数理解析研究所講究録988 巻 1997 年 120-131

120

ITERATION OF THE FUNCTION $c\exp[az+b/z]$
千葉大・理柳原 二郎 (NIRO YANAGIHARA)後藤 香代子 (KAYOKO GOTOH)
千葉大・自然科学

ABSTRACT. We will study iteration of the function $c\exp[az+b/z]$ . In
this note, we investigate orbits of the critical points, which are useful tofind conditions for real parametersand such that the Julia set of$c\exp[az+b/z]$ coincides with
.
$a,$

$b$

$c$

$\mathbb{C}^{*}$PART I
1. INTRODUCTION
There are many investigations on iteration of entire functions (analytic selfmappings on C), especially of exponential function. We shall study iteration of analytic functions on the punctured plane.
We denote by
the set of analytic self-mappings on . lt is known that$f\in E^{*}$ can be expressed as$ce^{az}$$\mathbb{C}^{*}=\mathbb{C}\backslash \{0\}$$E^{*}$$\mathbb{C}^{*}$$f(z)=z^{m}\exp[g(Z)+h(1/z)]$ ,where $m$ is an integer and , are entire functions. We call that belongs
to theclass, when both and are non-constant. The most simple
example of the function in this class is
$g$

$h$

$f$

$\mathrm{R}\circ \mathrm{a}\mathrm{d}\mathrm{s}\mathrm{t}\mathrm{r}\ddot{\mathrm{o}}\mathrm{m}$

$g$

$h$

$f(z)=c\exp[az+b/z],$$a,$ $b,$ $c\in \mathbb{R}\backslash \{0\}$,iteration of which will be investigated in this note.
We use the standard terminology in the iteration theory. We denote by
the n-th iteration of the function . Fatou set and Julia set of are denoted
by $F(f)$ and $J(f)$ , respctively. It is known that the set of repelling periodic
points of is dense in $J(f)[6]$ .
$f^{n}$

$f$

$f$

$f$

121
As seen below, the analytic map $w=c\exp[az+b/z],$
reduced to one of the following three canonical forms:
(1.1)
(1.2)
(1.3)

$a,$ $b,$ $c\in \mathbb{R}\backslash \{0\}$is$W=\lambda\exp[\mu(Z+1/Z)]$ ,if $ab>0$ ,$W=\lambda\exp[\mu(Z-1/Z)]$ , if $ab<0,$ $ac>0$ ,$W=\lambda\exp[-\mu(z-1/Z)]$ , if $ab<0,$ $ac<0$ ,where and are positive real numbers.
Case (1) : $ab>0$$\lambda$$\mu$(1-1) When $a>0,$ $b>0,$ $c>0$ . Put(1.1) with$\lambda=c\sqrt{\frac{a}{b}}>0,$$\mu=\sqrt{ab}>0$(1-2) When $a>0,$ $b>0,$ $c<0$ . Put

have (1.1) with$\lambda=-\frac{1}{c}\sqrt{\frac{b}{a}}>0,$$\lambda=\frac{1}{c}\sqrt{\frac{b}{a}}>0,$,, then we,$w= \sqrt{\frac{b}{a}}\cdot\frac{1}{W}$, then we$z=-\sqrt{\frac{b}{a}}Z,$$w=-\sqrt{\frac{b}{a}}W$, then we$\mu=\sqrt{ab}>0$.

$z= \sqrt{\frac{b}{a}}\cdot\frac{1}{Z}$$\mu=\sqrt{ab}>0$$\lambda=-c\sqrt{\frac{a}{b}}>0,$, then we have$w=- \sqrt{\frac{b}{a}}\cdot\frac{1}{W}$$z=- \sqrt{\frac{b}{a}}\cdot\frac{1}{Z}$(1-4) When $a<0,$ $b<0,$ $c<0$ . Put

have (1.1) with

$w=\sqrt{\frac{b}{a}}W$.

(1-3) When $a<0,$ $b<0,$ $c>0$ . Put

have (1.1) with

$z=\sqrt{\frac{b}{a}}Z,$.

$\mu=\sqrt{ab}>0$.

Case (2) : $ab<0,$ $ac>0$ .
(2-1) When $a>0,$ $b<0,$ $c>0$ . Put(1.2) with$\lambda=c\sqrt{\frac{a}{-b}}>0,$$z=\sqrt{\frac{-b}{a}}Z,$ $w=\sqrt{\frac{-b}{a}}W$$\mu=\sqrt{-ab}>0$, then we have.

(2-2) When $a<0,$ $b>0,$ $c<0$ . Put $z=-\sqrt{\frac{b}{-a}}Z,$ $w=-\sqrt{\frac{b}{-a}}W$ , then we

have (1.2) with$\lambda=-C\sqrt{\frac{-a}{b}}>0,$$\mu=\sqrt{-ab}>0$.

Case (3) : $ab<0,$ $ac<0$ .
(3-1) When $a<0,$ $b>0,$ $c>0$ . Put

have (1.3) with$\lambda=c\sqrt{\frac{-a}{b}}>0,$$z=\sqrt{\frac{b}{-a}}Z,$$\mu=\sqrt{-ab}>0$.

$w=\sqrt{\frac{b}{-a}}W$, then we122(3-2) When $a>0,$ $b<0,$ $c<0$ . Put $z=-\sqrt{\frac{-b}{a}}Z,$ $w=-\sqrt{\frac{-b}{a}}W$ , then we

have (1.3) withPut$\lambda=-c\sqrt{\frac{a}{-.b}}>0,$.

,$f_{2}(_{Z})=\lambda\exp[\mu(z-1/z)]$ ,

(1.4)
(1.5)
(1.6)

$f_{1}(z)=\lambda\exp[\mu(z+1/z)]$,$f_{3}(z)=\lambda\exp[-\mu(z-1/z)]$where$\pm i$$\mu=\sqrt{-ab}>0$$\lambda>0,$ $\mu>0$.

. Note that$f_{1}$

has critical points at$\pm 1,$$f_{2}$

We say that the function $f\in E^{*}$ belongs to the class,such that$\alpha_{1},$$\ldots$and$f_{3}$

have at

if there exist$S_{q}^{*}$$\alpha_{q}\in \mathbb{C}^{*}$$f$

is a covering map.
sing
Obviously all of:$\mathbb{C}^{*}\backslash f^{-}1(\{\alpha 1, \ldots, \alpha\}q)arrow \mathbb{C}^{*}\backslash \{\alpha_{1}, \ldots, \alpha_{q}\}$$\alpha_{1},$$\ldots,$$(f^{-1})=\{\alpha_{1}, \ldots, \alpha_{q}\}$$f_{1}(z),$and we write
are called finite singularities of. The union of all the classes
is denoted by .
andbelong to .
$f^{-1}$$\alpha_{q}$$f_{2}(z)$$S^{*}$$S_{q}^{*}$$S^{*}$$f_{3}(z)$Theorem 1. Let be a function in the classand $s\in$ sing $(f^{-1})$ . The
sequence $\{f^{n}(s)\}$ is said to satisfy the condition (S) provided that one of thefollowing conditions is satisfied:(S1) $|f^{n}(s)|arrow 0$ ,(S2)(S3) is a $preperiodic_{l}$ not a periodic point of .
Suppose that for every $s\in$ sing $(f^{-1})$ the sequence $\{f^{n}(s)\}$ satisfy thecondition (S). Then.
$f$

$S^{*},$$|f^{n}(s)|arrow\infty_{f}$

$f$

$s$

$J(f)=\mathbb{C}^{*}$In the case of entire functions, this theorem was given by Baker [1] andDevaney [4]. This theorem is a generalization of the results of them.
As seen by Theorem 1, it is important to observe the orbits of critical points.
We shall discribe the orbits of the critical points of$f_{1}(z)=\lambda\exp[\mu(z+1/z)]$,$\lambda>0$,$\mu>0$.

We write simply for .
Consider the funcion defind by$f$$f_{1}$$c_{1}$(1.7)$c_{1}=c_{1}( \mu)=\frac{1}{2\mu}(\sqrt{1+4\mu^{2}}-1)\exp[\sqrt{1+4\mu^{2}}-2\mu]$.

This funcion varies from to 1 as varies from to . Note that the equation$f(z)=z$ has at least one positive solution if and only if.
Indeed, $f(z)/z=\lambda\exp[\mu(z+1/z)]/z$ has the minimum value
at
$0$

$\mu$

$0$

$\infty$$\lambda e^{2\mu}\cdot c_{1}(\mu)\leqq 1$$\lambda e^{2\mu}\cdot c_{1}(\mu)$123$z=(1+\sqrt{1+4\mu^{2}})/2\mu>1$

positive solutions$A_{l},$. If$\lambda e^{2\mu}<1/c_{1}$, the equation $f(z)=z$ has two

such that$A_{u}$$A_{l}< \frac{1+\sqrt{1+4\mu^{2}}}{2\mu}<A_{u}$.

Furtherif$1<\lambda e^{2\mu}$,$A_{l}=1$

if$\lambda e^{2\mu}=1$,$A_{l}<1<A_{u}$if $0<\lambda e^{2\mu}<1$ ,$1<A_{l}$because has critical point at 1 andof 1 by iteration of . Let $a_{n}=f^{n}(1)$ , then$f(1)=\lambda e^{2\mu}$

$f$

$f$

(1.8)

. Firstly we observe the orbit$a_{1}=\lambda e^{2\mu}$and$a_{n}= \lambda\exp[\mu(a_{n-1}+\frac{1}{a_{n-1}})]=a_{1}\exp[\mu(a_{n-1}+\frac{1}{a_{n-1}}-2)]$Thus for any$n\in \mathrm{N}$

.

,

(1.9)

$a_{n}\geqq a_{1}$.

We consider $a_{2}=a_{1}\exp[\mu(a_{1}+1/a_{1}-2)]$ . Let(1.10)$K(x)=x \exp[\mu(x+\frac{1}{x}-2)]$ ,then If has a minimum at $c_{2}=(-1+\sqrt{1+4\mu^{2}})/2\mu<1$ , and decreases in$0<x<c_{2}$ , increases in $c_{2}<x$ . If
is the solution of the equation$c_{0}$(1.11)$c_{0}^{2} \exp[\mu(c_{0}+\frac{1}{c_{0}}-2)]=1$,$0<c_{0}<1$.

then $K(c\mathrm{o})=A_{u}$ and $c_{0}=1/A_{u}$ . So we shall prove existence of the solutionof the equation (1.11). Let$I \mathrm{f}_{0}(X)=x^{2}\exp[\mu(x+\frac{1}{x}-2)]$

Then

,

$0<x\leqq 1$.

has a minimum at $c=(-1+\sqrt{1+\mu^{2}})/\mu<1$ , and decreases in,$0<x<c$ , increases in $c<x\leqq 1$ . Since $\lim_{xarrow 0}IC_{0}(X)=\infty$ andthere exist only one $0<c_{0}<1$ such that
. We note that$I\mathrm{f}_{0}$$I\mathrm{f}_{0}(1)=1$$I\mathrm{f}_{0}(C\mathrm{o})=1$(1.12)$I\mathrm{f}_{0}(X)>1$,

that is$x \exp[\mu(x+\frac{1}{x}-2)]>\frac{1}{x}$if $0<x<c_{0}$ .

Theorem 2. Let$c_{1}$

and$c_{0}$

be as above. Then we have the following:

124, then $f^{n}(1)arrow\infty$ as $narrow\infty$ .
(1) If, then $\{f^{n}(1)\}$ is bounded from(2) If(3) If $c_{0}>\lambda e^{2\mu}>0$ , then $f^{n}(1)arrow\infty$ as $narrow\infty$ .
$\lambda e^{2\mu}>1/c_{1}$$1/c_{1}\geqq\lambda e^{2\mu}\geqq c_{0}$

$0$

and

Next we observe the orbit of-l. Let $b_{n}=f^{n}(-1)$ . Then$\infty$.

$b_{1}=\lambda e^{-2\mu}$$b_{n}= \lambda\exp[\mu(b_{n-1}+\frac{1}{b_{n-1}})]=b_{1}\exp[\mu(b_{n-1}+\frac{1}{b_{n-1}}+2)]$

Define two functions$L_{0}$

and

$F$

and

.

by

$L_{0}(x)=x \exp[\mu(x+\frac{1}{x}+2)]$,

and$F(x)= \frac{\sqrt{4x^{2}+1}-1}{2x}\exp[\sqrt{4x^{2}+1}+2x]$.

It is then easy to see that, $L_{0}(x)\geqq F(\mu)$ for any $x>0$ . Define two functions
and by$G$$L_{1}$$L_{1}(x)=x^{2} \exp[\mu(x+\frac{1}{x}+2)]$,

and$G(x)=( \frac{\sqrt{x^{2}+1}-1}{x})^{2}\exp[2\sqrt{x^{2}+1}+2x]$.

We easily get that $L_{1}(x)\geqq G(\mu)$ , for any $x>0$ . Note that $F(\mu),$ $G(\mu)$ increase
from toin $\mu>0$ .
Letbe positive numbers such that $F(\mu_{1})=1,$ $G(\mu_{2})=1$ , respectively.
Note that $0<\mu_{1}<\mu_{2}<1$ . Then we obtain the following.
$0$$\infty$$\mu_{1},$$\mu_{2}$Theorem 3. Let
be as in.
(1) Suppose that.
for any$0<\mu\leqq\mu_{2}$(2) Suppose that
. If
as
.
$c_{1},$$(\mathit{1}.7)_{f}(\mathit{1}.\mathit{1}\mathit{1})$$c_{0}$$\mu>\mu_{2}$$Then_{f}$$\lambda>0,$$\lambda e^{2\mu}>1/c_{1}$$\infty$$f^{n}(-1)arrow\infty$or$c_{1},$$c_{0}$

be as in (1.7),$(\mathit{1}.\mathit{1}\mathit{1})_{f}$and$c_{*}=c_{*}( \mu)=\frac{\sqrt{1+4\mu^{2}}-1}{2\mu}\exp[2\mu-\sqrt{1+4\mu^{2}}]$(1) If(2) If(3) If$\lambda e^{2\mu}>1/c_{1}$, then, then, then$c_{*}\leqq\lambda e^{2\mu}\leqq 1/c_{1}$$0<\lambda e^{2\mu}\leqq c_{0}$$J(f)=\mathbb{C}^{*}$.

$J(f)\neq \mathbb{C}^{*}$$J(f)=\mathbb{C}^{*}$$narrow\infty$.

$0<\lambda e^{2\mu}<c_{0_{\lambda}}f^{n}(-1)arrow$$narrow\infty$

Theorem 4. Let

as

.

125
2. PROOF or THEOREM 1
We need the lemma below.
Lemma. Suppose that $f\in S^{*}$ . Let
be the set of points of the form $f^{n}(s)$ ,. Then any constant limit of a$n=1,2,$in a component of $F(f)$ belongs to.
$E$$s\in \mathrm{s}\mathrm{i}\mathrm{n}\mathrm{g}(f^{-1}),$$\mathit{8}equence\{f^{n_{k}}(z)\}$$a$$\ldots$$L=\overline{E}\cup\{0\}\cup\{\infty\}$

Although this lemma was given by Baker [1, Theorem 2] for entire functions,his proof also works for analytic function :.
Suppose that $F(f)$ would be non-empty. By a result of Makienko [6, Lemma
4],
converges to a constantin a component of $F(f)$ . Since $F(f)$has no wandering component [6, Theorem 4], we may suppose that $f^{p}(D)\subset D$
for some . Thusalso converges to in , and hence $\alpha=f^{p}(a)$ .
The lennma above implies, that is $a=f^{m}(s)$ for someand$s\in$ sing $(f^{-1})$ .
Moreovercontains no finite singularities ofbyassumption. Thus
is attractive basin with super-attracting periodic point. Thus $(f^{p})’(a)=0$ hence $f’(\beta)=0$ for $\beta=f^{q}(a)$ with some $0\leqq q\leqq p-1$ .
Then $f(\beta)\in$ sing $(f^{-1})$ is periodic, which contradicts the assumption.
$f$$\mathbb{C}^{*}arrow \mathbb{C}^{*}$$D$$\alpha\in \mathbb{C}^{*}$$f^{n_{k}}$$f^{n_{k}+p}$

$p$

$a$

$D$

$\alpha\in L$$m\in \mathrm{N}$$f^{-1}$$D\backslash \{a\}$

$D$

$a$

$q,$

3. PROOF OF THEOREM 2
In the sequel, we frequently use the facts that
(3.1) the function$x+ \frac{1}{x}$increases on$x\geqq 1$, and decreases on $0<x<1$ ,

and
(3.2)

if and only if$x> \lambda\exp[\mu(x+\frac{1}{x})]$Case (1) :obtain thatsolution.
Case (2) :. By (3.1), $a_{n+1}>a_{n}$ for any.

, and hence we
as $narrow\infty$ , since the equation $f(z)=z$ has no positive$\lambda e^{2\mu}>1/c_{1}$$a_{n}arrow\infty$$A_{l}<x<A_{u}$$n\in \mathrm{N}$. By
and (3.1), $a_{2}<\lambda\exp[A\iota+A_{l}]=A_{l}$ ,Hence, by means of (3.2), it is shown inductively that$a_{1}<a_{n}<a_{n+1}$ .
$a_{1}<a_{n}<a_{n+1}<A_{l}$ , for any.
Note that $K(c_{2})=c_{1}<1$ and $K(1)=1$ . So we may take the value such
$1/c_{1}\geqq\lambda e^{2\mu}\geqq 1$$a_{1}\geqq 1$$n\in \mathrm{N}$$c_{3}$

that $0<c_{3}<c_{2}$ and,(3.3)

$IC(c_{3})=c_{3} \exp[\mu(c_{3}+\frac{1}{c_{3}}-2)]=1$

By $A_{u}>1$ and (3.1), we have$c_{0}<c_{3}$

.

,

126
Case (3) :so that,$1>\lambda e^{2\mu}>c_{3}$. We then have $c_{1}<K(a_{1})\leqq 1$ , and$a_{1}<a_{2}\leqq 1$,,$a_{3}= \lambda\exp[\mu(a_{2}+\frac{1}{a_{2}})]<\lambda\exp[\mu(a_{1}+\frac{1}{a_{1}})]=a_{2}$.

$a_{4}= \lambda\exp[\mu(a_{3}+\frac{1}{a_{3}})]>\lambda\exp[\mu(a_{2}+\frac{1}{a_{2}})]=a_{3}$and further,,$a_{3}-a_{1}= \lambda\exp[\mu(a_{2}+\frac{1}{a_{2}})]-\lambda e^{2\mu}\geqq\lambda e^{2\mu}-\lambda e^{2\mu}=0$$a_{4}-a_{2}= \lambda\{\exp[\mu(a_{3}+\frac{1}{a_{3}})]-\exp[\mu(a_{1}+\frac{1}{a_{1}})]\}\leqq 0$

Hence$a_{1}\leqq a_{3}\leqq a_{4}\leqq a_{2}\leqq 1$.

, and we obtain$a_{2m-1}\leqq a_{2m+1}\leqq a_{2m+2}\leqq a_{2m}\leqq 1$,

by induction.
Case (4) :. Then $IC’(a1)<0,$ $K(a_{1})>1$ . Since If $(c_{3})=1$$A_{u}\geqq
a_{2}>1>a_{1}$$K(c_{0})=A_{u},$and. Suppose $a_{1}<a_{n}\leqq A_{u}$ , then we have$c_{3}>\lambda e^{2\mu}\geqq c_{0}$$a_{n+1}= \lambda\exp[\mu(a_{n}+\frac{1}{a_{n}})]\leqq\lambda\exp[\mu(A_{u}+\frac{1}{A_{u}})]=A_{u}$,$a_{n+1}= \lambda\exp[\mu(a_{n}+\frac{1}{a_{n}})]\leqq\lambda\exp[\mu(a_{1}+\frac{1}{a_{1}}.)]=a_{2}\leqq A_{u}$if$1\leqq a_{n}$,if $1>a_{n}$ ,,$a_{1}<a_{n}\leqq A_{u}$ .
by (3.1). Hence for every
Case (5) : $c_{0}>\lambda e^{2\mu}>0$ . From $I\mathrm{f}’(a1)<0$ and If $(a_{1})>1$ , we have$IC(a_{1})>K(c_{0})=A_{u}$ . Thus we obtain from (3.1) and (3.2),$n\in \mathrm{N},$$a_{3}= \lambda\exp[\mu(a_{2}+\frac{1}{a_{2}})]>a_{2}$,.

$a_{4}= \lambda\exp[\mu(a_{3}+\frac{1}{a_{3}})]>\lambda\exp[\mu(a_{2}+\frac{1}{a_{2}})]=a_{3}$and $a_{n+1}>a_{n}>A_{u}$ , by induction. Hence$a_{n}arrow\infty$.

127
4. PROOF OF THEOREM 3

Case (1) : $\mu>\mu_{2}$ . $L_{0}$ andhave the minimum values $F(\mu)$ and$G(\mu)$ at $x=(\sqrt{1+4\mu^{2}}-1)/2\mu$ and $x=(\sqrt{1+\mu^{2}}-1)/\mu$ , respectively. Henceby definition ofand , for any $x>0$ ,$L_{1}(x)$$\mu_{1}$$\mu_{2}$$L_{0}(x)=x \exp[\mu(x+\frac{1}{x}+2)]\geqq F(\mu)>1$

,

(4.1)
$L_{1}(x)=x^{2} \exp[\mu(x+\frac{1}{x}+2)]\geqq G(\mu)>1$

First assume that$b_{1}=\lambda e^{-2\mu}\geqq 1$.

. Then$b_{2}= \lambda\exp[\mu(b_{1}+\frac{1}{b_{1}})]\geqq\lambda e^{2\mu}=a_{1}>b_{1}\geqq 1$,

and it is shown by induction that$b_{n+1}= \lambda\exp[\mu(b_{n}+\frac{1}{b_{n}})]>\lambda\exp[\mu(b_{n-1}+\frac{1}{b_{n-1}})]=b_{n}$$b_{n+1}= \lambda\exp[\mu(b_{n}+\frac{1}{b_{n}})]\geqq\lambda\exp[\mu(a_{n-1}+\frac{1}{a_{n-1}})]=a_{n}$,,for any $n\geqq 2$ . Note that $\lambda e^{2\mu}>1/c_{1}\geqq e^{4\mu}>e^{4\mu}/F(\mu)=1/c_{1}$ , by assumption. Thus Theorem 2 (1), implies.
Next assume that $b_{1}=\lambda e^{-2\mu}<1$ . Then by (4.1)$b_{n}\geqq a_{n-1}arrow\infty$(4.2)$b_{2}= \lambda\exp[\mu(b_{1}+\frac{1}{b_{1}})]>\frac{1}{b_{1}}>1$.

and hence by (3.1)$b_{3}= \lambda\exp[\mu(b_{2}+\frac{1}{b_{2}})]>\lambda\exp[\mu(\frac{1}{b_{1}}+b_{1})]=b_{2}$.

Further we have $b_{n+1}>b_{n}$ , for every $n\geqq 2$ by induction. Note that $1/b_{1}>A_{u}$because of (4.2) and (3.2). Therefore we obtain.
. Suppose that
Case (2) :. Then$b_{n}arrow\infty$$\mu_{2}\geqq\mu>\mu_{1}$$\lambda e^{2\mu}>1/c_{1}$$b_{n}= \lambda\exp[\mu(b_{n-1}+\frac{1}{b_{n-1}})]\geqq\lambda e^{2\mu}>\frac{1}{c_{1}}>1$

If$\lambda e^{-2\mu}\geqq 1$,, in exactly the same way as Case (1), we have$b_{n+1}>a_{n}$,$n\geqq 2$.

128
for any $n\geqq 2$ . By Theorem 2 (1),$b_{n}arrow\infty$. lf$\lambda e^{-2\mu}<1$, by (3.1),$b_{2}= \lambda\exp[\mu(b_{1}+\frac{1}{b_{1}})]>\lambda e^{2\mu}=a_{1}>\frac{1}{c_{1}}>1$

.

Hence

.

$b_{3}= \lambda\exp[\mu(b_{2}+\frac{1}{b_{2}})]>\lambda\exp[\mu(a_{1}+\frac{1}{a_{1}})]=a_{2}$By induction, $b_{n}>a_{n-1}$ for any $n\geqq 2$ . By Theorem 2 (1), we obtain
Suppose that
. Then by (1.12)$b_{n}arrow\infty$.

$1< \lambda^{2}\exp[\mu(\lambda e^{2\mu}+\frac{1}{\lambda e^{2\mu}}+2)]<\lambda e^{2\mu}\cdot\lambda\exp[\mu(\lambda e^{-2\mu}+\frac{1}{\lambda e^{-2\mu}})]=a_{1}\cdot b_{2}$.

$\lambda e^{2\mu}<c_{0}$$a_{2}= \lambda e^{2\mu}\exp[\mu(\lambda e^{2\mu}+\frac{1}{\lambda e^{2\mu}}-2)]>\frac{1}{\lambda e^{2\mu}}$By$\lambda e^{-2\mu}<\lambda e^{2\mu}<c_{0}<1$.

and (3.1),

Thus we have$a_{2}= \lambda\exp[\mu]<\lambda\exp[\mu(b_{2}+\frac{1}{b_{2}})]=b_{3}$and hence $a_{n}<b_{n+1}$ , for any $n\geqq 2$ , by induction. Therefore, as above.
Cas $e(3)$ : $\mu_{1}\geqq\mu>0$ . lf
, then,$b_{n}arrow\infty$as$narrow\infty$$\lambda e^{2\mu}>1/c_{1}$$\lambda e^{-2\mu}>\frac{2\mu}{\sqrt{4\mu^{2}+1}-1}\exp[-\sqrt{4\mu^{2}+1}-2\mu]=\frac{1}{F(\mu)}\geqq\frac{1}{F(\mu_{1})}=1$

$\ln$

.

exactly the same way as Case (1), we obtain.
If, we can prove that, in the same way as case (2).
$b_{n}arrow\infty$$\lambda e^{2\mu}<c_{0}$$b_{n}arrow\infty$

5. PROOF OF THEOREM 4
lf
or $0<\lambda e^{2\mu}<c_{0}$ , Theorem 2, Theorem 3 and Theorem
1 imply the assertion. So we consider the case.
this case, bydefinition of , it is shown that $a_{1}=c_{0},$ $a_{2}=1/c_{0}=A_{u}$ , thus we have $a_{n}--a_{2}$for $n\geqq 3,$ $b_{1}<a_{1}=c_{0}$ , and $b_{2}>A_{l}$ . Therefore
is a repelling fixed point,since $f’(1/c_{0})=\mu(A_{u}-1/A_{u})>1$ . Moreover
as $narrow\infty$ by Theorem
3(2). Hence both $\{f^{n}(1)\}$ and $\{f^{n}(-1)\}$ satisfy condition (S), and we haveby Theorem 1.
Next we shall prove (2). Let$\lambda e^{2\mu}>1/c_{1}$$\lambda e^{2\mu}=c_{0}$$c_{0}$$a_{2}$$b_{n}arrow\infty$$J(f)=\mathbb{C}^{*}$$e_{1}=e_{1}( \mu)=\frac{\sqrt{1+4\mu}+1}{2\mu}$

,

$\ln$

129

$e_{2}=e_{2}( \mu)=\frac{\sqrt{1+4\mu}-1}{2\mu}$.

Case (1) :. Then the equation $f(z)=z$ has only one positive
solution. Choose a positive real value 1 $<z_{0}<e_{1}$ nearand put$\rho=e_{1}-z_{0}$ .
Because of conformality of the disk $\Delta=\{|z-z_{0}|<\rho\}$is mapped by in a disk $\triangle’=\{|z-z_{0}’|<\rho’\}$ , where $z_{0}<z_{0}’<e_{1}$ and$e_{1}-z_{0}’=\rho’<\rho$ . Henceis normal in .
Case (2) :. Then $1\leqq A_{l}<e_{1}$ and $f’(A_{l})\in[0,1)$ . Thusis an attracting fixed point, and.
Case (3) :. Then $1>A_{l}>e_{2}$ and $f’(A_{l})\in(-1,0)$ . Thusis an attracting fixed point, and.
Case (4) :. Then $f^{f}(A_{l})=-1$ and so $(f^{2})’(Al)=1$ . Choose apositive real value $z_{0}<A_{l}$ nearand put $\rho=A_{l}-z_{0}$ . Since $0<(f^{2})’<1$
on, we can see thatis normal in the disk $\triangle=\{|z-z0|<\rho\}$ inthe same way as Case (1).
$\lambda e^{2\mu}=1/c_{1}$$e_{1}(\mu)$$e_{1}$$f$$f$$\{f^{n}\}$$\triangle$$1/c_{1}>\lambda e^{2\mu}\geqq 1$$A_{l}$$J(f)\neq \mathbb{C}^{*}$$1>\lambda e^{2\mu}>c_{*}$$A_{l}$$J(f)\neq \mathbb{C}^{*}$$\lambda e^{2\mu}=c_{*}$$A_{l}$$(z_{0}, A_{l})$$\{f^{2n}\}$PART II
The present Part II has been inspired by comments of attendants in the assembly 9.24-9.27, 1996, at the RIMS.

6. THE SITUATION WHEN$\lambda$

IS FIXEDIn Part I, we have investigated behaviors of the function
for varying
with fixed. In the present Part , we suppose that is fixed and willconsider the problem for varying . We write for simplicity for.
We consider only the case$\mu>0$ . Other cases can be treated similarly.
Put$f_{\lambda\mu}(z)$$\lambda$

$\Pi$

$\mu$

$\lambda$

$f$

$\mu$

$f_{\lambda\mu}$$\lambda>0,$(6.1)$M_{1}( \mu)=\frac{1}{2\mu}(\sqrt{1+4\mu^{2}}-1)\exp[\sqrt{1+4\mu^{2}}]$increases from to
be a number such that$\mathrm{O}\mathrm{b}_{\mathrm{V}}\mathrm{i}_{0}\mathrm{u}\mathrm{s}\mathrm{l}\mathrm{y}M_{1}(\mu)$$0$$\infty$

as

$\mu$

varies from to$M_{1}(\mu^{*})=1/\lambda$$0$.

$\infty$. Let$\mu^{*}=\mu^{*}(\lambda)$.

By Theorems 2 and 3 in Part I, we know that
(6.2)$\mu>\mu^{*}(\lambda)$

imples $a_{n}=f^{n}(1)arrow\infty,$ $b_{n}=f^{n}(-1)arrow\infty$ , as $narrow\infty$ .

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be the fixpoint of $f(x)$ , then $f’(X1)=\mu(.x_{1}-1/x_{1})$ . lf $f’(X1)=-1$ ,

Let
then

$x_{1}$.
$x_{1}=.$$\underline{\sqrt{1+4\mu^{2}}-1}$.

$2\mu$The value of$\lambda$with this fixpoint(6.3)$x_{1}$

must be$\lambda=\frac{1}{2\mu}(\sqrt{1+4\mu^{2}}-1)\exp[-\sqrt{1+4\mu^{2}}]$.

attains the. Then
We write the right hand side of (6.3) as,
maximum $\ell*=(\sqrt{2}-1)e-\sqrt{2}=$ 0.100702249. . . at $\mu=1/2$ . Thus if$f’(X_{1})=-1$, that is, $-1<f’(x)<1$ if
withthen there is no fixpoint$0<\mu<\mu^{*}(\lambda)$. Therefore we obtain the following theorem:$f(x)=x$ , provided$M_{2}(\mu)$$M_{2}(\mu)$$\lambda>l^{*}$$x_{1}$

Theorem 5.
, then.
(1) If(2) Suppose
0.100702249. . . , where $\ell*has$ been defined above.
) we have.
Then ifwill be defined $below_{f}$(3) If0.0322903204227226... , wherethen the case (3) in Theorem 4 does not occur. That is, there is no.
such that $J(f)=\mathbb{C}^{*}for$$\mu>\mu^{*}(\lambda)$$J(f)=\mathbb{C}^{*}$$\lambda\geqq\ell*=$$0<\mu\leqq\mu^{\langle}\lambda$$J(f)\neq \mathbb{C}^{*}$$\lambda>\lambda^{*}=$$\lambda^{*}$$c_{0}$$0<\lambda e^{2\mu}<c_{0}$We have only to prove the case (3) in Theorem 5. Put$M_{3}(x, \mu)=(xe^{2\mu})^{2}\exp[\mu(xe2\mu+1/xe^{2\mu}-2)]=IC_{0}(xe)2\mu$,is the function in (1.12) of Part 1. Let $x(\mu)$ be the implicit function
wheredetermined by$0<x(\mu)e^{2\mu}<1$ .
$M_{3}(x(\mu), \mu)=1$ ,$IC_{0}(X)$Then$\frac{dx}{d\mu}=-x\frac{(2\mu+1)_{X}e^{2}-\mu(2\mu-1)/(_{Xe)}2\mu+2}{\mu(_{Xe^{2\mu}}-1/(Xe\mu)2)+2}$.

Thus $x(\mu)$ takes the local maximum value $x=e^{-2\mu}(2\mu-1)/(2\mu+1)$ wheresatisfies$M_{3}(e^{-2\mu_{\frac{2\mu-1}{2\mu+1’}}} \mu)=(\frac{2\mu-1}{2\mu+1})^{2}\exp[\frac{4\mu}{4\mu^{2}-1}]=1$

$\mu$

,

from which we get $\mu=0.62783439300776\ldots$ . Then $x(\mu)=e^{-2\mu}(2\mu-1)/(2\mu+$$1)=$ 0.0322903204227226. . . , which number we denote as. Therefore if, then
can not take the valuein (1.11), for any . Hence, if, then $f^{n}(1)$ and $f^{n}(-1)$ are bounded from and.
if$\lambda^{*}$$\lambda>\lambda^{*}$$\lambda>\lambda^{*}$$\lambda e^{2\mu}$$c_{0}(\mu)$

$\mu$

$0$

$\infty$$\mu\leqq\mu^{*}(\lambda)$

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418-420.
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$\mathrm{R}^{\mathrm{o}}\mathrm{a}\mathrm{d}_{\mathrm{S}}\mathrm{t}\mathrm{r}\ddot{\mathrm{o}}\mathrm{m}$, On the iteration of analytic functions, Math. Scand. 1 (1953), 85-92.

NIRO YANAGIHARA
Department of Mathematics&Informatics
Faculty of Science
Chiba University-mail: yanagi@math.s.chiba-u.ac.jp$e$

KAYOKO GOTOH
Department of Mathematics&Informatics
Graduate School of Science and Technology
Chiba University-mail: kayoko@math.s.chiba-u.ac.jp$e$