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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 !

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 !

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 Green's identity (or 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 !

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 !

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 : $$\nabla_W F(\rho) = - (\nabla\rho.\nabla\delta_X + \nabla.\nabla\delta_X \rho)$$ which equals to $0$ after using Green's identity (or 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 !

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 Green's identity (or 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 !