Timeline for Derivative of distance function to a closed, rectifiable set
Current License: CC BY-SA 4.0
7 events
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| Dec 28, 2021 at 16:29 | comment | added | dohmatob | @MartinKell A bonus from your proof is that: If $\Gamma$ is a closed subset of $\mathbb R^d$, then $x \mapsto \mbox{dist}(x,\Gamma)$ is differentiable at almost-every $x \in \mathbb R^d \setminus \Gamma$. Indeed, see tzamfirescu.tricube.de/TZamfirescu-110.pdf | |
| Aug 19, 2019 at 16:10 | history | edited | Luc Guyot | CC BY-SA 4.0 |
Fixes typos: missing "s", $F$ --> $\Gamma$
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| Aug 18, 2019 at 19:38 | history | edited | Martin Kell | CC BY-SA 4.0 |
added 14 characters in body
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| Aug 18, 2019 at 19:37 | comment | added | Martin Kell | (1) I adjusted the claim to "..$x \notin \Gamma$ ...". The whole argument assumes $x \ne x_\Gamma$. (2) Note $f$ is $1$-Lipschitz and for general $1$-Lipschitz functions it holds that the gradient is equal to $e$ whenever a partial in direction $e$ is equal to $1$. Alternatively, observe that on the one hand $f(y)- f(x)$= -\|y-x\|$ by choice of $y$. On the other hand $f(y) = f(x) + v\cdot (x-y) + o(\|x-y\|) by differentiability in $x$. So if $y$ is close to $x$ then necessarily $v = e$ by the "equality case" of Cauchy-Schwarz. | |
| Aug 18, 2019 at 16:25 | comment | added | Luc Guyot | I would appreciate some hints regarding "But then $v=(x-z)/\|x-z\|$." I understand only that $f(y)$ and $\| y - z \|$ coincide for $y = (1 - \epsilon)x + \epsilon z, 0 \le \epsilon \le 1$ so that $\frac{\partial f}{\partial e}(x) = \frac{\partial \| . - z \|}{\partial e}(x)$ for $e = (x-z)/\|x-z\|$. | |
| Aug 18, 2019 at 16:10 | comment | added | Luc Guyot | I am confused by your second sentence: a point $\gamma \in \Gamma$ has a unique nearest point in $\Gamma$, that is itself. So $\text{dist}(., \Gamma)$ should be differentiable at $\gamma$ by the first sentence. | |
| Aug 16, 2019 at 11:09 | history | answered | Martin Kell | CC BY-SA 4.0 |