derivative in the Wasserstein space

Question

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 !

Answer 1

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).