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When I look for example of a uniformly convergent sequence which is not orbitally convergent, I found this example from this article.

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Then, I have some questions from the example above.
The first is about uniform convergence on $S^1$. But, I have not learned about $S^1$ yet. After read some notes about dynamical system, I think $S^1$ in Example 3.2 is $S^1=\{z\in \mathbb{C}:|z|=1\}=\{e^{2\pi i\theta}:0\leq \theta<1\}$ and the distance between two points on $S^1$ is induced by the arc length distance, i.e. the length of shortest arc connecting two points $\theta_1,\theta_2\in [0,1)$, which is defined by \begin{align*} d(e^{2\pi i\theta_1},e^{2\pi i\theta_2})= \begin{cases} |\theta_2-\theta_1|, \quad \text{if}\, |\theta_2-\theta_1|\leq \frac{1}{2},\\ 1-|\theta_2-\theta_1|,\quad \text{if}\, |\theta_2-\theta_1|\geq \frac{1}{2}. \end{cases} \end{align*} I try to prove that $f_n$ is uniformly convergent to $I_{S^1}$. This is my attempt.
Since $f_n(x)=xe^{2\pi i \alpha_n}$ for $x\in S^1$, then $f_n(e^{2\pi i\theta})=e^{2\pi i\theta}e^{2\pi i\alpha_n}=e^{2\pi i(\theta+\alpha_n)}$, and also $I_{S^1}=x=e^{2\pi i\theta}$. So, $$d(f_n(x),x)=d(e^{2\pi i(\theta+\alpha_n)},e^{2\pi i\theta})=\alpha_n.$$ Since $(\alpha_n)$ converges to 1, then I conclude that $f_n$ is uniformly convergent to $I_{S^1}$. Can I conclude like that?

And the second is, why the orbit $\{f_N^k(x):k\in\mathbb{N}\}$ is not dense in $S^1$?
Does it use a theorem that said "If $\alpha$ is irrational, then for every $z\in S^1$, the orbit $f_\alpha^n(z):n\in\mathbb{N})$ is dense in $S^1$"?

Sorry for the long questions, but I am confused because I am new to dynamical system and $S^1$. Thank you in advance.

user136524
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You have it right. Your argument for uniform convergence looks good to me. I think it might be easier to use the norm in $\mathbb C$ though. In that case, we have $$ |xe^{2\pi i \alpha_n} - x| = |x| |e^{2\pi i \alpha_n} - 1| = |e^{2\pi i \alpha_n} - 1| \to 0 \quad \text{ as } n\to\infty$$ and the convergence is uniform with respect to $x$.

Also, they assume orbital convergence, which implies $\{ f^k_N (x)\}_{k\in\mathbb N}$ is not dense for each $x\in S^1$. But this contradicts the theorem you cited, so we cannot have orbital convergence.

Trevor Norton
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  • Is the norm in $\mathbb{C}$ is $d: \mathbb{C}×\mathbb{C}\to \mathbb{R}$ with $d(z_1z_2)=|z_1-z_2|$? – user136524 Jan 10 '23 at 01:05
  • And why the orbital convergent assumption implies ${f_N^k(x):k\in\mathbb{N}}$ is not dense? Meanwhile, the definition of ${f_N^k(x):k\in\mathbb{N}}$ not dense is there exists $\epsilon_0>0$ and $x_1\in S^1$ such that for all $k\in \mathbb{N}$, $d(f_N^k(x),x_1)\geq \epsilon_0$. – user136524 Jan 10 '23 at 01:19
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    @user136524 The norm in $\mathbb C$ is $|z|$ which defines the metric you give. Orbital convergence gives ${f^k_N(x)}$ is not dense, since (for $\epsilon$ small enough) we will be restricted to an $\epsilon$-ball of $x$ and cannot have subsequences approach values outside the ball. For example, if $x = 1$ and $\epsilon =1$ then ${f^k_N(x)}$ will always be in the right-hand plane and $|f^k_N(x) - (-1)| \geq 1$ for all $k$. – Trevor Norton Jan 10 '23 at 18:17