Long time asymptotic of Fokker–Planck equation $\; \partial_tu-\nabla\!\!\cdot\!\left(\nabla u+xu\right) = 0$ Is it true that given a solution to the Fokker–Planck equation
$$\partial_tu-\nabla\!\!\cdot\!\left(\nabla u+x\hspace{0.2ex}u\right) = 0,$$
then we have 
$$
\left\|\frac{u-\rho}{\rho}\right\|_{\infty}\leq \varepsilon\left(t\right)
$$
with $\displaystyle\lim_{t\to \infty} \varepsilon\left(t\right) = 0$?
Where $\rho $ is the normalized Gaussian such that has the same mass of $u$.
I know that $\,\left\|u-\rho\right\|_{L^1}\lesssim e^{-t },$ but I have no idea how to prove or disprove this result. Can we prove the result using the rappresentation formula, obtained via a change of variable by heat equation?  

It is not true. Case $n=1$. Let $ w = \frac{v}{\rho} $ then $ w $ satisfies $$ \partial_t w -\rho^{-1}\nabla\cdot(\rho\nabla w) = 0. $$ and $$ \left\|\frac{v-\rho}{\rho}\right\|_{\infty} =\|w-1\|_{\infty}. $$ For this equation we have the following rapresentation formula for the solution with initial value f
$$ w(x,t) = \int f(xe^{-t}+y(1-e^{-2t})^{1/2})\rho(y)dy,$$
(The formula comes from a change of variables from the heat equation). 
Consider 
$$ f = \begin{cases}2 \quad \text{for} \quad x \in (-\alpha, \alpha) \\ 0 \quad \text{else} \end{cases}.$$
Choose $ \alpha $ such that $ \int f\rho = 1  $ then for $ x= -e^{t}1000 $ we have $ w(\bar{x},t) \leq 1/2 $ for every $ \bar{x} < x $ which prove that the fact is not true.
