Let $G\subseteq \mathbb{R^2}$ be a bounded convex domain. I need to find a bilipschitz map from $\partial G$ to $S^1$, i.e. a map $f:\partial G\rightarrow S^1$ that satisfies for all $x,y \in \partial G$ that
$$ \frac{|x-y|}{L}\leq|f(x)-f(y)|\leq L |x-y|$$ for some constant L.
Supposing without loss of generality that the origin is in the interior of $G$ I was thinking about the map given by $f(x)=x/|x|$. If we take $m=\min \{|z|:z \in \partial C \}$ then we have
\begin{align} \begin{split} |f(x)-f(y)|&=\left|\frac{x}{|x|}-\frac{y}{|y|}\right|\\&=\left|\frac{x|y|-y|x| + (x|x|-x|x|)}{|x| |y|}\right|\\& \leq \frac{|x|y|-x|x|| + |y|x|-x|x||}{|x||y|}\\&=\frac{||y|-|x|| + |y-x|}{|y|}\\& \leq \frac{2 |x-y| }{|y|}\\ & \leq \frac{2 |x-y| }{m}. \end{split} \end{align}
But I am having trouble proving that $m|x-y|/2 \leq |f(x)-f(y)|$
Any help would be highly appreciated.