155 (1998)
For almost every tent map, the turning point is typical
by
Henk B r u i n (Stockholm)
Abstract. Let T
abe the tent map with slope a. Let c be its turning point, and µ
athe absolutely continuous invariant probability measure. For an arbitrary, bounded, almost everywhere continuous function g, it is shown that for almost every a, T
g dµ
a= lim
n→∞ 1n
P
n−1i=0
g(T
ai(c)). As a corollary, we deduce that the critical point of a quadratic map is generically not typical for its absolutely continuous invariant probability measure, if it exists.
1. Introduction. Let T
a: I → I be the tent map with slope a. Brucks and Misiurewicz [BM] showed that for a.e. a ∈ [ √
2, 2], the orbit of the turning point is dense in the dynamical core. It is well known that for a > 1, the tent map T
ahas an absolutely continuous invariant probability measure (acip), µ
a, and that µ
ais ergodic. By Birkhoff’s Ergodic Theorem,
(1) \
g dµ
a= lim
n→∞
1 n
n−1
X
i=0
g(T
ai(x)) µ
a-a.e.
Here we take g ∈ G = {h : I → R | h is bounded and continuous a.e.}.
Because µ
ais absolutely continuous with respect to Lebesgue measure, (1) holds Lebesgue a.e. If (1) holds for a point x, then x is called typical with respect to g. Although most points are typical, it is very difficult to identify a typical point. It is natural to ask if the turning point c of T
ais typical.
We will prove
Theorem 1 (Main Theorem). Let g ∈ G. Then
(2) \
g dµ
a= lim
n→∞
1 n
n−1
X
i=0
g(T
ai(c)) for a.e. a ∈ [1, 2].
1991 Mathematics Subject Classification: 58F11, 58F03, 28D20.
Supported by the Deutsche Forschungsgemeinschaft (DFG). The research was carried out at the University of Erlangen-N¨ urnberg, Germany.
[215]
It follows that for a.e. a ∈ [1, 2], (2) holds for every bounded Riemann integrable function simultaneously. This answers a question of Brucks and Misiurewicz [BM]. Schmeling [Sc] recently obtained similar results for β- transformations. In our proof, as well as in [BM], the properties of the turn- ing point are used in a few arguments. We think, however, that Theorem 1 is true not only for c, but also for an arbitrary point y ∈ I.
The tent map T
ahas topological entropy log a. Hence one can state Theorem 1 as: For a.e. value of the topological entropy, the turning point of T
ais typical. Because the measure µ
aactually maximizes metric entropy [M], this has a striking consequence for unimodal maps in general:
Corollary 1. For a.e. h ∈ [0, log 2], if f is a unimodal map with h
top(f ) = h, then the turning point of f is typical for the measure of maximal entropy.
A result by Sands [Sa] states that for a.e. h ∈ [0, log 2], every S-unimodal map f with h
top(f ) = h satisfies the Collet–Eckmann condition, and there- fore has an acip. For an S-unimodal map, however, the acip in general does not maximize entropy, because if it did, and if f is conjugate to a tent map, the conjugacy ψ would be absolutely continuous. But then ψ has to be also C
1+αin a large neighbourhood of the critical point, as [MS, Ex- ercise 3.1, page 375] indicates. (In [M] an argument is given for unimodal maps with a nonrecurrent critical point.) As a consequence, all periodic points have to have the same Lyapunov exponent, which is very unlikely.
The only exception we are aware of is the full quadratic map x 7→ 4x(1 − x).
Hence combining Corollary 1 with Sands’ result, we obtain a large class of S-unimodal maps satisfying the Collet–Eckmann condition, but for which c is not typical for the acip. In contrast, Benedicks and Carleson [BC, Theo- rem 3] show that for the quadratic family f
a(x) = ax(1 − x) there is a set of parameters of positive Lebesgue measure for which f
ais Collet–Eckmann and c is typical for the acip (
1). Thus we are led to the conclusion that the entropy map a 7→ h
top(f
a), even when we disregard its flat pieces, has very bad absolute continuity properties.
The proof of the Main Theorem goes in short as follows. First we intro- duce some induced map of the tent map. We show that if a point is typical in some strong sense for this induced map, it is also typical for the original tent map (Proposition 1). In Sections 4 and 5 we prove certain properties of the induced map. Finally, we show, using a version of the Law of Large
(
1) Thunberg [T] showed another kind of typicality: for a positive measured set of
parameters, f
ahas an acip which can be approximated weakly by Dirac measures on
super-stable orbits of nearby maps.
Numbers (Lemma 8), that the turning point is indeed typical in this strong sense for a.e. parameter value (Sections 7 to 9).
Acknowledgments. I want to thank Karen Brucks for many discussions and Gerhard Keller for his help with Lemma 6. I am also grateful to the referee for the attentive comments.
2. Preliminaries. The tent map T
a: I = [0, 1] → I is defined as T
a(x) = min(ax, a(1 − x)). For a ≤ 1, the dynamics is uninteresting, and for a ∈ (1, √
2], T
ais finitely renormalizable. By considering the last renormalization instead of T
a, we reduce to the case a ∈ ( √
2, 2]. Let us only deal with a ∈ ( √
2, 2].
The point c = 1/2 is the turning point. We write c
n= c
n(a) = T
an(c).
Another notation is ϕ
n(a) = T
an(c). The core [c
2(a), c
1(a)] will be denoted as J(a).
For a ∈ [ √
2, 2], T
ahas an absolutely continuous invariant measure µ
a(acip for short). Its precise form can be found in [DGP], although we will not use that paper here. µ
a|
J(a)is equivalent to Lebesgue measure.
In the Main Theorem we considered g ∈ G. Using a well-known fact from measure theory (e.g. [P, p. 40]), it suffices to prove the following: Let B be the algebra of subsets of I whose boundaries have zero Lebesgue measure (or equivalently, µ
a-measure), and let B ∈ B. Then for a.e. a ∈ ( √
2, 2], µ
a(B) = lim
n→∞
1
n #{0 ≤ i < n | T
ai(c) ∈ B}.
It is this statement that we are going to prove.
The induced map that we will use is closely related to the Hofbauer tower (Markov extension) of the tent map. This object was introduced by Hofbauer (e.g. [H]). It is the disjoint union of the intervals {D
n}
n≥2, where D
2= [c
2, c
1] and for n ≥ 1,
D
n+1=
T
a(D
n) if D
n63 c, [c
n+1, c
1] if D
n3 c.
Hence the boundary points of D
nare forward images of c, one of which is c
n. If D
n3 c, then we call n a cutting time. We enumerate the cutting times by S
k: S
1= 2, and by abuse of notation S
0= 1. In this way we get D
Sk+1= [c
Sk+1, c
1] and an inductive argument shows that D
n= [c
n, c
n−Sk] if S
k< n ≤ S
k+1.
The action ˇ T
aon the tower is as follows. If x ∈ D
n, then T ˇ
a(x) = T
a(x) ∈
D
n+1if c 6∈ (c
n, x] or x = c = c
n, D
r+1if c ∈ (c
n, x],
where r is determined as follows: Clearly, c ∈ (c
n, x] implies that c ∈ D
n.
So n is a cutting time, say S
k. Then we set r = S
k− S
k−1. In fact, it is
not hard to show that r itself is a cutting time. One can define a function Q : N → N by
r = S
Q(k)= S
k− S
k−1.
The function Q is called the kneading map. For more details see [B2].
The tower can be viewed as a countable Markov chain with the intervals D
nas states. There is a transition from D
nto D
n+1for each n and a transition from D
Skto D
1+SQ(k)for each k. This will be used in Section 5 to estimate the number of branches of our induced map.
Another property of the tower is that if U is an interval in the tower, then ˇ T
an|
Uis continuous if and only if T
an|
Uis monotone.
3. The induced map F
aDefinition. Let ˇ F
abe the first return map to D
2in the Hofbauer tower.
The induced map F
ais the unique map such that π ◦ ˇ F
a= F
a◦ π.
For a.e. x we can define the transfer time s(x) as the integer such that F
a(x) = T
as(x). Then F
ahas the following properties:
• Each branch of F
ais linear.
• The image closure of each branch is D
2= [c
2, c
1] = J(a). If a < 2, then D
2is the only level in the tower that equals J(a). Hence s(x) is the smallest positive integer n such that there exists an interval H, x ∈ H ⊂ J(a), such that T
an(H) = J(a) and T
an|
His monotone.
• F
ahas countably many branches. The branch domain will be denoted by J
i(a). They form a partition of J(a). Lemma 1 below shows that
|J(a) \ S
i
J
i(a)| = 0.
• s|
Jiis constant. Let us denote this number by s
i. Let also
Φ
n(a) = F
an(c
3(a)).
The third iterate of c is chosen here, because F
anis well-defined in it for most parameter values (see Lemma 3).
Lemma 1. For every a ∈ [ √
2, 2] and every n ∈ N, F
anis well-defined for a.e. x ∈ J(a).
P r o o f. The tent map T
aadmits an acip µ
awith positive metric entropy log a. According to [K], µ can be lifted to an acip ˇ µ on the tower. Further- more, ˇ µ(D
2) > 0, and due to Birkhoff’s Ergodic Theorem, a.e. x in the tower visits D
2infinitely often. Hence for every n ∈ N, F
anis defined a.e.
Lemma 2. For each a
0∈ ( √
2, 2] there exists a neighbourhood U 3 a and
a constant C
1such that for all a ∈ U ,
X
i
s
i|J
i| = \
J
s(x) dx ≤ C
1. P r o o f. P
i
s
i|J
i| = T
J
s(x) dx < ∞ follows from the existence of the acip (see [B]). In our case, the uniform bound follows because there exist U 3 a
0, C
2> 0 and r ∈ (0, 1) such that for every a ∈ U ,
(3) X
si=n
|J
i| ≤ C
2r
n. We will prove this in Lemma 7.
The induced map F
apreserves Lebesgue measure, because every branch of F
ais linear and surjective. The invariant measure µ of T
acan be written as
µ(B) = C X
i s
X
i−1j=0
|T
a−j(B) ∩ J
i|,
where C is the normalizing factor. By Lemma 2, µ(I) = C P
i
s
i|J
i| < ∞, and the measure can indeed be normalized:
X
i
s
i|J
i| = 1 C .
Fix B ∈ B. We call x very typical with respect to B if (i) For all i ∈ N and 0 ≤ j < s
i,
n→∞
lim 1
n #{0 ≤ k < n | F
ak(x) ∈ T
a−j(B) ∩ J
i} = 1
|c
2− c
1| |T
a−j(B) ∩ J
i|.
In particular, this limit exists.
(ii) For every branch domain J
iof F
a, 1
|c
2− c
1| |J
i| = lim
n→∞
1
n #{0 ≤ j < n | F
aj(x) ∈ J
i}.
(iii) The following holds:
1
C = X
i
s
i|J
i| = lim
n→∞
1 n
n−1
X
i=0
s(F
ai(x)).
Proposition 1. If x is very typical with respect to B, then µ(B) = lim
n→∞
1
n #{0 ≤ k < n | T
ak(x) ∈ B}.
In other words, x is typical with respect to B for the original map.
P r o o f. Choose ε > 0. Let x be very typical. Because of (3), there exists L such that P
sj≥L
s
j|J
j| ≤ ε. Define N
k(x) = P
k−1i=0
s(F
ai(x)). By
condition (iii), lim
n→∞N
n(x)/n = 1/C. We abbreviate v(n, i) = #{(k, j) | 0 ≤ k < n, 0 ≤ j < s
i, F
ak(x) ∈ J
iand T
aj◦ F
ak(x) ∈ B}. Then
µ(B) = C X
i s
X
i−1j=0
|T
a−j(B) ∩ J
i|
= C X
i s
X
i−1j=0 n→∞
lim
1
n #{0 ≤ k < n | F
ak(x) ∈ T
a−j(B) ∩ J
i}
= C X
i n→∞
lim
1 n v(n, i)
≤ C X
si<L n→∞
lim
1
n v(n, i) + C X
si≥L
s
ilim
n→∞
#{0 ≤ k < n | F
ak(x) ∈ J
i}
≤ C lim
n→∞
1 n
X
si<L
v(n, i) + C X
si≥L
s
i|J
i|
≤ C lim sup
n→∞
1 n
X
i
v(n, i) + Cε
≤ C lim sup
n→∞
1
n #{0 ≤ k < N
n(x) | T
ak(x) ∈ B} + Cε
= C lim
n→∞
N
n(x)
n lim sup
n→∞
1
N
n(x) #{0 ≤ k < N
n(x) | T
ak(x) ∈ B} + Cε
= lim sup
n→∞
1
N
n(x) #{0 ≤ k < N
n(x) | T
ak(x) ∈ B} + Cε.
Because ε is arbitrary, and also lim
n→∞(N
n+1(x) − N
n(x))/n = 0, we obtain
µ(B) ≤ lim sup
N →∞
1
N #{0 ≤ k < N | T
ak(x) ∈ B}.
Combining properties (i) and (ii) gives
n→∞
lim 1
n #{0 ≤ k < n | F
ak(x) ∈ T
a−j(I \ B) ∩ J
i} = |T
a−j(I \ B) ∩ J
i|.
Therefore we can carry out the above computation for the complement I \ B as well. Because
N1#{0 ≤ k < N | T
ak(x) ∈ B ∪ (I \ B)} = 1, it follows that µ(B) = lim
N →∞ 1N
#{0 ≤ k < N | T
ak(x) ∈ B}, as asserted.
Remark. Since ˇ F
ais also an induced (in fact, first return) map over ˇ T
a, we can use the same argument to show that x ∈ D
2is typical with respect to B ⊂ F
n
D
nand lifted measure ˇ µ
aon the tower. It was shown in [B] that
many induced maps over (T
a, I) correspond to first return maps to some
subset A in the tower. As x is typical with respect to ˇ µ|
A/ˇ µ(A) and the first
return map to A, it immediately follows that x is typical for these induced maps.
In order to prove the Main Theorem, we need to show that c, or rather c
3, satisfies conditions (i)–(iii) for a.e. a. This will be done in Propositions 2 and 3.
4. Some more properties of J
i, ϕ
nand Φ
nLemma 3. If orb(c(a)) is dense in J(a), then Φ
n(a) is defined for every n ∈ N.
It immediately follows by [BM] that
Corollary 2. Φ
n(a) is defined for all n for a.e. a ∈ [ √ 2, 2].
P r o o f (of Lemma 3). Let k be such that there exists H, c
3∈ H ⊂ J(a), such that T
ak|
His monotone and T
ak(H) = J(a). Let p be the nonzero fixed point of T
a. Let
c
−v< c
−v−2< . . . < p < . . . < c
−v−3< c
−v−1be pre-turning points closest to p, where v > k. As orb(c(a)) is dense in J(a), there exists m such that c
m∈ (c
−v, c
−v−1). Take m minimal. Let H
03 c
3be the maximal interval such that T
am−3|
H0is monotone. Because
∂T
am−3(H
0) ⊂ orb(c(a)) and m is minimal, T
an−3(H
0) ⊃ [c
−v, c
−v−1]. Be- cause T
av+2([c
−v, c
−v−1]) = [c
2, c
1], we see for k
0= n − 3 + v + 2 > k that T
ak0|
H0is monotone and T
ak0(H
0) = J(a). It follows that Φ
n(a) is defined for all n ∈ N.
The previous lemmas showed that there exists a full-measure set A ⊂ [ √
2, 2] of parameters for which Φ
n(a) is defined for every n. In particular, c is not periodic for every a ∈ A. We assume from now on that a is always taken from A. The next lemma shows that all branches of Φ
n: ( √
2, 2] → J(a) are onto.
Lemma 4. Let a ∈ A, and suppose Φ
n(a) = T
am(c
3(a)). Then there exists an interval U = [a
1, a
2] 3 a such that ϕ
m+3maps U monotonically onto [c
1(a
1), c
2(a
2)] or [c
2(a
1), c
1(a
2)].
P r o o f. By definition π
−1◦ Φ
n(a) ∩ D
2is the nth return in the tower
of c
3∈ D
2to D
2. Suppose Φ
n(a) = ϕ
m+3(a) ∈ int J(a). Because any point
in π
−1(c) is mapped by ˇ T
ato a boundary point of some level in the tower,
and because boundary points are mapped to boundary points, it follows that
ϕ
j(a) 6= c for j < m+3. Hence ϕ
m+3is a diffeomorphism in a neighbourhood
of a. Since this is true for every point a
0such that Φ
n(a
0) ∈ int J(a
0), the
existence of the interval U follows.
For any C
1function f , let
dis(f, J) = sup
x,y∈J
|Df (x)|
|Df (y)|
be the distortion of f on J.
Lemma 5. Let U
n⊂ [ √
2, 2] be an interval on which ϕ
nis monotone.
Then
sup
Un⊂[√ 2,2]
dis(ϕ
n, U
n) → 1 as n → ∞.
Moreover ,
dadϕ
n(a) = O(a
n).
P r o o f. See [BM].
Corollary 3. There exists K > 0 with the following property. Let x = x(a) ∈ I be such that T
an(x) = c(a) for some n and T
aj(x) 6= c(a) for j < n.
Moreover , fix the itinerary of x up to entry n. Then
dx(a)da
≤ K.
P r o o f. Write G(a, x) = T
an(x) − c. Then 0 = d
da G(a, x) = ∂
∂a T
an(x) + ∂
∂x T
an(x) dx da = ∂
∂a T
an(x) + a
ndx da . As T
anis a degree n polynomial with coefficients in [0, 1],
∂∂a
T
an≤ Ka
n. The result follows.
The boundary points of J
i(a) are preimages of c. As long as J
i(a) persists,
|J
i(a)| = a
−si|J(a)| and J
i(a) moves with speed O(1) as a varies. Take n large and let U
nbe such that ϕ
n|
Unis monotone. By Lemma 5, dis(ϕ
n, U
n) is close to 1. There exists K (K → 1 as n → ∞) such that
|ϕ
−1n(J
i(a)) ∩ U
n|
|U
n| ≤ K|J
i(a)| = Ka
−si|J(a)|.
Let us now try to analyze how the branch domains J
i(a) are born and die if the parameter varies. As |J
i(a)| = a
−si|c
1(a) − c
2(a)|,
d
da |J
i(a)| = 1
2 a
−si(2a − 1 − s
i(a − 1)).
It is easy to see that for s
i≥ 5 and a ∈ [ √
2, 2],
dad|J
i(a)| < 0. These branch domains shrink as a increases, and therefore cannot be born in a point. The only way a branch domain can be created is by merging (countably) many smaller branch domains, with larger transfer times, into a new one. This happens whenever c is n-periodic, and the central branch of T
ancovers a point of T
a−1(c). This is the same moment at which the central branch of T
an+2covers (c
2, c
1).
As the kneading invariant (and topological entropy) of T
aincreases with
a, branch domains cannot disappear either, except in this merging process.
5. The proof of statement (3) Lemma 6. For every a
0∈ ( √
2, 2] for which c is not periodic under T
a0, there exist C
2, δ > 0 such that for every a ∈ (a
0− δ/2, a
0+ δ/2) and every n ≥ 1,
#{j | s
j(a) = n} ≤ C
2(a
0− δ)
n.
P r o o f. It is shown in [H] that a = exp h
top(T
a) is the exponential growth rate of the number of paths in the tower starting from D
2. Let G(a, n) =
#{j | s
j(a) = n} be the number of n-loops from D
2to D
2that do not visit D
2in between. We will choose δ > 0 below such that the combinatorics of the tower up to some level remains the same for all a ∈ (a
0− δ, a
0+ δ).
Then we argue that the exponential growth rate lim sup
nn1log G(a, n) for all a ∈ (a
0− δ/2, a
0+ δ/2) is smaller than h
top(T
a0−δ) = log(a
0− δ). From this the lemma follows. We will compute these exponential growth rates by means of the characteristic polynomials of well-chosen submatrices of the transition matrix corresponding to the tower.
Choice of δ. The assumption a
0> √
2 implies that c
3lies to the left of the non-zero fixed point of T
a0. It is easy to verify that for some integer u ≥ 0, c
3, . . . , c
2u+2lie to the right of c while c
2u+3lies to the left again.
This corresponds to the fact that T
a0is not renormalizable. In terms of the kneading map renormalizability is equivalent to the following statement ([B2, Proposition 1]): There exists k ≥ 1 such that
Q(k) = k − 1 and Q(k + j) ≥ k − 1 for all j ≥ 1.
Here S
kis the period of renormalization. In our case, this formula is false for S
k= S
1= 2. Therefore there exists u ≥ 0 such that
Q(1) = 0, Q(j) = 1 for 2 ≤ j ≤ u + 1, Q(u + 2) = 0.
Take δ maximal such that the cutting times S
0, . . . , S
u+2are the same for all a ∈ (a
0− δ, a
0+ δ). As c is not periodic under T
a0, δ is positive.
A lower bound for the entropy. The tower F
n≥2
D
ngives rise to a count- able transition matrix M = (m
i,j)
∞i,j=2, where m
i,j= 1 if and only if a transition D
i→ D
jis possible. Therefore m
i,i+1= 1 and m
Sk,1+SQ(k)= 1 for all i, k, and all other entries are zero. For a ∈ (a
0− δ, a
0+ δ) let M (u) be the (2u + 2) × (2u + 2) left upper submatrix of M . Denote the spectral radius of this matrix by %
0(u). For example,
M (2) =
1 1 0 0 0 0 0 0 1 0 0 0 0 1 0 1 0 0 0 0 0 0 1 0 0 1 0 0 0 1 1 0 0 0 0 0
.
Because M (u) is the transition matrix of F
2u+3n=2
D
n, we see that log %
0(u), the exponential growth rate of the paths from D
2in F
2u+3n=2
D
n, is less than or equal to the exponential growth rate of the paths from D
2in the whole tower. Therefore log %
0(u) ≤ inf{h
top(T
a) | a ∈ (a
0− δ, a
0+ δ)}
= log(a
0− δ).
An upper bound for G(a, n). In order to estimate G(a, n), we use a larger submatrix of M . Assume that S
u+3= S
u+2+ v = 2u + 3 + v. Let f M (u, v) be the (2u + 2 + v) × (2u + 2 + v) left upper submatrix of M in which we set e m
2,2= e m
2,3= 0 and e m
2u+3+v,2u+4= 1 + m
2u+3+v,2u+4. Denote the spectral radius by %
1(u, v). For example,
M (2, 4) = f
0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 1 0 0 0
.
We claim that for a ∈ (a
0− δ/2, a
0+ δ
2), i.e. u fixed, lim sup 1
n log G(a, n)
≤ max{log %
1(u, v) | v = 1, 2, 4, . . . , 2u, 2u + 2, 2u + 3}.
Clearly, G(a, 1) = 1 and, for n ≥ 2, G(a, n) is the number of paths of length n − 1 from D
3to D
2that do not visit D
2in between. The total number of paths of length n − 1 from D
3to D
2is m
n−13,2, the appropriate entry of the matrix M
n−1. By putting e m
2,2= e m
2,3= 0 we avoid counting the paths that visit D
2in between. The tower F
n≥2
D
ncan be pictured as a graph;
the branch points are the cutting levels D
Sk.
From D
Su+2there is a path D
Su+2→ D
2and a path upwards in the tower. This path splits again at D
Su+3into a path to D
1+SQ(u+3)and an- other to D
1+Su+3. This gives two paths D
Su+2] → D
1+Su+3and D
Su+2→ D
1+SQ(u+3), both of length v = S
Q(u+3)∈ {1, 2, 4, 6, . . . , 2u, 2u + 2, 2u + 3}.
At the branch point D
Su+3the same situation occurs: there are paths D
Su+3→ D
1+Su+4and D
Su+3→ D
1+SQ(u+4), both of length v
0= S
Q(u+4)∈ {1, 2, 4, 6, . . . , 2u, 2u + 2, 2u + 3, 2u + 3 + v}. The number of paths of length n from D
2increases if the path lengths between branch points decrease.
Therefore the choice v
0= 2u + 3 + v will give smaller values of G(a, n) for
large n than the choice v
0= 2u + 3. And if v is chosen such that G(a, n) is maximized (i.e. the largest values for G(a, n) are obtained for those a for which S
Q(u+3)= v), then choosing v
0= v (i.e. choosing a such that S
Q(u+4)= v) will also maximize G(a, n). By induction we should take the same value for S
Q(k)for each k ≥ u + 3. Therefore we can identify all branch points D
Sk, k ≥ u + 3. This gives rise to the transition matrix f M (u, v) and hence proves the claim.
The rome technique. To prove the lemma, it suffices to show that %
1(u, v)
≤ %
0(u). The spectral radius is the leading root of the characteristic polyno- mial. We will compute the characteristic polynomials of M (u) and f M (u, v) (denoted as cp
0and cp
1respectively) by means of the rome technique from [BGMY, Theorem 1.7]. Let M be some n × n matrix with nonnegative inte- ger entries. A path p is a sequence p
0. . . p
lof states such that m
pi−1,pi> 0 for all 1 ≤ i ≤ l. The length of the path is l(p) = l and w(p) = Q
l(p)i=1
m
pi−1,piis the width. A rome R = {r
1. . . r
k} (i.e. #(R) = k) is a subset of the states with the property that every closed path (i.e. p
0= p
l) contains at least one state from R. A path p = p
0. . . p
lis simple if p
0, p
l∈ R but p
i6∈ R for 1 ≤ i < l.
Theorem (Rome Theorem). The characteristic polynomial of M equals (−1)
n−kx
ndet(A
R(x) − I),
where I is the identity on R
kand A = (a
i,j)
ki,j=1is the matrix with entries a
i,j= P
p
w(p)x
−l(p). Here the sum runs over all simple paths from r
ito r
j. The characteristic polynomials. Let D
i→
kD
jstand for a path of length k from D
ito D
j. For M (u), the states D
2and D
3form a rome. The cor- responding simple paths are D
2→
1D
2, D
2→
1D
3, D
3→
2u+1D
2and D
3→
jD
3for j = 2, 4, . . . , 2u. Therefore the characteristic polynomial of M (u) is
cp
0(u) = x
2u+2det
1
x − 1 1
1 x x
2u+11
x
2+ . . . + 1 x
2u− 1
= x
2u+3− 2x
2u+1− 1
x + 1 .
For f M (u, v) we distinguish four cases.
(a) v = 1. In this case {D
2, D
3, D
2u+4} forms a rome and the simple
paths are D
3→
2u+1D
2, D
3→
2u+1D
2u+4, D
3→
jD
3for j = 2, 4, . . . , 2u,
D
2u+4→
1D
2and D
2u+4→
1D
2u+4. We give the characteristic polynomial
the sign that makes the leading coefficient positive:
−cp
1(u, 1) = − x
2u+3· det
−1 0 0
1 x
2u+11
x
2+ . . . + 1
x
2u− 1 1 x
2u+11
x 0 1
x − 1
= x
2(x
2u+2− 2x
2u+ 1)
(x + 1) .
Hence −
x1cp
1(u, v)−cp
0(u) = 1. As
x1is positive on (1, ∞), %
0(u) > %
1(u, 1).
(b) v = 2. In this case {D
2, D
3, D
2u+4} forms a rome and the simple paths are D
3→
2u+1D
2, D
3→
2u+1D
2u+4, D
3→
jD
3for j = 2, 4, . . . , 2u, D
2u+4→
2D
3and D
2u+4→
2D
2u+4. The characteristic polynomial is
cp
1(u, 2) = − x
2u+4· det
−1 0 0
1 x
2u+11
x
2+ . . . + 1
x
2u− 1 1 x
2u+10 1
x
21 x
2− 1
= x(x
2u+3− 2x
2u+1+ x − 1).
Therefore
1xcp
1(u, v)−(x+1)cp
0(u) = x, which is positive on (1, ∞). Because also
1xand x + 1 are positive on (1, ∞), %
0(u) > %
1(u, 2).
(c) v = 4, 6, . . . , 2u. Here {D
2, D
3, D
v+1, D
2u+4} forms a rome and the paths are D
3→
v−2D
v+1, D
3→
jD
3for j = 2, 4, . . . , v − 2, D
u+1→
jD
3for j = 2, . . . , 2u − v + 2, D
u+1→
2u−v+3D
2u+4, D
u+1→
2u−v+3D
2, D
2u+4→
vD
2u+4, D
2u+4→
vD
u+1and D
2u+4→
2D
2u+4. The charac- teristic polynomial is
cp
1(u, v) = x
2u+v+2× det
−1 0 0 0
0 1
x
2+ . . . + 1
x
v−2− 1 1
x
v−20
1 x
2u+3−v1
x
2+ . . . + 1
x
2u+2−v−1 1
x
2u+3−v0 0 1
x
v1 x
v− 1
= x(x
v− 1)(x
2u+3− 2x
2u+1+ x + 1) (x − 1)(x
2− 1) . It follows that
(x − 1)(x
2− 1)
x(x
v− 1) cp
1(u, v) − (x + 1)cp
0(u) = (x + 2),
which is positive on (1, ∞). Because (x − 1)(x
2− 1)/x(x
v− 1) and x + 1 are also positive in (1, ∞), %
0(u) > %
1(u, v).
(d) v = 2u + 3. Again {D
2, D
3, D
2u+4} forms a rome. The paths are D
3→
2u+1D
2, D
3→
2u+1D
2u+4, D
3→
jD
3for j = 2, 4, . . . , 2u and D
2u+4→
2u+3D
2u+4. This last path has width 2. We obtain
−cp
1(u, 2u + 3) = − x
4u+5× det
−1 0 0
1 x
2u+11
x
2+ . . . + 1
x
2u− 1 1 x
2u+10 0 2
x
2u+3− 1
= x(x
2u+3− 2)(x
2u+2− 2x
2u+ 1)
(x
2− 1) .
Therefore
− x − 1
x
2u+3− 2 cp
1(u, v) − cp
0(u) = 1.
Because 1/(x
2u+3− 2) and x − 1 are positive on (2
1/(2u+3), ∞) and cp
0(2
1/(2u+3)) < 0 it follows that %
0(u) > %
1(u, 2u + 3).
Hence in all cases %
0(u) > %
1(u, v). Therefore lim sup
n1log G(a, n) ≤ max{%
1(u, v) | v = 1, 2, 4, 6, . . . , 2u, 2u + 2, 2u + 3} < %
0(u) ≤ a
0− δ, proving the lemma.
Lemma 7. For every a
0∈ [ √
2, 2], there exist C
2, δ > 0 and r ∈ (0, 1) such that for every a ∈ (a
0− δ/2, a
0+ δ/2),
(3) X
sj=n
|J
i(a)| ≤ C
2r
n.
P r o o f. Because |J
i(a)| = |c
2(a) − c
1(a)|a
−si≤ a
−si, the statement follows immediately from Lemma 6. We can take δ and C
2as in Lemma 6 and r = (a
0− δ)/(a
0− δ/2) < 1.
6. Probabilistic lemmas. For each n ∈ N we consider the set of branch domains of the map Φ
nas a partition Z
nof the parameter space [ √
2, 2].
For m < n, Z
nis finer than Z
m, and W
n
Z
ncontains no nondegenerate intervals. An element of Z
nwill be denoted by Z
e1...en, where e
j= i if Φ
j−1(Z
e1...en) ⊂ J
i(a).
Lemma 8. Let {X
m} be a sequence of random variables with the following properties:
(a) There exists V < ∞ such that for every m ∈ N, Var(X
m| Z
e1...em) <
V for every branch domain Z
e1...em.
(b) X
m−1is constant on each interval Z
e1...em.
(c) There exist M ∈ R, N ∈ N and ε > 0 such that for every m > N ,
|M − E(X
m|Z
e1...em)| < ε.
Then
lim sup
m→∞
M − 1 m
m−1
X
i=0
X
i≤ ε a.s.
Notice that the random variables X
mare not independent, but only
“eventually almost independent”. We will use this lemma twice in the next two sections. In the next section, however, we will only consider a subse- quence of the branch domain partitions {Z
e1...en}. This does not affect the validity of the lemma.
P r o o f (of Lemma 8). Define Y
m= X
m− E(X
m|Z
e1...em). Then E(Y
m|Z
e1...em) = 0 and Var(Y
m|Z
e1...em) = E(Y
m2|Z
e1...em) < V for all m and all branch domains Z
e1...em. Let S
n= P
nm=1
Y
m, so E(S
12) = E(Y
12) ≤ V . By property (b), S
n−1is constant on each set Z
e1...en. Suppose by in- duction that E(S
n−12) ≤ (n − 1)V ; then
E(S
n2) = E(S
n−12) + E(Y
n2) + 2E(Y
nS
n−1)
≤ (n − 1)V + V + 2 X
Ze1...en
E(Y
nS
n−1|Z
e1...en)
≤ nV + 2 X
Ze1...en
S
n−1· E(Y
n|Z
e1...en) = nV.
By the Chebyshev inequality, P (S
n> nδ) ≤ nV /(n
2δ
2) = V /(nδ
2). In particular, P (S
n2> n
2δ) ≤ V /(n
2δ
2). Therefore P
n
P (S
n2> n
2δ
2) <
∞ and by the Borel–Cantelli Lemma, P (S
n2> n
2δ
2i.o.) = 0. As δ is arbitrary, S
n2/n
2→ 0 a.s. For the intermediate values of n, let D
n= max
n2<k<(n+1)2|S
k− S
n2|. Because |S
k− S
n2| = | P
kj=n2+1
X
j|, we have E(|S
k− S
n2|
2) ≤ (k − n
2)V ≤ 2nV . Hence
E(D
2n) ≤ E
(n+1)X
2−1k=n2+1
|S
k− S
n2|
2≤
(n+1)
X
2−1 k=n2+12nV = 4n
2V.
Using Chebyshev’s inequality again we obtain P (D
n≥ n
2δ) ≤ 4n
2V /(n
4δ
2)
= 4V /(n
2δ
2). By the Borel–Cantelli Lemma, P (D
n≥ n
2δ i.o.) = 0, and D
n/n
2→ 0 a.s. Combining things and taking n
2≤ k < (n + 1)
2, we get
S
kk ≤ S
n2+ D
nn
2→ 0 a.s.
Because X
m∈ Y
m+ [M − ε, M + ε] for m > N , lim sup
n→∞
1 n
X
n i=1X
i= lim sup
n→∞
1 n
X
N i=1X
i+ lim sup
n→∞
1 n
X
n i=N +1X
i≤ lim sup
n→∞
1
n S
N+ lim sup
n→∞
1 n
X
n i=N +1(Y
i+ M + ε)
≤ lim sup
n→∞
1
n S
N+ lim sup
n→∞
n − N
n (M + ε) ≤ M + ε.
The other inequality is proved similarly.
An additional lemma is needed to deal with the a-dependence of the acip.
Lemma 9. Let A be an interval, and let M : A → R and g
n: A → R be functions with the following properties:
(a) M is continuous a.e. on A.
(b) Let A(a
0, ε) = {a ∈ A | lim sup
n→∞|g
n(a) − M (a
0)| ≤ ε}. If ε > 0, then a.e. a
0∈ A is a density point of A(a
0, ε).
Then lim
n→∞g
n(a) = M (a) a.e.
P r o o f. Set B
k= {a ∈ A | lim sup
n→∞|g
n(a) − M (a)| ≥ 1/k}. Assume for a contradiction that there exists k such that |B
k| > 0. Take ε < 1/(3k) and let a
0∈ B
kbe a density point, both of B
kand of A(a
0, ε). Assume also that M is continuous at a
0. Let A
0be a neighbourhood of a
0so small that
• |M (a) − M (a
0)| ≤ ε for all a ∈ A
0,
• |A
0∩ A(a
0, ε)| ≥
34|A
0|, and
• |A
0∩ B
k| ≥
34|A
0|.
Then a ∈ A
0∩ A(a
0, ε) ∩ B
k6= ∅ and for all a ∈ A
0∩ A(a
0, ε) ∩ B
k, lim sup
n→∞
|g
n(a) − M (a)| ≤ lim sup
n→∞
|g
n(a) − M (a
0)| + |M (a) − M (a
0)|
≤ 2ε < 1/k.
This contradicts a ∈ B
k, proving the lemma.
7. Concerning condition (i). Choose B ∈ B. Hence ∂B is a closed zero-measure set.
Lemma 10. Choose ε > 0, a
0∈ A, k
1∈ N and 0 ≤ k
2< s
k1(a
0). For a close or equal to a
0let B
0(a) = T
a−k2(B) ∩ J
k1(a). Then there exists a neighbourhood A 3 a
0such that
lim sup
n→∞