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For J1 , J2 ∈ EA(J1 , J2 ) =1if J2 succeeds J10otherwiseThe next Proposition gives us (amongst other things) a nice characterization ofperiodic 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 forendpoints 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 Lemma4.12 (i) that each J ∈ En gives rise to at most one element J 0 ∈ En+1 . ThusCardEn+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 withJm+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) = xAssume 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+ agreefor 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 andq, 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 areonly 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.130XIII. MARKOV EXTENSIONS FOR INTERVAL MAPSCorollary 13.4.1. The number of periodic points T n (x) = x of period n is givenbyXCardF ix(T n ) :=An (J, J)J∈E13.2 The growth of periodic pointsIn this section we want apply the analysis from the previous section to studythe number of periodic points CardF ic(T n ) via the matrix A. Rather than tryto analyse infinite matrices A it is simpler to fix some large k1 and consider thetruncated (finite) matrix Ak (J, J 0 ) = A(J, J 0 ) if J, J 0 ∈ ∪kn=0 EnProposition 4.?? we know that every string of sucessors corresponding to a fixedpoint for T n : I → I must be of length n.

Moreover, since we can always choosea string starting in E0 then the uniqueness conclusion of Proposition 13.(ii) tell usthat the string contains an interval from E0 . Finally we see by 13.2 that all theelements of the string must therefore be in ∪ni=1 Ei . Thus by Corollary 13.4.1 we seethatXXCardF ix(T n ) =An (J, J) =Ank (J, J) = trace(Ank )J∈EJ∈Ewhenever k ≥ n.Since Ak can be thought of as a finite matrix, we have the standard matrixidentity!∞Xzmmtrace(Ak )Det(I − zAk ) = expmm=1mThe right hand-side carries information on the traces trace(Ak ).

To study theAN B Nleft land side, it helps to fix 1 ≤ N ≤ k and write Ak = CN DN where AN isthe square submatrix with indices ∪Ni=0 Ei , and DN is the square sub-matrix withindices ∪ki=N +1 Ei , etc.Lemma 13.5. For any k ≥ N ≥ 1, the sub-matrix DN has all of its eigenvalueswith modulus at most N 1/NWe shall postpone the proof until later and first consider the implications of thislemma.1Since DN − zI is invertable whenever |z| ≥ N N (by Lemma 13.5), we have thematrix identityAN − zCNBNDN − z=(AN − z) − BN (DN − z)−1 C0B(DN − z)−111C0D−zand taking determinants gives thatDet(Ak − zI) = Det (AN − z) − BN (DN − z)−1 CB(DN − z)−1 Det (DN − z) .The argument needs completingThe following theorem shows that the number of periodic points has a verydistinctive pattern..XIII. MARKOV EXTENSIONS FOR INTERVAL MAPS131Theorem 4.16.

For any β > 1 there exist constants(i) λ1 , . . . , λM , with |λi | > β (i = 1, . . . , M ), and(ii) C1 , . . . , CM ∈ Csuch thatMXN (n) =Ci λni + E(n)i=1where the error term E(n) satisfies |E(n)| ≤ Constantβ nProof. 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 Lemma4.15).The argument needs completingFinally,withsomeadditionalbounds we can let k → +∞ to see that the powerP∞ zmmseries m=1 m traceAk .It only remains to supply the proof of Lemma 4.15Proof of LemmanP4.15. We can define a norm on the matrixo Dk (and its powers) by0k||DN || = supD 0 ∈∪kEi DN (D, D ) : D ∈ ∪i=N +1 Ei .

By the spectral radiusi=N +111n ntheorem it suffices to show that lim supn→+∞ ||DN|| ≤ N NFor p ≥ 0 we let N (p) denote the number of possible strings of successorsJ1 → J2 → . . . → Jp−1 → Jpwith Ji ∈ ∪∞m=k+1 Em for 1 ≤ i ≤ p. If we set0 if r ≤ 0Tr =2k if r = 1and 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 thatN (r) ≤ Tr for r = 1, .

. . , p. Consider a typical string of successorsJ1 → 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 onlyl+1posibility 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 isbounded by N (p − k) ≤ Tn−k .(ii) Assume that J2 is a unique sucessor with J2 ∈ En+1 .

Then by the inductivehypothesis there are at most N (n) ≤ Tn choices for the remaining succesorsJi , 2 ≤ i ≤ n + 1(iii) Finally, the possibility J2 ∈ E0 is eliminated by our requirement that J2 ∈∪∞m=k+1 Em .132XIII. MARKOV EXTENSIONS FOR INTERVAL MAPSThus we see that Tp+1 ≤ Tp + Tp−k , completing the inductive step.We now see thatTp ≤ Tp−1 + Tp−k−1≤ (Tp−2 + Tp−k−2 ) + Tp−k−1≤ . . .

≤ Tp−k + (Tp−2k + . . . + Tp−k−2 ) + Tp−k−1 )≤ kTp−knand so Tmk ≤ k m T0 , for m ≥ 1. In particular, we see that ||DN|| ≤ N (n) for n ≥ 1.nSince ||(A − Ak ) || is sub-additive in n we see that1/nlim sup ||(A − Ak )n ||1n→+∞1= lim sup ||(A − Ak )mk ||1mkm→+∞1≤ lim sup N (mk) mkm→+∞11≤ lim sup(Tmk ) mk ≤ k km→+∞This completes the proof of the sublemma.13.3 Comments and referencesThe theory of Markov extensions was originated by Hofbauer [H]. The mainapplication we give to periodic points is based on work of Hofbauer and Keller.References1. F.

Hofbauer, ?????, Israel J. Math 34 ??? (19??), ???-???.2. F. Hofbauer and G. Keller, ?????, ????? ??? (??), ???-???..

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