Multiple mixing for a class of conservative surface flows
aa r X i v : . [ m a t h . D S ] S e p Multiple mixing for a class of conservative surface flows
Bassam Fayad and Adam Kanigowski11th July 2018
Abstract
Arnol’d and Kochergin mixing conservative flows on surfaces stand as the main and almostonly natural class of mixing transformations for which higher order mixing has not beenestablished, nor disproved.Under suitable arithmetic conditions on their unique rotation vector, of full Lebesguemeasure in the first case and of full Hausdorff dimension in the second, we show that theseflows are mixing of any order. For this, we show that they display a generalization of the socalled Ratner property on slow divergence of nearby orbits, that implies strong restrictions ontheir joinings, that in turn yield higher order mixing.This is the first case in which the Ratner property is used to prove multiple mixing outsideits original context of horocycle flows and we expect our approach will have further applica-tions.
A major open problem in ergodic theory is Rokhlin’s question on whether mixing implies mixingof all orders, also called multiple mixing [21]. In most of the known examples of mixing dynamicalsystems, multiple mixing is now known to hold. Moreover, a positive answer to Rokhlin’s questionis actually known to generally hold within various classes of mixing dynamical systems. The mostnoteworthy are K-systems where multiple mixing always holds [2], horocycle flows [18], mixingsystems with singular spectrum that display multiple mixing by a celebrated theorem of Host [10],and finite rank systems since Kalikow showed that rank one and mixing implies multiple mixing[11], a result that was extended to finite rank mixing systems by Ryzhikov [25].In the second half of the last century, Arnol’d and Kochergin introduced a major class ofconservative smooth mixing flows on surfaces, with non-degenerate saddle type singularities forthe first and degenerate ones for the second. Mixing of these flows was proved by Kochergin inthe case of degenerate power like singularities (see exact description below) in [13] and by Khaninand Sinai in a particular case of non-degenerate asymmetric saddle type singularities (see exactdescription below) [27]. The study of the mixing properties of these flows has known a revivalof interest since the beginning of the 2000’s, with results such as the computation of the speedof mixing [4] or extension of the Kahnin-Sinai mixing result to include all irrational translationvectors [17] (see also [15],[16]), or advances in the study of Arnol’d and Kochergin flows in thegeneral case where the Poincaré section return map is an interval exchange and not just a circularrotation [30, 31, 32].Arnol’d and Kochergin flows stand today as the main and almost only natural class of mixingtransformations for which higher order mixing has not been established, nor disproved. Our aimhere is to prove mixing of all orders for a rich subclass of these systems determined by the arithmet-ics of their unique rotation vector. For this, we use the representation of Arnol’d and Kochergin1flows as special flows above an irrational rotation R α of the circle and under a ceiling functionwith asymmetric logarithmic singularities for the first, and integrable power like singularities forthe second. Loosely speaking our main result is as follows (it will be made precise at the end ofthis introduction, see Corollaries 1.6 and 1.8). Theorem.
Arnol’d-Khanin-Sinai flows are mixing of all orders for a set of α ∈ (0 , of fullLebesgue measure. Kochergin flows are mixing of all orders for a set of α ∈ (0 , of full Hausdorffdimension. Similar mixing mechanisms due to orbit shear as in Kochergin and Arnol’d-Khanin-Sinai flowswere observed relatively recently such as in [5] or [1] and it should be possible to apply the tech-niques of the current paper to the study of higher order mixing for such parabolic systems.To explain our approach we need first to make a detour by Ratner’s study of horocycle flows.In the 1980’s, M. Ratner developed a rich machinery to study horocycle flows [22]-[24] and, inparticular, singled out a special way of controlled slow divergence of orbits of nearby points whichresulted in the notion of H p -property, later called R-property by J.-P. Thouvenot [28]. Thisproperty, to which we will come back with more detail in the sequel, has important dynamicalconsequences, mainly expressed by a restriction on the possible joining measures of a systemhaving the R-properties with other systems, and in particular with itself.A joining between two dynamical systems ( T, X, B , µ ) and ( S, Y, C , ν ) , ( X, B , µ ) and ( Y, C , ν ) being standard Borel probability spaces, is a measure ρ on X × Y invariant by T × S whosemarginals on X and Y are µ and ν . The definition for flows is similar. An important notion inRatner’s theory is that of finite extension joinings (FEJ). Definition 1.1.
An ergodic flow ( T t ) t ∈ R is said to have FEJ-property , acronym for finite extensionjoining , if for every ergodic flow ( S t ) t ∈ R acting on ( Y, C , ν ) and every ergodic joining ρ of ( T t ) t ∈ R and ( S t ) t ∈ R different from the product measure µ × ν , ρ yields a flow which is a finite extension of ( S t ) t ∈ R .It was shown in [26] that a mixing flow with FEJ-property is mixing of all orders. Moreover, itwas proved in [22] that a flow with R-property has the FEJ-property. It follows that mixing flowswith the R-property are mixing of all orders. Since the R-property for horocycle flows stemmedfrom polynomial shear along the orbits, and since Kochergin flows displayed a similar polynomialshear along the orbits, the idea that special flows over rotations may enjoy the R-property, andthus be multiple mixing, was then suggested by J-P. Thouvenot in the 1990’s (see p. 2 in [6]).However, whether natural classes of special flows (not necessarily mixing) over irrational rota-tions may have the R-property remained open until K. Fr¸aczek and M. Lemańczyk [6, 7] showed thata generalized R-property holds in some classes of special flows with roof functions of bounded vari-ation (which, by [12], are not mixing). More precisely, they have introduced a weaker notion thanthe R-property, called weak Ratner or WR-property that however still implies the FEJ-property(see Definition 2.1 and the comment after it) .Unfortunately, in the mixing examples of special flows under piecewise convex functions withsingularities such as Arnol’d and Kochergin flows, the shear may occur very abruptly as orbitsapproach the singularity and this may prevent them from having the weak Rather property. Indeed,we believe that these flows do not have the WR-property. That this should be true is corroboratedby the following result that shows that Kochergin flows, in the context of bounded type frequencyin the base (that is a priori favorable to controlling the shear), do not have the WR-property. Theorem 1.
Let α ∈ T be irrational with bounded partial quotients and f ( x ) = x γ + r , − < γ < , r > . Then the special flow above the circle rotation R α and under the ceiling function f does nothave the WR-property. We denote by T the circle R / Z . We refer to Section 3 for the exact definition of special flows.Theorem 1 has another consequence. It is known that every horocycle flow ( h t ) t ∈ R is looselyBernoulli [24]; therefore, for every irrational α , there exists a positive function in f ∈ L ( T ) such that ( h t ) t ∈ R is measurably isomorphic to ( T ft ) t ∈ R [20]. It follows from [12] and the fact that ( h t ) t ∈ R is mixing that f is of unbounded variation. Moreover, by [19], f can be made C except forone point. Since the R-property implies the WR-property and the R-property is an isomorphisminvariant, no special flow as in Theorem 1 is isomorphic to a horocycle flow. Actually, this line ofthought can be extended to show that horocycle flows are never isomorphic to special flows abovean irrational rotation and under a roof function that is convex and C except at one point. Forthe latter result, one needs to introduce the concept of strong Ratner property , which is also anisomorphism invariant, that specifies the occurrence of slow divergence of nearby orbits to the firsttime when the orbits do split apart. This will be dealt with in a future work.To bypass Theorem 1 and still use controlled divergence of orbits to show multiple mixing, ourapproach will be to further weaken the WR-property. Namely, we introduce the SWR-property,which stands for switchable weak Ratner property, that assumes that a pair of nearby pointsdisplays the WR-Property either under forward iteration in time or under backward iteration, andthis depending on the pair of points. We show that the SWR-property is sufficient to guaranteethe same FEJ consequences as the Ratner or the weak Ratner property. Consequently, a mixingflow enjoying the SWR-property is mixing of all orders.The main idea in showing that Arnold and Kocergin special flows may have the SWR-Propertyis the following. The main contribution to the shear between orbits is due to the visits of the flowlines to the neighborhood of the singularities. With the representation of these flows as specialflows above irrational rotations, the shear is translated into the divergence between the Birkhoffsums of the roof functions for nearby points, and this divergence is mainly due to the visits underthe base rotation to the neighborhoods of the points where the roof function has its singularities.If the base rotation angle α is of bounded type two nearby points will stay sufficiently far from thesingularity either when they are iterated forward or when they are iterated backward. In the caseof ceiling functions with only logarithmic singularities we are also able to exploit the progressivecontribution to the shear of these visits to the singularities to obtain multiple mixing for a fullmeasure set of numbers α .We will now describe the ceiling functions that will be considered in the sequel and state ourexact results on the SWR-property and multiple mixing. Definition 1.2.
Let h be a positive function h ∈ C ( T \ { } ) , decreasing on (0 , , lim x → + h ( x ) =+ ∞ , h ′ increasing on (0 , . Let f ∈ C ( T \ { a , ...., a k } ) for some numbers a , ..., a k ∈ T . We saythat f has singularities of type h at { a , ..., a k } if lim x → a + i f ′ ( x ) h ′ ( x − a i ) = A i and lim x → a − i f ′ ( x ) h ′ ( a i − x ) = − B i , (1)for some numbers A i , B i > , i = 1 , ..., k .Notice that in this definition h may only reflect a domination on the singularities of f since thecoefficients A i , B i may be equal to zero at some or at all i ’s.In all the sequel, we will consider α ∈ R \ Q and let ( q s ) be the sequence of denominators of thebest rational approximations of α . Namely ( q s ) is the unique increasing sequence such that q = 1 and k q s α k < k kα k for any k < q s +1 , k = q s . We recall that q s + q s +1 k q s α k q s +1 (2)Our results can deal with functions having several singularities but require a non resonancecondition of these singularities with α . Definition 1.3. [Badly approximable singularities] Given α ∈ R \ Q , we will say that { a , ...., a k } are badly approximable by α if there exists C > such that for every x ∈ T and every s ∈ N , thereexists at most one i ∈ { , ..., q s − } such that x + i α ∈ k [ i =1 (cid:20) − Cq s + a i , a i + 12 Cq s (cid:21) . (3) Remark 1.4.
It was shown in [8, Lemma 3] that if a i − a j ∈ ( Q + Q α ) \ ( Z + Z α ) whenever i = j then { a , ...., a k } are badly approximable by α .Note that if there is only one singularity, that is k = 1 , then by (2) it is always badly approxim-able by α . The following shows that for k > the set of singularities that are badly approximableby α is a thick set in [0 , k . Lemma 1.5.
Let α ∈ R \ Q . For any k ∈ N , the set E ⊂ [0 , k of k − tuples ( a , . . . , a k ) that arebadly approximable by α is a product of sets of full Hausdorff dimension in [0 , .Proof. Define B ( α ) := { b ∈ R : ∃ C > , ∀ k ∈ Z \ k kα − b k > C | k | } Then if ( a , . . . , a k ) are such that a i − a j ∈ B ( α ) for any i, j ∈ { , . . . , k } , i = j , then ( a , . . . , a k ) are badly approximable by α .But it was proven in [29] (see also [3]) that the set B ( α ) is a winning set in the sense of Schmidt(see [29, 3] and references therein). A winning set is of full Huasdorff dimension. Moreover, for awinning set B ∈ R we have that for any x , . . . , x n the set ∩ ns =1 ( x s + B ) is winning. So, if a , . . . , a l are such that a i − a j ∈ B ( α ) for any i, j ∈ { , . . . , l } , i = j , then the set of a ∈ [0 , such that a ∈ ∩ ls =1 ( a s + B ( α )) is winning which means that a , . . . , a l , a l +1 are badly approximable by α fora winning set of a l +1 . The statement of the Lemma follows then by induction and because a single a is always badly approximable by α .Our results deal with two types of singularities. Theorem 2 deals with logarithmic like singu-larities, while Theorem 3 deals with the case of at least one dominant power like singularities. In the case of logarithmic like singularities, the following theorem holds.
Theorem 2.
Let α ∈ R \ Q and f ∈ C ( T \ { a , ...., a k } ) with the singularities { a , . . . , a k } of type h and badly approximable by α , with some associated constant C > . Assume that P ki =1 A i = P ki =1 B i and that there exist a constant m > and a sequence ( x s ) such that for every s ∈ N , wehave x s < q s and lim s → + ∞ h ′ ( xs C ) q s h ( qs ) = 0 , lim s → + ∞ x s q s h ( q s ) = + ∞ ;2. P i/ ∈ K α q i x i < + ∞ , where K α := { s ∈ N : q s +1 < x s } ;3. h ( q s ) /h ( q s +1 ) > m .Then ( T ft ) t ∈ R has the SWR-property. To describe a consequence of Theorem 2 (see Corollary 1.6), set h ( x ) = − ln( x ) , for x ∈ (0 , .For α ∈ R \ Q , let K α := { n ∈ N : q n +1 < q n log ( q n ) } . We then define in view of . and . of Theorem 2 E := α ∈ T \ Q : X i/ ∈ K α q i < + ∞ . To have 3. of Theorem 2 it suffices to assume that α is Diophantine : for τ > define the setof Diophantine numbers α of exponent τ to be DC ( τ ) := { α ∈ T \ Q : ∃ r α ∈ N ∀ n ∈ N , q n +1 < r α q τn } . The set of Diophantine numbers is then D := { α ∈ T \ Q : ∃ τ > , α ∈ DC ( τ ) } . An equivalent definition of DC ( τ ) is that for any p, q ∈ Z × N ∗ we have that | α − pq | > C ( α ) q τ forsome C ( α ) > . Corollary 1.6.
Consider h ( x ) = − log( x ) . Let α ∈ D ∩ E . Let f ∈ C ( T \ { a , ...., a k } ) with thesingularities { a , . . . , a k } of type h and badly approximable by α , with some associated constant C > . Assume that P ki =1 A i = P ki =1 B i . Then ( T ft ) t ∈ R has the SWR-Property and is mixing ofall orders.Proof. We take for x s the sequence q s log q s and easily check the hypothesis of Theorem 2. There-fore ( T ft ) t ∈ R has the SWR-property. On the other hand, it was shown in [15] that ( T ft ) t ∈ R ismixing. Multiple mixing then follows from Theorem 4 and the FEJ-property.Corollary 1.6 covers a set of full Lebesgue measure of rotation angles α . Indeed, it is knownthat the set of Diophantine numbers D has full Lebesgue measure, and we will prove in AppendixA the following result. Denote by λ the Haar measure on T . Proposition 1.7.
It holds that λ ( E ) = 1 . Now, we will deal with power like singularities. We suppose f ∈ C ( T \ { a , ...., a k } ) with singu-larities { a , ...., a k } of type h . We divide the set { a , ..., a k } of singularities into two subsets : F the set of weak singularities, and E the set of strong singularities of type not less than h : namely,we suppose F = { a , ..., a v } and E = { a v +1 , ..., a k } ∈ E are such that, A i + B i > in (1) for i ∈ { v + 1 , . . . , k } , while each a i ∈ F is a singularity for f of type g i with g i a positive function in C ( T \ { } ) , decreasing on (0 , with g ′ i increasing and such that lim x → a + i (cid:12)(cid:12)(cid:12)(cid:12) f ′ ( x ) g ′ i ( x − a i ) (cid:12)(cid:12)(cid:12)(cid:12) and lim x → a − i (cid:12)(cid:12)(cid:12)(cid:12) f ′ ( x ) g ′ ( a i − x ) (cid:12)(cid:12)(cid:12)(cid:12) exist and are finite , (4) lim x → + g i ( x ) h ( x ) = 0 and for every s ∈ N there exists x i,s ∈ T , x i,s > q s h ( qs ) , such that lim s → + ∞ g ′ i ( x i,s ) h ′ ( qs ) =0 and P + ∞ i =1 q s x i,s < + ∞ .We always assume that E is not empty. Theorem 3.
Let α be irrational with bounded partial quotients, that is, α ∈ DC (0) . Assume that { a v +1 , ..., a k } are badly approximable by α with some constant C > . Assume that there existconstants D , D > such that for every s ∈ N D > − h ′ ( C q s ) q s h ( q s ) > D and h ( q s ) h ( q s +1 ) > D . (5) Then ( T ft ) t ∈ R has the SWR-property. Corollary 1.8.
Let α ∈ DC (0) . Let f ∈ C ( T \ { a , ...., a k } ) with all the singularities { a , ...., a k } of power-like type x γ i from the left and x δ i from the right, − < γ i , δ i < . Let γ = min i k { γ i , δ i } , E = { a i : min { γ i , δ i } = γ } . Then, if the points in E are badly approximable by α , we have that ( T ft ) t ∈ R has the SWR-Property and is mixing of all orders. Note that there are no combinatorial assumptions on the weak singularities a j / ∈ E . Proof of Corollary 1.8.
Take x s = s q s and easily check the hypothesis of Theorem 3. This givesthe SWR-Property . Mixing of ( T ft ) t ∈ R was established in [13]. Multiple mixing then follows fromTheorem 4 and the FEJ-property. In Section 2 we introduce the SWR-Property and study its joinings consequences. In Section 3 wegive a criterion involving the Birkhoff sums of the ceiling function that guarantees that a specialflow above an isometry has the SWR-property. The treatment of these sections is similar to [6, 7].In Section 4 we study the Birkhoff sums of logarithmic like and power like functions and proveTheorems 2 and 3. Section 5 is devoted to the proof of Theorem 1 on the absence of the SWR-Property for a subcalss of Kochergin flows. Finally Appendix A is devoted to the proof that theset of frequencies for which Theorem 2 holds has full Lebesgue measure.
Acknowledgments.
The second author would like to thank Professor Mariusz Lemańczyk for all his patience, helpand deep insight. The authors would also like to thank Krzysztof Fr¸aczek, Mariusz Lemańczykand Jean-Paul Thouvenot for valuable discussions on the subject.The results of Section 4 have been obtained by the two authors independently and the resultsof Section 5 by the second. The two authors decided to include Section 5 in this work because itis an integral part of the problems concerning Ratner’s property for this class of special flows.
Let ( X, B , µ ) be a probability standard Borel space. We additionally assume that X is a completemetric space with a metric d . Let ( T t ) t ∈ R be an ergodic flow acting on ( X, B , µ ) . Definition 2.1 (cf. [7], Definition 4) . Fix t ∈ R + and a compact set P ⊂ R \ { } . One says thatthe flow has the switchable R ( t , P ) -property if for every ǫ > and N ∈ N there exist κ = κ ( ǫ ) , δ = δ ( ǫ, N ) and a set Z = Z ( ǫ, N ) ⊂ B with µ ( Z ) > − ǫ such that for any x, y ∈ Z with d ( x, y ) < δ , x not in the orbit of y there exist M = M ( x, y ) , L = L ( x, y ) ∈ N with M, L > N and LM > κ and p = p ( x, y ) ∈ P such that L (cid:12)(cid:12) { n ∈ [ M, M + L ] : d ( T nt ( x ) , T nt + p ( y )) < ǫ } (cid:12)(cid:12) > − ǫ (6)or L (cid:12)(cid:12) { n ∈ [ M, M + L ] : d ( T n ( − t ) ( x ) , T n ( − t )+ p ( y )) < ǫ } (cid:12)(cid:12) > − ǫ. (7)If the set of t > such that the flow ( T t ) t ∈ R has the switchable R ( t , P ) -property is uncount-able, the flow is said to have SWR -property.For the sake of completeness, compare the SWR-property with the definition of the WR-property [7]. To have WR-property, we fix P ⊂ R \ { } and t ∈ R . ( T ft ) t ∈ R has R ( t , P ) propertyif in Definition 2.1, (6) holds (the condition (7) is not taken into account) and ( T ft ) t ∈ R has WR-property if the set of t ∈ R such that ( T ft ) t ∈ R has R ( t , P ) property is uncountable. Consequently,SWR-property is weaker than WR-property (and as Theorem 1 shows, it is strictly weaker).Now, again for sake of completeness, we will present a detailed proof (using some facts provedin [7]) of the fact under the “continuty” assumption on orbits (see below) that SWR-property hasFE-property as the original H p -property introduced by M. Ratner [22].We will state a lemma which is a simple consequence of Lemma 5.2. in [7]. Lemma 2.2.
Let
T, S : ( X, B , µ ) → ( X, B , µ ) be two ergodic automorphisms and let A ∈ B . Forany ǫ, δ, κ > there exist N = N ( ǫ, δ, κ ) and a measurable set Z = Z ( ǫ, δ, κ ) with µ ( Z ) > − δ such that for any M, L > N with LM > κ and any x ∈ Z we have (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) L M + L X i = M χ A ( T i x ) − µ ( A ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) < ǫ and (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) L M + L X i = M χ A ( S i x ) − µ ( A ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) < ǫ. We will add one more natural condition on the flow ( T t ) t ∈ R which can be viewed as “continuity”on orbits. The flow ( T t ) t ∈ R is called almost continuous [7] if for every ǫ > there exists a set X := X ( ǫ ) with µ ( X ) > − ǫ such that for every ǫ ′ > there exists δ ′ > such that for every x ∈ X , we have d ( T t ( x ) , T t ′ ( x )) < ǫ ′ for t, t ′ ∈ [ − δ, δ ] .For the definition and properties of joinings, we refer the reader to [28] or [9]. Our goal is nowto prove the following result. Theorem 4.
Let ( T t ) t ∈ R be a weakly mixing flow acting on a probability standard Borel space ( X, B , µ ) . Assume that ( T t ) t ∈ R satisfies the SWR-property. Let ( S t ) t ∈ R be an ergodic flow actingon a probability standard Borel space ( Y, C , ν ) and let ρ ∈ J (( T t ) t ∈ R , ( S t ) t ∈ R ) be an ergodic joining.Then either ρ is equal to µ ⊗ ν or is a finite extension of the measure ν . To prove this theorem we need some lemmas from [7].
Lemma 2.3.
Let ( T t ) t ∈ R be an ergodic almost continuous flow acting on ( X, B , µ ) , and ( S t ) t ∈ R be another ergodic flow acting on ( Y, C , ν ) . Let ρ ∈ J (( T t ) t ∈ R , ( S t ) t ∈ R ) be such that ρ is ergodicfor automorphisms T × S (hence, for T − × S − ). Let P ⊂ R be non-empty and compact. Let A ∈ B be such that µ ( ∂A ) = 0 and B ∈ C . Then, for every ǫ, δ, κ > there exist a natural number N = N ( ǫ, δ, κ ) and a set Z = Z ( ǫ, δ, κ ) ⊂ B ⊗ C with ρ ( Z ) > − δ such that for any N ∋ M, L > N with LM > κ and any p ∈ P , we have (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) L M + L X j = M χ T − p A × B ( T j x, S j y ) − ρ ( T − p A × B ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) < ǫ and (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) L M + L X j = M χ T − p A × B ( T − j x, S − j y ) − ρ ( T − p A × B ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) < ǫ for every ( x, y ) ∈ Z .Proof. The proof is a simple consequence of Lemma 5.4. in [7]. One uses this lemma first forthe flows ( T t ) t ∈ R and ( S t ) t ∈ R and ergodic joining ρ ∈ J (( T t ) t ∈ R , ( S t ) t ∈ R ) to get, for ǫ, δ , κ > , anatural number N + ∈ N and a set Z + ⊂ B ⊗ C with ρ ( Z + ) > − δ . Then, for flows ( T − t ) t ∈ R and ( S − t ) t ∈ R and the same ergodic joining ρ to get, for ǫ, δ , κ > , a natural number N − ∈ N anda set Z − ⊂ B ⊗ C with ρ ( Z − ) > − δ . To finish the proof one takes N := max( N + , N − ) and Z = Z + ∩ Z − .Next lemma is used in the proof of Theorem 3 in [22]. Lemma 2.4.
Let ( T t ) t ∈ R and ( S t ) t ∈ R be two ergodic flows. Let ρ ∈ J e (( T t ) t ∈ R , ( S t ) t ∈ R ) be anergodic joining. Then if there exists a set V with ρ ( V ) > such that for any points ( x, y ) , ( x ′ , y ) ∈ V either x is in the orbit of x ′ or d ( x, x ′ ) > c for some constant c > , then ρ is a finite extensionof ν . In what follows, we consider only ( X, d ) be a σ -compact metric space. Let A ∈ B . For η > we denote by V η ( A ) := { x ∈ X : d ( x, A ) < η } . Lemma 2.5. [cf. [7]] For any A ∈ B there exists R ⊂ (0 , + ∞ ) such that (0 , + ∞ ) \ R is countableand µ ( ∂V η ( A )) = 0 for η ∈ R . It particular, there exists a dense family ( B i ) i > in B with theproperty µ ( ∂B i ) = 0 for every i ∈ N .Proof of Theorem 4. Let ρ ∈ J (( T t ) t ∈ R , ( S t ) t ∈ R ) be an ergodic joining and ρ = µ × ν . Assume that ( T t ) t ∈ R has the switchable R ( t , P ) -property and ρ is ergodic for T t × S t (then ρ is ergodic for T − t × S − t ). Such t > always exists because an ergodic flow can have at most countably manynon-ergodic time automorphisms. For simplicity of notation, we assume t = 1 . Let { B i } i > and { C i } i > be two countable dense families in the σ -algebras B and C , respectively. Consider thefollowing real function: R ∋ t → k ( t ) := X i,j > (1 / i + j ) | ρ ( T t ( B i ) × C j ) − ρ ( B i × C j ) | . As in Lemma 5.4. in [7], we conclude that k : R → R is a continuous function and for any t ∈ R , k ( t ) > . Indeed, it follows by the fact that if for some r ∈ R \ { } we have for any i, j ∈ N ρ ( T r ( B i ) × C j ) = ρ ( B i × C j ) then ρ is product measure (recall that T t is assumed to be weakmixing hence every time r of the flow is ergodic).The set P ⊂ R \ { } is compact, therefore there exists ǫ > such that k ( p ) > ǫ for any p ∈ P . Itfollows by the definition of the function k that there exists a number R := R ( ǫ ) such that R X i,j > (1 / i + j ) | ρ ( T p ( B i ) × C j ) − ρ ( B i × C j ) | > ǫ/ for every p ∈ P . Therefore, for every p ∈ P , there exist i, j R such that | ρ ( T p ( B i ) × C j ) − ρ ( B i × C j ) | > ǫ .By Lemma 2.5, there exists ǫ ′ < ǫ such that for every i Rµ ( V ǫ ′ ( B i ) \ B i ) < ǫ, and µ ( ∂V ǫ ′ ( B i )) = 0 . It follows by the fact that ρ is a joining that | ρ ( V ǫ ′ ( B i ) × C j ) − ρ ( B i × C j ) | < ǫ and | ρ ( S − t V ǫ ′ ( B i ) × C j ) − ρ ( S − t B i × C j ) | < ǫ , (8)for i, j R and every t ∈ R . By the switchable R (1 , P ) -property, let κ := κ ( ǫ ′ ) . By Lemma 2.2applied to ǫ , , κ , the sets V ǫ ′ ( B i ) × C j , i, j R , and to automorphisms T × S and T − × S − ,we get N ∈ N and a set U ∈ B ⊗ C with ρ ( U ) > , such that for every L, M > N with LM > κ and every ( x, y ) ∈ U , we have (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) L M + L X k = M χ V ǫ ′ ( B i ) × C j ( T k x, S k y ) − ρ ( V ǫ ′ ( B i ) × C j ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) < ǫ (9) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) L M + L X k = M χ V ǫ ′ ( B i ) × C j ( T − k x, S − k y ) − ρ ( V ǫ ′ ( B i ) × C j ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) < ǫ . (10)Next, by Lemma 2.3 applied to ǫ , , κ > and the sets B i × C j , i, j R , there exist N ∈ N and a set U ⊂ B ⊗ C with ρ ( U ) > such that for every L, M > N with LM > κ and any p ∈ P ,we have (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) L M + L X k = M χ T − p B i × C j ( T k x, S k y ) − ρ ( T − p B i × C j ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) < ǫ (11)and (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) L M + L X k = M χ T − p B i × C j ( T − k x, S − k y ) − ρ ( T − p B i × C j ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) < ǫ . (12)It follows that if we set N := max( N , N ) and U := U ∩ U , then ρ ( U ) > and for every L, M > N with LM > κ , any p ∈ P , the equations (9), (10), (11), (12) are satisfied for every ( x, y ) ∈ U . Using the switchable R (1 , P ) -property with ǫ ′ > and N ∈ N , we obtain δ = δ ( ǫ ′ , N ) and Z = Z ( ǫ ′ , N ) with µ ( Z ) > − ǫ ′ . Now, we will use Lemma 2.4 with the set U := U ∩ ( Z × Y ) (then of course ρ ( U ) > ) and δ = δ ( ǫ ′ , N ) to prove that for every ( x, y ) , ( x ′ , y ) ∈ U , d ( x, x ′ ) > δ .Assume on the contrary that d ( x, x ′ ) < δ . Then by the switchable R(1,P)-property, there exist L , M > N with L M > κ and p ∈ P such that L (cid:12)(cid:12) { n ∈ [ M , M + L ] : d ( T n ( x ) , T n + p ( x ′ )) < ǫ ′ } (cid:12)(cid:12) > − ǫ ′ L (cid:12)(cid:12) { n ∈ [ M , M + L ] : d ( T − n ( x ) , T − n + p ( x ′ )) < ǫ ′ } (cid:12)(cid:12) > − ǫ ′ . Assume that the first inequality is satisfied. We will use equations (9) and (11) (in case the secondone is satisfied, we use equations (10) and (12)). Let i p , j p R be the numbers whichsatisfy | ρ ( T p ( B i p ) × C j p ) − ρ ( B i p × C j p ) | > ǫ . Let K = K ( x, x ′ , p ) := { n ∈ [ M , M + L ] : d ( T n ( x ) , T n + p ( x ′ )) < ǫ ′ } . It follows that if k ∈ K and T k + p x ′ ∈ A i then T k x ∈ V ǫ ′ ( A i ) . Therefore ρ ( T − p B i p × C j p ) L M + L X k = M χ T − p B ip × C jp ( T k x ′ , S k y ) + ǫ ǫ ′ L L + 1 L M + L X k = M χ V ǫ ′ ( B ip ) × C jp ( T k x, S k y ) + ǫ ǫ ρ ( V ǫ ′ ( B i p ) × C j p ) < ǫ + ρ ( B i p × C j p ) . (13)A similar arguments show that ρ ( B i p × C j p ) < ǫ + ρ ( T − p B i p × C j p ) and consequently, | ρ ( B i p × C j p ) − ρ ( T − p B i p × C j p ) | < ǫ . This contradicts our assumption that | ρ ( T p ( B i p ) × C j p ) − ρ ( B i p × C j p ) | > ǫ is satisfied. Therefore, for any ( x, y ) , ( x ′ , y ) ∈ U we have d ( x, x ′ ) > δ and an application ofLemma 2.4 completes the proof. In this section, we will prove a sufficient condition for SWR-property in the case of special flows overan ergodic isometry We start by recalling the definition of special flows. Let T be an automorphism ( X, B , µ ) . Let f ∈ L ( X, µ ) such that f > . The s pecial flow ( T ft ) t ∈ R defined above T and underthe ceiling function f is given by X × R / ∼ → X × R / ∼ ( x, s ) → ( x, s + t ) , where ∼ is the identification ( x, s + f ( x )) ∼ ( T ( x ) , s ) (14)Equivalently the flow ( T ft ) t ∈ R is defined for t + s > (with a similar definition for negative times)by T ft ( x, s ) = ( T n x, t + s − f ( n ) ( x )) where n is the unique integer such that f ( n ) ( x ) t + s < f ( n +1) ( x ) (15)and f ( n ) ( x ) = f ( x ) + . . . + f ( T n − x ) if n > if n = 0 − ( f ( T n x ) + . . . + f ( T − x )) if n < . If T preserves a unique probability measure µ then the special flow will preserve a unique prob-ability measure that is the normalized product measure of µ on the base and the Lebesgue measureon the fibers. If X is a matric space with a metric d , so is X f with the metric d f (( x, s ) , ( x ′ , s ′ )) := d ( x, x ′ ) + | s − s ′ | . Moreover, it is easy to show that if ( T ft ) t ∈ R is a special flow acting on X f , then ( T ft ) t ∈ R is almost continuous (see Section 2) with X ( ǫ ) = { ( x, s ) ∈ X f : x ∈ X, ǫ < s < f ( x ) − ǫ } .The following general lemma is a direct consequence of Birkhoff ergodic theorem. Lemma 3.1.
Let T be an ergodic automorphism ( X, B , µ ) . Let f ∈ L ( X, µ ) , R X f dµ = 0 . Forevery ǫ, κ > there exist N = N ( ǫ, κ ) and a set A = A ( ǫ, κ ) with µ ( A ) > − ǫ such that for every M > N (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) M M X i =1 f ( T i x ) − Z X f dµ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) κ (cid:12)(cid:12)(cid:12)(cid:12)Z X f dµ (cid:12)(cid:12)(cid:12)(cid:12) (16) for every x ∈ A . Remark 3.2.
Assume that additionally f is positive and bounded away from zero. Fix ǫ, κ > ( κ < | R X f dµ | < / ). It follows that there are constants r , r > such that if we take x ∈ A then for any M, L > N with LM > κ , we have r < f ( M ) ( x ) M < r and r < (1 − κ ) R f dµ ( M + L ) − (1 + κ ) R f dµ · ML f ( M + L ) ( x ) − f ( M ) ( x ) M < r . Proposition 3.3.
Let T : ( X, d ) → ( X, d ) be an ergodic isometry and f ∈ L ( X, B , µ ) a positivefunction bounded away from zero. Let ( T ft ) t ∈ R be the corresponding special flow. Let P ⊂ R \ { } be a compact set. Assume that for every ǫ > and N ∈ N there exist κ = κ ( ǫ ) , δ = δ ( ǫ, N ) and aset X ′ = X ′ ( ǫ, N ) with µ ( X ′ ) > − ǫ , such that for any x, y ∈ X ′ with < d ( x, y ) < δ there exist N ∋ M = M ( x, y ) , L = L ( x, y ) with M, L > N , LM > κ and p = p ( x, y ) ∈ P such that | f ( n ) ( x ) − f ( n ) ( y ) − p | < ǫ for every n ∈ [ M, M + L ] (17) or | f ( − n ) ( x ) − f ( − n ) ( y ) − p | < ǫ for every n ∈ [ M, M + L ] . (18) If γ > is such that the automorphism T fγ is ergodic, then ( T ft ) t ∈ R has the switchable R ( γ, P ) -property.Proof. Fix γ > such that T fγ is ergodic. Fix also k f k L > ǫ > . Apply Remark 3.2 with theconstants ǫ/ , κ to f and T , T − , respectively to obtain constants D , D > such that for x ∈ A , µ ( A ) > − ǫ/ (the set A is the intersection of two relevant sets), we have D < f ( M ) ( x ) M , f ( M + L ) ( x ) − f ( M ) ( x ) L , f ( − M ) ( x ) − M , f ( − M − L ) ( x ) − f ( − M ) ( x ) − L < D . (19)Fix N > D ǫ . Let ǫ ′ := min( D ǫ γ + D ) , ǫ ) . Let κ ′ := D D κ ( ǫ ′ ) . Let us consider the set X ( ǫ ) onwhich ( T ft ) t ∈ R is ǫ - “almost continuous”, that is X ( ǫ ) := { ( x, s ) ∈ X f : ǫ < s < f ( x ) − ǫ } . Now, we will use ergodicity of T fγ and T f − γ . It follows that there exist N := N ( ǫ ) and a set Z := Z ( ǫ ) with µ f ( Z ) > − ǫ and for every ( x, s ) ∈ Z and n > N (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n n X k =1 χ X ( ǫ ) T fki ( x, s ) − (1 − ǫ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) < κκ + 1 ǫ (20)2for i = γ, − γ . Moreover, since f ∈ L ( X, B , µ ) , there exists a set V = V ( ǫ ) ⊂ X with µ ( V ) > − ǫ and such that for every x ∈ V , f ( x ) < ǫ . Define the set Z ′ := Z ∩{ ( x, s ) ∈ X f : x ∈ V }∩{ ( x, s ) ∈ X f : x ∈ A } , then µ f ( Z ′ ) > − ǫ .Let δ ′ := δ ( ǫ ′ , γ max( N ,N ) D ) . Take two points ( x, s ) , ( x ′ , s ′ ) ∈ Z ′ , such that x = x ′ and d f (( x, s ) , ( x ′ , s ′ )) < δ ′ . It follows by definition of d f that d ( x, x ′ ) < δ ′ and therefore by our assump-tions there exist M, L > γ max( N ,N ) D with LM > κ , p ∈ P and such that for all n ∈ [ M, M + L ] either | f ( n ) ( x ) − f ( n ) ( y ) − p | < ǫ ′ or for all n ∈ [ M, M + L ] , | f ( − n ) ( x ) − f ( − n ) ( y ) − p | < ǫ ′ . We willconsider the second case (the proof in the first case goes along the same lines).Let us define M ′ := f ( − M ) ( x ) − s − γ and L ′ := f ( − M − L ) ( x ) − f ( − M ) ( x ) − γ . By (19) it follows that L ′ = f ( − L − M ) ( x ) − f ( − M ) ( x ) − L − L − γ > − LD − γ > N . Similarly, f ( − M ) ( x ) − s − γ > f ( − M ) ( x ) − γ > MD γ , so M ′ > N . Moreover, since ( x, s ) ∈ Z ′ , s < ǫ < N D M D D / (2 γ ) (by thechoice of N ) and therefore L ′ M ′ > LD γ − γf ( − M ) ( x ) − s > LD M D > κ ′ . It follows by the properties of M ′ , L ′ ∈ N that if ( x, s ) ∈ Z ′ ⊂ Z we have (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) L ′ M ′ + L ′ X k = M ′ χ X ( ǫ ) T f − kγ − (1 − ǫ ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) < ǫ . (21)Take any k ∈ [ M ′ , M ′ + L ′ ] such that T f − kγ ∈ X ( ǫ ) it follows that there exist a number m k ∈ [ M, M + L ] such that T f − kγ ( x, s ) = ( T m k x, − kγ + s − f ( − m k ) ( x )) , where, by the fact that T f − kγ ∈ X ( ǫ ) , f ( − m k − ( x ) + ǫ < − kγ + s < f ( − m k ) ( x ) − ǫ . Using additionally the inequality | s − s ′ | < δ ′ we hence obtain f ( − m k − ( x ′ ) f ( − m k − ( x ) + p − ǫ ′ f ( − m k − ( x ) + p + ǫ − δ ′ < − kγ + s ′ + p. A similar reasoning shows that − kγ + s ′ + p < − kγ + s + p + δ f ( − m k ) ( x ) + p − ǫ δ f ( − m k ) ( x ′ ) . Therefore, by the definition of the special flow, we have T f − kγ + p ( x ′ , s ′ ) = ( T m k x ′ , − kγ + s ′ + p − f ( − m k ) ( x ′ )) . Consequently, d f ( T f − kγ ( x, s ) , T f − kγ + p ( x ′ , s ′ )) = d f (( x, s ) , ( x ′ , s ′ )) + | f ( − m k ) ( x ) − f ( − m k ) ( x ′ ) − p | < ǫ. Now, the number of k ∈ [ M ′ , M ′ + L ′ ] such that T f − kγ ∈ X ( ǫ ) is, by (21), at least (1 − ǫ ) L ′ and forany such k we get that d f ( T f − kγ ( x, s ) , T f − kγ + p ( x ′ , s ′ )) < ǫ . Hence L ′ (cid:12)(cid:12)(cid:12) { k ∈ [ M ′ , M ′ + L ′ ] : d f ( T f − kγ ( x, s ) , T f − kγ + p ( x ′ , s ′ )) < ǫ } (cid:12)(cid:12)(cid:12) > − ǫ. This gives us the switchable R ( γ, P ) -property.Note that if a flow ( T ft ) t ∈ R is ergodic then the set of η ∈ R such that T fη is not ergodic, isat most countable and therefore, as a direct consequence of Proposition 3.3, we get that ( T ft ) t ∈ R enjoys SWR-property.3 In this section we will use Proposition 3.3 to prove SWR-property for special flows given by theassumptions in Theorem 2 and Theorem 3. In all the sequel we assume { a , .., a k } are badlyapproximable by α with a constant C > (see Definition 1.3). Lemma 4.1.
Let s ∈ N be such that q s +1 > Cq s and x ∈ T . Then { x + jα } [ qs +14 C ] j =0 ∩ k [ i =1 (cid:20) − Cq s + a i , a i + 14 Cq s (cid:21) ⊂ { x + rq s + i α } [ qs +14 Cqs ] r =0 , where i ∈ { , ..., q s − } is such that ρ ( { x + vα } q s − v =0 , { a i } ki =1 ) = ρ ( x + i α, { a i } ki =1 ) . For finite sets A, B ⊂ T , we use the notation ρ ( A, B ) = min a ∈ A,b ∈ B k a − b k . Proof: Take any < j < q s − , j = i . By (3), x + jα / ∈ S ki =1 [ − Cq s + a i , a i + Cq s ] . It followsthat for every r = 0 , ..., max { [ q s +1 Cq s ] , } , k x + rq s + jα − ( x + jα ) k = k rq s α k r k q s α k Cq s . Hence, we conclude by (3). (cid:3)
The following lemma is a simple consequence of the Denjoy-Koksma inequality.
Lemma 4.2.
Let h ∈ C ( T \ { } ) be positive and decreasing on (0 , with h ′ is increasing on (0 , and lim x → + h ( x ) = lim x → + ( − h ′ ( x )) = + ∞ . Denote by c := inf T h . Then for every x ∈ T and s ∈ N we have the following estimates: − q s (cid:18) h ( 12 q s ) − c (cid:19) − h ′ ( 12 q s ) > h ′ ( q s ) ( x ) > h ′ ( x + jα ) − q s (cid:18) h ( 12 q s ) − c (cid:19) + 2 h ′ ( 12 q s ) where j ∈ { , ..., q s − } is such that min ℓ ∈{ ,...,q s − } | x + ℓα | = x + jα .Proof. Fix s ∈ N . Consider ¯ h ( x ) = ( , if x ∈ [0 , q s ) h ′ ( x ) , otherwise.Then ¯ h ∈ BV ( T ) and we use the Denjoy-Koksma inequality to obtain | ¯ h ( q s ) ( x ) − q s R T ¯ h ( t ) dλ | < Var ¯ h . But R T ¯ h ( t ) dλ = h ( q s ) − c and Var ¯ h − h ′ ( q s ) . We then finish since h ′ ( q s ) ( x ) =¯ h ( q s ) ( x ) + h ′ ( x + jα ) χ [0 , qs ] ( x + jα ) and h ′ < . Lemma 4.3.
Let f ∈ C ( T \ { a , ..., a k } ) . Assume that for i = 1 , ..., k , lim x → a + i f ′ ( x ) r i ( x − a i ) and lim x → a − i f ′ ( x ) r i ( a i − x ) exist and are finite, where r i ∈ C ( T \ { } ) is decreasing on (0 , with r ′ i increasing on (0 , . Then there exists a constant H > such that | f ′ ( x ) | < H k X i =1 − r ′ i ( x − a i ) − r ′ i ( a i − x ) ! , for each x ∈ T . Proof.
By assumptions, there exists a constant z > such that for every i = 1 , ..., k and for every x ∈ [ − z + a i , a i ) | f ′ ( x ) | < − Kr ′ i ( a i − x ) and for every x ∈ ( a i , a i + z ] ; | f ′ ( x ) | < − Kr ′ i ( x − a i ) forsome constant K > .Moreover, since f ′ ∈ C ( T \ { a , ..., a k } ) ; it follows that there exists a constant R > such thatfor every x ∈ T \ S ki =1 [ − z + a i , a i + z ] , | f ′ ( x ) | < R . Denote by C := min i =1 ,...,k | sup T r ′ i | Now,the constant H := 2 max i =1 ,...,k { K, RC } satisfies the assertion of the lemma. We may assume WLOG that P ki =1 ( A i − B i ) > . Fix ≫ ǫ > and N ∈ N . Let d = P ki =1 ( A i − B i ) − min( , P ki =1 ( A i − B i )2 ) > . Define κ = κ ( ǫ ) := ǫm d d +1) Hk . By Lemma 3.1 for ǫ/ and κ , we get N = N ( ǫ/ , κ ) and a set A := A ( ǫ/ , κ ) with λ ( A ) > − ǫ such that (16) holds for x ∈ A . Define P := (cid:20) − d + 1) , − dm C (cid:21) ∪ (cid:20) dm C , d + 1) (cid:21) , (22)( C commes from Definition 1.3 of badly approximable singularities).In the sequel, we will assume s is an integer sufficiently large s > s , s = s ( ε, N ) to bedetermined later, for now assume that κq s > N .By assumptions 1. and 2. of Theorem 2, if s > s and s ( ε ) is sufficiently large we will have | h ′ ( x s C ) | q s h ( q s ) < ǫ , x s C q s h (cid:18) q s (cid:19) > ǫ , h (cid:18) q s (cid:19) > C, (23)and P s > s ,s/ ∈ K α x s q s < ǫ k . Set v s := x s C .Let B s := { x ∈ T : x − q s α, ..., x, ..., x + ( q s − α / ∈ k [ i =1 ( − v s + a i , a i + 4 v s ) } and Z ′ := T s > s ,s/ ∈ K α B s .Define Z := Z ′ ∩ A . Observe that λ ( Z ) > − ǫ ( λ ( B s ) > − kv s q s ). Set δ := q s h ( qs ) .Consider x, y ∈ Z with < k x − y k < δ (we assume that x < y ).The following proposition implies Theorem 2 due to Proposition 3.3. Proposition 4.4.
Consider x, y ∈ Z with < k x − y k < δ . Then there exists p ∈ P , M, L > κM > N such that either (17) holds for n ∈ [ M, M + L ] or (18) holds for n ∈ [ M, M + L ] . Proposition 4.4 can be deduced from the following main result on the drift of the Birkhoff sumsof a function with logarithmic like singularities. Let s := s ( x, y ) ( s > s ) be unique such that q s +1 h ( q s +1 ) < k x − y k q s h ( q s ) . (24)We will assume that q s +1 > q s . If not, then in (24), m q s h ( qs ) < k x − y k and we repeat theconsiderations below in the time interval [ q s − , q s ] . In other words, in this case we will see the driftbetween x and y before time q s . Proposition 4.5.
Consider x, y ∈ Z as in (24) . Part a
There exists n ∈ { , ..., max( q s +1 Cq s , } satisfying f ( n q s ) ( x ) − f ( n q s ) ( y ) ∈ P (25)5 or f ( − n q s ) ( x ) − f ( − n q s ) ( y ) ∈ P. (26) Part b
Let X = T n q s x and Y = T n q s y if (25) holds, and X = T − ( n q s +1) x and Y = T − ( n q s +1) y if (26) holds. For n = 1 , ..., [ κn q s ] + 1 the following holdsA. | f ( n ) ( X ) − f ( n ) ( Y ) | < ǫ or B. | f ( − n ) ( X ) − f ( − n ) ( Y ) | < ǫ. (27)The rest of this section is devoted to the proof of Proposition 4.5. But before this we show howit implies Proposition 4.4 and thus Theorem 2. Proof of Proposition 4.4.
Suppose (25) holds, the other case being similar. If A. from (27) holds,set M := n q s , L := [ κM ] + 1 and p := f ( n q s ) ( x ) − f ( n q s ) ( y ) ∈ P . If B. holds, we set M := [ n q s κ ] , L := [ κM ] + 1 and p := f ( n q s ) ( x ) − f ( n q s ) ( y ) ∈ P . Notice that in both cases M, L > κn q s > κq s > κq s > N . Finally, using A. or B. and the cocycle identity for the Birkhoff sums and thetriangular inequality shows that for n ∈ [ M, M + L ] , | f ( n ) ( x ) − f ( n ) ( y ) − p | < ε for some p ∈ P . ✷ For m ∈ N , we will often use the following non resonance conditions of a pair of points ( x, y ) withthe singularities { a , . . . , a k } . q m − [ j =0 T j [ x, y ] ∩ k [ i =1 [ − v m + a i , a i + 2 v m ] = ∅ (28) max( [ qm +14 C ] ,q m ) [ j =0 T j [ x, y ] ∩ k [ i =1 [ − v m + a i , a i + v m ] = ∅ . (29) max( [ qm +14 C ] ,q m ) [ j =0 T − j [ x, y ] ∩ k [ i =1 [ − v m + a i , a i + v m ] = ∅ . (30) Lemma 4.6.
Let x, y ∈ T as in (24) . Then for every m such that s m s , if we have at leastone of the following1. if m / ∈ K α and (28) is satisfied2. if m ∈ K α and q m +1 > q m ,then we have at least one of (29) or (30) .Proof. Assume m / ∈ K α . Since m > s we have that q m +1 > x m > kǫ q m > Cq m . Let t ∈ [0 , q m − ∩ Z and r ∈ { , ..., k } be such that ρ ( { x + jα } q m − j =0 , { a i } ki =1 ) = k x + t α − a r k . It follows by (3) that for t = j ∈ [0 , q m − ∩ Z , we have x + jα / ∈ k [ r =1 (cid:20) − Cq m + a r , a r + 12 Cq m (cid:21) . k x + ( − q m + t ) α − a r k < k x + t α − a r k ,we will show (29), if we have the reversed inequality,we will show the validity of (30). Suppose k x + ( − q m + t ) α − a r k < k x + t α − a r k (31)holds (the proof in the second case is analogous). For j / ∈ { t , t + q m , ..., t + [ q m +1 Cq m ] q m − } , byLemma 4.1 (for i = t ), x + jα / ∈ k [ i =1 (cid:20) − Cq m + a i , a i + 14 Cq m (cid:21) . By (24) and (23), for j = 0 , ..., [ q m +1 C ] − , k ( x + jα ) − ( y + jα ) k = k x − y k v s ǫ kq s Cq s Cq m , (32)hence (29) follows for j / ∈ { t , t + q m , ..., t + [ q m +1 Cq m ] q m − } . If j = rq m + t for some r =0 , ..., [ q m +1 Cq m ] − , then by (31) (and using t q m − ) ρ ( x + jα, { a i } ki =1 ) > r k q m α k + ρ ( x + t α, { a i } ki =1 ) > ρ ( x + t α, { a i } ki =1 ) (28) > v m . Therefore, since k ( x + jα ) − ( y + jα ) k = k x − y k v s v m (see (32)), (29) follows for j = rq m + t .If m ∈ K α (recall that, by assumption q m +1 > q m ).Denote by u m := [ q m +1 ] and consider x − u m α, ..., x, ..., x + ( u m − α (the length of this orbit issmaller than q m +1 ). By (3), there exists at most one j ∈ [ − u m , u m − such that x + j α ∈ k [ i =1 (cid:20) − Cq m +1 + a i , a i + 12 Cq m +1 (cid:21) . If j < , we will show (29), if not, we will show the validity of (30). Suppose j < . Then forevery j = 0 , ..., u m − , x + jα , y + jα / ∈ k [ i =1 [ − v m + a i , a i + v m ] . (33)Indeed, it follows that for j = 0 , ..., u m − , x + jα / ∈ S ki =1 [ − Cq m +1 + a i , a i + Cq m +1 ] . But m ∈ K α ,therefore Cq m +1 > x m C = 2 v m . Moreover, by (24), k x − y k v s v m . Therefore (29) follows. Lemma 4.7.
There exists s ′ ∈ N such that for every s > s ′ and any points x < y ∈ T such that (28) is satisfied for x, y and m = s , then ( d + 1) q s h (cid:18) q s (cid:19) k x − y k > f ( q s ) ( x ) − f ( q s ) ( y ) > dq s h (cid:18) q s (cid:19) k x − y k . (34) Proof.
By (28) and (24), f ( q s ) is differantiable on [ x, y ] . Therefore, there exists θ ∈ [ x, y ] such that f ( q s ) ( x ) − f ( q s ) ( y ) = k x − y k| f ′ ( q s ) ( θ ) | . It is enough to show that there exist d > and s ′ ∈ N such that for s > s ′ ( d + 1) q s h (cid:18) q s (cid:19) > | f ′ ( q s ) ( θ ) | > dq s h (cid:18) q s (cid:19) . s ∈ N , define ¯ f ′ s ( θ ) = ( , if θ ∈ S ki =1 [ − q s + a i , a i + q s ] f ′ ( θ ) , otherwise.It follows that ¯ f ′ s ∈ BV ( T ) and f ′ ( q s ) ( θ ) = ¯ f ′ s ( q s ) ( θ ) + X i ∈ J s f ′ ( θ + j i α ) + X i ∈ L s f ′ ( θ + l i α ) , (35)where J s = { i ∈ [1 , k ] : ∃ j i ∈ { , ..., q s − } : θ + j i α ∈ [ − q s + a i , a i ] } and L s := { i ∈ [1 , k ] : ∃ l i ∈{ , ..., q s − } : θ + l i α ∈ [ a i , a i + q s ] } . Note that for every i ∈ [1 , k ] there exists at most one j i ∈ { , ..., q s − } : θ + j i α ∈ [ − q s + a i , a i ] .We use the Denjoy-Koksma inequality to ¯ f ′ s , to get q s Z T ¯ f ′ s dλ − Var( ¯ f ′ s ) | ¯ f ′ s ( q s ) ( θ ) | q s Z T ¯ f ′ s dλ + Var( ¯ f ′ s ) . (36)We have Z T ¯ f ′ s dλ = k X i =1 f ( a i + 12 q s ) − f ( a i − q s ) and Var( ¯ f ′ s ) = 2 k X i =1 (cid:18) f ′ ( a i + 12 q s ) + f ′ ( a i − q s ) (cid:19) , (37)(if s ∈ N is sufficiently large). It follows by the assumptions on f ′ and h ′ and (28), that for every ǫ > there exists s ′ = s ′ ( ǫ ) such that for s > s ′ , we have for every i = 1 , ..., k : | f ′ ( θ + j i α ) | ( B i + 1) | h ′ ( θ + j i α ) | ( B i + 1) | h ′ ( x s C ) | ǫq s h ( 12 q s ) for i ∈ J s . (38)and similarly | f ′ ( θ + l i α ) | ǫq s h ( 12 q s ) for i ∈ L s . (39)On the other hand, by l’Hospital’s rule (( A i + ǫ ) − ( B i − ǫ )) h ( 12 q s ) > f ( a i + 12 q s ) − f ( a i − q s ) > (( A i − ǫ ) − ( B i + ǫ )) h ( 12 q s ) . (40) | f ′ ( a i + 12 q s ) | + | f ′ ( a i − q s ) | (( A i + 1) + ( B i + 1)) | h ′ ( 12 q s ) | ǫq s h ( 12 q s ) , (41)(by x s C < q s ).Now, using (35)-(41), we get q s h (cid:18) q s (cid:19) k X i =1 ( A i − B i ) ! − kǫ ! | f ′ ( q s ) ( θ ) | q s h (cid:18) q s (cid:19) k X i =1 ( A i − B i ) ! + 6 kǫ ! . which allows us to conclude if we assume WLOG that ε is sufficiently small.We will assume in the sequel that s > s ′ of Lemma 4.7. Lemma 4.8.
Let x, y ∈ T satisfy (24) . Assume x, y satisfy (29) with m = s , then there exists n ∈ { , ..., max( q s +1 Cq s , } such that (25) holds. Moreover, n q s h (cid:18) q s (cid:19) d + 1) d k x − y k . (42) If x, y satisfy satisfy (30) with m = s then there exists n ∈ { , ..., max( q s +1 Cq s , } such that (26) holds. for some n ∈ { , ..., max( q s +1 Cq s , } satisfying (42) . Proof.
We will use repeteadly Lemma 4.7 for x, y replaced by x + rq s α, y + rq s α respectively, r = 0 , , ..., max([ q s +1 Cq s ] − , (note that by (29) the points x + rq s , y + rq s satisfy the assumptionsof Lemma 4.7). If we fix R max([ q s +1 Cq s ] , , then using (34) for r = 0 , , ..., R , summing upthe obtained inequalities and using the cocycle identity, we obtain R k x − y k ( d + 1) q s h (cid:18) q s (cid:19) > f ( Rq s ) ( x ) − f ( Rq s ) ( y ) > R k x − y k dq s h (cid:18) q s (cid:19) . (43)Let e R := f ( Rq s ) ( x ) − f ( Rq s ) ( y ) . Then e R +1 − e R = f ( q s ) ( x + Rq s α ) − f ( q s ) ( y + Rq s α ) , so in viewof (43), (24), we obtain | e R +1 − e R | < d + 1 for R = 0 , ..., max([ q s +1 Cq s ] − , . (44)Moreover, by (43) and (24), e max([ qs +18 Cqs ] , > d max( q s +1 Cq s , q s h ( 12 q s ) k x − y k > dm max (cid:18) C − q s q s +1 , q s q s +1 (cid:19) > dm C . (45)Therefore, by (44), (45) and (22), there exists n ∈ { , ..., max([ q s +1 Cq s ] , } such that f ( n q s ) ( x ) − f ( n q s ) ( y ) = e n ∈ P Moreover, by (22) and (43) (for R = n ), n q s h (cid:18) q s (cid:19) d + 1) d k x − y k . (46)In case (30) is satisfied instead of (29), we show (26) in exactly the same fashion as we did for(25).We are ready now to finish the proof of Part a. of Proposition 4.5. If s / ∈ K α , then by the factthat x, y ∈ Z ⊂ B s , it follows that 1. in Lemma 4.6 is satisfied with m = s . If s ∈ K α then 2. inLemma 4.6 is satisfied with m = s . Therefore we can use Lemma 4.6 for x, y and m = s . Now, byLemma 4.7, if (29) holds we have (25), if (30) holds we have (26). Part a. of Proposition 4.5 issettled, we turn now to Part b. Let x, y ∈ T satisfy (29) for some m > s , then for every N ∪ { } ∋ l max( q m +1 Cq m − , for every n = 0 , ..., ( l + 1) q m , | f ( n ) ( x ) − f ( n ) ( y ) | < kH k x − y k ( l + 1) q m h ( 12 q m ) (47) If x, y ∈ T satisfy (30) for some m > s , thenfor every n = 0 , ..., ( l + 1) q m , | f ( − n ) ( x ) − f ( − n ) ( y ) | < kH k x − y k ( l + 1) q m h ( 12 q m ) . (48)9 Proof.
We only give the proof of the first case since the other is similar. For every n = 0 , ..., max( q m +1 C , q m ) , there exists θ n ∈ [ x, y ] such that | f ( n ) ( x ) − f ( n ) ( y ) = | f ′ ( n ) ( θ n ) |k x − y k .Therefore, using Lemma 4.3, for every n = 0 , ..., max( q m +1 C , q m ) , we have (cid:12)(cid:12)(cid:12) f ( n ) ( x ) − f ( n ) ( y ) (cid:12)(cid:12)(cid:12) H k x − y k k X i =1 − h ′ ( n ) ( θ n − a i ) − h ′ ( n ) ( a i − θ n ) ! . (49)Moreover, by monotonicity of h ′ , for every i = 1 , ..., k , − h ′ ( n ) ( θ n − a i ) − h ′ ( n ) ( x − a i ) and − h ′ ( n ) ( a i − θ n ) − h ′ ( n ) ( a i − y ) . Since − h ′ is positive, we get that − h ′ ( n ) ( θ n − a i ) − h ′ (( l +1) q m ) ( x − a i ) and − h ′ ( n ) ( a i − θ n ) < − h ′ (( l +1) q m ) ( a i − y ) . (50)It follows by Lemma 4.2, (29) and (23) that for every u = 0 , ..., l k x − y k| h ′ ( q m ) ( T uq m x − a i ) | k x − y k (cid:18) q m h ( 12 q m ) − h ′ ( 12 q m ) − h ′ ( v m ) (cid:19) k x − y k q m h ( 12 q m ) . Hence, summing up over u = 0 , ..., l , and using the cocycle identity, (49) implies (47).This finishesthe proof.To prove Proposition 4.5 Part b. , observe first that if s is sufficiently large, and up to even-tually changing κ to κ ′ = κ C , one of two possibilities holds : . There exists s m s , m ∈ K α , such that κn q s q m Cκn q s , or . There exist s m s and l > such that lq m κn q s ( l + 1) q m q m +1 C . Case 1. κn q s q m Cκn q s with s m s , m ∈ K α . Lemma 4.6 implies that either (29) or(30) holds for T n q s x, T n q s y, m . We then apply Lemma 4.9 to T n q s x, T n q s y, m with l = 0 , andaccording to whether we have (47) or (48) we will get A. or B. of Proposition 4.5 Part b. Indeed,suppose (47) holds then for n = 1 , . . . , [ κn q s ] + 1 , we have due to (46) | f ( n ) ( x ) − f ( n ) ( y ) | < kH k x − y k q m h ( 12 q m ) < CkHκn q s k x − y k h ( 12 q m ) < CkHκ d + 1) d < ε. Case 2.
There exist s m s and l > such that lq m κn q s ( l + 1) q m q m +1 C .If m ∈ K α , then Lemma 4.6 implies that either (29) or (30) holds for T n q s x, T n q s y, m .If m / ∈ K α , then we will first prove that T n q s x, T n q s y, m satisfy the hypothesis of Lemma 4.9.Due to Lemma 4.6, we just have to check (28) for T n q s x, T n q s y, m : q m − [ j =0 T j [ T n q s x, T n q s y ] ∩ k [ i =1 [ − v m + a i , a i + 2 v m ] = ∅ . (51)Indeed, let i and r be such that ρ ( { T n q s x + jα } q m − j =0 , { a i } ki =1 ) = k T n q s x + i α − a r k . It followsby (3), that for i = j ∈ { , ..., q m − } , T n q s x + jα / ∈ k [ i =1 [ − Cq m + a i , a i + 12 Cq m ] . (52)0Next, by the fact that m / ∈ K α and x ∈ B m ( m > s ), we get that k x + i α − a r k > v m , andtherefore k x + i α + n q s α − a r k > k x + i α − a r k − k n q s α k > v m − n q s +1 > v m − Cq s > v m , (recall that n q s +1 Cq s ) and (51) is thus proved ( k T n q s x − T n q s y k (24) < v m ).We are now able to apply Lemma 4.9 to T n q s x, T n q s y, m with l such that lq m κn q s ( l + 1) q m . Now and as in case 1., if (47) holds we get A., if (48) holds we get B. We may assume WLOG that A k + B k > . Let C k = max( A k , − B k ) > Let us define P := [ − Hk ( D + 2) , − C k D c ] ∪ [ C k D c , Hk ( D + 2) , ] where H is from Lemma 4.3, and c is such that for every s ∈ N q s +1 cq s . Fix C k D c > ǫ > and N ∈ N . Let κ := κ ( ǫ ) = ǫ D +2) kCH . We now use Lemma 3.1 to f and T to get, N = N ( ǫ/ , κ ) and a set A := A ( ǫ/ , κ ) with λ ( A ) > − ǫ such that (16) holds for x ∈ A . Denote for every s ∈ N , x s = min i =1 ,...,v x i,s (note that, by definition, x s > q s h ( qs ) ). Let s ∈ N be such that forevery i = 1 , ..., v and T ∋ x q s , g i ( x ) h ( x ) < ǫ kH ( D + 1) and g ′ i ( x i,s ) h ′ ( q s ) < ǫ min( 112 Hk ( D + 1) , for every s > s , (53) P + ∞ s > s q s x i,s < ǫ k for every i = 1 , . . . , v , h ( 12 q s ) > C, for s > s , (54)and for every i = 1 , ...k (cid:12)(cid:12)(cid:12)(cid:12) f ′ ( x ) h ′ ( x − a i ) (cid:12)(cid:12)(cid:12)(cid:12) > A i for x ∈ [ a i , a i + 1 q s − ] and (cid:12)(cid:12)(cid:12)(cid:12) f ′ ( x ) h ′ ( a i − x ) (cid:12)(cid:12)(cid:12)(cid:12) > B i , for x ∈ [ − q s − + a i , a i ] . (55)Define D s := { x ∈ T : x − q s α, ..., x, ..., x + ( q s − α / ∈ ( S vi =1 ( − x s + a i , a i + x s ) } ) . It followsthat λ ( B s ) > − vq s x s > − kq s x s . Define Z ′ := T s > s B s . It follows that λ ( Z ′ ) > − k P s > s q s x s > − ǫ . Now define Z := A ∩ Z ′ , λ ( Z ) > − ǫ .Let s ′ > s be such that q s ′ − > max { κ N, N } . Define δ := q s ′ h ( qs ′ ) .The following proposition implies Theorem 3 due to Proposition 3.3. Proposition 4.10.
Consider x, y ∈ Z with < k x − y k < δ . Then there exists p ∈ P , M, L > κM > N such that either (17) or (18) holds for n ∈ [ M, M + L ] . We can assume WLOG that x < y . Let s := s ( x, y ) be unique such that q s +1 h ( q s +1 ) k x − y k < q s h ( q s ) . (56)As in the precedent section, Proposition 4.10 follows from1 Proposition 4.11.
Consider x, y ∈ Z as in (56) . Part a.
There exists i ∈ { , ..., q s − − } , such that | f ( i ) ( x ) − f ( i ) ( y ) | ∈ P (57) or | f ( − i ) ( x ) − f ( − i ) ( y ) | ∈ P. (58) Part b.
Let X = T i x and Y = T i y if (57) holds, and X = T − i − x and Y = T − i − y if (58) holds, for n = 1 , ..., [ κi ] + 1 the following holdsA. | f ( n ) ( X ) − f ( n ) ( Y ) | < ǫ or B. | f ( − n ) ( X ) − f ( − n ) ( Y ) | < ǫ. (59)The rest of this section is devoted to the proof of Proposition 4.11.Consider the orbit x − q s − α, ..., x, ..., x + ( q s − − α (the length of this orbit is smaller than q s ). It follows by (3) that there exists at most one t s ∈ [ − q s − , q s − + 1] such that x + t s α ∈ S ki = v +1 [ − Cq s + a i , a i + Cq s ] . Hence at least one of the following two holds : q s − − [ j =0 T j [ x, y ] ∩ k [ i = v +1 [ − Cq s + a i , a i + 12 Cq s ] = ∅ (60)or q s − [ j =1 T − j [ x, y ] ∩ k [ i = v +1 [ − Cq s + a i , a i + 12 Cq s ] = ∅ . (61)The following Lemma directly implies the proof of Proposition 4.11. Lemma 4.12. If (60) then (57) and (59) hold. If (61) then (58) and (59) hold.Proof. We will suppose (60) holds, the proof of the other case being analogous.
Lemmata 4.13.
For n = 0 , ..., q s − − , | f ( n ) ( x ) − f ( n ) ( y ) | k (3 D + 2) . Proof.
By the choice of x, y ∈ Z and (56) we have for every i = 1 , . . . , k , a i / ∈ [ x + jα, y + jα ] with j ∈ { , . . . , q s − − } . It follows that for n = 0 , ..., q s − − , (cid:12)(cid:12) f ( n ) ( x ) − f ( n ) ( y ) (cid:12)(cid:12) = | f ′ ( θ n ) | k x − y k ,for some θ n ∈ [ x, y ] . Hence, using Lemma 4.3 and (4), for every n = 0 , ..., q s − we have (cid:12)(cid:12)(cid:12) f ( n ) ( x ) − f ( n ) ( y ) (cid:12)(cid:12)(cid:12) H k x − y k v X i =1 ( − g ′ ( n ) i ( θ n − a i ) − g ′ ( n ) i ( a i − θ n )) + k X i = v +1 ( − h ′ ( n ) ( θ n − a i ) − h ′ ( n ) ( a i − θ n )) ! . (62)Let φ stand for g i , i = 1 , ..., v and h . By the monotonicity of φ ′ on (0 , we obtain − φ ′ ( n ) ( θ n − a i ) − φ ′ ( n ) ( x − a i ) , − φ ′ ( n ) ( a i − θ n ) − φ ′ ( n ) ( a i − y ) . Since − φ ′ > , − φ ′ ( n ) ( x − a i ) − φ ′ ( n ) ( a i − y ) − φ ′ ( q s − ) ( x − a i ) − φ ′ ( q s − ) ( a i − y ) . (63)Using Lemma 4.2 (applied to x − a i , where j i ∈ [0 , q s − ] − is unique such that x + j i α ∈ [ a i , a i + q s − ] ), we obtain k x − y k (cid:16) − φ ′ ( q s − ) ( x − a i ) (cid:17) k x − y k (cid:18) q s − φ ( 12 q s − ) − φ ′ ( 12 q s − ) − φ ′ ( x + j i α ) (cid:19) . i ∈ E . It follows that for n = 0 , ..., q s − − we have x + nα, y + nα / ∈ S ki = v +1 [ − Cq s + a i , a i + Cq s ] , because by (54) and (56), k x − y k < Cq s and x + nα / ∈ S ki = v +1 [ − Cq s + a i , a i + Cq s ] .In this case, bo monotonicity of h ′ , − h ′ ( x + j i α ) < − h ′ ( Cq s ) and therefore by (5) and (56) k x − y k (cid:16) − h ′ ( q s − ) ( x − a i ) (cid:17) k x − y k q s − h ( 12 q s − ) − h ′ ( 1
12 1 q s − ) − h ′ ( 1 C q s ) ! q s − h ( q s − ) + 2 D q s − h ( q s − ) + D q s h ( q s ) q s h ( q s ) D q s h ( q s ) + q s − h ( q s − ) q s h ( q s ) D + 1 . (64)Similarly (replacing Cq s by Cq s ); we obtain k x − y k (cid:0) − h ′ ( q s − ) ( a i − y ) (cid:1) < D + 1 .For i ∈ F , by the fact that x, y ∈ Z , monotonicity of g ′ i and the choice of s , it follows that − g ′ i ( x + i j α − a i ) − g ′ i ( x i,s ) g ′ i ( x s ) ǫh ′ ( q s ) . Therefore, using (53), we get k x − y k (cid:16) − g ′ ( q s − ) i ( x − a i ) (cid:17) k x − y k q s − g i ( 12 q s − ) − g ′ i ( 1
12 1 q s − ) − ǫh ′ ( 1 C q s ) ! k x − y k ǫq s − h ( 12 q s − ) − ǫh ′ ( 1
12 1 q s − ) − ǫh ′ ( 1 C q s ) ! ǫ (3 D + 1) , (65)in the last inequality we use the last estimation in (64). Similarly we prove k x − y k (cid:16) − g ′ ( q s − ) i ( a i − y ) (cid:17) <ǫ (3 D + 1) . Therefore using (62) and the computations above, for n = 0 , ..., q s − − , | f ( n ) ( x ) − f ( n ) ( y ) | < H ( ǫ v (3 D + 1) + 2( k − v )(3 D + 1)) k (3 D + 2) , by the choice of ǫ . Lemmata 4.14.
There exists i ∈ { , ..., q s − − } , such that | f ( i ) ( x ) − f ( i ) ( y ) | > A k D c .Proof. Since q s − − q s − > q s − + 1 , there exists i ∈ [ q s − , q s − − such that x + i α ∈ [ a k , a k + q s − ] . We have assumed that A k + B k > . Suppose additionally A k > − B k (if A k − B k thenwe replace [ a k , a k + q s − ] by [ − q s − + a k , a k ] ). We claim that | ( f ( i +1) ( x ) − f ( i +1) ( y )) − ( f ( i ( x ) − f ( i ) ( y )) | > A k D c . Indeed, the LHS of this inequality is equal to | f ( x + i α ) − f ( y + i α ) | = | f ′ ( θ i ) |k x − y k , for some θ i ∈ [ x + i α, y + i α ] . Now, by (56), θ i ∈ [ a k , a k + q s − + q s h ( qs ) ] ⊂ [ a k , a k + q s − ] . By (55),monotonicity of h ′ , (5) twice (for s and s + 1 ) and (56) | f ′ ( θ i ) | > A k | h ′ ( θ i − a k ) | > A k | h ′ ( 2 q s − ) | > A k | h ′ (2 c q s ) | > A k D q s h ( 12 q s ) > A k D q s +1 c h ( 12 q s +1 ) > A k D c k x − y k ; (66)and the claim follows. Therefore, one of the numbers | f ( i +1) ( x ) − f ( i +1) ( y ) | or | f ( i ) ( x ) − f ( i ) ( y ) | is at least A k D c .As a consequence of the above lemmas, we obtain that at least one of the numbers f ( i +1) ( x ) − f ( i +1) ( y ) f ( i ) ( x ) − f ( i ) ( y ) belongs to the set P , and (57) is proved. The next result is the proofof (59).3 Lemmata 4.15.
The following hold: | f ( n ) ( T i +1 x ) − f ( n ) ( T i +1 y ) | < ǫ for all n κ ( i + 1) , (67) | f ( − n ) ( T i x ) − f ( − n ) ( T i y ) | < ǫ for all n κ ( i + 1) . (68) Proof.
First we show (67). Select (a unique) m ∈ N such that q m > κ ( i + 1) > q m − . By (3)applied to T i ( x ) , by the choice of i it follows that { T i x, ..., T i x + ( q m − α } ∩ k [ i = v +1 [ − Cq m + a i , a i + 12 Cq m ] = { T i x } . Therefore, using the same arguments which lead (62) we obtain (cf. (63)) for n = 0 , ..., κ ( i + 1) (cid:12)(cid:12)(cid:12) f ( n ) ( T i x ) − f ( n ) ( T i y ) (cid:12)(cid:12)(cid:12) H k x − y k ( v X i =1 ( − g ′ ( q m ) i ( T i +1 x − a i ) − g ′ ( q m ) i ( a i − T i +1 y ))+ k X i = v +1 − h ′ ( q m ) ( T i +1 x − a i ) − h ′ ( q m ) ( a i − T i +1 y )) . (69)Then for i ∈ E , again by repeating that lead to (64) we obtain k x − y k (cid:16) − h ′ ( q m ) ( T i +1 x − a i ) (cid:17) q m h ( q m ) + 3 D q m h ( q m ) q s h ( q s ) (3 D + 1) q m h ( q m ) q s h ( q s ) . But q m cκ ( i + 1) < cκq s − , thus (by the monotonicity of h ) k x − y k (cid:0) − h ′ ( q m ) ( T i +1 x − a i ) (cid:1) (3 D + 1) cκq s − q s = ǫ Hk , by the definition of κ . Similarly (replacing Cq m by Cq m ), we obtain k x − y k (cid:0) − h ′ ( q m ) ( a i − T i +1 y ) (cid:1) < ǫ Hk .If i ∈ F , then using monotonicity of g ′ , the choice of i and m , the fact that x, y ∈ Z and (56),we get − g ′ ( q m ) i ( T i +1 x − a i ) − g ′ ( q s − ) i ( x − a i ) and − g ′ ( q m ) i ( a i − T i +1 y ) − g ′ ( q s − ) i ( a i − y ) . Weproceed, repeating what lead to (65) (with q s − instead of q s − ) and using (56) and (53) k x − y k (cid:16) − g ′ ( q s − ) i ( x − a i ) (cid:17) q s − g ′ i ( q s − ) − g ′ i ( q s − ) − g ′ i ( x s ) q s h ( q s ) ǫ kH . Similarly k x − y k (cid:16) − g ′ ( q s − ) i ( a i − x ) (cid:17) < ǫ kH . Using this and (69) we get | f ( n ) ( T i +1 x ) − f ( n ) ( T i +1 y ) | < ǫ , which yields the first case of (67). To handle the second case,notice that { T i x − ( q m − α, ..., T i x } ∩ k [ i = v +1 [ − Cq m + a i , a i + 12 Cq m ] = { T i x } . We now proceed as before to obtain first | f ( − n ) ( T i x ) − f ( − n ) ( T i y ) | = k x − y k (cid:12)(cid:12) f ′ ( n ) ( θ n ) (cid:12)(cid:12) with θ n ∈ [ T i x − ( n − i ) α, T i y − ( n − i ) α ] and then estimating above by H k x − y k ( v X i =1 ( − g ′ ( q m ) i ( T i x − ( q m − α − a i ) − g ′ ( q m ) i ( a i − ( T i y − ( q m − α )))+ k X i = v +1 − h ′ ( q m ) ( T i x − ( q m − α − a i ) − h ′ ( q m ) ( a i − ( T i y − ( q m − α ))) . (70)We conclude exactly in the same way as in the first case.4We proceed to the proof of Lemma 4.12 in the case (60) is satisfied. If f ( i +1) ( x ) − f ( i +1) ( y ) ∈ P , then (67) gives A. in (59); if f ( i ) ( x ) − f ( i ) ( y ) ∈ P , then (68) gives B. in (59). The proof ofLemma 4.12 is thus completed since the case where (61) is satisfied is analogous.This finishes the proof of Proposition 4.11, thus of Theorem 3. In this section, we will prove Theorem 1. Let f be as in Theorem 1; for simplicity we assume that R T f = 1 . Let c > be such that for every s ∈ N , q s +1 cq s . Recall that C > is a constant fromDefinition 1.3 (such a constant exists, since k = 1 in our case); we may assume that C > c .Fix any compact P ⊂ R \ { } . We will prove that for any t ∈ R , ( T ft ) t ∈ R does not have R ( t , P ) property. For simplicity of the notations we will assume that t = 1 . Let d > c − γ be such that P ⊂ (cid:20) − | γ | d , − cd (cid:21) ∪ (cid:20) cd , | γ | d (cid:21) . (71)Let ǫ, κ > sufficiently small, smallness that will be determined in the course of the proof. We useLemma 3.1 for T x = x + α , to ǫ, κ to get a set A ⊂ T , λ ( A ) > − ǫ and N ∈ N , such that (16)holds for x ∈ A and n > N . Let N > max (cid:0) N , ǫ κ (cid:1) .We will hereafter assume that ( T ft ) t ∈ R has the R ( t , P ) property (see Definition 2.1) and obtaina contradiction. Thus, assume there exist a set Z ⊂ X f with λ f ( Z ) > − ǫ and < δ < ǫ suchthat for every ( x, s ) , ( y, s ′ ) ∈ Z with d f (( x, s ) , ( y, s ′ )) < δ , there exist M, L > N with LM > κ and p ∈ P such that L (cid:12)(cid:12)(cid:12) { n ∈ [ M, M + L ] : d f ( T fn ( x, s ) , T fn + p ( y, s ′ )) < ǫ } (cid:12)(cid:12)(cid:12) > − ǫ. (72)Consider V := { ( x, s ) ∈ Z : x ∈ A, s < ǫ } . (73)It follows that λ f ( V ) > − ǫ .The contradiction will come from the following two Propositions, the first one of which is aconsequence of (72) and (73). Proposition 5.1.
Let ( x, s ) , ( y, s ′ ) ∈ V with d f (( x, s ) , ( y, s ′ )) < δ . Then there exists an interval I = [ M ′ , M ′ + L ′ ] such that M ′ > N , L ′ M ′ > aκ ( a = a ( t ) > is a constant obtained in Lemma5.6), there exists p ∈ P and there exists m ∈ Z such that k x − y − mα k < ǫ and for every n ∈ [ M ′ , M ′ + L ′ ] , | f ( n ) ( x ) − f ( n + m ) ( y ) − p | < ǫ. (74) Remark 5.2.
For p ∈ P , n ∈ N , two points ( x, s ) , ( y, s ′ ) ∈ V are called p, n -close if d f ( T fn ( x, s ) , T fn + p ( y, s ′ )) < ǫ . Then ( x, s ) , ( y, s ′ ) have the WR-property (see (72)), if there exists atime interval [ M, M + L ] , such that they are p, n -close for a proportion − ǫ of n’s in [ M, M + L ] .In general, the set on which the points are p, n -close, can be any subset of [ M, M + L ] . Proposition5.1 says that in our context the property actually holds on a full interval of integers. This is whathappens also in the original case of horocycle flows, where, once the point are drifted after time R , they stay drifted for time εR .5 Proposition 5.3.
There exists a set W ⊂ T such that λ ( W ) > c ( d ) ( c = c ( d ) > being aconstant depending only on d ), and a number < δ < δ such that for every x ∈ W for every M > N , every k ∈ Z such that k x − ( x + δ ) − kα k < ǫ and every p ∈ P, if I = [ M, M + T ] is such that for every n ∈ I, | f ( n ) ( x ) − f ( n + k ) ( x + δ ) − p | < ǫ then TM < aκ . (75) Remark 5.4.
The points x ∈ W go too close to the singularity under iteration by R α , so thatpoints of the form ( x, s ) , ( x + δ , s ) split far apart before they get separated by a distance in P (Lemma 5.11 below). In other words, these points do not have the ’natural’ WR-property thatconsists of a controlled drift starting from the first time the points split. To make sure these pointscannot display the WR-Property in the future δ is chosen in such a way, that if for large MT fM ( x, s ) , T fM ( x + δ , s ) become close, then nevertheless d f ( T fM ( x, s ) , T fM ( x + δ , s )) ≫ M − γ , andLemma 5.5 then precludes the WR-property (see Lemma 5.10 below).Before we prove these propositions we will see how they imply Theorem 1. Proof of Theorem 1.
Take x ∈ W and s > such that ( x, s ) , ( x + δ , s ) ∈ V × V , which ispossible since the measure of V is arbitrarily close to if ε is sufficiently small. By Propostion 5.3, ( x, s ) , ( x + δ , s ) satisfy (75), hence they don’t satisfy (74), a contradiction. Lemma 5.5.
Let x, y ∈ T and let I be an integer interval such that for every n ∈ I , | f ( n ) ( x ) − f ( n ) ( y ) | < η (where η is a sufficiently small number). Then | I | < cη γ k x − y k − − γ .Proof. We assume that x < y . Let s ∈ N be unique such that q − γs +1 k x − y k < q − γs . (76)Denote I = [ a, b ] ∩ Z with a, b ∈ Z . Then, by the cocycle identity, the fact that a ∈ I , for n ∈ Z ,we have | f ( n ) ( x ) − f ( n ) ( y ) | > | f ( n − a ) ( T a x ) − f ( n − a ) ( T a y ) | − η. (77)Let k ∈ N be unique such that q k +1 > η γ q s +1 c > q k . (78)We will show that there exists n ∈ [0 , q k +1 ] such that | f (( n + a ) − a ) ( T a x ) − f (( n + a ) − a ) ( T a y ) | = | f ( n ) ( T a x ) − f ( n ) ( T a y ) | > η. (79)This, by (77), gives | f ( n + a ) ( x ) − f ( n + a ) ( y ) | > η and therefore n + a / ∈ I . It follows that | I | q k +1 cq k < η γ q s +1 (76) cη γ k x − y k − − γ which completes the proof. Now, we show(79). By (78) and η sufficiently small, we have s > k .Note that there exist n ∈ [0 , q k +1 ) such that T a x + n α ∈ [0 , q k +1 ] . By (76) and the fact that k + 1 s + 1 , we obtain T a y + n α ∈ [0 , q k +1 ] . Therefore (cid:12)(cid:12)(cid:12) ( f ( n +1) ( T a x ) − f ( n +1) ( T a y )) − ( f ( n ) ( T a x ) − f ( n ) ( T a y )) (cid:12)(cid:12)(cid:12) = | f ( T a x + n α ) − f ( T a y + n α ) | = | f ′ ( θ ) |k x − y k , (80)6for some θ ∈ [ T a x + n α, T a y + n α ] ⊂ [0 , q k +1 ] . Thus, by the monotonicity of f ′ and (78) | f ′ ( θ ) |k x − y k > | γ | ( 2 q k +1 ) − γ q − γs +1 = | γ | (cid:18) q k +1 q s +1 (cid:19) − γ > | γ | ( η γ c ) − γ > η, the last inequality by the fact that η is small enough. Therefore at least one of the numbers, | f ( n +1) ( T a x ) − f ( n +1) ( T a y ) | , | f ( n ) ( T a x ) − f ( n ) ( T a y ) | is bigger than η ; we set n either n , or n + 1 to obtain (79).The following lemma translates (72) into a property on the Birkhoff sums above R α of theceiling function f . Lemma 5.6.
Let ( x, s ) , ( y, s ′ ) ∈ V with d f (( x, s ) , ( y, s ′ )) < δ . There exist M , L > N with L M > κ such that L (cid:12)(cid:12)(cid:12) { r ∈ [ M , M + L ] : ∃ m r ∈ Z , s.t | x − y − m r α | < ǫ and | f ( r ) ( x ) − f ( r + m r ) ( y ) − p | < ǫ } (cid:12)(cid:12)(cid:12) > a. (81) Proof.
Assume WLOG that x < y . Let n ∈ [ M, M + L ] and r n be unique such that f ( r n ) ( x ) n + s < f ( r n +1) ( x ) . We will show that κ − κ M > r M >
11 + κ M − . (82)Indeed, first we show that r M > N . Indeed, if not, using Lemma 3.1 to N (we have N > N ) M M + s < f ( r M +1) ( x ) < f ( N ) ( x ) < (1 + κ ) N < N, a contradiction. Secondly, by the fact that ( x, s ) ∈ V (hence s < ǫ ) and r M > N , using Lemma3.1 to r M and the definition of N ( N > ǫ κ ), we get ( M > N ) (1 − κ ) r M < f ( r M ) ( x ) M + s < (1 + κ ) M and M M + s < f ( r M +1) ( x ) (1 + κ )( r M + 1) . Now, 82 follows.Analogously we prove that
11 + κ ( M + L ) − r M + L κ − κ ( M + L ) . (83)Set M := r M , L = r M + L − r M . It is easy to prove using (82) and (83) that M , L > N and L M > κ . Moreover, since f > c γ > , there exists a constant a = a ( t , γ ) > such that for every n ∈ [ M, M + L ] , | r n +1 − r n | a . It follows that the number of different r n ∈ [ M , M + L ] is atleast aL .Let n ∈ [ M, M + L ] be such that d f ( T fn ( x, s ) , T fn + p ( y, s ′ )) < ǫ . By (72), there are at least (1 − ǫ ) L of such n ∈ [ M, M + L ] . By the definition of d f and T f , there exist r n ∈ [ M , M + L ] and m n ∈ N such that | ( x + r n α ) − ( y + m n α ) | < ǫ and | f ( r n ) ( x ) − f ( m n ) ( y ) − p | < ǫ. We set m r = m r ( n ) := m n − r n ∈ Z to get | x − y − m r α | < ǫ and | f ( r n ) ( x ) − f ( r n + m r ) ( y ) − p | < ǫ .It follows that the number of different r n ∈ [ M , M + L ] is at least a (1 − ǫ ) L and hence (81)follows.7 Proof of Proposition 5.1.
Denote by U := { r ∈ [ M , M + L ] : ∃ m r | x − y − m r α | < ǫ and | f ( r ) ( x ) − f ( r + m r ) ( y ) − p | < ǫ } . It follows by (81) that | U | > aL . Let us choose in the integer interval [ M , M + L ] disjointsubintervals I = [ a , b ] , ..., I l = [ a l , b l ] such that U = I ∪ ... ∪ I l and for every i = 1 , ..., l thereexists m i ∈ Z such that | x − y − m i α | < ǫ and for r ∈ I i , | f ( r ) ( x ) − f ( r + m i ) ( y ) − p | < ǫ . Moreoverwe assume that for every i = 1 , ..., l , I i is maximal in the sense that | f ( h i ) ( x ) − f ( h i + m i ) ( y ) − p | > ǫ for h i = a i − , b i + 1 .We will show that there exists i = 1 , ..., l such that | I i | > aL . (84)This will obviously finish the proof of (74) with M ′ = a i , L ′ = | I i | , and m = m i ∈ Z .Let us show (84). If l there is nothing to prove. Assume l > .Notice that U is the set of n ′ s such that ( x, s ) , ( y, s ′ ) ∈ V are p, n -close. The next lemma impliesthat between any two disjoint integer intervals I j , I j +1 ⊂ U , on which ( x, s ) , ( y, s ′ ) are p, n -close,there will be an integer interval J j much longer than I j , such that for any n ∈ J j , ( x, s ) , ( y, s ′ ) arenot p, n -close. Lemmata 5.7.
Let i ∈ { , ..., l − } . There exist an interval [ c i , d i ] = J i ⊂ [ M , M + L ] suchthat for any r ∈ J i , C > | f ( r ) ( x ) − f ( r + m i ) ( y ) − p | > ǫ , c i > b i − , d i < a i +1 and | J i | > | I i | C ǫ γ (here C > commes from (3) ). Lemmata 5.7 will give (84). Indeed, by the definition of J i and I i , it follows that for i, j =2 , ..., l − with j = i − , i, i + 1 J i ∩ I i = ∅ and J i ∩ J j = ∅ . Hence, P l − i =2 | J i | L , and | I ∪ ... ∪ I l − | < C ǫ γ L < aL Therefore, by the fact that | U | > aL , we have | I ∪ I l | > aL and consequently, | I w | > aL forat least one of w = 0 or w = 1 .Hence to obtain (84) we just need to prove Lemma 5.7. Lemma 5.7.
Let v ∈ N be unique such that q − γv +1 < k x − ( y + m i α ) k q − γv . (85)Consider n ∈ I i = [ a i , b i ] . We have ǫ > | f ( n ) ( x ) − f ( n + m i ) ( y ) − p | = (cid:12)(cid:12)(cid:12) ( f ( a i ) ( x ) − f ( a i + m i ) ( y ) − p ) + ( f ( n − a i ) ( T a i x ) − f ( n − a i ) ( T a i + m i y )) (cid:12)(cid:12)(cid:12) > (cid:12)(cid:12)(cid:12) | f ( n − a i ) ( T a i x ) − f ( n − a i ) ( T a i + m i y ) | − ǫ (cid:12)(cid:12)(cid:12) . (86)8Hence, for n ∈ I i , | f ( n − a i ) ( T a i x ) − f ( n − a i ) ( T a i + m i y ) | < ǫ . It follows now by Lemma 5.5 appliedto η = 4 ǫ , the points T a i x, T a i + m i y and s = v that | I i | < ǫ ) γ k x − ( y + m i α ) k − − γ ǫ ) γ q v +1 . (87)Consider integer intervals K i = [ a i − q v − , a i ] and L i = [ a i , a i + q v − ] . It follows by (3) with k = 1 and for the point T a i x , similarly to the proof of Theorem 3, that there exist at most one t ∈ K i ∪ L i such that T a i x + t α ∈ [ − Cq v , Cq v ] . Assume t < . Then we consider L i . Moreover,we may assume that ǫ γ − γ > cC , and therefore, using (85) we obtain Cq v q − γv . It follows thatfor n ∈ [0 , q v − ] , / ∈ [ T a i + n x, T a i + m i + n y ] . Hence, n → sign ( f ( T a i + n x ) − f ( T a i + m i + n x )) is constant for n = 0 , ..., q v − , (88)(it may happen that T m i y < x ). It follows that | f ( n ) ( T a i x ) − f ( n ) ( T a i + m i y ) | q v − n =0 is increasing.Hence, for q v − > n > b i + 1 we have f ( n − a i ) ( T a i x ) − f ( n − a i ) ( T a i + m i y ) > ǫ. Moreover, by Lemma 4.2 (the RHS of the inequality) to h = f , x = θ (where f ( q v − ) ( T a i x ) − f ( q v − ) ( T a i + m i y ) = f ′ ( q v − ) ( θ ) k x − y − m i α k ) and s = v − , we obtain | f ( n ) ( T a i x ) − f ( n ) ( T a i + m i y ) | < | f ( q v − ) ( T a i x ) − f ( q v − ) ( T a i + m i y ) | C + 4 < C , (if necessery, to get the last inequality, we consider a bigger C ). We set J i = [ b i + 1 , a i + q v − ] ( b i a i + 2(4 ǫ ) γ q v +1 , by (87)). It follows that for n ∈ J i , we have by cocycle identity ǫ = 4 ǫ − ǫ < | f ( n − a i ) ( T a i x ) − f ( n − a i ) ( T a i + m i y ) | − | f ( a i ) ( x ) − f ( a i + m i ) ( y ) − p | | f ( n ) ( x ) − f ( n + m i ) ( y ) − p | | f ( a i ) ( x ) − f ( a i + m i ) ( y ) − p | + | f ( n − a i ) ( T a i x ) − f ( n − a i ) ( T a i + m i y ) | ǫ + 2 C < C . (89)Now, by (87) d i − c i = | J i | > q v − − ǫ ) γ q v +1 = 2(4 ǫ ) γ q v +1 (cid:18) q v − ǫ ) γ q v +1 − (cid:19) > | I i | C ǫ γ , (90)since q v − q v +1 > c > C and ǫ is small enough.Suppose n ∈ J i ∩ I i +1 . If m i +1 = m i , then immediately we have a contradiction. If m i +1 = m i then | f ( n ) ( x ) − f ( n + m i +1 ) ( y ) − p | = | ( f ( n ) ( x ) − f ( n + m i ) ( y ) − p ) + f ( m i +1 − m i ) ( T n + m i ) y > | m i +1 − m i | inf T f − C > ǫ, (91)since | m i +1 − m i | is of order ǫ because both k x − y − m i +1 α k and k x − y − m i α k are close to zero.Hence d i < a i +1 and Lemma 5.7 has been proved.This finishes the proof of Proposition 5.1.9 Lemmata 5.8.
Fix a number < ζ | γ | . For every v ∈ N , v > v and v sufficiently large, thereexists < δ v < δ , satisfying δ v ∈ [ 1 q − γv , cq − γv ] , (92) such that for every N ∋ | k | > ε − , k δ v − kα k > | k | ζ . (93) Lemmata 5.9.
For every w > w , w sufficiently large, there exists a set W w ⊂ A ∩ ( A − δ ) , (94) with λ ( W w ) > c ( c ( d ) will be specified in the proof ), such that the following holds for x ∈ W w and y := x + δ w : | f ( n ) ( x ) − f ( n ) ( y ) | < dc for every n = 0 , ..., q w − (95) there exists i ∈ { , ..., q w − l − } such that x + i α ∈ [ 12 dq w , dq w ] , (96) for some N ∋ l > depending on w , to be specified later, < f ( n ) ( x ) − f ( n ) ( y ) < cd for n i and f ( n ) ( x ) − f ( n ) ( y ) > | γ | d for i < n < wq w . (97)Before we proof the above Lemmas, let us first show, how they imply Proposition 5.3.Let w ∈ N be such that w > κ . (98)Denote W := W w and δ := δ w . By definition, δ ∈ [ 1 q − γw , cq − γw ] . (99) Lemma 5.10.
We have that (75) holds for k = 0 .Proof. Fix any M > N , any p ∈ P and any k = 0 such that k x − y − kα k < ǫ .Let I = [ M, M + R ] be such that for n ∈ [ M, M + R ] , we have | f ( n ) ( x ) − f ( n + k ) ( y ) − p | < ǫ .Note that since k x − y − kα k < ε and k x − y k < δ < ε , then | k | > c ǫ . By the cocycle identityand the triangle inequality, this implies that for every n ∈ [0 , R ] | f ( n ) ( x + M α ) − f ( n ) ( y + M α + kα ) | ǫ. Therefore, be Lemma 5.5,
R < c (4 ε ) γ k x − y + kα k − − γ (93) c (4 ε ) γ | k | ζ − γ | k | ζ − γ . (100)0On the other hand, since x ∈ W and | f ( n ) ( x ) − f ( n + k ) ( y ) − p | < ǫ we get that κM > | f ( M ) ( x ) − f ( M ) ( y ) | > | f ( k ) ( T M y ) |− ǫ − p > | k | inf T f − ǫ − p > | k | inf T f > inf T f c ε . (101)Hence, and using (95) ( ǫ is small enough), M > max( q w − , | k | ) . (102)Hence, by (100) and (102), RM | k | ζ − γ max( q w − , | k | ) q γ − ζ − γ w − (98) < aκ . This proves (75) for k = 0 . Lemma 5.11.
We have that (75) holds for k = 0 .Proof. Fix any M > N , p ∈ P and let I = [ M, M + R ] be such that for every n ∈ I , | f ( n ) ( x ) − f ( n ) ( y ) − p | < ǫ . By (97) and (71), for every n ∈ [0 , wq w ] and every p ∈ P , | f ( n ) ( x ) − f ( n ) ( y ) − p | > ǫ . Therefore, for M wq w , (75) holds. It follows by Lemma 5.5 applied to x and y (we have k x − y k = δ (99) > q − γw ), that R = | I | < ǫ ) γ k x − y k − − γ ǫ ) γ q w < q w . Therefore, by(98), for M > wq w , RM q w wq w < aκ . So, (75) holds for k = 0 .We thus proved (75) in Proposition 5.3. Let us now complete the proof by proving Lemmatas5.8 and 5.9. Proof of Lemmata 5.8.
Fix v ∈ N . To simplify the notations, we will write δ instead of δ v . Given u ∈ N , set B u := { η ∈ T : d (cid:16) η, { iα } q u − i = − q u (cid:17) > u q u } . (103)Let t ∈ N be unique such that q t +2 < q − γv q t +1 . (104)Let c = 4 c , then q t +1 > cq t − c (since t depends on v which is sufficiently large, t − c > v − by (104)). (cid:20) q t +1 , q t +1 (cid:21) ∩ \ i > t − c B i . (105)We will show below that this set is not empty. Now, let δ be any number in this set. This, by thedefinition of B i , will give (92) and (93). Indeed, (92) follows from (104) and (105). To show (93),note that for | k | < q t − c , k δ − kα k > k kα k − δ (92) , (104) > c | k | − q t +1 > c | k | − cq t − c > | k | ζ , | k | > ε − . If | k | > q t − c , let ℓ be unique, such that q ℓ +1 > | k | > q ℓ . By definition of B ℓ +1 k δ − kα k > sup | i | q ℓ +1 k δ − iα k > ℓ + 1) q ℓ +1 > | k | ζ , where the last inequality follows if v is sufficiently large (then by (104), t is large, so | k | is largeand therefore ℓ is large). Lemmata 5.12. (cid:20) q t +1 , q t +1 (cid:21) ∩ \ i > t − c B i = ∅ . (106) Proof.
The proof goes by induction. We will show that for every k > t − c there exists a closed in-terval E k ⊂ h q t +1 , q t +1 i ∩ (cid:16)T ki > t − c B i (cid:17) and E k +1 ⊂ E k . Moreover, we will show (for the inductionpurpose) that for every k > t − c , | E k | > c c q k . Indeed, we have min − q t − c +1 i q t − c − k iα k > q t − c > q t +1 . Set E t − c := [ q t +1 , q t +1 − t − c ) q t − c ] . It follows by the fact that t − c > v − and v is sufficiently large (by taking a bigger c , we may assume that c > ) that | E t − c | = 1 q t +1 − t − c ) q t − c > c c +2 q t − c (cid:18) c − c (cid:19) > c c +2 q t − c . Moreover, since t − c is sufficiently large, q t +1 > t − c ) q t − c . Moreover, c > and therefore q t +1 − t − c ) q t − c min i ∈{− q t − c ,...,q t − c − } k iα k − t − c ) q t − c . Therefore, by definition of E t − c and B t − c , E t − c ⊂ B t − c = t − c \ i = t − c B i . Suppose that for some k > t − c we have a closed interval E k ⊂ h q t +1 , q t +1 i ∩ (cid:16)T ki = t − c B i (cid:17) such that | E k | > c c q k . It follows that E k ∩ { iα } q k − i = − q k = ∅ . Let E k +1 ⊂ E k be the longest closed subinterval (in E k ) such that E k +1 ∩ B ck +1 = ∅ . (107)It follows that E k +1 ⊂ E k ⊂ h q t +1 , q t +1 i , and by (107), E k +1 ⊂ (cid:16)T k +1 i = t − c B i (cid:17) ( E k +1 ⊂ E k ). Itremains to prove that | E k +1 | > c c q k +1 . To do this note that | E k +1 | > | E k | [4 | E k | q k +1 + 1] − k + 1) q k +1 . Indeed, | E k ∩{ iα } q k +1 − − q k +1 | | E k | q k +1 and around each point of the form iα , i = − q k +1 , ..., q k +1 − ,we discard an interval of length k +1) q k +1 (see (103), for u = k + 1 ). We use the inductionassumption, the fact that k + 1 > t − c > v − (and v is sufficiently large) to obtain | E k | [4 | E k | q k +1 + 1] − k + 1) q k +1 > c c +2 q k +1 . Hence (106) is proved.2The proof of Lemmata 5.8 is thus finished.
Proof of Lemmata 5.9.
We will determine the set W w . To simplify notation, we will write W instead of W w , the dependence on w will be clear from the context. Let l ∈ N be such that (cid:18) q w − l q w (cid:19) − γ +1 d < (cid:18) q w − l +1 q w (cid:19) − γ +1 , (108)since w is large, l > . Set W , := { x ∈ T : { x, x + α, ..., x + ( wq w − α } ∩ [ − cq − γw , cq − γw ] = ∅} ,W , := { x ∈ T : { x, x + α, ..., x + ( q w − l − α } ∩ [ 12 dq w , dq w ] = ∅} . We have λ ( W , ) > − wq w cq − γw = 1 − cwq − γw .As l > , for i = 0 , ..., q w − l − , the sets T i [ dq w , dq w ] are pairwise disjoint. Therefore, by (108) λ ( W , ) = q w − l dq w > c q w − l +1 q w d > c (cid:18) d (cid:19) − γ d = 12 c (cid:18) d (cid:19) − γ − γ . Now we set W := W , ∩ W , . (109)Since w > w is sufficiently large, λ ( W ) > c , where c = c ( d ) > . We may assume that W ⊂ A ∩ ( A − δ ) , if not we take W := W ∩ A ∩ ( A − δ ) and use the fact that λ ( A ) > − ǫ , and δ < δ is small.This gives (94). Note that by (109) and the definition of W , , (96) follows. Let us show (95).By (96), we have { x, x + α, ..., x + ( q w − − α } ∩ [0 , dq w ] = ∅ . (110)Therefore, using (99) and k x − y k = δ , we have for every i = 0 , ..., q w − − T i ([ x, y ]) ∩ [0 , dq w ] = ∅ . (111)By (111), for i = 0 , ..., q w − − , / ∈ [ x + iα, y + iα ] , and therefore | f ( i ) ( x ) − f ( i ) ( y ) | | f ( q w − ) ( x ) − f ( q w − ) ( y ) | . Therefore and by (111), (99), monotonicity of f ′ , (111), Lemma 4.2 (to h = f ′ , andsome θ ∈ [ x, y ] ), it follows that for n q w − | f ( n ) ( x ) − f ( n ) ( y ) | | f ( q w − ) ( x ) − f ( q w − ) ( y ) | k x − y k| f ′ ( q w − ) ( θ ) | cq − γw dq − γw dc, (112)and (95) follows.Now we will show (97).Moreover, for x ∈ W ⊂ W , , and y := x + δ , / ∈ [ x + iα, y + iα ] for i = 0 , ..., wq w − f ( n ) ( x ) − f ( n ) ( y )) wq w − n =0 is an increasing sequence. (113)3By the fact that i < q w − l and (96) it follows that k x + jα − k > q w − l for j = i , j q w − l − . Moreover, since w > w is large enough, k x − y k = δ cq − γw < q w , and we obtain k x + jα k , k y + jα k > q w − l for j = 0 , ..., q w − l − , j = i . (114)Moreover, by (113), for n i < f ( n ) ( x ) − f ( n ) ( y ) ( f ( q w − l ) ( x ) − f ( q w − l ) ( y )) − ( f ( x + i α ) − f ( y + i α )) . (115)Let us consider ¯ f ( x ) = (cid:26) f ( x ) if x > q w − l ;0 if otherwise.Hence, by (114) f ( q w − l ) ( x ) − f ( x + i α ) = ¯ f ( q w − l ) ( x ) and f ( q w − l ) ( y ) − f ( y + i α ) = ¯ f ( q w − l ) ( y ) . (116)By / ∈ [ x + iα, y + iα ] for i = 0 , ..., wq w − there exists θ ∈ [ x, y ] such that ¯ f ( q w − l ) ( x ) − ¯ f ( q w − l ) ( y ) = k x − y k| ¯ f ′ ( q w − l ) ( θ ) | . But as in Lemma 4.2 we get the following: | ¯ f ′ ( q w − l ) ( θ ) | q w − l (2 q w − l ) − γ + 2 | γ | (2 q w − l ) − γ + | γ | (6 q w − l ) − γ q − γw − l . Therefore, using (115) and (116), (99) and (108), for every n i , we have < f ( n ) ( x ) − f ( n ) ( y ) k x − y k| ¯ f ′ ( q w − l ) ( θ ) | q − γw − l cq − γw < cd . For wq w > n > i , by (113) and monotonicity of f ′ , we have for some θ ∈ [ x + i α, y + i α ] ( ) , ( ) ⊂ [0 , dq w ] , and f ( n ) ( x ) − f ( n ) ( y ) > f ( x + i α ) − f ( y + i α ) = k x − y k| f ′ ( θ ) | > q − γw | γ | ( dq w − γ > | γ | d . This finishes the proof of (97).The proof of Lemmata 5.9 is complete.This finishes the proof of Theorem 1.
A Proof of Proposition 1.7
Consider the Gauss map T : [0 , → [0 , , T x := (cid:8) x (cid:9) , T (0) = 0 , and let µ be its invariantprobability measure given by its density with respect to the Lebesgue measure x dx . Lemma A.1.
There exists a constant
C > such that for every a ∈ T and for every k = l ∈ N µ (cid:0) T − k ((0 , a )) ∩ T − l ((0 , a )) (cid:1) Cµ ((0 , a )) . (117)4 Proof.
Assume that l > k . Then ( µ is T -invariant) µ (cid:0) T − k ((0 , a )) ∩ T − l ((0 , a )) (cid:1) = µ (cid:0) (0 , a ) ∩ T k − l (0 , a ) (cid:1) . Note that T k − l +1 (0 , a ) = S + ∞ i =1 ( c i , d i ) for some disjoint intervals ( c i , d i ) i = 1 , ..., + ∞ . We willprove that for every i ∈ N µ ( T − ( c i , d i ) ∩ (0 , a )) Cµ (( c i , d i )) µ ((0 , a )) (118)which implies (117) since µ (cid:0) (0 , a ) ∩ T k − l (0 , a ) (cid:1) Cµ ((0 , a )) µ ( + ∞ [ i =1 ( c i , d i )) = Cµ ( T k − l +1 (0 , a )) µ ((0 , a )) = Cµ ((0 , a )) . To prove (118), note that T − ( c i , d i ) = S + ∞ j =1 ( d i + j , c i + j ) . It follows that + ∞ X j =1 ( 1 d i + j , c i + j ) ∩ (0 , a ) ⊂ [ j > a − d i ( 1 d i + j , c i + j ) . Therefore, µ ( T − ( c i , d i ) ∩ (0 , a )) X j > a − d i µ (( 1 d i + j , c i + j )) C X j > a − d i c i − d i ( c i + j )( d i + j ) Cµ ((0 , a )) µ (( c i , d i )) for some constant C > (since the density function f ( x ) = x is bounded from above and belowon [0 , ). This completes the proof. Proposition A.2.
Let d > and set A := n x = [0; a , ... ] : ∃ N = N ( x ) ∀ n > N (cid:12)(cid:12)(cid:12) { k ∈ [ n , ( n + 1) ] : a k > dk } (cid:12)(cid:12)(cid:12) < o . Then λ ( A ) = 1 .Proof. We will prove that λ ( A c ) = 0 . To do this we will prove that µ ( A c ) = 0 ( λ and µ areequivalent.). Note that for k ∈ N if x = [0; a , ..., ] is the continued fraction of x , then T k ( x ) = a k + ak +1+ ··· . Therefore B ⊂ A , where B := (cid:26) x ∈ T : ∃ N = N ( x ) ∀ n > N (cid:12)(cid:12)(cid:12)(cid:12) { k ∈ [ n , ( n + 1) ] : T k x dk } (cid:12)(cid:12)(cid:12)(cid:12) < (cid:27) . We will prove that µ ( B c ) = 0 . To do this note that B c = + ∞ \ N =1 [ n > N B n , (119)where B n := n x ∈ T : (cid:12)(cid:12)(cid:12) { k ∈ [ n , ( n + 1) ] : T k x dk (cid:12)(cid:12)(cid:12) > o . Moreover, B n ⊂ [ i = i ∈ [ n , ( n +1) ] B ni ,i , (120)5where B ni ,i := { x ∈ T : T i x, T i x ∈ (0 , dn ] } . Let us note that B ni ,i = T − i ((0 , dn )) ∩ T − i ((0 , dn )) . By (117) from Lemma A.1, we get that µ ( B ni ,i ) Cn − . Therefore, using (120) and summingup over all i = i ∈ [ n , ( n + 1) ] , we get that µ ( B n ) Cn − . This and (119) yield µ ( B c ) = lim N → + ∞ µ [ n > N B n = 0 . This finishes the proof.
Lemma A.3.
Let α ∈ A . Then P s/ ∈ K α q s < + ∞ .Proof. Let N := N ( α ) be the number resulting from the fact that α ∈ A . We will prove that P s/ ∈ K α ,s > N ( q s ) < + ∞ . There exists a constant d > such that for any s ∈ N log( q s ) > (2 d ) s (121)(indeed, the sequence ( q s ) + ∞ s =1 grows exponentially fast). Let s / ∈ K α , s > N . Then a s +1 q s + q s − = q s +1 > q s log q s , and therefore, for s / ∈ K α , by (121) a s +1 > (ln( q s )) − > ds . (122)Since α ∈ A , for every k > in every interval of the form [( N + k ) , ( N + k + 1) ] there is atmost one s such that a s +1 > ds > ( d ( N + k ) ) . Therefore X s/ ∈ K α ,s > N q s d X s/ ∈ K α ,s > N s d X k > N + k ) = 2 d X k > N + k ) < + ∞ . This finishes the proof.Hence we proved that
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