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`International Journal of Mathematics and Mathematical SciencesVolume 2013 (2013), Article ID 929475, 5 pageshttp://dx.doi.org/10.1155/2013/929475`
Research Article

## On the Set of Fixed Points and Periodic Points of Continuously Differentiable Functions

Department of Mathematics, Berks College, Pennsylvania State University, Tulpehocken Road, P.O. Box 7009, Reading, PA 19610-6009, USA

Received 6 December 2012; Accepted 16 February 2013

Copyright © 2013 Aliasghar Alikhani-Koopaei. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

In recent years, researchers have studied the size of different sets related to the dynamics of self-maps of an interval. In this note we investigate the sets of fixed points and periodic points of continuously differentiable functions and show that typically such functions have a finite set of fixed points and a countable set of periodic points.

#### 1. Introduction and Notation

The set of periodic points of self-maps of intervals has been studied for different reasons. The functions with smaller sets of periodic points are more likely not to share a periodic point. Of course, one has to decide what “big” or “small” means and how to describe this notion. In this direction one would be interested in studying the size of the sets of periodic points of self-maps of an interval, in particular, and other sets arising in dynamical systems in general (see [15]). For example, typically continuous functions have a first category set of periodic points (see [1, 5]). This result was generalized in [2] for the set of chain recurrent points. At times, even the smallness of these sets in some sense could be useful. For example, in [6] we showed that two commuting continuous self-maps of an interval share a periodic point if one has a countable set of periodic points. Schwartz (see [7]) was able to show that if one of the two commuting continuous functions is also continuously differentiable, then it would necessarily follow that the functions share a periodic point. Schwartz's result along with the results given in [6] may suggest that continuously differentiable functions have a countable set of periodic points. This is not true in general. However, in this note we show that typically such functions have a finite set of fixed points and a countable set of periodic points. Here denotes the set of fixed points of . For and we define by induction: The orbit of under is given by the sequence . For , let and be the set of periodic points of order ; that is, Two functions on a given interval are said to be of the same monotone type if both are either strictly increasing or strictly decreasing on that interval. Here, for a partition of the interval , is the length of the largest subinterval of , is the open ball about with radius , denotes the set of interior points of , and is the length of the interval .

#### 2. Continuously Differentiable Functions

For , consider and to be the family of all continuous maps and continuously differentiable maps from into itself, respectively. Recall that the usual metrics and on and , respectively, are given by

It is well known that the metric spaces and are complete and hence Baire's category theorem holds in these spaces. We say that a typical function in or has a certain property if the set of those functions which does not have this property is of first category in or in . (Some authors prefer using the term generic instead of typical.) It is known that typically continuous self-maps of an interval have -perfect, measure zero sets of periodic points (see [2]). Here we show that typically members of have a finite set of fixed points and a countable set of periodic points.

Lemma 1. Let and so that is not finite. Then there exists so that and .

Proof. Suppose is not finite, and then we can choose a strictly monotone sequence in . Without loss of generality, assume so that for each . Let , and then, for each , . Thus by Rolle’s Theorem we have , so that . Let , then , and , and from the continuity of and it follows that and .

Lemma 2. For each and there is a polynomial with .

Proof. Let where . Take , , and . It is easy to see that , , and . Let and be a polynomial with . Then is a polynomial and . Thus we have hence . From and it follows that .

Lemma 3. Let be a positive integer and . If is a periodic point of order with , then .

Proof. We show that . The case follows similarly. Let with for . Since , there exist and so that . Since and , there exists so that and is strictly increasing or strictly decreasing on . Let be strictly increasing on , and then for we have ; hence, , implying that , a contradiction to . On the other hand when is strictly decreasing on , for we have ; hence, implying that , a contradiction to .

Lemma 4. For , the set is closed in .

Proof. Let and . Then there exists such that and . Without loss of generality, we may assume that , then . Let be arbitrary. Due to the uniform continuity of on there exists a positive integer such that, for , . Since , there exists a positive integer such that, for ,  . Choose the positive integer so that for , and let . Then for we have Implying that . We also have Thus we have and .

Theorem 5. There exists a residual subset of such that, for every , is finite.

Proof. If has infinitely many elements, then from Lemma 1 it follows that there exists a point so that . Let From Lemma 4 the set is closed in . To show that has no interior point, let and , and then, by Lemma 2, there exists a polynomial such that and for some . Let Choose so that for all and take . Then ,  ,  , and on , implying that is of first category and is residual.

Theorem 6. The set of functions with , , and is a nowhere dense subset of .

Proof. Let . It is easy to see that is a closed subset of . To show that is nowhere dense, let and . Choose and such that for . Let , where
It is clear that ,  ,  , and . It is easy to see that , , and .

Theorem 7. Let , , and be a finite set, and for . Then there exists a function with such that .

Proof. Let . Let be the elements of with distinct orbits, that is, for and . It is clear that each is a periodic point of with some period where is a factor of . This suggests that if one can construct a function that is sufficiently close to , and either or , then By repeating this process for each in a finite number of steps we can construct the desired function . Thus for convenience, we assume that, for ,  . Consider the partition of obtained by , , and choose a positive less than such that, for each , and each ,  , and if , then . Given that, for , , is continuous, and on , we may choose the nondegenerate closed intervals , of length less than and the positive numbers such that(i) is the midpoint of and for ,(ii) is strictly monotone on each for , , where ,(iii)the intervals are mutually disjoint for , ,(iv), for ,(v), where .
Let for some , , and be the associated and , respectively. Define We have for , , , , and . By considering four different cases, we construct a function so that, for Cases A and B, and for Cases C and D. From conditions (iv) and (v) we have , and from condition (v) it follows that is of the same monotone type as on each ,  .
Case A. Let for and .(i)If is strictly increasing on , then by taking we have on and   on and the function is also strictly increasing on , and as a result on . Thus .(ii)If is strictly decreasing on , then by taking we have on and on and the function is also strictly decreasing on and as a result, on . Thus .
Case B. Let for and .
If is strictly increasing on , take , and if is strictly decreasing on , take . Then similar to Case A, we may show that for ; hence, .
Case C. Let for ,   for , and .(i)If is strictly increasing on , then by taking we have for and for , and the function is also strictly increasing on . Thus for and for ; hence, .(ii)If is strictly decreasing on , then by taking we have for and   for , and the function is also decreasing on . Thus for and for ; hence, .
Case D. Let for ,   for , and .
If is strictly increasing on , take , and if is strictly decreasing on , take . Then similar to Case C, we may show that for and for . Thus .
Note that is strictly increasing on , so we have Thus for either or , of which in such case , , and for . Thus We have Also Hence .
Doing this recursively for each , , we get a function such that .

Theorem 8. Typically continuously differentiable self-maps of intervals have a countable set of periodic points.

Proof. Take
The sets and are first category sets (see Theorems 5 and 6). For from Lemmas 4 and 1 we have that is closed and . Without loss of generality we may assume that is neither a constant function nor a polynomial of first degree, and then by Lemma 2 we may choose a polynomial of degree such that and . Using Theorem 7 we can construct so that and the set , thus and , hence . This implies that for each the set is nowhere dense. Thus is a first category set, so is a residual set, and, for , is countable.

#### Acknowledgments

The author would like to thank professor David Preiss for his valuable comments and suggestions as well as the anonymous referees for their comments.

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