This post is based on a talk given by Keith Burns in the UH dynamical systems seminar yesterday, in which he presented a streamlined proof of Sharkovsky’s theorem due to him and Boris Hasselblatt.

**1. Background and statement of the theorem **

We recall the statement of Sharkovsky’s theorem. The setting is a continuous map , and one is interested in what dynamical properties must necessarily have if we know that has a periodic orbit of some given least period — for example, does existence of such an orbit force the existence of periodic points with least period for various values of ? Or does it force the map to have positive entropy?

The assumptions of continuity and one-dimensionality are vital: for discontinuous one-dimensional maps, or for continuous higher-dimensional maps, it is possible to have a point of least period without having points of any other period or having positive entropy.

In one dimension, on the other hand, one can begin with the following simple observation: if has a periodic point with least period 2, then there is such that , and so applying the intermediate value theorem to , the map must have a fixed point.

What if has a periodic point with least period 3? Suppose that (the other configurations are similar), and let and . Once again using the intermediate value theorem, we see that and . Thus we may consider the topological Markov chain given by the graph in Figure 1, and observe that it embeds into the dynamics of , so that in particular, has periodic points of *every* order.

Fig 1

Sharkovsky’s theorem generalises the above observations. Consider the following unorthodox total order on the integers:

Then if has a periodic point of least period , it has a point of least period for every .

The paper by Burns and Hasselblatt includes a discussion of the history of the theorem and the various proofs that have been given. Here we outline the most streamlined proof available at the present time.

**2. Sketch of the proof **

Given two intervals , write if . As in the case of a period 3 orbit above, the key will be to construct a collection of intervals (whose endpoints are certain points of the period orbit) such that the covering relations yield a directed graph with periodic paths of all the desired periods.

To deduce existence of various periodic orbits for from these covering relations, one makes the following observations.

- If , then contains a fixed point of .
- If , then .
- The above two observations guarantee existence of with . To guarantee that can be chosen with least period , it suffices to know that some to have disjoint interior from all the others, and that the endpoints of this do not follow the itinerary given by the sequence of covering relations above.

Detailed proofs of the above observations are in the Burns-Hasselblatt paper. With these tools in hand, the proof comes down to constructing the directed graph of covering relations referred to above.

Fig 2

To this end, consider a periodic orbit with points — the green points in Figure 2. These are permuted by ; some are moved to the left, some are moved to the right. Of all the points in the orbit that are moved to the right, let be the rightmost, so that all points in the orbit to the right of are moved to the left. Let be the next point to the right of , and let as in the figure.

Now consider the intervals . Note that has the following property: every point in lying to the left of moves to the right of , and vice versa. Let be the first interval that does not have this property, and let be a point that does not switch sides of under the action of . For concreteness we consider the case where is to the left of , as in Figure 2.

By construction, all the points between and are mapped to the right of . Of these, let be the point that is mapped furthest to the right.

Write for the interval , and consider the intervals for , together with . We claim that , , and for each , so that we have the covering relations shown in Figure 3, which can be used to establish existence of periodic points with all least periods greater than .

Fig 3

So why do the intervals have the covering relations shown?

- The fact that follows from the fact that and the endpoints of switch sides of .
- The definition of gives .
- Writing for the part of lying in or to the right of and for the part lying in or to the left, we see that and ; this is because for , all points of switch sides of under the action of . (This is the definition of .)
- Using similar notation for and , we see that and ; this is because was chosen to be the point that maps furthest to the right.
- This establishes that , and similarly .

For we see that lies to the left of , and to the right, so that .

The extra piece of information in the final item of the list above shows that the graph in Figure 3 can be replaced by the one in Figure 4.

Fig 4

Let us sum up what we know. One of the following two cases holds. *Case 1:* There is some point of the orbit such that and both lie on the same side of . *Case 2:* Every point of the orbit is mapped to the other side of by .

In Case 1, the arguments above show that for every and every even , there is an orbit of least period , which is at least as strong as the conclusion in Sharkovsky’s theorem.

In Case 2, one may observe that since permutes the elements of and every element is moved from one side of to the other, must be even. Moreover, by considering and separately, as invariant sets for , we can repeat the argument inductively to complete the proof of the theorem.