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CHAPTER XIII

MARKOV EXTENSIONS FOR INTERVAL MAPS

In section 4.3 we introduced piecewise expanding Markov interval maps and showed they can be effectively modelled by a (one sided) subshift of finite type. This was particularly useful in estimating numbers of periodic points. In this chapter we shall describe a more general construction which works for non-Markov piecewise expanding maps. 13.1 Basic constructions Assume that T : I → I is a interval map which is expanding on each of the intervals 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 image T (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.eps Figure 13.1. The image T (J) of J ∈ E gives rise to two more intervals DJ− and DJ+ in E Proof. 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 intervals We 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 ) is as 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 of intervals. We define collections of intervals En , n ≥ 1, (each a closed sub-interval of one of 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 can Typeset by AMS-TEX 125

126

XIII. MARKOV EXTENSIONS FOR INTERVAL MAPS

apply Lemma 13.1 to each interval J ∈ En−1 . We define the nth generation of intervals to be the collection of closed intervals En = {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) (in Lemma 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 Markov property by introducing more and more intervals so that the image of any J ∈ E is again the union of other intervals from E. The complications over and above those in the Markov case are that the larger family E will not in general be finite, and intervals 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 define an interval map T : I → I by ( βx if 0 ≤ x ≤ β1 Tx = βx − 1 if β1 < x ≤ 1 In 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, 2

then 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 not an algebraic number.) Simplifying Assumption. There are no choices 1 ≤ i ≤ k and n, m ≥ 1 such that T 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 successors to 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) or 0 T (J) = J 00 ∪ ∪s−1 i=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−1 i=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.eps Figure 13.2

XIII. MARKOV EXTENSIONS FOR INTERVAL MAPS

127

Proof. 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 for J ∈ 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 the order reversed) In case (a) we immediately have T (J) ∈ En+1 (i.e. the “either” case in part (1) of the statement and case (i) in part (2)). In case (b) we have by Lemma 13.1 that the sucessors to J are [T n+1 xs , xi ], Ii+1 , . . . , Ij , [xj , T n+1 xs+1 ], (i.e. the “or” case in part (1) of the statement). Moreover, we see that [T n+1 xs , xi ] ∈ En+1 is in the form of case (ii) of part (2). Alternatively, we can assume J = [a, b] = [T m (xj ), T n (xk )] ∈ En , say, with 0 ≤ 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 . In case (d) we take J 00 = [T (a), T (b)] ∈ En+1 . (The case xr < T (b) < T (a) < xr+1 being similar). These both correspond to the “Or” case of part (1) in the statement. Moreover, in both cases (c) and (d) J 00 ∈ En+1 This completes the proof of the Lemma Remarks. (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 most finitely many successors of the form T n (Ii ). (c) Assume that x ∈ J0 = [a, b] ∈ En . It is useful to enumerate the various possibilities J0 → J1 with T (x) ∈ J1 (cf. Figure 13.3): (1) (2) (3) (4)

If If If If

T a < xi < T x < T b < xi+1 T a < xi < T x < xi+1 < T b xi < T a < T x < xi+1 < T b xi < T a < T x < T b < xi+1

then then then then

J1 J1 J1 J1

= [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 endpoint of J0 ; and, (iii) in case (4) both of the endpoints of J1 are images of endpoints of J0 .

seventh.eps Figure 13.3

128

XIII. MARKOV EXTENSIONS FOR INTERVAL MAPS

Let (Jn )n∈Z+ be a sequence of sucessor intervals (i.e.J0 → J1 → J2 → . . . ) then we can associate a unique element π ((Jn )n∈Z+ ) ∈ I by π ((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 that π ((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 that π ((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, depending on analysing a number of similar cases. Here is a typical case (the others being variations on the same theme). The approach is to analyse the endpoints of the intervals 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 begin with a < c < x < b < d. By monotonicity of T on Ii the images of these points must have the same order (or be reversed). Assume that the order of the images is preserved and that T a < xj < T c < T x < T b < xj+1 < T d, say. The succesors to J0 , J00 containing x are then J1 = [xj , T (b)] and J10 = [T (c), xj+1 ] (by cases (1) and (3) above, respectively). 0 with a common endpoint): Let N be the smallest Step Two (JN and JN positive integar such that T N (c, b) ∩ {x0 , . . . , xk } 6= ∅. For 1 ≤ n ≤ N − 1 we see (by repeated application of cases (1),(3) and (4) above) that the succesors Jn each have one of their endpoints being equal to T n (b). Similarly, Jn0 will have an endpoint equal to T n (c). Assume that we arrive at T N (c) < xi < T N (x) < T N (b) < xi+1 , say, then xi 0 , by case (1) above. Furthermore, now usurps of T N (c) as the (left) endpoint of JN xi is also the (left) endpoint of JN . eighth.eps Figure 4.8 Step Three (Jn and Jn0 (n ≥ N ) with a common endpoint): We need the following easy sublemma. 0 Sublemma 13.3.1. If JN ⊂ JN with one (or both) of their endpoints the same, 0 then 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. 0 Four (JM and JM (M ≥ N ) with two common endpoints): Assume that M > N is the least integar for which T M (x, b) ∩ {x0 , . . . , xk } = 6 ∅. We see from

XIII. MARKOV EXTENSIONS FOR INTERVAL MAPS

129

sublemma 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 for 0 JM (by cases (1) or (2) above). Moreover, xj is also an end point for JM . We 0 0 conclude that JM = JM and so Jn = Jn for n ≥ M (by sublemma 4.13.1 now applied 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 Simplifying Assumption). We can define an infinite matrix A with entries 0 or 1, whose rows and columns are indexed by E, in the following way Definition. For J1 , J2 ∈ E A(J1 , J2 ) =



1

if J2 succeeds J1

0

otherwise

The next Proposition gives us (amongst other things) a nice characterization of periodic orbits for T : I → I. Proposition 13.4. (i) There exists C > 0 such that for all n ≥ 1 CardEn ≤ C (ii) There is a bijection between periodic points T n x = x and strings (i0 , i1 , . . . , in−1 ) such that A(i0 , i1 ) = A(i1 , i2 ) = . . . = A(in−1 , i0 ) = 1 (except possibly for endpoints of the original intervals Ii ) (iii) There exists N > 0 such that every J ∈ E has at most N succesors. Proof. (i) From Lemma 13.1 we see that each Ii (i = 1, . . . , k) gives rise to either two, one or no elements in E1 . Thus CardE1 ≤ 2k. For n ≥ 1 we see from Lemma 4.12 (i) that each J ∈ En gives rise to at most one element J 0 ∈ En+1 . Thus CardEn+1 ≤ CardEn ≤ . . . ≤ CardEN . Thus part (i) is proved with C = 2k. (ii) Let (Jn )∞ n=0 be a sequence of sucessors which is periodic (i.e. ∃n ≥ 1 with Jm+n = Jm , for k ≥ 0). Then if x ∈ I is the corresponding point (i.e. {x} = ∞ ∩m≥0 T −m Jm ) then T n (x) is the corresponding point for (Jn+m )∞ n=0 = (Jn )n=0 (i.e.{T m (x)} = ∩m≥0 T −m Jn+m ). Thus T n (x) = x Assume that T n (x) = x and choose a corresponding sequence of sucessors (Jn )n∈Z+ . We know by Lemma 13.3 that provided x 6∈ ∪n≥0 T −n (∪m≥0 T m {x1 , . . . , xk }) then the two sequences of sucesssors (for x = T n x) (Jn )n∈Z+ and (Jn+m )n∈Z+ agree for n ≥ N , say. If we choose nm ≥ N then the periodic sequence (Jn0 )n∈Z+ = (Jn+mN )n∈Z+ correspons to x since {x = T mN x} = ∩n∈Z+ T −n Jn0 . Assume that T n (x) = x and T m (x) = T q (xi ), for some 1 ≤ i ≤ k and q, m ≥ 0. In particular, if we choose (r − 1)n ≤ m < rn then we see x = T rn (x) = T q+(rn−m) (xi ). (i.e. x is characterised as the (unique) periodic orbit {x, T x, . . . , T n−1 x} into which the orbit of xi eventually falls). Since there are only k boundary points x1 , . . . , xk this means the bijection fails on at most k periodic orbits. (iii) This is an easy consequence of the definitions.

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Corollary 13.4.1. The number of periodic points T n (x) = x of period n is given by X CardF ix(T n ) := An (J, J) J∈E

13.2 The growth of periodic points In this section we want apply the analysis from the previous section to study the number of periodic points CardF ic(T n ) via the matrix A. Rather than try to analyse infinite matrices A it is simpler to fix some large k1 and consider the truncated (finite) matrix Ak (J, J 0 ) = A(J, J 0 ) if J, J 0 ∈ ∪kn=0 En Proposition 4.?? we know that every string of sucessors corresponding to a fixed point for T n : I → I must be of length n. Moreover, since we can always choose a string starting in E0 then the uniqueness conclusion of Proposition 13.(ii) tell us that the string contains an interval from E0 . Finally we see by 13.2 that all the elements of the string must therefore be in ∪ni=1 Ei . Thus by Corollary 13.4.1 we see that X X CardF ix(T n ) = An (J, J) = Ank (J, J) = trace(Ank ) J∈E

J∈E

whenever k ≥ n. Since Ak can be thought of as a finite matrix, we have the standard matrix identity ! ∞ X zm m trace(Ak ) Det(I − zAk ) = exp m m=1 m The right hand-side carries information on the traces trace(A k ).  To study the  AN B N left land side, it helps to fix 1 ≤ N ≤ k and write Ak = CN DN where AN is

the square submatrix with indices ∪N i=0 Ei , and DN is the square sub-matrix with indices ∪ki=N +1 Ei , etc. Lemma 13.5. For any k ≥ N ≥ 1, the sub-matrix DN has all of its eigenvalues with modulus at most N 1/N We shall postpone the proof until later and first consider the implications of this lemma. 1

Since DN − zI is invertable whenever |z| ≥ N N (by Lemma 13.5), we have the matrix identity 

AN − z CN

BN DN − z



=



(AN − z) − BN (DN − z)−1 C 0

B(DN − z)−1 1



1 C

0 D−z

and taking determinants gives that
Det(Ak − zI) = Det (AN − z) − BN (DN − z)−1 CB(DN − z)−1 Det (DN − z) . The argument needs completing The following theorem shows that the number of periodic points has a very distinctive pattern.



.

XIII. MARKOV EXTENSIONS FOR INTERVAL MAPS

131

Theorem 4.16. For any β > 1 there exist constants (i) λ1 , . . . , λM , with |λi | > β (i = 1, . . . , M ), and (ii) C1 , . . . , CM ∈ C such that M X N (n) = Ci λni + E(n) i=1

where the error term E(n) satisfies |E(n)| ≤ Constantβ n Proof. The above identity shows that provided N 1/n < β the eigenvalues of Ak ( for any k ≥ N ) are determined by the determinant of a N × N matrix (by Lemma 4.15). The argument needs completing Finally, with some additional bounds we can let k → +∞ to see that the power P∞ zm m series m=1 m traceAk . It only remains to supply the proof of Lemma 4.15 Proof of Lemma nP4.15. We can define a norm on the matrix o Dk (and its powers) by 0 k ||DN || = sup D 0 ∈∪k Ei DN (D, D ) : D ∈ ∪i=N +1 Ei . By the spectral radius i=N +1

1

1

n n theorem it suffices to show that lim supn→+∞ ||DN || ≤ N N For p ≥ 0 we let N (p) denote the number of possible strings of successors

J1 → J2 → . . . → Jp−1 → Jp with Ji ∈ ∪∞ m=k+1 Em for 1 ≤ i ≤ p. If we set  0 if r ≤ 0 Tr = 2k if r = 1 and then define inductively Tr+1 = Tr +Tr−k (r ≥ 1) then we claim that N (p) ≤ Tp . For p = 1 this is true by definition. We proceed by induction. and assume that N (r) ≤ Tr for r = 1, . . . , p. Consider a typical string of successors J1 → J2 → . . . → Jp → Jp+1 (of length p + 1) with Ji ∈ ∪∞ m=k+1 Em for 1 ≤ i ≤ p + 1. By Lemma 4.12 we see that there are three possible types of successor for J1 ∈ En (n ≥ k). There are three possibilities (i) If J1 = [T n xi , T l xj ] (with 0 ≤ l ≤ n − 1), say, then assume that J2 = [xk , T l xj ], say. To satisfy Ji ∈ ∪∞ m=k+1 Em for 2 ≤ i ≤ k + 3 the only l+1 posibility is that J3 = [T xq , T xj ], J4 = [T 2 xq , T l+2 xj ], . . . , Ji+2 = [T i xq , T l+i xj ], . . . , Jk+3 = [T k+1 xq , T l+k+1 xj ] (or with the endpoints reversed). Thus only the remaining intervals Jk+4 , . . . , Jp+1 are not determined, and by the inductive hypothesis the number of possible choices is bounded by N (p − k) ≤ Tn−k . (ii) Assume that J2 is a unique sucessor with J2 ∈ En+1 . Then by the inductive hypothesis there are at most N (n) ≤ Tn choices for the remaining succesors Ji , 2 ≤ i ≤ n + 1 (iii) Finally, the possibility J2 ∈ E0 is eliminated by our requirement that J2 ∈ ∪∞ m=k+1 Em .

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Thus we see that Tp+1 ≤ Tp + Tp−k , completing the inductive step. We now see that Tp ≤ Tp−1 + Tp−k−1 ≤ (Tp−2 + Tp−k−2 ) + Tp−k−1 ≤ . . . ≤ Tp−k + (Tp−2k + . . . + Tp−k−2 ) + Tp−k−1 ) ≤ kTp−k n and so Tmk ≤ k m T0 , for m ≥ 1. In particular, we see that ||DN || ≤ N (n) for n ≥ 1. n Since ||(A − Ak ) || is sub-additive in n we see that 1/n

lim sup ||(A − Ak )n ||1 n→+∞

1

= lim sup ||(A − Ak )mk ||1mk m→+∞

1

≤ lim sup N (mk) mk m→+∞

1

1

≤ lim sup(Tmk ) mk ≤ k k m→+∞

This completes the proof of the sublemma. 13.3 Comments and references The theory of Markov extensions was originated by Hofbauer [H]. The main application we give to periodic points is based on work of Hofbauer and Keller. References 1. F. Hofbauer, ?????, Israel J. Math 34 ??? (19??), ???-???. 2. F. Hofbauer and G. Keller, ?????, ????? ??? (??), ???-???.