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Onsemiconjugationofentirefunctions
WALTERBERGWEILERandA.HINKKANEN
MathematicalProceedingsoftheCambridgePhilosophicalSociety/Volume126/Issue03/May1999,pp565574
DOI:null,Publishedonline:08September2000
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Math. Proc. Camb. Phil. Soc. (1999), 126, 565
Printed in the United Kingdom
c
1999 Cambridge Philosophical Society
565
On semiconjugation of entire functions
By WALTER BERGWEILER
Mathematisches Seminar, Christian-Albrechts-Universit
¨
at zu Kiel,
Ludewig-Meyn-Str. 4, D-24098 Kiel, Germany
e-mail: bergweiler@math.uni-kiel.de
and A. HINKKANEN†
Department of Mathematics, University of Illinois at Urbana-Champaign, Urbana,
Illinois 61801, USA
e-mail: aimo@math.uiuc.edu
(Received 8 December 1997; revised 3 April 1998)
Abstract
Let f and h be transcendental entire functions and let g be a continuous and open
map of the complex plane into itself with g ◦ f = h ◦ g. We show that if f satisfies
a certain condition, which holds, in particular, if f has no wandering domains, then
g
−1
(J(h)) = J(f). Here J(·) denotes the Julia set of a function. We conclude that if f
has no wandering domains, then h has no wandering domains. Further, we show that
for given transcendental entire functions f and h, there are only countably many
entire functions g such that g ◦ f = h ◦ g.
1. Introduction and results
The Fatou set F (f) of an entire or rational function f is the set where the family
{f
n
} of iterates of f is normal and the Julia set J(f ) is its complement. These sets
play a fundamental role in complex dynamics (see [5, 10, 21, 23] for an introduction
to this theory).
Now let f and h be entire functions and let g : C → C be a non-constant continuous
function such that
g ◦ f = h ◦ g. (1)
Then we say that f and h are semiconjugated (by g) and call g a semiconjugacy.
Theorem 1. If f and h are transcendental entire functions, if g is a non-constant
continuous function and if (1) holds, then
g(J(f)) ⊂ J(h). (2)
If, in addition, g(C) is an open set and, in particular, if g is an open mapping, then
C \ g(C) contains at most one point.
† A.H. was partially supported by the U.S. National Science Foundation grant DMS 94-00999.
566 W. Bergweiler and A. Hinkkanen
The special case when g is entire is important and Theorem 1 is easy to prove in
this case. Theorem 1 is also easy to prove if we assume that g is open or discrete.
Even in the case when g is entire, however, it is not clear whether we also have
g
−1
(J(h)) ⊂ J(f) (3)
and thus
g
−1
(J(h)) = J(f). (4)
If g is a homeomorphism of C onto itself satisfying (1), then g is called a conjugacy.
In this case (4) clearly holds. It is also known that (4) holds if g(z)=e
z
(see [8]).
Here it suffices to assume that f is a holomorphic self-map of C\{0}.In[9]itwas
shown that (4) holds if g is entire and if there exists an entire function k such that
f = k ◦g and h = g ◦ k. Note that if f and h have this special form, then (1) is always
satisfied.
In order to state a further condition implying (4) we recall that a component U
of F (f) is called a wandering domain if f
m
(U) w f
n
(U)=6 for m n. A famous
theorem of Sullivan [24] says that rational functions do not have wandering domains.
Transcendental entire functions, however, may have wandering domains, but certain
classes of functions with no wandering domains are known (see [6, sections 4·5, 4·6]
for further discussion and references).
A further concept that we need is the set A(f) where the iterates of a transcendental
entire function f tend to ∞ about as fast as possible. Given such an f , we shall see
in Section 4 that
lim
n→∞
log log M(R, f
n
)
n
= ∞
for all large R>0 and that
A(f)={z ∈ C : there exists L ∈ N such that |f
n
(z)| >M(R, f
n−L
)forn>L}
is not empty for such R. For a further discussion of the set A(f ) we refer to Section 4,
but note that if f does not have wandering domains then A(f) ⊂ J(f); see Lemma 4
below.
Theorem 2. Let f and h be transcendental entire functions and let g : C → C be
open and continuous such that g ◦ f = h ◦ g.IfA(f ) ⊂ J(f) then g
−1
(J(h)) = J(f).In
particular, this is the case if f does not have wandering domains.
Consider the special case when f = h and g is entire. We then have f ◦ g = g ◦ f and
say that f and g commute. Theorem 2 implies that g
−1
(J(f)) = J(f ) which means
that J(f) is completely invariant under g. Now J(g) is known to be the smallest
closed completely invariant set with at least three points (see, for example [5,p.67]
for the special case of rational functions). We deduce that J(g) ⊂ J(f ). We thus have
the following result.
Corollary. Let f and g be commuting transcendental entire functions. If f does not
have wandering domains or, more generally, if A(f) ⊂ J(f) then J(g) ⊂ J(f ).
The conclusion that J(g) ⊂ J(f)iff does not have wandering domains was ob-
tained by Langley [20] under an additional growth restriction on g.
Of course the corollary implies that if neither f nor g has wandering domains, then
On semiconjugation of entire functions 567
J(f)=J(g). It is conjectured that this remains valid in general, i.e. even if f and g
are allowed to have wandering domains. It is known to be true for rational functions
(see [14, pp. 364–365], [19, p. 143] or [3, section 4]).
There are other results for commuting entire functions which have a generalization
to the situation of a semiconjugacy. One such result is:
Theorem 3. Let f and h be entire functions such that f is either transcendental or a
polynomial of degree at least 2 and h is not the identity mapping. Then there are only
countably many entire functions g such that (1) holds.
Of course, if h is the identity mapping then (1) reads g ◦ f = g, which is satisfied
by all constant maps g but not by any non-constant map g unless f(z)=ωz + c for
some root of unity ω and some c ∈ C (see, for example [17]). If f has this form, then
there are uncountably many non-constant solutions g of the equation g ◦ f = g.
Theorem 3 generalizes the result of Baker in [1, theorem 1, p. 244] where it was
proved that if f is a given entire function, either transcendental or a polynomial of
degree at least two, then there are only countably many entire functions g commuting
with f .
Theorem 4. Let f and h be transcendental entire functions and let g : C → C be
open and continuous such that (1) holds. If f has no wandering domains then h has no
wandering domains.
Theorem 4 generalizes the result in [9] where the conclusion was obtained if g is
entire and if f = k ◦g and h = g ◦ k for some entire function k. This was used in [9]to
exhibit certain new classes of entire functions with no wandering domains. If f = k◦g
and h = g ◦ k as in [9], then, by symmetry, f has wandering domains if and only if
h has wandering domains. In the situation of Theorem 4, however, it is possible that
f has wandering domains while h does not. An example is f(z)=z + e
z
+1+2πi,
g(z)=e
z
and h(z)=ze
z+1
.
2. An example
We give an example which shows that the non-constant continuous function g
need not be open or discrete in order to satisfy (1), even if f = h so that f and g
commute. The same example shows that for a given transcendental entire f there can
sometimes be uncountably many non-constant, continuous, and non-entire functions
g commuting with f (then also (1) holds with f = h).
First choose a number R ∈ (0, 1) and then a continuous function a:[R
2
,R] → (0, 1)
such that a(R
2
)=a(R)
2
. Note that then a is non-constant. The condition a(r
2
)=a(r)
2
extends the definition of a uniquely to r ∈ (0, 1). Define also a(0) = 0 and a(r)=r for
all r ∈ [1, +∞). Then a :[0, +∞) → [0, +∞) is continuous and satisfies a(r
2
)=a(r)
2
for all r > 0.
Next define G : C → C by G(re
iθ
)=a(r)e
iθ
. Then G(z
2
)=G(z)
2
for all z ∈ C.
Clearly a can be chosen so that G is neither discrete nor open. Now (1) holds with
f(z)=h(z)=z
2
and g = G.
To get an example with transcendental functions f and h, suppose that f = h is
transcendental, that f has a superattracting fixed point z
0
whose immediate basin
of attraction A is bounded and bounded by a Jordan curve, and whose basin of
568 W. Bergweiler and A. Hinkkanen
attraction contains no singularity of f
−1
other than z
0
, such that f(z)− z
0
has a zero
of order 2 at z
0
. Let ϕ be a conformal map of A onto the unit disk D with ϕ(z
0
)=0,
and note that ϕ conjugates f to z
2
.
For z ∈ A define g(z)=ϕ
−1
(G(ϕ(z))). Then g(f (z)) = ϕ
−1
(G(ϕ(f(z)))) =
ϕ
−1
(G(ϕ(z)
2
)) = ϕ
−1
(G(ϕ(z))
2
)=f(ϕ
−1
(G(ϕ(z)))) = f(g(z)) for z ∈ A.
If B is a component of the basin of attraction of z
0
with f
n
(B)=A where n is
minimal, define g(z)=f
−n
(g(f
n
(z))) ∈ B for all z ∈ B, with the appropriate branch
of f
−n
.
Since G coincides with the identity mapping on ∂D and since ϕ extends contin-
uously and bijectively to ∂A, we see that g extends continuously to ∂A and ∂B by
setting g(z)=z there. Finally, we set g(z)=z when z is outside the closure of the
basin of attraction of z
0
. Then g is a continuous non-constant function commuting
with f , hence satisfying (1) with h = f.
Obviously g need not be discrete or open and there are uncountably many possible
choices for g.
We still need to find a transcendental entire function f with the required prop-
erties. The function f(z)=c(e
z
2
− 1) satisfies all the conditions provided that c is a
sufficiently large positive number. Note that for any non-zero c, z
0
= 0 is a super-
attracting fixed point of f and that f has a zero or order 2 at the origin. The set
of singularities of f
−1
is {0, −c}.Ifc is large enough, we have |f (z)| > 1 whenever
|z| = 1. This implies that the component W of g
−1
(D) that contains 0 satisfies W ⊂ D
and that the immediate basin of attraction A of 0 satisfies
A ⊂ W . Moreover, f is
a proper map of degree 2 from W onto D. Thus, in the terminology of [10, section
VI·1] or [12], the triple (f; W, D) is a polynomial-like mapping of degree 2. The basic
result about polynomial-like mappings ([10, theorem VI·1·1] or [12, theorem 1]) says
that there exist a quasiconformal mapping ψ and a polynomial p of degree 2 such
that f(z)=ψ(p(ψ
−1
(z))) for z ∈ W . Since 0 is a superattracting fixed point of f we
deduce that p has a superattracting fixed point and by suitably normalizing ψ we
may assume that p(z)=z
2
. This implies that A = ψ(D) and ∂A = ψ(∂D). Thus ∂A is
not only a Jordan curve but a quasicircle, and f is conjugate to z 7→ z
2
in A. Similar
reasoning can be found in the proof of theorem 2 in [7, p. 529].
If c is large enough, we also have f
n
(−c) →∞as n →∞. Thus z
0
= 0 is then
the only singularity of f
−1
in the basin of attraction of z
0
= 0. This completes the
construction of the example.
3. Proof of Theorem 1
Let the assumptions of Theorem 1 be satisfied. Recall that, by Baker’s result [2],
J(f) is the closure of the repelling periodic points of f. Suppose that b ∈ J(f). Then
b = lim
n→∞
z
n
where f
k
n
(z
n
)=z
n
, say, and all the z
n
are distinct repelling periodic
points of f, and k
n
tends to infinity. Furthermore, g(b) = lim
n→∞
g(z
n
) and by (1),
it is seen that h
k
n
(g(z
n
)) = g(z
n
) for all n. We shall show that g(z
n
) ∈ J(h) for all n.
It then follows that g(b) = lim
n→∞
g(z
n
) ∈ J(h) and thus (2) holds.
To show that g(z
n
) ∈ J(h), suppose that g(z
n
) ∈ N(h). Since h
k
n
(g(z
n
)) = g(z
n
), the
point g(z
n
) is then an attracting, superattracting or Siegel fixed point of h
k
n
. Now
g(C) is a connected set containing more than one point. Hence we may choose a small
neighbourhood V
n
of g(z
n
) so that the set A = g(C)\ x
∞
p=0
h
pk
n
(V
n
) contains at least
two points. This implies that B = g
−1
(A) has at least two points. We then choose