Invariant measures for interval maps without Lyapunov exponents
aa r X i v : . [ m a t h . D S ] F e b INVARIANT MEASURES FOR INTERVAL MAPS WITHOUTLYAPUNOV EXPONENTS
JORGE OLIVARES-VINALES
Abstract.
We construct an invariant measure for a piecewise analytic intervalmap whose Lyapunov exponent is not defined. Moreover, for a set of fullmeasure, the pointwise Lyapunov exponent is not defined. This map has aLorenz-like singularity and non-flat critical points. Introduction
Lyapunov exponents play an important role in the study of the ergodic behaviorof dynamical systems. In particular, in the seminal work of Pesin (referred to as“Pesin Theory”), the existence and positivity of Lyapunov exponents were used tostudy the dynamics of non-uniformly hyperbolic systems, see for example [KH95,Supplement]. Using these ideas, Ledrappier [Led81] studied ergodic properties ofabsolutely continuous invariant measures for regular maps of the interval under theassumption that the Lyapunov exponent exists and is positive. Recently Dobbs[Dob14], [Dob15] developed the Pesin theory for noninvertible interval maps withLorenz-like singularities and non-flat critical points. Lima [Lim20] constructs asymbolic extension for these maps that code the measures with positive Lyapunovexponents.In the case of continuously differentiable interval maps, Przytycki proved thatergodic invariant measures have nonnegative Lyapunov exponent, or they are sup-ported on a strictly attracting periodic orbit of the system. Moreover, there existsa set of full measure for which the pointwise Lyapunov exponent exists and isnonnegative, see [Prz93], [RL20, Appendix A].In this paper, we show that the result above cannot be extended to continuouspiecewise differentiable interval maps with a finite number of non-flat critical pointsand Lorenz-like singularities. In particular, we construct a measure for a unimodalmap with a Lorenz-like singularity and two non-flat critical points for which theLyapunov exponent does not exist. Moreover, for this map, the pointwise Lyapunovexponent does not exist for a set of full measure. Thus, our example shows thatthe techniques developed by Dobbs [Dob14], [Dob15], and Lima, [Lim20], cannotbe extended to all maps with critical points and Lorenz-like singularities.Maps with Lorenz-like singularities are of interest since they appear in the studyof the Lorenz attractor, see [GW79], [LT99], and references therein. Apart fromthese motivations, these types of maps are of interest on their own since the pres-ence of these types of singularities create expansion, and hence enforce the chaoticbehavior of the system, see [ALV09], [LM13], [Dob14], and references therein.
This article was partially supported by ANID/CONICYT doctoral fellowship 21160715.
Additionally, the unimodal map that we consider has Fibonacci recurrence ofthe turning point (or just Fibonacci recurrence). Maps with Fibonacci recurrencefirst appeared in the work of Hofbauer and Keller [HK90] as possible interval mapshaving a wild attractor. Lyubich and Milnor [LM93] proved that unimodal mapswith a quadratic critical point and Fibonacci recurrence do not only have anyCantor attractor but also have a finite absolutely continuous invariant measure,see also [KN95]. Finally, Bruin, Keller, Nowicki, and van Strien [BKNvS96] provedthat a C − unimodal interval map with a critical point of order big enough and withFibonacci recurrence has a wild Cantor attractor. On the other hand, in the workof Branner and Hubbard [BH92], in the case of complex cubic polynomials, and thework of Yoccoz, in the case of complex quadratic polynomials, Fibonacci recurrenceappeared as the worst pattern of recurrence, see for example [Hub93], and [Mil00].Maps with Fibonacci recurrence also play an important role in the renormalizationtheory, see for example [Sma07], [LS12], [GS18], and references therein.1.1. Statement of results.
In order to state our main result, we need to recallsome definitions. A continuous map f : [ − , → [ − ,
1] is unimodal if there is c ∈ ( − ,
1) such that f | [ − ,c ) is increasing and f | ( c, is decreasing. We call c the turning point of f . For every A ⊂ [ − ,
1] and every x ∈ [ − , distance from x to A by dist( x, A ) := inf {| x − y | : y ∈ A } . We will use f ′ to denote the derivative of f . We will say that the point c ∈ [ − , Lorenz-like singularity if there exists ℓ + and ℓ − in (0 , L >
0, and δ > x ∈ ( c, c + δ )(1.1) 1 L | x − c | ℓ + ≤ | f ′ ( x ) | ≤ L | x − c | ℓ + , and for every x ∈ ( c − δ, c )(1.2) 1 L | x − c | ℓ − ≤ | f ′ ( x ) | ≤ L | x − c | ℓ − . We call ℓ + and ℓ − the right and left order of c respectively. For an interval map f ,a point ˆ c ∈ [ − ,
1] is called a critical point if f ′ (ˆ c ) = 0. We will say that a criticalpoint ˆ c is non-flat if there exist α + > α − > M >
0, and δ > x ∈ (ˆ c, ˆ c + δ )(1.3) (cid:12)(cid:12)(cid:12)(cid:12) log | f ′ ( x ) || x − ˆ c | α + (cid:12)(cid:12)(cid:12)(cid:12) ≤ M, and for every x ∈ (ˆ c − δ, ˆ c )(1.4) (cid:12)(cid:12)(cid:12)(cid:12) log | f ′ ( x ) || x − ˆ c | α − (cid:12)(cid:12)(cid:12)(cid:12) ≤ M. We call α + and α − the right and left order of ˆ c respectively. Let us denote byCrit( f ) the set of critical points of f . If f is a unimodal map with turning point c ,we will use the notation S ( f ) := Crit( f ) ∪ { c } . Let us denote by C ω the class ofanalytic maps. Here we will say that f is a C ω -unimodal map if it is of class C ω outside S ( f ). NVARIANT MEASURES FOR INTERVAL MAPS WITHOUT LYAPUNOV EXPONENTS 3
We denote the orbit of x ∈ [ − ,
1] under f by O f ( x ) := { f n ( x ) : n ≥ } . For a probability measure µ on [ − ,
1] that is invariant by f , we define the push-forward of µ by f as f ∗ µ := h ◦ f − . Denote by χ µ ( f ) := Z log | f ′ | dµ, its Lyapunov exponent , if the integral exists. Similarly, for every x ∈ [ − , O f ( x ) ∩ S ( f ) = ∅ , denote by χ f ( x ) := lim n →∞ n log | ( f n ) ′ ( x ) | , the pointwise Lyapunov exponent of f at x , if the limit exists.Let λ F ∈ (0 ,
2] be so that the map T λ F : [ − , → [ − , T λ F ( x ) := λ F (1 − | x | ) − , for every x ∈ [ − , µ P be the unique measurethat is ergodic, invariant by T λ F , and supported on O T λF (0), see § Theorem 1.
Let h : [ − , → [ − , be a homeomorphism of class C ω on [ − , \ { } with a unique non-flat critical point at , and put ˜ µ P := h ∗ µ P . Thenthe C ω -unimodal map f := h ◦ T λ F ◦ h − has a Lorenz-like singularity at ˜ c := h (0) and is so that:(1) χ ˜ µ P ( f ) is not defined.(2) For x ∈ O f (˜ c ) , the pointwise Lyapunov exponent of f at x does not exist if O f ( x ) ∩ S ( f ) = ∅ , and it is not defined if O f ( x ) ∩ S ( f ) = ∅ .(3) log(dist( · , S ( f ))) / ∈ L (˜ µ P ) . (4) f has exponential recurrence of the Lorenz-like singularity orbit, thus, lim sup n →∞ − log | f n (˜ c ) − ˜ c | n ∈ (0 , + ∞ ) . The map f in Theorem 1 has a Lorenz-like singularity at ˜ c and two non-flatcritical points, given by the preimages by f of the Lorenz-like singularity ˜ c , seeProposition 1.1. Since h is a homeomorphism of class C ω on [ − , \ { } , these arenon-flat critical points of inflection type.Dobbs constructed an example of a unimodal map with a flat critical pointand singularities at the boundary, for which the Lyapunov exponent of an invariantmeasure does not exist, see [Dob14, Proposition 43]. For interval maps with infiniteLyapunov exponent see [Ped20, Theorem A], and references therein.The negation of item (3) in Theorem 1 is considered in several works as a reg-ularity condition to study ergodic invariant measures. In [Lim20], Lima studiedmeasures satisfying this condition for interval maps with critical points and discon-tinuities, he called measures satisfying this condition f − adapted . By the Birkhoff JORGE OLIVARES-VINALES ergodic theorem, if log(dist( · , S ( f ))) ∈ L ( µ ), then for an ergodic invariant measure µ , we have lim n →∞ n log(dist( f n ( x ) , S ( f ))) = 0 ,µ − a.e. Ledrappier called measures satisfying this last condition non-degenerated ,for interval maps with a finite number of critical points, see [Led81]. The measure˜ µ P does not satisfy the non-degenerated condition. For more results related to thiscondition see [Lim20] and references therein.For continuously differentiable interval maps with a finite number of criticalpoints, every ergodic invariant measure that is not supported on an attractingperiodic point satisfies lim n →∞ n log(dist( f n ( x ) , S ( f ))) = 0, a.e., see [Prz93] and[RL20, Appendix]. Item (3) in Theorem 1 tells us that we cannot extend this topiecewise differentiable maps with a finite number of critical points and Lorenz-likesingularities.Item (4) in Theorem 1 stresses important information relative to the recurrence ofthe Lorenz-like singularity. This item represents a crucial difference between smoothinterval maps, and the case of interval maps with critical points and Lorenz-likesingularities. In the smooth case, certain conditions on the growth of the derivativerestrict the recurrence to the critical set, see for example [CE83], [Tsu93], [GS14],and references therein. Using the terminology in [DPU96], item (4) shows that RuleI is sharp for the map f in Theorem 1. Finally, we have that the map f in Theorem1 satisfies Tsuji’s weak regularity condition, due to the interaction between thecritical points and the Lorenz-like singularity.1.2. Example.
Now we will provide an example of a map f as in Theorem 1. Fix ℓ + and ℓ − in (0 , α + := − ℓ + and α − := − ℓ − . Define h α + ,α − : [ − , −→ [ − , h α + ,α − ( x ) = ( | x | α + if x ≥ −| x | α − if x < . So(1.7) h − α + ,α − ( x ) = ( | x | /α + if x ≥ −| x | /α − if x < . Then by (1.5), (1.6),(1.7), and the chain rule, we have f ′ ( x ) = λ F h ′ α + ,α − ( T λ F ( h − α + ,α − ( x ))) h ′ α + ,α − ( h − α + ,α − ( x ))for every x ∈ [ − , \{ h α + ,α − (0) } . The function h ′ α + ,α − ( T λ F ( h − α + ,α − ( x ))) is boundedfor x close enough to 0, see §
5. Then, by (1.6) and (1.7), there exists
L > x ∈ ( h (0) , h ( δ )), 1 L | x | ℓ + ≤ | f ′ ( x ) | ≤ L | x | ℓ + , and for every x ∈ ( h (0) , h ( − δ )),1 L | x | ℓ − ≤ | f ′ ( x ) | ≤ L | x | ℓ − . NVARIANT MEASURES FOR INTERVAL MAPS WITHOUT LYAPUNOV EXPONENTS 5
Thus, h (0) is a Lorenz-like singularity of f , see Figure 1. Also, by (1.6) and (1.7),if δ is small enough so that T − λ F (0) ∩ ( − δ, δ ) = ∅ , the two critical points of f arenon-flat. The one to the left of h (0) has right order α + and left order α − , and theone to the right of h (0) has right order α − and left order α + . (a) (b) (c) Figure 1.
Graphics of the functions T λ F ( x ) (Figure (A)), h α ( x )for α + = 2 and α − = 1 . f ( x ) (Figure (C)).1.3. Strategy and organization.
We now describe the strategy of the proof ofTheorem 1 and the organization of the paper.In § § § § O T λF (0), and following [LM93],we construct a partition of it that will allow us to estimate close return times tothe turning point and lower bounds for the distances of these close returns. In § f . In § µ P supported on O T λF (0),restricted to the partition constructed in § § f in terms of h − . Without loss of generality, we will assumethat h preserves orientations, Proposition 1.1.
Let h and f be as in Theorem 1. Then f has a Lorenz-likesingularity at ˜ c . Moreover there exists α + > , α − > , K > , and δ > suchthat the following property holds: For every x ∈ (˜ c, h ( δ )) , (1.8) K − | h − ( x ) | − α + ≤ | f ′ ( x ) | ≤ K | h − ( x ) | − α + , and for every x ∈ ( h ( − δ ) , ˜ c ) , (1.9) K − | h − ( x ) | − α − ≤ | f ′ ( x ) | ≤ K | h − ( x ) | − α − . JORGE OLIVARES-VINALES In § Proposition 1.2.
Let h and f be as in Theorem 1. Then(i) R log | f ′ | + d ˜ µ P = + ∞ ,(ii) R log | f ′ | − d ˜ µ P = −∞ , and(iii) R | log(dist( · , S ( f ))) | ˜ µ P = + ∞ . To prove the first part of Proposition 1.2, we use the fact that around the Lorenz-like singularity, the geometry of f grows at the same rate as the measure decreases.This implies that in a sequence of disjoint intervals that converges to the criticalpoint, the integral of log | f ′ | is bounded from below by a positive constant. For thesecond part, we use the fact that the two preimages of the turning point of f arecritical points and both belong to the set O f (˜ c ). The third part of the propositionis a consequence of the estimation that we get in the proof of the first part.In section § f is transitive on O f (˜ c ), will imply item (2) in Theorem 1. Recall that for x ∈ O f (˜ c )such that ˜ c ∈ O f ( x ) we have that the pointwise Lyapunov exponent is not defined,since for n large enough log | ( f n ) ′ ( x ) | is not defined. Proposition 1.3.
Let h and f be as in Theorem 1. Then for every x ∈ O f (˜ c ) with ˜ c / ∈ O f ( x ) , we have that (1.10) lim inf n →∞ n log | ( f n ) ′ ( x ) | ≤ (cid:18) − αϕ (cid:19) log λ < log λ ≤ lim sup n →∞ n log | ( f n ) ′ ( x ) | . To prove Proposition 1.3, we use the fact that f restricted to the set O f (˜ c )is minimal, so the orbit of every point accumulates points far from the turningpoint. In that case, the derivative is bounded so the limit of that subsequence mustbe the same as the one in T λ F . On the other hand, if we look at a subsequencethat accumulates at the Lorenz-like singularity, the growth of the derivative isexponential with respect to the return time, so the limit of this subsequence willbe bounded away from zero.In section § f is topologically conjugated to the Fibonacci tent map we knowthat h (0) is recurrent, and that the recurrence times are given by the Fibonaccinumbers. Then, to have an estimate on the recurrence of the turning point it isenough to estimate the decay of the distances | f S ( k ) (˜ c ) − ˜ c | , where S (0) = 1 , S (1) = 2 , S (2) = 3 , S (3) = 5 , . . . , are the Fibonacci numbers. Proposition 1.4.
There exist Θ , α ′ , α ′′ positive numbers, such that (1.11) λ − S ( k ) α ′′ Θ − ≤ | f S ( k ) (˜ c ) − ˜ c | ≤ λ − S ( k ) α ′ Θ , for every k ≥ . To prove Proposition 1.4 we estimate the diameter of certain symmetric inter-vals, whose closure are disjoint from the Lorenz-like singularity, and whose lengthsapproximate the left and right distance of the closest returns to the Lorenz-like sin-gularity. To do this we use the mean value theorem, the fact that h has a non-flat NVARIANT MEASURES FOR INTERVAL MAPS WITHOUT LYAPUNOV EXPONENTS 7 critical point at 0, and Lemma 3.5 that give us an estimate on the diameter of thepreimage of these intervals. The reason to use these intervals is because when wetry to make a direct estimation the distance | f S ( k ) (˜ c ) − ˜ c | we do not have controlon how close to zero is the derivative of h .1.4. Acknowledgements.
The author would like to thanks Juan Riverla-Letelierfor helpful discussions and encouragement for this work. The author is also gratefulto Daniel Coronel and Yuri Lima for helpful comments.2.
Preliminaries
Throughout the rest of this work, we will denote by I the closed interval [ − , ⊂ R . We use N to denote the set of integers that are greater than or equal to 1 andput N := N ∪ { } .We endow I with the distance induced by the absolute value | · | on R . For x ∈ R and r >
0, we denote by B ( x, r ) the open ball of I with center at x and radius r .For an interval J ⊂ I , we denote by | J | its length.For real numbers a, b we put [ a, b ] := [min { a, b } , max { a, b } ] in the same way( a, b ) := (min { a, b } , max { a, b } ).2.1. The kneading sequence.
Following [dMvS93], we will introduce the knead-ing invariant of a unimodal map and related properties. Let f : I → I be a unimodalwith turning point c ∈ (0 , c i := f i ( c ) for i ≥
1. Suppose f ( −
1) = f (1) = − c < c < c . Let Σ := { , , c } N be the space of sequences x = ( x , x , x , . . . ). In Σ we consider the topology generated by the cylinders[ a a . . . a n − ] k := { x ∈ Σ : x k + i = a i for all i = 0 , , . . . , n − } . With this topology, Σ is a compact space. Let us define i : I −→ Σ x ( i ( x ) , i ( x ) , . . . )where i n ( x ) = f n ( x ) ∈ [ − , c )1 if f n ( x ) ∈ ( c, c if f n ( x ) = c. The sequence i ( x ) is called the itinerary of x under f . Given n ∈ N and x ∈ I there exists δ > i n ( y ) ∈ { , } and is constant for every y ∈ ( x, x + δ ).Observe that this value is not the same as i n ( x ) if x is the turning point. It followsthat i ( x + ) := lim y ↓ x i ( y ) and i ( x − ) := lim y ↑ x i ( y )always exist. Notice that i ( x − ) and i ( x + ) belong to { , } N . The sequence e , e , e , . . . defined by e j := i j ( c +0 ) is called the kneading invariant of f . A se-quence a ∈ { , } N is admisible if there exists a unimodal map f : I → I withkneading invariant a .We say that Q : N → N defines a kneading map if Q ( k ) < k for all k ∈ N and( Q ( j )) k
1) + S ( Q ( k )) for k ≥
1. The kneading sequence { e j } j ≥ associated to Q is given by e = 1 and the relation(2.1) e S ( k − e S ( k − . . . e S ( k ) − e S ( k ) = e e . . . e S ( Q ( k )) − (1 − e S ( Q ( k )) ) , for k ≥
1. The length of each string in (2.1) is S ( Q ( k )), thus at the kth -step of theprocess we can construct S ( Q ( k )) symbols of the sequence. Since for every k ≥ Q ( k ) < k , we get that Q (1) = 0. So, for k = 1, each string in (2.1) has 1symbol. Then e = e S (0)+1 = 1 − e S (0) = 0 . Hence, c < c < c . The Fibonacci tent map.
We will say that a unimodal map f has Fibonaccirecurrence or it is a
Fibonacci unimodal map if the kneading map associated to itis given by Q (1) = 0 and Q ( k ) = k − k >
1. So the sequence { S ( n ) } n ≥ isgiven by the Fibonacci numbers S (0) = 1 , S (1) = 2 , S (2) = 3 , S (3) = 5 , . . . . For a Fibonacci unimodal map f we have that(2.2) | c S (0) − c | > | c S (1) − c | > . . . | c S ( n ) − c | > | c S ( n +1) − c | > . . . , and(2.3) | c − c | < | c − c | . See [LM93, Lemma 2.1] and references therein. The set O f ( c ) is a Cantor setand the restriction of f to this set is minimal and uniquely ergodic, see [Bru03,Proposition 1] or [CRL10, Proposition 4] and references therein. The kneadinginvariant for a Fibonacci unimodal map starts like100111011001010011100 . . . Let us consider the tent family T S : I → I defined by T s ( x ) = s (1 − | x | ) − x ∈ I and every s ∈ (0 , full , thus for every kneading map Q thereis a parameter s ∈ (0 ,
2] so that the kneading map of T s is Q , see [MT88], [dMvS93,Chapter 2]. So there exists λ F ∈ (0 ,
2] such that the kneading map associated to T λ F is given by Q ( k ) = max { , k − } .From now on we use the notations T := T λ F , λ := λ F , c := 0, and c i := T i ( c ).3. The set O T ( c )3.1. The combinatorics of the set O T ( c ) . In this section we will give an explicitdescription of the set O T ( c ) following [LM93].Put S ( −
2) = 0 and S ( −
1) = 1. From (2.1) we obtain that for every k ≥ c S ( k ) and c S ( k +2) are on opposite sides of c . Since c S (1) = c < c < c = c S (0) , we conclude that for k ≡ , c S ( k ) is to the right of c and if k ≡ c S ( k ) is to the left of c . Since we also know that c S (1) is to the left of c , we can conclude that for k ≡ c S ( k ) is to the left of c , and for k ≡ NVARIANT MEASURES FOR INTERVAL MAPS WITHOUT LYAPUNOV EXPONENTS 9 (mod 4), c S ( k ) is to the right of c . From this, we can conclude that if k is even thepoints c S ( k ) and c S ( k +1) are in opposite sides of c , and therefore[ c S ( k +1) , c S ( k ) ] ⊇ [ c S ( k +2) , c S ( k ) ] . In the case that k is odd, c S ( k ) and c S ( k +1) are on the same side with respect to c ,and therefore [ c S ( k +1) , c S ( k ) ] ⊆ [ c S ( k +2) , c S ( k ) ] . For each k ≥ I k be the smallest closed interval containing all of the points c S ( l ) for every l ≥ k . For each n ≥ I nk := T n ( I k ). By the above discussion(3.1) I k = ( [ c S ( k ) , c S ( k +1) ] if k is even,[ c S ( k ) , c S ( k +2) ] if k is odd. Lemma 3.1.
For every k ≥ , we have that T j is injective on [ c , c S ( k )+1 ] . Inparticular I j +1 k = [ c j +1 , c S ( k )+1+ j ] for every j ∈ { , . . . , S ( k − − } . Proof.
Since | c − c S ( k ) | > | c − c S ( m ) | and | T ([ c S ( k ) , c ]) | = λ | c − c S ( k ) | , for every0 ≤ k < m we get that c S ( k )+1 < c S ( m )+1 < c , in particular I k = [ c , c S ( k )+1 ]. Inthe case k ≥
1, by (2.1), with k replaced by k + 1, for every j ∈ { , . . . , S ( k − − } we have that c S ( k )+ j and c j are in the same side respect to c . Thus c / ∈ [ c S ( k )+ j , c j ] = T j − [ c S ( k )+1 , c ], and then the map T j is injective on [ c , c S ( k )+1 ]. In particular,for 1 < j ≤ S ( k − I jk = T j − ([ c , c S ( k )+1 ])(3.2) = [ c j , c S ( k )+ j ] (cid:3) Note that for k ≥
1, by Lemma 3.1, with j = S ( k − − I S ( k − k = [ c S ( k − , c S ( k )+ S ( k − ] = [ c S ( k − , c S ( k +1) ] . Then, by (2.1), c ∈ I S ( k − k and c / ∈ I nk for every 0 < n < S ( k − Lemma 3.2.
For all k ≥ we have that | c i − c | > | c S ( k − − c | , for all < i < S ( k ) , with i = S ( k − . Proof.
We will use induction on k . The cases k = 0 and 1 are vacuously true, thecases k = 2 and 3 are true by the definition of Fibonacci map and (2.3). Supposenow that it is true for k . We will prove that is true for k + 1. Case 1:
Since | c S ( k − − c | > | c S ( k ) − c | , we have that | c i − c | > | c S ( k ) − c | , for all 0 < i < S ( k ). Case 2:
Since c S ( k − < c S ( k )+1 < c
10 JORGE OLIVARES-VINALES and T i is injective on [ c S ( k − , c ] for 0 < i < S ( k −
2) by Lemma 3.1, we havethat c S ( k )+ i ∈ ( c S ( k − i , c i ), for 0 < i < S ( k − | c i − c | > | c S ( k − − c | > | c S ( k ) − c | and | c S ( k − i − c | > | c S ( k − − c | > | c S ( k ) − c | , for 0 < i < S ( k − . By 2.1 c i and c S ( k − i lie on the same side of c for 0 < i | c S ( k ) − c | for all 0 < i < S ( k − Case 3:
Since T S ( k − − is injective on [ c S ( k − , c ] we get that c S ( k )+ S ( k − ∈ ( c S ( k ) , c S ( k − ). Also, by (2.1), c S ( k )+ S ( k − and c S ( k − lie on the same side of c ,and opposite to c S ( k ) . Then | c S ( k )+ S ( k − − c | < | c S ( k − − c | . Hence c S ( k − < c S ( k )+ S ( k − < c . So by Lemma 3.1, c S ( k )+ S ( k − i ∈ ( c S ( k − i , c i ) for 0 < i < S ( k − | c i − c | > | c S ( k ) − c | and | c S ( k − i − c | > | c S ( k ) − c | , for 0 < i < S ( k − . Since, by (2.1), c S ( k − i and c i lie on the same side of c for0 < i < S ( k −
3) we get | c S ( k )+ S ( k − i − c | > | c S ( k ) − c | , for all 0 < i < S ( k − . Case 4:
It remains to prove that | c S ( k )+ S ( k − − c | > | c S ( k ) − c | . Suppose by contradiction that | c S ( k )+ S ( k − − c | < | c S ( k ) − c | . Then c S ( k )+1 < c S ( k )+ S ( k − < c . Since T S ( k − − is injective on [ c S ( k )+1 , c ]we get that c S ( k +1) ∈ ( c S ( k − , c S ( k )+ S ( k − ). Noting that by (2.1) c S ( k − and c S ( k )+ S ( k − are in the same side with respect to c we have either | c S ( k +1) − c | > | c S ( k )+ S ( k − − c | > | c S ( k ) − c | or | c S ( k +1) − c | > | c S ( k − − c | > | c S ( k ) − c | , a contradiction. So we must have | c S ( k )+ S ( k − − c | > | c S ( k ) − c | , and this conclude the proof. (cid:3) Let us denote J k := I S ( k − k +1 = [ c S ( k − , c S ( k +1)+ S ( k − ] , and put D k := [ c, c S ( k ) ] NVARIANT MEASURES FOR INTERVAL MAPS WITHOUT LYAPUNOV EXPONENTS 11 for every k ≥
1. For every n ≥ J nk := T n ( J k ) = I S ( k − nk +1 . Note that by definition D k ′ ⊂ [ c S ( k ) , c S ( k +2) ], for every k ′ ≥ k ≥ . Lemma 3.3.
For all < k < k ′ we have J k ′ ⊂ D k ′ − ⊂ I k and J k ∩ J k ′ = ∅ .Proof. First we will prove that J k +1 is contained in D k and c / ∈ J k +1 for every k ≥
0. Fix k ≥
0. By (2.1) with k replaced by k + 3 we have c S ( k ) and c S ( k +2)+ S ( k ) are in the same side of c . Since | c S ( k +2) − c | < | c S ( k +1) − c | , we have c S ( k +1)+1 < c S ( k +2)+1 < c . By Lemma 3.1, T S ( k ) − is injective on I k +1 . Then, c S ( k +2)+ S ( k ) ∈ ( c S ( k +2) , c S ( k ) ).By (2.1) with k replaced by k + 3, we thus conclude c S ( k +2)+ S ( k ) ∈ ( c S ( k ) , c ). Then J k +1 = [ c S ( k ) , c S ( k )+ S ( k +2) ] ⊂ [ c, c S ( k ) ] = D k ⊂ [ c S ( k ) , c S ( k +2) ] ⊆ I k and c / ∈ J k +1 . Since, by definition, for every k ′ > k we have D k ′ − ⊂ I k ′ − ⊂ I k ,we get J k ′ ⊂ D k ′ − ⊂ I k . Now we will prove that J k +1 and I k +1 are disjoint. If k is even then I k +1 =[ c S ( k +1) , c S ( k +3) ]. Since c S ( k ) and c S ( k +1) lie on opposite sides respect to c , we havethat c S ( k ) , c S ( k )+ S ( k +2) , and c S ( k +3) lie on the same side of c . By (3.2), with k replaced by k + 4, we get | c S ( k ) − c | > | c S ( k )+ S k +2 − c | > | c S ( k +3) − c | . So J k +1 ∩ I k +1 = ∅ . Now, if k is odd I k +1 = [ c S ( k +1) , c S ( k +2) ] and c S ( k ) , c s ( k +1) and c S ( k )+ S ( k +2) lie on the same side of c . Suppose that | c S ( k +1) − c | > | c S ( k )+ S ( k +2) − c | ,then [ c S ( k +1) , c S ( k )+ S ( k +2) ] ⊂ [ c S ( k ) , c S ( k )+ S ( k +2) ] . Since T S ( k − is injective on [ c S ( k ) , c S ( k )+ S ( k +2) ], then T S ( k − is injective on[ c S ( k +1) , c S ( k )+ S ( k +2) ]. So we get T S ( k − ([ c S ( k +1) , c S ( k )+ S ( k +2) ]) = [ c S ( k +1)+ S ( k − , c S ( k +3) ] . Since S ( k + 1) + S ( k − < S ( k + 4) , by Lemma 3.2, with k replaced by k + 4, weget | c S ( k +1)+ S ( k − − c | > | c S ( k +3) − c | . On the other hand, T S ( k − ( J k +1 ) = [ c S ( k +1) , c S ( k +3) ], then T S ( k − ( c S ( k +1) ) = c S ( k +1)+ S ( k − ∈ ( c S ( k +1) , c S ( k +3) )and by (2.1), with k replaced by k + 1, we have that c S ( k +1)+ S ( k − and c S ( k − are in the same side of c . Since k − ≡ k + 3 (mod 4), we have that c S ( k − and c S ( k +3) are on the same side of c , so c S ( k +1)+ S ( k − ∈ ( c, c S ( k +3) ). Thus | c S ( k +1)+ S ( k − − c | < | c S ( k +3) − c | , a contradiction. So we must have c S ( k )+ S ( k +2) ∈ ( c S ( k ) , c S ( k +1) ) and(3.5) J k +1 ∩ I k +1 = ∅ . This conclude the proof of the lemma. (cid:3)
Taking k ′ = k + 1 in Lemma (3.3) we get J k +1 ⊂ I k , and then(3.6) J k +1 ∪ I k +1 ⊂ I k ⊆ I S ( k − k , for every k ≥ Definition 3.4.
For k ≥ M k be the S ( k ) − fold union M k = [ ≤ n
In this section we will give an estimate on how thedistances | c S ( k ) − c | decrease as k → ∞ . Lemma 3.5.
The following limit lim k →∞ λ S ( k +1) | D k | exists and is strictly positive. NVARIANT MEASURES FOR INTERVAL MAPS WITHOUT LYAPUNOV EXPONENTS 13
Proof.
Since T ( D k ) = T ([ c, c S ( k ) ]) = I k , by (3.3) we have that T S ( k − ( D k ) = [ c S ( k − , c S ( k +1) ] . By (2.1), c S ( k − and c S ( k +1) are in opposite sides of c . Then D k − ∩ D k +1 = { c } ,so T S ( k − ( D k ) = D k − ∪ D k +1 . Since, by Lemma 3.1, T S ( k − is injective on I k and D k ⊂ I k we get that | T S ( k − ( D k ) | = λ S ( k − | D k | = | D k − | + | D k +1 | . For k ≥ ν k := | D k | / | D k +1 | . By the above we get λ S ( k − = ν k − + 1 ν k . By (2.2), ν k >
1, so 0 < ν − k <
1. Since λ >
1, we have λ S ( k − −→ ∞ as k −→ ∞ .Then ν k −→ ∞ as k −→ ∞ . So λ S ( k − − ν k − −→ k −→ ∞ . Then, if wedefine C k := ν k λ − S ( k ) , we have 0 < C k < C k ր n −→ ∞ . By definition of ν k , we have that | D || D k +1 | = k Y i =0 ν i = k Y i =0 λ S ( i ) C i = λ S ( k +2) − S (1) k Y i =0 C i . Then(3.8) | D k +1 | λ S ( k +2) − S (1) = | D | " k Y i =0 C i − . Since Q ki =0 C i converge to a strictly positive number as k −→ ∞ , the proof iscomplete. (cid:3) The invariant measure
Let us denote by µ P the unique ergodic invariant measure of T restricted to O T ( c ). As in the previous, section put S ( −
2) = 0 , S ( −
1) = 1 , S (0) = 1 , S (1) = 2 , . . . and put ϕ := √ . In this section we will estimate the value of µ P over theelements of M k for every k ≥ §
3, we know that the set O T ( c ) is contained in M k for very k ≥ I k , I k , . . . I S ( k − − k , J k , J k , . . . , J S ( k − − k , are disjoint. Then(4.1) S ( k − − X i =0 µ P ( I ik ) + S ( k − − X j =0 µ P ( J jk ) = 1 . Since T restricted to O T ( c ) is injective, except at the critical point that has twopreimages, we have that µ P ( I ik ) = µ P ( I jk )(4.2) µ P ( J pk ) = µ P ( J qk ) , (4.3)for every 0 ≤ i, j < S ( k −
1) and 0 ≤ p, q < S ( k − S ( k − µ P ( I k ) + S ( k − µ P ( J k ) = 1Since I k ⊔ J k ⊂ I k − , we have that(4.5) µ P ( I k ) + µ P ( J k ) = µ P ( I k − ) . And since J k − = I S ( k − k , using (4.2) with k replaced by k = 1, we have that(4.6) µ P ( J k − ) = µ P ( I k ) . Combining (4.5) and (4.6) we can write(4.7) (cid:20) (cid:21) (cid:20) µ P ( I k ) µ P ( J k ) (cid:21) = (cid:20) µ P ( I k − ) µ P ( J k − ) (cid:21) . Lemma 4.1.
For every m ≥ we have µ P ( I m ) = 1 ϕ m and µ P ( J m ) = 1 ϕ m +1 . Proof.
We will use induction to prove the lemma. For m = 1. We can apply k − (cid:20) µ P ( I k − ) µ P ( J k − ) (cid:21) , and we can write (4.7) as(4.8) (cid:20) (cid:21) k − (cid:20) µ P ( I k ) µ P ( J k ) (cid:21) = (cid:20) µ P ( I ) µ P ( J ) (cid:21) . Using that (cid:20) (cid:21) k − = (cid:20) S ( k − S ( k − S ( k − S ( k − (cid:21) , for k ≥
2. We can write(4.9) (cid:20) µ P ( I ) µ P ( J ) (cid:21) = (cid:20) S ( k − µ P ( I k ) + S ( k − µ P ( J k ) S ( k − µ P ( I k ) + S ( k − µ P ( J k ) (cid:21) . Multiplying the first equation in (4.9) by S ( k ) S ( k − we get(4.10) S ( k ) S ( k − µ P ( I ) = S ( k − S ( k − S ( k ) µ P ( I k ) + S ( k − S ( k − S ( k ) S ( k − S ( k − µ P ( J k ) . Using (4.4) we can write (4.10) as (4.11) S ( k ) S ( k − µ P ( I ) = S ( k ) µ P ( I k ) (cid:20) S ( k − S ( k − − S ( k − S ( k − (cid:21) + S ( k − S ( k − S ( k ) S ( k − . NVARIANT MEASURES FOR INTERVAL MAPS WITHOUT LYAPUNOV EXPONENTS 15
Since S ( k ) S ( k − −→ ϕ as k −→ ∞ , taking limit on (4.11) over k we get ϕµ P ( I ) = 1 , and then µ P ( I ) = 1 ϕ . Using (4.4) with k replaced by 1 we get that µ P ( J ) = 1 ϕ . So the lemma holds for m = 1.Suppose now that the result is true for m . By (4.6) we have that µ P ( I m +1 ) = µ P ( J m ) = 1 ϕ m +1 . By (4.5) we have that µ P ( J m +1 ) = µ P ( I m ) − µ P ( I m +1 )= 1 ϕ m − ϕ m +1 = 1 ϕ m (cid:18) − ϕ (cid:19) = 1 ϕ m ϕ = 1 ϕ m +2 , and we get the result. (cid:3) Proof of Proposition 1.1
In this section we will give the proof of Proposition 1.1. We use the same notationas in the previous section. Let h : [ − , → [ − ,
1] be as in Proposition 1.1 and f = h ◦ T ◦ h − . As in Theorem 1, put ˜ c := h (0).Since h has a non-flat critical point at 0, by (1.3) and (1.4) there are α + > α − >
0, and δ > e − M | ˆ x | α + ≤ | h ′ (ˆ x ) | ≤ e M | ˆ x | α + , for every ˆ x ∈ (0 , δ ) and(5.2) e − M | ˆ x | α − ≤ | h ′ (ˆ x ) | ≤ e M | ˆ x | α − , for every ˆ x ∈ ( − δ, c = 0 and c / ∈ I k = [ c c S ( k )+1 , c ] , we have that thereexist positive real numbers W and W such that for every x ∈ I k (5.3) W ≤ | h ′ ( T ( x )) | ≤ W . Proof of Proposition 1.1.
By the chain rule we have(5.4) f ′ ( x ) = λ h ′ ( T ( h − ( x ))) h ′ ( h − ( x )) , for every x ∈ ( h ( − δ ) , h ( δ )) \ { ˜ c } . Let K := max { λ − e M W − , λe M W } . Then by(5.3), (5.1), (5.2), and (5.4) we have that for every x ∈ (˜ c, h ( δ ))(5.5) 1 K | h − ( x ) | α + ≤ | f ′ ( x ) | ≤ K | h − ( x ) | α + , and for every x ∈ ( h ( − δ ) , ˜ c )(5.6) 1 K | h − ( x ) | α − ≤ | f ′ ( x ) | ≤ K | h − ( x ) | α − . Now, from (5.1) and (5.2) there exist M > M > x ∈ (0 , δ ) M − | x | α + +1 ≤ | h ( x ) | ≤ M | x | α + +1 , and for every x ∈ ( − δ, M − | x | α − +1 ≤ | h ( x ) | ≤ M | x | α − +1 . Since h is a homeomorphism, there exist constants M > M > x ∈ (˜ c, h ( δ ))(5.7) M − | x − ˜ c | α ++1 ≤ | h − ( x ) | ≤ M | x − ˜ c | α ++1 , and for every x ∈ ( h ( − δ ) , ˜ c )(5.8) M − | x − ˜ c | α − +1 ≤ | h − ( x ) | ≤ M | x − ˜ c | α − +1 . Then, by (5.5), (5.6), (5.7), and (5.8) we have that for every x ∈ (˜ c, h ( δ ))1 M | x − ˜ c | α + α ++1 ≤ | f ′ ( x ) | M | x − ˜ c | α + α ++1 , and for every x ∈ (˜ c, h ( − δ ))1 M | x − ˜ c | α − α − +1 ≤ | f ′ ( x ) | ≤ M | x − ˜ c | α − α − +1 . Thus, f has a Lorenz-like singularity at ˜ c . (cid:3) Proof of Proposition 1.2
In this section, we will give the proof of Proposition 1.2. We will use the samenotation as in the previous sections.First, take α := max { α + , α − } , where α + and α − are given in Proposition 1.1.From (1.8) and (1.9), we get that for every x ∈ ( h ( − δ ) , h ( δ )) \ { ˜ c } (6.1) 1 K | h − ( x ) | α ≤ | f ′ ( x ) | . Proof of Proposition 1.2.
First we prove ( i ). By Lemma 3.5, there is k ≥ h ( I k ) \ { ˜ c } and such that | f ′ | > h ( I k ) \ { ˜ c } . For n > k put L n := λ S ( n +1) | D n | . Definelog | f ′ | + := max { , log | f ′ |} and log | f ′ | − := { , − log | f ′ |} , NVARIANT MEASURES FOR INTERVAL MAPS WITHOUT LYAPUNOV EXPONENTS 17 on I \ { ˜ c } . By Lemma 3.3, for every n > k , we have J n ⊂ I k and for every k < n < n ′ , we have J n ∩ J n ′ = ∅ . So, since ˜ µ P ( { ˜ c } ) = 0 Z log | f ′ | + d ˜ µ P ≥ Z h ( I k ) log | f ′ | d ˜ µ P ≥ X n>k Z h ( J n ) log | f ′ | d ˜ µ P . Recall that ϕ := √ . Then for each n > k and x ∈ J n , we have by Lemma (3.3)and (6.1)(6.2) | f ′ ( h ( x )) | ≥ K − | D n − | α = K − λ αS ( n ) L − αn − . By the above together with Lemma 4.1 and the fact that S ( n ) ≥ ϕ n +2 Z h ( J n ) log | f ′ | d ˜ µ P ≥ ˜ µ P ( h ( J n )) log (cid:12)(cid:12)(cid:12) K − λ αS ( n ) L − αn − (cid:12)(cid:12)(cid:12) (6.3) ≥ (cid:18) ϕ (cid:19) n +1 h αS ( n ) log( λ ) + α log( K /α L n − ) − i ≥ ϕα λ ) + α (cid:18) ϕ (cid:19) n +1 log( K /α L n − ) − . By Lemma 3.5, (cid:16) ϕ (cid:17) n +1 log( K /α L n − ) − −→ n −→ ∞ . We get that(6.4) Z log | f ′ | + d ˜ µ P = + ∞ . Now we prove ( ii ). Suppose by contradiction that(6.5) Z log | f ′ | − d ˜ µ P < + ∞ . By the chain rule(6.6) log | f ′ ( x ) | = log( λ ) + log | h ′ ( T ( h − ( x ))) | − log | h ′ ( h − ( x )) | , on I \ { ˜ c } Since h ′ has a unique critical point, log | h ′ | is bounded away from thecritical point. In particular, is bounded from above in all I . Then − log | h ′ | isbounded from below on I . In particular, since ˜ µ P ( { ˜ c } ) = 0, the integral Z log | h ′ ◦ h − | d ˜ µ P , is defined. Since the only critical points of f are the points in f − ( { ˜ c } ), we havethat log | f ′ | is bounded away from { ˜ c } ∪ f − { ˜ c } . Let e V ⊂ I \ { ˜ c } be a neighborhoodof f − { ˜ c } such that log | f ′ ( x ) | < x ∈ e V , then by (6.5) and (6.6) −∞ < Z ˜ V log | f ′ | d ˜ µ P = log( λ ) d ˜ µ P ( ˜ V ) + Z ˜ V (cid:0) log | h ′ ◦ T ◦ h − | − log | h ′ ◦ h − | (cid:1) d ˜ µ P . Since h − ( e V ) is a neighborhood of T − ( c ), the function − log | h ′ ◦ h − | is boundedon e V . On the other hand, since h − ◦ f ( x ) = T ◦ h − ( x ) = c, we have h ′ ◦ T ◦ h − ( x ) = 0 if x ∈ f − (˜ c ). Thus h ′ ◦ T ◦ h − ( x ) = 0 for x ∈ I \ e V .Then log | h ′ ◦ T ◦ h − | is bounded in I \ e V . So(6.7) Z I \ e V log | h ′ ◦ T ◦ h − | d ˜ µ P > −∞ . Now, −∞ < Z ˜ V log | f ′ | d ˜ µ P ≤ (cid:18) log( λ ) + max x ∈ ˜ V {− log | h ′ ◦ h − ( x ) |} (cid:19) ˜ µ P ( ˜ V ) + Z ˜ V (cid:0) log | h ′ ◦ T ◦ h − | (cid:1) d ˜ µ P . So(6.8) Z ˜ V log | h ′ ◦ T ◦ h − | d ˜ µ P > −∞ . Together with (6.7) this implies that Z log | h ′ ◦ T ◦ h − | d ˜ µ P , is finite. Since the integral Z − log | h ′ ◦ h − | d ˜ µ P , is defined, we have Z log | f ′ | d ˜ µ P = log( λ ) + Z log | h ′ ◦ T ◦ h − | d ˜ µ P + Z − log | h ′ ◦ h − | d ˜ µ P = log( λ ) + Z log | h ′ ◦ h − ◦ f | d ˜ µ P + Z − log | h ′ ◦ h − | d ˜ µ P , and since ˜ µ P is f invariant we get Z log | f ′ | d ˜ µ P = log( λ ) , contradicting (6.4). This contradiction completes the proof of part ( ii ).Finally we prove ( iii ). By Proposition 1.1, f has a Lorenz-like singularity at ˜ c ,then there exist δ > ℓ + > , ℓ − > L > x ∈ (˜ c, ˜ c + δ ) and (1.2) holds for every x ∈ (˜ c − δ, ˜ c ). Let ℓ := max { ℓ + , ℓ − } , andchoose 0 < ˆ δ ≤ δ , so that log | f ′ ( x ) | > x, S ( f )) = | x − ˜ c | for every x ∈ (˜ c − ˆ δ, ˜ c + ˆ δ ) \ { ˜ c } . Let m ≥ I m ⊂ (˜ c − ˆ δ, ˜ c + ˆ δ ). Then, for every x ∈ I m \ { ˜ c } we havelog | f ′ ( x ) | ≤ log( L ) − ℓ log(dist( x, S ( f ))) . So, for every n ≥ m Z h ( J n ) log | f ′ | d ˜ µ P ≤ log( L )˜ µ P ( J n ) + ℓ Z h ( J n ) | log | x − ˜ c || d ˜ µ P ( x ) . By (6.3) and Lemma 3.5 we get that+ ∞ = Z h ( I m ) | log(dist( x, S ( f ))) | d ˜ µ P ( x ) , so log(dist( x, S ( f ))) / ∈ L (˜ µ P ). This conclude the proof of the proposition. (cid:3) NVARIANT MEASURES FOR INTERVAL MAPS WITHOUT LYAPUNOV EXPONENTS 19 Proof of Proposition 1.3
In this section, we will prove Proposition 1.3. We will use the same notation asin the previous sections. Recall that f ′ is not defined at ˜ c , so for x in I whose orbitcontains ˜ c the derivative ( f n ) ′ at x does not exist for large n .Let α + and α − be the right and left critical orders of 0 as the critical point of h , and let α := max { α + , α − } . Let M > k > h ( I k ), and (5.1), and (5.2) holds on I k .For every x ∈ I such that ˜ c / ∈ O f ( x ) we put(7.1) χ + f ( x ) := lim sup n →∞ n log | ( f n ) ′ ( x ) | , and(7.2) χ − f ( x ) := lim inf n →∞ n log | ( f n ) ′ ( x ) | . For any x ∈ h ( I k ) we put ˆ x := h − ( x ) ∈ I k . The proof of the Proposition 1.3 isgiven after the following lemma.
Lemma 7.1.
For every ˆ x ∈ I k ∩ O T ( c ) there exists an increasing sequence ofpositive integers { n i } i ≥ such that T n i (ˆ x ) ∈ I k + i and T m (ˆ x ) / ∈ I k + i +1 , for all i ≥ and all n i + 1 ≤ m < n i +1 . Moreover, (7.3) S ( k + i ) − S ( k ) ≤ n i ≤ S ( k + i + 2) − S ( k + 2) , for all i > .Proof. We will prove the lemma by induction. Let ˆ x ∈ I k ∩ O T ( c ). Recall that forany integer k ′ ≥
1, we have that I k ′ , I k ′ , . . . I S ( k ′ − − k ′ , J k ′ , J k ′ , . . . , J S ( k ′ − − k ′ , are pairwise disjoint. Now, by (3.7)ˆ x ∈ I k +1 or ˆ x ∈ J k +1 . If ˆ x ∈ J k +1 , for every 1 ≤ m < S ( k − T m (ˆ x ) ∈ J mk +1 , thus T m (ˆ x ) / ∈ I k +1 and T S ( k − (ˆ x ) ∈ I k +1 . In this case n = S ( k −
1) satisfies the desired properties. If ˆ x ∈ I k +1 , for every1 ≤ m < S ( k ) T m (ˆ x ) ∈ I mk +1 , thus T m (ˆ x ) / ∈ I k +1 , and T S ( k ) (ˆ x ) ∈ I k +1 or T S ( k ) (ˆ x ) ∈ J k +1 . In the former case n = S ( k ) satisfies the desired properties. In the later case wehave that for 1 ≤ m < S ( k ) + S ( k −
1) = S ( k + 1) T m (ˆ x ) / ∈ I k +1 and T S ( k +1) (ˆ x ) ∈ I k +1 . So n = S ( k + 1) satisfies the desired properties. So we have S ( k − ≤ n ≤ S ( k + 1) . Now suppose that for some i ≥ n i satisfying the conclusions of the lemma.Thus T n i (ˆ x ) ∈ I k + i and S ( k + i ) − S ( k ) ≤ n i ≤ S ( k + i + 2) − S ( k + 2) . By (3.7) T n i (ˆ x ) ∈ I k + i +1 or T n i (ˆ x ) ∈ J k + i +1 . If T n i (ˆ x ) ∈ J k + i +1 , for every 1 ≤ m < S ( k + i − T m + n i (ˆ x ) ∈ J mk + i +1 , thus T m + n i (ˆ x ) / ∈ I k + i +1 and T S ( k + i − n i (ˆ x ) ∈ I k + i +1 . In this case n i +1 = S ( k + i −
1) + n i satisfies the desired properties. If T n i (ˆ x ) ∈ I k + i +1 , for every 1 ≤ m < S ( k + i ) T m + n i (ˆ x ) ∈ I mk + i +1 , thus T m + n i (ˆ x ) / ∈ I k + i +1 , and T S ( k + i )+ n i (ˆ x ) ∈ I k + i +1 or T S ( k + i )+ n i (ˆ x ) ∈ J k + i +1 . In the former case n i +1 = S ( k + i ) + n i satisfies the desired properties. In the latercase we have that for 1 ≤ m < S ( k + i ) + S ( k + i −
1) = S ( k + i + 1) T m + n i (ˆ x ) / ∈ I k + i +1 and T S ( k + i +1)+ n i (ˆ x ) ∈ I k + i +1 . So n i +1 = S ( k + i + 1) + n i satisfies the desired properties. So we have S ( k + i − ≤ n n i +1 ≤ S ( k + i + 1) + n i . Since n i satisfies (7.3) S ( k + i + 1) − S ( k ) ≤ n i +1 ≤ S ( k + i + 3) − S ( k + 2) . This conclude the proof of the lemma. (cid:3)
NVARIANT MEASURES FOR INTERVAL MAPS WITHOUT LYAPUNOV EXPONENTS 21
Proof of Proposition 1.3.
Let x ∈ O f (˜ c ), with ˜ c / ∈ O f ( x ). Then ˆ x = h − ( x ) ∈O T ( c ), and c / ∈ O T (ˆ x ). By the chain rule, we have that for very n ≥ f n ) ′ ( x ) = n − Y i =0 λ h ′ ( T ( T i (ˆ x ))) h ′ ( T i (ˆ x )) = λ n h ′ ( T n (ˆ x )) h ′ (ˆ x ) . Then,(7.5) 1 n log | ( f n ) ′ ( x ) | = log λ + 1 n log | h ′ ( T n (ˆ x )) | − n log | h ′ (ˆ x ) | , for every n ≥
1. So, using (7.1) and (7.2), we get(7.6) χ + f ( x ) = log λ + lim sup n →∞ n log | h ′ ( T n (ˆ x )) | , and(7.7) χ − f ( x ) = log λ + lim inf n →∞ n log | h ′ ( T n (ˆ x )) | . Now, since ˆ x ∈ O T ( c ), we have that ˆ x belongs to one of the following sets I k , I k , . . . , I S ( k − − k , J k , . . . , J S ( k − − k . So there exists 0 ≤ l k ( x ) < S ( k ) such that T l k ( x ) (ˆ x ) ∈ I k . Let { n i } i ≥ be as inLemma 7.1, for T l k ( x ) ( x ). Note that for every i ≥ T n i + l k ( x )+1 (ˆ x ) ∈ I k + i ⊂ [ c S ( k )+1 , c ] . Then by (5.3) and (7.6) we have thatlog λ ≤ χ + f ( x ) . Now by (7.3) we have that(7.8) 1 S ( k + i + 2) ≤ n i . Also, since for every i ≥ T n i + l k ( x ) (ˆ x ) ∈ I k + i , by (2.2) and (3.1), we get(7.9) | T n i + l k ( x ) (ˆ x ) | ≤ | c S ( k + i ) | = | D k + i | . By (6.1) we have(7.10) 1 λK | T n i + l k ( x ) (ˆ x ) | α ≤ | h ′ ( T n i + l k ( x )+1 (ˆ x )) || h ′ ( T n i + l k ( x ) (ˆ x )) | . Combining (5.3), (7.9) and (7.10) we get(7.11) | h ′ ( T n i + l k ( x ) (ˆ x )) | ≤ λKW | D k + i | α . Since | D k + i | −→ i −→ ∞ , there exists i ′ ≥ i ≥ i ′ | D k + i | < (cid:18) λKW (cid:19) /α . Then for every i ≥ i ′ from (7.11) we getlog | h ′ ( T n i + l k ( x ) (ˆ x )) | ≤ log ( λKW | D k + i | α ) < . By the above and (7.8)1 n i log | h ′ ( T n i + l k ( x ) (ˆ x )) | ≤ S ( k + i + 2) log( λKW | D k + i | α ) . Taking limit as i −→ ∞ ,lim i →∞ n i log | h ′ ( T n i + l k ( x ) (ˆ x )) | ≤ lim i →∞ S ( k + i + 2) log | D k + i | α . Using Lemma 3.5lim i →∞ S ( k + i + 2) log | D k + i | α = lim i →∞ S ( k + i + 2) log λ − αS ( k + i +1) = − α log λ lim i →∞ S ( k + i + 1) S ( k + i + 2)= − αϕ log λ. Then by (7.7) χ − f ( x ) ≤ (1 − αϕ ) log λ < log λ. This conclude the proof of the proposition. (cid:3) Proof of Proposition 1.4
In this section we will give the proof of Proposition 1.4. We will use the samenotation as in the previous sections.For every k ≥
1, put D + k := ( | c S ( k ) | , c ) and D − k := ( −| c S ( k ) | , c ) A + k := D + k \ D + k +1 and A − k := D − k \ D − k +1 . Observe that(8.1) | A + k | = | D k | − | D k +1 | = | A − k | . Lemma 8.1.
There exist α ′′ + , α ′′− , α ′ + , α ′− , K , and Q positive real numbers suchthat (8.2) λ − S ( k ) α ′′ + Q − ≤ | h ( A + k ) | ≤ λ − S ( k ) α ′ + Q, and (8.3) λ − S ( k ) α ′′− Q − ≤ | h ( A − k ) | ≤ λ − S ( k ) α ′− Q, for every k ≥ K .Proof. By Lemma 3.5, there exists β > n →∞ λ S ( k +1) | D k | = β. Let ε > β − ε ) / ( β + ε ) ≥ /
2. Let
M >
K > k ≥ K , (5.1), (5.2) holds on A k , and the following holds:(8.4) λ − S ( k +1) ( β − ε ) ≤ | D k | ≤ λ − S ( k +1) ( β + ε ) , NVARIANT MEASURES FOR INTERVAL MAPS WITHOUT LYAPUNOV EXPONENTS 23 (8.5) λ − S ( k ) ≤ , and(8.6) S ( k + 1) S ( k ) < ϕ + ε. By (8.4), with k replaced by k + 1, we get(8.7) λ S ( k +2) β + ε ≤ | D k +1 | ≤ λ S ( k +2) β − ε . Combining (8.4) and (8.7), we get(8.8) λ S ( k ) β − εβ + ε ≤ | D k || D k +1 | ≤ λ S ( k ) β + εβ − ε . For k ≥ K , using the mean value theorem on the function h : A + k −→ h ( A + k ), thereexists γ + ∈ A + k such that(8.9) | h ( A + k ) || A + k | = | h ′ ( γ + ) | . Let α + , be the right order of 0 as a critical point of h . By (5.1), we have(8.10) e − M | γ + | α + ≤ | h ′ ( γ + ) | ≤ e M | γ + | α + . Since γ + ∈ A + k , we have that | D k +1 | ≤ | γ + | ≤ | D k | . Then, by (8.4), (8.5), (8.9) and (8.10) we have that(8.11) e − M | D k +1 | α + +1 (cid:18) | D k || D k +1 | − (cid:19) ≤ | h ( A + k ) | ≤ e M | D k | α + +1 (cid:18) − | D k +1 || D k | (cid:19) . Using (8.8) in (8.11), we obtain(8.12) e − M ( β − ε ) α + +1 λ − S ( k +2)( α + +1) (cid:18) λ S ( k ) β − εβ + ε − (cid:19) ≤ | h ( A + k ) | ≤ e M ( β + ε ) α + +1 λ − S ( k +1)( α + +1) (cid:18) − λ − S ( k ) β − εβ + ε (cid:19) . Put Q := e − M (cid:18) β (cid:19) α + +1
14 and Q := e M β. By (8.5) and since ( β − ε ) / ( β + ε ) ≥ /
2, we have that Q ≤ e − M ( β − ε ) α + +1 (cid:18) β − εβ + ε − λ − S ( k ) (cid:19) , and e M ( β + ε ) α + +1 (cid:18) − λ − S ( k ) β − εβ + ε (cid:19) ≤ Q , for every k ≥ K . Then(8.13) λ − S ( k +2)( α + +1) λ S ( k ) Q ≤ | h ( A + k ) | ≤ λ − S ( k +1)( α + +1) Q . Finally, put α ′ + := α + + 1 and α ′′ + := ( ϕ + ε ) ( α + + 1) − . Since S ( k ) = S ( k + 2) − S ( k − − S ( k + 2)( α + + 1) + S ( k ) = − S ( k ) (cid:18) S ( k + 2) S ( k ) ( α + + 1) − (cid:19) ≥ − S ( k ) α ′′ + . Then, taking Q := max { Q − , Q } we have λ − S ( k ) α ′′ + Q − ≤ | h ( A + k ) | ≤ λ S ( k ) α ′ + Q. In the same way we can prove (8.3). (cid:3)
Proof of Proposition 1.4.
Let α ′′ + , α ′′− , α ′ + , α ′− , K , and Q , be as in Lemma 8.1. By(8.1), we have that(8.14) n X m =0 | h ( A + k + m ) | = | h ( D + k ) | − | h ( D + k + n +1 ) | , for every n ≥
0. Then by (8.2) and (8.14) we get(8.15) Q − n X m =0 λ − S ( k + m ) α ′′ + ≤ | h ( D + k ) | − | h ( D + k + n +1 ) | ≤ Q n X m =0 λ − S ( k + m ) α ′ + , for every n ≥
0. Now, for m ≥ S ( k + m ) = S ( k ) + m − X j =0 S ( k + j − . Put F k + m := m − X j =0 S ( k + j − , and σ ′ ( k ) := 1 + ∞ X i =0 λ − α ′ + F k + i . Then, combining (8.15) and (8.16) we obtain(8.17) λ − S ( k ) α ′′ + Q − ≤ | h ( D + k ) | − | h ( D + k + n +1 ) | ≤ λ − S ( k ) α ′ + Qσ ′ ( k ) . If we put Θ := Q ∞ X i =0 λ − α ′′ + S ( i ) ! , then for every k ≥ K and every m ≥ Qσ ′ ( k ) ≤ Θ , and Θ − ≤ Q − . Then(8.18) λ − S ( k ) α ′′ + Θ − ≤ | h ( D + k ) | − | h ( D + k + n +1 ) | ≤ λ − S ( k ) α ′ + Θ . Since | D k + n +1 | −→ n −→ ∞ , and h is continuous, taking the limit in (8.18)as n −→ ∞ we obtain(8.19) λ − S ( k ) α ′′ + Θ − ≤ | h ( D + k ) | ≤ λ − S ( k ) α ′ + Θ . In the same way we can prove that(8.20) λ − S ( k ) α ′′− Θ − ≤ | h ( D − k ) | ≤ λ − S ( k ) α ′− Θ . NVARIANT MEASURES FOR INTERVAL MAPS WITHOUT LYAPUNOV EXPONENTS 25
Finally, put α ′′ := max { α ′′− , α ′′ + } , and α ′ := min { α ′− , α ′ + } . For any k ≥ K we havethat | f S ( k ) (˜ c ) − ˜ c | = | h ( D + k ) | or | f S ( k ) (˜ c ) − ˜ c | = | h ( D − k ) | . In any case, by (8.19) and (8.20) the result follows. (cid:3)
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Department of Mathematics, University of Rochester. Hylan Building, Rochester,NY 14627, U.S.A.Departamento de Ingenier´ıa Matem´atica, Universidad de Chile, Beauchef 851, San-tiago, Chile.
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