PII. S0161171203205366 http://ijmms.hindawi.com © Hindawi Publishing Corp.
ON COMMON FIXED POINTS, PERIODIC
POINTS, AND RECURRENT POINTS
OF CONTINUOUS FUNCTIONS
ALIASGHAR ALIKHANI-KOOPAEI
Received 15 May 2002 and in revised form 27 November 2002
It is known that two commuting continuous functions on an interval need not have a common fixed point. However, it is not known if such two functions have a common periodic point. We had conjectured that two commuting continuous functions on an interval will typically have disjoint sets of periodic points. In this paper, we first prove thatS is a nowhere dense subset of[0,1] if and only if {f∈C([0,1]):Fm(f )∩S= ∅}is a nowhere dense subset ofC([0,1]). We also give some results about the common fixed, periodic, and recurrent points of functions. We consider the class of functionsf with continuousωf studied by Bruckner and Ceder and show that the set of recurrent points of such functions are closed intervals.
2000 Mathematics Subject Classification: 54C50, 54C60, 54H25, 37E05, 26A21.
Conjecture 1.1and raised the possibility of using a Baire-category argument to settle the problem. It is also not difficult to show that commuting continuous self-maps of the intervals must have common recurrent points. However, this is not true in general for the case of metric spaces. In [1], we also constructed an example of a pair of commuting continuous self-maps of a compact metric space with no common periodic points, which also provided an answer to a question raised in [2].
Conjecture1.1. Two commuting continuous functions on an interval will
typically have disjoint sets of periodic points.
Steele [7] considered the above conjecture and investigated the likelihood of such pairf andgwhenP (f )is first category and pointed out the difficulties that one might face using a Baire-category argument.
In this paper, we first provide a natural setting for the study of this problem and give some results including a generalization of Steele’s main theorem. We also consider the class of functions with continuousωf(x)studied by Bruck-ner and Ceder [5] and show that the set of recurrent points of such functions is a closed interval. We begin with some preliminaries.
The class of continuous self-maps of a closed intervalIis denoted byC(I, I). Forf∈C(I, I)and any integern≥1,fndenotes thenth iterate off. The or-bit ofx underf (i.e., the set{fk(x):k≥0}) is denoted byO(f , x). The set of cluster points ofO(f , x)is denoted byω(f , x). We simply useω(x) in-stead ofω(f , x)when there is no room for confusion. We also useωf(x)for the function x→ω(f , x). A pointx is inR(f ) ifx∈ω(f , x). LetFn(f )= {x∈I:fn(x)=x}. We callF (f )=F
1(f )the set of fixed points off,P (f )= ∪∞
n=1Fn(f )the set of periodic points off, andR(f )the set of recurrent points of f, respectively. A subset Y ofI is called invariant under g ifg(Y )⊆Y. A closed, invariant, nonempty subset ofIis called minimal if it contains no proper subset that is also closed, invariant, and nonempty. Every closed, in-variant, and nonempty subset ofIcontains a minimal set. IfY is a minimal set, thenY⊆R(f ), and if it is not the orbit of a periodic point, then it is a perfect set. A minimal set is also nowhere dense.
SupposeᏴ= {(f , g):f , g∈C(I, I)}. Defineρ:Ᏼ→[0,∞)byρ((f1, g1),
its Bernstein polynomials are also inC(I, I). In the sequel, we useTheorem 1.2
given below from [3]. The proof ofTheorem 1.3is also trivial, and thus, it is omitted.
Theorem1.2. Iff∈C(I, I), thenP (f )=R(f ).
Theorem1.3. LetᏴandρbe as defined, then(Ᏼ, ρ)is a compact metric
space.
2. Main results
Lemma2.1. Letf∈C(I, I)and letS be anFσ subset ofI. ThenS is of first
category if and only ifSdoes not contain an interval.
Proof. SupposeSis of first category andS= ∪∞m
=1Smwith eachSmclosed. Then, from the Baire-category theorem, it follows thatS does not contain an interval. On the other hand, ifS does not contain an interval, then for each
m≥1,Sm is a nowhere dense set, implying thatS is a first-category subset ofI.
Corollary2.2. Letf∈C(I, I). ThenP (f )is of first category if and only if
P (f )does not contain an interval.
Theorem2.3. The setA= {(f , g):P (f )andP (g)are both first category}
is a residual subset ofᏴ.
Proof. Let A1 = {(f , g)∈ Ᏼ :P (f ) is not first category} and let A2 = {(f , g)∈Ᏼ:P (g) is not first category}, then we haveA=Ᏼ\(A1∪A2). We show that Ais a residual subset of Ᏼ by showing that A1and A2 are first-category subsets ofᏴ. We prove our claim for setA1; the case forA2similarly follows. Suppose(f , g)∈A1, thenP (f )is not of first category inI. Thus it contains an interval, hence we have
A1=(f , g)∈Ᏼ:P (f )is not first category
=(f , g)∈Ᏼ:P (f )contains an interval
= ∞
m=1
(f , g)∈Ᏼ:Fm(f )contains an interval.
(2.1)
Thus it suffices to show that, for each m ≥1, the set Bm = {(f , g)∈ Ᏼ :
Fm(f )contains an interval}is a first-category subset ofᏴ. This is achieved if we show thatBm,n= {(f , g)∈Ᏼ:Qn⊆Fm(f )} is a nowhere dense sub-set ofᏴ, where{Qn}is an enumeration of the rational intervals(a, b)with
a polynomialP, thusFm(P )does not containQn. This implies thatBm,n=Bm,n has no interior, henceBm,nis a nowhere dense subset ofᏴ.
Lemma2.4. LetS ⊂[0,1]be a nowhere dense set and letmbe a positive
integer. Then the setAm= {f∈C([0,1]):Fm(f )∩S= ∅}is a nowhere dense subset ofC([0,1]).
Proof. First we show thatAmis a closed subset ofC([0,1]). For this, let {gn} ⊆Am be a sequence and{gn}converges uniformly tog; we show that
g∈Am. SinceFm(gn)∩S= ∅, takeyn∈Fm(gn)∩S. Without loss of generality, we may assume limn→∞yn=y0, thusy0∈S. Let >0 be arbitrary. Choosek large enough so that|gm(y
k)−gm(y0)|< ,gkm−gm< , and|yk−y0|< . Then we have
gmy0
−y0≤gm
y0
−gmyk
+gmyk
−gmkyk
+gmkyk
−y0 ≤gmy0
−gmyk+gm
yk
−gmk
yk+gmk
yk
−y0 ≤gmy0−gmyk|+gm−gkm+yk−y0
≤3.
(2.2)
Sincewas arbitrary, we havegm(y
0)=y0, implying thaty0∈Fm(g)∩S, henceg∈Am. To show thatAmis nowhere dense subset ofC([0,1]), we show that for eachf∈Amand >0, there exists a functionh∈C([0,1])so that h−f< andFm(h)∩S= ∅. To this end, letPbe a Bernstein polynomial of
f so thatf−P< /4mand letE1=F1(P )∩S= {z1, z2, . . . , zk}. Due to the uniform continuity ofP, for >0, there existsδk>0 so that fort1, t2∈[0,1], with|t1−t2|<4δk, we have|P (t1)−P (t2)| ≤/4mk. Letδ
k=(1/4)min{|zi−
zj|fori=j, δk, /4}. For eachzi∈E1, we have one of the following cases. Case1. For allx∈(zi−δk, zi)∪(zi, zi+δk),P (x) > x.
Case2. For allx∈(zi−δk, zi)∪(zi, zi+δk),P (x) < x.
Case 3. For all x ∈(zi−δk, zi), P (x) > x, and for allx ∈ (zi, zi+δk),
P (x) < x.
Case 4. For all x ∈(zi−δk, zi), P (x) < x, and for allx ∈ (zi, zi+δk),
P (x) > x.
We only prove Cases1and3. Cases2and4are proved similarly. InCase 1, choosex1andx2so thatz1−δ
k< x1< z1< x2< z1+δ
kand take
P1(x)=P (x)forx≤x1orx≥x2, being linear on the interval[x1, x2]. InCase 3, choosex1,x2, andx3so thatz1−δ
k< x3< z1< x2< x1< z1+δ k,
Q1=Pk, we haveQ1−f ≤ Q1−Pk−1+Pk−1−Pk−2+···+P1−P+P−
f ≤k(/4mk)+/4m=/2mandF1(Q1)∩S= ∅. SinceQ1is a continuous piecewise polynomial,F2(Q1)has finitely many elements. Thus, by replacing the role of P with Q1 and the role of F1(P )with F2(Q1), similarly, we may choose a piecewise polynomialQ2inC([0,1])so thatQ2−Q1 ≤3/4mand
F2(Q2)∩S= ∅. Continuing this process, we may choose piecewise polynomials {Qi}m
i=1⊂C([0,1])so thatf−Q1 ≤3/4m,Qi−Qi−1 ≤3/4mfor 2≤i≤
m, andFi(Qi)∩S= ∅for 1≤i≤m. Thus we haveQm−f ≤ Qm−Qm−1+ Qm−1−Qm−2 + ··· + Q1−f ≤3/4m+3/4m+ ··· +3/4m+3/4m≤
m(3/4m)=3/4< , andFm(Qm)∩S= ∅. This implies thatQm∈Am and henceAmis a nowhere dense subset ofC([0,1]).
Lemma2.5. LetS ⊆[0,1]and letmbe a positive integer. If the setAm= {f∈C([0,1]):Fm(f )∩S= ∅}is a nowhere dense subset ofC([0,1]), thenS is a nowhere dense subset of[0,1].
Proof. On the contrary, supposeS is somewhere dense. ThenS contains
a closed interval[a, b]of positive length. Define the functionf asf (x)=x, forx ≤a or x ≥b, f (a+(b−a)/8)=1, f (a+(b−a)/4)=a+(b−a)/4,
f (x)=x, fora+(b−a)/4≤x≤a+3(b−a)/8,f (a+(b−a)/2)=0 and linear elsewhere. Take < (1/2)min{1−(a+(b−a)/4), a+(b−a)/2}. It is easy to see that for eachg∈C([0,1])withg−f< , there existx1, x2∈(a, b)so thatg(x1) > x1,g(x2) < x2. This implies the existence ofx3∈(a, b)∩F1(g)⊆
Fm(g)∩S, henceg∈Am, contrary to the assumption thatAm is a nowhere dense subset ofC([0,1]).
By combining Lemmas2.4and2.5, we have the following theorem.
Theorem2.6. LetS⊆[0,1]and letm be a positive integer. Then the set
Am= {f∈C([0,1]):Fm(f )∩S= ∅}is a nowhere dense subset ofC([0,1])if and only ifSis a nowhere dense subset of[0,1].
Theorem2.7. LetS⊂[0,1]be a first-category set. Then there exists a
resid-ual subsetMofC([0,1])so thatP (g)∩S= ∅for eachg∈M.
Proof. LetS= ∪∞n=1Sn, whereSnis nowhere dense for eachn. For each
pos-itive integermfromTheorem 2.6, we have that the setAm,n= {g∈C([0,1]):
Fm(g)∩Sn= ∅}is a nowhere dense subset ofC([0,1]). ThusB= ∪∞m=1∪∞n=1
Am,n is a first-category subset of C([0,1]). Let M=C([0,1])\B, then M is residual and ifg∈M, theng∈Am,nfor eachm≥1 andn≥1. Thus we have
Fm(g)∩Sn= ∅form≥1 andn≥1. This implies thatS∩P (g)= ∅for each
g∈M.
Theorem2.8. The functionh:Ᏼ→[0,∞)defined byh((f , g))=d(F (f ), F (g))
is a lower semicontinuous function.
Corollary2.9. LetᏭ= {(f , g)∈Ᏼ:fandghave a common fixed point}
and letᏮ= {(f , g)∈Ᏼ:f andghave a common periodic point}. ThenᏭand
Ꮾare closed andFσ subsets ofᏴ, respectively.
Now we turn our attention to commuting continuous functions on the inter-val and prove the following lemma that is used inTheorem 2.11.
Lemma2.10. Letf∈C(I, I)whereI=[0,1], and leta, b∈I,a=b. If >0
is a real number, then there exists a functionh∈C(I, I)such thath(a)=f (a),
h(b)=f (b), andh−f< .
Proof. We can assume <1. Then, forx∈I=[0,1], let
h(x)=f (x)+c−f (x)|x−a|c, (2.3)
withc=1/2 iff (b)=1/2 andc=1/3 iff (b)=1/2. Thenhis continuous,
h(a)=f (a),h(b)=f (b), and |h(x)−f (x)| = |(c−f (x))|x−a|c <|x−
a|(/2)≤/2 for any x∈I. Thus,h−f< . Moreover, h(x)∈[0,1]for
x∈I. Iff (x)≥c, 0≤h(x)≤f (x), andf (x)≤h(x)≤1 iff (x) < c. Thus
h∈C(I, I).
Theorem2.11. Letᐆ= {(f , g)∈Ᏼ:f◦g=g◦f}, thenᐆis a nonempty,
closed, and nowhere dense subset ofᏴ.
Proof. It is easy to see that(f , f )∈ᐆwhen f is the identity function,
soᐆis nonempty. Suppose(fn, gn)∈ᐆand (fn, gn)→(f , g). Then we have
fn →f, gn →g, and fn◦gn = gn◦fn for each n. We show that (f , g)∈ ᐆ. Let x ∈I and >0. Since f and g are uniformly continuous, we may choose δ, so that 0 < δ < and for every x1, x2 in I with |x1−x2|< δ, we have|f (x1)−f (x2)|< , |g(x1)−g(x2)|< . Sincefn→f and gn→g, there exists N ∈ N such that n > N implies fn−f < δ, gn−g< δ. Thus, if n > N, |f (g(x))−g(f (x))| ≤ |f (g(x))−f (gn(x))| + |f (gn(x))−
fn(gn(x))| + |gn(fn(x))−g(fn(x))| + |g(fn(x))−g(f (x))| ≤2δ+2≤4. Since >0 and x were arbitrary, we have f◦g=g◦f, so (f , g)∈ᐆ and
ᐆ is closed. Now we show that ᐆis nowhere dense. Let >0, (f , g)∈ᐆ. Suppose f or g is not the identity function, hence there exists x0 so that
f (x0)≠x0 or g(x0)≠x0. Supposeg(x0)≠x0, the case where f (x0)≠x0 is similar. Leth2(x)=g(x) for eachx ∈I, and let h1∈C(I, I) be a func-tion so thath1(x0)=f (x0), buth1(g(x0))≠f (g(x0))andf−h1< . From
Lemma 2.10, witha=x0andb=g(x0), it follows that suchh1∈C(I, I)exists. Then we have(h1, h2)∈B((f , g)), andh2(h1(x0))=h2(f (x0))=g(f (x0))=
f (g(x0))≠h1(g(x0))=h1(h2(x0)). Thusᐆis the union of a nowhere dense set and a singleton{(x, x)}, thus it is a nowhere dense subset ofᏴ.
Theorem2.12. Letf andgbe two commuting continuous self-maps of the
(i) f andghave a common periodic point;
(ii) for any two positive integersmandn, letAm,n= {x:fn(x)=gm(x)}. Then for allx∈Am,n,ω(f , x)is a perfect nowhere dense set.
Proof. The setAm,nis not empty as is shown in [2], however we give the
proof for completeness. IfAm,nis empty, the continuity offandgpermits us to assume without loss of generality that
fn(x) < gm(x) ∀x∈I. (2.4)
Sincegm(1)≤1, the set S= {x∈I:gm(x)≤x}is not empty. Thus, since
S is closed,S has a minimum elementc. Clearly,c=gm(c). Hencefn(c)=
fn(gm(c))=gm(fn(c))so thatfn(c)∈S. Consequently,fn(c)≥c=gm(c). Sincefn(c)≥gm(c)contradicts (2.4), the assertion thatA
m,nis empty is false. Now suppose that (i) does not hold, that is,f andgdo not have a common periodic point, and letx∈Am,n. Thenfn(x)=gm(x), and hencefn(f (x))=
f (gm(x))=gm(f (x)) and fn(g(x))=g(fn(x))=g(gm(x))=gm(g(x)). Thus ifx∈Am,n,f (x), g(x)∈Am,n, andfpn(x)=gpm(x)for each positive integerp. From this, it follows that O(g, x) is contained inAm,n whenever
x∈Am,n. From closeness ofAm,n, it follows thatω(g, x)⊆Am,n. Ifω(g, x) has an isolated point, then the isolated point is a periodic point ofgwhich is contained inAm,n, thus also a periodic point off, which is a contradiction. So for eachx∈Am,n,ω(g, x)is a perfect set. Suppose for somex∈Am,n,ω(g, x) is somewhere dense, then it contains an intervalI0. LetI1be the largest interval such thatI0⊆I1⊆Am,nandx0∈I1o, then there are positive integersk1,k2with
k1< k2so thatgk1(x)andgk2(x)are inI1. This implies thatgk2−k1(I1)⊆I1, and thusgshould have a periodic point inI1that is also a periodic point of
f, which is again a contradiction. It is also easy to see that the setω(g, x)is an invariant set underg. SupposeA1is a minimal set contained inω(g, x). Sincefandghave no common periodic points, the setA1cannot be the orbit of a periodic point. HenceA1is also a perfect set contained inAm,n∩ω(g, x). Thus, for eachx∈Am,n,ω(g, x)=ω(f , x) is a perfect nowhere dense set containing a perfect minimal set.
Remark2.13. Bruckner and Ceder [5] proved thatωf is Baire 1 if and only
if any infiniteω-limit set forf is perfect. Motivated by this result, one may extend the above Baire-1 classification ofωf to the closed subsets ofI. If so, the above theorem tells us either the two commuting continuous functions
f and g have a common periodic point or bothωf and ωg are Baire 1 on the closed setAm,nwithω(f , x)andω(g, x)being nowhere dense for each
x∈Am,n.
Remark2.14. An argument similar to the proof of the above theorem could
give the same result if the role of Am,n is replaced with any nonempty
particular,Theorem 2.12is also true if in its statement we replaceAm,nwith
F (f )orF (g).
The family of allω-limit sets of a continuous function with the Hausdorff metric forms a metric space. Bruckner and Ceder [5] introduced a kind of chaos in terms of the mapωf:x→ω(f , x). In the same paper, they proved Lemma 2.15and used it to proveTheorem 2.16which characterizes the con-tinuity ofωf.
Lemma2.15. Ifωf is continuous, thenF1(f )is connected.
Theorem2.16. The following conditions are equivalent:
(1) ωf is continuous; (2) {fn}∞
n=1is equicontinuous; (3) ωf2 is continuous;
(4) F1(f2)= ∩∞n=1fn(I);
(5) F1(f2)is connected and, for allx,{f2n(x)}∞n=1converges to a point of
F1(f2);
(6) F1(f2)is connected; (7) ωf is lower semicontinuous; (8) ωf is upper semicontinuous.
Using the above theorem, we show that the recurrent set of a continuous function defined on a closed interval with continuousωf is a closed interval.
Lemma2.17. For any natural numbern,ωfnis continuous whenωf is con-tinuous.
Proof. From the lower semicontinuity of ωf, it follows that {fk}∞k=1 is
equicontinuous, implying that {fnk}∞
k=1 is equicontinuous for eachn ∈N. Hence ωfn is also lower semicontinuous. Thus the lemma follows from
Theorem 2.16.
Theorem2.18. Supposefis a continuous self-map of the unit interval with
ωf:x→ω(f , x)continuous. ThenR(f )is a closed interval.
Proof. For eachm≥1,ωfmis continuous byLemma 2.17, thusF1(fm)=
Fm(f )is connected byLemma 2.15. We also haveF1(f )⊆Fn(f )for alln≥1, implying that∅ =F1(f )⊆∞n=1Fn(f ), thusP (f )= ∞n=1Fn(f )is connected. SinceP (f )⊆R(f )⊆P (f ), we haveR(f )connected. Now let{xn}∞
n=1be a se-quence inR(f )with limn→∞xn=x0. We show thatx0∈R(f ). Letbe an arbi-trary positive number. From the continuity ofωfand the fact that limn→∞xn=
x0, there existsN1∈Nso that|xn−x0|< and dH(ω(f , xn), ω(f , x0)) < for alln≥N1. Since{xn} ⊆R(f ), we havexn∈ω(f , xn)for eachn. Thus, for eachn≥N1, there exists zn ∈ω(f , x0) so that|xn−zn|< , implying that|zn−x0|<2. Henced(x0, ω(f , x0))≤2, for every >0, implying that
Remark 2.19. Due to the fact that R(f )= ∩∞m
=1∪∞n=m{x ∈I:|fn(x)−
x| ≤1/m}, we see that the recurrent set of continuous self-maps of a closed interval is anFσ δset. A question of considerable interest is the classification of such maps withFσ recurrent set. This class of functions contains nomadic functions. A continuous self-map of an interval is called nomadic if it has a dense orbit at some point. It is also known that ifR(f )is closed, thenf has topological entropy zero. ThusTheorem 2.18provides another subclass of the class of continuous self-maps of the interval with anFσ recurrent set as well as zero topological entropy.
Acknowledgment. I wish to thank the anonymous referee for his useful
and important suggestions as well as for proposing the present version of
Lemma 2.10which is more general than our original version.
References
[1] A. Alikhani-Koopaei,A counter example on common periodic points of functions, Int. J. Math. Math. Sci.21(1998), no. 4, 823–827.
[2] ,On common fixed and periodic points of commuting functions, Int. J. Math. Math. Sci.21(1998), no. 2, 269–276.
[3] L. S. Block and W. A. Coppel,Dynamics in One Dimension, Lecture Notes in Math-ematics, vol. 1513, Springer-Verlag, Berlin, 1992.
[4] W. M. Boyce,Commuting functions with no common fixed point, Trans. Amer. Math. Soc.137(1969), 77–92.
[5] A. M. Bruckner and J. Ceder,Chaos in terms of the mapx→ω(x, f ), Pacific J. Math.156(1992), no. 1, 63–96.
[6] J. P. Huneke,On common fixed points of commuting continuous functions on an interval, Trans. Amer. Math. Soc.139(1969), 371–381.
[7] T. H. Steele,A note on periodic points and commuting functions, Real Anal. Ex-change24(1998/99), no. 2, 781–789.
Aliasghar Alikhani-Koopaei: Berks-Lehigh Valley College, Pennsylvania State Univer-sity, Reading, PA 19610-6009, USA
