数理解析研究所講究録
1112 巻 1999 年 135-137

135

A NOTE ON YAMAKAWA’S QUESTIONTOSHIYUKI SUGAWA

須川敏幸・京都大学理学部

ABSTRACT. This short note gives an affirmative answer to a question raised by R. Yamakawa. This is included also in a result by Anisiu and Mocanu.

In the workshop on univalent functions held at RIMS during March 15-18, 1999, Professor Rikuo Yamakawa asked us the following question: Let $m_{0}=2.8329\ldots$ be the smallest

number $m$ satisfying $m>1$ and$\cos(\sqrt{m^{2}-1})+\sqrt{m^{2}-1}\sin(\sqrt{m^{2}-1})=e^{-1}$(cf. [2]). Is the analytic function $q(z)=m_{0}ze^{m}0z/(e^{m_{0}z}-1)$ subordinate to the funct\’ion$1+m_{0}z$ in the unit disk?
For analytic functions and in the unit disk we will say herethat is subordinate to if there is a holomorphic map from the unit disk into itself
such that $\omega(0)=0$ and.
The above number
is nowadays known as the possible smallest one such that thecondition $|f’’/f’|<m_{0}$ for an analytic function in the unit disk implies starlikeness of, i.e., ${\rm Re}(zf’(z)/f(z))>0$ (see [1], [3] and [4]). Yamakawa asserted that the affirmative
answer to the above question can be used in an important step of a proof of the abovementioned statement and its generalizations.
this short note, we will provide an elementary proof for the affirmative answer toYamakawa’s question. Actually, we will prove the following result.
$f$

$f$

$g$

$g$

$\omega$$f=g\mathrm{o}\omega$$m_{0}$$f$$f$$\dot{\mathrm{I}}\mathrm{n}$Proposition 1. For $0<m\leq 3\pi/2=4.71\ldots$ , the function $q_{m}(z)=mze^{mz}/(e^{mz}-1)$ issubordinate to $1+mz$ in the unit disk.

ark. The same result for $0<m\leq\pi$ was already known by Anisiu and Mocanu[1, Lemma 3]. This weaker result is sufficient to answer to Yamakawa’s question. Their
proof uses the Taylor expansion of $z/(e^{z}-1)$ in terms of the Bernoulli numbers.
${\rm Re} m$

By the maximum modulus principle, we can translate the above statement as follows.
$q_{m}(z)$is subordinate to $1+mz$ in the unit disk$\Leftrightarrow|\frac{ze^{z}}{e^{z}-1}-1|=|\frac{z}{1-e^{-z}}-1|<m$$\Leftrightarrow|\frac{z}{e^{z}-1}-1|<m$

in $|z|<m$$\Leftrightarrow|\frac{1}{e^{z}-1}-\frac{1}{z}|=|\frac{e^{z}-1-z}{z(e^{z}-1)}|<1$$\Leftrightarrow|e^{z}-1-z|\leq|z||e^{z}-1|$

Date: September 2, 1999.

in $|z|<m$in $|z|<m$in $|z|<m$ .

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Since the meromorphic function $F(z):=(e^{z}-1-Z)/z(e^{z}-1)$ is actually holomorphic
in the disk $|z|<2\pi$ , it suffices to show that $|F(z)|\leq 1$ in the boundary of the square$Q=\{z\in \mathbb{C};|{\rm Re} z|<3\pi/2, |{\rm Im} z|<3\pi/2\}$ by the maximum modulus principle again.
First we make rough estimates.
Lemma 2. For $|z|>1$ , the assertion $|F(z)|\leq 1$ holds $if|z|/(|z|-1)\leq|e^{z}-1|$ .
Proof. By assumption $(1-|z|)|e^{z}-1|+|z|\leq 0$ , so we obtain$|e^{z}-1-z|\leq|e^{z}-1|+|z|=|z||ez-1|+(1-|Z|)|e-z1|+|z|\leq|z||e^{z}-1|$ .

Lemma 3. If$z$with ${\rm Re} z<-1$ satisfies$|e^{\chi}|=e^{\mathrm{R}z} \mathrm{e}\leq\frac{-2{\rm Re} z-1}{4|z|^{2}}$then $|F(z)|\leq 1$ holds.

Proof. Note that $|z|>1$ and $|1+z|<|z|$ by assumption. We have$|z|-|1+z|= \frac{|z|^{2}-|1+\sim|^{2}\nu}{|z|+|1+z|}\geq\frac{-2{\rm Re} z-1}{2|z|}\geq 2|ze^{z}|\geq(1+|Z|)|e^{z}|$Therefore we obtain$|e^{z}-1-z|\leq|e^{z}|+|1+z|\leq|z|(1-|e^{z}|)\leq|z(1-e)z|$

The boundary of the squareand.

$\square$consists of three parts:. Note that every point$C_{1}= \{\frac{3\pi}{2}+it;|t|\leq\frac{3\pi}{2}\},$ $C_{2}=$$C_{3}= \{t\pm\frac{3\pi i}{2};|t|\leq\frac{3\pi}{2}\}$$\{-\frac{3\pi}{2}+it;|t|\leq\frac{3\pi}{2}\}$$3\pi/2\leq|z|\leq 3\pi/\sqrt{2}$$Q$.

$z$in$\partial Q$

satisfies.

the assumption of Lemma 2 is fulfilled because $|e^{z}-1|\geq e^{3\pi/2}-1>110>$$|z|/(|z|-1)$ for each $z\in C_{1}$ . For $z\in C_{2}$ , we haveOn

$C_{1}$$\frac{-2{\rm Re} z-1}{4|z|^{2}}\geq\frac{3\pi-1}{18\pi^{2}}>0.047>e^{-3\pi/2}=0.0089\ldots$

and hence the assumption in Lemma 3 is valid. Thus we see that $|F(Z)|\leq 1$ holds onslightly more. For
and . On the other hand, we have to be careful on
we consider the function$z=t \pm\frac{3}{2}\pi i\in C_{3}$$C_{3}$$C_{2}$$C_{1}$$f(t):=|Z(e^{z}-1)|^{2}-|e^{z}-1-Z|^{2}=(t^{2}+ \frac{9\pi^{2}}{4})(e^{2t}+1\mathrm{I}-((1+t)^{2}+(\frac{3\pi}{2}+e^{t})^{2})$$=(t^{2}+ \frac{9\pi^{2}}{4}-1)e^{2t}-3\pi e^{t}-(1+2t)$.

If $t\leq-1$ we have$f(t) \geq\frac{9\pi^{2}}{4}e^{2ti}-3\pi e-(1+2t)=(\frac{3\pi}{2}e^{t}-1)^{2}-(2+2t)\geq 0$

Otherwise, we have$f(t) \geq g(t):=(\frac{9\pi^{2}}{4}-1)e^{2t}-3\pi e-(t1+2t)$

,.

,

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and the function is increasing in the interval $[-1, +\infty)$ because $g^{\prime/}(t)\geq g^{\prime/}(-1)>21$
there and $g’(-1)=0.27\cdots>0$ . Since $g(-1)=0.40\cdot*\cdot>0$ , we have $f(t)\geq g(t)>0$ for$t>-1$ . These mean $|F|\leq 1$ on, too. Now the proof is complete.
$g$$C_{3}$

Remarks.

1. If we set $g_{m}(z)=(e^{mz}-1)/m$ for $m\neq 0$ , then we have$\frac{zg_{m}’(Z)}{g_{m}(z)}=\frac{mze^{mz}}{e^{mz}-1}=q_{m}(z)$and$1+, \frac{zg_{m}^{\prime/}(z)}{g_{m}(z)}=1+mz$.

Hence the statement thatis subordinate to $1+mz$ means thatis subordinate.
to2. By numerical experiments, we can see that the best possible value for $m$ in Proposition 1 is 4.813762. . . . The following figures indicate the images of the unit disk underthe mappings
and $1+mz$ for the values $m=\pi$ and $m=4.8$ respectively.
$zg_{m}’/g_{m}$$q_{m}$$1+zg_{m}^{\prime/}/g_{m}’$$q_{m}$FIGURE 1. $m=3.14$

FIGURE 2. $m=4.8$

REFERENCES
1. V. Anisiu and P. T. Mocanu, On a simple sufficient condition for starlikeness, Mathematica (Cluj) 31(54)’ 97-101.
2. P. T. Mocanu, On the radius of starlikeness of univalent functions (Romanian), Stud. Cerc. Mat.
(Cluj) 11 (1960), 337-341.
, On a Marx-Strohh\"acker differential subordination, Stud. Univ.
-Bolyai, Math. 36 (1991),
3.
$(1989)\backslash$$\mathrm{B}\mathrm{a}\mathrm{b}\mathrm{e}_{\mathrm{S}}$$\overline{77-84.}$

4. C.-F. Yuan, Univalence of analytic solutions to some first-order differential equations (Chinese), Acta
Math. Sinica 35 (1992), 483-491.
DEPARTMENT OF MATHEMATICS, KYOTO UNIVERSITY, 606-8502 KYOTO, JAPAN-mail address: sugawa@kusm.kyoto-u.ac.jp$E$