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

Authors: Aliasghar Alikhani-Koopaei
Journal: International Journal of Mathematics and Mathematical Sciences
 (2013)

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## 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.


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## Body

## 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 of f. For f and n∈ℕ we define fn(x) by induction:
(1)fn(x)=f(fn-1(x))withf0(x)=x,f1(x)=f(x).
The orbit of x under f is given by the sequence orb(x,f)={fi(x)}i=0∞. For n∈ℕ, let Fn(f)={x:fn(x)=x} and Pfn be the set of periodic points of order n; that is,
(2)Pfn=Fn(f)∖∪k<nFk(f).
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 T of the interval [a,b], ∥T∥ is the length of the largest subinterval of T, B(f,ϵ) is the open ball about f with radius ϵ, A∘ denotes the set of interior points of A, and λ(I) is the length of the interval I.


## 2. Continuously Differentiable Functions

ForI=[a,b], consider C(I,I) and C1(I,I) to be the family of all continuous maps and continuously differentiable maps from I into itself, respectively. Recall that the usual metrics ρ0 and ρ1 on C(I,I) and C1(I,I), respectively, are given by
(3)ρ0(f,g)=supx∈I|f(x)-g(x)|forf,g∈C(I,I),ρ1(f,g)=ρ0(f,g)+ρ0(f′,g′)forf,g∈C1(I,I).It is well known that the metric spaces(C(I,I);ρ0) and (C1(I,I);ρ1) are complete and hence Baire's category theorem holds in these spaces. We say that a typical function in C(I,I) or C1(I,I) has a certain property if the set of those functions which does not have this property is of first category in C(I,I) or in C1(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≥1 and f∈C1(I,I) so that Fn(f) is not finite. Then there exists x0∈I so that fn(x0)=x0 and (fn)′(x0)=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 have yi, xi<yi<xi+1 so that h′(yi)=0. Let x0=limi→∞xi, then x0∈I, and limi→∞yi=x0, and from the continuity of fn and (fn)′ it follows that fn(x0)=x0 and (fn)′(x0)=1.Lemma 2.
For eachf∈C1(I,I) and 0<ϵ<1 there is a polynomial p:I→Io with ρ1(f,p)<ϵ.Proof.
LetM=supx∈I{|f′(x)|,|a|,|b|,1} where I=[a,b]. Take β=ϵ/8M, α=β(a+b)/2, and g(x)=α+(1-β)f(x). It is easy to see that 0<β<1, g(I)⊂Io, and ρ1(f,g)<ϵ/4. Let
(4)ϵ1=12infx∈[a,b]{g(x)-a,b-g(x),ϵ8(b-a),ϵ8}
and s(t) be a polynomial with ρ0(g′,s)<ϵ1/(b-a+1). Then p(x)=g(a)+∫ax‍s(t)dt is a polynomial and g(x)-p(x)=∫ax‍(g′(t)-s(t))dt. Thus we have
(5)ρ1(g,p)=ρ0(g′,s)+ρ0(g,p)≤ϵ1(b-a)+1+ϵ1(b-a)(b-a)+1=ϵ1<ϵ4,
hence ρ1(f,p)≤ρ1(f,g)+ρ1(g,p)<ϵ/2. From g(I)⊂[a+2ϵ1,b-2ϵ1] and ρ1(g,p)≤ϵ1 it follows that p(I)⊂Io.Lemma 3.
Letn>2 be a positive integer and f∈C1(I,I). If x0 is a periodic point of order n with (fn)′(x0)=1, then orb(x0,f)∩{a,b}=∅.Proof.
We show thata∉orb(x0,f). The case b∉orb(x0,f) follows similarly. Let orb(x0,f)={a=x0,x1,…,xn=x0} with xi=f(xi-1)  for 1≤i≤n. Since n>2, there exist xi∈orb(x0,f)∩(a,b) and 1≤s≤n-1 so that fs(xi)=a. Since fs∈C1(I,I) and (fs)′(x0)≠0, there exists δ1>0 so that a<xi-δ1<xi<xi+δ1<b and fs is strictly increasing or strictly decreasing on J=[xi-δ1,xi+δ1]. Let fs be strictly increasing on J, and then for xi-δ1≤x≤xi we have fs(x)≤fs(xi)=a; hence, fs∣[xi-δ1,xi]=a, implying that (fs)′(xi)=0, a contradiction to (fn)′(x0)=1. On the other hand when fs is strictly decreasing on J, for xi≤x≤xi+δ1 we have fs(x)≤fs(xi)=a; hence, fs∣[xi,xi+δ1]=a implying that (fs)′(xi)=0, a contradiction to (fn)′(x0)=1.Lemma 4.
Fork≥1, the set
(6)𝔼k={f∈C1(I,I):Fk(f)∩{x:(fk)′(x)=1}≠∅}
is closed in C1(I,I).Proof.
Letgn∈𝔼k and ρ1(gn,g)→0. Then there exists {tn}n=1∞⊂I such that gnk(tn)=tn and (gnk)′(tn)=1. Without loss of generality, we may assume that limn→∞tn=t0, then t0∈I. Let ϵ>0 be arbitrary. Due to the uniform continuity of gk on I there exists a positive integer n1 such that, for n≥n1, |gk(tn)-gk(t0)|<ϵ. Since ρ0(gn,g)→0, there exists a positive integer n2 such that, for n≥n2,  ρ0(gnk,gk)<ϵ. Choose the positive integer n3 so that |tn-t0|<ϵ for n≥n3, and let m1=max(n1,n2,n3). Then for m≥m1 we have
(7)|gk(t0)-t0|≤|gk(t0)-gk(tm)|+|gk(tm)-gmk(tm)|+|gmk(tm)-t0|≤ϵ+ρ0(gk,gmk)+|tm-t0|≤3ϵ,
Implying that gk(t0)=t0. We also have
(8)|(gnk)′(tn)-(gk)′(tn)|≤limn→∞ρ1(gnk,gk)=0.
Thus we have limn→∞|1-(gk)′(tn)|=|1-(gk)′(t0)|=0 and g∈𝔼k.Theorem 5.
There exists a residual subsetM1 of C1(I,I) such that, for every f∈M1, F1(f) is finite.Proof.
IfF1(f) has infinitely many elements, then from Lemma 1 it follows that there exists a point x0∈F1(f) so that f′(x0)=1. Let
(9)𝔼1={f∈C1(I,I):∃x0∈I∩F1(f),f′(x0)=1}.
From Lemma 4 the set 𝔼1 is closed in C1(I,I). To show that 𝔼1 has no interior point, let f∈𝔼1 and ϵ>0, and then, by Lemma 2, there exists a polynomial p such that ρ1(f,p)<ϵ/2 and p(I)⊂[a+ϵ1,b-ϵ1] for some 0<ϵ1<ϵ/2. Let
(10)M(p)={x:p′(x)=1}={t1,t2,…,tm}.
Choose 0<ϵ2<ϵ1 so that p(x)-x≠ϵ2 for all x∈M(p) and take h=p-ϵ2. Then h∈C1(I,I),  ρ1(h,f)<ϵ,  M(h)={x:h′(x)=1}=M(p), and h(x)≠x on M(h), implying that 𝔼1 is of first category and M1=C1(I,I)∖𝔼1 is residual.Theorem 6.
The set of functionsf∈C1(I,I) with f(a)=b, f(b)=a, and (f2)′(a)=1 is a nowhere dense subset of C1(I,I).Proof.
LetA={f∈C1(I,I):f(a)=b,f(b)=a,and(f2)′(a)=1}. It is easy to see that A is a closed subset of C1(I,I). To show that A is nowhere dense, let f∈C1(I,I) and ϵ>0. Choose 0<γ<min{ϵ/(4(b-a)+4),1} and a<c<b such that |f(t)-f(a)|<(b-a)[1-ϵ/4] for t∈[a,c]. Let h(x)=f(x)+∫ax‍k(t)dt, where
(11)k(t)={-γ,t=a,0,t=a+c-a4,γ3,t=a+(c-a)2,0,t=c,t=b,linear,otherwise.
It is clear thath(a)=f(a)=b,  h(b)=f(b)=a,  h′(a)=f′(a)-γ, and h′(b)=f′(b). It is easy to see that h∈C1(I,I), (h2)′(a)=(h2)′(b)≠1, and ρ1(f,h)<ϵ/2.Theorem 7.
Letϵ>0, f∈C1(I,I), and An={x:(fn)′(x)=1}∩Fn(f) be a finite set, and orb(x,f)⊂(a,b) for x∈An. Then there exists a function g∈C1(I,I) with ρ1(f,g)<ϵ/4 such that {x:(gn)′(x)=1}∩Fn(g)=∅.Proof.
LetAn={x:(fn)′(x)=1}∩Fn(f)={zj}j=1l. Let B={tj}j=1k be the elements of An with distinct orbits, that is, orb(ti,f)∩orb(tj,f)=∅ for i≠j and ∪j=1korb(tj,f)⊇An. It is clear that each tj∈B is a periodic point of f with some period m where m is a factor of n. This suggests that if one can construct a function g that is sufficiently close to f, and either tj∉Pfm or (fm)′(tj)≠1, then
(12){x:(gn)′(x)=1}∩Fn(g)⊆An∖orb(tj,f).
By repeating this process for each tj∈B in a finite number of steps we can construct the desired function g. Thus for convenience, we assume that, for 1≤j≤k,  zj∈Pfn. Consider the partition T of [a,b] obtained by {fi(tj),1≤i≤n,1≤j≤k}, ϵ1=(1/4)min{ϵ,∥T∥}, and choose a positive δ less than ϵ1 such that, for each i, 1≤i≤n and each j,  1≤j≤k, and if |t-tj|<δ, then |fi(t)-fi(tj)|<ϵ1. Given that, for x∈An, orb(x,f)⊂(a,b), (fn)′ is continuous, and (fn)′(x)=1 on B, we may choose the nondegenerate closed intervals Hj, 1≤j≤k of length less than δ and the positive numbers γj such that(i)
tj∈(Hj)∘ is the midpoint of Hj and fi(Hj)∩orb(tj,f)={fi(tj)} for 0≤i≤n,(ii)
f is strictly monotone on each Hj,i=fi(Hj) for 0≤i≤n-1, 1≤j≤k, where Hj,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, for 0≤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=ti for some i, γ, and J=[c,d] be the associated γi and Hi, respectively. Define
(13)gJ,x0,γ(x)={-γ3,ifx=c+x0-c2,0,ifx=c+3(x0-c)4,γ,ifx=x0,0,ifx=x0+d-x04,-γ3,ifx=x0+d-x02,0,ifx∈[a,b]∖(c,d),linear,otherwise,rJ,x0,γ(x)={γ,ifx=c+x0-c4,0,ifx=x0,-γ(x0-c)d-x0ifx=x0+3(d-x0)4,0ifx∈[a,b]∖(c,d),linear,otherwise,h1(x)=∫ax‍gJ,x0,γ(t)dt,h2(x)=∫ax‍rJ,x0,γ(t)dt.
We have hm(c)=hm(d)=hm′(c)=hm′(d)=0 for m=1,2, h1(x0)=0, h1′(x0)=γ, h2(x0)=(x0-c)γ/2, and h2′(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=∅ and Fn(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 that h is of the same monotone type as f on each fs(J),  0≤s<n.
Case A. Let fn(x)<x for J∖{x0} and fn(x0)=x0.(i)
Iff is strictly increasing on J, then by taking h(x)=f(x)-h2(x) we have h(x)<f(x) on Jo and  h(x)=f(x) on I∖Jo and the function hn-1 is also strictly increasing on f(J), and as a result hn(x)<fn(x)≤x on Jo. Thus Fn(h)∩J=∅.(ii)
Iff is strictly decreasing on J, then by taking h(x)=f(x)+h2(x) we have h(x)>f(x) on Jo and h(x)=f(x) on I∖Jo and the function hn-1 is also strictly decreasing on f(J) and as a result, hn(x)<fn(x)≤x on Jo. Thus Fn(h)∩J=∅.
Case B. Let fn(x)>x for J∖{x0} and fn(x0)=x0.
Iff is strictly increasing on J, take h(x)=f(x)+h2(x), and if f is strictly decreasing on J, take h(x)=f(x)-h2(x). Then similar to Case A, we may show that hn(x)>fn(x)≥x for x∈Jo; hence, Fn(h)∩J=∅.
Case C. Let fn(x)<x for c<x<x0,  fn(x)>x for x0<x<d, and fn(x0)=x0.(i)
Iff is strictly increasing on J, then by taking h(x)=f(x)+h1(x) we have h(x)<f(x) for c<x<x0 and h(x)>f(x) for x0<x<d, and the function hn-1 is also strictly increasing on f(J). Thus hn(x)<fn(x)<x for c<x<x0 and hn(x)>fn(x)>x for x0<x<d; hence, Fn(h)∩J={x0}.(ii)
Iff is strictly decreasing on J, then by taking h(x)=f(x)-h1(x) we have h(x)>f(x) for c<x<x0 and  h(x)<f(x) for x0<x<d, and the function hn-1 is also decreasing on f(J). Thus hn(x)<fn(x)<x for c<x<x0 and hn(x)>fn(x)>x for x0<x<d; hence, Fn(h)∩J={x0}.
Case D. Let fn(x)>x for c<x<x0,  fn(x)<x for x0<x<d, and fn(x0)=x0.
Iff is strictly increasing on J, take h(x)=f(x)-h1(x), and if f is strictly decreasing on J, take h(x)=f(x)+h1(x). Then similar to Case C, we may show that hn(x)>fn(x)>x for c<x<x0 and hn(x)<fn(x)<x for x0<x<d. Thus Fn(h)∩J={x0}.
Note thathn is strictly increasing on J∪fn(J), so we have
(14)Fn(h)∩[(J∖fn(J))∪(fn(J)∖J)]=∅.
Thus for x∈I either orb(x,h)=orb(x,f) or orb(x,h)∩J≠∅, of which in such case Fn(h)∩J={x0}, h′(x0)=f′(x0)±γ, and h′(hi(x0))=f′(fi(x0)) for 1≤i≤n-1. Thus
(15)(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)′(x0)h′(x0)f′(x0)=f′(x0)±γf′(x0)=[1±γf′(x0)]≠1.
We have
(16)Fn(h)∩{x:(hn)′(x)=1}=[Fn(f)∩{x:(fn)′(x)=1}]∖{x0}.
Also
(17)ρ0(f′,h′)≤supx∈J{|g(x)|,|r(x)|}≤ϵ8k,ρ0(f,h)≤supx∈J{|h1(x)|,|h2(x)|}≤ϵ8k.
Hence ρ1(f,h)≤ρ0(f,h)+ρ0(f′,h′)≤ϵ/4k.
Doing this recursively for eachtj, 1≤j≤k, we get a function g∈B(f,ϵ/4) such that Fn(g)∩{x:(gn)′(x)=1}=∅.Theorem 8.
Typically continuously differentiable self-maps of intervals have a countable set of periodic points.Proof.
Take(18)Hn={f∈C1(I,I):Fn(f)isnotfinite},B1={f∈C1(I,I):a∈Pf2∩{x:(f2)′(x)=1}aaawithf(a)=bPf2{x:(f2)′(x)=1}},F1={f∈C1(I,I):∃x0∈Isuchthataaax0∈F1(f)andf′(x0)=1},𝔼n={f∈C1(I,I):Fn(f)∩{x:(fn)′(x)=1}≠∅}.
The setsF1 and B1 are first category sets (see Theorems 5 and 6). For n≥1 from Lemmas 4 and 1 we have that 𝔼n is closed and Hn⊆𝔼n. Without loss of generality we may assume that f∈C1(I,I) is neither a constant function nor a polynomial of first degree, and then by Lemma 2 we may choose a polynomial g∈C1(I,I) of degree n≥2 such that g(I)⊂Io and ρ1(f,g)<ϵ/4. Using Theorem 7 we can construct h∈C1(I,I) so that ρ1(g,h)<ϵ/4 and the set An=Fn(h)∩{x:(hn)′(x)=1}=∅, thus ρ1(f,h)<ϵ and h∉𝔼n, hence 𝔼no=∅. This implies that for each k≥2 the set 𝔼k is nowhere dense. Thus F=(∪k=2∞𝔼k)∪F1∪B1 is a first category set, so M=C1(I,I)∖F is a residual set, and, for f∈M, P(f)=∪n=1∞Fn(f) is countable.



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