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In an answer to this question, it is proved that the Riemann distance function is the same whether we use smooth paths or piecewise smooth paths.

What happens if we use $C^1$ paths? Can we approximate well enough any $C^1$ path with a smooth one?

This question is local: The image of any $C^1$ - path is compact, hence it can be covered by a finite number of coordinate balls; If we can approximate each segment which lies in a coordinate ball, we can approximate any path.

So, the question is reduced to the case of arbitrary metric on $\mathbb{R}^n$.

Attempted solution:

In this paper, it is proved that the Riemannian distance does not change if we use arbitrary absolutely continuous (a.c) paths. (See Corollary 3.13).

Thus, since "smooth $\subseteq C^1 \subseteq$ a.c" we get that

$$ d^{a.c}=d^{C^{\infty}} \ge d^{C^1} \ge d^{a.c},$$ so in particular $d^{C^1}=d^{C^{\infty}}$.

Perhaps there is an easier solution.

Community
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Asaf Shachar
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Yes, you can approximate any continuous map by $C^\infty$ maps. The cleanest way to do so is to embed $M$ in some $R^N$ and then use the fact that a small neighborhood $U$ of the image is diffeomorphic to the normal tangent bundle of $M$. In particular, you have a smooth projection $U\to M$. Now, given a continuous map $f: I\to M$, approximate $f$ by smooth functions $f_\epsilon: I\to R^N$ and then compose these approximates with $\pi$. The construction of $f_\epsilon$ is usually done via convolution with suitable bump-functions. See https://en.wikipedia.org/wiki/Mollifier.

Moishe Kohan
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  • Thanks for your answer. Actually by the argument I gave it's enough to consider $\mathbb{R}^n$, with arbitrary metric on it. However, if we take your approach, I am not sure I understand why after projecting the $f_\epsilon$ back to $M$ the "goodness" of the approximation is not damaged to much. – Asaf Shachar Oct 24 '15 at 10:38