Lipschitz constant of Polar parametrization If $p(t)=P(e^{it})=r(t)e^{it}$, where $r(t)>0$ is a mapping between $[0,2π]$ i.e. between the unit circle $T$ and a curve, then it seems that $\mathrm{Lip}(p)=\mathrm{Lip}(P)$ but I have't the proof.
 A: Turns out this is true. We may assume that $\mathrm{Lip}(p)=1$, that is, $\dot r^2+r^2\le 1$ where the dot indicates derivative with respect to $t$. Note that $r\le 1$. Our goal is to show that $|r(t)e^{it}-r(0)|\le |e^{it}-1|$ for all $t$. After squaring this simplifies to $$(1)\qquad \qquad 2\cos t\,(1-r(0)r(t))\le 2-r(0)^2-r(t)^2$$ which is trivially true when $\cos t\le 0$. So we only have to deal with $0<t<\pi/2$. 
We lose no generality in assuming that $r(t)\le r(0)$. It is convenient to write $r(0)=\cos \alpha$ where $0\le \alpha\le \pi/2$. Integrating differential inequality $\dot r\ge -\sqrt{1-r^2}$, we find that $r(t)\ge \cos(\alpha+t)$. The inequality (1) can be written as 
$$(2)\qquad\qquad r(t)^2-2r(t)\cos\alpha\cos t+\cos^2\alpha-2(1-\cos t)\le 0$$
Since the left hand side is a convex function of $r(t)$, it suffices to verify (2) at the endpoints $r(t)=\cos(\alpha+t)$ and $r(t)=\cos\alpha$. 
Plugging $r(t)=\cos(\alpha+t)$ into (2) we get $-(1-\cos t)^2\le 0$. Plugging $r(t)=\cos \alpha$ into (2) we get $-2\sin^2 \alpha  \, (1-\cos t)\le 0$. QED
There are somewhat similar lemmas in Radial extension of a bi-Lipschitz parametrization of a starlike Jordan curve by Kalaj and in On quasiconformal self-mappings of the unit disk satisfying Poisson's equation by Kalaj and Pavlović, although in both cases they are about modulus-preserving maps rather than argument-preserving. 
