Tip: Transform the function $f$ so it has one fixed point at $0$.
We have the unit disk $B_1(0):=\{x\in X \mid d(0,x)<1 \}$ with the metric space $d$.
A fixed piont is a $x\in B_1(0)$ such that $f(x)=x$.
I have problems to understand the function. To my understanding, for all $x \in B_1(0)$ applies $f(x)=x$ because $d(0,x)=|x-0|=x$ for all $x$. Do I have to transform the function so I have only one fixed point at zero? Because to my understanding $0$ is already a fixed point of $f$. I have no clue how to transform the function.
To show $f=id$, I could proof it with the identity theorem, which says:
Let $G\subseteq \mathbb{C}$ be a domain and $f,g$ holomorphic functions on that domain. The following are equivalent:
a) $f(z)=g(z)$ for all $z\in G$.
b) The set $\{z\in G \mid f(z)=g(z)\}$ has a limit point in G.
c) There exist a $z\in G$ such that $f^{(n)}(z)=g^{(n)}(z)$ for all $n\in \mathbb{N_0}$.
It is the first time I have to use the identity theorem and the task gives me a lot of problems. Any help appreciated.
Since it has two fixed points, it must have a fixed point $z_0$ with $z_0\neq0$. It follows then from the Schwarz lemma that, for some $\omega\in\mathbb C$ such that $|\omega|=1$, $f(z)=\omega z$. And now it follows from the fact that $f(z_0)=z_0$ that $\omega=1$ and therefore that $f=\operatorname{id}$.