Alsed\`a-Misiurewicz systems with place-dependent probabilities

We consider systems of two specific piecewise linear homeomorphisms of the unit interval, so called the Alsed\`a-Misiurewicz systems, and investigate the basic properties of Markov chains which arise when these two transformations are applied randomly with probabilities depending on the point of the interval. Though this iterated function system is not contracting in average and known methods do not apply, stability and the strong law of large numbers are proven.


Introduction
1.1. The main results. Let f 0 be an interval homeomorphisms such that its graph consists of two straight lines, the first one connecting (0, 0) with some point (x 0 , y 0 ) ∈ (1/2, 1) × [1/2, 1) under diagonal, and the second one connecting (x 0 , y 0 ) with (1,1). Next, let f 1 be the interval homeomorphisms defined by Figure 1). Setting a 0 = y0 x0 and a 1 = 1−y0 1−x0 we can write Fix two positive real functions p 0 , p 1 on [0, 1] with p 0 (x) + p 1 (x) = 1 for every x ∈ [0, 1]. It defines the following natural random process on the interval (0, 1): being at any point x ∈ (0, 1) we choose transformation f i with probability p i (x) and move to the point f i (x), i = 0, 1. To be more strict, we define the family of Markov chains with common transition probabilities given by the formula π(x, ·) := p 0 (x)δ f0(x) + p 1 (x)δ f1(x) , x ∈ (0, 1). As far as we know, stability of these Markov chains, i.e. arising from random application of transformations on the interval, were always proved under assumption that all transformations are contractions or are contracting in average. However, recently several papers have been published which established stability with dropping this assumptions. The first one and probably the most important for us was by Lluis Alsedà and Micha l Misiurewicz [1] in 2014 where the authors showed that if we consider two transformations defined in the first paragraph for y 0 = 1/2 and choose it randomly with constant and equal probabilities 1/2, 1/2 then the corresponding Markov chain is stable. After [2] we call the systems defined above the Alsedà-Misiurewicz systems (in [2] the only restriction for (x 0 , y 0 ) ∈ (0, 1) × (0, 1) is that it should be under diagonal). Later their results were generalized to the case of two C 2 diffeomorphisms ( [7]) or even arbitrary finite number of homeomorphisms ( [14]) satisfying additional assumptions that from each point we have positive probabilities of moving to the left and moving to the right, all functions are differentiable at 0 and 1 and the average Lyapunov exponents at these points are positive. In our setting we define the average Lyapunov exponents at 0 and 1 by the formulae (1) Λ 0 := p 0 (0) log(a 0 ) + p 1 (0) log(a 1 ), Λ 1 := p 0 (1) log(a 1 ) + p 1 (1) log(a 0 ).
In the general case coefficients a 0 , a 1 should be replaced by derivatives at 0 and 1, respectively. Note that all known results are proven under assumption that probability of choice of a transformation does not depend on the point of the interval. The most important papers concerning systems with place-dependent probabilities are probably [3], [9] where the stability of the corresponding Markov chains is proved under the most general assumptions in arbitrary locally compact metric spaces. However, one of them is contractivity in average which in our case is never satisfied, therefore we can neither apply the result, nor use the proof. In [3] and [13] one can find further references and historical comments.
The functions p 0 , p 1 are Dini continuous, which means that for every C ≥ 0 and t < 1 we have n β(Ct n ) < ∞, where β denotes the modulus of continuity of p 0 , p 1 , i.e.
We do not need any further assumptions on contractiveness of the system. Our two main results are the following theorems. Theorem 1. If (A1)-(A4) hold then there exists a unique Borel probability measure µ * ∈ M such that the Markov chain (X µ * n ) is stationary.
The last theorem was proved in the case of general Markov chains on compact spaces in [4]. Later it was proved in [5] in the case of iterated systems of contractions in R n with constant probabilities and in [6] in the case of systems contracting in average with place dependent probabilities on locally compact spaces. Our system does not satisfy assumptions of any of these theorems, however, using some ideas from the last paper we are still able to prove it. In its statement it is essential that it holds for every point x ∈ (0, 1), not only for µ * almost every.
The space of Borel probability measures on (0, 1) will be denoted by M 1 and the space of all positive Borel measures by M. Recall that the family of transition probabilities p(x, ·) ∈ M 1 , x ∈ (0, 1) by the formula Let us choose an initial distribution µ ∈ M 1 . Together with the transition probabilities it defines the Markov chain (X µ n ) on (0, 1). For simplicity of notation, we shall write (X x n ) when µ = δ x . Let us stress that values of this Markov chain are in the open interval (0, 1), not [0, 1].

U ϕdµ
for every µ ∈ M and ϕ ∈ C(0, 1). The operator P is linear, i.e. P (λ 1 µ 1 + λ 2 µ 2 ) = λ 1 P µ 1 + λ 2 P µ 2 for λ 1 , λ 2 ≥ 0, µ 1 , µ 2 ∈ M. It also preserves the total mass of a measure, i.e. P µ (0, 1) = µ (0, 1)) for µ ∈ M. We say that a measure µ * ∈ M is invariant for the operator P if P µ * = µ * . In that case we say that the operator P is asymptotically stable if P n ν → µ * weakly for every ν ∈ M 1 . The Markov-Feller operator P has the property that the distribution of X µ n is P n µ for all n ≥ 0 and µ ∈ M. Therefore one can choose an initial distribution µ ∈ M 1 in such a way that (X µ n ) is stationary if and only if µ is P -invariant and the Markov chain (X µ n ) is stable if and only if P is asymptotically stable. Following [7] we define By what we just mentioned, the theorem below is equivalent to the existence of a unique invariant probability measure for the Markov-Feller operator P .

The proof of Theorem 1
Proof of existence. The proof follows the lines of the proof from [7] with necessary changes. Namely, we shall show that there exist parameters M ≥ 1, α ∈ (0, 1) such that the class P M,α is invariant under the operator P . It is sufficient since in that case one can apply the standard Krylov-Bogoliubov technique, i.e. take any ν ∈ P M,α and define ν n = 1 n (ν + . . . + P n−1 ν). By the P -invariance of P M,α , all ν n 's are in P M,α , and by weak- * compactness of P M,α there exists an accumulation point µ * ∈ P M,α of this sequence which is an invariant measure. Details are left to the reader. What remains to show is the existence of parameters M, α with the desired property.
We are in position to show the invariance of P M,α for M, α chosen above. Take µ ∈ P M,α and x ∈ (0, 1).
We are now going to make some use of (A1). Take η 1 such that the following condition is satisfied There exists such η 1 . Indeed, since a 0 < 1 < a 1 , the linear function y −→ a 0 y − a 1 (y − a 1 η 1 ) is decreasing and, in consequence, it suffices to show that there exists such η 1 for y = 1 − y 0 . But since Let us also assume that η 1 is less than the length of the interval [1 − x 0 , 1 − y 0 ] and satisfies This is possible by the continuity of f 1 and To show this, however, we compute f 1 (y 0 ) = −y0(1−y0) x0 + 1 and obtain that (6) is equivalent to the condition y 0 (1 − y 0 ) > x 0 (1 − x 0 ). By the assumptions made on x 0 , y 0 we have 1/2 ≤ y 0 < x 0 , so our statement follows from the monotonicity of the function ψ(t) In order to simplify the reasoning we assume that x < y and ω such that f n ω (x) visits (0, 1 − x 0 ) infinitely often and f n ω (y) visits (x 0 , 1) infinitely often. In the end of the proof we will give a simple explanation that this assumption may be dropped.
, f s ω (y) again satisfy assumptions of the lemma, therefore we may assume s = 0. Next, define r to be the moment of the first visit of f r ω (y) in (1 − y 0 , 1). If we will show the claim for n ≤ r, then the points f r ω (x), f r ω (y) again satisfy the assumptions of the lemma, therefore we may assume r = u.
For this purpose observe that f n ω (y) = a n ω y and f n ω (x) = a n ω x for n ≤ r − 1, i.e. application of f 0 and f 1 is actually a multiplication by a 0 , a 1 , respectively. Indeed, assume contrary to our claim that f n−1 ω (y) > 1 − x 0 and ω n = 1. Then f n ω (y) = f 1 (f n−1 ω (y)) > f 1 (1 − x 0 ) = 1 − y 0 , hence r = n, which is a contradiction. Since f n ω (y) = a n ω y and f n ω (x) = a n ω x for n ≤ r − 1, we have for these n's (7) |f n ω (x) − f n ω (y)| = a n ω |x − y| But since f n ω (y) ≤ 1 − y 0 < y for n ≤ r − 1, we have a n ω < 1 which completes the proof in the case n ≤ r − 1.
The only point remaining now is to show that |f r x (this condition is equivalent to say that 1−x 0 ∈ kx, ky , thus the condition a r−1 x is equivalent to our case now). We assert that this function is nonincreasing. Indeed, hence the function is linear with the slope equal to a 0 y − a 1 x which is negative since |x − y| < a 1 η 1 and (4) holds for η 1 .
We compute now Combining that with the monotonicity of the considered function yields The proof is essentially the same as in the case of previous lemma. We define t and r in the same way and assume without loss of generality that t = 1, r = u (for n ≥ r we can apply Lemma 1). We again observe that f n ω (y) = a n ω y, f n ω (x) = a n ω x, and f n ω (y) ≤ 1 − y 0 for n ≤ r − 1. The difference is that y > 1 − y 0 is not true anymore. However, by the definition of a 1 we have 1 − y 0 = a 1 (1 − x 0 ) ≤ a 1 y, thus a n ω ≤ a 1 which proves the assertion for n ≤ r − 1 (cf. (7)).
If n = r then we have again two cases.
. Observation that f r ω (y) < a r ω y yields the assertion. Proof of Proposition 1. We can define the following infinite sequences of natural numbers To finish the proof notice that the statement for n ≤ u 1 follows from Lemma 2 (or its symmetric version) with u = u 1 . Hence, from the definition of (u k ), the points f u1 ω (x), f u1 ω (y) satisfy assumptions of Lemma 1 (or its symmetric version) with u = u 2 − u 1 . We continue in this fashion: for every k the points f u k ω (x), f u k ω (y) satisfy assumptions of Lemma 1 or its symmetric version with u = u k+1 − u k , and the conclusion follows.
To obtain the statement for any ω observe that for some k we cannot define t k+1 and in this case either Lemma 1 or 2 applies for f u k ω (x), f u k ω (y) with arbitrary large u.
From now on, M, α, ε, and p always stand for the quantities chosen in the proof of existence of a stationary measure. Fix x ∈ (0, 1) and define Lemma 3. If x < ε then P x (B x,n ) ≤ (ε/x) α p n for all n ≥ 0. The same estimation holds for P 1−x (B 1−x,n ).
Proof. Fix x ≤ ε and recall that we write a n ω = a ωn . . . a ω1 for n ≥ 1 and ω ∈ Σ. We first observe that E x 1 Bx,n−1 (a n ω ) −α ≤ p n . Indeed, by (2) we have Here (F n ) n≥1 stands for the natural filtration in (Σ, F ). Therefore Proceeding by induction yields E x 1 Bx,n−1 (a n ω ) −α ≤ p n . Observe that for all ω ∈ Σ with ω ∈ B x,n we have f j ω (x)(ω) = a j ω x and, in consequence, B x,n = {ω ∈ Σ : a j ω x < ε for all j ≤ n}. The Chebyshev inequality gives now Proof. First of all, by (6) is positive by (A3) and for any x ∈ [h, 1 − h] we can first take a sequence of length 2m ′′ with P x (f 2m ′′ ω (x) ∈ [1 − x 0 , y 0 + a 1 η]) > 0 (which may be less than 2m) and then apply the composition f 0 • f 1 exactly m ′ + (m − m ′′ ) many times.
Proof of Proposition 2. Let c be the point from Lemma 4. Take ρ > 0 to be any positive number less than distance from c to the boundary points of [1 − x 0 , y 0 ]. Take h = ε (recall that M, ε, α were the numbers given in the proof of existence of the stationary measure; see the comment under Proposition 2). Take n 1 to be the numbers given in Lemma 4. By the continuity of f 0 , f 1 and the compactness of Let n 2 be such that (a 0 ) n2 < 1/(2a 1 ). Put m := n 1 + n 2 and ξ := f m 0 (ε) (i.e. ξ is such a number that P Let us define the following optional times on Σ for u ∈ (0, 1): . From what has already been proved we conclude that By the strong Markov property for all ξ ≤ x ≤ 1 − ξ. By induction argument we get By Lemma 3 there exists C 1 > 0 and γ ∈ (0, 1) such that E x e γT1(x) ≤ C 1 for all x ∈ [ξ, 1 − ξ]. Induction argument applied below yields for every n. Fix κ ∈ (0, 1). We have again by the Chebyshev inequality for all x ∈ [ξ, 1 − ξ] provided that ρ ∈ (0, 1) was chosen sufficiently small. Again, conditioning argument yields E x e ρSn(x) ≤ C n 2 . Eventually, using again the Chebyshev inequality we obtain for such x, y and any λ ∈ (0, 1), Take λ such that C λ 2 e −ρ < 1 and put L = a 1 + 1, q = max{C λ 2 e −ρ , 1 2 λ } < 1. Then by the above we have E x |f n ω (x) − f n ω (y)| ≤ Lq n |x − y| for all natural n which is the desired conclusion.
Proof of uniqueness. Throughout the proof p i1,...,in (x) stands for First observe that for any x ∈ (0, 1) there exists a finite sequence (i 1 , .., i l ) ∈ {0, 1} l for some l such that (6) and f 1 • f 0 is a contraction on the interval [1 − x 0 , y 0 ], hence this composition has exactly one attractive fixed point c ∈ (1 − x 0 , y 0 ). Combining these facts yields c ∈ Γ µ for all P -invariant measures µ, where Γ µ denotes the topological support of this measure. The proof is completed by showing that the family (U n ϕ) is equicontinuous at c for any Lipschitz ϕ. Indeed, if there exist at least two different ergodic invariant measures µ 1 , µ 2 , then there exists a Lipschitz function ϕ such that ϕdµ 1 − ϕdµ 2 ≥ δ for some δ > 0. We consider the averages 1 n (ϕ(x) + U ϕ(x) + . . . + U n−1 ϕ(x)) which must differ from 1 n (ϕ(c) + U ϕ(c) + . . . + U n−1 ϕ(c)) at most δ/2, provided that x is sufficiently close to c. On the other hand, c ∈ Γ µ1 ∩ Γ µ2 , therefore in any neigbourhood of c we can find points x 1 , x 2 such that considered averages tend to ϕdµ 1 , ϕdµ 2 , respectively, which is a contradiction.

The proof of Theorem 2
Lemma 5. There exists 0 < h < 1/2 such that for all 0 < ξ < 1/2 there exists n 0 such that P n δ Proof. Recall that M , α, ε are the quantities given in the proof of existence of a stationary measure. The class P M,α is P -invariant. Take h > 0 such that M h α < 1/8. The definition of M yields the relation M ε α ≥ 1 which implies clearly P n δ x ∈ P M,α and thus P n δ x [h, 1 − h]) ≥ 3/4 for every x ∈ [ε, 1 − ε] and n ≥ 0.
Let n 0 be such that (ε/ξ) α p n < 1/8 for n ≥ n 0 . Take x ∈ [ε, 1 − ε] and x ∈ [ξ, 1 − ξ]. Denote by T the time of the first visit of x in [ε, 1 − ε]. Then by the strong Markov property, Lemma 3 and the first part of the proof we have From now on, h denotes the quantity given in Lemma 5.
We are in position to show Theorem 2. The idea is to apply the lower bound technique (cf. Theorem 4.1 in [9]).
Proof of Theorem 2. Take x < y, λ > 0 and a Lipschitz function ϕ. By the equicontinuity of (U n ϕ) at c there exists ρ > 0 such that Once again, by Lemma 6 there exists m 2 such that P x ⊗ P y (A m2 ) ≥ δ 2 . Put ξ 2 := min{f m1+m2 We continue in this fashion to construct a sequence m 1 , m 2 , . . . such that for all n's. It is easy to check that Hence for n ≥ m k we get By (9), (10), and the definition of A mj 's eventually we have provided that k was sufficiently large. Therefore lim n→∞ |U n ϕ(x) − U n ϕ(y)| → 0 for every x, y ∈ (0, 1). If µ * is the stationary probability measure and ν ∈ M 1 , then for any Lipschitz function ϕ we obtain by the Lebesgue Convergence Theorem. Thus P n ν → µ * weakly- * for every ν ∈ M 1 which is our assertion.

The proof of Theorem 3
Let c be the number given in Lemma 4. Recall that c is the unique attractive fixed point of the composition f 0 • f 1 on (1 − x 0 , y 0 ). For any ρ > 0 we will write S ρ (x) for the time of the first visit of the process (f n ω (x)) in (c − ρ, c + ρ).
We omit the proof, since it is an easy consequence of Lemma 3 and Lemma 4.
Proof. This is an immediate consequence of the Chebyshev inequality and the Borel-Cantelli lemma. Indeed, x) − f n ω (y)| ≥ r n } occurs only finitely many times P x -a.s which completes the proof.
The following lemma is proven in [6], Lemma 3. For the convenience of the reader we rewrite the proof here.
Lemma 9. There exists ρ > 0 such that for every x, y ∈ (c − ρ, c + ρ) the measures P x , P y on Σ are absolutely continuous.
Take i = (i 1 , . . . , i l ) ∈ Ξ. If Q and C i are not disjoint then also Q l and C i are not disjoint, hence we have be the above estimation Moreover, Recall here that the cylinders C i , i ∈ Ξ are disjoint. Combining that with two above inequalities yields which is the desired assertion.
Proof of Theorem 3. Let ϕ be any Lipschitz function. The statement for any continuous function follows from the density of the set of Lipschitz functions in C (0, 1) with the supremum norm. Let ρ be given in Lemma 9. There exists y ∈ (c − ρ, c + ρ) such that for P z -a.e. ω ∈ Σ where µ * is the unique P -invariant measure. It follows by the fact that c is in the support of µ * (see the beginning of the proof of uniqueness) and by the Birkhoff Ergodic Theorem. Take any z ∈ (c − ρ, c + ρ). For P z -a.e. ω ∈ Σ there exists n(ω) such that |f n ω (z) − f n ω (y)| ≤ r n for n ≥ n(ω), where q < r < 1 by Lemma 8. Therefore for P z -a.e. ω ∈ Σ. By Lemma 9 the measures P z , P y on Σ are absolutely continuous. Hence for ω ∈ D z , where D z ⊆ Σ is certain measurable set with P z (D z ) = 1.

Open problems
One cannot hope to prove uniqueness of a stationary measure with continuous probabilities without any additional assumptions on them. Indeed, in [13] it is proved that for two affine contractions of the interval there exists continuous probability functions p 1 , p 2 such that there is no uniqueness of a stationary measure for the corresponding Markov chain. It is reasonable to assume (A2) from two reasons. The first one is that it is exactly what we need to deduce from Proposition 2 the equicontinuity of the family (U n ) at some point of the interval [1 − x 0 , x 0 ]. The second is that this assumption appears also in papers [3] and [9], hence it seems to be natural.
The assumption (A3) is not very restrictive and without that the situation is more complicated. For example, admission of vanishing probabilities may easily create invariant intervals, i.e. disjoint intervals such that probability of getting from one to the another is zero. We used this assumption for example in the proof of Lemma 4.
The assumption (A4) was crucial to ensure the existence of a stationary measure and to show Lemma 3 which was a key ingredient in the proof of Proposition 2. Without it, different situations may happen. For example, X x n → 0 a.s., if Λ 0 < 0 and Λ 1 ≥ 0 (see [7]).
The assumption (A1) was also important in our reasoning. Essentially it was used only in the proof of Proposition 1, but we are not able to show Proposition 2 without Proposition 1. Proposition 1 is generally not true in the case of all Alsedà-Misiurewicz systems, so the following question is natural: 1. It is not possible (in general) to show Proposition 1 without (A1). However, is it possible to show Proposition 2 without this assumption? If not, then it is possible to show uniqueness of a stationary measure?
Our method of proving Theorem 2 does not provide any rate of convergence of U n ϕ to ϕdµ. However, if one assume that probabilities are Lipschitz continuous then we may expect that rate of convergence is exponential (see the main result in [12]), therefore sufficiently fast to provide the Central Limit Theorem (see [8], [11]).
2. Does the Central Limit Theorem hold for our Markov chains provided that probabilities are Lipschitz continuous?
The last question is connected with paper [2]. The authors prove there that under some assumptions, if y 0 < 1/2 then there are invariant Cantor sets for the iterated function system (f 0 , f 1 ). However, nothing is known in the case y 0 > 1/2 which is our case.

Acknowledgements
We are grateful to Tomasz Szarek for communicating the problem and reading the manuscript.