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In this article of Wikipedia it is stated that, if $\Omega$ is a subset of $\mathbb{R}^n$ with smooth boundary, then $$f(x)=\begin{cases} d(x,\partial\Omega),\;\;x\in\Omega\\ -d(x,\partial\Omega),\;\;x\notin \Omega \end{cases}$$ satisfies $|\nabla f(x)|=1$ for all $x\in\mathbb{R}^n$. Could you give me an outline of the proof?

I read an answer in this site solving in fact this question, but I do not understand it (maybe not enough details, maybe my level of geometry is not good...)

Motivation: Coarea formula says that, if $g\in L^1(\Omega)$, $u\in C^1(\bar{\Omega})$ and $|\nabla u|>0$, then $\int_{\Omega} g\,dx=\int_{\mathbb{R}}\int_{\{u=\lambda\}} g/|\nabla u|\,d\sigma\,d\lambda$. In the particular case $u(x)=d(x,\partial\Omega)$, I read that $\int_{\Omega} g\,dx=\int_{\mathbb{R}}\int_{\{u=\lambda\}} g\,d\sigma\,d\lambda$.

user39756
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1 Answers1

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Since $\partial\Omega$ is smooth, note that for any $x\notin\partial\Omega$ we'll have $d(x,\partial\Omega) = d(x,y)$ with $y\in\partial\Omega$ and $x-y$ orthogonal to $T_y\partial\Omega$. It follows that $\nabla f(x)$ will be a vector pointing toward $y$ and that $d(x,\partial\Omega)$ increases by one unit for every unit you move orthogonally away from $y$. From the directional derivative formula $D_v f(x) = \nabla f(x)\cdot v$ we immediately infer that $\|\nabla f(x)\|=1$.

Ted Shifrin
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  • Thank you for your answer. But I don't understand the fact that in the second line it is written "note that $\nabla f(x)$ will be a unit vector..." and in the last sentence "we infer that $|\nabla f(x)|=1$. Is it not the same? Why is $\nabla f(x)$ a unit vector pointing toward $y$? – user39756 Apr 24 '17 at 18:53
  • This follows immediately from what I said about following the normal line. For each step you take, the distance increases by one step. This means that the directional derivative in that direction is $1$. That, in turn, means that the magnitude of the gradient vector is $1$ for the reasons I gave. Figure this out first for the function $f(x)=|x|$ in $\Bbb R^n$. :) – Ted Shifrin Apr 24 '17 at 18:55
  • Ok, so I understand now why $\nabla f(x)\cdot v(x)=1$. Then $1=|\nabla f(x)||v(x)|\cos\theta=|\nabla f(x)|\cos\theta$, where $\theta(x)$ is the angle between $\nabla f(x)$ and $v(x)$. So my last question is: why do $\nabla f(x)$ and $v(x)$ point toward the same direction? I mean, why do we have your sentence "$|\nabla f(x)|$ pointing toward $y$..."? – user39756 Apr 24 '17 at 19:04
  • You really should understand the case of $f(x)=|x|$ first and then it generalizes. As you walk along the normal line to $\partial\Omega$ at $y$, locally these are all points for which $y$ is the closest point and the directional derivative is $1$ for the reason I said. $\nabla f$ points towards $y$ because moving along the normal line gives the direction in which $f$ increases (or decreases) at the greatest rate. You can compute this directly or see it intuitively (if you move locally "parallel" to $\partial\Omega$ at $x$ the distance instantaneously stays constant). – Ted Shifrin Apr 24 '17 at 19:10
  • I understand now :) Is it difficult to see that $f\in C^1(\mathbb{R}^n)$? Since $f$ is Lipschitz, $\nabla f(x)$ exists for a.e. $x\in\mathbb{R}^n$. Is it true that $\nabla f(x)$ exists for all $x\in\mathbb{R}^n$? – user39756 Apr 25 '17 at 07:45
  • @TedShifrin could you give some advice to proof the statement of your first line? – Valere Jan 18 '23 at 16:20
  • @JoséAntonio Think about Lagrange multipliers, writing $\partial\Omega$ locally as a level set. – Ted Shifrin Jan 18 '23 at 17:14