数理解析研究所講究録
第 1626 巻 2009 年 115-121

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Integral means inequality of certain analyticfunctions
Tadayuki Sekine, Kazuyuki Tsurumi and Shigeyoshi Owa
Abstract

We obtain the integral means inequality of $f(z)=z+ \sum_{n=2}^{\infty}a_{n}z$“ $(|z|<1)$
and $k_{\alpha}(z)= \frac{z}{(1-z)^{2(1-\alpha)}}(0\leqq\alpha<1, |z|<1)$.

Keywords: Analytic functions, Subordination, Integral means inequality.
2000 Malhematics Subject Classification. Primary $30C45$ .

1. Introduction
Let $A$ denote the class of functions $f(z)$ of the form$f(z)=z+ \sum_{n=2}^{\infty}a_{n}z^{n}$that are analytic in the open unit disk $\mathbb{U}=\{z\in \mathbb{C} : |z|<1\}$ .
$t(z),$
Let $g(z),$denote the analytic functions inand$p_{\alpha}(z),$$q_{\alpha}(z)$$k_{\alpha}(z)$$g(z)= \frac{zf’(z)}{f(z)}$

by
(1.1)

$(0\leqq\alpha<1)$$p_{\alpha}(z)= \frac{1+(1-2\alpha)z}{(1-z)}$$t(z)= \frac{f(z)}{z}$,$\mathbb{U}$

,,

(1.2)

(13)

$q_{\alpha}(z)= \frac{1}{(1-z)^{2(1-\alpha)}}$$(0\leqq\alpha<1)$(14)$k_{\alpha}(z)= \frac{z}{(1-z)^{2(1-\alpha)}}$$(0\leqq\alpha<1)$

,

(1.5)

and

respectively.

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In tliis paper, we obtain the integral means inequality of the functions $f(z)$ in $A$.
andHere we recall the concept of subordination between analytic functions. Letfunctions $f(z)$ and $g(z)$ be analytic in . We say that the function $f(z)$ is subordinate to $g(z)$ if there exists an analytic function $w(z)$ in satisfying $w(0)=0$
and $|w(z)|<1$ such that $f(z)=g(w(z))(|z|<1)$ . We denote this subordinationby $f(z)\prec g(z)$ . Let $g(z)$ be univalent in . Then $f(z)\prec g(z)$ if and only if$f(O)=g(O)$ and(see CH. Pommerenke [3]).
We need the following subordination theorem of J. E. Littlewood.
$k_{\alpha}(z)$$\mathbb{U}$$\mathbb{U}$$\mathbb{U}$$f(\mathbb{U})\subset g(\mathbb{U})$Lemma l.l(Littlewood [1]) If $f(z)$ and $g(z)$ are analytic in$g(z)$ , then, for $\mu>0$ and $z=re^{i\theta}(0<r<1)$$\int_{0}^{2\pi}|f(z)|^{\mu}d\theta\leqq\int_{0}^{2\pi}|g(z)|^{\mu}d\theta$$\mathbb{U}$rvith $f(z)\prec$.

Applying the lemma of Littlewood above, H. Silverman ([6]) showed the integral
means inequalities for univalent functions with negative coefficients. S. Owa andT. Sekine ([4]) proved integral means inequalities with coefficients inequalities fornormalized analytic functions and polynomials(see also Sekine et al. [5]).
In addition we need the following lemma of S. S. Miller and P. T. Mocanu.
Lemma 1.2(Miller and Mocanu [2]) Let $g(z)=g_{n}z^{n}+g_{n+1}z^{n+1}+\cdots$ be andytic
andin urith $g(z)\neq 0$ and $n\geqq 1$ . If$z_{0}=r_{0}e^{i\theta_{0}}(r_{0}<1)$$\mathbb{U}$$|g(z_{0})|= \max_{|z|\leqq|_{l0}|}|g(z)|$then$( i)\frac{z_{0}J(z_{0})}{g(z_{0})}=k$and(ii)${\rm Re}( \frac{z_{0}g’’(z_{0})}{g(z_{0})})+1\geqq k$,whem $k\geqq n\geqq 1$ .
2. Integral means inequality for $f(z)$ and$k_{\alpha}(z)$.

Lemma 2.1. Let $f(z)$ be in $A,$ $g(z)$ be the function given by (1.1) andthe function given by (1.2). If the function $f(z)$ satisfies${\rm Re} \{\beta\frac{zf^{t}(z)}{f(z)}+\gamma(1+\frac{zf^{l/}(z)}{f^{l}(z)})\}>\{\begin{array}{ll}\alpha(\beta+\frac{(1-2\alpha)\gamma}{2(1-\alpha)}), (0\leqq\alpha<1/2)\alpha\beta+\frac{(2\alpha+\alpha-1)\gamma}{2\alpha}, (1/2\leqq\alpha<1)\end{array}$for some real numbers $\beta>0$ and$\gamma>0$, then we have$g(z)\prec p_{\alpha}(z)$.

$p_{\alpha}(z)$be(2.1)

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Proof. First, we shall prove Lemma 2.1 for
function $u’(z)$ by$\alpha(0\leqq$a $<1/2)$ . Let us define the$g(z)= \frac{\sim^{r}f’(z)}{f(\hat{4,}\cdot)}=\frac{1+(1-2\alpha)t_{\sim}^{f};(z)}{1-w(\tilde{k})}(w(k*)\neq 1)$

.

(2.2)

Thus we have an analytic function $w(z)$ in such that $w(0)=0$ . Further, we provethat the analytic function $w(z)$ satisfies $|w(z)|<1(z\in \mathbb{U})$ for$\mathbb{U}$${\rm Re} \{\beta\frac{\sim\# f’(z)}{f(z)}+\gamma(1+\frac{zf’’.(z)}{f’(z)})\}$$={\rm Re} \{\beta(\frac{1+(1-2\alpha)w(z)}{1-w(z)})$$+ \gamma(\frac{1+(1-2\alpha)w(z)}{1-w(z)}+\frac{z(1-2\alpha)w^{l}(z)}{1+(1-2\alpha)w(z)}+\frac{zw^{f}(k*)}{1-w(z),-})\}$$> \alpha(\beta+\frac{(1-2\alpha)\gamma}{2(1-\alpha)})(\beta>0,$

If there exists$z_{0}\in \mathbb{U}$$\gamma>0,0\leqq\alpha<\frac{1}{2})$.

such that$| \approx|_{\simeq}|xo|\max_{<}|u\}(z)|=|w(z_{0})|=1$

,

(2.3)

then we have by Lemma 1.2,
$w(,\sim 0)=e^{*\theta},$

For such a point$z_{0}\in \mathbb{U}$$\frac{z_{0}w^{f}(z_{0})}{w(z_{0})}=k,$${\rm Re}( \frac{z_{0}w^{ll}(z_{0})}{w^{f}(z_{0})})+1\geqq k(k\geqq 1)$.

, we obtain that

距 $\{\beta\frac{z_{0}f’(0)}{f(z_{0})}+\gamma(1+\frac{z_{0}f^{\prime/}(z_{0})}{f’(z_{0})})\}$$={\rm Re} \{(\beta\frac{1+(1-2\alpha)w(\sim\gamma 0)}{1-w(z_{0})})$$+ \gamma(\frac{1+(1-2\alpha)w(z_{0})}{1-w(z_{0})}+\frac{z_{0}(1-2\alpha)w’(z_{0})}{1+(1-2\alpha)w(z_{0})}+\frac{z_{0}w^{f}(z_{0})}{1-w(z_{0})})\}$$={\rm Re} \{-(1-2\alpha)(\beta+\gamma)+\frac{2(1-\alpha)(\beta+\gamma)}{1-w(z_{0})}+\gamma\frac{(1-2\alpha)z_{0}w’(z_{0})}{1+(1-2\alpha)w(z_{0})}+\gamma\frac{z_{0}w’(z_{0})}{1-w(z_{0})}\}$$=-(1-2 \alpha)(\beta+\gamma)+2(1-\alpha)(\beta+\gamma){\rm Re}\{\frac{1}{1-w(z_{0})}\}$$+ \gamma(1-2\alpha){\rm Re}\{\frac{z_{0}w’(z_{0})}{1+(1-2\alpha)w(z_{0})}\}+\gamma{\rm Re} t\frac{z_{0}w’(z_{0})}{1-cv(z_{0})}\}$

118$=-(1-2 \alpha)(\beta+\gamma)+2(1-\alpha)(\beta+\gamma){\rm Re} t\frac{1}{1-w(z_{0})}\}$$+ \gamma k(1-2\alpha){\rm Re} t\frac{w(z_{0})}{1+(1-2\alpha)w(z_{0})}\}+\gamma k{\rm Re}\{\frac{w(z_{0})}{1-w(z_{0})}\}$$\leqq$$-(1-2 \alpha)(\beta+\gamma)+(1-\alpha)(\beta+\gamma)+\frac{\gamma k(1-2\alpha)}{2(1-\alpha)}-\frac{\gamma k}{2}$$= \alpha\beta+\alpha\gamma+\frac{\alpha\gamma}{2(1-\alpha)}(-k)$$\leqq\alpha\beta+\alpha\gamma+\frac{\alpha\gamma}{2(1-\alpha)}(-1)$$= \alpha(\beta+\frac{(1-2\alpha)\gamma}{2(1-\alpha)})$,which contradicts the hypothesis (2.1) for $\alpha(0\leqq\alpha<1/2)$ . Therefore there is nosuch that $|w(z_{0})|=1$ . This implies that $|w(z)|<1$ for all $z\in U$ . Thus wehave$z_{0}\in \mathbb{U}$$g(z)\prec p_{\alpha}(z)$.

Second, we shall prove Lemma 2.1 for $\alpha(1/2\leqq\alpha<1)$ in the same way. We
show that the analytic function $w(z)$ defined by (2.2) satisfies $|w(z)|<1(z\in \mathbb{U})$ for${\rm Re} \{\beta\frac{zf’(z)}{f(z)}+\gamma(1+\frac{f’’(z)}{f^{l}(z)})\}>\alpha\beta+\frac{(2\alpha^{2}+\alpha-1)\gamma}{2\alpha}(\beta>0,$

By Lemma 1.2, for the point$z_{0}$$\gamma>0,$$\frac{1}{2}\leqq\alpha<1)$satisfying (2.3), we have the following.

${\rm Re} \{\beta\frac{z_{0}f^{f}(z_{0})}{f(z_{0})}+\gamma(1+\frac{z_{0}f^{tt}(z_{0})}{f’(z_{0})})\}$$=$$-(1-2 \alpha)(\beta+\gamma)+2(1-\alpha)(\beta+\gamma){\rm Re}\{\frac{1}{1-w(z_{0})}\}$$+ \gamma k(1-2\alpha){\rm Re} t\frac{w(z_{0})}{1+(1-2\alpha)w(z_{0})}\}+\gamma kR\epsilon\{\frac{w(z_{0})}{1-w(z_{0})}\}$$\leqq$$-(1-2 \alpha)(\beta+\gamma)+(1-\alpha)(\beta+\gamma)+\gamma k(2\alpha-1)\frac{1}{2\alpha}-\frac{\gamma k}{2}$$= \alpha\beta+\alpha\gamma+\frac{(1-\alpha)\gamma}{2\alpha}(-k)$$\leqq\alpha\beta+\alpha\gamma+\frac{(\alpha-1)\gamma}{2\alpha}$$= \alpha\beta+\frac{(2\alpha^{2}+\alpha-1)\gamma}{2\alpha}$

,.

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which contradicts the hypothesis (2.1) for $\alpha(1/2\leqq\alpha<1)$ . Therefore there is nosuch that $|w(z_{0})|=1$ . This implies that $|w(z)|<1$ for all. Thus wehave$z_{0}\in \mathbb{U}$$z\in \mathbb{U}$$g(z)\prec p_{\alpha}(z)$.

Lemma 2.2. Let $f(z)$ be in $A,$ $g(z)$ be the function defined by (1.1) andbe the function defined by (1.2). If the function $f(z)$ satisfies$g(z)= \frac{zf’(z)}{f(z)}\prec p_{\alpha}(z)$$(\sim\in \mathbb{U})$then we have${\rm Re}\{g(z)\}>\alpha$$\mathbb{U}$,.

Proof. Since
is univalent in , we havethis lemma. Thus we have ${\rm Re}\{g(z)\}>\alpha$ , because$p_{\alpha}(z)$$p_{\epsilon r}(z)$by the assumption ofin U.

$g(\mathbb{U})\subset p_{\alpha}(\mathbb{U})$${\rm Re}\{p_{\alpha}(z)\}>\alpha$Lemma 2.3. Let $f(z)$ be in A. Let $g(z),$ $t(z)$ andbe the functions definedby (1.1), (1.3) and (1.4), respectively. If the flnction $f(z)$ satisfies$q_{\alpha}(z)$${\rm Re} \{g(z)\}={\rm Re}\{\frac{zf^{l}(z)}{f(z)}\}>\alpha$then we have$t(z)\prec q_{\alpha}(z)$$(\alpha\in \mathbb{U})$,.

Proof. Let us define the function $v(z)$ by.

$t(z)= \frac{f(\tilde{k})}{z}=\frac{1}{(1-v(z))^{2(1-\alpha)}}$(2.4)

Thus we have an analytic function $v(z)$ in such that $v(O)=0$ . Eirther, we provethat the analytic function $v(z)$ satisfies $|v(z)|<1(z\in \mathbb{U})$ for$\mathbb{U}$

.

(2.5)

${\rm Re} \{g(z)\}={\rm Re}\{\frac{zf’(z)}{f(\tilde{\sim})}\}>\alpha$If there exists$\sim 0\nu\in \mathbb{U}$such that$\max_{|z|\leqq|z_{0}|}|v(z)|=|v(z_{0})|=1$,

then we have by Lemma 1.2,$v(z_{0})=e^{i\theta},$$\frac{z_{0}v’(z_{0})}{\iota l(z_{0})}=k\geqq 1,$${\rm Re}( \frac{z_{0}v^{l/}(z_{0})}{v^{f}(z_{0})})+1\geqq k$.

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For such a point$z_{0}\in \mathbb{U}$, we obtain that$\frac{z_{0}f^{t}(\tilde{6}0)}{f(z_{0})}-1$$= \frac{2(1-\alpha)z_{0}(1-v(z_{0}))^{1-2\alpha}(v’(z_{0}))}{(1-v(z_{0}))^{2(1-\alpha)}}$$=$$\frac{2(1-\alpha)z_{0}v^{f}(z_{0})}{(1-v(z_{0}))}$$=$$\frac{2(1-\alpha)kv(z_{0})}{(1-v(z_{0}))}$.

Thus${\rm Re} \{\frac{z_{0}f^{l}(z_{0})}{f(z_{0})}\}$$=$$1+2(1- \alpha)k{\rm Re}\{\frac{v(z_{0})}{1-v(z_{0})}\}$$=$$1+2(1- \alpha)k(-\frac{1}{2})$$=1-(1-\alpha)k$$\leqq\alpha$,which contradicts the hypothesis (2.5). Therefore there is no$|v(z_{0})|=1$ . This implies that $|v(z)|<1$ for all. Thus we have$z_{0}\in \mathbb{U}$such that$z\in \mathbb{U}$$t(z)\prec q_{\alpha}(z)$.

We obtain the following result by using Lemmas 2.1, 2.2, 2.3 and 1.1.
be the function given by (1.5). IfTheorem 2.1. Let $f(z)$ be in A. Let$k_{\alpha}(z)$the function $f(z)$ satisfies${\rm Re} \{\beta\frac{\sim f^{f}(z)}{f(z)}+\gamma(1+\frac{zf’’(z)}{f’(z)})\}>\{\begin{array}{ll}\alpha(\beta+\frac{(1-2\alpha)\gamma}{2(1-\alpha)}), (0\leqq\alpha<1/2)\alpha\beta+\frac{(2\alpha^{2}+\alpha-1)\gamma}{2\alpha}, (1/2\leqq\alpha<1)\end{array}$for some real numbers $\beta>0$ andwe have$\gamma>0$, then for $\mu>0$ and $z=re^{i\theta}(0<r<1)$ ,$\int_{0}^{2\pi}|f(re^{i\theta})|^{\mu}d\theta\leqq\int_{0}^{2n}|k_{\alpha}(re^{:\theta})|^{\mu}d\theta$.

Proof. By Lemmas (2.1), (2.2) and (2.3), we have$t(z)\prec q_{\alpha}(z)$,under the hypothesis of the theorem. Then by Lemma 1.1, we have$\int_{0}^{2\pi}|t(re^{i\theta})|^{\mu}d\theta\leqq\int_{0}^{2\pi}|q_{\alpha}(re^{:\theta})|^{\mu}d\theta$

(2.6)

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for $\mu>0$ and $z=7^{\cdot}e^{i\theta}(0<r<1)$ , under the hypothesis of the theorem.
Thus we obtain the following integral means inequality by (2.6)$\int_{0}^{2\pi}|f(re^{i\theta})|^{\mu}d\theta\leqq\int_{0}^{2r}|k_{\alpha}(re^{i\theta})|^{\mu}d\theta$.

for $\mu>0$ and $z=re^{i\theta}(0<r<1)$ , under the hypothesis of the theorem.

References[1] J. E. Littlewood, On inequalities in the theory of functions, Proc. London Math.
Soc. , (2) 23 (1925), 481-519.

[2] S. S. Miller and P. T. Mocanu, Second order differential inequalities in thecomplex plane, J. Math. Anal. Appl., 65(1978), 289-305.
[3] CH. Pommerenke, Univalent Functions, Vandenhoeck 8 Ruprecht, Gottingen,

1975.

[4] S. Owa and T. Sekine, Integral means for analytic functions, J. Math. Anal.
A $ppl.,$ $304$ (2005), $772arrow 782$ .
[5] T. Sekine, S. Owa and R. Yamakawa, Integral means of certain analytic functions, General Math., $13(2005),99- 108$ .

[6] H. Silverman, Integral means for univalent functions with negative coefficients,
Houston J. Math., $23(1997),169- 174$ .

Research unit of Mathematics, College of Pharmacy, Nihon University
7-1 Narushinodai 7-chome, $fi\{mabashiarrow shi$ , Chiba $274- 8555_{f}$ Japan
E-mail: sekine.tadayuki@nihon-u.ac.jp
Tokyo Denki University$Nishib- cho_{j}$ Kanda Chiyoda-ku, Tokyo $101- 8457_{p}$ Japan$2arrow 2,$

Department of Mathematics, Kinki University
Higashi-Osaka, Osaka $577- 85\theta 2$, Japan

E-mail: owa@math.kindai.ac.jp