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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?

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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.

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  • $\begingroup$ 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? $\endgroup$ – Jason Mar 28 at 16:00
  • $\begingroup$ While I believe that "can you check this"-type questions are strongly discouraged on MO, your step-by-step derivation of a more formal statement of what I stated informally looks fine to me (but then I don't know what level of rigour you're aiming for). $\endgroup$ – gmvh Apr 4 at 9:29

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