# Distance function and its approximation

An easy and quick question:

Consider a function $$u\in C(\Omega)$$, where $$\Omega$$ is a bounded domain in $$\mathbb{R}^n$$.

Define a function $$Q$$ that measures the distance of a point $$(x,y) \in\mathbb{R}^{n+1}$$ from the graph of $$u$$. That is, consider the function $$Q(x,y)=\inf_{z\in \mathrm{graph}(u)} d_z (x,y)$$

where $$d_z(\zeta)=|\zeta - z|$$.

My question is:

• why the function $$Q$$ should be the limit of smooth approximations $$Q_p \geq Q$$ given by $$Q_p(x,y)=\bigg[\int_\Omega \{ d_{(\xi,u(\xi))}(x,y)\}^{-p} \mathrm{d}\xi \bigg]^{-1/p}\quad \textrm{for } (x,y) \notin \mathrm{graph}(u)$$
• why the derivative is $$DQ_p(x,y)=(Q_p(x,y) )^{p+1} \int_\Omega |(x-\xi, y-u(\xi)|^{-p-2}(x-\xi,y-u(\xi))\,\mathrm{d}\xi$$

I just tried to show the uniformly convergence. Can somebody check this following calculation?

Let $d_\xi (x,y) = d_{(\xi,u(\xi))}(x,y)\$. One have

$$d_\xi (x,y) \leq \sup_{\xi} d_\xi (x,y) \rightarrow \dfrac{1}{d_\xi (x,y) } \geq \dfrac{1}{\sup_{\xi} d_\xi (x,y)}$$ so applying the integral $$\int_\Omega \bigg( \dfrac{1}{d_\xi (x,y) } \bigg)^p \geq \bigg( \dfrac{1}{\sup_{\xi} d_\xi (x,y)} \bigg) ^p |\Omega|$$ and so we have $$\bigg[\int_\Omega \{ d_{(\xi,u(\xi))}(x,y)\}^{p} \mathrm{d}\xi \bigg]^{1/p}\geq \dfrac{1}{\sup_{\xi} d_\xi (x,y)} |\Omega|^{1/p}$$ and accordingly $$\dfrac{1}{\bigg[\int_\Omega \{ d_{(\xi,u(\xi))}(x,y)\}^{p} \mathrm{d}\xi \bigg]^{1/p}} \leq \dfrac{1}{\bigg( \dfrac{1}{\sup_{\xi} d_\xi (x,y)} \bigg) |\Omega|^{1/p}}$$ Now to prove the uniformly convergence, one need $$\lim_{p\rightarrow \infty } \sup_{(x,y)} | Q_p(x,y) - Q(x,y)| \rightarrow 0$$ By the calculation above, we have $$\lim_{p\rightarrow \infty } \sup_{(x,y)} | Q_p(x,y) - Q(x,y)|\leq \lim_{p\rightarrow \infty } \sup_{(x,y)} \bigg| \dfrac{1}{\bigg( \dfrac{1}{\sup_{\xi} d_\xi (x,y)} \bigg) |\Omega|^{1/p}} - \inf_\xi d_\xi(x,y)\bigg|$$ and using the fact that, in general, $$\dfrac{1}{\sup_x \dfrac{1}{f(x)}} \geq \inf_x f(x)$$ follows $$\lim_{p\rightarrow \infty } \sup_{(x,y)} | Q_p(x,y) - Q(x,y)|\leq \lim_{p\rightarrow \infty } \sup_{(x,y)} \bigg| \big(\dfrac{1}{|\Omega|^{1/p}} -1\big) \dfrac{1}{\sup_{\xi} d_\xi (x,y)}\bigg|$$

Now one can put $$\big(\dfrac{1}{|\Omega|^{1/p}} -1\big)$$ out of the supremum and calculete the limit for $$p\rightarrow \infty$$ the result is $$0$$.

Is this right?

The second identity appears to follow simply from the chain rule applied to $$Q_p(\eta) = f_p(g_p(\eta))$$, where \begin{align*} f_p(t) &=t^{-1/p},\\ g_p(x,y) &= \int_\Omega \left(d_{(\xi,u(\xi))}(x,y)\right)^{-p}\,\mathrm{d}\xi,\\ d_{(\xi,u(\xi))}(x,y) &= |(x-\xi,y-u(\xi))| \end{align*} by definition.
As for the first identity, it looks like $$Q_p$$ measures the inverse of the distance function $$d$$ in an $$L^p$$ norm, and thus the limit $$p\to\infty$$ should give the supremum norm for this inverse, hence the infimum of the distance itself.
• Thank you @gmvh . I wanted a more formal proof of the first question, for example show that $Q_p(x,y)$ converges uniformly to $Q(x,y)$...any idea? – Jason Mar 28 at 16:00