Given the hyperbolic Vlasov equation 
$$ \frac{\partial f }{\partial t} +v\nabla\_x f + F(t,x)\nabla\_vf =0$$ 
where $f=f(t,x,v)$ and $(t,x,v)\in \mathbb{ R}\times\mathbb{R}^{n}\times \mathbb{R}^{n} $. I wonder how can be proved that $$ \Vert f(t,x,v)\Vert\_{L^p(\mathbb{R}^{2n})} = \Vert f(0,x,v)\Vert\_{L^p(\mathbb{R}^{2n})}, \quad p\in [1,\infty] $$
Any hint is welcome. Thank in advance. 



Conservated quantity and hyperbolic equation
The "one phrase answer" is "divergence theorem".


Slightly wordier but a bit formally (for ease of typing I write $dz = dx~dv$ for the volume on phase space)


$$ \partial\_t \int f^p dz = \int \partial\_t f^p dz $$


Next, 


$$ 0 = \int \nabla\_x \cdot (vf^p) dz $$ 


assuming $f$ decays suitably fast at infinity, and similarly


$$ 0 = \int \nabla\_v \cdot (Ff^p) dz $$


Now, 


$$ \partial\_t f^p + \nabla\_x \cdot (vf^p) + \nabla\_v \cdot (F f^p) = p f^{p-1} \left[ \partial\_t + v\cdot \nabla\_x + F \cdot \nabla\_v \right]f = 0 $$


($\nabla\_x$ trivially acts on $v$ and $\nabla\_v$ trivially acts on $F(t,x)$) and the result follows. 


To be precise, one has to interpret $|f|^p = \lim\_{\epsilon\to 0} (\sqrt{\epsilon^2 + |f|^2}- \epsilon)^{p}$, and approximate your $f\in L^p$ with $f \in \mathcal{S}\cap L^p$ or similar. 


For the case $p = \infty$ it suffices to notice that $f$ is constant on the integral curves of the vector field $(1, v, F(t,x))$ on $\mathbb{R}\times\mathbb{R}^n \times \mathbb{R}^n$. 


Finally, something stronger is true: let $G:\mathbb{R}\to\mathbb{R}$ be smooth, then $\int G\circ f ~dz$ is invariant in time if $f$ solves Vlasov. 


