I am looking for an example of a simple discrete dynamical system with no equicontinuity points but not sensitive.

Found this example in the book of Kurka. But could not really understand it.

$X=\left\{ \left( x,y,z\right) \in \mathbb{R}^{3}:x^{2}+y^{2}\leq 1,z=0\right\} \cup \left\{ \left( x,y,z\right) \in \mathbb{R}^{3}:\left( x-1\right) ^{2}+z^{2}=1,y=0\right\} .$

The function $f$ is defined by : \begin{eqnarray*} f\left( r\cos \left( t\right) ,r\sin \left( t\right) ,0\right) &=&\left( r\cos \left( 2t\right) ,r\sin \left( 2t\right) ,0\right) \\ f\left( 1-\cos \left( t\right) ,0,\sin \left( t\right) \right) &=&\left( 1-\cos \left( 2t\right) ,0,\sin \left( 2t\right) \right) \end{eqnarray*}

Thank you.


1 Answer 1


That's about the simplest example I know, but I can explain it.

First, there's the doubling map, defined on any circle :

$$X_r = \{(x,y,z) \in \mathbb{R}^3 : x^2+y^2=r^2, z=0\},$$

$$f_r(r \cos(t), r \sin(t), 0) = (r \cos(2t), r \sin(2t), 0).$$

For any positive $r$, the system $(X_r, f_r)$ is sensitive, and has no equicontinuity point. Now, do the same on a disc:

$$X_D = \{(x,y,z) \in \mathbb{R}^3 : x^2+y^2\leq 1, z=0\},$$

$$f_D(r \cos(t), r \sin(t), 0) = (r \cos(2t), r \sin(2t), 0).$$

Then $(X_D, f_D)$ is not sensitive anymore: the point $(0,0,0)$ is stable. However, for the same reason, it is has an equicontinuity point, $(0,0,0)$.

The next step is to add a circle, on which we act via the doubling map, and glue it to $X_D$ at the point $(0,0,0)$. That's how you get $(X,f)$. Since all points belong to a circle on which we act via an angle-doubling map, $(X,f)$ has no equicontinuity point.

That said, $(0,0,0)$ is not stable anymore. The catch, however, is that $(X,f)$ is still not sensitive! The notion of sensitivity is global, i.e; uniform in $x$. I don't remember the definition exactly, but I think it is something like "For all $\varepsilon >0$, there exists $\delta >0$ such that, for all $x$...".

Now, the systems $(X_r, f_r)$ are each sensitive, but not uniformly so. If you start on $D$ and within a distance $\varepsilon$ from $(0,0,0)$, then you will always stay on $D$ and within a distance $\varepsilon$ from $(0,0,0)$, which contradicts sensitivity.


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