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This might be a very trivial question so bare with me. If $(X,d)$ is a length space we define a unit speed geodesic to be a path $\gamma:[0,1]\to X$ for which \begin{align*} d(\gamma(s),\gamma(t))=|t-s|d(\gamma(0),\gamma(1))\,\,\mathrm{for}\,\,\mathrm{all}\,\,s,t\in[0,1]. \end{align*} The author notes without further remarks that this is equivalent to if $\leq$ always a holds. I tried couple of tricks and I would like some fresh input if possible. I'm literally stuck and I believe there's a very simple way to see why this holds.

I know that we can find a reparameterization $\widetilde{\gamma}$ of $\gamma$ (i.e. $\gamma=\widetilde{\gamma}\circ\alpha$ for some non-decreasing continuous surjection $\alpha:[0,1]\to[0,1]$) for which the equation holds. From here to me it seems we can't do better than $d(\gamma(s),\gamma(t))=|\alpha(t)-\alpha(s)|d(\gamma(0),\gamma(1))$ for all $s,t\in[0,1]$. Or can we conclude something from here?

If someone is interested on the source, it's "A user's guide to optimal transportation", written by Ambrosio and Gigli. It's probably free to view through google. Page 34, equation (2.5).

Thanks in advance.

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I'd assume that the definition of 'length space', while seemingly not relevant to this question is not widely known. – user20266 Jun 6 '12 at 18:26
@Thomas: You're probably right. If $d(x,y)=\inf_{\gamma\in\mathscr{D}(x,y)}L_{d}(\gamma)$ for all $x,y\in X$: where $L_{d}$ is the length-function induced by $d$ and $\mathscr{D}(x,y)$ is the collection of those $\gamma\in X^{[0,1]}$ that are continuous, $\gamma(0)=x$ and $\gamma(1)=y$ and $L_{d}(\gamma)<\infty$ - we say that $(X,d)$ is a length space. In other words, if the distance of any pair of points is obtained as infimum over the lengths of rectifiable paths that join the points. – T. Eskin Jun 6 '12 at 18:35
@ThomasE. I retagged to replace (length-spaces) with the slightly more general (metric-geometry) – Willie Wong Jun 7 '12 at 13:43
@WillieWong: Thanks. – T. Eskin Jun 7 '12 at 14:32
up vote 3 down vote accepted

Clearly "$=$" $\Rightarrow$ "$\le$", so it suffices to consider "$\le$" $\Rightarrow$ "$=$". This follows from the triangle inequality.

Note, first, that, by assumption ($\le$) for $ 0 < s < t < 1$: $$d(\gamma(0), \gamma(s)) + d(\gamma(s), \gamma(t))+d(\gamma(t), \gamma(1))\le \left(s + (t-s) + (1-t)\right)\, d(\gamma(0), \gamma(1)) = d(\gamma(0), \gamma(1))$$ Now the rhs is $\le$ the lhs, by the triangle inequality. If there are $s, t$ such that strict inequality holds in the first inequality, you arrive at a contradiction.

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Thanks. I knew there existed a simple way to see this!:) – T. Eskin Jun 6 '12 at 18:39

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