The approximation of star-shaped domain Let $\Omega$ be a star-shaped domain in $\mathbb R^n$, that is, $\Omega$ is an open set such that for any $x\in \Omega$, $tx \in \Omega $ for $0 \leq t\leq 1$. Can we find a sequence of star-shaped domains $\{\Omega_i\}_{i\geq 1}$ such that $\Omega_i$ has the smooth boundary and $\bigcup\limits_{i \ge 1} {{\Omega _i}}  = \Omega $?
 A: A  star-shaped domain can be written as the union of bounded star-shaped domains, namely its intersections with open balls centered at the origin. Thus, we may assume $\Omega$ is bounded. 
Let $B_r$ be the closed ball of radius $r$ centered at the origin. Let $C_r(x)$ be the  convex hull of the union $B_r\cup \{x\}$. Define $\Omega_r=\{x : C_r(x)\subset \Omega\}$. Observations: 


*

*$\Omega_r$ is open. Indeed, $C_r(x)$ is a compact set, so if it's contained in $\Omega$, it is at positive distance from $\partial\Omega$. This gives $x$ some room to move about. 

*$\Omega_r$ is star-shaped, because $C_r(tx)\subset C_r(x)$ when $0\le t\le 1$.

*For every $x\in \Omega_r$ we have $C_r(x)\subset \Omega_r$. This is because $y\in C_r(x)$ implies $C_r(y)\subset C_r(x)\subset \Omega$. 

*$\Omega=\bigcup_{n=1}^\infty \Omega_{1/n}$. Indeed, for every $x\in \Omega$ the line segment from $0$ to $x$ is at positive distance from $\partial\Omega$, which implies $x\in \Omega_r$ for sufficiently small $r$.  


In view of 4, it suffices to approximate $\Omega_r$.
Define  $\rho: S^{n-1}\to (0,\infty)$ by $\rho(\xi )=\sup\{t:t\xi\in \Omega_r\}$. This function is Lipschitz continuous, being the supremum of the corresponding functions for the sets $C_r(x)$ (which are uniformly Lipschitz continuous, due to $x$ being bounded and $r$ fixed). Approximate $\rho$ uniformly by smooth functions $\rho_n$; scale them so that $\rho_n<\rho$ pointwise, and you have the desired approximation in the form $\{t\xi : 0\le t<\rho_n(\xi), \ \xi\in S^{n-1}\}$.     
