Let $G$ be an abelian group. We say $\Vert \cdot \Vert : G \rightarrow \Bbb R_{\ge0}$ is a norm if it satisfies
- $\Vert x \Vert =0 \Leftrightarrow x=0$
- $\Vert -x \Vert = \Vert x \Vert$
- $\Vert x+y \Vert \le \Vert x \Vert+\Vert y \Vert$
It's easy to prove $d(x,y)=\Vert x-y \Vert$ is a metric over $G$ and satisfies
- $d(x+z,y+z)=d(x,y)$
- $d(-x,-y)=d(x,y)$
In fact, any metric over $G$ satisfying this properties is induced by a Norm over $G$. We say two Norms over $G$ are topologically equivalent if they induce the same topology over $G$.
Let $G$ be an abelian group with a Norm, I've manage to prove the following function $$\Vert x \Vert_*= \frac{\Vert x \Vert}{1+\Vert x \Vert}$$ is a norm over $G$ and is topologically equivalent to the original norm. Also $\Vert x \Vert_*<1$ for all $x \in G$ and if $\Vert \cdot \Vert$ is unbounded then $\sup_{x \in G}{\Vert x \Vert}_*=1$. Now let $G$ be an abelian group with a norm such that $\Vert x \Vert<1 \ \ \forall \ x \in G$ and $\sup_{x \in G}{\Vert x \Vert}=1$, is the following function a norm over $G$? $$\Vert x \Vert^*= \frac{\Vert x \Vert}{1-\Vert x \Vert}$$ Is it topologically equivalent to the original norm? If this is true then the unbounded norms over an abelian group $G$ are in bijection with the bounded norms with $\Vert x \Vert<1 \ \ \forall \ x \in G$ and $\sup_{x \in G}{\Vert x \Vert}=1$, and the biyection "behaves well" with the topology (meaning it preserves the relation "being topologically equivalent"). This is what I'm trying to prove.