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Let $\Gamma \subset \mathbb{R}^2$ be a closed simple $C^1$ curve. For every $x \in \mathbb{R}^2\setminus\Gamma$ there exists some $p(x) \in \Gamma$ such that $$ (H) \quad \text{ dist}(x,\Gamma)=|x-p(x)|. $$ Of course $p(x)$ is not necessarily unique. E.g. for $\Gamma=\{x \in \mathbb{R}^2:\, |x|=r\}$ and $x=0$, any $p(0) \in \Gamma$ satisfies (H). However, if $0 < |x| < r$, then $p(x)=rx/|x|$ is the only point satisfying $(H)$.

The question is the following: Is it possible to characterize all closed simple $C^1$ curves $\Gamma \subset \mathbb{R}^2$ with the property:

there is some $\varepsilon_0 > 0$ such that for every $\varepsilon \in (0,\varepsilon_0)$, and every $x \in B_{\varepsilon}(\Gamma)$, there is a unique $p(x) \in \Gamma$ satisfying (H)? where, for $A \subset \mathbb{R}^2$ and $\varepsilon > 0$, $B_{\varepsilon}(A)$ denotes the set $\{x \in \mathbb{R}^2:\, \text{dist}(x,A) < \varepsilon \}$

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up vote 6 down vote accepted

The set of points which have more than one closest point to the curve is called the medial axis of that curve and it is the locus of the centers of maximal circles inside the region bounded by the curve. For example the medial axis of a circle is its center.

The distance from a compact set to its medial axis is called the reach of that set. Therefore in this terminology your question is asking for a criterion to determine curves of positive reach. Sets of positive reach were defined in this seminal paper of Federer "Curvature measures", mainly in order to be able to talk about curvature on curves, surfaces etc. which lack the desired differentiability properties. One thing to notice for example is that any curve of positive reach is $C^{1,1}$, i.e. the unit tangent vector is a Lipschitz function of arclength.

For some recent papers on characterizations of subsets of positive reach on Riemannian manifolds you can check out "Almost convex subsets" and "On the geometry of subsets of positive reach" by Lytchak, available from here. Federer had proved that a subset of positive reach that is a Lipschitz manifold is $C^{1,1}$, while Lytchak shows that it is enough to assume that the subset is a topological manifold.

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Notice that if an hypersurface $M$ of $R^n$ is of class $C^{1,1}$, then you can construct in a neighborhood of each point of $M$ a local solution to the so-called eikonal equation $|Du|=1$ with the boundary condition $u=0$ on $M$. This gives you that $u$ is indeed the signed distance function to $M$, and that $M$ has positive reach. A reference for local solutions to Hamilton-Jacobi equations with $C^^{1,1}$ boundary conditions may be found in a paper by Fathi in Ann. Fac. Sci. Toulouse.

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This holds for all smooth simple closed curves. You should look for tubular neighbourhoods in the literature.

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Think of curves of unbounded curvature ;). –  Gjergji Zaimi Dec 22 '11 at 12:53
    
Gjergji, you are right, I have not seen that the question was for $C^1$ curves. –  Sebastian Dec 22 '11 at 13:13
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