Critical paths for length cannot have kinks. This problem is in Spivak's Differential Geometry (Ch.9 #37), and he gives a sketch of a proof which I have been unable to finish.

So let's compute $\frac{dL(\overline{\alpha}(u))}{du}\mid_{u=0}$  where $L(\overline{\alpha}(u))=L_{0}^{t_{1}}(\gamma)+\int_{t_{1}}^{t_{0}+u}F(\,,\,)dt+\int_{t_{0}+u}^{1}F(\,,\,)dt$ , and of course $F(\alpha(u,t),\frac{\partial\alpha}{\partial t}(u,t))=\sqrt{\underset{i,j}{\sum}g_{ij}(\alpha)\frac{\partial\alpha^{i}}{\partial t}\frac{\partial\alpha^{j}}{\partial t}}$  (all the "$(u,t)$" omitted).
Well, $\frac{dL(\overline{\alpha}(u))}{du}=\frac{d}{du}\int_{t_{1}}^{t_{0}+u}F(\,,\,)dt+\frac{d}{du}\int_{t_{0}+u}^{1}F(\,,\,)dt$ . Now I apply the Leibniz integral rule, and the terms become 


*

*$F(\alpha(u,t_{0}+u),\frac{\partial\alpha}{\partial t}(u,t_{0}+u))+\int_{t_{1}}^{t_{0}+u}\frac{\partial}{\partial u}F(\alpha(u,t),\frac{\partial\alpha}{\partial t}(u,t))dt$ 


and


*

*$\int_{t_{0}+u}^{1}\frac{\partial}{\partial u}F(\alpha(u,t),\frac{\partial\alpha}{\partial t}(u,t))dt-F(\alpha(u,t_{0}+u),\frac{\partial\alpha}{\partial t}(u,t_{0}+u))$


, respectively. Evaluating their sum at $u=0$ , we just get $\int_{t_{1}}^{1}\frac{\partial}{\partial u}F(\alpha(0,t),\frac{\partial\alpha}{\partial t}(0,t))dt$ .
Note that $\alpha$  is not exactly a variation on $\gamma$  since $\alpha(0,t)$  has a piece of $\gamma$  replaced by a geodesic (*). But anyway, $\alpha$ is  a variation on the piecewise smooth curve $\alpha(0,t)$ , and the integral obtained yields the First Variation Formula for Length of $\alpha(0,t)\mid_{[t_{1},1]} .$ . For the moment ignore *, and assume this is the thing we want to show $\neq0$ . The integral term in the First Variation Formula will disappear, leaving 
$\left\langle \frac{\partial\alpha}{\partial u}(0,t_{0}),\frac{\partial\alpha}{\partial t}(0,t_{0}^{+})-\frac{\partial\alpha}{\partial t}(0,t_{0}^{-})\right\rangle$  .
This is sort of like $\left\langle \frac{\partial\alpha}{\partial u}(0,t_{0}),\Delta_{t_{0}}\frac{d\gamma}{dt}\right\rangle$  , where we know $\Delta_{t_{0}}\frac{d\gamma}{dt}\neq0$ . 
Assuming everything was correct so far, I have two questions:
1) Why (where) do we need $t_{1}$  to be sufficiently close to $t_{0}$, as hinted by Spivak ?
2) How can I conclude that the inner product term is not 0?
 A: If the metric changes sufficiently smoothly at the corner point ($C^1$ or something like that) then geodesics between points that approach the corner will closely approximate geodesics in the tangent space, where the answer follows from Minkowski or Cauchy inequalities applied to the tangent metric.  This is not affected by the infinitesimal amount of sewing at the ends needed to change the geodesics to variations.
A: Some more thoughts...
Can we just define $\alpha(0,t)$ to be $\gamma(t)$; in other words, let the geodesic "short-cuts" begin only when u>0?  Would this destroy the required $C^\infty$ properties of $\alpha$?  The geodesics are parametrized arclength, while the piece of $\alpha$ they approach can have an arbitrary parametrization.
Another note: By moving $t_1$ toward $t_0$ we can change the length of  $\frac{\partial\alpha}{\partial t}(0,t_{0}^{-})$, since the geodesic portion is parametrized by arclength.  This may be useful when we need to show the vectors in the inner product are not orthogonal.
