数理解析研究所講究録 1220 巻 2001 年 17-25

17Dynamics of structurally finite transcendental
entire functions
Masashi KISAKA(木坂 正史)
Graduate School of Humanand Environmental Studies,
Kyoto University, Kyoto 606-8501, Japane-mail:kisaka@math.h.kyoto-u.ac.jp

1Introduction
We investigate the dynamics of structurally finite transcendental entire functions, which was defined by Taniguchi ([2]). We will show that we can define
an itinerary for the points which remain in some region under iteration andthe set of all points which share the same itinerary forms acurve which goesto infinity. Also these curves belong to the Julia set and the points on thesecurves tend to infinity under iteration. This is ageneralization of the result
by Schleicher and Zimmer ([1]) for the exponential family.

2Structurally finite entire functions
In this section, we make abrief explanation of structurally finite entire
fucntions which was defined by Taniguchi ([2]).

Definition 1(Maskit surgery by connecting functions): $\mathbb{C}arrow \mathbb{C}(j=1,2)$ be two entire functions, andLet
be the set of
singular values of .
$f_{j}$$A_{j}$$f_{j}$

Assume that there is across cut$L$in$\mathbb{C}$such that

1. both ofandare either empty or consist of asingle point, which is an isolated point of each,$L\cap A_{1}$$L\cap A_{2}$$A_{j}$$z_{0}$

2.

$L$

separates$A_{1}\backslash \{z_{0}\}$from$A_{2}\backslash \{z_{0}\}$

, and

18
for
, thenis acritical value of each
3. if Ll”l,(U) hassuch that Un
asmall disk U with center
which contains acritical point forarelatively compact componenteach .
$A_{1}\ovalbox{\tt\small REJECT}$$L\mathrm{f}^{\ovalbox{\tt\small REJECT}}1A_{2}\ovalbox{\tt\small REJECT}$$\{\ovalbox{\tt\small REJECT} 0\}$$\{\ovalbox{\tt\small REJECT} \mathrm{z}_{0}\}$$f_{j}\ovalbox{\tt\small REJECT}$$A_{\ovalbox{\tt\small REJECT}}\ovalbox{\tt\small REJECT}$$z_{0}$$\{_{\ovalbox{\tt\small REJECT}}\ovalbox{\tt\small REJECT} \mathrm{s}_{0}\}$$f_{\ovalbox{\tt\small REJECT}}$$W_{\ovalbox{\tt\small REJECT}}$$f_{j}$:
is constructed from $fi$
Then we say that an entire function
and by Maskit surgery with respect to if the following assumptions are. Thencontaining
be the component ofsatisfied: Let
there exist$\mathbb{C}arrow \mathbb{C}$

$f$

$L$

$f_{2}$

1. components:
that
non-empty,$f_{j}$

and$\tilde{D}_{1}$$\tilde{D}_{j}arrow \mathrm{D}\-\mathrm{j}$2. across cut
and$\tilde{L}$in$\mathbb{C}$andof
is biholomorphic and$f_{1}^{-1}(D_{2})$$\tilde{D}_{2}$such that$f$$\phi_{j}$$\mathbb{C}\backslash \tilde{D}_{j}$$U_{j}$$U_{2}$$f_{2}^{-1}(D_{1})$, respectively, suchareif$\tilde{D}_{j}\cap W_{j}\neq \mathrm{G}5$

gives ahomeomorphism ofonto
3. aconformal map ofare components of
and
where
$U_{1}$$A_{j}\backslash \{z_{0}\}$$\mathbb{C}\backslash L$$D_{j}$

such that.

$f_{j}=f\circ\phi_{j}$$L\cap A_{j}$$\tilde{L}$ononto$L$,$\mathbb{C}\backslash \tilde{D}_{j}$,$\mathbb{C}\backslash \tilde{L}$Definition 2(structurally finite entire functions)
We say that an entire function is structurally finite if it can be constructedfrom afinite number of building blocks by Maskit surgeries. Here, abuilding
block is either aquadratic block$az^{2}+bz+c:\mathbb{C}arrow \mathbb{C}$$(a\neq 0)$

or an exponential block$a$ $\exp(bz)+c:\mathbb{C}arrow\mathbb{C}$$(ab\neq 0)$.

Then the following Representation Theorem holds:Theorem (Representation Theorem) Every structurally finite entire
function has the form$\int^{z}P(t)e^{Q(t)}dt$

with suitable polynomials$P$and$Q$.

19Note that if f is structurally finite, then f belongs to so called the Speiserclass, which is aclass of entire functions with only finite number of singularvalues.

3Statement of the result
Let $f(z)$ be astructurally finite transcendental entire function. Since we areinterested in the dynamics of , by the Representation Theorem and somesuitable linear conjugation, we can assume that $f(z)$ has the following form:$f$$f(z):=a \int_{0}^{z}P(t)e^{Q(t)}dt+b$,

where ,and $P$,
are monic polynomials with $\deg P=p\geq$, $\deg Q=q\geq 1$ . In what follows we consider only transcendental case, weassume here $q\geq 1$ . Then it is easy to see that has critical values andasymptotic directions which correspond to some finite asymptotic values.
In particular, has only finite number of singular values. So we take adisk$a$$b\in \mathbb{C}$

$Q$

$0$

$f$

$p$

$q$

$f$

$D:=\{z||z|<C.\}$

which contains all the singular values of . Thencomponents and each one by one lies in one of thedivided by the asymptotic paths$f$has exactly
domains which are$f^{-1}(\mathbb{C}\backslash D)$

$q$

$q$

$q$

$e^{\frac{(2k-1)\pi}{q}i}t$$(t\geq 0)$, $k=1,2$ ,$\cdots$, ,$q$which correspond to some finite asymptotic value. Let
paths, say$\Gamma(t):=e^{\frac{\pi}{q}i}t$$\Gamma$be one of these$(t\geq 0)$for example. Then each connected component ofis acurve whichtends toand its argument tends to one of $\frac{2k\pi}{q}(k=0,1, \cdots, q-1)$ , whichis the argument of asymptotic paths which corresponds to the asymptotic
value . Let be one of the components of. Then we make apartition of by using
so that$f^{-1}(\Gamma)$$\infty$$S$$\infty$$S$$f^{-1}(\mathbb{C}\backslash D)$$f^{-1}(\Gamma)$$S$$= \prod_{k\in \mathbb{Z}}R_{k}$.

20(See Figure 1). For apoint zE S such that $f^{n}(z)$ e S for any n CE N, wecan define its itinerary $S(z$ by$S(z):=\underline{s}=s_{0}s_{1}\cdots s_{n}\cdots$,

if$f^{n}(z)\in R_{s_{n}}$.

Main Theorem Let be astructurally finite transcendental entire funcsatisfying $|s_{n}|\leq F^{n}(x)$ , wheretion. For an itinerary$>0$ is some constant and$f$$\underline{s}=\mathrm{s}\mathrm{o}s1$$\ldots s_{n}\cdots$$x$$F(z):= \sum|a_{n}|z^{n}$,$f(z)= \sum a_{n}z^{n}$there exists acontinuous curve$h_{\underline{s}}(t)\subset S$$(t\geq t_{0})\exists$such thatfor fixed has the itinerary .
(1) All the points
for every .
(2)$t$$h_{\underline{s}}(t)$$f^{n}(h_{\underline{s}}(t))\in S$$n$$\underline{s}$

,

21
(3)

$f^{n}(h_{\underline{s}}(t))arrow\infty(narrow\infty)$(4)$\lim_{tarrow\infty}h_{\underline{s}}(t)=\infty$(5)$h_{\underline{s}}(t)$. In particular,$h_{\underline{s}}(t)\in J(f)$.

.

is injective with respect to .
$t$

4Preliminaries
Proposition 1If$z \in S_{0}:=\{z||\arg z-\frac{2k\pi}{q}|<\frac{\pi}{4q}$

and

$|z|$

(1)

,

$k=0$ ,

, $q-1\}$$\cdots$

is sufficiently large, then the following estimates hold:$f(z)= \frac{a}{q}z^{p-q+1}e^{z^{q}}(1+O(|z|^{-1}))$(2)$|f(z)| \underline{>}|\frac{a}{q}||z|^{p-q+1}\exp(\frac{|z|^{q}}{\sqrt{2}})$(3)$|f(z)| \leq 2|\frac{a}{q}|\exp(|z|^{q+\epsilon})$(4) Let$g_{s_{i}}$be the branch of.

.

for asmall$f^{-1}$$\epsilon>0$.

which takes values in$|g_{s}:(z)|\geq(\log|z|)^{\frac{1}{q}}-\epsilon$$R_{s}:$.

Define$h_{\underline{s}}^{n}(t):=g_{s_{1}}\mathrm{o}g_{s_{2}}\mathrm{o}\cdots \mathrm{o}g_{s_{n}}(\dot{F}^{n}(t))\in R_{s_{1}}$where$F(z):= \sum|a_{n}|z^{n}$,$f(z)= \sum a_{n}z^{n}$

Proposition 2$|h_{\underline{s}}^{n}(t)| \geq(\frac{1}{\sqrt{2}})^{\frac{1}{q}}t-\epsilon$

Proposition 3$|g_{s}’:(z)| \leq\frac{1}{|a|}|z|^{-\frac{1}{\sqrt{2}+\epsilon}}$

.

.

.

,. Then22
Now$h_{\underline{s}}^{n}(t)$can be written as follows:$h_{\underline{s}}^{n}(t)=h_{\underline{s}}^{1}(t)+ \sum_{k=1}^{n-1}(h_{\underline{s}}^{k+1}(t)-h_{\underline{s}}^{k}(t))$.

Then we have an estimate$|h_{\underline{s}}^{k+1}(t)-h_{\underline{s}}^{k}(t)| \leq\sup_{z\in L}|g_{s_{1}}’(z)||h_{\sigma(\underline{s})}^{k}(F(t))-h_{\sigma(\underline{s})}^{k-1}(F(t))|$where runs on the line segment between
Proposition 2, both points satisfy
$L$

$z$

$|*|$

Since all the components of$h_{\sigma(\underline{s})}^{k-1}(F(t))$:$( \frac{1}{\sqrt{2}})^{\frac{1}{q}}t_{1}-\epsilon$$f^{-1}(\Gamma)$.

in $S$ have the same asymptotics, we have$R_{s} \cap\{:|z|>M\}\subset\{z||\arg z|<\frac{\pi}{4q}\}$.

So we have$\min_{z\in L}|z|\geq((\frac{1}{\sqrt{2}})^{\frac{1}{q}}t_{1}-\epsilon)\cos\frac{\pi}{8q}$.

Prom the above estimate and Proposition 3, we have$\sup_{z\in L}|g_{s_{1}}’(z)|$$\leq\frac{1}{|a|}(((\frac{1}{\sqrt{2}})^{\frac{1}{q}}t_{1}-\epsilon)\cos\frac{\pi}{8q})^{-\frac{1}{\sqrt{2}+\epsilon}}$Repeating this procedure, we have$|h_{\underline{s}}^{n+1}(t)-h_{\underline{s}}^{n}(t)|$$\leq$$( \prod_{k=1}^{n}\frac{1}{|a|}(((\frac{1}{\sqrt{2}})^{\frac{1}{q}}t_{k}-\epsilon)\cos\frac{\pi}{8q})^{-\frac{1}{\sqrt{2}+\mathrm{g}}})$$\cross$

$=$

. By

$( \frac{1}{\sqrt{2}})^{\frac{1}{q}}F(t)-\epsilon$$\geq$$=$

and$h_{\sigma(\underline{s})}^{k}(F(t))$,$|h_{\sigma^{n}(\mathrm{g})}^{1}(F^{n}(t))-h_{\sigma^{n}(\mathfrak{g}}^{0}(F^{n}(t))|$$( \prod_{k=1}^{n}***)|g_{s_{n+1}}(F^{n+1}(t)))-F^{n}(t)|$.

23
In order to get an estimate for the term
the following propositions:

Proposition 4$\varphi_{ij}$:$R_{i}arrow R_{j}$

$z$

with

$|z|$

$F^{n}(\mathrm{t}\ovalbox{\tt\small REJECT}$, we$\mathrm{n}$is well-defined by the formula$f(\varphi_{ij}(z))=f(z)$for$|\mathrm{b}.+\mathrm{z}\mathrm{C}F^{n+1}(\mathrm{t})))$, $z\in R_{i}$large enough and satisfies$\varphi_{ij}(z)=z+\frac{2(j-i)\pi\sqrt{-1}}{q}\frac{1}{z^{q-1}}+O(\frac{1}{z^{2q-1}})$.

Proposition 5,$|g_{s}:(F(z))-z| \leq|\frac{2(s_{i}-l)\pi\sqrt{-1}}{q}+\epsilon|\frac{1}{|z|q-1}$

if $z\in R_{l}$ and $F(z)\in S$ .
By Propostion 5, we have the following estimate:$|h_{\underline{s}}^{n+1}(t)-h_{\underline{s}}^{n}(t)|$$\leq$$( \prod_{k=1}^{n}\frac{1}{|a|}(((\frac{1}{\sqrt{2}})^{\frac{1}{q}}t_{k}-\epsilon)\cos\frac{\pi}{8q})^{-\frac{1}{\sqrt{2}+\mathrm{g}}})\cross(\frac{2s_{n}\pi\sqrt{-1}}{q}+\epsilon)\frac{1}{t_{n}^{q-1}}$

where $t_{k}:=F^{k}(t)$ . Hence we have$g_{\underline{s}}^{n}(t)arrow\exists_{g_{\underline{s}}(t)}$locally uniformly for$t\geq\exists t_{0}(\geq x)$.

5Outline of proof of the Main Theorem
Now the proof of (1) and (2) are trivial.
(3) Since$f(h_{\underline{s}}^{n}(t))=h_{\sigma(\underline{s})}^{n-1}(F(t))$, we have$f(h_{\underline{s}}(t))=h_{\sigma(\underline{s})}(F(t))$

by taking alimit. In general we have.

$f^{n}(h_{\underline{s}}(t))=h_{\sigma^{n}(\underline{s})}(F^{n}(t))>( \frac{1}{\sqrt{2}})^{\frac{1}{q}}F^{n}(t)-\epsilon$

24
Hence we have,$f^{n}(h_{\underline{s}}(t))arrow\infty(narrow\infty)$.

Since it is well known that functions of finite type can have neither Bakerdomains nor wandering domains, this implies that$g_{\underline{s}}(t)\in J(f)$.

(4) Since$|h_{\underline{s}}^{n}(t)| \geq(\frac{1}{\sqrt{2}})^{\frac{1}{q}}t-\epsilon$

from Proposition 2, we have$|h_{\underline{s}}(t)| \geq(\frac{1}{\sqrt{2}})^{\frac{1}{q}}t-\epsilon$

Therefore$\lim_{tarrow\infty}h_{\underline{s}}(t)=\infty$(5) Suppose that$h_{\underline{s}}(t)$.

.

is not injective. Then$h_{\underline{s}}(t_{1})=h_{\underline{s}}(t_{2})$for some$t_{1}<t_{2}$. Hence we have$f^{n}(h_{\underline{s}}(t_{1}))=f^{n}(h_{\underline{s}}(t_{2}))$,that is,$h_{\sigma^{n}(\underline{s})}(F^{n}(t_{1}))=h_{\sigma^{n}(\underline{s})}(F^{n}(t_{2}))$.

On the other hand, it turns out that$|h_{\sigma^{n}(\mathrm{g})}(F^{n}(t_{k}))-F^{n}(t_{k})|<| \frac{2s_{n}\pi\sqrt{-1}}{q}+\epsilon|\frac{1}{|F^{n}(t_{k})|^{q-1}}$

for $k=1,2$ . Thenhence$g_{\sigma^{n}(\underline{s})}(F^{n}(t_{k}))-F^{n}(t_{k})$are bounded for $k=1,2$ and$(g_{\sigma^{n}(\underline{s})}(F^{n}(t_{1}))-F^{n}(t_{1}))-(g_{\sigma^{n}(\underline{s})}(F^{n}(t_{2}))-F^{n}(t_{2}))$$=F^{n}(t_{2})-F^{n}(t_{1})$is also bounded, which is impossible

25References[1] D. Schleicher and J. Zimmer, Dynamic Rays for Exponential Maps,
Preprint SUN Y Stony Brook, 1999/9.
[2] M. Taniguchi, Maskit surgery of entire functions, Preprin t.