I have to determine if $G = \{ g_a\}_{a \in \mathbb{R}}$ with : $\forall a \in \mathbb{R} \; g_a(x) = \sin(a+x)$ is a compact set of $\mathcal{C}([0;1])$. To do that, I use the Arzela-Ascoli's theorem, so :
- I proved that $G$ is equicontinuous ;
- $G(x) = \{ g_a(x) \; | \; a \in \mathbb{R} \} = [-1;1]$ is compact
And then we have : $\overline{G}$ compact on $\mathcal{C}([0;1])$. But I want to show that $G$ is compact on $\mathcal{C}([0;1])$. So, I've tried to show that $G$ is a close set of functions.
So, I've taken $(g_{a_n})_n \in G^{\mathbb{N}}$ which converges in $\mathcal{C}([0;1])$ towards $g \in \mathcal{C}([0;1])$, and I tried to prove that $\exists a \in \mathbb{R} \; g(x) = \sin(a+x)$.
If $a_n \rightarrow a \in \mathbb{R}$, then it's okay, we have the result. Otherwise, I use the fact that $(\sin a_n)_n$ converges in $\mathbb{R}$, so : $\exists l \in \mathbb{[-1;1]} \; \arcsin(l) = \lim_{n \rightarrow \infty} \arcsin(\sin(a_n))$ cause $\arcsin \in \mathbb{C}^0$.
But then, I'm stuck, cause $\arcsin(\sin(a_n)) = 2\pi k_n + a_n$ or $\arcsin(a_n) = 2\pi k_n + \pi - a_n = (2k_n+1) \pi - a_n$ depending of $a_n$ modulo $2\pi$. And so, I can't conclude.
So, I would like to know if someone can help me to figure out how to conclude !
Thank you !
Note that $g_{a+2\pi}=g_a$. If you can show that the map $a\mapsto g_a$ is continuous from $\mathbb R\to C([0,1])$, then $G$ is the image of $[0,2\pi]$, a compact set, under this continuous map, and thus $G$ is compact.