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If $\Omega$ is closed subset $\mathbb{R}$, then the distance function $$f\colon \mathbb{R}\rightarrow\mathbb{R}, \,\,\,\, x\mapsto d(\Omega,x),$$ is certainly a continuous function. An interesting property of this function is that, the points where $f$ vanishes is precisely the closed set $\Omega$. However, this function is not necessarily differentiable: when $I=\{0\}$, then $f(x)=|x|$, which is not differentiable at $0$.

My question is, can we find other non-closed nice sets (like open sets, or some others) $\Omega$, for which $f$ above will become differentiable function on $\mathbb{R}$? Can we characterize such sets?

(The obvious examples would be dense subsets of $\mathbb{R}$, for which $f$ will be constant function $0$.)

Groups
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  • It seems that $f$ is not differentiable at the "boundary" of $\Omega$. –  Oct 24 '15 at 09:03
  • you may be correct; for specific types of sets $\Omega$ it is true – Groups Oct 24 '15 at 09:42
  • If $\Omega$ is not closed then $d(x,\Omega)$ is the same as $d(x,\overline{\Omega})$ so asking about non closed sets reduces immediately to the closed case. If $\Omega$ is closed and not all of $R$ then in any component interval of $R\setminus \Omega$ it is abundantly clear what the function $d(x,\Omega)$ looks like and exactly where the corners are. So I think the problem is a dead end. – B. S. Thomson Oct 25 '15 at 00:29
  • Related to: Is the distance function differentiable?. For later readers who may be interested, this question is currently an active research area for infinite dimensional normed spaces. See Jon D. Vanderwerff's comments here, and more generally, see the google search "closed set" + differentiable + distance. – Dave L. Renfro Oct 26 '15 at 18:04
  • The "distance to a closed set" function is Lipschitz continuous (with Lipschitz constant $1),$ so the topic falls within the more general study of the differentiability properties of Lipschitz functions. However, I believe the special case of "distance to a closed set" includes many of the important results and counterexamples of the general theory, so the study of the differentiability properties of the distance to a closed set function is an important special case in the general theory of the differentiability properties of Lipschitz functions defined on normed spaces. – Dave L. Renfro Oct 26 '15 at 18:12

1 Answers1

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I try an answer.

Suppose that $f(x)=d(x,\Omega)$ is differentiable, and $\overline{\Omega}$ is not equal to $\mathbb{R}$. Let $U=\mathbb{R}-\overline{\Omega}$, and $]u,v[$ a connected component of $U$; we have $u,v\in \overline{\Omega}$.

Then for $x\in ]\frac{u+v}{2},v]$, we have $f(x)=v-x$, hence $f^{\prime}(x)=-1$. In particular, we have $f^{\prime}(v)=-1$. Then for $|x-v|$ small, $x\not =v$, we have $\displaystyle \frac{f(x)-f(v)}{x-v}\leq -\frac{1}{2}$. Hence as $f(v)=0$, for $x>v$, $x-v$ small, we get $f(x)<0$, a contradiction. The case of $]-\infty,v[$ and $]u,+\infty[$ can be shown to lead to a contradiction on the same way. So $\overline{\Omega}$ must be equal to $\mathbb{R}$

Kelenner
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  • Nice answer. I was not expecting myself that only dense sets would give differentiable functions. But, good to see this surprise. – Groups Oct 24 '15 at 10:41
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    You could simplify the last part of the argument by noting that $f$ has a global minimum in $v$ (in fact, in each point of $\overline{\Omega}$), so that $f'(v) = 0$. – PhoemueX Oct 24 '15 at 10:48
  • @phoemueX. Yes, you are true, thank you – Kelenner Oct 24 '15 at 13:14
  • Seems rather simpler to say that if $\Omega$ is a nonempty closed set that is not all of $R$ then there is a point that is isolated on at least one side. At that point $d(x,\Omega)$ cannot possibly be differentiable. (Closed or not is irrelevant since $d(x,\Omega)=d(x,\overline{\Omega})$ for any set.) – B. S. Thomson Oct 25 '15 at 12:16