# Boundary of star-shaped domain

Assume that $M\subset R^n$, $n\ge 3$, is a boundary of an open bounded set $D$ containing $0$, which is starlike w.r.t. 0, meaning that each ray $[0,x]$ from $x\in M$ to $0$ meets $M$ only once. Is $M$ smooth almost everywhere?

The unit circle is a boundary of an open bounded set $D$ containing $\mathbf{0}$,
which is starlike w.r.t. $\mathbf{0}$, and which is smooth almost everywhere.
When $f$ is the original Weierstrass function, $\;\; \left\{\begin{bmatrix} \left(\frac{1+a}{1+(-a)}+f(\theta)\right) \cdot \cos(\pi \cdot \theta) \\ \left(\frac{1+a}{1+(-a)}+f(\theta)\right) \cdot \sin(\pi \cdot \theta) \end{bmatrix} \: : \: \theta \in [-1,1]\right\}$
is a boundary of an open bounded set $D$ containing $\mathbf{0}$, which
is starlike w.r.t. $\mathbf{0}$, and which is not smooth almost everywhere.
• You should take $f$ to be positive (this can be arranged by adding a positive constant, it it needed). It is open because $D=\{z=re^{is}: r<f(s)\}$. Thanks. – djoke Oct 19 '12 at 21:07
• @Malota: Why can't you just increase the dimension of Ricky Demer's Weierstrass-function example by taking the product of his $D$ with the cube $(-1,1)^{n-2}$ (and increase the dimension of his round example by using an $n$-dimensional ball)? – Andreas Blass Oct 19 '12 at 21:50