Exercise :
Let $X$ be a Banach space and $C \subseteq X$ be closed, convex and bounded. Moreover, let $g:C \to C$ be a non-expansive operator, meaning that : $$\|g(u) - g(v) \| \leq \|u-v\| \; \forall u,v \in C$$ Show that : $$\inf\{\|u-g(u)\|:u \in C\} = 0$$
Attempt :
Since $X$ is Banach and $C$ is closed, this means that $C$ is also Banach.
Since it is $g : C \to C$, this means that $g(C) \subseteq C$. Since $C$ is convex, any such $g_n \in g(C)$ can be written as $$g_n(x) = (1-r_n)y + r_ng(x)$$ with $x,y \in C$ and $\{r_n\}_{n \geq 1} \in [0,1]$ with $r_n \xrightarrow{n \to \infty} 1$.
It also is :
$$\|g_n(x) - g_n(z) \| = r_n \|g(x) - g(z)\| \leq r_n\|x-z\|\; \forall x,z \in C$$
Let $\{x_n\}_{n \geq 1} \in C$ such that $g_n(x_n) = x_n$. Then : $$x_n = (1-r_n)y + r_ng(x_n)$$ But this leads to : $$\|x_n -g(x_n) \| = (1-r_n)\|y-g(x_n)\|$$ But since $C$ is also bounded, $\|y-g(x_n)\|$ will be a finite number and thus : $$\|x_n - g(x_n) \| = (1-r_n)\|y-g(x_n)\| \xrightarrow{n \to \infty} 0$$ $$\implies$$ $$\inf\{\|u-g(u)\|:u \in C\} = 0$$
I would really appreciate any feedback on this attempt, since it was completely based on intuitions (I have never handled something similar before). Any hints, tips or alternative elaborations will be most welcome.