Let $U$ be an open subset in $R^n$ and let $N$ be a $C^1$-submanifold. We have a family of geodesics $\gamma:[0,1]\rightarrow U$ in U with respect to the euclidian metric. Each geodesic is parametrized with constant speed and and intersects with $N$ in exactly one point $\gamma(\tau(\gamma))$. We have $sup_{t\in [0,1]}|\gamma(t),\tilde{\gamma}(t)|\leq C|\gamma(0),\tilde{\gamma}(0)|$ for all $\gamma$ and $\tilde{\gamma}$ and C is a constant.

Is the map $\gamma(0)\mapsto\gamma(\tau(\gamma))$ Lipschitz?

  • $\begingroup$ Would the statement be true if we assume that the intersections are transversal and replace Lipschitz by locally Lipschitz? $\endgroup$
    – Christian
    Aug 29, 2011 at 21:17

1 Answer 1


No, take $N = S^1$, $U = \mathbb{R}^2$ and a suitable collection of line-segments parallel to an axis (for example).

Edit: To respond to the question in the comment: even if we assume the intersections to be transversal we don't get a locally Lipschitz map. Consider for example the following open curve to be our manifold $N$ and the blue (and one red) vertical line-segments to indicate the set of geodesics. Then there is even a discontinuity for our mapping at the red line.

(The open ends of the curve are needed here for the example to work. A good starting point for proving something positive could be a closed curve $N$.)


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