Given a set $X$, a function $x \colon \mathbb{R} \to X$ is periodic if there exists $\tau>0$ such that $x(t+\tau)=x(t)$ for all $t \in \mathbb{R}$; and if $\tau$ is the smallest positive number with this property, we say that $\tau$ is the least period of $x$.

There obviously exist non-constant periodic functions without a least period - for example, taking $X=\{0,1\}$, the function $\mathbf{1}_\mathbb{Q} \colon \mathbb{R} \to \{0,1\}$ is clearly periodic, with every positive rational number being a period.

But now, let us consider "periodic orbits". Given a set $X$, a function $x \colon \mathbb{R} \to X$ is called a periodic orbit if

  • $x$ is a periodic function, and
  • there exists a family $(f^t)_{t > 0}$ of functions $f^t \colon X \to X$ such that $f^{s+t}=f^t \circ f^s$ for all $s,t>0$, and $x(s+t)=f^t(x(s))$ for all $s \in \mathbb{R}$ and $t>0$.

So, for example, taking $X=\{0,1\}$, it is easy to see that the periodic function $\mathbf{1}_\mathbb{Q}$ is not a periodic orbit: e.g., if the family $(f^t)$ as above exists, then $$ 0 = f^2(0) = f^\sqrt{2}(f^{2-\sqrt{2}}(0)) = f^\sqrt{2}(1) = f^\sqrt{2}(f^\sqrt{2}(0)) = f^{2\sqrt{2}}(0) = 1. $$

So my first question is:

Q1. Given a set $X$ (with cardinality at most that of the continuum) and a periodic orbit $x \colon \mathbb{R} \to X$, if $x$ is non-constant, does $x$ necessarily have a least period?

This question can alternatively be phrased more "directly" as follows: Given a family $(f^t)_{t > 0}$ of functions $f^t\colon X \to X$ satisfying $f^{s+t}=f^t \circ f^s$ for all $s,t > 0$, and a point $p \in X$, if there is a strictly decreasing sequence $t_n \to 0$ such that $f^{t_n}(p)=p$ for all $n$, does it necessarily follow that $f^t(p)=p$ for all $t>0$?

Q2. If the answer to Q1 is (in general) no, does it become yes if we additionally require $X$ to be a finite set?

  • $\begingroup$ If, e.g. $t_n=1/n$ then the condition does imply anything about $f^{\pi}$, right? $\endgroup$
    – Dirk
    Oct 14 '15 at 12:19
  • $\begingroup$ Yes, that's right. $\endgroup$ Oct 15 '15 at 12:53
  • $\begingroup$ Although I've already received an answer, I've reworded the question to help highlight the point of the question, in case that's helpful. $\endgroup$ Oct 21 '15 at 13:09
  • $\begingroup$ In brief, the answers are: no for Q1, since we can just take $X=\mathbb{R}/\mathbb{Q}$ and $f^t(x)=x+[t]$ (where $[t]$ is the element of $\mathbb{R}/\mathbb{Q}$ represented by t); but yes for Q2. (So for example, if $X=\{0,1\}$ and $(f^t)_{t>0}$ is a family of functions $f^t\colon \{0,1\} \to \{0,1\}$ satisfying $f^t \circ f^s$ for all $s,t>0$, if $f^{\frac{1}{n}}(0)=0$ for all $n$, it does follow that $f^\pi(0)=0$.) $\endgroup$ Oct 21 '15 at 13:26
  • $\begingroup$ (Well having said that, in the case that $X$ is finite, if there exists any $s>0$ such that $f^s(p)=p$, it follows that $f^t(p)=p$ for all $t>0$; we don't need a sequence of times tending to 0 at which $p$ is fixed.) $\endgroup$ Oct 21 '15 at 17:52

Two questions were asked

Concerning question I: Here's an application for my favorite bijection! Consider R the group of reals with addition and its subgroup Q of rationals. The group R mod Q has the same cardinality as R does (I guess you need the axiom of choice here). Let $\phi$ be a bijection from R mod Q to R. Let $f^t$ be the conjugate by $\phi$ of the translation by t in R mod Q. For all $t \in Q$ and all $x \in R$, $f^t(x)=x$.

Concerning question II: (Corrected from my previous wrong claim) For X finite, then your assumptions imply that all $f^t$ are the same map and are a projection (a solution $p$ of $p\circ p=p$). In particular, your claim is true if $X$ is finite.

Proof: let $n=|X|$, and notice that $f^t$ is for all k>1 the k-th iterate of a map $g:X\to X$ with $g=f^{t/k}$. Apply this to $k=n!$. I claim that $g^{n!}$ is a projection. Indeed consider any $x\in X$. Its orbit by $g$ consists in a tail of length $a\in\{0,\ldots,n-1\}$ followed by a cycle of length $b\in\{1,\ldots,n\}$, and $a+b\leq n$. So $g^{a}(x)$ has period $b$ dividing $n!$. As $n!\geq a$, the point $y=g^{n!}(x)$ is also a $g$-periodic point of period dividing $n!$, so it is fixed by $g^b$. Since $b$ divides $n!$, the point $y$ is fixed by $g^{n!}$. This proves the claim. Hence for all $t$, $f^t$ is a projection. Now consider two positive reals $s<t$. Then $f^t=f^s\circ f^{t-s}$ so the image of $f^t$ is contained in the image of $f^s$. But the image of $f^t$ is the same as the image of $f^{t/k}$ for all $k$ because the latter is also a projection. Since there is some $k$ so that $t/k<s$, we get that the image of $f^s$ is contained in the image of $f^{t/k}$ i.e. in the image of $f^t$. Since projections are characterized by their images, we are done.

  • $\begingroup$ Thank you very much for this! I now realise that in general, for any periodic point $x \in X$ of the dynamical system $(f^t)$, the dynamics of $(f^t)$ on the set $\{f^t(x):t > 0\}$ are conjugate to the unit-speed translation in the group $\mathbb{R}/G_x$, where $G_x:=\{0\} \cup \{t \in \mathbb{R} \setminus \{0\} : f^{|t|}(x)=x\}$. So then, your proof for part II is essentially a proof that for every proper subgroup $G$ of $\mathbb{R}$, $\mathbb{R}/G$ is infinite. $\endgroup$ Oct 13 '15 at 16:42

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