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Let $M$ be a riemannian manifold with riemannian metric $g$. The manifold $M$ can be made into a metric space with metric $d$, defined as follows for every couple of points $x,y\in M$ $$d(x,y):=\inf\limits_\gamma l_g(\gamma)$$ with $\gamma:[a,b]\rightarrow M$ with $\gamma(a)=x$ and $\gamma(b)=y$ and $l_g(\gamma)=\int_a^b\sqrt{g(\dot\gamma(t),\dot\gamma(t))}dt$.

With these definitions the curve $\gamma$ which minimizes the distance between $x$ and $y$ (i.e. such that $\gamma(0)=x$, $\gamma(1)=y$ and $l_g(\gamma)=d(x,y)$) is a minimizing geodesic.

I've read that the property of being a minimizing geodesic is stable by restriction: take $\gamma$ as before, then for any $k\in[0,1]$ it is true $l_g(x,\gamma(k))=d(x,\gamma(k))$. I can't find a proof of this fact: why is it true?

Now suppose that $g$ has some points of singularity (for example $g$ is a flat metric with conical singularities). Is it still true that the property of being a minimizing geodesic is stable by restriction?

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To see that the restriction of a minimizing geodesic is minimizing, it's probably easiest to consider the contrapositive: If a piecewise-$C^{1}$ path does not minimize length, then no extension minimizes length either, since the original segment can be shortened without changing its endpoints.

This is a feature of an arbitrary metric space in which distance is defined by minimizing (or infimizing) the arc lengths of paths joining the specified points, not just Riemannian manifolds.

(In case it matters, there's no well-defined notion of "geodesic" passing through a cone singularity: The obvious approach is to "cut and unroll", but the resulting path is cut-dependent.)

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    Thank you, but are you sure about your last sentence? I thought that in that case a geodesic is a straight line which can change direction at the conical singularity of total angle $2k\pi$, $k>1$, if both two angles at the conical point are greater than $\pi$. This seems well defined, isn't it? – 2017-01-23
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    What I had in mind was: If there is incident angle $\theta_{0} \neq 2\pi$ at the vertex, then the universal cover of the cone minus the vertex can be viewed as the open Euclidean half-plane $r > 0$, with $(r, \theta) \sim (r, \theta + \theta_{0})$. One might then view (the image of) $(r, \pm\pi)$ as straight-line continuations of $(r, 0)$, but (unless $\theta_{0} = 2\pi/n$ for some integer $n \geq 1$) $[\pi] \neq [-\pi]$ as elements of $\mathbf{R}/\theta_{0}\mathbf{Z}$. (And if $\theta_{0} < \pi$, a union of rays from the vertex is not length-minimizing between its endpoints.) – 2017-01-23