Radial Lipschitz Mapping Let $f:\mathbb{R}^2\rightarrow \mathbb{R}^2$ defined by
$$f(x,y)=(x\cos(y),x\sin(y)).$$
Is $f$ a Lipschitz function? Is it bi-Lipschitz? I tried this:
\begin{align*}|f(x_1 ,y_1) - f(x_2 , y_2)|^2&= x_1^2 +x_2^2 - 2x_1 x_2 (\cos(y_1)\cos(y_2)+\sin(y_1)\sin(y_2))\\&=x_1^2 +x_2^2 - 2x_1 x_2 \cos(y_1 + y_2),\end{align*}
but I don't know how to continue to make it appear $|(x_1 ,y_1)-(x_2 , y_2)|$. Any help is welcome.
 A: Useful scratch-work:\begin{aligned}d\left( f(x,y), f(u,v) \right)^2&=\left(x\cos(y)-u\cos(v)\right)^2+\left(x\sin(y)-u\sin(v)\right)^2
\\&=x^2\left(\cos^2(y)+\sin^2(y) \right)+u^2\left(\cos^2(v)+\sin^2(v)\right)\\&\quad-2xu\left(\cos(y)\cos(v) +\sin(y)\sin(v)\right)
\\&=x^2+u^2-2xu\cos(y-v).
\end{aligned}
Assume there is $K \geq 0$ so that
$$d\left( f(x,y), f(u,v) \right) \leq K d\left( (x,y), (u,v) \right) \;\text{ for all } (x,y), (u,v) \in \mathbb{R}^2.$$
Let $M=\max\{K, 1\}$.
Consider $(M\pi,\pi), (M\pi,0) \in \mathbb{R}^2$, and notice we have
\begin{aligned}d\left( f(M\pi,\pi), f(M\pi,0) \right)&=\sqrt{2(M\pi)^2-2(M\pi)^2\cos(\pi)}\\&=\sqrt{4(M\pi)^2}\\&=2M\pi >K\pi =Kd\left( (M\pi,\pi), (M\pi,0) \right).
\end{aligned}
Since our assumption leads to a contradiction, it follows that $f$ is not Lipschitz.
A: It will be Lipschitz continuous on any bounded rectangle since each of the component functions $f_1(x,y)=x \cos (y)$ and $f_2=x \sin (y)$ of the vector-valued function $f=(f_1,f_2)$ are $C^1$.
In particular, here we have $|\nabla f_1|^2=\cos^2(y)+x^2\sin^2(y)$  and $|\nabla f_2|^2=\sin^2(y)+x^2\cos^2(y)$. Hence $|\nabla f_1|,|\nabla f_2| \leq 1+x^2$. By the mean value theorem and the Cauchy-Schwarz inequality, we thus have
\begin{aligned}d\left( f(x,y), f(u,v) \right) &=\sqrt{\left(f_1(x,y)-f_1(u,v) \right)^2 +\left(f_2(x,y)-f_2(u,v) \right)^2 } \\& \leq  |f_1(x,y)-f_1(u,v)|+|f_2(x,y)-f_2(u,v)|
\\& \leq  2\sqrt{(x-u)^2+(y-v)^2}+2\sqrt{(x-u)^2+(y-v)^2}
\\&=4\sqrt{(x-u)^2+(y-v)^2} 
\\&=4d\left( (x,y), (u,v) \right) \quad \text { for all } (x,y), (u,v) \in [0,1] \times \mathbb{R} \end{aligned}
Note: Since our bound on each of $\nabla f_1$ and $\nabla f_2$ depends only on the first variable, $f$ is Lipschitz on every $G \subseteq  T \times \mathbb{R}$ whenever $T \subset \mathbb{R}$ is bounded.

Notice $f$ can't be bi-Lipschitz on any set on which it fails to be injective. In particular, for every $K \geq 1$ and any $x \in [0,1]$ we will have
$$\frac{1}{K}d\left((x, 0), (x,2\pi)\right)=\frac{2\pi}{K}\geq 0=d\left(f(x, 0), f(x,2\pi)\right).$$
Note: $f$ will fail to be injective on any set $S \subseteq \mathbb{R}^2$ where there exist either $(x_1, y_1), (-x_1,y_2 ) \in S$ such that $y_1 \equiv \pi +y_2\pmod {2\pi}$ or $(x_1, y_1), (x_1,y_2 ) \in S$ such that $y_1 \equiv y_2 \pmod {2\pi}$.
