arXiv:1503.01423v4 [math.DS] 22 Apr 2016
CENTRAL LIMIT THEOREM FOR THE MODULUS OF CONTINUITY OF AVERAGES OF OBSERVABLES ON TRANSVERSAL FAMILIES OF PIECEWISE EXPANDING UNIMODAL MAPS AMANDA DE LIMA AND DANIEL SMANIA Abstract. Consider a C 2 family of mixing C 4 piecewise expanding unimodal maps t ∈ [a, b] 7→ ft , with a critical point c, that is transversal to the topological classes of such maps. Given a Lipchitz observable φ consider the function Z Rφ (t) = φ dµt ,
where µt is the unique absolutely continuous invariant probability of ft . Suppose that σt > 0 for every t ∈ [a, b], where 2 Pn−1 R j Z φdµt j=0 φ ◦ ft − 2 2 dµt . σt = σt (φ) = lim √ n→∞ n m
We show that (
t ∈ [a, b] : t + h ∈ [a, b] and
converges to
1 p Ψ(t) − log |h|
Rφ (t + h) − Rφ (t) h
≤y
)
Z y s2 1 √ e− 2 ds, 2π −∞ where Ψ(t) is a dynamically defined function and m is the Lebesgue measure on [a, b], normalized in such way that m([a, b]) = 1. As a consequence we show that Rφ is not a Lipchitz function on any subset of [a, b] with positive Lebesgue measure.
Contents 1. 2. 3. 4. 5. 6. 7. 8. 9.
Introduction and statement of the main results Families of piecewise expanding unimodal maps Good transversal families Decomposition of the Newton quotient for good families Proof of the Central Limit Theorem for the modulus of continuity of Rφ Controlling how the orbit of the critical point moves Estimates for the Wild part Estimates for the Tame part The function Rφ is not Lipschitz on any subset of positive measure
2 3 4 9 12 20 36 45 53
Date: April 25, 2016. 2010 Mathematics Subject Classification. 37C30, 37C40, 37D50, 37E05, 37A05. Key words and phrases. linear response, dynamical systems, unimodal maps, expanding maps, ergodic theory, central limit theorem. A.L. was partially supported by FAPESP 2010/17419-6 and D.S. was partially supported by CNPq 305537/2012-1. 1
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AMANDA DE LIMA AND DANIEL SMANIA
55 55
Acknowledgments References
1. Introduction and statement of the main results Let ft be a smooth family of (piecewise) smooth maps on a manifold M , and let us suppose that for each ft there is a physical (or SBR) probability µt on M . Given an observable φ : M → R, we can ask if the function Rφ : [0, 1] −→ R R t 7−→ φ dµt
is differentiable and if we can find an explicit formula for its derivative. The study of this question is the so called linear response problem. D. Ruelle showed that Rφ is differentiable and also gave the formula for Rφ ′ , in the case of smooth uniformly hyperbolic dynamical systems (see Ruelle in [16] and [17], and Baladi and Smania in [4] for more details). In the setting of smooth families of piecewise expanding unimodal maps, Baladi and Smania (see [2]) proved that if we have a C 2 family of piecewise expanding unimodal maps of class C 3 , then Rφ is differentiable in t0 , with φ ∈ C 1+Lip , provided that the family ft is tangent to the topological class of ft0 at t = t0 . It turns out that the family s 7→ fs is tangent to the topological class of ft at the parameter t if and only if J(ft , vt ) =
M t −1 X k=0
vt (ftk (c)) = 0, Dftk (ft (c))
where vt = ∂s fs |s=t and Mt is either the period of the critical point c if c is periodic, or ∞, otherwise (see [3]). Now, let us consider a C 2 family of piecewise expanding unimodal maps of class C 4 that is transversal to the topological classes of piecewise unimodal maps, that is (1)
J(ft , vt ) =
M t −1 X k=0
vt (ftk (c)) 6= 0 Dftk (ft (c))
for every t. Baladi and Smania, [2] and [5], proved that Rφ is not differentiable, for most of the parameters t, even if φ is quite regular. One can ask what is the regularity of the function Rφ in this case. We know from Keller [9] (see also Mazzolena [14]) that Rφ has modulus of continuity |h|(log(1/|h|) + 1). We will show the Central Limit Theorem for the modulus of continuity of the function Rφ where φ is a lipschitzian observable. Let 2 Pn−1 R Z φ ◦ ftj − φdµt j=0 dµt 6= 0. √ σt2 = σt2 (φ) = lim n→∞ n Let t 7→ ft be a C 2 family of C 4 piecewise expanding unimodal maps. Note that each ft has a unique absolutely continuous invariant probability µt = ρt m, where
CLT for the modulus of continuity of averages of observables in transversal families
3
its density ρt has bounded variation. Let Z 1 (2) Lt = log |Dft | dµt > 0, ℓt = √ . Lt
Indeed ρt is continuous except on the forward orbit ftj (c) of the critical point (see Baladi [1]). Let St be the jump of ρt at the critical value, that is
(3)
St =
lim
x→ft (c)−
ρt (x) −
lim
x→ft (c)+
ρt (x) =
lim
x→ft (c)−
ρt (x) > 0.
Theorem 1.1. Let 2
t ∈ [a, b] 7→ ft ,
be a transversal C family of mixing C 4 piecewise expanding unimodal maps ft : [0, 1] → [0, 1].
If φ is a lipschitzian observable satisfying σt = 6 0 for every t ∈ [a, b], then for every y∈R (4) ) ( 1 Rφ (t + h) − Rφ (t) p ≤y lim m t ∈ [a, b] : t + h ∈ [a, b] and h→0 h Ψ(t) − log |h| converges to
1 √ 2π
where
Z
y
e−
s2 2
ds,
−∞
Ψ(t) = σt St Jt ℓt . and m is the Lebesgue measure normalized in such way that m([a, b]) = 1. Corollary 1.2. Under the same assumptions above, the function Rφ is not a lipschitzian function on any subset of [a, b] with positive Lebesgue measure. The proof of Corollary 1.2 will be given in the last section as a consequence of a stronger result (Corollary 9.1). 2. Families of piecewise expanding unimodal maps We begin this section by setting the one-parameter family of piecewise expanding unimodal maps. Definition 2.1. A piecewise expanding C r unimodal map f : [0, 1] → [0, 1] is a continuous map with a critical point c ∈ (0, 1), f (0) = f (1) = 0 and such that f |[0,c] and f |[c,1] are C r and 1 Df < 1. ∞
We say that f is mixing if f is topologically mixing on the interval [f 2 (c), f (c)]. For instance, if √ inf |Df (x)| > 2 x
then f is not renormalisable. In particular f is topologically mixing on [f 2 (c), f (c)]. We can see the set of all C r piecewise expanding unimodal maps that share the same critical point c ∈ (0, 1) as a convex subset of the affine subspace {f ∈
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AMANDA DE LIMA AND DANIEL SMANIA
B r : f (0) = f (1)} of the Banach space B r of all continuous functions f : [0, 1] → R that are C r on the intervals [0, c] and [c, 1], with the norm |f |r = |f |∞ + |f |[0,c] |C r + |f |[c,1]|C r . Let ft : [0, 1] → [0, 1], t ∈ [a, b] be a one-parameter family of piecewise expanding C 4 unimodal maps. We assume (1) For all t ∈ [a, b] the critical point of ft is c. (2) The maps ft are uniformly expanding, that is, there exist constants 1 < λ ≤ Λ < ∞ such that for all t ∈ [a, b], 1 1 Dft < λ and |Dft |∞ < Λ. ∞
(3) The map
is of class C 2 .
t ∈ [a, b] 7→ ft ∈ B 4
Each ft admits a unique absolutely continuous invariant probability measure µt and its density ρt has bounded variation (see [12]). By Keller (see [9]), (5)
|ρt+h − ρt |L1 ≤ C|h|(log
1 + 1). |h|
3. Good transversal families It turns out that we can cut the parameter interval of a transversal family ft in smaller intervals in such way that the family, when restricted to each one of those intervals satisfies stronger assumptions. Here, we introduce the notation of partitions following Schnellmann in [19]. Let us denote by K(t) = [ft2 (c), ft (c)] the support of ρt . Let Pj (t), j ≥ 1 be the partition on the dynamical interval composed by the maximal open intervals of smooth monotonicity for the map ftj : K(t) → K(t), where t is a fixed parameter value. Therefore, Pj (t) is the set of open intervals ω ⊂ K(t) such that ftj : ω → K(t) is C 4 and ω is maximal. We can also define analogous partitions on the parameter interval [a, b]. Let x0 : [a, b] −→ [0, 1] t 7−→ ft (c) be a C 2 map from the parameter interval into the dynamical interval. We will denote by xj (t) := ftj (x0 (t)), j ≥ 0, the orbit of the point x0 (t) under the map ft . Consider a interval J ⊂ [a, b]. Let us denote by Pj |J, j ≥ 1, the partition on the parameter interval composed by all open intervals ω in J such that xi (t) 6= c, for all i satisfying 0 ≤ i < j, that is fti (x0 (t)) = fti+1 (c) 6= c,
for all t ∈ ω, and such that ω is maximal, that is, if s ∈ ∂ω, then there exists 0 ≤ i < j such that xi (s) = c. The intervals ω ∈ Pj are also called cylinders.
CLT for the modulus of continuity of averages of observables in transversal families
5
We quote almost verbatim the definition of the Banach spaces Vα given in [19]. The spaces Vα were introduced by Keller [10]. Let m be the Lesbegue measure on the interval [0, 1] Definition 3.1. (Banach space Vα ) For every ψ : [0, 1] → R be a function in L1 (m) and γ > 0, we can define osc (ψ, γ, x) = ess sup ψ|(x−γ,x+γ) − ess inf ψ|(x−γ,x+γ). Given A > 0 and 0 < α ≤ 1 denote
1 α 0 0, with the following properties (I) There exists j0 > 0 with the following property. For every t ∈ [c, d] and for each j ≥ j0 there exists a neighborhood V of t such that for all t′ ∈ V \{t} and all 0 < i < j, we have fti′ (c) 6= c. In particular the one-sided limits lim+ ′
t →t
(6)
∂t′ ftj′ (c)
Dftj−1 (ft′ (c)) ′
and
lim− ′
t →t
∂t′ ftj′ (c)
Dftj−1 (ft′ (c)) ′
exist for every j ≥ j0 , and there is C ≥ 1 so that ∂t′ ftj′ (c) 1 ≤ lim ≤ C, t′ →t+ Dftj−1 C (ft′ (c)) ′
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AMANDA DE LIMA AND DANIEL SMANIA
and ∂t′ ftj′ (c) 1 ≤ ′lim− ≤ C, t →t Dftj−1 C (ft′ (c)) ′
(7)
for all j ≥ j0 and t ∈ [c − δ, d + δ].
(II) The map ft is mixing and there are constants δ > 0, L ≥ 1 and 0 < β˜ < 1 such that for all ψ ∈ Vα ||Lnt ψ||α ≤ Lβ˜n |ψ|α + L |ψ|L1 ,
(8)
for all t ∈ [c − δ, d + δ]. Here Lt is the Ruelle-Perron-Frobenious operator of ft given by X 1 ψ(y). (Lt ψ)(x) = |Dft (y)| ft (y)=x
(III) There is δ > 0 such that for every ζ > 0 there is a constant C˜ satisfying X 1 ˜ nζ ≤ Ce |x′n |ω |∞ ω∈Pn |[a−δ,b+δ]
for all n ≥ 1.
(IV) For all ϕ ∈ Vα such that σt (ϕ) > 0 the functions ξi : [c − δ, d + δ] → R i ≥ 1, defined by Z 1 i+1 ϕ(ft (c)) − ϕdµt ξi (t) = σt (ϕ) satisfy the ASIP for every exponent γ > 2/5. (V) There are positive constants C˜1 , C˜2 , C˜3 , C˜4 , C˜5 , C˜6 and β ∈ (0, 1) such that for every t ∈ [c−δ, d+δ] and its respective density ρt of the unique absolutely continuous invariant probability of ft A1 . The Perron-Frobenious operator Lt satisfies the Lasota-Yorke inequality in the space of bounded variation functions |Lkt φ|BV ≤ C˜6 β k |φ|BV + C˜5 |φ|L1 (m) . A2 . We have ρt ∈ BV and |ρt |BV < C˜1 . A3 . We have ρ′t ∈ BV and |ρ′t |BV < C˜2 . Moreover Z x M t −1 X ρt (x) = ρ′t (u) du + sk (t)Hftk (c) (x), 0
k=1
where Ha (x) = 0 if x < a and Ha (x) = −1 if x ≥ a, (9)
s1 (t) = and
ρt (c) ρt (c) + |Dft (c−)| |Dft (c+)|
sk (t) = Note that St = s1 (t).
s1 (t) . k−1 Dft (ft (c))
CLT for the modulus of continuity of averages of observables in transversal families
7
A4 . We have ρ′′t ∈ BV and |ρ′′t |BV < C˜3 . Moreover Z x M t −1 X ′ ′′ ρt (x) = ρt (u) du + s′k (t)Hftk (c) (x), 0
k=1
where
|s′k (t)| ≤
˜
C4 . |Dftk−1 (ft (c))|
(VI) Let j0 > 0 be the constant given by condition (I). For all i, j satisfying 0 ≤ i, j ≤ j0 and t ∈ [c, d], such that t + h ∈ [c − δ, d + δ] we have c∈ / Ii,j (t, h),
where Ii,j (t, h) is the smallest interval that contains the set i+j+1 i+1 i {ft+h (c), fti+j+1 (c), ft+h ◦ ftj+1 (c), ft+h ◦ ftj (c)}.
Remark 3.4. Conditions (I), (II) and (III) are exactly those that appears in Schnellmann [19], with obvious cosmetic modifications. Remark 3.5. If ft is a good transversal family then of course Eq. (4) converges to Z y s2 1 √ e− 2 ds 2π −∞ if and only if ) ( Rφ (t + h) − Rφ (t) 1 p ≤y (10) lim m t ∈ [a, b] : h→0 h Ψ(t) − log |h| converges to it as well.
Proposition 3.6. Let ft , t ∈ [a, b], be a transversal C 2 family of mixing C 4 piecewise expanding unimodal maps. Then there is a countable family of intervals [ci , di ] ⊂ [a, b], i ∈ ∆ ⊂ N, with pairwise disjoint interior and [ m([a, b] \ [ci , di ]) = 0, i∈∆
such that ft is a good transversal family on each [ci , di ], i ∈ ∆.
Proof. Since ft is transversal, there is just a countable subset Q of parameters where ft has a periodic critical point. It is easy to see that the subset Q′ ⊂ Q of parameters t such that ft is not good and it has a periodic critical point is finite, so without loss of generality we can assume that all maps ft are good. Consider Ω = [a, b] \ (Q ∪ {a, b}). It follows from the analysis in the proof of [4, Theorem 4.1] and [1, Proposition 3.3] that for every t′ ∈ Ω there exists ǫ1 = ǫ1 (t′ ) such that if [c, d] ⊂ (t′ − ǫ1 , t′ + ǫ1 ) then the family ft restricted to [c, d] satisfies condition (V ). By Schnellmann [19] for every t′ ∈ Ω there exists ǫ2 = ǫ2 (t′ ) such that if [c, d] ⊂ (t′ − ǫ2 , t′ + ǫ2 ) then the family ft restricted to [c, d] satisfies conditions (I), (II), (III) and (IV ). We claim that for every t′ ∈ Ω there is ǫ3 = ǫ3 (t′ ) such that if [c, d] ⊂ (t′ − ǫ3 , t′ + ǫ3 ) and δ > 0 is small enough then the family ft , with t ∈ [c, d], satisfies condition (V I). Indeed, since c is not a periodic point of ft′ , there is ǫ3 (t′ ) > 0 such that (11) η := min {|fti+j+1 (c) − c| : 0 ≤ j ≤ j0 and 0 < i ≤ j0 , t ∈ (t′ − ǫ3 , t′ + ǫ3 )} > 0,
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AMANDA DE LIMA AND DANIEL SMANIA
Since t ∈ [t′ − ǫ3 /2, t′ + ǫ3 /2] 7→ ft is a C 2 family the map i (t, h) 7→ ft+h (ftj (c))
is continuous for every 0 < i ≤ j0 and every j satisfying 0 ≤ j ≤ j0 . Therefore there is γ1 := γ1 (i, j) < ǫ3 /2 such that, if |h| ≤ γ1 and t ∈ [t′ − ǫ3 /2, t′ + ǫ3 /2], then i+1 j |ft+h (ft (c)) − fti+1 (ftj (c))| ≤ η,
and i |ft+h (ftj+1 (c)) − fti (ftj+1 (c))| ≤ η,
for all 0 ≤ j ≤ j0 and 0 < i ≤ j0 . Let γ := min{γ1 (i, j) : 0 ≤ j ≤ j0 and 0 < i ≤ j0 }. In particular if |h| ≤ γ1 and t ∈ [t′ − ǫ3 /2, t′ + ǫ3 /2] then c ∈ / Ii,j (t, h) for all 0 ≤ j ≤ j0 , 0 < i ≤ j0 . Let ǫ4 (t′ ) = min{ǫ1 (t′ ), ǫ2 (t′ ), γ}. Consider the family F of intervals [c, d] ⊂ [a, b] such that [c, d] ⊂ (t′ − ǫ4 (t′ ), t′ + ǫ4 (t′ )) for some t′ ∈ Ω. By the Vitali’s covering theorem there exists a countable family of intervals [ci , di ] ⊂ [a, b], [ci , di ] ∈ F , i ∈ ∆ ⊂ N, with pairwise disjoint interior and [ [ m([a, b] \ [ci , di ]) = m(Ω \ [ci , di ]) = 0. i∈∆
i∈∆
We will also need Lemma 3.7. Let 2
t ∈ [a, b] 7→ ft
be a good transversal C family of good and mixing C 4 piecewise expanding unimodal maps ft : [0, 1] → [0, 1].
If φ is a lipschitzian observable satisfying σt 6= 0 for every t ∈ [a, b] then
J = inf |J(ft , vt )|, σ = inf σt (φ), s = inf St , ℓ = inf ℓt , t∈[a,b]
t∈[a,b]
t∈[a,b]
t∈[a,b]
are positive, where St and ℓt are as defined in Eqs. (3) and (2) respectively. Moreover J(ft , vt ) does not changes signs for t ∈ [a, b]. In particular the function t → Ψ(t) = σt St Jt ℓt does not change signs for t ∈ [a, b] and satisfies inf |Ψ(t)| > 0. t
Proof. The function t 7→ J(ft , vt )
is not continuous in a transversal family (see [3]). Indeed, its points of discontinuity lie on the parameters t where the critical point c is periodic for ft , where this function have one-sided limits. However, in [3], Baladi and Smania showed that if vn converges to v and fn converges to f , then if J(fn , vn ) → 0 when n → ∞ we have J(f, v) = 0 and if J(f, v) 6= 0 then J(fn , vn ) has the same sign that J(f, v) for n large. From this it follows that J > 0 and that J(ft , vt ) does not changes signs for t ∈ [a, b]. In [19], Schnellmann proved that t 7→ σt is Hölder continuous.
CLT for the modulus of continuity of averages of observables in transversal families
9
Therefore, σ > 0. Note that St = s1 (t) > 0 everywhere, where s1 is as defined in Eq. (9). Suppose that limn s1 (tn ) = 0. Remember that (see [2] and [1]), Mtn −1
(12)
X
ρtn = ρabs,tn + ρsal,tn = ρabs,tn +
k=1
s1 (tn ) Hf k (c) (ftn (c)) tn Dftk−1 n
where ρabs,tn is absolutely continuous, (ρabs,tn )′ has bounded variation and |(ρabs,tn )′ |BV ≤ C.
(13)
Taking a subsequence, if necessary, we can assume that limn tn = t and that ρtn converges in L1 (m) to ρt . But if limn s1 (tn ) = 0 then by Eqs. (12) and (13) we conclude that ρt is a continuous function. But this is absurd since s1 (t) 6= 0 for every t. Remark 3.8. As an example, we have the family of tent maps defined by tx, if x < 1/2, ft (x) = t − tx, if x ≥ 1/2, t ∈ (1, 2). Tsujii [20] show that the family of tent maps satisfies J(ft0 , ∂t ft |t=t0 ) 6= 0 at every parameter t0 where ft0 has a periodic turning point. So the restriction of this family to a small neighborhood of such parameter t0 is a transversal family. We can observe that, since ft is a piecewise linear map for all t, the density ρt is purely a saltus function. 4. Decomposition of the Newton quotient for good families In this section we will assume that ft is a good family. In order to prove Theorem 1.1 we will decompose the quotient Rφ (t + h) − Rφ (t) h in two parts which will be called the Wild part and the Tame part of the decomposition. Definition 4.1. Let g : [0, 1] → R be a function of bounded variation and t ∈ [a, b]. We define the projection Πt : BV g
−→ BV R 7−→ g − ρt gdm.
Indeed Πt is also a well defined operator in L1 (m) and sup |Πt |BV < ∞ and sup |Πt |L1 (m) < ∞. t
t
R A function g ∈ L (m) belongs to Πt (BV ) if and only if g dm = 0. In particular the operator (I − Lt )−1 is well R defined on Πt (BV ).We are going to use the following observation quite often. If g dm = 0, and 1
g=
∞ X i=0
gi ,
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AMANDA DE LIMA AND DANIEL SMANIA
with gi ∈ BV and the convergence of the series is in the BV norm, then ∞ X (I − Lt )−1 Πt (gi ). (I − Lt )−1 g = i=0
Note also that
Πt ◦ Lt = Lt ◦ Πt . Proposition 4.2. Assume that ft is a family of piecewise expanding unimodal maps as defined in section 2 and let Lt be the Perron-Frobenius operator. Then ρt+h − ρt Lt+h (ρt ) − Lt (ρt ) . = (I − Lt+h )−1 h h Proof. Note that (I − Lt )−1 is well defined in Πt (BV ) and is given by (I − Lt )−1 (ρ) =
∞ X i=0
Lit (ρ),
for every ρ ∈ Πt (BV ). Therefore, the result follows as an immediate consequence of the identity (I − Lt+h )(ρt+h − ρt ) = (I − Lt+h )(ρt ) − (I − Lt )(ρt ). Proposition 4.3. Let ft be a C 2 family of good mixing C 4 piecewise expanding unimodal maps that satisfies property (V) in Definition 3.3. There exists C > 0 with the following property. For every t ∈ [a, b] such that the critical point of ft is not periodic, we can decompose Lt+h (ρt ) − Lt (ρt ) = Φh + rh h where ∞
1X sk+1 (t)Πt+h Hft+h (ftk (c)) − Hft (ftk (c)) Φh = h k=0
and rh satifies
Z
rh dm = 0
and
sup |rh |BV < C. h6=0
We will prove Proposition 4.3 in Section 8. We will call W(t, h) = (I −Lt+h )−1 Φh the Wild part and (I −Lt+h )−1 rh will be called the Tame part of the decomposition. Note that Z Z Rφ (t + h) − Rφ (t) = φW(t, h) dm + φ(I − Lt+h )−1 rh dm. h Definition 4.4. Given h 6= 0 and t ∈ [0, 1], let N := N (t, h) be the unique integer such that 1 1 ≤ |h| < (14) . N +1 N |Dft (ft (c))| |Dft (ft (c))|
There is some ambiguity in the definition of N (t, h) when ftk (c) = c for some k > 0. But since the family is transversal, there exists just a countable number of such parameters (see [3]).
CLT for the modulus of continuity of averages of observables in transversal families
Given a ∈ R define
11
⌊a⌋ = max{k ∈ Z : a ≥ k}.
The following proposition gives us a control on the orbit of the critical point. Proposition 4.5. For large K > 0 and every γ > 0 there exists δ > 0 such that for every small h0 there are sets Γδh′ ,h0 , Γδh0 ⊂ I = [a, b], with Γδh′ ,h0 ⊂ Γδh0 , for every h′ satisfying 0 < h′ < h0 , with the following properties A. limh′ →0 m(Γδh′ ,h0 ) = m(Γδh0 ) > 1 − γ. B. If t ∈ Γδh′ ,h0 and |h| ≤ h′ then there exists N3 (t, h) such that ⌊
(15)
K log N (t, h)⌋ ≤ N (t, h) − N3 (t, h) ≤ C5 K log N (t, h) 2
and (16)
c∈ / Ii,j for all 0 ≤ j < N3 (t, h) and 0 ≤ i < N3 (t, h) − j, where Ii,j is the smallest interval that contains the set i+j+1 i+1 i {ft+h (c), fti+j+1 (c), ft+h ◦ ftj+1 (c), ft+h ◦ ftj (c)}
C. For every t ∈ Γδh′ ,h0 the critical point of ft is not periodic. ˆ < h′ ≤ h0 then Γδ ′ ⊂ Γδ , D. If 0 < h h ,h0 ˆ h,h 0
where m is the normalized Lebesgue measure on I = [a, b]. We will prove Proposition 4.5 in Section 6. The following proposition is one of the most important results in this work. It relates the Birkhoff sum of the observable φ with the Wild part. This fact will allow us to use the almost sure invariance principle obtained by Schnellmann [19]. Proposition 4.6. Let ft be a good transversal family. Let φ : [0, 1] → R be a lipschitzian observable. If t ∈ Γδh,h0 , where Γδh,h0 is the set given by Proposition 4.5, then Z Z N3 (t,h) X 1 . φ(ftj (c)) − φdµt + O log log φ W(t, h)dm = s1 (t)J(ft , vt ) |h| j=0 We will prove Proposition 4.6 in Section 7. Proposition 4.7. Let ft be a good transversal family. Let φ : [0, 1] → R be a lipschitzian observable. If t ∈ Γδh,h0 , where Γδh,h0 is the set given by Proposition 4.5, then Z N3 (t,hn ) X 1 Rφ (t + hn ) − Rφ (t) j φ(ft (c)) − φdµt + O log log = s1 (t)J(ft , vt )hn |hn | j=0 Z 1 φ(I − Lt+h )−1 rh dm. + s1 (t)J(ft , vt ) The proof follows directly from Propositions 4.3 and 4.6.
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AMANDA DE LIMA AND DANIEL SMANIA
5. Proof of the Central Limit Theorem for the modulus of continuity of Rφ To simplify the notation in this section, given a transversal family t 7→ ft we will denote Stf = sf1 (t), Jtf = J(ft , ∂s fs |s=t ), σtf = σtf (φ). Moreover Z Lft = log |Dft |dµft ,
where µft is the unique absolutely continuous invariant probability of ft , and 1 ℓft = q . Lft
When there are not confusion with respect to which family we are dealing with, we will omit f in the notation. Lemma 5.1 (Functional Central Limit Theorem). Let ft be a good transversal C 2 family of C 4 unimodal maps and σt (φ) 6= 0 for every t. For each t ∈ [a, b] let us consider the continuous function θ 7→ XN (θ, t), where XN (θ, t) =
1 √
⌊N θ⌋−1
X
σt N
k=0
φ(ftk (c)) −
Z
(N θ − ⌊N θ⌋) ⌊N θ⌋ √ φ(ft (c)) − φ dµt + σt N
Z
φ dµt .
Considering the normalized Lebesgue measure on t ∈ [a, b], for each N the function t 7→ XN (·, t) induces a measure on the space of continuous functions and such D measures converges in distribution to the Wiener measure. We denote XN −→N W. Proof. By Schnellmann [19] we know that the sequence of functions Z 1 1 ξi (t) = φ(fti+1 (c)) − φdµt σt 0 satisfies the ASIP for every exponent error larger than 2/5. By [15, Theorem E], the ASIP implies the Functional Central Limit Theorem for XN (θ, t). We are going to need the following Proposition 5.2 ([6]). If νn P −→n L, an
(17)
where L is a positive constant and (an )n is a sequence such that an → ∞ when n → ∞, then D
XN −→N W
implies
D
Yn −→n W,
where Yn is 1 p σt νn (t)
⌊νn θ⌋−1
X
k=0
φ(ftk (c))
−
Z
φ dµt
(νn θ − ⌊νn θ⌋) ⌊ν θ⌋ p φ(ft n (c)) − + σt νn (t)
Z
φ dµt ,
CLT for the modulus of continuity of averages of observables in transversal families
13
Proof. See [6], page 152. From now on we will denote Z y s2 1 e− 2 ds. DN (y) = √ 2π −∞ The following lemma will be used many times Lemma 5.3 (A variation of Slutsky’s Theorem). Let An : [0, 1] → R be functions and Ωn ⊂ [0, 1] be such that lim inf m(Ωn ) > 1 − γ, n
and for every y ∈ R the sequence
an (y) = m(t ∈ Ωn : An (t) ≤ y)
eventually belongs to
O(y, ǫ) = (DN (y) − ǫ, DN (y) + ǫ),
that is, there is n0 = n0 (y) such that an (y) ∈ O(y, ǫ) for every n ≥ n0 . Then A. There exists δ > 0 such that if Bn : [0, 1] → R is a function such that lim inf m(t ∈ [0, 1] : |Bn (t) − 1| < δ) > 1 − γ, n
then the sequence bn (y) = m(t ∈ [0, 1] : An (t)Bn (t) ≤ y)
eventually belong to O(y, ǫ + 3γ). B. There exists δ > 0 such that if Bn : [0, 1] → R is a function such that lim inf m(t ∈ [0, 1] : |Bn (t)| < δ) > 1 − γ, n
then the sequence bn (y) = m(t ∈ [0, 1] : An (t) + Bn (t) ≤ y)
eventually belong to O(y, ǫ + 3γ). Proof of A. Define
n DA (y) = {t ∈ Ωn : An (t) ≤ y} n DB = {t ∈ [0, 1] : |Bn (t) − 1| < δ} n DAB (y) = {t ∈ [0, 1] : An (t)Bn (t) ≤ y} Choose δ > 0 such that
sup sup |DN (y) − DN (y(1 − δ ′ ))| < γ, y∈R |δ ′ | 0 there exists Q1 with the following property. Let ft be a good transversal C 2 family of C 4 piecewise expanding unimodal maps with σt (φ) 6= 0 for every t and Lt′ Q = sup 1 − < Q1 . Lt t,t′ ∈[c,d] Then for every h small enough we have ) ( 1 Rφ (t + h) − Rφ (t) 1 p ≤y m t ∈ [c, d] : m([c, d]) h σt ℓt St Jt − log |h|
CLT for the modulus of continuity of averages of observables in transversal families
15
belongs to O(y, 13γ). Proof. Without loss of generality we assume that [c, d] = [0, 1]. It is enough to prove the following claim: For every sequence hn →n 0
and every γ > 0, the sequence (
1 p sn = m t ∈ [0, 1] : σt ℓt St Jt − log |hn |
Rφ (t + hn ) − Rφ (t) hn
≤y
)
eventually belong to the interval O(y, 12γ). Fix a large K > 0. By Proposition 4.5, for every γ > 0 there exist δ > 0, h0 > 0 and sets Γδh,h0 , Γδh0 ⊂ I, with Γδh,h0 ⊂ Γδh0 , for every h 6= 0 satisfying |h| < h0 , such that A. limh→0 m(Γδh,h0 ) = m(Γδh0 ) > 1 − γ. B. If t ∈ Γδh,h0 then there exists N3 (t, h) such that ǫ ⌊ log N (t, h)⌋ ≤ N (t, h) − N3 (t, h) ≤ C5 K log N (t, h) 2 and i+1 i c∈ / [ft+h ◦ ftj+1 (c), ft+h ◦ ftj (c)]
for all 1 ≤ j < N3 (t, h) and 0 ≤ i < N3 (t, h) − j. For all h 6= 0 and t ∈ [0, 1], define N4 (t, h) = N3 (t, h) if t ∈ Γδh,h0 and |h| < h0 , and N4 (t, h) = N (t, h), otherwise. Therefore, for B we have (20) for every (t, h). Since
N (t, h) − N4 (t, h) ≤ C5 K log N (t, h) 1
N (t,h)+1 |Dft (ft (c))|
we have
≤ |h|
1 − γ.
In particular if Lt0 , t)| ≤ 2HQα } Lt then lim inf n m(Dn ) > 1 − γ. Let us apply Lemma 5.3.B with Ωn = Dn , An (t) = L Yn (1, t) and Bn (t) = Yn ( Ltt0 , t) − Yn (1, t). Observe that by Eq.(24) the sequence an (y) defined in Lemma 5.3 eventually belongs to O(y, ǫ) for all ǫ > 0. Hence, taking ǫ = γ, there exists δ1 = δ1 (γ) > 0 such that if 2HQα < δ we have Lt (26) m(t ∈ [0, 1] : Yn ( 0 , t) ≤ y) Lt eventually belongs to O(y, 4γ). Choose Q0 = Q0 γ > 0 such that if Q < Q0 then 2HQα < δ1 . Note that r Z ⌊N4 (t,hn )⌋−1 X Lt0 Lt0 1 k p , t) = (27) Yn ( φ(ft (c)) − φ dµt . Lt Lt σt N4 (t, hn ) k=0 Dn = {t ∈ [0, 1] : |Yn (1, t) − Yn (
CLT for the modulus of continuity of averages of observables in transversal families
17
By Eq. (21) and Lemma 5.3.A, the sequence p Z ⌊N4 (t,hn )⌋−1 X Lt0 k m(t ∈ [0, 1] : p φ(ft (c)) − φ dµt ≤ y) σt − log |hn | k=0
eventually belongs to O(y, 7γ). Applying again Lemma 5.3.A, with
p Z ⌊N4 (t,hn )⌋−1 X Lt0 An (t) = p φ(ftk (c)) − φ dµt , σt − log |hn | k=0
Ωn = [0, 1] and
Bn (t) =
s
Lt , Lt0
there exists δ2 = δ2 (γ) > 0 such that if s L t − 1 < δ2 (28) Lt0 for every t then
m(t ∈ [0, 1] :
√ Z ⌊N4 (t,hn )⌋−1 X Lt p φ(ftk (c)) − φ dµt ≤ y) σt − log |hn | k=0
eventually belong to O(y, 10γ). Choose Q1 = min{Q0 , δ2 } such that Q < Q1 implies Eq. (28). Finally by Propositions 4.6 and 4.7, if 0 < |hn | ≤ h0 and t ∈ Γδhn ,h0 we have Z N3 (t,hn ) X 1 Rφ (t + hn ) − Rφ (t) j φ(ft (c)) − φdµt + O log log = S t Jt h n |hn | j=0 Z 1 + φ(I − Lt+h )−1 rh dm. S t Jt Since log log |h1n | q →n 0 log |h1n |
and
sup |(I − Lt+h )−1 rh |L1 < ∞, we have Rφ (t + hn ) − Rφ (t) 1 p = p St σt Jt hn − log |hn | σt − log |hn |
N3 (t,hn )
X
φ(ftj (c))
j=1
where
lim
sup
n t∈Γδ hn ,h
0
|r(t, hn )| = 0.
−
Z
φdµt
+ r(t, hn ),
18
AMANDA DE LIMA AND DANIEL SMANIA
Hence, it is easy to conclude that Rφ (t + hn ) − Rφ (t) p St σt ℓt Jt hn − log |hn |
(29)
=
for every t ∈
1 p ℓt σt − log |hn |
Γδhn ,h0 ,
N3 (t,hn )
X
φ(ftj (c)) −
j=1
Z
φdµt
+ r′ (t, hn ),
where
1 ℓt = √ Lt
and lim
sup
n t∈Γδ hn ,h
|r′ (t, hn )| = 0.
0
Since m(Γδh,h0 ) > 1−γ, we can apply Lemma 5.3 (remember that N4 (t, h) = N3 (t, h) for t ∈ Γδh,h0 ) to conclude that the sequence m(t ∈ [0, 1] :
Rφ (t + hn ) − Rφ (t) p ≤ y) St σt ℓt Jt hn − log |hn |
eventually belong to the interval O(y, 13γ).
Lemma 5.7. Let [ci , di ] ⊂ [a, b], i ∈ ∆ ⊂ N, be intervals with pairwise disjoint interior and such that [ m([a, b] \ [ci , di ]) = 0. i∈∆
If t 7→ ft , with t ∈ [ci , di ], are good transversal families such that for all i ∈ ∆ and y ∈ R we have ) ( 1 1 Rφ (t + h) − Rφ (t) p ≤y m t ∈ [ci , di ] : m([ci , di ]) h σt ℓt St Jt − log |h|
eventually belongs to O(y, γ), then ) ( 1 1 Rφ (t + h) − Rφ (t) p ≤y m t ∈ [a, b] : t + h ∈ [a, b] and m([a, b]) h σt ℓt St Jt − log |h| eventually belongs to O(y, γ + ǫ), for every ǫ > 0.
Proof. Define ( Ω(h, y) = and Ωi (h, y) =
1 p t ∈ [a, b] : t + h ∈ [a, b] and σt ℓt St Jt − log |h|
(
1 p t ∈ [ci , di ] : t + h ∈ [a, b] and σt ℓt St Jt − log |h|
Rφ (t + h) − Rφ (t) h
Rφ (t + h) − Rφ (t) h
≤y
Of course Ωi (h, y) are pairwise disjoint up to a countable set, Ωi (h, y) ⊂ Ω(h, y) and m(Ω(h, y) \ ∪i Ωi (h, y)) = 0. Then X m(Ω(h, y)) = m(Ωi (h, y)). i∈∆
)
≤y
)
.
CLT for the modulus of continuity of averages of observables in transversal families
19
Given ǫ ∈ (0, 1), choose i0 such that m(∪i>i0 [ci , di ]) < ǫ m([a, b]). For every i ≤ i0 there exists hi > 0 such that for every |h| < hi we have m(Ωi (h, y)) m([ci , di ]) ˆ = mini≤i hi . Let belongs to O(y, γ + ǫ). Let h 0 Ui0 (h, y) = ∪i≤i0 Ωi (h, y), and Wi0 (h, y) = ∪i≤i0 [ci , di ]. ˆ we have Then for |h| < h X m([ci , di ]) m(Ωi (h, y)) m(Ui0 (h, y)) = m(Wi0 (h, y)) m(Wi0 (h, y)) m([ci , di ]) i≤i0
is a convex combination of elements of O(y, γ + ǫ), then it belongs to O(y, γ + ǫ). We conclude that (DN (y) − γ − 2ǫ)m([a, b])
≤ (DN (y) − γ − ǫ)(m([a, b]) − ǫ m([a, b]))
≤ (DN (y) − γ − ǫ)m(Wi0 (h, y)) ≤ m(Ui0 (h, y))
≤ m(Ω(h, y))
≤ m(Ui0 (h, y)) + ǫ m([a, b]) (30)
≤ (DN (y) + γ + ǫ)m(Wi0 (h, y)) + ǫ m([a, b]) ≤ (DN (y) + γ + 2ǫ)m([a, b]).
Proof of Theorem 1.1. Remember that t 7→ Lt is a continuous and positive function on [a, b]. Given γ > 0, let Q1 > 0 be as in Proposition 5.6. Then there are k > 0 and intervals [ci , di ], i ≤ k = k(γ), which forms a partition F of [a, b] and 1 − Lt′ < Q1 Lt t,t′ ∈[ci ,di ] sup
for every i ≤ k. Then the restrictions of the family ft to each one of the intervals [ci , di ] satisfy the assumptions of Proposition 5.6. Now it remains to apply Lemma 5.7 to the full family and the partition F . Since γ > 0 is arbitrary we completed the proof of Theorem 1.1.
20
AMANDA DE LIMA AND DANIEL SMANIA
6. Controlling how the orbit of the critical point moves The aim of this section is to prove Proposition 4.5. Let us denote by I = [0, 1] the interval of parameters. Remark 6.1. In Schnellmann [19, Lemma 4.4] it is proven that there is C1 > 0 such that if N ≥ 1, |t1 − t2 | < 1/N and if ω1 ∈ PN (t1 ) and ω2 ∈ PN (t2 ) have the same combinatorics up to the (N − 1)-th iteration then DftN1 (x1 ) Df N (x ) ≤ C1 , 2 t2
for all x1 ∈ ω1 and x2 ∈ ω2 . We also observe that if x, y ∈ ω ∈ PN (t), then by the bounded distortion lemma, there is C2 > 0 such that Df j (x) t ≤ C2 , Dftj (y) for every j ≤ N . Let
˜ = sup M
sup |∂t ftj (c)|,
0≤j≤j0 t∈[a,b]
and let us define (31)
˜ }, C3 = max{C, M
where C is the constant given by the transversality condition (see Eqs (6) and (7)) and C4 = sup sup |∂t ft (x)|. t∈[0,1] x∈[0,1]
To prove Proposition 4.5 we will need Lemma 6.2. Let N3 ∈ N and ω ∈ PN3 be such that 1 . |ω| ≤ N3 If t ∈ ω and
(32)
dist(t, ∂ω) > (M + 1)|h|,
where (33) Then (34)
M > max{C1 C3 C4 , C12 C2 C32 } c∈ / Ii,j (t, h)
for all 0 ≤ j < N3 and 0 ≤ i < N3 − j, where Ii,j (t, h) is the smallest interval that contains the set i+j+1 i+1 i ◦ ftj+1 (c), ft+h ◦ ftj (c)}. {ft+h (c), fti+j+1 (c), ft+h
Proof. Let j0 be as defined in condition (I) (see Definition 3.3). If j > j0 define i1 = 0 and if 0 ≤ j < j0 define i1 = j0 . First of all, we observe that if 0 ≤ j ≤ j0 and 0 ≤ i ≤ j0 then Eq. (34) follows from condition (V I). In particular (35)
c∈ / Ii,j (t, h) for every i < i1 .
Hence, it is left to consider the cases when
j0 < j < N3 and 0 < i ≤ N3 − j
CLT for the modulus of continuity of averages of observables in transversal families
21
and We claim that
0 < j < j0 and j0 < i ≤ N3 − j. c∈ / Ii1 ,j .
(36)
Indeed, if 0 ≤ j ≤ j0 , it follows from condition (V I). Now, if j0 < j < N3 , due to condition (I), Eqs. (6) and (7) the maps θ ∈ ω → fθk (c) ∈ [0, 1] are diffeomorphisms on their images for every j0 < k ≤ N3 and they do not contain the critical point in its image, for all j0 < k < N3 , θ ∈ ω. In particular if ω = (s1 , s2 ) then c∈ / {fθk (c) : θ ∈ ω} = (fsk1 (c), fsk2 (c))
(37)
for every j0 < k < N3 . Therefore, k c∈ / [ftk (c), ft+h (c)].
By the Mean Value Theorem and Remark 6.1, for every j < N3 j+1 |ftj+1 (c)−ft+h (c)| = |∂θ fθj+1 (c)|θ=θ1 ||h| ≤ C3 |Dfθj1 (fθ1 (c))||h| ≤ C3 C1 |Dftj (ft (c))||h|.
Moreover, |ft+h (ftj (c)) − ft (ftj (c))| ≤ |∂θ fθ (ftj (c))|θ=θ2 ||h| ≤ C4 |h|.
(38)
By assumption, d([t, t + h], ∂ω) > M |h|. Thus, |ω| ≥ (2M + 1)|h|.
(39)
If ∂ω = {s1 , s2 } and s ∈ [t, t + h] then (c) − fsk+1 (c)| |fsk+1 i
= ≥ ≥
(40)
|∂θ fθk+1 (c)|θ=θ3 ||si − s| 1 |Dfθk3 (fθ3 (c))|M |h| C3 1 |Dftk (ft (c))|M |h| C1 C3
for every k < N3 . Taking k = j we obtain |ft+h (ftj (c)) − ft (ftj (c))| (41)
M |h| C1 C3
≤
C4 |h| ≤
≤
M (c) − ftj+1 (c)|. |Dftj (ft (c))||h| ≤ |fsj+1 i C1 C3
Hence, (42)
(c)). (c), fsj+1 [ft+h (ftj (c)), ft (ftj (c))] ⊂ (fsj+1 2 1
In particular c∈ / I0,j (t, h) = Ii1 ,j (t, h).
We concluded the proof of our claim. Now fix 0 ≤ j < N3 . We are going to prove by induction on i that, for every i1 ≤ i < N3 − j, (43)
c∈ / Ii,j (t, h),
22
AMANDA DE LIMA AND DANIEL SMANIA
The case i = i1 follows from Eq. (36). Now suppose that Eq. (43) holds up to i. Provided that i ≥ i1 , we have i + j + 1 > j0 . Therefore, by Eq. (37), with k = i + j + 2, we obtain (i+1)+j
ft+h
(c)). (c), fs(i+1)+j+1 (ft+h (c)) ∈ (fs(i+1)+j+1 2 1
And as in Eq. (40) (44)
(i+1)+j
(fsi (c)) − ft+h |fs(i+1)+j i
(ft+h (c))| ≥
1 (i+1)+j |Dft+h (ft+h (c))|M |h| C3 C1
Moreover by induction assumption and Eq. (35) , we have for every 0 ≤ k ≤ i c 6∈ Ik,j (t, h)
Thus the points j+1 ft+h (c) and ftj+1 (c)
have the same combinatorics up to i iterations of the map ft+h . Then by Remark 6.1 j+1 i+1 j+1 i+1 j+1 i+1 j+1 |ft+h (ft (c)) − ft+h (ft+h (c))| ≤ C2 |Dft+h (ft+h (c))||ftj+1 (c) − ft+h (c)| i+1 j+1 ≤ C2 |Dft+h (ft+h (c))||∂θ fθj+1 (c)|θ=θ4 ||h|
i+1 j+1 ≤ C3 C2 |Dft+h (ft+h (c))||Dfθj4 (fθ4 (c))||h|
j i+1 j+1 ≤ C1 C2 C3 |Dft+h (ft+h (c))||Dft+h (ft+h (c))||h| (i+1)+j
≤ C1 C2 C3 |Dft+h
(45)
(ft+h (c))||h|
and j ftj (c) and ft+h (c)
have the same combinatorics up to i + 1 iterations of the map ft+h . Then by Remark 6.1 (i+1)+1
|ft+h
(i+1)+1
(ftj (c)) − ft+h
(i+1)+1
j (ft+h (c))| ≤ C2 |Dft+h
(i+1)+1
≤ C2 |Dft+h
j j (ft+h (c))||ftj (c) − ft+h (c)|
j (ft+h (c))||∂θ fθj (c)|θ=θ5 ||h|
(i+1)+1
≤ C2 C3 |Dft+h (46)
j (fθ5 (c))|||h| (ft+h (c))||Dfθj−1 5
≤ C1 C2 C3 |Dft+h
(i+1)+1
j j−1 (ft+h (c))||Dft+h (ft+h (c))||h|
≤ C1 C2 C3 |Dft+h
(i+1)+j
(ft+h (c))||h|.
Since C1 C2 C3
j0 . Observe that for each cylinder ω ∈ Pj j 1 , |ω| ≤ C3 λ
where C3 is the constant given by Eq. (31). Let N > 1 and define j = j(N ) as log(C3 N ) j= + 1. log λ Note that the cylinders of Pj divide the interval of parameters I in subintervals of length shorter than 1/N . Let J be one of these intervals in Pj . And we will denote by tR the right boundary point of J. Observe that, by defintion, there is an integer i, 0 ≤ i < j such that (c) = c. xi (tR ) = fti+1 R √ 1/τ Fix an integer τ such that 2 ≤ λ.
Definition 6.3 (The sets EN,J ). Let J ∈ Pj , j = j(N ). Let EN,J be the family of all intervals ω ∈ PN such that for every k satisfying K log N 0≤k≤ τ and for ω ˜ satisfying
ω ˜ = (a, b) ∈ PN −⌊K log N ⌋+q , with ω ⊂ ω ˜ ⊂ J,
where
q = min{(k + 1)τ, ⌊K log N ⌋}, one of the following statements holds: A. For every i satisfying we have
N − ⌊K log N ⌋ + kτ ≤ i < N − ⌊K log N ⌋ + q xi (a) 6= c.
B. For every i satisfying we have
N − ⌊K log N ⌋ + kτ ≤ i < N − ⌊K log N ⌋ + q xi (b) 6= c.
Define EN =
[
EN,J .
J∈Pj
Let us denote by |EN | the sum of the lengths of the intervals in this family. Given n ∈ N and ω ˜ ∈ Pn define δt := min{|fti (c) − ftj (c)| : fti (c) 6= ftj (c) i, j ≤ τ.} δω := Notice that if ω ˜ ⊃ ω then δω˜ ≤ δω .
mint∈ω δt . 2
24
AMANDA DE LIMA AND DANIEL SMANIA
Let CL be such that |fti (c) − fsi (c)| ≤ CL |t − s|
for all i ≤ τ , s, t ∈ [0, 1].
Lemma 6.4. There is C > 0 such that the following holds. If ω ˜ ∈ Pi , i > j0 , with |˜ ω| < 1/i and t ∈ ω ˜ then |xi (˜ ω )| |xi (˜ ω )| 1 ≤ |˜ ω| ≤ C . i i C |Dft (ft (c))| |Dft (ft (c))|
(47)
Moreover, if ω ∈ PN \ EN then there exists i satisfying such that ω ⊂ ω ˜ ∈ Pi and if
N − ⌊K log N ⌋ ≤ i ≤ N CL |˜ ω | < δω˜
then
|xi (˜ ω )| ≥ δω˜
and (48) for every t ∈ ω ˜.
δω˜ 1 1 ≤ |˜ ω| ≤ C C |Dfti (ft (c))| |Dfti (ft (c))|
Proof. If t ∈ ω ˜ ∈ Pk then by the Mean Value Theorem for some θ1 ∈ ω ˜ ω |, |xk (˜ ω )| = |∂θ fθk+1 (c)|θ=θ1 ||˜
then
|Dfθk1 (fθ1 (c))||˜ ω| |Dftk (ft (c))||˜ ω| ≤ ≤ |xk (˜ ω )| C1 C3 C3
and ω | ≤ C1 C3 |Dftk (ft (c))||˜ ω |, |xk (˜ ω )| ≤ C3 |Dfθk1 (fθ1 (c))||˜ therefore, Eq. (47) holds. Now assume ω ∈ PN \ EN . Then there exists k satisfying K log N 0≤k≤ τ
and
ω ˜ = (a, b) ∈ PN −⌊K log N ⌋+q , where q = min{(k + 1)τ, ⌊K log N ⌋}, such that xia (a) = c = xib (b), where in particular
N − ⌊K log N ⌋ + kτ ≤ ia , ib < N − ⌊K log N ⌋ + q, xN −⌊K log N ⌋+q (˜ ω) = (fana (c), fbnb (c)),
where Thus,
0 ≤ na , nb < τ, with na 6= nb . |xN −⌊K log N ⌋+q (˜ ω )| =
(49)
≥
≥
|fana (c) − fbnb (c)|
|fana (c) − fanb (c)| − |fanb (c) − fbnb (c)|
2δω˜ − CL |a − b| ≥ δω˜ .
CLT for the modulus of continuity of averages of observables in transversal families
25
Since δω˜ ≥ 0 depends only on a fixed finite number of iterations of the family ft , it will be easy to give positive lower bounds to it that hold for most of the intervals ω ˜ . Indeed define ΛδN0 = {t ∈ [0, 1] : for every N ≥ N0 if t ∈ ω ∈ PN −2⌊K log N ⌋ then δω > δ}. ′
Note that ΛδN0 ⊂ ΛδN0 +1 . Moreover δ ′ < δ implies ΛδN0 ⊃ ΛδN0 . Lemma 6.5. Given γ > 0 there exists δ > 0 such that lim |ΛδN0 | ≥ 1 − γ.
N0 →∞
Proof. Since ft is a transversal family, the set of parameters t such that fti (c) = ftj (c) for some i 6= j, with i, j ≤ τ + 1 is finite. Let t1 , . . . , tm be those parameters. The function t → δt is positive and continuous on O = [0, 1] \ {t1 , . . . , tm }.
Choose N0 large enough such that Thus,
#{ω ∈ PN0 −2⌊K log N0 ⌋ : ω ∩ {t1 , . . . , tm } 6= ∅} ≤ 2m. |{ω ∈ PN0 −2⌊K log N0 ⌋ : ω ⊂ O}| ≥ 1 −
2Cm > 1 − γ, λN0 −2⌊K log N0 ⌋
provided N0 is large enough. Let 1 δ := min{δω : ω ∈ PN −2⌊K log N ⌋ , ω ⊂ O}. 2 Note that δ > 0 and [ ΛδN ⊃ {ω ∈ PN −2⌊K log N ⌋ : ω ⊂ O} for every N ≥ N0 , provided that N0 is large.
Proposition 6.6. There exist Cˆ1 , Cˆ2 > 0, that do not depend on K, such that for every K′ < K there exists K = K(K′ ) > 0 such that ˆ
ˆ
′
|EN | ≤ KN C2 −C1 K .
(50)
The proof of this proposition follows from Lemma 6.7. There exists Cˆ1 > 0, that does not depend on K, such that for every K′ < K there exists K = K(K′ ) > 0 such that if J ∈ Pj , j =j(N), and EN,J is as defined before, then ˆ
′
|EN,J | ≤ KN −C1 K .
(51)
We will prove Lemma 6.7 later in this section. Proof of Proposition 6.6. We have EN =
[
EN,J .
J∈Pj
Since there are at most 2j cylinders of level j, we have by Lemma 6.7 that there exist Cˆ1 > 0 and K = K(K′ ) such that (52)
|EN | ≤ 2
log(C3 N ) log λ
ˆ
′
log 2
log 2
ˆ
′
KN −C1 K = KC3log λ N log λ −C1 K .
26
AMANDA DE LIMA AND DANIEL SMANIA
Define (53) ΩN0 = t ∈ [0, 1] : ∀N ≥ N0 ∃ ω ∈ PN −⌊K log N ⌋ \EN −⌊K log N ⌋ and t ∈ ω . Note that ΩN0 ⊂ ΩN0 +1 .
Corollary 6.8. If Cˆ2 − Cˆ1 K < −1 we have lim |ΩN0 | = 1.
(54)
N0 →∞
Proof. Notice that ΩN 0 =
\
[
ω.
N ≥N0 ω∈PN −⌊K log N ⌋ \EN −⌊K log N ⌋
If we choose K′ < K such that Cˆ2 − Cˆ1 K′ < −1 we have [ X [ ˆ ˆ ′ N0 →∞ |ΩcN0 | = K(N − ⌊K log N ⌋)C2 −C1 K −→ 0. ω ≤ N ≥N0 ω∈EN −⌊K log N ⌋
N ≥N0
From now on we choose and fix K > 0 satisfying Cˆ2 − Cˆ1 K < −1. Corollary 6.9. For every γ > 0 there exists δ > 0 such that lim m(ΛδN0 ∩ ΩN0 ) > 1 − γ.
N0 →∞
Definition 6.10. Given δ > 0 and h0 > 0, define Γδh0 as the set of all parameters t ∈ [0, 1] such that for every h, 0 < |h| ≤ h0 , there exists k satisfying N (t, h) − 2⌊ǫ log N (t, h)⌋ ≤ k ≤ N (t, h) − ⌊ǫ log N (t, h)⌋ such that if t ∈ ω ˆ ∈ Pk then |xk (ˆ ω )| > δ. Given t ∈ Γδh0 and h 6= 0, let N2 (t, h) be the largest k with this property. Definition 6.11. Given h and t ∈ [0, 1], define (55)
N1 (t, h) := N (t, h) − ⌊K log N (t, h)⌋,
and for h0 > 0 define ˆ1 (h0 ) := N Since lim max
N →∞ t∈[0,1]
we have
min
t∈I,|h|≤h0
N1 (t, h).
1 = 0, |DftN (ft (c))|
ˆ1 (h0 ) = +∞. lim N
h0 →0
Lemma 6.12. For every γ > 0 there exists δ > 0 such that lim m(Γδh0 ) > 1 − γ.
h0 →0
CLT for the modulus of continuity of averages of observables in transversal families
27
Proof. By Corollary 6.9 there exist δ > 0 and N0 such that m(ΛδN0 ∩ ΩN0 ) > 1 − γ.
Choose h0 such
ˆ1 (h0 ) > N0 . N
Let |h| ≤ h0 . Then
N (t, h) − ⌊K log N (t, h)⌋ ≥ N0 . If t ∈ ΛδN0 ∩ ΩN0 , choosing ω ˜ such that t ∈ ω ˜ ∈ PN (t,h)−⌊K log N (t,h)⌋ then ω ˜ 6∈ EN (t,h)−⌊K log N (t,h)⌋ .
Hence, by Lemma 6.4 there exists k satisfying (here N = N (t, h)) N − ⌊K log N ⌋ − ⌊K log(N − ⌊K log N ⌋)⌋ ≤ k ≤ N − ⌊K log N ⌋
such that if t ∈ ω ˜⊂ω ˆ ∈ Pk then since t ∈
ΛδN0 ,
|xk (ˆ ω )| ≥ δωˆ > δ
so that CL |ˆ ω| < δ < δω˜ . Therefore, Γδh0 ⊃ ΛδN0 ∩ ΩN0 .
Definition 6.13. Given h0 > 0 and δ > 0, for every h such that |h| ≤ h0 let Aδh,h0 be a covering of Γδh0 by intervals ω with the following properties P1 . There exists t ∈ Γδh0 such that t ∈ ω ∈ PN2 (t,h) . P2 . If t′ ∈ Γδh0 and t′ ∈ ω then ω ′ ⊂ ω, where t′ ∈ ω ′ ∈ PN2 (t′ ,h) . P3 . There does not exist t′′ ∈ Γδh0 such that t′′ ∈ ω ′′ ∈ PN2 (t′′ ,h) and ω $ ω ′′ .
One can check that one such collection Aδh,h0 does exist. Indeed, consider the covering of Γδh0 given by {ω : there exists t ∈ Γδh0 such that t ∈ ω ∈ PN2 (t,h) }.
Of course this covering satisfies property P1 . Remove from this covering all intervals ω that do not satisfy property P3 . Then the remaining collection is a covering of Γδh0 satisfying properties P1 , P2 and P3 . Note also that the distinct intervals in Aδh,h0 are pairwise disjoint. Indeed, if ω, ω ′ ∈ Aδh,h0 , with ω 6= ω ′ and ω ∩ ω ′ 6= ∅ then either ω $ ω ′ or ω ′ $ ω, which is in contradiction with property P3 . We note that |Aδh,h0 | ≥ m(Γδh0 ), since Aδh,h0 covers Γδh0 . Here |Aδh,h0 | denotes the Lebesgue measure of the union of the intervals in the family Aδh,h0 . Lemma 6.14. If h0 is small enough there are C5 > 0 and C6 > 0 such that the following holds. Given t′ ∈ Γδh0 , let ω be the unique interval in Aδh,h0 such that t′ ∈ ω. Let t ∈ Γδh0 be such that t ∈ ω ∈ PN2 (t,h) . Then ǫ (56) ⌊ log N (t′ , h)⌋ ≤ N (t′ , h) − N2 (t, h) ≤ C5 K log N (t′ , h) 2 and (57) ′
|ω| ≥ C6 δN (t′ , h)K
log λ 2
|h|.
Proof. Consider ω such that t′ ∈ ω ′ ∈ PN2 (t′ ,h) .
Then by property P2 we have ω ′ ⊂ ω. Since ′
N (t′ ,h)
δ ≤ |xN2 (t′ ,h) (ω ′ )| = |∂θ f N2 (t ,h) (c)||ω ′ | ≤ C1 C3 |Dft′ 2
(ft′ (c))|,
28
AMANDA DE LIMA AND DANIEL SMANIA
it follows that (58)
δ 1 C1 C3 . ≤ |ω ′ | ≤ |ω| ≤ N2 (t,h) C1 C3 |Df N′ 2 (t′ ,h) (ft′ (c))| |Dft (ft (c))| t
Since t, t′ ∈ ω, there is C1 > 1 such that
1 1 1 1 ≤ C1 ≤ . C1 |Dfti′ (ft′ (c))| |Dfti (ft (c))| |Dfti′ (ft′ (c))|
for every i ≤ N2 (t, h). Choose C¯ such that δ
(59)
C32 C13
>
1 . λC¯
Then ¯ N2 (t′ , h) ≥ N2 (t, h) − C,
otherwise C1 C3 δ 1 ≤ ′ ,h) N (t N (t,h) 2 2 C1 C3 |Df ′ |Dft (ft (c))| (ft′ (c))| t C1 C3 1 ≤ N2 (t,h)−N2 (t′ ,h) N2 (t′ ,h)+1 N2 (t′ ,h) |Dft (ft (c))| |Dft (ft (c))| C1 C1 C3 , ≤ ′ λC¯ |DftN′ 2 (t ,h) (ft′ (c))| which contradicts Eq. (59). In particular N (t′ , h) − N2 (t, h) ≥ N (t′ , h) − N2 (t′ , h) − C¯ ≥ ⌊ǫ log N (t′ , h)⌋ − C¯ ǫ ≥ ⌊ log N (t′ , h)⌋. 2 Note that the lower bound holds if h0 is small enough. Thus, N (t′ , h) > N2 (t, h). Moreover, |h| ≤ ≤ ≤
1 N (t′ ,h) |Dft′ (ft′ (c))|
1
1
N (t′ ,h)−N2 (t,h) N2 (t,h)+1 (c))| (ft′ |Dft′
N (t,h) (ft′ (c))| |Dft′ 2
1
N (t′ ,h)−N2 (t,h) N2 (t,h)+1 (c))| (ft′ |Dft′
On the other hand, |h| ≥ (60)
≥
C1 . N2 (t,h) |Dft (ft (c))|
1 N (t,h)+1 |Dft (ft (c))|
1
1
N (t,h)−N2 (t,h) N2 (t,h)+1 |Dft (ft (c))|
N (t,h) |Dft 2 (ft (c))|
1 . Λ
CLT for the modulus of continuity of averages of observables in transversal families
29
Then N (t′ ,h)−N2 (t,h)
log |Dft′ and consequently
≤
N (t,h)+1
(ft′ 2
(c))| − log C1
N (t,h)−N2 (t,h) N2 (t,h)+1 log |Dft (ft (c))|
+ log Λ
N (t′ , h) − N2 (t, h) ≤ Cˆ3 (N (t, h) − N2 (t, h)) + Cˆ4 .
In a similar way, we can obtain where
N (t, h) − N2 (t, h) ≤ Cˆ3 (N (t′ , h) − N2 (t, h)) + Cˆ4 , log Λ Cˆ3 = log λ
and
log C1 . Cˆ4 = log λ N (t, h) = N (t, h) − N2 (t, h) + N2 (t, h) ≤ 2ǫ log N (t, h) + N (t′ , h)
N (t, h) N (t′ , h) N (t, h) − 2ǫ log N (t, h) ≤ 2N (t′ , h), ≤
provided that h0 is small. Consequently N (t′ , h) − N2 (t, h) ≤ Cˆ3 (N (t, h) − N2 (t, h)) + Cˆ4 ≤ Cˆ3 2⌊K log N (t, h)⌋ + Cˆ4
≤ Cˆ3 2K log[2N (t′ , h)] + Cˆ4
≤ C5 K log N (t′ , h).
(61)
Here the last inequality holds if h0 is small enough. Moreover, by Eq. (58) N (t′ ,h)−N (t′ ,h)
N (t′ ,h)+1
2 1 δ |Dft′ (c))| (ft′ 2 δ |ω| ≥ = ′ ′ N (t ,h) C1 C3 |Df N′ 2 (t ,h) (ft′ (c))| C1 C3 (ft′ (c))| |Dft′ t (62) ′ ′ K log N (t′ ,h) log λ δ δ δ λN (t ,h)−N2 (t ,h) −1 2 ≥ |h| = λ N (t′ , h)K 2 |h|. ≥ ′ N (t ,h) C1 C3 |Df ′ C1 C3 C1 C3 λ (ft′ (c))|
t
Hence, we obtain Eq. (57). Choose ǫ > 0 such that
1 √ < 1 − ǫ. λ
Lemma 6.15. Given M > 0, define M +1 δ |h| . Bh,h = t : t ∈ ω ∈ Aδh,h0 and dist(t, ∂ω) > 0 ,M 1−ǫ Let hi = (1 − ǫ)i h0 . Given h satisfying 0 < |h| ≤ h0 , let i(h) = max{i ∈ N : |h| < (1 − ǫ)i−1 h0 }.
30
AMANDA DE LIMA AND DANIEL SMANIA
For every h > 0 define δ ˆδ Γ h,h0 = Γh0 ∩
\
i≥i(h)
Bhδ i ,h0 ,M .
Then ˆ < h then Γ ˆδ ˆδ , A. If 0 < h h,h0 ⊂ Γh,h ˆ 0 B. We have ˆ δ ) = m(Γδ ). lim m(Γ h0 h,h0 h→0
Proof. Note that min N (t, hi ) ≥ −
t∈[0,1]
log h0 i log(1 − ǫ) −1− , log Λ log Λ
where
i log(1 − ǫ) log h0 > 0 and − > 0. log Λ log Λ Therefore, if h0 is samall enough, there are K1 , K2 > 0, such that −
min N (t, hi ) ≥ K1 + iK2 .
t∈[0,1]
Define Ah =
[
ω.
ω∈Aδh,h 0
If ω ∈ Aδh,h0 then there is t ∈ Γδh0 such that t ∈ ω ∈ PN2 (t,h) . By Lemma 6.14 δ m(ω ∩ (Bh,h )c ) = m{t′ ∈ ω : dist(t′ , ∂ω) ≤ 0 ,M
M +1 |h|} 1−ǫ
M +1 |h| 1−ǫ 2(M + 1)|h| ≤ |ω| (1 − ǫ)|ω| 2C6 (M + 1) (63) ≤ log λ |ω|. δ(1 − ǫ)N (t, h)K 2 Choose K large enough such that K log λ > 2. Then √ ∞ ∞ X X 2C6 (M + 1) λ δ c (64) < ∞. m(Ahi ∩ (Bhi ,h0 ,M ) ) ≤ K log2 λ i=0 δ(K1 + iK2 ) i=0 ≤2
In particular m Γδh0 ∩
\
Bhδ i ,h0 ,M
i≥i(h)
= m(Γδh0 ) − m(Γδh0 ∩ ≥
m(Γδh0 )
−
≥ m(Γδh0 ) −
(65)
X
i≥i(h)
m(Γδh0
i≥i(h)
X
i≥i(h)
h→0
X
i≥i(h)
c Bhδ i ,h0 ,M )
∩ (Bhδ i ,h0 ,M )c )
m(Ahi ∩ (Bhδ i ,h0 ,M )c ).
Eq. (64) implies that lim
\
m(Ahi ∩ (Bhδ i ,h0 ,M )c ) = 0.
CLT for the modulus of continuity of averages of observables in transversal families
31
Proof of Proposition 4.5. By Lemma 6.12 for every γ > 0 there exists δ > 0 such that for every small h0 we have m(Γδh0 ) > 1 − γ. Choose M satisfying Eq. (33). Define ˆ δ \Q, Γδh,h0 = Γ h,h0 ˆδ where Γ h,h0 is the set defined in Lemma 6.15 and Q is the countable set of parameters where ft has a periodic critical point. By Lemma 6.15 Property A. holds. Let t′ ∈ Γδh,h0 , with |h| < h0 . There exists i ≥ i(h) such that hi+1 ≤ |h| ≤ hi ,
where hi = (1 − ǫ)i h0 . Thus, N (t′ , h) = N (t′ , hj ), for some j ∈ {i, i + 1}, and consequently N2 (t′ , h) = N2 (t′ , hj ). Then there exists a unique ω ∈ Aδhj ,h0 and t ∈ Γδh0 such that t, t′ ∈ ω ∈ PN2 (t,h) . Moreover, since t′ ∈ Bhδ j ,h0 ,M we have dist(t′ , ∂ω) ≥
M +1 hj ≥ (M + 1)|h|. 1−ǫ
Define N3 (t′ , h) = N2 (t, h). By Lemma 6.14 we have Eq. (15) holds. By Lemma 6.2, Eq. (16) holds. 6.1. Proof of Lemma 6.7. Let J be the interval as in the statement of Lemma 6.7. The sets EN,J ‘live’ in the parameter space. To estimate its measures we will compare them, following [18], with the measures of similarly defined sets in the phase space of the map ftR . ˆN,tR the set of ˆN,tR ). Let J = [tL , tR ]. Denote by E Definition 6.16 (The sets E all η ∈ PN (tR ) such that for all k satisfying
K log N 0≤k≤ τ
there is not where such that
η˜ ∈ PN −⌊K log N ⌋+j (tR ), η ⊂ η˜, j = min{(k + 1)τ, ⌊K log N ⌋}, N −⌊K log N ⌋+kτ
ftR
(˜ η ) ∈ Pj−kτ (tR ).
Using a strategy similar to the one applied in [18], we estimate the measure |EN,J | in terms of the measure |EˆN,tR |. To this end we need to define the map UJ . Recall that if F is a family of disjoint intervals then |F | denotes the sum of the measures of the intervals.
32
AMANDA DE LIMA AND DANIEL SMANIA
Definition 6.17 (The map UJ ). Let J = (tL , tR ). Consider the map UJ UJ : PN |J → PN (tR ) defined by Schnellmann [18, proof of Lemma 3.2] in the following way. Let ω ∈ PN |J and choose t ∈ ω. Since ω is a cylinder, it follows that xj (t) 6= c for all 0 ≤ j < N . Therefore, there is a cylinder ω(x0 (t)) in the partition PN (t) such that x0 (t) ∈ ω(x0 (t)). Let UJ (ω) = Ut,tR ,N (ω(x0 (t))), where Ut,tR ,N : PN (t) → PN (tR ) is such that for all η ∈ PN (t), the elements η and UJ (η) have the same combinatorics. symbt (fti (η)) = symbtR (ftiR (Ut,tR ,N (η)),
for 0 ≤ i < N . Schnellmann [18] proved that Ut,tR ,N is well defined when ft is a family of piecewise expanding unimodal maps satisfying our assumptions. In particular, if t < t′ and a certain symbolic dynamic appears in the dynamics of ft , then it also appears in the dynamics of ft′ . Therefore, the cylinder ω ′ = UJ (ω) = Ut,tR ,N (ω(x0 (t))) has the same combinatorics as ω, that is, symb(xj (ω)) = symbtR (ftjR (ω ′ )), when 0 ≤ j < N . Since there are not two cylinders in PN (tR ) with the same combinatorics, the element ω ′ does not depend on the choice of t ∈ ω. Therefore, UJ is well defined. ˆN,tR . Moreover, there exists C ′ ≥ 1 Lemma 6.18. If ω ∈ EN,J , then UJ (ω) ∈ E such that |ω| ≤ C ′ |UJ (ω)|.
(66) In particular
ˆN,tR |. |EN,J | ≤ C ′ |E
(67)
Proof. Note that UJ (ω) ∈ EˆN,tR follows from the fact that ω and UJ (ω) have the same combinatorics [18]. By [18, Lemma 3.2], there exists a constant C ′ ≥ 1 such that |ω| ≤ C ′ |UJ (ω)|. Thus, (68)
|EN,J | ≤
X
ω∈EN,J
|ω| ≤
X
ω∈EN,J
ˆN,tR |. C ′ |UJ (ω)| ≤ C ′ |E
Definition 6.19. For each η ′ ∈ PN −⌊K log N ⌋ (tR ), define the set ) ( ′ ˆ ˆ EN,tR ,η′ = η ∈ PN (tR ) : η ∈ EN,tR and η ⊂ η . Lemma 6.20. Let η ′ ∈ PN −⌊K log N ⌋ (tR ). Then (69)
ˆN,tR ,η′ ≤ 2⌊ #E
K log N τ
⌋+1 .
CLT for the modulus of continuity of averages of observables in transversal families
Proof. Define
33
K log N k0 = . τ
Notice that
N ≥ N − ⌊K log N ⌋ + k0 τ > N − τ.
If N = N − ⌊K log N ⌋ + k0 τ define k1 = k0 . Otherwise define k1 = k0 + 1. For every k satisfying 0 ≤ k ≤ k1 , define families of intervals Fk in the following way. If k ≤ k0 define (70) ˆN,tR ,η′ with η ⊂ ηˆ} Fk = {ˆ η ⊂ η ′ : ηˆ ∈ PN −⌊K log N ⌋+kτ (tR ) and there is η ∈ E otherwise k = k1 = k0 + 1 and ˆN,tR ,η′ . Fk1 = E
(71)
ˆN,tR ,η′ . We claim that Note that if k1 = k0 then we also have Fk1 = E #Fk ≤ 2k .
(72)
We observe that, taking k = k1 in Eq. (72) we obtain Eq. (69). Note that either F0 is the empty set or F0 = {η ′ }. Then #F0 ≤ 1. Moreover, it is easy to see that if ηˆk+1 ∈ Fk , with k < k1 , then there exists a unique ηˆk ∈ Fk such that ηˆk+1 ⊂ ηˆk . Therefore, it is enough to show that for each ηˆk ∈ Fk , with k < k1 , there are at most two intervals ηˆk+1 ∈ Fk+1 such that ηˆk+1 ⊂ ηˆk . Indeed, given k < k1 , for every ηˆk ∈ Fk we have ηˆk ∈ PN −⌊K log N ⌋+kτ (tR ). Moreover, there is j such that for every ηˆk+1 ∈ Fk+1 we have ηˆk+1 ∈ PN −⌊K log N ⌋+j (tR ), with kτ < j ≤ ⌊K log N ⌋, and j ≤ kτ + τ . Note that if the closure of ηˆk+1 = (a, b) is contained in the interior of ηˆk , then for every x ∈ ηˆk+1 we have ftpR (x) 6= c, for every p < N − ⌊K log N ⌋+ kτ . Furthermore, there are na , nb such that ftnRa (a) = c = ftnRb (b), where N − ⌊K log N ⌋ + kτ ≤ na , nb < N − ⌊K log N ⌋ + j.
We conclude that
N −⌊K log N ⌋+kτ
ftR
(ˆ ηk+1 ) ∈ Pj−kτ (tR ).
ˆN,tR ,η′ and where j − kτ ≤ τ . Therefore, if η ⊂ ηˆk+1 , with η ∈ PN (tR ), then η 6∈ E consequently ηˆk+1 6∈ Fk+1 . Since there are at most two intervals PN −⌊K log N ⌋+j (tR ) whose closure is not contained in the interior of ηˆk , we conclude that there are at most two intervals in Fk+1 that are contained in ηˆk . Lemma 6.21. Let η ′ , η ′′ ∈ PN −⌊K log N ⌋ (tR ) such that N −⌊K log N ⌋
ftR
N −⌊K log N ⌋
(η ′′ ).
N −⌊K log N ⌋
(EˆN,tR ,η′′ ).
(η ′ ) = ftR
Then N −⌊K log N ⌋
ftR
(EˆN,tR ,η′ ) = ftR
34
AMANDA DE LIMA AND DANIEL SMANIA
Proof. Let ω ′ = (y1′ , y2′ ) ∈ PN (tR ), with ω ′ ⊂ η ′ , be a cylinder in EˆN,tR ,η′ . Then N −⌊K log N ⌋
(73)
ftR
N −⌊K log N ⌋
(ω ′ ) ⊂ ftR
N −⌊K log N ⌋
(η ′ ) = ftR
(η ′′ ).
Remember that since ω ′ ∈ PN (tR ), it follows that for all x ∈ ω ′ ftiR (x) 6= c for all 0 ≤ i < N,
(74)
and if y ∈ ∂ω ′ , then there exists j, 0 ≤ j < N such that ftjR (y) = c. Define N −⌊K log N ⌋
ai = ftR
(yi′ ).
N −⌊K log N ⌋
(ω ′ ) = (a1 , a2 ) is an open interval and, by Eq. (73), we have Then ftR N −⌊K log N ⌋ ′′ (η ). Therefore, there is an open interval ω ′′ = (y1′′ , y2′′ ) ⊂ η ′′ (a1 , a2 ) ⊂ ftR N −⌊K log N ⌋ (ω ′′ ) = (a1 , a2 ) with such that ftR N −⌊K log N ⌋
ai = ftR
(yi′′ ).
We claim that ω ′′ is also a cylinder. Indeed, let x ∈ ω ′′ . Then, since ω ′′ ⊂ η ′′ and η ′′ is a cylinder of level N − ⌊K log N ⌋, it follows that ftiR (x) 6= c,
for all 1 ≤ i < N − ⌊K log N ⌋. On the other hand, N −⌊K log N ⌋
ftR
N −⌊K log N ⌋
(ω ′′ ) = ftR
(ω ′ ),
and by Eq. (74), we can conclude that ftiR (x) 6= c for all i satisfying N −⌊K log N ⌋ ≤ i < N . Therefore, for all x ∈ ω ′′ , we have ftiR (x) 6= c for all 0 ≤ i < N . Now, let yi′′ ∈ ∂ω2 . Since ω ′′ ⊂ η ′′ , we have two cases. Case 1: yi′′ ∈ ∂η ′′ . In this case, there is an integer j, 0 ≤ j < N − ⌊K log N ⌋, such that ftjR (yi′′ ) = c. Case 2: yi′′ ∈ / ∂η ′′ . In this case, ftjR (yi′′ ) 6= c for all 0 ≤ j < N − ⌊K log N ⌋. Then N −⌊K log N ⌋ ′′ N −⌊K log N ⌋ ′ N −⌊K log N ⌋ ′′ (η ) = (yi ) belongs to the interior of ftR (yi ) = ai = ftR ftR N −⌊K log N ⌋ ′ ′ ′ (η ). Thus, yi belongs to the interior of η , which implies that there ftR exists j such that N − ⌊K log N ⌋ ≤ j < N such that ftjR (yi′ ) = ftjR (yi′′ ) = c. Therefore ω ′′ ∈ PN (tR ). ˆN,tR ,η′ . Then for all 0 ≤ k ≤ ⌊ K log N ⌋, if By assumption, ω ′ ∈ E τ ′
ω ˜ k ∈ PN −⌊K log N ⌋+j(k) (tR ),
′
where ω ⊂ ω ˜ k ⊂ η and
j(k) = min{(k + 1)τ, ⌊K log N ⌋},
then there is zk′ ∈ ∂ ω ˜ satisfying (75)
q′
ftRk (zk′ ) = c, for some qk′ , 0 ≤ qk′ < N − ⌊K log N ⌋ + kτ.
In the same manner as for ω ′ , there exists a unique cylinder ω ˆ k ∈ PN −⌊K log N ⌋+j(k) , N −⌊K log N ⌋ N −⌊K log N ⌋ (ˆ ωk ). Note that ω ′′ ⊂ ω ˆ k . Let (˜ ωk ) = ftR ω ˆ k ⊂ η ′′ , such ftR ˆ k such that zk′′ ∈ ∂ ω N −⌊K log N ⌋
ftR
N −⌊K log N ⌋
(zk′ ) = ftR
(zk′′ ).
CLT for the modulus of continuity of averages of observables in transversal families
35
If zk′′ ∈ ∂η ′′ then there exists i < N − ⌊K log N ⌋ such that ftiR (zk′′ ) = c. Define qk′′ = i. If zk′′ 6∈ ∂η ′′ then zk′ 6∈ ∂η ′ . Thus, ftqR (zk′ ) 6= c for every q < N − ⌊K log N ⌋, which implies that q′
q′
N − ⌊K log N ⌋ ≤ qk′ < N − ⌊K log N ⌋ + kτ.
Then ftRk (zk′′ ) = ftRk (zk′′ ) = c. Define qk′′ = qk′ . ˆN,tR ,η′′ . In both cases we have 0 ≤ qk′′ < N − ⌊K log N ⌋ + kτ , then ω ′′ ∈ E N −⌊K log N ⌋
ftR
N −⌊K log N ⌋
(EˆN,tR ,η′′ ) ⊂ ftR
(EˆN,tR ,η′ ).
A similar argument shows that N −⌊K log N ⌋
ftR
N −⌊K log N ⌋
(EˆN,tR ,η′ ) ⊂ ftR
(EˆN,tR ,η′′ ).
Proof of Lemma 6.7. Due to Lemma 6.18 it is enough to show that for every K′ < K there exists C > 0 and K = K(K′ ) > 0 such that if J ∈ Pj , j = j(N ) then ′
|EˆN,tR | ≤ KN −CK .
(76) By Lemma 6.20 we have
#EˆN,tR ,η′ ≤ 2⌊
K log N τ
⌋+1
.
Let us define the set [
Ω=
N −⌊K log N ⌋
ftR
(EˆN,tR ,η′ ).
η ′ ∈PN −⌊K log N ⌋ (tR )
Note that ˆN,tR ⊂ f −(N −⌊K log N ⌋) (Ω). E tR Therefore, if µtR is the acip for ftR we have µtR (EˆN,tR ) ≤ µtR (Ω).
(77)
In [18, Section 6.2] it is shown that there is C1′ ≥ 1 such that for every density ρt of the unique acip of ft 1 ≤ ρt (x) ≤ C1′ , C1′ for µt -almost every x ∈ [0, 1], then 2
ˆN,tR | ≤ C1′ |Ω|. |E Since J ∈ Pj , j = j(N ), there exists an integer p, 0 ≤ p < j such that xp (tR ) = ftpR (ftR (c)) = c. In particular #{ftiR (c)}i≥0 = p + 1. Thus, N −⌊K log N ⌋
#{ftR
(η ′ ), η ′ ∈ PN −⌊K log N ⌋ (tR )} ≤ (p + 1)2 .
36
AMANDA DE LIMA AND DANIEL SMANIA
Therefore, by Lemma 6.21, 2
2
N −⌊K log N ⌋
|EˆN,tR | ≤ C1′ |Ω| = C1′ | ∪η′ ∈PN −⌊K log N ⌋ (tR ) ftR 2
≤ C1′ (p + 1)2
N −⌊K log N ⌋
max
η ′ ∈PN −⌊K log N ⌋ (tR )
|ftR
(Eˆη′ ))|
ˆη′ )| (E
⌊K log N ⌋ n o 1 # η ∈ PN (tR )|Eˆ ′ η λ N⌋ ⌊K log N ⌋ ⌊K log 2 N 1 1 ′2 2 ⌋+1 ′2 2 ⌊ K log τ ≤ C1 (p + 1) ≤ C1 (p + 1) 2 λ λ N⌋ ⌊K log N ⌋ ⌊K log 2 2 2 1 1 log(C3 N ) 2 2 ≤ C1′ j 2 ≤ C1′ λ log λ λ 2
≤ C1′ (p + 1)2
≤ KN −
log λ ′ 2 ǫ
where K = K(ǫ′ ).
7. Estimates for the Wild part
We start this section with a technical lemma. Lemma 7.1. Given a good transversal family ft there are constants L1 and L2 such that the following holds. Let ϕ : [0, 1] → R, |ϕ|L1 (m) > 0, be a function of bounded variation such that Z ϕdm = 0.
Then
(I − Lt )−1 (ϕ)
L1
|ϕ|BV + L2 |ϕ|L1 . ≤ L1 log |ϕ|L1
Proof. Let ˜ > 0 such that
Lβ˜˜ |ϕ|BV = |ϕ|L1 .
And let j0 the smallest integer such that j0 − 1 ≤ ˜ ≤ j0 . Hence, we have −1
(I − Lt )
(ϕ) =
j0 X i=0
Lit (ϕ) +
∞ X l=1
Llt (Ljt0 (ϕ)).
R ˜ l |ϕ|BV , when ϕ dm = 0 Observing Assumption (V )A1 , the fact that ≤ Lθ ˜ and θ uniform in t, as well as the elementary facts that |Lt |L1 = 1 with constants L and | · |L1 ≤ | · |BV , we see that |Llt ϕ|BV
˜ L |Lj0 ϕ|BV 1−θ t ˜ L C˜6 β j0 |ϕ|BV + C˜5 |ϕ|L1 + 1−θ + (j0 + c2 ) |ϕ|L1 .
|(I − Lt )−1 (ϕ)|L1 ≤ (j0 + 1)|ϕ|L1 + ≤ (j0 + 1)|ϕ|L1 |ϕ|BV ≤ c1 β j0 |ϕ|L1
By the choice of j0 , we have the desired estimate.
CLT for the modulus of continuity of averages of observables in transversal families
37
The following proposition will be quite important to study the Wild part of the decomposition. Denote supp(ψ) = {x ∈ [0, 1] : ψ(x) 6= 0}.
Proposition 7.2. There exist K, K1′ , K2′ > 0 such that the following holds. For all i, k ≥ 0, t ∈ [0, 1] and h 6= 0, let 1 ϕk,i,h = Lit+h Hft+h (ftk (c)) − Hft (ftk (c)) . h Then ϕk,i,h 1 ≤ K, (78) L
and
K ϕk,i,h ≤ . BV |h|
(79)
Furthermore, (I − Lt+h )−1 Πt+h (ϕk,i,h ) (80)
L1
≤ K1′ max{0, log |ϕk,i,h |BV } + K2′ .
Proof. Note that
i Lt+h Hf (f k (c)) − Hf (f k (c)) 1 t t+h L t t ≤ Hft+h (ftk (c)) − Hft (ftk (c)) L1 ≤ (sup |vt |)|h|,
(81)
t
and, by Assumption (V) in Definition 3.3 i Lt+h Hf (f k (c)) − Hf (f k (c)) t t+h BV t t ˆ ≤ 2C˜6 β i + C˜5 (sup |vt |)|h| ≤ C.
(82)
t
Thus, we have Eqs. (78) and (79). In particular |Πt+h (ϕk,i,h )|L1 (m) ≤ 2|ϕk,i,h |L1 (m) ≤ 2K,
and if h is small
|Πt+h (ϕk,i,h )|BV ≤ |ϕk,i,h |BV + |ϕk,i,h |BV sup |ρt |BV ≤ C|ϕk,i,h |BV , t∈[0,1]
where C ≥ 1. Now we can easily obtain Eq. (80) applying Lemma 7.1. Proposition 7.3. Let φ be a Lipchitz function. There exists K > 0 such that the following holds. Let t ∈ Γδh′ ,h0 and 0 < |h| ≤ h′ . Then 1 K (83) var Lit+h Hft+h (ftk (c)) − Hft (ftk (c)) ≤ , h |h||Dfti (ftk+1 (c))|
and
(84)
Z
φ(x)Lit+h
Hft+h (ftk (c)) − Hft (ftk (c)) h
(x) dx
= φ(fti+k+1 (c))vt (ftk (c)) + O(|Dfti (ftk+1 (c))||h|),
where 0 ≤ k ≤ N3 (t, h) and i < N3 (t, h) − k.
38
AMANDA DE LIMA AND DANIEL SMANIA
k+1 (c), ft+h (ftk (c)), ft (ftk (c)) belong to the same Proof. By Eq. (16), the points ft+h i interval of monotonicity of ft+h . Let
ζ : Dom(ζ) → Im(ζ) be an inverse branch associated to such interval of monotonicity, that is, ζ is a i diffeomorphism such that ft+h (ζ(y)) = y for every y ∈ Dom(ζ) and k+1 {ft+h (c), ft+h (ftk (c)), ft (ftk (c))} ⊂ Im(ζ).
Hence,
(85)
Lit+h Hft+h (ftk (c)) − Hft (ftk (c)) (x) 1 = 1 (x) H k (c)) (ζ(x)) − Hf (f k (c)) (ζ(x)) . Dom(ζ) f (f t i (ζ(x)) t+h t t Dft+h
There is a constant K ≥ 1 such that for all t ∈ [0, 1], h, and i, and every interval i of monotonicity Q of ft+h we have Df i (y ) 1 t+h 1 ≤ i (y ) ≤ K Dft+h K 2
for all y1 , y2 ∈ Q. Now we can estimate the variation of the function in Eq. (83) using familiar properties of the variation of functions (see Chapter 3 from Viana [21], for instance). var[0,1] Lit+h Hft+h (ftk (c)) − Hft (ftk (c))
1 = var[0,1] 1 (x) H k (ζ(x)) − H k (ζ(x)) Dom(ζ) ft+h (ft (c)) ft (ft (c)) i (ζ(x)) Dft+h ! 1 sup Hft+h (ftk (c)) − Hft (ftk (c)) = varDom(ζ) i Dft+h (ζ(x)) [0,1] ! 1 H k − H k 1 (x) sup + 2 sup Dom(ζ) ft+h (ft (c)) ft (ft (c)) i (ζ(x)) Dft+h [0,1] [0,1] ! 1 1Dom(ζ) (x) var[0,1] Hft+h (ftk (c)) − Hft (ftk (c)) + sup i Dft+h (ζ(x)) [0,1] ! 6K 1 ≤ 2varDom(ζ) + . i k+1 i Dft+h (ζ(x)) |Dft+h (ft+h (c))|
!
CLT for the modulus of continuity of averages of observables in transversal families
39
Now, note that since ζ is a diffeomorphism, it follows that ! ! 1 1 varDom(ζ) = varIm(ζ) i (ζ(x)) i (y) Dft+h Dft+h ! Z 1 dy = D i (y) Df Im(ζ) t+h Z j−1 i X 2 D ft+h (ft+h (y)) = − 2 dy i−j j j−1 Im(ζ) j=1 Dft+h (ft+h (y)) Dft+h (ft+h (y)) ≤ K1 |Im(ζ)| ≤ K1
≤
|Dom(ζ)| i (f k+1 (c))| |Dft+h t+h
CK2 . i (f k+1 (c))| |Dft+h t+h
Here we used that
≤
i j−1 X 2 D f (f (y)) t+h t+h − 2 i−j j j−1 j=1 Dft+h (ft+h (y)) Dft+h (ft+h (y)) i X C i−j λ j=1
≤ K1 ,
(86) and that
|Im(ζ)| ≤ K
1 |Dom(ζ)| ≤K . k+1 k+1 i i |Dft+h (ft+h (c))| |Dft+h (ft+h (c))|
Therefore, (87)
var[0,1] Lit+h Hft+h (ftk (c)) − Hft (ftk (c)) ≤
K3 . k+1 i |Dft+h (ft+h (c))|
k+1 Finally, by Eq. (16) note that the combinatorics up to i iterations of ft+h (c) by k+1 the map ft+h is the same as the combinatorics up to i iterations of ft (c) by the map ft . By Remark 6.1 we obtain
(88)
1 i (f k+1 (c))| |Dft+h t+h
≤ C1
1 |Dfti (ftk+1 (c))|
.
Eqs. (88) and (87) give us Eq. (83). Since 1 Hft+h (ftk (c)) − Hft (ftk (c)) = [ft+h (ftk (c)), ft (ftk (c))], supp h
by Eq. (85) we conclude that 1 i+1 k i Zi,k = supp Lit+h Hft+h (ftk (c)) − Hft (ftk (c)) = [ft+h (ft (c)), ft+h (ftk+1 (c))]. h
40
AMANDA DE LIMA AND DANIEL SMANIA
k+1 (c), ft+h (ftk (c)), ft (ftk (c)) belong to the same interval By Eq. (16), the points ft+h i of monotonicity of ft+h . Hence, 1 diam supp Lit+h Hft+h (ftk (c)) − Hft (ftk (c)) h i+1 k i = diam [ft+h (ft (c)), ft+h (ftk+1 (c))] i+1 k i = |ft+h (ft (c)) − ft+h (ftk+1 (c))|
i ≤ K|Dft+h (ftk+1 (c))||ft+h (ftk (c)) − ft (ftk (c))|
i ≤ K|Dft+h (ftk+1 (c))|| sup vt ||h| t
≤ C1 K|Dfti (ftk+1 (c))|| sup vt ||h|.
(89)
t
Therefore, Z
(90)
φ(x)Lit+h
Hft+h (ftk (c)) − Hft (ftk (c))
=
φ(fti+k+1 (c))
+
Z
Note that
φ(x) −
Z (91)
=
Lit+h Z
Z
Lit+h
(x) dx h Hft+h (ftk (c)) − Hft (ftk (c)) h
φ(fti+k+1 (c))
Lit+h
Hft+h (ftk (c)) − Hft (ftk (c))
Hft+h (ftk (c)) − Hft (ftk (c))
h Hft+h (ftk (c)) − Hft (ftk (c)) h
(x) dx
h
(x) dx.
(x) dx
(x)dx = vt (ftk (c)) + O(|h|).
Due to Eq (89) and the fact that φ is a lipschitzian function with Lipschitz constant L, and that fti+k+1 (c) ∈ Zi,k Z Hft+h (ftk (c)) − Hft (ftk (c)) φ(x) − φ(fti+k+1 (c)) Lit+h (x) dx h Z H ft+h (ftk (c)) − Hft (ftk (c)) (x) dx |φ(x) − φ(fti+k+1 (c))| Lit+h ≤ h Zi,k i Hft+h (ftk (c)) − Hft (ftk (c)) i k+1 1 ≤ LC1 K|Dft (ft (c))|| sup vt ||h| Lt+h L h t
(92)
≤ LC1 K|Dfti (ftk+1 (c))|| sup vt |2 |h|. t
Proof of Proposition 4.6. Let Φh be as in Proposition 4.3, that is ∞ 1X Φh = sk+1 (t)Πt+h Hft+h (ftk (c)) − Hft (ftk (c)) . h k=0
Γδh,h0 .
Given t ∈ Let N3 (t, h) be as in Proposition 4.5. Since t and h are fixed throughout this proof, we will write N3 instead of N3 (t, h) and N instead of N (t, h). Let us divide Φh as follows Φh = S 1 + S 2 .
CLT for the modulus of continuity of averages of observables in transversal families
41
Where N
S1 =
3 1X sk+1 (t)Πt+h Hft+h (ftk (c)) − Hft (ftk (c)) h
k=0
and S2 =
1 h
∞ X
k=N3 +1
sk+1 (t)Πt+h Hft+h (ftk (c)) − Hft (ftk (c)) .
Let us first estimate S2 . ∞ 1 X (I − Lt+h )−1 S2 = sk+1 (t)(I − Lt+h )−1 Πt+h Hft+h (ftk (c)) − Hft (ftk (c)) . h k=N3 +1
Thus,
(I − Lt+h )−1 S2 1 L ∞ X 1 −1 ≤ |sk+1 (t)| (I − Lt+h ) Πt+h Hft+h (ftk (c)) − Hft (ftk (c)) . h L1 k=N3 +1
By Proposition 7.2 and Lemma 7.1, taking 1 ϕ = Πt+h Hft+h (ftk (c)) − Hft (ftk (c)) , h we have, (I − Lt+h )−1 1 Πt+h Hf (f k (c)) − Hf (f k (c)) t t+h 1 t t h L 1 N +1 ≤ K1 log + K2 ≤ K1 log Λ + K2 |h| ≤ K1 (N + 1) log Λ + K2 ≤ K3 N + K4 .
Therefore,
(I − Lt+h )−1 S2
L1
∞ X
1 K5 N (K3 N + K4 ) ≤ N3 + K6 λk λ k=N3 +1 1+C5 K log λ 1 K5 N + K6 . ≤ N −C5 K log N + K6 ≤ K7 hK8 log λ log λ |h|
≤
It is left to analyze S1 . Applying the operator (I − Lt+h )−1 , ∞
−1
(I − Lt+h ) Then
N
3 1X i X Lt+h sk+1 (t)Πt+h Hft+h (ftk (c)) − Hft (ftk (c)) . (S1 ) = h i=0
k=0
(I − Lt+h )−1 (S1 ) =
N3 ∞ X 1X Lit+h Πt+h Hft+h (ftk (c)) − Hft (ftk (c)) sk+1 (t) h i=0 k=0
=S11 + S12 . Where S11 =
N3 X
k=0
sk+1 (t)
NX 3 −k i=0
1 i Lt+h Πt+h Hft+h (ftk (c)) − Hft (ftk (c)) h
42
AMANDA DE LIMA AND DANIEL SMANIA
and S12 =
N3 X
i=N3 −k+1
k=0
=
N3 X
k=0
∞ X
sk+1 (t)
1 i Lt+h Πt+h Hft+h (ftk (c)) − Hft (ftk (c)) h
1 3 −k Hft+h (ftk (c)) − Hft (ftk (c)) . sk+1 (t) Lt+h ◦ (I − Lt+h )−1 ◦ Πt+h ◦ LN t+h h
We observe that |S12 |L1 ≤ C
N3 X
k=0
3 −k |sk+1 (t)||(I − Lt+h )−1 ◦ Πt+h ◦ LN t+h
1 Hft+h (ftk (c)) − Hft (ftk (c)) |L1 . h
Let ϕk =
1 N3 −k Lt+h Hft+h (ftk (c)) − Hft (ftk (c)) , h
By Proposition 7.3 it follows that (93) (94)
|ϕk |BV = var(ϕk ) + |ϕk |L1 C ≤ + K1 N3 (t,h)−k k+1 |h||Dft (ft (c))| N (t,h)+1
≤
|Dft
(ft (c))| + K1 N3 (t,h)−k k+1 |Dft (ft (c))| N (t,h)+1−N3 (t,h) N3 (t,h)+1 C|Dft (ft (c))||Dftk (ft (c)))|
≤C
+ K1
≤ CΛN (t,h)+1−N3 (t,h)+k + K1 .
(95)
By Lemma 7.1 we have |(I − Lt+h )−1 ◦ Πt+h (ϕk )|L1
1 Hft+h (ftk (c)) − Hft (ftk (c)) |L1 h ≤ K1′ log(CΛN (t,h)+1−N3 (t,h)+k + K1 ) + K2′ 3 −k = |(I − Lt+h )−1 ◦ Πt+h ◦ LN t+h
≤ K1′ log(K2 ΛN (t,h)+1−N3 (t,h)+k ) + K2′ ≤ K3 (N − N3 + k + 1).
Therefore, |S12 |L1 ≤ K3
N3 X
k=0
|sk+1 (t)|(N − N3 + k + 1)
N3 N3 N3 X X X 1 k 1 ≤ K3 (N − N3 ) + K + 3 λk λk λk k=0 k=0 k=0 1 +1 . ≤ K4 K log N + K5 ≤ K log log |h|
!
CLT for the modulus of continuity of averages of observables in transversal families
43
We proceed to examine S11 . S11 =
N3 X
sk+1 (t)
i=0
k=0
−
Note that
|
|
N3 X
NX 3 −k
Lit+h
Hft+h (ftk (c)) − Hft (ftk (c)) h
{z
}
S111
sk+1 (t)
NX 3 −k i=0
k=0
Z Hft+h (ftk (c)) − Hft (ftk (c)) Lit+h ρt+h dm . h {z } S112
S112 = − =−
N3 X
sk+1 (t)
ρt+h
i=0
k=0 N3 X
NX 3 −k
sk+1 (t)
NX 3 −k i=0
k=0
Z
Hft+h (ftk (c)) − Hft (ftk (c)) h
dm
vt (ftk (c)) + O(h) ρt+h .
Adding and subtracting the sum N3 X
sk+1 (t)
NX 3 −k
vt (ftk (c))ρt ,
i=0
k=0
we obtain S112 = S1121 + S1122 , where S1121 = −
N3 X
sk+1 (t)
NX 3 −k
vt (ftk (c))ρt
i=0
k=0
and S1122 = −(ρt+h − ρt )
N3 X
k=0
sk+1 (t)(N3 − k)vt (ftk (c)) − O(h)
N3 X
k=0
sk+1 (t)(N3 − k)ρt+h .
By Eq. (5) N
|S1122 |L1 ≤ K1 sup |vt ||h| log t
+ |ρt+h |L1 |O(h)|
3 1 X |sk+1 (t)|(N3 − k) |h|
N3 X
k=0
k=0
|sk+1 (t)|(N3 − k)
N3 X 1 1 ≤ K2 |h| log + K3 |O(h)| N3 |h| λk k=0 1 1 1 |h| log + |O(h)| ≤ K log + |O(h)| . ≤ K4 N |h| log |h| |h| |h|
44
AMANDA DE LIMA AND DANIEL SMANIA
Therefore, taking φ : [0, 1] → R a lipschitzian observable, Z
Z φ(x)W(x) dx = φ(x)(I − Lt+h )−1 Φh (x) dx Z 1 = φ(x)(S111 + S1121 )(x) dx + O log log |h| Z N N −k 3 3 X X Hft+h (ftk (c)) − Hft (ftk (c)) i (x) dx = sk+1 (t) φ(x)Lt+h h i=0 k=0 Z N3 NX 3 −k X 1 k . vt (ft (c)) φ(x)ρt (x) dx + O log log − sk+1 (t) |h| i=0 k=0
By Eq. (84) we have Z
φ(x)Lit+h
Hft+h (ftk (c)) − Hft (ftk (c)) h
(x) dx
= φ(fti+k+1 (c))vt (ftk (c)) + O(|Dfti (ftk+1 (c))||h|)
= φ(fti+k+1 (c))vt (ftk (c)) + O(
|Dfti (ftk+1 (c))| ). |DftN (ft (c))|
Since N3 NX 3 −k X |Dfti (ftk+1 (c))| ) O( sk+1 (t) |DftN (ft (c))| i=0 k=0 N3 k X N N −i X 1 1 ≤ K1 < K, λ λ i=0
(96)
k=0
it follows that Z φ(x)W(x)dx N3 X
NX 3 −k
1 − φ dµt + O log log = |h| i=0 k=0 Z N3 NX 3 +1 X 1 = sk+1 (t)vt (ftk (c)) φ(ftj (c)) − φ dµt + O log log |h| sk+1 (t)vt (ftk (c))
k=0
φ(fti+k+1 (c))
Z
j=k+1
X Z j−1 NX 3 +1 1 j k φ(ft (c)) − φ dµt sk+1 (t)vt (ft (c)) + O log log = . |h| j=1 k=0
Adding and subtracting the series NX 3 +1 j=1
φ(ftj (c)) −
Z
φ dµt
X ∞ k=j
sk+1 (t)vt (ftk (c)),
CLT for the modulus of continuity of averages of observables in transversal families
we obtain X Z Z NX ∞ 3 +1 φ(ftj (c)) − φ dµt φ(x)W(x)dx = j=1
− |
k=0
NX 3 +1
φ(ftj (c)) −
j=1
Z
φ dµt
X ∞ k=j
{z I1
1 + O log log . |h|
Note that |I1 | < ∞. Indeed, |I1 | ≤ K1 ≤ K2
NX 3 +1 j=1
φ(ftj (c)) −
NX 3 +1 j=1
1 λ
j
Z
≤ K.
45
s1 (t) vt (ftk (c)) Dftk (ft (c)) s1 (t) vt (ftk (c)) Dftk (ft (c)) }
X ∞ 1 k φ dµt λ k=j
Therefore, Z Z NX 3 +1 1 φ(ftj (c)) − φ dµt + O log log φ(x)W(x)dx = s1 (t)J(ft , vt ) |h| j=1 (97)
= s1 (t)J(ft , vt )
Z N3 X 1 φ(ftj (c)) − φ dµt + O log log . |h| j=0
8. Estimates for the Tame part Let ν be a signed, finite and borelian measure on [0, 1]. Denote by |ν| the variation measure of ν and by ||ν|| the total variation of ν. Define the push-forward of ν by ft as the borelian measure (ft⋆ ν)(A) = ν(ft−1 (A)). Note that for every bounded borelian function g : [0, 1] → R Z Z g d(ft⋆ ν) = g ◦ ft dν.
It is also easy to see that
Suppose that ν has the form (98)
|ft⋆ ν| ≤ ft⋆ |ν|. ν = πm +
X
qx δx ,
ˆ x∈∆
ˆ ⊂ [0, 1] is where π ∈ L∞ (m) with support on [0, 1], m is the Lebesgue measure, ∆ a countable subset, qx ∈ R, with X |qx | < ∞, ˆ x∈∆
46
AMANDA DE LIMA AND DANIEL SMANIA
and δx is the Dirac measure supported on {x}. Then X |ν| = |π|m + |qx |δx , ˆ x∈∆
||ν|| = |π|L1 (m) + Furthermore, ft⋆ ν has the form ft⋆ ν = Lt (π)m +
X
ˆ x∈∆
X
|qx |.
qx δft (x) .
ˆ x∈∆
Proposition 8.1. Let ft be a C 1 family of C 1 piecewise expanding unimodal maps. Let ν be a signed, finite and borelian measure. Let ψt : [0, 1] 7→ R, t ∈ [0, 1] be such that ψt ∈ L∞ (ν) and t → ψt is a lipschitzian function with respect to the L∞ (|ν|) norm, that is, there exists L such that for all t, h we have |ψt+h − ψt |L∞ (ν) ≤ L|h|. Define ∆t,h (x) =
Z
x
0
⋆ dft+h (ψt+h ν) −
Z
x
0
dft⋆ (ψt ν).
Then there exist positive constants K1 , K2 such that |∆t,h |L1 (m) ≤ (L + K1 K2 )||ν|||h| for all t ∈ [0, 1], h, where K1 = sup |ψt |L∞ (ν) and K2 = sup |∂t ft (x)|. t
t,x
Proof. Observe that ∆t,h (x) =
Z
x
0
=
+
Therefore,
Z
x
Z
x
|
|
0
0
⋆ dft+h (ψt+h ν) − ⋆ dft+h (ψt+h ν) − {z
Z
x
0
Z
x
0
∆1
⋆ dft+h (ψt ν)
− {z
∆2
Z
0
x
dft⋆ (ψt ν) ⋆ dft+h (ψt ν) }
dft⋆ (ψt ν) . }
|∆t,h (x)| ≤ |∆1 (x)| + |∆2 (x)|.
We first estimate ∆1 . Z Z ⋆ ⋆ (ψt+h ν − ψt ν)| ≤ 1[0,x] d(ft+h (|ψt+h − ψt ||ν|)) |∆1 (x)| ≤ 1[0,x] d|ft+h Z ≤ 1[0,x] ◦ ft+h |ψt+h − ψt | d|ν| ≤ |ψt+h − ψt |L∞ (ν) ||ν|| ≤ L||ν|||h|. In particular
|∆1 |L1 (m) ≤ L||ν|||h|.
CLT for the modulus of continuity of averages of observables in transversal families
47
We now estimate ∆2 . ∆2 (x) = = = Therefore, |∆2 (x)| ≤ where
Z
Z
⋆ 1[0,x] dft+h (ψt ν) −
Z
(1f −1
Z
Z
1[0,x] ◦ ft+h d(ψt ν) − t+h ([0,x])
1[0,x] dft⋆ (ψt ν) Z
1[0,x] ◦ ftd(ψt ν)
− 1f −1 ([0,x]) ) d(ψt ν). t
|1f −1 ([0,x]) − 1f −1 ([0,x]) ||ψt | d|ν| ≤ K1 t
t+h
Z
|1f −1 ([0,x]) − 1f −1 ([0,x]) | d|ν| t+h
t
K1 = sup |ψt |L∞ (ν) . t
By the Fubini’s Theorem |∆2 |L1 (m) (99) Note that
≤
K1
≤
K1
Z Z
Z Z
|1f −1 ([0,x]) (y) − 1f −1 ([0,x]) (y)| d|ν|(y) dm(x) t
t+h
|1f −1 ([0,x]) (y) − 1f −1 ([0,x]) (y)| dm(x) d|ν|(y) t
t+h
|1f −1 ([0,x]) (y) − 1f −1 ([0,x]) (y)| = 1Uy (x), t
t+h
where
Uy = {x ∈ [0, 1] : ft+h (y) < x ≤ ft (y) or ft (y) < x ≤ ft+h (y)}.
Observe that Thus,
m(Uy ) = |ft+h (y) − ft (y)| ≤ K2 |h|. |∆2 |L1 (m)
(100)
Z Z
1Uy (x) dm(x) d|ν|(y)
≤
K1
≤
K1 K2 ||ν|||h|.
Remark 8.2. To avoid a cumbersome notation, in the Proof of Proposition 4.3 we will use the following notation. Whenever we take the supremum over all t ∈ [0, 1] we actually take the supremum over all t ∈ [0, 1] such that ft do not have a periodic critical point. And whenever we take the supremum over all h 6= 0 we indeed mean taking the supremum over all h 6= 0 such that 0 < |h| < δ, where δ > 0 is given by Definition 3.3. Proof of Proposition 4.3. We first examine 1 (Lt+h ρt − Lt ρt ). h As we have seen, the density ρt can be decomposed as ρt = (ρt )abs + (ρt )sal . We also have Lt+h ρt ∈ BV and
Lt+h ρt = (Lt+h ρt )abs + (Lt+h ρt )sal .
48
AMANDA DE LIMA AND DANIEL SMANIA
Therefore, (Lt+h ρt − Lt ρt ) = ((Lt+h ρt )abs − (Lt ρt )abs ) + ((Lt+h ρt )sal − (Lt ρt )sal ) .
Let us examine the absolutely continuous term 1 ((Lt+h ρt )abs − (Lt ρt )abs ). h Observe that for every t
(Lt ρt )(x) = (Lt ρt )abs (x) + (Lt ρt )sal (x). Differentiating with respect to x, ((Lt ρt )abs )′ (x) = (Lt ρt )′ (x) = ((Lt ρt )′ )abs (x) + ((Lt ρt )′ )sal (x). Then Z
(Lt ρt )abs (x) =
x
(Lt ρt )′ dm.
0
Similarly (Lt ρt+h )abs (x) =
Z
x
Z
x
(Lt+h ρt )′ (y) dm.
0
Therefore, (Lt+h ρt )abs (x) − (Lt ρt )abs (x) = = +
0
Z
x
Z 0x 0
We define (101)
At,h (x) =
Z
x
Z
x
0
and (102)
Bt,h (x) =
0
Our goal is to prove that At,h sup sup h t∈[0,1] h6=0
BV
(Lt+h ρt )′ − (Lt ρt )′ dm ((Lt+h ρt )′ )abs − ((Lt ρt )′ )abs dm
((Lt+h ρt )′ )sal − ((Lt ρt )′ )sal dm.
((Lt+h ρt )′ )abs − ((Lt ρt )′ )abs dm, ((Lt+h ρt )′ )sal − ((Lt ρt )′ )sal dm.
0 and K1 > 0 such that for every t ∈ Ω ˆ we have there exist Ω Rφ (t + h) − Rφ (t) p 1Ω (t + h) ≤ K1 . h − log |h| Since ft is a good transversal family, by Lemma 3.7 we have that inf t |Ψ(t)| > 0 and that Ψ(t) does not changes signs on t ∈ [a, b], so without loss of generality we can assume Ψ(t) > 0 for every t ∈ [a, b] and inf t Ψ(t) > 0 (otherwise replace the family ft by f−t ). So there exists K2 > 0 such that lim sup h→0+
lim sup h→0+
Rφ (t + h) − Rφ (t) p 1Ω (t + h) ≤ K2 Ψ(t)h − log |h|
ˆ Then there exists h0 > 0 and a set S ⊂ Ω ˆ with m(S) > 0 such for every t ∈ Ω. that for every t ∈ S we have Rφ (t + h) − Rφ (t) p 1Ω (t + h) ≤ K2 + 1 Ψ(t)h − log |h|
for every h satisfying 0 < h ≤ h0 . Let t0 ∈ (a, b) be a Lebesgue density point of S. Choose δ > 0 such that DN (K2 + 1) + δ < 1.
Then for every ǫ > 0 small enough,
m(S ∩ Iǫ ) > DN (K2 + 1) + δ, m(Iǫ ) where Iǫ = [t0 − ǫ, t0 + ǫ]. Let Sǫ = S ∩ Iǫ . It is a well-known fact that if Sǫ − h = {t − h : t ∈ Sǫ } then lim m(Sǫ ∩ (Sǫ − h)) = m(Sǫ ) > 0.
h→0
Note that for every t ∈ Sǫ ∩ (Sǫ − h), we have t, t + h ∈ Sǫ ⊂ S ⊂ Ω, then Rφ (t + h) − Rφ (t) p ≤ K2 + 1 Ψ(t)h − log |h|
for every 0 < h ≤ h0 . In particular
CLT for the modulus of continuity of averages of observables in transversal families
1 Rφ (t + h) − Rφ (t) 1 p m t ∈ Iǫ : ≤ K2 + 1 lim sup h Ψ(t)h − log |h| h→0+ m(Iǫ )
(111)
≥
55
!
m(Sǫ ) ≥ DN (K2 + 1) + δ. m(Iǫ )
On the other hand the restriction of ft to the interval Iǫ is a transversal family, then by Theorem 1.1 we obtain ! 1 1 Rφ (t + h) − Rφ (t) p m t ∈ Iǫ : ≤ K2 + 1 lim h→0+ m(Iǫ ) h Ψ(t)h − log |h| = DN (K2 + 1),
which contradicts Eq.(111).
Proof of Corollary 1.2. It follows from Corollary 9.1.
Remark 9.2. In Baladi and Smania [2][5] it is proven that for almost every t ∈ [a, b] there exists a sequence hn → 0 such that
Rφ (t + hn ) − Rφ (t) hn is not bounded. In particular Rφ is not a lipschitzian function on the whole interval [a, b]. Naturally Corollaries 9.1 and 1.2 do not follow from this when Ω is not an interval. Remark 9.3. Two weeks before this work be completed, Fabián Contreras sent us his Ph. D. Thesis [7] where he proves a result sharper than Corollary 9.1 when Ω = [a, b] and φ is a C 1 generic observable. He proves that for almost every t ∈ [a, b] the limit Rφ (t + h) − Rφ (t) (112) lim p h→0+ h | log h log log | log h||
exists and it is non zero. Note again that Corollaries 9.1 and 1.2 do not seem to follow from his result when Ω is not an interval. As in our case, the main difficult is to reduce the problem to Schnellmann’s main result in [19]. We are not completely familiar with his methods, but they seems to be quite different from our approach. Acknowledgments We would like to thank the anonymous referees for their valuable comments and suggestions. References [1] V. Baladi. On the susceptibility function of piecewise expanding interval maps. Comm. Math. Phys., 275(3):839–859, 2007. [2] V. Baladi and D. Smania. Linear response formula for piecewise expanding unimodal maps. Nonlinearity, 21(4):677–711, 2008. [3] V. Baladi and D. Smania. Smooth deformations of piecewise expanding unimodal maps. Discrete Contin. Dyn. Syst., 23(3):685–703, 2009. [4] V. Baladi and D. Smania. Alternative proofs of linear response for piecewise expanding unimodal maps. Ergodic Theory Dynam. Systems, 30(1):1–20, 2010.
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AMANDA DE LIMA AND DANIEL SMANIA
[5] V. Baladi and D. Smania. Corrigendum: Linear response formula for piecewise expanding unimodal maps. Nonlinearity, 25(7):2203–2205, 2012. [6] P. Billingsley. Convergence of probability measures. Wiley Series in Probability and Statistics: Probability and Statistics. John Wiley & Sons, Inc., New York, second edition, 1999. A WileyInterscience Publication. [7] F. Contreras. Regularity of absolutely continuous invariant measures for piecewise expanding unimodal maps. Ph.D’s Thesis. University of Maryland, 2015. [8] I. Karatzas and S. E. Shreve. Brownian motion and stochastic calculus, volume 113 of Graduate Texts in Mathematics. Springer-Verlag, New York, second edition, 1991. [9] G. Keller. Stochastic stability in some chaotic dynamical systems. Monatsh. Math., 94(4):313– 333, 1982. [10] G. Keller. Generalized bounded variation and applications to piecewise monotonic transformations. Z. Wahrsch. Verw. Gebiete, 69(3):461–478, 1985. [11] G. Keller, P. J. Howard, and R. Klages. Continuity properties of transport coefficients in simple maps. Nonlinearity, 21(8):1719–1743, 2008. [12] A. Lasota and J. A. Yorke. On the existence of invariant measures for piecewise monotonic transformations. Trans. Amer. Math. Soc., 186:481–488 (1974), 1973. [13] R. Leplaideur and B. Saussol. Central limit theorem for dimension of Gibbs measures in hyperbolic dynamics. Stoch. Dyn., 12(2):1150019, 24, 2012. [14] M. Mazzolena. Dinamiche espansive unidimensionali: dipendenza della misura invariante da un parametro. Master’s Thesis Roma 2, Tor Vergata, 2007. [15] W. Philipp and W. Stout. Almost sure invariance principles for partial sums of weakly dependent random variables. Mem. Amer. Math. Soc. 2, (issue 2, 161):iv+140, 1975. [16] D. Ruelle. Differentiation of SRB states. Comm. Math. Phys., 187(1):227–241, 1997. [17] D. Ruelle. Nonequilibrium statistical mechanics near equilibrium: computing higher-order terms. Nonlinearity, 11(1):5–18, 1998. [18] D. Schnellmann. Typical points for one-parameter families of piecewise expanding maps of the interval. Discrete Contin. Dyn. Syst., 31(3):877–911, 2011. [19] D. Schnellmann. Law of iterated logarithm and invariance principle for one-parameter families of interval maps. Probability Theory and Related Fields, pages 1–45, 2014. [20] M. Tsujii. A simple proof for monotonicity of entropy in the quadratic family. Ergodic Theory Dynam. Systems, 20(3):925–933, 2000. [21] M. Viana. Sthochastic dynamics of deterministic systems. Lecture notes XXI Colóquio Brasileiro de Matemática, IMPA, Rio de Janeiro, 1997. Departamento de Matemática, ICMC-USP, Caixa Postal 668, São Carlos-SP, CEP 13560-970 São Carlos-SP, Brazil E-mail address:
[email protected],
[email protected] URL: www.icmc.usp.br/∼smania/
