renormalization operator for affine dissipative lorenz maps colli

8/13/2019 Renormalization Operator for Affine Dissipative Lorenz Maps Colli

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What is striking about discontinuities is that intricate dynamics may appear even

without any expansion or criticality. This work is devoted to studying and charac-terizing a certain family of interval maps, which can be classied as Lorenz maps,described by three parameters inside the cube (0 , 1)3. To be specic, each map of thefamily is affine, and with derivative positive and smaller than one outside its singlediscontinuity point. One parameter ( b) describes the position of the discontinuitypoint into the interval and the two other parameters (here and ) are the slopesto the left and to the right of the discontinuity point.

In this family, the dynamical behaviour alternates between hyperbolic , i.e. exhibit-ing an attracting hyperbolic periodic point, and neutral , where there is an invariantset which is the complement of the orbit of a wandering interval, as in the Denjoycounter-examples for some circle diffeomorphisms with irrational rotation numbers

(see [5] for a general presentation).A key element in the study of the family is the renormalization operator R ,

dened for maps coming from parameters in a certain subset of the cube. Thesemaps are called renormalizable . A map f is said to be N renormalizable , N 1,if f, R f , . . . , R N 1f belong to the domain of R , and innitely renormalizable if N renormalizable for all N 1. It turns out that the neutral maps are exactly theinnitely renormalizable maps, and all the other maps (nonrenormalizable or nitelyrenormalizable) are hyperbolic.

We identify the cube (0 , 1)3 with the image of the one-to-one parametrization(, ,b ) f ,,b . In this way, we may say that a parameter is renormalizable ora parameter is innitely renormalizable, meaning that the corresponding functions

have this or the other property. The notation R will be also used for the operatorinduced in the parametrization cube.

Renormalization operators in dynamical systems came into the scene in the sev-enties, with the works of Feigenbaum ([6]) and Coullet-Tresser ([3]) on the quadraticfamily. Since then, intense research has been made aiming at proving the conjecturesposed by these and other authors.

The main advantage of working with this three-parameter family resides in thenite dimensionality of the domain of the renormalization operator, allied to thepossibility of explicitly writing its expression. In general, as a consequence of the notso good properties of the composition operator, the renormalization operator is noteven differentiable (see [4] for a recent account in the context of unimodal maps).

We will prove the following.

Theorem 1. The set of innitely renormalizable parameters (, ,b ) (0, 1)3 is a lamination, with C two-dimensional leaves, where each leaf is the graph of some function of (, ). The leaves vary continuously in the C r topology, for any r 0.The holonomies are H older continuous, but not uniformly H older continuous in (0, 1)3

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and not more than H older continuous, in general. Moreover, transversal bers to the

lamination intersect it in a zero Hausdorff dimension set.Differently from the renormalization operator of unimodal families, this one does

not present an attracting set: for every triple ( , ,b ) in the domain of R N for allN 1, the orbit {R N (, ,b )} accumulates in the set {(0, 0, 0), (0, 0, 1)}, which isoutside the open cube and represents trivial dynamics.

In this work we are concerned mostly with the parameter space. Some resultsabout the dynamics in the conguration space have been reported in previous works([1, 2, 8, 12, 11]). Our explanation on the conguration space will be restricted tothe assertions that are necessary to understand the parameter space, and this doesnot include, for example, involved combinatorial aspects.

Some of this works also study the renormalization operators for families like this([11, 14, 13]). But they are concerned more with the combinatorics, instead of dif-ferentiable structures of the parameter space.

In Section 2, we establish the dynamical dichotomy of this kind of maps: hiper-bolicity or wandering interval. Every assertion about the conguration space is con-tained in this Section. In Section 3, we list the main facts about the renormalizationoperator and its domain, explicitly writing its formula and calculating its derivative.The assertion that the set of innitely renormalizable maps is a lamination of contin-uous graphs is proved in Section 4. There we also show that this lamination may beextended to a foliation of the whole parameter space and that vertical bers intersectthe lamination in a zero Hausdorff dimension set. Section 5 is dedicated to studying

holonomies between vertical bers, which easily extend to arbitrary transversal berswith the results of the remaining sections. In Section 6 we show that the operatorhas a hyperbolic structure when ( , ) is not so far from (0 , 0), but uniform invariantunstable cone families containing the vertical direction are obtained for the wholecube. In Section 7 we prove that each leaf is C r , for any r 1, and as consequenceof the proof that leaves vary continuously in the C r topology, for every r 1.

2 Attracting periodic orbits versus wanderingintervals

Let {f ,,b },,b with (, ,b ) (0, 1)3 be the family of functions of the interval[b 1, b] given by

f ,,b (x) = b + x , x < 0b 1 + x , x > 0 (1)

(see Figure 1). This expression has the advantage of dealing with a unit intervaland putting the discontinuity point always at the origin. Up to a translation of the

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x-coordinate this is equivalent to a map of [0 , 1] into itself where the discontinuity

point is 1 b and the slopes are the same numbers and .

b

bb

1b 1

0

0

Figure 1: A typical function of the three-parameter family {f ,,b },,b .

We aim at describing the parameter space of this family accordingly to the dy-namical behavior, which is characterized by a dichotomy, as the following argumentsshow.

Accordingly to the value of b, take the smaller interval of [ b 1, 0) and (0 , b].Without loss of generality, we may assume b 12 and look at (0 , b]. Similar argumentswill be clearly valid for b > 12 with the interval [ b 1, 0) (one may think that if b >

12

then the conjugate map g = 1 f , where (x) = x , falls in the case b < 12).

Assuming b 12 , the rst consequence is that f ((0 , b]) [b 1, 0), since f < 1.If f has a negative root z then [z, 0) is a fundamental domain for f restricted to

the negative side. For all x (z, 0), f (x) > 0, and for every x [b 1, 0) there isi = i(x) 0 such that f i (x ) [z, 0). In fact, [0, b) is also a fundamental domain,since f ([z, 0)) = [0 , b).

If f has no negative root then there are two possibilities, although the rst isoutside the parameter space: (i) b = 0, where f i (x ) 0 for every x, or (ii) b > 0but f (x ) > 0 for all x [b 1, 0). In the latter case, f (x) (b 1, 0) for everyx (0, b] and f (x) (0, b) for every x [b 1, 0). This implies that the closureof f 2((0 , b]) is contained in (0 , b] and f 2 |(0 ,b] is a contraction. Hence f 2 |(0 ,b] has a(unique) xed point, and the original map f has a unique periodic orbit of periodtwo, which attracts every orbit.

Note that f < 1 implies that (ii) happens whenever b = 12 . In fact, for every(, ) (0, 1)2 there is = (, ) > 0 such that if b ( 12 ,

12 ] then the same

happens for f ,,b .If z = b 1 is a root of f then the return map to (0 , b] is also a contraction, but

the closure of f 2 |(0 ,b] is not anymore contained in (0 , b]. If we extend f 2 to [0, b] then

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0 is xed point, attracting all orbits. For the original map f , the orbits approach the

pair of points {0, b 1} as if they constituted a period two attracting periodic orbit .Note that 0 (and consequently also b 1) are approximated only by the right side.

We will never be free of this kind of behavior: a rst return map to an openinterval which is a contraction but the xed point is in one of the boundary points.But, under the dynamical point of view, this situation may be regarded as periodichyperbolic, even with no true attracting periodic orbit. We will call this kind of orbitas a false attracting periodic orbit , see below for a better characterization.

Now suppose again that f has a root z (b 1, 0). Then f (b 1) < 0 and[b 1, f (b 1)) is another fundamental domain. Moreover, as f is not onto (sincef < 1) the order b 1 < f (b) < f (b 1)) is always respected. We denote by W theinterval ( f (b), f (b 1)), exactly the maximal interval outside the image of f .

In order to obtain the return map to (0 , b) we follow the orbit of this interval.The rst iterate is ( b 1, f (b)), and let k be the rst integer such that

f k ((b 1, f (b))) (0, b) = (2)

(if f has no negative root then k = 1, but k may be equal to 1 even if f has a negativeroot). As each iterate of ( b 1, f (b)) is contained in some fundamental domain, and[0, b) is also a fundamental domain then there are only two possibilities (see Figures 2and 3):

1. f k ((b 1, f (b))) (0, b)

2. 0 f k ((b 1, f (b)))

b 1 0 b

W

f (b)

f (b 1) f 2(b 1)

f k (b 1)

f k 1(W )

Figure 2: First intersection of an iterate of (0 , b) with itself - rst case.

In case 1., the return map to (0 , b) is simply the restriction of f k +1 to this interval.Since f k+1 is a contraction, it has a unique xed point (at least for the extended map

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b 1 0 b

W

f (b)

f (b 1) f 2(b 1)

f k (b 1)

f k (W )

Figure 3: First intersection of an iterate of (0 , b) with itself - second case.

to [0, b]), and we easily conclude that f has an attracting periodic orbit (true or false)of period k + 1, attracting all orbits except the pre-orbit of 0.

To be more precise, there are essentially two subcases here. If the closure of f k ((b 1, f (b))) is contained in (0 , b) then there is a true attracting periodic orbit.In this case, 0 f k 1(W ) implies that the pre-orbit of 0 is nite, since there are nopre-images of points of W .

The other possibility is that 0 or b are in the closure of f k ((b 1, f (b))). In theformer case, the false attracting periodic orbit is the set of points {b 1, f (b1),. . . , f k 1(b 1), 0}. The cycle is interrupted at 0 since f is not well dened

at this point, but still b 1 = f (0+

). In the latter, the false orbit is the set{b, f (b), f 2(b), . . . , f k 1(b), 0}, and now b = f (0 ). Once again the pre-orbit of 0is nite.

In case 2., there is c (0, b) such that f k+1 (c) = 0. The return map R to (0, b), forpoints x (c, b), is given by R (x) = f k+1 (x) . But as the iterates of ( b 1, f (b)) arealways contained inside fundamental domains then R (x) = f k+2 (x ) for all x (0, c).Then we observe that lim x c+ R (x ) = 0 and lim x c R (x) = b, so that R has theaspect shown in Figure 4.

By an affine change of coordinates we rescale R to a function g dened in [b 1, b]with discontinuity at 0 and extend it to the boundary points b 1 and b. In this casewe say that f is renormalizable and call g the renormalization of f . The function

that takes f to g will be called the renormalization operator R .The renormalization reduces the dynamics to the new map R f , since, except for

the pre-orbit of 0 in the negative side, for every x [b 1, 0) there is n = n (x) suchthat f n x (0, b). Now the same analysis can be done for R f , leading to the samedichotomy: either there is an attracting periodic orbit or R f is renormalizable. Weconclude that either f is N times renormalizable ( N 0) and R N f has an attracting

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00

b

bc

R

f k+1

f k+2

Figure 4: Return map when 0 f k ((b 1, f (b)).

periodic orbit, true or false and attracting all orbits except the nite pre-orbit of 0,or f is innitely (many times) renormalizable.

In order to characterize in detail this dichotomy we introduce some new notationsand denitions. The return maps will not be rescaled, since we need to relate themto the original map f , but we will speak about renormalization even so.

Suppose that f is N times renormalizable, N 0. Let I 0 [a 0, b0] [b 1, b],c0 0 and R0 f . Let also W 0 be the interval W described above, which isthe maximal (non-trivial) interval inside [ b 1, b] such that f (x ) W 0 for all x [b 1, b]. This property immediately implies that the forward iterates of W 0 are pairwise disjoint .

Now suppose that for all 0 < i N the interval I i 1 = ( a i 1, bi 1), ci 1 I i 1,R i 1 : I i 1 \ ci 1 I i 1 (belonging, up to rescaling, to the three-parameter family{f ,,b }), and W i 1, the maximal interval outside the image of R i 1, have beendened. By induction, dene I i as the smaller interval of ( a i 1, ci 1) and ( ci 1, bi 1)(if these intervals were of equal size then, as remarked above, in the beginning of this Section, R i 1 would not be renormalizable, contradicting the hypothesis on N ).Then let ki 1 be the smallest integer such that ci 1 R

k i 1 +1i 1 (I i ) (see Figure 5). The

point ci is dened as the point of I i that satises Rk i 1 +1i 1 (ci ) = ci 1. The map R i is

the rst return map of R i 1 to I i , continuously extended to the boundary points of I i .Finally W i = R

k i 1i 1 (W i 1) is the (maximal) interval inside I i such that R i (x) W i

for any x I i .These denitions could be partially carried on one step further, even if f was

not N + 1 times renormalizable. The interval I N +1 is the smallest of ( a N , cN ) and(cN , bN ) and RN +1 is the rst return map of R N to I N +1 . But if R N +1 is a continuousmap (a contraction), then cN +1 and W N +1 are not dened.

We are ready to prove the following results. First we notice that every point yin the interval has at most one pre-image, but possibly none, since f is injective but

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a i 1 bi 1cici 1

W i 1

W i = Rk i 1i 1 (W i 1)

Figure 5: Induction scheme for return maps.

not onto. Hence every point y has a well dened past orbit y, f 1y, f 2y, . . ., butthis orbit may or not be nite.

Proposition 1. If there is n 0 such that 0 f n W 0 then the pre-orbit of 0 is nite.

Proof. The hypothesis implies f n (0) W 0. If f n (0) W 0 then f (n +1) (0) is notdened. Otherwise, f n (0) = f (b) or f n (0) = f (b 1), hence f (n +1) (0) = b orf (n +1) (0) = b 1, and this implies that f (n +2) (0) is not dened.

Proposition 2. If f is N but not N + 1 times renormalizable, for N < , then there is n 0 such that 0 f n (W 0), implying that 0 has a nite past orbit.

Proof. The hypothesis implies that RN is not renormalizable (accordingly to thenotations and denitions above). Then there is k 0 such that cN R kN (W N ). Nowwe observe, by the inductive construction above, that RkN (W N ) = f

l(W 0), for somel 0, and that f j (cN ) = 0, for some j 0. The assertion follows with n = j + l.

Proposition 3. If f is innitely renormalizable then 0 f n (W 0) for any n 0.

Proof. If N = then the points ci are dened for every i 0. As each ci is a pre-image of ci 1 then the past orbit of c0 = 0 is innite. If 0 belonged to f n (W 0), for somen 0, then the past orbit of 0 would be nite, by Proposition 1, contradiction.

Proposition 4. If f is N but not N + 1 times renormalizable, N < , then the (true or false) attracting periodic orbit is unique and attracts every orbit except the nite past orbit of 0.

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I 1 , I 2 , . . ., accumulating on b = 0, such that b I

k implies

f k (b 1) < 0 < f k (f (b)) . (4)

We denote the boundary points of I k by bE ,k and b

I ,k , where b

I ,k is the nearest to

b = 12 (I from internal, E from external). Therefore the parameter bI ,k is the

solution of the equationf k (b 1) = 0 (5)

and bE ,k is the solution of the equation

f k (f (b)) = 0 . (6)

0

12

b

I 1

I 2

I 3

I 4I

5

Figure 7: Renormalizable parameters for xed and .

These two equations can be explicitly stated for the piecewise linear family. Asb 1 and f (b) = b 1 + b are negative, the iterates are done only in the left side(with respect to the discontinuity point). It is easy to prove by induction that f k (x)(k 1), in the left side, is given by the expression

b

k 1

i=0

i+

kx . (7)

The solution of Equation (5) is then given by

bI ,k = k

k

i=0

i 1

(8)

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and of Equation (6) by

bE ,k = k

k

i=0

i + k 1

. (9)

The intervals where f is renormalizable, for b > 12 , are named I +1 , I

+2 , . . . , with

boundary points bI + ,k and bE + ,k . The value of the boundary points are obtained by

interchanging and , and subtracting the result from 1. Therefore

bI + ,k = 1 k

k

i=0

i 1

(10)

and

bE + ,k = 1 k

k

i=0

i + k 1

. (11)

Now we consider the domain of the renormalization operator as a whole. Let

D k =(, ) (0 ,1)2

{ } { } I k (, ) , (12)

where I k (, ) denotes the dependence of I k on (, ), as given in Equations (8),

(9), (10), and (11). These are the connected components of the domain of R . Un-

derstanding this set of connected components helps a lot in future estimates. InFigure 8, we show bI ,1, bE ,1, b

I + ,1 and b

E + ,1.

Below we list a collection of facts about these leaves that can be easily deducedfrom the formula. We let to the readers imagination the complete drawing of allleaves.

1. bI ,k does not depend on , as well as bI + ,k does not depend on .

2. bI ,k is an increasing function of , equal to 0 for = 0 and equal to 1k+1 for

= 1; bI + ,k is a decreasing function of , equal to 1 for = 0 and equal to1 1k+1 for = 1.

3. The derivative of bI,E ,k (resp. bI,E + ,k ) at = 0 (resp. = 0) is equal to zero for

all k 2 and equal to 1 (resp. 1) for k = 1.4. bI ,1 and b

I + ,1 coincide at ( , ) = (1 , 1) (and only there).

5. For all k 1, bE ,k and bI ,k +1 coincide at ( , ) = (1 , 1) (and only there).

6. For all k 1, bI ,k and bE ,k coincide at ( , 0) and (0 , ) (like two square sheets

of paper glued together by two adjacent edges).

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Figure 8: Boundaries of D 1 and D+1 .

7. The size of I k attains its maximum value

1(k + 1)( k + 2)

(13)

at ( , ) = (1 , 1).

The assertions above show, in particular, that the parameters regions for whichthere is an attracting periodic orbit disappear as ( , ) tends to (1 , 1). This is inagreement with the fact that rotations do not have attracting periodic orbits.

3.2 Distance between componentsIn Section 5 we shall need some lower and upper bounds on the distance betweenlayers of the renormalization operators domain. The distances are estimated alongvertical bers.

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Lemma 1. There is a positive function C (, ), monotonically increasing in both

arguments, such that

C (, ) 1 |b2 b1 |

l2 C (, ) , (14)

whenever (, ,b 1) D l1 , (, ,b 2) Dl2 and l1 > l 2 1, and

C (, ) 1 |b2 b1 |

l2 C (, ) , (15)

whenever (, ,b 1) D +l1 , (, ,b 2) D+l2 and l1 > l 2 1. Moreover, if l1, l2 1,

(, ,b 1) D l1 , and (, ,b 2) D+l2 then

C (, ) 1 | b1 b2 | 1 (16)

Proof. A straightforward calculation with Formulas (8), (9), (10), and (11) showsthat

1.

bE ,l 2 bI ,l 1

(1 )(1 )1 + 11

l2 , (17)

2.

bE + ,l 2

bI + ,l 1

(1 )(1 )

1 + 11 l2 , (18)

3.bI ,l2 b

E ,l 1

11

+ l2 , (19)

4.bI + ,l 2 b

E + ,l 1

11

+ l2 , (20)

and Inequalities (14) and (15) follow. For Inequality (16), |b1 b2 | is at least thedifference |bI + ,1 bI ,1 |, which is equal to

1 1 +

1 +

. (21)

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3.3 Formula and derivativeFirst we consider renormalization for a xed pair ( , ), for b inside an interval I k ,for some xed k 1. The renormalization R f is nothing more than the rescalingof the rst return map to the smaller interval of [ b 1, 0) and (0 , b]. The rescaledfunction must have discontinuity point at 0 and domain of size equal to one. As f ispiecewise linear then so is R f . The new slopes ( , ) are ( k+1 , k ), in the caseof b I k , and (

k , k+1 ), in the case of b I +k .In fact, it is easy to see that R sends the set { }{ } I k onto {

}{ } [0, 1].In order to obtain the explicit formula b b one must calculate the discontinuitypoint c1 of the return map. In the case that b I k , c1 is the solution of theimplicit equation f k (f (x )) = 0, for x > 0, where f (x) = b 1 + x , and f k (f (x)) =b k 1

i=0 i + k (b 1 + x ), accordingly to Equation 7. Then b = b c1

b , which after

manipulation gives

b = 1 1 b 1 (bI ,k ) 1 = 1 + ( k ) 1

k

i=0

i 1b 1 . (22)

If b I +k an analogous reasoning gives

b = 1 (1 b) 1 (1 bI + ,k ) 1 = ( k ) 1

k

i=0

i + 1(1 b) 1 (23)

In both cases the transformation is orientation-preserving, but not affine!

Denition 1. For = , k 1 and (, ) (0, 1)2 we call ,k, the one-variable function that sends b I k (, ) to b (0, 1), as above, where I

k (, ) indicates the

dependence of I k on and .

The derivative of R depends on the domain D k to which belongs. For Dk we

have

D R () =(k + 1) k k+1 0

k k 1 k 0 1 (b

I ,k )

1 2(b 1 (bI ,k ) 1) 1b 2

. (24)

By using symmetry, we will skip writing the explicit expression of D R () for D +k .It must be remarked that the derivative with respect to b of the third component

of R gives the expansion along vertical bers. Then D ,k, , the derivative of Rrestricted to {(, )} I k , is bounded between

1(bI ,k ) 2 and 1(bE ,k )

2 (inparticular greater than 4, since < 1 and bI ,k

12 ). By Equations (8) and (9), this

gives

1 2k D ,k, 1 2k

21

2. (25)

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(see Denition 1). In this way, we construct a sequence of nested intervals J 0 J 1

J 2 . . ., where

J n +1 = n ,k n n , n

0 ,k 0 0 , 0

n(I n +1kn +1 ( n +1 , n +1 )) . (39)

But by Inequation ( 33), |J n | 0, and the claim is proved.In particular, the intersection of { 0}{ 0} (0, 1) with the invariant lamination

is a totally disconnected set .

4.2 Foliation of the whole parameter spaceNow we aim at proving that the lamination just described can be extended to afoliation in the whole cube (0 , 1)3. Roughly speaking, the foliation will be dened bythe relative position of the xed point inside the interval which is the domain of thereturn map.

Let us rst dene the foliation outside the domain of the renormalization oper-ator. We start by the region comprised between the layers D 1 and D

+1 . For these

parameters there is an attracting periodic orbit of period 2. The position of therightmost point of this orbit is the solution of the equation

b + (b 1 + x ) = x . (40)

Dividing the solution x by b leads to the relative position of this point inside [0 , b]:

xb =

(1 + ) b 1

1 . (41)

We note that xb = 0 for b = bI ,1 and

xb = 1 for b = b

I + ,1. Moreover,

xb is strictly

increasing for b [bI ,1, bI + ,1], hence for xed (, ) and given 0 t 1 there is oneand only one b in this interval such that xb = t, dening, thus, for each t, a graphbt (, ).

Now we do the same for the region comprised between D k and Dk +1 , for k 1.

For these parameters there is an attracting periodic orbit of period k + 2, and theunique point x of this orbit in [0, b] satises

x

b = (1 k+1 ) 1

k+1

i=0

i k+1 b 1 . (42)

The same reasoning applies in this case (and also for the regions between D +k andD +k+1 ).

Inside the domain of R one proceeds as in the construction of the invariant lam-ination, by pulling back the foliation above under a nite number of iterates of R .This leaves are analytic since R is analytic.

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4.3 Bounded distortionWe recall that ,k, is always expanding (see Subsection 3.3). This fact will be usedto show that the composition of such functions has uniformly bounded distortion,where uniformly means independent of the composition. The bounded distortionwill be used in Subsection 4.4 and Section 5.

Here one must be careful because in general expansion almost automatically im-plies bounded distortion. But this is so since in many situations the logarithm of thederivative is Lipschitz or Holder continuous. In the present case, however, both arefalse.

In the next Lemma, we let m = m ,k m m , m and I m = I mkm , m 1, and =

n 0, and show that the distortion of is uniformly bounded.

Lemma 3. Let = n 0, as above, in such a way that maps J n onto (0, 1).Then log D (b)D (b) < 6 , (43) for any b,b J n .

Proof. Let b0 = b, b0 = b, bm = m 1 0(b) and bm = m 1 0(b), form = 1 , . . . , n + 1. We havelog

D (b)

D (b)

n

m =0

log D m (bm ) log D m (bm )

n

m =0max

I m|D log D m | |bm bm |

By the Mean Value Theorem there is, for each m , a point m (bm , bm ) such that

|bm bm | = |bn +1 bn +1 |

D (n m )( m ) min

I m +1D (n m +1 ) min

I mD m

1. (44)

As remarked in Subsection 3.3, the derivative of each m is always greater than 4,thus

minI m +1 D (n m +1 ) 4n m . (45)

Moreover if m = (the case + is analogous) then

minI m

D m = 1m (bI ,k m )

2 . (46)

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hence, by Lemma 3, |J | c|J | |I |, for some uniform constant c > 0 (Lemma 3

assures c = e6, but the value does not matter). Therefore

H ( U (n +1) ) =J U ( n +1)

|J | c

J U ( n )|J |

I = I k ( n +1 , n +1 )

|I | (51)

The pair ( n +1 , n +1 ) depends on ( 0, 0) (the position of the ber) and J (whichimplictly gives the combonatorics). But Inequation (29) gives the uniform estimate(independent of J ) max{ m , m } ( 0 0)m , for all m = 1 , . . . , n + 1. This impliesthat given > 0 there is n0 = n0( 0, 0) such that n n0 implies that n +1 and n +1 are smaller than (independently of J ).

On the other hand, if I k = I k (, ) then Inequation (26) implies |I

k |

2k

and |I +k | 2k . Therefore

{|I | ; I = I k (, )}

k=1

( 2k ) +

k=1

( 2k ) 2

1 2 +

2

1 2 . (52)

Now there is > 0 such that if and are smaller than then this sum is smalleror equal than 12 c

. Therefore if n n 0 then

H ( U n +1 ) 12

J U ( n )|J | =

12

H ( U (n )) . (53)

5 HolonomiesLet D be the set of innitely renormalizable parameters in the open cube, whichwe know to be formed by a collection of graphs of two-variable functions L(s )(, ),where each s is the combinatorics that denes the future orbit under iterations of the renormalization operator R . Let and be two (in principle topologically)transversal sections to the invariant lamination, with the additional condition thatthey intersect each leaf at one and only one point. The holonomy map from Dto

D is the function that takes ( , ,b ) in (, , b), with the following rule:

take the combinatorics s such that b = L(s )(, ) and let ( , , b) be the uniqueintersection point between and the graph of L(s).Transversality is not well dened at this point of the exposition, since we havenot proved that the functions L(, ) are differentiable. But the essential aspectsof this section can be discussed with vertical sections , which will be the most natu-ral transversal sections, since the vertical direction is invariant (it is the expandingdirection of the hyperbolic structure that will be constructed in Section 6).

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Hence we will discuss Holder continuity for vertical sections, and in fact it is not

so difficulty, by using arguments of a general nature, to generalize these results togeneral transversal sections, assuming for grant the results of Section 7.

5.1 H older continuity of holonomiesThe following Lemma says that holonomies between two given vertical bers areHlder continuous.

Proposition 6. Let ( 0, 0), ( 0, 0) (0, 1)2 be xed. There are constants M and , depending on 0, 0, 0, and

0 such that the following is true. For any pair of

combinatorics s1 and s2, if b1 = L(s1)( 0, 0), b2 = L(s2)( 0, 0), b1 = L(s1)( 0, 0),

and b2 = L(s2)( 0, 0), then

|b2 b1 | C |b2 b1 | . (54)

Proof. Let n 0 be such that s1 and s2 coincide up to the n -th position and let =n 0 (resp. = n 0) be the function that sends J U

(n )( 0, 0) (resp.J U (n ) ( 0, 0)) onto (0 , 1), following the notation of Section 4. Let ( m , m ) (resp.( m , m )) be the projection of the m -th iteration of J (resp. J ), m = 0 , 1, . . . , n + 1,and ( m , km ) the common symbols of the trajectories of J and J , m = 0 , 1, . . . , n .There is still the case where s1 and s2 differ already in the rst symbol, i.e. thepoints belong to different layers of the renormalization operators domain. In thiscase the reasoning is completely analogous and still much simpler. This will be calledthe n = 1 case.

Let > 0 be such that 0 2 0 and 0

2 0 . We claim that the same relation

is valid for the iterations: m 2 m and m

2 m . But this comes directly by

induction from the inductive denition of m , m , m , and m .We want to analyse |b1 b2 ||b1 b2 |

(55)

and prove that this quotient is bounded by a constant M depending only on 0, 0, 0,and

0. Observe that |(b1) (b2)| = |D ( )| | b1 b2 |, for some (b1, b2), and

|(b1) (b2)| = |D ( )| | b1 b2 |, for some

(b1, b2). Then, using the Chain Rule,

|b1 b2 ||b1 b2 |

|(b2) (b1)||(b2) (b1)|

(max |D n |) . . . (max |D 0 |)

(min |D n |) . . . (min |D 0 |)(56)

(in the n = 1 case the second factor of the right hand side disappears and the rstfactor is exactly equal to the left hand side).

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The derivative of each m is the bpartial derivative of the third component of

the renormalization operator, see Subsection 3.3. Thus, in the case that m = andkm = k,

(max |D m |)

min |D m |=

1m (bE ,k ) 2

1m (bI ,k ) 2, (57)

(where in this equation bE ,k = bE ,k ( m , m ) and b

I ,k = b

I ,k ( m , m )), and this, in

turn, is bounded by

m m

m m

2k 11 m

+ m m2

, (58)

using Formulas (8) and (9). If we call

M 1 = M 1( 0, 0) = (1 0) 1 + (1 0) 1 + 0 02 (59)

and pay attention to the choice of then we obtain the bound M 1 ( m 2m ) , which

is still bounded byM 1 ( 0 0)

m , (60)

by the remarks of Subsection 3.4. The fact that M 1 works also comes from theproperty, discussed in that Subsection, that m and m are smaller than 0 and 0.Moreover, the choice of M 1 is symmetric in and , in such a way that for m = +we obtain exactly the same bounds.

Now there is m 0 1 such that if m m 0 then M 1 ( 0 0)m 1, and m 0 depends

only on 0, 0 and . This implies that the second factor of the right hand side of Inequation (56) is bounded by a constant that depends only on 0, 0, 0, and 0.The rst factor of the right hand side of Inequation (56) has to be analysedaccordingly to the three following possibilities: (i) ( b1) < 12 and ( b2) < 12 ; (ii)(b1) > 12 and ( b2) >

12 ; (iii) (b1)

12 (b2)

12 < 0. The position with

respect to 12 of (bi ) follows the position of ( bi ), i = 1 , 2. In the rst case, as ( b1)and (b1) are in a layer that is different from the layer to which ( b2) and (b2)belong, Lemma 1 implies that there is l 1 such that|(b2) (b2)||(b2) (b1) |

C ( m +1 , m +1 )

lm +1

C ( m +1 , m +1 ) l m +1(61)

By the choice of , lm +1

2l m +1 . Therefore, using the monotonicity property of

C (, ) stated in Lemma 1,

|(b2) (b2) ||(b2) (b1)|

C ( 0, 0) C ( 0, 0) . (62)

In the second and third cases this bound works as well.

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In fact, the following is a corollary of the proof above.

Corollary 1. For any set S (0, 1)2 such that the closure of S is contained in (0, 1)2there are constants = (S ) > 0 and M = M (S ) > 0 such that the same statement of Lemma 6 holds for any pair of base points ( 0, 0) and ( 0, 0).

Observe that there is no chance of obtaining uniform H older continuity amongholonomies with base points in (0 , 1)2. Even uniform continuity is not allowed. Forexample, if ( 0, 0) is very near (1 , 1) and ( 0, 0) is very near (0 , 0), take b1 I 1 andb2 I +1 . Given > 0, there is a choice of ( 0, 0) and b1, b2 such that |b1 b2 | < .But if ( 0, 0) is sufficiently near (0 , 0), then |b1 b2 | > 1 .

In Section 7 we will show that each leaf of the lamination of innitely renormal-izable parameters is of class C 1, and also that leaves vary continuously in the C 1

topology. The former assertion allows considering transversal sections to the lamina-tion, from the differentiable point of view, while the latter says that near a transversalsection every crossing is also transversal.

As we will see in Section 6, the vertical bers are always transversal to the lamina-tion, since the unstable cone eld may be constructed in order to contain the verticaldirection, so they constitute a particular case of transversal sections.

With this is in mind, assuming for grant the results of Section 7, it is an exer-cise to show that the Holder continuity of holonomies may be extended to any pairof transversal sections. However, the statement must be done only locally, in theneighbourhood of the crossing.

In the same spirit, the result of Subsection 4.4 can be generalized to generaltransversal sections.

5.2 Not more than H olderThe results of the previous Subsection may be proved to be the best possible for thislamination. We will show that there are plenty of ber pairs where there is 0 < < 1such that the holonomy or its inverse is not ( , C )-Holder, for any constant C > 0.

For that purpose, we need lower bounds, in opposition to the previous Subsection.For example,

|b1 b2 ||b1 b2 |

|(b2) (b1) ||(b2) (b1)|

(min |D n |) . . . (min |D 0 |)

(max |D n |) . . . (max |D 0 |) (63)

is the analogous of Inequation (56). We have

(min |D m |)

max |D m |

m m

m m

2km

1 1

1 m + m m2 . (64)

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using Formulas (8) and (9). With the same constant M 1 there dened, we get

(min |D m |)

max |D m | M 11

m m

m m

2km. (65)

Once more, using Lemma 1,

|(b2) (b1) ||(b2) (b1)|

C 1 P (66)

where C 1 = C ( n +1 , n +1 ) 1C ( n +1 , n +1 ) and P is equal to 1, to ln +1

ln +1 for

some l 1, or to

ln +1

ln +1 , for some l 1. By putting everything together, taking

into account the remarks of Subsection 3.4, we conclude that there are powers l and j such that

|b1 b2 ||b1 b2 |

C 1M (n +1)1 0 0

l 0 0

j

. (67)

Also by Subsection 3.4 the powers l and j must be greater or equal than 2 n 1,therefore a uniform upper bound for |b1 b2 ||b1 b2 | cannot be found if, for example, 0 >

0

and 0 > 0 . These conditions can be achieved whenever 0 > 0, 0 > 0 and issufficiently near 1.

6 Invariant cones and hyperbolic structureNow we show that the renormalization operator has an invariant cone family withexpansion in the unstable cones, and for a certain restricted region (which is in thefuture of every orbit) there is in fact a hyperbolic structure, also with contraction inthe stable cones.

For each = ( , ,b ) belonging to the domain of the renormalization operatorwe associate a cone

C c() = {(v1, v2, v3); |v3 | c max{|v1 | , |v2 |}} , (68)

where c is a constant to be chosen. In fact the cones are equal everywhere, but we

keep this notation to remember that C c() is in the tangent space of .We will show that, for a suitable value of c, D R () C c() is strictly contained in

C c(R ), meaning that if |v3 | c max{|v1 | , |v2 |} and (v1, v 2, v 3) = D R () (v1, v2, v3)then |v3| > c max{|v1 | , |v2 |}.

Lemma 5. For any in the domain of R , if v belongs to C c() with c 6 then v = D R () v belongs to the interior of C c(R ).

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Proof. Firstly we observe that

max{|v1 | , |v2 |} (k + 2)max {|v1 | , |v2|} (69)

(this estimate could not be much better without using and ). Then

|v3 | 1b 2|v3 | 1

(bI ,k )

1 |v1 | 2 b 1 (bI ,k ) 1 |v2 | , (70)

that is to say,

|v3| 1b 2c 1

(bI ,k ) 1 2 b 1 (bI ,k )

1 max{|v1| , |v2 |}

k + 2 . (71)

As b I k we have

b 1 (bI ,k ) 1 (bE ,k )

1 (bI ,k ) 1 = , (72)

using Equations (8) and (9). Now

|v3 | 1 b 2c

(bI ,k ) 1 1

1k + 2

max{|v1 | , |v2 |} , (73)

but the factor 1 may be ignored, since < 1. Lemma 2 says that b 2 (b

I ,k )

1 , therefore it is enough to nd c such that b 2(c 1) 1 > c (k + 2),

for all k 1. Now b I k implies b 1k+1 , so that it suffices to nd c such that(k + 1) 2(c 1) 1 > c (k + 2), for all k 1. As any c 6 works well, this proves theLemma.

Now we examine how the derivative of the renormalization operator acts on vec-tors of the cones.

Lemma 6. For any in the domain of R and v C c(), with c 6,

v > 3 v . (74)

Proof. For c 6, if v C c() then v = |v3 | and as v C c(R ), by Lemma 5,then v = |v3 |. By Inequation (70) and Lemma 2,

|v3 | 1

cb 2(c 1) 1 |v3 | , (75)

using that |v1 | and |v2| are smaller than |v3 |. In particular, |v3 | > 3|v3 |, since b 12 .

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If we restrict our attention only to the intersection of the domain of R with the

regionQ {(, ,b ); , c . (79)

In this region b is at most equal to bI ,1 = 1+ , which is smaller than

110 , thus

b 2 > 100. It is sufficient to take c such that 4999 c > 5050, for example c 1.1 isenough.

The next Lemma assures that inverse branches expand vectors in the comple-ment of the unstable cone, in the restricted region Q, characterizing the hyperbolicstructure.

Lemma 8. If Q and v = D R () v belongs to the complement of C c(R ), with c = 1 .1, then

v 25

v . (80)

Proof. As v belongs to the complement of C c(R ), c = 1 .1, then

v < 1.1max{|v1 | , |v2 |} 1.

But for the induction on n we must rst do the case n > 0 and then the case n > 1.We will inductively dene the constant C r > 0 such that

( 0 0)k 1 p j (k) C r (88)

for all k 1 and j r .

7.4 Implicit equationOur aim is at obtaining a formula for n +1 ,w r n,w r .

The rst symbol s0 = ( 0, k0) of the combinatorics s determines the layer of the

domain of R in which (, ,

n (,

)) is contained, for all n

1. We denote k

= k

0for the sake of practicity. Remark that, once the rst symbol s 0 is xed n 1, thenthe point ( , , n (, )) is sent to the point ( , , n 1(, )), where =

k+1

(resp. = k ) and =

k (resp. = k+1 ), for 0 = (resp. 0 = +). Note

that the symbol s1 = ( 1, k1) = ( , k) may be anything, whatever be the assumptionon s0. If n 2 then n 1(, ) I

k

(, ).From now on we adopt the notation

R (, ,b ) = ( R 1(, ,b ), R 2(, ,b ), R 3(, ,b )) . (89)

This means that

= R 1(, , (, )) (90) = R 2(, , (, )) , (91)

but as the right hand side functions are independent of the third coordinates we maywrite

(, ) = ( R 1, R 2) = ( R 1(, ), R 2(, )) . (92)

Therefore the fundamental equation to begin with is the relation

n 1(R 1(, ), R 2(, )) = R 3(, , n (, )) . (93)

In order to prove Inequations (86) and (87) we need to estimate |n +1 ,w r n,w r |,for every w r W r . The starting point will be Equation (93).

We denote = n +1 and = n , = n and = n 1, and recall Equation(93):(R 1(, ), R 2(, )) = R 3(, , (, )) (94)(R 1(, ), R 2(, )) = R 3(, , (, )) (95)

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For the sake of convention let the left sides be written as

()(, ) = (R 1(, ), R 2(, )) (96)()(, ) = (R 1(, ), R 2(, )) . (97)By differentiating both sides of Equation (94) with respect to and we obtain() = 1 2 1(bI ,k )

2(bI ,k ) (98)

and() = 1 2 + 2 1 2(bI ,k )

1 . (99)

These equations can be rewritten as

()w = 2w + (w) 1 1 + , (100)

where = (bI ,k ) 2(bI ,k ) and (w) = 0 i f w = or =

1(bI ,k ) 1 and

(w) = 1 if w = . As 2w = ( 1)w , then

()w ()w = ( 1 1)w + (w) 1 1 1 1 . (101)

The difference w r w r will appear after writing wr = ww r 1, differentiating bothsides of Equation (101) with respect to wr 1, and isolating the desired term. Thedifferentiation gives

()w ()w w r 1 = ( 1

1

)w r + (wr

) ( 1

1

)w r 1 ( 1

1

)w r 1 ,(102)

where (wr ) = 0 if wr = w 2 . . . w r and (wr ) = 1 if w r = w2 . . . w r .Understanding ( 1)w j is a crucial step. We do not give a formula, but ratter

what it looks like. By induction, it may be easily proved that

( 1)w j = j

l=1

(l+1) w 1 ... w l = w j N (w j ; w1 , . . . , w l )w 1 . . . w l , (103)where N (w j ; w1 , . . . , w l ) is an integer and the indices 1, . . . , l are all strictly pos-itive. When l = 1 the decomposition of w j is unique and N (w j ; w j ) = 1, i.e. thecorresponding term in the sum is

2w j . (104)

In the subtraction ( 1 1)w r this term (for j = r ) gives

2w r 2w r = 2(w r w r ) + w r ( 2 2) . (105)

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Isolating w r w r we obtain

w r w r = T 1 + T 2 + T 3 + T 4 , (106)

where

T 1 = 2 ()w ()ww r 1

T 2 = 2w r 2 2

T 3 = 2r

l=2 w 1 ... w l = w rN (wr ; w1 , . . . , w l )( (l+1) w 1 ...w l (l+1) w 1 ...w l )

T 4 = (wr )2 1 1 w r 1 1 1 w r 1

The term T 3 is zero when r = 1.

7.5 Proof of the inductionThe following step is to give an estimate of |w r w r |, based on Equation (106),using the inductive hypotheses described in Subsection 7.3.

The proof will be done in the case that 0 = , remarking that the other situationis analogous. However, the induction hypotheses must be taken in the complete form,that is, with Inequalities (86) and (87), since s1 = ( 1, k1) = ( , k) may be anything,as previously remarked.

The case n = 0 . In this case = 0 is identically equal to 0 or 1, hence w r = 0,for all w r W r , r 1. Moreover = 1 G 1 and

|w r w r | = |w r | . (107)

The hypothesis (, ) I k (, ) implies = bI ,k or = b

E ,k , that is

(, ) = k gk (, ) , (108)

where gk (, ) = (1 + + . . . + k ) 1 or gk (, ) = (1 + + . . . + k + k ) 1,accordingly to Equations (8) and (9).

The series i=0

i has convergence radius equal to 1 and converges to (1 ) 1.It is a known fact from elementary theory of power series that the derivatives of apower series is the series of the derivatives and has the same convergence radius. Thisimplies, in particular, that

ki=0

i converges, uniformly in [0 , 0], to

(1 ) 1, for all 1. Also k uniformly converges to 0 in the C r topology, forany r 1, thefore in any case gk (, ) 1 uniformly converges to (1 ) 1 in the

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C r topology. This implies that gk (, ) uniformly converges to (1 ) in the C r

topology.From this, we conclude that there is a constant M r = M r, 0 , 0 such that

|gk,w j (, )| M r , (109)

for all w j W j , 0 j r , k 1, and (, ) [0, 0] [0, 0].Now we evaluate the derivatives of , using the product rule given by Equation

(83). We have

|w r | M rr

i=0 w i w r( k )w i . (110)

If wi has a then ( k )w i = 0 (this surely happens if i > r ). The same conclusionis true if i > k .

We examine the terms where wi is only made of s and i k, dividing into twocases. First, if k r then we see

( k )w i k(k 1) . . . (k i + 1) k i k! k r . (111)

On the other hand, if k > r then

( k )w i k(k 1) . . . (k i + 1) k i k(k 1) . . . (k r + 1) k r , (112)

Then ( k )w i q r (k) k r , where

q j (k) = k! if k jk(k 1) . . . (k j + 1) if k > j (113)

We observe that r r implies q r (k) q r (k) and that there is a total of 2 r termsin the sum (the sum of the coefficients in the binomial), in such a way that

|w r | 2r M r q r (k) k r . (114)

We call pr, 0 2r M r q r (k) . (115)

Inductive step: common estimates. If (, ) I k (, ) then

k , (116)

which directly comes from Equation (8). If in addition (, ) I k (, ) then, byEquations (8) and (9),

bI ,k

,bI ,k

,

,

bI ,kbE ,k

1 + k 2 . (117)

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On the other hand, (, ), (, ) I k (, ) implies , bE ,k , hence

1, 1 k (1 + + . . . + k + k ) (k + 2) k (118)

(taking into account that 0 and 0, we could obtain 1, 1 C k ,where C = C ( 0, 0) = 0 0 + (1 0) 1, but this would make any difference in ourarguments). If moreover


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where 1 denotes21 =1 and 1 , 2 denotes

21 =1

22 =1 . It may be easily proved

by induction that ( )w1 ...w r is given byr

l=1 1 ,..., l

1 ... l w 1 ... w l = w rN (wr ; w1 , . . . , w l )R 1 ,w 1 . . . R l ,w l , (125)where each N (w

r ; w1 , . . . , w l ) is an integer and each i is strictly positive. Theterm with l = r can be more explicitly written as

1 ,..., r

1 ... r R 1 ,w 1 R 2 ,w 2 . . . R r ,w r . (126)

As each term R 1 ,w 1 . . . R t ,w l depends only on and , it follows that()w r ()w r = (127)

r

l=1 1 ,..., l w 1 ... w l = w rN (wr ; w1 , . . . , w l )( )1 ... l R 1 ,w 1 . . . R t ,w l ,

which leads us to estimating the terms R 1 ,w 1 . . . R t ,w l , in conjunction with ( )1 ... l , bounded by the induction hypotheses. Recalling the notation r = # (wr ),r = # (wr ), we add the (local) notation = 1 . . . l and # () (resp. # ()) forthe number of s (resp. s) in the word .

In our notation, R 1(, ) =

= k+1 and R 2(, ) =

= k . We have

R 1,w i = 0 and R 2,w i = 0 if w i has at least two s. If each w i has at most one ,then exactly r of the w i s have a and l r do not have. Then R 1 ,w 1 . . . R l ,w lhas l r in its expression. The power of , in turn, has the following contributions:kl , coming from each one of the l terms; # (), coming from the i s that are equalto 1 (or ); r , coming from the r differentiations with respect to . Each differ-entiation with respect to also carries a multiplicative factor that is at most k + 1.Therefore we obtain

R 1 ,w 1 . . . R l ,w l (k + 1)r l r kl +# () r . (128)

The estimate of |(

)1 ... l | comes from induction. There are two options,

depending on . If = , then it is p(k)( k )n 1 k # () # () , (129)and if = + then it is p(k)( k )n 1 # () k # () , (130)

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2. Once pr,n (k) is dened, we are tasked with choosing a constant C r,n such that

( 0 0)k 1 pr,n (k) C r,n (160)

as a function of the previous constants C r,m , m n 1. Let C r be such that( 0 0)k 1 pr (k) C r (161)and C r be such that( 0 0)k 1(k + 1) r C r . (162)

Multiplying ( 0 0)k 1 by the expression of pr,n (k), given in (159), we get somethingsmaller than

C r,n ( 0 0)n 1C r C r,n 1 + 16n 1

m =0( 0 0)m C r,m + C r . (163)

3. Now assume that pr,n (k) and C r,n have been inductively dened by the expres-sions above, for all n. We will show that there is a constant C r such that

C r,n C r , n 0 . (164)

First, take n0 2 such that

0

0)n 0 1

max{C

r, 16 0 01 0 0 }

< 1

4 .

(165)

Take C r such that (i) C r < 14 C

r and (ii) 16

n 0 1m =0 C r,m

renormalization operator for affine dissipative lorenz maps colli

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