On the asymptotics of the scenery flow

Various notions of"zooming in"on measures exist in the literature and the scenery flow is one of them. It is of interest to describe the joint asymptotics of the scenery flows generated by a measure and the measure transported by a local diffeomorphism. We give both sufficient and necessary conditions for the scenery distributions to be asymptotic and provide some examples.


Introduction
Various notions of "zooming in" on measures exist in the literature and the scenery flow is one of them. A variant similar to the scenery flow was introduced by Bandt in [1] and studied by Graf in [4]. More recently, the scenery flow was defined by Gavish in [3] and studied by Hochman in [5,6]. One crucial step is to describe the joint asymptotics of the scenery flows generated by a measure and the measure transported by a local diffeomorphism, see Lemma 2.3 from [5] below. Unfortunately the lemma was stated without proof, and the assumptions are not sufficient to guarantee the conclusion. In this paper we give sufficient conditions for similar conclusions, and provide complete proofs. These statements show that the main results of [5] are not affected. We think however it is important on its own to have some more detailed insight to this crucial point in the properties of the scenery flow.
These results arose from discussions at the Lund dynamics seminar, and email correspondence with Michael Hochman. Hochman pointed out that he possesses a corrected statement similar to Proposition 2 (see Proposition 1.9 in [6]).
Let I = [−1, 1] and for any subset A of I and any t ≥ 0 let Let M(R) be the set of Radon measures on R and let M be the topological space of Borel probability measures on I, considered as a subspace of the dual of C(I) with the weak- * topology. (With this topology M is compact.) For any µ ∈ M(R), any x ∈ supp µ and any t ≥ 0, let µ x,t be the measure in M defined by Thus µ x,t tells what µ looks like in a window of radius e −t around x. The oneparameter family {µ x,t } t≥0 is called the scenery of µ at x. A Borel probability measure on M is called a distribution. The set of distributions is given the weak- * topology that comes from considering it as a subspace of the dual of C(M ). For any measure µ ∈ M(R), any x ∈ supp µ and any T ≥ 0, let µ x,T be the distribution defined by where λ is Lebesgue measure (one can show that t → µ x,t is left continuous, hence measurable, so that µ x,T is well defined. See for example Lemma 11 below). If P is a distribution and lim T →∞ µ x,T = P then µ is said to generate P at x. Thus µ generates P at x if and only if lim T →∞ 1 T T 0 g(µ x,t ) dt = g dP for every g ∈ C(M ).
If µ is a measure on X and f : X → Y is a measurable function, then f µ denotes the measure on Y given by for measurable subsets A of Y . The set of orientation preserving C k -diffeomorphisms R → R will be denoted by diff k + (R). The following appears as Lemma 2.3 in [5].
In particular µ generates P at x if and only if f µ generates P at f (x).
The idea is that locally around x, the diffeomorphism f acts as a translation that takes x to f (x) composed with a scaling around f (x) by the factor f ′ (x) = e s . However, no detailed proof is given in [5], and in fact it is possible to construct counterexamples to both of the statements in Lemma 2.3 (see Section 4). What can go wrong is that if µ has a lot of measure near the boundary of x + I t , then even the small distortion caused by f can push a relatively large amount of measure across the boundary, thus creating a big difference in normalisation between the measures µ x,t and (f µ) f (x),t−s . Under a certain (mild) condition, the set of t ≤ T for which this happens has small relative density in [0, T ] if T is large. Then the distributions µ x,T and f µ f (x),T will be asymptotic, even if µ x,t and (f µ) f (x),t−s are not.
2. An Auxiliary Result and its Consequences Lemma 1. Let µ ∈ M(R) and x ∈ supp µ, and suppose that Then µ x,T and f µ f (x),T are asymptotic for any f ∈ diff 1 + (R). Before we proceed to the proof of Lemma 1, we will state some consequences. The proofs of Propositions 2, 3 and 4 are given in Section 5.
The upper local dimension of a measure µ ∈ M(R) at x is defined to be It can be shown that if a Borel set E has positive µ-measure and D µ (x) ≥ c for all x ∈ E then the packing dimension of E is ≥ c (see for example Proposition 2.3 in [2]). This implies that D µ (x) ≤ 1 for µ-a.e. x, so the next proposition says in particular that µ-a.e. x is such that µ x,T and f µ f (x),T are asymptotic for all f ∈ diff 1 + (R).
If µ x,T has an ω-limit point P such that 0 ∈ supp ν for all ν ∈ supp P then the subset of M where S t : ν → ν 0,t is defined for all t ≥ 0 has full P -measure, and (M , S , P ) is a measure preserving flow 1 . A consequence of the next proposition is that in this situation, µ x,T and f µ f (x),T are asymptotic for all f ∈ diff 1 + (R).
Proposition 4. Let µ ∈ M(R) and x ∈ supp µ, and suppose that µ x,T has an ω-limit point P such that Then there exists a diffeomorphism f ∈ diff 1 + (R) such that µ x,T and f µ f (x),T are not asymptotic.

The Proof of Lemma 1
The proof of Lemma 1 goes via a few other lemmas.
using at * that λ is translation invariant. 1 Proof sketch: If µ x,T k → P , it can be shown that P ({ν; ν(c) > 0}) = 0 for any c ∈ (0, 1) \ {0}, and thus for fixed t > 0 the map S t is continuous P -a.e. (since S t is discontinuous at ν if and only if ν({±e −t }) > 0). Hence for any g ∈ C(M ), the composition g • S t is continuous P -a.e., so (see for example Theorem 3.1.5 in [7]). But this implies that S t µ x,T k → S t P , and it follows that P is S -invariant since S t µ x,T and µ x,T are asymptotic.
Lemma 6. Suppose that µ ∈ M(R) has 0 ∈ supp µ and satisfies the condition (1) at x = 0. Let A ⊂ I be an interval containing 0 and define for δ > 0 Then for any α > 0 Proof. Fix arbitrary γ > 1 and K > 1 and let Choose L = L(γ) > 0 and δ 0 = δ 0 (γ) > 0 so small that the inequalities are satisfied for all δ ∈ (0, δ 0 ). Take any δ ∈ (0, δ 0 ). By Chebyshev's inequality and the integral can be estimated by (the first inequality holds because e 2δ ≤ γ so that A t−2δ ⊂ e 2δ I t ⊂ γI t ). Since 0 ∈ A, the numerator in the last integral can be written as where a and b are the endpoints of A and J c (t) = [ce −(t±2δ) ] for c ∈ {a, b}. For each c, Suppose that c > 0 so that J c (t) = [ce −(t+2δ) , ce −(t−2δ) ] (if c < 0 then the endpoints come in the opposite order, but the computations are similar). Then for each k, where the last inequality is by Lemma 5. By the choice of L and δ 0 , the interval that appears in the last member is included in γI (k+1)L , so and since this holds for any γ > 1 and K > 1 the lemma follows from (1) when γ → 1 + and K → ∞.
Lemma 7. Suppose that µ ∈ M(R) satisfies the condition (1) at some x ∈ supp µ. Then for any ϕ ∈ C(I) and any α > 0, Proof. After a translation of the real line it may be assumed that it is enough to prove the lemma with f 0 in place of f . It will first be shown that (2) and thus Let ε > 0 and define h δ as in Lemma 6. Then there is a δ 1 > 0 such that Since lim t→∞ R(t) = 0, there is then a T 1 such that the right member of (3) is less than h δ1 (t) for t > T 1 . Thus This holds for any ε > 0, so (2) is proven for ϕ = χ A . It is not too hard to see that the set of ϕ that satisfy (2) for all α > 0 is a linear space. If A ⊂ I is an interval that does not contain 0, then A is the difference of two intervals that do, so (2) holds for ϕ = χ A for any interval A ⊂ I. Thus (2) holds whenever ϕ is a step function, and the lemma follows since any ϕ ∈ C(I) can be uniformly approximated by step functions.
The uniform structure of M will be used in the proof of Lemma 1 -therefore some facts about uniform spaces are recalled first. A uniform space is a set X with a notion of uniform closeness. This is represented by entourages, which are subsets of X × X satisfying some appropriate axioms. The idea is that an entourage V should contain all pairs of points that are "V -close" to each other. A fundamental system of entourages is a set Φ of entourages such that any entourage includes an entourage from Φ. As an example, if X has a metric d then a fundamental system of entourages for the metric uniformity on X is given by the sets such an entourage represents all pairs of points that are α-close. The uniform structure induces a topology on X, namely the one where the neighbourhood filter at x consists of the sets {y; (x, y) ∈ V } as V runs through all entourages. If X and Y are uniform spaces, then a function f : It can be shown that any uniformly continuous function is continuous, and that any continuous function from a compact uniform space to a uniform space is uniformly continuous.
The weak- * topology on M is the initial topology with respect to the maps This topology is also induced by the uniformity determined by the fundamental system of entourages consisting of all finite intersections of sets of the form where α > 0 and ϕ ∈ C(I).
Proof of Lemma 1. Take any g ∈ C(M ) -then g is bounded and uniformly continuous since M is compact. To prove the lemma, it is enough to show that for some α > 0 and some ϕ 1 , . . . , ϕ n ∈ C(I), since the sets of this form constitute a fundamental system of entourages of the uniformity on M . Let When T → ∞, the last expression goes to ε by Lemma 7, and since ε is arbitrary the lemma follows.

Motivating Examples
This section presents two examples of what can go wrong in Lemma 2.3 to motivate the altered formulation given in Lemma 1, where the conclusion is changed from ". . . then µ x,t and (f µ) f (x),t−s are asymptotic" to ". . . then µ x,T and f µ f (x),T are asymptotic" and an extra condition is added to the hypothesis. and Also for all n and all c in some small neighbourhood of 3/4, and thus ] for some small ε, one sees that µ 0,t and (f µ) 0,t are not asymptotic, since for t n = n log 2, The example still works if the weight of the k:th point mass is changed to w −k for any w > 1. The local dimension of µ at 0 then becomes log w/ log 2, which can take any value in (0, ∞). It would also be possible to replace the point masses by measures concentrated on the intervals (f −1 (2 −k ), 2 −k ). Thus one could give µ any prescribed ("global") dimension, for example by putting a suitable Cantor measure on each interval. In particular, it is possible to make µ non-atomic -in this case the map t → µ 0,t is continuous but changes rapidly near each t n .
In Example 1, the measures µ x,t and (f µ) f (x),t−s are far apart only for very short time periods, and it is still true that µ x,T and f µ f (x),T are asymptotic. The next example shows that even this can fail if µ x,t has a lot of mass near {±1} for all t. It is a special case of Proposition 4.
Define f ∈ diff 1 + (R) by letting then f (0) = 0 and f ′ (0) = 1. For c ∈ (0, 1), where the second inequality holds for large enough t. From Thus Since µ 0,t and (f µ) 0,t converge to different measures they are not asymptotic, and in fact they generate two different point mass distributions.

The Implications of Lemma 1
Here we will prove Propositions 2, 3 and 4 and then discuss the gap between Propositions 3 and 4.

Proof of Proposition 2. It follows from
If h : [0, ∞) → [0, ∞) is any decreasing function and ε, T and α are positive numbers, then Applying this to h(t) = log(µ(x + I t )) and ε = log γ gives (since γI t = I t−log γ ) Letting γ → 1 + shows that the condition (1) is satisfied, so the proposition follows by Lemma 1.
The following general fact about the weak- * topology will be used to prove Proposition 3 and Proposition 4. Proof. Suppose that ν(V ) > c. Any finite Borel measure on a locally compact space with a countable base is inner regular , so there is a compact set K ⊂ V such that ν(K) > c. By Urysohn's lemma, there is then a continuous function ϕ : X → R such that χ K ≤ ϕ ≤ χ V . Thus {ν; ν(ϕ) > c} is an open set that contains ν and is included in {ν; ν(V ) > c}, which proves the lemma.
Proof of Proposition 3. It will be shown that the condition (1) is satisfied, so that the proposition follows from Lemma 1. After a translation of the real line it may be assumed that x = 0.
Let Ω be the ω-limit set of µ 0,T . Then for every open set V that includes Ω, there is a T V such that µ 0,T ∈ V for all T ≥ T V . For if this was not the case, then there would be a sequence T k → ∞ such that µ 0,T k ∈ V c for all k. Since the space of distributions is compact there would be a converging subsequence, and the limit would lie in Ω ∩ V c , contradicting the assumption that Ω ⊂ V .
For any γ > 1 and K > 1, let this is a closed set by Lemma 8. Let ε K γ = sup P ∈Ω P (C K γ ). By Lemma 8 again, the set is an open set of distributions. Since it includes Ω (the reason for having the term 1 K is to make this true in case ε K γ = 0), it contains µ 0,T for all large enough T . Hence, 1) is replaced by the closed interval 1 γ I. This does not make any difference since the set of t for which µ 0,t gives positive measure to {± 1 γ } has Lebesgue measure 0). Thus to prove the proposition, it suffices to show that lim K→∞ lim sup γ→1 + ε K γ = 0. So take any ε > 0. Since C K γ decreases if γ decreases or K increases and P ( K,γ C K γ ) = 0 for any P ∈ Ω, the sets {{P ; P (C K γ ) < ε}} γ,K>1 cover Ω. Since they are open and Ω is compact there is a finite subcover, say Now, if γ decreases or K increases then {P ; P (C K γ ) < ε} increases. Thus Ω is in fact covered by {P ; P (C K γ ) < ε} whenever γ ≤ min k γ k and K ≥ max k K k . Hence and since ε is arbitrary this concludes the proof.
There is a discrepancy between the condition in Proposition 4 allowing µ x,T and f µ f (x),T not to be asymptotic, and the condition in Proposition 3 ensuring that they are. Namely, if P ({(1−β)δ −1 +βδ 1 ; 0 < β < 1}) = 0 for all ω-limit points P of µ x,T but there is some ω-limit point P 0 such that P 0 ({δ −1 , δ 1 }) > 0, then neither Proposition 4 nor Proposition 3 applies. In this case, it can happen either that µ x,T and f µ f (x),T are asymptotic for all f ∈ diff 1 + (R) or that there is some f for which they are not asymptotic, depending on µ and x. This is illustrated in the next two examples. As in example 2, for any c ∈ (0, 1). Take any f ∈ diff 1 + (R) such that f (0) = 0 and f ′ (0) = 1 -then f −1 has the form f −1 (x) = x(1 + r(x)) where lim x→0 r(x) = 0. Thus for any c ∈ (0, 1) where the inequality holds for all large enough t. Thus lim t→∞ (f µ) 0,t = δ 1 .
Since both G and H are continuous and increasing in (0, 1 2 since the inner parenthesis goes to e −1 − e so that the argument of the outer exponential function goes to −∞. For c ∈ (0, 1), Hence for any c ∈ (0, 1), so lim t→∞ µ 0,t = δ 1 . By the implicit function theorem, there is a diffeomorphism g defined in a neighbourhood (−ε, ε) of 0 such that g(0) = 0, g ′ (0) = 1 and
As Example 3 shows, the hypothesis in Proposition 3 is not the weakest possible. Proposition 3 was proved by showing that if P ({(1 − β)δ −1 + βδ 1 ; 0 ≤ β ≤ 1}) = 0 for every distribution P in the ω-limit set of µ x,T , then (1) is satisfied. The following proposition says that the opposite implication holds -thus to make the hypothesis of Proposition 3 weaker one would have to use something other than Lemma 1.
Proof. After a translation of the real line it may be assumed that x = 0. For any γ > 1 and any K > 1 let this is an open set by Lemma 8 and it includes {(1 − β)δ −1 + βδ 1 ; 0 ≤ β ≤ 1}. Let ε > 0. By (1) it is possible to choose K and γ > 1 such that Now let T 1 , T 2 , . . . be a sequence increasing to infinity such that µ 0,T k → P . Then where the inequality at * holds since V K γ is open. The proposition follows since ε is arbitrary.

Another example
In Example 2, 3 and 4 above, the measure µ generates a distribution P that is a point mass on a convex combination of δ −1 and δ 1 , so there is certainly ν ∈ supp P with 0 / ∈ supp ν. In all these examples this happens only at x = 0 -for any other point in the support of µ, the generated distribution would be a point mass on λ. This section will give an example of a measure µ that generates the same distribution P at µ-a.e. x, where P has 1 2 (δ −1 + δ 1 ) in its support. Let Λ be a countable set with at least two elements and let {[ l ]} l∈Λ be a collection of compact subintervals of I whose interiors are pairwise disjoint.
and suppose that T 1 − T 0 is µ-integrable. Then at µ-a.e. x, µ generates the distribution P given by Proof. It can be that T 1 (x) = T 0 (x) only if x lies in the left half of the leftmost basic interval or in the right half of the rightmost interval. Thus there is a basic interval where T 1 − T 0 > 0 if Λ has at least three elements, and if Λ has two elements, say Λ = {0, 1}, then T 1 − T 0 > 0 on at least one of the cylinders [01] and [10]. It follows that the denominator in the expression for P is positive.
Take any g ∈ C(M ) and fix some µ-typical x (meaning that the Birkhoff averages converge at x to the integral with respect to µ) such that [x 1 . . . x k ] is defined for all k. Since µ is a Bernoulli measure, there are constants c k such that σ k µ [x1,...,x k ] = c k µ 16MAGNUS ASPENBERG, FREDRIK EKSTRÖM, TOMAS PERSSON, AND JÖRG SCHMELING for all k, and thus Now, In particular, this implies that if n(T ) is the unique natural number such that T n(T ) (x) ≤ T < T n(T )+1 (x) then Using this at the first step together with the fact that g is bounded gives (5) and (6) The lemma follows from Birkhoff's theorem applied to the last member, since The support of the distribution P in Lemma 10 is E, where If ν / ∈ E then by Urysohn's lemma there is a non-negative function g ∈ C(M ) such that g(ν) > 0 and g is 0 on E. Then so the open neighbourhood {g > 0} of ν has P -measure 0. Thus supp P ⊂ E. To see the other inclusion, the next lemma is needed.
Take any ϕ ∈ C([−1, 1]). Again by the dominated convergence theorem, For each k, the function T k is finite and continuous in the interior of each cylinder of generation k, and if x does not lie in the interior of any such cylinder then T k (x) = ∞ and lim y→x T k (y) = ∞ -thus T k is continuous at every x ∈ I in the sense that lim y→x T k (y) = T k (x). Take any µ x0,t0 ∈ E and let V be an open set around µ x0,t0 . Because of Lemma 11 and the continuity of T 0 and T 1 , there is an Thus E ⊂ supp P and since supp P is closed it follows that E ⊂ supp P . and µ is symmetric around 0, Thus T 0 is integrable. Furthermore T 1 (x) = log 2 length of [x 1 ] + T 0 (σ(x)) = (|x 1 | + 1) log 2 + T 0 (σ(x)),

18MAGNUS ASPENBERG, FREDRIK EKSTRÖM, TOMAS PERSSON, AND JÖRG SCHMELING
where x 1 is the integer such that x ∈ [x 1 ], so T 1 dµ = 2 log 2 ∞ l=1 (l + 1)p l + T 0 dµ using that µ is σ-invariant. The sum converges because of (7), so T 1 is integrable as well. Thus by Lemma 10, µ generates a distribution P at µ-a.e. x with supp P = {µ x,t ; x ∈ supp µ, T 0 (x) ≤ t ≤ T 1 (x)}. p m 1 p n 2(p m 1 p n + ∞ l=n+1 p l ) → 1 2 , n → ∞ by (7). Thus for any ε > 0 there is a measure in supp P that gives measure > 1 2 − ε to each of the intervals ±[1 − ε, 1]. Since supp P is closed, it follows that 1 2 (δ −1 + δ 1 ) ∈ supp P . For µ-a.e. x, the local dimension of µ at x is where h µ (x) is the local entropy and u(x) is the Lyapunov exponent at x. By suitable choice of the probabilities (p l ), the dimension can be made to take any value in (0, 1). Consider for example where the constant c N is chosen so that (p N l ) sums to 1/2. These probabilities satisfy (7) for any finite N , and (this is not surprising since µ becomes Lebesgue measure if N = ∞). Thus, the dimension can be made arbitrarily close to 1. On the other hand, if two probabilities are swapped then the Lyapunov exponent changes but the entropy does not. In particular, it is possible to make the local dimension arbitrarily close to 0 by putting the largest probability at a high index. Thus for any ε > 0, there are (p ε l ) and (q ε l ) that sum to 1/2, satisfy (7) and give µ dimension at least 1 − ε and at most ε respectively. Any convex combination of (p ε l ) and (q ε l ) also sums to 1/2 and satisfies (7), and since the dimension varies continuously with the probabilities, it takes all values in (ε, 1 − ε) on the line segment between (p ε l ) and (q ε l ).