Does compact orbit imply periodicity?

A continuous dynamical system on a metric space $$X$$ is given by:

$$\varphi : \mathbb{R} \times X \rightarrow X$$ - continuous s.t.

$$\varphi (0,x) = x$$ for every $$x \in X$$

$$\varphi (t, \varphi(s,x) ) = \varphi(s+t, x)$$ for all $$s, t \in \mathbb{R}, \ \ x \in X$$

and

$$\gamma(x) = \{y \in X| \exists t\in \mathbb{R}; \varphi(t,x) = y \}$$ is the orbit of $$x$$.

$$x \in X$$ is periodic iff $$x =\varphi(T,x)$$ for some $$0 \neq T \in \mathbb R$$.

The question is the following:

Assume that $$x$$ is periodic. Show that $$\gamma(x)$$ is compact. Is the converse true?

For the first part, if $$x$$ is periodic of period $$T \ge 0$$ then $$\gamma(x)= \varphi (\{x\} \times[-T,T])$$ which is an image of a compact set, hence compact.

However, I am not sure about the converse. All I know is that for some sequence $$t_n \to \infty$$, $$\varphi(t_n,x) \to x'$$ for some $$x' \in \gamma(x)$$. Hence $$\exists t' \in \mathbb{R}$$ such that $$\varphi (t',x) = x'$$. On the other hand, we also have a continuous bijection from $$\mathbb R$$ to a compact set. At first sight this does not seem enough though.

Thank you.

P.S. The problem is from Bhattia & Szego: Stability Theory of Dynamical Systems

• Yes, see my answer here: math.stackexchange.com/questions/391315/… Nov 13, 2019 at 1:07
• also seems related math.stackexchange.com/questions/3169592/… Nov 13, 2019 at 1:16
• @MoisheKohan Thank you for the much more general answer. Because of my lack of topological group knowledge, in the meantime, I have come up with and written up a somewhat detailed solution to the particular case. Any feedback is welcome. Nov 18, 2019 at 15:02

1 Answer

While the answer of @Moishe Kohan answers an even more general question, in the meantime I was able to solve the particular problem (hopefully correctly). It makes use of the Baire category theorem and the following two lemmas:

Lemma 1. Assume that the $$x$$ is not periodic and $$\gamma(x)$$ is compact. Then for any $$x' \in\gamma(x)$$ there exists a real sequence $$(T_n) , T_n \to \infty$$ such that $$\varphi(T_n, x) \to x'$$.

Proof: Noticing that due to compactness, the sequence $$(\varphi(n,x))_n$$ converges up to a subsequence, denote its limit as $$x'' \in \gamma(x)$$ (i.e. $$\varphi(n(k), x) \to_{k \to \infty} x''$$ for some subsequence $$(n(k))_k$$ of $$(n)_n$$). Because $$x'' \in \gamma(x)$$, $$\exists T$$ such that $$x'= \varphi(T,x'')$$. Now setting $$(T_k)_k = (T+n(k))_k$$, one can notice that $$\varphi(T_k,x) \to x'$$. $$\square$$

Applying Lemma 1 and $$(\text{compactness} \implies \text{closedness}) + (\text{continuous image of a compact set is compact})$$ in metric spaces, one can obtain:

Lemma 2. Assume that the $$x$$ is not periodic and $$\gamma(x)$$ is compact. For any integer $$n$$, the set $$\varphi([n,n+1],x)$$ is closed and of empty interior in $$\gamma(x)$$.

Now for the sake of contradiction, assume that $$x$$ is not periodic. Remarking that $$\gamma(x) = \bigcup_{n \in \mathbb{Z}} \varphi([n,n+1],x),$$ according to Lemma 2 and the Baire category theorem (notice that $$\text{compactness of } \gamma(x) \implies \text{completeness of } \gamma(x)$$), the right hand side should be of empty interior in $$\gamma(x)$$. This is the desired contradiction.