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If $M$ is a graph of a $C^1$ function $f:\mathbb{R}^n\to\mathbb{R}$, is it true that the length minimizing geodesics on $M$ are $C^1$? I expect a counterexample.

For a related discussion see Metric angles in Riemannian manifolds of low regularity and Existence and uniqueness of geodesics in low regularity.

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    $\begingroup$ @LiviuNicolaescu I do not understand your comment. Geodesic equation requires Christoffel symbols which involve derivatives of $g_{ij}$, but the metric tensor is only continuous so it cannot be done. It seems that one cannot even write the reasonable equation for geodesics. $\endgroup$
    – Piotr Hajlasz
    Commented Jun 7, 2018 at 21:18
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    $\begingroup$ I think the geodesics is $C^1$, Otherwise there exists a point $x_0=(x,f(x))$ where the geodisic forms an angle or have fast oscilation. But around $x_0$ the graph is close to a plane (y,f(x)+df(y-x)) and therefore in both situation we should be able to find a shorter path. $\endgroup$
    – RaphaelB4
    Commented Jun 8, 2018 at 10:00
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    $\begingroup$ The condition that is is a graph is not needed --- any regular $C^1$ surface is a graph locally. It seems that geodesic might be not smooth at its end; if you are interested I might try to write it. $\endgroup$
    – Anton Petrunin
    Commented Jun 8, 2018 at 10:58
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    $\begingroup$ It seems that at the interior points geodesics have to be differentable --- RaphaelB4's answer seems to work (I do not see why it is $C^1$ tho). $\endgroup$
    – Anton Petrunin
    Commented Jun 8, 2018 at 11:04
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    $\begingroup$ I think a starting point is that the geodesic distance on $C^1$ surface verifies $d(x,y)=\|x-y\|(1+o(1))$ for $\|x-y\|\to0$. This should imply, for a geodesic parametrized by arc-lenght, that the quotients ${\gamma(s)-\gamma(t)\over s-t}$ are uniformly continuous on $\{(s,t)\in[0,1]\times[0,1] : s\neq t\}$, which implies $\gamma\in C^1$. $\endgroup$
    – Pietro Majer
    Commented Jun 8, 2018 at 11:47

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A comment: a length minimising geodesic $(u, f(u))$ on $M:={\rm graph}(f)$, which it is convenient to give by its projection $u:[0,1]\to\mathbb{R}^n $ on the domain of $f$, can be identified with a critical point $u\in H^1([0,1],\mathbb{R}^n)$ of $$E(u):={1\over 2}\int^1_0 \|\dot u \|^2+\big(\nabla f(u )\cdot \dot u \big)^2 dt= {1\over 2}\int^1_0 \|\dot u \|^2+\big(\partial_t f(u )\big)^2 dt.$$

There could be a simple argument to show regularity of the minimisers.

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  • $\begingroup$ (but not too simple) $\endgroup$
    – Pietro Majer
    Commented Jun 8, 2018 at 16:15
  • $\begingroup$ For a minimiser of $E$, the integrand must be constant, otherwise it can be improved by reparametrisation. $\endgroup$
    – Pietro Majer
    Commented Jun 8, 2018 at 16:54

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