I'm trying to read chapter 8 of the book on gradient flows by Ambrosio-Gigli-Savaré. In this context, I would like to better understand how the theory works for the following specific example. Take the family of probability measures on the real line $$\mu_t=(1-t)\delta_0+t\delta_1, \quad t\in(0,1),$$ where $\delta_x$ denotes the Dirac delta at $x$. It seems that this probability-valued curve is absolutely continuous with respect to the Wasserstein metric, but it does not seem to satisfy the conclusions of Theorem 8.3.1 in the book. In other words, there does not seem to be a vector field satisfying the continuity equation for this family of probabilities. As far as I can see the proof would already fail at equation (8.3.10), since it is possible to produce a test function with $\partial_t\phi\neq 0$ everywhere, yet $\partial_x\phi=0$ on the support of $\mu_t$, namely, $\{0,1\}$.

So my question is, what am I missing here?

Thanks a lot in advance.


This is the typical example of a curve of measures which is as nice as it gets for any "linear structure" (e.g. TV norm or whatever) but is somehow the worst case scenario for quadratic Wasserstein optimal transport.

In fact your curve is NOT absolutely continuous in the Wasserstein metric, contrarily to what you claim: indeed you can compute explicitly $$ W_2(\mu_t,\mu_s)=\sqrt{|t-s|}. $$ You can check this easily by moving a mass $|t-s|$ from $x=0$ to $x=1$, hence the cost $|t-s|$ for the squared distance $W_2^2$. More precisely, it is easy to check that the $W_2$ geodesic from $\mu_t$ to $\mu_s$ is of the form $$ (\tilde\mu_{\tau})_{\tau\in[0,1]}=s\delta_0+(t-s)\delta_{\tau}+(1-t)\delta_1, $$ say for $t>s$ (proving that this ansatz is optimal is a good exercise) As a consequence $$ \lim\limits_{h\to 0}\frac{W_2(\mu_{t+h},\mu_t)}{h}=+\infty \qquad\mbox{ for all }t\in(0,1), $$ and therefore your curve is clearly not absolutely continuous.

Heuristically this is due to the fact that, since your supports are fixed at distance $|1-0|>0$ from each other, you need to send mass infinitely fast at a positive distance thus with infinite cost (infinitesimally in time).

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    $\begingroup$ Very nice! The square root makes all the difference.. $\endgroup$ – S.Surace May 11 '19 at 21:47
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    $\begingroup$ Yes, people often underestimate the power of the square root ;-) $\endgroup$ – leo monsaingeon May 12 '19 at 2:27
  • $\begingroup$ Shouldn't the geodesic read $(1-t)\delta_0+|t-s|\delta_{\tau}+t\delta_1$? $\endgroup$ – S.Surace May 13 '19 at 15:19
  • $\begingroup$ Not exactly, but there was indeed a slight mistake which is now fixed (at least I hope). Thanks for pointing this out. $\endgroup$ – leo monsaingeon May 13 '19 at 18:31
  • $\begingroup$ Oh yes, of course. I think it is correct now. $\endgroup$ – S.Surace May 13 '19 at 19:37

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