Villani gives the following formula to find the gradient of a function $F$ of a probability density function $\rho$ in the Wasserstein space :
$$\nabla\_W F(\rho) = -\nabla.(\rho \nabla \frac{\delta F}{\delta \rho})$$
where, if $F$ is given as $F(\rho)=\int U(\rho) dx$, then $\frac{\delta F}{\delta \rho} = U'(\rho)$.


I am trying to find how much the value at a given (fixed) point $x$ of a continuous pdf varies along a displacement interpolation. I am thus trying to apply the formula above in the distributional sense where $U = \delta\_x$ (here, with $\delta$ the Dirac distribution), to get $F(\rho) = \rho(x) = <\delta\_x, \rho>$.


By using $\nabla.(f G) = (\nabla f) G + (\nabla.G) f$, I thus get : 
$$\nabla\_W F(\rho) = - (\nabla\rho.\nabla\delta\_x + \rho\triangle\delta\_x)$$
which equals to $0$ after using the definition of the derivatives in the distributional sense since both terms cancel each other.


So, what I get from it, is that the value at a fixed point $x$ (ie., not a point which gets advected with the function!) does not change when a function moves in the Wasserstein space... which I really don't believe to be true ! What went wrong ? 


Thanks !



derivative in the Wasserstein space
The dirac distribution $\delta\_x$ is not smooth enough to carry out this computation, which only works for functionals $F(\rho)=\int U(\rho)$...


One thing though: a displacement interpolation is nothing but a geodesic in the metric space of probabilities (endowed with the Wasserstein distance). If you look for example at Villani's book (chapter 8), we know that such a geodesic $(\rho\_t)\_{t\in[-\varepsilon,\varepsilon]}$ satisfies a continuity equation
$$
\partial\_t\rho\_t+\nabla\cdot(\rho\_tv\_t)=0
$$
for some particular vector-field $v\_t$ determining the tangent vector $s=-\nabla\cdot(\rho\_t v\_t)\in T\_{\rho\_t}\mathcal{P}$. This PDE is what you are looking for, since it allows to compute the variation $\partial\_t\rho\_t(x)$ for a given point $x$ (forgetting about smoothness issues, of course).


