Lipschitz Domain Let $U$ denote the open unit ball in the complex plane and $G=${$(z(1)+z(2),z(1)z(2))$ belongs to $C^2:z(1),z(2)$ belongs to U}.Is the boundary of $G$ Lipschitz?Justify.
Thanks for any help.
 A: To rephrase the question, we need to investigate if the boundary of 
$$
G=\{(z_1+z_2,z_1z_2)\in\mathbb{C}^2:z_1,z_2\in U\},
$$
is Lipschitz. Here $U=\{z\in\mathbb{C}:|z|<1\}$ is the unit disk. Let us call $w_1=z_1+z_2$ and $w_2=z_1z_2$.
First of all, it is obvious that the projection of $G$ onto the $w_1$-plane is $2U=\{z:|z|<2\}$, the disk of radius $2$. Let us fix $w\in2U$, and investigate the cross section $G_w=G\cap\{w_1=w\}$. Putting $w_1=2re^{i\theta}$, and $z_1=re^{i\theta}+\rho e^{i(\theta+\alpha)}$, we have $z_2=w_1-z_1=re^{i\theta}-\rho e^{i(\theta+\alpha)}$, hence
$$
w_2=z_1z_2=r^2e^{2i\theta}-\rho^2 e^{2i(\theta+\alpha)},
$$
where we assume $\alpha$ runs from $0$ to $\pi$, and $\rho$ from $-\rho^*$ to $\rho^*$, with $\rho^*=\rho^*(r,\alpha)$. We have $\rho^*(r,\alpha)>0$ for $0\leq r<0$. Also, $\rho^*(0,\cdot)\equiv1$ thus $G_0=U$, and $\rho^*\to0$ as $r\to1$, thus $G_w$ approaches the point $(2z,z^2)$ as $w\to2z$ with $z\in\partial U$. From continuity it is clear that 
$$
\partial G = \left(\bigcup_{w\in 2U}\partial G_w\right)\bigcup\left(\bigcup_{z\in\partial U}\{(2z,z^2)\}\right).
$$
For any fixed $r<1$, the function $\alpha\mapsto\rho^*(r,\alpha)$ is bi-Lipschitz. Moreover, the map $(r,\theta,\rho,\alpha)\mapsto(w_1,w_2)$ is locally a diffeomoprhism for $\rho\neq0$. Note that $\rho=0$ (with $r<1$) corresponds to points interior to $G$. So $\partial G$ is locally Lipschitz everywhere, except possibly near the "rim" $\{|w_1|=2\}$.
Let us look at how "sharp" the boundary is near the "rim". We fix again $w_1=2r$ (hence $\theta=0$ this time), with $r<1$ close to $1$, and consider two points $w_2=r^2\pm(1-r)^2$, i.e., $\rho=1-r$ and $\alpha\in\{0,\frac\pi2\}$, so that $(w_1,w_2)\in \partial G$. The distance between these two points is $2(1-r)^2$, which goes to $0$ much faster than $1-r$, as $r\nearrow1$.
Hence the boundary is not Lipschitz. We also see from the calculation that the boundary might be Hölder continuous with exponent $\frac12$.
