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I am trying to construct an example of a function $f: [0,1] \rightarrow [0,1]$ such that it has a periodic point of period 3 and NO other periodic points.

Any ideas? how can I even start envisioning this? I am pretty sure $f$ cannot be continuous, but how do I ensure the existence of just the periodic point with period 3?

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By Sharkovs’kiǐ’s theorem the function $f$ cannot be continuous, as you suspected. One simple approach is to let $\{x_\xi:\xi<2^\omega\}$ be an enumeration of $[0,1]$ and define

$$f:[0,1]\to[0,1]:x\mapsto\begin{cases} x_1,&\text{if }x=x_0\\ x_2,&\text{if }x=x_1\\ x_0,&\text{if }x=x_2\\ x_{\xi+1},&\text{if }x\notin\{x_0,x_1,x_2\}\;. \end{cases}$$

The orbit of each point of $[0,1]\setminus\{x_0,x_1,x_2\}$ is infinite, so there are no periodic points except $x_0,x_1$, and $x_2$. (Once you have one point with period $3$, you must of course have three such points, since the other two points in its orbit will also have period $3$.)

The enumeration of course uses the axiom of choice. You can avoid this while still using the same basic idea, but it takes more work. It’s well known that each irrational $x\in[0,1]$ has a unique continued fraction expansion $x=[0;a_1,a_2,\dots]$. It’s also well known that there is an effective enumeration $\{q_n:n\in\Bbb N\}$ of the rationals in $[0,1]$. Define $f:[0,1]\to[0,1]$ as follows:

$$\begin{align*} &f(q_0)=q_1\\ &f(q_1)=q_2\\ &f(q_2)=q_0\\ &f(q_n)=q_{n+1}\text{ for all }n\ge 3\\ &f\big([0;a_1,a_2,a_3,\dots]\big)=[0;a_1+1,a_2+1,a_3+1,\dots]\;. \end{align*}$$

Added: This paper by Bau-Sen Du is, according to the title, A collection of simple proofs of Sharkovsky's theorem.

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  • $\begingroup$ thanks a lot Brian! Just so that I am clear, you have given, not one, but two examples, am I right? What is the $w$ in the first example? A natural number I presume? Also, why was there a need to make is less than $2^w$? Moreover, I am having a harder time understanding the fourth piecewise in the first example, where you say $x= x_{\xi+1}$ if $x$ is not $x_0,x_1,x_2$. Can I visualize a circle of unit circumference when I try to visualize your functions? I am reading the paper listed and will get back with any questions. $\endgroup$ – user43901 Mar 30 '13 at 15:56
  • $\begingroup$ @user43901: Yes, it’s two examples. The $\omega$ in the first example is omega, the smallest infinite ordinal; if you’re not familiar with transfinite ordinals, you should probably just ignore the first example and go with the second. $\endgroup$ – Brian M. Scott Mar 30 '13 at 22:25
  • $\begingroup$ can I do the following? Let $f(0)= x+1/4$ when $x= 0, 1/4$ and let $f= 0$ for everything else. Won't that work? $\endgroup$ – user43901 Mar 30 '13 at 23:09
  • $\begingroup$ @user43901: Yes, that will work fine if you don’t care whether $f$ is $1$-$1$ or not. $\endgroup$ – Brian M. Scott Mar 30 '13 at 23:11
  • $\begingroup$ Awesome! Also, not related to current problem, I have two other questions posted. Would you mind taking a look at them? math.stackexchange.com/questions/346791/… math.stackexchange.com/questions/346684/… $\endgroup$ – user43901 Mar 30 '13 at 23:23
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The typical example of such a function is a rotación of the circle of angle $2\pi/3$. This suggests the following: $$ f(x)=x+\frac13\mod1. $$ All points are of minimal period three.

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  • $\begingroup$ And there aren't any other periodic points? $\endgroup$ – user43901 Mar 30 '13 at 23:33
  • $\begingroup$ @user43901 All the points are 3-periodic, as Julián explained. $\endgroup$ – Did Mar 31 '13 at 0:55
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Let $f(a)=b$, $f(b)=c$, $f(c)=a$, where $b\ne a$. What is $f(f(f(b)))$?

But presumably the question should be reworded to rule out this sort of example.

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  • $\begingroup$ Yes, I think you are right. We cannot have examples like this. What part of the question, do you think should be changed to avoid this? I am just having a hard time dealing with this thing. So an elaboration on the motivation of any future examples will be much appreciated. $\endgroup$ – user43901 Mar 30 '13 at 4:38
  • $\begingroup$ Look for Sharkovsky's (?) Theorem and related results. You can reword minimally by asking for a single orbit of three $3$-periodic points and no others. If you put no restrictions of the continuity kind, roughly anything can happen. The only restriction is that if there is a point of period $3$, there are at least $3$ such points. $\endgroup$ – André Nicolas Mar 30 '13 at 4:44

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