Backward volume contraction for endomorphisms with eventual volume expansion

C. R. Acad. Sci. Paris, Ser. I 342 (2006) 259–262 http://france.elsevier.com/direct/CRASS1/

Dynamical Systems

Backward volume contraction for endomorphisms with eventual volume expansion ✩ José F. Alves a , Armando Castro b , Vilton Pinheiro b a Centro de Matemática da Universidade do Porto, Rua do Campo Alegre 687, 4169-007 Porto, Portugal b Departamento de Matemática, Universidade Federal da Bahia, Av. Ademar de Barros s/n, 40170-110 Salvador, Brazil

Received 8 March 2004; accepted after revision 17 November 2005

Presented by Jean-Christophe Yoccoz

Abstract We consider smooth maps on compact Riemannian manifolds. We prove that under some mild condition of eventual volume expansion Lebesgue almost everywhere we have uniform backward volume contraction on every pre-orbit of Lebesgue almost every point. To cite this article: J.F. Alves et al., C. R. Acad. Sci. Paris, Ser. I 342 (2006).  2005 Académie des sciences. Published by Elsevier SAS. All rights reserved. Résumé Contraction en arrière pour des endomorphisms en expansion. Nous considérons des transformations différentiables sur des varietés Riemannienes compactes. Nous montrons que dans une certaine condition modérée d’expansion de volume nous pouvons déduire que pour Lebesgue presque chaque point nous avons contraction uniforme de volume en arrière de chaque pré-orbite. Pour citer cet article : J.F. Alves et al., C. R. Acad. Sci. Paris, Ser. I 342 (2006).  2005 Académie des sciences. Published by Elsevier SAS. All rights reserved.

1. Statement of results Let M be a compact Riemannian manifold and let Leb be a volume form on M that we call Lebesgue measure. We take f : M → M any smooth map. Let 0 < a1
a2
a3
· · · be a sequence converging to infinity. We define   
(1) h(x) = min n > 0: det Df n (x)  an , if this minimum exists, and h(x) = ∞, otherwise. For n  1, we take
 Γn = x ∈ M: h(x)  n .

(2)

Theorem 1.1. Assume that h ∈ Lp (Leb), for some p > 3, and take γ < (p − 3)/(p − 1). Choose any sequence 0 < b1
b2
b3
· · · such that bk bn  bk+n for every k, n ∈ N, and assume that there is n0 ∈ N such that ✩

Work carried out at the Federal University of Bahia. Partially supported by FCT through CMUP and UFBA. E-mail addresses: [email protected] (J.F. Alves), [email protected] (A. Castro), [email protected] (V. Pinheiro).

1631-073X/$ – see front matter  2005 Académie des sciences. Published by Elsevier SAS. All rights reserved. doi:10.1016/j.crma.2005.11.024

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bn
min{an , Leb(Γn )−γ } for every n  n0 . Then, for Leb almost every x ∈ M, there exists Cx > 0 such that |det Df n (y)| > Cx bn for every y ∈ f −n (x). We say that f : M → M is eventually volume expanding if there exists λ > 0 such that for Lebesgue almost every x∈M   1 (3) sup logdet Df n (x) > λ. n
1 n Let h and Γn be defined as in (1) and (2), associated to the sequence an = eλn . Corollary 1.2. Assume that f is eventually volume expanding. Given α > 0 there is β > 0 such that for Lebesgue almost every x ∈ M there are Cx > 0 such that for every n  1 and any y ∈ f −n (x) (i) if Leb(Γn )
O(e−αn ), then |det Df n (y)| > Cx eβn ; τ τ (ii) if Leb(Γn )
O(e−αn ) for some τ > 0, then |det Df n (y)| > Cx eβn ; (iii) if Leb(Γn )
O(n−α ) and α > 4, then |det Df n (y)| > Cx nβ . Specific rates will be obtained in Section 4 for some eventually volume expanding endomorphisms. In particular, non-uniformly expanding maps such as quadratic maps and Viana maps will be considered. 2. Concatenated collections Let (Un )n be a collection of measurable subsets of M whose union covers a full Lebesgue measure subset of M. We say that (Un )n is a concatenated collection if: x ∈ Un and f n (x) ∈ Um ⇒ x ∈ Un+m .  Given x ∈ n
1 Un , we define u(x) as the minimum n ∈ N for which x ∈ Un . Note that by definition we have  x ∈ Uu(x) . We define the chain generated by x ∈ n
1 Un as C(x) = {x, f (x), . . . , f u(x)−1 (x)}. n−1  j −1 Lemma 2.1. Let (Un )n be a concatenated collection. If n
1 j =0 Leb(f (u (n))) < ∞, then we have  sup{u(y): y ∈ n
1 Un and x ∈ C(y)} < ∞ for Lebesgue almost every x ∈ M. Assume that for a given x ∈ M there exists an infinite number of chains Cj = {yj , f (yj ), . . . , f sj −1 (yj )}, j  1, containing x with sj → ∞. For each j  1 let 1
rj < sj be such that x = f rj (yj ). First we verify that lim rj = ∞. If not, then replacing by a subsequence, we may assume that there is N > 0 such that rj < N for every j  1.  −i This implies that yj ∈ N i=1 f (x) for every j  1. At this point we need the smoothness of f . By compactness of M, the points x in M such that #(f −i (x)) = ∞ are singular values of f i , i ∈ N. By Sard’stheorem, the set of −i singular values of a C 1 map is a zero Lebesgue measure set. So, for almost all x ∈ M we have #( N i=1 f (x)) < ∞. r j As the number of chains is infinite, we obtain a contradiction. Since rj → ∞ and x = f (yj ) ∈ f rj (u−1 (sj )),   n−1  j −1 j −1 then we have x ∈ n
k n−1 n
1 j =0 f (u (n)), ∀k  1. The assumption j =0 Leb(f (u (n))) < ∞ implies that  n−1 j −1 Leb( n
k j =0 f (u (n))) → 0, when k → ∞. This completes the proof of Lemma 2.1. Lemma 2.2. Let (Un )n be a concatenated collection. If sup{u(y): y ∈ periodic point, then f −n (x) ⊂ Un ∪ · · · ∪ Un+N for all n  1.



n
1 Un

and x ∈ C(y)}
N and x is not a

 Assume that sup{u(y): y ∈ n
1 Un and x ∈ C(y)}
N , and take z ∈ f −n (x). Let zj = f j (z) for each j  0. We distinguish the cases x ∈ C(z) and x ∈ / C(z). If x ∈ C(z), and since x is not a periodic point, then n
#C(z) = u(z) by definition of u(·) and u(z)
N , since N is an upper bound for u(z), x ∈ C(z). Hence n
u(z)
N
n + N and we / C(z), then letting u0 = u(z) we must have u0 < n. Let u1 = u(zu0 ). conclude that z ∈ Uu(z) ⊂ Un ∪ · · · ∪ Un+N . If x ∈ If u0 + u1 < n we take u2 = u(zu0 +u1 ). We proceed in this way until we find the first s
n such that n
u0 + · · · + us .

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Note that us = u(zu0 +···+us−1 ), and by the choice of s we must have x ∈ C(zu0 +···+us−1 ). Our assumption implies that u(zu0 +···+us−1 )
N , and so u0 + · · · + us
n + N . By construction we have z ∈ Uu0 , f u0 (z) = zu0 ∈ Uu1 , f u0 +u1 (z) = zu0 +u1 ∈ Uu2 , . . . , f u0 +···+us−1 (z) = zu0 +···+us−1 ∈ Uus . By the definition of a concatenated collection we conclude that z ∈ Uu0 +u1 +···+us . 3. Proofs of main results  Let us now prove Theorem 1.2. Suppose that h ∈ Lp (Leb), for some p > 3. This implies that n
1 np Leb(h−1 (n)) < ∞, and so there exists some constant K > 0 such that Leb(h−1 (n))
Kn−p for every n  1. Now, taking 0 < γ < (p − 3)/(p − 1) we have for some K  > 0 ∞

1−γ ∞ ∞    1−γ  −1  n Leb h (k)
n K  /np−1 < ∞. n=1

k=n

n=1 n |det Df (x)|  bn },

Defining Un = {x ∈ M: then we have that {U1 , U2 , . . .} is a concatenated collection with respect  to the Lebesgue measure. Moreover, setting Un∗ = Un \ (U1 ∪ · · · ∪ Un−1 ) one observes that Un∗ ⊂ m
n h−1 (m), for otherwise there would be x ∈ Un∗ ∩ h−1 (m) with m < n, and so am  bm > |det Df m (x)|  am , which is not possible. As |det Df j (x)| < bj for every x ∈ Un∗ and j < n, we get Leb(f j (Un∗ ))
bj Leb(Un∗ ) for each j < n. Hence ∞  n−1 

∞  n−1     Leb f j Un∗
bj Leb Un∗

n=n0 +1 j =0

n=n0 +1 j =0




∞ n 0 −1 

∞ n−1     bj Leb Un∗ + bj Leb Un∗ n=n0 +1 j =n0

n=n0 +1 j =0




n 0 −1

∞ 

bj +

n−1 

 bj Leb Un∗ .

n=n0 +1 j =n0

j =0

Now, we just have to check that the last term in the sum above is finite. Indeed, ∞ n−1  

∞ ∞ ∞ ∞ n−1         bj Leb Un∗
bj Leb h−1 (k)
nbn Leb h−1 (k)

n=n0 +1 j =n0

n=n0 +1 j =n0




∞ 



n

n=n0 +1

=

∞  n=n0 +1

 n

∞ 

k=n

 Leb h−1 (k)

k=n ∞ 

−γ

n=n0 +1 ∞ 

k=n

 Leb h−1 (k)

k=n

 Leb h−1 (k)

1−γ

< ∞.

k=n

Using the fact that f is eventually volume expanding we deduce that the set of periodic points of f has zero Lebesgue measure. Otherwise, there would be some n for which Leb(Fix(f n )) > 0 and almost every x ∈ Fix(f n ) having an expanding direction, by eventual volume expansion. By an implicit theorem function argument we deduce that Fix(f n ) has zero Lebesgue measure in a neighborhood of x. Applying Lemmas 2.1 and 2.2, we get for each generic point x ∈ M a positive integer number Nx such that if y ∈ f −n (x) then y ∈ Un+s for some 0
s
Nx . Therefore, |det Df n+s (y)| > bn+s  bn . Taking Cx = K −Nx , where K = sup{|det Df (z)| : z ∈ M}, we obtain Theorem 1.1: n+s (y)|   det Df n (y) = |det Df > Cx b n . |det Df s (x)| Now we explain how we use Theorem 1.1 to prove Corollary 1.2. Recall that in Corollary 1.2 we have an = eλn  for each n ∈ N. Assume first that Leb(Γn )
O(e−c n ) for some c > 0. Then it is possible to choose c > 0 such that bn = ecn , for n  n0 . The other two cases are obtained under similar considerations.

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4. Examples: non-uniformly expanding maps An important class of dynamical systems where we can immediately apply our results is the class of non-uniformly expanding dynamical maps introduced in [2]. As particular examples of this kind of systems we present below onedimensional quadratic maps and the higher dimensional Viana maps. Quadratic maps. Let fa : [−1, 1] → [−1, 1] be given by fa (x) = 1 − ax 2 , for 0 < a
2. Results in [3,6] give that for a positive Lebesgue measure set of parameters fa in non-uniformly expanding. Freitas [5] proves that for Benedicks–Carleson parameters there are C, c > 0 such that Leb(Γn )
C e−cn for every n  1. Thus, it follows from Corollary 1.2 that there exists β > 0 such for Lebesgue almost every x ∈ I there is Cx > 0 such that |(f n ) (y)| > Cx eβn for every y ∈ f −n (x). Viana maps. Let a0 ∈ (1, 2) be such that the critical point x = 0 is pre-periodic for the quadratic map Q(x) = a0 − x 2 . Let S 1 = R/Z and b: S 1 → R given by b(s) = sin(2πs). For fixed small α > 0, consider the map fˆ from ˆ q(s, ˆ x)), where q(s, ˆ x) = a(s) − x 2 with a(s) = a0 + αb(s), and gˆ is the S 1 × R into itself given by fˆ(s, x) = (g(s), 1 ˆ = ds (mod Z) for some integer d  2. For α > 0 small enough there uniformly expanding map of S defined by g(s) is an interval I ⊂ (−2, 2) for which fˆ(S 1 × I ) is contained in the interior of S 1 × I . Thus, any map f sufficiently close to fˆ in the√C 0 topology has S 1 × I as a forward invariant region. Moreover, there are C, c > 0 such that that there exists β > 0 such for Leb(Γn )
C e−c n for every n  1; see [1,4,7]. Thus, it follows from Corollary 1.2 √ 1 n β n for every Y ∈ f −n (X). Lebesgue almost every X ∈ S × I there is CX > 0 such that |det Df (Y )| > CX e References [1] J.F. Alves, V. Araújo, Random perturbations of nonuniformly expanding maps, Astérisque 286 (2003) 25–62. [2] J.F. Alves, C. Bonatti, M. Viana, SRB measures for partially hyperbolic systems whose central direction is mostly expanding, Invent. Math. 140 (2000) 351–398. [3] M. Benedicks, L. Carleson, On iterations of 1 − ax 2 on (−1, 1), Ann. Math. 122 (1985) 1–25. [4] J. Buzzi, O. Sester, M. Tsujii, Weakly expanding skew-products of quadratic maps, Ergodic Theory Dynam. Systems 23 (5) (2003) 1401–1414. [5] J. Freitas, Continuity of SRB measure and entropy for Benedicks–Carleson quadratic maps, Preprint, CMUP, 2004. [6] M. Jakobson, Absolutely continuous invariant measures for one-parameter families of one-dimensional maps, Comm. Math. Phys. 81 (1981) 39–88. [7] M. Viana, Multidimensional non-hyperbolic attractors, Publ. Math. IHES 85 (1997) 63–96.