1. Introduction

Let be closed, bounded, and nondegenerate (i.e., neither of them consists of a single point) real intervals, and let be continuous functions. The aim of this paper is to discuss, under some additional assumptions on the maps and , the problem of (topological) conjugacy of and . More precisely, we investigate the existence and uniqueness of solutions of the following functional equation: Let us recall that a homeomorphism satisfying (1.1) is said to be a topological conjugacy between and ( and are then called topologically conjugate), whereas an arbitrary function fulfilling (1.1) is called a conjugacy between them (so the conjugacy needs not to be continuous, surjective, or injective).

A continuous function is said to be a horseshoe map (see [1]) if there exist an integer and a sequence of reals such that and for every , is a homeomorphism of the interval (which is called a lap of ) onto . We say that horseshoe maps and having that the same number of laps are of the same type if and are of the same type of monotonicity on their leftmost laps.

It is known (see [1, 2]) that two horseshoe maps of the same type and without homtervals (i.e., intervals on which all their iterates are monotone) are topologically conjugate. So are also transitive horseshoe maps having two laps each, and, in this case, topological conjugacy is only one (see [3]). Moreover, if is transitive and is an arbitrary function, then every increasing, continuous, and surjective solution of (1.1) is homeomorphic (see [4]). We will show (Example 3.7) that if we omit the assumption of the transitivity of , then such a solution needs not to be injective even if is continuous and transitive and is continuous and piecewise monotone.

However, the main purpose of this paper is to find some regularity conditions on and ensuring the uniqueness of conjugacy between them as well as implying that conjugacy is topological. The following fixed point principle plays a crucial role in the proof of our main result (Theorem 3.1).

Theorem 1.1 (see [5, Theorem , page ] and also [6, Theorem , page 12]). Let be a complete metric space and . If for a nondecreasing function such that then has a unique fixed point.

2. Preliminaries

We begin by recalling the basic definitions and introducing some notation.

Throughout the paper stands for the integer part function.

Let be a metric space. A function is called strictly contractive if Given a nondecreasing function such that for and a selfmap of , we say that is -contractive if it satisfies condition (1.3).

Given horseshoe maps having laps respectively, for every put A horseshoe map having laps is said to be piecewise expansive (resp., piecewise -expansive) if for every , is strictly contractive (resp., -contractive for a .

We start with the following.

Proposition 2.1. Let and be horseshoe maps having laps and , respectively. Assume also that is a monotone and surjective solution of (1.1).
If is increasing, then and for .
If is decreasing, then and for .

Proof. Let us first note that the fact that is monotone and surjective together with (1.1) gives for , and consequently Suppose that for an . Then, using the fact that is monotone, we see that consists of a single point, and therefore so does . But from (1.1) and the surjectivity of it follows that a contradiction. We have thus shown that
Assume that is increasing. Then, by (2.9), we obtain which together with (2.7) gives (2.5). Hence we immediately see that for any , .
The rest of the proof runs as before.

Proposition 2.2. Assume that
(H) and are horseshoe maps of the same type and having laps and , respectively,and is piecewise expansive. If is a continuous and nonconstant solution of (1.1), then is surjective. If, moreover, is even and is injective, then is strictly increasing.

Proof. Let with be such that . Then, by (1.1), we have This and the fact that for every , is strictly contractive shows that and are not in the same interval . Therefore, or , and consequently, by (2.11), or .
Assume that . Since , , (2.11) now gives Therefore, . Similarly, implies . We have thus shown that is a surjection.
Now, assume that is even (which obviously yields ). Suppose also, contrary to our claim, that is decreasing, and let us consider the case when . Then and , and (1.1) now gives which contradicts the fact that and are of the same type. Similar considerations apply to the case when .

Lemma 2.3. Let be odd, and let assumption (H) hold. If is a solution of (1.1) such that then If, moreover, is piecewise expansive, then (2.6) holds true.
Conversely, every function satisfying (2.15) is a solution of (1.1) fulfilling (2.14).

Proof. It is obvious that if fulfils (1.1) and (2.14), then (2.15) holds.
Assume that is piecewise expansive. If , then from (1.1) and (2.14) it follows that is a fixed point of , and the strict contractivity of gives .
Next, assume that , which clearly forces , , and . Since by (2.14) we obtain and , (1.1) implies and . If it were true that , we would conclude from the strict contractivity of and that a contradiction.
We have thus shown that . In the same manner we can see that .
Now, fix an . Suppose that is even and note that from (1.1), the equalities , and the fact that is odd, we get Since, by (2.14), , the injectivity of gives now . As the same conclusion can be drawn for odd , the proof of (2.6) is complete.
The rest of the proof is immediate.

Analysis similar to that in the proof of Lemma 2.3 (due to condition (2.18) it also applies to the case when is even) gives the following.

Lemma 2.4. Let assumption (H) hold. If is a solution of (1.1) for which then If, moreover, is piecewise expansive, then (2.5) holds true.
Conversely, every function satisfying (2.19) is a solution of (1.1) fulfilling (2.18).

3. Main Results

We can now formulate our main results.

Theorem 3.1. If assumption (H) holds and is piecewise -expansive, then there exists a unique function satisfying (1.1) and condition (2.18). This function is continuous, surjective and increasing. If, moreover, is piecewise expansive, then is also strictly increasing.

Proof. Put It is easily seen that all these spaces with the metric are complete.
Fix a , and set We will show that the above formula correctly defines a selfmap of . In order to do this let us first fix an and observe that we have In the same manner we can see that for . Thus, and, by Lemma 2.4, is a solution of (1.1) fulfilling (2.18) if and only if it is a fixed point of . Moreover, it is easily seen that and . Now, fix , and note that by the facts that for every , is -contractive and is increasing, we get Thus, the function is -contractive, and Theorem 1.1 now shows that has a unique fixed point . Similarly, has a unique fixed point , and has a unique fixed point . Therefore, and, in consequence, is a continuous, surjective and increasing solution of (1.1) fulfilling (2.18).
Assume, additionally, that is piecewise expansive. We will first show that is constant in no neighbourhood of or .
To do this, suppose that is constant on a neighbourhood of , and denote by , where , the maximal interval of constancy of . Since is surjective and satisfies (2.18), is a proper subset of . Therefore, and from (1.1) it follows that is constant on .
If , then But from the strict contractivity of we also get a contradiction.
Now, assume that . Then , so is constant on a neighbourhood of . Denote by , where , the maximal interval of constancy of . Since is surjective and satisfies (2.18), is a proper subset of . Therefore, and from (1.1) it follows that is constant on .
If , then which contradicts the strict contractivity of .
Finally, assume that . Then , so both and are intervals of constancy of , and therefore This together with the strict contractivity of and gives a contradiction.
Analysis similar to the above shows that is constant in no neighbourhood of .
Now, suppose that is constant on a neighbourhood of for an . Then from (1.1) it follows that is constant on a neighbourhood of or , which is impossible.
We have thus shown that if is not injective, then any interval of its constancy is contained in for an . Let be an interval of constancy of having the maximal length. By (1.1), is constant on . But since for an , from the strict contractivity of it follows that the interval is of greater length than , a contradiction.

Analysis similar to that in the proof of Theorem 3.1 with and application of Lemma 2.3 instead of Lemma 2.4 gives the following.

Theorem 3.2. If is odd, assumption (H) holds, and is piecewise -expansive, then there exists a unique function satisfying (1.1) and condition (2.14). This function is continuous, surjective, and decreasing. If, moreover, is piecewise expansive, then is also strictly decreasing.

Let us next note that an immediate consequence of Theorems 3.1 and 3.2 and Proposition 2.1 is what follows.

Corollary 3.3. If assumption (H) holds, is piecewise -expansive, and is odd, then (1.1) has exactly two monotone and surjective solutions. One of them is increasing, while the other is decreasing.

The following fact follows immediately from Propositions 2.1 and 2.2.

Remark 3.4. If assumption (H) holds, is piecewise expansive and is even, then (1.1) has no homeomorphic solution satisfying condition (2.14).

On the other hand, we also have the following.

Theorem 3.5. If assumption (H) holds and and are piecewise expansive, then there exists a unique function satisfying (1.1) and condition (2.18). This function is an increasing homeomorphism. If, moreover, is odd, then there is also exactly one map fulfilling (1.1) and condition (2.14). This map is a decreasing homeomorphism.

Proof. Put for . If , then we also set while for we put By Theorem 3.1 (for every , is -contractive with for , so its assumptions are satisfied) there exist increasing homeomorphisms such that Moreover, by Proposition 2.1, Let us observe that in the case when is odd from Theorem 3.2 and Proposition 2.1 it follows that we can also take which is a decreasing homeomorphism such that (3.14) holds true and Putting and using (3.13) and (3.14) we obtain Furthermore, from (3.15) (resp., (3.16) it follows that condition (2.18) (resp., (2.14) holds true.
We have thus shown that (1.1) has a homeomorphic, increasing solution fulfilling (2.18) and if, moreover, is odd, then it also has a homeomorphic, decreasing solution satisfying (2.14).
To prove the uniqueness, assume that is a solution of (1.1) fulfilling condition (2.18) (resp., (2.14). Then, by (3.14), we get which means that satisfies (3.13). Moreover, from (3.15) (resp., (3.16) it follows that for . Hence, by Theorem 3.1, and so .

As an immediate consequence of Theorem 3.5, Proposition 2.1, and Remark 3.4, we get the following.

Corollary 3.6. Let assumption (H) hold, and let and be piecewise expansive. If is even, then (1.1) has exactly one homeomorphic solution. This solution is strictly increasing. If is odd, then (1.1) has exactly two homeomorphic solutions. One of them is strictly increasing, while the other is strictly decreasing.

Finally, we give two examples. The first of them shows that if one replaces the assumption that is piecewise expansive by then the last assertion of Theorem 3.1 is no longer true.

Example 3.7. Let , and let be the standard tent map defined by It is well known (see, e.g., [7] which is also a good survey on transitive maps) that is transitive. It is also easily seen that fulfills the assumption of Theorem 3.1 with for and By Theorem 3.1 there exists a unique (continuous, surjective, and increasing) function satisfying equation (1.1) and condition (2.18). Put , and note that . Furthermore, from (1.1) it follows that . Therefore, has to be one of the intervals: . Consequently, is not injective, and thus (1.1) has no homeomorphic solution.

Example 3.8. Let , and let be the standard tent map. It is known (see, e.g., [7–9]) that the function given by is a topological conjugacy between and . From Theorem 3.1 it follows that is the unique conjugacy between these maps such that and . In particular, by Proposition 2.1, is the only continuous, increasing, and surjective conjugacy between and .

Corollary 3.9. The function given by (3.24) is the only one satisfying and such that for and for .

Example 3.10. If and is a horseshoe map, then for any positive integer the function is a continuous and surjective solution of (1.1). It is obvious that this solution does not satisfy neither (2.14) nor (2.18). We thus see that in this case (1.1) can even have infinitely many solutions.