TOPOLOGICAL CONDITIONS FOR THE EXISTENCE OF ABSORBING CANTOR SETS. Henk Bruin

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1 TOPOLOGICAL CONDITIONS FOR THE EXISTENCE OF ABSORBING CANTOR SETS Henk Bruin Abstract This paper deals with strange attractors of S-unimodal maps f It generalizes results from [BKNS] in the sense that very general topological conditions are given that either i) guarantee the existence of an absorbing Cantor set provided the critical point of f is sufficiently degenerate, or ii) prohibit the existence of an absorbing Cantor set altogether As a byproduct we obtain very weak topological conditions that imply the existence of an absolutely continuous invariant probability measure for f 1 Introduction Attractors are one of the central themes in dynamics, but a universal definition of attractor is hard to give Already in the context of interval maps one encounters ambiguities that one would hardly expect Let us illustrate this by the well-established classification of attractors for S-unimodal interval maps First we need to distinguish between a metric and a topological attractor Definition (cf [Mi1]) A closed set A is a metric (topological) attractor, if i) The basin B(A) = {x ω(x) A} has positive Lebesgue measure (is a residual set) ii) There is no proper closed subset A A such that B(A ) has positive Lebesgue measure (is a residual set) The following classification is due to Guckenheimer [G] for the topological part, and to Blokh and Lyubich [BL1] for the metric part Classification of Attractors Let f : I I be a non-flat S-unimodal (ie the Schwarzian derivative of f is negative) map of the interval Then f has a unique topological attractor, which is one of the following: i) An attracting periodic orbit ii) The union of n intervals which are cyclically permuted by f On each of these intervals f n is topological transitive iii) A Cantor set on which f acts as an adding machine This is the infinitely renormalizable case Furthermore, f has a unique metric attractor, which can be of type i), ii), iii) or 1991 Mathematics Subject Classification 58F13, 58F11 Key words and phrases attractors, unimodal maps, invariant measures Supported by the Netherlands Organization for Scientific Research (NWO) The research for this paper was carried out during the author s stay at the University of Erlangen-Nürnberg 1 Typeset by AMS-TEX

2 2 HENK BRUIN iv) A Cantor set, but iii) does not apply: f is finitely renormalizable An attractor of type iv) is called an absorbing Cantor set It is a metric but not a topological attractor It has been an open question for some years whether case iv) can occur If l = 2 absorbing Cantor sets do not exist Proofs were given by Lyubich [L1] and by Jakobson and Świ atek [JS1,JS2,JS3] However, as was proved in [BKNS], there are maps (with a degenerate critical point) that have an absorbing Cantor set This applies at least to maps with a special combinatorial structure, known as the Fibonacci dynamics The purpose of this paper is to generalize the results of [BKNS]: We want to indicate which topological constraints allow/prohibit the existence of absorbing Cantor sets The topology of absorbing Cantor sets is known to some extent [BL2,GJ,Ma1] If A is an absorbing Cantor set, then A = ω(c) (where c is the critical point of f), A is minimal and c must have a rigid recurrence behaviour, sometimes called persistent recurrence The precise formulation of this recurrence behaviour may change from one author to the next In section 3 we will discuss this in detail: The differences in formulation are very subtle, leading to intricate counter-examples (section 11) Let us discuss some of the ideas and tools used in this paper, and then state the main result in a simplified form Our main tool is a certain kind of induced map (section 4) It is a generalization of the induced map used in [BKNS] The dynamics of the induced map F can be considered as a random walk on a Markov chain with countably many states, {U k } k N From a topological viewpoint, F is well-understood The possible transitions from one state to another are given by the combinatorial structure of the original map f If f is non-renormalizable, the set of dense orbits on the chain is residual in the set of all orbits For the map f, this means that the set of points having a dense f-orbit is also residual From a measure theoretical viewpoint, one has to distinguish between recurrent and transient Markov chains In principle, if a point x escapes to infinity under iteration of F, then x tends to ω(c) under iteration of f Therefore a transient Markov chain corresponds to a map with an absorbing Cantor set (section 5) We will have to compute the probabilities to go from one state to another This involves complicated estimates, which makes up the hardest part of the proofs We will describe the combinatorics of a unimodal map by means of the kneading map Q This map has been developed by Hofbauer and Keller, eg [H,HK] It is a map on N with the property that Q(n) < n for all n > 0 For example, the Fibonacci map has the kneading map Q(k) = max{k 2, 0} The precise definition and some of the properties of Q will be given in section 2 Here we will point out how Q describes the dynamics on the Markov chain: From state U k one can reach U l if and only if l min{q(k) + 1, Q(Q(k 1) + 1) + 1} Therefore if k Q(k) is bounded (f is Fibonacci-like), then one can drop only a bounded number of states in the Markov chain This gives good chances to find a transient Markov chain We conjecture that the boundedness of k Q(k) is sufficient for the existence of an absorbing Cantor set (provided the critical order l is sufficiently large) Using an additional condition on Q, we can indeed prove this conjecture (section 6) On the other hand, we can show that in general no

3 ABSORBING CANTOR SETS 3 absorbing Cantor set can arise if k Q(k), irrespective the value of l (section 8) Related to the question of absorbing Cantor sets is the question whether f has an absolutely continuous (with respect to Lebesgue measure) invariant probability measure (acip for short) As was proved in [KN,LM], the Fibonacci map has an acip if the order l of the critical point is sufficiently small (l < 2 + ε) It is conjectured that the picture is as follows: Let f l be a family of Fibonacci maps with critical order l For l small, f l has an acip If l increases, the acip disappears, to be replaced by a conservative absolutely continuous infinite σ-finite invariant measure If l increases even more, then an absorbing Cantor set is created, and f l is therefore no longer conservative The same picture may be true for Fibonacci-like maps However, using the summability condition of Nowicki and van Strien [NS], we can give a large class of combinatorial structures that imply the existence of an acip, irrespective the value of l (section 9) In this case the stage of a σ-finite measure is never reached Let us summarize this exposition in the following, simplified, theorem: Theorem A Let f be a finitely renormalizable, non-flat S-unimodal map having critical order l < and kneading map Q Assume that Q is eventually nondecreasing a) If k Q(k) is bounded, then there exists l 0 (depending only on the upper bound of k Q(k)) such that f has an absorbing Cantor set, whenever l > l 0 b) If limk Q(k) =, then f has no absorbing Cantor set c) If lim k Q(k) log k =, then f has an absolutely continuous invariant probability measure The assumption that Q is eventually non-decreasing is strong, but it simplifies the metric estimates considerably Among other things, these metric estimates are needed to show that certain (dynamically defined) points u k accumulate exponentially fast on c In section 7 (and 10), we derive sufficient topological conditions for this behaviour These conditions are much weaker than that Q is eventually non-decreasing In fact, Theorem A is a special case of the Theorems 61, 81 and 91 We want to thank Gerhard Keller for the many fruitful discussions We also thank the referee for the useful remarks 2 Notation and Preliminaries f : I I S-unimodal map on the unit interval c, c n := f n (c) critical point and its iterates l order of the critical point ω(c), B(ω(c)) critical omega-limit set and its basin of attraction S k k-th cutting time Q kneading map H n (x) maximal interval of monotonicity of f n containing x M n (x) := f n (H n (x)) ˆx involution of x: f(ˆx) = f(x) x f := f(x) used for points close to c 1 d k := f Sk (c)

4 4 HENK BRUIN z k, A k := (z k 1, z k ) k-th closest precritical point u k, U k := (u k 1, u k ) generalized closest prefixed points F induced map F U k Û k := f Sk 1 U k Û k ϕ n (x) ϕ n (x) = k if and only if F n (x) U k y k := f S Q(k) 1 (dk 1 ) t f k right boundary point of H Sk 1(c 1 ) In some more detail: - f : [0, 1] [0, 1] is a unimodal map with f(0) = f(1) = 0 We assume that f is C 3 and has negative Schwarzian derivative (Sf := D3 f Df 3 2 ( D2 f Df )2 < 0 wherever it is defined) - f is called renormalizable of period n > 1 if there is an interval c J [c 2, c 1 ] such that f n (J) J The maximal interval with this property is called restrictive We assume throughout the paper that f admits no periodic attractor and is not renormalizable - denotes Lebesque measure or just the absolute value d(a, B) is the distance between sets or points - The order of the critical point l < Hence f(c) f(x) = O(1) c x l and Df(x) = O(l) x c l 1 It follows that the involution x ˆx (ˆx is such that ˆx x (if x c) and f(ˆx) = f(x)) is Lipschitz For simplicity we will assume that ˆx = 2c x - ω(x) is the set of accumulation points of orb(x), the forward orbit of x A forward invariant set A is minimal if every x A has a dense orbit in A - ω(x) is minimal if and only if x is uniformly recurrent [Got], ie for every neighbourhood U of x, there exists N = N(U) such that for every m for which f m (x) U, there exists n N such that also f m+n (x) U - Suppose that H n (x) = (a, b) x is the maximal interval on which f n is diffeomorphic, then r n (x) := min{ f n (x) f n (a), f n (x) f n (b) }, and R n (x) := max{ f n (x) f n (a), f n (x) f n (b) } Contraction Principle For every ε > 0 there exists δ > 0 such that for every n 0 and every interval of size J > ε, f n (J) δ In particular, f n J cannot be homeomorphic for every n The Contraction Principle holds if there are no wandering intervals or periodic attractors [BL2,MS] Let g : J R be C 1, then the distortion Dg(x) dis(g, J) := sup x,y J Dg(y) If g has negative Schwarzian derivative, g expands the cross ratio: Let j t be t j intervals, and let l, r be the components of t \ j Then is the cross-ratio of j l r and t

5 ABSORBING CANTOR SETS 5 Expansion of Cross-Ratios Let g : t T be a diffeomorphism with negative Schwarzian derivative Then g expands the cross-ratio, ie t j T J l r L R where the capital letters indicate the g-images Furthermore, we can use the Koebe Principle (see eg [MS]) Let T J again be intervals T is said to contain a δ-scaled neighbourhood of J if both components of T \ J have size δ J Koebe Principle Suppose Sg 0 Fix δ > 0 and let K = ( ) 1+δ 2 δ Then the following property holds: Let j t be intervals such that g t is monotone If T contains a δ-scaled neighbourhood of J, then dis(g, j) K A forward iterate c n is called a closest return if c j / [c n, ĉ n ] for 0 < j < n The closest precritical points z k and cutting times S k are defined as follows: S 0 := 1, z 0 := f 1 (c) (0, c) Inductively, and S k+1 := min{n > S k f n (c) (z k, c)} z k+1 := f Sk+1 (c) (z k, c) We will give a few properties of these notions More details can be found in [B2,B3] Let A k := (z k 1, z k ), and A 0 := (0, z 0 ) If S k < n S k+1, then (z k, c) and (c, ẑ k ) are maximal intervals on which f n is diffeomorphic c ẑ k ẑ k 1 f S k 1 z r 1 d k 1 = f S k 1(c) f S k 1(z k ) = z r c Figure 21 By construction, f Sk 1 (z k ) is again a closest precritical point, see figure 21 The kneading map Q : N N is defined such that It follows that f Sk 1 (z k ) = z Q(k) (21) S k = S k 1 + S Q(k) The kneading map determines the combinatorics of the map completely Define d k := f Sk (c) By figure 21 and the construction of the closest preimages, (22) d k 1 (z Q(k) 1, z Q(k) ) (ẑ Q(k), ẑ Q(k) 1 ) This is true for all k 1 If Q(k) = 0, then d k 1 I \ (z 0, ẑ 0 ) Notice also that f Sk maps (z k 1, c) diffeomorphically onto (d Q(k), d k ) and f Sk (z k ) = c Therefore d k and d Q(k) lie on different sides of c

6 6 HENK BRUIN Lemma 21 If there is no periodic attractor, then Q(k) < k for all k 1 Proof f Sk 1 maps both (z k 1, c) and (c, ẑ k 1 ) diffeomorphically onto (d k 1, c) If Q(k) k, then by (22), d k (z k 1, ẑ k 1 ) Hence f Sk 1 maps either (z k 1, c) or (c, ẑ k 1 ) diffeomorphically into itself, yielding a periodic attractor Lemma 22 There exists a unimodal map with Q as kneading map if and only if (23) {Q(k + j)} j 1 {Q(Q 2 (k) + j)} j 1, where denotes lexicographical order Proof See [H] or [B3] Formula (23) is the admissibility condition for kneading maps The geometric interpretation is that d k c d Q2 (k) c for all k z Q(k) 1 d k 1 z Q(k) c f S Q(k) d Q 2 (k) d k c d Q(k) Figure 22 This follows immediately from (22) by taking the S Q(k) -th iterate, see figure 22 Lemma 23 If f is renormalizable of period n, but has no n-periodic attractor, then there exists k such that S k = n and Q(k + j) k for all j 1 Proof Let [p, ˆp] be the restrictive interval of period n Then the situation is as in figure 23 f n z k 1 p z k c ẑ k ˆp ẑ k 1 p c d k ˆp Figure 23 Because f n ((p, c)) c and f n (p, c) is diffeomorphic, (p, c) z k for some k minimal, and n = S k Therefore p (z k 1, z k ) By Lemma 21, d k / (z k, ẑ k ), and as d k [p, ˆp], d k (ẑ k, ẑ k 1 ) Because f j ([p, ˆp]) c only if j is a multiple of n = S k (in which case f j ([p, ˆp]) [p, ˆp]), d m [p, ˆp] (z k 1, ẑ k 1 ) for every m k Therefore Q(m + 1) k for every m k We will often have to impose the following condition on the combinatorics: (24) Q(k + 1) > Q(Q 2 (k) + 1) for all sufficiently large k

7 ABSORBING CANTOR SETS 7 Geometrically, this means that there is at least one closest precritical point between d k and d Q 2(k), see figure 22 We have not been able to prove most of the metric estimates without (24) or a similar condition Condition (24) prohibits the existence of saddle-node like returns (see Section 7) Clearly (24) is true if Q is eventually non-decreasing Moreover Lemma 24 If Q is eventually non-decreasing and Q(k), then there exists k 0 such that for all k k 0, d k is a closest return and there is no other closest return past d k0 Proof Suppose that k 0 is such that Q(k + 1) Q(k) for all k k 0, and take k 0 such that Q(k) > k 1 for all k k 0 We will first show that d k+1 (d k, ˆd k ) for all k k 0 Suppose by contradiction that d k (d k+1, ˆd k+1 ) for some k k 0 Then Q(k + 2) Q(k + 1), so in fact Q(k + 2) = Q(k + 1) = n By (22), d k (d k+1, z n ) (ẑ n, ˆd k+1 ) As d k+2 = f Sn (d k+1 ), d k+1 = f Sn (d k ) and f Sn (z n ) = c, d k+1 (d k+2, c) It follows by induction that Q(k + j + 1) = Q(k + j) for all j 1, contradicting our assumption that Q(k) Next suppose that c n is a closest return for some S k < n < S k+1 Without loss of generality we can assume that c n (d k, c) and d k (z Q(k+1) 1, z Q(k+1) ) Write m := n S k < S Q(k+1) As c n is a closest return, M n 1 (c 1 ) c Hence also M m (d k ) c and H m (d k ) contains a preimage x of c As there are no points of f m (c) in (z Q(k+1) 1, c), f m (z Q(k+1) 1 ) [c, c n ] Because z Q(k+1) 1 is a closest precritical point, in fact f m (z Q(k+1) 1 ) = c, so m = S Q(k+1) 1 It follows that z Q(k+1) 1 f m ([d k, z Q(k+1) ]) = [z Q2 (k+1), c n ] See figure 24 z Q(k+1) 1 d k z Q(k+1) c f m = f S Q(k+1) 1 3 z Q 2 (k+1) z Q(k+1) 1 c n c Figure 24 Therefore r := n + S Q(k+1) 1 is a cutting time Furthermore r = S k + 2S Q(k+1) 1 = S k+1 + S Q(k+1) 1 S Q2 (k+1) < S k+1 + S Q(k+2) = S k+2 Hence r = S k+1 But then Q(Q(k + 1)) = Q(k + 1) 1, and because Q is nondecreasing, f is renormalizable with period S Q(k+1) 1 Passing to the renormalization (which has a non-decreasing kneading map) and repeating the arguments, we get the result Let t f k be such that H Sk 1(c f ) = (z f k 1, tf k ) Notice that t f k / [c 2, c 1 ], so t f k is not the image of any point t k The notation is just to indicate that t f k is close to cf

8 8 HENK BRUIN Lemma 25 Let κ := min{k Q(k) > 0} Then for all k > κ, M Sk 1(c f ) = f Sk 1 (z f k 1, tf k ) (d Q(k), d Q2 (k)) If (24) holds, then M Sk 1(c f ) = (d Q(k), d Q2 (k)), for k sufficiently large Proof By (21), at least f Sk 1 (z f k 1 ) = d Q(k) For k = κ + 1, f Sk 1 (t f k ) = c 1 = d Q 2(k), as one can verify by hand Now take k > κ + 1 arbitrary and suppose without loss of generality that d k 1 > c Therefore by (22), d k 1 (c, ẑ Q(k) 1 ) If (25) ẑ Q(k) 1 (d k 1, f Sk 1 1 (t f k 1 )), then by construction of H Sk 1(c f ), f Sk 1 1 (t f k ) = ẑ Q(k) 1 Hence f Sk 1 (t f k ) = f S Q(k) (ẑq(k) 1 ) = d Q2 (k) If (25) is false, so f Sk 1 1 (t f k 1 ) (d k 1, ẑ Q(k) 1 ), then t f k = tf k 1 and f Sk 1 (t f k ) f S Q(k) ((dk 1, ẑ Q(k) 1 )) = (d k, d Q 2(k)) This proves the first statement z f k 1 c f t f k t f k 1 f S k 1 1 c d k 1 ẑ Q(k) 1 d Q 2 (k 1) f S Q(k) 1 d Q(k) 1 y k c f S Q 2 (k) d Q(k) c d k d Q 2 (k) Figure 25 If f Sk 1 1 (t f k 1 ) = d Q 2 (k 1) and Q(Q 2 (k 1) + 1) < Q(k), then (25) is satisfied In this case all the relations of figure 25 (with y k := f S Q(k) 1 (dk 1 )) are true By induction the second statement follows too For the proofs of the following statements, we refer to [Ma2] - A point x is said to be nice if its forward orbit does not enter (x, ˆx) - An interval is called nice if it is symmetric and has nice boundary points - Let V be a nice interval and let m > 0 be the smallest integer such that c m V Then because f is not renormalizable, f m (V ) contains a boundary point x of V in its interior Pulling back this point along the orbit c, c 1,, c m we obtain two points y and ŷ in V y and ŷ are nice Define ψ by ψ(v ) = (y, ŷ) - Let V be a nice interval and x / V If n is the first visit time of x to V, then M n (x) V - Additionally, let x / ψ(v ), and n be the first visit time of x to ψ(v ) Then M n (x) V

9 ABSORBING CANTOR SETS 9 3 Persistent Recurrence and Related Notions In this section we discuss several notions concerning the recurrence behaviour of the critical point: - Johnson and Guckenheimer [GJ] introduced critical monotonicity as a sufficient condition for the non-existence of absorbing Cantor sets - Blokh and Lyubich [BL2] gave a necessary topological condition (r n (x) 0 for x ω(c)) for the existence of absorbing Cantor sets - In complex dynamics the notion of persistent recurrence has been used: it refers to Yoccoz τ-function Real interpretations of persistent recurrence appear in papers of Lyubich [L1-2] - The existence of absorbing Cantor sets was proven for sufficiently flat Fibonaccimaps In this paper we are showing that absorbing Cantor sets can also be found for more general, Fibonacci-like, maps It is not clear from the start which condition implies which, and in fact the difference is very subtle in some cases In this section we want to classify the several versions of recurrence and discuss their relations (knowing that some of these relations were proved in [BL2,L1,Mi2]) (C1) f is Fibonacci-like, ie k Q(k) is bounded (C2) Q(k) (C3) R n (c 1 ) 0 (C4) For every neighbourhood U c there exists N such that for every n > N, M n (c 1 ) does not cover a component of U \ {c} (C5) For every symmetric neighbourhood U c there exists N such that for every n > N, M n (c 1 ) U (C6) For every symmetric neighbourhood U c there exists N such that for every n > N: c n+1 / U or M n (c 1 ) U (ie f is not critically monotonic) (C7) For every nice interval U c, there exists N such that for every n > N: c n+1 / U or M n (c 1 ) U (C8) r n (c 1 ) 0 (C9) ω(c) is a minimal Cantor set containing c Proposition 31 The following implications hold: (C1) (C2) (C3) (C4) (C5) (C6) (C7) (C8) (C9) Here also means that the reverse implication is false Proof (C1) (C2): Trivial (C2) (C1): Trivial (C2) (C3): Lemma 25 states that M Sk 1(c 1 ) (d Q(k), d Q 2(k)) If Q(k), then R Sk 1(c 1 ) < d Q(k) d Q2 (k) < z Q(Q(k)+1) 1 ẑ Q(Q2 (k)+1) 1 0 as k By the Contraction Principle also R n (c 1 ) 0 (C3) (C2) lim n R n (c 1 ) 0 implies lim k R Sk 1(c 1 ) 0, whence d k c 0 As (d k, c) z Q(k+1) or ẑ Q(k+1), Q(k) (C3) (C4): Trivial (C4) (C3): Assume by contradiction that lim inf R n (c 1 ) ε > 0 Then there exist arbitrarily large integers n such that M n (c 1 ) ε Due to the Contraction Principle, there exists δ > 0 and arbitrarily large integers m such that M m (c 1 ) > δ and M m (c 1 ) c Therefore M m (c 1 ) contains a component of (c δ/2, c+δ/2)\{c}

10 10 HENK BRUIN (C4) (C5): Trivial (C5) (C4): See example 112 (C5) (C6): Trivial (C6) (C5): See example 113 (C6) (C7): Trivial (C7) (C6) See example 111 (C7) (C8): Assume (C7) holds but lim sup r n (c 1 ) = δ > 0 Choose a nice interval V so small that according to the Contraction Principle no interval of length 2δ can be mapped in a monotonic way into V Let n be arbitrary such that r n (c 1 ) δ Let m n be the smallest iterate such that c m+1 V As V is nice, M m n (c n+1 ) V, and by the choice of V, H m n (c n+1 ) f n m (V ) M n (c 1 ) Therefore also M m (c 1 ) V Since this can be done for infinitely many n, (C7) is false after all (C8) (C9): As r n (c 1 ) 0, c is recurrent Recall that f is assumed to have no periodic attractor, so #(orb(c)) = Assume by contradiction that ω(c) is not minimal Then c is not uniformly recurrent Therefore there exists a set V c such that b(n) := min{k n k > n, c k V } can be arbitrarily large Without loss of generality assume that V is nice If c n V, then we can pull-back V along the orbit c n, c n+1,, c n+b(n), obtaining an interval V n V As V is nice, V n V m = if b(n) b(m) (In particular, V 0 is the only interval of this kind on which f b(n) = f b(0) is not monotone) Because b(n) can be arbitrarily large, there are infinitely many sets V n Next take m(n) = min{k 1 c k V n } Then for each V n, there exists a set V n c 1 such that f m(n) 1 maps V n monotonically onto V n Now let U := ψ(v ) As f is not renormalizable, U V Let δ be the minimum length of the components of V \ U Let m (n) := min{k 0 c m(n)+k U} Then there exists V n c 1 such that f m (n)+m(n) 1 maps V n monotonically onto V, with c m (n)+m(n) U Therefore r m (n)+m(n) 1(c 1 ) δ Because this is true infinitely often, r k (c 1 ) 0 (C9) (C8): Counter-examples appear for example in [GT] (C8) (C7): Suppose (C8) is true, and, by contradiction, that (C7) is false Therefore there exists a sequence of iterates n i such that M ni (c 1 ) V c ni+1, where V is a nice interval Because (C8) is true, d(c ni+1, V ) 0 as i This means that V ω(c), and ω(c) is not minimal This contradicts the previous the (C8) (C9) implication The notion of persistence recurrence comes from complex dynamics: A quadratic map f(z) = z 2 + c 1 has a persistently recurrent critical point if τ(k) Here τ is Yoccoz τ-function in the critical tableau In order to relate this property to the ones we already have, we will assume in the next lemma that f(z) = z 2 + c 1, c 1 R, instead of just a smooth unimodal map Because of the notation c 1, we can maintain c n as the n-th iterate of the critical point Lemma 32 c is persistently recurrent if and only if r n (c 1 ) 0 Proof We will use the notation from [Mi2] P n (x) denotes the order n puzzle piece containing x First recall that the two boundary points of P n (c) R are preimages of the orientation reversing fixed point p They are symmetric two each other, and nice In other words P n (c) R is a nice interval for each n

11 ABSORBING CANTOR SETS 11 Because of Proposition 31, it suffices to show that persistent recurrence is equivalent to property (C7) Assume by contradiction that c is not persistently recurrent Therefore lim inf n τ(n) = k < Take n arbitrary such that f n k maps P n (c) by a two-fold covering onto P k (c) Then f n k 1 maps f(p n (c)) R to a fixed nice interval P k (c) R c n k Therefore (C7) is false Conversely, assume that (C7) is false, and that V is a nice interval such that M n (c 1 ) V c n+1 for arbitrarily large n Take k minimal such that P k (c) R V For n as above, let m 0 be the smallest integer such that c m+n+1 P k (c) R It follows that f n+m+1 maps P n+m+1 (c) by a two-fold covering onto P k (c) R Hence τ(n + m + 1) = k for arbitrarily large numbers n, and c is not persistently recurrent Lemma 33 We have r n (c 1 ) 0 if and only if r n (x) 0 for every x ω(c), in which case r n (x) 0 uniformly Proof Trivial Notice that it immediately follows that ω(c) is minimal Indeed, if for some x ω(c), c / ω(x), then r n (x) 0 Assume that there exists {x n } ω(c) such that lim sup n r n (x n ) δ > 0 Choose c U = ψ(v ) V so small that for every interval J of length 2δ, f n (J) V for every n 0 We already know that ω(c) is minimal, so c ω(x n ) Choose n arbitrarily large such that r n (x n ) δ and set m 1 (n) := min{k n f k (x n ) V } Pull-back V along the orbit x n, f(x n ),, f m1(n) (x n ), obtaining a neighbourhood of V n of x n Let m 2 (n) := min{k 1 c k V n } This gives an interval V n c 1 which is mapped monotonically onto V by f m1 (n)+m2(n) 1 Finally, let m 3 (n) := min{k 0 c m1(n)+m 2(n)+k U} Then there exists an interval V n c 1 which is mapped monotonically onto V by f m1 (n)+m2(n)+m3(n) 1 and moreover c m1(n)+m 2(n)+m 3(n) U Because this happens infinitely often, r k (c 1 ) 0, a contradiction Corollary 34 If r n (c 1 ) 0, then x B(ω(c)) if and only if r n (x) 0 Proof If x / B(ω(c)), then there exists δ > 0 and arbitrarily large iterates n such that d(f n (x), ω(c)) = d(f n (x), orb(c)) > δ Since M n (x) orb(c) {0, 1} it is clear that lim sup n r n (x) δ For the other direction, assume that lim sup n r n (x) = δ > 0 Since r n ω(c) 0 uniformly, there exists N such that r n ω(c) < δ 4 for all n N Let L := f(j) sup J J (supremum taken over all subintervals of [c 2, c 1 ]), and ε = δ Assume 4L N by contradiction that x B(ω(c)) Then there exists M such that d(f m (x), ω(c)) < ε for all m M Pick n N + M such that r n (x) δ 2 Then there exists neighbourhoods V U of f n N (x) such that f N V is monotone, f N (V ) = (f n (x) δ 2, f n (x) + δ 2 ), and f N (U) = (f n (x) δ 4, f n (x) + δ 4 ) By definition of ε, each component of U \ {f n N (x)} has size ε As n N M, there exists y U ω(c) But that means that r N (y) d(f N (y), f N ( V )) δ, contradicting the definition of N 4 4 The Induced Map F In this section we introduce the induced map used in sections 6 and 8 We will construct a countable interval partition, given by points {u k } {û k } These points

12 12 HENK BRUIN c ẑ k û k ẑ k 1 f S k 1 d k 1 z Q(k) u Q(k)+1 z Q(k)+1 c Figure 41 are defined as follows: û 1 = p is the orientation reversing fixed point of f and hence u 1 = ˆp Assume that u i is defined for every i < k The iterate f Sk 1 maps (z k 1, c) and (c, ẑ k 1 ) diffeomorphically onto (d k 1, c) Let u k (z k 1, z k ) be such that { uq(k)+1 (z Q(k), z Q(k)+1 ) if d k 1 < c, f Sk 1 (u k ) = û Q(k)+1 (ẑ Q(k)+1, ẑ Q(k) ) if d k 1 > c See figure 41 Then also f Sk (u k ) = f S Q(k) f Sk 1 (u k ) = f S Q(k) (u Q(k)+1 ) = u Q(Q(k)+1)+1 Let U = (ˆp, p) and U k = (u k 1, u k ) The induced map F is defined as F U k Ûk = f Sk 1 Hence { (uq(k)+1, û Q(Q(k 1)+1)+1 ) if d k 1 < c F (U k ) = F (Ûk) = (u Q(Q(k 1)+1)+1, û Q(k)+1 ) if d k 1 > c Clearly F is a Markov map, ie F preserves the partition {U k } F is also an extendible Markov map By this we mean that f Sk 1 (H Sk 1 (U k )) H Sr 1 (U r ) whenever F (U k ) U r Here H Sk 1 (U k ) is the maximal interval containing U k on which f Sk 1 Indeed, as H Sk 1 (U k ) = (z k 2, c), is monotone f Sk 1 ((z k 2, c) = (d Q(k 1), d k 1 ) H Sr 1 (U r ) = (z r 2, c), for each U r F (U k ) As F is a Markov map, there exist well-defined cylinder sets: U i0,i n 1 = {x F m (x) U im Û im for 0 m < n} Because F is extendible Markov, its iterates have the same property: Lemma 41 Let V = U i0,,i n 1 be an arbitrary cylinder Therefore F n 1 (V ) = U k, k = i n 1 and F n (V ) = f r (V ) = (u Q(k)+1, û Q(Q(k 1)+1)+1 ) or its involution Then f r (H r (V )) = f Sk 1 (H Sk 1 (U k )) = f Sk 1 ((z k 2, c)) = (d Q(k 1), d k 1 ) Proof Use induction

13 ABSORBING CANTOR SETS 13 5 Random Walks Governed by F The main tool for the proof of the (non-)existence of absorbing Cantor sets is a random walk on the states U k The transitions from one state to another are given by the induced map Write ϕ n (x) = k if F n (x) U k Ûk In order to prove that x B(ω(c)), one need to check the asymptotic behaviour of ϕ n (x) Lemma 51 i) If Q(k) and lim n ϕ n (x) =, then x B(ω(c)) ii) If r n (c 1 ) 0 and x B(ω(c)), then lim n ϕ n (x) = Proof i) As lim n ϕ n (x) =, F n 1 (x) c 0 By definition, F n 1 (x) U ϕn 1 (x) Because Q(k), also F n (x) d ϕn 1(x) 1 0 It follows from the Contraction Principle, that also for the intermediate iterates (ie for r 1 < r < r 2 where F n 1 = f r1 and F n = f r2 ), d(f r (x), f r r1 (c)) 0 as n Hence d(f r (x), ω(c)) 0 ii) Assume by contradiction that k 0 := lim inf n ϕ n (x) < Let δ := min k k 0 min{ z k 2 u k 1, u k c } Take n, s such that F n (x) = f s (x) U k for some k k 0 Then by Lemma 41, H s (x) (z k 2, c), whence r s (x) δ > 0 Since this holds for arbitrarily large s, lim sup s r s (x) > 0 According to Corollary 34, x / B(ω(c)) The asymptotic behaviour of ϕ n can be computed from the expectation E(ϕ n ϕ n 1 ), taken with respect to normalized Lebesgue measure on U If E(ϕ n ϕ n 1 ) ε > 0, then we expect that lim n ϕ n (x) = almost surely To prove this, we will use conditional expectations We also need boundedness of the variance For this reason, we will use functions ψ n ϕ n, which satisfy ϕ n if and only if ψ n, but also have bounded conditional variances Theorem 52 Let ψ n : U R, n N, be functions satisfying the following conditions: - ψ n 1 is constant on each cylinder U i0,i n 1 - There exist k 1 N and δ > 0 such that for every cylinder U i0,i n 1 with i n 1 k 1, E(ψ n ψ n 1 U i0,i n 1 ) δ - V ar(ψ n ψ n 1 U i0,i n 1 ) is uniformly bounded Then the set X = {x U ψ n (x) } has positive Lebesgue measure Proof We first restrict ourselves to X 0 = {x ϕ n (x) k 1 for all n} Define Ψ m = (ψ m ψ m 1 ) E((ψ m ψ m 1 ) U i0i m 1 ) Then E(Ψ m U i0i m 1 ) = 0 and there exists V such that V ar(ψ m U i0i m 1 ) = E(Ψ 2 m U i0i m 1 ) < V for all m and all cylinders U i0i m 1 Let T n := n m=1 Ψ m, so E(T1 2) = E(Ψ2 1 ) V By assumption T n 1 is constant on each set U i0i n 1 Suppose by induction that E(Tn 1) 2 (n 1)V, then E(Tn 2 ) = E(T n 1 2 ) + E(Ψ2 n ) + 2E(Ψ nt n 1 ) (n 1)V + V + 2 E(Ψ n T n 1 U i0i n 1 ) U i0 i n 1 nv + 2 T n 1 E(Ψ n U i0i n 1 ) = nv U i0 i n 1

14 14 HENK BRUIN = V By the Chebyshev inequality P ( T n > nε) nv n 2 ε 2 nε In particular P ( T 2 n 2 > n 2 ε) V Therefore n 2 ε 2 n P ( T n 2 > n2 ε) < and by the Borel-Cantelli Lemma, P ( T n 2 > n 2 ε infinitely often) = 0 As ε is arbitrary, T n 2 n 2 0 as Now for the intermediate values of n, let D n := max n 2 <k<(n+1) 2 T k T n 2 Because T k T n 2 = k j=n 2 +1 Ψ j, E( T k T n 2 2 ) (k n 2 )V 2nV Hence (n+1) 2 1 E(Dn) 2 E( T k T n 2 2 ) k=n 2 +1 (n+1) 2 1 k=n nV = 4n 2 V Using Chebyshev s inequality again, we obtain P (D n n 2 ε) 4n2 V n 4 ε = 4V 2 n 2 ε Hence 2 P (D n n 2 ε infinitely often) = 0, and Dn 0 as Combining things and taking n 2 n 2 k < (n + 1) 2, we get T k k T n 2 + D n n 2 0 as Because ψ m ψ m 1 = Ψ m + E(ψ m ψ m 1 U i0i m 1 ) Ψ m + δ, lim inf n 1 n n i=2 (ψ i ψ i 1 ) lim inf n 1 n T n + δ δ as Hence ψ n (x) for ae x X 0 Therefore X0 X Uk U k 1 as k, whence X > 0 6 Maps having Absorbing Cantor Sets Theorem 61 Let f be a non-renormalizable S-unimodal map with critical order l < Let its kneading map Q satisfy the properties: There exists N, k 1 N such that for all k k 1, (61) Q(k + 1) > Q 2 (k) + 1 and (62) Q(k) k N Then there exists l 0 = l 0 (N) such that, if l l 0, then f has an absorbing Cantor set Remark Condition (61) is rather unnatural It will be used effectively in Lemma 102 Note that (61) is satisfied if f is finitely renormalizable (use Lemma 23) and Q is eventually non-decreasing Note also that (61) immediately implies (24)

15 ABSORBING CANTOR SETS 15 Hence Lemma 25 applies We need (61) for the estimates, but we don t believe that it really affects the existence of an absorbing Cantor set According to Theorem 52 it suffices to check certain conditional expectations and variances This will be done in Proposition 62 This proposition uses the following metrical conditions: There exists C, l 0 1 and k 1 N such that for all l l 0 and k k 1 1 (63) Cl u k 1 u k C u k 1 c l and (64) 1 Cl d k 1 u Q(k)+1 C u Q(k)+1 c l The proofs of these statements are a generalization of the proofs in [BKNS] They are very technical; we put them in section 10 Proposition 62 Let f be a Fibonacci-like S-unimodal map having critical order l and kneading map Q Let k 1 be such that (63) and (64) hold Then there exists l 0 < such that if l l 0, and U ϕ0,ϕ n 1 is a cylinder with ϕ n 1 > k 1, E(ϕ n ϕ n 1 U ϕ0,ϕ n 1 ) > 1 and V ar(ϕ n ϕ n 1 U ϕ0,ϕ n 1 ) is uniformly bounded Proof Let V = U ϕ0,ϕ n 1 be any cylinder set such that ϕ n 1 k 1 Since F is extendible Markov, the Koebe space corresponding to F n V depends only on the last application of F in the composition In other words, suppose that F n = f s and F n 1 (V ) = U k, then f s (H s (V )) = f Sk 1 ((c, z k 2 )) = (d Q(k 1), d k 1 ) To be definite, assume that d Q(k 1) < c < d k 1 We will only calculate the conditional expectation and variance on the part V 0 V that is mapped to (u Q(Q(k 1)+1)+1, c) The estimates for the part V 1 of V which is mapped onto (c, û Q(k)+1 ) are similar Abbreviate q = Q(Q(k 1) + 1) + 1 > k 2N Let for r q, v r V be such that F n (v r ) = u r and F n (v ) = c Let also V r = (v r, v r 1 ), see figure 61 c v v q+[l] V r v r v r 1 v q F n V u q+[l] u r u q d Q(k 1) Figure 61

16 16 HENK BRUIN By (63), (1 C l ) ui+1 c u i c (1 1 ), whence Cl (1 C l )i j u i c u j c (1 1 Cl )i j Using also (64), we obtain that for l sufficiently large, and k < r q + l, there exist K = K(C, l) > 0 such that and d Q(k 1) u q d Q(k 1) u r d Q(k 1) u q u q c u q c u q 1 c 1 Cl 1 C l 1 (1 C l )r q K r q, 1 1 ur c u q 1 c U r u q u r 1 = U r u r 1 c 1 u r 1 c u q c 1 ur 1 c u q c 1 Cl (1 C l )r 1 q 1 (1 C l )r 1 q K r q These relations and the expansion of cross-ratios (with L = U r, J = (u r 1, u q ) and R = (u q, d Q(k 1) )) yield, for k < r q + l, V r v r 1 v q = l j R T Hence the conditional expectation: E(ϕ n k V 0 ) = 1 V 0 (j k) V j L J d Q(k 1) u q U r d Q(k 1) u r u q u r 1 K2 (r q) 2 j 1 V 0 { 2N v k v q + (l 2N) v q+[l] v + q+[l] j=k+1 K 2 j k (j q) 2 v j 1 v q } 1 V 0 {(K2 2 log l 2N) v k v q + (l 2N) v q+[l] v } for l sufficiently large Hence E(ϕ n k V 0 ) as l Let us compute the variance For j k, the interval (u j, c) has Koebe space (d Q(k 1), u j ) which is at least of order 1 l This gives a distortion bound of O(l2 ) V ar(ϕ n k V 0 ) E((ϕ n k) 2 V 0 ) = 1 V 0 4N 2 + (j k) 2 V j V 0 j>k (j k) 2 V j 4N 2 + O(l 2 ) (j k) 2 u j c u k c j>k 4N 2 + O(l 2 ) (j k) 2 (1 1 Cl )j k = 4N 2 + O(l 5 ) j>k j

17 ABSORBING CANTOR SETS 17 This concludes the proof Proof of Theorem 61 Combine Proposition 62, Theorem 52 and Lemma 51 7 Estimates on uk uk+1 u k c In this section we concentrate on the relative space between u k and u l (or similarly, the relative space between z k and z l ) for l > k Let ε > 0 be arbitrary We introduce the function ρ ε to measure how much z l c is smaller than z k c in factors 1 ε: z l c (1 ε) ρε(l) ρε(k) z k c More precisely, we define ρ = ρ ε : N N by: ρ(0) := 0, and if l k 1 is the largest integer such that ρ(l) = ρ(k 1), then { z ρ(k 1) + 1 if l z k z ρ(k) := l c ε, ρ(k 1) otherwise Hence ρ(k 1) ρ(k) k for each k 1, and for k l, zk c z l c (1 ε) ρ(k) ρ(l) Because z k (u k, u k 1 ) for all k, also uk+1 c u l c (1 ε) ρ(k) ρ(l) In this section we will try to estimate ρ using conditions on Q For instance, if Q is eventually non-decreasing, then ρ(k) = ρ(k) + 1 for k sufficiently large and ε sufficiently small (One should think of ε = O(l 2 )) Hence u k c decrease exponentially in this case z k c z k 1 c Remark One of the main difficulties in getting estimates for for arbitrary unimodal maps is the occurrence of almost saddle node bifurcations We speak of an almost saddle node bifurcation of period n if the graph of the central branch of f n is disjoint from, but almost tangent to the diagonal At the bifurcation itself, ie when the graph of the central branch is tangent, a neutrally attracting n-periodic point, say q, is created Without loss of generality, assume that q < c Then the precritical points z k accumulate on q By a continuity argument one can show that close before the bifurcation, the precritical points z k, z k+1,, z k+j0 cluster together around the spot where q is going to appear Here the period n = S k S Q2 (k) and j 0 can be arbitrarily large Therefore ρ ε (i) can be constant for arbitrarily long pieces However, we think that the occurrence of almost saddle node bifurcations does not affect the (non-)existence of absorbing Cantor sets (The existence of absolutely continuous invariant probability measures is a totally different matter, cf [B1]) As we cannot support this remark by rigid estimates, we will exclude almost saddle node bifurcations by assumption The estimates in this section rely on one technique developed by Martens [Ma1-2], and involve nice points We will use the following weaker version of niceness: (71) f j (x) / (x, ˆx) for every j < S k, where S k is minimal such that c f Sk ((x, c)) Proposition 71 There exists δ = δ(l) > 0 with the following property: Let x (z k 1, z k ) satisfy (71) If the interval J and n N are such that f n J is the monotone first visit of J to (x, ˆx), then there exists T J such that f n T is monotone and f n (T ) contains a δ-scaled neighbourhood of (x, ˆx) Proof Let M := (x f, c 1 ) Then f Sk 1 (M) c and by assumption f j (M) (x, ˆx) = for j < S k 1 Let m < S k be such that f m (M) f j (M) for all j < S k Let

18 18 HENK BRUIN r, l < S k be such that f r (M) and f l (M) are the closest right and left neighbours of f m (M) that are still disjoint from f m (M) Both neighbours need not exist, for instance if m {0, 1} In those cases define the disjoint right or left neighbour as the appropriate boundary point of I It follows that [f l (M), f r (M)] contains a 1-scaled neighbourhood of f m (M) (This holds true also if the closest disjoint left or right neighbour does not properly exist and x c is sufficiently small) Let H be the maximal interval such that f m H is monotone and f m (H) [f l (M), f r (M)] We claim that f m (H) = [f l (M), f r (M)] Suppose by contradiction that f m (H) [f l (M), f r (M)] Let L be a component of H \ M If I L, then also f m ( L) I and we already have a contradiction We can assume, by maximality of H, that there exist j < m such that c f j ( L) As f j (M) (x, ˆx) =, f m (L) f m j 1 (M) and f m j 1 (M) f m (M) = This contradicts the definition of closest disjoint neighbour, proving the claim Due to the expansion of cross-ratio, we can pull-back [f l (M), f r (M)] along the -scaled neighbourhood of orbit M, f(m),, f m (M) and show that H contains a 1 3 M Let H = f 1 (H), then H contains a δ-scaled neighbourhood of (x, ˆx), where δ = δ(l) = O( 1 l ) Now let J and n be as in the statement of the proposition To finish the proof, it suffices to show that there exists a neighbourhood T J such that f n maps T monotonically onto H Let T J be the maximal neighbourhood of J such that f n T is monotone and f n (T ) H Suppose by contradiction that f n (T ) H Let L be a component of T \J By maximality, there exists j < n such that c f j (L) By assumption f j (J) (x, ˆx) =, so f n (L) f n j 1 (M) and f n j 1 (M) is disjoint from (x, ˆx) As f m+1 is monotone on each component of H \{c}, also f n+m+1 L is monotone, and f n+m+1 (L) f n+m j (M) which is disjoint from f m (M) However f Sk ((x, ˆx)) c, so n + m j < S k Again, this contradicts the definition of the closest disjoint neighbours of f m (M) Remark In the second half of the proof it is not necessary that f j (J) is disjoint from (x, ˆx) for all j < n It suffices if f j (J) (x, ˆx) = when c f j (L) Corollary 72 There exist K(l) and ε(l) > 0 such that if c j / (d k, ˆd k ) for all 0 < j < S k+1, then dis(f Sk 1, (z f k, cf )) K(l) and zk 1 zk z k 1 c ε(l) In particular ρ ε (k) = ρ ε (k 1) + 1 If Q is (eventually) non-decreasing, then d k is a closest return for every k sufficiently large (Lemma 24) This implies that there exists k 0, R N such that ρ(k) = k R for all k k 0 Proof We use the Proposition 71 with x = d k Because c j / (d k, ˆd k ) for 0 < j < S k+1, f j ((d k, ˆd k )) (d k, ˆd k ) = for 0 < j < S Q(k+1) Let b be the middle point between c and ˆd k, and let J, (z f k, cf ) J (z f k 1, cf ), be such that f Sk 1 (J) = (b, d k ) Let T = H Sk 1(c f ), then z f k 1 T As f Sk 1 (z f k 1 ) = d Q(k) / (d k, ˆd k ), f Sk 1 (T ) contains a one-sided 1 -scaled neigh- 4 bourhood of (b, d k ) at the side of b Because c j / (d k, ˆd k ) for j < S k, one can show that f Sk 1 (T ) also contains the component of H \ {c} containing d k Here H is as in the proof of Proposition 71

19 ABSORBING CANTOR SETS 19 Hence f Sk 1 (T ) contains the necessary δ-scaled neighbourhood of (b, d k ), where δ = δ(l) is as in Proposition 71 Hence f Sk 1 J, and f Sk 1 (z f k, cf ), has bounded distortion Moreover, J (1 + O(δ)) z f k cf, so the second statement follows by non-flatness The next corollary gives a weaker condition implying ρ(k) = ρ(k 1) + 1 Let p k (z k 1, z k ) (ẑ k, ẑ k 1 ) be the orientation reversing S k 1 -periodic point of f By definition of closest precritical point, f j ((z k 1, z k )) (z k 1, ẑ k 1 ) = for all 0 < j < S k 1 Hence p k is a nice point Corollary 73 There exist δ(l), ε(l) > 0 with the following properties: For k arbitrary, let n be such that f n 1 (t f k ) = c Assume that f n (c) / (p Q(k+1) 1, ˆp Q(k+1) 1 ) Then f Sk 1 (z f k, cf ) has Koebe space δ on the side of d k If also d k c < d Q(k) c, then zk 1 zk z k 1 c ε The assumptions are such that the positions in figure 71 hold For example, if (24) holds, then Lemma 25 shows that n = S k S Q2 (k) for k sufficiently large One can verify that the assumptions of Corollary 73 hold if Q(k + 1) > Q 2 (k) + 2 and Q(k + 1) > Q(Q(k) + 1) for k sufficiently large f n (c) z Q(k+1) 2 p Q(k+1) 1 z Q(k+1) 1 d k c Figure 71 Proof The proof is similar to the proof of Corollary 72 Take x = p Q(k+1) 1 Therefore x is nice and d k (x, ˆx) Let J c f be the maximal neighbourhood of c f such that f Sk 1 (J) (x, ˆx) Let L (c f, 1) be the maximal interval adjacent to J on which f Sk 1 is monotone According to the remark below the proof of Proposition 71, it suffices to check that f j (J) (x, ˆx) = if f j (L) c By assumption j = n 1, and f n (c) / (x, ˆx) Hence f n 1 (J) (x, ˆx) = Therefore the (one-sided) Koebe space is sufficiently large If also d Q(k) c > d k c, then z k 1 z k z k 1 c ε follows by non-flatness 8 Maps Having no Absorbing Cantor Sets Theorem 81 Let Q be a kneading map satisfying (81) k Q(k) as k Let f be a non-flat S-unimodal map and have kneading map Q Assume also that there exists N N and ε > 0 such that (82) ρ ε (k + N) > ρ ε (k), for all k sufficiently large Then f has no absorbing Cantor set From the previous section one can derive topological conditions that imply (82) The simplest such condition is that Q is (eventually) non-decreasing Hence Theorem 81 yields that every non-flat S-unimodal map with k Q(k) and Q(k + 1) Q(k) for k sufficiently has no absorbing Cantor set

20 20 HENK BRUIN Instead of F we will used a reinduced map G, which has better distortion properties Let in this section ϕ G n (x) = k if Gn (x) = k Clearly ϕ G n (x) if ϕ n(x) For the proof of Theorem 81 we use Theorem 52 Therefore we need to check certain conditional expectations and variances But even the adjusted functions ϕ G n are not adequate First of all ϕ G n are so to say in the wrong direction: We need to prove that ϕ G n (x) does not tend to infinity for most points Secondly ϕ G n will not give us bounded variances Therefore we will use a certain truncation of ϕ G n Proof of Theorem 81 Let k 0 be such that ρ ε (k + N) > ρ ε (k) for all k k 0 Let K := ( 1+ε ε )2 be a distortion bound, and let M N be so large that Mε K j 0 j(1 ε) j/n 1 Let k 1 k 0 + M + N + 2 be so large that Q(k) k M N 2 for all k k 1 Now we define the reinduced map G Let x be such that F n (x) is defined for all x Let i = i(x) 1 be the smallest integer which satisfies one of the following properties (Take r such that F i 1 (x) U r Û r ) - r < k 1 - r k 1 and there exists a neighbourhood V x such that F i V is monotone and F i (V ) (u k0, û k0 ) - r k 1 and F i (x) (u r M N 1, û r M N 1 ) Then G(x) := F i (x) For s k 1, G U s consist of a middle branch G : V (u s M N 1, û s M N 1 ) coinciding with F : V (u s M N 1, û s M N 1 ) All other branches of G U k are longer Note that G is well-defined for ae x U: Either x k,j f j (u k û k ) (which applies to countably many points only), or i(x) is finite Indeed, if i(x) =, then ϕ n (x) < ϕ n 1 (x) for all n This is of course impossible The branches of G n have nice distortion properties First of all, as G is an induced map an extendible Markov map F, G is also extendible Markov, and it inherits the Koebe-spaces of F Before giving detailed distortion results, we introduce some more notation Suppose V U ϕ G 0,ϕ G is a branch-domain of Gn Let V + n 1 r V be such that G n (V n + ) = Ûr and Vr V be such that G n (Vn ) = U r Assume ϕ G n 1(V ) = s k 1 Hence G n (V ) contains an ε-scaled neighbourhood of (u s, û s ) In particular, G n (V ) contains an ε-scaled neighbourhood of r s Gn (V r + Vr ) By the Koebe Principle, it follows that (83) K Gn (V ) V r s Gn (V r + Vr ) r s (V r + Vr ) 1 K G n (V ) V Assume that G n (V ) = (u a, û b ) and let W V be the interval such that G n (W ) = (u a, u s M ) or (û s M, û b ) Then we can estimate W V from below Indeed, take x V such that G n (x) = c, and let W 0 W and W 1 be the components of

21 ABSORBING CANTOR SETS 21 l j r h G n W W 0 \ W x W 1 L u G n (W ) s M J R c Figure 81 V \ {x} Let h : I V be the (unique) surjective Möbius transformation such that W V = h 1 (W ) h 1 (V ) and h 1 (W 0\W ) h 1 (W 1) = 1 Then the situation is as in figure 81 As G n (V ) (u s M, û s M ), G n (W 1 ) G n (W 0 \ W ) By the expansion of cross-ratio (84) W V = l t j r L R T J = 1 Gn (W ) G n (W 1 ) G n (V ) G n (W 0 \ W ) Gn (W ) G n (V ) Define ψ 0 (x) = ϕ G 0 (x) = ϕ 0 (x) and ψ n (x) = ψ n 1 max{ϕ G n (x) ϕ G n 1(x), M} Clearly ψ n (x) ϕ G n (x) for every n We claim that for all cylinders U ϕ G 0,ϕG n 1 with ϕ G n 1 k 1 E(ψ n ψ n 1 U ϕ G 0,ϕ ) 1 and V ar(ψ G n ψ n 1 U n 1 ϕ G 0,ϕn 1) G is uniformly bounded Indeed, let again V U ϕ G 0,ϕ G be a branch-domain of Gn Assume that n 1 ϕ G n 1(V ) = s k 1, then G n (V ) = (u a, û b ) (u s M N 1, û s M N 1 ), whence G n (V ) contains an ε-scaled neighbourhood of (u s M, û s M ) Using (83), (84) and the choice of M, E(ψ n ψ n 1 V ) 1 V { r s M M V + r M u s a u s M + û s b û s M u a û b Mε r s Mε K r s K (r s) (1 ε) ρ(r) ρ(s) (r s)(1 ε) r s N 1 Vr (r s) V r + Vr } r s (r s)k u r û r u a û b r s

22 22 HENK BRUIN For the variance ( the second moment), V ar(ψ n ψ n 1 V ) 1 V { M 2 V r + r s M 2 + (r s) 2 K u r û r u a û b r s Vr + (r s) 2 V r + Vr } r s M 2 + (r s) 2 K(1 ε) ρ(r) ρ(s) r s M 2 + K r>s(r s) 2 (1 ε) (r s)/n < M 2 + O( 1 ε 5 ) Suppose f has an absorbing Cantor set Hence by Lemma 51, lim n ϕ G n (x) = lim n ϕ n (x) = ae Yet from the above computations and Theorem 52, it follows that ψ n on a positive measured set X B(ω(c)) Therefore for x X both lim n ϕ G n (x) = and lim n ϕ G n (x) lim n ψ n (x) = This contradiction shows, by means of Lemma 51, that B(ω(c)) cannot have full Lebesgue measure 9 Absolutely Continuous Invariant Probability Measures In this section we give a sufficient condition for the existence of an absolutely continuous (with respect to Lebesgue) invariant probability measure (acip) Theorem 91 Let f be non-flat S-unimodal If there exists ε > 0 and k 0 N such that for all k k 0 - ρ ε (Q(k + 1) 1) > ρ ε (Q(Q 2 (k) + 1)), - ρ ε (Q(Q(k) + 1) 1) > ρ ε (Q(Q 2 (k) + 1)), and - i (1 ε)ρε(i) ρε(q(i+1)) <, then f has an acip The first condition on ρ ε implies that (24) holds Using the results in section 7 we can derive Corollary 92 Let Q(k) be eventually non-decreasing and also lim k k Q(k) log k = Then every non-flat S-unimodal map with kneading map Q has an acip Proof of Theorem 91 We will prove the theorem by checking that the Nowicki-van Strien summability condition [NS] is satisfied: f has an acip if Df n (c 1 ) 1 l < n=1 Let T k be the partial sum S k n=1 Df n (c f ) 1 l Then by the chain-rule T k+1 = T k + Df Sk (c f ) 1 l S Q(k+1) n=1 Df n (d f k ) 1 l By the assumptions and Lemma 25, there exists k 1 k 0 such that for all k k 1, f Sk 1 (t f k ) = d Q 2 (k) and d k d Q2 (k) > ε d k c

23 ABSORBING CANTOR SETS 23 l j r z f k c f t f k f S k 1 L J R c ẑ k d k d Q 2 (k) Figure 91 Using non-flatness and the expansion of cross-ratios, it follows that for k k 1 (see figure 91), Df Sk (c f ) = Df(d k ) Df Sk 1 (c f l 1 J ) = O(l) d k c j l 1 t O(l) d k c r R T L l O(lε) d k c l 1 d k c z f k cf O(lε) d k c l z f k cf l O(lε)(1 ε)l{ρ(q(k+1)) ρ(k)} Now for the factor S Q(k+1) n=1 Df n (d f k ) 1 l, we compare Df n (d f k ) with Df n (c f ) Take again k k 1 and consider the points in figure 92 z f Q(k+1) d f k c f t f Q(k+1) f S Q(k+1) 1 By assumption d Q 2 (k+1) d k+1 c d Q(k+1) d Q 3 (k+1) Figure 92 d Q2 (k+1) d k+1, d Q3 (k+1) d Q(k+1) ε 2 d k d Q(k+1) Hence by the Koebe Principle, there exists K = O( 1 ε 2 ) such that 1 K Df SQ(k+1) 1 (d f k ) Df S Q(k+1) 1 (c f ) K But also for n < S Q(k+1) (because the Koebe spaces are the same), 1 K Df SQ(k+1) n 1 (f n (d f k )) Df S Q(k+1) n 1 (f n (c f )) K The chain-rule gives Df n (d f k ) 1 Df n (c f ) K 2 k k 1 T k+1 = T k + Df Sk (c f ) 1 l S Q(k+1) n=1 Df n (d f k ) 1 l T k + O(lε)(1 ε) ρ(k) ρ(q(k+1)) K 2 l Combining things, we get for S Q(k+1) n=1 T k {1 + O(lεK 2 l )(1 ε) ρ(k) ρ(q(k+1)) } Df n (c f ) 1 l