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...gue almost every point spends at most a fraction γ0 of time inside W ⊂ Bp+1 ∪· · ·∪Bp+q, we have that µ0(W) < γ0. As f is C1+α and µ0 is absolutely continuous, we may use Pesin’s entropy formula (see =-=[Pes77]-=-): ∫ hµ0(f) = log ‖ detDf‖dµ0 ≥ µ0(W)m1 + (1 − µ0(W))M1 As µ0(W) ≤ γ0 and m1 < M1, we conclude that By (5), γ0m1 + (1 − γ0)M1 ≤ hµ0(f). αm2 + (1 − α)M2 < γ0m1 + (1 − γ0)M1 − l log(1 + δ0). So, we can ...

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...t W ⊂ R1. We say that a system (f, µ) is Bernoulli (respectively Markov), if it is ergodically equivalent to a subshift of finite type endowed with a Bernoulli (respectively Markov) measure. See e.g. =-=[Mañ87]-=- for definitions. Theorem B. Assume hypotheses (H1), (H2), (H3), (H4) hold with δ0 and β sufficiently small. Then there exists a unique invariant measure µmax with hµmax(f) = htop(f). This measure als...

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... {µ ∈ I; µ(V ) ≤ α}. Since Lebesgue almost every point spends at most a fraction γ0 < α of time inside V ⊂ Bp+1 ∪ · · · ∪ Bp+q, all the ergodic absolutely continuous invariant measures constructed in =-=[ABV00]-=- belong to Kα. In particular, Kα is nonempty. We will see that Kα contains the equilibrium states of potentials with low variation. Let us recall the ergodic decomposition theorem, as it is proven in ...

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...erywhere. By the theorem of Oseledets [Ose68], ∫ log ‖ detDf‖dη = l∑ λi. (6) On the other hand, we have that λl > − log(1 + δ0), since by hyphotesis, ‖Df(x) −1 ‖ ≤ 1 + δ0. By Ruelle’s inequality (see =-=[Rue78]-=-) we have that hη(f) ≤ i=1 i=1 s∑ ∫ λi = log ‖ detDf(x)‖dη − l∑ i=s+1 λi. (7) Since m2 = sup log ‖ detDf(x)‖ < M2 = sup x∈V x∈V c log ‖ detDf(x)‖ and η(V ) > α we have: ∫ hη(f) ≤ log | det Df|dη ≤ η(V...

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...µ such µ(V ) ≤ α lies in K. Denoting λ1(x) ≥ λ2(x) ≥ . . .λs ≥ 0 > λs+1 · · · ≥ λl(x) the Lyapunov exponents in x, we know that λi = λi(x) is constant η-almost everywhere. By the theorem of Oseledets =-=[Ose68]-=-, ∫ log ‖ detDf‖dη = l∑ λi. (6) On the other hand, we have that λl > − log(1 + δ0), since by hyphotesis, ‖Df(x) −1 ‖ ≤ 1 + δ0. By Ruelle’s inequality (see [Rue78]) we have that hη(f) ≤ i=1 i=1 s∑ ∫ λi...

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...on of ergodic measures µx in Kα, by the previous case, µa(H) = 1 for µ almost every a, and this implies that µ(H) = ∫ µx(H)dµ = 1. Now we need the notion of hyperbolic time, first introduced by Alves =-=[Alv00]-=-. 9Definition 4.6. We say that n is a hyperbolic time for x with exponent c, if for every 1 ≤ j ≤ n: j−1 ∏ k=0 ‖Df(f n−k (x)) −1 ‖ ≤ e −cj . To prove that for an f-expanding measure almost every poin...

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...e from the dynamical behaviour certain facts valid at almost every point, uniform versions of which are taken as hypotheses in Yuri’s approach. For instance, we need no analogue of hypothesis (C5) in =-=[Yur99]-=-: in fact, for our examples in Section 3 the diameters of cylinders do not tend to zero. Besides, there a low variation potential may not satisfies the condition (C4) in [Yur99]. A combination of both...

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...ess property: there is a fixed hyperbolic saddle point p0 such that the stable manifold of p0 is contained in the unstable manifold of two other fixed points. For a discussion of related examples see =-=[Car93]-=- and [BV00]. 4 Expanding measures and hyperbolic times The proof of Theorem A occupies this section and the next one. Let us begin by detailing a bit more our strategy to prove the existence of equili...

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... states belong to K. In fact, if η = {ηx} is the ergodic decomposition of η, we should prove that the set S = {x ∈ M; ηx ∈ Kα} is a η-full measure set. Using the fact that ∫ hη(f) = hηx(f)dη(x), (see =-=[Rok67]-=-, for instance), we have Fφ(η) = ∫ Fφ(ηx)dη(x) Suppose by contradiction that η(Sc ) > 0. Observe that if y ∈ Sc , then ηy is in the hyphoteses of lemma 5.6 and thus: ∫ Fφ(ηy) = hηy + φdηy < ρhtop(f) +...

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...ov exponents of (f, ν) are all positive if and only if ν is f N -expanding for some iterate N. See [ABV00]. Also, if every invariant measure ν is f-expanding, then f is a uniformly expanding map. See =-=[AAS03]-=-. In fact, the same conclusion holds, more generally, if all invariant measures have only positive Lyapunov exponents. See [Cao]. The next statement proves that all measures in K are f-expanding, with...

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...or, independent of x and n. In this setting, the pressure P is given by ∫ { P = sup hµ(f) + φdµ µ∈I } . Several authors have been studying equilibrium states for non-hyperbolic systems: Bruin, Keller =-=[BK98]-=- and Denker, Urbanski [DU92, Urb98], for interval maps and rational functions on the sphere, and Buzzi, Maume, Sarig [Buz99, BMD02, BS, Sar03] and Yuri [Yur99, Yur00, Yur03], for countable Markov shif...

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... of (2), these measures are non-uniformly expanding, that is, µφ-almost every point has only positive Lyapunov exponents. Moreover, is possible to prove that µφ has a kind of weak Gibbs property (see =-=[Yur00]-=-) as in 2(1), with ∼ = meaning equality up to a factor with subexponential growth on the orbit of each x. The basic strategy for the construction is to find a subset K of invariant probability measur...

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...nt measure ν is f-expanding, then f is a uniformly expanding map. See [AAS03]. In fact, the same conclusion holds, more generally, if all invariant measures have only positive Lyapunov exponents. See =-=[Cao]-=-. The next statement proves that all measures in K are f-expanding, with uniform exponent. Let c > 0 be as in (4). Lemma 4.5. Every measure in µ ∈ K is f-expanding with exponent c: limsup n→+∞ n−1 1 ∑...