Gradient of the distance function Let $\Omega$ be open, bounded subset of $\mathbb{R}^n$. Let $d(x):=dist(x,\partial\Omega)$ denotes the distance of the point $x\in\Omega$ from the boundary $\partial\Omega$. Define function $$\xi:\overline{\Omega}\to[0,1]; \quad \xi(x):=\min\left\{\frac{d(x)}{\delta},1\right\},$$
where $\delta\in (0,1]$ is fixed.
I would like to prove that the following estimate for gradient holds: $$|\nabla\xi|\leqslant \frac{1}{\delta} \quad a.e.$$.
So far I concluded that when $d(x)\geqslant \delta$, then $\xi(x)=1$ and therefore $\nabla\xi(x)=0$ so the inequality holds. If $d(x)<\delta$ then $\xi(x)=\frac{d(x)}{\delta}$  and the only thing I know is that distance as a Lipschitz function is differentiable a.e. by Rademacher's Theorem so at least I am allowed to differentiate it. Of course if $x\in\partial\Omega$, then $\xi(x)=0$, so the natural conclusion is that $\xi$ must somehow 'get small' when $x$ pass through $\delta$-band in $\Omega$ to reach boundary. Still it does not help me to get the estimate.
I would be very pleased to see rigorous proof of this inequality.
Thank you very much,
K.
 A: OK, let’s try to prove the required inequality. I think you can find in books the conditions, when the following calculations are applicable 
(for instance, $\Omega$ should be non-empty). For the sake of simplicity I extend the function $\xi$ to the whole $\Bbb R^n$ by putting 
$$\xi(x):=\min\left\{\frac{dist(x,\Bbb R^n\setminus\Omega)}{\delta},1\right\}.$$
Now let $x_0\in\Bbb R^n$ be a point such that the function $\xi$ is differentiable at the point $x_0$. If $\nabla \xi(x_0)=0$ then there is nothing to prove. So, assume that $\nabla \xi(x_0)\ne 0$. For each $\lambda\in\Bbb R$ put  $g(\lambda)= \xi(x_0+\lambda\nabla \xi(x_0))$. Then the function $g$ has a derivative $g’(0)$ at the point $0$ and $g’(0)=\Delta\xi(x_0)=\|\nabla\xi(x_0)\|^2$ as a derivative of the composition of functions (I have a similar theorem in my analysis book). From the other side, a direct calculation yields 
$$|g’(0)|=\left|\lim_{\lambda\to 0}\frac{g(\lambda)-g(0)}{\lambda}\right|=
\left|\lim_{\lambda\to 0}\frac{\xi(x_0+\lambda\nabla \xi(x_0))- \xi(x_0)}{\lambda}\right|\le 
\lim_{\lambda\to 0}\left|\frac{\xi(x_0+\lambda\nabla \xi(x_0))- \xi(x_0)}{\lambda}\right|\le 
\lim_{\lambda\to 0}\frac{d(x_0+\lambda\nabla \xi(x_0),\xi(x_0))}{|\lambda\delta|}= 
\lim_{\lambda\to 0}\frac{\|\lambda\nabla\xi(x_0)\|}{|\lambda\delta|}=\frac{\|\nabla\xi(x_0)\|}{\delta}.$$ Comparing two expressions for $|g’(0)|$, we obtain the required inequality.  
