数理解析研究所講究録
第 1626 巻 2009 年 109-114

109

On geometric properties of certain multivalentfunctions with real coefficients
Hitoshi Saitoh
Department of Mathematics, Gunma National Collegeof Technology, Maebashi, Gunma 371-0837, Japansaitoh@nat.gunma-ct.ac.jp
AbstractLetbe the class of analytic functions with real $\infty efficients$ in the open unit disk .
Fbr $f(z)$ belonging to the class, some sufficient conditions for p-valently starlikenessand p-valently convexity are discussed.
$\mathcal{T}(p)$$\mathbb{U}$$\mathcal{T}(p)$1Introduction
Let$\mathcal{A}(p)$be the class of functions(1.1)$f(z)=z^{p}+ \sum_{n=1}^{\infty}a_{n+p^{Z^{n+p}}}$which are analytic in the open unit disk $\mathbb{U}=\{z : |z|<1\}$ .

We denote by $S^{*}(p)$ andthe subclasses ofare p-valently starlike and p-valently convex.
$\mathcal{K}(p)$A imction$f(z)\in \mathcal{A}(p)$$\mathcal{A}(p)$whose members map

is said to be p-valently starlike in${\rm Re} t\frac{zf’(z)}{f(z)}\}>0$$\mathbb{U}$${\rm Re} \{1+\frac{zf’’(z)}{f(z)}\}>0$

Let us define$\mathcal{T}(p)$.

(1.2)

if and only if$(z\in \mathbb{U})$

.

(1.3)

the class of analytic functions with real coefficients, that is,$\mathcal{T}(p)=\{f(z)\in \mathcal{A}(p)f(z)=z^{p}+\sum_{n=1}^{\infty}a_{n+p}z^{n+p},$

where is the set of real numbers. Then it follows that
Furthermore, let us defne the class of analytic fimctions in$\mathbb{R}$$a_{n+p}\in \mathbb{R}\}$$\mathcal{T}(p)\subset \mathcal{A}(p)$$\mathcal{P}$$\mathcal{P}=\{p(z)p(z)=1+\sum_{k=l}^{\infty}p_{k}z^{k}$$p(z)\in \mathcal{P}$onto domain which

if and only if$\mathbb{U}$$(z\in \mathbb{U})$Similarly, $f(z)\in A(p)$ is said to be pvalently convex in$\mathbb{U}$is called $Cara6odory$ function.

,$\mathbb{U}$(14).

, that is,${\rm Re} p(z)>0\}$.

(1.5)

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Preliminaries

2

For our results, we prepare the next lemmas.
Let(Numokawa [3])Lemma 1$p(z)\in \mathcal{P}$and suppose that there eststs a point${\rm Re} p(z)>0$

for${\rm Re} p(z_{0})=0$

and$\infty\in \mathbb{U}$such that$|z|<|z_{0}|$$p(\triangleleft)\neq 0$

(2.1)

.

Then we have
(2.2)$\frac{\triangleleft p’(\eta)}{p(\eta)}=ik$where$k$is real and$|k|\geqq 1$.
Let $p(z)=1+p_{1}z+p_{2}z^{2}+\cdots$ be analytic in

Lemma 2 (Saitoh [5])are $7ta$ numbers.
Suppose that$\alpha\geqq 1$

and $dl$ coefficients$p_{i}$$\mathbb{U}$

(2.3)

.

(2.4)

$p+{\rm Re} \frac{zf^{(p+l)}(z)}{f^{(p)}(z)}>0$

in U.

(2.5)

$k+{\rm Re} \frac{zf^{(k+1)}(z)}{f^{(k)}(z)}>0$

$in$

,

(2.6)

${\rm Re}\{p(z)+\alpha zp’(z)\}>0$

where$\mathbb{U}$$in$. Then we have
$1+{\rm Re} \{\frac{zp’(z)}{p(z)}\}>0$

Lemma 3

(Nunokawa [2])Then $f(z)$ is p-valent infor $k=0,1,2,$
Lenuna 4$p^{(n-1)}(0)=0$$\cdots$$\mathbb{U}$$in$$\mathbb{U}$Let $f(z)\in \mathcal{A}(p)$ and supposeand$\mathbb{U}$, $p-1$ . Zhis shows that $f(z)\in \mathcal{K}(p)$ and $f(z)\in S^{*}(p)$ .

(Owa-Nunokawa [4]). If

Let $p(z)$ be analytic in${\rm Re}\{p(z)+\alpha zp’(z)\}>\beta$$in$with $p(O)=1,$ $p’(O)=–$$\mathbb{U}$$\mathbb{U}$,$=$(2.7)

then${\rm Re} \{p(z)\}>\beta+(1-\beta)\{2\int_{0}^{1}\frac{1}{1+\rho^{nB\iota(\alpha)}}d\rho-1\}$$where\alpha\neq 0,$

3${\rm Re}(\alpha)\geqq 0$$in$$\mathbb{U}$

,

(2.8)

and $\beta<1$ .

Main results
First, we prove

Theorem 1Let

$f(z)\in \mathcal{A}\zeta p)$

and suppose that${\rm Re} \{f^{(p)}(z)+\alpha zf^{(p+1)}(z)\}>-\frac{p!}{2}\alpha$$(z\in \mathbb{U})$

(3.1)

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for some$\alpha(\alpha>0)$.

$\mathbb{R}en$

we have${\rm Re}\{f^{(p)}(z)\}>0$

Proof.

If there exists a point$z_{0}\in \mathbb{U}$$(z\in \mathbb{U})$

.

(3.2)

such that${\rm Re} \frac{f^{(p)}(z)}{p!}>0$$f\alpha r$$|z|<|z_{0}|$

and

and

${\rm Re} \frac{f^{(p)}(\triangleleft)}{p!}=0$$\frac{f^{(p)}(\infty)}{p!}\neq 0$,then $hom$ Lemma 1, we have$\triangleleft)f^{(p+1)}(\eta)\leqq-\frac{p!}{2}(1+|\frac{f^{(p)}(z_{0})}{p^{1}}|^{2})$.

This contradicts the assumption (3.1) and completes the proof.

ロNow, we prove
Let $f(z)\in \mathcal{T}(p)$ be analytic inTheorem 2
Suppose that$\mathbb{U}$.

${\rm Re} \{\frac{(1-\alpha p+\alpha j)f^{(j)}(z)+\alpha zf^{(j+1)}(z)}{z^{p-j}}\}>0$

where$\alpha\geqq 1$(3.3). Then we have$j+{\rm Re} \frac{zf^{(j+1)}(z)}{f^{(j)}(z)}>0$

for $j=0,1,2,$
Proof.

$(z\in \mathbb{U})$Let$\cdots,p$(3.4)$(z\in \mathbb{U})$.

$p(z)= \frac{(p-j)!f^{(j)}(z)}{p!z^{p-j}}$. Applying Lemma 2,$1+ \alpha{\rm Re}\frac{zf^{(j+1)}(z)-(p-j)f^{(j)}(z)}{f^{(j)}(z)}>0$$(z\in \mathbb{U})$$j+{\rm Re} \frac{zf^{(j+1)}(z)}{f^{(j)}(z)}>p-\frac{1}{\alpha}\geqq p-1>0$$(z\in \mathbb{U})$.

Therefore, we obtain.

Putting $j=0$ in Theorem 2, we have

Corollary 1
Suppose that

Let $f(z)\in \mathcal{T}(p)$ be andytic in$\mathbb{U}$.

${\rm Re} t\frac{(1-\alpha p)f(z)+\alpha zf’(z)}{z^{p}}\}>0$$(z\in \mathbb{U})$

(3.5)

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where

$\alpha\geqq 1$. Then we have$(z\in \mathbb{U})$${\rm Re} \frac{zf’(z)}{f(z)}>0$,

that is $f(z)\in S^{*}(p)$ .
Letting $j=1$ in Theorem 2, we have

Corollary 2
Suppose that

Let $f(z)\in \mathcal{T}(p)$ be andytic in$\mathbb{U}$.

where$\alpha\geqq 1$. Then we have
$1+{\rm Re} \frac{zf’’(z)}{f(z)}>0$

that is(3.6)$(z\in \mathbb{U})$${\rm Re} \{\frac{(1-\alpha p+\alpha)f^{l}(z)+\alpha zf^{\prime l}(z)}{z^{p-1}}\}>0$$f(z)\in \mathcal{K}(p)$$(z\in \mathbb{U})$,.

Next we prove
Let $f(z)\in \mathcal{T}(p)$ be analytic inTheorem 3Suppose that$\mathbb{U}$.

${\rm Re} \{\frac{(1-\alpha p+\alpha j)f^{(j)}(z)+\alpha zf^{(j+1)}(z)}{z^{p-j}}\}>0$

for $j=2,3,$$\cdots,p$, where$\alpha\geqq 1$$(z\in \mathbb{U})$(3.7). Then we have$k+{\rm Re} \frac{zf^{(k+1)}(z)}{f^{(k)}(z)}>0$

for $k=0,1,2,$ $\cdots,j-1$ . Therefore, we have $f(z)\in S^{*}(p)$ andProof. Rom Theorem 2,$j+{\rm Re} \frac{zf^{0+1)}(z)}{f^{(j)}(z)}>0$$f(z)\in \mathcal{K}(p)$.

$(z\in \mathbb{U})$for $j=0,1,2,$ $\cdots,p$ . If $j\geqq 2$ , using Lemma 3, we show that$k+{\rm Re} \frac{zf^{(k+1)}(z)}{f^{(k)}(z)}>0$$(z\in \mathbb{U})$for $k=0,1,2,$ $\cdots,j-1$ . In the case of $k=0$ and $k=1$ , we have $f(z)\in S^{*}(p)$ and $f(z)\in$$\mathcal{K}(p)$.

Putting $j=p$ in Theorem 3, we obtam
Corollary 3
Suppose that

Let $f(z)\in \mathcal{T}(p)$ be andytic in U.
${\rm Re}\{f^{(p)}(z)+\alpha zf^{(p+1)}(z)\}>0$

where$\alpha\geqq 1$. Then we have $f(z)\in S^{*}(p)$ and$f(z)\in \mathcal{K}(p)$$(z\in \mathbb{U})$(3.8).

Let us define generalized Libera-Bernardi integral operator$F(z)= \frac{c+p}{z^{c}}\int_{0}^{z}t^{c-1}f(t)dt$$(c>-p)$

(3.9)

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for $f(z)\in \mathcal{A}(p)$ .
Next, we prove the following theorem.

Theorem 4function

Let $f(z)\in \mathcal{T}(p)$ be analytic in$\mathbb{U}$and satisfies

Proof.

$(z\in \mathbb{U})$, then the$(c>-p)$$F(z)= \frac{c+p}{z^{c}}/o^{z}t^{c-1}f(t)dt$

belongs to $S^{*}(p)$ and${\rm Re} f^{(p)}(z)>0$

for all $c(p-1\leqq-c<p)$ .

$\mathcal{K}(p)$By differentiating (3.9), we have$F^{(p)}(z)+ \frac{1}{c+p}zF^{(p+1)}(z)=f^{(p)}(z)$.

Therefore,${\rm Re} \{F^{(p)}(z)+\frac{1}{c+p}zF^{CP+1)}(z)\}>0$

and$\frac{1}{c+p}\geqq 1$$(-p<c\leqq 1-p)$. Using Lemma 2 for$(z\in \mathbb{U})$$p(z)= \frac{F^{[p)}(z)}{p!}$$1+ \frac{1}{c+p}{\rm Re}\frac{zF^{(p+1)}(z)}{F(p)(z)}>0$$(z\in \mathbb{U})$, we obtain.

Then we have$p+{\rm Re} \frac{zF^{(p+1)}(z)}{F(p)(z)}>-c\geqq p-1>0$$(z\in \mathbb{U})$.

From Lemma 3, we have$k+{\rm Re} \frac{zF^{(p+1)}(z)}{F(p)(z)}>0$$(z\in \mathbb{U})$for $k=0,1,2,$ $\cdots,p-1$ .
Tffling $k=0$ , we have $F(z)\in S^{*}(p)$ , also letting $k=1$ , we obtain $F(z)\in \mathcal{K}(p)$ .

Applying $c=1-p$ in Theorem 4, we can prove
Corollary 4

Let $f(z)\in \mathcal{T}(p)$ be $and\phi c$ in$\mathbb{U}$and satisfies${\rm Re} f^{(p)}(z)>0$$(z\in \mathbb{U})$, then the$fi\iota nc\hslash on$$g(z)= \frac{1}{z^{1-p}}\int_{0}^{\epsilon}\frac{f(t)}{t^{p}}dt$

belongs to $S^{*}(p)$ and$\mathcal{K}(p)$.

Applying Lemma 4, we can prove

Theorem 5

If$f(z)\in \mathcal{T}(p)$be analytic in$\mathbb{U}$with${\rm Re} \frac{f^{(p)}(z)}{p1}>\beta$. If the function $F(z)$ given by(3.9), then${\rm Re} \frac{F^{(p)}(z)}{p^{1}}>\beta+(1-\beta)\{\int_{0}^{1}\frac{1}{1+\rho^{\frac{1}{c+p}}}d\rho-1\}$

where $\beta<1$ .

$(z\in \mathbb{U})$,(3.10)

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Proof.

By differentiatmg (3.9), we can show that$\frac{F^{(p)}(z)}{p!}+\frac{1}{c+p}\frac{zF^{(p+1)}(z)}{p!}=\frac{f^{(p)}(z)}{p!}$Letting $p(z)= \frac{F^{(p)}(z)}{p!}$ and $n=1,$$\alpha=\frac{1}{c+p}$.

in Lemma 4, we have 3.10 .
$)$

$($口

Putting $p=1$ in Theorem 5, we obtain

Corollary 5

If$f(z)\in \mathcal{T}(1)=\mathcal{T}$and ${\rm Re} f’(z)>0$ , let the function $F(z)$ given by$F(z)= \frac{c+1}{z^{c}}\int_{0}^{z}t^{c-1}f(t)dt$$(c>-1)$ ,(3.11)

then we have${\rm Re} F’(z)> \beta+(1-\beta)\{2\int_{0}^{1}\frac{1}{1+\rho^{\frac{1}{c+1}}}d\rho-1\}$.

References[1] Bernardi S. D., Convex and stariike univdent firnctions, Thrans. Amer. Math. Soc. 135 (1969),

429-446.

[2] Nunokawa M., On the theory of multivalent functions, Tsukuba Jour. Math. 11(2) (1987),273286.

[3] Nunokawa M., On properties of Non-Cambeodory flnctions, Proc. Japan Acad, 68 Ser.
A(6) (1992), 152-153.
[4] Owa S. and Nunokawa M., Applications of a subordination theorem, Jour. Math. Anal. Appl.
188(1) (1994), 219226.

[5] Saitoh H., $Geomet\sqrt c$ properties ofk\^oky\^uroku 1579 (2008), 101-109.

$ce\hslash ain$andytic functions urith red coefficients, RIMS