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I already asked a question about Lyapunov numbers , as I said , I'm working on a paper in which we introduce Lyapunov numbers and find relations between them, as I said in the previous post, the first and the third Lyapunov number are defined i, the following way:

The first :

$ L_{r} := \sup \lbrace \varepsilon : \forall x \in X , \forall U_{x} , \exists y \in U_{x} , \exists n \in \mathbb{Z}_{+} \text{such that}: d(f^{n}(x),f^{n}(y)) > \varepsilon \rbrace $

The third :

$ \overline{L_r} := \sup \lbrace \varepsilon : \forall x \in X , \forall U_{x} \quad \exists y \in U_{x} \quad \text{such that}: \limsup_{n \to \infty} d(f^{n}(x) , f^{n}(y) ) > \varepsilon \rbrace $

And a dynamical system $(X,f)$ is defined the way like this :

$(X,f)$ is a dynamical system , where $X$ is a compact metric space with metric $d$ and $ f : X \rightarrow X $ is a continuous map and $O_{f} = \lbrace f^{n}(x) \quad : n\geq 0 \rbrace $ is the orbit of the map $f$.

I have problem understanding the proof of this proposition :

There exists a topological dynamical system $(X, f)$ for which $L_{r}=2\overline{L_{r}}$.

Its proof is this :

Proof : We define the space $X$ as a compact surface in $R^{3}$ which is homeomorphic to a two-dimensional disk in $R^{2}$.
More precisely, the cylindrical coordinates of a point $(x, y, z) \in X$ have the form $(r,\varphi, z)$, where $r = \sqrt{x^{2} + y^{2}} $ and $\varphi$ is an angle, for which $x = r\cos \varphi$ and $y = r\sin \varphi$.
In other words, $(r,\varphi)$ are the polar coordinates of $(x, y)$, and $z$ remains unchanged.
Let $h(r) = 8r(1 − r)$.
Now, define $X$ as a set of points with cylindric coordinates $(r,\varphi, h(r))$, where $0 ≤ r ≤ 1$,$\varphi ∈ R$, and let the Euclidean metric (in $R^{3}$) $d$ be the metric on $X$.
Now we define a continuous map $f$ from $X$ to itself as follows $f : (r,\varphi, h(r)) → (g(r), 2\varphi, h(g(r)))$, where $g(x)$ is a continuous map $[0, 1] → [0, 1]$ with $g(0) = 0$, $g(1) = 1$ and $g(x) > x$ for all $x ∈ (0, 1)$.
From this properties one can easily deduce that $\lim_{n \to \infty} g^{n}(x) = 1$ for any $x ∈ (0, 1]$.
For example, let $g(x) = 2x − x2$.
Let $p ∈ X$ and $U$ be a neighborhood of $p$. If $p \neq (0, 0, 0)$, then for any $\delta > 0$ there are $n ∈ N$ and $q ∈ U$ such that $d(f^{n}(p), f^{n}(q)) > 2 − \delta $.
If $p = (0, 0, 0)$, then there are $n ∈ N$ and $q ∈ U$, for which $f^{n}(q)$ lies on a circumference of $X$ with the center $(0, 0, 2)$ (in $R^{3}$) and the radius $ \dfrac{1}{2}$ .
For these $n$ and $q$ we have $d(f^{n}(p), f^{n}(q)) > 2$ and so $L_{r} ≥ 2$.
Now, let $p = (0, 0, 0)$. The equality $\lim_{n \to \infty} d(f^{n}(p), f^{n}(q)) = 1$ holds for any $q \neq p$.
So $Lr ≤ 1$.
Since $L_{r} ≤ 2\overline{L_{r}}$ (by Proposition 2.1), it gives $Lr = 2\overline{L_{r}}$.

My questions :

1- How did they find out when $ p \neq (0,0,0)$ then for every $\delta > 0 $ there exists $ n \in \mathbb{N} $ and $ q \in U $ which $d(f^{n}(p) , f^{n}(q)) > 2 - \delta $ ?
2- How did they find out when $p=(0,0,0)$ then those things in the proof of the theorem happen ?

Here is the link of the paper On Lyapunov Numbers

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  • $\begingroup$ I guess near the end of the proof should be $\overline{L_{r}}≥ 2$ instead of $L_{r} ≥ 2$ and $L_r\le 1$ instead of $Lr\le 1$, also $g(x)$ should be $2x−x^2$ instead of $2x−x2$. $\endgroup$ – Alex Ravsky Jul 27 at 19:02
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Since $f(0,0,0)=(0,0,0)$, we have $f^n(0,0,0)=(0,0,0)$ for each $n\in\Bbb N$.

2- How did they find out when $p=(0,0,0)$ then those things in the proof of the theorem happen ?

Let $q=(r,\varphi,h(r))$, where $0\le r\le 1$ and $\varphi\in\Bbb R$. Since $q\ne p$, $r>0$. Then $f^n(q)=(g^n(r),n\varphi,h(g^n(r))$ for any $n\in\Bbb N$. Since $r>0$, $g^n(r)$ tends to $0$ when $n$ tends to infinity. Then $h(g^n(r)$ tends to zero. Thus for any $\delta>0$ there exists $N$ such that for each $n>N$, $f^n(q)$ belongs to $\delta$-neighborhood of the circle $C=\{(0, \psi,0): \psi\in \Bbb R\}$. Since the distance from $p$ to any point of $C$ equals $1$, the distance $d(p,f^n(q))$ tends to $1$.

1- How did they find out when $ p \neq (0,0,0)$ then for every $\delta > 0 $ there exists $ n \in \mathbb{N} $ and $ q \in U $ which $d(f^{n}(p) , f^{n}(q)) > 2 - \delta $ ?

Let $p=(r,\psi,h(r))$, where $0<r\le 1$ and $\psi\in\Bbb R$. There exists a number $m\in\Bbb N$ such that a point $q=(r,\psi+\pi/m,h(r))\in U$. Similarly to the answer to Question 2, we can show that for any $\varepsilon>0$ there exists a number $N$ such that for each odd $k>N$, $f^{km}(p)$ and $f^{km}(q)$ belong to $\delta/2$-neigborhood of the circle $C$. Since $km(\psi+\pi/m)-km\psi=k\pi$ and $k$ is odd, points $f^{km}(p)$ and $f^{km}(q)$ are in $\delta/2$-neigborhoods of endpoints of a diameter of the circle $C$. Then the triangle inequality implies that $d(f^{km}(p), f^{km}(q))> 2 - \delta$.

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