Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Assume that all prime periods of periodic orbits of a continuous map $f:[0:1]\to [0:1]$ are uniformly bounded (i.e. there exists N such that the prime period of every periodic orbit of f is smaller than N). What can you say about periods of periodic orbits of f? For example, can f have a periodic orbit of period 2007? Of period 2048?

This is just a problem I found online at in relation to what I am currently studying, so hopefully I am not missing anything to do it. I'm not sure how to approach this though. The only thing I can think of is that since $f([0,1])\subset [0,1]$ and is continuous, due to the intermediate value theorem, $f$ has a fixed point. But I am not sure on how to work this in, or if it is relevant. I figure the interval $[0,1]$ is somehow relevant, but am not positive.

share|cite|improve this question

I think this is an exercise in Sharkovskii's theorem. Suppose that $f$ has a periodic point $x$ with prime period $n$ which is not a power of $2$, so $n = 2^mk$, where $k\neq 1$ is odd. Then $y = f^{2^m}(x)$ has prime period $k$. By Sharkovskii's theorem, this implies that for every prime number $p>k$, there is a point $y$ with period $p$. This is, of course, necessarily the prime period of $y$. In particular, we see that $f$ cannot have uniformly bounded prime periods, since there are infinitely many prime numbers $>k$.

It follows that the only way that $f$ can have uniformly bounded prime periods is if all its periodic points have periods that are powers of $2$. So $f$ cannot have an orbit of period $2007$, but it can have a point of period $2048 = 2^{11}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.