We show that the nonwandering set consists of finitely many periodic orbits and an invariant setLwhich is topologically transitive and the disjoint union of finitely many closed intervals

Vol. LXXXI, 1 (2012), pp. 15–30

A TWO PARAMETER FAMILY OF PIECEWISE LINEAR TRANSFORMATIONS WITH NEGATIVE SLOPE

F. HOFBAUER

Abstract. We study a two parameter family of piecewise linear transformations on the interval [0,1] which have negative slope. We show that the nonwandering set consists of finitely many periodic orbits and an invariant setLwhich is topologically transitive and the disjoint union of finitely many closed intervals. We determine the number of these intervals.

1. Introduction

For a real numberβ >1, the beta transformation is defined byx7→βxmod 1 on the unit interval [0,1]. It can be used to generateβ-expansions of real numbers.

It is always topologically transitive and was first investigated in [9] and [8]. More recently, in [5] and [1], a beta transformation with negative slope was used to generate expansions with negative bases. It is defined on [0,1] byx7→ −βxmod 1.

It has more complicated dynamics as shown in [7]. Here we introduce a two parameter generalization of this negative beta transformation.

SetG= {(α, β) : α >1, β > 1, αβ−α−β <0}. We choose this set as the parameter space. We defineT : [0,1]→[0,1] by

T(x) =







1−αx if x∈M0=h 0,1

α i

, 1 +β

α−βx if x∈M1=1 α,1i

. (1)

Ifβ∈(0,1), then there is an attracting fixed point inM1, which attracts all orbits except the fixed point inM0. Ifβ = 1, thenT² is the identity onM1. Therefore, we assumeβ >1. Forα=β, we get the negative beta transformation.

We consider the nonwandering set Ω(T) of the transformationT defined by (1).

In particular, we are interested in the dependence of Ω(T) on the parametersαand β. This dependence was already investigated for other two parameter families of piecewise linear transformations, for tent maps in [4] and for the transformations x7→βx+αmod 1 in [2].

Received February 1, 2011.

2010Mathematics Subject Classification. Primary 37E05; Secondary 37B10.

Key words and phrases. Piecewise linear map; nonwandering set.

F. HOFBAUER

Before we state the results, we need some definitions. We setδn= 0 ifnis even andδn = 1 ifnis odd. We define sequences (cn)_n≥0and (dn)_n≥0by

(2) c0=d0= 1 and cn+1= 2cn−2δn, dn+1= 2dn+δn−1 for n≥0.

Using these sequences we define the sets

G0=G and Gn ={(α, β)∈G:α^cnβ^dn−α²−β <0} for n≥1.

We haveG_n+1⊂G_n and we define

Hn=Gn\Gn+1 for n≥0

which is a nonempty set. Then (H_n)_n≥0 is a partition of the parameter spaceG.

Furthermore, sets_n=c_n+d_n−1 forn≥0. It follows from (2) that s0= 1 and sn+1= 2sn−δn for n≥0.

(3)

We prove the following results.

The transformationT : [0,1]→[0,1] defined by (1) is topologically transitive, if (α, β)∈H0. Forn≥1 and (α, β)∈Hn, we have

Ω(T) =L∪

n

[

k=1

Pk

wherePk is a periodic orbit of periodsk−sk−1andLis a topologically transitive T-invariant subset of [0,1] which is the disjoint union ofsn closed intervals.

Figure 1 shows the parameter spaceGand the curvesα^cnβ^dn−α²−β= 0 for 1 ≤n ≤4. The parameter space Gis partitioned into the sets H₀, H₁, . . . by

Figure 1. Parameter space.

these curves. One sees thatH0,H1andH2are unbounded, whereasG3and hence also the setsHn withn≥3 are bounded.

It is well known that a piecewise linear transformation whose slopes have ab- solute values > 1, in particular, the transformation T defined by (1), has an

PIECEWISE LINEAR TRANSFORMATIONS WITH NEGATIVE SLOPE

absolutely continuous invariant measure whose support is a finite union of closed intervals (see e.g. [6]). Since this support isT-invariant and a subset of Ω(T), it must coincide withL. In the topologically transitive case it coincides with [0,1].

Hence the above result gives also the number of intervals of which the support of the absolutely continuous invariant measure consists.

The negative beta transformation is the special case of (1) where α=β. Set β0 = 2 and for n≥ 1, let βn be the largest solution of β^sn−β −1 = 0. Then (β, β) ∈ Hn is equivalent to βn+1 ≤ β < βn. Therefore, for the negative beta transformation the supportLof the absolutely continuous invariant measure con- sists ofs_ndisjoint closed intervals ifβ∈[β_n+1, β_n). This is proved in [7]. It is also proved there thatT|Lis topologically exact which implies topological transitivity.

The paper is organized as follows. In Section 2 we prove thatT is topologically transitive if (α, β)∈ H₀ and investigate Ω(T) for (α, β)∈H₁. For (α, β)∈ G₁, we find T³(0) < T²(0) < _α¹ and T²(0) < T⁴(0). The fixed point P₁ = {_1+α¹ } lies betweenT³(0) andT²(0) and all points in (T³(0), T²(0))\P₁ are wandering.

For (α, β)∈H₁ we show then that theT-invariant setL= [0, T³(0)]∪[T²(0),1], which is the disjoint union ofs1 = 2 closed intervals, is topologically transitive.

Hence Ω(T) =P1∪L.

If (α, β) ∈ G2, we also have _α¹ < T⁵(0) < T⁴(0) < T⁶(0). In particular, there exists a fixed point P2 which lies between T⁵(0) and T⁴(0), and the set L= [0, T³(0)]∪[T²(0), T⁵(0)]∪[T⁴(0),1] isT-invariant and the disjoint union of s2 = 3 closed intervals. If (α, β) moves from H2 to G3, then each of the three intervals of which L consists splits into two intervals such that L is then the disjoint union ofs3= 6 closed intervals, and there is a periodic orbitP3 of period s₃−s₂= 3 in the three gaps which emerge. If (α, β) moves fromH₃ toG₄, then five of the six intervals of whichLconsists split into two intervals, such thatLis then the disjoint union ofs₄= 11 closed intervals and there is a periodic orbitP₄ of periods₄−s₃= 5 in the five gaps which emerge. It continues in this way.

These further steps for (α, β)∈G_n withn≥2 are treated in Sections 3 and 4 using induction. In Section 3 the orbit of the point 0 is investigated, in particular, its dependence on the parametersαandβ. In Section 4 we findT-invariant subsets which are finite unions of intervals. This leads then to the proof of the results for Ω(T) stated above.

This proof is inspired by the Markov graph which was developed in [3], although we do not introduce it here. The intervals defined in Section 4 are those which occur as vertices in this graph and the course of the proof follows its recursive structure.

2. First steps

LetR: [0,1]→[0,1] be a transformation with two monotone pieces, which means that there isγ∈(0,1) such thatR|_[0,γ] andR|_(γ,1] are continuous and monotone.

We assume that R is expanding, which means that there existsκ >1 such that

|R(I)| ≥κ|I| holds for all intervalsI with γ /∈ I. Here |I| denotes the length of

F. HOFBAUER

the intervalI. Then we have

I⊂[0,1] a nonempty open interval ⇒ γ∈Rⁿ(I) for some n≥0 (4)

since otherwiseRⁿ(I) would be an interval satisfying|Rⁿ(I)| ≥κⁿ|I|for alln≥1, which is impossible asκ >1. IfR: [0,1]→[0,1] is not topologically transitive, it follows from general results for piecewise monotone transformations (see [3]), that there is a set A([0,1] which is R-invariant and a finite union of nondegenerate closed intervals. Using this we can show the following theorem.

Theorem 1. For(α, β)∈H0 the transformationT defined by (1)is topologi- cally transitive.

Proof. The transformation T has two monotone pieces and is expanding with κ= min(α, β). We assume thatT is not topologically transitive and hence there is a setA([0,1] which isT-invariant and a finite union of nondegenerate closed intervals. Letϑbe the fixed point ofT in [0,¹_α]. We haveϑ /∈A since otherwise theT-invariance ofAwould implyA= [0,1]. Using (4) we get also_α¹ ∈intA. Let J be the maximal subinterval ofA, which contains ¹_α. We haveJ = [_α¹−q,_α¹+p]

withp >0 andq >0. SetU = (_α¹,_α¹ +p] andV = [_α¹ −q,_α¹).

By (4) there is a minimaln≥1 satisfying _α¹ ∈Tⁿ(U). ThenTⁿ(U)⊂J since Ais T-invariant and J is the maximal subinterval ofA which contains _α¹. Since U ⊂ (_α¹,1], we have |T(U)| = β|U| and hence also |J| ≥ |Tⁿ(U)| ≥β|U|. This meansβp≤p+q.

Sinceϑ /∈ A, we have V ⊂(ϑ,_α¹) andT(V) ⊂(0, ϑ). Hence |T²(V)| =α²|V| and the left endpoint of T²(V) is larger than _α¹ −qsince α >1. If _α¹ ∈T²(V), thenT²(V)⊂J and we getα²|V|<|J|which meansα²q < p+q. If _α¹ ∈/ T²(V), then there is a minimaln ≥3 satisfying _α¹ ∈ Tⁿ(V) which implies Tⁿ(V) ⊂J. We getα²|V|=|T²(V)|<|Tⁿ(V)| ≤ |J|and again we have α²q < p+q.

We have shown (β−1)p≤qand (α²−1)q < p. This implies (β−1)(α²−1)<1, which is equivalent to (α, β)∈G1. It contradicts (α, β)∈H0and hence topological

transitivity ofT for (α, β)∈H0 is shown.

We need a similar result for tent maps. We use the same parameter spaceGas for the mapT in (1). For (α, β)∈Gwe define the tent mapS: [0,1]→[0,1] by (5) S(x) =







1−β+β

α+βx if x∈[0, γ] α−αx if x∈(γ,1]

where γ= 1− 1 α.

The pointγis called the critical point. This tent map has a unique fixed point in (γ,1] which we denote by%.

Proposition 1. Let S : [0,1] → [0,1] be a tent map as defined in (5) with (α, β)∈G. IfS(0)≤%, thenS is topologically transitive.

Proof. The tent mapSis expanding with κ= min(α, β). We proceed as in the last proof. We assume thatSis not topologically transitive and hence there is a set A([0,1] which isS-invariant and a finite union of nondegenerate closed intervals.

PIECEWISE LINEAR TRANSFORMATIONS WITH NEGATIVE SLOPE

We have% /∈Asince otherwise theS-invariance ofAwould implyA= [0,1]. Using (4) we get alsoγ∈intA. LetJ be the maximal subinterval ofAwhich contains γ. We have J = [γ−q, γ+p] with p > 0 and q > 0. Set U = [γ−q, γ) and V = (γ, γ+p].

We haveU ⊂[0, γ) and S(U) ⊂(%,1] since% /∈A. We get |S²(U)|=αβ|U|.

By (4) there is a minimal n≥ 2 with γ ∈ Sⁿ(U). Then Sⁿ(U)⊂ J. It follows thatαβ|U|=|S²(U)| ≤ |Sⁿ(U)| ≤ |J|. This meansαβq≤p+q.

Since% /∈A, we haveV ⊂(γ, %) andS(V)⊂(%,1). This gives|S²(V)|=α²|V| and the right endpoint of S²(V) is less than γ+p since α > 1. If γ ∈ S²(V), thenS²(V)⊂J and we getα²|V|<|J|which meansα²p < p+q. Ifγ /∈S²(V), then there is a minimaln≥3 withγ∈Sⁿ(V) which impliesSⁿ(V)⊂J. We get α²|V|=|S²(V)|<|Sⁿ(V)| ≤ |J|and again we haveα²p < p+q.

We have shown (αβ−1)q≤pand (α²−1)p < q. Hence (αβ−1)(α²−1)<1.

We assume S(0) ≤%which is equivalent to 1−β+^β_α ≤ _1+α^α . This contradicts (αβ−1)(α²−1)<1 and therefore, topological transitivity ofS is shown.

The behavior of the transformationT defined by (1) is determined by the orbit of the point 0. In order to find Ω(T) for (α, β)∈H1we need to know how an initial segment of this orbit is ordered. Notice thatG₁={(α, β)∈G:α²β−α²−β <0}

andG₂={(α, β)∈G:α²β²−α²−β <0}.

Lemma 1. If(α, β)∈G₁, thenT³(0)< T²(0)< _α¹ < T(0)andT²(0)< T⁴(0).

If(α, β)∈G2, then we haveT³(0)< T²(0)< _α¹ < T⁵(0)< T⁴(0)< T⁶(0)< T(0).

If(α, β)∈H1 andT⁴(0)> _α¹, then we have T⁵(0)≥T⁴(0).

Proof. For all (α, β)∈Gwe have T(0) = 1> 1

α, T²(0) = 1 +β

α−β < 1

α and T³(0) = 1−α−β+αβ.

Suppose that (α, β)∈G₁. Then α²β−α²−β <0. This implies T³(0)< T²(0) andT³(0)<_α¹ follows. Hence we get

T⁴(0) = 1−αT³(0) = 1−α+α²+αβ−α²β

and we haveT²(0)< T⁴(0) which is equivalent to (α−1)(α²β−α²−β)<0.

Additionally to (α, β)∈ G1 we assume now either (α, β)∈ G2 or T⁴(0) > _α¹. We have T⁴(0) > _α¹ also in the case, where (α, β) ∈G2, since this inequality is equivalent to (α−1)(α²β²−α²β−β)<0. Because T⁴(0)> _α¹ we get

T⁵(0) = 1 + β

α−βT⁴(0) = 1 + β

α−β+αβ−α²β−αβ²+α²β² andT⁵(0)>_α¹ holds since it is equivalent to (α²β−1)(α−1)(β−1)>0.

We observe that the inequalityT⁵(0)< T⁴(0) is equivalent to (α−1)(α²β²−α²−β)<0.

F. HOFBAUER

Therefore, we getT⁵(0)< T⁴(0) if (α, β)∈G2, andT⁵(0)≥T⁴(0) if (α, β)∈H1. BecauseT⁵(0)>_α¹, we getT⁶(0) = 1 + ^β_α−βT⁵(0)<1 and hence

T⁶(0) = 1 + β

α−β−β²

α +β²−αβ²+α²β²+αβ³−α²β³.

Therefore, for (α, β)∈G2, we getT⁴(0)< T⁶(0) since this inequality is equivalent

to (α−1)(β−1)(α²β²−α²−β)<0.

We use Lemma 1 to find Ω(T) for (α, β)∈H₁. To this end set K1=1

α,1i

, K2=h

T²(0),1 α i

, K3= [0, T³(0)] and U = (T³(0), T²(0)).

We assume (α, β)∈G1. Then by Lemma 1 these four intervals are disjoint and nonempty. Furthermore, setL1=K1∪K2∪K3which is the disjoint union of two closed intervals. Again by Lemma 1, we getT(L1)⊂L1 andT(U)⊃U. SinceT has slopeα < −1 on U, there is a fixed pointP1 in U and all other points inU are wandering.

Now assume (α, β)∈H1=G1\G2. It remains to show thatL1is topologically transitive. The first return map S to the interval K1∪K2 is T on K1 and T² onK2. HenceS is a tent map on the interval [T²(0),1] with critical point ¹_α and S(T²(0)) = T⁴(0). Since (α, β)∈ H1, we have either T⁴(0) ≤ _α¹ or T⁴(0) > _α¹ andT⁴(0)≤T⁵(0) by Lemma 1. In the second case we getT⁴(0)≤%, where%is the fixed point ofS =T in K1 = (_α¹,1]. Hence in both cases we haveT⁴(0)≤%.

Now Proposition 1 implies that there is a dense orbit inK₁∪K₂ under S. And this implies then that there is a dense orbit in L₁ under T proving that L₁ is topologically transitive. We have also shown that Ω(T) =P₁∪L₁.

3. The orbit of the point zero

In order to determine Ω(T) for (α, β)∈G2we need further properties of the orbit of the point 0. We introduce the kneading sequencee=e0e1e2· · · ∈ {0,1}^N0 of the transformation T. It is defined such that T^j(0) ∈ Me_j holds for all j ≥ 0, where M0 and M1 are as in (1). We analyze the symbolic sequences which can occur as initial segments of the kneading sequence.

Let B be a block consisting of the symbols 0 and 1. We call the number of symbols inB the length of the blockB. We defineB^∗ as follows. IfB ends with 1, then letB^∗ be the blockB with this 1 replaced by 00. IfB ends with 00, then letB^∗be the blockB with this 00 replaced by 1. In particular, we haveB^∗∗=B.

Set B₁ = 1 and for n ≥ 2 set B_n = B_n−1B^∗_n−1. We have then B₂ = 100, B₃ = 10011, B₄ = 10011100100 and so on. Lemma 1 implies thatebegins with 010011 = 0B2B₂^∗= 0B3if (α, β)∈G2.

Forn≥1, letan be the number of zeros andbn be the number of ones in the blockBn. Set rn =an+bn which is the length of the blockBn. In particular, we havea1= 0 andb1=r1= 1. We connect these numbers with those defined in (2) and in (3).

PIECEWISE LINEAR TRANSFORMATIONS WITH NEGATIVE SLOPE

Lemma 2. Forn ≥1, we have an+1 = 2an −2(−1)ⁿ, bn+1 = 2bn + (−1)ⁿ and rn+1 = 2rn−(−1)ⁿ. Furthermore, cn = an + 2δn, dn = bn + 1−δn and sn=rn+δn.

Proof. The recursion formulas for a_n, b_n and r_n follow from the definition B_n+1 =B_nB_n^∗ sinceB_n ends with 00 if n is even and with 1 if n is odd. The equation connecting a_n and c_n and that connecting b_n and d_n are then easily checked by induction using (2). Sinces_n =c_n+d_n−1 andr_n=a_n+b_n hold for allnby definition, also the equation connectingsn andrn follows.

Lemma 3. For n ≥ 1, we have sn+1 = rn +sn, rn+1 = 2sn −1, rn+1 = r_n+ 2r_n−1 andr_n+1−s_n+1=r_n−1−s_n−1.

Proof. These equations can be easily checked using rn+1 = 2rn−(−1)ⁿ and sn = rn +δn which are contained in Lemma 2, and sn+1 = 2sn −δn which is

contained in (3).

The next lemma investigates the orbit of the point 0. Set T0(x) = 1−αx and T1(x) = 1 + β

α−βx.

ThenT^j(0) =Tej−1◦ · · · ◦Te₁◦Te₀(0) for allj ≥1. LetC be a block containing pzeros andq ones, so that l =p+q is the length ofC. If ebegins with 0C1C, then we have

T^2l+2(0) = (−α)^p(−β)^q+1T^l+1(0) +β

α(−α)^p(−β)^q+T^l+1(0).

(6)

This can be shown by induction. Sinceel+1= 1 and henceT^l+1(0)∈M1, we have T^l+2(0) =T₁(T^l+1(0)) = 1 + β

α−βT^l+1(0) =T(0) +β

α−βT^l+1(0).

(7)

If 1≤m≤l and

T^l+m+1(0) =T^m(0) + (−α)^pm(−β)^qmβ

α−βT^l+1(0)

is already shown, wherepm is the number of zeros andqmis the number of ones ine1e2. . . e_m−1, we get

T^l+m+2(0) =







1−αT^m(0)+(−α)^pm+1(−β)^qmβ

α−βT^l+1(0)

ife_m= 0 1+β

α−βT^m(0)+(−α)^pm(−β)^qm+1β

α−βT^l+1(0)

ifem= 1.

In both cases we haveT^l+m+2(0) =T^m+1(0) + (−α)^pm+1(−β)^qm+1(^β_α−βT^l+1(0)).

Hence (6) is shown by induction sincep_l+1=pandq_l+1=q.

Now suppose thatebegins with 0C00C. In this case we have T^2l+3(0) = (−α)^p+2(−β)^qT^l+1(0) + (−α)^p+1(−β)^q+T^l+1(0).

(8)

F. HOFBAUER

This can be proved in the same way as (6), except that we have now T^l+2(0) =T0(T^l+1(0)) = 1−αT^l+1(0) and

T^l+3(0) =T0(T^l+2(0)) = 1−αT^l+2(0) =T(0)−α+α²T^l+1(0) instead of (7). Then we can proceed as above and get (8).

In the following proof we use equations like (6) and (8). All these equations can be proved in a similar way.

Lemma 4. Supposen≥2. If(α, β)∈G_n, thenebegins with0B_nB_n^∗ and T^rn−1+1(0)< T^rn+rn−1+1(0)< T^rn+1(0)< T^rn+1+1(0)< T(0).

Furthermore, T^j(0)6= ¹_α for 0 ≤j ≤rn+1. If (α, β)∈ H_n−1 and if T^sn−1−1(0) andT^{rn+sn−1−1}(0)are on the same side of _α¹, then T^rn+rn−1+1(0)≥T^rn+1(0).

Proof. We have r₁= 1,r₂= 3,r₃= 5 ands₁= 2. Hence forn= 2 the lemma is contained in Lemma 1. We proceed by induction. Suppose thatn≥3 and that the lemma is already proved forn−1 instead ofn. We assume (α, β)∈G_n−1and consider two cases.

First we assume that n−1 is even. Then the block B_n−1 ends with 00. Let C be the block B_n−1 with 00 removed. By induction hypothesis e begins with 0B_n−1B_n−1^∗ = 0C00C1. We set u = a_n−1 and v = b_n−1. Then the block C containsu−2 zeros andvones. Setk=u+v. This is the lengthr_n−1of the block Bn−1 andC has lengthk−2. We have cn−1=uanddn−1=v+ 1 by Lemma 2 and the recursions in Lemma 2 givean+ 2 = cn = 2uand bn =dn = 2v+ 1. It follows thatsn−1=rn−1=k andsn =rn+ 1 = 2k. By induction hypothesis we have also

T^j(0)6= 1

α for 0≤j≤2k−1 (9)

and T^k+1(0) < T^2k(0) < T(0) = 1. Since we have ek+1ek+2. . . e_2k−2 = C and e1e2. . . e_k−2 = C, for 0 ≤ j ≤ k−3, we get that T^2k+j(0) lies in the open interval with endpoints T^k+1+j(0) and T^1+j(0) which is contained either in M0

or M1. This impliesT^j(0)6= _α¹ for 2k≤j ≤3k−3 and e2ke2k+1. . . e_3k−3 =C.

Therefore,ebegins with 0C00C1C. We compute

T^3k−2(0) = (−α)^2u−2(−β)^2v+1T^k−1(0) + (−α)^2u−3(−β)^2v+1 + (−α)^u−2(−β)^v+1T^k−1(0) +β

α(−α)^u−2(−β)^v+T^k−1(0).

Sinceuis even andv is odd by Lemma 2, this implies 1

α−T^3k−2(0) =1

α−T^k−1(0)

(1 +α^u−2β^v+1−α^2u−2β^2v+1).

Either we assume thatT^sn−1−1(0) =T^k−1(0) and T^{rn+sn−1−1}(0) =T^3k−2(0) are on the same side of _α¹, or (α, β)∈G_n which means α^2uβ^2v+1−α²−β < 0. In both cases we get

1 +α^u−2β^v+1−α^2u−2β^2v+1>0.

PIECEWISE LINEAR TRANSFORMATIONS WITH NEGATIVE SLOPE

We always have 1 +α^u−2β^v+1−α^2u−2β^2v+1<1. Furthermore,T^k−1(0)< ¹_αholds becausee_k−1 = 0 and (9). This gives then T^k−1(0)< T^3k−2(0)< _α¹. Therefore, e3k−2= 0 is shown. ApplyingT we have also 0< T^3k−1(0)< T^k(0)< _α¹, where T^k(0)< _α¹ holds becauseek= 0 and (9). This givese_3k−1= 0. ApplyingT again, we getT^k+1(0)< T^3k(0)<1 which means

T^rn−1+1(0)< T^rn+rn−1+1(0)< T(0).

Now we know thatebegins with 0C00C1C00. We compute T^2k(0) = (−α)^u−1(−β)^v+1T^k(0) +β

αT^k(0) + 1 and T^3k(0) = (−α)^u(−β)^vT^2k(0)−(−α)^u(−β)^v−αT^k(0) + 1.

Sinceuis even andv is odd by Lemma 2, we get T^2k(0) =T^k(0)β

α−α^u−1β^v+1

+ 1 and

T^3k(0) =T^k(0)(α^2u−1β^2v+1−α^u−1β^v+1−α) + 1.

If now (α, β)∈G_n, thenα^2uβ^2v+1−α²−β <0 holds. We getT^3k(0)< T^2k(0) which meansT^rn+rn−1+1(0)< T^rn+1(0). On the other hand, if (α, β)∈H_n−1, we haveα^2uβ^2v+1−α²−β≥0 and we getT^rn+rn−1+1(0)≥T^rn+1(0).

From now on we assume (α, β)∈G_n. We have shownT^k+1(0)< T^3k(0)< T(0) above. Sincee_k+1e_k+2. . . e_2k−2=C ande₁e₂. . . e_k−2=C, for 0≤j ≤k−3, we get thatT^3k+j(0) lies in the open interval with endpointsT^k+1+j(0) andT^1+j(0) which is contained either inM0 or M1. Hence T^j(0) 6= _α¹ for 3k ≤j ≤4k−3.

This implies alsoe3ke3k+1. . . e_4k−3=C. Therefore,ebegins with 0C00C1C00C.

We setD=C00C. ThenDcontains 2u−2 zeros and 2vones. The length ofDis 2k−2 andBn=C00C1 =D1. Furthermore,ebegins with 0D1D. We compute

T^4k−2(0) = (−α)^2u−2(−β)^2v+1T^2k−1(0) + β

α(−α)^2u−2(−β)^2v+T^2k−1(0).

This implies 1

α−T^4k−2(0) =

T^2k−1(0)− 1 α

(α^2u−2β^2v+1−1).

Since e_2k−1 = 1, we getT^2k−1(0)> _α¹ using (9). This gives _α¹ −T^4k−2(0)> 0.

Since (α, β)∈Gn, we haveα^2u−2β^2v+1−1<_α^β2 and we get 0< 1

α−T^4k−2(0)<

1− 1 α

β

α² < 1 α(α+ 1).

The last inequality is equivalent toα²β−α²−β <0 which holds because (α, β)∈ G_n ⊂G₂. We have shown _1+α¹ < T^4k−2(0)< _α¹ which gives e_4k−2= 0. Applying T we get T^4k−1(0) < _1+α¹ < _α¹ which gives e_4k−1 = 0. By Lemma 2, we have r_n+1= 2r_n+ 1 = 4k−1. Therefore,T^j(0)6= _α¹ is shown for allj≤r_n+1.

Now we know thatebegins with 0D1D00 = 0B_nB_n^∗. We compute T^4k(0) = (−α)^2u(−β)^2vT^2k(0)−(−α)^2u(−β)^2v−αα

βT^2k(0)−α β

+ 1.

F. HOFBAUER

This implies

1−T^4k(0) = (1−T^2k(0))

α^2uβ^2v−α² β

.

We have 0< α^2uβ^2v−^α_β² <1 since (α, β)∈Gn. Furthermore,T^2k(0)<1 holds by the induction hypothesis. Hence we get T^2k(0) < T^4k(0) < 1 which means T^rn+1(0) < T^rn+1+1(0) < T(0). The lemma is completely proved in the case wheren−1 is even.

Now we assume that n−1 is odd. We proceed as above, but the details are different. In particular, the blockB_n−1ends with 1. LetCbe the blockB_n−1with this 1 removed. By induction hypothesis ebegins with 0B_n−1B_n−1^∗ = 0C1C00.

We setu=a_n−1andv=b_n−1. Then the blockCcontainsuzeros andv−1 ones.

Setk=u+v. This is the lengthr_n−1 of the blockB_n−1 andChas lengthk−1.

We havec_n−1=u+ 2 andd_n−1=v by Lemma 2 and the recursions in Lemma 2 givean =cn= 2u+ 2 and bn+ 1 =dn= 2v. It follows thatsn−1−1 =rn−1=k andrn=sn= 2k+ 1. By induction hypothesis, we have also

T^j(0)6= 1

α for 0≤j≤2k+ 1 (10)

andT^k+1(0)< T^2k+2(0) < T(0) = 1. Since we haveek+1ek+2. . . e_2k−1=C and e1e2. . . e_k−1 = C, for 1 ≤ j ≤ k−1, we get that T^2k+1+j(0) lies in the open interval with endpointsT^k+j(0) andT^j(0) which is contained either inM0orM1. This impliesT^j(0)6= _α¹ for 2k+ 2≤j≤3kande2k+2e2k+3. . . e3k=C. Therefore, ebegins with 0C1C00C. We compute

T^3k+1(0) = (−α)^2u+2(−β)^2v−1T^k(0) +β

α(−α)^2u+2(−β)^2v−2 + (−α)^u+2(−β)^v−1T^k(0) + (−α)^u+1(−β)^v−1+T^k(0).

Sinceuis even andv is odd by Lemma 2, this implies T^3k+1(0)− 1

α=

T^k(0)− 1 α

(1 +α^u+2β^v−1−α^2u+2β^2v−1).

Either we assume thatT^sn−1−1(0) =T^k(0) andT^{rn+sn−1−1}(0) =T^3k+1(0) are on the same side of _α¹, or (α, β)∈G_n which meansα^2u+2β^2v−α²−β <0. In both cases we get

1 +α^u+2β^v−1−α^2u+2β^2v−1>0.

We always have 1 +α^u+2β^v−1−α^2u+2β^2v−1<1. Furthermore, we get _α¹ < T^k(0) usingek = 1 and (10). This together implies _α¹ < T^3k+1(0)< T^k(0). Therefore, we have e3k+1 = 1. Applying T we have also T^k+1(0) < T^3k+2(0) < 1 which means

T^rn−1+1(0)< T^rn+rn−1+1(0)< T(0).

Now we have shown thatebegins with 0C1C00C1. We compute T^2k+2(0) = (−α)^u+2(−β)^vT^k(0) +β

α(−α)^u+2(−β)^v−1+ (−α)²T^k(0)−α+ 1, T^3k+2(0) = (−α)^u(−β)^vT^2k+2(0)−(−α)^u(−β)^v−βT^k(0) + 1 +β

α.

PIECEWISE LINEAR TRANSFORMATIONS WITH NEGATIVE SLOPE

Sinceuis even andv is odd by Lemma 2, we get 1−T^2k+2(0) =

T^k(0)− 1 α

(α^u+2β^v−α²) and 1−T^3k+2(0) =

T^k(0)− 1 α

(α^u+2β^v−α^2u+2β^2v+β).

If now (α, β)∈Gn, thenα^2u+2β^2v−α²−β <0 holds. We getT^3k+2(0)< T^2k+2(0) which means T^rn+rn−1+1(0) < T^rn+1(0). On the other hand, if (α, β) ∈ H_n−1, thenα^2u+2β^2v−α²−β ≥0 and we getT^rn+rn−1+1(0)≥T^rn+1(0).

From now on assume (α, β)∈Gn. We have shownT^k+1(0)< T^3k+2(0)< T(0) above. Sinceek+1ek+2. . . e_2k−1=C ande1e2. . . e_k−1=C, for 1≤j ≤k−1, we get that T^3k+1+j(0) lies in the open interval with endpoints T^k+j(0) and T^j(0) which is contained either inM0 orM1. This givesT^j(0)6= _α¹ for 3k+ 2≤j≤4k ande3k+2e3k+3. . . e4k =C. Henceebegins with 0C1C00C1C. We setD=C1C.

ThenD contains 2uzeros and 2v−1 ones. The length ofD is 2k−1 and we have Bn=C1C00 =D00. Furthermore,ebegins with 0D00D. We compute

T^4k+1(0) = (−α)^2u+2(−β)^2v−1T^2k(0) + (−α)^2u+1(−β)^2v−1+T^2k(0).

This gives

T^4k+1(0)− 1 α=

1

α−T^2k(0)

(α^2u+2β^2v−1−1).

Since e_2k = 0, we get T^2k(0) < _α¹ using (10). This implies T^4k+1(0) > _α¹ and e4k+1= 1. By Lemma 2 we havern+1= 2rn−1 = 4k+ 1. Therefore,T^j(0)6= _α¹ is shown for allj≤rn+1.

Now we know thatebegins with 0D00D1 = 0BnB_n^∗. We compute T^4k+2(0) = (−α)^2u(−β)^2vT^2k+2(0)−(−α)^2u(−β)^2v−β

α 1

α(T^2k+2(0)−1) + 1.

This implies

1−T^4k+2(0) = (1−T^2k+2(0))

α^2uβ^2v− β α²

.

Since (α, β) ∈ Gn, we have 0 < α^2uβ^2v − _α^β2 < 1, and T^2k+2(0) < 1 holds by induction hypothesis. Hence we get T^2k+2(0) < T^4k+2(0) < 1 which means T^rn+1(0)< T^rn+1+1(0)< T(0). The lemma is completely proved also in the case

wheren−1 is odd.

4. The nonwandering set

We define the intervals which are used to constructT-invariant sets. For n≥1, we define

Ks_n−1=





 h

T^sn−1(0),1 α i

, if T^sn−1(0)≤ _α¹, 1

α, T^sn−1(0)i

, if T^sn−1(0)> _α¹.

For all (α, β) ∈ G, we have K0 =Ks₀−1 = [0,_α¹], K1 =Ks₁−1 = (_α¹, T(0)] and K2=Ks₂−1= [T²(0),_α¹].

F. HOFBAUER

Suppose thatn≥2 and (α, β)∈G_n−1. Thenebegins with 0B_n−1B_n−1^∗ = 0Bn

andT^j(0)6= _α¹ forj ≤rn by Lemmas 1 and 4. If n is even, thenBn ends with 00 and has lengthr_n=s_n. Hence e_sn−1=e_sn= 0 and we get T^sn−1(0)< ¹_α and T^sn(0)< _α¹. We have thenKs_n−1= [T^sn−1(0),_α¹]⊂M0. We define

Ks_n =T(Ks_n−1) = [0, T^sn(0)]⊂M0 and Kr_n+1=Ks_n+1=T(Ks_n) = [T^sn+1(0), T(0)].

Ifnis odd, thenBn ends with 1 and has lengthrn=sn−1. Hencees_n−1= 1 and we getT^sn−1(0)>_α¹. We have thenKs_n−1= (_α¹, T^sn−1(0)]⊂M1and we define

Kr_n+1=Ks_n=T(Ks_n−1) = [T^sn(0),1] = [T^sn(0), T(0)].

Notice that in both cases, for even and oddn, we haveKr_n+1= [T^rn+1(0), T(0)].

Suppose now (α, β)∈ Gn ⊂Gn−1. Then e begins with 0BnB^∗_n and we have T^j(0)6= _α¹ forj≤r_n+1by Lemma 4. We continue to define the intervalsK_j. The blockB_n has lengthr_n and the initial segment which the blocksB_n andB^∗_n have in common, has lengthr_n−2 +δ_n which is equal to s_n−2 by Lemma 2. Using this and Lemma 3 we get

e_j =e_rn+j for 1≤j≤s_n−2 and e_sn−16=e_rn+sn−1=e_sn+1−1. For 1 ≤ j ≤ s_n −2, we set K_rn+j = T^j−1(K_rn+1) which is a closed interval contained either inM₀or inM₁ and has endpointsT^rn+j(0) andT^j(0). NowK_j is defined forj ≤s_n+1−2. Furthermore, T(K_sn+1−2) has endpoints T^sn+1−1(0) andT^sn−1(0) which are on different sides of _α¹. Hence bothT(K_sn+1−2)∩M₀and T(K_sn+1−2)∩M₁ are nonempty. One of these intervals isK_sn+1−1 and the other one isK_sn−1. Hence the interval T(K_sn+1−2) is the disjoint union of the intervals K_sn+1−1 and K_sn−1. Now we can continue to define intervals K_j forj≥s_n+1 as in the previous paragraph.

If (α, β) ∈ Gn, then the intervals Kj for 0 ≤j ≤ rn+1+ 1 are defined. The intervalKjis mapped monotonically ontoKj+1ifj /∈ {si−2 : 1≤i≤n+ 1}, and the intervalKs_i−2is mapped monotonically ontoKs_i−1∪Ksi−1−1for 1≤i≤n+1.

SetL0= [0,1] andLn =Sr_n+1

j=s_n−1Kj. We can prove now the following results.

Proposition 2. Suppose that n≥2 and(α, β)∈Gn.

(a) The intervalsKj forsn−1≤j≤rn+1are disjoint andKs_n−1andKs_n+1−1

have the common endpoint _α¹. (b) Ln isT-invariant andLn⊂L_n−1.

(c) L_n−1\Ln is the union of disjoint open intervalsUj with1≤j≤r_n−1 and we have T(Uj) =Uj+1 for1≤j≤rn−1−1 andT(Urn−1)⊃U1.

Proof. We have r2 =s2 = 3 and r3 =s3−1 = 5. For (α, β)∈ G2, we have K2=Ks₂−1= [T²(0),_α¹],K3= [0, T³(0)], K4= [T⁴(0), T(0)] andK5=Ks₃−1= Kr₃ = (_α¹, T⁵(0)]. Hence forn= 2 we get (a) from Lemma 1.

We proceed by induction. Supposen≥3 and (α, β)∈Gn ⊂G_n−1. We assume that (a) is already shown forn−1 instead ofn, this means that the intervalsKj

PIECEWISE LINEAR TRANSFORMATIONS WITH NEGATIVE SLOPE

fors_n−1−1 ≤j ≤rn are disjoint. In the following we write] for the union of disjoint sets.

We start withKrn−1+1= [T^rn−1+1(0), T(0)]⊂M1. By Lemma 4 we have T^rn−1+1(0)< T^rn+rn−1+1(0)< T^rn+1(0)< T(0).

It follows that the intervals

Kr_n+r_n−1+1= [T^rn−1+1(0), T^rn+rn−1+1(0)] and Kr_n+1= [T^rn+1(0), T(0)]

are disjoint and the nonempty open intervalU1 = (T^rn+rn−1+1(0), T^rn+1(0)) lies between them. Therefore, we have

K_rn+rn−1+1]U₁]K_rn+1=K_rn−1+1.

We setU_j=T^j−1(U₁) forj≥2. Furthermore, forr_n−1+ 2≤j≤s_n−2 we have that K_j =T(K_j−1) is contained either in M₀ or M₁. SinceU₁ ⊂K_rn−1+1, the setsU_j for 1≤j≤s_n−r_n−1−1 are intervals and we get

K_rn+j]U_j−rn−1]K_{rn−rn−1+j}=K_j for r_n−1+ 1≤j≤s_n−2 (11)

andT(Kr_n+s_n−2)]Us_n−rn−1−1]Kr_n−rn−1+s_n−1=T(Ks_n−2). By the first equa- tion of Lemma 3 this means

T(Ks_n+1−2)]Usn−1−1]Kr_n+sn−1−1=T(Ks_n−2).

We have T(Ks_n+1−2) = Ks_n+1−1]Ks_n−1 and T(Ks_n−2) = Ks_n−1 ]Ks_n−1−1. Therefore, we get

K_sn+1−1]U_sn−1−1]K_{rn+sn−1−1}=K_sn−1−1.

SinceT(K_j) =K_j+1 holds fors_n−1−1≤j≤r_n−1 (notice thatr_n−1=s_n−1−1 or r_n−1 =s_n−1), the sets U_j fors_n−1 =s_n−r_n−1 ≤j ≤r_n−1+ 1 are intervals and

Ksn+1−sn−1+j]Uj]Krn+j =Kj for sn−1−1≤j ≤rn−1. (12)

By induction hypothesis, the intervals Kj for s_n−1−1 ≤ j ≤ rn are disjoint.

By (11) and (12), each of the intervals Kj for sn−1−1 ≤ j ≤ sn−2 contains three disjoint intervals. Hence the intervals on the left hand sides of (11) and (12) are disjoint and they are also disjoint fromKj for sn−1 ≤j ≤rn (notice that r_n =s_n−1 orr_n=s_n). The intervals on the left hand sides of (11) and (12) are U_j for 1≤j ≤r_n−1 and

Kr_n+j for sn−1−1≤j≤sn−2 =sn+1−2−rn

Kr_n+l for 1≤l≤sn−2−r_n−1=s_n−1−2

Ks_n+1−2+l for 1≤l≤rn−1−sn−1+ 2 =rn+1−sn+1+ 2

where we have used the equations of Lemma 3. This list contains the intervalsKj

forrn+ 1≤j≤rn+1. Therefore, these intervals are disjoint and they are also dis- joint from the intervalsKjforsn−1≤j≤rn. By definition the intervalsKs_n+1−1

andKs_n−1have the common endpoint _α¹. Hence (a) is shown by induction.

F. HOFBAUER

The intervalsUj are disjoint. We haveT(Uj) =Uj+1 for 1≤j ≤r_n−1−1 and

r_n

[

j=sn−1−1

Kj\

r_n+1

[

j=sn−1

Kj=

s_n−2

[

j=sn−1−1

Kj\

r_n+1

[

j=rn+1

Kj=

rn−1

[

j=1

Uj

(13)

by (11) and (12). Sincern−1is odd by Lemma 2 andrn+ 2rn−1=rn+1holds by Lemma 3, we get

T(U_rn−1) =T^rn−1(U₁) = (T^rn+rn−1+1(0), T^rn+1+1(0)).

We haveT^rn+1(0)< T^rn+1+1(0) by Lemma 4 and therefore, T(U_rn−1)⊃U₁= (T^rn+rn−1+1(0), T^rn+1(0)).

We show (b). We get L_n ⊂L_n−1 from Lemma 1 if n= 2, and from (11) and (12) if n ≥ 3. We show T(L_n) ⊂ L_n. For s_n −1 ≤ j ≤ s_n+1 −3, we have T(K_j) =K_j+1. Furthermore, T(K_sn+1−2) =K_sn+1−1∪K_sn−1. If n+ 1 is even, then

sn+1=rn+1, T(Ks_n+1−1) =Ks_n+1=Kr_n+1 and T(Kr_n+1) =Kr_n+1+1. Ifn+ 1 is odd, then

sn+1=rn+1+ 1 and T(Ks_n+1−1)⊂Ks_n+1=Kr_n+1+1. SinceT^rn+1(0)< T^rn+1+1(0) holds by Lemma 4, we have also

Kr_n+1+1= [T^rn+1+1(0), T(0)]⊂[T^rn+1(0), T(0)] =Kr_n+1

andT(Ln)⊂Ln is shown.

We show (c). Ifn= 2, we haver_n−1=r1= 1 andU1= (T⁵(0), T⁴(0)). In this caseT(U1)⊃U1 and L1\L2 =U1 follow from Lemma 1. Forn ≥3, letUj for 1 ≤j ≤r_n−1 be as above. We have shown T(Uj) =Uj+1 for 1≤j ≤r_n−1−1 andT(Urn−1)⊃U1. Furthermore, we haveL_n−1\Ln=Srn−1

j=1 Uj by (13).

Finally we are able to determine the nonwandering set Ω(T).

Theorem 2. Forn≥1and(α, β)∈H_n, we haveΩ(T) =L_n∪Sn

k=1P_k, where P_k is a periodic orbit of period s_k−s_k−1. The setL_n is topologically transitive and the disjoint union ofsn closed intervals.

Proof. Forn= 1, this is already shown at the end of Section 2. Therefore, we supposen≥2. We first considerLn. We haveLn=Srn+1

j=sn−1Kj and the intervals in this union are disjoint by Proposition 2. Since two of these intervals have a common endpoint,Ln is the disjoint union ofrn+1−sn+ 1 =sn closed intervals.

Forr_n+ 2≤j≤s_n+1−2, the intervalK_j−1is mapped monotonically onto the intervalK_j. The intervalK_sn+1−2is mapped monotonically ontoK_sn+1−1∪Ksn−1

which contains _α¹ in its interior. Let δ be the inverse image of _α¹ in the interval K_rn+1 = [T^rn+1(0), T(0)] under the composition of these maps. The interval K_sn−1is mapped monotonically ontoK_rn+1andK_sn+1−1is mapped monotonically ontoKr_n+1+1= [T^rn+1+1(0), T(0)], one underTand the other one underT². Since