EQUICONTINUITY OF ITERATES OF CIRCLE MAPS

Let f be a continuous map of the circle to itsel Necessary and sufficient conditions are given for the family ofiterates {f"}= to be equicontinuous.

THEOREM.Let f 6 C(S1,S1).Then {f"}=, is equicontinuous if and only if one of the following holds: (1) f is conjugate to a rotation.
(2) F consists ofexactly two distinct points and every other point on S has period two.

PRELIMINARIES.
Let f 6 C(S,S).We think of the circle S as R/Z and for z,F 6 S with x :/: F we denote by [z,y] the closed interval from z counterclockwise to y.Let d(x,9) denote the where Ix, Y]l is the length of the interval Ix, ].For any nor, negative integer n define f inductively by f f o f"-*, where f0 is the identity map on S*.A point z 6 S is a periodic point of f if there is a positive integer n such that f"(z) z.The least inch n is called the period of z.A point of period one is called a fixed point.Let F, denote the fixed point set of f", Vn >_ 1 and P(f) the set of periodic points of f.If x 6 S then the trajectory of x is the sequence 7(x, f) {f"(z)} ,_>o and the w-limit set of z, w(x, f) N ,,_>0 U ,_>,nf" (x).Equivalently, 6 w(x, f) if and only if $/is a limit point of the trajectory 7(x, f), i.e. f(x) --, y for some sequence ofintegers nk --* co.Let ."{f, f, f3 }.The family of functions/" is said to be equicontinuous ifgiven e > 0 there exists a 6 > 0 such that d(f(z),J(y)) < whenever d(x, 9) < 6 for all z, 9 6 S and all > 1.
The following theorem is proved by J. Cano [5]: THEOREM A. Let f .6'(I,I)such that {f}n=l is equicontinuous.Then F1 is connected and ifit is non-degenerate then F P(f).
The next theorem which is given in [4] and is due to A. M. Bruckner and Thalcyin Hu (only if) and W. Boy (if): THEOREM B. Let f .Co(I,I).Then {f}= is equicontinuous if and only if .=f (.q F2.Combining these two theorems we get the following corollary: COROLLARY.Let f E C (I, I).If f has a periodic point of period n > 2, then {f }.= cannot be equicontinuous.

RESULTS
Let f E C (S , ,S ') such that {ff},oo__l is equicontinuous.We consider three cases" (I) f has a fixed point on S1. (II) the smallest period ofthe periodic points of f on S is n >_ 2. (III) f has no periodic points on 51.
We start with case (1).The basic result ofthis case is Theorem 1.We first show the following four LEMMA 1.Let f E C(,5'1, 51) such that {f}= is equicontinuous.Suppose that there is a fixed point p on S1, and let J be the component of F2 containing p.If J is either {p or a proper closed interval containing p then there exists an open interval K containing J such that w(z, f) c_ J, for every in K.
PROOF.First suppose that d' {p}.Let ISll/4 > 0. By equicontinuity of {f} there is an open interval K containing p such that for every z in K and for every r _> 1, d(f(z), p) < f.Define L--n .of(K).Then L is a closed, proper, invariant interval.By previous results on the interval (Theorems A and B), the fixed point set of flL and f2lL is connected and therefore, it is Moreover, by the above corollary all periodic points of flL have period or 2. But the fixed point p is the only periodic point of f in L. Therefore P(f) F and by [2] the w-limit points coincide with the fixed points.Hence p is the only limit point of f in L and thus ,,(z, f) {p} J, for every z in L.
Since K C L, w(z, f) {p} J, for every z in K. Now suppose that J is a proper closed interval containing p. Let ql and 92 be the endpoints of J, which are fixed points under f.Let [$1 JI/4 > 0. By equicontinuity of {f},, there is an open interval Klaround q such that for every z in K and for every n _ 1,'d(ff(z),f(q)) Similarly there is an open interval K around 92 such that for every z in K and for every n _> 1, d(f(z),f(92)) < .Define L---U.ofJ(K1U2UKz).Then L is a closed, proper, invariant interval.By previous results on the interval (Theorems A and B), the fixed point set of flL and f2[L is connected and therefore it is J.Moreover, by the Corollary all periodic points of f[L have period or 2, which we know that lie in J. Since P(f) is closed, by [2], it coincides with the set of limit points.Therefore w(z, f) C J, for every z in L. Let K KI U J u K. Then K C L and w(z, f) C J, for every a: in K.
LEMMA 2. Let f E C"(S1,S1) such that {f} is equicontinuous.Suppose that there is a fixed point p on $1, and let J be the component of F containing p. Define PROOF.There are three cases" (i) J {p), (ii) J is a poper closed interval containing p and (iii) J S1. If(iii) holds then obviously S J S1. Therefore assume (i) and (ii) hold.Then by 1, there exists an open interval K containing J such that w(:, f) C_ J, for every in K.Note that S is nonempty since S _D K. First we show that S is open: Let x E S. Then w(x, f) C_ J, by definition of S. Choose N large enough such that fv(x) E K. By continuity of fv there is a neighborhood U of z such that if V U then fv(v) K.But then w(f (V), f) to(V, f) c_ J and V S. Therefore S is open.
Let T be the component of S containing J and therefore K, as well.Then T is open and connected.
We will show that T S1.Suppose T :f-S 1.Then S 1-T is a closed interval or a point.Let J [q, q2] where possibly q q2 P.
Suppose first that S T is a closed interval.Let zl and z2 be the endpoints ofthis closed interval such that [z2, zl] NJ 0. Let rnind(q,zl),d(q2, z2)).By equicontinuity of(f') at zl, there is an open interval V around zl inch that for every x in V and for every n _> 1, d(f(x),f(zl)) < .Let x T such that d(x, zl) < .Since w(x, f) C_ J and the orbit of zl stays by definition out of T, there exists a positive integer k such that d(f'(x), fk(zl)) > , which is a contradiction.Now suppose that S T {z}.Let 1/2rnin{d(z,q),d(q2,z)}.By equicontinuity of{f} at z there is an open interval V around z such that for every x in V and for every n _> 1, d(f'(x), f(z)) < .Since w(x, f) C_ J for every x E T and f(z) z is a fixed point of f, we get a contradiction.
PROOF.Suppose that there is an f .C(S1,S)s uch that F2 S and F1 {p}.Let z be a point on S {p} of period two.Let K be the closed interval with endpoints z and f(z) which contains p and let L be the closed interval with the same endpoints that does not contain p.Since f is a homeomorphism, we have two cases: (i) y() and y(L) L or () f(K) L and f(L) K.
If (i) holds then, since f(L)= L, there would be another fixed point of f in L, which is a contradiction since F1 C K.
If(ii) holds then f(K) L implies that p cannot be a fixed point which is again a contradiction, i"i LEMMA 4. Let f E C o ($1, $1).If F2 S and F consists ofmore than two distinct points then f is the identity on S. PROOF.Assume that F consists of exactly k > 2 distinct fixed points Pl,P2,...,Pk-Let Li i, Pi+l] for 1, 2 k 1 and Lk [p,/h] so that the interior of each Li does not contain any fixed points.Then we have two cases: (i) f(Li) Li and (ii) f(Li) S Li.
If (i) holds then pick x in the interior of Li.Note that f(x) is a point in the interior of Li and denote by Mz the dosed interval with endpoints x and f(x) which is free of fixed points.If f(Mz) M= then there would be another fixed point in Mx contradicting that the interior of Li contains no fixed points.
Thus the only choice is x f(x) for every x E Li and f[Li is the identity map.The same argument applied to every Li shows that f is the identity map on S1.
If (ii) holds then there are points in L which map onto the other fixed points contradicting that F=S 1.El THEOREM 1.Let f .C(S,S1 ) such that {f}l is equicontinuous.Suppose that there is a fixed point p on S. Then f has periodic points ofperiod at most two and F2 is'connected.Furthermore F1 is either connected or it consists of exactly two distinct points and every other point on S has period two.Moreover if F is a nondegenerate interval then F1 P(f).
PROOF.Let J be the component of F containing p.There are three cases: (i) J {p}, (ii) J is a proper closed interval containing p and (iii) J S1. Asinine that (i) holds.Then, by Lemma 2, w(x, f) {p} for every x E ,.ql.Thus the fixed point p is the only periodic point of f on S and hence P(f) F1 F2 {p} is connected.
Assume that (ii) holds.Then, by Lemma 2, w(x, f) C_ J for every x S and the periodic points of f on S lie in J.By results on the interval applied to f[J, either p is the unique fixed point of f on S or F is a nondegenerate interval and the fixed points are the only periodic points of f on S1.In particular, both F1 and F2 are connected.
Assume that (iii) holds.Then all of the points of Slare periodic with period or 2 and F2 is connected.By Lemma 3, F1 cannot consist of one point and by Lemma 4 if F1 consists of more than two points then f is the identity map.Otherwise F1 consists of exactly two distinct points and every other point on 5 'l has period two.
We now investisate case (lI) where the smallest period of the periodic points of / on S is n The ma result here is Theorem 2. We use Lemma 5 in the proofofthe main theorem.
TIIEOREM 2. Let f E U(S1,S1) such that f,l is equicontinuous.Suppose that the smallest period of the periodic points of f on $1 is n _ 2. Then eve point on S is periodic with PROOF.Let p be a periodic point of period n on S1.Then f' p) p and therefore p is a fixed point of f'.Applying Theorem to f", we conclude that Fn is ether connected or it consists ofexly two distinct pos and every other point on S has period We lahn that there is no continuous map of the drcle having two points of period two and every other point periodic of period four.Otherwise, if g is such a map, let p, g(p) be the two points of period two and let K [p, g(p)] and L [g(p), p].Since g is a homeomorpldsm, we have two cases: (i) g() K and g(L) L or (ii) g(K) L and g(L) K.In both cases g (K) K. Hence if x E K is a point of period four then g (x) :/: and g () E K. If M is the closed interval with endpoints and g() lying in K then g(M) M. Therefore M contains a periodic point ofperiod two, contradicting the assumption that p and g(p) are the only points ofperiod two and every other point has period four.
Hence Fn is connected.Suppose that F : S1.Then F is a proper closed interval contahn8 the orbit ofp under f.Moreover f(F,) C F,.This implies that / has a fixed point on $1, contradicting the hypothesis that the smallest possle period ofthe periodic points is n 1. Hence Fn S1.
LEMMA .Let / U(S1, S1).Suppose that there exists a positive integer n _ 2 such that every point on S is periodic with period n.Then f is conjugate to a rational rotation.Now we consider case (I where f U(S1,S1) has no periodic points and (f is quicontinuous.The main result here is listed in Theorem 3. Note that f must be onto, since otherwise f(S1) I is homeomorphic to a closed interval and f(/') C I, so f has a fixed point.We shall adapt the techniques and use results due to L Auslander and Y. Katznelson Ill.
Let x E S. In If], I is defined to be the largest interval cong x such that ff'() J, Vm _ 1. Denote by zl and z the endpoints of , where possibly z x.The following are showed in [l l: 5 is closed and zl, z f () for k _ 1. Ifx, y S then w(, f) if and only if is an endpoint of I.If z and are the endpoints of I, then (z) and () are the endpoints of f().Also f(5)--and f(5)= 5, Vm _ 1.The intervals (m 0,1,2, ...) are pahvise disjoint and if/() f(x'), then I e.The sets (5 form a partition of S (that is, ifx, F S then 5 or 5 5 and U s J S1). Finally, at most countably many ofthe sets I are non-degenerate ( Before we show our result, we state the following theorem proved in [6] wlfieh concerns homeomorphisms. TtlEOREM C. Let f be an orientation preserving homeomorphism of S to itself.For x S , let R() / a (rood 1) denote irrational rotation by u.Then f is conjugate to some R f and only some (all) orbits of are dense on 81.