数理解析研究所講究録 1341 巻 2003 年 45-51

45PROPERTIES OF CERTAIN INTEGRALOPERATOR
JIN LIN LIU
Department of Mathematics,
Yangzhou University,
Yangzhou 225002, Jiangsu,
People’s Republic of China
E-Mail:Jlliucn@yahoo.com.cn-andShigeyoshi Owa
Department of Mathematics
Kinki University

Higashi-Osaka Osaka 577-8502
Japan-Mail:owa@math.kindai.ac.jp$\mathrm{E}$

Abstractalent
Let $A(p)$ denote the class of functions $f(z)$ which are analytic andof $A(p)$ consisting of analytic andin the unit disk . Anew subclasswhich-valent functions $f(z)$ associated with the certain integral operator
is the generalization of the integral operator studied by I.B.Jung, Y.C.Kim and
H.M.Srivastava (J. Math. Anal. AppL 248(2000), 475-481) is introduced. Somefor functions $f(z)$ belonging to $A(p)$ areinteresting properties of the operator
investigated.
$\mathrm{p}-$$U$$\Omega(\alpha,\beta;\gamma)$$p$$Q_{\beta}^{\alpha}$$Q_{\beta}^{\alpha}$

Key Words and phrases: Integral operator, extreme point, multivalent.

2000 Mathematics Subject Classification: Primary 30C45

461. Introduction.
Let $A(p)$ denote the class of functions of the form

(1.1)$(p\in N=\{1,2,3, \cdots\})$$f(z)=z^{p}+ \sum_{n=1}^{\infty}a_{p+n}z^{\mathrm{p}+n}$$<1$ }. Let
which are analytic and -valent in the unit disk $U=$ { : $\in C$ anddenote the class of functions $f(z)$ of the form (1.1) which satisfy the condition
$z$

$p$

$|z|$

$z$

$S_{p}^{*}(\gamma)$$Re \{\frac{zf’(z)}{f(z)}\}>p\gamma$for $0\leq\gamma<1$ and $\in U$. Afunction inis called -valent starlike of order 7in U.
Let $f(z)$ and $g(z)$ be analytic in . Then we say that the function $g(z)$ is subordinate
to $f(z)$ if there exists an analytic function $w(z)$ in such that $|w(z)|<1(z\in U)$ and$g(z)=f(w(z))$ . For this relation the symbol $g(z)$ $\prec f(z)$ is used. In case $f(z)$ isunivalent in we have that the subordination $g(z)\prec f(z)$ is equivalent toand $g(U)\subset f(U)$ .
Recently, Jung, Kim and Srivastava [3] introduced the following integral operator:$z$$S_{p}^{*}(\gamma)$

$p$

$U$

$U$

$U$

$g(0)=\mathrm{f}(\mathrm{z})$$Q_{\beta}^{\alpha}f(z)$$=(\begin{array}{l}\alpha+\beta\beta\end{array})$$\frac{\alpha}{z^{\beta}}\int_{0}^{z}(1-\frac{t}{z})^{\alpha-1}t^{\beta-1}f(t)dt$$(\alpha>0,\beta>-1;f\in A(1))$

.

(1.2)

They also showed that
$Q_{\beta}^{a}f(z)=z+ \sum_{n=2}^{\infty}\frac{\Gamma(\beta+n)\Gamma(\alpha+\beta+1)}{\Gamma(\beta+\alpha+n)\Gamma(\beta+1)}a_{n}z^{n}$.

It follows from (1.3) that one can define the operator
forand $\beta>-1$ . Someinteresting subclasses of analytic function, associated with the operator , have been
considered recently by Jung et a1.[3], Aouf et al.[l], Li[5], Liu[6] and others.
Motivated by Jung, Kim and Srivastava’s work [3], we now consider alinear operator: $A(p)arrow A(p)$ as following:$Q_{\beta}^{\alpha}$$\alpha\geq 0$$Q_{\beta}^{\alpha}$$Q_{\beta}^{\alpha}$$Q_{\beta}^{\alpha}f(z)=(\begin{array}{ll}p+\alpha+\beta -1p+\beta -\mathrm{l}\end{array})$$\frac{\alpha}{z^{\beta}}\int_{0}^{z}(1-\frac{t}{z})^{\alpha-1}t^{\beta-1}f(t)dt$$(\alpha\geq 0,\beta>-1;f\in A(p))$.

We note that
$Q_{\beta}^{\alpha}f(z)=z^{p}+ \sum_{n=1}^{\infty}\frac{\Gamma(p+n+\beta)\Gamma(p+\alpha+\beta)}{\Gamma(p+n+\alpha+\beta)\Gamma(p+\beta)}a_{p+n}z^{p+n}$(1.3)

47(a $\geq 0,\beta>-1;$ f $\in A(p)$ ).

(1.4)

It is easily verified from the definition (1.4) that$z(Q_{\beta}^{\alpha}f(z))’=(\alpha+\beta+p-1)Q_{\beta}^{\alpha-1}f(z)$ $-(\alpha+\beta-1)Q_{\beta}^{\alpha}f(z)$$Q_{\beta}^{\alpha}$.

(1.5)

When $p=1$ , the identity (1.5) is given in [3]. One can easily see that the operator
has an inverse operator
andis an unit operator.
$f(z)\inA(p)$if it satisfies the condition
Afunctionis said to be in the class$Q_{\beta+\alpha}^{-\alpha}$$Q_{\beta}^{0}$$\Omega(\alpha,\beta;\gamma)$$\frac{z(Q_{\beta}^{\alpha}f(z))}{Q_{\beta}^{\alpha}f(z)},$(1.6)$+ \frac{pz^{p}}{1-z^{p}}\prec\frac{p+p(1-2\gamma)z}{1-z}$

for all $z\in U$ and $0\leq\gamma<1$ .
In this paper, we shall show the extreme points of the closed convex hull of the class. It is then used to determine the coefficient bounds.
In the sequel, we denote the closed convex hull of aclass $H$ by $coH$ . Also, let$E(coH)$ denote the set of all extreme points of $H$ .
$\Omega(\alpha,\beta;\gamma)$

2. Main Results.
In order to derive our main results, we shall need the following lemmas.
Lemma 1([4]).
consists of the functions given by$E(coS_{p}^{*}(\alpha))$$\frac{z^{p}}{(1-xz)^{2p(1-\gamma)}}=z^{p}+\sum_{n=1}^{\infty}\frac{(2p-2p\gamma)_{n}}{n!}x^{n}z^{p+n}(z\in U)$,

(2.1)

where $(a)_{n}=a(a+1)\cdots(a+n-1),x\in C$ and $|x|=1$ .
Lemma 2([9]). The function $(1-z)^{\rho}\equiv e^{\rho\log(1-z)},\rho\neq 0$ , is univalent in if andonly if is either in the closed disk $|\rho-1|\leq 1$ or in the closed disk $|\rho+1|\leq 1$ .
be anaandand letLemma 3([7]). Let $q(z)$ be univalent in$D$lytic in adomain
containing $q(U)$ with $\phi(w)\neq 0$ when $w\in q(U)$ . Set $Q(z)=$$zq’(z)\phi(q(z))$ , $h(z)=\theta(q(z))+Q(z)$ and suppose that(1) $Q(z)$ is starlike (univalent) in ;$+zB’[perp] z[perp]\}Q(z)>0(z\in U)$ .
(2)
If $p(z)$ is analytic in , with $p(0)=q(0),p(U)\subset D$ and$U$$\rho$$U$$\phi(w)$$\theta(w)$$U$$Re\{_{Q(z)}^{\underline{z}h\lrcorner’z[perp]}\}=Re\{_{\phi(q(z))}^{\theta}\lrcorner’q\mathrm{u}z1$$U$$\theta(p(z))+zp’(z)\phi(p(z))\prec\theta(q(z))+zq’(z)\phi(q(z))=h(z)$then $p(z)\prec q(z)$ and $q(z)$ is the best dominant,(2.2)

48
Theorem 1. Afunction $f(z)$ $\in A(p)$ is inexpressed as$\Omega(\alpha,\beta;\gamma)$

if and only if $f(z)$ can be$f(z)=Q_{\beta+\alpha}^{-\alpha} \{z^{p}(1-z^{p})exp[-2p(1-\gamma)\int_{X}\log(1-xz)d\mu(x)]\}$

,

(2.3)

where is aprobability measure defined on the unit circle $X=\{x:|x|=1\}$ .
. Then by Herglotz formula [2], we have
Proof. Let$\mu$$f(z)\in\Omega(\alpha,\beta;\gamma)$$\frac{z(Q_{\beta}^{\alpha}f(z))’}{Q_{\beta}^{\alpha}f(z)}+\frac{pz^{p}}{1-z^{p}}=p(1-\gamma)\int_{X}\frac{1+xz}{1-xz}d\mu(x)+p\gamma$

where is aprobability measure defined on the unit circle $X=\{x :of the identity$\mu$$\frac{d}{dz}\log\frac{Q_{\beta}^{\alpha}f(z)}{z^{p}(1-z^{p})}=\frac{1}{z}[\frac{z(Q_{\beta}^{\alpha}f(z))’}{Q_{\beta}^{\alpha}f(z)}+\frac{pz^{p}}{1-z^{p}}-p]$

,

(2.4)

|x|=1\}$,. By means(2.5)(2.4) yields$Q_{\beta}^{\alpha}f(z)=z^{p}(1-z^{p}) \exp[-2p(1-\gamma)\int_{X}\log(1-xz)d\mu(x)]$

.

(2.6)

Thus
$f(z)=Q_{\beta+\alpha}^{-\alpha} \{z^{p}(1-z^{p})\exp[-2p(1-\gamma)\int_{X}\log(1-xz)d\mu(x)]\}$

Now the proof is complete.
, thenTheorem 2. Let
Proof. We define alinear operator on$0\leq\gamma_{1}<\gamma_{2}<1$$\Omega(\alpha,\beta;\gamma_{2})\subset\Omega(\alpha,\beta;\gamma_{1})$$\Omega(\alpha,\beta;\gamma)$$T_{\gamma}(f)= \frac{Q_{\beta}^{\alpha}f(z)}{1-z^{p}}$as following:$(z\in U)$.

.

.

(2.7)

is alinear homeomorphism fromto. It is well-known thatfor. The result follows immediately.
Theorem 3. (i) The extreme points ofare given by the functions

Then$\Omega(\alpha,\beta;\gamma)$$T_{\gamma}$$S_{p}^{*}(\gamma_{2})\subset S_{p}^{*}(\gamma_{1})$$S_{p}^{*}(\gamma)$$0\leq\gamma_{1}<\gamma_{2}<1$$co\Omega(\alpha,\beta;\gamma)$$f_{l}(z)=Q_{\beta+\alpha}^{-\alpha} \{\frac{z^{p}(1-z^{p})}{(1-xz)^{2p(1-\gamma)}}\}$$(x\in C, |x|=1;z\in U)$

.

(2.8)

49

$(i_{\dot{i}})$Co:$\Omega(\alpha,\beta;\gamma)=\{f$ $\mathrm{f}(\mathrm{z})=J_{X}f_{x}(z)d\mu(x)\}$

(2.9)

,

where varies over the probability measures defined on the unit circle $X$ .
toProof. Since defined by (2.7) is alinear homeomorphism fromit preserves extreme points. By making use of Lemma 1, the results follow at once.
According to Theorem 3and Lemma 1, we have the following corollaries.
$\mu$$\Omega(\alpha,,\theta;\gamma)$$T_{\gamma}$

Corollary 1. Let$f(z)=z^{\mathrm{p}}+ \sum_{n=1}^{\infty}a_{p+n}z^{p+n}\in\Omega(\alpha,\beta;\gamma)$$S_{p}^{*}(\gamma)$,. Then’$|a_{p+n}|\leq\{$$1\leq n<pn\geq p.$$\frac{(2p-2p\gamma)_{n-p}|\prod_{k=1}^{\mathrm{p}}(2p-2p\gamma+n-k)\mathrm{m}_{n!}2-2_{\hslash_{\frac{\Gamma\{p+n+\alpha+\beta)\Gamma(p+\beta\lrcorner}{\Gamma(p+n+\beta)\Gamma(p+\alpha+\beta)-\prod_{k=1}^{p}(n-p+k)|}}}}{n},.\cdot.’\frac{\Gamma(p+n+\alpha+\beta)\Gamma\{p+\beta)}{\overline{\Gamma(\mathrm{p}+n+\beta})\Gamma(p+\alpha+\beta)}$,
The result is sharp.
Corollary 2. Let $f(z)$$=z^{p}+ \sum_{n=1}^{\infty}a_{p+n}z^{p+n}\in\Omega(\alpha,\beta;\gamma)$. Then for $|z|=r<1$ .

$|f(z)| \leq r^{p}+\sum_{n=1}^{p-1}\frac{(2p-2p\gamma)_{n}}{n!}\cdot\frac{\Gamma(p+n+\alpha+\beta)\Gamma(p+\beta)}{\Gamma(p+n+\beta)\Gamma(p+\alpha+\beta)}r^{p+n}$$+ \sum_{n=p}^{\infty}\frac{(2p-2p\gamma)_{n-p}|\prod_{k=1}^{p}(2p-2p\gamma+n-k)-\prod_{k=1}^{p}(n-p+k)|}{n!}\cdot\frac{\Gamma(p+n+\alpha+\beta)\Gamma(p+\beta)}{\Gamma(p+n+\beta)\Gamma(p+\alpha+\sqrt)}\mathrm{r}^{p+n}$

The result is sharp.
. Let be acomplex number withTheorem 4. Let
satisfy either $|2p\rho(1-\gamma)+1|\leq 1$ or $|2p\rho(1-\gamma)-1|\leq 1$ . Then$f(z)\in\Omega(\alpha,\beta;\gamma)$$\rho$$( \frac{Q_{\beta}^{\alpha}f(z)}{z^{p}(1-z^{p})})^{\rho}\prec\frac{1}{(1-z)^{2p\rho(1-\gamma)}}=\mathrm{q}(\mathrm{z})$$(z\in U)$,$\rho\neq 0$

and(2.11)

where $q(z)$ is the best dominant.
Proof. Let$p(z)=( \frac{Q_{\beta}^{\alpha}f(z)}{z^{p}(1-z^{p})})^{\rho}$then $p(z)$ in analytic is$U$,(2.11)

with $p(0)=1$ . Differentiating (2.11) logarithmically we have$\frac{zp’(z)}{p(z)}=\rho(\frac{z(Q_{\beta}^{\alpha}f(z))’}{Q_{\beta}^{\alpha}f(z)}+\frac{pz^{p}}{1-z^{p}}-p)$.

(2.12)

50
Since$f(z)\in\Omega(\alpha, \beta;\gamma)\}$(2.12) is equivalent to$p+ \frac{zp’(z)}{\rho p(z)}\prec\frac{p+p(1-2\gamma)z}{1-z}=h(z)$.

(2.13)

If we take$q(z)= \frac{1}{(1-z)^{2p\rho(1-\gamma)}}$, $\theta(w)=p$ and$\phi(w)=\frac{1}{\rho w}$,(2.14)

then $q(z)$ is univalent by the condition of the theorem and Lemma 2. It is easy to showthat $q(z)$ ,satisfy the conditions of Lemma 3. Since
and$\theta(w)$$\phi(w)$(2.15)$Q(z)=zq’(z) \phi(q(z))=\frac{2p(1-\gamma)z}{1-z}$

is univalent starlike in$U$and$h(z)= \theta(q(z))+Q(z)=\frac{p+p(1-2\gamma)z}{1-z}$,(2.16)

it may be readily checked that the conditions (1) and (2) of Lemma 3are satisfied.
Thus the result follows from (2.13) immediately.

Acknowledgement
The research is partly supported by Jiangsu Gaoxiao Natural Science Foundation(OIKJB110009).

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