Annals of Mathematics Decay of geometry for
unimodal maps:
An elementary proof

By Weixiao Shen

Annals of Mathematics, 163 (2006), 383–404
Decay of geometry for unimodal maps:
An elementary proof
By Weixiao Shen
Abstract
We prove that a nonrenormalizable smooth unimodal interval map with
critical order between 1 and 2 displays decay of geometry, by an elementary
and purely “real” argument. This completes a “real” approach to Milnor’s
attractor problem for smooth unimodal maps with critical order not greater
than 2.
1. Introduction
The dynamical properties of unimodal interval maps have been extensively
studied recently. A major breakthrough is a complete solution of Milnor’s
attractor problem for smooth unimodal maps with quadratic critical points.
Let f be a unimodal map. Following [19], let us define a (minimal)
measure-theoretical attractor to be an invariant compact set A such that
{x : ω(x) ⊂ A} has positive Lebesgue measure, but no invariant compact
proper subset of A has this property. Similarly, we define a topological attrac-
tor by replacing “has positive Lebesgue measure” with “is a residual set”. By
a wild attractor we mean a measure-theoretical attractor which fails to be a
topological one. In [19], Milnor asked if wild attractors can exist.
For smooth unimodal maps with nonflat critical points, this problem was

Let us state the result more precisely. By a unimodal map, we mean
a C
1
map f :[−1, 1] → [−1, 1] with a unique critical point 0, such that
f(−1) = f(1) = −1. We shall assume that f is C
3
except at 0, and there
are C
3
local diffeomorphisms φ, ψ such that f(x)=ψ(|φ(x)|

) for x close to 0,
where >1 is a constant, called the critical order. We shall refer to such a map
as a C
3
unimodal map with critical order . Recall that f is renormalizable
if there exist an interval I which contains the critical point 0 in its interior,
and a positive integer s>1, such that the intervals I,f(I), ··· ,f
s−1
(I) have
pairwise disjoint interiors, f
s
(I) ⊂ I, and f
s
(∂I) ⊂ ∂I.
Main Theorem. Let f :[−1, 1] → [−1, 1] be a nonrenormalizable C
3
unimodal map with critical order  ∈ (1, 2]. Assume that f has a nonperiodic
recurrent critical point. Then f displays decay of geometry.
Corollary 1.1. A C

DECAY GEOMETRY
385
Definition. We say that f displays decay of geometry if there are constants
C>0 and λ>1 such that
|I
m(k)
|
|I
m(k)+1
|
≥ Cλ
k
.
According to [8], [12], for any 1 <≤ 2, there is a constant  = () > 0,
such that f displays decay of geometry if
lim inf
n
|I
n+1
|
|I
n
|
≤ .
The last inequality is called the starting condition.
Prior to this work, real methods were known to work for some special ex-
amples. The so-called “essentially unbounded” combinatorics admits a rather
simple argument ([8], [12]). The more difficult cases, namely the Fibonacci
combinatorics and the so-called “rotation-like” combinatorics, are also resolved
in [10] and [5] respectively. Those arguments are again complicated and seem

=
|f(b)|−|f(f
s
n+1
(0))|
|f(b)|−|f(f
s
n
(0))|
,B
n
=

|f
s
n
(0)|
|f
s
n+1
(0)|

/2
,
where b is an endpoint of I
m(k+1)−1
. It is not difficult to show that the Main
Theorem follows from the following:
Main Lemma. There exists a universal constant σ>0 such that for all
n sufficiently large,

|T is a diffeomorphism,
386 WEIXIAO SHEN
and if f
n
(T ) is contained in a small neighborhood of the critical point, then
for any interval J  T , C(f
n
(T ),f
n
(J))/C(T,J) is bounded from below by a
constant close to 1. (See §2.4.) In particular, for any x ∈ T , this gives us a
lower bound on |(f
n
)

(x)| in terms of the length of the intervals T −{x} and
their images under f
n
.
We shall choose an appropriate neighborhood T
n
of f
s
n
(0), such that
f
s
n+1
−s
n

n
) not to exceed I
m(k)−1
.
Our proof can be modified to deal with a nonrenormalizable C
3
uni-
modal map with critical order 2 + , with >0 sufficiently small. In gen-
eral, the decay of geometry property does not hold, but we can show that
lim inf |I
m(k)
|/|I
m(k)+1
| is bounded from below by a universal constant C(),
and C() →∞as  → 0. The argument in [12] is still valid to show that such
a map does not have a wild attractor as well. It is also possible to weaken
the smoothness condition to be C
2
. These (minor) issues will not be discussed
further in this paper.
In Section 2, we shall give the necessary definitions and recall some known
facts which will be used in our argument. These facts include Martens’ real
bounds ([16]) and Kozlovski’s result on cross-ratio distortion ([11]). We shall
deduce the Main Theorem from the Main Lemma. In Section 3, we shall
define the intervals T
n
and investigate the location of the boundary points
of T
n
and f

2. Preliminaries
2.1. Pull back, nice intervals. Given an open interval I ⊂ [−1, 1], and an
orbit x, f(x), ···, f
n
(x) with f
n
(x) ∈ I,bypulling back I along {f
i
(x)}
n
i=0
,we
get a sequence of intervals I
i
 f
i
(x) such that I
n
= I, and I
i
is a component
of f
−1
(I
i+1
) for each 0 ≤ i ≤ n − 1. The interval I
0
is produced by this pull
back procedure, and will be denoted by I(n; x). The pull back is monotone if
none of these intervals I

→ I defined by
x → f
k(x)
(x). The first return map to I is the restriction of R
I
on D
I
∩ I.
For any given x ∈D
I
, the pull back of I along the orbit x, f(x), ,f
k(x)
(x)
is either unimodal or monotone, according to whether I(k(x); x)  0 or not.
This follows from the basic property of a nice interval that any two intervals
obtained by pulling back this interval are either disjoint, or nested, i.e., one
contains the other.
2.2. The principal nest. Let q denote the orientation-reversing fixed point
of f. Let I
0
=(−q,q), and for all n ≥ 1, let I
n
be the return domain to I
n−1
which contains the critical point. All these intervals I
n
are nice. The sequence
I
0
⊃ I

, if |g
m(k)
(z)|≤|z|,
then z ∈ I
m(k)
− I
m(k)+1
.
Proof.Ifz ∈ I
m(k)+i
−I
m(k)+i+1
for some 1 ≤ i ≤ m(k+1)−m(k)−1, then
g
m(k)
(z) ∈ I
m(k)+i−1
− I
m(k)+i
, and hence |g
m(k)
(z)| > |z|, which contradicts
the hypothesis of this lemma.
388 WEIXIAO SHEN
Lemma 2.2. Let J ⊂ I
m(k)−1
−I
m(k)
be a return domain to I
m(k)−1

by Lemma 2.1. For any 0 ≤ i ≤ p − 1, let 0 ≤ j
i
≤ m(k) − m(k − 1) − 1be
such that g
i
m(k−1)
(J) ⊂ I
m(k−1)+j
i
−I
m(k−1)+j
i
+1
and let P
i
be the component
of I
m(k−1)+j
i
−I
m(k−1)+j
i
+1
which contains g
i
m(k−1)
(J). Then it is easy to see
that for any 0 ≤ i ≤ p − 2, g
m(k−1)
maps a neighborhood of g

diffeomorphism.
Proof. Let s

be the return time of 0 to I
m(k−1)
. We pull back the nice
interval I
m(k−1)
along {f
i
(0)}
s
i=s

and denote by P  f
s

(0) the interval pro-
duced. By the previous lemma, this pull back is monotone and P is contained
in I
m(k)−1
. The pull back of P along {f
i
(0)}
s

i=0
is certainly unimodal, and the
interval produced is contained in D
I

,
DECAY GEOMETRY
389
where L, R are the components of T − J.Ifh : T → R is a homeomorphism
onto its image, we write
C(h; T,J)=
C(h(T ),h(J))
C(T,J)
.
A diffeomorphism with negative Schwarzian always expands the cross-ratio. In
general, a smooth map does not expand the cross-ratio, but in small scales,
cross-ratios are still “almost expanded” by the dynamics of f.
Lemma 2.5 (Theorem C, [11]). For each k sufficiently large, there is a
positive number O
k
, with O
k
→ 1 as k →∞and with the following property.
Let T ⊂ [−1, 1] be an interval and let n be a positive integer. Assume that
f
n
|T is monotone and f
n
(T ) ⊂ I
m(k−1)
. Then for any interval J  T ,
C(f
n
; T,J) ≥O
k

where T
+
,T
−
are the components of T −{z}. To see this, we just apply the
lemma to J

=(z −, z + ), and let  go to 0.
The estimate on cross-ratio distortion enables us to apply the following
lemma, called the real Koebe principle. This lemma is well-known, and a proof
can be found, for example, in [18].
Lemma 2.6. Let τ>0 and 0 <C≤ 1 be constants. Let I be an interval,
and let h : I → h(I)=(−τ, 1+τ ) be a diffeomorphism. Assume that for
any intervals J  T ⊂ I, there exists C(h; T,J) ≥ C. Then for any x, y ∈
h
−1
([0, 1]),
h

(x)
h

(y)
≤
1
C
6
(1 + τ)
2
τ

≥|(f
s
n
)

(c
1
)|B
n−1
/A
n−1
≥ Cλ
n
,
where we use the fact A
n−1
> 1.
390 WEIXIAO SHEN
For any k ≥ 0, consider the first return of the critical point to I
m(k)
, which
is a closest return, denoted by f
s
n
k
(0). Obviously, n
k
≥ k, and thus
|(f
s

By Corollary 2.3, there is an interval J  f(0) with f
−1
(J) ⊂ I
m(k)
and such that f
s
n
k
−1
: J → I
m(k−1)
is a diffeomorphism. By Lemma 2.4,
f
s
n
k
(I
m(k)+1
) ⊂ I
m(k)
is well inside I
m(k−1)
, and so by Lemmas 2.5 and 2.6,
the map f
s
n
k
−1
|f(I
m(k)+1

k
)

(c
1
)|≤K
|I
m(k)
|
|f(I
m(k+1)
)|
|c
s
n
k
|
−1
.
Since c
s
n
k
∈ I
m(k)
, this implies
|(f
s
n
k

n
k
−1
≤

|I
m(k−1)
|
|I
m(k+1)
|

/2
.(2.3)
These inequalities (2.2) and (2.3) imply the claim.
Let us consider again the map f
s
n
k
−1
|J as above. Applying Lemma 2.5,
we have
C(J, f (I
m(k)+1
))
−1
≥O
k
C(I
m(k−1)

|I
m(k)
|

1/
.
This inequality, together with the claim above, implies that |I
m(k)
|/|I
m(k)+1
|
grows exponentially fast. The proof of the Main Theorem is completed.
2.6. Two elementary lemmas. We shall need the following two elementary
lemmas to deal with the case <2.
DECAY GEOMETRY
391
Lemma 2.7. For any α ∈ (0, 1), the function
 → φ(α, )=α
1−

2

1
α
t
−1
dt
is a monotone increasing function on (0, ∞).
Proof. Direct computation shows:
∂φ

−log
√
α
0
t(e
t
− e
−t
)dt
> 0.
Lemma 2.8. For any 1 >a>b, and any 1 ≤  ≤ 2,
1 −b

1 −a

≥
1 −b
2
1 −a
2
.
Proof. By a continuity argument, it suffices to prove the lemma when  is
rational. Let  = m/n, with m, n ∈ N, and let x = b
1/n
, y = a
1/n
. Then
1 >
x
y

2n−1
.
Multiplying by (1 −x)/(1 −y) on both sides, we obtain the desired inequality.
3. The closest critical returns
Let s
1
<s
2
< ··· be all the closest return times. Let n
0
be such that
s
n
0
is the return time of 0 to I
m(1)
. For any n ≥ n
0
, let k = k(n) be so that
c
s
n
∈ I
m(k)
− I
m(k+1)
. Note that we have c
s
n
∈ I

n
(T
n
) ⊂ I
m(k)−1
.
392 WEIXIAO SHEN
We shall use the cross-ratio estimate to get a lower bound for
|(f
s
n+1
−s
n
−1
)

(f(c
s
n
))|.
To do this, it will be necessary to know the location of the boundary points of
T
n
and their images under f
s
n+1
−s
n
.
Note. Let u

n
), so organized that |x
n
|≤|y
n
|.
Lemma 3.1. T
n
⊂ I
m(k+1)−1
.
Proof. Arguing by contradiction, assume T
n
⊂ I
m(k+1)−1
. Then there
exists z ∈ T
n
∩ ∂I
m(k+1)−1
. Clearly, g
i
m(k)
(z) ∈ ∂I
m(k+1)−i−1
for all 0 ≤ i ≤
m(k +1)− m(k) −1. In particular,
w = g
m(k+1)−m(k)−1
m(k)

n
) ∈
(c
s
n
, −c
s
n
), we have ν ≥ m(k +1)−m(k). So the forward orbit of w intersects
I
m(k)−1
, i.e., w ∈ D
I
m(k)−1
. But this is absurd since I
m(k)
is a return domain
to the nice interval I
m(k)−1
.
Definition. We say that s
n
is of type I if f
s
n+1
−s
n
(T
n
) ⊃ (c

n
,y
n
)  0.
Moreover, if s
n
is of type II, then
• p ≥ 2 and g
m(k)
(I
m(k)+1
)  0,
• c
s
n
is the first return of 0 to I
m(k)
,
• If q ∈ N is minimal such that g
q
m(k)−1
(0) ∈ I
m(k)
, then there exist 1 ≤
q

<q,1≤ p

≤ p − 1 such that x
n

n−1
for all n ≥ 2. In this case, c
s
n
is the first return of the critical point to I
n−3
, and
c
s
n+1
= g
n−2
(c)=g
2
n−3
(c)=g
n−3
(c
s
n
),
for all n ≥ 3. Thus for all n ≥ 4, T
n
is the component of I
n−3
−{c} which
contains c
s
n
, and f

m(k)−1
along the orbit {f
i
(z)}
r
i=0
, and denote by J
i
 f
i
(z) the
intervals obtained. Then this pull back is unimodal or monotone according to
J = J
0
 0 or not. We say that z is good if 0 ∈ J, and bad otherwise.
Lemma 3.3. Let z, r, J, and J
1
be as above. Then the following hold.
1) If z is good, then J ∩I
m(k)+1
= ∅;
2) If z is bad, then f
r
is monotone on (0,z), and f
i
(0) ∈ I
m(k)
for all
1 ≤ i ≤ r.
Moreover, in either case, there is a closest return c

m(k)−1
, which
is necessarily not in I
m(k)
, to verify the last statement of the lemma.
Assume now that z is bad. As f
r
is monotone on each component of
J −{0} and (0,z) ⊂ J, f
r
|(0,z) is monotone. As we noted above, f
i
(J
1
)is
disjoint from I
m(k)
for all 0 ≤ i ≤ r − 2, which proves that f
i
(0) ∈ I
m(k)
for
all 1 ≤ i ≤ r − 1. The statement f
r
(0) ∈ I
m(k)
is obvious. We proved 2). To
verify the last statement of the lemma in this case, we just take c
s
to be the

m(k)
. As above, let J  z denote the interval
obtained by pulling back I
m(k)−1
along {f
i
(z)}
r
i=0
, and let J

be the component
of J −{0} which contains z. Then J

⊂ I
m(k)
, f
r
|J

is monotone, and f
r
(J

) ⊃
(c
s
n−1
, −c
s

i
m(k)
|γ
i
is well-defined and monotone,
• g
i
m(k)
(γ
i
) ⊂ I
m(k)
−{0}.
Moreover, let 0 ≤ j
i
≤ p−1 be such that g
i
m(k)
(c
s
n
) ∈ I
m(k)+j
i
− I
m(k)+j
i
+1
.
Note that T

(T
n
)=f
r
(J

) ⊃ (c
s
n−1
, −c
s
n−1
),
and s
n
is of type I. In particular, this is the case if g
ν−1
m(k)
(γ
ν−1
) is a component
of I
m(k)
−{0}.
If p =1org
m(k)
(I
m(k)+1
)  0, then for each 0 ≤ i ≤ ν −2, g
m(k)

Now we assume that p ≥ 2 and g
m(k)
(I
m(k)+1
)  0. We claim that for
each 1 ≤ i ≤ ν − 1, there is 0 ≤ p
i
≤ p − 1 with g
p
i
m(k)
(0) ∈ I
m(k)+j
i
+1
, such
that g
i
m(k)
(γ
i
) is the component of I
m(k)
−{g
p
i
m(k)
(0)} which does not contain
the critical point.
Let us prove this claim by induction on i.Fori = 1, the claim is true with

I
m(k)+1
; so the claim is true with p
i+1
= p
i
+1.
If j
i
= 0, then g
i
m(k)
(γ
i
) contains a component I
m(k)
− I
m(k)+1
, and thus it
contains the return domain to I
m(k)
which contains g
i
m(k)
(c
s
n
). Therefore,
g
i+1

m(k)
.To
prove the last term of the lemma, we first notice that p

:= p
ν−1
= 0 and that
z = g
ν−1
m(k)
(c
s
n
) is bad, for otherwise, g
ν−1
m(k)
(γ
ν−1
) ⊃ J

which implies that s
n
is
of type I by our previous remark. Let q

∈ N be such that g
q

m(k)−1
= f


. In particular, q

<q. Since c
s
n−1
is exactly
DECAY GEOMETRY
395
the point in {g
i
m(k)−1
(0), 1 ≤ i ≤ q − 1} which is closest to the critical point
and since the intervals g
i
m(k)−1
(0,g
p

m(k)
(0)), 1 ≤ i ≤ q, are pairwise disjoint, we
have c
s
n−1
= g
q

m(k)−1
(0).
Remark 3.2. In the case that p = 1, we see from the above proof that

s
n
∈ I
m(k)
−I
m(k+1)
, and let b
n
be an
endpoint of I
m(k+1)−1
. Recall that
A
n
=
|b
n
|

−|c
s
n+1
|

|b
n
|

−|c
s

n
))|≥(1 + σ)
A
n
A
n−1
B
n−1
B
n
.
The proof is organized as follows. First of all, by means of cross-ratio, we
prove
|(f
s
n+1
−s
n
)

(f(c
s
n
))|
A
n−1
B
n
A
n

s
n
|)
=1+
|y
n
|−|x
n
|
|y
n
| + |x
n
|
|x
n
|−|c
s
n
|
|x
n
| + |c
s
n
|
,
and
W
n

constant greater than 1. In this case, we prove that |x
n
|/|c
s
n
| is strictly bigger
than 1, and then the desired estimate follows from easy observations.
396 WEIXIAO SHEN
Case 2. s
n
is of type I, and |I
m(k)−1
|/|I
m(k)
| is close to 1. In this case, by
Martens’ real bounds, we have m(k −1) −m(k) > 1 and that g
m(k−1)
displays
a high return, i.e., g
m(k−1)
(I
m(k−1)+1
)  0. According to the relative position
of c
s
n
with the orientation-preserving fixed point of g
m(k−1)
|I
m(k−1)+1

we can prove the Main Lemma under the assumption that f does not satisfy
the starting condition. From this point of view, the second case above is not
necessary. We include an argument for this case as well so that we can prove
the decay of geometry property without reference to the starting condition.
4.1. Proof of (4.1). Applying Lemma 2.5 to the map f
s
n+1
−s
n
−1
: f(T
n
) →
(x
n
,y
n
), we obtain
C(f
s
n+1
−s
n
−1
; f(T
n
), {f(c
s
n
)}) ≥O

n
|
|f(L
n
)| + |f(R
n
)|
|f(L
n
)||f(R
n
)|
.(4.2)
By Lemma 3.1, T
n
is contained in a component of I
m(k+1)−1
−{0}. So for all
n sufficiently large, we have
|f(L
n
)|≤|c
s
n
|

, |f(R
n
)|≤|b
n

n
|

)|c
s
n
|

≥ A
n
|b
n
|

(|b
n
|

−|c
s
n+1
|

)|c
s
n
|

.(4.3)
Since |x

s
n
|

.(4.4)
Since |y
n
|≥|x
n
|,wehave
|y
n
− c
s
n+1
||x
n
− c
s
n+1
|≥(|y
n
| + |c
s
n+1
|)(|x
n
|−|c
s
n+1


(f(c
s
n
))||c
s
n+1
|
−1
≥O
k
(|y
n
| + |c
s
n+1
|)(|x
n
|−|c
s
n+1
|)
|y
n
| + |x
n
|
A
n
|x

s
n+1
|
|c
s
n
|
2

/2
(|y
n
| + |c
s
n+1
|)(|x
n
|−|c
s
n+1
|)
(|y
n
| + |x
n
|)|x
n
|
1
φ(|c

|−|c
s
n+1
|)
(|y
n
| + |x
n
|)|x
n
|
1
φ(|c
s
n+1
/x
n
|, 2)
= O
k
A
n

|x
n
||c
s
n+1
|
|c

n
||c
s
n+1
|
|c
s
n
|
2

/2
2|x
n
|(|y
n
| + |c
s
n
|)
(|y
n
| + |x
n
|)(|x
n
| + |c
s
n
|)

1
> 1.
Lemma 4.1. There exists a constant ρ
2
> 1, such that |x
n
|/|c
s
n
| >ρ
2
.
Proof. First notice that
|y
n
|
|c
s
n
|
≥
|I
m(k)−1
|
|I
m(k)
|
≥ ρ
1
,

| so
that x
n
∈ I
m(k)−1
.
As before, let ν ∈ N be such that c
s
n+1
= g
ν
m(k)
(c
s
n
), and let z = g
ν−1
m(k)
(c
s
n
).
Then |z|≥|c
s
n
|.Ifm(k +1)>m(k) + 1, then c
s
n
∈ I
m(k)+1

: J
1
→ I
m(k)−1
is a diffeomorphism. From
the fact x
n
∈ I
m(k)−1
, by Remark 3.2, it follows that z is bad (so J
1
 c
1
), and
398 WEIXIAO SHEN
that x
n
= f
r−1
(c
1
). Since f
r−1
(f(0,z)) = (x
n
,c
s
n+1
) is well inside I
m(k)−1

| or |x
n
|/|c
s
n
| is close
to 1. Lemma 4.1 shows that we are in the former case. If W
n
is also close
to 1, then so is |y
n
|/|c
s
n−1
|.As|I
m(k)−1
|/|I
m(k)
|≥ρ
1
, it follows that c
s
n−1
∈
I
m(k)−1
− I
m(k)
and A
n−1

− I
m(k)+1
.
Lemma 4.2. |g

m(k−1)
(ζ)| > 1 is uniformly bounded from above, and uni-
formly bounded away from 1.
Proof. Let s be the return time of 0 to I
m(k)−1
. Let M be the component
of I
m(k−1)+1
−{0} containing ζ. By Lemma 2.5, C(f
s
,M,{ζ}) ≥O
k
. Observe
that for each component J of M −{ζ}, |g
m(k−1)
(J)|/|J| is uniformly bounded
below from 1. It follows that |g

m(k−1)
(ζ)| is bounded below from 1. Since the
pull back of I
m(k−1)
along the orbit {f
i
(0)}

≤ C

|I
m(k)−1
|
|I
m(k)
)|


,
and hence the derivative is uniformly bounded.
Case 2.1. c
s
n
∈(ζ,−ζ). Since c
s
n
∈I
m(k+1)−1
−I
m(k+1)
,wehavem(k +1)
= m(k) + 1. We claim that there is an interval T

n
with I
m(k)
⊃ T


m(k)−1
(c
s
n
)=c
s
n+1
. Since g
m(k)−1
pushes points in
I
m(k)
−(−ζ,ζ) farther away from 0, g
ν

−1
m(k)−1
(c
s
n
) is contained in a component
P of I
m(k)−1
−I
m(k)
. Note that g
ν

−1
m(k)−1

m(k−1)
has uniformly bounded
distortion on [ζ,b], and thus by Lemma 4.2, |g

m(k−1)
| is uniformly bounded
from above on [ζ,b]. Consequently, (|b|−|ζ|)/(|y
n
|−|b|) is bounded from
zero. Since A
n−1
is close to 1 and |c
s
n
| < |ζ|, it follows that c
s
n−1
∈ I
m(k)
and (|c
s
n−1
|−|c
s
n
|)/(|b|−|c
s
n−1
|) is very small. Since we are also assuming
that |x

s
n+1
|. Note that |c
s
n−1
|≥|ξ|, and hence |c
s
n−1
|−|c
s
n
| is not much
smaller than |c
s
n
|−|c
s
n+1
|. Therefore, A
n
and B
n
/B
n−1
are both close to 1 as
well. Moreover, |(f
s
n+1
−s
n

and g
m(k)
displays a low return, g
m(k)
(I
m(k)+1
)  0, the following hold:
•|c
s
n
| < |w
n
| < |x
n
| < |y
n
|;
• the points g
i
m(k)
(0), 1 ≤ i ≤ p, lie on the same side of the critical point.
We no longer have W
n
≥ 1, but instead, Lemma 3.2 provides more com-
binatorial information to apply. We claim

|x
n
|
|c

n
|
.(4.6)
To see this, let J be the entry domain to I
m(k)
which contains c
s
n−1
=
g
q

m(k)−1
(0). Then g
q−q

m(k)−1
|J : J → I
m(k)
is a diffeomorphism and
I
m(k)−1
− I
m(k)
⊃ J ⊃ g
q

m(k)−1
(I
m(k)+1

n
|, we have by Lemma 2.5 that
|x
n
|−|w
n
|
|c
s
n−1
|−|x
n
|
≥ C(J, (x
n
,c
s
n−1
))
−1
≥O
k
C(g
q−q

m(k)−1
(J),g
q−q

m(k)−1

n
|−|w
n
|
2|w
n
|
,
which implies (4.6) by rearranging.
Note. Set
U
n
= A
n−1
W
n
=

x
n
c
s
n−1

/2
|y
n
|

−|c

|
.
Then, (4.1) becomes
|(f
s
n+1
−s
n
)

(f(c
s
n
))|
A
n−1
A
n
B
n
B
n−1
≥O
k
U
n
V
n
,
where V

|

−|c
s
n−1
|

≥
|y
n
|
2
−|c
s
n
|
2
|y
n
|
2
−|c
s
n−1
|
2
≥
|y
n
|

2
−|c
s
n−1
|
2
|y
n
|−|w
n
| +2|w
n
|
2
/|c
s
n−1
|
|y
n
| + |w
n
|
=
(λµ −1)(λµ
2
− µ +2)
µ
3
(λ

1, and A
n−1
=(|y
n
|

−|c
s
n
|

)/(|y
n
|

−|c
s
n−1
|

) is very big. The Main Lemma
follows.
Let us assume that V
n
> 1 is uniformly bounded away from 1. To show
that U
n
V
n
is bounded from below by a constant greater than 1, we may assume

DECAY GEOMETRY
401
Moreover,
1
1 −U
n
≥
Q
Q −P
≥
µ
3
(λ
2
− 1)
(µ −1)(2λµ −µ
2
− µ − 2)
.(4.8)
Let θ =2λµ − (µ
2
+ µ + 2). Then λ =(µ
2
+ µ +2+θ)/2µ. Substituting
this equality to (4.8), we obtain
1
1 −U
n
≥
θ

n
is the first return of 0 to I
m(k)
. So there
is an interval J  c
1
such that f
s
n
−1
: J → I
m(k)−1
is a diffeomorphism, and
f
−1
(J) ⊂ I
m(k)
.Asf
s
n
((0,c
s
n
)) ⊂ (c
s
n
,w
n
), and x
n

n
)))
−1
≥ 1+C(I
m(k)−1
,f
s
n
((0,c
s
n
)))
−1
O
k
≥ 1+C(I
m(k)−1
, (c
s
n
,w
n
))
−1
O
k
=1+
|y
n
|−|w

|
|w
n
|−|c
s
n
|
O
k
≥ 1+
λµ −1
2
|x
n
|
|x
n
|−|c
s
n
|
O
k
≥ 1+
λµ −1
2
α
α −1
O
k

(µ −1)
5
+7(µ − 1)
4
+ 21(µ − 1)
3
+ 37(µ − 1)
2
+ 43(µ − 1)+21
4(µ −1)
≥
37(µ −1)
2
+ 43(µ − 1) + 21
4(µ −1)
≥
2
√
37 ×21 + 43
4
≥ 24,
which implies U
n
≥ 23/24.
Since λµ =(µ
2
+ µ +2+θ)/2 ≥ 2, we have α
2
≥ 1+O
k

α −1
α +1
≥
0.9
2.9
·
0.5
2.5
>
1
20
,
and consequently,
U
n
V
n
>
23
24
·
21
20
> 1.
Case 3.2. θ>1. By (4.9),
1
1 −U
n
≥
θ

5+4
√
2)µ +2}
≥
µ(µ
2
+4.2µ +2)
2(µ −1)
≥
7.2(µ −1)
2
+13.4(µ − 1)+7.2
2(µ −1)
≥ 13.
Thus, U
n
≥ 12/13. Moreover,
λ =
µ
2
+ µ +2+θ
2µ
≥
µ
2
+ µ +3
2µ
≥
√
3+

V
n
is bounded from below by a constant greater than 1.
University of Science and Technology of China, Hefei, P. R. China
E-mail address:
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(Received September 25, 2002)
(Revised August 10, 2004)