- Research Article
- Open Access
On Uniqueness of Conjugacy of Continuous and Piecewise Monotone Functions
© K. Ciepliński and M. C. Zdun. 2009
- Received: 23 December 2008
- Accepted: 24 June 2009
- Published: 10 August 2009
We investigate the existence and uniqueness of solutions of the functional equation , , where are closed intervals, and , are some continuous piecewise monotone functions. A fixed point principle plays a crucial role in the proof of our main result.
- Continuous Function
- Functional Equation
- Single Point
- Differential Geometry
- Arbitrary Function
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).
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 ). Moreover, if is transitive and is an arbitrary function, then every increasing, continuous, and surjective solution of (1.1) is homeomorphic (see ). 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).
then has a unique fixed point.
We begin by recalling the basic definitions and introducing some notation.
Throughout the paper stands for the integer part function.
for and a selfmap of , we say that is -contractive if it satisfies condition (1.3).
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.
Let and be horseshoe maps having laps and , respectively. Assume also that is a monotone and surjective solution of (1.1).
and for .
and for .
which together with (2.7) gives (2.5). Hence we immediately see that for any , .
The rest of the proof runs as before.
(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.
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 .
Therefore, . Similarly, implies . We have thus shown that is a surjection.
which contradicts the fact that and are of the same type. Similar considerations apply to the case when .
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).
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 .
We have thus shown that . In the same manner we can see that .
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.
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).
We can now formulate our main results.
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.
It is easily seen that all these spaces with the metric are complete.
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 .
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 .
which contradicts the strict contractivity of .
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.
and application of Lemma 2.3 instead of Lemma 2.4 gives the following.
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.
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.
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.
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.
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).
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.
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.
then the last assertion of Theorem 3.1 is no longer true.
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.
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 .
and such that for and for .
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.
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