Suppose we are given a continuous function . For example, the graph of may look like this:
One interesting thing to do is to randomly pick a point , keep applying , and see where we end up. We can visualize this process via a cobweb diagram, drawn as follows:
We see that we converge towards the point . In particular, if we start at , we stay there all the time — is a fixed point.
Some choices of give more complicated behaviour. For example, in the following diagram
we see that in the limit, the point oscillates between and . In this case, and are -periodic points. Perhaps if we want to care about the limiting behaviour, we should understand these periodic points.
Let be a function. A point is -periodic if and for .
For example, for the following choice of , we have a -periodic point:
We are now ready to state the desired theorem.
Let be a continuous function. If has a -periodic point, then has an -periodic point for all .
Let's try to understand this. A good start might be to find a fixed point, i.e. a -periodic point. We can achieve this using the intermediate value theorem. From now on, we fix a single continuous function .
Let be a (closed) subinterval. If , then contains a fixed point.
So we just need to find an interval such that . We suppose the three points in the -periodic orbit are . We may wlog assume that , and .
Then is a closed subinterval containing and . So it in particular contains . So we have found a fixed point.
How about periodic points of larger period? If we want to find an -periodic point, we might try to look for fixed points of . This is in some sense a good strategy, but it is not good enough. If we are given a fixed point , we know it has period at most . Perhaps it is secretly a fixed point of . So we need to be a bit more delicate than that.
We first introduce a convenient notation:
If are subintervals, we write if .
We have seen that if , then contains a fixed point of . The following is the desired strengthening of the lemma.
If , then there exists an such that for and .
This is trivial if we don't require , since , and we can apply the lemma. The (slightly) hard part is getting the points to lie in the .
Now if , then we can find some such that . We still have , so we have a sequence
We can now replace with a as above, and keep going on to obtain a new sequence
such that , and for all . Then since , it follows that we can find some such that . By assumption, . So we are done.
We are essentially done with the proof. We set and . Then we have , and . If we feel like drawing diagrams, we can draw it as
Now we have a sequence of the form
for any number of copies of . So for each , we get some such that , and crucially, for all . To see that is actually of period , we need to check that for . To do so, we simply have to note that . Thus if for some , then must be . We can check manually that does not satisfy the required property, since . This concludes the proof.
This result is a special case of a more general theorem called Sharkovskii's theorem. The statement is as follows:
Define an ordering on the naturals by
Then if has an -periodic point and , then there is an -periodic point.
One way to prove this is to use adapt the above strategy with some clever choices of the intervals involved.
As a concluding remark, recall that we led ourselves into looking at periodic points when we tried to understand the limiting behaviour as we keep iterating . In the first two examples, the fixed point and the -periodic points were stable, namely if we start somewhere near the periodic points and keep applying , the point converges towards the fixed point or cycle. In general, there is no reason to believe that the periodic points given by our theorem will be stable. This makes it slightly less interesting — there is no way we can "naturally" discover these periodic points as we did at the beginning. We must have started at exactly the periodic point to discover the periodicity.