Volume 2013, Article ID 929475,5pages http://dx.doi.org/10.1155/2013/929475
Research Article
On the Set of Fixed Points and Periodic Points of
Continuously Differentiable Functions
Aliasghar Alikhani-Koopaei
Department of Mathematics, Berks College, Pennsylvania State University, Tulpehocken Road, P.O. Box 7009, Reading, PA 19610-6009, USA
Correspondence should be addressed to Aliasghar Alikhani-Koopaei; [email protected]
Received 6 December 2012; Accepted 16 February 2013
Academic Editor: Paolo Ricci
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.
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 [1–5]). 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
𝐹1(𝑓) = {𝑥 : 𝑓(𝑥) = 𝑥}denotes the set of fixed points of𝑓.
For𝑓and𝑛 ∈Nwe define𝑓𝑛(𝑥)by induction:
𝑓𝑛(𝑥) = 𝑓 (𝑓𝑛−1(𝑥))
with𝑓0(𝑥) = 𝑥, 𝑓1(𝑥) = 𝑓 (𝑥) .
(1)
The orbit of𝑥under𝑓is given by the sequence orb(𝑥, 𝑓) =
{𝑓𝑖(𝑥)}∞
𝑖=0. For𝑛 ∈N, let𝐹𝑛(𝑓) = {𝑥 : 𝑓𝑛(𝑥) = 𝑥}and𝑃𝑓𝑛be
the set of periodic points of order𝑛; that is,
𝑃𝑓𝑛= 𝐹𝑛(𝑓) \ ∪𝑘<𝑛𝐹𝑘(𝑓) . (2)
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 𝑇of the
interval[𝑎, 𝑏],‖𝑇‖is the length of the largest subinterval of
𝑇,𝐵(𝑓, 𝜖)is the open ball about𝑓with radius𝜖,𝐴∘denotes
the set of interior points of𝐴, and𝜆(𝐼)is the length of the
interval𝐼.
2. Continuously Differentiable Functions
from𝐼into itself, respectively. Recall that the usual metrics𝜌0 and𝜌1on𝐶(𝐼, 𝐼)and𝐶1(𝐼, 𝐼), respectively, are given by
𝜌0(𝑓, 𝑔) =sup
𝑥∈𝐼𝑓(𝑥) − 𝑔(𝑥) for𝑓, 𝑔 ∈ 𝐶 (𝐼, 𝐼) ,
𝜌1(𝑓, 𝑔) = 𝜌0(𝑓, 𝑔) + 𝜌0(𝑓, 𝑔) for𝑓, 𝑔 ∈ 𝐶1(𝐼, 𝐼) . (3)
It is well known that the metric spaces (𝐶(𝐼, 𝐼); 𝜌0)and
(𝐶1(𝐼, 𝐼); 𝜌
1)are complete and hence Baire’s category theorem
holds in these spaces. We say that a typical function in𝐶(𝐼, 𝐼)
or𝐶1(𝐼, 𝐼)has a certain property if the set of those functions
which does not have this property is of first category in𝐶(𝐼, 𝐼)
or in𝐶1(𝐼, 𝐼). (Some authors prefer using the term generic
instead of typical.) It is known that typically continuous
self-maps of an interval have 𝜎-perfect, measure zero sets
of periodic points (see [2]). Here we show that typically
members of(𝐶1(𝐼, 𝐼), 𝜌1)have a finite set of fixed points and
a countable set of periodic points.
Lemma 1. Let𝑛 ≥ 1and𝑓 ∈ 𝐶1(𝐼, 𝐼)so that𝐹𝑛(𝑓)is not
finite. Then there exists 𝑥0 ∈ 𝐼 so that𝑓𝑛(𝑥0) = 𝑥0 and
(𝑓𝑛)(𝑥 0) = 1.
Proof. Suppose𝐹𝑛(𝑓)is not finite, and then we can choose
a strictly monotone sequence in 𝐹𝑛(𝑓). Without loss of
generality, assume {𝑥𝑖}∞𝑖=1 ⊂ 𝐹𝑛(𝑓)so that 𝑥𝑖 < 𝑥𝑖+1 for
each 𝑖 ≥ 1. Let ℎ(𝑥) = 𝑓𝑛(𝑥) − 𝑥, and then, for each 𝑖,
ℎ(𝑥𝑖) = ℎ(𝑥𝑖+1) = 0. Thus by Rolle’s Theorem we have𝑦𝑖,
𝑥𝑖 < 𝑦𝑖 < 𝑥𝑖+1so thatℎ(𝑦𝑖) = 0. Let𝑥0 = lim𝑖 → ∞𝑥𝑖, then
𝑥0 ∈ 𝐼, and lim𝑖 → ∞𝑦𝑖 = 𝑥0, and from the continuity of𝑓𝑛
and(𝑓𝑛)it follows that𝑓𝑛(𝑥0) = 𝑥0and(𝑓𝑛)(𝑥0) = 1.
Lemma 2. For each𝑓 ∈ 𝐶1(𝐼, 𝐼)and0 < 𝜖 < 1there is a
polynomial𝑝 : 𝐼 → 𝐼𝑜with𝜌1(𝑓, 𝑝) < 𝜖.
Proof. Let𝑀 = sup𝑥∈𝐼{|𝑓(𝑥)|, |𝑎|, |𝑏|, 1}where𝐼 = [𝑎, 𝑏].
Take𝛽 = 𝜖/8𝑀,𝛼 = 𝛽(𝑎 + 𝑏)/2, and𝑔(𝑥) = 𝛼 + (1 − 𝛽)𝑓(𝑥).
It is easy to see that0 < 𝛽 < 1,𝑔(𝐼) ⊂ 𝐼𝑜, and𝜌1(𝑓, 𝑔) < 𝜖/4.
Let
𝜖1= 12𝑥∈[𝑎,𝑏]inf {𝑔 (𝑥) − 𝑎, 𝑏 − 𝑔 (𝑥) ,8 (𝑏 − 𝑎)𝜖 ,8𝜖} (4)
and𝑠(𝑡)be a polynomial with𝜌0(𝑔, 𝑠) < 𝜖1/(𝑏 − 𝑎 + 1). Then
𝑝(𝑥) = 𝑔(𝑎) + ∫𝑎𝑥𝑠(𝑡)𝑑𝑡is a polynomial and𝑔(𝑥) − 𝑝(𝑥) =
∫𝑎𝑥(𝑔(𝑡) − 𝑠(𝑡))𝑑𝑡. Thus we have
𝜌1(𝑔, 𝑝) = 𝜌0(𝑔, 𝑠) + 𝜌0(𝑔, 𝑝) ≤ 𝜖1
(𝑏 − 𝑎) + 1+
𝜖1(𝑏 − 𝑎) (𝑏 − 𝑎) + 1
= 𝜖1< 𝜖4,
(5)
hence𝜌1(𝑓, 𝑝) ≤ 𝜌1(𝑓, 𝑔) + 𝜌1(𝑔, 𝑝) < 𝜖/2. From𝑔(𝐼) ⊂ [𝑎 +
2𝜖1, 𝑏 − 2𝜖1]and𝜌1(𝑔, 𝑝) ≤ 𝜖1it follows that𝑝(𝐼) ⊂ 𝐼𝑜.
Lemma 3. Let𝑛 > 2be a positive integer and𝑓 ∈ 𝐶1(𝐼, 𝐼).
If𝑥0is a periodic point of order 𝑛with(𝑓𝑛)(𝑥0) = 1, then
orb(𝑥0, 𝑓) ∩ {𝑎, 𝑏} = 0.
Proof. We show that𝑎 ∉orb(𝑥0, 𝑓). The case𝑏 ∉ orb(𝑥0, 𝑓)
follows similarly. Let orb(𝑥0, 𝑓) = {𝑎 = 𝑥0, 𝑥1, . . . , 𝑥𝑛 = 𝑥0}
with𝑥𝑖 = 𝑓(𝑥𝑖−1)for1 ≤ 𝑖 ≤ 𝑛. Since𝑛 > 2, there exist
𝑥𝑖 ∈ orb(𝑥0, 𝑓) ∩ (𝑎, 𝑏)and1 ≤ 𝑠 ≤ 𝑛 − 1so that𝑓𝑠(𝑥𝑖) =
𝑎. Since𝑓𝑠 ∈ 𝐶1(𝐼, 𝐼)and(𝑓𝑠)(𝑥0) ̸= 0, there exists𝛿1 > 0
so that𝑎 < 𝑥𝑖− 𝛿1 < 𝑥𝑖 < 𝑥𝑖+ 𝛿1 < 𝑏and 𝑓𝑠 is strictly
increasing or strictly decreasing on𝐽 = [𝑥𝑖− 𝛿1, 𝑥𝑖+ 𝛿1]. Let
𝑓𝑠be strictly increasing on𝐽, and then for𝑥
𝑖−𝛿1≤ 𝑥 ≤ 𝑥𝑖we
have𝑓𝑠(𝑥) ≤ 𝑓𝑠(𝑥𝑖) = 𝑎; hence,𝑓𝑠|[𝑥𝑖−𝛿1,𝑥𝑖]= 𝑎, implying that
(𝑓𝑠)(𝑥
𝑖) = 0, a contradiction to(𝑓𝑛)(𝑥0) = 1. On the other
hand when𝑓𝑠is strictly decreasing on𝐽, for𝑥𝑖≤ 𝑥 ≤ 𝑥𝑖+ 𝛿1
we have𝑓𝑠(𝑥) ≤ 𝑓𝑠(𝑥𝑖) = 𝑎; hence,𝑓𝑠|[𝑥𝑖,𝑥𝑖+𝛿1]= 𝑎implying
that(𝑓𝑠)(𝑥𝑖) = 0, a contradiction to(𝑓𝑛)(𝑥0) = 1.
Lemma 4. For𝑘 ≥ 1, the set
E𝑘= {𝑓 ∈ 𝐶1(𝐼, 𝐼) : 𝐹𝑘(𝑓) ∩ {𝑥 : (𝑓𝑘)(𝑥) = 1} ̸= 0}
(6)
is closed in𝐶1(𝐼, 𝐼).
Proof. Let𝑔𝑛 ∈ E𝑘 and 𝜌1(𝑔𝑛, 𝑔) → 0. Then there exists
{𝑡𝑛}∞𝑛=1 ⊂ 𝐼such that𝑔𝑘𝑛(𝑡𝑛) = 𝑡𝑛and(𝑔𝑛𝑘)(𝑡𝑛) = 1. Without
loss of generality, we may assume that lim𝑛 → ∞𝑡𝑛 = 𝑡0, then
𝑡0∈ 𝐼. Let𝜖 > 0be arbitrary. Due to the uniform continuity of
𝑔𝑘on𝐼there exists a positive integer𝑛
1such that, for𝑛 ≥ 𝑛1,
|𝑔𝑘(𝑡𝑛) − 𝑔𝑘(𝑡0)| < 𝜖. Since 𝜌0(𝑔𝑛, 𝑔) → 0, there exists a
positive integer𝑛2 such that, for 𝑛 ≥ 𝑛2,𝜌0(𝑔𝑛𝑘, 𝑔𝑘) < 𝜖.
Choose the positive integer𝑛3so that|𝑡𝑛− 𝑡0| < 𝜖for𝑛 ≥ 𝑛3,
and let𝑚1=max(𝑛1, 𝑛2, 𝑛3). Then for𝑚 ≥ 𝑚1we have
𝑔𝑘(𝑡0) − 𝑡0
≤ 𝑔𝑘(𝑡0) − 𝑔𝑘(𝑡𝑚) +𝑔 𝑘(𝑡𝑚) − 𝑔𝑚𝑘 (𝑡𝑚) + 𝑔𝑘𝑚(𝑡𝑚) − 𝑡0
≤ 𝜖 + 𝜌0(𝑔𝑘, 𝑔𝑘𝑚) + 𝑡𝑚− 𝑡0 ≤ 3𝜖,
(7)
Implying that𝑔𝑘(𝑡0) = 𝑡0. We also have
(𝑔𝑘𝑛)
(𝑡𝑛) − (𝑔𝑘)(𝑡𝑛) ≤𝑛 → ∞lim𝜌1(𝑔𝑛𝑘, 𝑔𝑘) = 0. (8)
Thus we have lim𝑛 → ∞|1 − (𝑔𝑘)(𝑡𝑛)| = |1 − (𝑔𝑘)(𝑡0)| = 0and
𝑔 ∈E𝑘.
Theorem 5. There exists a residual subset𝑀1of𝐶1(𝐼, 𝐼)such
that, for every𝑓 ∈ 𝑀1,𝐹1(𝑓)is finite.
Proof. If 𝐹1(𝑓) has infinitely many elements, then from
Lemma1it follows that there exists a point𝑥0 ∈ 𝐹1(𝑓)so
that𝑓(𝑥0) = 1. Let
E1= {𝑓 ∈ 𝐶1(𝐼, 𝐼) : ∃𝑥0∈ 𝐼 ∩ 𝐹1(𝑓) , 𝑓(𝑥0) = 1} . (9)
From Lemma4the setE1is closed in𝐶1(𝐼, 𝐼). To show that
Lemma2, there exists a polynomial𝑝such that𝜌1(𝑓, 𝑝) < 𝜖/2 and𝑝(𝐼) ⊂ [𝑎 + 𝜖1, 𝑏 − 𝜖1]for some0 < 𝜖1< 𝜖/2. Let
𝑀 (𝑝 ) = {𝑥 : 𝑝(𝑥) = 1} = {𝑡1, 𝑡2, . . . , 𝑡𝑚} . (10)
Choose0 < 𝜖2< 𝜖1so that𝑝(𝑥) − 𝑥 ̸= 𝜖2for all𝑥 ∈ 𝑀(𝑝 )and
takeℎ = 𝑝 − 𝜖2. Thenℎ ∈ 𝐶1(𝐼, 𝐼),𝜌1(ℎ, 𝑓) < 𝜖,𝑀(ℎ) = {𝑥 :
ℎ(𝑥) = 1} = 𝑀(𝑝), andℎ(𝑥) ̸= 𝑥on𝑀(ℎ), implying thatE 1
is of first category and𝑀1= 𝐶1(𝐼, 𝐼) \E1is residual.
Theorem 6. The set of functions𝑓 ∈ 𝐶1(𝐼, 𝐼)with 𝑓(𝑎) =
𝑏, 𝑓(𝑏) = 𝑎, and(𝑓2)(𝑎) = 1is a nowhere dense subset of
𝐶1(𝐼, 𝐼).
Proof. Let 𝐴 = {𝑓 ∈ 𝐶1(𝐼, 𝐼) : 𝑓(𝑎) = 𝑏, 𝑓(𝑏) =
𝑎, and (𝑓2)(𝑎) = 1}. It is easy to see that𝐴is a closed subset of𝐶1(𝐼, 𝐼). To show that𝐴is nowhere dense, let𝑓 ∈ 𝐶1(𝐼, 𝐼) and𝜖 > 0. Choose0 < 𝛾 < min{𝜖/(4(𝑏 − 𝑎) + 4), 1}and
𝑎 < 𝑐 < 𝑏such that|𝑓(𝑡) − 𝑓(𝑎)| < (𝑏 − 𝑎)[1 − 𝜖/4] for
𝑡 ∈ [𝑎, 𝑐]. Letℎ(𝑥) = 𝑓(𝑥) + ∫𝑎𝑥𝑘(𝑡)𝑑𝑡, where
𝑘 (𝑡) = { { { { { { { { { { { { { { { { { { { { { { {
−𝛾, 𝑡 = 𝑎,
0, 𝑡 = 𝑎 +𝑐 − 𝑎 4 , 𝛾
3, 𝑡 = 𝑎 + (𝑐 − 𝑎)2 , 0, 𝑡 = 𝑐, 𝑡 = 𝑏,
linear, otherwise.
(11)
It is clear thatℎ(𝑎) = 𝑓(𝑎) = 𝑏,ℎ(𝑏) = 𝑓(𝑏) = 𝑎,ℎ(𝑎) =
𝑓(𝑎) − 𝛾, andℎ(𝑏) = 𝑓(𝑏). It is easy to see thatℎ ∈ 𝐶1(𝐼, 𝐼),
(ℎ2)(𝑎) = (ℎ2)(𝑏) ̸= 1, and𝜌
1(𝑓, ℎ) < 𝜖/2.
Theorem 7. Let𝜖 > 0,𝑓 ∈ 𝐶1(𝐼, 𝐼), and𝐴𝑛 = {𝑥 : (𝑓𝑛)(𝑥) =
1} ∩ 𝐹𝑛(𝑓)be a finite set, and𝑜𝑟𝑏(𝑥, 𝑓) ⊂ (𝑎, 𝑏)for𝑥 ∈ 𝐴𝑛.
Then there exists a function𝑔 ∈ 𝐶1(𝐼, 𝐼)with𝜌1(𝑓, 𝑔) < 𝜖/4
such that{𝑥 : (𝑔𝑛)(𝑥) = 1} ∩ 𝐹𝑛(𝑔) = 0.
Proof. Let𝐴𝑛 = {𝑥 : (𝑓𝑛)(𝑥) = 1} ∩ 𝐹𝑛(𝑓) = {𝑧𝑗}𝑙𝑗=1. Let
𝐵 = {𝑡𝑗}𝑘𝑗=1be the elements of𝐴𝑛with distinct orbits, that is,
orb(𝑡𝑖, 𝑓) ∩orb(𝑡𝑗, 𝑓) = 0for𝑖 ̸= 𝑗and∪𝑘𝑗=1orb(𝑡𝑗, 𝑓) ⊇ 𝐴𝑛.
It is clear that each𝑡𝑗 ∈ 𝐵is a periodic point of𝑓with some
period𝑚where𝑚is a factor of𝑛. This suggests that if one
can construct a function𝑔that is sufficiently close to𝑓, and
either𝑡𝑗∉ 𝑃𝑓𝑚or(𝑓𝑚)(𝑡𝑗) ̸= 1, then
{𝑥 : (𝑔𝑛)(𝑥) = 1} ∩ 𝐹𝑛(𝑔) ⊆ 𝐴𝑛\orb(𝑡𝑗, 𝑓) . (12)
By repeating this process for each𝑡𝑗 ∈ 𝐵in a finite number
of steps we can construct the desired function𝑔. Thus for
convenience, we assume that, for1 ≤ 𝑗 ≤ 𝑘,𝑧𝑗 ∈ 𝑃𝑓𝑛.
Consider the partition𝑇of [𝑎, 𝑏]obtained by {𝑓𝑖(𝑡𝑗), 1 ≤
𝑖 ≤ 𝑛, 1 ≤ 𝑗 ≤ 𝑘},𝜖1 = (1/4)min{𝜖, ‖ 𝑇 ‖}, and choose a
positive𝛿less than𝜖1such that, for each𝑖, 1 ≤ 𝑖 ≤ 𝑛and each
𝑗, 1 ≤ 𝑗 ≤ 𝑘, and if|𝑡−𝑡𝑗| < 𝛿, then|𝑓𝑖(𝑡)−𝑓𝑖(𝑡𝑗)| < 𝜖1. Given
that, for𝑥 ∈ 𝐴𝑛, orb(𝑥, 𝑓) ⊂ (𝑎, 𝑏),(𝑓𝑛)is continuous, and
(𝑓𝑛)(𝑥) = 1on𝐵, we may choose the nondegenerate closed
intervals𝐻𝑗, 1 ≤ 𝑗 ≤ 𝑘of length less than𝛿and the positive
numbers𝛾𝑗such that
(i)𝑡𝑗 ∈ (𝐻𝑗)∘ is the midpoint of 𝐻𝑗 and 𝑓𝑖(𝐻𝑗) ∩
orb(𝑡𝑗, 𝑓) = {𝑓𝑖(𝑡𝑗)}for0 ≤ 𝑖 ≤ 𝑛,
(ii)𝑓is strictly monotone on each𝐻𝑗,𝑖 = 𝑓𝑖(𝐻𝑗)for0 ≤
𝑖 ≤ 𝑛 − 1,1 ≤ 𝑗 ≤ 𝑘, where𝐻𝑗,0= 𝐻𝑗,
(iii) the intervals𝐻𝑗,𝑖 = 𝑓𝑖(𝐻𝑗)are mutually disjoint for
1 ≤ 𝑖 ≤ 𝑛,1 ≤ 𝑗 ≤ 𝑘,
(iv)𝑎 <min(𝐻𝑗,𝑖) − 𝛾𝑗𝜆(𝐻𝑗,𝑖) <max(𝐻𝑗,𝑖) + 𝛾𝑗𝜆(𝐻𝑗,𝑖) < 𝑏, for0 ≤ 𝑖 ≤ 𝑛 − 1,
(v) max1≤𝑗≤𝑘{𝛾𝑗, (𝑏 − 𝑎)𝛾𝑗} < (1/8)[min0≤𝑖≤𝑛, 1≤𝑗≤𝑘{𝜖/𝑘,
𝜆(𝑓𝑖(𝐻𝑗)), 𝛽𝑗,𝑖}], where 𝛽𝑗,𝑖 = inf{|𝑓(𝑥)| : 𝑥 ∈
𝑓𝑖(𝐻𝑗)}.
Let𝑥0 = 𝑡𝑖for some𝑖,𝛾, and𝐽 = [𝑐, 𝑑]be the associated
𝛾𝑖and𝐻𝑖, respectively. Define
𝑔𝐽,𝑥0,𝛾(𝑥) = { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { −𝛾
3, if𝑥 = 𝑐 + 𝑥0− 𝑐
2 ,
0, if𝑥 = 𝑐 +3 (𝑥0− 𝑐)
4 , 𝛾, if𝑥 = 𝑥0,
0, if𝑥 = 𝑥0+𝑑 − 𝑥0
4 ,
−𝛾3, if𝑥 = 𝑥0+𝑑 − 𝑥0
2 , 0, if𝑥 ∈ [𝑎, 𝑏] \ (𝑐, 𝑑) ,
linear, otherwise,
𝑟𝐽,𝑥0,𝛾(𝑥) = { { { { { { { { { { { { { { { { { { { { { { {
𝛾, if 𝑥 = 𝑐 +𝑥0− 𝑐
4 , 0, if 𝑥 = 𝑥0,
−𝛾 (𝑥0− 𝑐)
𝑑 − 𝑥0 if 𝑥 = 𝑥0+3 (𝑑 − 𝑥4 0),
0 if 𝑥 ∈ [𝑎, 𝑏] \ (𝑐, 𝑑) ,
linear, otherwise,
ℎ1(𝑥) = ∫ 𝑥
𝑎 𝑔𝐽,𝑥0,𝛾(𝑡) 𝑑𝑡,
ℎ2(𝑥) = ∫ 𝑥
𝑎 𝑟𝐽,𝑥0,𝛾(𝑡) 𝑑𝑡.
(13)
We haveℎ𝑚(𝑐) = ℎ𝑚(𝑑) = ℎ𝑚(𝑐) = ℎ𝑚(𝑑) = 0for𝑚 = 1, 2,
ℎ1(𝑥0) = 0,ℎ1(𝑥0) = 𝛾,ℎ2(𝑥0) = (𝑥0− 𝑐)𝛾/2, andℎ2(𝑥0) = 0. By considering four different cases, we construct a function
ℎ ∈ 𝐶1(𝐼, 𝐼) ∩ 𝐵(𝑓, 𝜖)so that, for Cases A and B,𝐹𝑛(ℎ) ∩ {𝑥 :
(ℎ𝑛)(𝑥) = 1} ∩ 𝐽 = 0and𝐹
𝑛(ℎ) ∩ {𝑥 : (ℎ𝑛)(𝑥) = 1} ∩ 𝐽 =
{𝑥0}for Cases C and D. From conditions (iv) and (v) we have
ℎ(𝐼) ⊂ 𝐼, and from condition (v) it follows thatℎis of the
same monotone type as𝑓on each𝑓𝑠(𝐽), 0 ≤ 𝑠 < 𝑛.
Case A. Let𝑓𝑛(𝑥) < 𝑥for𝐽 \ {𝑥0}and𝑓𝑛(𝑥0) = 𝑥0.
(i) If 𝑓 is strictly increasing on 𝐽, then by taking
andℎ(𝑥) = 𝑓(𝑥)on𝐼 \ 𝐽𝑜 and the function ℎ𝑛−1 is
also strictly increasing on𝑓(𝐽), and as a resultℎ𝑛(𝑥) <
𝑓𝑛(𝑥) ≤ 𝑥on𝐽𝑜. Thus𝐹
𝑛(ℎ) ∩ 𝐽 = 0.
(ii) If𝑓is strictly decreasing on𝐽, then by takingℎ(𝑥) =
𝑓(𝑥) + ℎ2(𝑥)we haveℎ(𝑥) > 𝑓(𝑥)on𝐽𝑜andℎ(𝑥) =
𝑓(𝑥)on𝐼 \ 𝐽𝑜 and the functionℎ𝑛−1 is also strictly
decreasing on𝑓(𝐽)and as a result,ℎ𝑛(𝑥) < 𝑓𝑛(𝑥) ≤ 𝑥
on𝐽𝑜. Thus𝐹𝑛(ℎ) ∩ 𝐽 = 0.
Case B. Let𝑓𝑛(𝑥) > 𝑥for𝐽 \ {𝑥0}and𝑓𝑛(𝑥0) = 𝑥0.
If𝑓is strictly increasing on𝐽, takeℎ(𝑥) = 𝑓(𝑥) + ℎ2(𝑥),
and if𝑓is strictly decreasing on𝐽, takeℎ(𝑥) = 𝑓(𝑥) − ℎ2(𝑥).
Then similar to Case A, we may show thatℎ𝑛(𝑥) > 𝑓𝑛(𝑥) ≥ 𝑥
for𝑥 ∈ 𝐽𝑜; hence,𝐹𝑛(ℎ) ∩ 𝐽 = 0.
Case C. Let𝑓𝑛(𝑥) < 𝑥for𝑐 < 𝑥 < 𝑥0,𝑓𝑛(𝑥) > 𝑥for𝑥0< 𝑥 <
𝑑, and𝑓𝑛(𝑥0) = 𝑥0.
(i) If𝑓is strictly increasing on𝐽, then by takingℎ(𝑥) =
𝑓(𝑥) + ℎ1(𝑥)we haveℎ(𝑥) < 𝑓(𝑥)for𝑐 < 𝑥 < 𝑥0 andℎ(𝑥) > 𝑓(𝑥)for𝑥0 < 𝑥 < 𝑑, and the function
ℎ𝑛−1is also strictly increasing on𝑓(𝐽). Thusℎ𝑛(𝑥) <
𝑓𝑛(𝑥) < 𝑥for𝑐 < 𝑥 < 𝑥
0andℎ𝑛(𝑥) > 𝑓𝑛(𝑥) > 𝑥for
𝑥0< 𝑥 < 𝑑; hence,𝐹𝑛(ℎ) ∩ 𝐽 = {𝑥0}.
(ii) If𝑓is strictly decreasing on𝐽, then by takingℎ(𝑥) =
𝑓(𝑥) − ℎ1(𝑥)we haveℎ(𝑥) > 𝑓(𝑥)for𝑐 < 𝑥 < 𝑥0 andℎ(𝑥) < 𝑓(𝑥)for𝑥0 < 𝑥 < 𝑑, and the function
ℎ𝑛−1is also decreasing on𝑓(𝐽). Thusℎ𝑛(𝑥) < 𝑓𝑛(𝑥) <
𝑥for𝑐 < 𝑥 < 𝑥0andℎ𝑛(𝑥) > 𝑓𝑛(𝑥) > 𝑥for𝑥0< 𝑥 <
𝑑; hence,𝐹𝑛(ℎ) ∩ 𝐽 = {𝑥0}.
Case D. Let𝑓𝑛(𝑥) > 𝑥for𝑐 < 𝑥 < 𝑥0,𝑓𝑛(𝑥) < 𝑥for𝑥0< 𝑥 <
𝑑, and𝑓𝑛(𝑥0) = 𝑥0.
If𝑓is strictly increasing on𝐽, takeℎ(𝑥) = 𝑓(𝑥) − ℎ1(𝑥),
and if𝑓is strictly decreasing on𝐽, takeℎ(𝑥) = 𝑓(𝑥) + ℎ1(𝑥).
Then similar to Case C, we may show thatℎ𝑛(𝑥) > 𝑓𝑛(𝑥) > 𝑥
for𝑐 < 𝑥 < 𝑥0andℎ𝑛(𝑥) < 𝑓𝑛(𝑥) < 𝑥for𝑥0 < 𝑥 < 𝑑. Thus
𝐹𝑛(ℎ) ∩ 𝐽 = {𝑥0}.
Note thatℎ𝑛is strictly increasing on𝐽 ∪ 𝑓𝑛(𝐽), so we have
𝐹𝑛(ℎ) ∩ [(𝐽 \ 𝑓𝑛(𝐽)) ∪ (𝑓𝑛(𝐽) \ 𝐽)] = 0. (14)
Thus for𝑥 ∈ 𝐼either orb(𝑥, ℎ) = orb(𝑥, 𝑓)or orb(𝑥, ℎ) ∩
𝐽 ̸= 0, of which in such case𝐹𝑛(ℎ)∩𝐽 = {𝑥0},ℎ(𝑥0) = 𝑓(𝑥0)±
𝛾, andℎ(ℎ𝑖(𝑥0)) = 𝑓(𝑓𝑖(𝑥0))for1 ≤ 𝑖 ≤ 𝑛 − 1. Thus
(ℎ𝑛)(𝑥0)
= ℎ(ℎ𝑛−1(𝑥0)) ℎ(ℎ𝑛−2(𝑥0)) ⋅ ⋅ ⋅ ℎ(ℎ (𝑥0)) ℎ(𝑥0)
= 𝑓(𝑓𝑛−1(𝑥0)) 𝑓(𝑓𝑛−2(𝑥0))
× 𝑓(𝑓𝑛−3(𝑥0)) ⋅ ⋅ ⋅ 𝑓(𝑓 (𝑥0)) ℎ(𝑥0)
= (𝑓𝑛)
(𝑥
0) ℎ(𝑥0)
𝑓(𝑥
0) =
𝑓(𝑥 0) ± 𝛾
𝑓(𝑥 0)
= [1 ± 𝛾 𝑓(𝑥
0)] ̸= 1.
(15)
We have
𝐹𝑛(ℎ) ∩ {𝑥 : (ℎ𝑛)(𝑥) = 1}
= [𝐹𝑛(𝑓) ∩ {𝑥 : (𝑓𝑛)(𝑥) = 1}] \ {𝑥0} . (16)
Also
𝜌0(𝑓, ℎ) ≤sup
𝑥∈𝐽{𝑔 (𝑥),|𝑟(𝑥)|} ≤ 𝜖8𝑘,
𝜌0(𝑓, ℎ) ≤sup
𝑥∈𝐽{ℎ1(𝑥) ,ℎ2(𝑥)} ≤
𝜖 8𝑘.
(17)
Hence𝜌1(𝑓, ℎ) ≤ 𝜌0(𝑓, ℎ) + 𝜌0(𝑓, ℎ) ≤ 𝜖/4𝑘.
Doing this recursively for each𝑡𝑗,1 ≤ 𝑗 ≤ 𝑘, we get a
function𝑔 ∈ 𝐵(𝑓, 𝜖/4)such that𝐹𝑛(𝑔) ∩ {𝑥 : (𝑔𝑛)(𝑥) = 1} =
0.
Theorem 8. Typically continuously differentiable self-maps of intervals have a countable set of periodic points.
Proof. Take
𝐻𝑛= {𝑓 ∈ 𝐶1(𝐼, 𝐼) : 𝐹𝑛(𝑓) is not finite} ,
𝐵1= {𝑓 ∈ 𝐶1(𝐼, 𝐼) : 𝑎 ∈ 𝑃𝑓2∩ {𝑥 : (𝑓2)
(𝑥) = 1}
with𝑓 (𝑎) = 𝑏} ,
𝐹1= {𝑓 ∈ 𝐶1(𝐼, 𝐼) : ∃𝑥0∈ 𝐼 such that
𝑥0∈ 𝐹1(𝑓) and𝑓(𝑥0) = 1} ,
E𝑛= {𝑓 ∈ 𝐶1(𝐼, 𝐼) : 𝐹
𝑛(𝑓) ∩ {𝑥 : (𝑓𝑛)(𝑥) = 1} ̸= 0} .
(18)
The sets𝐹1and𝐵1are first category sets (see Theorems
5and6). For𝑛 ≥ 1from Lemmas4and1we have thatE𝑛
is closed and𝐻𝑛 ⊆ E𝑛. Without loss of generality we may
assume that𝑓 ∈ 𝐶1(𝐼, 𝐼)is neither a constant function nor
a polynomial of first degree, and then by Lemma2we may
choose a polynomial 𝑔 ∈ 𝐶1(𝐼, 𝐼) of degree𝑛 ≥ 2 such
that𝑔(𝐼) ⊂ 𝐼𝑜and𝜌1(𝑓, 𝑔) < 𝜖/4. Using Theorem7we can
construct ℎ ∈ 𝐶1(𝐼, 𝐼)so that 𝜌1(𝑔, ℎ) < 𝜖/4and the set
𝐴𝑛 = 𝐹𝑛(ℎ) ∩ {𝑥 : (ℎ𝑛)(𝑥) = 1} = 0, thus𝜌1(𝑓, ℎ) < 𝜖
andℎ ∉ E𝑛, henceE𝑜𝑛 = 0. This implies that for each𝑘 ≥ 2
the setE𝑘is nowhere dense. Thus𝐹 = (∪∞𝑘=2E𝑘) ∪ 𝐹1∪ 𝐵1is
a first category set, so𝑀 = 𝐶1(𝐼, 𝐼) \ 𝐹is a residual set, and,
for𝑓 ∈ 𝑀,𝑃(𝑓) = ∪∞𝑛=1𝐹𝑛(𝑓)is countable.
Acknowledgments
References
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[3] A. A. Alikhani-Koopaei, “On periodic and recurrent points
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