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CHAPTER XIIIMARKOV EXTENSIONS FOR INTERVAL MAPSIn section 4.3 we introduced piecewise expanding Markov interval maps andshowed they can be effectively modelled by a (one sided) subshift of finite type. Thiswas particularly useful in estimating numbers of periodic points. In this chapter weshall describe a more general construction which works for non-Markov piecewiseexpanding maps.13.1 Basic constructionsAssume that T : I → I is a interval map which is expanding on each of theintervals I = I1 ∪ .
. . ∪ Ik . We shall denote their endpoints by {x0 , x1 , . . . , xk } (i.e.I1 = [x0 , x1 ], . . . , Ik = [xk−1 , xk ]).The following simple lemma will be used several times.Lemma 13.1. Let J be a sub interval of one of the intervals I1 , . . . , Ik . The imageT (J) will be a union either of the form:(a)(b)(c)(d)T (J) = DJ ⊂ Ii ;T (J) = ∪tj=s Ij ;T (J) = DJ− ∪ ∪tj=s Ij or T (J) = ∪tj=s Ij ∪ DJ+T (J) = DJ− ∪ ∪tj=s Ij ∪ DJ+where DJ , DJ− ⊂ Is−1 and DJ+ ⊂ It+1 are each closed non-degenerate (strict)sub-intervals.fifth.epsFigure 13.1. The image T (J) of J ∈ E gives rise to two more intervalsDJ− and DJ+ in EProof.
Since T (J) is a closed interval in I it is clear that this exhausts all possibilities.Definition. We call E0 = {I1 , . . . , Ik } the zeroth generation of intervalsWe can apply Lemma 13.1 to each J = Ii (i = 1, . . . , n) to get new sub-intervals.For example, DIi (when T (Ii ) is as in case (a) or (c)) and DI−i , DI+i (when T (Ii ) isas in case (d)). If T (Ii ) corresponds to case (b) then no new intervals are created.Definition. We call E1 = {DI−i , DI+i or DIi : i = 1, . . . , k} the first generation ofintervals.We define collections of intervals En , n ≥ 1, (each a closed sub-interval of oneof the original intervals I1 , .
. . , Ik ) inductively. Given a collection of intervals En−1(each a closed sub-interval of one of the original intervals I1 , . . . , Ik ) then we canTypeset by AMS-TEX125126XIII. MARKOV EXTENSIONS FOR INTERVAL MAPSapply Lemma 13.1 to each interval J ∈ En−1 . We define the nth generation ofintervals to be the collection of closed intervalsEn = {DJ− , DJ+ or DJ : J ∈ En−1 }We write E = ∪∞n=0 En .If J ∈ En , say, then each of the sets J 0 in the appropriate union for T (J) (inLemma 13.1) is called a successor to J.
We write J → J 0 .The essence of this construction is to compensate for the absence of the Markovproperty by introducing more and more intervals so that the image of any J ∈ E isagain the union of other intervals from E. The complications over and above thosein the Markov case are that the larger family E will not in general be finite, andintervals in E will not in general have disjoint interiors.The following simple example illustrates this construction.Example (β-transformation). Let I = [0, 1] and choose 2 > β > 1.
We can definean interval map T : I → I by(βxif 0 ≤ x ≤ β1Tx =βx − 1 if β1 < x ≤ 1In this case, we have that E0 = {[0, β1 ], [ β1 , 1]}. Since β < 2 we have that T [0, β1 ] =[0, β1 ] ∪ [ β1 , 1] and T [ β1 , 1] = [0, β1 ] ∪ [ β1 , β − 1]. Thus E1 = {[ β1 , 1 − β]}. If β <51/2 −1,say,2then T [ β1 , β − 1] = [0, β 2 − β − 1] ⊂ [0, β1 ] and so E2 = {[0, β 2 − β − 1]}.For the purposes of exposition we want to make the following simplifying assumption (which, for example, holds for the β-transformation when β > 1 is notan algebraic number.)Simplifying Assumption.
There are no choices 1 ≤ i ≤ k and n, m ≥ 1 such thatT n (T m xi ) = T m xi (i.e. none of the endpoints is pre-periodic).The next lemma shows that there are some restrictions on the possible successorsto a given interval in En .Lemma 13.2 (Restrictions on successors).(1) If J ∈ En (n ≥ 1) then there are two possibilities.
Either J 0 = T (J) or0T (J) = J 00 ∪ ∪s−1i=r Ii ∪ J where(i) J 00 ∈ Ej , for some 0 ≤ j ≤ n (perhaps with J 00 = ∅),(ii) Ir , . . . , Is−1 are original intervals (perhaps with ∪s−1i=r Ii = ∅), and(ii) J 0 ∈ En+1 ,(2) Each element J ∈ En must be of one of the two types:(i) J = T n (Ii ), for some 1 ≤ i ≤ k,(ii) J has endpoints T m xi , T n xj with n − 1 ≥ m ≥ 0 (for some 1 ≤ i, j ≤ k)(i.e.
J = [T m xi , T n xj ] or J = [T m xi , T n xj ])sixth.epsFigure 13.2XIII. MARKOV EXTENSIONS FOR INTERVAL MAPS127Proof. We prove the lemma by induction on n.For n = 1 both parts of the lemma are immediate from Lemma 4.11 and Assumption (2). Assume that both part (1) and part (2) of the lemma are know forJ ∈ En .If we first assume that J = T n Ir = [T n xs , T n xs+1 ], say, then either(a) T (J) = T n+1 Ir ; or(b) xi−1 ≤ T n+1 xs < xi < . . . < xj < T n+1 xs+1 ≤ xj+1 , say (perhaps with theorder reversed)In case (a) we immediately have T (J) ∈ En+1 (i.e. the “either” case in part (1) ofthe statement and case (i) in part (2)).
In case (b) we have by Lemma 13.1 thatthe sucessors to J are [T n+1 xs , xi ], Ii+1 , . . . , Ij , [xj , T n+1 xs+1 ], (i.e. the “or” casein part (1) of the statement). Moreover, we see that [T n+1 xs , xi ] ∈ En+1 is in theform of case (ii) of part (2).Alternatively, we can assume J = [a, b] = [T m (xj ), T n (xk )] ∈ En , say, with0 ≤ m ≤ n − 1 (the case with the endpoints reversed being similar).There are now two possibilities.(c) xr−1 < T (a) < xr < .
. . < xs < T (b) < xs+1 , say.(d) xr < T (a) < T (b) < xr+1(where in each case T (a) and T (b) might be in the reverse order).In case (c) we can take J 0 = [T (a), xr ] ∈ En+1 and J 00 = [xs , T (xj )] ∈ Em+1 . Incase (d) we take J 00 = [T (a), T (b)] ∈ En+1 . (The case xr < T (b) < T (a) < xr+1being similar). These both correspond to the “Or” case of part (1) in the statement.Moreover, in both cases (c) and (d) J 00 ∈ En+1This completes the proof of the LemmaRemarks.(a) If we don’t assume the simplifying Assumption then there are yet more possibilities with J 0 ∈ Di , 1 ≤ i ≤ n.(b) Since the map T : I → I is clearly piecewise expanding, there are at mostfinitely many successors of the form T n (Ii ).(c) Assume that x ∈ J0 = [a, b] ∈ En .
It is useful to enumerate the variouspossibilities J0 → J1 with T (x) ∈ J1 (cf. Figure 13.3):(1)(2)(3)(4)IfIfIfIfT a < xi < T x < T b < xi+1T a < xi < T x < xi+1 < T bxi < T a < T x < xi+1 < T bxi < T a < T x < T b < xi+1thenthenthenthenJ1J1J1J1= [xi , T b];= [xi , xi+1 ];= [T a, xi+1 ]= [T a, T b]and four more cases if T |J reverses orientation.What characterises these cases is:(i) In case (2) none of the endpoints of J1 are images of J0 ;(ii) in cases (1) and (3) one of the end points of J1 is the image of an endpointof J0 ; and,(iii) in case (4) both of the endpoints of J1 are images of endpoints of J0 .seventh.epsFigure 13.3128XIII.
MARKOV EXTENSIONS FOR INTERVAL MAPSLet (Jn )n∈Z+ be a sequence of sucessor intervals (i.e.J0 → J1 → J2 → . . . ) thenwe can associate a unique element π ((Jn )n∈Z+ ) ∈ I byπ ((Jn )n∈Z+ ) = ∩n∈Z+ T −n Jn(in complete analogy with the Markov case).Although x ∈ I will not in general have a unique sequence (Jn )n∈Z+ such thatπ ((Jn )n∈Z+ ), the next lemma shows the remarkable fact that (generically) the “tail”of such a sequence is uniquely determined.Lemma 13.3. Let (Jn )n∈Z+ and (Jn0 )n∈Z+ be two sequences of sucessors such thatπ ((Jn )n∈Z+ ) = π ((Jn )n∈Z+ ) = x ∈ I. = ∩n∈Z+ T −n Jn0 ). If ∀m ≥ 0, T m (x) 6∈∪n≥0 T n {x1 , .
. . , xk } then ∃N ≥ 0 with Jn = Jn0 for n ≥ N .Proof.Let us denote J0 = [a, b] and J00 = [c, d]. The proof is a little involved, dependingon analysing a number of similar cases. Here is a typical case (the others beingvariations on the same theme). The approach is to analyse the endpoints of theintervals Jn and Jn0 and to show that that for sufficiently large n they must agree.Step One (Comparing J0 and J00 ): Assume (for definiteness) that we beginwith a < c < x < b < d. By monotonicity of T on Ii the images of these pointsmust have the same order (or be reversed).
Assume that the order of the images ispreserved and that T a < xj < T c < T x < T b < xj+1 < T d, say. The succesors toJ0 , J00 containing x are then J1 = [xj , T (b)] and J10 = [T (c), xj+1 ] (by cases (1) and(3) above, respectively).0with a common endpoint): Let N be the smallestStep Two (JN and JNpositive integar such that T N (c, b) ∩ {x0 , . . . , xk } 6= ∅. For 1 ≤ n ≤ N − 1 wesee (by repeated application of cases (1),(3) and (4) above) that the succesors Jneach have one of their endpoints being equal to T n (b).
Similarly, Jn0 will have anendpoint equal to T n (c).Assume that we arrive at T N (c) < xi < T N (x) < T N (b) < xi+1 , say, then xi0, by case (1) above. Furthermore,now usurps of T N (c) as the (left) endpoint of JNxi is also the (left) endpoint of JN .eighth.epsFigure 4.8Step Three (Jn and Jn0 (n ≥ N ) with a common endpoint): We need thefollowing easy sublemma.0Sublemma 13.3.1. If JN ⊂ JNwith one (or both) of their endpoints the same,0then the same is true for Jn , Jn , n ≥ N .Proof of sublemma 13.3.1. This is something which one sees from the four cases(1) to (4) above.0Four (JM and JM(M ≥ N ) with two common endpoints): Assume thatM > N is the least integar for which T M (x, b) ∩ {x0 , .
. . , xk } =6 ∅. We see fromXIII. MARKOV EXTENSIONS FOR INTERVAL MAPS129sublemma 4.13.1 that for N ≤ n < M the point T n−N xi is an endpoint for Jn , Jn0 .Assume that T M (x) < xj < T M (b), say, then xj usurps T M (b) as an endpoint for0JM (by cases (1) or (2) above). Moreover, xj is also an end point for JM. We00conclude that JM = JM and so Jn = Jn for n ≥ M (by sublemma 4.13.1 nowapplied to both endpoints).The other cases are similar. This completes the proof ♣If T : I → I is not Markov then E = ∪∞n=0 En is infinite (by the SimplifyingAssumption). We can define an infinite matrix A with entries 0 or 1, whose rowsand columns are indexed by E, in the following wayDefinition.
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