Today the 17 October 2019 I discussed a very remarkable fixed point theorem discovered by the Ukrainian mathematician Oleksandr Micholayovych Sharkovsky.
We recall that a periodic point of period for a function is a point such that . With this definition, a periodic point of period is also periodic of period for every which is a multiple of . If but for every from 1 to , we say that is the least period of .
Theorem 1. (Sharkovsky’s “little” theorem) Let be an interval and let be a continuous function su. If has a point of least period 3, then it has points of arbitrary least period; in particular, it has a fixed point.
Note that no hypothesis is made on being open or closed, bounded or unbounded.
Our proof of Sharkovsky’s “little” theorem follows the one given in (Sternberg, 2010), and could even be given in a Calculus 1 course: the most advanced result will be the intermediate value theorem.
Lemma 1. Let be a compact interval of the real line, let be a continuous function. Suppose that for some compact interval it is . Then has a fixed point in .
Proof. Let and be the minimum and the maximum of in , respectively. As , it is and . Choose such that and . Then is nonpositive at and nonnegative at . By the intermediate value theorem applied to , must have a fixed point in the closed and bounded interval (possibly reduced to a single point) delimited by and , which is a subset of .
Lemma 2. In the hypotheses of Lemma 1, let be a closed and bounded interval contained in . Then there exists a closed and bounded subinterval of such that .
Proof. Let . We may suppose , otherwise the statement is trivial. Let be the largest such that . Two cases are possible.
- There exists such that . Let be the smallest such , and let . Then surely , but if for some we had either or , then by the intermediate value theorem, for some we would also have either or , against our choice of and .
- for every . Let then be the largest such that , and let . Then for reasons similar to those of the previous point.
Proof of Sharkovsky’s “little” theorem. Let be such that , , and . Up to cycling between these three values and replacing with , we may suppose . Fix a positive integer : we will prove that there exists such that and for every .
Let and be the “left” and “right” side of the closed and bounded interval : then and by the intermediate value theorem. In particular, , and Lemma 1 immediately tells us that has a fixed point in . Also, , so also has a point of period 2 in , again by Lemma 1: call it . This point cannot be a fixed point, because then it would also belong to as , but which has period 3. As we can obviously take , we only need to consider the case .
By Lemma 2, there exists a closed and bounded subinterval of such that . In turn, as , there also exists a closed and bounded subinterval of such that , again by Lemma 2: but then, . By iterating the procedure, we find a sequence of closed and bounded intervals such that, for every , and .
We stop at and recall that : we are still in the situation of Lemma 2, with in the role of . So we choose as a closed and bounded subinterval not of , but of , such that . In turn, as , there exists a closed and bounded subinterval of such that . Following the chain of inclusions we obtain . By Lemma 1, has a fixed point in , which is a periodic point of period for .
Can the least period of for be smaller than ? No, it cannot, for the following reason. If has period , then so has , and in addition is divisible by . But while for every : consequently, if has period , then . But this is impossible, because by construction as , while .
Theorem 1 is a special case of a much more general, and complex, result also due to Sharkovsky. Before stating it, we need to define a special ordering on positive integers.
Definition. The Sharkovsky ordering between positive integers is defined as follows:
- Identify the number , with odd integer, with the pair .
- Sort the pairs with in lexicographic order.
That is: first, list all the odd numbers, in increasing order; then, all the doubles of the odd numbers, in increasing order; then, all the quadruples of the odd numbers, in increasing order; and so on.
For example, and
- Set for every and .
That is: the powers of 2 follow, in the Sharkovskii ordering, any number which has an odd factor.
For example, .
- Sort the pairs of the form —i.e., the powers of 2—in reverse order.
The set of positive integer with the Sharkowsky ordering has then the form:
Note that is a total ordering.
Theorem 2. (Sharkovsky’s “great” theorem) Let be an interval on the real line and let be a continuous function.
- If has a point of least period , and , then has a point of least period . In particular, if has a periodic point, then it has a fixed point.
- For every integer it is possible to choose and so that has a point of minimum period and no points of minimum period for any . In particular, there are functions whose only periodic points are fixed.
- Keith Burns and Boris Hasselblatt. The Sharkovsky theorem: A natural direct proof. The American Mathematical Monthly 118(3) (2011), 229–244. doi:10.4169/amer.math.monthly.118.03.229
- Robert L. Devaney, An Introduction to Chaotic Dynamical Systems, Second Edition, Westview Press 2003.
- Shlomo Sternberg, Dynamical Systems, Dover 2010.