Let $\Omega\subset\mathbb{R}^n$ be a bounded "smooth" domain and $\Omega_\delta=\{x\in \Omega:\ d(x,\partial\Omega)<\delta\}$. Let $U_\delta$ denote the part of the boundary of $\Omega_\delta$ which does not contains points of $\partial\Omega$. Choose $\delta>0$ in such a way that for each point $p\in U_\delta$, there is only one point $T(p)\in \partial \Omega$ which minimizes the distance between $p$ and $\partial\Omega$.
Define $u:\Omega\to\mathbb{R}$ by $u(x)=1$ if $x\in \Omega\setminus\Omega_\delta$ and if $x(t)=(1-t)p+tT(p)$ for $t\in(0,1)\in \Omega_\delta$, we define $u(x(t))=(1-t)$.
My question is: Does $u$ has a weak derivative in $\Omega$? Maybe we can use absolutely continuity on lines, but I can't see a straightforward way to do it. Also, it seems to me that this function is $C^1$ excpet on $U_\delta$. Is this true?
The function is
$$u(x) = \min \left\lbrace 1, \frac{1}{\delta}\cdot d(x,\partial\Omega)\right\rbrace,$$
which is Lipschitz-continuous. If I remember correctly, that already implies weak differentiability. [By a theorem of Rademacher, a Lipschitz function is differentiable almost everywhere in the classical sense, and the a.e. defined partial derivatives are the distributional derivatives, they belong to $L^\infty$, so for bounded $\Omega$ to $L^p(\Omega)$ for all $p\in [1,\infty]$.]
Also, in a neighbourhood of the boundary, the $d(x,\partial\Omega)$ is at least $k-1$ times continuously differentiable if the boundary is $C^k$, where points of non-differentiability occur only when several boundary points have the same minimal distance to the point in the interior. $\delta$ was chosen so small that that doesn't occur in $\Omega_\delta \cup U_\delta$, so $u$ is continuously differentiable in $\Omega\setminus U_\delta$ (assuming $\partial\Omega\in C^2$).