I learned that if we are given a $C^0$ Riemannian metric on a smooth manifold $M$, geodesics (i.e. length minimizing curves) are absolutely continuous, and if the metrics is $C^{0,\alpha}$, then the geodesics are even $C^{1, \beta}$ where $\beta = \alpha/(2-\alpha)$.

However, I did not find any results on the existence of geodesics. A priori, it could be possible that no geodesics exist at all, or am I wrong?

Are there results that there exist "enough" geodesics in some sense, like when two points are close enough, or for almost any two points (suppose maybe that $M$ is compact)?

A problem in considering the geodesics equation seems to be that it involves the Christoffel symbols which are not continuous. If we have a solution to the geodesic equation, is it length minimizing for two points close enough?

It is known that given a continuous metric $g$, there exists a sequence of smooth Riemannian metrics that converges to $g_n$ such that the corresponding distance functions converge to the one associated to $g$. What happens with the geodesics (i.e. solutions to the geodesics equation) in this case? Do they converge in some sense?

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    $\begingroup$ this paper seems relevant: arxiv.org/abs/1212.6962 $\endgroup$
    – johndoe
    Jun 14, 2013 at 9:11
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    $\begingroup$ Yes, indeed, this paper answers the main question raised by OP: Take results of section 4 and combine them with Arzela-Ascoli theorem. $\endgroup$
    – Misha
    Jun 14, 2013 at 10:26
  • $\begingroup$ Could you elaborate on this? $\endgroup$ Jun 14, 2013 at 10:44
  • $\begingroup$ Kofi: Did you read section 4 of this paper, more specifically, proof of theorem 4.5? $\endgroup$
    – Misha
    Jun 14, 2013 at 11:00
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    $\begingroup$ In any compact metric space, any two points that can be joined by a rectifiable curve, can be joined by a length minimizer. This follows immediately from Arzela-Askoli and lower semi-continuity of length. As a corollary, the same holds in a locally compact space and two points sufficiently close to each other. $\endgroup$ Jun 14, 2013 at 16:47

1 Answer 1


Ok, thank you Misha for the comments, let me try to fill out the hints you gave myself. I try to prove the following:

Let $g_n$ be a sequence of complete smooth metrics that converge in $C^0$ against the continuous metric $g$. Let $p$, $q \in M$ such that they are not cutpoints of each other w.r.t. any $g_n$. Let $\gamma_n$ be a $g_n$-geodesic connecting $p$ and $q$, then $\gamma_n$ converges to a $g$-geodesic in $C^0$.

By theorem 4.5 in the paper quoted above, for each $\eta>0$, there exists a number $N \in \mathbb{N}$ such that $$ (1-\eta)d_n(p, q) \leq d(p, q) \leq (1+\eta) d_n(p, q) ~~~~ \forall n\geq N$$ Hence for $\delta = \varepsilon/(1+\eta)$, $|s-t| < \delta$ implies $$ \varepsilon > (1+\eta)|s-t| = (1+\eta)d_n(\gamma_n(s), \gamma_n(t)) \geq d(\gamma_n(s), \gamma_n(t))$$ Hence the set of $\gamma_n$ with $n \geq N$ is equicontinuous and contained in some compact set by the same reasining as in the proof of the quoted thm. 4.5. Hence it subconverges in $C^0$ to a path $\gamma$. By the assumption that $p$ and $q$ are not cutpoints for any $g_n$, $\gamma_n$ is uniquely determined, so $(\gamma_n)$ converges itself (they fulfill a differential equation depending smoothly on the metric and hence are $C^0$ close if the metrics are $C^0$ close).

I do not yet see why $\gamma$ is a $g$-geodesic though.

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    $\begingroup$ As Sergei Ivanov points out, there is no need to approximate the metric by smooth ones. You can construct the existence (but not the uniqueness) of length-minimizing curves directly using a length-minimizing sequence of curves using Arzela-Ascoli, which also shows that the curve is absolutely continuous. You get everything in one shot. $\endgroup$
    – Deane Yang
    Oct 12, 2013 at 15:57

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