,1]}Q(x)>0$

Using this fact, (3.23) gives us that$\frac{1-|z|^{2{\rm Re}(\alpha)}}{{\rm Re}(\alpha)}|z\frac{f’(z)}{f’(z)}|\leqq\frac{1}{|\alpha|}\max_{|z|\leqq 1}\{(\frac{1-|z|^{2\ (\alpha)}}{{\rm Re}(\alpha)})|z|( \frac{|z|+|a_{2}|}{1+|a_{2}||z|})\}$.

(3.25)

Prom (3.25) and (3.15), we obtain$\frac{1-|z|^{2\mathrm{R}\epsilon(\alpha)}}{{\rm Re}(\alpha)}|\frac{zf’(z)}{f’(z)}|\leqq 1$$(z\in \mathrm{U})$.

(3.26)

Using (3.26) and Theorem 2.2, we obtain that the integral operator(3.27)$G_{\beta}(z)= \{\beta\int_{0}^{z}u^{\beta-1}f’(u)du\}^{\#}$

belongs to the class
$S$

$S$

.

Corollary 3.4
sat. Therefore, it follows from (3.27) and (3.18), thatLet$\alpha$$F_{a_{1}}\rho(z)$is in the classbe a complex number with ${\rm Re}(\alpha)>0$ and the function $g(z)$ $\in A$$\dot{u}h$$| \frac{zg’(z)-g(z)}{zg(z)}|\leqq 1$$(z\in \mathrm{U})$.

(3.28)

Then, for$\max_{|z|\leqq 1}\{(\frac{1-|z|^{2\ (\alpha)}}{{\rm Re}(\alpha)})|z|( \frac{|z|+|a_{2}|}{1+|a_{2}||z|})\}\leqq|\alpha|\leqq 1$$\theta\iota e$

integral operator,(3.29)

101101(3.30)$F_{\alpha}(z)= \{\frac{1}{\alpha}\int_{0}^{z}g(u)^{\frac{1}{a}}u^{-1}du\}^{\alpha}$

is in the class .
Proof Prom Theorem 3.3 for$|\alpha|<1$ and we have$S$$\beta=\frac{1}{\alpha}$$F_{\alpha,\beta}(z)=F_{\alpha}(z)$, the condition${\rm Re}(\beta)\geqq{\rm Re}(\alpha)>0$, is identical with.

References[1] S.S.Miller and P.T. Mocanu, Differential subordinations ,Pure and Applied Mathematics
225, Marcel Dekker, 2000.

[2] Z. Nehari, Conformed mapping, Me Graw-Hill Book Comp., New York, Toronto, London,

1952.

[3] N.N. Pascu, On a univalence criterion II, Inter. Seminar on Functional Equations Approximation and Convexity, Cluj -Napoca, Preprint No. 6, 1985, 153-154.
[4] N.N.Pascu, An improvement of Becker’s univalence criterion. Proceedings of the Commemorative Simion Stoilow (Bra\S ov, 1987) Univ. Bra50V, 1987, 43-48.
[5] V. Pescar, New univalence criteria, Transilvania University of Bra\S ov, 2002.
[6] Ch. Pommerenke, Univalent functions, Vandenhoeck Ruprecht in Gottingen, 1975.

V.Pescar: Department of Mathematics
Faculty of Mathematics and Computer Sciencen $Runsd.vania^{\chi}$ University of,
2200 Bra\S ov
Romania$\mathrm{B}\mathrm{r}\Re \mathrm{o}\mathrm{v}$S.Owa:Department of Mathematics
Kinki University
Higashi -Osaka, Osaka 577-8502
Japa