Discrete Holomorphic Local Dynamical Systems

300 Dierk Schleicher
of f are discrete and thus f is constant. If c �∈ {0,1}, then differentiation yields
( f ′ ◦ f ) f ′ −1 = cf ′ − c or f ′ ( f ′ ◦ f − c)=1 − c.Sincec �= 1, it follows that f ′ omits
the value 0 and f ′ ◦ f omits the value c �= 0, so f ′ can assume the value c only at
the omitted values of f . By Picard’s Theorem A.4, it follows that f ′ is constant and
thus f is a polynomial of degree at most 1.
In our case, f is not a polynomial of degree 0 or 1 by hypothesis, so g is a
non-constant meromorphic function. Suppose p ∈ C is such that g(p) ∈{0,1,∞}.
If g(p)=∞, thenf (p) =p; ifg(p)=0, then f ( f (p)) = p; andifg(p)=1, then
f ( f (p)) = f (p).
If g is transcendental, then by Picard’s theorem there are infinitely many p ∈ C
with g(p) ∈{0,1,∞}. Ifg is a non-constant rational map (which in fact never
happens), then there are p0, p1, p∞ ∈ C with g(pi) =i, and at least two of them
are in C. ⊓⊔
As an example, the map f (z) =ez + z has no fixed points; in this case g(z) =
eez + 1hasnop∈Cwith g(p) ∈{1,∞}, but of course infinitely many p with
g(p)=0 (corresponding to periodic points of period 2).
Proof of Theorem 1.4. Let f be an entire function other than a polynomial of degree
0or1.ByLemma1.5, f has (at least) two periodic points of period 1 or 2; replacing
f by f ◦ f if necessary, we may suppose that f has two fixed points, say at p, p ′ ∈ C
(note that J( f )=J( f ◦ f )). If | f ′ (p)| > 1, then p ∈ J( f ). In order to show that
J( f ) �= /0, we may assume that p ∈ F( f ) and in particular that | f ′ (p)|≤1. Let W be
the Fatou component containing p.
If | f ′ (p)| = 1, then any limit function of the family of iterates { f ◦n |W } is nonconstant.
So let f ◦n j be a subsequence that converges to a non-constant limit function;
then f ◦(n j+1−n j) converges to the identity on W, and this implies that f |W is
injective. If W = C, then this is a contradiction to the choice of f , hence F( f ) �= C.
Thefinalcasewehavetoconsideris| f ′ (p)| < 1; for convenience, we may suppose
that p = 0. If F( f )=C, then f ◦n → 0 uniformly on compacts in C. We will
show that f is a polynomial of degree at most 1. Let D be an open disk centered at p
such that f (D) ⊂ D. ForN ∈ N, letDn := f ◦(−n) (D); sinceD is simply connected,
it follows that also Dn is simply connected. Let rn be maximal so that the round cir-
cle ∂Drn (0) ⊂ Dn;thisimpliesthatDrn ⊂ Dn. WehaveDn ⊂ Dn+1 and �
n Dn = C,
hence rn+1 ≥ rn and rn → ∞.
As usual, for r > 0, define M(r; f ) := max{| f (z)|: |z| = r}. For each n ∈ N,
define maps hn : D → C via hn(z) =M(rn/2; f ) −1 f (rnz). Allhn satisfy hn(0)=0
and M(1/2;hn)=1. Let
cn := sup{r > 0: ∂Dr(0) ⊂ hn(D)}.
If some cn satisfies cn < 1, then there are points an,bn ∈ C with |an| = 1, |bn| = 2so
that hn(D) ∩{an,bn} = /0. If this happens for infinitely many n, then we can extract
a subsequence with cn < 1, and it follows easily from Montel’s theorem that the hn
form a normal family. After extracting another subsequence, we may suppose that
the hn converge to a holomorphic limit function h: D → C that inherits from the hn
the properties that h(0)=0andM(1/2;h)=1. Thus h is non-constant and there is