Absolutely Continuous Invariant Measures for Nonuniformly Expanding Maps

Huyi Hu ^{∗}  Sandro Vaienti ^{†}

February 19, 2012

Abstract For a large class of nonuniformly expanding maps of ℝ^{m}, with indifferent fixed points and unbounded distortion and non necessarily Markovian, we construct an absolutely continuous invariant measure. We extend to our case techniques previously used for expanding maps on quasi-Hölder spaces. We give general conditions and provide examples to which apply our result.

0 Introduction

The aim of this paper is to treat a class of multidimensional nonsingular transformations with indifferent fixed points which do not enjoy any Markov property. These maps exhibit two major difficulties. First the presence of discontinuities (the boundaries of the domains of local injectivity); second the nonuniformity caused by the presence of the indifferent fixed points. While there are several techniques to handle with the former point (see e.g. [C], [Bu], [S]), there is an essential difficulty for the latter one: unbounded distortion.

It is well known that for a nonuniformly expanding map T on the unit interval with an indifferent fixed point p, unbounded distortion occurs at the fixed point (see for instance the examples treated in [Pi] and [Th]). That is, for any neighborhood U of p, there are points x ∈ U such that |(T_{1}^{−n})′(p)/(T_{1}^{−n})′(x)| is unbounded as n increases, where T_{1} denote the restriction of T to some neighborhood of p. However, this only happens at the indifferent fixed point p; if we remove an arbitrarily small neighborhood of p, and we consider the first return map with respect to its complement, then the reduced system has bounded distortion. Unfortunately, this is not the case for nonuniformly expanding maps T in higher dimensional spaces with an indifferent fixed point p (we just consider only one indifferent fixed point for convenience). The system has unbounded distortion in the following sense: there are uncountably many points z such that for any neighborhood V of z, we can find ẑ ∈ V such that the ratio

∗Mathematics Department, Michigan State University, East Lansing, MI 48824, USA. email: <hu@math.msu.edu>.

†UMR-6207 Centre de Physique Théorique, CNRS, Universités d’Aix-Marseille I, II,Université du Sud Toulon-Var and FRUMAM, Fédéderation de Recherche des Unitiés de Mathématiques de Marseille; address: CPT, Luminy Case 907, F-13288 Marseille Cedex 9 e-mail: <vaienti@cpt.univ-mrs.fr>.

| det DT_{1}^{−n}(z)|/| det DT_{1}^{−n}(ẑ)|

is unbounded as n → ∞ (see Example 1 in Section 2). The points with unbounded distortion are not given explicitly by the map. Therefore, the methods for the one dimensional case cannot be applied directly here, since the first return map with respect to the complement of any neighborhood of p still have unbounded distortion.

Distortion estimates play an important role for the existence of absolutely continuous invariant measures in the case of hyperbolic or expanding maps. This is because the bounds of distortion give the bounds of the ratio of the density function. In many works, bounded distortion are either assumed or proved (e.g. [Yo1, Yo2], [ABV], [FJ], [Yu1, Yu2], [BPS]). However, for many systems the density function h(x) may be only an L^{1} function and, the ratio h(x)/h(y) may be unbounded as well on close points x and y. Therefore we need some techniques to handle these situations.

This work is an attempt toward this direction: we will prove the existence of absolutely continuous invariant measures for maps with indifferent fixed points in higher dimensions and in presence of unbounded distortion.

Existence of absolutely continuous invariant measures for expanding systems with an indifferent fixed point was proved for one dimensional cases in 1980 ([Pi], [Th]). However, there is no corresponding results for higher dimensional cases, except for some special examples ([Yu2], [H]).

In this paper we are able to cover an open set of maps in the space of expanding systems with an indifferent fixed point p whose local expression is an isometry plus homogeneous terms and higher order terms (see Example 3 and Remark 2.1 thereafter). We could also deal with maps whose differential has at least one eigenvalue greater than one at the indifferent fixed point (see Example 2). Actually our assumptions (1 to 4) are formulated in a general way with the attempt to capture and control the delicate behavior around the indifferent fixed point due to the lack of bounded distortion. We provide in Theorem B and C sufficient conditions to check those assumptions and we successively apply them to examples in Section 2.

What we get here could not be derived easily from other existing results. Since the distortions are unbounded, Young’s results [Yo1, Yo2] do not follow directly. Neither do Yuri’s techniques [Yu1, Yu2] for the same reasons, even if we model our map to give it a Markov or a finite range structure. Also, the condition lim sup_{n→∞} \frac{1}{n} ∑_{i=0}^{n−1} log ||DT (T^{i}(x))^{−1}|| < 0 in [ABV] cannot be obtained in our case since T may admit a σ-finite absolutely continuous invariant measure. If we study the first return map T̂ instead, then ∥DT̂_{x}(v)∥ can be arbitrary large for x close to the discontinuity set, and therefore the assumptions on the critical set in [ABV] are not satisfied.

Our construction consists of the following steps. We first replace the transformation T by a first return map T̂ with respect to a domain outside a neighborhood of the indifferent fixed point to get a uniformly expanding map with a countable number of discontinuity surfaces. Then we prove a Lasota-Yorke [LY] inequality on the induced system by acting the Perron-Frobenius (PF) operator on the space of “quasi-Hölder” functions to obtain a density function ĥ. This result is interesting in itself since it extends the work of [S] to piecewise expanding maps with a countable number of branches. The density function ĥ defines an absolutely continuous probability measure μ̂ invariant under T̂; then we extend μ̂ to an absolutely continuous invariant measure μ for T. The measure μ has finitely many ergodic components, and these could be finite or infinite, depending on the behavior of T near the fixed point. Moreover these maps can be arranged in such a way that the absolutely continuous invariant measure has both finite and infinite components that lie side by side (see Example 1).

The space of quasi-Hölder functions, introduced by Keller [K], developed by Blank [Bl] and successfully applied by Saussol [S] and successively by Buzzi [Bu] (see also [BK]) and Tsujii [Ts] to the multidimensional expanding case, reveals to be very useful to control the oscillations of a function under the iteration of the PF operator across the discontinuities of the map ^{∗}. Our result shows that it is also useful for unbounded distortion caused by nonuniform expansion, since as we point out in Remark 1.9, the oscillations of test functions are produced under iteration, not only by the propagation of discontinuities, but also by the distortion of the determinant of the map.

The plan of the paper is the following: in Section 1 we state the assumptions and the main theorem (Theorem A) about existence of absolutely continuous invariant measures, and give sufficient conditions (Theorems B and C) to produce a wide class of maps which fit the preceding assumptions. In Section 2 we study carefully some concrete examples. The proofs of Theorem B and C are respectively in Section 3 and 4. Section 5 contains the proof of Theorem A, while the last section, Section 6, deals with the proof of the Lasota-Yorke inequality for the induced system.

∗The use of the more standard space of bounded variation functions (in the sense of distribution) allowed as well to get absolute invariant measures for some class of piecewise uniformly expanding maps, see for instance [BG, PGB, Ad, C], but they need a stronger control of the geometrical shape of the discontinuity surfaces.

1 Assumptions and statements of results

Let M ⊂ ℝ^{m} be a compact subset with \overline{int M} = M and d be the Euclidean distance. Let ν be the Lebesgue measure on M. We assume νM = 1.

Let T : M → M be a map satisfying the following assumptions.

For A ⊂ M and ε > 0, denote B_{ε}(A) = {x ∈ ℝ^{m}; d(x, A) ≤ ε}.

Assumption 1. (Piecewise smoothness) There are finitely many disjoint open sets U_{1}, ⋯, U_{K} with M = ⋃_{i=1}^{K} \overline{U_{i}} such that for each i,

(a) T_{i} := T |_{U_{i}} : U_{i} → M is C^{1+α};

(b) T_{i} can be extended to a C^{1+α} map T_{i} : Ũ_{i} → M such that T_{i}Ũ_{i} ⊃ B_{ε_{1}}(T_{i}U_{i}) for some ε_{1} > 0, where Ũ_{i} is a neighborhood of U_{i}.

Assumption 2. (Fixed point) There is a point p ∈ U_{1} such that:

(a) T p = p;

(b) T^{−1}{p} ∉ ∂U_{j} for any j.

Since M ⊂ ℝ^{m}, we may take a coordinate system such that p = 0. Hence, we write |x| = d(x, p) if x ∈ M.

For any x ∈ U_{i}, we define s(x) = s(x, T) as the inverse of the slowest expansion near x by:

s(x, T) = min{s : d(x, y) ≤ sd(T x, T y), y ∈ U_{i}, d(x, y) ≤ min{ε_{1}, 0.1|x|} }.

where the factor 0.1 makes the ball away from the origin, though any other small factor would work as well.

Denote by γ_{m} the volume of the unit ball in ℝ^{m}.

Assumption 3. (Expanding Rates) There exists an open connected region R bounded by a smooth surface with p ∈ R, \overline{R} ⊂ T R, \overline{T R} ⊂ U_{1} and with either \overline{R} ⊂ T U_{j} or \overline{R} ∩ T U_{j} = ∅ such that:

(a) ∀x ∈ M \ {p} we have 0 < s(x) ≤ 1 , and if s(x) = 1 then x ∈ R and |T x| > |x|;

(b) there exist constants η_{0} ∈ (0, 1), ε_{2} > 0 such that

s^{α} + λ ≤ η_{0} < 1,

where

s := max{s(x) : x ∈ M\R},

λ = max\{2 sup_{ε_{0}≤ε_{2}} sup_{ε≤ε_{0}} \frac{G_{U}(ε, ε_{0})}{ε^{α}}ε_{0}^{α}, \frac{3sγ_{m−1}}{(1 − s)γ_{m}} \},  (1.1)

G_{U}(ε, ε_{0}) = sup_{x∈M} G_{U}(x, ε, ε_{0}),  (1.2)

and

G_{U}(x, ε, ε_{0}) = ∑_{j=1}^{K} \frac{ν(T_{j}^{−1}B_{ε}(∂T U_{j}) ∩ B_{(1−s)ε_{0}}(x))}{ν(B_{(1−s)ε_{0}}(x))};

(c) there exists N = N_{s} > 0 and ε_{3} > 0 such that for all x ∈ B_{ε_{3}}(T R\R),

s(T_{1}^{−N}(x), T_{1}^{N}) ≤ \frac{s}{5m}(\frac{λ(1 − s)^{m}}{2C_{p}I^{2}} )^{1/α}

for λ given by (1.1) and where I and C_{p} are constants defined below in Assumption 4(c).

We say that T : M → M is an almost expanding piecewise smooth map with an indifferent fixed point p if it verifies Assumption 1, 2 and 3(a), and s(p) = 1.

Remark 1.1. We first observe that Assumption 1 does not require any Markov property of the partition {U_{1}, ⋯, U_{K}}. Moreover, by Assumption 3(a), the map T_{j} : U_{j} → T_{j}(U_{j}) is noncontracting for each j, and therefore it is a local diffeomorphisms. Also, by the assumption, for any x ∈ U_{1}, T_{1}^{−n}x → p, because the set of limit points of {T_{1}^{−n}x} cannot contain any other point but p.

Remark 1.2. By Assumption 1(b), |detDT (x)| is bounded. This is because the map x → |detDT (x)| is continuous on \overline{U}_{i} for each i, and \overline{U}_{i} is compact.

Remark 1.3. Assumption 3(b) is the main assumption that requires uniform dilation outside R and gives condition on the relations between expanding rates and discontinuity. It is proved in [S] (see the proof of Lemma 2.1) that if the boundary of U_{i} consists of piecewise C^{1} codimension one embedded compact submanifolds, then G_{U}(ε, ε_{0}) ≤ 2Y \frac{γ_{m−1}}{γ_{m}} \frac{sε}{(1 − s)ε_{0}} (1 + o(1)), where Y is the

maximal number of smooth components of the boundary of all U_{i} that meet in one point. We refer to [S] for more details about the meaning of G_{U}(ε, ε_{0}).

Remark 1.4. We do not require the boundary of U_{i} to be piecewise smooth. In fact, they could be fractals as analyzed in [S]. However, Assumption 3(b) implies ν(∂U_{j}) = 0 for any j = 1, ⋯, K. ^{†}

Δ = \{x ∈ M : lim_{ε→0} \frac{ν(B_{ε}(x) ∩ ∂U_{j})}{νB_{ε}(x)} = 1 \} .

G_{U} (x, ε, ε_{0}) ≥ \frac{ν(T_{j}^{−1}B_{ε}(∂T U_{j}) ∩ B_{(1−s)ε_{0}}(x))}{ν(B_{(1−s)ε_{0}}(x))} ≥ \frac{ν(∂U_{j} ∩ B_{ε}(x))}{ν(B_{ε}(x))}

†In fact, if ν(∂U_{j}) > 0 for some j, then we take the set of the density points By the Lebesgue-Vitali Theorem (see, e.g. [SG], Chapter 10), νΔ = ν(∂U_{j}) > 0. In particular, Δ ≠ ∅. Therefore for any x ∈ Δ, if ε_{0} is sufficiently small and ε = (1 − s)ε_{0}, then is sufficiently close to 1, which contradicts the assumption.

Remark 1.5. We allow that s(x, T) = 1 for some x other than p. However we still need some expanding rate inside R. This is given by Assumption 3(c). If s(T_{1}^{−N}(x), T_{1}^{N}) can be arbitrarily small by taking N sufficiently large, then Assumption 3(c) is always true.

Denote R_{0} = T R\R. Clearly, R_{0} ⊂ U_{1} because of the choice of R.

Assumption 4. (Distortions)

(a) There exists c > 0 such that for any x, y ∈ T U_{j} with d(x, y) ≤ ε_{1},

|det DT_{j}^{−1}(x) − det DT_{j}^{−1}(y)| ≤ c| det DT_{j}^{−1}(x)|d(x, y)^{α},

where ε_{1} is given by Assumption 1(b);

(b) For any b > 0, there exist J > 0, ε_{4} > 0 such that for any ε ∈ (0, ε_{4}], we can find 0 < N = N(ε) ≤ ∞ with

\frac{| det DT_{1}^{−n}(y)|}{| det DT_{1}^{−n}(x)|} ≤ 1 + Jε^{α} ∀y ∈ B_{ε}(x), x ∈ B_{ε_{4}}(R_{0}), n ∈ (0, N],

and

∑_{n=N}^{∞} sup_{y∈B_{ε}(x)} | det DT_{1}^{−n}(y)| ≤ bε^{m+α} ∀x ∈ B_{ε_{4}}(R_{0});

(c) There exist constants I > 1, C_{p} > 0, ε_{5} > 0 such that for any 0 < ε_{0} ≤ ε_{5}, n > 0, there is a finite or countable partition ξ = ξ_{n} of B_{ε_{0}}(R_{0}) such that ∀A ∈ ξ, 0 < ε ≤ ε_{0}, diam(A ∩ B_{ε_{0}}(∂R_{0})) ≤ 5mε_{0},

\frac{ν(B_{ε}(∂R_{0}) ∩ A))}{ν(B_{ε_{0}}(∂R_{0}) ∩ A)} ≤ C_{p} (\frac{ε}{ε_{0}} )^{α} ,  (1.3)

whenever ν(T_{1}^{−n}(B_{ε_{0}}(∂R_{0})) ∩ A) ≠ 0, and for any x, y ∈ A,

\frac{| det DT_{1}^{−n}(y)|}{| det DT_{1}^{−n}(x)|} ≤ I.  (1.4)

Remark 1.6. In fact, Assumption 4(a) is a consequence of Assumption 1(a) since |detDT (x)| is bounded from above as we mentioned in Remark 1.2. However we state it here independently due to its importance for our arguments.

Remark 1.7. If T_{1}^{−1} has bounded distortion in B_{ε_{5}}(R_{0}) in the sense that for any J_{0} > 1, there is ε > 0 such that for any x, y ∈ B_{ε_{5}}(R_{0}) with d(x, y) ≤ ε and

for any n > 0, \frac{| det DT_{1}^{−n}(y)|}{| det DT_{1}^{−n}(x)|} ≤ J_{0}d(x, y)^{α}, then Assumption 4(b) and (c) are

true with ε_{4} = ε_{5} = ε_{0}.

Remark 1.8. Actually, by our proof the condition diam(A∩B_{ε_{0}}(∂R_{0})) ≤ 5mε_{0} in Assumption 4(c) can be replaced by

diam T_{1}^{−n}(A ∩ B_{ε_{0}}(∂R_{0})) ≤ s(\frac{λ(1 − s)^{m}}{2C_{p}I^{2}} )^{1/α}

for all n ≥ N_{s}, where s and N_{s} are given by Assumption 3(b) and (c) respectively (see (6.4)).

Remark 1.9. When we iterate the system, oscillations of the test functions are produced by both discontinuities ∂U_{j} and distortion of | det DT |. It is very common for an expanding system in a multidimensional space with an indifferent fixed point to have unbounded distortion near the fixed point, (see Example 1 in the next section). Assumption 4(b) requires that the distortion (namely the ratio \frac{| det DT_{1}^{−n}(y)|}{| det DT_{1}^{−n}(x)|}) is uniformly bounded only up to some n = N(ϵ) whenever

the points x and y are ϵ-close and | det DT_{1}^{−n}(y)| is summable for n ≥ N(ϵ) with the sum which is small (of the order of a power of ϵ).

Remark 1.10. Assumption 4(c) is a usual bounded distortion estimate since it controls uniformly the ratio \frac{| det DT_{1}^{−n}(y)|}{| det DT_{1}^{−n}(x)|} when the points x and y are chosen close enough in an element of a partition depending on n.

We are now ready to state our main result.

Theorem A. Suppose T : M → M satisfies Assumption 1 to 4. Then T admits an absolutely continuous invariant measure μ with at most finitely many ergodic components μ_{1}, ⋯, μ_{s} that are either finite or σ-finite, and the density functions of μ_{i} are bounded on any compact set away from p. Hence,

· μ is finite if ∑_{n=1}^{∞} ν(T_{1}^{−n}R) < ∞.

Moreover, if for any ball B_{ε}(x) in M, there exists Ñ = Ñ(x, ε) > 0 such that T ÑB_{ε}(x) ⊃ M, then the density function is bounded below by a positive number. Hence

· μ is σ-finite if ∑_{n=1}^{∞} ν(T_{1}^{−n}R) = ∞.

Remark 1.11. We will give an example in the next section showing that it is possible for μ to have both finite and σ-finite ergodic components simultaneously, and both contain the same indifferent fixed point p in their supports.

Since Assumption 4(b) and 4(c) are difficult to verify, we give some sufficient conditions in the next theorems.

The conditions satisfied by the maps studied in the second part of both Theorem B and C are the following:

There are constants γ′ > γ > 0, C_{i}, C′_{i} > 0, i = 0, 1, 2, such that in a neighborhood of the indifferent fixed point p = 0,

|x|(1 − C′_{0}|x|^{γ} + O(|x|^{γ′})) ≤ |T_{1}^{−1}x| ≤ |x|(1 − C_{0}|x|^{γ} + O(|x|^{γ′})), (1.5)
1 − C′_{1}|x|^{γ} ≤ ∥DT_{1}^{−1}(x)∥ ≤ 1 − C_{1}|x|^{γ},  (1.6)
C′_{2}|x|^{γ−1} ≤ ∥D^{2}T_{1}^{−1}(x)∥ ≤ C_{2}|x|^{γ−1}.  (1.7)

Remark 1.12. If a map T_{1} satisfies all of these inequalities, then ∥DT_{1}(p)∥ = ∥DT_{1}(p)^{−1}∥ = 1, and DT_{1}(p) is an isometry. If γ > 1, then D^{2}T_{1}(p) = 0.

Further, consider the space of C^{3} maps that is expanding at every point x in a neighborhood of p except at p itself and whose differential DT_{1}(p) is an isometry. Then it is easy to see that the maps satisfying (1.5) to (1.7) form a generic set in the C^{3} topology.

We would like to point out that the local behaviors given by the inequalities (1.5) to (1.7) allow us to apply the useful Lemmas 3.1 and 3.2 (see below), which permit a good control of the iterates of the Jacobian of the map. This will allow us to check assumptions 4(b) and 4(c) on the examples of the next section.

Theorem B. Suppose T : M → M verifies Assumption 1 to 3 and 4(a). Then the Assumption 4(b) is satisfied if one of the following two conditions holds:

i) There exists a constant κ ∈ (0, 1) such that | det DT | ≥ κ^{−1} > 1, and a constant α̂ > α such that T is C^{1+α̂} in a neighborhood of p. In this case, μ is finite if Assumption 4(c) also holds.

ii) There exists an open region R̃ ⊂ R containing p with T_{1}^{−L}R ⊂ R̃ for some L > 0, and constants γ′ > γ > 0, C_{0}, C_{1}, C_{2} > 0 such that the second inequalities in (1.5) to (1.7) hold; and there exist constants δ, τ > 0, C_{δ}, C_{τ} > 0 with

\frac{1}{γ(1 − α)} − τ < \frac{δ − 1}{m + α}  (1.8)

such that for any x ∈ R_{0}, n ≥ L,

| det DT_{1}^{−n}(x)| ≤ \frac{C_{δ}}{n^{δ}} ,  ∥DT_{1}^{−n}(x)∥ ≤ \frac{C_{τ}}{n^{τ}} .  (1.9)

In Theorem C, part i) below, we use the partial order x ≺ y between two points x and y if |x_{j}| < |y_{j}| for every j = 1, ⋯, m, where x_{j} and y_{j} are the jth component of x and y in ℝ^{m} respectively. In part ii) we denote by E(v_{1}, ⋯, v_{k}) the subspace spanned by vectors v_{1}, ⋯, v_{k}, and by E_{x}(S) the tangent space of a submanifold S at a point x ∈ S. Also, we may use a coordinate system (t, ϕ) near p where t = |x| and ϕ ∈ 𝕊^{m−1}, the m − 1 dimensional sphere.

Theorem C. Suppose T : M → M verifies Assumption 1 to 3 and 4(a). Then the Assumption 4(c) is satisfied if one of the following two conditions holds: i) There is a partition of T R into finite or countable number of regions {D_{i}}

such that ν(R \ ∪_{i}D_{i}) = 0, T (D_{i} ∩ R) = D_{i} ∩ T R and on each D_{i} ∩ T R, with the partial order decribed as above, T satisfies the following: x ≺ y implies | det DT (x)| < | det DT (y)|; if x ≺ y, then T_{1}^{−1}x ≺ T_{1}^{−1}y; x ≺ T x for any x ∈ D_{i} ∩ R; and for any y ∈ R_{0} ∩ D_{i} there is x ∈ ∂R ∩ D_{i} such that x ≺ y ≺ T x.

ii) Suppose T is C^{1+γ} and satisfies (1.5) to (1.7) near p. There are two families of cones {C_{x}} and {C′_{x}}, uniformly continuous in (t, ϕ) in the tangent bundle over the set T R, where t ≥ 0 and ϕ = 𝕊^{m−1} with (t, ϕ) ∈ T R, such that (a) DT_{x}(C_{x}) ⊂ C_{T x} and DT_{x}(C′_{x}) ⊃ C′_{T x} ∀x ∈ R; (b) there exists a positive angle θ_{0} such that for any x ∈ T R and v ∈ C_{x} and v′ ∈ C′_{x}, the angle between these two vectors is bounded from below by θ_{0};

(c) ∃d > 0, such that

\frac{| det DT_{x}|_{E(v,v′)}|}{∥DT_{x}|_{E(v)}∥ · ∥DT_{x}|_{E(v′)}∥} ≤ 1 − d|x|^{γ}  (1.10)

for any v, v′ ∈ C_{x}; and (d) C_{x} contains the position vector from p to x for all x ∈ T R, C′_{x} contains E_{x}(∂B_{ε}(R_{0})) for all x ∈ ∂(B_{ε}(R_{0})), 0 < ε ≤ ε_{5}, and

∥DT_{x}|_{E(∂(T_{1}^{−n}R))}∥ ≤ \frac{|T x|^{1/(1−θ)}}{|x|^{1/(1−θ)}} ∀x ∈ ∂(T_{1}^{−n}R), n > 0  (1.11)

for some θ with (1 + γ)(1 − θ) > 1.

Remark 1.13. The conditions in Theorem B.i) mean that DT_{p} has at least one eigenvalue with absolute value greater than 1.

Remark 1.14. In Theorem C. ii), the cones are not in general continuous with x at the point p, though they are continuous in (t, ϕ). (See Example 1.)

Remark 1.15. The condition in Theorem C.ii) part (c) implies that under DT, vectors in the cone C_{x} expands faster than that in C′_{x}. To see this, let us first recall that if a 2 × 2 matrix has two eigenvectors v_{1} and v_{2} with corresponding eigenvalues λ_{1} > λ_{2} > 0 respectively, then under iterations all vectors, except for v_{2}, move toward v_{1}, up to a coefficient. So if two vectors are close to v_{1}, then the angle between these two vectors decreases under iterations of A, and if we replace v_{1} by v_{2}, then the angle between them increases. In our case, cone invariance in part (a) corresponds the fact that DT_{x} has two eigenvectors and (1.10) implies that the angle between v and v′ becomes smaller under DT_{x}. (See Lemma 4.2.)

Remark 1.16. If we write DT (x) = T_{0}(x) + T_{γ}(x) + T_{h}(x), where T_{0} = DT_{p}, T_{γ} satisfies T_{γ}(tx) = t^{γ}T_{γ}(x) ∀t > 0 and |T_{h}(x)| = O(|x|^{γ′}), γ′ > γ, then the construction of the cones {C_{x}} and {C′_{x}} depends substantially on T_{γ} as x near p: this is explicitly seen in the proof of Lemma 4.3. So it is easy to get uniformity near t = 0.

2 Examples

Now we give three examples of maps for which all the assumptions of Theorem A can be checked. Since we allow discontinuities, it is easy to construct a map that satisfies Assumption 1 to 3 and 4(a). So our examples concentrate mostly on the local behavior of the map in the neighborhood of the indifferent fixed point p = 0, although a complete description is provided for example 1.

In the latter we show that the map has unbounded distortion, and the conditions in part ii) of Theorem B and C can be verified, and then that the map can have both finite and infinite components for the absolutely continuous invariant measure.

Example 1. We let M ⊂ ℝ^{2} and near the fixed point p = (0, 0), the map T has the form

T (x, y) = (x(1 + x^{2} + y^{2}), y(1 + x^{2} + y^{2})^{2})  (2.1)

up to order O(|z|^{4}), where z = (x, y) and |z| = \sqrt{x^{2} + y^{2}}.

It is easy to see that

DT (x, y) = ( \begin{matrix}1 + 3x^{2} + y^{2} + O(|z|^{4}) & 2xy + O(|z|^{4}) \\ 4xy + O(|z|^{4}) & 1 + 2x^{2} + 6y^{2} + O(|z|^{4}) \end{matrix} ) , (2.2)

and

det DT (x, y) = 1 + 5x^{2} + 7y^{2} + O(|z|^{4}),  (2.3)

Note that in this example, T is locally injective and T^{−1} will denote its inverse.

Unbounded distortion We begin to show that the distortion is unbounded even away from p = 0 in the sense that there are uncountably many points z such that for any neighborhood V of z, we can find ẑ ∈ V such that the ratio

| det DT_{1}^{−n}(z)|/| det DT_{1}^{−n}(ẑ)|  (2.4)

is unbounded as n → ∞. Take z′ = (x_{0}, 0) and denote z′_{n} = T^{−n}z′. By Lemma 3.1 in the next section,

we have |z′_{n}| ∼ \frac{1}{\sqrt{2n}}, where a_{n} ∼ b_{n} means lim_{n→∞} \frac{a_{n}}{b_{n}} = 1. Hence by (2.3) and

Lemma 3.2 with t_{i} = z_{i}, r(t_{i}) = |detDT (z_{i})|, C′ = 5 and γC = 2, we get

| det DT^{−n}(z′)| ≤ \frac{D′}{n^{5/2}} for some D′ > 0. On the other hand if we take z′′ =

(0, y_{0}) and denote z′′_{n} = T^{−n}z′′, then |z′′_{n}| ∼ \frac{1}{\sqrt{4n}} and | det DT^{−n}(z′′)| ≥ \frac{D′′}{n^{7/4}}

for some D′′ > 0. So \frac{| det DT^{−n}(z′′)|}{| det DT^{−n}(z′)|} → ∞ as n → ∞.

We take a curve from z′ to z′′ that does not contain the origin. If for every z on the curve, there is a neighborhood V such that for all ẑ ∈ V, the ratio in (2.4) is bounded for all n > 0, then the ratio | det DT^{−n}(z′′)|/| det DT^{−n}(z′)| should be bounded. This contradicts the above fact. So we know that there are some points on the curve at which distortion is unbounded. By moving z′ and z′′, we can get uncountably many pairwise disjoint curves and therefore we get what we need.

Validity of Assumption 4(b) For any z = (x, y), we put again z_{n} = T^{−n}z.

Note that

|z|(1 + |z|^{2} + O(|z|^{4})) ≤ |T z| ≤ |z|(1 + 2|z|^{2} + O(|z|^{4})),

or

|z_{n}|(1 + |z_{n}|^{2} + O(|z_{n}|^{4})) ≤ |z_{n−1}| ≤ |z_{n}|(1 + 2|z_{n}|^{2} + O(|z_{n}|^{4})).

So by Lemma 3.1, we have

\frac{1}{\sqrt{4(n + k)}} + O(n^{−β′}) ≤ |z_{n}| ≤ \frac{1}{\sqrt{2(n + k)}} + O(n^{−β′}),  (2.5)

for some integer k, where β′ > 1/2.

Since (2.3) implies that | det DT (z)|^{−1} ≤ 1 − 5|z|^{2} + O(|z|^{4}), by (2.5) and Lemma 3.2 we get

| det DT^{−n}(z)| ≤ Dn^{−5/2}.  (2.6)

Also by (2.2),

DT^{−1}(x, y) = ( \begin{matrix}1 − 3x^{2} − y^{2} + O(r^{4}) & −2xy + O(r^{4}) \\ −4xy + O(r^{4}) & 1 − 2x^{2} − 6y^{2} + O(r^{4}) \end{matrix} ) .

So ∥DT^{−1}(z)∥ ≤ 1 − |z|^{2} + O(|z|^{4}), hence by Lemma 3.2,

∥DT^{−n}(z)∥ ≤ D′n^{−1/2}  (2.7)

for some D′ > 0. Now by (2.6), (2.7) and (1.9), we know that δ = 5/2 and τ = 1/2. Since m = 2 and γ = 2, we have (1.8) if α = 1/2. By Theorem B.ii), T satisfies Assumption 4(b).

Validity of Assumption 4(c) It is easy to check that we can apply Theorem C.i) as we will do in Example 2 below. However, we use this map to show how to apply Theorem C.ii).

Notice that if we take two vectors v_{0} = (x, y)^{∗} and v′_{0} = (y, −x)^{∗} at the tangent plane of z = (x, y), where the asterisk denotes transpose, then by (2.2) we have

DT_{z}(v_{0}) = ( \begin{matrix}x + 3x^{3} + 3xy^{2} + O(|z|^{5}) \\ y + 6x^{2}y + 6y^{3} + O(|z|^{5}) \end{matrix} ) ,

DT_{z}(v′_{0}) = ( \begin{matrix}y + x^{2}y + y^{3} + O(|z|^{5}) \\ −x − 2x^{3} − 2xy^{2} + O(|z|^{5}) \end{matrix} ) .

This means that |DT_{z}(v′_{0})| < |DT_{z}(v_{0})|. We define C_{z} at each point z as the cone bounded by lines generated by vectors 3v_{0} + 2v′_{0} and 3v_{0} − 2v′_{0} and containing v_{0}, and define C′_{z} as the cone bounded by lines generated by vectors 3v′_{0} + 2v_{0} and 3v′_{0} − 2v_{0} and disjoint with C_{z}. We can check that part (a) and (b) in Theorem B.ii) are satisfied. Also we can check that for all unit vector v′ ∈ C′_{z}, |DT_{z}(v′)| ≤ |T z|^{2.5}/|z|^{2.5}. So if we take R in such a way that the tangent lines of ∂(T_{1}^{−n_{0}}R) are in the cones C′ for some n_{0} ≥ 0, then we use the fact DT^{−1}(C′) ⊂ C′ to get that part (c) is satisfied for all n ≥ n_{0} with 1 − θ = 2/5.

Coexistence of finite and σ-finite components We now arrange this map in such a way that the absolute continuous invariant measure μ has a finite and a σ-finite ergodic components simultaneously, and both contain the same indifferent fixed point p = 0 in their supports. We assume that our map satisfies Assumptions 1 to 3 and 4(a), besides the condition 4(b) and 4(c) checked above. We also assume that there is a partition of M = {M_{1}, M_{2}} such that for i = 1, 2, T M_{i} = M_{i} and for any ball B_{ε}(x) in M_{i}, there exists an integer N such that T^{N}B_{ε}(x) = M_{i}, and

{z = (x, y) ∈ B_{1}(p) : y < x^{2}} ⊂ M_{1}, {z = (x, y) ∈ B_{1}(p) : y > x^{2}} ⊂ M_{2}.

This is possible since it is easy to check that T Γ ∩ B_{1}(p) = Γ, where Γ = {(x, y) ∈ B_{1}(p) : y = x^{2}}. The assumptions just required insure the applicability of Theorem A. Since both M_{1} and M_{2} are invariant sets, T has absolutely continuous invariant measures μ_{1} and μ_{2} with respect to the Lebesgue measure restricted to M_{1} and M_{2} respectively. Now we show μ_{1}M_{1} < ∞ and μ_{2}M_{2} = ∞.

For this purpose we may assume that R = B_{1}(p). By (2.5), we know that T_{1}^{−n}R ⊂ B_{2/\sqrt{2n}}(p) for all large n. So

ν(T_{1}^{−n}R ∩ M_{1}) ≤ ν\{(x, y) : x^{2} + y^{2} ≤ \frac{4}{2n}, |y| ≤ |x|^{2}\} ≤ C(\frac{4}{2n})^{3/2}

for some C > 0. It follows that ∑_{n=1}^{∞} ν(T_{1}^{−n}R ∩ M_{1}) < ∞. Applying Theorem A

to the system T : M_{1} → M_{1}, we get that μ_{1}M_{1} ≤ ∞.

Also, by (2.5), we have that T_{1}^{−n}R ⊃ B_{1/2\sqrt{4n}}(p) for all large n. Hence it

is easy to see that ν(T_{1}^{−n}R) ≥ π/16n and therefore ∑_{n=1}^{∞} ν(T_{1}^{−n}R) = ∞. Since

ν(T_{1}^{−n}R ∩ M_{1}) + ν(T_{1}^{−n}R ∩ M_{2}) = ν(T_{1}^{−n}R), we get ∑_{n=1}^{∞} ν(T_{1}^{−n}R ∩ M_{2}) = ∞.

So we have μ_{2}M_{2} = ∞.

Full construction of the map We now show how to arrange this map in order to verify all the assumptions required by Theorem A.

Let M = {(x, y) ∈ ℝ^{2} : |x|, |y| ≤ 1000}. Take a partition {U_{1}, ⋯, U_{K}} of M such that

(i) U_{1} = {(x, y) ∈ ℝ^{2} : |x|, |y| ≤ 10}; (ii) diam U_{i} ≤ 10 if i ≠ 1; (iii) ∂U_{i} are piecewise smooth curves; and (iv) every point x ∈ M is contained in at most 3 ∂U_{i}s. Let R = {(x, y) ∈ ℝ^{2} : x^{2} + y^{2} ≤ (3 + ϵ_{1})^{2}}, where ϵ_{1} was used to define s(x)

in Section 1.

We define T : M → M such that

(a) on the set {(x, y) ∈ ℝ^{2} : x^{2} + y^{2} ≤ 4^{2}}, T has the form

T (x, y) = (x(1 + x^{2} + y^{2}), y(1 + x^{2} + y^{2})^{τ(x,y)})

where τ is a smooth function decreasing with x^{2} + y^{2} such that τ(x, y) = 2 if x^{2} + y^{2} ≤ 1; τ(x, y) = 1 if 2 ≤ x^{2} + y^{2} ≤ 3.5. So in the neighborhood of (0, 0), T has the form (2.1);

(b) T_{i} = T |_{U_{i}} is a C^{2} map for all i, and can be extended to an ϵ neighborhood of U_{i};

(c) s(x) < 0.1, ∀x ∉ R;

(d) R ∉ T (∂U_{i}) for any i. Note that on the boundary of R, s(x)^{−1} ≥ 10. In fact, observe that near the

boundary of R,

T (x, y) = (x(1 + x^{2} + y^{2}), y(1 + x^{2} + y^{2})).

It is easy to see by the choice of ϵ_{1} that at the point x = (3+ϵ_{1}, 0), s(x)^{−1} ≥ 10. Observe that DT is the same at every point of ∂R, up to a rotation. So we have that s(x)^{−1} ≥ 10, ∀x ∈ ∂R. This shows that such systems exist.

It is obvious that Assumption 1 to 2 are satisfied. Note that T R = {(x, y) : x^{2} + y^{2} ≤ (3 + ε_{1})(1 + (3 + ε_{1})^{2})} ⊂ U_{1}. So by (d)

the requirements for R in Assumption 3 are satisfied.

Assumption 3(a) is obvious. Since for x ∈ T R \ R, |T_{1}^{−n}x| has the order n^{−1/2}, we get ∥DT_{1}^{−1}(T_{1}^{−n}x)∥ ≤

1 − c/n for some c > 0. So we have ∥DT_{1}^{−n}(x)∥ → 0. By Remark 1.5, Assumption 3(c) follows.

Assumption 4(a) follows from (b). Now we verify Assumption 3(b). By (c) we have s ≤ 0.1. Note that γ_{2} = π,

γ_{1} = 2. By (iv), we can choose Y = 3. Then

\frac{4s}{1 − s}Y \frac{γ_{1}}{γ_{2}} ≤ \frac{0.4}{0.9}3\frac{2}{π} = \frac{8}{3π} ≤ 0.85.

So by Remark 1.3, we make Assumption 3(b) holds.

The next example shows how to use part i) in Theorem B and C to verify Assumption 4(b) and 4(c).

Example 2. Let T : ℝ^{3} → ℝ^{3} be given by

T (x, y, z)= (x(1 + x^{2} + y^{2} + z^{2}), y(1 + x^{2} + y^{2} + z^{2})^{2}, z(2 + x^{2} + y^{2} + z^{2})^{3})

as (x, y, z) near the origin.

Note that by an argument similar to that given in the preceding example, we can prove that for this map the distortion is unbounded away from the origin.

Since det DT_{(0,0,0)} = 2 and T is C^{∞} near the origin, by Theorem B.i), Assumption 4(b) is satisfied.

Let D_{i}, i = 1, ⋯, 8, be the eight octants in ℝ^{3}. Clearly all the requirements in Theorem C.i) are satisfied. So we get Assumption 4(c) as well.

The last example shows that our results cover an open set in the space of piecewise expanding maps T with an indifferent fixed point p such that DT (p) is an isometry, and s(x) > 1 ∀x ≠ p.

Example 3. Let T : ℝ^{m} → ℝ^{m} be given by

T (x)= Ex + H(x, x, x) + O(|x|^{4}),  (2.8)

where x ∈ ℝ^{m} is a column vector, E is an m × m orthogonal matrix, H : ℝ^{m} × ℝ^{m} × ℝ^{m} → ℝ^{m} is a linear map close to the map H_{0}(x, x, x) = x^{T} x(Ex), and x^{T} denote the transpose of x.

We first consider the case H = H_{0}. Then

T (x)= Ex(1 + x^{T} x) + O(|x|^{4}),  (2.9)

and

DT (x) = E(I + |x|^{2}I + 2xx^{T}) + O(|x|^{3}),

where I denote the m × m identity matrix. It is easy to see that

| det DT (x)| = 1 + (m + 2)|x|^{2} + O(|x|^{3}),
∥DT^{−1}(x)∥^{−1} = 1 + |x|^{2} + O(|x|^{3}).

So if H is close to H_{0}, then

| det DT_{p}(x)|^{−1} ≤ 1 − C̄′_{1}|x|^{2} + O(|x|^{3}),
∥DT_{p}^{−1}p(x)∥ = 1 − Ĉ_{1}^{′|}x|^{2} + O(|x|^{3})

for some C̄′_{1} close to m + 2 and Ĉ′_{1} close to 1.

Also note that if H is close to H_{0}, then

|T^{−1}(x)| ≥ |x|(1 − C′|x|^{2} + O(|x|^{3})),

for some C′ close to 1. So if we take x in a neighborhood of p and denote x_{n} = T^{−n}x, then by Lemma 3.1,

|x_{n−1}| ≥ \frac{1}{(2C(n + k))^{1/2}} + O(\frac{1}{(n + k)^{β′}} )

for some k > 0 and β > 1/2. By Lemma 3.2, we get that

| det DT_{1}^{−n}(x)| ≤ D̄(\frac{k}{n + k} )^{C̄′_{1}/2C}, ∥DT_{1}^{−n}(x)∥ ≤ D̂(\frac{k}{n + k} )^{Ĉ′_{1}/2C} (2.10)

for some D̄, D̂ > 0. Hence, we get (1.9) for δ = C̄′_{1}/2C and τ = Ĉ′_{1}/2C which are close to 1+m/2 and 1/2 respectively. Since γ = 2, we get (1.8) if α is close to 0. The requirements of Thereom B.ii) are satisfied. This verifies Assumption 4(b).

For each x we let E_{x} denote the one dimensional subspace spanned by the vector x, and E′_{x} be the orthogonal complement of E. Then we define the cones

C_{x} = {u + u′ : u ∈ E_{x}, u′ ∈ E′_{x}, |u′| ≤ |u|/2}
C′_{x} = {u + u′ : u ∈ E_{x}, u ∈ E′_{x}, |u| ≤ |u′|/2}

Note that if H = H_{0}, then we have |DT (x)(u)| = 1 + 3|x|^{2}|u| for u ∈ E_{x} and |DT (x)(u′)| = 1 + |x|^{2}|u′| for u′ ∈ E′_{x}. Using this fact we can check that all requirements in Theorem C.ii) are satisfied. Also note that DT change continuously with the third order term H of T. So if H is close to H_{0}, then the requirements of Theorem C.ii) are also satisfied. This verifies Assumption 4(c) for these T.

Remark 2.1. Consider the set S of all C^{4} maps T defined in a neighborhood of 0 in ℝ^{m} satisfying T (0) = 0, DT (0) = E, an isometry map, and s(x) < 1 if x ≠ 0, where s(x) is defined before Assumption 3. The condition s(x) < 1 implies D^{2}T (p) = 0. Hence the Taylor expression of T has the form (2.8), where H = 6D^{3}T.

If T_{a}, T_{b} ∈ S are close in C^{3} topology, then the corresponding H_{a} and H_{b} are close. So the example implies that our results cover an open subset of S that containing the map given in (2.9).

3 Proof of Theorem B

We first prove some lemmas.

For γ > 0, let β = 1/γ.

Lemma 3.1. If

t_{n−1} ≥ t_{n} + Ct_{n}^{1+γ} + O(t_{n}^{1+γ′})  ∀n > 0,  (3.1)

where γ′ > γ, then for all large n,

t_{n} ≤ \frac{1}{(γC(n + k))^{β}} + O(\frac{1}{(n + k)^{β′}} )  ∀n > 0  (3.2)

for some β′ > β and k ∈ ℤ. The result remains true if we exchange “≤” and “≥”. Therefore, if (3.1) becomes an equality, then so does (3.2).

Proof: We claim that if

t_{n−1} ≥ t_{n} + Ct_{n}^{1+γ} + C′t_{n}^{1+γ′},  (3.3)

for some large n and

t_{n}^{γ} ≥ \frac{1}{γCn}(1 + \frac{1}{n^{δ′}} )  (3.4)

for some δ′ > 0, then

t_{n−1}^{γ} ≥ \frac{1}{γC(n − 1)}(1 + \frac{1}{(n − 1)^{δ′}} ).

This gives the results since we can choose an integer k such that for some large n_{0} > 0,

t_{n}^{γ} ≤ \frac{1}{γC(n_{0} + k)}(1 + \frac{1}{(n_{0} + k)^{δ′}} ).

By relabelling the indices, the claim implies (3.2) for all n ≥ n_{0}.

Now we prove the claim. Denote γ_{n} = γ(1 + n^{−δ′})^{−1}. By (3.3) and (3.4),

t_{n−1}^{γ} ≥ t_{n}^{γ}(1 + Ct_{n}^{γ} + C′t_{n}^{γ′})^{γ} ≥ \frac{1}{Cnγ_{n}} (1 + \frac{C}{Cnγ_{n}} + \frac{C′}{(Cnγ_{n})^{γ′/γ}} )^{γ}.

To prove the lemma we only need to show that

\frac{1}{nγ_{n}} (1 + \frac{1}{nγ_{n}} + \frac{C′}{(Cnγ_{n})^{γ′/γ}} )^{γ} ≥ \frac{1}{(n − 1)γ_{n−1}} ,

or, equivalently,

\frac{n − 1}{n}(1 + \frac{1}{nγ} + \frac{1}{n^{1+δ′}γ} + \frac{C′}{(Cnγ_{n})^{γ′/γ}} )^{γ} ≥ \frac{γ_{n}}{γ_{n−1}} = \frac{1 + (n − 1)^{−δ′}}{1 + n^{−δ′}} .

Take δ′ < min{1, γ′/γ − 1}. Then (nγ_{n})^{−(γ′/γ)} is of higher order. We can check that as n → ∞, the left side of the inequality is like 1 + n^{−(1+δ′)} and the right side is like 1 + δ′n^{−(1+δ′)}. Since δ′ < 1, the right side is smaller as n is large. □

Lemma 3.2. If for all n > 0, t_{n} satisfies (3.2), and r(t_{n}) ≤ 1−C′t_{n}^{γ}+O(t_{n}^{1+γ′}), where C′ > 0, then there exists D > 0 such that for all k_{0} ≥ k,

∏_{i=k_{0}−k}^{n+k_{0}−k} r(t_{i}) ≤ D(\frac{k}{n + k} )^{C′/γC}.  (3.5)

The result remains true if we replace “≤” by “≥” in all three inequalities.

Proof: Note that

r(t_{n}) ≤ 1 − \frac{C′}{γCn} + O(\frac{1}{n^{1+γ′}} ) = (1 − \frac{1}{n})^{\frac{C′}{γC}} · (1 + O(\frac{1}{n^{1+γ′}} )),

where γ′ > 0. Then we take product. □

Lemma 3.3. Let θ ∈ (0, 1) and C̄′_{1}, C̄_{2}, D̄_{1} > 0, and let R̃ ⊂ ℝ^{m} be a bounded region containing the origin. Suppose the map T : R̃ → ℝ^{m} is injective with T^{−1}R̃ ⊂ R̃ and satisfies

d(T x, T y) ≥ (1 + C̄′_{1}|x|^{γ})d(x, y),  (3.6)
log|\frac{det DT (x)}{det DT (y)}| ≤ C̄_{2}|x|^{γ−1}d(x, y)  (3.7)

for all x, y ∈ R̃ with d(x, y) ≤ |x|/2. Then there exists J′ > 0 such that for all x, y ∈ T R̃ with

d(x_{i}, y_{i})^{1−θ} ≤ D̄_{1}|x_{i}|, i = 1, ⋯, n,  (3.8)

where x_{i} = T^{−i}x and y_{i} = T^{−i}y, we have

log|\frac{det DT^{n}(x_{n})}{det DT^{n}(y_{n})}| ≤ J′d(x, y)^{θ}.  (3.9)

Proof: We prove by induction that for all i = 1, ⋯, n,

log|\frac{det DT^{i}(x_{n})}{det DT^{i}(y_{n})}| ≤ J′d(x_{n−i}, y_{n−i})^{θ}.  (3.10)

For i = 1, by (3.7), (3.8) and (3.6), we have

log|\frac{det DT (x_{n})}{det DT (y_{n})}| ≤ C̄_{2}D̄_{1}|x_{n}|^{γ}d(x_{n}, y_{n})^{θ} ≤ C̄_{2}D̄_{1}|x_{n−1}|^{γ}d(x_{n−1}, y_{n−1})^{θ}.

So if J′ ≥ sup{C̄_{2}D̄_{1}|x|^{γ} : x ∈ R̃} then the right side of the inequality is less than J′d(x_{n}, y_{n})^{θ} because |x_{n}| ≤ |x|.

Suppose (3.10) is true up to i = k − 1. Then similarly we have

log|\frac{det DT^{k}(x_{n})}{det DT^{k}(y_{n})}| ≤ log|\frac{det DT^{k−1}(x_{n})}{det DT^{k−1}(y_{n})}| + log|\frac{det DT (x_{n−k+1})}{det DT (y_{n−k+1})}|

≤ J′d(x_{n−k+1}, y_{n−k+1})^{θ} + C̄_{2}|x_{n−k+1}|^{γ−1}d(x_{n−k+1}, y_{n−k+1})
= J′(1 + \frac{C̄_{2}D̄_{1}}{J′} |x_{n−k+1}|^{γ}) · \frac{d(x_{n−k+1}, y_{n−k+1})^{θ}}{d(x_{n−k}, y_{n−k})^{θ}} · d(x_{n−k}, y_{n−k})^{θ}
≤ J′(1 + \frac{C̄_{2}D̄_{1}}{J′} |x_{n−k+1}|^{γ}) · \frac{1}{(1 + C̄′_{1}|x_{n−k+1}|^{γ})^{θ}} d(x_{n−k}, y_{n−k})^{θ}.

Clearly if J′ is large enough, then the right side is bounded by J′d(x_{n−k}, y_{n−k})^{θ}. We get (3.10) for i = k. □

We are finally ready to prove the theorem of this section.

Proof of Theorem B: Part i). We assume that T is C^{1+α̂} and | det DT | ≥ κ^{−1} > 1 on T R. We also regard α̂ ≤ 1. So there exist c_{1} > 0 such that

\frac{| det DT_{1}^{−1}(y)|}{| det DT_{1}^{−1}(x)|} ≤ 1 + c_{1}d(x, y)α̂

for all x, y ∈ T R. Let x_{i} = T_{1}^{−i}x and y_{i} = T_{1}^{−i}y. Clearly, d(x_{i}, y_{i}) ≤ d(x, y). So if d(x, y) ≤ ε and 0 < n ≤ N, then

\frac{| det DT_{1}^{−n}(y)|}{| det DT_{1}^{−n}(x)|} ≤ (1 + c_{1}d(x, y)α̂)^{n} ≤ (1 + c_{1}εα̂)^{N}.  (3.11)

Also, there exists C > 0 such that for any y ∈ B_{ε}(R_{0}), | det DT_{1}^{−n}(y)| ≤ Cκ^{n}. Hence,

∑_{n=N}^{∞} sup_{y∈B_{ε}(x)} | det DT_{1}^{−n}(y)| ≤ \frac{Cκ^{N}}{1 − κ}.

Let b > 0 be given.

Consider the function

σ(ε) = \frac{(1 + c_{1}εα̂)^{N_{0}−c_{2} log ε}}{1 + Jε^{α}} ,

where N_{0} = 1 + log(C^{−1}b(1 − κ))/ log κ and c_{2} = −(m + α)/ log κ. Since lim_{ε→0}(1 + c_{1}εα̂)^{N_{0}−c_{2} log ε} = 1, we have lim_{ε→0} σ(ε) = 1. Note that if

(N_{0} − c_{2} log ε)α̂c_{1}ε^{α̂−1} · (1 + Jε^{α}) − αJε^{α−1} · (1 + c_{1}εα̂) < 0,  (3.12)

then σ′(ε) < 0. Since α̂ > α, the first term in (3.12) is of higher order. So we can choose J > 0 and ε_{4} > 0 such that for all ε ∈ (0, ε_{4}], (3.12) holds and therefore σ(ε) ≤ 1.

Now for each ε ∈ (0, ε_{4}], we take N = N(ε) as the integer part of N_{0}−c_{2} log ε. Clearly, for such N we have

\frac{Cκ^{N}}{1 − κ} ≤ bε^{m+α}.

So the second inequality in Assumption 4(b) is true. For the first inequality, note that

(1 + c_{1}εα̂)^{N} ≤ (1 + c_{1}εα̂)^{N_{0}−c_{2} log ε} ≤ 1 + Jε^{α}.

Then by (3.11) we get what we need.

Part ii). Let us put β = 1/γ and θ = α. Take ρ > 0 such that

\frac{β}{1 − θ} − τ < ρ < \frac{δ − 1}{m + α}.  (3.13)

Let b > 0 be given.

Note that by Lemma 3.1, (1.5) implies that there exists C̄_{0} > 0 such that

for any x ∈ R_{0}, |x_{n}| ≥ \frac{1}{(C̄_{0}n)^{β}}. Take N_{b} ≥ L such that for all n ≥ N_{b},

b^{− \frac{1}{m+α}} (∑_{k=n}^{∞} \frac{C_{δ}}{k^{δ}} )^{\frac{1}{m+α}} < \frac{1}{n^{ρ}} < \frac{1}{(n − 1)^{ρ}} < \frac{1}{2C_{τ} C̄0^{\frac{β}{1−θ}} n^{\frac{β}{1−θ} −τ}} ,  (3.14)

where C_{δ} and C_{τ} are as in (1.9). The inequality is possible because of (3.13).

Note that (1.6) and (1.7) imply (3.6) and (3.7) respectively. By Lemma 3.3 we can take J′ > 0 such that (3.9) holds for any x ∈ R_{0}, n > 0 whenever (3.8) holds with D̄_{1} = 1 for all x_{i}, y_{i}, i = 1, ⋯, n.

Take ε′_{4} > 0 such that for all x, y with x ∈ R_{0}, d(x, y) ≤ ε′_{4}, n = 1, ⋯, N_{b}, we have d(x_{n}, y_{n})^{1−θ} ≤ |x_{n}|. By the choice of J′, (3.9) holds for all 1 ≤ n ≤ N_{b}.

Then we take ε_{4} = min{ε′_{4}, 1/N_{b}^{ρ}}, and J > 0 such that e^{J′ε_{4}^{θ}} ≤ 1 + Jε_{4}^{θ}. We show that J and ε_{4} satisfies the requirement. Let ε ∈ (0, ε_{4}]. Take

N = N(ε) > N_{b} such that

\frac{1}{N^{ρ}} ≤ ε < \frac{1}{(N − 1)^{ρ}} .

By the first inequality of (1.9) and (3.14),

∑_{k=N}^{∞} sup_{y∈B_{ε}(x)} | det DT^{−k}(y)| ≤ ∑_{k=N}^{∞} \frac{C_{δ}}{k^{δ}} ≤ b · \frac{1}{N^{ρ(m+α)}} ≤ bε^{m+α}.

On the other hand, if x ∈ R_{0} and d(x, y) ≤ ε, then by the last inequality of (1.9) and (3.14), for any N_{b} < n ≤ N,

d(x_{n}, y_{n}) ≤ \frac{2C_{τ}}{n^{τ}} ε ≤ \frac{2C_{τ}}{n^{τ}} \frac{1}{(N − 1)^{ρ}} ≤ \frac{1}{C̄0^{\frac{β}{1−θ}} n^{\frac{β}{1−θ}}} ≤ |x_{n}|^{\frac{1}{1−θ}} .

So we know that (3.9) holds for all 0 ≤ n ≤ N. Then by the choice of J and the fact θ = α,

|\frac{det DT^{n}(x_{n})}{det DT^{n}(y_{n})}| ≤ e^{J′d(x,y)^{θ}} ≤ e^{J′ε^{α}} ≤ 1 + Jε^{α}.

This is what we need. □

4 Proof of Theorem C

The proof consists again of two parts. Proof of Theorem C: Part i). For x ∈ ∂R, denote

D(x) = {z ∈ R_{0} : x ≺ z ≺ T x}.

Clearly the collection {D(x) : x ∈ ∂R ∩ D_{i}} form a cover of D_{i} ∩ R_{0}. Note that by the meaning of x ≺ T x, every D(x) contains an open set in D_{i} ∩ R_{0}. So we can construct a finite or countable partition ξ of R_{0} such that every element of ξ belongs to the closure \overline{D(x)} of some D(x).

Note that for any x,

\frac{| det DT_{1}^{−n}(x)|}{| det DT_{1}^{−n}(T x)|} = \frac{| det DT (x)|}{| det DT (x_{n})|} ≤ | det DT (x)|

is always bounded. So for any y, z ∈ \overline{D(x)}, we have

\frac{| det DT_{1}^{−n}(y)|}{| det DT_{1}^{−n}(z)|} ≤ \frac{| det DT_{1}^{−n}(x)|}{| det DT_{1}^{−n}(T x)|} ≤ | det DT (x)|.

Hence (1.4) follows. Obviously we can arrange the partition ξ in such a way that (1.3) also holds. Therefore ξ is a desired partition for any n. Part ii). First, we take θ > 0 such that

DT_{x}(v′) ≤ (|T x|/|x|)^{1/(1−θ)}

for all x ∈ ∂(T_{1}^{−n}R) and v′ ∈ E_{x}(∂(T_{1}^{−n}R)). This is possible because of the assumption stated in part (d) of Theorem C.ii). So for any n > 0, if we take x, y ∈ ∂R_{0} such that d(x_{n}, y_{n}) ≤ D̄_{1}|x_{n}|^{1/(1−θ)}, we have

d(x_{i}, y_{i}) ≤ D̄_{1}|x_{i}|^{1/(1−θ)}  ∀i = 1, ⋯, n.  (4.1)

By Lemma 3.3, we get that there exists I_{1} > 0 such that

\frac{| det DT_{1}^{−n}(y)|}{| det DT_{1}^{−n}(x)|} ≤ I_{1}.  (4.2)

That is, (1.4) holds for all such x, y.

We construct ξ = ξ_{n}. Note that we only need do it for n sufficiently large. Since the family of cones C′_{x} are uniformly continuous in x = (t, ϕ), we can find t_{0} > 0 such that for any x, y ∈ T R with d(x, y) ≤ t_{0}, the Hausdorff distance between C′_{x} and C′_{y} is less than θ_{0}/2. Then we take N > 0 large enough such that for any x ∈ R_{0} and n > N, |x_{n}| ≤ t_{0}. Note that for any x, the position vector from p to x, denoted by u_{x}, is contained in C_{x}. By part (a) and (d) in the conditions of the theorem we know that at x ∈ T_{1}^{−n}(∂R_{0}), C′_{x} contains the tangent plane of the surface. Hence, if v′ ∈ E_{x}(T_{1}^{−n}(∂R_{0}), then the angle between u_{x} and v′, denoted by ∠(u_{x}, v′), is larger than θ_{0}, and therefore for any v′ ∈ E_{y}(T_{1}^{−n}(∂R_{0}), we have ∠(u_{x}, v′) ≥ θ_{0}/2, whenever y ∈ T_{1}^{−n}(∂R_{0}) with d(x, y) ≤ t_{0}. So for any x, y ∈ T_{1}^{−n}(∂R_{0}) with d(x, y) ≤ t_{0}, we have d_{S}(x, y) ≤ d(x, y)/ sin(θ_{0}/2), where d_{S}(·, ·) is the distance restricted to the surfaces {T_{1}^{−n}(∂R_{0})}. This means that we can take a partition ξ^{(n)} on T_{1}^{−n}(∂R_{0}) such that every element of ξ^{(n)} is contained in a ball of radius |x_{n}|^{1/(1−θ)} and containing a ball of radius |x_{n}|^{1/(1−θ)}/10m sin(θ_{0}/2), with respect to the metric on T_{1}^{−n}(∂R_{0}), and these elements are close to (m−1) dimensional disks. Denote ξ′ = T^{n}ξ^{(n)}. Clearly, it is a partition of ∂R_{0}. Then we can take a partition ξ of R_{0} whose elements has the form ∪_{x∈A′}F_{x} ∩ R_{0}, where A′ is an element of ξ′, and F_{x} is given in Lemma 4.1.

Now we prove that ξ satisfies (1.3) and (1.4). Condition (1.6) implies ∥DT (p)∥ = 1. We first consider the case that DT (p) = id.

By (1.5), we know that d(x, T x) ≤ C|x|^{1+γ} for some C > 0. So the “width” of the annulus T_{1}^{−i}(B_{ε_{5}}(R_{0})) is bounded by C′|T_{1}^{−i}x|^{1+γ} for some C′ > 0. By part (b) and (d) of the condition in the theorem, for 0 < ε ≤ ε_{5}, x ∈ T_{1}^{−i}(∂B_{ε}(R_{0})), the angle between the tangent space of T_{1}^{−i}(∂B_{ε}(R)) and the position vector u_{x} is larger than θ_{0}. So the length of the curve F_{T_{1}^{−i}x}∩T_{1}^{−i}B_{ε_{5}}(R_{0}) is bounded by C|T_{1}^{−i}x|^{1+γ} for some C ≥ C′. Hence, for any x, y ∈ B_{ε_{5}}(R_{0}) with y ∈ F_{x}, we can get

d(x_{i}, y_{i}) ≤ C|x_{i}|^{1+γ}  (4.3)

and therefore by applying Lemma 3.3 we get

\frac{| det DT_{1}^{−n}(y)|}{| det DT_{1}^{−n}(x)|} ≤ I_{2}  (4.4)

for some I_{2} > 0. Also, the construction of ξ′ implies (4.1) and therefore (4.2) for any x, y ∈ A′, where A′ ∈ ξ′. So by the construction of ξ, we get (1.4) with I = I_{1}I_{2}^{2} for any x, y ∈ A.

On the other hand, for any x, y ∈ B_{ε_{5}}(R_{0}) with y ∈ F_{x}, we have (4.3). So we can apply Lemma 4.4 to get that inside A, distortion of |DT |_{E(F)}| is bounded. It means that for each x ∈ A, the ratio of the length of T_{1}^{−n}(B_{ε}(∂R_{0}) ∩ A)) ∩ F_{x_{n}} and the length of T_{1}^{−n}(B_{ε_{0}}(∂R_{0}) ∩ A)) ∩ F_{x_{n}} is uniformly bounded by ε/ε_{0} multiplied by a constant. Notice that the angle between the tangent vectors of F and the tangent space of T_{1}^{−n}(∂B_{ε}(R_{0})) are greater than θ_{0}. Also notice that by the construction of ξ′, for any A ∈ ξ, the size of the set T_{1}^{−n}A along the fiber direction is much smaller than the size of T_{1}^{−n}A′. Hence, the ratio between ν(T_{1}^{−n}B_{ε}(∂R_{0}) ∩ A)) and ν(T_{1}^{−n}B_{ε_{0}}(∂R_{0}) ∩ A)) is bounded by a constant times ε/ε_{0} ≤ (ε/ε_{0})^{α} for some α ∈ (0, 1]. Now we use (1.4) to get (1.3).^{‡}

‡Let us make this argument more precise. We denote with A′_{n} and A_{n}(ε) respectively the backward iterates T_{1}^{−n}A′ of some A′ ∈ ξ′ and of the set A∩B_{ε}(∂R_{0}) where A = ∪_{x∈A′}F_{x}∩R_{0}.

If DT_{p} ≠ id, then it is a rotation, say S. Hence near p we can write T x = Sx+T_{r}(x) where |T_{r}(x)| ≤ C|x|^{1+γ}. If we write T^{(i)} = id +S^{−i} ∘T_{r} ∘S^{i−1}, then T^{n} = S^{n} ∘ T^{(n)} ∘ ⋯ ∘ T^{(1)}. It implies that the “width” of the annulus T_{1}^{−i}R_{0} is bounded by C|T_{1}^{−i}x|^{1+γ}. Then we apply the same arguments to get (1.4) and (1.3). □

Lemma 4.1. There is a foliation on {F_{x}} on T R \ {p} consisting of curves from p to points on ∂(T R) such that for any x ∈ T R, the tangent line of F_{x} lies in C_{x}, and T F_{x} ∩ T R = F_{T x}.

Proof: Denote E_{x} = ∩_{n≥0}DT_{T_{1}^{−n}x}^{n}(C_{T_{1}^{−n}x}) for all x ∈ T R\{p}. By Lemma 4.2, we know that sine of the angle between any two vectors in DT_{T_{1}^{−n}x}^{n}(C_{T_{1}^{−n}x}) is less than (1 − d|x_{n}|^{γ}) ⋯ (1 − d|x_{1}|^{γ}). By (1.5) and Lemma 3.1, the product diverges as n → ∞. So {E_{x}} is a subbundle of the tangent bundle over T R\{p}. Further, we have DT_{x}(E_{x}) = E_{T x} for all x ∈ R. By Lemma 4.3, we know that {E_{x}}

Since the angles between the tangent spaces of the curves F_{x} and the tangent spaces of the ϵ-neighborhood of the boundary of R_{0} are uniformly bounded away from zero, the length of the curve F_{x} ∩ B_{ε}(∂R_{0}), when x ∈ A′, is of order ϵ. Its n-backward iterate in A_{n}(ε) will be therefore bounded by a constant times ϵ times d_{n,M}^{1+γ}, where d_{n,M} is the maximum over the ϵ-compact neighborhood of R_{0} of |T_{1}^{−i}x| (see above; equivalently we set d_{n,m} the minimum of |T_{1}^{−i}x| over the ϵ-compact neighborhood of R_{0}). Let us call this upper bound l_{n,ϵ}. We construct then the l_{n,ϵ}-neighborhood of A′_{n}, B_{l_{n,ϵ}}(A′_{n}). Clearly

ν(A_{n}(ε)) ≤ ν(B_{l_{n,ϵ}}(A′_{n})) ≤ ν′(A′_{n,ε})l_{n,ϵ}

where A′_{n,ε} = {z ∈ T_{1}^{−n}A′; d(z, A′_{n}) ≤ l_{n,ϵ}} and ν′ denotes the Riemannian volume on

T_{1}^{−n}∂R_{0}. Since A′_{n,ε} is contained in a ball of radius dn,M^{\frac{1}{1−θ}} + l_{n,ϵ} and A′_{n} by construction

contains a ball of radius ^{\frac{d \frac{1}{1−θ} ; n,m sin θ_{0}/2}{10m}}, we have that ν′(A′_{n,ε}) ≤ const(dn,M^{\frac{1}{1−θ}} + l_{n,ϵ})^{m−1}γ_{m−1}

and ν′(A′_{n}) ≥ (^{\frac{d \frac{1}{1−θ} ; n,m sin θ_{0}/2}{10m}})^{m−1}γ_{m−1}. But d_{n,M} , d_{n,m} are of order n^{−β}, with β = 1/γ (see

Lemma 3.1), and since (1 + γ)(1 − θ) > 1, we see immediately that for large n:

ν(A_{n}(ε)) ≤ C′ν′(A′_{n})l_{n,ϵ}

where C′ is a suitable constant, depending on m. Let us now define the following objects: A_{n}(ε_{0}): the backward iterate of A ∩ B_{ε_{0}}(∂R_{0}), l′_{n,ϵ_{0}}: the minimum length of the backward images of the curves F_{x} ∩ B_{ε_{0}}(∂R_{0}), when x ∈ A′; A′_{n,ε_{0}} = {z ∈ A′_{n}; d(z, ∂A′_{n}) ≥ l′_{n,ϵ_{0}}} and B_{l′′_{n,ε0}} (A′_{n,ε_{0}}) the l′′_{n,ϵ_{0}}-neighborhood of A′_{n,ε_{0}}, being l′′_{n,ϵ_{0}} = l′_{n,ϵ_{0}} sin θ_{0}. Moreover by what we already said above and which follows from Lemma 4.3, the bounded distortion property along the points of the backward images of the curves A ∩ B_{ε_{0}}(∂R_{0}), will imply that l′_{n,ϵ_{0}} will be of the same order as l_{n,ϵ_{0}} (the maximum length of the backward images of the curves). Taking this into account we get:

ν(A_{n}(ε_{0})) ≥ ν(B_{l′′_{n,ε0}} (A′_{n,ε_{0}})) ≥ (( \frac{dn,m^{\frac{1}{1−θ}} sin θ_{0}/2}{10m} − l′′_{n,ϵ_{0}})^{m−1}γ_{m−1}l′′_{n,ϵ_{0}}

By using as above the uniform bounds on d_{n,M} , d_{n,m} when n is large, we see that ν(A_{n}(ε_{0})) ≥ C′′ν′(A′_{n})l_{n,ϵ_{0}}, where C′′ is a suitable constant depending on m. By dividing ν(A_{n}(ε)) and ν(A_{n}(ε_{0})), we get the desired result.

satisfies the Hölder condition near each x with Hölder constants depending on x. Note that {E_{x}} determines a vector field. We can integrate it to get a family of curves {F_{x}} from p to boundary points of T R that satisfies T F_{x} ∩T R = F_{T x}. By our assumption, {F_{x}} is the “strong unstable manifold” at x.

It is easy to see that the curve passing through x is unique, and therefore {F_{x}} forms a foliation. In fact, if there are two such curves F_{x} and F′_{x} that pass through x, then we can take a curve Γ close to x joining y ∈ F_{x} and y′ ∈ F′_{x} such that the tangent line of Γ is in C′. Let us denote by A_{n} the area of the “triangle” bounded by the curves T_{1}^{−n}Γ, T_{1}^{−n}F_{x,y} and T_{1}^{−n}F′_{x,y′}, and by L_{n} and L′_{n} the lengths of the curves T_{1}^{−n}F_{x,y} and T_{1}^{−n}F′_{x,y′} respectively, where F_{x,y} is the part of the curve in F_{x} between x and y, and F′_{x,y′} is understood in a similar way. By the assumption stated in part (c), the ratio between A_{n} and L_{n} · L′_{n} tends to infinity, which is a contradiction. □

Lemma 4.2. For any v, v′ ∈ C_{x},

sin ∠(DT_{x}(v), DT_{x}(v′)) ≤ (1 − d|x|^{γ}) sin ∠(v, v′),

where the symbol ∠(v, v′) denotes the angle between the vectors v and v′.

Proof: Note that

| det DT_{x}|_{E(v,v′)}| = \frac{|DT_{x}(v)| · |DT_{x}(v′)| · sin ∠(DT_{x}(v), DT_{x}(v′))}{|v| · |v′| · sin ∠(v, v)}

and

||DT_{x}|_{E(v)}|| = \frac{|DT_{x}(v)|}{|v|}, ||DT_{x}|_{E(v′)}|| = \frac{|DT_{x}(v)|}{|v′|}.

Then the results follows from (1.10). □

Lemma 4.3. There exist constants H > 0, a > 0, and τ_{1} ∈ (0, 1), such that for all x ∈ T R\{p},

d(E_{x}, E_{y}) ≤ \frac{Hd(x, y)^{τ_{1}}}{|x|^{τ_{1}}}  ∀y ∈ B(x, a|x|),  (4.5)

where d(E_{x}, E_{y}) is defined by d(E_{x}, E_{y}) = sin ∠(v_{x}, v_{y}), v_{x} and v_{y} are the tangent vectors of F_{x} and F_{y} at x and y respectively chosen in the way that 0 ≤ ∠(v_{x}, v_{y}) < π/2.

Proof: We note that we only need prove (4.5) for all x in a small neighborhood R̃ ⊂ R of p, because DT_{x}(E_{x}) = E_{T x}, and then the results can be extended to T R.

Take d̃ ∈ (0, d). Then for each x we can extend C_{x} to C̃_{x} such that (1.10) holds with d̃ for all v ∈ C_{x} and v′ ∈ C̃_{x}. By (1.5) and the fact that T is C^{1+γ}, we can write DT (x) = T_{0}(x) + T_{γ}(x) + T_{h}(x), where T_{0} = DT_{p}, T_{γ} satisfies T_{γ}(tx) = t^{γ}T_{γ}(x) ∀t > 0 and |T_{h}(x)| = o(|x|^{γ}). So it is easy to see that we can find ε_{a} > 0 such that C̃_{x} ∩ 𝕊^{m−1} contains an ε_{a}-neighborhood of C_{x} ∩ 𝕊^{m−1} in 𝕊^{m−1} for all x with |x| small. Moreover, since C_{x} is uniformly continuous in (t, ϕ), we can take a > 0 and R̃ small such that for all x ∈ R̃, with d(x, y) ≤ a|x|^{γ}, C_{y} ⊂ C̃_{x}. So if v ∈ C_{x} and v′ ∈ C_{y}, we have

\frac{| det DT_{x}|_{E(v,v′)}|}{∥DT_{x}|_{E(v)}∥ · ∥DT_{x}|_{E(v′)}∥} ≤ 1 − d̃|x|^{γ}.

Hence, by the same arguments used in Lemma 4.2 we have

sin ∠(DT_{x}(v), DT_{x}(v′)) ≤ (1 − d̃|x|^{γ}) sin ∠(v, v′).  (4.6)

Take τ_{1} ∈ (0, 1) such that

(1 − \frac{d̃}{2}|x|^{γ})(\frac{|T x|}{|x|} · \frac{d(x, y)}{d(T x, T y)})^{τ_{1}} ≤ 1  (4.7)

for all x ∈ R̃ close to p with d(x, y) ≤ a|x|.

Take 0 < a_{1} ≤ a such that if d(x, y) ≤ a_{1}|x|, then

∥DT (x) − DT (y)∥ ≤ C̄_{2}|x|^{γ−1}d(x, y)^{τ_{1}}  (4.8)

for some C̄_{2} > 0. This is possible because of (1.7).

Take H > 0 such that Hd̃> 2C̄_{2}.

Let L = {L_{x} : x ∈ R̃\{p} } be the set of all line bundles in the tangent bundle over R̃. Clearly DT induces a map D : L → L given by (DL)_{x} = DT_{x}(L_{T_{1}^{−1}x}), and E = {E_{x}} is the unique fixed point of D contained in C. Denote

H = \{\{L_{x}\} ∈ L ∩ C : d(L_{x}, L_{y}) ≤ \frac{Hd(x, y)^{τ_{1}}}{|x|^{τ_{1}}} ∀y ∈ B(x, a_{1}|x|) \}. (4.9)

We show that D(H) ⊂ H. This implies the result since {E_{x}} = ∩_{n≥0}D^{n}C.

Take {L_{x}} ∈ H. Let x, y ∈ R̃ with d(x, y) ≤ a_{1}|x|. Take unit vectors e_{x} ∈ L_{x}, e_{y} ∈ L_{y}. So sin ∠(e_{x}, e_{y}) ≤ H|x|^{−τ_{1}}d(x, y)^{τ_{1}}. By (4.6) and (4.8),

sin ∠(DT_{x}(e_{x}), DT_{y}(e_{y}))
≤ sin ∠(DT_{x}(e_{x}), DT_{x}(e_{y})) + sin ∠(DT_{x}(e_{y}), DT_{y}(e_{y}))
≤ (1 − d̃|x|^{γ}) sin ∠(e_{x}, e_{y}) + |DT_{x}(e_{y}) − DT_{y}(e_{y})|
≤ (1 − d̃|x|^{γ})\frac{Hd(x, y)^{τ_{1}}}{|x|^{τ_{1}}} + C̄_{2}|x|^{γ−1}d(x, y)^{τ_{1}}
= [(1 − d̃|x|^{γ})H + C̄_{2}|x|^{γ}]\frac{d(T x, T y)^{τ_{1}}}{|T x|^{τ_{1}}} · \frac{d(x, y)^{τ_{1}}}{d(T x, T y)^{τ_{1}}} \frac{|T x|^{τ_{1}}}{|x|^{τ_{1}}} .

By the choice of H, the quantity in the bracket is less than 1 − d̃|x|^{γ}/2. Then by (4.7) the right side of the inequality is less than H|T x|^{−τ_{1}}d(T x, T y)^{τ_{1}}. We get the desired results. □

Lemma 4.4. There exists J^{∗} > 0 such that for any x, y with d(x_{i}, y_{i}) ≤ |x_{i}|γ̄ for some γ̄ > 1, i = 1, ⋯, n,

\frac{|DT_{1}^{−n}(y)|_{E_{y}(F)}|}{|DT_{1}^{−n}(x)|_{E_{x}(F)}|} ≤ J^{∗}.  (4.10)

Proof: Take an integer r̄ ≥ 2C′_{0}/C_{0}, where C_{0} and C′_{0} are as in (1.6). We assume that x_{0} ≤ 1/(γC′_{0}k_{0})^{β} for some k_{0} ≥ 1. Then we take k_{i} = (r̄^{i} − 1)k_{0} for i = 1, ⋯, ℓ − 1, where ℓ − 1 is the largest number j such that k_{j} < n. Let k_{ℓ} = n. By Lemma 3.1, we know that

|x_{j}|^{γ} ≤ 2/(γC′_{0}(k_{0} + j)).  (4.11)

Hence, (1.6) implies

∥DT_{x_{k_{i}}}^{k_{i}−k_{i−1}}∥ ≤ ∏_{j=k_{i−1}}^{k_{i}−1} ∥DT_{x_{j}}∥ ≤ ∏_{j=k_{i−1}}^{k_{i}−1}(1 + \frac{2C_{1}}{γC′_{0}(k_{0} + j)}) ≤ ∏_{j=k_{i−1}}^{k_{i}−1}(1 + \frac{1}{k_{0} + j} )^{C}

for some C larger than 2C/γC′_{0} if k_{i} is large enough. So the choice of r̄ gives

∥DT_{x_{k_{i}}}^{k_{i}−k_{i−1}}∥ ≤ (\frac{k_{0} + k_{i}}{k_{0} + k_{i−1}} )^{C} ≤ r̄^{C}  (4.12)

for all i ≥ 0.

Let e_{x} be the unit tangent vector of F at x. We have

\frac{|DT_{1}^{−n}(y)|_{E_{y}(F)}|}{|DT_{1}^{−n}(x)|_{E_{x}(F)}|} = \frac{|DT_{x_{n}}^{n}(e_{x_{n}})|}{|DT_{y_{n}}^{n}(e_{y_{n}})|} = \frac{|DT_{x_{n}}^{n}(e_{x_{n}})|}{|DT_{x_{n}}^{n}(e_{y_{n}})|} · \frac{|DT_{x_{n}}^{n}(e_{y_{n}})|}{|DT_{y_{n}}^{n}(e_{y_{n}})|}

= ∏_{i=1}^{ℓ} \frac{|DT_{x_{k_{i}}}^{k_{i}−k_{i−1}}(e_{x_{k_{i}}})|}{|DT_{x_{k_{i}}}^{k_{i}−k_{i−1}}(e_{y_{k_{i}}})|} · ∏_{j=1}^{n} \frac{|DT_{x_{j}}(e_{y_{j}})|}{|DT_{y_{j}}(e_{y_{j}})|}.

By the results of Lemma 4.3 and (4.12), each factor in the first product is bounded by

1 + \frac{|DT_{x_{k_{i}}}^{k_{i}−k_{i−1}}(e_{x_{k_{i}}})| − |DT_{x_{k_{i}}}^{k_{i}−k_{i−1}}(e_{y_{k_{i}}})|}{|DT_{x_{k_{i}}}^{k_{i}−k_{i−1}}(e_{y_{k_{i}}})|}

≤ 1 + \frac{|DT_{x_{k_{i}}}^{k_{i}−k_{i−1}}(e_{x_{k_{i}}} − e_{y_{k_{i}}})|}{|DT_{x_{k_{i}}}^{k_{i}−k_{i−1}}(e_{y_{k_{i}}})|} ≤ 1 + \frac{∥DT_{x_{k_{i}}}^{k_{i}−k_{i−1}}∥ · |e_{x_{k_{i}}} − e_{y_{k_{i}}} |}{|DT_{x_{k_{i}}}^{k_{i}−k_{i−1}}(e_{y_{k_{i}}})|}

≤ 1 + \frac{r̄^{C} · BHd(x_{k_{i}}, y_{k_{i}})^{τ_{1}}}{|x_{k_{i}}|} ≤ 1 + r̄^{C}BH|x_{k_{i}}|^{τ_{1}(γ̄−1)},

where we use the fact that |e_{x_{k_{i}}} − e_{y_{k_{i}}} | ≤ B sin ∠(e_{x_{k_{i}}} , e_{y_{k_{i}}}) for some B > 0. Also note that by (4.11) and the choice of k_{i}, {|x_{k_{i}}|} decreases exponentially fast as i → ∞. Since γ̄ > 1, the first product in above equality is convergent.

For the second product, by (1.7) each factor is bounded by

1 + \frac{|DT_{x_{j}}(e_{y_{j}})| − |DT_{y_{j}}(e_{y_{j}})|}{|DT_{y_{j}}(e_{y_{j}})|} ≤ 1 + \frac{C|x_{j}|^{γ−1}d(x, y)}{|DT_{y_{j}}(e_{y_{j}})|} ≤ 1 + \frac{C|x_{j}|^{γ̄+γ−1}}{|DT_{y_{j}}(e_{y_{j}})|} .

By (4.11) and the fact γ̄ > 1, we know that ∑_{j} |x_{j}|^{γ̄+γ−1} converges. So the product is also bounded. We get the result. □

5 Proof of Theorem A

In this section we first introduce a subspace V_{α} of L^{1} ≡ L^{1}(ℝ^{m}, ν) with compact unit ball that contains the density function of the invariant measures of the induced map of T with respect to the relatively compact subspace M\R. Here we only give a brief description and list some properties we use. We refer to [S] and [K] for more details.

Let f be an L^{1}(ℝ^{m}, ν) function. If Ω is a Borel subset of ℝ^{m}, we define the oscillation of f over Ω by the difference of essential supremum and essential infimum of f over Ω:

osc(f, Ω) = Esup_{Ω} f − Einf_{Ω} f.

If B_{ϵ}(x) denotes the ball of radius ϵ about the point x, then we get a measurable function x → osc(f, B_{ϵ}(x)). The function have the following properties.

Proposition 5.1. Let f, f_{i}, g ∈ L^{∞}(ℝ^{m}, ν) with g ≥ 0, ε > 0, and S be a Borel subset of ℝ^{m}. Then

(i) osc(∑_{i} f_{i}, B_{ε}(·)) ≤ ∑_{i} osc(f_{i}, B_{ε}(·)),

(ii) osc(fχ_{S}, B_{ε}(·)) ≤ osc(f, S ∩ B_{ε}(·))χ_{S}(·) + 2[ Esup_{S∩B_{ε}(·)} f]χ_{B_{ε}(S)∩B_{ε}(S^{c})},

(iii) osc(fg, S)≤ osc(f, S)Esup_{S} g + osc(g, S)Einf_{S} f.

Proof: See [S] Proposition 3.2. □

Take 0 < α < 1 and ε_{0} > 0. We define the α-seminorm of f as

|f|_{α} = sup_{0<ϵ≤ϵ_{0}} ϵ^{−α} ∫_{ℝ^{m}} osc(f, B_{ϵ}(x))dν(x).  (5.1)

We will consider the space of the functions f with bounded α-seminorm, namely,

V_{α} = {f ∈ L^{1} : |f|_{α} < ∞} ,  (5.2)

and equip V_{α} with the norm

∥ · ∥_{α}=∥ · ∥_{1} +| · |_{α},  (5.3)

where ∥ · ∥_{1} denotes the L^{1} norm. This space does not depend on the choice of ϵ_{0}. With the ∥ · ∥_{α} norm, V_{α} is a Banach space; moreover according to Theorem 1.13 in [K], the unit ball in V_{α} is compact in L^{1}.

Proposition 5.2. Let f ∈ V_{α}. Then

(i) ∥f∥_{∞} ≤ \frac{1}{γ_{m}ε_{0}^{m}} ∥f∥_{α} provided ε_{0} ≤ 1,

(ii) There exists a ball B_{ε}(x) such that Einf_{B_{ε}(x)} f > 0.

Proof: See [S] Proposition 3.4 and Lemma 3.1. □

To prove Theorem A we need one more ingredient, the so-called Lasota-Yorke’s inequality, which will be proved in Section 6. This inequality provides an upper bound on the action of the Perron-Frobenius operator on the elements in V_{α}. Such an operator will be defined on the subspace M\R with a potential given by the inverse of the determinant of the induced map. We denote it by P̂f. We will prove

|P̂f|_{α} ≤ η|f|_{α} + D∥f∥_{1}

for some η < 1 and D < ∞. This, plus the compactness of the unit ball of V_{α} in L^{1}, allow us to invoke the ergodic theorem of Ionescu-Tulcea and Marinescu ([IM], see also [K], Theorem 3.3,) to obtain an invariant probability measure μ absolutely continuous with respect to ν on M\R. The measure μ has finite number of ergodic components, and is “unique greatest” in the sense that any other measure absolutely continuous with respect to ν is absolutely continuous with respect to μ.

Proof of Theorem A:

Recall that R is given in Assumption 3. We construct an induced system (M̂, T̂). Denote M̂ = M\R. Let T̂ : M̂ → M̂ be the first return map of T, so that T̂(x) = T (x) if x ∉ T^{−1}R, otherwise T̂(x) = T^{i+1}(x) = T_{1}^{i}T_{j}(x) if x ∈ T_{j}^{−1}R, where i is the smallest positive integer such that T_{1}^{i}T_{j}(x) ∉ R. We denote g(x) = | det DT (x)|^{−1}, and similarly ĝ(x) = g(x) if x ∉ T^{−1}R and ĝ(x) = | det DT̂(x)|^{−1} = | det DT^{i+1}(x)|^{−1} if otherwise. Let ν̂ be the renormalization of the Lebesgue measure ν restricted to M̂.

Let P be the Perron-Frobenius operator of T with the potential function log g(x), i.e.

P f(x) = ∑_{T y=x} f(y)g(y).

Then let P̂ be the Perron-Frobenius operator of T̂ with the potential function log ĝ(x), i.e.

P̂f(x) = ∑_{j=1}^{K} ∑_{i=0}^{∞} f(T_{j}^{−1}T_{1}^{−i}x)ĝ(T_{j}^{−1}T_{1}^{−i}x).  (5.4)

By the definition of the induced system, we know that if x ∈ M \ T R, then i = 0, and if x ∈ T R \ R, then j ≠ 1.

By Proposition 6.2 in the next section we have the Lasota-Yorke’s inequality for the induced system (M̂, T̂). So T̂ has an absolutely continuous invariant probability measure μ̂ on M̂ with density function ĥ that has finitely many ergodic components.

We extend ĥ to R\{p} to get a density function h on M\{p}. That is, if x ∈ R\{p}, we let

h(x) = ∑_{j=2}^{K} h(T_{j}^{−1}x)g(T_{j}^{−1}x)

+ ∑_{j=2}^{K} ∑_{i=1}^{∞} h(T_{j}^{−1}T_{1}^{−i}x)g(T_{j}^{−1}T_{1}^{−i}x)g(T_{1}^{−i}x) ⋯ g(T_{1}^{−1}x). (5.5)

It is clear that h is well defined and nonnegative. Also, by this definition, for x ∈ T R\{p},

h(T_{1}^{−1}x) = ∑_{j=2}^{K} h(T_{j}^{−1}T_{1}^{−1}x)g(T_{j}^{−1}T_{1}^{−1}x)

+ ∑_{j=2}^{K} ∑_{i=2}^{∞} h(T_{j}^{−1}T_{1}^{−i}x)g(T_{j}^{−1}T_{1}^{−i}x)g(T_{1}^{−i}x) ⋯ g(T_{1}^{−2}x). (5.6)

Note that

P h(x) = h(T_{1}^{−1}x)g(T_{1}^{−1}x) + ∑_{j=2}^{K} h(T_{j}^{−1}x)g(T_{j}^{−1}x).

So if x ∈ R\{p}, we substitute h(T_{1}^{−1}x) in (5.6) and then compare it with (5.5) to get that the right side is equal to h(x). If x ∈ T R\R, we substitute h(T_{1}^{−1}x) and then compare it with (5.4), using the fact j ≠ 1 and ĝ(T_{j}^{−1}T_{1}^{−i}x) = g(T_{j}^{−1}T_{1}^{−i}x)g(T_{1}^{−i}x) ⋯ g(T_{1}^{−1}x), to get P h(x) = P̂h(x), which is also equal to ĥ(x) = h(x). Since outside T R, P̂f = P f for any f and ĥ = h, we obtain P h = h on M\{p}.

Let μ be the measure on M with density h. Clearly, μ is invariant under T and has the same number of ergodic components as μ̂ does.

Next, we show that μM is finite if ∑_{i=1}^{∞} ν(T_{1}^{−i}R) < ∞. Recall R_{0} = T R\R,

and let R_{n} = T_{1}^{−n}R_{0} for n > 0. By Remark 1.1, diam R_{n} → 0. So we have R = ∑_{n=1}^{∞} R_{n} ∪ {p}. Since μ is invariant, we have

μR_{i} = μR_{i+1} + ∑_{j=2}^{K′} μ(T_{j}^{−1}R_{i}),

where we assume that in addition to T_{1}^{−1}R ⊂ U_{1}, R has K′ − 1 preimages in U_{2}, ⋯, U_{K′}, where K′ ≤ K. Take summation from i = n to infinity, we get

μR_{n} = ∑_{j=2}^{K′} μ(T_{j}^{−1} ⋃_{i=n}^{∞} R_{i}) = ∑_{j=2}^{K′} μ(T_{j}^{−1}T_{1}^{−n}R).

Note that ∥ĥ∥_{∞} ≤ ∞ since ĥ ∈ V_{α}, and then note that the Jacobian of T_{j}^{−1} is less than or equal to 1. We have

μ(T_{j}^{−1}T_{1}^{−n}R) ≤ ∥ĥ∥_{∞}ν(T_{j}^{−1}T_{1}^{−n}R) ≤ ∥ĥ∥_{∞}ν(T_{1}^{−n}R).

Hence

μR = ∑_{n=1}^{∞} μR_{n} ≤ ∥ĥ∥_{∞}(K′ − 1) ∑_{n=1}^{∞} ν(T_{1}^{−n}R) < ∞.  (5.7)

Now we prove the last part of the theorem. By Proposition 5.2(ii), there is a ball B_{ε}(z) ⊂ M \ R such that Einf_{B_{ε}(x)} ĥ ≥ h_{∗} > 0 for some constant h_{∗}. By

the assumption in the statement of Theorem A, there exists Ñ > 0 such that T ÑB_{ε}(z) ⊃ M. So for any x ∈ M, there is y_{0} ∈ B_{ε}(z) such that T Ñy_{0} = x. Since | det DT | is bounded above, we have g_{∗} := inf{g(y) : y ∈ M} > 0. Hence, for every x,

h(x) = (P Ñh)(x) = ∑_{T Ñ y=x} h(y) ∏_{i=0}^{Ñ−1} g(T^{i}y) ≥ h(y_{0}) ∏_{i=0}^{Ñ−1} g(T^{i}y_{0}) ≥ h_{∗}gÑ ; ∗.

By splitting R over the disjoint union (mod 0) of the R_{n} as in (5.7), we get

μR = ∑_{n=1}^{∞} μR_{n} ≥ (h_{∗}gÑ ; ∗)g_{∗} ∑_{j=2}^{K′} ν(T_{1}^{−n}R) = ∞.

This ends the proof. □

6 A Lasota-Yorke type inequality

Let R be as in Assumption 3. Denote T̂_{ij} = T_{1}^{i}T_{j} and U_{ij} = T̂_{ij}^{−1}(R_{0}) = T_{j}^{−1}R_{i} for i > 1 and U_{0j} = U_{j}\T_{j}^{−1}R. So if T U_{l} ∌ p, then U_{il} is undefined for any i > 0 and U_{0l} = U_{l}. Clearly, U_{ij} ⊂ U_{j} for all i > 0 and {U_{ij}, i ≥ 0} are pairwise disjoint.

Lemma 6.1. There exists 0 < ε_{6} ≤ ε_{5} such that for any ε_{0} ≤ ε_{6}, ε ≤ ε_{0} and ∀x ∈ M̂,

2 ∑_{j=1}^{K} ∑_{i=0}^{∞} \frac{ν(T̂_{ij}^{−1}B_{ε}(∂R_{0}) ∩ B_{(1−s)ε_{0}}(x))}{ν(B_{(1−s)ε_{0}}(x))} ≤ \frac{λε^{α}}{ε_{0}^{α}} ,  (6.1)

where λ is given by Assumption 3(b).

Proof: Note that the sets ∪_{i=1}^{∞}∂U_{ij}, j = 1, ⋯, K, are pairwise separated. So by Assumption 3(b) and the definition of λ in (1.1) we only need prove that there exists ε_{6} > 0 such that for any given j, neighborhood if 0 < ε ≤ ε_{0} ≤ ε_{6}, then

2 ∑_{i=0}^{∞} \frac{ν(T̂_{ij}^{−1}B_{ε}(∂R_{0}) ∩ B_{(1−s)ε_{6}}(x))}{ν(B_{(1−s)ε_{6}}(x))} ≤ \frac{λε^{α}}{ε_{0}^{α}} .  (6.2)

Take

ε_{6} ≤ min\{ε_{5}, ε_{3}\} · (\frac{λ(1 − s)^{m}}{2C_{p}I^{2}} )^{1/α},

where ε_{3} is given by Assumption 3(c).

Recall that N_{s} is also given by Assumption 3(c). Reduce ε_{6} if necessary such that for any x, the ball B_{(1−s)ε_{6}}(x) intersects at most one connected component of the set {T̂_{ij}^{−1}B_{ε_{6}}(∂R_{0}), 0 < i ≤ N_{s}, 1 < j ≤ K}, which, we remember, are pairwise disjoint. We also require ε_{6} small enough such that for any 1 < j ≤ K, 1 ≤ i ≤ N_{s}, the part T̂_{ij}^{−1}∂R_{0} ∩ B_{ε_{6}}(x) are close to an (m − 1) dimensional plane.

Take ε and ε_{0} such that 0 < ε ≤ ε_{0} ≤ ε_{6} .

We first consider the case 1 ≤ i ≤ N_{s}. Note that T̂_{ij}^{−1}B_{ε}(∂R_{0})∩B_{(1−s)ε_{0}}(x) ⊂ B_{sε}(T_{ij}^{−1}∂R_{0}) ∩ B_{(1−s)ε_{0}}(x). The volume of the latter is close to γ_{m−1}((1 −

s)ε_{0})^{m−1} · 2sε = 2sγ_{m−1}ε(1 − s)^{m−1}ε_{0}^{m−1}. So \frac{ν(T̂_{ij}^{−1}B_{ε}(∂R) ∩ B_{(1−s)ε_{0}}(x))}{ν(B_{(1−s)ε_{0}}(x))} is

close to \frac{2sγ_{m−1}ε(1 − s)^{m−1}ε_{0}^{m−1}}{γ_{m}(1 − s)^{m}ε_{0}^{m}} = \frac{2sγ_{m−1}ε}{(1 − s)γ_{m}ε_{0}} . Hence, by Assumption 3(b),

we know that it is less than λε^{α}/ε_{0}^{α}.

Now we consider the case that i ≥ N_{s}.

Let ε̃ = ε_{0}(\frac{2C_{p}I^{2}}{λ(1 − s)^{m}} )^{1/α}. We have ε̃ ≤ ε_{5}.

For each i, we take a partition ξ_{i} = {Ã_{i1}, Ã_{i2}, ⋯, } satisfying Assumption 4(c) with n = i and ε̃ ≤ ε_{5}. Denote A_{ik} = Ã_{ik} ∩ Bε̃(∂R_{0}), A′_{ik} = Ã_{ik} ∩ B_{ε}(∂R_{0}), A_{ijk} = T̂_{ij}^{−1}A_{ik} and A′_{ijk} = T̂_{ij}^{−1}A′_{ik}. Then we let

A = {A_{ijk} : A′_{ijk} ∩ B_{(1−s)ε_{0}}(x) ≠ ∅}, A′ = {A′_{ijk} : A_{ijk} ∈ A}.

By abusing notations, we may also think that A and A′ are the unions of the sets they contain.

By the fact

νA_{ijk} = ∫_{A_{ik}} | det DT̂_{ij}^{−1}(x)|dν(x)

and Assumption 4(c), we know that

\frac{νA′_{ijk}}{νA_{ijk}} ≤ \frac{C_{p}ε^{α}}{ε̃^{α}} · I^{2} = \frac{C_{p}I^{2}ε^{α}λ(1 − s)^{m}}{2C_{p}I^{2}ε_{0}^{α}} = \frac{ε^{α}λ(1 − s)^{m}}{2ε_{0}^{α}} .  (6.3)

Denote s^{∗} = sup{s(T_{1}^{−N_{s}}(z), T_{1}^{N_{s}}) : z ∈ Bε̃(R_{0})}. Note that by Assumption

4(c), diam A_{ik} ≤ 5mε̃ ≤ 5mε_{0}(\frac{2C_{p}I^{2}}{λ(1 − s)^{m}} )^{1/α}. Since i ≥ N_{s}, by Assumption

3(c), we have

diam A_{ijk} ≤ 5mε_{0}(\frac{2C_{p}I^{2}}{λ(1 − s)^{m}} )^{1/α} · s^{∗} = sε_{0}.  (6.4)

So if A_{ijk} ∈ A, then A_{ijk} ∩ B_{(1−s)ε_{0}}(x) ≠ ∅, and therefore A_{ijk} ⊂ B_{ε_{0}}(x). That is,

A ⊂ B_{ε_{0}}(x).  (6.5)

Note that

⋃_{i=0}^{∞} T̂_{ij}^{−1}B_{ε}(∂R) ∩ B_{(1−s)ε_{0}}(x) ⊂ A′.  (6.6)

By (6.3) to (6.6), we get

2 ∑_{i=0}^{∞} \frac{ν(T̂_{ij}^{−1}B_{ε}(∂R) ∩ B_{(1−s)ε_{0}}(x))}{ν(B_{(1−s)ε_{0}}(x))} ≤ 2 · \frac{νA′}{νA} · \frac{νA}{νB_{ε_{0}}(x)} · \frac{μB_{ε_{0}}(x)}{ν(B_{(1−s)ε_{0}}(x)}

≤ 2 · \frac{ε^{α}λ(1 − s)^{m}}{2ε_{0}^{α}} · 1 · \frac{γ_{m}ε_{0}^{m}}{γ_{m}(1 − s)^{m}ε_{0}^{m}} = λ\frac{ε^{α}}{ε_{0}^{α}} .

This is (6.2), the formula we need to show. □

Proposition 6.2. Assume that T : M → M satisfies Assumption 1 to 4, and T̂ : M̂ → M̂ is the reduced system with respect to M̂ = M\R. Then there exist η < 1 and D < ∞ such that for any f ∈ V_{α} = V_{α}(ε_{0}), we have P f ∈ V_{α} and

|P̂f|_{α} ≤ η|f|_{α} + D∥f∥_{1}

for all ε_{0} sufficiently small.

Proof: Take ζ > 0 such that for any ε ≤ ε_{4},

(1 + Js^{α}ε^{a})(1 + cs^{a}ε^{α}) ≤ 1 + ζε^{α},  (6.7)

where c, J and ε_{4} are given in Assumption 4(a) and (b) respectively.

Recall that by Assumption 3(b), s^{α} + λ ≤ η_{0} < 1. Take b > 0 such that (s^{α} + λ) + 3K′bγ_{m}^{−1} < 1, where K′ is the number of preimages of p for the map T. Recall also that ε_{1}, ε_{2}, ε_{4} and ε_{6} are given in Assumption 1(b), 3(b), 4(b) and Lemma 6.1 respectively. Take ε_{0} ≤ min{ε_{1}, ε_{2}, ε_{4}, ε_{6}} such that

η := (1 + ζε_{0}^{α})(s^{α} + λ) + 3K′bγ_{m}^{−1} < 1.  (6.8)

Denote

G_{R}(x, ε, ε_{0}) = 2 ∑_{j=1}^{K} ∑_{i=0}^{N(ε)} \frac{ν(T̂_{ij}^{−1}B_{ε}(∂R_{0}) ∩ B_{(1−s)ε_{0}}(x))}{ν(B_{(1−s)ε_{0}}(x))}.

Recall that G_{U}(x, ε, ε_{0}) is given by (1.2) in Assumption 3(b). Note that if ε_{0} is small, then supp G_{U}(·, ε, ε_{0}) and supp G_{R}(·, ε, ε_{0}) are disjoint. Also, by (1.1) and Lemma 6.1, we know that

G(ε, ε_{0}) = sup_{x∈M}\{G_{U}(x, ε, ε_{0}), G_{R}(x, ε, ε_{0})\} ≤ \frac{λε^{α}}{ε_{0}^{α}} .  (6.9)

Then we take

D = 2ζ + 2(1 + ζε^{α}) sup_{ε≤ε_{0}} G(ε, ε_{0})ε^{−α} + K′bγ_{m}^{−1}.  (6.10)

By (6.9), G(ε, ε_{0})ε^{−α} ≤ λε_{0}^{−α}. We have D < ∞.

Let ε ≤ ε_{0}.

By Proposition 5.1,

osc(P̂f, B_{ε}(x)) ≤ ∑_{j=1}^{K} ∑_{i=0}^{∞} osc((fĝ) ∘ T̂_{ij}^{−1}χ_{T̂U_{ij}}, B_{ε}(x))

≤ ∑_{j=1}^{K} ∑_{i=0}^{∞}(osc((fĝ) ∘ T̂_{ij}^{−1}, B_{ε}(x))χ_{T̂U_{ij}}(x) + [2 Esup_{B_{ε}(x)}(fĝ) ∘ T̂_{ij}^{−1}]χ_{B_{ε}(∂T̂U_{ij})}(x))

=:∑_{j=1}^{K} ∑_{i=0}^{∞}(R_{ij}^{(1)}(x)χ_{T̂U_{ij}}(x) + R_{ij}^{(2)}(x)χ_{B_{ε}(∂T̂U_{ij})}(x)).  (6.11)

Denote y_{ij} = T̂_{ij}^{−1}x. We can choose N = N(ε) > 0 for each 0 < ε ≤ ε_{0} according to Assumption 4(b).

For R_{ij}^{(1)}(x) with x ∈ T̂U_{ij}, we first consider the case i ≤ N(ε). By Assumption 4(a), (b) and (6.7), we have ĝ(y′_{ij})/ĝ(y_{ij}) ≤ (1 + Js^{α}ε^{α})(1 + cs^{α}ε^{α}) ≤ 1 + ζε^{α} if d(T^{i+1}y_{ij}, T^{i+1}y′_{ij}) ≤ sε. Hence ĝ(y′_{ij}) ≤ (1 + ζε^{α})ĝ(y_{ij}) and osc(ĝ ∘ T̂_{ij}^{−1}, B_{ε}(x)) ≤ osc(ĝ, B_{sε}(y_{ij})) ≤ 2ζε^{α}ĝ(y_{ij}). So we get for almost every x,

R_{ij}^{(1)}(x) = osc(fĝ, T̂_{ij}^{−1}B_{ε}(x) ∩ U_{ij})
≤ osc(f, B_{sε}(y_{ij}) ∩ U_{ij}) Esup_{B_{sε}(y_{ij})∩U_{ij}} ĝ + osc(ĝ, B_{sε}(y_{ij}) ∩ U_{ij}) Einf_{B_{sε}(y_{ij})∩U_{ij}} f
≤ (1 + ζε^{α}) osc(f, B_{sε}(y_{ij}) ∩ U_{ij})ĝ(y_{ij}) + 2ζε^{α}f(y_{ij})ĝ(y_{ij}).

If i > N(ε), then we must have x ∈ R_{0}, and therefore for almost every x,

R_{ij}^{(1)}(x) = osc(fĝ, T̂_{ij}^{−1}B_{ε}(x) ∩ U_{ij})
≤ osc(f, T̂_{ij}^{−1}B_{ε}(x) ∩ U_{ij})) Einf_{T̂_{ij}^{−1}B_{ε}(x)∩U_{ij}} ĝ + osc(ĝ, T̂_{ij}^{−1}B_{ε}(x) ∩ U_{ij}) Esup_{T̂_{ij}^{−1}B_{ε}(x)} f
≤ osc(f, B_{sε}(y_{ij}) ∩ U_{ij})ĝ(y_{ij}) + ∥f∥_{∞} sup_{T̂_{ij}^{−1}B_{ε}(x)} ĝ.

By Assumption 4(b), for any x ∈ R_{0}, ∑_{i=N}^{∞}( sup_{T̂_{ij}^{−1}B_{ε}(x)} ĝ) ≤ bε^{m+α}. Hence,

∑_{j=1}^{K} ∑_{i=0}^{∞} R_{ij}^{(1)}(x)χ_{T̂U_{ij}}(x) ≤ K′bε^{m+α}∥f∥_{∞}χ_{R_{0}}(x)

+ ∑_{j=1}^{K} ∑_{i=0}^{∞}((1 + ζε^{α}) osc(f, B_{sε}(y_{ij}) ∩ U_{ij})ĝ(y_{ij}) + 2ζε^{α}f(y_{ij})ĝ(y_{ij}))
≤ K′bε^{m+α}∥f∥_{∞}χ_{R_{0}}(x)+ (1 + ζε^{α})[P̂ osc(f, B_{sε}(·))](x)+ 2ζε^{α}(P̂f)(x).

Since ∫M̂ P̂fdν̂ = ∫M̂ fdν̂ for any integrable function f, we have

∫M̂ ∑_{j=1}^{K} ∑_{i=0}^{∞} R_{ij}^{(1)}χ_{T̂U_{ij}}dν̂

≤ K′bε^{m+α}∥f∥_{∞}ν̂R_{0} + (1 + ζε^{α}) ∫M̂ osc(f, B_{sε}(·))dν̂ + 2ζε^{α} ∫M̂ fdν̂
≤ (1 + ζε^{α})s^{α}ε^{α}|f|_{α} + 2ζε^{α}∥f∥_{1} + K′bε^{m+α}∥f∥_{∞}ν̂R_{0}.  (6.12)

As for R_{ij}^{(2)}(x), if i ≤ N(ε), then we have

Esup_{B_{ε}(x)}(fĝ) ∘ T̂_{ij}^{−1} ≤ [ Esup_{B_{sε}(y_{ij})} |f|]ĝ(y_{ij})(1 + ζε^{α}).

Hence by the same method as in [S], we get that

∫M̂ ∑_{j=1}^{K} ∑_{i=0}^{N(ε)} R_{ij}^{(2)}χ_{B_{ε}(∂T̂U_{ij})}dν̂ ≤ 2(1 + ζε^{α})G(ε, ε_{0})(ε_{0}^{α}|f|_{α} + ∥f∥_{1}).

If i ≥ N(ε), then Esup_{B_{ε}(x)}(fĝ) ∘ T̂_{ij}^{−1} ≤ ∥f∥_{∞} sup_{T̂_{ij}^{−1}B_{ε}(x)} ĝ, and

∑_{j=1}^{K} ∑_{i=N(ε)}^{∞} R_{ij}^{(2)}χ_{B_{ε}(∂T̂U_{ij})} ≤ 2K′∥f∥_{∞} ∑_{i=N(ε)}^{∞} sup_{T̂_{ij}^{−1}B_{ε}(x)} ĝ.

Again, by Assumption 4(b) it is bounded by 2K′bε^{m+α}∥f∥_{∞}. So we have

∫M̂ ∑_{j=1}^{K} ∑_{i=0}^{∞} R_{ij}^{(2)}χ_{B_{ε}(∂T̂U_{ij})}dν̂

≤ 2(1 + ζε^{α})G(ε, ε_{0})(ε_{0}^{α}|f|_{α} + ∥f∥_{1}) + 2K′bε^{m+α}∥f∥_{∞}ν̂B_{ε}(∂R_{0}).(6.13)

We may assume that ν̂R_{0} + ν̂B_{ε}(∂R_{0}) ≤ 1. By Proposition 5.2(i) and (5.3) we have that ε^{m+α}∥f∥_{∞} ≤ γ_{m}^{−1}ε^{α}∥f∥_{α} and ∥f∥_{α} = |f|_{α} + ∥f∥_{1} respectively. So by (6.11), (6.12) and (6.13), we get

∫M̂ osc(P f, B_{ε}(·))dν̂ ≤ [(1 + ζε^{α})(s^{α}ε^{α} + 2G(ε, ε_{0})ε_{0}^{α}) + 3K′bγ_{m}^{−1}ε^{α}]|f|_{α}
+ [2ζε^{α} + 2(1 + ζε^{α})G(ε, ε_{0}) + 3K′bγ_{m}^{−1}ε^{α}]∥f∥_{1}.

Now the result follows by the choice of η and D in (6.8) and (6.10). □

Acknowledgment During the paper was written, H. Hu was partially supported by the National Science Foundation grant DMS-9970646 and DMS-0240097. He also wishes to acknowledge the hospitality and support of the Centre de Physique Théorique, CNRS, Marseille, where part of work was done. S. Vaienti thanks the Department of Mathematics at Michigan State University for the kind hospitality during the completion of this work. We thank the anonymous referee for the careful reading the paper and the comments and suggestions which helped us to considerably improve the paper. We finally thank F. Ledrappier for useful discussions and suggestions.

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