数理解析研究所講究録
第 1824 巻 2013 年 82-85

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ON THE REPRESENTATION AND THE RESIDUE
OF CONCAVE FUNCTIONSRINTARO OHNO
ABSTRACT. In [2] we introduced several integral representation formulas for concave
functions. Using those, we gave a general formula to describe the residue offunctions with a pole at $p\in(0,1)$ . In the present article we will present alternate
versions of the formulas, as weil as a shortcut for the calculation to obtain the range ofthe residue.
Key words: concave univalent functions, integral representations$\infty ncave$

1. INTRODUCTION
be theLet be the complex plane, the Riemann sphere andis concave, i.e.
is said to be concave, ifunit disk.
univalent function :is convex. Commonly there are several types of concave functions, which map
conformally onto a simply connected, concave domain in :(1) meromorphic, univalent functions with a simple pole at the origin and the nor, said to belong to the classmalization(2) meromorphic, univalent functions with a simple pole at the point $p\in(O, 1)$ andandthe normalization $f(z)=z+ \sum_{n=2}^{\infty}a_{m}z^{n}$ , said to belong to the class(3) analytic, univalent functions satisfying $f(1)=\infty$ with the normalizations $f(z)=$withatless or equal toand an opening angle of$\alpha\in(1,2], said to$ belong $to the$ class.
A detailed discussion of these classes has already been done in [2]. We therefore concenfor the present article.
trate on the class$\mathbb{D}=\{z\in \mathbb{C} :$\hat{\mathbb{C}}$$\mathbb{C}$

$A$

$f$

$\mathbb{D}arrow\hat{\mathbb{C}}$|z|<1\}$$f(\mathbb{D})$$\mathbb{D}$$\mathbb{C}\backslash f(\mathbb{D})$$\hat{\mathbb{C}}$$f$$f(z)= \frac{1}{z}+\sum_{n=0}^{\infty}a_{r}z^{n}$$\mathcal{C}0_{0},$$f$$Co_{p}$$f$$f(\mathbb{D})$$z+ \sum_{n=2}^{\infty}a_{n}z^{n}$$\alpha\pi$$\infty$$Co(\alpha)$$Co_{p}$2. ALTERNATIVE FORMULAS
In [2] we introduced the following integral representation fomula for functions of$Co_{p}.$Theorem 1. [2] Let $p\in(0,1)$ . For a meromorphic function :of class, holomorphic in utth $\varphi(p)=p$ , such that the concave
there exists a function :junction can be represented as$\mathbb{D}arrow\hat{\mathbb{C}}$$f$$\mathbb{D}arrow \mathbb{D}$$\mathcal{C}0_{p},$$\mathbb{D}$$\varphi$

(1)

for

$f’(z)= \frac{p^{2}}{(z-p)^{2}(1-zp)^{2}}\exp\int_{0}^{z}\frac{-2\varphi(\zeta)}{1-\zeta\varphi(\zeta)}d\zeta$$z\in \mathbb{D}$. Conversely, for any holomorphic function

there enists a concave function of class$\mathcal{C}0_{p}$mapping
described by (1).
$\varphi$$\mathbb{D}arrow \mathbb{D}$with $\varphi(p)=p,$

Howcvcr, a fixcd point of thc function at is not very useful for further discussions.
Using several transformations we obtain an altemate version of Theorem 1.
$\varphi$$p$2010 Mathcmatics Subjcct $ClassiJicatio;\iota.$ $30C45.$
Key words and phrases. Concave functions; Integral representations.

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Corollary 2. Let $p\in(O, 1)$ . For a meromorphic function :of class , there, holomorphic in$\Psi(0)=0$
exists a function:
withsuch that the $\omega ncave$function can be represented as$f$$\Psi$(2)$\mathbb{D}arrow \mathbb{D}$$\mathbb{D}arrow\hat{\mathbb{C}}$$\mathcal{C}0_{p}$$\mathbb{D}$$f’(z)= \frac{p^{2}}{(z-p)^{2}(1-zp)^{2}}\exp(2\int_{p}^{\succ_{-pz}^{-\iota}}\frac{p}{1-\gamma)\zeta}-\frac{\Psi(\zeta)}{1-\zeta\Psi(\zeta)}d\zeta)$. Conversely, for any holomorphic function mapping
there exists a concave function of classdescribed by (2).
. Applying the transformation
Proof. Let $p\in(O, 1)$ andfor$z\in \mathbb{D}$$\Psi$$\mathbb{D}arrow \mathbb{D}$with $\Psi(0)=0,$$Co_{p}$$z\in \mathbb{D}$$\zeta=L_{\frac{x}{px}}^{-}1-$

we obtain

and$\Phi(x)=\varphi(\zeta)$$\int_{0}^{z}\frac{-2\varphi(\zeta)}{1-\zeta\prime p(\zeta)}d\zeta=l^{\frac{p-z}{1-pz}}\frac{-2\Phi(x)}{1-RI\frac{x}{px}\Phi(x)}\cdot\frac{p^{2}-1}{(1-px)^{2}}dx$$= \int_{p^{-pz}}^{\succ^{-}}\frac{-2\Phi(x)(p^{2}-1)}{(1-px)^{2}-(p-x)\Phi(x)(1-px)}dx.$

Here the function is holomorphic in with $\Phi(0)=p$ . Therefore there exists a function:
holomorphic in with $\Psi(0)=0$ , such that. Then$\Phi$$\Psi$$\mathbb{D}arrow \mathbb{D}$$\mathbb{D}$$\mathbb{D}$$\Phi(x)=\frac{p-\Psi(x)}{1-p\Psi(x)}$$\int_{0}^{z}\frac{-2\varphi(\zeta)}{i-\zeta\varphi(\zeta)}d\zeta=\int_{p^{\underline{A=}\frac{z}{pz}}}^{1}\frac{-2\frac{p-\Psi(x)}{1-p\Psi(x)}(p^{2}-1)}{(1-px)^{2}-(p-x)\frac{p-\Psi(x)}{1-p\Psi(\omega)}(1-px)}dx$$= \int_{p}^{\frac{p-z}{1-pz}}\frac{-2(p-\Psi(x))(p^{2}-1)}{(1-px)((1-p^{2})-x\Psi(x)(1-p^{2}))}dx$$= \int_{p}^{\frac{p-z}{1-pz}}\frac{-2(\Psi(x)-p)}{(1-px)(1-x\Psi(x)}dx$$=2 \int_{p}^{\frac{r-z}{1-pz}}\frac{p}{1-px}-\frac{\Psi(x)}{1-x\Psi(x)}dx.$

Changing the variable inside the integration and replacing the integral in (1) leads to thestatement.
The formula for the residue derived from the integral representation in [2] was g\’iven as$\square$follows.
Theorem 3. [2] Let $f(z)\in Co_{p}$ be a concave function with a simple pole at some point$p\in(O, i)$ . Then the residueof this function can be described by some function :, holomorphic in and $\varphi(p)=p$ , such that$f$$\mathbb{D}$$\mathbb{D}arrow$$\varphi$$\mathbb{D}$(3)${\rm Res}_{p}f=- \frac{p^{2}}{(1-p^{2})^{2}}exp.\int_{0}^{p}\frac{-2\varphi(z)}{1-x\varphi(z)}dz.$

Applying the alternative representation from Corollary 2, we obtain
Corollary 4. Let $f(z)\in Co_{p}$ be a concave function with a simple pole at some point $p\in$$(0,1)$ . Then the residueof this function $f$ can be described by some function :holomorphic in and $\Psi(0)=0$ , such that$\Psi$$\mathbb{D}$(4)${\rm Res}_{p}f=- \frac{p^{2}}{(1-p^{2})^{2}}\exp 2\int_{0}^{p}\frac{\Psi(x)}{1-x\Psi(x)}-\frac{p}{i-px}dx.$$\mathbb{D}arrow \mathbb{D},$84
the original presentation is the fixed point of
The advantage of Corollary 4means for the construction, than a fixed point atprovide
easiermuchthe origin. This. Furthermore, the Schwarz Lemma can be applied directly without any complicatedanalysis, giving a way for the estimate of special values. We will show an application in
the next section.
$\Psi a\dagger$$ov\epsilon r$$p$

3. RANGE OF THE RESIDUE
Wirths proved the following statement in [3] using the inequality$| \frac{1}{f(z)}-\frac{1}{z}+\frac{1+p^{2}}{p}|\leq 1$

provided by Miller in [1].
Theorem 5. [3] Let $p\in(0,1)$ . For$a={\rm Res}_{p}f$if and only if(5)$a\in \mathbb{C}$there exists a function $f\in Co_{p}$ such that$|a+ \frac{p^{2}}{1-p^{4}}|\leq\frac{p^{4}}{1-p^{4}}.$Let $\theta\in[0,2\pi).$ $A\mu$nctionf$\in \mathcal{C}0_{p}$has the residue$a=- \frac{p^{2}}{1-p^{4}}+e^{:\theta}\frac{p^{4}}{1-p^{4}}$if and only if(6)$f_{\theta}(z)= \frac{z_{1+p}-+(1+e^{i\theta})z^{2}}{(1-\frac{z}{p})(1-pz)}.$The established representation formula for the residue can be useri for a differen approach of the same statement as described in [2]. For the present discussion we will use
Corollary 4, which provides a shortcut for the proof. We also present some details, omitted
in [2]
be holomorphic in with fixed point at the origin.
Proof. Let $p\in(0,1)$ and :$f\in
Co_{p}$$a={\rmRes}_{p}f$we obtain with the use of Corollary 4withFor$\dagger$$\Psi$$\mathbb{D}arrow \mathbb{D}$$\mathbb{D}$$|a+ \frac{p^{2}}{1-p^{4}}|^{(4)}=\frac{p^{2}}{1-p^{4}}|\frac{1+p^{2}}{1-p^{2}}\exp(2\int_{0}^{p}\frac{\Psi(x)}{1-x\Psi(x)}-\frac{p}{1-px}dx)-1|.$

Some basic calculations yield$\frac{1+p^{2}}{1-p^{2}}=\exp 2(\frac{1}{2}\log\frac{1+\emptyset}{1-p^{2}})=\exp\int_{0}^{p}\frac{2p}{1-p^{2}x^{2}}dx$

and therefore$|a+ \frac{p^{2}}{1-p^{4}}|$$=$$\frac{p^{2}}{1-p^{4}}|\exp\int_{0}^{p}2(\frac{p}{1-p^{2}x^{2}}-\frac{p}{1-px}+\frac{\Psi(x)}{1-x\Psi(x)})dx-1|$$= \frac{\emptyset}{1-p^{4}}|\exp\int_{0}^{p}2\frac{\Psi(x)-p^{2_{X}}}{(1-x\Psi(x))(1-p^{2}x^{2})}dx-1|.$From the triangle inequality, we know that$|e^{w}-1|=| \sum_{n=1}^{\infty}\frac{w^{n}}{n!}|\leq\sum_{n=1}^{\infty}\frac{|w|^{n}}{n!}=e^{|w|}-1.$

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Hence
$|a+ \frac{p^{2}}{1-p^{4}}|\leq\frac{p^{2}}{1-p^{4}}(\exp\int_{0}^{p}2|\frac{\Psi(x)-p^{2_{X}}}{(1-x\Psi(x))(i-p^{2}x^{2})}|dx-1)$

Due tofixed point atorigiil, we$|\Psi(x)|\leq x$ for$0<x<p$ . Furthermore, since(7)$t1_{1}e$$t1_{1}e$$CdJ\downarrow$.

apply the $Sd_{lWalZ}$ Lemma and havefor $|w|\leq x$ , we have$| \frac{w-p^{2}x}{i-xw}|\leq\frac{(1-p^{2})x}{1-x^{2}}$$| \frac{\Psi(x)-p^{2_{X}}}{1-x\Psi(x)}|\leq\frac{(1-p^{2})x}{1-x^{2}}.$Using the above, we finally obtain$|a+ \frac{p^{2}}{1-p^{4}}| (7)\leq \frac{p^{2}}{i-p^{4}}(\exp\int_{0}^{p}2\frac{(1-p^{2})x}{(1-x^{2})(1-p^{2}x^{2})}dx-1)$$= \frac{p^{2}}{1-p^{4}}(\exp(\log(1+p^{2}))-1)$$= \frac{p^{4}}{1-p^{4}}.$

The rest of the proof goes according to the way described in [2].
$\square$

REFERENCES[1] J. Miller, Convex and starlike meromorphic functions, Proc. Amer. Math. Soc. 80 (1980), 607-613.
[2] R. Ohno, Characterizations for concave functions and integral representations, Proceedings of the19th ICFIDCAA (2013), Tohoku Univ. Press, to appear.
[3] K.-J. Wirths, On the residuum of concave univalent functions, Serdica Math. J. 32 (2006), 209-214.

GRADUATE SCHOOL OF INFORMATION SCIENCES,TOHOKU UNIVERSITY, SENDAI, 980-8579, JAPAN.
-mail address: rohnoQims. is. tohoku. ac. jp$E$