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Suppose I have a bounded set $\Omega$ with a Lipschitz boundary. Is it possible to equip this boundary with a Riemannian distance function $d$ that satisfies: $\lambda |x-y| \leq d(x,y) \leq \mu|x-y|$ for all points in $\partial \Omega$? If not, what is the smallest regularity of the boundary one needs for this property to hold?

Edit: Ultimately what I want to do is the following. I have a set $\Omega \subset \mathbb R^d$ that is bounded, and $\partial \Omega$ is connected. Then, I define the following distance function, $d(x,y) = \inf ( \int_0^1 \|\gamma'\| dt \colon \gamma(0)=x, \gamma(1)=y \text{ and }\gamma \subset\partial \Omega )$. I am interested in conditions where there exist $\mu > 0$ such that $d(x,y) \leq \mu |x-y|$ for all points $x,y \in \partial \Omega$.

I suspect that this is true when $\partial \Omega$ is Lipschitz, in the sense that there exists a finite cover of $\partial \Omega$ where in each component of the cover, say $U_j$ it holds that $U_j \cap \partial \Omega$ is the graph of a Lipschitz function.

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    $\begingroup$ @dsk24: Please specify what do you mean by Lipschitz boundary and Riemannian metric in it. Both have at least two reasonable interpretations. One would be to work in the category of closed Lipschitz-tamed submanifolds and Riemannian metrics on charts which are a.e. preserved by transition maps. $\endgroup$
    – Misha
    Oct 17 '12 at 13:20
  • $\begingroup$ Thank you for your comment. I have edited the above to include my definition of a Lipschitz boundary and what I am trying to achieve. $\endgroup$
    – dcs24
    Oct 17 '12 at 15:05
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    $\begingroup$ Your "Lipschitz boundary" might have topological singularities; in this case there is no hope for a "Riemannian metric". $\endgroup$
    – ε-δ
    Oct 17 '12 at 18:24
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By compactness, there exists $r>0$ so that if points in $S=\partial \Omega$ are within Euclidean distance $\le r$, then they belong to an open subset $U\subset {\mathbb R}^n$ where $S$ is a graph of a Lipschitz function.

Let $f: D\to {\mathbb R}$ be an $L$-Lipschitz function defined on a domain in ${\mathbb R}^{n-1}$ with the graph $G_f$ equipped with the path-metric induced from ${\mathbb R}^n$. Then you define a $L'$-Lipschitz retraction $D\times {\mathbb R}\to G_f$ by $(x,t)\to (x,f(x))$. (Here $L'=\sqrt{L^2+1}$.) It follows that the path-metric on $G_f$ is bi-Lipschitz equivalent to the restriction of the distance function from ${\mathbb R}^n$: $$ |p-q|\le d(p,q)\le L |p-q|. $$ By compactness and connectivity of $S$, it follows that the diameter of $S$ (with the path-metric $d$) is finite (say, $D$). Therefore, if $p, q\in S$ satisfy $|p-q|>r$, then $$ |p-q|\le d(p,q)\le (D/r) |p-q| $$ so you have the required bi-Lipschitz inequality between the distances in this case. If $|p-q|<r$, then by the above observation you also have the required bi-Lipschitz inequality.

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  • $\begingroup$ A very nice proof indeed. Thank you for your help. $\endgroup$
    – dcs24
    Oct 17 '12 at 21:36
  • $\begingroup$ Your first sentence is incorrect; it seems you use a definition different from the one given by dcs24. $\endgroup$
    – ε-δ
    Oct 19 '12 at 17:40
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    $\begingroup$ epsilon-delta: No, everything is correct, my number $r$ is the Lebesgue number of the covering described by dcs34. See en.wikipedia.org/wiki/Lebesgue's_number_lemma $\endgroup$
    – Misha
    Oct 19 '12 at 18:54

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