When considering a subset $\Omega \subset \mathbb{R}^{n}$. If we consider $\nu$, the outward unit surface normal to $\partial \Omega$, what are the requirements of $\partial \Omega$ which will guarantee that $\nu$ is continuous and why?
Does $\partial \Omega$ have to maybe be $C^{1}$ or a Lipschitz continuous?
Thanks.
On
It requires some level of smoothness, consider the following: If for example you consider the case in $3$ dimensional space, then $\Omega \subset \mathbb{R}^{3}$ and by definition of '$C^{1}$ boundary' you have $\gamma \in C^{1}$ for each $x \in \partial \Omega$, taking $\gamma(x,y) = z$ you would define the implicit equation $F(x,y,z): = \gamma - z = 0$ and then use the theorem that states " If a function $F$ is differentiable, the gradient of $F$ at a point is either zero, or perpendicular to the level set of $F$ at that point." Since the surface(curve in $2$ dimensions) is precisely the level set it follows that we can now define the unit normal vector as: $$\textbf{n} := \frac{\nabla F(x,y,z)}{\Vert \nabla F(x,y,z) \Vert}$$
As you can see the continuous differentiability of $\gamma$ is used in defining the unit normal vector $\textbf{n}$.
Suppose $\Omega$ is such that there is a regular parametrization $\gamma$ of its border, i.e. $\gamma \in C^1$ and $\nabla\gamma$ never vanish (so for example your parametrization don't get stuck a while on a point somewhere). Then you can "build" the exterior normal (at least in the case $n = 2$ and $n=3$) out of $\nabla \gamma $ and $\|\nabla \gamma \|^{-1}$.