Sarkovskii's theorem and the islands of stability in the cascade diagram

We will give an indication of why period 3 seems to occur at the end of all the occurrences of finite attractors for quadratic maps. This is just a very small piece of a fantastic result called Sarkovskii's theorem. Our discussion is a rehashing of more detailed material found in [Dev89b].

Let be a continuous map. Suppose f has a 3-cycle. Then f has a periodic point of prime period n for all integers .

In short, period 3 implies periodic points of all periods. This doesn't say anything about the stability of these periodic points.

The proof of this theorem begins by assuming we have a 3-cycle for f(x). This would consist of three distinct points a,b,c such that f(a)=b, f(b)=c, and f(c)=a. By symmetry, we may assume that a is the smallest of the three: a< b,c. Then there are two cases b<c and c<b. We shall assume b<c and leave the second case to the reader to carry out.

Assuming a<b<c, we define the intervals A=[a,b] and B=[b,c]. Since f(a)=b and f(b)=c and f is continuous, we have

(The Intermediate Value Theorem says that on a closed interval [a,b] a continuous function assumes all values between f(a) and f(b).) Similarly, since f(b)=c and f(c)=a, we have

Question: What do these relations remind you of? See here.

The key idea is to iterate these subset relations to prove the existence of periodic points of a given period. To do this, we require two lemmas.

Lemma 1: Let I be a closed interval such that . Then f has a fixed point in I.

Let the interval be I=[c,d]. Let S be the subset of such that f(x)=c, and let T be the subset of such that f(x)= d. Since f is continuous, both S and T are closed subsets of I, and hence compact. Since , S and T are disjoint. Therefore, we can consider the distance between these two sets

Since S and T are compact, this minimum is achieved for some points and . If , then f(c)=c and we have found our fixed point. Similarly, if , d is a fixed point. Now assume and . Then , while . By the Intermediate Value Theorem, there is a point x between and such that f(x)-x=0. This is our fixed point.

Lemma 2: Suppose I and J are closed intervals such that . Then there is a closed interval such that .

This is proved by basically the same proof as the first lemma. Let I=[a,b] and J=[c,d]. Let S be the set of such that f(x)=c, and T the set of x with f(x)=d. Once again we find , in I such that . Let be the interval with endpoints equal to and . We claim that . The Intermediate Value Theorem (IVT) certainly implies that . If equality doesn't happen, then there is a point such that f(x)>d or f(x)<c. Suppose the former happens (the latter case may be argued similarly). Then again by the IVT there is a point between and x such that . However, . This is a contradiction since and . Therefore, no such point x can exist, and we have proved our lemma.

Now we may return to the 3-cycle a<b<c and the proof of our theorem. Since , by Lemma 1, f has a fixed point in B. That takes care of the theorem for n=1.

Since , there is an interval such that , by Lemma 2. Since , there is an interval such that . Then . By Lemma 1, has a fixed point p in . Since , we cannot have f(p)=p, unless p=b the common boundary of A and B. But b is clearly not a fixed point of f, since f(b)=c. Thus, p is a periodic point of prime period 2.

Having been supplied with a periodic point of prime period 3, we suppose now that n>3. We define a sequence of intervals as follows:

• .
• Since , by Lemma 2 there is an interval such that .
• Since , by Lemma 2 there is an interval such that .
• Continue the last step all the ways to creating such that .
• Then . So . Choose so that .

All told, we have . Thus, by Lemma 1, has a fixed point p in . We claim p is a periodic point of f with prime period n. All we have to do is show that p cannot have any smaller period k<n.

Suppose . Then the iterates of f on p cycle through the points

forever. By our construction, these points are all contained in , and so all the iterates are forever contained in B. On the other hand, (this was the clever part of our construction). That can only happen if , the endpoint shared by A and B. Then , and , a contradiction. This proves that the period of p is no smaller than n.

This theorem is dramatically visualized in the bifurcation diagram, although that's a story for another day.