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155 (1998)

For almost every tent map, the turning point is typical

by

Henk B r u i n (Stockholm)

Abstract. Let T

a

be the tent map with slope a. Let c be its turning point, and µ

a

the 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→∞ 1

n

P

n−1

i=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

a

has an absolutely continuous invariant probability measure (acip), µ

a

, and that µ

a

is 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

a

is 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

a

is 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]

(2)

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

a

has topological entropy log a. Hence one can state Theorem 1 as: For a.e. value of the topological entropy, the turning point of T

a

is typical. Because the measure µ

a

actually 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+α

in 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

a

is 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

a

has an acip which can be approximated weakly by Dirac measures on

super-stable orbits of nearby maps.

(3)

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

a

is 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

a

has 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

n

63 c, [c

n+1

, c

1

] if D

n

3 c.

Hence the boundary points of D

n

are forward images of c, one of which is c

n

. If D

n

3 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

a

on the tower is as follows. If x ∈ D

n

, then T ˇ

a

(x) = T

a

(x) ∈

 D

n+1

if c 6∈ (c

n

, x] or x = c = c

n

, D

r+1

if 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

(4)

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

n

as states. There is a transition from D

n

to D

n+1

for each n and a transition from D

Sk

to 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

|

U

is continuous if and only if T

an

|

U

is monotone.

3. The induced map F

a

Definition. Let ˇ F

a

be the first return map to D

2

in the Hofbauer tower.

The induced map F

a

is 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

a

has the following properties:

• Each branch of F

a

is linear.

• The image closure of each branch is D

2

= [c

2

, c

1

] = J(a). If a < 2, then D

2

is 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

|

H

is monotone.

• F

a

has 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|

Ji

is 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

an

is well-defined in it for most parameter values (see Lemma 3).

Lemma 1. For every a ∈ [

2, 2] and every n ∈ N, F

an

is well-defined for a.e. x ∈ J(a).

P r o o f. The tent map T

a

admits an acip µ

a

with 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

2

infinitely often. Hence for every n ∈ N, F

an

is defined a.e.

Lemma 2. For each a

0

∈ (

2, 2] there exists a neighbourhood U 3 a and

a constant C

1

such that for all a ∈ U ,

(5)

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

2

r

n

. We will prove this in Lemma 7.

The induced map F

a

preserves Lebesgue measure, because every branch of F

a

is linear and surjective. The invariant measure µ of T

a

can be written as

µ(B) = C X

i s

X

i−1

j=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

i

of 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−1

i=0

s(F

ai

(x)). By

(6)

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

i

and T

aj

◦ F

ak

(x) ∈ B}. Then

µ(B) = C X

i s

X

i−1

j=0

|T

a−j

(B) ∩ J

i

|

= C X

i s

X

i−1

j=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

i

lim

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 →∞ 1

N

#{0 ≤ k < N | T

ak

(x) ∈ B}, as asserted.

Remark. Since ˇ F

a

is also an induced (in fact, first return) map over ˇ T

a

, we can use the same argument to show that x ∈ D

2

is typical with respect to B ⊂ F

n

D

n

and lifted measure ˇ µ

a

on 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

µ(A) and the first

(7)

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

n

and Φ

n

Lemma 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

|

H

is 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−1

be 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

0

3 c

3

be the maximal interval such that T

am−3

|

H0

is 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

|

H0

is 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+3

maps 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

2

is the nth return in the tower

of c

3

∈ D

2

to 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

a

to 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+3

is a diffeomorphism in a neighbourhood

of a. Since this is true for every point a

0

such that Φ

n

(a

0

) ∈ int J(a

0

), the

existence of the interval U follows.

(8)

For any C

1

function 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

n

is monotone.

Then

sup

Un⊂[√ 2,2]

dis(ϕ

n

, U

n

) → 1 as n → ∞.

Moreover ,

dad

ϕ

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

n

dx da . As T

an

is 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

n

be such that ϕ

n

|

Un

is monotone. By Lemma 5, dis(ϕ

n

, U

n

) is close to 1. There exists K (K → 1 as n → ∞) such that

−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

an

covers a point of T

a−1

(c). This is the same moment at which the central branch of T

an+2

covers (c

2

, c

1

).

As the kneading invariant (and topological entropy) of T

a

increases with

a, branch domains cannot disappear either, except in this merging process.

(9)

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

2

to D

2

that do not visit D

2

in 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

nn1

log 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

3

lies 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+2

lie to the right of c while c

2u+3

lies to the left again.

This corresponds to the fact that T

a0

is 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

k

is 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+2

are 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

n

gives 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

j

is 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

 

 

 

.

(10)

Because M (u) is the transition matrix of F

2u+3

n=2

D

n

, we see that log %

0

(u), the exponential growth rate of the paths from D

2

in F

2u+3

n=2

D

n

, is less than or equal to the exponential growth rate of the paths from D

2

in 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

3

to D

2

that do not visit D

2

in between. The total number of paths of length n − 1 from D

3

to D

2

is 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

2

in between. The tower F

n≥2

D

n

can be pictured as a graph;

the branch points are the cutting levels D

Sk

.

From D

Su+2

there is a path D

Su+2

→ D

2

and a path upwards in the tower. This path splits again at D

Su+3

into 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+3

and 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+3

the same situation occurs: there are paths D

Su+3

→ D

1+Su+4

and 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

2

increases 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

(11)

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

0

and cp

1

respectively) 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

l

of 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,pi

is 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

l

is simple if p

0

, p

l

∈ R but p

i

6∈ R for 1 ≤ i < l.

Theorem (Rome Theorem). The characteristic polynomial of M equals (−1)

n−k

x

n

det(A

R

(x) − I),

where I is the identity on R

k

and A = (a

i,j

)

ki,j=1

is the matrix with entries a

i,j

= P

p

w(p)x

−l(p)

. Here the sum runs over all simple paths from r

i

to r

j

. The characteristic polynomials. Let D

i

k

D

j

stand for a path of length k from D

i

to D

j

. For M (u), the states D

2

and D

3

form a rome. The cor- responding simple paths are D

2

1

D

2

, D

2

1

D

3

, D

3

2u+1

D

2

and D

3

j

D

3

for j = 2, 4, . . . , 2u. Therefore the characteristic polynomial of M (u) is

cp

0

(u) = x

2u+2

det

  1

x − 1 1

1 x x

2u+1

1

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+1

D

2

, D

3

2u+1

D

2u+4

, D

3

j

D

3

for j = 2, 4, . . . , 2u,

D

2u+4

1

D

2

and D

2u+4

1

D

2u+4

. We give the characteristic polynomial

(12)

the sign that makes the leading coefficient positive:

−cp

1

(u, 1) = − x

2u+3

· det

 

 

−1 0 0

1 x

2u+1

1

x

2

+ . . . + 1

x

2u

− 1 1 x

2u+1

1

x 0 1

x − 1

 

 

= x

2

(x

2u+2

− 2x

2u

+ 1)

(x + 1) .

Hence −

x1

cp

1

(u, v)−cp

0

(u) = 1. As

x1

is 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+1

D

2

, D

3

2u+1

D

2u+4

, D

3

j

D

3

for j = 2, 4, . . . , 2u, D

2u+4

2

D

3

and D

2u+4

2

D

2u+4

. The characteristic polynomial is

cp

1

(u, 2) = − x

2u+4

· det

 

 

−1 0 0

1 x

2u+1

1

x

2

+ . . . + 1

x

2u

− 1 1 x

2u+1

0 1

x

2

1 x

2

− 1

 

 

= x(x

2u+3

− 2x

2u+1

+ x − 1).

Therefore

1x

cp

1

(u, v)−(x+1)cp

0

(u) = x, which is positive on (1, ∞). Because also

1x

and 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−2

D

v+1

, D

3

j

D

3

for j = 2, 4, . . . , v − 2, D

u+1

j

D

3

for j = 2, . . . , 2u − v + 2, D

u+1

2u−v+3

D

2u+4

, D

u+1

2u−v+3

D

2

, D

2u+4

v

D

2u+4

, D

2u+4

v

D

u+1

and D

2u+4

2

D

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−2

0

1 x

2u+3−v

1

x

2

+ . . . + 1

x

2u+2−v

−1 1

x

2u+3−v

0 0 1

x

v

1 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),

(13)

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+1

D

2

, D

3

2u+1

D

2u+4

, D

3

j

D

3

for j = 2, 4, . . . , 2u and D

2u+4

2u+3

D

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+1

1

x

2

+ . . . + 1

x

2u

− 1 1 x

2u+1

0 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

n1

log 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

2

r

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

2

as 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

n

as a partition Z

n

of the parameter space [

2, 2].

For m < n, Z

n

is finer than Z

m

, and W

n

Z

n

contains no nondegenerate intervals. An element of Z

n

will 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−1

is constant on each interval Z

e1...em

.

(14)

(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

m

are 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

n

m=1

Y

m

, so E(S

12

) = E(Y

12

) ≤ V . By property (b), S

n−1

is 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

n

S

n−1

)

≤ (n − 1)V + V + 2 X

Ze1...en

E(Y

n

S

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

2

δ

2

) = V /(nδ

2

). In particular, P (S

n2

> n

2

δ) ≤ V /(n

2

δ

2

). Therefore P

n

P (S

n2

> n

2

δ

2

) <

∞ and by the Borel–Cantelli Lemma, P (S

n2

> n

2

δ

2

i.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

k

j=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

2−1

k=n2+1

|S

k

− S

n2

|

2



(n+1)

X

2−1 k=n2+1

2nV = 4n

2

V.

Using Chebyshev’s inequality again we obtain P (D

n

≥ n

2

δ) ≤ 4n

2

V /(n

4

δ

2

)

= 4V /(n

2

δ

2

). By the Borel–Cantelli Lemma, P (D

n

≥ n

2

δ i.o.) = 0, and D

n

/n

2

→ 0 a.s. Combining things and taking n

2

≤ k < (n + 1)

2

, we get

S

k

k S

n2

+ D

n

n

2

→ 0 a.s.

(15)

Because X

m

∈ Y

m

+ [M − ε, M + ε] for m > N , lim sup

n→∞

1 n

X

n i=1

X

i

= lim sup

n→∞

1 n

X

N i=1

X

i

+ lim sup

n→∞

1 n

X

n i=N +1

X

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

k

be a density point, both of B

k

and of A(a

0

, ε). Assume also that M is continuous at a

0

. Let A

0

be a neighbourhood of a

0

so 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

k

6= ∅ 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

0

let B

0

(a) = T

a−k2

(B) ∩ J

k1

(a). Then there exists a neighbourhood A 3 a

0

such that

lim sup

n→∞

1

n #{0 ≤ i < n | Φ

i

(a) ∈ B

0

(a)} − |B

0

(a

0

)|

|J(a

0

)|

≤ ε.

(16)

P r o o f. Suppose we have chosen a

0

∈ A and ε > 0. Let J = {J

i

}

i

be the partition of J(a

0

) into branch domains of F

a0

. The partition J ∨ F

a−10

J ∨ F

a−20

J ∨ . . . contains no nondegenerate intervals. Furthermore, as B ∈ B, also ∂B

0

(a

0

) is a closed set of zero Lebesgue measure. Therefore we can find N and a neighbourhood U of ∂B

0

(a

0

) with the following properties:

• |U | ≤ ε/8.

• U consists of a finite number of intervals, say U

i

, i = 1, . . . , L.

• The boundary points of each U

i

are boundary points of cylinder sets in J ∨ F

a−10

J ∨ . . . ∨ F

a−N0

J .

In this way, we have chosen at most 2L cylinder sets, say K

i

, i = 1, . . . , 2L, which determine the neighbourhood U in a topological way: U can be defined persistently under small changes of the parameter. Let us write U = U (a).

Let Z

e1...en

⊂ A denote a branch domain of Φ

n

. Fix R ∈ N and an interval A 3 a

0

such that

• J

k1

(a) persists as a varies in A.

• dis(Φ

r

, Z

e1...er

) ≤ 1 + ε/4 for every r ≥ R and every branch domain Z

e1...er

such that Z

e1...er

∩ A 6= ∅.

• The intervals K

i

, i = 1, . . . , 2L, persist as a varies in A, and |U (a)| ≤ ε/4 for all a ∈ A.

|B

a0

|

|J(a)| |B

a00

|

|J(a

0

)|

ε

4 for all a ∈ A.

Let

X e

r+

=

 1 if Φ

r

(a) ∈ B

0

(a) ∪ U (a), 0 otherwise,

and

X e

r

=

n 1 if Φ

r

(a) ∈ B

0

(a) \ U (a), 0 otherwise.

Hence e X

r±

are constant on Z

e1...er+N

. We claim that for any set Z

e1...er

⊂ A, E( e X

r+

|Z

e1...er

) ≤ |B

a00

|

|J(a

0

)| + ε.

Here the expectation is taken with respect to normalized Lebesgue measure on A. Indeed, we have

E( e X

r+

|Z

e1...er

) ≤

 1 + ε

4

 |B

0

(a) ∪ U (a)|

|J(a)|

 1 + ε

4

 |B

a0

|

|J(a)| + ε 4



 1 + ε

4

 |B

a00

|

|J(a

0

)| + ε 2



|B

a00

|

|J(a

0

)| + ε.

Cytaty

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