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?



Existence of Geodesics in continuous metrics
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.


