Selection of measures for a potential with two maxima at the zero temperature limit
aa r X i v : . [ m a t h . D S ] A p r Selection of measures for a potential with two maxima at the zerotemperature limit
A. T. Baraviera ∗ , R. Leplaideur † , A. O. Lopes ‡ October 30, 2018
Abstract
For the subshift of finite type Σ = { , , } N we study the convergence at temperaturezero of the Gibbs measure associated to a non-locally constant H¨older potential which admitsonly two maximizing measures. These measures are Dirac measures at two different fixedpoints. The potential is flattest at one of these two fixed points.The question we are interested is: which of these probabilities the invariant Gibbs statewill select when temperature goes to zero?We prove that on the one hand the Gibbs measure converges, and at the other hand itdoes not necessarily converge to the measures where the potential is the flattest.We consider a family of potentials of the above form; for some of them there is the selectionof a convex combination of the two Dirac measures, and for others there is a selection of theDirac measure associated to the flattest point. In the first case this is contrary to what wasexpected if we consider the analogous problem in Aubry-Mather theory [1]. Keywords: selection of measures, transfer operator, Gibbs measures, ergodic optmization
The topic of optimization in Ergodic Theory deals with the study of maximizing or minimizingmeasures. Considering a dynamical system (
X, T ) and A : X → R , a A -maximizing measure isa T -invariant probability measure µ such that Z A dµ = max ν T − inv (cid:26)Z A dν (cid:27) . Existence of maximizing measures is for instance ensured when X is compact, and T and A arecontinuous. ∗ [email protected], Instituto de Matem´atica - UFRGS - Partially supported by DynEuroBraz † [email protected], D´epartement de Math´ematiques - Universit´e de Brest- Partially supportedby DynEuroBraz and Convenio Brasil-Franca ‡ [email protected], Instituto de Matem´atica - UFRGS - Partially supported by DynEuroBraz,CNPq, PRONEX – Sistemas Dinamicos, INCT, Convenio Brasil-Franca, and beneficiary of CAPES financialsupport µ is an equilibrium state for A if it satisfies h µ ( T ) + Z A dµ = sup ν T − inv (cid:26) h ν ( T ) + Z A dν (cid:27) , where h ν ( T ) is the usual Kolmogorov entropy. It is well-known that any accumulation point forthe equilibrium state µ β associated to βA , where β is a large positive real parameter, as β goesto + ∞ is a A -maximizing measure. In Statistical Mechanics the parameter β is the inverse ofthe temperature. The study of selection is to consider the following question: which maximizingmeasure is obtained as the limit of the equilibrium state associated to βA , when β → ∞ ? Insome cases there is no convergence (see [3]). When the maximizing probability is unique thereis convergence. Therefore, the interesting situation to analyze is when there is more than one A -maximizing probability.In [1], Anantharaman and al. study one example of selection for Lagrangian dynamics.They consider an external parameter ǫ , and for each ǫ there is a natural probability which canbe associated to an eigen-function problem. There, they show, among other things, that if thepotential has only two points of maxima, then this natural probability converges, when ǫ → , to the Dirac measure concentrated in the point (which is maxima of the potential) were thepotential is the flattest.In the present paper we study the same kind of problem but for the dynamics of the shiftwith three symbols. The main difference between these two problems (Euler-Lagrange flow andthe shift) is that for the case of the dynamics of the shift, every choice of A is possible and makessense to be analyzed. In Aubry-Mather theory the dynamics (the Euler-Lagrange flow) dependsof the Lagrangian (or, potential) considered. In our case there is no relation of the potentialwith the dynamics. In [1] the parameter ǫ , such that ǫ goes to infinity, is related to viscositysolutions, and here the parameter β is the inverse of temperature. The question of selectionmakes sense in both settings. There is a natural hope, that every result in one theory has itsdual version for the other theory. This was the first motivation for this paper: considering inΣ := { , , } N the Holder potential A ( x ) = − d ( x, ∞ ) if x ∈ [0] − d ( x, ∞ ) if x ∈ [1] − α otherwise . It is reasonable to say that the potential A is more flat at 0 ∞ . We initially expected thatthe Gibbs measure for the potential βA converges to the δ ∞ , as the temperature goes to 0, andwe wanted to study how the selection occurs.For our surprise, for the case α <
1, we find out that the Gibbs measure always converges,but not to δ ∞ ; it can select in the limit another convex combination of δ ∞ and δ ∞ .Moreover, the convex combination is not continuous on α . Nevertheless it takes quite sur-prising values as a function of α . For this reason we believe that it would be very difficult toestablish a global and general selection theory for the class of all subshifts of finite type (withfinite alphabet) and any potential, not only due to this unexpected selection behavior but alsobecause even convergence does not always occur, see [3].The invariant probability is obtained by the junction of the eigen-function and the eigen-probability [7]. A curious phenomena that happens in our examples is that the eigen-measure2nd the eigen-function (see section (1.3) for definitions) have opposite behavior. When β → ∞ ,the eigen-measure became exponential bigger around 0 ∞ , when compared to points around 1 ∞ .For the eigen-function the opposite happens. Therefore, we need a very fine analysis of thecontrol of the invariant Gibbs state.The terminology ”selection” was borrowed from the theory of viscosity solutions (see forinstance [1] for references). We work here with a full shift Σ over the alphabet { , , } . Points in Σ are sequences x =( x , x , . . . ) with x i ∈ { , , } . We will consider the usual terminology and the usual topologyin Σ. Hence, we recall that a cylinder [ X , . . . X k ] is the set of points x = ( x n ) such that x i = X i for every i ∈ J , k K . We also recall that the distance between x = ( x n ) and y = ( y n ) is definedby d ( x, y ) = 12 min { n, x n = y n } . The two special points 0 ∞ and 1 ∞ respectively denote the points (0 , , . . . ) and (1 , , . . . ). Theyare fixed points for the shift σ over Σ.As we said above, we consider over this shift the Lipschitz potential A defined as follows: A ( x ) = − d ( x, ∞ ) if x ∈ [0] − d ( x, ∞ ) if x ∈ [1] − α otherwisefor some α >
0. Then this potential is always non-positive. There are only two maximizingmeasures, respectively δ ∞ and δ ∞ . We point out that the potential is flattest close to 0 ∞ .It is well-known (see e.g. [2]) that there exists a unique equilibrium state for βA (for all β ∈ R ). It is also a Gibbs measure (see also Subsection 1.3).Our main result is: Theorem
Let µ β be the unique Gibbs measure associated to βA . Let ρ be the golden mean ρ := 1 + √ . Then1. for α > , µ β converges to ( δ ∞ + δ ∞ ) as β goes to + ∞ ,2. for α = 1 , µ β converges to ρ ( ρ δ ∞ + δ ∞ ) as β goes to + ∞ ,3. for α < , µ β converges to δ ∞ as β goes to + ∞ . As we already said it above, this result is surprising because it was expected that in every cases µ β would converge to δ ∞ . Discontinuity of the limit measure as a function of α is of course lesssurprising. Nevertheless, the values which appear in function of α , and in particular for α = 1,are quite surprising.For α = 0, it is expected that µ β converges to δ ∞ (the flattest one !). Then, we could haveexpected the exact inverse situation between α < α > for α < , µ β would converge to
12 ( δ ∞ + δ ∞ ) and for α > , µ β would converge to δ ∞ , the measure 12 ( δ ∞ + δ ∞ ) being a kind3f “smooth” transition with the limit case δ ∞ for the case α = 0. It turns out that this is notthe case.On the other hand, if α goes to + ∞ , the system looks, in some sense, the full-shift with twosymbols { , } N . In that case, it is not so surprising that the limit measure is 12 ( δ ∞ + δ ∞ ),whatever the slopes are. Indeed, for { , } N , every µ β typical orbit is an alternation of stringsof 0’s and 1’s. Following [6], the convex combination would be given by the costs between thetwo maximizing zones, δ ∞ and δ ∞ . Hence, every typical orbit sees the two symbols and thisis an heuristic argument which in some sense justifies that µ β converges to 12 ( δ ∞ + δ ∞ ).We emphasize that one simple generalization of our theorem would be to replace − d ( x, ∞ )with some − Γ d ( x, ∞ ), with Γ >
1. In this case the same result holds and our method caneasily be adapted. Nevertheless, the computation would be a little bit more complicate and theformulas less convenient to be used. If y = ( y , y , . . . ) is a point in Σ and if a = 0 , ,
2, we denote by ay the point ( a, y , y , . . . ) in Σ.We recall that the main tool is the transfer operator defined as follows: L β ϕ ( x ) = X y ∈ σ − ( x ) e βA ( y ) ϕ ( y )= e − βd (0 x, ∞ ) ϕ (0 x ) + e − βd (1 x, ∞ ) ϕ (1 x ) + e − αβ ϕ (2 x ) . where β is the inverse of the temperature. It acts on continuous functions and its dual operator,denoted by L ∗ β , acts on probability measures. Most of the time we will omit the subscript β .We know that there exists some function H and some probability measure ν such that L ( H ) = e P H and L ∗ ( ν ) = e P ν . Then, the probability measure dµ = Hdν is σ − invariant and isthe unique equilibrium state. It is the so called Gibbs measure associated to βA .Throughout, they will thus be referred to as the eigen-measure and the eigen-function.The plan of the proof of the main result of the paper is the following:In Section 2 we give the exponential asymptotics for the eigen-function (obtaining what iscalled a calibrated subaction) and the pressure.In Section 3 we prove the convergence of the eigen-measure to δ ∞ . For this we give precisevalues for the ν -measures of rings.In Section 4 we compute the exact values of the eigen-function on the same rings consideredbefore in Section 3.In Section 5 we finish the proof of our Theorem. We first recall a usual definition in that theory.
Definition 2.1.
We say that u : Ω → R is a calibrated subaction for A if for any y we have u ( y ) = sup σ ( x )= y { A ( x ) + u ( x ) − m ( A ) } .
4e denote by V any accumulation point for 1 β log H β as β goes to + ∞ . It is clearly acalibrated subaction, see [4]. If we add a constant to a calibrated subaction, it will be also acalibrated subaction.We remind that the Peierls’ barrier is given by h ( x, y ) = lim ǫ → lim sup n { n − X j =0 A ( σ j ( z )) − m ( A ) , n ≥ , σ n ( z ) = y, d ( z, x ) < ǫ } . Remark 1.
We let the reader check that for every x = 0 ∞ , ∞ both numbers h (0 ∞ , x ) and h (1 ∞ , x ) are negative.It is known that if u is a calibrated subaction then it satisfies u ( y ) = sup x ∈ Ω [ h ( x , y ) + u ( x )] , where h is the Peierls barrier and Ω is the Aubry-set [4] [5] (Theorem 10).In the present case the Aubry-set is the union of the two fixed points p = 0 ∞ and q = 1 ∞ .In this way, any calibrated subaction is determined by its values on p and q . Lemma 2.2.
The functions defined by u ( x ) = − d ( x, ∞ ) and u ( x ) = − d ( x, ∞ ) are bothcalibrated subactions.Proof. The proof is only done for u , the other case being similar. We consider y ∈ Σ and wewant to prove − d (0 ∞ , y ) =: u ( y ) = max { A (0 y ) + u (0 y ) , A (1 y ) + u (1 y ) , A (2 y ) + u (2 y ) } . (1)We set y = ( y , y , y , . . . ). We first assume that y = 0. Note that both A (1 y ) and A (2 y ) arenegative and u (1 y ) = u (2 y ) = −
1. Hence u ( y ) = − A (1 y )+ u (1 y )and A (2 y ) + u (2 y ). Now A (0 y ) = − and u ( y ) = − . Hence (1) holds in that case. Assumenow that y belong to the cylinder 0 n . Then u ( y ) = − n . Again, note that u ( y ) is bigger thanboth terms A (1 y ) + u (1 y ) and A (2 y ) + u (2 y ). We also get − n = − n +1 + − n +1 = A (0 y ) + u (0 y ) . Hence, (1) holds in that case too.Using Lemma 2.2 we can get a more simple formulation for V . Lemma 2.3. V ( x ) = sup { [ V (0 ∞ ) − d (0 ∞ , x )] , [ V (1 ∞ ) − d (1 ∞ , x )] } Proof.
This follows from the fact that such V is calibrated and from the expression of the Peierlsbarrier. Indeed, we claim that we have h (0 ∞ , y ) = u ( y ) and h (1 ∞ , y ) = u ( y ) . Again, we only prove that we get h (0 ∞ , x ) = u ( x ) = − d ( x, ∞ ), the other equality beingsimilar. 5et x = ( x , x , . . . ) be in Σ. We get u ( x ) = max( h (0 ∞ , x ) + u (0 ∞ ) , h (1 ∞ , x ) + u (1 ∞ )) = max( h (0 ∞ , x ) , h (1 ∞ , x ) − . Note that u ( x ) ≥ − u ( x ) = h (0 ∞ , x ) . Now, we use properties of the eigenfunction H β to obtain some relations satisfied by V . Acalibrated subaction, in the present situation, is determined by its values 0 ∞ and 1 ∞ . We justneed the relative values of V at these points. Proposition 2.4.
For α > , we get V (1 ∞ ) = V (0 ∞ ) + 1 and lim β → + ∞ β log P ( β ) = − .For < α ≤ , we get V (1 ∞ ) = V (0 ∞ ) + α and lim β → + ∞ β log P ( β ) = − (1 + α ) .Proof. Up to the fact that we consider a sub-family we assume that lim β → + ∞ β log P ( β ) exists andis equal to real number γ .From the equation for the eigenfunction we get the pair of equations( e P ( β ) − H β (0 ∞ ) = e − αβ H β (2) + e − β H β (1 0 ∞ ) , (2a)( e P ( β ) − H β (1 ∞ ) = e − αβ H β (2) + e − β H β (0 1 ∞ ) . (2b)Remember that by Lemma 2.2 we get V (10 ∞ ) = max { [ V (0 ∞ ) − , [ V (1 ∞ ) −
32 ] } ,V (2 x x .. ) = max { [ V (0 ∞ ) − , [ V (1 ∞ ) − } ,V (01 ∞ ) = max { [ V (0 ∞ ) −
12 ] , [ V (1 ∞ ) − } . Then, taking β log in Equation (2a) and making β go to + ∞ we get γ + V (0 ∞ ) = max { [ V (0 ∞ ) − − α ] , [ V (1 ∞ ) − − α ] , [ V (0 ∞ ) − −
32 ] , [ V (1 ∞ ) − −
32 ] } = max { [ V (0 ∞ ) − − α ] , [ V (1 ∞ ) − − α ] , [ V (0 ∞ ) −
52 ] , [ V (1 ∞ ) − } = max { [ V (0 ∞ ) − − α ] , [ V (0 ∞ ) −
52 ] , [ V (1 ∞ ) − } . (3)Similarly with (2b) we finally get γ + V (1 ∞ ) = max { [ V (0 ∞ ) − , [ V (1 ∞ ) − / , [ V (1 ∞ ) − − α ] } . (4) We first deal with the case α >
1. We will show that V (1 ∞ ) = V (0 ∞ ) + 1. We dividethe analysis in two cases: 6) if α > /
2, then, we have to solve γ + V (0 ∞ ) = max { [ V (0 ∞ ) −
52 ] , [ V (1 ∞ ) − } , (5a) γ + V (1 ∞ ) = max { [ V (0 ∞ ) − , [ V (1 ∞ ) − / } . (5b)Now, we show that this system of equation is solvable if and only if V (0 ∞ ) − ≤ V (1 ∞ ) − V (0 ∞ ) − ≥ V (1 ∞ ) − / V (0 ∞ ) − > V (1 ∞ ) −
3. Then, we get γ + V (0 ∞ ) = V (0 ∞ ) − /
2, which showsthat we have γ = − /
2. Thus, we must have V (0 ∞ ) − ≥ V (1 ∞ ) − / γ = − ), and we get V (0 ∞ ) − γ + V (1 ∞ ) = − / V (1 ∞ ) . From this follows that V (1 ∞ ) = 3 / V (0 ∞ ). This yields V (1 ∞ ) − / V (0 ∞ )) − V (0 ∞ ) − > V (0 ∞ ) − , which produces a contradiction.Then, we have γ + V (0 ∞ ) = V (1 ∞ ) − γ ≥ − . If V (0 ∞ ) − ≤ V (1 ∞ ) − /
2, then (5b)shows that γ is equal to − which is impossible. Hence we must get γ + V (1 ∞ ) = V (0 ∞ ) − . (7)Finally, (6) and (7) yield γ = −
2, and V (1 ∞ ) = V (0 ∞ ) + 1.2) The case 1 < α ≤ . The proof is similar. It is explicitly reproduced here, but the readercan skip it in a first reading.The new system to solve is γ + V (0 ∞ ) = max { [ V (0 ∞ ) − (1 + α )] , [ V (1 ∞ ) − } , (8a) γ + V (1 ∞ ) = max { [ V (0 ∞ ) − , [ V (1 ∞ ) − / } . (8b)Again, we show that this system of equation is solvable if, and only if, V (0 ∞ ) − (1+ α ) ≤ V (1 ∞ ) − V (0 ∞ ) − ≥ V (1 ∞ ) − / V (0 ∞ ) − (1 + α ) > V (1 ∞ ) −
3. Then, we get γ + V (0 ∞ ) = V (0 ∞ ) − (1 + α ),which shows that we have γ = − (1 + α ) > − . Thus, we must have V (0 ∞ ) − ≥ V (1 ∞ ) − / γ = − ), and we get V (0 ∞ ) − γ + V (1 ∞ ) = − (1 + α ) + V (1 ∞ ) . From this follows that V (1 ∞ ) = α + V (0 ∞ ). This yields V (1 ∞ ) − α + V (0 ∞ )) − V (0 ∞ ) − > V (0 ∞ ) − , which produces a contradiction. 7hen, we have γ + V (0 ∞ ) = V (1 ∞ ) − γ ≥ − (1 + α ) > − . If V (0 ∞ ) − ≤ V (1 ∞ ) − / γ is equal to − which is impossible. Hence we must get γ + V (1 ∞ ) = V (0 ∞ ) − . (10)Finally, (9) and (10) yield γ = −
2, and V (1 ∞ ) = V (0 ∞ ) + 1.We point out here that the above discussion can be done for every sub-family of β ’s. Inparticular, this shows that 1 β log P ( β ) can have only one accumulation point. In other words, itconverges to γ = − Now, we deal with the case α ≤
1. We will show that V (1 ∞ ) = V (0 ∞ ) + α . The systemwe have to solve is γ + V (0 ∞ ) = max { [ V (0 ∞ ) − (1 + α )] , [ V (1 ∞ ) − } , (11a) γ + V (1 ∞ ) = max { [ V (0 ∞ ) − , [ V (1 ∞ ) − / , [ V (1 ∞ ) − − α ] } . (11b)We show that, whatever is the case α ≤ or α ≥ , the system can be solved if, and only if, V (0 ∞ ) − (1 + α ) ≥ V (1 ∞ ) − V (0 ∞ ) − ≥ V (1 ∞ ) − / , V (1 ∞ ) − − α .Let us proceed by contradiction and assume we get V (0 ∞ ) − (1 + α ) < V (1 ∞ ) −
3. In thatcase, if we assume that we get V (0 ∞ ) − ≥ V (1 ∞ ) − / , V (1 ∞ ) − − α , then the system tosolve is exactly given by equations (6) and (7). This yields γ = −
2, and V (1 ∞ ) = V (0 ∞ ) + 1.Then, we get V (1 ∞ ) − V (0 ∞ ) − ≤ V (0 ∞ ) − (1 + α ) which produced a contradictionwith our assumption V (0 ∞ ) − (1 + α ) < V (1 ∞ ) − V (0 ∞ ) − ≤ V (1 ∞ ) − / , V (1 ∞ ) − − α , and the bigger term only dependson the relative position of α with respect to . Depending of this position, we get γ = −
72 or γ = − − α . Then (11a) would give in both case V (0 ∞ ) − γ > V (0 ∞ ) − (1 + α ) , which produces a contradiction. Hence, we must get V (0 ∞ ) − (1 + α ) ≥ V (1 ∞ ) − γ = − (1 + α ) . (12)If V (0 ∞ ) − ≥ V (1 ∞ ) − / , V (1 ∞ ) − − α does not hold, then we would get γ = −
72 or γ = − − α , which is impossible. Thus we must get V (0 ∞ ) − ≥ V (1 ∞ ) − / , V (1 ∞ ) − − α and we finally get V (1 ∞ ) + γ = V (1 ∞ ) − (1 + α ) = V (0 ∞ ) − . (13)This finishes the proof of the proposition (again γ is the unique possible accumulation point for1 β log P ( β )). ν In this section we study the eigen-measure ν βA . We prove it converges to the Dirac measure δ ∞ . We also show exact limit values on special sets.8 .1 A useful function We define and study a function F depending on the pressure P ( β ) and on the parameter β . Definition 3.1.
For Z ≥ and β ≥ F ( Z, β ) := ∞ X k =0 e − kZ e β k +1 and its partial sum F n ( Z, β ) := n X k =0 e − kZ e β k +1 . Clearly, F n ( Z, β ) → F ( Z, β ) when n → ∞ .We remind that as β goes to + ∞ , P goes exponentially fast to 0. The asymptotic behaviorof F (for β very large) can be obtained as follows: Lemma 3.2.
For every β > we get (cid:12)(cid:12)(cid:12)(cid:12) F ( P, β ) − P (cid:12)(cid:12)(cid:12)(cid:12) ≤ βe β/ X n ≥ ( P ln 2 ) n ) . Proof.
Let us consider a positive Z . Note that the function x
7→ − Zx + β . x is decreasing on R + . We can thus compare the sum and the integral: Z + ∞ Ze − xZ e β x dx ≤ ZF ( Z, β ) ≤ Z + ∞ Ze − xZ e β x dx + Ze β . Let us study the integral. We get Z + ∞ Ze − xZ e β x dx = h − e − xZ e β x i + ∞ − Z + ∞ β e − xZ ln 22 x e β x dx. = e β − Z + ∞ β e − xZ ln 22 x e β x dx. Let us set u = x in this last integral. We get Z + ∞ Ze − xZ e β x dx = e β − Z β e − Z ln u ln 2 e β u du. Writing e − Z ln u ln 2 = + ∞ X n =0 n ! (cid:18) − Z ln u ln 2 (cid:19) n we get Z + ∞ Ze − xZ e β x dx = e β − Z β + ∞ X n =0 n ! (cid:18) − Z ln u ln 2 (cid:19) n e β u du. To get the inverse of the two sums we remind that Z | ln u | n du = Z + ∞ v n e − v dv = n !. Thenfor Z < ln 2 we get 9 + ∞ Ze − xZ e β x dx = e β − + ∞ X n =0 n ! (cid:18) − Z ln 2 (cid:19) n Z β u ) n e β u du = 1 − + ∞ X n =1 n ! (cid:18) − Z ln 2 (cid:19) n Z β u ) n e β u du. Now, note that (cid:12)(cid:12)(cid:12)(cid:12) n ! (cid:18) − Z ln 2 (cid:19) n Z β u ) n e β u du (cid:12)(cid:12)(cid:12)(cid:12) ≤ n ! (cid:18) Z ln 2 (cid:19) n β e β Z | ln u | n du = (cid:18) Z ln 2 (cid:19) n β e β . We also recall that for positive β , the pressure is strictly smaller than the topological entropyln 2. This shows the lemma. [0] and [1] We remind that the eigen-probability for βA , ν β , is a conformal measure: for any cylinder set B where σ is injective ν β ( σ ( B )) = Z B e P ( β ) − βA ( x ) d ν β ( x ) . We shall use this simple relation to compute exact values for ν β of some special cylinders.For simplicity we drop the subscribe β in ν β and simply write ν . We shall also use thenotation ∗ for the pair of symbols which are not 0 and ∗ for the pair of symbols which are not1. Then [0 ∗ ] = [01] ⊔ [02] and [1 ∗ ] = [10] ⊔ [12](and the unions are disjoint).We can now estimate the measures of the cylinders [0] and [1]. Lemma 3.3. ν [0] = e − β F ( P, β ) ν [0 ∗ ] ν [1] = e − β F ( P, β ) ν [1 ∗ ] Proof.
Conformality yields ν [0 ∗ ] = ν [ σ (00 ∗ )] = e P + β ν [00 ∗ ] = e P + β + β ν [000 ∗ ] , and so on. By induction we get ν [0 ∗ ] = e ( n − P + β ( + ... + n ) ν [00 . . . | {z } n ∗ ] . (14)Hence, we get ν [0] = ∞ X n =1 ν [00 . . . | {z } n ∗ ] = ∞ X n =1 e − ( n − P e − β e β n ν [0 ∗ ] = e − β F ( P, β ) ν [0 ∗ ] . Similarly we get ν [1] = e − β F ( P, β ) ν [1 ∗ ].10sing [0 ∗ ] = [01] ⊔ [02] and[1 ∗ ] = [10] ⊔ [12] and the conformal property of ν we obtain thefollowing system: ν [1 ∗ ] = ν [2] e − P − β + ν [0] e − P − β . (15a) ν [0 ∗ ] = ν [2] e − P − β + ν [1] e − P − β . (15b)This system is the key point to determine the convergence of the eigen-measure. Proposition 3.4.
The ratio ν [0] ν [1] goes exponentially fast to + ∞ as β goes to + ∞ .Proof. By Lemma 3.3 the system (15) can be transformed into a system in ν [0], ν [1], and ν [2]: ν [0] = e − β/ F ( P, β ) { ν [2] e − P − β + ν [1] e − P − β } ν [1] = e − (3 β ) / F ( P, β ) { ν [2] e − P − β + ν [0] e − P − β } This yields ν [0] ν [1] = e β F ( P, β ) ( 1 + e − P − β F ( P, β ) ) F ( P, β ) ( 1 + e − P − β F ( P, β ) ) (16)Finally, when β → ∞ , Proposition 2.4 and Lemma 3.2 show that ν [0] ν [1] goes to + ∞ exponen-tially fast. The exponential speed is larger than 1 − ε for every positive ε .We point out that Lemma 3.3 also allow to transform the system (15) into a system in ν ([0 ∗ ]), ν ([1 ∗ ]), and ν (2). From this system we get ν [0 ∗ ] ν [1 ∗ ] = e β ( 1 + e − P − β F ( P, β ) )( 1 + e − P − β F ( P, β ) ) . (17)Nevertheless, at that point of the proof we do not have enough information on P to con-clude which is the limit of the ratio. Proposition 2.4 and Lemma 3.2 just allow to ensure that1 β log ν [0 ∗ ] ν [1 ∗ ] goes to 0. However, we can get ratios for other rings: Corollary 3.5.
For every n ≥ , ν [0 n ∗ ] ν [1 n ∗ ] = e β (1 − n − ) ν [0 ∗ ] ν [1 ∗ ] . The ratio ν [0 n ∗ ] ν [1 n ∗ ] goes to + ∞ as β goes to + ∞ with exponential speed larger than (1 − n − ) − ε for every positive ε . .3 Convergence of the eigen-measure In this subsection we get a finer estimate for P ( β ) and conclude that ν goes to the Dirac measure δ ∞ .The conformal property yields ν ([2]) = ν ([20]) + ν ([21]) + ν ([22]) = e − P − αβ ( ν [0] + ν [1] + ν [2]) = e − P − αβ . (18)On the other hand the solution of the system obtained in the proof of Proposition 3.4 showsthat we have ν ([0]) = 1 + e − P − β F ( P, β )1 − e − P F ( P, β ) F ( P, β ) e − β F ( P, β ) e − P − β ν ([2]) ,ν ([1]) = 1 + e − P − β F ( P, β )1 − e − P F ( P, β ) F ( P, β ) e − β F ( P, β ) e − P − β ν ([2]) . Using the formula ν ([0]) + ν ([1]) + ν ([2]) = 1 we get another expression for ν ([2]):1 = ν ([2]) (cid:18) e − P − β F ( P, β )1 − e − P F ( P, β ) F ( P, β ) e − β F ( P, β ) e − P − β +1 + e − P − β F ( P, β )1 − e − P F ( P, β ) F ( P, β ) e − β F ( P, β ) e − P − β (cid:19) = ν ([2]) (cid:18) e − P − β F ( P, β ) + e − P − β F ( P, β ) + e − P − β F ( P, β ) F ( P, β )1 − e − P F ( P, β ) F ( P, β ) e − β (cid:19) . (19)Lemma 3.2 and Proposition 2.4 show that whatever the value of α is, e − P − β F ( P, β ) goes to0 as β goes to + ∞ . On the other hand, e − P − β F ( P, β ) is exponentially big (of order e β if α isbigger than 1 and e αβ if α is smaller than 1). Remember that Equation (18) shows that ν ([2])goes exponentially fast to 0 with exponential speed − αβ . Lemma 3.6. If α > , we get lim β → + ∞ P ( β ) e β = 1 . For α = 1 , P ( β ) e β goes to √ .Proof. We first do the case α >
1. As we said above, the numerator in the right hand side of(19) has order e β . On the other hand ν ([2]) has order e − αβ . Therefore, the denominator of theright hand side of (19) goes to 0 with exponential speed e (1 − α ) β . Then, Lemma 3.2 shows that P ( β ) e − β goes to 1.Let us now deal with the case α = 1. Copying what we did above we get e P = e − β P ε ( β )1 − e − P (cid:16) e − β P (cid:17) (1 + ε ( β )) , with ε i ( β ) going to 0 as β goes to + ∞ . Let l be any accumulation point for P e β . We thus get= l − l = ll − . This yields l = √ . 12 orollary 3.7. As β goes to + ∞ , the ratio ν [0 ∗ ] ν [1 ∗ ] goes to 1 for α > to √ for α = 1 andto + ∞ for α < . The convergence is non-exponential for α ≥ and has exponential speed − α if α < .Proof. We remind that Equation (17) gives ν [0 ∗ ] ν [1 ∗ ] = e β ( 1 + e − P − β F ( P, β ) )( 1 + e − P − β F ( P, β ) ) . We already know that e − β F ( P, β ) goes to 0 as β goes to + ∞ . The denominator has fordominating term e − β P . For α < ν [0 ∗ ] ν [1 ∗ ] goes to + ∞ . For α ≥ ν ([2]) goes to 0 as β goes to + ∞ . Then Proposition 3.4 yields: Corollary 3.8.
The measure ν goes to the Dirac measure δ ∞ as β goes to + ∞ . H In this section we get estimates at the non-exponential scale for the asymptotic behavior of theeigenfunction H β . In what follows, for simplicity, we will drop the subindex β . We know that H ( x ) = lim N →∞ N N − X k =0 L k (1I)( x ) e kP (20)where L is the transfer operator (see Subsection 1.3). We recall that ∗ (resp. ∗ ) denotes anysymbol different to 0 (resp. to 1). We start with the following result. Lemma 4.1.
The eigen-function is constant on cylinders [0 n ∗ ] , [1 n ∗ ] and [2] .Proof. Owing to Equation 20, it is sufficient to prove that for every k , L k (1I) is constant oncylinders [0 n ∗ ], [1 n ∗ ] and [2]. For x in Σ, we get L kβ (1I)( x ) = X z ∈{ , , } k e β.S k ( A )( zx ) , where S k ( A ) is the Birkhoff sum A + A ◦ σ + . . . + A ◦ k − σ . Now, note that the potential isconstant on the cylinders [0 m ∗ ], [1 m ∗ ] (whatever m ≥ H ) and on the eigen-measure are not yet sufficient to conclude the proof. Indeed,one important fact is that the eigen-measure and the eigen-function have opposite behavior: Lemma 4.2.
For α ≥ and for every integer n ≥ , lim β → + ∞ β log µ ([0 n ∗ ]) µ ([1 n ∗ ]) = 0 . roof. By definition we get µ ([0 n ∗ ]) µ ([1 n ∗ ]) = H (0 n ∗ ) ν ([0 n ∗ ]) H (1 n ∗ ) ν ([1 n ∗ ]) . Using Corollaries 3.5 and 3.7, weget that 1 β log ν ([0 n ∗ ]) ν ([1 n ∗ ]) goes to 1 − n − as β goes to + ∞ .On the other hand, Lemma 2.3 and Proposition 2.4 shows that 1 β log H (0 n ∗ ) H (1 n ∗ ) goes to − n − as β goes to + ∞ . Remark 2.
For α < β → + ∞ β log µ ([0 n ∗ ]) µ ([1 n ∗ ]) =2 − α .Lemma 4.2 shows that the convergence and the study of selection for µ cannot be obtainedat the exponential scale. We thus must get more precise estimates. We recall that the functions F ( P, β ) and F n ( P, β ) were defined in Definition 3.1.
Lemma 4.3.
For every n ≥ we get H (0 n ∗ ) = e ( n − P − β n ( e P − e P + e − αβ h e P + β H (1 ∞ ) − (cid:16) F n − ( P, β )(1 + e − P − αβ ) + e (1 − α ) β (cid:17) H (0 ∞ ) i , (21) H (1 n ∗ ) = e ( n − P − β n ( e P − e P + e − αβ h e P +3 β H (0 ∞ ) − (cid:16) F n − ( P, β )(1 + e − P − αβ ) + e (3 − α ) β (cid:17) H (1 ∞ ) i , (22) where F − ≡ .Proof. Using the equality L ( H ) = e P H we get the following system of equations e − β H (0 ∗ ) + e − αβ H (2) = ( e P − H (1 ∞ ) ,e − β H (1 ∗ ) + e − αβ H (2) = ( e P − H (0 ∞ ) ,e − β H (0 ∗ ) + e − β H (1 ∗ ) +( e − αβ − e P ) H (2) = 0 . (23)Solving this system in terms of H (1 ∞ ) and H (0 ∞ ) we find: H (0 ∗ ) = e β ( e P − e P + e − αβ h e P H (1 ∞ ) − e − αβ H (0 ∞ ) i (24) H (1 ∗ ) = e β ( e P − e P + e − αβ h e P H (0 ∞ ) − e − αβ H (1 ∞ ) i (25)Again, the equality L ( H ) = e p H yields e P H (0 n ∗ ) = e − β n +1 H (0 n +1 ∗ ) + e − β H (1 ∗ ) + e − αβ H (2) . Introducing the second equation in (23), we get H (0 n +1 ∗ ) = e P + β n +1 H (0 n ∗ ) + e β n +1 ( e P − H (0 ∞ ) .
14y induction, we get for every n ≥ H (0 n ∗ ) in function of H (0 ∞ ) and H (0 ∗ ).Then, introducing (24) in this expression, we let the reader check that we get (21). The proofof (22) is similar.As we said above, the exponential scale is not sufficient to determine the limit and theselection for the Gibbs measure. Due to the values of the subactions, the good parameter toestimate is e β H (0 ∞ ) H (1 ∞ ) . Lemma 4.3 allows us to solve that problem. Proposition 4.4. As β goes to + ∞ we get the following limits:(i) if α > , then, lim β → + ∞ e β H (0 ∞ ) H (1 ∞ ) = 1 ,(ii) if α = 1 , then, lim β → + ∞ e β H (0 ∞ ) H (1 ∞ ) = 1 + √ ,(iii) if < α < , then, lim β → + ∞ e β H (0 ∞ ) H (1 ∞ ) = + ∞ .Proof. Equalities (21) and (22) yield for any fixed ne β − β n − H (0 n ∗ ) H (1 n ∗ ) = e P − [ F n − ( P, β ) (1 + e − P − α β ) e − β + e − (1+ α ) β ] ( e β H (0) H (1) ) e P ( e β H (0) H (1) ) − [ F n − ( P, β ) (1 + e − P − α β ) e − β + e (1 − α ) β ] . (26)For, β fixed, we set x = x β = e β H (0) H (1) . Then, taking the limit as n goes to + ∞ we get x = e P − [ F ( P, β ) (1 + e − P − α β ) e − β + e − (1+ α ) β ] xe P x − [ F ( P, β ) (1 + e − P − α β ) e − β + e (1 − α ) β ] , (the eigen-function is continuous). Let us set a = d = e P and b = − [ F ( P, β ) (1 + e − P − α β ) e − β + e − (1+ α ) β ] ,c = − [ F ( P, β ) (1 + e − P − α β ) e − β + e (1 − α ) β ] . We can write the above equation in the form x = a + b xd x + c . As x is positive we can solve this equation and we get x = ( b − c ) + p ( c − b ) + 4 a d d . (27)Note that( b − c ) = ( F ( P, β ) − F ( P, β ) ) e − β (1 + e − P − α β ) + e − α β ( e β − e − β ) . Now, Lemma 3.2 shows that e − β ( F ( P, β ) − F ( P, β ) ) → β goes to + ∞ . On the otherhand we get, 15or α > e − α β ( e β − e − β ) → . for α < e − α β ( e β − e − β ) → + ∞ ,for α = 1, e − α β ( e β − e − β ) → β goes to + ∞ . From this, we get that for the three cases of possiblevalues of α , the corresponding limits for ( b − c ) are the same:for α > b − c → . for α < b − c → + ∞ ,for α = 1, b − c → α > β → + ∞ e β H (0 ∞ ) H (1 ∞ ) = 1 , for α = 1, lim β → + ∞ e β H (0 ∞ ) H (1 ∞ ) = 1 + √ , and for 0 < α <
1, lim β → + ∞ e β H (0 ∞ ) H (1 ∞ ) = + ∞ . Now, we can finish the proof of our Main Theorem. We recall that any accumulation point for µ β is a A -maximizing measure. Hence, such an accumulation point is a convex combinationof the two Dirac measures δ ∞ and δ ∞ . This convex combination can be found if we get anestimate for lim β → + ∞ µ ([0]) µ ([1]) . We get µ ([0]) µ ([1]) = P + ∞ n =1 µ ([0 n ∗ ]) P + ∞ n =1 µ ([1 n ∗ ])= P + ∞ n =1 H (0 n ∗ ) ν ([0 n ∗ ]) P + ∞ n =1 H (1 n ∗ ) ν ([1 n ∗ ])= P + ∞ n =1 H (0 n ∗ ) e − ( n − P − β ( + ... + n ) P + ∞ n =1 H (1 n ∗ ) e − ( n − P − β ( + ... + n ) ν ([0 ∗ ]) ν ([1 ∗ ])= P + ∞ n =1 e β (1 − n − ) H (0 n ∗ ) H (1 n ∗ ) µ ([1 n ∗ ]) P + ∞ n =1 µ ([1 n ∗ ]) ν ([0 ∗ ]) ν ([1 ∗ ]) . (28)The proof will follow from the next technical lemma:16 emma 5.1. There exists β such that for every n ≥ , for every β ≥ β and for every α (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) e β (1 − n − ) H (0 n ∗ ) H (1 n ∗ ) × e β H (0 ∞ ) H (1 ∞ ) − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ e − β . Proof.
We re-employ notations from the proof of Proposition 4.4. We denote by R n − (1) thetail R n − (1) = F ( P, β ) − F n − ( P, β ) = ∞ X k = n − e − k P + β k +1 ,R n − (3) the tail R n − (3) = F ( P, β ) − F n − ( P, β ) = ∞ X k = n − e − k P + β k +1 and ∆ n − = R n − (1) − R n − (3) = e − ( n − P ( e β n − e β n ) + ... . Then, e β − β n − H (0 n ∗ ) H (1 n ∗ ) = a + bx + x ∆ n − e − β (1 + e − P − α β ) + xR n − (3) e − β (1 + e − P − α β ) c + dx + R n − (3) e − β (1 + e − P − α β )= x + x ∆ n − e − β (1 + e − P − α β ) c + dx + R n − (3) e − β (1 + e − P − α β ) . (29)Remember that by definition we have x = e β H (0 ∞ ) H (1 ∞ ) . Now Equation (22) yields H (1 n ∗ ) H (1 ∞ ) e P + e − αβ ( e P − e − ( n − P + β n − β = dx + c + R n − (3) e − β (1 + e − P − α β ) . If n goes to + ∞ the right hand side term of this equality goes to c + dx . On the other side it isalways non-negative. This shows that c + dx is always non-negative. Therefore (29) yields (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) e β (1 − n − ) H (0 n ∗ ) H (1 n ∗ ) × e β H (0 ∞ ) H (1 ∞ ) − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ | ∆ n − | R n − (3) . Now, note that R n − (1) = F ( P, β n − ) and R n − (3) = F ( P, β n − ). Then, Lemma 3.2 shows that | ∆ n − | R n − (3) is of order P ( β ) β n e β n − . Remember that P converges to 0 at least in e − β . For n ≥ β sufficiently big, P ( β ) β n e β n − is lower than e − β .Now Equation (28) and Lemma 5.1 show that we get for every β ≥ β e β H (0 ∞ ) H (1 ∞ ) (1 − e − β ) ν ([0 ∗ ]) ν ([1 ∗ ≤ µ ([0]) µ ([1]) ≤ e β H (0 ∞ ) H (1 ∞ ) (1 + e − β ) ν ([0 ∗ ]) ν ([1 ∗ , (for β big the terms µ ([0 k ∗ ]) and µ ([1 k ∗ ]), k = 1 , µ β “goes” to acombination of δ ∞ and δ ∞ ). Then Corollary 3.7 and Proposition 4.4 conclude the proof.17 eferences [1] N. Anantharaman, R. Iturriaga, P. Padilla and H. Sanchez-Morgado Physical solutions ofthe Hamilton–Jacobi equation. Disc. Cont. Dyn. Syst. Ser. B no. 3, 513–528 (2005).[2] R. Bowen. Equilibrium States and the Ergodic Theory of Anosov Diffeomorphisms , volume470 of
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