Suppose that $g$ is a $C^{1,1}$ (i.e., continuously differentiable with locally Lipschitz first derivative) Riemannian metric on a smooth manifold $M$. It seems to be known that locally the exponential map $\exp_p$ corresponding to $g$ at some point $p$ of the manifold is still a local homeomorphism (and Lipschitz). I am looking for a proof of this result (if it is true). The standard proof for $C^2$ metrics uses the fact that the tangent map of $exp_p$ at $0\in T_pM$ is the identity, so that the inverse function theorem gives that $exp_p$ is a local diffeomorphism. However, for a $C^{1,1}$ metric, $\exp_p$ is only (locally) Lipschitz, so it is not clear that there even exists a tangent map of $exp_p$ at $0$ and the inverse function theorem does not apply. Is there some argument that can replace the use of the inverse function theorem in this situation?
Note that $C^{1,1}$ metric admits an approximation by $C^2$metrics with uniformly bounded $C^2$norm. In particular the curvature is bounded, hence we get a bounds for the distortion of the exponential map depending on the size of the neighborhood of $0$. It remains to pass to the limit. 


More generally the exponential map of any Lipschitz connection or spray is a Lipeomorphism (biLipschitz homeomorphism) in a neighborhood of the origin. The proof uses Leach's inverse function theorem which holds under strong differentiability. The nontrivial part is to show that the exponential map is strongly differentiable at the origin. Details are given here http://arxiv.org/abs/1308.6675 

