Dynamical obstruction to the existence of continuous sub-actions for interval maps with regularly varying property
DDynamical obstructionto the existence of continuous sub-actionsfor interval maps with regularly varying property
Eduardo Garibaldi ∗ Department of Mathematics, University of Campinas, 13083-859 Campinas, Brazil(email: [email protected])
Irene Inoquio-Renteria † ICFM, Universidad Austral de Chile, casilla 567 Valdivia, Chile(email: [email protected])
January 23, 2019
Abstract
In ergodic optimization theory, the existence of sub-actions is an importanttool in the study of the so-called optimizing measures. For transformationswith regularly varying property, we highlight a class of moduli of continuitywhich is not compatible with the existence of continuous sub-actions. Ourresult relies fundamentally on the local behavior of the dynamics near a fixedpoint and applies to interval maps that are expanding outside an neutral fixedpoint, including Manneville-Pomeau and Farey maps.
Keywords: ergodic optimization, sub-actions, modulus of continuity, non-uniformly expanding dynamics, intermittent maps.
Mathematical subject classification:
Let T : X → X be a continuous surjective map on a compact metric space X .Suppose that f : X → R is a continuous function (called potential ). Let M ( X, T )denote the set of T -invariant Borel probability measures on X . As usual the max-imum ergodic average is defined as m ( f, T ) := max µ ∈ M ( X,T ) (cid:90) f dµ. ∗ Supported by CNPq grant 304792/2017-9. † Supported by FONDECYT 11130341 and BCH-CONICYT postdoctoral fellowship 74170014. a r X i v : . [ m a t h . D S ] J a n Given a potential f : X → R , a function u : X → R is said to be a sub-action for f if it satisfies the cohomological inequality f + u − u ◦ T ≤ m ( f, T ) . The existence of sub-actions for a potential f plays an important role in the study ofmeasures µ in M ( X, T ) that maximize (or minimize) the average (cid:82) X f dµ . The studyof these measures gave rise to the ergodic optimization (see [Jen06, Jen18, Gar17]and references therein).The existence of continuous sub-actions is guaranteed when the map is uniformlyexpanding and the potentials have H¨older modulus of continuity (see [CLT01] forthe context of expanding transformations of the circle). For related studies onthe existence of sub-actions, see [LT03, LT05, LRR07, GLT09], and see also [Sou03,Bra07, Bra08, Mor09] for results in one-dimensional dynamics.For transitive expanding dynamics, generic continuous potentials do not admitbounded measurable sub-actions (see [BJ02, Theorem C] and for details [Gar17,Appendix]). Surprisingly there are few cases in the literature about specific ex-amples of non-existence of continuous sub-actions. An example is provided byMorris [Mor07, Proposition 2] in the context of shift spaces.Our theorem highlights a dynamical obstruction on the existence of continuoussub-actions. It seems that Morris [Mor09] was the first to notice this kind ofphenomenon. Although our result holds for interval dynamics, we are convincedthat such an obstruction must occur in a similar way for multidimensional settings.Precisely, we deal with interval maps with a regularly varying property and weidentify an associated class of moduli of continuity whose members do not alwaysadmit continuous sub-actions. We present our theorem in the following subsection.In the Appendix, we address the natural question of the existence of sub-actions insuch a setting. Let [0 ,
1] be endowed with the standard metric on R . Our dynamical setting willbe interval maps T : [0 , → [0 , , defined for x close enough to 0 as an invertiblefunction of the form T ( x ) := x (1 ± V ( x )), where for some σ >
0, the continuousand increasing function V : [0 , + ∞ ) → (0 ,
1) satisfieslim x → V ( tx ) V ( x ) = t σ , for all t > . (1)The function V is said to be regularly varying at with index σ .By a modulus of continuity , we mean a continuous and non-decreasing function ω : [0 , + ∞ ) → [0 , + ∞ ) satisfying lim (cid:15) → ω ( (cid:15) ) = ω (0) = 0. Let M denote the familyof concave modulus of continuity. For a given ω ∈ M , we denote by C ω ([0 , ϕ : [0 , → R with a multiple of ω as modulus of continuity: | ϕ ( x ) − ϕ ( y ) | ≤ Cω ( d ( x, y )) for some constant C >
0, for all x, y ∈ [0 , Theorem 1.
Let T : [0 , → [0 , be an interval map such that, for x close to , T is invertible and has the form T ( x ) := x (1 ± V ( x )) , where the continuous and increasing function V : [0 , + ∞ ) → (0 , is regularly varying at with index σ > .Suppose that ω ∈ M satisfies lim inf x → ω ( x ) V ( x ) > . (2) Then there exists a function f ∈ C ω ([0 , , with m ( f, T ) = (cid:82) f dδ = f (0) , thatdoes not admit continuous sub-action. The main novelty of our general result is the clear identification of condition (2)as an obstruction to the existence of continuous sub-actions. An immediate questionis whether the opposite condition, that is, a null limit inferior would be sufficientto ensure existence. As a complement of discussion as well as an initial answer,we provide an example of existence result in the Appendix (see Theorem 2): byconsidering certain maps with an indifferent fixed point and a stronger assumptionthan a null limit inferior, we show that sub-actions do exist and we highlight theirassociated regularity.In the following Subsection, we give examples of applications of Theorem 1.We gather in Section 2 preliminary results. In Section 3, we present the proof ofTheorem 1. In the Appendix, we detail the context and the proof of an existenceresult inspired by a construction due to Contreras, Lopes and Thieullen [CLT01,Proposition 11].
A trivial example of elements of M are the functions ω ( h ) = Ch α with α ∈ (0 , α -H¨older continuous functions. The family M also includesthe minimal concave majorants ω of non-decreasing subadditive functions ω :[0 , + ∞ ) → [0 , + ∞ ), with lim h → ω ( h ) = ω (0) = 0 . Following [Med01] these con-cave majorants are infinitely differentiable on (0 , + ∞ ). Moreover, if ω (cid:48) (0) < ∞ then ω ( h ) = ω (cid:48) (0) h on some neighborhood of 0.Another example of members of M are the functions ω ( h ) = h (cid:0) log (cid:0) h k (cid:1) + 1 (cid:1) (for k > h small enough), which describe locally H¨older continuous functions. Amore general class of modulus of continuity in M is defined as follows: for 0 ≤ α < β ≥ α + β >
0, consider ω α,β : [0 , + ∞ ) → [0 , + ∞ ) given as ω α,β ( h ) := (cid:26) h α ( − log h ) − β , < h < h ,h α ( − log h ) − β , h ≥ h , (3)where h is taken small enough so that ω α,β is concave. Note that ω α, is reducedto the H¨older continuity, and ω ,β for β > Remark 1.
Let ω α,β : [0 , + ∞ ) → [0 , + ∞ ) be the modulus of continuity definedin (3). It is easy to see that for every (cid:15) > α , lim h → ω α,β ( h ) h (cid:15) = + ∞ . (4) Note that M includes many functions besides the previous examples for thesimple fact that for each pair ω , ω ∈ M , we have ω ◦ ω ∈ M . However, weare interested in a class of modulus of continuity whose behavior near 0 satisfiescondition (2), which is dictated by the dynamics.Let V : [0 , + ∞ ) → (0 ,
1) be a continuous and increasing function which is regu-larly varying at 0 with index σ >
0. Consider the modulus of continuity ω α,β definedin (3) with 0 ≤ α < min { σ, } and β ≥ α + β >
0. Thanks to prop-erty (4), the condition lim inf x → ω α,β ( x ) V ( x ) > x → x σ V ( x ) > Corollary 1.
Let T : [0 , → [0 , be an interval map such that in a neighbor-hood of the origin T is invertible and has the form T ( x ) = x (1 ± V ( x )) , where V : [0 , + ∞ ) → (0 , is a continuous, increasing and regularly varying functionat with index σ > that satisfies lim inf x → x σ V ( x ) > . Let ω α,β ( x ) be defined asin (3). Then, for α = σ and β = 0 or for ≤ α < min { σ, } and β ≥ with α + β > , there is a function f ∈ C ω α,β ([0 , which does not admit continuoussub-action. Examples of this kind of dynamics include Manneville-Pomeau interval map:for a given s > T s : [0 , → [0 ,
1] is defined as T s ( x ) := x (1 + x s ) mod 1 . Note that T (cid:48) s ( x ) ≥ x with equality only at x = 0. Let c be the unique pointin (0 ,
1) such that T s ( c ) = 1 and T s | [0 ,c ] : [0 , c ] → [0 ,
1] is a diffeomorphism. Let usdenote U s : [0 , → [0 , c ] the corresponding inverse branch. Note that U (cid:48) s ( x ) ≤ x and U s is concave, so that cx ≤ U s ( x ) ≤ x . If we write U s ( x ) = x (1 − V ( x )),then 0 ≤ V ( x ) ≤ − c. Moreover, by using the identity T s ◦ U s = Id, we have V ( x ) = x s (1 − V ( x )) s +1 for all x (cid:54) = 0 . Hence lim x → V ( x ) = 0,lim x → V ( tx ) V ( x ) = lim x → t s (cid:18) − V ( tx )1 − V ( x ) (cid:19) s +1 = t s and lim x → x s V ( x ) = lim x → − V ( x )) s +1 = 1 . It is not difficult to argue that V is increasing. Then Corollary 1 applies to U s aswell. Corollary 2.
Let s ∈ (0 , and T s ( x ) = x + x s for x close enough to . Denote U s the corresponding inverse branch. Let ω α,β ( x ) be defined as in (3), where either α ∈ [0 , min { s, } ) and β ≥ with α + β > or α = s and β = 0 . Then thereare functions f, g ∈ C ω α,β ([0 , which do not admit continuous sub-actions withrespect to T s and U s , respectively. The above corollary is an extension of Morris’ result [Mor09], which establishedthat for T s ( x ) = x + x s mod 1, there is f ∈ C ω s, ([0 , ,
1] with indifferentfixed point at x = 0 is defined as follows: for ρ ∈ (0 , F ρ : [0 , → [0 ,
1] begiven as F ρ ( x ) = (cid:40) x (1 − x ρ ) /ρ if 0 ≤ x ≤ − /ρ (1 − x ρ ) /ρ x if 2 − /ρ < x ≤ . Note that Farey map corresponds to the special case ρ = 1. For any ρ ∈ (0 , G ρ ( x ) = x (1+ x ρ ) /ρ . Note thenthat the functions V ( x ) = − x ρ ) /ρ − W ( x ) = 1 − x ρ ) /ρ are continuous,increasing, regularly varying with index ρ , and satisfy lim x → x ρ V ( x ) = lim x → x ρ W ( x ) = ρ > . Clearly, F ρ ( x ) = x (1 + V ( x )) and G ρ ( x ) = x (1 − W ( x )). Corollary 3.
For ρ ∈ (0 , , let F ρ ( x ) = x (1 − x ρ ) /ρ and G ρ ( x ) = x (1+ x ρ ) /ρ for x close to . Let ω α,β ( x ) be defined as in (3), where either α ∈ [0 , ρ ) and β ≥ with α + β > or α = ρ and β = 0 . Then there are functions f, g ∈ C ω α,β ([0 , whichdo not admit continuous sub-actions with respect to F ρ and G ρ , respectively. As a final example of application of our theorem, let T ( x ) = x = 0 x + x | log x | if 0 < x ≤ / x − / < x ≤ . Note that V ( x ) = x | log x | , x >
0, is a regularly varying function with index 1.For k >
0, the concave modulus of continuity defined for h sufficiently small as ω ( h ) = h (cid:0) log (cid:0) h k (cid:1) + 1 (cid:1) clearly satisfies lim x → ω ( x ) V ( x ) = k log 2 >
0. Recalling thatsuch a modulus describes locally H¨older continuous functions, we have the followingresult.
Corollary 4.
With respect to a dynamics that behaves as T ( x ) = x + x | log x | for x > sufficiently small, there exist locally H¨older continuous functions that donot admit continuous sub-actions. Recall that M denotes the family of concave modulus of continuity. Note that,given a non-identically null ω ∈ M , then ([0 , , ω ◦ d ) is a metric space. Indeed, thesubadditivity of ω follows from its concavity and thus, since ω is non-decreasing,we obtain the triangle inequality: ω ( d ( x, y )) ≤ ω ( d ( x, z )) + ω ( d ( z, y )) ∀ x, y, z ∈ [0 , . In particular, a function ϕ : [0 , → R with modulus of continuity ω ∈ M is nothingelse than a Lipschitz function with respect to the metric ω ◦ d. We will use the following property.
Lemma 1.
Let ω ∈ M . For any positive constant χ, we have χ χ ω ( h ) ≤ ω ( χh ) ≤ ( χ + 1) ω ( h ) . Proof.
Since ω is subadditive, we have for all positive integer n ≥ ω ( nh ) ≤ nω ( h ).For a positive constant χ , by monotonicity of ω , we see that ω ( χh ) ≤ ω ( (cid:100) χ (cid:101) h ) ≤ (cid:100) χ (cid:101) ω ( h ) ≤ ( χ + 1) ω ( h ) , where (cid:100)·(cid:101) denotes the ceiling function. Then, we also obtain ω ( χh ) ≥ χ + 1 ω ( h ) = χ χ ω ( h ) . Given σ >
0, a measurable function V : [0 , + ∞ ) → (0 , + ∞ ) is said to be regularlyvarying at with index σ if condition (1) holds. A regularly varying functioncan be represented in the form V ( x ) = x σ V ( x ), where the function V satisfieslim x → V ( tx ) V ( x ) = 1, for all t >
0. Similarly a measurable function V : [0 , + ∞ ) → (0 , + ∞ ) is regularly varying at ∞ with index σ ∈ R if the function x (cid:55)→ V ( x )is regularly varying at 0 . For properties of regularly varying functions, we referto [Sen76] and [Aar97]. See also [Kar33] for details concerning the original literature.Recall that near to origin the dynamics is supposed invertible and defined as T ( x ) = x (1 ± V ( x )). Let ( w n ) + ∞ n =0 ⊂ [0 ,
1] be a sequence of points obtained bychoosing w close enough to 0 and by defining w n +1 = T ∓ ( w n ), n ≥
0. In clearterms, for x (cid:55)→ x (1+ V ( x )) we take pre-images, and for x (cid:55)→ x (1 − V ( x )) we considerfuture iterates. Note that in both cases w n → n → ∞ . A sequence of iterationtimes will also play a central role in our construction. More precisely, let ( n k ) k ≥ be an increasing sequence of positive integers such that for some γ ∈ (0 , k →∞ n k n k +1 = γ. (5)The study of the behavior close to 0 can be done in a similar way for both x (cid:55)→ x (1 + V ( x )) and x (cid:55)→ x (1 − V ( x )). From now on in this subsection, we look at thecase T ( x ) = x (1 − V ( x )). We will point out in the end similarities and particularitiesto the other caseWe write α j ∼ β j whenever α j β j → j → ∞ . The next lemma summarizes themain properties concerning the asymptotic behavior of the sequences ( w n = T ( w n − ))and ( n k ). Lemma 2.
The following properties hold(i) w n ∼ σ /σ b ( n ) , where b − ( x ) := 1 V ( x ) ; (6) (ii) d ( w n , w n +1 ) ∼ σ /σ nb ( n ) ; (7) (iii) n k n k +1 ∼ γ /σ b ( n k +1 ) b ( n k ) . (8) Proof.
To verify Part (iii), we first note that b − ( tx ) b − ( x ) = V (1 /x ) V (1 /tx ) → /t ) σ = t σ as x → ∞ , which means that b − is regularly varying at ∞ with index σ . Hence, itsinverse, the increasing function b , is regularly varying at ∞ with index 1 /σ (fordetails, see [Sen76]).We set b ( y ) = y /σ B ( y ), where lim y →∞ B ( ty ) B ( y ) = 1, for every t > . The function B has the following representation (for a proof, see [Sen76, Theorem 1 . Y > Y, ∞ ) → R , ε : [ Y, ∞ ) → ( − σ , σ ), withΘ( y ) → θ ∈ R + as y → ∞ and ε ( t ) → t → ∞ , such that B ( y ) = Θ( y ) e (cid:82) yY ε ( t ) t dt ∀ y ≥ Y. Then log B ( n k ) B ( n k +1 ) = log Θ( n k )Θ( n k +1 ) + (cid:90) n k n k +1 ε ( t ) t dt and (cid:0) sup [ n k , + ∞ ) ε (cid:1) log n k n k +1 ≤ (cid:90) n k n k +1 ε ( t ) t dt ≤ (cid:0) inf [ n k , + ∞ ) ε (cid:1) log n k n k +1 ensure that B ( n k ) B ( n k +1 ) → k → + ∞ . Therefore n k b ( n k ) n k +1 b ( n k +1 ) = (cid:16) n k n k +1 (cid:17) /σ B ( n k ) B ( n k +1 ) → γ /σ as k → ∞ . Part (i) follows from [Aar97, Lemma 4.8.6] which is deduced using that b − (cid:16) w n (cid:17) ∼ nσ. (9)The asymptotic equivalence (9) implies that V ( w n ) = 1 /b − (cid:0) w n (cid:1) ∼ nσ , so it followsthat d ( w n , w n +1 ) = w n V ( w n ) ∼ σ /σ nb ( n ) and therefore Part (ii) holds. Remark 2.
Since b is a continuous and increasing function and since we considerthe standard metric on R , by the asymptotic equivalence (7), there exists a constant C > such that for every i ≤ j , ( j − i ) C − σ /σ j b ( j ) ≤ d ( w i , w j ) ≤ ( j − i ) C σ /σ i b ( i ) . (10)The next lemma provides us estimates on the cardinality of future iterates thatstay within suitable intervals. Lemma 3.
Let us consider ( w n k ) + ∞ k =1 a subsequence of ( w n ) + ∞ n =0 , where ( n k ) k ≥ is anincreasing sequence satisfying (5) and T n k − n k − ( w n k − ) = w n k . For k ≥ , denote R k := 13 C n k − b ( n k − ) n k b ( n k ) d ( w n k , w n k − ) . Then, for z ∈ [ w n k + R k , w n k − ] and k large enough, (cid:8) ≤ j < n k − n k − : R k ≤ d ( T j ( z ) , w n k ) ≤ d ( w n k , w n k − ) (cid:9) ≥≥ C n k − b ( n k − ) d ( w n k , w n k − ) , where C := ( C − − C − ) σ /σ > . In particular, there is C > such that, for k sufficiently large, (cid:8) ≤ j < n k − n k − : R k ≤ d ( T j ( w n k − ) , w n k ) ≤ d ( w n k , w n k − ) (cid:9) ≥ C V ( w n k ) . Proof.
Let (cid:96) ≥ w n k − + (cid:96) < z ≤ w n k − +( (cid:96) − . Note that a nonnegativeinteger j such that R k ≤ d ( w n k − + (cid:96) + j , w n k ) and d ( w n k − +( (cid:96) − j , w n k ) ≤ d ( w n k , w n k − ) (11)belongs to (cid:8) j : R k ≤ d ( T j ( z ) , w n k ) ≤ d ( w n k , w n k − ) (cid:9) . Moreover, thanks to (10),any j ≥ R k ≤ ( n k − n k − − (cid:96) − j ) C − σ /σ n k b ( n k ) and( n k − n k − − ( (cid:96) − − j ) C σ /σ n k − b ( n k − ) ≤ d ( w n k , w n k − ) (12)satisfies (11). Denoting κ := n k − n k − − (cid:96) , there are exactly (cid:98) κ − C σ /σ n k b ( n k ) R k (cid:99) − (cid:100) κ + 1 − C − σ /σ n k − b ( n k − ) d ( w n k , w n k − ) (cid:101) + 1nonnegative integers j that fulfill (12). Therefore, we have (cid:8) j : R k ≤ d ( T j ( z ) , w n k ) ≤ d ( w n k , w n k − ) (cid:9) ≥≥ C − σ /σ n k − b ( n k − ) d ( w n k , w n k − ) − C σ /σ n k b ( n k ) R k −
2= 13 ( C − − C − ) σ /σ n k − b ( n k − ) d ( w n k , w n k − ) − . Note that, from Remark 2 and Lemma 2, as k → ∞ σ /σ n k − b ( n k − ) d ( w n k , w n k − ) ≥ C − n k (cid:16) − n k − n k (cid:17) n k − b ( n k − ) n k b ( n k ) → ∞ . Hence, ignoring at most finitely many initial terms of ( n k ) if necessary, we obtain (cid:8) j : R k ≤ d ( T j ( z ) , w n k ) ≤ d ( w n k , w n k − ) (cid:9) ≥ C n k − b ( n k − ) d ( w n k , w n k − ) . In particular, for z = w n k − , from (10) we have d ( w n k , w n k − ) σ (cid:8) j : R k ≤ d ( T j ( w n k − ) , w n k ) ≤ d ( w n k , w n k − ) (cid:9) ≥≥ C d ( w n k , w n k − ) σ +1 n k − b ( n k − ) ≥ C (cid:104) ( n k − n k − ) C − σ /σ n k b ( n k ) (cid:105) σ +1 n k − b ( n k − )= C C σ +10 σ ( σ +1) /σ (cid:16) − n k − n k (cid:17) σ +1 n k − b ( n k − ) n k b ( n k ) n k b ( n k ) σ . Note now that, from (6) and (9), nb ( n ) σ ∼ σnw σn ∼ w σn V ( w n ) . Denote thus C (cid:48) := C C σ +10 σ ( σ +1)2 /σ (1 − γ ) σ +1 γ /σ >
0. Following the previousestimate and the above asymptotic equivalence, from (5) and (8), for k large enough, (cid:8) j : R k ≤ d ( T j ( w n k − ) , w n k ) ≤ d ( w n k − , w n k ) (cid:9) ≥ C (cid:48) V ( w n k ) w σn k d ( w n k , w n k − ) σ . Note now that, from Remark 2 and Lemma 2, for k sufficiently large, d ( w n k , w n k − ) ≤ (cid:18) − n k − n k (cid:19) C σ n k b ( n k ) n k − b ( n k − ) 1 σ /σ b ( n k ) ≤ − γ ) C σ γ /σ w n k . We obtain thus a constant C (cid:48)(cid:48) > w σnk d ( w nk ,w nk − ) σ ≥ C (cid:48)(cid:48) whenever k is largeenough, which completes the proof with C := C (cid:48) C (cid:48)(cid:48) . Comments on local behavior near to origin for x (cid:55)→ x (1 + V ( x )) . In thiscase, we deal with a sequence of past iterates ( w n = T ( w n +1 )), where T ( x ) = x (1+ V ( x )) in a neighborhood of 0. It is not a surprise that asymptotic equivalencesare exactly the same as in the statement of Lemma 2. One may show easily sucha fact with minor adjustments in the proof and an appropriate version of [Aar97,Lemma 4.8.6], which can be obtained repeating almost verbatim original arguments.The statement of Lemma 3 for this case obviously requires contextual changes sincethe sequences are now related by T n k − n k − ( w n k ) = w n k − . If one follows the samelines of proof, one will conclude that for z ∈ [ w n k , w n k − − R k ] and k large enough, (cid:8) ≤ j < n k − n k − : R k ≤ d ( T j ( z ) , w n k − ) ≤ d ( w n k , w n k − ) (cid:9) ≥≥ C n k − b ( n k − ) d ( w n k , w n k − ) , and in particular for k sufficiently large, (cid:8) ≤ j < n k − n k − : R k ≤ d ( T j ( w n k ) , w n k − ) ≤ d ( w n k , w n k − ) (cid:9) ≥ C V ( w n k ) . (13)0 We will present in details the proof of Theorem 1 when T ( x ) = x (1 − V ( x )) for x close to 0. In the end, we will comment on the small changes of arguments requiredto prove the theorem in the case x (cid:55)→ x (1 + V ( x )). Hence, let ( w n k ) + ∞ k =1 be asubsequence of future iterates ( w n = T n ( w )) + ∞ n =0 , where w ∈ (0 ,
1) is a point closeenough to 0 and ( n k ) k ≥ is an increasing sequence such that lim k → + ∞ n k n k +1 = γ forsome γ ∈ (0 , S := { w n k } + ∞ k =1 ∪ { } . For every k >
1, set I k = (cid:16)
15 (3 w n k + 2 w n k +1 ) ,
15 (3 w n k + 2 w n k − ) (cid:17) and J k = (cid:16)
13 ( w n k + 2 w n k +1 ) ,
13 (2 w n k + w n k +1 ) (cid:17) , and denote Y := ( w n , ∪ (cid:83) k J k . Since { Y, I k ( k > } is an open cover of((0 , , ω ◦ d ), we may consider a partition of unity subordinate to it (see Figure 1).Precisely, let { ϕ Y , ϕ k : ((0 , , ω ◦ d ) → [0 ,
1] ( k > } be a family of Lips-chitz continuous functions such that ϕ Y + (cid:80) k ϕ k = 1, with Supp( ϕ Y ) ⊂ Y andSupp( ϕ k ) ⊂ I k . In particular, ω is a modulus of continuity of ϕ Y and of ϕ k ( k > d − := d ( w n k , w n k − ), d + := d ( w n k , w n k +1 ) J k − J k I k +1 I k I k − w n k w n k +1 w n k − / d + / d − w nk + w nk +1 w nk − + w nk For ξ >
0, defineΦ( x ) := ϕ k ( x ) , x ∈ I k , k = 1 mod 3 − ξϕ k ( x ) , x ∈ I k , k = 2 mod 30 , otherwise , and consider f : [0 , → R given as f ( x ) := Φ( x ) ω ( d ( x, S )) . (14)This function clearly vanishes on S . Moreover, f has ω as modulus of continuity.We will show that, for ξ large enough, f does not admit a continuous sub-action.1We have T m k ( w n k − ) = w n k , where m k := n k − n k − , and S m k f ( w n k − ) = m k − (cid:88) j =0 f (cid:0) T j (cid:0) w n k − (cid:1)(cid:1) = m k − (cid:88) j =0 Φ( w n k − + j ) ω ( d ( w n k − + j , S )) . Recall the definition of R k in the statement of Lemma 3. Note that, for k largeenough, [ w n k , w n k + R k ) ⊂ (cid:2) w n k , (2 w n k + w n k − ) (cid:1) ⊂ I k . Besides, by construction ϕ k ≡ (cid:2) (2 w n k + w n k +1 ) , (2 w n k + w n k − ) (cid:3) . Therefore, if k = 1 mod 3 issufficiently large, from Lemma 3 we get S m k f ( w n k − ) ≥ (cid:8) j : R k ≤ d ( w n k − + j , w n k ) ≤ d ( w n k , w n k − ) (cid:9) ω ( R k ) ≥ C V ( w n k ) ω ( R k ) . We will show that for k sufficiently large, ω ( R k ) V ( w nk ) is bounded from below by a positiveconstant. As a matter of fact, by the definition of R k and (8),lim k →∞ R k d ( w n k , w n k − ) = 13 γ /σ C . For C := γ /σ C >
0, using the monotonicity of ω and Lemma 1, we have thatfor a sufficiently large k , ω ( R k ) ≥ C C ω ( d ( w n k , w n k − )) . Moreover, from Remark 2 and Lemma 2, we see that for k sufficiently large, d ( w n k , w n k − ) ≥ C − σ (cid:18) − n k − n k (cid:19) σ /σ b ( n k ) ≥ C − σ (1 − γ ) w n k . Then, for C := C −
10 1 σ (1 − γ ) >
0, we obtain ω ( R k ) V ( w n k ) ≥ C C C C ω ( w n k ) V ( w n k ) . Therefore, thanks to hypothesis (2), we conclude that there exists a constant C > k = 1 mod 3 large enough, S m k f ( w n k − ) > C . We will show in Subsection 3.1 that m ( f, T ) = 0 for ξ large enough. Let usassume this fact for a moment and argue that the inequality f ≤ u ◦ T − u is impossible for every continuous function u : [0 , → R . Suppose the oppositehappens. Then, if k = 1 mod 3 is sufficiently large, we have shown that u ( w n k ) = u (cid:0) T m k (cid:0) w n k − (cid:1)(cid:1) ≥ S m k f ( w n k − ) + u ( w n k − ) > C + u ( w n k − ) . Since u is continuous at 0, by letting k → + ∞ , we get a contradiction.2 m ( f, T ) = 0 It remains to argue that, for ξ large enough, m ( f, T ) = 0. Since f (0) = 0 and δ is T -invariant, clearly m ( f, T ) ≥ (cid:82) f dδ = f (0) = 0. If ξ is sufficiently large, by choosinga suitable constant γ ∈ (0 ,
1) and an appropriate initial point w close enough to 0,we will show that for each x there is n ( x ) such that S n ( x ) f ( x ) ≤
0. From Birkhoff’sergodic theorem, we thus conclude that m ( f, T ) ≤
0, which completes the proof.We first choose γ ∈ (0 ,
1) satisfying γ /σ > . (15)Note now that, replacing w by w n with n large enough, we may assume that theconstant C in Remark 2 is as close as we want to 1. Thus, we suppose henceforththat 1 < C ≤ γ /σ . (16)Furthermore, thanks to (8), if n is sufficiently large, we may also assume that12 γ /σ ≤ n k b ( n k ) n k +1 b ( n k +1 ) ∀ k ≥ . (17)If x ∈ [0 , \ (cid:91) k =1 mod 3 I k , just take n ( x ) = 1, since f ( x ) ≤ . Suppose then x ∈ I k for some k = 1 mod 3. Define p ( x ) := min { p ≥ T p ( x ) / ∈ I k } . Note that S p ( x ) f ( x ) ≤ { j ≥ T j ( x ) ∈ I k } ω (cid:16)
25 max { d ( w n k +1 , w n k ) , d ( w n k , w n k − ) } (cid:17) . Let us estimate the cardinality in the right term. Denote L k := (cid:108) C σ /σ n k b ( n k ) d ( w n k , w n k − ) (cid:109) . From Remark 2, we have d ( w n k , w n k − L k ) ≥ L k C −
10 1 σ /σ n k b ( n k ) > d ( w n k , w n k − ),which means that w n k − L k is greater than the right endpoint of I k . Thanks to (15),(16) and (17),37 C σ /σ n k +1 b ( n k +1 ) d ( w n k +1 , w n k ) ≤ C n k +1 b ( n k +1 ) n k b ( n k ) ( n k +1 − n k ) ≤ n k +1 − n k , so that L k +1 ≤ n k +1 − n k . Hence, a similar reasoning shows that w n k + L k +1 is smallerthan the left endpoint of I k . Therefore, by the monotonicity of T , we obtain { j : T j ( x ) ∈ I k } ≤ ( L k −
1) + ( L k +1 − ≤ C σ /σ n k +1 b ( n k +1 ) d ( w n k +1 , w n k − ) . S p ( x ) f ( x ) ≤ C σ /σ n k +1 b ( n k +1 ) d ( w n k +1 , w n k − ) ω (cid:0) d ( w n k +1 , w n k − ) (cid:1) . (18)Now, for y ∈ (cid:2) w n k +1 + R k +1 , (3 w n k + 2 w n k +1 ) (cid:3) , denote q ( y ) := min { q ≥ d ( T q ( y ) , w n k +1 ) < R k +1 } . Clearly, S q ( y ) f ( y ) ≤ − ξ (cid:8) j ≥ R k +1 ≤ d ( T j ( y ) , w n k +1 ) ≤ d ( w n k +1 , w n k ) (cid:9) ω ( R k +1 ) . Thanks to Lemma 3, we obtain that S q ( y ) f ( y ) ≤ − ξ C n k b ( n k ) d ( w n k , w n k +1 ) ω ( R k +1 ) . (19)We claim that, whenever ξ is sufficiently large, for n ( x ) := p ( x ) + q ( T p ( x ) ( x )) onehas S n ( x ) f ( x ) ≤ k n k +1 b ( n k +1 ) d ( w n k +1 , w n k − ) ω (cid:0) d ( w n k +1 , w n k − ) (cid:1) n k b ( n k ) d ( w n k , w n k +1 ) ω ( R k +1 ) < ∞ . Recalling the asymptotic equivalence (8), we just have to show that both supremasup k d ( w n k +1 , w n k − ) d ( w n k , w n k +1 ) and sup k ω (cid:0) d ( w n k +1 , w n k − ) (cid:1) ω ( R k +1 )are finite. With respect to the first one, from (10) it is immediate that d ( w n k , w n k − ) d ( w n k +1 , w n k ) ≤ C ( n k − n k − )1 / (cid:2) σ /σ n k − b ( n k − ) (cid:3) C − ( n k +1 − n k )1 / (cid:2) σ /σ n k +1 b ( n k +1 ) (cid:3) = C − n k − n k n k +1 n k − n k +1 b ( n k +1 ) n k b ( n k ) n k b ( n k ) n k − b ( n k − ) , (20)which ensures d ( w nk +1 ,w nk − ) d ( w nk ,w nk +1 ) = 1 + d ( w nk ,w nk − ) d ( w nk +1 ,w nk ) is bounded from above. Withrespect to the second one, note first that, thanks to (20), d ( w n k +1 , w n k − ) R k +1 = 3 C n k +1 b ( n k +1 ) n k b ( n k ) d ( w n k +1 , w n k − ) d ( w n k , w n k +1 )is bounded from above. Hence, there exists a positive constant C such that d ( w n k +1 , w n k − ) ≤ C R k +1 . By the monotonicity of ω and Lemma 1, we obtain ω (cid:0) d ( w n k +1 , w n k − ) (cid:1) ω ( R k +1 ) ≤ C + 1 < ∞ . The proof is complete.4
Comments on the proof of Theorem 1 for x (cid:55)→ x (1+ V ( x )) . We consider nowa subsequence ( w n k ) that fulfills w n k − = T n k − n k − ( w n k ), where T ( x ) = x (1+ V ( x ))in a neighborhood of 0. Note that orbits are moving monotonically away fromthe origin, that is, they are moving to the right instead of to the left as in theprevious case. This merely produces a, let us say, reflexive effect on our arguments,exchanging the roles of indices k = 1 mod 3 and k = 2 mod 3. In practical terms,we define Φ for this case asΦ( x ) := − ξϕ k ( x ) , x ∈ I k , k = 1 mod 3 ϕ k ( x ) , x ∈ I k , k = 2 mod 30 , otherwise . Introducing f as in (14) and supposing by a moment that m ( f, T ) = 0, we applythe same strategy to show that f does not admit continuous sub-action. In fact, for k = 2 mod 3 sufficiently large, using (13) one estimates the number of iterates thatremain in the interval [ (2 w n k + w n k +1 ) , w n k − R k +1 ] to conclude that S m k +1 f ( w n k +1 )is bounded from below by a positive constant and thus to reach a contradiction.In order to show that, for the same choice of parameters (15), (16), and (17), m ( f, T ) = 0 whenever ξ is sufficiently large, suitable adjustments are required toobtain that for x ∈ I k with k = 2 mod 3, there is n ( x ) such that S n ( x ) f ( x ) ≤ . Similarly to the previous case, the key observation is that such a Birkhoff summay be bounded from above by the difference of two terms, the first one takes intoaccount the iterates that remain in I k , the second one considers iterates that remainin [ (2 w n k − + w n k ) , w n k − − R k ], and their ratio is uniformly bounded. Appendix: On the existence of sub-actions
Since the analysis of the existence of sub-actions is a global issue, we fix a particularclass of dynamics with intermittent behavior. Our working class of maps with twobranches provides an example of situation in which one can guarantee the existenceof sub-actions for potentials with various moduli of continuity, highlighting clearlythe associate regularity of these sub-actions. Similar arguments are feasible forintermittent dynamics with more inverse branches.At the best of our knowledge, there are no previous works at such a level ofgenerality about the regularity of potentials and sub-actions.Throughout this section we consider a class J of one-dimensional maps, sothat each T ∈ J is a piecewise two to one interval map defined on ([0 , , d ) withdiscontinuity c ∈ (0 ,
1) such that lim x → c − T ( x ) = 1 and lim x → c + T ( x ) = 0 . Moreover, T takes the form T ( x ) := x (1 + V ( x )) on [0 , c ] , where for some σ >
0, the continuousand increasing function V : [0 , + ∞ ) → [0 ,
1) is regularly varying with index σ (recall (1)). Finally, we assume that there is λ > x, y ∈ ( c, d ( T ( x ) , T ( y )) ≥ λd ( x, y ) . As in § M denotes the set of continuous, non-decreasing, concave modulusof continuity. For a given function V as above, we consider an appropriate ω ∈ M satisfying the following assumption:5[A] There exist constants γ > ξ > η ∈ (0 ,
1) such that ω ( ξh ) V ( ξh ) ≥ ξ γ ω ( h ) V ( h ) , ∀ h ∈ (0 , η ) , ∀ ξ ∈ (1 , ξ ] . (21)One can easily verify that, for V and ω fulfilling (21),lim h → ω ( h ) V ( h ) = 0 . (22)The converse statement is not satisfied in general, see Remark 3.From Assumption A, we define a modulus of continuity Ω ∈ M so that potentialswith modulus of continuity ω admit sub-actions with modulus of continuity Ω.Before we state this result, we first provide examples of maps in J for whichcondition (21) holds. Examples
A prototypical example in J is the Manneville-Pomeau interval map defined forsome s ∈ (0 ,
1) as T s ( x ) := x (1 + x s ) mod 1 . Consider the class of modulus ofcontinuity ω α,β as in (3). For s < α <
1, condition (21) follows immediately with γ = α − s : for h sufficiently small, ω α,β ( ξh )( ξh ) s ≥ ξ α − s h α ( − log h ) − β h s = ξ α − s ω α,β ( h ) h s . Another interesting family of interval maps in J is given by H ρ : [0 , → [0 , ρ ∈ (0 , H ρ ( x ) = (cid:40) x (1 − x ρ ) /ρ if 0 ≤ x ≤ − /ρ , /ρ x − /ρ − if 2 − /ρ < x ≤ . The function V ( h ) = − h ρ ) /ρ − ρ . For ρ < α <
1, we have that ω α,β and V satisfy condition (21), since ω α,β ( ξh ) V ( ξh ) V ( h ) ω α,β ( h ) = ξ α (cid:18) log( ξh )log h (cid:19) − β V ( h ) V ( ξh )implies that lim h → ω α,β ( ξh ) V ( ξh ) V ( h ) ω α,β ( h ) = ξ α lim h → V ( h ) V ( ξh ) = ξ α − ρ . As another example,following [Hol05], consider a family defined for 0 < τ < θ > T τ,θ ( x ) = (cid:40) x + τ (log 2) θ +1 x τ | log x | θ +1 if 0 ≤ x ≤ / , x − / < x ≤ . In this case, the function V τ,θ ( h ) = τ (log 2) θ +1 h τ | log h | θ +1 is regularly varying withindex τ . Condition (21) is satisfied, for instance, with the modulus of continuity ω k ( h ) = h (cid:0) log (cid:0) h k (cid:1) + 1 (cid:1) for k ≥ h sufficiently small. Indeed, one has ω k ( ξh ) V τ,θ ( ξh ) V τ,θ ( h ) ω k ( h ) = ξ − τ (cid:12)(cid:12)(cid:12)(cid:12) log h log( ξh ) (cid:12)(cid:12)(cid:12)(cid:12) θ +1 − k log( ξh )1 − k log h , h → ω k ( ξh ) V τ,θ ( ξh ) V τ,θ ( h ) ω k ( h ) = ξ − τ lim h → − k log( ξh )1 − k log h = ξ − τ . Remark 3. [Condition (21) is more restricted than (22) .] For θ > and k ≥ , consider T ,θ and ω k as above. It is easy to see that ω k ( h ) V ,θ ( h ) → as h → . However, from ω k ( ξh ) V ,θ ( ξh ) V ,θ ( h ) ω k ( h ) = (cid:12)(cid:12)(cid:12)(cid:12) log h log( ξh ) (cid:12)(cid:12)(cid:12)(cid:12) θ +1 − k log( ξh )1 − k log h , we get lim h → ω k ( ξh ) V ,θ ( ξh ) V ,θ ( h ) ω k ( h ) = 1 . Hence, property (22) is satisfied, however (21) fails.
Defining a continuous increasing concave modulus of continuity
For V and ω fulfilling (21), let ϑ : [0 , ∞ ) → [0 , ∞ ) be the continuous functiondefined as ϑ ( x ) := (cid:40) ω ( x ) V ( x ) , x > , , x = 0 , (23)and let ϑ : [0 , ∞ ) → [0 , ∞ ) be the continuous increasing function given as ϑ ( x ) = max ≤ y ≤ x ϑ ( y ) , ≤ x ≤ , max [0 , ϑ , x ≥ , (24)Denote then ϑ ∗ the concave conjugate Legendre transform of ϑ , defined as ϑ ∗ ( x ) = min y ∈ [0 , ∞ ) [ xy − ϑ ( y )] , ∀ x ≥ . (25)By the very definition, ϑ ∗ is concave, increasing and continuous on (0 , ∞ ) . To seethat ϑ ∗ is continuous at 0, note that ϑ ∗ (0) = − max [0 , ϑ and ϑ ∗ (0) ≤ ϑ ∗ ( (cid:15) ) ≤ (cid:15) − ϑ (1) = (cid:15) + ϑ ∗ (0) . For the continuous concave increasing function ϑ ( x ) = min { ϑ ∗ ( x ) , ϑ ∗ (1) } , (26)a similar reasoning shows that its concave conjugate Legendre transform , ϑ ∗ ( x ) = min y ∈ [0 , ∞ ) [ xy − ϑ ( y )] , ∀ x ≥ , (27)is also a continuous concave increasing function. Moreover ϑ ( x ) ≤ ϑ ( x ) ≤ ϑ ∗ ( x )for all x ∈ [0 , . Actually, ϑ ∗ is the smallest concave function that lies above ϑ on [0 , . Note that ϑ ∗ (0) = − ϑ ∗ (1) . We have obtained a function Ω := ϑ ∗ + ϑ ∗ (1) that belongs to M . Theorem 2.
Let T : [0 , → [0 , be a map in J with discontinuity c ∈ (0 , such that T ( x ) = x (1 + V ( x )) for all x ∈ [0 , c ] , where V is regularly varying at .Let ω be a modulus of continuity in M for which Assumption A holds. Then, every f ∈ C ω ([0 , admits continuous sub-actions in C Ω ([0 , , where Ω is defined bythe process (23) - (27) . Proof of the theorem
In the following results we will assume the hypotheses of Theorem 2. In particular,we keep in mind all the constants of Assumption A.
Lemma 4.
There are constants (cid:37) T > and C ∈ (0 , min { ξ , η − } − such thatfor all x, y ∈ [0 , , with d ( x, y ) < (cid:37) T , we have d ( T ( x ) , T ( y )) ≥ d ( x, y ) (cid:0) C V ( d ( x, y )) (cid:1) . (28) Proof.
Let x, y ∈ [0 , c ] with x < y . Since V and T are increasing, note that d ( T ( x ) , T ( y )) = d ( x, y ) + d ( x, y ) V ( y ) + x ( V ( y ) − V ( x )) ≥ d ( x, y ) (cid:0) V ( d ( x, y )) (cid:1) . Consider now x, y ∈ ( c, V ([0 , ⊂ [0 , d ( T ( x ) , T ( y )) ≥ λ d ( x, y ) ≥ d ( x, y ) (cid:0) λ − V ( d ( x, y )) (cid:1) . Fix (cid:37) > x ∈ [ c − (cid:37)/ , c ) and y ∈ ( c, c + (cid:37)/
2] it follows that d ( T ( x ) , T ( y )) ≥ / . We choose (cid:37) T ∈ (0 , (cid:37) ) such that V (cid:0) h (cid:1) ≥ σ +1 V ( h ) for all h ∈ [0 , (cid:37) T ]. Then for c − (cid:37) T / ≤ x < c < y ≤ c + (cid:37) T / d ( T ( x ) , T ( y )) ≥ − d ( T ( x ) , T ( y )) = lim t → c − d ( T ( t ) , T ( x )) + lim t → c + d ( T ( y ) , T ( t )) ≥ lim t → c − d ( t, x ) (cid:0) V ( d ( t, x )) (cid:1) + lim t → c + d ( y, t ) (cid:0) λ − V ( d ( y, t )) (cid:1) = d ( x, y ) + d ( c, x ) V ( d ( c, x )) + ( λ − d ( y, c ) V ( d ( y, c )) . Suppose that d ( c, x ) ≥ d ( y, c ), then 2 d ( c, x ) ≥ d ( x, y ) and d ( T ( x ) , T ( y )) ≥ d ( x, y ) + 12 d ( x, y ) V (cid:16) d ( x, y ) (cid:17) ≥ d ( x, y ) + 12 σ +2 d ( x, y ) V ( d ( x, y )) . Similarly, if d ( c, y ) ≥ d ( x, c ), then 2 d ( c, y ) ≥ d ( x, y ) and d ( T ( x ) , T ( y )) ≥ d ( x, y ) + ( λ − σ +2 d ( x, y ) V ( d ( x, y )) . Take C := min (cid:26) σ +2 , λ − σ +2 , ξ − , η − (cid:27) . Proposition 3.
There are constants (cid:37)
T,ω > and C > such that, given asequence { x k } k ≥ in [0 , , with T ( x k +1 ) = x k for k ≥ , and a point y ∈ [0 , with d ( x , y ) < (cid:37) T,ω , there is { y k } k ≥ ⊂ [0 , , with T ( y k +1 ) = y k for k ≥ , satisfying Ω (cid:0) d ( x k , y k ) (cid:1) + C k (cid:88) j =1 ω (cid:0) d ( x j , y j ) (cid:1) ≤ Ω (cid:0) d ( x , y ) (cid:1) ∀ k ≥ . (29) Proof.
Let (cid:37)
T,ω = min { (cid:37) T , η } , where (cid:37) T is as in the statement of Lemma 4. For x , x , y ∈ [0 ,
1] with T ( x ) = x and d ( x , y ) < (cid:37) T,ω , we can choose y ∈ T − ( y )with d ( x , y ) ≤ d ( x , y ) < (cid:37) T,ω . Then from Lemma 4, d ( x , y ) = d ( T ( x ) , T ( y )) ≥ d ( x , y ) (cid:0) C V ( d ( x , y )) (cid:1) . (cid:0) d ( x , y ) (cid:1) ≥ Ω (cid:0) d ( x , y ) (cid:0) C V ( d ( x , y ) (cid:1)(cid:1) . For h = d ( x , y ), we can writeΩ (cid:0) h (1 + C V ( h )) (cid:1) = Ω (cid:0) (1 − V ( h )) h + V ( h ) (1 + C ) h ) (cid:1) . As Ω = ϑ ∗ + ϑ ∗ (1) is concave, we see thatΩ (cid:0) h (1 + C V ( h )) (cid:1) ≥ (1 − V ( h )) Ω( h ) + V ( h ) Ω (cid:0) (1 + C ) h (cid:1) = Ω( h ) + V ( h ) (cid:16) ϑ ∗ ((1 + C ) h ) − ϑ ∗ ( h ) (cid:17) . Recalling that ϑ ∗ ≥ ϑ , we haveΩ (cid:0) h (1 + C V ( h )) (cid:1) ≥ Ω( h ) + V ( h ) ϑ ∗ ( h ) (cid:16) ϑ ∗ (cid:0) (1 + C ) h (cid:1) ϑ ∗ ( h ) − (cid:17) ≥ Ω( h ) + ω ( h ) (cid:16) ϑ ∗ (cid:0) (1 + C ) h (cid:1) ϑ ∗ ( h ) − (cid:17) . We claim that ϑ ∗ ((1+ C ) h ) ϑ ∗ ( h ) ≥ (1+ C ) γ . As a matter of fact, following Assumption A,for 1 + C ≤ ξ , since h = d ( x , y ) < (cid:37) T,ω ≤ η ,ϑ ((1 + C ) h ) ϑ ( h ) ≥ (1 + C ) γ , and thus ϑ ((1 + C ) h ) ϑ ( h ) ≥ (1 + C ) γ . Write ξ = 1 + C and recall that the transform Legendre is order reversing, then ϑ (cid:16) hξ (cid:17) = ϑ ∗ (cid:16) hξ (cid:17) = ( ϑ ( ξ h )) ∗ ≤ ( ξ γ ϑ ( h )) ∗ = ξ γ ϑ ∗ (cid:16) hξ (cid:17) = ξ γ ϑ (cid:16) hξ γ (cid:17) . Applying again the concave conjugate, we get ϑ ∗ ( ξh ) = (cid:16) ϑ (cid:16) hξ (cid:17)(cid:17) ∗ ≥ (cid:16) ξ γ ϑ (cid:16) hξ γ (cid:17)(cid:17) ∗ = ξ γ ϑ ∗ ( h ) . Therefore, for C := (1 + C ) γ −
1, we have shown that, for x , x , y ∈ [0 ,
1] with T ( x ) = x and d ( x , y ) < (cid:37) T,ω , there is y ∈ T − ( y ), with d ( x , y ) ≤ d ( x , y ) <(cid:37) T,ω , such that Ω ( d ( x , y )) ≥ Ω( d ( x , y )) + C ω ( d ( x , y )) . Inequality (29) follows straightforward from the above inequality.For ω ∈ M and ϕ ∈ C ω ([0 , | ϕ | ω = sup x (cid:54) = y | ϕ ( x ) − ϕ ( y ) | ω ( d ( x, y )) . Lemma 5.
Let g k ( x ) := sup T k ( y )= x S k (cid:0) f − m ( f, T ) (cid:1) ( y ) , for k ≥ . Then, there is L = L ( (cid:37) T,ω ) > such that for every k ≥ , | g k ( x ) − g k ( y ) | ≤ L C − | f | ω Ω( d ( x, y )) , ∀ x, y ∈ [0 , and | g k ( x ) | ≤ L C − | f | ω Ω(1) , ∀ x ∈ [0 , , where (cid:37) T,ω and C are as in the statement of Proposition 3. Proof.
Without loss of generality, we suppose that m ( f, T ) = 0. Let x , y ∈ [0 , d ( x , y ) < (cid:37) T,ω . Fix k ≥ g k ( x ) ≥ g k ( y ). Given (cid:15) >
0, there exists x k ∈ T − k ( x ) with g k ( x ) − (cid:15) < S k f ( x k ). We apply the previousproposition and consider y k ∈ T − k ( y ) so that k − (cid:88) j =0 ω (cid:0) d (cid:0) T j ( x k ) , T j ( y k ) (cid:1)(cid:1) ≤ C − (cid:16) Ω (cid:0) d ( x , y ) (cid:1) − Ω (cid:0) d ( x k , y k ) (cid:1)(cid:17) ≤ C − Ω (cid:0) d ( x , y ) (cid:1) . Thus, | g k ( x ) − g k ( y ) | − (cid:15) < S k f ( x k ) − S k f ( y k ) ≤ | f | ω k − (cid:88) j =0 ω (cid:0) d (cid:0) T j ( x k ) , T j ( y k ) (cid:1)(cid:1) ≤ C − | f | ω Ω( d ( x , y )) . Therefore, as (cid:15) > d ( x , y ) < (cid:37) T,ω and k ≥ | g k ( x ) − g k ( y ) | ≤ C − | f | ω Ω( d ( x , y )) . For z ∈ [0 , I z = ( z − (cid:37) T,ω / , z + (cid:37) T,ω / ∩ [0 , z i ∈ [0 , ≤ i ≤ L −
1, which are assumed ordered, such that { I z i } L − i =1 isan open cover of [0 , x + (cid:37) T,ω ≤ y in [0 , i x < i y for which x ∈ I z ix and y ∈ I z iy . Note that, as Ω is increasing, the above localproperty provides | g k ( x ) − g k ( y ) | ≤| g k ( x ) − g k ( z i x ) | + (cid:88) i x ≤ i
1] and k ≥
1, one has | g k (˜ x ) | > C . By the previous discussion, we wouldhave | g k (˜ x ) − g k ( x ) | ≤ C for all x ∈ [0 , , so that | g k | > C everywhere. Thenthere would be a sequence (˜ x (cid:96) ) (cid:96) ≥ such that T (cid:96)k (˜ x (cid:96) ) = ˜ x and S (cid:96)k f (˜ x (cid:96) ) > (cid:96) C ,hence 1 (cid:96) k S (cid:96)k f (˜ x (cid:96) ) > C k > . This contradicts the fact that m ( f, T ) = 0 . Indeed, it is easy to see that the Borelprobabilities ν (cid:96) = (cid:96)k (cid:0) δ ˜ x (cid:96) + δ T (˜ x (cid:96) ) + . . . + δ T (cid:96)k − (˜ x (cid:96) ) (cid:1) have, with respect to theweak-star topology, T -invariant measures as accumulation probabilities as (cid:96) → ∞ .Hence, if ν ∞ is any one of these accumulation probabilities, then m ( f, T ) ≥ (cid:90) f dν ∞ = lim j →∞ (cid:96) j k S (cid:96) j k f (˜ x k (cid:96)j ) ≥ C k . Proof of Theorem 2.
Following [CLT01, Proposition 11], denote g ≡ x ∈ [0 , U f ( x ) := sup k ≥ g k ( x ) = sup (cid:110) S k (cid:0) f − m ( f, T ) (cid:1) : k ≥ T k ( y ) = x (cid:111) . Thanks to Lemma 5, U f is a well-defined real function and actually U f ∈ C Ω ([0 , U f ◦ T ≥ U f + f − m ( f, T )holds and therefore U f is a sub-action. Acknowledgment:
We are indebted to J. T. A. Gomes for his attentive readingof this appendix.
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